Introduction to Whole Numbers

As we begin our study of elementary algebra, we need to refresh some of our skills and vocabulary. This chapter will focus on whole numbers, integers, fractions, decimals, and real numbers. We will also begin our use of algebraic notation and vocabulary.

Use Place Value with Whole Numbers

The most basic numbers used in algebra are the numbers we use to count objects in our world: 1, 2, 3, 4, and so on. These are called the counting numbers. Counting numbers are also called natural numbers. If we add zero to the counting numbers, we get the set of whole numbers.

Counting Numbers: 1, 2, 3, …

Whole Numbers: 0, 1, 2, 3, …

The notation “…” is called ellipsis and means “and so on,” or that the pattern continues endlessly.

We can visualize counting numbers and whole numbers on a number line (see Figure 1).

A horizontal number line with arrows on each end and values of zero to six runs along the bottom of the diagram. A second horizontal line with a left-facing arrow lies above the first and extend from zero to three. This line is labled “smaller”. A third horizontal line with a right-facing arrow lies above the first two, but runs from three to six and is labeled “larger”. — The numbers on the number line get larger as they go from left to right, and smaller as they go from right to left. While this number line shows only the whole numbers 0 through 6, the numbers keep going without end.

Our number system is called a place value system, because the value of a digit depends on its position in a number. Figure 2 shows the place values. The place values are separated into groups of three, which are called periods. The periods are ones, thousands, millions, billions, trillions, and so on. In a written number, commas separate the periods.

This figure is a table illustrating the number 5,278,194 within the place value system. The table is shown with a header row, labeled “Place Value”, divided into a second header row labeled “Trillions”, “Billions”, “Millions”, “Thousands” and “Ones”. Under the header “Trillions” are three labeled columns, written from bottom to top, that read “Hundred trillions”, “Ten trillions” and “Trillions”. Under the header “Billions” are three labeled columns, written from bottom to top, that read “Hundred billions”, “Ten billions” and “Billions”. Under the header “Millions” are three labeled columns, written from bottom to top, that read “Hundred millions”, “Ten millions” and “Millions”. Under the header “Thousands” are three labeled columns, written from bottom to top, that read “Hundred thousands”, “Ten thousands” and “Thousands”. Under the header “Ones” are three labeled columns, written from bottom to top, that read “Hundreds”, “Tens” and “Ones”. From left to right, below the columns labeled “Millions”, “Hundred thousands”, “Ten thousands”, “Thousands”, “Hundreds”, “Tens”, and “Ones”, are the following values: 5, 2, 7, 8, 1, 9, 4. This means there are 5 millions, 2 hundred thousands, 7 ten thousands, 8 thousands, 1 hundreds, 9 tens, and 4 ones in the number five million two hundred seventy-nine thousand one hundred ninety-four. — The number 5,278,194 is shown in the chart. The digit 5 is in the millions place. The digit 2 is in the hundred-thousands place. The digit 7 is in the ten-thousands place. The digit 8 is in the thousands place. The digit 1 is in the hundreds place. The digit 9 is in the tens place. The digit 4 is in the ones place.

In the number 63,407,218, find the place value of each digit:

ⓐ 7
ⓑ 0
ⓒ 1
ⓓ 6
ⓔ 3

Solution

Place the number in the place value chart:

ⓐ 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.

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 at the left, where the periods have the largest value. The ones period is not named. The commas separate the periods, so wherever there is a comma in the number, put a comma between the words (see Figure 3). The number 74,218,369 is written as seventy-four million, two hundred eighteen thousand, three hundred sixty-nine.

In this figure, the numbers 74, 218 and 369 are listed in a row, separated by commas. Each number has a curly bracket beneath it with the word “millions” written below the number 74, “thousands” written below the number 218, and “ones” written below the number 369. A left-facing arrow points at these three words, labeling them “periods”. One row down is the number “74”, a right-facing arrow and the words “Seventy-four million” followed by a comma. The next row below is the number “218”, a right-facing arrow and the words “two hundred eighteen thousand” followed by a comma. On the bottom row is the number “369”, a right-facing arrow and the words “three hundred sixty-nine”.

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

Solution

Name the number in each period, followed by the period name.

Put the commas in to separate the periods.

So, $8, 165, 432, 098, 710$ is named as eight trillion, one hundred sixty-five billion, four hundred thirty-two million, ninety-eight thousand, seven hundred ten.

We are now going to reverse the process by writing the digits from the name of the number. To write the number in digits, we first look for the clue words that indicate the periods. It is helpful to draw three blanks for the needed periods and then fill in the blanks with the numbers, separating the periods with commas.

Write nine billion, two hundred forty-six million, seventy-three thousand, one hundred eighty-nine as a whole number using digits.

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.

The number is 9,246,073,189.

In 2013, the U.S. Census Bureau estimated the population of the state of New York as 19,651,127. We could say the population of New York was approximately 20 million. In many cases, you don’t need the exact value; an approximate number is good enough.

The process of approximating a number is called rounding. Numbers are rounded to a specific place value, depending on how much accuracy is needed. Saying that the population of New York is approximately 20 million means that we rounded to the millions place.

How to Round Whole Numbers

Round 23,658 to the nearest hundred.

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains the numbers corresponding with the written steps and instructions. In the top row, the first cell says: “Step 1. Locate the given place value with an arrow. All digits to the left do not change.” In the the second cell, the instructions say: “Locate the hundreds place in 23,658.” In the third cell, there is the number 23,658 with an arrow pointing to the digit 6, labeling it “hundreds place.”

One row down, the instructions in the first cell say: “Step 2. Underline the digit to the right of the given place value.” In the second cell, the instructions say: “Underline the 5, which is to the right of the hundreds place.” In the third cell, there is the number 23,658 again, the same arrow pointing to the digit 6, labeling it the hundreds place. The 5 is also underlined in this cell.

One row down, the first cell says: “Step 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.” In the second cell, the instructions say: “Add 1 to the 6 in the hundreds place, since 5 is greater than or equal to 5.” The third cell contains the number 23,658 again, with an arrow pointing at the digit 6 and the text “add 1”. There is also a curly bracket under the digits 5 and 8, with an arrow pointing at them and the text “replace with 0s.”

In the bottom row, the first cell says: “Step 4. Replace all digits to the right of the given place value with zeros. So, 23,700 is rounded to the nearest hundred.” In the second cell, the instructions say: “Replace all digits to the right of the hundreds place with zeros.” The third cell contains the number 23,700, which we have reached by rounding the number 23,658 to the nearest hundred.

Round $103,978$ to the nearest:

ⓐ hundred
ⓑ thousand
ⓒ ten thousand

Solution

ⓐ

Locate the hundreds place in 103,978.
Underline the digit to the right of the hundreds place.
Since 7 is greater than or equal to 5, add 1 to the 9. Replace all digits to the right of the hundreds place with zeros.
	So, 104,000 is 103,978 rounded to the nearest hundred.

ⓑ

Locate the thousands place and underline the digit to the right of the thousands place.
Since 9 is greater than or equal to 5, add 1 to the 3. Replace all digits to the right of the hundreds place with zeros.
	So, 104,000 is 103,978 rounded to the nearest thousand.

ⓒ

Locate the ten thousands place and underline the digit to the right of the ten thousands place.
Since 3 is less than 5, we leave the 0 as is, and then replace the digits to the right with zeros.
	So, 100,000 is 103,978 rounded to the nearest ten thousand.

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

Identify Multiples and Apply Divisibility Tests

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

A diagram made up of two rows of numbers. The top row reads “2, 4, 6, 8, 10, 12,” followed by an elipsis. Below 2 is 2 times 1, below 4 is 2 times 2, below 6 is 2 times 3, below 8 is 2 times 4, below 10 is 2 times 5, and below 12 is 2 times 6.

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

A diagram made up of two rows of numbers. The top row reads “3, 6, 9, 12, 15, 18,” followed by an elipsis. Below 3 is 3 times 1, below 6 is 3 times 2, below 9 is 3 times 3, below 12 is 3 times 4, below 15 is 3 times 5, and below 18 is 3 times 6.

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

Table 4 shows the multiples of 2 through 9 for the first 12 counting numbers.

Counting Number	1	2	3	4	5	6	7	8	9	10	11	12
Multiples of 2	2	4	6	8	10	12	14	16	18	20	22	24
Multiples of 3	3	6	9	12	15	18	21	24	27	30	33	36
Multiples of 4	4	8	12	16	20	24	28	32	36	40	44	48
Multiples of 5	5	10	15	20	25	30	35	40	45	50	55	60
Multiples of 6	6	12	18	24	30	36	42	48	54	60	66	72
Multiples of 7	7	14	21	28	35	42	49	56	63	70	77	84
Multiples of 8	8	16	24	32	40	48	56	64	72	80	88	96
Multiples of 9	9	18	27	36	45	54	63	72	81	90	99	108
Multiples of 10	10	20	30	40	50	60	70	80	90	100	110	120

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

Look at the multiples of 5 in Table 4. They all end in 5 or 0. Numbers with last digit of 5 or 0 are divisible by 5. Looking for other patterns in Table 4 that shows multiples of the numbers 2 through 9, we can discover the following divisibility tests:

Is 5,625 divisible by 2? By 3? By 5? By 6? By 10?

Solution

This table demonstrates divisibility rules for 5,625 by 2, 3, 5, 10, and 6, detailing the criteria, steps, and outcomes for each test.
Is 5,625 divisible by 2?
Does it end in 0,2,4,6, or 8?	No. 5,625 is not divisible by 2.
Is 5,625 divisible by 3?
What is the sum of the digits?	$5 + 6 + 2 + 5 = 18$
Is the sum divisible by 3?	Yes. 5,625 is divisble by 3.
Is 5,625 divisible by 5 or 10?
What is the last digit? It is 5.	5,625 is divisble by 5 but not by 10.
Is 5,625 divisible by 6?
Is it divisible by both 2 and 3?	No, 5,625 is not divisible by 2, so 5,625 is not divisible by 6.

Find Prime Factorizations and Least Common Multiples

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

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

An image shows the equation 8 times 9 equals 72. Written below the expression 8 times 9 is a curly bracket and the word “factors” while written below 72 is a horizontal bracket and the word “product”.

Other ways to factor 72 are $1 \cdot 72, 2 \cdot 36, 3 \cdot 24, 4 \cdot 18, and 6 \cdot 12 .$ Seventy-two has many factors: 1, 2, 3, 4, 6, 8, 9, 12, 18, 36, and 72.

Some numbers, like 72, have many factors. Other numbers have only two factors.

The counting numbers from 2 to 19 are listed in Figure 4, with their factors. Make sure to agree with the “prime” or “composite” label for each!

A table is shown with eleven rows and seven columns. The first row is a header row, and each cell labels the contents of the column below it. In the header row, the first three cells read from left to right “Number”, “Factors”, and “Prime or Composite?” The entire fourth column is blank. The last three cells read from left to right “Number”, “Factor”, and “Prime or Composite?” again. In each subsequent row, the first cell contains a number, the second contains its factors, and the third indicates whether the number is prime or composite. The three columns to the left of the blank middle column contain this information for the number 2 through 10, and the three columns to the right of the blank middle column contain this information for the number 11 through 19. On the left side of the blank column, in the first row below the header row, the cells read from left to right: “2”, “1,2”, and “Prime”. In the next row, the cells read from left to right: “3”, “1,3”, and “Prime”. In the next row, the cells read from left to right: “4”, “1,2,4”, and “Composite”. In the next row, the cells read from left to right: “5”, “1,5”, and “Prime”. In the next row, the cells read from left to right: “6”, “1,2,3,6” and “Composite”. In the next row, the cells read from left to right: “7”, “1,7”, and “Prime”. In the next row, the cells read from left to right: “8”, “1,2,4,8”, and “Composite”. In the next row, the cells read from left to right: “9”, “1,3,9”, and “Composite”. In the bottom row, the cells read from left to right: “10”, “1,2,5,10”, and “Composite”. On the right side of the blank column, in the first row below the header row, the cells read from left to right: “11”, “1,11”, and “Prime”. In the next row, the cells read from left to right: “12”, “1,2,3,4,6,12”, and “Composite”. In the next row, the cells read from left to right: “13”, “1,13”, and “Prime”. In the next row, the cells read from left to right “14”, “1,2,7,14”, and “Composite”. In the next row, the cells read from left to right: “15”, “1,3,5,15”, and “Composite”. In the next row, the cells read from left to right: “16”, “1,2,4,8,16”, and “Composite”. In the next row, the cells read from left to right, “17”, “1,17”, and “Prime”. In the next row, the cells read from left to right, “18”, “1,2,3,6,9,18”, and “Composite”. In the bottom row, the cells read from left to right: “19”, “1,19”, and “Prime”.

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

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

To find the prime factorization of a composite number, find any two factors of the number and use them to create two branches. If a factor is prime, that branch is complete. Circle that prime!

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

How to Find the Prime Factorization of a Composite Number

Factor 48.

Solution

One row down, the instructions in the first cell say: “Step 2. If a factor is prime, that branch is complete. Circle the prime.” In the second cell, the instructions say: “2 is prime. Circle the prime.” In the third cell, the factor tree from step 1 is repeated, but the 2 at the bottom of the tree is now circled.

One row down, the first cell says: “Step 3. If a factor is not prime, write it as the product of two factors and continue the process.” In the second cell, the instructions say: “24 is not prime. Break it into 2 more factors.” The third cell contains the original factor tree, with 48 at the top and two downward-pointing branches terminating at 2, which is underlined, and 24. Two more branches descend from 24 and terminate at 4 and 6 respectively. One line down, the instructions in the middle of the cell say “4 and 6 are not prime. Break them each into two factors.” In the cell on the right, the factor tree is repeated once more. Two branches descend from the 4 and terminate at 2 and 2. Both 2s are circled. Two more branches descend from 6 and terminate at a 2 and a 3, which are both circled. The instructions on the left say “2 and 3 are prime, so circle them.”

In the bottom row, the first cell says: “Step 4. Write the composite number as the product of all the circled primes.” The second cell is left blank. The third cell contains the algebraic equation 48 equals 2 times 2 times 2 times 2 times 3.

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

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

Find the prime factorization of 252.

Solution

Step 1. Find two factors whose product is 252. 12 and 21 are not prime. Break 12 and 21 into two more factors. Continue until all primes are factored.
Step 2. Write 252 as the product of all the circled primes.	$252 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7$

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. Two methods are used most often to find the least common multiple and we will look at both of them.

The first method is the Listing Multiples Method. To find the least common multiple of 12 and 18, we list the first few multiples of 12 and 18:

Two rows of numbers are shown. The first row begins with 12, followed by a colon, then 12, 24, 36, 48, 60, 72, 84, 96, 108, and an elipsis. 36, 72, and 108 are bolded written in red. The second row begins with 18, followed by a colon, then 18, 36, 54, 72, 90, 108, and an elipsis. Again, the numbers 36, 72, and 108 are bolded written in red. On the line below is the phrase “Common Multiples”, a colon and the numbers 36, 72, and 108, written in red. One line below is the phrase “Least Common Multiple”, a colon and the number 36, written in blue.

Notice that some numbers appear in both lists. They are the common multiples of 12 and 18.

We see that the first few common multiples of 12 and 18 are 36, 72, and 108. Since 36 is the smallest of the common multiples, we call it the least common multiple. We often use the abbreviation LCM.

The procedure box lists the steps to take to find the LCM using the prime factors method we used above for 12 and 18.

Find the least common multiple of 15 and 20 by listing multiples.

Solution

Make lists of the first few multiples of 15 and of 20, and use them to find the least common multiple.
Look for the smallest number that appears in both lists.	The first number to appear on both lists is 60, so 60 is the least common multiple of 15 and 20.

Notice that 120 is in both lists, too. It is a common multiple, but it is not the least common multiple.

Our second method to find the least common multiple of two numbers is to use The Prime Factors Method. Let’s find the LCM of 12 and 18 again, this time using their prime factors.

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

Find the Least Common Multiple (LCM) of 12 and 18 using the prime factors method.

Solution

One row down, the instructions in the first cell say: “Step 2. List the primes of each number. Match primes vertically when possible.” In the second cell, the instructions say: “List the primes of 12. List the primes of 18. Line up with the primes of 12 when possible. If not create a new column.” The third cell contains 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.

One row down, the instructions in the first cell say: “Bring down the number from each column.” The second cell is blank. The third cell contains the prime factorizations of 12 and 18 again, illustrated as two equations aligned just as they were before. This time, 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.

In the bottom row of the table, the first cell says: “Step 4: Multiply the factors.” The second cell is bank. The third cell contains the equation LCM equals 36.

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

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

Find the Least Common Multiple (LCM) of 24 and 36 using the prime factors method.

Solution

Find the primes of 24 and 36. Match primes vertically when possible. Bring down all columns.
Multiply the factors.
	The LCM of 24 and 36 is 72.

Key Concepts

Place Value as in Figure 2.
Name a Whole Number in Words
1. Start at the left and name the number in each period, followed by the period name.
2. Put commas in the number to separate the periods.
3. Do not name the ones period.
Write a Whole Number Using Digits
1. Identify the words that indicate periods. (Remember the ones period is never named.)
2. Draw 3 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.
Round Whole Numbers
1. Locate the given place value and mark it with an arrow. All digits to the left of the arrow do not change.
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. Replace all digits to the right of the given place value with zeros.
Divisibility Tests: A number is divisible by:
- 2 if the last digit is 0, 2, 4, 6, or 8.
- 3 if the sum of the digits is divisible by 3.
- 5 if the last digit is 5 or 0.
- 6 if it is divisible by both 2 and 3.
- 10 if it ends with 0.
Find the Prime Factorization of a Composite Number
1. Find two factors whose product is the given number, and use these numbers to create two branches.
2. If a factor is prime, that branch is complete. Circle the prime, like a bud on the tree.
3. If a factor is not prime, write it as the product of two factors and continue the process.
4. Write the composite number as the product of all the circled primes.
Find the Least Common Multiple by Listing Multiples
1. List several multiples of each number.
2. Look for the smallest number that appears on both lists.
3. This number is the LCM.
Find the Least Common Multiple Using the Prime Factors Method
1. Write each number as a product of primes.
2. List the primes of each number. Match primes vertically when possible.
3. Bring down the columns.
4. Multiply the factors.

Practice Makes Perfect

Use Place Value with Whole Numbers

In the following exercises, find the place value of each digit in the given numbers.

51,493 ⓐ 1, ⓑ 4, ⓒ 9, ⓓ 5, ⓔ 3

Solution

ⓐ thousands ⓑ hundreds ⓒ tens ⓓ ten thousands ⓔ ones

87,210 ⓐ 2 ⓑ 8 ⓒ 0 ⓓ 7 ⓔ 1

164,285 ⓐ 5, ⓑ 6, ⓒ 1, ⓓ 8, ⓔ 2

Solution

ⓐ ones ⓑ ten thousands ⓒ hundred thousands ⓓ tens ⓔ hundreds

395,076 ⓐ 5 ⓑ 3 ⓒ 7 ⓓ 0 ⓔ 9

93,285,170 ⓐ 9 ⓑ 8 ⓒ 7 ⓓ 5 ⓔ 3

Solution

ⓐ ten millions ⓑ ten thousands ⓒ tens ⓓ thousands ⓔ millions

36,084,215 ⓐ 8 ⓑ 6 ⓒ 5 ⓓ 4 ⓔ 3

7,284,915,860,132 ⓐ 7 ⓑ 4 ⓒ 5 ⓓ 3 ⓔ 0

Solution

ⓐ trillions ⓑ billions ⓒ millions ⓓ tens ⓔ thousands

2,850,361,159,433
ⓐ 9
ⓑ 8
ⓒ 6
ⓓ 4
ⓔ 2

In the following exercises, name each number using 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

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

In the following, round to the indicated place value.

Round to the nearest ten.

ⓐ 386 ⓑ 2,931

Solution

ⓐ 390 ⓑ 2,930

Round to the nearest ten.

ⓐ 792 ⓑ 5,647

Round to the nearest hundred.

ⓐ 13,748 ⓑ 391,794

Solution

ⓐ 13,700 ⓑ 391,800

Round to the nearest hundred.

ⓐ 28,166 ⓑ 481,628

Round to the nearest ten.

ⓐ 1,492 ⓑ 1,497

Solution

ⓐ 1,490 ⓑ 1,500

Round to the nearest ten.

ⓐ 2,791 ⓑ 2,795

Round to the nearest hundred.

ⓐ 63,994 ⓑ 63,940

Solution

ⓐ 64,000 ⓑ 63,900

Round to the nearest hundred.

ⓐ 49,584 ⓑ 49,548

In the following exercises, round each number to the nearest ⓐ hundred, ⓑ thousand, ⓒ ten thousand.

392,546

Solution

ⓐ 392,500 ⓑ 393,000 ⓒ 390,000

619,348

2,586,991

Solution

ⓐ 2,587,000 ⓑ 2,587,000 ⓒ 2,590,000

4,287,965

Identify Multiples and Factors

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

Solution

divisible by 2, 3, and 6

9,696

Solution

divisible by 3 and 5

900

Solution

divisible by 2, 3, 5, 6, and 10

800

986

Solution

divisible by 2

942

350

Solution

divisible by 2, 5, and 10

550

22,335

Solution

divisible by 3 and 5

39,075

Find Prime Factorizations and Least Common Multiples

In the following exercises, find the prime factorization.

Solution

$2 \cdot 43$

132

Solution

$2 \cdot 2 \cdot 3 \cdot 11$

455

693

Solution

$3 \cdot 3 \cdot 7 \cdot 11$

400

432

Solution

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3$

627

2,160

Solution

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5$

2,520

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

8, 12

Solution

4, 3

12, 16

Solution

30, 40

20, 30

Solution

44, 55

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

8, 12

Solution

12, 16

28, 40

Solution

280

84, 90

55, 88

Solution

440

60, 72

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 ⓐ ten ⓑ hundred ⓒ thousand; and ⓓ ten-thousand.

Solution

ⓐ $24,490 ⓑ $24,500 ⓒ $24,000 ⓓ $20,000

Remodeling a Kitchen Marissa’s kitchen remodeling cost $18,549, Round the cost to the nearest ⓐ ten ⓑ hundred ⓒ thousand and ⓓ ten-thousand.

Population The population of China was 1,339,724,852 on November 1, 2010. Round the population to the nearest ⓐ billion ⓑ hundred-million; and ⓒ million.

Solution

ⓐ 1,000,000,000 ⓑ 1,300,000,000 ⓒ 1,340,000,000

Astronomy The average distance between Earth and the sun is 149,597,888 kilometers. Round the distance to the nearest ⓐ hundred-million ⓑ ten-million; and ⓒ million.

Grocery Shopping Hot dogs are sold in packages of 10, but hot dog buns come in packs of eight. What is the smallest number that makes the hot dogs and buns come out even?

Solution

Grocery Shopping Paper plates are sold in packages of 12 and party cups come in packs of eight. What is the smallest number that makes the plates and cups come out even?

Writing Exercises

Give an everyday example where it helps to round numbers.

Solution

Answers may vary.

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?

Solution

Answers may vary.

Explain in your own words how to find the prime factorization of a composite number, using any method you prefer.

Self Check

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

A table with four columns and four rows is shown. The columns are titled “I can …”, “Confidently”, “With some help”, and “No – I don’t get it!”. The first column has three rows of text that read “use place value with whole numbers”, “identify multiples and apply divisibility rules” and “find prime factorization and least common multiples”. All other spaces on the table are blank.

ⓑ If most of your checks were:

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

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

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

composite number

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

counting numbers

The counting numbers are the numbers 1, 2, 3, …

divisible by a number

If a number

m

is a multiple of

n

, then

m

is divisible by

n

. (If 6 is a multiple of 3, then 6 is divisible by 3.)

factors

a \cdot b = m

, then

a and b

are factors of

m

. Since 3 · 4 = 12, then 3 and 4 are factors of 12.

least common multiple

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

multiple of a number

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

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.

prime factorization

The prime factorization of a number is the product of prime numbers that equals the number.

prime number

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

whole numbers

The whole numbers are the numbers 0, 1, 2, 3, ....

Use the Language of Algebra

Use Variables and Algebraic Symbols

Suppose this year Greg is 20 years old and Alex is 23. You know that 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 variables and the 3 is a constant. The ages change (“vary”) but the 3 years between them always stays the same (“constant”). Since Greg’s age and Alex’s age will always differ by 3 years, 3 is the constant.

In algebra, we use letters of the alphabet to represent variables. So if 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
$12$	$15$
$20$	$23$
$35$	$38$
$g$	$g + 3$

The letters used to represent these changing ages are called variables. The letters most commonly used for variables are x, y, a, b, and c.

To write algebraically, we need some operation symbols as well as numbers and variables. There are several types of symbols we will be using.

There are four basic arithmetic operations: addition, subtraction, multiplication, and division. We’ll list the symbols used to indicate these operations in the table below. You’ll probably recognize some of them.

Operation	Notation	Say:	The result is…
Addition	$a + b$	a plus b	the sum of a and b
Subtraction	$a - b$	a minus b	the difference of a and b
Multiplication	$a \cdot b, a b, (a) (b),$ $(a) b, a (b)$	a times b	the product of a and b
Division	$a \div b, a / b, \frac{a}{b}, b a$	a divided by b	the quotient of a and b, a is called the dividend, and b is called the divisor

We perform these operations on two numbers. When translating from symbolic form to English, or from English to symbolic form, pay attention to the words “of” and “and.”

The difference of 9 and 2 means subtract 9 and 2, in other words, 9 minus 2, which we write symbolically as $9 - 2 .$
The product of 4 and 8 means multiply 4 and 8, in other words 4 times 8, which we write symbolically as $4 \cdot 8 .$

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

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

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

The expressions a b can be read from left to right or right to left, though in English we usually read from left to right (Table 3). In general, a a. For example 7 < 11 is equivalent to 11 > 7. And a > b is equivalent to b < a. For example 17 > 4 is equivalent to 4 < 17.

Inequality Symbols	Words
$a \neq b$	a is not equal to b
a < b	a is less than b
$a \leq b$	a is less than or equal to b
a > b	a is greater than b
$a \geq b$	a is greater than or equal to b

Translate from algebra into English:

ⓐ $17 \leq 26$ ⓑ $8 \neq 17 - 3$ ⓒ $12 > 27 \div 3$ ⓓ $y + 7 < 19$

Solution

ⓐ $17 \leq 26$
17 is less than or equal to 26

ⓑ $8 \neq 17 - 3$
8 is not equal to 17 minus 3

ⓒ $12 > 27 \div 3$
12 is greater than 27 divided by 3

ⓓ $y + 7 < 19$
y plus 7 is less than 19

Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in English. They help to make clear which expressions are to be kept together and separate from other expressions. We will introduce three types now.

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

\begin{matrix} 8 (14 - 8) & 21 - 3 [2 + 4 (9 - 8)] & 24 \div {13 - 2 [1 (6 - 5) + 4]} \end{matrix}

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 an English phrase. Here are some examples of expressions:

Expression	Words	English Phrase
$3 + 5$	3 plus 5	the sum of three and five
$n - 1$	n minus one	the difference of n and one
$6 \cdot 7$	6 times 7	the product of six and seven
$\frac{x}{y}$	x divided by y	the quotient of x and y

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

An equation is two expressions linked 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	English Sentence
$3 + 5 = 8$	The sum of three and five is equal to eight.
$n - 1 = 14$	n minus one equals fourteen.
$6 \cdot 7 = 42$	The product of six and seven is equal to forty-two.
$x = 53$	x is equal to fifty-three.
$y + 9 = 2 y - 3$	y plus nine is equal to two y minus three.

Determine if each is an expression or an equation:

ⓐ $2 (x + 3) = 10$ ⓑ $4 (y - 1) + 1$ ⓒ $x \div 25$ ⓓ $y + 8 = 40$

Solution

This table differentiates between mathematical expressions and equations, providing examples and explanations for each type.
ⓐ $2 (x + 3) = 10$	This is an equation—two expressions are connected with an equal sign.
ⓑ $4 (y - 1) + 1$	This is an expression—no equal sign.
ⓒ $x \div 25$	This is an expression—no equal sign.
ⓓ $y + 8 = 40$	This is an equation—two expressions are connected with an equal sign.

Suppose we need to multiply 2 nine times. We could write this as $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 .$ This is tedious and it can be hard to keep track of all those 2s, so we use exponents. We write $2 \cdot 2 \cdot 2$ as $2^{3}$ and $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ as $2^{9} .$ In expressions such as $2^{3},$ the 2 is called the base and the 3 is called the exponent. The exponent tells us how many times we need to multiply the base.

The number two is shown with a superscipted number three to the right of it. an arrow is drawn to the number two and labeled “base” while another arrow is drawn to the superscripted three and labeled “exponent”. This means multiply 2 by itself, three times, as in 2 times 2 times 2.

We read $2^{3}$ as “two to the third power” or “two cubed.”

We say $2^{3}$ is in exponential notation and $2 \cdot 2 \cdot 2$ is in expanded notation.

While we read $a^{n}$ as “a to the $n^{t h}$ power,” we usually read:

$a^{2}$ “a squared”
$a^{3}$ “a cubed”

We’ll see later why $a^{2}$ and $a^{3}$ have special names.

Table 7 shows how we read some expressions with exponents.

Expression	In Words
$7^{2}$	7 to the second power or 7 squared
$5^{3}$	5 to the third power or 5 cubed
$9^{4}$	9 to the fourth power
$12^{5}$	$12$ to the fifth power

Simplify: $3^{4} .$

Solution

Step-by-step evaluation of the exponential expression 3^4.
	$3^{4}$
Expand the expression.	$3 \cdot 3 \cdot 3 \cdot 3$
Multiply left to right.	$9 \cdot 3 \cdot 3$
Multiply.	$27 \cdot 3$
Multiply.	81

Simplify Expressions Using the Order of Operations

To simplify an expression means to do all the math possible. For example, to simplify $4 \cdot 2 + 1$ we’d first multiply $4 \cdot 2$ 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:

\begin{matrix} 4 \cdot 2 + 1 \\ 8 + 1 \\ 9 \end{matrix}

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

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 \cdot 7

If you simplify this expression, what do you get?

Some students say 49,

\begin{matrix} 4 + 3 \cdot 7 \\ Since 4 + 3 gives 7 . & 7 \cdot 7 \\ And 7 \cdot 7 is 49 . & 49 \end{matrix}

Others say 25,

\begin{matrix} 4 + 3 \cdot 7 \\ Since 3 \cdot 7 is 21 . & 4 + 21 \\ And 21 + 4 makes 25 . & 25 \end{matrix}

Imagine the confusion in our banking system if every problem had several different correct answers!

The same expression should give the same result. So mathematicians early on established some guidelines that are called the Order of Operations.

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.”

\begin{matrix} P arentheses & P lease \\ E xponents & E xcuse \\ M ultiplication D ivision & M y D ear \\ A ddition S ubtraction & A unt S ally \end{matrix}

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.

Let’s try an example.

Simplify: ⓐ $4 + 3 \cdot 7$ ⓑ $(4 + 3) \cdot 7 .$

Solution

ⓐ


Are there any parentheses? No.
Are there any exponents? No.
Is there any multiplication or division? Yes.
Multiply first.
Add.

ⓑ


Are there any parentheses? Yes.
Simplify inside the parentheses.
Are there any exponents? No.
Is there any multiplication or division? Yes.
Multiply.

Simplify: $18 \div 6 + 4 (5 - 2) .$

Solution

Parentheses? Yes, subtract first.	$18 \div 6 + 4 (5 - 2)$
Exponents? No.
Multiplication or division? Yes.
Divide first because we multiply and divide left to right.
Any other multiplication or division? Yes.
Multiply.
Any other multiplication or division? No.
Any addition or subtraction? Yes.

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

Simplify: $5 + 2^{3} + 3 [6 - 3 (4 - 2)] .$

Solution


Are there any parentheses (or other grouping symbol)? Yes.
Focus on the parentheses that are inside the brackets.
Subtract.
Continue inside the brackets and multiply.
Continue inside the brackets and subtract.
The expression inside the brackets requires no further simplification.
Are there any exponents? Yes.
Simplify exponents.
Is there any multiplication or division? Yes.
Multiply.
Is there any addition or subtraction? Yes.
Add.
Add.

Evaluate an Expression

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

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

Evaluate $7 x - 4,$ when ⓐ $x = 5$ and ⓑ $x = 1 .$

Solution

ⓐ



Multiply.
Subtract.

ⓑ



Multiply.
Subtract.

Evaluate the following for $x = 4,$ when ⓐ $x^{2}$ ⓑ $3^{x} .$

Solution

ⓐ

	$x^{2}$

Use definition of exponent.	$4 \cdot 4$
Simplify.	$16$

ⓑ

	$3^{x}$

Use definition of exponent.	$3 \cdot 3 \cdot 3 \cdot 3$
Simplify.	$81$

Evaluate $2 x^{2} + 3 x + 8$ when $x = 4 .$

Solution

	$2 x^{2} + 3 x + 8$

Follow the order of operations.	$2 (16) + 3 (4) + 8$
	$32 + 12 + 8$
	$52$

Identify and Combine Like Terms

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

Examples of terms are $7, y, 5 x^{2}, 9 a, and b^{5} .$

The constant that multiplies the variable is called the coefficient.

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 $x = 1 \cdot x .$

Identify the coefficient of each term: ⓐ 14y ⓑ $15 x^{2}$ ⓒ a.

Solution

ⓐ The coefficient of 14y is 14.

ⓑ The coefficient of $15 x^{2}$ is 15.

ⓒ The coefficient of a is 1 since $a = 1 a .$

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

\begin{matrix} 5 x & 7 & n^{2} & 4 & 3 x & 9 n^{2} \end{matrix}

The 7 and the 4 are both constant terms.

The 5x and the 3x are both terms with x.

The $n^{2}$ and the $9 n^{2}$ are both terms with $n^{2} .$

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

7 and 4 are like terms.
5x and 3x are like terms.
$x^{2}$ and $9 x^{2}$ are like terms.

Identify the like terms: $y^{3},$ $7 x^{2},$ 14, 23, $4 y^{3},$ 9x, $5 x^{2} .$

Solution

$y^{3}$ and $4 y^{3}$ are like terms because both have $y^{3};$ the variable and the exponent match.

$7 x^{2}$ and $5 x^{2}$ are like terms because both have $x^{2};$ the variable and the exponent match.

14 and 23 are like terms because both are constants.

There is no other term like 9x.

Adding or subtracting terms forms an expression. In the expression $2 x^{2} + 3 x + 8,$ from Example 9, the three terms are $2 x^{2}, 3 x,$ and 8.

Identify the terms in each expression.

ⓐ $9 x^{2} + 7 x + 12$
ⓑ $8 x + 3 y$

Solution

ⓐ The terms of $9 x^{2} + 7 x + 12$ are $9 x^{2},$ 7x, and 12.
ⓑ The terms of $8 x + 3 y$ are 8x and 3y.

If there are like terms in an expression, you can simplify the expression by combining the like terms. What do you think $4 x + 7 x + x$ would simplify to? If you thought 12x, you would be right!

\begin{matrix} 4 x + 7 x + x \\ x + x + x + x + x + x + x + x + x + x + x + x \\ 12 x \end{matrix}

Add the coefficients and keep the same variable. It doesn’t matter what x is—if you have 4 of something and add 7 more of the same thing and then add 1 more, the result is 12 of them. For example, 4 oranges plus 7 oranges plus 1 orange is 12 oranges. We will discuss the mathematical properties behind this later.

Simplify: $4 x + 7 x + x .$

Add the coefficients. 12x

How To Combine Like Terms

Simplify: $2 x^{2} + 3 x + 7 + x^{2} + 4 x + 5 .$

Solution

Three lines of instructions are listed in a column on the left side of the image while four algebraic expressions are listed on the right. The first line of instruction on the left says: “Step 1. Identify like terms.” Across from step 1 in the right column is the algebraic expression: 2x squared plus 3x plus 7 plus x squared plus 4x plus 5. One line down on the right, the same algebraic expression is repeated, except each of the terms appears in one of three colors to illustrate that these are like terms: 2x squared and x squared appear as red, illustrating that these are like terms; 3x and 4x appear as blue, illustrating that these are also like terms; 7 and 5 appear as green, illustrating that these are like terms as well.

The second line of instruction on the left says: “Step 2. Rearrange the expression so the like terms are together. Across from step 2 in the right column is the original algebraic expression with terms reordered so that like terms appear side by side: 2x squared plus x2, both written in red, plus 3x plus 4x, both written n blue, plus 7 plus 5, both written in green.

The third line of instruction on the left says: “Step 3. Combine like terms.” Across from step 3 in the right column is the algebraic expression with like terms combined: 3x squared in red, plus 7x in blue, plus 12 in green.

Translate an English Phrase to an Algebraic Expression

In the last section, we listed many operation symbols that are used in algebra, then we translated expressions and equations into English phrases and sentences. Now we’ll reverse the process. We’ll translate English phrases into algebraic expressions. The symbols and variables we’ve talked about will help us do that. Table 18 summarizes them.

Operation	Phrase	Expression
Addition	a plus b the sum of $a$ and b $a$ increased by b b more than $a$ the total of $a$ and b b added to $a$	$a + b$
Subtraction	$a$ minus b the difference of $a$ and b $a$ decreased by b b less than $a$ b subtracted from $a$	$a - b$
Multiplication	$a$ times b the product of $a$ and b twice $a$	$a \cdot b, a b, a (b), (a) (b)$ 2a
Division	$a$ divided by b the quotient of $a$ and b the ratio of $a$ and b b divided into $a$	$a \div b, a / b, \frac{a}{b}, b a$

Look closely at these phrases using the four operations:

Four phrases are shown. The first reads “the sum of a and b”, where the words “of” and “and” are written in red. The second reads “the difference of a and b”, where the words “of” and “and” are written in red. The third reads “the product of a and b”, where the words “of” and “and” are written in red. The fourth reads “the quotient of a and b”, where the words “of” and “and” are written in red.

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

Translate each English phrase into an algebraic expression: ⓐ the difference of $17 x$ and $5$ ⓑ the quotient of $10 x^{2}$ and $7 .$

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 key word is “quotient,” which tells us the operation is division.

The phrase “the quotient of 10x squared and 7”, where the words “of” and “and” are written in red, is written above the expression “divide 10x squared by 7”. The expression written below reads “10x squared, division sign,v7”.

This can also be written $10 x^{2} / 7 or \frac{10 x^{2}}{7} .$

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 8 added to your present age. How old were you seven years ago? This is 7 years less than your age now. You subtract 7 from your present age. Seven “less than” means 7 subtracted from your present age.

Translate the English phrase into an algebraic expression: ⓐ Seventeen more than y ⓑ Nine less than $9 x^{2} .$

Solution

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

$\begin{matrix} Seventeen more than y \\ Seventeen added to y \\ y + 17 \end{matrix}$
ⓑ The key words are less than. They tell us to subtract. Less than means “subtracted from.”

$\begin{matrix} Nine less than 9 x^{2} \\ Nine subtracted from 9 x^{2} \\ 9 x^{2} - 9 \end{matrix}$

Translate the English phrase into an algebraic expression: ⓐ five times the sum of m and n ⓑ the sum of five times m and n.

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, $(m + n) .$ This forces us to determine the sum first. (Remember the order of operations.)

$\begin{matrix} five times the sum of m and n \\ 5 (m + n) \end{matrix}$
ⓑ 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.

$\begin{matrix} the sum of five times m and n \\ 5 m + n \end{matrix}$

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

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

Solution

This table illustrates the step-by-step process of translating the verbal phrase "6 less than the width" into its algebraic expression, w-6.
Write a phrase about the width of the rectangle.	6 less than the length
Substitute $l$ for "the length."	6 less than $l$
Rewrite "less than" as "subtracted from."	6 subtracted from $l$
Translate the phrase into algebra.	$l - 6$

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

Solution

Step-by-step translation of a verbal phrase about dimes into an algebraic expression.
Write a phrase about the number of dimes.	three less than four times the number of quarters
Substitute $q$ for the number of quarters.	3 less than 4 times $q$
Translate "4 times $q$ ."	3 less than $4 q$
Translate the phase into algebra.	$4 q - 3$

Key Concepts

Notation The result is…

$\begin{matrix} \circ a + b & the sum of a and b \\ \circ a - b & the difference of a and b \\ \circ a \cdot b, a b, (a) (b) (a) b, a (b) & the product of a and b \\ \circ a \div b, a / b, \frac{a}{b}, b a & the quotient of a and b \end{matrix}$
Inequality

$\begin{matrix} \circ a b is read “ a is greater than b ” & a is to the right of b on the number line \end{matrix}$
Inequality Symbols Words

$\begin{matrix} \circ a \neq b & a is not equal to b \\ \circ a b & a is greater than b \\ \circ a \geq b & a is greater than or equal to b \end{matrix}$
Grouping Symbols
- Parentheses $()$
- Brackets $[]$
- Braces ${}$
Exponential Notation
- $a^{n}$ means multiply $a$ by itself, $n$ times. The expression $a^{n}$ is read $a$ to the $n^{t h}$ power.
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.
Combine Like Terms
1. Identify like terms.
2. Rearrange the expression so like terms are together.
3. Add or subtract the coefficients and keep the same variable for each group of like terms.

Practice Makes Perfect

Use Variables and Algebraic Symbols

In the following exercises, translate from algebra to English.

$16 - 9$

Solution

16 minus 9, the difference of sixteen and nine

$3 \cdot 9$

$28 \div 4$

Solution

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

$x + 11$

$(2) (7)$

Solution

2 times 7, the product of two and seven

$(4) (8)$

$14 < 21$

Solution

fourteen is less than twenty-one

$17 < 35$

$36 \geq 19$

Solution

thirty-six is greater than or equal to nineteen

$6 n = 36$

$y - 1 > 6$

Solution

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

$y - 4 > 8$

$2 \leq 18 \div 6$

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

$a \neq 1 \cdot 12$

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

$9 \cdot 6 = 54$

Solution

equation

$7 \cdot 9 = 63$

$5 \cdot 4 + 3$

Solution

expression

$x + 7$

$x + 9$

Solution

expression

$y - 5 = 25$

Simplify Expressions Using the Order of Operations

In the following exercises, simplify each expression.

$5^{3}$

Solution

125

$8^{3}$

$2^{8}$

Solution

256

$10^{5}$

In the following exercises, simplify using the order of operations.

ⓐ $3 + 8 \cdot 5$ ⓑ $(3 + 8) \cdot 5$

Solution

ⓐ 43 ⓑ 55

ⓐ $2 + 6 \cdot 3$ ⓑ $(2 + 6) \cdot 3$

$2^{3} - 12 \div (9 - 5)$

Solution

$3^{2} - 18 \div (11 - 5)$

$3 \cdot 8 + 5 \cdot 2$

Solution

$4 \cdot 7 + 3 \cdot 5$

$2 + 8 (6 + 1)$

Solution

$4 + 6 (3 + 6)$

$4 \cdot 12 / 8$

Solution

$2 \cdot 36 / 6$

$(6 + 10) \div (2 + 2)$

Solution

$(9 + 12) \div (3 + 4)$

$20 \div 4 + 6 \cdot 5$

Solution

$33 \div 3 + 8 \cdot 2$

$3^{2} + 7^{2}$

Solution

${(3 + 7)}^{2}$

$3 (1 + 9 \cdot 6) - 4^{2}$

Solution

149

$5 (2 + 8 \cdot 4) - 7^{2}$

$2 [1 + 3 (10 - 2)]$

Solution

$5 [2 + 4 (3 - 2)]$

Evaluate an Expression

In the following exercises, evaluate the following expressions.

$7 x + 8$ when $x = 2$

Solution

$8 x - 6$ when $x = 7$

$x^{2}$ when $x = 12$

Solution

144

$x^{3}$ when $x = 5$

$x^{5}$ when $x = 2$

Solution

$4^{x}$ when $x = 2$

$x^{2} + 3 x - 7$ when $x = 4$

Solution

$6 x + 3 y - 9$ when
$x = 6, y = 9$

${(x - y)}^{2}$ when
$x = 10, y = 7$

Solution

${(x + y)}^{2}$ when $x = 6, y = 9$

$a^{2} + b^{2}$ when $a = 3, b = 8$

Solution

$r^{2} - s^{2}$ when $r = 12, s = 5$

$2 l + 2 w$ when
$l = 15, w = 12$

Solution

$2 l + 2 w$ when
$l = 18, w = 14$

Simplify Expressions by Combining Like Terms

In the following exercises, identify the coefficient of each term.

Solution

13m

$5 r^{2}$

Solution

$6 x^{3}$

In the following exercises, identify the like terms.

$x^{3}, 8 x, 14, 8 y, 5, 8 x^{3}$

Solution

$x^{3} and 8 x^{3}, 14 and 5$

$6 z, 3 w^{2}, 1, 6 z^{2}, 4 z, w^{2}$

$9 a, a^{2}, 16, 16 b^{2}, 4, 9 b^{2}$

Solution

$16 and 4, 16 b^{2} and 9 b^{2}$

$3, 25 r^{2}, 10 s, 10 r, 4 r^{2}, 3 s$

In the following exercises, identify the terms in each expression.

$15 x^{2} + 6 x + 2$

Solution

$15 x^{2}, 6 x, 2$

$11 x^{2} + 8 x + 5$

$10 y^{3} + y + 2$

Solution

$10 y^{3}, y, 2$

$9 y^{3} + y + 5$

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

$10 x + 3 x$

Solution

13x

$15 x + 4 x$

$4 c + 2 c + c$

Solution

$6 y + 4 y + y$

$7 u + 2 + 3 u + 1$

Solution

$10 u + 3$

$8 d + 6 + 2 d + 5$

$10 a + 7 + 5 a - 2 + 7 a - 4$

Solution

$22 a + 1$

$7 c + 4 + 6 c - 3 + 9 c - 1$

$3 x^{2} + 12 x + 11 + 14 x^{2} + 8 x + 5$

Solution

$17 x^{2} + 20 x + 16$

$5 b^{2} + 9 b + 10 + 2 b^{2} + 3 b - 4$

Translate an English Phrase to an Algebraic Expression

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

the difference of 14 and 9

Solution

$14 - 9$

the difference of 19 and 8

the product of 9 and 7

Solution

$9 \cdot 7$

the product of 8 and 7

the quotient of 36 and 9

Solution

$36 \div 9$

the quotient of 42 and 7

the sum of 8x and 3x

Solution

$8 x + 3 x$

the sum of 13x and 3x

the quotient of y and 3

Solution

$\frac{y}{3}$

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

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.

Solution

$c + 3$

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.

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

$2 n - 7$

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

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?
ⓑ how much will his insurance company pay?

Solution

ⓐ $750 ⓑ $1,350

Home insurance Armando’s home insurance has a $2,500 deductible per incident. This means that he pays $2,500 and the insurance company will pay all costs beyond $2,500. If Armando files a claim for $19,400.

ⓐ how much will he pay?
ⓑ how much will the insurance company pay?

Writing Exercises

Explain the difference between an expression and an equation.

Solution

Answers may vary

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

Explain how you identify the like terms in the expression $8 a^{2} + 4 a + 9 - a^{2} - 1 .$

Solution

Answers may vary

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

Self Check

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

A table is shown that is composed of four columns and six rows. The header row reads, from left to right, “I can …”, “Confidently”, “With some help” and “No – I don’t get it!”. The phrases in the first column read “use variables and algebraic symbols.”, “simplify expressions using the order of operations.”, “evaluate an expression.”, “identify and combine like terms.”, and “translate English phrases to algebraic expressions.”

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

coefficient

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

constant

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

equality symbol

The symbol “

=

” is called the equal sign. We read

a = b

as “

a

is equal to

b

.”

equation

An equation is two expressions connected by an equal sign.

evaluate an expression

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

expression

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

like terms

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

simplify an expression

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

term

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

variable

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

Add and Subtract Integers

Use Negatives and Opposites

Our work so far has only included the counting numbers and the whole numbers. But if you have ever experienced a temperature below zero or accidentally overdrawn your checking account, you are already familiar with negative numbers. Negative numbers are numbers less than $0 .$ The negative numbers are to the left of zero on the number line. See Figure 1.

A number line extends from negative 4 to 4. A bracket is under the values “negative 4” to “0” and is labeled “Negative numbers”. Another bracket is under the values 0 to 4 and labeled “positive numbers”. There is an arrow in between both brackets pointing upward to zero. — The number line shows the location of positive and negative numbers.

The arrows on the ends of the number line indicate that the numbers keep going forever. There is no biggest positive number, and there is no smallest negative number.

Is zero a positive or a negative number? Numbers larger than zero are positive, and numbers smaller than zero are negative. Zero is neither positive nor negative.

Consider how numbers are ordered on the number line. Going from left to right, the numbers increase in value. Going from right to left, the numbers decrease in value. See Figure 2.

A number line ranges from negative 4 to 4. An arrow above the number line extends from negative 1 towards 4 and is labeled “larger”. An arrow below the number line extends from 1 towards negative 4 and is labeled “smaller”. — The numbers on a number line increase in value going from left to right and decrease in value going from right to left.

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.

a > b (read “a is greater than b”) when a is to the right of b on the number line.

Now we need to extend the number line which showed the whole numbers to include negative numbers, too. The numbers marked by points in Figure 3 are called the integers. The integers are the numbers $… - 3, −2, −1, 0, 1, 2, 3 …$

A number line extends from negative four to four. Points are plotted at negative four, negative three, negative two, negative one, zero, one, two, 3, and four. — All the marked numbers are called *integers*.

Order each of the following pairs of numbers, using < or >: ⓐ $14___6$ ⓑ $−1___9$ ⓒ $−1___−4$ ⓓ $2___−20 .$

Solution

It may be helpful to refer to the number line shown.
A number line ranges from negative twenty to fifteen with ticks marks between numbers. Every fifth tick mark is labeled a number. Points are plotted at points negative twenty, negative 4, negative 1, 2, 6, 9 and 14.

This table demonstrates the correspondence between the relative positions of numbers on a number line and their mathematical inequalities.
ⓐ 14 is to the right of 6 on the number line.	$\begin{matrix} 14___6 \\ 14 > 6 \end{matrix}$
ⓑ −1 is to the left of 9 on the number line.	$\begin{matrix} −1___9 \\ −1 < 9 \end{matrix}$
ⓒ −1 is to the right of −4 on the number line.	$\begin{matrix} −1___−4 \\ −1 > −4 \end{matrix}$
ⓓ 2 is to the right of −20 on the number line.	$\begin{matrix} 2___−20 \\ 2 > −20 \end{matrix}$

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

Figure 4 illustrates the definition.

A number line ranges from negative 4 to 4. There are two brackets above the number line. The bracket on the left spans from negative three to 0. The bracket on the right spans from zero to three. Points are plotted on both negative three and three. — The opposite of 3 is $−3 .$

Sometimes in algebra the same symbol has different meanings. Just like some words in English, the specific meaning becomes clear by looking at how it is used. You have seen the symbol “−” used in three different ways.

\begin{matrix} 10 - 4 & Between two numbers, it indicates the operation of subtraction . \\ We read 10 - 4 as “ 10 minus 4. ” \\ −8 & In front of a number, it indicates a negative number. \\ We read −8 as “negative eight.” \\ − x & In front of a variable, it indicates the opposite . We read − x as “the opposite of x . ” \\ − (−2) & Here there are two “ - ” signs. The one in the parentheses tells us the number is \\ negative 2 . The one outside the parentheses tells us to take the opposite of −2 . \\ We read − (−2) as “the opposite of negative two.” \end{matrix}

This table illustrates the diverse meanings and interpretations of the "−" symbol in mathematical expressions, including subtraction, negative numbers, and opposites.
$10 - 4$	Between two numbers, it indicates the operation of subtraction. We read $10 - 4$ as "10 minus 4."
$−8$	In front of a number, it indicates a negative number. We read −8 as "negative eight."
$− x$	In front of a variable, it indicates the opposite. We read $− x$ as "the opposite of $x$ ."
$− (−2)$	Here there are two "−" signs. The one in the parentheses tells us the number is negative 2. The one outside the parentheses tells us to take the opposite of −2. We read $− (−2)$ as "the opposite of negative two."

Find: ⓐ the opposite of 7 ⓑ the opposite of $−10$ ⓒ $− (−6) .$

Solution

ⓐ −7 is the same distance from 0 as 7, but on the opposite side of 0.	The opposite of 7 is −7.
ⓑ 10 is the same distance from 0 as −10, but on the opposite side of 0.	The opposite of −10 is 10.
ⓒ −(−6) (the opposite of –6).	The opposite of −(−6) is 6.

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

When evaluating the opposite of a variable, we must be very careful. Without knowing whether the variable represents a positive or negative number, we don’t know whether $− x$ is positive or negative. We can see this in Example 3.

Evaluate ⓐ $− x,$ when $x = 8$ ⓑ $− x,$ when $x = −8 .$

Solution

ⓐ

−x

Write the opposite of 8.
ⓑ

−x

Write the opposite of −8. 8

Simplify: Expressions with Absolute Value

We saw that numbers such as $2 and −2$ are opposites because they are the same distance from 0 on the number line. They are both two units from 0. The distance between 0 and any number on the number line is called the absolute value of that number.

For example,

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

Figure 5 illustrates this idea.

A number line is shown ranging from negative 5 to 5. A bracket labeled “5 units” lies above the points negative 5 to 0. An arrow labeled “negative 5 is 5 units from 0, so absolute value of negative 5 equals 5.” is written above the labeled bracket. A bracket labeled “5 units” lies above the points “0” to “5”. An arrow labeled “5 is 5 units from 0, so absolute value of 5 equals 5.” and is written above the labeled bracket. — The integers $5 and are 5$ units away from $0 .$

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

Mathematicians say it more precisely, “absolute values are always non-negative.” Non-negative means greater than or equal to zero.

Simplify: ⓐ $| 3 |$ ⓑ $| −44 |$ ⓒ $| 0 |$ .

Solution

The absolute value of a number is the distance between the number and zero. Distance is never negative, so the absolute value is never negative.

ⓐ $| 3 |$
$3$

ⓑ $| −44 |$
$44$

ⓒ $| 0 |$
$0$

In the next example, we’ll order expressions with absolute values. Remember, positive numbers are always greater than negative numbers!

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

ⓐ $| −5 |___- | −5 |$ ⓑ $8___- | −8 |$ ⓒ $−9___- | −9 |$ ⓓ $− (−16)___- | −16 |$

Solution

This table demonstrates the simplification of expressions involving absolute values and negative numbers, followed by their comparison (ordering) using inequality or equality signs.
ⓐ Simplify. Order.	$\begin{matrix} \| −5 \| & ___ & - \| −5 \| \\ 5 & ___ & - 5 \\ 5 & > & −5 \\ \| −5 \| & > & - \| −5 \| \end{matrix}$
ⓑ Simplify. Order.	$\begin{matrix} 8 & ___ & - \| −8 \| \\ 8 & ___ & - 8 \\ 8 & > & −8 \\ 8 & > & - \| −8 \| \end{matrix}$
ⓒ Simplify. Order.	$\begin{matrix} −9 & ___ & - \| −9 \| \\ −9 & ___ & - 9 \\ −9 & = & −9 \\ −9 & = & - \| −9 \| \end{matrix}$
ⓓ Simplify. Order.	$\begin{matrix} − (−16) & ___ & - \| −16 \| \\ 16 & ___ & - 16 \\ 16 & > & −16 \\ − (−16) & > & - \| −16 \| \end{matrix}$

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

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

Simplify: $24 - | 19 - 3 (6 - 2) | .$

Solution

Step-by-step solution demonstrating the order of operations for simplifying a mathematical expression.
	$24 - \| 19 - 3 (6 - 2) \|$
Work inside parentheses first: subtract 2 from 6.	$24 - \| 19 - 3 (4) \|$
Multiply 3(4).	$24 - \| 19 - 12 \|$
Subtract inside the absolute value bars.	$24 - \| 7 \|$
Take the absolute value.	$24 - 7$
Subtract.	$17$

Evaluate: ⓐ $| x | when x = −35$ ⓑ $| − y | when y = −20$ ⓒ $− | u | when u = 12$ ⓓ $− | p | when p = −14 .$

Solution

ⓐ

| x | when x = −35

	$\| x \|$

Take the absolute value.	35

ⓑ

| − y | when y = −20

	$\| - y \|$

Simplify.	$\| 20 \|$
Take the absolute value.	20

ⓒ

− | u | when u = 12

	$- \| u \|$

Take the absolute value.	$- 12$

ⓓ

− | p | when p = −14

	$- \| p \|$

Take the absolute value.	$- 14$

Add 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 challenging.

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

We let one color (blue) represent positive. The other color (red) will represent the negatives. If we have one positive counter and one negative counter, the value of the pair is zero. They form a neutral pair. The value of this neutral pair is zero.

In this image we have a blue counter above a red counter with a circle around both. The equation to the right is 1 plus negative 1 equals 0.

We will use the counters to show how to add the four addition facts using the numbers $5, −5$ and $3, −3 .$

\begin{matrix} 5 + 3 & −5 + (−3) & −5 + 3 & 5 + (−3) \end{matrix}

To add $5 + 3,$ we realize that $5 + 3$ means the sum of 5 and 3.

We start with 5 positives.
And then we add 3 positives.
We now have 8 positives. The sum of 5 and 3 is 8.

Now we will add $−5 + (−3) .$ Watch for similarities to the last example $5 + 3 = 8 .$

To add $−5 + (−3),$ we realize this means the sum of $−5 and - 3 .$

We start with 5 negatives.
And then we add 3 negatives.
We now have 8 negatives. The sum of −5 and −3 is −8.

In what ways were these first two examples 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 8—either 8 positives or 8 negatives.

When the signs were the same, the counters were all the same color, and so we added them.

This figure is divided into two columns. In the left column there are eight blue counters in a horizontal row. Under them is the text “8 positives.” Centered under this is the equation 5 plus 3 equals 8. In the right column are eight red counters in a horizontal row which are labled below with the phrase “8 negatives”. Centered under this is the equation negative 5 plus negative 3 equals negative 8, where negative 3 is in parentheses.

Add: ⓐ $1 + 4$ ⓑ $−1 + (−4) .$

Solution

ⓐ

Five light blue circles arranged horizontally, illustrating the sum of 1 and 4, with '1 + 4' and '5' written next to them, representing a basic addition problem.

1 positive plus 4 positives is 5 positives.

ⓑ

Five red circles visually represent the sum of -1 and -4, equaling -5, illustrating the addition of negative numbers.

1 negative plus 4 negatives is 5 negatives.

So what happens when the signs are different? Let’s add $−5 + 3 .$ We realize this means the sum of $−5$ and 3. When the counters were the same color, we put them in a row. When the counters are a different color, we line them up under each other.

	−5 + 3 means the sum of −5 and 3.
We start with 5 negatives.
And then we add 3 positives.
We remove any neutral pairs.
We have 2 negatives left.
The sum of −5 and 3 is −2.	−5 + 3 = −2

Notice that there were more negatives than positives, so the result was negative.

Let’s now add the last combination, $5 + (− 3) .$

	5 + (−3) means the sum of 5 and −3.
We start with 5 positives.
And then we add 3 negatives.
We remove any neutral pairs.
We have 2 positives left.
The sum of 5 and −3 is 2.	5 + (−3) = 2

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

Two images are shown and labeled. The left image shows five red counters in a horizontal row drawn above three blue counters in a horizontal row, where the first three pairs of red and blue counters are circled. Above this diagram is written “negative 5 plus 3” and below is written “More negatives – the sum is negative.” The right image shows five blue counters in a horizontal row drawn above three red counters in a horizontal row, where the first three pairs of red and blue counters are circled. Above this diagram is written “5 plus negative 3” and below is written “More positives – the sum is positive.”

Add: ⓐ $−1 + 5$ ⓑ $1 + (−5) .$

Solution

ⓐ

	−1 + 5

There are more positives, so the sum is positive.	4

ⓑ

	1 + (−5)

There are more negatives, so the sum is negative.	−4

Now that we have added small positive and negative integers with a model, we can visualize the model in our minds to simplify problems with any numbers.

When you need to add numbers such as $37 + (− 53),$ you really don’t want to have to count out 37 blue counters and 53 red counters. With the model in your mind, can you visualize what you would do to solve the problem?

Picture 37 blue counters with 53 red counters lined up underneath. Since there would be more red (negative) counters than blue (positive) counters, the sum would be negative. How many more red counters would there be? Because $53 - 37 = 16,$ there are 16 more red counters.

Therefore, the sum of $37 + (− 53)$ is $−16 .$

37 + (− 53) = −16

Let’s try another one. We’ll add $−74 + (− 27) .$ Again, imagine 74 red counters and 27 more red counters, so we’d have 101 red counters. This means the sum is $−101 .$

−74 + (− 27) = −101

Let’s look again at the results of adding the different combinations of $5, −5$ and $3, −3 .$

Visualize the model as you simplify the expressions in the following examples.

Simplify: ⓐ $19 + (−47)$ ⓑ $−14 + (−36) .$

Solution

ⓐ Since the signs are different, we subtract $19 from 47 .$ The answer will be negative because there are more negatives than positives.
$\begin{matrix} 19 + (−47) \\ Add. & −28 \end{matrix}$
ⓑ Since the signs are the same, we add. The answer will be negative because there are only negatives.
$\begin{matrix} −14 + (−36) \\ Add. & −50 \end{matrix}$

The techniques used up to now extend to more complicated problems, like the ones we’ve seen before. Remember to follow the order of operations!

Simplify: $−5 + 3 (−2 + 7) .$

Solution

Step-by-step simplification of the expression -5+3(-2+7) demonstrating the order of operations, resulting in 10.
	$−5 + 3 (−2 + 7)$
Simplify inside the parentheses.	$−5 + 3 (5)$
Multiply.	$−5 + 15$
Add left to right.	$10$

Subtract Integers

We will continue to use counters to model the 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 $“ 5$ take away $3. ”$ When you use counters, you can think of subtraction the same way!

We will model the four subtraction facts using the numbers $5$ and $3 .$

\begin{matrix} 5 - 3 & −5 - (−3) & −5 - 3 & 5 - (−3) \end{matrix}

To subtract $5 - 3,$ we restate the problem as $“ 5$ take away $3. ”$

We start with 5 positives.
We ‘take away’ 3 positives.
We have 2 positives left.
The difference of 5 and 3 is 2.	2

Now we will subtract $−5 - (−3) .$ Watch for similarities to the last example $5 - 3 = 2 .$

To subtract $−5 - (−3),$ we restate this as $“ –5$ take away $–3 ”$

We start with 5 negatives.
We ‘take away’ 3 negatives.
We have 2 negatives left.
The difference of −5 and −3 is −2.	−2

Notice that these two examples are much alike: The first example, we subtract 3 positives from 5 positives and end up with 2 positives.

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

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

Two images are shown and labeled. The first image shows five blue counters, three of which are circled with an arrow. Above the counters is the equation “5 minus 3 equals 2.” The second image shows five red counters, three of which are circled with an arrow. Above the counters is the equation “negative 5, minus, negative 3, equals negative 2.”

Subtract: ⓐ $7 - 5$ ⓑ $−7 - (−5) .$

Solution

Illustrates subtraction scenarios for positive and negative integers with both verbal and mathematical forms.
ⓐ Take 5 positive from 7 positives and get 2 positives.	$\begin{matrix} 7 - 5 \\ 2 \end{matrix}$
ⓑ Take 5 negatives from 7 negatives and get 2 negatives.	$\begin{matrix} −7 - (−5) \\ −2 \end{matrix}$

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

To subtract $−5 - 3,$ we restate it as $−5$ take away 3.

We start with 5 negatives. We need to take away 3 positives, but we do not have any positives to take away.

Remember, a neutral pair has value zero. If we add 0 to 5 its value is still 5. We add neutral pairs to the 5 negatives until we get 3 positives to take away.

	−5 − 3 means −5 take away 3.
We start with 5 negatives.
We now add the neutrals needed to get 3 positives.
We remove the 3 positives.
We are left with 8 negatives.
The difference of −5 and 3 is −8.	−5 − 3 = −8

And now, the fourth case, $5 - (−3) .$ We start with 5 positives. We need to take away 3 negatives, but there are no negatives to take away. So we add neutral pairs until we have 3 negatives to take away.

	5 − (−3) means 5 take away −3.
We start with 5 positives.
We now add the needed neutrals pairs.
We remove the 3 negatives.
We are left with 8 positives.
The difference of 5 and −3 is 8.	5 − (−3) = 8

Subtract: ⓐ $−3 - 1$ ⓑ $3 - (−1) .$

Solution

ⓐ

Take 1 positive from the one added neutral pair.

A row of five pale orange circles with red outlines on a white background, with a gap between the fourth and fifth circles.

A light blue circle is encircled by a purple elliptical shape with an arrow indicating a clockwise rotation, all set against a white background.

−3 − 1

−4

ⓑ

Take 1 negative from the one added neutral pair.

A row of four light blue circles with a thin, darker blue outline, set against a plain white background. The circles are evenly spaced and identical in appearance.

Red circle with orange fill and a purple elliptical arrow around it, pointing clockwise.

3 − (−1)

4

Have you noticed that subtraction of signed numbers can be done by adding the opposite? In Example 13, $−3 - 1$ is the same as $−3 + (−1)$ and $3 - (−1)$ is the same as $3 + 1 .$ You will often see this idea, the subtraction property, written as follows:

Look at these two examples.

Two images are shown and labeled. The first image shows four gray spheres drawn next to two gray spheres, where the four are circled in red, with a red arrow leading away to the lower left. This drawing is labeled above as “6 minus 4” and below as “2.” The second image shows four gray spheres and four red spheres, drawn one above the other and circled in red, with a red arrow leading away to the lower left, and two gray spheres drawn to the side of the four gray spheres. This drawing is labeled above as “6 plus, open parenthesis, negative 4, close parenthesis” and below as “2.”

6 - 4 gives the same answer as 6 + (−4) .

Of course, when you have a subtraction problem that has only positive numbers, like $6 - 4,$ you just do the subtraction. You already knew how to subtract $6 - 4$ long ago. But knowing that $6 - 4$ gives the same answer as $6 + (−4)$ helps when you are subtracting negative numbers. Make sure that you understand how $6 - 4$ and $6 + (−4)$ give the same results!

Simplify: ⓐ $13 - 8$ and $13 + (−8)$ ⓑ $−17 - 9$ and $−17 + (−9) .$

Solution

Examples demonstrating that subtracting a number is equivalent to adding its opposite, showing original subtraction problems and their corresponding addition forms.
ⓐ Subtract.	$\begin{matrix} 13 - 8 \\ 5 \end{matrix}$	$\begin{matrix} 13 + (−8) \\ 5 \end{matrix}$
ⓑ Subtract.	$\begin{matrix} −17 - 9 \\ −26 \end{matrix}$	$\begin{matrix} −17 + (−9) \\ −26 \end{matrix}$

Look at what happens when we subtract a negative.

This figure is divided vertically into two halves. The left part of the figure contains the expression 8 minus negative 5, where negative 5 is in parentheses. The expression sits above a group of 8 blue counters next to a group of five blue counters in a row, with a space between the two groups. Underneath the group of five blue counters is a group of five red counters, which are circled. The circle has an arrow pointing away toward bottom left of the image, symbolizing subtraction. Below the counters is the number 13. The right part of the figure contains the expression 8 plus 5. The expression sits above a group of 8 blue counters next to a group of five blue counters in a row, with a space between the two groups. Underneath the counters is the number 13.

8 - (−5) gives the same answer as 8 + 5

Subtracting a negative number is like adding a positive!

You will often see this written as $a - (− b) = a + b .$

Does that work for other numbers, too? Let’s do the following example and see.

Simplify: ⓐ $9 - (−15)$ and $9 + 15$ ⓑ $−7 - (−4)$ and $−7 + 4 .$

Solution

This table illustrates integer subtraction by converting problems into equivalent addition problems with the same solution.
ⓐ Subtract.	$\begin{matrix} 9 - (−15) \\ 24 \end{matrix}$	$\begin{matrix} 9 + 15 \\ 24 \end{matrix}$
ⓑ Subtract.	$\begin{matrix} −7 - (−4) \\ −3 \end{matrix}$	$\begin{matrix} −7 + 4 \\ −3 \end{matrix}$

Let’s look again at the results of subtracting the different combinations of $5, −5$ and $3, −3 .$

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

Simplify: $7 - (−4 - 3) - 9 .$

Solution

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

Key Concepts

Addition of Positive and Negative Integers
$\begin{matrix} 5 + 3 & - 5 + (−3) \\ 8 & −8 \\ both positive, & both negative, \\ sum positive & sum negative \\ −5 + 3 & 5 + (−3) \\ −2 & 2 \\ different signs, & different signs, \\ more negatives & more positives \\ sum negative & sum positive \end{matrix}$
Property of Absolute Value: $| n | \geq 0$ for all numbers. Absolute values are always greater than or equal to zero!
Subtraction of Integers
$\begin{matrix} 5 - 3 & −5 - (−3) \\ 2 & −2 \\ 5 positives & 5 negatives \\ take away 3 positives & take away 3 negatives \\ 2 positives & 2 negatives \\ −5 - 3 & 5 - (−3) \\ −8 & 8 \\ 5 negatives, want to & 5 positives, want to \\ subtract 3 positives & subtract 3 negatives \\ need neutral pairs & need neutral pairs \end{matrix}$
Subtraction Property: Subtracting a number is the same as adding its opposite.

Practice Makes Perfect

Use Negatives and Opposites of Integers

In the following exercises, order each of the following pairs of numbers, using < or >.

ⓐ $9___4$
ⓑ $−3___6$
ⓒ $−8___−2$
ⓓ $1___−10$

Solution

ⓐ > ⓑ < ⓒ < ⓓ >

ⓐ $−7___3$
ⓑ $−10___−5$
ⓒ $2___−6$
ⓓ $8___9$

In the following exercises, find the opposite of each number.

ⓐ 2 ⓑ $−6$

Solution

ⓐ $−2$ ⓑ 6

ⓐ 9ⓑ $−4$

In the following exercises, simplify.

$− (−4)$

Solution

$− (−8)$

$− (−15)$

Solution

$− (−11)$

In the following exercises, evaluate.

$− c$ whenⓐ $c = 12$ ⓑ $c = −12$

Solution

ⓐ $−12$ ⓑ 12

$− d$ when
ⓐ $d = 21$
ⓑ $d = −21$

Simplify Expressions with Absolute Value

In the following exercises, simplify.

ⓐ $| −32 |$ ⓑ $| 0 |$ ⓒ $| 16 |$

Solution

ⓐ 32 ⓑ 0 ⓒ 16

ⓐ $| 0 |$
ⓑ $| −40 |$ ⓒ $| 22 |$

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

ⓐ $−6___| −6 |$ ⓑ $− | −3 |___−3$

Solution

ⓐ < ⓑ $=$

ⓐ $| −5 |___− | −5 |$ ⓑ $9___− | −9 |$

In the following exercises, simplify.

$− (−5) and − | −5 |$

Solution

$5, −5$

$− | −9 | and − (−9)$

$8 | −7 |$

Solution

$5 | −5 |$

$| 15 - 7 | - | 14 - 6 |$

Solution

$| 17 - 8 | - | 13 - 4 |$

$18 - | 2 (8 - 3) |$

Solution

$18 - | 3 (8 - 5) |$

In the following exercises, evaluate.

ⓐ $− | p | when p = 19$
ⓑ $− | q | when q = −33$

Solution

ⓐ $−19$ ⓑ $−33$

ⓐ $− | a | when a = 60$
ⓑ $− | b | when b = −12$

Add Integers

In the following exercises, simplify each expression.

$−21 + (−59)$

Solution

$−80$

$−35 + (−47)$

$48 + (−16)$

Solution

$34 + (−19)$

$−14 + (−12) + 4$

Solution

$−22$

$−17 + (−18) + 6$

$135 + (−110) + 83$

Solution

108

$−38 + 27 + (−8) + 12$

$19 + 2 (−3 + 8)$

Solution

$24 + 3 (−5 + 9)$

Subtract Integers

In the following exercises, simplify.

$8 - 2$

Solution

$−6 - (−4)$

$−5 - 4$

Solution

$−9$

$−7 - 2$

$8 - (−4)$

Solution

$7 - (−3)$

ⓐ $44 - 28$ ⓑ $44 + (−28)$

Solution

ⓐ 16 ⓑ 16

ⓐ $35 - 16$ ⓑ $35 + (−16)$

ⓐ $27 - (−18)$ ⓑ $27 + 18$

Solution

ⓐ 45 ⓑ 45

ⓐ $46 - (−37)$ ⓑ $46 + 37$

In the following exercises, simplify each expression.

$15 - (−12)$

Solution

$14 - (−11)$

$48 - 87$

Solution

$−39$

$45 - 69$

$−17 - 42$

Solution

$−59$

$−19 - 46$

$−103 - (−52)$

Solution

$−51$

$−105 - (−68)$

$−45 - (–54)$

Solution

$−58 - (−67)$

$8 - 3 - 7$

Solution

$−2$

$9 - 6 - 5$

$−5 - 4 + 7$

Solution

$−2$

$−3 - 8 + 4$

$−14 - (−27) + 9$

Solution

$64 + (−17) - 9$

$(2 - 7) - (3 - 8)$

Solution

$(1 - 8) - (2 - 9)$

$− (6 - 8) - (2 - 4)$

Solution

$− (4 - 5) - (7 - 8)$

$25 - [10 - (3 - 12)]$

Solution

$32 - [5 - (15 - 20)]$

$6 \cdot 3 - 4 \cdot 3 - 7 \cdot 2$

Solution

$−8$

$5 \cdot 7 - 8 \cdot 2 - 4 \cdot 9$

$5^{2} - 6^{2}$

Solution

$−11$

$6^{2} - 7^{2}$

Everyday Math

Elevation The highest elevation in the United States is Mount McKinley, Alaska, at 20,320 feet above sea level. The lowest elevation is Death Valley, California, at 282 feet below sea level.

Use integers to write the elevation of:

ⓐ Mount McKinley. ⓑ Death Valley.

Solution

ⓐ 20,320 feet ⓑ −282 feet

Extreme temperatures The highest recorded temperature on Earth was $57 °$ Celsius. The lowest recorded temperature was $90 ° below 0 °$ Celsius.

Use integers to write the:

ⓐ highest recorded temperature.
ⓑ 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 of:

ⓐ Pennsylvania.
ⓑ Texas.

Solution

ⓐ $540 million ⓑ $− $27$ billion

College enrollments Across the United States, community college enrollment grew by 1,400,000 students from Fall 2007 to Fall 2010. In California, community college enrollment declined by 110,171 students from Fall 2009 to Fall 2010.

Use integers to write the change in enrollment:

ⓐ in the U.S. from Fall 2007 to Fall 2010.
ⓑ in California from Fall 2009 to Fall 2010.

Stock Market The week of September 15, 2008 was one of the most volatile weeks ever for the US stock market. The closing numbers of the Dow Jones Industrial Average each day were:

Monday	$−504$
Tuesday	$+ 142$
Wednesday	$−449$
Thursday	$+ 410$
Friday	$+ 369$

What was the overall change for the week? Was it positive or negative?

Solution

32, negative

Stock Market During the week of June 22, 2009, the closing numbers of the Dow Jones Industrial Average each day were:

Monday	$−201$
Tuesday	$−16$
Wednesday	$−23$
Thursday	$+ 172$
Friday	$−34$

What was the overall change for the week? Was it positive or negative?

Writing Exercises

Give an example of a negative number from your life experience.

Solution

Answers may vary

What are the three uses of the $“ - ”$ sign in algebra? Explain how they differ.

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

Solution

Answers may vary

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 table is shown with four columns and five rows. The column titles, from left to right, are “I can …”, “Confidently”, “With some help” and “No – I don’t get it!” The first column includes the phrases “use negatives and opposites of integers.”, “Simplify: expressions with absolute value.”, “add integers.” and “subtract integers.”

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

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 |

integers

The whole numbers and their opposites are called the integers: ...−3, −2, −1, 0, 1, 2, 3...

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:

− a

means the opposite of the number. The notation

− a

is read “the opposite of

a

.”

Multiply and Divide Integers

Multiply Integers

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

We remember that $a \cdot b$ means add a, b times.

Two images are shown side-by-side. The image on the left has the equation five times three at the top. Below this it reads “add 5, 3 times.” Below this depicts three rows of blue counters, with five counters in each row. Under this, it says “15 positives.” Under thisis the equation“5 times 3 equals 15.” The image on the right reads “negative 5 times three. The three is in parentheses. Below this it reads, “add negative five, three times.” Under this are fifteen red counters in three rows of five. Below this it reads” “15 negatives”. Below this is the equation negative five times 3 equals negative 15.”

The next two examples are more interesting.

What does it mean to multiply 5 by $−3 ?$ It means subtract 5, 3 times. Looking at subtraction as “taking away,” it means to take away 5, 3 times. But there is nothing to take away, so we start by adding neutral pairs on the workspace. Then we take away 5 three times.

This figure has two columns. In the top row, the left column contains the expression 5 times negative 3. This means take away 5, three times. Below this, there are three groups of five red negative counters, and below each group of red counters is an identical group of five blue positive counters. What are left are fifteen negatives, represented by 15 red counters. Underneath the counters is the equation 5 times negative 3 equals negative 15. In the top row, the right column contains the expression negative 5 times negative 3. This means take away negative 5, three times. Below this, there are three groups of five blue positive counters, and below each group of blue counters is an identical group of five red negative counters. What are left are fifteen positives, represented by 15 blue counters. Underneath the blue counters is the equation negative 5 times negative 3 equals 15.

In summary:

\begin{matrix} 5 \cdot 3 & = & 15 & −5 (3) & = & −15 \\ 5 (−3) & = & −15 & (−5) (−3) & = & 15 \end{matrix}

Notice that for multiplication of two signed numbers, when the:

signs are the same, the product is positive.
signs are different, the product is negative.

We’ll put this all together in the chart below.

Multiplication of Signed Numbers

For multiplication of two signed numbers:

Same signs	Product	Example
Two positives Two negatives	Positive Positive	$\begin{matrix} 7 \cdot 4 & = & 28 \\ −8 (−6) & = & 48 \end{matrix}$

Different signs	Product	Example
Positive · negative Negative · positive	Negative Negative	$\begin{matrix} 7 (−9) & = & −63 \\ −5 \cdot 10 & = & −50 \end{matrix}$

Multiply: ⓐ $−9 \cdot 3$ ⓑ $−2 (−5)$ ⓒ $4 (−8)$ ⓓ $7 \cdot 6 .$

Solution

Examples of integer multiplication, illustrating how the signs of numbers determine the product's sign.
ⓐ Multiply, noting that the signs are different so the product is negative.	$\begin{matrix} −9 \cdot 3 \\ −27 \end{matrix}$
ⓑ Multiply, noting that the signs are the same so the product is positive.	$\begin{matrix} −2 (−5) \\ 10 \end{matrix}$
ⓒ Multiply, with different signs.	$\begin{matrix} 4 (−8) \\ −32 \end{matrix}$
ⓓ Multiply, with same signs.	$\begin{matrix} 7 \cdot 6 \\ 42 \end{matrix}$

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.

\begin{matrix} −1 \cdot 4 & −1 (−3) \\ Multiply. & −4 & 3 \\ −4 is the opposite of 4 . & 3 is the opposite of −3 . \end{matrix}

Each time we multiply a number by $−1,$ we get its opposite!

Multiply: ⓐ $−1 \cdot 7$ ⓑ $−1 (−11) .$

Solution

Examples demonstrating multiplication by -1, showing how the product's sign is determined and always results in the opposite value of the original number.
ⓐ Multiply, noting that the signs are different so the product is negative.	$\begin{matrix} −1 \cdot 7 \\ −7 \\ −7 is the opposite of 7 . \end{matrix}$
ⓑ Multiply, noting that the signs are the same so the product is positive.	$\begin{matrix} −1 (−11) \\ 11 \\ 11 is the opposite of −11 . \end{matrix}$

Divide Integers

What about division? Division is the inverse operation of multiplication. So, $15 \div 3 = 5$ because $5 \cdot 3 = 15 .$ In words, this expression says that 15 can be divided into three groups of five each because adding five three times gives 15. Look at some examples of multiplying integers, to figure out the rules for dividing integers.

\begin{matrix} 5 \cdot 3 & = & 15 so 15 \div 3 & = & 5 & −5 (3) & = & −15 so −15 \div 3 & = & −5 \\ (−5) (−3) & = & 15 so 15 \div (−3) & = & −5 & 5 (−3) & = & −15 so −15 \div (−3) & = & 5 \end{matrix}

Division follows the same rules as multiplication!

For division of two signed numbers, when the:

signs are the same, the quotient is positive.
signs are different, the quotient is negative.

And remember that we can always check the answer of a division problem by multiplying.

Multiplication and Division of Signed Numbers

For multiplication and division of two signed numbers:

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

Same signs	Result
Two positives Two negatives	Positive Positive
If the signs are the same, the result is positive.

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

Divide: ⓐ $−27 \div 3$ ⓑ $−100 \div (−4) .$

Solution

Examples demonstrating division rules for integers with different and same signs.
ⓐ Divide. With different signs, the quotient is negative.	$\begin{matrix} - 27 \div 3 \\ −9 \end{matrix}$
ⓑ Divide. With signs that are the same, the quotient is positive.	$\begin{matrix} - 100 \div (−4) \\ 25 \end{matrix}$

Simplify Expressions with Integers

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

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

Simplify: $7 (−2) + 4 (−7) - 6 .$

Solution

Step-by-step evaluation of the mathematical expression 7(-2) + 4(-7) - 6, showing the progression of operations.
	$7 (−2) + 4 (−7) - 6$
Multiply first.	$−14 + (−28) - 6$
Add.	$−42 - 6$
Subtract.	$−48$

Simplify: ⓐ ${(−2)}^{4}$ ⓑ $− 2^{4} .$

Solution

Step-by-step evaluation of exponential expressions, differentiating between a negative base raised to a power and the negative of a positive base.
ⓐ Write in expanded form. Multiply. Multiply. Multiply.	$\begin{matrix} {(−2)}^{4} \\ (−2) (−2) (−2) (−2) \\ 4 (−2) (−2) \\ −8 (−2) \\ 16 \end{matrix}$
ⓑ Write in expanded form. We are asked to find the opposite of $2^{4} .$ Multiply. Multiply. Multiply.	$\begin{matrix} − 2^{4} \\ − (2 \cdot 2 \cdot 2 \cdot 2) \\ − (4 \cdot 2 \cdot 2) \\ − (8 \cdot 2) \\ -16 \end{matrix}$

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

The next example reminds us to simplify inside parentheses first.

Simplify: $12 - 3 (9 - 12) .$

Solution

Step-by-step evaluation of a mathematical expression.
	$12 - 3 (9 - 12)$
Subtract in parentheses first.	$12 - 3 (−3)$
Multiply.	$12 - (−9)$
Subtract.	$21$

Simplify: $8 (−9) \div {(−2)}^{3} .$

Solution

Demonstrates the step-by-step evaluation of the mathematical expression 8(-9) ÷ (-2)^3, illustrating the order of operations.
	$8 (−9) \div {(−2)}^{3}$
Exponents first.	$8 (−9) \div (−8)$
Multiply.	$−72 \div (−8)$
Divide.	$9$

Simplify: $−30 \div 2 + (−3) (−7) .$

Solution

Step-by-step evaluation of a mathematical expression, illustrating the application of the order of operations to reach the final numerical result.
	$−30 \div 2 + (−3) (−7)$
Multiply and divide left to right, so divide first.	$−15 + (−3) (−7)$
Multiply.	$−15 + 21$
Add.	$6$

Evaluate Variable Expressions with Integers

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

When $n = −5,$ evaluate: ⓐ $n + 1$ ⓑ $− n + 1 .$

Solution

ⓐ



Simplify.	−4

ⓑ



Simplify.
Add.	6

Evaluate ${(x + y)}^{2}$ when $x = −18$ and $y = 24 .$

Solution



Add inside parenthesis.	(6)²
Simplify.	36

Evaluate $20 - z$ when ⓐ $z = 12$ and ⓑ $z = −12 .$

Solution

ⓐ



Subtract.	8

ⓑ



Subtract.	32

Evaluate: $2 x^{2} + 3 x + 8$ when $x = 4 .$

Solution

Substitute

4 for x .

Use parentheses to show multiplication.


Substitute.
Evaluate exponents.
Multiply.
Add.	52

Translate Phrases to Expressions with Integers

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

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

Solution

Step-by-step translation and solution of a mathematical word problem involving sums and integers.
	the sum of 8 and $−12,$ increased by 3.
Translate.	$[8 + (−12)] + 3$
Simplify. Be careful not to confuse the brackets with an absolute value sign.	$(−4) + 3$
Add.	$−1$

When we first introduced the operation symbols, we saw that the expression may be read in several ways. They are listed in the chart below.

$a - b$
$a$ minus $b$ the difference of $a$ and $b$ $b$ subtracted from $a$ $b$ less than $a$

Be careful to get a and b in the right order!

Translate and then simplify ⓐ the difference of 13 and $−21$ ⓑ subtract 24 from $−19 .$

Solution

Demonstrates translating verbal math problems involving integers into expressions and finding their solutions.
ⓐ Translate. Simplify.	$\begin{matrix} the difference of 13 and - 21 \\ 13 - (−21) \\ 34 \end{matrix}$
ⓑ Translate. Remember, "subtract $b$ from $a$ means $a - b$ . Simplify.	$\begin{matrix} subtract 24 from - 19 \\ - 19 - 24 \\ - 43 \end{matrix}$

Once again, our prior work translating English 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

Demonstrates translation and simplification of a mathematical product from text to numerical result.
	$the product of −2 and 14$
Translate.	$(−2) (14)$
Simplify.	$−28$

Translate to an algebraic expression and simplify if possible: the quotient of $−56$ and $−7 .$

Solution

Steps to translate a verbal phrase into a mathematical expression and simplify the quotient of two negative integers.
	$the quotient of −56 and −7$
Translate.	$−56 \div (−7)$
Simplify.	$8$

Use Integers in Applications

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

How to Apply a Strategy to Solve Applications with Integers

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

Solution

This is a table with two columns. The left column includes steps to solve the problem. The right column includes the math to solve the problem. In the first row, the left column says “Step 1. Read the problem. Make sure all the words and ideas are understood.” The right column is blank.

In the second row, the left column says “Step 2. Identify what we are asked to find”. The right column says, “the difference of the morning and afternoon temperatures.”

In the third row, the left column says, “Step 3. Write a phrase that gives the information to find it.” Next to this in the right column, it says “the difference of 11 and negative 9.”

In the fourth row, the left column says, “Step 4. Translate the phrase to an expression.” The right column contains 11 minus negative 9.

In the fifth row, the left column says, “Step 5. Simplify the expression.” The right column contains 20.

The final row says, “Step five. Write a complete sentence that answers the question.” Next to this in the right column, it says “the difference in temperatures was 20 degrees.”

The Mustangs football team received three penalties in the third quarter. Each penalty gave them a loss of fifteen yards. What is the number of yards lost?

Solution

Steps for solving a word problem involving a 15-yard penalty, leading to a total loss of 45 yards.
Step 1. Read the problem. Make sure all the words and ideas are understood.
Step 2. Identify what we are asked to find.	the number of yards lost
Step 3. Write a phrase that gives the information to find it.	three times a 15-yard penalty
Step 4. Translate the phrase to an expression.	$3 (−15)$
Step 5. Simplify the expression.	$−45$
Step 6. Answer the question with a complete sentence.	The team lost 45 yards.

Key Concepts

Multiplication and Division of Two Signed Numbers
- Same signs—Product is positive
- Different signs—Product is negative
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

Multiply Integers

In the following exercises, multiply.

$−4 \cdot 8$

Solution

$−32$

$−3 \cdot 9$

$9 (−7)$

Solution

$−63$

$13 (−5)$

$- 1 \cdot 6$

Solution

$−6$

$- 1 \cdot 3$

$−1 (−14)$

Solution

$−1 (−19)$

Divide Integers

In the following exercises, divide.

$−24 \div 6$

Solution

$−4$

$35 \div (−7)$

$−52 \div (−4)$

Solution

$−84 \div (−6)$

$−180 \div 15$

Solution

$−12$

$−192 \div 12$

Simplify Expressions with Integers

In the following exercises, simplify each expression.

$5 (−6) + 7 (−2) - 3$

Solution

$−47$

$8 (−4) + 5 (−4) - 6$

${(−2)}^{6}$

Solution

${(−3)}^{5}$

$− 4^{2}$

Solution

$−16$

$− 6^{2}$

$−3 (−5) (6)$

Solution

$−4 (−6) (3)$

$(8 - 11) (9 - 12)$

Solution

$(6 - 11) (8 - 13)$

$26 - 3 (2 - 7)$

Solution

$23 - 2 (4 - 6)$

$65 \div (−5) + (−28) \div (−7)$

Solution

$−9$

$52 \div (−4) + (−32) \div (−8)$

$9 - 2 [3 - 8 (−2)]$

Solution

$−29$

$11 - 3 [7 - 4 (−2)]$

${(−3)}^{2} - 24 \div (8 - 2)$

Solution

${(−4)}^{2} - 32 \div (12 - 4)$

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

$y + (−14)$ when ⓐ $y = −33$ ⓑ $y = 30$

Solution

ⓐ $−47$ ⓑ 16

$x + (−21)$ whenⓐ $x = −27$ ⓑ $x = 44$

ⓐ $a + 3$ when $a = −7$
ⓑ $− a + 3$ when $a = −7$

Solution

ⓐ $−4$ ⓑ 10

ⓐ $d + (−9)$ when $d = −8$
ⓑ $− d + (−9)$ when $d = −8$

$m + n$ when
$m = −15, n = 7$

Solution

$−8$

$p + q$ when
$p = −9, q = 17$

$r + s$ when $r = −9, s = −7$

Solution

$−16$

$t + u$ when $t = −6, u = −5$

${(x + y)}^{2}$ when
$x = −3, y = 14$

Solution

121

${(y + z)}^{2}$ when
$y = −3, z = 15$

$−2 x + 17$ when

ⓐ $x = 8$
ⓑ $x = −8$

Solution

ⓐ 1
ⓑ 33

$−5 y + 14$ when
ⓐ $y = 9$
ⓑ $y = −9$

$10 - 3 m$ when
ⓐ $m = 5$
ⓑ $m = −5$

Solution

ⓐ $−5$ ⓑ 25

$18 - 4 n$ when
ⓐ $n = 3$
ⓑ $n = −3$

$2 w^{2} - 3 w + 7$ when
$w = −2$

Solution

$3 u^{2} - 4 u + 5$ when $u = −3$

$9 a - 2 b - 8$ when
$a = −6 and b = −3$

Solution

$−56$

$7 m - 4 n - 2$ when
$m = −4 and n = −9$

Translate English Phrases to Algebraic Expressions

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

the sum of 3 and $−15,$ increased by 7

Solution

$(3 + (−15)) + 7; - 5$

the sum of $−8$ and $−9,$ increased by 23

the difference of 10 and $−18$

Solution

$10 - (−18); 28$

subtract 11 from $−25$

the difference of $−5$ and $−30$

Solution

$−5 - (−30); 25$

subtract $−6$ from $−13$

the product of $−3 and 15$

Solution

$−3 \cdot 15; - 45$

the product of $−4 and 16$

the quotient of $−60$ and $−20$

Solution

$−60 \div (−20); 3$

the quotient of $−40$ and $−20$

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

Solution

$\frac{−6}{a + b}$

the quotient of $−7$ and the sum of m and n

the product of $−10$ and the difference of $p and q$

Solution

$−10 (p - q)$

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

Use Integers in Applications

In the following exercises, solve.

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

Solution

$96 °$

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

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

Solution

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

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

Solution

$− $28$

Checking Account Selina has $165 in her checking account. She writes a check for $207. What is the new balance in her checking account?

Checking Account Diontre has a balance of $− $38$ in his checking account. He deposits $225 to the account. What is the new balance?

Solution

$187

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

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

$− $3600$

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 integers.

Solution

Answers may vary

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

Why is $− 2^{4} \neq {(−2)}^{4} ?$

Solution

Answers may vary

Why is $− 4^{3} = {(−4)}^{3} ?$

Self Check

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

A table is shown that is composed of four columns and seven rows. The titles of the columns are “I can …”, “Confidently”, “With some help” and “No – I don’t get it!”. The first column reads “multiple integers.”, “divide integers.”, “simplify expressions with integers.”, “evaluate variable expressions with integers.”, “translate English phrases to algebraic expressions.” and “use integers in applications.”

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

Visualize Fractions

Find Equivalent Fractions

Fractions are a way to represent parts of a whole. The fraction $\frac{1}{3}$ means that one whole has been divided into 3 equal parts and each part is one of the three equal parts. See Figure 1. The fraction $\frac{2}{3}$ represents two of three equal parts. In the fraction $\frac{2}{3},$ the 2 is called the numerator and the 3 is called the denominator.

Two circles are shown, each divided into three equal pieces by lines. The left hand circle is labeled “one third” in each section. Each section is shaded. The circle on the right is shaded in two of its three sections. — The circle on the left has been divided into 3 equal parts. Each part is $\frac{1}{3}$ of the 3 equal parts. In the circle on the right, $\frac{2}{3}$ of the circle is shaded (2 of the 3 equal parts).

If a whole pie has been cut into 6 pieces and we eat all 6 pieces, we ate $\frac{6}{6}$ pieces, or, in other words, one whole pie.

A circle is shown and is divided into six section. All sections are shaded.

So $\frac{6}{6} = 1 .$ This leads us to the property of one that tells us that any number, except zero, divided by itself is 1.

If a pie was cut in $6$ pieces and we ate all 6, we ate $\frac{6}{6}$ pieces, or, in other words, one whole pie. If the pie was cut into 8 pieces and we ate all 8, we ate $\frac{8}{8}$ pieces, or one whole pie. We ate the same amount—one whole pie.

The fractions $\frac{6}{6}$ and $\frac{8}{8}$ have the same value, 1, and so they are called equivalent fractions. Equivalent fractions are fractions that have the same value.

Let’s think of pizzas this time. Figure 2 shows two images: a single pizza on the left, cut into two equal pieces, and a second pizza of the same size, cut into eight pieces on the right. This is a way to show that $\frac{1}{2}$ is equivalent to $\frac{4}{8} .$ In other words, they are equivalent fractions.

A circle is shown that is divided into eight equal wedges by lines. The left side of the circle is a pizza with four sections making up the pizza slices. The right side has four shaded sections. Below the diagram is the fraction four eighths. — Since the same amount is of each pizza is shaded, we see that $\frac{1}{2}$ is equivalent to $\frac{4}{8} .$ They are equivalent fractions.

How can we use mathematics to change $\frac{1}{2}$ into $\frac{4}{8} ?$ How could we take a pizza that is cut into 2 pieces and cut it into 8 pieces? We could cut each of the 2 larger pieces into 4 smaller pieces! The whole pizza would then be cut into $8$ pieces instead of just 2. Mathematically, what we’ve described could be written like this as $\frac{1 \cdot 4}{2 \cdot 4} = \frac{4}{8} .$ See Figure 3.

A circle is shown and is divided in half by a vertical black line. It is further divided into eighths by the addition of dotted red lines. — Cutting each half of the pizza into $4$ pieces, gives us pizza cut into 8 pieces: $\frac{1 \cdot 4}{2 \cdot 4} = \frac{4}{8} .$

This model leads to the following property:

If we had cut the pizza differently, we could get

$An image shows three rows of fractions. In the first row are the fractions “1, times 2, divided by 2, times 2, equals two fourths”. Next to this is the word “so” and the fraction “one half, equals two fourths. The second row reads “1, times 3, divided by 2 times 3, equals three sixths”. Next to this is the word “so” and the fraction “one half equals, three sixths”. The third row reads “1 times 10, divided by 2 times 10, ten twentieths”. Next to this is the word “so” and the fraction “one half equals, ten twentieths”.$

So, we say $\frac{1}{2}, \frac{2}{4}, \frac{3}{6}, and \frac{10}{20}$ are equivalent fractions.

Find three fractions equivalent to $\frac{2}{5} .$

Solution

To find a fraction equivalent to $\frac{2}{5},$ we multiply the numerator and denominator by the same number. We can choose any number, except for zero. Let’s multiply them by 2, 3, and then 5.
$A row of fractions reads “2 times 2, divided by 5 times 2, equals four tenths”. Next to this is “2, times 3, divided by 5 times 3, equals six fifteenths”. Next to this is “2 times 5, divided by 5 times 5, equals ten twenty-fifths”.$
So, $\frac{4}{10}, \frac{6}{15}, and \frac{10}{25}$ are equivalent to $\frac{2}{5} .$

Simplify Fractions

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

For example,

$\frac{2}{3}$ is simplified because there are no common factors of 2 and 3.
$\frac{10}{15}$ is not simplified because $5$ is a common factor of 10 and 15.

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

In Example 1, we used the equivalent fractions property to find equivalent fractions. Now we’ll use the equivalent fractions property in reverse to simplify fractions. We can rewrite the property to show both forms together.

Simplify: $- \frac{32}{56} .$

Solution

	$- \frac{32}{56}$
Rewrite the numerator and denominator showing the common factors.	$A mathematical fraction displaying negative (4 multiplied by 8) over (7 multiplied by 8), with the common factor '8' highlighted in red in both numerator and denominator.$
Simplify using the equivalent fractions property.	$- \frac{4}{7}$

Notice that the fraction

- \frac{4}{7}

is simplified because there are no more common factors.

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

How to Simplify a Fraction

Simplify: $- \frac{210}{385} .$

Solution

A table is shown with three columns and three rows. The first row of the left column reads “Step 1. Rewrite the numerator and denominator to show the common factors. If needed, use a factor tree”. Next to this in the middle column, it reads “rewrite 210 and 285 as the product of the primes. Next to this in the right column, it reads “negative 210 divided by 385.” Under this, is the equation “two times three times five times seven.” The five and 7 are blue and red respectively.

$The next row down reads “Step 2. Simplify using the equivalent fractions property by dividing out common factors.” Next to this in the middle column, it reads, “Mark the common factors 5 and 7.” Next to this in the right column, it has the equation 2 times, three times five, times seven over 5 times seven times 11. Both the 5 and the 7 are crossed out as common factors. Under this is the equation “negative two times 3 divided by 11.”$ The next row reads, “Step 3. Multiply the remaining factors, if necessary.” Next to this in the right column is negative six elevenths.

The next row reads, “Step 3. Multiply the remaining factors, if necessary.” Next to this in the right column is negative six elevenths.

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

Simplify: $\frac{5 x}{5 y} .$

Solution

	$\frac{5 x}{5 y}$
Rewrite showing the common factors, then divide out the common factors.	$The image illustrates the simplification of the fraction 5x/5y. It shows how to rewrite the expression to identify common factors (5), divide them out, and then simplify to x/y.$
Simplify.	$\frac{x}{y}$

Multiply Fractions

Many people find multiplying and dividing fractions easier than adding and subtracting fractions. So we will start with fraction multiplication.

We’ll use a model to show you how to multiply two fractions and to help you remember the procedure. Let’s start with $\frac{3}{4} .$

A rectangle made up of four squares in a row. The first three squares are shaded.

Now we’ll take $\frac{1}{2}$ of $\frac{3}{4} .$

A rectangle made up of four squares in a row. The first three squares are shaded. The bottom halves of the first three squares are shaded darker with diagonal lines.

Notice that now, the whole is divided into 8 equal parts. So $\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8} .$

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

When multiplying fractions, the properties of positive and negative numbers still apply, of course. It is a good idea to determine the sign of the product as the first step. In Example 5, we will multiply negative and a positive, so the product will be negative.

Multiply: $- \frac{11}{12} \cdot \frac{5}{7} .$

Solution

The first step is to find the sign of the product. Since the signs are the different, the product is negative.

This table illustrates the step-by-step process of multiplying two fractions, showing each mathematical operation and the resulting expression.
	$- \frac{11}{12} \cdot \frac{5}{7}$
Determine the sign of the product; multiply.	$- \frac{11 \cdot 5}{12 \cdot 7}$
Are there any common factors in the numerator and the demoninator? No.	$- \frac{55}{84}$

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 $\frac{a}{1} .$ So, for example, $3 = \frac{3}{1} .$

Multiply: $- \frac{12}{5} (−20 x) .$

Solution

Determine the sign of the product. The signs are the same, so the product is positive.

	$- \frac{12}{5} (−20 x)$
Write $20 x$ as a fraction.	$\frac{12}{5} (\frac{20 x}{1})$
Multiply.
Rewrite 20 to show the common factor 5 and divide it out.	$A mathematical expression showing the fraction (12 * 4 * 5x) divided by (5 * 1), with the '5' in the numerator and denominator canceled out.$
Simplify.	$48 x$

Divide Fractions

Now that we know how to multiply fractions, we are almost ready to divide. Before we can do that, we need some vocabulary.

The reciprocal of a fraction is found by inverting the fraction, placing the numerator in the denominator and the denominator in the numerator. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2} .$

Notice that $\frac{2}{3} \cdot \frac{3}{2} = 1 .$ A number and its reciprocal multiply to 1.

To get a product of positive 1 when multiplying two numbers, the numbers must have the same sign. So reciprocals must have the same sign.

The reciprocal of $- \frac{10}{7}$ is $- \frac{7}{10},$ since $- \frac{10}{7} (- \frac{7}{10}) = 1 .$

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

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

Divide: $- \frac{2}{3} \div \frac{n}{5} .$

Solution

Step-by-step guide on dividing a numerical fraction by an algebraic fraction using the reciprocal method.
	$- \frac{2}{3} \div \frac{n}{5}$
To divide, multiply the first fraction by the reciprocal of the second.	$- \frac{2}{3} \cdot \frac{5}{n}$
Multiply.	$- \frac{10}{3 n}$

Find the quotient: $- \frac{7}{18} \div (- \frac{14}{27}) .$

Solution

	$- \frac{7}{18} \div (− \frac{14}{27})$
To divide, multiply the first fraction by the reciprocal of the second.	$- \frac{7}{18} \cdot − \frac{27}{14}$
Determine the sign of the product, and then multiply..	$\frac{7 \cdot 27}{18 \cdot 14}$
Rewrite showing common factors.	$A mathematical fraction is shown where common factors (7 and 9) in the numerator (793) and denominator (9272) are crossed out, indicating cancellation to simplify the expression. The remaining terms are 3 in the numerator and 22 in the denominator.$
Remove common factors.	$\frac{3}{2 \cdot 2}$
Simplify.	$\frac{3}{4}$

There are several ways to remember which steps to take to multiply or divide fractions. One way is to repeat the call outs to yourself. If you do this each time you do an exercise, you will have the steps memorized.

“To multiply fractions, multiply the numerators and multiply the denominators.”
“To divide fractions, multiply the first fraction by the reciprocal of the second.”

Another way is to keep two examples in mind:

$This is an image with two columns. The first column reads “One fourth of two pizzas is one half of a pizza. Below this are two pizzas side-by-side with a line down the center of each one representing one half. The halves are labeled “one half”. Under this is the equation “2 times 1 fourth”. Under this is another equation “two over 1 times 1 fourth.” Under this is the fraction two fourths and under this is the fraction one half. The next column reads “there are eight quarters in two dollars.” Under this are eight quarters in two rows of four. Under this is the fraction equation 2 divided by one fourth. Under this is the equation “two over one divided by one fourth.” Under this is two over one times four over one. Under this is the answer “8”.$

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

Some examples of complex fractions are:

\frac{\frac{6}{7}}{3} \frac{\frac{3}{4}}{\frac{5}{8}} \frac{\frac{x}{2}}{\frac{5}{6}}

To simplify a complex fraction, we remember that the fraction bar means division. For example, the complex fraction $\frac{\frac{3}{4}}{\frac{5}{8}}$ means $\frac{3}{4} \div \frac{5}{8} .$

Simplify: $\frac{\frac{3}{4}}{\frac{5}{8}} .$

Solution

	$\frac{\frac{3}{4}}{\frac{5}{8}}$
Rewrite as division.	$\frac{3}{4} \div \frac{5}{8}$
Multiply the first fraction by the reciprocal of the second.	$\frac{3}{4} \cdot \frac{8}{5}$
Multiply.	$\frac{3 \cdot 8}{4 \cdot 5}$
Look for common factors.	$An example of fraction simplification, where the common factor '4' is canceled from both the numerator and denominator of (342)/(4*5).$
Divide out common factors and simplify.	$\frac{6}{5}$

Simplify: $\frac{\frac{x}{2}}{\frac{x y}{6}} .$

Solution

	$\frac{\frac{x}{2}}{\frac{x y}{6}}$
Rewrite as division.	$\frac{x}{2} \div \frac{x y}{6}$
Multiply the first fraction by the reciprocal of the second.	$\frac{x}{2} \cdot \frac{6}{x y}$
Multiply.	$\frac{x \cdot 6}{2 \cdot x y}$
Look for common factors.	$Simplifying a fraction by canceling out common variables 'x' and 'z' from the numerator and denominator, resulting in 3 over y.$
Divide out common factors and simplify.	$\frac{3}{y}$

Simplify Expressions with a Fraction Bar

The line that separates the numerator from the denominator in a fraction is called a fraction bar. A fraction bar acts as grouping symbol. The order of operations then tells us to simplify the numerator and then the denominator. Then we divide.

To simplify the expression $\frac{5 - 3}{7 + 1},$ we first simplify the numerator and the denominator separately. Then we divide.

\frac{5 - 3}{7 + 1}

\frac{2}{8}

\frac{1}{4}

Simplify: $\frac{4 - 2 (3)}{2^{2} + 2} .$

Solution

Step-by-step simplification of a mathematical expression using the order of operations.
	$\frac{4 - 2 (3)}{2^{2} + 2}$
Use the order of operations to simpliy the numerator and the denominator.	$\frac{4 - 6}{4 + 2}$
Simplify the numerator and the denominator.	$\frac{−2}{6}$
Simplify. A negative divided by a positive is negative.	$- \frac{1}{3}$

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

\begin{matrix} \frac{−1}{3} = - \frac{1}{3} & \frac{negative}{positive} = negative \\ \frac{1}{−3} = - \frac{1}{3} & \frac{positive}{negative} = negative \end{matrix}

Simplify: $\frac{4 (−3) + 6 (−2)}{−3 (2) - 2} .$

Solution

This table demonstrates the step-by-step simplification of a complex rational expression involving integer arithmetic.
	$\frac{4 (−3) + 6 (−2)}{−3 (2) - 2}$
Multiply.	$\frac{−12 + (−12)}{−6 - 2}$
Simplify.	$\frac{−24}{−8}$
Divide.	$3$

Translate Phrases to Expressions with Fractions

Now that we have done some work with fractions, we are ready to translate phrases that would result in expressions with fractions.

The English words quotient and ratio are often used to describe fractions. Remember that “quotient” means division. The quotient of $a$ and $b$ is the result we get from dividing $a$ by $b,$ or $\frac{a}{b} .$

Translate the English phrase into an algebraic expression: the quotient of the difference of m and n, and p.

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 $m and n by p .$

\frac{m - n}{p}

Key Concepts

Equivalent Fractions Property: If $a, b, c$ are numbers where $b \neq 0, c \neq 0,$ then
$\frac{a}{b} = \frac{a \cdot c}{b \cdot c}$ and $\frac{a \cdot c}{b \cdot c} = \frac{a}{b} .$
Fraction Division: If $a, b, c and d$ are numbers where $b \neq 0, c \neq 0, and d \neq 0,$ then $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} .$ To divide fractions, multiply the first fraction by the reciprocal of the second.
Fraction Multiplication: If $a, b, c and d$ are numbers where $b \neq 0, and d \neq 0,$ then $\frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d} .$ To multiply fractions, multiply the numerators and multiply the denominators.
Placement of Negative Sign in a Fraction: For any positive numbers $a and b,$ $\frac{- a}{b} = \frac{a}{- b} = - \frac{a}{b} .$
Property of One: $\frac{a}{a} = 1;$ Any number, except zero, divided by itself is one.
Simplify a Fraction
1. Rewrite the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers first.
2. Simplify using the equivalent fractions property by dividing out common factors.
3. Multiply any remaining factors.
Simplify an Expression with a Fraction Bar
1. Simplify the expression in the numerator. Simplify the expression in the denominator.
2. Simplify the fraction.

Practice Makes Perfect

Find Equivalent Fractions

In the following exercises, find three fractions equivalent to the given fraction. Show your work, using figures or algebra.

$\frac{3}{8}$

Solution

$\frac{6}{16}, \frac{9}{24}, \frac{12}{32}$ answers may vary

$\frac{5}{8}$

$\frac{5}{9}$

Solution

$\frac{10}{18}, \frac{15}{27}, \frac{20}{36}$ answers may vary

$\frac{1}{8}$

Simplify Fractions

In the following exercises, simplify.

$- \frac{40}{88}$

Solution

$- \frac{5}{11}$

$- \frac{63}{99}$

$- \frac{108}{63}$

Solution

$- \frac{12}{7}$

$- \frac{104}{48}$

$\frac{120}{252}$

Solution

$\frac{10}{21}$

$\frac{182}{294}$

$- \frac{3 x}{12 y}$

Solution

$- \frac{x}{4 y}$

$- \frac{4 x}{32 y}$

$\frac{14 x^{2}}{21 y}$

Solution

$\frac{2 x^{2}}{3 y}$

$\frac{24 a}{32 b^{2}}$

Multiply Fractions

In the following exercises, multiply.

$\frac{3}{4} \cdot \frac{9}{10}$

Solution

$\frac{27}{40}$

$\frac{4}{5} \cdot \frac{2}{7}$

$- \frac{2}{3} (- \frac{3}{8})$

Solution

$\frac{1}{4}$

$- \frac{3}{4} (- \frac{4}{9})$

$- \frac{5}{9} \cdot \frac{3}{10}$

Solution

$- \frac{1}{6}$

$- \frac{3}{8} \cdot \frac{4}{15}$

$(- \frac{14}{15}) (\frac{9}{20})$

Solution

$- \frac{21}{50}$

$(- \frac{9}{10}) (\frac{25}{33})$

$(- \frac{63}{84}) (- \frac{44}{90})$

Solution

$\frac{11}{30}$

$(- \frac{33}{60}) (- \frac{40}{88})$

$4 \cdot \frac{5}{11}$

Solution

$\frac{20}{11}$

$5 \cdot \frac{8}{3}$

$\frac{3}{7} \cdot 21 n$

Solution

$\frac{5}{6} \cdot 30 m$

$−8 (\frac{17}{4})$

Solution

$−34$

$(−1) (- \frac{6}{7})$

Divide Fractions

In the following exercises, divide.

$\frac{3}{4} \div \frac{2}{3}$

Solution

$\frac{9}{8}$

$\frac{4}{5} \div \frac{3}{4}$

$- \frac{7}{9} \div (- \frac{7}{4})$

Solution

$\frac{4}{9}$

$- \frac{5}{6} \div (- \frac{5}{6})$

$\frac{3}{4} \div \frac{x}{11}$

Solution

$\frac{33}{4 x}$

$\frac{2}{5} \div \frac{y}{9}$

$\frac{5}{18} \div (- \frac{15}{24})$

Solution

$- \frac{4}{9}$

$\frac{7}{18} \div (- \frac{14}{27})$

$\frac{8 u}{15} \div \frac{12 v}{25}$

Solution

$\frac{10 u}{9 v}$

$\frac{12 r}{25} \div \frac{18 s}{35}$

$−5 \div \frac{1}{2}$

Solution

$−10$

$−3 \div \frac{1}{4}$

$\frac{3}{4} \div (−12)$

Solution

$- \frac{1}{16}$

$−15 \div (- \frac{5}{3})$

In the following exercises, simplify.

$\frac{- \frac{8}{21}}{\frac{12}{35}}$

Solution

$- \frac{10}{9}$

$\frac{- \frac{9}{16}}{\frac{33}{40}}$

$\frac{- \frac{4}{5}}{2}$

Solution

$- \frac{2}{5}$

$\frac{5}{\frac{3}{10}}$

$\frac{\frac{m}{3}}{\frac{n}{2}}$

Solution

$\frac{2 m}{3 n}$

$\frac{- \frac{3}{8}}{- \frac{y}{12}}$

Simplify Expressions Written with a Fraction Bar

In the following exercises, simplify.

$\frac{22 + 3}{10}$

Solution

$\frac{5}{2}$

$\frac{19 - 4}{6}$

$\frac{48}{24 - 15}$

Solution

$\frac{16}{3}$

$\frac{46}{4 + 4}$

$\frac{−6 + 6}{8 + 4}$

Solution

$\frac{−6 + 3}{17 - 8}$

$\frac{4 \cdot 3}{6 \cdot 6}$

Solution

$\frac{1}{3}$

$\frac{6 \cdot 6}{9 \cdot 2}$

$\frac{4^{2} - 1}{25}$

Solution

$\frac{3}{5}$

$\frac{7^{2} + 1}{60}$

$\frac{8 \cdot 3 + 2 \cdot 9}{14 + 3}$

Solution

$2 \frac{8}{17}$

$\frac{9 \cdot 6 - 4 \cdot 7}{22 + 3}$

$\frac{5 \cdot 6 - 3 \cdot 4}{4 \cdot 5 - 2 \cdot 3}$

Solution

$\frac{9}{7}$

$\frac{8 \cdot 9 - 7 \cdot 6}{5 \cdot 6 - 9 \cdot 2}$

$\frac{5^{2} - 3^{2}}{3 - 5}$

Solution

$−8$

$\frac{6^{2} - 4^{2}}{4 - 6}$

$\frac{7 \cdot 4 - 2 (8 - 5)}{9 \cdot 3 - 3 \cdot 5}$

Solution

$\frac{11}{6}$

$\frac{9 \cdot 7 - 3 (12 - 8)}{8 \cdot 7 - 6 \cdot 6}$

$\frac{9 (8 - 2) - 3 (15 - 7)}{6 (7 - 1) - 3 (17 - 9)}$

Solution

$\frac{5}{2}$

$\frac{8 (9 - 2) - 4 (14 - 9)}{7 (8 - 3) - 3 (16 - 9)}$

Translate Phrases to Expressions with Fractions

In the following exercises, translate each English phrase into an algebraic expression.

the quotient of r and the sum of s and 10

Solution

$\frac{r}{s + 10}$

the quotient of A and the difference of 3 and B

the quotient of the difference of $x and y, and - 3$

Solution

$\frac{x - y}{−3}$

the quotient of the sum of $m and n, and 4 q$

Everyday Math

Baking. A recipe for chocolate chip cookies calls for $\frac{3}{4}$ cup brown sugar. Imelda wants to double the recipe. ⓐ How much brown sugar will Imelda need? Show your calculation. ⓑ Measuring cups usually come in sets of $\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, and 1$ cup. Draw a diagram to show two different ways that Imelda could measure the brown sugar needed to double the cookie recipe.

Solution

ⓐ $(2) (\frac{3}{4}); \frac{3}{2}$ cups ⓑ answers will vary

Baking. Nina is making 4 pans of fudge to serve after a music recital. For each pan, she needs $\frac{2}{3}$ cup of condensed milk. ⓐ How much condensed milk will Nina need? Show your calculation. ⓑ Measuring cups usually come in sets of $\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, and 1$ cup. Draw a diagram to show two different ways that Nina could measure the condensed milk needed for $4$ pans of fudge.

Portions Don purchased a bulk package of candy that weighs $5$ pounds. He wants to sell the candy in little bags that hold $\frac{1}{4}$ pound. How many little bags of candy can he fill from the bulk package?

Solution

20 bags

Portions Kristen has $\frac{3}{4}$ yards of ribbon that she wants to cut into $6$ equal parts to make hair ribbons for her daughter’s 6 dolls. How long will each doll’s hair ribbon be?

Writing Exercises

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.

Solution

Answers may vary

Give an example from everyday life that demonstrates how $\frac{1}{2} \cdot \frac{2}{3} is \frac{1}{3} .$

Explain how you find the reciprocal of a fraction.

Solution

Answers may vary

Explain how you find the reciprocal of a negative number.

Self Check

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

$A table is shown that is made up of four columns and seven rows. The first row reads “I can…” in the first column, “Confidently” in the second column, “With some help” in the third column and “No – I don’t get it” in the last column. The next row down in the first column reads “find equivalent fractions”, under this reads “simplify fractions”, under this reads “multiply fractions”, under this reads “divide fractions”, under this reads “Simplify expressions written with a fraction bar” and under this reads “translate phrases to expressions with fractions.”$

ⓑ After looking at the checklist, do you think you are well prepared for the next section? Why or why not?

complex fraction

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

denominator

The denominator is the value on the bottom part of the fraction that indicates the number of equal parts into which the whole has been divided.

equivalent fractions

Equivalent fractions are fractions that have the same value.

fraction

A fraction is written

\frac{a}{b}

, where

b \neq 0

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.

numerator

The numerator is the value on the top part of the fraction that indicates how many parts of the whole are included.

reciprocal

The reciprocal of

\frac{a}{b}

\frac{b}{a}

. A number and its reciprocal multiply to one:

\frac{a}{b} \cdot \frac{b}{a} = 1

simplified fraction

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

Add and Subtract Fractions

Add or Subtract Fractions with a Common Denominator

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

Find the sum: $\frac{x}{3} + \frac{2}{3} .$

Solution

Illustrates adding fractions with a common denominator, showing the operation and the resulting expression.
	$\frac{x}{3} + \frac{2}{3}$
Add the numerators and place the sum over the common denominator.	$\frac{x + 2}{3}$

Find the difference: $- \frac{23}{24} - \frac{13}{24} .$

Solution

This table illustrates the step-by-step process of subtracting fractions with common denominators and simplifying the final result.
	$- \frac{23}{24} - \frac{13}{24}$
Subtract the numerators and place the difference over the common denominator.	$\frac{−23 - 13}{24}$
Simplify.	$\frac{−36}{24}$
Simplify. Remember, $- \frac{a}{b} = \frac{− a}{b}$ .	$- \frac{3}{2}$

Simplify: $- \frac{10}{x} - \frac{4}{x} .$

Solution

Step-by-step simplification of algebraic fractions with a common denominator.
	$- \frac{10}{x} - \frac{4}{x}$
Subtract the numerators and place the difference over the common denominator.	$\frac{−14}{x}$
Rewrite with the sign in front of the fraction.	$- \frac{14}{x}$

Now we will do an example that has both addition and subtraction.

Simplify: $\frac{3}{8} + (- \frac{5}{8}) - \frac{1}{8} .$

Solution

This table illustrates the step-by-step process for adding and subtracting fractions that share a common denominator.
Add and subtract fractions—do they have a common denominator? Yes.	$\frac{3}{8} + (- \frac{5}{8}) - \frac{1}{8}$
Add and subtract the numerators and place the difference over the common denominator.	$\frac{3 + (−5) - 1}{8}$
Simplify left to right.	$\frac{−2 - 1}{8}$
Simplify.	$- \frac{3}{8}$

Add or Subtract Fractions with Different Denominators

As we have seen, to add or subtract fractions, their denominators must be the same. The least common denominator (LCD) of two fractions is the smallest number that can be used as a common denominator of the fractions. The LCD of the two fractions is the least common multiple (LCM) of their denominators.

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

How to Add or Subtract Fractions

Add: $\frac{7}{12} + \frac{5}{18} .$

Solution

$In this figure, we have a table with directions on the left, hints or explanations in the middle, and mathematical statements on the right. On the first line, we have “Step 1. Do they have a common denominator? No – rewrite each fraction with the LCD (least common denominator).” To the right of this, we have the statement “No. Find the LCD 12, 18.” To the right of this, we have 12 equals 2 times 2 times 3 and 18 equals 2 times 3 times 3. The LCD is hence 2 times 2 times 3 times 3, which equals 36. As another hint, we have “Change into equivalent fractions with the LCD,. Do not simplify the equivalent fractions! If you do, you’ll get back to the original fractions and lose the common denominator!” To the right of this, we have 7/12 plus 5/18, which becomes the quantity (7 times 3) over the quantity (12 times 3) plus the quantity (5 times 2) over the quantity (18 times 2), which becomes 21/36 plus 10/36.$ $The next step reads “Step 2. Add or subtract the fractions.” The hint reads “Add.” And we have 31/36.$ The final step reads “Step 3. Simplify, if possible.” The explanation reads “Because 31 is a prime number, it has no factors in common with 36. The answer is simplified.”

The final step reads “Step 3. Simplify, if possible.” The explanation reads “Because 31 is a prime number, it has no factors in common with 36. The answer is simplified.”

When finding the equivalent fractions needed to create the common denominators, there is a quick way to find the number we need to multiply both the numerator and denominator. This method works if we found the LCD by factoring into primes.

Look at the factors of the LCD and then at each column above those factors. The “missing” factors of each denominator are the numbers we need.

The number 12 is factored into 2 times 2 times 3 with an extra space after the 3, and the number 18 is factored into 2 times 3 times 3 with an extra space between the 2 and the first 3. There are arrows pointing to these extra spaces that are marked “missing factors.” The LCD is marked as 2 times 2 times 3 times 3, which is equal to 36. The numbers that create the LCD are the factors from 12 and 18, with the common factors counted only once (namely, the first 2 and the first 3).

In Example 5, the LCD, 36, has two factors of 2 and two factors of $3 .$

The numerator 12 has two factors of 2 but only one of 3—so it is “missing” one 3—we multiply the numerator and denominator by 3.

The numerator 18 is missing one factor of 2—so we multiply the numerator and denominator by 2.

We will apply this method as we subtract the fractions in Example 6.

Subtract: $\frac{7}{15} - \frac{19}{24} .$

Solution

Do the fractions have a common denominator? No, so we need to find the LCD.

Find the LCD. $The image shows the calculation of the Least Common Denominator (LCD) for the fractions 7/15 and 19/24. It details the prime factorization of 15 (35) and 24 (2223), leading to an LCD of 120.$
Notice, 15 is “missing” three factors of 2 and 24 is “missing” the 5 from the factors of the LCD. So we multiply 8 in the first fraction and 5 in the second fraction to get the LCD.
Rewrite as equivalent fractions with the LCD.	$A mathematical expression showing the subtraction of two fractions: (7 multiplied by 8) divided by (15 multiplied by 8), minus (19 multiplied by 5) divided by (24 multiplied by 5). The numbers 8 and 5 are highlighted in red.$
Simplify.	$A mathematical expression showing the subtraction of two fractions with a common denominator: 56/120 - 95/120.$
Subtract.	$− \frac{39}{120}$
Check to see if the answer can be simplified.	$− \frac{13 \cdot 3}{40 \cdot 3}$
Both 39 and 120 have a factor of 3.
Simplify.	$− \frac{13}{40}$

Do not simplify the equivalent fractions! If you do, you’ll get back to the original fractions and lose the common denominator!

In the next example, one of the fractions has a variable in its numerator. Notice that we do the same steps as when both numerators are numbers.

Add: $\frac{3}{5} + \frac{x}{8} .$

Solution

The fractions have different denominators.

		$A mathematical expression showing the sum of two fractions: 3/5 plus x/8. The numbers and variable are in black text on a white background.$
Find the LCD.
Rewrite as equivalent fractions with the LCD.		$A mathematical expression shows the addition of two fractions: (3 times 8) divided by (5 times 8) plus (x times 5) divided by (8 times 5). The numbers 8 and 5 in red highlight common factors.$
Simplify.		$A mathematical expression shows the sum of two fractions, 24/40 and 5x/40. Both fractions share a common denominator of 40.$
Add.		$A mathematical expression showing the fraction (24 + 5x) / 40.$

Remember, we can only add like terms: 24 and 5x are not like terms.

We now have all four operations for fractions. Table 7 summarizes fraction operations.

Fraction Multiplication	Fraction Division
$\frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d}$ Multiply the numerators and multiply the denominators	$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ Multiply the first fraction by the reciprocal of the second.
Fraction Addition	Fraction Subtraction
$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$ Add the numerators and place the sum over the common denominator.	$\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$ Subtract the numerators and place the difference over the common denominator.
To multiply or divide fractions, an LCD is NOT needed. To add or subtract fractions, an LCD is needed.

Simplify: ⓐ $\frac{5 x}{6} - \frac{3}{10}$ ⓑ $\frac{5 x}{6} \cdot \frac{3}{10} .$

Solution

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

Demonstrates fraction operations (subtraction, multiplication) with verbal explanations and step-by-step mathematical solutions.
ⓐ What is the operation? The operation is subtraction. Do the fractions have a common denominator? No. Rewrite each fraction as an equivalent fraction with the LCD. Subtract the numerators and place the difference over the common denominators. Simplify, if possible. There are no common factors. The fraction is simplified.	$\begin{matrix} \frac{5 x}{6} - \frac{3}{10} \\ \frac{5 x \cdot 5}{6 \cdot 5} - \frac{3 \cdot 3}{10 \cdot 3} \\ \frac{25 x}{30} - \frac{9}{30} \\ \frac{25 x - 9}{30} \end{matrix}$
ⓑ What is the operation? Multiplication. To multiply fractions, multiply the numerators and multiply the denominators. Rewrite, showing common factors. Remove common factors. Simplify.	$\begin{matrix} \frac{5 x}{6} \cdot \frac{3}{10} \\ \frac{5 x \cdot 3}{6 \cdot 10} \\ \frac{5 x \cdot 3}{2 \cdot 3 \cdot 2 \cdot 5} \\ \frac{x}{4} \end{matrix}$

Notice we needed an LCD to add $\frac{5 x}{6} - \frac{3}{10},$ but not to multiply $\frac{5 x}{6} \cdot \frac{3}{10} .$

Use the Order of Operations to Simplify Complex Fractions

We have seen that a complex fraction is a fraction in which the numerator or denominator contains a fraction. The fraction bar indicates division. We simplified the complex fraction $\frac{\frac{3}{4}}{\frac{5}{8}}$ by dividing $\frac{3}{4}$ by $\frac{5}{8} .$

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

How to Simplify Complex Fractions

Simplify: $\frac{{(\frac{1}{2})}^{2}}{4 + 3^{2}} .$

Solution

In this figure, we have a table with directions on the left and mathematical statements on the right. On the first line, we have “Step 1. Simplify the numerator. Remember one half squared means one half times one half.” To the right of this, we have the quantity (1/2) squared all over the quantity (4 plus 3 squared). Then, we have 1/4 over the quantity (4 plus 3 squared).

The next line’s direction reads “Step 2. Simplify the denominator.” To the right of this, we have 1/4 over the quantity (4 plus 9), under which we have 1/4 over 13.

The final step is “Step 3. Divide the numerator by the denominator. Simplify if possible. Remember, thirteen equals thirteen over 1.” To the right we have 1/4 divided by 13. Then we have 1/4 times 1/13, which equals 1/52.

Simplify: $\frac{\frac{1}{2} + \frac{2}{3}}{\frac{3}{4} - \frac{1}{6}} .$

Solution

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

This table illustrates the step-by-step process of simplifying a complex mathematical fraction, showing the progressive transformation of the expression.
	$\frac{(\frac{1}{2} + \frac{2}{3})}{(\frac{3}{4} - \frac{1}{6})}$
Simplify the numerator (LCD = 6) and simplify the denominator (LCD = 12).	$\frac{(\frac{3}{6} + \frac{4}{6})}{(\frac{9}{12} - \frac{2}{12})}$
Simplify.	$\frac{(\frac{7}{6})}{(\frac{7}{12})}$
Divide the numerator by the denominator.	$\frac{7}{6} \div \frac{7}{12}$
Simplify.	$\frac{7}{6} \cdot \frac{12}{7}$
Divide out common factors.	$\frac{7 \cdot 6 \cdot 2}{6 \cdot 7}$
Simplify.	$2$

Evaluate Variable Expressions with Fractions

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

Evaluate $x + \frac{1}{3}$ when ⓐ $x = - \frac{1}{3}$ ⓑ $x = - \frac{3}{4} .$

Solution

ⓐ To evaluate $x + \frac{1}{3}$ when $x = - \frac{1}{3},$ substitute $- \frac{1}{3}$ for $x$ in the expression.

$A mathematical expression showing 'x' plus the fraction '1/3'.$

Simplify. 0
ⓑ To evaluate $x + \frac{1}{3}$ when $x = - \frac{3}{4},$ we substitute $- \frac{3}{4}$ for x in the expression.

$Text: Substitute -3/4 for x. The fraction -3/4 is highlighted in red.$ $A mathematical expression showing the addition of two fractions: negative three-fourths plus one-third. The number three and four are in red color, while the negative sign and fractions for one-third are in black.$

Rewrite as equivalent fractions with the LCD, 12. $An equation for adding -3/4 and 1/3, showing the fractions multiplied by factors to achieve a common denominator of 12: -(3*3)/(4*3) + (1*4)/(3*4).$

Simplify. $A mathematical expression showing the addition of two fractions with the same denominator: -9/12 + 4/12.$

Add. $- \frac{5}{12}$

Evaluate $- \frac{5}{6} - y$ when $y = - \frac{2}{3} .$

Solution


$The image displays the mathematical instruction 'Substitute -2/3 for y.' in a gray sans-serif font, with the fraction '-2/3' highlighted in red.$	$A mathematical expression showing the subtraction of two negative fractions: -5/6 - (-2/3), where the second fraction -2/3 is enclosed in parentheses and highlighted in red.$
Rewrite as equivalent fractions with the LCD, 6.	$A mathematical expression showing the subtraction of negative five-sixths and negative four-sixths, written as -5/6 - (-4/6), where the second fraction is enclosed in parentheses and in red.$
Subtract.	$A mathematical expression showing the fraction: negative 5 minus negative 4, all divided by 6.$
Simplify.	$- \frac{1}{6}$

Evaluate $2 x^{2} y$ when $x = \frac{1}{4}$ and $y = - \frac{2}{3} .$

Solution

Substitute the values into the expression.

	$2 x^{2} y$
$The image shows the instruction: 'Substitute 1/4 for x and -2/3 for y.' The fraction 1/4 is colored red, and -2/3 is colored light blue, making them visually distinct from the black text.$	$A mathematical expression showing the product of 2, the square of 1/4, and -2/3. The fraction 1/4 is colored red and -2/3 is colored blue, highlighting different parts of the expression.$
Simplify exponents first.	$2 (\frac{1}{16}) (- \frac{2}{3})$
Multiply. Divide out the common factors. Notice we write 16 as $2 \cdot 2 \cdot 4$ to make it easy to remove common factors.	$- \frac{2 \cdot 1 \cdot 2}{2 \cdot 2 \cdot 4 \cdot 3}$
Simplify.	$- \frac{1}{12}$

The next example will have only variables, no constants.

Evaluate $\frac{p + q}{r}$ when $p = −4, q = −2, and r = 8 .$

Solution

To evaluate

\frac{p + q}{r}

when

p = −4, q = −2, and r = 8,

we substitute the values into the expression.

	$\frac{p + q}{r}$
	$A mathematical fraction is displayed, with the numerator being '-4 + (-2)' and the denominator being '8'.$
Add in the numerator first.	$\frac{−6}{8}$
Simplify.	$- \frac{3}{4}$

Key Concepts

Fraction Addition and Subtraction: If $a, b, and c$ are numbers where $c \neq 0,$ then
$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$ and $\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c} .$
To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.
Strategy for Adding or Subtracting Fractions
1. Do they have a common denominator?
  Yes—go to step 2.
  No—Rewrite each fraction with the LCD (Least Common Denominator). Find the LCD. Change each fraction into an equivalent fraction with the LCD as its denominator.
2. Add or subtract the fractions.
3. Simplify, if possible. To multiply or divide fractions, an LCD IS NOT needed. To add or subtract fractions, an LCD IS needed.
Simplify Complex Fractions
1. Simplify the numerator.
2. Simplify the denominator.
3. Divide the numerator by the denominator. Simplify if possible.

Practice Makes Perfect

Add and Subtract Fractions with a Common Denominator

In the following exercises, add.

$\frac{6}{13} + \frac{5}{13}$

Solution

$\frac{11}{13}$

$\frac{4}{15} + \frac{7}{15}$

$\frac{x}{4} + \frac{3}{4}$

Solution

$\frac{x + 3}{4}$

$\frac{8}{q} + \frac{6}{q}$

$- \frac{3}{16} + (- \frac{7}{16})$

Solution

$- \frac{5}{8}$

$- \frac{5}{16} + (- \frac{9}{16})$

$- \frac{8}{17} + \frac{15}{17}$

Solution

$\frac{7}{17}$

$- \frac{9}{19} + \frac{17}{19}$

$\frac{6}{13} + (- \frac{10}{13}) + (- \frac{12}{13})$

Solution

$- \frac{16}{13}$

$\frac{5}{12} + (- \frac{7}{12}) + (- \frac{11}{12})$

In the following exercises, subtract.

$\frac{11}{15} - \frac{7}{15}$

Solution

$\frac{4}{15}$

$\frac{9}{13} - \frac{4}{13}$

$\frac{11}{12} - \frac{5}{12}$

Solution

$\frac{1}{2}$

$\frac{7}{12} - \frac{5}{12}$

$\frac{19}{21} - \frac{4}{21}$

Solution

$\frac{5}{7}$

$\frac{17}{21} - \frac{8}{21}$

$\frac{5 y}{8} - \frac{7}{8}$

Solution

$\frac{5 y - 7}{8}$

$\frac{11 z}{13} - \frac{8}{13}$

$- \frac{23}{u} - \frac{15}{u}$

Solution

$- \frac{38}{u}$

$- \frac{29}{v} - \frac{26}{v}$

$- \frac{3}{5} - (- \frac{4}{5})$

Solution

$\frac{1}{5}$

$- \frac{3}{7} - (- \frac{5}{7})$

$- \frac{7}{9} - (- \frac{5}{9})$

Solution

$- \frac{2}{9}$

$- \frac{8}{11} - (- \frac{5}{11})$

Mixed Practice

In the following exercises, simplify.

$- \frac{5}{18} \cdot \frac{9}{10}$

Solution

$- \frac{1}{4}$

$- \frac{3}{14} \cdot \frac{7}{12}$

$\frac{n}{5} - \frac{4}{5}$

Solution

$\frac{n - 4}{5}$

$\frac{6}{11} - \frac{s}{11}$

$- \frac{7}{24} + \frac{2}{24}$

Solution

$- \frac{5}{24}$

$- \frac{5}{18} + \frac{1}{18}$

$\frac{8}{15} \div \frac{12}{5}$

Solution

$\frac{2}{9}$

$\frac{7}{12} \div \frac{9}{28}$

Add or Subtract Fractions with Different Denominators

In the following exercises, add or subtract.

$\frac{1}{2} + \frac{1}{7}$

Solution

$\frac{9}{14}$

$\frac{1}{3} + \frac{1}{8}$

$\frac{1}{3} - (- \frac{1}{9})$

Solution

$\frac{4}{9}$

$\frac{1}{4} - (- \frac{1}{8})$

$\frac{7}{12} + \frac{5}{8}$

Solution

$\frac{29}{24}$

$\frac{5}{12} + \frac{3}{8}$

$\frac{7}{12} - \frac{9}{16}$

Solution

$\frac{1}{48}$

$\frac{7}{16} - \frac{5}{12}$

$\frac{2}{3} - \frac{3}{8}$

Solution

$\frac{7}{24}$

$\frac{5}{6} - \frac{3}{4}$

$- \frac{11}{30} + \frac{27}{40}$

Solution

$\frac{37}{120}$

$- \frac{9}{20} + \frac{17}{30}$

$- \frac{13}{30} + \frac{25}{42}$

Solution

$\frac{17}{105}$

$- \frac{23}{30} + \frac{5}{48}$

$- \frac{39}{56} - \frac{22}{35}$

Solution

$- \frac{53}{40}$

$- \frac{33}{49} - \frac{18}{35}$

$- \frac{2}{3} - (- \frac{3}{4})$

Solution

$\frac{1}{12}$

$- \frac{3}{4} - (- \frac{4}{5})$

$1 + \frac{7}{8}$

Solution

$\frac{15}{8}$

$1 - \frac{3}{10}$

$\frac{x}{3} + \frac{1}{4}$

Solution

$\frac{4 x + 3}{12}$

$\frac{y}{2} + \frac{2}{3}$

$\frac{y}{4} - \frac{3}{5}$

Solution

$\frac{5 y - 12}{20}$

$\frac{x}{5} - \frac{1}{4}$

Mixed Practice

In the following exercises, simplify.

ⓐ $\frac{2}{3} + \frac{1}{6}$ ⓑ $\frac{2}{3} \div \frac{1}{6}$

Solution

ⓐ $\frac{5}{6}$ ⓑ 4

ⓐ $- \frac{2}{5} - \frac{1}{8}$ ⓑ $- \frac{2}{5} \cdot \frac{1}{8}$

ⓐ $\frac{5 n}{6} \div \frac{8}{15}$ ⓑ $\frac{5 n}{6} - \frac{8}{15}$

Solution

ⓐ $\frac{25 n}{16}$ ⓑ $\frac{25 n - 16}{30}$

ⓐ $\frac{3 a}{8} \div \frac{7}{12}$ ⓑ $\frac{3 a}{8} - \frac{7}{12}$

$- \frac{3}{8} \div (- \frac{3}{10})$

Solution

$\frac{5}{4}$

$- \frac{5}{12} \div (- \frac{5}{9})$

$- \frac{3}{8} + \frac{5}{12}$

Solution

$\frac{1}{24}$

$- \frac{1}{8} + \frac{7}{12}$

$\frac{5}{6} - \frac{1}{9}$

Solution

$\frac{13}{18}$

$\frac{5}{9} - \frac{1}{6}$

$- \frac{7}{15} - \frac{y}{4}$

Solution

$\frac{−28 - 15 y}{60}$

$- \frac{3}{8} - \frac{x}{11}$

$\frac{11}{12 a} \cdot \frac{9 a}{16}$

Solution

$\frac{33}{64}$

$\frac{10 y}{13} \cdot \frac{8}{15 y}$

Use the Order of Operations to Simplify Complex Fractions

In the following exercises, simplify.

$\frac{2^{3} + 4^{2}}{{(\frac{2}{3})}^{2}}$

Solution

$\frac{3^{3} - 3^{2}}{{(\frac{3}{4})}^{2}}$

$\frac{{(\frac{3}{5})}^{2}}{{(\frac{3}{7})}^{2}}$

Solution

$\frac{49}{25}$

$\frac{{(\frac{3}{4})}^{2}}{{(\frac{5}{8})}^{2}}$

$\frac{2}{\frac{1}{3} + \frac{1}{5}}$

Solution

$\frac{15}{4}$

$\frac{5}{\frac{1}{4} + \frac{1}{3}}$

$\frac{\frac{7}{8} - \frac{2}{3}}{\frac{1}{2} + \frac{3}{8}}$

Solution

$\frac{5}{21}$

$\frac{\frac{3}{4} - \frac{3}{5}}{\frac{1}{4} + \frac{2}{5}}$

$\frac{1}{2} + \frac{2}{3} \cdot \frac{5}{12}$

Solution

$\frac{7}{9}$

$\frac{1}{3} + \frac{2}{5} \cdot \frac{3}{4}$

$1 - \frac{3}{5} \div \frac{1}{10}$

Solution

$−5$

$1 - \frac{5}{6} \div \frac{1}{12}$

$\frac{2}{3} + \frac{1}{6} + \frac{3}{4}$

Solution

$\frac{19}{12}$

$\frac{2}{3} + \frac{1}{4} + \frac{3}{5}$

$\frac{3}{8} - \frac{1}{6} + \frac{3}{4}$

Solution

$\frac{23}{24}$

$\frac{2}{5} + \frac{5}{8} - \frac{3}{4}$

$12 (\frac{9}{20} - \frac{4}{15})$

Solution

$\frac{11}{5}$

$8 (\frac{15}{16} - \frac{5}{6})$

$\frac{\frac{5}{8} + \frac{1}{6}}{\frac{19}{24}}$

Solution

$\frac{\frac{1}{6} + \frac{3}{10}}{\frac{14}{30}}$

$(\frac{5}{9} + \frac{1}{6}) \div (\frac{2}{3} - \frac{1}{2})$

Solution

$\frac{13}{3}$

$(\frac{3}{4} + \frac{1}{6}) \div (\frac{5}{8} - \frac{1}{3})$

Evaluate Variable Expressions with Fractions

In the following exercises, evaluate.

$x + (- \frac{5}{6})$ when
ⓐ $x = \frac{1}{3}$
ⓑ $x = - \frac{1}{6}$

Solution

ⓐ $- \frac{1}{2}$ ⓑ $−1$

$x + (- \frac{11}{12})$ when
ⓐ $x = \frac{11}{12}$ ⓑ $x = \frac{3}{4}$

$x - \frac{2}{5}$ when ⓐ $x = \frac{3}{5}$ ⓑ $x = - \frac{3}{5}$

Solution

ⓐ $\frac{1}{5}$ ⓑ $−1$

$x - \frac{1}{3}$ when ⓐ $x = \frac{2}{3}$ ⓑ $x = - \frac{2}{3}$

$\frac{7}{10} - w$ when ⓐ $w = \frac{1}{2}$ ⓑ $w = - \frac{1}{2}$

Solution

ⓐ $\frac{1}{5}$ ⓑ $\frac{6}{5}$

$\frac{5}{12} - w$ when ⓐ $w = \frac{1}{4}$ ⓑ $w = - \frac{1}{4}$

$2 x^{2} y^{3}$ when $x = - \frac{2}{3}$ and $y = - \frac{1}{2}$

Solution

$- \frac{1}{9}$

$8 u^{2} v^{3}$ when $u = - \frac{3}{4}$ and $v = - \frac{1}{2}$

$\frac{a + b}{a - b}$ when $a = −3, b = 8$

Solution

$- \frac{5}{11}$

$\frac{r - s}{r + s}$ when $r = 10, s = −5$

Everyday Math

Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs $\frac{1}{2}$ yard of print fabric and $\frac{3}{8}$ yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover?

Solution

$\frac{7}{8}$ yard

Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs $\frac{1}{2}$ cup of sugar for the chocolate chip cookies and $\frac{1}{4}$ of sugar for the oatmeal cookies. How much sugar does she need altogether?

Writing Exercises

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

Solution

Answers may vary

How do you find the LCD of 2 fractions?

Self Check

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

$This is a table that has five rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “add and subtract fractions with different denominators,” “identify and use fraction operations,” “use the order of operations to simplify complex fractions,” and “evaluate variable expressions with fractions.” The rest of the cells are blank.$

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

least common denominator

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

Decimals

Name and Write Decimals

Decimals are another way of writing fractions whose denominators are powers of 10.

\begin{matrix} 0.1 & = & \frac{1}{10} & 0.1 is “one tenth” \\ 0.01 & = & \frac{1}{100} & 0.01 is “one hundredth” \\ 0.001 & = & \frac{1}{1,000} & 0.001 is “one thousandth” \\ 0.0001 & = & \frac{1}{10,000} & 0.0001 is “one ten-thousandth” \end{matrix}

Notice that “ten thousand” is a number larger than one, but “one ten-thousandth” is a number smaller than one. The “th” at the end of the name tells you that the number is smaller than one.

When we name a whole number, the name corresponds to the place value based on the powers of ten. We read 10,000 as “ten thousand” and 10,000,000 as “ten million.” Likewise, the names of the decimal places correspond to their fraction values. Figure 1 shows the names of the place values to the left and right of the decimal point.

A table is shown with the title Place Value. From left to right the row reads “Hundred thousands,” “Ten thousands,” “Thousands,” “Hundreds,” “Tens,” and “Ones.” Then there is a blank cell and below it is a decimal point. To the right of this, the cells read “Tenths,” “Hundredths,” “Thousandths,” “Ten-thousandths,” and “Hundred-thousandths.” — Place value of decimal numbers are shown to the left and right of the decimal point.

How to Name Decimals

Name the decimal 4.3.

Solution

A table is given with four steps. Additionally, the number 4.3 is given. The first step reads “Step 1. Name the number to the left of the decimal point.” To the right of this, it is noted that “4 is to the left of the decimal point.” To the right of this, it reads “four” followed by a large blank space.

The second step reads “Step 2. Write ‘and’ for the decimal point.” To the right of this it reads “four and” followed by a blank space.

The third step reads “Step 3. Name the ‘number’ part to the right of the decimal point as if it were a whole number.” To the right of this, it reads “3 is to the right of the decimal point.” To the right of this, it reads “four and three” followed by a blank.

Finally, the last step reads “Step 4. Name the decimal place.” To the right of this, it reads “four and three tenths.”

We summarize the steps needed to name a decimal below.

Name the decimal: $−15.571 .$

Solution

Illustrates the step-by-step process of converting the decimal -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.	negative fifteen and five hundred seventy-one __________
The 1 is in the thousandths place.	negative fifteen and five hundred seventy-one thousandths

When we write a check we write both the numerals and the name of the number. Let’s see how to write the decimal from the name.

How to Write Decimals

Write “fourteen and twenty-four thousandths” as a decimal.

Solution

A table is given with four steps. The first step reads “Step 1. Look for the work ‘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.” To the right of this, we have the words “fourteen and twenty-four thousandths.” Below this word, we have “fourteen and twenty-four thousandths” with the word “and” underlined. Below this word, we have a small blank space separated from a larger blank space by a decimal point. Under this, we have 14 in the small blank space followed by the decimal point and the larger blank space.

The second step reads “Step 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.” To the right of this it reads “The last word is thousandths.” To the right of this there is the number 14 followed by a decimal point and three small blank spaces. Under the blank spaces, the words “tenths,” “hundredths,” and “thousandths” are written.

The third step reads “Step 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.” To the right of this, we have 14 followed by a decimal followed by a blank space followed by 2 and 4 on the other two previously blank spaces.

Finally, the last step reads “Step 4. Fill in zeros for empty place holders as needed.” To the right of this, it reads “Zeros are needed in the tenths place.” To the right of this, we have 14 followed by a decimal point followed by 0, 2, and 4, respectively, on the blank spaces. Below this, we have “fourteen and twenty-four thousandths is written 14.024.”

We summarize the steps to writing a decimal.

Round Decimals

Rounding decimals is very much like rounding whole numbers. We will round decimals with a method based on the one we used to round whole numbers.

How to Round Decimals

Round 18.379 to the nearest hundredth.

Solution

A table is given with four steps. The first step reads “Step 1: Locate the given place value and mark it with an arrow.” To the right of this, we have the number 18.379; above it, are the words hundreds place, which has an arrow pointing to the 7.

The second step reads “Step 2. Underline the digit to the right of the given place value.” To the right of this, we have 18.379 with the 9 underlined.

The third step reads “Step 3. Is this digit greater than or equal to 5? Below this reads, “Yes: add 1 to the digit in the given place value.” Below this reads, “No: do not change the digit in the given place value.” To the right of this, it says “Because 9 is greater than or equal to ” To the right of this, we have the number 18.379 with the 9 marked “delete” and the 7 marked “add 1.”

Finally, the last step reads “Step 4. Rewrite the number, removing all digits to the right of the rounding digit.” To the right of this, we have 18.38 followed by “18.38 is 18.379 rounded to the nearest hundredth.”

We summarize the steps for rounding a decimal here.

Round 18.379 to the nearest ⓐ tenth ⓑ whole number.

Solution

Round 18.379

ⓐ to the nearest tenth

Locate the tenths place with an arrow.
Underline the digit to the right of the given place value.
Because 7 is greater than or equal to 5, add 1 to the 3.
Rewrite the number, deleting all digits to the right of the rounding digit.
Notice that the deleted digits were NOT replaced with zeros.	So, 18.379 rounded to the nearest tenth is 18.4.

ⓑ to the nearest whole number

Locate the ones place with an arrow.
Underline the digit to the right of the given place value.
Since 3 is not greater than or equal to 5, do not add 1 to the 8.
Rewrite the number, deleting all digits to the right of the rounding digit.
	So, 18.379 rounded to the nearest whole number is 18.

Add and Subtract Decimals

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

Add: $23.5 + 41.38 .$

Solution

Step-by-step guide on adding decimal numbers, demonstrating alignment, placeholders, and the final sum.
Write the numbers so the decimal points line up vertically.	$\underset{______}{\begin{matrix} 23.5 \\ + 41.38 \end{matrix}}$
Put 0 as a placeholder after the 5 in 23.5. Remember, $\frac{5}{10} = \frac{50}{100} so 0.5 = 0.50$ .	$\underset{______}{\begin{matrix} 23.50 \\ + 41.38 \end{matrix}}$
Add the numbers as if they were whole numbers. Then place the decimal point in the sum.	$\begin{matrix} \underset{______}{\begin{matrix} 23.50 \\ + 41.38 \end{matrix}} \\ 64.88 \end{matrix}$

Subtract: $20 - 14.65 .$

Solution

Step-by-step process for subtracting a decimal from a whole number, exemplified by 20 - 14.65.
	$20 - 14.65$
Write the numbers so the decimal points line up vertically. Remember, 20 is a whole number, so place the decimal point after the 0.	$\underset{______}{\begin{matrix} 20. \\ - 14.65 \end{matrix}}$
Put in zeros to the right as placeholders.	$\begin{matrix} 20.00 \\ \underset{______}{−14.65} \end{matrix}$
Subtract and place the decimal point in the answer.	$\begin{matrix} \underset{__________}{\begin{matrix} \overset{1}{2} \overset{\overset{9}{10}}{0} . \overset{\overset{9}{10}}{0} \overset{10}{0} \\ - 1 4 . 6 5 \end{matrix}} \\ 5 . 3 5 \end{matrix}$

Multiply and Divide 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 convert them to fractions and then multiply.

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. Look for a pattern!


Convert to fractions.	$Two mathematical expressions demonstrating the multiplication of fractions: 3/10 multiplied by 7/10, and 2/10 multiplied by 46/100.$
Multiply.	$Two fractions are displayed side by side: '21/100' and '92/1000'. The numbers are black against a white background.$
Convert to decimals.

Notice, in the first example, we multiplied two numbers that each had one digit after the decimal point and the product had two decimal places. In the second example, we multiplied a number with one decimal place by a number with two decimal places and the product had three decimal places.

We multiply the numbers just as we do whole numbers, temporarily ignoring the decimal point. We then count the number of decimal points in the factors and that sum tells us the number of decimal places in the product.

The rules for multiplying positive and negative numbers apply to decimals, too, of course!

When multiplying two numbers,

if their signs are the same the product is positive.
if their signs are different the product is negative.

When we multiply signed decimals, first we determine the sign of the product and then multiply as if the numbers were both positive. Finally, we write the product with the appropriate sign.

Multiply: $(−3.9) (4.075) .$

Solution

	(−3.9)(4.075)
The signs are different. The product will be negative.
Write in vertical format, lining up the numbers on the right.
Multiply.
Add the number of decimal places in the factors (1 + 3). Place the decimal point 4 places from the right.
The signs are different, so the product is negative.	(−3.9)(4.075) = −15.8925

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

Multiply 5.63 ⓐ by 10 ⓑ by 100 ⓒ by 1,000.

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.

ⓐ

	5.63(10)
There is 1 zero in 10, so move the decimal point 1 place to the right.

ⓑ

	5.63(100)
There are 2 zeros in 100, so move the decimal point 2 places to the right.

ⓒ

	5.63(1,000)
There are 3 zeros in 1,000, so move the decimal point 3 places to the right.
A zero must be added at the end.

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 divide decimals, determine what power of 10 to multiply the denominator by to make it a whole number. Then multiply the numerator by that same power of $10 .$ 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. For example:

\begin{matrix} \frac{0.8}{0.4} \\ \frac{0.8 (10)}{0.4 (10)} \\ \frac{8}{4} \end{matrix}

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.

We review the notation and vocabulary for division:

\underset{dividend}{a} \div \underset{divisor}{b} = \underset{quotient}{c} \underset{divisor}{b} \begin{matrix} \underset{quotient}{c} \\ \underset{dividend}{a} \end{matrix}

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

Divide: $−25.65 \div (−0.06) .$

Solution

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

	$−25.65 \div (−0.06)$
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.
Divide. Place the decimal point in the quotient above the decimal point in the dividend.
Write the quotient with the appropriate sign.	$−25.65 \div (−0.06) = 427.5$

A common application of dividing whole numbers into decimals is when we want to find the price of one item that is sold as part of a multi-pack. For example, suppose a case of 24 water bottles costs $3.99. To find the price of one water bottle, we would divide $3.99 by 24. We show this division in Example 11. In calculations with money, we will round the answer to the nearest cent (hundredth).

Divide: $$3.99 \div 24 .$

Solution

	$$3.99 \div 24$
Place the decimal point in the quotient above the decimal point in the dividend.
Divide as usual. When do we stop? Since this division involves money, we round it to the nearest cent (hundredth.) To do this, we must carry the division to the thousandths place.
Round to the nearest cent.	$$0.166 \approx $0.17$ $$3.99 \div 24 \approx $0.17$

Convert Decimals, Fractions, and Percents

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

0 0.03 = \frac{3}{100}

Notice, when the number to the left of the decimal is zero, we get a fraction whose numerator is less than its denominator. Fractions like this are called proper fractions.

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

Write 0.374 as a fraction.

Solution

	$0.374$
Determine the place value of the final digit.
Write the fraction for 0.374: The numerator is 374. The denominator is 1,000.	$\frac{374}{1000}$
Simplify the fraction.	$\frac{2 \cdot 187}{2 \cdot 500}$
Divide out the common factors.	$\frac{187}{500}$ so, $0.374 = \frac{187}{500}$

Did you notice that the number of zeros in the denominator of $\frac{374}{1,000}$ is the same as the number of decimal places in 0.374?

We’ve learned to convert decimals to fractions. Now we will do the reverse—convert fractions to decimals. Remember that the fraction bar means division. So $\frac{4}{5}$ can be written $4 \div 5$ or $54 .$ This leads to the following method for converting a fraction to a decimal.

Write $- \frac{5}{8}$ as a decimal.

Solution

Since a fraction bar means division, we begin by writing $\frac{5}{8}$ as $85 .$ Now divide.

This is a long division problem with 8 dividing 5.000 and 0.625 as the quotient. Below 5.000 we have 48, a solid horizontal line, 20, 16, a solid horizontal line, 40, 40, and a final horizontal line. So five eighths equals 0.625.

When we divide, we will not always get a zero remainder. Sometimes the quotient ends up with a decimal that repeats. A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly. A bar is placed over the repeating block of digits to indicate it repeats.

A bar is placed over the repeating block of digits to indicate it repeats.

Write $\frac{43}{22}$ as a decimal.

Solution

Sometimes we may have to simplify expressions with fractions and decimals together.

Simplify: $\frac{7}{8} + 6.4 .$

Solution

First we must change one number so both numbers are in the same form. We can change the fraction to a decimal, or change the decimal to a fraction. Usually it is easier to change the fraction to a decimal.

		$\frac{7}{8} + 6.4$
Change $\frac{7}{8}$ to a decimal.
Add.		$0.875 + 6.4$
		$7.275$
		So, $\frac{7}{8} + 6.4 = 7.275$

A percent is a ratio whose denominator is 100. Percent means per hundred. We use the percent symbol, %, to show percent.

Since a percent is a ratio, it can easily be expressed as a fraction. Percent means per 100, so the denominator of the fraction is 100. We then change the fraction to a decimal by dividing the numerator by the denominator.

Demonstrates converting percentages (6%, 78%, 135%) into fractions with a denominator of 100 and subsequently into decimal form.
	$6%$	$78%$	$135%$
Write as a ratio with denominator 100.	$\frac{6}{100}$	$\frac{78}{100}$	$\frac{135}{100}$
Change the fraction to a decimal by dividing the numerator by the denominator.	$0.06$	$0.78$	$1.35$

Do you see the pattern? To convert a percent number to a decimal number, we move the decimal point two places to the left.

The first part of this figure shows 6% with an arrow drawn from between the 6 and the percentage sign to the space to the left of 6 and then to the space further to the left of that space. Below this, the number 0.06 is given. The second part of this figure shows 78% with an arrow drawn from between the 8 and the percentage sign to the space between the 7 and the 8 and then to the space to the left of the 7. Below this, the number 0.78 is given. The third part of this figure shows 2.7% with an arrow drawn from the decimal point to the space to the left of the 2 and then to the space further to the left of that space. Below this, the number 0.027 is given. The fourth part of this figure shows 135% with an arrow drawn from between the 5 and the percentage sign to the space between 3 and 5 and then to the space between 1 and 3. Below this, the number 1.35 is given.

Convert each percent to a decimal: ⓐ 62% ⓑ 135% ⓒ 35.7%.

Solution

ⓐ

Move the decimal point two places to the left.	$0.62$

ⓑ

Move the decimal point two places to the left.	$1.35$

ⓒ

Move the decimal point two places to the left.	$0.357$

Converting a decimal to a percent makes sense if we remember the definition of percent and keep place value in mind.

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.

This table illustrates the conversion of three decimal numbers (0.83, 1.05, and 0.075) into their equivalent fraction and percentage forms.
	$0.83$	$1.05$	$0.075$
Write as a fraction.	$\frac{83}{100}$	$1 \frac{5}{100}$	$\frac{75}{1000}$
The denominator is 100.		$\frac{105}{100}$	$\frac{7.5}{100}$
Write the ratio as a percent.	$83%$	$105%$	$7.5%$

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

The first part of this figure shows 0.05 with an arrow drawn from the decimal point to the space between 0 and 5 and then to the space after 5. Below this, the number 5% is given. The second part of this figure shows 0.83 with an arrow drawn from the decimal point to the space between 8 and 3 and then to the space after 3. Below this, the number 83% is given. The third part of this figure shows 1.05 with an arrow drawn from the decimal point to the space between 0 and 5 and then to the space after 5. Below this, the number 105% is given. The fourth part of this figure shows 0.075 with an arrow drawn from the decimal point to the space between 0 and 7 and then to the space between 7 and 5. Below this, the number 7.5% is given. The fifth part of this figure shows 0.3 with an arrow drawn from the decimal point to the space after 3 and then to space further to the right of that 3. Below this, the number 30% is given.

Convert each decimal to a percent: ⓐ 0.51 ⓑ 1.25 ⓒ 0.093.

Solution

ⓐ

Move the decimal point two places to the right.	$51%$

ⓑ

Move the decimal point two places to the right.	$125%$

ⓒ

Move the decimal point two places to the right.	$9.3%$

Key Concepts

Name a Decimal
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
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.
Round a Decimal
1. Locate the given place value and mark it with an arrow.
2. Underline the digit to the right of the place value.
3. Is 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, deleting all digits to the right of the rounding digit.
Add or Subtract Decimals
1. Write the numbers so the decimal points line up vertically.
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 Decimals
1. Determine the sign of the product.
2. Write in vertical format, lining up the numbers on the right. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
3. Place the decimal point. The number of decimal places in the product is the sum of the decimal places in the factors.
4. Write the product with the appropriate sign.
Multiply a Decimal by a Power of Ten
1. Move the decimal point to the right the same number of places as the number of zeros in the power of 10.
2. Add zeros at the end of the number as needed.
Divide Decimals
1. Determine the sign of the quotient.
2. Make the divisor a whole number by “moving” the decimal point all the way to the right. “Move” the decimal point in the dividend the same number of places - adding zeros as needed.
3. Divide. Place the decimal point in the quotient above the decimal point in the dividend.
4. Write the quotient with the appropriate sign.
Convert a Decimal to a Proper Fraction
1. Determine the place value of the final digit.
2. Write the fraction: numerator—the ‘numbers’ to the right of the decimal point; denominator—the place value corresponding to the final digit.
Convert a Fraction to a Decimal Divide the numerator of the fraction by the denominator.

Practice Makes Perfect

Name and Write Decimals

In the following exercises, write as a decimal.

Twenty-nine and eighty-one hundredths

Solution

29.81

Sixty-one and seventy-four hundredths

Seven tenths

Solution

0.7

Six tenths

Twenty-nine thousandth

Solution

0.029

Thirty-five thousandths

Negative eleven and nine ten-thousandths

Solution

$−11.0009$

Negative fifty-nine and two ten-thousandths

In the following exercises, name each decimal.

5.5

Solution

five and five tenths

14.02

8.71

Solution

eight and seventy-one hundredths

2.64

0.002

Solution

two thousandths

0.479

$− 17 .9$

Solution

negative seventeen and nine tenths

$− 31 .4$

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

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

ⓐ 5.78 ⓑ 5.8 ⓒ 6

1.6381

63.479

Solution

ⓐ 63.48 ⓑ 63.5 ⓒ 63

$84 .281$

Add and Subtract Decimals

In the following exercises, add or subtract.

$16.92 + 7.56$

Solution

24.48

$248.25 - 91.29$

$21.76 - 30.99$

Solution

$−9.23$

$38.6 + 13.67$

$−16.53 - 24.38$

Solution

$−40.91$

$−19.47 - 32.58$

$−38.69 + 31.47$

Solution

$−7.22$

$29.83 + 19.76$

$72.5 - 100$

Solution

$−27.5$

$86.2 - 100$

$15 + 0.73$

Solution

15.73

$27 + 0.87$

$91.95 - (−10.462)$

Solution

102.412

$94.69 - (−12.678)$

$55.01 - 3.7$

Solution

51.31

$59.08 - 4.6$

$2.51 - 7.4$

Solution

$−4.89$

$3.84 - 6.1$

Multiply and Divide Decimals

In the following exercises, multiply.

$(0.24) (0.6)$

Solution

0.144

$(0.81) (0.3)$

$(5.9) (7.12)$

Solution

42.008

$(2.3) (9.41)$

$(−4.3) (2.71)$

Solution

$−11.653$

$(−8.5) (1.69)$

$(−5.18) (−65.23)$

Solution

337.8914

$(−9.16) (−68.34)$

$(0.06) (21.75)$

Solution

1.305

$(0.08) (52.45)$

$(9.24) (10)$

Solution

92.4

$(6.531) (10)$

$(55.2) (1000)$

Solution

55,200

$(99.4) (1000)$

In the following exercises, divide.

$4.75 \div 25$

Solution

0.19

$12.04 \div 43$

$$117.25 \div 48$

Solution

$2.44

$$109.24 \div 36$

$0.6 \div 0.2$

Solution

$0.8 \div 0.4$

$1.44 \div (−0.3)$

Solution

$−4.8$

$1.25 \div (−0.5)$

$−1.75 \div (−0.05)$

Solution

$−1.15 \div (−0.05)$

$5.2 \div 2.5$

Solution

2.08

$6.5 \div 3.25$

$11 \div 0.55$

Solution

$14 \div 0.35$

Convert Decimals, Fractions and Percents

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

0.04

Solution

$\frac{1}{25}$

0.19

0.52

Solution

$\frac{13}{25}$

0.78

1.25

Solution

$\frac{5}{4}$

1.35

0.375

Solution

$\frac{3}{8}$

0.464

0.095

Solution

$\frac{19}{200}$

0.085

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

$\frac{17}{20}$

Solution

0.85

$\frac{13}{20}$

$\frac{11}{4}$

Solution

2.75

$\frac{17}{4}$

$- \frac{310}{25}$

Solution

$−12.4$

$- \frac{284}{25}$

$\frac{15}{11}$

Solution

$1. \overset{—}{36}$

$\frac{18}{11}$

$\frac{15}{111}$

Solution

$0. \overset{—}{135}$

$\frac{25}{111}$

$2.4 + \frac{5}{8}$

Solution

3.025

$3.9 + \frac{9}{20}$

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

Solution

0.01

63%

Solution

0.63

71%

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 decimal to a percent.

0.01

Solution

0.03

1.35

Solution

135%

1.56

Solution

300%

0.0875

Solution

8.75%

0.0625

2.254

Solution

225.4%

2.317

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

ⓐ $58,966 ⓑ $59,000 ⓒ $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 $1,624.99 and when the clerk calculated the sales tax it came out to exactly $142.186625. Round the sales tax to the nearest ⓐ penny and ⓑ dollar.

Solution

ⓐ $142.19; ⓑ $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 and ⓑ dollar.

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

ⓐ $243.57 ⓑ $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

How does knowing about US money help you learn about decimals?

Solution

Answers may vary

Explain how you write “three and nine hundredths” as a decimal.

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

Answers may vary

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

Self Check

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

$This is a table that has six rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “name and write decimals,” “round decimals,” “add and subtract decimals,” “multiply and divide decimals,” and “convert decimals, fractions and percents.” The rest of the cells are blank.$

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

decimal

A decimal is another way of writing a fraction whose denominator is a power of ten.

percent

A percent is a ratio whose denominator is 100.

repeating decimal

A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly.

The Real Numbers

Simplify Expressions with Square Roots

Remember that when a number n is multiplied by itself, we write $n^{2}$ and read it “n squared.” The result is called the square of n. For example,

\begin{matrix} 8^{2} & read ‘ 8 squared’ \\ 64 & 64 is called the square of 8 . \end{matrix}

Similarly, 121 is the square of 11, because $11^{2}$ is 121.

Complete the following table to show the squares of the counting numbers 1 through 15.

There is a table with two rows and 17 columns. The first row reads from left to right Number, n, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15. The second row reads from left to right Square, n squared, blank, blank, blank, blank, blank, blank, blank, 64, blank, blank, 121, blank, blank, blank, and blank.

The numbers in the second row are called perfect square numbers. It will be helpful to learn to recognize the perfect square numbers.

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

\begin{matrix} {(−3)}^{2} = 9 & {(−8)}^{2} = 64 & {(−11)}^{2} = 121 & {(−15)}^{2} = 225 \end{matrix}

Did you notice that these squares are the same as the squares of the positive numbers?

Sometimes we will need to look at the relationship between numbers and their squares in reverse. Because $10^{2} = 100,$ we say 100 is the square of 10. We also say that 10 is a square root of 100. A number whose square is $m$ is called a square root of 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 wanted the positive square root of a positive number? The radical sign, $\sqrt{m},$ denotes the positive square root. The positive square root is called the principal square root. When we use the radical sign that always means we want the principal square root.

We also use the radical sign for the square root of zero. Because $0^{2} = 0,$ $\sqrt{0} = 0 .$ Notice that zero has only one square root.

Since 10 is the principal square root of 100, we write $\sqrt{100} = 10 .$ You may want to complete the following table to help you recognize square roots.

There is a table with two rows and 15 columns. The first row reads from left to right 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 row consists of all blanks except for the tenth cell under the square root of 100, which reads 10.

Simplify: ⓐ $\sqrt{25}$ ⓑ $\sqrt{121} .$

Solution

Demonstrates finding square roots using known squared values.
ⓐ Since $5^{2} = 25$	$\begin{matrix} \sqrt{25} \\ 5 \end{matrix}$
ⓑ Since $11^{2} = 121$	$\begin{matrix} \sqrt{121} \\ 11 \end{matrix}$

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

Simplify: ⓐ $− \sqrt{9}$ ⓑ $− \sqrt{144} .$

Solution

This table illustrates examples of evaluating mathematical expressions where a negative sign precedes a square root.
ⓐ The negative is in front of the radical sign.	$\begin{matrix} - \sqrt{9} \\ - 3 \end{matrix}$
ⓑ The negative is in front of the radical sign.	$\begin{matrix} - \sqrt{144} \\ - 12 \end{matrix}$

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

We have already described numbers as counting numbers, whole numbers, and integers. What is the difference between these types of numbers?

\begin{matrix} Counting numbers & 1, 2, 3, 4, … \\ Whole numbers & 0, 1, 2, 3, 4, … \\ Integers & … −3, −2, −1, 0, 1, 2, 3, … \end{matrix}

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

All signed fractions, such as $\frac{4}{5}, - \frac{7}{8}, \frac{13}{4}, - \frac{20}{3}$ are rational numbers. Each numerator and each denominator is an integer.

Are integers rational numbers? To decide if an integer is a rational number, we try to write it as a ratio of two integers. Each integer can be written as a ratio of integers in many ways. For example, 3 is equivalent to $\frac{3}{1}, \frac{6}{2}, \frac{9}{3}, \frac{12}{4}, \frac{15}{5} …$

An easy way to write an integer as a ratio of integers is to write it as a fraction with denominator one.

\begin{matrix} 3 = \frac{3}{1} & −8 = - \frac{8}{1} & 0 = \frac{0}{1} \end{matrix}

Since any integer can be written as the ratio of two integers, all integers are rational numbers! Remember that the counting numbers and 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 $7 \frac{3}{10},$ we can write it as an improper fraction, $\frac{73}{10} .$ 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 as the ratio of two integers: ⓐ $−27$ ⓑ 7.31.

Solution

Examples demonstrating the conversion of integers and decimals into various fractional forms.
ⓐ Write it as a fraction with denominator 1.	$\begin{matrix} −27 \\ \frac{−27}{1} \end{matrix}$
ⓑ Write it as a mixed number. Remember, 7 is the whole number and the decimal part, 0.31, indicates hundredths. Convert to an improper fraction.	$\begin{matrix} 7.31 \\ 7 \frac{31}{100} \\ \frac{731}{100} \end{matrix}$

So we see that $−27$ and 7.31 are both rational numbers, since they can be written as the ratio of two integers.

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 $a = \frac{a}{1}$ for any integer, a. We can also change any integer to a decimal by adding a decimal point and a zero.

\begin{matrix} Integer & −2 & −1 & 0 & 1 & 2 & 3 \\ Decimal form & −2.0 & −1.0 & 0.0 & 1.0 & 2.0 & 3.0 \end{matrix}

These decimal numbers stop.

We have also seen that every fraction is a rational number. Look at the decimal form of the fractions we considered above.

\begin{matrix} Ratio of integers & \frac{4}{5} & - \frac{7}{8} & \frac{13}{4} & - \frac{20}{3} \\ The decimal form & 0.8 & −0.875 & 3.25 & \begin{matrix} - 6.666 … \\ - 6. \overset{–}{6} \end{matrix} \end{matrix}

These decimals either stop or repeat.

What do these examples tell us?

Every rational number can be written both as a ratio of integers, $(\frac{p}{q},$ where p and q are integers and $q \neq 0),$ and as a decimal that either stops or repeats.

Here are the numbers we looked at above expressed as a ratio of integers and as a decimal:

Fractions					Integers
Number	$\frac{4}{5}$	$- \frac{7}{8}$	$\frac{13}{4}$	$- \frac{20}{3}$	$−2$	$−1$	$0$	$1$	$2$	$3$
Ratio of Integers	$\frac{4}{5}$	$- \frac{7}{8}$	$\frac{13}{4}$	$- \frac{20}{3}$	$- \frac{2}{1}$	$- \frac{1}{1}$	$\frac{0}{1}$	$\frac{1}{1}$	$\frac{2}{1}$	$\frac{3}{1}$
Decimal Form	$0.8$	$−0.875$	$3.25$	$−6. \overset{–}{6}$	$−2.0$	$−1.0$	$0.0$	$1.0$	$2.0$	$3.0$

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 . . .

We can even create a decimal pattern that does not stop or repeat, such as

2.01001000100001…

Numbers whose decimal form does not stop or repeat cannot be written as a fraction of integers. We call these numbers irrational.

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

Given the numbers $0.58 \overset{–}{3}, 0.47, 3.605551275 . . .$ list the ⓐ rational numbers ⓑ irrational numbers.

Solution

This table illustrates how to classify numbers as rational or irrational by examining their decimal representations, focusing on whether decimals repeat, stop, or neither.
ⓐ Look for decimals that repeat or stop.	The 3 repeats in $0.58 \overset{–}{3}$ . The decimal 0.47 stops after the 7. So $0.58 \overset{–}{3}$ and 0.47 are rational.
ⓑ Look for decimals that neither stop nor repeat.	$3.605551275 …$ has no repeating block of digits and it does not stop. So $3.605551275 …$ is irrational.

For each number given, identify whether it is rational or irrational: ⓐ $\sqrt{36}$ ⓑ $\sqrt{44} .$

Solution

ⓐ Recognize that 36 is a perfect square, since $6^{2} = 36 .$ So $\sqrt{36} = 6,$ therefore $\sqrt{36}$ is rational.
ⓑ Remember that $6^{2} = 36$ and $7^{2} = 49,$ so 44 is not a perfect square. Therefore, the decimal form of $\sqrt{44}$ will never repeat and never stop, so $\sqrt{44}$ is irrational.

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

All the numbers we use in elementary algebra are real numbers. Figure 1 illustrates how the number sets we’ve discussed in this section fit together.

This figure consists of a Venn diagram. To start there is a large rectangle marked Real Numbers. The right half of the rectangle consists of Irrational Numbers. The left half consists of Rational Numbers. Within the Rational Numbers rectangle, there are Integers …, negative 2, negative 1, 0, 1, 2, …. Within the Integers rectangle, there are Whole Numbers 0, 1, 2, 3, … Within the Whole Numbers rectangle, there are Counting Numbers 1, 2, 3, … — This chart shows the number sets that make up the set of real numbers. Does the term “real numbers” seem strange to you? Are there any numbers that are not “real,” and, if so, what could they be?

Can we simplify $\sqrt{−25} ?$ Is there a number whose square is $−25 ?$

{()}^{2} = −25 ?

None of the numbers that we have dealt with so far has a square that is $−25 .$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $\sqrt{−25} .$

The square root of a negative number is not a real number.

For each number given, identify whether it is a real number or not a real number: ⓐ $\sqrt{−169}$ ⓑ $− \sqrt{64} .$

Solution

ⓐ There is no real number whose square is $−169 .$ Therefore, $\sqrt{−169}$ is not a real number.
ⓑ Since the negative is in front of the radical, $− \sqrt{64}$ is $−8,$ Since $−8$ is a real number, $− \sqrt{64}$ is a real number.

Given the numbers $−7, \frac{14}{5}, 8, \sqrt{5}, 5.9, − \sqrt{64},$ list the ⓐ whole numbers ⓑ integers ⓒ rational numbers ⓓ irrational numbers ⓔ real numbers.

Solution

ⓐ Remember, the whole numbers are 0, 1, 2, 3, … and 8 is the only whole number given.
ⓑ The integers are the whole numbers, their opposites, and 0. So the whole number 8 is an integer, and $−7$ is the opposite of a whole number so it is an integer, too. Also, notice that 64 is the square of 8 so $− \sqrt{64} = −8 .$ So the integers are $−7, 8, − \sqrt{64} .$
ⓒ Since all integers are rational, then $−7, 8, − \sqrt{64}$ are rational. Rational numbers also include fractions and decimals that repeat or stop, so $\frac{14}{5} and 5.9$ are rational. So the list of rational numbers is $−7, \frac{14}{5}, 8, 5.9, - \sqrt{64} .$
ⓓ Remember that 5 is not a perfect square, so $\sqrt{5}$ is irrational.
ⓔ All the numbers listed are real numbers.

Locate Fractions on the Number Line

The last time we looked at the number line, it only had positive and negative integers on it. We now want to include fractions and decimals on it.

Let’s start with fractions and locate $\frac{1}{5}, - \frac{4}{5}, 3, \frac{7}{4}, - \frac{9}{2}, −5, and \frac{8}{3}$ on the number line.

We’ll start with the whole numbers $3$ and $−5 .$ because they are the easiest to plot. See Figure 2.

The proper fractions listed are $\frac{1}{5} and - \frac{4}{5} .$ We know the proper fraction $\frac{1}{5}$ has value less than one and so would be located between $0 and 1.$ The denominator is 5, so we divide the unit from 0 to 1 into 5 equal parts $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} .$ We plot $\frac{1}{5} .$ See Figure 2.

Similarly, $- \frac{4}{5}$ is between 0 and $−1 .$ After dividing the unit into 5 equal parts we plot $- \frac{4}{5} .$ See Figure 2.

Finally, look at the improper fractions $\frac{7}{4}, - \frac{9}{2}, \frac{8}{3} .$ These are fractions in which the numerator is greater than the denominator. Locating these points may be easier if you change each of them to a mixed number. See Figure 2.

\begin{matrix} \frac{7}{4} = 1 \frac{3}{4} & - \frac{9}{2} = −4 \frac{1}{2} & \frac{8}{3} = 2 \frac{2}{3} \end{matrix}

Figure 2 shows the number line with all the points plotted.

There is a number line shown that runs from negative 6 to positive 6. From left to right, the numbers marked are negative 5, negative 9/2, negative 4/5, 1/5, 4/5, 8/3, and 3. The number negative 9/2 is halfway between negative 5 and negative 4. The number negative 4/5 is slightly to the right of negative 1. The number 1/5 is slightly to the right of 0. The number 4/5 is slightly to the left of 1. The number 8/3 is between 2 and 3, but a little closer to 3.

Locate and label the following on a number line: $4, \frac{3}{4}, - \frac{1}{4}, −3, \frac{6}{5}, - \frac{5}{2}, and \frac{7}{3} .$

Solution

Locate and plot the integers, $4, −3 .$

Locate the proper fraction $\frac{3}{4}$ first. The fraction $\frac{3}{4}$ is between 0 and 1. Divide the distance between 0 and 1 into four equal parts then, we plot $\frac{3}{4} .$ Similarly plot $- \frac{1}{4} .$

Now locate the improper fractions $\frac{6}{5}, - \frac{5}{2}, \frac{7}{3} .$ It is easier to plot them if we convert them to mixed numbers and then plot them as described above: $\frac{6}{5} = 1 \frac{1}{5}, - \frac{5}{2} = −2 \frac{1}{2}, \frac{7}{3} = 2 \frac{1}{3} .$

There is a number line shown that runs from negative 6 to positive 6. From left to right, the numbers marked are negative 3, negative 5/2, negative 1/4, 3/4, 6/5, 7/3, and 4. The number negative 5/2 is halfway between negative 3 and negative 2. The number negative 1/4 is slightly to the left of 0. The number 3/4 is slightly to the left of 1. The number 6/5 is slightly to the right of 1. The number 7/3 is between 2 and 3, but a little closer to 2.

In Example 9, we’ll use the inequality symbols to order fractions. In previous chapters we used the number line to order numbers.

a < b “a is less than b” when a is to the left of b on the number line
a > b “a is greater than b” when 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 >. It may be helpful to refer Figure 3.

ⓐ $- \frac{2}{3}___−1$ ⓑ $−3 \frac{1}{2}___−3$ ⓒ $- \frac{3}{4}___- \frac{1}{4}$ ⓓ $−2___- \frac{8}{3}$

There is a number line shown that runs from negative 4 to positive 4. From left to right, the numbers marked are negative 3 and 1/2, negative 3, negative 8/3, negative 2, negative 1, negative 3/4, negative 2/3, and negative 1/4. The number negative 3 and 1/2 is between negative 4 and negative 3 The number negative 8/3 is between negative 3 and negative 2, but closer to negative 3. The numbers negative 3/4, negative 2/3, and negative 1/4 are all between negative 1 and 0.

Solution

This table translates verbal descriptions of number positions on a number line into their corresponding mathematical inequalities, illustrating concepts of greater than and less than.
ⓐ $- \frac{2}{3}$ is to the right of $−1$ on the number line.	$\begin{matrix} - \frac{2}{3}___−1 \\ - \frac{2}{3} > −1 \end{matrix}$
ⓑ $−3 \frac{1}{2}$ is to the right of $−3$ on the number line.	$\begin{matrix} −3 \frac{1}{2}___−3 \\ −3 \frac{1}{2} < −3 \end{matrix}$
ⓒ $- \frac{3}{4}$ is to the right of $- \frac{1}{4}$ on the number line.	$\begin{matrix} - \frac{3}{4}___- \frac{1}{4} \\ - \frac{3}{4} < - \frac{1}{4} \end{matrix}$
ⓓ $−2$ is to the right of $- \frac{8}{3}$ on the number line.	$\begin{matrix} −2___- \frac{8}{3} \\ −2 > - \frac{8}{3} \end{matrix}$

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 the number line.

Solution

A proper fraction has value less than one. The decimal number 0.4 is equivalent to $\frac{4}{10},$ a proper fraction, so 0.4 is located between 0 and 1. On a number line, divide the interval between 0 and 1 into 10 equal parts. Now label the parts 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0. We write 0 as 0.0 and 1 and 1.0, so that the numbers are consistently in tenths. Finally, mark 0.4 on the number line. See Figure 4.

There is a number line shown that runs from 0.0 to 1. The only point given is 0.4, which is between 0.3 and 0.5.

Locate $−0.74$ on the number line.

Solution

The decimal $−0.74$ is equivalent to $- \frac{74}{100},$ so it is located between 0 and $−1 .$ On a number line, mark off and label the hundredths in the interval between 0 and $−1 .$ See Figure 5.

There is a number line shown that runs from negative 1.00 to 0.00. The only point given is negative 0.74, which is between negative 0.8 and negative 0.7.

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,

$0.40 > 0.04$

Again, we can use the number line to order numbers.

a < b “a is less than b” when a is to the left of b on the number line
a > b “a is greater than b” when a is to the right of b on the number line

Where are 0.04 and 0.40 located on the number line? See Figure 6.

There is a number line shown that runs from negative 0.0 to 1.0. From left to right, there are points 0.04 and 0.4 marked. The point 0.04 is between 0.0 and 0.1. The point 0.4 is between 0.3 and 0.5.

We see that 0.40 is to the right of 0.04 on the number line. This is another way to demonstrate that 0.40 > 0.04.

How does 0.31 compare to 0.308? This doesn’t translate into money to make it easy to compare. But if we convert 0.31 and 0.308 into fractions, we can tell which is larger.

	0.31	0.308
Convert to fractions.	$\frac{31}{100}$	$\frac{308}{1000}$
We need a common denominator to compare them.	$A mathematical expression showing the fraction (31 * 10) / (100 * 10), illustrating the multiplication of both the numerator and denominator by 10, highlighted in red.$	$The mathematical fraction 308 over 1000, representing three hundred eight thousandths.$
	$\frac{310}{1000}$	$\frac{308}{1000}$

Because 310 > 308, we know that $\frac{310}{1000} > \frac{308}{1000} .$ Therefore, 0.31 > 0.308.

Notice what we did in converting 0.31 to a fraction—we started with the fraction $\frac{31}{100}$ and ended with the equivalent fraction $\frac{310}{1000} .$ Converting $\frac{310}{1000}$ 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!

\frac{31}{100} = \frac{310}{1000} and 0.31 = 0.310

We say 0.31 and 0.310 are equivalent decimals.

We use equivalent decimals when we order decimals.

The steps we take to order decimals are summarized here.

Order $0.64___0.6$ using $<$ or $> .$

Solution

Demonstrates step-by-step comparison of decimal numbers (0.64 and 0.6) by aligning decimal points and equalizing decimal places.
Write the numbers one under the other, lining up the decimal points.	$\begin{matrix} 0.64 \\ 0.6 \end{matrix}$
Add a zero to 0.6 to make it a decimal with 2 decimal places. Now they are both hundredths.	$\begin{matrix} 0.64 \\ 0.60 \end{matrix}$
64 is greater than 60.	$64 > 60$
64 hundredths is greater than 60 hundredths.	$0.64 > 0.60$
	$0.64 > 0.6$

Order $0.83___0.803$ using $<$ or $> .$

Solution

Step-by-step guide for comparing decimal numbers, using 0.83 and 0.803 as an example to show alignment and zero padding.
	$0.83___0.803$
Write the numbers one under the other, lining up the decimals.	$\begin{matrix} 0.83 \\ 0.803 \end{matrix}$
They do not have the same number of digits. Write one zero at the end of 0.83.	$\begin{matrix} 0.830 \\ 0.803 \end{matrix}$
Since $830 > 803$ , 830 thousandths is greater than 803 thousandths.	$0.830 > 0.803$
	$0.83 > 0.803$

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 $−9 < −6 .$ See Figure 7.

There is a number line shown that runs from negative 10 to 0. There are not points given and the hashmarks exist at every integer between negative 10 and 0.

If we zoomed in on the interval between 0 and $−1,$ as shown in Example 14, we would see in the same way that $−0.2 > −0.3 and - 0.9 < −0.6 .$

Use $<$ or $>$ to order $−0.1___−0.8 .$

Solution

Steps demonstrating how to compare two negative decimal numbers, -0.1 and -0.8, by aligning their decimal points and comparing their values.
	$−0.1___−0.8$
Write the numbers one under the other, lining up the decimal points. They have the same number of digits.	$\begin{matrix} −0.1 \\ −0.8 \end{matrix}$
Since $−1 > −8$ , −1 tenth is greater than −8 tenths.	$−0.1 > −0.8$

Key Concepts

Square Root Notation
$\sqrt{m}$ is read ‘the square root of m.’ If $m = n^{2},$ then $\sqrt{m} = n,$ for $n \geq 0 .$
Order Decimals
1. Write the numbers one under the other, lining up the decimal points.
2. Check to see if both numbers have the same number of digits. If not, write zeros at the end of the one with fewer digits to make them match.
3. Compare the numbers as if they were whole numbers.
4. Order the numbers using the appropriate inequality sign.

Practice Makes Perfect

Simplify Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{36}$

Solution

$\sqrt{4}$

$\sqrt{64}$

Solution

$\sqrt{169}$

$\sqrt{9}$

Solution

$\sqrt{16}$

$\sqrt{100}$

Solution

$\sqrt{144}$

$− \sqrt{4}$

Solution

$−2$

$− \sqrt{100}$

$− \sqrt{1}$

Solution

$−1$

$− \sqrt{121}$

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

In the following exercises, write as the ratio of two integers.

ⓐ 5 ⓑ 3.19

Solution

ⓐ $\frac{5}{1}$ ⓑ $\frac{319}{100}$

ⓐ 8 ⓑ 1.61

ⓐ $− 12$ ⓑ 9.279

Solution

ⓐ $\frac{−12}{1}$ ⓑ $\frac{9279}{1000}$

ⓐ $− 16$ ⓑ 4.399

In the following exercises, list the ⓐ rational numbers, ⓑ irrational numbers

$0.75, 0.22 \overset{–}{3}, 1.39174 …$

Solution

ⓐ $0.75, 0.22 \overset{–}{3}$ ⓑ $1.39174 …$

$0.36, 0.94729 …, 2.52 \overset{–}{8}$

$0.4 \overset{–}{5}, 1.919293 …, 3.59$

Solution

ⓐ $0.4 \overset{–}{5}, 3.59$ ⓑ $1.919293 …$

$0.1 \overset{–}{3}, 0.42982 …, 1.875$

In the following exercises, identify whether each number is rational or irrational.

ⓐ $\sqrt{25}$ ⓑ $\sqrt{30}$

Solution

ⓐ rational ⓑ irrational

ⓐ $\sqrt{44}$ ⓑ $\sqrt{49}$

ⓐ $\sqrt{164}$ ⓑ $\sqrt{169}$

Solution

ⓐ irrational ⓑ rational

ⓐ $\sqrt{225}$ ⓑ $\sqrt{216}$

In the following exercises, identify whether each number is a real number or not a real number.

ⓐ $− \sqrt{81}$ ⓑ $\sqrt{−121}$

Solution

ⓐ real number ⓑ not a real number

ⓐ $− \sqrt{64}$ ⓑ $\sqrt{−9}$

ⓐ $\sqrt{−36}$ ⓑ $− \sqrt{144}$

Solution

ⓐ not a real number ⓑ real number

ⓐ $\sqrt{−49}$ ⓑ $− \sqrt{144}$

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

$−8, 0, 1.95286 …, \frac{12}{5}, \sqrt{36}, 9$

Solution

ⓐ $0, \sqrt{36}, 9$ ⓑ $−8, 0, \sqrt{36}, 9$ ⓒ $−8, 0, \frac{12}{5}, \sqrt{36}, 9$ ⓓ $1.95286 …$ ⓔ $−8, 0, 1.95286 …, \frac{12}{5}, \sqrt{36}, 9$

$−9, −3 \frac{4}{9}, − \sqrt{9}, 0.40 \overset{–}{9}, \frac{11}{6}, 7$

$− \sqrt{100}, −7, - \frac{8}{3}, −1, 0.77, 3 \frac{1}{4}$

Solution

ⓐ none ⓑ $− \sqrt{100}, −7, −1$ ⓒ $− \sqrt{100}, −7, - \frac{8}{3}, −1, 0.77, 3 \frac{1}{4}$ ⓓ none ⓔ $− \sqrt{100}, −7, - \frac{8}{3}, −1, 0.77, 3 \frac{1}{4}$

$−6, - \frac{5}{2}, 0, 0. \overset{———}{714285}, 2 \frac{1}{5}, \sqrt{14}$

Locate Fractions on the Number Line

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

$\frac{3}{4}, \frac{8}{5}, \frac{10}{3}$

Solution

There is a number line shown that runs from 0 to 6. From left to right the points read 3/4, 8/5, and 10/3. The point for 3/4 is between 0 and 1. The point for 8/5 is between 1 and 2. The point for 10/3 is between 3 and 4.

$\frac{1}{4}, \frac{9}{5}, \frac{11}{3}$

$\frac{3}{10}, \frac{7}{2}, \frac{11}{6}, 4$

Solution

There is a number line shown that runs from 0 to 6. From left to right the points read 3/10, 11/6, 7/2, and 4. The point for 3/10 is between 0 and 1. The point for 11/6 is between 1 and 2. The point for 7/2 is between 3 and 4.

$\frac{7}{10}, \frac{5}{2}, \frac{13}{8}, 3$

$\frac{2}{5}, - \frac{2}{5}$

Solution

There is a number line shown that runs from negative 1 to 1. From left to right the points read negative 2/5 and 2/5. The point for negative 2/5 is between negative 1 and 0. The point for 2/5 is between 0 and 1.

$\frac{3}{4}, - \frac{3}{4}$

$\frac{3}{4}, - \frac{3}{4}, 1 \frac{2}{3}, −1 \frac{2}{3}, \frac{5}{2}, - \frac{5}{2}$

Solution

$\frac{2}{5}, - \frac{2}{5}, 1 \frac{3}{4}, −1 \frac{3}{4}, \frac{8}{3}, - \frac{8}{3}$

In the following exercises, order each of the pairs of numbers, using < or >.

$−1___- \frac{1}{4}$

Solution

$−1___- \frac{1}{3}$

$−2 \frac{1}{2}___−3$

Solution

$−1 \frac{3}{4}___−2$

$- \frac{5}{12}___- \frac{7}{12}$

Solution

$- \frac{9}{10}___- \frac{3}{10}$

$−3___- \frac{13}{5}$

Solution

$−4___- \frac{23}{6}$

Locate Decimals on the Number Line In the following exercises, locate the number on the number line.

0.8

Solution

There is a number line shown that runs from negative 4 to 4. The point 0.8 is between 0 and 1.

$−0.9$

$−1.6$

Solution

There is a number line shown that runs from negative 4 to 4. The point negative 1.6 is between negative 2 and negative 1.

3.1

In the following exercises, order each pair of numbers, using < or >.

$0.37___0.63$

Solution

$0.86___0.69$

$0.91___0.901$

Solution

$0.415___0.41$

$−0.5___−0.3$

Solution

$−0.1___−0.4$

$−0.62___−0.619$

Solution

$−7.31___−7.3$

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 busses will be needed?
ⓑ Why must the answer be a whole number?
ⓒ Why shouldn’t you round the answer the usual way, by choosing the whole number closest to the exact answer?

Solution

ⓐ 4 busses ⓑ answers may vary ⓒ answers may vary

Child care Serena wants to open a licensed child care center. Her state requires there be no more than 12 children for each teacher. She would like her child care center to serve 40 children.

ⓐ How many teachers will be needed? ⓑ Why must the answer be a whole number? ⓒ Why shouldn’t you round the answer the usual way, by choosing the whole number closest to the exact answer?

Writing Exercises

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

Solution

Answers may vary

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

Self Check

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

$This is a table that has five rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “simplify expressions with square roots,” “identify integers, rational numbers, irrational numbers and real numbers,” locate fractions on the number line,” and “locate decimals on the number line.” The rest of the cells are blank$

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

equivalent decimals

Two decimals are equivalent if they convert to equivalent fractions.

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 of the form

\frac{p}{q}

, where p and q are integers and

q \neq 0

. A rational number can be written as the ratio of two integers. Its decimal form stops or repeats.

radical sign

A radical sign is the symbol

\sqrt{m}

that denotes the positive square root.

real number

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

square and square root

n^{2} = m

, then

m

is the square of

n

and

n

is a square root of

m

Properties of Real Numbers

Use the Commutative and Associative Properties

Think about adding two numbers, say 5 and 3. The order we add them doesn’t affect the result, does it?

\begin{matrix} 5 + 3 & 3 + 5 \\ 8 & 8 \end{matrix}

5 + 3 = 3 + 5

The results are the same.

As we can see, the order in which we add does not matter!

What about multiplying $5 and 3 ?$

\begin{matrix} 5 \cdot 3 & 3 \cdot 5 \\ 15 & 15 \end{matrix}

5 \cdot 3 = 3 \cdot 5

Again, the results are the same!

The order in which we multiply does not matter!

These examples illustrate the commutative property. When adding or multiplying, changing the order gives the same result.

The commutative property has to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.

What about subtraction? Does order matter when we subtract numbers? Does $7 - 3$ give the same result as $3 - 7 ?$

\begin{matrix} \begin{matrix} 7 - 3 & 3 - 7 \\ 4 & −4 \end{matrix} \\ 4 \neq − 4 \\ 7 - 3 \neq 3 - 7 \end{matrix}

The results are not the same.

Since changing the order of the subtraction did not give the same result, we know that subtraction is not commutative.

Let’s see what happens when we divide two numbers. Is division commutative?

\begin{matrix} \begin{matrix} 12 \div 4 & 4 \div 12 \\ \frac{12}{4} & \frac{4}{12} \\ 3 & \frac{1}{3} \end{matrix} \\ 3 \neq \frac{1}{3} \\ 12 \div 4 \neq 4 \div 12 \end{matrix}

The results are not the same.

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

Addition and multiplication are commutative.
Subtraction and Division are not commutative.

If you were asked to simplify this expression, how would you do it and what would your answer be?

7 + 8 + 2

Some people would think $7 + 8 is 15$ and then $15 + 2 is 17 .$ Others might start with $8 + 2 makes 10$ and then $7 + 10 makes 17 .$

Either way gives the same result. Remember, we use parentheses as grouping symbols to indicate which operation should be done first.

Demonstrates the associative property of addition using the numbers 7, 8, and 2, showing that different groupings result in the same total.
Add $7 + 8$ . Add.	$\begin{matrix} (7 + 8) + 2 \\ 15 + 2 \\ 17 \end{matrix}$
Add $8 + 2$ . Add.	$\begin{matrix} 7 + (8 + 2) \\ 7 + 10 \\ 17 \end{matrix}$
	$(7 + 8) + 2 = 7 + (8 + 2)$

When adding three numbers, changing the grouping of the numbers gives the same result.

This is true for multiplication, too.

This table demonstrates the associative property of multiplication with fractions, showing how grouping factors differently does not change the product.
Multiply. $5 \cdot \frac{1}{3}$ Multiply.	$\begin{matrix} (5 \cdot \frac{1}{3}) \cdot 3 \\ \frac{5}{3} \cdot 3 \\ 5 \end{matrix}$
Multiply. $\frac{1}{3} \cdot 3$ . Multiply.	$\begin{matrix} 5 \cdot (\frac{1}{3} \cdot 3) \\ 5 \cdot 1 \\ 5 \end{matrix}$
	$(5 \cdot \frac{1}{3}) \cdot 3 = 5 \cdot (\frac{1}{3} \cdot 3)$

When multiplying three numbers, changing the grouping of the numbers gives the same result.

You probably know this, but the terminology may be new to you. These examples illustrate the associative property.

Let’s think again about multiplying $5 \cdot \frac{1}{3} \cdot 3 .$ We got the same result both ways, but which way was easier? Multiplying $\frac{1}{3}$ and $3$ first, as shown above on the right side, eliminates the fraction in the first step. Using the associative property can make the math easier!

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

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

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

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

Solution

Step-by-step simplification of an algebraic expression by combining like terms using the commutative property of addition.
	$18 p + 6 q + 15 p + 5 q$
Use the commutative property of addition to re-order so that like terms are together.	$18 p + 15 p + 6 q + 5 q$
Add like terms.	$33 p + 11 q$

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. When adding or subtracting fractions, combine those with a common denominator first.

Simplify: $(\frac{5}{13} + \frac{3}{4}) + \frac{1}{4} .$

Solution

Step-by-step solution simplifying a fractional expression by grouping terms with common denominators for easier calculation.
	$(\frac{5}{13} + \frac{3}{4}) + \frac{1}{4}$
Notice that the last 2 terms have a common denominator, so change the grouping.	$\frac{5}{13} + (\frac{3}{4} + \frac{1}{4})$
Add in parentheses first.	$\frac{5}{13} + (\frac{4}{4})$
Simplify the fraction.	$\frac{5}{13} + 1$
Add.	$1 \frac{5}{13}$
Convert to an improper fraction.	$\frac{18}{13}$

Use the associative property to simplify $6 (3 x) .$

Solution

Steps to simplify the algebraic expression 6(3x) using the associative property of multiplication.
	$6 (3 x)$
Change the grouping.	$(6 \cdot 3) x$
Multiply in the parentheses.	$18 x$

Notice that we can multiply $6 \cdot 3$ but we could not multiply 3x without having a value for x.

Use the Identity and Inverse Properties of Addition and Multiplication

What happens when we add 0 to any number? Adding 0 doesn’t change the value. For this reason, we call 0 the additive identity.

For example,

\begin{matrix} 13 + 0 & −14 + 0 & 0 + (−8) \\ 13 & - 14 & - 8 \end{matrix}

These examples illustrate the Identity Property of Addition that states that for any real number $a,$ $a + 0 = a$ and $0 + a = a .$

What happens when we multiply any number by one? Multiplying by 1 doesn’t change the value. So we call 1 the multiplicative identity.

For example,

\begin{matrix} 43 \cdot 1 & - 27 \cdot 1 & 1 \cdot \frac{3}{5} \\ 43 & - 27 & \frac{3}{5} \end{matrix}

These examples illustrate the Identity Property of Multiplication that states that for any real number $a,$ $a \cdot 1 = a$ and $1 \cdot a = a .$

We summarize the Identity Properties below.

In the top line of this figure, we have the question “What number added to 5 gives the additive identity, 0?” On the following line, we have 5 plus a blank space equals 0. Then it is stated that “We know 5 plus negative 5 equals 0.” On the following line, we have the question “What number added to negative 6 gives the additive identity, 0?” On the following line, we have negative 6 plus a blank space equals 0. Then it is stated that “We know negative 6 plus 6 equals 0.”

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

What number multiplied by $\frac{2}{3}$ gives the multiplicative identity, 1? In other words, $\frac{2}{3}$ times what results in 1?

We have the statement that 2/3 times a blank space equals 1. Then it is stated that “We know 2/3 times 3/2 equals 1.”

What number multiplied by 2 gives the multiplicative identity, 1? In other words 2 times what results in 1?

We have the statement that 2 times a blank space equals 1. Then it is stated that “We know 2 times 1/2 equals 1.”

Notice that in each case, the missing number was the reciprocal of the number!

We call $\frac{1}{a}$ the multiplicative inverse of a. The reciprocal of $a$ number is its multiplicative inverse. A number and its reciprocal multiply to one, which is the multiplicative identity. This leads to the Inverse Property of Multiplication that states that for any real number $a, a \neq 0, a \cdot \frac{1}{a} = 1 .$

We’ll formally state the inverse properties here:

Inverse Property

This table defines additive and multiplicative inverses for real numbers, providing their descriptions and mathematical expressions.
of addition	For any real number $a$ , $− a$ is the additive inverse of $a$ . A number and its opposite add to zero.	$a + (− a) = 0$
of multiplication	For any real number $a$ , $\frac{1}{a}$ is the multiplicative inverse of $a$ . A number and its reciprocal multiply to one.	$a \cdot \frac{1}{a} = 1$

Find the additive inverse of ⓐ $\frac{5}{8}$ ⓑ $0.6$ ⓒ $−8$ ⓓ $- \frac{4}{3} .$

Solution

To find the additive inverse, we find the opposite.

ⓐ The additive inverse of $\frac{5}{8}$ is the opposite of $\frac{5}{8} .$ The additive inverse of $\frac{5}{8}$ is $- \frac{5}{8} .$
ⓑ The additive inverse of 0.6 is the opposite of 0.6. The additive inverse of 0.6 is $−0.6 .$
ⓒ The additive inverse of $−8$ is the opposite of $−8 .$ We write the opposite of $−8$ as $− (−8),$ and then simplify it to 8. Therefore, the additive inverse of $−8$ is 8.
ⓓ The additive inverse of $- \frac{4}{3}$ is the opposite of $- \frac{4}{3} .$ We write this as $− (- \frac{4}{3}),$ and then simplify to $\frac{4}{3} .$ Thus, the additive inverse of $- \frac{4}{3}$ is $\frac{4}{3} .$

Find the multiplicative inverse of ⓐ $9$ ⓑ $- \frac{1}{9}$ ⓒ $0.9 .$

Solution

To find the multiplicative inverse, we find the reciprocal.

ⓐ The multiplicative inverse of 9 is the reciprocal of 9, which is $\frac{1}{9} .$ Therefore, the multiplicative inverse of 9 is $\frac{1}{9} .$
ⓑ The multiplicative inverse of $- \frac{1}{9}$ is the reciprocal of $- \frac{1}{9},$ which is $−9 .$ Thus, the multiplicative inverse of $- \frac{1}{9}$ is $−9 .$
ⓒ To find the multiplicative inverse of 0.9, we first convert 0.9 to a fraction, $\frac{9}{10} .$ Then we find the reciprocal of the fraction. The reciprocal of $\frac{9}{10}$ is $\frac{10}{9} .$ So the multiplicative inverse of 0.9 is $\frac{10}{9} .$

Use the Properties of Zero

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

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

0 \div 3 = 0

We can check division with the related multiplication fact.

12 \div 6 = 2 because 2 \cdot 6 = 12 .

So we know $0 \div 3 = 0$ because $0 \cdot 3 = 0 .$

Now think about dividing by zero. What is the result of dividing 4 by 0? Think about the related multiplication fact: $4 \div 0 = ?$ means $? \cdot 0 = 4 .$ Is there a number that multiplied by 0 gives 4? Since any real number multiplied by 0 gives 0, there is no real number that can be multiplied by 0 to obtain 4.

We conclude that there is no answer to $4 \div 0$ and so we say that division by 0 is undefined.

We summarize the properties of zero below.

Properties of Zero

Multiplication by Zero: For any real number a,

This table illustrates the multiplication property of zero, showing that the product of any number and zero is always zero.
$a \cdot 0 = 0 0 \cdot a = 0$	The product of any number and 0 is 0.

Division of Zero, Division by Zero: For any real number $a, a \neq 0$

This table presents mathematical rules for division involving zero, clarifying cases where the result is zero or undefined, along with their verbal descriptions.
$\frac{0}{a} = 0$	Zero divided by any real number except itself is zero.
$\frac{a}{0} is undefined$	Division by zero is undefined.

Simplify: ⓐ $−8 \cdot 0$ ⓑ $\frac{0}{−2}$ ⓒ $\frac{−32}{0} .$

Solution

This table illustrates fundamental mathematical rules for operations involving zero, specifically multiplication by zero and division by zero, with corresponding examples.
ⓐ The product of any real number and 0 is 0.	$\begin{matrix} −8 \cdot 0 \\ 0 \end{matrix}$
ⓑ The product of any real number and 0 is 0.	$\begin{matrix} \frac{0}{−2} \\ 0 \end{matrix}$
ⓒ Division by 0 is undefined.	$\begin{matrix} \frac{−32}{0} \\ Undefined \end{matrix}$

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

Simplify: ⓐ $\frac{0}{n + 5},$ where $n \neq − 5$ ⓑ $\frac{10 - 3 p}{0},$ where $10 - 3 p \neq 0 .$

Solution

Fundamental rules of division involving zero: Zero divided by a non-zero number is zero; division by zero is undefined, shown with mathematical examples.
ⓐ Zero divided by any real number except itself is 0.	$\begin{matrix} \frac{0}{n + 5} \\ 0 \end{matrix}$
ⓑ Division by 0 is undefined.	$\begin{matrix} \frac{10 - 3 p}{0} \\ Undefined \end{matrix}$

Simplify: $−84 n + (−73 n) + 84 n .$

Solution

Step-by-step simplification of the algebraic expression -84n + (-73n) + 84n using the commutative property to combine like terms.
	$−84 n + (−73 n) + 84 n$
Notice that the first and third terms are opposites; use the commutative property of addition to re-order the terms.	$−84 n + 84 n + (−73 n)$
Add left to right.	$0 + (−73 n)$
Add.	$−73 n$

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

Simplify: $\frac{7}{15} \cdot \frac{8}{23} \cdot \frac{15}{7} .$

Solution

Demonstrates simplifying a fractional multiplication by reordering factors using the commutative property.
	$\frac{7}{15} \cdot \frac{8}{23} \cdot \frac{15}{7}$
Notice that the first and third terms are reciprocals, so use the commutative property of multiplication to re-order the factors.	$\frac{7}{15} \cdot \frac{15}{7} \cdot \frac{8}{23}$
Multiply left to right.	$1 \cdot \frac{8}{23}$
Multiply.	$\frac{8}{23}$

Simplify: $\frac{3}{4} \cdot \frac{4}{3} (6 x + 12) .$

Solution

This table illustrates the step-by-step simplification of a mathematical expression using reciprocals and the multiplicative identity.
	$\frac{3}{4} \cdot \frac{4}{3} (6 x + 12)$
There is nothing to do in the parentheses, so multiply the two fractions first—notice, they are reciprocals.	$1 (6 x + 12)$
Simplify by recognizing the multiplicative identity.	$6 x + 12$

Simplify Expressions Using the Distributive Property

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

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

Back to our friends at the movies, we could find the total amount of money they need like this:

\begin{matrix} 3 (9.25) \\ \begin{matrix} 3 (9 & + & 0.25) \\ 3 (9) & + & 3 (0.25) \\ 27 & + & 0.75 \end{matrix} \\ 27.75 \end{matrix}

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 11.

Simplify: $3 (x + 4) .$

Solution

This table illustrates the step-by-step simplification of the algebraic expression 3(x+4) using the distributive property.
	$3 (x + 4)$
Distribute.	$3 \cdot x + 3 \cdot 4$
Multiply.	$3 x + 12$

Some students find it helpful to draw in arrows to remind them how to use the distributive property. Then the first step in Example 11 would look like this:

We have the expression 3 times (x plus 4) with two arrows coming from the 3. One arrow points to the x, and the other arrow points to the 4.

Simplify: $8 (\frac{3}{8} x + \frac{1}{4}) .$

Solution


Distribute.
Multiply.

Using the distributive property as shown in Example 13 will be very useful when we solve money applications in later chapters.

Simplify: $100 (0.3 + 0.25 q) .$

Solution


Distribute.
Multiply.

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

Simplify: $−2 (4 y + 1) .$

Solution


Distribute.
Multiply.

Simplify: $−11 (4 - 3 a) .$

Solution

Distribute.
Multiply.
Simplify.

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

Example 16 will show how to use the distributive property to find the opposite of an expression.

Simplify: $− (y + 5) .$

Solution

Step-by-step demonstration of multiplying (y+5) by -1, showing distribution and simplification.
	$(y + 5)$
Multiplying by −1 results in the opposite.	$−1 (y + 5)$
Distribute.	$−1 \cdot y + (−1) \cdot 5$
Simplify.	$− y + (−5)$
	$− y - 5$

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

Simplify: $8 - 2 (x + 3) .$

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

Solution

Step-by-step simplification of the algebraic expression 8 - 2(x + 3).
	$8 - 2 (x + 3)$
Distribute.	$8 - 2 \cdot x - 2 \cdot 3$
Multiply.	$8 - 2 x - 6$
Combine like terms.	$−2 x + 2$

Simplify: $4 (x - 8) - (x + 3) .$

Solution

Demonstrates the step-by-step simplification of an algebraic expression using distribution and combining like terms.
	$4 (x - 8) - (x + 3)$
Distribute.	$4 x - 32 - x - 3$
Combine like terms.	$3 x - 35$

All the properties of real numbers we have used in this chapter are summarized in Table 22.

Commutative Property
of addition If $a, b$ are real numbers, then of multiplication If $a, b$ are real numbers, then	$a + b = b + a$ $a \cdot b = b \cdot a$
Associative Property
of addition If $a, b, c$ are real numbers, then of multiplication If $a, b, c$ are real numbers, then	$(a + b) + c = a + (b + c)$ $(a \cdot b) \cdot c = a \cdot (b \cdot c)$
Distributive Property
If $a, b, c$ are real numbers, then	$a (b + c) = a b + a c$
Identity Property
of addition For any real number $a :$ 0 is the additive identity of multiplication For any real number $a :$ $1$ is the multiplicative identity	$\begin{matrix} a + 0 = a \\ 0 + a = a \end{matrix}$ $\begin{matrix} a \cdot 1 = a \\ 1 \cdot a = a \end{matrix}$
Inverse Property
of addition For any real number $a,$ $− a$ is the additive inverse of $a$ of multiplication For any real number $a, a \neq 0$ $\frac{1}{a}$ is the multiplicative inverse of $a .$	$a + (− a) = 0$ $a \cdot \frac{1}{a} = 1$
Properties of Zero
For any real number a, For any real number $a, a \neq 0$ For any real number $a, a \neq 0$	$\begin{matrix} a \cdot 0 = 0 \\ 0 \cdot a = 0 \end{matrix}$ $\frac{0}{a} = 0$ $\frac{a}{0}$ is undefined

Key Concepts

Commutative Property of
- Addition: If $a, b$ are real numbers, then $a + b = b + a .$
- Multiplication: If $a, b$ are real numbers, then $a \cdot b = b \cdot a .$ When adding or multiplying, changing the order gives the same result.
Associative Property of
- Addition: If $a, b, c$ are real numbers, then $(a + b) + c = a + (b + c) .$
- Multiplication: If $a, b, c$ are real numbers, then $(a \cdot b) \cdot c = a \cdot (b \cdot c) .$
  When adding or multiplying, changing the grouping gives the same result.
Distributive Property: If a,b,c are real numbers, then
- $a (b + c) = a b + a c$
- $(b + c) a = b a + c a$
- $a (b - c) = a b - a c$
- $(b - c) a = b a - c a$
Identity Property
- of Addition: For any real number $a : a + 0 = a 0 + a = a$
  0 is the additive identity
- of Multiplication: For any real number $a : a \cdot 1 = a 1 \cdot a = a$
  $1$ is the multiplicative identity
Inverse Property
- of Addition: For any real number $a, a + (− a) = 0 .$ A number and its opposite add to zero. $− a$ is the additive inverse of $a .$
- of Multiplication: For any real number $a, (a \neq 0) a \cdot \frac{1}{a} = 1 .$ A number and its reciprocal multiply to one. $\frac{1}{a}$ is the multiplicative inverse of $a .$
Properties of Zero
- For any real number $a,$
  $a \cdot 0 = 0 0 \cdot a = 0$ – The product of any real number and 0 is 0.
- $\frac{0}{a} = 0$ for $a \neq 0$ – Zero divided by any real number except zero is zero.
- $\frac{a}{0}$ is undefined – Division by zero is undefined.

Practice Makes Perfect

Use the Commutative and Associative Properties

In the following exercises, use the associative property to simplify.

3(4x)

Solution

12x

4(7m)

$(y + 12) + 28$

Solution

$y + 40$

$(n + 17) + 33$

In the following exercises, simplify.

$\frac{1}{2} + \frac{7}{8} + (- \frac{1}{2})$

Solution

$\frac{7}{8}$

$\frac{2}{5} + \frac{5}{12} + (- \frac{2}{5})$

$\frac{3}{20} \cdot \frac{49}{11} \cdot \frac{20}{3}$

Solution

$\frac{49}{11}$

$\frac{13}{18} \cdot \frac{25}{7} \cdot \frac{18}{13}$

$−24 \cdot 7 \cdot \frac{3}{8}$

Solution

$−63$

$−36 \cdot 11 \cdot \frac{4}{9}$

$(\frac{5}{6} + \frac{8}{15}) + \frac{7}{15}$

Solution

$1 \frac{5}{6}$

$(\frac{11}{12} + \frac{4}{9}) + \frac{5}{9}$

17(0.25)(4)

Solution

36(0.2)(5)

[2.48(12)](0.5)

Solution

14.88

[9.731(4)](0.75)

7(4a)

Solution

28a

9(8w)

$−15 (5 m)$

Solution

$−75 m$

$−23 (2 n)$

$12 (\frac{5}{6} p)$

Solution

10p

$20 (\frac{3}{5} q)$

$43 m + (−12 n) + (−16 m) + (−9 n)$

Solution

$27 m + (−21 n)$

$−22 p + 17 q + (−35 p) + (−27 q)$

$\frac{3}{8} g + \frac{1}{12} h + \frac{7}{8} g + \frac{5}{12} h$

Solution

$\frac{5}{4} g + \frac{1}{2} h$

$\frac{5}{6} a + \frac{3}{10} b + \frac{1}{6} a + \frac{9}{10} b$

$6.8 p + 9.14 q + (−4.37 p) + (−0.88 q)$

Solution

$2.43 p + 8.26 q$

$9.6 m + 7.22 n + (−2.19 m) + (−0.65 n)$

Use the Identity and Inverse Properties of Addition and Multiplication

In the following exercises, find the additive inverse of each number.

