### Learning Objectives

- 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

Before you get started, take this readiness quiz.

Simplify: $4y-9+9.$

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

### Example 2.27

Solve: $7x+8=\mathrm{-13}.$

Solve: $3x+4=\mathrm{-8}.$

Solve: $5a+3=\mathrm{-37}.$

### Example 2.28

Solve: $8y-9=31.$

Solve: $5y-9=16.$

Solve: $3m-8=19.$

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

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

Solve: $\mathrm{-6}c=\mathrm{-7}c-1.$

### Example 2.30

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

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

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

### Example 2.31

Solve: $12x=\text{\u2212}x+26.$

Solve: $12j=\mathrm{-4}j+32.$

Solve: $8h=\mathrm{-4}h+12.$

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

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

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

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.

- 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: $8n-4=\mathrm{-2}n+6.$

Solve: $8q-5=\mathrm{-4}q+7.$

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

### Example 2.34

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

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

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

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

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

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

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

### Example 2.36

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

Solve: $2.8x+12=\mathrm{-1.4}x-9.$

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

### 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.

$9x-3=60$

$14w+5=117$

$2a+8=\mathrm{-28}$

$\mathrm{-62}=8n-6$

$35=\mathrm{-13}y+9$

$\mathrm{-12}p-9=9$

**Solve Equations with Variables on Both Sides**

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

$19z=18z-7$

$9x+36=15x$

$c=\mathrm{-3}c-20$

$9q=44-2q$

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

$\mathrm{-18}a-8=\mathrm{-22}a$

**Solve Equations with Variables and Constants on Both Sides**

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

$8x-15=7x+3$

$26+13d=14d+11$

$2p-1=4p-33$

$4a+5=\text{\u2212}a-40$

$5y-30=\mathrm{-5}y+30$

$7s+12=5+4s$

$2z-6=23-z$

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

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

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

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

$14n+8.25=9n+19.60$

$2.4w-100=0.8w+28$

$5.6r+13.1=3.5r+57.2$

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

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

#### 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 $10x+14=\mathrm{-2}x+38$ explaining all the steps of your solution as in the examples in this section.

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?

#### 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?