3.5 Solve Equations Using Integers; The Division Property of Equality - Prealgebra

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 3.5

Before you get started, take this readiness quiz.

$Evaluate x + 4 when x = −4 .$
If you missed this problem, review Example 3.22.
$Solve: y - 6 = 10 .$
If you missed this problem, review Example 2.33.
Translate into an algebraic expression $5$ less than $x .$
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.

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

Example 3.60

Determine whether each of the following is a solution of $2 x - 5 = −13 :$

ⓐ $x = 4$
ⓑ $x = −4$
ⓒ $x = −9 .$

Solution

ⓐ Substitute 4 for x in the equation to determine if it is true.


Multiply.
Subtract.

Since $x = 4$ does not result in a true equation, $4$ is not a solution to the equation.

ⓑ Substitute −4 for x in the equation to determine if it is true.

Multiply.
Subtract.

Since $x = −4$ results in a true equation, $−4$ is a solution to the equation.

ⓒ Substitute −9 for x in the equation to determine if it is true.

Substitute −9 for x.
Multiply.
Subtract.

Since $x = −9$ does not result in a true equation, $−9$ is not a solution to the equation.

Try It 3.119

Determine whether each of the following is a solution of $2 x - 8 = −14 :$

ⓐ $x = −11$
ⓑ $x = 11$
ⓒ $x = −3$

Try It 3.120

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

ⓐ $y = 4$
ⓑ $y = −4$
ⓒ $y = −7$

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
$For any numbers a, b, c,$ $if a = b then a - c = b - c .$	$For any numbers a, b, c,$ $if a = b then a + c = b + c .$

Example 3.61

Solve: $y + 9 = 5 .$

Solution


Subtract 9 from each side to undo the addition.
Simplify.

Check the result by substituting $−4$ into the original equation.

	$y + 9 = 5$
Substitute −4 for y	$−4 + 9 \overset{?}{=} 5$
	$5 = 5 ✓$

Since $y = −4$ makes $y + 9 = 5$ a true statement, we found the solution to this equation.

Try It 3.121

Solve:

$y + 11 = 7$

Try It 3.122

Solve:

$y + 15 = −4$

Example 3.62

Solve: $a - 6 = −8$

Solution


Add 6 to each side to undo the subtraction.
Simplify.
Check the result by substituting $−2$ into the original equation:
Substitute $−2$ for $a$

The solution to $a - 6 = −8$ is $−2 .$

Since $a = −2$ makes $a - 6 = −8$ a true statement, we found the solution to this equation.

Try It 3.123

Solve:

$a - 2 = −8$

Try It 3.124

Solve:

$n - 4 = −8$

Model the Division Property of Equality

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

We will model an equation with envelopes and counters in Figure 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 $2$ groups of the same size. So $6$ counters divided into $2$ groups means there must be $3$ counters in each group (since $6 \div 2 = 3) .$

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

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 $2$ as we did with the envelopes and counters.

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

We found that each envelope contains $3 counters.$ Does this check? We know $2 \cdot 3 = 6,$ 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 $3$ identical envelopes and $12 counters.$ How many counters are in each envelope? We have to separate the $12 counters$ into $3 groups.$ Since $12 \div 3 = 4,$ there must be $4 counters$ 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 $3 x = 12 .$ We can divide both sides of the equation by $3 .$

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

Does this check? It does because $3 \cdot 4 = 12 .$

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.

Solution

There are $4 envelopes,$ or $4$ unknown values, on the left that match the $8 counters$ on the right. Let’s call the unknown quantity in the envelopes $x .$

Write the equation.
Divide both sides by 4.
Simplify.

There are $2 counters$ in each envelope.

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

\begin{matrix} For any numbers & a, b, c, and & c \neq 0, \\ If & a = b then & \frac{a}{c} = \frac{b}{c} . \end{matrix}

Example 3.64

$Solve: 7 x = −49 .$

Solution

To isolate $x,$ we need to undo multiplication.


Divide each side by 7.
Simplify.

Check the solution.

	$7 x = −49$
Substitute −7 for x.	$7 (−7) \overset{?}{=} −49$
	$−49 = −49 ✓$

Therefore, $−7$ is the solution to the equation.

Try It 3.127

Solve:

$8 a = 56$

Try It 3.128

Solve:

$11 n = 121$

Example 3.65

Solve: $−3 y = 63 .$

Solution

To isolate $y,$ we need to undo the multiplication.


Divide each side by −3.
Simplify

Check the solution.

	$−3 y = 63$
Substitute −21 for y.	$−3 (−21) \overset{?}{=} 63$
	$63 = 63 ✓$

Since this is a true statement, $y = −21$ is the solution to the equation.

Try It 3.129

Solve:

$−8 p = 96$

Try It 3.130

Solve:

$−12 m = 108$

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 $x$ is equal to $−3 .$

Solution

	five more than $x$ is equal to $−3$
Translate	$x + 5 = −3$
Subtract $5$ from both sides.	$x + 5 - 5 = −3 - 5$
Simplify.	$x = −8$

Check the answer by substituting it into the original equation.

