Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
Prealgebra 2e

3.5 Solve Equations Using Integers; The Division Property of Equality

Prealgebra 2e3.5 Solve Equations Using Integers; The Division Property of Equality

Learning Objectives

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

  • Determine whether an integer is a solution of an equation
  • Solve equations with integers using the Addition and Subtraction Properties of Equality
  • Model the Division Property of Equality
  • Solve equations using the Division Property of Equality
  • Translate to an equation and solve

Be Prepared 3.10

Before you get started, take this readiness quiz.

Evaluatex+4whenx=−4.Evaluatex+4whenx=−4.
If you missed this problem, review Example 3.22.

Be Prepared 3.11

Solve:y6=10.Solve:y6=10.
If you missed this problem, review Example 2.33.

Be Prepared 3.12

Translate into an algebraic expression 55 less than x.x.
If you missed this problem, review Table 1.3.

Determine Whether a Number is a Solution of an Equation

In Solve Equations with the Subtraction and Addition Properties of Equality, we saw that a solution of an equation is a value of a variable that makes a true statement when substituted into that equation. In that section, we found solutions that were whole numbers. Now that we’ve worked with integers, we’ll find integer solutions to equations.

The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number or an integer.

How To

How to determine whether a number is a solution to an equation.

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

Example 3.60

Determine whether each of the following is a solution of 2x5=−13:2x5=−13:

  1. x=4x=4
  2. x=−4x=−4
  3. x=−9.x=−9.

Try It 3.119

Determine whether each of the following is a solution of 2x8=−14:2x8=−14:

  1. x=−11x=−11
  2. x=11x=11
  3. x=−3x=−3

Try It 3.120

Determine whether each of the following is a solution of 2y+3=−11:2y+3=−11:

  1. y=4y=4
  2. y=−4y=−4
  3. y=−7y=−7

Solve Equations with Integers Using the Addition and Subtraction Properties of Equality

In Solve Equations with the Subtraction and Addition Properties of Equality, we solved equations similar to the two shown here using the Subtraction and Addition Properties of Equality. Now we can use them again with integers.

This figure has two columns. The first column has the equation x plus 4 equals 12. Underneath there is x plus 4 minus 4 equals 12 minus 4. Under this there is x equals 8. The second column has the equation y minus 5 equals 9. Underneath there is the equation y minus 5 plus 5 equals 9 plus 5. Under this there is y equals 14.

When you add or subtract the same quantity from both sides of an equation, you still have equality.

Properties of Equalities

Subtraction Property of Equality Addition Property of Equality
For any numbersa,b,c,For any numbersa,b,c,
ifa=bthenac=bc.ifa=bthenac=bc.
For any numbersa,b,c,For any numbersa,b,c,
ifa=bthena+c=b+c.ifa=bthena+c=b+c.

Example 3.61

Solve: y+9=5.y+9=5.

Try It 3.121

Solve:

y+11=7y+11=7

Try It 3.122

Solve:

y+15=−4y+15=−4

Example 3.62

Solve: a6=−8a6=−8

Try It 3.123

Solve:

a2=−8a2=−8

Try It 3.124

Solve:

n4=−8n4=−8

Model the Division Property of Equality

All of the equations we have solved so far have been of the form x+a=bx+a=b or xa=b.xa=b. We were able to isolate the variable by adding or subtracting the constant term. Now we’ll see how to solve equations that involve division.

We will model an equation with envelopes and counters in Figure 3.21.

This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed.
Figure 3.21

Here, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?

To determine the number, separate the counters on the right side into 22 groups of the same size. So 66 counters divided into 22 groups means there must be 33 counters in each group (since 6÷2=3).6÷2=3).

What equation models the situation shown in Figure 3.22? There are two envelopes, and each contains xx counters. Together, the two envelopes must contain a total of 66 counters. So the equation that models the situation is 2x=6.2x=6.

This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed. Under the figure is two times x equals 6.
Figure 3.22

We can divide both sides of the equation by 22 as we did with the envelopes and counters.

This figure has two rows. The first row has the equation 2x divided by 2 equals 6 divided by 2. The second row has the equation x equals 3.

