Contents

Solve Equations with Variables and Constants on Both Sides

Learning Objectives

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

Solve an equation with constants on both sides
Solve an equation with variables on both sides
Solve an equation with variables and constants on both sides

Be Prepared 2.3

Before you get started, take this readiness quiz.

Simplify: $4 y - 9 + 9 . 4 y - 9 + 9 .$
If you missed this problem, review Example 1.129.

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 $a x = b . a x = b .$ We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Example 2.27

Solve: $7 x + 8 = −13 . 7 x + 8 = −13 .$

Show solution

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.

Since the left side is the “ $x x$ ”, 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 $x = −3 x = −3$ .

Try It 2.53

Solve: $3 x + 4 = −8 . 3 x + 4 = −8 .$

Try It 2.54

Solve: $5 a + 3 = −37 . 5 a + 3 = −37 .$

Example 2.28

Solve: $8 y - 9 = 31 . 8 y - 9 = 31 .$

Show solution

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 $8 y 8 y$ , 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 $y = 5 y = 5$ .

Try It 2.55

Solve: $5 y - 9 = 16 . 5 y - 9 = 16 .$

Try It 2.56

Solve: $3 m - 8 = 19 . 3 m - 8 = 19 .$

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.

Example 2.29

Solve: $9 x = 8 x - 6 . 9 x = 8 x - 6 .$

Show solution

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 $x x$ ’s on the right, so subtract the $8 x 8 x$ 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 $x = −6 x = −6$ .

Try It 2.57

Solve: $6 n = 5 n - 10 . 6 n = 5 n - 10 .$

Try It 2.58

Solve: $−6 c = −7 c - 1 . −6 c = −7 c - 1 .$

Example 2.30

Solve: $5 y - 9 = 8 y . 5 y - 9 = 8 y .$

Show solution

Solution

The only constant is on the left and the $y y$ ’s are on both sides. Let’s leave the constant on the left and get the variables to the right.


Subtract $5 y 5 y$ 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 $y = −3 . y = −3 .$

Try It 2.59

Solve: $3 p - 14 = 5 p . 3 p - 14 = 5 p .$

Try It 2.60

Solve: $8 m + 9 = 5 m . 8 m + 9 = 5 m .$

Example 2.31

Solve: $12 x = − x + 26 . 12 x = − x + 26 .$

Show solution

Solution

The only constant is on the right, so let the left side be the “variable” side.


Remove the $− x − x$ from the right side by adding $x x$ to both sides.
Simplify.
All the $x x$ ’s are on the left and the constants are on the right. Divide both sides by 13.
Simplify.

Try It 2.61

Solve: $12 j = −4 j + 32 . 12 j = −4 j + 32 .$

Try It 2.62

Solve: $8 h = −4 h + 12 . 8 h = −4 h + 12 .$

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.

Example 2.32

How to Solve Equations with Variables and Constants on Both Sides

Solve: $7 x + 5 = 6 x + 2 . 7 x + 5 = 6 x + 2 .$

Show solution

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 math. On the top row of the table, the first cell on the left reads: “Step 1. Choose which side will the “variable” side—the other side will be the “constant” side.” The text in the second cell reads: “The variable terms are 7 x and 6 x. Since 7 is greater than 6, we will make the left side the “x” side and so the right side will be the “constant” side.” The third cell contains the equation 7 x plus 5 equals 6 x plus 2, and the left side of the equation is labeled “variable” written in red, and the right side of the equation is labeled “constant” written in red.

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.

Try It 2.63

Solve: $12 x + 8 = 6 x + 2 . 12 x + 8 = 6 x + 2 .$

Try It 2.64

Solve: $9 y + 4 = 7 y + 12 . 9 y + 4 = 7 y + 12 .$

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.

How To

Beginning Strategy for Solving Equations with Variables and Constants on Both Sides of the Equation.

Step 1. Choose which side will be the “variable” side—the other side will be the “constant” side.
Step 2. Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
Step 3. Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
Step 4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
Step 5. Check the solution by substituting it into the original equation.

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.

Example 2.33

Solve: $8 n - 4 = −2 n + 6 . 8 n - 4 = −2 n + 6 .$

Show solution

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since $8 > −2 8 > −2$ , make the left side the “variable” side.
We don’t want variable terms on the right side—add $2 n 2 n$ 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 $44$ to both sides.
Simplify.
The variable term is on the left and the constant term is on the right. To get the coefficient of $n n$ to be one, divide both sides by 10.
Simplify.
Check:
Let $n = 1 n = 1$ .

Try It 2.65

Solve: $8 q - 5 = −4 q + 7 . 8 q - 5 = −4 q + 7 .$

Try It 2.66

Solve: $7 n - 3 = n + 3 . 7 n - 3 = n + 3 .$

Example 2.34

Solve: $7 a - 3 = 13 a + 7 . 7 a - 3 = 13 a + 7 .$

Show solution

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since $13 > 7 13 > 7$ , make the right side the “variable” side and the left side the “constant” side.