ⓐ $\frac{2}{5}$ ⓑ 4.3
ⓒ $−8$
ⓓ $- \frac{10}{3}$

Solution

ⓐ $- \frac{2}{5}$ ⓑ $−4.3$ ⓒ 8 ⓓ $\frac{10}{3}$

ⓐ $\frac{5}{9}$
ⓑ 2.1 ⓒ $−3$ ⓓ $- \frac{9}{5}$

ⓐ $- \frac{7}{6}$ ⓑ $−0.075$ ⓒ 23 ⓓ $\frac{1}{4}$

Solution

ⓐ $\frac{7}{6}$ ⓑ 0.075 ⓒ $−23$ ⓓ $- \frac{1}{4}$

ⓐ $- \frac{8}{3}$ ⓑ $−0.019$ ⓒ 52 ⓓ $\frac{5}{6}$

In the following exercises, find the multiplicative inverse of each number.

ⓐ 6 ⓑ $- \frac{3}{4}$ ⓒ 0.7

Solution

ⓐ $\frac{1}{6}$ ⓑ $- \frac{4}{3}$ ⓒ $\frac{10}{7}$

ⓐ 12 ⓑ $- \frac{9}{2}$ ⓒ 0.13

ⓐ $\frac{11}{12}$ ⓑ $−1.1$ ⓒ $−4$

Solution

ⓐ $\frac{12}{11}$ ⓑ $- \frac{10}{11}$ ⓒ $- \frac{1}{4}$

ⓐ $\frac{17}{20}$ ⓑ $−1.5$ ⓒ $−3$

Use the Properties of Zero

In the following exercises, simplify.

$\frac{0}{6}$

Solution

$\frac{3}{0}$

$0 \div \frac{11}{12}$

Solution

$\frac{6}{0}$

$\frac{0}{3}$

Solution

$0 \cdot \frac{8}{15}$

$(−3.14) (0)$

Solution

$\frac{\frac{1}{10}}{0}$

Mixed Practice

In the following exercises, simplify.

$19 a + 44 - 19 a$

Solution

$27 c + 16 - 27 c$

$10 (0.1 d)$

Solution

$100 (0.01 p)$

$\frac{0}{u - 4.99},$ where $u \neq 4.99$

Solution

$\frac{0}{v - 65.1},$ where $v \neq 65.1$

$0 \div (x - \frac{1}{2}),$ where $x \neq \frac{1}{2}$

Solution

$0 \div (y - \frac{1}{6}),$ where $y \neq \frac{1}{6}$

$\frac{32 - 5 a}{0},$ where
$32 - 5 a \neq 0$

Solution

undefined

$\frac{28 - 9 b}{0},$ where
$28 - 9 b \neq 0$

$(\frac{3}{4} + \frac{9}{10} m) \div 0$ where
$\frac{3}{4} + \frac{9}{10} m \neq 0$

Solution

undefined

$(\frac{5}{16} n - \frac{3}{7}) \div 0$ where
$\frac{5}{16} n - \frac{3}{7} \neq 0$

$15 \cdot \frac{3}{5} (4 d + 10)$

Solution

$36 d + 90$

$18 \cdot \frac{5}{6} (15 h + 24)$

Simplify Expressions Using the Distributive Property

In the following exercises, simplify using the distributive property.

$8 (4 y + 9)$

Solution

$32 y + 72$

$9 (3 w + 7)$

$6 (c - 13)$

Solution

$6 c - 78$

$7 (y - 13)$

$\frac{1}{4} (3 q + 12)$

Solution

$\frac{3}{4} q + 3$

$\frac{1}{5} (4 m + 20)$

$9 (\frac{5}{9} y - \frac{1}{3})$

Solution

$5 y - 3$

$10 (\frac{3}{10} x - \frac{2}{5})$

$12 (\frac{1}{4} + \frac{2}{3} r)$

Solution

$3 + 8 r$

$12 (\frac{1}{6} + \frac{3}{4} s)$

$r (s - 18)$

Solution

$r s - 18 r$

$u (v - 10)$

$(y + 4) p$

Solution

$y p + 4 p$

$(a + 7) x$

$−7 (4 p + 1)$

Solution

$−28 p - 7$

$−9 (9 a + 4)$

$−3 (x - 6)$

Solution

$−3 x + 18$

$−4 (q - 7)$

$− (3 x - 7)$

Solution

$−3 x + 7$

$− (5 p - 4)$

$16 - 3 (y + 8)$

Solution

$−3 y - 8$

$18 - 4 (x + 2)$

$4 - 11 (3 c - 2)$

Solution

$−33 c + 26$

$9 - 6 (7 n - 5)$

$22 - (a + 3)$

Solution

$− a + 19$

$8 - (r - 7)$

$(5 m - 3) - (m + 7)$

Solution

$4 m - 10$

$(4 y - 1) - (y - 2)$

$5 (2 n + 9) + 12 (n - 3)$

Solution

$22 n + 9$

$9 (5 u + 8) + 2 (u - 6)$

$9 (8 x - 3) - (−2)$

Solution

$72 x - 25$

$4 (6 x - 1) - (−8)$

$14 (c - 1) - 8 (c - 6)$

Solution

$6 c + 34$

$11 (n - 7) - 5 (n - 1)$

$6 (7 y + 8) - (30 y - 15)$

Solution

$12 y + 63$

$7 (3 n + 9) - (4 n - 13)$

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 as a copay. Calculate Carrie’s copay:

ⓐ First, by multiplying 0.20 by 80 to find her copay for each filling and then multiplying your answer by 5 to find her total copay for 5 fillings.
ⓑ Next, by multiplying [5(0.20)](80)
ⓒ Which of the properties of real numbers says that your answers to parts (a), where you multiplied 5[(0.20)(80)] and (b), where you multiplied [5(0.20)](80), should be equal?

Solution

ⓐ $80 ⓑ $80 ⓒ answers will vary

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 to the oven. She wants to allow 20 minutes per pound cooking time. Calculate the length of time needed to roast the turkey:

ⓐ First, by multiplying $24 \cdot 20$ to find the total number of minutes and then multiplying the answer by $\frac{1}{60}$ to convert minutes into hours.
ⓑ Next, by multiplying $24 (20 \cdot \frac{1}{60}) .$
ⓒ Which of the properties of real numbers says that your answers to parts (a), where you multiplied $(24 \cdot 20) \frac{1}{60},$ and (b), where you multiplied $24 (20 \cdot \frac{1}{60}),$ should be equal?

Buying by the case Trader Joe’s grocery stores sold a bottle of wine they called “Two Buck Chuck” for $1.99. They sold a case of 12 bottles for $23.88. To find the cost of 12 bottles at $1.99, notice that 1.99 is $2 - 0.01 .$

ⓐ Multiply 12(1.99) by using the distributive property to multiply $12 (2 - 0.01) .$
ⓑ Was it a bargain to buy “Two Buck Chuck” by the case?

Solution

ⓐ $23.88 ⓑ no, the price is the same

Multi-pack purchase Adele’s shampoo sells for $3.99 per bottle at the grocery store. At the warehouse store, the same shampoo is sold as a 3 pack for $10.49. To find the cost of 3 bottles at $3.99, notice that 3.99 is $4 - 0.01 .$

ⓐ Multiply 3(3.99) by using the distributive property to multiply $3 (4 - 0.01) .$
ⓑ How much would Adele save by buying 3 bottles at the warehouse store instead of at the grocery store?

Writing Exercises

In your own words, state the commutative property of addition.

Solution

Answers may vary

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

Simplify $8 (x - \frac{1}{4})$ using the distributive property and explain each step.

Solution

Answers may vary

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

Self Check

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

This is a table that has five rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “use the commutative and associative properties,” “use the identity and inverse properties of addition and multiplication,” “use the properties of zero,” and “simplify expressions using the distributive property.” The rest of the cells are blank.

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

additive identity

The additive identity is the number 0; adding 0 to any number does not change its value.

additive inverse

The opposite of a number is its additive inverse. A number and its additive inverse add to 0.

multiplicative identity

The multiplicative identity is the number 1; multiplying 1 by any number does not change the value of the number.

multiplicative inverse

The reciprocal of a number is its multiplicative inverse. A number and its multiplicative inverse multiply to one.

Systems of Measurement

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 U.S. 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, and hours.

The equivalencies of measurements are shown in Table 1. The table also shows, in parentheses, the common abbreviations for each measurement.

U.S. System of Measurement
$\begin{matrix} Length & \begin{matrix} 1 foot (ft.) & = & 12 inches (in.) \\ 1 yard (yd.) & = & 3 feet (ft.) \\ 1 mile (mi.) & = & 5,280 feet (ft.) \end{matrix} \end{matrix}$	$\begin{matrix} Volume & \begin{matrix} 3 teaspoons (t) & = & 1 tablespoon (T) \\ 16 tablespoons (T) & = & 1 cup (C) \\ 1 cup (C) & = & 8 fluid ounces (fl. oz.) \\ 1 pint (pt.) & = & 2 cups (C) \\ 1 quart (qt.) & = & 2 pints (pt.) \\ 1 gallon (gal) & = & 4 quarts (qt.) \end{matrix} \end{matrix}$
$\begin{matrix} Weight & \begin{matrix} 1 pound (lb.) & = & 16 ounces (oz.) \\ 1 ton & = & 2000 pounds (lb.) \end{matrix} \end{matrix}$	$\begin{matrix} Time & \begin{matrix} 1 minute (min) & = & 60 seconds (sec) \\ 1 hour (hr) & = & 60 minutes (min) \\ 1 day & = & 24 hours (hr) \\ 1 week (wk) & = & 7 days \\ 1 year (yr) & = & 365 days \end{matrix} \end{matrix}$

In many real-life applications, we need to convert between units of measurement, such as feet and yards, minutes and seconds, quarts and gallons, etc. We will use the identity property of multiplication to do these conversions. We’ll restate the identity property of multiplication here for easy reference.

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 change inches to feet. We know that 1 foot is equal to 12 inches, so we will write 1 as the fraction $\frac{1 foot}{12 inches} .$ When we multiply by this fraction we do not change the value, but just change the units.

But $\frac{12 inches}{1 foot}$ also equals 1. How do we decide whether to multiply by $\frac{1 foot}{12 inches}$ or $\frac{12 inches}{1 foot} ?$ We choose the fraction that will make the units we want to convert from divide out. Treat the unit words like factors and “divide out” common units like we do common factors. If we want to convert $66$ inches to feet, which multiplication will eliminate the inches?

$Two expressions are given: 66 inches times the fraction (1 foot) over (12 inches), and 66 inches times the fraction (12 inches) over (1 foot). This second expression is crossed out. Below this, it is stated that “The first form works since 66 inches times the fraction (1 foot) over (12 inches), with inches crossed off in both instances.$

The inches divide out and leave only feet. The second form does not have any units that will divide out and so will not help us.

How to Make Unit Conversions

MaryAnne is 66 inches tall. Convert her height into feet.

Solution

$A table is given with three columns. In the first column are directions. The second column has exposition, and the third column has the mathematical steps. In the first row, the direction is “Step 1. Multiply the measurement to be converted by; write as a fraction relating the units given and the units needed.” The exposition is “Multiply inches by, writing as a fraction relating inches and feet. We need inches in the denominator so that the inches will divide out!” The mathematical step is 66 inches times the fraction (1 foot) over (12 inches).$ $In the following row, we have “Step 2. Multiply.” The hint is “Think of 66 inches as the quantity 66 inches divided by 1.” The math portion is the fraction (66 inches times 1 foot) over 12 inches.$ $In the following row, we have “Step 3. Simplify the fraction.” The hint is that “Notice: inches divide out.” We obtain 66 feet divided by 12.$ Then the last step is “Step 4. Simplify.” The hint is “Divide 66 by 12.” Hence, our final mathematical statement is 5.5 feet.

Then the last step is “Step 4. Simplify.” The hint is “Divide 66 by 12.” Hence, our final mathematical statement is 5.5 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.

Solution

We will convert 3.2 tons into pounds. We will use the identity property of multiplication, writing 1 as the fraction

\frac{2000 pounds}{1 ton} .

	$3.2 tons$
Multiply the measurement to be converted, by 1.	$3.2 tons \cdot 1$
Write 1 as a fraction relating tons and pounds.	$3.2 tons \cdot \frac{2,000 pounds}{1 ton}$
Simplify.
Multiply.	$6,400 pounds$
	Ndula weighs almost 6,400 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 from her boyfriend for 9 weeks. Convert the time to minutes.

Solution

To convert weeks into minutes we will convert weeks into days, days into hours, and then hours into minutes. To do this we will multiply by conversion factors of 1.

	9 weeks
Write $1$ as $\frac{7 days}{1 week}$ , $\frac{24 hours}{1 day}$ , and $\frac{60 minutes}{1 hour}$ .	$\frac{9 wk}{1} \cdot \frac{7 days}{1 wk} \cdot \frac{24 hr}{1 day} \cdot \frac{60 min}{1 hr}$
Divide out the common units.	$A dimensional analysis calculation converting 9 weeks into minutes. The series of fractions shows how weeks are converted to days, days to hours, and hours to minutes, with units canceled at each step.$
Multiply.	$\frac{9 \cdot 7 \cdot 24 \cdot 60 min}{1 \cdot 1 \cdot 1 \cdot 1}$
Multiply.	$90,720 min$

Juliet and her boyfriend will be apart for 90,720 minutes (although it may seem like an eternity!).

How many ounces are in 1 gallon?

Solution

We will convert gallons to ounces by multiplying by several conversion factors. Refer to Table 1.

	1 gallon
Multiply the measurement to be converted by 1.	$\frac{1 gallon}{1} \cdot \frac{4 quarts}{1 gallon} \cdot \frac{2 pints}{1 quart} \cdot \frac{2 cups}{1 pint} \cdot \frac{8 ounces}{1 cup}$
Use conversion factors to get to the right unit. Simplify.
Multiply.	$\frac{1 \cdot 4 \cdot 2 \cdot 2 \cdot 8 ounces}{1 \cdot 1 \cdot 1 \cdot 1 \cdot 1}$
Simplify.	$128 ounces$ There are 128 ounces in a gallon.

Use Mixed Units of Measurement in the U.S. System

We often use mixed units of measurement in everyday situations. Suppose Joe is 5 feet 10 inches tall, stays at work for 7 hours and 45 minutes, and then eats a 1 pound 2 ounce steak for dinner—all these measurements have mixed units.

Performing arithmetic operations on measurements with mixed units of measures requires care. Be sure to add or subtract like units!

Seymour 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?

Solution

We will add the weights of the steaks to find the total weight of the steaks.

Add the ounces. Then add the pounds.
Convert 22 ounces to pounds and ounces.	2 pounds + 1 pound, 6 ounces
Add the pounds.	3 pounds, 6 ounces
	Seymour bought 3 pounds 6 ounces of steak.

Anthony bought four planks of wood that were each 6 feet 4 inches long. What is the total length of the wood he purchased?

Solution

We will multiply the length of one plank to find the total length.

Multiply the inches and then the feet.
Convert the 16 inches to feet. Add the feet.
	Anthony bought 25 feet and 4 inches of wood.

Make Unit Conversions in the Metric System

In the metric system, units are related by powers of 10. The roots words of their names reflect this relation. For example, the basic unit for measuring length is a meter. One kilometer is 1,000 meters; the prefix kilo means thousand. One centimeter is $\frac{1}{100}$ of a meter, just like one cent is $\frac{1}{100}$ of one dollar.

The equivalencies of measurements in the metric system are shown in Table 7. The common abbreviations for each measurement are given in parentheses.

Metric System of Measurement
Length	Mass	Capacity
1 kilometer (km) = 1,000 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) = 1,000 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) = 1,000 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 = 1,000 millimeters	1 gram = 100 centigrams 1 gram = 1,000 milligrams	1 liter = 100 centiliters 1 liter = 1,000 milliliters

To make conversions in the metric system, we will use the same technique we did in the US 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 5K or 10K race? The length 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 10K race. How many meters did he run?

Solution

We will convert kilometers to meters using the identity property of multiplication.

	10 kilometers
Multiply the measurement to be converted by 1.
Write 1 as a fraction relating kilometers and meters.
Simplify.
Multiply.	10,000 meters
	Nick ran 10,000 meters.

Eleanor’s newborn baby weighed 3,200 grams. How many kilograms did the baby weigh?

Solution

We will convert grams into kilograms.


Multiply the measurement to be converted by 1.
Write 1 as a function relating kilograms and grams.	$A mathematical expression shows a unit conversion: 3,200 grams multiplied by the fraction 1 kg divided by 1,000 grams. The word 'grams' is highlighted in red in both the initial quantity and the denominator.$
Simplify.
Multiply.	$\frac{3,200 kilograms}{1,000}$
Divide.	3.2 kilograms The baby weighed 3.2 kilograms.

As you become familiar with the metric system you may see a pattern. Since the system is based on multiples of ten, the calculations involve multiplying by multiples of ten. We have learned how to simplify these calculations by just moving the decimal.

To multiply by 10, 100, or 1,000, we move the decimal to the right one, two, or three places, respectively. To multiply by 0.1, 0.01, or 0.001, we move the decimal to the left one, two, or three places, respectively.

We can apply this pattern when we make measurement conversions in the metric system. In Example 8, we changed 3,200 grams to kilograms by multiplying by $\frac{1}{1000}$ (or 0.001). This is the same as moving the decimal three places to the left.

$We have the statement 3200 g times the fraction 1 kg over 1000 g, with the g’s crossed out. Below this, we have 3.2. We also have the statement 3200 times 1/1000, with an arrow drawn from the right of the final 0 in 3200 to the space between the 0’s, to the space between the 2 and the 0, and then to the space between the 3 and the 2. Below this, we have 3.2.$

Convert ⓐ 350 L to kiloliters ⓑ 4.1 L to milliliters.

Solution

ⓐ We will convert liters to kiloliters. In Table 7, we see that $1 kiloliter = 1,000 liters.$

$350 L$

Multiply by 1, writing 1 as a fraction relating liters to kiloliters. $350 L \cdot \frac{1 kL}{1,000 L}$

Simplify. $350 L \cdot \frac{1 kL}{1,000 L}$

$0.35 kL$
ⓑ We will convert liters to milliliters. From Table 7 we see that $1 liter = 1,000 milliliters.$

Multiply by 1, writing 1 as a fraction relating liters to milliliters.

Simplify. $A mathematical expression showing the conversion of 4.1 liters to milliliters, written as 4.1 L multiplied by the fraction 1,000 mL divided by 1 L.$

Move the decimal 3 units to the right.

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 US system. But it may be easier because of the relation of the units to the powers of 10. 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

We can convert both measurements to either centimeters or meters. Since meters is the larger unit, we will subtract the lengths in meters. We convert 85 centimeters to meters by moving the decimal 2 places to the left.

This table provides a text-based problem statement and a distinct mathematical calculation example.
Write the 85 centimeters as meters.	$\begin{matrix} 1.60 m \\ \underset{_______}{−0.85 m} \\ 0.75 m \end{matrix}$

Ryland is $0.75 m$ taller than his brother.

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

We will find the amount of olive oil in millileters then convert to liters.

This table illustrates the step-by-step calculation and conversion of 3 x 150 mL to liters.
	Triple $150 mL$
Translate to algebra.	$3 \cdot 150 mL$
Multiply.	$450 mL$
Convert to liters.	$450 \cdot \frac{0.001 L}{1 mL}$
Simplify.	$0.45 L$
	Dena needs 0.45 liters of olive oil.

Convert Between the U.S. and the Metric Systems of Measurement

Many measurements in the United States are made in metric units. Our soda may come in 2-liter bottles, our calcium may come in 500-mg capsules, and we may run a 5K 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	Mass	Capacity
$\begin{matrix} \begin{matrix} 1 in. & = & 2.54 cm \\ 1 ft. & = & 0.305 m \\ 1 yd. & = & 0.914 m \\ 1 mi. & = & 1.61 km \\ 1 m & = & 3.28 ft. \end{matrix} \end{matrix}$	$\begin{matrix} \begin{matrix} 1 lb. & = & 0.45 kg \\ 1 oz. & = & 28 g \\ 1 kg & = & 2.2 lb. \end{matrix} \end{matrix}$	$\begin{matrix} \begin{matrix} 1 qt. & = & 0.95 L \\ 1 fl. oz. & = & 30 mL \\ 1 L & = & 1.06 qt. \end{matrix} \end{matrix}$

Figure 1 shows how inches and centimeters are related on a ruler.

A ruler with inches and centimeters. — This ruler shows inches and centimeters.

Figure 2 shows the ounce and milliliter markings on a measuring cup.

A measuring cup showing milliliters and ounces. — This measuring cup shows ounces and milliliters.

Figure 3 shows how pounds and kilograms marked on a bathroom scale.

We are given an image of a bathroom scale showing pounds. — This scale shows pounds and kilograms.

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 ounces are in the bottle? Round to the nearest tenth of an ounce.

Solution

Step-by-step conversion of 500 milliliters to ounces using a unit conversion factor, resulting in 16.7 ounces.
	$500 mL$
Multiply by a unit conversion factor relating mL and ounces.	$500 milliliters \cdot \frac{1 ounce}{30 milliliters}$
Simplify.	$\frac{500 ounce}{30}$
Divide.	$16.7 ounces.$
	The water bottle has 16.7 ounces.

Soleil was on a road trip and saw a sign that said the next rest stop was in 100 kilometers. How many miles until the next rest stop?

Solution

Step-by-step guide converting 100 kilometers to miles, detailing the calculation process and final distance.
	$100 kilometers$
Multiply by a unit conversion factor relating km and mi.	$100 kilometers \cdot \frac{1 mile}{1.61 kilometer}$
Simplify.	$\frac{100 miles}{1.61}$
Divide.	$62 miles.$
	Soleil will travel 62 miles.

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 4 shows the relationship between the two systems.

Two thermometers are shown, one in Celsius (°C) and another in Fahrenheit (°F). They are marked “Water boils” at 100°C and 212°F. They are marked “Normal body temperature” at 37°C and 98.6°F. They are marked “Water freezes” at 0°C and 32°F. — The diagram shows normal body temperature, along with the freezing and boiling temperatures of water in degrees Fahrenheit and degrees Celsius.

Convert $50 °$ Fahrenheit into degrees Celsius.

Solution

We will substitute

50 ° F

into the formula to find C.



Simplify in parentheses.
Multiply.
	So we found that 50°F is equivalent to 10°C.

While visiting Paris, Woody saw the temperature was $20 °$ Celsius. Convert the temperature into degrees Fahrenheit.

Solution

We will substitute

20 ° C

into the formula to find F.



Multiply.
Add.
	So we found that 20°C is equivalent to 68°F.

Key Concepts

Metric System of Measurement
- Length
  $\begin{matrix} 1 kilometer (km) & = & 1,000 m \\ 1 hectometer (hm) & = & 100 m \\ 1 dekameter (dam) & = & 10 m \\ 1 meter (m) & = & 1 m \\ 1 decimeter (dm) & = & 0.1 m \\ 1 centimeter (cm) & = & 0.01 m \\ 1 millimeter (mm) & = & 0.001 m \\ 1 meter & = & 100 centimeters \\ 1 meter & = & 1,000 millimeters \end{matrix}$
- Mass
  $\begin{matrix} 1 kilogram (kg) & = & 1,000 g \\ 1 hectogram (hg) & = & 100 g \\ 1 dekagram (dag) & = & 10 g \\ 1 gram (g) & = & 1 g \\ 1 decigram (dg) & = & 0.1 g \\ 1 centigram (cg) & = & 0.01 g \\ 1 milligram (mg) & = & 0.001 g \\ 1 gram & = & 100 centigrams \\ 1 gram & = & 1,000 milligrams \end{matrix}$
- Capacity
  $\begin{matrix} 1 kiloliter (kL) & = & 1,000 L \\ 1 hectoliter (hL) & = & 100 L \\ 1 dekaliter (daL) & = & 10 L \\ 1 liter (L) & = & 1 L \\ 1 deciliter (dL) & = & 0.1 L \\ 1 centiliter (cL) & = & 0.01 L \\ 1 milliliter (mL) & = & 0.001 L \\ 1 liter & = & 100 centiliters \\ 1 liter & = & 1,000 milliliters \end{matrix}$
Temperature Conversion
- To convert from Fahrenheit temperature, F, to Celsius temperature, C, use the formula $C = \frac{5}{9} (F - 32)$
- To convert from Celsius temperature, C, to Fahrenheit temperature, F, use the formula $F = \frac{9}{5} C + 32$

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.

Solution

72 inches

A floor tile is 2 feet wide. Convert the width to inches.

A ribbon is 18 inches long. Convert the length to feet.

Solution

1.5 feet

Carson is 45 inches tall. Convert his height to feet.

A football field is 160 feet wide. Convert the width to yards.

Solution

$53 \frac{1}{3}$ yards

On a baseball diamond, the distance from home plate to first base is 30 yards. Convert the distance to feet.

Ulises lives 1.5 miles from school. Convert the distance to feet.

Solution

7,920 feet

Denver, Colorado, is 5,183 feet above sea level. Convert the height to miles.

A killer whale weighs 4.6 tons. Convert the weight to pounds.

Solution

9,200 pounds

Blue whales can weigh as much as 150 tons. Convert the weight to pounds.

An empty bus weighs 35,000 pounds. Convert the weight to tons.

Solution

$17 \frac{1}{2}$ tons

At take-off, an airplane weighs 220,000 pounds. Convert the weight to tons.

Rocco waited $1 \frac{1}{2}$ hours for his appointment. Convert the time to seconds.

Solution

5,400 s

Misty’s surgery lasted $2 \frac{1}{4}$ hours. Convert the time to seconds.

How many teaspoons are in a pint?

Solution

$96$ teaspoons

How many tablespoons are in a gallon?

JJ’s cat, Posy, weighs 14 pounds. Convert her weight to ounces.

Solution

$224$ ounces

April’s dog, Beans, weighs 8 pounds. Convert his weight to ounces.

Crista will serve 20 cups of juice at her son’s party. Convert the volume to gallons.

Solution

$1 \frac{1}{4}$ gallons

Lance needs 50 cups of water for the runners in a race. Convert the volume to gallons.

Jon is 6 feet 4 inches tall. Convert his height to inches.

Solution

76 in.

Faye is 4 feet 10 inches tall. Convert her height to inches.

The voyage of the Mayflower took 2 months and 5 days. Convert the time to days.

Solution

65 days

Lynn’s cruise lasted 6 days and 18 hours. Convert the time to hours.

Baby Preston weighed 7 pounds 3 ounces at birth. Convert his weight to ounces.

Solution

115 ounces

Baby Audrey weighted 6 pounds 15 ounces at birth. Convert her weight to ounces.

Use Mixed Units of Measurement in the U.S. System

In the following exercises, solve.

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?

Solution

8 lbs. 13 oz.

Judy bought 1 pound 6 ounces of almonds, 2 pounds 3 ounces of walnuts, and 8 ounces of cashews. How many pounds of nuts did Judy buy?

One day Anya kept track of the number of minutes she spent driving. She recorded 45, 10, 8, 65, 20, and 35. How many hours did Anya spend driving?

Solution

3.05 hours

Last year Eric went on 6 business trips. The number of days of each was 5, 2, 8, 12, 6, and 3. How many weeks did Eric spend on business trips last year?

Renee attached a 6 feet 6 inch extension cord to her computer’s 3 feet 8 inch power cord. What was the total length of the cords?

Solution

10 ft. 2 in.

Fawzi’s SUV is 6 feet 4 inches tall. If he puts a 2 feet 10 inch box on top of his SUV, what is the total height of the SUV and the box?

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?

Solution

4 yards

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?

Make Unit Conversions in the Metric System

In the following exercises, convert the units.

Ghalib ran 5 kilometers. Convert the length to meters.

Solution

5,000 meters

Kitaka hiked 8 kilometers. Convert the length to meters.

Estrella is 1.55 meters tall. Convert her height to centimeters.

Solution

155 centimeters

The width of the wading pool is 2.45 meters. Convert the width to centimeters.

Mount Whitney is 3,072 meters tall. Convert the height to kilometers.

Solution

3.072 kilometers

The depth of the Mariana Trench is 10,911 meters. Convert the depth to kilometers.

June’s multivitamin contains 1,500 milligrams of calcium. Convert this to grams.

Solution

1.5 grams

A typical ruby-throated hummingbird weights 3 grams. Convert this to milligrams.

One stick of butter contains 91.6 grams of fat. Convert this to milligrams.

Solution

91,600 milligrams

One serving of gourmet ice cream has 25 grams of fat. Convert this to milligrams.

The maximum mass of an airmail letter is 2 kilograms. Convert this to grams.

Solution

2,000 grams

Dimitri’s daughter weighed 3.8 kilograms at birth. Convert this to grams.

A bottle of wine contained 750 milliliters. Convert this to liters.

Solution

0.75 liters

A bottle of medicine contained 300 milliliters. Convert this to liters.

Use Mixed Units of Measurement in the Metric System

In the following exercises, solve.

Matthias is 1.8 meters tall. His son is 89 centimeters tall. How much taller is Matthias than his son?

Solution

91 centimeters

Stavros is 1.6 meters tall. His sister is 95 centimeters tall. How much taller is Stavros than his sister?

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?

Solution

855 grams

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?

Harry mailed 5 packages that weighed 420 grams each. What was the total weight of the packages in kilograms?

Solution

2.1 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?

Jonas drinks 200 milliliters of water 8 times a day. How many liters of water does Jonas drink in a day?

Solution

1.6 liters

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?

Convert Between the U.S. and the Metric Systems of Measurement

In the following exercises, make the unit conversions. Round to the nearest tenth.

Bill is 75 inches tall. Convert his height to centimeters.

Solution

190.5 centimeters

Frankie is 42 inches tall. Convert his height to centimeters.

Marcus passed a football 24 yards. Convert the pass length to meters

Solution

21.9 meters

Connie bought 9 yards of fabric to make drapes. Convert the fabric length to meters.

Each American throws out an average of 1,650 pounds of garbage per year. Convert this weight to kilograms.

Solution

750 kilograms

An average American will throw away 90,000 pounds of trash over his or her lifetime. Convert this weight to kilograms.

A 5K run is 5 kilometers long. Convert this length to miles.

Solution

3.1 miles

Kathryn is 1.6 meters tall. Convert her height to feet.

Dawn’s suitcase weighed 20 kilograms. Convert the weight to pounds.

Solution

44 pounds

Jackson’s backpack weighed 15 kilograms. Convert the weight to pounds.

Ozzie put 14 gallons of gas in his truck. Convert the volume to liters.

Solution

53.2 liters

Bernard bought 8 gallons of paint. Convert the volume to liters.

Convert between Fahrenheit and Celsius Temperatures

In the following exercises, convert the Fahrenheit temperatures to degrees Celsius. Round to the nearest tenth.

$86 °$ Fahrenheit

Solution

$30 ° C$

$77 °$ Fahrenheit

$104 °$ Fahrenheit

Solution

$40 ° C$

$14 °$ Fahrenheit

$72 °$ Fahrenheit

Solution

$22.2 ° C$

$4 °$ Fahrenheit

$0 °$ Fahrenheit

Solution

$− 17.8 ° C$

$120 °$ Fahrenheit

In the following exercises, convert the Celsius temperatures to degrees Fahrenheit. Round to the nearest tenth.

$5 °$ Celsius

Solution

$41 ° F$

$25 °$ Celsius

$− 10 °$ Celsius

Solution

$14 ° F$

$− 15 °$ Celsius

$22 °$ Celsius

Solution

$71.6 ° F$

$8 °$ Celsius

$43 °$ Celsius

Solution

$109.4 ° F$

$16 °$ Celsius

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?

Solution

14.6 kilograms

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 lane-marking stripe?

Writing Exercises

Some people think that $65 ° to 75 °$ Fahrenheit is the ideal temperature range.

ⓐ What is your ideal temperature range? Why do you think so?
ⓑ Convert your ideal temperatures from Fahrenheit to Celsius.

Solution

Answers may vary.

ⓐ Did you grow up using the U.S. or the metric system of measurement?
ⓑ Describe two examples in your life when you had to convert between the two systems of measurement.

Self Check

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

This is a table that has seven rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “define US units of measurement and convert from one unit to another,” “use US units of measurement,” “define metric units of measurement and convert from one unit to another,” “use metric units of measurement,” “convert between the US and the metric system of measurement,” and “convert between Fahrenheit and Celsius temperatures.” The rest of the cells are blank.

ⓑ 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

Use Place Value with Whole Number

In the following exercises find the place value of each digit.

26,915

ⓐ 1 ⓑ 2 ⓒ 9 ⓓ 5 ⓔ 6

Solution

ⓐ tens ⓑ ten thousands ⓒ hundreds ⓓ ones ⓔ thousands

359,417

ⓐ 9 ⓑ 3 ⓒ 4 ⓓ 7 ⓔ 1

58,129,304

ⓐ 5 ⓑ 0 ⓒ 1 ⓓ 8 ⓔ 2

Solution

ⓐ ten millions ⓑ tens ⓒ hundred thousands ⓓ millions ⓔ ten thousands

9,430,286,157

ⓐ 6 ⓑ 4 ⓒ 9 ⓓ 0 ⓔ 5

In the following exercises, name each number.

6,104

Solution

six thousand, one hundred four

493,068

3,975,284

Solution

three million, nine hundred seventy-five thousand, two hundred eighty-four

85,620,435

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

three hundred fifteen

Solution

$315$

sixty-five thousand, nine hundred twelve

ninety million, four hundred twenty-five thousand, sixteen

Solution

$90,425,016$

one billion, forty-three million, nine hundred twenty-two thousand, three hundred eleven

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

Round to the nearest ten.

ⓐ 407 ⓑ 8,564

Solution

ⓐ $410$ ⓑ $8,560$

Round to the nearest hundred.

ⓐ 25,846 ⓑ 25,864

In the following exercises, round each number to the nearest ⓐ hundred ⓑ thousand ⓒ ten thousand.

864,951

Solution

ⓐ $865,000$ ⓑ $865,000$ ⓒ $860,000$

3,972,849

Identify Multiples and Factors

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

168

Solution

$by 2, 3, 6$

264

375

Solution

$by 3, 5$

750

1430

Solution

$by 2, 5, 10$

1080

Find Prime Factorizations and Least Common Multiples

In the following exercises, find the prime factorization.

420

Solution

$2 \cdot 2 \cdot 3 \cdot 5 \cdot 7$

115

225

Solution

$3 \cdot 3 \cdot 5 \cdot 5$

2475

1560

Solution

$2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 13$

Solution

$2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$

168

252

Solution

$2 \cdot 2 \cdot 3 \cdot 3 \cdot 7$

391

In the following exercises, find the least common multiple of the following numbers using the multiples method.

6,15

Solution

$30$

60, 75

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

24, 30

Solution

120

70, 84

Use the Language of Algebra

Use Variables and Algebraic Symbols

In the following exercises, translate the following from algebra to English.

$25 - 7$

Solution

25 minus 7, the difference of twenty-five and seven

$5 \cdot 6$

$45 \div 5$

Solution

45 divided by 5, the quotient of forty-five and five

$x + 8$

$42 \geq 27$

Solution

forty-two is greater than or equal to twenty-seven

$3 n = 24$

$3 \leq 20 \div 4$

Solution

3 is less than or equal to 20 divided by 4, three is less than or equal to the quotient of twenty and four

$a \neq 7 \cdot 4$

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

$6 \cdot 3 + 5$

Solution

expression

$y - 8 = 32$

Simplify Expressions Using the Order of Operations

In the following exercises, simplify each expression.

$3^{5}$

Solution

243

$10^{8}$

In the following exercises, simplify

$6 + 10 / 2 + 2$

Solution

$9 + 12 / 3 + 4$

$20 \div (4 + 6) \cdot 5$

Solution

$33 \div (3 + 8) \cdot 2$

${(4^{2} + 5^{2})}^{2}$

Solution

1681

${(4 + 5)}^{2}$

Evaluate an Expression

In the following exercises, evaluate the following expressions.

$9 x + 7$ when $x = 3$

Solution

$5 x - 4$ when $x = 6$

$x^{4}$ when $x = 3$

Solution

$3^{x}$ when $x = 3$

$x^{2} + 5 x - 8$ when $x = 6$

Solution

$2 x + 4 y - 5$ when
$x = 7, y = 8$

Simplify Expressions by Combining Like Terms

In the following exercises, identify the coefficient of each term.

$12 n$

Solution

$9 x^{2}$

In the following exercises, identify the like terms.

$3 n, n^{2}, 12, 12 p^{2}, 3, 3 n^{2}$

Solution

$12 and 3, n^{2} and 3 n^{2}$

$5, 18 r^{2}, 9 s, 9 r, 5 r^{2}, 5 s$

In the following exercises, identify the terms in each expression.

$11 x^{2} + 3 x + 6$

Solution

$11 x^{2}, 3 x, 6$

$22 y^{3} + y + 15$

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

$17 a + 9 a$

Solution

$26 a$

$18 z + 9 z$

$9 x + 3 x + 8$

Solution

$12 x + 8$

$8 a + 5 a + 9$

$7 p + 6 + 5 p - 4$

Solution

$12 p + 2$

$8 x + 7 + 4 x - 5$

Translate an English Phrase to an Algebraic Expression

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

the sum of 8 and 12

Solution

$8 + 12$

the sum of 9 and 1

the difference of $x$ and 4

Solution

$x - 4$

the difference of $x$ and 3

the product of 6 and $y$

Solution

$6 y$

the product of 9 and $y$

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

$b + 15$

Marcella has 6 fewer boy cousins than girl cousins. Let $g$ represent the number of girl cousins. Write an expression for the number of boy cousins.

Add and Subtract Integers

Use Negatives and Opposites of Integers

In the following exercises, order each of the following pairs of numbers, using < or >.

ⓐ $6___2$
ⓑ $−7___4$
ⓒ $−9___−1$
ⓓ $9___−3$

Solution

ⓐ > ⓑ < ⓒ < ⓓ >

ⓐ $−5___1$
ⓑ $−4___−9$
ⓒ $6___10$
ⓓ $3___−8$

In the following exercises,, find the opposite of each number.

ⓐ $−8$ ⓑ 1

Solution

ⓐ 8 ⓑ $−1$

ⓐ $−2$ ⓑ $6$

In the following exercises, simplify.

$− (−19)$

Solution

$− (−53)$

In the following exercises, simplify.

$− m$ when
ⓐ $m = 3$
ⓑ $m = −3$

Solution

ⓐ $−3$ ⓑ 3

$− p$ when
ⓐ $p = 6$
ⓑ $p = −6$

Simplify Expressions with Absolute Value

In the following exercises,, simplify.

ⓐ $| 7 |$ ⓑ $| −25 |$ ⓒ $| 0 |$

Solution

ⓐ 7 ⓑ 25 ⓒ 0

ⓐ $| 5 |$ ⓑ $| 0 |$ ⓒ $| −19 |$

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

ⓐ $−8___| −8 |$
ⓑ $− | −2 |___−2$

Solution

ⓐ < ⓑ =

ⓐ $| −3 |___- | −3 |$
ⓑ $4___- | −4 |$

In the following exercises, simplify.

$| 8 - 4 |$

Solution

$| 9 - 6 |$

$8 (14 - 2 | −2 |)$

Solution

$6 (13 - 4 | −2 |)$

In the following exercises, evaluate.

ⓐ $| x |$ when $x = −28$ ⓑ $| x |$ when $x = −15$

Solution

ⓐ 28 ⓑ 15

ⓐ $| y |$ when $y = −37$
ⓑ $| − z |$ when $z = −24$

Add Integers

In the following exercises, simplify each expression.

$−200 + 65$

Solution

$−135$

$−150 + 45$

$2 + (−8) + 6$

Solution

$4 + (−9) + 7$

$140 + (−75) + 67$

Solution

132

$−32 + 24 + (−6) + 10$

Subtract Integers

In the following exercises, simplify.

$9 - 3$

Solution

$−5 - (−1)$

ⓐ $15 - 6$ ⓑ $15 + (−6)$

Solution

ⓐ 9 ⓑ 9

ⓐ $12 - 9$ ⓑ $12 + (−9)$

ⓐ $8 - (−9)$ ⓑ $8 + 9$

Solution

ⓐ 17 ⓑ 17

ⓐ $4 - (−4)$ ⓑ $4 + 4$

In the following exercises, simplify each expression.

$10 - (−19)$

Solution

$11 - (−18)$

$31 - 79$

Solution

$−48$

$39 - 81$

$−31 - 11$

Solution

$−42$

$−32 - 18$

$−15 - (−28) + 5$

Solution

$71 + (−10) - 8$

$−16 - (−4 + 1) - 7$

Solution

$−20$

$−15 - (−6 + 4) - 3$

Multiply Integers

In the following exercises, multiply.

$−5 (7)$

Solution

$−35$

$−8 (6)$

$−18 (−2)$

Solution

$36$

$−10 (−6)$

Divide Integers

In the following exercises, divide.

$−28 \div 7$

Solution

$−4$

$56 \div (−7)$

$−120 \div (−20)$

Solution

$−200 \div 25$

Simplify Expressions with Integers

In the following exercises, simplify each expression.

$−8 (−2) - 3 (−9)$

Solution

$−7 (−4) - 5 (−3)$

${(−5)}^{3}$

Solution

$−125$

${(−4)}^{3}$

$−4 \cdot 2 \cdot 11$

Solution

$−88$

$−5 \cdot 3 \cdot 10$

$−10 (−4) \div (−8)$

Solution

$−5$

$−8 (−6) \div (−4)$

$31 - 4 (3 - 9)$

Solution

$24 - 3 (2 - 10)$

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

$x + 8$ when ⓐ $x = −26$ ⓑ $x = −95$

Solution

ⓐ $−18$ ⓑ $−87$

$y + 9$ when ⓐ $y = −29$ ⓑ $y = −84$

When $b = −11$ , evaluate:ⓐ $b + 6$ ⓑ $− b + 6$

Solution

ⓐ $−5$ ⓑ 17

When $c = −9$ , evaluate:
ⓐ $c + (−4)$ >ⓑ $− c + (−4)$

$p^{2} - 5 p + 2$ when
$p = −1$

Solution

$q^{2} - 2 q + 9$ when $q = −2$

$6 x - 5 y + 15$ when $x = 3$ and $y = −1$

Solution

$3 p - 2 q + 9$ when $p = 8$ and $q = −2$

Translate English Phrases to Algebraic Expressions

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

the sum of $−4$ and $−17$ , increased by 32

Solution

$(−4 + (−17)) + 32; 11$

ⓐ the difference of 15 and $−7$ ⓑ subtract 15 from $−7$

the quotient of $−45$ and $−9$

Solution

$\frac{−45}{−9}; 5$

the product of $−12$ and the difference of $c and d$

Use Integers in Applications

In the following exercises, solve.

Temperature The high temperature one day in Miami Beach, Florida, was $76 °$ . That same day, the high temperature in Buffalo, New York was $− 8 °$ . What was the difference between the temperature in Miami Beach and the temperature in Buffalo?

Solution

84 degrees

Checking Account Adrianne has a balance of $− $22$ in her checking account. She deposits $301 to the account. What is the new balance?

Visualize Fractions

Find Equivalent Fractions

In the following exercises, find three fractions equivalent to the given fraction. Show your work, using figures or algebra.

$\frac{1}{4}$

Solution

$\frac{2}{8}, \frac{3}{12}, \frac{4}{16}$ answers may vary

$\frac{1}{3}$

$\frac{5}{6}$

Solution

$\frac{10}{12}, \frac{15}{18}, \frac{20}{24}$ answers may vary

$\frac{2}{7}$

Simplify Fractions

In the following exercises, simplify.

$\frac{7}{21}$

Solution

$\frac{1}{3}$

$\frac{8}{24}$

$\frac{15}{20}$

Solution

$\frac{3}{4}$

$\frac{12}{18}$

$- \frac{168}{192}$

Solution

$- \frac{7}{8}$

$- \frac{140}{224}$

$\frac{11 x}{11 y}$

Solution

$\frac{x}{y}$

$\frac{15 a}{15 b}$

Multiply Fractions

In the following exercises, multiply.

$\frac{2}{5} \cdot \frac{1}{3}$

Solution

$\frac{2}{15}$

$\frac{1}{2} \cdot \frac{3}{8}$

$\frac{7}{12} (- \frac{8}{21})$

Solution

$- \frac{2}{9}$

$\frac{5}{12} (- \frac{8}{15})$

$−28 p (- \frac{1}{4})$

Solution

$7 p$

$−51 q (- \frac{1}{3})$

$\frac{14}{5} (−15)$

Solution

$−42$

$−1 (- \frac{3}{8})$

Divide Fractions

In the following exercises, divide.

$\frac{1}{2} \div \frac{1}{4}$

Solution

$\frac{1}{2} \div \frac{1}{8}$

$- \frac{4}{5} \div \frac{4}{7}$

Solution

$- \frac{7}{5}$

$- \frac{3}{4} \div \frac{3}{5}$

$\frac{5}{8} \div \frac{a}{10}$

Solution

$\frac{25}{4 a}$

$\frac{5}{6} \div \frac{c}{15}$

$\frac{7 p}{12} \div \frac{21 p}{8}$

Solution

$\frac{2}{9}$

$\frac{5 q}{12} \div \frac{15 q}{8}$

$\frac{2}{5} \div (−10)$

Solution

$- \frac{1}{25}$

$−18 \div - (\frac{9}{2})$

In the following exercises, simplify.

$\frac{\frac{2}{3}}{\frac{8}{9}}$

Solution

$\frac{3}{4}$

$\frac{\frac{4}{5}}{\frac{8}{15}}$

$\frac{- \frac{9}{10}}{3}$

Solution

$- \frac{3}{10}$

$\frac{2}{\frac{5}{8}}$

$\frac{\frac{r}{5}}{\frac{s}{3}}$

Solution

$\frac{3 r}{5 s}$

$\frac{- \frac{x}{6}}{- \frac{8}{9}}$

Simplify Expressions Written with a Fraction Bar

In the following exercises, simplify.

$\frac{4 + 11}{8}$

Solution

$\frac{15}{8}$

$\frac{9 + 3}{7}$

$\frac{30}{7 - 12}$

Solution

$−6$

$\frac{15}{4 - 9}$

$\frac{22 - 14}{19 - 13}$

Solution

$\frac{4}{3}$

$\frac{15 + 9}{18 + 12}$

$\frac{5 \cdot 8}{−10}$

Solution

$−4$

$\frac{3 \cdot 4}{−24}$

$\frac{15 \cdot 5 - 5^{2}}{2 \cdot 10}$

Solution

$\frac{5}{2}$

$\frac{12 \cdot 9 - 3^{2}}{3 \cdot 18}$

$\frac{2 + 4 (3)}{−3 - 2^{2}}$

Solution

$−2$

$\frac{7 + 3 (5)}{−2 - 3^{2}}$

Translate Phrases to Expressions with Fractions

In the following exercises, translate each English phrase into an algebraic expression.

the quotient of c and the sum of d and 9.

Solution

$\frac{c}{d + 9}$

the quotient of the difference of h and k, and $−5$ .

Add and Subtract Fractions

Add and Subtract Fractions with a Common Denominator

In the following exercises, add.

$\frac{4}{9} + \frac{1}{9}$

Solution

$\frac{5}{9}$

$\frac{2}{9} + \frac{5}{9}$

$\frac{y}{3} + \frac{2}{3}$

Solution

$\frac{y + 2}{3}$

$\frac{7}{p} + \frac{9}{p}$

$- \frac{1}{8} + (- \frac{3}{8})$

Solution

$- \frac{1}{2}$

$- \frac{1}{8} + (- \frac{5}{8})$

In the following exercises, subtract.

$\frac{4}{5} - \frac{1}{5}$

Solution

$\frac{3}{5}$

$\frac{4}{5} - \frac{3}{5}$

$\frac{y}{17} - \frac{9}{17}$

Solution

$\frac{y - 9}{17}$

$\frac{x}{19} - \frac{8}{19}$

$- \frac{8}{d} - \frac{3}{d}$

Solution

$- \frac{11}{d}$

$- \frac{7}{c} - \frac{7}{c}$

Add or Subtract Fractions with Different Denominators

In the following exercises, add or subtract.

$\frac{1}{3} + \frac{1}{5}$

Solution

$\frac{8}{15}$

$\frac{1}{4} + \frac{1}{5}$

$\frac{1}{5} - (- \frac{1}{10})$

Solution

$\frac{3}{10}$

$\frac{1}{2} - (- \frac{1}{6})$

$\frac{2}{3} + \frac{3}{4}$

Solution

$\frac{17}{12}$

$\frac{3}{4} + \frac{2}{5}$

$\frac{11}{12} - \frac{3}{8}$

Solution

$\frac{13}{24}$

$\frac{5}{8} - \frac{7}{12}$

$- \frac{9}{16} - (- \frac{4}{5})$

Solution

$\frac{19}{80}$

$- \frac{7}{20} - (- \frac{5}{8})$

$1 + \frac{5}{6}$

Solution

$\frac{11}{6}$

$1 - \frac{5}{9}$

Use the Order of Operations to Simplify Complex Fractions

In the following exercises, simplify.

$\frac{{(\frac{1}{5})}^{2}}{2 + 3^{2}}$

Solution

$\frac{1}{275}$

$\frac{{(\frac{1}{3})}^{2}}{5 + 2^{2}}$

$\frac{\frac{2}{3} + \frac{1}{2}}{\frac{3}{4} - \frac{2}{3}}$

Solution

$\frac{\frac{3}{4} + \frac{1}{2}}{\frac{5}{6} - \frac{2}{3}}$

Evaluate Variable Expressions with Fractions

In the following exercises, evaluate.

$x + \frac{1}{2}$ when
ⓐ $x = - \frac{1}{8}$
ⓑ $x = - \frac{1}{2}$

Solution

ⓐ $\frac{3}{8}$ ⓑ 0

$x + \frac{2}{3}$ when
ⓐ $x = - \frac{1}{6}$
ⓑ $x = - \frac{5}{3}$

$4 p^{2} q$ when $p = - \frac{1}{2}$ and $q = \frac{5}{9}$

Solution

$\frac{5}{9}$

$5 m^{2} n$ when $m = - \frac{2}{5}$ and $n = \frac{1}{3}$

$\frac{u + v}{w}$ when
$u = −4, v = −8, w = 2$

Solution

$−6$

$\frac{m + n}{p}$ when
$m = −6, n = −2, p = 4$

Decimals

Name and Write Decimals

In the following exercises, write as a decimal.

Eight and three hundredths

Solution

8.03

Nine and seven hundredths

One thousandth

Solution

0.001

Nine thousandths

In the following exercises, name each decimal.

7.8

Solution

seven and eight tenths

5.01

0.005

Solution

five thousandths

0.381

Round Decimals

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

5.7932

Solution

ⓐ 5.79 ⓑ 5.8 ⓒ 6

3.6284

12.4768

Solution

ⓐ 12.48 ⓑ 12.5 ⓒ 12

25.8449

Add and Subtract Decimals

In the following exercises, add or subtract.

$18.37 + 9.36$

Solution

27.73

$256.37 - 85.49$

$15.35 - 20.88$

Solution

−5.53

$37.5 + 12.23$

$−4.2 + (−9.3)$

Solution

−13.5

$−8.6 + (−8.6)$

$100 - 64.2$

Solution

35.8

$100 - 65.83$

$2.51 + 40$

Solution

42.51

$9.38 + 60$

Multiply and Divide Decimals

In the following exercises, multiply.

$(0.3) (0.4)$

Solution

0.12

$(0.6) (0.7)$

$(8.52) (3.14)$

Solution

26.7528

$(5.32) (4.86)$

$(0.09) (24.78)$

Solution

2.2302

$(0.04) (36.89)$

In the following exercises, divide.

$0.15 \div 5$

Solution

0.03

$0.27 \div 3$

$$8.49 \div 12$

Solution

$0.71

$$16.99 \div 9$

$12 \div 0.08$

Solution

150

$5 \div 0.04$

Convert Decimals, Fractions, and Percents

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

0.08

Solution

$\frac{2}{25}$

0.17

0.425

Solution

$\frac{17}{40}$

0.184

1.75

Solution

$\frac{7}{4}$

0.035

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

$\frac{2}{5}$

Solution

0.4

$\frac{4}{5}$

$- \frac{3}{8}$

Solution

$−0.375$

$- \frac{5}{8}$

$\frac{5}{9}$

Solution

$0. \overset{–}{5}$

$\frac{2}{9}$

$\frac{1}{2} + 6.5$

Solution

$\frac{1}{4} + 10.75$

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

Solution

0.05

40%

Solution

0.4

50%

115%

Solution

1.15

125%

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

0.18

Solution

18%

0.15

0.009

Solution

0.9%

0.008

1.5

Solution

150%

2.2

The Real Numbers

Simplify Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{64}$

Solution

$\sqrt{144}$

$− \sqrt{25}$

Solution

−5

$− \sqrt{81}$

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

In the following exercises, write as the ratio of two integers.

ⓐ 9 ⓑ 8.47

Solution

ⓐ $\frac{9}{1}$ ⓑ $\frac{847}{100}$

ⓐ $−15$ ⓑ 3.591

In the following exercises, list the ⓐ rational numbers, ⓑ irrational numbers.

$0.84, 0.79132 …, 1. \overset{–}{3}$

Solution

ⓐ $0.84, 1. \overset{–}{3}$ ⓑ $0.79132 …,$

$2.3 \overset{–}{8}, 0.572, 4.93814 …$

In the following exercises, identify whether each number is rational or irrational.

ⓐ $\sqrt{121}$ ⓑ $\sqrt{48}$

Solution

ⓐ rational ⓑ irrational

ⓐ $\sqrt{56}$ ⓑ $\sqrt{16}$

In the following exercises, identify whether each number is a real number or not a real number.

ⓐ $\sqrt{−9}$ ⓑ $− \sqrt{169}$

Solution

ⓐ not a real number ⓑ real number

ⓐ $\sqrt{−64}$ ⓑ $− \sqrt{81}$

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

$−4, 0, \frac{5}{6}, \sqrt{16}, \sqrt{18}, 5.2537 …$

Solution

ⓐ $0, \sqrt{16}$ ⓑ $−4, 0, \sqrt{16}$ ⓒ $−4, 0, \frac{5}{6}, \sqrt{16}$ ⓓ $\sqrt{18}, 5.2537 …$ ⓔ $−4, 0, \frac{5}{6}, \sqrt{16}, \sqrt{18}, 5.2537 …$

$− \sqrt{4}, 0. \overset{—}{36}, \frac{13}{3}, 6.9152 …, \sqrt{48}, 10 \frac{1}{2}$

Locate Fractions on the Number Line

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

$\frac{2}{3}, \frac{5}{4}, \frac{12}{5}$

Solution

This figure is a number line ranging from 0 to 6 with tick marks for each integer. 2 thirds, 5 fourths, and 12 fifths are plotted.

$\frac{1}{3}, \frac{7}{4}, \frac{13}{5}$

$2 \frac{1}{3}, −2 \frac{1}{3}$

Solution

This figure is a number line ranging from negative 4 to 4 with tick marks for each integer. Negative 2 and 1 third, and 2 and 1 third are plotted.

$1 \frac{3}{5}, −1 \frac{3}{5}$

In the following exercises, order each of the following pairs of numbers, using < or >.

$−1___- \frac{1}{8}$

Solution

$−3 \frac{1}{4}___−4$

$- \frac{7}{9}___- \frac{4}{9}$

Solution

$−2___- \frac{19}{8}$

Locate Decimals on the Number Line

In the following exercises, locate on the number line.

$0.3$

Solution

This figure is a number line ranging from 0 to 1 with tick marks for each tenth of an integer. 0.3 is plotted.

$−0.2$

$−2.5$

Solution

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. Negative 2.5 is plotted.

2.7

In the following exercises, order each of the following pairs of numbers, using < or >.

$0.9___0.6$

Solution

$0.7___0.8$

$−0.6___−0.59$

Solution

$−0.27___−0.3$

Properties of Real Numbers

Use the Commutative and Associative Properties

In the following exercises, use the Associative Property to simplify.

$−12 (4 m)$

Solution

$−48 m$

$30 (\frac{5}{6} q)$

$(a + 16) + 31$

Solution

$a + 47$

$(c + 0.2) + 0.7$

In the following exercises, simplify.

$6 y + 37 + (−6 y)$

Solution

$\frac{1}{4} + \frac{11}{15} + (- \frac{1}{4})$

$\frac{14}{11} + \frac{35}{9} + (- \frac{14}{11})$

Solution

$\frac{35}{9}$

$−18 \cdot 15 \cdot \frac{2}{9}$

$(\frac{7}{12} + \frac{4}{5}) + \frac{1}{5}$

Solution

$1 \frac{7}{12}$

$(3.98 d + 0.75 d) + 1.25 d$

$11 x + 8 y + 16 x + 15 y$

Solution

$27 x + 23 y$

$52 m + (−20 n) + (−18 m) + (−5 n)$

Use the Identity and Inverse Properties of Addition and Multiplication

In the following exercises, find the additive inverse of each number.

ⓐ $\frac{1}{3}$
ⓑ $5.1$
ⓒ $−14$
ⓓ $- \frac{8}{5}$

Solution

ⓐ $- \frac{1}{3}$ ⓑ $−5.1$ ⓒ 14 ⓓ $\frac{8}{5}$

ⓐ $- \frac{7}{8}$
ⓑ $−0.03$
ⓒ $17$
ⓓ $\frac{12}{5}$

In the following exercises, find the multiplicative inverse of each number.

ⓐ 10 ⓑ $- \frac{4}{9}$ ⓒ 0.6

Solution

ⓐ $\frac{1}{10}$ ⓑ $- \frac{9}{4}$ ⓒ $\frac{5}{3}$

ⓐ $- \frac{9}{2}$ ⓑ $−7$ ⓒ 2.1

Use the Properties of Zero

In the following exercises, simplify.

$83 \cdot 0$

Solution

$\frac{0}{9}$

$\frac{5}{0}$

Solution

undefined

$0 \div \frac{2}{3}$

In the following exercises, simplify.

$43 + 39 + (−43)$

Solution

$(n + 6.75) + 0.25$

$\frac{5}{13} \cdot 57 \cdot \frac{13}{5}$

Solution

$\frac{1}{6} \cdot 17 \cdot 12$

$\frac{2}{3} \cdot 28 \cdot \frac{3}{7}$

Solution

$9 (6 x - 11) + 15$

Simplify Expressions Using the Distributive Property

In the following exercises, simplify using the Distributive Property.

$7 (x + 9)$

Solution

$7 x + 63$

$9 (u - 4)$

$−3 (6 m - 1)$

Solution

$−18 m + 3$

$−8 (−7 a - 12)$

$\frac{1}{3} (15 n - 6)$

Solution

$5 n - 2$

$(y + 10) \cdot p$

$(a - 4) - (6 a + 9)$

Solution

$−5 a - 13$

$4 (x + 3) - 8 (x - 7)$

Systems of Measurement

1.1 Define U.S. Units of Measurement and Convert from One Unit to Another

In the following exercises, convert the units. Round to the nearest tenth.

A floral arbor is 7 feet tall. Convert the height to inches.

Solution

84 inches

A picture frame is 42 inches wide. Convert the width to feet.

Kelly is 5 feet 4 inches tall. Convert her height to inches.

Solution

64 inches

A playground is 45 feet wide. Convert the width to yards.

The height of Mount Shasta is 14,179 feet. Convert the height to miles.

Solution

2.7 miles

Shamu weights 4.5 tons. Convert the weight to pounds.

The play lasted $1 \frac{3}{4}$ hours. Convert the time to minutes.

Solution

105 minutes

How many tablespoons are in a quart?

Naomi’s baby weighed 5 pounds 14 ounces at birth. Convert the weight to ounces.

Solution

94 ounces

Trinh needs 30 cups of paint for her class art project. Convert the volume to gallons.

Use Mixed Units of Measurement in the U.S. System.

In the following exercises, solve.

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?

Solution

10 lbs. 6 oz.

Every day last week Pedro recorded the number of minutes he spent reading. The number of minutes were 50, 25, 83, 45, 32, 60, 135. How many hours did Pedro spend reading?

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?

Solution

15 feet

Dalila wants to make throw pillow covers. Each cover takes 30 inches of fabric. How many yards of fabric does she need for 4 covers?

Make Unit Conversions in the Metric System

In the following exercises, convert the units.

Donna is 1.7 meters tall. Convert her height to centimeters.

Solution

170 centimeters

Mount Everest is 8,850 meters tall. Convert the height to kilometers.

One cup of yogurt contains 488 milligrams of calcium. Convert this to grams.

Solution

0.488 grams

One cup of yogurt contains 13 grams of protein. Convert this to milligrams.

Sergio weighed 2.9 kilograms at birth. Convert this to grams.

Solution

2,900 grams

A bottle of water contained 650 milliliters. Convert this to liters.

Use Mixed Units of Measurement in the Metric System

In the following exerices, solve.

Minh is 2 meters tall. His daughter is 88 centimeters tall. How much taller is Minh than his daughter?

Solution

1.12 meter

Selma had a 1 liter bottle of water. If she drank 145 milliliters, how much water was left in the bottle?

One serving of cranberry juice contains 30 grams of sugar. How many kilograms of sugar are in 30 servings of cranberry juice?

Solution

0.9 kilograms

One ounce of tofu provided 2 grams of protein. How many milligrams of protein are provided by 5 ounces of tofu?

Convert between the U.S. and the Metric Systems of Measurement

In the following exercises, make the unit conversions. Round to the nearest tenth.

Majid is 69 inches tall. Convert his height to centimeters.

Solution

175.3 centimeters

A college basketball court is 84 feet long. Convert this length to meters.

Caroline walked 2.5 kilometers. Convert this length to miles.

Solution

1.6 miles

Lucas weighs 78 kilograms. Convert his weight to pounds.

Steve’s car holds 55 liters of gas. Convert this to gallons.

Solution

14.6 gallons

A box of books weighs 25 pounds. Convert the weight to kilograms.

Convert between Fahrenheit and Celsius Temperatures

In the following exercises, convert the Fahrenheit temperatures to degrees Celsius. Round to the nearest tenth.

95° Fahrenheit

Solution

35° C

23° Fahrenheit

20° Fahrenheit

Solution

–6.7° C

64° Fahrenheit

In the following exercises, convert the Celsius temperatures to degrees Fahrenheit. Round to the nearest tenth.

30° Celsius

Solution

86° F

–5° Celsius

–12° Celsius

Solution

10.4° F

24° Celsius

Chapter Practice Test

Write as a whole number using digits: two hundred five thousand, six hundred seventeen.

Solution

205,617

Find the prime factorization of 504.

Find the Least Common Multiple of 18 and 24.

Solution

Combine like terms: $5 n + 8 + 2 n - 1 .$

In the following exercises, evaluate.

$− | x |$ when $x = −2$

Solution

$−2$

$11 - a$ when $a = −3$

Translate to an algebraic expression and simplify: twenty less than negative 7.

Solution

$−7 - 20; - 27$

Monique has a balance of $− $18$ in her checking account. She deposits $152 to the account. What is the new balance?

Round 677.1348 to the nearest hundredth.

Solution

677.13

Convert $\frac{4}{5}$ to a decimal.

Convert 1.85 to a percent.

Solution

185%

Locate $\frac{2}{3}, −1.5, and \frac{9}{4}$ on a number line.

In the following exercises, simplify each expression.

$4 + 10 (3 + 9) - 5^{2}$

Solution

$−85 + 42$

$−19 - 25$

Solution

$−44$

${(−2)}^{4}$

$−5 (−9) \div 15$

Solution

$\frac{3}{8} \cdot \frac{11}{12}$

$\frac{4}{5} \div \frac{9}{20}$

Solution

$\frac{16}{9}$

$\frac{12 + 3 \cdot 5}{15 - 6}$

$\frac{m}{7} + \frac{10}{7}$

Solution

$\frac{m + 10}{7}$

$\frac{7}{12} - \frac{3}{8}$

$−5.8 + (−4.7)$

Solution

$−10.5$

$100 - 64.25$

$(0.07) (31.95)$

Solution

2.2365

$9 \div 0.05$

$−14 (\frac{5}{7} p)$

Solution

$−10 p$

$(u + 8) - 9$

$6 x + (−4 y) + 9 x + 8 y$

Solution

$15 x + 4 y$

$\frac{0}{23}$

$\frac{75}{0}$

Solution

undefined

$−2 (13 q - 5)$

A movie lasted $1 \frac{2}{3}$ hours. How many minutes did it last? (1 hour = 60 minutes)

Solution

100 minutes

Mike’s SUV is 5 feet 11 inches tall. He wants to put a rooftop cargo bag on the the SUV. The cargo bag is 1 foot 6 inches tall. What will the total height be of the SUV with the cargo bag on the roof? (1 foot = 12 inches)

Jennifer ran 2.8 miles. Convert this length to kilometers. (1 mile = 1.61 kilometers)

Solution

4.508 km

Introduction

This is a photo of several rocks carefully stacked to achieve balance. — The rocks in this formation must remain perfectly balanced around the center for the formation to hold its shape.

If we carefully placed more rocks of equal weight on both sides of this formation, it would still balance. Similarly, the expressions in an equation remain balanced when we add the same quantity to both sides of the equation. In this chapter, we will solve equations, remembering that what we do to one side of the equation, we must also do to the other side.

Solve Equations Using the Subtraction and Addition Properties of Equality

Verify a Solution of an Equation

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 – so that we end up with a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

Determine whether $x = \frac{3}{2}$ is a solution of $4 x - 2 = 2 x + 1$ .

Solution

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


$The image shows the text 'Substitute 3/2 for x.' The fraction 3/2 is written with the numeral 3 in red at the top and the numeral 2 in red at the bottom, separated by a horizontal line.$	$A mathematical equation asks whether 4 times (3/2) minus 2 is equal to 2 times (3/2) plus 1. The fraction 3/2 is highlighted in red.$
Multiply.
Subtract.

Since $x = \frac{3}{2}$ results in a true equation (4 is in fact equal to 4), $\frac{3}{2}$ is a solution to the equation $4 x - 2 = 2 x + 1$ .

Solve Equations Using the Subtraction and Addition Properties of Equality

We are going to use a model to clarify the process of solving an equation. An envelope represents the variable – since its contents are unknown – and each counter represents one. We will set out one envelope and some counters on our workspace, as shown in Figure 1. Both sides of the workspace have the same number of counters, but some counters are “hidden” in the envelope. Can you tell how many counters are in the envelope?

This image illustrates a workspace divided into two sides. The content of the left side is equal to the content of the right side. On the left side, there are three circular counters and an envelope containing an unknown number of counters. On the right side are eight counters. — The illustration shows a model of an equation with one variable. On the left side of the workspace is an unknown (envelope) and three counters, while on the right side of the workspace are eight counters.

What are you thinking? 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 at the bottom left to get the envelope by itself. The 3 counters on the left can be matched with 3 on the right and so I can take them away from both sides. That leaves five on the right—so there must be 5 counters in the envelope.” See Figure 2 for an illustration of this process.

This figure contains two illustrations of workspaces, divided each into two sides. On the left side of the first workspace there are three counters circled in purple and an envelope containing an unknown number of counters. On the right side are eight counters, three of which are also circled in purple. An arrow to the right of the workspace points to the second workspace. On the left side of the second workspace, there is just an envelope. On the right side are five counters. This workspace is identical to the first workspace, except that the three counters circled in purple have been removed from both sides. — The illustration shows a model for solving an equation with one variable. On both sides of the workspace remove three counters, leaving only the unknown (envelope) and five counters on the right side. The unknown is equal to five counters.

What algebraic equation would match this situation? In Figure 3 each side of the workspace represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope $x$ .

Let’s write algebraically the steps we took to discover how many counters were in the envelope:


First, we took away three from each side.
Then we were left with five.

Check:

Five in the envelope plus three more does equal eight!

5 + 3 = 8

Our model has given us an idea of what we need to do to solve one kind of equation. The goal is to isolate the variable by itself on one side of the equation. To solve equations such as these mathematically, we use the Subtraction Property of Equality.

Let’s see how to use this property to solve an equation. Remember, the goal is 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: $y + 37 = −13 .$

Solution

To get y by itself, we will undo the addition of 37 by using the Subtraction Property of Equality.


Subtract 37 from each side to ‘undo’ the addition.
Simplify.
Check:
Substitute $y = −50$

Since $y = −50$ makes $y + 37 = −13$ a true statement, we have the solution to this equation.

What happens when an equation has a number subtracted from the variable, as in the equation $x - 5 = 8$ ? We use another property of equations to solve equations where a number is subtracted from the variable. 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.

In Example 2, 37 was added to the y and so we subtracted 37 to ‘undo’ the addition. In Example 3, we will need to ‘undo’ subtraction by using the Addition Property of Equality.

Solve: $a - 28 = −37 .$

Solution


Add 28 to each side to ‘undo’ the subtraction.
Simplify.
Check:
Substitute $a = −9$

		The solution to $a - 28 = −37$ is $a = −9 .$

Solve: $x - \frac{5}{8} = \frac{3}{4} .$

Solution


Use the Addition Property of Equality.		$An algebraic equation is shown: x - 5/8 + 5/8 = 3/4 + 5/8. The fractions 5/8 appear in red text.$
Find the LCD to add the fractions on the right.		$A mathematical equation showing x minus 5/8 plus 5/8 equals 6/8 plus 5/8. This algebraic expression requires solving for the variable x, involving basic fraction addition and subtraction.$
Simplify.
Check:
Substitute $x = \frac{11}{8} .$
Subtract.	$A mathematical expression displaying the fraction 6/8 and 3/4, with a question mark over an equals sign, asking if the two fractions are equivalent. Both fractions are indeed equal.$
Simplify.	$A mathematical expression shows the fraction 3/4 equals 3/4, followed by a checkmark, indicating correctness or approval.$
		The solution to $x - \frac{5}{8} = \frac{3}{4}$ is $x = \frac{11}{8} .$

The next example will be an equation with decimals.

Solve: $n - 0.63 = −4.2 .$

Solution


Use the Addition Property of Equality.
Add.
Check:
Let $n = −3.57$ .

Solve Equations That Require Simplification

In the previous examples, we were able to isolate the variable with just one operation. Most 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 you try to isolate the variable. Remember that to simplify an expression means to do all the operations in the expression. Simplify one side of the equation at a time. Note that simplification is different from the process used to solve an equation in which we apply an operation to both sides.

How to Solve Equations That Require Simplification

Solve: $9 x - 5 - 8 x - 6 = 7 .$

Solution

In the second row of the table, the first cell says “Step 2. Isolate the variable.” In the second cell, the instructions say “Now isolate x. Undo subtraction by adding 11 to both sides.” The third cell contains the equation x minus 11 plus 11 equals 7 plus 11, with “plus 11” written in red on both sides.

In the third row of the table, the first cell says: “Step 3. Simplify the equation on both sides of the equation.” The second cell is left blank. The third cell contains x equals 18.

Solve: $5 (n - 4) - 4 n = −8 .$

Solution

We simplify both sides of the equation as much as possible before we try to isolate the variable.


Distribute on the left.
Use the Commutative Property to rearrange terms.
Combine like terms.
Each side is as simplified as possible. Next, isolate $n$ .
Undo subtraction by using the Addition Property of Equality.
Add.
Check. Substitute $n = 12$ .
	The solution to $5 (n - 4) - 4 n = −8$ is $n = 12 .$

Solve: $3 (2 y - 1) - 5 y = 2 (y + 1) - 2 (y + 3) .$

Solution

We simplify both sides of the equation before we isolate the variable.


Distribute on both sides.
Use the Commutative Property of Addition.
Combine like terms.
Each side is as simplified as possible. Next, isolate $y$ .
Undo subtraction by using the Addition Property of Equality.
Add.
Check. Let $y = −1$ .
	The solution to $3 (2 y - 1) - 5 y = 2 (y + 1) - 2 (y + 3)$ is $y = −1 .$

Translate to an Equation and Solve

To solve applications algebraically, we will begin by translating from English sentences into equations. Our first step is to look for the word (or words) that would translate to the equals sign. Table 9 shows us some of the words that are commonly used.

Equals =
is is equal to is the same as the result is gives was will be

The steps we use to translate a sentence into an equation are listed below.

Translate and solve: Eleven more than x is equal to 54.

Solution

Translate.
Subtract 11 from both sides.
Simplify.
Check: Is 54 eleven more than 43? $\begin{matrix} 43 + 11 & \overset{?}{=} & 54 \\ 54 & = & 54 ✓ \end{matrix}$

Translate and solve: The difference of $12 t$ and $11 t$ is $−14$ .

Solution

Translate.
Simplify.
Check: $\begin{matrix} 12 (−14) - 11 (−14) & \overset{?}{=} & −14 \\ −168 + 154 & \overset{?}{=} & −14 \\ −14 & = & −14 ✓ \end{matrix}$

Translate and Solve Applications

Most of the time a question that requires an algebraic solution comes out of a real life question. To begin with that question is asked in English (or the language of the person asking) and not in math symbols. Because of this, it is an important skill to be able to translate an everyday situation into algebraic language.

We will 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. For example, you might use q for the number of quarters if you were solving a problem about coins.

How to Solve Translate and Solve Applications

The MacIntyre family recycled newspapers for two months. The two months of newspapers weighed a total of 57 pounds. The second month, the newspapers weighed 28 pounds. How much did the newspapers weigh the first month?

Solution

In the second row, the first cell says “Step 2. Identify what we are asked to find.” The second cell says “What are we asked to find?” The third cell says: “How much did the newspapers weigh the 2nd month?”

In the third row, the first cell says “Step 3. Name what we are looking for. Choose a variable to represent that quantity.” The second cell says “Choose a variable.” The third cell says “Let w equal weight of the newspapers the 1st month.”

In the fourth row, the first cell says “Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.” The second cell says “Restate the problem. We know that the weight of the newspapers the second month is 28 pounds.” The third cell says “Weight of newspapers the 1st month plus the weight of the newspapers the 2nd month equals 57 pounds. Weight from 1st month plus 28 equals 57.” One line down, the second cell says “Translate into an equation using the variable w.” The third cell contains the equation w plus 28 equals 57.

In the fifth row, the first cell says “Step 5. Solve the equation using good algebra techniques.” The second cell says “Solve.” The third cell contains the equation with 28 being subtracted from both sides: w plus 28 minus 28 equals 57 minus 28, with minus 28 written in red. Below this is w equals 29.

In the sixth row, the first cell says “Step 6. Check the answer and make sure it makes sense.” The second cell says “Does 1st month’s weight plus 2nd month’s weight equal 57 pounds?” The third cell contains the equation 29 plus 28 might equal 57. Below this is 57 equals 57 with a check mark next to it.

In the seventh and final row, the first cell says ‘Step 7. Answer the question with a complete sentence.” The second cell says “Write a sentence to answer ‘How much did the newspapers weigh the 2nd month?’” The third cell contains the sentence “The 2nd month the newspapers weighed 29 pounds.”

Randell paid $28,675 for his new car. This was $875 less than the sticker price. What was the sticker price of the car?

Solution

This table illustrates a seven-step process for solving a word problem, exemplified by calculating a car's sticker price.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	"What was the sticker price of the car?"
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $s =$ the sticker price of the car.
Step 4. Translate into an equation. Restate the problem in one sentence.	$28,675 is $875 less than the sticker price
Step 5. Solve the equation.	$$28,675 is $875 less than s$ $\begin{matrix} 28,675 & = & s - 875 \\ 28,675 + 875 & = & s - 875 + 875 \\ 29,550 & = & s \end{matrix}$
Step 6. Check the answer. Is $875 less than $29,550 equal to $28,675? $\begin{matrix} \begin{matrix} 29,550 - 875 & \overset{?}{=} & 28,675 \\ 28,675 & = & 28,675 ✓ \end{matrix} \end{matrix}$
Step 7. Answer the question with a complete sentence.	The sticker price of the car was $29,550.

Key Concepts

To Determine Whether a Number is a Solution to an Equation
1. Substitute the number in for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting statement is true.
  - If it is true, the number is a solution.
  - If it is not true, the number is not a solution.
Addition Property of Equality
- For any numbers a, b, and c, if $a = b$ , then $a + c = b + c$ .
Subtraction Property of Equality
- For any numbers a, b, and c, if $a = b$ , then $a - c = b - c$ .
To Translate a Sentence to an 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.
To Solve an Application
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important 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.

Practice Makes Perfect

Verify a Solution of an Equation

In the following exercises, determine whether the given value is a solution to the equation.

Is $y = \frac{5}{3}$ a solution of
$6 y + 10 = 12 y$ ?

Solution

yes

Is $x = \frac{9}{4}$ a solution of
$4 x + 9 = 8 x$ ?

Is $u = - \frac{1}{2}$ a solution of
$8 u - 1 = 6 u$ ?

Solution

Is $v = - \frac{1}{3}$ a solution of
$9 v - 2 = 3 v$ ?

Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, solve each equation using the Subtraction and Addition Properties of Equality.

$x + 24 = 35$

Solution

$x = 11$

$x + 17 = 22$

$y + 45 = −66$

Solution

$y = −111$

$y + 39 = −83$

$b + \frac{1}{4} = \frac{3}{4}$

Solution

$b = \frac{1}{2}$

$a + \frac{2}{5} = \frac{4}{5}$

$p + 2.4 = −9.3$

Solution

$p = −11.7$

$m + 7.9 = 11.6$

$a - 45 = 76$

Solution

$a = 121$

$a - 30 = 57$

$m - 18 = −200$

Solution

$m = −182$

$m - 12 = −12$

$x - \frac{1}{3} = 2$

Solution

$x = \frac{7}{3}$

$x - \frac{1}{5} = 4$

$y - 3.8 = 10$

Solution

$y = 13.8$

$y - 7.2 = 5$

$x - 165 = −420$

Solution

$x = −255$

$z - 101 = −314$

$z + 0.52 = −8.5$

Solution

$z = −9.02$

$x + 0.93 = −4.1$

$q + \frac{3}{4} = \frac{1}{2}$

Solution

$q = - \frac{1}{4}$

$p + \frac{1}{3} = \frac{5}{6}$

$p - \frac{2}{5} = \frac{2}{3}$

Solution

$p = \frac{16}{15}$

$y - \frac{3}{4} = \frac{3}{5}$

Solve Equations that Require Simplification

In the following exercises, solve each equation.

$c + 31 - 10 = 46$

Solution

$c = 25$

$m + 16 - 28 = 5$

$9 x + 5 - 8 x + 14 = 20$

Solution

$x = 1$

$6 x + 8 - 5 x + 16 = 32$

$−6 x - 11 + 7 x - 5 = −16$

Solution

$x = 0$

$−8 n - 17 + 9 n - 4 = −41$

$5 (y - 6) - 4 y = −6$

Solution

$y = 24$

$9 (y - 2) - 8 y = −16$

$8 (u + 1.5) - 7 u = 4.9$

Solution

$u = −7.1$

$5 (w + 2.2) - 4 w = 9.3$

$6 a - 5 (a - 2) + 9 = −11$

Solution

$a = −30$

$8 c - 7 (c - 3) + 4 = −16$

$6 (y - 2) - 5 y = 4 (y + 3)$
$- 4 (y - 1)$

Solution

$y = 28$

$9 (x - 1) - 8 x = −3 (x + 5)$
$+ 3 (x - 5)$

$3 (5 n - 1) - 14 n + 9$
$= 10 (n - 4) - 6 n - 4 (n + 1)$

Solution

$n = −50$

$2 (8 m + 3) - 15 m - 4$
$= 9 (m + 6) - 2 (m - 1) - 7 m$

$− (j + 2) + 2 j - 1 = 5$

Solution

$j = 8$

$− (k + 7) + 2 k + 8 = 7$

$− (\frac{1}{4} a - \frac{3}{4}) + \frac{5}{4} a = −2$

Solution

$a = - \frac{11}{4}$

$− (\frac{2}{3} d - \frac{1}{3}) + \frac{5}{3} d = −4$

$8 (4 x + 5) - 5 (6 x) - x$
$= 53 - 6 (x + 1) + 3 (2 x + 2)$

Solution

$x = 13$

$6 (9 y - 1) - 10 (5 y) - 3 y$
$= 22 - 4 (2 y - 12) + 8 (y - 6)$

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve it.

Nine more than $x$ is equal to 52.

Solution

$x + 9 = 52; x = 43$

The sum of x and $−15$ is 23.

Ten less than m is $−14$ .

Solution

$m - 10 = −14; m = −4$

Three less than y is $−19$ .

The sum of y and $−30$ is 40.

Solution

$y + (−30) = 40; y = 70$

Twelve more than p is equal to 67.

The difference of $9 x and 8 x$ is 107.

Solution

$9 x - 8 x = 107; 107$

The difference of $5 c and 4 c$ is 602.

The difference of $n$ and $\frac{1}{6}$ is $\frac{1}{2}$ .

Solution

$n - \frac{1}{6} = \frac{1}{2}; \frac{2}{3}$

The difference of $f$ and $\frac{1}{3}$ is $\frac{1}{12}$ .

The sum of $−4 n$ and $5 n$ is $−82$ .

Solution

$−4 n + 5 n = −82; - 82$

The sum of $−9 m$ and $10 m$ is $−95$ .

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Distance Avril rode her bike a total of 18 miles, from home to the library and then to the beach. The distance from Avril’s house to the library is 7 miles. What is the distance from the library to the beach?

Solution

11 miles

Reading Jeff read a total of 54 pages in his History and Sociology textbooks. He read 41 pages in his History textbook. How many pages did he read in his Sociology textbook?

Age Eva’s daughter is 15 years younger than her son. Eva’s son is 22 years old. How old is her daughter?

Solution

7 years old

Age Pablo’s father is 3 years older than his mother. Pablo’s mother is 42 years old. How old is his father?

Groceries For a family birthday dinner, Celeste bought a turkey that weighed 5 pounds less than the one she bought for Thanksgiving. The birthday turkey weighed 16 pounds. How much did the Thanksgiving turkey weigh?

Solution

21 pounds

Weight Allie weighs 8 pounds less than her twin sister Lorrie. Allie weighs 124 pounds. How much does Lorrie weigh?

Health 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

Health 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?

Salary 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

Textbooks 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?

Everyday Math

Construction Miguel wants to drill a hole for a $\frac{5}{8}$ inch screw. The hole should be $\frac{1}{12}$ inch smaller than the screw. Let $d$ equal the size of the hole he should drill. Solve the equation $d + \frac{1}{12} = \frac{5}{8}$ to see what size the hole should be.

Solution

$d = \frac{13}{24} inch$

Baking Kelsey needs $\frac{2}{3}$ cup of sugar for the cookie recipe she wants to make. She only has $\frac{3}{8}$ cup of sugar and will borrow the rest from her neighbor. Let $s$ equal the amount of sugar she will borrow. Solve the equation $\frac{3}{8} + s = \frac{2}{3}$ to find the amount of sugar she should ask to borrow.

Writing Exercises

Is $−8$ a solution to the equation $3 x = 16 - 5 x$ ? How do you know?

Solution

No. Justifications will vary.

What is the first step in your solution to the equation $10 x + 2 = 4 x + 26$ ?

Self Check

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

This is a table that has six rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “verify a solution of an equation,” “solve equations using the subtraction and addition properties of equality,” “solve equations that require simplification,” “translate to an equation and solve,” and “translate and solve applications.” The rest of the cells are blank.

ⓑ If most of your checks were:

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

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

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

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

Solve Equations Using the Division and Multiplication Properties of Equality

You may have noticed that 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 on the side of the equation with the variable. Now we will see how to solve equations that have a variable multiplied by a constant and so will require division to isolate the variable.

Let’s look at our puzzle again with the envelopes and counters in Figure 1.

In the illustration 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?

How do we determine the number? We have to separate the counters on the right side into two groups of the same size to correspond with the two envelopes on the left side. The 6 counters divided into 2 equal groups gives 3 counters in each group (since $6 \div 2 = 3$ ).

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.


If we divide both sides of the equation by 2, as we did with the envelopes and counters,
we get:

We found that each envelope contains 3 counters. Does this check? We know $2 \cdot 3 = 6$ , so it works! Three counters in each of two envelopes does equal six!

This example leads to the Division Property of Equality.

The goal in solving an equation is to ‘undo’ the operation on the variable. In the next example, the variable is multiplied by 5, so we will divide both sides by 5 to ‘undo’ the multiplication.

Solve: $5 x = −27 .$

Solution

To isolate $x$ , “undo” the multiplication by 5.
Divide to ‘undo’ the multiplication.
Simplify.
Check:
Substitute $- \frac{27}{5}$ for $x .$	$This image displays the equation 5(-27/5) =? -27, a math problem testing basic multiplication with fractions and negative numbers. The statement is true as 5 * (-27/5) simplifies to -27.$

		Since this is a true statement, $x = - \frac{27}{5}$ is the solution to $5 x = −27$ .

Consider the equation $\frac{x}{4} = 3$ . 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.

Solve: $\frac{y}{−7} = −14 .$

Solution

Here $y$ is divided by $−7$ . We must multiply by $−7$ to isolate $y$ .


Multiply both sides by $−7$ .		$A mathematical equation shows -7 multiplied by the fraction y over -7, which is equal to -7 multiplied by -14. The -7 values on both sides of the equation are highlighted in red.$
Multiply.
Simplify.
Check: $\frac{y}{−7} = −14$
Substitute $y = 98$ .
Divide.

Solve: $− n = 9 .$

Solution


Remember $− n$ is equivalent to $−1 n$ .
Divide both sides by $−1$ .
Divide.
Notice that there are two other ways to solve $− n = 9$ . We can also solve this equation by multiplying both sides by $−1$ and also by taking the opposite of both sides.
Check:
Substitute $n = −9$ .
Simplify.

Solve: $\frac{3}{4} x = 12 .$

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 $\frac{3}{4}$ .


Multiply by the reciprocal of $\frac{3}{4}$ .
Reciprocals multiply to 1.
Multiply.
Notice that we could have divided both sides of the equation $\frac{3}{4} x = 12$ by $\frac{3}{4}$ to isolate $x$ . While this would work, most people would find multiplying by the reciprocal easier.
Check:
Substitute $x = 16$ .

In the next example, all the variable terms are on the right side of the equation. As always, our goal in solving the equation is to isolate the variable.

Solve: $\frac{8}{15} = - \frac{4}{5} x .$

Solution


Multiply by the reciprocal of $- \frac{4}{5}$ .		$A mathematical equation is displayed, showing the product of two fractions, (-5/4) and (8/15), on the left side, equated to the product of (-5/4) and (-4/5x) on the right.$
Reciprocals multiply to 1.		$A mathematical equation shows the simplification of a fraction: -(542)/(435) = 1x. Common factors '5' and '4' are crossed out from both the numerator and the denominator, leading to further calculation.$
Multiply.		$A mathematical equation shows a negative fraction, '-2/3,' which is set equal to the variable 'x.' The expression is centered against a plain white background.$
Check:
Let $x = - \frac{2}{3}$ .	$A mathematical equation showing the fraction 8/15 equals the product of -4/5 and -2/3. The numbers 2 and 3 in the second fraction are highlighted in red.$
	$A mathematical equation showing the fraction 8 over 15 is equal to 8 over 15, followed by a checkmark indicating correctness.$

Solve Equations That Require Simplification

Many equations start out more complicated than the ones we have been working with.

With these more complicated equations the first step is to simplify both sides of the equation as much as possible. This usually involves combining like terms or using the distributive property.

Solve: $14 - 23 = 12 y - 4 y - 5 y .$

Solution

Begin by simplifying each side of the equation.


Simplify each side.
Divide both sides by $3$ .
Check:
Substitute $y = −3$ .

Solve: $−4 (a - 3) - 7 = 25 .$

Solution

Here we will simplify each side of the equation by using the distributive property first.


Distribute.
Simplify.
Simplify.
Divide both sides by $−4$ to isolate $a$ .
Divide.
Check:
Substitute $a = −5$ .

Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.

Translate to an Equation and Solve

In the next few examples, we will translate sentences into equations and then solve the equations. You might want to review the translation table in the previous chapter.

Translate and solve: The number 143 is the product of $−11$ and y.

Solution

Begin by translating the sentence into an equation.

Translate.
Divide by $−11$ .
Simplify.
Check: $\begin{matrix} 143 & = & −11 y \\ 143 & \overset{?}{=} & −11 (−13) \\ 143 & = & 143 ✓ \end{matrix}$

Translate and solve: $n$ divided by 8 is $−32$ .

Solution

Begin by translating the sentence into an equation. Translate.
Multiple both sides by 8.
Simplify.
Check:	Is $n$ divided by 8 equal to −32?
Let $n = −256$ .	Is $−256$ divided by $8$ equal to $−32$ ?
Translate.	$\frac{−256}{8} \overset{?}{=} −32$
Simplify.	$−32 = −32 ✓$

Translate and solve: The quotient of $y$ and $−4$ is $68$ .

Solution

Begin by translating the sentence into an equation.

Translate.
Multiply both sides by $−4$ .		$A mathematical equation shows -4 multiplied by the fraction y over -4, which equals -4 multiplied by 68. The numbers -4 and 68 are in black, while the multiplication signs are implied by parentheses.$
Simplify.
Check:	Is the quotient of $y$ and $−4$ equal to $68$ ?
Let $y = −272$ .	Is the quotient of $−272$ and $−4$ equal to $68$ ?
Translate.	$\frac{−272}{−4} \overset{?}{=} 68$
Simplify.	$68 = 68 ✓$

Translate and solve: Three-fourths of $p$ is 18.

Solution

Begin by translating the sentence into an equation. Remember, “of” translates into multiplication.

Translate.
Multiply both sides by $\frac{4}{3} .$
Simplify.
Check:	Is three-fourths of p equal to 18?
Let $p = 24 .$	Is three-fourths of 24 equal to 18?
Translate.	$\frac{3}{4} \cdot 24 \overset{?}{=} 18$
Simplify.	$18 = 18 ✓$

Translate and solve: The sum of three-eighths and $x$ is one-half.

Solution

Begin by translating the sentence into an equation.

Translate.
Subtract $\frac{3}{8}$ from each side.		$A mathematical equation is displayed on a white background, reading '3/8 - 3/8 + x = 1/2 - 3/8'. The fractions on the right side of the equals sign are partially in red.$
Simplify and rewrite fractions with common denominators.		$A mathematical equation displays 'x = 4/8 - 3/8' on a white background, representing a subtraction problem with fractions sharing a common denominator.$
Simplify.
Check:	Is the sum of three-eighths and $x$ equal to one-half?
$Let x = \frac{1}{8} .$	Is the sum of three-eighths and one-eighth equal to one-half?
Translate.	$\frac{3}{8} + \frac{1}{8} \overset{?}{=} \frac{1}{2}$
Simplify.	$\frac{4}{8} \overset{?}{=} \frac{1}{2}$
Simplify.	$\frac{1}{2} = \frac{1}{2} ✓$

Translate and Solve Applications

To solve applications using the Division and Multiplication Properties of Equality, we will follow the same steps we used in the last section. We will restate the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve.

Denae bought 6 pounds of grapes for $10.74. What was the cost of one pound of grapes?

Solution

Steps for solving a word problem to find the unit cost of grapes, from identifying the unknown to checking the solution.
What are you asked to find?	The cost of 1 pound of grapes
Assign a variable.	Let $c$ = the cost of one pound.
Write a sentence that gives the information to find it.	The cost of 6 pounds is $10.74.
Translate into an equation.	$6 c = 10.74$
Solve.	$\begin{matrix} \frac{6 c}{6} = \frac{10.74}{6} \\ c = 1.79 \end{matrix}$
	The grapes cost $1.79 per pound.
Check: If one pound costs $1.79, do 6 pounds cost #10.74? $\begin{matrix} 6 (1.79) & \overset{?}{=} & 10.74 \\ 10.74 & = & 10.74 ✓ \end{matrix}$

Andreas bought a used car for $12,000. Because the car was 4-years old, its price was $\frac{3}{4}$ of the original price, when the car was new. What was the original price of the car?

Solution

This table demonstrates the step-by-step process for solving a word problem involving fractions, from identifying the unknown to checking the solution.
What are you asked to find?	The original price of the car
Assign a variable.	Let $p$ = the original price.
Write a sentence that gives the information to find it.	$12,000 is $\frac{3}{4}$ of the original price.
Translate into an equation.	$12,000 = \frac{3}{4} p$
Solve.	$\begin{matrix} \frac{4}{3} (12,000) = \frac{4}{3} \cdot \frac{3}{4} p \\ 16,000 = p \end{matrix}$
	The original cost of the car was $16,000.
Check: Is $\frac{3}{4}$ of $16,000 equal to $12,000? $\begin{matrix} \frac{3}{4} \cdot 16,000 & \overset{?}{=} & 12,000 \\ 12,000 & = & 12,000 ✓ \end{matrix}$

Key Concepts

The Division Property of Equality—For any numbers a, b, and c, and $c \neq 0$ , if $a = b$ , then $\frac{a}{c} = \frac{b}{c}$ .
When you divide both sides of an equation by any non-zero number, you still have equality.
The Multiplication Property of Equality—For any numbers a, b, and c, if $a = b$ , then $a c = b c$ .
If you multiply both sides of an equation by the same number, you still have equality.

Practice Makes Perfect

Solve Equations Using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation using the Division and Multiplication Properties of Equality and check the solution.

$8 x = 56$

Solution

$x = 7$

$7 p = 63$

$−5 c = 55$

Solution

$c = −11$

$−9 x = −27$

$−809 = 15 y$

Solution

$y = - \frac{809}{15}$

$−731 = 19 y$

$−37 p = −541$

Solution

$p = \frac{541}{37}$

$−19 m = −586$

$0.25 z = 3.25$

Solution

$z = 13$

$0.75 a = 11.25$

$−13 x = 0$

Solution

$x = 0$

$24 x = 0$

$\frac{x}{4} = 35$

Solution

$x = 140$

$\frac{z}{2} = 54$

$−20 = \frac{q}{−5}$

Solution

$q = 100$

$\frac{c}{−3} = −12$

$\frac{y}{9} = −16$

Solution

$y = −144$

$\frac{q}{6} = −38$

$\frac{m}{−12} = 45$

Solution

$m = −540$

$−24 = \frac{p}{−20}$

$− y = 6$

Solution

$y = −6$

$− u = 15$

$− v = −72$

Solution

$v = 72$

$− x = −39$

$\frac{2}{3} y = 48$

Solution

$y = 72$

$\frac{3}{5} r = 75$

$- \frac{5}{8} w = 40$

Solution

$w = −64$

$24 = - \frac{3}{4} x$

$- \frac{2}{5} = \frac{1}{10} a$

Solution

$a = −4$

$- \frac{1}{3} q = - \frac{5}{6}$

$- \frac{7}{10} x = - \frac{14}{3}$

Solution

$x = \frac{20}{3}$

$\frac{3}{8} y = - \frac{1}{4}$

$\frac{7}{12} = - \frac{3}{4} p$

Solution

$p = - \frac{7}{9}$

$\frac{11}{18} = - \frac{5}{6} q$

$- \frac{5}{18} = - \frac{10}{9} u$

Solution

$u = \frac{1}{4}$

$- \frac{7}{20} = - \frac{7}{4} v$

Solve Equations That Require Simplification

In the following exercises, solve each equation requiring simplification.

$100 - 16 = 4 p - 10 p - p$

Solution

$p = −12$

$−18 - 7 = 5 t - 9 t - 6 t$

$\frac{7}{8} n - \frac{3}{4} n = 9 + 2$

Solution

$n = 88$

$\frac{5}{12} q + \frac{1}{2} q = 25 - 3$

$0.25 d + 0.10 d = 6 - 0.75$

Solution

$d = 15$

$0.05 p - 0.01 p = 2 + 0.24$

$−10 (q - 4) - 57 = 93$

Solution

$q = −11$

$−12 (d - 5) - 29 = 43$

$−10 (x + 4) - 19 = 85$

Solution

$x = - \frac{72}{5}$

$−15 (z + 9) - 11 = 75$

Mixed Practice

In the following exercises, solve each equation.

$\frac{9}{10} x = 90$

Solution

$x = 100$

$\frac{5}{12} y = 60$

$y + 46 = 55$

Solution

$y = 9$

$x + 33 = 41$

$\frac{w}{−2} = 99$

Solution

$w = −198$

$\frac{s}{−3} = −60$

$27 = 6 a$

Solution

$a = \frac{9}{2}$

$− a = 7$

$− x = 2$

Solution

$x = −2$

$z - 16 = −59$

$m - 41 = −14$

Solution

$m = 27$

$0.04 r = 52.60$

$63.90 = 0.03 p$

Solution

$p = 2130$

$−15 x = −120$

$84 = −12 z$

Solution

$z = −7$

$19.36 = x - 0.2 x$

$c - 0.3 c = 35.70$

Solution

$c = 51$

$− y = −9$

$− x = −8$

Solution

$x = 8$

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve.

187 is the product of $−17$ and m.

133 is the product of $−19$ and n.

Solution

$133 = −19 n; n = −7$

$−184$ is the product of 23 and p.

$−152$ is the product of 8 and q.

Solution

$−152 = 8 q; q = −19$

u divided by 7 is equal to $−49$ .

r divided by 12 is equal to $−48$ .

Solution

$\frac{r}{12} = −48; r = −576$

h divided by $−13$ is equal to $−65$ .

j divided by $−20$ is equal to $−80$ .

Solution

$\frac{j}{−20} = −80; j = 1,600$

The quotient $c$ and $−19$ is 38.

The quotient of $b$ and $−6$ is 18.

Solution

$\frac{b}{−6} = 18; b = −108$

The quotient of $h$ and 26 is $−52$ .

The quotient $k$ and 22 is $−66$ .

Solution

$\frac{k}{22} = −66; k = −1,452$

Five-sixths of y is 15.

Three-tenths of x is 15.

Solution

$\frac{3}{10} x = 15; x = 50$

Four-thirds of w is 36.

Five-halves of v is 50.

Solution

$\frac{5}{2} v = 50; v = 20$

The sum of nine-tenths and g is two-thirds.

The sum of two-fifths and f is one-half.

Solution

$\frac{2}{5} + f = \frac{1}{2}; f = \frac{1}{10}$

The difference of p and one-sixth is two-thirds.

The difference of q and one-eighth is three-fourths.

Solution

$q - \frac{1}{8} = \frac{3}{4}; q = \frac{7}{8}$

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Kindergarten Connie’s kindergarten class has 24 children. She wants them to get into 4 equal groups. How many children will she put in each group?

Balloons Ramona bought 18 balloons for a party. She wants to make 3 equal bunches. How many balloons did she use in each bunch?

Solution

6 balloons

Tickets Mollie paid $36.25 for 5 movie tickets. What was the price of each ticket?

Shopping Serena paid $12.96 for a pack of 12 pairs of sport socks. What was the price of pair of sport socks?

Solution

$1.08

Sewing Nancy used 14 yards of fabric to make flags for one-third of the drill team. How much fabric, would Nancy need to make flags for the whole team?

MPG John’s SUV gets 18 miles per gallon (mpg). This is half as many mpg as his wife’s hybrid car. How many miles per gallon does the hybrid car get?

Solution

36 mpg

Height Aiden is 27 inches tall. He is $\frac{3}{8}$ as tall as his father. How tall is his father?

Real estate Bea earned $11,700 commission for selling a house, calculated as $\frac{6}{100}$ of the selling price. What was the selling price of the house?

Solution

$195,000

Everyday Math

Commission Every week Perry gets paid $150 plus 12% of his total sales amount over $1,250. Solve the equation $840 = 150 + 0.12 (a - 1250)$ for a, to find the total amount Perry must sell in order to be paid $840 one week.

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 $0.49 s + 0.21 (s - 5) = 9.45$ for s, to find the number of 49-cent stamps Travis bought.

Solution

15 49-cent stamps

Writing Exercises

Frida started to solve the equation $−3 x = 36$ by adding 3 to both sides. Explain why Frida’s method will not solve the equation.

Emiliano thinks $x = 40$ is the solution to the equation $\frac{1}{2} x = 80$ . Explain why he 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.

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

Solve Equations with Variables and Constants on Both Sides

Solve Equations with Constants on Both Sides

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 will learn to solve equations in which the variable terms, or constant terms, or both 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 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 began with variables and constants on both sides into the form $a x = b .$ We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Solve: $7 x + 8 = −13 .$

Solution

In this equation, the variable is found 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 will write the labels above the equation to help us remember what goes where.

This figure shows the equation 7x plus 8 equals negative 13, with the left side of the equation labeled “variable”, written in red, and the right side of the equation labeled “constant”, written in red.

Since the left side is the “ $x$ ”, or variable side, the 8 is out of place. We must “undo” adding 8 by subtracting 8, and to keep the equality we must subtract 8 from both sides.


Use the Subtraction Property of Equality.
Simplify.
Now all the variables are on the left and the constant on the right. The equation looks like those you learned to solve earlier.
Use the Division Property of Equality.
Simplify.
Check:
Let $x = −3$ .

Solve: $8 y - 9 = 31 .$

Solution

Notice, the variable is only on the left side of the equation, so we will call this side the “variable” side, and the right side will be the “constant” side. Since the left side is the “variable” side, the 9 is out of place. It is subtracted from the $8 y$ , so to “undo” subtraction, add 9 to both sides. Remember, whatever you do to the left, you must do to the right.


Add 9 to both sides.
Simplify.
		The variables are now on one side and the constants on the other. We continue from here as we did earlier.
Divide both sides by 8.
Simplify.
Check:
Let $y = 5$ .

Solve Equations with Variables on Both Sides

What if there are variables on both sides of the equation? For equations like this, begin as we did above—choose 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.

Solve: $9 x = 8 x - 6 .$

Solution

Here the variable is on both sides, but the constants only appear on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.


We don’t want any $x$ ’s on the right, so subtract the $8 x$ from both sides.
Simplify.
We succeeded in getting the variables on one side and the constants on the other, and have obtained the solution.
Check:
Let $x = −6$ .

Solve: $5 y - 9 = 8 y .$

Solution

The only constant is on the left and the $y$ ’s are on both sides. Let’s leave the constant on the left and get the variables to the right.


Subtract $5 y$ from both sides.
Simplify.
We have the y’s on the right and the constants on the left. Divide both sides by 3.
Simplify.
Check:
Let $y = −3 .$

Solve: $12 x = − x + 26 .$

Solution

The only constant is on the right, so let the left side be the “variable” side.


Remove the $− x$ from the right side by adding $x$ to both sides.
Simplify.
All the $x$ ’s are on the left and the constants are on the right. Divide both sides by 13.
Simplify.

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. It may take several steps to solve this equation, so we need a clear and organized strategy.

How to Solve Equations with Variables and Constants on Both Sides

Solve: $7 x + 5 = 6 x + 2 .$

Solution

In the second row of the table, the first cell says: “Step 2. Collect the variable terms to the “variable” side of the equation, using the addition or subtraction property of equality.” In the second cell, the instructions say: “ With the right side as the “constant” side, the 6x is out of place, so subtract 6x from both sides. Combine like terms. Now the variable is only on the left side!” The third cell contains the original equation with 6x subtracted from both sides: 7 x minus 6 x plus 5 equals 6 x minus 6 x plus 2, with “minus 6 x” written in red on both sides. Below this is the same equation with like terms combined: x plus 5 equals 2.

In the third row of the table, the first cell says: “Step 3. Collect all the constants to the other side of the equation, using the addition or subtraction property of equality.” In the second cell, the instructions say: “The right side is the “constant” side, so the 5 is out of place. Subtract 5 from both sides. Simplify.” The third cell contains the equation x plus 5 minus 5 equals 2 minus 5, with “minus 5” written in red on both sides. Below this is the answer to the equation: x equals negative 3.

In the fourth row of the table, the first cell says: “Step 4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.” In the second cell, the instructions say: “The coefficient of x is one. The equation is solved.” The third cell is blank.

In the fifth row of the table, the first cell says: “Step 5. Check.” The instructions in the second cell say: “Check. Let x equal negative 3. Simplify. Add.” In the third cell is the original equation again: 7 x plus 5 equals 6x plus 2. Below this is the same equation with negative 3 substituted in for x: 7 times negative 3 (in paretheses) plus 5 might equal 6 times negative 3 (in parentheses) plus 2, with the “times negative 3” written in red on both sides of the equation. Below this is the equation negative 21 plus 5 might equal negative 18 plus 2. On the last line is the equation negative 16 equals negative 16, with a check mark next to it.

We’ll list the steps below so you can easily refer to them. But we’ll call this the ‘Beginning Strategy’ because we’ll be adding some steps later in this chapter.

In Step 1, a helpful approach is to make the “variable” side the side that has the variable with the larger coefficient. This usually makes the arithmetic easier.

Solve: $8 n - 4 = −2 n + 6 .$

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since $8 > −2$ , make the left side the “variable” side.
We don’t want variable terms on the right side—add $2 n$ to both sides to leave only constants on the right.
Combine like terms.
We don’t want any constants on the left side, so add $4$ to both sides.
Simplify.
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 10.
Simplify.
Check:
Let $n = 1$ .

Solve: $7 a - 3 = 13 a + 7 .$

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since $13 > 7$ , make the right side the “variable” side and the left side the “constant” side.


Subtract $7 a$ from both sides to remove the variable term from the left.
Combine like terms.
Subtract $7$ from both sides to remove the constant from the right.
Simplify.
Divide both sides by $6$ to make $1$ the coefficient of $a$ .		$A mathematical equation shows a fraction -10/6 equal to the fraction 6a/6. The denominators are both 6, colored in red, while the numerators -10 and 6a are in black.$
Simplify.
Check:
Let $a = - \frac{5}{3}$ .
	$A mathematical equation is displayed: -35/3 - 9/3 = ? - 65/3 + 21/3. The '?' indicates a missing term, and the fractions all share a common denominator of 3, involving subtraction and addition operations.$

In the last example, we could have made the left side the “variable” side, but it would have led to a negative coefficient on the variable term. (Try it!) While we could work with the negative, there is less chance of errors when working with positives. The strategy outlined above helps avoid the negatives!

To solve an equation with fractions, we just follow the steps of our strategy to get the solution!

Solve: $\frac{5}{4} x + 6 = \frac{1}{4} x - 2 .$

Solution

Since $\frac{5}{4} > \frac{1}{4}$ , make the left side the “variable” side and the right side the “constant” side.


Subtract $\frac{1}{4} x$ from both sides.
Combine like terms.
Subtract $6$ from both sides.
Simplify.
$\begin{matrix} Check: & \frac{5}{4} x + 6 & = & \frac{1}{4} x - 2 \\ Let x = −8 . & \frac{5}{4} (−8) + 6 & \overset{?}{=} & \frac{1}{4} (−8) - 2 \\ −10 + 6 & \overset{?}{=} & −2 - 2 \\ −4 & = & −4 ✓ \end{matrix}$

We will use the same strategy to find the solution for an equation with decimals.

Solve: $7.8 x + 4 = 5.4 x - 8 .$

Solution

Since $7.8 > 5.4$ , make the left side the “variable” side and the right side the “constant” side.


Subtract $5.4 x$ from both sides.
Combine like terms.
Subtract $4$ from both sides.
Simplify.
Use the Division Propery of Equality.
Simplify.
Check:
Let $x = −5$ .

Key Concepts

Beginning Strategy for Solving an Equation with Variables and Constants on Both Sides of the Equation
1. Choose which side will be the “variable” side—the other side will be the “constant” side.
2. Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
3. Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
5. Check the solution by substituting it into the original equation.

Practice Makes Perfect

Solve Equations with Constants on Both Sides

In the following exercises, solve the following equations with constants on both sides.

$9 x - 3 = 60$

$12 x - 8 = 64$

Solution

$x = 6$

$14 w + 5 = 117$

$15 y + 7 = 97$

Solution

$y = 6$

$2 a + 8 = −28$

$3 m + 9 = −15$

Solution

$m = −8$

$−62 = 8 n - 6$

$−77 = 9 b - 5$

Solution

$b = −8$

$35 = −13 y + 9$

$60 = −21 x - 24$

Solution

$x = −4$

$−12 p - 9 = 9$

$−14 q - 2 = 16$

Solution

$q = - \frac{9}{7}$

Solve Equations with Variables on Both Sides

In the following exercises, solve the following equations with variables on both sides.

$19 z = 18 z - 7$

$21 k = 20 k - 11$

Solution

$k = −11$

$9 x + 36 = 15 x$

$8 x + 27 = 11 x$

Solution

$x = 9$

$c = −3 c - 20$

$b = −4 b - 15$

Solution

$b = −3$

$9 q = 44 - 2 q$

$5 z = 39 - 8 z$

Solution

$z = 3$

$6 y + \frac{1}{2} = 5 y$

$4 x + \frac{3}{4} = 3 x$

Solution

$x = - \frac{3}{4}$

$−18 a - 8 = −22 a$

$−11 r - 8 = −7 r$

Solution

$r = −2$

Solve Equations with Variables and Constants on Both Sides

In the following exercises, solve the following equations with variables and constants on both sides.

$8 x - 15 = 7 x + 3$

$6 x - 17 = 5 x + 2$

Solution

$x = 19$

$26 + 13 d = 14 d + 11$

$21 + 18 f = 19 f + 14$

Solution

$f = 7$

$2 p - 1 = 4 p - 33$

$12 q - 5 = 9 q - 20$

Solution

$q = −5$

$4 a + 5 = − a - 40$

$8 c + 7 = −3 c - 37$

Solution

$c = −4$

$5 y - 30 = −5 y + 30$

$7 x - 17 = −8 x + 13$

Solution

$x = 2$

$7 s + 12 = 5 + 4 s$

$9 p + 14 = 6 + 4 p$

Solution

$p = - \frac{8}{5}$

$2 z - 6 = 23 - z$

$3 y - 4 = 12 - y$

Solution

$y = 4$

$\frac{5}{3} c - 3 = \frac{2}{3} c - 16$

$\frac{7}{4} m - 7 = \frac{3}{4} m - 13$

Solution

$m = −6$

$8 - \frac{2}{5} q = \frac{3}{5} q + 6$

$11 - \frac{1}{5} a = \frac{4}{5} a + 4$

Solution

$a = 7$

$\frac{4}{3} n + 9 = \frac{1}{3} n - 9$

$\frac{5}{4} a + 15 = \frac{3}{4} a - 5$

Solution

$a = −40$

$\frac{1}{4} y + 7 = \frac{3}{4} y - 3$

$\frac{3}{5} p + 2 = \frac{4}{5} p - 1$

Solution

$p = 15$

$14 n + 8.25 = 9 n + 19.60$

$13 z + 6.45 = 8 z + 23.75$

Solution

$z = 3.46$

$2.4 w - 100 = 0.8 w + 28$

$2.7 w - 80 = 1.2 w + 10$

Solution

$w = 60$

$5.6 r + 13.1 = 3.5 r + 57.2$

$6.6 x - 18.9 = 3.4 x + 54.7$

Solution

$x = 23$

Everyday Math

Concert tickets At a school concert the total value of tickets sold was $1506. 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 $6 s + 27 s - 45 = 1506$ .

Making a fence Jovani has 150 feet of fencing to make a rectangular garden in his backyard. He wants the length to be 15 feet more than the width. Find the width, w, by solving the equation $150 = 2 w + 30 + 2 w$ .

Solution

30 feet

Writing Exercises

Solve the equation $\frac{6}{5} y - 8 = \frac{1}{5} y + 7$ explaining all the steps of your solution as in the examples in this section.

Solve the equation $10 x + 14 = −2 x + 38$ explaining all the steps of your solution as in the examples in this section.

Solution

$x = 2$ Justifications will vary.

When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient of $x$ to be the “variable” side?

Is $x = −2$ a solution to the equation $5 - 2 x = −4 x + 1$ ? How do you know?

Solution

Yes. Justifications will vary.

Self Check

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

This is a table that has four rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can...,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can...” reads: “solve an equation with constants on both sides,” “solve an equation with variables on both sides,” and “solve an equation with variables and constants on both sides. ” The rest of the cells are blank.

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

Use a General Strategy to Solve Linear Equations

Solve Equations Using the General Strategy

Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.

Beginning by simplifying each side of the equation makes the remaining steps easier.

How to Solve Linear Equations Using the General Strategy

Solve: $−6 (x + 3) = 24 .$

Solution

This figure is a table that has three columns and five rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Simplify each side of the equation as much as possible.” The text in the second cell reads: “Use the Distributive Property. Notice that each side of the equation is simplified as much as possible.” The third cell contains the equation negative 6 times x plus 3, where x plus 3 is in parentheses, equals 24. Below this is the same equation with the negative 6 distributed across the parentheses: negative 6x minus 18 equals 24.

In the second row of the table, the first cell says: “Step 2. Collect all variable terms on one side of the equation.” In the second cell, the instructions say: “Nothing to do—all x’s are on the left side. The third cell is blank.

In the third row of the table, the first cell says: “Step 3. Collect constant terms on the other side of the equation. In the second cell, the instructions say: “To get constants only on the right, add 18 to each side. Simplify.” The third cell contains the same equation with 18 added to both sides: negative 6x minus 18 plus 18 equals 24 plus 18. Below this is the equation negative 6x equals 42.

In the fourth row of the table, the first cell says: “Step 4. Make the coefficient of the variable term equal to 1.” In the second cell, the instructions say: “Divide each side by negative 6. Simplify. The third cell contains the same equation divided by negative 6 on both sides: negative 6x over negative 6 equals 42 over negative 6, with “divided by negative 6” written in red on both sides. Below this is the answer to the equation: x equals negative 7.

Solve: $− (y + 9) = 8 .$

Solution


Simplify each side of the equation as much as possible by distributing.
The only $y$ term is on the left side, so all variable terms are on the left side of the equation.
Add $9$ to both sides to get all constant terms on the right side of the equation.
Simplify.
Rewrite $- y$ as $−1 y$ .
Make the coefficient of the variable term to equal to $1$ by dividing both sides by $−1$ .
Simplify.
Check:
Let $y = −17$ .

Solve: $5 (a - 3) + 5 = −10$ .

Solution


Simplify each side of the equation as much as possible.
Distribute.
Combine like terms.
The only $a$ term is on the left side, so all variable terms are on one side of the equation.
Add $10$ to both sides to get all constant terms on the other side of the equation.
Simplify.
Make the coefficient of the variable term to equal to $1$ by dividing both sides by $5$ .
Simplify.
Check:
Let $a = 0$ .

Solve: $\frac{2}{3} (6 m - 3) = 8 - m$ .

Solution


Distribute.
Add $m$ to get the variables only to the left.
Simplify.
Add $2$ to get constants only on the right.
Simplify.
Divide by $5$ .
Simplify.
Check:	$A mathematical equation is displayed: 2/3(6m - 3) = 8 - m. It shows a linear equation with a variable 'm' to be solved, involving a fraction, parentheses, and distribution.$
Let $m = 2$ .

Solve: $8 - 2 (3 y + 5) = 0$ .

Solution


Simplify—use the Distributive Property.
Combine like terms.
Add $2$ to both sides to collect constants on the right.
Simplify.
Divide both sides by $−6$ .
Simplify.
Check: Let $y = - \frac{1}{3} .$

Solve: $4 (x - 1) - 2 = 5 (2 x + 3) + 6$ .

Solution


Distribute.
Combine like terms.
Subtract $4 x$ to get the variables only on the right side since $10 > 4$ .
Simplify.
Subtract $21$ to get the constants on left.
Simplify.
Divide by 6.		$A mathematical equation displaying two fractions set equal to each other: -27/6 = 6x/6. The denominators in both fractions are 6.$
Simplify.
Check:
Let $x = - \frac{9}{2}$ .

Solve: $10 [3 - 8 (2 s - 5)] = 15 (40 - 5 s)$ .

Solution


Simplify from the innermost parentheses first.
Combine like terms in the brackets.
Distribute.
Add $160 s$ to get the s’s to the right.
Simplify.
Subtract 600 to get the constants to the left.
Simplify.
Divide.
Simplify.
Check:
Substitute $s = −2$ .

Solve: $0.36 (100 n + 5) = 0.6 (30 n + 15)$ .

Solution


Distribute.
Subtract $18 n$ to get the variables to the left.
Simplify.
Subtract $1.8$ to get the constants to the right.
Simplify.
Divide.
Simplify.
Check:
Let $n = 0.4$ .

Classify Equations

Consider the equation we solved at the start of the last section, $7 x + 8 = −13$ . The solution we found was $x = −3$ . This means the equation $7 x + 8 = −13$ is true when we replace the variable, x, with the value $−3$ . We showed this when we checked the solution $x = −3$ and evaluated $7 x + 8 = −13$ for $x = −3$ .

This figure shows why we can say the equation 7x plus 8 equals negative 13 is true when the variable x is replaced with the value negative 3. The first line shows the equation with negative 3 substituted in for x: 7 times negative 3 plus 8 might equal negative 13. Below this is the equation negative 21 plus 8 might equal negative 13. Below this is the equation negative 13 equals negative 13, with a check mark next to it.

If we evaluate $7 x + 8$ for a different value of x, the left side will not be $−13$ .

The equation $7 x + 8 = −13$ is true when we replace the variable, x, with the value $−3$ , but not true when we replace x with any other value. Whether or not the equation $7 x + 8 = −13$ is true depends on the value of the variable. Equations like this are called conditional equations.

All the equations we have solved so far are conditional equations.

Now let’s consider the equation $2 y + 6 = 2 (y + 3)$ . Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.


Distribute.
Subtract $2 y$ to get the $y$ ’s to one side.
Simplify—the $y$ ’s are gone!

But $6 = 6$ is true.

This means that the equation $2 y + 6 = 2 (y + 3)$ is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity.

What happens when we solve the equation $5 z = 5 z - 1$ ?


Subtract $5 z$ to get the constant alone on the right.
Simplify—the $z$ ’s are gone!

But $0 \neq − 1$ .

Solving the equation $5 z = 5 z - 1$ led to the false statement $0 = −1$ . The equation $5 z = 5 z - 1$ will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$6 (2 n - 1) + 3 = 2 n - 8 + 5 (2 n + 1)$

Solution


Distribute.
Combine like terms.
Subtract $12 n$ to get the $n$ ’s to one side.
Simplify.
This is a true statement.	The equation is an identity. The solution is all real numbers.

Classify as a conditional equation, an identity, or a contradiction. Then state the solution.

$10 + 4 (p - 5) = 0$

Solution


Distribute.
Combine like terms.
Add $10$ to both sides.
Simplify.
Divide.
Simplify.
The equation is true when $p = \frac{5}{2}$ .	This is a conditional equation. The solution is $p = \frac{5}{2} .$

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

$5 m + 3 (9 + 3 m) = 2 (7 m - 11)$

Solution


Distribute.
Combine like terms.
Subtract $14 m$ from both sides.
Simplify.
But $27 \neq −22$ .	The equation is a contradiction. It has no solution.

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

Key Concepts

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 on one side of the equation.
  Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms on the other side of the equation.
  Use the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable term 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.

Practice Makes Perfect

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

$15 (y - 9) = −60$

$21 (y - 5) = −42$

Solution

$y = 3$

$−9 (2 n + 1) = 36$

$−16 (3 n + 4) = 32$

Solution

$n = −2$

$8 (22 + 11 r) = 0$

$5 (8 + 6 p) = 0$

Solution

$p = - \frac{4}{3}$

$− (w - 12) = 30$

$− (t - 19) = 28$

Solution

$t = −9$

$9 (6 a + 8) + 9 = 81$

$8 (9 b - 4) - 12 = 100$

Solution

$b = 2$

$32 + 3 (z + 4) = 41$

$21 + 2 (m - 4) = 25$

Solution

$m = 6$

$51 + 5 (4 - q) = 56$

$−6 + 6 (5 - k) = 15$

Solution

$k = \frac{3}{2}$

$2 (9 s - 6) - 62 = 16$

$8 (6 t - 5) - 35 = −27$

Solution

$t = 1$

$3 (10 - 2 x) + 54 = 0$

$−2 (11 - 7 x) + 54 = 4$

Solution

$x = −2$

$\frac{2}{3} (9 c - 3) = 22$

$\frac{3}{5} (10 x - 5) = 27$

Solution

$x = 5$

$\frac{1}{5} (15 c + 10) = c + 7$

$\frac{1}{4} (20 d + 12) = d + 7$

Solution

$d = 1$

$18 - (9 r + 7) = −16$

$15 - (3 r + 8) = 28$

Solution

$r = −7$

$5 - (n - 1) = 19$

$−3 - (m - 1) = 13$

Solution

$m = −15$

$11 - 4 (y - 8) = 43$

$18 - 2 (y - 3) = 32$

Solution

$y = −4$

$24 - 8 (3 v + 6) = 0$

$35 - 5 (2 w + 8) = −10$

Solution

$w = \frac{1}{2}$

$4 (a - 12) = 3 (a + 5)$

$−2 (a - 6) = 4 (a - 3)$

Solution

$a = 4$

$2 (5 - u) = −3 (2 u + 6)$

$5 (8 - r) = −2 (2 r - 16)$

Solution

$r = 8$

$3 (4 n - 1) - 2 = 8 n + 3$

$9 (2 m - 3) - 8 = 4 m + 7$

Solution

$m = 3$

$12 + 2 (5 - 3 y) = −9 (y - 1) - 2$

$−15 + 4 (2 - 5 y) = −7 (y - 4) + 4$

Solution

$y = −3$

$8 (x - 4) - 7 x = 14$

$5 (x - 4) - 4 x = 14$

Solution

$x = 34$

$5 + 6 (3 s - 5) = −3 + 2 (8 s - 1)$

$−12 + 8 (x - 5) = −4 + 3 (5 x - 2)$

Solution

$x = −6$

$4 (u - 1) - 8 = 6 (3 u - 2) - 7$

$7 (2 n - 5) = 8 (4 n - 1) - 9$

Solution

$n = −1$

$4 (p - 4) - (p + 7) = 5 (p - 3)$

$3 (a - 2) - (a + 6) = 4 (a - 1)$

Solution

$a = −4$

$− (9 y + 5) - (3 y - 7)$
$= 16 - (4 y - 2)$

$− (7 m + 4) - (2 m - 5)$
$= 14 - (5 m - 3)$

Solution

$m = −4$

$4 [5 - 8 (4 c - 3)]$
$= 12 (1 - 13 c) - 8$

$5 [9 - 2 (6 d - 1)]$
$= 11 (4 - 10 d) - 139$

Solution

$d = −3$

$3 [−9 + 8 (4 h - 3)]$
$= 2 (5 - 12 h) - 19$

$3 [−14 + 2 (15 k - 6)]$
$= 8 (3 - 5 k) - 24$

Solution

$k = \frac{3}{5}$

$5 [2 (m + 4) + 8 (m - 7)]$
$= 2 [3 (5 + m) - (21 - 3 m)]$

$10 [5 (n + 1) + 4 (n - 1)]$
$= 11 [7 (5 + n) - (25 - 3 n)]$

Solution

$n = −5$

$5 (1.2 u - 4.8) = −12$

$4 (2.5 v - 0.6) = 7.6$

Solution

$v = 1$

$0.25 (q - 6) = 0.1 (q + 18)$

$0.2 (p - 6) = 0.4 (p + 14)$

Solution

$p = −34$

$0.2 (30 n + 50) = 28$

$0.5 (16 m + 34) = −15$

Solution

$m = −4$

Classify Equations

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

$23 z + 19 = 3 (5 z - 9) + 8 z + 46$

$15 y + 32 = 2 (10 y - 7) - 5 y + 46$

Solution

identity; all real numbers

$5 (b - 9) + 4 (3 b + 9) = 6 (4 b - 5) - 7 b + 21$

$9 (a - 4) + 3 (2 a + 5) = 7 (3 a - 4) - 6 a + 7$

Solution

identity; all real numbers

$18 (5 j - 1) + 29 = 47$

$24 (3 d - 4) + 100 = 52$

Solution

conditional equation; $d = \frac{2}{3}$

$22 (3 m - 4) = 8 (2 m + 9)$

$30 (2 n - 1) = 5 (10 n + 8)$

Solution

conditional equation; $n = 7$

$7 v + 42 = 11 (3 v + 8) - 2 (13 v - 1)$

$18 u - 51 = 9 (4 u + 5) - 6 (3 u - 10)$

Solution

contradiction; no solution

$3 (6 q - 9) + 7 (q + 4) = 5 (6 q + 8) - 5 (q + 1)$

$5 (p + 4) + 8 (2 p - 1) = 9 (3 p - 5) - 6 (p - 2)$

Solution

contradiction; no solution

$12 (6 h - 1) = 8 (8 h + 5) - 4$

$9 (4 k - 7) = 11 (3 k + 1) + 4$

Solution

conditional equation; $k = 26$

$45 (3 y - 2) = 9 (15 y - 6)$

$60 (2 x - 1) = 15 (8 x + 5)$

Solution

contradiction; no solution

$16 (6 n + 15) = 48 (2 n + 5)$

$36 (4 m + 5) = 12 (12 m + 15)$

Solution

identity; all real numbers

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

$11 (8 c + 5) - 8 c = 2 (40 c + 25) + 5$

Solution

identity; all real numbers

Everyday Math

Fencing Micah has 44 feet of fencing to make a dog run in his yard. He wants the length to be 2.5 feet more than the width. Find the length, L, by solving the equation $2 L + 2 (L - 2.5) = 44$ .

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 $0.05 n + 0.10 (2 n - 1) = 1.90$ .

Solution

8 nickels

Writing Exercises

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

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.

Solution

Answers will vary.

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

Solve the equation $\frac{1}{4} (8 x + 20) = 3 x - 4$ explaining all the steps of your solution as in the examples in this section.

Solution

Answers will vary.

Self Check

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

This is a table that has three rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads: “solve equations using the general strategy for solving linear equations,” and “classify equations.” The rest of the cells are blank.

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

conditional equation

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

contradiction

An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.

identity

An equation that is true for any value of the variable is called an identity. The solution of an identity is all real numbers.

Solve Equations with Fractions or Decimals

Solve Equations with Fraction Coefficients

Let’s use the general strategy for solving linear equations introduced earlier to solve the equation, $\frac{1}{8} x + \frac{1}{2} = \frac{1}{4}$ .

	$A mathematical equation is displayed: 1/8x + 1/2 = 1/4. The equation involves fractions, a variable 'x', addition, and an equality sign, set against a plain white background.$
To isolate the $x$ term, subtract $\frac{1}{2}$ from both sides.
Simplify the left side.
Change the constants to equivalent fractions with the LCD.
Subtract.
Multiply both sides by the reciprocal of $\frac{1}{8}$ .
Simplify.

This method worked fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

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

Let’s solve a similar equation, but this time use the method that eliminates the fractions.

How to Solve Equations with Fraction Coefficients

Solve: $\frac{1}{6} y - \frac{1}{3} = \frac{5}{6}$ .

Solution

$This figure is a table that has three columns and three rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Find the least common denominator of all the fractions in the equation.” The text in the second cell reads: “What is the LCD of 1/6, 1/3, and 5/6?” The third cell contains the equation one-sixth y minus 1/3 equals 5/6, with LCD equals 6 written next to it.$ $In the second row of the table, the first cell says: “Step 2. Multiply both sides of the equation by that LCD. This clears the fractions.” In the second cell, the instructions say: “Multiply both sides of the equation by the LCD 6. Use the Distributive Property. Simplify—and notice, no more fractions!” The third cell contains the equation 6 times one-sixth y minus 1/3, with one-sixth y minus 1/3 in brackets, equals 6 times 5/6, with “6 times” written in red on both sides. Below this is the same equation with the 6 distributed on both sides: 6 times one-sixth y minus 6 times 1/3 equals 6 times 5/6. Below this is the equation y minus 2 equals 5.$ In the third row of the table, the first cell says: “Step 3. Solve using the General Strategy for Solving Linear Equations.” In the second cell, the instructions say: “Isolate the x term, add 2. Simplify.” The third cell contains the equation with 2 added to both sides: y minus 2 plus 2 equals 5 plus 2, with “plus 2” written in red on both sides. Below this is the equation y equals 7.

In the third row of the table, the first cell says: “Step 3. Solve using the General Strategy for Solving Linear Equations.” In the second cell, the instructions say: “Isolate the x term, add 2. Simplify.” The third cell contains the equation with 2 added to both sides: y minus 2 plus 2 equals 5 plus 2, with “plus 2” written in red on both sides. Below this is the equation y equals 7.

Notice in Example 1, 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: $6 = \frac{1}{2} v + \frac{2}{5} v - \frac{3}{4} v$ .

Solution

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

Find the LCD of all fractions in the equation.		$A mathematical equation shown is 6 = 1/2v + 2/5v - 3/4v, displaying a linear equation with fractions where the constant 6 is set equal to a sum and difference of terms involving the variable 'v' with fractional coefficients.$
The LCD is 20.
Multiply both sides of the equation by 20.
Distribute.
Simplify—notice, no more fractions!
Combine like terms.
Divide by 3.
Simplify.
Check:
Let $v = 40$ .

In the next example, we again have variables on both sides of the equation.

Solve: $a + \frac{3}{4} = \frac{3}{8} a - \frac{1}{2}$ .

Solution


Find the LCD of all fractions in the equation. The LCD is 8.
Multiply both sides by the LCD.		$A mathematical equation is displayed: 8(a + 3/4) = 8(3/8a - 1/2). The equation involves a variable 'a', integers, and fractions, with both sides multiplied by the integer 8.$
Distribute.
Simplify—no more fractions.
Subtract $3 a$ from both sides.
Simplify.
Subtract 6 from both sides.
Simplify.
Divide by 5.		$A mathematical equation shows '5a divided by 5 equals -10 divided by 5'. The number 5 in the denominator of both fractions is highlighted in red, indicating division by 5 on both sides of the equation.$
Simplify.
Check:
Let $a = −2$ .	$A mathematical problem presented as an equation to be verified: -2 + 3/4 =? (3/8)(-2) - 1/2, involving integers and fractions.$
	$A mathematical equation featuring fractions: -8/4 + 3/4 = ? 16/8 - 4/8. The question mark indicates an unknown value or a query regarding the equality of the expressions.$
	$A mathematical equation shows that the negative fraction -5/4 is equal to the negative fraction -10/8, demonstrating the concept of equivalent fractions with negative values.$
	$A mathematical equation shows '-5/4 = -5/4' with a checkmark, indicating the equality is correct. The expression represents a negative fraction equal to itself, often used in math education.$

In the next example, we start by using the Distributive Property. This step clears the fractions right away.

Solve: $−5 = \frac{1}{4} (8 x + 4)$ .

Solution


Distribute.
Simplify. Now there are no fractions.
Subtract 1 from both sides.
Simplify.
Divide by 2.
Simplify.
Check:	$A mathematical equation is displayed on a white background: -5 = 1/4(8x + 4). The equation shows -5 on the left side of the equals sign. On the right, the fraction 1/4 is multiplied by the quantity (8x + 4).$
Let $x = −3$ .

In the next example, even after distributing, we still have fractions to clear.

Solve: $\frac{1}{2} (y - 5) = \frac{1}{4} (y - 1)$ .

Solution


Distribute.		$The image shows a linear equation involving fractions and the variable 'y': (1/2) * y - (1/2) * 5 = (1/4) * y - (1/4) * 1.$
Simplify.
Multiply by the LCD, 4.
Distribute.		$A linear equation involving fractions and the variable 'y' is presented as 4(1/2)y - 4(5/2) = 4(1/4)y - 4(1/4).$
Simplify.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
Check:
Let $y = 9$ .
Finish the check on your own.

Solve: $\frac{5 x - 3}{4} = \frac{x}{2}$ .

Solution

		$An algebraic equation is shown, with a fraction on each side of the equals sign. On the left, the numerator is 5x - 3 and the denominator is 4. On the right, the numerator is x and the denominator is 2.$
Multiply by the LCD, 4.		$A mathematical equation shows '4 multiplied by the fraction (5x minus 3) over 4' equals '4 multiplied by the fraction x over 2.' The number 4 is highlighted in red on both sides of the equation.$
Simplify.
Collect the variables to the right.
Simplify.
Divide.
Simplify.
Check:
Let $x = 1$ .
	$A mathematical equation questions whether the fraction 2/4 is equivalent to 1/2, illustrating a common concept in comparing and simplifying fractions.$

Solve: $\frac{a}{6} + 2 = \frac{a}{4} + 3$ .

Solution


Multiply by the LCD, 12.
Distribute.		$The mathematical equation 12 * (d/6) + 12 * 2 = 12 * (d/4) + 12 * 3 is displayed, demonstrating an algebraic problem with fractions and a variable 'd'.$
Simplify.
Collect the variables to the right.
Simplify.
Collect the constants to the left.
Simplify.
Check:
Let $a = −12$ .

Solve: $\frac{4 q + 3}{2} + 6 = \frac{3 q + 5}{4}$ .

Solution


Multiply by the LCD, 4.
Distribute.		$A mathematical equation showing 4 multiplied by the fraction (4q+3)/2, added to the product of 4 and 6, which equals 4 multiplied by the fraction (3q+5)/4.$
Simplify.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
Divide by 5.
Simplify.
Check:
Let $q = −5$ .
Finish the check on your own.

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, $0.3 = \frac{3}{10}$ and $0.17 = \frac{17}{100}$ . So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.

Solve: $0.06 x + 0.02 = 0.25 x - 1.5$ .

Solution

Look at the decimals and think of the equivalent fractions.

0.06 = \frac{6}{100} 0.02 = \frac{2}{100} 0.25 = \frac{25}{100} 1.5 = 1 \frac{5}{10}

Notice, the LCD is 100.

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


Multiply both sides by 100.
Distribute.
Multiply, and now we have no more decimals.
Collect the variables to the right.
Simplify.
Collect the constants to the left.
Simplify.
Divide by 19.
Simplify.
Check:	Let $x = 8$ .

The next example uses an equation that is typical of the money applications in the next chapter. Notice that we distribute the decimal before we clear all the decimals.

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

Solution


Distribute first.
Combine like terms.
To clear decimals, multiply by 100.
Distribute.
Subtract 15 from both sides.
Simplify.
Divide by 30.
Simplify.
Check it yourself by substituting $x = 9$ into the original equation.

Key Concepts

Strategy to Solve an Equation with Fraction Coefficients
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.

Practice Makes Perfect

Solve Equations with Fraction Coefficients

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

$\frac{1}{4} x - \frac{1}{2} = - \frac{3}{4}$

$\frac{3}{4} x - \frac{1}{2} = \frac{1}{4}$

Solution

$x = 1$

$\frac{5}{6} y - \frac{2}{3} = - \frac{3}{2}$

$\frac{5}{6} y - \frac{1}{3} = - \frac{7}{6}$

Solution

$y = −1$

$\frac{1}{2} a + \frac{3}{8} = \frac{3}{4}$

$\frac{5}{8} b + \frac{1}{2} = - \frac{3}{4}$

Solution

$b = −2$

$2 = \frac{1}{3} x - \frac{1}{2} x + \frac{2}{3} x$

$2 = \frac{3}{5} x - \frac{1}{3} x + \frac{2}{5} x$

Solution

$x = 3$

$\frac{1}{4} m - \frac{4}{5} m + \frac{1}{2} m = −1$

$\frac{5}{6} n - \frac{1}{4} n - \frac{1}{2} n = −2$

Solution

$n = −24$

$x + \frac{1}{2} = \frac{2}{3} x - \frac{1}{2}$

$x + \frac{3}{4} = \frac{1}{2} x - \frac{5}{4}$

Solution

$x = −4$

$\frac{1}{3} w + \frac{5}{4} = w - \frac{1}{4}$

$\frac{3}{2} z + \frac{1}{3} = z - \frac{2}{3}$

Solution

$z = −2$

$\frac{1}{2} x - \frac{1}{4} = \frac{1}{12} x + \frac{1}{6}$

$\frac{1}{2} a - \frac{1}{4} = \frac{1}{6} a + \frac{1}{12}$

Solution

$a = 1$

$\frac{1}{3} b + \frac{1}{5} = \frac{2}{5} b - \frac{3}{5}$

$\frac{1}{3} x + \frac{2}{5} = \frac{1}{5} x - \frac{2}{5}$

Solution

$x = −6$

$1 = \frac{1}{6} (12 x - 6)$

$1 = \frac{1}{5} (15 x - 10)$

Solution

$x = 1$

$\frac{1}{4} (p - 7) = \frac{1}{3} (p + 5)$

$\frac{1}{5} (q + 3) = \frac{1}{2} (q - 3)$

Solution

$q = 7$

$\frac{1}{2} (x + 4) = \frac{3}{4}$

$\frac{1}{3} (x + 5) = \frac{5}{6}$

Solution

$x = - \frac{5}{2}$

$\frac{5 q - 8}{5} = \frac{2 q}{10}$

$\frac{4 m + 2}{6} = \frac{m}{3}$

Solution

$m = −1$

$\frac{4 n + 8}{4} = \frac{n}{3}$

$\frac{3 p + 6}{3} = \frac{p}{2}$

Solution

$p = −4$

$\frac{u}{3} - 4 = \frac{u}{2} - 3$

$\frac{v}{10} + 1 = \frac{v}{4} - 2$

Solution

$v = 20$

$\frac{c}{15} + 1 = \frac{c}{10} - 1$

$\frac{d}{6} + 3 = \frac{d}{8} + 2$

Solution

$d = −24$

$\frac{3 x + 4}{2} + 1 = \frac{5 x + 10}{8}$

$\frac{10 y - 2}{3} + 3 = \frac{10 y + 1}{9}$

Solution

$y = −1$

$\frac{7 u - 1}{4} - 1 = \frac{4 u + 8}{5}$

$\frac{3 v - 6}{2} + 5 = \frac{11 v - 4}{5}$

Solution

$v = 4$

Solve Equations with Decimal Coefficients

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

$0.6 y + 3 = 9$

$0.4 y - 4 = 2$

Solution

$y = 15$

$3.6 j - 2 = 5.2$

$2.1 k + 3 = 7.2$

Solution

$k = 2$

$0.4 x + 0.6 = 0.5 x - 1.2$

$0.7 x + 0.4 = 0.6 x + 2.4$

Solution

$x = 20$

$0.23 x + 1.47 = 0.37 x - 1.05$

$0.48 x + 1.56 = 0.58 x - 0.64$

Solution

$x = 22$

$0.9 x - 1.25 = 0.75 x + 1.75$

$1.2 x - 0.91 = 0.8 x + 2.29$

Solution

$x = 8$

$0.05 n + 0.10 (n + 8) = 2.15$

$0.05 n + 0.10 (n + 7) = 3.55$

Solution

$n = 19$

$0.10 d + 0.25 (d + 5) = 4.05$

$0.10 d + 0.25 (d + 7) = 5.25$

Solution

$d = 10$

$0.05 (q - 5) + 0.25 q = 3.05$

$0.05 (q - 8) + 0.25 q = 4.10$

Solution

$q = 15$

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 $0.10 d + 0.01 (d + 2) = 2$ for $d$ , the number of dimes.

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

Solution

$s = 35$

Writing Exercises

Explain how you find the least common denominator of $\frac{3}{8}$ , $\frac{1}{6}$ , and $\frac{2}{3}$ .

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

Solution

Answers will vary.

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

In the equation $0.35 x + 2.1 = 3.85$ what is the LCD? How do you know?

Solution

100. Justifications will vary.

Self Check

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

$This is a table that has three rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads: “solve equations with fraction coefficients,” and “solve equations with decimal coefficients.” The rest of the cells are blank.$

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

Solve a Formula for a Specific Variable

Use the Distance, Rate, and Time Formula

One formula you will use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant rate. Rate is an equivalent word for “speed.” The basic idea of rate may already familiar to you. Do you know what distance you travel if you drive 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 highway.) If you said 120 miles, you already know how to use this formula!

We will use the Strategy for Solving Applications that we used earlier in this chapter. When our problem requires a formula, we change Step 4. In place of writing a sentence, we write the appropriate formula. We write the revised steps here for reference.

You may want to create a mini-chart to summarize the information in the problem. See the chart in this first example.

Jamal rides his bike at a uniform rate of 12 miles per hour for $3 \frac{1}{2}$ hours. What distance has he traveled?

Solution

Step 1. Read the problem.
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.		$d = r t$

Substitute in the given information.		$d = 12 \cdot 3 \frac{1}{2}$
Step 5. Solve the equation.		$d = 42$ miles
Step 6. Check
Does 42 miles make sense?
Jamal rides:

Step 7. Answer the question with a complete sentence.		Jamal rode 42 miles.

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

Step 1. Read the problem.
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.		$d = r t$
Substitute in the given information.		$520 = 65 t$
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.
	$\begin{matrix} d & = & r t \\ 520 & \overset{?}{=} & 65 \cdot 8 \\ 520 & = & 520 ✓ \end{matrix}$
Step 7. Answer the question with a complete sentence. Rey’s trip will take 8 hours.

Solve a Formula for a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. 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 $d = r t$ . This formula gives the value of $d$ , distance, when you substitute in the values of $r and t$ , the rate and time. 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 for $t$ . If you had to do this often, you might wonder why there is not a formula that gives the value of $t$ when you substitute in the values of $d and r$ . We can make a formula like this by solving the formula $d = r t$ for $t$ .

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

Solve the formula $d = r t$ for $t$ :

ⓐ when $d = 520$ and $r = 65$
ⓑ in general

Solution

We will write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.

ⓐ when $d = 520$ and $r = 65$		ⓑ in general
Write the formula.	$d = r t$	Write the formula.	$d = r t$
Substitute.	$520 = 65 t$
Divide, to isolate $t$ .	$\frac{520}{65} = \frac{65 t}{65}$	Divide, to isolate $t$ .	$\frac{d}{r} = \frac{r t}{r}$
Simplify.	$8 = t$	Simplify.	$\frac{d}{r} = t$

We say the formula $t = \frac{d}{r}$ is solved for $t$ .

Solve the formula $A = \frac{1}{2} b h$ for $h$ :

ⓐ when $A = 90$ and $b = 15$ ⓑ in general

Solution

ⓐ when $A = 90$ and $b = 15$	ⓑ in general
Write the formula.	Write the formula.
Substitute.
Clear the fractions.	Clear the fractions.	$A mathematical equation shows both sides being multiplied by 2, represented in red text, to simplify the expression for the area of a triangle, A = (1/2)bh, by removing the fraction.$
Simplify.	Simplify.
Solve for $h$ .	Solve for $h$ .

We can now find the height of a triangle, if we know the area and the base, by using the formula $h = \frac{2 A}{b}$ .

The formula $I = P r t$ is used to calculate simple interest, I, for a principal, P, invested at rate, r, for t years.

Solve the formula $I = P r t$ to find the principal, $P$ :

ⓐ when $I = $ 5,600, r = 4 %, t = 7 years$ ⓑ in general

Solution

ⓐ $I = $5,600$ , $r = 4%$ , $t = 7 years$	ⓑ in general
Write the formula.	Write the formula.
Substitute.
Simplify.	Simplify.
Divide, to isolate P.	Divide, to isolate P.
Simplify.	Simplify.
The principal is		$A mathematical formula is shown: P = I / rt. The letter P is equal to a fraction where I is the numerator, and the product of r and t is the denominator. The variables are P, I, r, and t.$

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.

Solve the formula $3 x + 2 y = 18$ for y:

ⓐ when $x = 4$ ⓑ in general

Solution

ⓐ when $x = 4$	ⓑ in general

Substitute.
Subtract to isolate the $y$ -term.	Subtract to isolate the $y$ -term.
Divide.	Divide.	$Equation showing two y over two equals eighteen over two minus three x over two. The number two in the denominator of each fraction is highlighted.$
Simplify.	Simplify.

In Examples 1.60 through 1.64 we used the numbers in part ⓐ as a guide to solving in general in part ⓑ. Now we will solve a formula in general without using numbers as a guide.

Solve the formula $P = a + b + c$ for $a$ .

Solution

We will isolate $a$ on one side of the equation.
Both $b$ and $c$ are added to $a$ , so we subtract them from both sides of the equation.
Simplify.

Solve the formula $6 x + 5 y = 13$ for y.

Solution


Subtract $6 x$ from both sides to isolate the term with $y$ .
Simplify.
Divide by 5 to make the coefficient 1.	$A mathematical equation is shown with fractions. The left side is 5y over 5, and the right side is 13 minus 6x, all over 5. The denominator 5 on both sides is highlighted in red.$
Simplify.	$A mathematical equation is displayed, showing y equals the fraction with numerator 13 minus 6x and denominator 5.$

The fraction is simplified. We cannot divide $13 - 6 x$ by 5.

Key Concepts

To Solve an Application (with a formula)
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. Write the appropriate formula 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.
Distance, Rate and Time
For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula: $d = r t$ where d = distance, r = rate, t = time.
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 other variables and constants on the other side.

Practice Makes Perfect

Use the Distance, Rate, and Time Formula

In the following exercises, solve.

Steve drove for $8 \frac{1}{2}$ hours at 72 miles per hour. How much distance did he travel?

Socorro drove for $4 \frac{5}{6}$ hours at 60 miles per hour. How much distance did she travel?

Solution

290 miles

Yuki walked for $1 \frac{3}{4}$ hours at 4 miles per hour. How far did she walk?

Francie rode her bike for $2 \frac{1}{2}$ hours at 12 miles per hour. How far did she ride?

Solution

30 miles

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?

Megan is taking the bus from New York City to Montreal. The distance is 380 miles and the bus travels at a steady rate of 76 miles per hour. How long will the bus ride be?

Solution

5 hours

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?

Kareem wants to ride his bike from St. Louis 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?

Solution

11.25 hours

Javier is driving to Bangor, 240 miles away. If he needs to be in Bangor in 4 hours, at what rate does he need to drive?

Alejandra is driving to Cincinnati, 450 miles away. If she wants to be there in 6 hours, at what rate does she need to drive?

Solution

75 mph

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?

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?

Solution

75 mph

Solve a Formula for a Specific Variable

In the following exercises, use the formula $d = r t$ .

Solve for $t$ ⓐ when $d = 350$ and $r = 70$ ⓑ in general

Solve for $t$ ⓐ when $d = 240 and r = 60$ ⓑ in general

Solution

ⓐ $t = 4$ ⓑ $t = \frac{d}{r}$

Solve for $t$ ⓐ when $d = 510 and r = 60$ ⓑ in general

Solve for $t$
ⓐ when $d = 175 and r = 50$
ⓑ in general

Solution

ⓐ $t = 3.5$ ⓑ $t = \frac{d}{r}$

Solve for $r$
ⓐ when $d = 204 and t = 3$ ⓑ in general

Solve for $r$ ⓐ when $d = 420 and t = 6$ ⓑ in general

Solution

ⓐ $r = 70$ ⓑ $r = \frac{d}{t}$

Solve for $r$ ⓐ when $d = 160 and t = 2.5$ ⓑ in general

Solve for $r$ ⓐ when $d = 180 and t = 4.5$ ⓑ in general

Solution

ⓐ $r = 40$ ⓑ $r = \frac{d}{t}$

In the following exercises, use the formula $A = \frac{1}{2} b h$ .

Solve for $b$ ⓐ when $A = 126 and h = 18$ ⓑ in general

Solve for $h$
ⓐ when $A = 176 and b = 22$ ⓑ in general

Solution

ⓐ $h = 16$ ⓑ $h = \frac{2 A}{b}$

Solve for $h$ ⓐ when $A = 375 and b = 25$ ⓑ in general

Solve for $b$ ⓐ when $A = 65 and h = 13$ ⓑ in general

Solution

ⓐ $b = 10$ ⓑ $b = \frac{2 A}{h}$

In the following exercises, use the formula I = Prt.

Solve for the principal, P forⓐ $I = $ 5,480, r = 4 %,$ $t = 7 years$ ⓑ in general

Solve for the principal, P for
ⓐ $I = $ 3,950, r = 6 %,$ $t = 5 years$ ⓑ in general

Solution

ⓐ $P = $ 13, 166.67$ ⓑ $P = \frac{I}{r t}$

Solve for the time, t for ⓐ $I = $ 2,376, P = $ 9,000,$ $r = 4.4 %$ ⓑ in general

Solve for the time, t for
ⓐ $I = $ 624, P = $ 6,000,$ $r = 5.2 %$ ⓑ in general

Solution

ⓐ $t = 2$ years ⓑ $t = \frac{I}{\Pr}$

In the following exercises, solve.

Solve the formula $2 x + 3 y = 12$ for yⓐ when $x = 3$ ⓑ in general

Solve the formula $5 x + 2 y = 10$ for yⓐ when $x = 4$ ⓑ in general

Solution

ⓐ $y = −5$ ⓑ $y = \frac{10 - 5 x}{2}$

Solve the formula $3 x - y = 7$ for yⓐ when $x = −2$ ⓑ in general

Solve the formula $4 x + y = 5$ for yⓐ when $x = −3$ ⓑ in general

Solution

ⓐ $y = 17$ ⓑ $y = 5 - 4 x$

Solve $a + b = 90$ for $b$ .

Solve $a + b = 90$ for $a$ .

Solution

$a = 90 - b$

Solve $180 = a + b + c$ for $a$ .

Solve $180 = a + b + c$ for $c$ .

Solution

$c = 180 - a - b$

Solve the formula $8 x + y = 15$ for y.

Solve the formula $9 x + y = 13$ for y.

Solution

$y = 13 - 9 x$

Solve the formula $−4 x + y = −6$ for y.

Solve the formula $−5 x + y = −1$ for y.

Solution

$y = −1 + 5 x$

Solve the formula $4 x + 3 y = 7$ for y.

Solve the formula $3 x + 2 y = 11$ for y.

Solution

$y = \frac{11 - 3 x}{2}$

Solve the formula $x - y = −4$ for y.

Solve the formula $x - y = −3$ for y.

Solution

$y = 3 + x$

Solve the formula $P = 2 L + 2 W$ for $L$ .

Solve the formula $P = 2 L + 2 W$ for $W$ .

Solution

$W = \frac{P - 2 L}{2}$

Solve the formula $C = π d$ for $d$ .

Solve the formula $C = π d$ for $π$ .

Solution

$π = \frac{C}{d}$

Solve the formula $V = L W H$ for $L$ .

Solve the formula $V = L W H$ for $H$ .

Solution

$H = \frac{V}{L W}$

Everyday Math

Converting temperature While on a tour in Greece, Tatyana saw that the temperature was 40^o Celsius. Solve for F in the formula $C = \frac{5}{9} (F - 32)$ to find the Fahrenheit temperature.

Converting temperature Yon was visiting the United States and he saw that the temperature in Seattle one day was 50^o Fahrenheit. Solve for C in the formula $F = \frac{9}{5} C + 32$ to find the Celsius temperature.

Solution

10°C

Writing Exercises

Solve the equation $2 x + 3 y = 6$ for $y$ ⓐ when $x = −3$ ⓑ in general ⓒ Which solution is easier for you, ⓐ or ⓑ ? Why?

Solve the equation $5 x - 2 y = 10$ for $x$ ⓐ when $y = 10$ ⓑ in general
ⓒ Which solution is easier for you, ⓐ or ⓑ ? Why?

Solution

Answers will vary.

Self Check

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

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

Solve Linear Inequalities

Graph Inequalities on the Number Line

Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

What about the solution of an inequality? What number would make the inequality $x > 3$ true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality $x > 3$ .

We show the solutions to the inequality $x > 3$ on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of $x > 3$ is shown in Figure 1. Please note that the following convention is used: thick arrows point in the positive direction and thin arrows point in the negative direction.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 3 is graphed on the number line, with an open parenthesis at x equals 3, and a red line extending to the right of the parenthesis. — The inequality $x > 3$ is graphed on this number line.

The graph of the inequality $x \geq 3$ is very much like the graph of $x > 3$ , but now we need to show that 3 is a solution, too. We do that by putting a bracket at $x = 3$ , as shown in Figure 2.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than or equal to 3 is graphed on the number line, with an open bracket at x equals 3, and a red line extending to the right of the bracket. — The inequality $x \geq 3$ is graphed on this number line.

Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.

Graph on the number line:

ⓐ $x \leq 1$ ⓑ $x < 5$ ⓒ $x > - 1$

Solution

ⓐ $x \leq 1$
This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at $x = 1$ to show that it is included.
ⓑ $x < 5$
This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put a parenthesis at $x = 5$ to show it is not included.
ⓒ $x > - 1$
This means all numbers greater than $−1$ , but not including $−1$ . We shade in all the numbers on the number line to the right of $−1$ , then put a parenthesis at $x = −1$ to show it is not included.

We can also represent inequalities using interval notation. As we saw above, the inequality $x > 3$ means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation, we express $x > 3$ as $(3, \infty) .$ The symbol $\infty$ is read as ‘infinity’. It is not an actual number. Figure 3 shows both the number line and the interval notation.

The inequality $x \leq 1$ means all numbers less than or equal to 1. There is no lower end to those numbers. We write $x \leq 1$ in interval notation as $(− \infty, 1]$ . The symbol $− \infty$ is read as ‘negative infinity’. Figure 4 shows both the number line and interval notation.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to 1 is graphed on the number line, with an open bracket at x equals 1, and a red line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 1, bracket. — The inequality $x \leq 1$ is graphed on this number line and written in interval notation.

Did you notice how the parenthesis or bracket in the interval notation matches the symbol at the endpoint of the arrow? These relationships are shown in Figure 5.

This figure shows the same four number lines as above, with the same interval notation labels. Below the interval notation for each number line, there is text indicating how the notation on the number lines is similar to the interval notation. The first number line is a graph of x is greater than a, and the interval notation is parenthesis, a comma infinity, parenthesis. The text below reads: “Both have a left parenthesis.” The second number line is a graph of x is greater than or equal to a, and the interval notation is bracket, a comma infinity, parenthesis. The text below reads: “Both have a left bracket.” The third number line is a graph of x is less than a, and the interval notation is parenthesis, negative infinity comma a, parenthesis. The text below reads: “Both have a right parenthesis.” The last number line is a graph of x is less than or equal to a, and the interval notation is parenthesis, negative infinity comma a, bracket. The text below reads: “Both have a right bracket.” — The notation for inequalities on a number line and in interval notation use similar symbols to express the endpoints of intervals.

Graph on the number line and write in interval notation.

ⓐ $x \geq −3$ ⓑ $x < 2.5$ ⓒ $x \leq - \frac{3}{5}$

Solution

ⓐ

Shade to the right of $−3$ , and put a bracket at $−3$ .

Write in interval notation.
ⓑ

Shade to the left of $2.5$ , and put a parenthesis at $2.5$ .

Write in interval notation.
ⓒ

Shade to the left of $- \frac{3}{5}$ , and put a bracket at $- \frac{3}{5}$ .

Write in interval notation.

Solve Inequalities using the Subtraction and Addition Properties of Inequality

The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.

Similar properties hold true for inequalities.

For example, we know that −4 is less than 2.
If we subtract 5 from both quantities, is the left side still less than the right side?
We get −9 on the left and −3 on the right.
And we know −9 is less than −3.
	The inequality sign stayed the same.

Similarly we could show that the inequality also stays the same for addition.

This leads us to the Subtraction and Addition Properties of Inequality.

We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality $x + 5 > 9$ , the steps would look like this:

Steps to solve the inequality x + 5 > 9, demonstrating the process of isolating x.
	$x + 5 > 9$
Subtract 5 from both sides to isolate $x$ .	$x + 5 - 5 > 9 - 5$
Simplify.	$x > 4$

Any number greater than 4 is a solution to this inequality.

Solve the inequality $n - \frac{1}{2} \leq \frac{5}{8}$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Add $\frac{1}{2}$ to both sides of the inequality.
Simplify.	$A mathematical inequality shows 'n' is less than or equal to the fraction 9/8, presented in black font on a white background.$
Graph the solution on the number line.
Write the solution in interval notation.	$(- \infty, \frac{9}{8}]$

Solve Inequalities using the Division and Multiplication Properties of Inequality

The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).

Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?

Consider some numerical examples.


Divide both sides by 5.	$A math problem showing two fractions, 10/5 and 15/5, with a red question mark between them, implying a comparison or operation needs to be determined.$	Multiply both sides by 5.
Simplify.
Fill in the inequality signs.

The inequality signs stayed the same.

Does the inequality stay the same when we divide or multiply by a negative number?


Divide both sides by −5.	$A mathematical problem comparing two fractions, 10/-5 and 15/-5, with a red question mark positioned between them, indicating the need for a relational operator.$	Multiply both sides by −5.
Simplify.
Fill in the inequality signs.

The inequality signs reversed their direction.

When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.

Here are the Division and Multiplication Properties of Inequality for easy reference.

When we divide or multiply an inequality by a:

positive number, the inequality stays the same.
negative number, the inequality reverses.

Solve the inequality $7 y < 42$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Divide both sides of the inequality by 7. Since $7 > 0$ , the inequality stays the same.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $−10 a \geq 50$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Divide both sides of the inequality by −10. Since $−10 < 0$ , the inequality reverses.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $−20 < \frac{4}{5} u$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides of the inequality by $\frac{5}{4}$ . Since $\frac{5}{4} > 0$ , the inequality stays the same.
Simplify.
Rewrite the variable on the left.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $\frac{t}{−2} \geq 8$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides of the inequality by $−2$ . Since $−2 < 0$ , the inequality reverses.	$An algebraic inequality is displayed, showing -2 multiplied by the fraction t/-2, which is less than or equal to -2 multiplied by 8.$
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve Inequalities That Require Simplification

Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.

Solve the inequality $4 m \leq 9 m + 17$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Subtract $9 m$ from both sides to collect the variables on the left.
Simplify.
Divide both sides of the inequality by −5, and reverse the inequality.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $8 p + 3 (p - 12) > 7 p - 28$ , graph the solution on the number line, and write the solution in interval notation.

Solution

Simplify each side as much as possible.	$8 p + 3 (p - 12) > 7 p - 28$
Distribute.	$8 p + 3 p - 36 > 7 p - 28$
Combine like terms.	$11 p - 36 > 7 p - 28$
Subtract $7 p$ from both sides to collect the variables on the left.	$11 p - 36 - 7 p > 7 p - 28 - 7 p$
Simplify.	$4 p - 36 > −28$
Add 36 to both sides to collect the constants on the right.	$4 p - 36 + 36 > −28 + 36$
Simplify.	$4 p > 8$
Divide both sides of the inequality by 4; the inequality stays the same.	$\frac{4 p}{4} > \frac{8}{4}$
Simplify.	$p > 2$
Graph the solution on the number line.
Write the solution in interal notation.	$(2, ∞)$

Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.

Solve the inequality $8 x - 2 (5 - x) < 4 (x + 9) + 6 x$ , graph the solution on the number line, and write the solution in interval notation.

Solution

Simplify each side as much as possible.	$8 x - 2 (5 - x) < 4 (x + 9) + 6 x$
Distribute.	$8 x - 10 + 2 x < 4 x + 36 + 6 x$
Combine like terms.	$10 x - 10 < 10 x + 36$
Subtract $10 x$ from both sides to collect the variables on the left.	$10 x - 10 - 10 x < 10 x + 36 - 10 x$
Simplify.	$−10 < 36$
The $x$ ’s are gone, and we have a true statement.	The inequality is an identity. The solution is all real numbers.
Graph the solution on the number line.
Write the solution in interval notation.	$(−∞, ∞)$

Solve the inequality $\frac{1}{3} a - \frac{1}{8} a > \frac{5}{24} a + \frac{3}{4}$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides by the LCD, 24, to clear the fractions.
Simplify.
Combine like terms.
Subtract $5 a$ from both sides to collect the variables on the left.
Simplify.
The statement is false!	The inequality is a contradiction.
	There is no solution.
Graph the solution on the number line.
Write the solution in interval notation.	There is no solution.

Translate to an Inequality and Solve

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.

Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.

Table 17 shows some common phrases that indicate inequalities.

$>$	$\geq$	$<$	$\leq$
is greater than	is greater than or equal to	is less than	is less than or equal to
is more than	is at least	is smaller than	is at most
is larger than	is no less than	has fewer than	is no more than
exceeds	is the minimum	is lower than	is the maximum

Translate and solve. Then write the solution in interval notation and graph on the number line.

Twelve times c is no more than 96.

Solution

Translate.
Solve—divide both sides by 12.
Simplify.
Write in interval notation.
Graph on the number line.

Translate and solve. Then write the solution in interval notation and graph on the number line.

Thirty less than x is at least 45.

Solution

Translate.
Solve—add 30 to both sides.
Simplify.
Write in interval notation.
Graph on the number line.

Key Concepts

Subtraction Property of Inequality
For any numbers a, b, and c,
if $a < b$ then $a - c b$ then $a - c > b - c .$
Addition Property of Inequality
For any numbers a, b, and c,
if $a < b$ then $a + c b$ then $a + c > b + c .$
Division and Multiplication Properties of Inequality
For any numbers a, b, and c,
if $a < b$ and $c > 0$ , then $\frac{a}{c} < \frac{b}{c}$ and $a c > b c$ .
if $a > b$ and $c > 0$ , then $\frac{a}{c} > \frac{b}{c}$ and $a c > b c$ .
if $a < b$ and $c < 0$ , then $\frac{a}{c} > \frac{b}{c}$ and $a c > b c$ .
if $a > b$ and $c < 0$ , then $\frac{a}{c} < \frac{b}{c}$ and $a c < b c$ .
When we divide or multiply an inequality by a:
- positive number, the inequality stays the same.
- negative number, the inequality reverses.

Section Exercises

Practice Makes Perfect

Graph Inequalities on the Number Line

In the following exercises, graph each inequality on the number line.

ⓐ $x \leq - 2$
ⓑ $x > - 1$ ⓒ $x < 0$

ⓐ $x > 1$ ⓑ $x < - 2$
ⓒ $x \geq −3$

Solution

ⓐ $x \geq −3$ ⓑ $x < 4$ ⓒ $x \leq - 2$

ⓐ $x \leq 0$ ⓑ $x > - 4$ ⓒ $x \geq −1$

Solution

In the following exercises, graph each inequality on the number line and write in interval notation.

ⓐ $x < - 2$ ⓑ $x \geq −3.5$ ⓒ $x \leq \frac{2}{3}$

ⓐ $x > 3$ ⓑ $x \leq - 0.5$ ⓒ $x \geq \frac{1}{3}$

Solution

ⓐ $x \geq −4$ ⓑ $x < 2.5$ ⓒ $x > - \frac{3}{2}$

ⓐ $x \leq 5$ ⓑ $x \geq −1.5$ ⓒ $x < - \frac{7}{3}$

Solution

Solve Inequalities using the Subtraction and Addition Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$n - 11 < 33$

$m - 45 \leq 62$

Solution

$u + 25 > 21$

$v + 12 > 3$

Solution

$a + \frac{3}{4} \geq \frac{7}{10}$

$b + \frac{7}{8} \geq \frac{1}{6}$

Solution

$f - \frac{13}{20} < - \frac{5}{12}$

$g - \frac{11}{12} < - \frac{5}{18}$

Solution

Solve Inequalities using the Division and Multiplication Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$8 x > 72$

$6 y < 48$

Solution

$7 r \leq 56$

$9 s \geq 81$

Solution

$−5 u \geq 65$

$−8 v \leq 96$

Solution

$−9 c < 126$

$−7 d > 105$

Solution

$20 > \frac{2}{5} h$

$40 < \frac{5}{8} k$

Solution

$\frac{7}{6} j \geq 42$

$\frac{9}{4} g \leq 36$

Solution

$\frac{a}{−3} \leq 9$

$\frac{b}{−10} \geq 30$

Solution

$−25 < \frac{p}{−5}$

$−18 > \frac{q}{−6}$

Solution

$9 t \geq −27$

$7 s < - 28$

Solution

$\frac{2}{3} y > - 36$

$\frac{3}{5} x \leq - 45$

Solution

Solve Inequalities That Require Simplification

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$4 v \geq 9 v - 40$

$5 u \leq 8 u - 21$

Solution

$13 q < 7 q - 29$

$9 p > 14 p - 18$

Solution

$12 x + 3 (x + 7) > 10 x - 24$

$9 y + 5 (y + 3) < 4 y - 35$

Solution

$6 h - 4 (h - 1) \leq 7 h - 11$

$4 k - (k - 2) \geq 7 k - 26$

Solution

$8 m - 2 (14 - m) \geq 7 (m - 4) + 3 m$

$6 n - 12 (3 - n) \leq 9 (n - 4) + 9 n$

Solution

$\frac{3}{4} b - \frac{1}{3} b < \frac{5}{12} b - \frac{1}{2}$

$9 u + 5 (2 u - 5) \geq 12 (u - 1) + 7 u$

Solution

At the top of this figure is the result of the inequality: the inequality is a contradiction. Below this is a number line ranging from negative 2 to 2 with tick marks for each integer. Because this is a contradiction, no inequality is graphed on the number line. Below the number line is the statement: “No solution”.

$\frac{2}{3} g - \frac{1}{2} (g - 14) \leq \frac{1}{6} (g + 42)$

$\frac{5}{6} a - \frac{1}{4} a > \frac{7}{12} a + \frac{2}{3}$

Solution

$\frac{4}{5} h - \frac{2}{3} (h - 9) \geq \frac{1}{15} (2 h + 90)$

$12 v + 3 (4 v - 1) \leq 19 (v - 2) + 5 v$

Solution

Mixed practice

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$15 k \leq - 40$

$35 k \geq −77$

Solution

$23 p - 2 (6 - 5 p) > 3 (11 p - 4)$

$18 q - 4 (10 - 3 q) < 5 (6 q - 8)$

Solution

$- \frac{9}{4} x \geq - \frac{5}{12}$

$- \frac{21}{8} y \leq - \frac{15}{28}$

Solution

$c + 34 < - 99$

$d + 29 > - 61$

Solution

$\frac{m}{18} \geq −4$

$\frac{n}{13} \leq - 6$

Solution

Translate to an Inequality and Solve

In the following exercises, translate and solve .Then write the solution in interval notation and graph on the number line.

Fourteen times d is greater than 56.

Ninety times c is less than 450.

Solution

Eight times z is smaller than $−40$ .

Ten times y is at most $−110$ .

Solution

Three more than h is no less than 25.

Six more than k exceeds 25.

Solution

Ten less than w is at least 39.

Twelve less than x is no less than 21.

Solution

Negative five times r is no more than 95.

Negative two times s is lower than 56.

Solution

Nineteen less than b is at most $−22$ .

Fifteen less than a is at least $−7$ .

Solution

Everyday Math

Safety A child’s height, h, must be at least 57 inches for the child to safely ride in the front seat of a car. Write this as an inequality.

Fighter pilots The maximum height, h, of a fighter pilot is 77 inches. Write this as an inequality.

Solution

$h \leq 77$

Elevators The total weight, w, of an elevator’s passengers can be no more than 1,200 pounds. Write this as an inequality.

Shopping The number of items, n, a shopper can have in the express check-out lane is at most 8. Write this as an inequality.

Solution

$n \leq 8$

Writing Exercises

Give an example from your life using the phrase ‘at least’.

Give an example from your life using the phrase ‘at most’.

Solution

Answers will vary.

Explain why it is necessary to reverse the inequality when solving $−5 x > 10$ .

Explain why it is necessary to reverse the inequality when solving $\frac{n}{−3} < 12$ .

Solution

Answers will vary.

Self Check

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

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

Chapter 2 Review Exercises

Solve Equations using the Subtraction and Addition Properties of Equality

Verify a Solution of an Equation

In the following exercises, determine whether each number is a solution to the equation.

$10 x - 1 = 5 x; x = \frac{1}{5}$

$w + 2 = \frac{5}{8}; w = \frac{3}{8}$

Solution

$−12 n + 5 = 8 n; n = - \frac{5}{4}$

$6 a - 3 = −7 a, a = \frac{3}{13}$

Solution

yes

Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, solve each equation using the Subtraction Property of Equality.

$x + 7 = 19$

$y + 2 = −6$

Solution

$y = −8$

$a + \frac{1}{3} = \frac{5}{3}$

$n + 3.6 = 5.1$

Solution

$n = 1.5$

In the following exercises, solve each equation using the Addition Property of Equality.

$u - 7 = 10$

$x - 9 = −4$

Solution

$x = 5$

$c - \frac{3}{11} = \frac{9}{11}$

$p - 4.8 = 14$

Solution

$p = 18.8$

In the following exercises, solve each equation.

$n - 12 = 32$

$y + 16 = −9$

Solution

$y = −25$

$f + \frac{2}{3} = 4$

$d - 3.9 = 8.2$

Solution

$d = 12.1$

Solve Equations That Require Simplification

In the following exercises, solve each equation.

$y + 8 - 15 = −3$

$7 x + 10 - 6 x + 3 = 5$

Solution

$x = −8$

$6 (n - 1) - 5 n = −14$

$8 (3 p + 5) - 23 (p - 1) = 35$

Solution

$p = −28$

Translate to an Equation and Solve

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

The sum of $−6$ and $m$ is 25.

Four less than $n$ is 13.

Solution

$n - 4 = 13; n = 17$

Translate and Solve Applications

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?

Tan weighs 146 pounds. Minh weighs 15 pounds more than Tan. How much does Minh weigh?

Solution

161 pounds

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?

Elissa earned $152.84 this week, which was $21.65 more than she earned last week. How much did she earn last week?

Solution

$131.19

Solve Equations using the Division and Multiplication Properties of Equality

Solve Equations Using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation using the division and multiplication properties of equality and check the solution.

$8 x = 72$

$13 a = −65$

Solution

$a = −5$

$0.25 p = 5.25$

$− y = 4$

Solution

$y = −4$

$\frac{n}{6} = 18$

$\frac{y}{−10} = 30$

Solution

$y = −300$

$36 = \frac{3}{4} x$

$\frac{5}{8} u = \frac{15}{16}$

Solution

$u = \frac{3}{2}$

$−18 m = −72$

$\frac{c}{9} = 36$

Solution

$c = 324$

$0.45 x = 6.75$

$\frac{11}{12} = \frac{2}{3} y$

Solution

$y = \frac{11}{8}$

Solve Equations That Require Simplification

In the following exercises, solve each equation requiring simplification.

$5 r - 3 r + 9 r = 35 - 2$

$24 x + 8 x - 11 x = −7 - 14$

Solution

$x = −1$

$\frac{11}{12} n - \frac{5}{6} n = 9 - 5$

$−9 (d - 2) - 15 = −24$

Solution

$d = 3$

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve.

143 is the product of $−11$ and y.

The quotient of b and 9 is $−27$ .

Solution

$\frac{b}{9} = −27; b = −243$

The sum of q and one-fourth is one.

The difference of s and one-twelfth is one fourth.

Solution

$s - \frac{1}{12} = \frac{1}{4}; s = \frac{1}{3}$

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Ray paid $21 for 12 tickets at the county fair. What was the price of each ticket?

Janet gets paid $24 per hour. She heard that this is $\frac{3}{4}$ of what Adam is paid. How much is Adam paid per hour?

Solution

$32

Solve Equations with Variables and Constants on Both Sides

Solve an Equation with Constants on Both Sides

In the following exercises, solve the following equations with constants on both sides.

$8 p + 7 = 47$

$10 w - 5 = 65$

Solution

$w = 7$

$3 x + 19 = −47$

$32 = −4 - 9 n$

Solution

$n = −4$

Solve an Equation with Variables on Both Sides

In the following exercises, solve the following equations with variables on both sides.

$7 y = 6 y - 13$

$5 a + 21 = 2 a$

Solution

$a = −7$

$k = −6 k - 35$

$4 x - \frac{3}{8} = 3 x$

Solution

$x = \frac{3}{8}$

Solve an Equation with Variables and Constants on Both Sides

In the following exercises, solve the following equations with variables and constants on both sides.

$12 x - 9 = 3 x + 45$

$5 n - 20 = −7 n - 80$

Solution

$n = −5$

$4 u + 16 = −19 - u$

$\frac{5}{8} c - 4 = \frac{3}{8} c + 4$

Solution

$c = 32$

Use a General Strategy for Solving Linear Equations

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

$6 (x + 6) = 24$

$9 (2 p - 5) = 72$

Solution

$p = \frac{13}{2}$

$− (s + 4) = 18$

$8 + 3 (n - 9) = 17$

Solution

$n = 12$

$23 - 3 (y - 7) = 8$

$\frac{1}{3} (6 m + 21) = m - 7$

Solution

$m = −14$

$4 (3.5 y + 0.25) = 365$

$0.25 (q - 8) = 0.1 (q + 7)$

Solution

$q = 18$

$8 (r - 2) = 6 (r + 10)$

$5 + 7 (2 - 5 x) = 2 (9 x + 1)$
$- (13 x - 57)$

Solution

$x = −1$

$(9 n + 5) - (3 n - 7)$
$= 20 - (4 n - 2)$

$2 [−16 + 5 (8 k - 6)]$
$= 8 (3 - 4 k) - 32$

Solution

$k = \frac{3}{4}$

Classify Equations

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

$17 y - 3 (4 - 2 y) = 11 (y - 1)$
$+ 12 y - 1$

$9 u + 32 = 15 (u - 4)$
$- 3 (2 u + 21)$

Solution

contradiction; no solution

$−8 (7 m + 4) = −6 (8 m + 9)$

$21 (c - 1) - 19 (c + 1)$
$= 2 (c - 20)$

Solution

identity; all real numbers

Solve Equations with Fractions and Decimals

Solve Equations with Fraction Coefficients

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

$\frac{2}{5} n - \frac{1}{10} = \frac{7}{10}$

$\frac{1}{3} x + \frac{1}{5} x = 8$

Solution

$x = 15$

$\frac{3}{4} a - \frac{1}{3} = \frac{1}{2} a - \frac{5}{6}$

$\frac{1}{2} (k - 3) = \frac{1}{3} (k + 16)$

Solution

$k = 41$

$\frac{3 x - 2}{5} = \frac{3 x + 4}{8}$

$\frac{5 y - 1}{3} + 4 = \frac{−8 y + 4}{6}$

Solution

$y = −1$

Solve Equations with Decimal Coefficients

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

$0.8 x - 0.3 = 0.7 x + 0.2$

$0.36 u + 2.55 = 0.41 u + 6.8$

Solution

$u = −85$

$0.6 p - 1.9 = 0.78 p + 1.7$

$0.7 y + 2.5 = 0.95 y - 9.25$

Solution

$y = 47$

Solve a Formula for a Specific Variable

Use the Distance, Rate, and Time Formula

In the following exercises, solve.

Natalie drove for $7 \frac{1}{2}$ hours at 60 miles per hour. How much distance did she travel?

Mallory is taking the bus from St. Louis to Chicago. The distance is 300 miles and the bus travels at a steady rate of 60 miles per hour. How long will the bus ride be?

Solution

5 hours

Aaron’s friend drove him from Buffalo to Cleveland. The distance is 187 miles and the trip took 2.75 hours. How fast was Aaron’s friend driving?

Link rode his bike at a steady rate of 15 miles per hour for $2 \frac{1}{2}$ hours. How much distance did he travel?

Solution

37.5 miles

Solve a Formula for a Specific Variable

In the following exercises, solve.

Use the formula. $d = r t$ to solve for t
ⓐ when $d = 510$ and $r = 60$
ⓑ in general

Use the formula. $d = r t$ to solve for $r$
ⓐ when when $d = 451$ and $t = 5.5$
ⓑ in general

Solution

ⓐ $r = 82 mph$ ; ⓑ $r = \frac{D}{t}$

Use the formula $A = \frac{1}{2} b h$ to solve for $b$
ⓐ when $A = 390$ and $h = 26$
ⓑ in general

Use the formula $A = \frac{1}{2} b h$ to solve for $h$
ⓐ when $A = 153$ and $b = 18$
ⓑ in general

Solution

ⓐ $h = 17$ ⓑ $h = \frac{2 A}{b}$

Use the formula $I = P r t$ to solve for the principal, P for
ⓐ $I = $ 2, 501, r = 4.1 %,$
$t = 5 years$
ⓑ in general

Solve the formula $4 x + 3 y = 6$ for y
ⓐ when $x = −2$
ⓑ in general

Solution

ⓐ $y = \frac{14}{3}$ ⓑ $y = \frac{6 - 4 x}{3}$

Solve $180 = a + b + c$ for $c$ .

Solve the formula $V = L W H$ for $H$ .

Solution

$H = \frac{V}{L W}$

Solve Linear Inequalities

Graph Inequalities on the Number Line

In the following exercises, graph each inequality on the number line.

ⓐ $x \leq 4$
ⓑ $x > - 2$
ⓒ $x < 1$

ⓐ $x > 0$
ⓑ $x < - 3$
ⓒ $x \geq −1$

Solution

In the following exercises, graph each inequality on the number line and write in interval notation.

ⓐ $x < - 1$
ⓑ $x \geq −2.5$
ⓒ $x \leq \frac{5}{4}$

ⓐ $x > 2$
ⓑ $x \leq - 1.5$
ⓒ $x \geq \frac{5}{3}$

Solution

Solve Inequalities using the Subtraction and Addition Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$n - 12 \leq 23$

$m + 14 \leq 56$

Solution

$a + \frac{2}{3} \geq \frac{7}{12}$

$b - \frac{7}{8} \geq - \frac{1}{2}$

Solution

Solve Inequalities using the Division and Multiplication Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$9 x > 54$

$−12 d \leq 108$

Solution

$\frac{5}{6} j < - 60$

$\frac{q}{- 2} \geq - 24$

Solution

Solve Inequalities That Require Simplification

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$6 p > 15 p - 30$

$9 h - 7 (h - 1) \leq 4 h - 23$

Solution

$5 n - 15 (4 - n) < 10 (n - 6) + 10 n$

$\frac{3}{8} a - \frac{1}{12} a > \frac{5}{12} a + \frac{3}{4}$

Solution

Translate to an Inequality and Solve

In the following exercises, translate and solve. Then write the solution in interval notation and graph on the number line.

Five more than z is at most 19.

Three less than c is at least 360.

Solution

Nine times n exceeds 42.

Negative two times a is no more than 8.

Solution

Everyday Math

Describe how you have used two topics from this chapter in your life outside of your math class during the past month.

Chapter 2 Practice Test

Determine whether each number is a solution to the equation $6 x - 3 = x + 20$ .

ⓐ 5ⓑ $\frac{23}{5}$

Solution

ⓐ no ⓑ yes

In the following exercises, solve each equation.

$n - \frac{2}{3} = \frac{1}{4}$

$\frac{9}{2} c = 144$

Solution

$c = 32$

$4 y - 8 = 16$

$−8 x - 15 + 9 x - 1 = −21$

Solution

$x = −5$

$−15 a = 120$

$\frac{2}{3} x = 6$

Solution

$x = 9$

$x - 3.8 = 8.2$

$10 y = −5 y - 60$

Solution

$y = −4$

$8 n - 2 = 6 n - 12$

$9 m - 2 - 4 m - m = 42 - 8$

Solution

$m = 9$

$−5 (2 x - 1) = 45$

$− (d - 9) = 23$

Solution

$d = −14$

$\frac{1}{4} (12 m - 28) = 6 - 2 (3 m - 1)$

$2 (6 x - 5) - 8 = −22$

Solution

$x = - \frac{1}{3}$

$8 (3 a - 5) - 7 (4 a - 3) = 20 - 3 a$

$\frac{1}{4} p - \frac{1}{3} = \frac{1}{2}$

Solution

$p = \frac{10}{3}$

$0.1 d + 0.25 (d + 8) = 4.1$

$14 n - 3 (4 n + 5) = −9 + 2 (n - 8)$

Solution

contradiction; no solution

$9 (3 u - 2) - 4 [6 - 8 (u - 1)]$ $= 3 (u - 2)$

Solve the formula $x - 2 y = 5$ for y
ⓐ when $x = −3$
ⓑ in general

Solution

ⓐ $y = 4$ ⓑ $y = \frac{5 - x}{2}$

In the following exercises, graph on the number line and write in interval notation.

$x \geq −3.5$

$x < \frac{11}{4}$

Solution

This figure is a number line ranging from 1 to 5 with tick marks for each integer. The inequality x is less than 11/4 is graphed on the number line, with an open parenthesis at x equals 11/4, and a dark line extending to the left of the parenthesis. Below the number line is the solution written in interval notation: parenthesis, negative infinity comma 11/4, parenthesis.

In the following exercises,, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$8 k \geq 5 k - 120$

$3 c - 10 (c - 2) < 5 c + 16$

Solution

In the following exercises, translate to an equation or inequality and solve.

4 less than twice x is 16.

Fifteen more than n is at least 48.

Solution

$n + 15 \geq 48; n \geq 33$

Samuel paid $25.82 for gas this week, which was $3.47 less than he paid last week. How much had he paid last week?

Jenna bought a coat on sale for $120, which was $\frac{2}{3}$ of the original price. What was the original price of the coat?

Solution

$120 = \frac{2}{3} p$ ; The original price was $180.

Sean took the bus from Seattle to Boise, a distance of 506 miles. If the trip took $7 \frac{2}{3}$ hours, what was the speed of the bus?

Introduction

A photo of cars on the highway. — Sophisticated mathematical models are used to predict traffic patterns on our nation’s highways.

Mathematical formulas model phenomena in every facet of our lives. They are used to explain events and predict outcomes in fields such as transportation, business, economics, medicine, chemistry, engineering, and many more. In this chapter, we will apply our skills in solving equations to solve problems in a variety of situations.

Use a Problem-Solving Strategy

Approach Word Problems with a Positive Attitude

“If you think you can… or think you can’t… you’re right.”—Henry Ford

The world is full of word problems! Will my income qualify me to rent that apartment? How much punch do I need to make for the party? What size diamond can I afford to buy my girlfriend? Should I fly or drive to my family reunion?

How much money do I need to fill the car with gas? How much tip should I leave at a restaurant? How many socks should I pack for vacation? What size turkey do I need to buy for Thanksgiving dinner, and then what time do I need to put it in the oven? If my sister and I buy our mother a present, how much does 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 below?

A student is shown with thought bubbles saying “I don’t know whether to add, subtract, multiply, or divide!,” “I don’t understand word problems!,” “My teachers never explained this!,” “If I just skip all the word problems, I can probably still pass the class,” and “I just can’t do this!” — Negative thoughts 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. If we take control and believe we can be successful, we will be able to master word problems! Read the positive thoughts in Figure 2 and say them out loud.

A student is shown with thought bubbles saying “While word problems were hard in the past, I think I can try them now,” “I am better prepared now. I think I will begin to understand word problems,” “I think I can! I think I can!,” and “It may take time, but I can begin to solve word problems.” — Thinking positive thoughts is a first step towards success.

Think of something, outside of school, that you can do now but couldn’t do 3 years ago. Is it driving a car? Snowboarding? Cooking a gourmet meal? Speaking a new language? Your past experiences with word problems happened when you were younger—now you’re older and ready to succeed!

Use a Problem-Solving Strategy for Word Problems

We have reviewed translating English phrases into algebraic expressions, using some basic mathematical vocabulary and symbols. We have also translated English sentences into algebraic equations and solved some word problems. The word problems applied math to everyday situations. We restated the situation in one sentence, assigned a variable, and then wrote an equation to solve the problem. This method works as long as the situation is familiar and the math is not too complicated.

Now, we’ll expand our strategy so we can use it to successfully solve any word problem. We’ll list the strategy here, and then we’ll use it to solve some problems. We summarize below an effective strategy for problem solving.

Pilar bought a purse on sale for $18, which is one-half of the original price. What was the original price of the purse?

Solution

Step 1. Read the problem. Read the problem two or more times if necessary. Look up any unfamiliar words in a dictionary or on the internet.

In this problem, is it clear what is being discussed? Is every word familiar?

Step 2. Identify what you are looking for. Did you ever go into your bedroom to get something and then forget what you were 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 purse” tell us what we need to find.

Step 3. Name what we are looking for. Choose a variable to represent that quantity. We can use any letter for the variable, but choose one that makes it easy to remember what it represents.

Let $p =$ the original price of the purse.

Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Translate the English sentence into an algebraic equation.

Reread the problem carefully to see how the given information is related. Often, there is one sentence that gives this information, or it may help to write one sentence with all the important information. Look for clue words to help translate the sentence into algebra. Translate the sentence into an equation.

Restate the problem in one sentence with all the important information.
Translate into an equation.

Step 5. Solve the equation using good algebraic techniques. Even if you know the solution right away, using good algebraic techniques here will better prepare you to solve problems that do not have obvious answers.

Solve the equation.
Multiply both sides by 2.
Simplify.

Step 6. Check the answer in the problem to make sure it makes sense. We solved the equation and 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 purse 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 purse?”

The answer to the question is: “The original price of the purse was $36.”

If this were a homework exercise, our work might look like this:

Pilar bought a purse on sale for $18, which is one-half the original price. What was the original price of the purse?

	Let $p =$ the original price.
	18 is one-half the original price.

Multiply both sides by 2.
Simplify.
Check. Is $36 a reasonable price for a purse?
Yes.
Is 18 one half of 36?
$18 \overset{?}{=} \frac{1}{2} \cdot 36$
$18 = 18 ✓$
	The original price of the purse was $36.

Let’s try this approach with another example.

Ginny and her classmates formed a study group. The number of girls in the study group was three more than twice the number of boys. There were 11 girls in the study group. How many boys were in the study group?

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	How many boys were in the study group?
Step 3. Name. Choose a variable to represent the number of boys.	Let $n =$ the number of boys.
Step 4. Translate. Restate the problem in one sentence with all the important information.
Translate into an equation.
Step 5. Solve the equation.
Subtract 3 from each side.
Simplify.
Divide each side by 2.
Simplify.
Step 6. Check. First, is our answer reasonable? Yes, having 4 boys in a study group seems OK. The problem says the number of girls was 3 more than twice the number of boys. If there are four boys, does that make eleven girls? Twice 4 boys is 8. Three more than 8 is 11.
Step 7. Answer the question.	There were 4 boys in the study group.

Solve Number Problems

Now that we have a problem solving strategy, we will use it on several different types of word problems. The first type we will work on is “number problems.” Number problems give some clues about one or more numbers. We use these clues to write an equation. Number problems don’t usually arise on an everyday basis, but they provide a good introduction to practicing the problem solving strategy outlined above.

The difference of a number and six is 13. Find the number.

Solution

Step 1. Read the problem. Are all the words familiar?
Step 2. Identify what we are looking for.	the number
Step 3. Name. Choose a variable to represent the number.	Let $n =$ the number.
Step 4. Translate. Remember to look for clue words like "difference... of... and..."
Restate the problem as one sentence.
Translate into an equation.
Step 5. Solve the equation.
Simplify.
Step 6. Check.
The difference of 19 and 6 is 13. It checks!
Step 7. Answer the question.	The number is 19.

The sum of twice a number and seven is 15. Find the number.

Solution

Step 1. Read the problem.
Step 2. Identify what we 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.
Step 5. Solve the equation.
Subtract 7 from each side and simplify.
Divide each side by 2 and simplify.
Step 6. Check.
Is the sum of twice 4 and 7 equal to 15?
$\begin{matrix} 2 \cdot 4 + 7 & ≟ & 15 \\ 15 & = & 15 ✓ \end{matrix}$
Step 7. Answer the question.	The number is 4.

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

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

One number is five more than another. The sum of the numbers is 21. Find the numbers.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		We are looking for two numbers.
Step 3. Name. We have two numbers to name and need a name for each.
Choose a variable to represent the first number.		Let $n = 1^{st}$ number.
What do we know about the second number?		One number is five more than another.
		$n + 5 = 2^{nd}$ number
Step 4. Translate. Restate the problem as one sentence with all the important information.		The sum of the 1^st number and the 2^nd number is 21.
Translate into an equation.
Substitute the variable expressions.
Step 5. Solve the equation.
Combine like terms.
Subtract 5 from both sides and simplify.
Divide by 2 and simplify.
Find the second number, too.


Step 6. Check.
Do these numbers check in the problem?
Is one number 5 more than the other?	$13 ≟ 8 + 5$
Is thirteen 5 more than 8? Yes.	$13 = 13 ✓$
Is the sum of the two numbers 21?	$8 + 13 ≟ 21$
	$21 = 21 ✓$
Step 7. Answer the question.		The numbers are 8 and 13.

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		We are looking for two numbers.
Step 3. Name.
Choose a variable.		Let $n = 1^{st}$ number.
One number is 4 less than the other.		$n - 4 = 2^{nd}$ number
Step 4. Translate.
Write as one sentence.		The sum of the 2 numbers is negative 14.
Translate into an equation.
Step 5. Solve the equation.
Combine like terms.
Add 4 to each side and simplify.
Simplify.




Step 6. Check.
Is −9 four less than −5?	$−5 - 4 ≟ −9$
	$−9 = −9 ✓$
Is their sum −14?	$−5 + (−9) ≟ −14$
	$−14 = −14 ✓$
Step 7. Answer the question.		The numbers are −5 and −9.

One number is ten more than twice another. Their sum is one. Find the numbers.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.		We are looking for two numbers.
Step 3. Name.
Choose a variable.		Let $x = 1^{st}$ number.
One number is 10 more than twice another.		$2 x + 10 = 2^{nd}$ number
Step 4. Translate.
Restate as one sentence.		Their sum is one.
		The sum of the two numbers is 1.
Translate into an equation.
Step 5. Solve the equation.
Combine like terms.
Subtract 10 from each side.
Divide each side by 3.




Step 6. Check.
Is ten more than twice −3 equal to 4?	$2 (−3) + 10 ≟ 4$
	$−6 + 10 ≟ 4$
	$4 = 4 ✓$
Is their sum 1?	$−3 + 4 ≟ 1$
	$1 = 1 ✓$
Step 7. Answer the question.		The numbers are −3 and 4.

Some number problems involve consecutive integers. Consecutive integers are integers that immediately follow each other. Examples of consecutive integers are:

\begin{matrix} 1, 2, 3, 4 \\ - 10, −9, −8, −7 \\ 150, 151, 152, 153 \end{matrix}

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 $n + 1 + 1,$ which is $n + 2 .$

\begin{matrix} n & 1^{st} integer \\ n + 1 & 2^{nd} consecutive integer \\ n + 2 & 3^{rd} consecutive integer . . . etc. \end{matrix}

The sum of two consecutive integers is 47. Find the numbers.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.		two consecutive integers
Step 3. Name each number.		Let $n = 1^{st}$ integer.
		$n + 1 =$ next consecutive integer
Step 4. Translate.
Restate as one sentence.		The sum of the integers is 47.
Translate into an equation.
Step 5. Solve the equation.
Combine like terms.
Subtract 1 from each side.
Divide each side by 2.



Step 6. Check.
$\begin{matrix} 23 + 24 & ≟ & 47 \\ 47 & = & 47 ✓ \end{matrix}$
Step 7. Answer the question.		The two consecutive integers are 23 and 24.

Find three consecutive integers whose sum is $−42 .$

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		three consecutive integers
Step 3. Name each of the three numbers.		Let $n = 1^{st}$ integer.
		$n + 1 =$ 2^nd consecutive integer
		$n + 2 =$ 3^rd consecutive integer
Step 4. Translate.
Restate as one sentence.		The sum of the three integers is −42.
Translate into an equation.
Step 5. Solve the equation.
Combine like terms.
Subtract 3 from each side.
Divide each side by 3.


		$The image displays the fraction '-1/4' in black text on a plain white background, appearing as if written or printed directly on the surface.$



Step 6. Check.
$\begin{matrix} −13 + (−14) + (−15) & ≟ & −42 \\ −42 & = & −42 ✓ \end{matrix}$
Step 7. Answer the question.		The three consecutive integers are −13, −14, and −15.

Now that we have worked with consecutive integers, we will expand our work to include consecutive even integers and consecutive odd integers. Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

\begin{matrix} 18, 20, 22 \\ 64, 66, 68 \\ −12, −10, −8 \end{matrix}

Notice each integer is 2 more than the number preceding it. If we call the first one n, then the next one is $n + 2 .$ The next one would be $n + 2 + 2$ or $n + 4 .$

\begin{matrix} n & 1^{st} even integer \\ n + 2 & 2^{nd} consecutive even integer \\ n + 4 & 3^{rd} consecutive even integer . . . etc. \end{matrix}

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers 77, 79, and 81.

\begin{matrix} 77, 79, 81 \\ n, n + 2, n + 4 \end{matrix}

\begin{matrix} n & 1^{st} odd integer \\ n + 2 & 2^{nd} consecutive odd integer \\ n + 4 & 3^{rd} consecutive odd integer . . . etc. \end{matrix}

Does it seem strange to add 2 (an even number) to get from one odd integer to the next? Do you get an odd number or an even number when we add 2 to 3? to 11? to 47?

Whether the problem asks for consecutive even numbers or odd numbers, you don’t have to do anything different. The pattern is still the same—to get from one odd or one even integer to the next, add 2.

Find three consecutive even integers whose sum is 84.

Solution

Step-by-step solution for finding three consecutive even integers whose sum is 84, illustrating the process from problem understanding to final answer.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	three consecutive even integers
Step 3. Name the integers.	Let $n = 1^{st}$ even integer. $n + 2 = 2^{nd}$ consecutive even integer $n + 4 = 3^{rd}$ consecutive even integer
Step 4. Translate.
Restate as one sentence.	The sume of the three even integers is 84.
Translate into an equation.	$n + n + 2 + n + 4 = 84$
Step 5. Solve the equation.
Combine like terms.	$n + n + 2 + n + 4 = 84$
Subtract 6 from each side.	$3 n + 6 = 84$
Divide each side by 3.	$3 n = 78$
	$\begin{matrix} n = 26 & 1^{st} integer \\ n + 2 & 2^{nd} integer \\ 26 + 2 \\ 28 \\ n + 4 & 3^{rd} integer \\ 26 + 4 \\ 30 \end{matrix}$
Step 6. Check.
$\begin{matrix} 26 + 28 + 30 \overset{?}{=} 84 \\ 84 = 84 ✓ \end{matrix}$
Step 7. Answer the question.	The three consecutive integers are 26, 28, and 30.

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		How much does the husband earn?
Step 3. Name.
Choose a variable to represent the amount the husband earns.		Let $h =$ the amount the husband earns.
The wife earns $16,000 less than twice that.		$2 h - 16,000$ the amount the wife earns.
Step 4. Translate.		Together the husband and wife earn $110,000.
Restate the problem in one sentence with all the important information.
Translate into an equation.
Step 5. Solve the equation.		h + 2h − 16,000 = 110,000
Combine like terms.		3h − 16,000 = 110,000
Add 16,000 to both sides and simplify.		3h = 126,000
Divide each side by 3.		h = 42,000
		$42,000 amount husband earns
		2h − 16,000 amount wife earns
		2(42,000) − 16,000
		84,000 − 16,000
		68,000
Step 6. Check.
If the wife earns $68,000 and the husband earns $42,000 is the total $110,000? Yes!
Step 7. Answer the question.		The husband earns $42,000 a year.

Key Concepts

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

$\begin{matrix} n & 1^{st} integer \\ n + 1 & 2^{nd} integer consecutive integer \\ n + 2 & 3^{rd} consecutive integer . . . etc. \end{matrix}$

Consecutive even integers are even integers that immediately follow one another.

$\begin{matrix} n & 1^{st} integer \\ n + 2 & 2^{nd} integer consecutive integer \\ n + 4 & 3^{rd} consecutive integer . . . etc. \end{matrix}$

Consecutive odd integers are odd integers that immediately follow one another.

$\begin{matrix} n & 1^{st} integer \\ n + 2 & 2^{nd} integer consecutive integer \\ n + 4 & 3^{rd} consecutive integer . . . etc. \end{matrix}$

Practice Makes Perfect

Use the Approach Word Problems with a Positive Attitude

In the following exercises, prepare the lists described.

List five positive thoughts you can say to yourself that will help you approach word problems with a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often.

Solution

Answers will vary

List five negative thoughts that you have said to yourself in the past that will hinder your progress on word problems. You may want to write each one on a small piece of paper and rip it up to symbolically destroy the negative thoughts.

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.

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

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

125

One-fourth of the candies in a bag of M&M’s 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 four more than twice the number of boys. Find the number of boys.

Solution

There are 18 Cub Scouts in Pack 645. The number of scouts is three more than five times the number of adult leaders. Find the number of adult leaders.

Huong is organizing paperback and hardback books for her club’s used book sale. The number of paperbacks is 12 less than three times the number of hardbacks. Huong had 162 paperbacks. How many hardback books were there?

Solution

Jeff is lining up children’s and adult bicycles at the bike shop where he works. The number of children’s bicycles is nine less than three times the number of adult bicycles. There are 42 adult bicycles. How many children’s bicycles are there?

Philip pays $1,620 in rent every month. This amount is $120 more than twice what his brother Paul pays for rent. How much does Paul pay for rent?

Solution

$750

Marc just bought an SUV for $54,000. This is $7,400 less than twice what his wife paid for her car last year. How much did his wife pay for her car?

Laurie has $46,000 invested in stocks and bonds. The amount invested in stocks is $8,000 less than three times the amount invested in bonds. How much does Laurie have invested in bonds?

Solution

$13,500

Erica earned a total of $50,450 last year from her two jobs. The amount she earned from her job at the store was $1,250 more than three times the amount she earned from her job at the college. How much did she earn from her job at the college?

Solve Number Problems

In the following exercises, solve each number word problem.

The sum of a number and eight is 12. Find the number.

Solution

The sum of a number and nine is 17. Find the number.

The difference of a number and 12 is three. Find the number.

Solution

The difference of a number and eight is four. Find the number.

The sum of three times a number and eight is 23. Find the number.

Solution

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

The difference of four times a number and seven is 21. Find the number.

Three times the sum of a number and nine is 12. Find the number.

Solution

$−5$

Six times the sum of a number and eight is 30. Find the number.

One number is six more than the other. Their sum is 42. Find the numbers.

Solution

18, 24

One number is five more than the other. Their sum is 33. Find the numbers.

The sum of two numbers is 20. One number is four less than the other. Find the numbers.

Solution

8, 12

The sum of two numbers is 27. One number is seven less than the other. Find the numbers.

The sum of two numbers is $−45 .$ One number is nine more than the other. Find the numbers.

Solution

$−18, −27$

The sum of two numbers is $−61 .$ One number is 35 more than the other. Find the numbers.

The sum of two numbers is $−316 .$ One number is 94 less than the other. Find the numbers.

Solution

$−111, −205$

The sum of two numbers is $−284 .$ One number is 62 less than the other. Find the numbers.

One number is 14 less than another. If their sum is increased by seven, the result is 85. Find the numbers.

Solution

32, 46

One number is 11 less than another. If their sum is increased by eight, the result is 71. Find the numbers.

One number is five more than another. If their sum is increased by nine, the result is 60. Find the numbers.

Solution

23, 28

One number is eight more than another. If their sum is increased by 17, the result is 95. Find the numbers.

One number is one more than twice another. Their sum is $−5 .$ 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 14. One number is two less than three times the other. Find the numbers.

Solution

4, 10

The sum of two numbers is zero. One number is nine less than twice the other. Find the numbers.

The sum of two consecutive integers is 77. Find the integers.

Solution

38, 39

The sum of two consecutive integers is 89. Find the integers.

The sum of 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 $−3 .$

Find three consecutive even integers whose sum is 258.

Solution

84, 86, 88

Find three consecutive even integers whose sum is 222.

Find three consecutive odd integers whose sum is 171.

Solution

55, 57, 59

Find three consecutive odd integers whose sum is 291.

Find three consecutive even integers whose sum is $−36 .$

Solution

$−10, −12, −14$

Find three consecutive even integers whose sum is $−84 .$

Find three consecutive odd integers whose sum is $−213 .$

Solution

$−69, −71, −73$

Find three consecutive odd integers whose sum is $−267 .$

Everyday Math

Sale Price Patty paid $35 for a purse on sale for $10 off the original price. What was the original price of the purse?

Solution

$45

Sale Price 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?

Buying in Bulk Minh spent $6.25 on five sticker books to give his nephews. Find the cost of each sticker book.

Solution

$1.25

Buying in Bulk Alicia bought a package of eight peaches for $3.20. Find the cost of each peach.

Price before Sales Tax Tom paid $1,166.40 for a new refrigerator, including $86.40 tax. What was the price of the refrigerator?

Solution

$1080

Price before Sales Tax Kenji paid $2,279 for a new living room set, including $129 tax. What was the price of the living room set?

Writing Exercises

What has been your past experience solving word problems?

Solution

answers will vary

When you start to solve a word problem, how do you decide what to let the variable represent?

What are consecutive odd integers? Name three consecutive odd integers between 50 and 60.

Solution

Consecutive odd integers are odd numbers that immediately follow each other. An example of three consecutive odd integers between 50 and 60 would be 51, 53, and 55.

What are consecutive even integers? Name three consecutive even integers between $−50$ and $−40 .$

Self Check

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

ⓑ If most of your checks were:

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

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

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

Solve Percent Applications

Translate and Solve Basic Percent Equations

We will solve percent equations using the methods we used to solve equations with fractions or decimals. Without the tools of algebra, the best method available to solve percent problems was by setting them up as proportions. Now as an algebra student, you can just translate English sentences into algebraic equations and then solve the equations.

We can use any letter you like as a variable, but it is a good idea to choose a letter that will remind us of what you are looking for. We must be sure to change the given percent to a decimal when we put it in the equation.

Translate and solve: What number is 35% of 90?

Solution


Translate into algebra. Let $n$ = the number.
Remember "of" means multiply, "is" means equals.
Multiply.
	$31.5$ is $35%$ of $90$

We must be very careful when we translate the words in the next example. The unknown quantity will not be isolated at first, like it was in Example 1. We will again use direct translation to write the equation.

Translate and solve: 6.5% of what number is $1.17?

Solution


Translate. Let $n =$ the number.
Multiply.
Divide both sides by 0.065 and simplify.
	$6.5%$ of $$18$ is $$1.17$

In the next example, we are looking for the percent.

Translate and solve: 144 is what percent of 96?

Solution


Translate into algebra. Let $p =$ the percent.
Multiply.
Divide by 96 and simplify.
Convert to percent.
	$144$ is $150%$ of $96$

Note that we are asked to find percent, so we must have our final result in percent form.

Solve Applications of Percent

Many applications of percent—such as tips, sales tax, discounts, and interest—occur in our daily lives. To solve these applications we’ll translate to a basic percent equation, just like those we solved in previous examples. Once we translate the sentence into a percent equation, we know how to solve it.

We will restate the problem solving strategy we used earlier for easy reference.

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 will involve everyday situations, you can rely on your own experience.

Dezohn and his girlfriend enjoyed a nice dinner at a restaurant and his bill was $68.50. He wants to leave an 18% tip. If the tip will be 18% of the total bill, how much tip should he leave?

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		the amount of tip should Dezohn leave
Step 3. Name what we are looking for.
Choose a variable to represent it.		Let t = amount of tip.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it.
Translate the sentence into an equation.		$A mathematical equation is displayed with the variable 't' set equal to the product of 0.18 and 68.50, demonstrating a calculation likely for a percentage or fraction of a given value.$
Step 5. Solve the equation. Multiply.
Step 6. Check. Does this make sense?
Yes, 20% of $70 is $14.
Step 7. Answer the question with a complete sentence.		Dezohn should leave a tip of $12.33.

Notice that we used t to represent the unknown tip.

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?

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		the total amount of potassium that is recommended
Step 3. Name what we are looking for.
Choose a variable to represent it.		Let $a =$ total amount of potassium.
Step 4. Translate. Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Step 6. Check. Does this make sense?
Yes, 2% is a small percent and 85 is a small part of 4,250.
Step 7. Answer the question with a complete sentence.		The amount of potassium that is recommended is 4,250 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

Step 1. Read the problem.
Step 2. Identify what we are looking for.		the percent of the total calories from fat
Step 3. Name what we are looking for.
Choose a variable to represent it.		Let $p =$ percent of fat.
Step 4. Translate. Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Divide by 480.
Put in a percent form.
Step 6. Check. Does this make sense?
Yes, 240 is half of 480, so 50% makes sense.
Step 7. Answer the question with a complete sentence.		Of the total calories in each brownie, 50% is fat.

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, the difference of the new amount and the original amount. Then we find what percent the amount of increase is of the original amount.

In 2011, the California governor proposed raising community college fees from $26 a unit to $36 a unit. Find the percent increase. (Round to the nearest tenth of a percent.)

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		the percent increase
Step 3. Name what we are looking for.
Choose a variable to represent it.		Let $p =$ the percent.
Step 4. Translate. Write a sentence that gives the information to find it.
First find the amount of increase.		new amount − original amount = increase
		$36 - 26 = 10$
Find the percent.		Increase is what percent of the original amount?

Translate into an equation.
Step 5. Solve the equation.
Divide by 26.
Change to percent form; round to the nearest tenth.
Step 6. Check. Does this make sense?
Yes, 38.4% is close to $\frac{1}{3}$ , and 10 is close to $\frac{1}{3}$ of 26.
Step 7. Answer the question with a complete sentence.		The new fees represent a 38.5% increase over the old fees.

Notice that we rounded the division to the nearest thousandth in order to round the percent to the nearest tenth.

Finding the percent decrease is very similar to finding the percent increase, but now the amount of decrease is the difference of the original amount and the new amount. Then we find what percent the amount of decrease is 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

Step 1. Read the problem.
Step 2. Identify what we are looking for.		the percent decrease
Step 3. Name what we are looking for.
Choose a variable to represent that quantity.		Let $p =$ the percent decrease.
Step 4. Translate. Write a sentence that gives the information to find it.
First find the amount of decrease.		$3.71 - 3.64 = 0.07$
Find the percent.		Decrease is what percent of the original amaount?

Translate into an equation.
Step 5. Solve the equation.
Divide by 3.71.
Change to percent form; round to the nearest tenth.
Step 6. Check. Does this make sense?
Yes, if the original price was $4, a 2% decrease would be 8 cents.
Step 7. Answer the question with a complete sentence.		The price of gas decreased 1.9%.

Solve Simple Interest Applications

Do you know that banks pay you to keep your money? The money a customer puts in the bank is called the principal, P, and the money the bank pays the customer is called the interest. The interest is computed as a certain percent of the principal; called the rate of interest, r. We usually express rate of interest as a percent per year, and we calculate it by using the decimal equivalent of the percent. The variable t, (for time) represents the number of years the money is in the account.

To find the interest we use the simple interest formula, $I = P r t .$

Interest may also be calculated another way, called compound interest. This type of interest will be covered in later math classes.

The formula we use to calculate simple interest is $I = P r t .$ To use the formula, we substitute in the values the problem gives us for the variables, and then solve for the unknown variable. It may be helpful to organize the information in a chart.

Nathaly deposited $12,500 in her bank account where it will earn 4% interest. How much interest will Nathaly earn in 5 years?

\begin{matrix} I & = & ? \\ P & = & $12,500 \\ r & = & 4 % \\ t & = & 5 years \end{matrix}

Solution

A seven-step method for solving problems, demonstrated by calculating simple interest.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the amount of interest earned
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $I =$ the amount of interest.
Step 4. Translate into an equation. Write the formula. Substitute in the given information.	$\begin{matrix} I = P r t \\ I = (12,500) (.04) (5) \end{matrix}$
Step 5. Solve the equation.	$I = 2,500$
Step 6. Check: Does this make sense? Is $2,500 a reasonable interest on $12,500? Yes.
Step 7. Answer the question with a complete sentence.	The interest is $2,500.

There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we don’t know the rate. To find the rate, we use the simple interest formula, substitute in the given values for the principal and time, and then solve for the rate.

Loren loaned 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?

\begin{matrix} I & = & $660 \\ P & = & $3,000 \\ r & = & ? \\ t & = & 4 years \end{matrix}

Solution

Detailed 7-step process for solving a simple interest problem to determine the interest rate.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the rate of interest
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $r =$ the rate of interest.
Step 4. Translate into an equation. Write the formula. Substitute in the given information.	$\begin{matrix} I = P r t \\ 660 = (3,000) r (4) \end{matrix}$
Step 5. Solve the equation. Divide. Change to percent form.	$\begin{matrix} 660 = (12,000) r \\ 0.055 = r \\ 5.5% = r \end{matrix}$
Step 6. Check: Does this make sense? $\begin{matrix} I = P r t \\ 660 \overset{?}{=} (3,000) (0.055) (4) \\ 660 = 660 ✓ \end{matrix}$
Step 7. Answer the question with a complete sentence.	The rate of interest was 5.5%.

Notice that in this example, Loren’s brother paid Loren interest, just like a bank would have paid interest if Loren invested his money there.

Eduardo noticed that his new car loan papers stated that with a 7.5% interest rate, he would pay $6,596.25 in interest over 5 years. How much did he borrow to pay for his car?

Solution

Steps for solving a word problem to find the principal amount using the simple interest formula.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the amount borrowed (the principal)
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $P =$ principal borrowed.
Step 4. Translate into an equation. Write the formula. Substitute in the given information.	$\begin{matrix} I & = & P r t \\ 6,596.25 & = & P (0.075) (5) \end{matrix}$
Step 5. Solve the equation. Divide.	$\begin{matrix} 6,596.25 & = & 0.375 P \\ 17,590 & = & P \end{matrix}$
Step 6. Check: Does this make sense? $\begin{matrix} I & = & P r t \\ 6,596.25 & \overset{?}{=} & (17,590) (0.075) (5) \\ 6,596.25 & = & 6,596.25 ✓ \end{matrix}$
Step 7. Answer the question with a complete sentence.	The principal was $17,590.

Solve Applications with Discount or Mark-up

Applications of discount are very common in retail settings. When you buy an item on sale, the original price has been discounted by some dollar amount. The discount rate, usually given as a percent, is used to determine the amount of the discount. To determine the amount of discount, we multiply the discount rate by the original price.

We summarize the discount model in the box below.

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

This table provides a step-by-step walkthrough for calculating a discount, demonstrating the process from problem setup to the final solution.
ⓐ $\begin{matrix} Original price & = & $ 140 \\ Discount rate & = & 35 % \\ Discount & = & ? \end{matrix}$
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the amount of discount
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $d =$ the amount of discount.
Step 4. Translate into an equation. Write a sentence that gives the information to find it. Translate into an equation.	The discount is 35% of $140. $d = 0.35 (140)$
Step 5. Solve the equation.	$d = 49$
Step 6. Check: Does this make sense? Is a $49 discount reasonable for a $140 dress? Yes.
Step 7. Write a complete sentence to answer the question.	The amount of discount was $49.

ⓑ
Read the problem again.

Step 1. Identify what we are looking for.		the sale price of the dress
Step 2. Name what we are looking for.
Choose a variable to represent that quantity.		Let $s =$ the sale price.
Step 3. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 4. Solve the equation.
Step 5. Check. Does this make sense?
Is the sale price less than the original price?
Yes, $91 is less than $140.
Step 6. Answer the question with a complete sentence.		The sale price of the dress was $91.

There may be times when we know the original price and the sale price, and we want to know the discount rate. To find the discount rate, first we will find the amount of discount and then use it to compute the rate as a percent of the original price. Example 13 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

This table illustrates a seven-step process for solving a word problem to calculate the discount amount, from identifying the unknown to providing the final answer.
ⓐ $\begin{matrix} Original price & = & $ 31 \\ Discount & = & ? \\ Sale Price & = & $ 13.95 \end{matrix}$
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the amount of discount
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $d =$ the amount of discount.
Step 4. Translate into an equation. Write a sentence that gives the information to find it. Translate into an equation.	The discount is the difference between the original price and the sale price. $d = 31 - 13.95$
Step 5. Solve the equation.	$d = 17.05$
Step 6. Check: Does this make sense? Is 17.05 less than 31? Yes.
Step 7. Answer the question with a complete sentence.	The amount of discount was $17.05.

ⓑ
Read the problem again.

Step 1. Identify what we are looking for.		the discount rate
Step 2. Name what we are looking for.
Choose a variable to represent it.		Let $r =$ the discount rate.
Step 3. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 4. Solve the equation.
Divide both sides by 31.
Change to percent form.
Step 5. Check. Does this make sense?
Is $17.05 equal to 55% of $31?
$17.05 ≟ 0.55 (31)$
$17.05 = 17.05 ✓$
Step 6. Answer the question with a complete sentence.		The rate of discount was 55%.

Applications of mark-up are very common in retail settings. The price a retailer pays for an item is called the original cost. The retailer then adds a mark-up to the original cost to get the list price, the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost.

We summarize the mark-up model in the box below.

Adam’s art gallery bought a photograph at original cost $250. Adam marked the price up 40%. Find the ⓐ amount of mark-up and ⓑ the list price of the photograph.

Solution

ⓐ

Step 1. Read the problem.
Step 2. Identify what we are looking for.		the amount of mark-up
Step 3. Name what we are looking for.
Choose a variable to represent it.		Let $m =$ the amount of markup.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Step 6. Check. Does this make sense?
Yes, 40% is less than one-half and 100 is less than half of 250.
Step 7. Answer the question with a complete sentence.		The mark-up on the phtograph was $100.

ⓑ

Step 1. Read the problem again.
Step 2. Identify what we are looking for.		the list price
Step 3. Name what we are looking for.
Choose a variable to represent it.		Let $p =$ the list price.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Step 6. Check. Does this make sense?
Is the list price more than the net price? Is $350 more than $250? Yes.
Step 7. Answer the question with a complete sentence.		The list price of the photograph was $350.

Key Concepts

Percent Increase To find the percent increase:
1. Find the amount of increase. $increase = new amount - original amount$
2. Find the percent increase. Increase is what percent of the original amount?
Percent Decrease To find the percent decrease:
1. Find the amount of decrease. $decrease = original amount - new amount$
2. Find the percent decrease. Decrease is what percent of the original amount?
Simple Interest If an amount of money, P, called the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is

$\begin{matrix} I & = & P r t \\ where I & = & interest \\ P & = & principal \\ r & = & rate \\ t & = & time \end{matrix}$
Discount
- amount of discount is discount rate $\cdot$ original price
- sale price is original price – discount
Mark-up
- amount of mark-up is mark-up rate $\cdot$ original cost
- list price is original cost + mark up

Practice Makes Perfect

Translate and Solve Basic Percent Equations

In the following exercises, translate and solve.