We found that each envelope contains 3 counters.3 counters. Does this check? We know 2·3=6,2·3=6, so it works. Three counters in each of two envelopes does equal six.

Figure 3.23 shows another example.

This image has two columns. In the first column are three envelopes. In the second column there are four rows of  three blue circles. Underneath the image is the equation 3x equals 12.
Figure 3.23

Now we have 33 identical envelopes and 12 counters.12 counters. How many counters are in each envelope? We have to separate the 12 counters12 counters into 3 groups.3 groups. Since 12÷3=4,12÷3=4, there must be 4 counters4 counters in each envelope. See Figure 3.24.

This image has two columns. In the first column are four envelopes. In the second column there are twelve blue circles.
Figure 3.24

The equation that models the situation is 3x=12.3x=12. We can divide both sides of the equation by 3.3.

This image shows the equation 3x divided by 3 equals 12 divided by 3. Below this equation is the equation x equals 4.

Does this check? It does because 3·4=12.3·4=12.

Manipulative Mathematics

Doing the Manipulative Mathematics activity “Division Property of Equality” will help you develop a better understanding of how to solve equations using the Division Property of Equality.

Example 3.63

Write an equation modeled by the envelopes and counters, and then solve it.

This image has two columns. In the first column are four envelopes. In the second column there are 8 blue circles.

Try It 3.125

Write the equation modeled by the envelopes and counters. Then solve it.

This image has two columns. In the first column are four envelopes. In the second column there are 12 blue circles.

Try It 3.126

Write the equation modeled by the envelopes and counters. Then solve it.

This image has two columns. In the first column are three envelopes. In the second column there are six blue circles.

Solve Equations Using the Division Property of Equality

The previous examples lead to the Division Property of Equality. When you divide both sides of an equation by any nonzero number, you still have equality.

Division Property of Equality

For any numbers a,b,c,and c0, If a=bthen ac=bc. For any numbers a,b,c,and c0, If a=bthen ac=bc.

Example 3.64

Solve:7x=−49.Solve:7x=−49.

Try It 3.127

Solve:

8a=568a=56

Try It 3.128

Solve:

11n=12111n=121

Example 3.65

Solve: −3y=63.−3y=63.

Try It 3.129

Solve:

−8p=96−8p=96

Try It 3.130

Solve:

−12m=108−12m=108

Translate to an Equation and Solve

In the past several examples, we were given an equation containing a variable. In the next few examples, we’ll have to first translate word sentences into equations with variables and then we will solve the equations.

Example 3.66

Translate and solve: five more than xx is equal to −3.−3.

Try It 3.131

Translate and solve:

Seven more than xx is equal to −2−2.

Try It 3.132

Translate and solve:

Eleven more thanyis equal to 2.Eleven more thanyis equal to 2.

Example 3.67

Translate and solve: the difference of nn and 66 is −10.−10.

Try It 3.133

Translate and solve:

The difference of pp and 22 is −4−4.

Try It 3.134

Translate and solve:

The difference of qq and 77 is −3−3.

Example 3.68

Translate and solve: the number 108108 is the product of −9−9 and y.y.

Try It 3.135

Translate and solve:

The number 132132 is the product of −12−12 and yy.

Try It 3.136

Translate and solve:

The number 117117 is the product of −13−13 and zz.

Section 3.5 Exercises

Practice Makes Perfect

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each number is a solution of the given equation.

285.

4x2=64x2=6

  1. x=−2x=−2
  2. x=−1x=−1
  3. x=2x=2
286.

4y10=−144y10=−14

  1. y=−6y=−6
  2. y=−1y=−1
  3. y=1y=1
287.

9a+27=−639a+27=−63

  1. a=6a=6
  2. a=−6a=−6
  3. a=−10a=−10
288.

7c+42=−567c+42=−56

  1. c=2c=2
  2. c=−2c=−2
  3. c=−14c=−14

Solve Equations Using the Addition and Subtraction Properties of Equality

In the following exercises, solve for the unknown.