Subtract $7 a 7 a$ from both sides to remove the variable term from the left.
Combine like terms.
Subtract $77$ from both sides to remove the constant from the right.
Simplify.
Divide both sides by $66$ to make $11$ the coefficient of $a a$ .
Simplify.
Check:
Let $a = - \frac{5}{3} a = - \frac{5}{3}$ .

Try It 2.67

Solve: $2 a - 2 = 6 a + 18 . 2 a - 2 = 6 a + 18 .$

Try It 2.68

Solve: $4 k - 1 = 7 k + 17 . 4 k - 1 = 7 k + 17 .$

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!

Example 2.35

Solve: $\frac{5}{4} x + 6 = \frac{1}{4} x - 2 . \frac{5}{4} x + 6 = \frac{1}{4} x - 2 .$

Show solution

Solution

Since $\frac{5}{4} > \frac{1}{4} \frac{5}{4} > \frac{1}{4}$ , make the left side the “variable” side and the right side the “constant” side.


Subtract $\frac{1}{4} x \frac{1}{4} x$ from both sides.
Combine like terms.
Subtract $66$ from both sides.
Simplify.
$\begin{matrix} Check: & \frac{5}{4} x + 6 & = & \frac{1}{4} x - 2 \\ Let x = −8 . & \frac{5}{4} (−8) + 6 & \overset{?}{=} & \frac{1}{4} (−8) - 2 \\ −10 + 6 & \overset{?}{=} & −2 - 2 \\ −4 & = & −4 ✓ \end{matrix} \begin{matrix} Check: & \frac{5}{4} x + 6 = \frac{1}{4} x - 2 \\ Let & \frac{5}{4} (−8) + 6 \overset{?}{=} \frac{1}{4} (−8) - 2 \\ −10 + 6 \overset{?}{=} −2 - 2 \\ −4 = −4 ✓ \end{matrix}$

Try It 2.69

Solve: $\frac{7}{8} x - 12 = - \frac{1}{8} x - 2 . \frac{7}{8} x - 12 = - \frac{1}{8} x - 2 .$

Try It 2.70

Solve: $\frac{7}{6} y + 11 = \frac{1}{6} y + 8 . \frac{7}{6} y + 11 = \frac{1}{6} y + 8 .$

We will use the same strategy to find the solution for an equation with decimals.

Example 2.36

Solve: $7.8 x + 4 = 5.4 x - 8 . 7.8 x + 4 = 5.4 x - 8 .$

Show solution

Solution

Since $7.8 > 5.4 7.8 > 5.4$ , make the left side the “variable” side and the right side the “constant” side.


Subtract $5.4 x 5.4 x$ from both sides.
Combine like terms.
Subtract $44$ from both sides.
Simplify.
Use the Division Propery of Equality.
Simplify.
Check:
Let $x = −5 x = −5$ .

Try It 2.71

Solve: $2.8 x + 12 = −1.4 x - 9 . 2.8 x + 12 = −1.4 x - 9 .$

Try It 2.72

Solve: $3.6 y + 8 = 1.2 y - 4 . 3.6 y + 8 = 1.2 y - 4 .$

Practice Makes Perfect

Solve Equations with Constants on Both Sides

In the following exercises, solve the following equations with constants on both sides.

174.

$9 x - 3 = 60 9 x - 3 = 60$

175.

$12 x - 8 = 64 12 x - 8 = 64$

176.

$14 w + 5 = 117 14 w + 5 = 117$

177.

$15 y + 7 = 97 15 y + 7 = 97$

178.

$2 a + 8 = −28 2 a + 8 = −28$

179.

$3 m + 9 = −15 3 m + 9 = −15$

180.

$−62 = 8 n - 6 −62 = 8 n - 6$

181.

$−77 = 9 b - 5 −77 = 9 b - 5$

182.

$35 = −13 y + 9 35 = −13 y + 9$

183.

$60 = −21 x - 24 60 = −21 x - 24$

184.

$−12 p - 9 = 9 −12 p - 9 = 9$

185.

$−14 q - 2 = 16 −14 q - 2 = 16$

Solve Equations with Variables on Both Sides

In the following exercises, solve the following equations with variables on both sides.

186.

$19 z = 18 z - 7 19 z = 18 z - 7$

187.

$21 k = 20 k - 11 21 k = 20 k - 11$

188.

$9 x + 36 = 15 x 9 x + 36 = 15 x$

189.

$8 x + 27 = 11 x 8 x + 27 = 11 x$

190.

$c = −3 c - 20 c = −3 c - 20$

191.