What number is 45% of 120?

Solution

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 2250 is what number?

Solution

18,000

600% of 1740 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 1500 is 540?

Solution

36%

What percent of 1800 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 Percent Applications

In the following exercises, solve.

Geneva treated her parents to dinner at their favorite restaurant. The bill was $74.25. Geneva wants to leave 16% of the total bill as a tip. How much should the tip be?

Solution

$11.88

When Hiro and his co-workers had lunch at a restaurant near their work, the bill was $90.50. They want to leave 18% of the total bill as a tip. How much should the tip be?

Trong has 12% of each paycheck automatically deposited to his savings account. His last paycheck was $2165. 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 eight grams of fiber, which is 33% of the recommended daily amount. What is the total recommended daily amount of fiber?

Solution

24.2 g

One serving of trail mix has 67 grams of carbohydrates, which is 22% of the recommended daily amount. What is the total recommended daily amount of carbohydrates?

A bacon cheeseburger at a popular fast food restaurant contains 2070 milligrams (mg) of sodium, which is 86% of the recommended daily amount. What is the total recommended daily amount of sodium?

Solution

2407 mg

A grilled chicken salad at a popular fast food restaurant contains 650 milligrams (mg) of sodium, which is 27% of the recommended daily amount. What is the total recommended daily amount of sodium?

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

175 lb.

Tricia got a 6% raise on her weekly salary. The raise was $30 per week. What was her original salary?

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

$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?

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?

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, solve.

Tamanika got a raise in her hourly pay, from $15.50 to $17.36. Find the percent increase.

Solution

12%

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 $12,000 in 2010. Find the percent increase.

Solution

200%

The price of a share of one stock rose from $12.50 to $50. Find the percent increase.

According to Time magazine annual global seafood consumption rose from 22 pounds per person in the 1960s to 38 pounds per person in 2011. 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%

In 10 years, the population of Detroit fell from 950,000 to about 712,500. Find the percent decrease.

In 1 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 1 year. Find the percent decrease. (Round to the nearest tenth of a percent.)

Solve Simple Interest Applications

In the following exercises, solve.

Casey deposited $1,450 in a bank account with interest rate 4%. How much interest was earned in two 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

Kenneth loaned 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 loaned 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?

Solve Applications with Discount or Mark-up

In the following exercises, find the sale price.

Perla bought a cell phone that was on sale for $50 off. The original price of the cell phone was $189.

Solution

$139

Sophie saw a dress she liked on sale for $15 off. The original price of the dress was $96.

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 televisions. One television, with original price $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

ⓐ $26.97 ⓑ $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

ⓐ $128.37 ⓑ $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 (round to the nearest cent.)

Solution

ⓐ $332.48 ⓑ $617.47

Jay bought a guitar on sale at 45% off. The original price was $514.75 (round to the nearest cent.)

In the following exercises, find ⓐ the amount of discount and ⓑ the discount rate. (Round to the nearest tenth of a percent if needed.)

Larry and Donna bought a sofa at the sale price of $1,344. The original price of the sofa was $1,920.

Solution

ⓐ $576 ⓑ 30%

Hiroshi bought a lawnmower at the sale price of $240. The original price of the lawnmower is $300.

Patty bought a baby stroller on sale for $301.75. The original price of the stroller was $355.

Solution

ⓐ $53.25 ⓑ 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. To the nearest tenth of a percent, what was the rate of discount?

Solution

ⓐ $370 ⓑ 43.5%

Stella bought a dinette set on sale for $725. The original price was $1,299. To the nearest tenth of a percent, what was the rate of discount?

In the following exercises, find ⓐ the amount of the mark-up and ⓑ the list price.

Daria bought a bracelet at original cost $16 to sell in her handicraft store. She marked the price up 45%.

Solution

ⓐ $7.20 ⓑ $23.20

Regina bought a handmade quilt at original cost $120 to sell in her quilt store. She marked the price up 55%.

Tom paid $0.60 a pound for tomatoes to sell at his produce store. He added a 33% mark-up.

Solution

ⓐ $0.20 ⓑ $0.80

Flora paid her supplier $0.74 a stem for roses to sell at her flower shop. She added an 85% mark-up.

Alan bought a used bicycle for $115. After re-conditioning it, he added 225% mark-up and then advertised it for sale.

Solution

ⓐ $258.75 ⓑ $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

Leaving a Tip 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%

Splitting a Bill Four friends went out to lunch and the bill came to $53.75. They decided to add enough tip to make a total of $64, so that they could easily split the bill evenly among themselves. What percent tip did they leave?

Writing Exercises

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 number should be greater than 44. Since 80% equals 0.8 in decimal form, 0.8 is less than one, and we must multiply the number by 0.8 to get 44, the number must be greater than 44.

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

He meant that he should have packed half the shorts and twice the shirts.

Because of road construction in one city, commuters were advised to plan that their Monday morning commute would take 150% of their usual commuting time. Explain what this means.

Self Check

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

This is a table that has two rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “translate and solve basic percent equations,” “solve percent applications,” “find percent increase and percent decrease,” “solve simple interest applications,” and “solve applications with discount or mark-up.” The rest of the cells are blank.

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

amount of discount

The amount of discount is the amount resulting when a discount rate is multiplied by the original price of an item.

discount rate

The discount rate is the percent used to determine the amount of a discount, common in retail settings.

interest

Interest is the money that a bank pays its customers for keeping their money in the bank.

list price

The list price is the price a retailer sells an item for.

mark-up

A mark-up is a percentage of the original cost used to increase the price of an item.

original cost

The original cost in a retail setting, is the price that a retailer pays for an item.

principal

The principal is the original amount of money invested or borrowed for a period of time at a specific interest rate.

rate of interest

The rate of interest is a percent of the principal, usually expressed as a percent per year.

simple interest

Simple interest is the interest earned according to the formula

I = P r t .

Solve Mixture Applications

Solve Coin Word Problems

In mixture problems, we will have two or more items with different values to combine together. The mixture model is used by grocers and bartenders to make sure they set fair prices for the products they sell. Many other professionals, like chemists, investment bankers, and landscapers also use the mixture model.

We will start by looking at an application everyone is familiar with—money!

Imagine that we take a handful of coins from a pocket or purse and place them on a desk. How would we determine the value of that pile of coins? If we can form a step-by-step plan for finding the total value of the coins, it will help us as we begin solving coin word problems.

So what would we do? To get some order to the mess of coins, we could separate the coins into piles 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, we would add the total value of each pile.

Piles of pennies, nickels, dimes, and quarters

How would we determine the value of each pile? Think about the dime pile—how much is it worth? If we count the number of dimes, we’ll know how many we have—the number of dimes.

But this does not tell us the value of all the dimes. Say we counted 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. This method leads to the following model.

The number of dimes times the value of each dime equals the total value of the dimes.

\begin{matrix} n u m b e r \cdot v a l u e & = & t o t a l v a l u e \\ 17 \cdot $0.10 & = & $ 1.70 \end{matrix}

We could continue this process for each type of coin, and then we would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.

Let’s look at a specific case. Suppose there are 14 quarters, 17 dimes, 21 nickels, and 39 pennies.

This table has five rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Quarters, 14, 0.25, and 3.50. The third row reads Dimes, 17, 0.10, and 1.70. The fourth row reads Nickels, 21, 0.05, and 1.05. The fifth row reads Pennies, 39, 0.01, and 0.39. The extra cell reads 6.64.

The total value of all the coins is $6.64.

Notice how the chart helps organize all the information! Let’s see how we use this method 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

Step 1. Read the problem. Make sure all the words and ideas are understood.

Determine the types of coins involved.
Think about the strategy we used to find the value of the handful of coins. The first thing we need is to notice what types of coins are involved. Adalberto has dimes and nickels.
Create a table to organize the information. See chart below.
- 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. The table below shows this information.

Step 2. Identify what we are looking for.

We are asked to find the number of dimes and nickels Adalberto has.

Step 3. Name what we 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.

Next we counted the number of each type of coin. In this problem we cannot count each type of coin—that is what you are looking for—but we have a clue. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.

\begin{matrix} Let d & = & number of dimes. \\ d + 9 & = & number of nickels \end{matrix}

Fill in the “number” column in the table to help get everything organized.

This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Dimes, d, 0.10, and blank. The third row reads Nickels, d plus 9, 0.05, and blank. The extra cell reads 2.25.

Now we have all the information we need from the problem!

We multiply the number times the value to get the total value of each type of coin. While we do not know the actual number, we do have an expression to represent it.

And so now multiply $n u m b e r \cdot v a l u e = t o t a l v a l u e .$ See how this is done in the table below.

Notice that we made the heading of the table show the model.

Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence. Translate the English sentence into an algebraic equation.

Write the equation by adding the total values of all the types of coins.

The sentence, “value of dimes plus value of nickels equals total value of coins,” can be translated to an equation. Translate “value of dimes” to 0.10d, translate “value of nickles” to 0.05d, and translate “total value of coins” to 2.25. The full equation is 0.10d plus 0.05 times the quantity d plus 9 equals 2.25.

Step 5. Solve the equation using good algebra techniques.

Now solve this equation.
Distribute.
Combine like terms.
Subtract 0.45 from each side.
Divide.
So there are 12 dimes.
The number of nickels is $d + 9$ .

	21

Step 6. Check the answer in the problem and make sure it makes sense.

Does this check?

Calculates the total monetary value for given quantities of dimes and nickels.
12 dimes	$12 (0.10) = 1.20$
21 nickels	$\begin{matrix} 21 (0.05) = \underset{____}{1.05} \\ $2.25 ✓ \end{matrix}$

Step 7. Answer the question with a complete sentence.

Adalberto has twelve dimes and twenty-one nickels.

If this were a homework exercise, our work might look like the following.

Word problem: Adalberto has $2.25 in dimes and nickels. He has nine more nickels than dimes. A table organizes the number, value, and total value of each coin type. The solution shows the equations used to determine he has twelve dimes and twenty-one nickels, with a check confirming the total amount.

Solve Coin Word Problems.

Read the problem. Make sure all the words and ideas are understood.
- Determine the types of coins involved.
- Create a table to organize the information.
- Label the columns “type,” “number,” “value,” “total value.”
- List the types of coins.
- Write in the value of each type of coin.
- Write in the total value of all the coins.
Identify what we are looking for.
Name what we 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.
Translate into an equation.
It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.
Write the equation by adding the total values of all the types of coins.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

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

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.

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.
Let q represent the number of quarters.
Then the number of pennies is 2q.

Multiply the ‘number’ and the ‘value’ to get the ‘total value’ of each type of coin.

A table calculates the total value of coins. It shows 'q' quarters (each worth $0.25) and '2q' pennies (each worth $0.01), resulting in total values of $0.25q and $0.01(2q) respectively. The overall sum is $2.43.

Step 4. Translate. Write the equation by adding the ‘total value’ of all the types of coins.

Step-by-step solution for a coin problem involving quarters and pennies, including verification of the answer.
Step 5. Solve the equation.	$\begin{matrix} 0.25 q + 0.01 (2 q) & = & 2.43 \end{matrix}$
Multiply.	$\begin{matrix} 0.25 q + 0.02 q & = & 2.43 \end{matrix}$
Combine like terms.	$\begin{matrix} 0.27 q & = & 2.43 \end{matrix}$
Divide by 0.27.	$\begin{matrix} q & = & 9 quarters \end{matrix}$
The number of pennies is $2 q$ .	$\begin{matrix} 2 q \\ 2 \cdot 9 \\ 18 pennies \end{matrix}$
Step 6. Check the answer in the problem.
Maria has 9 quarters and 18 pennies. Does this make $2.43?
$\begin{matrix} 9 quarters & 9 (0.25) & = & 2.25 \\ 18 pennies & 18 (0.01) & = & \underset{____}{0.18} \\ Total & $2.43 ✓ \end{matrix}$
Step 7. Answer the question.	Maria has nine quarters and eighteen pennies.

In the next example, we’ll show only the completed table—remember the steps we take to fill in the table.

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

Step 1. Read the problem.
Determine the types of coins involved.	pennies and nickels
Create a table.
Write in the value of each type of coin.	Pennies are worth $0.01. Nickels are worth $0.05.
Step 2. Identify what we are looking for.	the number of pennies and nickels
Step 3. Name. Represent the number of each type of coin using variables.
The number of nickels is defined in terms of the number of pennies, so start with pennies.	Let $p =$ number of pennies.
The number of nickels is two more than ten times the number of pennies.	$10 p + 2 =$ number of nickels.
Multiply the number and the value to get the total value of each type of coin.

Step 4. Translate. Write the equation by adding the total value of all the types of coins.
Step 5. Solve the equation.
How many nickels?
Step 6. Check the answer in the problem and make sure it makes sense Danny has four pennies and 42 nickels. Is the total value $2.14? $\begin{matrix} 4 (0.01) + 42 (0.05) & ≟ & 2.14 \\ 2.14 & = & 2.14 ✓ \end{matrix}$
Step 7. Answer the question.	Danny has four pennies and 42 nickels.

Solve Ticket and Stamp Word Problems

Problems involving tickets or stamps are very much like coin problems. Each type of ticket and stamp has a value, just like each type of coin does. So to solve these problems, we will follow the same steps we used to solve coin problems.

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 five less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

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.

Step 2. Identify what we 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 was five less than three times the number of student tickets sold.

Let s be the number of student tickets.
Then $3 s - 5$ is the number of adult tickets

Multiply the number times the value to get the total value of each type of ticket.

Step 4. Translate. Write the equation by adding the total values of each type of ticket.

6 s + 9 (3 s - 5) = 1506

Step 5. Solve the equation.

This table presents the step-by-step calculation for determining the number of student and adult tickets.
$\begin{matrix} 6 s + 27 s - 45 & = & 1506 \\ 33 s - 45 & = & 1506 \\ 33 s & = & 1551 \\ s & = & 47 student tickets \end{matrix}$
$\begin{matrix} 3 s - 5 \\ 3 (47) - 5 \end{matrix}$
136 adult tickets

Step 6. Check the answer.

There were 47 student tickets at $6 each and 136 adult tickets at $9 each. Is the total value $1,506? We find the total value of each type of ticket by multiplying the number of tickets times its value then add to get the total value of all the tickets sold.

\begin{matrix} 47 \cdot 6 & = & 282 \\ 136 \cdot 9 & = & \underset{_____}{1,224} \\ 1,506 ✓ \end{matrix}

Step 7. Answer the question. They sold 47 student tickets and 136 adult tickets.

We have learned how to find the total number of tickets when the number of one type of ticket is based on the number of the other type. Next, we’ll look at an example where we know the total number of tickets and have to figure out how the two types of tickets relate.

Suppose Bianca sold a total of 100 tickets. Each ticket was either an adult ticket or a child ticket. If she sold 20 child tickets, how many adult tickets did she sell?

Did you say ‘80’? How did you figure that out? Did you subtract 20 from 100?

If she sold 45 child tickets, how many adult tickets did she sell?

Did you say ‘55’? How did you find it? By subtracting 45 from 100?

What if she sold 75 child tickets? How many adult tickets did she sell?

The number of adult tickets must be $100 - 75 .$ She sold 25 adult tickets.

Now, suppose Bianca sold x child tickets. Then how many adult tickets did she sell? To find out, we would follow the same logic we used above. In each case, we subtracted the number of child tickets from 100 to get the number of adult tickets. We now do the same with x.

We have summarized this below.

This table has five rows and two columns. The top row is a header row that reads from left to right Child tickets and Adult tickets. The second row reads 20 and 80. The third row reads 45 and 55. The fourth row reads 75 and 25. The fifth row reads x and 100 plus x.

We can apply these techniques to other examples

Galen sold 810 tickets for his church’s carnival for a total of $2,820. Children’s tickets cost $3 each and adult tickets cost $5 each. How many children’s tickets and how many adult tickets did he sell?

Solution

Step 1. Read the problem.

Determine the types of tickets involved. There are children tickets and adult tickets.
Create a table to organize the information.

Step 2. Identify what we are looking for.

We are looking for the number of children and adult tickets.

Step 3. Name. Represent the number of each type of ticket using variables.

We know the total number of tickets sold was 810. This means the number of children’s tickets plus the number of adult tickets must add up to 810.
Let c be the number of children tickets.
Then $810 - c$ is the number of adult tickets.
Multiply the number times the value to get the total value of each type of ticket.

Step 4. Translate.

Write the equation by adding the total values of each type of ticket.

Step 5. Solve the equation.

\begin{matrix} 3 c + 5 (810 - c) & = & 2,820 \\ 3 c + 4,050 - 5 c & = & 2,820 \\ - 2 c & = & −1,230 \\ c & = & 615 children tickets \end{matrix}

How many adults?

810 - c

810 - 615

195 adult tickets

Step 6. Check the answer. There were 615 children’s tickets at $3 each and 195 adult tickets at $5 each. Is the total value $2,820?

\begin{matrix} 615 \cdot 3 & = & 1845 \\ 195 \cdot 5 & = & \underset{____}{975} \\ 2,820 ✓ \end{matrix}

Step 7. Answer the question. Galen sold 615 children’s tickets and 195 adult tickets.

Now, we’ll do one where we fill in the table all at once.

Monica paid $8.36 for stamps. The number of 41-cent stamps was four more than twice the number of two-cent stamps. How many 41-cent stamps and how many two-cent stamps did Monica buy?

Solution

The types of stamps are 41-cent stamps and two-cent stamps. Their names also give the value!

“The number of 41-cent stamps was four more than twice the number of two-cent stamps.”

\begin{matrix} Let x = number of 2-cent stamps. \\ 2 x + 4 = number of 41-cent stamps \end{matrix}

Step-by-step solution to a word problem, demonstrating how to set up, solve, and check a linear equation to determine the number of two-cent and 41-cent stamps purchased.
Write the equation from the total values.	$\begin{matrix} 0.41 (2 x + 4) + 0.02 x = 8.36 \end{matrix}$
Solve the equation.	$\begin{matrix} 0.82 x + 1.64 + 0.02 x & = & 8.36 \\ 0.84 x + 1.64 & = & 8.36 \\ 0.84 x & = & 6.72 \\ x & = & 8 \end{matrix}$
Monica bought eight two-cent stamps.	$2 x + 4 for x = 8 .$
Find the number of 41-cent stamps she bought be evaluating.	$\begin{matrix} 2 x + 4 \\ 2 (8) + 4 \\ 20 \end{matrix}$
Check.	$\begin{matrix} 8 (0.02) + 20 (0.41) & \overset{?}{=} & 8.36 \\ 0.16 + 8.20 & \overset{?}{=} & 8.36 \\ 8.36 & = & 8.36 ✓ \end{matrix}$
	Monica bought eight two-cent stamps and 20 41-cent stamps.

Solve Mixture Word Problems

Now we’ll solve some more general applications of the mixture model. Grocers and bartenders use the mixture model to set a fair price for a product made from mixing two or more ingredients. Financial planners use the mixture model when they invest money in a variety of accounts and want to find the overall interest rate. Landscape designers use the mixture model when they have an assortment of plants and a fixed budget, and event coordinators do the same when choosing appetizers and entrees for a banquet.

Our first mixture word problem will be making trail mix from raisins and nuts.

Henning is mixing raisins and nuts to make 10 pounds of trail mix. Raisins cost $2 a pound and nuts cost $6 a pound. If Henning wants his cost for the trail mix to be $5.20 a pound, how many pounds of raisins and how many pounds of nuts should he use?

Solution

As before, we fill in a chart to organize our information.

The 10 pounds of trail mix will come from mixing raisins and nuts.

\begin{matrix} Let x = number of pounds of raisins. \\ 10 - x = number of pounds of nuts \end{matrix}

We enter the price per pound for each item.

We multiply the number times the value to get the total value.

This table has four rows and four columns. The top row is a header row that reads from left to right Type, Number of pounds, Price per pound ($), and Total Value ($). The second row reads raisins, x, 2, and 2x. The third row reads nuts, 10 minus x, 6, and 6 times the quantity (10 minus x). The fourth row reads trail mix, 10, 5.20, and 10 times 5.20.

Notice that the last line in the table gives the information for the total amount of the mixture.

We know the value of the raisins plus the value of the nuts will be the value of the trail mix.

Write the equation from the total values.
Solve the equation.


Find the number of pounds of nuts.

	8 pounds of nuts
Check. $\begin{matrix} 2 ($2) + 8 ($6) & \overset{?}{=} & 10 ($5.20) \\ $4 + $48 & \overset{?}{=} & $52 \\ $52 & = & $52 ✓ \end{matrix}$
	Henning mixed two pounds of raisins with eight pounds of nuts.

We can also use the mixture model to solve investment problems using simple interest. We have used the simple interest formula, $I = P r t,$ where $t$ represented the number of years. When we just need to find the interest for one year, $t = 1,$ so then $I = P r .$

Stacey has $20,000 to invest in two different bank accounts. One account pays interest at 3% per year and the other account pays interest at 5% per year. How much should she invest in each account if she wants to earn 4.5% interest per year on the total amount?

Solution

We will fill in a chart to organize our information. We will use the simple interest formula to find the interest earned in the different accounts.

The interest on the mixed investment will come from adding the interest from the account earning 3% and the interest from the account earning 5% to get the total interest on the $20,000.

\begin{matrix} Let x & = & amount invested at 3%. \\ 20,000 - x & = & amount invested at 5% \end{matrix}

The amount invested is the principal for each account.

We enter the interest rate for each account.

We multiply the amount invested times the rate to get the interest.

This table has four rows and four columns. The top row is a header row that reads from left to right Type, Amount invested, Rate, and Interest. The second row reads 3%, x, 0.03, and 0.03x. The third row reads 5%, 20,000 minus x, 0.05, and 0.05 times the quantity (20,000 minus x). The fourth row reads 4.5%, 20,000, 0.045, and 0.045 times 20,000.

Notice that the total amount invested, 20,000, is the sum of the amount invested at 3% and the amount invested at 5%. And the total interest, $0.045 (20,000),$ is the sum of the interest earned in the 3% account and the interest earned in the 5% account.

As with the other mixture applications, the last column in the table gives us the equation to solve.

Write the equation from the interest earned. Solve the equation.	$\begin{matrix} 0.03 x + 0.05 (20,000 - x) & = & 0.045 (20,000) \\ 0.03 x + 1,000 - 0.05 x & = & 900 \\ - 0.02 x + 1,000 & = & 900 \\ - 0.02 x & = & - 100 \\ x & = & 5,000 \end{matrix}$ amount invested at 3%
Find the amount invested at 5%.
Check. $\begin{matrix} 0.03 x + 0.05 (15,000 + x) & ≟ & 0.045 (20,000) \\ 150 + 750 & ≟ & 900 \\ 900 & = & 900 ✓ \end{matrix}$
	Stacey should invest $5,000 in the account that earns 3% and $15,000 in the account that earns 5%.

Key Concepts

Total Value of Coins For the same type of coin, the total value of a number of coins is found by using the model.
$n u m b e r \cdot v a l u e = t o t a l v a l u e$ where number is the number of coins and value is the value of each coin; total value is the total value of all the coins
Problem-Solving Strategy—Coin Word Problems
1. Read the problem. Make all the words and ideas are understood. Determine the types of coins involved.
  - Create a table to organize the information.
  - Label the columns type, number, value, total value.
  - List the types of coins.
  - Write in the value of each type of coin.
  - Write in the total value of all the coins.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
  Use variable expressions to represent the number of each type of coin and write them in the table.
  Multiply the number times the value to get the total value of each type of coin.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.
  Write the equation by adding the total values of all the types of coins.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Practice Makes Perfect

Solve Coin Word Problems

In the following exercises, solve each coin word problem.

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 seven 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?

A cash box of $1 and $5 bills is worth $45. The number of $1 bills is three more than the number of $5 bills. How many of each bill does it contain?

Solution

10 at $1, 7 at $5

Joe’s wallet contains $1 and $5 bills worth $47. The number of $1 bills is five more than the number of $5 bills. How many of each bill does he have?

Rachelle has $6.30 in nickels and quarters in her coin purse. The number of nickels is twice the number of quarters. How many coins of each type does she have?

Solution

18 quarters, 36 nickels

Deloise has $1.20 in pennies and nickels in a jar on her desk. The number of pennies is three times the number of nickels. How many coins of each type does she have?

Harrison has $9.30 in his coin collection, all in pennies and dimes. The number of dimes is three times the number of pennies. How many coins of each type does he have?

Solution

30 pennies, 90 dimes

Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type 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 at $10, 5 at $5

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?

Carolyn has $2.55 in her purse in nickels and dimes. The number of nickels is nine 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 three more than three times the number of quarters. How many of each are there?

Solution

63 dimes, 20 quarters

Tyler has $9.70 in dimes and quarters. The number of quarters is eight more than four times the number of dimes. How many of each coin does he have?

Mukul has $3.75 in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and nine more nickels than quarters. How many of each coin are in his pocket?

Solution

16 nickels, 12 dimes, 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 each ticket or stamp word problem.

The school play sold $550 in tickets 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 sold $1,340 in tickets 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?

The ice rink sold 95 tickets for the afternoon skating session, for a total of $828. General admission tickets cost $10 each and youth tickets cost $8 each. How many general admission tickets and how many youth tickets were sold?

Solution

34 general, 61 youth

For the 7:30 show time, 140 movie tickets were sold. Receipts from the $13 adult tickets and the $10 senior tickets totaled $1,664. How many adult tickets and how many senior tickets were sold?

The box office sold 360 tickets to a concert at the college. The total receipts were $4170. General admission tickets cost $15 and student tickets cost $10. How many of each kind of ticket was sold?

Solution

114 general, 246 student

Last Saturday, the museum box office sold 281 tickets for a total of $3954. Adult tickets cost $15 and student tickets cost $12. How many of each kind of ticket was sold?

Julie went to the post office and bought both $0.41 stamps and $0.26 postcards. She spent $51.40. The number of stamps was 20 more than twice the number of postcards. How many of each did she buy?

Solution

40 postcards, 100 stamps

Jason went to the post office and bought both $0.41 stamps and $0.26 postcards and spent $10.28. The number of stamps was four more than twice the number of postcards. How many of each did he buy?

Maria spent $12.50 at the post office. She bought three times as many $0.41 stamps as $0.02 stamps. How many of each did she buy?

Solution

30 at $0.41, 10 at $0.02

Hector spent $33.20 at the post office. He bought four times as many $0.41 stamps as $0.02 stamps. How many of each did he buy?

Hilda has $210 worth of $10 and $12 stock shares. The numbers of $10 shares is five more than twice the number of $12 shares. How many of each does she have?

Solution

15 $10 shares, 5 $12 shares

Mario invested $475 in $45 and $25 stock shares. The number of $25 shares was five less than three times the number of $45 shares. How many of each type of share did he buy?

Solve Mixture Word Problems

In the following exercises, solve each mixture word problem.

Lauren in making 15 liters of mimosas for a brunch banquet. Orange juice costs her $1.50 per liter and champagne costs her $12 per liter. How many liters of orange juice and how many liters of champagne should she use for the mimosas to cost Lauren $5 per liter?

Solution

5 liters champagne, 10 liters orange juice

Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. Macadamia nuts cost $9 per pound and almonds cost $5.25 per pound. How many pounds of macadamia nuts and how many pounds of almonds should Macario use for the mixture to cost $6.50 per pound to make?

Kaapo is mixing Kona beans and Maui beans to make 25 pounds of coffee blend. Kona beans cost Kaapo $15 per pound and Maui beans cost $24 per pound. How many pounds of each coffee bean should Kaapo use for his blend to cost him $17.70 per pound?

Solution

7.5 lbs Maui beans, 17.5 Kona beans

Estelle is making 30 pounds of fruit salad from strawberries and blueberries. Strawberries cost $1.80 per pound and blueberries cost $4.50 per pound. If Estelle wants the fruit salad to cost her $2.52 per pound, how many pounds of each berry should she use?

Carmen wants to tile the floor of his house. He will need 1000 square feet of tile. He will do most of the floor with a tile that costs $1.50 per square foot, but also wants to use an accent tile that costs $9.00 per square foot. How many square feet of each tile should he plan to use if he wants the overall cost to be $3 per square foot?

Solution

800 at $1.50, 200 at $9.00

Riley is planning to plant a lawn in his yard. He will need nine pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs $3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound?

Vartan was paid $25,000 for a cell phone app that he wrote and wants to invest it to save for his son’s education. He wants to put some of the money into a bond that pays 4% annual interest and the rest into stocks that pay 9% annual interest. If he wants to earn 7.4% annual interest on the total amount, how much money should he invest in each account?

Solution

$8000 at 4%, $17,000 at 9%

Vern sold his 1964 Ford Mustang for $55,000 and wants to invest the money to earn him 5.8% interest per year. He will put some of the money into Fund A that earns 3% per year and the rest in Fund B that earns 10% per year. How much should he invest into each fund if he wants to earn 5.8% interest per year on the total amount?

Stephanie inherited $40,000. She wants to put some of the money in a certificate of deposit that pays 2.1% interest per year and the rest in a mutual fund account that pays 6.5% per year. How much should she invest in each account if she wants to earn 5.4% interest per year on the total amount?

Solution

$10,000 in CD, $30,000 in mutual fund

Avery and Caden have saved $27,000 towards a down payment on a house. They want to keep some of the money in a bank account that pays 2.4% annual interest and the rest in a stock fund that pays 7.2% annual interest. How much should they put into each account so that they earn 6% interest per year?

Dominic pays 7% interest on his $15,000 college loan and 12% interest on his $11,000 car loan. What average interest rate does he pay on the total $26,000 he owes? (Round your answer to the nearest tenth of a percent.)

Solution

9.1%

Liam borrowed a total of $35,000 to pay for college. He pays his parents 3% interest on the $8,000 he borrowed from them and pays the bank 6.8% on the rest. What average interest rate does he pay on the total $35,000? (Round your answer to the nearest tenth of a percent.)

Everyday Math

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

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 $15 full-year registration fee and how many had paid the $10 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 $250 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 six quarters, nine dimes, and four 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.

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

mixture problems

Mixture problems combine two or more items with different values together.

Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

Solve Applications Using Properties of Triangles

In this section we will use some common geometry formulas. We will adapt our problem-solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve. In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

We will start geometry applications by looking at the properties of triangles. Let’s review some basic facts about triangles. Triangles have three sides and three interior angles. Usually each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

The plural of the word vertex is vertices. All triangles have three vertices. Triangles are named by their vertices: The triangle in Figure 1 is called $△ A B C .$

A triangle with vertices A, B, and C. The sides opposite these vertices are marked a, b, and c, respectively. — Triangle ABC has vertices A, B, and C. The lengths of the sides are a, b, and c.

The three angles of a triangle are related in a special way. The sum of their measures is $180 ° .$ Note that we read $m ∠ A$ as “the measure of angle A.” So in $△ A B C$ in Figure 1,

m ∠ A + m ∠ B + m ∠ C = 180 °

Because the perimeter of a figure is the length of its boundary, the perimeter of $△ A B C$ is the sum of the lengths of its three sides.

P = a + b + c

To find the area of a triangle, we need to know its base and height. The height is a line that connects the base to the opposite vertex and makes a $90 °$ angle with the base. We will draw $△ A B C$ again, and now show the height, h. See Figure 2.

A triangle with vertices A, B, and C. The sides opposite these vertices are marked a, b, and c, respectively. The side b is parallel to the bottom of the page, and it has a dashed line drawn from vertex B to it. This line is marked h and makes a right angle with side b. — The formula for the area of $△ A B C$ is $A = \frac{1}{2} b h,$ where b is the base and h is the height.

The measures of two angles of a triangle are 55 and 82 degrees. Find the measure of the third angle.

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
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.	Let $x =$ the measure of the angle.
Step 4. Translate.
Write the appropriate formula and substitute.	$m ∠ A + m ∠ B + m ∠ C = 180$
Step 5. Solve the equation.	$\begin{matrix} 55 + 82 + x & = & 180 \\ 137 + x & = & 180 \\ x & = & 43 \end{matrix}$
Step 6. Check. $\begin{matrix} 55 + 82 + 43 & ≟ & 180 \\ 180 & = & 180 ✓ \end{matrix}$
Step 7. Answer the question.	The measure of the third angle is 43 degrees.

The perimeter of a triangular garden is 24 feet. The lengths of two sides are four feet and nine feet. How long is the third side?

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
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 and substitute.
Substitute in the given information.
Step 5. Solve the equation.
Step 6. Check. $\begin{matrix} P & = & a + b + c \\ 24 & ≟ & 4 + 9 + 11 \\ 24 & = & 24 ✓ \end{matrix}$
Step 7. Answer the question.	The third side is 11 feet long.

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

Step 1. Read the problem. Draw the figure and label it with the given information.	Area $= 90 m^{2}$
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.
Step 5. Solve the equation.
Step 6. Check. $\begin{matrix} A & = & \frac{1}{2} b h \\ 90 & ≟ & \frac{1}{2} \cdot 15 \cdot 12 \\ 90 & = & 90 ✓ \end{matrix}$
Step 7. Answer the question.	The height of the triangle is 12 meters.

The triangle properties we used so far apply to all triangles. Now we will look at one specific type of triangle—a right triangle. A right triangle has one $90 °$ angle, which we usually mark with a small square in the corner.

A right triangle with the largest angle marked 90 degrees.

One angle of a right triangle measures $28 ° .$ What is the measure of the third angle?

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the measure of an angle
Step 3. Name. Choose a variable to represent it.	Let $x =$ the measure of an angle.
Step 4. Translate.	$m ∠ A + m ∠ B + m ∠ C = 180$
Write the appropriate formula and substitute.	$x + 90 + 28 = 180$
Step 5. Solve the equation.	$\begin{matrix} x + 118 & = & 180 \\ x & = & 62 \end{matrix}$
Step 6. Check. $\begin{matrix} 180 & ≟ & 90 + 28 + 62 \\ 180 & = & 180 ✓ \end{matrix}$
Step 7. Answer the question.	The measure of the third angle is 62^°.

In the examples we have seen 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. 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 degrees more than the measure of the smallest angle. Find the measures of all three angles.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the measures of all three angles
Step 3. Name. Choose a variable to represent it.	Let $a = 1^{st}$ angle. $a + 20 = 2^{nd}$ angle $90 = 3^{rd}$ angle (the right angle)
Draw the figure and label it with the given information
Step 4. Translate
Write the appropriate formula. Substitute into the formula.
Step 5. Solve the equation.	55 90 third angle
Step 6. Check. $\begin{matrix} 35 + 55 + 90 & ≟ & 180 \\ 180 & = & 180 ✓ \end{matrix}$
Step 7. Answer the question.	The three angles measure 35^°, 55^°, and 90^°.

Use the Pythagorean Theorem

We have learned how the measures of the angles of a triangle relate to each other. Now, we will learn how the lengths of the sides relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem. This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.

Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a $90 °$ angle, marked with a small square in the corner. The side of the triangle opposite the $90 °$ angle is called the hypotenuse and each of the other sides are called legs.

Three right triangles with different orientations. The right angles are marked with two small lines that make a small square with the angle. Opposite these angles, hypotenuse is written. The other sides are marked “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 lengths of the two legs equals the square of the length of the hypotenuse. In symbols we say: in any right triangle, $a^{2} + b^{2} = c^{2},$ where $a and b$ are the lengths of the legs and $c$ is the length of the hypotenuse.

Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.

To solve exercises that use the Pythagorean Theorem, we will need to find square roots. We have used the notation $\sqrt{m}$ and the definition:

If $m = n^{2},$ then $\sqrt{m} = n,$ for $n \geq 0 .$

For example, we found that $\sqrt{25}$ is 5 because $25 = 5^{2} .$

Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.

Use the Pythagorean Theorem to find the length of the hypotenuse shown below.

A right triangle with legs marked 3 and 4.

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. Label side c on the figure.	Let c = the length of the hypotenuse.
Step 4. Translate.
Write the appropriate formula.	$a^{2} + b^{2} = c^{2}$
Substitute.	$3^{2} + 4^{2} = c^{2}$
Step 5. Solve the equation.	$9 + 16 = c^{2}$
Simplify.	$25 = c^{2}$
Use the definition of square root.	$\sqrt{25} = c$
Simplify.	$5 = c$
Step 6. Check.
Step 7. Answer the question.	The length of the hypotenuse is 5.

Use the Pythagorean Theorem to find the length of the leg shown below.

A right angle with one leg marked 5. The hypotenuse is labeled 13.

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.
Lable side b.
Step 4. Translate
Write the appropriate formula.		$a^{2} + b^{2} = c^{2}$
Substitute.		$5^{2} + b^{2} = 13^{2}$
Step 5. Solve the equation.		$25 + b^{2} = 169$
Isolate the variable term.		$b^{2} = 144$
Use the definition of square root.		$b^{2} = \sqrt{144}$
Simplify.		$b = 12$
Step 6. Check.
Step 7. Answer the question.		The length of the leg is 12.

A gazebo is shown. In one of its corners, a triangle is made with the wood. The hypotenuse is marked 10 inches, and one of the legs is marked x

Kelvin is building a gazebo and wants to brace each corner by placing a $10 ″$ piece of wood diagonally as shown above.

If he fastens the wood so that the ends of the brace are the same distance from the corner, what is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.

Solution

This table outlines the methodical steps for solving a mathematical problem, from reading the question to checking the answer.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the distance from the corner that the bracket should be attached
Step 3. Name. Choose a variable to represent it.	Let $x =$ the distance from the corner.
Step 4. Translate Write the appropriate formula and substitute.	$\begin{matrix} a^{2} + b^{2} & = & c^{2} \\ x^{2} + x^{2} & = & 10^{2} \end{matrix}$
Step 5. Solve the equation. Isolate the variable. Use the definition of square root. Simplify. Approximate to the nearest tenth.	$\begin{matrix} 2 x^{2} & = & 100 \\ x^{2} & = & 50 \\ x & = & \sqrt{50} \\ x & \approx & 7.1 \end{matrix}$
Step 6. Check. $\begin{matrix} a^{2} + b^{2} & = & c^{2} \\ {(7.1)}^{2} + {(7.1)}^{2} & \approx & 10^{2} Yes. \end{matrix}$
Step 7. Answer the question.	Kelvin should fasten each piece of wood approximately 7.1" from the corner.

Solve Applications Using Rectangle Properties

You may already be familiar with the properties of rectangles. Rectangles have four sides and four right $(90 °)$ angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, L, and its adjacent side as the width, W.

The distance around this rectangle is $L + W + L + W,$ or $2 L + 2 W .$ This is the perimeter, P, of the rectangle.

P = 2 L + 2 W

What about the area of a rectangle? Imagine a rectangular rug that is 2-feet long by 3-feet wide. Its area is 6 square feet. There are six squares in the figure.

A rectangles composed of 6 squares that is three high and two wide. The height is marked 3 and the width is marked 2.

\begin{matrix} A = 6 \\ A = 2 \cdot 3 \\ A = L \cdot W \end{matrix}

The area is the length times the width.

The formula for the area of a rectangle is $A = L W .$

The length of a rectangle is 32 meters and the width is 20 meters. What is the perimeter?

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
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.
Step 5. Solve the equation.
Step 6. Check. $\begin{matrix} P & ≟ & 104 \\ 20 + 32 + 20 + 32 & ≟ & 104 \\ 104 & = & 104 ✓ \end{matrix}$
Step 7. Answer the question.	The perimeter of the rectangle is 104 meters.

The area of a rectangular room is 168 square feet. The length is 14 feet. What is the width?

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
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 = the width.
Step 4. Translate.
Write the appropriate formula.	$A = L W$
Substitute.	$168 = 14 W$
Step 5. Solve the equation.	$\frac{168}{14} = \frac{14 W}{14}$ $12 = W$
Step 6. Check. $\begin{matrix} A & = & L W \\ 168 & ≟ & 14 \cdot 12 \\ 168 & = & 168 ✓ \end{matrix}$
Step 7. Answer the question.	The width of the room is 12 feet.

Find the length of a rectangle with perimeter 50 inches and width 10 inches.

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.
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.
Step 5. Solve the equation.	$The mathematical equation 30/2 = 2L/2 is shown on a white background, demonstrating a basic algebraic problem involving fractions.$
Step 6. Check. $\begin{matrix} P & = & 50 \\ 15 + 10 + 15 + 10 & ≟ & 50 \\ 50 & = & 50 ✓ \end{matrix}$
Step 7. Answer the question.	The length is 15 inches.

We have solved problems where either the length or width was given, along with the perimeter or area; now we will learn how to solve problems in which the width is defined in terms of the length. We will 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 feet less than the length. The perimeter is 52 feet. Find the length and width.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length and width of a rectangle
Step 3. Name. Choose a variable to represent it. Since the width is defined in terms of the length, we let L = length. The width is two feet less than the length, so we let L − 2 = width.	$P = 52$ ft
Step 4. Translate.
Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information.	$P = 2 L + 2 W$
Substitute in the given information.	$52 = 2 L + 2 (L - 2)$
Step 5. Solve the equation.	$52 = 2 L + 2 L - 4$
Combine like terms.	$52 = 4 L - 4$
Add 4 to each side.	$56 = 4 L$
Divide by 4.	$\frac{56}{4} = \frac{4 L}{4}$ $14 = L$ The length is 14 feet.
Now we need to find the width.	The width is $L - 2$ . The width is 12 feet.
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 length of a rectangle is four centimeters more than twice the width. The perimeter is 32 centimeters. Find the length and width.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length and the width
Step 3. Name. Choose a variable to represent the width.
The length is four more than twice the width.
Step 4. Translate
Write the appropriate formula.
Substitute in the given information.
Step 5. Solve the equation.	12 The length is 12 cm.
Step 6. Check. $\begin{matrix} P & = & 2 L + 2 W \\ 32 & ≟ & 2 \cdot 12 + 2 \cdot 4 \\ 32 & = & 32 ✓ \end{matrix}$
Step 7. Answer the question.	The length is 12 cm and the width is 4 cm.

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

Step 1. Read the problem. Draw the figure and label it with the given information.	$P = 150$ ft
Step 2. Identify what you are looking for.	the length and the width of the pool
Step 3. Name. Choose a variable to represent the width. The length is 15 feet more than the width.
Step 4. Translate
Write the appropriate formula.
Substitute.
Step 5. Solve the equation.
Step 6. Check. $\begin{matrix} P & = & 2 L + 2 W \\ 150 & ≟ & 2 (45) + 2 (30) \\ 150 & = & 150 ✓ \end{matrix}$
Step 7. Answer the question.	The length of the pool is 45 feet and the width is 30 feet.

Key Concepts

Problem-Solving Strategy for Geometry Applications
1. Read the problem and make all the words and ideas are understood. Draw the figure and label it with the given information.
2. Identify what we are looking for.
3. Name what we are looking for by choosing a variable to represent it.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
Triangle Properties For $△ A B C$
Angle measures:
- $m ∠ A + m ∠ B + m ∠ C = 180$
Perimeter:
- $P = a + b + c$
Area:
- $A = \frac{1}{2} b h, b = base, h = height$
A right triangle has one 90° angle.
The Pythagorean Theorem In any right triangle, $a^{2} + b^{2} = c^{2}$ where c is the length of the hypotenuse and a and b are the lengths of the legs.
Properties of Rectangles
- Rectangles have four sides and four right (90°) angles.
- The lengths of opposite sides are equal.
- The perimeter of a rectangle is the sum of twice the length and twice the width: $P = 2 L + 2 W .$ The area of a rectangle is the length times the width: $A = L W .$

Practice Makes Perfect

Solving Applications Using Triangle Properties

In the following exercises, solve using triangle properties.

The measures of two angles of a triangle are 26 and 98 degrees. Find the measure of the third angle.

Solution

56 degrees

The measures of two angles of a triangle are 61 and 84 degrees. Find the measure of the third angle.

The measures of two angles of a triangle are 105 and 31 degrees. Find the measure of the third angle.

Solution

44 degrees

The measures of two angles of a triangle are 47 and 72 degrees. Find the measure of the third angle.

The perimeter of a triangular pool is 36 yards. The lengths of two sides are 10 yards and 15 yards. How long is the third side?

Solution

11 feet

A triangular courtyard has perimeter 120 meters. The lengths of two sides are 30 meters and 50 meters. How long is the third side?

If a triangle has sides 6 feet and 9 feet and the perimeter is 23 feet, how long is the third side?

Solution

8 feet

If a triangle has sides 14 centimeters and 18 centimeters and the perimeter is 49 centimeters, how long is the third side?

A triangular flag has base one foot and height 1.5 foot. What is its area?

Solution

0.75 sq. ft.

A triangular window has base eight feet and height six feet. What is its area?

What is the base of a triangle with area 207 square inches and height 18 inches?

Solution

23 inches

What is the height of a triangle with area 893 square inches and base 38 inches?

One angle of a right triangle measures 33 degrees. What is the measure of the other small angle?

Solution

One angle of a right triangle measures 51 degrees. What is the measure of the other small angle?

One angle of a right triangle measures 22.5 degrees. What is the measure of the other small angle?

Solution

67.5

One angle of a right triangle measures 36.5 degrees. What is the measure of the other small angle?

The perimeter of a triangle is 39 feet. One side of the triangle is one foot longer than the second side. The third side is two feet longer than the second side. Find the length of each side.

Solution

13 ft., 12 ft., 14 ft.

The perimeter of a triangle is 35 feet. One side of the triangle is five feet longer than the second side. The third side is three feet longer than the second side. Find the length of each side.

One side of a triangle is twice the shortest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.

Solution

3 ft., 6 ft., 8 ft.

One side of a triangle is three times the shortest side. The third side is three feet more than the shortest side. The perimeter is 13 feet. Find the lengths of all three sides.

The two smaller angles of a right triangle have equal measures. Find the measures of all three angles.

Solution

$45 °, 45 °, 90 °$

The measure of the smallest angle of a right triangle is 20° less than the measure of the next larger angle. Find the measures of all three angles.

The angles in a triangle are such that one angle is twice the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

Solution

$30 °, 60 °, 90 °$

The angles in a triangle are such that one angle is 20° more than the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

Use the Pythagorean Theorem

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

A right triangle with legs marked 9 and 12.

Solution

A right triangle with legs marked 16 and 12.

A right triangle with legs marked 15 and 20.

Solution

A right triangle with legs marked 5 and 12.

In the following exercises, use the Pythagorean Theorem to find the length of the leg. Round to the nearest tenth, if necessary.

A right triangle with one leg marked 6 and hypotenuse marked 10.

Solution

A right triangle with one leg marked 8 and hypotenuse marked 17.

A right triangle with one leg marked 5 and hypotenuse marked 13.

Solution

A right triangle with one leg marked 16 and hypotenuse marked 20.

A right triangle with one leg marked 8 and hypotenuse marked 13.

Solution

10.2

A right triangle with both legs marked 6.

A right-angled triangle with a horizontal base of length 5 and a hypotenuse of length 11. The square symbol indicates the right angle between the base and the vertical side.

Solution

9.8

A right triangle with legs marked 5 and 7.

In the following exercises, solve using the Pythagorean Theorem. 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, as shown below. How far from the base of the pole should the end of the string of lights be anchored?

A right triangle with one leg marked 12 and hypotenuse marked 13.

Solution

5 feet

Pam wants to put a banner across her garage door, as shown below, to congratulate her son for 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 house is shown with a banner over the garage door. The garage door is marked 16 ft wide and 12 ft high.

Chi is planning to put a path of paving stones through her flower garden, as shown below. The flower garden is a square with side 10 feet. What will the length of the path be?

A square garden is shown that is marked 10’ on the side. There is a path of stones along the diagonal of the square.

Solution

14.1 feet

Brian borrowed a 20 foot extension ladder to use when he paints his house. If he sets the base of the ladder 6 feet from the house, as shown below, how far up will the top of the ladder reach?

A house is shown with a ladder leaning against it. The ladder is marked 20’, and the distance from the house to the base of the ladder is marked 6’.

Solve Applications Using Rectangle Properties

In the following exercises, solve using rectangle properties.

The length of a rectangle is 85 feet and the width is 45 feet. What is the perimeter?

Solution

260 feet

The length of a rectangle is 26 inches and the width is 58 inches. What is the perimeter?

A rectangular room is 15 feet wide by 14 feet long. What is its perimeter?

Solution

58 feet

A driveway is in the shape of a rectangle 20 feet wide by 35 feet long. What is its perimeter?

The area of a rectangle is 414 square meters. The length is 18 meters. What is the width?

Solution

23 meters

The area of a rectangle is 782 square centimeters. The width is 17 centimeters. What is the length?

The width of a rectangular window is 24 inches. The area is 624 square inches. What is the length?

Solution

26 inches

The length of a rectangular poster is 28 inches. The area is 1316 square inches. What is the width?

Find the length of a rectangle with perimeter 124 and width 38.

Solution

Find the width of a rectangle with perimeter 92 and length 19.

Find the width of a rectangle with perimeter 16.2 and length 3.2.

Solution

4.9

Find the length of a rectangle with perimeter 20.2 and width 7.8.

The length of a rectangle is nine inches more than the width. The perimeter is 46 inches. Find the length and the width.

Solution

16 in., 7 in.

The width of a rectangle is eight 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 five 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 seven 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 length, 12.8 m width

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 is 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.

Solution

50 ft., 25 ft.

The length of a rectangle is three times the width. The perimeter of the rectangle is 72 feet. Find the length and width of the rectangle.

The length of a rectangle is three meters less than twice the width. The perimeter of the rectangle is 36 meters. Find the dimensions of the rectangle.

Solution

7 m width, 11 m length

The length of a rectangle is five inches more than twice the width. The perimeter is 34 inches. Find the length and width.

The perimeter of a rectangular field is 560 yards. The length is 40 yards more than the width. Find the length and width of the field.

Solution

160 yd., 120 yd.

The perimeter of a rectangular atrium is 160 feet. The length is 16 feet more than the width. Find the length and width of the atrium.

A rectangular parking lot has perimeter 250 feet. The length is five feet more than twice the width. Find the length and width of the parking lot.

Solution

85 ft., 40 ft.

A rectangular rug has perimeter 240 inches. The length is 12 inches more than twice the width. Find the length and width of the rug.

Everyday Math

Christa wants to put a fence around her triangular flowerbed. The sides of the flowerbed are six feet, eight feet and 10 feet. How many feet of fencing will she need to enclose her flowerbed?

Solution

240 feet

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 length?

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

area; answers will vary

If you need to put a fence around your backyard, do you need to know the perimeter or the area of the backyard? Explain your reasoning.

Look at the two figures below.

On the left, we have a rectangle with height 2 and width 8. On the right, we have a square with height 4 and width 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.
ⓑ Answers will vary.
ⓒ Answers will vary.
ⓓ The areas are the same.
ⓔ The 2x8 rectangle has a larger perimeter than the 4x4 square.

Write a geometry word problem that relates to your life experience, then solve it and explain all your steps.

Self Check

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

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

Solve Uniform Motion Applications

When planning a road trip, it often helps to know how long it will take to reach the destination or how far to travel each day. We would use the distance, rate, and time formula, $D = r t,$ which we have already seen.

In this section, we will use this formula in situations that require a little more algebra to solve than the ones we saw earlier. Generally, we will be looking at comparing two scenarios, such as two vehicles travelling at different rates or in opposite directions. When the speed of each vehicle is constant, we call applications like this uniform motion problems.

Our problem-solving strategies will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the money applications.

The steps are listed here for easy reference:

Use a Problem-Solving Strategy in Distance, Rate, and Time Applications.

Read the problem. Make sure all the words and ideas are understood.
- Draw a diagram to illustrate what it happening.
- Create a table to organize the information.
- Label the columns rate, time, distance.
- List the two scenarios.
- Write in the information you know.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
- Complete the chart.
- Use variable expressions to represent that quantity in each row.
- Multiply the rate times the time to get the distance.
Translate into an equation.
- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in 4 hours and the local train takes 5 hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

Draw a diagram to illustrate what it happening. Shown below is a sketch of what is happening in the example.
Create a table to organize the information.
Label the columns “Rate,” “Time,” and “Distance.”
List the two scenarios.
Write in the information you know.

Step 2. Identify what we are looking for.

We are asked to find the speed of both trains.
Notice that the distance formula uses the word “rate,” but it is more common to use “speed” when we talk about vehicles in everyday English.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

Complete the chart
Use variable expressions to represent that quantity in each row.
We are looking for the speed of the trains. Let’s let r represent the speed of the local train. Since the speed of the express train is 12 mph faster, we represent that as $r + 12 .$

$\begin{matrix} r & = & speed of the local train \\ r + 12 & = & speed of the express train \end{matrix}$

Fill in the speeds into the chart.

A table with three rows and four columns. The first row is a header row and reads from left to right _____, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have Express and then Local. Below the Rate header cell, we have r plus 12 and then r. Below the Time header cell, we have 4 and then 5. The rest of the cells are blank.

Multiply the rate times the time to get the distance.

Step 4. Translate into an equation.

Restate the problem in one sentence with all the important information.
Then, translate the sentence into an equation.
The equation to model this situation will come from the relation between the distances. Look at the diagram we drew above. How is the distance travelled by the express train related to the distance travelled by the local train?
Since both trains leave from Pittsburgh and travel to Washington, D.C. they travel the same distance. So we write:

The sentence, “The distance traveled by the express train equals the distance traveled by the local train,” can be translated to an equation. Translate “distance traveled by the express train” to 4 times the quantity r plus 12, and translate “distance traveled by the local train” to 5r. The full equation is 4 times the quantity r plus 12 equals 5r.

Step 5. Solve the equation using good algebra techniques.

Now solve this equation.	So the speed of the local train is 48 mph.
Find the speed of the express train.	The speed of the express train is 60 mph.

Step 6. Check the answer in the problem and make sure it makes sense.

Comparison of express and local train travel, detailing speed, time, and how both achieve a 240-mile distance.
express train	$60 mph (4 hours) = 240 miles$
local train	$48 mph (5 hours) = 240 miles ✓$

Step 7. Answer the question with a complete sentence.

The speed of the local train is 48 mph and the speed of the express train is 60 mph.

In Example 1, the last example, we had two trains traveling the same distance. The diagram and the chart helped us write the equation we solved. Let’s see how this works in another case.

Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove 1.5 hours while his parents drove 1 hour to get to the restaurant. Christopher’s average speed was 10 miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

Draw a diagram to illustrate what it happening. Below shows a sketch of what is happening in the example.
Create a table to organize the information.
Label the columns rate, time, distance.
List the two scenarios.
Write in the information you know.

A table with three rows and four columns and an extra cell at the bottom of the fourth column. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have Christopher and Parents. Below the time header cell, we have 1.5 and 1. The extra cell contains 115. The rest of the cells are blank.

Step 2. Identify what we are looking for.

We are asked to find the average speeds of Christopher and his parents.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

Complete the chart.
Use variable expressions to represent that quantity in each row.
We are looking for their average speeds. Let’s let r represent the average speed of the parents. Since the Christopher’s speed is 10 mph faster, we represent that as $r + 10 .$

Fill in the speeds into the chart.

Multiply the rate times the time to get the distance.

Step 4. Translate into an equation.

Restate the problem in one sentence with all the important information.
Then, translate the sentence into an equation.
Again, we need to identify a relationship between the distances in order to write an equation. Look at the diagram we created above and notice the relationship between the distance Christopher traveled and the distance his parents traveled.

The distance Christopher travelled plus the distance his parents travel must add up to 115 miles. So we write:

The sentence, “The distance traveled by Christopher plus the distance traveled by his parents equals 115 miles,” can be translated to an equation. Translate “distance traveled by Christopher” to 1.5 times the quantity r plus 10, and translate “distance traveled by his parents” to r. The full equation is 1.5 times the quantity r plus 10, plus r equals 115.

Step 5. Solve the equation using good algebra techniques.

Step-by-step solution for a word problem calculating parent's and Christopher's speeds using an algebraic equation.
Now solve this equation.	$\begin{matrix} 1.5 (r + 10) + r & = & 115 \\ 1.5 r + 15 + r & = & 115 \\ 2.5 r + 15 & = & 115 \\ 2.5 r & = & 100 \\ r & = & 40 \end{matrix}$
	So the parents' speed was 40 mph.
Christopher's speed is $r + 10$ .	$\begin{matrix} r + 10 \\ 40 + 10 \\ 50 \end{matrix}$
	Christopher's speed was 50 mph.

Step 6. Check the answer in the problem and make sure it makes sense.

Calculated distances traveled by Christopher and his parents, summing up to a total journey distance.
Christopher drove	$\begin{matrix} 50 mph (1.5 hours) & = & 75 miles \end{matrix}$
His parents drove	$\begin{matrix} 40 mph (1 hours) & = & \underset{_______}{40 miles} \\ 115 miles \end{matrix}$

This table outlines a problem-solving step, specifically providing the calculated speeds for Christopher and his parents.
Step 7. Answer the question with a complete sentence.	Christopher's speed was 50 mph. His parents' speed was 40 mph.

As you read the next example, think about the relationship of the distances traveled. Which of the previous two examples is more similar to this situation?

Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at 70 mph and the truck traveling east has an average speed of 60 mph. How long will they travel before they are 325 miles apart?

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

Draw a diagram to illustrate what it happening.
Create a table to organize the information.

A table with three rows and four columns and an extra cell at the bottom of the fourth column. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have West and East. Below the rate header cell, we have 70 and 60. The extra cell contains 325. The rest of the cells are blank.

Step 2. Identify what we are looking for.

We are asked to find the amount of time the trucks will travel until they are 325 miles apart.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

We are looking for the time travelled. Both trucks will travel the same amount of time. Let’s call the time t. Since their speeds are different, they will travel different distances.
Complete the chart.

Step 4. Translate into an equation.

We need to find a relation between the distances in order to write an equation. Looking at the diagram, what is the relationship between the distance each of the trucks will travel?
The distance traveled by the truck going west plus the distance travelled by the truck going east must add up to 325 miles. So we write:

Distance traveled by westbound truck plus distance traveled by eastbound truck equals 325. The first part corresponds to 70t and the second part corresponds to 60.

Step 5. Solve the equation using good algebra techniques.

Step-by-step solution for the algebraic equation 70t + 60t = 325, demonstrating the process to find the value of 't'.
Now solve this equation.	$\begin{matrix} 70 t + 60 t & = & 325 \\ 130 t & = & 325 \\ t & = & 2.5 \end{matrix}$

So it will take the trucks 2.5 hours to be 325 miles apart.

Step 6. Check the answer in the problem and make sure it makes sense.

Distances covered by two trucks traveling in opposite directions over 2.5 hours, including total distance.
Truck going West	$\begin{matrix} 70 mph (2.5 hours) & = & 175 miles \end{matrix}$
Truck going East	$\begin{matrix} 60 mph (2.5 hours) & = & \underset{________}{150 miles} \\ 325 miles \end{matrix}$

This table presents a problem-solving step and its corresponding answer.
Step 7. Answer the question with a complete sentence.	It will take the trucks 2.5 hours to be 325 miles apart.

When Katie Mae walks to school, it takes her 30 minutes. If she rides her bike, it takes her 15 minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What are her walking speed and her speed riding her bike?

Solution

First, we draw a diagram that represents the situation to help us see what is happening.

A house and a school are represented by two separate lines. There is a line marked walking from the house to the school that takes 30 minutes. There is a line marked biking from the house to the school that take 15 minutes and is 3 mph faster. The space between the house and school is marked distance.

We are asked to find her speed walking and riding her bike. Let’s call her walking speed r. Since her biking speed is three miles per hour faster, we will call that speed $r + 3 .$ We write the speeds in the chart.

The speed is in miles per hour, so we need to express the times in hours, too, in order for the units to be the same. Remember, one hour is 60 minutes. So:

\begin{matrix} 30 minutes is \frac{30}{60} or \frac{1}{2} hour \\ 15 minutes is \frac{15}{60} or \frac{1}{4} hour \end{matrix}

Next, we multiply rate times time to fill in the distance column.

A table with three rows and four columns. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have walk and bike. Below the rate header cell, we have r and r plus 3. Below the time header cell, we have 1/2 and 1/4. Below the distance cell we have 1/2 times r and 1/4 times the quantity (r plus 3).

The equation will come from the fact that the distance from Katie Mae’s home to her school is the same whether she is walking or riding her bike.

So we say:


Translate into an equation.
Solve this equation.
Clear the fractions by multiplying by the LCD of all the fractions in the equation.
Simplify.	6 mph (Katie Mae's biking speed)
Let's check if this works. Walk 3 mph (0.5 hour) = 1.5 miles Bike 6 mph (0.25 hour) = 1.5 miles
Yes, either way Katie Mae travels 1.5 miles to school.	Katie Mae’s walking speed is 3 mph. Her speed riding her bike is 6 mph.

In the distance, rate, and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.

Hamilton loves to travel to Las Vegas, 255 miles from his home in Orange County. On his last trip, he left his house at 2:00 pm. The first part of his trip was on congested city freeways. At 4:00 pm, the traffic cleared and he was able to drive through the desert at a speed 1.75 times as fast as when he drove in the congested area. He arrived in Las Vegas at 6:30 pm. How fast was he driving during each part of his trip?

Solution

A diagram will help us model this trip.

Home (2:00 pm) and Las Vegas (6:30 pm) are represented by two separate lines. The space between home and Las Vegas is marked 255 miles. There is an arrow marked city driving from Home/2:00 pm to 4:00 pm. Then there is an arrow marked desert driving from the tip of the previous one at 4:00 pm to Las Vegas/6:30 pm.

Next, we create a table to organize the information.

We know the total distance is 255 miles. We are looking for the rate of speed for each part of the trip. The rate in the desert is 1.75 times the rate in the city. If we let $r =$ the rate in the city, then the rate in the desert is $1.75 r .$

The times here are given as clock times. Hamilton started from home at 2:00 pm and entered the desert at 4:30 pm. So he spent two hours driving the congested freeways in the city. Then he drove faster from 4:00 pm until 6:30 pm in the desert. So he drove 2.5 hours in the desert.

Now, we multiply the rates by the times.

A table with three rows and four columns and an extra cell at the bottom of the fourth column. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have city and desert. Below the rate header cell, we have r and 1.75r. Below the time head cell, we have 2 and 2.5. Below the Distance header cell we have 2r, 2.5 times 1.75r, and 255.

By looking at the diagram below, we can see that the sum of the distance driven in the city and the distance driven in the desert is 255 miles.


Translate into an equation.
Solve this equation.
Check.
	Hamilton drove 40 mph in the city and 70 mph in the desert.

Key Concepts

Distance, Rate, and Time
- D = rt where D = distance, r = rate, t = time
Problem-Solving Strategy—Distance, Rate, and Time Applications
1. Read the problem. Make sure all the words and ideas are understood.
  Draw a diagram to illustrate what it happening.
  Create a table to organize the information: Label the columns rate, time, distance. List the two scenarios. Write in the information you know.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
  Complete the chart.
  Use variable expressions to represent that quantity in each row.
  Multiply the rate times the time to get the distance.
4. Translate into an equation.
  Restate the problem in one sentence with all the important information.
  Then, translate the sentence into an equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Practice Makes Perfect

Solve Uniform Motion Applications

In the following exercises, solve.

Lilah is moving from Portland to Seattle. It takes her three hours to go by train. Mason leaves the train station in Portland and drives to the train station in Seattle with all Lilah’s boxes in his car. It takes him 2.4 hours to get to Seattle, driving at 15 miles per hour faster than the speed of the train. Find Mason’s speed and the speed of the train.

Solution

Mason 75 mph, train 60 mph

Kathy and Cheryl are walking in a fundraiser. Kathy completes the course in 4.8 hours and Cheryl completes the course in 8 hours. Kathy walks two miles per hour faster than Cheryl. Find Kathy’s speed and Cheryl’s speed.

Two busses go from Sacramento for San Diego. The express bus makes the trip in 6.8 hours and the local bus takes 10.2 hours for the trip. The speed of the express bus is 25 mph faster than the speed of the local bus. Find the speed of both busses.

Solution

express bus 75mph, local 50 mph

A commercial jet and a private airplane fly from Denver to Phoenix. It takes the commercial jet 1.1 hours for the flight, and it takes the private airplane 1.8 hours. The speed of the commercial jet is 210 miles per hour faster than the speed of the private airplane. Find the speed of both airplanes.

Saul drove his truck 3 hours from Dallas towards Kansas City and stopped at a truck stop to get dinner. At the truck stop he met Erwin, who had driven 4 hours from Kansas City towards Dallas. The distance between Dallas and Kansas City is 542 miles, and Erwin’s speed was eight miles per hour slower than Saul’s speed. Find the speed of the two truckers.

Solution

Saul 82 mph, Erwin 74 mph

Charlie and Violet met for lunch at a restaurant between Memphis and New Orleans. Charlie had left Memphis and drove 4.8 hours towards New Orleans. Violet had left New Orleans and drove 2 hours towards Memphis, at a speed 10 miles per hour faster than Charlie’s speed. The distance between Memphis and New Orleans is 394 miles. Find the speed of the two drivers.

Sisters Helen and Anne live 332 miles apart. For Thanksgiving, they met at their other sister’s house partway between their homes. Helen drove 3.2 hours and Anne drove 2.8 hours. Helen’s average speed was four miles per hour faster than Anne’s. Find Helen’s average speed and Anne’s average speed.

Solution

Helen 57.2 mph, Anne 53.2 mph

Ethan and Leo start riding their bikes at the opposite ends of a 65-mile bike path. After Ethan has ridden 1.5 hours and Leo has ridden 2 hours, they meet on the path. Ethan’s speed is six miles per hour faster than Leo’s speed. Find the speed of the two bikers.

Elvira and Aletheia live 3.1 miles apart on the same street. They are in a study group that meets at a coffee shop between their houses. It took Elvira half an hour and Aletheia two-thirds of an hour to walk to the coffee shop. Aletheia’s speed is 0.6 miles per hour slower than Elvira’s speed. Find both women’s walking speeds.

Solution

Aletheia 2.4 mph, Elvira 3 mph

DaMarcus and Fabian live 23 miles apart and play soccer at a park between their homes. DaMarcus rode his bike for three-quarters of an hour and Fabian rode his bike for half an hour to get to the park. Fabian’s speed was six miles per hour faster than DaMarcus’ speed. Find the speed of both soccer players.

Cindy and Richard leave their dorm in Charleston at the same time. Cindy rides her bicycle north at a speed of 18 miles per hour. Richard rides his bicycle south at a speed of 14 miles per hour. How long will it take them to be 96 miles apart?

Solution

3 hours

Matt and Chris leave their uncle’s house in Phoenix at the same time. Matt drives west on I-60 at a speed of 76 miles per hour. Chris drives east on I-60 at a speed of 82 miles per hour. How many hours will it take them to be 632 miles apart?

Two busses leave Billings at the same time. The Seattle bus heads west on I-90 at a speed of 73 miles per hour while the Chicago bus heads east at a speed of 79 miles an hour. How many hours will it take them to be 532 miles apart?

Solution

3.5 hours

Two boats leave the same dock in Cairo at the same time. One heads north on the Mississippi River while the other heads south. The northbound boat travels four miles per hour. The southbound boat goes eight miles per hour. How long will it take them to be 54 miles apart?

Lorena walks the path around the park in 30 minutes. If she jogs, it takes her 20 minutes. Her jogging speed is 1.5 miles per hour faster than her walking speed. Find Lorena’s walking speed and jogging speed.

Solution

walking 3 mph, jogging 4.5 mph

Julian rides his bike uphill for 45 minutes, then turns around and rides back downhill. It takes him 15 minutes to get back to where he started. His uphill speed is 3.2 miles per hour slower than his downhill speed. Find Julian’s uphill and downhill speed.

Cassius drives his boat upstream for 45 minutes. It takes him 30 minutes to return downstream. His speed going upstream is three miles per hour slower than his speed going downstream. Find his upstream and downstream speeds.

Solution

upstream 6 mph, downstream 9 mph

It takes Darline 20 minutes to drive to work in light traffic. To come home, when there is heavy traffic, it takes her 36 minutes. Her speed in light traffic is 24 miles per hour faster than her speed in heavy traffic. Find her speed in light traffic and in heavy traffic.

At 1:30 Marlon left his house to go to the beach, a distance of 7.6 miles. He rode his skateboard until 2:15, then walked the rest of the way. He arrived at the beach at 3:00. Marlon’s speed on his skateboard is 2.5 times his walking speed. Find his speed when skateboarding and when walking.

Solution

skateboarding 7.2 mph, walking 2.9 mph

Aaron left at 9:15 to drive to his mountain cabin 108 miles away. He drove on the freeway until 10:45, and then he drove on the mountain road. He arrived at 11:05. His speed on the freeway was three times his speed on the mountain road. Find Aaron’s speed on the freeway and on the mountain road.

Marisol left Los Angeles at 2:30 to drive to Santa Barbara, a distance of 95 miles. The traffic was heavy until 3:20. She drove the rest of the way in very light traffic and arrived at 4:20. Her speed in heavy traffic was 40 miles per hour slower than her speed in light traffic. Find her speed in heavy traffic and in light traffic.

Solution

heavy traffic 30 mph, light traffic 70 mph

Lizette is training for a marathon. At 7:00 she left her house and ran until 8:15, then she walked until 11:15. She covered a total distance of 19 miles. Her running speed was five miles per hour faster than her walking speed. Find her running and walking speeds.

Everyday Math

John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50. At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18.

ⓐ What was his average speed on the drive from Irvine to Los Angeles?
ⓑ What was his average speed on the drive from Los Angeles to Irvine?
ⓒ What was the total time he spent driving to and from this meeting?
ⓓ John drove a total of 90 miles roundtrip. Find his average speed. (Round to the nearest tenth.)

Solution

ⓐ 36 mph ⓑ 25 mph ⓒ 3.05 hours ⓓ 29.5 mph

Sarah wants to arrive at her friend’s wedding at 3:00. The distance from Sarah’s house to the wedding is 95 miles. Based on usual traffic patterns, Sarah predicts she can drive the first 15 miles at 60 miles per hour, the next 10 miles at 30 miles per hour, and the remainder of the drive at 70 miles per hour.

ⓐ How long will it take Sarah to drive the first 15 miles?
ⓑ How long will it take Sarah to drive the next 10 miles?
ⓒ How long will it take Sarah to drive the rest of the trip?
ⓓ What time should Sarah leave her house?

Writing Exercises

When solving a uniform motion problem, how does drawing a diagram of the situation help you?

Solution

Answers will vary.

When solving a uniform motion problem, how does creating a table help you?

Self Check

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

A table with two rows and four columns. The first row reads: 'I can...', 'Confidently', 'With Some Help', and 'No-I don't get it!' The second row reads: 'solve uniform motion applications' with the rest of the cells blank for students to fill out.

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

Solve Applications with Linear Inequalities

Many real-life situations require us to solve inequalities. In fact, inequality applications are so common that we often do not even realize we are doing algebra. For example, how many gallons of gas can be put in the car for $20? Is the rent on an apartment affordable? Is there enough time before class to go get lunch, eat it, and return? How much money should each family member’s holiday gift cost without going over budget?

The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations. We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality.

Emma got a new job and will have to move. Her monthly income will be $5,625. To qualify to rent an apartment, Emma’s monthly income must be at least three times as much as the rent. What is the highest rent Emma will qualify for?

Solution

This table details a 7-step process for solving word problems with inequalities, exemplified by determining the maximum affordable rent.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the highest rent Emma will qualify for
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $r =$ the rent.
Step 4. Translate into an inequality. First write a sentence that gives the information to find it.	Emma's monthly income must be at least three times the rent.
Step 5. Solve the inequality. Remember, $a > x$ has the same meaning as $x < a$ .	$\begin{matrix} 5,625 & \geq & 3 r \\ 1,875 & \geq & r \\ r & \leq & 1,875 \end{matrix}$
Step 6. Check the answer in the problem and make sure it makes sense. A maximum rent of $1,875 seems reasonable for an income of $5,625.
Step 7. Answer the question with a complete sentence.	The maximum rent is $1,875.

Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down. To check applications like this, we will round our answer to a number that is easy to compute with and make sure that number makes the inequality true.

Dawn won a mini-grant of $4,000 to buy tablet computers for her classroom. The tablets she would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets Dawn can buy?

Solution

Steps to solve a word problem using inequalities, exemplified by determining the maximum number of tablets Dawn can buy within a budget.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the maximum number of tablets Dawn can buy
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $n =$ the number of tablets.
Step 4. Translate. Write a sentence that gives the information to find it. Translate into an inequality.	$254.12 times the number of tablets is no more than $4,000. $254.12 n \leq 4,000$
Step 5. Solve the inequality. But $n$ must be a whole number of tablets, so round to 15.	$n \leq 15.74$ $n \leq 15$
Step 6. Check the answer in the problem and make sure it makes sense. Rounding down the price to $250, 15 tablets would cost $3,750, while 16 tablets would be $4,000. So a maximum of 15 tablets at $254.12 seems reasonable.
Step 7. Answer the question with a complete sentence.	Dawn can buy a maximum of 15 tablets.

Pete works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925?

Solution

A step-by-step guide demonstrating how to solve a word problem involving an inequality for sales commission.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the total sales needed for his variable pay option to exceed the fixed amount of $925
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $s =$ the total sales.
Step 4. Translate. Write a sentence that gives the information to find it. Translate into an inequality. Remember to convert the percent to a decimal.	$500 + 0.12 s > 925$
Step 5. Solve the inequality.	$\begin{matrix} 0.12 s & > & 425 \\ s & > & 3,541. \overset{—}{66} \end{matrix}$
Step 6. Check the answer in the problem and make sure it makes sense. If we round the total sales up to $4,000, we see that $500 + 0.12 (4,000) = 980$ , which is more than $925.
Step 7. Answer the question with a complete sentence.	The total sales must be more than $3,541.67.

Sergio and Lizeth have a very tight vacation budget. They plan to rent a car from a company that charges $75 a week plus $0.25 a mile. How many miles can they travel and still keep within their $200 budget?

Solution

Step-by-step solution for an inequality word problem calculating maximum travel distance within a budget.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the number of miles Sergio and Lizeth can travel
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $m =$ the number of miles.
Step 4. Translate. Write a sentence that gives the information to find it. Translate into an inequality.	$75 plus 0.25 times the number of miles is less than or equal to $200. $75 + 0.25 m \leq 200$
Step 5. Solve the inequality.	$\begin{matrix} 0.25 m & \leq & 125 \\ m & \leq & 500 miles \end{matrix}$
Step 6. Check the answer in the problem and make sure it makes sense. Yes, $75 + 0.25 (500) = 200$ .
Step 7. Write a sentence that answers the question.	Sergio and Lizeth can travel 500 miles and still stay on budget.

A common goal of most businesses is to make a profit. Profit is the money that remains when the expenses have been subtracted from the money earned. In the next example, we will find the number of jobs a small businessman needs to do every month in order to make a certain amount of profit.

Elliot has a landscape maintenance business. His monthly expenses are $1,100. If he charges $60 per job, how many jobs must he do to earn a profit of at least $4,000 a month?

Solution

This table provides a step-by-step guide on how to solve a word problem by setting up and solving an inequality, using Elliot's job earnings as an example.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the number of jobs Elliot needs
Step 3. Name what we are looking for. Choose a variable to represent it.	Let $j =$ the number of jobs.
Step 4. Translate Write a sentence that gives the information to find it.	$60 times the number of jobs minus $1,100 is at least $4,000.
Translate into an inequality.

Step 5. Solve the inequality.
Step 6. Check the answer in the problem and make sure it makes sense. If Elliot did 90 jobs, his profit would be $60 (90) - 1,100$ , or $$4,300$ . This is more than $$4,000$ .
Step 7. Write a sentence that answer the question.	Elliot must work at least 85 jobs.

Sometimes life gets complicated! There are many situations in which several quantities contribute to the total expense. We must make sure to account for all the individual expenses when we solve problems like this.

Brenda’s best friend is having a destination wedding and the event will require 3 nights in a hotel. Brenda has $500 in savings and can earn $15 an hour babysitting. She expects to pay $350 airfare, $375 for food and entertainment and $60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip?

Solution

This table outlines a 7-step process for solving an inequality word problem, demonstrating each step with an example about babysitting hours.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	the number of hours Brenda must babysit
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $h =$ the number of hours.
Step 4. Translate. Write a sentence that gives the information to find it. Translate into an inequality.	The expenses must be less than or equal to the income. The cost of airfare plus the cost of food and entertainment and the hotel bill must be less than or equal to the savings plus the amount earned babysitting. $$ 350 + $ 375 + $ 60 (3) \leq $ 500 + $ 15 h$
Step 5. Solve the inequality.	$\begin{matrix} 905 & \leq & 500 + 15 h \\ 405 & \leq & 15 h \\ 27 & \leq & h \\ h & \geq & 27 \end{matrix}$
Step 6. Check the answer in the problem and make sure it makes sense. We substitute 27 into the inequality. $\begin{matrix} 905 \leq 500 + 15 h \\ 905 \leq 500 + 15 (27) \\ 905 \leq 905 \end{matrix}$
Step 7. Write a sentence that answers the question.	Brenda must babysit at least 27 hours.

Key Concepts

Solving inequalities
1. Read the problem.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate. Write a sentence that gives the information to find it. Translate into an inequality.
5. Solve the inequality.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Section Exercises

Practice Makes Perfect

Solve Applications with Linear Inequalities

In the following exercises, solve.

Mona is planning her son’s birthday party and has a budget of $285. The Fun Zone charges $19 per child. How many children can she have at the party and stay within her budget?

Solution

15 children

Carlos is looking at apartments with three of his friends. They want the monthly rent to be no more than $2360. If the roommates split the rent evenly among the four of them, what is the maximum rent each will pay?

A water taxi has a maximum load of 1,800 pounds. If the average weight of one person is 150 pounds, how many people can safely ride in the water taxi?

Solution

12 people

Marcela is registering for her college classes, which cost $105 per unit. How many units can she take to have a maximum cost of $1,365?

Arleen got a $20 gift card for the coffee shop. Her favorite iced drink costs $3.79. What is the maximum number of drinks she can buy with the gift card?

Solution

five drinks

Teegan likes to play golf. He has budgeted $60 next month for the driving range. It costs him $10.55 for a bucket of balls each time he goes. What is the maximum number of times he can go to the driving range next month?

Joni sells kitchen aprons online for $32.50 each. How many aprons must she sell next month if she wants to earn at least $1,000?

Solution

31 aprons

Ryan charges his neighbors $17.50 to wash their car. How many cars must he wash next summer if his goal is to earn at least $1,500?

Keshad gets paid $2,400 per month plus 6% of his sales. His brother earns $3,300 per month. For what amount of total sales will Keshad’s monthly pay be higher than his brother’s monthly pay?

Solution

$15,000

Kimuyen needs to earn $4,150 per month in order to pay all her expenses. Her job pays her $3,475 per month plus 4% of her total sales. What is the minimum Kimuyen’s total sales must be in order for her to pay all her expenses?

Andre has been offered an entry-level job. The company offered him $48,000 per year plus 3.5% of his total sales. Andre knows that the average pay for this job is $62,000. What would Andre’s total sales need to be for his pay to be at least as high as the average pay for this job?

Solution

$400,000

Nataly is considering two job offers. The first job would pay her $83,000 per year. The second would pay her $66,500 plus 15% of her total sales. What would her total sales need to be for her salary on the second offer be higher than the first?

Jake’s water bill is $24.80 per month plus $2.20 per ccf (hundred cubic feet) of water. What is the maximum number of ccf Jake can use if he wants his bill to be no more than $60?

Solution

16 ccf

Kiyoshi’s phone plan costs $17.50 per month plus $0.15 per text message. What is the maximum number of text messages Kiyoshi can use so the phone bill is no more than $56.50?

Marlon’s TV plan costs $49.99 per month plus $5.49 per first-run movie. How many first-run movies can he watch if he wants to keep his monthly bill to be a maximum of $100?

Solution

nine movies

Kellen wants to rent a banquet room in a restaurant for her cousin’s baby shower. The restaurant charges $350 for the banquet room plus $32.50 per person for lunch. How many people can Kellen have at the shower if she wants the maximum cost to be $1,500?

Moshde runs a hairstyling business from her house. She charges $45 for a haircut and style. Her monthly expenses are $960. She wants to be able to put at least $1,200 per month into her savings account order to open her own salon. How many “cut & styles” must she do to save at least $1,200 per month?

Solution

48 cut & styles

Noe installs and configures software on home computers. He charges $125 per job. His monthly expenses are $1,600. How many jobs must he work in order to make a profit of at least $2,400?

Katherine is a personal chef. She charges $115 per four-person meal. Her monthly expenses are $3,150. How many four-person meals must she sell in order to make a profit of at least $1,900?

Solution

44 meals

Melissa makes necklaces and sells them online. She charges $88 per necklace. Her monthly expenses are $3745. How many necklaces must she sell if she wants to make a profit of at least $1,650?

Five student government officers want to go to the state convention. It will cost them $110 for registration, $375 for transportation and food, and $42 per person for the hotel. There is $450 budgeted for the convention in the student government savings account. They can earn the rest of the money they need by having a car wash. If they charge $5 per car, how many cars must they wash in order to have enough money to pay for the trip?

Solution

49 cars

Cesar is planning a 4-day trip to visit his friend at a college in another state. It will cost him $198 for airfare, $56 for local transportation, and $45 per day for food. He has $189 in savings and can earn $35 for each lawn he mows. How many lawns must he mow to have enough money to pay for the trip?

Alonzo works as a car detailer. He charges $175 per car. He is planning to move out of his parents’ house and rent his first apartment. He will need to pay $120 for application fees, $950 for security deposit, and first and last months’ rent at $1,140 per month. He has $1,810 in savings. How many cars must he detail to have enough money to rent the apartment?

Solution

9 cars

Eun-Kyung works as a tutor and earns $60 per hour. She has $792 in savings. She is planning an anniversary party for her parents. She would like to invite 40 guests. The party will cost her $1,520 for food and drinks and $150 for the photographer. She will also have a favor for each of the guests, and each favor will cost $7.50. How many hours must she tutor to have enough money for the party?

Everyday Math

Maximum Load on a Stage In 2014, a high school stage collapsed in Fullerton, California, when 250 students got on stage for the finale of a musical production. Two dozen students were injured. The stage could support a maximum of 12,750 pounds. If the average weight of a student is assumed to be 140 pounds, what is the maximum number of students who could safely be on the stage?

Solution

91 students

Maximum Weight on a Boat In 2004, a water taxi sank in Baltimore harbor and five people drowned. The water taxi had a maximum capacity of 3,500 pounds (25 people with average weight 140 pounds). The average weight of the 25 people on the water taxi when it sank was 168 pounds per person. What should the maximum number of people of this weight have been?

Wedding Budget Adele and Walter found the perfect venue for their wedding reception. The cost is $9,850 for up to 100 guests, plus $38 for each additional guest. How many guests can attend if Adele and Walter want the total cost to be no more than $12,500?

Solution

169 guests

Shower Budget Penny is planning a baby shower for her daughter-in-law. The restaurant charges $950 for up to 25 guests, plus $31.95 for each additional guest. How many guests can attend if Penny wants the total cost to be no more than $1,500?

Writing Exercises

Find your last month’s phone bill and the hourly salary you are paid at your job. (If you do not have a job, use the hourly salary you would realistically be paid if you had a job.) Calculate the number of hours of work it would take you to earn at least enough money to pay your phone bill by writing an appropriate inequality and then solving it.

Solution

Answers will vary.

Find out how many units you have left, after this term, to achieve your college goal and estimate the number of units you can take each term in college. Calculate the number of terms it will take you to achieve your college goal by writing an appropriate inequality and then solving it.

Self Check

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

A self-assessment chart with the statement 'I can... solve applications with linear inequalities.' and three options for evaluating understanding: '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?

Chapter 3 Review Exercises

3.1 Using a Problem Solving Strategy

Approach Word Problems with a Positive Attitude

In the following exercises, reflect on your approach to word problems.

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

116

There are nine saxophone players in the band. The number of saxophone players is one less than twice the number of tuba players. Find the number of tuba players.

Solve Number Problems

In the following exercises, solve each number word problem.

The sum of a number and three is forty-one. Find the number.

Solution

Twice the difference of a number and ten is fifty-four. Find the number.

One number is nine less than another. Their sum is negative twenty-seven. Find the numbers.

Solution

$−18, −9$

One number is eleven more than another. If their sum is increased by seventeen, the result is 90. Find the numbers.

One number is two more than four times another. Their sum is $−13 .$ Find the numbers.

Solution

$−3, −10$

The sum of two consecutive integers is $−135 .$ Find the numbers.

Find three consecutive integers whose sum is $−141 .$

Solution

$−48, −47, −46$

Find three consecutive even integers whose sum is 234.

Find three consecutive odd integers whose sum is 51.

Solution

15, 17, 19

Koji has $5,502 in his savings account. This is $30 less than six times the amount in his checking account. How much money does Koji have in his checking account?

3.2 Solve Percent Applications

Translate and Solve Basic Percent Equations

In the following exercises, translate and solve.

What number is 67% of 250?

Solution

167.5

300% of 82 is what number?

12.5% of what number is 20?

Solution

160

72 is 30% of what number?

What percent of 125 is 150?

Solution

120%

127.5 is what percent of 850?

Solve Percent Applications

In the following exercises, solve.

The bill for Dino’s lunch was $19.45. He wanted to leave 20% of the total bill as a tip. How much should the tip be?

Solution

$3.89

Reza was very sick and lost 15% of his original weight. He lost 27 pounds. What was his original weight?

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?

Solution

$875

Jaden earns $2,680 per month. He pays $938 a month for rent. What percent of his monthly pay goes to rent?

Find Percent Increase and Percent Decrease

In the following exercises, solve.

Angel’s got a raise in his annual salary from $55,400 to $56,785. Find the percent increase.

Solution

2.5%

Rowena’s monthly gasoline bill dropped from $83.75 last month to $56.95 this month. Find the percent decrease.

Solve Simple Interest Applications

In the following exercises, solve.

Winston deposited $3,294 in a bank account with interest rate 2.6%. How much interest was earned in 5 years?

Solution

$428.22

Moira borrowed $4,500 from her grandfather to pay for her first year of college. Three years later, she repaid the $4,500 plus $243 interest. What was the rate of interest?

Jaime’s refrigerator loan statement said he would pay $1,026 in interest for a 4-year loan at 13.5%. How much did Jaime borrow to buy the refrigerator?

Solution

$1,900

In 12 years, a bond that paid 6.35% interest earned $7,620 interest. What was the principal of the bond?

Solve Applications with Discount or Mark-up

In the following exercises, find the sale price.

The original price of a handbag was $84. Carole bought it on sale for $21 off.

Solution

$63

Marian wants to buy a coffee table that costs $495. Next week the coffee table will be on sale for $149 off.

In the following exercises, find ⓐ the amount of discount and ⓑ the sale price.

Emmett bought a pair of shoes on sale at 40% off from an original price of $138.

Solution

ⓐ $55.20 ⓑ $82.80

Anastasia bought a dress on sale at 75% off from an original price of $280.

In the following exercises, find ⓐ the amount of discount and ⓑ the discount rate. (Round to the nearest tenth of a percent, if needed.)

Zack bought a printer for his office that was on sale for $380. The original price of the printer was $450.

Solution

ⓐ $70 ⓑ 15.6%

Lacey bought a pair of boots on sale for $95. The original price of the boots was $200.

In the following exercises, find ⓐ the amount of the mark-up and ⓑ the list price.

Nga and Lauren bought a chest at a flea market for $50. They re-finished it and then added a 350% mark-up.

Solution

ⓐ $175 ⓑ $225

Carly bought bottled water for $0.24 per bottle at the discount store. She added a 75% mark-up before selling them at the football game.

3.3 Solve Mixture 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 five 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 eight 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

six $5 bills, 11 $10 bills

Lenny has $3.69 in pennies, dimes, and quarters. The number of pennies is three more than the number of dimes. The number of quarters is twice the number of dimes. How many of each coin does he have?

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 three less than 10 times the number of adults. The total amount of money from ticket sales was $619. How many of each ticket were sold?

125 tickets were sold for the jazz band concert for a total of $1,022. Student tickets cost $6 each and general admission tickets cost $10 each. How many of each kind of ticket were sold?

Solution

57 students, 68 adults

One afternoon the water park sold 525 tickets for a total of $13,545. Child tickets cost $19 each and adult tickets cost $40 each. How many of each kind of ticket were sold?

Ana spent $4.06 buying stamps. The number of $0.41 stamps she bought was five more than the number of $0.26 stamps. How many of each did she buy?

Solution

three $0.26 stamps, eight $0.41 stamps

Yumi spent $34.15 buying stamps. The number of $0.56 stamps she bought was 10 less than four times the number of $0.41 stamps. How many of each did she buy?

Solve Mixture Word Problems

In the following exercises, solve each mixture word problem.

Marquese is making 10 pounds of trail mix from raisins and nuts. Raisins cost $3.45 per pound and nuts cost $7.95 per pound. How many pounds of raisins and how many pounds of nuts should Marquese use for the trail mix to cost him $6.96 per pound?

Solution

2.2 lb. of raisins, 7.8 lb. of nuts

Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tile. She will use basic tiles that cost $8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be $10 per square foot?

Shawn has $15,000 to invest. She will put some of it into a fund that pays 4.5% annual interest and the rest in a certificate of deposit that pays 1.8% annual interest. How much should she invest in each account if she wants to earn 4.05% annual interest on the total amount?

Solution

$12,500 at 4.5%, $2,500 at 1.8%

Enrique borrowed $23,500 to buy a car. He pays his uncle 2% interest on the $4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total $23,500? (Round your answer to the nearest tenth of a percent.)

3.4 Solve Geometry Applications: Triangles, Rectangles and the Pythagorean Theorem

Solve Applications Using Triangle Properties

In the following exercises, solve using triangle properties.

The measures of two angles of a triangle are 22 and 85 degrees. Find the measure of the third angle.

Solution

$73 °$

The playground at a shopping mall is a triangle with perimeter 48 feet. The lengths of two sides are 19 feet and 14 feet. How long is the third side?

A triangular road sign has base 30 inches and height 40 inches. What is its area?

Solution

600 square inches

What is the height of a triangle with area 67.5 square meters and base 9 meters?

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 right triangle measures $58 ° .$ What is the measure of the other angles of the triangle?

The measure of the smallest angle in a right triangle is $45 °$ less than the measure of the next larger angle. Find the measures of all three angles.

Solution

$22.5 °, 67.5 °, 90 °$

The perimeter of a triangle is 97 feet. One side of the triangle is eleven feet more than the smallest side. The third side is six feet more than twice the smallest side. Find the lengths of all sides.

Use the Pythagorean Theorem

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

A right triangle with one leg labeled 10 and the other labeled 24.

Solution

A right-angled triangle with legs measuring 6 and 8 units, featuring a square symbol indicating the 90-degree angle.

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-angled triangle showing two sides with lengths 15 and 17. The right angle is indicated, suggesting a Pythagorean theorem problem to find the third side.

Solution

A right triangle with one leg labeled 15 and the hypotenuse labeled 25.

A right-angled triangle showing legs of length 4 and 7 units, indicating a geometry problem or concept.

Solution

8.1

A right triangle with one leg labeled 10 and the other labeled 11.

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?

A right-angled triangle with one leg measuring 8 feet and the hypotenuse measuring 10 feet. The right angle is indicated at the bottom left vertex of the triangle.

Solution

$6^{'}$

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 right triangle with 36 inches labeled on one leg and 15 inches labeled on the other. A dotted line is drawn parallel to the two triangle legs, forming a rectangle.

Solve Applications Using Rectangle Properties

In the following exercises, solve using rectangle properties.

The length of a rectangle is 36 feet and the width is 19 feet. Find the ⓐ perimeter ⓑ area.

Solution

ⓐ 110 ft. ⓑ 684 sq. ft.

A sidewalk in front of Kathy’s house is in the shape of a rectangle four feet wide by 45 feet long. Find the ⓐ perimeter ⓑ area.

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 2,700 square centimeters. What is the length?

The length of a rectangle is 12 cm more than the width. The perimeter is 74 cm. Find the length and the width.

Solution

24.5 cm, 12.5 cm

The width of a rectangle is three more than twice the length. The perimeter is 96 inches. Find the length and the width.

3.5 Solve Uniform Motion Applications

Solve Uniform Motion Applications

In the following exercises, solve.

When Gabe drives from Sacramento to Redding it takes him 2.2 hours. It takes Elsa 2 hours to drive the same distance. Elsa’s speed is seven miles per hour faster than Gabe’s speed. Find Gabe’s speed and Elsa’s speed.

Solution

Gabe 70 mph, Elsa 77 mph

Louellen and Tracy met at a restaurant on the road between Chicago and Nashville. Louellen had left Chicago and drove 3.2 hours towards Nashville. Tracy had left Nashville and drove 4 hours towards Chicago, at a speed one mile per hour faster than Louellen’s speed. The distance between Chicago and Nashville is 472 miles. Find Louellen’s speed and Tracy’s speed.

Two busses leave Amarillo at the same time. The Albuquerque bus heads west on the I-40 at a speed of 72 miles per hour, and the Oklahoma City bus heads east on the I-40 at a speed of 78 miles per hour. How many hours will it take them to be 375 miles apart?

Solution

2.5 hours

Kyle rowed his boat upstream for 50 minutes. It took him 30 minutes to row back downstream. His speed going upstream is two miles per hour slower than his speed going downstream. Find Kyle’s upstream and downstream speeds.

At 6:30, Devon left her house and rode her bike on the flat road until 7:30. Then she started riding uphill and rode until 8:00. She rode a total of 15 miles. Her speed on the flat road was three miles per hour faster than her speed going uphill. Find Devon’s speed on the flat road and riding uphill.

Solution

flat road 11 mph, uphill 8 mph

Anthony drove from New York City to Baltimore, a distance of 192 miles. He left at 3:45 and had heavy traffic until 5:30. Traffic was light for the rest of the drive, and he arrived at 7:30. His speed in light traffic was four miles per hour more than twice his speed in heavy traffic. Find Anthony’s driving speed in heavy traffic and light traffic.

3.6 Solve Applications with Linear Inequalities

Solve Applications with Linear Inequalities

In the following exercises, solve.

Julianne has a weekly food budget of $231 for her family. If she plans to budget the same amount for each of the seven days of the week, what is the maximum amount she can spend on food each day?

Solution

$33 per day

Rogelio paints watercolors. He got a $100 gift card to the art supply store and wants to use it to buy $12 ″ \times 16 ″$ canvases. Each canvas costs $10.99. What is the maximum number of canvases he can buy with his gift card?

Briana has been offered a sales job in another city. The offer was for $42,500 plus 8% of her total sales. In order to make it worth the move, Briana needs to have an annual salary of at least $66,500. What would her total sales need to be for her to move?

Solution

at least $300,000

Renee’s car costs her $195 per month plus $0.09 per mile. How many miles can Renee drive so that her monthly car expenses are no more than $250?

Costa is an accountant. During tax season, he charges $125 to do a simple tax return. His expenses for buying software, renting an office, and advertising are $6,000. How many tax returns must he do if he wants to make a profit of at least $8,000?

Solution

at least 112 jobs

Jenna is planning a 5-day resort vacation with three of her friends. It will cost her $279 for airfare, $300 for food and entertainment, and $65 per day for her share of the hotel. She has $550 saved towards her vacation and can earn $25 per hour as an assistant in her uncle’s photography studio. How many hours must she work in order to have enough money for her vacation?

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?

Solution

One number is three more than twice another. Their sum is $−63 .$ Find the numbers.

The sum of two consecutive odd integers is $−96 .$ Find the numbers.

Solution

$−49, −47$

Marla’s breakfast was 525 calories. This was 35% of her total calories for the day. How many calories did she have that day?

Humberto’s hourly pay increased from $16.25 to $17.55. Find the percent increase.

Solution

Melinda deposited $5,985 in a bank account with an interest rate of 1.9%. How much interest was earned in 2 years?

Dotty bought a freezer on sale for $486.50. The original price of the freezer was $695. Find ⓐ the amount of discount and ⓑ the discount rate.

Solution

ⓐ $208.50 ⓑ 30%

Bonita has $2.95 in dimes and quarters in her pocket. If she has five more dimes than quarters, how many of each coin does she have?

At a concert, $1,600 in tickets were sold. Adult tickets were $9 each and children’s tickets were $4 each. If the number of adult tickets was 30 less than twice the number of children’s tickets, how many of each kind were sold?

Solution

140 adult, 85 children

Kim is making eight gallons of punch from fruit juice and soda. The fruit juice costs $6.04 per gallon and the soda costs $4.28 per gallon. How much fruit juice and how much soda should she use so that the punch costs $5.71 per gallon?

The measure of one angle of a triangle is twice the measure of the smallest angle. The measure of the third angle is 14 more than the measure of the smallest angle. Find the measures of all three angles.

Solution

$41.5 °, 55.5 °, 83 °$

What is the height of a triangle with area 277.2 square inches and base 44 inches?

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-angled triangle with one leg measuring 24 units and the hypotenuse measuring 26 units.

Solution

A right triangle with one leg labeled 6 and the other leg labeled 9.

A baseball diamond is really a square with sides of 90 feet. How far is it from home plate to second base, as shown?

The image shows a simple baseball field laid out as a perfect square. Home, 1st, 2nd, and 3rd base are labeled. A dotted line runs diagonally from home to 2nd base.

Solution

127.3 ft.

The length of a rectangle is two feet more than five times the width. The perimeter is 40 feet. Find the dimensions of the rectangle.

Two planes leave Dallas at the same time. One heads east at a speed of 428 miles per hour. The other plane heads west at a speed of 382 miles per hour. How many hours will it take them to be 2,025 miles apart?

Solution

2.5 hours

Leon drove from his house in Cincinnati to his sister’s house in Cleveland, a distance of 252 miles. It took him $4 \frac{1}{2}$ hours. For the first half hour he had heavy traffic, and the rest of the time his speed was five miles per hour less than twice his speed in heavy traffic. What was his speed in heavy traffic?

Chloe has a budget of $800 for costumes for the 18 members of her musical theater group. If all the costumes are the same price, what is the maximum she can spend for each costume?

Solution

at most $44.44 per costume

Frank found a rental car deal online for $49 per week plus $0.24 per mile. How many miles could he drive if he wants the total cost for one week to be no more than $150?

Introduction

This is a drawing of a graph. The x-axis ranges from 1994 through 2010 in two-year increments. The y-axis is labeled 0 to 30 million in increments of 5 millon per year. The y-axis is labeled “Annual Vehicle Sales (MM/year)” There are three line graphs. The first shows the annual sale of gas motorcyles from 5 million in 1994 to about 15 million in 2010. The next line is a green line labled EV for electric vehicles. It shows sales were null from 1994 through 2002, but they quickly rose to more than 25 million in sales per year. The last line is labeled gas cars and starts at 0 in 1994 and slowly rises from 2002 to 2010 to just over 10 million. — This graph illustrates the annual vehicle sales of gas motorcycles, gas cars, and electric vehicles from 1994 to 2010. It is a line graph with x- and y-axes, one of the most common types of graphs. (credit: Steve Jurvetson, Flickr)

Graphs are found in all areas of our lives—from commercials showing you which cell phone carrier provides the best coverage, to bank statements and news articles, to the boardroom of major corporations. In this chapter, we will study the rectangular coordinate system, which is the basis for most consumer graphs. We will look at linear graphs, slopes of lines, equations of lines, and linear inequalities.

Use the Rectangular Coordinate System

Plot Points on a Rectangular Coordinate System

Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the xy-plane or the ‘coordinate plane’.

The horizontal number line is called the x-axis. The vertical number line is called the y-axis. The x-axis and the y-axis together form the rectangular coordinate system. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See Figure 1.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 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 labeled "III" and the bottom-right portion of the plane is labeled "IV". — ‘Quadrant’ has the root ‘quad,’ which means ‘four.’

In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the x-coordinate of the point, and the second number is the y-coordinate of the point.

The phrase ‘ordered pair’ means the order is important. What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is $(0, 0)$ . The point $(0, 0)$ has a special name. It is called the origin.

We use the coordinates to locate a point on the xy-plane. Let’s plot the point $(1, 3)$ as an example. First, locate 1 on the x-axis and lightly sketch a vertical line through $x = 1$ . Then, locate 3 on the y-axis and sketch a horizontal line through $y = 3$ . Now, find the point where these two lines meet—that is the point with coordinates $(1, 3)$ .

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. An arrow starts at the origin and extends right to the number 2 on the x-axis. The point (1, 3) is plotted and labeled. Two dotted lines, one parallel to the x-axis, the other parallel to the y-axis, meet perpendicularly at 1, 3. The dotted line parallel to the x-axis intercepts the y-axis at 3. The dotted line parallel to the y-axis intercepts the x-axis at 1.

Notice that the vertical line through $x = 1$ and the horizontal line through $y = 3$ are not part of the graph. We just used them to help us locate the point $(1, 3)$ .

Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:

ⓐ $(−5, 4)$ ⓑ $(−3, −4)$ ⓒ $(2, −3)$ ⓓ $(−2, 3)$ ⓔ $(3, \frac{5}{2})$ .

Solution

The first number of the coordinate pair is the x-coordinate, and the second number is the y-coordinate.

ⓐ Since $x = −5$ , the point is to the left of the y-axis. Also, since $y = 4$ , the point is above the x-axis. The point $(−5, 4)$ is in Quadrant II.
ⓑ Since $x = −3$ , the point is to the left of the y-axis. Also, since $y = −4$ , the point is below the x-axis. The point $(−3, −4)$ is in Quadrant III.
ⓒ Since $x = 2$ , the point is to the right of the y-axis. Since $y = −3$ , the point is below the x-axis. The point $(2, −3)$ is in Quadrant lV.
ⓓ Since $x = −2$ , the point is to the left of the y-axis. Since $y = 3$ , the point is above the x-axis. The point $(−2, 3)$ is in Quadrant II.
ⓔ Since $x = 3$ , the point is to the right of the y-axis. Since $y = \frac{5}{2}$ , the point is above the x-axis. (It may be helpful to write $\frac{5}{2}$ as a mixed number or decimal.) The point $(3, \frac{5}{2})$ is in Quadrant I.

How do the signs affect the location of the points? You may have noticed some patterns as you graphed the points in the previous example.

For the point in Figure 2 in Quadrant IV, what do you notice about the signs of the coordinates? What about the signs of the coordinates of 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?

What if one coordinate is zero as shown in Figure 3? Where is the point $(0, 4)$ located? Where is the point $(−2, 0)$ located?

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. Points (0, 4) and (negative 2, 0) are plotted and labeled.

The point $(0, 4)$ is on the y-axis and the point $(−2, 0)$ is on the x-axis.

Plot each point:

ⓐ $(0, 5)$ ⓑ $(4, 0)$ ⓒ $(−3, 0)$ ⓓ $(0, 0)$ ⓔ $(0, −1)$ .

Solution

ⓐ Since $x = 0$ , the point whose coordinates are $(0, 5)$ is on the y-axis.
ⓑ Since $y = 0$ , the point whose coordinates are $(4, 0)$ is on the x-axis.
ⓒ Since $y = 0$ , the point whose coordinates are $(−3, 0)$ is on the x-axis.
ⓓ Since $x = 0$ and $y = 0$ , the point whose coordinates are $(0, 0)$ is the origin.
ⓔ Since $x = 0$ , the point whose coordinates are $(0, −1)$ is on the y-axis.

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, when you write the ordered pair use the correct order, $(x, y)$ .

Name the ordered pair of each point shown in the rectangular coordinate system.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The points (4, 0), (negative 2, 0), (0, 0), (0, 2), and (0, negative 3) are plotted and labeled A, B, C, D, and E, respectively.

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 $(4, −4)$ .

Point E is on the y-axis at $y = −2$ . The coordinates of point E are $(0, −2) .$

Point F is on the x-axis at $x = 3$ . The coordinates of point F are $(3, 0) .$

Verify Solutions to an Equation in Two Variables

Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation you got exactly one solution. The process of solving an equation ended with a statement like $x = 4$ . (Then, you checked the solution by substituting back into the equation.)

Here’s an example of an equation in one variable, and its one solution.

\begin{matrix} 3 x + 5 & = & 17 \\ 3 x & = & 12 \\ x & = & 4 \end{matrix}

But equations can have more than one variable. Equations with two variables may be of the form $A x + B y = C$ . Equations of this form are called linear equations in two variables.

Notice the word line in linear. Here is an example of a linear equation in two variables, $x$ and $y$ .

In this figure, we see the linear equation Ax plus By equals C. Below this is the equation x plus 4y equals 8. Below this are the values A equals 1, B equals 4, and C equals 8.

The equation $y = −3 x + 5$ is also a linear equation. But it does not appear to be in the form $A x + B y = C$ . We can use the Addition Property of Equality and rewrite it in $A x + B y = C$ form.

Illustrates the step-by-step conversion of a linear equation from slope-intercept form to standard form.
	$y = −3 x + 5$
Add to both sides.	$y + 3 x = −3 x + 5 + 3 x$
Simplify.	$y + 3 x = 5$
Use the Commutative Property to put it in $A x + B y = C$ form.	$3 x + y = 5$

By rewriting $y = −3 x + 5$ as $3 x + y = 5$ , we can easily see that it is a linear equation in two variables because it is of the form $A x + B y = C$ . When an equation is in the form $A x + B y = C$ , we say it is in standard form.

Most people prefer to have $A$ , $B$ , and $C$ be integers and $A \geq 0$ when writing a linear equation in standard form, although it is not strictly necessary.

Linear equations have infinitely many solutions. For every number that is substituted for $x$ there is a corresponding $y$ value. This pair of values is a solution to the linear equation and is represented by the ordered pair $(x, y)$ . When we substitute these values of $x$ and $y$ into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.

Determine which ordered pairs are solutions to the equation $x + 4 y = 8$ .

ⓐ $(0, 2)$ ⓑ $(2, −4)$ ⓒ $(−4, 3)$

Solution

Substitute the x- and y-values from each ordered pair into the equation and determine if the result is a true statement.
The equation x plus 4y equals 8 is checked for three ordered pairs. The pair (0, 2) satisfies the equation; substituting gives 8 equals 8. The pair (2, negative 4) does not satisfy the equation; substituting gives negative 14 does not equal 8. The pair (negative 4, 3) satisfies the equation; substituting gives 8 equals 8.

Which of the following ordered pairs are solutions to the equation $y = 5 x - 1$ ?

ⓐ $(0, −1)$ ⓑ $(1, 4)$ ⓒ $(−2, −7)$

Solution

Substitute the x- and y-values from each ordered pair into the equation and determine if it results in a true statement.
The graph shows three ordered pairs and checks if they are solutions to y equals 5x minus 1. The first pair (0, -1) satisfies the equation. The second pair (1, 4) also satisfies the equation. The third pair (-2, -7) does not satisfy the equation.

Complete a Table of Solutions to a Linear Equation in Two Variables

In the examples above, we substituted the x- and y-values of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do you find the ordered pairs if they are not given? It’s easier than you might think—you can just pick a value for $x$ and then solve the equation for $y$ . Or, pick a value for $y$ and then solve for $x$ .

We’ll start by looking at the solutions to the equation $y = 5 x - 1$ that we found in Example 5. We can summarize this information in a table of solutions, as shown in Table 2.

$y = 5 x - 1$
$x$	$y$	$(x, y)$
0	$−1$	$(0, −1)$
1	4	$(1, 4)$

To find a third solution, we’ll let $x = 2$ and solve for $y$ .

The figure shows the steps to solve for y when x equals 2 in the equation y equals 5 x minus 1. The equation y equals 5 x minus 1 is shown. Below it is the equation with 2 substituted in for x which is y equals 5 times 2 minus 1. To solve for y first multiply so that the equation becomes y equals 10 minus 1 then subtract so that the equation is y equals 9.

The ordered pair $(2, 9)$ is a solution to $y = 5 x - 1$ . We will add it to Table 3.

$y = 5 x - 1$
$x$	$y$	$(x, y)$
0	$−1$	$(0, −1)$
1	4	$(1, 4)$
2	9	$(2, 9)$

We can find more solutions to the equation by substituting in 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 infinitely many solutions of this equation.

Complete Table 4 to find three solutions to the equation $y = 4 x - 2$ .

$y = 4 x - 2$
$x$	$y$	$(x, y)$
0
$−1$
2

Solution

Substitute $x = 0$ , $x = −1$ , and $x = 2$ into $y = 4 x - 2$ .
Table showing three points on the line y equals 4x minus 2. The first point is x equals 0, y equals negative 2, or the ordered pair 0, negative 2. The second point is x equals negative 1, y equals negative 6, or the ordered pair negative 1, negative 6. The third point is x equals 2, y equals 6, or the ordered pair 2, 6.

The results are summarized in Table 5.

$y = 4 x - 2$
$x$	$y$	$(x, y)$
0	$−2$	$(0, −2)$
$−1$	$−6$	$(−1, −6)$
2	6	$(2, 6)$

Complete Table 10 to find three solutions to the equation $5 x - 4 y = 20$ .

$5 x - 4 y = 20$
$x$	$y$	$(x, y)$
0
	0
	5

Solution

Substitute the given value into the equation $5 x - 4 y = 20$ and solve for the other variable. Then, fill in the values in the table.
The graph of the equation 5x minus 4y equals 20 is shown for three cases. When x equals 0, y equals negative 5, and the point is (0, negative 5). When y equals 0, x equals 4, and the point is (4, 0). When y equals 5, x equals 8, and the point is (8, 5).

The results are summarized in Table 11.

$5 x - 4 y = 20$
$x$	$y$	$(x, y)$
0	$−5$	$(0, −5)$
4	0	$(4, 0)$
8	5	$(8, 5)$

Find Solutions to a Linear Equation

To find a solution to a linear equation, you really can pick any number you want to substitute into the equation for $x$ or $y .$ But since you’ll need to use that number to solve for the other variable it’s a good idea to choose a number that’s easy to work with.

When the equation is in y-form, with the y by itself on one side of the equation, it is usually easier to choose values of $x$ and then solve for $y$ .

Find three solutions to the equation $y = −3 x + 2$ .

Solution

We can substitute any value we want for

x

or any value for

y

. Since the equation is in y-form, it will be easier to substitute in values of

x

. Let’s pick

x = 0

x = 1

, and

x = −1



Substitute the value into the equation.
Simplify.
Simplify.
Write the ordered pair.			(0, 2)	(1, −1)	(−1, 5)
Check.
$y = −3 x + 2$	$y = −3 x + 2$	$y = −3 x + 2$
$2 ≟ −3 \cdot 0 + 2$	$−1 ≟ −3 \cdot 1 + 2$	$5 ≟ −3 (−1) + 2$
$2 ≟ 0 + 2$	$−1 ≟ −3 + 2$	$5 ≟ 3 + 2$
$2 = 2 ✓$	$−1 = −1 ✓$	$5 = 5 ✓$

So,

(0, 2)

(1, −1)

and

(−1, 5)

are all solutions to

y = −3 x + 2

. We show them in Table 17.

$y = −3 x + 2$
$x$	$y$	$(x, y)$
0	2	$(0, 2)$
1	$−1$	$(1, −1)$
$−1$	5	$(−1, 5)$

We have seen how using zero as one value of $x$ makes finding the value of $y$ easy. When an equation is in standard form, with both the $x$ and $y$ on the same side of the equation, it is usually easier to first find one solution when $x = 0$ find a second solution when $y = 0$ , and then find a third solution.

Find three solutions to the equation $3 x + 2 y = 6$ .

Solution

We can substitute any value we want for

x

or any value for

y

. Since the equation is in standard form, let’s pick first

x = 0

, then

y = 0

, and then find a third point.

Step-by-step solutions for three algebraic equations, showing substitution, simplification, and final ordered pairs.


Substitute the value into the equation.
Simplify.
Solve.

Write the ordered pair.	(0, 3)	(2, 0)	$(1, \frac{3}{2})$

Step-by-step verification of the linear equation 3x + 2y = 6, demonstrating checks for three different (x,y) solution pairs.
Check.
$3 x + 2 y = 6$	$3 x + 2 y = 6$	$3 x + 2 y = 6$
$3 \cdot 0 + 2 \cdot 3 ≟ 6$	$3 \cdot 2 + 2 \cdot 0 ≟ 6$	$3 \cdot 1 + 2 \cdot \frac{3}{2} ≟ 6$
$0 + 6 ≟ 6$	$6 + 0 ≟ 6$	$3 + 3 ≟ 6$
$6 = 6 ✓$	$6 = 6 ✓$	$6 = 6 ✓$

(0, 3)

(2, 0)

, and

(1, \frac{3}{2})

are all solutions to the equation

3 x + 2 y = 6

. We can list these three solutions in Table 20.

$3 x + 2 y = 6$
$x$	$y$	$(x, y)$
0	3	$(0, 3)$
2	0	$(2, 0)$
1	$\frac{3}{2}$	$(1, \frac{3}{2})$

Key Concepts

Sign Patterns of the Quadrants
$\begin{matrix} Quadrant I & Quadrant II & Quadrant III & Quadrant IV \\ (x, y) & (x, y) & (x, y) & (x, y) \\ (+, +) & (−, +) & (−, −) & (+, −) \end{matrix}$
Points on the Axes
- On the x-axis, $y = 0$ . Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates $(a, 0)$ .
- On the y-axis, $x = 0$ . Points with an x-coordinate equal to 0 are on the y-axis, and have coordinates $(0, b) .$
Solution of a Linear Equation
- An ordered pair $(x, y)$ is a solution of the linear equation $A x + B y = C$ , if the equation is a true statement when the x- and y- values of the ordered pair are substituted into the equation.

Practice Makes Perfect

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.

ⓐ $(−4, 2)$
ⓑ $(−1, −2)$
ⓒ $(3, −5)$
ⓓ $(−3, 5)$
ⓔ $(\frac{5}{3}, 2)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (negative 4, 2) is plotted and labeled "a". The point (negative 1, negative 2) is plotted and labeled "b". The point (3, negative 5) is plotted and labeled "c". The point (negative 3, 5) is plotted and labeled “d”. The point (5 thirds, 2) is plotted and labeled “e”.

ⓐ $(−2, −3)$
ⓑ $(3, −3)$
ⓒ $(−4, 1)$
ⓓ $(4, −1)$
ⓔ $(\frac{3}{2}, 1)$

ⓐ $(3, −1)$
ⓑ $(−3, 1)$
ⓒ $(−2, 2)$
ⓓ $(−4, −3)$
ⓔ $(1, \frac{14}{5})$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (3, negative 1) is plotted and labeled "a". The point (negative 3, 1) is plotted and labeled "b". The point (negative 2, 2) is plotted and labeled "c". The point (negative 4, negative 3) is plotted and labeled “d”. The point (1, 14 fifths) is plotted and labeled “e”.

ⓐ $(−1, 1)$
ⓑ $(−2, −1)$
ⓒ $(2, 1)$
ⓓ $(1, −4)$
ⓔ $(3, \frac{7}{2})$

In the following exercises, plot each point in a rectangular coordinate system.

ⓐ $(−2, 0)$
ⓑ $(−3, 0)$
ⓒ $(0, 0)$
ⓓ $(0, 4)$
ⓔ $(0, 2)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (negative 2, 0) is plotted and labeled "a". The point (negative 3, 0) is plotted and labeled "b". The point (0, 0) is plotted and labeled "c". The point (0, 4) is plotted and labeled “d”. The point (0, 3) is plotted and labeled “e”.

ⓐ $(0, 1)$
ⓑ $(0, −4)$
ⓒ $(−1, 0)$
ⓓ $(0, 0)$
ⓔ $(5, 0)$

ⓐ $(0, 0)$
ⓑ $(0, −3)$
ⓒ $(−4, 0)$
ⓓ $(1, 0)$
ⓔ $(0, −2)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (0, 0) is plotted and labeled "a". The point (0, negative 3) is plotted and labeled "b". The point (negative 4, 0) is plotted and labeled "c". The point (1, 0) is plotted and labeled “d”. The point (0, negative 2) is plotted and labeled “e”.

ⓐ $(−3, 0)$
ⓑ $(0, 5)$
ⓒ $(0, −2)$
ⓓ $(2, 0)$
ⓔ $(0, 0)$

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 x- and y-axes each run from negative 6 to 6. The point (negative 4, 1) is plotted and labeled “A”. The point (negative 3, negative 4) is plotted and labeled “B”. The point (1, negative 3) is plotted and labeled “C”. The point (4, 3) is plotted and labeled “D”.

Solution

A: $(−4, 1)$ B: $(−3, −4)$ C: $(1, −3)$ D: $(4, 3)$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The point (negative 4, 2) is plotted and labeled “A”. The point (3, 5) is plotted and labeled “B”. The point (negative 3, negative 2) is plotted and labeled “C”. The point (5, negative 1) is plotted and labeled “D”.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (0, negative 2) is plotted and labeled “A”. The point (negative 2, 0) is plotted and labeled “B”. The point (0, 5) is plotted and labeled “C”. The point (5, 0) is plotted and labeled “D”.

Solution

A: $(0, −2)$ B: $(−2, 0)$ C: $(0, 5)$ D: $(5, 0)$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (0, negative 1) is plotted and labeled “A”. The point (negative 1, 0) is plotted and labeled “B”. The point (4, 0) is plotted and labeled “C”. The point (0, 4) is plotted and labeled “D”.

Verify Solutions to an Equation in Two Variables

In the following exercises, which ordered pairs are solutions to the given equations?

$2 x + y = 6$

ⓐ $(1, 4)$
ⓑ $(3, 0)$
ⓒ $(2, 3)$

Solution

a, b

$x + 3 y = 9$

ⓐ $(0, 3)$
ⓑ $(6, 1)$
ⓒ $(−3, −3)$

$4 x - 2 y = 8$

ⓐ $(3, 2)$
ⓑ $(1, 4)$
ⓒ $(0, −4)$

Solution

a, c

$3 x - 2 y = 12$

ⓐ $(4, 0)$
ⓑ $(2, −3)$
ⓒ $(1, 6)$

$y = 4 x + 3$

ⓐ $(4, 3)$
ⓑ $(−1, −1)$
ⓒ $(\frac{1}{2}, 5)$

Solution

b, c

$y = 2 x - 5$

ⓐ $(0, −5)$
ⓑ $(2, 1)$
ⓒ $(\frac{1}{2}, −4)$

$y = \frac{1}{2} x - 1$

ⓐ $(2, 0)$
ⓑ $(−6, −4)$
ⓒ $(−4, −1)$

Solution

a, b

$y = \frac{1}{3} x + 1$

ⓐ $(−3, 0)$
ⓑ $(9, 4)$
ⓒ $(−6, −1)$

Complete a Table of Solutions to a Linear Equation

In the following exercises, complete the table to find solutions to each linear equation.

$y = 2 x - 4$

$x$	$y$	$(x, y)$
0
2
$−1$

Solution

$x$	$y$	$(x, y)$
0	$−4$	$(0, −4)$
2	0	$(2, 0)$
$−1$	$−6$	$(−1, −6)$

$y = 3 x - 1$

$x$	$y$	$(x, y)$
0
2
$−1$

$y = − x + 5$

$x$	$y$	$(x, y)$
0
3
$−2$

Solution

$x$	$y$	$(x, y)$
0	5	$(0, 5)$
3	2	$(3, 2)$
$−2$	7	$(−2, 7)$

$y = − x + 2$

$x$	$y$	$(x, y)$
0
3
$−2$

$y = \frac{1}{3} x + 1$

$x$	$y$	$(x, y)$
0
3
6

Solution

$x$	$y$	$(x, y)$
0	1	$(0, 1)$
3	2	$(3, 2)$
6	3	$(6, 3)$

$y = \frac{1}{2} x + 4$

$x$	$y$	$(x, y)$
0
2
4

$y = - \frac{3}{2} x - 2$

$x$	$y$	$(x, y)$
0
2
$−2$

Solution

$x$	$y$	$(x, y)$
0	$−2$	$(0, −2)$
2	$−5$	$(2, −5)$
$−2$	1	$(−2, 1)$

$y = - \frac{2}{3} x - 1$

$x$	$y$	$(x, y)$
0
3
$−3$

$x + 3 y = 6$

$x$	$y$	$(x, y)$
0
3
	0

Solution

$x$	$y$	$(x, y)$
0	2	$(0, 2)$
3	4	$(3, 1)$
6	0	$(6, 0)$

$x + 2 y = 8$

$x$	$y$	$(x, y)$
0
4
	0

$2 x - 5 y = 10$

$x$	$y$	$(x, y)$
0
10
	0

Solution

$x$	$y$	$(x, y)$
0	$−2$	$(0, −2)$
10	2	$(10, 2)$
5	0	$(5, 0)$

$3 x - 4 y = 12$

$x$	$y$	$(x, y)$
0
8
	0

Find Solutions to a Linear Equation

In the following exercises, find three solutions to each linear equation.

$y = 5 x - 8$

Solution

Answers will vary.

$y = 3 x - 9$

$y = −4 x + 5$

Solution

Answers will vary.

$y = −2 x + 7$

$x + y = 8$

Solution

Answers will vary.

$x + y = 6$

$x + y = −2$

Solution

Answers will vary.

$x + y = −1$

$3 x + y = 5$

Solution

Answers will vary.

$2 x + y = 3$

$4 x - y = 8$

Solution

Answers will vary.

$5 x - y = 10$

$2 x + 4 y = 8$

Solution

Answers will vary.

$3 x + 2 y = 6$

$5 x - 2 y = 10$

Solution

Answers will vary.

$4 x - 3 y = 12$

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 below, and shown as an ordered pair in the third column.

ⓐ Plot the points on a coordinate plane.

A blank Cartesian coordinate plane featuring a grid. The x-axis is labeled from 0 to 12, and the y-axis is labeled from 0 to 25, ready for data plotting.

ⓑ Why is only Quadrant I needed?

Age $x$	Weight $y$	$(x, y)$
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)

Solution

ⓐ
The graph shows the x y-coordinate plane. The x- and y-axes each run from 0 to 25. The points (0, 7), (2, 11), (4, 15), (6, 16), (8, 19), (10, 20) and (12, 21) are plotted and labeled.
ⓑ 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 below, and shown as an ordered pair in the third column.

ⓐ Plot the points on a coordinate plane.

An empty 5x5 grid on a Cartesian coordinate plane with x and y axes marked from 0 to 50, showing increments of 10.

ⓑ Why is only Quadrant I needed?

Height $x$	Weight $y$	$(x, y)$
28	22	(28, 22)
31	27	(31, 27)
33	33	(33, 33)
37	35	(37, 35)
40	41	(40, 41)
42	45	(42, 45)

Writing Exercises

Explain in words how you plot the point $(4, −2)$ in a rectangular coordinate system.

Solution

Answers will vary.

How do you determine if an ordered pair is a solution to a given equation?

Is the point $(−3, 0)$ on the x-axis or y-axis? How do you know?

Solution

Answers will vary.

Is the point $(0, 8)$ on the x-axis or y-axis? How do you know?

Self Check

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

ⓑ 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

A linear equation is of the form

A x + B y = C

, where A and B are not both zero, is called a linear equation in two variables.

ordered pair

An ordered pair

(x, y)

gives the coordinates of a point in a rectangular coordinate system.

origin

The point

(0, 0)

is called the origin. It is the point where the x-axis and y-axis intersect.

quadrant

The x-axis and the y-axis divide a plane into four regions, called quadrants.

rectangular coordinate system

A grid system is used in algebra to show a relationship between two variables; also called the xy-plane or the ‘coordinate plane’.

x-coordinate

The first number in an ordered pair

(x, y)

y-coordinate

The second number in an ordered pair

(x, y)

Graph Linear Equations in Two Variables

Recognize the Relationship Between the Solutions of an Equation and its Graph

In the previous section, we found several solutions to the equation $3 x + 2 y = 6$ . They are listed in Table 1. So, the ordered pairs $(0, 3)$ , $(2, 0)$ , and $(1, \frac{3}{2})$ are some solutions to the equation $3 x + 2 y = 6$ . We can plot these solutions in the rectangular coordinate system as shown in Figure 1.

$3 x + 2 y = 6$
$x$	$y$	$(x, y)$
0	3	$(0, 3)$
2	0	$(2, 0)$
1	$\frac{3}{2}$	$(1, \frac{3}{2})$

The figure shows four points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the four points at (0, 3), (1, three halves), (2, 0), and (4, negative 3). The four points appear to line up along a straight line.

Notice how the points line up perfectly? We connect the points with a line to get the graph of the equation $3 x + 2 y = 6$ . See Figure 2. Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

The figure shows a straight line drawn through four points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the four points at (0, 3), (1, three halves), (2, 0), and (4, negative 3). A straight line with a negative slope goes through all four points. The line has arrows on both ends pointing to the edge of the figure. The line is labeled with the equation 3x plus 2y equals 6.

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 3. If you substitute $x = −2$ and $y = 6$ into the equation, you find that it is a solution to the equation.

The figure shows a straight line and two points and on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the two points and are labeled by the coordinates “(negative 2, 6)” and “(4, 1)”. The straight line goes through the point (negative 2, 6) but does not go through the point (4, 1).

Checking if ordered pair negative 2, 6 solves 3x + 2y = 6. Substituting gives 3 times negative 2 plus 2 times 6 equals 6. Simplifying yields negative 6 plus 12 equals 6, then 6 equals 6. A checkmark indicates the solution is correct.

So the point $(−2, 6)$ is a solution to the equation $3 x + 2 y = 6$ . (The phrase “the point whose coordinates are $(−2, 6)$ ” is often shortened to “the point $(−2, 6)$ .”)

Checking if (4, 1) solves 3x + 2y = 6. Substituting gives 3(4) + 2(1) = 6, which simplifies to 12 + 2 = 6. This further simplifies to 14 does not equal 6, so the ordered pair is not a solution.

So $(4, 1)$ is not a solution to the equation $3 x + 2 y = 6$ . Therefore, the point $(4, 1)$ is not on the line. See Figure 2. 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 $3 x + 2 y = 6$ .

The graph of $y = 2 x - 3$ is shown.

The figure shows a straight line on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line has a positive slope and goes through the y-axis at the (0, negative 3). The line is labeled with the equation y equals 2x negative 3.

For each ordered pair, decide:

ⓐ Is the ordered pair a solution to the equation?
ⓑ Is the point on the line?

A $(0, −3)$ B $(3, 3)$ C $(2, −3)$ D $(−1, −5)$

Solution

Substitute the x- and y- values into the equation to check if the ordered pair is a solution to the equation.

ⓐ
ⓑ 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 $y = 2 x - 3$ , and the point $(2, −3)$ is not on the line.

The points that are solutions to $y = 2 x - 3$ are on the line, but the point that is not a solution is not on the line.

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The method we used to graph $3 x + 2 y = 6$ is called plotting points, or the Point–Plotting Method.

How To Graph an Equation By Plotting Points

Graph the equation $y = 2 x + 1$ by plotting points.

Solution

The figure shows the three step procedure for graphing a line from the equation using the example equation y equals 2x minus 1. The first step is to “Find three points whose coordinates are solutions to the equation. Organize the solutions in a table”. The remark is made that “You can choose any values for x or y. In this case, since y is isolated on the left side of the equation, it is easier to choose values for x”. The work for the first step of the example is shown through a series of equations aligned vertically. From the top down, the equations are y equals 2x plus 1, x equals 0 (where the 0 is blue), y equals 2x plus 1, y equals 2(0) plus 1 (where the 0 is blue), y equals 0 plus 1, y equals 1, x equals 1 (where the 1 is blue), y equals 2x plus 1, y equals 2(1) plus 1 (where the 1 is blue), y equals 2 plus 1, y equals 3, x equals negative 2 (where the negative 2 is blue), y equals 2x plus 1, y equals 2(negative 2) plus 1 (where the negative 2 is blue), y equals negative 4 plus 1, y equals negative 3. The work is then organized in a table. The table has 5 rows and 3 columns. The first row is a title row with the equation y equals 2x plus 1. The second row is a header row and it labels each column. The first column header is “x”, the second is “y” and the third is “(x, y)”. Under the first column are the numbers 0, 1, and negative 2. Under the second column are the numbers 1, 3, and negative 3. Under the third column are the ordered pairs (0, 1), (1, 3), and (negative 2, negative 3).

The second step is to “Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work!” For the example the points are (0, 1), (1, 3), and (negative 2, negative 3). A graph shows the three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points at (0, 1), (1, 3), and (negative 2, negative 3). The question “Do the points line up?” is stated and followed with the answer “Yes, the points line up.”

The third step of the procedure is “Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.” A graph shows a straight line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points at (0, 1), (1, 3), and (negative 2, negative 3). A straight line goes through all three points. The line has arrows on both ends pointing to the edge of the figure. The line is labeled with the equation y equals 2x plus 1. The statement “This line is the graph of y equals 2x plus 1” is included next to the graph.

The steps to take when graphing a linear equation by plotting points are summarized below.

It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you only plot two points and one of them is incorrect, you can still draw a line but it will not represent the solutions to the equation. It will be the wrong line.

If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. Look at the difference between part (a) and part (b) in Figure 4.

Figure a shows three points with a straight line going through them. Figure b shows three points that do not lie on the same line.

Let’s do another example. This time, we’ll show the last two steps all on one grid.

Graph the equation $y = −3 x$ .

Solution

Find three points that are solutions to the equation. Here, again, it’s easier to choose values for $x$ . Do you see why?
Three sets of equations demonstrate finding ordered pairs for y equals negative three times x. The first set uses x equals zero, resulting in y equals zero. The second uses x equals one, yielding y equals negative three. The third uses x equals negative two, giving y equals six.

We list the points in Table 2.

$y = −3 x$
$x$	$y$	$(x, y)$
0	0	$(0, 0)$
1	$−3$	$(1, −3)$
$−2$	6	$(−2, 6)$

Plot the points, check that they line up, and draw the line.

When an equation includes a fraction as the coefficient of $x$ , we can still 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 $y = \frac{1}{2} x + 3$ .

Solution

Find three points that are solutions to the equation. Since this equation has the fraction $\frac{1}{2}$ as a coefficient of $x,$ we will choose values of $x$ carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of 2 a good choice for values of $x$ ?
Three sets of equations demonstrate finding ordered pairs for the equation y equals one-half x plus three. The first set uses x equals zero, resulting in y equals three. The second uses x equals two, yielding y equals four. The third uses x equals four, giving y equals five.

The points are shown in Table 3.

$y = \frac{1}{2} x + 3$
$x$	$y$	$(x, y)$
0	3	$(0, 3)$
2	4	$(2, 4)$
4	5	$(4, 5)$

Plot the points, check that they line up, and draw the line.

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. Let’s see what happens in the equation $2 x + y = 3$ . If $y = 0$ what is the value of $x$ ?

The figure shows a set of equations used to determine an ordered pair from the equation 2x plus y equals 3. The first equation is y equals 0 (where the 0 is red). The second equation is the two- variable equation 2x plus y equals 3. The third equation is the onenegative variable equation 2x plus 0 equals 3 (where the 0 is red). The fourth equation is 2x equals 3. The fifth equation is x equals three halves. The last line is the ordered pair (three halves, 0).

This point has a fraction for the x- coordinate and, while we could graph this point, it is hard to be precise graphing fractions. Remember in the example $y = \frac{1}{2} x + 3$ , we carefully chose values for $x$ so as not to graph fractions at all. If we solve the equation $2 x + y = 3$ for $y$ , it will be easier to find three solutions to the equation.

\begin{matrix} 2 x + y & = & 3 \\ y & = & −2 x + 3 \end{matrix}

The solutions for $x = 0$ , $x = 1$ , and $x = −1$ are shown in the Table 4. The graph is shown in Figure 5.

$2 x + y = 3$
$x$	$y$	$(x, y)$
0	3	$(0, 3)$
1	1	$(1, 1)$
$−1$	5	$(−1, 5)$

The figure shows a straight line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points which are labeled by their ordered pairs (negative 1, 5), (0, 3), and (1, 1). A straight line goes through all three points. The line has arrows on both ends pointing to the outside of the figure. The line is labeled with the equation 2x plus y equals 3.

Can you locate the point $(\frac{3}{2}, 0)$ , which we found by letting $y = 0$ , on the line?

Graph the equation $3 x + y = −1$ .

Solution

This table demonstrates steps to solve the linear equation 3x + y = -1 for y.
Find three points that are solutions to the equation.	$3 x + y = −1$
First, solve the equation for $y$ .	$y = −3 x - 1$

We’ll let $x$ be 0, 1, and $−1$ to find 3 points. The ordered pairs are shown in Table 6. Plot the points, check that they line up, and draw the line. See Figure 6.

$3 x + y = −1$
$x$	$y$	$(x, y)$
0	$−1$	$(0, −1)$
1	$−4$	$(1, −4)$
$−1$	2	$(−1, 2)$

A straight line passes through the points negative 1 comma 2, 0 comma negative 1, and 1 comma negative 4 on a coordinate plane. The line's equation is three x plus y equals negative one, indicating it continues infinitely in both directions.

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!

The equation in Example 5 was written in standard form, with both $x$ and $y$ on the same side. We solved that equation for $y$ in just one step. But for other equations in standard form it is not that easy to solve for $y$ , so we will leave them in standard form. We can still find a first point to plot by letting $x = 0$ and solving for $y$ . We can plot a second point by letting $y = 0$ and then solving for $x$ . Then we will plot a third point by using some other value for $x$ or $y$ .

Graph the equation $2 x - 3 y = 6$ .

Solution

Step-by-step solution demonstrating how to find three points that satisfy the linear equation 2x - 3y = 6 by substituting values for x and y.
Find three points that are solutions to the equation.	$\begin{matrix} 2 x - 3 y & = & 6 \end{matrix}$
First, let $x = 0$ .	$\begin{matrix} 2 (0) - 3 y & = & 6 \end{matrix}$
Solve for $y$ .	$\begin{matrix} −3 y & = & 6 \\ y & = & −2 \end{matrix}$
Now let $y = 0$ .	$\begin{matrix} 2 x - 3 (0) & = & 6 \end{matrix}$
Solve for $x$ .	$\begin{matrix} 2 x & = & 6 \\ x & = & 3 \end{matrix}$
We need a third point. Remember, we can choose any value for $x$ or $y$ . We'll let $x = 6$ .	$\begin{matrix} 2 (6) - 3 y & = & 6 \end{matrix}$
Solve for $y$ .	$\begin{matrix} 12 - 3 y & = & 6 \\ −3 y & = & −6 \\ y & = & 2 \end{matrix}$

We list the ordered pairs in Table 8. Plot the points, check that they line up, and draw the line. See Figure 7.

$2 x - 3 y = 6$
$x$	$y$	$(x, y)$
0	$−2$	$(0, −2)$
3	0	$(3, 0)$
6	2	$(6, 2)$

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 $x = −3$ . This equation has only one variable, $x$ . The equation says that $x$ is always equal to $−3$ , so its value does not depend on $y$ . No matter what $y$ is, the value of $x$ is always $−3$ .

So to make a table of values, write $−3$ in for all the $x$ values. Then choose any values for $y$ . Since $x$ does not depend on $y$ , you can choose any numbers you like. But to fit the points on our coordinate graph, we’ll use 1, 2, and 3 for the y-coordinates. See Table 9.

$x = −3$
$x$	$y$	$(x, y)$
$−3$	1	$(−3, 1)$
$−3$	2	$(−3, 2)$
$−3$	3	$(−3, 3)$

Plot the points from Table 9 and connect them with a straight line. Notice in Figure 8 that we have graphed a vertical line.

A vertical line passes through the points negative 3 comma 1, negative 3 comma 2, and negative 3 comma 3 on a coordinate plane. The line represents x equals negative 3.

Graph the equation $x = 2$ .

Solution

The equation has only one variable, $x$ , and $x$ is always equal to 2. We create Table 10 where $x$ is always 2 and then put in any values for $y$ . The graph is a vertical line passing through the x-axis at 2. See Figure 9.

$x = 2$
$x$	$y$	$(x, y)$
2	1	$(2, 1)$
2	2	$(2, 2)$
2	3	$(2, 3)$

The figure shows a straight vertical line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points which are labeled by their ordered pairs (2, 1), (2, 2), and (2, 3). A vertical straight line goes through all three points. The line has arrows on both ends pointing to the outside of the figure. The line is labeled with the equation x equals 2.

What if the equation has $y$ but no $x$ ? Let’s graph the equation $y = 4$ . This time the y- value is a constant, so in this equation, $y$ does not depend on $x$ . Fill in 4 for all the $y$ ’s in Table 11 and then choose any values for $x$ . We’ll use 0, 2, and 4 for the x-coordinates.

$y = 4$
$x$	$y$	$(x, y)$
0	4	$(0, 4)$
2	4	$(2, 4)$
4	4	$(4, 4)$

The graph is a horizontal line passing through the y-axis at 4. See Figure 10.

The figure shows a straight horizontal line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points which are labeled by their ordered pairs (0, 4), (2, 4), and (4, 4). A straight horizontal line goes through all three points. The line has arrows on both ends pointing to the outside of the figure. The line is labeled with the equation y equals 4.

Graph the equation $y = −1 .$

Solution

The equation $y = −1$ has only one variable, $y$ . The value of $y$ is constant. All the ordered pairs in Table 12 have the same y-coordinate. The graph is a horizontal line passing through the y-axis at $−1$ , as shown in Figure 11.

$y = −1$
$x$	$y$	$(x, y)$
0	$−1$	$(0, −1)$
3	$−1$	$(3, −1)$
$−3$	$−1$	$(−3, −1)$

A graph shows a horizontal line at y equals negative 1. Three points are marked on the line: negative 3, negative 1; 0, negative 1; and 3, negative 1. The x and y axes range from negative 7 to positive 7.

The equations for vertical and horizontal lines look very similar to equations like $y = 4 x .$ What is the difference between the equations $y = 4 x$ and $y = 4$ ?

The equation $y = 4 x$ 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 $x$ . See Table 13.

$y = 4 x$			$y = 4$
$x$	$y$	$(x, y)$	$x$	$y$	$(x, y)$
0	0	$(0, 0)$	0	4	$(0, 4)$
1	4	$(1, 4)$	1	4	$(1, 4)$
2	8	$(2, 8)$	2	4	$(2, 4)$

The figure shows a two straight lines drawn on the same x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. One line is a straight horizontal line labeled with the equation y equals 4. The other line is a slanted line labeled with the equation y equals 4x.

Notice, in Figure 12, the equation $y = 4 x$ gives a slanted line, while $y = 4$ gives a horizontal line.

Graph $y = −3 x$ and $y = −3$ in the same rectangular coordinate system.

Solution

Notice that the first equation has the variable $x$ , while the second does not. See Table 14. The two graphs are shown in Figure 13.

$y = −3 x$			$y = −3$
$x$	$y$	$(x, y)$	$x$	$y$	$(x, y)$
0	0	$(0, 0)$	0	$−3$	$(0, −3)$
1	$−3$	$(1, −3)$	1	$−3$	$(1, −3)$
2	$−6$	$(2, −6)$	2	$−3$	$(2, −3)$

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 in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work!
3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.

Practice Makes Perfect

Recognize the Relationship Between the Solutions of an Equation and its Graph

In the following exercises, for each ordered pair, decide:

ⓐ Is the ordered pair a solution to the equation? ⓑ Is the point on the line?

$y = x + 2$

ⓐ $(0, 2)$
ⓑ $(1, 2)$
ⓒ $(−1, 1)$
ⓓ $(−3, −1)$

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 6, negative 4), (negative 5, negative 3), (negative 4, negative 2), (negative 3, negative 1), (negative 2, 0), (negative 1, 1), (0, 2), (1, 3), (2, 4), (3, 5), (4, 6), and (5, 7).

Solution

ⓐ yes; yes ⓑ no; no ⓒ yes; yes ⓓ yes; yes

$y = x - 4$

ⓐ $(0, −4)$
ⓑ $(3, −1)$
ⓒ $(2, 2)$
ⓓ $(1, −5)$

$y = \frac{1}{2} x - 3$

ⓐ $(0, −3)$
ⓑ $(2, −2)$
ⓒ $(−2, −4)$
ⓓ $(4, 1)$

Solution

ⓐ yes; yes ⓑ yes; yes ⓒ yes; yes ⓓ no; no

$y = \frac{1}{3} x + 2$

ⓐ $(0, 2)$
ⓑ $(3, 3)$
ⓒ $(−3, 2)$
ⓓ $(−6, 0)$

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

$y = 3 x - 1$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 3, negative 10), (negative 2, negative 7), (negative 1, negative 4), (0, negative 1), (1, 2), (2, 5), and (3, 8).

$y = 2 x + 3$

$y = −2 x + 2$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 4, 10), (negative 3, 8), (negative 2, 6), (negative 1, 4), (0, 2), (1, 0), (2, negative 2), (3, negative 4), (4, negative 6), and (5, negative 8).

$y = −3 x + 1$

$y = x + 2$

Solution

$y = x - 3$

$y = − x - 3$

Solution

$y = − x - 2$

$y = 2 x$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 5, negative 10), (negative 4, negative 8), (negative 3, negative 6), (negative 2, negative 4), (negative 1, negative 2), (0, 0), (1, 2), (2, 4), (3, 6), (4, 8), and (5, 10).

$y = 3 x$

$y = −4 x$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 3, 12), (negative 2, 8), (negative 1, 4), (0, 0), (1, negative 4), (2, negative 8), and (3, negative 12).

$y = −2 x$

$y = \frac{1}{2} x + 2$

Solution

$y = \frac{1}{3} x - 1$

$y = \frac{4}{3} x - 5$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 3, negative 9), (0, negative 5), (3, negative 1), (6, 3), and (9, 7).

$y = \frac{3}{2} x - 3$

$y = - \frac{2}{5} x + 1$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 10, 5), (negative 5, 3), (0, 1), (5, negative 1), and (10, negative 3).

$y = - \frac{4}{5} x - 1$

$y = - \frac{3}{2} x + 2$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 6, 11), (negative 4, 8), (negative 2, 5), (0, 2), (2, negative 1), (4, negative 4), (6, negative 7), and (8, negative 10).

$y = - \frac{5}{3} x + 4$

$x + y = 6$

Solution

$x + y = 4$

$x + y = −3$

Solution

$x + y = −2$

$x - y = 2$

Solution

$x - y = 1$

$x - y = −1$

Solution

$x - y = −3$

$3 x + y = 7$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to -7. The equation 3 x plus y equals 7 is graphed.

$5 x + y = 6$

$2 x + y = −3$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 5, 7), (negative 4, 5), (negative 3, 3), (negative 2, 1), (negative 1, negative 1), (0, negative 3), (1, negative 5), and (2, negative 7).

$4 x + y = −5$

$\frac{1}{3} x + y = 2$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 6, 4), (negative 3, 3), (0, 2), (3, 1), and (6, 0).

$\frac{1}{2} x + y = 3$

$\frac{2}{5} x - y = 4$

Solution

$\frac{3}{4} x - y = 6$

$2 x + 3 y = 12$

Solution

$4 x + 2 y = 12$

$3 x - 4 y = 12$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 4, negative 6), (0, negative 3), (4, 0), and (8, 3).

$2 x - 5 y = 10$

$x - 6 y = 3$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 6, negative three halves), (negative 3, negative 1), (0, negative one half), (3, 0), and (6, one half).

$x - 4 y = 2$

$5 x + 2 y = 4$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 2, 7), (0, 2), (2, negative 3), and (4, negative 8).

$3 x + 5 y = 5$

Graph Vertical and Horizontal Lines

In the following exercises, graph each equation.

$x = 4$

Solution

The figure shows a straight vertical line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The vertical line goes through the points (4, 0), (4, 1), (4, 2) and all points with first coordinate 4.

$x = 3$

$x = −2$

Solution

The figure shows a straight vertical line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The vertical line goes through the points (negative 2, 0), (negative 2, 1), (negative 2, 2) and all points with first coordinate negative 2.

$x = −5$

$y = 3$

Solution

The figure shows a straight horizontal line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The horizontal line goes through the points (0, 3), (1, 3), (2, 3) and all points with second coordinate 3.

$y = 1$

$y = −5$

Solution

The figure shows a straight horizontal line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The horizontal line goes through the points (0, negative 5), (1, negative 5), (2, negative 5) and all points with second coordinate negative 5.

$y = −2$

$x = \frac{7}{3}$

Solution

The figure shows a straight vertical line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The vertical line goes through the points (7/3, 0), (7/3, 1), (7/3, 2) and all points with first coordinate 7/3.

$x = \frac{5}{4}$

$y = - \frac{15}{4}$

Solution

The figure shows a straight horizontal line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The horizontal line goes through the points (0, negative 15/4), (1, negative 15/4), (2, negative 15/4) and all points with second coordinate negative 15/4.

$y = - \frac{5}{3}$

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

$y = 2 x$ and $y = 2$

Solution

$y = 5 x$ and $y = 5$

$y = - \frac{1}{2} x$ and $y = - \frac{1}{2}$

Solution

$y = - \frac{1}{3} x$ and $y = - \frac{1}{3}$

Mixed Practice

In the following exercises, graph each equation.

$y = 4 x$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 2, negative 8), (negative 1, negative 4), (0, 0), (1, 4), and (2, 8).

$y = 2 x$

$y = - \frac{1}{2} x + 3$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 6, 6), (negative 4, 5), (negative 2, 4), (0, 3), (2, 2), (4, 1), and (6, 0).

$y = \frac{1}{4} x - 2$

$y = − x$

Solution

$y = x$

$x - y = 3$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 3, negative 7), (negative 2, negative 6), (negative 1, negative 4), (0, negative 3), (1, negative 2), (2, negative 1), (3, 0), (4, 1), (5, 2), and (6, 3).

$x + y = −5$

$4 x + y = 2$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 2, 6), (negative 1, 4), (0, 2), (1, negative 2), and (2, negative 6).

$2 x + y = 6$

$y = −1$

Solution

The figure shows a straight horizontal line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The horizontal line goes through the points (0, negative 1), (1, negative 1), (2, negative 1) and all points with second coordinate negative 1.

$y = 5$

$2 x + 6 y = 12$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 6, 4), (negative 3, 3), (0, 2), (3, 1), and (6, 0).

$5 x + 2 y = 10$

$x = 3$

Solution

The figure shows a straight vertical line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The vertical line goes through the points (3, 0), (3, 1), (3, 2) and all points with first coordinate 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 $y = 594 + 0.32 x$ gives the cost, $y$ , for driving $x$ miles. Calculate the rental cost for driving 400, 800, and 1200 miles, and then graph the line.

Solution

$722, $850, $978
The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from 0 to 1200 in increments of 100. The y-axis of the plane runs from 0 to 1000 in increments of 100. The straight line starts at the point (0, 594) and goes through the points (400, 722), (800, 850), and (1200, 978). The right end of the line has an arrow pointing up and to the right.

Weekly earnings. At the art gallery where he works, Salvador gets paid $200 per week plus 15% of the sales he makes, so the equation $y = 200 + 0.15 x$ gives the amount, $y$ , he earns for selling $x$ dollars of artwork. Calculate the amount Salvador earns for selling $900, $1600, and $2000, and then graph the line.

Writing Exercises

Explain how you would choose three x- values to make a table to graph the line $y = \frac{1}{5} x - 2$ .

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.

This table has 4 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “…recognize the relation between the solutions of an equation and its graph.”, “…graph a linear equation by plotting points.”, and “…graph vertical and horizontal lines.”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

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

graph of a linear equation

The graph of a linear equation

A x + B y = 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.

horizontal line

A horizontal line is the graph of an equation of the form

y = b

. The line passes through the y-axis at

(0, b)

vertical line

A vertical line is the graph of an equation of the form

x = a

. The line passes through the x-axis at

(a, 0)

Graph with Intercepts

Identify the x- and y- Intercepts on a Graph

Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.

At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x- axis and the y- axis. These points are called the intercepts of the line.

Let’s look at the graphs of the lines in Figure 1.

Four figures, each showing a different straight line on the x y- coordinate plane. The x- axis of the planes runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Figure a shows a straight line crossing the x- axis at the point (3, 0) and crossing the y- axis at the point (0, 6). The graph is labeled with the equation 2x plus y equals 6. Figure b shows a straight line crossing the x- axis at the point (4, 0) and crossing the y- axis at the point (0, negative 3). The graph is labeled with the equation 3x minus 4y equals 12. Figure c shows a straight line crossing the x- axis at the point (5, 0) and crossing the y- axis at the point (0, negative 5). The graph is labeled with the equation x minus y equals 5. Figure d shows a straight line crossing the x- axis and y- axis at the point (0, 0). The graph is labeled with the equation y equals negative 2x. — Examples of graphs crossing the x-axis.

First, notice where each of these lines crosses the $x$ -axis. See Figure 1.

Figure	The line crosses the x- axis at:	Ordered pair of this point
Figure (a)	3	$(3, 0)$
Figure (b)	4	$(4, 0)$
Figure (c)	5	$(5, 0)$
Figure (d)	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 a line. The x- intercept occurs when $y$ is zero.

Now, let’s look at the points where these lines cross the y- axis. See Table 2.

Figure	The line crosses the y-axis at:	Ordered pair for this point
Figure (a)	6	$(0, 6)$
Figure (b)	$−3$	$(0, −3)$
Figure (c)	$−5$	$(0, −5)$
Figure (d)	0	$(0, 0)$

What is the pattern here?

In each row, the x- coordinate of the point where the line crosses the y- axis is zero. The point where the line crosses the y- axis has the form $(0, b)$ and is called the y- intercept of the line. The y- intercept occurs when $x$ is zero.

Find the x- and y- intercepts on each graph.

Three figures, each showing a different straight line on the x y- coordinate plane. The x- axis of the planes runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Figure a shows a straight line going through the points (negative 6, 5), (negative 4, 4), (negative 2, 3), (0, 2), (2, 1), (4, 0), and (6, negative 1). Figure b shows a straight line going through the points (0, negative 6), (1, negative 3), (2, 0), (3, 3), and (4, 6). Figure c shows a straight line going through the points (negative 6, 1), (negative 5, 0), (negative 4, negative 1), (negative 3, negative 2), (negative 2, negative 3), (negative 1, negative 4), (0, negative 5), and (1, negative 6).

Solution

ⓐ The graph crosses the x- axis at the point $(4, 0)$ . The x- intercept is $(4, 0)$ .
The graph crosses the y- axis at the point $(0, 2)$ . The y- intercept is $(0, 2)$ .
ⓑ The graph crosses the x- axis at the point $(2, 0)$ . The x- intercept is $(2, 0)$
The graph crosses the y- axis at the point $(0, −6)$ . The y- intercept is $(0, −6)$ .
ⓒ The graph crosses the x- axis at the point $(−5, 0)$ . The x- intercept is $(−5, 0)$ .
The graph crosses the y- axis at the point $(0, −5)$ . The y- intercept is $(0, −5)$ .

Find the x- and y- Intercepts 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 intercepts of $2 x + y = 6$ .

Solution

We will let $y = 0$ to find the x- intercept, and let $x = 0$ to find the y- intercept. We will fill in the table, which reminds us of what we need to find.

The figure shows a table with four rows and two columns. The first row is a title row and it labels the table with the equation 2 x plus y equals 6. The second row is a header row and it labels each column. The first column header is “x” and the second is "y". The third row is labeled “x- intercept” and has the first column blank and a 0 in the second column. The fourth row is labeled “y- intercept” and has a 0 in the first column with the second column blank.

To find the x- intercept, let $y = 0$ .

A step-by-step demonstration of finding x and y intercepts for an equation, showing the algebraic operations and resulting mathematical expressions and coordinates.

Let y = 0.
Simplify.

The x-intercept is	(3, 0)
To find the y-intercept, let x = 0.

Let x = 0.
Simplify.

The y-intercept is	(0, 6)

The intercepts are the points $(3, 0)$ and $(0, 6)$ as shown in Table 4.

$2 x + y = 6$
$x$	$y$
3	0
0	6

Find the intercepts of $4 x - 3 y = 12$ .

Solution

To find the x-intercept, let y = 0.

Let y = 0.
Simplify.


The x-intercept is	(3, 0)
To find the y-intercept, let x = 0.

Let x = 0.
Simplify.


The y-intercept is	(0, −4)

The intercepts are the points (3, 0) and (0, −4) as shown in the following table.

$4 x - 3 y = 12$
$x$	$y$
3	0
0	$−4$

Graph a Line Using the Intercepts

To graph a linear equation by plotting points, you need to find three points whose coordinates are solutions to the equation. You can use the x- and y- intercepts as two of your three points. Find the intercepts, and then find a third point to ensure accuracy. Make sure the points line up—then draw the line. This method is often the quickest way to graph a line.

How to Graph a Line Using Intercepts

Graph $- x + 2 y = 6$ using the intercepts.

Solution

The figure shows a table with the general procedure for graphing a line using the intercepts along with a specific example using the equation negative x plus 2y equals 6. Step 1 of the general procedure is “Find the x and y- intercepts of the line. Let y equals 0 and solve for x. Let x equals 0 and solve for y”. Step 1 for the example is a series of statements and equations: “Find the x- intercept. Let y equals 0”, negative x plus 2y equals 6, negative x plus 2(0) equals 6 (where the 0 is red), negative x equals 6, x equals negative 6, “The x- intercept is (negative 6, 0)”, “Find the y- intercept. Let x equals 0”, negative x plus 2y equals 6, negative 0 plus 2y equals 6 (where the 0 is red), 2y equals 6, y equals 3, and “The y- intercept is (0, 3)”.

Step 3 for the example is a table and a graph. The table has four rows and three columns. The first row is a header row and it labels each column. The first column header is “x”, the second is "y", and the third is “(x,y)”. Under the first column are the numbers negative 6, 0 and 2. Under the second column are the numbers 0, 3, and 4. Under the third column are the ordered pairs (negative 6, 0), (0, 3), and (2, 4). The graph has three points on the x- y coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (negative 6, 0), (0, 3), and (2, 4).

Step 4 of the general procedure is “Draw the line.” For the specific example, there is the statement “See the graph” and a graph of a straight line going through three points on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (negative 6, 0), (0, 3), and (2, 4). The straight line is drawn through the points (negative 6, 0), (negative 4, 1), (negative 2, 2), (0, 3), (2, 4), (4, 5), and (6, 6).

The steps to graph a linear equation using the intercepts are summarized below.

Graph $4 x - 3 y = 12$ using the intercepts.

Solution

Find the intercepts and a third point.

We list the points in Table 7 and show the graph below.

$4 x - 3 y = 12$
$x$	$y$	$(x, y)$
3	0	$(3, 0)$
0	$−4$	$(0, −4)$
6	4	$(6, 4)$

The figure shows the graph of a straight line going through three points on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (0, negative 4), (3, 0), and (6, 4). The straight line is drawn through the points (0, negative 4), (3, 0), and (6, 4).

Graph $y = 5 x$ using the intercepts.

Solution

The figure shows two sets of statements and equations to find the intercepts from an equation. The first set of statements and equations is “x- intercept”, “let y equals 0”, y equals 5x, 0 equals 5x (where the 0 is red), 0 equals x, (0, 0). The second set of statements and equations is “y- intercept”, “let x equals 0”, y equals 5x, y equals 5(0) (where the 0 is red), y equals 0, (0, 0).

This line has only one intercept. It is the point $(0, 0)$ .

To ensure accuracy we need to plot three points. Since the x- and y- intercepts are the same point, we need two more points to graph the line.

The figure shows two sets of statements and equations to find two points from an equation. The first set of statements and equations is “Let x equals 1”, y equals 5x, y equals 5(1) (where the 1 is red), y equals 5. The second set of statements and equations is “Let x equals negative 1”, y equals 5x, y equals 5(negative 1) (where the negative 1 is red), y equals negative 5.

See Table 8.

$y = 5 x$
$x$	$y$	$(x, y)$
0	0	$(0, 0)$
1	5	$(1, 5)$
$−1$	$−5$	$(−1, −5)$

Plot the three points, check that they line up, and draw the line.

The figure shows the graph of a straight line going through three points on the x y- coordinate plane. The x- axis of the plane runs from negative 10 to 10. The y- axis of the planes runs from negative 10 to 10. Three points are marked and labeled with their coordinates at (negative 1, negative 5), (0, 0), and (1, 5). The straight line is drawn through the points (negative 1, negative 5), (0, 0), and (1, 5).

Key Concepts

Find the x- and y- Intercepts from the Equation of a Line
- Use the equation of the line to find the x- intercept of the line, let $y = 0$ and solve for x.
- Use the equation of the line to find the y- intercept of the line, let $x = 0$ and solve for y.
Graph a Linear Equation using the Intercepts
1. Find the x- and y- intercepts of the line.
  Let $y = 0$ and solve for x.
  Let $x = 0$ and solve for y.
2. Find a third solution to the equation.
3. Plot the three points and then check that they line up.
4. Draw the line.
Strategy for Choosing the Most Convenient Method to Graph a Line:
- Consider the form of the equation.
- If it only has one variable, it is a vertical or horizontal line.
  $x = a$ is a vertical line passing through the x- axis at $a$
  $y = b$ is a horizontal line passing through the y- axis at $b$ .
- If y is isolated on one side of the equation, graph by plotting points.
- Choose any three values for x and then solve for the corresponding y- values.
- If the equation is of the form $a x + b y = c$ , find the intercepts. Find the x- and y- intercepts and then a third point.

Practice Makes Perfect

Identify the x- and y- Intercepts on a Graph

In the following exercises, find the x- and y- intercepts on each graph.

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 10 to 10. The y- axis of the planes runs from negative 10 to 10. The straight line goes through the points (negative 5, 8), (negative 4, 7), (negative 3, 6), (negative 2, 5), (negative 1, 4), (0, 3), (1, 2), (2, 1), (3, 0), (4, negative 1), (5, negative 2) and (6, negative 3).

Solution

$(3, 0), (0, 3)$

Solution

$(5, 0), (0, −5)$

Solution

$(−2, 0), (0, −2)$

Solution

$(−1, 0), (0, 1)$

Solution

$(6, 0), (0, 3)$

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. The straight line goes through the points (negative 6, 5), (negative 4, 4), (negative 2, 3), (0, 2), (2, 1), (4, 0), and (6, negative 1).

Solution

$(0, 0)$

Find the x- and y- Intercepts from an Equation of a Line

In the following exercises, find the intercepts for each equation.

$x + y = 4$

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

$(−3, 0), (0, 3)$

$x - y = −4$

$x + 2 y = 8$

Solution

$(8, 0), (0, 4)$

$x + 2 y = 10$

$3 x + y = 6$

Solution

$(2, 0), (0, 6)$

$3 x + y = 9$

$x - 3 y = 12$

Solution

$(12, 0), (0, −4)$

$x - 2 y = 8$

$4 x - y = 8$

Solution

$(2, 0), (0, −8)$

$5 x - y = 5$

$2 x + 5 y = 10$

Solution

$(5, 0), (0, 2)$

$2 x + 3 y = 6$

$3 x - 2 y = 12$

Solution

$(4, 0), (0, −6)$

$3 x - 5 y = 30$

$y = \frac{1}{3} x + 1$

Solution

$(−3, 0), (0, 1)$

$y = \frac{1}{4} x - 1$

$y = \frac{1}{5} x + 2$

Solution

$(−10, 0), (0, 2)$

$y = \frac{1}{3} x + 4$

$y = 3 x$

Solution

$(0, 0)$

$y = −2 x$

$y = −4 x$

Solution

$(0, 0)$

$y = 5 x$

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

$- x + 5 y = 10$

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 12 to 12. The y- axis of the planes runs from negative 12 to 12. The line graphed is negative x plus 5 y equals 10.

$- x + 4 y = 8$

$x + 2 y = 4$

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 12 to 12. The y- axis of the planes runs from negative 12 to 12. The straight line goes through the points (negative 8, 6), (negative 6, 5), (negative 4, 4), (negative 2, 3), (0, 2), (2, 1), (4, 0), (6, negative 1), (8, negative 2), and (10, negative 3).

$x + 2 y = 6$

$x + y = 2$

Solution

$x + y = 5$

$x + y = −3$

Solution

$x + y = −1$

$x - y = 1$

Solution

$x - y = 2$

$x - y = −4$

Solution

$x - y = −3$

$4 x + y = 4$

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 12 to 12. The y- axis of the planes runs from negative 12 to 12. The straight line goes through the points (negative 2, 12), (negative 1, 8), (0, 4), (1, 0), (2, negative 4), (3, negative 8), and (4, negative 12).

$3 x + y = 3$

$2 x + 4 y = 12$

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. The straight line goes through the points (negative 6, 6), (negative 4, 5), (negative 2, 4), (0, 3), (2, 2), (4, 1), and (6, 0).

$3 x + 2 y = 12$

$3 x - 2 y = 6$

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 12 to 12. The y- axis of the planes runs from negative 12 to 12. The straight line goes through the points (negative 6, negative 12), (negative 4, negative 9), (negative 2, negative 6), (0, negative 3), (2, 0), (4, 3), (6, 6), (8, 9), and (10, 12).

$5 x - 2 y = 10$

$2 x - 5 y = −20$

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 12 to 12. The y- axis of the planes runs from negative 12 to 12. The straight line goes through the points (negative 10, 0), (negative 5, 2), (0, 4), (5, 6), and (10, 8).

$3 x - 4 y = −12$

$3 x - y = −6$

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 12 to 12. The y- axis of the planes runs from negative 12 to 12. The straight line goes through the points (negative 6, negative 12), (negative 5, negative 9), (negative 4, negative 6), (negative 3, negative 3), (negative 2, 0), (1, 3), (2, 6), (3, 9), and (4, 12).

$2 x - y = −8$

$y = −2 x$

Solution

$y = −4 x$

$y = x$

Solution

$y = 3 x$

Everyday Math

Road trip. Damien is driving from Chicago to Denver, a distance of 1000 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 figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from 0 to 16. The y- axis of the planes runs from 0 to 1200 in increments of 200. The straight line goes through the points (0, 1000), (3, 800), (6, 600), (9, 400), (12, 200), and (15, 0). The points (0, 1000) and (15, 0) are marked and labeled with their coordinates.

ⓐ Find the x- and y- intercepts.
ⓑ Explain what the x- and y- intercepts mean for Damien.

Solution

ⓐ $(0, 1000), (15, 0)$
ⓑ At $(0, 1000)$ , he has been gone 0 hours and has 1000 miles left. At $(15, 0)$ , he has been gone 15 hours and has 0 miles left to go.

Road trip. Ozzie filled up the gas tank of his truck and headed out on a road trip. The x- axis on the graph below 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 figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from 0 to 350 in increments of 50. The y- axis of the planes runs from 0 to 18 in increments of 2. The straight line goes through the points (0, 16), (150, 8), and (300, 0). The points (0, 16) and (300, 0) are marked and labeled with their coordinates

ⓐ 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 $3 x - 2 y = 6$ ?

Solution

Answers will vary.

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $4 x + y = −4$ ? Why?

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $y = \frac{2}{3} x - 2$ ? Why?

Solution

Answers will vary.

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $y = 6$ ? Why?

Self Check

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

The figure shows a table with four rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is "confidently", the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “identify the x and y intercepts of a graph”, “find the x and y intercepts from an equation of a line”, and “graph a line using intercepts”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved.

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

intercepts of a line

The points where a line crosses the x- axis and the y- axis are called the intercepts of the line.

x- intercept

The point

(a, 0)

where the line crosses the x- axis; the x- intercept occurs when

y

is zero.

y-intercept

The point

(0, b)

where the line crosses the y- axis; the y- intercept occurs when

x

is zero.

Understand Slope of a Line

When you graph linear equations, you may notice that some lines tilt up as they go from left to right and some lines tilt down. Some lines are very steep and some lines are flatter. What determines whether a line tilts up or down or if it is steep or flat?

In mathematics, the ‘tilt’ 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, grade of a highway, and a ramp for a wheelchair are some examples where you literally see slopes. And when you ride a bicycle, you feel the slope as you pump uphill or coast downhill.

In this section, we will explore the concept of slope.

Use Geoboards to Model Slope

A geoboard is a board with a grid of pegs on it. Using rubber bands on a geoboard gives us 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.

We’ll start by stretching a rubber band between two pegs as shown in Figure 1.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 4 and the peg in column 4, row 2, forming a line.

Doesn’t 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

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 2, the peg in column 1, row 4, and the peg in column 4, row 2, forming a right triangle. The 1, 2 peg is the vertex of the 90 degree angle, while the line between the 1, 4 and 4, 2 pegs forms the hypotenuse of the triangle.

We carefully make a 90º angle around the third peg, so one of the newly formed lines is vertical and the other is horizontal.

To find the slope of the line, we measure the distance along the vertical and horizontal sides of the triangle. The vertical distance is called the rise and the horizontal distance is called the run, as shown in Figure 3.

In this illustration, there are two perpendicular lines with arrows. The first line extends straight upward and is labeled “rise”. The second arrow extends straight rightward and is labeled “run”.

If our geoboard and rubber band look just like the one shown in Figure 4, the rise is 2. The rubber band goes up 2 units. (Each space is one unit.)

What is the run?

The rubber band goes across 3 units. The run is 3 (see Figure 4).

The slope of a line is the ratio of the rise to the run. In mathematics, it is always referred to with the letter $m$ .

What is the slope of the line on the geoboard in Figure 4?

\begin{matrix} m & = & \frac{rise}{run} \\ m & = & \frac{2}{3} \end{matrix}

The line has slope $\frac{2}{3}$ . This means that the line rises 2 units for every 3 units of run.

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. If the rise goes up it is positive and if it goes down it is negative. The run will go from left to right and be positive.

What is the slope of the line on the geoboard shown?
The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 5 and the peg in column 5, row 2, forming a line.

Solution

Use the definition of slope: $m = \frac{rise}{run} .$

Start at the left peg and count the spaces up and to the right to reach the second peg.

This table demonstrates the calculation of a line's slope (m) using its defined rise and run components.
The rise is 3.	$m = \frac{3}{run}$
The run is 4.	$m = \frac{3}{4}$
	The slope is $\frac{3}{4}$ .

This means that the line rises 3 units for every 4 units of run.

Solution

Use the definition of slope: $m = \frac{rise}{run} .$

Start at the left peg and count the units down and to the right to reach the second peg.

This table demonstrates the step-by-step calculation of slope (m) from given rise and run values, concluding with m = -1/3.
The rise is −1.	$= \frac{−1}{run}$
The run is 3.	$\begin{matrix} m = \frac{−1}{3} \\ m = - \frac{1}{3} \end{matrix}$
	The slope is $- \frac{1}{3}$ .

This means that the line drops 1 unit for every 3 units of run.

Notice that in Example 1 the slope is positive and in Example 2 the slope is negative. Do you notice any difference in the two lines shown in Figure 5(a) and Figure 5(b)?

The figure shows two grids of evenly spaced pegs, one labeled (a) and one labeled (b). There are 5 columns and 5 rows of pegs in each grid. In the (a) grid, a rubber band is stretched between the peg in column 1, row 5 and the peg in column 5, row 2, forming a line. Below this grid is the slope of a line defined as m equals 3 fourths. In the (b) grid, a rubber band is stretched between the peg in column 1, row 3 and the peg in column 4, row 4, forming a line. Below this grid is the slope of a line defined as m equals negative 1 third.

We ‘read’ a line from left to right just like we read words in English. As you read from left to right, the line in Figure 5(a) is going up; it has positive slope. The line in Figure 5(b) is going down; it has negative slope.

Use a geoboard to model a line with slope $\frac{1}{2}$ .

Solution

To model a line on a geoboard, we need the rise and the run.

Demonstrates steps to use the slope formula, showing the action and the resulting mathematical expression for each step.
Use the slope formula.	$m = \frac{rise}{run}$
Replace $m$ with $\frac{1}{2}$ .	$\frac{1}{2} = \frac{rise}{run}$

So, the rise is 1 and the run is 2.

Start at a peg in the lower left of the geoboard.

Stretch the rubber band up 1 unit, and then right 2 units.

The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is $\frac{1}{2}$ .

Use a geoboard to model a line with slope $\frac{−1}{4} .$

Solution

Steps and corresponding mathematical expressions for applying the slope formula.
Use the slope formula.	$m = \frac{rise}{run}$
Replace $m$ with $\frac{−1}{4}$ .	$\frac{−1}{4} = \frac{rise}{run}$

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 go to the right 4 units, as shown.

The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is $\frac{−1}{4}$ .

Use $m = \frac{rise}{run}$ to Find the Slope of a Line from its Graph

Now, we’ll look at some graphs on the $x y$ -coordinate plane and see how to find their slopes. The method will be very similar to what we just modeled on our geoboards.

To find the slope, we must count out the rise and the run. But where do we start?

We locate two points on the line whose coordinates are integers. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.

How to Use $m = \frac{rise}{run}$ to Find the Slope of a Line from its Graph

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 6 and the y-axis runs from negative 4 to 2. A line passes through the points (0, negative 3) and (5, 1).

Solution

This table has three columns and four rows. The first row says, “Step 1. Locate two points on the graph whose coordinates are integers. Mark (0, negative 3) and (5, 1).” To the right is a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 1 to 6. The y-axis of the plane runs from negative 4 to 2. The points (0, negative 3) and (5, 1) are plotted.

The second row says, “Step 2. Starting with the point on the left, sketch a right triangle, with the hypotenuse going from the first point to the second point. Starting at (0, negative 3), sketch a right triangle to (5, 1).” In the graph on the right, an additional point is plotted at (0, 1). The three points form a right triangle, with the line from (0, negative 3) to (5, 1) forming the hypotenuse and the lines from (0, negative 3) to (0, 1) and (0, 1) to (5, 1) forming the legs.

The third row then says, “Step 3. Count the rise and the run on the legs of the triangle.” The rise is 4 and the run is 5.

The fourth row says, “Step 4. Take the ratio of the rise to run to find the slope. Use the slope formula. Substitute the values of the rise and run.” To the right is the slope formula, m equals rise divided by run. The slope of the line is 4 divided by 5, or four fifths. This means that y increases 4 units as x increases 5 units.

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 9 and the y-axis runs from negative 1 to 7. A line passes through the points (0, 5), (3, 3), and (6, 1).

Solution

Locate two points on the graph whose coordinates are integers.	$(0, 5)$ and $(3, 3)$
Which point is on the left?	$(0, 5)$
Starting at $(0, 5)$ , sketch a right triangle to $(3, 3)$ .
Count the rise—it is negative.	The rise is $−2$ .
Count the run.	The run is 3.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values of the rise and run.	$m = \frac{−2}{3}$
Simplify.	$m = - \frac{2}{3}$
	The slope of the line is $- \frac{2}{3}$ .

So $y$ decreases by 2 units as $x$ increases by 3 units.

What if we used the points $(−3, 7)$ and $(6, 1)$ to find the slope of the line?

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 3, 7) and (6, 1). An additional point is plotted at (negative 3, 1). The three points form a right triangle, with the line from (negative 3, 7) to (6, 1) forming the hypotenuse and the lines from (negative 3, 7) to negative 1, 7) and from (negative 1, 7) to (6, 1) forming the legs.

The rise would be $−6$ and the run would be 9. Then $m = \frac{−6}{9}$ , and that simplifies to $m = - \frac{2}{3}$ . Remember, it does not matter which points you use—the slope of the line is always the same.

In the last two examples, the lines had y-intercepts with integer values, so it was convenient to use the y-intercept as one of the points to find the slope. In the next example, the y-intercept is a fraction. Instead of using that point, we’ll look for two other points whose coordinates are integers. This will make the slope calculations easier.

Find the slope of the line shown.

The graph shows the x y coordinate plane. The x-axis runs from 0 to 8 and the y-axis runs from 0 to 7. A line passes through the points (2, 3) and (7, 6).

Solution

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 triangle to $(7, 6)$ .
Count the rise.	The rise is 3.
Count the run.	The run is 5.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values of the rise and run.	$m = \frac{3}{5}$
	The slope of the line is $\frac{3}{5}$ .

This means that $y$ increases 5 units as $x$ increases 3 units.

When we used geoboards to introduce the concept of slope, we said that we would always start with the point on the left and count the rise and the run to get to the point on the right. That way the run was always positive and the rise determined whether the slope was positive or negative.

What would happen if we started with the point on the right?

Let’s use the points $(2, 3)$ and $(7, 6)$ again, but now we’ll start at $(7, 6)$ .

The graph shows the x y coordinate plane. The x -axis runs from 0 to 8. The y -axis runs from 0 to 7. A line passes through the points (2, 3) and (7, 6). An additional point is plotted at (7, 3). The three points form a right triangle, with the line from (2, 3) to (7, 6) forming the hypotenuse and the lines from (2, 3) to (7, 3) and from (7, 3) to (7, 6) forming the legs.

Detailed steps to calculate the slope of a line (m) using the rise over run method, presenting both actions and their numerical outcomes.
Count the rise.	The rise is $−3$ .
Count the run. It goes from right to left, so it is negative.	The run is $−5$ .
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values of the rise and run.	$m = \frac{−3}{−5}$
	The slope of the line is $\frac{−3}{−5}$ .

It does not matter where you start—the slope of the line is always the same.

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.

\begin{matrix} Horizontal line y = b & Vertical line x = a \\ y -coordinates are the same. & x -coordinates are the same. \end{matrix}

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 when we do this.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 5 and the y-axis runs from negative 1 to 7. A line passes through the points (0, 4) and (3, 4).

Calculates the slope of a horizontal line (y=4) by identifying the rise (0) and run (3), demonstrating that the resulting slope is 0.
What is the rise?	The rise is $0$ .
Count the run.	The run is $3$ .
What is the slope?	$\begin{matrix} m = \frac{rise}{run} \\ m = \frac{0}{3} \\ m = 0 \end{matrix}$
	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.

The floor of your room is horizontal. Its slope is 0. If you carefully placed a ball on the floor, it would not roll away.

Now, we’ll consider a vertical line, the line.

The graph shows the x y coordinate plane. The x-axis runs from negative 1 to 5 and the y-axis runs from negative 2 to 2. A line passes through the points (3, 0) and (3, 2).

Illustrates slope calculation with a zero run, resulting in an undefined value.
What is the rise?	The rise is $2$ .
Count the run.	The run is $0$ .
What is the slope?	$\begin{matrix} m = \frac{rise}{run} \\ m = \frac{2}{0} \end{matrix}$

But we can’t divide by 0. Division by 0 is not defined. So we say that the slope of the vertical line $x = 3$ is undefined.

The slope of any vertical line is undefined. When the x-coordinates of a line are all the same, the run is 0.

Find the slope of each line:

ⓐ $x = 8$ ⓑ $y = −5$ .

Solution

ⓐ $x = 8$
This is a vertical line.
Its slope is undefined.
ⓑ $y = −5$
This is a horizontal line.
It has slope 0.

Remember, we ‘read’ a line from left to right, just like we read written words in English.

Use the Slope Formula to find the Slope of a Line Between Two Points

Sometimes we’ll need to find the slope of a line between two points when we don’t have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but as we’ll see, there is a way to find the slope without graphing. Before we get to it, we need to introduce some algebraic notation.

We have seen that an ordered pair $(x, y)$ gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol $(x, y)$ be used to represent two different points? Mathematicians use subscripts to distinguish the points.

\begin{matrix} (x_{1}, y_{1}) & read ‘ x sub 1, y sub 1’ \\ (x_{2}, y_{2}) & read ‘ x sub 2, y sub 2’ \end{matrix}

The use of subscripts in math is very much like the use of last name initials in elementary school. Maybe you remember Laura C. and Laura M. in your third grade class?

We will use $(x_{1}, y_{1})$ to identify the first point and $(x_{2}, y_{2})$ to identify the second point.

If we had more than two points, we could use $(x_{3}, y_{3})$ , $(x_{4}, y_{4})$ , and so on.

Let’s see how the rise and run relate to the coordinates of the two points by taking another look at the slope of the line between the points $(2, 3)$ and $(7, 6)$ .

The graph shows the x y coordinate plane. The x and y-axes run from 0 to 7. A line passes through the points (2, 3) and (7, 6), which are plotted and labeled. The ordered pair (2, 3) is labeled (x subscript 1, y subscript 1). The ordered pair (7, 6) is labeled (x subscript 2, y subscript 2). An additional point is plotted at (2, 6). The three points form a right triangle, with the line from (2, 3) to (7, 6) forming the hypotenuse and the lines from (2, 3) to (2, 6) and from (2, 6) to (7, 6) forming the legs. The first leg, from (2, 3) to (2, 6) is labeled y subscript 2 minus y subscript 1, 6 minus 3, and 3. The second leg, from (2, 3) to (7, 6), is labeled x subscript 2 minus x subscript 1, y minus 2, and 5.

Since we have two points, we will use subscript notation, $(\overset{x_{1},}{2,} \overset{y_{1}}{3})$ $(\overset{x_{2}, y_{2}}{7, 6})$ .

On the graph, we counted the rise of 3 and the run of 5.

Notice that the rise of 3 can be found by subtracting the y-coordinates 6 and 3.

3 = 6 - 3

And the run of 5 can be found by subtracting the x-coordinates 7 and 2.

5 = 7 - 2

We know $m = \frac{rise}{run}$ . So $m = \frac{3}{5}$ .

We rewrite the rise and run by putting in the coordinates $m = \frac{6 - 3}{7 - 2}$ .

But 6 is $y_{2}$ , the y-coordinate of the second point and 3 is $y_{1}$ , the y-coordinate of the first point.

So we can rewrite the slope using subscript notation. $m = \frac{y_{2} - y_{1}}{7 - 2}$

Also, 7 is $x_{2}$ , the x-coordinate of the second point and 2 is $x_{1}$ , the x-coordinate of the first point.

So, again, we rewrite the slope using subscript notation. $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

We’ve shown that $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is really another version of $m = \frac{rise}{run}$ . We can use this formula to find the slope of a line when we have two points on the line.

Use the slope formula to find the slope of the line between the points $(1, 2)$ and $(4, 5)$ .

Solution

Step-by-step calculation of the slope of a line between two given points (1,2) and (4,5) using the slope formula.
We'll call $(1, 2)$ point #1 and $(4, 5)$ point #2.	$(\overset{x_{1}, y_{1}}{1, 2}) {(\overset{x_{2}, y_{2}}{4, 5})}_{}$
Use the slope formula.	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
Substitute the values.
$y$ of the second point minus $y$ of the first point	$m = \frac{5 - 2}{x_{2} - x_{1}}$ .
$x$ of the second point minus $x$ of the first point	$m = \frac{5 - 2}{4 - 1}$ .
Simplify the numerator and the denominator.	$m = \frac{3}{3}$ .
Simplify.	$m = 1$ .

Let’s confirm this by counting out the slope on a graph using $m = \frac{rise}{run}$ .

The graph shows the x y-coordinate plane. The x and y-axes of the plane run from 0 to 7. A line passes through the points (1, 2) and (4, 5), which are plotted. An additional point is plotted at (1, 5). The three points form a right triangle, with the line from (1, 2) to (4, 5) forming the hypotenuse and the lines from (1, 2) to (1, 5) and from (1, 5) to (4, 5) forming the legs. The leg from (1, 2) to (1, 5) is labeled “rise” and the leg from (1, 5) to (4, 5) is labeled “run”.

It doesn’t matter which point you call point #1 and which one you call point #2. The slope will be the same. Try the calculation yourself.

Use the slope formula to find the slope of the line through the points $(−2, −3)$ and $(−7, 4)$ .

Solution

This table outlines the step-by-step process and mathematical expressions used to calculate the slope between two specific points: (-2,-3) and (-7,4).
We'll call $(−2, −3)$ point #1 and $(−7, 4)$ point #2.	$(\overset{x_{1}, y_{1}}{−2, −3}) {(\overset{x_{2}, y_{2}}{−7, 4})}_{}$
Use the slope formula.	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
Substitute the values.
$y$ of the second point minus $y$ of the first point	$m = \frac{4 - (−3)}{x_{2} - x_{1}}$ .
$x$ of the second point minus $x$ of the first point	$m = \frac{4 - (−3)}{−7 - (−2)}$ .
Simplify.	$\begin{matrix} m = \frac{7}{−5} \\ m = - \frac{7}{5} \end{matrix}$

Let’s verify this slope on the graph shown.

The graph shows the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 2 and the y-axis of the plane runs from negative 6 to 5. A line passes through the points (negative 7, 4) and (negative 2, negative 3), which are plotted and labeled. An additional point is plotted at (negative 7, negative 3). The three points form a right triangle, with the line from (negative 7, 4) to (negative 2, negative 3) forming the hypotenuse and the lines from (negative 7, 4) to (negative 7, negative 3) and from (negative 7, negative 3) to (negative 2, negative 3) forming the legs. The leg from (negative 7, 4) to (negative 7, negative 3) is labeled “rise” and the leg from (negative 7, negative 3) to (negative 2, negative 3) is labeled “run”.

\begin{matrix} m & = & \frac{rise}{run} \\ m & = & \frac{−7}{5} \\ m & = & - \frac{7}{5} \end{matrix}

Graph a Line Given a Point and the Slope

Up to now, in this chapter, we have graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.

One other method we can use to graph lines is called the point–slope method. We will use this method when we know one point and the slope of the line. We will start by plotting the point and then use the definition of slope to draw the graph of the line.

How To Graph a Line Given a Point and The Slope

Graph the line passing through the point $(1, −1)$ whose slope is $m = \frac{3}{4}$ .

Solution

This table has three columns and four rows. The first row says, “Step 1. Plot the given point. Plot (1, negative 1).” To the right is a graph of the x y-coordinate plane. The x-axis of the plane runs from negative 1 to 7. The y-axis of the plane runs from negative 3 to 4. The point (0, negative 1) is plotted.

The second row says, “Step 2. Use the slope formula m equals rise divided by run to identify the rise and the run.” The rise and run are 3 and 4, so m equals 3 divided by 4.

The third row says “Step 3. Starting at the given point, count out the rise and run to mark the second point.” We start at (1, negative 1) and count the rise and run. Up three units and right 4 units. In the graph on the right, an additional two points are plotted: (1, 2), which is 3 units up from (1, negative 1), and (5, 2), which is 3 units up and 4 units right from (1, negative 1).

The fourth row says “Step 4. Connect the points with a line.” On the graph to the right, a line is drawn through the points (1, negative 1) and (5, 2). This line is also the hypotenuse of the right triangle formed by the three points, (1, negative 1), (1, 2) and (5, 2).

Graph the line with y-intercept 2 whose slope is $m = - \frac{2}{3}$ .

Solution

Plot the given point, the y-intercept, $(0, 2)$ .

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. The point (0, 2) is plotted.

This table demonstrates how to identify the rise and run from a given slope (m) by breaking down the fraction into its numerator and denominator components.
Identify the rise and the run.	$m = - \frac{2}{3}$
	$\frac{rise}{run} = \frac{−2}{3}$
	$rise = −2$
	$run = 3$

Count the rise and the run. Mark the second point.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. The points (0, 2), (0, 0), and (3,0) are plotted and labeled. The line from (0, 2) to (0, 0) is labeled “down 2” and the line from (0, 0) to (3, 0) is labeled “right 3”.

Connect the two points with a line.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. A line passes through the plotted points (0, 2) and (3,0).

You can check your work by finding a third point. Since the slope is $m = - \frac{2}{3}$ , it can be written as $m = \frac{2}{−3}$ . Go back to $(0, 2)$ and count out the rise, 2, and the run, $−3$ .

Graph the line passing through the point $(−1, −3)$ whose slope is $m = 4 .$

Solution

Plot the given point.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. The point (negative 1, negative 3) is plotted and labeled.

Steps demonstrating how to identify the 'rise' and 'run' components from a given slope 'm' expressed as an integer.
Identify the rise and the run.	$m = 4$
Write 4 as a fraction.	$\frac{rise}{run} = \frac{4}{1}$
	$rise = 4, run = 1$

Count the rise and run and mark the second point.

This figure shows how to graph the line passing through the point (negative 1, negative 3) whose slope is 4. The first step is to identify the rise and run. The rise is 4 and the run is 1. 4 divided by 1 is 4, so the slope is 4. Next we count the rise and run and mark the second point. To the right is a graph of the x y-coordinate plane. The x and y-axes run from negative 5 to 5. We start at the plotted point (negative 1, negative 3) and count the rise, 4. We reach the point negative 1, 1, which we plot. We then count the run from this point, which is 1. We reach the point (0, 1), which is plotted. The last step is to connect the two points with a line. We draw a line which passes through the points (negative 1, negative 3) and (0, 1).

Connect the two points with a line.

The graph shows the x y coordinate plane. The x and y-axes run from negative 5 to 5. A line passes through the plotted points (-1, -3) and (1,0).

You can check your work by finding a third point. Since the slope is $m = 4$ , it can be written as $m = \frac{−4}{−1}$ . Go back to $(−1, −3)$ and count out the rise, $−4$ , and the run, $−1$ .

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 equals 9 feet". There is a line segment with arrows at each end measuring the horizontal length of the root and is labeled "run equals 18 feet".

Solution

This table illustrates the step-by-step calculation and interpretation of a roof's slope.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values for rise and run.	$m = \frac{9}{18}$
Simplify.	$m = \frac{1}{2}$
The slope of the roof is $\frac{1}{2}$ .
	The roof rises 1 foot for every 2 feet of horizontal run.

Have you ever thought about the sewage pipes going from your house to the street? They must slope down $\frac{1}{4}$ inch per foot in order to drain properly. What is the required slope?

This figure is a right triangle. One leg is negative one quarter inch and the other leg is one foot.

Solution

Steps to calculate the slope of a pipe, showing formula application, unit conversion, and simplification to arrive at the final slope value.
Use the slope formula.	$\begin{matrix} m = \frac{rise}{run} \\ m = \frac{- \frac{1}{4} inch}{1 foot} \\ m = \frac{- \frac{1}{4} inch}{12 inches} \end{matrix}$
Simplify.	$m = - \frac{1}{48}$
	The slope of the pipe is $- \frac{1}{48}$ .

The pipe drops 1 inch for every 48 inches of horizontal run.

Key Concepts

Find the Slope of a Line from its Graph using m=riserun
1. Locate two points on the line whose coordinates are integers.
2. Starting with 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.
Graph a Line Given a Point and the Slope
1. Plot the given point.
2. Use the slope formula $m = \frac{rise}{run}$ to identify the rise and the run.
3. Starting at the given point, count out the rise and run to mark the second point.
4. Connect the points with a line.
Slope 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

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 pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 3 and the peg in column 5, row 2, forming a line.

Solution

$\frac{1}{4}$

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 4 and the peg in column 5, row 2, forming a line.

Solution

$\frac{2}{3}$

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 3, row 4 and the peg in column 5, row 1, forming a line.

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 1 and the peg in column 4, row 4, forming a line.

Solution

$\frac{−3}{2} = - \frac{3}{2}$

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 1 and the peg in column 5, row 4, forming a line.

Solution

$- \frac{3}{4}$

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 2 and the peg in column 4, row 5, forming a line.

In the following exercises, model each slope. Draw a picture to show your results.

$\frac{2}{3}$

Solution

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 2, row 5 and the peg in column 5, row 3, forming a line.

$\frac{3}{4}$

$\frac{1}{4}$

Solution

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 4 and the peg in column 5, row 3, forming a line.

$\frac{4}{3}$

$- \frac{1}{2}$

Solution

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 4 and the peg in column 3, row 5, forming a line.

$- \frac{3}{4}$

$- \frac{2}{3}$

Solution

The figure shows a grid of evenly spaced pegs. There are 5 columns and 5 rows of pegs. A rubber band is stretched between the peg in column 1, row 2 and the peg in column 4, row 4, forming a line.

$- \frac{3}{2}$

Use $m = \frac{rise}{run}$ to 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 and y-axes run from negative 10 to 10. A line passes through the points (negative 10, negative 8), (0, negative 4), and (10, 0).

Solution

$\frac{2}{5}$

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, negative 8) and (2, negative 2).

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 4, negative 6) and (4, 4).

Solution

$\frac{5}{4}$

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line intercepts the y-axis at (0, negative 2) and passes through the point (3, 3).

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 3, 3) and (3, 1).

Solution

$- \frac{1}{3}$

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, 4) and (2, 2).

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line intercepts the y-axis at (0, 6) and passes through the point (4, 3).

Solution

$- \frac{3}{4}$

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the point (negative 3, 1) and intercepts the y-axis at (0, negative 1).

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, 1) and (2, 4).

Solution

$\frac{3}{4}$

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 1, 1) and (2, 3).

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 1, 6) and (1, 1).

Solution

$−3$

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the point (negative 1, 3) and intercepts the x-axis at (3, 0).

The graph shows the x y coordinate plane. The x and y-axes run from negative 7 to 7. A line passes through the points (negative 2, 6) and (1, 4).

Solution

$- \frac{2}{3}$

The graph shows the x y coordinate plane. The x and y-axes run from negative 10 to 10. A line passes through the points (negative 1, 3) and (1, 2).

The graph shows the x y coordinate plane. The x and y-axes run from negative 10 to 10. A line intercepts the x-axis at (negative 2, 0) and passes through the point (2, 1).

Solution

$\frac{1}{4}$

The graph shows the x y coordinate plane. The x and y-axes run from negative 10 to 10. A line passes through the points (4, 2) and (7, 3).

Find the Slope of Horizontal and Vertical Lines

In the following exercises, find the slope of each line.

$y = 3$

Solution

$y = 1$

$x = 4$

Solution

undefined

$x = 2$

$y = −2$

Solution

$y = −3$

$x = −5$

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

$\frac{5}{2}$

$(2, 3), (5, 7)$

$(0, 3), (4, 6)$

Solution

$\frac{3}{4}$

$(0, 1), (5, 4)$

$(2, 5), (4, 0)$

Solution

$- \frac{5}{2}$

$(3, 6), (8, 0)$

$(−3, 3), (4, −5)$

Solution

$- \frac{8}{7}$

$(−2, 4), (3, −1)$

$(−1, −2), (2, 5)$

Solution

$\frac{7}{3}$

$(−2, −1), (6, 5)$

$(4, −5), (1, −2)$

Solution

$−1$

$(3, −6), (2, −2)$

Graph a Line Given a Point and the Slope

In the following exercises, graph each line with the given point and slope.

$(1, −2)$ ; $m = \frac{3}{4}$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (1, negative 2) and (5, 1).

$(1, −1)$ ; $m = \frac{2}{3}$

$(2, 5)$ ; $m = - \frac{1}{3}$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (2, 5) and (5, 4).

$(1, 4)$ ; $m = - \frac{1}{2}$

$(−3, 4)$ ; $m = - \frac{3}{2}$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 3, 4) and (negative 1, 1).

$(−2, 5)$ ; $m = - \frac{5}{4}$

$(−1, −4)$ ; $m = \frac{4}{3}$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 1, negative 4) and intercepts the x-axis at (2, 0).

$(−3, −5)$ ; $m = \frac{3}{2}$

y-intercept 3; $m = - \frac{2}{5}$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line intercepts the y-axis at (0, 3) and passes through the point (5, 1).

y-intercept 5; $m = - \frac{4}{3}$

x-intercept $−2$ ; $m = \frac{3}{4}$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line intercepts the x-axis at (negative 2, 0) and passes through the point (2, 3).

x-intercept $−1$ ; $m = \frac{1}{5}$

$(−3, 3)$ ; $m = 2$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (negative 3, 3) and (negative 2, 5).

$(−4, 2)$ ; $m = 4$

$(1, 5)$ ; $m = −3$

Solution

The graph shows the x y coordinate plane. The x and y-axes run from negative 12 to 12. A line passes through the points (1, 5) and (2, 2).

$(2, 3)$ ; $m = −1$

Everyday Math

Slope of a roof. An easy way to determine the slope of a roof is to set one end of a 12 inch level on the roof surface and hold it level. Then take a tape measure or ruler and measure from the other end of the level down to the roof surface. This will give you the slope of the roof. Builders, sometimes, refer to this as pitch and state it as an “ $x$ 12 pitch” meaning $\frac{x}{12}$ , where $x$ is the measurement from the roof to the level—the rise. It is also sometimes stated as an “ $x$ -in-12 pitch”.

ⓐ What is the slope of the roof in this picture?
ⓑ What is the pitch in construction terms?

Solution

ⓐ $\frac{1}{3}$ ⓑ 4 12 pitch or 4-in-12 pitch

The slope of the roof shown here is measured with a 12” level and a ruler. What is the slope of this roof?

This figure shows one side of a sloped roof of a house. The rise of the roof is measured with a ruler and shown to be 7 inches. The run of the roof is measured with a twelve inch level and shown to be 12 inches.

Road grade. A local road has a grade of 6%. The grade of a road is its slope expressed as a percent. Find the slope of the road as a fraction and then simplify. What rise and run would reflect this slope or grade?

Solution

$\frac{3}{50}$ ; rise = 3, run = 50

Highway grade. A local road rises 2 feet for every 50 feet of highway.

ⓐ What is the slope of the highway?
ⓑ The grade of a highway is its slope expressed as a percent. What is the grade of this highway?

Wheelchair ramp. The rules for wheelchair ramps require a maximum 1-inch rise for a 12-inch run.

ⓐ How long must the ramp be to accommodate a 24-inch rise to the door?
ⓑ Create a model of this ramp.

Solution

ⓐ 288 inches (24 feet) ⓑ Models will vary.

Wheelchair ramp. A 1-inch rise for a 16-inch run makes it easier for the wheelchair rider to ascend a ramp.

ⓐ How long must a ramp be to easily accommodate a 24-inch rise to the door?
ⓑ Create a model of this ramp.

Writing Exercises

What does the sign of the slope tell you about a line?

Solution

When the slope is a positive number the line goes up from left to right. When the slope is a negative number the line goes down from left to right.

How does the graph of a line with slope $m = \frac{1}{2}$ differ from the graph of a line with slope $m = 2$ ?

Why is the slope of a vertical line “undefined”?

Solution

A vertical line has 0 run and since division by 0 is undefined the slope is undefined.

Self Check

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

This table has seven rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the next row reads “use geoboards to model slope.” The third row reads “use m equals rise divided by run to find the slope of a line from its graph.” The fourth row reads “find the slope of horizontal and vertical lines.” The fifth row reads “use the slope formula to find the slope of a line between two points.” The sixth row reads “graph a line given a point and the slope.” The last row reads “solve slope applications.” The remaining columns are blank.

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

geoboard

A geoboard is a board with a grid of pegs on it.

negative slope

A negative slope of a line goes down as you read from left to right.

positive slope

A positive slope of a line goes up as you read from left to right.

rise

The rise of a line is its vertical change.

run

The run of a line is its horizontal change.

slope formula

The slope of the line between two points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

slope of a line

The slope of a line is

m = \frac{rise}{run}

. The rise measures the vertical change and the run measures the horizontal change.

Use the Slope-Intercept Form of an Equation of a Line

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

We have graphed linear equations by plotting points, using intercepts, recognizing horizontal and vertical lines, and using the point–slope method. Once we see how an equation in slope–intercept form and its graph are related, we’ll have one more method we can use to graph lines.

In Graph Linear Equations in Two Variables, we graphed the line of the equation $y = \frac{1}{2} x + 3$ by plotting points. See Figure 1. Let’s find the slope of this line.

This figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The line is labeled with the equation y equals one half x, plus 3. The points (0, 3), (2, 4) and (4, 5) are labeled also. A red vertical line begins at the point (2, 4) and ends one unit above the point. It is labeled “Rise equals 1”. A red horizontal line begins at the end of the vertical line and ends at the point (4, 5). It is labeled “Run equals 2. The red lines create a right triangle with the line y equals one half x, plus 3 as the hypotenuse.

The red lines show us the rise is 1 and the run is 2. Substituting into the slope formula:

\begin{matrix} m & = & \frac{rise}{run} \\ m & = & \frac{1}{2} \end{matrix}

What is the y-intercept of the line? The y-intercept is where the line crosses the y-axis, so y-intercept is $(0, 3)$ . The equation of this line is:

$The figure shows the equation y equals one half x, plus 3. The fraction one half is colored red and the number 3 is colored blue.$

Notice, the line has:

The figure shows the statement “slope m equals one half and y-intercept (0, 3). The slope, one half, is colored red and the number 3 in the y-intercept is colored blue.

When a linear equation is solved for $y$ , the coefficient of the $x$ term is the slope and the constant term is the y-coordinate of the y-intercept. We say that the equation $y = \frac{1}{2} x + 3$ is in slope–intercept form.

$The figure shows the statement “m equals one half; y-intercept is (0, 3). The slope, one half, is colored red and the number 3 in the y-intercept is colored blue. Below that statement is the equation y equals one half x, plus 3. The fraction one half is colored red and the number 3 is colored blue. Below the equation is another equation y equals m x, plus b. The variable m is colored red and the variable b is colored blue.$

Sometimes the slope–intercept form is called the “y-form.”

Use the graph to find the slope and y-intercept of the line, $y = 2 x + 1$ .

Compare these values to the equation $y = m x + b$ .

Solution

To find the slope of the line, we need to choose two points on the line. We’ll use the points $(0, 1)$ and $(1, 3)$ .


Find the rise and run.


Find the y-intercept of the line.	The y-intercept is the point (0, 1).

The slope is the same as the coefficient of $x$ and the y-coordinate of the y-intercept is the same as the constant term.

Identify the Slope and y-Intercept From an Equation of a Line

In Understand Slope of a Line, we graphed a line using the slope and a point. When we are given an equation in slope–intercept form, we can use the y-intercept as the point, and then count out the slope from there. Let’s practice finding the values of the slope and y-intercept from the equation of a line.

Identify the slope and y-intercept of the line with equation $y = −3 x + 5$ .

Solution

We compare our equation to the slope–intercept form of the equation.


Write the equation of the line.
Identify the slope.
Identify the y-intercept.

When an equation of a line is not given in slope–intercept form, our first step will be to solve the equation for $y$ .

Identify the slope and y-intercept of the line with equation $x + 2 y = 6$ .

Solution

This equation is not in slope–intercept form. In order to compare it to the slope–intercept form we must first solve the equation for $y$ .

Solve for y.	$x + 2 y = 6$
Subtract x from each side.
Divide both sides by 2.
Simplify.
$(Remember: \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c})$
Simplify.
Write the slope–intercept form of the equation of the line.
Write the equation of the line.	$A mathematical equation, y = -1/2x + 3, is displayed on a white background. The fraction '-1/2' is written in red, and the number '3' is in blue, while the rest of the equation is in black.$
Identify the slope.
Identify the y-intercept.

Graph a Line Using its Slope and Intercept

Now that we know how to find the slope and y-intercept of a line from its equation, we can graph the line by plotting the y-intercept and then using the slope to find another point.

How to Graph a Line Using its Slope and Intercept

Graph the line of the equation $y = 4 x - 2$ using its slope and y-intercept.

Solution

The figure shows the steps to graph the equation y equals 4x minus 2. Step 1 is to find the slope intercept form of the equation. The equation is already in slope intercept form.

Step 2 is to identify the slope and y-intercept. Use the equation y equals m x, plus b. The equation y equals m x, plus b is shown with the variable m colored red and the variable b colored blue. Below that is the equation y equals 4 x, plus -2. The number 4 is colored red and -2 is colored blue. From this equation we can see that m equals 4 and b equals -2 so the slope is 4 and the y-intercept is the point (0, negative 2).

Step 3 is to plot the y-intercept. An x y-coordinate plane is shown with the x-axis of the plane running from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The point (0, negative 2) is plotted.

Step 4 is to use the slope formula m equals rise over run to identify the rise and the run. Since m equals 4, rise over run equals 4 over 1. From this we can determine that the rise is 4 and the run is 1.

Step 5 is to start at they-intercept, count out the rise and run to mark the second point. So start at the point (0, negative 2) and count the rise and the run. The rise is up 4 and the run is right 1. On the x y-coordinate plane is a red vertical line starts at the point (0, negative 2) and rises 4 units at its end a red horizontal line runs 1 unit to end at the point (1, 2). The point (1, 2) is plotted.

Step 6 is to connect the points with a line. On the x y-coordinate plane the points (0, negative 2) and (1, 2) are plotted and a line runs through the two points. The line is the graph of y equals 4 x, minus 2.

Graph the line of the equation $y = − x + 4$ using its slope and y-intercept.

Solution

	$y = m x + b$
The equation is in slope–intercept form.	$y = − x + 4$
Identify the slope and y-intercept.	$m = −1$
	y-intercept is (0, 4)
Plot the y-intercept.	See graph below.
Identify the rise and the run.	$m = \frac{−1}{1}$
Count out the rise and run to mark the second point.	rise −1, run 1
Draw the line.
To check your work, you can find another point on the line and make sure it is a solution of the equation. In the graph we see the line goes through (4, 0).
Check. $\begin{matrix} y & = & − x + 4 \\ 0 & \overset{?}{=} & - 4 + 4 \\ 0 & = & 0 ✓ \end{matrix}$

Graph the line of the equation $y = - \frac{2}{3} x - 3$ using its slope and y-intercept.

Solution

	$y = m x + b$
The equation is in slope–intercept form.	$y = - \frac{2}{3} x - 3$
Identify the slope and y-intercept.	$m = - \frac{2}{3}$ ; y-intercept is (0, −3)
Plot the y-intercept.	See graph below.
Identify the rise and the run.
Count out the rise and run to mark the second point.
Draw the line.

Graph the line of the equation $4 x - 3 y = 12$ using its slope and y-intercept.

Solution

	$4 x - 3 y = 12$
Find the slope–intercept form of the equation.	$−3 y = −4 x + 12$
	$- \frac{3 y}{3} = \frac{−4 x + 12}{−3}$
The equation is now in slope–intercept form.	$y = \frac{4}{3} x - 4$
Identify the slope and y-intercept.	$m = \frac{4}{3}$
	y-intercept is (0, −4)
Plot the y-intercept.	See graph below.
Identify the rise and the run; count out the rise and run to mark the second point.
Draw the line.

We have used a grid with $x$ and $y$ both going from about $−10$ to 10 for all the equations we’ve graphed so far. Not all linear equations can be graphed on this small grid. Often, especially in applications with real-world data, we’ll need to extend the axes to bigger positive or smaller negative numbers.

Graph the line of the equation $y = 0.2 x + 45$ using its slope and y-intercept.

Solution

We’ll use a grid with the axes going from about $−80$ to 80.

	$y = m x + b$
The equation is in slope–intercept form.	$y = 0.2 x + 45$
Identify the slope and y-intercept.	$m = 0.2$
	The y-intercept is (0, 45)
Plot the y-intercept.	See graph below.
Count out the rise and run to mark the second point. The slope is $m = 0.2$ ; in fraction form this means $m = \frac{2}{10}$ . Given the scale of our graph, it would be easier to use the equivalent fraction $m = \frac{10}{50}$ .
Draw the line.

Now that we have graphed lines by using the slope and y-intercept, let’s summarize all the methods we have used to graph lines. See Figure 2.

The table has two rows and four columns. The first row spans all four columns and is a header row. The header is “Methods to Graph Lines”. The second row is made up of four columns. The first column is labeled “Plotting Points” and shows a smaller table with four rows and two columns. The first row is a header row with the first column labeled “x” and the second labeled “y”. The rest of the table is blank. Below the table it reads “Find three points. Plot the points, make sure they line up, then draw the line.” The Second column is labeled “Slope–Intercept” and shows the equation y equals m x, plus b. Below the equation it reads “Find the slope and y-intercept. Start at the y-intercept, then count the slope to get a second point.” The third column is labeled “Intercepts” and shows a smaller table with four rows and two columns. The first row is a header row with the first column labeled “x” and the second labeled “y”. The second row has a 0 in the “x” column and the “y” column is blank. The second row is blank in the “x” column and has a 0 in the “y” column. The third row is blank. Below the table it reads “Find the intercepts and a third point. Plot the points, make sure they line up, then draw the line.” The fourth column is labeled “Recognize Vertical and Horizontal Lines”. Below that it reads “The equation has only one variable.” The equation x equals a is a vertical line and the equation y equals b is a horizontal line.

Choose the Most Convenient Method to Graph a Line

Now that we have seen several methods we can use to graph lines, how do we know which method to use for a given equation?

While we could plot points, use the slope–intercept form, or find the intercepts for any equation, if we recognize the most convenient way to graph a certain type of equation, our work will be easier. Generally, plotting points is not the most efficient way to graph a line. We saw better methods in sections 4.3, 4.4, and earlier in this section. Let’s look for some patterns to help determine the most convenient method to graph a line.

Here are six equations we graphed in this chapter, and the method we used to graph each of them.

\begin{matrix} Equation & Method \\ #1 & x = 2 & Vertical line \\ #2 & y = 4 & Horizontal line \\ #3 & − x + 2 y = 6 & Intercepts \\ #4 & 4 x - 3 y = 12 & Intercepts \\ #5 & y = 4 x - 2 & Slope–intercept \\ #6 & y = − x + 4 & Slope–intercept \end{matrix}

Equations #1 and #2 each have just one variable. Remember, in equations of this form the value of that one variable is constant; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.

In equations #3 and #4, both $x$ and $y$ are on the same side of the equation. These two equations are of the form $A x + B y = C$ . We substituted $y = 0$ to find the x-intercept and $x = 0$ to find the y-intercept, and then found a third point by choosing another value for $x$ or $y$ .

Equations #5 and #6 are written in slope–intercept form. After identifying the slope and y-intercept from the equation we used them to graph the line.

This leads to the following strategy.

Determine the most convenient method to graph each line.

ⓐ $y = −6$ ⓑ $5 x - 3 y = 15$ ⓒ $x = 7$ ⓓ $y = \frac{2}{5} x - 1$ .

Solution

ⓐ $y = −6$
This equation has only one variable, $y$ . Its graph is a horizontal line crossing the y-axis at $−6$ .
ⓑ $5 x - 3 y = 15$
This equation is of the form $A x + B y = C$ . The easiest way to graph it will be to find the intercepts and one more point.
ⓒ $x = 7$
There is only one variable, $x$ . The graph is a vertical line crossing the x-axis at 7.
ⓓ $y = \frac{2}{5} x - 1$
Since this equation is in $y = m x + b$ form, it will be easiest to graph this line by using the slope and y-intercept.

Graph and Interpret Applications of Slope–Intercept

Many real-world applications are modeled by linear equations. We will take a look at a few applications here so you can see how equations written in slope–intercept form relate to real-world situations.

Usually when a linear equation models a real-world situation, different letters are used for the variables, instead of x and y. The variable names remind us of what quantities are being measured.

The equation $F = \frac{9}{5} C + 32$ is used to convert temperatures, $C$ , on the Celsius scale to temperatures, $F$ , on the Fahrenheit scale.

ⓐ Find the Fahrenheit temperature for a Celsius temperature of 0.
ⓑ Find the Fahrenheit temperature for a Celsius temperature of 20.
ⓒ Interpret the slope and F-intercept of the equation.
ⓓ Graph the equation.

Solution

Examples of converting Celsius temperatures to Fahrenheit, showing problem statements and step-by-step calculations.
ⓐ Find the Fahrenheit temperature for a Celsius temperature of 0. Find $F$ when $C = 0$ . Simplify.	$\begin{matrix} F = \frac{9}{5} C + 32 \\ F = \frac{9}{5} (0) + 32 \\ F = 32 \end{matrix}$
ⓑ Find the Fahrenheit temperature for a Celsius temperature of 20. Find $F$ when $C = 20$ . Simplify. Simplify.	$\begin{matrix} F = \frac{9}{5} C + 32 \\ F = \frac{9}{5} (20) + 32 \\ F = 36 + 32 \\ F = 68 \end{matrix}$

Even though this equation uses $F$ and $C$ , it is still in slope–intercept form.

This image shows three lines of equations. The first line reads y equals m x plus b. The second line reads F equals m C plus b and the third line reads F equals nine fifths times C plus 32.

The slope, $\frac{9}{5}$ , means that the temperature Fahrenheit (F) increases 9 degrees when the temperature Celsius (C) increases 5 degrees.

The F-intercept means that when the temperature is $0 °$ on the Celsius scale, it is $32 °$ on the Fahrenheit scale.

ⓓ Graph the equation.

We’ll need to use a larger scale than our usual. Start at the F-intercept $(0, 32)$ then count out the rise of 9 and the run of 5 to get a second point. See Figure 3.

A graph with x-axis ranging from –20 to 50, and y-axis ranging from –20 to 50. A line runs upwards, intercepting at points (0, 32) and a second point which the student must find.

The cost of running some types business has two components—a fixed cost and a variable cost. The fixed cost is always the same regardless of how many units are produced. This is the cost of rent, insurance, equipment, advertising, and other items that must be paid regularly. The variable cost depends on the number of units produced. It is for the material and labor needed to produce each item.

Stella has a home business selling gourmet pizzas. The equation $C = 4 p + 25$ models the relation between her weekly cost, C, in dollars and the number of pizzas, p, that she sells.

ⓐ Find Stella’s cost for a week when she sells no pizzas.
ⓑ Find the cost for a week when she sells 15 pizzas.
ⓒ Interpret the slope and C-intercept of the equation.
ⓓ Graph the equation.

Solution

ⓐ Find Stella's cost for a week when she sells no pizzas.
Find C when $p = 0$ .
Simplify.
	Stella's fixed cost is $25 when she sells no pizzas.
ⓑ Find the cost for a week when she sells 15 pizzas.
Find C when $p = 15$ .
Simplify.

	Stella's costs are $85 when she sells 15 pizzas.
ⓒ Interpret the slope and C-intercept of the equation.
	The slope, 4, means that the cost increases by $4 for each pizza Stella sells. The C-intercept means that even when Stella sells no pizzas, her costs for the week are $25.
ⓓ Graph the equation. We'll need to use a larger scale than our usual. Start at the C-intercept (0, 25) then count out the rise of 4 and the run of 1 to get a second point.

Use Slopes to Identify Parallel Lines

The slope of a line indicates how steep the line is and whether it rises or falls as we read it from left to right. Two lines that have the same slope are called parallel lines. Parallel lines never intersect.

The figure shows three pairs of lines side-by-side. The pair of lines on the left run diagonally rising from left to right. The pair run side-by-side, not crossing. The pair of lines in the middle run diagonally dropping from left to right. The pair run side-by-side, not crossing. The pair of lines on the right run diagonally also dropping from left to right, but with a lesser slope. The pair run side-by-side, not crossing.

We say this more formally in terms of the rectangular coordinate system. Two lines that have the same slope and different y-intercepts are called parallel lines. See Figure 4.

The figure shows two lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line goes through the points (negative 5,1) and (5,5). The other line goes through the points (negative 5, negative 4) and (5,0). — Verify that both lines have the same slope, $m = \frac{2}{5}$ , and different y-intercepts.

What about vertical lines? The slope of a vertical line is undefined, so vertical lines don’t fit in the definition above. We say that vertical lines that have different x-intercepts are parallel. See Figure 5.

The figure shows two vertical lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line goes through the points (2,1) and (2,5). The other line goes through the points (5, negative 4) and (5,0). — Vertical lines with diferent x-intercepts are parallel.

Let’s graph the equations $y = −2 x + 3$ and $2 x + y = −1$ on the same grid. The first equation is already in slope–intercept form: $y = −2 x + 3$ . We solve the second equation for $y$ :

\begin{matrix} 2 x + y & = & −1 \\ y & = & −2 x - 1 \end{matrix}

Graph the lines.

Notice the lines look parallel. What is the slope of each line? What is the y-intercept of each line?

\begin{matrix} y & = & m x + b & y & = & m x + b \\ y & = & −2 x + 3 & y & = & −2 x - 1 \\ m & = & −2 & m & = & −2 \\ b & = & 3, (0, 3) & b & = & −1, (0, −1) \end{matrix}

The slopes of the lines are the same and the y-intercept of each line is different. So we know these lines are parallel.

Since parallel lines have the same slope and different y-intercepts, we can now just look at the slope–intercept form of the equations of lines and decide if the lines are parallel.

Use slopes and y-intercepts to determine if the lines $3 x - 2 y = 6$ and $y = \frac{3}{2} x + 1$ are parallel.

Solution

Step-by-step analysis of two linear equations, demonstrating how to find their slopes and y-intercepts by converting them to slope-intercept form.
Solve the first equation for $y$ .	$\begin{matrix} 3 x - 2 y & = & 6 \\ - 2 y & = & −3 x + 6 \\ \frac{−2 y}{−2} & = & \frac{−3 x + 6}{−2} \end{matrix}$	and	$y = \frac{3}{2} x + 1$
The equation is now in slope-intercept form.	$y = \frac{3}{2} x - 3$
The equation of the second line is already in slope-intercept form.			$y = \frac{3}{2} x + 1$
Identify the slope and $y$ -intercept of both lines.	$\begin{matrix} y & = & \frac{3}{2} x - 3 \\ y & = & m x + b \\ m & = & \frac{3}{2} \end{matrix}$		$\begin{matrix} y = \frac{3}{2} x + 1 \\ y = m x + b \\ m = \frac{3}{2} \end{matrix}$
	$y -intercept is (0, −3)$		$y -intercept is (0, 1)$

The lines have the same slope and different y-intercepts and so they are parallel. You may want to graph the lines to confirm whether they are parallel.

Use slopes and y-intercepts to determine if the lines $y = −4$ and $y = 3$ are parallel.