289.

n + 12 = 5 n + 12 = 5

290.

m + 16 = 2 m + 16 = 2

291.

p + 9 = −8 p + 9 = −8

292.

q + 5 = −6 q + 5 = −6

293.

u 3 = −7 u 3 = −7

294.

v 7 = −8 v 7 = −8

295.

h 10 = −4 h 10 = −4

296.

k 9 = −5 k 9 = −5

297.

x + ( −2 ) = −18 x + ( −2 ) = −18

298.

y + ( −3 ) = −10 y + ( −3 ) = −10

299.

r ( −5 ) = −9 r ( −5 ) = −9

300.

s ( −2 ) = −11 s ( −2 ) = −11

Model the Division Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve it.

301.
No Alt Text
302.
No Alt Text
303.
No Alt Text
304.
No Alt Text

Solve Equations Using the Division Property of Equality

In the following exercises, solve each equation using the division property of equality and check the solution.

305.

5 x = 45 5 x = 45

306.

4 p = 64 4 p = 64

307.

−7 c = 56 −7 c = 56

308.

−9 x = 54 −9 x = 54

309.

−14 p = −42 −14 p = −42

310.

−8 m = −40 −8 m = −40

311.

−120 = 10 q −120 = 10 q

312.

−75 = 15 y −75 = 15 y

313.

24 x = 480 24 x = 480

314.

18 n = 540 18 n = 540

315.

−3 z = 0 −3 z = 0

316.

4 u = 0 4 u = 0

Translate to an Equation and Solve

In the following exercises, translate and solve.

317.

Four more than nn is equal to 1.

318.

Nine more than mm is equal to 5.

319.

The sum of eight and pp is −3−3.

320.

The sum of two and qq is −7−7.

321.

The difference of aa and three is −14−14.

322.

The difference of bb and 55 is −2−2.

323.

The number −42 is the product of −7 and xx.

324.

The number −54 is the product of −9 and yy.

325.

The product of -15 and ff is 75.

326.

The product of −18 and gg is 36.

327.

−6 plus cc is equal to 4.

328.

−2 plus dd is equal to 1.

329.

Nine less than mm is −4.

330.

Thirteen less than nn is −10−10.

Mixed Practice

In the following exercises, solve.

331.
  1. x+2=10x+2=10
  2. 2x=102x=10
332.
  1. y+6=12y+6=12
  2. 6y=126y=12
333.
  1. −3p=27−3p=27
  2. p3=27p3=27
334.
  1. −2q=34−2q=34
  2. q2=34q2=34
335.

a 4 = 16 a 4 = 16

336.

b 1 = 11 b 1 = 11

337.

−8 m = −56 −8 m = −56

338.

−6 n = −48 −6 n = −48

339.

−39 = u + 13 −39 = u + 13

340.

−100 = v + 25 −100 = v + 25

341.

11 r = −99 11 r = −99

342.

15 s = −300 15 s = −300

343.

100 = 20 d 100 = 20 d

344.

250 = 25 n 250 = 25 n

345.

−49 = x 7 −49 = x 7

346.

64 = y 4 64 = y 4

Everyday Math

347.

Cookie packaging A package of 51 cookies51 cookies has 33 equal rows of cookies. Find the number of cookies in each row, c,c, by solving the equation 3c=51.3c=51.

348.

Kindergarten class Connie’s kindergarten class has 24 children.24 children. She wants them to get into 44 equal groups. Find the number of children in each group, g,g, by solving the equation 4g=24.4g=24.

Writing Exercises

349.

Is modeling the Division Property of Equality with envelopes and counters helpful to understanding how to solve the equation 3x=15?3x=15? Explain why or why not.

350.

Suppose you are using envelopes and counters to model solving the equations x+4=12x+4=12 and 4x=12.4x=12. Explain how you would solve each equation.

351.

Frida started to solve the equation −3x=36−3x=36 by adding 33 to both sides. Explain why Frida’s method will not solve the equation.

352.

Raoul started to solve the equation 4y=404y=40 by subtracting 44 from both sides. Explain why Raoul’s method will not solve the equation.

Self Check

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

.

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

Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/prealgebra-2e/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/prealgebra-2e/pages/1-introduction
Citation information

© Jul 24, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.