$\begin{array}{l} x + 5 = −3 \\ −8 + 5 \overset{?}{=} −3 \\ −3 = −3 ✓ \end{array}$

Try It 3.131

Translate and solve:

Seven more than $x$ is equal to $−2$ .

Try It 3.132

Translate and solve:

$Eleven more than y is equal to 2.$

Example 3.67

Translate and solve: the difference of $n$ and $6$ is $−10 .$

Solution

	the difference of $n$ and $6$ is $−10$
Translate.	$n - 6 = −10$
Add $6$ to each side.	$n - 6 + 6 = −10 + 6$
Simplify.	$n = −4$

Check the answer by substituting it into the original equation.

$\begin{matrix} n - 6 = −10 \\ −4 - 6 \overset{?}{=} −10 \\ −10 = −10 ✓ \end{matrix}$

Try It 3.133

Translate and solve:

The difference of $p$ and $2$ is $−4$ .

Try It 3.134

Translate and solve:

The difference of $q$ and $7$ is $−3$ .

Example 3.68

Translate and solve: the number $108$ is the product of $−9$ and $y .$

Solution

	the number of $108$ is the product of $−9$ and $y$
Translate.	$108 = −9 y$
Divide by $−9$ .	$\frac{108}{−9} = \frac{−9 y}{−9}$
Simplify.	$−12 = y$

Check the answer by substituting it into the original equation.

$\begin{matrix} 108 = −9 y \\ 108 \overset{?}{=} −9 (−12) \\ 108 = 108 ✓ \end{matrix}$

Try It 3.135

Translate and solve:

The number $132$ is the product of $−12$ and $y$ .

Try It 3.136

Translate and solve:

The number $117$ is the product of $−13$ and $z$ .

Media Access Additional Online Resources

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.

$4 x - 2 = 6$

ⓐ $x = −2$
ⓑ $x = −1$
ⓒ $x = 2$

286.

$4 y - 10 = −14$

ⓐ $y = −6$
ⓑ $y = −1$
ⓒ $y = 1$

287.

$9 a + 27 = −63$

ⓐ $a = 6$
ⓑ $a = −6$
ⓒ $a = −10$

288.

$7 c + 42 = −56$

ⓐ $c = 2$
ⓑ $c = −2$
ⓒ $c = −14$

Solve Equations Using the Addition and Subtraction Properties of Equality

In the following exercises, solve for the unknown.

289.

$n + 12 = 5$

290.

$m + 16 = 2$

291.

$p + 9 = −8$

292.

$q + 5 = −6$

293.

$u - 3 = −7$

294.

$v - 7 = −8$

295.

$h - 10 = −4$

296.

$k - 9 = −5$

297.

$x + (−2) = −18$

298.

$y + (−3) = −10$

299.

$r - (−5) = −9$

300.

$s - (−2) = −11$

Model the Division Property of Equality

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

301.

302.

303.

304.

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.

$5 x = 45$

306.

$4 p = 64$

307.

$−7 c = 56$

308.

$−9 x = 54$

309.

$−14 p = −42$

310.

$−8 m = −40$

311.

$−120 = 10 q$

312.

$−75 = 15 y$

313.

$24 x = 480$

314.

$18 n = 540$

315.

$−3 z = 0$

316.

$4 u = 0$

Translate to an Equation and Solve

In the following exercises, translate and solve.

317.

Four more than $n$ is equal to 1.

318.

Nine more than $m$ is equal to 5.

319.

The sum of eight and $p$ is $−3$ .

320.

The sum of two and $q$ is $−7$ .

321.

The difference of $a$ and three is $−14$ .

322.

The difference of $b$ and $5$ is $−2$ .

323.

The number −42 is the product of −7 and $x$ .

324.

The number −54 is the product of −9 and $y$ .

325.

The product of $f$ and −15 is 75.

326.

The product of $g$ and −18 is 36.

327.

−6 plus $c$ is equal to 4.

328.

−2 plus $d$ is equal to 1.

329.

Nine less than $n$ is −4.

330.

Thirteen less than $n$ is $−10$ .

Mixed Practice

In the following exercises, solve.

331.

ⓐ $x + 2 = 10$
ⓑ $2 x = 10$

332.

ⓐ $y + 6 = 12$
ⓑ $6 y = 12$

333.

ⓐ $−3 p = 27$
ⓑ $p - 3 = 27$

334.

ⓐ $−2 q = 34$
ⓑ $q - 2 = 34$

335.

$a - 4 = 16$

336.

$b - 1 = 11$

337.

$−8 m = −56$

338.

$−6 n = −48$

339.

$−39 = u + 13$

340.

$−100 = v + 25$

341.

$11 r = −99$

342.

$15 s = −300$

343.

$100 = 20 d$

344.

$250 = 25 n$

345.

$−49 = x - 7$

346.

$64 = y - 4$

Everyday Math

347.

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

348.

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

Writing Exercises

349.

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

350.

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

351.

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.

352.

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

Self Check

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

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