Solution

This table demonstrates converting horizontal line equations (y = -4, y = 3) to slope-intercept form and identifying their slopes and y-intercepts.
	$\begin{matrix} y = −4 \\ y = 0 x - 4 \end{matrix}$	and	$\begin{matrix} y = 3 \\ y = 0 x + 3 \end{matrix}$
Write each equation in slope-intercept form.	$y = 0 x - 4$		$y = 0 x + 3$
Since there is no $x$ term we write $0 x$ .	$y = m x + b$		$y = m x + b$
Identify the slope and $y$ -intercept of both lines.	$m = 0$		$m = 0$
	$y -intercept is (0, 4)$		$y -intercept is (0, 3)$

The lines have the same slope and different y-intercepts and so they are parallel.

There is another way you can look at this example. If you recognize right away from the equations that these are horizontal lines, you know their slopes are both 0. Since the horizontal lines cross the y-axis at $y = −4$ and at $y = 3$ , we know the y-intercepts are $(0, −4)$ and $(0, 3)$ . The lines have the same slope and different y-intercepts and so they are parallel.

Use slopes and y-intercepts to determine if the lines $x = −2$ and $x = −5$ are parallel.

Solution

x = −2 and x = −5

Since there is no $y$ , the equations cannot be put in slope–intercept form. But we recognize them as equations of vertical lines. Their x-intercepts are $−2$ and $−5$ . Since their x-intercepts are different, the vertical lines are parallel.

Use slopes and y-intercepts to determine if the lines $y = 2 x - 3$ and $−6 x + 3 y = −9$ are parallel. You may want to graph these lines, too, to see what they look like.

Solution

Steps to demonstrate the equivalence of y=2x-3 and -6x+3y=-9 by converting to slope-intercept form and comparing properties.
	$y = 2 x - 3$	and	$−6 x + 3 y = −9$
The first equation is already in slope-intercept form.	$y = 2 x - 3$
Solve the second equation for $y$ .			$\begin{matrix} - 6 x + 3 y & = & −9 \\ 3 y & = & 6 x - 9 \\ \frac{3 y}{3} & = & \frac{6 x - 9}{3} \\ y & = & 2 x - 3 \end{matrix}$
The second equation is now in slope-intercept form.	$y = 2 x - 3$
Identify the slope and $y$ -intercept of both lines.	$\begin{matrix} y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix}$		$\begin{matrix} y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix}$
	$y -intercept is (0, −3)$		$y -intercept is (0, −3)$

The lines have the same slope, but they also have the same y-intercepts. Their equations represent the same line. They are not parallel; they are the same line.

Use Slopes to Identify Perpendicular Lines

Let’s look at the lines whose equations are $y = \frac{1}{4} x - 1$ and $y = −4 x + 2$ , shown in Figure 6.

These lines lie in the same plane and intersect in right angles. We call these lines perpendicular.

What do you notice about the slopes of these two lines? As we read from left to right, the line $y = \frac{1}{4} x - 1$ rises, so its slope is positive. The line $y = −4 x + 2$ drops from left to right, so it has a negative slope. Does it make sense to you that the slopes of two perpendicular lines will have opposite signs?

If we look at the slope of the first line, $m_{1} = \frac{1}{4}$ , and the slope of the second line, $m_{2} = −4$ , we can see that they are negative reciprocals of each other. If we multiply them, their product is $−1 .$

\begin{matrix} m_{1} \cdot m_{2} \\ \frac{1}{4} (−4) \\ - 1 \end{matrix}

This is always true for perpendicular lines and leads us to this definition.

We were able to look at the slope–intercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.

We find the slope–intercept form of the equation, and then see if the slopes are negative reciprocals. If the product of the slopes is $−1$ , the lines are perpendicular. Perpendicular lines may have the same y-intercepts.

Use slopes to determine if the lines, $y = −5 x - 4$ and $x - 5 y = 5$ are perpendicular.

Solution

Steps to convert two linear equations to slope-intercept form and identify their respective slopes (m values).
The first equation is already in slope-intercept form.	$\begin{matrix} y & = & −5 x - 4 \end{matrix}$
Solve the second equation for $y$ .	$\begin{matrix} x - 5 y & = & 5 \\ - 5 y & = & − x + 5 \\ \frac{−5 y}{−5} & = & \frac{− x + 5}{−5} \\ y & = & \frac{1}{5} x - 1 \end{matrix}$
Identify the slope of each line.	$\begin{matrix} y & = & −5 x - 4 \\ y & = & m x + b \\ m_{1} & = & −5 \end{matrix}$	$\begin{matrix} y & = & \frac{1}{5} x - 1 \\ y & = & m x + b \\ m_{2} & = & \frac{1}{5} \end{matrix}$

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes,

\begin{matrix} m_{1} \cdot m_{2} \\ - 5 (\frac{1}{5}) \\ - 1 ✓ \end{matrix}

Use slopes to determine if the lines, $7 x + 2 y = 3$ and $2 x + 7 y = 5$ are perpendicular.

Solution

This table demonstrates the step-by-step process of solving two linear equations for y and identifying their respective slopes.
Solve the equations for $y$ .	$\begin{matrix} 7 x + 2 y & = & 3 \\ 2 y & = & −7 x + 3 \\ \frac{2 y}{2} & = & \frac{−7 x + 3}{2} \\ y & = & - \frac{7}{2} x + \frac{3}{2} \end{matrix}$	$\begin{matrix} 2 x + 7 y & = & 5 \\ 7 y & = & −2 x + 5 \\ \frac{7 y}{7} & = & \frac{−2 x + 5}{7} \\ y & = & - \frac{2}{7} x + \frac{5}{7} \end{matrix}$
Identify the slope of each line.	$\begin{matrix} y & = & m x + b \\ m_{1} & = & - \frac{7}{2} \end{matrix}$	$\begin{matrix} y & = & m x + b \\ m_{2} & = & - \frac{2}{7} \end{matrix}$

The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.

Key Concepts

The slope–intercept form of an equation of a line with slope $m$ and y-intercept, $(0, b)$ is, $y = m x + b$ .
Graph a Line Using its Slope and y-Intercept
1. Find the slope-intercept form of the equation of the line.
2. Identify the slope and y-intercept.
3. Plot the y-intercept.
4. Use the slope formula $m = \frac{rise}{run}$ to identify the rise and the run.
5. Starting at the y-intercept, count out the rise and run to mark the second point.
6. Connect the points with a line.
Strategy for Choosing the Most Convenient Method to Graph a Line: Consider the form of the equation.
- If it only has one variable, it is a vertical or horizontal line.
  $x = a$ is a vertical line passing through the x-axis at $a$ .
  $y = b$ is a horizontal line passing through the y-axis at $b$ .
- If $y$ is isolated on one side of the equation, in the form $y = m x + b$ , graph by using the slope and y-intercept.
  Identify the slope and y-intercept and then graph.
- If the equation is of the form $A x + B y = C$ , find the intercepts.
  Find the x- and y-intercepts, a third point, and then graph.
Parallel lines are lines in the same plane that do not intersect.
- Parallel lines have the same slope and different y-intercepts.
- If m₁ and m₂ are the slopes of two parallel lines then $m_{1} = m_{2} .$
- Parallel vertical lines have different x-intercepts.
Perpendicular lines are lines in the same plane that form a right angle.
- If $m_{1} and m_{2}$ are the slopes of two perpendicular lines, then $m_{1} \cdot m_{2} = −1$ and $m_{1} = \frac{−1}{m_{2}}$ .
- Vertical lines and horizontal lines are always perpendicular to each other.

Practice Makes Perfect

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

In the following exercises, use the graph to find the slope and y-intercept of each line. Compare the values to the equation $y = m x + b$ .

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 5) and (1, negative 2).

$y = 3 x - 5$

$y = 4 x - 2$

Solution

slope $m = 4$ and y-intercept $(0, −2)$

$y = − x + 4$

$y = −3 x + 1$

Solution

slope $m = −3$ and y-intercept $(0, 1)$

$y = - \frac{4}{3} x + 1$

$y = - \frac{2}{5} x + 3$

Solution

slope $m = - \frac{2}{5}$ and y-intercept $(0, 3)$

Identify the Slope and y-Intercept From an Equation of a Line

In the following exercises, identify the slope and y-intercept of each line.

$y = −7 x + 3$

$y = −9 x + 7$

Solution

$−9; (0, 7)$

$y = 6 x - 8$

$y = 4 x - 10$

Solution

$4; (0, −10)$

$3 x + y = 5$

$4 x + y = 8$

Solution

$−4; (0, 8)$

$6 x + 4 y = 12$

$8 x + 3 y = 12$

Solution

$- \frac{8}{3}; (0, 4)$

$5 x - 2 y = 6$

$7 x - 3 y = 9$

Solution

$\frac{7}{3}; (0, −3)$

Graph a Line Using Its Slope and Intercept

In the following exercises, graph the line of each equation using its slope and y-intercept.

$y = x + 3$

$y = x + 4$

Solution

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, 4) and (1, 5).

$y = 3 x - 1$

$y = 2 x - 3$

Solution

$y = − x + 2$

$y = − x + 3$

Solution

$y = − x - 4$

$y = − x - 2$

Solution

$y = - \frac{3}{4} x - 1$

$y = - \frac{2}{5} x - 3$

Solution

$y = - \frac{3}{5} x + 2$

$y = - \frac{2}{3} x + 1$

Solution

$3 x - 4 y = 8$

$4 x - 3 y = 6$

Solution

$y = 0.1 x + 15$

$y = 0.3 x + 25$

Solution

Choose the Most Convenient Method to Graph a Line

In the following exercises, determine the most convenient method to graph each line.

$x = 2$

$y = 4$

Solution

horizontal line

$y = 5$

$x = −3$

Solution

vertical line

$y = −3 x + 4$

$y = −5 x + 2$

Solution

slope–intercept

$x - y = 5$

$x - y = 1$

Solution

intercepts

$y = \frac{2}{3} x - 1$

$y = \frac{4}{5} x - 3$

Solution

slope–intercept

$y = −3$

$y = −1$

Solution

horizontal line

$3 x - 2 y = −12$

$2 x - 5 y = −10$

Solution

intercepts

$y = - \frac{1}{4} x + 3$

$y = - \frac{1}{3} x + 5$

Solution

slope–intercept

Graph and Interpret Applications of Slope–Intercept

The equation $P = 31 + 1.75 w$ models the relation between the amount of Tuyet’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

ⓐ Find Tuyet’s payment for a month when 0 units of water are used.
ⓑ Find Tuyet’s payment for a month when 12 units of water are used.
ⓒ Interpret the slope and P-intercept of the equation.
ⓓ Graph the equation.

The equation $P = 28 + 2.54 w$ models the relation between the amount of Randy’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

ⓐ Find the payment for a month when Randy used 0 units of water.
ⓑ Find the payment for a month when Randy used 15 units of water.
ⓒ Interpret the slope and P-intercept of the equation.
ⓓ Graph the equation.

Solution

ⓐ $28
ⓑ $66.10
ⓒ The slope, 2.54, means that Randy’s payment, P, increases by $2.54 when the number of units of water he used, w, increases by 1. The P–intercept means that if the number units of water Randy used was 0, the payment would be $28.
ⓓ

Bruce drives his car for his job. The equation $R = 0.575 m + 42$ models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day.

ⓐ Find the amount Bruce is reimbursed on a day when he drives 0 miles.
ⓑ Find the amount Bruce is reimbursed on a day when he drives 220 miles.
ⓒ Interpret the slope and R-intercept of the equation.
ⓓ Graph the equation.

Janelle is planning to rent a car while on vacation. The equation $C = 0.32 m + 15$ models the relation between the cost in dollars, C, per day and the number of miles, m, she drives in one day.

ⓐ Find the cost if Janelle drives the car 0 miles one day.
ⓑ Find the cost on a day when Janelle drives the car 400 miles.
ⓒ Interpret the slope and C–intercept of the equation.
ⓓ Graph the equation.

Solution

ⓐ $15
ⓑ $143
ⓒ The slope, 0.32, means that the cost, C, increases by $0.32 when the number of miles driven, m, increases by 1. The C-intercept means that if Janelle drives 0 miles one day, the cost would be $15.
ⓓ

Cherie works in retail and her weekly salary includes commission for the amount she sells. The equation $S = 400 + 0.15 c$ models the relation between her weekly salary, S, in dollars and the amount of her sales, c, in dollars.

ⓐ Find Cherie’s salary for a week when her sales were 0.
ⓑ Find Cherie’s salary for a week when her sales were 3600.
ⓒ Interpret the slope and S–intercept of the equation.
ⓓ Graph the equation.

Patel’s weekly salary includes a base pay plus commission on his sales. The equation $S = 750 + 0.09 c$ models the relation between his weekly salary, S, in dollars and the amount of his sales, c, in dollars.

ⓐ Find Patel’s salary for a week when his sales were 0.
ⓑ Find Patel’s salary for a week when his sales were 18,540.
ⓒ Interpret the slope and S-intercept of the equation.
ⓓ Graph the equation.

Solution

ⓐ $750
ⓑ $2418.60
ⓒ The slope, 0.09, means that Patel’s salary, S, increases by $0.09 for every $1 increase in his sales. The S-intercept means that when his sales are $0, his salary is $750.
ⓓ

Costa is planning a lunch banquet. The equation $C = 450 + 28 g$ models the relation between the cost in dollars, C, of the banquet and the number of guests, g.

ⓐ Find the cost if the number of guests is 40.
ⓑ Find the cost if the number of guests is 80.
ⓒ Interpret the slope and C-intercept of the equation.
ⓓ Graph the equation.

Margie is planning a dinner banquet. The equation $C = 750 + 42 g$ models the relation between the cost in dollars, C of the banquet and the number of guests, g.

ⓐ Find the cost if the number of guests is 50.
ⓑ Find the cost if the number of guests is 100.
ⓒ Interpret the slope and C–intercept of the equation.
ⓓ Graph the equation.

Solution

ⓐ $2850
ⓑ $4950
ⓒ The slope, 42, means that the cost, C, increases by $42 for when the number of guests increases by 1. The C-intercept means that when the number of guests is 0, the cost would be $750.
ⓓ

Use Slopes to Identify Parallel Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel.

$y = \frac{3}{4} x - 3; 3 x - 4 y = - 2$

$y = \frac{2}{3} x - 1; 2 x - 3 y = - 2$

Solution

parallel

$2 x - 5 y = - 3; y = \frac{2}{5} x + 1$

$3 x - 4 y = - 2; y = \frac{3}{4} x - 3$

Solution

parallel

$2 x - 4 y = 6; x - 2 y = 3$

$6 x - 3 y = 9; 2 x - y = 3$

Solution

not parallel

$4 x + 2 y = 6; 6 x + 3 y = 3$

$8 x + 6 y = 6; 12 x + 9 y = 12$

Solution

parallel

$x = 5; x = - 6$

$x = 7; x = - 8$

Solution

parallel

$x = - 4; x = - 1$

$x = - 3; x = - 2$

Solution

parallel

$y = 2; y = 6$

$y = 5; y = 1$

Solution

parallel

$y = - 4; y = 3$

$y = - 1; y = 2$

Solution

parallel

$x - y = 2; 2 x - 2 y = 4$

$4 x + 4 y = 8; x + y = 2$

Solution

not parallel

$x - 3 y = 6; 2 x - 6 y = 12$

$5 x - 2 y = 11; 5 x - y = 7$

Solution

not parallel

$3 x - 6 y = 12; 6 x - 3 y = 3$

$4 x - 8 y = 16; x - 2 y = 4$

Solution

not parallel

$9 x - 3 y = 6; 3 x - y = 2$

$x - 5 y = 10; 5 x - y = - 10$

Solution

not parallel

$7 x - 4 y = 8; 4 x + 7 y = 14$

$9 x - 5 y = 4; 5 x + 9 y = - 1$

Solution

not parallel

Use Slopes to Identify Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are perpendicular.

$3 x - 2 y = 8; 2 x + 3 y = 6$

$x - 4 y = 8; 4 x + y = 2$

Solution

perpendicular

$2 x + 5 y = 3; 5 x - 2 y = 6$

$2 x + 3 y = 5; 3 x - 2 y = 7$

Solution

perpendicular

$3 x - 2 y = 1; 2 x - 3 y = 2$

$3 x - 4 y = 8; 4 x - 3 y = 6$

Solution

not perpendicular

$5 x + 2 y = 6; 2 x + 5 y = 8$

$2 x + 4 y = 3; 6 x + 3 y = 2$

Solution

not perpendicular

$4 x - 2 y = 5; 3 x + 6 y = 8$

$2 x - 6 y = 4; 12 x + 4 y = 9$

Solution

perpendicular

$6 x - 4 y = 5; 8 x + 12 y = 3$

$8 x - 2 y = 7; 3 x + 12 y = 9$

Solution

perpendicular

Everyday Math

The equation $C = \frac{5}{9} F - 17.8$ can be used to convert temperatures F, on the Fahrenheit scale to temperatures, C, on the Celsius scale.

ⓐ Explain what the slope of the equation means.
ⓑ Explain what the C–intercept of the equation means.

The equation $n = 4 T - 160$ is used to estimate the number of cricket chirps, n, in one minute based on the temperature in degrees Fahrenheit, T.

ⓐ Explain what the slope of the equation means.
ⓑ Explain what the n–intercept of the equation means. Is this a realistic situation?

Solution

ⓐ For every increase of one degree Fahrenheit, the number of chirps increases by four.
ⓑ There would be $−160$ chirps when the Fahrenheit temperature is $0 °$ . (Notice that this does not make sense; this model cannot be used for all possible temperatures.)

Writing Exercises

Explain in your own words how to decide which method to use to graph a line.

Why are all horizontal lines parallel?

Solution

Answers will vary.

Self Check

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

This table has eight rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the next row reads “recognize the relation between the graph and the slope-intercept form of an equation of a line.” The third row reads “identify the Slope and y-intercept from an equation of a line”. The fourth row reads “graph a line using its slope and intercept”. The fifth row reads “choose the most convenient method to graph a line.” The sixth row reads “graph and interpret applications of slope-intercept”. The seventh row reads “use slopes to identify parallel lines” and the last row reads “use slopes to identify perpendicular lines.” The remaining columns are blank.

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

parallel lines

Lines in the same plane that do not intersect.

perpendicular lines

Lines in the same plane that form a right angle.

slope-intercept form of an equation of a line

The slope–intercept form of an equation of a line with slope

m

and y-intercept,

(0, b)

is,

y = m x + b

Find the Equation of a Line

How do online retailers know that ‘you may also like’ a particular item based on something you just ordered? How can economists know how a rise in the minimum wage will affect the unemployment rate? How do medical researchers create drugs to target cancer cells? How can traffic engineers predict the effect on your commuting time of an increase or decrease in gas prices? It’s all mathematics.

You are at an exciting point in your mathematical journey as the mathematics you are studying has interesting applications in the real world.

The physical sciences, social sciences, and the business world are full of situations that can be modeled with linear equations relating two variables. Data is collected and graphed. If the data points appear to form a straight line, an equation of that line can be used to predict the value of one variable based on the value of the other variable.

To create a mathematical model of a linear relation between two variables, we must be able to find the equation of the line. In this section we will look at several ways to write the equation of a line. The specific method we use will be determined by what information we are given.

Find an Equation of the Line Given the Slope and y-Intercept

We can easily determine the slope and intercept of a line if the equation was written in slope–intercept form, $y = m x + b .$ Now, we will do the reverse—we will start with the slope and y-intercept and use them to find the equation of the line.

Find an equation of a line with slope $−7$ and y-intercept $(0, −1)$ .

Solution

Since we are given the slope and y-intercept of the line, we can substitute the needed values into the slope–intercept form, $y = m x + b$ .

Name the slope.
Name the y-intercept.
Substitute the values into $y = m x + b .$

Sometimes, the slope and intercept need to be determined from the graph.

Find the equation of the line shown.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. A line intercepts the y-axis at (0, negative 4), passes through the plotted point (3, negative 2), and intercepts the x-axis at (4, 0).

Solution

We need to find the slope and y-intercept of the line from the graph so we can substitute the needed values into the slope–intercept form, $y = m x + b$ .

To find the slope, we choose two points on the graph.

The y-intercept is $(0, −4)$ and the graph passes through $(3, −2)$ .

Find the slope by counting the rise and run.
	$A mathematical equation is displayed against a white background, showing 'm = 2/3'. The variable 'm' is in black, and the fraction '2/3' is in red, with '2' above a horizontal line and '3' below.$
Find the y-intercept.
Substitute the values into $y = m x + b .$
	$The image displays the linear equation y = (2/3)x - 4. The fraction 2/3 is colored red, indicating the slope, while the number 4, part of the y-intercept, is colored blue, with a minus sign preceding it.$

Find an Equation of the Line Given the Slope and a Point

Finding an equation of a line using the slope–intercept form of the equation works well when you are given the slope and y-intercept or when you read them off a graph. But what happens when you have another point instead of the y-intercept?

We are going to use the slope formula to derive another form of an equation of the line. Suppose we have a line that has slope $m$ and that contains some specific point $(x_{1}, y_{1})$ and some other point, which we will just call $(x, y)$ . We can write the slope of this line and then change it to a different form.

Illustrates the step-by-step derivation of the point-slope form of a linear equation from the slope formula.
	$m = \frac{y - y_{1}}{x - x_{1}}$
Multiply both sides of the equation by $x - x_{1}$ .	$\begin{matrix} m (x - x_{1}) & = & (\frac{y - y_{1}}{x - x_{1}}) (x - x_{1}) \end{matrix}$
Simplify.	$\begin{matrix} m (x - x_{1}) & = & y - y_{1} \end{matrix}$
Rewrite the equation with the $y$ terms on the left.	$\begin{matrix} y - y_{1} & = & m (x - x_{1}) \end{matrix}$

This format is called the point–slope form of an equation of a line.

We can use the point–slope form of an equation to find an equation of a line when we are given the slope and one point. Then we will rewrite the equation in slope–intercept form. Most applications of linear equations use the the slope–intercept form.

Find an Equation of a Line Given the Slope and a Point

Find an equation of a line with slope $m = \frac{2}{5}$ that contains the point $(10, 3)$ . Write the equation in slope–intercept form.

Solution

In the second row, the first cell reads: “Step 2. Identify the point.” The second cell reads: “The point is given.” The third cell contains the ordered pair (10, 3). A superscript x subscript 1 is written over 10, and a superscript y subscript 1 is written over 3.

Step 3 substitutes values into point-slope form: y minus y sub one equals m times x minus x sub one. The equation becomes y minus 3 equals 2/5 times x minus 10. Simplify to get y minus 3 equals 2/5 x minus 4.

In the fourth row, the first cell reads: “Write the equation in slope-intercept form.” The second cell is blank. In the third cell is y equals 2 fifths x minus 1.

Find an equation of a line with slope $m = - \frac{1}{3}$ that contains the point $(6, −4)$ . Write the equation in slope–intercept form.

Solution

Since we are given a point and the slope of the line, we can substitute the needed values into the point–slope form, $y - y_{1} = m (x - x_{1})$ .

Identify the slope.	$The equation m = -1/3, displaying a blue 'm' and an equal sign, followed by a negative sign and the fraction 1/3 in red, all on a white background.$
Identify the point.
Substitute the values into $y - y_{1} = m (x - x_{1}) .$
	$A mathematical equation in point-slope form: y - (-4) = -1/3(x - 6). The numbers -4 and -6 are highlighted in red, and the fraction 1/3 is highlighted in blue.$
Simplify.
Write in slope–intercept form.

Find an equation of a horizontal line that contains the point $(−1, 2)$ . Write the equation in slope–intercept form.

Solution

Every horizontal line has slope 0. We can substitute the slope and points into the point–slope form, $y - y_{1} = m (x - x_{1})$ .

Identify the slope.
Identify the point.
Substitute the values into $y - y_{1} = m (x - x_{1}) .$

Simplify.


Write in slope–intercept form.	It is in y-form, but could be written $y = 0 x + 2$ .

Did we end up with the form of a horizontal line, $y = a$ ?

Find an Equation of the Line Given Two Points

When real-world data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when just two points are given.

We have two options so far for finding an equation of a line: slope–intercept or point–slope. Since we will know two points, it will make more sense to use the point–slope form.

But then we need the slope. Can we find the slope with just two points? Yes. Then, once we have the slope, we can use it and one of the given points to find the equation.

Find an Equation of a Line Given Two Points

Find an equation of a line that contains the points $(5, 4)$ and $(3, 6)$ . Write the equation in slope–intercept form.

Solution

Table showing steps to find the slope of a line. Step 1: Find the slope using the formula y two minus y one, divided by x two minus x one. Calculation shows the slope, m, as negative one.

In the second row, the first cell reads: “Step 2. Choose one point.” The second cell reads: “Choose either point.” The third cell contains the ordered pair (5, 4) with a superscript x subscript 1 over 5 and a superscript y subscript 1 over 4.

In the third row, the first cell reads: “Step 3. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top line of the second cell is left blank. The third cell contains the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the point-slope form with 5 substituted for x subscript 1, 4 substituted for y subscript 1, and negative 1 substituted for m: y minus 4 equals negative 1 times x minus 5 in parentheses. Below this is y minus 4 equals negative x plus 5.

In the fourth row, the first cell reads: “Step 4. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals negative x plus 9.

Use the point $(3, 6)$ and see that you get the same equation.

Find an equation of a line that contains the points $(−3, −1)$ and $(2, −2)$ . Write the equation in slope–intercept form.

Solution

Since we have two points, we will find an equation of the line using the point–slope form. The first step will be to find the slope.

Find the slope of the line through (−3, −1) and (2, −2).

	$A mathematical equation is displayed on a white background, showing 'm = -1/5'. The letter 'm' is followed by an equals sign, then a fraction where -1 is the numerator and 5 is the denominator.$
	$A mathematical equation displays 'm = -1/5' in a clean, white background. The variable 'm' is shown in light blue, the equals sign and negative sign are black, and the fraction '1/5' is in red.$
Choose either point.
Substitute the values into $y - y_{1} = m (x - x_{1}) .$


Write in slope–intercept form.

Find an equation of a line that contains the points $(−2, 4)$ and $(−2, −3)$ . Write the equation in slope–intercept form.

Solution

Again, the first step will be to find the slope.

Demonstration of finding the slope of a line between two points, illustrating a case where the slope is undefined.
Find the slope of the line through $(−2, 4) and (−2, −3)$ .	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
	$m = \frac{−3 - 4}{−2 - (−2)}$
	$m = \frac{−7}{0}$
	The slope is undefined.

This tells us it is a vertical line. Both of our points have an x-coordinate of $−2$ . So our equation of the line is $x = −2$ . Since there is no $y$ , we cannot write it in slope–intercept form.

You may want to sketch a graph using the two given points. Does the graph agree with our conclusion that this is a vertical line?

We have seen that we can use either the slope–intercept form or the point–slope form to find an equation of a line. Which form we use will depend on the information we are given. This is summarized in Table 8.

To Write an Equation of a Line
If given:	Use:	Form:
Slope and y-intercept	slope–intercept	$y = m x + b$
Slope and a point	point–slope	$y - y_{1} = m (x - x_{1})$
Two points	point–slope	$y - y_{1} = m (x - x_{1})$

Find an Equation of a Line Parallel to a Given Line

Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point–slope equation.

First let’s look at this graphically.

The graph shows the graph of $y = 2 x - 3$ . We want to graph a line parallel to this line and passing through the point $(−2, 1)$ .

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). Elsewhere on the graph, the point (negative 2, 1) is plotted.

We know that parallel lines have the same slope. So the second line will have the same slope as $y = 2 x - 3$ . That slope is $m_{∥} = 2$ . We’ll use the notation $m_{∥}$ to represent the slope of a line parallel to a line with slope $m$ . (Notice that the subscript $∥$ looks like two parallel lines.)

The second line will pass through $(−2, 1)$ and have $m = 2$ . To graph the line, we start at $(−2, 1)$ and count out the rise and run. With $m = 2$ (or $m = \frac{2}{1}$ ), we count out the rise 2 and the run 1. We draw the line.

Do the lines appear parallel? Does the second line pass through $(−2, 1)$ ?

Now, let’s see how to do this algebraically.

We can use either the slope–intercept form or the point–slope form to find an equation of a line. Here we know one point and can find the slope. So we will use the point–slope form.

How to Find an Equation of a Line Parallel to a Given Line

Find an equation of a line parallel to $y = 2 x - 3$ that contains the point $(−2, 1)$ . Write the equation in slope–intercept form.

Solution

In the second row, the first cell reads: “Step 2. Find the slope of the parallel line.” The second cell reads “Parallel lines have the same slope.” The third cell contains the slope of the parallel line, defined as m parallel equals 2.

In the third row, the first cell reads “Step 3. Identify the point.” The second cell reads “The given point is (negative 2, 1).” The third cell contains the ordered pair (negative 2, 1) with a superscript x subscript 1 above negative 2 and a superscript y subscript 1 above 1.

In the fourth row, the first cell reads “Step 4. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top of the second cell is blank. The third cell contains the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the form with negative 2 substituted for x subscript 1, 1 substituted for y subscript 1, and 2 substituted for m: y minus 1 equals 2 times x minus negative 2 in parentheses. One line down, the text in the second cell says “Simplify.” The right column contains y minus 1 equals 2 times x plus 2. Below this is y minus 1 equals 2x plus 4.

In the fifth row, the first cell says “Step 5. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals 2x plus 5.

Does this equation make sense? What is the y-intercept of the line? What is the slope?

Find an Equation of a Line Perpendicular to a Given Line

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point–slope equation, like we did with parallel lines.

The graph shows the graph of $y = 2 x - 3$ . Now, we want to graph a line perpendicular to this line and passing through $(−2, 1)$ .

We know that perpendicular lines have slopes that are negative reciprocals. We’ll use the notation $m_{⊥}$ to represent the slope of a line perpendicular to a line with slope $m$ . (Notice that the subscript ⊥ looks like the right angles made by two perpendicular lines.)

\begin{matrix} y = 2 x - 3 & perpendicular line \\ m = 2 & m_{⊥} = - \frac{1}{2} \end{matrix}

We now know the perpendicular line will pass through $(−2, 1)$ with $m_{⊥} = - \frac{1}{2}$ .

To graph the line, we will start at $(−2, 1)$ and count out the rise $−1$ and the run 2. Then we draw the line.

Graph of two lines in x y-plane. The first line, y equals 2x minus 3, intersects the y-axis at negative 3 and the x-axis at one and a half, zero. A second line passes through negative 2, 1, forming a right triangle with the first line.

Do the lines appear perpendicular? Does the second line pass through $(−2, 1)$ ?

Now, let’s see how to do this algebraically. We can use either the slope–intercept form or the point–slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point–slope form.

How to Find an Equation of a Line Perpendicular to a Given Line

Find an equation of a line perpendicular to $y = 2 x - 3$ that contains the point $(−2, 1)$ . Write the equation in slope–intercept form.

Solution

In the second row, the first cell reads: “Step 2. Find the slope of the perpendicular line.” The second cell reads “The slopes of perpendicular lines are negative reciprocals.” The third cell contains the slope of the perpendicular line, defined as m perpendicular equals negative 1 half.

In the fifth row, the first cell says “Step 5. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals negative 1 half x.

Find an equation of a line perpendicular to $x = 5$ that contains the point $(3, −2)$ . Write the equation in slope–intercept form.

Solution

Again, since we know one point, the point–slope option seems more promising than the slope–intercept option. We need the slope to use this form, and we know the new line will be perpendicular to $x = 5$ . This line is vertical, so its perpendicular will be horizontal. This tells us the $m_{⊥} = 0$ .

Step-by-step derivation of the equation of a perpendicular line given a point and its slope.
Identify the point.	$(3, −2)$
Identify the slope of the perpendicular line.	$m_{⊥} = 0$
Substitute the values into $y - y_{1} = m (x - x_{1})$ .	$\begin{matrix} y - y_{1} = m (x - x_{1}) \\ y - (−2) = 0 (x - 3) \\ y + 2 = 0 \end{matrix}$
Simplify.	$y = −2$

Sketch the graph of both lines. Do they appear to be perpendicular?

In Example 11, we used the point–slope form to find the equation. We could have looked at this in a different way.

We want to find a line that is perpendicular to $x = 5$ that contains the point $(3, −2)$ . The graph shows us the line $x = 5$ and the point $(3, −2)$ .

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is x equals 5 intercepts the x-axis at (5, 0) and runs parallel to the y-axis. Elsewhere on the graph, the point (3, negative 2) is plotted.

We know every line perpendicular to a vetical line is horizontal, so we will sketch the horizontal line through $(3, −2)$ .

Do the lines appear perpendicular?

If we look at a few points on this horizontal line, we notice they all have y-coordinates of $−2$ . So, the equation of the line perpendicular to the vertical line $x = 5$ is $y = −2$ .

Find an equation of a line that is perpendicular to $y = −4$ that contains the point $(−4, 2)$ . Write the equation in slope–intercept form.

Solution

The line $y = −4$ is a horizontal line. Any line perpendicular to it must be vertical, in the form $x = a$ . Since the perpendicular line is vertical and passes through $(−4, 2)$ , every point on it has an x-coordinate of $−4$ . The equation of the perpendicular line is $x = −4$ . You may want to sketch the lines. Do they appear perpendicular?

Key Concepts

To Find an Equation of a Line Given the Slope and a Point
1. Identify the slope.
2. Identify the point.
3. Substitute the values into the point-slope form, $y - y_{1} = m (x - x_{1})$ .
4. Write the equation in slope-intercept form.
To Find an Equation of a Line Given Two Points
1. Find the slope using the given points.
2. Choose one point.
3. Substitute the values into the point-slope form, $y - y_{1} = m (x - x_{1})$ .
4. Write the equation in slope-intercept form.
To Write and Equation of a Line
- If given slope and y-intercept, use slope–intercept form $y = m x + b$ .
- If given slope and a point, use point–slope form $y - y_{1} = m (x - x_{1})$ .
- If given two points, use point–slope form $y - y_{1} = m (x - x_{1})$ .
To Find an Equation of a Line Parallel to a Given Line
1. Find the slope of the given line.
2. Find the slope of the parallel line.
3. Identify the point.
4. Substitute the values into the point-slope form, $y - y_{1} = m (x - x_{1})$ .
5. Write the equation in slope-intercept form.
To Find an Equation of a Line Perpendicular to a Given Line
1. Find the slope of the given line.
2. Find the slope of the perpendicular line.
3. Identify the point.
4. Substitute the values into the point-slope form, $y - y_{1} = m (x - x_{1})$ .
5. Write the equation in slope-intercept form.

Practice Makes Perfect

Find an Equation of the Line Given the Slope and y-Intercept

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope–intercept form.

slope 3 and y-intercept $(0, 5)$

slope 4 and y-intercept $(0, 1)$

Solution

$y = 4 x + 1$

slope 6 and y-intercept $(0, −4)$

slope 8 and y-intercept $(0, −6)$

Solution

$y = 8 x - 6$

slope $−1$ and y-intercept $(0, 3)$

slope $−1$ and y-intercept $(0, 7)$

Solution

$y = − x + 7$

slope $−2$ and y-intercept $(0, −3)$

slope $−3$ and y-intercept $(0, −1)$

Solution

$y = −3 x - 1$

slope $\frac{3}{5}$ and y-intercept $(0, −1)$

slope $\frac{1}{5}$ and y-intercept $(0, −5)$

Solution

$y = \frac{1}{5} x - 5$

slope $- \frac{3}{4}$ and y-intercept $(0, −2)$

slope $- \frac{2}{3}$ and y-intercept $(0, −3)$

Solution

$y = - \frac{2}{3} x - 3$

slope 0 and y-intercept $(0, −1)$

slope 0 and y-intercept $(0, 2)$

Solution

$y = 2$

slope $−3$ and y-intercept $(0, 0)$

slope $−4$ and y-intercept $(0, 0)$

Solution

$y = −4 x$

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope–intercept form.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (1, negative 2) is plotted. A line intercepts the y-axis at (0, negative 5), passes through the point (1, negative 2), and intercepts the x-axis at (5 thirds, 0).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (2, 0) is plotted. A line intercepts the y-axis at (0, 4) and intercepts the x-axis at (2, 0).

Solution

$y = −2 x + 4$

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (6, 0) is plotted. A line intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (6, 0).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (4, 5) is plotted. A line intercepts the x-axis at (negative 8 thirds, 0), intercepts the y-axis at (0, 2), and passes through the point (4, 5).

Solution

$y = \frac{3}{4} x + 2$

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (3, negative 1) is plotted. A line intercepts the y-axis at (0, 2), intercepts the x-axis at (9 fourths, 0), and passes through the point (3, negative 1).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (2, negative 4) is plotted. A line intercepts the x-axis at (negative 2 thirds, 0), intercepts the y-axis at (0, negative 1), and passes through the point (2, negative 4).

Solution

$y = - \frac{3}{2} x - 1$

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (2, negative 2) is plotted. A line running parallel to the x-axis intercepts the y-axis at (0, negative 2) and passes through the point (2, negative 2).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 9 to 9. The point (negative 3, 6) is plotted. A line running parallel to the x-axis passes through (negative 3, 6) and intercepts the y-axis at (0, 6).

Solution

$y = 6$

Find an Equation of the Line Given the Slope and a Point

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope–intercept form.

$m = \frac{5}{8}$ , point $(8, 3)$

$m = \frac{3}{8}$ , point $(8, 2)$

Solution

$y = \frac{3}{8} x - 1$

$m = \frac{1}{6}$ , point $(6, 1)$

$m = \frac{5}{6}$ , point $(6, 7)$

Solution

$y = \frac{5}{6} x + 2$

$m = - \frac{3}{4}$ , point $(8, −5)$

$m = - \frac{3}{5}$ , point $(10, −5)$

Solution

$y = - \frac{3}{5} x + 1$

$m = - \frac{1}{4}$ , point $(−12, −6)$

$m = - \frac{1}{3}$ , point $(−9, −8)$

Solution

$y = - \frac{1}{3} x - 11$

Horizontal line containing $(−2, 5)$

Horizontal line containing $(−1, 4)$

Solution

$y = 4$

Horizontal line containing $(−2, −3)$

Horizontal line containing $(−1, −7)$

Solution

$y = −7$

$m = - \frac{3}{2}$ , point $(−4, −3)$

$m = - \frac{5}{2}$ , point $(−8, −2)$

Solution

$y = - \frac{5}{2} x - 22$

$m = −7$ , point $(−1, −3)$

$m = −4$ , point $(−2, −3)$

Solution

$y = −4 x - 11$

Horizontal line containing $(2, −3)$

Horizontal line containing $(4, −8)$

Solution

$y = −8$

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope–intercept form.

$(2, 6)$ and $(5, 3)$

$(3, 1)$ and $(2, 5)$

Solution

$y = −4 x + 13$

$(4, 3)$ and $(8, 1)$

$(2, 7)$ and $(3, 8)$

Solution

$y = x + 5$

$(−3, −4)$ and $(5 - 2)$

$(−5, −3)$ and $(4, −6)$

Solution

$y = - \frac{1}{3} x - \frac{14}{3}$

$(−1, 3)$ and $(−6, −7)$

$(−2, 8)$ and $(−4, −6)$

Solution

$y = 7 x + 22$

$(6, −4)$ and $(−2, 5)$

$(3, −2)$ and $(−4, 4)$

Solution

$y = - \frac{6}{7} x + \frac{4}{7}$

$(0, 4)$ and $(2, −3)$

$(0, −2)$ and $(−5, −3)$

Solution

$y = \frac{1}{5} x - 2$

$(7, 2)$ and $(7, −2)$

$(4, 2)$ and $(4, −3)$

Solution

$x = 4$

$(−7, −1)$ and $(−7, −4)$

$(−2, 1)$ and $(−2, −4)$

Solution

$x = −2$

$(6, 1)$ and $(0, 1)$

$(6, 2)$ and $(−3, 2)$

Solution

$y = 2$

$(3, −4)$ and $(5, −4)$

$(−6, −3)$ and $(−1, −3)$

Solution

$y = −3$

$(4, 3)$ and $(8, 0)$

$(0, 0)$ and $(1, 4)$

Solution

$y = 4 x$

$(−2, −3)$ and $(−5, −6)$

$(−3, 0)$ and $(−7, −2)$

Solution

$y = \frac{1}{2} x + \frac{3}{2}$

$(8, −1)$ and $(8, −5)$

$(3, 5)$ and $(−7, 5)$

Solution

$y = 5$

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope–intercept form.

line $y = 4 x + 2$ , point $(1, 2)$

line $y = 3 x + 4$ , point $(2, 5)$

Solution

$y = 3 x - 1$

line $y = −2 x - 3$ , point $(−1, 3)$

line $y = −3 x - 1$ , point $(2, −3)$

Solution

$y = −3 x + 3$

line $3 x - y = 4$ , point $(3, 1)$

line $2 x - y = 6$ , point $(3, 0)$

Solution

$y = 2 x - 6$

line $4 x + 3 y = 6$ , point $(0, −3)$

line $2 x + 3 y = 6$ , point $(0, 5)$

Solution

$y = - \frac{2}{3} x + 5$

line $x = −3$ , point $(−2, −1)$

line $x = −4$ , point $(−3, −5)$

Solution

$x = −3$

line $x - 2 = 0$ , point $(1, −2)$

line $x - 6 = 0$ , point $(4, −3)$

Solution

$x = 4$

line $y = 5$ , point $(2, −2)$

line $y = 1$ , point $(3, −4)$

Solution

$y = −4$

line $y + 2 = 0$ , point $(3, −3)$

line $y + 7 = 0$ , point $(1, −1)$

Solution

$y = −1$

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope–intercept form.

line $y = −2 x + 3$ , point $(2, 2)$

line $y = − x + 5$ , point $(3, 3)$

Solution

$y = x$

line $y = \frac{3}{4} x - 2$ , point $(−3, 4)$

line $y = \frac{2}{3} x - 4$ , point $(2, −4)$

Solution

$y = - \frac{3}{2} x - 1$

line $2 x - 3 y = 8$ , point $(4, −1)$

line $4 x - 3 y = 5$ , point $(−3, 2)$

Solution

$y = - \frac{3}{4} x - \frac{1}{4}$

line $2 x + 5 y = 6$ , point $(0, 0)$

line $4 x + 5 y = −3$ , point $(0, 0)$

Solution

$y = \frac{5}{4} x$

line $y - 3 = 0$ , point $(−2, −4)$

line $y - 6 = 0$ , point $(−5, −3)$

Solution

$x = −5$

line y-axis, point $(3, 4)$

line y-axis, point $(2, 1)$

Solution

$y = 1$

Mixed Practice

In the following exercises, find the equation of each line. Write the equation in slope–intercept form.

Containing the points $(4, 3)$ and $(8, 1)$

Containing the points $(2, 7)$ and $(3, 8)$

Solution

$y = x + 5$

$m = \frac{1}{6}$ , containing point $(6, 1)$

$m = \frac{5}{6}$ , containing point $(6, 7)$

Solution

$y = \frac{5}{6} x + 2$

Parallel to the line $4 x + 3 y = 6$ , containing point $(0, −3)$

Parallel to the line $2 x + 3 y = 6$ , containing point $(0, 5)$

Solution

$y = - \frac{2}{3} x + 5$

$m = - \frac{3}{4}$ , containing point $(8, −5)$

$m = - \frac{3}{5}$ , containing point $(10, −5)$

Solution

$y = - \frac{3}{5} x + 1$

Perpendicular to the line $y - 1 = 0$ , point $(−2, 6)$

Perpendicular to the line y-axis, point $(−6, 2)$

Solution

$y = 2$

Containing the points $(4, 3)$ and $(8, 1)$

Containing the points $(−2, 0)$ and $(−3, −2)$

Solution

$y = 2 x + 4$

Parallel to the line $x = −3$ , containing point $(−2, −1)$

Parallel to the line $x = −4$ , containing point $(−3, −5)$

Solution

$x = −3$

Containing the points $(−3, −4)$ and $(2, −5)$

Containing the points $(−5, −3)$ and $(4, −6)$

Solution

$y = - \frac{1}{3} x - \frac{14}{3}$

Perpendicular to the line $x - 2 y = 5$ , containing point $(−2, 2)$

Perpendicular to the line $4 x + 3 y = 1$ , containing point $(0, 0)$

Solution

$y = \frac{3}{4} x$

Everyday Math

Cholesterol. The age, $x$ , and LDL cholesterol level, $y$ , of two men are given by the points $(18, 68)$ and $(27, 122)$ . Find a linear equation that models the relationship between age and LDL cholesterol level.

Fuel consumption. The city mpg, $x$ , and highway mpg, $y$ , of two cars are given by the points $(29, 40)$ and $(19, 28)$ . Find a linear equation that models the relationship between city mpg and highway mpg.

Solution

$y = 1.2 x + 5.2$

Writing Exercises

Why are all horizontal lines parallel?

Explain in your own words why the slopes of two perpendicular lines must have opposite signs.

Solution

Answers will vary.

Self Check

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

Table assessing student understanding of finding linear equations. Columns: "I can...", "confidently", "with some help", "no-I don’t get it!". Rows describe skills, including finding equations from slope/intercept, slope/point, two points, parallel lines, and perpendicular lines.

ⓑ 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?

point–slope form

The point–slope form of an equation of a line with slope

m

and containing the point

(x_{1}, y_{1})

y - y_{1} = m (x - x_{1})

Graphs of Linear Inequalities

Verify Solutions to an Inequality in Two Variables

We have learned how to solve inequalities in one variable. Now, we will look at inequalities in two variables. Inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs—so that your business would make a profit.

Do you remember that an inequality with one variable had many solutions? The solution to the inequality $x > 3$ is any number greater than 3. We showed this on the number line by shading in the number line to the right of 3, and putting an open parenthesis at 3. See Figure 1.

The figure shows a number line extending from negative 5 to 5. A parenthesis is shown at positive 3 and an arrow extends form positive 3 to positive infinity.

Similarly, inequalities in two variables have many solutions. Any ordered pair $(x, y)$ that makes the inequality true when we substitute in the values is a solution of the inequality.

Determine whether each ordered pair is a solution to the inequality $y > x + 4$ :

ⓐ $(0, 0)$ ⓑ $(1, 6)$ ⓒ $(2, 6)$ ⓓ $(−5, −15)$ ⓔ $(−8, 12)$

Solution

ⓐ

$(0, 0)$

Simplify.
So, $(0, 0)$ is not a solution to $y > x + 4$ .
ⓑ

$(1, 6)$

Simplify.
So, $(1, 6)$ is a solution to $y > x + 4$ .
ⓒ

$(2, 6)$

Simplify.
So, $(2, 6)$ is not a solution to $y > x + 4$ .
ⓓ

$(−5, −15)$

Simplify.
So, $(−5, −15)$ is not a solution to $y > x + 4$ .
ⓔ

$(−8, 12)$

Simplify.
So, $(−8, 12)$ is a solution to $y > x + 4$ .

Recognize the Relation Between the Solutions of an Inequality and its Graph

Now, we will look at how the solutions of an inequality relate to its graph.

Let’s think about the number line in Figure 1 again. The point $x = 3$ separated that number line into two parts. On one side of 3 are all the numbers less than 3. On the other side of 3 all the numbers are greater than 3. See Figure 2.

The solution to $x > 3$ is the shaded part of the number line to the right of $x = 3$ .

Similarly, the line $y = x + 4$ separates the plane into two regions. On one side of the line are points with $y < x + 4$ . On the other side of the line are the points with $y > x + 4$ . We call the line $y = x + 4$ a boundary line.

For an inequality in one variable, the endpoint is shown with a parenthesis or a bracket depending on whether or not $a$ is included in the solution:

The figure shows two number lines. The number line on the left is labeled x is less than a. The number line shows a parenthesis at a and an arrow that points to the left. The number line on the right is labeled x is less than or equal to a. The number line shows a bracket at a and an arrow that points to the left.

Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to indicate whether or not it the line is included in the solution. This is summarized in Table 6

$A x + B y < C$	$A x + B y \leq C$
$A x + B y > C$	$A x + B y \geq C$
Boundary line is not included in solution.	Boundary line is included in solution.
Boundary line is dashed.	Boundary line is solid.

Now, let’s take a look at what we found in Example 1. We’ll start by graphing the line $y = x + 4$ , and then we’ll plot the five points we tested. See Figure 3.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals x plus 4 is plotted as an arrow extending from the bottom left toward the upper right. The following points are plotted and labeled (negative 8, 12), (1, 6), (2, 6), (0, 0), and (negative 5, negative 15).

In Example 1 we found that some of the points were solutions to the inequality $y > x + 4$ and some were not.

Which of the points we plotted are solutions to the inequality $y > x + 4$ ? The points $(1, 6)$ and $(−8, 12)$ are solutions to the inequality $y > x + 4$ . Notice that they are both on the same side of the boundary line $y = x + 4$ .

The two points $(0, 0)$ and $(−5, −15)$ are on the other side of the boundary line $y = x + 4$ , and they are not solutions to the inequality $y > x + 4$ . For those two points, $y < x + 4$ .

What about the point $(2, 6)$ ? Because $6 = 2 + 4$ , the point is a solution to the equation $y = x + 4$ . So the point $(2, 6)$ is on the boundary line.

Let’s take another point on the left side of the boundary line and test whether or not it is a solution to the inequality $y > x + 4$ . The point $(0, 10)$ clearly looks to be to the left of the boundary line, doesn’t it? Is it a solution to the inequality?

\begin{matrix} y > x + 4 \\ 10 \overset{?}{>} 0 + 4 \\ 10 > 4 & So, (0, 10) is a solution to y > x + 4 . \end{matrix}

Any point you choose on the left side of the boundary line is a solution to the inequality $y > x + 4$ . All points on the left are solutions.

Similarly, all points on the right side of the boundary line, the side with $(0, 0)$ and $(−5, −15)$ , are not solutions to $y > x + 4$ . See Figure 4.

The graph of the inequality $y > x + 4$ is shown in Figure 5 below. The line $y = x + 4$ divides the plane into two regions. The shaded side shows the solutions to the inequality $y > x + 4$ .

The points on the boundary line, those where $y = x + 4$ , are not solutions to the inequality $y > x + 4$ , so the line itself is not part of the solution. We show that by making the line dashed, not solid.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals x plus 4 is plotted as a dashed arrow extending from the bottom left toward the upper right. The coordinate plane to the upper left of the line is shaded. — The graph of the inequality $y > x + 4$ .

The boundary line shown is $y = 2 x - 1$ . Write the inequality shown by the graph.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals 2 x minus 1 is plotted as a solid arrow extending from the bottom left toward the upper right. The coordinate plane to the left of the line is shaded

Solution

The line $y = 2 x - 1$ is the boundary line. On one side of the line are the points with $y > 2 x - 1$ and on the other side of the line are the points with $y < 2 x - 1$ .

Let’s test the point $(0, 0)$ and see which inequality describes its side of the boundary line.

At $(0, 0)$ , which inequality is true:

\begin{matrix} y > 2 x - 1 & or & y < 2 x - 1 ? \\ y > 2 x - 1 & y < 2 x - 1 \\ 0 \overset{?}{>} 2 \cdot 0 - 1 & 0 \overset{?}{<} 2 \cdot 0 - 1 \\ 0 > −1 True & 0 < −1 False \end{matrix}

Since, $y > 2 x - 1$ is true, the side of the line with $(0, 0)$ , is the solution. The shaded region shows the solution of the inequality $y > 2 x - 1$ .

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality $y \geq 2 x - 1$ .

We could use any point as a test point, provided it is not on the line. Why did we choose $(0, 0)$ ? Because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that $y < 2 x - 1$ .

The boundary line shown is $2 x + 3 y = 6$ . Write the inequality shown by the graph.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line 2 x plus 3 y equals 6 is plotted as a dashed arrow extending from the top left toward the bottom right. The coordinate plane to the bottom of the line is shaded.

Solution

The line $2 x + 3 y = 6$ is the boundary line. On one side of the line are the points with $2 x + 3 y > 6$ and on the other side of the line are the points with $2 x + 3 y < 6$ .

Let’s test the point $(0, 0)$ and see which inequality describes its side of the boundary line.

At $(0, 0)$ , which inequality is true:

\begin{matrix} 2 x + 3 y & > & 6 & or & 2 x + 3 y & < & 6 ? \\ 2 x + 3 y & > & 6 & 2 x + 3 y & < & 6 \\ 2 (0) + 3 (0) & \overset{?}{>} & 6 & 2 (0) + 3 (0) & \overset{?}{<} & 6 \\ 0 & > & 6 False & 0 & < & 6 True \end{matrix}

So the side with $(0, 0)$ is the side where $2 x + 3 y < 6$ .

(You may want to pick a point on the other side of the boundary line and check that $2 x + 3 y > 6$ .)

Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.

The graph shows the solution to the inequality $2 x + 3 y < 6$ .

Graph Linear Inequalities

Now, we’re ready to put all this together to graph linear inequalities.

How to Graph Linear Inequalities

Graph the linear inequality $y \geq \frac{3}{4} x - 2$ .

Solution

This figure is a table that has three columns and three rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Identify and graph the boundary line. If the inequality is less than or equal to or greater than or equal to, the boundary line is solid. If the inequality is less than or greater than, the boundary line is dashed. The text in the second cell reads: “Replace the inequality sign with an equal sign to find the boundary line. Graph the boundary line y equals three-fourths x minus 2. The inequality sign is greater than or equal to, so we draw a solid line. The third cell contains the graph of the line three-fourths x minus 2 on a coordinate plane.

In the third row of the table, the first cell says: “Step 3. Shade in one side of the boundary line. If the test point is a solution, shade in the side that includes the point. If the test point is not a solution, shade in the opposite side. In the second cell, the instructions say: The test point (0, 0) is a solution to y is greater than or equal to three-fourths x minus 2. So we shade in that side.” In the third cell is the graph of the line three-fourths x minus 2 on a coordinate plane with the region above the line shaded.

The steps we take to graph a linear inequality are summarized here.

Graph the linear inequality $x - 2 y < 5$ .

Solution

First we graph the boundary line $x - 2 y = 5$ . The inequality is $<$ so we draw a dashed line.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x minus 2 y equals 5 is plotted as a dashed arrow extending from the bottom left toward the top right.

Then we test a point. We’ll use $(0, 0)$ again because it is easy to evaluate and it is not on the boundary line.

Is $(0, 0)$ a solution of $x - 2 y < 5$ ?
The figure shows the inequality 0 minus 2 times 0 in parentheses is less than 5, with a question mark above the inequality symbol. The next line shows 0 minus 0 is less than 5, with a question mark above the inequality symbol. The third line shows 0 is less than 5.

The point $(0, 0)$ is a solution of $x - 2 y < 5$ , so we shade in that side of the boundary line.
The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x minus 2 y equals 5 is plotted as a dashed arrow extending from the bottom left toward the top right. The point (0, 0) is plotted, but not labeled. The region above the line is shaded.

What if the boundary line goes through the origin? Then we won’t be able to use $(0, 0)$ as a test point. No problem—we’ll just choose some other point that is not on the boundary line.

Graph the linear inequality $y \leq −4 x$ .

Solution

First we graph the boundary line $y = −4 x$ . It is in slope–intercept form, with $m = −4 and b = 0$ . The inequality is $\leq$ so we draw a solid line.
The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line s y equals negative 4 x is plotted as a solid arrow extending from the top left toward the bottom right.

Now, we need a test point. We can see that the point $(1, 0)$ is not on the boundary line.

Is $(1, 0)$ a solution of $y \leq −4 x$ ?
The figure shows 0 is less than or equal to negative 4 times 1 in parentheses, with a question mark above the inequality symbol. The next line shows 0 is not less than or equal to negative 4.

The point $(1, 0)$ is not a solution to $y \leq −4 x$ , so we shade in the opposite side of the boundary line. See Figure 6.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals negative 4 x is plotted as a solid arrow extending from the top left toward the bottom right. The point (1, 0) is plotted, but not labeled. The region to the left of the line is shaded.

Some linear inequalities have only one variable. They may have an x but no y, or a y but no x. In these cases, the boundary line will be either a vertical or a horizontal line. Do you remember?

\begin{matrix} x = a & vertical line \\ y = b & horizontal line \end{matrix}

Graph the linear inequality $y > 3$ .

Solution

First we graph the boundary line $y = 3$ . It is a horizontal line. The inequality is > so we draw a dashed line.

We test the point $(0, 0)$ .

\begin{matrix} y > 3 \\ 0 > 3 \end{matrix}

$(0, 0)$ is not a solution to $y > 3$ .

So we shade the side that does not include (0, 0).
The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals 3 is plotted as a dashed arrow horizontally across the plane. The region above the line is shaded.

Key Concepts

To Graph a Linear Inequality
1. Identify and graph the boundary line.
 If the inequality is $\leq or \geq$ , the boundary line is solid.
 If the inequality is < or >, the boundary line is dashed.
2. Test a point that is not on the boundary line. Is it a solution of the inequality?
3. Shade in one side of the boundary line.
 If the test point is a solution, shade in the side that includes the point.
 If the test point is not a solution, shade in the opposite side.

Section Exercises

Practice Makes Perfect

Verify Solutions to an Inequality in Two Variables

In the following exercises, determine whether each ordered pair is a solution to the given inequality.

Determine whether each ordered pair is a solution to the inequality $y > x - 1$ :

ⓐ $(0, 1)$
ⓑ $(−4, −1)$
ⓒ $(4, 2)$
ⓓ $(3, 0)$
ⓔ $(−2, −3)$

Determine whether each ordered pair is a solution to the inequality $y > x - 3$ :

ⓐ $(0, 0)$
ⓑ $(2, 1)$
ⓒ $(−1, −5)$
ⓓ $(−6, −3)$
ⓔ $(1, 0)$

Solution

ⓐ yes ⓑ yes ⓒ no ⓓ yes ⓔ yes

Determine whether each ordered pair is a solution to the inequality $y < x + 2$ :

ⓐ $(0, 3)$
ⓑ $(−3, −2)$
ⓒ $(−2, 0)$
ⓓ $(0, 0)$
ⓔ $(−1, 4)$

Determine whether each ordered pair is a solution to the inequality $y < x + 5$ :

ⓐ $(−3, 0)$
ⓑ $(1, 6)$
ⓒ $(−6, −2)$
ⓓ $(0, 1)$
ⓔ $(5, −4)$

Solution

ⓐ yes ⓑ no ⓒ yes ⓓ yes ⓔ yes

Determine whether each ordered pair is a solution to the inequality $x + y > 4$ :

ⓐ $(5, 1)$
ⓑ $(−2, 6)$
ⓒ $(3, 2)$
ⓓ $(10, −5)$
ⓔ $(0, 0)$

Determine whether each ordered pair is a solution to the inequality $x + y > 2$ :

ⓐ $(1, 1)$
ⓑ $(4, −3)$
ⓒ $(0, 0)$
ⓓ $(−8, 12)$
ⓔ $(3, 0)$

Solution

ⓐ no ⓑ no ⓒ no ⓓ yes ⓔ yes

Recognize the Relation Between the Solutions of an Inequality and its Graph

In the following exercises, write the inequality shown by the shaded region.

Write the inequality shown by the graph with the boundary line $y = 3 x - 4 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals 3x minus 4 is plotted as a dashed line extending from the bottom left toward the top right. The region to the right of the line is shaded.

Write the inequality shown by the graph with the boundary line $y = 2 x - 4 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals 2x minus 4 is plotted as a solid line extending from the bottom left toward the top right. The region below the line is shaded.

Solution

$y < 2 x - 4$

Write the inequality shown by the graph with the boundary line $y = \frac{1}{2} x + 1 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals negative one-half x plus 1 is plotted as a solid line extending from the bottom left toward the top right. The region below the line is shaded.

Write the inequality shown by the graph with the boundary line $y = - \frac{1}{3} x - 2 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals negative one-third x minus 2 is plotted as a solid line extending from the top left toward the bottom right. The region below the line is shaded.

Solution

$y \leq - \frac{1}{3} x - 2$

Write the inequality shown by the shaded region in the graph with the boundary line $x + y = 5 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x plus y equals 5 is plotted as a solid line extending from the top left toward the bottom right. The region above the line is shaded.

Write the inequality shown by the shaded region in the graph with the boundary line $x + y = 3 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x plus y equals 3 is plotted as a solid line extending from the top left toward the bottom right. The region above the line is shaded.

Solution

$x + y \geq 3$

Write the inequality shown by the shaded region in the graph with the boundary line $2 x + y = −4 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line 2 x plus y equals negative 4 is plotted as a solid line extending from the top left toward the bottom right. The region below the line is shaded.

Write the inequality shown by the shaded region in the graph with the boundary line $x + 2 y = −2 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x plus 2 y equals negative 2 is plotted as a solid line extending from the top left toward the bottom right. The region below the line is shaded.

Solution

$x + 2 y \leq −2$

Write the inequality shown by the shaded region in the graph with the boundary line $3 x - y = 6 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line 3 x minus y equals 6 is plotted as a dashed line extending from the bottom left toward the top right. The region to the left of the line is shaded.

Write the inequality shown by the shaded region in the graph with the boundary line $2 x - y = 4 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line 2 x minus y equals 4 is plotted as a dashed line extending from the bottom left toward the top right. The region to the left of the line is shaded.

Solution

$2 x - y < 4$

Write the inequality shown by the shaded region in the graph with the boundary line $2 x - 5 y = 10 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line 2 x minus 5 y equals 10 is plotted as a dashed line extending from the bottom left toward the top right. The region below the line is shaded.

Write the inequality shown by the shaded region in the graph with the boundary line $4 x - 3 y = 12 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line 4 x minus 3 y equals 12 is plotted as a dashed line extending from the bottom left toward the top right. The region below the line is shaded.

Solution

$4 x - 3 y > 12$

Graph Linear Inequalities

In the following exercises, graph each linear inequality.

Graph the linear inequality $y > \frac{2}{3} x - 1$ .

Graph the linear inequality $y < \frac{3}{5} x + 2$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals three-fifths x plus 2 is plotted as a dashed line extending from the bottom left toward the top right. The region below the line is shaded.

Graph the linear inequality $y \leq - \frac{1}{2} x + 4$ .

Graph the linear inequality $y \geq - \frac{1}{3} x - 2$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals negative one-third x minus 2 is plotted as a solid line extending from the top left toward the bottom right. The region below the line is shaded.

Graph the linear inequality $x - y \leq 3$ .

Graph the linear inequality $x - y \geq −2$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x minus y equals negative 2 is plotted as a solid line extending from the bottom left toward the top right. The region below the line is shaded.

Graph the linear inequality $4 x + y > −4$ .

Graph the linear inequality $x + 5 y < −5$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x plus 5 y equals negative 5 is plotted as a dashed line extending from the top left toward the bottom right. The region below the line is shaded.

Graph the linear inequality $3 x + 2 y \geq −6$ .

Graph the linear inequality $4 x + 2 y \geq −8$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line 4 x plus 2 y equals negative 8 is plotted as a solid line extending from the top left toward the bottom right. The region to the right of the line is shaded.

Graph the linear inequality $y > 4 x$ .

Graph the linear inequality $y > x$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals x is plotted as a solid line extending from the bottom left toward the top right. The region above the line is shaded.

Graph the linear inequality $y \leq − x$ .

Graph the linear inequality $y \leq −3 x$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals negative 3 x is plotted as a solid line extending from the top left toward the bottom right. The region to the left of the line is shaded.

Graph the linear inequality $y \geq −2$ .

Graph the linear inequality $y < −1$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals negative 1 is plotted as a dashed horizontal line. The region below the line is shaded.

Graph the linear inequality $y < 4$ .

Graph the linear inequality $y \geq 2$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals 2 is plotted as a solid horizontal line. The region above the line is shaded.

Graph the linear inequality $x \leq 5$ .

Graph the linear inequality $x > −2$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x equals negative 2 is plotted as a dashed vertical line. The region to the right of the line is shaded.

Graph the linear inequality $x > −3$ .

Graph the linear inequality $x \leq 4$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x equals 4 is plotted as a solid vertical line. The region to the left of the line is shaded.

Graph the linear inequality $x - y < 4$ .

Graph the linear inequality $x - y < −3$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x minus y equals negative 3 is plotted as a dashed line extending from the bottom left toward the top right. The region above the line is shaded.

Graph the linear inequality $y \geq \frac{3}{2} x$ .

Graph the linear inequality $y \leq \frac{5}{4} x$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals five-fourths x is plotted as a solid line extending from the bottom left toward the top right. The region below the line is shaded.

Graph the linear inequality $y > −2 x + 1$ .

Graph the linear inequality $y < −3 x - 4$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line y equals negative 3 x minus 4 is plotted as a dashed line extending from the top left toward the bottom right. The region to the left of the line is shaded.

Graph the linear inequality $x \leq −1$ .

Graph the linear inequality $x \geq 0$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The line x equals negative 0 is plotted as a solid vertical line along the y-axis. The region to the right of the line is shaded.

Everyday Math

Money. Gerry wants to have a maximum of $100 cash at the ticket booth when his church carnival opens. He will have $1 bills and $5 bills. If x is the number of $1 bills and y is the number of $5 bills, the inequality $x + 5 y \leq 100$ models the situation.

ⓐ Graph the inequality.
ⓑ List three solutions to the inequality $x + 5 y \leq 100$ where both x and y are integers.

Shopping. Tula has $20 to spend at the used book sale. Hardcover books cost $2 each and paperback books cost $0.50 each. If x is the number of hardcover books Tula can buy and y is the number of paperback books she can buy, the inequality $2 x + \frac{1}{2} y \leq 20$ models the situation.

ⓐ Graph the inequality.
ⓑ List three solutions to the inequality $2 x + \frac{1}{2} y \leq 20$ where both x and y are whole numbers.

Solution

ⓐ
ⓑ Answers will vary.

Writing Exercises

Lester thinks that the solution of any inequality with a > sign is the region above the line and the solution of any inequality with a < sign is the region below the line. Is Lester correct? Explain why or why not.

Explain why in some graphs of linear inequalities the boundary line is solid but in other graphs it is dashed.

Self Check

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

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

Chapter 4 Review Exercises

Rectangular Coordinate System

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system.

ⓐ $(−1, −5)$
ⓑ $(−3, 4)$
ⓒ $(2, −3)$
ⓓ $(1, \frac{5}{2})$

ⓐ $(4, 3)$
ⓑ $(−4, 3)$
ⓒ $(−4, −3)$
ⓓ $(4, −3)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (4, 3) is plotted and labeled "a". The point (negative 4, 3) is plotted and labeled "b". The point (negative 4, negative 3) is plotted and labeled "c". The point (4, negative 3) is plotted and labeled “d”.

ⓐ $(−2, 0)$
ⓑ $(0, −4)$
ⓒ $(0, 5)$
ⓓ $(3, 0)$

ⓐ $(2, \frac{3}{2})$
ⓑ $(3, \frac{4}{3})$
ⓒ $(\frac{1}{3}, −4)$
ⓓ $(\frac{1}{2}, −5)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (2, three halves) is plotted and labeled "a". The point (3, four thirds) is plotted and labeled "b". The point (one third, negative 4) is plotted and labeled "c". The point (one-half, negative 5) is plotted and labeled “d”.

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 x- and y-axes each run from negative 6 to 6. The point (5, 3) is plotted and labeled "a". The point (2, negative 1) is plotted and labeled "b". The point (negative 3, negative 2) is plotted and labeled "c". The point (negative 1, 4) is plotted and labeled “d”.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (2, 0) is plotted and labeled "a". The point (0, negative 5) is plotted and labeled "b". The point (negative 4, 0) is plotted and labeled "c". The point (0, 3) is plotted and labeled “d”.

Solution

ⓐ $(2, 0)$ ⓑ $(0, −5)$ ⓒ $(−4.0)$ ⓓ $(0, 3)$

Verify Solutions to an Equation in Two Variables

In the following exercises, which ordered pairs are solutions to the given equations?

$5 x + y = 10$

ⓐ $(5, 1)$
ⓑ $(2, 0)$
ⓒ $(4, −10)$

$y = 6 x - 2$

ⓐ $(1, 4)$
ⓑ $(\frac{1}{3}, 0)$
ⓒ $(6, −2)$

Solution

a, b

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.

$y = 4 x - 1$

$x$	$y$	$(x, y)$
0
1
$−2$

$y = - \frac{1}{2} x + 3$

$x$	$y$	$(x, y)$
0
4
$−2$

Solution

$x$	$y$	$(x, y)$
0	3	$(0, 3)$
4	1	(4, 1)
$−2$	4	$(−2, 4)$

$x + 2 y = 5$

$x$	$y$	$(x, y)$
	0
1
$−1$

$3 x + 2 y = 6$

$x$	$y$	$(x, y)$
0
	0
$−2$

Solution

$x$	$y$	$(x, y)$
0	$3$	$(0, 3)$
2	0	$(2, 0)$
$−2$	$6$	$(−2, 6)$

Find Solutions to a Linear Equation in Two Variables

In the following exercises, find three solutions to each linear equation.

$x + y = 3$

$x + y = −4$

Solution

Answers will vary.

$y = 3 x + 1$

$y = − x - 1$

Solution

Answers will vary.

Graphing Linear Equations

Recognize the Relation Between the Solutions of an Equation and its Graph

In the following exercises, for each ordered pair, decide:

ⓐ Is the ordered pair a solution to the equation?
ⓑ Is the point on the line?

$y = − x + 4$

ⓐ $(0, 4)$

ⓑ $(−1, 3)$

ⓓ $(−2, 6)$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative x plus 4 is plotted as an arrow extending from the top left toward the bottom right.

$y = \frac{2}{3} x - 1$

ⓐ $(0, −1)$

ⓑ $(3, 1)$

ⓓ $(6, 4)$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals two-thirds x minus 1 is plotted as an arrow extending from the bottom left toward the top right.

Solution

ⓐ yes; yes ⓑ yes; yes ⓒ yes; yes ⓓ no; no

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

$y = 4 x - 3$

$y = −3 x$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative 3 x is plotted as an arrow extending from the top left toward the bottom right.

$y = \frac{1}{2} x + 3$

$x - y = 6$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x minus y equals 6 is plotted as an arrow extending from the bottom left toward the top right.

$2 x + y = 7$

$3 x - 2 y = 6$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line 3 x minus 2 y equals 6 is plotted as an arrow extending from the bottom left toward the top right.

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

$y = −2$

$x = 3$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x equals 3 is plotted as a vertical line.

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

$y = −2 x$ and $y = −2$

$y = \frac{4}{3} x$ and $y = \frac{4}{3}$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals four-thirds x is plotted as an arrow extending from the bottom left toward the top right. The line y equals four-thirds is plotted as a horizontal line.

Graphing with Intercepts

Identify the x- and y-Intercepts on a Graph

In the following exercises, find the x- and y-intercepts.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 4, 0) and (0, 4) is plotted.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (3, 0) and (0, 3) is plotted.

Solution

$(3, 0), (0, 3)$

Find the x- and y-Intercepts from an Equation of a Line

In the following exercises, find the intercepts of each equation.

$x + y = 5$

$x - y = −1$

Solution

$(−1, 0), (0, 1)$

$x + 2 y = 6$

$2 x + 3 y = 12$

Solution

$(6, 0), (0, 4)$

$y = \frac{3}{4} x - 12$

$y = 3 x$

Solution

$(0, 0)$

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

$− x + 3 y = 3$

$x + y = −2$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x plus y equals negative 2 is plotted as an arrow extending from the top left toward the bottom right.

$x - y = 4$

$2 x - y = 5$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line 2 x minus y equals 5 is plotted as an arrow extending from the bottom left toward the top right.

$2 x - 4 y = 8$

$y = 2 x$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 2 x is plotted as an arrow extending from the bottom left toward the top right.

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

$\frac{4}{3}$

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

$- \frac{2}{3}$

In the following exercises, model each slope. Draw a picture to show your results.

$\frac{1}{3}$

$\frac{3}{2}$

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.

$- \frac{2}{3}$

$- \frac{1}{2}$

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 2 and the point in column 3 row 3.

Use $m = \frac{rise}{run}$ to 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- and y-axes each run from negative 7 to 7. A line passing through the points (negative 1, 3), (0, 0), and (1, negative 3) is plotted.

Solution

A graph on a Cartesian coordinate plane displays a straight line with a positive slope, passing through the x-axis at (5, 0) and the y-axis at (0, -2.5), extending infinitely.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 3, 6) and (5, 2) is plotted.

Solution

$- \frac{1}{2}$

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

Use the Slope Formula to find the Slope of a Line between Two Points

In the following exercises, use the slope formula to find the slope of the line between each pair of points.

$(−1, −1), (0, 5)$

$(3, 5), (4, −1)$

Solution

$−6$

$(−5, −2), (3, 2)$

$(2, 1), (4, 6)$

Solution

$\frac{5}{2}$

Graph a Line Given a Point and the Slope

In the following exercises, graph each line with the given point and slope.

$(2, −2)$ ; $m = \frac{5}{2}$

$(−3, 4)$ ; $m = - \frac{1}{3}$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 3, 4) and (0, 3) is plotted.

x-intercept $−4$ ; $m = 3$

y-intercept 1; $m = - \frac{3}{4}$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (0, 1) and (4, negative 2) is plotted.

Solve Slope Applications

In the following exercises, solve these slope applications.

The roof pictured below has a rise of 10 feet and a run of 15 feet. What is its slope?

The figure shows a person on a ladder using a hammer on the roof of a building.

A mountain road rises 50 feet for a 500-foot run. What is its slope?

Solution

$\frac{1}{10}$

Intercept Form of an Equation of a Line

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

In the following exercises, use the graph to find the slope and y-intercept of each line. Compare the values to the equation $y = m x + b$ .

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 4 x minus 1 is plotted from the lower left to the top right.

$y = 4 x - 1$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals two-thirds x plus 4 is plotted from the top left to the bottom right.

$y = - \frac{2}{3} x + 4$

Solution

slope $m = - \frac{2}{3}$ and y-intercept $(0, 4)$

Identify the Slope and y-Intercept from an Equation of a Line

In the following exercises, identify the slope and y-intercept of each line.

$y = −4 x + 9$

$y = \frac{5}{3} x - 6$

Solution

$\frac{5}{3}; (0, −6)$

$5 x + y = 10$

$4 x - 5 y = 8$

Solution

$\frac{4}{5}; (0, - \frac{8}{5})$

Graph a Line Using Its Slope and Intercept

In the following exercises, graph the line of each equation using its slope and y-intercept.

$y = 2 x + 3$

$y = − x - 1$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative x minus 1 is plotted from the top left to the bottom right.

$y = - \frac{2}{5} x + 3$

$4 x - 3 y = 12$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line 4 x minus 3 y equals 12 is plotted from the bottom left to the top right.

In the following exercises, determine the most convenient method to graph each line.

$x = 5$

$y = −3$

Solution

horizontal line

$2 x + y = 5$

$x - y = 2$

Solution

intercepts

$y = x + 2$

$y = \frac{3}{4} x - 1$

Solution

plotting points

Graph and Interpret Applications of Slope–Intercept

Katherine is a private chef. The equation $C = 6.5 m + 42$ models the relation between her weekly cost, C, in dollars and the number of meals, m, that she serves.

ⓐ Find Katherine’s cost for a week when she serves no meals.
ⓑ Find the cost for a week when she serves 14 meals.
ⓒ Interpret the slope and C-intercept of the equation.
ⓓ Graph the equation.

Marjorie teaches piano. The equation $P = 35 s - 250$ models the relation between her weekly profit, P, in dollars and the number of student lessons, s, that she teaches.

ⓐ Find Marjorie’s profit for a week when she teaches no student lessons.
ⓑ Find the profit for a week when she teaches 20 student lessons.
ⓒ Interpret the slope and P–intercept of the equation.
ⓓ Graph the equation.

Solution

ⓐ −$250 ⓑ $450 ⓒ The slope, 35, means that Marjorie’s weekly profit, P, increases by $35 for each additional student lesson she teaches. The P–intercept means that when the number of lessons is 0, Marjorie loses $250. ⓓ
The graph shows the x y-coordinate plane where h is plotted along the x-axis and P is potted along the y-axis. The x-axis runs from 0 to 24. The y-axis runs from negative 300 to 500. The line P equals 35 h minus 250 is plotted from the bottom left to the top right.

Use Slopes to Identify Parallel Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel.

$4 x - 3 y = −1; y = \frac{4}{3} x - 3$

$2 x - y = 8; x - 2 y = 4$

Solution

not parallel

Use Slopes to Identify Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are perpendicular.

$y = 5 x - 1; 10 x + 2 y = 0$

$3 x - 2 y = 5; 2 x + 3 y = 6$

Solution

perpendicular

Find the Equation of a Line

Find an Equation of the Line Given the Slope and y-Intercept

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope–intercept form.

slope $\frac{1}{3}$ and $y -intercept$ $(0, −6)$

slope $−5$ and $y -intercept$ $(0, −3)$

Solution

$y = −5 x - 3$

slope $0$ and $y -intercept$ $(0, 4)$

slope $−2$ and $y -intercept$ $(0, 0)$

Solution

$y = −2 x$

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope–intercept form.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 2 x plus 1 is plotted from the bottom left to the top right.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative 3 x plus 5 is plotted from the top left to the bottom right.

Solution

$y = −3 x + 5$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals three-fourths x minus 2 is plotted from the bottom left to the top right.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative 4 is plotted as a horizontal line.

Solution

$y = −4$

Find an Equation of the Line Given the Slope and a Point

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope–intercept form.

$m = - \frac{1}{4}$ , point $(−8, 3)$

$m = \frac{3}{5}$ , point $(10, 6)$

Solution

$y = \frac{3}{5} x$

Horizontal line containing $(−2, 7)$

$m = −2$ , point $(−1, −3)$

Solution

$y = −2 x - 5$

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope–intercept form.

$(2, 10)$ and $(−2, −2)$

$(7, 1)$ and $(5, 0)$

Solution

$y = \frac{1}{2} x - \frac{5}{2}$

$(3, 8)$ and $(3, −4)$ .

$(5, 2)$ and $(−1, 2)$

Solution

$y = 2$

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope–intercept form.

line $y = −3 x + 6$ , point $(1, −5)$

line $2 x + 5 y = −10$ , point $(10, 4)$

Solution

$y = - \frac{2}{5} x + 8$

line $x = 4$ , point $(−2, −1)$

line $y = −5$ , point $(−4, 3)$

Solution

$y = 3$

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope–intercept form.

line $y = - \frac{4}{5} x + 2$ , point $(8, 9)$

line $2 x - 3 y = 9$ , point $(−4, 0)$

Solution

$y = - \frac{3}{2} x - 6$

line $y = 3$ , point $(−1, −3)$

line $x = −5$ point $(2, 1)$

Solution

$y = 1$

Graph Linear Inequalities

Verify Solutions to an Inequality in Two Variables

In the following exercises, determine whether each ordered pair is a solution to the given inequality.

Determine whether each ordered pair is a solution to the inequality $y < x - 3$ :

ⓐ $(0, 1)$
ⓑ $(−2, −4)$
ⓒ $(5, 2)$
ⓓ $(3, −1)$
ⓔ $(−1, −5)$

Determine whether each ordered pair is a solution to the inequality $x + y > 4$ :

ⓐ $(6, 1)$
ⓑ $(−3, 6)$
ⓒ $(3, 2)$
ⓓ $(−5, 10)$
ⓔ $(0, 0)$

Solution

ⓐ yes ⓑ no ⓒ yes ⓓ yes ⓔ no

Recognize the Relation Between the Solutions of an Inequality and its Graph

In the following exercises, write the inequality shown by the shaded region.

Write the inequality shown by the graph with the boundary line $y = − x + 2$ .

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative x plus 2 is plotted as a solid line extending from the top left toward the bottom right. The region below the line is shaded.

Write the inequality shown by the graph with the boundary line $y = \frac{2}{3} x - 3$ .

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals two-thirds x minus 3 is plotted as a dashed line extending from the bottom left toward the top right. The region above the line is shaded.

Solution

$y \geq \frac{2}{3} x - 3$

Write the inequality shown by the shaded region in the graph with the boundary line $x + y = −4$ .

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x plus y equals negative 4 is plotted as a dashed line extending from the top left toward the bottom right. The region above the line is shaded.

Write the inequality shown by the shaded region in the graph with the boundary line $x - 2 y = 6 .$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x minus 2 y equals 6 is plotted as a solid line extending from the bottom left toward the top right. The region below the line is shaded.

Solution

$x - 2 y \geq 6$

Graph Linear Inequalities

In the following exercises, graph each linear inequality.

Graph the linear inequality $y > \frac{2}{5} x - 4$ .

Graph the linear inequality $y < - \frac{1}{4} x + 3$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative one-fourth x plus 3 is plotted as a solid line extending from the top left toward the bottom right. The region below the line is shaded.

Graph the linear inequality $x - y \leq 5$ .

Graph the linear inequality $3 x + 2 y > 10$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line 3 x plus 2 y equals 10 is plotted as a dashed line extending from the top left toward the bottom right. The region above the line is shaded.

Graph the linear inequality $y \leq −3 x$ .

Graph the linear inequality $y < 6$ .

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 6 is plotted as a dashed, horizontal line. The region below the line is shaded.

Practice Test

Plot each point in a rectangular coordinate system.

ⓐ $(2, 5)$
ⓑ $(−1, −3)$
ⓒ $(0, 2)$
ⓓ $(−4, \frac{3}{2})$
ⓔ $(5, 0)$

Which of the given ordered pairs are solutions to the equation $3 x - y = 6$ ?

ⓐ $(3, 3)$
ⓑ $(2, 0)$
ⓒ $(4, −6)$

Solution

ⓐ yes ⓑ yes ⓒ no

Find three solutions to the linear equation $y = −2 x - 4$ .

Find the x- and y-intercepts of the equation $4 x - 3 y = 12$ .

Solution

$(3, 0), (0, −4)$

Find the slope of each line shown.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A line passing through the points (negative 5, 2) and (0, negative 1) is plotted from the top left toward the bottom right.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A vertical line passing through the point (2, 0) is plotted.

Solution

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The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. A horizontal line passing through the point (0, 5) is plotted.

Find the slope of the line between the points $(5, 2)$ and $(−1, −4)$ .

Solution

$1$

Graph the line with slope $\frac{1}{2}$ containing the point $(−3, −4)$ .

Graph the line for each of the following equations.

$y = \frac{5}{3} x - 1$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals five-thirds x minus 1 is plotted. The line passes through the points (0, negative 1) and (three-fifths, 0).

$y = − x$

$x - y = 2$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line x minus y equals 2 is plotted. The line passes through the points (0, negative 2) and (2, 0).

$4 x + 2 y = −8$

$y = 2$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals 2 is plotted as a horizontal line passing through the point (0, 2).

$x = −3$

Find the equation of each line. Write the equation in slope–intercept form.

slope $- \frac{3}{4}$ and y-intercept $(0, −2)$

Solution

$y = - \frac{3}{4} x - 2$

$m = 2$ , point $(−3, −1)$

containing $(10, 1)$ and $(6, −1)$

Solution

$y = \frac{1}{2} x - 4$

parallel to the line $y = - \frac{2}{3} x - 1$ , containing the point $(−3, 8)$

perpendicular to the line $y = \frac{5}{4} x + 2$ , containing the point $(−10, 3)$

Solution

$y = - \frac{4}{5} x - 5$

Write the inequality shown by the graph with the boundary line $y = − x - 3$ .

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative x minus 3 is plotted. The solid line passes through the points (negative 3, 0) and (0, negative 3).

Graph each linear inequality.

$y > \frac{3}{2} x + 5$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals three-halves x plus 5 is plotted. The dashed line passes through the points (0, 5) and (2, 8).

$x - y \geq −4$

$y \leq −5 x$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The line y equals negative 5 x is plotted. The solid line passes through the points (0, 0) and (1, negative 5).

$y < 3$

boundary line

The line with equation

A x + B y = C

that separates the region where

A x + B y > C

from the region where

A x + B y < C

linear inequality

An inequality that can be written in one of the following forms:

\begin{matrix} A x + B y > C & A x + B y \geq C & A x + B y < C & A x + B y \leq C \end{matrix}

where

A and B

are not both zero.

solution of a linear inequality

An ordered pair

(x, y)

is a solution to a linear inequality the inequality is true when we substitute the values of x and y.

Introduction

This is a photo of a two story home with several large windows. — Designing the number and sizes of windows in a home can pose challenges for an architect.

An architect designing a home may have restrictions on both the area and perimeter of the windows because of energy and structural concerns. The length and width chosen for each window would have to satisfy two equations: one for the area and the other for the perimeter. Similarly, a banker may have a fixed amount of money to put into two investment funds. A restaurant owner may want to increase profits, but in order to do that he will need to hire more staff. A job applicant may compare salary and costs of commuting for two job offers.

In this chapter, we will look at methods to solve situations like these using equations with two variables.

Solve Systems of Equations by Graphing

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations and Inequalities we learned how to solve linear equations with one variable. Remember that the solution of an equation is a value of the variable that makes a true statement when substituted into the equation.

Now we will work with systems of linear equations, two or more linear equations grouped together.

We will focus our work here on systems of two linear equations in two unknowns. Later, you may solve larger systems of equations.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

{\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix}

A linear equation in two variables, like 2x + y = 7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs (x, y) that make both equations true. These are called the solutions to a system of equations.

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Let’s consider the system below:

{\begin{matrix} 3 x - y = 7 \\ x - 2 y = 4 \end{matrix}

Is the ordered pair $(2, −1)$ a solution?

This figure begins with a sentence, “We substitute x =2 and y = -1 into both equations.” The first equation shows that 3x minus y equals 7. Then 3 times 2 minus negative, in parentheses, equals 7. Then 7 equals 7 is true. The second equation reads x minus 2y equals 4. Then 2 minus 2 times negative one in parentheses equals 4. Then 4 = 4 is true.

The ordered pair (2, −1) made both equations true. Therefore (2, −1) is a solution to this system.

Let’s try another ordered pair. Is the ordered pair (3, 2) a solution?

This figure begins with the sentence, “We substitute x equals 3 and y equals 2 into both equations.” The first equation reads 3 times x minus 7equals 7. Then, 3 times 3 minus 2 equals 7. Then 7 = 7 is true. The second equation reads x minus 2y equals 4. The n times 2 minus 2 times 2 = 4. Then negative 1 = 4 is false.

The ordered pair (3, 2) made one equation true, but it made the other equation false. Since it is not a solution to both equations, it is not a solution to this system.

Determine whether the ordered pair is a solution to the system: ${\begin{matrix} x - y = −1 \\ 2 x - y = −5 \end{matrix}$

ⓐ $(−2, −1)$ ⓑ $(−4, −3)$

Solution

ⓐ

(–2, –1) does not make both equations true. (–2, –1) is not a solution.

ⓑ

(–4, –3) does make both equations true. (–4, –3) is a solution.

Solve a System of Linear Equations by Graphing

In this chapter we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown in Figure 1:

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

For the first example of solving a system of linear equations in this section and in the next two sections, we will solve the same system of two linear equations. But we’ll use a different method in each section. After seeing the third method, you’ll decide which method was the most convenient way to solve this system.

How to Solve a System of Linear Equations by Graphing

Solve the system by graphing: ${\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix} .$

Solution

This table has four rows and three columns. The first column acts as the header column. The first row reads, “Step 1. Graph the first equation.” Then it reads, “To graph the first line, write the equation in slope-intercept form.” The equation reads 2x + y = 7 and becomes y = -2x + 7 where m = -2 and b = 7. Then it shows a graph of the equations 2x + y = 7. The equation x – 2y = 6 is also listed.

The second row reads, “Step 2. Graph the second equation on the same rectangular coordinate system.” Then it says, “To graph the second line, use intercepts.” This is followed by the equation x – 2y = 6 and the ordered pairs (0, -3) and (6, 0). The last column of this row shows a graph of the two equations.

The third row reads, “Step 3. Determine whether the lines intersect, are parallel, or are the same line.” Then “Look at the graph of the lines.” Finally it reads, “The lines intersect.”

The fourth row reads, “Step 4. Identify the solution to the system. If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an infinite number of solutions.” Then it reads, “Since the lines intersect, find the point of intersection. Check the point in both equations.” Finally it reads, “The lines intersect at (4, -1). It then uses substitution to show that, “The solution is (4, -1).”

The steps to use to solve a system of linear equations by graphing are shown below.

Solve the system by graphing: ${\begin{matrix} y = 2 x + 1 \\ y = 4 x - 1 \end{matrix} .$

Solution

Both of the equations in this system are in slope-intercept form, so we will use their slopes and y-intercepts to graph them.

{\begin{matrix} y = 2 x + 1 \\ y = 4 x - 1 \end{matrix}

Find the slope and y-intercept of the first equation.
Find the slope and y-intercept of the first equation.
Graph the two lines.
Determine the point of intersection.	The lines intersect at (1, 3).

Check the solution in both equations.	$\begin{matrix} \begin{matrix} y & = & 2 x + 1 \\ 3 & \overset{?}{=} & 2 \cdot 1 + 1 \\ 3 & = & 3 ✓ \end{matrix} & \begin{matrix} y & = & 4 x - 1 \\ 3 & \overset{?}{=} & 4 \cdot 1 - 1 \\ 3 & = & 3 ✓ \end{matrix} \end{matrix}$
	The solution is (1, 3).

Both equations in Example 3 were given in slope–intercept form. This made it easy for us to quickly graph the lines. In the next example, we’ll first re-write the equations into slope–intercept form.

Solve the system by graphing: ${\begin{matrix} 3 x + y = −1 \\ 2 x + y = 0 \end{matrix} .$

Solution

We’ll solve both of these equations for

y

so that we can easily graph them using their slopes and y-intercepts.

{\begin{matrix} 3 x + y = −1 \\ 2 x + y = 0 \end{matrix}

Solve the first equation for y. Find the slope and y-intercept. Solve the second equation for y. Find the slope and y-intercept.	$\begin{matrix} \begin{matrix} 3 x + y & = & - 1 \\ y & = & - 3 x - 1 \\ m & = & - 3 \\ b & = & - 1 \\ 2 x + y & = & 0 \\ y & = & - 2 x \\ m & = & - 2 \\ b & = & 0 \end{matrix} \end{matrix}$
Graph the lines.
Determine the point of intersection.	The lines intersect at (−1, 2).
Check the solution in both equations.	$\begin{matrix} \begin{matrix} 3 x + y & = & - 1 \\ 3 (- 1) + 2 & \overset{?}{=} & - 1 \\ - 1 & = & - 1 ✓ \end{matrix} & \begin{matrix} 2 x + y & = & 0 \\ 2 (- 1) + 2 & \overset{?}{=} & 0 \\ 0 & = & 0 ✓ \end{matrix} \end{matrix}$
	The solution is (−1, 2).

Usually when equations are given in standard form, the most convenient way to graph them is by using the intercepts. We’ll do this in Example 5.

Solve the system by graphing: ${\begin{matrix} x + y = 2 \\ x - y = 4 \end{matrix} .$

Solution

We will find the x- and y-intercepts of both equations and use them to graph the lines.


To find the intercepts, let x = 0 and solve for y, then let y = 0 and solve for x.	$\begin{matrix} \begin{matrix} x + y & = & 2 \\ 0 + y & = & 2 \\ y & = & 2 \end{matrix} & \begin{matrix} x + y & = & 2 \\ x + 0 & = & 2 \\ x & = & 2 \end{matrix} \end{matrix}$

To find the intercepts, let x = 0 then let y = 0.	$\begin{matrix} \begin{matrix} x - y & = & 4 \\ 0 - y & = & 4 \\ - y & = & 4 \\ y & = & - 4 \end{matrix} & \begin{matrix} x - y & = & 4 \\ x - 0 & = & 4 \\ x & = & 4 \end{matrix} \end{matrix}$
Graph the line.
Determine the point of intersection.	The lines intersect at (3, −1).
Check the solution in both equations.	$\begin{matrix} x + y & = & 2 & x - y & = & 4 \\ 3 + (−1) & \overset{?}{=} & 2 & 3 - (−1) & \overset{?}{=} & 4 \\ 2 & = & 2 ✓ & 4 & = & 4 ✓ \end{matrix}$ The solution is (3, −1).

Do you remember how to graph a linear equation with just one variable? It will be either a vertical or a horizontal line.

Solve the system by graphing: ${\begin{matrix} y = 6 \\ 2 x + 3 y = 12 \end{matrix} .$

Solution


We know the first equation represents a horizontal line whose y-intercept is 6.
The second equation is most conveniently graphed using intercepts.
To find the intercepts, let x = 0 and then y = 0.
Graph the lines.
Determine the point of intersection.	The lines intersect at (−3, 6).
Check the solution to both equations.	$\begin{matrix} y & = & 6 & 2 x + 3 y & = & 12 \\ 6 & \overset{?}{=} & 6 ✓ & 2 (−3) + 3 (6) & \overset{?}{=} & 12 \\ −6 + 18 & \overset{?}{=} & 12 \\ 12 & = & 12 ✓ \end{matrix}$
	The solution is (−3, 6).

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Solve the system by graphing: ${\begin{matrix} y = \frac{1}{2} x - 3 \\ x - 2 y = 4 \end{matrix} .$

Solution


To graph the first equation, we will use its slope and y-intercept.


To graph the second equation, we will use the intercepts.

Graph the lines.
Determine the point of intersection.	The lines are parallel.
	Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system.

Solve the system by graphing: ${\begin{matrix} y = 2 x - 3 \\ −6 x + 3 y = - 9 \end{matrix} .$

Solution


Find the slope and y-intercept of the first equation.
Find the intercepts of the second equation.

Graph the lines.
Determine the point of intersection.	The lines are the same!
	Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true.
	There are infinitely many solutions to this system.

If you write the second equation in Example 8 in slope-intercept form, you may recognize that the equations have the same slope and same y-intercept.

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident. Coincident lines have the same slope and same y-intercept.

Determine the Number of Solutions of a Linear System

There will be times when we will want to know how many solutions there will be to a system of linear equations, but we might not actually have to find the solution. It will be helpful to determine this without graphing.

We have seen that two lines in the same plane must either intersect or are parallel. The systems of equations in Example 2 through Example 6 all had two intersecting lines. Each system had one solution.

A system with parallel lines, like Example 7, has no solution. What happened in Example 8? The equations have coincident lines, and so the system had infinitely many solutions.

We’ll organize these results in Figure 2 below:

This table has two columns and four rows. The first row labels each column “Graph” and “Number of solutions.” Under “Graph” are “2 intersecting lines,” “Parallel lines,” and “Same line.” Under “Number of solutions” are “1,” “None,” and “Infinitely many.”

Parallel lines have the same slope but different y-intercepts. So, if we write both equations in a system of linear equations in slope–intercept form, we can see how many solutions there will be without graphing! Look at the system we solved in Example 7.

\begin{matrix} {\begin{matrix} y & = & \frac{1}{2} x - 3 \\ x - 2 y & = & 4 \end{matrix} \\ The first line is in slope–intercept form. & If we solve the second equation for y, we get \\ y = \frac{1}{2} x - 3 & \begin{matrix} x - 2 y & = & 4 \\ −2 y & = & − x + 4 \\ y & = & \frac{1}{2} x - 2 \end{matrix} \\ m = \frac{1}{2}, b = −3 & m = \frac{1}{2}, b = −2 \end{matrix}

The two lines have the same slope but different y-intercepts. They are parallel lines.

Figure 3 shows how to determine the number of solutions of a linear system by looking at the slopes and intercepts.

This table is entitled “Number of Solutions of a Linear System of Equations.” There are four columns. The columns are labeled, “Slopes,” “Intercepts,” “Type of Lines,” “Number of Solutions.” Under “Slopes” are “Different,” “Same,” and “Same.” Under “Intercepts,” the first cell is blank, then the words “Different” and “Same” appear. Under “Types of Lines” are the words, “Intersecting,” “Parallel,” and “Coincident.” Under “Number of Solutions” are “1 point,” “No Solution,” and “Infinitely many solutions.”

Let’s take one more look at our equations in Example 7 that gave us parallel lines.

{\begin{matrix} y = \frac{1}{2} x - 3 \\ x - 2 y = 4 \end{matrix}

When both lines were in slope-intercept form we had:

y = \frac{1}{2} x - 3 y = \frac{1}{2} x - 2

Do you recognize that it is impossible to have a single ordered pair $(x, y)$ that is a solution to both of those equations?

We call a system of equations like this an inconsistent system. It has no solution.

A system of equations that has at least one solution is called a consistent system.

We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent equations, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See Figure 4 and Figure 5.

This figure shows three x y coordinate planes in a horizontal row. The first shows two lines intersecting. The second shows two parallel lines. The third shows two coincident lines.

This table has four columns and four rows. The columns are labeled, “Lines,” “Intersecting,” “Parallel,” and “Coincident.” In the first row under the labeled column “lines” it reads “Number of solutions.” Reading across, it tell us that an intersecting line contains 1 point, a parallel line provides no solution, and a coincident line has infinitely many solutions. A consistent/inconsistent line has consistent lines if they are intersecting, inconsistent lines if they are parallel and consistent if the lines are coincident. Finally, dependent and independent lines are considered independent if the lines intersect, they are also independent if the lines are parallel, and they are dependent if the lines are coincident.

Without graphing, determine the number of solutions and then classify the system of equations: ${\begin{matrix} y = 3 x - 1 \\ 6 x - 2 y = 12 \end{matrix} .$

Solution

This table demonstrates the step-by-step process of comparing two linear equations by converting them to slope-intercept form to determine their relationship.
We will compare the slopes and intercepts of the two lines.	${\begin{matrix} y = 3 x - 1 \\ 6 x - 2 y = 12 \end{matrix} .$
The first equation is already in slope-intercept form.	$y = 3 x - 1$
Write the second equation in slope-intercept form.	$\begin{matrix} 6 x - 2 y & = & 12 \\ −2 y & = & −6 x + 12 \\ \frac{−2 y}{−2} & = & \frac{−6 x + 12}{−2} \\ y & = & 3 x - 6 \end{matrix}$
Find the slope and intercept of each line.	$\begin{matrix} y & = & 3 x - 1 & y & = & 3 x - 6 \\ m & = & 3 & m & = & 3 \\ b & = & −1 & b & = & −6 \end{matrix}$
	Since the slopes are the same and $y$ -intercepts are different, the lines are parallel.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

Without graphing, determine the number of solutions and then classify the system of equations: ${\begin{matrix} 2 x + y = - 3 \\ x - 5 y = 5 \end{matrix} .$

Solution

This table demonstrates a step-by-step solution for analyzing a system of two linear equations by converting them to slope-intercept form to compare their slopes and intercepts.
We will compare the slope and intercepts of the two lines.	${\begin{matrix} 2 x + y = - 3 \\ x - 5 y = 5 \end{matrix}$
Write both equations in slope-intercept form.	$\begin{matrix} 2 x + y & = & −3 \\ y & = & −2 x −3 \end{matrix}$	$\begin{matrix} x - 5 y = 5 & = & 5 \\ −5 y & = & − x + 5 \\ \frac{−5 y}{−5} & = & \frac{− x + 5}{−5} \\ y & = & \frac{1}{5} x - 1 \end{matrix}$
Find the slope and intercept of each line.	$\begin{matrix} y & = & −2 x - 3 \\ m & = & −2 \\ b & = & −3 \end{matrix}$	$\begin{matrix} y & = & \frac{1}{5} x - 1 \\ m & = & \frac{1}{5} \\ b & = & −1 \end{matrix}$
	Since the slopes are different, the lines intersect.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Without graphing, determine the number of solutions and then classify the system of equations. ${\begin{matrix} 3 x - 2 y = 4 \\ y = \frac{3}{2} x - 2 \end{matrix}$

Solution

This table demonstrates analyzing a system of two linear equations by converting them to slope-intercept form and comparing slopes/intercepts to determine line relationships.
We will compare the slopes and intercepts of the two lines.	${\begin{matrix} 3 x - 2 y = 4 \\ y = \frac{3}{2} x - 2 \end{matrix}$
Write the first equation in slope-intercept form.	$\begin{matrix} 3 x - 2 y & = & 4 \\ −2 y & = & −3 x + 4 \\ \frac{−2 y}{−2} & = & \frac{−3 x + 4}{−2} \\ y & = & \frac{3}{2} x - 2 \end{matrix}$
The second equation is already in slope-intercept form.	$\begin{matrix} y = \frac{3}{2} x - 2 \end{matrix}$
	Since the slopes are the same, they have the same slope and same $y$ -intercept and so the lines are coincident.

A system of equations whose graphs are coincident lines has infinitely many solutions and is consistent and dependent.

Solve Applications of Systems of Equations by Graphing

We will use the same problem solving strategy we used in Math Models to set up and solve applications of systems of linear equations. We’ll modify the strategy slightly here to make it appropriate for systems of equations.

Step 5 is where we will use the method introduced in this section. We will graph the equations and find the solution.

Sondra is making 10 quarts of punch from fruit juice and club soda. The number of quarts of fruit juice is 4 times the number of quarts of club soda. How many quarts of fruit juice and how many quarts of club soda does Sondra need?

Solution

Step 1. Read the problem.

Step 2. Identify what we are looking for.

We are looking for the number of quarts of fruit juice and the number of quarts of club soda that Sondra will need.

Step 3. Name what we are looking for. Choose variables to represent those quantities.

Let $f =$ number of quarts of fruit juice.
$c =$ number of quarts of club soda

Step 4. Translate into a system of equations.

We now have the system. ${\begin{matrix} f + c = 10 \\ f = 4 c \end{matrix}$

Step 5. Solve the system of equations using good algebra techniques.
This figure shows two equations and their graph. The first equation is f = 4c where b = 4 and b = 0. The second equation is f + c = 10. f = negative c +10 where b = negative 1 and b = 10. The x y coordinate plane shows a graph of these two lines which intersect at (2, 8).

The point of intersection (2, 8) is the solution. This means Sondra needs 2 quarts of club soda and 8 quarts of fruit juice.

Step 6. Check the answer in the problem and make sure it makes sense.

Does this make sense in the problem?

Yes, the number of quarts of fruit juice, 8 is 4 times the number of quarts of club soda, 2.

Yes, 10 quarts of punch is 8 quarts of fruit juice plus 2 quarts of club soda.

Step 7. Answer the question with a complete sentence.

Sondra needs 8 quarts of fruit juice and 2 quarts of soda.

Key Concepts

To solve a system of linear equations by graphing
1. Graph the first equation.
2. Graph the second equation on the same rectangular coordinate system.
3. Determine whether the lines intersect, are parallel, or are the same line.
4. Identify the solution to the system.
  If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system.
  If the lines are parallel, the system has no solution.
  If the lines are the same, the system has an infinite number of solutions.
5. Check the solution in both equations.
Determine the number of solutions from the graph of a linear system
Determine the number of solutions of a linear system by looking at the slopes and intercepts
Determine the number of solutions and how to classify a system of equations
Problem Solving Strategy for Systems of Linear Equations
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose variables to represent those quantities.
4. Translate into a system of equations.
5. Solve the system of equations using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Practice Makes Perfect

Determine Whether an Ordered Pair is a Solution of a System of Equations. In the following exercises, determine if the following points are solutions to the given system of equations.

${\begin{matrix} 2 x - 6 y = 0 \\ 3 x - 4 y = 5 \end{matrix}$

ⓐ $(3, 1)$ ⓑ $(−3, 4)$

Solution

ⓐ yes ⓑ no

${\begin{matrix} 7 x - 4 y = −1 \\ −3 x - 2 y = 1 \end{matrix}$

ⓐ $(1, 2)$ ⓑ $(1, −2)$

${\begin{matrix} 2 x + y = 5 \\ x + y = 1 \end{matrix}$

ⓐ $(4, −3)$ ⓑ $(2, 0)$

Solution

ⓐ yes ⓑ no

${\begin{matrix} −3 x + y = 8 \\ - x + 2 y = −9 \end{matrix}$

ⓐ $(−5, −7)$ ⓑ $(−5, 7)$

${\begin{matrix} x + y = 2 \\ y = \frac{3}{4} x \end{matrix}$

ⓐ $(\frac{8}{7}, \frac{6}{7})$ ⓑ $(1, \frac{3}{4})$

Solution

ⓐ yes ⓑ no

${\begin{matrix} x + y = 1 \\ y = \frac{2}{5} x \end{matrix}$

ⓐ $(\frac{5}{7}, \frac{2}{7})$ ⓑ $(5, 2)$

${\begin{matrix} x + 5 y = 10 \\ y = \frac{3}{5} x + 1 \end{matrix}$

ⓐ $(−10, 4)$ ⓑ $(\frac{5}{4}, \frac{7}{4})$

Solution

ⓐ no ⓑ yes

${\begin{matrix} x + 3 y = 9 \\ y = \frac{2}{3} x - 2 \end{matrix}$

ⓐ $(−6, 5)$ ⓑ $(5, \frac{4}{3})$

Solve a System of Linear Equations by Graphing In the following exercises, solve the following systems of equations by graphing.

${\begin{matrix} 3 x + y = −3 \\ 2 x + 3 y = 5 \end{matrix}$

Solution

$(−2, 3)$

${\begin{matrix} − x + y = 2 \\ 2 x + y = −4 \end{matrix}$

${\begin{matrix} −3 x + y = −1 \\ 2 x + y = 4 \end{matrix}$

Solution

$(1, 2)$

${\begin{matrix} −2 x + 3 y = −3 \\ x + y = 4 \end{matrix}$

${\begin{matrix} y = x + 2 \\ y = −2 x + 2 \end{matrix}$

Solution

$(0, 2)$

${\begin{matrix} y = x - 2 \\ y = −3 x + 2 \end{matrix}$

${\begin{matrix} y = \frac{3}{2} x + 1 \\ y = - \frac{1}{2} x + 5 \end{matrix}$

Solution

$(2, 4)$

${\begin{matrix} y = \frac{2}{3} x - 2 \\ y = - \frac{1}{3} x - 5 \end{matrix}$

${\begin{matrix} - x + y = −3 \\ 4 x + 4 y = 4 \end{matrix}$

Solution

$(2, −1)$

${\begin{matrix} x - y = 3 \\ 2 x - y = 4 \end{matrix}$

${\begin{matrix} −3 x + y = −1 \\ 2 x + y = 4 \end{matrix}$

Solution

$(1, 2)$

${\begin{matrix} −3 x + y = −2 \\ 4 x - 2 y = 6 \end{matrix}$

${\begin{matrix} x + y = 5 \\ 2 x - y = 4 \end{matrix}$

Solution

$(3, 2)$

${\begin{matrix} x - y = 2 \\ 2 x - y = 6 \end{matrix}$

${\begin{matrix} x + y = 2 \\ x - y = 0 \end{matrix}$

Solution

$(1, 1)$

${\begin{matrix} x + y = 6 \\ x - y = −8 \end{matrix}$

${\begin{matrix} x + y = −5 \\ x - y = 3 \end{matrix}$

Solution

$(−1, −4)$

${\begin{matrix} x + y = 4 \\ x - y = 0 \end{matrix}$

${\begin{matrix} x + y = −4 \\ - x + 2 y = −2 \end{matrix}$

Solution

$(- 2, - 2)$

${\begin{matrix} - x + 3 y = 3 \\ x + 3 y = 3 \end{matrix}$

${\begin{matrix} −2 x + 3 y = 3 \\ x + 3 y = 12 \end{matrix}$

Solution

$(3, 3)$

${\begin{matrix} 2 x - y = 4 \\ 2 x + 3 y = 12 \end{matrix}$

${\begin{matrix} 2 x + 3 y = 6 \\ y = −2 \end{matrix}$

Solution

$(6, −2)$

${\begin{matrix} −2 x + y = 2 \\ y = 4 \end{matrix}$

${\begin{matrix} x - 3 y = −3 \\ y = 2 \end{matrix}$

Solution

$(3, 2)$

${\begin{matrix} 2 x - 2 y = 8 \\ y = −3 \end{matrix}$

${\begin{matrix} 2 x - y = −1 \\ x = 1 \end{matrix}$

Solution

$(1, 3)$

${\begin{matrix} x + 2 y = 2 \\ x = −2 \end{matrix}$

${\begin{matrix} x - 3 y = −6 \\ x = −3 \end{matrix}$

Solution

$(−3, 1)$

${\begin{matrix} x + y = 4 \\ x = 1 \end{matrix}$

${\begin{matrix} 4 x - 3 y = 8 \\ 8 x - 6 y = 14 \end{matrix}$

Solution

no solution

${\begin{matrix} x + 3 y = 4 \\ −2 x - 6 y = 3 \end{matrix}$

${\begin{matrix} −2 x + 4 y = 4 \\ y = \frac{1}{2} x \end{matrix}$

Solution

no solution

${\begin{matrix} 3 x + 5 y = 10 \\ y = - \frac{3}{5} x + 1 \end{matrix}$

${\begin{matrix} x = −3 y + 4 \\ 2 x + 6 y = 8 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} 4 x = 3 y + 7 \\ 8 x - 6 y = 14 \end{matrix}$

${\begin{matrix} 2 x + y = 6 \\ −8 x - 4 y = −24 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} 5 x + 2 y = 7 \\ −10 x - 4 y = −14 \end{matrix}$

${\begin{matrix} x + 3 y = −6 \\ 4 y = - \frac{4}{3} x - 8 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} − x + 2 y = −6 \\ y = - \frac{1}{2} x - 1 \end{matrix}$

${\begin{matrix} −3 x + 2 y = −2 \\ y = − x + 4 \end{matrix}$

Solution

$(2, 2)$

${\begin{matrix} - x + 2 y = −2 \\ y = − x - 1 \end{matrix}$

Determine the Number of Solutions of a Linear System Without graphing the following systems of equations, determine the number of solutions and then classify the system of equations.

${\begin{matrix} y = \frac{2}{3} x + 1 \\ −2 x + 3 y = 5 \end{matrix}$

Solution

0 solutions

${\begin{matrix} y = \frac{1}{3} x + 2 \\ x - 3 y = 9 \end{matrix}$

${\begin{matrix} y = −2 x + 1 \\ 4 x + 2 y = 8 \end{matrix}$

Solution

0 solutions

${\begin{matrix} y = 3 x + 4 \\ 9 x - 3 y = 18 \end{matrix}$

${\begin{matrix} y = \frac{2}{3} x + 1 \\ 2 x - 3 y = 7 \end{matrix}$

Solution

no solutions, inconsistent, independent

${\begin{matrix} 3 x + 4 y = 12 \\ y = −3 x - 1 \end{matrix}$

${\begin{matrix} 4 x + 2 y = 10 \\ 4 x - 2 y = −6 \end{matrix}$

Solution

consistent, 1 solution

${\begin{matrix} 5 x + 3 y = 4 \\ 2 x - 3 y = 5 \end{matrix}$

${\begin{matrix} y = - \frac{1}{2} x + 5 \\ x + 2 y = 10 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} y = x + 1 \\ − x + y = 1 \end{matrix}$

${\begin{matrix} y = 2 x + 3 \\ 2 x - y = −3 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} 5 x - 2 y = 10 \\ y = \frac{5}{2} x - 5 \end{matrix}$

Solve Applications of Systems of Equations by Graphing In the following exercises, solve.

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Solution

Molly needs 16 ounces of strawberry juice and 48 ounces of water.

Jamal is making a snack mix that contains only pretzels and nuts. For every ounce of nuts, he will use 2 ounces of pretzels. How many ounces of pretzels and how many ounces of nuts does he need to make 45 ounces of snack mix?

Enrique is making a party mix that contains raisins and nuts. For each ounce of nuts, he uses twice the amount of raisins. How many ounces of nuts and how many ounces of raisins does he need to make 24 ounces of party mix?

Solution

Enrique needs 8 ounces of nuts and 16 ounces of water.

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

Everyday Math

Leo is planning his spring flower garden. He wants to plant tulip and daffodil bulbs. He will plant 6 times as many daffodil bulbs as tulip bulbs. If he wants to plant 350 bulbs, how many tulip bulbs and how many daffodil bulbs should he plant?

Solution

Leo should plant 50 tulips and 300 daffodils.

A marketing company surveys 1,200 people. They surveyed twice as many females as males. How many males and females did they survey?

Writing Exercises

In a system of linear equations, the two equations have the same slope. Describe the possible solutions to the system.

Solution

Given that it is only known that the slopes of both linear equations are the same, there are either no solutions (the graphs of the equations are parallel) or infinitely many.

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Self Check

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

This figure shows a table with four rows and four columns. The columns are labeled, “I can…,” “Confidently.” “With some help.” and “No - I don’t get it.” The only column with filled in cells below it is labeled “I can…” It reads, “determine whether an ordered pair is a solution of a system of equations.” “solve a system of linear equations by graphing.” “determine the number of solutions of a linear system.” and “solve applications of systems of equations by graphing.”

If most of your checks were:

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

coincident lines

Coincident lines are lines that have the same slope and same y-intercept.

consistent system

A consistent system of equations is a system of equations with at least one solution.

dependent equations

Two equations are dependent if all the solutions of one equation are also solutions of the other equation.

inconsistent system

An inconsistent system of equations is a system of equations with no solution.

independent equations

Two equations are independent if they have different solutions.

solutions of a system of equations

Solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x, y).

system of linear equations

When two or more linear equations are grouped together, they form a system of linear equations.

Solving Systems of Equations by Substitution

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

{\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix}

We will first solve one of the equations for either x or y. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Example 1.

How to Solve a System of Equations by Substitution

Solve the system by substitution. ${\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix}$

Solution

This figure has three columns and six rows. The first row says, “Step 1. Solve one of the equations for either variable.” To the right of this, the middl row reads, “We’ll solve the first equation for y.” The third column shows the two equations: 2x + y = 7 and x – 2y = 6. It shows that 2x + y = 7 becomes y = 7 – 2x.

The second row reads, “Step 2. Substitute the expression from Step 1 into the other equation.” Then, “We replace y in the second equation with the expression 7 – 2x.” It then shows the x – 2y = 6 becomes x – 2(7 – 2x) = 6.

The third row says, “Step 3: Solve the resulting equation.” Then “Now we have an equation with just 1 variable. WE know how to solve this!” It then shows that x – 2(7 – 2x) = 6 becomes x – 14 + 4x = 6 which becomes 5x = 20. Thus x = 4.

The fourth row says, “Step 4. Substitute the solution in Step 3 into one of the original quaitons to find the other variable.” Then, “We’ll use the first equation and replace x with 4.” Then it shows that 2x + y = 7 becomes 2(4) + y = 7. This becomes 8 + y = 7, and thus y = −1.

The fifth row reads, “Step 5. Write the solution as an ordered pair.” Then “The ordered air is (x, y).” Then (4, −1).

The sixth row reads, “Step 6. Check that the order pair is a solution to both original equations.” Then, “Substitute (4, −1) into both equations and make sure they are both true.” It then shows that 2x + y = 7 becomxe 2(4) + −1 = 7, and thus 7 = 7. It also shows that x – 2y = 6 becomes 4 – 2(−1) = 6, and thus 6−6. It also states, “Both equations are ture. (4, −1) is the solution to the system.”

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Example 2.

Solve the system by substitution.

${\begin{matrix} x + y = −1 \\ y = x + 5 \end{matrix}$

Solution

The second equation is already solved for y. We will substitute the expression in place of y in the first equation.


The second equation is already solved for y. We will substitute into the first equation.
Replace the y with x + 5.
Solve the resulting equation for x.


Substitute x = −3 into y = x + 5 to find y.

The ordered pair is (−3, 2).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} x + y & = & - 1 \\ - 3 + 2 & \overset{?}{=} & - 1 \\ - 1 & = & - 1 ✓ \end{matrix} & \begin{matrix} y & = & x + 5 \\ 2 & \overset{?}{=} & - 3 + 5 \\ 2 & = & 2 ✓ \end{matrix} \end{matrix}$
	The solution is (−3, 2).

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y.

Solve the system by substitution. ${\begin{matrix} 3 x + y = 5 \\ 2 x + 4 y = −10 \end{matrix}$

Solution

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Solve for y. Substitute into the other equation.
Replace the y with −3x + 5.
Solve the resulting equation for x.

Substitute x = 3 into 3x + y = 5 to find y.

The ordered pair is (3, −4).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} 3 x + y & = & 5 \\ 3 \cdot 3 + (- 4) & \overset{?}{=} & 5 \\ 9 - 4 & \overset{?}{=} & 5 \\ 5 & = & 5 ✓ \end{matrix} & \begin{matrix} 2 x + 4 y & = & - 10 \\ 2 \cdot 3 + 4 (- 4) & = & - 10 \\ 6 - 16 & \overset{?}{=} & - 10 \\ - 10 & = & - 10 ✓ \end{matrix} \end{matrix}$
	The solution is (3, −4).

In Example 3 it was easiest to solve for y in the first equation because it had a coefficient of 1. In Example 4 it will be easier to solve for x.

Solve the system by substitution. ${\begin{matrix} x - 2 y = −2 \\ 3 x + 2 y = 34 \end{matrix}$

Solution

We will solve the first equation for $x$ and then substitute the expression into the second equation.


Solve for x. Substitute into the other equation.
Replace the x with 2y − 2.
Solve the resulting equation for y.
Substitute y = 5 into x − 2y = −2 to find x.
The ordered pair is (8, 5).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} x - 2 y & = & - 2 \\ 8 - 2 \cdot 5 & \overset{?}{=} & - 2 \\ 8 - 10 & \overset{?}{=} & - 2 \\ - 2 & = & - 2 ✓ \end{matrix} & \begin{matrix} 3 x + 2 y & = & 34 \\ 3 \cdot 8 + 2 \cdot 5 & \overset{?}{=} & 34 \\ 24 + 10 & \overset{?}{=} & 34 \\ 34 & = & 34 ✓ \end{matrix} \end{matrix}$
	The solution is (8, 5).

When both equations are already solved for the same variable, it is easy to substitute!

Solve the system by substitution. ${\begin{matrix} y = −2 x + 5 \\ y = \frac{1}{2} x \end{matrix}$

Solution

Since both equations are solved for y, we can substitute one into the other.

Substitute $\frac{1}{2} x$ for y in the first equation.
Replace the y with $\frac{1}{2} x .$
Solve the resulting equation. Start by clearing the fraction.
Solve for x.

Substitute x = 2 into y = $\frac{1}{2} x$ to find y.
The ordered pair is (2,1).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} y & = & \frac{1}{2} x \\ 1 & \overset{?}{=} & \frac{1}{2} \cdot 2 \\ 1 & = & 1 ✓ \end{matrix} & \begin{matrix} y & = & - 2 x + 5 \\ 1 & \overset{?}{=} & - 2 \cdot 2 + 5 \\ 1 & = & - 4 + 5 \\ 1 & = & 1 ✓ \end{matrix} \end{matrix}$
	The solution is (2,1).

Be very careful with the signs in the next example.

Solve the system by substitution. ${\begin{matrix} 4 x + 2 y = 4 \\ 6 x - y = 8 \end{matrix}$

Solution

We need to solve one equation for one variable. We will solve the first equation for y.


Solve the first equation for y.
Substitute −2x + 2 for y in the second equation.
Replace the y with −2x + 2.
Solve the equation for x.

Substitute $x = \frac{5}{4}$ into 4x + 2y = 4 to find y.	$A mathematical equation is displayed against a white background: 4(5/4) + 2y = 4. The numerator and denominator of the fraction, 5 and 4 respectively, are highlighted in red.$
The ordered pair is $(\frac{5}{4}, - \frac{1}{2}) .$
Check the ordered pair in both equations. $\begin{matrix} \begin{matrix} 4 x + 2 y & = & 4 \\ 4 (\frac{5}{4}) + 2 (- \frac{1}{2}) & \overset{?}{=} & 4 \\ 5 - 1 & \overset{?}{=} & 4 \\ 4 & = & 4 ✓ \end{matrix} & \begin{matrix} 6 x - y & = & 8 \\ 6 (\frac{5}{4}) - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{15}{4} - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{16}{2} & \overset{?}{=} & 8 \\ 8 & = & 8 ✓ \end{matrix} \end{matrix}$
	The solution is $(\frac{5}{4}, - \frac{1}{2}) .$

In Example 7, it will take a little more work to solve one equation for x or y.

Solve the system by substitution. ${\begin{matrix} 4 x - 3 y = 6 \\ 15 y - 20 x = −30 \end{matrix}$

Solution

We need to solve one equation for one variable. We will solve the first equation for x.


Solve the first equation for x.
Substitute $\frac{3}{4} y + \frac{3}{2}$ for x in the second equation.
Replace the x with $\frac{3}{4} y + \frac{3}{2} .$	$A mathematical equation is displayed: 15y - 20(3/4y + 3/2) = -30. The fractions 3/4 and 3/2 are highlighted in red.$
Solve for y.

Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.

Look back at the equations in Example 7. Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Solve the system by substitution. ${\begin{matrix} 5 x - 2 y = −10 \\ y = \frac{5}{2} x \end{matrix}$

Solution

The second equation is already solved for y, so we can substitute for y in the first equation.

Substitute x for y in the first equation.
Replace the y with $\frac{5}{2} x .$	$A mathematical equation is displayed, reading '5x minus 2 multiplied by (5 over 2 multiplied by x) equals -10'. The fraction '5 over 2' and the 'x' immediately following it are colored red.$
Solve for x.

Since 0 = −10 is a false statement the equations are inconsistent. The graphs of the two equation would be parallel lines. The system has no solutions.

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

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

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for two numbers.
Step 3. Name what we are looking for.	Let $n =$ the first number Let $m =$ the second number
Step 4. Translate into a system of equations.	The sum of two numbers is zero.

	One number is nine less than the other.

The system is:
Step 5. Solve the system of equations. We will use substitution since the second equation is solved for n.
Substitute m − 9 for n in the first equation.
Solve for m.


Substitute $m = \frac{9}{2}$ into the second equation and then solve for n.	$A white background displays two mathematical expressions. The top expression reads 'm = 9/2,' with the fraction '9/2' encircled in red. A red arrow points from this circled fraction to the variable 'm' in the lower expression, 'n = m - 9.' This visually indicates the substitution of the value of m into the second equation.$
	$A mathematical equation is displayed, showing 'm' equals the fraction nine over two, minus nine. The fraction '9/2' is highlighted in red, while 'm =' and '- 9' are in black.$

	$A mathematical expression shows 'n' equals negative nine-halves, written as 'n = -9/2' on a white background. The variable 'n' is isolated, indicating its value as a negative fraction or decimal.$
Step 6. Check the answer in the problem.	Do these numbers make sense in the problem? We will leave this to you!
Step 7. Answer the question.	The numbers are $\frac{9}{2}$ and $- \frac{9}{2} .$

In the Example 10, we’ll use the formula for the perimeter of a rectangle, P = 2L + 2W.

The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and the width.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the length and width.
Step 3. Name what we are looking for.	Let $L =$ the length $W =$ the width
Step 4. Translate into a system of equations.	The perimeter of a rectangle is 88.
	2L + 2W = P
	The length is five more than twice the width.

The system is:
Step 5. Solve the system of equations. We will use substitution since the second equation is solved for L. Substitute 2W + 5 for L in the first equation.
Solve for W.



Substitute W = 13 into the second equation and then solve for L.


Step 6. Check the answer in the problem.	Does a rectangle with length 31 and width 13 have perimeter 88? Yes.
Step 7. Answer the equation.	The length is 31 and the width is 13.

For Example 11 we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

Solution

We will draw and label a figure.

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the measures of the angles.
Step 3. Name what we are looking for.	Let $a =$ the measure of the 1^st angle $b =$ the measure of the 2^nd angle
Step 4. Translate into a system of equations.	The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle.

	The sum of the measures of the angles of a triangle is 180.

The system is:
Step 5. Solve the system of equations. We will use substitution since the first equation is solved for a.
Substitute 3b + 10 for a in the second equation.
Solve for b.


Substitute b = 20 into the first equation and then solve for a.
Step 6. Check the answer in the problem.	We will leave this to you!
Step 7. Answer the question.	The measures of the small angles are 20 and 70.

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the number of training sessions that would make the pay equal.
Step 3. Name what we are looking for.	Let $s =$ Heather’s salary. $n =$ the number of training sessions
Step 4. Translate into a system of equations.	Option A would pay her $25,000 plus $15 for each training session.

	Option B would pay her $10,000 + $40 for each training session

The system is:
Step 5. Solve the system of equations. We will use substitution.
Substitute 25,000 + 15n for s in the second equation.
Solve for n.


Step 6. Check the answer.	Are 600 training sessions a year reasonable? Are the two options equal when n = 600?
Step 7. Answer the question.	The salary options would be equal for 600 training sessions.

Key Concepts

Solve a system of equations by substitution
1. Solve one of the equations for either variable.
2. Substitute the expression from Step 1 into the other equation.
3. Solve the resulting equation.
4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
5. Write the solution as an ordered pair.
6. Check that the ordered pair is a solution to both original equations.

Practice Makes Perfect

Solve a System of Equations by Substitution

In the following exercises, solve the systems of equations by substitution.

${\begin{matrix} 2 x + y = −4 \\ 3 x - 2 y = −6 \end{matrix}$

Solution

$(−2, 0)$

${\begin{matrix} 2 x + y = −2 \\ 3 x - y = 7 \end{matrix}$

${\begin{matrix} x - 2 y = −5 \\ 2 x - 3 y = −4 \end{matrix}$

Solution

$(7, 6)$

${\begin{matrix} x - 3 y = −9 \\ 2 x + 5 y = 4 \end{matrix}$

${\begin{matrix} 5 x - 2 y = −6 \\ y = 3 x + 3 \end{matrix}$

Solution

$(0, 3)$

${\begin{matrix} −2 x + 2 y = 6 \\ y = −3 x + 1 \end{matrix}$

${\begin{matrix} 2 x + 3 y = 3 \\ y = − x + 3 \end{matrix}$

Solution

$(6, −3)$

${\begin{matrix} 2 x + 5 y = −14 \\ y = −2 x + 2 \end{matrix}$

${\begin{matrix} 2 x + 5 y = 1 \\ y = \frac{1}{3} x - 2 \end{matrix}$

Solution

$(3, −1)$

${\begin{matrix} 3 x + 4 y = 1 \\ y = - \frac{2}{5} x + 2 \end{matrix}$

${\begin{matrix} 3 x - 2 y = 6 \\ y = \frac{2}{3} x + 2 \end{matrix}$

Solution

$(6, 6)$

${\begin{matrix} −3 x - 5 y = 3 \\ y = \frac{1}{2} x - 5 \end{matrix}$

${\begin{matrix} 2 x + y = 10 \\ - x + y = −5 \end{matrix}$

Solution

$(5, 0)$

${\begin{matrix} −2 x + y = 10 \\ - x + 2 y = 16 \end{matrix}$

${\begin{matrix} 3 x + y = 1 \\ −4 x + y = 15 \end{matrix}$

Solution

$(−2, 7)$

${\begin{matrix} x + y = 0 \\ 2 x + 3 y = −4 \end{matrix}$

${\begin{matrix} x + 3 y = 1 \\ 3 x + 5 y = −5 \end{matrix}$

Solution

$(−5, 2)$

${\begin{matrix} x + 2 y = −1 \\ 2 x + 3 y = 1 \end{matrix}$

${\begin{matrix} 2 x + y = 5 \\ x - 2 y = −15 \end{matrix}$

Solution

$(−1, 7)$

${\begin{matrix} 4 x + y = 10 \\ x - 2 y = −20 \end{matrix}$

${\begin{matrix} y = −2 x - 1 \\ y = - \frac{1}{3} x + 4 \end{matrix}$

Solution

$(−3, 5)$

${\begin{matrix} y = x - 6 \\ y = - \frac{3}{2} x + 4 \end{matrix}$

${\begin{matrix} y = 2 x - 8 \\ y = \frac{3}{5} x + 6 \end{matrix}$

Solution

(10, 12)

${\begin{matrix} y = − x - 1 \\ y = x + 7 \end{matrix}$

${\begin{matrix} 4 x + 2 y = 8 \\ 8 x - y = 1 \end{matrix}$

Solution

$(\frac{1}{2}, 3)$

${\begin{matrix} - x - 12 y = −1 \\ 2 x - 8 y = −6 \end{matrix}$

${\begin{matrix} 15 x + 2 y = 6 \\ −5 x + 2 y = −4 \end{matrix}$

Solution

$(\frac{1}{2}, - \frac{3}{4})$

${\begin{matrix} 2 x - 15 y = 7 \\ 12 x + 2 y = −4 \end{matrix}$

${\begin{matrix} y = 3 x \\ 6 x - 2 y = 0 \end{matrix}$

Solution

Infinitely many solutions

${\begin{matrix} x = 2 y \\ 4 x - 8 y = 0 \end{matrix}$

${\begin{matrix} 2 x + 16 y = 8 \\ - x - 8 y = −4 \end{matrix}$

Solution

Infinitely many solutions

${\begin{matrix} 15 x + 4 y = 6 \\ −30 x - 8 y = −12 \end{matrix}$

${\begin{matrix} y = −4 x \\ 4 x + y = 1 \end{matrix}$

Solution

No solution

${\begin{matrix} y = - \frac{1}{4} x \\ x + 4 y = 8 \end{matrix}$

${\begin{matrix} y = \frac{7}{8} x + 4 \\ −7 x + 8 y = 6 \end{matrix}$

Solution

No solution

${\begin{matrix} y = - \frac{2}{3} x + 5 \\ 2 x + 3 y = 11 \end{matrix}$

Solve Applications of Systems of Equations by Substitution

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is 15. One number is 3 less than the other. Find the numbers.

Solution

The numbers are 6 and 9.

The sum of two numbers is 30. One number is 4 less than the other. Find the numbers.

The sum of two numbers is −26. One number is 12 less than the other. Find the numbers.

Solution

The numbers are −7 and −19.

The perimeter of a rectangle is 50. The length is 5 more than the width. Find the length and width.

The perimeter of a rectangle is 60. The length is 10 more than the width. Find the length and width.

Solution

The length is 20 and the width is 10.

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width.

The perimeter of a rectangle is 84. The length is 10 more than three times the width. Find the length and width.

Solution

The length is 34 and the width is 8.

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

Solution

The measures are 16° and 74°.

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Solution

The measures are 45° and 45°.

Maxim has been offered positions by two car dealers. The first company pays a salary of $10,000 plus a commission of $1,000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $ 14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Solution

80 cable packages would need to be sold.

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Solution

Mitchell would need to sell 120 stoves.

Everyday Math

When Gloria spent 15 minutes on the elliptical trainer and then did circuit training for 30 minutes, her fitness app says she burned 435 calories. When she spent 30 minutes on the elliptical trainer and 40 minutes circuit training she burned 690 calories. Solve the system ${\begin{matrix} 15 e + 30 c = 435 \\ 30 e + 40 c = 690 \end{matrix}$ for $e$ , the number of calories she burns for each minute on the elliptical trainer, and $c$ , the number of calories she burns for each minute of circuit training.

Stephanie left Riverside, California, driving her motorhome north on Interstate 15 towards Salt Lake City at a speed of 56 miles per hour. Half an hour later, Tina left Riverside in her car on the same route as Stephanie, driving 70 miles per hour. Solve the system ${\begin{matrix} 56 s = 70 t \\ s = t + \frac{1}{2} \end{matrix}$ .

ⓐ for $t$ to find out how long it will take Tina to catch up to Stephanie.
ⓑ what is the value of $s$ , the number of hours Stephanie will have driven before Tina catches up to her?

Solution

ⓐ $t = 2$ hours ⓑ $s = 2 \frac{1}{2}$ hours

Writing Exercises

Solve the system of equations
${\begin{matrix} x + y = 10 \\ x - y = 6 \end{matrix}$

ⓐ by graphing. ⓑ by substitution. ⓒ Which method do you prefer? Why?

Solve the system of equations
${\begin{matrix} 3 x + y = 12 \\ x = y - 8 \end{matrix}$ by substitution and explain all your steps in words.

Solution

Answers will vary.

Self Check

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

This figure shows a table with three rows and four columns. The columns are labeled, “I can…,” “Confidently.” “With some help.” and “No - I don’t get it.” The only column with filled in cells below it is labeled “I can…” It reads, “solve a system of equations by substitution.” “solve applications of systems of equations by substitution.”

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

Solve Systems of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

Solve a System of Equations by Elimination

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

\begin{matrix} if & a & = & b \\ and & c & = & d \\ then & a + c & = & b + d \end{matrix}

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

\begin{matrix} 3 x + y = 5 \\ \underset{_________}{2 x - y = 0} \\ 5 x = 5 \end{matrix}

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

{\begin{matrix} x + 4 y = 2 \\ 2 x + 5 y = −2 \end{matrix}

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

This figure shows two equations. The first is negative 2 times x plus 4y in parentheses equals negative 2 times 2. The second is 2x + 5y = negative 2. This figure shows two equations. The first is negative 2x minus 8y = negative 4. The second is 2x + 5y = -negative 2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

This figure shows two equations being added together. The first is negative 2x – 8y = −4 and 2x plus 5y = negative 2. The answer is negative 3y = negative 6.

We’ll do one more:

{\begin{matrix} 4 x - 3 y = 10 \\ 3 x + 5 y = −7 \end{matrix}

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12x and −12x.

This figure shows two equations. The first is 3 times 4x minus 3y in parentheses equals 3 times 10. The second is negative 4 times 3x plus 5y in parentheses equals negative 4 times negative 7.

This gives us these two new equations:

{\begin{matrix} 12 x - 9 y = 30 \\ −12 x - 20 y = 28 \end{matrix}

When we add these equations,

\begin{matrix} {\begin{matrix} 12 x - 9 y = 30 \\ \underset{_____________}{−12 x - 20 y = 28} \end{matrix} \\ - 29 y = 58 \end{matrix}

the x’s are eliminated and we just have −29y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

How to Solve a System of Equations by Elimination

Solve the system by elimination. ${\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix}$

Solution

$This figure has seven rows and three columns. The first row reads, “Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.” It also says, “Both equations are in standard form, A x + B y = C. There are no fractions.” It also gives the two equations as 2x + y = 7 and x – 2y = 6.$

The second row reads, “Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.” It also says, “We can eliminate the y’s by multiplying the first equation by 2. Multiply both sides of 2x + y = 7 by 2.” It also shows the steps with equations. Initially the equations are ex + y = 7 and x – 2y = 6. Then they become 2(2x + y) = 2 times 7 and x – 2y = 6. They then become 4x + 2y = 14 and x – 2y = 6.

The third row says, “Step 3: Add the equations resulting from step 2 to eliminate one variable.” It also says, “We add the x’s, y’s, and constants.” It then gives the equation as 5x = 20.

The fourth row says, “Step 4: Solve for the remaining variable.” It also says, “Solve for x.” It gives the equation as x = 4.

The fifth row says, “Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.” It also says, “Substitute x = 4 into the second equation, x – 2y = 6. Then solve for y.” It then gives the equations as x – 2y = 6 which becomes 4 – 2y = 6. This is then −2y = 2, and thus, y = −1.

The sixth row says, “Step 6: Write the solution as an order pair.” It also says, “Write it as (x, y).” It gives the ordered pair as (4, −1).

The seventh row says, “Step 7: Check that the ordered pair is a solution to both original equations.” It also says, “Substitute (4, −1) into 2x + y = 7 and x – 2y = 6. Do they make both equations true? Yes!” It then gives the equations. 2x + y = 7 becomes 2 times 4 + −1 = 7 which is 7 = 7. x – 2y = 6 becomes 4 – 2 times −1 = 6 which is 6 = 6. The row then says, “The solution is (4, −1).”

The steps are listed below for easy reference.

First we’ll do an example where we can eliminate one variable right away.

Solve the system by elimination. ${\begin{matrix} x + y = 10 \\ x - y = 12 \end{matrix}$

Solution


Both equations are in standard form.
The coefficients of y are already opposites.
Add the two equations to eliminate y. The resulting equation has only 1 variable, x.
Solve for x, the remaining variable. Substitute x = 11 into one of the original equations.

Solve for the other variable, y.
Write the solution as an ordered pair.	The ordered pair is (11, −1).
Check that the ordered pair is a solution to both original equations. $\begin{matrix} \begin{matrix} x + y & = & 10 \\ 11 + (- 1) & \overset{?}{=} & 10 \\ 10 & = & 10 ✓ \end{matrix} & \begin{matrix} x - y & = & 12 \\ 11 - (- 1) & \overset{?}{=} & 12 \\ 12 & = & 12 ✓ \end{matrix} \end{matrix}$
	The solution is (11, −1).

In Example 3, we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

Solve the system by elimination. ${\begin{matrix} 3 x - 2 y = −2 \\ 5 x - 6 y = 10 \end{matrix}$

Solution


Both equations are in standard form.
None of the coefficients are opposites.
We can make the coefficients of y opposites by multiplying the first equation by −3.
Simplify.
Add the two equations to eliminate y.
Solve for the remaining variable, x. Substitute x = −4 into one of the original equations.

Solve for y.
Write the solution as an ordered pair.	The ordered pair is (−4, −5).
Check that the ordered pair is a solution to both original equations. $\begin{matrix} \begin{matrix} 3 x - 2 y & = & - 2 \\ 3 (- 4) - 2 (- 5) & \overset{?}{=} & - 2 \\ - 12 + 10 & \overset{?}{=} & - 2 \\ - 2 y & = & - 2 ✓ \end{matrix} & \begin{matrix} 5 x - 6 y & = & 10 \\ 3 (- 4) - 6 (- 5) & \overset{?}{=} & 10 \\ - 20 + 30 & \overset{?}{=} & 10 \\ 10 & = & 10 ✓ \end{matrix} \end{matrix}$
	The solution is (−4, −5).

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Solve the system by elimination. ${\begin{matrix} 4 x - 3 y = 9 \\ 7 x + 2 y = −6 \end{matrix}$

Solution

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.


Both equations are in standard form. To get opposite coefficients of y, we will multiply the first equation by 2 and the second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x. Substitute x = 0 into one of the original equations.

Solve for y.

Write the solution as an ordered pair.	The ordered pair is (0, −3).
Check that the ordered pair is a solution to both original equations. $\begin{matrix} \begin{matrix} 4 x - 3 y & = & 9 \\ 4 (0) - 3 (- 3) & \overset{?}{=} & 9 \\ 9 & = & 9 ✓ \end{matrix} & \begin{matrix} 7 x + 2 y & = & - 6 \\ 7 (0) + 2 (- 3) & \overset{?}{=} & - 6 \\ - 6 & = & - 6 ✓ \end{matrix} \end{matrix}$
	The solution is (0, −3).

What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.

Solve the system by elimination. ${\begin{matrix} x + \frac{1}{2} y = 6 \\ \frac{3}{2} x + \frac{2}{3} y = \frac{17}{2} \end{matrix}$

Solution

In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.

	$A system of two linear equations with fractional coefficients is displayed, enclosed by a brace on the left side. The equations are x + 1/2y = 6 and 3/2x + 2/3y = 17/2.$
To clear the fractions, multiply each equation by its LCD.	$The image displays a system of two linear equations. In the first equation, both sides are multiplied by 2, and in the second equation, both sides are multiplied by 6. This operation is performed to eliminate the fractional coefficients, specifically the denominators of 2 in the first equation and 2 and 3 in the second equation, thereby simplifying the system for further algebraic manipulation.$
Simplify.
Now we are ready to eliminate one of the variables. Notice that both equations are in standard form.
We can eliminate y multiplying the top equation by −4.
Simplify and add. Substitute x = 3 into one of the original equations.
Solve for y.
	$A mathematical equation is displayed on a white background, reading '1/2y = 3'. The fraction 1/2 is vertically aligned, followed by the variable 'y', an equals sign, and the number '3'.$

Write the solution as an ordered pair.	The ordered pair is (3, 6).
Check that the ordered pair is a solution to both original equations. $\begin{matrix} \begin{matrix} x + \frac{1}{2} y & = & 6 \\ 3 + \frac{1}{2} (6) & \overset{?}{=} & 6 \\ 3 + 3 & \overset{?}{=} & 6 \\ 6 & = & 6 ✓ \end{matrix} & \begin{matrix} \frac{3}{2} x + \frac{2}{3} y & = & \frac{17}{2} \\ \frac{3}{2} (3) + \frac{2}{3} (6) & \overset{?}{=} & \frac{17}{2} \\ \frac{9}{2} + 4 & \overset{?}{=} & \frac{17}{2} \\ \frac{9}{2} + \frac{8}{2} & \overset{?}{=} & \frac{17}{2} \\ \frac{17}{2} & = & \frac{17}{2} ✓ \end{matrix} \end{matrix}$
	The solution is (3, 6).

In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

Solve the system by elimination. ${\begin{matrix} 3 x + 4 y = 12 \\ y = 3 - \frac{3}{4} x \end{matrix}$

Solution

Steps to solve a system of dependent linear equations, illustrating the algebraic transformations that result in the identity 0=0.
	${\begin{matrix} 3 x + 4 y = 12 \\ y = 3 - \frac{3}{4} x \end{matrix}$
Write the second equation in standard form.	${\begin{matrix} 3 x + 4 y & = & 12 \\ \frac{3}{4} x + y & = & 3 \end{matrix}$
Clear the fractions by multiplying the second equation by 4.	${\begin{matrix} 3 x + 4 y & = & 12 \\ 4 (\frac{3}{4} x + y) & = & 4 (3) \end{matrix}$
Simplify.	${\begin{matrix} 3 x + 4 y & = & 12 \\ 3 x + 4 y & = & 12 \end{matrix}$
To eliminate a variable, we multiply the second equation by $−1$ . Simplify and add.	$\begin{matrix} \underset{________________}{{\begin{matrix} 3 x + 4 y & = & 12 \\ - 3 x - 4 y & = & −12 \end{matrix}} \\ 0 = 0 \end{matrix}$

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Solve the system by elimination. ${\begin{matrix} −6 x + 15 y = 10 \\ 2 x - 5 y = −5 \end{matrix}$

Solution

Steps demonstrating the elimination method to solve a system of linear equations, illustrating a case with no solution.
The equations are in standard form.	${\begin{matrix} - 6 x + 15 y & = & 10 \\ 2 x - 5 y & = & −5 \end{matrix}$
Multiply the second equation by 3 to eliminate a variable.	${\begin{matrix} - 6 x + 15 y & = & 10 \\ 3 (2 x - 5 y) & = & 3 (−5) \end{matrix}$
Simplify and add.	$\begin{matrix} \underset{__________________}{{\begin{matrix} - 6 x + 15 y & = & 10 \\ 6 x - 15 y & = & −15 \end{matrix}} \\ 0 \neq −5 \end{matrix}$

This statement is false. The equations are inconsistent and so their graphs would be parallel lines.

The system does not have a solution.

Solve Applications of Systems of Equations by Elimination

Some applications problems translate directly into equations in standard form, so we will use the elimination method to solve them. As before, we use our Problem Solving Strategy to help us stay focused and organized.

The sum of two numbers is 39. Their difference is 9. Find the numbers.

Solution

A 7-step guide demonstrating how to solve a word problem by setting up and solving a system of linear equations.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for two numbers.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $n =$ the first number. $m =$ the second number.
Step 4. Translate into a system of equations. The system is:	The sum of two numbers is 39. $n + m = 39$ Their difference is 9. $\begin{matrix} n - m = 9 \\ {\begin{matrix} n + m = 39 \\ n - m = 9 \end{matrix} \end{matrix}$
Step 5. Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form and the coefficients of $m$ are opposites. Add. Solve for $n$ . Substitute $n = 24$ into one of the original equations and solve for $m$ .	$\begin{matrix} \underset{____________}{{\begin{matrix} n + m = 39 \\ n - m = 9 \end{matrix}} \\ 2 n = 48 \\ n = 24 \\ n + m = 39 \\ 24 + m = 39 \\ m = 15 \end{matrix}$
Step 6. Check the answer.	Since $24 + 15 = 39$ and $24 - 15 = 9$ , the answers check.
Step 7. Answer the question.	The numbers are 24 and 15.

Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the number of calories in one order of medium fries and in one small soda.
Step 3. Name what we are looking for.	Let f = the number of calories in 1 order of medium fries. s = the number of calories in 1 small soda.
Step 4. Translate into a system of equations:	one medium fries and two small sodas had a total of 620 calories

	two medium fries and one small soda had a total of 820 calories.

Our system is:
Step 5. Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form. To get opposite coefficients of f, multiply the top equation by −2.
Simplify and add.
Solve for s.
Substitute s = 140 into one of the original equations and then solve for f.



Step 6. Check the answer.	Verify that these numbers make sense in the problem and that they are solutions to both equations. We leave this to you!
Step 7. Answer the question.	The small soda has 140 calories and the fries have 340 calories.

Choose the Most Convenient Method to Solve a System of Linear Equations

When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

This table has two rows and three columns. The first row labels the columns as “Graphing,” “Substitution,” and “Elimination.” Under “Graphing” it says, “Use when you need a picture of the situation.” Under “Substitution” it says, “Use when one equation is already solved for one variable.” Under “Elimination” it says, “Use when the equations are in standard form.”

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ ${\begin{matrix} 3 x + 8 y = 40 \\ 7 x - 4 y = −32 \end{matrix}$ ⓑ ${\begin{matrix} 5 x + 6 y = 12 \\ y = \frac{2}{3} x - 1 \end{matrix}$

Solution

ⓐ $\begin{matrix} {\begin{matrix} 3 x + 8 y = 40 \\ 7 x - 4 y = −32 \end{matrix} \end{matrix}$
Since both equations are in standard form, using elimination will be most convenient.
ⓑ $\begin{matrix} {\begin{matrix} 5 x + 6 y = 12 \\ y = \frac{2}{3} x - 1 \end{matrix} \end{matrix}$

>Since one equation is already solved for y, using substitution will be most convenient.

Key Concepts

To Solve a System of Equations by Elimination
1. Write both equations in standard form. If any coefficients are fractions, clear them.
2. Make the coefficients of one variable opposites.
  - Decide which variable you will eliminate.
  - Multiply one or both equations so that the coefficients of that variable are opposites.
3. Add the equations resulting from Step 2 to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
6. Write the solution as an ordered pair.
7. Check that the ordered pair is a solution to both original equations.

Practice Makes Perfect

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

${\begin{matrix} 5 x + 2 y = 2 \\ −3 x - y = 0 \end{matrix}$

${\begin{matrix} −3 x + y = −9 \\ x - 2 y = −12 \end{matrix}$

Solution

(6, 9)

${\begin{matrix} 6 x - 5 y = −1 \\ 2 x + y = 13 \end{matrix}$

${\begin{matrix} 3 x - y = −7 \\ 4 x + 2 y = −6 \end{matrix}$

Solution

$(−2, 1)$

${\begin{matrix} x + y = −1 \\ x - y = −5 \end{matrix}$

${\begin{matrix} x + y = −8 \\ x - y = −6 \end{matrix}$

Solution

$(−7, −1)$

${\begin{matrix} 3 x - 2 y = 1 \\ - x + 2 y = 9 \end{matrix}$

${\begin{matrix} −7 x + 6 y = −10 \\ x - 6 y = 22 \end{matrix}$

Solution

$(−2, −4)$

${\begin{matrix} 3 x + 2 y = −3 \\ - x - 2 y = −19 \end{matrix}$

${\begin{matrix} 5 x + 2 y = 1 \\ −5 x - 4 y = −7 \end{matrix}$

Solution

$(−1, 3)$

${\begin{matrix} 6 x + 4 y = −4 \\ −6 x - 5 y = 8 \end{matrix}$

${\begin{matrix} 3 x - 4 y = −11 \\ x - 2 y = −5 \end{matrix}$

Solution

$(−1, 2)$

${\begin{matrix} 5 x - 7 y = 29 \\ x + 3 y = −3 \end{matrix}$

${\begin{matrix} 6 x - 5 y = −75 \\ - x - 2 y = −13 \end{matrix}$

Solution

$(−5, 9)$

${\begin{matrix} - x + 4 y = 8 \\ 3 x + 5 y = 10 \end{matrix}$

${\begin{matrix} 2 x - 5 y = 7 \\ 3 x - y = 17 \end{matrix}$

Solution

(6, 1)

${\begin{matrix} 5 x - 3 y = −1 \\ 2 x - y = 2 \end{matrix}$

${\begin{matrix} 7 x + y = −4 \\ 13 x + 3 y = 4 \end{matrix}$

Solution

$(−2, 10)$

${\begin{matrix} −3 x + 5 y = −13 \\ 2 x + y = −26 \end{matrix}$

${\begin{matrix} 3 x - 5 y = −9 \\ 5 x + 2 y = 16 \end{matrix}$

Solution

(2, 3)

${\begin{matrix} 4 x - 3 y = 3 \\ 2 x + 5 y = −31 \end{matrix}$

${\begin{matrix} 4 x + 7 y = 14 \\ −2 x + 3 y = 32 \end{matrix}$

Solution

$(−7, 6)$

${\begin{matrix} 5 x + 2 y = 21 \\ 7 x - 4 y = 9 \end{matrix}$

${\begin{matrix} 3 x + 8 y = −3 \\ 2 x + 5 y = −3 \end{matrix}$

Solution

$(−9, 3)$

${\begin{matrix} 11 x + 9 y = −5 \\ 7 x + 5 y = −1 \end{matrix}$

${\begin{matrix} 3 x + 8 y = 67 \\ 5 x + 3 y = 60 \end{matrix}$

Solution

(9, 5)

${\begin{matrix} 2 x + 9 y = −4 \\ 3 x + 13 y = −7 \end{matrix}$

${\begin{matrix} \frac{1}{3} x - y = −3 \\ x + \frac{5}{2} y = 2 \end{matrix}$

Solution

$(−3, 2)$

${\begin{matrix} x + \frac{1}{2} y = \frac{3}{2} \\ \frac{1}{5} x - \frac{1}{5} y = 3 \end{matrix}$

${\begin{matrix} x + \frac{1}{3} y = −1 \\ \frac{1}{2} x - \frac{1}{3} y = −2 \end{matrix}$

Solution

$(−2, 3)$

${\begin{matrix} \frac{1}{3} x - y = −3 \\ \frac{2}{3} x + \frac{5}{2} y = 3 \end{matrix}$

${\begin{matrix} 2 x + y = 3 \\ 6 x + 3 y = 9 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} x - 4 y = −1 \\ −3 x + 12 y = 3 \end{matrix}$

${\begin{matrix} −3 x - y = 8 \\ 6 x + 2 y = −16 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} 4 x + 3 y = 2 \\ 20 x + 15 y = 10 \end{matrix}$

${\begin{matrix} 3 x + 2 y = 6 \\ −6 x - 4 y = −12 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} 5 x - 8 y = 12 \\ 10 x - 16 y = 20 \end{matrix}$

${\begin{matrix} −11 x + 12 y = 60 \\ −22 x + 24 y = 90 \end{matrix}$

Solution

inconsistent, no solution

${\begin{matrix} 7 x - 9 y = 16 \\ −21 x + 27 y = −24 \end{matrix}$

${\begin{matrix} 5 x - 3 y = 15 \\ y = \frac{5}{3} x - 2 \end{matrix}$

Solution

inconsistent, no solution

${\begin{matrix} 2 x + 4 y = 7 \\ y = - \frac{1}{2} x - 4 \end{matrix}$

Solve Applications of Systems of Equations by Elimination

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is 65. Their difference is 25. Find the numbers.

Solution

The numbers are 20 and 45.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

The sum of two numbers is −27. Their difference is −59. Find the numbers.

Solution

The numbers are 16 and −43.

The sum of two numbers is −45. Their difference is −89. Find the numbers.

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Solution

A shirt costs $16 and a sweater costs $33.

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

Solution

There are 860 mg in a hot dog. There are 1,000 mg in a cup of cottage cheese.

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

Choose the Most Convenient Method to Solve a System of Linear Equations

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ ${\begin{matrix} 8 x - 15 y = −32 \\ 6 x + 3 y = −5 \end{matrix}$ ⓑ ${\begin{matrix} x = 4 y - 3 \\ 4 x - 2 y = −6 \end{matrix}$

Solution

ⓐ elimination ⓑ substitution

ⓐ ${\begin{matrix} y = 7 x - 5 \\ 3 x - 2 y = 16 \end{matrix}$ ⓑ ${\begin{matrix} 12 x - 5 y = −42 \\ 3 x + 7 y = −15 \end{matrix}$

ⓐ ${\begin{matrix} y = 4 x + 9 \\ 5 x - 2 y = −21 \end{matrix}$ ⓑ ${\begin{matrix} 9 x - 4 y = 24 \\ 3 x + 5 y = −14 \end{matrix}$

Solution

ⓐ substitution ⓑ elimination

ⓐ ${\begin{matrix} 14 x - 15 y = −30 \\ 7 x + 2 y = 10 \end{matrix}$ ⓑ ${\begin{matrix} x = 9 y - 11 \\ 2 x - 7 y = −27 \end{matrix}$

Everyday Math

Norris can row 3 miles upstream against the current in 1 hour, the same amount of time it takes him to row 5 miles downstream, with the current. Solve the system. ${\begin{matrix} r - c = 3 \\ r + c = 5 \end{matrix}$

ⓐ for $r$ , his rowing speed in still water.
ⓑ Then solve for $c$ , the speed of the river current.

Solution

ⓐ $r = 4$ ⓑ $c = 1$

Josie wants to make 10 pounds of trail mix using nuts and raisins, and she wants the total cost of the trail mix to be $54. Nuts cost $6 per pound and raisins cost $3 per pound. Solve the system ${\begin{matrix} n + r = 10 \\ 6 n + 3 r = 54 \end{matrix}$ to find $n$ , the number of pounds of nuts, and $r$ , the number of pounds of raisins she should use.

Writing Exercises

Solve the system
${\begin{matrix} x + y = 10 \\ 5 x + 8 y = 56 \end{matrix}$

ⓐ by substitution ⓑ by graphing ⓒ Which method do you prefer? Why?

Solution

ⓐ (8, 2)
ⓑ
ⓒ Answers will vary.

Solve the system
${\begin{matrix} x + y = −12 \\ y = 4 - \frac{1}{2} x \end{matrix}$

ⓐ by substitution ⓑ by graphing ⓒ Which method do you prefer? Why?

Self Check

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

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

Solve Applications with Systems of Equations

Previously in this chapter we solved several applications with systems of linear equations. In this section, we’ll look at some specific types of applications that relate two quantities. We’ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.

We will use our Problem Solving Strategy for Systems of Linear Equations.

Translate to a System of Equations

Many of the problems we solved in earlier applications related two quantities. Here are two of the examples from the chapter on Math Models.

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

In that chapter we translated each situation into one equation using only one variable. Sometimes it was a bit of a challenge figuring out how to name the two quantities, wasn’t it?

Let’s see how we can translate these two problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem Solving Strategy.

How to Translate to a System of Equations

Translate to a system of equations:

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

Solution

This figure has four rows and three columns. The first row reads, “Step 1: Read the problem. Make sure you understand all the words and ideas. This is a number problem. The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.”

The second row reads, “Step 2: Identify what you are looking for. ‘Find the numbers.’ We are looking for 2 numbers.”

The third row reads, “Step 3: Name what you are looking for. Choose variables to represent those quantities. We will use two variables, m and n. Let me = one number n = second number.”

The fourth row reads, “Step 4: Translate into a system of equations. We will write one equation for each sentence.” The figure then shows how, “The sum of the numbers is -14” becomes m + n = -14 and “One number is four less than the other” becomes m = n – 4. The figure then says, “The system is m + n = -14 and m = n – 4.”

We’ll do another example where we stop after we write the system of equations.

Translate to a system of equations:

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

Solution

This table demonstrates translating a word problem about a couple's earnings into a system of linear equations, showing variable definition and equation formation.
We are looking for the amount that the husband and wife each earn.	Let $h =$ the amount the husband earns. $w =$ the amount the wife earns.
Translate.	A married couple together earns $110,000.
	$w + h = 110,000$
	The wife earns $16,000 less than twice what husband earns.
	$w = 2 h - 16,000$
The system of equations is:	${\begin{matrix} w + h = 110,000 \\ w = 2 h - 16,000 \end{matrix}$

Solve Direct Translation Applications

We set up, but did not solve, the systems of equations in Example 1 and Example 2 Now we’ll translate a situation to a system of equations and then solve it.

Translate to a system of equations and then solve:

Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the ages of Devon and Cooper.
Step 3. Name what we are looking for.	Let $d =$ Devon’s age. $c =$ Cooper’s age
Step 4. Translate into a system of equations.	Devon is 26 years older than Cooper.

	The sum of their ages is 50.

The system is:
Step 5. Solve the system of equations. Solve by substitution.
Substitute c + 26 into the second equation.
Solve for c.


Substitute c = 12 into the first equation and then solve for d.

Step 6. Check the answer in the problem.	Is Devon’s age 26 more than Cooper’s? Yes, 38 is 26 more than 12. Is the sum of their ages 50? Yes, 38 plus 12 is 50.
Step 7. Answer the question.	Devon is 38 and Cooper is 12 years old.

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the number of calories burned each minute on the elliptical trainer and each minute of circuit training.
Step 3. Name what we are looking for.	Let $e =$ number of calories burned per minute on the elliptical trainer. $c =$ number of calories burned per minute while circuit training
Step 4. Translate into a system of equations.	10 minutes on the elliptical and circuit training for 20 minutes, burned 278 calories

	20 minutes on the elliptical and 30 minutes of circuit training burned 473 calories

The system is:
Step 5. Solve the system of equations.
Multiply the first equation by −2 to get opposite coefficients of e.
Simplify and add the equations. Solve for c.
Substitute c = 8.3 into one of the original equations to solve for e.




Step 6. Check the answer in the problem.	Check the math on your own.

Step 7. Answer the question.	Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute while on the elliptical trainer.

Solve Geometry Applications

When we learned about Math Models, we solved geometry applications using properties of triangles and rectangles. Now we’ll add to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Translate to a system of equations and then solve:

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

Solution

This table outlines the step-by-step process for solving a system of equations to find two complementary angles with a given difference.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the measure of each angle.
Step 3. Name what we are looking for.	Let $x =$ the measure of the first angle $x =$ . $m =$ the measure of the second angle.
Step 4. Translate into a system of equations.	The angles are complementary. $x + y = 90$
	The difference of the two angles is 26 degrees. $x - y = 26$
The system is	${\begin{matrix} x + y = 90 \\ x - y = 26 \end{matrix}$
Step 5. Solve the system of equations by elimination.	$\begin{matrix} \underset{_________}{{\begin{cases} x + y = 90 \\ x - y = 26 \end{cases}} \\ 2 x = 116 \end{matrix}$
Substitute $x = 58$ into the first equation.	$\begin{matrix} x + y & = & 90 \\ 58 + y & = & 90 \\ y & = & 32 \end{matrix}$
Step 6. Check the answer in the problem. $\begin{matrix} 58 + 32 = 90 ✓ \\ 58 - 32 = 26 ✓ \end{matrix}$
Step 7. Answer the question.	The angle measures are 58 degrees and 32 degrees.

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the measure of each angle.
Step 3. Name what we are looking for.	Let $x =$ the measure of the first angle. $y =$ the measure of the second angle
Step 4. Translate into a system of equations.	The angles are supplementary.

	The larger angle is twelve less than five times the smaller angle

The system is: Step 5. Solve the system of equations substitution.
Substitute 5x − 12 for y in the first equation.
Solve for x.


Substitute 32 for in the second equation, then solve for y.


Step 6. Check the answer in the problem. $\begin{matrix} 32 + 158 & = & 180 ✓ \\ 5 \cdot 32 - 12 & = & 147 ✓ \end{matrix}$
Step 7. Answer the question.	The angle measures are 148 and 32.

Translate to a system of equations and then solve:

Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the length and width.

Step 3. Name what we are looking for.	Let $L =$ the length of the fenced yard. $W =$ the width of the fenced yard
Step 4. Translate into a system of equations.	One length and two widths equal 125.

	The length will be 5 feet more than four times the width.

The system is: Step 5. Solve the system of equations by substitution.
Substitute L = 4W + 5 into the first equation, then solve for W.


Substitute 20 for W in the second equation, then solve for L.



Step 6. Check the answer in the problem. $\begin{matrix} 20 + 28 + 20 & = & 125 ✓ \\ 85 & = & 4 \cdot 20 + 5 ✓ \end{matrix}$
Step 7. Answer the equation.	The length is 85 feet and the width is 20 feet.

Solve Uniform Motion Applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = rt where D is the distance travelled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Translate to a system of equations and then solve:

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

Solution

A diagram is useful in helping us visualize the situation.
This figure shows a diagram. Denver is on the left and St. Louis is on the right. There is a ray stretching from St. Louis to Denver. It is labeled “Joni” and “65 m p h.” There is another ray stretching from St. Louis to Denver. It is labeled “Kelly (1/2 hour later)” and “78 m p h.”

This figure shows a diagram. Denver is on the left and St. Louis is on the right. There is a ray stretching from St. Louis to Denver. It is labeled “Joni” and “65 m p h.” There is another ray stretching from St. Louis to Denver. It is labeled “Kelly (1/2 hour later)” and “78 m p h.”

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart.
We are looking for the length of time Kelly, k, and Joni, j, will each drive. Since $D = r \cdot t$ we can fill in the Distance column.
Translate into a system of equations. To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So, $65 j = 78 k .$ Also, since Kelly left later, her time will be $\frac{1}{2}$ hour less than Joni’s time. So, $k = j - \frac{1}{2} .$
Now we have the system.
Solve the system of equations by substitution.
Substitute $k = j - \frac{1}{2}$ into the second equation, then solve for j.



To find Kelly’s time, substitute j = 3 into the first equation, then solve for k.	$The image shows a mathematical equation centered on a white background. The equation is 'k = j - 1/2', where k and j are variables and 1/2 is a fraction being subtracted from j.$

	$A mathematical expression showing the variable k equals 5/2 or 2 1/2. The two forms represent the same value, with 5/2 as an improper fraction and 2 1/2 as a mixed number.$
Check the answer in the problem. Joni 3 hours (65 mph) = 195 miles. Kelly $2 \frac{1}{2}$ hours (78 mph) = 195 miles. Yes, they will have traveled the same distance when they meet.
Answer the question.	Kelly will catch up to Joni in $2 \frac{1}{2}$ hours. By then, Joni will have traveled 3 hours.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

Figure 1 and Figure 2 show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c.

In Figure 1 the boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c.

This figure shows a boat floating in water. On the right, there is an arrow pointing towards the boat. It is labeled “c.” On the left, there is an arrow pointing away from the boat. It is labeled “b.”

In Figure 2 the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is $b - c$ .

This figure shows a boat floating in water. To the left is an arrow pointing away from the boat labeled “b,” and an arrow pointing towards the boat labeled “c.”

We’ll put some numbers to this situation in Example 9.

Translate to a system of equations and then solve:

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Solution

Read the problem.

This is a uniform motion problem and a picture will help us visualize the situation.
This figure shows an arrow labeled “c” which continues to the right, representing the wave. Under the wave is a ray that points to the right and is labeled “four hours.” Under this ray is another ray pointing to the left labeled “five hours.” It is the same length as the ray labeled “four hours.” There is a bracket under the ray labeled “five hours.” The bracket is labeled “60 miles.”

Identify what we are looking for.	We are looking for the speed of the ship in still water and the speed of the current.
Name what we are looking for.	Let $s =$ the rate of the ship in still water. $c =$ the rate of the current
A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship; therefore, the ship’s actual rate is s + c. Going upstream, the current slows the ship; therefore, the actual rate is s − c.
Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles.
Translate into a system of equations. Since rate times time is distance, we can write the system of equations.
Solve the system of equations. Distribute to put both equations in standard form, then solve by elimination.
Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for s.
Substitute s = 13.5 into one of the original equations.




Check the answer in the problem. The downstream rate would be 13.5 + 1.5 = 15 mph. In 4 hours the ship would travel 15 · 4 = 60 miles. The upstream rate would be 13.5 − 1.5 = 12 mph. In 5 hours the ship would travel 12 · 5 = 60 miles.
Answer the question.	The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in Example 10. A wind current in the same direction as the plane is flying is called a tailwind. A wind current blowing against the direction of the plane is called a headwind.

Translate to a system of equations and then solve:

A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

Read the problem.

This is a uniform motion problem and a picture will help us visualize.
This figure shows an arrow labeled “3 hours” which continues to the right, representing the wind. Under the wave is a ray that points to the right and is labeled “j plus w equals 365” and “1,095 miles”. Under this ray is another ray pointing to the left labeled “j minus w equals 329” and “987 miles.”

This figure shows an arrow labeled “3 hours” which continues to the right, representing the wind. Under the wave is a ray that points to the right and is labeled “j plus w equals 365” and “1,095 miles”. Under this ray is another ray pointing to the left labeled “j minus w equals 329” and “987 miles.”

Identify what we are looking for.	We are looking for the speed of the jet in still air and the speed of the wind.
Name what we are looking for.	Let $j =$ the speed of the jet in still air. $w =$ the speed of the wind
A chart will help us organize the information. The jet makes two trips-one in a tailwind and one in a headwind. In a tailwind, the wind helps the jet and so the rate is j + w. In a headwind, the wind slows the jet and so the rate is j − w.
Each trip takes 3 hours. In a tailwind the jet flies 1095 miles. In a headwind the jet flies 987 miles.
Translate into a system of equations. Since rate times time is distance, we get the system of equations.
Solve the system of equations. Distribute, then solve by elimination.
Add, and solve for j. Substitute j = 347 into one of the original equations, then solve for w.




Check the answer in the problem. With the tailwind, the actual rate of the jet would be 347 + 18 = 365 mph. In 3 hours the jet would travel 365 · 3 = 1095 miles. Going into the headwind, the jet’s actual rate would be 347 − 18 = 329 mph. In 3 hours the jet would travel 329 · 3 = 987 miles.
Answer the question.	The rate of the jet is 347 mph and the rate of the wind is 18 mph.

Practice Makes Perfect

Translate to a System of Equations

In the following exercises, translate to a system of equations and solve the system.

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

Solution

The numbers are 6 and 9.

The sum of two numbers is twenty-five. One number is five less than the other. Find the numbers.

The sum of two numbers is negative thirty. One number is five times the other. Find the numbers.

Solution

The numbers are −5 and −25.

The sum of two numbers is negative sixteen. One number is seven times the other. Find the numbers.

Twice a number plus three times a second number is twenty-two. Three times the first number plus four times the second is thirty-one. Find the numbers.

Solution

The numbers are 5 and 4.

Six times a number plus twice a second number is four. Twice the first number plus four times the second number is eighteen. Find the numbers.

Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen. Find the numbers.

Solution

The numbers are 2 and 3.

Twice a number plus three times a second number is negative one. The first number plus four times the second number is two. Find the numbers.

A married couple together earn $75,000. The husband earns $15,000 more than five times what his wife earns. What does the wife earn?

Solution

$10,000

During two years in college, a student earned $9,500. The second year she earned $500 more than twice the amount she earned the first year. How much did she earn the first year?

Daniela invested a total of $50,000, some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was $5000 more than twice the amount she put into the CD. How much did she invest in each account?

Solution

She put $15,000 into a CD and $35,000 in bonds.

Jorge invested $28,000 into two accounts. The amount he put in his money market account was $2,000 less than twice what he put into a CD. How much did he invest in each account?

In her last two years in college, Marlene received $42,000 in loans. The first year she received a loan that was $6,000 less than three times the amount of the second year’s loan. What was the amount of her loan for each year?

Solution

The amount of the first year’s loan was $30,000 and the amount of the second year’s loan was $12,000.

Jen and David owe $22,000 in loans for their two cars. The amount of the loan for Jen’s car is $2000 less than twice the amount of the loan for David’s car. How much is each car loan?

Solve Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

Alyssa is twelve years older than her sister, Bethany. The sum of their ages is forty-four. Find their ages.

Solution

Bethany is 16 years old and Alyssa is 28 years old.

Robert is 15 years older than his sister, Helen. The sum of their ages is sixty-three. Find their ages.

The age of Noelle’s dad is six less than three times Noelle’s age. The sum of their ages is seventy-four. Find their ages.

Solution

Noelle is 20 years old and her dad is 54 years old.

The age of Mark’s dad is 4 less than twice Marks’s age. The sum of their ages is ninety-five. Find their ages.

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

Solution

The small container holds 20 gallons and the large container holds 30 gallons.

June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Solution

There were 10 calories burned jogging and 10 calories burned cycling.

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball?

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.

Solution

Notebooks are $4 and thumb drives are $20.

Nancy bought seven pounds of oranges and three pounds of bananas for $17. Her husband later bought three pounds of oranges and six pounds of bananas for $12. What was the cost per pound of the oranges and the bananas?

Solve Geometry Applications In the following exercises, translate to a system of equations and solve.

The difference of two complementary angles is 30 degrees. Find the measures of the angles.

Solution

The measures are 60 degrees and 30 degrees.

The difference of two complementary angles is 68 degrees. Find the measures of the angles.

The difference of two supplementary angles is 70 degrees. Find the measures of the angles.

Solution

The measures are 125 degrees and 55 degrees.

The difference of two supplementary angles is 24 degrees. Find the measure of the angles.

The difference of two supplementary angles is 8 degrees. Find the measures of the angles.

Solution

94 degrees and 86 degrees

The difference of two supplementary angles is 88 degrees. Find the measures of the angles.

The difference of two complementary angles is 55 degrees. Find the measures of the angles.

Solution

72.5 degrees and 17.5 degrees

The difference of two complementary angles is 17 degrees. Find the measures of the angles.

Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.

Solution

The measures are 44 degrees and 136 degrees.

Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.

Solution

The measures are 34 degrees and 56 degrees.

Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Solution

The width is 10 feet and the length is 25 feet.

Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his rectangular front yard. The length, the side along the house, is five feet less than three times the width. Find the length and width of the fencing.

A frame around a rectangular family portrait has a perimeter of 60 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

Solution

The width is 15 feet and the length is 15 feet.

The perimeter of a rectangular toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.

Solve Uniform Motion Applications In the following exercises, translate to a system of equations and solve.

Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?

Solution

It took Sarah’s sister 12 hours.

College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?

At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?

Solution

It took Lucy’s friend 2 hours.

Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

The Jones family took a 12 mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

Solution

The canoe rate is 5 mph and the current rate is 1 mph.

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current.

Solution

The boat rate is 6.60 mph and the current rate is 2.50 mph.

A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current.

A small jet can fly 1,072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

The jet rate is 240 mph and the wind speed is 28 mph.

A small jet can fly 1,435 miles in 5 hours with a tailwind but only 1215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

The jet rate is 415 mph and the wind speed is 19 mph.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1,170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Everyday Math

At a school concert, 425 tickets were sold. Student tickets cost $5 each and adult tickets cost $8 each. The total receipts for the concert were $2,851. Solve the system

${\begin{matrix} s + a = 425 \\ 5 s + 8 a = 2,851 \end{matrix}$

to find $s$ , the number of student tickets and $a$ , the number of adult tickets.

Solution

$s = 183, a = 242$

The first graders at one school went on a field trip to the zoo. The total number of children and adults who went on the field trip was 115. The number of adults was $\frac{1}{4}$ the number of children. Solve the system

${\begin{matrix} c + a = 115 \\ a = \frac{1}{4} c \end{matrix}$

to find $c$ , the number of children and $a$ , the number of adults.

Writing Exercises

Write an application problem similar to Example 3 using the ages of two of your friends or family members. Then translate to a system of equations and solve it.

Solution

Answers will vary.

Write a uniform motion problem similar to Example 8 that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.

Self Check

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

ⓑ 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?

complementary angles

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

supplementary angles

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

Solve Mixture Applications with Systems of Equations

Solve Mixture Applications

When we solved mixture applications with coins and tickets earlier, we started by creating a table so we could organize the information. For a coin example with nickels and dimes, the table looked like this:

This is a table with three rows and four columns. The first row of the table is a header row, and each cell names the column or columns below it. The first cell from the left is named “Type.” The second cell contains the equation “Number” times “Value” equals “Total Value,” with one column corresponding to “Number,” one column corresponding to “Value,” and one column corresponding to total value. Hence the content of the “Number” column times the content of the “Value” column equals the content of the “Total Value” column. In the second row of the table, the “Type” column contains “nickels,” the “Number” column is blank, the “Value” column contains 0.05, and the “Total Value” column is blank. In the third row of the table, the “Type” column contains “dimes,” the “Number” column is blank, the “Value column contains 0.10, and the “Total Value” column is blank.

Using one variable meant that we had to relate the number of nickels and the number of dimes. We had to decide if we were going to let n be the number of nickels and then write the number of dimes in terms of n, or if we would let d be the number of dimes and write the number of nickels in terms of d.

Now that we know how to solve systems of equations with two variables, we’ll just let n be the number of nickels and d be the number of dimes. We’ll write one equation based on the total value column, like we did before, and the other equation will come from the number column.

For the first example, we’ll do a ticket problem where the ticket prices are in whole dollars, so we won’t need to use decimals just yet.

Translate to a system of equations and solve:

The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled $1,302. How many $11 adult and how many $8 child tickets were sold?

Solution

Step 1. Read the problem.	We will create a table to organize the information.
Step 2. Identify what we are looking for.	We are looking for the number of adult tickets and the number of child tickets sold.
Step 3. Name what we are looking for.	Let $a =$ the number of adult tickets. $c =$ the number of child tickets
A table will help us organize the data. We have two types of tickets: adult and child.	Write a and c for the number of tickets.
Write the total number of tickets sold at the bottom of the Number column.	Altogether 147 were sold.
Write the value of each type of ticket in the Value column.	The value of each adult ticket is $11. The value of each child tickets is $8.
The number times the value gives the total value, so the total value of adult tickets is $a \cdot 11 = 11 a$ , and the total value of child tickets is $c \cdot 8 = 8 c .$
Altogether the total value of the tickets was $1,302.	Fill in the Total Value column.
Step 4. Translate into a system of equations.
The Number column and the Total Value column give us the system of equations. We will use the elimination method to solve this system.
Multiply the first equation by −8.
Simplify and add, then solve for a.

Substitute a = 42 into the first equation, then solve for c.

Step 5. Check the answer in the problem. 42 adult tickets at $11 per ticket makes $462 105 child tickets at $8 per ticket makes $840. The total receipts are $1,302.✓
Step 6. Answer the question.	The movie theater sold 42 adult tickets and 105 child tickets.

In Example 2 we’ll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the ‘number’ column will be easy.

Translate to a system of equations and solve:

Priam has a collection of nickels and quarters, with a total value of $7.30. The number of nickels is six less than three times the number of quarters. How many nickels and how many quarters does he have?

Solution

Step 1. Read the problem.	We will create a table to organize the information.
Step 2. Identify what we are looking for.	We are looking for the number of nickels and the number of quarters.
Step 3. Name what we are looking for.	Let $n =$ the number of nickels. $q =$ the number of quarters
A table will help us organize the data. We have two types of coins, nickels and quarters.	Write n and q for the number of each type of coin.
Fill in the Value column with the value of each type of coin.	The value of each nickel is $0.05. The value of each quarter is $0.25.
The number times the value gives the total value, so, the total value of the nickels is n (0.05) = 0.05n and the total value of quarters is q(0.25) = 0.25q. Altogether the total value of the coins is $7.30.
Step 4. Translate into a system of equations.
The Total value column gives one equation.
We also know the number of nickels is six less than three times the number of quarters. Translate to get the second equation.
Now we have the system to solve.
Step 5. Solve the system of equations We will use the substitution method. Substitute n = 3q − 6 into the first equation. Simplify and solve for q.





To find the number of nickels, substitute q = 19 into the second equation.


Step 6. Check the answer in the problem. $\begin{matrix} 19 quarters at $ 0.25 & = & $ 4.75 \\ 51 nickels at $ 0.05 & = & $ 2.55 \\ Total & = & $ 7.30 ✓ \\ 3 \cdot 19 - 16 & = & 51 ✓ \end{matrix}$
Step 7. Answer the question.	Priam has 19 quarters and 51 nickels.

Some mixture applications involve combining foods or drinks. Example situations might include combining raisins and nuts to make a trail mix or using two types of coffee beans to make a blend.

Translate to a system of equations and solve:

Carson wants to make 20 pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix costs him $7.60 per pound. Nuts cost $9.00 per pound and chocolate chips cost $2.00 per pound. How many pounds of nuts and how many pounds of chocolate chips should he use?

Solution

Step 1. Read the problem.	We will create a table to organize the information.
Step 2. Identify what we are looking for.	We are looking for the number of pounds of nuts and the number of pounds of chocolate chips.
Step 3. Name what we are looking for.	Let $n =$ the number of pound of nuts. $c =$ the number of pounds of chips
Carson will mix nuts and chocolate chips to get trail mix. Write in n and c for the number of pounds of nuts and chocolate chips. There will be 20 pounds of trail mix. Put the price per pound of each item in the Value column. Fill in the last column using
Number · Value = Total Value
Step 4. Translate into a system of equations. We get the equations from the Number and Total Value columns.
Step 5. Solve the system of equations We will use elimination to solve the system.
Multiply the first equation by −2 to eliminate c.
Simplify and add. Solve for n.

To find the number of pounds of chocolate chips, substitute n = 16 into the first equation, then solve for c.
	$c = 4$
Step 6. Check the answer in the problem. $\begin{matrix} 16 + 4 & = & 20 ✓ \\ 9 \cdot 16 + 2 \cdot 4 & = & 152 ✓ \end{matrix}$
Step 7. Answer the question.	Carson should mix 16 pounds of nuts with 4 pounds of chocolate chips to create the trail mix.

Another application of mixture problems relates to concentrated cleaning supplies, other chemicals, and mixed drinks. The concentration is given as a percent. For example, a 20% concentrated household cleanser means that 20% of the total amount is cleanser, and the rest is water. To make 35 ounces of a 20% concentration, you mix 7 ounces (20% of 35) of the cleanser with 28 ounces of water.

For these kinds of mixture problems, we’ll use percent instead of value for one of the columns in our table.

Translate to a system of equations and solve:

Sasheena is a lab assistant at her community college. She needs to make 200 milliliters of a 40% solution of sulfuric acid for a lab experiment. The lab has only 25% and 50% solutions in the storeroom. How much should she mix of the 25% and the 50% solutions to make the 40% solution?

Solution

Step 1. Read the problem.	A figure may help us visualize the situation, then we will create a table to organize the information.
Sasheena must mix some of the 25% solution and some of the 50% solution together to get 200 ml of the 40% solution.
Step 2. Identify what we are looking for.	We are looking for how much of each solution she needs.
Step 3. Name what we are looking for.	Let $x =$ number of ml of 25% solution. $y =$ number of ml of 50% solution
A table will help us organize the data. She will mix x ml of 25% with y ml of 50% to get 200 ml of 40% solution. We write the percents as decimals in the chart. We multiply the number of units times the concentration to get the total amount of sulfuric acid in each solution.
Step 4. Translate into a system of equations. We get the equations from the Number column and the Amount column.
Now we have the system.
Step 5. Solve the system of equations. We will solve the system by elimination. Multiply the first equation by −0.5 to eliminate y.
Simplify and add to solve for x.
To solve for y, substitute x = 80 into the first equation.


Step 6. Check the answer in the problem. $\begin{matrix} 80 + 120 & = & 120 ✓ \\ 0.25 (80) + 0.50 (120) & = & 80 ✓ \\ Yes! \end{matrix}$
Step 7. Answer the question.	Sasheena should mix 80 ml of the 25% solution with 120 ml of the 50% solution to get the 200 ml of the 40% solution.

Solve Interest Applications

The formula to model interest applications is I = Prt. Interest, I, is the product of the principal, P, the rate, r, and the time, t. In our work here, we will calculate the interest earned in one year, so t will be 1.

We modify the column titles in the mixture table to show the formula for interest, as you’ll see in Example 5.

Translate to a system of equations and solve:

Adnan has $40,000 to invest and hopes to earn 7.1% interest per year. He will put some of the money into a stock fund that earns 8% per year and the rest into bonds that earns 3% per year. How much money should he put into each fund?

Solution

Step 1. Read the problem.	A chart will help us organize the information.
Step 2. Identify what we are looking for.	We are looking for the amount to invest in each fund.
Step 3. Name what we are looking for.	Let $s =$ the amount invested in stocks. $b =$ the amount invested in bonds.
Write the interest rate as a decimal for each fund. Multiply: Principal · Rate · Time to get the Interest.
Step 4. Translate into a system of equations. We get our system of equations from the Principal column and the Interest column.
Step 5. Solve the system of equations Solve by elimination. Multiply the top equation by −0.03.
Simplify and add to solve for s.

To find b, substitute s = 32,800 into the first equation.

Step 6. Check the answer in the problem.	We leave the check to you.
Step 7. Answer the question.	Adnan should invest $32,800 in stock and $7,200 in bonds.

Did you notice that the Principal column represents the total amount of money invested while the Interest column represents only the interest earned? Likewise, the first equation in our system, s + b = 40,000, represents the total amount of money invested and the second equation, 0.08s + 0.03b = 0.071(40,000), represents the interest earned.

Translate to a system of equations and solve:

Rosie owes $21,540 on her two student loans. The interest rate on her bank loan is 10.5% and the interest rate on the federal loan is 5.9%. The total amount of interest she paid last year was $1,669.68. What was the principal for each loan?

Solution

Step 1. Read the problem.	A chart will help us organize the information.
Step 2. Identify what we are looking for.	We are looking for the principal of each loan.
Step 3. Name what we are looking for.	Let $b =$ the principal for the bank loan. $f =$ the principal on the federal loan
The total loans are $21,540.
Record the interest rates as decimals in the chart.
Multiply using the formula l = Pr t to get the Interest.
Step 4. Translate into a system of equations. The system of equations comes from the Principal column and the Interest column.
Step 5. Solve the system of equations We will use substitution to solve. Solve the first equation for b.
Substitute b = −f + 21,540 into the second equation.
Simplify and solve for f.



To find b, substitute f = 12,870 into the first equation.

Step 6. Check the answer in the problem.	We leave the check to you.
Step 7. Answer the question.	The principal of the bank loan is $12,870 and the principal for the federal loan is $8,670.

Key Concepts

Table for coin and mixture applications
Table for concentration applications
Table for interest applications

Practice Makes Perfect

Solve Mixture Applications

In the following exercises, translate to a system of equations and solve.

Tickets to a Broadway show cost $35 for adults and $15 for children. The total receipts for 1650 tickets at one performance were $47,150. How many adult and how many child tickets were sold?

Solution

There 1120 adult tickets and 530 child tickets sold.

Tickets for a show are $70 for adults and $50 for children. One evening performance had a total of 300 tickets sold and the receipts totaled $17,200. How many adult and how many child tickets were sold?

Tickets for a train cost $10 for children and $22 for adults. Josie paid $1,200 for a total of 72 tickets. How many children’s tickets and how many adult tickets did Josie buy?

Solution

Josie bought 40 adult tickets and 32 children tickets.

Tickets for a baseball game are $69 for Main Level seats and $39 for Terrace Level seats. A group of sixteen friends went to the game and spent a total of $804 for the tickets. How many of Main Level and how many Terrace Level tickets did they buy?

Tickets for a dance recital cost $15 for adults and $7 for children. The dance company sold 253 tickets and the total receipts were $2,771. How many adult tickets and how many child tickets were sold?

Solution

There were 125 adult tickets and 128 children tickets sold.

Tickets for the community fair cost $12 for adults and $5 dollars for children. On the first day of the fair, 312 tickets were sold for a total of $2,204. How many adult tickets and how many child tickets were sold?

Brandon has a cup of quarters and dimes with a total value of $3.80. The number of quarters is four less than twice the number of dimes. How many quarters and how many dimes does Brandon have?

Solution

Brandon has 12 quarters and 8 dimes.

Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is $0.95. The number of nickels is two less than five times the number of dimes. How many nickels and how many dimes are in the coin purse?

Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value $13.10. The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have?

Solution

Peter had 11 dimes and 48 quarters.

Lucinda had a pocketful of dimes and quarters with a value of $ $6.20. The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have?

A cashier has 30 bills, all of which are $10 or $20 bills. The total value of the money is $460. How many of each type of bill does the cashier have?

Solution

The cashier has fourteen $10 bills and sixteen $20 bills.

A cashier has 54 bills, all of which are $10 or $20 bills. The total value of the money is $910. How many of each type of bill does the cashier have?

Marissa wants to blend candy selling for $1.80 per pound with candy costing $1.20 per pound to get a mixture that costs her $1.40 per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use?

Solution

Marissa should use 60 pounds of the $1.20/lb candy and 30 pounds of the $1.80/lb candy.

How many pounds of nuts selling for $6 per pound and raisins selling for $3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him $5 per pound?

Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is $1.79 per gallon and the cost of the fruit drink is $2.49 per gallon. Hannah’s budget requires that the punch cost $2.21 per gallon. How many gallons of soda and how many gallons of fruit drink does she need?

Solution

Hannah needs 10 gallons of soda and 15 gallons of fruit drink.

Joseph would like to make 12 pounds of a coffee blend at a cost of $6.25 per pound. He blends Ground Chicory at $4.40 a pound with Jamaican Blue Mountain at $8.84 per pound. How much of each type of coffee should he use?

Julia and her husband own a coffee shop. They experimented with mixing a City Roast Columbian coffee that cost $7.80 per pound with French Roast Columbian coffee that cost $8.10 per pound to make a 20 pound blend. Their blend should cost them $7.92 per pound. How much of each type of coffee should they buy?

Solution

Julia and her husband should buy 12 pounds of City Roast Columbian coffee and 8 pounds of French Roast Columbian coffee.

Melody wants to sell bags of mixed candy at her lemonade stand. She will mix chocolate pieces that cost $4.89 per bag with peanut butter pieces that cost $3.79 per bag to get a total of twenty-five bags of mixed candy. Melody wants the bags of mixed candy to cost her $4.23 a bag to make. How many bags of chocolate pieces and how many bags of peanut butter pieces should she use?

Jotham needs 70 liters of a 50% alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution?

Solution

Jotham should mix 42 liters of the 30% solution and 28 liters of the 80% solution.

Joy is preparing 15 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution?

A scientist needs 65 liters of a 15% alcohol solution. She has available a 25% and a 12% solution. How many liters of the 25% and how many liters of the 12% solutions should she mix to make the 15% solution?

Solution

The scientist should mix 15 liters of the 25% solution and 50 liters of the 12% solution.

A scientist needs 120 liters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many liters of the 25% and how many liters of the 10% solutions should the scientist mix to make the 20% solution?

A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used?

Solution

160 liters of the 40% solution and 80 liters of the 70% solution will be used.

A 90% antifreeze solution is to be mixed with a 75% antifreeze solution to get 360 liters of a 85% solution. How many liters of the 90% and how many liters of the 75% solutions will be used?

Solve Interest Applications

In the following exercises, translate to a system of equations and solve.

Hattie had $3,000 to invest and wants to earn 10.6% interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year. How much money should she put into each account?

Solution

Hattie should invest $900 at 12% and $2,100 at 10%.

Carol invested $2,560 into two accounts. One account paid 8% interest and the other paid 6% interest. She earned 7.25% interest on the total investment. How much money did she put in each account?

Sam invested $48,000, some at 6% interest and the rest at 10%. How much did he invest at each rate if he received $4,000 in interest in one year?

Solution

Sam invested $28,000 at 10% and $20,000 at 6%.

Arnold invested $64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received $4,500 in interest in one year?

After four years in college, Josie owes $65,800 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owed for one year was $2,878.50. What is the amount of each loan?

Solution

The federal loan is $62,500 and the bank loan is $3,300.

Mark wants to invest $10,000 to pay for his daughter’s wedding next year. He will invest some of the money in a short term CD that pays 12% interest and the rest in a money market savings account that pays 5% interest. How much should he invest at each rate if he wants to earn $1,095 in interest in one year?

A trust fund worth $25,000 is invested in two different portfolios. This year, one portfolio is expected to earn 5.25% interest and the other is expected to earn 4%. Plans are for the total interest on the fund to be $1150 in one year. How much money should be invested at each rate?

Solution

$12,000 should be invested at 5.25% and $13,000 should be invested at 4%.

A business has two loans totaling $85,000. One loan has a rate of 6% and the other has a rate of 4.5%. This year, the business expects to pay $4650 in interest on the two loans. How much is each loan?

Everyday Math

In the following exercises, translate to a system of equations and solve.

Solution

14 boys paid the full-year fee. 4 boys paid the partial-year fee,

As the treasurer of her daughter’s Girl Scout troop, Laney collected money for some girls and adults to go to a three-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?

Writing Exercises

Take a handful of two types of coins, and write a problem similar to Example 2 relating the total number of coins and their total value. Set up a system of equations to describe your situation and then solve it.

Solution

Answers will vary.

In Example 6 we solved the system of equations ${\begin{matrix} b + f = 21,540 \\ 0.105 b + 0.059 f = 1669.68 \end{matrix}$ by substitution. Would you have used substitution or elimination to solve this system? Why?

Self Check

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

After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Graphing Systems of Linear Inequalities

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations.

A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown below.

{\begin{matrix} x + 4 y \geq 10 \\ 3 x - 2 y < 12 \end{matrix}

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs $(x, y)$ that make both inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Determine whether the ordered pair is a solution to the system. ${\begin{matrix} x + 4 y \geq 10 \\ 3 x - 2 y < 12 \end{matrix}$

ⓐ (−2, 4) ⓑ (3,1)

Solution

ⓐ Is the ordered pair (−2, 4) a solution?

The ordered pair (−2, 4) made both inequalities true. Therefore (−2, 4) is a solution to this system.

ⓑ Is the ordered pair (3,1) a solution?

The ordered pair (3,1) made one inequality true, but the other one false. Therefore (3,1) is not a solution to this system.

Solve a System of Linear Inequalities by Graphing

The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

How to Solve a System of Linear inequalities

Solve the system by graphing.

${\begin{matrix} y \geq 2 x - 1 \\ y < x + 1 \end{matrix}$

Solution

This is a table with three columns and several rows. The first row says, “Step 1: Graph the first inequality. We will graph y is greater than or equal to 2x – 1.” There are two equations givens, y is greater than or equal to 2x – 1 and y is less than x + 1. The table then reads, “Graph the boundary line. We graph the line y = 2x – 1. It is a solid line because the inequality sign is greater than or equal to. Shade in the side of the boundary line where the inequality is true. We choose (0, 0) as a test point. It is a solution to y is greater than or equal to 2x – 1, so we shad in the left side of the boundary line.” There is a figure of a line graphed on an x y coordinate plane. The area to the left of the line is shaded.

The second row then says, “Step 2: On the same grid, graph the second inequality. We will graph y is less than x + 1 on the same grid. Grph the boundary line. We graph the lin y = x + 1. It is a dashed line because the inequality sign is less than. There is a graph which shows two lines graphed on an x y coordinate plane. The area to the left of one line is shaded. The area to the right of the second line is shaded. There is a small area where the shaded areas overlap. The table then says, “Shade in the side of that boundary line where the inequality is true. Again we use (0, 0) as a test point. It is a solution so we shade in that side of the line y = x + 1.

The third row then says, “Step 3: The solution is the region where the shading overlaps. The poing where the boundary lines intersect is not a solution because it is not a solution to y is less than x + 1. The solution is all points in the purple shaded region.”

Solve the system by graphing. ${\begin{matrix} x - y > 3 \\ y < - \frac{1}{5} x + 4 \end{matrix}$

Solution

Graph x − y > 3, by graphing x − y = 3 and
testing a point.

The intercepts are x = 3 and y = −3 and the boundary
line will be dashed.

Test (0, 0). It makes the inequality false. So,
shade the side that does not contain (0, 0) red.

Graph

y < - \frac{1}{5} x + 4

by graphing

y = - \frac{1}{5} x + 4

using the slope

m = - \frac{1}{5}

and y−intercept
b = 4. The boundary line will be dashed.

Test (0, 0). It makes the inequality true, so shade the side that contains (0, 0) blue.

Choose a test point in the solution and verify that it is a solution to both inequalities.

A graph on a coordinate plane displays two linear inequalities using dashed lines. The solution sets for each inequality are shown by a light blue and a dark grey shaded region.

The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice which is the darker-shaded region.

Solve the system by graphing. ${\begin{matrix} x - 2 y < 5 \\ y > −4 \end{matrix}$

Solution

Graph

x - 2 y < 5

, by graphing

x - 2 y = 5

and testing a point.
The intercepts are x = 5 and y = −2.5 and the boundary line will be dashed.

Test (0, 0). It makes the inequality true. So, shade the side
that contains (0, 0) red.

A coordinate plane showing a dashed red line representing the equation y = 1/2x - 2. The region above this line is shaded orange, illustrating the solution set for the inequality y > 1/2x - 2.

Graph y > −4, by graphing y = −4 and recognizing that it is a
horizontal line through y = −4. The boundary line will be dashed.

Test (0, 0). It makes the inequality true. So, shade (blue)
the side that contains (0, 0) blue.

A coordinate plane displays two dashed lines intersecting. A red dashed line slopes upwards, and a black dashed horizontal line is at y=-4. Three distinct regions are shaded: orange, blue, and gray, indicating different solution sets.

The point (0, 0) is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed.

The solution is the area shaded twice which is the darker-shaded region.

Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in Example 5.

Solve the system by graphing. ${\begin{matrix} 4 x + 3 y \geq 12 \\ y < - \frac{4}{3} x + 1 \end{matrix}$

Solution

Graph $4 x + 3 y \geq 12$ , by graphing $4 x + 3 y = 12$ and testing a point. The intercepts are x = 3 and y = 4 and the boundary line will be solid. Test (0, 0). It makes the inequality false. So, shade the side that does not contain (0, 0) red.
Graph $y < - \frac{4}{3} x + 1$ by graphing $y = - \frac{4}{3} x + 1$ using the slope $m = \frac{4}{3}$ and the y-intercept b = 1. The boundary line will be dashed. Test (0, 0). It makes the inequality true. So, shade the side that contains (0, 0) blue.

There is no point in both shaded regions, so the system has no solution. This system has no solution.

Solve the system by graphing. ${\begin{matrix} y > \frac{1}{2} x - 4 \\ x - 2 y < −4 \end{matrix}$

Solution

Graph $y > \frac{1}{2} x - 4$ by graphing $y = \frac{1}{2} x - 4$ using the slope $m = \frac{1}{2}$ and the intercept b = −4. The boundary line will be dashed. Test (0, 0). It makes the inequality true. So, shade the side that contains (0, 0) red.
Graph $x - 2 y < - 4$ by graphing $x - 2 y = - 4$ and testing a point. The intercepts are x = −4 and y = 2 and the boundary line will be dashed. Choose a test point in the solution and verify that it is a solution to both inequalities.

No point on the boundary lines is included in the solution as both lines are dashed.

The solution is the region that is shaded twice, which is also the solution to $x - 2 y < −4$ .

Solve Applications of Systems of Inequalities

The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system as we did above to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so their graphs will only show Quadrant I.

Christy sells her photographs at a booth at a street fair. At the start of the day, she wants to have at least 25 photos to display at her booth. Each small photo she displays costs her $4 and each large photo costs her $10. She doesn’t want to spend more than $200 on photos to display.

ⓐ Write a system of inequalities to model this situation.

ⓑ Graph the system.

ⓓ Could she display 3 large and 22 small photos?

Solution

ⓐ Let $x =$ the number of small photos.
$y =$ the number of large photos
To find the system of inequalities, translate the information.
$\begin{matrix} She wants to have at least 25 photos. \\ The number of small plus the number of large should be at least 25. \\ x + y \geq 25 \\ $4 for each small and $10 for each large must be no more than $200 \\ 4 x + 10 y \leq 200 \end{matrix}$
We have our system of inequalities. ${\begin{matrix} x + y \geq 25 \\ 4 x + 10 y \leq 200 \end{matrix}$

ⓑ

To graph

x + y \geq 25

, graph x + y = 25 as a solid line.
Choose (0, 0) as a test point. Since it does not make the inequality
true, shade the side that does not include the point (0, 0) red.

To graph

4 x + 10 y \leq 200

, graph 4x + 10y = 200 as a solid line.
Choose (0, 0) as a test point. Since it does not make the inequality
true, shade the side that includes the point (0, 0) blue.

A coordinate plane displays the graphs of two linear inequalities, with a red line and a blue line, and their respective shaded regions demonstrating the solution set.

The solution of the system is the region of the graph that is double shaded and so is shaded darker.

ⓒ To determine if 10 small and 20 large photos would work, we see if the point (10, 20) is in the solution region. It is not. Christy would not display 10 small and 20 large photos.
ⓓ To determine if 20 small and 10 large photos would work, we see if the point (20, 10) is in the solution region. It is. Christy could choose to display 20 small and 10 large photos.

Notice that we could also test the possible solutions by substituting the values into each inequality.

Mary needs to purchase supplies of answer sheets and pencils for a standardized test to be given to the juniors at her high school. The number of the answer sheets needed is at least 5 more than the number of pencils. The pencils cost $2 and the answer sheets cost $1. Mary’s budget for these supplies allows for a maximum cost of $400.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could Mary purchase 100 pencils and 100 answer sheets?
ⓓ Could Mary purchase 150 pencils and 150 answer sheets?

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could Mary purchase 100 pencils and 100 answer sheets?
ⓓ Could Mary purchase 150 pencils and 150 answer sheets?

Solution

ⓐ ${\begin{matrix} a \geq p + 5 \\ a + 2 p \leq 400 \end{matrix}$
ⓑ
ⓒ no
ⓓ no

Omar needs to eat at least 800 calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than $5. At the hamburger restaurant near his college, each hamburger has 240 calories and costs $1.40. Each cookie has 160 calories and costs $0.50.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he eat 3 hamburgers and 1 cookie?
ⓓ Could he eat 2 hamburgers and 4 cookies?

Solution

ⓐ Let $h =$ the number of hamburgers.
$c =$ the number of cookies
To find the system of inequalities, translate the information.
The calories from hamburgers at 240 calories each, plus the calories from cookies at 160 calories each must be more that 800.

240 h + 160 c \geq 800

The amount spent on hamburgers at $1.40 each, plus the amount spent on cookies at $0.50 each must be no more than $5.00.

1.40 h + 0.50 c \leq 5

We have our system of inequalities. ${\begin{matrix} 240 h + 160 c \geq 800 \\ 1.40 h + 0.50 c \leq 5 \end{matrix}$

ⓑ

To graph

240 h + 160 c \geq 800

graph

240 h + 160 c = 800

as a solid line.
Choose (0, 0) as a test point. it does not make the inequality true.
So, shade (red) the side that does not include the point (0, 0).

To graph

1.40 h + 0.50 c \leq 5

, graph

1.40 h + 0.50 c = 5

as a solid line.
Choose (0,0) as a test point. It makes the inequality true. So, shade
(blue) the side that includes the point.

A coordinate graph illustrates two linear boundaries, one red and one blue, defining a common shaded region. The axes are labeled 'h' and 'c', indicating a system of inequalities.

The solution of the system is the region of the graph that is double shaded and so is shaded darker.

ⓒ To determine if 3 hamburgers and 1 cookie would meet Omar’s criteria, we see if the point (3, 1) is in the solution region. It is. He might choose to eat 3 hamburgers and 1 cookie.
ⓓ To determine if 2 hamburgers and 4 cookies would meet Omar’s criteria, we see if the point (2, 4) is in the solution region. It is. He might choose to eat 2 hamburgers and 4 cookies.

We could also test the possible solutions by substituting the values into each inequality.

Key Concepts

To Solve a System of Linear Inequalities by Graphing
1. Graph the first inequality.
  - Graph the boundary line.
  - Shade in the side of the boundary line where the inequality is true.
2. On the same grid, graph the second inequality.
  - Graph the boundary line.
  - Shade in the side of that boundary line where the inequality is true.
3. The solution is the region where the shading overlaps.
4. Check by choosing a test point.

Section Exercises

Practice Makes Perfect

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

In the following exercises, determine whether each ordered pair is a solution to the system.

${\begin{matrix} 3 x + y > 5 \\ 2 x - y \leq 10 \end{matrix}$

ⓐ $(3, −3)$ ⓑ $(7, 1)$

Solution

ⓐ true ⓑ false

${\begin{matrix} 4 x - y < 10 \\ −2 x + 2 y > −8 \end{matrix}$

ⓐ $(5, −2)$ ⓑ $(−1, 3)$

${\begin{matrix} y > \frac{2}{3} x - 5 \\ x + \frac{1}{2} y \leq 4 \end{matrix}$

ⓐ $(6, −4)$ ⓑ $(3, 0)$

Solution

ⓐ false ⓑ true

${\begin{matrix} y < \frac{3}{2} x + 3 \\ \frac{3}{4} x - 2 y < 5 \end{matrix}$

ⓐ $(−4, −1)$ ⓑ $(8, 3)$

${\begin{matrix} 7 x + 2 y > 14 \\ 5 x - y \leq 8 \end{matrix}$

ⓐ $(2, 3)$ ⓑ $(7, −1)$

Solution

ⓐ true ⓑ false

${\begin{matrix} 6 x - 5 y < 20 \\ −2 x + 7 y > −8 \end{matrix}$

ⓐ $(1, −3)$ ⓑ $(−4, 4)$

${\begin{matrix} 2 x + 3 y \geq 2 \\ 4 x - 6 y < −1 \end{matrix}$

ⓐ $(\frac{3}{2}, \frac{4}{3})$ ⓑ $(\frac{1}{4}, \frac{7}{6})$

Solution

ⓐ true ⓑ true

${\begin{matrix} 5 x - 3 y < −2 \\ 10 x + 6 y > 4 \end{matrix}$

ⓐ $(\frac{1}{5}, \frac{2}{3})$ ⓑ $(- \frac{3}{10}, \frac{7}{6})$

Solve a System of Linear Inequalities by Graphing

In the following exercises, solve each system by graphing.

${\begin{matrix} y \leq 3 x + 2 \\ y > x - 1 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of y is less than or equal to 3x + 2 and y is greater than x – 1. The area to the left or right of each line is shaded different colors with the overlapping area also shaded a different color. Both lines are dotted.

${\begin{matrix} y < −2 x + 2 \\ y \geq - x - 1 \end{matrix}$

${\begin{matrix} y < 2 x - 1 \\ y \leq - \frac{1}{2} x + 4 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of y is less than 2x - 1 and y is less than or equal to -(1/2)x + 4. The area to the left or below each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

${\begin{matrix} y \geq - \frac{2}{3} x + 2 \\ y > 2 x - 3 \end{matrix}$

${\begin{matrix} x - y > 1 \\ y < - \frac{1}{4} x + 3 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of x – y is greater than 1 and y is less than –(1/4)x + 3. The area to the right or below each line is shaded different colors with the overlapping area also shaded a different color. Both lines are dotted.

${\begin{matrix} x + 2 y < 4 \\ y < x - 2 \end{matrix}$

${\begin{matrix} 3 x - y \leq 6 \\ y \geq - \frac{1}{2} x \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of 3x – y is less than or equal to 6 and y is greater than or equal to –(1/2)x. The area to the right or above each line is shaded different colors with the overlapping area also shaded a different color.

${\begin{matrix} 2 x + 4 y \geq 8 \\ y \leq \frac{3}{4} x \end{matrix}$

${\begin{matrix} 2 x - 5 y < 10 \\ 3 x + 4 y \geq 12 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of 2x – 5y is less than 10 and 3x +4y is greater than or equal to 12. The area to the right above each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

${\begin{matrix} 3 x - 2 y \leq 6 \\ −4 x - 2 y > 8 \end{matrix}$

${\begin{matrix} 2 x + 2 y > −4 \\ - x + 3 y \geq 9 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of 2x + 2y is greater than -4 and –x + 3y is greater than or equal to 9. The area to the right or above each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

${\begin{matrix} 2 x + y > −6 \\ - x + 2 y \geq −4 \end{matrix}$

${\begin{matrix} x - 2 y < 3 \\ y \leq 1 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of x – 2y is less than 3 and y is less than or equal to 1. The area to the left or below each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

${\begin{matrix} x - 3 y > 4 \\ y \leq - 1 \end{matrix}$

${\begin{matrix} y \geq - \frac{1}{2} x - 3 \\ x \leq 2 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of y is greater than or equal to (-1/2)x - 3 and x is less than or equal to 2. The area to the left or right of each line is shaded different colors with the overlapping area also shaded a different color.

${\begin{matrix} y \leq - \frac{2}{3} x + 5 \\ x \geq 3 \end{matrix}$

${\begin{matrix} y \geq \frac{3}{4} x - 2 \\ y < 2 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of y is greater than or equal to (3/4)x - 2 and y is less than 2. The area to the left or below each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

${\begin{matrix} y \leq - \frac{1}{2} x + 3 \\ y < 1 \end{matrix}$

${\begin{matrix} 3 x - 4 y < 8 \\ x < 1 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of 3x – 4y is less than 8 and x is less than 1. The area to the left of each line is shaded different colors with the overlapping area also shaded a different color. Both lines are dotted.

${\begin{matrix} −3 x + 5 y > 10 \\ x > −1 \end{matrix}$

${\begin{matrix} x \geq 3 \\ y \leq 2 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of x is greater than or equal to 3 and y less than or equal to 2. The area to the right or below each line is shaded different colors with the overlapping area also shaded a different color.

${\begin{matrix} x \leq - 1 \\ y \geq 3 \end{matrix}$

${\begin{matrix} 2 x + 4 y > 4 \\ y \leq - \frac{1}{2} x - 2 \end{matrix}$

Solution

No solution
This figure shows a graph on an x y-coordinate plane of 2x + 4y is greater than 4 and y is less than or equal to (-1/2)x - 2. The area to the left or right of each line is shaded different colors. There is no area where the shaded areas overlap. One line is dotted.

${\begin{matrix} x - 3 y \geq 6 \\ y > \frac{1}{3} x + 1 \end{matrix}$

${\begin{matrix} −2 x + 6 y < 0 \\ 6 y > 2 x + 4 \end{matrix}$

Solution

No solution
This figure shows a graph on an x y-coordinate plane of -2x + 6y is less than 0 and 6y is greater than 2x + 4. The area to the left or right of each line is shaded different colors. There is no area where the shaded areas overlap. Both lines are dotted.

${\begin{matrix} −3 x + 6 y > 12 \\ 4 y \leq 2 x - 4 \end{matrix}$

${\begin{matrix} y \geq −3 x + 2 \\ 3 x + y > 5 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of y is greater than or equal to -3x + 2 and 3x + y is greater than 5. The area to the right of each line is shaded different colors. One line is within the shaded area of the other. One line is dotted.

${\begin{matrix} y \geq \frac{1}{2} x - 1 \\ −2 x + 4 y \geq 4 \end{matrix}$

${\begin{matrix} y \leq - \frac{1}{4} x - 2 \\ x + 4 y < 6 \end{matrix}$

Solution

$x + 4 y < 6$
This figure shows a graph on an x y-coordinate plane of y is less than or equal to (negative 1/4)x – 2 and x + 4y is less than 6. The area below each line is shaded different colors. One line is within the shaded area of the other. One line is dotted.

${\begin{matrix} y \geq 3 x - 1 \\ −3 x + y > −4 \end{matrix}$

${\begin{matrix} 3 y > x + 2 \\ −2 x + 6 y > 8 \end{matrix}$

Solution

$−2 x + 6 y > 8$
This figure shows a graph on an x y-coordinate plane of 3y is greater than x + 2 and -2x + 6y is greater than 8. The area above each line is shaded different colors. One line is within the shaded area of the other. Both lines are dotted.

${\begin{matrix} y < \frac{3}{4} x - 2 \\ −3 x + 4 y < 7 \end{matrix}$

Solve Applications of Systems of Inequalities

In the following exercises, translate to a system of inequalities and solve.

Caitlyn sells her drawings at the county fair. She wants to sell at least 60 drawings and has portraits and landscapes. She sells the portraits for $15 and the landscapes for $10. She needs to sell at least $800 worth of drawings in order to earn a profit.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Will she make a profit if she sells 20 portraits and 35 landscapes?
ⓓ Will she make a profit if she sells 50 portraits and 20 landscapes?

Solution

ⓐ ${\begin{matrix} p + l \geq 60 \\ 15 p + 10 l \geq 800 \end{matrix}$
ⓑ
ⓒ No
ⓓ Yes

Jake does not want to spend more than $50 on bags of fertilizer and peat moss for his garden. Fertilizer costs $2 a bag and peat moss costs $5 a bag. Jake’s van can hold at most 20 bags.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can he buy 15 bags of fertilizer and 4 bags of peat moss?
ⓓ Can he buy 10 bags of fertilizer and 10 bags of peat moss?

Reiko needs to mail her Christmas cards and packages and wants to keep her mailing costs to no more than $500. The number of cards is at least 4 more than twice the number of packages. The cost of mailing a card (with pictures enclosed) is $3 and for a package the cost is $7.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can she mail 60 cards and 26 packages?
ⓓ Can she mail 90 cards and 40 packages?

Solution

ⓐ ${\begin{matrix} 7 p + 3 c \leq 500 \\ p \geq 2 c + 4 \end{matrix}$
ⓑ
ⓒ Yes
ⓓ No

Juan is studying for his final exams in Chemistry and Algebra. He knows he only has 24 hours to study, and it will take him at least three times as long to study for Algebra than Chemistry.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can he spend 4 hours on Chemistry and 20 hours on Algebra?
ⓓ Can he spend 6 hours on Chemistry and 18 hours on Algebra?

Jocelyn is pregnant and needs to eat at least 500 more calories a day than usual. When buying groceries one day with a budget of $15 for the extra food, she buys bananas that have 90 calories each and chocolate granola bars that have 150 calories each. The bananas cost $0.35 each and the granola bars cost $2.50 each.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could she buy 5 bananas and 6 granola bars?
ⓓ Could she buy 3 bananas and 4 granola bars?

Solution

ⓐ ${\begin{matrix} 90 b + 150 g \geq 500 \\ 0.35 b + 2.50 g \leq 15 \end{matrix}$
ⓑ
ⓒ No
ⓓ Yes

Mark is attempting to build muscle mass and so he needs to eat at least an additional 80 grams of protein a day. A bottle of protein water costs $3.20 and a protein bar costs $1.75. The protein water supplies 27 grams of protein and the bar supplies 16 gram. If he has $ 10 dollars to spend

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he buy 3 bottles of protein water and 1 protein bar?
ⓓ Could he buy no bottles of protein water and 5 protein bars?

Jocelyn desires to increase both her protein consumption and caloric intake. She desires to have at least 35 more grams of protein each day and no more than an additional 200 calories daily. An ounce of cheddar cheese has 7 grams of protein and 110 calories. An ounce of parmesan cheese has 11 grams of protein and 22 calories.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could she eat 1 ounce of cheddar cheese and 3 ounces of parmesan cheese?
ⓓ Could she eat 2 ounces of cheddar cheese and 1 ounce of parmesan cheese?

Solution

ⓐ ${\begin{matrix} 7 c + 11 p \geq 35 \\ 110 c + 22 p \leq 200 \end{matrix}$
ⓑ
ⓒ Yes
ⓓ No

Mark is increasing his exercise routine by running and walking at least 4 miles each day. His goal is to burn a minimum of 1,500 calories from this exercise. Walking burns 270 calories/mile and running burns 650 calories.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he meet his goal by walking 3 miles and running 1 mile?
ⓓ Could he meet his goal by walking 2 miles and running 2 mile?

Everyday Math

Tickets for an American Baseball League game for 3 adults and 3 children cost less than $75, while tickets for 2 adults and 4 children cost less than $62.

ⓐ Write a system of inequalities to model this problem.
ⓑ Graph the system.
ⓒ Could the tickets cost $20 for adults and $8 for children?
ⓓ Could the tickets cost $15 for adults and $5 for children?

Solution

ⓐ ${\begin{matrix} 3 a + 3 c < 75 \\ 2 a + 4 c < 62 \end{matrix}$
ⓑ
ⓒ No
ⓓ Yes

Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and $4 per adult. Evening tickets cost $6 per child and $8 per adult. They plan on spending no more than $80 on the matinee tickets and no more than $100 on the evening tickets.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could they take 9 children and 4 adults to both shows?
ⓓ Could they take 8 children and 5 adults to both shows?

Writing Exercises

Graph the inequality $x - y \geq 3$ . How do you know which side of the line $x - y = 3$ should be shaded?

Solution

Answers will vary.

Graph the system ${\begin{matrix} x + 2 y \leq 6 \\ y \geq - \frac{1}{2} x - 4 \end{matrix}$ . What does the solution mean?

Self Check

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

This table has four columsn and four rows. The columns are labeled, “I can………,” “confidently.” “with some help.” “no – I don’t get it!” The only rows filled in are under the “I can……...” column. The rows say, “determine whether an ordered pair is a solution of a system of linear inequalities.” “solve a system of linear inequalities by graphing.” and “solving applications of systmes of inequalities.”

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

Chapter 5 Review Exercises

Solve Systems of Equations by Graphing

Determine Whether an Ordered Pair is a Solution of a System of Equations.

In the following exercises, determine if the following points are solutions to the given system of equations.

${\begin{matrix} x + 3 y = −9 \\ 2 x - 4 y = 12 \end{matrix}$

ⓐ $(−3, −2)$ ⓑ $(0, −3)$

Solution

ⓐ no ⓑ yes

${\begin{matrix} x + y = 8 \\ y = x - 4 \end{matrix}$

ⓐ $(6, 2)$ ⓑ $(9, −1)$

Solve a System of Linear Equations by Graphing

In the following exercises, solve the following systems of equations by graphing.

${\begin{matrix} 3 x + y = 6 \\ x + 3 y = −6 \end{matrix}$

Solution

$(3, −3)$
This figure shows a graph on an x y-coordinate plane of 3x plus y = 6 and x plus 3y = negative 6.

