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Elementary Algebra

2.3 Solve Equations with Variables and Constants on Both Sides

Elementary Algebra2.3 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.

  1. Simplify: 4y9+9.4y9+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 ax=b.ax=b. We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Example 2.27

Solve: 7x+8=−13.7x+8=−13.

Try It 2.53

Solve: 3x+4=−8.3x+4=−8.

Try It 2.54

Solve: 5a+3=−37.5a+3=−37.

Example 2.28

Solve: 8y9=31.8y9=31.

Try It 2.55

Solve: 5y9=16.5y9=16.

Try It 2.56

Solve: 3m8=19.3m8=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: 9x=8x6.9x=8x6.

Try It 2.57

Solve: 6n=5n10.6n=5n10.

Try It 2.58

Solve: −6c=−7c1.−6c=−7c1.

Example 2.30

Solve: 5y9=8y.5y9=8y.

Try It 2.59

Solve: 3p14=5p.3p14=5p.

Try It 2.60

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

Example 2.31

Solve: 12x=x+26.12x=x+26.

Try It 2.61

Solve: 12j=−4j+32.12j=−4j+32.

Try It 2.62

Solve: 8h=−4h+12.8h=−4h+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: 7x+5=6x+2.7x+5=6x+2.

Try It 2.63

Solve: 12x+8=6x+2.12x+8=6x+2.

Try It 2.64

Solve: 9y+4=7y+12.9y+4=7y+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.

  1. Step 1. Choose which side will be the “variable” side—the other side will be the “constant” side.
  2. Step 2. Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
  3. Step 3. Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
  4. Step 4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
  5. 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: 8n4=−2n+6.8n4=−2n+6.

Try It 2.65

Solve: 8q5=−4q+7.8q5=−4q+7.

Try It 2.66

Solve: 7n3=n+3.7n3=n+3.

Example 2.34

Solve: 7a3=13a+7.7a3=13a+7.

Try It 2.67

Solve: 2a2=6a+18.2a2=6a+18.

Try It 2.68

Solve: 4k1=7k+17.4k1=7k+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: 54x+6=14x2.54x+6=14x2.

Try It 2.69

Solve: 78x12=18x2.78x12=18x2.

Try It 2.70

Solve: 76y+11=16y+8.76y+11=16y+8.

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

Example 2.36

Solve: 7.8x+4=5.4x8.7.8x+4=5.4x8.

Try It 2.71

Solve: 2.8x+12=−1.4x9.2.8x+12=−1.4x9.

Try It 2.72

Solve: 3.6y+8=1.2y4.3.6y+8=1.2y4.

Section 2.3 Exercises

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 + 1 2 = 5 y 6 y + 1 2 = 5 y

195.

4 x + 3 4 = 3 x 4 x + 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.

5 3 c 3 = 2 3 c 16 5 3 c 3 = 2 3 c 16

213.

7 4 m 7 = 3 4 m 13 7 4 m 7 = 3 4 m 13

214.

8 2 5 q = 3 5 q + 6 8 2 5 q = 3 5 q + 6

215.

11 1 5 a = 4 5 a + 4 11 1 5 a = 4 5 a + 4

216.

4 3 n + 9 = 1 3 n 9 4 3 n + 9 = 1 3 n 9

217.

5 4 a + 15 = 3 4 a 5 5 4 a + 15 = 3 4 a 5

218.

1 4 y + 7 = 3 4 y 3 1 4 y + 7 = 3 4 y 3

219.

3 5 p + 2 = 4 5 p 1 3 5 p + 2 = 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 6s+27s45=15066s+27s45=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=2w+30+2w150=2w+30+2w.

Writing Exercises

228.

Solve the equation 65y8=15y+765y8=15y+7 explaining all the steps of your solution as in the examples in this section.

229.

Solve the equation 10x+14=−2x+3810x+14=−2x+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 xx to be the “variable” side?

231.

Is x=−2x=−2 a solution to the equation 52x=−4x+152x=−4x+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?

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