$b = −4 b - 15 b = −4 b - 15$

192.

$9 q = 44 - 2 q 9 q = 44 - 2 q$

193.

$5 z = 39 - 8 z 5 z = 39 - 8 z$

194.

$6 y + \frac{1}{2} = 5 y 6 y + \frac{1}{2} = 5 y$

195.

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

196.

$−18 a - 8 = −22 a −18 a - 8 = −22 a$

197.

$−11 r - 8 = −7 r −11 r - 8 = −7 r$

Solve Equations with Variables and Constants on Both Sides

In the following exercises, solve the following equations with variables and constants on both sides.

198.

$8 x - 15 = 7 x + 3 8 x - 15 = 7 x + 3$

199.

$6 x - 17 = 5 x + 2 6 x - 17 = 5 x + 2$

200.

$26 + 13 d = 14 d + 11 26 + 13 d = 14 d + 11$

201.

$21 + 18 f = 19 f + 14 21 + 18 f = 19 f + 14$

202.

$2 p - 1 = 4 p - 33 2 p - 1 = 4 p - 33$

203.

$12 q - 5 = 9 q - 20 12 q - 5 = 9 q - 20$

204.

$4 a + 5 = − a - 40 4 a + 5 = − a - 40$

205.

$8 c + 7 = −3 c - 37 8 c + 7 = −3 c - 37$

206.

$5 y - 30 = −5 y + 30 5 y - 30 = −5 y + 30$

207.

$7 x - 17 = −8 x + 13 7 x - 17 = −8 x + 13$

208.

$7 s + 12 = 5 + 4 s 7 s + 12 = 5 + 4 s$

209.

$9 p + 14 = 6 + 4 p 9 p + 14 = 6 + 4 p$

210.

$2 z - 6 = 23 - z 2 z - 6 = 23 - z$

211.

$3 y - 4 = 12 - y 3 y - 4 = 12 - y$

212.

$\frac{5}{3} c - 3 = \frac{2}{3} c - 16 \frac{5}{3} c - 3 = \frac{2}{3} c - 16$

213.

$\frac{7}{4} m - 7 = \frac{3}{4} m - 13 \frac{7}{4} m - 7 = \frac{3}{4} m - 13$

214.

$8 - \frac{2}{5} q = \frac{3}{5} q + 6 8 - \frac{2}{5} q = \frac{3}{5} q + 6$

215.

$11 - \frac{1}{5} a = \frac{4}{5} a + 4 11 - \frac{1}{5} a = \frac{4}{5} a + 4$

216.

$\frac{4}{3} n + 9 = \frac{1}{3} n - 9 \frac{4}{3} n + 9 = \frac{1}{3} n - 9$

217.

$\frac{5}{4} a + 15 = \frac{3}{4} a - 5 \frac{5}{4} a + 15 = \frac{3}{4} a - 5$

218.

$\frac{1}{4} y + 7 = \frac{3}{4} y - 3 \frac{1}{4} y + 7 = \frac{3}{4} y - 3$

219.

$\frac{3}{5} p + 2 = \frac{4}{5} p - 1 \frac{3}{5} p + 2 = \frac{4}{5} p - 1$

220.

$14 n + 8.25 = 9 n + 19.60 14 n + 8.25 = 9 n + 19.60$

221.

$13 z + 6.45 = 8 z + 23.75 13 z + 6.45 = 8 z + 23.75$

222.

$2.4 w - 100 = 0.8 w + 28 2.4 w - 100 = 0.8 w + 28$

223.

$2.7 w - 80 = 1.2 w + 10 2.7 w - 80 = 1.2 w + 10$

224.

$5.6 r + 13.1 = 3.5 r + 57.2 5.6 r + 13.1 = 3.5 r + 57.2$

225.

$6.6 x - 18.9 = 3.4 x + 54.7 6.6 x - 18.9 = 3.4 x + 54.7$

Everyday Math

226.

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 $6 s + 27 s - 45 = 1506 6 s + 27 s - 45 = 1506$ .

227.

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 $150 = 2 w + 30 + 2 w 150 = 2 w + 30 + 2 w$ .

Writing Exercises

228.

Solve the equation $\frac{6}{5} y - 8 = \frac{1}{5} y + 7 \frac{6}{5} y - 8 = \frac{1}{5} y + 7$ explaining all the steps of your solution as in the examples in this section.

229.

Solve the equation $10 x + 14 = −2 x + 38 10 x + 14 = −2 x + 38$ explaining all the steps of your solution as in the examples in this section.

230.

When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient of $x x$ to be the “variable” side?

231.

Is $x = −2 x = −2$ a solution to the equation $5 - 2 x = −4 x + 1 5 - 2 x = −4 x + 1$ ? How do you know?

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?