${\begin{matrix} y = x - 2 \\ y = −2 x - 2 \end{matrix}$

${\begin{matrix} 2 x - y = 6 \\ y = 4 \end{matrix}$

Solution

$(5, 4)$
This figure shows a graph on an x y-coordinate plane of 2x – y = 6 and y = 4.

${\begin{matrix} x + 4 y = −1 \\ x = 3 \end{matrix}$

${\begin{matrix} 2 x - y = 5 \\ 4 x - 2 y = 10 \end{matrix}$

Solution

coincident lines
This figure shows a graph on an x y-coordinate plane of 2x – y = 5 and 4x – 2y = 10.

${\begin{matrix} - x + 2 y = 4 \\ y = \frac{1}{2} x - 3 \end{matrix}$

Determine the Number of Solutions of a Linear System

In the following exercises, without graphing determine the number of solutions and then classify the system of equations.

${\begin{matrix} y = \frac{2}{5} x + 2 \\ −2 x + 5 y = 10 \end{matrix}$

Solution

infinitely many solutions, consistent system, dependent equations

${\begin{matrix} 3 x + 2 y = 6 \\ y = −3 x + 4 \end{matrix}$

${\begin{matrix} 5 x - 4 y = 0 \\ y = \frac{5}{4} x - 5 \end{matrix}$

Solution

no solutions, inconsistent system, independent equations

${\begin{matrix} y = - \frac{3}{4} x + 1 \\ 6 x + 8 y = 8 \end{matrix}$

Solve Applications of Systems of Equations by Graphing

LaVelle is making a pitcher of caffe mocha. For each ounce of chocolate syrup, she uses five ounces of coffee. How many ounces of chocolate syrup and how many ounces of coffee does she need to make 48 ounces of caffe mocha?

Solution

LaVelle needs 8 ounces of chocolate syrup and 40 ounces of coffee.

Eli is making a party mix that contains pretzels and chex. For each cup of pretzels, he uses three cups of chex. How many cups of pretzels and how many cups of chex does he need to make 12 cups of party mix?

Solve Systems of Equations by Substitution

Solve a System of Equations by Substitution

In the following exercises, solve the systems of equations by substitution.

${\begin{matrix} 3 x - y = −5 \\ y = 2 x + 4 \end{matrix}$

Solution

$(−1, 2)$

${\begin{matrix} 3 x - 2 y = 2 \\ y = \frac{1}{2} x + 3 \end{matrix}$

${\begin{matrix} x - y = 0 \\ 2 x + 5 y = −14 \end{matrix}$

Solution

$(−2, −2)$

${\begin{matrix} y = −2 x + 7 \\ y = \frac{2}{3} x - 1 \end{matrix}$

${\begin{matrix} y = −5 x \\ 5 x + y = 6 \end{matrix}$

Solution

no solution

${\begin{matrix} y = - \frac{1}{3} x + 2 \\ x + 3 y = 6 \end{matrix}$

Solve Applications of Systems of Equations by Substitution

In the following exercises, translate to a system of equations and solve.

The sum of two number is 55. One number is 11 less than the other. Find the numbers.

Solution

The numbers are 22 and 33.

The perimeter of a rectangle is 128. The length is 16 more than the width. Find the length and width.

The measure of one of the small angles of a right triangle is 2 less than 3 times the measure of the other small angle. Find the measure of both angles.

Solution

The measures are 23 degrees and 67 degrees.

Gabriela works for an insurance company that pays her a salary of $32,000 plus a commission of $100 for each policy she sells. She is considering changing jobs to a company that would pay a salary of $40,000 plus a commission of $80 for each policy sold. How many policies would Gabriela need to sell to make the total pay the same?

Solve Systems of Equations by Elimination

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

${\begin{matrix} x + y = 12 \\ x - y = −10 \end{matrix}$

Solution

$(1, 11)$

${\begin{matrix} 4 x + 2 y = 2 \\ −4 x - 3 y = −9 \end{matrix}$

${\begin{matrix} 3 x - 8 y = 20 \\ x + 3 y = 1 \end{matrix}$

Solution

$(4, −1)$

${\begin{matrix} 3 x - 2 y = 6 \\ 4 x + 3 y = 8 \end{matrix}$

${\begin{matrix} 9 x + 4 y = 2 \\ 5 x + 3 y = 5 \end{matrix}$

Solution

$(−2, 5)$

${\begin{matrix} - x + 3 y = 8 \\ 2 x - 6 y = −20 \end{matrix}$

Solve Applications of Systems of Equations by Elimination

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is $−90$ . Their difference is $16$ . Find the numbers.

Solution

The numbers are $−37$ and $−53$ .

Omar stops at a donut shop every day on his way to work. Last week he had 8 donuts and 5 cappuccinos, which gave him a total of 3,000 calories. This week he had 6 donuts and 3 cappuccinos, which was a total of 2,160 calories. How many calories are in one donut? How many calories are in one cappuccino?

Choose the Most Convenient Method to Solve a System of Linear Equations

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

${\begin{matrix} 6 x - 5 y = 27 \\ 3 x + 10 y = −24 \end{matrix}$

Solution

elimination

${\begin{matrix} y = 3 x - 9 \\ 4 x - 5 y = 23 \end{matrix}$

Solve Applications with Systems of Equations

Translate to a System of Equations

In the following exercises, translate to a system of equations. Do not solve the system.

The sum of two numbers is $−32$ . One number is two less than twice the other. Find the numbers.

Solution

${\begin{matrix} x + y = −32 \\ x = 2 y - 2 \end{matrix}$ The numbers are –10 and –22

Four times a number plus three times a second number is $−9$ . Twice the first number plus the second number is three. Find the numbers.

Last month Jim and Debbie earned $7,200. Debbie earned $1,600 more than Jim earned. How much did they each earn?

Solution

${\begin{matrix} j + d = 7200 \\ d = j + 1600 \end{matrix}$ Debbie earned $4400 and Jim earned $2800

Henri has $24,000 invested in stocks and bonds. The amount in stocks is $6,000 more than three times the amount in bonds. How much is each investment?

Solve Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

Pam is 3 years older than her sister, Jan. The sum of their ages is 99. Find their ages.

Solution

Pam is 51 and Jan is 48.

Mollie wants to plant 200 bulbs in her garden. She wantsall irises and tulips. She wants to plant three times as many tulips as irises. How many irises and how many tulips should she plant?

Solve Geometry Applications

In the following exercises, translate to a system of equations and solve.

The difference of two supplementary angles is 58 degrees. Find the measures of the angles.

Solution

The measures are 119 degrees and 61 degrees.

Two angles are complementary. The measure of the larger angle is five more than four times the measure of the smaller angle. Find the measures of both angles.

Becca is hanging a 28 foot floral garland on the two sides and top of a pergola to prepare for a wedding. The height is four feet less than the width. Find the height and width of the pergola.

Solution

The pergola is 8 feet high and 12 feet wide.

The perimeter of a city rectangular park is 1428 feet. The length is 78 feet more than twice the width. Find the length and width of the park.

Solve Uniform Motion Applications

In the following exercises, translate to a system of equations and solve.

Sheila and Lenore were driving to their grandmother’s house. Lenore left one hour after Sheila. Sheila drove at a rate of 45 mph, and Lenore drove at a rate of 60 mph. How long will it take for Lenore to catch up to Sheila?

Solution

It will take Lenore 3 hours.

Bob left home, riding his bike at a rate of 10 miles per hour to go to the lake. Cheryl, his wife, left 45 minutes $(\frac{3}{4}$ hour) later, driving her car at a rate of 25 miles per hour. How long will it take Cheryl to catch up to Bob?

Marcus can drive his boat 36 miles down the river in three hours but takes four hours to return upstream. Find the rate of the boat in still water and the rate of the current.

Solution

The rate of the boat is 10.5 mph. The rate of the current is 1.5 mph.

A passenger jet can fly 804 miles in 2 hours with a tailwind but only 776 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solve Mixture Applications with Systems of Equations

Solve Mixture Applications

In the following exercises, translate to a system of equations and solve.

Lynn paid a total of $2,780 for 261 tickets to the theater. Student tickets cost $10 and adult tickets cost $15. How many student tickets and how many adult tickets did Lynn buy?

Solution

Lynn bought 227 student tickets and 34 adult tickets.

Priam has dimes and pennies in a cup holder in his car. The total value of the coins is $4.21. The number of dimes is three less than four times the number of pennies. How many dimes and how many pennies are in the cup?

Yumi wants to make 12 cups of party mix using candies and nuts. Her budget requires the party mix to cost her $1.29 per cup. The candies are $2.49 per cup and the nuts are $0.69 per cup. How many cups of candies and how many cups of nuts should she use?

Solution

Yumi should use 4 cups of candies and 8 cups of nuts.

A scientist needs 70 liters of a 40% solution of alcohol. He has a 30% and a 60% solution available. How many liters of the 30% and how many liters of the 60% solutions should he mix to make the 40% solution?

Solve Interest Applications

In the following exercises, translate to a system of equations and solve.

Jack has $12,000 to invest and wants to earn 7.5% interest per year. He will put some of the money into a savings account that earns 4% per year and the rest into CD account that earns 9% per year. How much money should he put into each account?

Solution

Jack should put $3600 into savings and $8400 into the CD.

When she graduates college, Linda will owe $43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was $1585. What is the amount of each loan?

Graphing Systems of Linear Inequalities

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

In the following exercises, determine whether each ordered pair is a solution to the system.

${\begin{matrix} 4 x + y > 6 \\ 3 x - y \leq 12 \end{matrix}$

ⓐ $(2, −1)$ ⓑ $(3, −2)$

Solution

ⓐ yes ⓑ yes

${\begin{matrix} y > \frac{1}{3} x + 2 \\ x - \frac{1}{4} y \leq 10 \end{matrix}$

ⓐ $(6, 5)$ ⓑ $(15, 8)$

Solve a System of Linear Inequalities by Graphing

In the following exercises, solve each system by graphing.

${\begin{matrix} y < 3 x + 1 \\ y \geq - x - 2 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of y is less than 3x + 1 and y is greater than or equal to -x - 2. The area to the right of each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

${\begin{matrix} x - y > −1 \\ y < \frac{1}{3} x - 2 \end{matrix}$

${\begin{matrix} 2 x - 3 y < 6 \\ 3 x + 4 y \geq 12 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane of 2x – 3y is less than 6 and 3x + 4y is greater than or equal to 12. The area to the left or right of each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

${\begin{matrix} y \leq - \frac{3}{4} x + 1 \\ x \geq −5 \end{matrix}$

${\begin{matrix} x + 3 y < 5 \\ y \geq - \frac{1}{3} x + 6 \end{matrix}$

Solution

No solution
This figure shows a graph on an x y-coordinate plane of x + 3y is less than 5 and y is greater than or equal to -(1/3)x + 6. The area to the above or below each line is shaded different colors. There is no overlapping shaded area. One line is dotted.

${\begin{matrix} y \geq 2 x - 5 \\ −6 x + 3 y > −4 \end{matrix}$

Solve Applications of Systems of Inequalities

In the following exercises, translate to a system of inequalities and solve.

Roxana makes bracelets and necklaces and sells them at the farmers’ market. She sells the bracelets for $12 each and the necklaces for $18 each. At the market next weekend she will have room to display no more than 40 pieces, and she needs to sell at least $500 worth in order to earn a profit.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Should she display 26 bracelets and 14 necklaces?
ⓓ Should she display 39 bracelets and 1 necklace?

Solution

ⓐ ${\begin{matrix} b + n \leq 40 \\ 12 b + 18 n \geq 500 \end{matrix}$
ⓑ
This figure shows a graph on an x y-coordinate plane of b + n is less than or equal to 40 and 12b + 18n is greater than or equal to 500. The area to the left or right of each line is shaded different colors with the overlapping area also shaded a different color.
ⓒ yes
ⓓ no

Annie has a budget of $600 to purchase paperback books and hardcover books for her classroom. She wants the number of hardcover to be at least 5 more than three times the number of paperback books. Paperback books cost $4 each and hardcover books cost $15 each.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can she buy 8 paperback books and 40 hardcover books?
ⓓ Can she buy 10 paperback books and 37 hardcover books?

Practice Test

For the following set of equations, determine if each ordered pair is a solution.

${\begin{matrix} x - 4 y = −8 \\ 2 x + 5 y = 10 \end{matrix}$

ⓐ $(0, 2)$ ⓑ $(4, 3)$

Solution

ⓐ yes ⓑ no

In the following exercises, solve the following systems by graphing.

${\begin{matrix} x - y = 5 \\ x + 2 y = −4 \end{matrix}$

${\begin{matrix} x - y > −2 \\ y \leq 3 x + 1 \end{matrix}$

Solution

This figure shows a graph on an x y-coordinate plane x – y is greater than -2 and y is less than or equal to 3x + 1. The area to the left of each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

In the following exercises, solve each system of equations. Use either substitution or elimination.

${\begin{matrix} 3 x - 2 y = 3 \\ y = 2 x - 1 \end{matrix}$

${\begin{matrix} x + y = −3 \\ x - y = 11 \end{matrix}$

Solution

$(4, −7)$

${\begin{matrix} 4 x - 3 y = 7 \\ 5 x - 2 y = 0 \end{matrix}$

${\begin{matrix} y = - \frac{4}{5} x + 1 \\ 8 x + 10 y = 10 \end{matrix}$

Solution

infinitely many solutions

${\begin{matrix} 2 x + 3 y = 12 \\ −4 x + 6 y = −16 \end{matrix}$

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is −24. One number is 104 less than the other. Find the numbers.

Solution

The numbers are 40 and –64

Ramon wants to plant cucumbers and tomatoes in his garden. He has room for 16 plants, and he wants to plant three times as many cucumbers as tomatoes. How many cucumbers and how many tomatoes should he plant?

Two angles are complementary. The measure of the larger angle is six more than twice the measure of the smaller angle. Find the measures of both angles.

Solution

The measures of the angles are 28 degrees and 62 degrees.

On Monday, Lance ran for 30 minutes and swam for 20 minutes. His fitness app told him he had burned 610 calories. On Wednesday, the fitness app told him he burned 695 calories when he ran for 25 minutes and swam for 40 minutes. How many calories did he burn for one minute of running? How many calories did he burn for one minute of swimming?

Kathy left home to walk to the mall, walking quickly at a rate of 4 miles per hour. Her sister Abby left home 15 minutes later and rode her bike to the mall at a rate of 10 miles per hour. How long will it take Abby to catch up to Kathy?

Solution

It will take Kathy $\frac{1}{6}$ of an hour (or 10 minutes).

It takes $5 \frac{1}{2}$ hours for a jet to fly 2,475 miles with a headwind from San Jose, California to Lihue, Hawaii. The return flight from Lihue to San Jose with a tailwind, takes 5 hours. Find the speed of the jet in still air and the speed of the wind.

Liz paid $160 for 28 tickets to take the Brownie troop to the science museum. Children’s tickets cost $5 and adult tickets cost $9. How many children’s tickets and how many adult tickets did Liz buy?

Solution

Liz bought 23 children’s tickets and 5 adult tickets.

A pharmacist needs 20 liters of a 2% saline solution. He has a 1% and a 5% solution available. How many liters of the 1% and how many liters of the 5% solutions should she mix to make the 2% solution?

Translate to a system of inequalities and solve.

Andi wants to spend no more than $50 on Halloween treats. She wants to buy candy bars that cost $1 each and lollipops that cost $0.50 each, and she wants the number of lollipops to be at least three times the number of candy bars.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can she buy 20 candy bars and 70 lollipops?
ⓓ Can she buy 15 candy bars and 65 lollipops?

Solution

ⓐ ${\begin{matrix} C + 0.5 L \leq 50 \\ L \geq 3 C \end{matrix}$
ⓑ
This figure shows a graph on an x y-coordinate plane of C + 0.5L is less than or equal to 50 and L is greater than or equal to 3C. The area to the left or right of each line is shaded different colors with the overlapping area also shaded a different color.
ⓒ No
ⓓ Yes

system of linear inequalities

Two or more linear inequalities grouped together form a system of linear inequalities.

Introduction

This is a photo of a bridge at sunset. — Architects use polynomials to design curved shapes such as this suspension bridge, the Silver Jubilee bridge in Halton, England.

We have seen that the graphs of linear equations are straight lines. Graphs of other types of equations, called polynomial equations, are curves, like the outline of this suspension bridge. Architects use polynomials to design the shape of a bridge like this and to draw the blueprints for it. Engineers use polynomials to calculate the stress on the bridge’s supports to ensure they are strong enough for the intended load. In this chapter, you will explore operations with and properties of polynomials.

Add and Subtract Polynomials

Identify Polynomials, Monomials, Binomials and Trinomials

You have 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 $a x^{m}$ , where $a$ is a constant and $m$ is a whole number, it is called a monomial. Some examples of monomial are $8, −2 x^{2}, 4 y^{3}, and 11 z^{7}$ .

A monomial, or two or more monomials combined by addition or subtraction, is a polynomial. Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term. A binomial has exactly two terms, and a trinomial has exactly three terms. There are no special names for polynomials with more than three terms.

Here are some examples of polynomials.

$\begin{matrix} Polynomial & b + 1 & 4 y^{2} - 7 y + 2 & 4 x^{4} + x^{3} + 8 x^{2} - 9 x + 1 \\ Monomial & 14 & 8 y^{2} & −9 x^{3} y^{5} & −13 \\ Binomial & a + 7 & 4 b - 5 & y^{2} - 16 & 3 x^{3} - 9 x^{2} \\ Trinomial & x^{2} - 7 x + 12 & 9 y^{2} + 2 y - 8 & 6 m^{4} - m^{3} + 8 m & z^{4} + 3 z^{2} - 1 \end{matrix}$

Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials.

Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial.

ⓐ $4 y^{2} - 8 y - 6$
ⓑ $−5 a^{4} b^{2}$
ⓒ $2 x^{5} - 5 x^{3} - 9 x^{2} + 3 x + 4$
ⓓ $13 - 5 m^{3}$
ⓔ $q$

Solution

Examples of polynomials, showing their number of terms and classification type.
	Polynomial	Number of terms	Type
ⓐ	$4 y^{2} - 8 y - 6$	$3$	Trinomial
ⓑ	$−5 a^{4} b^{2}$	$1$	Monomial
ⓒ	$2 x^{5} - 5 x^{3} - 9 x^{2} + 3 x + 4$	$5$	Polynomial
ⓓ	$13 - 5 m^{3}$	$2$	Binomial
ⓔ	$q$	$1$	Monomial

Determine the Degree of Polynomials

The degree of a polynomial and the degree of its terms are determined by the exponents of the variable.

A monomial that has no variable, just a constant, is a special case. The degree of a constant is 0—it has no variable.

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.

This table has 11 rows and 5 columns. The first column is a header column, and it names each row. The first row is named “Monomial,” and each cell in this row contains a different monomial. The second row is named “Degree,” and each cell in this row contains the degree of the monomial above it. The degree of 14 is 0, the degree of 8y squared is 2, the degree of negative 9x cubed y to the fifth power is 8, and the degree of negative 13a is 1. The third row is named “Binomial,” and each cell in this row contains a different binomial. The fourth row is named “Degree of each term,” and each cell contains the degrees of the two terms in the binomial above it. The fifth row is named “Degree of polynomial,” and each cell contains the degree of the binomial as a whole.” The degrees of the terms in a plus 7 are 0 and 1, and the degree of the whole binomial is 1. The degrees of the terms in 4b squared minus 5b are 2 and 1, and the degree of the whole binomial is 2. The degrees of the terms in x squared y squared minus 16 are 4 and 0, and the degree of the whole binomial is 4. The degrees of the terms in 3n cubed minus 9n squared are 3 and 2, and the degree of the whole binomial is 3. The sixth row is named “Trinomial,” and each cell in this row contains a different trinomial. The seventh row is named “Degree of each term,” and each cell contains the degrees of the three terms in the trinomial above it. The eighth row is named “Degree of polynomial,” and each cell contains the degree of the trinomial as a whole. The degrees of the terms in x squared minus 7x plus 12 are 2, 1, and 0, and the degree of the whole trinomial is 2. The degrees of the terms in 9a squared plus 6ab plus b squared are 2, 2, and 2, and the degree of the trinomial as a whole is 2. The degrees of the terms in 6m to the fourth power minus m cubed n squared plus 8mn to the fifth power are 4, 5, and 6, and the degree of the whole trinomial is 6. The degrees of the terms in z to the fourth power plus 3z squared minus 1 are 4, 2, and 0, and the degree of the whole trinomial is 4. The ninth row is named “Polynomial,” and each cell contains a different polynomial. The tenth row is named “Degree of each term,” and the eleventh row is named “Degree of polynomial.” The degrees of the terms in b plus 1 are 1 and 0, and the degree of the whole polynomial is 1. The degrees of the terms in 4y squared minus 7y plus 2 are 2, 1, and 0, and the degree of the whole polynomial is 2. The degrees of the terms in 4x to the fourth power plus x cubed plus 8x squared minus 9x plus 1 are 4, 3, 2, 1, and 0, and the degree of the whole polynomial is 4.

A polynomial is in standard form when the terms of a polynomial are written in descending order of degrees. Get in the habit of writing the term with the highest degree first.

Find the degree of the following polynomials.

ⓐ $10 y$
ⓑ $4 x^{3} - 7 x + 5$
ⓒ $−15$
ⓓ $−8 b^{2} + 9 b - 2$
ⓔ $8 x y^{2} + 2 y$

Solution

This table demonstrates how to determine the degree of various mathematical expressions, including monomials, polynomials, and constants.
ⓐ The exponent of $y$ is one. $y = y^{1}$	$10 y$ The degree is 1.
ⓑ The highest degree of all the terms is 3.	$4 x^{3} - 7 x + 5$ The degree is 3.
ⓒ The degree of a constant is 0.	$−15$ The degree is 0.
ⓓ The highest degree of all the terms is 2.	$−8 b^{2} + 9 b - 2$ The degree is 2.
ⓔ The highest degree of all the terms is 3.	$8 x y^{2} + 2 y$ The degree is 3.

Add and Subtract Monomials

You have learned how to simplify expressions by combining like terms. Remember, like terms must have the same variables with the same exponent. Since monomials are terms, adding and subtracting monomials is the same as combining like terms. If the monomials are like terms, we just combine them by adding or subtracting the coefficient.

Add: $25 y^{2} + 15 y^{2}$ .

Solution

This table demonstrates the process of combining like terms in an algebraic expression.
	$25 y^{2} + 15 y^{2}$
Combine like terms.	$40 y^{2}$

Subtract: $16 p - (−7 p)$ .

Solution

This table demonstrates the step-by-step simplification of an algebraic expression by combining like terms.
	$16 p - (−7 p)$
Combine like terms.	$23 p$

Remember that like terms must have the same variables with the same exponents.

Simplify: $c^{2} + 7 d^{2} - 6 c^{2}$ .

Solution

Demonstrates simplifying an algebraic expression by combining like terms.
	$c^{2} + 7 d^{2} - 6 c^{2}$
Combine like terms.	$−5 c^{2} + 7 d^{2}$

Simplify: $u^{2} v + 5 u^{2} - 3 v^{2}$ .

Solution

An algebraic expression and a note on its like terms.
	$u^{2} v + 5 u^{2} - 3 v^{2}$
There are no like terms to combine.	$u^{2} v + 5 u^{2} - 3 v^{2}$

Add and Subtract Polynomials

We can think of adding and subtracting polynomials as just adding and subtracting a series of monomials. Look for the like terms—those with the same variables and the same exponent. The Commutative Property allows us to rearrange the terms to put like terms together.

Find the sum: $(5 y^{2} - 3 y + 15) + (3 y^{2} - 4 y - 11) .$

Solution

Identify like terms.
Rearrange to get the like terms together.
Combine like terms.

Find the difference: $(9 w^{2} - 7 w + 5) - (2 w^{2} - 4) .$

Solution


Distribute and identify like terms.
Rearrange the terms.
Combine like terms.

Subtract: $(c^{2} - 4 c + 7)$ from $(7 c^{2} - 5 c + 3)$ .

Solution



Distribute and identify like terms.
Rearrange the terms.
Combine like terms.

Find the sum: $(u^{2} - 6 u v + 5 v^{2}) + (3 u^{2} + 2 u v)$ .

Solution

Step-by-step addition of two polynomial expressions, showing distribution, rearrangement, and combining like terms for simplification.
	$(u^{2} - 6 u v + 5 v^{2}) + (3 u^{2} + 2 u v)$
Distribute.	$u^{2} - 6 u v + 5 v^{2} + 3 u^{2} + 2 u v$
Rearrange the terms, to put like terms together.	$u^{2} + 3 u^{2} - 6 u v + 2 u v + 5 v^{2}$
Combine like terms.	$4 u^{2} - 4 u v + 5 v^{2}$

Find the difference: $(p^{2} + q^{2}) - (p^{2} + 10 p q - 2 q^{2})$ .

Solution

Step-by-step algebraic simplification of the expression (p^2 + q^2) - (p^2 + 10pq - 2q^2).
	$(p^{2} + q^{2}) - (p^{2} + 10 p q - 2 q^{2})$
Distribute.	$p^{2} + q^{2} - p^{2} - 10 p q + 2 q^{2}$
Rearrange the terms, to put like terms together.	$p^{2} - p^{2} - 10 p q + q^{2} + 2 q^{2}$
Combine like terms.	$−10 p q + 3 q^{2}$

Simplify: $(a^{3} - a^{2} b) - (a b^{2} + b^{3}) + (a^{2} b + a b^{2})$ .

Solution

Step-by-step simplification of an algebraic expression through distribution, rearrangement, and combining like terms.
	$(a^{3} - a^{2} b) - (a b^{2} + b^{3}) + (a^{2} b + a b^{2})$
Distribute.	$a^{3} - a^{2} b - a b^{2} - b^{3} + a^{2} b + a b^{2}$
Rearrange the terms, to put like terms together.	$a^{3} - a^{2} b + a^{2} b - a b^{2} + a b^{2} - b^{3}$
Combine like terms.	$a^{3} - b^{3}$

Evaluate a Polynomial for a Given Value

We have already learned how to evaluate expressions. Since polynomials are expressions, we’ll follow the same procedures to evaluate a polynomial. We will substitute the given value for the variable and then simplify using the order of operations.

Evaluate $5 x^{2} - 8 x + 4$ when

ⓐ $x = 4$
ⓑ $x = −2$
ⓒ $x = 0$

Solution

ⓐ $x = 4$


Simplify the exponents.
Multiply.
Simplify.

ⓑ $x = −2$


Simplify the exponents.
Multiply.
Simplify.

ⓒ $x = 0$


Simplify the exponents.
Multiply.
Simplify.

The polynomial $−16 t^{2} + 250$ gives the height of a ball $t$ seconds after it is dropped from a 250 foot tall building. Find the height after $t = 2$ seconds.

Solution

Step-by-step evaluation of the expression -16t^2 + 250 at t=2, determining the height of a ball.
	$−16 t^{2} + 250$
Substitute $t = 2$ .	$−16 {(2)}^{2} + 250$
Simplify.	$−16 \cdot 4 + 250$
Simplify.	$−64 + 250$
Simplify.	$186$
	After 2 seconds the height of the ball is 186 feet.

The polynomial $6 x^{2} + 15 x y$ gives the cost, in dollars, of producing a rectangular container whose top and bottom are squares with side x feet and sides of height y feet. Find the cost of producing a box with $x = 4$ feet and $y = 6$ feet.

Solution



Simplify.
Simplify.
Simplify.
	The cost of producing the box is $456.

Key Concepts

Monomials
- A monomial is a term of the form $a x^{m}$ , where $a$ is a constant and $m$ is a whole number
Polynomials
- polynomial—A monomial, or two or more monomials combined by addition or subtraction is a polynomial.
- monomial—A polynomial with exactly one term is called a monomial.
- binomial—A polynomial with exactly two terms is called a binomial.
- trinomial—A polynomial with exactly three terms is called a trinomial.
Degree of a Polynomial
- The degree of a term is the sum of the exponents of its variables.
- The degree of a constant is 0.
- The degree of a polynomial is the highest degree of all its terms.

Practice Makes Perfect

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.

ⓐ $81 b^{5} - 24 b^{3} + 1$ ⓑ $5 c^{3} + 11 c^{2} - c - 8$ ⓒ $\frac{14}{15} y + \frac{1}{7}$
ⓓ 5
ⓔ $4 y + 17$

Solution

ⓐ trinomial ⓑ polynomial ⓒ binomial ⓓ monomial ⓔ binomial

ⓐ $x^{2} - y^{2}$ ⓑ $−13 c^{4}$ ⓒ $x^{2} + 5 x - 7$ ⓓ $x^{2} y^{2} - 2 x y + 8$ ⓔ 19

ⓐ $8 - 3 x$ ⓑ $z^{2} - 5 z - 6$ ⓒ $y^{3} - 8 y^{2} + 2 y - 16$ ⓓ $81 b^{5} - 24 b^{3} + 1$ ⓔ $−18$

Solution

ⓐ binomial ⓑ trinomial ⓒ polynomial ⓓ trinomial ⓔ monomial

ⓐ $11 y^{2}$ ⓑ $−73$ ⓒ $6 x^{2} - 3 x y + 4 x - 2 y + y^{2}$ ⓓ $4 y + 17$ ⓔ $5 c^{3} + 11 c^{2} - c - 8$

Determine the Degree of Polynomials

In the following exercises, determine the degree of each polynomial.

ⓐ $6 a^{2} + 12 a + 14$ ⓑ $18 x y^{2} z$ ⓒ $5 x + 2$ ⓓ $y^{3} - 8 y^{2} + 2 y - 16$ ⓔ $−24$

Solution

ⓐ 2 ⓑ 4 ⓒ 1 ⓓ 3 ⓔ 0

ⓐ $9 y^{3} - 10 y^{2} + 2 y - 6$ ⓑ $−12 p^{4}$ ⓒ $a^{2} + 9 a + 18$ ⓓ $20 x^{2} y^{2} - 10 a^{2} b^{2} + 30$ ⓔ 17

ⓐ $14 - 29 x$ ⓑ $z^{2} - 5 z - 6$ ⓒ $y^{3} - 8 y^{2} + 2 y - 16$ ⓓ $23 a b^{2} - 14$ ⓔ $−3$

Solution

ⓐ 1 ⓑ 2 ⓒ 3 ⓓ 3 ⓔ 0

ⓐ $62 y^{2}$ ⓑ 15 ⓒ $6 x^{2} - 3 x y + 4 x - 2 y + y^{2}$ ⓓ $10 - 9 x$ ⓔ $m^{4} + 4 m^{3} + 6 m^{2} + 4 m + 1$

Add and Subtract Monomials

In the following exercises, add or subtract the monomials.

${7x}^{2} + 5 x^{2}$

Solution

$12 x^{2}$

${4y}^{3} + 6 y^{3}$

$−12 w + 18 w$

Solution

$6 w$

$−3 m + 9 m$

$4a - 9 a$

Solution

$−5 a$

$− y - 5 y$

$28 x - (−12 x)$

Solution

$40 x$

$13 z - (−4 z)$

$−5 b - 17 b$

Solution

$−22 b$

$−10 x - 35 x$

$12 a + 5 b - 22 a$

Solution

$−10a + 5 b$

$14x - 3 y - 13 x$

$2 a^{2} + b^{2} - 6 a^{2}$

Solution

$−4 a^{2} + b^{2}$

$5 u^{2} + 4 v^{2} - 6 u^{2}$

$x y^{2} - 5 x - 5 y^{2}$

Solution

$x y^{2} - 5 x - 5 y^{2}$

$p q^{2} - 4 p - 3 q^{2}$

$a^{2} b - 4 a - 5 a b^{2}$

Solution

$a^{2} b - 4 a - 5 a b^{2}$

$x^{2} y - 3 x + 7 x y^{2}$

$12a + 8 b$

Solution

$12a + 8 b$

$19y + 5 z$

Add: $4 a, −3 b, −8 a$

Solution

$−4 a - 3 b$

Add: $4x, 3 y, −3 x$

Subtract $5 x^{6} from - 12 x^{6}$ .

Solution

$−17 x^{6}$

Subtract $2 p^{4} from - 7 p^{4}$ .

Add and Subtract Polynomials

In the following exercises, add or subtract the polynomials.

$(5 y^{2} + 12 y + 4) + (6 y^{2} - 8 y + 7)$

Solution

$11 y^{2} + 4 y + 11$

$(4 y^{2} + 10 y + 3) + (8 y^{2} - 6 y + 5)$

$(x^{2} + 6 x + 8) + (−4 x^{2} + 11 x - 9)$

Solution

$−3 x^{2} + 17 x - 1$

$(y^{2} + 9 y + 4) + (−2 y^{2} - 5 y - 1)$

$(8 x^{2} - 5 x + 2) + (3 x^{2} + 3)$

Solution

$11 x^{2} - 5 x + 5$

$(7 x^{2} - 9 x + 2) + (6 x^{2} - 4)$

$(5 a^{2} + 8) + (a^{2} - 4 a - 9)$

Solution

$6 a^{2} - 4 a - 1$

$(p^{2} - 6 p - 18) + (2 p^{2} + 11)$

$(4 m^{2} - 6 m - 3) - (2 m^{2} + m - 7)$

Solution

$2 m^{2} - 7 m + 4$

$(3 b^{2} - 4 b + 1) - (5 b^{2} - b - 2)$

$(a^{2} + 8 a + 5) - (a^{2} - 3 a + 2)$

Solution

$11 a + 3$

$(b^{2} - 7 b + 5) - (b^{2} - 2 b + 9)$

$(12 s^{2} - 15 s) - (s - 9)$

Solution

$12 s^{2} - 16 s + 9$

$(10 r^{2} - 20 r) - (r - 8)$

Subtract $(9 x^{2} + 2)$ from $(12 x^{2} - x + 6)$ .

Solution

$3 x^{2} - x + 4$

Subtract $(5 y^{2} - y + 12)$ from $(10 y^{2} - 8 y - 20)$ .

Subtract $(7 w^{2} - 4 w + 2)$ from $(8 w^{2} - w + 6)$ .

Solution

$w^{2} + 3 w + 4$

Subtract $(5 x^{2} - x + 12)$ from $(9 x^{2} - 6 x - 20)$ .

Find the sum of $(2 p^{3} - 8)$ and $(p^{2} + 9 p + 18)$ .

Solution

$2 p^{3} + p^{2} + 9 p + 10$

Find the sum of
$(q^{2} + 4 q + 13)$ and $(7 q^{3} - 3)$ .

Find the sum of $(8 a^{3} - 8 a)$ and $(a^{2} + 6 a + 12)$ .

Solution

$8 a^{3} + a^{2} - 2 a + 12$

Find the sum of
$(b^{2} + 5 b + 13)$ and $(4 b^{3} - 6)$ .

Find the difference of
$(w^{2} + w - 42)$ and
$(w^{2} - 10 w + 24)$ .

Solution

$11 w - 66$

Find the difference of
$(z^{2} - 3 z - 18)$ and
$(z^{2} + 5 z - 20)$ .

Find the difference of
$(c^{2} + 4 c - 33)$ and
$(c^{2} - 8 c + 12)$ .

Solution

$12 c - 45$

Find the difference of
$(t^{2} - 5 t - 15)$ and
$(t^{2} + 4 t - 17)$ .

$(7 x^{2} - 2 x y + 6 y^{2}) + (3 x^{2} - 5 x y)$

Solution

$10 x^{2} - 7 x y + 6 y^{2}$

$(−5 x^{2} - 4 x y - 3 y^{2}) + (2 x^{2} - 7 x y)$

$(7 m^{2} + m n - 8 n^{2}) + (3 m^{2} + 2 m n)$

Solution

$10 m^{2} + 3 m n - 8 n^{2}$

$(2 r^{2} - 3 r s - 2 s^{2}) + (5 r^{2} - 3 r s)$

$(a^{2} - b^{2}) - (a^{2} + 3 a b - 4 b^{2})$

Solution

$−3 a b + 3 b^{2}$

$(m^{2} + 2 n^{2}) - (m^{2} - 8 m n - n^{2})$

$(u^{2} - v^{2}) - (u^{2} - 4 u v - 3 v^{2})$

Solution

$4 u v + 2 v^{2}$

$(j^{2} - k^{2}) - (j^{2} - 8 j k - 5 k^{2})$

$(p^{3} - 3 p^{2} q) + (2 p q^{2} + 4 q^{3}) - (3 p^{2} q + p q^{2})$

Solution

$p^{3} - 6 p^{2} q + p q^{2} + 4 q^{3}$

$(a^{3} - 2 a^{2} b) + (a b^{2} + b^{3}) - (3 a^{2} b + 4 a b^{2})$

$(x^{3} - x^{2} y) - (4 x y^{2} - y^{3}) + (3 x^{2} y - x y^{2})$

Solution

$x^{3} + 2 x^{2} y - 5 x y^{2} + y^{3}$

$(x^{3} - 2 x^{2} y) - (x y^{2} - 3 y^{3}) - (x^{2} y - 4 x y^{2})$

Evaluate a Polynomial for a Given Value

In the following exercises, evaluate each polynomial for the given value.

Evaluate $8 y^{2} - 3 y + 2$ when:

ⓐ $y = 5$ ⓑ $y = −2$ ⓒ $y = 0$

Solution

ⓐ 187 ⓑ 40 ⓒ 2

Evaluate $5 y^{2} - y - 7$ when:

ⓐ $y = −4$ ⓑ $y = 1$ ⓒ $y = 0$

Evaluate $4 - 36 x$ when:

ⓐ $x = 3$ ⓑ $x = 0$ ⓒ $x = −1$

Solution

ⓐ −104 ⓑ 4 ⓒ 40

Evaluate $16 - 36 x^{2}$ when:

ⓐ $x = −1$ ⓑ $x = 0$ ⓒ $x = 2$

A painter drops a brush from a platform 75 feet high. The polynomial $−16 t^{2} + 75$ gives the height of the brush $t$ seconds after it was dropped. Find the height after $t = 2$ seconds.

Solution

A girl drops a ball off a cliff into the ocean. The polynomial $−16 t^{2} + 250$ gives the height of a ball $t$ seconds after it is dropped from a 250-foot tall cliff. Find the height after $t = 2$ seconds.

A manufacturer of stereo sound speakers has found that the revenue received from selling the speakers at a cost of p dollars each is given by the polynomial $−4 p^{2} + 420 p .$ Find the revenue received when $p = 60$ dollars.

Solution

$10,800

A manufacturer of the latest basketball shoes has found that the revenue received from selling the shoes at a cost of p dollars each is given by the polynomial $−4 p^{2} + 420 p .$ Find the revenue received when $p = 90$ dollars.

Everyday Math

Fuel Efficiency The fuel efficiency (in miles per gallon) of a car going at a speed of $x$ miles per hour is given by the polynomial $- \frac{1}{150} x^{2} + \frac{1}{3} x$ . Find the fuel efficiency when $x = 30 mph$ .

Solution

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 $0.06 x^{2} + 1.1 x$ . Find the stopping distance when $x = 40 mph$ .

Rental Cost The cost to rent a rug cleaner for $d$ days is given by the polynomial $5.50 d + 25$ . Find the cost to rent the cleaner for 6 days.

Solution

$58

Height of Projectile The height (in feet) of an object projected upward is given by the polynomial $−16 t^{2} + 60 t + 90$ where $t$ represents time in seconds. Find the height after $t = 2.5$ seconds.

Temperature Conversion The temperature in degrees Fahrenheit is given by the polynomial $\frac{9}{5} c + 32$ where $c$ represents the temperature in degrees Celsius. Find the temperature in degrees Fahrenheit when $c = 65 ° .$

Solution

149

Writing Exercises

Using your own words, explain the difference between a monomial, a binomial, and a trinomial.

Using your own words, explain the difference between a polynomial with five terms and a polynomial with a degree of 5.

Solution

Answers will vary.

Ariana thinks the sum $6 y^{2} + 5 y^{4}$ is $11 y^{6}$ . What is wrong with her reasoning?

Jonathan thinks that $\frac{1}{3}$ and $\frac{1}{x}$ are both monomials. What is wrong with his reasoning?

Solution

Answers will vary.

Self Check

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

ⓑ If most of your checks were:

binomial

A binomial is a polynomial with exactly two terms.

degree of a constant

The degree of any constant is 0.

degree of a polynomial

The degree of a polynomial is the highest degree of all its terms.

degree of a term

The degree of a term is the exponent of its variable.

monomial

A monomial is a term of the form

a x^{m}

, where

a

is a constant and

m

is a whole number; a monomial has exactly one term.

polynomial

A polynomial is a monomial, or two or more monomials combined by addition or subtraction.

standard form

A polynomial is in standard form when the terms of a polynomial are written in descending order of degrees.

trinomial

A trinomial is a polynomial with exactly three terms.

Use Multiplication Properties of Exponents

Simplify Expressions with Exponents

Remember that an exponent indicates repeated multiplication of the same quantity. For example, $2^{4}$ means to multiply 2 by itself 4 times, so $2^{4}$ means $2 \cdot 2 \cdot 2 \cdot 2$ .

Let’s review the vocabulary for expressions with exponents.

Before we begin working with variable expressions containing exponents, let’s simplify a few expressions involving only numbers.

Simplify: ⓐ $4^{3}$ ⓑ $7^{1}$ ⓒ ${(\frac{5}{6})}^{2}$ ⓓ ${(0.63)}^{2} .$

Solution

Examples demonstrating the evaluation of exponential expressions, showing the expansion into factors and the simplified numerical result.
ⓐ	$4^{3}$
Multiply three factors of 4.	$4 \cdot 4 \cdot 4$
Simplify.	$64$
ⓑ	$7^{1}$
Multiply one factor of 7.	$7$
ⓒ	${(\frac{5}{6})}^{2}$
Multiply two factors.	$(\frac{5}{6}) (\frac{5}{6})$
Simplify.	$\frac{25}{36}$
ⓓ	${(0.63)}^{2}$
Multiply two factors.	$(0.63) (0.63)$
Simplify.	$0.3969$

Simplify: ⓐ ${(−5)}^{4}$ ⓑ $− 5^{4} .$

Solution

Compares the evaluation of exponential expressions with a negative base vs. a negative sign outside the base, illustrating the impact of parentheses.
ⓐ	${(−5)}^{4}$
Multiply four factors of $−5$ .	$(−5) (−5) (−5) (−5)$
Simplify.	$625$
ⓑ	$− 5^{4}$
Multiply four factors of 5.	$− (5 \cdot 5 \cdot 5 \cdot 5)$
Simplify.	$−625$

Notice the similarities and differences in Example 2ⓐ and Example 2ⓑ ! Why are the answers different? As we follow the order of operations in part ⓐ the parentheses tell us to raise the $(−5)$ to the 4^th power. In part ⓑ we raise just the 5 to the 4^th power and then take the opposite.

Simplify Expressions Using the Product Property for 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.

First, we will look at an example that leads to the Product Property.


What does this mean? How many factors altogether?
So, we have
Notice that 5 is the sum of the exponents, 2 and 3.

We write:

\begin{matrix} x^{2} \cdot x^{3} \\ x^{2 + 3} \\ x^{5} \end{matrix}

The base stayed the same and we added the exponents. This leads to the Product Property for Exponents.

An example with numbers helps to verify this property.

\begin{matrix} 2^{2} \cdot 2^{3} & \overset{?}{=} & 2^{2 + 3} \\ 4 \cdot 8 & \overset{?}{=} & 2^{5} \\ 32 & = & 32 ✓ \end{matrix}

Simplify: $y^{5} \cdot y^{6} .$

Solution


Use the product property, a^m · aⁿ = a^m+n.
Simplify.

Simplify: ⓐ $2^{5} \cdot 2^{9}$ ⓑ $3 \cdot 3^{4} .$

Solution

ⓐ

Use the product property, a^m · aⁿ = a^m+n.

Simplify.
ⓑ

Use the product property, a^m · aⁿ = a^m+n.

Simplify.

Simplify: ⓐ $a^{7} \cdot a$ ⓑ $x^{27} \cdot x^{13} .$

Solution

ⓐ

Rewrite, a = a¹.

Use the product property, a^m · aⁿ = a^m+n.

Simplify.
ⓑ

Notice, the bases are the same, so add the exponents.

Simplify.

We can extend the Product Property for Exponents to more than two factors.

Simplify: $d^{4} \cdot d^{5} \cdot d^{2} .$

Solution


Add the exponents, since bases are the same.
Simplify.

Simplify Expressions Using the Power Property for 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.


What does this mean? How many factors altogether?
So we have
Notice that 6 is the product of the exponents, 2 and 3.

We write:

\begin{matrix} {(x^{2})}^{3} \\ x^{2 \cdot 3} \\ x^{6} \end{matrix}

We multiplied the exponents. This leads to the Power Property for Exponents.

An example with numbers helps to verify this property.

\begin{matrix} {(3^{2})}^{3} & \overset{?}{=} & 3^{2 \cdot 3} \\ {(9)}^{3} & \overset{?}{=} & 3^{6} \\ 729 & = & 729 ✓ \end{matrix}

Simplify: ⓐ ${(y^{5})}^{9}$ ⓑ ${(4^{4})}^{7} .$

Solution

ⓐ


Use the power property, (a^m)ⁿ = a^m·n.
Simplify.

ⓑ


Use the power property.
Simplify.

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. Can you find this pattern?

This table demonstrates the step-by-step expansion and simplification of the algebraic expression (2x) to 2 x .
	${(2 x)}^{3}$
What does this mean?	$2 x \cdot 2 x \cdot 2 x$
We group the like factors together.	$2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot x$
How many factors of 2 and of $x$ ?	$2^{3} \cdot x^{3}$

Notice that each factor was raised to the power and ${(2 x)}^{3}$ is $2^{3} \cdot x^{3}$ .

Illustrates the power of a product rule, showing (2x)^3 expanded as 2^3 * x^3.
We write:	${(2 x)}^{3}$
	$2^{3} \cdot x^{3}$

The exponent applies to each of the factors! This leads to the Product to a Power Property for Exponents.

An example with numbers helps to verify this property:

\begin{matrix} {(2 \cdot 3)}^{2} & \overset{?}{=} & 2^{2} \cdot 3^{2} \\ 6^{2} & \overset{?}{=} & 4 \cdot 9 \\ 36 & = & 36 ✓ \end{matrix}

Simplify: ⓐ ${(−9 d)}^{2}$ ⓑ ${(3 m n)}^{3} .$

Solution

ⓐ

Use Power of a Product Property, (ab)^m = a^mb^m.

Simplify.
ⓑ

Use Power of a Product Property, (ab)^m = a^mb^m.

Simplify.

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 $a and b$ are real numbers, and $m and n$ are whole numbers, then

Common exponent properties and their corresponding mathematical formulas.
Product Property	$\begin{matrix} a^{m} \cdot a^{n} & = & a^{m + n} \end{matrix}$
Power Property	$\begin{matrix} {(a^{m})}^{n} & = & a^{m \cdot n} \end{matrix}$
Product to a Power	$\begin{matrix} {(a b)}^{m} & = & a^{m} b^{m} \end{matrix}$

All exponent properties hold true for any real numbers $m and n$ . Right now, we only use whole number exponents.

Simplify: ⓐ ${(y^{3})}^{6} {(y^{5})}^{4}$ ⓑ ${(−6 x^{4} y^{5})}^{2} .$

Solution

Steps demonstrating the simplification of algebraic expressions with exponents using various exponent properties.
ⓐ	${(y^{3})}^{6} {(y^{5})}^{4}$
Use the Power Property.	$y^{18} \cdot y^{20}$
Add the exponents.	$y^{38}$
ⓑ	${(−6 x^{4} y^{5})}^{2}$
Use the Product to a Power Property.	${(−6)}^{2} {(x^{4})}^{2} {(y^{5})}^{2}$
Use the Power Property.	${(−6)}^{2}$
Simplify.	$36 x^{8} y^{10}$

Simplify: ⓐ ${(5 m)}^{2} (3 m^{3})$ ⓑ ${(3 x^{2} y)}^{4} {(2 x y^{2})}^{3} .$

Solution

Step-by-step simplification of two algebraic expressions involving exponents and their properties.
ⓐ	${(5 m)}^{2} (3 m^{3})$
Raise $5 m$ to the second power.	$5^{2} m^{2} \cdot 3 m^{3}$
Simplify.	$25 m^{2} \cdot 3 m^{3}$
Use the Commutative Property.	$25 \cdot 3 \cdot m^{2} \cdot m^{3}$
Multiply the constants and add the exponents.	$75 m^{5}$
ⓑ	${(3 x^{2} y)}^{4} {(2 x y^{2})}^{3}$
Use the Product to a Power Property.	$(3^{4} x^{8} y^{4}) (2^{3} x^{3} y^{6})$
Simplify.	$(81 x^{8} y^{4}) (8 x^{3} y^{6})$
Use the Commutative Property.	$81 \cdot 8 \cdot x^{8} \cdot x^{3} \cdot y^{4} \cdot y^{6}$
Multiply the constants and add the exponents.	$648 x^{11} y^{10}$

Multiply Monomials

Since a monomial is an algebraic expression, we can use the properties of exponents to multiply monomials.

Multiply: $(3 x^{2}) (−4 x^{3}) .$

Solution

Step-by-step solution for multiplying monomial expressions using the Commutative Property.
	$(3 x^{2}) (−4 x^{3})$
Use the Commutative Property to rearrange the terms.	$3 \cdot (−4) \cdot x^{2} \cdot x^{3}$
Multiply.	$−12 x^{5}$

Multiply: $(\frac{5}{6} x^{3} y) (12 x y^{2}) .$

Solution

Simplification of an algebraic expression, showing steps including the Commutative Property to arrive at the final product.
	$(\frac{5}{6} x^{3} y) (12 x y^{2})$
Use the Commutative Property to rearrange the terms.	$\frac{5}{6} \cdot 12 \cdot x^{3} \cdot x \cdot y \cdot y^{2}$
Multiply.	$10 x^{4} y^{3}$

Key Concepts

Exponential Notation
Properties of Exponents
- If $a, b$ are real numbers and $m, n$ are whole numbers, then
  $\begin{matrix} Product Property & a^{m} \cdot a^{n} & = & a^{m + n} \\ Power Property & {(a^{m})}^{n} & = & a^{m \cdot n} \\ Product to a Power & {(a b)}^{m} & = & a^{m} b^{m} \end{matrix}$

Practice Makes Perfect

Simplify Expressions with Exponents

In the following exercises, simplify each expression with exponents.

ⓐ $3^{5}$ ⓑ $9^{1}$ ⓒ ${(\frac{1}{3})}^{2}$ ⓓ ${(0.2)}^{4}$

ⓐ $10^{4}$ ⓑ $17^{1}$ ⓒ ${(\frac{2}{9})}^{2}$ ⓓ ${(0.5)}^{3}$

Solution

ⓐ 10,000 ⓑ 17 ⓒ $\frac{4}{81}$ ⓓ 0.125

ⓐ $2^{6}$ ⓑ $14^{1}$ ⓒ ${(\frac{2}{5})}^{3}$ ⓓ ${(0.7)}^{2}$

ⓐ $8^{3}$ ⓑ $8^{1}$ ⓒ ${(\frac{3}{4})}^{3}$ ⓓ ${(0.4)}^{3}$

Solution

ⓐ 512 ⓑ 8 ⓒ $\frac{27}{64}$
ⓓ 0.064

ⓐ ${(−6)}^{4}$ ⓑ $− 6^{4}$

ⓐ ${(−2)}^{6}$ ⓑ $− 2^{6}$

Solution

ⓐ 64 ⓑ $−64$

ⓐ $− {(\frac{1}{4})}^{4}$ ⓑ ${(- \frac{1}{4})}^{4}$

ⓐ $− {(\frac{2}{3})}^{2}$ ⓑ ${(- \frac{2}{3})}^{2}$

Solution

ⓐ $- \frac{4}{9}$ ⓑ $\frac{4}{9}$

ⓐ $− {0.5}^{2}$ ⓑ ${(−0.5)}^{2}$

ⓐ $− {0.1}^{4}$ ⓑ ${(−0.1)}^{4}$

Solution

ⓐ $−0.0001$ ⓑ 0.0001

Simplify Expressions Using the Product Property for Exponents

In the following exercises, simplify each expression using the Product Property for Exponents.

$d^{3} \cdot d^{6}$

$x^{4} \cdot x^{2}$

Solution

$x^{6}$

$n^{19} \cdot n^{12}$

$q^{27} \cdot q^{15}$

Solution

$q^{42}$

ⓐ $4^{5} \cdot 4^{9}$ ⓑ $8^{9} \cdot 8$

ⓐ $3^{10} \cdot 3^{6}$ ⓑ $5 \cdot 5^{4}$

Solution

ⓐ $3^{16}$ ⓑ $5^{5}$

ⓐ $y \cdot y^{3}$ ⓑ $z^{25} \cdot z^{8}$

ⓐ $w^{5} \cdot w^{}$ ⓑ $u^{41} \cdot u^{53}$

Solution

ⓐ $w^{6}$ ⓑ $u^{94}$

$w \cdot w^{2} \cdot w^{3}$

$y \cdot y^{3} \cdot y^{5}$

Solution

$y^{9}$

$a^{4} \cdot a^{3} \cdot a^{9}$

$c^{5} \cdot c^{11} \cdot c^{2}$

Solution

$c^{18}$

$m^{x} \cdot m^{3}$

$n^{y} \cdot n^{2}$

Solution

$n^{y + 2}$

$y^{a} \cdot y^{b}$

$x^{p} \cdot x^{q}$

Solution

$x^{p + q}$

Simplify Expressions Using the Power Property for Exponents

In the following exercises, simplify each expression using the Power Property for Exponents.

ⓐ ${(m^{4})}^{2}$ ⓑ ${(10^{3})}^{6}$

ⓐ ${(b^{2})}^{7}$ ⓑ ${(3^{8})}^{2}$

Solution

ⓐ $b^{14}$ ⓑ $3^{16}$

ⓐ ${(y^{3})}^{x}$ ⓑ ${(5^{x})}^{y}$

ⓐ ${(x^{2})}^{y}$ ⓑ ${(7^{a})}^{b}$

Solution

ⓐ $x^{2 y}$ ⓑ $7^{a b}$

Simplify Expressions Using the Product to a Power Property

In the following exercises, simplify each expression using the Product to a Power Property.

ⓐ ${(6 a)}^{2}$ ⓑ ${(3 x y)}^{2}$

ⓐ ${(5 x)}^{2}$ ⓑ ${(4 a b)}^{2}$

Solution

ⓐ $25 x^{2}$ ⓑ $16 a^{2} b^{2}$

ⓐ ${(−4 m)}^{3}$ ⓑ ${(5 a b)}^{3}$

ⓐ ${(−7 n)}^{3}$ ⓑ ${(3 x y z)}^{4}$

Solution

ⓐ $−343 n^{3}$ ⓑ $81 x^{4} y^{4} z^{4}$

Simplify Expressions by Applying Several Properties

In the following exercises, simplify each expression.

ⓐ ${(y^{2})}^{4} \cdot {(y^{3})}^{2}$ ⓑ ${(10 a^{2} b)}^{3}$

ⓐ ${(w^{4})}^{3} \cdot {(w^{5})}^{2}$ ⓑ ${(2 x y^{4})}^{5}$

Solution

ⓐ $w^{22}$ ⓑ $32 x^{5} y^{20}$

ⓐ ${(−2 r^{3} s^{2})}^{4}$ ⓑ ${(m^{5})}^{3} \cdot {(m^{9})}^{4}$

ⓐ ${(−10 q^{2} p^{4})}^{3}$ ⓑ ${(n^{3})}^{10} \cdot {(n^{5})}^{2}$

Solution

ⓐ $−1000 q^{6} p^{12}$ ⓑ $n^{40}$

ⓐ ${(3 x)}^{2} (5 x)$ ⓑ ${(5 t^{2})}^{3} {(3 t)}^{2}$

ⓐ ${(2 y)}^{3} (6 y)$ ⓑ ${(10 k^{4})}^{3} {(5 k^{6})}^{2}$

Solution

ⓐ $48 y^{4}$ ⓑ $25,000 k^{24}$

ⓐ ${(5 a)}^{2} {(2 a)}^{3}$ ⓑ ${(\frac{1}{2} y^{2})}^{3} {(\frac{2}{3} y)}^{2}$

ⓐ ${(4 b)}^{2} {(3 b)}^{3}$ ⓑ ${(\frac{1}{2} j^{2})}^{5} {(\frac{2}{5} j^{3})}^{2}$

Solution

ⓐ $432 b^{5}$ ⓑ $\frac{1}{200} j^{16}$

ⓐ ${(\frac{2}{5} x^{2} y)}^{3}$ ⓑ ${(\frac{8}{9} x y^{4})}^{2}$

ⓐ ${(2 r^{2})}^{3} {(4 r)}^{2}$ ⓑ ${(3 x^{3})}^{3} {(x^{5})}^{4}$

Solution

ⓐ $128 r^{8}$ ⓑ ${27 x}^{29}$

ⓐ ${(m^{2} n)}^{2} {(2 m n^{5})}^{4}$ ⓑ ${(3 p q^{4})}^{2} {(6 p^{6} q)}^{2}$

Multiply Monomials

In the following exercises, multiply the monomials.

$(6 y^{7}) (−3 y^{4})$

Solution

$−18 y^{11}$

$(−10 x^{5}) (−3 x^{3})$

$(−8 u^{6}) (−9 u)$

Solution

$72 u^{7}$

$(−6 c^{4}) (−12 c)$

$(\frac{1}{5} f^{8}) (20 f^{3})$

Solution

$4 f^{11}$

$(\frac{1}{4} d^{5}) (36 d^{2})$

$(4 a^{3} b) (9 a^{2} b^{6})$

Solution

$36 a^{5} b^{7}$

$(6 m^{4} n^{3}) (7 m n^{5})$

$(\frac{4}{7} r s^{2}) (14 r s^{3})$

Solution

$8 r^{2} s^{5}$

$(\frac{5}{8} x^{3} y) (24 x^{5} y)$

$(\frac{2}{3} x^{2} y) (\frac{3}{4} x y^{2})$

Solution

$\frac{1}{2} x^{3} y^{3}$

$(\frac{3}{5} m^{3} n^{2}) (\frac{5}{9} m^{2} n^{3})$

Mixed Practice

In the following exercises, simplify each expression.

${(x^{2})}^{4} \cdot {(x^{3})}^{2}$

Solution

$x^{14}$

${(y^{4})}^{3} \cdot {(y^{5})}^{2}$

${(a^{2})}^{6} \cdot {(a^{3})}^{8}$

Solution

$a^{36}$

${(b^{7})}^{5} \cdot {(b^{2})}^{6}$

${(2 m^{6})}^{3}$

Solution

$8 m^{18}$

${(3 y^{2})}^{4}$

${(10 x^{2} y)}^{3}$

Solution

$1000 x^{6} y^{3}$

${(2 m n^{4})}^{5}$

${(−2 a^{3} b^{2})}^{4}$

Solution

$16 a^{12} b^{8}$

${(−10 u^{2} v^{4})}^{3}$

${(\frac{2}{3} x^{2} y)}^{3}$

Solution

$\frac{8}{27} x^{6} y^{3}$

${(\frac{7}{9} p q^{4})}^{2}$

${(8 a^{3})}^{2} {(2 a)}^{4}$

Solution

$1024 a^{10}$

${(5 r^{2})}^{3} {(3 r)}^{2}$

${(10 p^{4})}^{3} {(5 p^{6})}^{2}$

Solution

$25000 p^{24}$

${(4 x^{3})}^{3} {(2 x^{5})}^{4}$

${(\frac{1}{2} x^{2} y^{3})}^{4} {(4 x^{5} y^{3})}^{2}$

Solution

$x^{18} y^{18}$

${(\frac{1}{3} m^{3} n^{2})}^{4} {(9 m^{8} n^{3})}^{2}$

${(3 m^{2} n)}^{2} {(2 m n^{5})}^{4}$

Solution

$144 m^{8} n^{22}$

${(2 p q^{4})}^{3} {(5 p^{6} q)}^{2}$

Everyday Math

Email Kate emails a flyer to ten of her friends and tells them to forward it to ten of their friends, who forward it to ten of their friends, and so on. The number of people who receive the email on the second round is $10^{2}$ , on the third round is $10^{3}$ , as shown in the table below. How many people will receive the email on the sixth round? Simplify the expression to show the number of people who receive the email.

Round	Number of people
1	10
2	$10^{2}$
3	$10^{3}$
…	…
6	?

Solution

$1,000,000$

Salary Jamal’s boss gives him a 3% raise every year on his birthday. This means that each year, Jamal’s salary is 1.03 times his last year’s salary. If his original salary was $35,000, his salary after 1 year was $$ 35,000 (1.03)$ , after 2 years was $$ 35,000 {(1.03)}^{2}$ , after 3 years was $$ 35,000 {(1.03)}^{3}$ , as shown in the table below. What will Jamal’s salary be after 10 years? Simplify the expression, to show Jamal’s salary in dollars.

Year	Salary
1	$$ 35,000 (1.03)$
2	$$ 35,000 {(1.03)}^{2}$
3	$$ 35,000 {(1.03)}^{3}$
…	…
10	?

Clearance A department store is clearing out merchandise in order to make room for new inventory. The plan is to mark down items by 30% each week. This means that each week the cost of an item is 70% of the previous week’s cost. If the original cost of a sofa was $1,000, the cost for the first week would be $$ 1,000 (0.70)$ and the cost of the item during the second week would be $$ 1,000 {(0.70)}^{2}$ . Complete the table shown below. What will be the cost of the sofa during the fifth week? Simplify the expression, to show the cost in dollars.

Week	Cost
1	$$ 1,000 (0.70)$
2	$$ 1,000 {(0.70)}^{2}$
3
4	…
5	?

Solution

$168.07

Depreciation Once a new car is driven away from the dealer, it begins to lose value. Each year, a car loses 10% of its value. This means that each year the value of a car is 90% of the previous year’s value. If a new car was purchased for $20,000, the value at the end of the first year would be $$ 20,000 (0.90)$ and the value of the car after the end of the second year would be $$ 20,000 {(0.90)}^{2}$ . Complete the table shown below. What will be the value of the car at the end of the eighth year? Simplify the expression, to show the value in dollars.

Year	Cost
1	$$ 20,000 (0.90)$
2	$$ 20,000 {(0.90)}^{2}$
3
…	…
8	?

Writing Exercises

Use the Product Property for Exponents to explain why $x \cdot x = x^{2}$ .

Solution

Answers will vary.

Explain why $− 5^{3} = {(−5)}^{3}$ but $− 5^{4} \neq {(−5)}^{4}$ .

Jorge thinks ${(\frac{1}{2})}^{2}$ is 1. What is wrong with his reasoning?

Solution

Answers will vary.

Explain why $x^{3} \cdot x^{5}$ is $x^{8}$ , and not $x^{15}$ .

Self Check

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

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

Multiply Polynomials

Multiply a Polynomial by a Monomial

We have used the Distributive Property to simplify expressions like $2 (x - 3)$ . You multiplied both terms in the parentheses, $x and 3$ , by 2, to get $2 x - 6$ . 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! Here’s an example:

Multiply: $4 (x + 3) .$

Solution


Distribute.
Simplify.

Multiply: $y (y - 2) .$

Solution


Distribute.
Simplify.

Multiply: $7 x (2 x + y) .$

Solution


Distribute.
Simplify.

Multiply: $−2 y (4 y^{2} + 3 y - 5) .$

Solution


Distribute.
Simplify.

Multiply: $2 x^{3} (x^{2} - 8 x + 1) .$

Solution


Distribute.
Simplify.

Multiply: $(x + 3) p .$

Solution

The monomial is the second factor.
Distribute.
Simplify.

Multiply a Binomial by a Binomial

Just like there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a binomial times a binomial. We will start by using the Distributive Property.

Multiply a Binomial by a Binomial Using the Distributive Property

Look at Example 6, where we multiplied a binomial by a monomial.


We distributed the p to get:
What if we have (x + 7) instead of p?
Distribute (x + 7).
Distribute again.
Combine like terms.

Notice that before combining like terms, you had four terms. You multiplied the two terms of the first binomial by the two terms of the second binomial—four multiplications.

Multiply: $(y + 5) (y + 8) .$

Solution


Distribute (y + 8).
Distribute again
Combine like terms.

Multiply: $(2 y + 5) (3 y + 4) .$

Solution


Distribute (3y + 4).
Distribute again
Combine like terms.

Multiply: $(4 y + 3) (2 y - 5) .$

Solution


Distribute.
Distribute again.
Combine like terms.

Multiply: $(x - 2) (x - y) .$

Solution


Distribute.
Distribute again.
There are no like terms to combine.

Multiply a Binomial by a Binomial 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, like in Example 10, 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.

\begin{matrix} (x - 2) (x - y) \\ x^{2} - x y - 2 x + 2 y \end{matrix}

Where did the first term, $x^{2}$ , come from?

This figure explains how to multiply a binomial using the FOIL method. It has two columns, with written instructions on the left and math on the right. At the top of the figure, the text in the left column says “It is the product of x and x, the first terms in x minus 2 and x minus y.” In the right column is the product of x minus 2 and x minus y. An arrow extends from the x in x minus 2, and terminates at the x in x minus y. Below this is the word “First.” One row down, the text in the left column says “The next terms, negative xy, is the product of x and negative y, the two outer terms.” In the right column is the product of x minus 2 and x minus y, with another arrow extending from the x in x minus 2 to the y in x minus y. Below this is the word “Outer.” One row down, the text in the left column says “The third term, negative 2 x, is the product of negative 2 and x, the two inner terms.” In the right column is the product of x minus 2 and x minus y with a third arrow extending from minus 2 in x minus 2 and terminating at the x in x minus y. Below this is the word “Inner.” In the last row, the text in the left column says “And the last term, plus 2y, came from multiplying the two last terms, negative 2 and negative y.” In the right column is the product of x minus 2 and x minus y, with a fourth arrow extending from the minus 2 in x minus 2 to the minus y in x minus y. Below this is the word “Last.”

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.

\begin{matrix} (x - 2) (x - y) \\ \underset{F O I L}{x^{2} - x y - 2 x + 2 y} \end{matrix}

Let’s look at $(x + 3) (x + 7)$ .

Distibutive Property	FOIL

Notice how the terms in third line fit the FOIL pattern.

Now we will do an example where we use the FOIL pattern to multiply two binomials.

How to Multiply a Binomial by a Binomial using the FOIL Method

Multiply using the FOIL method: $(x + 5) (x + 9) .$

Solution

In the second row, the first cell reads “Step 2. Multiply the outer terms.” In the second cell is the product of x plus 5 and x plus 9 again, with an arrow extending from x in the first binomial to the 9 in the second binomial. The third cell contains x squared plus 9x plus blank plus blank, with the letter F under the x squared, O under the 9x, and I and L beneath the two blanks.

In the third row, the first cell reads “Step 3. Multiply the inner terms.” The second cell contains the product of x plus 5 and x plus 9 again, with an arrow extending from 5 in the first binomial to the x in the second binomial. The third cell contains x squared plus 9x plus 5x plus blank, with F beneath x squared, O beneath 9x, I beneath 5x, and L beneath the blank.

In the fourth row, the first cell reads “Step 4. Multiply the last terms.” In the second cell is the product of x plus 5 and x plus 9 again, with an arrow extending from 5 in the first binomial to 9 in the second binomial. The third cell contains x squared plus 9x plus 6x plus 45, with F beneath x squared, O beneath 9x, I beneath 6x, and L beneath 45.

In the final row, the first cell reads “Step 5. Combine like terms, when possible.” The second cell is blank. The third cell contains the final expression: x squared plus 15x plus 45.

We summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!

When you multiply by the FOIL method, drawing the lines will help your brain focus on the pattern and make it easier to apply.

Multiply: $(y - 7) (y + 4) .$

Solution

Multiply: $(4 x + 3) (2 x - 5) .$

Solution

The final products in the last four examples were trinomials because we could combine the two middle terms. This is not always the case.

Multiply: $(3 x - y) (2 x - 5) .$

Solution



Multiply the First.
Multiply the Outer.
Multiply the Inner.
Multiply the Last.
Combine like terms—there are none.

Be careful of the exponents in the next example.

Multiply: $(n^{2} + 4) (n - 1) .$

Solution



Multiply the First.
Multiply the Outer.
Multiply the Inner.
Multiply the Last.
Combine like terms—there are none.

Multiply: $(3 p q + 5) (6 p q - 11) .$

Solution


Multiply the First.
Multiply the Outer.
Multiply the Inner.
Multiply the Last.
Combine like terms—there are none.

Multiply a Binomial by a Binomial Using the Vertical Method

The FOIL method is usually the quickest method for multiplying two binomials, but it only works for binomials. You can use the Distributive Property to find the product of any two polynomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.

This figure shows the vertical multiplication of 23 and 46. The number 23 is above the number 46. Below this, there is the partial product 138 over the partial product 92. The final product is at the bottom and is 1058. Text on the right side of the image says “Start by multiplying 23 by 6 to get 138. Next, 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: $(3 y - 1) (2 y - 6) .$

Solution

It does not matter which binomial goes on the top.

$\begin{matrix} \begin{matrix} Multiply 3 y - 1 by −6 . \\ Multiply 3 y - 1 by 2y. \\ Add like terms. \end{matrix} & \begin{matrix} 3 y - 1 \\ \underset{________}{\times 2 y - 6} \\ −18 y + 6 \\ \underset{_____________}{6 y^{2} - 2 y} \\ 6 y^{2} - 20 y + 6 \end{matrix} & \begin{matrix} partial product \\ partial product \\ product \end{matrix} \end{matrix}$

Notice the partial products are the same as the terms in the FOIL method.

We have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The methods are listed here all together, to help you remember them.

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, FOIL 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: $(b + 3) (2 b^{2} - 5 b + 8) .$

Solution


Distribute.
Multiply.
Combine like terms.

Now let’s do this same multiplication using the Vertical Method.

Multiply using the Vertical Method: $(b + 3) (2 b^{2} - 5 b + 8) .$

Solution

It is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products this way.

Multiply (2b² − 5b + 8) by 3.
Multiply (2b² − 5b + 8) by b.
Add like terms.

We have now seen two methods you can use to multiply a trinomial by a binomial. After you practice each method, you’ll probably find you prefer one way over the other. We list both methods are listed here, for easy reference.

Key Concepts

FOIL Method for Multiplying Two Binomials—To multiply two binomials:
1. Multiply the First terms.
2. Multiply the Outer terms.
3. Multiply the Inner terms.
4. Multiply the Last terms.
Multiplying Two Binomials—To multiply binomials, use the:
- Distributive Property (Example 7)
- FOIL Method (Example 12)
- Vertical Method (Example 17)
Multiplying a Trinomial by a Binomial—To multiply a trinomial by a binomial, use the:
- Distributive Property (Example 18)
- Vertical Method (Example 19)

Practice Makes Perfect

Multiply a Polynomial by a Monomial

In the following exercises, multiply.

$4 (w + 10)$

Solution

$4 w + 40$

$6 (b + 8)$

$−3 (a + 7)$

Solution

$−3 a - 21$

$−5 (p + 9)$

$2 (x - 7)$

Solution

$2 x - 14$

$7 (y - 4)$

$−3 (k - 4)$

Solution

$−3 k + 12$

$−8 (j - 5)$

$q (q + 5)$

Solution

$q^{2} + 5 q$

$k (k + 7)$

$− b (b + 9)$

Solution

$− b^{2} - 9 b$

$− y (y + 3)$

$− x (x - 10)$

Solution

$− x^{2} + 10 x$

$− p (p - 15)$

$6 r (4 r + s)$

Solution

$24 r^{2} + 6 r s$

$5 c (9 c + d)$

$12 x (x - 10)$

Solution

$12 x^{2} - 120 x$

$9 m (m - 11)$

$−9 a (3 a + 5)$

Solution

$−27 a^{2} - 45 a$

$−4 p (2 p + 7)$

$3 (p^{2} + 10 p + 25)$

Solution

$3 p^{2} + 30 p + 75$

$6 (y^{2} + 8 y + 16)$

$−8 x (x^{2} + 2 x - 15)$

Solution

$−8 x^{3} - 16 x^{2} + 120 x$

$−5 t (t^{2} + 3 t - 18)$

$5 q^{3} (q^{3} - 2 q + 6)$

Solution

$5 q^{6} - 10 q^{4} + 30 q^{3}$

$4 x^{3} (x^{4} - 3 x + 7)$

$−8 y (y^{2} + 2 y - 15)$

Solution

$−8 y^{3} - 16 y^{2} + 120 y$

$−5 m (m^{2} + 3 m - 18)$

$5 q^{3} (q^{2} - 2 q + 6)$

Solution

$5 q^{5} - 10 q^{4} + 30 q^{3}$

$9 r^{3} (r^{2} - 3 r + 5)$

$−4 z^{2} (3 z^{2} + 12 z - 1)$

Solution

$−12 z^{4} - 48 z^{3} + 4 z^{2}$

$−3 x^{2} (7 x^{2} + 10 x - 1)$

$(2 m - 9) m$

Solution

$2 m^{2} - 9 m$

$(8 j - 1) j$

$(w - 6) \cdot 8$

Solution

$8 w - 48$

$(k - 4) \cdot 5$

$4 (x + 10)$

Solution

$4 x + 40$

$6 (y + 8)$

$15 (r - 24)$

Solution

$15 r - 360$

$12 (v - 30)$

$−3 (m + 11)$

Solution

$−3 m - 33$

$−4 (p + 15)$

$−8 (z - 5)$

Solution

$−8 z + 40$

$−3 (x - 9)$

$u (u + 5)$

Solution

$u^{2} + 5 u$

$q (q + 7)$

$n (n^{2} - 3 n)$

Solution

$n^{3} - 3 n^{2}$

$s (s^{2} - 6 s)$

$6 x (4 x + y)$

Solution

$24 x^{2} + 6 x y$

$5 a (9 a + b)$

$5 p (11 p - 5 q)$

Solution

$55 p^{2} - 25 p q$

$12 u (3 u - 4 v)$

$3 (v^{2} + 10 v + 25)$

Solution

$3 v^{2} + 30 v + 75$

$6 (x^{2} + 8 x + 16)$

$2 n (4 n^{2} - 4 n + 1)$

Solution

$8 n^{3} - 8 n^{2} + 2 n$

$3 r (2 r^{2} - 6 r + 2)$

$−8 y (y^{2} + 2 y - 15)$

Solution

$−8 y^{3} - 16 y^{2} + 120 y$

$−5 m (m^{2} + 3 m - 18)$

$5 q^{3} (q^{2} - 2 q + 6)$

Solution

$5 q^{5} - 10 q^{4} + 30 q^{3}$

$9 r^{3} (r^{2} - 3 r + 5)$

$−4 z^{2} (3 z^{2} + 12 z - 1)$

Solution

$−12 z^{4} - 48 z^{3} + 4 z^{2}$

$−3 x^{2} (7 x^{2} + 10 x - 1)$

$(2 y - 9) y$

Solution

$2 y^{2} - 9 y$

$(8 b - 1) b$

Multiply a Binomial by a Binomial

In the following exercises, multiply the following binomials using: ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical Method.

$(w + 5) (w + 7)$

Solution

$w^{2} + 12 w + 35$

$(y + 9) (y + 3)$

$(p + 11) (p - 4)$

Solution

$p^{2} + 7 p - 44$

$(q + 4) (q - 8)$

In the following exercises, multiply the binomials. Use any method.

$(x + 8) (x + 3)$

Solution

$x^{2} + 11 x + 24$

$(y + 7) (y + 4)$

$(y - 6) (y - 2)$

Solution

$y^{2} - 8 y + 12$

$(x - 7) (x - 2)$

$(w - 4) (w + 7)$

Solution

$w^{2} + 3 w - 28$

$(q - 5) (q + 8)$

$(p + 12) (p - 5)$

Solution

$p^{2} + 7 p - 60$

$(m + 11) (m - 4)$

$(6 p + 5) (p + 1)$

Solution

$6 p^{2} + 11 p + 5$

$(7 m + 1) (m + 3)$

$(2 t - 9) (10 t + 1)$

Solution

$20 t^{2} - 88 t - 9$

$(3 r - 8) (11 r + 1)$

$(5 x - y) (3 x - 6)$

Solution

$15 x^{2} - 3 x y - 30 x + 6 y$

$(10 a - b) (3 a - 4)$

$(a + b) (2 a + 3 b)$

Solution

$2 a^{2} + 5 a b + 3 b^{2}$

$(r + s) (3 r + 2 s)$

$(4 z - y) (z - 6)$

Solution

$4 z^{2} - 24 z - z y + 6 y$

$(5 x - y) (x - 4)$

$(x^{2} + 3) (x + 2)$

Solution

$x^{3} + 2 x^{2} + 3 x + 6$

$(y^{2} - 4) (y + 3)$

$(x^{2} + 8) (x^{2} - 5)$

Solution

$x^{4} + 3 x^{2} - 40$

$(y^{2} - 7) (y^{2} - 4)$

$(5 a b - 1) (2 a b + 3)$

Solution

$10 a^{2} b^{2} + 13 a b - 3$

$(2 x y + 3) (3 x y + 2)$

$(6 p q - 3) (4 p q - 5)$

Solution

$24 p^{2} q^{2} - 42 p q + 15$

$(3 r s - 7) (3 r s - 4)$

Multiply a Trinomial by a Binomial

In the following exercises, multiply using ⓐ the Distributive Property ⓑ the Vertical Method.

$(x + 5) (x^{2} + 4 x + 3)$

Solution

$x^{3} + 9 x^{2} + 23 x + 15$

$(u + 4) (u^{2} + 3 u + 2)$

$(y + 8) (4 y^{2} + y - 7)$

Solution

$4 y^{3} + 33 y^{2} + y - 56$

$(a + 10) (3 a^{2} + a - 5)$

In the following exercises, multiply. Use either method.

$(w - 7) (w^{2} - 9 w + 10)$

Solution

$w^{3} - 16 w^{2} + 73 w - 70$

$(p - 4) (p^{2} - 6 p + 9)$

$(3 q + 1) (q^{2} - 4 q - 5)$

Solution

$3 q^{3} - 11 q^{2} - 19 q - 5$

$(6 r + 1) (r^{2} - 7 r - 9)$

Mixed Practice

$(10 y - 6) + (4 y - 7)$

Solution

$14 y - 13$

$(15 p - 4) + (3 p - 5)$

$(x^{2} - 4 x - 34) - (x^{2} + 7 x - 6)$

Solution

$−11 x - 28$

$(j^{2} - 8 j - 27) - (j^{2} + 2 j - 12)$

$5 q (3 q^{2} - 6 q + 11)$

Solution

$15 q^{3} - 30 q^{2} + 55 q$

$8 t (2 t^{2} - 5 t + 6)$

$(s - 7) (s + 9)$

Solution

$s^{2} + 2 s - 63$

$(x - 5) (x + 13)$

$(y^{2} - 2 y) (y + 1)$

Solution

$y^{3} - y^{2} - 2 y$

$(a^{2} - 3 a) (4 a + 5)$

$(3 n - 4) (n^{2} + n - 7)$

Solution

$3 n^{3} - n^{2} - 25 n + 28$

$(6 k - 1) (k^{2} + 2 k - 4)$

$(7 p + 10) (7 p - 10)$

Solution

$49 p^{2} - 100$

$(3 y + 8) (3 y - 8)$

$(4 m^{2} - 3 m - 7) m^{2}$

Solution

$4 m^{4} - 3 m^{3} - 7 m^{2}$

$(15 c^{2} - 4 c + 5) c^{4}$

$(5 a + 7 b) (5 a + 7 b)$

Solution

$25 a^{2} + 70 a b + 49 b^{2}$

$(3 x - 11 y) (3 x - 11 y)$

$(4 y + 12 z) (4 y - 12 z)$

Solution

$16 y^{2} - 144 z^{2}$

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 $10 + 5$ .

ⓐ Multiply $(10 + 3) (10 + 5)$ by the FOIL method.
ⓑ Multiply $13 \cdot 15$ without using a calculator.
ⓒ Which way is easier for you? Why?

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 $20 - 3$ .

ⓐ Multiply $(20 - 2) (20 - 3)$ by the FOIL method.
ⓑ Multiply $18 \cdot 17$ without using a calculator.
ⓒ Which way is easier for you? Why?

Solution

ⓐ 306 ⓑ 306 ⓒ Answers will vary.

Writing Exercises

Which method do you prefer to use when multiplying two binomials: the Distributive Property, the FOIL method, or the Vertical Method? Why?

Which method do you prefer to use when multiplying a trinomial by a binomial: the Distributive Property or the Vertical Method? Why?

Solution

Answers will vary.

Multiply the following:

$\begin{matrix} (x + 2) (x - 2) \\ (y + 7) (y - 7) \\ (w + 5) (w - 5) \end{matrix}$

Explain the pattern that you see in your answers.

Multiply the following:

$\begin{matrix} (m - 3) (m + 3) \\ (n - 10) (n + 10) \\ (p - 8) (p + 8) \end{matrix}$

Explain the pattern that you see in your answers.

Solution

Answers may vary.

Multiply the following:

$\begin{matrix} (p + 3) (p + 3) \\ (q + 6) (q + 6) \\ (r + 1) (r + 1) \end{matrix}$

Explain the pattern that you see in your answers.

Multiply the following:

$\begin{matrix} (x - 4) (x - 4) \\ (y - 1) (y - 1) \\ (z - 7) (z - 7) \end{matrix}$

Explain the pattern that you see in your answers.

Solution

Answers may vary.

Self Check

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

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

Special Products

Square a Binomial Using the Binomial Squares Pattern

Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and using the methods of the last section, there is less work to do if you learn to use a pattern.

Step-by-step expansion of the binomial (x+9)^2 using the FOIL method, from definition to simplified quadratic expression.
Let's start by looking at ${(x + 9)}^{2}$ .
What does this mean?	${(x + 9)}^{2}$
It means to multiply $(x + 9)$ by itself.	$(x + 9) (x + 9)$
Then, using FOIL, we get:	$x^{2} + 9 x + 9 x + 81$
Combining like terms gives:	$x^{2} + 18 x + 81$

Step-by-step algebraic expansion of (y-7)^2, illustrating the FOIL method and simplification of terms.
Here's another one:	${(y - 7)}^{2}$
Multiply $(y - 7)$ by itself.	$(y - 7) (y - 7)$
Using FOIL, we get:	$y^{2} - 7 y - 7 y + 49$
And combining like terms:	$y^{2} - 14 y + 49$

This table demonstrates the step-by-step algebraic expansion of a binomial squared, illustrating the process including the FOIL method.
And one more:	${(2 x + 3)}^{2}$
Multiply.	$(2 x + 3) (2 x + 3)$
Use FOIL:	$4 x^{2} + 6 x + 6 x + 9$
Combine like terms.	$4 x^{2} + 12 x + 9$

Look at these results. Do you see any patterns?

What about the number of terms? In each example we squared a binomial and the result was a trinomial.

{(a + b)}^{2} =____+____+____

Now look at the first term in each result. Where did it come from?

This figure has three columns. The first column contains the expression x plus 9, in parentheses, squared. Below this is the product of x plus 9 and x plus 9. Below this is x squared plus 9x plus 9x plus 81. Below this is x squared plus 18x plus 81. The second column contains the expression y minus 7, in parentheses, squared. Below this is the product of y minus 7 and y minus 7. Below this is y squared minus 7y minus 7y plus 49. Below this is the expression y squared minus 14y plus 49. The third column contains the expression 2x plus 3, in parentheses, squared. Below this is the product of 2x plus 3 and 2x plus 3. Below this is 4x squared plus 6x plus 6x plus 9. Below this is 4x squared plus 12x plus 9.

The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!

{(a + b)}^{2} = a^{2} +____+____

To get the first term of the product, square the first term.

Where did the last term come from? Look at the examples and find the pattern.

The last term is the product of the last terms, which is the square of the last term.

{(a + b)}^{2} =____+____+ b^{2}

To get the last term of the product, square the last term.

Finally, look at the middle term. Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial.

\begin{matrix} {(a + b)}^{2} =____+ 2 a b +____ \\ {(a - b)}^{2} =____- 2 a b +____ \end{matrix}

To get the middle term of the product, multiply the terms and double their product.

Putting it all together:

To square a binomial:

square the first term
square the last term
double their product

A number example helps verify the pattern.

Steps for expanding and simplifying the binomial expression (10+4)^2 using algebraic identity.
	${(10 + 4)}^{2}$
Square the first term.	$10^{2} +___+$
Square the last term.	$10^{2} +___+ 4^{2}$
Double their product.	$10^{2} + 2 \cdot 10 \cdot 4 + 4^{2}$
Simplify.	$100 + 80 + 16$
Simplify.	$196$

To multiply ${(10 + 4)}^{2}$ usually you’d follow the Order of Operations.

\begin{matrix} {(10 + 4)}^{2} \\ {(14)}^{2} \\ 196 \end{matrix}

The pattern works!

Multiply: ${(x + 5)}^{2} .$

Solution


Square the first term.
Square the last term.
Double the product.
Simplify.

Multiply: ${(y - 3)}^{2} .$

Solution


Square the first term.
Square the last term.
Double the product.
Simplify.

Multiply: ${(4 x + 6)}^{2} .$

Solution


Use the pattern.
Simplify.

Multiply: ${(2 x - 3 y)}^{2} .$

Solution


Use the pattern.
Simplify.

Multiply: ${(4 u^{3} + 1)}^{2} .$

Solution


Use the pattern.
Simplify.

Multiply Conjugates Using the Product of Conjugates Pattern

We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

What do you notice about these pairs of binomials?

\begin{matrix} (x - 9) (x + 9) & (y - 8) (y + 8) & (2 x - 5) (2 x + 5) \end{matrix}

Look at the first term of each binomial in each pair.

This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. The second is y minus 8, in parentheses, times y plus 8, in parentheses. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses

Notice the first terms are the same in each pair.

Look at the last terms of each binomial in each pair.

Notice the last terms are the same in each pair.

Notice how each pair has one sum and one difference.

This figure has three products. The first is x minus 9, in parentheses, times x plus 9, in parentheses. Below the x minus 9 is the word “difference”. Below x plus 9 is the word “sum”. The second is y minus 8, in parentheses, times y plus 8, in parentheses. Below y minus 8 is the word “difference”. Below y plus 8 is the word “sum”. The last is 2x minus 5, in parentheses, times 2x plus 5, in parentheses. Below the 2x minus 5 is the word “difference” and below 2x plus 5 is the word “sum”.

A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference has a special name. It is called a conjugate pair and is of the form $(a - b), (a + b)$ .

There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier.

Let’s look for the pattern by using FOIL to multiply some conjugate pairs.

\begin{matrix} (x - 9) (x + 9) & (y - 8) (y + 8) & (2 x - 5) (2 x + 5) \\ x^{2} + 9 x - 9 x - 81 & y^{2} + 8 y - 8 y - 64 & 4 x^{2} + 10 x - 10 x - 25 \\ x^{2} - 81 & y^{2} - 64 & 4 x^{2} - 25 \end{matrix}

This figure has three columns. The first column contains the product of x plus 9 and x minus 9. Below this is the expression x squared minus 9x plus 9x minus 81. Below this is x squared minus 81. The second column contains the product of y minus 8 and y plus 8. Below this is the expression y squared plus 8y minus 8y minus 64. Below this is y squared minus 64. The third column contains the product of 2x minus 5 and 2x plus 5. Below this is the expression 4x squared plus 10x minus 10x minus 25. Below this is 4x squared minus 25.

Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term.

\begin{matrix} (a + b) (a - b) = a^{2} -____ \\ To get the first term, square the first term . \end{matrix}

The last term came from multiplying the last terms, the square of the last term.

\begin{matrix} (a + b) (a - b) = a^{2} - b^{2} \\ To get the last term, square the last term . \end{matrix}

What do you observe about the products?

The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.

Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.

The product of conjugates is always of the form $a^{2} - b^{2}$ . This is called a difference of squares.

This leads to the pattern:

Let’s test this pattern with a numerical example.

Comparison of two methods (difference of squares vs. order of operations) to simplify the expression (10-2)(10+2).
	$(10 - 2) (10 + 2)$
It is the product of conjudgates, so the result will be the difference of two squares.	$____-____$
Square the first term.	$10^{2} -____$
Square the last term.	$10^{2} - 2^{2}$
Simplify.	$100 - 4$
Simplify.	$96$
What do you get using the order of operations?
	$\begin{matrix} (10 - 2) (10 + 2) \\ (8) (12) \\ 96 \end{matrix}$

Notice, the result is the same!

Multiply: $(x - 8) (x + 8) .$

Solution

First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference.

It fits the pattern.
Square the first term, x.
Square the last term, 8.
The product is a difference of squares.

Multiply: $(2 x + 5) (2 x - 5) .$

Solution

Are the binomials conjugates?

It is the product of conjugates.
Square the first term, 2x.
Square the last term, 5.
Simplify. The product is a difference of squares.

The binomials in the next example may look backwards – the variable is in the second term. But the two binomials are still conjugates, so we use the same pattern to multiply them.

Find the product: $(3 + 5 x) (3 - 5 x) .$

Solution

It is the product of conjugates.
Use the pattern.
Simplify.

Now we’ll multiply conjugates that have two variables.

Find the product: $(5 m - 9 n) (5 m + 9 n) .$

Solution

This fits the pattern.
Use the pattern.
Simplify.

Find the product: $(c d - 8) (c d + 8) .$

Solution

This fits the pattern.
Use the pattern.
Simplify.

Find the product: $(6 u^{2} - 11 v^{5}) (6 u^{2} + 11 v^{5}) .$

Solution

This fits the pattern.
Use the pattern.
Simplify.

Recognize and Use the Appropriate Special Product Pattern

We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.

Comparing the Special Product Patterns

This table summarizes formulas and key characteristics for binomial squares and products of conjugates, highlighting their algebraic properties and differences.
Binomial Squares	Product of Conjugates
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$	$(a - b) (a + b) = a^{2} - b^{2}$
${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
- Squaring a binomial	- Multiplying conjugates
- Product is a trinomial	- Product is a binomial
- Inner and outer terms with FOIL are the same.	- Inner and outer terms with FOIL are opposites.
- Middle term is double the product of the terms.	- There is no middle term.

Choose the appropriate pattern and use it to find the product:

ⓐ $(2 x - 3) (2 x + 3)$ ⓑ ${(8 x - 5)}^{2}$ ⓒ ${(6 m + 7)}^{2}$ ⓓ $(5 x - 6) (6 x + 5)$

Solution

ⓐ $(2 x - 3) (2 x + 3)$ These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.

This fits the pattern.

Use the pattern.

Simplify.
ⓑ ${(8 x - 5)}^{2}$ We are asked to square a binomial. It fits the binomial squares pattern.

Use the pattern.

Simplify.
ⓒ ${(6 m + 7)}^{2}$ Again, we will square a binomial so we use the binomial squares pattern.

Use the pattern.

Simplify.

ⓓ

(5 x - 6) (6 x + 5)

This product does not fit the patterns, so we will use FOIL.

This table illustrates the step-by-step multiplication of two binomials using the FOIL method, resulting in a simplified quadratic expression.
	$(5 x - 6) (6 x + 5)$
Use FOIL.	$30 x^{2} + 25 x - 36 x - 30$
Simplify.	$30 x^{2} - 11 x - 30$

Key Concepts

Binomial Squares Pattern
- If $a, b$ are real numbers,
- ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
- ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
- To square a binomial: square the first term, square the last term, double their product.
Product of Conjugates Pattern
- If $a, b$ are real numbers,
- $(a - b) (a + b) = a^{2} - b^{2}$
- The product is called a difference of squares.
To multiply conjugates:
- square the first term square the last term write it as a difference of squares

Practice Makes Perfect

Square a Binomial Using the Binomial Squares Pattern

In the following exercises, square each binomial using the Binomial Squares Pattern.

${(w + 4)}^{2}$

${(q + 12)}^{2}$

Solution

$q^{2} + 24 q + 144$

${(y + \frac{1}{4})}^{2}$

${(x + \frac{2}{3})}^{2}$

Solution

$x^{2} + \frac{4}{3} x + \frac{4}{9}$

${(b - 7)}^{2}$

${(y - 6)}^{2}$

Solution

$y^{2} - 12 y + 36$

${(m - 15)}^{2}$

${(p - 13)}^{2}$

Solution

$p^{2} - 26 p + 169$

${(3 d + 1)}^{2}$

${(4 a + 10)}^{2}$

Solution

$16 a^{2} + 80 a + 100$

${(2 q + \frac{1}{3})}^{2}$

${(3 z + \frac{1}{5})}^{2}$

Solution

$9 z^{2} + \frac{6}{5} z + \frac{1}{25}$

${(3 x - y)}^{2}$

${(2 y - 3 z)}^{2}$

Solution

$4 y^{2} - 12 y z + 9 z^{2}$

${(\frac{1}{5} x - \frac{1}{7} y)}^{2}$

${(\frac{1}{8} x - \frac{1}{9} y)}^{2}$

Solution

$\frac{1}{64} x^{2} - \frac{1}{36} x y + \frac{1}{81} y^{2}$

${(3 x^{2} + 2)}^{2}$

${(5 u^{2} + 9)}^{2}$

Solution

$25 u^{4} + 90 u^{2} + 81$

${(4 y^{3} - 2)}^{2}$

${(8 p^{3} - 3)}^{2}$

Solution

$64 p^{6} - 48 p^{3} + 9$

Multiply Conjugates Using the Product of Conjugates Pattern

In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern.

$(m - 7) (m + 7)$

$(c - 5) (c + 5)$

Solution

$c^{2} - 25$

$(x + \frac{3}{4}) (x - \frac{3}{4})$

$(b + \frac{6}{7}) (b - \frac{6}{7})$

Solution

$b^{2} - \frac{36}{49}$

$(5 k + 6) (5 k - 6)$

$(8 j + 4) (8 j - 4)$

Solution

$64 j^{2} - 16$

$(11 k + 4) (11 k - 4)$

$(9 c + 5) (9 c - 5)$

Solution

$81 c^{2} - 25$

$(11 - b) (11 + b)$

$(13 - q) (13 + q)$

Solution

$169 - q^{2}$

$(5 - 3 x) (5 + 3 x)$

$(4 - 6 y) (4 + 6 y)$

Solution

$16 - 36 y^{2}$

$(9 c - 2 d) (9 c + 2 d)$

$(7 w + 10 x) (7 w - 10 x)$

Solution

$49 w^{2} - 100 x^{2}$

$(m + \frac{2}{3} n) (m - \frac{2}{3} n)$

$(p + \frac{4}{5} q) (p - \frac{4}{5} q)$

Solution

$p^{2} - \frac{16}{25} q^{2}$

$(a b - 4) (a b + 4)$

$(x y - 9) (x y + 9)$

Solution

$x^{2} y^{2} - 81$

$(u v - \frac{3}{5}) (u v + \frac{3}{5})$

$(r s - \frac{2}{7}) (r s + \frac{2}{7})$

Solution

$r^{2} s^{2} - \frac{4}{49}$

$(2 x^{2} - 3 y^{4}) (2 x^{2} + 3 y^{4})$

$(6 m^{3} - 4 n^{5}) (6 m^{3} + 4 n^{5})$

Solution

$36 m^{6} - 16 n^{10}$

$(12 p^{3} - 11 q^{2}) (12 p^{3} + 11 q^{2})$

$(15 m^{2} - 8 n^{4}) (15 m^{2} + 8 n^{4})$

Solution

$225 m^{4} - 64 n^{8}$

Recognize and Use the Appropriate Special Product Pattern

In the following exercises, find each product.

ⓐ $(p - 3) (p + 3)$ ⓑ ${(t - 9)}^{2}$ ⓒ ${(m + n)}^{2}$ ⓓ $(2 x + y) (x - 2 y)$

ⓐ ${(2 r + 12)}^{2}$ ⓑ $(3 p + 8) (3 p - 8)$ ⓒ $(7 a + b) (a - 7 b)$ ⓓ ${(k - 6)}^{2}$

Solution

ⓐ $4 r^{2} + 48 r + 144$ ⓑ $9 p^{2} - 64$ ⓒ $7 a^{2} - 48 a b - 7 b^{2}$ ⓓ $k^{2} - 12 k + 36$

ⓐ ${(a^{5} - 7 b)}^{2}$ ⓑ $(x^{2} + 8 y) (8 x - y^{2})$ ⓒ $(r^{6} + s^{6}) (r^{6} - s^{6})$ ⓓ ${(y^{4} + 2 z)}^{2}$

ⓐ $(x^{5} + y^{5}) (x^{5} - y^{5})$ ⓑ ${(m^{3} - 8 n)}^{2}$ ⓒ ${(9 p + 8 q)}^{2}$ ⓓ $(r^{2} - s^{3}) (r^{3} + s^{2})$

Solution

ⓐ $x^{10} - y^{10}$ ⓑ $m^{6} - 16 m^{3} n + 64 n^{2}$ ⓒ $81 p^{2} + 144 p q + 64 q^{2}$ ⓓ $r^{5} + r^{2} s^{2} - r^{3} s^{3} - s^{5}$

Everyday Math

Mental math You can use the product of conjugates pattern to multiply numbers without a calculator. Say you need to multiply 47 times 53. Think of 47 as $50 - 3$ and 53 as $50 + 3$ .

ⓐ Multiply $(50 - 3) (50 + 3)$ by using the product of conjugates pattern, $(a - b) (a + b) = a^{2} - b^{2}$ .
ⓑ Multiply $47 \cdot 53$ without using a calculator.
ⓒ Which way is easier for you? Why?

Mental math You can use the binomial squares pattern to multiply numbers without a calculator. Say you need to square 65. Think of 65 as $60 + 5$ .

ⓐ Multiply ${(60 + 5)}^{2}$ by using the binomial squares pattern, ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ .
ⓑ Square 65 without using a calculator.
ⓒ Which way is easier for you? Why?

Solution

ⓐ 4,225 ⓑ 4,225 ⓒ Answers will vary.

Writing Exercises

How do you decide which pattern to use?

Why does ${(a + b)}^{2}$ result in a trinomial, but $(a - b) (a + b)$ result in a binomial?

Solution

Answers will vary.

Marta did the following work on her homework paper:

\begin{matrix} {(3 - y)}^{2} \\ 3^{2} - y^{2} \\ 9 - y^{2} \end{matrix}

Explain what is wrong with Marta’s work.

Use the order of operations to show that ${(3 + 5)}^{2}$ is 64, and then use that numerical example to explain why ${(a + b)}^{2} \neq a^{2} + b^{2}$ .

Solution

Answers will vary.

Self Check

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

ⓑ 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?

conjugate pair

A conjugate pair is two binomials of the form

(a - b), (a + b)

; the pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.

Divide Monomials

Simplify Expressions Using the Quotient Property for Exponents

Earlier in this chapter, we developed the properties of exponents for multiplication. We summarize these properties below.

Summary of Exponent Properties for Multiplication

If $a and b$ are real numbers, and $m and n$ are whole numbers, then

This table presents fundamental properties of exponents, including product, power, and product-to-a-power rules, along with their corresponding mathematical formulas.
Product Property	$a^{m} \cdot a^{n} = a^{m + n}$
Power Property	${(a^{m})}^{n} = a^{m \cdot n}$
Product to a Power	${(a b)}^{m} = a^{m} b^{m}$

Now we will look at the exponent properties for division. A quick memory refresher may help before we get started. You have learned to simplify fractions by dividing out common factors from the numerator and denominator using the Equivalent Fractions Property. This property will also help you work with algebraic fractions—which are also quotients.

As before, we’ll try to discover a property by looking at some examples.

Step-by-step simplification of rational expressions with exponents using the Equivalent Fractions Property.
Consider	$\frac{x^{5}}{x^{2}}$	and	$\frac{x^{2}}{x^{3}}$
What do they mean?	$\frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x}$		$\frac{x \cdot x}{x \cdot x \cdot x}$
Use the Equivalent Fractions Property.	$\frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x}$		$\frac{x \cdot x \cdot 1}{x \cdot x \cdot x}$
Simplify.	$x^{3}$		$\frac{1}{x}$

Notice, in each case the bases were the same and we subtracted exponents.

When the larger exponent was in the numerator, we were left with factors in the numerator.

When the larger exponent was in the denominator, we were left with factors in the denominator—notice the numerator of 1.

We write:

\begin{matrix} \frac{x^{5}}{x^{2}} & \frac{x^{2}}{x^{3}} \\ x^{5 - 2} & \frac{1}{x^{3 - 2}} \\ x^{3} & \frac{1}{x} \end{matrix}

This leads to the Quotient Property for Exponents.

A couple of examples with numbers may help to verify this property.

\begin{matrix} \frac{3^{4}}{3^{2}} & = & 3^{4 - 2} & \frac{5^{2}}{5^{3}} & = & \frac{1}{5^{3 - 2}} \\ \frac{81}{9} & = & 3^{2} & \frac{25}{125} & = & \frac{1}{5^{1}} \\ 9 & = & 9 ✓ & \frac{1}{5} & = & \frac{1}{5} ✓ \end{matrix}

Simplify: ⓐ $\frac{x^{9}}{x^{7}}$ ⓑ $\frac{3^{10}}{3^{2}} .$

Solution

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

ⓐ

Since 9 > 7, there are more factors of x in the numerator.

Use the Quotient Property, $\frac{a^{m}}{a^{n}} = a^{m - n}$ .

Simplify.
ⓑ

Since 10 > 2, there are more factors of x in the numerator.

Use the Quotient Property, $\frac{a^{m}}{a^{n}} = a^{m - n}$ .

Simplify.

Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.

Simplify: ⓐ $\frac{b^{8}}{b^{12}}$ ⓑ $\frac{7^{3}}{7^{5}} .$

Solution

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

ⓐ

Since 12 > 8, there are more factors of b in the denominator.

Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .

Simplify.
ⓑ

Since 5 > 3, there are more factors of 3 in the denominator.

Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .

Simplify.

Simplify. $The fraction one forty-ninth is displayed, represented as 1 over 49 with a horizontal line separating the numerator and the denominator.$

Notice that when the larger exponent is in the denominator, we are left with factors in the denominator.

Notice the difference in the two previous examples:

If we start with more factors in the numerator, we will end up with factors in the numerator.
If we start with more factors in the denominator, we will end up with factors in the denominator.

The first step in simplifying an expression using the Quotient Property for Exponents is to determine whether the exponent is larger in the numerator or the denominator.

Simplify: ⓐ $\frac{a^{5}}{a^{9}}$ ⓑ $\frac{x^{11}}{x^{7}} .$

Solution

ⓐ Is the exponent of $a$ larger in the numerator or denominator? Since 9 > 5, there are more $a' s$ in the denominator and so we will end up with factors in the denominator.

Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .

Simplify.
ⓑ Notice there are more factors of $x$ in the numerator, since 11 > 7. So we will end up with factors in the numerator.

Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .

Simplify.

Simplify Expressions with an Exponent of Zero

A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like $\frac{a^{m}}{a^{m}}$ . From your earlier work with fractions, you know that:

\frac{2}{2} = 1 \frac{17}{17} = 1 \frac{−43}{−43} = 1

In words, a number divided by itself is 1. So, $\frac{x}{x} = 1$ , for any $x (x \neq 0)$ , since any number divided by itself is 1.

The Quotient Property for Exponents shows us how to simplify $\frac{a^{m}}{a^{n}}$ when $m > n$ and when $n < m$ by subtracting exponents. What if $m = n$ ?

Consider $\frac{8}{8}$ , which we know is 1.

This table illustrates the derivation of the zero exponent rule, showing step-by-step how any non-zero number raised to the power of zero equals one.
	$\begin{matrix} \frac{8}{8} = 1 \end{matrix}$
Write $8$ as $2^{3}$ .	$\begin{matrix} \frac{2^{3}}{2^{3}} = 1 \end{matrix}$
Subtract exponents.	$\begin{matrix} 2^{3 - 3} = 1 \end{matrix}$
Simplify.	$\begin{matrix} 2^{0} = 1 \end{matrix}$

Now we will simplify $\frac{a^{m}}{a^{m}}$ in two ways to lead us to the definition of the zero exponent. In general, for $a \neq 0$ :

$This figure is divided into two columns. At the top of the figure, the left and right columns both contain a to the m power divided by a to the m power. In the next row, the left column contains a to the m minus m power. The right column contains the fraction m factors of a divided by m factors of a, represented in the numerator and denominator by a times a followed by an ellipsis. All the as in the numerator and denominator are canceled out. In the bottom row, the left column contains a to the zero power. The right column contains 1.$

We see $\frac{a^{m}}{a^{m}}$ simplifies to $a^{0}$ and to 1. So $a^{0} = 1$ .

In this text, we assume any variable that we raise to the zero power is not zero.

Simplify: ⓐ $9^{0}$ ⓑ $n^{0} .$

Solution

The definition says any non-zero number raised to the zero power is 1.

This table demonstrates the zero exponent rule with examples, showing that any non-zero base raised to the power of zero equals one.
ⓐ Use the definition of the zero exponent.	$\begin{matrix} 9^{0} \\ 1 \end{matrix}$
ⓑ Use the definition of the zero exponent.	$\begin{matrix} n^{0} \\ 1 \end{matrix}$

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 ${(2 x)}^{0}$ . We can use the product to a power rule to rewrite this expression.

Step-by-step simplification of (2x)^0, illustrating the product to a power rule and zero exponent property.
	${(2 x)}^{0}$
Use the product to a power rule.	$2^{0} x^{0}$
Use the zero exponent property.	$1 \cdot 1$
Simplify.	$1$

This tells us that any nonzero expression raised to the zero power is one.

Simplify: ⓐ ${(5 b)}^{0}$ ⓑ ${(−4 a^{2} b)}^{0} .$

Solution

Examples demonstrating the zero exponent rule, where any non-zero base raised to the power of zero equals one.
ⓐ	${(5 b)}^{0}$
Use the definition of the zero exponent.	$1$
ⓑ	${(−4 a^{2} b)}^{0}$
Use the definition of the zero exponent.	$1$

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.

Steps demonstrating how to expand and simplify a fraction raised to a power, showing (x/y)^3 equals x^3/y^3.
	${(\frac{x}{y})}^{3}$
This means:	$\frac{x}{y} \cdot \frac{x}{y} \cdot \frac{x}{y}$
Multiply the fractions.	$\frac{x \cdot x \cdot x}{y \cdot y \cdot y}$
Write with exponents.	$\frac{x^{3}}{y^{3}}$

Notice that the exponent applies to both the numerator and the denominator.

Illustration of the power of a quotient rule: (x/y)3 = x3/y3.
We write:	${(\frac{x}{y})}^{3}$
	$\frac{x^{3}}{y^{3}}$

This leads to the Quotient to a Power Property for Exponents.

An example with numbers may help you understand this property:

\begin{matrix} {(\frac{2}{3})}^{3} & = & \frac{2^{3}}{3^{3}} \\ \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} & = & \frac{8}{27} \\ \frac{8}{27} & = & \frac{8}{27} ✓ \end{matrix}

Simplify: ⓐ ${(\frac{3}{7})}^{2}$ ⓑ ${(\frac{b}{3})}^{4}$ ⓒ ${(\frac{k}{j})}^{3} .$

Solution

ⓐ


Use the Quotient Property, ${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$ .
Simplify.

ⓑ


Use the Quotient Property, ${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$ .
Simplify.

ⓒ


Raise the numerator and denominator to the third power.

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 $a and b$ are real numbers, and $m and n$ are whole numbers, then

This table summarizes fundamental exponent properties, including product, power, quotient, and zero exponent rules, presented with their corresponding mathematical formulas.
Product Property	$a^{m} \cdot a^{n} = a^{m + n}$
Power Property	${(a^{m})}^{n} = a^{m \cdot n}$
Product to a Power	${(a b)}^{m} = a^{m} b^{m}$
Quotient Property	$\begin{matrix} \frac{a^{m}}{a^{n}} & = & a^{m - n}, a \neq 0, m > n \\ \frac{a^{m}}{a^{n}} & = & \frac{1}{a^{n - m}}, a \neq 0, n > m \end{matrix}$
Zero Exponent Definition	$a^{o} = 1, a \neq 0$
Quotient to a Power Property	${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}, b \neq 0$

Simplify: $\frac{{(y^{4})}^{2}}{y^{6}} .$

Solution

Steps to simplify the exponential expression (y^4)^2 / y^6 using exponent rules.
	$\frac{{(y^{4})}^{2}}{y^{6}}$
Multiply the exponents in the numerator.	$\frac{y^{8}}{y^{6}}$
Subtract the exponents.	$y^{2}$

Simplify: $\frac{b^{12}}{{(b^{2})}^{6}} .$

Solution

This table illustrates the step-by-step simplification of the exponential expression b^12 / (b^2)^6 using exponent rules, resulting in a final value of 1.
	$\frac{b^{12}}{{(b^{2})}^{6}}$
Multiply the exponents in the numerator.	$\frac{b^{12}}{b^{12}}$
Subtract the exponents.	$b^{0}$
Simplify.	$1$

Simplify: ${(\frac{y^{9}}{y^{4}})}^{2} .$

Solution

This table demonstrates the step-by-step simplification of the algebraic expression (y^9 / y^4)^2, applying exponent rules for division and power of a power.
	${(\frac{y^{9}}{y^{4}})}^{2}$
Remember parentheses come before exponents. Notice the bases are the same, so we can simplify inside the parentheses. Subtract the exponents.	${(y^{5})}^{2}$
Multiply the exponents.	$y^{10}$

Simplify: ${(\frac{j^{2}}{k^{3}})}^{4} .$

Solution

Here we cannot simplify inside the parentheses first, since the bases are not the same.

Step-by-step simplification of a rational expression using the Quotient to a Power Property and other exponent rules.
	${(\frac{j^{2}}{k^{3}})}^{4}$
Raise the numerator and denominator to the fourth power using the Quotient to a Power Property, ${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$ .
Use the Power Property and simplify.	$\frac{j^{8}}{k^{12}}$

Simplify: ${(\frac{2 m^{2}}{5 n})}^{4} .$

Solution

Step-by-step simplification of a rational expression raised to a power using exponent properties.
	${(\frac{2 m^{2}}{5 n})}^{4}$
Raise the numberator and denominator to the fourth power, using the Quotient to a Power Property, ${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$ .	$\frac{{(2 m^{2})}^{4}}{{(5 n)}^{4}}$
Raise each factor to the fourth power.	$\frac{{(2 m^{2})}^{4}}{{(5 n)}^{4}}$
Use the Power Property and simplify.	$\frac{16 m^{8}}{625 n^{4}}$

Simplify: $\frac{{(x^{3})}^{4} {(x^{2})}^{5}}{{(x^{6})}^{5}} .$

Solution

Step-by-step simplification of an algebraic expression using power and quotient properties of exponents.
	$\frac{{(x^{3})}^{4} {(x^{2})}^{5}}{{(x^{6})}^{5}}$
Use the Power Property, ${(a^{m})}^{n} = a^{m \cdot n}$ .	$\frac{(x^{12}) (x^{10})}{(x^{30})}$
Add the exponents in the numerator.	$\frac{x^{22}}{x^{30}}$
Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .	$\frac{1}{x^{8}}$

Simplify: $\frac{{(10 p^{3})}^{2}}{{(5 p)}^{3} {(2 p^{5})}^{4}} .$

Solution

Step-by-step simplification of a complex algebraic expression using properties of exponents.
	$\frac{{(10 p^{3})}^{2}}{{(5 p)}^{3} {(2 p^{5})}^{4}}$
Use the Product to a Power Property, ${(a b)}^{m} = a^{m} b^{m}$ .	$\frac{{(10)}^{2} {(p^{3})}^{2}}{{(5)}^{3} {(p)}^{3} {(2)}^{4} {(p^{5})}^{4}}$
Use the Power Property, ${(a^{m})}^{n} = a^{m \cdot n}$ .	$\frac{100 p^{6}}{125 p^{3} \cdot 16 p^{20}}$
Add the exponents in the denominator.	$\frac{100 p^{6}}{125 \cdot 16 p^{23}}$
Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .	$\frac{100}{125 \cdot 16 p^{17}}$
Simplify.	$\frac{1}{20 p^{17}}$

Divide Monomials

You have now been introduced to all the properties of exponents and used them to simplify expressions. Next, you’ll see how to use these properties to divide monomials. Later, you’ll use them to divide polynomials.

Find the quotient: $56 x^{7} \div 8 x^{3} .$

Solution

Steps for simplifying 56x^7 8x^3 to 7x^4, illustrating algebraic division and exponent properties.
	$56 x^{7} \div 8 x^{3}$
Rewrite as a fraction.	$\frac{56 x^{7}}{8 x^{3}}$
Use fraction multiplication.	$\frac{56}{8} \cdot \frac{x^{7}}{x^{3}}$
Simplify and use the Quotient Property.	$7 x^{4}$

Find the quotient: $\frac{45 a^{2} b^{3}}{−5 a b^{5}} .$

Solution

Steps to simplify a rational algebraic expression using fraction multiplication and quotient properties.
	$\frac{45 a^{2} b^{3}}{−5 a b^{5}}$
Use fraction multiplication.	$\frac{45}{−5} \cdot \frac{a^{2}}{a} \cdot \frac{b^{3}}{b^{5}}$
Simplify and use the Quotient Property.	$−9 \cdot a \cdot \frac{1}{b^{2}}$
Multiply.	$- \frac{9 a}{b^{2}}$

Find the quotient: $\frac{24 a^{5} b^{3}}{48 a b^{4}} .$

Solution

Step-by-step simplification of an algebraic rational expression using fraction multiplication and quotient properties.
	$\frac{24 a^{5} b^{3}}{48 a b^{4}}$
Use fraction multiplication.	$\frac{24}{48} \cdot \frac{a^{5}}{a} \cdot \frac{b^{3}}{b^{4}}$
Simplify and use the Quotient Property.	$\frac{1}{2} \cdot a^{4} \cdot \frac{1}{b}$
Multiply.	$\frac{a^{4}}{2 b}$

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: $\frac{14 x^{7} y^{12}}{21 x^{11} y^{6}} .$

Solution

Be very careful to simplify $\frac{14}{21}$ by dividing out a common factor, and to simplify the variables by subtracting their exponents.

This table demonstrates the simplification of a rational expression using the Quotient Property, showing the original expression and its simplified form.
	$\frac{14 x^{7} y^{12}}{21 x^{11} y^{6}}$
Simplify and use the Quotient Property.	$\frac{2 y^{6}}{3 x^{4}}$

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. This follows the order of operations. Remember, a fraction bar is a grouping symbol.

Find the quotient: $\frac{(6 x^{2} y^{3}) (5 x^{3} y^{2})}{(3 x^{4} y^{5})} .$

Solution

Step-by-step simplification of a rational algebraic expression.
	$\frac{(6 x^{2} y^{3}) (5 x^{3} y^{2})}{(3 x^{4} y^{5})}$
Simplify the numerator.	$\frac{30 x^{5} y^{5}}{3 x^{4} y^{5}}$
Simplify.	$10 x$

Key Concepts

Quotient Property for Exponents:
- If $a$ is a real number, $a \neq 0$ , and $m, n$ are whole numbers, then:
  $\frac{a^{m}}{a^{n}} = a^{m - n}, m > n and \frac{a^{m}}{a^{n}} = \frac{1}{a^{m - n}}, n > m$
Zero Exponent
- If $a$ is a non-zero number, then $a^{0} = 1$ .
Quotient to a Power Property for Exponents:
- If $a$ and $b$ are real numbers, $b \neq 0,$ and $m$ is a counting number, then:
  ${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}$
- To raise a fraction to a power, raise the numerator and denominator to that power.
Summary of Exponent Properties
- If $a, b$ are real numbers and $m, n$ are whole numbers, then
  $\begin{matrix} Product Property & a^{m} \cdot a^{n} & = & a^{m + n} \\ Power Property & {(a^{m})}^{n} & = & a^{m \cdot n} \\ Product to a Power & {(a b)}^{m} & = & a^{m} b^{m} \\ Quotient Property & \frac{a^{m}}{a^{n}} & = & a^{m - n}, a \neq 0, m > n \\ \frac{a^{m}}{a^{n}} & = & \frac{1}{a^{n - m}}, a \neq 0, n > m \\ Zero Exponent Definition & a^{o} & = & 1, a \neq 0 \\ Quotient to a Power Property & {(\frac{a}{b})}^{m} & = & \frac{a^{m}}{b^{m}}, b \neq 0 \end{matrix}$

Practice Makes Perfect

Simplify Expressions Using the Quotient Property for Exponents

In the following exercises, simplify.

ⓐ $\frac{x^{18}}{x^{3}}$ ⓑ $\frac{5^{12}}{5^{3}}$

ⓐ $\frac{y^{20}}{y^{10}}$ ⓑ $\frac{7^{16}}{7^{2}}$

Solution

ⓐ $y^{10}$ ⓑ $7^{14}$

ⓐ $\frac{p^{21}}{p^{7}}$ ⓑ $\frac{4^{16}}{4^{4}}$

ⓐ $\frac{u^{24}}{u^{3}}$ ⓑ $\frac{9^{15}}{9^{5}}$

Solution

ⓐ $u^{21}$ ⓑ $9^{10}$

ⓐ $\frac{q^{18}}{q^{36}}$ ⓑ $\frac{10^{2}}{10^{3}}$

ⓐ $\frac{t^{10}}{t^{40}}$ ⓑ $\frac{8^{3}}{8^{5}}$

Solution

ⓐ $\frac{1}{t^{30}}$ ⓑ $\frac{1}{64}$

ⓐ $\frac{b}{b^{9}}$ ⓑ $\frac{4}{4^{6}}$

ⓐ $\frac{x}{x^{7}}$ ⓑ $\frac{10}{10^{3}}$

Solution

ⓐ $\frac{1}{x^{6}}$ ⓑ $\frac{1}{100}$

Simplify Expressions with Zero Exponents

In the following exercises, simplify.

ⓐ $20^{0}$ ⓑ $b^{0}$

ⓐ $13^{0}$ ⓑ $k^{0}$

Solution

ⓐ 1 ⓑ 1

ⓐ $− 27^{0}$ ⓑ $− (27^{0})$

ⓐ $− 15^{0}$ ⓑ $− (15^{0})$

Solution

ⓐ $−1$ ⓑ $−1$

ⓐ ${(25 x)}^{0}$ ⓑ $25 x^{0}$

ⓐ ${(6 y)}^{0}$ ⓑ $6 y^{0}$

Solution

ⓐ 1 ⓑ 6

ⓐ ${(12 x)}^{0}$ ⓑ ${(−56 p^{4} q^{3})}^{0}$

ⓐ $7 y^{0}$ ${(17 y)}^{0}$ ⓑ ${(−93 c^{7} d^{15})}^{0}$

Solution

ⓐ 7 ⓑ 1

ⓐ $12 n^{0} - 18 m^{0}$ ⓑ ${(12 n)}^{0} - {(18 m)}^{0}$

ⓐ $15 r^{0} - 22 s^{0}$ ⓑ ${(15 r)}^{0} - {(22 s)}^{0}$

Solution

ⓐ $−7$ ⓑ 0

Simplify Expressions Using the Quotient to a Power Property

In the following exercises, simplify.

ⓐ ${(\frac{3}{4})}^{3}$ ⓑ ${(\frac{p}{2})}^{5}$ ⓒ ${(\frac{x}{y})}^{6}$

ⓐ ${(\frac{2}{5})}^{2}$ ⓑ ${(\frac{x}{3})}^{4}$ ⓒ ${(\frac{a}{b})}^{5}$

Solution

ⓐ $\frac{4}{25}$ ⓑ $\frac{x^{4}}{81}$ ⓒ $\frac{a^{5}}{b^{5}}$

ⓐ ${(\frac{a}{3 b})}^{4}$ ⓑ ${(\frac{5}{4 m})}^{2}$

ⓐ ${(\frac{x}{2 y})}^{3}$ ⓑ ${(\frac{10}{3 q})}^{4}$

Solution

ⓐ $\frac{x^{3}}{8 y^{3}}$ ⓑ $\frac{10,000}{81 q^{4}}$

Simplify Expressions by Applying Several Properties

In the following exercises, simplify.

$\frac{{(a^{2})}^{3}}{a^{4}}$

$\frac{{(p^{3})}^{4}}{p^{5}}$

Solution

$p^{7}$

$\frac{{(y^{3})}^{4}}{y^{10}}$

$\frac{{(x^{4})}^{5}}{x^{15}}$

Solution

$x^{5}$

$\frac{u^{6}}{{(u^{3})}^{2}}$

$\frac{v^{20}}{{(v^{4})}^{5}}$

Solution

$\frac{m^{12}}{{(m^{8})}^{3}}$

$\frac{n^{8}}{{(n^{6})}^{4}}$

Solution

$\frac{1}{n^{16}}$

${(\frac{p^{9}}{p^{3}})}^{5}$

${(\frac{q^{8}}{q^{2}})}^{3}$

Solution

$q^{18}$

${(\frac{r^{2}}{r^{6}})}^{3}$

${(\frac{m^{4}}{m^{7}})}^{4}$

Solution

$\frac{1}{m^{12}}$

${(\frac{p}{r^{11}})}^{2}$

${(\frac{a}{b^{6}})}^{3}$

Solution

$\frac{a^{3}}{b^{18}}$

${(\frac{w^{5}}{x^{3}})}^{8}$

${(\frac{y^{4}}{z^{10}})}^{5}$

Solution

$\frac{y^{20}}{z^{50}}$

${(\frac{2 j^{3}}{3 k})}^{4}$

${(\frac{3 m^{5}}{5 n})}^{3}$

Solution

$\frac{27 m^{15}}{125 n^{3}}$

${(\frac{3 c^{2}}{4 d^{6}})}^{3}$

${(\frac{5 u^{7}}{2 v^{3}})}^{4}$

Solution

$\frac{625 u^{28}}{16 v^{^{12}}}$

${(\frac{k^{2} k^{8}}{k^{3}})}^{2}$

${(\frac{j^{2} j^{5}}{j^{4}})}^{3}$

Solution

$j^{9}$

$\frac{{(t^{2})}^{5} {(t^{4})}^{2}}{{(t^{3})}^{7}}$

$\frac{{(q^{3})}^{6} {(q^{2})}^{3}}{{(q^{4})}^{8}}$

Solution

$\frac{1}{q^{8}}$

$\frac{{(−2 p^{2})}^{4} {(3 p^{4})}^{2}}{{(−6 p^{3})}^{2}}$

$\frac{{(−2 k^{3})}^{2} {(6 k^{2})}^{4}}{{(9 k^{4})}^{2}}$

Solution

$64 k^{6}$

$\frac{{(−4 m^{3})}^{2} {(5 m^{4})}^{3}}{{(−10 m^{6})}^{3}}$

$\frac{{(−10 n^{2})}^{3} {(4 n^{5})}^{2}}{{(2 n^{8})}^{2}}$

Solution

$−4,000$

Divide Monomials

In the following exercises, divide the monomials.

$56 b^{8} \div 7 b^{2}$

$63 v^{10} \div 9 v^{2}$

Solution

$7 v^{8}$

$−88 y^{15} \div 8 y^{3}$

$−72 u^{12} \div 12 u^{4}$

Solution

$−6 u^{8}$

$\frac{45 a^{6} b^{8}}{−15 a^{10} b^{2}}$

$\frac{54 x^{9} y^{3}}{−18 x^{6} y^{15}}$

Solution

$- \frac{3 x^{3}}{y^{12}}$

$\frac{15 r^{4} s^{9}}{18 r^{9} s^{2}}$

$\frac{20 m^{8} n^{4}}{30 m^{5} n^{9}}$

Solution

$\frac{2 m^{3}}{3 n^{5}}$

$\frac{18 a^{4} b^{8}}{−27 a^{9} b^{5}}$

$\frac{45 x^{5} y^{9}}{−60 x^{8} y^{6}}$

Solution

$\frac{−3 y^{3}}{4 x^{3}}$

$\frac{64 q^{11} r^{9} s^{3}}{48 q^{6} r^{8} s^{5}}$

$\frac{65 a^{10} b^{8} c^{5}}{42 a^{7} b^{6} c^{8}}$

Solution

$\frac{65 a^{3} b^{2}}{42 c^{3}}$

$\frac{(10 m^{5} n^{4}) (5 m^{3} n^{6})}{25 m^{7} n^{5}}$

$\frac{(−18 p^{4} q^{7}) (−6 p^{3} q^{8})}{−36 p^{12} q^{10}}$

Solution

$\frac{−3 q^{5}}{p^{5}}$

$\frac{(6 a^{4} b^{3}) (4 a b^{5})}{(12 a^{2} b) (a^{3} b)}$

$\frac{(4 u^{2} v^{5}) (15 u^{3} v)}{(12 u^{3} v) (u^{4} v)}$

Solution

$\frac{5 v^{4}}{u^{2}}$

Mixed Practice

ⓐ $24 a^{5} + 2 a^{5}$ ⓑ $24 a^{5} - 2 a^{5}$ ⓒ $24 a^{5} \cdot 2 a^{5}$ ⓓ $24 a^{5} \div 2 a^{5}$

ⓐ $15 n^{10} + 3 n^{10}$ ⓑ $15 n^{10} - 3 n^{10}$ ⓒ $15 n^{10} \cdot 3 n^{10}$ ⓓ $15 n^{10} \div 3 n^{10}$

Solution

ⓐ $18 n^{10}$ ⓑ $12 n^{10}$ ⓒ $45 n^{20}$ ⓓ 5

ⓐ $p^{4} \cdot p^{6}$ ⓑ ${(p^{4})}^{6}$

ⓐ $q^{5} \cdot q^{3}$ ⓑ ${(q^{5})}^{3}$

Solution

ⓐ $q^{8}$ ⓑ $q^{15}$

ⓐ $\frac{y^{3}}{y}$ ⓑ $\frac{y}{y^{3}}$

ⓐ $\frac{z^{6}}{z^{5}}$ ⓑ $\frac{z^{5}}{z^{6}}$

Solution

ⓐ $z$ ⓑ $\frac{1}{z}$

$(8 x^{5}) (9 x) \div 6 x^{3}$

$(4 y) (12 y^{7}) \div 8 y^{2}$

Solution

$6 y^{6}$

$\frac{27 a^{7}}{3 a^{3}} + \frac{54 a^{9}}{9 a^{5}}$

$\frac{32 c^{11}}{4 c^{5}} + \frac{42 c^{9}}{6 c^{3}}$

Solution

$15 c^{6}$

$\frac{32 y^{5}}{8 y^{2}} - \frac{60 y^{10}}{5 y^{7}}$

$\frac{48 x^{6}}{6 x^{4}} - \frac{35 x^{9}}{7 x^{7}}$

Solution

$3 x^{2}$

$\frac{63 r^{6} s^{3}}{9 r^{4} s^{2}} - \frac{72 r^{2} s^{2}}{6 s}$

$\frac{56 y^{4} z^{5}}{7 y^{3} z^{3}} - \frac{45 y^{2} z^{2}}{5 y}$

Solution

$− y z^{2}$

Everyday Math

Memory One megabyte is approximately $10^{6}$ bytes. One gigabyte is approximately $10^{9}$ bytes. How many megabytes are in one gigabyte?

Memory One gigabyte is approximately $10^{9}$ bytes. One terabyte is approximately $10^{12}$ bytes. How many gigabytes are in one terabyte?

Solution

$10^{3}$

Writing Exercises

Jennifer thinks the quotient $\frac{a^{24}}{a^{6}}$ simplifies to $a^{4}$ . What is wrong with her reasoning?

Maurice simplifies the quotient $\frac{d^{7}}{d}$ by writing $\frac{d^{7}}{d} = 7$ . What is wrong with his reasoning?

Solution

Answers will vary.

When Drake simplified $− 3^{0}$ and ${(−3)}^{0}$ he got the same answer. Explain how using the Order of Operations correctly gives different answers.

Robert thinks $x^{0}$ simplifies to 0. What would you say to convince Robert he 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.

ⓑ 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?

Divide Polynomials

Divide a Polynomial by a Monomial

In the last section, you learned how to divide a monomial by a monomial. As you continue to build up your knowledge of polynomials the next procedure is to divide a polynomial of two or more terms by a monomial.

The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition.

Demonstrates the algebraic simplification of adding two fractions with a common denominator.
The sum,	$\frac{y}{5} + \frac{2}{5},$
simplifies to	$\frac{y + 2}{5} .$

Now we will do this in reverse to split a single fraction into separate fractions.

We’ll state the fraction addition property here just as you learned it and in reverse.

We use the form on the left to add fractions and we use the form on the right to divide a polynomial by a monomial.

Demonstrates splitting a fraction with a sum in the numerator into a sum of two fractions, illustrating a fundamental algebraic property.
For example,	$\frac{y + 2}{5}$
can be written	$\frac{y}{5} + \frac{2}{5} .$

We use this form of fraction addition to divide polynomials by monomials.

Find the quotient: $\frac{7 y^{2} + 21}{7} .$

Solution

Steps to simplify the rational expression (7y^2 + 21) / 7 by dividing each term of the numerator by the denominator.
	$\frac{7 y^{2} + 21}{7}$
Divide each term of the numerator by the denominator.	$\frac{7 y^{2}}{7} + \frac{21}{7}$
Simplify each fraction.	$y^{2} + 3$

Remember that division can be represented as a fraction. When you are asked to divide a polynomial by a monomial and it is not already in fraction form, write a fraction with the polynomial in the numerator and the monomial in the denominator.

Find the quotient: $(18 x^{3} - 36 x^{2}) \div 6 x .$

Solution

Step-by-step example of dividing a polynomial by a monomial.
	$(18 x^{3} - 36 x^{2}) \div 6 x$
Rewrite as a fraction.	$\frac{18 x^{3} - 36 x^{2}}{6 x}$
Divide each term of the numerator by the denominator.	$\frac{18 x^{3}}{6 x} - \frac{36 x^{2}}{6 x}$
Simplify.	$3 x^{2} - 6 x$

When we divide by a negative, we must be extra careful with the signs.

Find the quotient: $\frac{12 d^{2} - 16 d}{−4} .$

Solution

Step-by-step simplification of a polynomial division problem, illustrating the process of dividing each term by the denominator.
	$\frac{12 d^{2} - 16 d}{−4}$
Divide each term of the numerator by the denominator.	$\frac{12 d^{2}}{−4} - \frac{16 d}{−4}$
Simplify. Remember, subtracting a negative is like adding a positive!	$−3 d^{2} + 4 d$

Find the quotient: $\frac{105 y^{5} + 75 y^{3}}{5 y^{2}} .$

Solution

This table illustrates the step-by-step process of dividing a polynomial by a monomial, including separating terms and simplifying the resulting expressions.
	$\frac{105 y^{5} + 75 y^{3}}{5 y^{2}}$
Separate the terms.	$\frac{105 y^{5}}{5 y^{2}} + \frac{75 y^{3}}{5 y^{2}}$
Simplify.	$21 y^{3} + 15 y$

Find the quotient: $(15 x^{3} y - 35 x y^{2}) \div (−5 x y) .$

Solution

Step-by-step solution demonstrating the division of a polynomial by a monomial.
	$(15 x^{3} y - 35 x y^{2}) \div (−5 x y)$
Rewrite as a fraction.	$\frac{15 x^{3} y - 35 x y^{2}}{−5 x y}$
Separate the terms.	$\frac{15 x^{3} y}{−5 x y} - \frac{35 x y^{2}}{−5 x y}$
Simplify.	$−3 x^{2} + 7 y$

Find the quotient: $\frac{36 x^{3} y^{2} + 27 x^{2} y^{2} - 9 x^{2} y^{3}}{9 x^{2} y} .$

Solution

Step-by-step simplification of a polynomial expression by dividing each term by a common monomial denominator.
	$\frac{36 x^{3} y^{2} + 27 x^{2} y^{2} - 9 x^{2} y^{3}}{9 x^{2} y}$
Separate the terms.	$\frac{36 x^{3} y^{2}}{9 x^{2} y} + \frac{27 x^{2} y^{2}}{9 x^{2} y} - \frac{9 x^{2} y^{3}}{9 x^{2} y}$
Simplify.	$4 x y + 3 y - y^{2}$

Find the quotient: $\frac{10 x^{2} + 5 x - 20}{5 x} .$

Solution

Steps for simplifying a rational algebraic expression by separating terms into individual fractions.
	$\frac{10 x^{2} + 5 x - 20}{5 x}$
Separate the terms.	$\frac{10 x^{2}}{5 x} + \frac{5 x}{5 x} - \frac{20}{5 x}$
Simplify.	$2 x + 1 - \frac{4}{x}$

Divide a Polynomial by a Binomial

To divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.

We write the long division
We divide the first two digits, 87, by 25.
We multiply 3 times 25 and write the product under the 87.
Now we subtract 75 from 87.
Then we bring down the third digit of the dividend, 5.
Repeat the process, dividing 25 into 125.

We check division by multiplying the quotient by the divisor.

If we did the division correctly, the product should equal the dividend.

\begin{matrix} 35 \cdot 25 \\ 875 ✓ \end{matrix}

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

Find the quotient: $(x^{2} + 9 x + 20) \div (x + 5) .$

Solution


Write it as a long division problem.
Be sure the dividend is in standard form.
Divide x² by x. It may help to ask yourself, "What do I need to multiply x by to get x²?"
Put the answer, x, in the quotient over the x term.
Multiply x times x + 5. Line up the like terms under the dividend.
Subtract x² + 5x from x² + 9x.
Then bring down the last term, 20.
Divide 4x by x. It may help to ask yourself, "What do I need to multiply x by to get 4x?"
Put the answer, 4, in the quotient over the constant term.
Multiply 4 times x + 5.
Subtract 4x + 20 from 4x + 20.
Check:
Multiply the quotient by the divisor.
(x + 4)(x + 5)
You should get the dividend.
x² + 9x + 20✓

When the divisor has subtraction sign, we must be extra careful when we multiply the partial quotient and then subtract. It may be safer to show that we change the signs and then add.

Find the quotient: $(2 x^{2} - 5 x - 3) \div (x - 3) .$

Solution


Write it as a long division problem.
Be sure the dividend is in standard form.
Divide 2x² by x. Put the answer, 2x, in the quotient over the x term.
Multiply 2x times x − 3. Line up the like terms under the dividend.
Subtract 2x² − 6x from 2x² − 5x. Change the signs and then add. Then bring down the last term.
Divide x by x. Put the answer, 1, in the quotient over the constant term.
Multiply 1 times x − 3.
Subtract x − 3 from x − 3 by changing the signs and adding.
To check, multiply (x − 3)(2x + 1).
The result should be 2x² − 5x − 3.

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In Example 10, we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Find the quotient: $(x^{3} - x^{2} + x + 4) \div (x + 1) .$

Solution


Write it as a long division problem.
Be sure the dividend is in standard form.
Divide x³ by x. Put the answer, x², in the quotient over the x² term. Multiply x² times x + 1. Line up the like terms under the dividend.
Subtract x³ + x² from x³ − x² by changing the signs and adding. Then bring down the next term.
Divide −2x² by x. Put the answer, −2x, in the quotient over the x term. Multiply −2x times x + 1. Line up the like terms under the dividend.
Subtract −2x² − 2x from −2x² + x by changing the signs and adding. Then bring down the last term.
Divide 3x by x. Put the answer, 3, in the quotient over the constant term. Multiply 3 times x + 1. Line up the like terms under the dividend.
Subtract 3x + 3 from 3x + 4 by changing the signs and adding. Write the remainder as a fraction with the divisor as the denominator.	$The sum of 3 x plus 4 and negative 3 x plus negative 3 is 1. Therefore, the polynomial x cubed minus x squared plus x plus 4, divided by the binomial x plus 1, equals x squared minus 2 x plus the fraction 1 over x plus 1.$
To check, multiply $(x + 1) (x^{2} - 2 x + 3 + \frac{1}{x + 1}) .$ The result should be $x^{3} - x^{2} + x + 4$ .

Look back at the dividends in Example 8, Example 9, and Example 10. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in Example 11 will be $x^{4} - x^{2} + 5 x - 2$ . It is missing an $x^{3}$ term. We will add in $0 x^{3}$ as a placeholder.

Find the quotient: $(x^{4} - x^{2} + 5 x - 2) \div (x + 2) .$

Solution

Notice that there is no $x^{3}$ term in the dividend. We will add $0 x^{3}$ as a placeholder.


Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms.
Divide x⁴ by x. Put the answer, x³, in the quotient over the x³ term. Multiply x³ times x + 2. Line up the like terms. Subtract and then bring down the next term.
Divide −2x³ by x. Put the answer, −2x², in the quotient over the x² term. Multiply −2x² times x + 1. Line up the like terms. Subtract and bring down the next term.
Divide 3x² by x. Put the answer, 3x, in the quotient over the x term. Multiply 3x times x + 1. Line up the like terms. Subtract and bring down the next term.
Divide −x by x. Put the answer, −1, in the quotient over the constant term. Multiply −1 times x + 1. Line up the like terms. Change the signs, add.
To check, multiply $(x + 2) (x^{3} - 2 x^{2} + 3 x - 1)$ .
The result should be $x^{4} - x^{2} + 5 x - 2$ .

In Example 12, we will divide by $2 a - 3$ . As we divide we will have to consider the constants as well as the variables.

Find the quotient: $(8 a^{3} + 27) \div (2 a + 3) .$

Solution

This time we will show the division all in one step. We need to add two placeholders in order to divide.

To check, multiply $(2 a + 3) (4 a^{2} - 6 a + 9)$ .

The result should be $8 a^{3} + 27$ .

Key Concepts

Fraction Addition
- If $a, b, and c$ are numbers where $c \neq 0$ , then
  $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c} and \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$
Division of a Polynomial by a Monomial
- To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Practice Makes Perfect

In the following exercises, divide each polynomial by the monomial.

$\frac{45 y + 36}{9}$

$\frac{30 b + 75}{5}$

Solution

$6 b + 15$

$\frac{8 d^{2} - 4 d}{2}$

$\frac{42 x^{2} - 14 x}{7}$

Solution

$6 x^{2} - 2 x$

$(16 y^{2} - 20 y) \div 4 y$

$(55 w^{2} - 10 w) \div 5 w$

Solution

$11 w - 2$

$(9 n^{4} + 6 n^{3}) \div 3 n$

$(8 x^{3} + 6 x^{2}) \div 2 x$

Solution

$4 x^{2} + 3 x$

$\frac{18 y^{2} - 12 y}{−6}$

$\frac{20 b^{2} - 12 b}{−4}$

Solution

$−5 b^{2} + 3 b$

$\frac{35 a^{4} + 65 a^{2}}{−5}$

$\frac{51 m^{4} + 72 m^{3}}{−3}$

Solution

$−17 m^{4} - 24 m^{3}$

$\frac{310 y^{4} - 200 y^{3}}{5 y^{2}}$

$\frac{412 z^{8} - 48 z^{5}}{4 z^{3}}$

Solution

$103 z^{5} - 12 z^{2}$

$\frac{46 x^{3} + 38 x^{2}}{2 x^{2}}$

$\frac{51 y^{4} + 42 y^{2}}{3 y^{2}}$

Solution

$17 y^{2} + 14$

$(24 p^{2} - 33 p) \div (−3 p)$

$(35 x^{4} - 21 x) \div (−7 x)$

Solution

$−5 x^{3} + 3$

$(63 m^{4} - 42 m^{3}) \div (−7 m^{2})$

$(48 y^{4} - 24 y^{3}) \div (−8 y^{2})$

Solution

$−6 y^{2} + 3 y$

$(63 a^{2} b^{3} + 72 a b^{4}) \div (9 a b)$

$(45 x^{3} y^{4} + 60 x y^{2}) \div (5 x y)$

Solution

$9 x^{2} y^{3} + 12 y$

$\frac{52 p^{5} q^{4} + 36 p^{4} q^{3} - 64 p^{3} q^{2}}{4 p^{2} q}$

$\frac{49 c^{2} d^{2} - 70 c^{3} d^{3} - 35 c^{2} d^{4}}{7 c d^{2}}$

Solution

$7 c - 10 c^{2} d - 5 c d^{2}$

$\frac{66 x^{3} y^{2} - 110 x^{2} y^{3} - 44 x^{4} y^{3}}{11 x^{2} y^{2}}$

$\frac{72 r^{5} s^{2} + 132 r^{4} s^{3} - 96 r^{3} s^{5}}{12 r^{2} s^{2}}$

Solution

$6 r^{3} + 11 r^{2} s - 8 r s^{3}$

$\frac{4 w^{2} + 2 w - 5}{2 w}$

$\frac{12 q^{2} + 3 q - 1}{3 q}$

Solution

$4 q + 1 - \frac{1}{3 q}$

$\frac{10 x^{2} + 5 x - 4}{−5 x}$

$\frac{20 y^{2} + 12 y - 1}{−4 y}$

Solution

$−5 y - 3 + \frac{1}{4 y}$

$\frac{36 p^{3} + 18 p^{2} - 12 p}{6 p^{2}}$

$\frac{63 a^{3} - 108 a^{2} + 99 a}{9 a^{2}}$

Solution

$7 a - 12 + \frac{11}{a}$

Divide a Polynomial by a Binomial

In the following exercises, divide each polynomial by the binomial.

$(y^{2} + 7 y + 12) \div (y + 3)$

$(d^{2} + 8 d + 12) \div (d + 2)$

Solution

$d + 6$

$(x^{2} - 3 x - 10) \div (x + 2)$

$(a^{2} - 2 a - 35) \div (a + 5)$

Solution

$a - 7$

$(t^{2} - 12 t + 36) \div (t - 6)$

$(x^{2} - 14 x + 49) \div (x - 7)$

Solution

$x - 7$

$(6 m^{2} - 19 m - 20) \div (m - 4)$

$(4 x^{2} - 17 x - 15) \div (x - 5)$

Solution

$4 x + 3$

$(q^{2} + 2 q + 20) \div (q + 6)$

$(p^{2} + 11 p + 16) \div (p + 8)$

Solution

$p + 3 - \frac{8}{p + 8}$

$(y^{2} - 3 y - 15) \div (y - 8)$

$(x^{2} + 2 x - 30) \div (x - 5)$

Solution

$x + 7 + \frac{5}{x - 5}$

$(3 b^{3} + b^{2} + 2) \div (b + 1)$

$(2 n^{3} - 10 n + 24) \div (n + 3)$

Solution

$2 n^{2} - 6 n + 8$

$(2 y^{3} - 6 y - 36) \div (y - 3)$

$(7 q^{3} - 5 q - 2) \div (q - 1)$

Solution

$7 q^{2} + 7 q + 2$

$(z^{3} + 1) \div (z + 1)$

$(m^{3} + 1000) \div (m + 10)$

Solution

$m^{2} - 10 m + 100$

$(a^{3} - 125) \div (a - 5)$

$(x^{3} - 216) \div (x - 6)$

Solution

$x^{2} + 6 x + 36$

$(64 x^{3} - 27) \div (4 x - 3)$

$(125 y^{3} - 64) \div (5 y - 4)$

Solution

$25 y^{2} + 20 x + 16$

Everyday Math

Average cost Pictures Plus produces digital albums. The company’s average cost (in dollars) to make $x$ albums is given by the expression $\frac{7 x + 500}{x}$ .

ⓐ Find the quotient by dividing the numerator by the denominator.
ⓑ What will the average cost (in dollars) be to produce 20 albums?

Handshakes At a company meeting, every employee shakes hands with every other employee. The number of handshakes is given by the expression $\frac{n^{2} - n}{2}$ , where $n$ represents the number of employees. How many handshakes will there be if there are 10 employees at the meeting?

Solution

Writing Exercises

James divides $48 y + 6$ by 6 this way: $\frac{48 y + 6}{6} = 48 y$ . What is wrong with his reasoning?

Divide $\frac{10 x^{2} + x - 12}{2 x}$ and explain with words how you get each term of the quotient.

Solution

Answers will vary.

Self Check

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

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

Integer Exponents and Scientific Notation

Use the Definition of a Negative Exponent

We saw that the Quotient Property for Exponents introduced earlier in this chapter, has two forms depending on whether the exponent is larger in the numerator or the denominator.

What if we just subtract exponents regardless of which is larger?

Let’s consider $\frac{x^{2}}{x^{5}}$ .

We subtract the exponent in the denominator from the exponent in the numerator.

\begin{matrix} \frac{x^{2}}{x^{5}} \\ x^{2 - 5} \\ x^{−3} \end{matrix}

We can also simplify $\frac{x^{2}}{x^{5}}$ by dividing out common factors:

Illustrated in this figure is x times x divided by x times x times x times x times x. Two xes cancel out in the numerator and denominator. Below this is the simplified term: 1 divided by x cubed.

This implies that $x^{−3} = \frac{1}{x^{3}}$ and it leads us to the definition of a negative exponent.

The negative exponent tells us we can 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 the expression with only positive exponents.

For example, if after simplifying an expression we end up with the expression $x^{−3}$ , we will take one more step and write $\frac{1}{x^{3}}$ . The answer is considered to be in simplest form when it has only positive exponents.

Simplify: ⓐ $4^{−2}$ ⓑ $10^{−3} .$

Solution

Step-by-step examples demonstrating the simplification of mathematical expressions with negative exponents.
ⓐ	$4^{−2}$
Use the definition of a negative exponent, $a^{− n} = \frac{1}{a^{n}}$ .	$\frac{1}{4^{2}}$
Simplify.	$\frac{1}{16}$
ⓑ	$10^{−3}$
Use the definition of a negative exponent, $a^{− n} = \frac{1}{a^{n}}$ .	$\frac{1}{10^{3}}$
Simplify.	$\frac{1}{1000}$

In Example 1 we raised an integer to a negative exponent. What happens when we raise a fraction to a negative exponent? We’ll start by looking at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.

Steps demonstrating the simplification of the expression 1/(a^-n) using the definition of negative exponents, resulting in a^n.
	$\frac{1}{a^{− n}}$
Use the definition of a negative exponent, $a^{− n} = \frac{1}{a^{n}}$ .	$\frac{1}{\frac{1}{a^{n}}}$
Simplify the complex fraction.	$1 \cdot \frac{a^{n}}{1}$
Multiply.	$a^{n}$

This leads to the Property of Negative Exponents.

Simplify: ⓐ $\frac{1}{y^{−4}}$ ⓑ $\frac{1}{3^{−2}} .$

Solution

Examples demonstrating step-by-step simplification of mathematical expressions using the property of negative exponents.
ⓐ	$\frac{1}{y^{−4}}$
Use the property of a negative exponent, $\frac{1}{a^{− n}} = a^{n}$ .	$y^{4}$
ⓑ	$\frac{1}{3^{−2}}$
Use the property of a negative exponent, $\frac{1}{a^{− n}} = a^{n}$ .	$3^{2}$
Simplify.	$9$

Suppose now we have a fraction raised to a negative exponent. Let’s use our definition of negative exponents to lead us to a new property.

Illustrates the step-by-step simplification of a fractional expression with a negative exponent, demonstrating the rule for inverses.
	${(\frac{3}{4})}^{−2}$
Use the definition of a negative exponent, $a^{− n} = \frac{1}{a^{n}}$ .	$\frac{1}{{(\frac{3}{4})}^{2}}$
Simplify the denominator.	$\frac{1}{\frac{9}{16}}$
Simplify the complex fraction.	$\frac{16}{9}$
But we know that $\frac{16}{9}$ is ${(\frac{4}{3})}^{2}$ .
This tells us that:	${(\frac{3}{4})}^{−2} = {(\frac{4}{3})}^{2}$

To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base—the fraction—and changed the sign of the exponent.

This leads us to the Quotient to a Negative Power Property.

Simplify: ⓐ ${(\frac{5}{7})}^{−2}$ ⓑ ${(- \frac{2 x}{y})}^{−3} .$

Solution

Step-by-step simplification of expressions using the Quotient to a Negative Exponent Property.
ⓐ	${(\frac{5}{7})}^{−2}$
Use the Quotient to a Negative Exponent Property, ${(\frac{a}{b})}^{− n} = {(\frac{b}{a})}^{n}$ .
Take the reciprocal of the fraction and change the sign of the exponent.	${(\frac{7}{5})}^{2}$
Simplify.	$\frac{49}{25}$
ⓑ	${(- \frac{2 x}{y})}^{−3}$
Use the Quotient to a Negative Exponent Property, ${(\frac{a}{b})}^{− n} = {(\frac{b}{a})}^{n}$ .
Take the reciprocal of the fraction and change the sign of the exponent.	${(- \frac{y}{2 x})}^{3}$
Simplify.	$- \frac{y^{3}}{8 x^{3}}$

When simplifying an expression with exponents, we must be careful to correctly identify the base.

Simplify: ⓐ ${(−3)}^{−2}$ ⓑ $− 3^{−2}$ ⓒ ${(- \frac{1}{3})}^{−2}$ ⓓ $− {(\frac{1}{3})}^{−2} .$

Solution

Step-by-step evaluation of mathematical expressions involving negative exponents, demonstrating their application to various bases and handling of negative signs.
ⓐ Here the exponent applies to the base $−3$ .	${(−3)}^{−2}$
Take the reciprocal of the base and change the sign of the exponent.	$\frac{1}{{(−3)}^{−2}}$
Simplify.	$\frac{1}{9}$
ⓑ The expression $− 3^{−2}$ means "find the opposite of $3^{−2}$ ." Here the exponent applies to the base ${(- \frac{1}{3})}^{}$ .	$− 3^{−2}$
Rewrite as a product with $-1$ .	$−1 \cdot 3^{−2}$
Take the reciprocal of the base and change the sign of the exponent.	$−1 \cdot \frac{1}{3^{2}}$
Simplify.	$- \frac{1}{9}$
ⓒ Here the exponent applies to the base ${(- \frac{1}{3})}^{}$ .	${(- \frac{1}{3})}^{−2}$
Take the reciprocal of the base and change the sign of the exponent.	${(- \frac{3}{1})}^{2}$
Simplify.	$9$
ⓓ The expression $− {(\frac{1}{3})}^{−2}$ means "find the opposite of ${(\frac{1}{3})}^{−2}$ ." Here the exponent applies to the base $(\frac{1}{3})$ .
Rewrite as a product with $-1$ .	$−1 \cdot {(\frac{1}{3})}^{−2}$
Take the reciprocal of the base and change the sign of the exponent.	$−1 \cdot {(\frac{3}{1})}^{2}$
Simplify.	$−9$

We must be careful to follow the Order of Operations. In the next example, parts (a) and (b) look similar, but the results are different.

Simplify: ⓐ $4 \cdot 2^{−1}$ ⓑ ${(4 \cdot 2)}^{−1} .$

Solution

Step-by-step simplification of expressions involving exponents and multiplication, demonstrating the order of operations.
ⓐ Do exponents before multiplication.	$4 \cdot 2^{−1}$
Use $a^{− n} = \frac{1}{a^{n}}$ .	$4 \cdot \frac{1}{2^{1}}$
Simplify.	$2$
ⓑ	${(4 \cdot 2)}^{−1}$
Simplify inside the parentheses first.	${(8)}^{−1}$
Use $a^{− n} = \frac{1}{a^{n}}$ .	$\frac{1}{8^{1}}$
Simplify.	$\frac{1}{8}$

When a variable is raised to a negative exponent, we apply the definition the same way we did with numbers. We will assume all variables are non-zero.

Simplify: ⓐ $x^{−6}$ ⓑ ${(u^{4})}^{−3} .$

Solution

ⓐ
$\begin{matrix} x^{−6} \\ Use the definition of a negative exponent, a^{− n} = \frac{1}{a^{n}} . & \frac{1}{x^{6}} \end{matrix}$
ⓑ
$\begin{matrix} {(u^{4})}^{−3} \\ Use the definition of a negative exponent, a^{− n} = \frac{1}{a^{n}} . & \frac{1}{{(u^{4})}^{3}} \\ Simplify. & \frac{1}{u^{12}} \end{matrix}$

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, we simplify expressions in parentheses before applying exponents. We’ll see how this works in the next example.

Simplify: ⓐ $5 y^{−1}$ ⓑ ${(5 y)}^{−1}$ ⓒ ${(−5 y)}^{−1} .$

Solution

ⓐ

The table demonstrates the steps to simplify the expression 5y⁻¹, illustrating the rule for negative exponents by taking the reciprocal.
	$5 y^{−1}$
Notice the exponent applies to just the base y. Take the reciprocal of y and change the sign of the exponent.	$5 \cdot \frac{1}{y^{1}}$
Simplify.	$\frac{5}{y}$

ⓑ

Steps to simplify an algebraic expression with a negative exponent, detailing each transformation.
	${(5 y)}^{−1}$
Here the parentheses make the exponent apply to the base 5y. Take the reciprocal of 5y and change the sign of the exponent.	$\frac{1}{{(5 y)}^{1}}$
Simplify.	$\frac{1}{5 y}$

ⓒ

Step-by-step simplification of the algebraic expression (-5y)^-1.
	${(−5 y)}^{−1}$
The base here is −5y. Take the reciprocal of −5y and change the sign of the exponent.	$\frac{1}{{(−5 y)}^{1}}$
Simplify.	$\frac{1}{−5 y}$
Use $\frac{a}{− b} = - \frac{a}{b} .$	$- \frac{1}{5 y}$

With negative exponents, the Quotient Rule needs only one form $\frac{a^{m}}{a^{n}} = a^{m - n}$ , for $a \neq 0$ . When the exponent in the denominator is larger than the exponent in the numerator, the exponent of the quotient will be negative.

Simplify Expressions with Integer Exponents

All of the exponent properties we developed earlier in the chapter with whole number exponents apply to integer exponents, too. We restate them here for reference.

Simplify: ⓐ $x^{−4} \cdot x^{6}$ ⓑ $y^{−6} \cdot y^{4}$ ⓒ $z^{−5} \cdot z^{−3} .$

Solution

ⓐ

Steps to simplify the exponential expression x^-4 * x^6 using the product property of exponents.
	$x^{−4} \cdot x^{6}$
Use the Product Property, $a^{m} \cdot a^{n} = a^{m + n} .$	$x^{−4 + 6}$
Simplify.	$x^{2}$

ⓑ

Step-by-step simplification of an algebraic expression involving negative exponents using exponent rules.
	$y^{−6} \cdot y^{4}$
Notice the same bases, so add the exponents.	$y^{−6 + 4}$
Simplify.	$y^{−2}$
Use the definition of a negative exponent, $a^{− n} = \frac{1}{a^{n}} .$	$\frac{1}{y^{2}}$

ⓒ

Step-by-step simplification of an expression involving negative exponents: z^-5 * z^-3.
	$z^{−5} \cdot z^{−3}$
Add the exponents, since the bases are the same.	$z^{−5 - 3}$
Simplify.	$z^{−8}$
Take the reciprocal and change the sign of the exponent, using the definition of a negative exponent.	$\frac{1}{z^{8}}$

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.

Simplify: $(m^{4} n^{−3}) (m^{−5} n^{−2}) .$

Solution

Step-by-step guide to simplifying an algebraic expression with negative exponents using exponent properties.
	$(m^{4} n^{−3}) (m^{−5} n^{−2})$
Use the Commutative Property to get like bases together.	$m^{4} m^{−5} \cdot n^{−2} n^{−3}$
Add the exponents for each base.	$m^{−1} \cdot n^{−5}$
Take reciprocals and change the signs of the exponents.	$\frac{1}{m^{1}} \cdot \frac{1}{n^{5}}$
Simplify.	$\frac{1}{m n^{5}}$

In the next two examples, we’ll use the Power Property and the Product to a Power Property.

Simplify: ${(6 k^{3})}^{−2} .$

Solution

Demonstration of simplifying the algebraic expression (6k^3)^-2 using various exponent properties.
	${(6 k^{3})}^{−2}$
Use the Product to a Power Property, ${(a b)}^{m} = a^{m} b^{m} .$	${(6)}^{−2} {(k^{3})}^{−2}$
Use the Power Property, ${(a^{m})}^{n} = a^{m \cdot n} .$	$6^{−2} k^{−6}$
Use the Definition of a Negative Exponent, $a^{− n} = \frac{1}{a^{n}} .$	$\frac{1}{6^{2}} \cdot \frac{1}{k^{6}}$
Simplify.	$\frac{1}{36 k^{6}}$

Simplify: ${(5 x^{−3})}^{2} .$

Solution

Step-by-step simplification of an algebraic expression using exponent properties.
	${(5 x^{−3})}^{2}$
Use the Product to a Power Property, ${(a b)}^{m} = a^{m} b^{m} .$	$5^{2} {(x^{−3})}^{2}$
Simplify 5² and multiply the exponents of x using the Power Property, ${(a^{m})}^{n} = a^{m \cdot n} .$	$25 \cdot x^{−6}$
Rewrite x⁻⁶ by using the Definition of a Negative Exponent, $a^{− n} = \frac{1}{a^{n}} .$	$25 \cdot \frac{1}{x^{6}}$
Simplify.	$\frac{25}{x^{6}}$

To simplify a fraction, we use the Quotient Property and subtract the exponents.

Simplify: $\frac{r^{5}}{r^{−4}} .$

Solution

Step-by-step simplification of an exponential expression, r^5 / r^-4, using the Quotient Property of Exponents.
	$\frac{r^{5}}{r^{−4}}$
Use the Quotient Property, $\frac{a^{m}}{a^{n}} = a^{m - n} .$	$r^{5 - (−4)}$
Simplify.	$r^{9}$

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 4,000 and $0.004$ . We know that 4,000 means $4 \times 1,000$ and 0.004 means $4 \times \frac{1}{1,000}$ .

If we write the 1000 as a power of ten in exponential form, we can rewrite these numbers in this way:

\begin{matrix} 4,000 & 0.004 \\ 4 \times 1,000 & 4 \times \frac{1}{1,000} \\ 4 \times 10^{3} & 4 \times \frac{1}{10^{3}} \\ 4 \times 10^{−3} \end{matrix}

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.

It is customary in scientific notation to use as the $\times$ multiplication sign, even though we avoid using this sign elsewhere in algebra.

If we look at what happened to the decimal point, we can see a method to easily convert from decimal notation to scientific notation.

This figure illustrates how to convert a number to scientific notation. It has two columns. In the first column is 4000 equals 4 times 10 to the third power. Below this, the equation is repeated, with an arrow demonstrating that the decimal point at the end of 4000 has moved three places to the left, so that 4000 becomes 4.000. The second column has 0.004 equals 4 times 10 to the negative third power. Below this, the equation is repeated, with an arrow demonstrating how the decimal point in 0.004 is moved three places to the right to produce 4.

In both cases, the decimal was moved 3 places to get the first factor between 1 and 10.

$\begin{matrix} The power of 10 is positive when the number is larger than 1: & 4,000 = 4 \times 10^{3} \\ The power of 10 is negative when the number is between 0 and 1: & 0.004 = 4 \times 10^{−3} \end{matrix}$

How to Convert from Decimal Notation to Scientific Notation

Write in scientific notation: 37,000.

Solution

In the second row, the first cell reads “Step 2. Count the number of decimal places, n, that the decimal place was moved. The second cell reads “The decimal point was moved 4 places to the left.” The third cell contains 370000 again, with an arrow showing the decimal point jumping places to the left from the end of the number until it ends up between the 3 and the 7.

In the third row, the first cell reads “Step 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 10 to the n power. If it’s between 0 and 1, the power of 10 will be 10 to the negative n power.” The second cell reads “37,000 is greater than 1, so the power of 10 will have exponent 4.” The third cell contains 3.7 times 10 to the fourth power.

In the fourth row, the first cell reads “Step 4. Check.” The second cell reads “Check to see if your answer makes sense.” The third cell reads “10 to the fourth power is 10,000 and 10,000 times 3.7 will be 37,000.” Below this is 37,000 equals 3.7 times 10 to the fourth power.

Write in scientific notation: $0.0052 .$

Solution

The original number, $0.0052$ , is between 0 and 1 so we will have a negative power of 10.


Move the decimal point to get 5.2, a number between 1 and 10.
Count the number of decimal places the point was moved.
Write as a product with a power of 10.
Check.
$\begin{matrix} 5.2 \times 10^{−3} \\ 5.2 \times \frac{1}{10^{3}} \\ 5.2 \times \frac{1}{1000} \\ 5.2 \times 0.001 \end{matrix}$
$0.0052$

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.

\begin{matrix} 9.12 \times 10^{4} & 9.12 \times 10^{−4} \\ 9.12 \times 10,000 & 9.12 \times 0.0001 \\ 91,200 & 0.000912 \end{matrix}

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.

\begin{matrix} 9.12 \times 10^{4} = 91,200 & 9.12 \times 10^{−4} = 0.000912 \end{matrix}

This figure has two columns. In the left column is 9.12 times 10 to the fourth power equals 91,200. Below this, the same scientific notation is repeated, with an arrow showing the decimal point in 9.12 being moved four places to the right. Because there are no digits after 2, the final two places are represented by blank spaces. Below this is the text “Move the decimal point four places to the right.” In the right column is 9.12 times 10 to the negative fourth power equals 0.000912. Below this, the same scientific notation is repeated, with an arrow showing the decimal point in 9.12 being moved four places to the left. Because there are no digits before 9, the remaining three places are represented by spaces. Below this is the text “Move the decimal point 4 places to the left.”

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.

How to Convert Scientific Notation to Decimal Form

Convert to decimal form: $6.2 \times 10^{3} .$

Solution

In the second row, the first cell reads “Step 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 absolute value of n places to the left.” The second cell reads “The exponent is positive so move the decimal point 3 places to the right. We need to add two zeros as placeholders.” The third cell contains 6.200, with an arrow showing the decimal point jumping places to the right, from between the 6 and 2 to after the second 00 in 6.200. Below this is the number 6,200.

In the third row, the first cell reads “Step 3. Check to see if your answer makes sense.” The second cell is blank. The third reads “10 cubed is 1000 and 1000 times 6.2 will be 6,200.” Beneath this is 6.2 times 10 cubed equals 6,200.

The steps are summarized below.

Convert to decimal form: $8.9 \times 10^{−2} .$

Solution


Determine the exponent, n, on the factor 10.
Since the exponent is negative, move the decimal point 2 places to the left.
Add zeros as needed for placeholders.

Multiply and Divide Using Scientific Notation

Astronomers use very large numbers to describe distances in the universe and ages of stars and planets. Chemists use very small numbers to describe the size of an atom or the charge on an electron. When scientists perform calculations with very large or very small numbers, they use scientific notation. Scientific notation provides a way for the calculations to be done without writing a lot of zeros. We will see how the Properties of Exponents are used to multiply and divide numbers in scientific notation.

Multiply. Write answers in decimal form: $(4 \times 10^{5}) (2 \times 10^{−7}) .$

Solution

$\begin{matrix} (4 \times 10^{5}) (2 \times 10^{−7}) \\ Use the Commutative Property to rearrange the factors. & 4 \cdot 2 \cdot 10^{5} \cdot 10^{−7} \\ Multiply. & 8 \times 10^{−2} \\ Change to decimal form by moving the decimal two places left. & 0.08 \end{matrix}$

Divide. Write answers in decimal form: $\frac{9 \times 10^{3}}{3 \times 10^{−2}} .$

Solution

Illustrates the step-by-step division of numbers in scientific notation, converting the final answer to decimal form.
	$\frac{9 \times 10^{3}}{3 \times 10^{−2}}$
Separate the factors, rewriting as the product of two fractions.	$\frac{9}{3} \times \frac{10^{3}}{10^{−2}}$
Divide.	$3 \times 10^{5}$
Change to decimal form by moving the decimal five places right.	300,000

Key Concepts

Property of Negative Exponents
- If $n$ is a positive integer and $a \neq 0$ , then $\frac{1}{a^{− n}} = a^{n}$
Quotient to a Negative Exponent
- If $a, b$ are real numbers, $b \neq 0$ and $n$ is an integer , then ${(\frac{a}{b})}^{− n} = {(\frac{b}{a})}^{n}$
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 $10^{n}$
  - between 0 and 1, the power of 10 will be $10^{− n}$
4. Check.
To convert scientific notation to decimal form:
1. Determine the exponent, $n$ , on the factor 10.
2. Move the decimal nplaces, 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.

Section Exercises

Practice Makes Perfect

Use the Definition of a Negative Exponent

In the following exercises, simplify.

ⓐ $4^{−2}$ ⓑ $10^{−3}$

ⓐ $3^{−4}$ ⓑ $10^{−2}$

Solution

ⓐ $\frac{1}{81}$ ⓑ $\frac{1}{100}$

ⓐ $5^{- 3}$ ⓑ $10^{−5}$

ⓐ $2^{−8}$ ⓑ $10^{−2}$

Solution

ⓐ $\frac{1}{256}$ ⓑ $\frac{1}{100}$

ⓐ $\frac{1}{c^{−5}}$ ⓑ $\frac{1}{3^{−2}}$

ⓐ $\frac{1}{c^{−5}}$ ⓑ $\frac{1}{5^{−2}}$

Solution

ⓐ $c^{5}$ ⓑ 25

ⓐ $\frac{1}{q^{−10}}$ ⓑ $\frac{1}{10^{−3}}$

ⓐ $\frac{1}{t^{−9}}$ ⓑ $\frac{1}{10^{−4}}$

Solution

ⓐ $t^{9}$ ⓑ 10000

ⓐ ${(\frac{5}{8})}^{−2}$ ⓑ ${(- \frac{3 m}{n})}^{−2}$

ⓐ ${(\frac{3}{10})}^{−2}$ ⓑ ${(- \frac{2}{c d})}^{−3}$

Solution

ⓐ $\frac{100}{9}$ ⓑ $- \frac{c^{3} d^{3}}{8}$

ⓐ ${(\frac{4}{9})}^{−3}$ ⓑ ${(- \frac{u^{2}}{2 v})}^{−5}$

ⓐ ${(\frac{7}{2})}^{−3}$ ⓑ ${(- \frac{3}{x y^{2}})}^{−3}$

Solution

ⓐ $\frac{8}{343}$ ⓑ $- \frac{x^{3} y^{6}}{27}$

ⓐ ${(- 5)}^{- 2}$ ⓑ $− 5^{−2}$ ⓒ ${(- \frac{1}{5})}^{−2}$ ⓓ $− {(\frac{1}{5})}^{−2}$

ⓐ ${(−7)}^{−2}$ ⓑ $- 7^{- 2}$ ⓒ ${(- \frac{1}{7})}^{−2}$ ⓓ $− {(\frac{1}{7})}^{−2}$

Solution

ⓐ $\frac{1}{49}$ ⓑ $- \frac{1}{49}$ ⓒ 49 ⓓ $−49$

ⓐ $− 3^{−3}$ ⓑ ${(- \frac{1}{3})}^{−3}$ ⓒ $− {(\frac{1}{3})}^{−3}$ ⓓ ${(−3)}^{−3}$

ⓐ $− 5^{−3}$ ⓑ ${(- \frac{1}{5})}^{−3}$ ⓒ $− {(\frac{1}{5})}^{−3}$ ⓓ ${(−5)}^{−3}$

Solution

ⓐ $- \frac{1}{125}$ ⓑ $−125$ ⓒ $−125$ ⓓ $- \frac{1}{125}$

ⓐ $3 \cdot 5^{−1}$ ⓑ ${(3 \cdot 5)}^{−1}$

ⓐ $2 \cdot 5^{−1}$ ⓑ ${(2 \cdot 5)}^{−1}$

Solution

ⓐ $\frac{2}{5}$ ⓑ $\frac{1}{10}$

ⓐ $4 \cdot 5^{−2}$ ⓑ ${(4 \cdot 5)}^{−2}$

ⓐ $3 \cdot 4^{−2}$ ⓑ ${(3 \cdot 4)}^{−2}$

Solution

ⓐ $\frac{3}{16}$ ⓑ $\frac{1}{144}$

ⓐ $m^{−4}$ ⓑ ${(x^{3})}^{−4}$

ⓐ $b^{−5}$ ⓑ ${(k^{2})}^{−5}$

Solution

ⓐ $\frac{1}{b^{5}}$ ⓑ $\frac{1}{k^{10}}$

ⓐ $p^{−10}$ ⓑ ${(q^{6})}^{−8}$

ⓐ $s^{−8}$ ⓑ ${(a^{9})}^{−10}$

Solution

ⓐ $\frac{1}{s^{8}}$ ⓑ $\frac{1}{a^{90}}$

ⓐ $7 n^{−1}$ ⓑ ${(7 n)}^{−1}$ ⓒ ${(−7 n)}^{−1}$

ⓐ $6 r^{−1}$ ⓑ ${(6 r)}^{−1}$ ⓒ ${(−6 r)}^{−1}$

Solution

ⓐ $\frac{6}{r}$ ⓑ $\frac{1}{6 r}$ ⓒ $- \frac{1}{6 r}$

ⓐ ${(3 p)}^{−2}$ ⓑ $3 p^{−2}$ ⓒ $−3 p^{−2}$

ⓐ ${(2 q)}^{−4}$ ⓑ $2 q^{−4}$ ⓒ $−2 q^{−4}$

Solution

ⓐ $\frac{1}{16 q^{4}}$ ⓑ $\frac{2}{q^{4}}$ ⓒ $- \frac{2}{q^{4}}$

Simplify Expressions with Integer Exponents

In the following exercises, simplify.

ⓐ $b^{4} b^{−8}$ ⓑ $r^{−2} r^{5}$ ⓒ $x^{−7} x^{−3}$

ⓐ $s^{3} \cdot s^{−7}$ ⓑ $q^{−8} \cdot q^{3}$ ⓒ $y^{−2} \cdot y^{−5}$

Solution

ⓐ $\frac{1}{s^{4}}$ ⓑ $\frac{1}{q^{5}}$ ⓒ $\frac{1}{y^{7}}$

ⓐ $a^{3} \cdot a^{−3}$ ⓑ $a \cdot a^{3}$ ⓒ $a \cdot a^{−3}$

ⓐ $y^{5} \cdot y^{−5}$ ⓑ $y \cdot y^{5}$ ⓒ $y \cdot y^{−5}$

Solution

ⓐ 1 ⓑ $y^{6}$ ⓒ $\frac{1}{y^{4}}$

$p^{5} \cdot p^{−2} \cdot p^{−4}$

$x^{4} \cdot x^{−2} \cdot x^{−3}$

Solution

$\frac{1}{x}$

$(w^{4} x^{−5}) (w^{−2} x^{−4})$

$(m^{3} n^{−3}) (m^{−5} n^{−1})$

Solution

$\frac{1}{m^{2} n^{4}}$

$(u v^{−2}) (u^{−5} v^{−3})$

$(p q^{−4}) (p^{−6} q^{−3})$

Solution

$\frac{1}{p^{5} q^{7}}$

$(−6 c^{−3} d^{9}) (2 c^{4} d^{−5})$

$(−2 j^{−5} k^{8}) (7 j^{2} k^{−3})$

Solution

$- \frac{14 k^{5}}{j^{3}}$

$(−4 r^{−2} s^{−8}) (9 r^{4} s^{3})$

$(−5 m^{4} n^{6}) (8 m^{−5} n^{−3})$

Solution

$- \frac{40 n^{3}}{m}$

${(5 x^{2})}^{−2}$

${(4 y^{3})}^{−3}$

Solution

$\frac{1}{64 y^{9}}$

${(3 z^{−3})}^{2}$

${(2 p^{−5})}^{2}$

Solution

$\frac{4}{p^{10}}$

$\frac{t^{9}}{t^{−3}}$

$\frac{n^{5}}{n^{−2}}$

Solution

$n^{7}$

$\frac{x^{−7}}{x^{−3}}$

$\frac{y^{−5}}{y^{−10}}$

Solution

$y^{5}$

Convert from Decimal Notation to Scientific Notation

In the following exercises, write each number in scientific notation.

57,000

340,000

Solution

$3.4 \times 10^{5}$

8,750,000

1,290,000

Solution

$1.29 \times 10^{6}$

0.026

0.041

Solution

$4.1 \times 10^{−2}$

0.00000871

0.00000103

Solution

$1.03 \times 10^{−6}$

Convert Scientific Notation to Decimal Form

In the following exercises, convert each number to decimal form.

$5.2 \times 10^{2}$

$8.3 \times 10^{2}$

Solution

830

$7.5 \times 10^{6}$

$1.6 \times 10^{10}$

Solution

16,000,000,000

$2.5 \times 10^{−2}$

$3.8 \times 10^{−2}$

Solution

0.038

$4.13 \times 10^{−5}$

$1.93 \times 10^{−5}$

Solution

0.0000193

Multiply and Divide Using Scientific Notation

In the following exercises, multiply. Write your answer in decimal form.

$(3 \times 10^{−5}) (3 \times 10^{9})$

$(2 \times 10^{2}) (1 \times 10^{−4})$

Solution

0.02

$(7.1 \times 10^{−2}) (2.4 \times 10^{−4})$

$(3.5 \times 10^{−4}) (1.6 \times 10^{−2})$

Solution

$0.0000056$

In the following exercises, divide. Write your answer in decimal form.

$\frac{7 \times 10^{−3}}{1 \times 10^{−7}}$

$\frac{5 \times 10^{−2}}{1 \times 10^{−10}}$

Solution

500,000,000

$\frac{6 \times 10^{4}}{3 \times 10^{−2}}$

$\frac{8 \times 10^{6}}{4 \times 10^{−1}}$

Solution

20,000,000

Everyday Math

The population of the United States on July 4, 2010 was almost 310,000,000. Write the number in scientific notation.

The population of the world on July 4, 2010 was more than 6,850,000,000. Write the number in scientific notation

Solution

$6.85 \times 10^{9}$ .

The average width of a human hair is 0.0018 centimeters. Write the number in scientific notation.

The probability of winning the 2010 Megamillions lottery was about 0.0000000057. Write the number in scientific notation.

Solution

$5.7 \times 10^{−9}$

In 2010, the number of Facebook users each day who changed their status to ‘engaged’ was $2 \times 10^{4}$ . Convert this number to decimal form.

At the start of 2012, the US federal budget had a deficit of more than $$ 1.5 \times 10^{13}$ . Convert this number to decimal form.

Solution

15,000,000,000,000

The concentration of carbon dioxide in the atmosphere is $3.9 \times 10^{−4}$ . Convert this number to decimal form.

The width of a proton is $1 \times 10^{−5}$ of the width of an atom. Convert this number to decimal form.

Solution

0.00001

Health care costs The Centers for Medicare and Medicaid projects that consumers will spend more than $4 trillion on health care by 2017.

ⓐ Write 4 trillion in decimal notation.
ⓑ Write 4 trillion in scientific notation.

Coin production In 1942, the U.S. Mint produced 154,500,000 nickels. Write 154,500,000 in scientific notation.

Solution

$1.545 \times 10^{8}$

Distance The distance between Earth and one of the brightest stars in the night star is 33.7 light years. One light year is about 6,000,000,000,000 (6 trillion), miles.

ⓐ Write the number of miles in one light year in scientific notation.
ⓑ Use scientific notation to find the distance between Earth and the star in miles. Write the answer in scientific notation.

Debt At the end of fiscal year 2015 the gross United States federal government debt was estimated to be approximately $18,600,000,000,000 ($18.6 trillion), according to the Federal Budget. The population of the United States was approximately 300,000,000 people at the end of fiscal year 2015.

ⓐ Write the debt in scientific notation.
ⓑ Write the population in scientific notation.
ⓒ Find the amount of debt per person by using scientific notation to divide the debt by the population. Write the answer in scientific notation.

Solution

ⓐ $1.86 \times 10^{13}$ ⓑ $3 \times 10^{8}$ ⓒ $6.2 \times 10^{4}$

Writing Exercises

ⓐ Explain the meaning of the exponent in the expression $2^{3}$ .
ⓑ 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.

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

Chapter 6 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.

ⓐ $11 c^{4} - 23 c^{2} + 1$
ⓑ $9 p^{3} + 6 p^{2} - p - 5$
ⓒ $\frac{3}{7} x + \frac{5}{14}$
ⓓ 10
ⓔ $2 y - 12$

ⓐ $a^{2} - b^{2}$
ⓑ $24 d^{3}$
ⓒ $x^{2} + 8 x - 10$
ⓓ $m^{2} n^{2} - 2 m n + 6$
ⓔ $7 y^{3} + y^{2} - 2 y - 4$

Solution

ⓐ binomial ⓑ monomial ⓒ trinomial ⓓ trinomial ⓔ other polynomial

Determine the Degree of Polynomials

In the following exercises, determine the degree of each polynomial.

ⓐ $3 x^{2} + 9 x + 10$
ⓑ $14 a^{2} b c$
ⓒ $6 y + 1$
ⓓ $n^{3} - 4 n^{2} + 2 n - 8$
ⓔ $−19$

ⓐ $5 p^{3} - 8 p^{2} + 10 p - 4$
ⓑ $−20 q^{4}$
ⓒ $x^{2} + 6 x + 12$
ⓓ $23 r^{2} s^{2} - 4 r s + 5$
ⓔ 100

Solution

ⓐ 3 ⓑ 4 ⓒ 2 ⓓ 4 ⓔ 0

Add and Subtract Monomials

In the following exercises, add or subtract the monomials.

${5y}^{3} + 8 y^{3}$

$−14 k + 19 k$

Solution

$5 k$

$12 q - (−6 q)$

$−9 c - 18 c$

Solution

$−27 c$

$12x - 4 y - 9 x$

$3 m^{2} + 7 n^{2} - 3 m^{2}$

Solution

$7 n^{2}$

$6 x^{2} y - 4 x + 8 x y^{2}$

$13a + b$

Solution

$13a + b$

Add and Subtract Polynomials

In the following exercises, add or subtract the polynomials.

$(5 x^{2} + 12 x + 1) + (6 x^{2} - 8 x + 3)$

$(9 p^{2} - 5 p + 3) + (4 p^{2} - 4)$

Solution

$13 p^{2} - 5 p - 1$

$(10 m^{2} - 8 m - 1) - (5 m^{2} + m - 2)$

$(7 y^{2} - 8 y) - (y - 4)$

Solution

$7 y^{2} - 9 y + 4$

Subtract
$(3 s^{2} + 10) from (15 s^{2} - 2 s + 8)$

Find the sum of $(a^{2} + 6 a + 9) and (5 a^{3} - 7)$

Solution

$5 a^{3} + a^{2} + 6 a + 2$

Evaluate a Polynomial for a Given Value of the Variable

In the following exercises, evaluate each polynomial for the given value.

Evaluate $3 y^{2} - y + 1$ when:

ⓐ $y = 5$
ⓑ $y = −1$
ⓒ $y = 0$

Evaluate $10 - 12 x$ when:

ⓐ $x = 3$
ⓑ $x = 0$
ⓒ $x = −1$

Solution

ⓐ $−26$ ⓑ 10 ⓒ 22

Randee drops a stone off the 200 foot high cliff into the ocean. The polynomial $−16 t^{2} + 200$ gives the height of a stone $t$ seconds after it is dropped from the cliff. Find the height after $t = 3$ seconds.

A manufacturer of stereo sound speakers has found that the revenue received from selling the speakers at a cost of p dollars each is given by the polynomial $−4 p^{2} + 460 p .$ Find the revenue received when $p = 75$ dollars.

Solution

12,000

Use Multiplication Properties of Exponents

Simplify Expressions with Exponents

In the following exercises, simplify.

$10^{4}$

$17^{1}$

Solution

${(\frac{2}{9})}^{2}$

${(0.5)}^{3}$

Solution

0.125

${(−2)}^{6}$

$− 2^{6}$

Solution

$−64$

Simplify Expressions Using the Product Property for Exponents

In the following exercises, simplify each expression.

$x^{4} \cdot x^{3}$

$p^{15} \cdot p^{16}$

Solution

$p^{31}$

$4^{10} \cdot 4^{6}$

$8 \cdot 8^{5}$

Solution

$8^{6}$

$n \cdot n^{2} \cdot n^{4}$

$y^{c} \cdot y^{3}$

Solution

$y^{c + 3}$

Simplify Expressions Using the Power Property for Exponents

In the following exercises, simplify each expression.

${(m^{3})}^{5}$

${(5^{3})}^{2}$

Solution

$5^{6}$

${(y^{4})}^{x}$

${(3^{r})}^{s}$

Solution

$3^{r s}$

Simplify Expressions Using the Product to a Power Property

In the following exercises, simplify each expression.

${(4 a)}^{2}$

${(−5 y)}^{3}$

Solution

$−125 y^{3}$

${(2 m n)}^{5}$

${(10 x y z)}^{3}$

Solution

$1000 x^{3} y^{3} z^{3}$

Simplify Expressions by Applying Several Properties

In the following exercises, simplify each expression.

${(p^{2})}^{5} \cdot {(p^{3})}^{6}$

${(4 a^{3} b^{2})}^{3}$

Solution

$64 a^{9} b^{6}$

${(5 x)}^{2} (7 x)$

${(2 q^{3})}^{4} {(3 q)}^{2}$

Solution

$144 q^{14}$

${(\frac{1}{3} x^{2})}^{2} {(\frac{1}{2} x)}^{3}$

${(\frac{2}{5} m^{2} n)}^{3}$

Solution

$\frac{8}{125} m^{6} n^{3}$

Multiply Monomials

In the following exercises 8, multiply the monomials.

$(−15 x^{2}) (6 x^{4})$

$(−9 n^{7}) (−16 n)$

Solution

$144 n^{8}$

$(7 p^{5} q^{3}) (8 p q^{9})$

$(\frac{5}{9} a b^{2}) (27 a b^{3})$

Solution

$15 a^{2} b^{5}$

Multiply Polynomials

Multiply a Polynomial by a Monomial

In the following exercises, multiply.

$7 (a + 9)$

$−4 (y + 13)$

Solution

$−4 y - 52$

$−5 (r - 2)$

$p (p + 3)$

Solution

$p^{2} + 3 p$

$− m (m + 15)$

$−6 u (2 u + 7)$

Solution

$−12 u^{2} - 42 u$

$9 (b^{2} + 6 b + 8)$

$3 q^{2} (q^{2} - 7 q + 6)$ 3

Solution

$9 q^{4} - 63 q^{3} + 54 q^{2}$

$(5 z - 1) z$

$(b - 4) \cdot 11$

Solution

$11 b - 44$

Multiply a Binomial by a Binomial

In the following exercises, multiply the binomials using: ⓐ the Distributive Property, ⓑ the FOIL method, ⓒ the Vertical Method.

$(x - 4) (x + 10)$

$(6 y - 7) (2 y - 5)$

Solution

ⓐ $12 y^{2} - 44 y + 35$ ⓑ $12 y^{2} - 44 y + 35$ ⓒ $12 y^{2} - 44 y + 35$

In the following exercises, multiply the binomials. Use any method.

$(x + 3) (x + 9)$

$(y - 4) (y - 8)$

Solution

$y^{2} - 12 y + 32$

$(p - 7) (p + 4)$

$(q + 16) (q - 3)$

Solution

$q^{2} + 13 q - 48$

$(5 m - 8) (12 m + 1)$

$(u^{2} + 6) (u^{2} - 5)$

Solution

$u^{4} + u^{2} - 30$

$(9 x - y) (6 x - 5)$

$(8 m n + 3) (2 m n - 1)$

Solution

$16 m^{2} n^{2} - 2 m n - 3$

Multiply a Trinomial by a Binomial

In the following exercises, multiply using ⓐ the Distributive Property, ⓑ the Vertical Method.

$(n + 1) (n^{2} + 5 n - 2)$

$(3 x - 4) (6 x^{2} + x - 10)$

Solution

ⓐ $18 x^{3} - 21 x^{2} - 34 x + 40$ ⓑ $18 x^{3} - 21 x^{2} - 34 x + 40$

In the following exercises, multiply. Use either method.

$(y - 2) (y^{2} - 8 y + 9)$

$(7 m + 1) (m^{2} - 10 m - 3)$

Solution

$7 m^{3} - 69 m^{2} - 31 m - 3$

Special Products

Square a Binomial Using the Binomial Squares Pattern

In the following exercises, square each binomial using the Binomial Squares Pattern.

${(c + 11)}^{2}$

${(q - 15)}^{2}$

Solution

$q^{2} - 30 q + 225$

${(x + \frac{1}{3})}^{2}$

${(8 u + 1)}^{2}$

Solution

$64 u^{2} + 16 u + 1$

${(3 n^{3} - 2)}^{2}$

${(4 a - 3 b)}^{2}$

Solution

$16 a^{2} - 24 a b + 9 b^{2}$

Multiply Conjugates Using the Product of Conjugates Pattern

In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern.

$(s - 7) (s + 7)$

$(y + \frac{2}{5}) (y - \frac{2}{5})$

Solution

$y^{2} - \frac{4}{25}$

$(12 c + 13) (12 c - 13)$

$(6 - r) (6 + r)$

Solution

$36 - r^{2}$

$(u + \frac{3}{4} v) (u - \frac{3}{4} v)$

$(5 p^{4} - 4 q^{3}) (5 p^{4} + 4 q^{3})$

Solution

$25 p^{8} - 16 q^{6}$

Recognize and Use the Appropriate Special Product Pattern

In the following exercises, find each product.

${(3 m + 10)}^{2}$

$(6 a + 11) (6 a - 11)$

Solution

$36 a^{2} - 121$

$(5 x + y) (x - 5 y)$

${(c^{4} + 9 d)}^{2}$

Solution

$c^{8} + 18 c^{4} d + 81 d^{2}$

$(p^{5} + q^{5}) (p^{5} - q^{5})$

$(a^{2} + 4 b) (4 a - b^{2})$

Solution

$4 a^{3} - a^{2} b^{2} + 16 a b - 4 b^{3}$

Divide Monomials

Simplify Expressions Using the Quotient Property for Exponents

In the following exercises, simplify.

$\frac{u^{24}}{u^{6}}$

$\frac{10^{25}}{10^{5}}$

Solution

$10^{20}$

$\frac{3^{4}}{3^{6}}$

$\frac{v^{12}}{v^{48}}$

Solution

$\frac{1}{v^{36}}$

$\frac{x}{x^{5}}$

$\frac{5}{5^{8}}$

Solution

$\frac{1}{5^{7}}$

Simplify Expressions with Zero Exponents

In the following exercises, simplify.

$75^{0}$

$x^{0}$

Solution

$− 12^{0}$

$(− 12^{0})$ ${(−12)}^{0}$

Solution

−1

$25 x^{0}$

${(25 x)}^{0}$

Solution

$19 n^{0} - 25 m^{0}$

${(19 n)}^{0} - {(25 m)}^{0}$

Solution

Simplify Expressions Using the Quotient to a Power Property

In the following exercises, simplify.

${(\frac{2}{5})}^{3}$

${(\frac{m}{3})}^{4}$

Solution

$\frac{m^{4}}{81}$

${(\frac{r}{s})}^{8}$

${(\frac{x}{2 y})}^{6}$

Solution

$\frac{x^{6}}{64 y^{6}}$

Simplify Expressions by Applying Several Properties

In the following exercises, simplify.

$\frac{{(x^{3})}^{5}}{x^{9}}$

$\frac{n^{10}}{{(n^{5})}^{2}}$

Solution

${(\frac{q^{6}}{q^{8}})}^{3}$

${(\frac{r^{8}}{r^{3}})}^{4}$

Solution

$r^{20}$

${(\frac{c^{2}}{d^{5}})}^{9}$

${(\frac{3 x^{4}}{2 y^{2}})}^{5}$

Solution

$\frac{243 x^{20}}{32 y^{10}}$

${(\frac{v^{3} v^{9}}{v^{6}})}^{4}$

$\frac{{(3 n^{2})}^{4} {(−5 n^{4})}^{3}}{{(−2 n^{5})}^{2}}$

Solution

$- \frac{10,125 n^{10}}{4}$

Divide Monomials

In the following exercises, divide the monomials.

$−65 y^{14} \div 5 y^{2}$

$\frac{64 a^{5} b^{9}}{−16 a^{10} b^{3}}$

Solution

$- \frac{4 b^{6}}{a^{5}}$

$\frac{144 x^{15} y^{8} z^{3}}{18 x^{10} y^{2} z^{12}}$

$\frac{(8 p^{6} q^{2}) (9 p^{3} q^{5})}{16 p^{8} q^{7}}$

Solution

$\frac{9 p}{2}$

Divide Polynomials

Divide a Polynomial by a Monomial

In the following exercises, divide each polynomial by the monomial.

$\frac{42 z^{2} - 18 z}{6}$

$(35 x^{2} - 75 x) \div 5 x$

Solution

$7 x - 15$

$\frac{81 n^{4} + 105 n^{2}}{−3}$

$\frac{550 p^{6} - 300 p^{4}}{10 p^{3}}$

Solution

$55 p^{3} - 30 p$

$(63 x y^{3} + 56 x^{2} y^{4}) \div (7 x y)$

$\frac{96 a^{5} b^{2} - 48 a^{4} b^{3} - 56 a^{2} b^{4}}{8 a b^{2}}$

Solution

$12 a^{4} - 6 a^{3} b - 7 a b^{2}$

$\frac{57 m^{2} - 12 m + 1}{−3 m}$

$\frac{105 y^{5} + 50 y^{3} - 5 y}{5 y^{3}}$

Solution

$21 y^{2} + 10 - \frac{1}{y^{2}}$

Divide a Polynomial by a Binomial

In the following exercises, divide each polynomial by the binomial.

$(k^{2} - 2 k - 99) \div (k + 9)$

$(v^{2} - 16 v + 64) \div (v - 8)$

Solution

$v - 8$

$(3 x^{2} - 8 x - 35) \div (x - 5)$

$(n^{2} - 3 n - 14) \div (n + 3)$

Solution

$n - 6 + \frac{4}{n + 3}$

$(4 m^{3} + m - 5) \div (m - 1)$

$(u^{3} - 8) \div (u - 2)$

Solution

$u^{2} + 2 u + 4$

Integer Exponents and Scientific Notation

Use the Definition of a Negative Exponent

In the following exercises, simplify.

$9^{−2}$

${(−5)}^{−3}$

Solution

$- \frac{1}{125}$

$3 \cdot 4^{−3}$

${(6 u)}^{−3}$

Solution

$\frac{1}{216 u^{3}}$

${(\frac{2}{5})}^{−1}$

${(\frac{3}{4})}^{−2}$

Solution

$\frac{16}{9}$

Simplify Expressions with Integer Exponents

In the following exercises, simplify.

$p^{−2} \cdot p^{8}$

$q^{−6} \cdot q^{−5}$

Solution

$\frac{1}{q^{11}}$

$(c^{−2} d) (c^{−3} d^{−2})$

${(y^{8})}^{−1}$

Solution

$\frac{1}{y^{8}}$

${(q^{−4})}^{−3}$

$\frac{a^{8}}{a^{12}}$

Solution

$\frac{1}{a^{4}}$

$\frac{n^{5}}{n^{−4}}$

$\frac{r^{−2}}{r^{−3}}$

Solution

$r$

Convert from Decimal Notation to Scientific Notation

In the following exercises, write each number in scientific notation.

8,500,000

0.00429

Solution

$4.29 \times 10^{−3}$

The thickness of a dime is about 0.053 inches.

In 2015, the population of the world was about 7,200,000,000 people.

Solution

$7.2 \times 10^{9}$

Convert Scientific Notation to Decimal Form

In the following exercises, convert each number to decimal form.

$3.8 \times 10^{5}$

$1.5 \times 10^{10}$

Solution

$15,000,000,000$

$9.1 \times 10^{−7}$

$5.5 \times 10^{−1}$

Solution

$0.55$

Multiply and Divide Using Scientific Notation

In the following exercises, multiply and write your answer in decimal form.

$(2 \times 10^{5}) (4 \times 10^{−3})$

$(3.5 \times 10^{−2}) (6.2 \times 10^{−1})$

Solution

$0.0217$

In the following exercises, divide and write your answer in decimal form.

$\frac{8 \times 10^{5}}{4 \times 10^{−1}}$

$\frac{9 \times 10^{−5}}{3 \times 10^{2}}$

Solution

$0.0000003$

Chapter Practice Test

For the polynomial $10 x^{4} + 9 y^{2} - 1$
ⓐ Is it a monomial, binomial, or trinomial?
ⓑ What is its degree?

In the following exercises, simplify each expression.

$(12 a^{2} - 7 a + 4) + (3 a^{2} + 8 a - 10)$

Solution

$15 a^{2} + a - 6$

$(9 p^{2} - 5 p + 1) - (2 p^{2} - 6)$

${(- \frac{2}{5})}^{3}$

Solution

$- \frac{8}{125}$

$u \cdot u^{4}$

${(4 a^{3} b^{5})}^{2}$

Solution

$16 a^{6} b^{10}$

$(−9 r^{4} s^{5}) (4 r s^{7})$

$3 k (k^{2} - 7 k + 13)$

Solution

$3 k^{3} - 21 k^{2} + 39 k$

$(m + 6) (m + 12)$

$(v - 9) (9 v - 5)$

Solution

$9 v^{2} - 86 v + 45$

$(4 c - 11) (3 c - 8)$

$(n - 6) (n^{2} - 5 n + 4)$

Solution

$n^{3} - 11 n^{2} + 34 n - 24$

$(2 x - 15 y) (5 x + 7 y)$

$(7 p - 5) (7 p + 5)$

Solution

$49 p^{2} - 25$

${(9 v - 2)}^{2}$

$\frac{3^{8}}{3^{10}}$

Solution

$\frac{1}{9}$

${(\frac{m^{4} \cdot m}{m^{3}})}^{6}$

${(87 x^{15} y^{3} z^{22})}^{0}$

Solution

$1$

$\frac{80 c^{8} d^{2}}{16 c d^{10}}$

$\frac{12 x^{2} + 42 x - 6}{2 x}$

Solution

$6 x + 21 - \frac{3}{x}$

$(70 x y^{4} + 95 x^{3} y) \div 5 x y$

$\frac{64 x^{3} - 1}{4 x - 1}$

Solution

$16 x^{2} + 4 x + 1$

$(y^{2} - 5 y - 18) \div (y + 3)$

$5^{−2}$

Solution

$\frac{1}{25}$

${(4 m)}^{−3}$

$q^{−4} \cdot q^{−5}$

Solution

$\frac{1}{q^{9}}$

$\frac{n^{−2}}{n^{−10}}$

Convert 83,000,000 to scientific notation.

Solution

$8.3 \times 10^{7}$

Convert $6.91 \times 10^{−5}$ to decimal form.

In the following exercises, simplify, and write your answer in decimal form.

$(3.4 \times 10^{9}) (2.2 \times 10^{−5})$

Solution

74,800

$\frac{8.4 \times 10^{−3}}{4 \times 10^{3}}$

A helicopter flying at an altitude of 1000 feet drops a rescue package. The polynomial $−16 t^{2} + 1000$ gives the height of the package $t$ seconds a after it was dropped. Find the height when $t = 6$ seconds.

Solution

424 feet

negative exponent

n

is a positive integer and

a \neq 0

, then

a^{− n} = \frac{1}{a^{n}}

scientific notation

A number is expressed in scientific notation when it is of the form

a \times 10^{n}

where

a \geq 1 and a < 10

and

n

is an integer.

Introduction

This figure is a photo of the Sydney Harbor bridge in Sydney, Australia. — The Sydney Harbor Bridge is one of Australia’s most photographed landmarks. It is the world’s largest steel arch bridge with the top of the bridge standing 134 meters above the harbor. Can you see why it is known by the locals as the “Coathanger”?

Quadratic expressions may be used to model physical properties of a large bridge, the trajectory of a baseball or rocket, and revenue and profit of a business. By factoring these expressions, specific characteristics of the model can be identified. In this chapter, you will explore the process of factoring expressions and see how factoring is used to solve certain types of equations.

Greatest Common Factor and Factor by Grouping

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.

This figure has two factors being multiplied. They are 8 and 7. Beside this equation there are other factors multiplied. They are 2x and (x+3). The product is given as 2x^2 plus 6x. Above the figure is an arrow towards the right with multiply inside. Below the figure is an arrow to the left with factor inside.

We have learned how to factor numbers to find the least common multiple (LCM) of two or more numbers. Now we will factor expressions and find the greatest common factor of two or more expressions. The method we use is similar to what we used to find the LCM.

First we’ll find the GCF of two numbers.

How to Find the Greatest Common Factor of Two or More Expressions

Find the GCF of 54 and 36.

Solution

This table has three columns. In the first column are the steps for factoring. The first row has the first step, factor each coefficient into primes and write all variables with exponents in expanded form. The second column in the first row has “factor 54 and 36”. The third column in the first row has 54 and 36 factored with factor trees. The prime factors of 54 are circled and are 3, 3, 2, and3. The prime factors of 36 are circled and are 2,3,2,3.

The second row has the second step of “in each column, circle the common factors. The second column in the second row has the statement “circle the 2, 3 and 3 that are shared by both numbers”. The third column in the second row has the prime factors of 36 and 54 in rows above each other. The common factors of 2, 3, and 3 are circled.

The third row has the step “bring down the common factors that all expressions share”. The second column in the third row has “bring down the 2,3, and 3 then multiply”. The third column in the third row has “GCF = 2 times 3 times 3”.

The fourth row has the fourth step “multiply the factors”. The second column in the fourth row is blank. The third column in the fourth row has “GCF = 18” and “the GCF of 54 and 36 is 18”.

Notice that, because the GCF is a factor of both numbers, 54 and 36 can be written as multiples of 18.

\begin{matrix} 54 = 18 \cdot 3 \\ 36 = 18 \cdot 2 \end{matrix}

We summarize the steps we use to find the GCF below.

In the first example, the GCF was a constant. In the next two examples, we will get variables in the greatest common factor.

Find the greatest common factor of $27 x^{3} and 18 x^{4}$ .

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.
	The GCF of $27 x^{3}$ and $18 x^{4}$ is $9 x^{3} .$

Find the GCF of $4 x^{2} y, 6 x y^{3}$ .

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.
	The GCF of $4 x^{2} y$ and $6 x y^{3}$ is $2 xy$ .

Find the GCF of: $21 x^{3}, 9 x^{2}, 15 x$ .

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.
	The GCF of $21 x^{3}$ , $9 x^{2}$ and $15 x$ is $3 x .$

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 $2 \cdot 6 or 3 \cdot 4),$ in algebra, it can be useful to represent a polynomial in factored form. One way to do this is by finding the GCF of all the terms. Remember, we multiply a polynomial by a monomial as follows:

\begin{matrix} 2 (x + 7) & factors \\ 2 \cdot x + 2 \cdot 7 \\ 2 x + 14 & product \end{matrix}

Now we will start with a product, like $2 x + 14$ , and end with its factors, $2 (x + 7)$ . To do this we apply the Distributive Property “in reverse.”

We state the Distributive Property here just as you saw it in earlier chapters and “in reverse.”

So how do you use the Distributive Property to factor a polynomial? You just find the GCF of all the terms and write the polynomial as a product!

How to Factor the Greatest Common Factor from a Polynomial

Factor: $4 x + 12$ .

Solution

This table has three columns. In the first column are the steps for factoring. The first row has the first step, “Find the G C F of all the terms of the polynomial”. The second column in the first row has “find the G C F of 4 x and 12”. The third column in the first row has 4 x factored as 2 times 2 times x and below it 18 factored as 2 times 2 times 3. Then, below the factors are the statements, “G C F = 2 times 2” and “G C F = 4”.

The second row has the second step “rewrite each term as a product using the G C F”. The second column in the second row has the statement “Rewrite 4 x and 12 as products of their G C F, 4” Then the two equations 4 x = 4 times x and 12 = 4 times 3. The third column in the second row has the expressions 4x + 12 and below this 4 times x + 4 times 3.

The third row has the step “Use the reverse distributive property to factor the expression”. The second column in the third row is blank. The third column in the third row has “4(x + 3)”.

The fourth row has the fourth step “check by multiplying the factors”. The second column in the fourth row is blank. The third column in the fourth row has three expressions. The first is 4(x + 3), the second is 4 times x + 4 times 3. The third is 4 x + 12.

Factor: $5 a + 5$ .

Solution

Find the GCF of 5a and 5.

Rewrite each term as a product using the GCF.
Use the Distributive Property "in reverse" to factor the GCF.
Check by mulitplying the factors to get the orginal polynomial.
$5 (a + 1)$
$5 \cdot a + 5 \cdot 1$
$5 a + 5 ✓$

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: $12 x - 60$ .

Solution

Find the GCF of 12x and 60.

Rewrite each term as a product using the GCF.
Factor the GCF.
Check by mulitplying the factors.
$12 (x - 5)$
$12 \cdot x - 12 \cdot 5$
$12 x - 60 ✓$

Now we’ll factor the greatest common factor from a trinomial. We start by finding the GCF of all three terms.

Factor: $4 y^{2} + 24 y + 28$ .

Solution

We start by finding the GCF of all three terms.

Find the GCF of $4 y^{2}$ , $24 y$ and 28.

Rewrite each term as a product using the GCF.
Factor the GCF.
Check by mulitplying.
$4 (y^{2} + 6 y + 7)$
$4 \cdot y^{2} + 4 \cdot 6 y + 4 \cdot 7$
$4 y^{2} + 24 y + 28 ✓$

Factor: $5 x^{3} - 25 x^{2}$ .

Solution

Find the GCF of $5 x^{3}$ and $25 x^{2} .$

Rewrite each term.
Factor the GCF.
Check.
$5 x^{2} (x - 5)$
$5 x^{2} \cdot x - 5 x^{2} \cdot 5$
$5 x^{3} - 25 x^{2} ✓$

Factor: $21 x^{3} - 9 x^{2} + 15 x$ .

Solution

In a previous example we found the GCF of $21 x^{3}, 9 x^{2}, 15 x$ to be $3 x$ .


Rewrite each term using the GCF, 3x.
Factor the GCF.
Check.
$3 x (7 x^{2} - 3 x + 5)$
$3 x \cdot 7 x^{2} - 3 x \cdot 3 x + 3 x \cdot 5$
$21 x^{3} - 9 x^{2} + 15 x ✓$

Factor: $8 m^{3} - 12 m^{2} n + 20 m n^{2}$ .

Solution

Find the GCF of $8 m^{3}$ , $12 m^{2} n$ , $20 m n^{2}$ .

Rewrite each term.
Factor the GCF.
Check.
$4 m (2 m^{2} - 3 m n + 5 n^{2})$
$4 m \cdot 2 m^{2} - 4 m \cdot 3 m n + 4 m \cdot 5 n^{2}$
$8 m^{3} - 12 m^{2} n + 20 m n^{2} ✓$

When the leading coefficient is negative, we factor the negative out as part of the GCF.

Factor: $−8 y - 24$ .

Solution

When the leading coefficient is negative, the GCF will be negative.

Ignoring the signs of the terms, we first find the GCF of 8y and 24 is 8. Since the expression −8y − 24 has a negative leading coefficient, we use −8 as the GCF.
Rewrite each term using the GCF.
Factor the GCF.
Check.
$−8 (y + 3)$
$−8 \cdot y + (−8) \cdot 3$
$−8 y - 24 ✓$

Factor: $−6 a^{2} + 36 a$ .

Solution

The leading coefficient is negative, so the GCF will be negative?

Since the leading coefficient is negative, the GCF is negative, −6a.
Rewrite each term using the GCF.
Factor the GCF.
Check.
$−6 a (a - 6)$
$−6 a \cdot a + (−6 a) (−6)$
$−6 a^{2} + 36 a ✓$

Factor: $5 q (q + 7) - 6 (q + 7)$ .

Solution

The GCF is the binomial $q + 7$ .


Factor the GCF, (q + 7).
Check on your own by multiplying.

Factor by Grouping

When there is no common factor of all the terms of a polynomial, look for a common factor in just some of the terms. When there are four terms, a good way to start is by separating the polynomial into two parts with two terms in each part. Then look for the GCF in each part. If the polynomial can be factored, you will find a common factor emerges from both parts.

(Not all polynomials can be factored. Just like some numbers are prime, some polynomials are prime.)

How to Factor by Grouping

Factor: $x y + 3 y + 2 x + 6$ .

Solution

This table gives the steps for factoring x y + 3 y + 2 x + 6. In the first row there is the statement, “group terms with common factors”. In the next column, there is the statement of no common factors of all 4 terms. The last column shows the first two terms grouped and the last two terms grouped.

The second row has the statement, “factor out the common factor from each group”. The second column in the second row states to factor out the GCF from the two separate groups. The third column in the second row has the expression y(x + 3) + 2(x + 3).

The third row has the statement, “factor the common factor from the expression”. The second column in this row points out there is a common factor of (x + 3). The third column in the third row shows the factor of (x + 3) factored from the two groups, (x + 3) times (y + 2).

The last row has the statement, “check”. The second column in this row states to multiply (x + 3)(y + 2). The product is shown in the last column of the original polynomial x y + 3 y + 2 x + 6.

Factor: $x^{2} + 3 x - 2 x - 6$ .

Solution

$\begin{matrix} There is no GCF in all four terms. & x^{2} + 3 x −2 x - 6 \\ Separate into two parts. & \underset{⎵}{x^{2} + 3 x} \underset{⎵}{−2 x - 6} \\ \begin{matrix} Factor the GCF from both parts. Be careful \\ with the signs when factoring the GCF from \\ the last two terms. \end{matrix} & \begin{matrix} x (x + 3) - 2 (x + 3) \\ (x + 3) (x - 2) \end{matrix} \\ Check on your own by multiplying. \end{matrix}$

Key Concepts

Finding the Greatest Common Factor (GCF): To find the GCF of two expressions:
1. Factor each coefficient into primes. Write all variables with exponents in expanded form.
2. List all factors—matching common factors in a column. In each column, circle the common factors.
3. Bring down the common factors that all expressions share.
4. Multiply the factors as in Example 2.
Factor the Greatest Common Factor from a Polynomial: To factor a greatest common factor from a polynomial:
1. Find the GCF of all the terms of the polynomial.
2. Rewrite each term as a product using the GCF.
3. Use the ‘reverse’ Distributive Property to factor the expression.
4. Check by multiplying the factors as in Example 5.
Factor by Grouping: To factor a polynomial with 4 four or more terms
1. Group terms with common factors.
2. Factor out the common factor in each group.
3. Factor the common factor from the expression.
4. Check by multiplying the factors as in Example 15.

Practice Makes Perfect

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

8, 18

Solution

24, 40

72, 162

Solution

150, 275

10a, 50

Solution

5b, 30

$3 x, 10 x^{2}$

Solution

$x$

$21 b^{2}, 14 b$

$8 w^{2}, 24 w^{3}$

Solution

$8 w^{2}$

$30 x^{2}, 18 x^{3}$

$10 p^{3} q, 12 p q^{2}$

Solution

$2 p q$

$8 a^{2} b^{3}, 10 a b^{2}$

$12 m^{2} n^{3}, 30 m^{5} n^{3}$

Solution

$6 m^{2} n^{3}$

$28 x^{2} y^{4}, 42 x^{4} y^{4}$

$10 a^{3}, 12 a^{2}, 14 a$

Solution

$2 a$

$20 y^{3}, 28 y^{2}, 40 y$

$35 x^{3}, 10 x^{4}, 5 x^{5}$

Solution

$5 x^{3}$

$27 p^{2}, 45 p^{3}, 9 p^{4}$

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

$4 x + 20$

Solution

$4 (x + 5)$

$8 y + 16$

$6 m + 9$

Solution

$3 (2 m + 3)$

$14 p + 35$

$9 q + 9$

Solution

$9 (q + 1)$

$7 r + 7$

$8 m - 8$

Solution

$8 (m - 1)$

$4 n - 4$

$9 n - 63$

Solution

$9 (n - 7)$

$45 b - 18$

$3 x^{2} + 6 x - 9$

Solution

$3 (x^{2} + 2 x - 3)$

$4 y^{2} + 8 y - 4$

$8 p^{2} + 4 p + 2$

Solution

$2 (4 p^{2} + 2 p + 1)$

$10 q^{2} + 14 q + 20$

$8 y^{3} + 16 y^{2}$

Solution

$8 y^{2} (y + 2)$

$12 x^{3} - 10 x$

$5 x^{3} - 15 x^{2} + 20 x$

Solution

$5 x (x^{2} - 3 x + 4)$

$8 m^{2} - 40 m + 16$

$12 x y^{2} + 18 x^{2} y^{2} - 30 y^{3}$

Solution

$6 y^{2} (2 x + 3 x^{2} - 5 y)$

$21 p q^{2} + 35 p^{2} q^{2} - 28 q^{3}$

$−2 x - 4$

Solution

$−2 (x + 2)$

$−3 b + 12$

$5 x (x + 1) + 3 (x + 1)$

Solution

$(x + 1) (5 x + 3)$

$2 x (x - 1) + 9 (x - 1)$

$3 b (b - 2) - 13 (b - 2)$

Solution

$(b - 2) (3 b - 13)$

$6 m (m - 5) - 7 (m - 5)$

Factor by Grouping

In the following exercises, factor by grouping.

$x y + 2 y + 3 x + 6$

Solution

$(y + 3) (x + 2)$

$m n + 4 n + 6 m + 24$

$u v - 9 u + 2 v - 18$

Solution

$(u + 2) (v - 9)$

$p q - 10 p + 8 q - 80$

$b^{2} + 5 b - 4 b - 20$

Solution

$(b - 4) (b + 5)$

$m^{2} + 6 m - 12 m - 72$

$p^{2} + 4 p - 9 p - 36$

Solution

$(p - 9) (p + 4)$

$x^{2} + 5 x - 3 x - 15$

Mixed Practice

In the following exercises, factor.

$−20 x - 10$

Solution

$−10 (2 x + 1)$

$5 x^{3} - x^{2} + x$

$3 x^{3} - 7 x^{2} + 6 x - 14$

Solution

$(x^{2} + 2) (3 x - 7)$

$x^{3} + x^{2} - x - 1$

$x^{2} + x y + 5 x + 5 y$

Solution

$(x + y) (x + 5)$

$5 x^{3} + 3 x^{2} - 5 x - 3$

Everyday Math

Area of a rectangle The area of a rectangle with length 6 less than the width is given by the expression $w^{2} - 6 w$ , where $w =$ width. Factor the greatest common factor from the polynomial.

Solution

$w (w - 6)$

Height of a baseball The height of a baseball t seconds after it is hit is given by the expression $−16 t^{2} + 80 t + 4$ . Factor the greatest common factor from the polynomial.

Writing Exercises

The greatest common factor of 36 and 60 is 12. Explain what this means.

Solution

Answers will vary.

What is the GCF of $y^{4}, y^{5}, and y^{10}$ ? Write a general rule that tells you how to find the GCF of $y^{a}, y^{b}, and y^{c}$ .

Self Check

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

This table has the following statements all to be preceded by “I can…”. The first is “find the greatest common factor of two or more expressions”. The second is “factor the greatest common factor from a polynomial”. The third is “factor by grouping”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.

ⓑ If most of your checks were:

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

factoring

Factoring is splitting a product into factors; in other words, it is the reverse process of multiplying.

greatest common factor

The greatest common factor is the largest expression that is a factor of two or more expressions is the greatest common factor (GCF).

Factor Trinomials of the Form x²+bx+c

Factor Trinomials of the Form x² + bx + c

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication—to start with the product and end up with the factors. Let’s look at an example of multiplying binomials to refresh your memory.

This figure shows the steps of multiplying the factors (x + 2) times (x + 3). The multiplying is completed using FOIL to demonstrate. The first term is x squared and is below F. The second term is 3 x below “O”. The third term is 2 x below “I”. The fourth term is 6 below L. The simplified product is then given as x 2 plus 5 x + 6.

To factor the trinomial means to start with the product, $x^{2} + 5 x + 6$ , and end with the factors, $(x + 2) (x + 3)$ . You need to think about where each of the terms in the trinomial came from.

The first term came from multiplying the first term in each binomial. So to get $x^{2}$ in the product, each binomial must start with an x.

\begin{matrix} x^{2} + 5 x + 6 \\ (x) (x) \end{matrix}

The last term in the trinomial came from multiplying the last term in each binomial. So the last terms must multiply to 6.

What two numbers multiply to 6?

The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use?

Consider the middle term. It came from adding the outer and inner terms.

So the numbers that must have a product of 6 will need a sum of 5. We’ll test both possibilities and summarize the results in Table 1—the table will be very helpful when you work with numbers that can be factored in many different ways.

Factors of $6$	Sum of factors
$1, 6$	$1 + 6 = 7$
$2, 3$	$2 + 3 = 5$

We see that 2 and 3 are the numbers that multiply to 6 and add to 5. So we have the factors of $x^{2} + 5 x + 6$ . They are $(x + 2) (x + 3)$ .

\begin{matrix} x^{2} + 5 x + 6 & product \\ (x + 2) (x + 3) & factors \end{matrix}

You should check this by multiplying.

Looking back, we started with $x^{2} + 5 x + 6$ , which is of the form $x^{2} + b x + c$ , where $b = 5$ and $c = 6$ . We factored it into two binomials of the form $(x + m) and (x + n)$ .

\begin{matrix} x^{2} + 5 x + 6 & x^{2} + b x + c \\ (x + 2) (x + 3) & (x + m) (x + n) \end{matrix}

To get the correct factors, we found two numbers m and n whose product is c and sum is b.

How to Factor Trinomials of the Form $x^{2} + b x + c$

Factor: $x^{2} + 7 x + 12$ .

Solution

This table gives the steps for factoring x squared + 7 x + 12. The first row states the first step “write the factors as two binomials with first terms x”. In the second column of the first row it states, “write two sets of parentheses and put x as the first term”. In the third column, it has the expression x squared + 7 x +12. Below the expression are two sets of parentheses with x as the first term.

The second row states the second step “find two numbers m and n that multiply to c, m times n = c and add to b, m + n = b”. In the second column of the second row are the factors of 12 and their sums. 1,12 with sum 1 + 12 = 13. 2, 6 with sum 2 + 6 =8. 3, 4 with sum 3 + 4 = 7.

The third row states “use m and n as the last terms of the factors”. The second column states “use 3 and 4 as the last terms of the binomials”. The third column in this row has the product (x + 3)(x + 4).

In the fourth row the statement is “check by multiplying the factors”. The product of (x + 3)(x +4) is shown to be x 2 + 7 x + 12.

Let’s summarize the steps we used to find the factors.

Factor: $u^{2} + 11 u + 24$ .

Solution

Notice that the variable is u, so the factors will have first terms u.

$\begin{matrix} u^{2} + 11 u + 24 \\ Write the factors as two binomials with first terms u . & (u) (u) \end{matrix}$

Find two numbers that: multiply to 24 and add to 11.

Factors of $24$	Sum of factors
$1, 24$	$1 + 24 = 25$
$2, 12$	$2 + 12 = 14$
$3, 8$	$3 + 8 = 11 *$
$4, 6$	$4 + 6 = 10$

$\begin{matrix} Use 3 and 8 as the last terms of the binomials. & (u + 3) (u + 8) \\ Check. \\ (u + 3) (u + 8) \\ u^{2} + 3 u + 8 u + 24 \\ u^{2} + 11 u + 24 ✓ \end{matrix}$

Factor: $y^{2} + 17 y + 60$ .

Solution

$\begin{matrix} y^{2} + 17 y + 60 \\ Write the factors as two binomials with first terms y . & (y) (y) \end{matrix}$

Find two numbers that multiply to 60 and add to 17.

Factors of $60$	Sum of factors
$1, 60$	$1 + 60 = 61$
$2, 30$	$2 + 30 = 32$
$3, 20$	$3 + 20 = 23$
$4, 15$	$4 + 15 = 19$
$5, 12$	$5 + 12 = 17 *$
$6, 10$	$6 + 10 = 16$

$\begin{matrix} Use 5 and 12 as the last terms. & (y + 5) (y + 12) \\ Check. \\ (y + 5) (y + 12) \\ (y^{2} + 12 y + 5 y + 60) \\ (y^{2} + 17 y + 60) ✓ \end{matrix}$

Factor Trinomials of the Form x² + bx + c with b Negative, c Positive

In the examples so far, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

Remember: To get a negative sum and a positive product, the numbers must both be negative.

Again, think about FOIL and where each term in the trinomial came from. Just as before,

the first term, $x^{2}$ , comes from the product of the two first terms in each binomial factor, x and y;
the positive last term is the product of the two last terms
the negative middle term is the sum of the outer and inner terms.

How do you get a positive product and a negative sum? With two negative numbers.

Factor: $t^{2} - 11 t + 28$ .

Solution

Again, with the positive last term, 28, and the negative middle term, $−11 t$ , we need two negative factors. Find two numbers that multiply 28 and add to $−11$ .

$\begin{matrix} t^{2} - 11 t + 28 \\ Write the factors as two binomials with first terms t . & (t) (t) \end{matrix}$

Find two numbers that: multiply to 28 and add to $−11$ .

Factors of $28$	Sum of factors
$−1, −28$	$−1 + (−28) = −29$
$−2, −14$	$−2 + (−14) = −16$
$−4, −7$	$−4 + (−7) = −11 *$

$\begin{matrix} Use −4, −7 as the last terms of the binomials. & (t - 4) (t - 7) \\ Check. \\ (t - 4) (t - 7) \\ t^{2} - 7 t - 4 t + 28 \\ t^{2} - 11 t + 28 ✓ \end{matrix}$

Factor Trinomials of the Form $x^{2} + b x + c$ with c Negative

Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

Remember: To get a negative product, the numbers must have different signs.

Factor: $z^{2} + 4 z - 5$ .

Solution

To get a negative last term, multiply one positive and one negative. We need factors of $−5$ that add to positive 4.

Factors of $−5$	Sum of factors
$1, −5$	$1 + (−5) = −4$
$−1, 5$	$−1 + 5 = 4 *$

Notice: We listed both $1, −5 and - 1, 5$ to make sure we got the sign of the middle term correct.

Steps for factoring the quadratic expression z^2 + 4z - 5 into binomials.
	$z^{2} + 4 z - 5$
Factors will be two binomials with first terms z.	$(z) (z)$
Use −1, 5 as the last terms of the binomials.	$(z - 1) (z + 5)$
Check. $\begin{matrix} (z - 1) (z + 5) \\ z^{2} + 5 z - 1 z - 5 \\ z^{2} + 4 z - 5 ✓ \end{matrix}$

Let’s make a minor change to the last trinomial and see what effect it has on the factors.

Factor: $z^{2} - 4 z - 5$ .

Solution

This time, we need factors of $−5$ that add to $−4$ .

Factors of $−5$	Sum of factors
$1, −5$	$1 + (−5) = −4 *$
$−1, 5$	$−1 + 5 = 4$

Step-by-step guide on factoring the quadratic expression z^2 - 4z - 5, including intermediate mathematical forms and a check.
	$z^{2} - 4 z - 5$
Factors will be two binomials with first terms z.	$(z) (z)$
Use 1, −5 as the last terms of the binomials.	$(z + 1) (z - 5)$
Check. $\begin{matrix} (z + 1) (z - 5) \\ z^{2} - 5 z + 1 z - 5 \\ z^{2} - 4 z - 5 ✓ \end{matrix}$

Notice that the factors of $z^{2} - 4 z - 5$ are very similar to the factors of $z^{2} + 4 z - 5$ . It is very important to make sure you choose the factor pair that results in the correct sign of the middle term.

Factor: $q^{2} - 2 q - 15$ .

Solution

$\begin{matrix} q^{2} - 2 q - 15 \\ Factors will be two binomials with first terms q . & (q) (q) \\ You can use 3, −5 as the last terms of the & (q + 3) (q - 5) \\ binomials. \end{matrix}$

Factors of $−15$	Sum of factors
$1, −15$	$1 + (−15) = −14$
$−1, 15$	$−1 + 15 = 14$
$3, −5$	$3 + (−5) = −2 *$
$−3, 5$	$−3 + 5 = 2$

Check.

$(q + 3) (q - 5)$

$q^{2} - 5 q + 3 q - 15$

$q^{2} - 2 q - 15 ✓$

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

Factor: $y^{2} - 6 y + 15$ .

Solution

Initial step in factoring a quadratic expression: outlining the binomial structure for y^2 - 6y + 15.
	$y^{2} - 6 y + 15$
Factors will be two binomials with first terms y.	$(y) (y)$

Factors of 15	Sum of factors
$−1, −15$	$−1 + (−15) = −16$
$−3, −5$	$−3 + (−5) = −8$

As shown in the table, none of the factors add to $−6$ ; therefore, the expression is prime.

Factor: $2 x + x^{2} - 48$ .

Solution

This table illustrates the initial steps in factoring a quadratic expression, showing the reordering of terms and the setup for binomial factors.
	$2 x + x^{2} - 48$
First we put the terms in decreasing degree order.	$x^{2} + 2 x - 48$
Factors will be two binomials with first terms x.	$(x) (x)$

As shown in the table, you can use $−6, 8$ as the last terms of the binomials.

(x - 6) (x + 8)

Factors of $−48$	Sum of factors
$−1, 48$	$−1 + 48 = 47$
$−2, 24$ $−3, 16$ $−4, 12$ $−6, 8$	$−2 + 24 = 22$ $−3 + 16 = 13$ $−4 + 12 = 8$ $−6 + 8 = 2$

Check.

$(x - 6) (x + 8)$

$x^{2} - 6 x + 8 x - 48$

$x^{2} + 2 x - 48 ✓$

Let’s summarize the method we just developed to factor trinomials of the form $x^{2} + b x + c$ .

Factor Trinomials of the Form x² + bxy + cy²

Sometimes you’ll need to factor trinomials of the form $x^{2} + b x y + c y^{2}$ with two variables, such as $x^{2} + 12 x y + 36 y^{2} .$ The first term, $x^{2}$ , is the product of the first terms of the binomial factors, $x \cdot x$ . The $y^{2}$ in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in the previous objective.

Factor: $x^{2} + 12 x y + 36 y^{2}$ .

Solution

Demonstrates an algebraic expression (x^2 + 12xy + 36y^2) with notes, indicating steps for its manipulation or factorization.
	$x^{2} + 12 x y + 36 y^{2}$
Note that the first terms are x, last terms contain y.	$(x_y) (x_y)$

Find the numbers that multiply to 36 and add to 12.

Factors of $36$	Sum of factors
1, 36	$1 + 36 = 37$
2, 18	$2 + 18 = 20$
3, 12	$3 + 12 = 15$
4, 9	$4 + 9 = 13$
6, 6	$6 + 6 = 12 *$

Steps to expand and verify the algebraic expression (x+6y)(x+6y).
Use 6 and 6 as the coefficients of the last terms.	$(x + 6 y) (x + 6 y)$
Check your answer. $\begin{matrix} (x + 6 y) (x + 6 y) \\ x^{2} + 6 x y + 6 x y + 36 y^{2} \\ x^{2} + 12 x y + 36 y^{2} ✓ \end{matrix}$

Factor: $r^{2} - 8 r s - 9 s^{2}$ .

Solution

We need $r$ in the first term of each binomial and $s$ in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

$\begin{matrix} r^{2} - 8 r s - 9 s^{2} \\ Note that the first terms are r, last terms contain s . & (r_s) (r_s) \end{matrix}$

Find the numbers that multiply to $−9$ and add to $−8$ .

Factors of $−9$	Sum of factors
$1, −9$	$1 + (−9) = −8 *$
$−1, 9$	$−1 + 9 = 8$
$3, −3$	$3 + (−3) = 0$

$\begin{matrix} Use 1, −9 as coefficients of the last terms. & (r + s) (r - 9 s) \\ Check your answer. \\ (r - 9 s) (r + s) \\ r^{2} + r s - 9 r s - 9 s^{2} \\ r^{2} - 8 r s - 9 s^{2} ✓ \end{matrix}$

Factor: $u^{2} - 9 u v - 12 v^{2}$ .

Solution

We need u in the first term of each binomial and $v$ in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

Illustrates the factorization of a quadratic expression involving two variables.
	$u^{2} - 9 u v - 12 v^{2}$
Note that the first terms are u, last terms contain v.	$(u_v) (u_v)$

Find the numbers that multiply to $−12$ and add to $−9$ .

Factors of $−12$	Sum of factors
$1, −12$	$1 + (−12) = −11$
$−1, 12$	$−1 + 12 = 11$
$2, −6$	$2 + (−6) = −4$
$−2, 6$	$−2 + 6 = 4$
$3, −4$	$3 + (−4) = −1$
$−3, 4$	$−3 + 4 = 1$

Note there are no factor pairs that give us $−9$ as a sum. The trinomial is prime.

Key Concepts

Factor trinomials of the form $x^{2} + b x + c$
1. Write the factors as two binomials with first terms x: $(x) (x)$ .
2. Find two numbers m and n that
  Multiply to c, $m \cdot n = c$
  Add to b, $m + n = b$
3. Use m and n as the last terms of the factors: $(x + m) (x + n)$ .
4. Check by multiplying the factors.

Practice Makes Perfect

Factor Trinomials of the Form $x^{2} + b x + c$

In the following exercises, factor each trinomial of the form $x^{2} + b x + c$ .

$x^{2} + 4 x + 3$

Solution

$(x + 1) (x + 3)$

$y^{2} + 8 y + 7$

$m^{2} + 12 m + 11$

Solution

$(m + 1) (m + 11)$

$b^{2} + 14 b + 13$

$a^{2} + 9 a + 20$

Solution

$(a + 4) (a + 5)$

$m^{2} + 7 m + 12$

$p^{2} + 11 p + 30$

Solution

$(p + 5) (p + 6)$

$w^{2} + 10 w + 21$

$n^{2} + 19 n + 48$

Solution

$(n + 3) (n + 16)$

$b^{2} + 14 b + 48$

$a^{2} + 25 a + 100$

Solution

$(a + 5) (a + 20)$

$u^{2} + 101 u + 100$

$x^{2} - 8 x + 12$

Solution

$(x - 2) (x - 6)$

$q^{2} - 13 q + 36$

$y^{2} - 18 y + 45$

Solution

$(y - 3) (y - 15)$

$m^{2} - 13 m + 30$

$x^{2} - 8 x + 7$

Solution

$(x - 1) (x - 7)$

$y^{2} - 5 y + 6$

$p^{2} + 5 p - 6$

Solution

$(p - 1) (p + 6)$

$n^{2} + 6 n - 7$

$y^{2} - 6 y - 7$

Solution

$(y + 1) (y - 7)$

$v^{2} - 2 v - 3$

$x^{2} - x - 12$

Solution

$(x - 4) (x + 3)$

$r^{2} - 2 r - 8$

$a^{2} - 3 a - 28$

Solution

$(a - 7) (a + 4)$

$b^{2} - 13 b - 30$

$w^{2} - 5 w - 36$

Solution

$(w - 9) (w + 4)$

$t^{2} - 3 t - 54$

$x^{2} + x + 5$

Solution

prime

$x^{2} - 3 x - 9$

$8 - 6 x + x^{2}$

Solution

$(x - 4) (x - 2)$

$7 x + x^{2} + 6$

$x^{2} - 12 - 11 x$

Solution

$(x - 12) (x + 1)$

$−11 - 10 x + x^{2}$

Factor Trinomials of the Form $x^{2} + b x y + c y^{2}$

In the following exercises, factor each trinomial of the form $x^{2} + b x y + c y^{2}$ . If the trinomial cannot be factored, answer “Prime.”

$p^{2} + 3 p q + 2 q^{2}$

Solution

$(p + q) (p + 2 q)$

$m^{2} + 6 m n + 5 n^{2}$

$r^{2} + 15 r s + 36 s^{2}$

Solution

$(r + 3 s) (r + 12 s)$

$u^{2} + 10 u v + 24 v^{2}$

$m^{2} - 12 m n + 20 n^{2}$

Solution

$(m - 2 n) (m - 10 n)$

$p^{2} - 16 p q + 63 q^{2}$

$x^{2} - 2 x y - 80 y^{2}$

Solution

$(x + 8 y) (x - 10 y)$

$p^{2} - 8 p q - 65 q^{2}$

$m^{2} - 64 m n - 65 n^{2}$

Solution

$(m + n) (m - 65 n)$

$p^{2} - 2 p q - 35 q^{2}$

$a^{2} + 5 a b - 24 b^{2}$

Solution

$(a + 8 b) (a - 3 b)$

$r^{2} + 3 r s - 28 s^{2}$

$x^{2} - 3 x y - 14 y^{2}$

Solution

prime

$u^{2} - 8 u v - 24 v^{2}$

$m^{2} - 5 m n + 30 n^{2}$

Solution

prime

$c^{2} - 7 c d + 18 d^{2}$

Mixed Practice

In the following exercises, factor each expression.

$u^{2} - 12 u + 36$

Solution

$(u - 6) (u - 6)$

$w^{2} + 4 w - 32$

$x^{2} - 14 x - 32$

Solution

$(x + 2) (x - 16)$

$y^{2} + 41 y + 40$

$r^{2} - 20 r s + 64 s^{2}$

Solution

$(r - 4 s) (r - 16 s)$

$x^{2} - 16 x y + 64 y^{2}$

$k^{2} + 34 k + 120$

Solution

$(k + 4) (k + 30)$

$m^{2} + 29 m + 120$

$y^{2} + 10 y + 15$

Solution

prime

$z^{2} - 3 z + 28$

$m^{2} + m n - 56 n^{2}$

Solution

$(m + 8 n) (m - 7 n)$

$q^{2} - 29 q r - 96 r^{2}$

$u^{2} - 17 u v + 30 v^{2}$

Solution

$(u - 15 v) (u - 2 v)$

$m^{2} - 31 m n + 30 n^{2}$

$c^{2} - 8 c d + 26 d^{2}$

Solution

prime

$r^{2} + 11 r s + 36 s^{2}$

Everyday Math

Consecutive integers Deirdre is thinking of two consecutive integers whose product is 56. The trinomial $x^{2} + x - 56$ describes how these numbers are related. Factor the trinomial.

Solution

$(x + 8) (x - 7)$

Consecutive integers Deshawn is thinking of two consecutive integers whose product is 182. The trinomial $x^{2} + x - 182$ describes how these numbers are related. Factor the trinomial.

Writing Exercises

Many trinomials of the form $x^{2} + b x + c$ factor into the product of two binomials $(x + m) (x + n)$ . Explain how you find the values of m and n.

Solution

Answers may vary

How do you determine whether to use plus or minus signs in the binomial factors of a trinomial of the form $x^{2} + b x + c$ where $b$ and $c$ may be positive or negative numbers?

Will factored $x^{2} - x - 20$ as $(x + 5) (x - 4)$ . Bill factored it as $(x + 4) (x - 5)$ . Phil factored it as $(x - 5) (x - 4)$ . Who is correct? Explain why the other two are wrong.

Solution

Answers may vary

Look at Example 3, where we factored $y^{2} + 17 y + 60$ . We made a table listing all pairs of factors of 60 and their sums. Do you find this kind of table helpful? Why or why not?

Self Check

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

This table has the following statements all to be preceded by “I can…”. The first is “factor trinomials of the form x ^ 2 +b x + c”. The second is “factor trinomials of the form x^2 + b x y + c y ^ 2”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.

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

Factor Trinomials of the Form ax²+bx+c

Recognize a Preliminary Strategy for Factoring

Let’s summarize where we are so far with factoring polynomials. In the first two sections of this chapter, we used three methods of factoring: factoring the GCF, factoring by grouping, and factoring a trinomial by “undoing” FOIL. More methods will follow as you continue in this chapter, as well as later in your studies of algebra.

How will you know when to use each factoring method? As you learn more methods of factoring, how will you know when to apply each method and not get them confused? It will help to organize the factoring methods into a strategy that can guide you to use the correct method.

As you start to factor a polynomial, always ask first, “Is there a greatest common factor?” If there is, factor it first.

The next thing to consider is the type of polynomial. How many terms does it have? Is it a binomial? A trinomial? Or does it have more than three terms?

If it is a trinomial where the leading coefficient is one, $x^{2} + b x + c$ , use the “undo FOIL” method.

If it has more than three terms, try the grouping method. This is the only method to use for polynomials of more than three terms.

Some polynomials cannot be factored. They are called “prime.”

Below we summarize the methods we have so far. These are detailed in Choose a strategy to factor polynomials completely.

This figure lists strategies for factoring polynomials. At the top of the figure is G C F, where factoring always starts. From there, the figure has three branches. The first is binomial, the second is trinomial with the form x ^ 2 + b x +c, and the third is “more than three terms”, which is labeled with grouping.

Use the preliminary strategy to completely factor a polynomial. A polynomial is factored completely if, other than monomials, all of its factors are prime.

Identify the best method to use to factor each polynomial.

ⓐ $6 y^{2} - 72$
ⓑ $r^{2} - 10 r - 24$
ⓒ $p^{2} + 5 p + p q + 5 q$

Solution

ⓐ

This table illustrates the step-by-step process of factoring the expression 6y^2 - 72, including identifying the greatest common factor and recognizing the resulting binomial.
	$6 y^{2} - 72$
Is there a greatest common factor?	Yes, 6.
Factor out the 6.	$6 (y^{2} - 12)$
Is it a binomial, trinomial, or are there more than 3 terms?	Binomial, we have no method to factor binomials yet.

ⓑ

Steps for factoring the trinomial polynomial r^2 - 10r - 24, including checking for common factors and identifying its type for method selection.
	$r^{2} - 10 r - 24$
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	Trinomial, with leading coefficient 1, so “undo” FOIL.

ⓒ

Guide for selecting a factoring method by evaluating a polynomial's common factors and number of terms.
	$p^{2} + 5 p + p q + 5 q$
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	More than three terms, so factor using grouping.

Factor Trinomials of the form ax² + bx + c with a GCF

Now that we have organized what we’ve covered so far, we are ready to factor trinomials whose leading coefficient is not 1, trinomials of the form $a x^{2} + b x + c$ .

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods in the last section. Let’s do a few examples to see how this works.

Watch out for the signs in the next two examples.

Factor completely: $2 n^{2} - 8 n - 42$ .

Solution

Use the preliminary strategy.

Illustrates finding and factoring out the Greatest Common Factor (GCF) from an algebraic expression.
Is there a greatest common factor?	$2 n^{2} - 8 n - 42$
Yes, GCF = 2. Factor it out.	$2 (n^{2} - 4 n - 21)$

Inside the parentheses, is it a binomial, trinomial, or are there more than three terms?

This table illustrates the step-by-step process of factoring a trinomial, showing both the procedural description and the corresponding mathematical expressions.
It is a trinomial whose coefficient is 1, so undo FOIL.	$2 (n) (n)$
Use 3 and −7 as the last terms of the binomials.	$2 (n + 3) (n - 7)$

Factors of $−21$	Sum of factors
$1, −21$	$1 + (−21) = −20$
$3, −7$	$3 + (−7) = −4 *$

Check.

$2 (n + 3) (n - 7)$

$2 (n^{2} - 7 n + 3 n - 21)$

$2 (n^{2} - 4 n - 21)$

$2 n^{2} - 8 n - 42 ✓$

Factor completely: $4 y^{2} - 36 y + 56$ .

Solution

Use the preliminary strategy.

This table illustrates the step-by-step process of factoring a quadratic trinomial, including identifying the GCF and undoing FOIL to find the final factored form.
Is there a greatest common factor?	$4 y^{2} - 36 y + 56$
Yes, GCF = 4. Factor it.	$4 (y^{2} - 9 y + 14)$
Inside the parentheses, is it a binomial, trinomial, or are there more than three terms?
It is a trinomial whose coefficient is 1. So undo FOIL.	$4 (y) (y)$
Use a table like the one below to find two numbers that multiply to 14 and add to −9.
Both factors of 14 must be negative.	$4 (y - 2) (y - 7)$

Factors of $14$	Sum of factors
$−1, −14$	$−1 + (−14) = −15$
$−2, −7$	$−2 + (−7) = −9 *$

Check.

$4 (y - 2) (y - 7)$

$4 (y^{2} - 7 y - 2 y + 14)$

$4 (y^{2} - 9 y + 14)$

$4 y^{2} - 36 y + 56 ✓$

In the next example the GCF will include a variable.

Factor completely: $4 u^{3} + 16 u^{2} - 20 u$ .

Solution

Use the preliminary strategy.

This table illustrates the step-by-step process of factoring a polynomial expression, including finding the greatest common factor and binomial factors.
Is there a greatest common factor?	$4 u^{3} + 16 u^{2} - 20 u$
Yes, GCF = 4u. Factor it.	$4 u (u^{2} + 4 u - 5)$
Binomial, trinomial, or more than three terms?
It is a trinomial. So “undo FOIL.”	$4 u (u) (u)$
Use a table like the table below to find two numbers that multiply to −5 and add to 4.	$4 u (u - 1) (u + 5)$

Factors of $−5$	Sum of factors
$−1, 5$	$−1 + 5 = 4 *$
$1, −5$	$1 + (−5) = −4$

Check.

$4 u (u - 1) (u + 5)$

$4 u (u^{2} + 5 u - u - 5)$

$4 u (u^{2} + 4 u - 5)$

$4 u^{3} + 16 u^{2} - 20 u ✓$

Factor Trinomials using Trial and Error

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial $3 x^{2} + 5 x + 2$ .

From our earlier work we expect this will factor into two binomials.

\begin{matrix} 3 x^{2} + 5 x + 2 \\ () () \end{matrix}

We know the first terms of the binomial factors will multiply to give us $3 x^{2}$ . The only factors of $3 x^{2}$ are $1 x, 3 x$ . We can place them in the binomials.

This figure has the polynomial 3 x^ 2 +5 x +2. Underneath there are two terms, 1 x, and 3 x. Below these are the two factors x and (3 x) being shown multiplied.

Check. Does $1 x \cdot 3 x = 3 x^{2}$ ?

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1 and 2. But we now have two cases to consider as it will make a difference if we write 1, 2, or 2, 1.

This figure demonstrates the possible factors of the polynomial 3x^2 +5x +2. The polynomial is written twice. Underneath both, there are the terms 1x, 3x under the 3x^2. Also, there are the factors 1,2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3x + 2) and the next is (x + 2)(3x + 1).

Which factors are correct? To decide that, we multiply the inner and outer terms.

Since the middle term of the trinomial is 5x, the factors in the first case will work. Let’s FOIL to check.

\begin{matrix} (x + 1) (3 x + 2) \\ 3 x^{2} + 2 x + 3 x + 2 \\ 3 x^{2} + 5 x + 2 ✓ \end{matrix}

Our result of the factoring is:

\begin{matrix} 3 x^{2} + 5 x + 2 \\ (x + 1) (3 x + 2) \end{matrix}

How to Factor Trinomials of the Form $a x^{2} + b x + c$ Using Trial and Error

Factor completely: $3 y^{2} + 22 y + 7$ .

Solution

This table summarizes the steps for factoring 3 y ^ 2 + 22 y + 7. The first row states write the trinomial in descending order. The polynomial is written 3 y ^ 2 +22 y + 7.

The second row states find all the factor pairs of the first term. The only pairs listed are 1 y, 3 y. Then, since there is only one pair, they are in the parentheses written (1 y ) and (3 y ).

The third row states “find all the factored pairs of the third term”. It also states the only factors of 7 are 1 and 7.

The fourth row states test all the possible combinations of the factors until the correct product is found. The possible factors are shown (y + 1)(3 y + 7) and (y + 7)(3y + 1). Under each factor is the products of the outer terms and the inner terms. For the first it is 7y and 3y. For the second it is 21 y and y. The combination (y + 7)(3 y + 1) is the correct factoring.

The last row states to check by multiplying. The product of (y + 7)(3 y + 1) is shown as 3 y ^ 2 + 22 y + 7.

When the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

Factor completely: $6 b^{2} - 13 b + 5$ .

Solution

The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

$6 b^{2} - 13 b + 5$
Possible factors	Product
$(b - 1) (6 b - 5)$	$6 b^{2} - 11 b + 5$
$(b - 5) (6 b - 1)$	$6 b^{2} - 31 b + 5$
$(2 b - 1) (3 b - 5)$	$6 b^{2} - 13 b + 5$ *
$(2 b - 5) (3 b - 1)$	$6 b^{2} - 17 b + 5$

This table demonstrates how to check if given factors are correct by multiplying them to obtain the original polynomial, including a step-by-step example.
The correct factors are those whose product is the original trinomial.	$(2 b - 1) (3 b - 5)$
Check by multiplying. $\begin{matrix} (2 b - 1) (3 b - 5) \\ 6 b^{2} - 10 b - 3 b + 5 \\ 6 b^{2} - 13 b + 5 ✓ \end{matrix}$

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Factor completely: $14 x^{2} - 47 x - 7$ .

Solution

The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since it is negative, one factor must be positive and one negative.

Consider all the combinations of factors. We use each pair of the factors of $14 x^{2}$ with each pair of factors of $−7 .$

Factors of $14 x^{2}$	Pair with	Factors of $−7$
$x$ , $14 x$		$1$ , $−7$ $−7$ , $1$ (reverse order)
$x$ , $14 x$		$−1$ , $7$ $7$ , $−1$ (reverse order)
$2 x, 7 x$		$1$ , $−7$ $−7$ , $1$ (reverse order)
$2 x, 7 x$		$−1$ , $7$ $7$ , $−1$ (reverse order)

These pairings lead to the following eight combinations.

This table has the heading 14 x ^ 2 – 47 x minus 7. This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. In the first row under “possible factors” it reads (x+1) and (14 x minus 7). Under product, in the next column, it says “not an option”. In the next row down, it shows (x minus 7) and (14 x plus 1). In the next row down, it shows (x minus 1) and (14 x plus 7). Next to this in the product column, it says “not an option.” The next row down under “possible factors”, it has the equation (x plus 7 and 14 x minus 1. Next to this in the product column it has 14 x ^2 plus 97 x minus 7. The next row down under possible factors, it has 2 x plus 1 and 7 x minus 7. Next to this under the product column, is says “not an option”. The next row down reads 2 x minus 7 and 7x plus 1. Next to this under the product column, it has 14 x ^2 minus 47 x minus 7 with the asterisk following the 7. The next row down reads 2 x minus 1 and 7 x plus 7. Next to this in the product column it reads “not an option”. The final row reads 2 x plus 7 and 7 x minus 1. Next to this in the product column it reads 14, x, ^ 2 plus 47 x minus 7. Next to the table is a box with four arrows point to each “not an option” row. The reason given in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor. That means that each of these combinations is not an option.”

This table demonstrates how to verify factors of a trinomial by multiplying the binomial expressions.
The correct factors are those whose product is the original trinomial.	$(2 x - 7) (7 x + 1)$
Check by multiplying. $\begin{matrix} (2 x - 7) (7 x + 1) \\ 14 x^{2} + 2 x - 49 x - 7 \\ 14 x^{2} - 47 x - 7 ✓ \end{matrix}$

Factor completely: $18 n^{2} - 37 n + 15$ .

Solution

The trinomial is already in descending order.	$18 n^{2} - 37 n + 15$
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative facotrs.

Consider all the combinations of factors.

Illustrates factoring a trinomial, showing how to find correct factors and verify them through multiplication.
The correct factors are those whose product is the original trinomial.	$(2 n - 3) (9 n - 5)$
Check by multiplying. $\begin{matrix} (2 n - 3) (9 n - 5) \\ 18 n^{2} - 10 n - 27 n + 15 \\ 18 n^{2} - 37 n + 15 ✓ \end{matrix}$

Don’t forget to look for a GCF first.

Factor completely: $10 y^{4} + 55 y^{3} + 60 y^{2}$ .

Solution

	$10 y^{4} + 55 y^{3} + 60 y^{2}$
Notice the greatest common factor, and factor it first.	$15 y^{2} (2 y^{2} + 11 y + 12)$
Factor the trinomial.

Consider all the combinations.

Illustrates the factored form of a trinomial and the verification steps through multiplication.
The correct factors are those whose product is the original trinomial. Remember to include the factor 5y².	$5 y^{2} (y + 4) (2 y + 3)$
Check by multiplying. $\begin{matrix} 5 y^{2} (y + 4) (2 y + 3) \\ 5 y^{2} (2 y^{2} + 8 y + 3 y + 12) \\ 10 y^{4} + 55 y^{3} + 60 y^{2} ✓ \end{matrix}$

Factor Trinomials using the “ac” Method

Another way to factor trinomials of the form $a x^{2} + b x + c$ is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!

How to Factor Trinomials Using the “ac” Method

Factor: $6 x^{2} + 7 x + 2$ .

Solution

This table lists the steps for factoring 6 x ^ 2 + 7 x + 2. The first step is to factor the GCF. This polynomial has none.

The second row states to find the product a c. Then, it lists a c as 6 times 2 = 12.

The third step is to find two numbers m and n in which m times n = a c and m + n = b. The middle column reads, “find two numbers that add to 7. Both factors must be positive”. The numbers are 3 and 4. 3 times 4 is 12 and 3 + 4 is 7.

The next step is to split the middle term using m and n. That is, to write 7 x as 3 x + 4 x. Therefore, 6 x ^ 2 + 7 x + 2 is rewritten as 6 x ^ 2 +3 x + 4 x + 2.

The next step is to factor by grouping. 3 x(2 x + 1) + 2(2 x + 1) then factor again (2 x + 1)(3 x + 2).

The last step is to check by multiplying. Multiply the factors (2 x + 1)(3 x + 2) to get 6 x ^ 2 + 7 x + 2.

When the third term of the trinomial is negative, the factors of the third term will have opposite signs.

Factor: $8 u^{2} - 17 u - 21$ .

Solution

Is there a greatest common factor? No.
Find $a \cdot c$ .	$a \cdot c$
	$8 (−21)$
	$−168$

Find two numbers that multiply to $−168$ and add to $−17$ . The larger factor must be negative.

Factors of $−168$	Sum of factors
$1, −168$	$1 + (−168) = −167$
$2, −84$	$2 + (−84) = −82$
$3, −56$	$3 + (−56) = −53$
$4, −42$	$4 + (−42) = −38$
$6, −28$	$6 + (−28) = −22$
$7, −24$	$7 + (−24) = −17 *$
$8, −21$	$8 + (−21) = −13$

Steps to factor a quadratic expression by grouping, including checking the result.
Split the middle term using 7u and −24u.	$\begin{matrix} 8 u^{2} - \underset{↙}{17} \underset{↘}{u} - 21 \\ \underset{⎵}{8 u^{2} + 7 u} \underset{⎵}{−24 u - 21} \end{matrix}$
Factor by grouping.	$\begin{matrix} u (8 u + 7) - 3 (8 u + 7) \\ (8 u + 7) (u - 3) \end{matrix}$
Check by multiplying. $\begin{matrix} (8 u + 7) (u - 3) \\ 8 u^{2} - 24 u + 7 u - 21 \\ 8 u^{2} - 17 u - 21 ✓ \end{matrix}$

Factor: $2 x^{2} + 6 x + 5$ .

Solution

Is there a greatest common factor? No.
Find $a \cdot c$ .	$a \cdot c$
	$2 (5)$
	$10$

Find two numbers that multiply to 10 and add to 6.

Factors of $10$	Sum of factors
$1, 10$	$1 + 10 = 11$
2, 5	$2 + 5 = 7$

There are no factors that multiply to 10 and add to 6. The polynomial is prime.

Don’t forget to look for a common factor!

Factor: $10 y^{2} - 55 y + 70$ .

Solution

Is there a greatest common factor? Yes. The GCF is 5.
Factor it. Be careful to keep the factor of 5 all the way through the solution!
The trinomial inside the parentheses has a leading coefficient that is not 1.
Factor the trinomial.
Check by mulitplying all three factors.
$5 (2 y^{2} - 2 y - 4 y + 14)$
$5 (2 y^{2} - 11 y + 14)$
$10 y^{2} - 55 y + 70 ✓$

We can now update the Preliminary Factoring Strategy, as shown in Figure 1 and detailed in Choose a strategy to factor polynomials completely (updated), to include trinomials of the form $a x^{2} + b x + c$ . Remember, some polynomials are prime and so they cannot be factored.

This figure has the strategy for factoring polynomials. At the top of the figure is GCF. Below this, there are three options. The first is binomial. The second is trinomial. Under trinomial there are x squared + b x + c and a x squared + b x +c. The two methods here are trial and error and the “a c” method. The third option is for more than three terms. It is grouping.

Key Concepts

Factor Trinomials of the Form $a x^{2} + b x + c$ using Trial and Error: See Example 5.
1. Write the trinomial in descending order of degrees.
2. Find all the factor pairs of the first term.
3. Find all the factor pairs of the third term.
4. Test all the possible combinations of the factors until the correct product is found.
5. Check by multiplying.
Factor Trinomials of the Form $a x^{2} + b x + c$ Using the “ac” Method: See Example 10.
1. Factor any GCF.
2. Find the product ac.
3. Find two numbers m and n that:
  $\begin{matrix} Multiply to a c & m \cdot n = a \cdot c \\ Add to b & m + n = b \end{matrix}$
4. Split the middle term using m and n:
5. Factor by grouping.
6. Check by multiplying the factors.
Choose a strategy to factor polynomials completely (updated):
1. Is there a greatest common factor? Factor it.
2. Is the polynomial a binomial, trinomial, or are there more than three terms?
  If it is a binomial, right now we have no method to factor it.
  If it is a trinomial of the form $x^{2} + b x + c$
     Undo FOIL $(x) (x)$ .
  If it is a trinomial of the form $a x^{2} + b x + c$
     Use Trial and Error or the “ac” method.
  If it has more than three terms
     Use the grouping method.
3. Check by multiplying the factors.

Practice Makes Perfect

Recognize a Preliminary Strategy to Factor Polynomials Completely

In the following exercises, identify the best method to use to factor each polynomial.

ⓐ $10 q^{2} + 50$
ⓑ $a^{2} - 5 a - 14$
ⓒ $u v + 2 u + 3 v + 6$

Solution

ⓐ factor the GCF, binomial ⓑ Undo FOIL ⓒ factor by grouping

ⓐ $n^{2} + 10 n + 24$
ⓑ $8 u^{2} + 16$
ⓒ $p q + 5 p + 2 q + 10$

ⓐ $x^{2} + 4 x - 21$
ⓑ $a b + 10 b + 4 a + 40$
ⓒ $6 c^{2} + 24$

Solution

ⓐ undo FOIL ⓑ factor by grouping ⓒ factor the GCF, binomial

ⓐ $20 x^{2} + 100$
ⓑ $u v + 6 u + 4 v + 24$
ⓒ $y^{2} - 8 y + 15$

Factor Trinomials of the form $a x^{2} + b x + c$ with a GCF

In the following exercises, factor completely.

$5 x^{2} + 35 x + 30$

Solution

$5 (x + 1) (x + 6)$

$12 s^{2} + 24 s + 12$

$2 z^{2} - 2 z - 24$

Solution

$2 (z - 4) (z + 3)$

$3 u^{2} - 12 u - 36$

$7 v^{2} - 63 v + 56$

Solution

$7 (v - 1) (v - 8)$

$5 w^{2} - 30 w + 45$

$p^{3} - 8 p^{2} - 20 p$

Solution

$p (p - 10) (p + 2)$

$q^{3} - 5 q^{2} - 24 q$

$3 m^{3} - 21 m^{2} + 30 m$

Solution

$3 m (m - 5) (m - 2)$

$11 n^{3} - 55 n^{2} + 44 n$

$5 x^{4} + 10 x^{3} - 75 x^{2}$

Solution

$5 x^{2} (x - 3) (x + 5)$

$6 y^{4} + 12 y^{3} - 48 y^{2}$

Factor Trinomials Using Trial and Error

In the following exercises, factor.

$2 t^{2} + 7 t + 5$

Solution

$(2 t + 5) (t + 1)$

$5 y^{2} + 16 y + 11$

$11 x^{2} + 34 x + 3$

Solution

$(11 x + 1) (x + 3)$

$7 b^{2} + 50 b + 7$

$4 w^{2} - 5 w + 1$

Solution

$(4 w - 1) (w - 1)$

$5 x^{2} - 17 x + 6$

$6 p^{2} - 19 p + 10$

Solution

$(3 p - 2) (2 p - 5)$

$21 m^{2} - 29 m + 10$

$4 q^{2} - 7 q - 2$

Solution

$(4 q + 1) (q - 2)$

$10 y^{2} - 53 y - 11$

$4 p^{2} + 17 p - 15$

Solution

$(4 p - 3) (p + 5)$

$6 u^{2} + 5 u - 14$

$16 x^{2} - 32 x + 16$

Solution

$16 (x - 1) (x - 1)$

$81 a^{2} + 153 a - 18$

$30 q^{3} + 140 q^{2} + 80 q$

Solution

$10 q (3 q + 2) (q + 4)$

$5 y^{3} + 30 y^{2} - 35 y$

Factor Trinomials using the ‘ac’ Method

In the following exercises, factor.

$5 n^{2} + 21 n + 4$

Solution

$(5 n + 1) (n + 4)$

$8 w^{2} + 25 w + 3$

$9 z^{2} + 15 z + 4$

Solution

$(3 z + 1) (3 z + 4)$

$3 m^{2} + 26 m + 48$

$4 k^{2} - 16 k + 15$

Solution

$(2 k - 3) (2 k - 5)$

$4 q^{2} - 9 q + 5$

$5 s^{2} - 9 s + 4$

Solution

$(5 s - 4) (s - 1)$

$4 r^{2} - 20 r + 25$

$6 y^{2} + y - 15$

Solution

$(3 y + 5) (2 y - 3)$

$6 p^{2} + p - 22$

$2 n^{2} - 27 n - 45$

Solution

$(2 n + 3) (n - 15)$

$12 z^{2} - 41 z - 11$

$3 x^{2} + 5 x + 4$

Solution

prime

$4 y^{2} + 15 y + 6$

$60 y^{2} + 290 y - 50$

Solution

$10 (6 y - 1) (y + 5)$

$6 u^{2} - 46 u - 16$

$48 z^{3} - 102 z^{2} - 45 z$

Solution

$3 z (8 z + 3) (2 z - 5)$

$90 n^{3} + 42 n^{2} - 216 n$

$16 s^{2} + 40 s + 24$

Solution

$8 (2 s + 3) (s + 1)$

$24 p^{2} + 160 p + 96$

$48 y^{2} + 12 y - 36$

Solution

$12 (4 y - 3) (y + 1)$

$30 x^{2} + 105 x - 60$

Mixed Practice

In the following exercises, factor.

$12 y^{2} - 29 y + 14$

Solution

$(4 y - 7) (3 y - 2)$

$12 x^{2} + 36 y - 24 z$

$a^{2} - a - 20$

Solution

$(a - 5) (a + 4)$

$m^{2} - m - 12$

$6 n^{2} + 5 n - 4$

Solution

$(2 n - 1) (3 n + 4)$

$12 y^{2} - 37 y + 21$

$2 p^{2} + 4 p + 3$

Solution

prime

$3 q^{2} + 6 q + 2$

$13 z^{2} + 39 z - 26$

Solution

$13 (z^{2} + 3 z - 2)$

$5 r^{2} + 25 r + 30$

$x^{2} + 3 x - 28$

Solution

$(x + 7) (x - 4)$

$6 u^{2} + 7 u - 5$

$3 p^{2} + 21 p$

Solution

$3 p (p + 7)$

$7 x^{2} - 21 x$

$6 r^{2} + 30 r + 36$

Solution

$6 (r + 2) (r + 3)$

$18 m^{2} + 15 m + 3$

$24 n^{2} + 20 n + 4$

Solution

$4 (2 n + 1) (3 n + 1)$

$4 a^{2} + 5 a + 2$

$x^{2} + 2 x - 24$

Solution

$(x + 6) (x - 4)$

$2 b^{2} - 7 b + 4$

Everyday Math

Height of a toy rocket The height of a toy rocket launched with an initial speed of 80 feet per second from the balcony of an apartment building is related to the number of seconds, t, since it is launched by the trinomial $−16 t^{2} + 80 t + 96$ . Completely factor the trinomial.

Solution

$−16 (t - 6) (t + 1)$

Height of a beach ball The height of a beach ball tossed up with an initial speed of 12 feet per second from a height of 4 feet is related to the number of seconds, t, since it is tossed by the trinomial $−16 t^{2} + 12 t + 4$ . Completely factor the trinomial.

Writing Exercises

List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form $a x^{2} + b x + c .$

Solution

Answers may vary.

How is the “ac” method similar to the “undo FOIL” method? How is it different?

What are the questions, in order, that you ask yourself as you start to factor a polynomial? What do you need to do as a result of the answer to each question?

Solution

Answers may vary.

On your paper draw the chart that summarizes the factoring strategy. Try to do it without looking at the book. When you are done, look back at the book to finish it or verify it.

Self Check

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

This table has the following statements all to be preceded by “I can…”. The first row is “recognize a preliminary strategy to factor polynomials completely”. The second row is “factor trinomials of the form a x ^ 2 + b x + c with a GCF”. The third row is “factor trinomials using trial and error”. And the fourth row is “factor trinomials using the “ac” method”. In the columns beside these statements are the headers, “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?

prime polynomials

Polynomials that cannot be factored are prime polynomials.

Factor Special Products

The strategy for factoring we developed in the last section will guide you as you factor most binomials, trinomials, and polynomials with more than three terms. We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.

Factor Perfect Square Trinomials

Some trinomials are perfect squares. They result from multiplying a binomial times itself. You can square a binomial by using FOIL, but using the Binomial Squares pattern you saw in a previous chapter saves you a step. Let’s review the Binomial Squares pattern by squaring a binomial using FOIL.

This image shows the FOIL procedure for multiplying (3x + 4) squared. The polynomial is written with two factors (3x + 4)(3x + 4). Then, the terms are 9 x squared + 12 x + 12 x + 16, demonstrating first, outer, inner, last. Finally, the product is written, 9 x squared + 24 x + 16.

The first term is the square of the first term of the binomial and the last term is the square of the last. The middle term is twice the product of the two terms of the binomial.

\begin{matrix} {(3 x)}^{2} + 2 (3 x \cdot 4) + 4^{2} \\ 9 x^{2} + 24 x + 16 \end{matrix}

The trinomial 9x² + 24 +16 is called a perfect square trinomial. It is the square of the binomial 3x+4.

We’ll repeat the Binomial Squares Pattern here to use as a reference in factoring.

When you square a binomial, the product is a perfect square trinomial. In this chapter, you are learning to factor—now, you will start with a perfect square trinomial and factor it into its prime factors.

You could factor this trinomial using the methods described in the last section, since it is of the form ax² + bx + c. But if you recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, you will save yourself a lot of work.

Here is the pattern—the reverse of the binomial squares pattern.

To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, $a^{2}$ . Next check that the last term is a perfect square, $b^{2}$ . Then check the middle term—is it twice the product, 2ab? If everything checks, you can easily write the factors.

How to Factor Perfect Square Trinomials

Factor: $9 x^{2} + 12 x + 4$ .

Solution

This table gives the steps for factoring 9 x squared +12 x +4. The first step is recognizing the perfect square pattern “a” squared + 2 a b + b squared. This includes, is the first term a perfect square and is the last term a perfect square. The first term can be written as (3 x) squared and the last term can be written as 2 squared. Also, in the first step, the middle term has to be twice “a” times b. This is verified by 2 times 3 x times 2 being 12 x.

The second step is writing the square of the binomial. The polynomial is written as (3 x) squared + 2 times 3 x times 2 + 2 squared. This is factored as (3 x + 2) squared.

The last step is to check with multiplication.

The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern $a^{2} - 2 a b + b^{2}$ , which factors to ${(a - b)}^{2}$ .

The steps are summarized here.

We’ll work one now where the middle term is negative.

Factor: $81 y^{2} - 72 y + 16$ .

Solution

The first and last terms are squares. See if the middle term fits the pattern of a perfect square trinomial. The middle term is negative, so the binomial square would be ${(a - b)}^{2}$ .


Are the first and last terms perfect squares?
Check the middle term.
Does is match ${(a - b)}^{2}$ ? Yes.
Write the square of a binomial.
Check by mulitplying.
${(9 y - 4)}^{2}$
${(9 y)}^{2} - 2 \cdot 9 y \cdot 4 + 4^{2}$
$81 y^{2} - 72 y + 16 ✓$

The next example will be a perfect square trinomial with two variables.

Factor: $36 x^{2} + 84 x y + 49 y^{2}$ .

Solution


Test each term to verify the pattern.
Factor.
Check by mulitplying.
${(6 x + 7 y)}^{2}$
${(6 x)}^{2} + 2 \cdot 6 x \cdot 7 y + {(7 y)}^{2}$
$36 x^{2} + 84 x y + 49 y^{2} ✓$

Factor: $9 x^{2} + 50 x + 25$ .

Solution

Steps to factor 9x^2 + 50x + 25, showing an initial attempt at perfect square trinomials and then successful factorization using the 'ac' method.
	$\begin{matrix} 9 x^{2} + 50 x + 25 \end{matrix}$
Are the first and last terms perfect squares?	$\begin{matrix} {(3 x)}^{2} {(5)}^{2} \end{matrix}$
Check the middle term—is it $2 a b ?$	$\begin{matrix} {(3 x)}^{2}_{↘} \underset{\underset{30 x}{2 (3 x) (5)}}{}_{↙} {(5)}^{2} \end{matrix}$
No! $30 x \neq 50 x$	This does not fit the pattern!
Factor using the “ac” method.	$9 x^{2} + 50 x + 25$
$\begin{matrix} Notice: \begin{matrix} a c \\ 9 \cdot 25 \\ 225 \end{matrix} and \begin{matrix} 5 \cdot 45 = 225 \\ 5 + 45 = 50 \end{matrix} \end{matrix}$
Split the middle term. Factor by grouping.	$\begin{matrix} 9 x^{2} + 5 x + 45 x + 25 \\ x (9 x + 5) + 5 (9 x + 5) \\ (9 x + 5) (x + 5) \end{matrix}$
Check. $\begin{matrix} (9 x + 5) (x + 5) \\ 9 x^{2} + 45 x + 5 x + 25 \\ 9 x^{2} + 50 x + 25 ✓ \end{matrix}$

Remember the very first step in our Strategy for Factoring Polynomials? It was to ask “is there a greatest common factor?” and, if there was, you factor the GCF before going any further. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.

Factor: $36 x^{2} y - 48 x y + 16 y$ .

Solution

	$36 x^{2} y - 48 x y + 16 y$
Is there a GCF? Yes, 4y, so factor it out.	$4 y (9 x^{2} - 12 x + 4)$
Is this a perfect square trinomial?
Verify the pattern.
Factor.	$4 y {(3 x - 2)}^{2}$
Remember: Keep the factor 4y in the final product.
Check.
$4 y {(3 x - 2)}^{2}$
$4 y [{(3 x)}^{2} - 2 \cdot 3 x \cdot 2 + 2^{2}]$
$4 y {(9 x)}^{2} - 12 x + 4$
$36 x^{2} y - 48 x y + 16 y ✓$

Factor Differences of Squares

The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

\begin{matrix} (3 x - 4) (3 x + 4) \\ 9 x^{2} - 16 \end{matrix}

Remember, when you multiply conjugate binomials, the middle terms of the product add to 0. All you have left is a binomial, the difference of squares.

Multiplying conjugates is the only way to get a binomial from the product of two binomials.

To factor, we will use the product pattern “in reverse” to factor the difference of squares. A difference of squares factors to a product of conjugates.

Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.

How to Factor Differences of Squares

Factor: $x^{2} - 4$ .

Solution

This table gives the steps for factoring x squared minus 4. The first step is identifying the pattern in the binomial including it is a difference. Also, the first and last terms are verified as squares.

The second step is writing the two terms as squares, x squared and 2 squared.

The second step is writing the two terms as squares, x squared and 2 squared. The third step is to write the factoring as a product of the conjugates (x – 2)(x + 2).

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression $a^{2} + b^{2}$ is prime!

Don’t forget that 1 is a perfect square. We’ll need to use that fact in the next example.

Factor: $64 y^{2} - 1$ .

Solution


Is this a difference? Yes.
Are the first and last terms perfect squares?
Yes - write them as squares.
Factor as the product of conjugates.
Check by multiplying.
$(8 y - 1) (8 y + 1)$
$64 y^{2} - 1 ✓$

Factor: $121 x^{2} - 49 y^{2}$ .

Solution

$\begin{matrix} 121 x^{2} - 49 y^{2} \\ Is this a difference of squares? Yes. & {(11 x)}^{2} - {(7 y)}^{2} \\ Factor as the product of conjugates. & (11 x - 7 y) (11 x + 7 y) \\ Check by multiplying. \\ (11 x - 7 y) (11 x + 7 y) \\ 121 x^{2} - 49 y^{2} ✓ \end{matrix}$

The binomial in the next example may look “backwards,” but it’s still the difference of squares.

Factor: $100 - h^{2}$ .

Solution

This table demonstrates the step-by-step process of factoring the difference of squares expression 100 - h^2, including a caution.
	$100 - h^{2}$
Is this a difference of squares? Yes.	${(10)}^{2} - {(h)}^{2}$
Factor as the product of conjugates.	$(10 - h) (10 + h)$
Check by multiplying. $\begin{matrix} (10 - h) (10 + h) \\ 100 - h^{2} ✓ \end{matrix}$
Be careful not to rewrite the original expression as $h^{2} - 100$ .

Factor $h^{2} - 100$ on your own and then notice how the result differs from $(10 - h) (10 + h)$ .

To completely factor the binomial in the next example, we’ll factor a difference of squares twice!

Factor: $x^{4} - y^{4}$ .

Solution

This table illustrates the step-by-step factorization of the algebraic expression x^4 - y^4 into (x - y)(x + y)(x^2 + y^2) using the difference of squares method, complete with a verification.
	$x^{4} - y^{4}$
Is this a difference of squares? Yes.	${(x^{2})}^{2} - {(y^{2})}^{2}$
Factor it as the product of conjugates.	$(x^{2} - y^{2}) (x^{2} + y^{2})$
Notice the first binomial is also a difference of squares!	$({(x)}^{2} - {(y)}^{2}) (x^{2} + y^{2})$
Factor it as the product of conjugates. The last factor, the sum of squares, cannot be factored.	$(x - y) (x + y) (x^{2} + y^{2})$
Check by multiplying. $\begin{matrix} (x - y) (x + y) (x^{2} + y^{2}) \\ [(x - y) (x + y)] (x^{2} + y^{2}) \\ (x^{2} - y^{2}) (x^{2} + y^{2}) \\ x^{4} - y^{4} ✓ \end{matrix}$

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.

Factor: $8 x^{2} y - 98 y$ .

Solution

Step-by-step factorization of the algebraic expression 8x^2y - 98y, demonstrating GCF and difference of squares methods.
	$8 x^{2} y - 98 y$
Is there a GCF? Yes, 2y—factor it out!	$2 y (4 x^{2} - 49)$
Is the binomial a difference of squares? Yes.	$2 y ({(2 x)}^{2} - {(7)}^{2})$
Factor as a product of conjugates.	$2 y (2 x - 7) (2 x + 7)$
Check by multiplying. $\begin{matrix} 2 y (2 x - 7) (2 x + 7) \\ 2 y [(2 x - 7) (2 x + 7)] \\ 2 y (4 x^{2} - 49) \\ 8 x^{2} y - 98 y ✓ \end{matrix}$

Factor: $6 x^{2} + 96$ .

Solution

Steps demonstrating the factoring of the polynomial 6x^2 + 96, including GCF extraction and identifying unfactorable sums of squares.
	$6 x^{2} + 96$
Is there a GCF? Yes, 6—factor it out!	$6 (x^{2} + 16)$
Is the binomial a difference of squares? No, it is a sum of squares. Sums of squares do not factor!
Check by multiplying. $\begin{matrix} 6 (x^{2} + 16) \\ 6 x^{2} + 96 ✓ \end{matrix}$

Factor Sums and Differences of Cubes

There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.

\begin{matrix} a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \end{matrix}

We’ll check the first pattern and leave the second to you.


Distribute.
Multiply.	$a^{3} - a^{2} b + {a b}^{2} + a^{2} b - {a b}^{2} + b^{3}$
Combine like terms.	$a^{3} + b^{3}$

The two patterns look very similar, don’t they? But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. If you recognize the pattern of the signs, it may help you memorize the patterns.

This figure demonstrates the sign patterns in the sum and difference of two cubes. For the sum of two cubes, this figure shows the first two signs are plus and the first and the third signs are opposite, plus minus. The difference of two cubes has the first two signs the same, minus. The first and the third sign are minus plus.

The trinomial factor in the sum and difference of cubes pattern cannot be factored.

It can be very helpful if you learn to recognize the cubes of the integers from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers from 1 to 10 in Table 11.

n	1	2	3	4	5	6	7	8	9	10
$n^{3}$	1	8	27	64	125	216	343	512	729	1000

How to Factor the Sum or Difference of Cubes

Factor: $x^{3} + 64$ .

Solution

This table gives the steps for factoring x cubed + 64. The first step is to verify the binomial fits the pattern. Also, to check the sign for a sum or difference. This binomial is a sum that fits the pattern.

The second step is to write the terms as cubes, x cubed + 4 cubed.

The third step is follow the pattern for the sum of two cubes, (x + 4)(x squared minus x times 4 + 4 squared).

The fourth step is to simplify, (x + 4)(x squared minus 4 x +16).

The last step is to check the answer with multiplication.

Factor: $x^{3} - 1000$ .

Solution


This binomial is a difference. The first and last terms are perfect cubes.
Write the terms as cubes.
Use the difference of cubes pattern.
Simplify.
Check by multiplying.

Be careful to use the correct signs in the factors of the sum and difference of cubes.

Factor: $512 - 125 p^{3}$ .

Solution


This binomial is a difference. The first and last terms are perfect cubes.
Write the terms as cubes.
Use the difference of cubes pattern.
Simplify.
Check by multiplying.	We'll leave the check to you.

Factor: $27 u^{3} - 125 v^{3}$ .

Solution


This binomial is a difference. The first and last terms are perfect cubes.
Write the terms as cubes.
Use the difference of cubes pattern.
Simplify.
Check by multiplying.	We'll leave the check to you.

In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.

Factor: $5 m^{3} + 40 n^{3}$ .

Solution


Factor the common factor.
This binomial is a sum. The first and last terms are perfect cubes.
Write the terms as cubes.
Use the sum of cubes pattern.
Simplify.

Check. To check, you may find it easier to multiply the sum of cubes factors first, then multiply that product by 5. We’ll leave the multiplication for you.

$5 (\overset{}{m + 2 n}) (\overset{}{m^{2} - 2 m n + 4 n^{2}})$

Key Concepts

Factor perfect square trinomials See Example 1.
$\begin{matrix} Step 1. Does the trinomial fit the pattern? & a^{2} + 2 a b + b^{2} & a^{2} - 2 a b + b^{2} \\ Is the first term a perfect square? & {(a)}^{2} & {(a)}^{2} \\ Write it as a square. \\ Is the last term a perfect square? & {(a)}^{2} {(b)}^{2} & {(a)}^{2} {(b)}^{2} \\ Write it as a square. \\ Check the middle term. Is it 2 a b ? & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot b}{}_{↙} {(b)}^{2} & {(a)}^{2}_{↘} \underset{2 \cdot a \cdot b}{}_{↙} {(b)}^{2} \\ Step 2. Write the square of the binomial. & {(a + b)}^{2} & {(a - b)}^{2} \\ Step 3. Check by multiplying. \end{matrix}$
Factor differences of squares See Example 6.
$\begin{matrix} Step 1. Does the binomial fit the pattern? & a^{2} - b^{2} \\ Is this a difference? & ____-____ \\ Are the first and last terms perfect squares? \\ Step 2. Write them as squares. & {(a)}^{2} - {(b)}^{2} \\ Step 3. Write the product of conjugates. & (a - b) (a + b) \\ Step 4. Check by multiplying. \end{matrix}$
Factor sum and difference of cubes To factor the sum or difference of cubes: See Example 13.
1. Does the binomial fit the sum or difference of cubes pattern? Is it a sum or difference? Are the first and last terms perfect cubes?
2. Write them as cubes.
3. Use either the sum or difference of cubes pattern.
4. Simplify inside the parentheses
5. Check by multiplying the factors.

Practice Makes Perfect

Factor Perfect Square Trinomials

In the following exercises, factor.

$16 y^{2} + 24 y + 9$

Solution

${(4 y + 3)}^{2}$

$25 v^{2} + 20 v + 4$

$36 s^{2} + 84 s + 49$

Solution

${(6 s + 7)}^{2}$

$49 s^{2} + 154 s + 121$

$100 x^{2} - 20 x + 1$

Solution

${(10 x - 1)}^{2}$

$64 z^{2} - 16 z + 1$

$25 n^{2} - 120 n + 144$

Solution

${(5 n - 12)}^{2}$

$4 p^{2} - 52 p + 169$

$49 x^{2} - 28 x y + 4 y^{2}$

Solution

${(7 x - 2 y)}^{2}$

$25 r^{2} - 60 r s + 36 s^{2}$

$25 n^{2} + 25 n + 4$

Solution

$(5 n + 4) (5 n + 1)$

$100 y^{2} - 20 y + 1$

$64 m^{2} - 16 m + 1$

Solution

$(8 m - 1)^{2}$

$100 x^{2} - 25 x + 1$

$10 k^{2} + 80 k + 160$

Solution

$10 {(k + 4)}^{2}$

$64 x^{2} - 96 x + 36$

$75 u^{3} - 30 u^{2} v + 3 u v^{2}$

Solution

$3 u {(5 u - v)}^{2}$

$90 p^{3} + 300 p^{2} q + 250 p q^{2}$

Factor Differences of Squares

In the following exercises, factor.

$x^{2} - 16$

Solution

$(x - 4) (x + 4)$

$n^{2} - 9$

$25 v^{2} - 1$

Solution

$(5 v - 1) (5 v + 1)$

$169 q^{2} - 1$

$121 x^{2} - 144 y^{2}$

Solution

$(11 x - 12 y) (11 x + 12 y)$

$49 x^{2} - 81 y^{2}$

$169 c^{2} - 36 d^{2}$

Solution

$(13 c - 6 d) (13 c + 6 d)$

$36 p^{2} - 49 q^{2}$

$4 - 49 x^{2}$

Solution

$(2 - 7 x) (2 + 7 x)$

$121 - 25 s^{2}$

$16 z^{4} - 1$

Solution

$(2 z - 1) (2 z + 1) (4 z^{2} + 1)$

$m^{4} - n^{4}$

$5 q^{2} - 45$

Solution

$5 (q - 3) (q + 3)$

$98 r^{3} - 72 r$

$24 p^{2} + 54$

Solution

$6 (4 p^{2} + 9)$

$20 b^{2} + 140$

Factor Sums and Differences of Cubes

In the following exercises, factor.

$x^{3} + 125$

Solution

$(x + 5) (x^{2} - 5 x + 25)$

$n^{3} + 512$

$z^{3} - 27$

Solution

$(z - 3) (z^{2} + 3 z + 9)$

$v^{3} - 216$

$8 - 343 t^{3}$

Solution

$(2 - 7 t) (4 + 14 t + 49 t^{2})$

$125 - 27 w^{3}$

$8 y^{3} - 125 z^{3}$

Solution

$(2 y - 5 z) (4 y^{2} + 10 y z + 25 z^{2})$

$27 x^{3} - 64 y^{3}$

$7 k^{3} + 56$

Solution

$7 (k + 2) (k^{2} - 2 k + 4)$

$6 x^{3} - 48 y^{3}$

$2 - 16 y^{3}$

Solution

$2 (1 - 2 y) (1 + 2 y + 4 y^{2})$

$−2 x^{3} - 16 y^{3}$

Mixed Practice

In the following exercises, factor.

$64 a^{2} - 25$

Solution

$(8 a - 5) (8 a + 5)$

$121 x^{2} - 144$

$27 q^{2} - 3$

Solution

$3 (3 q - 1) (3 q + 1)$

$4 p^{2} - 100$

$16 x^{2} - 72 x + 81$

Solution

${(4 x - 9)}^{2}$

$36 y^{2} + 12 y + 1$

$8 p^{2} + 2$

Solution

$2 (4 p^{2} + 1)$

$81 x^{2} + 169$

$125 - 8 y^{3}$

Solution

$(5 - 2 y) (25 + 10 y + 4 y^{2})$

$27 u^{3} + 1000$

$45 n^{2} + 60 n + 20$

Solution

$5 {(3 n + 2)}^{2}$

$48 q^{3} - 24 q^{2} + 3 q$

Everyday Math

Landscaping Sue and Alan are planning to put a 15 foot square swimming pool in their backyard. They will surround the pool with a tiled deck, the same width on all sides. If the width of the deck is w, the total area of the pool and deck is given by the trinomial $4 w^{2} + 60 w + 225$ . Factor the trinomial.

Solution

${(2 w + 15)}^{2}$

Home repair The height a twelve foot ladder can reach up the side of a building if the ladder’s base is b feet from the building is the square root of the binomial $144 - b^{2}$ . Factor the binomial.

Writing Exercises

Why was it important to practice using the binomial squares pattern in the chapter on multiplying polynomials?

Solution

Answers may vary.

How do you recognize the binomial squares pattern?

Explain why $n^{2} + 25 \neq {(n + 5)}^{2}$ . Use algebra, words, or pictures.

Solution

Answers may vary.

Maribel factored $y^{2} - 30 y + 81$ as ${(y - 9)}^{2}$ . Was she right or wrong? How do you know?

Self Check

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

This table has the following statements all to be preceded by “I can…”. The first row is “factor perfect square trinomials”. The second row is “factor differences of squares”. The third row is “factor sums and differences of cubes”. In the columns beside these statements are the headers, “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?

perfect square trinomials pattern

If a and b are real numbers,

\begin{matrix} a^{2} + 2 a b + b^{2} = {(a + b)}^{2} & a^{2} - 2 a b + b^{2} = {(a - b)}^{2} \end{matrix}

difference of squares pattern

If a and b are real numbers,
This image shows the difference of two squares formula, a squared – b squared = (a – b)(a + b). Also, the squares are labeled, a squared and b squared. The difference is shown between the two terms. Finally, the factoring (a – b)(a + b) are labeled as conjugates.

This image shows the difference of two squares formula, a squared – b squared = (a – b)(a + b). Also, the squares are labeled, a squared and b squared. The difference is shown between the two terms. Finally, the factoring (a – b)(a + b) are labeled as conjugates.

sum and difference of cubes pattern

\begin{matrix} a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) \\ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \end{matrix}

General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

You have now become acquainted with all the methods of factoring that you will need in this course. (In your next algebra course, more methods will be added to your repertoire.) The figure below summarizes all the factoring methods we have covered. Note 5 outlines a strategy you should use when factoring polynomials.

This figure presents a general strategy for factoring polynomials. First, at the top, there is GCF, which is where factoring starts. Below this, there are three options, binomial, trinomial, and more than three terms. For binomial, there are the difference of two squares, the sum of squares, the sum of cubes, and the difference of cubes. For trinomials, there are two forms, x squared plus bx plus c and ax squared 2 plus b x plus c. There are also the sum and difference of two squares formulas as well as the “a c” method. Finally, for more than three terms, the method is grouping.

Remember, a polynomial is completely factored if, other than monomials, its factors are prime!

Factor completely: $4 x^{5} + 12 x^{4}$ .

Solution

$\begin{matrix} Is there a GCF? & Yes, 4 x^{4} . & 4 x^{5} + 12 x^{4} \\ Factor out the GCF. & 4 x^{4} (x + 3) \\ In the parentheses, is it a binomial, a \\ trinomial, or are there more than three terms? & Binomial. \\ Is it a sum? & Yes. \\ Of squares? Of cubes? & No. \\ Check. \\ Is the expression factored completely? & Yes. \\ Multiply. \\ 4 x^{4} (x + 3) \\ 4 x^{4} \cdot x + 4 x^{4} \cdot 3 \\ 4 x^{5} + 12 x^{4} ✓ \end{matrix}$

Factor completely: $12 x^{2} - 11 x + 2$ .

Solution


Is there a GCF?	No.
Is it a binomial, trinomial, or are there more than three terms?	Trinomial.
Are a and c perfect squares?	No, a = 12, not a perfect square.
Use trial and error or the “ac” method. We will use trial and error here.

This table has the heading of 12 x squared minus 11 x plus 2 and gives the possible factors. The first column is labeled possible factors and the second column is labeled product. Four rows have not an option in the product column. This is explained by the text, “if the trinomial has no common factors, then neither factor can contain a common factor”. The last factors, 3 x - 2 in parentheses and 4 x - 1 in parentheses, give the product of 12 x squared minus 11 x plus 2.

Check.

$(3 x - 2) (4 x - 1)$

$12 x^{2} - 3 x - 8 x + 2$

$12 x^{2} - 11 x + 2 ✓$

Factor completely: $g^{3} + 25 g$ .

Solution

A step-by-step guide to factoring an algebraic expression, illustrating the process with questions, outcomes, and mathematical forms.
Is there a GCF?	Yes, g.	$g^{3} + 25 g$
Factor out the GCF.		$g (g^{2} + 25)$
In the parentheses, is it a binomial, trinomial, or are there more than three terms?	Binomial.
Is it a sum ? Of squares?	Yes.	Sums of squares are prime.
Check.
Is the expression factored completely?	Yes.
Multiply. $\begin{matrix} g (g^{2} + 25) \\ g^{3} + 25 g ✓ \end{matrix}$

Factor completely: $12 y^{2} - 75$ .

Solution

Detailed steps for factoring the algebraic expression 12y^2 - 75, showing identification of GCF and difference of squares.
Is there a GCF?	Yes, 3.	$12 y^{2} - 75$
Factor out the GCF.		$3 (4 y^{2} - 25)$
In the parentheses, is it a binomial, trinomial, or are there more than three terms?	Binomial.
Is it a sum?	No.
Is it a difference? Of squares or cubes?	Yes, squares.	$3 ({(2 y)}^{2} - {(5)}^{2})$
Write as a product of conjugates.		$3 (2 y - 5) (2 y + 5)$
Check.
Is the expression factored completely?	Yes.
Neither binomial is a difference of squares.
Multiply. $\begin{matrix} 3 (2 y - 5) (2 y + 5) \\ 3 (4 y^{2} - 25) \\ 12 y^{2} - 75 ✓ \end{matrix}$

Factor completely: $4 a^{2} - 12 a b + 9 b^{2}$ .

Solution

Is there a GCF?	No.
Is it a binomial, trinomial, or are there more terms?
Trinomial with $a \neq 1$ . But the first term is a perfect square.
Is the last term a perfect square?	Yes.
Does it fit the pattern, $a^{2} - 2 a b + b^{2} ?$	Yes.
Write it as a square.
Check your answer.
Is the expression factored completely?
Yes.
The binomial is not a difference of squares.
Multiply.
${(2 a - 3 b)}^{2}$
${(2 a)}^{2} - 2 \cdot 2 a \cdot 3 b + {(3 b)}^{2}$
$4 a^{2} - 12 a b + 9 b^{2} ✓$

Factor completely: $6 y^{2} - 18 y - 60$ .

Solution

Step-by-step guide demonstrating how to factor a trinomial polynomial, from identifying the GCF to checking the final factored form.
Is there a GCF?	Yes, 6.	$6 y^{2} - 18 y - 60$
Factor out the GCF.		$6 (y^{2} - 3 y - 10)$
In the parentheses, is it a binomial, trinomial, or are there more terms?	Trinomial with leading coefficient 1.
“Undo” FOIL.	$6 (y) (y)$	$6 (y + 2) (y - 5)$
Check your answer.
Is the expression factored completely?		Yes.
Neither binomial is a difference of squares.
Multiply. $\begin{matrix} 6 (y + 2) (y - 5) \\ 6 (y^{2} - 5 y + 2 y - 10) \\ 6 (y^{2} - 3 y - 10) \\ 6 y^{2} - 18 y - 60 ✓ \end{matrix}$

Factor completely: $24 x^{3} + 81$ .

Solution

Is there a GCF?	Yes, 3.	$24 x^{3} + 81$
Factor it out.		$3 (8 x^{3} + 27)$
In the parentheses, is it a binomial, trinomial, or are there more than three terms?	Binomial.
Is it a sum or difference?	Sum.
Of squares or cubes?	Sum of cubes.
Write it using the sum of cubes pattern.
Is the expression factored completely?	Yes.	$3 (2 x + 3) (4 x^{2} - 6 x + 9)$
Check by multiplying.		We leave the check to you.

Factor completely: $2 x^{4} - 32$ .

Solution

A step-by-step guide demonstrating the process of factoring the polynomial 2x^4 - 32, including questions and corresponding expressions.
Is there a GCF?	Yes, 2.	$2 x^{4} - 32$
Factor it out.		$2 (x^{4} - 16)$
In the parentheses, is it a binomial, trinomial, or are there more than three terms?	Binomial.
Is it a sum or difference?	Yes.
Of squares or cubes?	Difference of squares.	$2 ({(x^{2})}^{2} - {(4)}^{2})$
Write it as a product of conjugates.		$2 (x^{2} - 4) (x^{2} + 4)$
The first binomial is again a difference of squares.		$2 ({(x)}^{2} - {(2)}^{2}) (x^{2} + 4)$
Write it as a product of conjugates.		$2 (x - 2) (x + 2) (x^{2} + 4)$
Is the expression factored completely?	Yes.
None of these binomials is a difference of squares.
Check your answer.
Multiply. $\begin{matrix} 2 (x - 2) (x + 2) (x^{2} + 4) \\ 2 (x^{2} - 4) (x^{2} + 4) \\ 2 (x^{4} - 16) \\ 2 x^{4} - 32 ✓ \end{matrix}$

Factor completely: $3 x^{2} + 6 b x - 3 a x - 6 a b$ .

Solution

This table demonstrates the step-by-step process of factoring a polynomial expression by grouping, including GCF identification and checking the solution.
Is there a GCF?	Yes, 3.	$3 x^{2} + 6 b x - 3 a x - 6 a b$
Factor out the GCF.		$3 (x^{2} + 2 b x - a x - 2 a b)$
In the parentheses, is it a binomial, trinomial, or are there more terms?	More than 3 terms.
Use grouping.		$\begin{matrix} 3 [x (x + 2 b) - a (x + 2 b)] \\ 3 (x + 2 b) (x - a) \end{matrix}$
Check your answer.
Is the expression factored completely? Yes.
Multiply. $\begin{matrix} 3 (x + 2 b) (x - a) \\ 3 (x^{2} - a x + 2 b x - 2 a b) \\ 3 x^{2} - 3 a x + 6 b x - 6 a b ✓ \end{matrix}$

Factor completely: $10 x^{2} - 34 x - 24$ .

Solution

Step-by-step method for factoring a quadratic trinomial, illustrating GCF, classification, trial/error, and verification.
Is there a GCF?	Yes, 2.	$10 x^{2} - 34 x - 24$
Factor out the GCF.		$2 (5 x^{2} - 17 x - 12)$
In the parentheses, is it a binomial, trinomial, or are there more than three terms?	Trinomial with $a \neq 1 .$
Use trial and error or the “ac” method.		$\begin{matrix} 2 (5 x^{2} - 17 x −12) \\ 2 (5 x + 3) (x - 4) \end{matrix}$
Check your answer. Is the expression factored completely? Yes.
Multiply. $\begin{matrix} 2 (5 x + 3) (x - 4) \\ 2 (5 x^{2} - 20 x + 3 x - 12) \\ 2 (5 x^{2} - 17 x - 12) \\ 10 x^{2} - 34 x - 24 ✓ \end{matrix}$

Key Concepts

General Strategy for Factoring Polynomials See Figure 1.
How to Factor Polynomials
1. Is there a greatest common factor? Factor it out.
2. Is the polynomial a binomial, trinomial, or are there more than three terms?
  - If it is a binomial:
    Is it a sum?
    - Of squares? Sums of squares do not factor.
    - Of cubes? Use the sum of cubes pattern.
    Is it a difference?
    - Of squares? Factor as the product of conjugates.
    - Of cubes? Use the difference of cubes pattern.
  - If it is a trinomial:
    Is it of the form x2+bx+c? Undo FOIL.
    Is it of the form ax2+bx+c?
    - If ‘a’ and ‘c’ are squares, check if it fits the trinomial square pattern.
    - Use the trial and error or ‘ac’ method.
  - If it has more than three terms:
    Use the grouping method.
3. Check. Is it factored completely? Do the factors multiply back to the original polynomial?

Practice Makes Perfect

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

$10 x^{4} + 35 x^{3}$

Solution

$5 x^{3} (2 x + 7)$

$18 p^{6} + 24 p^{3}$

$y^{2} + 10 y - 39$

Solution

$(y - 3) (y + 13)$

$b^{2} - 17 b + 60$

$2 n^{2} + 13 n - 7$

Solution

$(2 n - 1) (n + 7)$

$8 x^{2} - 9 x - 3$

$a^{5} + 9 a^{3}$

Solution

$a^{3} (a^{2} + 9)$

$75 m^{3} + 12 m$

$121 r^{2} - s^{2}$

Solution

$(11 r - s) (11 r + s)$

$49 b^{2} - 36 a^{2}$

$8 m^{2} - 32$

Solution

$8 (m - 2) (m + 2)$

$36 q^{2} - 100$

$25 w^{2} - 60 w + 36$

Solution

${(5 w - 6)}^{2}$

$49 b^{2} - 112 b + 64$

$m^{2} + 14 m n + 49 n^{2}$

Solution

${(m + 7 n)}^{2}$

$64 x^{2} + 16 x y + y^{2}$

$7 b^{2} + 7 b - 42$

Solution

$7 (b + 3) (b - 2)$

$3 n^{2} + 30 n + 72$

$3 x^{3} - 81$

Solution

$3 (x - 3) (x^{2} + 3 x + 9)$

$5 t^{3} - 40$

$k^{4} - 16$

Solution

$(k - 2) (k + 2) (k^{2} + 4)$

$m^{4} - 81$

$15 p q - 15 p + 12 q - 12$

Solution

$3 (5 p + 4) (q - 1)$

$12 a b - 6 a + 10 b - 5$

$4 x^{2} + 40 x + 84$

Solution

$4 (x + 3) (x + 7)$

$5 q^{2} - 15 q - 90$

$u^{5} + u^{2}$

Solution

$u^{2} (u + 1) (u^{2} - u + 1)$

$5 n^{3} + 320$

$4 c^{2} + 20 c d + 81 d^{2}$

Solution

prime

$25 x^{2} + 35 x y + 49 y^{2}$

$10 m^{4} - 6250$

Solution

$10 (m - 5) (m + 5) (m^{2} + 25)$

$3 v^{4} - 768$

Everyday Math

Watermelon drop A springtime tradition at the University of California San Diego is the Watermelon Drop, where a watermelon is dropped from the seventh story of Urey Hall.

ⓐ The binomial $−16 t^{2} + 80$ gives the height of the watermelon $t$ seconds after it is dropped. Factor the greatest common factor from this binomial.
ⓑ If the watermelon is thrown down with initial velocity 8 feet per second, its height after $t$ seconds is given by the trinomial $−16 t^{2} - 8 t + 80$ . Completely factor this trinomial.

Solution

ⓐ $−16 (t^{2} - 5)$ ⓑ $−8 (2 t + 5) (t - 2)$

Pumpkin drop A fall tradition at the University of California San Diego is the Pumpkin Drop, where a pumpkin is dropped from the eleventh story of Tioga Hall.

ⓐ The binomial $−16 t^{2} + 128$ gives the height of the pumpkin t seconds after it is dropped. Factor the greatest common factor from this binomial.
ⓑ If the pumpkin is thrown down with initial velocity 32 feet per second, its height after $t$ seconds is given by the trinomial $−16 t^{2} - 32 t + 128$ . Completely factor this trinomial.

Writing Exercises

The difference of squares $y^{4} - 625$ can be factored as $(y^{2} - 25) (y^{2} + 25)$ . But it is not completely factored. What more must be done to completely factor it?

Solution

Answer may vary.

Of all the factoring methods covered in this chapter (GCF, grouping, undo FOIL, ‘ac’ method, special products) which is the easiest for you? Which is the hardest? Explain your answers.

Self Check

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

This table has the following statements all to be preceded by “I can…”. The row states “recognize and use the appropriate method to factor a polynomial completely”. In the columns beside these statements are the headers, “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 section? Why or why not?

Quadratic Equations

We have already solved linear equations, equations of the form $a x + b y = c$ . In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. Listed below are some examples of quadratic equations:

\begin{matrix} x^{2} + 5 x + 6 = 0 & 3 y^{2} + 4 y = 10 & 64 u^{2} - 81 = 0 & n (n + 1) = 42 \end{matrix}

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get $n^{2} + n$ .

The general form of a quadratic equation is $a x^{2} + b x + c = 0, with a \neq 0$ .

To solve quadratic equations we need methods different than the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

Solve Quadratic Equations Using the Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

We will now use the Zero Product Property, to solve a quadratic equation.

How to Use the Zero Product Property to Solve a Quadratic Equation

Solve: $(x + 1) (x - 4) = 0$ .

Solution

This table gives the steps for solving (x + 1)(x – 4) = 0. The first step is to set each factor equal to 0. Since it is a product equal to 0, at least one factor must equal 0. x + 1 = 0 or x – 4 = 0.

The last step is to check both answers by substituting the values for x into the original equation.

We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property.

Solve: $(5 n - 2) (6 n - 1) = 0$ .

Solution

	$(5 n - 2) (6 n - 1) = 0$
Use the Zero Product Property to set each factor to 0.	$5 n - 2 = 0$	$6 n - 1 = 0$
Solve the equations.	$n = \frac{2}{5}$	$n = \frac{1}{6}$
Check your answers.

Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.

Solve: $3 p (10 p + 7) = 0$ .

Solution

	$3 p (10 p + 7) = 0$
Use the Zero Product Property to set each factor to 0.	$3 p = 0$	$10 p + 7 = 0$
Solve the equations.	$p = 0$	$10 p = −7$
		$p = - \frac{7}{10}$
Check your answers.

It may appear that there is only one factor in the next example. Remember, however, that ${(y - 8)}^{2}$ means $(y - 8) (y - 8)$ .

Solve: ${(y - 8)}^{2} = 0$ .

Solution

	${(y - 8)}^{2} = 0$
Rewrite the left side as a product.	$(y - 8) (y - 8) = 0$
Use the Zero Product Property and set each factor to 0.	$y - 8 = 0$	$y - 8 = 0$
Solve the equations.	$y = 8$	$y = 8$
When a solution repeats, we call it a double root.
Check your answer.

Solve Quadratic Equations by Factoring

Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we must be sure to start with the quadratic equation in standard form, $a x^{2} + b x + c = 0$ . Then we can factor the expression on the left.

How to Solve a Quadratic Equation by Factoring

Solve: $x^{2} + 2 x - 8 = 0$ .

Solution

This table gives the steps for solving the equation x squared + 2 x – 8 = 0. The first step is writing the equation in standard quadratic form, which it is.

The second step is factoring the quadratic expression x squared + 2 x – 8. The factors are (x + 4), (x – 2).

The next step is using the zero product property and set each factor equal to 0, x + 4 = 0 and x – 2 = 0.

The next step is solving both linear equations, x = −4 or x = 2.

The last step is checking both solutions by substituting them into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form.

Solve: $2 y^{2} = 13 y + 45$ .

Solution

	$2 y^{2} = 13 y + 45$
Write the quadratic equation in standard form.	$2 y^{2} - 13 y - 45 = 0$
Factor the quadratic expression.	$(2 y + 5) (y - 9) = 0$
Use the Zero Product Property to set each factor to 0.	$2 y + 5 = 0$	$y - 9 = 0$
Solve each equation.	$y = - \frac{5}{2}$	$y = 9$
Check your answers.

Solve: $5 x^{2} - 13 x = 7 x$ .

Solution

	$5 x^{2} - 13 x = 7 x$
Write the quadratic equation in standard form.	$5 x^{2} - 20 x = 0$
Factor the left side of the equation.	$5 x (x - 4) = 0$
Use the Zero Product Property to set each factor to 0.	$5 x = 0$	$x - 4 = 0$
Solve each equation.	$x = 0$	$x = 4$
Check your answers.

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Solve: $144 q^{2} = 25$ .

Solution


Write the quadratic equation in standard form.
Factor. It is a difference of squares.
Use the Zero Product Property to set each factor to 0.	$\begin{matrix} 12 q - 5 = 0 \end{matrix}$	$\begin{matrix} 12 q + 5 = 0 \end{matrix}$
Solve each equation.	$\begin{matrix} 12 q & = & 5 \\ q & = & \frac{5}{12} \end{matrix}$	$\begin{matrix} 12 q & = & –5 \\ q & = & - \frac{5}{12} \end{matrix}$
Check your answers.

The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Solve: $(3 x - 8) (x - 1) = 3 x$ .

Solution

Step-by-step solution of a quadratic equation by factoring, demonstrating transformations to find the roots.
	$\begin{matrix} (3 x - 8) (x - 1) & = & 3 x \end{matrix}$
Multiply the binomials.	$\begin{matrix} 3 x^{2} - 11 x + 8 & = & 3 x \end{matrix}$
Write the quadratic equation in standard form.	$\begin{matrix} 3 x^{2} - 14 x + 8 & = & 0 \end{matrix}$
Factor the trinomial.	$\begin{matrix} (3 x - 2) (x - 4) & = & 0 \end{matrix}$
Use the Zero Product Property to set each factor to 0. Solve each equation.	$\begin{matrix} 3 x - 2 & = & 0 & x - 4 & = & 0 \\ 3 x & = & 2 & x & = & 4 \\ x & = & \frac{2}{3} \end{matrix}$
Check your answers.	The check is left to you!

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations.

Solve: $9 m^{3} + 100 m = 60 m^{2}$ .

Solution

Steps for solving algebraic equations by factoring, showing each action and its mathematical representation.

Bring all the terms to one side so that the other side is zero.
Factor the greatest common factor first.
Factor the trinomial.
Use the Zero Product Property to set each factor to 0.
Solve each equation.	$Three mathematical expressions are shown: m = 0, m = 10/3, and m = 10/3. The letter 'm' is equal to zero in the first expression and to the fraction 10 over 3 in the other two.$
Check your answers.	The check is left to you.

When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Solve: $4 x^{2} = 16 x + 84$ .

Solution

Steps demonstrating how to solve a quadratic equation by factoring, from the initial equation to the final solutions.
	$\begin{matrix} 4 x^{2} & = & 16 x + 84 \end{matrix}$
Write the quadratic equation in standard form.	$\begin{matrix} 4 x^{2} - 16 x - 84 & = & 0 \end{matrix}$
Factor the greatest common factor first.	$\begin{matrix} 4 (x^{2} - 4 x - 21) & = & 0 \end{matrix}$
Factor the trinomial.	$\begin{matrix} 4 (x - 7) (x + 3) & = & 0 \end{matrix}$
Use the Zero Product Property to set each factor to 0. Solve each equation.	$\begin{matrix} 4 & \neq & 0 & x - 7 & = & 0 & x + 3 & = & 0 \\ 4 & \neq & 0 & x & = & 7 & x & = & −3 \end{matrix}$
Check your answers.	The check is left to you.

Solve Applications Modeled by Quadratic Equations

The problem solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We will copy the problem solving strategy here so we can use it for reference.

We will start with a number problem to get practice translating words into a quadratic equation.

The product of two consecutive integers is 132. Find the integers.

Solution

This table outlines a 7-step process for solving a word problem to find consecutive integers whose product is 132, from reading the problem to checking the final answer.
Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for two consecutive integers.
Step 3. Name what we are looking for.	$\begin{matrix} Let n = the first integer \\ n + 1 = the next consecutive integer \end{matrix}$
Step 4. Translate into an equation. Restate the problem in a sentence.	The product of the two consecutive integers is 132.
	The first integer times the next integer is 132.
Translate to an equation.	$\begin{matrix} n (n + 1) & = & 132 \end{matrix}$
Step 5. Solve the equation.	$\begin{matrix} n^{2} + n & = & 132 \end{matrix}$
Bring all the terms to one side.	$\begin{matrix} n^{2} + n - 132 & = & 0 \end{matrix}$
Factor the trinomial.	$\begin{matrix} (n - 11) (n + 12) & = & 0 \end{matrix}$
Use the zero product property. Solve the equations.	$\begin{matrix} n - 11 & = & 0 & n + 12 & = & 0 \\ n & = & 11 & n & = & −12 \end{matrix}$
There are two values for $n$ that are solutions to this problem. So there are two sets of consecutive integers that will work.
$\begin{matrix} If the first integer is n = 11 & If the first integer is n = −12 \\ then the next integer is n + 1 & then the next integer is n + 1 \\ 11 + 1 & −12 + 1 \\ 12 & −11 \end{matrix}$
Step 6. Check the answer.
The consecutive integers are $11, 12$ and $−11, −12$ . The product $11 \cdot 12 = 132$ and the product $−11 (−12) = 132$ . Both pairs of consecutive integers are solutions.
Step 7. Answer the question. The consecutive integers are $11, 12$ and $−11, −12$ .

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give 132.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

A rectangular garden has an area $15$ square feet. The length of the garden is two feet more than the width. Find the length and width of the garden.

Solution

Step 1. Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation.
Step 2. Identify what you are looking for.	We are looking for the length and width.
Step 3. Name what you are looking for. The length is two feet more than width.	Let W = the width of the garden. W + 2 = the length of the garden
Step 4. Translate into an equation. Restate the important information in a sentence.	The area of the rectangular garden is 15 square feet.
Use the formula for the area of a rectangle.	$A = L \cdot W$
Substitute in the variables.	$15 = (W + 2) W$
Step 5. Solve the equation. Distribute first.	$15 = W^{2} + 2 W$
Get zero on one side.	$0 = W^{2} + 2 W - 15$
Factor the trinomial.	$0 = (W + 5) (W - 3)$
Use the Zero Product Property.	$0 = W + 5$	$0 = W - 3$
Solve each equation.	$−5 = W$	$3 = W$
Since W is the width of the garden, it does not make sense for it to be negative. We eliminate that value for W.	$−5 = W$ $W = 3$	$3 = W$ Width is 3 feet.
Find the value of the length.	$W + 2 = length$
	$3 + 2$
	$5$	Length is 5 feet.
Step 6. Check the answer. Does the answer make sense?

	Yes, this makes sense.
Step 7. Answer the question.	The width of the garden is 3 feet and the length is 5 feet.

In an earlier chapter, we used the Pythagorean Theorem $(a^{2} + b^{2} = c^{2})$ . It gave the relation between the legs and the hypotenuse of a right triangle.

We will use this formula in the next example.

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle, as shown below. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the deck.

This figure is a right triangle. The vertical leg is labeled “x – 7”. the horizontal leg, the base, is labeled “x”. The hypotenuse is labeled “17”.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the lengths of the sides of the deck.
Step 3. Name what you are looking for. One side is 7 less than the other.	Let x = length of a side of the deck x − 7 = length of other side
Step 4. Translate into an equation. Since this is a right triangle we can use the Pythagorean Theorem.	$a^{2} + b^{2} = c^{2}$
Substitute in the variables.	$x^{2} + {(x - 7)}^{2} = 17^{2}$
Step 5. Solve the equation.	$x^{2} + x^{2} - 14 x + 49 = 289$
Simplify.	$2 x^{2} - 14 x + 49 = 289$
It is a quadratic equation, so get zero on one side.	$2 x^{2} - 14 x - 240 = 0$
Factor the greatest common factor.	$2 (x^{2} - 7 x - 120) = 0$
Factor the trinomial.	$2 (x - 15) (x + 8) = 0$
Use the Zero Product Property.	$2 \neq 0$	$x - 15 = 0$	$x + 8 = 0$
Solve.	$2 \neq 0$	$x = 15$	$x = −8$
Since x is a side of the triangle, $x = −8$ does not make sense.	$2 \neq 0$	$x = 15$	$x = −8$
Find the length of the other side.
If the length of one side is
then the length of the other side is

	8 is the length of the other side.
Step 6. Check the answer. Do these numbers make sense?

Step 7. Answer the question.	The sides of the deck are 8, 15, and 17 feet.

Key Concepts

Zero Product Property If $a \cdot b = 0$ , then either $a = 0$ or $b = 0$ or both. See Example 1.
Solve a quadratic equation by factoring To solve a quadratic equation by factoring: See Example 5.
1. Write the quadratic equation in standard form, $a x^{2} + b x + c = 0$ .
2. Factor the quadratic expression.
3. Use the Zero Product Property.
4. Solve the linear equations.
5. Check.
Use a problem solving strategy to solve word problems See Example 12.
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Section Exercises

Practice Makes Perfect

Use the Zero Product Property

In the following exercises, solve.

$(x - 3) (x + 7) = 0$

Solution

$x = 3, x = −7$

$(y - 11) (y + 1) = 0$

$(3 a - 10) (2 a - 7) = 0$

Solution

$a = 10 / 3, a = 7 / 2$

$(5 b + 1) (6 b + 1) = 0$

$6 m (12 m - 5) = 0$

Solution

$m = 0, m = 5 / 12$

$2 x (6 x - 3) = 0$

${(y - 3)}^{2} = 0$

Solution

$y = 3$

${(b + 10)}^{2} = 0$

${(2 x - 1)}^{2} = 0$

Solution

$x = 1 / 2$

${(3 y + 5)}^{2} = 0$

Solve Quadratic Equations by Factoring

In the following exercises, solve.

$x^{2} + 7 x + 12 = 0$

Solution

$x = −3, x = −4$

$y^{2} - 8 y + 15 = 0$

$5 a^{2} - 26 a = 24$

Solution

$a = −4 / 5, a = 6$

$4 b^{2} + 7 b = −3$

$4 m^{2} = 17 m - 15$

Solution

$m = 5 / 4, m = 3$

$n^{2} = 5 n - 6$

$7 a^{2} + 14 a = 7 a$

Solution

$a = −1, a = 0$

$12 b^{2} - 15 b = −9 b$

$49 m^{2} = 144$

Solution

$m = 12 / 7, m = −12 / 7$

$625 = x^{2}$

$(y - 3) (y + 2) = 4 y$

Solution

$y = −1, y = 6$

$(p - 5) (p + 3) = −7$

$(2 x + 1) (x - 3) = −4 x$

Solution

$x = 3 / 2, x = −1$

$(x + 6) (x - 3) = −8$

$16 p^{3} = 24 p^{2} - 9 p$

Solution

$p = 0, p = ¾$

$m^{3} - 2 m^{2} = − m$

$20 x^{2} - 60 x = −45$

Solution

$x = 3 / 2$

$3 y^{2} - 18 y = −27$

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

The product of two consecutive integers is 56. Find the integers.

Solution

$7 and 8; - 8 and −7$

The product of two consecutive integers is 42. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

Solution

$4 feet and 7 feet$

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

Solution

$6 feet and 8 feet$

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

Mixed Practice

In the following exercises, solve.

$(x + 8) (x - 3) = 0$

Solution

$x = −8, x = 3$

$(3 y - 5) (y + 7) = 0$

$p^{2} + 12 p + 11 = 0$

Solution

$p = −1, p = −11$

$q^{2} - 12 q - 13 = 0$

$m^{2} = 6 m + 16$

Solution

$m = −2, m = 8$

$4 n^{2} + 19 n = 5$

$a^{3} - a^{2} - 42 a = 0$

Solution

$a = 0, a = −6, a = 7$

$4 b^{2} - 60 b + 224 = 0$

The product of two consecutive integers is 110. Find the integers.

Solution

$10 and 11; - 11 and −10$

The length of one leg of a right triangle is three feet more than the other leg. If the hypotenuse is 15 feet, find the lengths of the two legs.

Everyday Math

Area of a patio If each side of a square patio is increased by 4 feet, the area of the patio would be 196 square feet. Solve the equation ${(s + 4)}^{2} = 196$ for s to find the length of a side of the patio.

Solution

10 feet

Watermelon drop A watermelon is dropped from the tenth story of a building. Solve the equation $−16 t^{2} + 144 = 0$ for $t$ to find the number of seconds it takes the watermelon to reach the ground.

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Solution

Answers may vary.

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

Self Check

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

This table has the following statements all to be preceded by “I can…”. The first row is “solve quadratic equations by using the zero product property”. The second row is “solve quadratic equations by factoring”. The third row is “solve applications modeled by quadratic equations”. In the columns beside these statements are the headers, “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 section? Why or why not?

Chapter 7 Review Exercises

7.1 Greatest Common Factor and Factor by Grouping

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

$42, 60$

Solution

$450, 420$

$90, 150, 105$

Solution

$15$

$60, 294, 630$

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

$24 x - 42$

Solution

$6 (4 x - 7)$

$35 y + 84$

$15 m^{4} + 6 m^{2} n$

Solution

$3 m^{2} (5 m^{2} + 2 n)$

$24 p t^{4} + 16 t^{7}$

Factor by Grouping

In the following exercises, factor by grouping.

$a x - a y + b x - b y$

Solution

$(a + b) (x - y)$

$x^{2} y - x y^{2} + 2 x - 2 y$

$x^{2} + 7 x - 3 x - 21$

Solution

$(x - 3) (x + 7)$

$4 x^{2} - 16 x + 3 x - 12$

$m^{3} + m^{2} + m + 1$

Solution

$(m^{2} + 1) (m + 1)$

$5 x - 5 y - y + x$

7.2 Factor Trinomials of the form $x^{2} + b x + c$

Factor Trinomials of the Form $x^{2} + b x + c$

In the following exercises, factor each trinomial of the form $x^{2} + b x + c$ .

$u^{2} + 17 u + 72$

Solution

$(u + 8) (u + 9)$

$a^{2} + 14 a + 33$

$k^{2} - 16 k + 60$

Solution

$(k - 6) (k - 10)$

$r^{2} - 11 r + 28$

$y^{2} + 6 y - 7$

Solution

$(y + 7) (y - 1)$

$m^{2} + 3 m - 54$

$s^{2} - 2 s - 8$

Solution

$(s - 4) (s + 2)$

$x^{2} - 3 x - 10$

Factor Trinomials of the Form $x^{2} + b x y + c y^{2}$

In the following examples, factor each trinomial of the form $x^{2} + b x y + c y^{2}$ .

$x^{2} + 12 x y + 35 y^{2}$

Solution

$(x + 5 y) (x + 7 y)$

$u^{2} + 14 u v + 48 v^{2}$

$a^{2} + 4 a b - 21 b^{2}$

Solution

$(a + 7 b) (a - 3 b)$

$p^{2} - 5 p q - 36 q^{2}$

7.3 Factoring Trinomials of the form $a x^{2} + b x + c$

Recognize a Preliminary Strategy to Factor Polynomials Completely

In the following exercises, identify the best method to use to factor each polynomial.

$y^{2} - 17 y + 42$

Solution

Undo FOIL

$12 r^{2} + 32 r + 5$

$8 a^{3} + 72 a$

Solution

Factor the GCF

$4 m - m n - 3 n + 12$

Factor Trinomials of the Form $a x^{2} + b x + c$ with a GCF

In the following exercises, factor completely.

$6 x^{2} + 42 x + 60$

Solution

$6 (x + 2) (x + 5)$

$8 a^{2} + 32 a + 24$

$3 n^{4} - 12 n^{3} - 96 n^{2}$

Solution

$3 n^{2} (n - 8) (n + 4)$

$5 y^{3} + 25 y^{2} - 70 y$

Factor Trinomials Using the “ac” Method

In the following exercises, factor.

$2 x^{2} + 9 x + 4$

Solution

$(x + 4) (2 x + 1)$

$3 y^{2} + 17 y + 10$

$18 a^{2} - 9 a + 1$

Solution

$(3 a - 1) (6 a - 1)$

$8 u^{2} - 14 u + 3$

$15 p^{2} + 2 p - 8$

Solution

$(5 p + 4) (3 p - 2)$

$15 x^{2} + x - 2$

$40 s^{2} - s - 6$

Solution

$(5 s - 2) (8 s + 3)$

$20 n^{2} - 7 n - 3$

Factor Trinomials with a GCF Using the “ac” Method

In the following exercises, factor.

$3 x^{2} + 3 x - 36$

Solution

$3 (x + 4) (x - 3)$

$4 x^{2} + 4 x - 8$

$60 y^{2} - 85 y - 25$

Solution

$5 (4 y + 1) (3 y - 5)$

$18 a^{2} - 57 a - 21$

7.4 Factoring Special Products

Factor Perfect Square Trinomials

In the following exercises, factor.

$25 x^{2} + 30 x + 9$

Solution

${(5 x + 3)}^{2}$

$16 y^{2} + 72 y + 81$

$36 a^{2} - 84 a b + 49 b^{2}$

Solution

${(6 a - 7 b)}^{2}$

$64 r^{2} - 176 r s + 121 s^{2}$

$40 x^{2} + 360 x + 810$

Solution

$10 {(2 x + 9)}^{2}$

$75 u^{2} + 180 u + 108$

$2 y^{3} - 16 y^{2} + 32 y$

Solution

$2 y {(y - 4)}^{2}$

$5 k^{3} - 70 k^{2} + 245 k$

Factor Differences of Squares

In the following exercises, factor.

$81 r^{2} - 25$

Solution

$(9 r - 5) (9 r + 5)$

$49 a^{2} - 144$

$169 m^{2} - n^{2}$

Solution

$(13 m + n) (13 m - n)$

$64 x^{2} - y^{2}$

$25 p^{2} - 1$

Solution

$(5 p - 1) (5 p + 1)$

$1 - 16 s^{2}$

$9 - 121 y^{2}$

Solution

$(3 + 11 y) (3 - 11 y)$

$100 k^{2} - 81$

$20 x^{2} - 125$

Solution

$5 (2 x - 5) (2 x + 5)$

$18 y^{2} - 98$

$49 u^{3} - 9 u$

Solution

$u (7 u + 3) (7 u - 3)$

$169 n^{3} - n$

Factor Sums and Differences of Cubes

In the following exercises, factor.

$a^{3} - 125$

Solution

$(a - 5) (a^{2} + 5 a + 25)$

$b^{3} - 216$

$2 m^{3} + 54$

Solution

$2 (m + 3) (m^{2} - 3 m + 9)$

$81 x^{3} + 3$

7.5 General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

$24 x^{3} + 44 x^{2}$

Solution

$4 x^{2} (6 x + 11)$

$24 a^{4} - 9 a^{3}$

$16 n^{2} - 56 m n + 49 m^{2}$

Solution

${(4 n - 7 m)}^{2}$

$6 a^{2} - 25 a - 9$

$5 r^{2} + 22 r - 48$

Solution

$(r + 6) (5 r - 8)$

$5 u^{4} - 45 u^{2}$

$n^{4} - 81$

Solution

$(n^{2} + 9) (n + 3) (n - 3)$

$64 j^{2} + 225$

$5 x^{2} + 5 x - 60$

Solution

$5 (x - 3) (x + 4)$

$b^{3} - 64$

$m^{3} + 125$

Solution

$(m + 5) (m^{2} - 5 m + 25)$

$2 b^{2} - 2 b c + 5 c b - 5 c^{2}$

7.6 Quadratic Equations

Use the Zero Product Property

In the following exercises, solve.

$(a - 3) (a + 7) = 0$

Solution

$a = 3, a = −7$

$(b - 3) (b + 10) = 0$

$3 m (2 m - 5) (m + 6) = 0$

Solution

$m = 0 m = –6 m = \frac{5}{2}$

$7 n (3 n + 8) (n - 5) = 0$

Solve Quadratic Equations by Factoring

In the following exercises, solve.

$x^{2} + 9 x + 20 = 0$

Solution

$x = −4, x = −5$

$y^{2} - y - 72 = 0$

$2 p^{2} - 11 p = 40$

Solution

$p = - \frac{5}{2}, p = 8$

$q^{3} + 3 q^{2} + 2 q = 0$

$144 m^{2} - 25 = 0$

Solution

$m = \frac{5}{12}, m = - \frac{5}{12}$

$4 n^{2} = 36$

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

The product of two consecutive numbers is $462$ . Find the numbers.

Solution

$−21 and −22; 21 and 22$

The area of a rectangular shaped patio $400$ square feet. The length of the patio is $9$ feet more than its width. Find the length and width.

Practice Test

In the following exercises, find the Greatest Common Factor in each expression.

$14 y - 42$

Solution

$14 (y - 3)$

$−6 x^{2} - 30 x$

$80 a^{2} + 120 a^{3}$

Solution

$40 a^{2} (2 + 3 a)$

$5 m (m - 1) + 3 (m - 1)$

In the following exercises, factor completely.

$x^{2} + 13 x + 42$

Solution

$(x + 7) (x + 6)$

$p^{2} + p q - 12 q^{2}$

$3 a^{3} - 6 a^{2} - 72 a$

Solution

$3 a (a - 6) (a + 4)$

$s^{2} - 25 s + 84$

$5 n^{2} + 30 n + 45$

Solution

$5 (n + 3)^{2}$

$64 y^{2} - 49$

$x y - 8 y + 7 x - 56$

Solution

$(x - 8) (y + 7)$

$40 r^{2} + 810$

$9 s^{2} - 12 s + 4$

Solution

${(3 s - 2)}^{2}$

$n^{2} + 12 n + 36$

$100 - a^{2}$

Solution

$(10 - a) (10 + a)$

$6 x^{2} - 11 x - 10$

$3 x^{2} - 75 y^{2}$

Solution

$3 (x + 5 y) (x - 5 y)$

$c^{3} - 1000 d^{3}$

$a b - 3 b - 2 a + 6$

Solution

$(a - 3) (b - 2)$

$6 u^{2} + 3 u - 18$

$8 m^{2} + 22 m + 5$

Solution

$(4 m + 1) (2 m + 5)$

In the following exercises, solve.

$x^{2} + 9 x + 20 = 0$

$y^{2} = y + 132$

Solution

$y = −11, y = 12$

$5 a^{2} + 26 a = 24$

$9 b^{2} - 9 = 0$

Solution

$b = 1, b = −1$

$16 - m^{2} = 0$

$4 n^{2} + 19 n + 21 = 0$

Solution

$n = - \frac{7}{4}, n = −3$

$(x - 3) (x + 2) = 6$

The product of two consecutive integers is 156. Find the integers.

Solution

$12 and 13; - 13 and −12$

The area of a rectangular place mat is $168$ square inches. Its length is two inches longer than the width. Find the length and width of the place mat.

quadratic equations

are equations in which the variable is squared.

Zero Product Property

The Zero Product Property states that, if the product of two quantities is zero, at least one of the quantities is zero.

Introduction

This is a photo of two people rowing a boat in a canal. — Rowing a boat downstream can be very relaxing, but it takes much more effort to row the boat upstream.

Like rowing a boat, riding a bicycle is a situation in which going in one direction, downhill, is easy, but going in the opposite direction, uphill, can be more work. The trip to reach a destination may be quick, but the return trip whether upstream or uphill will take longer.

Rational equations are used to model situations like these. In this chapter, we will work with rational expressions, solve rational equations, and use them to solve problems in a variety of applications.

Simplify Rational Expressions

In Chapter 1, we reviewed the properties of fractions and their operations. We introduced rational numbers, which are just fractions where the numerators and denominators are integers, and the denominator is not zero.

In this chapter, we will work with fractions whose numerators and denominators are polynomials. We call these rational expressions.

Remember, division by 0 is undefined.

Here are some examples of rational expressions:

- \frac{13}{42} \frac{7 y}{8 z} \frac{5 x + 2}{x^{2} - 7} \frac{4 x^{2} + 3 x - 1}{2 x - 8}

Notice that the first rational expression listed above, $- \frac{13}{42},$ is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero.

We will perform the same operations with rational expressions that we do with fractions. We will simplify, add, subtract, multiply, divide, and use them in applications.

Determine the Values for Which a Rational Expression is Undefined

When we work with a numerical fraction, it is easy to avoid dividing by zero, because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero.

If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator.

So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero. That way, when we solve a rational equation for example, we will know whether the algebraic solutions we find are allowed or not.

Determine the values for which the rational expression is undefined:

Solution

The expression will be undefined when the denominator is zero.

ⓐ

This table demonstrates the steps to determine when a rational expression is undefined by setting the denominator to zero.
	$\frac{9 y}{x}$
Set the denominator equal to zero. Solve for the variable.	$x = 0$
	$\frac{9 y}{x} is undefined for x = 0 .$

ⓑ

Steps to determine where a rational expression is undefined by setting its denominator to zero and solving for the variable.
	$\frac{4 b - 3}{2 b + 5}$
Set the denominator equal to zero. Solve for the variable.	$\begin{matrix} 2 b + 5 & = & 0 \\ 2 b & = & −5 \\ b & = & - \frac{5}{2} \end{matrix}$
	$\frac{4 b - 3}{2 b + 5}$ is undefined for $b = - \frac{5}{2} .$

ⓒ

This table illustrates the step-by-step procedure to determine the values for which a given rational expression is undefined by solving its denominator for zero.
	$\frac{x + 4}{x^{2} + 5 x + 6}$
Set the denominator equal to zero. Solve for the variable.	$\begin{matrix} x^{2} + 5 x + 6 & = & 0 \\ (x + 2) (x + 3) & = & 0 \\ x + 2 = 0 or x + 3 & = & 0 \\ x = −2 or x & = & −3 \end{matrix}$
	$\frac{x + 4}{x^{2} + 5 x + 6}$ is undefined for $x = −2 or x = −3 .$

Saying that the rational expression $\frac{x + 4}{x^{2} + 5 x + 6}$ is undefined for $x = −2 or x = −3$ is similar to writing the phrase “void where prohibited” in contest rules.

Evaluate Rational Expressions

To evaluate a rational expression, we substitute values of the variables into the expression and simplify, just as we have for many other expressions in this book.

Evaluate $\frac{2 x + 3}{3 x - 5}$ for each value:

Solution

ⓐ

	$A mathematical expression displaying the fraction (2x + 3) over (3x - 5), where 2x + 3 is the numerator and 3x - 5 is the denominator.$
	$A mathematical fraction showing the substitution of 0 for a variable in both the numerator and denominator, with the zeroes highlighted in red: (2(0)+3) / (3(0)-5).$
Simplify.	$The image displays the fraction -3/5, representing a negative value where three is divided by five.$

ⓑ

	$A mathematical fraction with the expression 2x + 3 in the numerator and 3x - 5 in the denominator, presented in a clear, standard algebraic format.$
	$A mathematical expression displaying a fraction where the variable in the numerator 2(x) + 3 and denominator 3(x) - 5 has been substituted with the number 2, highlighted in red. The expression is 2(2) + 3 / 3(2) - 5.$
Simplify.	$A mathematical fraction showing the expression (4+3) in the numerator and (6-5) in the denominator.$
	$A mathematical fraction displays the number 7 over the number 1, representing the value of seven divided by one. The numeral 7 is positioned as the numerator above a horizontal line, with the numeral 1 as the denominator below it.$

ⓒ

	$A mathematical expression displaying a fraction: the numerator is 2x + 3, and the denominator is 3x - 5.$
	$A mathematical fraction is shown, with 2(-3) + 3 in the numerator and 3(-3) - 5 in the denominator. The number -3 is highlighted in red in both parts of the expression.$
Simplify.	$A mathematical expression showing a fraction with numerator '-6 + 3' and denominator '-9 - 5'.$
	$The mathematical expression shows the fraction -3 divided by -14.$
	$A fraction shows 3 over 14, represented as 3/14, centered on a white background.$

Evaluate $\frac{x^{2} + 8 x + 7}{x^{2} - 4}$ for each value:

Solution

ⓐ

	$A mathematical expression showing the fraction (x^2 + 8x + 7) divided by (x^2 - 4). The numerator is a quadratic trinomial, and the denominator is a difference of squares.$
	$The evaluation of an algebraic fraction where the variable is replaced by zero, represented as ((0)^2 + 8(0) + 7) / ((0)^2 - 4).$
Simplify.	$A white background displays the mathematical fraction seven over negative four, representing the value -7/4.$
	$The mathematical expression displays a negative fraction, written as minus seven over four (-7/4).$

ⓑ

	$A mathematical expression showing the fraction (x^2 + 8x + 7) divided by (x^2 - 4). The numerator is a quadratic trinomial, and the denominator is a difference of squares.$
	$A mathematical fraction with (2)^2 + 8(2) + 7 in the numerator and (2)^2 - 4 in the denominator, where the number 2 is highlighted in red.$
Simplify.	$A mathematical expression showing the fraction (4 + 16 + 7) / (4 - 4), which results in division by zero, rendering the expression undefined.$

This rational expression is undefined for x = 2.

ⓒ

	$A mathematical expression showing a fraction. The numerator is x squared plus 8x plus 7. The denominator is x squared minus 4.$
	$A mathematical fraction showing the evaluation of (x^2 + 8x + 7) / (x^2 - 4) with x = -1, where the -1 is highlighted in red.$
Simplify.	$A mathematical expression showing a fraction. The numerator is 1 - 8 + 7 and the denominator is 1 - 4.$
	$A mathematical expression showing the fraction with numerator -7 + 7 and denominator -3.$
	$A mathematical expression displaying the fraction 0 over -3, which simplifies to 0. This image shows a fundamental concept in division where zero divided by any non-zero number is zero.$

Remember that a fraction is simplified when it has no common factors, other than 1, in its numerator and denominator. When we evaluate a rational expression, we make sure to simplify the resulting fraction.

Evaluate $\frac{a^{2} + 2 a b + b^{2}}{3 a b^{2}}$ for each value:

Solution

ⓐ

	$\frac{a^{2} + 2 a b + b^{2}}{3 a b^{2}}$ when $a = 1, b = 2$ .
	$A mathematical fraction with a numerator expanded as (1)^2 + 2(1)(2) + (2)^2 and a denominator as 3(1)(2)^2, featuring red and blue colored numbers.$
Simplify.	$A mathematical expression showing the fraction (1 + 4 + 4) over 3(4).$
	$The fraction 9/12 is displayed on a white background, with a horizontal line separating the numerator '9' from the denominator '12'.$
	$The image displays the fraction '3/4' centered on a plain white background. The numerator '3' is positioned above a horizontal fraction bar, with the denominator '4' directly below the bar. The text is black and clear.$

ⓑ

	$\frac{a^{2} + 2 a b + b^{2}}{3 a b^{2}}$ when $a = −2, b = −1$ .

Simplify.	$A mathematical expression showing the fraction (4 + 4 + 1) divided by -6, presented in a clean, white background.$
	$A mathematical expression displaying the fraction -9/6.$
	$A mathematical expression displaying the negative fraction -3/2, set against a plain white background.$

ⓒ

	$\frac{a^{2} + 2 a b + b^{2}}{3 a b^{2}}$ when $a = \frac{1}{3}, b = 0$ .
	$A mathematical expression showing a fraction. The numerator is (1/3)^2 + 2(1/3)(0) + (0)^2, and the denominator is 3(1/3)(0)^2. The number 1/3 is in red, and 0 is in blue.$
Simplify.	$A mathematical expression showing a fraction 1/9 plus two zeros in the numerator, all divided by zero, resulting in an undefined expression due to division by zero.$
	$A mathematical expression showing the fraction 1 over 9, which is then divided by 0, resulting in an undefined or impossible mathematical operation, commonly known as division by zero.$
	The expression is undefined.

Simplify Rational Expressions

Just like a fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator, a rational expression is simplified if it has no common factors, other than 1, in its numerator and denominator.

For example:

$\frac{2}{3}$ is simplified because there are no common factors of 2 and 3.
$\frac{2 x}{3 x}$ is not simplified because x is a common factor of 2x and 3x.

We use the Equivalent Fractions Property to simplify numerical fractions. We restate it here as we will also use it to simplify rational expressions.

Notice that in the Equivalent Fractions Property, the values that would make the denominators zero are specifically disallowed. We see $b \neq 0, c \neq 0$ clearly stated. Every time we write a rational expression, we should make a similar statement disallowing values that would make a denominator zero. However, to let us focus on the work at hand, we will omit writing it in the examples.

Let’s start by reviewing how we simplify numerical fractions.

Simplify: $- \frac{36}{63} .$

Solution

	$The image displays the fraction -36/63, represented with a horizontal line separating the numerator -36 and the denominator 63.$
Rewrite the numerator and denominator showing the common factors.	$A mathematical fraction displaying -(49)/(79), with the common multiplier '9' highlighted in red, indicating its cancellation for simplification to -4/7.$
Simplify using the Equivalent Fractions Property.	$The negative fraction -4/7, showing a minus sign preceding the fraction bar with 4 as the numerator and 7 as the denominator.$

Notice that the fraction $- \frac{4}{7}$ is simplified because there are no more common factors.

Throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, $x \neq 0$ and $y \neq 0$ .

Simplify: $\frac{3 x y}{18 x^{2} y^{2}} .$

Solution

	$A fraction with 3xy in the numerator and 18x²y² in the denominator. This algebraic expression can be simplified by canceling common terms.$
Rewrite the numerator and denominator showing the common factors.	$A fraction with 1 multiplied by 3xy in the numerator and 6xy multiplied by 3xy in the denominator. The 3xy term is highlighted in red, indicating a common factor for simplification.$
Simplify using the Equivalent Fractions Property.	$The image displays the mathematical expression one divided by six xy, written as a fraction with 1 as the numerator and 6xy as the denominator.$

Did you notice that these are the same steps we took when we divided monomials in Polynomials?

To simplify rational expressions we first write the numerator and denominator in factored form. Then we remove the common factors using the Equivalent Fractions Property.

Be very careful as you remove common factors. Factors are multiplied to make a product. You can remove a factor from a product. You cannot remove a term from a sum.

$This figure contains three columns. The first column, shows the numerator and denominator in factored form. The numerator has 2 times 3 times 7. The denominator has 3 times 5 times 7. The common factors, 3 and 7 are crossed out. The second row, first column shows what remains after the threes and sevens are crossed out, which is 2 over 5 in fraction form. The last row in the first column reads “We removed the common factors of 3 and 7. They are the factors of the product.” The first row of the middle column shows 3 x and then x minus 9 in parentheses in the numerator. The denominator shows 5 and then x-9 in parentheses. The common factors x minus 9 are crossed out. The second row of the middle column shows what remains after removing the common factors, which is 3 x over 5 in fraction form. The last row in the middle column reads, “We removed the common factor x minus 9. It is a factor of the product.” The first row of the third column shows x plus 5 in the numerator and x in the denominator. The second row says “No common factors” and the third row reads, “While there is an x in both the numerator and the denominator, the x in the numerator is a term of a sum”.$

Note that removing the x’s from $\frac{x + 5}{x}$ would be like cancelling the 2’s in the fraction $\frac{2 + 5}{2}$ !

How to Simplify Rational Binomials

Simplify: $\frac{2 x + 8}{5 x + 20} .$

Solution

This figure is a table with three columns and two rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell says “Step 1. Factor the numerator and denominator completely.” The second cell says “Factor 2x plus 8 and 5x minus 20.” The third cell contains 2x plus 8, divided by 5x plus 20. Below this is 2 times x plus 4 divided by 5 times x plus 4.

In the second row, the first cell says “Step 2. Simplify by dividing out common factors.” The second cell says “Divide out the common factors.” The third cell contains 2 times x plus 4 divided by 5 times x plus 4, where x plus 4 cancels out in the numerator and the denominator. It simplifies to 2 fifths.

We now summarize the steps you should follow to simplify rational expressions.

Usually, we leave the simplified rational expression in factored form. This way it is easy to check that we have removed all the common factors!

We’ll use the methods we covered in Factoring to factor the polynomials in the numerators and denominators in the following examples.

Simplify: $\frac{x^{2} + 5 x + 6}{x^{2} + 8 x + 12} .$

Solution

$\begin{matrix} \frac{x^{2} + 5 x + 6}{x^{2} + 8 x + 12} \\ Factor the numerator and denominator. & \frac{(x + 2) (x + 3)}{(x + 2) (x + 6)} \\ \begin{matrix} Remove the common factor x + 2 from \\ the numerator and the denominator. \end{matrix} & \frac{(x + 2) (x + 3)}{(x + 2) (x + 6)} \\ \frac{x + 3}{x + 6} \end{matrix}$

Can you tell which values of x must be excluded in this example?

Simplify: $\frac{y^{2} + y - 42}{y^{2} - 36} .$

Solution

$\begin{matrix} \frac{y^{2} + y - 42}{y^{2} - 36} \\ Factor the numerator and denominator. & \frac{(y + 7) (y - 6)}{(y + 6) (y - 6)} \\ \begin{matrix} Remove the common factor y - 6 from \\ the numerator and the denominator. \end{matrix} & \frac{(y + 7) (y - 6)}{(y + 6) (y - 6)} \\ \frac{y + 7}{y + 6} \end{matrix}$

Simplify: $\frac{p^{3} - 2 p^{2} + 2 p - 4}{p^{2} - 7 p + 10} .$

Solution

$\begin{matrix} \frac{p^{3} - 2 p^{2} + 2 p - 4}{p^{2} - 7 p + 10} \\ \begin{matrix} Factor the numerator and denominator, \\ using grouping to factor the numerator. \end{matrix} & \frac{p^{2} (p - 2) + 2 (p - 2)}{(p - 5) (p - 2)} \\ \frac{(p^{2} + 2) (p - 2)}{(p - 5) (p - 2)} \\ \begin{matrix} Remove the common factor of p - 2 \\ from the numerator and the denominator. \end{matrix} & \frac{(p^{2} + 2) (p - 2)}{(p - 5) (p - 2)} \\ \frac{p^{2} + 2}{p - 5} \end{matrix}$

Simplify: $\frac{2 n^{2} - 14 n}{4 n^{2} - 16 n - 48} .$

Solution

$\begin{matrix} \frac{2 n^{2} - 14 n}{4 n^{2} - 16 n - 48} \\ \begin{matrix} Factor the numerator and denominator, \\ first factoring out the GCF. \end{matrix} & \frac{2 n (n - 7)}{4 (n^{2} - 4 n - 12)} \\ \frac{2 n (n - 7)}{4 (n - 6) (n + 2)} \\ Remove the common factor, 2. & \frac{2 n (n - 7)}{2 \cdot 2 (n - 6) (n + 2)} \\ \frac{n (n - 7)}{2 (n - 6) (n + 2)} \end{matrix}$

Simplify: $\frac{3 b^{2} - 12 b + 12}{6 b^{2} - 24} .$

Solution

$\begin{matrix} \frac{3 b^{2} - 12 b + 12}{6 b^{2} - 24} \\ \begin{matrix} Factor the numerator and denominator, \\ first factoring out the GCF. \end{matrix} & \frac{3 (b^{2} - 4 b + 4)}{6 (b^{2} - 4)} \\ \frac{3 (b - 2) (b - 2)}{6 (b + 2) (b - 2)} \\ Remove the common factors of b - 2 and 3 . & \frac{3 (b - 2) (b - 2)}{3 \cdot 2 (b + 2) (b - 2)} \\ \frac{b - 2}{2 (b + 2)} \end{matrix}$

Simplify: $\frac{m^{3} + 8}{m^{2} - 4} .$

Solution

$\begin{matrix} \frac{m^{3} + 8}{m^{2} - 4} \\ \begin{matrix} Factor the numerator and denominator, \\ using the formulas for sum of cubes and \\ difference of squares. \end{matrix} & \frac{(m + 2) (m^{2} - 2 m + 4)}{(m + 2) (m - 2)} \\ Remove the common factor of m + 2 . & \frac{(m + 2) (m^{2} - 2 m + 4)}{(m + 2) (m - 2)} \\ \frac{m^{2} - 2 m + 4}{m - 2} \end{matrix}$

Simplify Rational Expressions with Opposite Factors

Now we will see how to simplify a rational expression whose numerator and denominator have opposite factors. Let’s start with a numerical fraction, say $\frac{7}{−7}$ . We know this fraction simplifies to $−1$ . We also recognize that the numerator and denominator are opposites.

In Foundations, we introduced opposite notation: the opposite of $a$ is $− a$ . We remember, too, that $− a = −1 \cdot a$ .

We simplify the fraction $\frac{a}{− a}$ , whose numerator and denominator are opposites, in this way:

$\begin{matrix} \frac{a}{− a} \\ We could rewrite this. & \frac{1 \cdot a}{−1 \cdot a} \\ Remove the common factors. & \frac{1}{−1} \\ Simplify. & −1 \end{matrix}$

So, in the same way, we can simplify the fraction $\frac{x - 3}{− (x - 3)}$ :

$\begin{matrix} We could rewrite this. & \frac{1 \cdot (x - 3)}{−1 \cdot (x - 3)} \\ Remove the common factors. & \frac{1}{−1} \\ Simplify. & −1 \end{matrix}$

But the opposite of $x - 3$ could be written differently:

$\begin{matrix} − (x - 3) \\ Distribute. & − x + 3 \\ Rewrite. & 3 - x \end{matrix}$

This means the fraction $\frac{x - 3}{3 - x}$ simplifies to $−1$ .

In general, we could write the opposite of $a - b$ as $b - a$ . So the rational expression $\frac{a - b}{b - a}$ simplifies to $−1$ .

We will use this property to simplify rational expressions that contain opposites in their numerators and denominators.

Simplify: $\frac{x - 8}{8 - x} .$

Solution

$\begin{matrix} \frac{x - 8}{8 - x} \\ Recognize that x - 8 and 8 - x are opposites. & −1 \end{matrix}$

Remember, the first step in simplifying a rational expression is to factor the numerator and denominator completely.

Simplify: $\frac{14 - 2 x}{x^{2} - 49} .$

Solution

	$A mathematical expression displaying the fraction (14 - 2x) over (x^2 - 49).$
Factor the numerator and denominator.	$A mathematical expression showing the fraction 2(7-x) over (x+7)(x-7).$
Recognize that $7 - x and x - 7 are opposites$ .	$A math expression showing the fraction 2(7-x) over (x+7)(x-7). Diagonal lines on (7-x) in the numerator and (x-7) in the denominator indicate their cancellation. The fraction is multiplied by a red (-1).$
Simplify.	$A mathematical expression displaying a fraction: negative 2 over the quantity of x plus 7. The fraction is written with a horizontal line separating the numerator 2 from the denominator x+7, with a minus sign to the left of the fraction line.$

Simplify: $\frac{x^{2} - 4 x - 32}{64 - x^{2}} .$

Solution

	$A fraction with the numerator x squared minus 4x minus 32, and the denominator 64 minus x squared.$
Factor the numerator and denominator.	$A mathematical fraction is displayed with (x-8)(x+4) in the numerator and (8-x)(8+x) in the denominator.$
Recognize the factors that are opposites.
Simplify.	$A mathematical expression showing a negative fraction. The numerator is x plus 4, and the denominator is x plus 8. The entire fraction is negated.$

Key Concepts

Determine the Values for Which a Rational Expression is Undefined
1. Set the denominator equal to zero.
2. Solve the equation, if possible.
Simplified Rational Expression
- A rational expression is considered simplified if there are no common factors in its numerator and denominator.
Simplify a Rational Expression
1. Factor the numerator and denominator completely.
2. Simplify by dividing out common factors.
Opposites in a Rational Expression
- The opposite of $a - b$ is $b - a$ .
  $\begin{matrix} \frac{a - b}{b - a} = −1 & a \neq 0, b \neq 0, a \neq b \end{matrix}$

Practice Makes Perfect

In the following exercises, determine the values for which the rational expression is undefined.

Solution

Evaluate Rational Expressions

In the following exercises, evaluate the rational expression for the given values.

$\frac{2 x}{x - 1}$

Solution

$\frac{4 y - 1}{5 y - 3}$

$\frac{2 p + 3}{p^{2} + 1}$

Solution

$\frac{x + 3}{2 - 3 x}$

$\frac{y^{2} + 5 y + 6}{y^{2} - 1}$

Solution

$\frac{z^{2} + 3 z - 10}{z^{2} - 1}$

$\frac{a^{2} - 4}{a^{2} + 5 a + 4}$

Solution

$\frac{b^{2} + 2}{b^{2} - 3 b - 4}$

$\frac{x^{2} + 3 x y + 2 y^{2}}{2 x^{3} y}$

ⓐ $x = 1, y = −1$
ⓑ $x = 2, y = 1$

Solution

$\frac{c^{2} + c d - 2 d^{2}}{c d^{3}}$

ⓐ $c = 2, d = −1$
ⓑ $c = 1, d = −1$

$\frac{m^{2} - 4 n^{2}}{5 m n^{3}}$

ⓐ $m = 2, n = 1$
ⓑ $m = −1, n = −1$

Solution

$\frac{2 s^{2} t}{s^{2} - 9 t^{2}}$

ⓐ $s = 4, t = 1$
ⓑ $s = −1, t = −1$

Simplify Rational Expressions

In the following exercises, simplify.

$- \frac{4}{52}$

Solution

$- \frac{1}{13}$

$- \frac{44}{55}$

$\frac{56}{63}$

Solution

$\frac{8}{9}$

$\frac{65}{104}$

$\frac{6 a b^{2}}{12 a^{2} b}$

Solution

$\frac{b}{2 a}$

$\frac{15 x y}{3 x^{3} y^{3}}$

$\frac{8 m^{3} n}{12 m n^{2}}$

Solution

$\frac{2 m^{2}}{3 n}$

$\frac{36 v^{3} w^{2}}{27 v w^{3}}$

$\frac{3 a + 6}{4 a + 8}$

Solution

$\frac{3}{4}$

$\frac{5 b + 5}{6 b + 6}$

$\frac{3 c - 9}{5 c - 15}$

Solution

$\frac{3}{5}$

$\frac{4 d + 8}{9 d + 18}$

$\frac{7 m + 63}{5 m + 45}$

Solution

$\frac{7}{5}$

$\frac{8 n - 96}{3 n - 36}$

$\frac{12 p - 240}{5 p - 100}$

Solution

$\frac{12}{5}$

$\frac{6 q + 210}{5 q + 175}$

$\frac{a^{2} - a - 12}{a^{2} - 8 a + 16}$

Solution

$\frac{a + 3}{a - 4}$

$\frac{x^{2} + 4 x - 5}{x^{2} - 2 x + 1}$

$\frac{y^{2} + 3 y - 4}{y^{2} - 6 y + 5}$

Solution

$\frac{y + 4}{y - 5}$

$\frac{v^{2} + 8 v + 15}{v^{2} - v - 12}$

$\frac{x^{2} - 25}{x^{2} + 2 x - 15}$

Solution

$\frac{x - 5}{x - 3}$

$\frac{a^{2} - 4}{a^{2} + 6 a - 16}$

$\frac{y^{2} - 2 y - 3}{y^{2} - 9}$

Solution

$\frac{y + 1}{y + 3}$

$\frac{b^{2} + 9 b + 18}{b^{2} - 36}$

$\frac{y^{3} + y^{2} + y + 1}{y^{2} + 2 y + 1}$

Solution

$\frac{y^{2} + 1}{y + 1}$

$\frac{p^{3} + 3 p^{2} + 4 p + 12}{p^{2} + p - 6}$

$\frac{x^{3} - 2 x^{2} - 25 x + 50}{x^{2} - 25}$

Solution

$x - 2$

$\frac{q^{3} + 3 q^{2} - 4 q - 12}{q^{2} - 4}$

$\frac{3 a^{2} + 15 a}{6 a^{2} + 6 a - 36}$

Solution

$\frac{a (a + 5)}{2 (a + 3) (a - 2)}$

$\frac{8 b^{2} - 32 b}{2 b^{2} - 6 b - 80}$

$\frac{−5 c^{2} - 10 c}{−10 c^{2} + 30 c + 100}$

Solution

$\frac{c}{2 (c - 5)}$

$\frac{4 d^{2} - 24 d}{2 d^{2} - 4 d - 48}$

$\frac{3 m^{2} + 30 m + 75}{4 m^{2} - 100}$

Solution

$\frac{3 (m + 5)}{4 (m - 5)}$

$\frac{5 n^{2} + 30 n + 45}{2 n^{2} - 18}$

$\frac{5 r^{2} + 30 r - 35}{r^{2} - 49}$

Solution

$\frac{5 (r - 1)}{r - 7}$

$\frac{3 s^{2} + 30 s + 72}{3 s^{2} - 48}$

$\frac{t^{3} - 27}{t^{2} - 9}$

Solution

$\frac{t^{2} + 3 t + 9}{t + 3}$

$\frac{v^{3} - 1}{v^{2} - 1}$

$\frac{w^{3} + 216}{w^{2} - 36}$

Solution

$\frac{w^{2} - 6 w + 36}{w - 6}$

$\frac{v^{3} + 125}{v^{2} - 25}$

Simplify Rational Expressions with Opposite Factors

In the following exercises, simplify each rational expression.

$\frac{a - 5}{5 - a}$

Solution

$−1$

$\frac{b - 12}{12 - b}$

$\frac{11 - c}{c - 11}$

Solution

$−1$

$\frac{5 - d}{d - 5}$

$\frac{12 - 2 x}{x^{2} - 36}$

Solution

$- \frac{2}{x + 6}$

$\frac{20 - 5 y}{y^{2} - 16}$

$\frac{4 v - 32}{64 - v^{2}}$

Solution

$- \frac{4}{8 + v}$

$\frac{7 w - 21}{9 - w^{2}}$

$\frac{y^{2} - 11 y + 24}{9 - y^{2}}$

Solution

$- \frac{(y - 8)}{3 + y}$

$\frac{z^{2} - 9 z + 20}{16 - z^{2}}$

$\frac{a^{2} - 5 a - 36}{81 - a^{2}}$

Solution

$- \frac{a + 4}{9 + a}$

$\frac{b^{2} + b - 42}{36 - b^{2}}$

Everyday Math

Tax Rates For the tax year 2015, the amount of tax owed by a single person earning between $37,450 and $90,750, can be found by evaluating the formula $0.25 x - 4206.25,$ where x is income. The average tax rate for this income can be found by evaluating the formula $\frac{0.25 x - 4206.25}{x} .$ What would be the average tax rate for a single person earning $50,000?

Solution

16.6%

Work The length of time it takes for two people for perform the same task if they work together can be found by evaluating the formula $\frac{x y}{x + y} .$ If Tom can paint the den in $x =$ 45 minutes and his brother Bobby can paint it in $y =$ 60 minutes, how many minutes will it take them if they work together?

Writing Exercises

Explain how you find the values of x for which the rational expression $\frac{x^{2} - x - 20}{x^{2} - 4}$ is undefined.

Solution

Answers will vary, but all should reference setting the denominator function to zero.

Explain all the steps you take to simplify the rational expression $\frac{p^{2} + 4 p - 21}{9 - p^{2}} .$

Self Check

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

This figure shows a table with four columns and five rows. The first row is a header row and each column is labeled. The first column header is labeled “I can…”, the second is labeled “Confidently”, the third is labeled “With some help”, and the fourth is labeled “No—I don’t get it!” In the first column under “I can”, the cells read “determine the values for which a rational expression is undefined,” “evaluate rational expressions,” “simplify rational expressions,” and “simplify rational expressions with opposite factors.” The rest of the cells are blank.

ⓑ If most of your checks were:

rational expression

A rational expression is an expression of the form

\frac{p}{q}

, where p and q are polynomials and

q \neq 0 .

Multiply and Divide Rational Expressions

Multiply Rational Expressions

To multiply rational expressions, we do just what we did with numerical fractions. We multiply the numerators and multiply the denominators. Then, if there are any common factors, we remove them to simplify the result.

We’ll do the first example with numerical fractions to remind us of how we multiplied fractions without variables.

Multiply: $\frac{10}{28} \cdot \frac{8}{15} .$

Solution

	$The image displays the multiplication of two fractions: 10/28 multiplied by 8/15, represented as (10/28) (8/15).$
Multiply the numerators and denominators.	$A mathematical expression displaying the fraction (10 multiplied by 8) divided by (28 multiplied by 15).$
Look for common factors, and then remove them.	$A mathematical fraction is shown with terms being cancelled out. In the numerator, 5 and 4 are crossed out, leaving 2 multiplied by 2. In the denominator, 4 and 5 are crossed out, leaving 7 multiplied by 3.$
Simplify.	$The fraction 4/21 is displayed on a white background, with 4 as the numerator and 21 as the denominator, separated by a horizontal line.$

Remember, throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, $x \neq 0$ and $y \neq 0$ .

Mulitply: $\frac{2 x}{3 y^{2}} \cdot \frac{6 x y^{3}}{x^{2} y} .$

Solution

	$A mathematical expression showing the product of two fractions: (2x / 3y^2) multiplied by (6xy^3 / x^2y).$
Multiply.	$An algebraic fraction is displayed with the numerator '2x multiplied by 6xy^3' and the denominator '3y^2 multiplied by x^2y'.$
Factor the numerator and denominator completely, and then remove common factors.	$A mathematical fraction showing simplification. Common factors like 'x', 'B', and 'y' are struck through in red from both the numerator and denominator, illustrating term cancellation.$
Simplify.

How to Multiply Rational Expressions

Mulitply: $\frac{2 x}{x^{2} - 7 x + 12} \cdot \frac{x^{2} - 9}{6 x^{2}} .$

Solution

The above image has three columns and three rows to show how to multiply rational expressions. Step one is to factor each numerator and denominator completely. Factor x squared minus 9 and x squared minus 7 x plus 12. The rational equation is 2x divided by x squared plus x plus 12 times x squared minus 9 divided by 6x squared, then to 2x divided by x minus 3 times x minus 4 times x minus 3 times x plus 3 divided by 6x squared.

Step 2 is to multiply the numerators and denominators. It is helpful to multiply the monomials first. Multiply 2x times x minus 3 times x plus 3 divided by 6x squared times x minus 3 times x minus 4.

Step 3 is to divide out the common factors, canceling out 2, x, and x minus 3 in the numerator and 2, x and x minus 3 in the denominator. Leave the denominator in factored form to get x plus 3 divided by 3x times x minus 4.

Multiply: $\frac{n^{2} - 7 n}{n^{2} + 2 n + 1} \cdot \frac{n + 1}{2 n} .$

Solution

Step-by-step simplification of a rational algebraic expression.
	$\frac{n^{2} - 7 n}{n^{2} + 2 n + 1} \cdot \frac{n + 1}{2 n}$
Factor each numerator and denominator.	$\frac{n (n - 7)}{(n + 1) (n + 1)} \cdot \frac{n + 1}{2 n}$
Multiply the numerators and the denominators.	$\frac{n (n - 7) (n + 1)}{(n + 1) (n + 1) 2 n}$
Remove common factors.	$\frac{n (n - 7) (n + 1)}{(n + 1) (n + 1) 2 n}$
Simplify.	$\frac{n - 7}{2 (n + 1)}$

Multiply: $\frac{16 - 4 x}{2 x - 12} \cdot \frac{x^{2} - 5 x - 6}{x^{2} - 16} .$

Solution

This table illustrates the step-by-step simplification of a rational algebraic expression through factoring and canceling common terms.
	$\frac{16 - 4 x}{2 x - 12} \cdot \frac{x^{2} - 5 x - 6}{x^{2} - 16}$
Factor each numerator and denominator.	$\frac{4 (4 - x)}{2 (x - 6)} \cdot \frac{(x - 6) (x + 1)}{(x - 4) (x + 4)}$
Multiply the numerators and the denominators.	$\frac{4 (4 - x) (x - 6) (x + 1)}{2 (x - 6) (x - 4) (x + 4)}$
Remove common factors.	$(−1) \frac{2 \cdot 2 (4 - x) (x - 6) (x + 1)}{2 (x - 6) (x - 4) (x + 4)}$
Simplify.	$- \frac{2 (x + 1)}{(x + 4)}$

Multiply: $\frac{2 x - 6}{x^{2} - 8 x + 15} \cdot \frac{x^{2} - 25}{2 x + 10} .$

Solution

	$A mathematical expression showing the multiplication of two algebraic fractions: (2x-6)/(x^2-8x+15) * (x^2-25)/(2x+10).$
Factor each numerator and denominator.	$An algebraic expression illustrating the multiplication of two rational fractions: 2(x-3) over (x-3)(x-5) multiplied by (x-5)(x+5) over 2(x+5).$
Multiply the numerators and denominators.
Remove common factors.	$A fraction displays identical algebraic expressions in the numerator and denominator, with Z, (x-3), (x-5), and (x+5) terms all struck through, demonstrating cancellation.$
Simplify.

Divide Rational Expressions

To divide rational expressions we multiply the first fraction by the reciprocal of the second, just like we did for numerical fractions.

Remember, the reciprocal of $\frac{a}{b}$ is $\frac{b}{a}$ . To find the reciprocal we simply put the numerator in the denominator and the denominator in the numerator. We “flip” the fraction.

How to Divide Rational Expressions

Divide: $\frac{x + 9}{6 - x} \div \frac{x^{2} - 81}{x - 6} .$

Solution

$The above image has three columns. It shows the steps to divide rational expressions. Step one is to rewrite the division as the product of the first rational expression and the reciprocal of the second for x plus 9 divided by 6 minus x divided by x squared minus 81 divided by x minus 6. “Flip” the second fraction and change the division sign to multiplication to get x plus 9 divided by 6 minus x times x minus 6 divided by x squared minus 81.$ Step two is to factor the numerators and denominators completely. Factor x squared minus 81 to get x plus 9 divided by 6 minus x times x minus 6 divided by x minus 9 times x plus 9.

Step two is to factor the numerators and denominators completely. Factor x squared minus 81 to get x plus 9 divided by 6 minus x times x minus 6 divided by x minus 9 times x plus 9.

Step three is to multiply the numerators and denominators to get x plus 9 times x minus 6 divided by 6 minus x times x minus 9 times x plus 9.

Step four is to simplify by dividing out common factors. Divide out the common factors x plus 9, x minus 6 from the numerator and 6 minus x and x plus 9 from the denominator. Remember opposites divide to negative 1. This simplifies to negative 1 divided by x minus 9.

Divide: $\frac{3 n^{2}}{n^{2} - 4 n} \div \frac{9 n^{2} - 45 n}{n^{2} - 7 n + 10} .$

Solution

	$A mathematical expression showing the division of two algebraic fractions. The first fraction is 3n^2 over n^2 - 4n, and the second is 9n^2 - 45n over n^2 - 7n + 10.$
Rewrite the division as the product of the first rational expression and the reciprocal of the second.
Factor the numerators and denominators and then multiply.	$A mathematical expression showing a fraction. The numerator is 3 * n * n * (n - 5)(n - 2). The denominator is n(n - 4) * 3 * 3 * n * (n - 5).$
Simplify by dividing out common factors.	$An algebraic fraction undergoing simplification, with identical terms 'B', 'n', and 'n(n-5)' struck through in red in both the numerator and denominator.$
	$A mathematical expression showing the fraction (n-2) divided by 3(n-4).$

Remember, first rewrite the division as multiplication of the first expression by the reciprocal of the second. Then factor everything and look for common factors.

Divide: $\frac{2 x^{2} + 5 x - 12}{x^{2} - 16} \div \frac{2 x^{2} - 13 x + 15}{x^{2} - 8 x + 16} .$

Solution

Step-by-step simplification of the division of two rational expressions through factoring and canceling common terms.
	$\frac{2 x^{2} + 5 x - 12}{x^{2} - 16} \div \frac{2 x^{2} - 13 x + 15}{x^{2} - 8 x + 16}$
Rewrite the division as multiplication of the first expression by the reciprocal of the second.	$\frac{2 x^{2} + 5 x - 12}{x^{2} - 16} \cdot \frac{x^{2} - 8 x + 16}{2 x^{2} - 13 x + 15}$
Factor the numerators and denominators and then multiply.	$\frac{(2 x - 3) (x + 4) (x - 4) (x - 4)}{(x - 4) (x + 4) (2 x - 3) (x - 5)}$
Simplify by dividing out common factors.	$\frac{(2 x - 3) (x + 4) (x - 4) (x - 4)}{(x - 4) (x + 4) (2 x - 3) (x - 5)}$
Simplify.	$\frac{x - 4}{x - 5}$

Divide: $\frac{p^{3} + q^{3}}{2 p^{2} + 2 p q + 2 q^{2}} \div \frac{p^{2} - q^{2}}{6} .$

Solution

Detailed steps for simplifying a rational algebraic expression through division, multiplication, factoring, and canceling common terms.
	$\frac{p^{3} + q^{3}}{2 p^{2} + 2 p q + 2 q^{2}} \div \frac{p^{2} - q^{2}}{6}$
Rewrite the division as a multiplication of the first expression times the reciprocal of the second.	$\frac{p^{3} + q^{3}}{2 p^{2} + 2 p q + 2 q^{2}} \cdot \frac{6}{p^{2} - q^{2}}$
Factor the numerators and denominators and then multiply.	$\frac{(p + q) (p^{2} - p q + q^{2}) 6}{2 (p^{2} + p q + q^{2}) (p - q) (p + q)}$
Simplify by dividing out common factors.	$\frac{(p + q) (p^{2} - p q + q^{2}) 6^{3}}{2 (p^{2} + p q + q^{2}) (p - q) (p + q)}$
Simplify.	$\frac{3 (p^{2} - p q + q^{2})}{(p - q) (p^{2} + p q + q^{2})}$

Before doing the next example, let’s look at how we divide a fraction by a whole number. When we divide $\frac{3}{5} \div 4$ , we first write 4 as a fraction so that we can find its reciprocal.

\begin{matrix} \frac{3}{5} \div 4 \\ \frac{3}{5} \div \frac{4}{1} \\ \frac{3}{5} \cdot \frac{1}{4} \end{matrix}

We do the same thing when we divide rational expressions.

Divide: $\frac{a^{2} - b^{2}}{3 a b} \div (a^{2} + 2 a b + b^{2}) .$

Solution

Step-by-step simplification of an algebraic rational expression using division, factoring, and common factor cancellation.
	$\frac{a^{2} - b^{2}}{3 a b} \div (a^{2} + 2 a b + b^{2})$
Write the second expression as a fraction.	$\frac{a^{2} - b^{2}}{3 a b} \div \frac{a^{2} + 2 a b + b^{2}}{1}$
Rewrite the division as the first expression times the reciprocal of the second expression.	$\frac{a^{2} - b^{2}}{3 a b} \cdot \frac{1}{a^{2} + 2 a b + b^{2}}$
Factor the numerators and the denominators, and then multiply.	$\frac{(a - b) (a + b) \cdot 1}{3 a b \cdot (a + b) (a + b)}$
Simplify by dividing out common factors.	$\frac{(a - b) (a + b)}{3 a b \cdot (a + b) (a + b)}$
Simplify.	$\frac{(a - b)}{3 a b (a + b)}$

Remember a fraction bar means division. A complex fraction is another way of writing division of two fractions.

Divide: $\frac{\frac{6 x^{2} - 7 x + 2}{4 x - 8}}{\frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}} .$

Solution

Step-by-step simplification of a complex rational algebraic expression, detailing the process from initial division to the final simplified fraction.
	$\frac{\frac{6 x^{2} - 7 x + 2}{4 x - 8}}{\frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}}$
Rewrite with a division sign.	$\frac{6 x^{2} - 7 x + 2}{4 x - 8} \div \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$
Rewrite as product of first times reciprocal of second.	$\frac{6 x^{2} - 7 x + 2}{4 x - 8} \cdot \frac{x^{2} - 5 x + 6}{2 x^{2} - 7 x + 3}$
Factor the numerators and the denominators, and then multiply.	$\frac{(2 x - 1) (3 x - 2) (x - 2) (x - 3)}{4 (x - 2) (2 x - 1) (x - 3)}$
Simplify by dividing out common factors.	$\frac{(2 x - 1) (3 x - 2) (x - 2) (x - 3)}{4 (x - 2) (2 x - 1) (x - 3)}$
Simplify.	$\frac{3 x - 2}{4}$

If we have more than two rational expressions to work with, we still follow the same procedure. The first step will be to rewrite any division as multiplication by the reciprocal. Then we factor and multiply.

Divide: $\frac{3 x - 6}{4 x - 4} \cdot \frac{x^{2} + 2 x - 3}{x^{2} - 3 x - 10} \div \frac{2 x + 12}{8 x + 16} .$

Solution


Rewrite the division as multiplication by the reciprocal.	$A mathematical expression showing the multiplication of three rational algebraic fractions: (3x-6)/(4x-4), (x^2+2x-3)/(x^2-3x-10), and (8x+16)/(2x+12), with the last fraction in red.$
Factor the numerators and the denominators, and then multiply.	$A complex algebraic fraction displays a numerator of 3 * 8 * (x-2)(x+3)(x-1)(x+2) and a denominator of 4 * 2 * (x-1)(x+2)(x-5)(x+6), set up for simplification.$
Simplify by dividing out common factors.
Simplify.	$A mathematical expression showing a fraction. The numerator is 3(x-2)(x+3) and the denominator is (x-5)(x+6).$

Key Concepts

Multiplication of Rational Expressions
- If $p, q, r, s$ are polynomials where $q \neq 0, s \neq 0$ , then $\frac{p}{q} \cdot \frac{r}{s} = \frac{p r}{q s}$ .
- To multiply rational expressions, multiply the numerators and multiply the denominators
Multiply a Rational Expression
1. Factor each numerator and denominator completely.
2. Multiply the numerators and denominators.
3. Simplify by dividing out common factors.
Division of Rational Expressions
- If $p, q, r, s$ are polynomials where $q \neq 0, r \neq 0, s \neq 0$ , then $\frac{p}{q} \div \frac{r}{s} = \frac{p}{q} \cdot \frac{s}{r}$ .
- To divide rational expressions multiply the first fraction by the reciprocal of the second.
Divide Rational Expressions
1. Rewrite the division as the product of the first rational expression and the reciprocal of the second.
2. Factor the numerators and denominators completely.
3. Multiply the numerators and denominators together.
4. Simplify by dividing out common factors.

Practice Makes Perfect

Multiply Rational Expressions

In the following exercises, multiply.

$\frac{12}{16} \cdot \frac{4}{10}$

Solution

$\frac{3}{10}$

$\frac{32}{5} \cdot \frac{16}{24}$

$\frac{18}{10} \cdot \frac{4}{30}$

Solution

$\frac{6}{25}$

$\frac{21}{36} \cdot \frac{45}{24}$

$\frac{5 x^{2} y^{4}}{12 x y^{3}} \cdot \frac{6 x^{2}}{20 y^{2}}$

Solution

$\frac{x^{3}}{8 y}$

$\frac{8 w^{3} y}{9 y^{2}} \cdot \frac{3 y}{4 w^{4}}$

$\frac{12 a^{3} b}{b^{2}} \cdot \frac{2 a b^{2}}{9 b^{3}}$

Solution

$\frac{8 a^{4}}{3 b^{2}}$

$\frac{4 m n^{2}}{5 n^{3}} \cdot \frac{m n^{3}}{8 m^{2} n^{2}}$

$\frac{5 p^{2}}{p^{2} - 5 p - 36} \cdot \frac{p^{2} - 16}{10 p}$

Solution

$\frac{p (p - 4)}{2 (p - 9)}$

$\frac{3 q^{2}}{q^{2} + q - 6} \cdot \frac{q^{2} - 9}{9 q}$

$\frac{4 r}{r^{2} - 3 r - 10} \cdot \frac{r^{2} - 25}{8 r^{2}}$

Solution

$\frac{r + 5}{2 r (r + 2)}$

$\frac{s}{s^{2} - 9 s + 14} \cdot \frac{s^{2} - 49}{7 s^{2}}$

$\frac{x^{2} - 7 x}{x^{2} + 6 x + 9} \cdot \frac{x + 3}{4 x}$

Solution

$\frac{x - 7}{4 (x + 3)}$

$\frac{2 y^{2} - 10 y}{y^{2} + 10 y + 25} \cdot \frac{y + 5}{6 y}$

$\frac{z^{2} + 3 z}{z^{2} - 3 z - 4} \cdot \frac{z - 4}{z^{2}}$

Solution

$\frac{z + 3}{z (z + 1)}$

$\frac{2 a^{2} + 8 a}{a^{2} - 9 a + 20} \cdot \frac{a - 5}{a^{2}}$

$\frac{28 - 4 b}{3 b - 3} \cdot \frac{b^{2} + 8 b - 9}{b^{2} - 49}$

Solution

$- \frac{4 (b + 9)}{3 (b + 7)}$

$\frac{18 c - 2 c^{2}}{6 c + 30} \cdot \frac{c^{2} + 7 c + 10}{c^{2} - 81}$

$\frac{35 d - 7 d^{2}}{d^{2} + 7 d} \cdot \frac{d^{2} + 12 d + 35}{d^{2} - 25}$

Solution

$−7$

$\frac{72 m - 12 m^{2}}{8 m + 32} \cdot \frac{m^{2} + 10 m + 24}{m^{2} - 36}$

$\frac{4 n + 20}{n^{2} + n - 20} \cdot \frac{n^{2} - 16}{4 n + 16}$

Solution

$1$

$\frac{6 p^{2} - 6 p}{p^{2} + 7 p - 18} \cdot \frac{p^{2} - 81}{3 p^{2} - 27 p}$

$\frac{q^{2} - 2 q}{q^{2} + 6 q - 16} \cdot \frac{q^{2} - 64}{q^{2} - 8 q}$

Solution

$1$

$\frac{2 r^{2} - 2 r}{r^{2} + 4 r - 5} \cdot \frac{r^{2} - 25}{2 r^{2} - 10 r}$

Divide Rational Expressions

In the following exercises, divide.

$\frac{t - 6}{3 - t} \div \frac{t - 5}{t^{2} - 9}$

Solution

$\frac{(6 - t) (t + 3)}{t - 5}$

$\frac{v - 5}{11 - v} \div \frac{v^{2} - 25}{v - 11}$

$\frac{10 + w}{w - 8} \div \frac{100 - w^{2}}{8 - w}$

Solution

$- \frac{1}{10 - w}$

$\frac{7 + x}{x - 6} \div {\frac{49 - x}{x + 6}}^{2}$

$\frac{27 y^{2}}{3 y - 21} \div \frac{3 y^{2} + 18}{y^{2} + 13 y + 42}$

Solution

$\frac{3 y^{2} (y + 6) (y + 7)}{(y - 7) (y^{2} + 6)}$

$\frac{24 z^{2}}{2 z - 8} \div \frac{4 z - 28}{z^{2} - 11 z + 28}$

$\frac{16 a^{2}}{4 a + 36} \div \frac{4 a^{2} - 24 a}{a^{2} + 4 a - 45}$

Solution

$\frac{a (a - 5)}{a - 6}$

$\frac{24 b^{2}}{2 b - 4} \div \frac{12 b^{2} + 36 b}{b^{2} - 11 b + 18}$

$\frac{5 c^{2} + 9 c - 2}{c^{2} - 4} \div \frac{5 c^{2} - 16 c + 3}{c^{2} + 4 c + 4}$

Solution

$\frac{(c + 2) (c + 2)}{(c - 2) (c - 3)}$

$\frac{2 d^{2} + d - 3}{d^{2} - 16} \div \frac{2 d^{2} - 9 d - 18}{d^{2} - 8 d + 16}$

$\frac{6 m^{2} - 11 m - 2}{9 - m^{2}} \div \frac{6 m^{2} + 25 m + 4}{m^{2} - 6 m + 9}$

Solution

$- \frac{(m - 2) (m - 3)}{(3 + m) (m + 4)}$

$\frac{2 n^{2} - 3 n - 14}{25 - n^{2}} \div \frac{2 n^{2} - 13 n + 21}{n^{2} - 10 n + 25}$

$\frac{3 s^{2}}{s^{2} - 16} \div \frac{s^{3} + 4 s^{2} + 16 s}{s^{3} - 64}$

Solution

$\frac{3 s}{s + 4}$

$\frac{r^{2} - 9}{15} \div \frac{r^{3} - 27}{5 r^{2} + 15 r + 45}$

$\frac{p^{3} + q^{3}}{3 p^{2} + 3 p q + 3 q^{2}} \div \frac{p^{2} - q^{2}}{12}$

Solution

$\frac{4 (p^{2} - p q + q^{2})}{(p - q) (p^{2} + p q + q^{2})}$

$\frac{v^{3} - 8 w^{3}}{2 v^{2} + 4 v w + 8 w^{2}} \div \frac{v^{2} - 4 w^{2}}{4}$

$\frac{t^{2} - 9}{2 t} \div (t^{2} - 6 t + 9)$

Solution

$\frac{t + 3}{2 t (t - 3)}$

$\frac{x^{2} + 3 x - 10}{4 x} \div (2 x^{2} + 20 x + 50)$

$\frac{2 y^{2} - 10 y z - 48 z^{2}}{2 y - 1} \div (4 y^{2} - 32 y z)$

Solution

$\frac{y + 3 z}{2 y (2 y - 1)}$

$\frac{2 m^{2} - 98 n^{2}}{2 m + 6} \div (m^{2} - 7 m n)$

$\frac{\frac{2 a^{2} - a - 21}{5 a + 20}}{\frac{a^{2} + 7 a + 12}{a^{2} + 8 a + 16}}$

Solution

$\frac{2 a - 7}{5}$

$\frac{\frac{3 b^{2} + 2 b - 8}{12 b + 18}}{\frac{3 b^{2} + 2 b - 8}{2 b^{2} - 7 b - 15}}$

$\frac{\frac{12 c^{2} - 12}{2 c^{2} - 3 c + 1}}{\frac{4 c + 4}{6 c^{2} - 13 c + 5}}$

Solution

$3 (3 c - 5)$

$\frac{\frac{4 d^{2} + 7 d - 2}{35 d + 10}}{\frac{d^{2} - 4}{7 d^{2} - 12 d - 4}}$

$\frac{10 m^{2} + 80 m}{3 m - 9} \cdot \frac{m^{2} + 4 m - 21}{m^{2} - 9 m + 20}$
$\div \frac{5 m^{2} + 10 m}{2 m - 10}$

Solution

$\frac{4 (m + 8) (m + 7)}{3 (m - 4) (m + 2)}$

$\frac{4 n^{2} + 32 n}{3 n + 2} \cdot \frac{3 n^{2} - n - 2}{n^{2} + n - 30}$
$\div \frac{108 n^{2} - 24 n}{n + 6}$

$\frac{12 p^{2} + 3 p}{p + 3} \div \frac{p^{2} + 2 p - 63}{p^{2} - p - 12}$
$\cdot \frac{p - 7}{9 p^{3} - 9 p^{2}}$

Solution

$\frac{(4 p + 1) (p - 4)}{3 p (p + 9) (p - 1)}$

$\frac{6 q + 3}{9 q^{2} - 9 q} \div \frac{q^{2} + 14 q + 33}{q^{2} + 4 q - 5}$
$\cdot \frac{4 q^{2} + 12 q}{12 q + 6}$

Everyday Math

Probability The director of large company is interviewing applicants for two identical jobs. If $w =$ the number of women applicants and $m =$ the number of men applicants, then the probability that two women are selected for the jobs is $\frac{w}{w + m} \cdot \frac{w - 1}{w + m - 1} .$

ⓐ Simplify the probability by multiplying the two rational expressions.
ⓑ Find the probability that two women are selected when $w = 5$ and $m = 10$ .

Solution

ⓐ $\frac{w (w - 1)}{(w + m) (w + m - 1)}$
ⓑ $\frac{2}{21}$

Area of a triangle The area of a triangle with base b and height h is $\frac{b h}{2} .$ If the triangle is stretched to make a new triangle with base and height three times as much as in the original triangle, the area is $\frac{9 b h}{2} .$ Calculate how the area of the new triangle compares to the area of the original triangle by dividing $\frac{9 b h}{2}$ by $\frac{b h}{2}$ .

Writing Exercises

ⓐ Multiply $\frac{7}{4} \cdot \frac{9}{10}$ and explain all your steps.
ⓑ Multiply $\frac{n}{n - 3} \cdot \frac{9}{n + 3}$ and explain all your steps.

Solution

Answers will vary.

ⓐ Divide $\frac{24}{5} \div 6$ and explain all your steps.
ⓑ Divide $\frac{x^{2} - 1}{x} \div (x + 1)$ and explain all your steps.

Self Check

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

The above image is a table with four columns and four rows. The first row is the header row. The first header is labeled “I can…”, the second “Confidently”, the third, “With some help”, and the fourth “No – I don’t get it!”. In the first column under “I can”, the next row reads multiply rational expressions.”, the next row reads “divide rational expressions.”, the last row reads “after reviewing this checklist, what will you do to become confident for all objectives?” The remaining columns are blank.

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

Add and Subtract Rational Expressions with a Common Denominator

Add Rational Expressions with a Common Denominator

What is the first step you take when you add numerical fractions? You check if they have a common denominator. If they do, you add the numerators and place the sum over the common denominator. If they do not have a common denominator, you find one before you add.

It is the same with rational expressions. To add rational expressions, they must have a common denominator. When the denominators are the same, you add the numerators and place the sum over the common denominator.

We will add two numerical fractions first, to remind us of how this is done.

Add: $\frac{5}{18} + \frac{7}{18} .$

Solution

Step-by-step process for adding two fractions with a common denominator and simplifying the result.
	$\frac{5}{18} + \frac{7}{18}$
The fractions have a common denominator, so add the numerators and place the sum over the common denominator.	$\frac{5 + 7}{18}$
Add in the numerator.	$\frac{12}{18}$
Factor the numerator and denominator to show the common factors.	$\frac{6 \cdot 2}{6 \cdot 3}$
Remove common factors.	$\frac{6 \cdot 2}{6 \cdot 3}$
Simplify.	$\frac{2}{3}$

Remember, we do not allow values that would make the denominator zero. What value of $y$ should be excluded in the next example?

Add: $\frac{3 y}{4 y - 3} + \frac{7}{4 y - 3} .$

Solution

Demonstrates adding rational expressions with a common denominator by combining numerators over the shared denominator.
	$\frac{3 y}{4 y - 3} + \frac{7}{4 y - 3}$
The fractions have a common denominator, so add the numerators and place the sum over the common denominator.	$\frac{3 y + 7}{4 y - 3}$

The numerator and denominator cannot be factored. The fraction is simplified.

Add: $\frac{7 x + 12}{x + 3} + \frac{x^{2}}{x + 3} .$

Solution

Step-by-step solution for adding and simplifying rational expressions.
	$\frac{7 x + 12}{x + 3} + \frac{x^{2}}{x + 3}$
The fractions have a common denominator, so add the numerators and place the sum over the common denominator.	$\frac{7 x + 12 + x^{2}}{x + 3}$
Write the degrees in descending order.	$\frac{x^{2} + 7 x + 12}{x + 3}$
Factor the numerator.	$\frac{(x + 3) (x + 4)}{x + 3}$
Simplify by removing common factors.	$\frac{(x + 3) (x + 4)}{x + 3}$
Simplify.	$x + 4$

Subtract Rational Expressions with a Common Denominator

To subtract rational expressions, they must also have a common denominator. When the denominators are the same, you subtract the numerators and place the difference over the common denominator.

We always simplify rational expressions. Be sure to factor, if possible, after you subtract the numerators so you can identify any common factors.

Subtract: $\frac{n^{2}}{n - 10} - \frac{100}{n - 10} .$

Solution

Step-by-step simplification of a rational algebraic expression using common denominators and factoring.
	$\frac{n^{2}}{n - 10} - \frac{100}{n - 10}$
The fractions have a common denominator, so subtract the numerators and place the difference over the common denominator.	$\frac{n^{2} - 100}{n - 10}$
Factor the numerator.	$\frac{(n - 10) (n + 10)}{n - 10}$
Simplify by removing common factors.	$\frac{(n - 10) (n + 10)}{n - 10}$
Simplify.	$n + 10$

Be careful of the signs when you subtract a binomial!

Subtract: $\frac{y^{2}}{y - 6} - \frac{2 y + 24}{y - 6} .$

Solution

Step-by-step simplification of a rational algebraic expression involving subtraction and factorization.
	$\frac{y^{2}}{y - 6} - \frac{2 y + 24}{y - 6}$
The fractions have a common denominator, so subtract the numerators and place the difference over the common denominator.	$\frac{y^{2} - (2 y + 24)}{y - 6}$
Distribute the sign in the numerator.	$\frac{y^{2} - 2 y - 24}{y - 6}$
Factor the numerator.	$\frac{(y - 6) (y + 4)}{y - 6}$
Remove common factors.	$\frac{(y - 6) (y + 4)}{y - 6}$
Simplify.	$y + 4$

Subtract: $\frac{5 x^{2} - 7 x + 3}{x^{2} - 3 x - 18} - \frac{4 x^{2} + x - 9}{x^{2} - 3 x - 18} .$

Solution

This table illustrates the step-by-step process of subtracting two rational expressions, showing each mathematical transformation from initial problem to simplified solution.
	$\frac{5 x^{2} - 7 x + 3}{x^{2} - 3 x - 18} - \frac{4 x^{2} + x - 9}{x^{2} - 3 x - 18}$
Subtract the numerators and place the difference over the common denominator.	$\frac{5 x^{2} - 7 x + 3 - (4 x^{2} + x - 9)}{x^{2} - 3 x - 18}$
Distribute the sign in the numerator.	$\frac{5 x^{2} - 7 x + 3 - 4 x^{2} - x + 9}{x^{2} - 3 x - 18}$
Combine like terms.	$\frac{x^{2} - 8 x + 12}{x^{2} - 3 x - 18}$
Factor the numerator and the denominator.	$\frac{(x - 2) (x - 6)}{(x + 3) (x - 6)}$
Simplify by removing common factors.	$\frac{(x - 2) (x - 6)}{(x + 3) (x - 6)}$
Simplify.	$\frac{(x - 2)}{(x + 3)}$

Add and Subtract Rational Expressions whose Denominators are Opposites

When the denominators of two rational expressions are opposites, it is easy to get a common denominator. We just have to multiply one of the fractions by $\frac{−1}{−1}$ .

Let’s see how this works.

	$A mathematical expression showing the sum of two fractions: 7 over d plus 5 over negative d.$
Multiply the second fraction by $\frac{−1}{−1}$ .	$An algebraic expression showing the sum of two fractions. The first term is 7/d, and the second term is a fraction with a numerator of (-1) to the power of 5 and a denominator of (-1)(-d).$
The denominators are the same.	$The image shows the mathematical expression 7/d + -5/d, representing the addition of two fractions with a common denominator 'd'.$
Simplify.	$A mathematical expression displaying the fraction 2 over d.$

Add: $\frac{4 u - 1}{3 u - 1} + \frac{u}{1 - 3 u} .$

Solution

	$Algebraic expression: (4u-1)/(3u-1) + u/(1-3u). It's a sum of two fractions with denominators that are additive inverses of each other.$
The denominators are opposites, so multiply the second fraction by $\frac{−1}{−1}$ .	$A mathematical expression representing the sum of two algebraic fractions. The first term is (4u-1)/(3u-1), and the second is ((-1)u)/((-1)(1-3u)), illustrating a step in simplifying the expression.$
Simplify the second fraction.	$The image displays the sum of two algebraic fractions with a common denominator, written as (4u-1)/(3u-1) + (-u)/(3u-1).$
The denominators are the same. Add the numerators.	$A mathematical fraction displays '4u - 1 - u' as the numerator and '3u - 1' as the denominator, which simplifies to (3u - 1) / (3u - 1).$
Simplify.	$The mathematical expression showing the fraction (3u-1) divided by (3u-1). This expression simplifies to 1, assuming that (3u-1) is not equal to zero.$
Simplify.

Subtract: $\frac{m^{2} - 6 m}{m^{2} - 1} - \frac{3 m + 2}{1 - m^{2}} .$

Solution

	$A mathematical expression showing the subtraction of two algebraic fractions: (m^2 - 6m) / (m^2 - 1) - (3m + 2) / (1 - m^2).$
The denominators are opposites, so multiply the second fraction by $\frac{−1}{−1}$ .	$A mathematical expression displaying the subtraction of two algebraic fractions. The first fraction is (m^2 - 6m)/(m^2 - 1), and the second is -1(3m + 2)/-1(1 - m^2).$
Simplify the second fraction.	$A mathematical expression showing the subtraction of two algebraic fractions with a common denominator of m^2 - 1. The first fraction is (m^2 - 6m) / (m^2 - 1) and the second is (-3m - 2) / (m^2 - 1).$
The denominators are the same. Subtract the numerators.	$A mathematical expression showing the fraction (m^2 - 6m - (-3m - 2)) / (m^2 - 1).$
Distribute.	$\frac{m^{2} - 6 m + 3 m + 2}{m^{2} - 1}$
Combine like terms.	$A mathematical expression showing a fraction with the numerator m squared minus 3m plus 2, and the denominator m squared minus 1.$
Factor the numerator and denominator.	$A mathematical fraction displaying (m-1)(m-2) in the numerator and (m-1)(m+1) in the denominator, illustrating an algebraic expression suitable for simplification.$
Simplify by removing common factors.	$A fraction showing the expression (m-1)(m-2) divided by (m-1)(m+1), with the (m-1) terms crossed out in both the numerator and denominator, implying simplification to (m-2)/(m+1).$
Simplify.	$A mathematical fraction displays 'm minus 2' in the numerator and 'm plus 1' in the denominator, set against a plain white background.$

Key Concepts

Rational Expression Addition
- If $p, q, and r$ are polynomials where $r \neq 0$ , then
  
  $\frac{p}{r} + \frac{q}{r} = \frac{p + q}{r}$
- To add rational expressions with a common denominator, add the numerators and place the sum over the common denominator.
Rational Expression Subtraction
- If $p, q, and r$ are polynomials where $r \neq 0$ , then
  
  $\frac{p}{r} - \frac{q}{r} = \frac{p - q}{r}$
- To subtract rational expressions, subtract the numerators and place the difference over the common denominator.

Practice Makes Perfect

Add Rational Expressions with a Common Denominator

In the following exercises, add.

$\frac{2}{15} + \frac{7}{15}$

Solution

$\frac{3}{5}$

$\frac{4}{21} + \frac{3}{21}$

$\frac{7}{24} + \frac{11}{24}$

Solution

$\frac{3}{4}$

$\frac{7}{36} + \frac{13}{36}$

$\frac{3 a}{a - b} + \frac{1}{a - b}$

Solution

$\frac{3 a + 1}{a - b}$

$\frac{3 c}{4 c - 5} + \frac{5}{4 c - 5}$

$\frac{d}{d + 8} + \frac{5}{d + 8}$

Solution

$\frac{d + 5}{d + 8}$

$\frac{7 m}{2 m + n} + \frac{4}{2 m + n}$

$\frac{p^{2} + 10 p}{p + 2} + \frac{16}{p + 2}$

Solution

$p + 8$

$\frac{q^{2} + 12 q}{q + 3} + \frac{27}{q + 3}$

$\frac{2 r^{2}}{2 r - 1} + \frac{15 r - 8}{2 r - 1}$

Solution

$r + 8$

$\frac{3 s^{2}}{3 s - 2} + \frac{13 s - 10}{3 s - 2}$

$\frac{8 t^{2}}{t + 4} + \frac{32 t}{t + 4}$

Solution

$8 t$

$\frac{6 v^{2}}{v + 5} + \frac{30 v}{v + 5}$

$\frac{2 w^{2}}{w^{2} - 16} + \frac{8 w}{w^{2} - 16}$

Solution

$\frac{2 w}{w - 4}$

$\frac{7 x^{2}}{x^{2} - 9} + \frac{21 x}{x^{2} - 9}$

Subtract Rational Expressions with a Common Denominator

In the following exercises, subtract.

$\frac{y^{2}}{y + 8} - \frac{64}{y + 8}$

Solution

$y - 8$

$\frac{z^{2}}{z + 2} - \frac{4}{z + 2}$

$\frac{9 a^{2}}{3 a - 7} - \frac{49}{3 a - 7}$

Solution

$3 a + 7$

$\frac{25 b^{2}}{5 b - 6} - \frac{36}{5 b - 6}$

$\frac{c^{2}}{c - 8} - \frac{6 c + 16}{c - 8}$

Solution

$c + 2$

$\frac{d^{2}}{d - 9} - \frac{6 d + 27}{d - 9}$

$\frac{3 m^{2}}{6 m - 30} - \frac{21 m - 30}{6 m - 30}$

Solution

$\frac{m - 2}{2}$

$\frac{2 n^{2}}{4 n - 32} - \frac{18 n - 16}{4 n - 32}$

$\frac{6 p^{2} + 3 p + 4}{p^{2} + 4 p - 5} - \frac{5 p^{2} + p + 7}{p^{2} + 4 p - 5}$

Solution

$\frac{p + 3}{p + 5}$

$\frac{5 q^{2} + 3 q - 9}{q^{2} + 6 q + 8} - \frac{4 q^{2} + 9 q + 7}{q^{2} + 6 q + 8}$

$\frac{5 r^{2} + 7 r - 33}{r^{2} - 49} - \frac{4 r^{2} + 5 r + 30}{r^{2} - 49}$

Solution

$\frac{r + 9}{r + 7}$

$\frac{7 t^{2} - t - 4}{t^{2} - 25} - \frac{6 t^{2} + 12 t - 44}{t^{2} - 25}$

Add and Subtract Rational Expressions whose Denominators are Opposites

In the following exercises, add.

$\frac{10 v}{2 v - 1} + \frac{2 v + 4}{1 - 2 v}$

Solution

$4$

$\frac{20 w}{5 w - 2} + \frac{5 w + 6}{2 - 5 w}$

$\frac{10 x^{2} + 16 x - 7}{8 x - 3} + \frac{2 x^{2} + 3 x - 1}{3 - 8 x}$

Solution

$x + 2$

$\frac{6 y^{2} + 2 y - 11}{3 y - 7} + \frac{3 y^{2} - 3 y + 17}{7 - 3 y}$

In the following exercises, subtract.

$\frac{z^{2} + 6 z}{z^{2} - 25} - \frac{3 z + 20}{25 - z^{2}}$

Solution

$\frac{z + 4}{z - 5}$

$\frac{a^{2} + 3 a}{a^{2} - 9} - \frac{3 a - 27}{9 - a^{2}}$

$\frac{2 b^{2} + 30 b - 13}{b^{2} - 49} - \frac{2 b^{2} - 5 b - 8}{49 - b^{2}}$

Solution

$\frac{4 b - 3}{b - 7}$

$\frac{c^{2} + 5 c - 10}{c^{2} - 16} - \frac{c^{2} - 8 c - 10}{16 - c^{2}}$

Everyday Math

Sarah ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. If $r$ represents Sarah’s speed when she ran, then her running time is modeled by the expression $\frac{8}{r}$ and her biking time is modeled by the expression $\frac{24}{r + 4} .$ Add the rational expressions $\frac{8}{r} + \frac{24}{r + 4}$ to get an expression for the total amount of time Sarah ran and biked.

Solution

$\frac{32 (r + 1)}{r (r + 4)}$

If Pete can paint a wall in $p$ hours, then in one hour he can paint $\frac{1}{p}$ of the wall. It would take Penelope 3 hours longer than Pete to paint the wall, so in one hour she can paint $\frac{1}{p + 3}$ of the wall. Add the rational expressions $\frac{1}{p} + \frac{1}{p + 3}$ to get an expression for the part of the wall Pete and Penelope would paint in one hour if they worked together.

Writing Exercises

Donald thinks that $\frac{3}{x} + \frac{4}{x}$ is $\frac{7}{2 x} .$ Is Donald correct? Explain.

Explain how you find the Least Common Denominator of $x^{2} + 5 x + 4$ and $x^{2} - 16 .$

Self Check

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

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

Add and Subtract Rational Expressions with Unlike Denominators

Find the Least Common Denominator of Rational Expressions

When we add or subtract rational expressions with unlike denominators we will need to get common denominators. If we review the procedure we used with numerical fractions, we will know what to do with rational expressions.

Let’s look at the example $\frac{7}{12} + \frac{5}{18}$ from Foundations. Since the denominators are not the same, the first step was to find the least common denominator (LCD). Remember, the LCD is the least common multiple of the denominators. It is the smallest number we can use as a common denominator.

To find the LCD of 12 and 18, we factored each number into primes, lining up any common primes in columns. Then we “brought down” one prime from each column. Finally, we multiplied the factors to find the LCD.

\frac{\begin{matrix} 12 = 2 \cdot 2 \cdot 3 \\ 18 = 2 \cdot 3 \cdot 3 \end{matrix}}{\begin{matrix} LCD = 2 \cdot 2 \cdot 3 \cdot 3 \\ LCD = 36 \end{matrix}}

We do the same thing for rational expressions. However, we leave the LCD in factored form.

Find the LCD for $\frac{8}{x^{2} - 2 x - 3}, \frac{3 x}{x^{2} + 4 x + 3}$ .

Solution

Steps to find the Least Common Denominator (LCD) for given rational expressions by factoring their denominators.
	$Find the LCD for \frac{8}{x^{2} - 2 x - 3}, \frac{3 x}{x^{2} + 4 x + 3} .$
Factor each expression completely, lining up common factors. Bring down the columns.	$\frac{\begin{matrix} x^{2} - 2 x - 3 = (x + 1) (x - 3) \\ x^{2} + 4 x + 3 = (x + 1) (x + 3) \end{matrix}}{LCD = (x + 1) (x - 3) (x + 3)}$
Multiply the factors.	$The LCD is (x + 1) (x - 3) (x + 3) .$

Find Equivalent Rational Expressions

When we add numerical fractions, once we find the LCD, we rewrite each fraction as an equivalent fraction with the LCD.

$The above image shows how to find the LCD (least common denominator) when adding numerical fractions in the example seven-twelfths plus five-eighteenths. The image shows 7 times 3 divided by 12 times 3 plus 5 times 2 plus 18 times 2. Below this is 21 divided by 36 plus 10 divided by 36. The image next to this shows that 12 equals 2 times 2 times 3. Below this shows 18 equals 2 times 3 times 3. A line is drawn. Below it is LCD equals 2 times 2 times 3 times 3. The line below this shows that the LCD equals 36.$

We will do the same thing for rational expressions.

Rewrite as equivalent rational expressions with denominator $(x + 1) (x - 3) (x + 3)$ : $\frac{8}{x^{2} - 2 x - 3}, \frac{3 x}{x^{2} + 4 x + 3} .$

Solution

	$Two algebraic fractions are shown: 8 over (x^2 - 2x - 3) and 3x over (x^2 + 4x + 3).$
Factor each denominator.	$Two algebraic fractions are shown: 8 over (x+1)(x-3) and 3x over (x+1)(x+3).$
Find the LCD.
Multiply each denominator by the 'missing' factor and multiply each numerator by the same factor.
Simplify the numerators.	$Two algebraic fractions displayed. The first is 8x+24 over (x+1)(x-3)(x+3), and the second is 3x^2-9x over (x+1)(x+3)(x-3). Both fractions have common denominators.$

Add Rational Expressions with Different Denominators

Now we have all the steps we need to add rational expressions with different denominators. As we have done previously, we will do one example of adding numerical fractions first.

Add: $\frac{7}{12} + \frac{5}{18} .$

Solution

	$The image shows the fraction addition problem '7 over 12 plus 5 over 18'.$
Find the LCD of 12 and 18.
Rewrite each fraction as an equivalent fraction with the LCD.	$A math problem displaying the sum of two fractions: (7 * 3) / (12 * 3) + (5 * 2) / (18 * 2). The numbers 3 and 2 are highlighted in red, suggesting multiplication to achieve common denominators.$
Add the fractions.	$A mathematical expression showing the sum of two fractions with a common denominator: 21/36 + 10/36.$
The fraction cannot be simplified.	$A fraction is displayed with 31 as the numerator, a horizontal line, and 36 as the denominator.$

Now we will add rational expressions whose denominators are monomials.

Add: $\frac{5}{12 x^{2} y} + \frac{4}{21 x y^{2}} .$

Solution

		$An algebraic expression showing the sum of two fractions: 5 over 12x squared y, plus 4 over 21xy squared.$
Find the LCD of $12 x^{2} y$ and $21 x y^{2}$ .
		$A mathematical expression showing the sum of two algebraic fractions: 5/(12x^2y) + 4/(21xy^2).$
Rewrite each rational expression as an equivalent fraction with the LCD.		$An algebraic expression displaying the addition of two rational terms: (5 * 7y) / (12x^2y * 7y) and (4 * 4x) / (21xy^2 * 4x), illustrating a step in finding a common denominator for two fractions.$
Simplify.		$A mathematical expression showing the sum of two fractions: 35y over 84x^2y^2 plus 16x over 84x^2y^2. Both fractions share a common denominator.$
Add the rational expressions.		$A fraction with the numerator 16x + 35y and the denominator 84x^2y^2 is displayed on a white background, representing a mathematical algebraic expression.$
There are no factors common to the numerator and denominator. The fraction cannot be simplified.

Now we are ready to tackle polynomial denominators.

How to Add Rational Expressions with Different Denominators

Add: $\frac{3}{x - 3} + \frac{2}{x - 2} .$

Solution

$The above image shows the steps to add fractions whose denominators are monomials for the example 5 divided by 12 x squared y plus 4 divided by 21 x y squared. Find the LCD of 12 x squared y and 21 x y squared. To the right of this expression is 12 x squared y equals 2 times 2 times 3 times x times x times y. Below that is 21 x y squared equals 3 times 7 times x times y times y. A line is drawn. Below that is LCD equals 2 times 2 times 3 times 7 times x times x times y times y. Below that is LCD equals 84 x squared y squared. Rewrite each rational expression as an equivalent fraction with the LCD. The original equation is shown. Below that is 5 times 7 y divided by 12 x squared y times 7 y plus 4 times 4 x divided by 21 x y squared times 4 x. Simplify to get 35 y divided by 84 x squared y squared plus 16 x divided by x squared y squared. Add the rational expressions 16 x plus 35 y divided by 84 x squared y squared. There are no factors common to the numeration and denominator. The fraction cannot be simplified.$ Step 2 is to add the rational expression. Then, add the numerators and place the sum over the common denominator to get 3 x minus 6 plus 2 x minus 6 divided by x minus 3 times x minus 2.

Step 2 is to add the rational expression. Then, add the numerators and place the sum over the common denominator to get 3 x minus 6 plus 2 x minus 6 divided by x minus 3 times x minus 2.

Step 3 is to simplify, if possible. Because 5 x minus 12 cannot be factored, the answer is simplified to 5 x minus 12 divided by x minus 3 times x minus 2.

The steps to use to add rational expressions are summarized in the following procedure box.

Add: $\frac{2 a}{2 a b + b^{2}} + \frac{3 a}{4 a^{2} - b^{2}} .$

Solution

	$A mathematical expression showing the sum of two fractions: 2a divided by (2ab + b^2) plus 3a divided by (4a^2 - b^2).$
Do the expressions have a common denominator? No. Rewrite each expression with the LCD.
Find the LCD.
Rewrite each rational expression as an equivalent rational expression with the LCD.	$An algebraic expression displaying the sum of two fractions with a common denominator, prior to simplification. Numerators are 2a(2a-b) and 3ab, while the common denominator is b(2a+b)(2a-b).$
Simplify the numerators.	$An algebraic expression showing the sum of two fractions with a common denominator of b(2a+b)(2a-b). The first numerator is 4a^2-2ab, and the second numerator is 3ab.$
Add the rational expressions.	$A mathematical expression showing a fraction with 4a^2 - 2ab + 3ab in the numerator and b(2a + b)(2a - b) in the denominator.$
Simplify the numerator.	$A mathematical fraction is shown, with the numerator as 4a^2 + ab and the denominator as b(2a + b)(2a - b).$
Factor the numerator.	$A mathematical fraction with the numerator a(4a + b) and the denominator b(2a + b)(2a - b).$
There are no factors common to the numerator and denominator. The fraction cannot be simplified.

Avoid the temptation to simplify too soon! In the example above, we must leave the first rational expression as $\frac{2 a (2 a - b)}{b (2 a + b) (2 a - b)}$ to be able to add it to $\frac{3 a \cdot b}{(2 a + b) (2 a - b) \cdot b}$ . Simplify only after you have combined the numerators.

Add: $\frac{8}{x^{2} - 2 x - 3} + \frac{3 x}{x^{2} + 4 x + 3} .$

Solution

	$An algebraic expression featuring the sum of two rational fractions: 8/(x^2 - 2x - 3) + 3x/(x^2 + 4x + 3).$
Do the expressions have a common denominator? No. Rewrite each expression with the LCD.
Find the LCD.
Rewrite each rational expression as an equivalent fraction with the LCD.	$An algebraic expression showing the sum of two rational fractions, where common factors in the numerators and denominators are highlighted in red, indicating potential simplification steps.$
Simplify the numerators.
Add the rational expressions.	$A mathematical fraction displaying the expression (3x^2 - x + 24) divided by (x + 1)(x - 3)(x + 3).$
Simplify the numerator.	$A mathematical fraction displaying the expression (3x^2 - x + 24) divided by (x + 1)(x - 3)(x + 3).$
The numerator is prime, so there are no common factors.

Subtract Rational Expressions with Different Denominators

The process we use to subtract rational expressions with different denominators is the same as for addition. We just have to be very careful of the signs when subtracting the numerators.

How to Subtract Rational Expressions with Different Denominators

Subtract: $\frac{x}{x - 3} - \frac{x - 2}{x + 3} .$

Solution

The above image has 3 columns. It shows the steps on how to subtract rational expressions with different denominators for x divided by x minus three minus x plus x minus 3. Step 1 is to Determine if the expressions have a common denominator. Yes – go to step 2. No – Rewrite each rational expression with the LCD. Find the LCD. Rewrite each rational expression as an equivalent rational expression with the LCD. In the above expression, the answer is no. Find the LCD of x minus 3, x plus 3. To the right of this is x – 3: x – 3. Below that is x – 2: x – 2. A line is drawn. Below that is written the LCD is x – 3 times x plus 3. Rewrite as x times x plus 3 divided by x minus 3 times x plus 3 minus x minus 2 times x minus 3 divided by x plus 3 times x minus 3. Keep the denominators factored! Factor to get x squared plus 3 x divided by x minus 3 times x plus 3 minus x squared minus 5 x plus 6 divided by x minus 3 times x plus 3.

Step 2 is to subtract the rational expressions. Subtract the numerators and place the difference over the common denominator to get x 2 plus 3 x minus x squared minus 5 x plus 6 divided by x minus 3 times x plus 3. Then to x squared plus 3 x minus x squared plus 5 x minus 6 divided by x minus 3 times x plus 3. Be careful with the signs! Then to 8 x minus 6 divided by x minus 3 times x plus 3.

Step 3 is to simplify, if possible. The numerator and denominator have no factors in common. The answer is simplified to 2 times 4 x minus 3 divided by x minus 3 times x plus 3.

The steps to take to subtract rational expressions are listed below.

Subtract: $\frac{8 y}{y^{2} - 16} - \frac{4}{y - 4} .$

Solution

	$A mathematical expression showing the subtraction of two algebraic fractions: 8y divided by the quantity y squared minus 16, minus 4 divided by the quantity y minus 4.$
Do the expressions have a common denominator? No. Rewrite each expression with the LCD.
Find the LCD.
Rewrite each rational expression as an equivalent rational expression with the LCD.	$A mathematical expression showing the subtraction of two fractions with a common denominator of (y-4)(y+4). The numerators are 8y and 4(y+4), with parts of the second fraction highlighted in red.$
Simplify the numerators.	$An algebraic expression showing the subtraction of two fractions with a common denominator of (y-4)(y+4). The numerators are 8y and 4y+16.$
Subtract the rational expressions.	$An algebraic expression shown as a fraction with 8y - 4y - 16 in the numerator and (y - 4)(y + 4) in the denominator.$
Simplify the numerators.	$A mathematical expression showing the fraction (4y - 16) / ((y - 4)(y + 4)).$
Factor the numerator to look for common factors.	$A mathematical expression showing the fraction 4(y-4) over (y-4)(y+4).$
Remove common factors.	$A mathematical expression showing the fraction 4(y-4) over (y-4)(y+4), with the common factor (y-4) in both the numerator and denominator crossed out, indicating cancellation.$
Simplify.	$A mathematical expression shows the fraction 4 over (y + 4).$

There are lots of negative signs in the next example. Be extra careful!

Subtract: $\frac{−3 n - 9}{n^{2} + n - 6} - \frac{n + 3}{2 - n} .$

Solution


Factor the denominator.
Since $n - 2$ and $2 - n$ are opposites, we will mutliply the second rational expression by $\frac{−1}{−1}$ .
Simplify.
Do the expressions have a common denominator? No.
Find the LCD.
Rewrite each rational expression as an equivalent rational expression with the LCD.	$An algebraic expression showing the sum of two fractions with a common denominator of (n-2)(n+3). The numerators are (-3n - 9) and (n+3)(n+3), with an (n+3) term highlighted in red.$
Simplify the numerators.	$A mathematical expression showing the addition of two rational algebraic expressions. Both fractions have a common denominator of (n-2)(n+3). The numerators are -3n-9 and n^2+6n+9 respectively.$
Simplify the rational expressions.	$A mathematical fraction. The numerator is -3n - 9 + n^2 + 6n + 9, and the denominator is the product of (n-2) and (n+3).$
Simplify the numerator.	$A mathematical expression showing a fraction with n squared plus 3n in the numerator and the product of (n minus 2) and (n plus 3) in the denominator.$
Factor the numerator to look for common factors.	$A mathematical expression showing the fraction n(p+3) / ((n-2)(p+3)), where the (p+3) terms in both the numerator and denominator are crossed out, indicating they are being canceled.$
Simplify.	$A mathematical expression displaying the fraction n over (n-2).$

When one expression is not in fraction form, we can write it as a fraction with denominator 1.

Subtract: $\frac{5 c + 4}{c - 2} - 3 .$

Solution

	$A mathematical expression displays a fraction '5c + 4' over 'c - 2', followed by a subtraction of 3. The expression is (5c + 4)/(c - 2) - 3.$
Write $3$ as $\frac{3}{1}$ to have 2 rational expressions.	$A mathematical expression showing the subtraction of two fractions: (5c + 4) / (c - 2) - 3/1.$
Do the rational expressions have a common denominator? No.
Find the LCD of $c - 2$ and $1 .$ LCD = $c - 2 .$
Rewrite $\frac{3}{1}$ as an equivalent rational expression with the LCD.	$A mathematical expression displaying the subtraction of two algebraic fractions. The first term is (5c+4)/(c-2), and the second term is 3(c-2) over 1(c-2), with factors in red.$
Simplify.	$A mathematical expression showing the subtraction of two fractions with a common denominator (c-2): (5c+4)/(c-2) - (3c-6)/(c-2).$
Subtract the rational expressions.	$A mathematical expression showing the fraction (5c + 4 - (3c - 6)) divided by (c - 2).$
Simplify.	$A fraction with a numerator of 2c + 10 and a denominator of c - 2 is shown on a white background.$
Factor to check for common factors.	$A fraction with 2(c + 5) in the numerator and c - 2 in the denominator.$
There are no common factors; the rational expression is simplified.

We follow the same steps as before to find the LCD when we have more than two rational expressions. In the next example we will start by factoring all three denominators to find their LCD.

Simplify: $\frac{2 u}{u - 1} + \frac{1}{u} - \frac{2 u - 1}{u^{2} - u} .$

Solution


Do the rational expressions have a common denominator? No.
Find the LCD.
Rewrite each rational expression as an equivalent rational expression with the LCD.	$A multi-term mathematical expression involving fractions with the variable 'u'. It shows three rational terms being added and subtracted, all sharing a common denominator of u(u-1) or (u-1)u.$

Write as one rational expression.	$A mathematical expression displaying a fraction. The numerator is 2u^2 + u - 1 - 2u + 1, and the denominator is u(u - 1).$
Simplify.	$A mathematical expression displaying the fraction 2u squared minus u over u multiplied by the quantity u minus 1, all in black font against a white background.$
Factor the numerator, and remove common factors.	$A mathematical fraction with mu(2u-1) in the numerator and mu(u-1) in the denominator.$
Simplify.	$A mathematical expression displaying the fraction (2u-1) divided by (u-1).$

Key Concepts

Find the Least Common Denominator of Rational Expressions
1. Factor each expression completely.
2. List the factors of each expression. Match factors vertically when possible.
3. Bring down the columns.
4. Multiply the factors.
Add or Subtract Rational Expressions
1. Determine if the expressions have a common denominator.
  Yes – go to step 2.
  No – Rewrite each rational expression with the LCD.
  - Find the LCD.
  - Rewrite each rational expression as an equivalent rational expression with the LCD.
2. Add or subtract the rational expressions.
3. Simplify, if possible.

Practice Makes Perfect

In the following exercises, find the LCD.

$\frac{5}{x^{2} - 2 x - 8}, \frac{2 x}{x^{2} - x - 12}$

Solution

$(x - 4) (x + 2) (x + 3)$

$\frac{8}{y^{2} + 12 y + 35}, \frac{3 y}{y^{2} + y - 42}$

$\frac{9}{z^{2} + 2 z - 8}, \frac{4 z}{z^{2} - 4}$

Solution

$(z - 2) (z + 4) (z + 2)$

$\frac{6}{a^{2} + 14 a + 45}, \frac{5 a}{a^{2} - 81}$

$\frac{4}{b^{2} + 6 b + 9}, \frac{2 b}{b^{2} - 2 b - 15}$

Solution

$(b + 3) (b + 3) (b - 5)$

$\frac{5}{c^{2} - 4 c + 4}, \frac{3 c}{c^{2} - 10 c + 16}$

$\frac{2}{3 d^{2} + 14 d - 5}, \frac{5 d}{3 d^{2} - 19 d + 6}$

Solution

$(3 d - 1) (d + 5) (d - 6)$

$\frac{3}{5 m^{2} - 3 m - 2}, \frac{6 m}{5 m^{2} + 17 m + 6}$

In the following exercises, write as equivalent rational expressions with the given LCD.

$\frac{5}{x^{2} - 2 x - 8}, \frac{2 x}{x^{2} - x - 12}$
LCD $(x - 4) (x + 2) (x + 3)$

Solution

$\frac{5 x + 15}{(x - 4) (x + 2) (x + 3)},$
$\frac{2 x^{2} + 4 x}{(x - 4) (x + 2) (x + 3)}$

$\frac{8}{y^{2} + 12 y + 35}, \frac{3 y}{y^{2} + y - 42}$
LCD $(y + 7) (y + 5) (y - 6)$

$\frac{9}{z^{2} + 2 z - 8}, \frac{4 z}{z^{2} - 4}$
LCD $(z - 2) (z + 4) (z + 2)$

Solution

$\frac{9 z + 18}{(z - 2) (z + 4) (z + 2)},$
$\frac{4 z^{2} + 16 z}{(z - 2) (z + 4) (z + 2)}$

$\frac{6}{a^{2} + 14 a + 45}, \frac{5 a}{a^{2} - 81}$
LCD $(a + 9) (a + 5) (a - 9)$

$\frac{4}{b^{2} + 6 b + 9}, \frac{2 b}{b^{2} - 2 b - 15}$
LCD $(b + 3) (b + 3) (b - 5)$

Solution

$\frac{4 b - 20}{(b + 3) (b + 3) (b - 5)},$
$\frac{2 b^{2} + 6 b}{(b + 3) (b + 3) (b - 5)}$

$\frac{5}{c^{2} - 4 c + 4}, \frac{3 c}{c^{2} - 10 c + 16}$
LCD $(c - 2) (c - 2) (c - 8)$

$\frac{2}{3 d^{2} + 14 d - 5}, \frac{5 d}{3 d^{2} - 19 d + 6}$
LCD $(3 d - 1) (d + 5) (d - 6)$

Solution

$\frac{2 d - 12}{(3 d - 1) (d + 5) (d - 6)},$
$\frac{5 d^{2} + 25 d}{(3 d - 1) (d + 5) (d - 6)}$

$\frac{3}{5 m^{2} - 3 m - 2}, \frac{6 m}{5 m^{2} + 17 m + 6}$
LCD $(5 m + 2) (m - 1) (m + 3)$

In the following exercises, add.

$\frac{5}{24} + \frac{11}{36}$

Solution

$\frac{37}{72}$

$\frac{7}{30} + \frac{13}{45}$

$\frac{9}{20} + \frac{11}{30}$

Solution

$\frac{49}{60}$

$\frac{8}{27} + \frac{7}{18}$

$\frac{7}{10 x^{2} y} + \frac{4}{15 x y^{2}}$

Solution

$\frac{21 y + 8 x}{30 x^{2} y^{2}}$

$\frac{1}{12 a^{3} b^{2}} + \frac{5}{9 a^{2} b^{3}}$

$\frac{1}{2 m} + \frac{7}{8 m^{2} n}$

Solution

$\frac{4 m n + 7}{8 m^{2} n}$

$\frac{5}{6 p^{2} q} + \frac{1}{4 p}$

$\frac{3}{r + 4} + \frac{2}{r - 5}$

Solution

$\frac{5 r - 7}{(r + 4) (r - 5)}$

$\frac{4}{s - 7} + \frac{5}{s + 3}$

$\frac{8}{t + 5} + \frac{6}{t - 5}$

Solution

$\frac{14 t - 10}{(t + 5) (t - 5)}$

$\frac{7}{v + 5} + \frac{9}{v - 5}$

$\frac{5}{3 w - 2} + \frac{2}{w + 1}$

Solution

$\frac{11 w + 1}{(3 w - 2) (w + 1)}$

$\frac{4}{2 x + 5} + \frac{2}{x - 1}$

$\frac{2 y}{y + 3} + \frac{3}{y - 1}$

Solution

$\frac{2 y^{2} + y + 9}{(y + 3) (y - 1)}$

$\frac{3 z}{z - 2} + \frac{1}{z + 5}$

$\frac{5 b}{a^{2} b - 2 a^{2}} + \frac{2 b}{b^{2} - 4}$

Solution

$\frac{b (5 b + 10 + 2 a^{2})}{a^{2} (b - 2) (b + 2)}$

$\frac{4}{c d + 3 c} + \frac{1}{d^{2} - 9}$

$\frac{2 m}{3 m - 3} + \frac{5 m}{m^{2} + 3 m - 4}$

Solution

$\frac{2 m^{2} + 23 m}{3 (m - 1) (m + 4)}$

$\frac{3}{4 n + 4} + \frac{6}{n^{2} - n - 2}$

$\frac{3}{n^{2} + 3 n - 18} + \frac{4 n}{n^{2} + 8 n + 12}$

Solution

$\frac{4 n^{2} - 9 n + 6}{(n - 3) (n + 6) (n + 2)}$

$\frac{6}{q^{2} - 3 q - 10} + \frac{5 q}{q^{2} - 8 q + 15}$

$\frac{3 r}{r^{2} + 7 r + 6} + \frac{9}{r^{2} + 4 r + 3}$

Solution

$\frac{3 (r^{2} + 6 r + 18)}{(r + 1) (r + 6) (r + 3)}$

$\frac{2 s}{s^{2} + 2 s - 8} + \frac{4}{s^{2} + 3 s - 10}$

In the following exercises, subtract.

$\frac{t}{t - 6} - \frac{t - 2}{t + 6}$

Solution

$\frac{2 (7 t - 6)}{(t - 6) (t + 6)}$

$\frac{v}{v - 3} - \frac{v - 6}{v + 1}$

$\frac{w + 2}{w + 4} - \frac{w}{w - 2}$

Solution

$\frac{−4 (1 + w)}{(w + 4) (w - 2)}$

$\frac{x - 3}{x + 6} - \frac{x}{x + 3}$

$\frac{y - 4}{y + 1} - \frac{1}{y + 7}$

Solution

$\frac{y^{2} + 2 y - 29}{(y + 1) (y + 7)}$

$\frac{z + 8}{z - 3} - \frac{z}{z - 2}$

$\frac{5 a}{a + 3} - \frac{a + 2}{a + 6}$

Solution

$\frac{4 a^{2} + 25 a - 6}{(a + 3) (a + 6)}$

$\frac{3 b}{b - 2} - \frac{b - 6}{b - 8}$

$\frac{6 c}{c^{2} - 25} - \frac{3}{c + 5}$

Solution

$\frac{3}{c - 5}$

$\frac{4 d}{d^{2} - 81} - \frac{2}{d + 9}$

$\frac{6}{m + 6} - \frac{12 m}{m^{2} - 36}$

Solution

$\frac{−6}{m - 6}$

$\frac{4}{n + 4} - \frac{8 n}{n^{2} - 16}$

$\frac{−9 p - 17}{p^{2} - 4 p - 21} - \frac{p + 1}{7 - p}$

Solution

$\frac{p + 2}{p + 3}$

$\frac{- 13 q - 8}{q^{2} + 2 q - 24} - \frac{q + 2}{4 - q}$

$\frac{−2 r - 16}{r^{2} + 6 r - 16} - \frac{5}{2 - r}$

Solution

$\frac{3}{r - 2}$

$\frac{2 t - 30}{t^{2} + 6 t - 27} - \frac{2}{3 - t}$

$\frac{5 v - 2}{v + 3} - 4$

Solution

$\frac{v - 14}{v + 3}$

$\frac{6 w + 5}{w - 1} + 2$

$\frac{2 x + 7}{10 x - 1} + 3$

Solution

$\frac{4 (8 x + 1)}{10 x - 1}$

$\frac{8 y - 4}{5 y + 2} - 6$

In the following exercises, add and subtract.

$\frac{5 a}{a - 2} + \frac{9}{a} - \frac{2 a + 18}{a^{2} - 2 a}$

Solution

$\frac{5 a^{2} + 7 a - 36}{a (a - 2)}$

$\frac{2 b}{b - 5} + \frac{3}{2 b} - \frac{2 b - 15}{2 b^{2} - 10 b}$

$\frac{c}{c + 2} + \frac{5}{c - 2} - \frac{10 c}{c^{2} - 4}$

Solution

$\frac{c - 5}{c + 2}$

$\frac{6 d}{d - 5} + \frac{1}{d + 4} - \frac{7 d - 5}{d^{2} - d - 20}$

In the following exercises, simplify.

$\frac{6 a}{3 a b + b^{2}} + \frac{3 a}{9 a^{2} - b^{2}}$

Solution

$\frac{3 a (6 a - b)}{b (3 a + b) (3 a - b)}$

$\frac{2 c}{2 c + 10} + \frac{7 c}{c^{2} + 9 c + 20}$

$\frac{6 d}{d^{2} - 64} - \frac{3}{d - 8}$

Solution

$\frac{3}{d + 8}$

$\frac{5}{n + 7} - \frac{10 n}{n^{2} - 49}$

$\frac{4 m}{m^{2} + 6 m - 7} + \frac{2}{m^{2} + 10 m + 21}$

Solution

$\frac{2 (2 m^{2} + 7 m - 1)}{(m + 7) (m - 1) (m + 3)}$

$\frac{3 p}{p^{2} + 4 p - 12} + \frac{1}{p^{2} + p - 30}$

$\frac{−5 n - 5}{n^{2} + n - 6} + \frac{n + 1}{2 - n}$

Solution

$- \frac{(n + 1) (n + 8)}{(n + 3) (n - 2)}$

$\frac{−4 b - 24}{b^{2} + b - 30} + \frac{b + 7}{5 - b}$

$\frac{7}{15 p} + \frac{5}{18 p q}$

Solution

$\frac{42 q + 25}{90 p q}$

$\frac{3}{20 a^{2}} + \frac{11}{12 a b^{2}}$

$\frac{4}{x - 2} + \frac{3}{x + 5}$

Solution

$\frac{7 (x + 2)}{(x - 2) (x + 5)}$

$\frac{6}{m + 4} + \frac{9}{m - 8}$

$\frac{2 q + 7}{q + 4} - 2$

Solution

$\frac{- 1}{q + 4}$

$\frac{3 y - 1}{y + 4} - 2$

$\frac{z + 2}{z - 5} - \frac{z}{z + 1}$

Solution

$\frac{2 (4 z + 1)}{(z - 5) (z + 1)}$

$\frac{t}{t - 5} - \frac{t - 1}{t + 5}$

$\frac{3 d}{d + 2} + \frac{4}{d} - \frac{d + 8}{d^{2} + 2 d}$

Solution

$\frac{3 (d + 1)}{d + 2}$

$\frac{2 q}{q + 5} + \frac{3}{q - 3} - \frac{13 q + 15}{q^{2} + 2 q - 15}$

Everyday Math

Decorating cupcakes Victoria can decorate an order of cupcakes for a wedding in $t$ hours, so in 1 hour she can decorate $\frac{1}{t}$ of the cupcakes. It would take her sister 3 hours longer to decorate the same order of cupcakes, so in 1 hour she can decorate $\frac{1}{t + 3}$ of the cupcakes.

ⓐ Find the fraction of the decorating job that Victoria and her sister, working together, would complete in one hour by adding the rational expressions $\frac{1}{t} + \frac{1}{t + 3} .$
ⓑ Evaluate your answer to part (a) when $t = 5 .$

Solution

ⓐ $\frac{2 t + 3}{t (t + 3)}$ ⓑ $\frac{13}{40}$

Kayaking When Trina kayaks upriver, it takes her $\frac{5}{3 - c}$ hours to go 5 miles, where $c$ is the speed of the river current. It takes her $\frac{5}{3 + c}$ hours to kayak 5 miles down the river.

ⓐ Find an expression for the number of hours it would take Trina to kayak 5 miles up the river and then return by adding $\frac{5}{3 - c} + \frac{5}{3 + c} .$
ⓑ Evaluate your answer to part (a) when $c = 1$ to find the number of hours it would take Trina if the speed of the river current is 1 mile per hour.

Writing Exercises

Felipe thinks $\frac{1}{x} + \frac{1}{y}$ is $\frac{2}{x + y} .$

ⓐ Choose numerical values for x and y and evaluate $\frac{1}{x} + \frac{1}{y} .$
ⓑ Evaluate $\frac{2}{x + y}$ for the same values of x and y you used in part (a).
ⓓ Find the correct expression for $\frac{1}{x} + \frac{1}{y} .$

Solution

Answers may vary.

Simplify the expression $\frac{4}{n^{2} + 6 n + 9} - \frac{1}{n^{2} - 9}$ and explain all your steps.

Self Check

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

ⓑ 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?

Simplify Complex Rational Expressions

Complex fractions are fractions in which the numerator or denominator contains a fraction. In Chapter 1 we simplified complex fractions like these:

\frac{\frac{3}{4}}{\frac{5}{8}} \frac{\frac{x}{2}}{\frac{x y}{6}}

In this section we will simplify complex rational expressions, which are rational expressions with rational expressions in the numerator or denominator.

Here are a few complex rational expressions:

\frac{\frac{4}{y - 3}}{\frac{8}{y^{2} - 9}} \frac{\frac{1}{x} + \frac{1}{y}}{\frac{x}{y} - \frac{y}{x}} \frac{\frac{2}{x + 6}}{\frac{4}{x - 6} - \frac{4}{x^{2} - 36}}

Remember, we always exclude values that would make any denominator zero.

We will use two methods to simplify complex rational expressions.

Simplify a Complex Rational Expression by Writing it as Division

We have already seen this complex rational expression earlier in this chapter.

\frac{\frac{6 x^{2} - 7 x + 2}{4 x - 8}}{\frac{2 x^{2} - 8 x + 3}{x^{2} - 5 x + 6}}

We noted that fraction bars tell us to divide, so rewrote it as the division problem

(\frac{6 x^{2} - 7 x + 2}{4 x - 8}) \div (\frac{2 x^{2} - 8 x + 3}{x^{2} - 5 x + 6})

Then we multiplied the first rational expression by the reciprocal of the second, just like we do when we divide two fractions.

This is one method to simplify rational expressions. We write it as if we were dividing two fractions.

Simplify: $\frac{\frac{4}{y - 3}}{\frac{8}{y^{2} - 9}} .$

Solution

Step-by-step process for simplifying a complex rational expression into its most reduced form.
	$\frac{\frac{4}{y - 3}}{\frac{8}{y^{2} - 9}}$
Rewrite the complex fraction as division.	$\frac{4}{y - 3} \div \frac{8}{y^{2} - 9}$
Rewrite as the product of first times the reciprocal of the second.	$\frac{4}{y - 3} \cdot \frac{y^{2} - 9}{8}$
Multiply.	$\frac{4 (y^{2} - 9)}{8 (y - 3)}$
Factor to look for common factors.	$\frac{4 (y - 3) (y + 3)}{4 \cdot 2 (y - 3)}$
Remove common factors.	$\frac{4 (y - 3) (y + 3)}{4 \cdot 2 (y - 3)}$
Simplify.	$\frac{y + 3}{2}$

Are there any value(s) of $y$ that should not be allowed? The simplified rational expression has just a constant in the denominator. But the original complex rational expression had denominators of $y - 3$ and $y^{2} - 9$ . This expression would be undefined if $y = 3$ or $y = −3$ .

Fraction bars act as grouping symbols. So to follow the Order of Operations, we simplify the numerator and denominator as much as possible before we can do the division.

Simplify: $\frac{\frac{1}{3} + \frac{1}{6}}{\frac{1}{2} - \frac{1}{3}} .$

Solution

	$A mathematical expression featuring a complex fraction. The numerator shows the sum of one-third and one-sixth, while the denominator displays the difference between one-half and one-third.$
Simplify the numerator and denominator.
Find the LCD and add the fractions in the numerator. Find the LCD and add the fractions in the denominator.	$A complex fraction showing addition and subtraction of fractions. Red numbers highlight the multiplication steps to find common denominators in both the numerator and denominator.$
Simplify the numerator and denominator.	$A complex fraction mathematical problem showing (2/6 + 1/6) divided by (3/6 - 2/6).$
Simplify the numerator and denominator, again.	$A mathematical expression showing the division of two fractions, (3/6) by (1/6), which simplifies to 3. This illustrates how to divide fractions with a common denominator.$
Rewrite the complex rational expression as a division problem.	$A mathematical expression showing the division of two fractions: three-sixths divided by one-sixth.$
Multiply the first times by the reciprocal of the second.	$A mathematical expression showing the multiplication of two fractions: 3/6 multiplied by 6/1.$
Simplify.

How to Simplify a Complex Rational Expression by Writing it as Division

Simplify: $\frac{\frac{1}{x} + \frac{1}{y}}{\frac{x}{y} - \frac{y}{x}} .$

Solution

$The above image has three columns. The image shows steps on how to divide complex rational expressions in three steps. Step one is to simplify the numerator and denominator. We will simplify the sum in the numerator and difference in the denominator for the example 1 divided by x plus 1 divided by y divided by x divided by y minus y divided by x. Find a common denominator and add the fractions in the numerator and find a common denominator and subtract the fractions in the numerator to get 1 times y divided by x times y plus 1 times x divided by y times x divided by x times x divided by y times x minus y times y divided by x times y. Then, we get y divided by x y plus x plus x y divided by x squared divided by x y minus y squared divided by x y. We now have just one rational expression in the numerator and one in the denominator, y plus x divided by x y divided by x squared minus y squared divided by x y.$ Step two is to rewrite the complex rational expression as a division problem. We write the numerator divided by the denominator.

Step two is to rewrite the complex rational expression as a division problem. We write the numerator divided by the denominator.

Step three is to divide the expressions. Multiply the first by the reciprocal of the second to get y plus x divided by x y times x y divided by x squared minus y squared. Factor any expressions if possible. We now have x y times y plus x divided by x y times x minus y times x plus y. Remove common factors. Cross out x, y and y plus x from the numerator. Cross out x, y and x plus y from the denominator. Simplify to get 1 divided by x minus y.

Simplify: $\frac{n - \frac{4 n}{n + 5}}{\frac{1}{n + 5} + \frac{1}{n - 5}} .$

Solution

	$A complex algebraic fraction involving variables 'n' and constants, with a numerator of (n - 4n/(n+5)) and a denominator of (1/(n+5) + 1/(n-5)).$
Simplify the numerator and denominator.
Find the LCD and add the fractions in the numerator. Find the LCD and add the fractions in the denominator.	$An algebraic expression presented as a complex fraction. The numerator involves the subtraction of two rational terms, and the denominator involves the addition of two rational terms, all featuring the variable 'n'.$
Simplify the numerators.	$A complex algebraic fraction involving the variable 'n'. The numerator is the subtraction of (n^2+5n)/(n+5) and 4n/(n+5), all divided by the sum of (n-5)/((n+5)(n-5)) and (n+5)/((n-5)(n+5)).$
Subtract the rational expressions in the numerator and add in the denominator. Simplify.	$A complex fraction with algebraic expressions, where the numerator is (n^2 + n)/(n + 5) and the denominator is 2n/((n + 5)(n - 5)).$
Rewrite as fraction division.	$The image displays a mathematical expression: the fraction (n^2 + n) / (n + 5) divided by the fraction 2n / ((n + 5)(n - 5)).$
Multiply the first times the reciprocal of the second.	$An algebraic expression showing the product of two fractions. The first fraction is (n^2 + n) divided by (n + 5). The second fraction is ((n + 5)(n - 5)) divided by (2n).$
Factor any expressions if possible.	$A mathematical expression showing a fraction with numerator n(n+1)(n+5)(n-5) and denominator (n+5)2n.$
Remove common factors.	$A fractional algebraic expression with factors n, (n+1), (n+5), and (n-5) in the numerator, and (n+5) and 2n in the denominator, featuring cancellation marks for (n+5) and n.$
Simplify.	$A mathematical expression showing the fraction (n+1)(n-5) divided by 2. The numerator consists of the product of two binomials, (n+1) and (n-5), and the denominator is the number 2.$

Simplify a Complex Rational Expression by Using the LCD

We “cleared” the fractions by multiplying by the LCD when we solved equations with fractions. We can use that strategy here to simplify complex rational expressions. We will multiply the numerator and denominator by LCD of all the rational expressions.

Let’s look at the complex rational expression we simplified one way in Example 2. We will simplify it here by multiplying the numerator and denominator by the LCD. When we multiply by $\frac{LCD}{LCD}$ we are multiplying by 1, so the value stays the same.

Simplify: $\frac{\frac{1}{3} + \frac{1}{6}}{\frac{1}{2} - \frac{1}{3}} .$

Solution

	$A complex fraction mathematical expression: the sum of one-third and one-sixth, divided by the difference between one-half and one-third. It represents a multi-step fraction calculation.$
The LCD of all the fractions in the whole expression is 6.
Clear the fractions by multiplying the numerator and denominator by that LCD.
Distribute.	$A complex fraction where the numerator is (61/3 + 61/6) and the denominator is (61/2 - 61/3), with the number 6 highlighted in red.$
Simplify.	$A mathematical fraction displays the expression (2 + 1) divided by (3 - 2) in black text on a white background.$
	$The image displays the fraction 3/1, with the number 3 positioned above a horizontal fraction bar and the number 1 directly below the bar, set against a plain white background.$

How to Simplify a Complex Rational Expression by Using the LCD

Simplify: $\frac{\frac{1}{x} + \frac{1}{y}}{\frac{x}{y} - \frac{y}{x}} .$

Solution

$The above image has 3 columns. It shows the steps on how to simplify a complex rational expression using the LCD for 1 divided by x plus 1 divided by y divided by x divided by y minus y divided by x. Step one is to find the LCD of all fractions in the complex rational expression. The LCD of all the fractions is x y. Multiply the numerator and denominator by the LCD.$ Step two is to multiply both the numerator and denominator by x y to get x y times 1 divided by x plus 1 divided by y divided x y times x divided by y minus y divided by x.

Step two is to multiply both the numerator and denominator by x y to get x y times 1 divided by x plus 1 divided by y divided x y times x divided by y minus y divided by x.

Step three is to simplify the expression. Distribute to get x y times 1 divided by x plus x y times 1 divided y divided by x y times x divided by y minus x y times y divided by x. Simplify to get y plus x divided by x squared minus y squared. Remove common factors. Cross out y plus x in the numerator. Cross out x plus y in the numerator. Simplify to get 1 divided by x minus y.

Be sure to start by factoring all the denominators so you can find the LCD.

Simplify: $\frac{\frac{2}{x + 6}}{\frac{4}{x - 6} - \frac{4}{x^{2} - 36}} .$

Solution

	$A complex algebraic fraction with 2/(x+6) in the numerator and a difference of two fractions, 4/(x-6) - 4/(x^2-36), in the denominator.$
Find the LCD of all fractions in the complex rational expression. The LCD is $(x + 6) (x - 6)$ .
Multiply the numerator and denominator by the LCD.	$A complex algebraic fraction demonstrating a step where both numerator and denominator are multiplied by the common factor (x+6)(x-6), highlighted in red, to simplify the expression.$
Simplify the expression.
Distribute in the denominator.	$A step in simplifying a rational expression, illustrating the multiplication of the numerator and denominator by the common factor (x+6)(x-6) to remove inner fractions.$
Simplify.	$Algebraic expression being simplified, demonstrating cancellation of common factors (x+6) and (x-6) in numerator and denominator to reduce the complex fraction.$
Simplify.	$A mathematical fraction is shown. The numerator is 2(x - 6) and the denominator is 4(x + 6) - 4.$
To simplify the denominator, distribute and combine like terms.	$A mathematical expression shown as a fraction with 2(x-6) in the numerator and 4x+20 in the denominator, set against a plain white background.$
Remove common factors.	$A mathematical expression showing a fraction with 2(x-6) in the numerator and 2(2x+10) in the denominator.$
Simplify.	$A mathematical expression showing the fraction (x-6) over (2x+10).$
Notice that there are no more factors common to the numerator and denominator.

Simplify: $\frac{\frac{4}{m^{2} - 7 m + 12}}{\frac{3}{m - 3} - \frac{2}{m - 4}} .$

Solution

	$A complex algebraic fraction. The numerator is 4, and the denominator is another fraction with m^2 - 7m + 12 as its numerator and (3/(m-3) - 2/(m-4)) as its denominator.$
Find the LCD of all fractions in the complex rational expression. The LCD is $(m - 3) (m - 4)$ .
Multiply the numerator and denominator by the LCD.	$A step in simplifying a complex algebraic fraction by multiplying by the common denominator (m-3)(m-4) to clear fractions in both the numerator and denominator.$
Simplify.
Simplify.	$A mathematical expression featuring a fraction with '4' as the numerator and '3(m-4) - 2(m-3)' as the denominator, suitable for algebraic simplification.$
Distribute.	$The image shows a mathematical fraction where the numerator is 4 and the denominator is 3m - 12 - 2m + 6.$
Combine like terms.	$A mathematical expression showing the fraction 4 over (m minus 6).$

Simplify: $\frac{\frac{y}{y + 1}}{1 + \frac{1}{y - 1}} .$

Solution

	$A complex fraction with the numerator as y divided by (y+1) and the denominator as 1 plus 1 divided by (y-1).$
Find the LCD of all fractions in the complex rational expression.
The LCD is $(y + 1) (y - 1)$ .
Multiply the numerator and denominator by the LCD.	$An algebraic fraction where both numerator and denominator are multiplied by the red terms (y+1)(y-1). The expression involves a fraction y/(y+1) in the numerator and (1 + 1/(y-1)) in the denominator.$
Distribute in the denominator and simplify.
Simplify.	$A mathematical expression showing a fraction. The numerator is (y-1)y. The denominator is (y+1)(y-1) + (y+1).$
Simplify the denominator, and leave the numerator factored.	$A mathematical fraction with y(y-1) in the numerator and y^2 - 1 + y + 1 in the denominator, set against a plain white background.$
	$A mathematical expression showing a fraction with y(y-1) in the numerator and y^2 + y in the denominator, written in black text on a white background.$
Factor the denominator, and remove factors common with the numerator.	$A mathematical expression showing the fraction x(y-1) over x(y+1), where the 'x' terms are crossed out, indicating simplification to (y-1) over (y+1).$
Simplify.	$A mathematical expression showing the fraction (y - 1) / (y + 1).$

Key Concepts

To Simplify a Rational Expression by Writing it as Division
1. Simplify the numerator and denominator.
2. Rewrite the complex rational expression as a division problem.
3. Divide the expressions.
To Simplify a Complex Rational Expression by Using the LCD
1. Find the LCD of all fractions in the complex rational expression.
2. Multiply the numerator and denominator by the LCD.
3. Simplify the expression.

Practice Makes Perfect

Simplify a Complex Rational Expression by Writing It as Division

In the following exercises, simplify.

$\frac{\frac{2 a}{a + 4}}{\frac{4 a^{2}}{a^{2} - 16}}$

Solution

$\frac{a - 4}{2 a}$

$\frac{\frac{3 b}{b - 5}}{\frac{b^{2}}{b^{2} - 25}}$

$\frac{\frac{5}{c^{2} + 5 c - 14}}{\frac{10}{c + 7}}$

Solution

$\frac{1}{2 (c - 2)}$

$\frac{\frac{8}{d^{2} + 9 d + 18}}{\frac{12}{d + 6}}$

$\frac{\frac{1}{2} + \frac{5}{6}}{\frac{2}{3} + \frac{7}{9}}$

Solution

$\frac{12}{13}$

$\frac{\frac{1}{2} + \frac{3}{4}}{\frac{3}{5} + \frac{7}{10}}$

$\frac{\frac{2}{3} - \frac{1}{9}}{\frac{3}{4} + \frac{5}{6}}$

Solution

$\frac{20}{57}$

$\frac{\frac{1}{2} - \frac{1}{6}}{\frac{2}{3} + \frac{3}{4}}$

$\frac{\frac{n}{m} + \frac{1}{n}}{\frac{1}{n} - \frac{n}{m}}$

Solution

$\frac{n^{2} + m}{m - n^{2}}$

$\frac{\frac{1}{p} + \frac{p}{q}}{\frac{q}{p} - \frac{1}{q}}$

$\frac{\frac{1}{r} + \frac{1}{t}}{\frac{1}{r^{2}} - \frac{1}{t^{2}}}$

Solution

$\frac{r t}{t - r}$

$\frac{\frac{2}{v} + \frac{2}{w}}{\frac{1}{v^{2}} - \frac{1}{w^{2}}}$

$\frac{x - \frac{2 x}{x + 3}}{\frac{1}{x + 3} + \frac{1}{x - 3}}$

Solution

$\frac{(x + 1) (x - 3)}{2}$

$\frac{y - \frac{2 y}{y - 4}}{\frac{2}{y - 4} - \frac{2}{y + 4}}$

$\frac{2 - \frac{2}{a + 3}}{\frac{1}{a + 3} + \frac{a}{2}}$

Solution

$\frac{4}{a + 1}$

$\frac{4 - \frac{4}{b - 5}}{\frac{1}{b - 5} + \frac{b}{4}}$

Simplify a Complex Rational Expression by Using the LCD

In the following exercises, simplify.

$\frac{\frac{1}{3} + \frac{1}{8}}{\frac{1}{4} + \frac{1}{12}}$

Solution

$\frac{11}{8}$

$\frac{\frac{1}{4} + \frac{1}{9}}{\frac{1}{6} + \frac{1}{12}}$

$\frac{\frac{5}{6} + \frac{2}{9}}{\frac{7}{18} - \frac{1}{3}}$

Solution

$19$

$\frac{\frac{1}{6} + \frac{4}{15}}{\frac{3}{5} - \frac{1}{2}}$

$\frac{\frac{c}{d} + \frac{1}{d}}{\frac{1}{d} - \frac{d}{c}}$

Solution

$\frac{c^{2} + c}{c - d^{2}}$

$\frac{\frac{1}{m} + \frac{m}{n}}{\frac{n}{m} - \frac{1}{n}}$

$\frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p^{2}} - \frac{1}{q^{2}}}$

Solution

$\frac{p q}{q - p}$

$\frac{\frac{2}{r} + \frac{2}{t}}{\frac{1}{r^{2}} - \frac{1}{t^{2}}}$

$\frac{\frac{2}{x + 5}}{\frac{3}{x - 5} + \frac{1}{x^{2} - 25}}$

Solution

$\frac{2 x - 10}{3 x + 16}$

$\frac{\frac{5}{y - 4}}{\frac{3}{y + 4} + \frac{2}{y^{2} - 16}}$

$\frac{\frac{5}{z^{2} - 64} + \frac{3}{z + 8}}{\frac{1}{z + 8} + \frac{2}{z - 8}}$

Solution

$\frac{3 z - 19}{3 z + 8}$

$\frac{\frac{3}{s + 6} + \frac{5}{s - 6}}{\frac{1}{s^{2} - 36} + \frac{4}{s + 6}}$

$\frac{\frac{4}{a^{2} - 2 a - 15}}{\frac{1}{a - 5} + \frac{2}{a + 3}}$

Solution

$\frac{4}{3 a - 7}$

$\frac{\frac{5}{b^{2} - 6 b - 27}}{\frac{3}{b - 9} + \frac{1}{b + 3}}$

$\frac{\frac{5}{c + 2} - \frac{3}{c + 7}}{\frac{5 c}{c^{2} + 9 c + 14}}$

Solution

$\frac{2 c + 29}{5 c}$

$\frac{\frac{6}{d - 4} - \frac{2}{d + 7}}{\frac{2 d}{d^{2} + 3 d - 28}}$

$\frac{2 + \frac{1}{p - 3}}{\frac{5}{p - 3}}$

Solution

$\frac{(2 p - 5)}{5}$

$\frac{\frac{n}{n - 2}}{3 + \frac{5}{n - 2}}$

$\frac{\frac{m}{m + 5}}{4 + \frac{1}{m - 5}}$

Solution

$\frac{m (m - 5)}{4 m^{2} + m - 95}$

$\frac{7 + \frac{2}{q - 2}}{\frac{1}{q + 2}}$

Simplify

In the following exercises, use either method.

$\frac{\frac{3}{4} - \frac{2}{7}}{\frac{1}{2} + \frac{5}{14}}$

Solution

$\frac{13}{24}$

$\frac{\frac{v}{w} + \frac{1}{v}}{\frac{1}{v} - \frac{v}{w}}$

$\frac{\frac{2}{a + 4}}{\frac{1}{a^{2} - 16}}$

Solution

$2 (a - 4)$

$\frac{\frac{3}{b^{2} - 3 b - 40}}{\frac{5}{b + 5} - \frac{2}{b - 8}}$

$\frac{\frac{3}{m} + \frac{3}{n}}{\frac{1}{m^{2}} - \frac{1}{n^{2}}}$

Solution

$\frac{3 m n}{n - m}$

$\frac{\frac{2}{r - 9}}{\frac{1}{r + 9} + \frac{3}{r^{2} - 81}}$

$\frac{x - \frac{3 x}{x + 2}}{\frac{3}{x + 2} + \frac{3}{x - 2}}$

Solution

$\frac{(x - 1) (x - 2)}{6}$

$\frac{\frac{y}{y + 3}}{2 + \frac{1}{y - 3}}$

Everyday Math

Electronics The resistance of a circuit formed by connecting two resistors in parallel is $\frac{1}{\frac{1}{R_{1}} + \frac{1}{R_{2}}}$ .

ⓐ Simplify the complex fraction $\frac{1}{\frac{1}{R_{1}} + \frac{1}{R_{2}}}$ .
ⓑ Find the resistance of the circuit when $R_{1} = 8$ and $R_{2} = 12$ .

Solution

ⓐ $\frac{R_{1} R_{2}}{R_{2} + R_{1}}$ ⓑ $\frac{24}{5}$

Ironing Lenore can do the ironing for her family’s business in $h$ hours. Her daughter would take $h + 2$ hours to get the ironing done. If Lenore and her daughter work together, using 2 irons, the number of hours it would take them to do all the ironing is $\frac{1}{\frac{1}{h} + \frac{1}{h + 2}}$ .

ⓐ Simplify the complex fraction $\frac{1}{\frac{1}{h} + \frac{1}{h + 2}}$ .
ⓑ Find the number of hours it would take Lenore and her daughter, working together, to get the ironing done if $h = 4$ .

Writing Exercises

In this section, you learned to simplify the complex fraction $\frac{\frac{3}{x + 2}}{\frac{x}{x^{2} - 4}}$ two ways:

rewriting it as a division problem

multiplying the numerator and denominator by the LCD

Which method do you prefer? Why?

Solution

Answers will vary.

Efraim wants to start simplifying the complex fraction $\frac{\frac{1}{a} + \frac{1}{b}}{\frac{1}{a} - \frac{1}{b}}$ by cancelling the variables from the numerator and denominator. Explain what is wrong with Efraim’s plan.

Self Check

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

The above image is four columns and three rows. The first row is the header row. The first header is labeled “I can…”, the second “Confidently”, the third, “With some help”, and the fourth “No – I don’t get it!”. In the first column under “I can”, the next row reads “simplify a complex rational expression by writing it as division.”, the next row reads “simplify a complex rational expression by using the LCD.” The remaining columns are blank.

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

complex rational expression

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

Solve Rational Equations

After defining the terms expression and equation early in Foundations, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations. We have simplified many rational expressions so far in this chapter. Now we will solve rational equations.

The definition of a rational equation is similar to the definition of equation we used in Foundations.

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

\begin{matrix} Rational Expression & Rational Equation \\ \frac{1}{8} x + \frac{1}{2} & \frac{1}{8} x + \frac{1}{2} = \frac{1}{4} \\ \frac{y + 6}{y^{2} - 36} & \frac{y + 6}{y^{2} - 36} = y + 1 \\ \frac{1}{n - 3} + \frac{1}{n + 4} & \frac{1}{n - 3} + \frac{1}{n + 4} = \frac{15}{n^{2} + n - 12} \end{matrix}

Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

Here is an example we did when we worked with linear equations:


We multiplied both sides by the LCD.
Then we distributed.
We simplified—and then we had an equation with no fractions.
Finally, we solved that equation.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then we will have an equation that does not contain rational expressions and thus is much easier for us to solve.

But because the original equation may have a variable in a denominator we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution.

We note any possible extraneous solutions, c, by writing $x \neq c$ next to the equation.

How to Solve Equations with Rational Expressions

Solve: $\frac{1}{x} + \frac{1}{3} = \frac{5}{6} .$

Solution

Step two is to find the least common denominator of all denominators in the equation. Find the LCD of 1 divided by x one-third, and five-sixths. The x is 6 x.

$Step three is to clear the fractions by multiplying both sides of the equation by the LCD. Multiply both sides of the equation by the LCD, 6 x to get 6 times 1 divided by x plus one-third equals 6 x times five-sixths. Use the Distributive Property to get 6 x times 1 divided by x plus 6 x times one-third equals 6 x times five-sixths. Simplify – and notice, no more fractions and we have 6 plus 2 x equals 5 x.$ Step 4 is to solve the resulting equation. Simplify to get 6 equals 3 x and 2 equals x.

Step 4 is to solve the resulting equation. Simplify to get 6 equals 3 x and 2 equals x.

Step 5 is to check. If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation. We did not get 0 as an algebraic solution. We substitute x equals 2 into the original equation to get one-half plus one-third equals five-sixths, then three-sixths plus two-sixths equals five-sixths and finally, five-sixths equal five-sixths.

The steps of this method are shown below.

We always start by noting the values that would cause any denominators to be zero.

Solve: $1 - \frac{5}{y} = - \frac{6}{y^{2}} .$

Solution


Note any value of the variable that would make any denominator zero.
Find the least common denominator of all denominators in the equation. The LCD is $y^{2}$ .
Clear the fractions by multiplying both sides of the equation by the LCD.
Distribute.
Multiply.
Solve the resulting equation. First write the quadratic equation in standard form.
Factor.
Use the Zero Product Property.
Solve.
Check.
We did not get 0 as an algebraic solution.

Solve: $\frac{5}{3 u - 2} = \frac{3}{2 u} .$

Solution

	$A mathematical equation is displayed, showing the fraction 5 over the expression 3u minus 2, set equal to the fraction 3 over 2u.$
Note any value of the variable that would make any denominator zero.
Find the least common denominator of all denominators in the equation. The LCD is $2 u (3 u - 2)$ .
Clear the fractions by multiplying both sides of the equation by the LCD.	$An algebraic equation showing 2u(3u-2) multiplied by a fraction on both sides of the equality, with the fractions being 5/(3u-2) on the left and 3/(2u) on the right.$
Remove common factors.
Simplify.
Multiply.
Solve the resulting equation.
We did not get 0 or $\frac{2}{3}$ as algebraic solutions.

When one of the denominators is a quadratic, remember to factor it first to find the LCD.

Solve: $\frac{2}{p + 2} + \frac{4}{p - 2} = \frac{p - 1}{p^{2} - 4} .$

Solution

	$A mathematical equation is displayed, showing a sum of two fractions on the left side equal to a fraction on the right side. The equation is (2/(p+2)) + (4/(p-2)) = (p-1)/(p^2-4).$
Note any value of the variable that would make any denominator zero.	$An algebraic equation is shown, featuring the sum of two fractions, 2/(p+2) and 4/(p-2), equaling a third fraction, (p-1)/((p+2)(p+2)), with the conditions p not equal to -2, p not equal to 2.$
Find the least common denominator of all denominators in the equation. The LCD is $(p + 2) (p - 2)$ .
Clear the fractions by multiplying both sides of the equation by the LCD.	$An algebraic equation showing (p+2)(p-2) multiplied by the sum of two fractions on the left side, equaling (p+2)(p-2) multiplied by a single fraction on the right side. The common multiplier (p+2)(p-2) is in red.$
Distribute.	$A mathematical equation where a common factor, (p+2)(p-2), highlighted in red, is multiplied across three terms involving fractions. This step is typically used to clear denominators in rational equations.$
Remove common factors.
Simplify.
Distribute.
Solve.


We did not get $2 or - 2$ as algebraic solutions.

Solve: $\frac{4}{q - 4} - \frac{3}{q - 3} = 1 .$

Solution


Note any value of the variable that would make any denominator zero.	$A mathematical equation showing the sum of two fractions equaling one, with specified restrictions for the variable q. The equation is 4/(q-4) + 3/(q-3) = 1, where q is not equal to 4 or 3.$
Find the least common denominator of all denominators in the equation. The LCD is $(q - 4) (q - 3)$ .
Clear the fractions by multiplying both sides of the equation by the LCD.	$An algebraic equation showing a step where a rational expression is multiplied by the common denominator (q-4)(q-3) on both sides to clear the fractions.$
Distribute.
Remove common factors.	$A mathematical equation illustrating the simplification of algebraic expressions by canceling common factors. The image shows (q-4) and (q-3) being canceled out in the numerators and denominators of the fractional terms.$
Simplify.
Simplify.
Combine like terms.
Solve. First write in standard form.
Factor.
Use the Zero Product Property.
We did not get 4 or 3 as algebraic solutions.

Solve: $\frac{m + 11}{m^{2} - 5 m + 4} = \frac{5}{m - 4} - \frac{3}{m - 1} .$

Solution


Factor all the denominators, so we can note any value of the variable the would make any denominator zero.	$The image displays a complex algebraic fraction (m+11)/((m-4)(m-1)) rewritten as the difference of two simpler fractions: 5/(m-4) - 3/(m-1), with restrictions m!=4 and m!=1.$
Find the least common denominator of all denominators in the equation. The LCD is $(m - 4) (m - 1)$ .
Clear the fractions.	$An algebraic equation showing both sides multiplied by (m-4)(m-1) to clear denominators, simplifying the expression involving fractions with variables m.$
Distribute.
Remove common factors.
Simplify.
Solve the resulting equation.

Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution. There is no solution to this equation.

The equation we solved in Example 6 had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. Some equations have no solution.

Solve: $\frac{n}{12} + \frac{n + 3}{3 n} = \frac{1}{n} .$

Solution

	$A mathematical equation is displayed on a white background. The equation is n/12 + (n+3)/(3n) = 1/n, featuring fractions with variables in both numerators and denominators.$
Note any value of the variable that would make any denominator zero.
Find the least common denominator of all denominators in the equation. The LCD is $12 n$ .
Clear the fractions by multiplying both sides of the equation by the LCD.	$A mathematical equation is shown with 12n multiplying a sum of fractions (n/12 + (n+3)/(3n)) on the left, and 12n multiplying a fraction (1/n) on the right, all in black text with '12n' in red.$
Distribute.
Remove common factors.
Simplify.
Solve the resulting equation.



Check.
$n = 0$ is an extraneous solution.

Solve: $\frac{y}{y + 6} = \frac{72}{y^{2} - 36} + 4 .$

Solution


Factor all the denominators, so we can note any value of the variable that would make any denominator zero.
Find the least common denominator. The LCD is $(y - 6) (y + 6)$ .
Clear the fractions.	$A mathematical equation demonstrating the process of clearing fractions by multiplying both sides of the equation by the least common denominator, (y-6)(y+6).$
Simplify.
Simplify.
Solve the resulting equation.




Check.
$y = - 6$ is an extraneous solution.

Solve: $\frac{x}{2 x - 2} - \frac{2}{3 x + 3} = \frac{5 x^{2} - 2 x + 9}{12 x^{2} - 12} .$

Solution


We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD.	$A mathematical equation shows the subtraction of two algebraic fractions set equal to a third algebraic fraction. The equation is x over 2(x-1) minus 2 over 3(x+1) equals (5x^2 - 2x + 9) over 12(x-1)(x+1).$
Note any value of the variable that would make any denominator zero.	$A mathematical equation featuring fractions: x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)), with the conditions x ≠ 1 and x ≠ -1.$
Find the least common denominator.The LCD is $12 (x - 1) (x + 1)$
Clear the fractions.	$An algebraic equation showing the multiplication of both sides by 12(x-1)(x+1) to eliminate fractional terms, a common step in solving rational equations.$
Simplify.
Simplify.
Solve the resulting equation.



Check.
$x = 1$ and $x = - 1$ are extraneous solutions. The equation has no solution.

Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

We’ll start with a formula relating distance, rate, and time. We have used it many times before, but not usually in this form.

Solve: $\frac{D}{T} = R for T .$

Solution


Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equations by the LCD, T.
Simplify.
Divide both sides by R to isolate T.	$A mathematical equation shows D/R = RT/R, with the variable R in the denominator of both fractions highlighted in red, indicating a step in solving or manipulating the equation.$
Simplify.

Example 11 uses the formula for slope that we used to get the point-slope form of an equation of a line.

Solve: $m = \frac{x - 2}{y - 3} for y .$

Solution


Note any value of the variable that would make any denominator zero.	$A mathematical equation displays the slope m as a fraction: m = (x-2) / (y-3), with the condition that y is not equal to 3.$
Clear the fractions by multiplying both sides of the equations by the LCD, $y - 3$ .
Simplify.
Isolate the term with y.
Divide both sides by m to isolate y.	$A mathematical equation shows 'ym over m equals x minus 2 plus 3m over m' with the variable 'm' in the denominator highlighted in red. The numerator of the first fraction is 'ym', and the denominator is 'm'. The numerator of the second fraction is 'x - 2 + 3m', and the denominator is 'm'.$
Simplify.

Be sure to follow all the steps in Example 12. It may look like a very simple formula, but we cannot solve it instantly for either denominator.

Solve $\frac{1}{c} + \frac{1}{m} = 1 for c .$

Solution

	$A mathematical equation showing the sum of two fractions, 1 over c plus 1 over m, equaling 1, with the instruction to solve for c.$
Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equations by the LCD, $c m$ .
Distribute.
Simplify.
Collect the terms with c to the right.
Factor the expression on the right.
To isolate c, divide both sides by $m - 1$ .
Simplify by removing common factors.

Notice that even though we excluded $c = 0 and m = 0$ from the original equation, we must also now state that $m \neq 1$ .

Key Concepts

Strategy to Solve Equations with Rational Expressions
1. Note any value of the variable that would make any denominator zero.
2. Find the least common denominator of all denominators in the equation.
3. Clear the fractions by multiplying both sides of the equation by the LCD.
4. Solve the resulting equation.
5. Check.
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.

Practice Makes Perfect

Solve Rational Equations

In the following exercises, solve.

$\frac{1}{a} + \frac{2}{5} = \frac{1}{2}$

Solution

$10$

$\frac{5}{6} + \frac{3}{b} = \frac{1}{3}$

$\frac{5}{2} - \frac{1}{c} = \frac{3}{4}$

Solution

$\frac{4}{7}$

$\frac{6}{3} - \frac{2}{d} = \frac{4}{9}$

$\frac{4}{5} + \frac{1}{4} = \frac{2}{v}$

Solution

$\frac{40}{21}$

$\frac{3}{7} + \frac{2}{3} = \frac{1}{w}$

$\frac{7}{9} + \frac{1}{x} = \frac{2}{3}$

Solution

$−9$

$\frac{3}{8} + \frac{2}{y} = \frac{1}{4}$

$1 - \frac{2}{m} = \frac{8}{m^{2}}$

Solution

$−2, 4$

$1 + \frac{4}{n} = \frac{21}{n^{2}}$

$1 + \frac{9}{p} = \frac{−20}{p^{2}}$

Solution

$−5, −4$

$1 - \frac{7}{q} = \frac{−6}{q^{2}}$

$\frac{1}{r + 3} = \frac{4}{2 r}$

Solution

$−6$

$\frac{3}{t - 6} = \frac{1}{t}$

$\frac{5}{3 v - 2} = \frac{7}{4 v}$

Solution

$14$

$\frac{8}{2 w + 1} = \frac{3}{w}$

$\frac{3}{x + 4} + \frac{7}{x - 4} = \frac{8}{x^{2} - 16}$

Solution

$\frac{−4}{5}$

$\frac{5}{y - 9} + \frac{1}{y + 9} = \frac{18}{y^{2} - 81}$

$\frac{8}{z - 10} + \frac{7}{z + 10} = \frac{5}{z^{2} - 100}$

Solution

$- \frac{1}{3}$

$\frac{9}{a + 11} + \frac{6}{a - 11} = \frac{7}{a^{2} - 121}$

$\frac{1}{q + 4} - \frac{2}{q - 2} = 1$

Solution

$- 2, - 1$

$\frac{3}{r + 10} - \frac{4}{r - 4} = 1$

$\frac{1}{t + 7} - \frac{5}{t - 5} = 1$

Solution

$−5, −1$

$\frac{2}{s + 7} - \frac{3}{s - 3} = 1$

$\frac{v - 10}{v^{2} - 5 v + 4} = \frac{3}{v - 1} - \frac{6}{v - 4}$

Solution

$no solution$

$\frac{w + 8}{w^{2} - 11 w + 28} = \frac{5}{w - 7} + \frac{2}{w - 4}$

$\frac{x - 10}{x^{2} + 8 x + 12} = \frac{3}{x + 2} + \frac{4}{x + 6}$

Solution

$no solution$

$\frac{y - 3}{y^{2} - 4 y - 5} = \frac{1}{y + 1} + \frac{8}{y - 5}$

$\frac{z}{16} + \frac{z + 2}{4 z} = \frac{1}{2 z}$

Solution

$−4$

$\frac{a}{9} + \frac{a + 3}{3 a} = \frac{1}{a}$

$\frac{b + 3}{3 b} + \frac{b}{24} = \frac{1}{b}$

Solution

$−8$

$\frac{c + 3}{12 c} + \frac{c}{36} = \frac{1}{4 c}$

$\frac{d}{d + 3} = \frac{18}{d^{2} - 9} + 4$

Solution

$2$

$\frac{m}{m + 5} = \frac{50}{m^{2} - 25} + 6$

$\frac{n}{n + 2} = \frac{8}{n^{2} - 4} + 3$

Solution

$1$

$\frac{p}{p + 7} = \frac{98}{p^{2} - 49} + 8$

$\frac{q}{3 q - 9} - \frac{3}{4 q + 12}$
$= \frac{7 q^{2} + 6 q + 63}{24 q^{2} - 216}$

Solution

$no solution$

$\frac{r}{3 r - 15} - \frac{1}{4 r + 20}$
$= \frac{3 r^{2} + 17 r + 40}{12 r^{2} - 300}$

$\frac{s}{2 s + 6} - \frac{2}{5 s + 5}$
$= \frac{5 s^{2} - s - 18}{10 s^{2} + 40 s + 30}$

Solution

$no solution$

$\frac{t}{6 t - 12} - \frac{5}{2 t + 10}$
$= \frac{t^{2} - 23 t + 70}{12 t^{2} + 36 t - 120}$

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

$\frac{C}{r} = 2 π for r$

Solution

$r = \frac{C}{2 π}$

$\frac{I}{r} = P for r$

$\frac{V}{h} = l w for h$

Solution

$h = \frac{v}{l w}$

$\frac{2 A}{b} = h for b$

$\frac{v + 3}{w - 1} = \frac{1}{2} for w$

Solution

$w = 2 v + 7$

$\frac{x + 5}{2 - y} = \frac{4}{3} for y$

$a = \frac{b + 3}{c - 2} for c$

Solution

$c = \frac{b + 3 + 2 a}{a}$

$m = \frac{n}{2 - n} for n$

$\frac{1}{p} + \frac{2}{q} = 4 for p$

Solution

$p = \frac{q}{4 q - 2}$

$\frac{3}{s} + \frac{1}{t} = 2 for s$

$\frac{2}{v} + \frac{1}{5} = \frac{3}{w} for w$

Solution

$w = \frac{15 v}{10 + v}$

$\frac{6}{x} + \frac{2}{3} = \frac{1}{y} for y$

$\frac{m + 3}{n - 2} = \frac{4}{5} for n$

Solution

$n = \frac{5 m + 23}{4}$

$\frac{E}{c} = m^{2} for c$

$\frac{3}{x} - \frac{5}{y} = \frac{1}{4} for y$

Solution

$y = \frac{20 x}{12 - x}$

$\frac{R}{T} = W for T$

$r = \frac{s}{3 - t} for t$

Solution

$t = \frac{3 r - s}{r}$

$c = \frac{2}{a} + \frac{b}{5} for a$

Everyday Math

House Painting Alain can paint a house in 4 days. Spiro would take 7 days to paint the same house. Solve the equation $\frac{1}{4} + \frac{1}{7} = \frac{1}{t}$ for t to find the number of days it would take them to paint the house if they worked together.

Solution

$2 \frac{6}{11}$ days

Boating Ari can drive his boat 18 miles with the current in the same amount of time it takes to drive 10 miles against the current. If the speed of the boat is 7 knots, solve the equation $\frac{18}{7 + c} = \frac{10}{7 - c}$ for c to find the speed of the current.

Writing Exercises

Why is there no solution to the equation $\frac{3}{x - 2} = \frac{5}{x - 2}$ ?

Solution

Answers will vary.

Pete thinks the equation $\frac{y}{y + 6} = \frac{72}{y^{2} - 36} + 4$ has two solutions, $y = −6 and y = 4$ . Explain why Pete is wrong.

Self Check

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

This table has three rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the next row reads “solve rational equations”. The next row reads, “solve rational equations for a specific variable”. The remaining columns are blank.

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

rational equation

A rational equation is two rational expressions connected by an equal sign.

extraneous solution to a rational equation

An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

Solve Proportion and Similar Figure Applications

Solve Proportions

When two rational expressions are equal, the equation relating them is called a proportion.

The equation $\frac{1}{2} = \frac{4}{8}$ is a proportion because the two fractions are equal. The proportion $\frac{1}{2} = \frac{4}{8}$ is read “1 is to 2 as 4 is to 8.”

Proportions are used in many applications to ‘scale up’ quantities. We’ll start with a very simple example so you can see how proportions work. Even if you can figure out the answer to the example right away, make sure you also learn to solve it using proportions.

Suppose a school principal wants to have 1 teacher for 20 students. She could use proportions to find the number of teachers for 60 students. We let x be the number of teachers for 60 students and then set up the proportion:

\frac{1 teacher}{20 students} = \frac{x teachers}{60 students}

We are careful to match the units of the numerators and the units of the denominators—teachers in the numerators, students in the denominators.

Since a proportion is an equation with rational expressions, we will solve proportions the same way we solved equations in Solve Rational Equations. We’ll multiply both sides of the equation by the LCD to clear the fractions and then solve the resulting equation.

So let’s finish solving the principal’s problem now. We will omit writing the units until the last step.

	$A mathematical equation is displayed against a white background. The equation shows the fraction 1/20 on the left side, an equals sign in the middle, and the fraction x/60 on the right side.$
Multiply both sides by the LCD, 60.
Simplify.
	The principal needs 3 teachers for 60 students.

Now we’ll do a few examples of solving numerical proportions without any units. Then we will solve applications using proportions.

Solve the proportion: $\frac{x}{63} = \frac{4}{7} .$

Solution

		$A mathematical equation is displayed, showing a variable 'x' divided by 63, which equals the fraction 4/7. The equation is x/63 = 4/7.$
To isolate $x$ , multiply both sides by the LCD, 63.
Simplify.
Divide the common factors.
Check. To check our answer, we substitute into the original proportion.
	$A mathematical equation shows a fraction with x over 63, set equal to the fraction 4 over 7, likely intended for solving for x.$
	$A math problem displays the fractions 36/63 and 4/7, with a question mark next to an equals sign in between, asking if they are equivalent.$
Show common factors.	$A mathematical expression displays the fraction (4 * 9) / (7 * 9) with a question mark over an equals sign, followed by the simplified fraction 4/7, illustrating the simplification of fractions.$
Simplify.

When we work with proportions, we exclude values that would make either denominator zero, just like we do for all rational expressions. What value(s) should be excluded for the proportion in the next example?

Solve the proportion: $\frac{144}{a} = \frac{9}{4} .$

Solution


Multiply both sides by the LCD.
Remove common factors on each side.
Simplify.
Divide both sides by 9.		$A mathematical equation shows '576/9 = 9a/9'. The left side of the equation is the fraction 576 over 9, and the right side is the fraction 9a over 9, with an equals sign in between.$
Simplify.
Check.
	$A mathematical equation is displayed, showing the fraction 144 over 'a' set equal to the fraction 9 over 4, representing a problem to solve for the variable 'a'.$
	$A mathematical problem is displayed, comparing the fractions '144/64' and '9/4' with a question mark over an equality sign, asking if they are equivalent. The denominator '64' is highlighted in red.$
Show common factors.	$A mathematical problem asking if the fraction (9 * 16) / (4 * 16) is equal to 9/4. This visually demonstrates the principle of simplifying fractions by canceling common factors in the numerator and denominator.$
Simplify.

Solve the proportion: $\frac{n}{n + 14} = \frac{5}{7} .$

Solution

		$A mathematical equation shows a fraction n over the sum of n and 14, which is set equal to the fraction 5 over 7, appearing in a clean, isolated format on a white background.$
Multiply both sides by the LCD.
Remove common factors on each side.
Simplify.
Solve for $n$ .

Check.
	$A mathematical equation shows a fraction with 'n' in the numerator and 'n + 14' in the denominator, set equal to the fraction '5/7'.$
	$A math problem showing the fraction 35 over (35 plus 14) on the left side, which is equal to an unknown numerator (?) over 7 on the right side. The problem asks to solve for the missing numerator.$
Simplify.	$A mathematical problem displaying the fractions 35/49 and 5/7, with a question mark positioned above the equals sign, prompting verification of their equivalence.$
Show common factors.	$A mathematical expression showing the fraction (5 multiplied by 7) divided by (7 multiplied by 7), with a question mark above an equals sign, followed by the fraction 5/7. It asks if (57)/(77) is equal to 5/7.$
Simplify.

Solve: $\frac{p + 12}{9} = \frac{p - 12}{6} .$

Solution

		$A mathematical equation is displayed: the fraction p plus 12 over 9 equals the fraction p minus 12 over 6. This is a linear equation with one variable 'p'.$
Multiply both sides by the LCD, 18.		$A mathematical equation is displayed, showing 18 multiplied by the fraction (p + 12)/9 on the left side, equal to 18 multiplied by the fraction (p - 12)/6 on the right side.$
Simplify.
Distribute.
Solve for $p$ .
Check.
	$An algebraic equation is shown, with the fraction (p+12)/9 on the left side and the fraction (p-12)/6 on the right side, separated by an equals sign.$
	$A math problem comparing two fractions: (60+12)/9 and (60-12)/6, with a question mark over an equals sign between them. Both sides evaluate to 8, indicating they are equal.$
Simplify.	$A math problem asking if the fraction 72/9 is equal to 48/6. The expression shows 72 over 9, followed by an equal sign with a question mark on top, and then 48 over 6.$
Divide.

To solve applications with proportions, we will follow our usual strategy for solving applications. But when we set up the proportion, we must make sure to have the units correct—the units in the numerators must match and the units in the denominators must 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

Identify what we are asked to find, and choose a variable to represent it.		How many ml of acetaminophen will the doctor prescribe?
		Let $a = ml$ of acetaminophen.
Write a sentence that gives the information to find it.		If 5 ml is prescribed for every 25 pounds, how much will be prescribed for 80 pounds?
Translate into a proportion–be careful of the units. $\frac{ml}{pounds} = \frac{ml}{pounds}$		$A mathematical equation is shown with the fraction 5/25 equal to the fraction a/80, where 'a' is a variable.$
Multiply both sides by the LCD, 400.
Remove common factors on each side.
Simplify, but don't multiply on the left. Notice what the next step will be.
Solve for $a$ .		$A mathematical equation is displayed, showing the fraction 16 multiplied by 5, all divided by 5, which equals 5a divided by 5. The denominator '5' is highlighted in red on both sides of the equation.$
Check.
Is the answer reasonable?
Yes, since 80 is about 3 times 25, the medicine should be about 3 times 5. So 16 ml makes sense.

Write a complete sentence.		The pediatrician would prescribe 16 ml of acetaminophen to Zoe.

A 16-ounce iced caramel macchiato has 230 calories. How many calories are there in a 24-ounce iced caramel macchiato?

Solution

Identify what we are asked to find, and choose a variable to represent it.		How many calories are in a 24 ounce iced caramel macchiato?
		Let $c = calories$ in 24 ounces.
Write a sentence that gives the information to find it.		If there are 230 calories in 16 ounces, then how many calories are in 24 ounces?
Translate into a proportion–be careful of the units. $\frac{calories}{ounce} = \frac{calories}{ounce}$
Multiply both sides by the LCD, 48.		$A mathematical equation shows '48(230/16) = 48(c/24)' where 48 is in red, illustrating the multiplication of 48 by two different fractions set equal to each other.$
Remove common factors on each side.
Simplify.
Solve for $c$ .

Check.
Is the answer reasonable?
Yes, 345 calories for 24 ounces is more than 290 calories for 16 ounces, but not too much more.

Write a complete sentence.		There are 345 calories in a 24-ounce iced caramel macchiato.

Josiah went to Mexico for spring break and changed $325 dollars into Mexican pesos. At that time, the exchange rate had $1 US is equal to 12.54 Mexican pesos. How many Mexican pesos did he get for his trip?

Solution

What are you asked to find?		How many Mexican pesos did Josiah get?
Assign a variable.		Let $p = the number of Mexican pesos.$
Write a sentence that gives the information to find it.		If $1 US is equal to 12.54 Mexican pesos, then $325 is how many pesos?
Translate into a proportion–be careful of the units.
$\frac{$}{pesos} = \frac{$}{pesos}$
Multiply both sides by the LCD, $12.54 p$ .
Remove common factors on each side.
Simplify.
Check.
Is the answer reasonable?
Yes, $100 would be 1,254 pesos. $325 is a little more than 3 times this amount, so our answer of 4075.5 pesos makes sense.

Write a complete sentence.		Josiah got 4075.5 pesos for his spring break trip.

In the example above, we related the number of pesos to the number of dollars by using a proportion. We could say the number of pesos is proportional to the number of dollars. If two quantities are related by a proportion, we say that they are proportional.

Solve Similar Figure Applications

When you shrink or enlarge a photo on a phone or tablet, figure out a distance on a map, or use a pattern to build a bookcase or sew a dress, you are working with similar figures. If two figures have exactly the same shape, but different sizes, they are said to be similar. One is a scale model of the other. All their corresponding angles have the same measures and their corresponding sides are in the same ratio.

For example, the two triangles in Figure 1 are similar. Each side of $Δ A B C$ is 4 times the length of the corresponding side of $Δ X Y Z$ .

Diagram of two triangles, ABC and XYZ. Triangle ABC has side lengths AB equal to 12, BC equal to 16, and AC equal to 20. Triangle XYZ has side lengths XY equal to 3, YZ equal to 4, and XZ equal to 5. The image indicates that angle A is congruent to angle X, angle B is congruent to angle Y, and angle C is congruent to angle Z. Ratios of corresponding sides are shown: AB/XY equals 12/3, BC/YZ equals 16/4, and AC/XZ equals 20/5.

This is summed up in the Property of Similar Triangles.

To solve applications with similar figures we will follow the Problem-Solving Strategy for Geometry Applications we used earlier.

$Δ A B C$ is similar to $Δ X Y Z$ . The lengths of two sides of each triangle are given. Find the lengths of the third sides.

The above image shows two similar triangles. Two sides are given for each triangle. The larger triangle is labeled A B C. The length of A to B is 4. The length from B to C is a. The length from C to A is 3.2. The smaller triangle is labeled X Y Z. The length from X to Y is 3. The length from Y to Z is 4.5. The length from Z to X is y.

Solution

Step 1. Read the problem. Draw the figure and label it with the given information.	Figure is given.
Step 2. Identify what we are looking for.	the length of the sides of similar triangles
Step 3. Name the variables.	Let $a =$ length of the third side of $Δ A B C .$ $y =$ length of the third side of $Δ X Y Z$
Step 4. Translate.	Since the triangles are similar, the corresponding sides are proportional.
We need to write an equation that compares the side we are looking for to a known ratio. Since the side AB = 4 corresponds to the side XY = 3 we know $\frac{A B}{X Y} = \frac{4}{3}$ . So we write equations with $\frac{A B}{X Y}$ to find the sides we are looking for. Be careful to match up corresponding sides correctly.	$\frac{A B}{X Y} = \frac{B C}{Y Z} = \frac{A C}{X Z} .$
Substitute.	$A mathematical equation shows a fraction 4/3 equal to another fraction a/4.5. This represents a proportion where 'a' is an unknown variable to be solved.$
Step 5. Solve the equation.

Step 6. Check. $\begin{matrix} \begin{matrix} \frac{4}{3} & \overset{?}{=} & \frac{6}{4.5} \\ 4 (4.5) & \overset{?}{=} & 6 (3) \\ 18 & = & 18 ✓ \end{matrix} & \begin{matrix} \frac{4}{3} & \overset{?}{=} & \frac{3.2}{2.4} \\ 4 (2.4) & \overset{?}{=} & 3.2 (3) \\ 9.6 & = & 9.6 ✓ \end{matrix} \end{matrix}$
Step 7. Answer the question.	The third side of $Δ A B C$ is 6 and the third side of $Δ X Y Z$ is 2.4.

The next example shows how similar triangles are used with maps.

On a map, San Francisco, Las Vegas, and Los Angeles form a triangle whose sides are shown in the figure below. If the actual distance from Los Angeles to Las Vegas is 270 miles find the distance from Los Angeles to San Francisco.

The above image shows two similar triangles and how they are used with maps. The smaller triangle on the left shows San Francisco, Las Vegas and Los Angeles on the three points. San Francisco to Los Angeles is 1.3 inches. Los Angeles to Las Vegas is 1 inch. Las Vegas to San Francisco is 2.1 inches. The second larger triangle shows the same points. The distance from San Francisco to Los Angeles is x. The distance from Los Angeles to Las Vegas is 270 miles. The distance from Las Vegas to San Francisco is not noted.

Solution

Read the problem. Draw the figures and label with the given information.	The figures are shown above.
Identify what we are looking for.	The actual distance from Los Angeles to San Francisco.
Name the variables.	Let $x =$ distance from Los Angeles to San Francisco.
Translate into an equation. Since the triangles are similar, the corresponding sides are proportional. We'll make the numerators "miles" and the denominators "inches."
Solve the equation.

Check.
On the map, the distance from Los Angeles to San Francisco is more than the distance from Los Angeles to Las Vegas. Since 351 is more than 270 the answer makes sense.

Answer the question.	The distance from Los Angeles to San Francisco is 351 miles.

We can use similar figures to find heights that we cannot directly measure.

Tyler is 6 feet tall. Late one afternoon, his shadow was 8 feet long. At the same time, the shadow of a tree was 24 feet long. Find the height of the tree.

Solution

Read the problem and draw a figure.
We are looking for h, the height of the tree.
We will use similar triangles to write an equation.
The small triangle is similar to the large triangle.
Solve the proportion.
Simplify.
Check.
Tyler's height is less than his shadow's length so it makes sense that the tree's height is less than the length of its shadow.

Key Concepts

Property of Similar Triangles
- If $Δ A B C$ is similar to $Δ X Y Z$ , then their corresponding angle measures are equal and their corresponding sides are in the same ratio.
Problem Solving Strategy for Geometry Applications
1. Read the problem and make sure all the words and ideas are understood. Draw the figure and label it with the given information.
2. Identify what we are looking for.
3. Name what we are looking for by choosing a variable to represent it.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Practice Makes Perfect

Solve Proportions

In the following exercises, solve.

$\frac{x}{56} = \frac{7}{8}$

Solution

$49$

$\frac{n}{91} = \frac{8}{13}$

$\frac{49}{63} = \frac{z}{9}$

Solution

$7$

$\frac{56}{72} = \frac{y}{9}$

$\frac{5}{a} = \frac{65}{117}$

Solution

$9$

$\frac{4}{b} = \frac{64}{144}$

$\frac{98}{154} = \frac{−7}{p}$

Solution

$−11$

$\frac{72}{156} = \frac{−6}{q}$

$\frac{a}{−8} = \frac{−42}{48}$

Solution

$7$

$\frac{b}{−7} = \frac{−30}{42}$

$\frac{2.7}{j} = \frac{0.9}{0.2}$

Solution

$0.6$

$\frac{2.8}{k} = \frac{2.1}{1.5}$

$\frac{a}{a + 12} = \frac{4}{7}$

Solution

$16$

$\frac{b}{b - 16} = \frac{11}{9}$

$\frac{c}{c - 104} = - \frac{5}{8}$

Solution

$40$

$\frac{d}{d - 48} = - \frac{13}{3}$

$\frac{m + 90}{25} = \frac{m + 30}{15}$

Solution

$60$

$\frac{n + 10}{4} = \frac{40 - n}{6}$

$\frac{2 p + 4}{8} = \frac{p + 18}{6}$

Solution

$30$

$\frac{q - 2}{2} = \frac{2 q - 7}{18}$

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?

A veterinarian prescribed Sunny, a 65 pound dog, an antibacterial medicine in case an infection emerges after her teeth were cleaned. If the dosage is 5 mg for every pound, how much medicine was Sunny given?

Solution

$325 mg$

Belle, a 13 pound cat, is suffering from joint pain. How much medicine should the veterinarian prescribe if the dosage is 1.8 mg per pound?

A new energy drink advertises 106 calories for 8 ounces. How many calories are in 12 ounces of the drink?

Solution

$159 calories$

One 12 ounce can of soda has 150 calories. If Josiah drinks the big 32 ounce size from the local mini-mart, how many calories does he get?

A new 7 ounce lemon ice drink is advertised for having only 140 calories. How many ounces could Sally drink if she wanted to drink just 100 calories?

Solution

$5 oz$

Reese loves to drink healthy green smoothies. A 16 ounce serving of smoothie has 170 calories. Reese drinks 24 ounces of these smoothies in one day. How many calories of smoothie is he consuming in one day?

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

$252.5 Canadian dollars$

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 for each US dollar?

Solution

$0.80 Euros$

Martha changed $350 US into 385 Australian dollars. How many Australian dollars did she receive for each US dollar?

When traveling to Great Britain, Bethany exchanged her $900 into 570 British pounds. How many pounds did she receive for each American dollar?

Solution

$0.63 British pounds$

A missionary commissioned to South Africa had to exchange his $500 for the South African Rand which is worth 12.63 for every dollar. How many Rand did he have after the exchange?

Ronald needs a morning breakfast drink that will give him at least 390 calories. Orange juice has 130 calories in one cup. How many cups does he need to drink to reach his calorie goal?

Solution

$3 cups$

Sarah drinks a 32-ounce energy drink containing 80 calories per 12 ounce. How many calories did she drink?

Elizabeth is returning to the United States from Canada. She changes the remaining 300 Canadian dollars she has to $230.05 in American dollars. What was $1 worth in Canadian dollars?

Solution

$1.30 Canadian dollars$

Ben needs to convert $1000 to the Japanese Yen. One American dollar is worth 123.3 Yen. How much Yen will he have?

A golden retriever weighing 85 pounds has diarrhea. His medicine is prescribed as 1 teaspoon per 5 pounds. How much medicine should he be given?

Solution

$17 tsp$

Five-year-old Lacy was stung by a bee. The dosage for the anti-itch liquid is 150 mg for her weight of 40 pounds. What is the dosage per pound?

Karen eats $\frac{1}{2}$ 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

$\frac{3}{4}$ cup

An oatmeal cookie recipe calls for $\frac{1}{2}$ cup of butter to make 4 dozen cookies. Hilda needs to make 10 dozen cookies for the bake sale. How many cups of butter will she need?

Solve Similar Figure Applications

In the following exercises, $Δ A B C$ is similar to $Δ X Y Z$ . Find the length of the indicated side.

The above image shows two triangles. The larger triangle is labeled A B C. The length from A to B is 15. The length from A to C is b. The length from B to C is 9. The smaller triangle is labeled X Y Z. The length from X to Y is 10. The length from X to Z is 8. The length from Y to Z is x.

side b

Solution

$12$

side x

In the following exercises, $Δ D E F$ is similar to $Δ N P Q$ .

The above image shows two triangles side by side. The smaller triangle is labeled D E F. The length from E to D is five-halves. The length from D to F is 7. The length from E to F is d. The second triangle is labeled N P Q. the length from P to N is zero. The length from N to Q is 9. The length from Q to P is eleven-halves.

Find the length of side d.

Solution

$\frac{77}{18}$

Find the length of side q.

In the following two exercises, use the map shown. On the map, New York City, Chicago, and Memphis form a triangle whose sides are shown in the figure below. The actual distance from New York to Chicago is 800 miles.

The above image shows a triangle. Each angle is labeled, clockwise, “Chicago”, “New York”, and “Memphis”. The side that extends from Chicago to New York is labeled 8 inches. The side that extends from New York to Memphis is labeled 9.5 inches and the side extending from Memphis to Chicago is labeled 5.4 inches.

Find the actual distance from New York to Memphis.

Solution

950 miles

Find the actual distance from Chicago to Memphis.

In the following two exercises, use the map shown. On the map, Atlanta, Miami, and New Orleans form a triangle whose sides are shown in the figure below. The actual distance from Atlanta to New Orleans is 420 miles.

The above image shows a triangle. Each angle is labeled, clockwise, “Atlanta”, “Miami”, and “New Orleans”. The side that extends from Atlanta to Miami is labeled 3 inches. The side that extends from Miami to New Orleans is labeled 3.4 inches and the side extending from New Orleans to Atlanta is labeled 2.1 inches.

Find the actual distance from New Orleans to Miami.

Solution

680 miles

Find the actual distance from Atlanta to Miami.

A 2 foot tall dog casts a 3 foot shadow at the same time a cat casts a one foot shadow. How tall is the cat?

Solution

$\frac{2}{3}$ foot $(8 in)$

Larry and Tom were standing next to each other in the backyard when Tom challenged Larry to guess how tall he was. Larry knew his own height is 6.5 feet and when they measured their shadows, Larry’s shadow was 8 feet and Tom’s was 7.75 feet long. What is Tom’s height?

The tower portion of a windmill is 212 feet tall. A six foot tall person standing next to the tower casts a seven foot shadow. How long is the windmill’s shadow?

Solution

$247.3 feet$

The height of the Statue of Liberty is 305 feet. Nicole, who is standing next to the statue, casts a 6 foot shadow and she is 5 feet tall. How long should the shadow of the statue be?

Everyday Math

Heart Rate At the gym, Carol takes her pulse for 10 seconds 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 per minute
no

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?

Cost of a Road Trip 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

Cost of a Road Trip 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?

Lawn Fertilizer 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

House Paint 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?

Cooking Natalia’s pasta recipe calls for 2 pounds of pasta for 1 quart of sauce. How many pounds of pasta should Natalia cook if she has 2.5 quarts of sauce?

Solution

5 pounds

Heating Oil A 275 gallon oil tank costs $400 to fill. How much would it cost to fill a 180 gallon oil tank?

Writing Exercises

Marisol solves the proportion $\frac{144}{a} = \frac{9}{4}$ by ‘cross multiplying’, so her first step looks like $4 \cdot 144 = 9 \cdot a .$ Explain how this differs from the method of solution shown in Example 2.

Solution

Answers will vary.

Find a printed map and then write and solve an application problem similar to Example 9.

Self Check

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

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

proportion

A proportion is an equation of the form

\frac{a}{b} = \frac{c}{d}

, where

b \neq 0, d \neq 0 .

The proportion is read “

a

is to

b

, as

c

is to

d .

”

similar figures

Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides are in the same ratio.

Solve Uniform Motion and Work Applications

Solve Uniform Motion Applications

We have solved uniform motion problems using the formula $D = r t$ in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The above image is a table with 4 columns and three rows. The first row is the header row. The second column in the header row has the word “rate”. The third column has the word, “Time”. The fourth column says “Distance”. The rest of the spaces are blank.

The formula $D = r t$ assumes we know r and t and use them to find D. If we know D and r and need to find t, we would solve the equation for t and get the formula $t = \frac{D}{r}$ .

We have also explained how flying with or against a current affects the speed of a vehicle. We will revisit that idea in the next example.

An airplane can fly 200 miles into a 30 mph headwind in the same amount of time it takes to fly 300 miles with a 30 mph tailwind. What is the speed of the airplane?

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

The above image has two parallel arrows. The first arrow has its tip pointing to the right The words,”300 miles with the wind r plus 30” above the arrow tip. Below that is a squiggly line. To the left of the squiggly line it says, “Wind 30 miles per hour”. Below that is an arrow with its tip pointing to the left. Below that are the words, “200 miles against the wind r minus 30”.

We fill in the chart to organize the information.

We are looking for the speed of the airplane.	Let $r =$ the speed of the airplane.
When the plane flies with the wind, the wind increases its speed and the rate is $r + 30$ .
When the plane flies against the wind, the wind decreases its speed and the rate is $r - 30$ .
Write in the rates. Write in the distances. Since $D = r ∙ t$ , we solve for t and get $\frac{D}{r}$ . We divide the distance by the rate in each row, and place the expression in the time column.
We know the times are equal and so we write our equation.	$\frac{200}{r - 30} = \frac{300}{r + 30}$
We multiply both sides by the LCD. $200 (r + 30) = 300 (r - 30)$	$(r + 30) (r - 30) (\frac{200}{r - 30}) = (r + 30) (r - 30) (\frac{300}{r + 30})$
Simplify.	$(r + 30) (200) = (r - 30) (300)$
	$200 r + 6000 = 300 r - 9000$
Solve.	$15000 = 100 r$ $150 = r$
Check.
Is 150 mph a reasonable speed for an airplane? Yes. If the plane is traveling 150 mph and the wind is 30 mph:
Tailwind $150 + 30 = 180 mph \frac{300}{180} = \frac{5}{3}$ hours
Headwind $150 - 30 = 120 mph \frac{200}{120} = \frac{5}{3}$ hours
The times are equal, so it checks.	The plane was traveling 150 mph.

In the next example, we will know the total time resulting from travelling different distances at different speeds.

Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

We fill in the chart to organize the information.

We are looking for Jazmine’s running speed.	Let $r =$ Jazmine’s running speed.
Her biking speed is 4 miles faster than her running speed.	$r + 4 =$ her biking speed
The distances are given, enter them into the chart.
Since $D = r ∙ t$ , we solve for t and get $t = \frac{D}{r}$ . We divide the distance by the rate in each row, and place the expression in the time column.
Write a word sentence.	Her time plus the time biking is 3 hours.
Translate the sentence to get the equation.	$\frac{8}{r} + \frac{24}{r + 4} = 3$
Solve.	$\begin{matrix} r (r + 4) (\frac{8}{r} + \frac{24}{r + 4}) & = & 3 ∙ r (r + 4) \\ 8 (r + 4) + 24 r & = & 3 r (r + 4) \\ 8 r + 32 + 24 r & = & 3 r^{2} + 12 r \\ 32 + 32 r & = & 3 r^{2} + 12 r \\ 0 & = & 3 r^{2} - 20 r - 32 \\ 0 & = & (3 r + 4) (r - 8) \end{matrix}$
	$(3 r + 4) = 0 (r - 8) = 0$
	$r = - \frac{4}{3} r = 8$
Check. $r = - \frac{4}{3}$ $r = 8$
A negative speed does not make sense in this problem, so $r = 8$ is the solution.
Is 8 mph a reasonable running speed? Yes.
$\begin{matrix} Run 8 mph & \frac{8 miles}{8 mph} = 1 hour \\ Bike 12 mph & \frac{24 miles}{12 mph} = 2 hours \\ Total 3 hours & Jazmine’s running speed is 8 mph. \end{matrix}$

Once again, we will use the uniform motion formula solved for the variable t.

Hamilton rode his bike downhill 12 miles on the river trail from his house to the ocean and then rode uphill to return home. His uphill speed was 8 miles per hour slower than his downhill speed. It took him 2 hours longer to get home than it took him to get to the ocean. Find Hamilton’s downhill speed.

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

The above figure show 2 diagonal, parallel lines pointing in opposite directions. The top line points to the right, and downward and has “12 miles” written beneath it. The bottom line points to the left and upward, and has, “ 8 miles per hour slower, 2 hours longer” written beneath it.

We fill in the chart to organize the information.

We are looking for Hamilton’s downhill speed.	Let $r =$ Hamilton’s downhill speed.
His uphill speed is 8 miles per hour slower. Enter the rates into the chart.	$r - 8 =$ Hamilton’s uphill speed
The distance is the same in both directions, 12 miles. Since $D = r ∙ t$ , we solve for t and get $t = \frac{D}{r}$ . We divide the distance by the rate in each row, and place the expression in the time column.
Write a word sentence about the time.	He took 2 hours longer uphill than downhill. The uphill time is 2 more than the downhill time.
Translate the sentence to get the equation. Solve.	$\begin{matrix} \frac{12}{r - 8} & = & \frac{12}{r} + 2 \\ r (r - 8) (\frac{12}{r - 8}) & = & r (r - 8) (\frac{12}{r} + 2) \\ 12 r & = & 12 (r - 8) + 2 r (r - 8) \\ 12 r & = & 12 r - 96 + 2 r^{2} - 16 r \\ 0 & = & 2 r^{2} - 16 r - 96 \\ 0 & = & 2 (r^{2} - 8 r - 48) \\ 0 & = & 2 (r - 12) (r + 4) \\ r - 12 & = & 0 r + 4 = 0 \\ r & = & 12 r = −4 \end{matrix}$
Check. Is 12 mph a reasonable speed for biking downhill? Yes.
Downhill $12 mph \frac{12 miles}{12 mph} = 1 hour$
Uphill $12 - 8 = 4 mph \frac{12 miles}{4 mph} = 3 hours$
	The uphill time is 2 hours more than the downhill time. Hamilton’s downhill speed is 12 mph.

Solve Work Applications

Suppose Pete can paint a room in 10 hours. If he works at a steady pace, in 1 hour he would paint $\frac{1}{10}$ of the room. If Alicia would take 8 hours to paint the same room, then in 1 hour she would paint $\frac{1}{8}$ of the room. How long would it take Pete and Alicia to paint the room if they worked together (and didn’t interfere with each other’s progress)?

This is a typical ‘work’ application. There are three quantities involved here – the time it would take each of the two people to do the job alone and the time it would take for them to do the job together.

Let’s get back to Pete and Alicia painting the room. We will let t be the number of hours it would take them to paint the room together. So in 1 hour working together they have completed $\frac{1}{t}$ of the job.

In one hour Pete did $\frac{1}{10}$ of the job. Alicia did $\frac{1}{8}$ of the job. And together they did $\frac{1}{t}$ of the job.

We can model this with the word equation and then translate to a rational equation. To find the time it would take them if they worked together, we solve for t.



Multiply by the LCD, $40 t$ .
Distribute.
Simplify and solve.

	$A mathematical equation is displayed on a white background, showing t = 40/9. The variable 't' is equal to the fraction forty over nine.$
We’ll write as a mixed number so that we can convert it to hours and minutes.
Remember, 1 hour = 60 minutes.	$A mathematical equation shows 't = 4 hours + 4/9(60 minutes)', calculating a total time by adding 4 hours to a fraction of 60 minutes.$
Multiply, and then round to the nearest minute.
	It would take Pete and Alica about 4 hours and 27 minutes to paint the room.

Keep in mind, it should take less time for two people to complete a job working together than for either person to do it alone.

The weekly gossip magazine has a big story about the Princess’ baby and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours to do the job and Press #2 takes 12 hours to do the job. How long will it take the printer to get the magazine printed with both presses running together?

Solution

This is a work problem. A chart will help us organize the information.

Let $t =$ the number of hours needed to complete the job together.
Enter the hours per job for Press #1, Press #2 and when they work together. If a job on Press #1 takes 6 hours, then in 1 hour $\frac{1}{6}$ of the job is completed. Similarly find the part of the job completed/hours for Press #2 and when they both work together.
Write a word sentence.
The part completed by Press #1 plus the part completed by Press #2 equals the amount completed together.
Translate to an equation.
Solve.	$The image shows a mathematical equation with fractions: one-sixth plus one-twelfth equals one over t (1/6 + 1/12 = 1/t).$
Multiply by the LCD, $12 t$ .
Simplify.


	When both presses are running it takes 4 hours to do the job.

Corey can shovel all the snow from the sidewalk and driveway in 4 hours. If he and his twin Casey work together, they can finish shoveling the snow in 2 hours. How many hours would it take Casey to do the job by himself?

Solution

This is a work application. A chart will help us organize the information.
We are looking for how many hours it would take Casey to complete the job by himself.
Let $t =$ the number of hours needed for Casey to complete.
Enter the hours per job for Corey, Casey, and when they work together. If Corey takes 4 hours, then in 1 hour $\frac{1}{4}$ of the job is completed. Similarly find the part of the job completed/hours for Casey and when they both work together.
Write a word sentence.
The part completed by Corey plus the part completed by Casey equals the amount completed together.
Translate to an equation:
Solve.	$A mathematical equation is displayed, showing '1/4 + 1/t = 1/2'. This is an algebraic expression involving fractions and a variable 't', likely requiring solving for 't'.$
Multiply by the LCD, $4 t$ .
Simplify.

	It would take Casey 4 hours to do the job alone.

Practice Makes Perfect

Solve Uniform Motion Applications

In the following exercises, solve uniform motion applications

Mary takes a sightseeing tour on a helicopter that can fly 450 miles against a 35 mph headwind in the same amount of time it can travel 702 miles with a 35 mph tailwind. Find the speed of the helicopter.

Solution

$160 mph$

A private jet can fly 1210 miles against a 25 mph headwind in the same amount of time it can fly 1694 miles with a 25 mph tailwind. Find the speed of the jet.

A boat travels 140 miles downstream in the same time as it travels 92 miles upstream. The speed of the current is 6mph. What is the speed of the boat?

Solution

$29 mph$

Darrin can skateboard 2 miles against a 4 mph wind in the same amount of time he skateboards 6 miles with a 4 mph wind. Find the speed Darrin skateboards with no wind.

Jane spent 2 hours exploring a mountain with a dirt bike. When she rode the 40 miles uphill, she went 5 mph slower than when she reached the peak and rode for 12 miles along the summit. What was her rate along the summit?

Solution

$30 mph$

Jill wanted to lose some weight so she planned a day of exercising. She spent a total of 2 hours riding her bike and jogging. She biked for 12 miles and jogged for 6 miles. Her rate for jogging was 10 mph less than biking rate. What was her rate when jogging?

Bill wanted to try out different water craft. He went 62 miles downstream in a motor boat and 27 miles downstream on a jet ski. His speed on the jet ski was 10 mph faster than in the motor boat. Bill spent a total of 4 hours on the water. What was his rate of speed in the motor boat?

Solution

$20 mph$

Nancy took a 3 hour drive. She went 50 miles before she got caught in a storm. Then she drove 68 miles at 9 mph less than she had driven when the weather was good. What was her speed driving in the storm?

Chester rode his bike uphill 24 miles and then back downhill at 2 mph faster than his uphill. If it took him 2 hours longer to ride uphill than downhill, l, what was his uphill rate?

Solution

$4 mph$

Matthew jogged to his friend’s house 12 miles away and then got a ride back home. It took him 2 hours longer to jog there than ride back. His jogging rate was 25 mph slower than the rate when he was riding. What was his jogging rate?

Hudson travels 1080 miles in a jet and then 240 miles by car to get to a business meeting. The jet goes 300 mph faster than the rate of the car, and the car ride takes 1 hour longer than the jet. What is the speed of the car?

Solution

$60 mph$

Nathan walked on an asphalt pathway for 12 miles. He walked the 12 miles back to his car on a gravel road through the forest. On the asphalt he walked 2 miles per hour faster than on the gravel. The walk on the gravel took one hour longer than the walk on the asphalt. How fast did he walk on the gravel?

John can fly his airplane 2800 miles with a wind speed of 50 mph in the same time he can travel 2400 miles against the wind. If the speed of the wind is 50 mph, find the speed of his airplane.

Solution

$650 mph$

Jim’s speedboat can travel 20 miles upstream against a 3 mph current in the same amount of time it travels 22 miles downstream with a 3 mph current speed. Find the speed of the Jim’s boat.

Hazel needs to get to her granddaughter’s house by taking an airplane and a rental car. She travels 900 miles by plane and 250 miles by car. The plane travels 250 mph faster than the car. If she drives the rental car for 2 hours more than she rode the plane, find the speed of the car.

Solution

$50 mph$

Stu trained for 3 hours yesterday. He ran 14 miles and then biked 40 miles. His biking speed is 6 mph faster than his running speed. What is his running speed?

When driving the 9 hour trip home, Sharon drove 390 miles on the interstate and 150 miles on country roads. Her speed on the interstate was 15 more than on country roads. What was her speed on country roads?

Solution

$50 mph$

Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?

Solve Work Applications

In the following exercises, solve work applications.

Mike, an experienced bricklayer, can build a wall in 3 hours, while his son, who is learning, can do the job in 6 hours. How long does it take for them to build a wall together?

Solution

$2 hours$

It takes Sam 4 hours to rake the front lawn while his brother, Dave, can rake the lawn in 2 hours. How long will it take them to rake the lawn working together?

Mary can clean her apartment in 6 hours while her roommate can clean the apartment in 5 hours. If they work together, how long would it take them to clean the apartment?

Solution

$2 hours and 44 minutes$

Brian can lay a slab of concrete in 6 hours, while Greg can do it in 4 hours. If Brian and Greg work together, how long will it take?

Leeson can proofread a newspaper copy in 4 hours. If Ryan helps, they can do the job in 3 hours. How long would it take for Ryan to do his job alone?

Solution

$12 hours$

Paul can clean a classroom floor in 3 hours. When his assistant helps him, the job takes 2 hours. How long would it take the assistant to do it alone?

Josephine can correct her students’ test papers in 5 hours, but if her teacher’s assistant helps, it would take them 3 hours. How long would it take the assistant to do it alone?

Solution

$7 hours and 30 minutes$

Washing his dad’s car alone, eight year old Levi takes 2.5 hours. If his dad helps him, then it takes 1 hour. How long does it take the Levi’s dad to wash the car by himself?

Jackson can remove the shingles off of a house in 7 hours, while Martin can remove the shingles in 5 hours. How long will it take them to remove the shingles if they work together?

Solution

$2 hours and 55 minutes$

At the end of the day Dodie can clean her hair salon in 15 minutes. Ann, who works with her, can clean the salon in 30 minutes. How long would it take them to clean the shop if they work together?

Ronald can shovel the driveway in 4 hours, but if his brother Donald helps it would take 2 hours. How long would it take Donald to shovel the driveway alone?

Solution

$4 hours$

It takes Tina 3 hours to frost her holiday cookies, but if Candy helps her it takes 2 hours. How long would it take Candy to frost the holiday cookies by herself?

Everyday Math

Dana enjoys taking her dog for a walk, but sometimes her dog gets away and she has to run after him. Dana walked her dog for 7 miles but then had to run for 1 mile, spending a total time of 2.5 hours with her dog. Her running speed was 3 mph faster than her walking speed. Find her walking speed.

Solution

3 mph

Ken and Joe leave their apartment to go to a football game 45 miles away. Ken drives his car 30 mph faster Joe can ride his bike. If it takes Joe 2 hours longer than Ken to get to the game, what is Joe’s speed?

Writing Exercises

In Example 3, the solution $h = −4$ is crossed out. Explain why.

Paula and Yuki are roommates. It takes Paula 3 hours to clean their apartment. It takes Yuki 4 hours to clean the apartment. The equation $\frac{1}{3} + \frac{1}{4} = \frac{1}{t}$ can be used to find t, the number of hours it would take both of them, working together, to clean their apartment. Explain how this equation models the situation.

Self Check

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

ⓑ 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?

Use Direct and Inverse Variation

When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities. We will discuss direct variation and inverse variation in this section.

Solve Direct Variation Problems

Lindsay gets paid $15 per hour at her job. If we let s be her salary and h be the number of hours she has worked, we could model this situation with the equation

s = 15 h

Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates x and y. Then we can use that equation to find values of y for other values of x.

How to Solve Direct Variation Problems

If y varies directly with x and $y = 20$ when $x = 8$ , find the equation that relates x and y.

Solution

The above image has 3 columns. The table shows the steps to solve direct variation problems. Step one is to write the formula for the direct variation. The direct variation formula is y equals k x. Then we get y equals k times x.

Step two is substitute the given values for the variables. We are given y equals 20 and x equals 8. Then we have 20 equals k times 8.

Step three is to solve for the constant variation. Divide both sides of the equation by 8, then multiply. We now get 20 divided by 8 equals k. K equals 2.5.

Step four is to write the equation that relates x and y. Rewrite the general equation with the value we found k to get y equals 2 and five-tenths times x.

We’ll list the steps below.

Now we’ll solve a few applications of direct variation.

When Raoul runs on the treadmill at the gym, the number of calories, c, he burns varies directly with the number of minutes, m, he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

ⓐ Write the equation that relates c and m.
ⓑ How many calories would he burn if he ran on the treadmill for 25 minutes?

Solution

ⓐ

	The number of calories, $c$ , varies directly with the number of minutes, $m$ , on the treadmill, and $c = 315$ when $m = 18$ .
Write the formula for direct variation.
We will use $c$ in place of $y$ and $m$ in place of $x$ .
Substitute the given values for the variables.
Solve for the constant of variation.

Write the equation that relates $c$ and $m$ .
Substitute in the constant of variation.

ⓑ

	Find $c$ when $m = 25$ .
Write the equation that relates $c$ and $m$ .
Substitute the given value for $m$ .
Simplify.
	Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.

In the previous example, the variables c and m were named in the problem. Usually that is not the case. We will have to name the variables in the next example as part of the solution, just like we do in most applied problems.

The number of gallons of gas Eunice’s car uses varies directly with the number of miles she drives. Last week she drove 469.8 miles and used 14.5 gallons of gas.

ⓐ Write the equation that relates the number of gallons of gas used to the number of miles driven.
ⓑ How many gallons of gas would Eunice’s car use if she drove 1000 miles?

Solution

	The number of gallons of gas varies directly with the number of miles driven.
First we will name the variables.	Let $g =$ number of gallons of gas. $m =$ number of miles driven
Write the formula for direct variation.
We will use $g$ in place of $y$ and $m$ in place of $x$ .
Substitute the given values for the variables.

Solve for the constant of variation.
We will round to the nearest thousandth.
Write the equation that relates $g$ and $m$ .
Substitute in the constant of variation.

ⓑ

\begin{matrix} Find g when m = 1000 . \\ Write the equation that relates g and m . & g = 0.031 m \\ Substitute the given value for m . & g = 0.031 (1000) \\ Simplify. & g = 31 \\ Eunice’s car would use 31 gallons of gas if she drove it 1,000 miles. \end{matrix}

Notice that in this example, the units on the constant of variation are gallons/mile. In everyday life, we usually talk about miles/gallon.

In some situations, one variable varies directly with the square of the other variable. When that happens, the equation of direct variation is $y = k x^{2}$ . We solve these applications just as we did the previous ones, by substituting the given values into the equation to solve for k.

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 4” will support a maximum load of 75 pounds.

ⓐ Write the equation that relates the maximum load to the cross-section.
ⓑ What is the maximum load that can be supported by a beam with diagonal 8”?

Solution

	The maximum load varies directly with the square of the diagonal of the cross-section.
Name the variables.	Let $L =$ maximum load. $c =$ the diagonal of the cross-section
Write the formula for direct variation, where $y$ varies directly with the square of $x$ .
We will use $L$ in place of $y$ and $c$ in place of $x$ .
Substitute the given values for the variables.

Solve for the constant of variation.	$A mathematical equation shows a fraction 75/16 on the left side, equated to a fraction (k * 16)/16 on the right side, indicating an algebraic step to solve for k.$

Write the equation that relates $L$ and $c$ .
Substitute in the constant of variation.

ⓑ

\begin{matrix} Find L when c = 8 . \\ Write the equation that relates L and c . & L = 4.6875 c^{2} \\ Substitute the given value for c . & L = 4.6875 {(8)}^{2} \\ Simplify. & L = 300 \\ \begin{matrix} A beam with diagonal 8” could support \\ a maximum load of 300 pounds. \end{matrix} \end{matrix}

Solve Inverse Variation Problems

Many applications involve two variable that vary inversely. As one variable increases, the other decreases. The equation that relates them is $y = \frac{k}{x}$ .

The word ‘inverse’ in inverse variation refers to the multiplicative inverse. The multiplicative inverse of x is $\frac{1}{x}$ .

We solve inverse variation problems in the same way we solved direct variation problems. Only the general form of the equation has changed. We will copy the procedure box here and just change ‘direct’ to ‘inverse’.

If y varies inversely with $x$ and $y = 20$ when $x = 8$ , find the equation that relates x and y.

Solution

Write the formula for inverse variation.
Substitute the given values for the variables.

Solve for the constant of variation.	$A mathematical equation is displayed, showing 8 multiplied by 20 on the left side, equaling 8 multiplied by the fraction k over 8 on the right side: 8(20) = 8(k/8).$

Write the equation that relates $x$ and $y$ .
Substitute in the constant of variation.

The fuel consumption (mpg) of a car varies inversely with its weight. A car that weighs 3100 pounds gets 26 mpg on the highway.

ⓐ Write the equation of variation.
ⓑ What would be the fuel consumption of a car that weighs 4030 pounds?

Solution

ⓐ

	The fuel consumption varies inversely with the weight.
First we will name the variables.	Let $f =$ fuel consumption. $w =$ weight
Write the formula for inverse variation.
We will use $f$ in place of $y$ and $w$ in place of $x$ .
Substitute the given values for the variables.

Solve for the constant of variation.	$A mathematical equation is displayed, showing 3100 multiplied by 26 on the left side, which equals 3100 multiplied by the fraction k over 3100 on the right side.$

Write the equation that relates $f$ and $w$ .
Substitute in the constant of variation.

ⓑ

Calculation of car fuel consumption (mpg) based on its weight (pounds) using an inverse relationship.
	$Find f when w = 4030 .$
Write the equation that relates f and w.	$f = \frac{80,600}{w}$
Substitute the given value for w.	$f = \frac{80,600}{4030}$
Simplify.	$f = 20$ A car that weighs 4030 pounds would have fuel consumption of 20 mpg.

The frequency of a guitar string varies inversely with its length. A 26” long string has a frequency of 440 vibrations per second.

ⓐ Write the equation of variation.
ⓑ How many vibrations per second will there be if the string’s length is reduced to 20” by putting a finger on a fret?

Solution

ⓐ

	The frequency varies inversely with the length.
Name the variables.	Let $f =$ frequency. $L =$ length
Write the formula for inverse variation.
We will use $f$ in place of $y$ and $L$ in place of $x$ .
Substitute the given values for the variables.

Solve for the constant of variation.	$A mathematical equation shows 26 multiplied by 440 equals 26 multiplied by the fraction k over 26.$

Write the equation that relates $f$ and $L$ .
Substitute in the constant of variation.	$A mathematical equation shows f equals 11,440 divided by L. The variable 'f' is on the left side of the equation, and a fraction with '11,440' as the numerator and 'L' as the denominator is on the right.$

ⓑ

This table demonstrates the step-by-step calculation of a guitar string's frequency (f) when its length (L) is 20 units.
	$Find f when L = 20 .$
Write the equation that relates f and L.	$f = \frac{11,440}{L}$
Substitute the given value for L.	$f = \frac{11,440}{20}$
Simplify.	$f = 572$ A 20” guitar string has frequency 572 vibrations per second.

Section Exercises

Practice Makes Perfect

Solve Direct Variation Problems

In the following exercises, solve.

If $y$ varies directly as $x$ and $y = 14, when x = 3$ , find the equation that relates $x and y$ .

Solution

$y = \frac{14}{3} x$

If $p$ varies directly as $q$ and $p = 5, when q = 2$ , find the equation that relates $p and q$ .

If $v$ varies directly as $w$ and $v = 24, when w = 8$ , find the equation that relates $v and w .$

Solution

$v = 3 w$

If $a$ varies directly as $b$ and $a = 16, when b = 4$ , find the equation that relates $a and b .$

If $p$ varies directly as $q$ and $p = 9.6, when q = 3$ , find the equation that relates $p and q .$

Solution

$p = 3.2 q$

If $y$ varies directly as $x$ and $y = 12.4, when x = 4,$ find the equation that relates $x and y$

If $a$ varies directly as $b$ and $a = 6, when b = \frac{1}{3}$ , find the equation that relates $a and b .$

Solution

$a = 18 b$

If $v$ varies directly as $w$ and $v = 8, when w = \frac{1}{2}$ , find the equation that relates $v and w .$

The amount of money Sally earns, P, varies directly with the number, n, of necklaces she sells. When Sally sells 15 necklaces she earns $150.

ⓐ Write the equation that relates P and n.
ⓑ How much money would she earn if she sold 4 necklaces?

Solution

ⓐ $P = 10 n$ ⓑ $$ 40$

The price, P, that Eric pays for gas varies directly with the number of gallons, g, he buys. It costs him $50 to buy 20 gallons of gas.

ⓐ Write the equation that relates P and g.
ⓑ How much would 33 gallons cost Eric?

Terri needs to make some pies for a fundraiser. The number of apples, a, varies directly with number of pies, p. It takes nine apples to make two pies.

ⓐ Write the equation that relates a and p.
ⓑ How many apples would Terri need for six pies?

Solution

ⓐ $a = 4.5 p$ ⓑ 27 apples

Joseph is traveling on a road trip. The distance, d, he travels before stopping for lunch varies directly with the speed, v, he travels. He can travel 120 miles at a speed of 60 mph.

ⓐ Write the equation that relates d and v.
ⓑ How far would he travel before stopping for lunch at a rate of 65 mph?

The price of gas that Jesse purchased varies directly to how many gallons he purchased. He purchased 10 gallons of gas for $39.80.

ⓐ Write the equation that relates the price to the number of gallons.
ⓑ How much will it cost Jesse for 15 gallons of gas?

Solution

ⓐ $p = 3.98 g$ ⓑ $$ 59.70$

The distance that Sarah travels varies directly to how long she drives. She travels 440 miles in 8 hours.

ⓐ Write the equation that relates the distance to the number of hours.
ⓑ How far can Sally travel in 6 hours?

The mass of a liquid varies directly with its volume. A liquid with mass 16 kilograms has a volume of 2 liters.

ⓐ Write the equation that relates the mass to the volume.
ⓑ What is the volume of this liquid if its mass is 128 kilograms?

Solution

ⓐ $m = 8 v$ ⓑ $16 liters$

The length that a spring stretches varies directly with a weight placed at the end of the spring. When Sarah placed a 10 pound watermelon on a hanging scale, the spring stretched 5 inches.

ⓐ Write the equation that relates the length of the spring to the weight.
ⓑ What weight of watermelon would stretch the spring 6 inches?

The distance an object falls varies directly to the square of the time it falls. A ball falls 45 feet in 3 seconds.

ⓐ Write the equation that relates the distance to the time.
ⓑ How far will the ball fall in 7 seconds?

Solution

ⓐ $d = 5 t^{2}$ ⓑ 245 feet

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 6 inch will support a maximum load of 108 pounds.

ⓐ Write the equation that relates the load to the diagonal of the cross-section.
ⓑ What load will a beam with a 10 inch diagonal support?

The area of a circle varies directly as the square of the radius. A circular pizza with a radius of 6 inches has an area of 113.04 square inches.

ⓐ Write the equation that relates the area to the radius.
ⓑ What is the area of a personal pizza with a radius 4 inches?

Solution

ⓐ $A = 3.14 r^{2}$ ⓑ $50.24 sq. in.$

The distance an object falls varies directly to the square of the time it falls. A ball falls 72 feet in 3 seconds,

ⓐ Write the equation that relates the distance to the time.
ⓑ How far will the ball have fallen in 8 seconds?

Solve Inverse Variation Problems

In the following exercises, solve.

If $y$ varies inversely with $x$ and $y = 5$ when $x = 4$ find the equation that relates $x$ and $y .$

Solution

$y = \frac{20}{x}$

If $p$ varies inversely with $q$ and $p = 2$ when $q = 1$ find the equation that relates $p$ and $q .$

If $v$ varies inversely with $w$ and $v = 6$ when $w = \frac{1}{2}$ find the equation that relates $v$ and $w .$

Solution

$v = \frac{3}{w}$

If $a$ varies inversely with $b$ and $a = 12$ when $b = \frac{1}{3}$ find the equation that relates $a$ and $b .$

Write an inverse variation equation to solve the following problems.

The fuel consumption (mpg) of a car varies inversely with its weight. A Toyota Corolla weighs 2800 pounds and gets 33 mpg on the highway.

ⓐ Write the equation that relates the mpg to the car’s weight.
ⓑ What would the fuel consumption be for a Toyota Sequoia that weighs 5500 pounds?

Solution

ⓐ $g = \frac{92,400}{w}$ ⓑ 16.8 mpg

A car’s value varies inversely with its age. Jackie bought a 10 year old car for $2,400.

ⓐ Write the equation that relates the car’s value to its age.
ⓑ What will be the value of Jackie’s car when it is 15 years old ?

The time required to empty a tank varies inversely as the rate of pumping. It took Janet 5 hours to pump her flooded basement using a pump that was rated at 200 gpm (gallons per minute),

ⓐ Write the equation that relates the number of hours to the pump rate.
ⓑ How long would it take Janet to pump her basement if she used a pump rated at 400 gpm?

Solution

ⓐ $t = \frac{1000}{r}$ ⓑ $2.5 hours$

The volume of a gas in a container varies inversely as pressure on the gas. A container of helium has a volume of 370 cubic inches under a pressure of 15 psi.

ⓐ Write the equation that relates the volume to the pressure.
ⓑ What would be the volume of this gas if the pressure was increased to 20 psi?

On a string instrument, the length of a string varies inversely as the frequency of its vibrations. An 11-inch string on a violin has a frequency of 400 cycles per second.

ⓐ Write the equation that relates the string length to its frequency.
ⓑ What is the frequency of a 10-inch string?

Solution

ⓐ $L = \frac{4,400}{f}$ ⓑ 440 cycles per second

Paul, a dentist, determined that the number of cavities that develops in his patient’s mouth each year varies inversely to the number of minutes spent brushing each night. His patient, Lori, had 4 cavities when brushing her teeth 30 seconds (0.5 minutes) each night.

ⓐ Write the equation that relates the number of cavities to the time spent brushing.
ⓑ How many cavities would Paul expect Lori to have if she had brushed her teeth for 2 minutes each night?

The number of tickets for a sports fundraiser varies inversely to the price of each ticket. Brianna can buy 25 tickets at $5each.

ⓐ Write the equation that relates the number of tickets to the price of each ticket.
ⓑ How many tickets could Brianna buy if the price of each ticket was $2.50?

Solution

ⓐ $t = \frac{125}{p}$ ⓑ 50 tickets

Boyle’s Law states that if the temperature of a gas stays constant, then the pressure varies inversely to the volume of the gas. Braydon, a scuba diver, has a tank that holds 6 liters of air under a pressure of 220 psi.

ⓐ Write the equation that relates pressure to volume.
ⓑ If the pressure increases to 330 psi, how much air can Braydon’s tank hold?

Mixed Practice

If $y$ varies directly as $x$ and $y = 5, when x = 3 .$ , find the equation that relates $x and y$ .

Solution

$y = \frac{5}{3} x$

If $v$ varies directly as $w$ and $v = 21, when w = 8 .$ find the equation that relates $v and w .$

If $p$ varies inversely with $q$ and $p = 5$ when $q = 6$ , find the equation that relates $p$ and $q .$

Solution

$p = \frac{30}{q}$

If $y$ varies inversely with $x$ and $y = 11$ when $x = 3$ find the equation that relates $x$ and $y .$

If $p$ varies directly as $q$ and $p = 10, when q = 2 .$ find the equation that relates $p and q$ .

Solution

$p = 5 q$

If $v$ varies inversely with $w$ and $v = 18$ when $w = \frac{1}{3}$ find the equation that relates $v$ and $w .$

The force needed to break a board varies inversely with its length. If Tom uses 20 pounds of pressure to break a 1.5-foot long board, how many pounds of pressure would he need to use to break a 6 foot long board?

Solution

$5 pounds$

The number of hours it takes for ice to melt varies inversely with the air temperature. A block of ice melts in 2.5 hours when the temperature is 54 degrees. How long would it take for the same block of ice to melt if the temperature was 45 degrees?

The length a spring stretches varies directly with a weight placed at the end of the spring. When Meredith placed a 6-pound cantaloupe on a hanging scale, the spring stretched 2 inches. How far would the spring stretch if the cantaloupe weighed 9 pounds?

Solution

$3 inches$

The amount that June gets paid varies directly the number of hours she works. When she worked 15 hours, she got paid $111. How much will she be paid for working 18 hours?

The fuel consumption (mpg) of a car varies inversely with its weight. A Ford Focus weighs 3000 pounds and gets 28.7 mpg on the highway. What would the fuel consumption be for a Ford Expedition that weighs 5,500 pounds? Round to the nearest tenth.

Solution

$15.6 mpg$

The volume of a gas in a container varies inversely as the pressure on the gas. If a container of argon has a volume of 336 cubic inches under a pressure of 2,500 psi, what will be its volume if the pressure is decreased to 2,000 psi?

The distance an object falls varies directly to the square of the time it falls. If an object falls 52.8 feet in 4 seconds, how far will it fall in 9 seconds?

Solution

$267.3 feet$

The area of the face of a Ferris wheel varies directly with the square of its radius. If the area of one face of a Ferris wheel with diameter 150 feet is 70,650 square feet, what is the area of one face of a Ferris wheel with diameter of 16 feet?

Everyday Math

Ride Service It costs $35 for a ride from the city center to the airport, 14 miles away.

ⓐ Write the equation that relates the cost, c, with the number of miles, m.
ⓑ What would it cost to travel 22 miles with this service?

Solution

ⓐ $c = 2.5 m$ ⓑ $55

Road Trip The number of hours it takes Jack to drive from Boston to Bangor is inversely proportional to his average driving speed. When he drives at an average speed of 40 miles per hour, it takes him 6 hours for the trip.

ⓐ Write the equation that relates the number of hours, h, with the speed, s.
ⓑ How long would the trip take if his average speed was 75 miles per hour?

Writing Exercises

In your own words, explain the difference between direct variation and inverse variation.

Solution

Answers will vary.

Make up an example from your life experience of inverse variation.

Self Check

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

This image is four columns and three rows. The first row is the header row. The first header is labeled “I can…”, the second “Confidently”, the third, “With some help”, and the fourth “No – I don’t get it!”. In the first column under “I can”, the next row reads “solve direct variation problems.”, the next row reads “solve direct variation problems.” The remaining columns are blank.

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

Chapter 8 Review Exercises

Simplify Rational Expressions

Determine the Values for Which a Rational Expression is Undefined

In the following exercises, determine the values for which the rational expression is undefined.

$\frac{2 a + 1}{3 a - 2}$

Solution

$a \neq \frac{2}{3}$

$\frac{b - 3}{b^{2} - 16}$

$\frac{3 x y^{2}}{5 y}$

Solution

$y \neq 0$

$\frac{u - 3}{u^{2} - u - 30}$

Evaluate Rational Expressions

In the following exercises, evaluate the rational expressions for the given values.

$\frac{4 p - 1}{p^{2} + 5} when p = −1$

Solution

$- \frac{5}{6}$

$\frac{q^{2} - 5}{q + 3} when q = 7$

$\frac{y^{2} - 8}{y^{2} - y - 2} when y = 1$

Solution

$\frac{7}{2}$

$\frac{z^{2} + 2}{4 z - z^{2}} when z = 3$

Simplify Rational Expressions

In the following exercises, simplify.

$\frac{10}{24}$

Solution

$\frac{5}{12}$

$\frac{8 m^{4}}{16 m n^{3}}$

$\frac{14 a - 14}{a - 1}$

Solution

$14$

$\frac{b^{2} + 7 b + 12}{b^{2} + 8 b + 16}$

Simplify Rational Expressions with Opposite Factors

In the following exercises, simplify.

$\frac{c^{2} - c - 2}{4 - c^{2}}$

Solution

$- \frac{c + 1}{c + 2}$

$\frac{d - 16}{16 - d}$

$\frac{7 v - 35}{25 - v^{2}}$

Solution

$- \frac{7}{5 + v}$

$\frac{w^{2} - 3 w - 28}{49 - w^{2}}$

Multiply and Divide Rational Expressions

Multiply Rational Expressions

In the following exercises, multiply.

$\frac{3}{8} \cdot \frac{2}{15}$

Solution

$\frac{1}{20}$

$\frac{2 x y^{2}}{8 y^{3}} \cdot \frac{16 y}{24 x}$

$\frac{3 a^{2} + 21 a}{a^{2} + 6 a - 7} \cdot \frac{a - 1}{a b}$

Solution

$\frac{3}{b}$

$\frac{5 z^{2}}{5 z^{2} + 40 z + 35} \cdot \frac{z^{2} - 1}{3 z}$

Divide Rational Expressions

In the following exercises, divide.

$\frac{t^{2} - 4 t - 12}{t^{2} + 8 t + 12} \div \frac{t^{2} - 36}{6 t}$

Solution

$\frac{6 t}{(t + 6)^{2}}$

$\frac{r^{2} - 16}{4} \div \frac{r^{3} - 64}{2 r^{2} + 8 r + 32}$

$\frac{11 + w}{w - 9} \div \frac{121 - w^{2}}{9 - w}$

Solution

$\frac{−1}{11 - w}$

$\frac{3 y^{2} - 12 y - 63}{4 y + 3} \div (6 y^{2} - 42 y)$

$\frac{\frac{c^{2} - 64}{3 c^{2} + 26 c + 16}}{\frac{c^{2} - 4 c - 32}{15 c + 10}}$

Solution

$\frac{5}{c + 4}$

$\frac{8 m^{2} - 8 m}{m - 4} \cdot \frac{m^{2} + 2 m - 24}{m^{2} + 7 m + 10} \div \frac{2 m^{2} - 6 m}{m + 5}$

Add and Subtract Rational Expressions with a Common Denominator

Add Rational Expressions with a Common Denominator

In the following exercises, add.

$\frac{3}{5} + \frac{2}{5}$

Solution

$1$

$\frac{4 a^{2}}{2 a - 1} - \frac{1}{2 a - 1}$

$\frac{p^{2} + 10 p}{p + 5} + \frac{25}{p + 5}$

Solution

$p + 5$

$\frac{3 x}{x - 1} + \frac{2}{x - 1}$

Subtract Rational Expressions with a Common Denominator

In the following exercises, subtract.

$\frac{d^{2}}{d + 4} - \frac{3 d + 28}{d + 4}$

Solution

$d - 7$

$\frac{z^{2}}{z + 10} - \frac{100}{z + 10}$

$\frac{4 q^{2} - q + 3}{q^{2} + 6 q + 5} - \frac{3 q^{2} + q + 6}{q^{2} + 6 q + 5}$

Solution

$\frac{q - 3}{q + 5}$

$\frac{5 t^{2} + 4 t + 3}{t^{2} - 25} - \frac{4 t^{2} - 8 t - 32}{t^{2} - 25}$

Add and Subtract Rational Expressions whose Denominators are Opposites

In the following exercises, add and subtract.

$\frac{18 w}{6 w - 1} + \frac{3 w - 2}{1 - 6 w}$

Solution

$\frac{15 w + 2}{6 w - 1}$

$\frac{a^{2} + 3 a}{a^{2} - 16} - \frac{3 a + 8}{16 - a^{2}}$

$\frac{2 b^{2} + 3 b - 15}{b^{2} - 49} - \frac{b^{2} + 16 b - 1}{49 - b^{2}}$

Solution

$\frac{3 b^{2} + 19 b - 16}{(b - 7) (b + 7)}$

$\frac{8 y^{2} - 10 y + 7}{2 y - 5} + \frac{2 y^{2} + 7 y + 2}{5 - 2 y}$

Add and Subtract Rational Expressions With Unlike Denominators

Find the Least Common Denominator of Rational Expressions

In the following exercises, find the LCD.

$\frac{4}{m^{2} - 3 m - 10}, \frac{2 m}{m^{2} - m - 20}$

Solution

$(m + 2) (m - 5) (m + 4)$

$\frac{6}{n^{2} - 4}, \frac{2 n}{n^{2} - 4 n + 4}$

$\frac{5}{3 p^{2} + 19 p + 6}, \frac{2 p}{3 p^{2} + 25 p + 8}$

Solution

$(3 p + 1) (p + 6) (p + 8)$

Find Equivalent Rational Expressions

In the following exercises, rewrite as equivalent rational expressions with the given denominator.

Rewrite as equivalent rational expressions with denominator $(m + 2) (m - 5) (m + 4)$ :

\frac{4}{m^{2} - 3 m - 10}, \frac{2 m}{m^{2} - m - 20} .

Rewrite as equivalent rational expressions with denominator $(n - 2) (n - 2) (n + 2)$ :

\frac{6}{n^{2} - 4 n + 4}, \frac{2 n}{n^{2} - 4} .

Solution

$\frac{6 n + 12}{(n - 2) (n - 2) (n + 2)},$
$\frac{2 n^{2} - 4 n}{(n - 2) (n - 2) (n + 2)}$

Rewrite as equivalent rational expressions with denominator $(3 p + 1) (p + 6) (p + 8)$ :

\frac{5}{3 p^{2} + 19 p + 6}, \frac{7 p}{3 p^{2} + 25 p + 8}

Add Rational Expressions with Different Denominators

In the following exercises, add.

$\frac{2}{3} + \frac{3}{5}$

Solution

$\frac{19}{15}$

$\frac{7}{5 a} + \frac{3}{2 b}$

$\frac{2}{c - 2} + \frac{9}{c + 3}$

Solution

$\frac{11 c - 12}{(c - 2) (c + 3)}$

$\frac{3 d}{d^{2} - 9} + \frac{5}{d^{2} + 6 d + 9}$

$\frac{2 x}{x^{2} + 10 x + 24} + \frac{3 x}{x^{2} + 8 x + 16}$

Solution

$\frac{5 x^{2} + 26 x}{(x + 4) (x + 4) (x + 6)}$

$\frac{5 q}{p^{2} q - p^{2}} + \frac{4 q}{q^{2} - 1}$

Subtract Rational Expressions with Different Denominators

In the following exercises, subtract and add.

$\frac{3 v}{v + 2} - \frac{v + 2}{v + 8}$

Solution

$\frac{2 (v^{2} + 10 v - 2)}{(v + 2) (v + 8)}$

$\frac{−3 w - 15}{w^{2} + w - 20} - \frac{w + 2}{4 - w}$

$\frac{7 m + 3}{m + 2} - 5$

Solution

$\frac{2 m - 7}{m + 2}$

$\frac{n}{n + 3} + \frac{2}{n - 3} - \frac{n - 9}{n^{2} - 9}$

$\frac{8 d}{d^{2} - 64} - \frac{4}{d + 8}$

Solution

$\frac{4}{d - 8}$

$\frac{5}{12 x^{2} y} + \frac{7}{20 x y^{3}}$

Simplify Complex Rational Expressions

Simplify a Complex Rational Expression by Writing it as Division

In the following exercises, simplify.

$\frac{\frac{5 a}{a + 2}}{\frac{10 a^{2}}{a^{2} - 4}}$

Solution

$\frac{a - 2}{2 a}$

$\frac{\frac{2}{5} + \frac{5}{6}}{\frac{1}{3} + \frac{1}{4}}$

$\frac{x - \frac{3 x}{x + 5}}{\frac{1}{x + 5} + \frac{1}{x - 5}}$

Solution

$\frac{(x + 2) (x - 5)}{2}$

$\frac{\frac{2}{m} + \frac{m}{n}}{\frac{n}{m} - \frac{1}{n}}$

Simplify a Complex Rational Expression by Using the LCD

In the following exercises, simplify.

$\frac{6 + \frac{2}{q - 4}}{\frac{5}{q + 4}}$

Solution

$\frac{2 (3 q - 11) (q + 4)}{5 (q - 4)}$

$\frac{\frac{3}{a^{2}} - \frac{1}{b}}{\frac{1}{a} + \frac{1}{b^{2}}}$

$\frac{\frac{2}{z^{2} - 49} + \frac{1}{z + 7}}{\frac{9}{z + 7} + \frac{12}{z - 7}}$

Solution

$\frac{z - 5}{21 z + 21}$

$\frac{\frac{3}{y^{2} - 4 y - 32}}{\frac{2}{y - 8} + \frac{1}{y + 4}}$

Solve Rational Equations

Solve Rational Equations

In the following exercises, solve.

$\frac{1}{2} + \frac{2}{3} = \frac{1}{x}$

Solution

$\frac{6}{7}$

$1 - \frac{2}{m} = \frac{8}{m^{2}}$

$\frac{1}{b - 2} + \frac{1}{b + 2} = \frac{3}{b^{2} - 4}$

Solution

$\frac{3}{2}$

$\frac{3}{q + 8} - \frac{2}{q - 2} = 1$

$\frac{v - 15}{v^{2} - 9 v + 18} = \frac{4}{v - 3} + \frac{2}{v - 6}$

Solution

$no solution$

$\frac{z}{12} + \frac{z + 3}{3 z} = \frac{1}{z}$

Solve a Rational Equation for a Specific Variable

In the following exercises, solve for the indicated variable.

$\frac{V}{l} = h w for l$

Solution

$l = \frac{V}{h w}$

$\frac{1}{x} - \frac{2}{y} = 5 for y$

$x = \frac{y + 5}{z - 7} for z$

Solution

$z = \frac{y + 5 + 7 x}{x}$

$P = \frac{k}{V} for V$

Solve Proportion and Similar Figure Applications Similarity

Solve Proportions

In the following exercises, solve.

$\frac{x}{4} = \frac{3}{5}$

Solution

$\frac{12}{5}$

$\frac{3}{y} = \frac{9}{5}$

$\frac{s}{s + 20} = \frac{3}{7}$

Solution

$15$

$\frac{t - 3}{5} = \frac{t + 2}{9}$

In the following exercises, solve using proportions.

Rachael had a 21 ounce strawberry shake that has 739 calories. How many calories are there in a 32 ounce shake?

Solution

$1126 calories$

Leo went to Mexico over Christmas break and changed $525 dollars into Mexican pesos. At that time, the exchange rate had $1 US is equal to 16.25 Mexican pesos. How many Mexican pesos did he get for his trip?

Solve Similar Figure Applications

In the following exercises, solve.

∆ABC is similar to ∆XYZ. The lengths of two sides of each triangle are given in the figure. Find the lengths of the third sides.

This image shows two triangles. The large triangle is labeled A B C. The length from A to B is labeled 8. The length from B to C is labeled 7. The length from C to A is labeled b. The smaller triangle is triangle x y z. The length from x to y is labeled 2 and two-thirds. The length from y to z is labeled x. The length from x to z is labeled 3.

Solution

$b = 9; x = 2 \frac{1}{3}$

On a map of Europe, Paris, Rome, and Vienna form a triangle whose sides are shown in the figure below. If the actual distance from Rome to Vienna is 700 miles, find the distance from

ⓐ Paris to Rome
ⓑ Paris to Vienna

This is an image of a triangle. Clockwise beginning at the top, each vertex is labeled. The top vertex is labeled “Paris”, the next vertex is labeled “Vienna”, and the next vertex is labeled “Rome”. The distance from Paris to Vienna is 7.7 centimeters. The distance from Vienna to Rome is 7 centimeters. The distance from Rome to Paris is 8.9 centimeters.

Tony is 5.75 feet tall. Late one afternoon, his shadow was 8 feet long. At the same time, the shadow of a nearby tree was 32 feet long. Find the height of the tree.

Solution

23 feet

The height of a lighthouse in Pensacola, Florida is 150 feet. Standing next to the statue, 5.5 foot tall Natalie cast a 1.1 foot shadow How long would the shadow of the lighthouse be?

Solve Uniform Motion and Work Applications Problems

Solve Uniform Motion Applications

In the following exercises, solve.

When making the 5-hour drive home from visiting her parents, Lisa ran into bad weather. She was able to drive 176 miles while the weather was good, but then driving 10 mph slower, went 81 miles in the bad weather. How fast did she drive when the weather was bad?

Solution

$45 mph$

Mark is riding on a plane that can fly 490 miles with a headwind of 20 mph in the same time that it can fly 350 miles against a tailwind of 20 mph. What is the speed of the plane?

John can ride his bicycle 8 mph faster than Luke can ride his bike. It takes Luke 3 hours longer than John to ride 48 miles. How fast can John ride his bike?

Solution

$16 mph$

Mark was training for a triathlon. He ran 8 kilometers and biked 32 kilometers in a total of 3 hours. His running speed was 8 kilometers per hour less than his biking speed. What was his running speed?

Solve Work Applications

In the following exercises, solve.

Jerry can frame a room in 1 hour, while Jake takes 4 hours. How long could they frame a room working together?

Solution

$\frac{4}{5} hour$

Lisa takes 3 hours to mow the lawn while her cousin, Barb, takes 2 hours. How long will it take them working together?

Jeffrey can paint a house in 6 days, but if he gets a helper he can do it in 4 days. How long would it take the helper to paint the house alone?

Solution

$12 days$

Sue and Deb work together writing a book that takes them 90 days. If Sue worked alone it would take her 120 days. How long would it take Deb to write the book alone?

Use Direct and Inverse Variation

Solve Direct Variation Problems

In the following exercises, solve.

If $y$ varies directly as $x$ , when $y = 9$ and $x = 3$ , find $x$ when $y = 21 .$

Solution

$7$

If $y$ varies inversely as $x$ , when $y = 20$ and $x = 2$ find $y$ when $x = 4 .$

If $m$ varies inversely with the square of $n$ , when $m = 4$ and $n = 6$ find $m$ when $n = 2 .$

Solution

$36$

Vanessa is traveling to see her fiancé. The distance, d, varies directly with the speed, v, she drives. If she travels 258 miles driving 60 mph, how far would she travel going 70 mph?

If the cost of a pizza varies directly with its diameter, and if an 8” diameter pizza costs $12, how much would a 6” diameter pizza cost?

Solution

$$ 9$

The distance to stop a car varies directly with the square of its speed. It takes 200 feet to stop a car going 50 mph. How many feet would it take to stop a car going 60 mph?

Solve Inverse Variation Problems

In the following exercises, solve.

The number of tickets for a music fundraiser varies inversely with the price of the tickets. If Madelyn has just enough money to purchase 12 tickets for $6 each, how many tickets can Madelyn afford to buy if the price increased to $8?

Solution

$9 tickets$

On a string instrument, the length of a string varies inversely with the frequency of its vibrations. If an 11-inch string on a violin has a frequency of 360 cycles per second, what frequency does a 12 inch string have?

Practice Test

In the following exercises, simplify.

$\frac{3 a^{2} b}{6 a b^{2}}$

Solution

$\frac{a}{2 b}$

$\frac{5 b - 25}{b^{2} - 25}$

In the following exercises, perform the indicated operation and simplify.

$\frac{4 x}{x + 2} \cdot \frac{x^{2} + 5 x + 6}{12 x^{2}}$

Solution

$\frac{x + 3}{3 x}$

$\frac{5 y}{4 y - 8} \cdot \frac{y^{2} - 4}{10}$

$\frac{4}{p q} + \frac{5}{p}$

Solution

$\frac{4 + 5 q}{p q}$

$\frac{1}{z - 9} - \frac{3}{z + 9}$

$\frac{\frac{2}{3} + \frac{3}{5}}{\frac{2}{5}}$

Solution

$\frac{19}{6}$

$\frac{\frac{1}{m} - \frac{1}{n}}{\frac{1}{n} + \frac{1}{m}}$

In the following exercises, solve each equation.

$\frac{1}{2} + \frac{2}{7} = \frac{1}{x}$

Solution

$\frac{14}{11}$

$\frac{5}{y - 6} = \frac{3}{y + 6}$

$\frac{1}{z - 5} + \frac{1}{z + 5} = \frac{1}{z^{2} - 25}$

Solution

$\frac{1}{2}$

$\frac{t}{4} = \frac{3}{5}$

$\frac{2}{r - 2} = \frac{3}{r - 1}$

Solution

In the following exercises, solve.

If $y$ varies directly with $x,$ and $x = 5$ when $y = 30,$ find $x$ when $y = 42 .$

If $y$ varies inversely with $x$ and $x = 6$ when $y = 20,$ find $y$ when $x = 2 .$

Solution

If $y$ varies inversely with the square of $x$ and $x = 3$ when $y = 9,$ find $y$ when $x = 4 .$

The recommended erythromycin dosage for dogs, is 5 mg for every pound the dog weighs. If Daisy weighs 25 pounds, how many milligrams of erythromycin should her veterinarian prescribe?

Solution

$125 mg$

Julia spent 4 hours Sunday afternoon exercising at the gym. She ran on the treadmill for 10 miles and then biked for 20 miles. Her biking speed was 5 mph faster than her running speed on the treadmill. What was her running speed?

Kurt can ride his bike for 30 miles with the wind in the same amount of time that he can go 21 miles against the wind. If the wind’s speed is 6 mph, what is Kurt’s speed on his bike?

Solution

$34 mph$

Amanda jogs to the park 8 miles using one route and then returns via a 14-mile route. The return trip takes her 1 hour longer than her jog to the park. Find her jogging rate.

An experienced window washer can wash all the windows in Mike’s house in 2 hours, while a new trainee can wash all the windows in 7 hours. How long would it take them working together?

Solution

$1 \frac{5}{9} hours$

Josh can split a truckload of logs in 8 hours, but working with his dad they can get it done in 3 hours. How long would it take Josh’s dad working alone to split the logs?

The price that Tyler pays for gas varies directly with the number of gallons he buys. If 24 gallons cost him $59.76, what would 30 gallons cost?

Solution

$$ 74.70$

The volume of a gas in a container varies inversely with the pressure on the gas. If a container of nitrogen has a volume of 29.5 liters with 2000 psi, what is the volume if the tank has a 14.7 psi rating? Round to the nearest whole number.

The cities of Dayton, Columbus, and Cincinnati form a triangle in southern Ohio, as shown on the figure below, that gives the map distances between these cities in inches.

This is an image of a triangle. Clockwise beginning at the top, each vertex is labeled. The top vertex is labeled “Dayton”, the next vertex is labeled “Columbus”, and the next vertex is labeled “Cincinnati”. The distance from Dayton to Columbus is 3.2 inches. The distance from Columbus to Cincinnati is 5.3 inches. The distance from Cincinnati to Dayton is 2.4 inches.

The actual distance from Dayton to Cincinnati is 48 miles. What is the actual distance between Dayton and Columbus?

Solution

64 miles

Introduction

This figure shows a rock cliff. — Square roots are used to determine the time it would take for a stone falling from the edge of this cliff to hit the land below.

Suppose a stone falls from the edge of a cliff. The number of feet the stone has dropped after $t$ seconds can be found by multiplying 16 times the square of $t$ . But to calculate the number of seconds it would take the stone to hit the land below, we need to use a square root. In this chapter, we will introduce and apply the properties of square roots, and extend these concepts to higher order roots and rational exponents.

Simplify and Use Square Roots

Simplify Expressions with Square Roots

Remember that when a number $n$ is multiplied by itself, we write $n^{2}$ and read it “n squared.” For example, $15^{2}$ reads as “15 squared,” and 225 is called the square of 15, since $15^{2} = 225$ .

Sometimes we will need to look at the relationship between numbers and their squares in reverse. Because 225 is the square of 15, we can also say that 15 is a square root of 225. A number whose square is $m$ is called a square root of $m$ .

Notice ${(−15)}^{2} = 225$ also, so $−15$ is also a square root of 225. Therefore, both 15 and $−15$ are square roots of 225.

So, every positive number has two square roots—one positive and one negative. What if we only wanted the positive square root of a positive number? The radical sign, $\sqrt{m}$ , denotes the positive square root. The positive square root is also called the principal square root.

We also use the radical sign for the square root of zero. Because $0^{2} = 0$ , $\sqrt{0} = 0$ . Notice that zero has only one square root.

Since 15 is the positive square root of 225, we write $\sqrt{225} = 15$ . Fill in Figure 1 to make a table of square roots you can refer to as you work this chapter.

This table has fifteen columns and two rows. The first row contains the following numbers: the square root of 1, the square root of 4, the square root of 9, the square root of 16, the square root of 25, the square root of 36, the square root of 49, the square root of 64, the square root of 81, the square root of 100, the square root of 121, the square root of 144, the square root of 169, the square root of 196, and the square root of 225. The second row is completely empty except for the last column. The number 15 is in the last column.

We know that every positive number has two square roots and the radical sign indicates the positive one. We write $\sqrt{225} = 15$ . If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, $− \sqrt{225} = −15$ .

Solution

ⓐ

This table illustrates the calculation of the square root of 36, showing the expression, its result, and the underlying mathematical justification.
	$\sqrt{36}$
Since $6^{2} = 36$	6

ⓑ

Calculation demonstrating that the square root of 196 equals 14.
	$\sqrt{196}$
Since $14^{2} = 196$	14

ⓒ

Illustrates the evaluation of -sqrt(81), emphasizing the role of the negative sign's position.
	$− \sqrt{81}$
The negative is in front of the radical sign.	−9

ⓓ

Demonstrates the evaluation of -sqrt(289), clarifying the role of the negative sign and presenting the final value of -17.
	$− \sqrt{289}$
The negative is in front of the radical sign.	−17

Simplify: ⓐ $\sqrt{−169}$ ⓑ $− \sqrt{64}$ .

Solution

ⓐ

This table demonstrates that the square root of a negative number, such as -169, is not a real number, providing both explanation and mathematical context.
	$\sqrt{−169}$
There is no real number whose square is −169.	$\sqrt{−169}$ is not a real number.

ⓑ

Evaluation of the expression -sqrt(64), showing the impact of the leading negative sign on the final result of -8.
	$− \sqrt{64}$
The negative is in front of the radical.	−8

When using the order of operations to simplify an expression that has square roots, we treat the radical as a grouping symbol.

Simplify: ⓐ $\sqrt{25} + \sqrt{144}$ ⓑ $\sqrt{25 + 144}$ .

Solution

ⓐ

Illustrates the step-by-step simplification of a mathematical expression involving square roots using the order of operations.
	$\sqrt{25} + \sqrt{144}$
Use the order of operations.	$5 + 12$
Simplify.	17

ⓑ

Steps illustrating the simplification of a mathematical square root expression.
	$\sqrt{25 + 144}$
Simplify under the radical sign.	$\sqrt{169}$
Simplify.	13
Notice the different answers in parts ⓐ and ⓑ !

Estimate Square Roots

So far we have only considered square roots of perfect square numbers. The square roots of other numbers are not whole numbers. Look at Table 9 below.

Number	Square Root
4	$\sqrt{4}$ = 2
5	$\sqrt{5}$
6	$\sqrt{6}$
7	$\sqrt{7}$
8	$\sqrt{8}$
9	$\sqrt{9}$ = 3

The square roots of numbers between 4 and 9 must be between the two consecutive whole numbers 2 and 3, and they are not whole numbers. Based on the pattern in the table above, we could say that $\sqrt{5}$ must be between 2 and 3. Using inequality symbols, we write:

2 < \sqrt{5} < 3

Estimate $\sqrt{60}$ between two consecutive whole numbers.

Solution

Think of the perfect square numbers closest to 60. Make a small table of these perfect squares and their squares roots.


Locate 60 between two consecutive perfect squares.
$\sqrt{60}$ is between their square roots.

Approximate Square Roots

There are mathematical methods to approximate square roots, but nowadays most people use a calculator to find them. Find the $\sqrt{x}$ key on your calculator. You will 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 square root. It is an approximation, accurate to the number of digits shown on your calculator’s display. The symbol for an approximation is $\approx$ and it is read ‘approximately.’

Suppose your calculator has a 10-digit display. You would see that

\sqrt{5} \approx 2.236067978

If we wanted to round $\sqrt{5}$ to two decimal places, we would say

\sqrt{5} \approx 2.24

How do we know these values are approximations and not the exact values? Look at what happens when we square them:

\begin{matrix} {(2.236067978)}^{2} & = & 5.000000002 \\ {(2.24)}^{2} & = & 5.0176 \end{matrix}

Their squares are close to 5, but are not exactly equal to 5.

Using the square root key on a calculator and then rounding to two decimal places, we can find:

\begin{matrix} \sqrt{4} & = & 2 \\ \sqrt{5} & \approx & 2.24 \\ \sqrt{6} & \approx & 2.45 \\ \sqrt{7} & \approx & 2.65 \\ \sqrt{8} & \approx & 2.83 \\ \sqrt{9} & = & 3 \end{matrix}

Round $\sqrt{17}$ to two decimal places.

Solution

Steps to approximate the square root of 17, demonstrating calculator use and rounding to two decimal places, resulting in 4.12.
	$\sqrt{17}$
Use the calculator square root key.	4.123105626...
Round to two decimal places.	4.12
	$\sqrt{17} \approx 4.12$

Simplify Variable Expressions with Square Roots

What if we have to find a square root of an expression with a variable? Consider $\sqrt{9 x^{2}}$ . Can you think of an expression whose square is $9 x^{2}$ ?

\begin{matrix} {(?)}^{2} & = & 9 x^{2} \\ {(3 x)}^{2} & = & 9 x^{2}, & so \sqrt{9 x^{2}} = 3 x \end{matrix}

When we use the radical sign to take the square root of a variable expression, we should specify that $x \geq 0$ to make sure we get the principal square root.

However, in this chapter we will assume that each variable in a square-root expression represents a non-negative number and so we will not write $x \geq 0$ next to every radical.

What about square roots of higher powers of variables? Think about the Power Property of Exponents we used in Chapter 6.

{(a^{m})}^{n} = a^{m \cdot n}

If we square $a^{m}$ , the exponent will become $2 m$ .

{(a^{m})}^{2} = a^{2 m}

How does this help us take square roots? Let’s look at a few:

\begin{matrix} \sqrt{25 u^{8}} = 5 u^{4} & because {(5 u^{4})}^{2} = 25 u^{8} \\ \sqrt{16 r^{20}} = 4 r^{10} & because {(4 r^{10})}^{2} = 16 r^{20} \\ \sqrt{196 q^{36}} = 14 q^{18} & because {(14 q^{18})}^{2} = 196 q^{36} \end{matrix}

Simplify: ⓐ $\sqrt{x^{6}}$ ⓑ $\sqrt{y^{16}}$ .

Solution

ⓐ

Simplification of the mathematical expression <m:math><m:msqrt><m:mrow><m:msup><m:mi>x</m:mi><m:mn>6</m:mn></m:msup></m:mrow></m:msqrt></m:math> with justification.
	$\sqrt{x^{6}}$
$Since {(x^{3})}^{2} = x^{6} .$	$x^{3}$

ⓑ

Simplification of the square root of y to the power of 16, demonstrating the result y to the power of 8 using exponent properties.
	$\sqrt{y^{16}}$
$Since {(y^{8})}^{2} = y^{16} .$	$y^{8}$

Simplify: $\sqrt{16 n^{2}}$ .

Solution

This table demonstrates the simplification of the square root expression sqrt(16n^2) to 4n, including its mathematical justification.
	$\sqrt{16 n^{2}}$
$Since {(4 n)}^{2} = 16 n^{2} .$	$4 n$

Simplify: $− \sqrt{81 c^{2}}$ .

Solution

This table demonstrates the simplification of the mathematical expression −√(81c^2) along with the reasoning for the steps.
	$− \sqrt{81 c^{2}}$
$Since {(9 c)}^{2} = 81 c^{2} .$	$−9 c$

Simplify: $\sqrt{36 x^{2} y^{2}}$ .

Solution

An example demonstrating the simplification of the square root expression `sqrt(36x^2y^2)` to `6xy`, including mathematical justification.
	$\sqrt{36 x^{2} y^{2}}$
$Since {(6 x y)}^{2} = 36 x^{2} y^{2} .$	$6 x y$

Simplify: $\sqrt{64 p^{64}}$ .

Solution

This table illustrates the simplification of a square root expression, showing the original term, its explanation, and the simplified result.
	$\sqrt{64 p^{64}}$
$Since {(8 p^{32})}^{2} = 64 p^{64} .$	$8 p^{32}$

Simplify: $\sqrt{121 a^{6} b^{8}}$

Solution

This table demonstrates the simplification of an algebraic square root, presenting the original expression, the rationale, and the final simplified form.
	$\sqrt{121 a^{6} b^{8}}$
$Since {(11 a^{3} b^{4})}^{2} = 121 a^{6} b^{8} .$	$11 a^{3} b^{4}$

Key Concepts

Note that the square root of a negative number is not a real number.
Every positive number has two square roots, one positive and one negative. The positive square root of a positive number is the principal square root.
We can estimate square roots using nearby perfect squares.
We can approximate square roots using a calculator.
When we use the radical sign to take the square root of a variable expression, we should specify that $x \geq 0$ to make sure we get the principal square root.

Practice Makes Perfect

Simplify Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{36}$

Solution

$\sqrt{4}$

$\sqrt{64}$

Solution

$\sqrt{169}$

$\sqrt{9}$

Solution

$\sqrt{16}$

$\sqrt{100}$

Solution

$\sqrt{144}$

$− \sqrt{4}$

Solution

$−2$

$− \sqrt{100}$

$− \sqrt{1}$

Solution

$−1$

$− \sqrt{121}$

$\sqrt{−121}$

Solution

not a real number

$\sqrt{−36}$

$\sqrt{−9}$

Solution

not a real number

$\sqrt{−49}$

$\sqrt{9 + 16}$

Solution

$\sqrt{25 + 144}$

$\sqrt{9} + \sqrt{16}$

Solution

$\sqrt{25} + \sqrt{144}$

Estimate Square Roots

In the following exercises, estimate each square root between two consecutive whole numbers.

$\sqrt{70}$

Solution

$8 < \sqrt{70} < 9$

$\sqrt{55}$

$\sqrt{200}$

Solution

$14 < \sqrt{200} < 15$

$\sqrt{172}$

Approximate Square Roots

In the following exercises, approximate each square root and round to two decimal places.

$\sqrt{19}$

Solution

4.36

$\sqrt{21}$

$\sqrt{53}$

Solution

7.28

$\sqrt{47}$

Simplify Variable Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{y^{2}}$

Solution

$y$

$\sqrt{b^{2}}$

$\sqrt{a^{14}}$

Solution

$a^{7}$

$\sqrt{w^{24}}$

$\sqrt{49 x^{2}}$

Solution

$7 x$

$\sqrt{100 y^{2}}$

$\sqrt{121 m^{20}}$

Solution

$11 m^{10}$

$\sqrt{25 h^{44}}$

$\sqrt{81 x^{36}}$

Solution

$9 x^{18}$

$\sqrt{144 z^{84}}$

$− \sqrt{81 x^{18}}$

Solution

$−9 x^{9}$

$− \sqrt{100 m^{32}}$

$− \sqrt{64 a^{2}}$

Solution

$−8 a$

$− \sqrt{25 x^{2}}$

$\sqrt{144 x^{2} y^{2}}$

Solution

$12 x y$

$\sqrt{196 a^{2} b^{2}}$

$\sqrt{169 w^{8} y^{10}}$

Solution

$13 w^{4} y^{5}$

$\sqrt{81 p^{24} q^{6}}$

$\sqrt{9 c^{8} d^{12}}$

Solution

$3 c^{4} d^{6}$

$\sqrt{36 r^{6} s^{20}}$

Everyday Math

Decorating Denise wants to have 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?

Solution

25 centimeters

Decorating Morris wants to have a square mosaic inlaid in his new patio. His budget allows for 2025 square inch tiles. How long can a side of the mosaic be?

Writing Exercises

Why is there no real number equal to $\sqrt{−64}$ ?

Solution

Answers will vary.

What is the difference between $9^{2}$ and $\sqrt{9}$ ?

Self Check

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

This table has four columns and five rows. The columns are labeled, “I can…,” “Confidentally,” “With some help,” and “No – I don’t get it!” Under the “I can…,” column are, “simplify expressions with square roots.,” “estimate square roots.,” “approximate square roots.,” and “4) simplify variable expressions with square roots.” All the other rows under the different columns are empty.

ⓑ 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?

square of a number

If $n^{2} = m$ , then $m$ is the square of $n$

square root of a number

If $n^{2} = m$ , then $n$ is a square root of $m$

square root notation

If $m = n^{2}$ , then $\sqrt{m} = n$ . We read $\sqrt{m}$ as ‘the square root of $m$ .’

Simplify Square Roots

In the last section, we estimated the square root of a number between two consecutive whole numbers. We can say that $\sqrt{50}$ is between 7 and 8. This is fairly easy to do when the numbers are small enough that we can use Figure 1 in Simplify and Use Square Roots.

But what if we want to estimate $\sqrt{500}$ ? If we simplify the square root first, we’ll be able to estimate it easily. There are other reasons, too, to simplify square roots as you’ll see later in this chapter.

A square root is considered simplified if its radicand contains no perfect square factors.

So $\sqrt{31}$ is simplified. But $\sqrt{32}$ is not simplified, because 16 is a perfect square factor of 32.

Use the Product Property to Simplify Square Roots

The properties we will use to simplify expressions with square roots are similar to the properties of exponents. We know that ${(a b)}^{m} = a^{m} b^{m}$ . The corresponding property of square roots says that $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$ .

We use the Product Property of Square Roots to remove all perfect square factors from a radical. We will show how to do this in Example 1.

How To Use the Product Property to Simplify a Square Root

Simplify: $\sqrt{50}$ .

Solution

This figure has three columns and three rows. The first row says, “Step 1. Find the largest perfect square factor of the radicand. Rewrite the radicand as a product using the perfect square factor.” It then says, “25 is the largest perfect square factor of 50. 50 equals 25 times 2. Always write the perfect square factor first.” Then it shows the square root of 50 and the square root of 25 times 2.

The second row says, “Step 2. Use the product rule to rewrite the radical as the product of two radicals.” The second column is empty, but the third column shows the square root of 25 times the square root of 2.

The third row says, “Step 3. Simplify the square root of the perfect square.” The second column is empty, but the third column shows 5 times the square root of 2.

Notice in the previous example that the simplified form of $\sqrt{50}$ is $5 \sqrt{2}$ , which is the product of an integer and a square root. We always write the integer in front of the square root.

Simplify: $\sqrt{500}$ .

Solution

Illustrates the step-by-step process of simplifying the square root of 500 by factoring out perfect squares.
	$\sqrt{500}$
Rewrite the radicand as a product using the largest perfect square factor.	$\sqrt{100 \cdot 5}$
Rewrite the radical as the product of two radicals.	$\sqrt{100} \cdot \sqrt{5}$
Simplify.	$10 \sqrt{5}$

We could use the simplified form $10 \sqrt{5}$ to estimate $\sqrt{500}$ . We know 5 is between 2 and 3, and $\sqrt{500}$ is $10 \sqrt{5}$ . So $\sqrt{500}$ is between 20 and 30.

The next example is much like the previous examples, but with variables.

Simplify: $\sqrt{x^{3}}$ .

Solution

This table illustrates the step-by-step simplification of the square root expression sqrt(x^3), showing the procedure and the evolving mathematical form.
	$\sqrt{x^{3}}$
Rewrite the radicand as a product using the largest perfect square factor.	$\sqrt{x^{2} \cdot x}$
Rewrite the radical as the product of two radicals.	$\sqrt{x^{2}} \cdot \sqrt{x}$
Simplify.	$x \sqrt{x}$

We follow the same procedure when there is a coefficient in the radical, too.

Simplify: $\sqrt{25 y^{5} .}$

Solution

Illustrates the step-by-step process of simplifying the square root expression sqrt(25y^5).
	$\sqrt{25 y^{5}}$
Rewrite the radicand as a product using the largest perfect square factor.	$\sqrt{25 y^{4} \cdot y}$
Rewrite the radical as the product of two radicals.	$\sqrt{25 y^{4}} \cdot \sqrt{y}$
Simplify.	$5 y^{2} \sqrt{y}$

In the next example both the constant and the variable have perfect square factors.

Simplify: $\sqrt{72 n^{7}}$ .

Solution

Step-by-step simplification of a square root expression (sqrt(72n^7)), showing the process of extracting perfect square factors.
	$\sqrt{72 n^{7}}$
Rewrite the radicand as a product using the largest perfect square factor.	$\sqrt{36 n^{6} \cdot 2 n}$
Rewrite the radical as the product of two radicals.	$\sqrt{36 n^{6}} \cdot \sqrt{2 n}$
Simplify.	$6 n^{3} \sqrt{2 n}$

Simplify: $\sqrt{63 u^{3} v^{5}}$ .

Solution

This table illustrates the step-by-step process of simplifying the radical expression sqrt(63u^3v^5).
	$\sqrt{63 u^{3} v^{5}}$
Rewrite the radicand as a product using the largest perfect square factor.	$\sqrt{9 u^{2} v^{4} \cdot 7 u v}$
Rewrite the radical as the product of two radicals.	$\sqrt{9 u^{2} v^{4}} \cdot \sqrt{7 u v}$
Simplify.	$3 u v^{2} \sqrt{7 u v}$

We have seen how to use the Order of Operations to simplify some expressions with radicals. To simplify $\sqrt{25} + \sqrt{144}$ we must simplify each square root separately first, then add to get the sum of 17.

The expression $\sqrt{17} + \sqrt{7}$ cannot be simplified—to begin we’d need to simplify each square root, but neither 17 nor 7 contains a perfect square factor.

In the next example, we have the sum of an integer and a square root. We simplify the square root but cannot add the resulting expression to the integer.

Simplify: $3 + \sqrt{32}$ .

Solution

This table demonstrates the step-by-step simplification of the radical expression 3 + sqrt(32).
	$3 + \sqrt{32}$
Rewrite the radicand as a product using the largest perfect square factor.	$3 + \sqrt{16 \cdot 2}$
Rewrite the radical as the product of two radicals.	$3 + \sqrt{16} \cdot \sqrt{2}$
Simplify.	$3 + 4 \sqrt{2}$

The terms are not like and so we cannot add them. Trying to add an integer and a radical is like trying to add an integer and a variable—they are not like terms!

The next example includes a fraction with a radical in the numerator. Remember that in order to simplify a fraction you need a common factor in the numerator and denominator.

Simplify: $\frac{4 - \sqrt{48}}{2}$ .

Solution

Step-by-step simplification of the radical expression (4 - 48)/2 to its simplified form 2(1 - 3).
	$\frac{4 - \sqrt{48}}{2}$
Rewrite the radicand as a product using the largest perfect square factor.	$\frac{4 - \sqrt{16 \cdot 3}}{2}$
Rewrite the radical as the product of two radicals.	$\frac{4 - \sqrt{16} \cdot \sqrt{3}}{2}$
Simplify.	$\frac{4 - 4 \sqrt{3}}{2}$
Factor the common factor from the numerator.	$\frac{4 (1 - \sqrt{3})}{2}$
Remove the common factor, 2, from the numerator and denominator.	$\frac{2 \cdot 2 (1 - \sqrt{3})}{2}$
Simplify.	$2 (1 - \sqrt{3})$

Use the Quotient Property to Simplify Square Roots

Whenever you have to simplify a square root, the first step you should take is to determine whether the radicand is a perfect square. A perfect square fraction is a fraction in which both the numerator and the denominator are perfect squares.

Simplify: $\sqrt{\frac{9}{64}}$ .

Solution

$\begin{matrix} \sqrt{\frac{9}{64}} \\ Since {(\frac{3}{8})}^{2} = \frac{9}{64} & \frac{3}{8} \end{matrix}$

If the numerator and denominator have any common factors, remove them. You may find a perfect square fraction!

Simplify: $\sqrt{\frac{45}{80}}$ .

Solution

This table demonstrates the step-by-step process of simplifying a square root of a fraction, showing both the instructions and the corresponding mathematical expressions.
	$\sqrt{\frac{45}{80}}$
Simplify inside the radical first. Rewrite showing the common factors of the numerator and denominator.	$\sqrt{\frac{5 \cdot 9}{5 \cdot 16}}$
Simplify the fraction by removing common factors.	$\sqrt{\frac{9}{16}}$
$Simplify. {(\frac{3}{4})}^{2} = \frac{9}{16}$	$\frac{3}{4}$

In the last example, our first step was to simplify the fraction under the radical by removing common factors. In the next example we will use the Quotient Property to simplify under the radical. We divide the like bases by subtracting their exponents, $\frac{a^{m}}{a^{n}} = a^{m - n}, a \neq 0$ .

Simplify: $\sqrt{\frac{m^{6}}{m^{4}}}$ .

Solution

Step-by-step simplification of the radical expression involving variables and exponents.
	$\sqrt{\frac{m^{6}}{m^{4}}}$
Simplify the fraction inside the radical first.
Divide the like bases by subtracting the exponents.	$\sqrt{m^{2}}$
Simplify.	$m$

Simplify: $\sqrt{\frac{48 p^{7}}{3 p^{3}}}$ .

Solution

This table illustrates the step-by-step simplification of a radical algebraic expression.
	$\sqrt{\frac{48 p^{7}}{3 p^{3}}}$
Simplify the fraction inside the radical first.	$\sqrt{16 p^{4}}$
Simplify.	$4 p^{2}$

Remember the Quotient to a Power Property? It said we could raise a fraction to a power by raising the numerator and denominator to the power separately.

{(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}, b \neq 0

We can use a similar property to simplify a square root of a fraction. After removing all common factors from the numerator and denominator, if the fraction is not a perfect square we simplify the numerator and denominator separately.

Simplify: $\sqrt{\frac{21}{64}}$ .

Solution

Step-by-step simplification of the radical expression sqrt(21/64) using the quotient property of square roots.
	$\sqrt{\frac{21}{64}}$
We cannot simplify the fraction inside the radical. Rewrite using the quotient property.	$\frac{\sqrt{21}}{\sqrt{64}}$
Simplify the square root of 64. The numerator cannot be simplified.	$\frac{\sqrt{21}}{8}$

How to Use the Quotient Property to Simplify a Square Root

Simplify: $\sqrt{\frac{27 m^{3}}{196}}$ .

Solution

$This table has three columns and three rows. The first row reads, “Step 1. Simplify the fraction in the radicand, if possible.” Then it shows that 27 m cubed over 196 cannot be simplified. Then it shows the square root of 27 m cubed over 196.$ The second row says, “Step 2. Use the Quotient Property to rewrite the radical as the quotient of two radicals.” Then it says, “We rewrite the square root of 27 m cubed over 196 as the quotient of the square root of 27 m cubed and the square root of 196.” Then it shows the square root of 27 m cubed over the square root of 196.

The second row says, “Step 2. Use the Quotient Property to rewrite the radical as the quotient of two radicals.” Then it says, “We rewrite the square root of 27 m cubed over 196 as the quotient of the square root of 27 m cubed and the square root of 196.” Then it shows the square root of 27 m cubed over the square root of 196.

The third row says, “Step 3. Simplify the radicals in the numerator and the denominator.” Then it says, “9 m squared and 196 are perfect squares.” It then shows the square root of 9 m squared time the square root of 3 m over the square root of 196. It then shows 3 m times the square root of 3 m over 14.

Simplify: $\sqrt{\frac{45 x^{5}}{y^{4}}}$ .

Solution

Step-by-step simplification of a radical expression involving a fraction, illustrating the application of the Quotient Property of radicals.
	$\sqrt{\frac{45 x^{5}}{y^{4}}}$
We cannot simplify the fraction in the radicand. Rewrite using the Quotient Property.	$\frac{\sqrt{45 x^{5}}}{\sqrt{y^{4}}}$
Simplify the radicals in the numerator and the denominator.	$\frac{\sqrt{9 x^{4}} \cdot \sqrt{5 x}}{y^{2}}$
Simplify.	$\frac{3 x^{2} \sqrt{5 x}}{y^{2}}$

Be sure to simplify the fraction in the radicand first, if possible.

Simplify: $\sqrt{\frac{81 d^{9}}{25 d^{4}}}$ .

Solution

Step-by-step simplification of a radical expression involving variables and fractions.
	$\sqrt{\frac{81 d^{9}}{25 d^{4}}}$
Simplify the fraction in the radicand.	$\sqrt{\frac{81 d^{5}}{25}}$
Rewrite using the Quotient Property.	$\frac{\sqrt{81 d^{5}}}{\sqrt{25}}$
Simplify the radicals in the numerator and the denominator.	$\frac{\sqrt{81 d^{4}} \cdot \sqrt{d}}{5}$
Simplify.	$\frac{9 d^{2} \sqrt{d}}{5}$

Simplify: $\sqrt{\frac{18 p^{5} q^{7}}{32 p q^{2}}}$ .

Solution

Step-by-step simplification of a radical expression with fractions and variables using algebraic properties.
	$\sqrt{\frac{18 p^{5} q^{7}}{32 p q^{2}}}$
Simplify the fraction in the radicand, if possible.	$\sqrt{\frac{9 p^{4} q^{5}}{16}}$
Rewrite using the Quotient Property.	$\frac{\sqrt{9 p^{4} q^{5}}}{\sqrt{16}}$
Simplify the radicals in the numerator and the denominator.	$\frac{\sqrt{9 p^{4} q^{4}} \cdot \sqrt{q}}{4}$
Simplify.	$\frac{3 p^{2} q^{2} \sqrt{q}}{4}$

Key Concepts

Simplified Square Root $\sqrt{a}$ is considered simplified if $a$ has no perfect-square factors.
Product Property of Square Roots If a, b are non-negative real numbers, then

$\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$
Simplify a Square Root Using the Product Property To simplify a square root using the Product Property:
1. Find the largest perfect square factor of the radicand. Rewrite the radicand as a product using the perfect square factor.
2. Use the product rule to rewrite the radical as the product of two radicals.
3. Simplify the square root of the perfect square.
Quotient Property of Square Roots If a, b are non-negative real numbers and $b \neq 0$ , then

$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$
Simplify a Square Root Using the Quotient Property To simplify a square root using the Quotient Property:
1. Simplify the fraction in the radicand, if possible.
2. Use the Quotient Rule to rewrite the radical as the quotient of two radicals.
3. Simplify the radicals in the numerator and the denominator.

Practice Makes Perfect

Use the Product Property to Simplify Square Roots

In the following exercises, simplify.

$\sqrt{27}$

Solution

$3 \sqrt{3}$

$\sqrt{80}$

$\sqrt{125}$

Solution

$5 \sqrt{5}$

$\sqrt{96}$

$\sqrt{200}$

Solution

$10 \sqrt{2}$

$\sqrt{147}$

$\sqrt{450}$

Solution

$15 \sqrt{2}$

$\sqrt{252}$

$\sqrt{800}$

Solution

$20 \sqrt{2}$

$\sqrt{288}$

$\sqrt{675}$

Solution

$15 \sqrt{3}$

$\sqrt{1250}$

$\sqrt{x^{7}}$

Solution

$x^{3} \sqrt{x}$

$\sqrt{y^{11}}$

$\sqrt{p^{3}}$

Solution

$p \sqrt{p}$

$\sqrt{q^{5}}$

$\sqrt{m^{13}}$

Solution

$m^{6} \sqrt{m}$

$\sqrt{n^{21}}$

$\sqrt{r^{25}}$

Solution

$r^{12} \sqrt{r}$

$\sqrt{s^{33}}$

$\sqrt{49 n^{17}}$

Solution

$7 n^{8} \sqrt{n}$

$\sqrt{25 m^{9}}$

$\sqrt{81 r^{15}}$

Solution

$9 r^{7} \sqrt{r}$

$\sqrt{100 s^{19}}$

$\sqrt{98 m^{5}}$

Solution

$7 m^{2} \sqrt{2 m}$

$\sqrt{32 n^{11}}$

$\sqrt{125 r^{13}}$

Solution

$5 r^{6} \sqrt{5 r}$

$\sqrt{80 s^{15}}$

$\sqrt{200 p^{13}}$

Solution

$10 p^{6} \sqrt{2 p}$

$\sqrt{128 q^{3}}$

$\sqrt{242 m^{23}}$

Solution

$11 m^{11} \sqrt{2 m}$

$\sqrt{175 n^{13}}$

$\sqrt{147 m^{7} n^{11}}$

Solution

$7 m^{3} n^{5} \sqrt{3 m n}$

$\sqrt{48 m^{7} n^{5}}$

$\sqrt{75 r^{13} s^{9}}$

Solution

$5 r^{6} s^{4} \sqrt{3 r s}$

$\sqrt{96 r^{3} s^{3}}$

$\sqrt{300 p^{9} q^{11}}$

Solution

$10 p^{4} q^{5} \sqrt{3 p q}$

$\sqrt{192 q^{3} r^{7}}$

$\sqrt{242 m^{13} n^{21}}$

Solution

$11 m^{6} n^{10} \sqrt{2 m n}$

$\sqrt{150 m^{9} n^{3}}$

$5 + \sqrt{12}$

Solution

$5 + 2 \sqrt{3}$

$8 + \sqrt{96}$

$1 + \sqrt{45}$

Solution

$1 + 3 \sqrt{5}$

$3 + \sqrt{125}$

$\frac{10 - \sqrt{24}}{2}$

Solution

$5 - \sqrt{6}$

$\frac{8 - \sqrt{80}}{4}$

$\frac{3 + \sqrt{90}}{3}$

Solution

$1 + \sqrt{10}$

$\frac{15 + \sqrt{75}}{5}$

Use the Quotient Property to Simplify Square Roots

In the following exercises, simplify.

$\sqrt{\frac{49}{64}}$

Solution

$\frac{7}{8}$

$\sqrt{\frac{100}{36}}$

$\sqrt{\frac{121}{16}}$

Solution

$\frac{11}{4}$

$\sqrt{\frac{144}{169}}$

$\sqrt{\frac{72}{98}}$

Solution

$\frac{6}{7}$

$\sqrt{\frac{75}{12}}$

$\sqrt{\frac{45}{125}}$

Solution

$\frac{3}{5}$

$\sqrt{\frac{300}{243}}$

$\sqrt{\frac{x^{10}}{x^{6}}}$

Solution

$x^{2}$

$\sqrt{\frac{p^{20}}{p^{10}}}$

$\sqrt{\frac{y^{4}}{y^{8}}}$

Solution

$\frac{1}{y^{2}}$

$\sqrt{\frac{q^{8}}{q^{14}}}$

$\sqrt{\frac{200 x^{7}}{2 x^{3}}}$

Solution

$10 x^{2}$

$\sqrt{\frac{98 y^{11}}{2 y^{5}}}$

$\sqrt{\frac{96 p^{9}}{6 p}}$

Solution

$4 p^{4}$

$\sqrt{\frac{108 q^{10}}{3 q^{2}}}$

$\sqrt{\frac{36}{35}}$

Solution

$\frac{6}{\sqrt{35}}$

$\sqrt{\frac{144}{65}}$

$\sqrt{\frac{20}{81}}$

Solution

$\frac{2 \sqrt{5}}{9}$

$\sqrt{\frac{21}{196}}$

$\sqrt{\frac{96 x^{7}}{121}}$

Solution

$\frac{4 x^{3} \sqrt{6 x}}{11}$

$\sqrt{\frac{108 y^{4}}{49}}$

$\sqrt{\frac{300 m^{5}}{64}}$

Solution

$\frac{5 m^{2} \sqrt{3 m}}{4}$

$\sqrt{\frac{125 n^{7}}{169}}$

$\sqrt{\frac{98 r^{5}}{100}}$

Solution

$\frac{7 r^{2} \sqrt{2 r}}{10}$

$\sqrt{\frac{180 s^{10}}{144}}$

$\sqrt{\frac{28 q^{6}}{225}}$

Solution

$\frac{2 q^{3} \sqrt{7}}{15}$

$\sqrt{\frac{150 r^{3}}{256}}$

$\sqrt{\frac{75 r^{9}}{s^{8}}}$

Solution

$\frac{5 r^{4} \sqrt{3 r}}{s^{4}}$

$\sqrt{\frac{72 x^{5}}{y^{6}}}$

$\sqrt{\frac{28 p^{7}}{q^{2}}}$

Solution

$\frac{2 p^{3} \sqrt{7 p}}{q}$

$\sqrt{\frac{45 r^{3}}{s^{10}}}$

$\sqrt{\frac{100 x^{5}}{36 x^{3}}}$

Solution

$\frac{5 x}{3}$

$\sqrt{\frac{49 r^{12}}{16 r^{6}}}$

$\sqrt{\frac{121 p^{5}}{81 p^{2}}}$

Solution

$\frac{11 p \sqrt{p}}{9}$

$\sqrt{\frac{25 r^{8}}{64 r}}$

$\sqrt{\frac{32 x^{5} y^{3}}{18 x^{3} y}}$

Solution

$\frac{4 x y}{3}$

$\sqrt{\frac{75 r^{6} s^{8}}{48 r s^{4}}}$

$\sqrt{\frac{27 p^{2} q}{108 p^{5} q^{3}}}$

Solution

$\frac{1}{2 p q \sqrt{p}}$

$\sqrt{\frac{50 r^{5} s^{2}}{128 r^{2} s^{5}}}$

Everyday Math

ⓐ Elliott decides to construct a square garden that will take up 288 square feet of his yard. Simplify $\sqrt{288}$ to determine the length and the width of his garden. Round to the nearest tenth of a foot.
ⓑ Suppose Elliott decides to reduce the size of his square garden so that he can create a 5-foot-wide walking path on the north and east sides of the garden. Simplify $\sqrt{288} - 5$ to determine the length and width of the new garden. Round to the nearest tenth of a foot.

Solution

ⓐ $17.0 feet$ ⓑ $12.0 feet$

ⓐ Melissa accidentally drops a pair of sunglasses from the top of a roller coaster, 64 feet above the ground. Simplify $\sqrt{\frac{64}{16}}$ to determine the number of seconds it takes for the sunglasses to reach the ground.
ⓑ Suppose the sunglasses in the previous example were dropped from a height of 144 feet. Simplify $\sqrt{\frac{144}{16}}$ to determine the number of seconds it takes for the sunglasses to reach the ground.

Writing Exercises

Explain why $\sqrt{x^{4}} = x^{2}$ . Then explain why $\sqrt{x^{16}} = x^{8}$ .

Solution

Answers will vary.

Explain why $7 + \sqrt{9}$ is not equal to $\sqrt{7 + 9}$ .

Self Check

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

This table has four columns and three rows. The columns are labeled, “I can…,” “confidently,” “with some help,” and “no—I don’t get it!” The rows under “I can…” Read, “use the Product Property to simplify square roots.,” and “use the Quotient Property to simplify square roots.” The other rows unders the other columns are blank.

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

Add and Subtract Square Roots

We know that we must follow the order of operations to simplify expressions with square roots. The radical is a grouping symbol, so we work inside the radical first. We simplify $\sqrt{2 + 7}$ in this way:

$\begin{matrix} \sqrt{2 + 7} \\ Add inside the radical. & \sqrt{9} \\ Simplify. & 3 \end{matrix}$

So if we have to add $\sqrt{2} + \sqrt{7}$ , we must not combine them into one radical.

\sqrt{2} + \sqrt{7} \neq \sqrt{2 + 7}

Trying to add square roots with different radicands is like trying to add unlike terms.

$\begin{matrix} But, just like we can add & x + x, & we can add & \sqrt{3} + \sqrt{3} . \\ x + x = 2 x & \sqrt{3} + \sqrt{3} = 2 \sqrt{3} \end{matrix}$

Adding square roots with the same radicand is just like adding like terms. We call square roots with the same radicand like square roots to remind us they work the same as like terms.

We add and subtract like square roots in the same way we add and subtract like terms. We know that $3 x + 8 x$ is $11 x$ . Similarly we add $3 \sqrt{x} + 8 \sqrt{x}$ and the result is $11 \sqrt{x} .$

Add and Subtract Like Square Roots

Think about adding like terms with variables as you do the next few examples. When you have like radicands, you just add or subtract the coefficients. When the radicands are not like, you cannot combine the terms.

Simplify: $2 \sqrt{2} - 7 \sqrt{2}$ .

Solution

Example demonstrating the subtraction of like radicals.
	$2 \sqrt{2} - 7 \sqrt{2}$
Since the radicals are like, we subtract the coefficients.	$−5 \sqrt{2}$

Simplify: $3 \sqrt{y} + 4 \sqrt{y}$ .

Solution

Demonstrates simplifying like radical expressions by adding their coefficients to obtain a combined radical term.
	$3 \sqrt{y} + 4 \sqrt{y}$
Since the radicals are like, we add the coefficients.	$7 \sqrt{y}$

Simplify: $4 \sqrt{x} - 2 \sqrt{y}$ .

Solution

This table demonstrates why expressions with unlike radicals cannot be subtracted, illustrating the final form of such an expression.
	$4 \sqrt{x} - 2 \sqrt{y}$
Since the radicals are not like, we cannot subtract them. We leave the expression as is.	$4 \sqrt{x} - 2 \sqrt{y}$

Simplify: $5 \sqrt{13} + 4 \sqrt{13} + 2 \sqrt{13}$ .

Solution

This table illustrates the process of adding like radicals, showing the initial expression, the rule applied, and the simplified result.
	$5 \sqrt{13} + 4 \sqrt{13} + 2 \sqrt{13}$
Since the radicals are like, we add the coefficients.	$11 \sqrt{13}$

Simplify: $2 \sqrt{6} - 6 \sqrt{6} + 3 \sqrt{3}$ .

Solution

This table illustrates the step-by-step simplification of a radical expression by combining like terms, showing the operation and the resulting expression.
	$2 \sqrt{6} - 6 \sqrt{6} + 3 \sqrt{3}$
Since the first two radicals are like, we subtract their coefficients.	$- 4 \sqrt{6} + 3 \sqrt{3}$

Simplify: $2 \sqrt{5 n} - 6 \sqrt{5 n} + 4 \sqrt{5 n}$ .

Solution

Step-by-step simplification of an algebraic expression involving like radicals, demonstrating its reduction to zero.
	$2 \sqrt{5 n} - 6 \sqrt{5 n} + 4 \sqrt{5 n}$
Since the radicals are like, we combine them.	$0 \sqrt{5 n}$
Simplify.	0

When radicals contain more than one variable, as long as all the variables and their exponents are identical, the radicals are like.

Simplify: $\sqrt{3 x y} + 5 \sqrt{3 x y} - 4 \sqrt{3 x y}$ .

Solution

This table demonstrates the step-by-step simplification of a mathematical expression involving like radicals.
	$\sqrt{3 x y} + 5 \sqrt{3 x y} - 4 \sqrt{3 x y}$
Since the radicals are like, we combine them.	$2 \sqrt{3 x y}$

Add and Subtract Square Roots that Need Simplification

Remember that we always simplify square roots by removing the largest perfect-square factor. Sometimes when we have to add or subtract square roots that do not appear to have like radicals, we find like radicals after simplifying the square roots.

Simplify: $\sqrt{20} + 3 \sqrt{5}$ .

Solution

This table illustrates the step-by-step process of simplifying and combining radical expressions.
	$\sqrt{20} + 3 \sqrt{5}$
Simplify the radicals, when possible.	$\sqrt{4} \cdot \sqrt{5} + 3 \sqrt{5}$
	$2 \sqrt{5} + 3 \sqrt{5}$
Combine the like radicals.	$5 \sqrt{5}$

Simplify: $\sqrt{48} - \sqrt{75}$ .

Solution

Steps to simplify and combine square root expressions.
	$\sqrt{48} - \sqrt{75}$
Simplify the radicals.	$\sqrt{16} \cdot \sqrt{3} - \sqrt{25} \cdot \sqrt{3}$
	$4 \sqrt{3} - 5 \sqrt{3}$
Combine the like radicals.	$− \sqrt{3}$

Just like we use the Associative Property of Multiplication to simplify $5 (3 x)$ and get $15 x$ , we can simplify $5 (3 \sqrt{x})$ and get $15 \sqrt{x}$ . We will use the Associative Property to do this in the next example.

Simplify: $5 \sqrt{18} - 2 \sqrt{8}$ .

Solution

This table demonstrates the step-by-step simplification and combination of radical expressions.
	$5 \sqrt{18} - 2 \sqrt{8}$
Simplify the radicals.	$5 \cdot \sqrt{9} \cdot \sqrt{2} - 2 \cdot \sqrt{4} \cdot \sqrt{2}$
	$5 \cdot 3 \cdot \sqrt{2} - 2 \cdot 2 \cdot \sqrt{2}$
	$15 \sqrt{2} - 4 \sqrt{2}$
Combine the like radicals.	$11 \sqrt{2}$

Simplify: $\frac{3}{4} \sqrt{192} - \frac{5}{6} \sqrt{108}$ .

Solution

Step-by-step simplification of an algebraic expression involving radicals.
	$\frac{3}{4} \sqrt{192} - \frac{5}{6} \sqrt{108}$
Simplify the radicals.	$\frac{3}{4} \sqrt{64} \cdot \sqrt{3} - \frac{5}{6} \sqrt{36} \cdot \sqrt{3}$
	$\frac{3}{4} \cdot 8 \cdot \sqrt{3} - \frac{5}{6} \cdot 6 \cdot \sqrt{3}$
	$6 \sqrt{3} - 5 \sqrt{3}$
Combine the like radicals.	$\sqrt{3}$

Simplify: $\frac{2}{3} \sqrt{48} - \frac{3}{4} \sqrt{12}$ .

Solution

Step-by-step simplification of a radical mathematical expression.
	$\frac{2}{3} \sqrt{48} - \frac{3}{4} \sqrt{12}$
Simplify the radicals.	$\frac{2}{3} \sqrt{16} \cdot \sqrt{3} - \frac{3}{4} \sqrt{4} \cdot \sqrt{3}$
	$\frac{2}{3} \cdot 4 \cdot \sqrt{3} - \frac{3}{4} \cdot 2 \cdot \sqrt{3}$
	$\frac{8}{3} \sqrt{3} - \frac{3}{2} \sqrt{3}$
Find a common denominator to subtract the coefficients of the like radicals.	$\frac{16}{6} \sqrt{3} - \frac{9}{6} \sqrt{3}$
Simplify.	$\frac{7}{6} \sqrt{3}$

In the next example, we will remove constant and variable factors from the square roots.

Simplify: $\sqrt{18 n^{5}} - \sqrt{32 n^{5}}$ .

Solution

Illustrates the step-by-step simplification of an algebraic expression involving the subtraction of radical terms.
	$\sqrt{18 n^{5}} - \sqrt{32 n^{5}}$
Simplify the radicals.	$\sqrt{9 n^{4}} \cdot \sqrt{2 n} - \sqrt{16 n^{4}} \cdot \sqrt{2 n}$
	$3 n^{2} \sqrt{2 n} - 4 n^{2} \sqrt{2 n}$
Combine the like radicals.	$− n^{2} \sqrt{2 n}$

Simplify: $9 \sqrt{50 m^{2}} - 6 \sqrt{48 m^{2}}$ .

Solution

Step-by-step simplification of an algebraic expression with radicals, showing reduction and the condition for combining terms.
	$9 \sqrt{50 m^{2}} - 6 \sqrt{48 m^{2}}$
Simplify the radicals.	$9 \sqrt{25 m^{2}} \cdot \sqrt{2} - 6 \sqrt{16 m^{2}} \cdot \sqrt{3}$
	$9 \cdot 5 m \cdot \sqrt{2} - 6 \cdot 4 m \cdot \sqrt{3}$
	$45 m \sqrt{2} - 24 m \sqrt{3}$
The radicals are not like and so cannot be combined.

Simplify: $2 \sqrt{8 x^{2}} - 5 x \sqrt{32} + 5 \sqrt{18 x^{2}}$ .

Solution

Demonstrates the step-by-step simplification of an algebraic expression containing radicals.
	$2 \sqrt{8 x^{2}} - 5 x \sqrt{32} + 5 \sqrt{18 x^{2}}$
Simplify the radicals.	$2 \sqrt{4 x^{2}} \cdot \sqrt{2} - 5 x \sqrt{16} \cdot \sqrt{2} + 5 \sqrt{9 x^{2}} \cdot \sqrt{2}$
	$2 \cdot 2 x \cdot \sqrt{2} - 5 x \cdot 4 \cdot \sqrt{2} + 5 \cdot 3 x \cdot \sqrt{2}$
	$4 x \sqrt{2} - 20 x \sqrt{2} + 15 x \sqrt{2}$
Combine the like radicals.	$− x \sqrt{2}$

Key Concepts

To add or subtract like square roots, add or subtract the coefficients and keep the like square root.
Sometimes when we have to add or subtract square roots that do not appear to have like radicals, we find like radicals after simplifying the square roots.

Practice Makes Perfect

Add and Subtract Like Square Roots

In the following exercises, simplify.

$8 \sqrt{2} - 5 \sqrt{2}$

Solution

$3 \sqrt{2}$

$7 \sqrt{2} - 3 \sqrt{2}$

$3 \sqrt{5} + 6 \sqrt{5}$

Solution

$9 \sqrt{5}$

$4 \sqrt{5} + 8 \sqrt{5}$

$9 \sqrt{7} - 10 \sqrt{7}$

Solution

$− \sqrt{7}$

$11 \sqrt{7} - 12 \sqrt{7}$

$7 \sqrt{y} + 2 \sqrt{y}$

Solution

$9 \sqrt{y}$

$9 \sqrt{n} + 3 \sqrt{n}$

$\sqrt{a} - 4 \sqrt{a}$

Solution

$−3 \sqrt{a}$

$\sqrt{b} - 6 \sqrt{b}$

$5 \sqrt{c} + 2 \sqrt{c}$

Solution

$7 \sqrt{c}$

$7 \sqrt{d} + 2 \sqrt{d}$

$8 \sqrt{a} - 2 \sqrt{b}$

Solution

$8 \sqrt{a} - 2 \sqrt{b}$

$5 \sqrt{c} - 3 \sqrt{d}$

$5 \sqrt{m} + \sqrt{n}$

Solution

$5 \sqrt{m} + \sqrt{n}$

$\sqrt{n} + 3 \sqrt{p}$

$8 \sqrt{7} + 2 \sqrt{7} + 3 \sqrt{7}$

Solution

$13 \sqrt{7}$

$6 \sqrt{5} + 3 \sqrt{5} + \sqrt{5}$

$3 \sqrt{11} + 2 \sqrt{11} - 8 \sqrt{11}$

Solution

$−3 \sqrt{11}$

$2 \sqrt{15} + 5 \sqrt{15} - 9 \sqrt{15}$

$3 \sqrt{3} - 8 \sqrt{3} + 7 \sqrt{5}$

Solution

$−5 \sqrt{3} + 7 \sqrt{5}$

$5 \sqrt{7} - 8 \sqrt{7} + 6 \sqrt{3}$

$6 \sqrt{2} + 2 \sqrt{2} - 3 \sqrt{5}$

Solution

$8 \sqrt{2} - 3 \sqrt{5}$

$7 \sqrt{5} + \sqrt{5} - 8 \sqrt{10}$

$3 \sqrt{2 a} - 4 \sqrt{2 a} + 5 \sqrt{2 a}$

Solution

$4 \sqrt{2 a}$

$\sqrt{11 b} - 5 \sqrt{11 b} + 3 \sqrt{11 b}$

$8 \sqrt{3 c} + 2 \sqrt{3 c} - 9 \sqrt{3 c}$

Solution

$\sqrt{3 c}$

$3 \sqrt{5 d} + 8 \sqrt{5 d} - 11 \sqrt{5 d}$

$5 \sqrt{3 a b} + \sqrt{3 a b} - 2 \sqrt{3 a b}$

Solution

$4 \sqrt{3 a b}$

$8 \sqrt{11 c d} + 5 \sqrt{11 c d} - 9 \sqrt{11 c d}$

$2 \sqrt{p q} - 5 \sqrt{p q} + 4 \sqrt{p q}$

Solution

$\sqrt{p q}$

$11 \sqrt{2 r s} - 9 \sqrt{2 r s} + 3 \sqrt{2 r s}$

Add and Subtract Square Roots that Need Simplification

In the following exercises, simplify.

$\sqrt{50} + 4 \sqrt{2}$

Solution

$9 \sqrt{2}$

$\sqrt{48} + 2 \sqrt{3}$

$\sqrt{80} - 3 \sqrt{5}$

Solution

$\sqrt{5}$

$\sqrt{28} - 4 \sqrt{7}$

$\sqrt{27} - \sqrt{75}$

Solution

$−2 \sqrt{3}$

$\sqrt{72} - \sqrt{98}$

$\sqrt{48} + \sqrt{27}$

Solution

$7 \sqrt{3}$

$\sqrt{45} + \sqrt{80}$

$2 \sqrt{50} - 3 \sqrt{72}$

Solution

$−8 \sqrt{2}$

$3 \sqrt{98} - \sqrt{128}$

$2 \sqrt{12} + 3 \sqrt{48}$

Solution

$16 \sqrt{3}$

$4 \sqrt{75} + 2 \sqrt{108}$

$\frac{2}{3} \sqrt{72} + \frac{1}{5} \sqrt{50}$

Solution

$5 \sqrt{2}$

$\frac{2}{5} \sqrt{75} + \frac{3}{4} \sqrt{48}$

$\frac{1}{2} \sqrt{20} - \frac{2}{3} \sqrt{45}$

Solution

$− \sqrt{5}$

$\frac{2}{3} \sqrt{54} - \frac{3}{4} \sqrt{96}$

$\frac{1}{6} \sqrt{27} - \frac{3}{8} \sqrt{48}$

Solution

$− \sqrt{3}$

$\frac{1}{8} \sqrt{32} - \frac{1}{10} \sqrt{50}$

$\frac{1}{4} \sqrt{98} - \frac{1}{3} \sqrt{128}$

Solution

$- \frac{11}{12} \sqrt{2}$

$\frac{1}{3} \sqrt{24} + \frac{1}{4} \sqrt{54}$

$\sqrt{72 a^{5}} - \sqrt{50 a^{5}}$

Solution

$a^{2} \sqrt{2 a}$

$\sqrt{48 b^{5}} - \sqrt{75 b^{5}}$

$\sqrt{80 c^{7}} - \sqrt{20 c^{7}}$

Solution

$2 c^{3} \sqrt{5 c}$

$\sqrt{96 d^{9}} - \sqrt{24 d^{9}}$

$9 \sqrt{80 p^{4}} - 6 \sqrt{98 p^{4}}$

Solution

$36 p^{2} \sqrt{5} - 42 p^{2} \sqrt{2}$

$8 \sqrt{72 q^{6}} - 3 \sqrt{75 q^{6}}$

$2 \sqrt{50 r^{8}} + 4 \sqrt{54 r^{8}}$

Solution

$10 r^{4} \sqrt{2} + 12 r^{4} \sqrt{6}$

$5 \sqrt{27 s^{6}} + 2 \sqrt{20 s^{6}}$

$3 \sqrt{20 x^{2}} - 4 \sqrt{45 x^{2}} + 5 x \sqrt{80}$

Solution

$14 x \sqrt{5}$

$2 \sqrt{28 x^{2}} - \sqrt{63 x^{2}} + 6 x \sqrt{7}$

$3 \sqrt{128 y^{2}} + 4 y \sqrt{162} - 8 \sqrt{98 y^{2}}$

Solution

$4 y \sqrt{2}$

$3 \sqrt{75 y^{2}} + 8 y \sqrt{48} - \sqrt{300 y^{2}}$

Mixed Practice

$2 \sqrt{8} + 6 \sqrt{8} - 5 \sqrt{8}$

Solution

$6 \sqrt{2}$

$\frac{2}{3} \sqrt{27} + \frac{3}{4} \sqrt{48}$

$\sqrt{175 k^{4}} - \sqrt{63 k^{4}}$

Solution

$2 k^{2} \sqrt{7}$

$\frac{5}{6} \sqrt{162} + \frac{3}{16} \sqrt{128}$

$2 \sqrt{363} - 2 \sqrt{300}$

Solution

$2 \sqrt{3}$

$\sqrt{150} + 4 \sqrt{6}$

$9 \sqrt{2} - 8 \sqrt{2}$

Solution

$\sqrt{2}$

$5 \sqrt{x} - 8 \sqrt{y}$

$8 \sqrt{13} - 4 \sqrt{13} - 3 \sqrt{13}$

Solution

$\sqrt{13}$

$5 \sqrt{12 c^{4}} - 3 \sqrt{27 c^{6}}$

$\sqrt{80 a^{5}} - \sqrt{45 a^{5}}$

Solution

$a^{2} \sqrt{5 a}$

$\frac{3}{5} \sqrt{75} - \frac{1}{4} \sqrt{48}$

$21 \sqrt{19} - 2 \sqrt{19}$

Solution

$19 \sqrt{19}$

$\sqrt{500} + \sqrt{405}$

$\frac{5}{6} \sqrt{27} + \frac{5}{8} \sqrt{48}$

Solution

$5 \sqrt{3}$

$11 \sqrt{11} - 10 \sqrt{11}$

$\sqrt{75} - \sqrt{108}$

Solution

$− \sqrt{3}$

$2 \sqrt{98} - 4 \sqrt{72}$

$4 \sqrt{24 x^{2}} - \sqrt{54 x^{2}} + 3 x \sqrt{6}$

Solution

$8 x \sqrt{6}$

$8 \sqrt{80 y^{6}} - 6 \sqrt{48 y^{6}}$

Everyday Math

A decorator decides to use square tiles as an accent strip in the design of a new shower, but she wants to rotate the tiles to look like diamonds. She will use 9 large tiles that measure 8 inches on a side and 8 small tiles that measure 2 inches on a side. Determine the width of the accent strip by simplifying the expression $9 (8 \sqrt{2}) + 8 (2 \sqrt{2})$ . (Round to the nearest tenth of an inch.)

Solution

$124.5 inches$

Suzy wants to use square tiles on the border of a spa she is installing in her backyard. She will use large tiles that have area of 12 square inches, medium tiles that have area of 8 square inches, and small tiles that have area of 4 square inches. Once section of the border will require 4 large tiles, 8 medium tiles, and 10 small tiles to cover the width of the wall. Simplify the expression $4 \sqrt{12} + 8 \sqrt{8} + 10 \sqrt{4}$ to determine the width of the wall. (Round to the nearest tenth of an inch.)

Writing Exercises

Explain the difference between like radicals and unlike radicals. Make sure your answer makes sense for radicals containing both numbers and variables.

Solution

Answers will vary.

Explain the process for determining whether two radicals are like or unlike. Make sure your answer makes sense for radicals containing both numbers and variables.

Self Check

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

This table has four columns and three rows. The columns are labeled, “I can…,” “Confidently,” “With some help,” and “No – I don’t get it!” Under the “I can…” column the rows read, “add and subtract like square roots.,” and “add and subtract square roots that need simplification.” The other rows under the other columns are empty.

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

like square roots

Square roots with the same radicand are called like square roots.

Multiply Square Roots

We have used the Product Property of Square Roots to simplify square roots by removing the perfect square factors. The Product Property of Square Roots says

\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}

We can use the Product Property of Square Roots ‘in reverse’ to multiply square roots.

\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}

Remember, we assume all variables are greater than or equal to zero.

We will rewrite the Product Property of Square Roots so we see both ways together.

So we can multiply $\sqrt{3} \cdot \sqrt{5}$ in this way:

\begin{matrix} \sqrt{3} \cdot \sqrt{5} \\ \sqrt{3 \cdot 5} \\ \sqrt{15} \end{matrix}

Sometimes the product gives us a perfect square:

\begin{matrix} \sqrt{2} \cdot \sqrt{8} \\ \sqrt{2 \cdot 8} \\ \sqrt{16} \\ 4 \end{matrix}

Even when the product is not a perfect square, we must look for perfect-square factors and simplify the radical whenever possible.

Multiplying radicals with coefficients is much like multiplying variables with coefficients. To multiply $4 x \cdot 3 y$ we multiply the coefficients together and then the variables. The result is $12 x y$ . Keep this in mind as you do these examples.

Simplify: ⓐ $\sqrt{2} \cdot \sqrt{6}$ ⓑ $(4 \sqrt{3}) (2 \sqrt{12})$ .

Solution

ⓐ

This table illustrates the step-by-step process of simplifying the product of square roots, from initial multiplication to the final simplified radical form.
	$\sqrt{2} \cdot \sqrt{6}$
Multiply using the Product Property.	$\sqrt{12}$
Simplify the radical.	$\sqrt{4} \cdot \sqrt{3}$
Simplify.	$2 \sqrt{3}$

ⓑ

This table demonstrates the step-by-step simplification of a radical expression, from multiplication to a final integer result.
	$(4 \sqrt{3}) (2 \sqrt{12})$
Multiply using the Product Property.	$8 \sqrt{36}$
Simplify the radical.	$8 \cdot 6$
Simplify.	$48$

Notice that in (b) we multiplied the coefficients and multiplied the radicals. Also, we did not simplify $\sqrt{12}$ . We waited to get the product and then simplified.

Simplify: $(6 \sqrt{2}) (3 \sqrt{10})$ .

Solution

Demonstration of simplifying a radical expression by multiplication, showing each procedural step and its corresponding mathematical form.
	$(6 \sqrt{2}) (3 \sqrt{10})$
Multiply using the Product Property.	$18 \sqrt{20}$
Simplify the radical.	$18 \sqrt{4} \cdot \sqrt{5}$
Simplify.	$18 \cdot 2 \cdot \sqrt{5}$
	$36 \sqrt{5}$

When we have to multiply square roots, we first find the product and then remove any perfect square factors.

Simplify: ⓐ $(\sqrt{8 x^{3}}) (\sqrt{3 x})$ ⓑ $(\sqrt{20 y^{2}}) (\sqrt{5 y^{3}})$ .

Solution

ⓐ

Illustrates the step-by-step simplification of a product involving square root expressions.
	$(\sqrt{8 x^{3}}) (\sqrt{3 x})$
Multiply using the Product Property.	$\sqrt{24 x^{4}}$
Simplify the radical.	$\sqrt{4 x^{4}} \cdot \sqrt{6}$
Simplify.	$2 x^{2} \sqrt{6}$

ⓑ

Step-by-step simplification of a product of square root expressions.
	$(\sqrt{20 y^{2}}) (\sqrt{5 y^{3}})$
Multiply using the Product Property.	$\sqrt{100 y^{5}}$
Simplify the radical.	$10 y^{2} \sqrt{y}$

Simplify: $(10 \sqrt{6 p^{3}}) (3 \sqrt{18 p})$ .

Solution

Step-by-step simplification of the radical expression (10sqrt(6p^3))(3sqrt(18p)) to its simplified form 180p^2sqrt(3).
	$(10 \sqrt{6 p^{3}}) (3 \sqrt{18 p})$
Multiply.	$30 \sqrt{108 p^{4}}$
Simplify the radical.	$30 \sqrt{36 p^{4}} \cdot \sqrt{3}$
	$30 \cdot 6 p^{2} \cdot \sqrt{3}$
	$180 p^{2} \sqrt{3}$

Simplify: ⓐ ${(\sqrt{2})}^{2}$ ⓑ ${(− \sqrt{11})}^{2}$ .

Solution

ⓐ

Step-by-step simplification of (sqrt(2))^2, illustrating how squaring a square root yields the base number 2.
	${(\sqrt{2})}^{2}$
Rewrite as a product.	$(\sqrt{2}) (\sqrt{2})$
Multiply.	$\sqrt{4}$
Simplify.	$2$

ⓑ

Steps to simplify a squared square root expression, showing rewriting, multiplication, and final integer result.
	${(− \sqrt{11})}^{2}$
Rewrite as a product.	$(− \sqrt{11}) (− \sqrt{11})$
Multiply.	$\sqrt{121}$
Simplify.	$11$

The results of the previous example lead us to this property.

By realizing that squaring and taking a square root are ‘opposite’ operations, we can simplify ${(\sqrt{2})}^{2}$ and get 2 right away. When we multiply the two like square roots in part (a) of the next example, it is the same as squaring.

Simplify: ⓐ $(2 \sqrt{3}) (8 \sqrt{3})$ ⓑ ${(3 \sqrt{6})}^{2}$ .

Solution

ⓐ

Step-by-step simplification of the product of radical expressions, (23)(83), to its integer value.
	$(2 \sqrt{3}) (8 \sqrt{3})$
Multiply. Remember, ${(\sqrt{3})}^{2} = 3 .$	$16 \cdot 3$
Simplify.	$48$

ⓑ

Illustrates the step-by-step simplification of the mathematical expression (36) to its final value, 54.
	${(3 \sqrt{6})}^{2}$
Multiply.	$9 \cdot 6$
Simplify.	$54$

Use Polynomial Multiplication to Multiply Square Roots

In the next few examples, we will use the Distributive Property to multiply expressions with square roots.

We will first distribute and then simplify the square roots when possible.

Simplify: ⓐ $3 (5 - \sqrt{2})$ ⓑ $\sqrt{2} (4 - \sqrt{10})$ .

Solution

ⓐ

This table illustrates the distributive property by showing an algebraic expression and its expanded form after distribution.
	$3 (5 - \sqrt{2})$
Distribute.	$15 - 3 \sqrt{2}$

ⓑ

Step-by-step simplification of a mathematical expression, illustrating distribution and reduction of square roots.
	$\sqrt{2} (4 - \sqrt{10})$
Distribute.	$4 \sqrt{2} - \sqrt{20}$
	$4 \sqrt{2} - 2 \sqrt{5}$

Simplify: ⓐ $\sqrt{5} (7 + 2 \sqrt{5})$ ⓑ $\sqrt{6} (\sqrt{2} + \sqrt{18})$ .

Solution

ⓐ

This table illustrates the step-by-step simplification of the mathematical expression $\\sqrt{5}(7 + 2\\sqrt{5})$, detailing the multiplication and final simplified form.
	$\sqrt{5} (7 + 2 \sqrt{5})$
Multiply.	$7 \sqrt{5} + 2 \cdot 5$
Simplify.	$7 \sqrt{5} + 10$
	$10 + 7 \sqrt{5}$

ⓑ

Step-by-step simplification of the radical expression sqrt(6)*(sqrt(2) + sqrt(18)).
	$\sqrt{6} (\sqrt{2} + \sqrt{18})$
Multiply.	$\sqrt{12} + \sqrt{108}$
Simplify.	$\sqrt{4} \cdot \sqrt{3} + \sqrt{36} \cdot \sqrt{3}$
	$2 \sqrt{3} + 6 \sqrt{3}$
Combine like radicals.	$8 \sqrt{3}$

When we worked with polynomials, we multiplied binomials by binomials. Remember, this gave us four products before we combined any like terms. To be sure to get all four products, we organized our work—usually by the FOIL method.

Simplify: $(2 + \sqrt{3}) (4 - \sqrt{3})$ .

Solution

Steps to multiply and simplify an expression involving binomials with square roots.
	$(2 + \sqrt{3}) (4 - \sqrt{3})$
Multiply.	$8 - 2 \sqrt{3} + 4 \sqrt{3} - 3$
Combine like terms.	$5 + 2 \sqrt{3}$

Simplify: $(3 - 2 \sqrt{7}) (4 - 2 \sqrt{7})$ .

Solution

Step-by-step multiplication and simplification of an algebraic expression involving square roots.
	$(3 - 2 \sqrt{7}) (4 - 2 \sqrt{7})$
Multiply.	$12 - 6 \sqrt{7} - 8 \sqrt{7} + 4 \cdot 7$
Simplify.	$12 - 6 \sqrt{7} - 8 \sqrt{7} + 28$
Combine like terms.	$40 - 14 \sqrt{7}$

Simplify: $(3 \sqrt{2} - \sqrt{5}) (\sqrt{2} + 4 \sqrt{5})$ .

Solution

This table demonstrates the step-by-step multiplication and simplification of an algebraic expression involving square roots, showing each operation and the resulting expression.
	$(3 \sqrt{2} - \sqrt{5}) (\sqrt{2} + 4 \sqrt{5})$
Multiply.	$3 \cdot 2 + 12 \sqrt{10} - \sqrt{10} - 4 \cdot 5$
Simplify.	$6 + 12 \sqrt{10} - \sqrt{10} - 20$
Combine like terms.	$−14 + 11 \sqrt{10}$

Simplify: $(4 - 2 \sqrt{x}) (1 + 3 \sqrt{x})$ .

Solution

This table demonstrates the step-by-step simplification of the algebraic expression (4-2x)(1+3x) by multiplication and combining like terms.
	$(4 - 2 \sqrt{x}) (1 + 3 \sqrt{x})$
Multiply.	$4 + 12 \sqrt{x} - 2 \sqrt{x} - 6 x$
Combine like terms.	$4 + 10 \sqrt{x} - 6 x$

Note that some special products made our work easier when we multiplied binomials earlier. This is true when we multiply square roots, too. The special product formulas we used are shown below.

We will use the special product formulas in the next few examples. We will start with the Binomial Squares formula.

Simplify: ⓐ ${(2 + \sqrt{3})}^{2}$ ⓑ ${(4 - 2 \sqrt{5})}^{2}$ .

Solution

Be sure to include the $2 a b$ term when squaring a binomial.

ⓐ

Multiply using the binomial square pattern.

Simplify.

Combine like terms.
ⓑ

Multiply using the binomial square pattern.

Simplify.

Combine like terms.

Simplify: ${(1 + 3 \sqrt{x})}^{2}$ .

Solution


Multiply using the binomial square pattern.
Simplify.

In the next two examples, we will find the product of conjugates.

Simplify: $(4 - \sqrt{2}) (4 + \sqrt{2})$ .

Solution


Multiply using the binomial square pattern.
Simplify.

Simplify: $(5 - 2 \sqrt{3}) (5 + 2 \sqrt{3})$ .

Solution


Multiply using the binomial square pattern.
Simplify.	$A fraction displaying the operation (25 - 4 * 3) divided by 13. Following the order of operations, the numerator simplifies to 25 - 12 = 13, making the fraction 13/13, which equals 1.$

Key Concepts

Product Property of Square Roots If a, b are nonnegative real numbers, then

$\sqrt{a b} = \sqrt{a} \cdot \sqrt{b} and \sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$
Special formulas for multiplying binomials and conjugates:

$\begin{matrix} {(a + b)}^{2} = a^{2} + 2 a b + b^{2} & (a - b) (a + b) = a^{2} - b^{2} \\ {(a - b)}^{2} = a^{2} - 2 a b + b^{2} \end{matrix}$
The FOIL method can be used to multiply binomials containing radicals.

Practice Makes Perfect

Multiply Square Roots

In the following exercises, simplify.

ⓐ $\sqrt{2} \cdot \sqrt{8}$ ⓑ $(3 \sqrt{3}) (2 \sqrt{18})$

Solution

ⓐ $4$ ⓑ $18 \sqrt{6}$

ⓐ $\sqrt{6} \cdot \sqrt{6}$ ⓑ $(3 \sqrt{2}) (2 \sqrt{32})$

ⓐ $\sqrt{7} \cdot \sqrt{14}$ ⓑ $(4 \sqrt{8}) (5 \sqrt{8})$

Solution

ⓐ $7 \sqrt{2}$ ⓑ 160

ⓐ $\sqrt{6} \cdot \sqrt{12}$ ⓑ $(2 \sqrt{5}) (2 \sqrt{10})$

$(5 \sqrt{2}) (3 \sqrt{6})$

Solution

$30 \sqrt{3}$

$(2 \sqrt{3}) (4 \sqrt{6})$

$(−2 \sqrt{3}) (3 \sqrt{18})$

Solution

$−18 \sqrt{6}$

$(−4 \sqrt{5}) (5 \sqrt{10})$

$(5 \sqrt{6}) (− \sqrt{12})$

Solution

$−30 \sqrt{2}$

$(6 \sqrt{2}) (− \sqrt{10})$

$(−2 \sqrt{7}) (−2 \sqrt{14})$

Solution

$28 \sqrt{2}$

$(−2 \sqrt{11}) (−4 \sqrt{22})$

ⓐ $(\sqrt{15 y}) (\sqrt{5 y^{3}})$ ⓑ $(\sqrt{2 n^{2}}) (\sqrt{18 n^{3}})$

Solution

ⓐ $5 y^{2} \sqrt{3}$ ⓑ $6 n^{2} \sqrt{n}$

ⓐ $(\sqrt{14 x^{3}}) (\sqrt{7 x^{3}})$ ⓑ $(\sqrt{3 q^{2}}) (\sqrt{48 q^{3}})$

ⓐ $(\sqrt{16 y^{2}}) (\sqrt{8 y^{4}})$ ⓑ $(\sqrt{11 s^{6}}) (\sqrt{11 s})$

Solution

ⓐ $8 y^{3} \sqrt{2}$ ⓑ $11 s^{3} \sqrt{s}$

ⓐ $(\sqrt{8 x^{3}}) (\sqrt{3 x})$ ⓑ $(\sqrt{7 r}) (\sqrt{7 r^{8}})$

$(2 \sqrt{5 b^{3}}) (4 \sqrt{15 b})$

Solution

$40 b^{2} \sqrt{3}$

$(3 \sqrt{8 c^{5}}) (2 \sqrt{6 c^{3}})$

$(5 \sqrt{2 d^{7}}) (3 \sqrt{50 d^{3}})$

Solution

$150 d^{5}$

$(4 \sqrt{6 t^{2}}) (3 \sqrt{3 t^{2}})$

$(3 \sqrt{4 y^{4}}) (3 \sqrt{9 y^{5}})$

Solution

$54 y^{4} \sqrt{y}$

$(−2 \sqrt{7 z^{3}}) (3 \sqrt{14 z^{8}})$

$(4 \sqrt{2 k^{5}}) (−3 \sqrt{32 k^{6}})$

Solution

$−96 k^{5} \sqrt{k}$

ⓐ ${(\sqrt{7})}^{2}$ ⓑ ${(− \sqrt{15})}^{2}$

ⓐ ${(\sqrt{11})}^{2}$ ⓑ ${(− \sqrt{21})}^{2}$

Solution

ⓐ 11 ⓑ 21

ⓐ ${(\sqrt{19})}^{2}$ ⓑ ${(− \sqrt{5})}^{2}$

ⓐ ${(\sqrt{23})}^{2}$
ⓑ ${(− \sqrt{3})}^{2}$

Solution

ⓐ 23 ⓑ 3

ⓐ $(4 \sqrt{11}) (−3 \sqrt{11})$ ⓑ ${(5 \sqrt{3})}^{2}$

ⓐ $(2 \sqrt{13}) (−9 \sqrt{13})$ ⓑ ${(6 \sqrt{5})}^{2}$

Solution

ⓐ −234 ⓑ 180

ⓐ $(−3 \sqrt{12}) (−2 \sqrt{6})$ ⓑ ${(−4 \sqrt{10})}^{2}$

ⓐ $(−7 \sqrt{5}) (−3 \sqrt{10})$ ⓑ ${(−2 \sqrt{14})}^{2}$

Solution

ⓐ $105 \sqrt{2}$ ⓑ 56

Use Polynomial Multiplication to Multiply Square Roots

In the following exercises, simplify.

ⓐ $3 (4 - \sqrt{3})$ ⓑ $\sqrt{2} (4 - \sqrt{6})$

ⓐ $4 (6 - \sqrt{11})$ ⓑ $\sqrt{2} (5 - \sqrt{12})$

Solution

ⓐ $24 - 4 \sqrt{11}$ ⓑ $5 \sqrt{2} - 2 \sqrt{6}$

ⓐ $5 (3 - \sqrt{7})$ ⓑ $\sqrt{3} (4 - \sqrt{15})$

ⓐ $7 (−2 - \sqrt{11})$ ⓑ $\sqrt{7} (6 - \sqrt{14})$

Solution

ⓐ $−14 - 7 \sqrt{11}$ ⓑ $6 \sqrt{7} - 7 \sqrt{2}$

ⓐ $\sqrt{7} (5 + 2 \sqrt{7})$ ⓑ $\sqrt{5} (\sqrt{10} + \sqrt{18})$

ⓐ $\sqrt{11} (8 + 4 \sqrt{11})$ ⓑ $\sqrt{3} (\sqrt{12} + \sqrt{27})$

Solution

ⓐ $44 + 8 \sqrt{11}$ ⓑ $15$

ⓐ $\sqrt{11} (−3 + 4 \sqrt{11})$ ⓑ $\sqrt{3} (\sqrt{15} - \sqrt{18})$

ⓐ $\sqrt{2} (−5 + 9 \sqrt{2})$ ⓑ $\sqrt{7} (\sqrt{3} - \sqrt{21})$

Solution

ⓐ $18 - 5 \sqrt{2}$ ⓑ $\sqrt{21} - 7 \sqrt{3}$

$(8 + \sqrt{3}) (2 - \sqrt{3})$

$(7 + \sqrt{3}) (9 - \sqrt{3})$

Solution

$60 + 2 \sqrt{3}$

$(8 - \sqrt{2}) (3 + \sqrt{2})$

$(9 - \sqrt{2}) (6 + \sqrt{2})$

Solution

$52 + 3 \sqrt{2}$

$(3 - \sqrt{7}) (5 - \sqrt{7})$

$(5 - \sqrt{7}) (4 - \sqrt{7})$

Solution

$27 - 9 \sqrt{7}$

$(1 + 3 \sqrt{10}) (5 - 2 \sqrt{10})$

$(7 - 2 \sqrt{5}) (4 + 9 \sqrt{5})$

Solution

$−62 + 55 \sqrt{5}$

$(\sqrt{3} + \sqrt{10}) (\sqrt{3} + 2 \sqrt{10})$

$(\sqrt{11} + \sqrt{5}) (\sqrt{11} + 6 \sqrt{5})$

Solution

$41 + 7 \sqrt{55}$

$(2 \sqrt{7} - 5 \sqrt{11}) (4 \sqrt{7} + 9 \sqrt{11})$

$(4 \sqrt{6} + 7 \sqrt{13}) (8 \sqrt{6} - 3 \sqrt{13})$

Solution

$−81 + 44 \sqrt{78}$

$(5 - \sqrt{u}) (3 + \sqrt{u})$

$(9 - \sqrt{w}) (2 + \sqrt{w})$

Solution

$18 + 7 \sqrt{w} - w$

$(7 + 2 \sqrt{m}) (4 + 9 \sqrt{m})$

$(6 + 5 \sqrt{n}) (11 + 3 \sqrt{n})$

Solution

$66 + 73 \sqrt{n} + 15 n$

ⓐ ${(3 + \sqrt{5})}^{2}$
ⓑ ${(2 - 5 \sqrt{3})}^{2}$

ⓐ ${(4 + \sqrt{11})}^{2}$ ⓑ ${(3 - 2 \sqrt{5})}^{2}$

Solution

ⓐ $27 + 8 \sqrt{11}$ ⓑ $29 - 12 \sqrt{5}$

ⓐ ${(9 - \sqrt{6})}^{2}$ ⓑ ${(10 + 3 \sqrt{7})}^{2}$

ⓐ ${(5 - \sqrt{10})}^{2}$ ⓑ ${(8 + 3 \sqrt{2})}^{2}$

Solution

ⓐ $35 - 10 \sqrt{10}$ ⓑ $82 + 48 \sqrt{2}$

$(3 - \sqrt{5}) (3 + \sqrt{5})$

$(10 - \sqrt{3}) (10 + \sqrt{3})$

Solution

$(4 + \sqrt{2}) (4 - \sqrt{2})$

$(7 + \sqrt{10}) (7 - \sqrt{10})$

Solution

$(4 + 9 \sqrt{3}) (4 - 9 \sqrt{3})$

$(1 + 8 \sqrt{2}) (1 - 8 \sqrt{2})$

Solution

$−127$

$(12 - 5 \sqrt{5}) (12 + 5 \sqrt{5})$

$(9 - 4 \sqrt{3}) (9 + 4 \sqrt{3})$

Solution

Mixed Practice

In the following exercises, simplify.

$\sqrt{3} \cdot \sqrt{21}$

$(4 \sqrt{6}) (− \sqrt{18})$

Solution

$−24 \sqrt{3}$

$(−5 + \sqrt{7}) (6 + \sqrt{21})$

$(−5 \sqrt{7}) (6 \sqrt{21})$

Solution

$−210 \sqrt{3}$

$(−4 \sqrt{2}) (2 \sqrt{18})$

$(\sqrt{35 y^{3}}) (\sqrt{7 y^{3}})$

Solution

$7 y^{3} \sqrt{5}$

$(4 \sqrt{12 x^{5}}) (2 \sqrt{6 x^{3}})$

${(\sqrt{29})}^{2}$

Solution

$29$

$(−4 \sqrt{17}) (−3 \sqrt{17})$

$(−4 + \sqrt{17}) (−3 + \sqrt{17})$

Solution

$29 - 7 \sqrt{17}$

Everyday Math

A landscaper wants to put a square reflecting pool next to a triangular deck, as shown below. The triangular deck is a right triangle, with legs of length 9 feet and 11 feet, and the pool will be adjacent to the hypotenuse.

ⓐ Use the Pythagorean Theorem to find the length of a side of the pool. Round your answer to the nearest tenth of a foot.
ⓑ Find the exact area of the pool.

This figure is an illustration of a square pool with a deck in the shape of a right triangle. the pool's sides are x inches long while the deck's hypotenuse is x inches long and its legs are nine and eleven inches long.

An artist wants to make a small monument in the shape of a square base topped by a right triangle, as shown below. The square base will be adjacent to one leg of the triangle. The other leg of the triangle will measure 2 feet and the hypotenuse will be 5 feet.

ⓐ Use the Pythagorean Theorem to find the length of a side of the square base. Round your answer to the nearest tenth of a foot.
ⓑ Find the exact area of the face of the square base.

Solution

ⓐ $4.6 feet$ ⓑ $21 sq. feet$

A square garden will be made with a stone border on one edge. If only $3 + \sqrt{10}$ feet of stone are available, simplify ${(3 + \sqrt{10})}^{2}$ to determine the area of the largest such garden. Round your answer to the nearest tenth of a foot.

A garden will be made so as to contain two square sections, one section with side length $\sqrt{5} + \sqrt{6}$ yards and one section with side length $\sqrt{2} + \sqrt{3}$ yards. Simplify ${(\sqrt{5} + \sqrt{6})}^{2} + {(\sqrt{2} + \sqrt{3})}^{2}$ to determine the total area of the garden. Round your answer to the nearest tenth.

Solution

$31.9 square yards$

Suppose a third section will be added to the garden in the previous exercise. The third section is to have a width of $\sqrt{432}$ yards. Write an expression that gives the total area of the garden.

Writing Exercises

ⓐ Explain why ${(− \sqrt{n})}^{2}$ is always positive, for $n \geq 0$ .
ⓑ Explain why $- {(\sqrt{n})}^{2}$ is always negative, for $n \geq 0$ .

Solution

ⓐ when squaring a negative, it becomes a positive ⓑ since the negative is not included in the parenthesis, it is not squared, and remains negative

Use the binomial square pattern to simplify ${(3 + \sqrt{2})}^{2}$ . Explain all your steps.

Self Check

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

This table has four columns and three rows. The columns are labeled, “I can…,” “confidently.,” “with some help.,” and “no minus I don’t get it!” The rows under the “I can…” column read, “multiply square roots.,” and “use polynomial multiplication to multiply square roots.” The other rows under the other columns are empty.

ⓑ 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?

Divide Square Roots

We know that we simplify fractions by removing factors common to the numerator and the denominator. When we have a fraction with a square root in the numerator, we first simplify the square root. Then we can look for common factors.

This figure shows two columns. The first is labeled “Common Factors” and has 3 times the square root of 2 over 3 times 5 beneath it. Both number threes are red. The second column is labeled “No common factors” and has 2 times the square root of 3 over 3 times 5.

Simplify: $\frac{\sqrt{54}}{6}$ .

Solution

This table illustrates the step-by-step simplification of the radical expression sqrt(54)/6, demonstrating the process from its initial to its final simplified form.
	$\frac{\sqrt{54}}{6}$
Simplify the radical.	$\frac{\sqrt{9} \cdot \sqrt{6}}{6}$
Simplify.	$\frac{3 \sqrt{6}}{6}$
Remove the common factors.	$\frac{3 \sqrt{6}}{3 \cdot 2}$
Simplify.	$\frac{\sqrt{6}}{2}$

Simplify: $\frac{6 - \sqrt{24}}{12}$ .

Solution

Step-by-step guide to simplifying the radical expression (6 - sqrt(24)) / 12, demonstrating each transformation to arrive at (3 - sqrt(6)) / 6.
	$\frac{6 - \sqrt{24}}{12}$
Simplify the radical.	$\frac{6 - \sqrt{4} \cdot \sqrt{6}}{12}$
Simplify.	$\frac{6 - 2 \sqrt{6}}{12}$
Factor the common factor from the numerator.	$\frac{2 (3 - \sqrt{6})}{2 \cdot 6}$
Remove the common factors.	$\frac{2 (3 - \sqrt{6})}{2 \cdot 6}$
Simplify.	$\frac{3 - \sqrt{6}}{6}$

We have used the Quotient Property of Square Roots to simplify square roots of fractions. The Quotient Property of Square Roots says

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, b \neq 0

Sometimes we will need to use the Quotient Property of Square Roots ‘in reverse’ to simplify a fraction with square roots.

\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}, b \neq 0

We will rewrite the Quotient Property of Square Roots so we see both ways together. Remember: we assume all variables are greater than or equal to zero so that their square roots are real numbers.

We will use the Quotient Property of Square Roots ‘in reverse’ when the fraction we start with is the quotient of two square roots, and neither radicand is a perfect square. When we write the fraction in a single square root, we may find common factors in the numerator and denominator.

Simplify: $\frac{\sqrt{27}}{\sqrt{75}}$ .

Solution

Step-by-step guide on simplifying the rational expression sqrt(27)/sqrt(75), detailing each procedural step and the corresponding mathematical expression.
	$\frac{\sqrt{27}}{\sqrt{75}}$
Neither radicand is a perfect square, so rewrite using the quotient property of square roots.	$\sqrt{\frac{27}{75}}$
Remove common factors in the numerator and denominator.	$\sqrt{\frac{3 \cdot 9}{3 \cdot 25}}$
Simplify.	$\sqrt{\frac{9}{25}}$
	$\frac{3}{5}$

We will use the Quotient Property for Exponents, $\frac{a^{m}}{a^{n}} = a^{m - n}$ , when we have variables with exponents in the radicands.

Simplify: $\frac{\sqrt{6 y^{5}}}{\sqrt{2 y}}$ .

Solution

Steps to simplify the quotient of two square root expressions involving variables.
	$\frac{\sqrt{6 y^{5}}}{\sqrt{2 y}}$
Neither radicand is a perfect square, so rewrite using the quotient property of square roots.	$\sqrt{\frac{6 y^{5}}{2 y}}$
Remove common factors in the numerator and denominator.	$\sqrt{\frac{2 \cdot 3 \cdot y^{4} \cdot y}{2 \cdot y}}$
Simplify.	$\sqrt{3 y^{4}}$
Simplify the radical.	$y^{2} \sqrt{3}$

Simplify: $\frac{\sqrt{72 x^{3}}}{\sqrt{162 x}}$ .

Solution

This table illustrates the step-by-step simplification of a radical expression involving a quotient of square roots.
	$\frac{\sqrt{72 x^{3}}}{\sqrt{162 x}}$
Rewrite using the quotient property of square roots.	$\sqrt{\frac{72 x^{3}}{162 x}}$
Remove common factors.	$\sqrt{\frac{18 \cdot 4 \cdot x^{2} \cdot x}{18 \cdot 9 \cdot x}}$
Simplify.	$\sqrt{\frac{4 x^{2}}{9}}$
Simplify the radical.	$\frac{2 x}{3}$

Simplify: $\frac{\sqrt{147 a b^{8}}}{\sqrt{3 a^{3} b^{4}}}$ .

Solution

Step-by-step simplification of a rational expression with square roots, demonstrating algebraic properties.
	$\frac{\sqrt{147 a b^{8}}}{\sqrt{3 a^{3} b^{4}}}$
Rewrite using the quotient property of square roots.	$\sqrt{\frac{147 a b^{8}}{3 a^{3} b^{4}}}$
Remove common factors.	$\sqrt{\frac{49 b^{4}}{a^{2}}}$
Simplify the radical.	$\frac{7 b^{2}}{a}$

Rationalize a One Term Denominator

Before the calculator became a tool of everyday life, tables of square roots were used to find approximate values of square roots. Figure 1 shows a portion of a table of squares and square roots. Square roots are approximated to five decimal places in this table.

This table has three solumn and eleven rows. The columns are labeled, “n,” “n squared,” and “the square root of n.” Under the column labeled “n” are the following numbers: 200; 201; 202; 203; 204; 205; 206; 207; 208; 209; and 210. Under the column labeled, “n squared” are the following numbers: 40,000; 40,401; 40,804; 41,209; 41,616; 42,025; 42,436; 42,849; 43,264; 43,681; 44,100. Under the column labeled, “the square root of n” are the following numbers: 14.14214; 14.17745; 14.21267; 14.24781; 14.28286; 14.31782; 14.35270; 14.38749; 14.42221; 14.45683; 14.49138. — A table of square roots was used to find approximate values of square roots before there were calculators.

If someone needed to approximate a fraction with a square root in the denominator, it meant doing long division with a five decimal-place divisor. This was a very cumbersome process.

For this reason, a process called rationalizing the denominator was developed. A fraction with a radical in the denominator is converted to an equivalent fraction whose denominator is an integer. This process is still used today and is useful in other areas of mathematics, too.

Square roots of numbers that are not perfect squares are irrational numbers. When we rationalize the denominator, we write an equivalent fraction with a rational number in the denominator.

Let’s look at a numerical example.

$\begin{matrix} Suppose we need an approximate value for the fraction. & \frac{1}{\sqrt{2}} \\ A five decimal place approximation to \sqrt{2} is 1.41421 . & \frac{1}{1.41421} \\ Without a calculator, would you want to do this division? & 1.41421 1.0 \end{matrix}$

But we can find a fraction equivalent to $\frac{1}{\sqrt{2}}$ by multiplying the numerator and denominator by $\sqrt{2}$ .

$This figure shows three fractions. The first fraction is 1 over the square root of 2. The second is 1 times the square root of 2 over the square root of 2 times the square root of 2. The third shows the square root of 2 over 2.$

Now if we need an approximate value, we divide $2 1.41421$ . This is much easier.

Even though we have calculators available nearly everywhere, a fraction with a radical in the denominator still must be rationalized. It is not considered simplified if the denominator contains a square root.

Similarly, a square root is not considered simplified if the radicand contains a fraction.

To rationalize a denominator, we use the property that ${(\sqrt{a})}^{2} = a$ . If we square an irrational square root, we get a rational number.

We will use this property to rationalize the denominator in the next example.

Simplify: $\frac{4}{\sqrt{3}}$ .

Solution

To rationalize a denominator, we can multiply a square root by itself. To keep the fraction equivalent, we multiply both the numerator and denominator by the same factor.

This table illustrates the step-by-step process of rationalizing the denominator of a fraction containing a square root.
	$\frac{4}{\sqrt{3}}$
Multiply both the numerator and denominator by $\sqrt{3} .$	$\frac{4 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}$
Simplify.	$\frac{4 \sqrt{3}}{3}$

Simplify: $- \frac{8}{3 \sqrt{6}}$ .

Solution

To remove the square root from the denominator, we multiply it by itself. To keep the fractions equivalent, we multiply both the numerator and denominator by $\sqrt{6}$ .

	$A mathematical expression displaying a fraction with a negative sign, where the numerator is 8 and the denominator is 3 multiplied by the square root of 6, represented as -8 / (3sqrt(6)).$
Multiply both the numerator and the denominator by $\sqrt{6}$ .
Simplify.	$A mathematical expression showing a negative fraction: - (8 times the square root of 6) divided by (3 multiplied by 6).$
Remove common factors.	$A mathematical expression showing a negative fraction where the factor 2 is crossed out in both the numerator and denominator. The numerator is 4 times the cancelled 2 times the square root of 6, and the denominator is 3 times the cancelled 2 times 3.$
Simplify.

Always simplify the radical in the denominator first, before you rationalize it. This way the numbers stay smaller and easier to work with.

Simplify: $\sqrt{\frac{5}{12}}$ .

Solution

	$A mathematical expression displays the square root of the fraction 5/12, with the numerator 5 and the denominator 12 clearly visible under the radical symbol.$
The fraction is not a perfect square, so rewrite using the Quotient Property.
Simplify the denominator	$A mathematical fraction with the square root of 5 in the numerator and 2 times the square root of 3 in the denominator.$
Rationalize the denominator.
Simplify.	$A mathematical fraction with the square root of 15 in the numerator and 2 multiplied by 3 in the denominator.$
Simplify.

Simplify: $\sqrt{\frac{11}{28}}$ .

Solution

	$A mathematical expression showing the square root of the fraction 11 over 28.$
Rewrite using the Quotient Property.	$A mathematical expression showing the square root of 11 divided by the square root of 28, presented as a fraction.$
Simplify the denominator.
Rationalize the denominator.
Simplify.
Simplify.

Rationalize a Two-Term Denominator

When the denominator of a fraction is a sum or difference with square roots, we use the Product of Conjugates pattern to rationalize the denominator.

\begin{matrix} (a - b) (a + b) & (2 - \sqrt{5}) (2 + \sqrt{5}) \\ a^{2} - b^{2} & 2^{2} - {(\sqrt{5})}^{2} \\ 4 - 5 \\ −1 \end{matrix}

When we multiply a binomial that includes a square root by its conjugate, the product has no square roots.

Simplify: $\frac{4}{4 + \sqrt{2}}$ .

Solution

	$A fraction with 4 in the numerator and 4 + square root of 2 in the denominator.$
Multiply the numerator and denominator by the conjugate of the denominator.	$A mathematical expression showing the fraction 4(4-sqrt(2)) divided by the product of (4+sqrt(2)) and (4-sqrt(2)), with (4-sqrt(2)) highlighted in red in both the numerator and denominator.$
Multiply the conjugates in the denominator.	$A mathematical fraction is shown. The numerator is 4(4 - square root of 2). The denominator is 4 squared - (square root of 2) squared.$
Simplify the denominator.	$A mathematical expression showing a fraction. The numerator is 4(4 - square root of 2), and the denominator is 16 - 2.$
Simplify the denominator.	$A mathematical expression showing the fraction 4 multiplied by (4 minus the square root of 2), all divided by 14.$
Remove common factors from the numerator and denominator.	$A mathematical expression representing the fraction 2 multiplied by the quantity 4 minus the square root of 2, all divided by 7.$
We leave the numerator in factored form to make it easier to look for common factors after we have simplified the denominator.

Simplify: $\frac{5}{2 - \sqrt{3}}$ .

Solution

	$A mathematical fraction is displayed, with 5 as the numerator and (2 - sqrt(3)) as the denominator. This represents the expression 5 / (2 minus the square root of 3).$
Multiply the numerator and denominator by the conjugate of the denominator.	$A fraction: 5(2 + sqrt(3)) / ((2 - sqrt(3))(2 + sqrt(3))). The common term (2 + sqrt(3)) is highlighted in red in both the numerator and denominator, indicating a step in simplification.$
Multiply the conjugates in the denominator.	$A mathematical expression showing a fraction. The numerator is 5 times (2 plus the square root of 3). The denominator is 2 squared minus the square root of 3 squared.$
Simplify the denominator.	$A mathematical expression showing a fraction with 5 multiplied by (2 plus the square root of 3) in the numerator, and (4 minus 3) in the denominator, on a white background.$
Simplify the denominator.	$A mathematical expression showing the fraction 5(2 + 3)/1.$
Simplify.

Simplify: $\frac{\sqrt{3}}{\sqrt{u} - \sqrt{6}}$ .

Solution

	$A mathematical expression showing a fraction with the square root of 3 in the numerator and the square root of u minus the square root of 6 in the denominator.$
Multiply the numerator and denominator by the conjugate of the denominator.	$A fraction with a square root of three times the quantity of the square root of u plus the square root of six in the numerator. The denominator is the product of the quantity of the square root of u minus the square root of six and the quantity of the square root of u plus the square root of six.$
Multiply the conjugates in the denominator.	$A mathematical expression showing the fraction: square root of 3 multiplied by the sum of square root of 'u' and square root of 6, all divided by 'u' minus 6.$
Simplify the denominator.	$A fraction with a numerator of 'square root 3 times the quantity square root u plus square root 6' and a denominator of 'u minus 6'.$

Simplify: $\frac{\sqrt{x} + \sqrt{7}}{\sqrt{x} - \sqrt{7}}$ .

Solution

	$A mathematical fraction is displayed, with the numerator as the square root of x plus the square root of 7, and the denominator as the square root of x minus the square root of 7.$
Multiply the numerator and denominator by the conjugate of the denominator.	$An algebraic fraction showing (sqrt(x)+sqrt(7))^2 divided by (sqrt(x)-sqrt(7))(sqrt(x)+sqrt(7)), illustrating a step in simplifying radical expressions.$
Multiply the conjugates in the denominator.	$A mathematical fraction with the numerator as the product of (square root x + square root 7) and (square root x + square root 7), and the denominator as (square root x) squared minus (square root 7) squared.$
Simplify the denominator.	$A fraction with numerator (sqrt(x) + sqrt(7))^2 and denominator (x - 7).$
We do not square the numerator. In factored form, we can see there are no common factors to remove from the numerator and denominator.

Key Concepts

Quotient Property of Square Roots
- If a, b are non-negative real numbers and $b \neq 0$ , then
  $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} and \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
Simplified Square Roots
A square root is considered simplified if there are
- no perfect square factors in the radicand
- no fractions in the radicand
- no square roots in the denominator of a fraction

Practice Makes Perfect

Divide Square Roots

In the following exercises, simplify.

$\frac{\sqrt{27}}{6}$

Solution

$\frac{\sqrt{3}}{2}$

$\frac{\sqrt{50}}{10}$

$\frac{\sqrt{72}}{9}$

Solution

$\frac{2 \sqrt{2}}{3}$

$\frac{\sqrt{243}}{6}$

$\frac{2 - \sqrt{32}}{8}$

Solution

$\frac{1 - 2 \sqrt{2}}{4}$

$\frac{3 + \sqrt{27}}{9}$

$\frac{6 + \sqrt{45}}{6}$

Solution

$\frac{2 + \sqrt{5}}{2}$

$\frac{10 - \sqrt{200}}{20}$

$\frac{\sqrt{80}}{\sqrt{125}}$

Solution

$\frac{4}{5}$

$\frac{\sqrt{72}}{\sqrt{200}}$

$\frac{\sqrt{128}}{\sqrt{72}}$

Solution

$\frac{4}{3}$

$\frac{\sqrt{48}}{\sqrt{75}}$

ⓐ $\frac{\sqrt{8 x^{6}}}{\sqrt{2 x^{2}}}$ ⓑ $\frac{\sqrt{200 m^{5}}}{\sqrt{98 m}}$

Solution

ⓐ $2 x^{2}$ ⓑ $\frac{10 m^{2}}{7}$

ⓐ $\frac{\sqrt{10 y^{3}}}{\sqrt{5 y}}$ ⓑ $\frac{\sqrt{108 n^{7}}}{\sqrt{243 n^{3}}}$

$\frac{\sqrt{75 r^{3}}}{\sqrt{108 r}}$

Solution

$\frac{5 r}{6}$

$\frac{\sqrt{196 q^{5}}}{\sqrt{484 q}}$

$\frac{\sqrt{108 p^{5} q^{2}}}{\sqrt{3 p^{3} q^{6}}}$

Solution

$\frac{6 p \sqrt{102}}{q^{2}}$

$\frac{\sqrt{98 r s^{10}}}{\sqrt{2 r^{3} s^{4}}}$

$\frac{\sqrt{320 m n^{5}}}{\sqrt{45 m^{7} n^{3}}}$

Solution

$\frac{8 n}{3 m^{3}}$

$\frac{\sqrt{810 c^{3} d^{7}}}{\sqrt{1000 c^{5} d}}$

$\frac{\sqrt{98}}{14}$

Solution

$\frac{\sqrt{2}}{2}$

$\frac{\sqrt{72}}{18}$

$\frac{5 + \sqrt{125}}{15}$

Solution

$\frac{1 + \sqrt{5}}{3}$

$\frac{6 - \sqrt{45}}{12}$

$\frac{\sqrt{96}}{\sqrt{150}}$

Solution

$\frac{4}{5}$

$\frac{\sqrt{28}}{\sqrt{63}}$

$\frac{\sqrt{26 y^{7}}}{\sqrt{2 y}}$

Solution

$y^{3} \sqrt{13}$

$\frac{\sqrt{15 x^{3}}}{\sqrt{3 x}}$

Rationalize a One-Term Denominator

In the following exercises, simplify and rationalize the denominator.

$\frac{10}{\sqrt{6}}$

Solution

$\frac{5 \sqrt{6}}{3}$

$\frac{8}{\sqrt{3}}$

$\frac{6}{\sqrt{7}}$

Solution

$\frac{6 \sqrt{7}}{7}$

$\frac{4}{\sqrt{5}}$

$\frac{3}{\sqrt{13}}$

Solution

$\frac{3 \sqrt{13}}{13}$

$\frac{10}{\sqrt{11}}$

$\frac{10}{3 \sqrt{10}}$

Solution

$\frac{\sqrt{10}}{3}$

$\frac{2}{5 \sqrt{2}}$

$\frac{4}{9 \sqrt{5}}$

Solution

$\frac{4 \sqrt{5}}{45}$

$\frac{9}{2 \sqrt{7}}$

$- \frac{9}{2 \sqrt{3}}$

Solution

$- \frac{3 \sqrt{3}}{2}$

$- \frac{8}{3 \sqrt{6}}$

$\sqrt{\frac{3}{20}}$

Solution

$\frac{\sqrt{15}}{10}$

$\sqrt{\frac{4}{27}}$

$\sqrt{\frac{7}{40}}$

Solution

$\frac{\sqrt{70}}{20}$

$\sqrt{\frac{8}{45}}$

$\sqrt{\frac{19}{175}}$

Solution

$\frac{\sqrt{133}}{35}$

$\sqrt{\frac{17}{192}}$

Rationalize a Two-Term Denominator

In the following exercises, simplify by rationalizing the denominator.

ⓐ $\frac{3}{3 + \sqrt{11}}$ ⓑ $\frac{8}{1 - \sqrt{5}}$

Solution

ⓐ $\frac{3 (3 - \sqrt{11})}{−2}$ ⓑ $−2 (1 + \sqrt{5})$

ⓐ $\frac{4}{4 + \sqrt{7}}$ ⓑ $\frac{7}{2 - \sqrt{6}}$

ⓐ $\frac{5}{5 + \sqrt{6}}$ ⓑ $\frac{6}{3 - \sqrt{7}}$

Solution

ⓐ $\frac{5 (5 - \sqrt{6})}{19}$ ⓑ $3 (3 + \sqrt{7})$

ⓐ $\frac{6}{6 + \sqrt{5}}$ ⓑ $\frac{5}{4 - \sqrt{11}}$

$\frac{\sqrt{3}}{\sqrt{m} - \sqrt{5}}$

Solution

$\frac{\sqrt{3} (\sqrt{m} + \sqrt{5})}{m - 5}$

$\frac{\sqrt{5}}{\sqrt{n} - \sqrt{7}}$

$\frac{\sqrt{2}}{\sqrt{x} - \sqrt{6}}$

Solution

$\frac{\sqrt{2} (\sqrt{x} + \sqrt{6})}{x - 6}$

$\frac{\sqrt{7}}{\sqrt{y} + \sqrt{3}}$

$\frac{\sqrt{r} + \sqrt{5}}{\sqrt{r} - \sqrt{5}}$

Solution

${\frac{(\sqrt{r} + \sqrt{5})}{r - 5}}^{2}$

$\frac{\sqrt{s} - \sqrt{6}}{\sqrt{s} + \sqrt{6}}$

$\frac{\sqrt{150 x^{2} y^{6}}}{\sqrt{6 x^{4} y^{2}}}$

Solution

$\frac{5 y^{2}}{x}$

$\frac{\sqrt{80 p^{3} q}}{\sqrt{5 p q^{5}}}$

$\frac{15}{\sqrt{5}}$

Solution

$3 \sqrt{5}$

$\frac{3}{5 \sqrt{8}}$

$\sqrt{\frac{8}{54}}$

Solution

$\frac{2 \sqrt{3}}{9}$

$\sqrt{\frac{12}{20}}$

$\frac{3}{5 + \sqrt{5}}$

Solution

$\frac{3 (5 - \sqrt{5})}{20}$

$\frac{20}{4 - \sqrt{3}}$

$\frac{\sqrt{2}}{\sqrt{x} - \sqrt{3}}$

Solution

$\frac{\sqrt{2} (\sqrt{x} + \sqrt{3})}{x - 3}$

$\frac{\sqrt{5}}{\sqrt{y} - \sqrt{7}}$

$\frac{\sqrt{x} + \sqrt{8}}{\sqrt{x} - \sqrt{8}}$

Solution

${\frac{(\sqrt{x} + 2 \sqrt{2})}{x - 8}}^{2}$

$\frac{\sqrt{m} - \sqrt{3}}{\sqrt{m} + \sqrt{3}}$

Everyday Math

A supply kit is dropped from an airplane flying at an altitude of 250 feet. Simplify $\sqrt{\frac{250}{16}}$ to determine how many seconds it takes for the supply kit to reach the ground.

Solution

$\frac{5 \sqrt{10}}{4} seconds$

A flare is dropped into the ocean from an airplane flying at an altitude of 1,200 feet. Simplify $\sqrt{\frac{1200}{16}}$ to determine how many seconds it takes for the flare to reach the ocean.

Writing Exercises

ⓐ Simplify $\sqrt{\frac{27}{3}}$ and explain all your steps.
ⓑ Simplify $\sqrt{\frac{27}{5}}$ and explain all your steps.

Solution

Answers will vary.

ⓐ Approximate $\frac{1}{\sqrt{2}}$ by dividing $\frac{1}{1.414}$ using long division without a calculator.
ⓑ Rationalizing the denominator of $\frac{1}{\sqrt{2}}$ gives $\frac{\sqrt{2}}{2} .$ Approximate $\frac{\sqrt{2}}{2}$ by dividing $\frac{1.414}{2}$ using long division without a calculator.

Self Check

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

This table has four columns and four rows. The columns are labeled, “I can…,” “confidently.,” “with some help.,” and “no – I don’t get it!” The rows under the column “I can…” read, “divide square roots,” “rationalize a one term denominator.,” and “rationalize a two term denominator.” All the other rows under the columns are empty.

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

rationalizing the denominator

The process of converting a fraction with a radical in the denominator to an equivalent fraction whose denominator is an integer is called rationalizing the denominator.

Solve Equations with Square Roots

Solve Radical Equations

In this section we will solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.

As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Since squaring a quantity and taking a square root are ‘opposite’ operations, we will square both sides in order to remove the radical sign and solve for the variable inside.

But remember that when we write $\sqrt{a}$ we mean the principal square root. So $\sqrt{a} \geq 0$ always. When we solve radical equations by squaring both sides we may get an algebraic solution that would make $\sqrt{a}$ negative. This algebraic solution would not be a solution to the original radical equation; it is an extraneous solution. We saw extraneous solutions when we solved rational equations, too.

For the equation $\sqrt{x + 2} = x$ :

ⓐ Is $x = 2$ a solution? ⓑ Is $x = −1$ a solution?

Solution

ⓐ Is

x = 2

a solution?


Let x = 2.
Simplify.

	2 is a solution.

ⓑ Is

x = −1

a solution?


Let x = −1.
Simplify.

	−1 is not a solution.
	−1 is an extraneous solution to the equation.

Now we will see how to solve a radical equation. Our strategy is based on the relation between taking a square root and squaring.

For a \geq 0, {(\sqrt{a})}^{2} = a

How to Solve Radical Equations

Solve: $\sqrt{2 x - 1} = 7$ .

Solution

This table has three columns and four rows. The first row says, “Step 1. Isolate the radical on one side of equation. The square root of (2x minus 1) is already isolated on the left side.” It then shows the equation: the square root of (2x minus 1) equals 7.

The second row says, “Step 2. Square both sides of the equation. Remember, the square root of a squared equals a.” It then shows the equation: the square root of (2x minus 1) squared equals 7 squared.

The third row then says, “Step 3. Solve the new equation.” It indicates that 2x minus 1 equals 49 or 2x equals 50 which means that x equals 25.

The fourth row says, “Step 4. Check the answer. Check:” It then indicates the square root of (2x minus 1) equals 7. This becomes the square root of (2 times 25 minus 1) equals 7. This becomes the square root of (50 minus 1) equals 7. This becomes the square root of 49 equals 7, and thus 7 equals 7. The figure then states, “The solutions is x equals 25.”

Solve: $\sqrt{5 n - 4} - 9 = 0$ .

Solution


To isolate the radical, add 9 to both sides.
Simplify.
Square both sides of the equation.
Solve the new equation.


Check the answer.

	The solution is n = 17.

Solve: $\sqrt{3 y + 5} + 2 = 5$ .

Solution


To isolate the radical, subtract 2 from both sides.
Simplify.
Square both sides of the equation.
Solve the new equation.


Check the answer.

	The solution is $y = \frac{4}{3}$ .

When we use a radical sign, we mean the principal or positive root. If an equation has a square root equal to a negative number, that equation will have no solution.

Solve: $\sqrt{9 k - 2} + 1 = 0$ .

Solution


To isolate the radical, subtract 1 from both sides.
Simplify.
Since the square root is equal to a negative number, the equation has no solution.

If one side of the equation is a binomial, we use the binomial squares formula when we square it.

Don’t forget the middle term!

Solve: $\sqrt{p - 1} + 1 = p$ .

Solution


To isolate the radical, subtract 1 from both sides.
Simplify.
Square both sides of the equation.
Simplify, then solve the new equation.
It is a quadratic equation, so get zero on one side.
Factor the right side.
Use the zero product property.
Solve each equation.
Check the answers.

	The solutions are p = 1, p = 2.

Solve: $\sqrt{r + 4} - r + 2 = 0$ .

Solution

	$\sqrt{r + 4} - r + 2 = 0$
Isolate the radical.	$\sqrt{r + 4} = r - 2$
Square both sides of the equation.	${(\sqrt{r + 4})}^{2} = {(r - 2)}^{2}$
Solve the new equation.	$r + 4 = r^{2} - 4 r + 4$
It is a quadratic equation, so get zero on one side.	$0 = r^{2} - 5 r$
Factor the right side.	$0 = r (r - 5)$
Use the zero product property.	$0 = r 0 = r - 5$
Solve the equation.	$r = 0 r = 5$
Check the answer.
	The solution is $r = 5$ .
	$r = 0$ is an extraneous solution.

When there is a coefficient in front of the radical, we must square it, too.

Solve: $3 \sqrt{3 x - 5} - 8 = 4$ .

Solution

	$3 \sqrt{3 x - 5} - 8 = 4$
Isolate the radical.	$3 \sqrt{3 x - 5} = 12$
Square both sides of the equation.	${(3 \sqrt{3 x - 5})}^{2} = {(12)}^{2}$
Simplify, then solve the new equation.	$9 (3 x - 5) = 144$
Distribute.	$27 x - 45 = 144$
Solve the equation.	$27 x = 189$
	$x = 7$
Check the answer.
	The solution is $x = 7$ .

Solve: $\sqrt{4 z - 3} = \sqrt{3 z + 2}$ .

Solution

Step-by-step solution demonstrating how to solve a radical equation by isolating and squaring radical terms.
	$\begin{matrix} \sqrt{4 z - 3} = \sqrt{3 z + 2} \end{matrix}$
The radical terms are isolated.	$\begin{matrix} \sqrt{4 z - 3} = \sqrt{3 z + 2} \end{matrix}$
Square both sides of the equation.	$\begin{matrix} {(\sqrt{4 z - 3})}^{2} = {(\sqrt{3 z + 2})}^{2} \end{matrix}$
Simplify, then solve the new equation.	$\begin{matrix} \begin{matrix} 4 z - 3 = 3 z + 2 \\ z - 3 = 2 \\ z = 5 \end{matrix} \end{matrix}$
Check the answer.
We leave it to you to show that 5 checks!	The solution is $z = 5 .$

Sometimes after squaring both sides of an equation, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and square both sides of the equation again.

Solve: $\sqrt{m} + 1 = \sqrt{m + 9}$ .

Solution

Step-by-step solution for the radical equation "sqrt(m) + 1 = sqrt(m + 9)", illustrating the process of isolating and squaring radicals to find the value of 'm'.
	$\sqrt{m} + 1 = \sqrt{m + 9}$
The radical on the right side is isolated. Square both sides.	${(\sqrt{m} + 1)}^{2} = {(\sqrt{m + 9})}^{2}$
Simplify—be very careful as you multiply!	$m + 2 \sqrt{m} + 1 = m + 9$
There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical.	$2 \sqrt{m} = 8$
Square both sides.	${(2 \sqrt{m})}^{2} = {(8)}^{2}$
Simplify, then solve the new equation.	$4 m = 64$
	$m = 16$
	Check the answer.
We leave it to you to show that $m = 16$ checks!	The solution is $m = 16 .$

Solve: $\sqrt{q - 2} + 3 = \sqrt{4 q + 1}$ .

Solution

Step-by-step solution demonstrating how to solve a radical equation by isolating and squaring radical terms to find the values of q.
	$\sqrt{q - 2} + 3 = \sqrt{4 q + 1}$
The radical on the right side is isolated. Square both sides.	${(\sqrt{q - 2} + 3)}^{2} = {(\sqrt{4 q + 1})}^{2}$
Simplify.	$q - 2 + 6 \sqrt{q - 2} + 9 = 4 q + 1$
There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical.	$6 \sqrt{q - 2} = 3 q - 6$
Square both sides.	${(6 \sqrt{q - 2})}^{2} = {(3 q - 6)}^{2}$
Simplify, then solve the new equation.	$36 (q - 2) = 9 q^{2} - 36 q + 36$
Distribute.	$36 q - 72 = 9 q^{2} - 36 q + 36$
It is a quadratic equation, so get zero on one side.	$0 = 9 q^{2} - 72 q + 108$
Factor the right side.	$0 = 9 (q^{2} - 8 q + 12)$
	$0 = 9 (q - 6) (q - 2)$
Use the zero product property.	$\begin{matrix} q - 6 = 0 & q - 2 = 0 \\ q = 6 & q = 2 \end{matrix}$
The checks are left to you. (Both solutions should work.)	$The solutions are q = 6 and q = 2 .$

Use Square Roots in Applications

As you progress through your college courses, you’ll encounter formulas that include square roots in many disciplines. We have already used formulas to solve geometry applications.

We will use our Problem Solving Strategy for Geometry Applications, with slight modifications, to give us a plan for solving applications with formulas from any discipline.

We used the formula $A = L \cdot W$ to find the area of a rectangle with length L and width W. A square is a rectangle in which the length and width are equal. If we let s be the length of a side of a square, the area of the square is $s^{2}$ .

This figure shows a square with two sides labeled s. It also indicates that A equals s squared.

The formula $A = s^{2}$ gives us the area of a square if we know the length of a side. What if we want to find the length of a side for a given area? Then we need to solve the equation for s.

$\begin{matrix} A = s^{2} \\ Take the square root of both sides. & \sqrt{A} = \sqrt{s^{2}} \\ Simplify. & \sqrt{A} = s \end{matrix}$

We can use the formula $s = \sqrt{A}$ to find the length of a side of a square for a given area.

We will show an example of this in the next example.

Mike and Lychelle want to make a square patio. They have enough concrete to pave an area of 200 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of the patio. Round your answer to the nearest tenth of a foot.

Solution

Step 1. Read the problem. Draw a figure and label it with the given information.
	A = 200 square feet
Step 2. Identify what you are looking for.	The length of a side of the square patio.
Step 3. Name what you are looking for by choosing a variable to represent it.	Let s = the length of a side.
Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute the given information.
Step 5. Solve the equation using good algebra techniques. Round to one decimal place.
Step 6. Check the answer in the problem and make sure it makes sense.

This is close enough because we rounded the square root. Is a patio with side 14.1 feet reasonable? Yes.
Step 7. Answer the question with a complete sentence.	Each side of the patio should be 14.1 feet.

Another application of square roots has to do with gravity.

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting $h = 64$ into the formula.



Take the square root of 64.	$A mathematical equation displays 't = 8/4' on a white background, representing the variable t being equal to the fraction eight divided by four.$
Simplify the fraction.

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. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the sunglasses to reach the river.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	The time it takes for the sunglasses to reach the river.
Step 3. Name what you are looking for by choosing a variable to represent it.	Let t = time.
Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
$A math equation displays 5 followed by a question mark over an equals sign, then the fraction 20 over 4, asking if 5 is equal to 20 divided by 4.$ $5 = 5 ✓$
Does 5 seconds seem reasonable? Yes.
Step 7. Answer the question with a complete sentence.	It will take 5 seconds for the sunglasses to hit the water.

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.

After a car accident, the skid marks for one car measured 190 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	The speed of a car.
Step 3. Name what we are looking for.	Let s = the speed.
Step 4. Translate into an equation by writing the appropriate formula.
Substitute the given information.
Step 5. Solve the equation.

Round to 1 decimal place.
Step 6. Check the answer in the problem. $67.5 \overset{?}{\approx} \sqrt{24 (190)}$ $67.5 \overset{?}{\approx} \sqrt{4560}$ $67.5 \overset{?}{\approx} 67.5277...$
Is 67.5 mph a reasonable speed?	Yes.
Step 7. Answer the question with a complete sentence.	The speed of the car was approximately 67.5 miles per hour.

Key Concepts

To Solve a Radical Equation:
1. Isolate the radical on one side of the equation.
2. Square both sides of the equation.
3. Solve the new equation.
4. Check the answer. Some solutions obtained may not work in the original equation.
Solving Applications with Formulas
1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
2. Identify what we are looking for.
3. Name what we are looking for by choosing a variable to represent it.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
Area of a Square
Falling Objects
- On Earth, if an object is dropped from a height of $h$ feet, the time in seconds it will take to reach the ground is found by using the formula $t = \frac{\sqrt{h}}{4}$ .
Skid Marks and Speed of a Car
- If the length of the skid marks is d feet, then the speed, s, of the car before the brakes were applied can be found by using the formula $s = \sqrt{24 d}$ .

Practice Makes Perfect

Solve Radical Equations

In the following exercises, check whether the given values are solutions.

For the equation $\sqrt{x + 12} = x$ : ⓐ Is $x = 4$ a solution? ⓑ Is $x = −3$ a solution?

Solution

ⓐ $yes$ ⓑ $no$

For the equation $\sqrt{− y + 20} = y$ : ⓐ Is $y = 4$ a solution? ⓑ Is $y = −5$ a solution?

For the equation $\sqrt{t + 6} = t$ : ⓐ Is $t = −2$ a solution? ⓑ Is $t = 3$ a solution?

Solution

ⓐ no ⓑ yes

For the equation $\sqrt{u + 42} = u$ : ⓐ Is $u = −6$ a solution? ⓑ Is $u = 7$ a solution?

In the following exercises, solve.

$\sqrt{5 y + 1} = 4$

Solution

$\sqrt{7 z + 15} = 6$

$\sqrt{5 x - 6} = 8$

Solution

$\sqrt{4 x - 3} = 7$

$\sqrt{2 m - 3} - 5 = 0$

Solution

$\sqrt{2 n - 1} - 3 = 0$

$\sqrt{6 v - 2} - 10 = 0$

Solution

$17$

$\sqrt{4 u + 2} - 6 = 0$

$\sqrt{5 q + 3} - 4 = 0$

Solution

$\frac{13}{5}$

$\sqrt{4 m + 2} + 2 = 6$

$\sqrt{6 n + 1} + 4 = 8$

Solution

$\frac{5}{2}$

$\sqrt{2 u - 3} + 2 = 0$

$\sqrt{5 v - 2} + 5 = 0$

Solution

$no solution$

$\sqrt{3 z - 5} + 2 = 0$

$\sqrt{2 m + 1} + 4 = 0$

Solution

$no solution$

ⓐ $\sqrt{u - 3} + 3 = u$
ⓑ $\sqrt{x + 1} - x + 1 = 0$

ⓐ $\sqrt{v - 10} + 10 = v$
ⓑ $\sqrt{y + 4} - y + 2 = 0$

Solution

ⓐ $10, 11$ ⓑ $5$

ⓐ $\sqrt{r - 1} - r = −1$
ⓑ $\sqrt{z + 100} - z + 10 = 0$

ⓐ $\sqrt{s - 8} - s = −8$
ⓑ $\sqrt{w + 25} - w + 5 = 0$

Solution

ⓐ $8, 9$ ⓑ $11$

$3 \sqrt{2 x - 3} - 20 = 7$

$2 \sqrt{5 x + 1} - 8 = 0$

Solution

$3$

$2 \sqrt{8 r + 1} - 8 = 2$

$3 \sqrt{7 y + 1} - 10 = 8$

Solution

$5$

$\sqrt{3 u - 2} = \sqrt{5 u + 1}$

$\sqrt{4 v + 3} = \sqrt{v - 6}$

Solution

not a real number

$\sqrt{8 + 2 r} = \sqrt{3 r + 10}$

$\sqrt{12 c + 6} = \sqrt{10 - 4 c}$

Solution

$\frac{1}{4}$

ⓐ $\sqrt{a} + 2 = \sqrt{a + 4}$
ⓑ $\sqrt{b - 2} + 1 = \sqrt{3 b + 2}$

ⓐ $\sqrt{r} + 6 = \sqrt{r + 8}$
ⓑ $\sqrt{s - 3} + 2 = \sqrt{s + 4}$

Solution

ⓐ $no solution$ ⓑ $\frac{57}{16}$

ⓐ $\sqrt{u} + 1 = \sqrt{u + 4}$
ⓑ $\sqrt{n - 5} + 4 = \sqrt{3 n + 7}$

ⓐ $\sqrt{x} + 10 = \sqrt{x + 2}$
ⓑ $\sqrt{y - 2} + 2 = \sqrt{2 y + 4}$

Solution

ⓐ $no solution$ ⓑ 6

$\sqrt{2 y + 4} + 6 = 0$

$\sqrt{8 u + 1} + 9 = 0$

Solution

no solution

$\sqrt{a} + 1 = \sqrt{a + 5}$

$\sqrt{d} - 2 = \sqrt{d - 20}$

Solution

$\sqrt{6 s + 4} = \sqrt{8 s - 28}$

$\sqrt{9 p + 9} = \sqrt{10 p - 6}$

Solution

Use Square Roots in Applications

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Solution

$11.4 feet$

Gravity While putting up holiday decorations, Renee dropped a light bulb from the top of a 64 foot tall tree. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the light bulb to reach the ground.

Gravity An airplane dropped a flare from a height of 1024 feet above a lake. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the flare to reach the water.

Solution

$8 seconds$

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the cell phone to reach the ground.

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the hammer to reach the river.

Solution

$15.8 seconds$

Accident investigation The skid marks for a car involved in an accident measured 54 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Solution

$72 miles per hour$

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

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. Use the formula $s = \sqrt{24 d}$ to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Solution

$53.0 miles per hour$

Writing Exercises

Explain why an equation of the form $\sqrt{x} + 1 = 0$ has no solution.

ⓐ Solve the equation $\sqrt{r + 4} - r + 2 = 0$ .
ⓑ Explain why one of the “solutions” that was found was not actually a solution to the equation.

Solution

Answers will vary.

Self Check

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

This table has two rows and four columns. The first row labels each column, “I can…,” “Confidently,” “With some help,” and “No minus I don’t get it!” The row under “I can…,” reads, “use square roots in applications.” All the other rows are empty.

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

radical equation

An equation in which the variable is in the radicand of a square root is called a radical equation

Higher Roots

Simplify Expressions with Higher Roots

Up to now, in this chapter we have worked with squares and square roots. We will now extend our work to include higher powers and higher roots.

Let’s review some vocabulary first.

\begin{matrix} We write: & We say: \\ n^{2} & n squared \\ n^{3} & n cubed \\ n^{4} & n to the fourth \\ n^{5} & n to the fifth \end{matrix}

The terms ‘squared’ and ‘cubed’ come from the formulas for area of a square and volume of a cube.

It will be helpful to have a table of the powers of the integers from $−5 to 5$ . See Figure 1.

This figure consists of two tables. The first table shows the results of raising the numbers 1, 2, 3, 4, 5, x, and x squared to the second, third, fourth, and fifth powers. The second table shows the results of raising the numbers negative one through negative five to the second, third, fourth, and fifth powers. The table first has five columns and nine rows. The second has five columns and seven rows. The columns in both tables are labeled, “Number,” “Square,” “Cube,” “Fourth power,” “Fifth power,” nothing, “Number,” “Square,” “Cube,” “Fourth power,” and “Fifth power.” In both tables, the next row reads: n, n squared, n cubed, n to the fourth power, n to the fifth power, nothing, n, n squared, n cubed, n to the fourth power, and n to the fifth power. In the first table, 1 squared, 1 cubed, 1 to the fourth power, and 1 to the fifth power are all shown to be 1. In the next row, 2 squared is 4, 2 cubed is 8, 2 to the fourth power is 16, and 2 to the fifth power is 32. In the next row, 3 squared is 9, 3 cubed is 27, 3 to the fourth power is 81, and 3 to the fifth power is 243. In the next row, 4 squared is 16, 4 cubed is 64, 4 to the fourth power is 246, and 4 to the fifth power is 1024. In the next row, 5 squared is 25, 5 cubed is 125, 5 to the fourth power is 625, and 5 to the fifth power is 3125. In the next row, x squared, x cubed, x to the fourth power, and x to the fifth power are listed. In the next row, x squared squared is x to the fourth power, x cubed squared is x to the fifth power, x squared to the fourth power is x to the eighth power, and x squared to the fifth power is x to the tenth power. In the second table, negative 1 squared is 1, negative 1 cubed is negative 1, negative 1 to the fourth power is 1, and negative 1 to the fifth power is negative 1. In the next row, negative 2 squared is 4, negative 2 cubed is negative 8, negative 2 to the fourth power is 16, and negative 2 to the fifth power is negative 32. In the next row, negative 4 squared is 16, negative 4 cubed is negative 64, negative 4 to the fourth power is 256, and negative 4 to the fifth power is negative 1024. In the next row, negative 5 squared is 25, negative 5 cubed is negative 125, negative 5 to the fourth power is 625, and negative 5 to the fifth power is negative 3125. — First through fifth powers of integers from $−5$ to $5 .$

Notice the signs in Figure 1. All powers of positive numbers are positive, of course. But when we have a negative number, the even powers are positive and the odd powers are negative. We’ll copy the row with the powers of $−2$ below to help you see this.

This figure has five columns and two rows. The first row labels each column: n, n squared, n cubed, n to the fourth power, and n to the fifth power. The second row reads: negative 2, 4, negative 8, 16, and negative 32.

Earlier in this chapter we defined the square root of a number.

If n^{2} = m, then n is a square root of m .

And we have used the notation $\sqrt{m}$ to denote the principal square root. So $\sqrt{m} \geq 0$ always.

We will now extend the definition to higher roots.

We do not write the index for a square root. Just like we use the word ‘cubed’ for $b^{3}$ , we use the term ‘cube root’ for $\sqrt[3]{a}$ .

We refer to Figure 1 to help us find higher roots.

\begin{matrix} 4^{3} & = & 64 & \sqrt[3]{64} & = & 4 \\ 3^{4} & = & 81 & \sqrt[4]{81} & = & 3 \\ {(−2)}^{5} & = & −32 & \sqrt[5]{−32} & = & −2 \end{matrix}

Could we have an even root of a negative number? No. We know that the square root of a negative number is not a real number. The same is true for any even root. Even roots of negative numbers are not real numbers. Odd roots of negative numbers are real numbers.

Solution

ⓐ

This table illustrates finding the cube root of 8, demonstrating its value and the mathematical reason.
	$\sqrt[3]{8}$
Since ${(2)}^{3} = 8 .$	$2$

ⓑ

This table presents the fourth root of 81, its resulting value, and the mathematical explanation for the solution.
	$\sqrt[4]{81}$
Since ${(3)}^{4} = 81 .$	$3$

ⓒ

Example showing the evaluation of the fifth root of 32, with its mathematical justification.
	$\sqrt[5]{32}$
Since ${(2)}^{5} = 32 .$	$2$

Solution

ⓐ

Demonstrates the calculation of the cube root of -64 with a step-by-step explanation.
	$\sqrt[3]{−64}$
Since ${(−4)}^{3} = −64 .$	$−4$

ⓑ

This table demonstrates why the fourth root of -16 is not a real number, providing the mathematical expression, the reasoning, and the conclusion.
	$\sqrt[4]{−16}$
Think, ${(?)}^{4} = −16 .$ No real number raised to the fourth power is negative.	Not a real number.

ⓒ

Calculation of the fifth root of -243, including the expression, explanation, and final solution.
	$\sqrt[5]{−243}$
Since ${(−3)}^{5} = −243 .$	$−3$

When we worked with square roots that had variables in the radicand, we restricted the variables to non-negative values. Now we will remove this restriction.

The odd root of a number can be either positive or negative. We have seen that $\sqrt[3]{−64} = −4$ .

But the even root of a non-negative number is always non-negative, because we take the principal nth root.

Suppose we start with $a = −5$ .

\begin{matrix} {(−5)}^{4} = 625 & \sqrt[4]{625} = 5 \end{matrix}

How can we make sure the fourth root of −5 raised to the fourth power, ${(−5)}^{4}$ is 5? We will see in the following property.

Solution

We use the absolute value to be sure to get the positive root.

ⓐ

Illustration of simplifying sqrt(x^2) to |x| and the underlying mathematical justification.
	$\sqrt{x^{2}}$
Since ${(x)}^{2} = x^{2}$ and we want the positive root.	$\| x \|$

ⓑ

This table illustrates the simplification of the cube root of n cubed, providing the mathematical reasoning for the derivation.
	$\sqrt[3]{n^{3}}$
Since ${(n)}^{3} = n^{3} .$ It is an odd root so there is no need for an absolute value sign.	$n$

ⓒ

Illustrates the simplification of the fourth root of p to the fourth power to the absolute value of p, with an explanation.
	$\sqrt[4]{p^{4}}$
Since ${(p)}^{4} = p^{4}$ and we want the positive root.	$\| p \|$

ⓓ

This table demonstrates the simplification of the fifth root of y^5 to y, explaining why an absolute value is not required for odd roots.
	$\sqrt[5]{y^{5}}$
Since ${(y)}^{5} = y^{5} .$ It is an odd root so there is no need for an absolute value sign.	$y$

Simplify: ⓐ $\sqrt[3]{y^{18}}$ ⓑ $\sqrt[4]{z^{8}}$ .

Solution

ⓐ

This table demonstrates the step-by-step algebraic simplification of the cube root of y to the power of 18.
	$\sqrt[3]{y^{18}}$
Since ${(y^{6})}^{3} = y^{18} .$	$\sqrt[3]{{(y^{6})}^{3}}$
	$y^{6}$

ⓑ

Step-by-step simplification of the radical expression 4th root of z to the power of 8, showing mathematical reasons.
	$\sqrt[4]{z^{8}}$
Since ${(z^{2})}^{4} = z^{8} .$	$\sqrt[4]{{(z^{2})}^{4}}$
Since $z^{2}$ is positive, we do not need an absolute value sign.	$z^{2}$

Simplify: ⓐ $\sqrt[3]{64 p^{6}}$ ⓑ $\sqrt[4]{16 q^{12}}$ .

Solution

ⓐ

Illustrates the step-by-step process of simplifying a cube root expression with a perfect cube radicand.
	$\sqrt[3]{64 p^{6}}$
Rewrite $64 p^{6} as {(4 p^{2})}^{3} .$	$\sqrt[3]{{(4 p^{2})}^{3}}$
Take the cube root.	$4 p^{2}$

ⓑ

Steps to simplify the fourth root of 16q^12, showing intermediate algebraic expressions for each operation.
	$\sqrt[4]{16 q^{12}}$
Rewrite the radicand as a fourth power.	$\sqrt[4]{{(2 q^{3})}^{4}}$
Take the fourth root.	$2 \| q^{3} \|$

Use the Product Property to Simplify Expressions with Higher Roots

We will simplify expressions with higher roots in much the same way as we simplified expressions with square roots. An nth root is considered simplified if it has no factors of $m^{n}$ .

We will generalize the Product Property of Square Roots to include any integer root $n \geq 2$ .

Simplify: ⓐ $\sqrt[3]{x^{4}}$ ⓑ $\sqrt[4]{x^{7}}$ .

Solution

ⓐ

Steps for simplifying the cube root of x^4 by factoring the radicand and extracting perfect cubes.
	$\sqrt[3]{x^{4}}$
Rewrite the radicand as a product using the largest perfect cube factor.	$\sqrt[3]{x^{3} \cdot x}$
Rewrite the radical as the product of two radicals.	$\sqrt[3]{x^{3}} \cdot \sqrt[3]{x}$
Simplify.	$x \sqrt[3]{x}$

ⓑ

Illustrates the step-by-step simplification of the radical expression fourth root of x to the power of seven.
	$\sqrt[4]{x^{7}}$
Rewrite the radicand as a product using the greatest perfect fourth power factor.	$\sqrt[4]{x^{4} \cdot x^{3}}$
Rewrite the radical as the product of two radicals.	$\sqrt[4]{x^{4}} \cdot \sqrt[4]{x^{3}}$
Simplify.	$\| x \| \sqrt[4]{x^{3}}$

Simplify: ⓐ $\sqrt[3]{16}$ ⓑ $\sqrt[4]{243}$ .

Solution

ⓐ

Step-by-step process demonstrating the simplification of the cube root of 16.
	$\sqrt[3]{16}$
	$\sqrt[3]{2^{4}}$
Rewrite the radicand as a product using the greatest perfect cube factor.	$\sqrt[3]{2^{3} \cdot 2}$
Rewrite the radical as the product of two radicals.	$\sqrt[3]{2^{3}} \cdot \sqrt[3]{2}$
Simplify.	$2 \sqrt[3]{2}$

ⓑ

Step-by-step simplification of the fourth root of 243.
	$\sqrt[4]{243}$
	$\sqrt[4]{3^{5}}$
Rewrite the radicand as a product using the greatest perfect fourth power factor.	$\sqrt[4]{3^{4} \cdot 3}$
Rewrite the radical as the product of two radicals.	$\sqrt[4]{3^{4}} \cdot \sqrt[4]{3}$
Simplify.	$3 \sqrt[4]{3}$

Don’t forget to use the absolute value signs when taking an even root of an expression with a variable in the radical.

Simplify: ⓐ $\sqrt[3]{24 x^{7}}$ ⓑ $\sqrt[4]{80 y^{14}}$ .

Solution

ⓐ

This table details the step-by-step simplification of the cube root expression (24x^7)^(1/3) to its simplified form 2x^2 * (3x)^(1/3).
	$\sqrt[3]{24 x^{7}}$
Rewrite the radicand as a product using perfect cube factors.	$\sqrt[3]{2^{3} x^{6} \cdot 3 x}$
Rewrite the radical as the product of two radicals.	$\sqrt[3]{2^{3} x^{6}} \cdot \sqrt[3]{3 x}$
Rewrite the first radicand as ${(2 x^{2})}^{3} .$	$\sqrt[3]{{(2 x^{2})}^{3}} \cdot \sqrt[3]{3 x}$
Simplify.	$2 x^{2} \sqrt[3]{3 x}$

ⓑ

This table illustrates the step-by-step process of simplifying a fourth root radical expression.
	$\sqrt[4]{80 y^{14}}$
Rewrite the radicand as a product using perfect fourth power factors.	$\sqrt[4]{2^{4} y^{12} \cdot 5 y^{2}}$
Rewrite the radical as the product of two radicals.	$\sqrt[4]{2^{4} y^{12}} \cdot \sqrt[4]{5 y^{2}}$
Rewrite the first radicand as ${(2 y^{3})}^{4} .$	$\sqrt[4]{{(2 y^{3})}^{4}} \cdot \sqrt[4]{5 y^{2}}$
Simplify.	$2 \| y^{3} \| \sqrt[4]{5 y^{2}}$

Simplify: ⓐ $\sqrt[3]{−27}$ ⓑ $\sqrt[4]{−16}$ .

Solution

ⓐ

Steps to calculate the cube root of -27.
	$\sqrt[3]{−27}$
Rewrite the radicand as a product using perfect cube factors.	$\sqrt[3]{{(−3)}^{3}}$
Take the cube root.	$−3$

ⓑ

This table demonstrates why the fourth root of -16 is not considered a real number.
	$\sqrt[4]{−16}$
There is no real number $n$ where $n^{4} = −16 .$	Not a real number.

Use the Quotient Property to Simplify Expressions with Higher Roots

We can simplify higher roots with quotients in the same way we simplified square roots. First we simplify any fractions inside the radical.

Simplify: ⓐ $\sqrt[3]{\frac{a^{8}}{a^{5}}}$ ⓑ $\sqrt[4]{\frac{a^{10}}{a^{2}}}$ .

Solution

ⓐ

Step-by-step simplification of the radical expression ³√(a⁸/a⁵) to 'a', illustrating algebraic simplification.
	$\sqrt[3]{\frac{a^{8}}{a^{5}}}$
Simplify the fraction under the radical first.	$\sqrt[3]{a^{3}}$
Simplify.	$a$

ⓑ

Step-by-step process demonstrating the simplification of a radical expression involving variables.
	$\sqrt[4]{\frac{a^{10}}{a^{2}}}$
Simplify the fraction under the radical first.	$\sqrt[4]{a^{8}}$
Rewrite the radicand using perfect fourth power factors.	$\sqrt[4]{{(a^{2})}^{4}}$
Simplify.	$a^{2}$

Previously, we used the Quotient Property ‘in reverse’ to simplify square roots. Now we will generalize the formula to include higher roots.

Simplify: ⓐ $\frac{\sqrt[3]{−108}}{\sqrt[3]{2}}$ ⓑ $\frac{\sqrt[4]{96 x^{7}}}{\sqrt[4]{3 x^{2}}}$ .

Solution

ⓐ

Step-by-step simplification of a cube root expression using radical properties.
	$\frac{\sqrt[3]{−108}}{\sqrt[3]{2}}$
Neither radicand is a perfect cube, so use the Quotient Property to write as one radical.	$\sqrt[3]{\frac{−108}{2}}$
Simplify the fraction under the radical.	$\sqrt[3]{−54}$
Rewrite the radicand as a product using perfect cube factors.	$\sqrt[3]{{(−3)}^{3} \cdot 2}$
Rewrite the radical as the product of two radicals.	$\sqrt[3]{{(−3)}^{3}} \cdot \sqrt[3]{2}$
Simplify.	$−3 \sqrt[3]{2}$

ⓑ

Step-by-step simplification of a radical expression involving the division of fourth roots.
	$\frac{\sqrt[4]{96 x^{7}}}{\sqrt[4]{3 x^{2}}}$
Neither radicand is a perfect fourth power, so use the Quotient Property to write as one radical.	$\sqrt[4]{\frac{96 x^{7}}{3 x^{2}}}$
Simplify the fraction under the radical.	$\sqrt[4]{32 x^{5}}$
Rewrite the radicand as a product using perfect fourth power factors.	$\sqrt[4]{2^{4} x^{4} \cdot 2 x}$
Rewrite the radical as the product of two radicals.	$\sqrt[4]{{(2 x)}^{4}} \cdot \sqrt[4]{2 x}$
Simplify.	$2 \| x \| \sqrt[4]{2 x}$

If the fraction inside the radical cannot be simplified, we use the first form of the Quotient Property to rewrite the expression as the quotient of two radicals.

Simplify: ⓐ $\sqrt[3]{\frac{24 x^{7}}{y^{3}}}$ ⓑ $\sqrt[4]{\frac{48 x^{10}}{y^{8}}}$ .

Solution

ⓐ

Step-by-step process for simplifying a cube root expression involving a fraction and variables.
	$\sqrt[3]{\frac{24 x^{7}}{y^{3}}}$
The fraction in the radicand cannot be simplified. Use the Quotient Property to write as two radicals.	$\frac{\sqrt[3]{24 x^{7}}}{\sqrt[3]{y^{3}}}$
Rewrite each radicand as a product using perfect cube factors.	$\frac{\sqrt[3]{8 x^{6} \cdot 3 x}}{\sqrt[3]{y^{3}}}$
Rewrite the numerator as the product of two radicals.	$\frac{\sqrt[3]{{(2 x^{2})}^{3}} \sqrt[3]{3 x}}{\sqrt[3]{y^{3}}}$
Simplify.	$\frac{2 x^{2} \sqrt[3]{3 x}}{y}$

ⓑ

Step-by-step process demonstrating the simplification of a fourth root algebraic expression using radical properties.
	$\sqrt[4]{\frac{48 x^{10}}{y^{8}}}$
The fraction in the radicand cannot be simplified. Use the Quotient Property to write as two radicals.	$\frac{\sqrt[4]{48 x^{10}}}{\sqrt[4]{y^{8}}}$
Rewrite each radicand as a product using perfect fourth power factors.	$\frac{\sqrt[4]{16 x^{8} \cdot 3 x^{2}}}{\sqrt[4]{y^{8}}}$
Rewrite the numerator as the product of two radicals.	$\frac{\sqrt[4]{{(2 x^{2})}^{4}} \sqrt[4]{3 x^{2}}}{\sqrt[4]{{(y^{2})}^{4}}}$
Simplify.	$\frac{2 x^{2} \sqrt[4]{3 x^{2}}}{y^{2}}$

Add and Subtract Higher Roots

We can add and subtract higher roots like we added and subtracted square roots. First we provide a formal definition of like radicals.

Like radicals have the same index and the same radicand.

$9 \sqrt[4]{42 x}$ and $−2 \sqrt[4]{42 x}$ are like radicals.
$5 \sqrt[3]{125 x}$ and $6 \sqrt[3]{125 y}$ are not like radicals. The radicands are different.
$2 \sqrt[5]{1000 q}$ and $−4 \sqrt[4]{1000 q}$ are not like radicals. The indices are different.

We add and subtract like radicals in the same way we add and subtract like terms. We can add $9 \sqrt[4]{42 x} + (−2 \sqrt[4]{42 x})$ and the result is $7 \sqrt[4]{42 x}$ .

Simplify: ⓐ $\sqrt[3]{4 x} + \sqrt[3]{4 x}$ ⓑ $4 \sqrt[4]{8} - 2 \sqrt[4]{8}$ .

Solution

ⓐ

Example illustrating the addition of like radicals, including an explanation and the simplified expression.
	$\sqrt[3]{4 x} + \sqrt[3]{4 x}$
The radicals are like, so we add the coefficients.	$2 \sqrt[3]{4 x}$

ⓑ

Simplification of like radical expressions, demonstrating the subtraction of coefficients with an accompanying explanation.
	$4 \sqrt[4]{8} - 2 \sqrt[4]{8}$
The radicals are like, so we subtract the coefficients.	$2 \sqrt[4]{8}$

When an expression does not appear to have like radicals, we will simplify each radical first. Sometimes this leads to an expression with like radicals.

Simplify: ⓐ $\sqrt[3]{54} - \sqrt[3]{16}$ ⓑ $\sqrt[4]{48} + \sqrt[4]{243}$ .

Solution

ⓐ

Step-by-step simplification of an expression involving cube roots.
	$\sqrt[3]{54} - \sqrt[3]{16}$
Rewrite each radicand using perfect cube factors.	$\sqrt[3]{27} \cdot \sqrt[3]{2} - \sqrt[3]{8} \cdot \sqrt[3]{2}$
Rewrite the perfect cubes.	$\sqrt[3]{{(3)}^{3}} \sqrt[3]{2} - \sqrt[3]{{(2)}^{3}} \sqrt[3]{2}$
Simplify the radicals where possible.	$3 \sqrt[3]{2} - 2 \sqrt[3]{2}$
Combine like radicals.	$\sqrt[3]{2}$

ⓑ

Step-by-step simplification of a sum of fourth roots by factoring out perfect fourth powers and combining like radicals.
	$\sqrt[4]{48} + \sqrt[4]{243}$
Rewrite using perfect fourth power factors.	$\sqrt[4]{16} \cdot \sqrt[4]{3} + \sqrt[4]{81} \cdot \sqrt[4]{3}$
Rewrite the perfect fourth powers.	$\sqrt[4]{{(2)}^{4}} \sqrt[4]{3} + \sqrt[4]{{(3)}^{4}} \sqrt[4]{3}$
Simplify the radicals where possible.	$2 \sqrt[4]{3} + 3 \sqrt[4]{3}$
Combine like radicals.	$5 \sqrt[4]{3}$

Simplify: ⓐ $\sqrt[3]{24 x^{4}} - \sqrt[3]{−81 x^{7}}$ ⓑ $\sqrt[4]{162 y^{9}} + \sqrt[4]{512 y^{5}}$ .

Solution

ⓐ

Step-by-step simplification of an algebraic expression involving cube roots.
	$\sqrt[3]{24 x^{4}} - \sqrt[3]{−81 x^{7}}$
Rewrite each radicand using perfect cube factors.	$\sqrt[3]{8 x^{3}} \cdot \sqrt[3]{3 x} - \sqrt[3]{−27 x^{6}} \cdot \sqrt[3]{3 x}$
Rewrite the perfect cubes.	$\sqrt[3]{{(2 x)}^{3}} \sqrt[3]{3 x} - \sqrt[3]{{(−3 x^{2})}^{3}} \sqrt[3]{3 x}$
Simplify the radicals where possible.	$2 x \sqrt[3]{3 x} - (−3 x^{2} \sqrt[3]{3 x})$

ⓑ

Step-by-step simplification of a sum involving fourth-root radical expressions.
	$\sqrt[4]{162 y^{9}} + \sqrt[4]{512 y^{5}}$
Rewrite each radicand using perfect fourth power factors.	$\sqrt[4]{81 y^{8}} \cdot \sqrt[4]{2 y} + \sqrt[4]{256 y^{4}} \cdot \sqrt[4]{2 y}$
Rewrite the perfect fourth powers.	$\sqrt[4]{{(3 y^{2})}^{4}} \cdot \sqrt[4]{2 y} + \sqrt[4]{{(4 y)}^{4}} \cdot \sqrt[4]{2 y}$
Simplify the radicals where possible.	$3 y^{2} \sqrt[4]{2 y} + 4 \| y \| \sqrt[4]{2 y}$

Key Concepts

Properties of
an when n is an even number and
- $a \geq 0$ , then $\sqrt[n]{a}$ is a real number
- $a < 0$ , then $\sqrt[n]{a}$ is not a real number
- When $n$ is an odd number, $\sqrt[n]{a}$ is a real number for all values of a.
- For any integer $n \geq 2$ , when n is odd $\sqrt[n]{a^{n}} = a$
- For any integer $n \geq 2$ , when n is even $\sqrt[n]{a^{n}} = | a |$
$\sqrt[n]{a}$ is considered simplified if a has no factors of $m^{n}$ .
Product Property of nth Roots

$\sqrt[n]{a b} = \sqrt[n]{a} \cdot \sqrt[n]{b} and \sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a b}$
Quotient Property of nth Roots

$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} and \frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$
To combine like radicals, simply add or subtract the coefficients while keeping the radical the same.

Practice Makes Perfect

Simplify Expressions with Higher Roots

In the following exercises, simplify.

Solution

ⓐ $\sqrt[5]{u^{5}}$ ⓑ $\sqrt[8]{v^{8}}$

ⓐ $\sqrt[3]{a^{3}}$
ⓑ

Solution

ⓐ $a$ ⓑ $| b |$

ⓐ $\sqrt[4]{y^{4}}$ ⓑ $\sqrt[7]{m^{7}}$

ⓐ $\sqrt[8]{k^{8}}$ ⓑ $\sqrt[6]{p^{6}}$

Solution

ⓐ $| k |$ ⓑ $| p |$

ⓐ $\sqrt[3]{x^{9}}$ ⓑ $\sqrt[4]{y^{12}}$

ⓐ $\sqrt[5]{a^{10}}$ ⓑ $\sqrt[3]{b^{27}}$

Solution

ⓐ $a^{2}$ ⓑ $b^{9}$

ⓐ $\sqrt[4]{m^{8}}$ ⓑ $\sqrt[5]{n^{20}}$

ⓐ $\sqrt[6]{r^{12}}$ ⓑ $\sqrt[3]{s^{30}}$

Solution

ⓐ $r^{2}$ ⓑ $s^{10}$

ⓐ $\sqrt[4]{16 x^{8}}$ ⓑ $\sqrt[6]{64 y^{12}}$

ⓐ $\sqrt[3]{−8 c^{9}}$ ⓑ $\sqrt[3]{125 d^{15}}$

Solution

ⓐ $−2 c^{3}$ ⓑ $5 d^{5}$

ⓐ $\sqrt[3]{216 a^{6}}$ ⓑ $\sqrt[5]{32 b^{20}}$

ⓐ $\sqrt[7]{128 r^{14}}$ ⓑ $\sqrt[4]{81 s^{24}}$

Solution

ⓐ $2 r^{2}$ ⓑ $3 s^{6}$

Use the Product Property to Simplify Expressions with Higher Roots

In the following exercises, simplify.

ⓐ $\sqrt[3]{r^{5}}$ ⓑ $\sqrt[4]{s^{10}}$

ⓐ $\sqrt[5]{u^{7}}$ ⓑ $\sqrt[6]{v^{11}}$

Solution

ⓐ $u \sqrt[5]{u^{2}}$ ⓑ $|v| \sqrt[6]{v^{5}}$

ⓐ $\sqrt[4]{m^{5}}$ ⓑ $\sqrt[8]{n^{10}}$

ⓐ $\sqrt[5]{p^{8}}$ ⓑ $\sqrt[3]{q^{8}}$

Solution

ⓐ $p \sqrt[5]{p^{3}}$ ⓑ $q^{2} \sqrt[3]{q^{2}}$

ⓐ $\sqrt[4]{32}$ ⓑ $\sqrt[7]{64}$

ⓐ $\sqrt[3]{625}$ ⓑ $\sqrt[6]{128}$

Solution

ⓐ $5 \sqrt[3]{5}$ ⓑ $2 \sqrt[6]{2}$

ⓐ $\sqrt[5]{64}$ ⓑ $\sqrt[3]{256}$

ⓐ $\sqrt[4]{3125}$ ⓑ $\sqrt[3]{81}$

Solution

ⓐ $5 \sqrt[4]{5}$ ⓑ $3 \sqrt[3]{3}$

ⓐ $\sqrt[3]{108 x^{5}}$ ⓑ $\sqrt[4]{48 y^{6}}$

ⓐ $\sqrt[5]{96 a^{7}}$ ⓑ $\sqrt[3]{375 b^{4}}$

Solution

ⓐ $2 a \sqrt[5]{3 a^{2}}$ ⓑ $5 b \sqrt[3]{3 b}$

ⓐ $\sqrt[4]{405 m^{10}}$ ⓑ $\sqrt[5]{160 n^{8}}$

ⓐ $\sqrt[3]{512 p^{5}}$ ⓑ $\sqrt[4]{324 q^{7}}$

Solution

ⓐ $8 p \sqrt[3]{p^{2}}$ ⓑ $3 |q| \sqrt[4]{4 q^{3}}$

ⓐ $\sqrt[3]{−864}$ ⓑ $\sqrt[4]{−256}$

ⓐ $\sqrt[5]{−486}$ ⓑ $\sqrt[6]{−64}$

Solution

ⓐ $−3 \sqrt[5]{2}$ ⓑ $not real$

ⓐ $\sqrt[5]{−32}$ ⓑ $\sqrt[8]{−1}$

ⓐ $\sqrt[3]{−8}$ ⓑ $\sqrt[4]{−16}$

Solution

ⓐ $−2$ ⓑ $not real$

Use the Quotient Property to Simplify Expressions with Higher Roots

In the following exercises, simplify.

ⓐ $\sqrt[3]{\frac{p^{11}}{p^{2}}}$ ⓑ $\sqrt[4]{\frac{q^{17}}{q^{13}}}$

ⓐ $\sqrt[5]{\frac{d^{12}}{d^{7}}}$ ⓑ $\sqrt[8]{\frac{m^{12}}{m^{4}}}$

Solution

ⓐ $d$ ⓑ $| m |$

ⓐ $\sqrt[5]{\frac{u^{21}}{u^{11}}}$ ⓑ $\sqrt[6]{\frac{v^{30}}{v^{12}}}$

ⓐ $\sqrt[3]{\frac{r^{14}}{r^{5}}}$ ⓑ $\sqrt[4]{\frac{c^{21}}{c^{9}}}$

Solution

ⓐ $r^{3}$ ⓑ $| c^{3} |$

ⓐ $\frac{\sqrt[4]{64}}{\sqrt[4]{2}}$ ⓑ $\frac{\sqrt[5]{128 x^{8}}}{\sqrt[5]{2 x^{2}}}$

ⓐ $\frac{\sqrt[3]{−625}}{\sqrt[3]{5}}$ ⓑ $\frac{\sqrt[4]{80 m^{7}}}{\sqrt[4]{5 m}}$

Solution

ⓐ $−5$ ⓑ $2 |m| \sqrt[4]{m^{2}}$

ⓐ $\sqrt[3]{\frac{1250}{2}}$ ⓑ $\sqrt[4]{\frac{486 y^{9}}{2 y^{3}}}$

ⓐ $\sqrt[3]{\frac{162}{6}}$ ⓑ $\sqrt[4]{\frac{160 r^{10}}{5 r^{3}}}$

Solution

ⓐ $3 \sqrt[3]{6}$ ⓑ $2 | r | \sqrt[4]{2 r^{3}}$

ⓐ $\sqrt[3]{\frac{54 a^{8}}{b^{3}}}$ ⓑ $\sqrt[4]{\frac{64 c^{5}}{d^{2}}}$

ⓐ $\sqrt[5]{\frac{96 r^{11}}{s^{3}}}$ ⓑ $\sqrt[6]{\frac{128 u^{7}}{v^{3}}}$

Solution

ⓐ $2 r^{2} \sqrt[5]{\frac{3 r}{s^{3}}}$ ⓑ $2 |u| \sqrt[6]{\frac{2 u}{v^{3}}}$

ⓐ $\sqrt[3]{\frac{81 s^{8}}{t^{3}}}$ ⓑ $\sqrt[4]{\frac{64 p^{15}}{q^{12}}}$

ⓐ $\sqrt[3]{\frac{625 u^{10}}{v^{3}}}$ ⓑ $\sqrt[4]{\frac{729 c^{21}}{d^{8}}}$

Solution

ⓐ $\frac{5 u^{3} \sqrt[3]{5 u}}{v}$ ⓑ $\frac{3 |c^{5}| \sqrt[4]{9 c}}{d^{2}}$

Add and Subtract Higher Roots

In the following exercises, simplify.

ⓐ $\sqrt[7]{8 p} + \sqrt[7]{8 p}$ ⓑ $3 \sqrt[3]{25} - \sqrt[3]{25}$

ⓐ $\sqrt[3]{15 q} + \sqrt[3]{15 q}$ ⓑ $2 \sqrt[4]{27} - 6 \sqrt[4]{27}$

Solution

ⓐ $2 \sqrt[3]{15 q}$ ⓑ $−4 \sqrt[4]{27}$

ⓐ $3 \sqrt[5]{9 x} + 7 \sqrt[5]{9 x}$ ⓑ $8 \sqrt[7]{3 q} - 2 \sqrt[7]{3 q}$

ⓐ An algebraic expression showing the sum of 23 times the 12th root of 4y and 19 times the 12th root of 4y. The expression is 23(12th root of 4y) + 19(12th root of 4y). ⓑ The mathematical expression 31 times the 10th root of 5z minus 17 times the 10th root of 5z is displayed.

Solution

ⓐ The image displays the mathematical expression 42 times the 12th root of 4y. ⓑ The image shows the mathematical expression 14 times the 10th root of 5z.

ⓐ $\sqrt[3]{81} - \sqrt[3]{192}$ ⓑ $\sqrt[4]{512} - \sqrt[4]{32}$

ⓐ $\sqrt[3]{250} - \sqrt[3]{54}$ ⓑ $\sqrt[4]{243} - \sqrt[4]{1875}$

Solution

ⓐ $2 \sqrt[3]{2}$ ⓑ $−2 \sqrt[4]{3}$

ⓐ $\sqrt[3]{128} + \sqrt[3]{250}$ ⓑ $\sqrt[5]{729} + \sqrt[5]{96}$

ⓐ $\sqrt[4]{243} + \sqrt[4]{1250}$ ⓑ $\sqrt[3]{2000} + \sqrt[3]{54}$

Solution

ⓐ $3 \sqrt[4]{3} + 5 \sqrt[4]{2}$ ⓑ $13 \sqrt[3]{2}$

ⓐ $\sqrt[3]{64 a^{10}} - \sqrt[3]{−216 a^{12}}$ ⓑ $\sqrt[4]{486 u^{7}} + \sqrt[4]{768 u^{3}}$

ⓐ $\sqrt[3]{80 b^{5}} - \sqrt[3]{−270 b^{3}}$ ⓑ $\sqrt[4]{160 v^{10}} - \sqrt[4]{1280 v^{3}}$

Solution

ⓐ $2 b \sqrt[3]{10 b^{2}} + 3 b \sqrt[3]{10}$ ⓑ $2 v^{2} \sqrt[4]{10 v^{2}} - 4 \sqrt[4]{5 v^{3}}$

Mixed Practice

In the following exercises, simplify.

$\sqrt[4]{16}$

$\sqrt[6]{64}$

Solution

$2$

$\sqrt[3]{a^{3}}$

A mathematical expression featuring the 12th root of b raised to the power of 12. This expression simplifies to 'b'.

Solution

$| b |$

$\sqrt[3]{−8 c^{9}}$

$\sqrt[3]{125 d^{15}}$

Solution

$5 d^{5}$

$\sqrt[3]{r^{5}}$

$\sqrt[4]{s^{10}}$

Solution

$s^{2} \sqrt[4]{s^{2}}$

$\sqrt[3]{108 x^{5}}$

$\sqrt[4]{48 y^{6}}$

Solution

$2 |y| \sqrt[4]{3 y^{2}}$

$\sqrt[5]{−486}$

$\sqrt[6]{−64}$

Solution

$not real$

$\frac{\sqrt[4]{64}}{\sqrt[4]{2}}$

$\frac{\sqrt[5]{128 x^{8}}}{\sqrt[5]{2 x^{2}}}$

Solution

$2 x \sqrt[5]{2 x}$

$\sqrt[5]{\frac{96 r^{11}}{s^{3}}}$

$\sqrt[6]{\frac{128 u^{7}}{v^{3}}}$

Solution

$2 |u| \sqrt[6]{\frac{2 u}{v^{3}}}$

$\sqrt[3]{81} - \sqrt[3]{192}$

$\sqrt[4]{512} - \sqrt[4]{32}$

Solution

$2 \sqrt[4]{2}$

$\sqrt[3]{64 a^{10}} - \sqrt[3]{−216 a^{12}}$

$\sqrt[4]{486 u^{7}} + \sqrt[4]{768 u^{3}}$

Solution

$3 │ u │ \sqrt[4]{6 u^{3}} + 4 \sqrt[4]{3 u^{3}}$

Everyday Math

Population growth The expression $10 \cdot x^{n}$ models the growth of a mold population after $n$ generations. There were 10 spores at the start, and each had $x$ offspring. So $10 \cdot x^{5}$ is the number of offspring at the fifth generation. At the fifth generation there were 10,240 offspring. Simplify the expression $\sqrt[5]{\frac{10,240}{10}}$ to determine the number of offspring of each spore.

Spread of a virus The expression $3 \cdot x^{n}$ models the spread of a virus after $n$ cycles. There were three people originally infected with the virus, and each of them infected $x$ people. So $3 \cdot x^{4}$ is the number of people infected on the fourth cycle. At the fourth cycle 1875 people were infected. Simplify the expression $\sqrt[4]{\frac{1875}{3}}$ to determine the number of people each person infected.

Solution

$5$

Writing Exercises

Explain how you know that $\sqrt[5]{x^{10}} = x^{2}$ .

Explain why $\sqrt[4]{−64}$ is not a real number but $\sqrt[3]{−64}$ is.

Solution

Answers may vary.

Self Check

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

This table has four columns and five rows. The first row labels each column: “I can…,” “Confidentaly,” “With some help,” and “No – I don’t get it!” The rows under the “I can…,” column read, “simplify expressions with hither roots.,” “use the product property to simplify expressions with higher roots.,” “use the quotient property to simplify expressions with higher roots.,” and “add and subtract higher roots.” The rest of the rows under the columns are empty.

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

nth root of a number

b^{n} = a

, then

b

is an nth root of

a

principal nth root

The principal nth root of

a

is written

\sqrt[n]{a}

index

\sqrt[n]{a}

n is called the index of the radical.

like radicals

Radicals with the same index and same radicand are called like radicals.

Rational Exponents

Simplify Expressions with $a^{\frac{1}{n}}$

Rational exponents are another way of writing expressions with radicals. When we use rational exponents, we can apply the properties of exponents to simplify expressions.

The Power Property for Exponents says that ${(a^{m})}^{n} = a^{m \cdot n}$ when m and n are whole numbers. Let’s assume we are now not limited to whole numbers.

Suppose we want to find a number p such that ${(8^{p})}^{3} = 8$ . We will use the Power Property of Exponents to find the value of p.

Step-by-step solution for an exponential equation, illustrating the relationship between fractional exponents and roots.
	$\begin{matrix} {(8^{p})}^{3} & = & 8 \end{matrix}$
Multiply the exponents on the left.	$\begin{matrix} 8^{3 p} & = & 8 \end{matrix}$
Write the exponent 1 on the right.	$\begin{matrix} 8^{3 p} & = & 8^{1} \end{matrix}$
The exponents must be equal.	$\begin{matrix} 3 p & = & 1 \end{matrix}$
Solve for $p .$	$\begin{matrix} p & = & \frac{1}{3} \end{matrix}$
	$\begin{matrix} So {(8^{\frac{1}{3}})}^{3} & = & 8 . \end{matrix}$
But we know also ${(\sqrt[3]{8})}^{3} = 8$ . Then it must be that $8^{\frac{1}{3}} = \sqrt[3]{8}$ .

But we know also ${(\sqrt[3]{8})}^{3} = 8$ . Then it must be that $8^{\frac{1}{3}} = \sqrt[3]{8}$ .

This same logic can be used for any positive integer exponent n to show that $a^{\frac{1}{n}} = \sqrt[n]{a}$ .

There will be times when working with expressions will be easier if you use rational exponents and times when it will be easier if you use radicals. In the first few examples, you’ll practice converting expressions between these two notations.

Solution

We want to write each expression in the form $\sqrt[n]{a}$ .

ⓐ

This table illustrates the conversion of a fractional exponent (x^(1/2)) to its equivalent radical form (sqrt(x)), including a rule explanation.
	$x^{\frac{1}{2}}$
The denominator of the exponent is 2, so the index of the radical is 2. We do not show the index when it is 2.	$\sqrt{x}$

ⓑ

This table demonstrates the relationship between fractional exponents and their radical equivalents, showing how the exponent's denominator determines the radical's index.
	$y^{\frac{1}{3}}$
The denominator of the exponent is 3, so the index is 3.	$\sqrt[3]{y}$

ⓒ

Conversion of fractional exponents to radical form, showing the exponent's denominator as the radical's index.
	$z^{\frac{1}{4}}$
The denominator of the exponent is 4, so the index is 4.	$\sqrt[4]{z}$

Solution

We want to write each radical in the form $a^{\frac{1}{n}}$ .

ⓐ

Illustrates the equivalence between square root notation and fractional exponent notation, with explanations for the transformation.
	$\sqrt{x}$
No index is shown, so it is 2. The denominator of the exponent will be 2.	$x^{\frac{1}{2}}$

ⓑ

Conversion of cube roots to fractional exponents, illustrating how the root's index determines the exponent's denominator.
	$\sqrt[3]{y}$
The index is 3, so the denominator of the exponent is 3.	$y^{\frac{1}{3}}$

ⓒ

Illustrates the conversion of a fourth root expression to its equivalent exponential form, detailing the rule for the exponent's denominator.
	$\sqrt[4]{z}$
The index is 4, so the denominator of the exponent is 4.	$z^{\frac{1}{4}}$

Solution

We want to write each radical in the form $a^{\frac{1}{n}}$ .

ⓐ

Illustrates converting radical expressions to rational exponent form, showing how the root's index determines the fractional exponent.
	$\sqrt{5 y}$
No index is shown, so it is 2. The denominator of the exponent will be 2.	${(5 y)}^{\frac{1}{2}}$

ⓑ

Illustrates converting the cube root of 4x to its equivalent exponential form with a fractional exponent, explaining the role of the index.
	$\sqrt[3]{4 x}$
The index is 3, so the denominator of the exponent is 3.	${(4 x)}^{\frac{1}{3}}$

ⓒ

Conversion of a radical expression to an equivalent exponential form, showing an example with an explanation.
	$3 \sqrt[4]{5 z}$
The index is 4, so the denominator of the exponent is 4.	$3 {(5 z)}^{\frac{1}{4}}$

In the next example, you may find it easier to simplify the expressions if you rewrite them as radicals first.

Solution

ⓐ

This table illustrates the step-by-step simplification of a numerical expression with a fractional exponent.
	$25^{\frac{1}{2}}$
Rewrite as a square root.	$\sqrt{25}$
Simplify.	$5$

ⓑ

This table illustrates the step-by-step simplification of a numerical expression with a fractional exponent, converting it to a cube root for evaluation.
	$64^{\frac{1}{3}}$
Rewrite as a cube root.	$\sqrt[3]{64}$
Recognize 64 is a perfect cube.	$\sqrt[3]{4^{3}}$
Simplify.	$4$

ⓒ

Step-by-step simplification of 256^(1/4), demonstrating rewriting as a root and final simplification.
	$256^{\frac{1}{4}}$
Rewrite as a fourth root.	$\sqrt[4]{256}$
Recognize 256 is a perfect fourth power.	$\sqrt[4]{4^{4}}$
Simplify.	$4$

Be careful of the placement of the negative signs in the next example. We will need to use the property $a^{− n} = \frac{1}{a^{n}}$ in one case.

Solution

ⓐ

Illustrates the step-by-step evaluation of a fractional exponent expression by conversion to a cube root and simplification.
	${(−64)}^{\frac{1}{3}}$
Rewrite as a cube root.	$\sqrt[3]{−64}$
Rewrite $−64$ as a perfect cube.	$\sqrt[3]{{(−4)}^{3}}$
Simplify.	$−4$

ⓑ

Detailed steps for simplifying the mathematical expression -64^(1/3) by rewriting it as a cube root and evaluating the final result.
	$− 64^{\frac{1}{3}}$
The exponent applies only to the 64.	$− (64^{\frac{1}{3}})$
Rewrite as a cube root.	$− \sqrt[3]{64}$
Rewrite 64 as $4^{3} .$	$− \sqrt[3]{4^{3}}$
Simplify.	$−4$

ⓒ

This table demonstrates the step-by-step process for simplifying a mathematical expression with a negative fractional exponent.
	${(64)}^{- \frac{1}{3}}$
Rewrite as a fraction with a positive exponent, using the property, $a^{− n} = \frac{1}{a^{n}} .$	$\frac{1}{\sqrt[3]{64}}$
Write as a cube root.
Rewrite 64 as $4^{3} .$	$\frac{1}{\sqrt[3]{4^{3}}}$
Simplify.	$\frac{1}{4}$

Solution

ⓐ

Evaluation of (-16)^(1/4), illustrating it has no real number solution.
	${(−16)}^{\frac{1}{4}}$
Rewrite as a fourth root.	$\sqrt[4]{−16}$
There is no real number whose fourth power is $−16 .$

ⓑ

Step-by-step simplification of the mathematical expression -16^(1/4) to its final integer value.
	$− 16^{\frac{1}{4}}$
The exponent only applies to the 16. Rewrite as a fourth root.	$− \sqrt[4]{16}$
Rewrite $16$ as $2^{4} .$	$− \sqrt[4]{2^{4}}$
Simplify.	$−2$

ⓒ

This table shows the step-by-step simplification of the mathematical expression (16)^(-1/4), transforming it into its simplified fractional form of 1/2.
	${(16)}^{- \frac{1}{4}}$
Rewrite using the property $a^{− n} = \frac{1}{a^{n}} .$	$\frac{1}{{(16)}^{\frac{1}{4}}}$
Rewrite as a fourth root.	$\frac{1}{\sqrt[4]{16}}$
Rewrite $16$ as $2^{4} .$	$\frac{1}{\sqrt[4]{2^{4}}}$
Simplify.	$\frac{1}{2}$

Simplify Expressions with $a^{\frac{m}{n}}$

Let’s work with the Power Property for Exponents some more.

Suppose we raise $a^{\frac{1}{n}}$ to the power m.

Steps to simplify an exponential expression with a fractional power, demonstrating its equivalence to a radical form.
	${(a^{\frac{1}{n}})}^{m}$
Multiply the exponents.	$a^{\frac{1}{n}}^{\cdot m}$
Simplify.	$a^{\frac{m}{n}}$
So $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m} .$

Now suppose we take $a^{m}$ to the $\frac{1}{n}$ power.

This table demonstrates the steps to simplify (a^m)^(1/n) to a^(m/n), illustrating its equivalence with the n-th root of (a^m).
	${(a^{m})}^{\frac{1}{n}}$
Multiply the exponents.	$a^{m \cdot \frac{1}{n}}$
Simplify.	$a^{\frac{m}{n}}$
So $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ also.

Which form do we use to simplify an expression? We usually take the root first—that way we keep the numbers in the radicand smaller.

Solution

We want to use $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ to write each radical in the form $a^{\frac{m}{n}}$ .

Solution

We will rewrite each expression as a radical first using the property, $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m}$ . This form lets us take the root first and so we keep the numbers in the radicand smaller than if we used the other form.

ⓐ

This table illustrates the step-by-step evaluation of the exponential expression 9^(3/2) by converting it to radical form.
	$9^{\frac{3}{2}}$
The power of the radical is the numerator of the exponent, 3. Since the denominator of the exponent is 2, this is a square root.	${(\sqrt{9})}^{3}$
Simplify.	${(3)}^{3}$
	$27$

ⓑ

Steps to simplify an expression with a rational exponent by converting it to radical form.
	$125^{\frac{2}{3}}$
The power of the radical is the numerator of the exponent, 2. The index of the radical is the denominator of the exponent, 3.	${(\sqrt[3]{125})}^{2}$
Simplify.	${(5)}^{2}$
	$25$

ⓒ

Steps to simplify an expression with a rational exponent by converting it to radical form and calculating the result.
	$81^{\frac{3}{4}}$
The power of the radical is the numerator of the exponent, 3. The index of the radical is the denominator of the exponent, 4.	${(\sqrt[4]{81})}^{3}$
Simplify.	${(3)}^{3}$
	$27$

Remember that $b^{− p} = \frac{1}{b^{p}}$ . The negative sign in the exponent does not change the sign of the expression.

Solution

We will rewrite each expression first using $b^{− p} = \frac{1}{b^{p}}$ and then change to radical form.

ⓐ

Step-by-step simplification of an expression with a negative fractional exponent, showing the application of exponent and radical rules.
	$16^{- \frac{3}{2}}$
Rewrite using $b^{− p} = \frac{1}{b^{p}} .$	$\frac{1}{16^{\frac{3}{2}}}$
Change to radical form. The power of the radical is the numerator of the exponent, 3. The index is the denominator of the exponent, 2.	$\frac{1}{{(\sqrt{16})}^{3}}$
Simplify.	$\frac{1}{4^{3}}$
	$\frac{1}{64}$

ⓑ

Detailed steps to simplify 32^(-2/5) by converting to a fraction, radical form, and evaluating to 1/4.
	$32^{- \frac{2}{5}}$
Rewrite using $b^{− p} = \frac{1}{b^{p}} .$	$\frac{1}{32^{\frac{2}{5}}}$
Change to radical form.	$\frac{1}{{(\sqrt[5]{32})}^{2}}$
Rewrite the radicand as a power.	$\frac{1}{{(\sqrt[5]{2^{5}})}^{2}}$
Simplify.	$\frac{1}{2^{2}}$
	$\frac{1}{4}$

ⓒ

Step-by-step simplification of an expression with a negative fractional exponent, demonstrating conversions to positive exponents and radical form.
	$4^{- \frac{5}{2}}$
Rewrite using $b^{− p} = \frac{1}{b^{p}} .$	$\frac{1}{4^{\frac{5}{2}}}$
Change to radical form.	$\frac{1}{{(\sqrt{4})}^{5}}$
Simplify.	$\frac{1}{2^{5}}$
	$\frac{1}{32}$

Solution

ⓐ

Step-by-step simplification of the expression -25^(3/2) by rewriting it in radical form and evaluating it to -125.
	$− 25^{\frac{3}{2}}$
Rewrite in radical form.	$− {(\sqrt{25})}^{3}$
Simplify the radical.	$− {(5)}^{3}$
Simplify.	$−125$

ⓑ

This table illustrates the step-by-step simplification of the mathematical expression -25^(-3/2) by applying exponent rules and converting to radical form.
	$− 25^{- \frac{3}{2}}$
Rewrite using $b^{− p} = \frac{1}{b^{p}} .$	$− (\frac{1}{25^{\frac{3}{2}}})$
Rewrite in radical form.	$− (\frac{1}{{(\sqrt{25})}^{3}})$
Simplify the radical.	$− (\frac{1}{{(5)}^{3}})$
Simplify.	$- \frac{1}{125}$

ⓒ

This table evaluates the expression (-25)^(3/2), demonstrating its transformation into radical form and concluding it is not a real number.
	${(−25)}^{\frac{3}{2}}$
Rewrite in radical form.	${(\sqrt{−25})}^{3}$
There is no real number whose square root is $−25 .$	Not a real number.

Use the Laws of Exponents to Simplify Expressions with Rational Exponents

The same laws of exponents that we already used apply to rational exponents, too. We will list the Exponent Properties here to have them for reference as we simplify expressions.

When we multiply the same base, we add the exponents.

Simplify: ⓐ $2^{\frac{1}{2}} \cdot 2^{\frac{5}{2}}$ ⓑ $x^{\frac{2}{3}} \cdot x^{\frac{4}{3}}$ ⓒ $z^{\frac{3}{4}} \cdot z^{\frac{5}{4}}$ .

Solution

ⓐ

Step-by-step simplification of the exponential expression 2^(1/2) * 2^(5/2) to 8.
	$2^{\frac{1}{2}} \cdot 2^{\frac{5}{2}}$
The bases are the same, so we add the exponents.	$2^{\frac{1}{2} + \frac{5}{2}}$
Add the fractions.	$2^{\frac{6}{2}}$
Simplify the exponent.	$2^{3}$
Simplify.	$8$

ⓑ

Steps demonstrating the simplification of an algebraic expression involving the multiplication of exponential terms with the same base.
	$x^{\frac{2}{3}} \cdot x^{\frac{4}{3}}$
The bases are the same, so we add the exponents.	$x^{\frac{2}{3} + \frac{4}{3}}$
Add the fractions.	$x^{\frac{6}{3}}$
Simplify.	$x^{2}$

ⓒ

Step-by-step simplification of an algebraic expression with fractional exponents using exponent rules.
	$z^{\frac{3}{4}} \cdot z^{\frac{5}{4}}$
The bases are the same, so we add the exponents.	$z^{\frac{3}{4} + \frac{5}{4}}$
Add the fractions.	$z^{\frac{8}{4}}$
Simplify.	$z^{2}$

We will use the Power Property in the next example.

Solution

ⓐ

Simplifying an expression with a power raised to a power by multiplying exponents, with steps and examples.
	${(x^{4})}^{\frac{1}{2}}$
To raise a power to a power, we multiply the exponents.	$x^{4 \cdot \frac{1}{2}}$
Simplify.	$x^{2}$

ⓑ

This table illustrates the step-by-step simplification of an algebraic expression involving a power raised to a fractional power, applying the rule for multiplying exponents.
	${(y^{6})}^{\frac{1}{3}}$
To raise a power to a power, we multiply the exponents.	$y^{6 \cdot \frac{1}{3}}$
Simplify.	$y^{2}$

ⓒ

Demonstration of simplifying the exponential expression (z^9)^(2/3) by applying the power to a power rule.
	${(z^{9})}^{\frac{2}{3}}$
To raise a power to a power, we multiply the exponents.	$z^{9 \cdot \frac{2}{3}}$
Simplify.	$z^{6}$

The Quotient Property tells us that when we divide with the same base, we subtract the exponents.

Simplify: ⓐ $\frac{x^{\frac{4}{3}}}{x^{\frac{1}{3}}}$ ⓑ $\frac{y^{\frac{3}{4}}}{y^{\frac{1}{4}}}$ ⓒ $\frac{z^{\frac{2}{3}}}{z^{\frac{5}{3}}}$ .

Solution

ⓐ

Step-by-step simplification of an exponential expression involving division with the same base by subtracting fractional exponents.
	$\frac{x^{\frac{4}{3}}}{x^{\frac{1}{3}}}$
To divide with the same base, we subtract the exponents.	$x^{\frac{4}{3} - \frac{1}{3}}$
Simplify.	$x$

ⓑ

Steps to simplify an exponential expression by dividing terms with the same base and subtracting their fractional exponents.
	$\frac{y^{\frac{3}{4}}}{y^{\frac{1}{4}}}$
To divide with the same base, we subtract the exponents.	$y^{\frac{3}{4} - \frac{1}{4}}$
Simplify.	$y^{\frac{1}{2}}$

ⓒ

This table demonstrates the simplification of an exponential expression by applying rules for dividing with the same base and rewriting negative exponents.
	$\frac{z^{\frac{2}{3}}}{z^{\frac{5}{3}}}$
To divide with the same base, we subtract the exponents.	$z^{\frac{2}{3} - \frac{5}{3}}$
Rewrite without a negative exponent.	$\frac{1}{z}$

Sometimes we need to use more than one property. In the next two examples, we will use both the Product to a Power Property and then the Power Property.

Simplify: ⓐ ${(27 u^{\frac{1}{2}})}^{\frac{2}{3}}$ ⓑ ${(8 v^{\frac{1}{4}})}^{\frac{2}{3}}$ .

Solution

ⓐ

Demonstrates the step-by-step simplification of the expression (27u^(1/2))^(2/3) to 9u^(1/3) using exponent rules.
	${(27 u^{\frac{1}{2}})}^{\frac{2}{3}}$
First we use the Product to a Power Property.	${(27)}^{\frac{2}{3}} {(u^{\frac{1}{2}})}^{\frac{2}{3}}$
Rewrite 27 as a power of 3.	${(3^{3})}^{\frac{2}{3}} {(u^{\frac{1}{2}})}^{\frac{2}{3}}$
To raise a power to a power, we multiply the exponents.	$(3^{2}) (u^{\frac{1}{3}})$
Simplify.	$9 u^{\frac{1}{3}}$

ⓑ

This table demonstrates the step-by-step simplification of an algebraic expression involving rational exponents.
	${(8 v^{\frac{1}{4}})}^{\frac{2}{3}}$
First we use the Product to a Power Property.	${(8)}^{\frac{2}{3}} {(v^{\frac{1}{4}})}^{\frac{2}{3}}$
Rewrite 8 as a power of 2.	${(2^{3})}^{\frac{2}{3}} {(v^{\frac{1}{4}})}^{\frac{2}{3}}$
To raise a power to a power, we multiply the exponents.	$(2^{2}) (v^{\frac{1}{6}})$
Simplify.	$4 v^{\frac{1}{6}}$

Simplify: ⓐ ${(m^{3} n^{9})}^{\frac{1}{3}}$ ⓑ ${(p^{4} q^{8})}^{\frac{1}{4}}$ .

Solution

ⓐ

Step-by-step simplification of a mathematical expression involving powers and roots using the Product to a Power Property.
	${(m^{3} n^{9})}^{\frac{1}{3}}$
First we use the Product to a Power Property.	${(m^{3})}^{\frac{1}{3}} {(n^{9})}^{\frac{1}{3}}$
To raise a power to a power, we multiply the exponents.	$m n^{3}$

ⓑ

Step-by-step example illustrating the simplification of an exponential expression (p^4q^8)^(1/4) using exponent properties.
	${(p^{4} q^{8})}^{\frac{1}{4}}$
First we use the Product to a Power Property.	${(p^{4})}^{\frac{1}{4}} {(q^{8})}^{\frac{1}{4}}$
To raise a power to a power, we multiply the exponents.	$p q^{2}$

We will use both the Product and Quotient Properties in the next example.

Simplify: ⓐ $\frac{x^{\frac{3}{4}} \cdot x^{- \frac{1}{4}}}{x^{- \frac{6}{4}}}$ ⓑ $\frac{y^{\frac{4}{3}} \cdot y}{y^{- \frac{2}{3}}}$ .

Solution

ⓐ

This table illustrates the step-by-step simplification of an algebraic expression involving fractional exponents using properties of exponents.
	$\frac{x^{\frac{3}{4}} \cdot x^{- \frac{1}{4}}}{x^{- \frac{6}{4}}}$
Use the Product Property in the numerator, add the exponents.	$\frac{x^{\frac{2}{4}}}{x^{- \frac{6}{4}}}$
Use the Quotient Property, subtract the exponents.	$x^{\frac{8}{4}}$
Simplify.	$x^{2}$

ⓑ

Step-by-step simplification of an algebraic expression using product and quotient properties of exponents.
	$\frac{y^{\frac{4}{3}} \cdot y}{y^{- \frac{2}{3}}}$
Use the Product Property in the numerator, add the exponents.	$\frac{y^{\frac{7}{3}}}{y^{- \frac{2}{3}}}$
Use the Quotient Property, subtract the exponents.	$y^{\frac{9}{3}}$
Simplify.	$y^{3}$

Key Concepts

Summary of Exponent Properties
If a,b are real numbers and m,n are rational numbers, then
- Product Property $a^{m} \cdot a^{n} = a^{m + n}$
- Power Property ${(a^{m})}^{n} = a^{m \cdot n}$
- Product to a Power ${(a b)}^{m} = a^{m} b^{m}$
- Quotient Property:
  
  $\frac{a^{m}}{a^{n}} = a^{m - n}, a \neq 0, m > n$
  
  $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}, a \neq 0, n > m$
- Zero Exponent Definition $a^{0} = 1$ , $a \neq 0$
- Quotient to a Power Property ${(\frac{a}{b})}^{m} = \frac{a^{m}}{b^{m}}, b \neq 0$

Section Exercises

Practice Makes Perfect

Simplify Expressions with $a^{\frac{1}{n}}$

In the following exercises, write as a radical expression.

Solution

In the following exercises, write with a rational exponent.

ⓐ $\sqrt[8]{r}$ ⓑ A mathematical radical symbol is displayed, with '19' as the index above the left arm of the root and '5' as the radicand inside, indicating the 19th root of 5. ⓒ $\sqrt[4]{t}$

Solution

ⓐ $\sqrt[5]{u}$ ⓑ $\sqrt{v}$ ⓒ A close-up view of the mathematical expression '16th root of W', stylized as a radical sign with '16' as the index and 'W' as the radicand.

Solution

ⓐ $\sqrt[3]{25 a}$ ⓑ $\sqrt{3 b}$ ⓒ A mathematical expression showing the tenth root of 40c, represented as an n-th root symbol with 10 as the index and 40c as the radicand.

Solution

In the following exercises, simplify.

Solution

ⓐ 5 ⓑ 3 ⓒ 2

Solution

ⓐ 6 ⓑ 2 ⓒ 3

Solution

Simplify Expressions with $a^{\frac{m}{n}}$

In the following exercises, write with a rational exponent.

Solution

In the following exercises, simplify.

Solution

Use the Laws of Exponents to Simplify Expressions with Rational Exponents

In the following exercises, simplify.

ⓐ $4^{\frac{5}{8}} \cdot 4^{\frac{11}{8}}$ ⓑ $m^{\frac{7}{12}} \cdot m^{\frac{17}{12}}$ ⓒ $p^{\frac{3}{7}} \cdot p^{\frac{18}{7}}$

ⓐ $6^{\frac{5}{2}} \cdot 6^{\frac{1}{2}}$ ⓑ $n^{\frac{2}{10}} \cdot n^{\frac{8}{10}}$ ⓒ $q^{\frac{2}{5}} \cdot q^{\frac{13}{5}}$

Solution

ⓐ $5^{\frac{1}{2}} \cdot 5^{\frac{7}{2}}$ ⓑ $c^{\frac{3}{4}} \cdot c^{\frac{9}{4}}$ ⓒ $d^{\frac{3}{5}} \cdot d^{\frac{2}{5}}$

ⓐ $10^{\frac{1}{3}} \cdot 10^{\frac{5}{3}}$ ⓑ $x^{\frac{5}{6}} \cdot x^{\frac{7}{6}}$ ⓒ $y^{\frac{11}{8}} \cdot y^{\frac{21}{8}}$

Solution

ⓐ $\frac{x^{\frac{7}{2}}}{x^{\frac{5}{2}}}$ ⓑ $\frac{y^{\frac{5}{2}}}{y^{\frac{1}{2}}}$ ⓒ $\frac{r^{\frac{4}{5}}}{r^{\frac{9}{5}}}$

ⓐ $\frac{s^{\frac{11}{5}}}{s^{\frac{6}{5}}}$ ⓑ $\frac{z^{\frac{7}{3}}}{z^{\frac{1}{3}}}$ ⓒ $\frac{w^{\frac{2}{7}}}{w^{\frac{9}{7}}}$

Solution

ⓐ $\frac{t^{\frac{12}{5}}}{t^{\frac{7}{5}}}$ ⓑ $\frac{x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$ ⓒ $\frac{m^{\frac{13}{8}}}{m^{\frac{5}{8}}}$

ⓐ $\frac{u^{\frac{13}{9}}}{u^{\frac{4}{9}}}$ ⓑ $\frac{r^{\frac{15}{7}}}{r^{\frac{8}{7}}}$ ⓒ $\frac{n^{\frac{3}{5}}}{n^{\frac{8}{5}}}$

Solution

ⓐ ${(9 p^{\frac{2}{3}})}^{\frac{5}{2}}$ ⓑ ${(27 q^{\frac{3}{2}})}^{\frac{4}{3}}$

ⓐ ${(81 r^{\frac{4}{5}})}^{\frac{1}{4}}$ ⓑ ${(64 s^{\frac{3}{7}})}^{\frac{1}{6}}$

Solution

ⓐ $3 r^{\frac{1}{5}}$ ⓑ $2 s^{\frac{1}{14}}$

ⓐ ${(16 u^{\frac{1}{3}})}^{\frac{3}{4}}$ ⓑ ${(100 v^{\frac{2}{5}})}^{\frac{3}{2}}$

ⓐ ${(27 m^{\frac{3}{4}})}^{\frac{2}{3}}$ ⓑ ${(625 n^{\frac{8}{3}})}^{\frac{3}{4}}$

Solution

ⓐ $9 m^{\frac{1}{2}}$ ⓑ $125 n^{2}$

ⓐ ${(x^{8} y^{10})}^{\frac{1}{2}}$ ⓑ ${(a^{9} b^{12})}^{\frac{1}{3}}$

ⓐ ${(r^{8} s^{4})}^{\frac{1}{4}}$ ⓑ ${(u^{15} v^{20})}^{\frac{1}{5}}$

Solution

ⓐ $r^{2} s$ ⓑ $u^{3} v^{4}$

ⓐ ${(a^{6} b^{16})}^{\frac{1}{2}}$ ⓑ ${(j^{9} k^{6})}^{\frac{2}{3}}$

ⓐ ${(r^{16} s^{10})}^{\frac{1}{2}}$ ⓑ ${(u^{10} v^{5})}^{\frac{4}{5}}$

Solution

ⓐ $r^{8} s^{5}$ ⓑ $u^{8} v^{4}$

ⓐ $\frac{r^{\frac{5}{2}} \cdot r^{- \frac{1}{2}}}{r^{- \frac{3}{2}}}$ ⓑ $\frac{s^{\frac{1}{5}} \cdot s}{s^{- \frac{9}{5}}}$

ⓐ $\frac{a^{\frac{3}{4}} \cdot a^{- \frac{1}{4}}}{a^{- \frac{10}{4}}}$ ⓑ $\frac{b^{\frac{2}{3}} \cdot b}{b^{- \frac{7}{3}}}$

Solution

ⓐ $a^{3}$ ⓑ $b^{4}$

ⓐ $\frac{c^{\frac{5}{3}} \cdot c^{- \frac{1}{3}}}{c^{- \frac{2}{3}}}$ ⓑ $\frac{d^{\frac{3}{5}} \cdot d}{d^{- \frac{2}{5}}}$

ⓐ $\frac{m^{\frac{7}{4}} \cdot m^{- \frac{5}{4}}}{m^{- \frac{2}{4}}}$ ⓑ $\frac{n^{\frac{3}{7}} \cdot n}{n^{- \frac{4}{7}}}$

Solution

ⓐ $m$ ⓑ $n^{2}$

$4^{\frac{5}{2}} \cdot 4^{\frac{1}{2}}$

$n^{\frac{2}{6}} \cdot n^{\frac{4}{6}}$

Solution

$n$

${(a^{24})}^{\frac{1}{6}}$

${(b^{10})}^{\frac{3}{5}}$

Solution

$b^{6}$

$\frac{w^{\frac{2}{5}}}{w^{\frac{7}{5}}}$

$\frac{z^{\frac{2}{3}}}{z^{\frac{8}{3}}}$

Solution

$\frac{1}{z^{2}}$

${(27 r^{\frac{3}{5}})}^{\frac{1}{3}}$

${(64 s^{\frac{3}{5}})}^{\frac{1}{6}}$

Solution

$2 s^{\frac{1}{10}}$

${(r^{9} s^{12})}^{\frac{1}{3}}$

${(u^{12} v^{18})}^{\frac{1}{6}}$

Solution

$u^{2} v^{3}$

Everyday Math

Landscaping Joe wants to have a square garden plot in his backyard. He has enough compost to cover an area of 144 square feet. Simplify $144^{\frac{1}{2}}$ to find the length of each side of his garden.

Landscaping Elliott wants to make a square patio in his yard. He has enough concrete to pave an area of 242 square feet. Simplify $242^{\frac{1}{2}}$ to find the length of each side of his patio.Round to the nearest tenth of a foot.

Solution

15.6 feet

Gravity While putting up holiday decorations, Bob dropped a decoration from the top of a tree that is 12 feet tall. Simplify $\frac{12^{\frac{1}{2}}}{16^{\frac{1}{2}}}$ to find how many seconds it took for the decoration to reach the ground. Round to the nearest tenth of a second.

Gravity An airplane dropped a flare from a height of 1024 feet above a lake. Simplify $\frac{1024^{\frac{1}{2}}}{16^{\frac{1}{2}}}$ to find how many seconds it took for the flare to reach the water.

Solution

8 seconds

Writing Exercises

Show two different algebraic methods to simplify $4^{\frac{3}{2}} .$ Explain all your steps.

Explain why the expression ${(−16)}^{\frac{3}{2}}$ cannot be evaluated.

Solution

Answers will vary.

Chapter 9 Review Exercises

Simplify and Use Square Roots

Simplify Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{64}$

$\sqrt{144}$

Solution

$- \sqrt{25}$

$- \sqrt{81}$

Solution

$−9$

$\sqrt{−9}$

$\sqrt{−36}$

Solution

not a real number

$\sqrt{64} + \sqrt{225}$

$\sqrt{64 + 225}$

Solution

Estimate Square Roots

In the following exercises, estimate each square root between two consecutive whole numbers.

$\sqrt{28}$

$\sqrt{155}$

Solution

$12 < \sqrt{155} < 13$

Approximate Square Roots

In the following exercises, approximate each square root and round to two decimal places.

$\sqrt{15}$

$\sqrt{57}$

Solution

7.55

Simplify Variable Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{q^{2}}$

$\sqrt{64 b^{2}}$

Solution

$8 |b|$

$− \sqrt{121 a^{2}}$

$\sqrt{225 m^{2} n^{2}}$

Solution

$15 |m| |n|$

$− \sqrt{100 q^{2}}$

$\sqrt{49 y^{2}}$

Solution

$7 |y|$

$\sqrt{4 a^{2} b^{2}}$

$\sqrt{121 c^{2} d^{2}}$

Solution

$11 |c| |d|$

Simplify Square Roots

Use the Product Property to Simplify Square Roots

In the following exercises, simplify.

$\sqrt{300}$

$\sqrt{98}$

Solution

$7 \sqrt{2}$

$\sqrt{x^{13}}$

$\sqrt{y^{19}}$

Solution

$|y^{9}| \sqrt{y}$

$\sqrt{16 m^{4}}$

$\sqrt{36 n^{13}}$

Solution

$6 n^{6} \sqrt{n}$

$\sqrt{288 m^{21}}$

$\sqrt{150 n^{7}}$

Solution

$5 |n^{3}| \sqrt{6 n}$

$\sqrt{48 r^{5} s^{4}}$

$\sqrt{108 r^{5} s^{3}}$

Solution

$6 r^{2} |s| \sqrt{3 r s}$

$\frac{10 - \sqrt{50}}{5}$

$\frac{6 + \sqrt{72}}{6}$

Solution

$1 + \sqrt{2}$

Use the Quotient Property to Simplify Square Roots

In the following exercises, simplify.

$\sqrt{\frac{16}{25}}$

$\sqrt{\frac{81}{36}}$

Solution

$\frac{3}{2}$

$\sqrt{\frac{x^{8}}{x^{4}}}$

$\sqrt{\frac{y^{6}}{y^{2}}}$

Solution

$y^{2}$

$\sqrt{\frac{98 p^{6}}{2 p^{2}}}$

$\sqrt{\frac{72 q^{8}}{2 q^{4}}}$

Solution

$6 q^{2}$

$\sqrt{\frac{65}{121}}$

$\sqrt{\frac{26}{169}}$

Solution

$\frac{\sqrt{26}}{13}$

$\sqrt{\frac{64 x^{4}}{25 x^{2}}}$

$\sqrt{\frac{36 r^{10}}{16 r^{5}}}$

Solution

$\frac{3 r^{2} \sqrt{r}}{2}$

$\sqrt{\frac{48 p^{3} q^{5}}{27 p q}}$

$\sqrt{\frac{12 r^{5} s^{7}}{75 r^{2} s}}$

Solution

$\frac{2 |r| |s^{3}| \sqrt{r}}{5}$

Add and Subtract Square Roots

Add and Subtract Like Square Roots

In the following exercises, simplify.

$3 \sqrt{2} + \sqrt{2}$

$5 \sqrt{5} + 7 \sqrt{5}$

Solution

$12 \sqrt{5}$

$4 \sqrt{y} + 4 \sqrt{y}$

$6 \sqrt{m} - 2 \sqrt{m}$

Solution

$4 \sqrt{m}$

$−3 \sqrt{7} + 2 \sqrt{7} - \sqrt{7}$

$8 \sqrt{13} + 2 \sqrt{3} + 3 \sqrt{13}$

Solution

$11 \sqrt{13} + 2 \sqrt{3}$

$3 \sqrt{5 x y} - \sqrt{5 x y} + 3 \sqrt{5 x y}$

$2 \sqrt{3 r s} + \sqrt{3 r s} - 5 \sqrt{r s}$

Solution

$3 \sqrt{3 r s} - 5 \sqrt{r s}$

Add and Subtract Square Roots that Need Simplification

In the following exercises, simplify.

$\sqrt{32} + 3 \sqrt{2}$

$\sqrt{8} + 3 \sqrt{2}$

Solution

$5 \sqrt{2}$

$\sqrt{72} + \sqrt{50}$

$\sqrt{48} + \sqrt{75}$

Solution

$9 \sqrt{3}$

$3 \sqrt{32} + \sqrt{98}$

$\frac{1}{3} \sqrt{27} - \frac{1}{8} \sqrt{192}$

Solution

$0$

$\sqrt{50 y^{5}} - \sqrt{72 y^{5}}$

$6 \sqrt{18 n^{4}} - 3 \sqrt{8 n^{4}} + n^{2} \sqrt{50}$

Solution

$17 n^{2} \sqrt{2}$

Multiply Square Roots

Multiply Square Roots

In the following exercises, simplify.

$\sqrt{2} \cdot \sqrt{20}$

$2 \sqrt{2} \cdot 6 \sqrt{14}$

Solution

$24 \sqrt{7}$

$\sqrt{2 m^{2}} \cdot \sqrt{20 m^{4}}$

$(6 \sqrt{2 y}) (3 \sqrt{50 y^{3}})$

Solution

$180 y^{2}$

$(6 \sqrt{3 v^{4}}) (5 \sqrt{30 v})$

${(\sqrt{8})}^{2}$

Solution

${(− \sqrt{10})}^{2}$

$(2 \sqrt{5}) (5 \sqrt{5})$

Solution

$(−3 \sqrt{3}) (5 \sqrt{18})$

Use Polynomial Multiplication to Multiply Square Roots

In the following exercises, simplify.

$10 (2 - \sqrt{7})$

Solution

$20 - 10 \sqrt{7}$

$\sqrt{3} (4 + \sqrt{12})$

$(5 + \sqrt{2}) (3 - \sqrt{2})$

Solution

$13 - 2 \sqrt{2}$

$(5 - 3 \sqrt{7}) (1 - 2 \sqrt{7})$

$(1 - 3 \sqrt{x}) (5 + 2 \sqrt{x})$

Solution

$5 - 13 \sqrt{x} - 6 x$

$(3 + 4 \sqrt{y}) (10 - \sqrt{y})$

${(1 + 6 \sqrt{p})}^{2}$

Solution

$1 + 12 \sqrt{p} + 36 p$

${(2 - 6 \sqrt{5})}^{2}$

$(3 + 2 \sqrt{7}) (3 - 2 \sqrt{7})$

Solution

$−19$

$(6 - \sqrt{11}) (6 + \sqrt{11})$

Divide Square Roots

Divide Square Roots

In the following exercises, simplify.

$\frac{\sqrt{75}}{10}$

Solution

$\frac{\sqrt{3}}{2}$

$\frac{2 - \sqrt{12}}{6}$

$\frac{\sqrt{48}}{\sqrt{27}}$

Solution

$\frac{4}{3}$

$\frac{\sqrt{75 x^{7}}}{\sqrt{3 x^{3}}}$

$\frac{\sqrt{20 y^{5}}}{\sqrt{2 y}}$

Solution

$y^{2} \sqrt{10}$

$\frac{\sqrt{98 p^{6} q^{4}}}{\sqrt{2 p^{4} q^{8}}}$

Rationalize a One Term Denominator

In the following exercises, rationalize the denominator.

$\frac{10}{\sqrt{15}}$

Solution

$\frac{2 \sqrt{15}}{3}$

$\frac{6}{\sqrt{6}}$

$\frac{5}{3 \sqrt{5}}$

Solution

$\frac{\sqrt{5}}{3}$

$\frac{10}{2 \sqrt{6}}$

$\sqrt{\frac{3}{28}}$

Solution

$\frac{\sqrt{21}}{14}$

$\sqrt{\frac{9}{75}}$

Rationalize a Two Term Denominator

In the following exercises, rationalize the denominator.

$\frac{4}{4 + \sqrt{27}}$

Solution

$\frac{16 - 12 \sqrt{3}}{−11}$

$\frac{5}{2 - \sqrt{10}}$

$\frac{4}{2 - \sqrt{5}}$

Solution

$−8 - 4 \sqrt{5}$

$\frac{5}{4 - \sqrt{8}}$

$\frac{\sqrt{2}}{\sqrt{p} + \sqrt{3}}$

Solution

$\frac{\sqrt{2 p} - \sqrt{6}}{p - 3}$

$\frac{\sqrt{x} - \sqrt{2}}{\sqrt{x} + \sqrt{2}}$

Solve Equations with Square Roots

Solve Radical Equations

In the following exercises, solve the equation.

$\sqrt{7 z + 1} = 6$

Solution

$\sqrt{4 u - 2} - 4 = 0$

$\sqrt{6 m + 4} - 5 = 0$

Solution

$\frac{7}{2}$

$\sqrt{2 u - 3} + 2 = 0$

$\sqrt{u - 4} + 4 = u$

Solution

4 and 5

$\sqrt{v - 9} + 9 = 0$

$\sqrt{r - 4} - r = −10$

Solution

$\sqrt{s - 9} - s = −9$

$2 \sqrt{2 x - 7} - 4 = 8$

Solution

$\frac{43}{2}$

$\sqrt{2 - x} = \sqrt{2 x - 7}$

$\sqrt{a} + 3 = \sqrt{a + 9}$

Solution

$\sqrt{r} + 3 = \sqrt{r + 4}$

$\sqrt{u} + 2 = \sqrt{u + 5}$

Solution

$\frac{1}{16}$

$\sqrt{n + 11} - 1 = \sqrt{n + 4}$

$\sqrt{y + 5} + 1 = \sqrt{2 y + 3}$

Solution

Use Square Roots in Applications

In the following exercises, solve. Round approximations to one decimal place.

A pallet of sod will cover an area of about 600 square feet. Trinh wants to order a pallet of sod to make a square lawn in his backyard. Use the formula $s = \sqrt{A}$ to find the length of each side of his lawn.

A helicopter dropped a package from a height of 900 feet above a stranded hiker. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the package to reach the hiker.

Solution

7.5 seconds

Officer Morales measured the skid marks of one of the cars involved in an accident. The length of the skid marks was 245 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied.

Higher Roots

Simplify Expressions with Higher Roots

In the following exercises, simplify.

ⓐ $\sqrt[6]{64}$
ⓑ $\sqrt[3]{64}$

Solution

ⓐ 2 ⓑ 4

ⓐ $\sqrt[3]{−27}$
ⓑ $\sqrt[4]{−64}$

ⓐ $\sqrt[9]{d^{9}}$
ⓑ $\sqrt[8]{v^{8}}$

Solution

ⓐ $d$ ⓑ $| v |$

ⓐ $\sqrt[5]{a^{10}}$
ⓑ $\sqrt[3]{b^{27}}$

ⓐ $\sqrt[4]{16 x^{8}}$
ⓑ $\sqrt[6]{64 y^{12}}$

Solution

ⓐ $2 x^{2}$ ⓑ $2 y^{2}$

ⓐ $\sqrt[7]{128 r^{14}}$
ⓑ $\sqrt[4]{81 s^{24}}$

Use the Product Property to Simplify Expressions with Higher Roots

In the following exercises, simplify.

ⓐ $\sqrt[9]{d^{9}}$
ⓑ A mathematical expression showing the 11th root of m raised to the power of 17, written as '11th root of m^17'.

Solution

ⓐ $d$ ⓑ $m \sqrt[11]{m^{6}}$

ⓐ $\sqrt[3]{54}$
ⓑ $\sqrt[4]{128}$

ⓐ $\sqrt[5]{64 c^{8}}$
ⓑ $\sqrt[4]{48 d^{7}}$

Solution

ⓐ $2 c \sqrt[5]{2 c^{3}}$ ⓑ $2 |d| \sqrt[4]{3 d^{3}}$

ⓐ $\sqrt[3]{343 q^{7}}$
ⓑ $\sqrt[6]{192 r^{9}}$

ⓐ $\sqrt[3]{−500}$
ⓑ $\sqrt[4]{−16}$

Solution

ⓐ $−5 \sqrt[3]{4}$ ⓑ not a real number

Use the Quotient Property to Simplify Expressions with Higher Roots

In the following exercises, simplify.

$\sqrt[5]{\frac{r^{10}}{r^{5}}}$

$\sqrt[3]{\frac{w^{12}}{w^{2}}}$

Solution

$w^{3} \sqrt[3]{w}$

$\sqrt[4]{\frac{64 y^{8}}{4 y^{5}}}$

$\sqrt[3]{\frac{54 z^{9}}{2 z^{3}}}$

Solution

$3 z^{2}$

$\sqrt[6]{\frac{64 a^{7}}{b^{2}}}$

Add and Subtract Higher Roots

In the following exercises, simplify.

$4 \sqrt[5]{20} - 2 \sqrt[5]{20}$

Solution

$2 \sqrt[5]{20}$

$4 \sqrt[3]{18} + 3 \sqrt[3]{18}$

$\sqrt[4]{1250} - \sqrt[4]{162}$

Solution

$2 \sqrt[4]{2}$

$\sqrt[3]{640 c^{5}} - \sqrt[3]{−80 c^{3}}$

$\sqrt[5]{96 t^{8}} + \sqrt[5]{486 t^{4}}$

Solution

$2 t \sqrt[5]{3 t^{3}} + 3 \sqrt[5]{2 t^{4}}$

Rational Exponents

Simplify Expressions with $a^{\frac{1}{n}}$

In the following exercises, write as a radical expression.

$r^{\frac{1}{8}}$

$s^{\frac{1}{10}}$

Solution

A close-up image showing a mathematical expression: the square root of nineteen divided by S. It appears to be a variable or a value S under the radical symbol, with 19 also inside.

In the following exercises, write with a rational exponent.

$\sqrt[5]{u}$

$\sqrt[6]{v}$

Solution

$v^{\frac{1}{6}}$

$\sqrt[3]{9 m}$

$\sqrt[6]{10 z}$

Solution

${(10 z)}^{\frac{1}{6}}$

In the following exercises, simplify.

$16^{\frac{1}{4}}$

$32^{\frac{1}{5}}$

Solution

${(−125)}^{\frac{1}{3}}$

${(125)}^{- \frac{1}{3}}$

Solution

$\frac{1}{5}$

${(−9)}^{\frac{1}{2}}$

${(36)}^{- \frac{1}{2}}$

Solution

$\frac{1}{6}$

Simplify Expressions with $a^{\frac{m}{n}}$

In the following exercises, write with a rational exponent.

$\sqrt[3]{q^{5}}$

$\sqrt[5]{n^{8}}$

Solution

$n^{\frac{8}{5}}$

In the following exercises, simplify.

$27^{- \frac{2}{3}}$

$64^{\frac{5}{2}}$

Solution

32,768

$36^{\frac{3}{2}}$

$81^{- \frac{5}{2}}$

Solution

$\frac{1}{59,049}$

Use the Laws of Exponents to Simplify Expressions with Rational Exponents

In the following exercises, simplify.

$3^{\frac{4}{5}} \cdot 3^{\frac{6}{5}}$

${(x^{6})}^{\frac{4}{3}}$

Solution

$x^{8}$

$\frac{z^{\frac{5}{2}}}{z^{\frac{7}{5}}}$

${(16 s^{\frac{9}{4}})}^{\frac{1}{4}}$

Solution

$2 s^{\frac{9}{16}}$

${(m^{8} n^{12})}^{\frac{1}{4}}$

$\frac{z^{\frac{2}{3}} \cdot z^{- \frac{1}{3}}}{z^{- \frac{5}{3}}}$

Solution

$z^{2}$

Practice Test

In the following exercises, simplify.

$\sqrt{81 + 144}$

$\sqrt{169 m^{4} n^{2}}$

Solution

$13 m^{2} | n |$

$\sqrt{36 n^{13}}$

$3 \sqrt{13} + 5 \sqrt{2} + \sqrt{13}$

Solution

$4 \sqrt{13} + 5 \sqrt{2}$

$5 \sqrt{20} + 2 \sqrt{125}$

$(3 \sqrt{6 y}) (2 \sqrt{50 y^{3}})$

Solution

$60 y^{2} \sqrt{3}$

$(2 - 5 \sqrt{x}) (3 + \sqrt{x})$

${(1 - 2 \sqrt{q})}^{2}$

Solution

$1 - 4 \sqrt{q} + 4 q$

ⓐ $\sqrt[4]{a^{12}}$
ⓑ $\sqrt[3]{b^{21}}$

ⓐ $\sqrt[4]{81 x^{12}}$
ⓑ $\sqrt[6]{64 y^{18}}$

Solution

ⓐ $3 |x^{3}|$ ⓑ $2 |y^{3}|$

$\sqrt{\frac{64 r^{12}}{25 r^{6}}}$

$\sqrt{\frac{14 y^{3}}{7 y}}$

Solution

$|y| \sqrt{2}$

$\frac{\sqrt[5]{256 x^{7}}}{\sqrt[5]{4 x^{2}}}$

$\sqrt[4]{512} - 2 \sqrt[4]{32}$

Solution

$0$

ⓐ $256^{\frac{1}{4}}$
ⓑ $243^{\frac{1}{5}}$

$49^{\frac{3}{2}}$

Solution

343

$25^{- \frac{5}{2}}$

$\frac{w^{\frac{3}{4}}}{w^{\frac{7}{4}}}$

Solution

$\frac{1}{w}$

${(27 s^{\frac{3}{5}})}^{\frac{1}{3}}$

In the following exercises, rationalize the denominator.

$\frac{3}{2 \sqrt{6}}$

Solution

$\frac{\sqrt{6}}{4}$

$\frac{\sqrt{3}}{\sqrt{x} + \sqrt{5}}$

In the following exercises, solve.

$3 \sqrt{2 x - 3} - 20 = 7$

Solution

$\sqrt{3 u - 2} = \sqrt{5 u + 1}$

In the following exercise, solve.

A helicopter flying at an altitude of 600 feet dropped a package to a lifeboat. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the package to reach the hiker. Round your answer to the nearest tenth of a second.

Solution

6.1 seconds

rational exponents

If $\sqrt[n]{a}$ is a real number and $n \geq 2$ , $a^{\frac{1}{n}} = \sqrt[n]{a}$ .
For any positive integers m and n, $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m}$ and $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Introduction

A photo of multi-colored fireworks exploding in a night sky. — Fireworks accompany festive celebrations around the world. (Credit: modification of work by tlc, Flickr)

The trajectories of fireworks are modeled by quadratic equations. The equations can be used to predict the maximum height of a firework and the number of seconds it will take from launch to explosion. In this chapter, we will study the properties of quadratic equations, solve them, graph them, and see how they are applied as models of various situations.

Solve Quadratic Equations Using the Square Root Property

Quadratic equations are equations of the form $a x^{2} + b x + c = 0$ , where $a \neq 0$ . They differ from linear equations by including a term with the variable raised to the second power. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will use three other methods to solve quadratic equations.

Solve Quadratic Equations of the Form ax² = k Using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation $x^{2} = 9$ .

\begin{matrix} x^{2} & = & 9 \\ Put the equation in standard form. & x^{2} - 9 & = & 0 \\ Factor the left side. & (x - 3) (x + 3) & = & 0 \\ Use the Zero Product Property. & (x - 3) = 0, (x + 3) & = & 0 \\ Solve each equation. & x = 3, x & = & −3 \\ Combine the two solutions into \pm form. & x & = & \pm 3 \\ (The solution is read ‘ x is equal to positive or negative 3.’) \end{matrix}

We can easily use factoring to find the solutions of similar equations, like $x^{2} = 16$ and $x^{2} = 25$ , because 16 and 25 are perfect squares. But what happens when we have an equation like $x^{2} = 7$ ? Since 7 is not a perfect square, we cannot solve the equation by factoring.

These equations are all of the form $x^{2} = k$ .
We defined the square root of a number in this way:

If n^{2} = m, then n is a square root of m .

This leads to the Square Root Property.

Notice that the Square Root Property gives two solutions to an equation of the form $x^{2} = k$ : the principal square root of $k$ and its opposite. We could also write the solution as $x = \pm \sqrt{k}$ .

Now, we will solve the equation $x^{2} = 9$ again, this time using the Square Root Property.

\begin{matrix} x^{2} & = & 9 \\ Use the Square Root Property. & x & = & \pm \sqrt{9} \\ Simplify the radical. & x & = & \pm 3 \\ Rewrite to show the two solutions. & x = 3, x & = & −3 \end{matrix}

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation $x^{2} = 7$ .

\begin{matrix} \begin{matrix} Use the Square Root Property. \end{matrix} & \begin{matrix} x^{2} & = & 7 \\ x & = & \pm \sqrt{7} \end{matrix} \\ Rewrite to show two solutions. & x = \sqrt{7}, x = − \sqrt{7} \\ We cannot simplify \sqrt{7}, so we leave the answer as a radical. \end{matrix}

Solve: $x^{2} = 169$ .

Solution

$\begin{matrix} \begin{matrix} Use the Square Root Property. \\ Simplify the radical. \end{matrix} & \begin{matrix} x^{2} & = & 169 \\ x & = & \pm \sqrt{169} \\ x & = & \pm 13 \end{matrix} \\ Rewrite to show two solutions. & x = 13, x = −13 \end{matrix}$

How to Solve a Quadratic Equation of the Form ${a x}^{2} = k$ Using the Square Root Property

Solve: $x^{2} - 48 = 0$ .

Solution

The image shows the given equation, x squared minus 48 equals zero. Step one is to isolate the quadratic term and make its coefficient one so add 48 to both sides of the equation to get x squared by itself.

Step two is to use the Square Root Property to get x equals plus or minus the square root of 48.

Step three, simplify the square root of 48 by writing 48 as the product of 16 and three. The square root of 16 is four. The simplified solution is x equals plus or minus four square root of three.

Step four, check the solutions by substituting each solution into the original equation. When x equals four square root of three, replace x in the original equation with four square root of three to get four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, a true statement. When x equals negative four square root of three, replace x in the original equation with negative four square root of three to get negative four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, also a true statement.

To use the Square Root Property, the coefficient of the variable term must equal 1. In the next example, we must divide both sides of the equation by 5 before using the Square Root Property.

Solve: $5 m^{2} = 80$ .

Solution

The quadratic term is isolated.		$5 m^{2} = 80$
Divide by 5 to make its cofficient 1.		$\frac{5 m^{2}}{5} = \frac{80}{5}$
Simplify.		$m^{2} = 16$
Use the Square Root Property.		$m = \pm \sqrt{16}$
Simplify the radical.		$m = \pm 4$
Rewrite to show two solutions.		$m = 4, m = − 4$
Check the solutions.

The Square Root Property started by stating, ‘If $x^{2} = k$ , and $k \geq 0$ ’. What will happen if $k < 0$ ? This will be the case in the next example.

Solve: $q^{2} + 24 = 0$ .

Solution

Demonstrates solving the quadratic equation q^2 + 24 = 0, illustrating steps that lead to the conclusion of no real solution.
	$q^{2} + 24 = 0$
Isolate the quadratic term.	$q^{2} = −24$
Use the Square Root Property.	$q = \pm \sqrt{−24}$
The $\sqrt{−24}$ is not a real number.	There is no real solution.

Remember, we first isolate the quadratic term and then make the coefficient equal to one.

Solve: $\frac{2}{3} u^{2} + 5 = 17$ .

Solution

		$\frac{2}{3} u^{2} + 5 = 17$
Isolate the quadratic term.		$\frac{2}{3} u^{2} = 12$
Multiply by $\frac{3}{2}$ to make the coefficient 1.		$\frac{3}{2} \cdot \frac{2}{3} u^{2} = \frac{3}{2} \cdot 12$
Simplify.		$u^{2} = 18$
Use the Square Root Property.		$u = \pm \sqrt{18}$
Simplify the radical.		$u = \pm \sqrt{9} \sqrt{2}$
Simplify.		$u = \pm 3 \sqrt{2}$
Rewrite to show two solutions.		$u = 3 \sqrt{2}, u = − 3 \sqrt{2}$
Check.

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

Solve: $2 c^{2} - 4 = 45$ .

Solution

Step-by-step solution of a quadratic equation using the Square Root Property.
Isolate the quadratic term. Divide by 2 to make the coefficient 1. Simplify. Use the Square Root Property. Simplify the radical. Rationalize the denominator. Simplify.	$\begin{matrix} 2 c^{2} - 4 & = & 45 \\ 2 c^{2} & = & 49 \\ \frac{2 c^{2}}{2} & = & \frac{49}{2} \\ c^{2} & = & \frac{49}{2} \\ c & = & \pm \sqrt{\frac{49}{2}} \\ c & = & \pm \frac{\sqrt{49}}{\sqrt{2}} \\ c & = & \pm \frac{\sqrt{49} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} \\ c & = & \pm \frac{7 \sqrt{2}}{2} \end{matrix}$
Rewrite to show two solutions.	$c = \frac{7 \sqrt{2}}{2}, c = - \frac{7 \sqrt{2}}{2}$
Check. We leave the check for you.

Solve Quadratic Equations of the Form a(x − h)² = k Using the Square Root Property

We can use the Square Root Property to solve an equation like ${(x - 3)}^{2} = 16$ , too. We will treat the whole binomial, $(x - 3)$ , as the quadratic term.

Solve: ${(x - 3)}^{2} = 16$ .

Solution

		${(x - 3)}^{2} = 16$
Use the Square Root Property.		$x - 3 = \pm \sqrt{16}$
Simplify.		$x - 3 = \pm 4$
Write as two equations.		$x - 3 = 4, x - 3 = − 4$
Solve.		$x = 7, x = − 1$
Check.

Solve: ${(y - 7)}^{2} = 12$ .

Solution

		${(y - 7)}^{2} = 12$
Use the Square Root Property.		$y - 7 = \pm \sqrt{12}$
Simplify the radical.		$y - 7 = \pm 2 \sqrt{3}$
Solve for y.		$y = 7 \pm 2 \sqrt{3}$
Rewrite to show two solutions.		$y = 7 + 2 \sqrt{3}, y = 7 - 2 \sqrt{3}$
Check.

Remember, when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Solve: ${(x - \frac{1}{2})}^{2} = \frac{5}{4} .$

Solution

This table illustrates the step-by-step solution of a quadratic equation using the Square Root Property.
Use the Square Root Property. Rewrite the radical as a fraction of square roots. Simplify the radical. Solve for x.	$\begin{matrix} {(x - \frac{1}{2})}^{2} & = & \frac{5}{4} \\ x - \frac{1}{2} & = & \pm \sqrt{\frac{5}{4}} \\ x - \frac{1}{2} & = & \pm \frac{\sqrt{5}}{\sqrt{4}} \\ x - \frac{1}{2} & = & \pm \frac{\sqrt{5}}{2} \\ x & = & \frac{1}{2} \pm \frac{\sqrt{5}}{2} \end{matrix}$
Rewrite to show two solutions.	$x = \frac{1}{2} + \frac{\sqrt{5}}{2}, x = \frac{1}{2} - \frac{\sqrt{5}}{2}$
Check. We leave the check for you.

We will start the solution to the next example by isolating the binomial.

Solve: ${(x - 2)}^{2} + 3 = 30$ .

Solution

Step-by-step solution to a quadratic equation using the Square Root Property.
	$\begin{matrix} {(x - 2)}^{2} + 3 & = & 30 \end{matrix}$
Isolate the binomial term.	$\begin{matrix} {(x - 2)}^{2} & = & 27 \end{matrix}$
Use the Square Root Property.	$\begin{matrix} x - 2 & = & \pm \sqrt{27} \end{matrix}$
Simplify the radical.	$\begin{matrix} x - 2 & = & \pm 3 \sqrt{3} \end{matrix}$
Solve for x.	$\begin{matrix} x & = & 2 \pm 3 \sqrt{3} \end{matrix}$
Rewrite to show two solutions.	$x = 2 + 3 \sqrt{3}, x = 2 - 3 \sqrt{3}$
Check. We leave the check for you.

Solve: ${(3 v - 7)}^{2} = −12$ .

Solution

Demonstration of applying the square root property to solve (3v-7)^2 = -12, resulting in no real solution.
Use the Square Root Property.	$\begin{matrix} {(3 v - 7)}^{2} & = & −12 \\ 3 v - 7 & = & \pm \sqrt{−12} \end{matrix}$
The $\sqrt{−12}$ is not a real number.	There is no real solution.

The left sides of the equations in the next two examples do not seem to be of the form $a {(x - h)}^{2}$ . But they are perfect square trinomials, so we will factor to put them in the form we need.

Solve: $p^{2} - 10 p + 25 = 18$ .

Solution

The left side of the equation is a perfect square trinomial. We will factor it first.

Demonstrates solving a perfect square trinomial using the Square Root Property, showing steps like factoring, simplifying radicals, and presenting two solutions for p.
Factor the perfect square trinomial. Use the Square Root Property. Simplify the radical. Solve for p.	$\begin{matrix} p^{2} - 10 p + 25 & = & 18 \\ {(p - 5)}^{2} & = & 18 \\ p - 5 & = & \pm \sqrt{18} \\ p - 5 & = & \pm 3 \sqrt{2} \\ p & = & 5 \pm 3 \sqrt{2} \end{matrix}$
Rewrite to show two solutions.	$p = 5 + 3 \sqrt{2,} p = 5 - 3 \sqrt{2}$
Check. We leave the check for you.

Solve: $4 n^{2} + 4 n + 1 = 16$ .

Solution

Again, we notice the left side of the equation is a perfect square trinomial. We will factor it first.

		$4 n^{2} + 4 n + 1 = 16$
Factor the perfect square trinomial.		${(2 n + 1)}^{2} = 16$
Use the Square Root Property.		$2 n + 1 = \pm \sqrt{16}$
Simplify the radical.		$2 n + 1 = \pm 4$
Solve for n.		$2 n = − 1 \pm 4$
Divide each side by 2.		$\begin{matrix} \frac{2 n}{2} & = & \frac{- 1 \pm 4}{2} \\ n & = & \frac{- 1 \pm 4}{2} \end{matrix}$
Rewrite to show two solutions.		$n = \frac{- 1 + 4}{2}, n = \frac{- 1 - 4}{2}$
Simplify each equation.		$n = \frac{3}{2}, n = - \frac{5}{2}$
Check.

Key Concepts

Square Root Property
If $x^{2} = k$ , and $k \geq 0$ , then $x = \sqrt{k} or x = − \sqrt{k}$ .

Practice Makes Perfect

Solve Quadratic Equations of the form $a x^{2} = k$ Using the Square Root Property

In the following exercises, solve the following quadratic equations.

$a^{2} = 49$

Solution

$a = \pm 7$

$b^{2} = 144$

$r^{2} - 24 = 0$

Solution

$r = \pm 2 \sqrt{6}$

$t^{2} - 75 = 0$

$u^{2} - 300 = 0$

Solution

$u = \pm 10 \sqrt{3}$

$v^{2} - 80 = 0$

$4 m^{2} = 36$

Solution

$m = \pm 3$

$3 n^{2} = 48$

$x^{2} + 20 = 0$

Solution

no real solution

$y^{2} + 64 = 0$

$\frac{2}{5} a^{2} + 3 = 11$

Solution

$a = \pm 2 \sqrt{5}$

$\frac{3}{2} b^{2} - 7 = 41$

$7 p^{2} + 10 = 26$

Solution

$p = \pm \frac{4 \sqrt{7}}{7}$

$2 q^{2} + 5 = 30$

Solve Quadratic Equations of the Form $a {(x - h)}^{2} = k$ Using the Square Root Property

In the following exercises, solve the following quadratic equations.

${(x + 2)}^{2} = 9$

Solution

$x = 1, x = −5$

${(y - 5)}^{2} = 36$

${(u - 6)}^{2} = 64$

Solution

$u = 14, u = −2$

${(v + 10)}^{2} = 121$

${(m - 6)}^{2} = 20$

Solution

$m = 6 \pm 2 \sqrt{5}$

${(n + 5)}^{2} = 32$

${(r - \frac{1}{2})}^{2} = \frac{3}{4}$

Solution

$r = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$

${(t - \frac{5}{6})}^{2} = \frac{11}{25}$

${(a - 7)}^{2} + 5 = 55$

Solution

$a = 7 \pm 5 \sqrt{2}$

${(b - 1)}^{2} - 9 = 39$

${(5 c + 1)}^{2} = −27$

Solution

no real solution

${(8 d - 6)}^{2} = −24$

$m^{2} - 4 m + 4 = 8$

Solution

$m = 2 \pm 2 \sqrt{2}$

$n^{2} + 8 n + 16 = 27$

$25 x^{2} - 30 x + 9 = 36$

Solution

$x = - \frac{3}{5}, x = \frac{9}{5}$

$9 y^{2} + 12 y + 4 = 9$

Mixed Practice

In the following exercises, solve using the Square Root Property.

$2 r^{2} = 32$

Solution

$r = \pm 4$

$4 t^{2} = 16$

${(a - 4)}^{2} = 28$

Solution

$a = 4 \pm 2 \sqrt{7}$

${(b + 7)}^{2} = 8$

$9 w^{2} - 24 w + 16 = 1$

Solution

$w = 1, w = \frac{5}{3}$

$4 z^{2} + 4 z + 1 = 49$

$a^{2} - 18 = 0$

Solution

$a = \pm 3 \sqrt{2}$

$b^{2} - 108 = 0$

${(p - \frac{1}{3})}^{2} = \frac{7}{9}$

Solution

$p = \frac{1}{3} \pm \frac{\sqrt{7}}{3}$

${(q - \frac{3}{5})}^{2} = \frac{3}{4}$

$m^{2} + 12 = 0$

Solution

no real solution

$n^{2} + 48 = 0$

$u^{2} - 14 u + 49 = 72$

Solution

$u = 7 \pm 6 \sqrt{2}$

$v^{2} + 18 v + 81 = 50$

${(m - 4)}^{2} + 3 = 15$

Solution

$m = 4 \pm 2 \sqrt{3}$

${(n - 7)}^{2} - 8 = 64$

${(x + 5)}^{2} = 4$

Solution

$x = −3, x = −7$

${(y - 4)}^{2} = 64$

$6 c^{2} + 4 = 29$

Solution

$c = \pm \frac{5 \sqrt{6}}{6}$

$2 d^{2} - 4 = 77$

${(x - 6)}^{2} + 7 = 3$

Solution

no real solution

${(y - 4)}^{2} + 10 = 9$

Everyday Math

Paola has enough mulch to cover 48 square feet. She wants to use it to make three square vegetable gardens of equal sizes. Solve the equation $3 s^{2} = 48$ to find $s$ , the length of each garden side.

Solution

4 feet

Kathy is drawing up the blueprints for a house she is designing. She wants to have four square windows of equal size in the living room, with a total area of 64 square feet. Solve the equation $4 s^{2} = 64$ to find $s$ , the length of the sides of the windows.

Writing Exercises

Explain why the equation $x^{2} + 12 = 8$ has no solution.

Solution

Answers will vary.

Explain why the equation $y^{2} + 8 = 12$ has two solutions.

Self Check

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

ⓑ 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.

quadratic equation

A quadratic equation is an equation of the form

a x^{2} + b x + c = 0

, where

a \neq 0 .

Square Root Property

The Square Root Property states that, if

x^{2} = k

and

k \geq 0

, then

x = \sqrt{k} or x = − \sqrt{k .}

Solve Quadratic Equations by Completing the Square

So far, we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called ‘completing the square.’

Complete The Square of a Binomial Expression

In the last section, we were able to use the Square Root Property to solve the equation ${(y - 7)}^{2} = 12$ because the left side was a perfect square.

\begin{matrix} {(y - 7)}^{2} & = & 12 \\ y - 7 & = & \pm \sqrt{12} \\ y - 7 & = & \pm 2 \sqrt{3} \\ y & = & 7 \pm 2 \sqrt{3} \end{matrix}

We also solved an equation in which the left side was a perfect square trinomial, but we had to rewrite it the form ${(x - k)}^{2}$ in order to use the square root property.

\begin{matrix} x^{2} - 10 x + 25 & = & 18 \\ {(x - 5)}^{2} & = & 18 \end{matrix}

What happens if the variable is not part of a perfect square? Can we use algebra to make a perfect square?

Let’s study the binomial square pattern we have used many times. We will look at two examples.

\begin{matrix} \begin{matrix} {(x + 9)}^{2} \\ (x + 9) (x + 9) \\ x^{2} + 9 x + 9 x + 81 \\ x^{2} + 18 x + 81 \end{matrix} & \begin{matrix} {(y - 7)}^{2} \\ (y - 7) (y - 7) \\ y^{2} - 7 y - 7 y + 49 \\ y^{2} - 14 y + 49 \end{matrix} \end{matrix}

We can use this pattern to “make” a perfect square.

We will start with the expression $x^{2} + 6 x$ . Since there is a plus sign between the two terms, we will use the ${(a + b)}^{2}$ pattern.

a^{2} + 2 a b + b^{2} = {(a + b)}^{2}

Notice that the first term of $x^{2} + 6 x$ is a square, $x^{2}$ .

We now know $a = x$ .

What number can we add to $x^{2} + 6 x$ to make a perfect square trinomial?

The image shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus six x plus a blank space. The x squared is below the a squared, the six x is below two a b and the blank is below the b squared.

The middle term of the Binomial Squares Pattern, $2 a b$ , is twice the product of the two terms of the binomial. This means twice the product of $x$ and some number is $6 x$ . So, two times some number must be six. The number we need is $\frac{1}{2} \cdot 6 = 3 .$ The second term in the binomial, $b,$ must be 3.

The image is similar to the image above. It shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus two times three times x plus a blank space. The x squared is below the a squared, the two times three times x is below two a b and the blank is below the b squared.

We now know $b = 3$ .

Now, we just square the second term of the binomial to get the last term of the perfect square trinomial, so we square three to get the last term, nine.

The image shows the expression a squared plus two a b plus b squared. Below it is the expression x squared plus six x plus nine.

We can now factor to

The image shows the expression quantity a plus b squared. Below it is the expression quantity x plus three squared.

So, we found that adding nine to $x^{2} + 6 x$ ‘completes the square,’ and we write it as ${(x + 3)}^{2}$ .

Complete the square to make a perfect square trinomial. Then, write the result as a binomial square.

$x^{2} + 14 x$

Solution

The coefficient of x is 14.
$\begin{matrix} Find {(\frac{1}{2} b)}^{2} . \\ {(\frac{1}{2} \cdot 14)}^{2} \\ {(7)}^{2} \\ 49 \end{matrix}$
Add 49 to the binomial to complete the square.	$x^{2} + 14 x + 49$
Rewrite as a binomial square.	${(x + 7)}^{2}$

Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. $m^{2} - 26 m$

Solution

The coefficient of m is −26.
$\begin{matrix} Find {(\frac{1}{2} b)}^{2} . \\ {(\frac{1}{2} \cdot (− 26))}^{2} \\ {(− 13)}^{2} \\ 169 \end{matrix}$
Add 169 to the binomial to complete the square.	$m^{2} - 26 m + 169$
Rewrite as a binomial square.	${(m - 13)}^{2}$

Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.

$u^{2} - 9 u$

Solution

The coefficient of u is −9.
$\begin{matrix} Find {(\frac{1}{2} b)}^{2} . \\ {(\frac{1}{2} \cdot (− 9))}^{2} \\ {(- \frac{9}{2})}^{2} \\ \frac{81}{4} \end{matrix}$
Add $\frac{81}{4}$ to the binomial to complete the square.	$u^{2} - 9 u + \frac{81}{4}$
Rewrite as a binomial square.	${(u - \frac{9}{2})}^{2}$

Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.

$p^{2} + \frac{1}{2} p$

Solution

The coefficient of p is $\frac{1}{2}$ .
$\begin{matrix} Find {(\frac{1}{2} b)}^{2} . \\ {(\frac{1}{2} \cdot \frac{1}{2})}^{2} \\ {(\frac{1}{4})}^{2} \\ \frac{1}{16} \end{matrix}$
Add $\frac{1}{16}$ to the binomial to complete the square.	$p^{2} + \frac{1}{2} p + \frac{1}{16}$
Rewrite as a binomial square.	${(p + \frac{1}{4})}^{2}$

Solve Quadratic Equations of the Form x² + bx + c = 0 by completing the square

In solving equations, we must always do the same thing to both sides of the equation. This is true, of course, when we solve a quadratic equation by completing the square, too. When we add a term to one side of the equation to make a perfect square trinomial, we must also add the same term to the other side of the equation.

For example, if we start with the equation $x^{2} + 6 x = 40$ and we want to complete the square on the left, we will add nine to both sides of the equation.

The image shows the equation x squared plus six x equals 40. Below that the equation is rewritten as x squared plus six x plus blank space equals 40 plus blank space. Below that the equation is rewritten again as x squared plus six x plus nine equals 40 plus nine.

Then, we factor on the left and simplify on the right.

{(x + 3)}^{2} = 49

Now the equation is in the form to solve using the Square Root Property. Completing the square is a way to transform an equation into the form we need to be able to use the Square Root Property.

How To Solve a Quadratic Equation of the Form $x^{2} + b x + c = 0$ by Completing the Square

Solve $x^{2} + 8 x = 48$ by completing the square.

Solution

The image shows the steps to solve the equation x squared plus eight x equals 48. Step one is to isolate the variable terms on one side and the constant terms on the other. The equation already has all the variables on the left.

Step two is to find the quantity half of b squared, the number to complete the square and add it to both sides of the equation. The coefficient of x is eight so b is eight. Take half of eight, which is four and square it to get 16. Add 16 to both sides of the equation to get x squared plus eight x plus 16 equals 48 plus 16.

Step three is to factor the perfect square trinomial as a binomial square. The left side is the perfect square trinomial x squared plus eight x plus 16 which factors to the quantity x plus four squared. Adding on the right side 48 plus 16 is 64. The equation is now the quantity x plus four squared equals 64.

Step four is to use the square root property to make the equation x plus four equals plus or minus the square root of 64.

Step five is to simplify the radical and then solve the two resulting equations. The square root of 64 is eight. The equation can be written as two equations: x plus four equals eight and x plus four equals negative eight. Solving each equation gives x equals four or negative 12.

Solve $y^{2} - 6 y = 16$ by completing the square.

Solution

The variable terms are on the left side.
Take half of $- 6$ and square it. $(\frac{1}{2} (− 6))^{2} = 9$
Add 9 to both sides.
Factor the perfect square trinomial as a binomial square.
Use the Square Root Property.
Simplify the radical.
Solve for y.
Rewrite to show two solutions.
Solve the equations.
Check.

Solve $x^{2} + 4 x = −21$ by completing the square.

Solution

The variable terms are on the left side.
Take half of $4$ and square it. $(\frac{1}{2} (4))^{2} = 4$
Add 4 to both sides.
Factor the perfect square trinomial as a binomial square.
Use the Square Root Property.
We cannot take the square root of a negative number.		There is no real solution.

In the previous example, there was no real solution because ${(x + k)}^{2}$ was equal to a negative number.

Solve $p^{2} - 18 p = −6$ by completing the square.

Solution

The variable terms are on the left side.
Take half of $− 18$ and square it. $(\frac{1}{2} (− 18))^{2} = 81$
Add 81 to both sides.
Factor the perfect square trinomial as a binomial square.
Use the Square Root Property.
Simplify the radical.
Solve for p.
Rewrite to show two solutions.
Check.

Another way to check this would be to use a calculator. Evaluate $p^{2} - 18 p$ for both of the solutions. The answer should be $−6$ .

We will start the next example by isolating the variable terms on the left side of the equation.

Solve $x^{2} + 10 x + 4 = 15$ by completing the square.

Solution

The variable terms are on the left side.
Subtract $4$ to get the constant terms on the right side.
Take half of 10 and square it. $(\frac{1}{2} (10))^{2} = 25$
Add 25 to both sides.
Factor the perfect square trinomial as a binomial square.
Use the Square Root Property.
Simplify the radical.
Solve for x.
Rewrite to show two equations.
Solve the equations.
Check.

To solve the next equation, we must first collect all the variable terms to the left side of the equation. Then, we proceed as we did in the previous examples.

Solve $n^{2} = 3 n + 11$ by completing the square.

Solution


Subtract 3n to get the variable terms on the left side.
Take half of $− 3$ and square it. $(\frac{1}{2} (− 3))^{2} = \frac{9}{4}$
Add $\frac{9}{4}$ to both sides.		$A mathematical equation showing n squared minus 3n plus the fraction 9/4 equals 11 plus the fraction 9/4. The fraction 9/4 is highlighted in red on both sides of the equation.$
Factor the perfect square trinomial as a binomial square.		$A mathematical equation shows (n - 3/2)^2 = 44/4 + 9/4. The equation involves a variable 'n', fractions, and an exponent.$
Add the fractions on the right side.
Use the Square Root Property.
Simplify the radical.
Solve for n.
Rewrite to show two equations.
Check. We leave the check for you!

Notice that the left side of the next equation is in factored form. But the right side is not zero, so we cannot use the Zero Product Property. Instead, we multiply the factors and then put the equation into the standard form to solve by completing the square.

Solve $(x - 3) (x + 5) = 9$ by completing the square.

Solution


We multiply binomials on the left.
Add 15 to get the variable terms on the left side.
Take half of 2 and square it. $(\frac{1}{2} (2))^{2} = 1$
Add 1 to both sides.
Factor the perfect square trinomial as a binomial square.
Use the Square Root Property.
Solve for x.
Rewite to show two solutions.
Simplify.
Check. We leave the check for you!

Solve Quadratic Equations of the form ax² + bx + c = 0 by completing the square

The process of completing the square works best when the leading coefficient is one, so the left side of the equation is of the form $x^{2} + b x + c$ . If the $x^{2}$ term has a coefficient, we take some preliminary steps to make the coefficient equal to one.

Sometimes the coefficient can be factored from all three terms of the trinomial. This will be our strategy in the next example.

Solve $3 x^{2} - 12 x - 15 = 0$ by completing the square.

Solution

To complete the square, we need the coefficient of $x^{2}$ to be one. If we factor out the coefficient of $x^{2}$ as a common factor, we can continue with solving the equation by completing the square.


Factor out the greatest common factor.
Divide both sides by 3 to isolate the trinomial.
Simplify.
Subtract 5 to get the constant terms on the right.
Take half of 4 and square it. $(\frac{1}{2} (4))^{2} = 4$
Add 4 to both sides.
Factor the perfect square trinomial as a binomial square.
Use the Square Root Property.
Solve for x.
Rewrite to show 2 solutions.
Simplify.
Check.

To complete the square, the leading coefficient must be one. When the leading coefficient is not a factor of all the terms, we will divide both sides of the equation by the leading coefficient. This will give us a fraction for the second coefficient. We have already seen how to complete the square with fractions in this section.

Solve $2 x^{2} - 3 x = 20$ by completing the square.

Solution

Again, our first step will be to make the coefficient of $x^{2}$ be one. By dividing both sides of the equation by the coefficient of $x^{2}$ , we can then continue with solving the equation by completing the square.


Divide both sides by 2 to get the coefficient of $x^{2}$ to be 1.		$A mathematical equation is displayed: the fraction 2x squared minus 3x all over 2, is equal to the fraction 20 over 2.$
Simplify.
Take half of $- \frac{3}{2}$ and square it. $(\frac{1}{2} (- \frac{3}{2}))^{2} = \frac{9}{16}$
Add $\frac{9}{16}$ to both sides.
Factor the perfect square trinomial as a binomial square.		$A mathematical equation is displayed against a white background: (x - 3/4)^2 = 160/16 + 9/16. It shows a binomial squared on the left and a sum of fractions on the right.$
Add the fractions on the right side.
Use the Square Root Property.
Simplify the radical.
Solve for x.
Rewrite to show 2 solutions.
Simplify.
Check. We leave the check for you.

Solve $3 x^{2} + 2 x = 4$ by completing the square.

Solution


Divide both sides by 3 to make the coefficient of $x^{2}$ equal 1.		$A mathematical equation displays (3x^2 + 2x) / 3 = 4/3, representing a quadratic expression set equal to a fraction.$
Simplify.
Take half of $\frac{2}{3}$ and square it. $(\frac{1}{2} \cdot \frac{2}{3})^{2} = \frac{1}{9}$
Add $\frac{1}{9}$ to both sides.
Factor the perfect square trinomial as a binomial square.
Use the Square Root Property.
Simplify the radical.
Solve for x.
Rewrite to show 2 solutions.		$The image shows two solutions for x: x = -1/3 + sqrt(13)/3 and x = -1/3 - sqrt(13)/3, expressed as fractions with a common denominator.$
Check. We leave the check for you.

Key Concepts

Binomial Squares Pattern If $a, b$ are real numbers,
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
Complete a Square
To complete the square of x2+bx:
1. Identify $b$ , the coefficient of $x$ .
2. Find ${(\frac{1}{2} b)}^{2}$ , the number to complete the square.
3. Add the ${(\frac{1}{2} b)}^{2}$ to $x^{2} + b x$ .

Practice Makes Perfect

Complete the Square of a Binomial Expression

In the following exercises, complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.

$a^{2} + 10 a$

Solution

${(a + 5)}^{2}$

$b^{2} + 12 b$

$m^{2} + 18 m$

Solution

${(m + 9)}^{2}$

$n^{2} + 16 n$

$m^{2} - 24 m$

Solution

${(m - 12)}^{2}$

$n^{2} - 16 n$

$p^{2} - 22 p$

Solution

${(p - 11)}^{2}$

$q^{2} - 6 q$

$x^{2} - 9 x$

Solution

${(x - \frac{9}{2})}^{2}$

$y^{2} + 11 y$

$p^{2} - \frac{1}{3} p$

Solution

${(p - \frac{1}{6})}^{2}$

$q^{2} + \frac{3}{4} q$

Solve Quadratic Equations of the Form $x^{2} + b x + c = 0$ by Completing the Square

In the following exercises, solve by completing the square.

$v^{2} + 6 v = 40$

Solution

$v = −10, v = 4$

$w^{2} + 8 w = 65$

$u^{2} + 2 u = 3$

Solution

$u = −3, u = 1$

$z^{2} + 12 z = −11$

$c^{2} - 12 c = 13$

Solution

$c = −1, c = 13$

$d^{2} - 8 d = 9$

$x^{2} - 20 x = 21$

Solution

$x = −1, x = 21$

$y^{2} - 2 y = 8$

$m^{2} + 4 m = −44$

Solution

no real solution

$n^{2} - 2 n = −3$

$r^{2} + 6 r = −11$

Solution

no real solution

$t^{2} - 14 t = −50$

$a^{2} - 10 a = −5$

Solution

$a = 5 \pm 2 \sqrt{5}$

$b^{2} + 6 b = 41$

$u^{2} - 14 u + 12 = −1$

Solution

$u = 1, u = 13$

$z^{2} + 2 z - 5 = 2$

$v^{2} = 9 v + 2$

Solution

$v = \frac{9}{2} \pm \frac{\sqrt{89}}{2}$

$w^{2} = 5 w - 1$

$(x + 6) (x - 2) = 9$

Solution

$x = −7, x = 3$

$(y + 9) (y + 7) = 79$

Solve Quadratic Equations of the Form $a x^{2} + b x + c = 0$ by Completing the Square

In the following exercises, solve by completing the square.

$3 m^{2} + 30 m - 27 = 6$

Solution

$m = −11, m = 1$

$2 n^{2} + 4 n - 26 = 0$

$2 c^{2} + c = 6$

Solution

$c = −2, c = \frac{3}{2}$

$3 d^{2} - 4 d = 15$

$2 p^{2} + 7 p = 14$

Solution

$p = - \frac{7}{4} \pm \frac{\sqrt{161}}{4}$

$3 q^{2} - 5 q = 9$

Everyday Math

Rafi is designing a rectangular playground to have an area of 320 square feet. He wants one side of the playground to be four feet longer than the other side. Solve the equation $p^{2} + 4 p = 320$ for $p$ , the length of one side of the playground. What is the length of the other side?

Solution

16 feet, 20 feet

Yvette wants to put a square swimming pool in the corner of her backyard. She will have a 3 foot deck on the south side of the pool and a 9 foot deck on the west side of the pool. She has a total area of 1080 square feet for the pool and two decks. Solve the equation $(s + 3) (s + 9) = 1080$ for $s$ , the length of a side of the pool.

Writing Exercises

Solve the equation $x^{2} + 10 x = −25$ ⓐ by using the Square Root Property and ⓑ by completing the square. ⓒ Which method do you prefer? Why?

Solution

Solve the equation $y^{2} + 8 y = 48$ by completing the square and explain all your steps.

Self Check

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

This table has four rows and four columns. The first row is a header row and it labels each column. The first column is labeled “I can ...”, the second “Confidently”, the third “With some help” and the last “No–I don’t get it”. In the “I can...” column the next row reads “complete the square of a binomial expression.” The next row reads “solve quadratic equations of the form x squared plus b x plus c equals zero by completing the square.” and the last row reads “solve quadratic equations of the form a x squared plus b x plus c equals zero by completing the square.” The remaining columns are blank.

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

completing the square

Completing the square is a method used to solve quadratic equations.

Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes.’ In this section, we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’ so that we would do the algebraic steps only once and then use the new formula to find the value of the specific variable. Now, we will go through the steps of completing the square in general to solve a quadratic equation for x. It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form $a x^{2} + b x + c = 0$ as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

Step-by-step derivation of the quadratic formula using the method of completing the square, detailing each algebraic transformation.
We start with the standard form of a quadratic equation and solve it for x by completing the square.	$a x^{2} + b x + c = 0 a \neq 0$
Isolate the variable terms on one side.	$\begin{matrix} a x^{2} + b x & = & − c \end{matrix}$
Make leading coefficient 1, by dividing by a.	$\begin{matrix} \frac{a x^{2}}{a} + \frac{b}{a} x & = & - \frac{c}{a} \end{matrix}$
Simplify.	$\begin{matrix} x^{2} + \frac{b}{a} x & = & - \frac{c}{a} \end{matrix}$
To complete the square, find ${(\frac{1}{2} \cdot \frac{b}{a})}^{2}$ and add it to both sides of the equation. ${(\frac{1}{2} \frac{b}{a})}^{2} = \frac{b^{2}}{4 a^{2}}$	$\begin{matrix} x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}} & = & - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} \end{matrix}$
The left side is a perfect square, factor it.	$\begin{matrix} {(x + \frac{b}{2 a})}^{2} & = & - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} \end{matrix}$
Find the common denominator of the right side and write equivalent fractions with the common denominator.	$\begin{matrix} {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2}}{4 a^{2}} - \frac{c \cdot 4 a}{a \cdot 4 a} \end{matrix}$
Simplify.	$\begin{matrix} {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2}}{4 a^{2}} - \frac{4 a c}{4 a^{2}} \end{matrix}$
Combine to one fraction.	$\begin{matrix} {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2} - 4 a c}{4 a^{2}} \end{matrix}$
Use the square root property.	$\begin{matrix} x + \frac{b}{2 a} & = & \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} \end{matrix}$
Simplify.	$\begin{matrix} x + \frac{b}{2 a} & = & \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{matrix}$
Add $- \frac{b}{2 a}$ to both sides of the equation.	$\begin{matrix} x & = & - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{matrix}$
Combine the terms on the right side.	$\begin{matrix} x & = & \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{matrix}$

This last equation is the Quadratic Formula.

To use the Quadratic Formula, we substitute the values of $a, b, and c$ into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

How to Solve a Quadratic Equation Using the Quadratic Formula

Solve $2 x^{2} + 9 x - 5 = 0$ by using the Quadratic Formula.

Solution

The image shows the steps to solve the quadratic equation two x squared plus nine x minus five equals zero. Step one is to write the quadratic equation in standard form and identify the a, b, and c values. This equation is already in standard for. The value of a is two, the value of b is nine and the value of c is negative five.

Step two is to write the quadratic formula. Then substitute in the values of a, b, and c. Substitute two for a, nine for b and negative five for c in the formula x equals the quantity negative b plus or minus the square root of b squared minus four times a times c divided by two times a. The formula becomes x equals negative nine plus or minus the square root of negative nine squared minus four time two times negative five all divided by two times two.

Step three is to simplify the formula. Squaring negative nine and performing the multiplication to get negative nine plus or minus the square root of 81 minus negative 40 all divided by four. This simplifies further to negative nine plus or minus the square root of 121 all divided by four which reduces to negative nine plus or minus 11 all divided by four. Negative nine plus 11 divided by four is two fourths which reduces to one half. Negative nine minus 11 divided by four is negative 20 fourths which reduces to negative five.

Step four is to check the solutions by putting each answer in the original equation to check. Replace x in two x squared plus nine x minus five equals zero with one half to get two times one half squared plus nine times one half minus five. Simplify to get one half plus nine halves minus five which is zero. Replace x in two x squared plus nine x minus five equals zero with negative five to get two times negative five squared plus nine times negative five minus five. Simplify to get 50 minus 45 minus five which is zero.

If you say the formula as you write it in each problem, you’ll have it memorized in no time. And remember, the Quadratic Formula is an equation. Be sure you start with ‘ $x =$ ’.

Solve $x^{2} - 6 x + 5 = 0$ by using the Quadratic Formula.

Solution


This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.		$A mathematical equation is displayed on a white background: x equals a fraction where the numerator is 6 plus or minus the square root of 36 minus 20, and the denominator is 2.$ $A mathematical equation shows X equals the fraction of (6 plus or minus 4) over 2, indicating two possible solutions for X.$
Rewrite to show two solutions.
Simplify.
Check.

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

Solve $4 y^{2} - 5 y - 3 = 0$ by using the Quadratic Formula.

Solution

We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named ‘x’.


This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.		$A mathematical equation for y is shown on a white background, which states y equals a fraction with 5 plus or minus the square root of 73 in the numerator, and 8 in the denominator.$
Rewrite to show two solutions.
Check. We leave the check to you.

Solve $2 x^{2} + 10 x + 11 = 0$ by using the Quadratic Formula.

Solution


This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Simplify the radical.
Factor out the common factor in the numerator.
Remove the common factors.		$The equation shows x equals a fraction where the numerator is -5 plus or minus the square root of 3, and the denominator is 2, representing solutions from a quadratic formula.$
Rewrite to show two solutions.
Check. We leave the check to you.

We cannot take the square root of a negative number. So, when we substitute $a$ , $b$ , and $c$ into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example.

Solve $3 p^{2} + 2 p + 9 = 0$ by using the Quadratic Formula.

Solution

This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Simplify the radical.		$A mathematical equation shows p equals a fraction. The numerator is -2 plus or minus the square root of -104, and the denominator is 6.$
We cannot take the square root of a negative number.		There is no real solution.

The quadratic equations we have solved so far in this section were all written in standard form, $a x^{2} + b x + c = 0$ . Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Solve $x (x + 6) + 4 = 0$ by using the Quadratic Formula.

Solution


Distribute to get the equation in standard form.
This equation is now in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.		$A mathematical equation shows X equals a fraction with a numerator of -6 plus or minus the square root of 36 minus 16, all divided by 2. This is a step in solving a quadratic equation.$
Simplify inside the radical.
Simplify the radical.		$A mathematical equation for x is displayed as a fraction: x equals negative six plus or minus two times the square root of five, all divided by two.$
Factor out the common factor in the numerator.
Remove the common factors.
Rewrite to show two solutions.
Check. We leave the check to you.

When we solved linear equations, if an equation had too many fractions we ‘cleared the fractions’ by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions—to solve. We can use the same strategy with quadratic equations.

Solve $\frac{1}{2} u^{2} + \frac{2}{3} u = \frac{1}{3}$ by using the Quadratic Formula.

Solution


Multiply both sides by the LCD, 6, to clear the fractions.
Multiply.
Subtract 2 to get the equation in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.		$A mathematical equation for 'u' is shown, calculated as the fraction of '-4 plus or minus the square root of (16 + 24)' all divided by '6'.$
Simplify the radical.
Factor out the common factor in the numerator.		$A mathematical equation shows u equals a fraction: numerator is 2 multiplied by the quantity -2 plus or minus the square root of 10, and the denominator is 6.$
Remove the common factors.		$The image displays a mathematical equation, expressing the variable 'u' as a fraction: the numerator is -2 plus or minus the square root of 10, and the denominator is 3.$
Rewrite to show two solutions.
Check. We leave the check to you.

Think about the equation ${(x - 3)}^{2} = 0$ . We know from the Zero Products Principle that this equation has only one solution: $x = 3$ .

We will see in the next example how using the Quadratic Formula to solve an equation with a perfect square also gives just one solution.

Solve $4 x^{2} - 20 x = −25$ by using the Quadratic Formula.

Solution


Add 25 to get the equation in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.		$A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 400 minus 400, and the denominator is 8.$ $A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 0, and the denominator is 8.$
Simplify the radical.
Simplify the fraction.
Check. We leave the check to you.

Did you recognize that $4 x^{2} - 20 x + 25$ is a perfect square?

Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation?

Yes, the quantity inside the radical of the Quadratic Formula makes it easy for us to determine the number of solutions. This quantity is called the discriminant.

Let’s look at the discriminant of the equations in Example 1, Example 5, and Example 8, and the number of solutions to those quadratic equations.

	Quadratic Equation (in standard form)	Discriminant $b^{2} - 4 a c$	Sign of the Discriminant	Number of real solutions
Example 1	$2 x^{2} + 9 x - 5 = 0$	$9^{2} - 4 \cdot 2 (−5) = 121$	+	2
Example 8	$4 x^{2} - 20 x + 25 = 0$	${(−20)}^{2} - 4 \cdot 4 \cdot 25 = 0$	0	1
Example 5	$3 p^{2} + 2 p + 9 = 0$	$2^{2} - 4 \cdot 3 \cdot 9 = −104$	−	0

When the discriminant is positive $(x = \frac{− b \pm \sqrt{+}}{2 a})$ the quadratic equation has two solutions.

When the discriminant is zero $(x = \frac{− b \pm \sqrt{0}}{2 a})$ the quadratic equation has one solution.

When the discriminant is negative $(x = \frac{− b \pm \sqrt{-}}{2 a})$ the quadratic equation has no real solutions.

Determine the number of solutions to each quadratic equation:

Solution

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

ⓐ

Step-by-step calculation of the discriminant for a quadratic equation (2v^2 - 3v + 6 = 0) to determine that it has no real solutions.
	$\begin{matrix} 2 v^{2} - 3 v + 6 & = & 0 \end{matrix}$
The equation is in standard form, identify a, b, c.	$\begin{matrix} a = 2, b = −3, c & = & 6 \end{matrix}$
Write the discriminant.	$b^{2} - 4 a c$
Substitute in the values of a, b, c.	${(3)}^{2} - 4 \cdot 2 \cdot 6$
Simplify.	$\begin{matrix} 9 - 48 \\ −39 \end{matrix}$
Because the discriminant is negative, there are no real solutions to the equation.

ⓑ

Step-by-step calculation of the discriminant for the quadratic equation 3x^2 + 7x - 9 = 0 to determine the number of solutions.
	$3 x^{2} + 7 x - 9 = 0$
The equation is in standard form, identify a, b, c.	$a = 3, b = 7, c = −9$
Write the discriminant.	$b^{2} - 4 a c$
Substitute in the values of a, b, c.	${(7)}^{2} - 4 \cdot 3 \cdot (−9)$
Simplify.	$\begin{matrix} 49 + 108 \\ 157 \end{matrix}$
Because the discriminant is positive, there are two solutions to the equation.

ⓒ

Steps to calculate the discriminant of the quadratic equation 5n^2 + n + 4 = 0, showing it has no real solutions.
	$\begin{matrix} 5 n^{2} + n + 4 & = & 0 \end{matrix}$
The equation is in standard form, identify a, b, c.	$\begin{matrix} a = 5, b = 1, c & = & 4 \end{matrix}$
Write the discriminant.	$b^{2} - 4 a c$
Substitute in the values of a, b, c.	${(1)}^{2} - 4 \cdot 5 \cdot 4$
Simplify.	$\begin{matrix} 1 - 80 \\ −79 \end{matrix}$
Because the discriminant is negative, there are no real solutions to the equation.

ⓓ

Demonstrates the step-by-step process of calculating the discriminant for a quadratic equation and interpreting its result.
	$\begin{matrix} 9 y^{2} - 6 y + 1 & = & 0 \end{matrix}$
The equation is in standard form, identify a, b, c.	$\begin{matrix} a = 9, b = −6, c & = & 1 \end{matrix}$
Write the discriminant.	$b^{2} - 4 a c$
Substitute in the values of a, b, c.	${(−6)}^{2} - 4 \cdot 9 \cdot 1$
Simplify.	$\begin{matrix} 36 - 36 \\ 0 \end{matrix}$
Because the discriminant is 0, there is one solution to the equation.

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We have used four methods to solve quadratic equations:

Factoring
Square Root Property
Completing the Square
Quadratic Formula

You can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method to use.

What about the method of completing the square? Most people find that method cumbersome and prefer not to use it. We needed to include it in this chapter because we completed the square in general to derive the Quadratic Formula. You will also use the process of completing the square in other areas of algebra.

Identify the most appropriate method to use to solve each quadratic equation:

Solution

ⓐ $5 z^{2} = 17$

Since the equation is in the $a x^{2} = k$ , the most appropriate method is to use the Square Root Property.

ⓑ $4 x^{2} - 12 x + 9 = 0$

We recognize that the left side of the equation is a perfect square trinomial, and so Factoring will be the most appropriate method.

Put the equation in standard form. $8 u^{2} + 6 u - 11 = 0$

While our first thought may be to try Factoring, thinking about all the possibilities for trial and error leads us to choose the Quadratic Formula as the most appropriate method

Key Concepts

Quadratic Formula The solutions to a quadratic equation of the form $a x^{2} + b x + c = 0,$ $a \neq 0$ are given by the formula:

$x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Solve a Quadratic Equation Using the Quadratic Formula
To solve a quadratic equation using the Quadratic Formula.
1. Write the quadratic formula in standard form. Identify the $a, b, c$ values.
2. Write the quadratic formula. Then substitute in the values of $a, b, c .$
3. Simplify.
4. Check the solutions.
Using the Discriminant, $b^{2} - 4 a c$ , to Determine the Number of Solutions of a Quadratic Equation
For a quadratic equation of the form ax2+bx+c=0, a≠0,
- if $b^{2} - 4 a c > 0$ , the equation has 2 solutions.
- if $b^{2} - 4 a c = 0$ , the equation has 1 solution.
- if $b^{2} - 4 a c < 0$ , the equation has no real solutions.
To identify the most appropriate method to solve a quadratic equation:
1. Try Factoring first. If the quadratic factors easily this method is very quick.
2. Try the Square Root Property next. If the equation fits the form $a x^{2} = k$ or $a {(x - h)}^{2} = k$ , it can easily be solved by using the Square Root Property.
3. Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.

Practice Makes Perfect

Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

$4 m^{2} + m - 3 = 0$

Solution

$m = −1, m = \frac{3}{4}$

$4 n^{2} - 9 n + 5 = 0$

$2 p^{2} - 7 p + 3 = 0$

Solution

$p = \frac{1}{2}, p = 3$

$3 q^{2} + 8 q - 3 = 0$

$p^{2} + 7 p + 12 = 0$

Solution

$p = −4, p = −3$

$q^{2} + 3 q - 18 = 0$

$r^{2} - 8 r - 33 = 0$

Solution

$r = −3, r = 11$

$t^{2} + 13 t + 40 = 0$

$3 u^{2} + 7 u - 2 = 0$

Solution

$u = \frac{−7 \pm \sqrt{73}}{6}$

$6 z^{2} - 9 z + 1 = 0$

$2 a^{2} - 6 a + 3 = 0$

Solution

$a = \frac{3 \pm \sqrt{3}}{2}$

$5 b^{2} + 2 b - 4 = 0$

$2 x^{2} + 3 x + 9 = 0$

Solution

no real solution

$6 y^{2} - 5 y + 2 = 0$

$v (v + 5) - 10 = 0$

Solution

$v = \frac{−5 \pm \sqrt{65}}{2}$

$3 w (w - 2) - 8 = 0$

$\frac{1}{3} m^{2} + \frac{1}{12} m = \frac{1}{4}$

Solution

$m = −1, m = \frac{3}{4}$

$\frac{1}{3} n^{2} + n = - \frac{1}{2}$

$16 c^{2} + 24 c + 9 = 0$

Solution

$c = - \frac{3}{4}$

$25 d^{2} - 60 d + 36 = 0$

$5 m^{2} + 2 m - 7 = 0$

Solution

$m = - \frac{7}{5}, m = 1$

$8 n^{2} - 3 n + 3 = 0$

$p^{2} - 6 p - 27 = 0$

Solution

$p = −3, p = 9$

$25 q^{2} + 30 q + 9 = 0$

$4 r^{2} + 3 r - 5 = 0$

Solution

$r = \frac{−3 \pm \sqrt{89}}{8}$

$3 t (t - 2) = 2$

$2 a^{2} + 12 a + 5 = 0$

Solution

$a = \frac{−6 \pm \sqrt{26}}{2}$

$4 d^{2} - 7 d + 2 = 0$

$\frac{3}{4} b^{2} + \frac{1}{2} b = \frac{3}{8}$

Solution

$b = \frac{−2 \pm \sqrt{22}}{6}$

$\frac{1}{9} c^{2} + \frac{2}{3} c = 3$

$2 x^{2} + 12 x - 3 = 0$

Solution

$x = \frac{−6 \pm \sqrt{42}}{4}$

$16 y^{2} + 8 y + 1 = 0$

Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation

In the following exercises, determine the number of solutions to each quadratic equation.

ⓐ $4 x^{2} - 5 x + 16 = 0$
ⓑ $36 y^{2} + 36 y + 9 = 0$
ⓓ $18 n^{2} - 7 n + 3 = 0$

Solution

ⓐ $9 v^{2} - 15 v + 25 = 0$
ⓑ $100 w^{2} + 60 w + 9 = 0$
ⓓ $15 d^{2} - 4 d + 8 = 0$

ⓐ $r^{2} + 12 r + 36 = 0$
ⓑ $8 t^{2} - 11 t + 5 = 0$
ⓓ $3 v^{2} - 5 v - 1 = 0$

Solution

ⓐ $25 p^{2} + 10 p + 1 = 0$
ⓑ $7 q^{2} - 3 q - 6 = 0$
ⓓ $25 z^{2} - 60 z + 36 = 0$

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

Solution

Everyday Math

A flare is fired straight up from a ship at sea. Solve the equation $16 (t^{2} - 13 t + 40) = 0$ for $t$ , the number of seconds it will take for the flare to be at an altitude of 640 feet.

Solution

5 seconds, 8 seconds

An architect is designing a hotel lobby. She wants to have a triangular window looking out to an atrium, with the width of the window 6 feet more than the height. Due to energy restrictions, the area of the window must be 140 square feet. Solve the equation $\frac{1}{2} h^{2} + 3 h = 140$ for $h$ , the height of the window.

Writing Exercises

Solve the equation $x^{2} + 10 x = 200$
ⓐ by completing the square
ⓑ using the Quadratic Formula
ⓒ Which method do you prefer? Why?

Solution

Solve the equation $12 y^{2} + 23 y = 24$
ⓐ by completing the square
ⓑ using the Quadratic Formula
ⓒ Which method do you prefer? Why?

Self Check

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

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

discriminant

In the Quadratic Formula,

x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a}

the quantity

b^{2} - 4 a c

is called the discriminant.

Solve Applications Modeled by Quadratic Equations

Solve Applications of the Quadratic Formula

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications. To get us started, we will copy our usual Problem Solving Strategy here so we can follow the steps.

We have solved number applications that involved consecutive even integers and consecutive odd integers by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one n, then the next one is $n + 2$ . The next one would be $n + 2 + 2$ or $n + 4$ . This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

\begin{matrix} Consecutive even integers & Consecutive odd integers \\ 64, 66, 68 & 77, 79, 81 \\ \begin{matrix} n & 1^{st} even integer \\ n + 2 & 2^{nd} consecutive even integer \\ n + 4 & 3^{rd} consecutive even integer \end{matrix} & \begin{matrix} n & 1^{st} odd integer \\ n + 2 & 2^{nd} consecutive odd integer \\ n + 4 & 3^{rd} consecutive odd integer \end{matrix} \end{matrix}

Some applications of consecutive odd integers or consecutive even integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

The product of two consecutive odd integers is 195. Find the integers.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		We are looking for two consecutive odd integers.
Step 3. Name what we are looking for.		Let $n =$ the first odd integer. $n + 2 =$ the next odd integer
Step 4. Translate into an equation. State the problem in one sentence.		"The product of two consecutive odd integers is 195." The product of the first odd integer and the second odd integer is 195.
Translate into an equation
Step 5. Solve the equation. Distribute.
Subtract 195 to get the equation in standard form.
Identify the a, b, c values.
Write the quadratic equation.
Then substitute in the values of a, b, c..
Simplify.		$A mathematical equation is displayed, showing n equals a fraction where the numerator is -2 plus or minus the square root of 4 plus 780, all divided by 2.$
Simplify the radical.		$A mathematical equation shows 'n' is equal to a fraction where the numerator is '-2 plus or minus 28' and the denominator is '2'.$
Rewrite to show two solutions.
Solve each equation.		$Two mathematical equations are displayed horizontally on a white background: 'n = 26/2' and 'n = -30/2'. The equations show the variable 'n' being defined by fractions.$
There are two values of n that are solutions. This will give us two pairs of consecutive odd integers for our solution.		First odd integer $n = 13$ next odd integer $n + 2$ $13 + 2$ $15$
		First odd integer $n = −15$ next odd integer $n + 2$ $−15 + 2$ $−13$
Step 6. Check the answer. Do these pairs work? Are they consecutive odd integers? Is their product 195?		$\begin{matrix} 13, 15, yes & −13, −15, yes \\ 13 \cdot 15 = 195, yes & −13 (−15) = 195, yes \end{matrix}$
Step 7. Answer the question.		The two consecutive odd integers whose product is 195 are 13, 15, and −13, −15.

We will use the formula for the area of a triangle to solve the next example.

Recall that, when we solve geometry applications, it is helpful to draw the figure.

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can have an area of 120 square feet and the architect wants the width to be 4 feet more than twice the height. Find the height and width of the window.

Solution

Step 1. Read the problem. Draw a picture.
Step 2. Identify what we are looking for.		We are looking for the height and width.
Step 3. Name what we are looking for.		Let $h =$ the height of the triangle. $2 h + 4 =$ the width of the triangle
Step 4. Translate.		We know the area. Write the formula for the area of a triangle.
Step 5. Solve the equation. Substitute in the values.
Distribute.
This is a quadratic equation, rewrite it in standard form.
Solve the equation using the Quadratic Formula. Identify the a, b, c values.
Write the quadratic equation.
Then substitute in the values of a, b, c..
Simplify.		$A mathematical equation for 'h' is shown: h equals a fraction where the numerator is -2 plus or minus the square root of (4 plus 480), and the denominator is 2.$
Simplify the radical.
Rewrite to show two solutions.
Simplify.
Since h is the height of a window, a value of $h = − 12$ does not make sense.
		The height of the triangle: $h = 10$ The width of the triangle: $2 h + 4$ $2 \cdot 10 + 4$ $24$
Step 6. Check the answer. Does a triangle with a height 10 and width 24 have area 120? Yes.
Step 7. Answer the question.		The height of the triangular window is 10 feet and the width is 24 feet.

Notice that the solutions were integers. That tells us that we could have solved the equation by factoring.

When we wrote the equation in standard form, $h^{2} + 2 h - 120 = 0$ , we could have factored it. If we did, we would have solved the equation $(h + 12) (h - 10) = 0$ .

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Solution

Step 1. Read the problem. Draw a picture.
Step 2. Identify what we are looking for.		We are looking for the height of the pole.
Step 3. Name what we are looking for.		The distance from the base of the pole to either stake is the same as the height of the pole. Let $x =$ the height of the pole. $x =$ the distance from the pole to stake
Each side is a right triangle. We draw a picture of one of them.
Step 4. Translate into an equation. We can use the Pythagorean Theorem to solve for x.
Write the Pythagorean Theorem.		$a^{2} + b^{2} = c^{2}$
Step 5. Solve the equation. Substitute.		$x^{2} + x^{2} = 10^{2}$
Simplify.		$2 x^{2} = 100$
Divide by 2 to isolate the variable.		$\frac{2 x^{2}}{2} = \frac{100}{2}$
Simplify.		$x^{2} = 50$
Use the Square Root Property.		$x = \pm \sqrt{50}$
Simplify the radical.		$x = \pm 5 \sqrt{2}$
Rewrite to show two solutions.		$x = 5 \sqrt{2}$ $x = − 5 \sqrt{2}$
Approximate this number to the nearest tenth with a calculator.		$x \approx 7.1$
Step 6. Check the answer. Check on your own in the Pythagorean Theorem.
Step 7. Answer the question.		The pole should be about 7.1 feet tall.

Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than three times the width. Find the length and width. Round to the nearest tenth of a foot.

Solution

Step 1. Read the problem. Draw a picture.
Step 2. Identify what we are looking for.		We are looking for the length and width.
Step 3. Name what we are looking for.		Let $w =$ the width of the rectangle. $3 w - 1 =$ the length of the rectangle
Step 4. Translate into an equation. We know the area. Write the formula for the area of a rectangle.
Step 5. Solve the equation. Substitute in the values.
Distribute.
This is a quadratic equation, rewrite it in standard form.
Solve the equation using the Quadratic Formula.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.		$The image shows a mathematical equation where W is calculated as the fraction of (1 plus or minus the square root of (1 + 1800)) all divided by 6.$ $A mathematical equation shows W equals the fraction of one plus or minus the square root of 1801, all divided by six.$
Rewrite to show two solutions.
Approximate the answers using a calculator. We eliminate the negative solution for the width.
Step 6. Check the answer. Make sure that the answers make sense.
Step 7. Answer the question.		The width of the rectangle is approximately 7.2 feet and the length 20.6 feet.

The height of a projectile shot upwards is modeled by a quadratic equation. The initial velocity, $v_{0}$ , propels the object up until gravity causes the object to fall back down.

We can use the formula for projectile motion to find how many seconds it will take for a firework to reach a specific height.

A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.		We are looking for the number of seconds, which is time.
Step 3. Name what we are looking for.		Let $t =$ the number of seconds.
Step 4. Translate into an equation.		Use the formula.
		$h = −16 t^{2} + v_{0} t$
Step 5. Solve the equation. We know the velocity $v_{0}$ is 130 feet per second.
The height is 260 feet. Substitute the values.
This is a quadratic equation, rewrite it in standard form.
Solve the equation using the Quadratic Formula.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Rewrite to show two solutions.		$Two mathematical solutions for 't' are presented, each as a fraction: (130 + sqrt(260))/32 and (130 - sqrt(260))/32.$
Approximate the answers with a calculator.		$t \approx 4.6$ seconds, $t \approx 3.6$
Step 6. Check the answer. The check is left to you.
Step 7. Answer the question.		The firework will go up and then fall back down. As the firework goes up, it will reach 260 feet after approximately 3.6 seconds. It will also pass that height on the way down at 4.6 seconds.

Key Concepts

Area of a Triangle For a triangle with base, $b$ , and height, $h$ , the area, $A$ , is given by the formula: $A = \frac{1}{2} b h$
Pythagorean Theorem In any right triangle, where $a$ and $b$ are the lengths of the legs, and $c$ is the length of the hypothenuse, $a^{2} + b^{2} = c^{2}$
Projectile motion The height in feet, $h$ , of an object shot upwards into the air with initial velocity, $v_{0}$ , after $t$ seconds can be modeled by the formula:

$h = −16 t^{2} + v_{0} t$

Practice Makes Perfect

Solve Applications of the Quadratic Formula

In the following exercises, solve by using methods of factoring, the square root principle, or the Quadratic Formula. Round your answers to the nearest tenth.

The product of two consecutive odd numbers is 255. Find the numbers.

Solution

Two consecutive odd numbers whose product is 255 are 15 and 17, and $−15$ and $−17$ .

The product of two consecutive even numbers is 360. Find the numbers.

The product of two consecutive even numbers is 624. Find the numbers.

Solution

Two consecutive even numbers whose product is 624 are 24 and 26, and $−26$ and $−24$ .

The product of two consecutive odd numbers is 1023. Find the numbers.

The product of two consecutive odd numbers is 483. Find the numbers.

Solution

Two consecutive odd numbers whose product is 483 are 21 and 23, and $−21$ and $−23$ .

The product of two consecutive even numbers is 528. Find the numbers.

A triangle with area 45 square inches has a height that is two less than four times the width. Find the height and width of the triangle.

Solution

The width of the triangle is 5 inches and the height is 18 inches.

The width of a triangle is six more than twice the height. The area of the triangle is 88 square yards. Find the height and width of the triangle.

The hypotenuse of a right triangle is twice the length of one of its legs. The length of the other leg is three feet. Find the lengths of the three sides of the triangle. Round to the nearest tenth.

Solution

The leg of the right triangle is 1.7 feet and the hypotenuse is 3.4 feet.

The hypotenuse of a right triangle is 10 cm long. One of the triangle’s legs is three times the length of the other leg. Find the lengths of the three sides of the triangle. Round to the nearest tenth.

A farmer plans to fence off sections of a rectangular corral. The diagonal distance from one corner of the corral to the opposite corner is five yards longer than the width of the corral. The length of the corral is three times the width. Find the length of the diagonal of the corral. Round to the nearest tenth.

The image shows rectangle with the long sides horizontal. A diagonal line runs from the top left corner of the rectangle to the bottom right corner.

Solution

The length of the diagonal of the fence is 7.3 yards.

Nautical flags are used to represent letters of the alphabet. The flag for the letter O consists of a yellow right triangle and a red right triangle which are sewn together along their hypotenuse to form a square. The adjoining side of the two triangles is three inches longer than a side of the flag. Find the length of the side of the flag.

The image shows a square with a diagonal line running from the top left corner to the bottom right corner. The diagonal splits the square into two right triangles. The lower triangle is red and the upper triangle is yellow.

The length of a rectangular driveway is five feet more than three times the width. The area is 350 square feet. Find the length and width of the driveway.

Solution

The width of the driveway is 10 feet and its length is 35 feet.

A rectangular lawn has area 140 square yards. Its width that is six less than twice the length. What are the length and width of the lawn?

A firework rocket is shot upward at a rate of 640 ft/sec. Use the projectile formula $h = −16 t^{2} + v_{0} t$ to determine when the height of the firework rocket will be 1200 feet.

Solution

The rocket will reach 1,200 feet on its way up in 2 seconds and on the way down in 38 seconds.

An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula $h = −16 t^{2} + v_{0} t$ to determine when height of the arrow will be 400 feet.

Everyday Math

A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial height of 8 feet. Use the formula $h = −16 t^{2} + v_{0} t + 8$ to determine how many seconds it will take for the bullet to hit the ground. (That is, when will $h = 0$ ?)

Solution

70 seconds

A city planner wants to build a bridge across a lake in a park. To find the length of the bridge, he makes a right triangle with one leg and the hypotenuse on land and the bridge as the other leg. The length of the hypotenuse is 340 feet and the leg is 160 feet. Find the length of the bridge.

The image shows a right triangle with a horizontal side stretching across a lake, a vertical side on the left labeled a and the hypotenuse connecting the two.

Writing Exercises

Make up a problem involving the product of two consecutive odd integers. Start by choosing two consecutive odd integers. ⓐ What are your integers? ⓑ What is the product of your integers? ⓒ Solve the equation $n (n + 2) = p$ , where $p$ is the product you found in part (b). ⓓ Did you get the numbers you started with?

Solution

Make up a problem involving the product of two consecutive even integers. Start by choosing two consecutive even integers. ⓐ What are your integers? ⓑ What is the product of your integers? ⓒ Solve the equation $n (n + 2) = p$ , where $p$ is the product you found in part (b). ⓓ Did you get the numbers you started with?

Self Check

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

This table has two rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the next row reads “solve applications of the quadratic formula.” The remaining columns are blank.

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

consecutive even integers

Consecutive even integers are even integers that follow right after one another. If an even integer is represented by

n

, the next consecutive even integer is

n + 2

, and the next after that is

n + 4

consecutive odd integers

Consecutive odd integers are odd integers that follow right after one another. If an odd integer is represented by

n

, the next consecutive odd integer is

n + 2

, and the next after that is

n + 4

Graphing Quadratic Equations in Two Variables

Recognize the Graph of a Quadratic Equation in Two Variables

We have graphed equations of the form $A x + B y = C$ . We called equations like this linear equations because their graphs are straight lines.

Now, we will graph equations of the form $y = a x^{2} + b x + c$ . We call this kind of equation a quadratic equation in two variables.

Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations.

Let’s look first at graphing the quadratic equation $y = x^{2}$ . We will choose integer values of $x$ between $−2$ and 2 and find their $y$ values. See Table 1.

$y = x^{2}$
$x$	$y$
0	0
1	1
$−1$	1
2	4
$−2$	4

Notice when we let $x = 1$ and $x = −1$ , we got the same value for $y$ .

\begin{matrix} y = x^{2} & y = x^{2} \\ y = 1^{2} & y = {(−1)}^{2} \\ y = 1 & y = 1 \end{matrix}

The same thing happened when we let $x = 2$ and $x = −2$ .

Now, we will plot the points to show the graph of $y = x^{2}$ . See Figure 1.

This figure shows an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (0, 0). Other points on the curve are located at (-2, 4), (-1, 1), (1, 1) and (2, 4).

The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this.

In Example 1 you will practice graphing a parabola by plotting a few points.

Graph $y = x^{2} - 1$ .

Solution

We will graph the equation by plotting points.

Choose integers values for x, substitute them into the equation and solve for y.
Record the values of the ordered pairs in the chart.
Plot the points, and then connect them with a smooth curve. The result will be the graph of the equation $y = x^{2} - 1$ .

How do the equations $y = x^{2}$ and $y = x^{2} - 1$ differ? What is the difference between their graphs? How are their graphs the same?

All parabolas of the form $y = a x^{2} + b x + c$ open upwards or downwards. See Figure 2.

This figure shows two graphs side by side. The graph on the left side shows an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the inequality a greater than 0 which means the parabola opens upwards. The graph on the right side shows a downward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the inequality a less than 0 which means the parabola opens downwards.

Notice that the only difference in the two equations is the negative sign before the $x^{2}$ in the equation of the second graph in Figure 2. When the $x^{2}$ term is positive, the parabola opens upward, and when the $x^{2}$ term is negative, the parabola opens downward.

Determine whether each parabola opens upward or downward:

ⓐ $y = −3 x^{2} + 2 x - 4$ ⓑ $y = 6 x^{2} + 7 x - 9$

Solution

ⓐ Find the value of "a".		Since the “a” is negative, the parabola will open downward.
ⓑ Find the value of "a".		Since the “a” is positive, the parabola will open upward.

Find the Axis of Symmetry and Vertex of a Parabola

Look again at Figure 2. Do you see that we could fold each parabola in half and that one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

We show the same two graphs again with the axis of symmetry in blue. See Figure 3.

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (-2, -1). Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (2, 7). Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3.

The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of $y = a x^{2} + b x + c$ is $x = - \frac{b}{2 a} .$

So, to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula $x = - \frac{b}{2 a}$ .

The figure shows the steps to find the axis of symmetry for two parabolas. On the left side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the quantity 2 times 1, which simplifies to x equals negative 2. On the right side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals negative x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the quantity 2 times -1, which simplifies to x equals 2.

Look back at Figure 3. Are these the equations of the dashed red lines?

The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. This point is called the vertex of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its x-coordinate is $- \frac{b}{2 a}$ . To find the y-coordinate of the vertex, we substitute the value of the x-coordinate into the quadratic equation.

The figure shows the steps to find the vertex for two parabolas. On the left side is the given equation y equals x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals -2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of -2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with -2 plugged in for the x value which is y equals -2 squared plus 4 times -2 plus 3. This simplifies to y equals -1. Below this is the statement “vertex is (-2, -1)”. On the right side is the given equation y equals negative x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals 2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of 2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with 2 plugged in for the x value which is y equals negative the quantity 2 squared, plus 4 times 2 plus 3. This simplifies to y equals 7. Below this is the statement “vertex is (2, 7)”.

For the parabola $y = 3 x^{2} - 6 x + 2$ find: ⓐ the axis of symmetry and ⓑ the vertex.

Solution

ⓐ
The axis of symmetry is the line $x = - \frac{b}{2 a}$ .
Substitute the values of a, b into the equation.		$A mathematical equation shows X equals the negative of a fraction, where the numerator is -6 and the denominator is 2 multiplied by 3.$
Simplify.		$x = 1$
		The axis of symmetry is the line $x = 1$ .
ⓑ
The vertex is on the line of symmetry, so its x-coordinate will be $x = 1$ .
Substitute $x = 1$ into the equation and solve for y.
Simplify.
This is the y-coordinate.		$y = −1$ The vertex is $(1, − 1) .$

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the x- and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y-intercept the value of $x$ is zero. So, to find the y-intercept, we substitute $x = 0$ into the equation.

Let’s find the y-intercepts of the two parabolas shown in the figure below.

At an x-intercept, the value of $y$ is zero. To find an x-intercept, we substitute $y = 0$ into the equation. In other words, we will need to solve the equation $0 = a x^{2} + b x + c$ for $x$ .

\begin{matrix} y = a x^{2} + b x + c \\ 0 = a x^{2} + b x + c \end{matrix}

But solving quadratic equations like this is exactly what we have done earlier in this chapter.

We can now find the x-intercepts of the two parabolas shown in Figure 4.

First, we will find the x-intercepts of a parabola with equation $y = x^{2} + 4 x + 3$ .


Let $y = 0$ .
Factor.
Use the zero product property.
Solve.
		The x intercepts are $(− 1, 0)$ and $(− 3, 0) .$

Now, we will find the x-intercepts of the parabola with equation $y = - x^{2} + 4 x + 3$ .


Let $y = 0$ .
This quadratic does not factor, so we use the Quadratic Formula.
$a = −1$ , $b = 4$ , $c = 3$
Simplify.		$A mathematical equation where X is equal to a fraction. The numerator of the fraction is -4 plus or minus the square root of 28, and the denominator is -2.$ $A mathematical equation for X, showing X equals a fraction where the numerator is -4 plus or minus 2 times the square root of 7, and the denominator is -2.$ $A mathematical equation showing x equals the fraction where the numerator is -2 times the quantity of 2 plus or minus the square root of 7, and the denominator is -2.$
		The x intercepts are $(2 + \sqrt{7}, 0)$ and $(2 - \sqrt{7}, 0)$ .

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph.

\begin{matrix} (2 + \sqrt{7}, 0) \approx (4.6, 0) & (2 - \sqrt{7}, 0) \approx (−0.6, 0) \end{matrix}

Do these results agree with our graphs? See Figure 5.

Find the intercepts of the parabola $y = x^{2} - 2 x - 8$ .

Solution


To find the y-intercept, let $x = 0$ and solve for y.
		When $x = 0$ , then $y = −8$ . The y-intercept is the point $(0, −8)$ .

To find the x-intercept, let $y = 0$ and solve for x.
Solve by factoring.

When $y = 0$ , then $x = 4 or x = −2$ . The x-intercepts are the points $(4, 0)$ and $(−2, 0)$ .

In this chapter, we have been solving quadratic equations of the form $a x^{2} + b x + c = 0$ . We solved for $x$ and the results were the solutions to the equation.

We are now looking at quadratic equations in two variables of the form $y = a x^{2} + b x + c$ . The graphs of these equations are parabolas. The x-intercepts of the parabolas occur where $y = 0$ .

For example:

\begin{matrix} Quadratic equation & Quadratic equation in two variables \\ y = x^{2} - 2 x - 15 \\ \begin{matrix} x^{2} - 2 x - 15 & = & 0 \\ (x - 5) (x + 3) & = & 0 \end{matrix} & let y = 0 \begin{matrix} 0 = x^{2} - 2 x - 15 \\ 0 = (x - 5) (x + 3) \end{matrix} \\ \begin{matrix} x - 5 & = & 0 & x + 3 & = & 0 \\ x & = & 5 & x & = & −3 \end{matrix} & \begin{matrix} x - 5 & = & 0 & x + 3 & = & 0 \\ x & = & 5 & x & = & −3 \end{matrix} \\ (5, 0) and (−3, 0) \\ x -intercepts \end{matrix}

The solutions of the quadratic equation are the $x$ values of the x-intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the equations give the x-intercepts of the graphs, the number of x-intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic equation of the form $a x^{2} + b x + c = 0$ . Now, we can use the discriminant to tell us how many x-intercepts there are on the graph.

This figure shows three graphs side by side. The leftmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. Below the graph is the inequality b squared minus 4 a c greater than 0. Below that is the statement “Two solutions”. Below that is the statement “ Two x-intercepts”. The middle graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is on the x-axis. Below the graph is the equation b squared minus 4 a c equals 0. Below that is the statement “One solution”. Below that is the statement “ One x-intercept”. The rightmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper left quadrant. Below the graph is the inequality b squared minus 4 a c less than 0. Below that is the statement “No real solutions”. Below that is the statement “ No x-intercept”.

Before you start solving the quadratic equation to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Find the intercepts of the parabola $y = 5 x^{2} + x + 4$ .

Solution


To find the y-intercept, let $x = 0$ and solve for y.		When $x = 0$ , then $y = 4$ . The y-intercept is the point $(0, 4)$ .

To find the x-intercept, let $y = 0$ and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.		$\begin{matrix} \begin{matrix} b^{2} & - & 4 a c \\ 1^{2} & - & 4 \cdot 5 \cdot 4 \\ 1 & - & 80 \end{matrix} \\ −79 \end{matrix}$
Since the value of the discriminant is negative, there is no real solution to the equation.		There are no x-intercepts.

Find the intercepts of the parabola $y = 4 x^{2} - 12 x + 9$ .

Solution


To find the y-intercept, let $x = 0$ and solve for y.
		When $x = 0$ , then $y = 9$ . The y-intercept is the point $(0, 9)$ .

To find the x-intercept, let $y = 0$ and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.		$\begin{matrix} \begin{matrix} b^{2} & - & 4 a c \\ 2^{2} & - & 4 \cdot 4 \cdot 9 \\ 144 & - & 144 \end{matrix} \\ 0 \end{matrix}$
		Since the value of the discriminant is 0, there is only one real solution to the equation. Therefore, there is only one x-intercept.
Solve the equation by factoring the perfect square trinomial.
Use the Zero Product Property.
Solve for x.
		When $y = 0$ , then $\frac{3}{2} = x .$
		The x-intercept is the point $(\frac{3}{2}, 0) .$

Graph Quadratic Equations in Two Variables

Now, we have all the pieces we need in order to graph a quadratic equation in two variables. We just need to put them together. In the next example, we will see how to do this.

How To Graph a Quadratic Equation in Two Variables

Graph $y = x^{2} - 6 x + 8$ .

Solution

The image shows the steps to graph the quadratic equation y equals x squared minus 6 x plus 8. Step 1 is to write the quadratic equation with y on one side. This equation has y on one side already. The value of a is one, the value of b is -6 and the value of c is 8.

Step 2 is to determine whether the parabola opens upward or downward. Since a is positive, the parabola opens upward.

Step 3 is to find the axis of symmetry. The axis of symmetry is the line x equals negative b divided by the quantity 2 a. Plugging in the values of b and a the formula becomes x equals negative -6 divided by the quantity 2 times 1 which simplifies to x equals 3. The axis of symmetry is the line x equals 3.

Step 4 is to find the vertex. The vertex is on the axis of symmetry. Substitute x equals 3 into the equation and solve for y. The equation is y equals x squared minus 6 x plus 8. Replacing x with 3 it becomes y equals 3 squared minus 6 times 3 plus 8 which simplifies to y equals -1. The vertex is (3, -1).

Step 5 is to find the y-intercept and find the point symmetric to the y-intercept across the axis of symmetry. We substitute x equals 0 into the equation. The equation is y equals x squared minus 6 x plus 8. Replacing x with 0 it becomes y equals 0 squared minus 6 times 0 plus 8 which simplifies to y equals 8. The y-intercept is (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. The point symmetric to the y-intercept is (6, 8).

Step 6 is to find the x-intercepts. We substitute y equals 0 into the equation. The equation becomes 0 equals x squared minus 6 x plus 8. We can solve this quadratic equation by factoring to get 0 equals the quantity x minus 2 times the quantity x minus 4. Solve each equation to get x equals 2 and x equals 4. The x-intercepts are (2, 0) and (4, 0).

Step 7 is to graph the parabola. We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these five points to sketch the parabola. The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -2 to 10. The y-axis of the plane runs from -3 to 10. The vertex is at the point (3, -1). Four points are plotted on the curve at (0, 8), (6, 8), (2, 0) and (4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3.

We were able to find the x-intercepts in the last example by factoring. We find the x-intercepts in the next example by factoring, too.

Graph $y = − x^{2} + 6 x - 9$ .

Solution

The equation y has on one side.
Since a is $- 1$ , the parabola opens downward. To find the axis of symmetry, find $x = - \frac{b}{2 a}$ .		The axis of symmetry is $x = 3 .$ The vertex is on the line $x = 3 .$
Find y when $x = 3 .$		The vertex is $(3, 0) .$
The y-intercept occurs when $x = 0 .$ Substitute $x = 0 .$ Simplify. The point $(0, −9)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $(6, −9) .$ Point symmetric to the y-intercept is $(6, −9)$		The y-intercept is $(0, −9) .$
The x-intercept occurs when $y = 0 .$
Substitute $y = 0 .$
Factor the GCF.
Factor the trinomial.
Solve for x.
Connect the points to graph the parabola.

For the graph of $y = - x^{2} + 6 x - 9$ , the vertex and the x-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation $0 = − x^{2} + 6 x - 9$ is 0, so there is only one solution. That means there is only one x-intercept, and it is the vertex of the parabola.

How many x-intercepts would you expect to see on the graph of $y = x^{2} + 4 x + 5$ ?

Graph $y = x^{2} + 4 x + 5$ .

Solution

The equation has y on one side.
Since a is 1, the parabola opens upward.
To find the axis of symmetry, find $x = - \frac{b}{2 a} .$		The axis of symmetry is $x = −2 .$
The vertex is on the line $x = −2 .$
Find y when $x = −2 .$		The vertex is $(−2, 1) .$
The y-intercept occurs when $x = 0 .$ Substitute $x = 0 .$ Simplify. The point $(0, 5)$ is two units to the right of the line of symmetry. The point two units to the left of the line of symmetry is $(−4, 5) .$		The y-intercept is $(0, 5) .$ Point symmetric to the y- intercept is $(−4, 5)$ .
The x- intercept occurs when $y = 0 .$
Substitute $y = 0 .$ Test the discriminant.
		$b^{2} - 4 a c$ $4^{2} - 4 \cdot 15$ $16 - 20$ $−4$
Since the value of the discriminant is negative, there is no solution and so no x- intercept. Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.

Finding the y-intercept by substituting $x = 0$ into the equation is easy, isn’t it? But we needed to use the Quadratic Formula to find the x-intercepts in Example 9. We will use the Quadratic Formula again in the next example.

Graph $y = 2 x^{2} - 4 x - 3$ .

Solution


The equation y has one side. Since a is 2, the parabola opens upward.
To find the axis of symmetry, find $x = - \frac{b}{2 a}$ .		The axis of symmetry is $x = 1$ .
The vertex on the line $x = 1 .$
Find y when $x = 1$ .		The vertex is $(1, − 5)$ .
The y-intercept occurs when $x = 0 .$
Substitute $x = 0 .$
Simplify.		The y-intercept is $(0, −3)$ .
The point $(0, −3)$ is one unit to the left of the line of symmetry. The point one unit to the right of the line of symmetry is $(2, −3)$		Point symmetric to the y-intercept is $(2, −3) .$
The x-intercept occurs when $y = 0$ .
Substitute $y = 0$ .
Use the Quadratic Formula.
Substitute in the values of a, b, c.
Simplify.		$A mathematical equation shows x equals a fraction where the numerator is 4 plus or minus the square root of (16 plus 24), all divided by 4.$
Simplify inside the radical.
Simplify the radical.		$A mathematical equation is displayed, showing x equals the fraction 4 plus or minus 2 times the square root of 10, all divided by 4.$
Factor the GCF.		$A mathematical equation shows x equals the fraction with a numerator of two times the quantity of two plus or minus the square root of ten, and a denominator of four.$
Remove common factors.		$A mathematical equation displays x equals a fraction where the numerator is 2 plus or minus the square root of 10, and the denominator is 2.$
Write as two equations.
Approximate the values.
		The approximate values of the x-intercepts are $(2.5, 0)$ and $(−0.6, 0)$ .
Graph the parabola using the points found.

Solve Maximum and Minimum Applications

Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic equation. The y-coordinate of the vertex is the minimum y-value of a parabola that opens upward. It is the maximum y-value of a parabola that opens downward. See Figure 6.

This figure shows two graphs side by side. The left graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper right quadrant. The vertex is labeled “maximum”. The right graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. The vertex is labeled “minimum”.

Find the minimum value of the quadratic equation $y = x^{2} + 2 x - 8$ .

Solution


Since a is positive, the parabola opens upward.
The quadratic equation has a minimum.
Find the axis of symmetry.		The axis of symmetry is $x = −1$ .
The vertex is on the line $x = −1 .$
Find y when $x = −1 .$		The vertex is $(−1, −9)$ .
Since the parabola has a minimum, the y-coordinate of the vertex is the minimum y-value of the quadratic equation.
The minimum value of the quadratic is $−9$ and it occurs when $x = −1$ .
Show the graph to verify the result.

We have used the formula

h = −16 t^{2} + v_{0} t + h_{0}

to calculate the height in feet, $h$ , of an object shot upwards into the air with initial velocity, $v_{0}$ , after $t$ seconds.

This formula is a quadratic equation in the variable $t$ , so its graph is a parabola. By solving for the coordinates of the vertex, we can find how long it will take the object to reach its maximum height. Then, we can calculate the maximum height.

The quadratic equation $h = −16 t^{2} + v_{0} t + h_{0}$ models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

ⓐ How many seconds will it take the volleyball to reach its maximum height?
ⓑ Find the maximum height of the volleyball.

Solution

$h = −16 t^{2} + 176 t + 4$

Since a is negative, the parabola opens downward.

The quadratic equation has a maximum.

ⓐ
$\begin{matrix} Find the axis of symmetry. & \begin{matrix} t = - \frac{b}{2 a} \\ t = - \frac{176}{2 (−16)} \\ t = 5.5 \end{matrix} \\ The axis of symmetry is t = 5.5 . \\ The vertex is on the line t = 5.5 . & The maximum occurs when t = 5.5 seconds. \end{matrix}$

ⓑ

Find h when $t = 5.5$ .
Use a calculator to simplify.
		The vertex is $(5.5, 488)$ .
Since the parabola has a maximum, the h-coordinate of the vertex is the maximum y-value of the quadratic equation.		The maximum value of the quadratic is 488 feet and it occurs when $t = 5.5$ seconds.

Key Concepts

The graph of every quadratic equation is a parabola.
Parabola Orientation For the quadratic equation y=ax2+bx+c, if
- $a > 0$ , the parabola opens upward.
- $a < 0$ , the parabola opens downward.
Axis of Symmetry and Vertex of a Parabola For a parabola with equation y=ax2+bx+c:
- The axis of symmetry of a parabola is the line $x = - \frac{b}{2 a}$ .
- The vertex is on the axis of symmetry, so its x-coordinate is $- \frac{b}{2 a}$ .
- To find the y-coordinate of the vertex we substitute $x = - \frac{b}{2 a}$ into the quadratic equation.
Find the Intercepts of a Parabola To find the intercepts of a parabola with equation $y = a x^{2} + b x + c$ :
$\begin{matrix} y -intercept & x -intercepts \\ Let x = 0 and solve for y . & Let y = 0 and solve for x . \end{matrix}$
To Graph a Quadratic Equation in Two Variables
1. Write the quadratic equation with $y$ on one side.
2. Determine whether the parabola opens upward or downward.
3. Find the axis of symmetry.
4. Find the vertex.
5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
6. Find the x-intercepts.
7. Graph the parabola.
Minimum or Maximum Values of a Quadratic Equation
- The y-coordinate of the vertex of the graph of a quadratic equation is the
- minimum value of the quadratic equation if the parabola opens upward.
- maximum value of the quadratic equation if the parabola opens downward.

Section Exercises

Practice Makes Perfect

Recognize the Graph of a Quadratic Equation in Two Variables

In the following exercises, graph:

$y = x^{2} + 3$

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has a vertex at (0, 3) and goes through the point (1, 4).

$y = − x^{2} + 1$

In the following exercises, determine if the parabola opens up or down.

$y = −2 x^{2} - 6 x - 7$

Solution

down

$y = 6 x^{2} + 2 x + 3$

$y = 4 x^{2} + x - 4$

Solution

$y = −9 x^{2} - 24 x - 16$

Find the Axis of Symmetry and Vertex of a Parabola

In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.

$y = x^{2} + 8 x - 1$

Solution

ⓐ $x = −4$ ⓑ $(−4, −17)$

$y = x^{2} + 10 x + 25$

$y = − x^{2} + 2 x + 5$

Solution

ⓐ $x = 1$ ⓑ $(1, 6)$

$y = −2 x^{2} - 8 x - 3$

Find the Intercepts of a Parabola

In the following exercises, find the x- and y-intercepts.

$y = x^{2} + 7 x + 6$

Solution

$y : (0, 6); x : (−1, 0), (−6, 0)$

$y = x^{2} + 10 x - 11$

$y = − x^{2} + 8 x - 19$

Solution

$y : (0, - 19); x : none$

$y = x^{2} + 6 x + 13$

$y = 4 x^{2} - 20 x + 25$

Solution

$y : (0, 25); x : (\frac{5}{2}, 0)$

$y = − x^{2} - 14 x - 49$

Graph Quadratic Equations in Two Variables

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$y = x^{2} + 6 x + 5$

Solution

$y : (0, 5); x : (−1, 0), (−5, 0);$
axis: $x = −3; vertex : (−3, −4)$

$y = x^{2} + 4 x - 12$

$y = x^{2} + 4 x + 3$

Solution

$y : (0, 3); x : (−1, 0), (−3, 0);$
axis: $x = −2; vertex: (−2, −1)$

$y = x^{2} - 6 x + 8$

$y = 9 x^{2} + 12 x + 4$

Solution

$y : (0, 4) x : (- \frac{2}{3}, 0);$
axis: $x = - \frac{2}{3}; vertex: (- \frac{2}{3}, 0)$
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 5. The parabola has points plotted at the vertex (-2 thirds, 0) and the intercept (0, 4). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2 thirds.

$y = − x^{2} + 8 x - 16$

$y = − x^{2} + 2 x - 7$

Solution

$y : (0, −7); x : none;$
axis: $x = 1; vertex : (1, −6)$
This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -15 to 5. The parabola has points plotted at the vertex (1, -6) and the intercept (0, -7). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

$y = 5 x^{2} + 2$

$y = 2 x^{2} - 4 x + 1$

Solution

$y : (0, 1); x : (1.7, 0), (0.3, 0);$
axis: $x = 1; vertex : (1, −1)$

$y = 3 x^{2} - 6 x - 1$

$y = 2 x^{2} - 4 x + 2$

Solution

$y : (0, 2) x : (1, 0);$
axis: $x = 1; vertex: (1, 0)$
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (1, 0) and the intercept (0, 2). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

$y = −4 x^{2} - 6 x - 2$

$y = − x^{2} - 4 x + 2$

Solution

$y : (0, 2) x : (−4.4, 0), (0.4, 0);$
axis: $x = −2; vertex: (−2, 6)$

$y = x^{2} + 6 x + 8$

$y = 5 x^{2} - 10 x + 8$

Solution

$y : (0, 8); x : none;$
axis: $x = 1; vertex : (1, 3)$
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (1, 3) and the intercept(0, 8). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

$y = −16 x^{2} + 24 x - 9$

$y = 3 x^{2} + 18 x + 20$

Solution

$y : (0, 20) x : (−4.5, 0), (−1.5, 0);$
axis: $x = −3; vertex: (−3, −7)$
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (-3, -7) and the intercepts (-4.5, 0) and (-1.5, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -3.

$y = −2 x^{2} + 8 x - 10$

Solve Maximum and Minimum Applications

In the following exercises, find the maximum or minimum value.

$y = 2 x^{2} + x - 1$

Solution

The minimum value is $- \frac{9}{8}$ when $x = - \frac{1}{4}$ .

$y = −4 x^{2} + 12 x - 5$

$y = x^{2} - 6 x + 15$

Solution

The minimum value is 6 when $x = 3$ .

$y = − x^{2} + 4 x - 5$

$y = −9 x^{2} + 16$

Solution

The maximum value is 16 when $x = 0$ .

$y = 4 x^{2} - 49$

In the following exercises, solve. Round answers to the nearest tenth.

An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic equation $h = −16 t^{2} + 168 t + 45$ to find how long it will take the arrow to reach its maximum height, and then find the maximum height.

Solution

In 5.3 sec the arrow will reach maximum height of 486 ft.

A stone is thrown vertically upward from a platform that is 20 feet high at a rate of 160 ft/sec. Use the quadratic equation $h = −16 t^{2} + 160 t + 20$ to find how long it will take the stone to reach its maximum height, and then find the maximum height.

A computer store owner estimates that by charging $x$ dollars each for a certain computer, he can sell $40 - x$ computers each week. The quadratic equation $R = − x^{2} + 40 x$ is used to find the revenue, $R$ , received when the selling price of a computer is $x$ . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

Solution

Charging $20 for each computer will give the maximum revenue of $400.

A retailer who sells backpacks estimates that, by selling them for $x$ dollars each, he will be able to sell $100 - x$ backpacks a month. The quadratic equation $R = − x^{2} + 100 x$ is used to find the $R$ received when the selling price of a backpack is $x$ . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation $A = x (240 - 2 x)$ gives the area of the corral, $A$ , for the length, $x,$ of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.

Solution

The length of the side along the river of the corral is 60 feet and the maximum area is 7,200 sq ft.

A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic equation $A = x (100 - 2 x)$ gives the area, $A$ , of the dog run for the length, $x$ , of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.

Everyday Math

In the previous set of exercises, you worked with the quadratic equation $R = − x^{2} + 40 x$ that modeled the revenue received from selling computers at a price of $x$ dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model.
ⓐ Graph the equation $R = − x^{2} + 40 x$ . ⓑ Find the values of the x-intercepts.

Solution

ⓐ
ⓑ $(0, 0), (40, 0)$

In the previous set of exercises, you worked with the quadratic equation $R = − x^{2} + 100 x$ that modeled the revenue received from selling backpacks at a price of $x$ dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model.
ⓐ Graph the equation $R = − x^{2} + 100 x$ . ⓑ Find the values of the x-intercepts.

Writing Exercises

For the revenue model in Exercise 82 and Exercise 86, explain what the x-intercepts mean to the computer store owner.

Solution

Answers will vary.

For the revenue model in Exercise 83 and Exercise 87, explain what the x-intercepts mean to the backpack retailer.

Self Check

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

This table has six rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the second row reads “solve quadratic equations using the quadratic for recognize the graph of a quadratic equation in two variables.” The third row reads “find the axis of symmetry and vertex of a parabola.” The fourth row reads “find the intercepts of a parabola.” The fifth row reads “graph quadratic equations in two variables.” and the last row reads “solve maximum and minimum applications.” The remaining columns are blank.

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

Chapter 10 Review Exercises

10.1 Solve Quadratic Equations Using the Square Root Property

In the following exercises, solve using the Square Root Property.

$x^{2} = 100$

Solution

$x = \pm 10$

$y^{2} = 144$

$m^{2} - 40 = 0$

Solution

$m = \pm 2 \sqrt{10}$

$n^{2} - 80 = 0$

$4 a^{2} = 100$

Solution

$a = \pm 5$

$2 b^{2} = 72$

$r^{2} + 32 = 0$

Solution

no solution

$t^{2} + 18 = 0$

$\frac{4}{3} v^{2} + 4 = 28$

Solution

$v = \pm 3 \sqrt{2}$

$\frac{2}{3} w^{2} - 20 = 30$

$5 c^{2} + 3 = 19$

Solution

$c = \pm \frac{4 \sqrt{5}}{5}$

$3 d^{2} - 6 = 43$

In the following exercises, solve using the Square Root Property.

${(p - 5)}^{2} + 3 = 19$

Solution

$p = 1, 9$

${(q + 4)}^{2} = 9$

${(u + 1)}^{2} = 45$

Solution

$u = −1 \pm 3 \sqrt{5}$

${(z - 5)}^{2} = 50$

${(x - \frac{1}{4})}^{2} = \frac{3}{16}$

Solution

$x = \frac{1}{4} \pm \frac{\sqrt{3}}{4}$

${(y - \frac{2}{3})}^{2} = \frac{2}{9}$

${(m - 7)}^{2} + 6 = 30$

Solution

$m = 7 \pm 2 \sqrt{6}$

${(n - 4)}^{2} - 50 = 150$

${(5 c + 3)}^{2} = −20$

Solution

no solution

${(4 c - 1)}^{2} = −18$

$m^{2} - 6 m + 9 = 48$

Solution

$m = 3 \pm 4 \sqrt{3}$

$n^{2} + 10 n + 25 = 12$

$64 a^{2} + 48 a + 9 = 81$

Solution

$a = - \frac{3}{2}, \frac{3}{4}$

$4 b^{2} - 28 b + 49 = 25$

10.2 Solve Quadratic Equations Using Completing the Square

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

$x^{2} + 22 x$

Solution

${(x + 11)}^{2}$

$y^{2} + 6 y$

$m^{2} - 8 m$

Solution

${(m - 4)}^{2}$

$n^{2} - 10 n$

$a^{2} - 3 a$

Solution

${(a - \frac{3}{2})}^{2}$

$b^{2} + 13 b$

$p^{2} + \frac{4}{5} p$

Solution

${(p + \frac{2}{5})}^{2}$

$q^{2} - \frac{1}{3} q$

In the following exercises, solve by completing the square.

$c^{2} + 20 c = 21$

Solution

$c = 1, −21$

$d^{2} + 14 d = −13$

$x^{2} - 4 x = 32$

Solution

$x = −4, 8$

$y^{2} - 16 y = 36$

$r^{2} + 6 r = −100$

Solution

no solution

$t^{2} - 12 t = −40$

$v^{2} - 14 v = −31$

Solution

$v = 7 \pm 3 \sqrt{2}$

$w^{2} - 20 w = 100$

$m^{2} + 10 m - 4 = −13$

Solution

$m = −9, −1$

$n^{2} - 6 n + 11 = 34$

$a^{2} = 3 a + 8$

Solution

$a = \frac{3}{2} \pm \frac{\sqrt{41}}{2}$

$b^{2} = 11 b - 5$

$(u + 8) (u + 4) = 14$

Solution

$u = −6 \pm 3 \sqrt{2}$

$(z - 10) (z + 2) = 28$

$3 p^{2} - 18 p + 15 = 15$

Solution

$p = 0, 6$

$5 q^{2} + 70 q + 20 = 0$

$4 y^{2} - 6 y = 4$

Solution

$y = - \frac{1}{2}, 2$

$2 x^{2} + 2 x = 4$

$3 c^{2} + 2 c = 9$

Solution

$c = - \frac{1}{3} \pm \frac{2 \sqrt{7}}{3}$

$4 d^{2} - 2 d = 8$

10.3 Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

$4 x^{2} - 5 x + 1 = 0$

Solution

$x = \frac{1}{4}, 1$

$7 y^{2} + 4 y - 3 = 0$

$r^{2} - r - 42 = 0$

Solution

$r = −6, 7$

$t^{2} + 13 t + 22 = 0$

$4 v^{2} + v - 5 = 0$

Solution

$v = - \frac{5}{4}, 1$

$2 w^{2} + 9 w + 2 = 0$

$3 m^{2} + 8 m + 2 = 0$

Solution

$m = \frac{−4 \pm \sqrt{10}}{3}$

$5 n^{2} + 2 n - 1 = 0$

$6 a^{2} - 5 a + 2 = 0$

Solution

no real solution

$4 b^{2} - b + 8 = 0$

$u (u - 10) + 3 = 0$

Solution

$u = 5 \pm \sqrt{22}$

$5 z (z - 2) = 3$

$\frac{1}{8} p^{2} - \frac{1}{5} p = - \frac{1}{20}$

Solution

$p = \frac{4 \pm \sqrt{6}}{5}$

$\frac{2}{5} q^{2} + \frac{3}{10} q = \frac{1}{10}$

$4 c^{2} + 4 c + 1 = 0$

Solution

$c = - \frac{1}{2}$

$9 d^{2} - 12 d = −4$

In the following exercises, determine the number of solutions to each quadratic equation.

ⓐ $9 x^{2} - 6 x + 1 = 0$
ⓑ $3 y^{2} - 8 y + 1 = 0$
ⓓ $5 n^{2} - n + 1 = 0$

Solution

ⓐ $5 x^{2} - 7 x - 8 = 0$
ⓑ $7 x^{2} - 10 x + 5 = 0$
ⓓ $15 x^{2} - 8 x + 4 = 0$

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation.

ⓐ $16 r^{2} - 8 r + 1 = 0$
ⓑ $5 t^{2} - 8 t + 3 = 9$

Solution

ⓐ $4 d^{2} + 10 d - 5 = 21$
ⓑ $25 x^{2} - 60 x + 36 = 0$

10.4 Solve Applications Modeled by Quadratic Equations

In the following exercises, solve by using methods of factoring, the square root principle, or the quadratic formula.

Find two consecutive odd numbers whose product is 323.

Solution

Two consecutive odd numbers whose product is 323 are 17 and 19, and $−17$ and $−19 .$

Find two consecutive even numbers whose product is 624.

A triangular banner has an area of 351 square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.

Solution

The height of the banner is 13 cm and the length of the side is 54 cm.

Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is 70 square inches. Find the height and width of the case.

A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is 5 feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.

The image shows a rectangular pathway with a right inlaid in the lower left corner. The right angle of the triangle overlays the lower left corner of the rectangle. The left leg of the right triangle overlays the left side of the rectangle and the hypotenuse of the right triangle runs from the upper left corner of the rectangle to a point on the bottom of the rectangle.

Solution

The lengths of the sides of the mosaic are 2.2 and 4.4 feet.

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.

The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.

Solution

The width of the front walk is 8.2 feet and its length is 30.6 feet.

For Sophia’s graduation party, several tables of the same width will be arranged end to end to give a serving table with a total area of 75 square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth. Round answer to the nearest tenth.

The image shows four rectangular tables placed side by side to create one large table.

A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula $h = −16 t^{2} + v_{0} t$ to determine when the ball will be 384 feet from the ground. Round to the nearest tenth.

Solution

The ball will reach 384 feet on its way up in 4 seconds and on the way down in 6 seconds.

A bullet is fired straight up from the ground at a velocity of 320 ft/sec. Use the formula $h = −16 t^{2} + v_{0} t$ to determine when the bullet will reach 800 feet. Round to the nearest tenth.

10.5 Graphing Quadratic Equations in Two Variables

In the following exercises, graph by plotting point.

Graph $y = x^{2} - 2$

Solution

Graph $y = − x^{2} + 3$

In the following exercises, determine if the following parabolas open up or down.

$y = −3 x^{2} + 3 x - 1$

Solution

down

$y = 5 x^{2} + 6 x + 3$

$y = x^{2} + 8 x - 1$

Solution

$y = −4 x^{2} - 7 x + 1$

In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.

$y = − x^{2} + 6 x + 8$

Solution

ⓐ $x = 3$ ⓑ $(3, 17)$

$y = 2 x^{2} - 8 x + 1$

In the following exercises, find the x- and y-intercepts.

$y = x^{2} - 4 x - 5$

Solution

$y : (0, 5); x : (5, 0), (−1, 0)$

$y = x^{2} - 8 x + 15$

$y = x^{2} - 4 x + 10$

Solution

$y : (0, 10); x : none$

$y = −5 x^{2} - 30 x - 46$

$y = 16 x^{2} - 8 x + 1$

Solution

$y : (0, 1); x : (\frac{1}{4}, 0)$

$y = x^{2} + 16 x + 64$

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$y = x^{2} + 8 x + 15$

Solution

$y : (0, 15); x : (−3, 0), (−5, 0);$
axis: $x = −4; vertex : (−4, −1)$

$y = x^{2} - 2 x - 3$

$y = − x^{2} + 8 x - 16$

Solution

$y : (0, −16); x : (4, 0);$
axis: $x = 4; vertex : (4, 0)$
This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -15 to 12. The y-axis of the plane runs from -20 to 2. The parabola has points plotted at the vertex (4, 0) and the intercept (0, -16). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 4.

$y = 4 x^{2} - 4 x + 1$

$y = x^{2} + 6 x + 13$

Solution

$y : (0, 13); x : none;$
axis: $x = −3; vertex : (−3, 4)$
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -2 to 18. The parabola has points plotted at the vertex (-3, 4) and the intercept (0, 13). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -3.

$y = −2 x^{2} - 8 x - 12$

$y = −4 x^{2} + 16 x - 11$

Solution

$y : (0, −11) x : (3.1, 0), (0.9, 0);$
axis: $x = 2; vertex: (2, 5)$
This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (2, 5) and the intercepts (3.1, 0) and (0.9, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2.

$y = x^{2} + 8 x + 10$

In the following exercises, find the minimum or maximum value.

$y = 7 x^{2} + 14 x + 6$

Solution

The minimum value is $−1$ when $x = −1$ .

$y = −3 x^{2} + 12 x - 10$

In the following exercises, solve. Rounding answers to the nearest tenth.

A ball is thrown upward from the ground with an initial velocity of 112 ft/sec. Use the quadratic equation $h = −16 t^{2} + 112 t$ to find how long it will take the ball to reach maximum height, and then find the maximum height.

Solution

In 3.5 seconds the ball is at its maximum height of 196 feet.

A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using 180 feet of fencing on three sides of the yard. The quadratic equation $A = −2 x^{2} + 180 x$ gives the area, $A$ , of the yard for the length, $x$ , of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.

Practice Test

Use the Square Root Property to solve the quadratic equation: $3 {(w + 5)}^{2} = 27$ .

Solution

$w = −2, −8$

Use Completing the Square to solve the quadratic equation: $a^{2} - 8 a + 7 = 23$ .

Use the Quadratic Formula to solve the quadratic equation: $2 m^{2} - 5 m + 3 = 0$ .

Solution

$m = 1, \frac{3}{2}$

Solve the following quadratic equations. Use any method.

$8 v^{2} + 3 = 35$

$3 n^{2} + 8 n + 3 = 0$

Solution

$n = \frac{−4 \pm \sqrt{7}}{3}$

$2 b^{2} + 6 b - 8 = 0$

$x (x + 3) + 12 = 0$

Solution

no real solution

$\frac{4}{3} y^{2} - 4 y + 3 = 0$

Use the discriminant to determine the number of solutions of each quadratic equation.

$6 p^{2} - 13 p + 7 = 0$

Solution

$3 q^{2} - 10 q + 12 = 0$

Solve by factoring, the Square Root Property, or the Quadratic Formula.

Find two consecutive even numbers whose product is 360.

Solution

Two consecutive even number are $−20$ and $−18$ and 18 and 20.

The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)

For each parabola, find ⓐ which ways it opens, ⓑ the axis of symmetry, ⓒ the vertex, ⓓ the x- and y-intercepts, and ⓔ the maximum or minimum value.

$y = 3 x^{2} + 6 x + 8$

Solution

$y = x^{2} - 4$

$y = x^{2} + 10 x + 24$

Solution

$y = −3 x^{2} + 12 x - 8$

$y = − x^{2} - 8 x + 16$

Solution

Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry.

$y = 2 x^{2} + 6 x + 2$

$y = 16 x^{2} + 24 x + 9$

Solution

$y : (0, 9); x : (- \frac{3}{4}, 0);$
axis: $x = - \frac{3}{4}; vertex : (- \frac{3}{4}, 0)$
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (3 fourths, 0) and the intercept (0, 9). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3 fourths.

Solve.

A water balloon is launched upward at the rate of 86 ft/sec. Using the formula $h = −16 t^{2} + 86 t$ , find how long it will take the balloon to reach the maximum height and then find the maximum height. Round to the nearest tenth.

axis of symmetry

The axis of symmetry is the vertical line passing through the middle of the parabola

y = a x^{2} + b x + c .

parabola

The graph of a quadratic equation in two variables is a parabola.

quadratic equation in two variables

A quadratic equation in two variables, where

a, b, and c

are real numbers and

a \neq 0

is an equation of the form

y = a x^{2} + b x + c .

vertex

The point on the parabola that is on the axis of symmetry is called the vertex of the parabola; it is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards.

x-intercepts of a parabola

The x-intercepts are the points on the parabola where

y = 0 .

y-intercept of a parabola

The y-intercept is the point on the parabola where

x = 0 .

	$A mathematical expression showing 'x' plus the fraction '1/3'.$

Simplify.	0


$Text: Substitute -3/4 for x. The fraction -3/4 is highlighted in red.$	$A mathematical expression showing the addition of two fractions: negative three-fourths plus one-third. The number three and four are in red color, while the negative sign and fractions for one-third are in black.$
Rewrite as equivalent fractions with the LCD, 12.	$An equation for adding -3/4 and 1/3, showing the fractions multiplied by factors to achieve a common denominator of 12: -(33)/(43) + (14)/(34).$
Simplify.	$A mathematical expression showing the addition of two fractions with the same denominator: -9/12 + 4/12.$
Add.	$- \frac{5}{12}$

	$350 L$
Multiply by 1, writing 1 as a fraction relating liters to kiloliters.	$350 L \cdot \frac{1 kL}{1,000 L}$
Simplify.	$350 L \cdot \frac{1 kL}{1,000 L}$
	$0.35 kL$


Multiply by 1, writing 1 as a fraction relating liters to milliliters.
Simplify.	$A mathematical expression showing the conversion of 4.1 liters to milliliters, written as 4.1 L multiplied by the fraction 1,000 mL divided by 1 L.$
Move the decimal 3 units to the right.


Shade to the right of $−3$ , and put a bracket at $−3$ .
Write in interval notation.


Shade to the left of $2.5$ , and put a parenthesis at $2.5$ .
Write in interval notation.


Shade to the left of $- \frac{3}{5}$ , and put a bracket at $- \frac{3}{5}$ .
Write in interval notation.

$y = 3 x - 1$
$x$	$y$	$(x, y)$
0	$−1$	$(0, −1)$
$−1$	$−4$	$(−1, −4)$
2	5	$(2, 5)$

$y = 6 x + 1$
$x$	$y$	$(x, y)$
0	1	$(0, 1)$
1	7	$(1, 7)$
$−2$	$−11$	$(−2, −11)$

$2 x - 5 y = 20$
$x$	$y$	$(x, y)$
0	$−4$	$(0, −4)$
10	0	$(10, 0)$
$−5$	$−6$	$(−5, −6)$

$3 x - 4 y = 12$
$x$	$y$	$(x, y)$
0	$−3$	$(0, −3)$
4	0	$(4, 0)$
$−4$	$−6$	$(−4, −6)$

$(0, 0)$

Simplify.		So, $(0, 0)$ is not a solution to $y > x + 4$ .

$(1, 6)$

Simplify.		So, $(1, 6)$ is a solution to $y > x + 4$ .

$(2, 6)$

Simplify.		So, $(2, 6)$ is not a solution to $y > x + 4$ .

$(−5, −15)$

Simplify.		So, $(−5, −15)$ is not a solution to $y > x + 4$ .

$(−8, 12)$

Simplify.		So, $(−8, 12)$ is a solution to $y > x + 4$ .


Rewrite, a = a¹.
Use the product property, a^m · aⁿ = a^m+n.
Simplify.


Notice, the bases are the same, so add the exponents.
Simplify.


Use Power of a Product Property, (ab)^m = a^mb^m.
Simplify.

Since 9 > 7, there are more factors of x in the numerator.
Use the Quotient Property, $\frac{a^{m}}{a^{n}} = a^{m - n}$ .
Simplify.

Elementary Algebra 2e

Contents

Preface

About OpenStax

About OpenStax Resources

Customization

Art attribution in Elementary Algebra 2e

Errata

Format

About Elementary Algebra

Coverage and Scope

Changes to the Second Edition

Key Features and Boxes

Examples

Be Prepared!

Try It

How To

Media

Art Program

Section Exercises

Chapter Review Features

Answers to Questions in the Book

Additional Resources

Student and Instructor Resources

Partner Resources

About the Authors

Senior Contributing Authors

Reviewers

Introduction

Introduction to Whole Numbers

Use Place Value with Whole Numbers

Solution

Solution

Solution

How to Round Whole Numbers

Solution

Solution

Identify Multiples and Apply Divisibility Tests

Solution

Find Prime Factorizations and Least Common Multiples

How to Find the Prime Factorization of a Composite Number

Solution

Solution

Solution

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

Solution

Solution

Key Concepts

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

Use the Language of Algebra

Use Variables and Algebraic Symbols

Solution

Solution

Solution

Simplify Expressions Using the Order of Operations

Solution

Solution

Solution

Evaluate an Expression

Solution

Solution

Solution

Identify and Combine Like Terms

Solution

Solution

Solution

How To Combine Like Terms

Solution

Translate an English Phrase to an Algebraic Expression

Solution

Solution

Solution

Solution

Solution

Key Concepts

Practice Makes Perfect

Everyday Math

Since 10 > 2, there are more factors of x in the numerator.
Use the Quotient Property, $\frac{a^{m}}{a^{n}} = a^{m - n}$ .
Simplify.

Since 12 > 8, there are more factors of b in the denominator.
Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .
Simplify.

Since 5 > 3, there are more factors of 3 in the denominator.
Use the Quotient Property, $\frac{a^{m}}{a^{n}} = \frac{1}{a^{n - m}}$ .
Simplify.
Simplify.	$The fraction one forty-ninth is displayed, represented as 1 over 49 with a horizontal line separating the numerator and the denominator.$