# 3.2 Solve Equations using the Division and Multiplication Properties of Equality

## Learning Objectives

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

• Solve equations using the Division and Multiplication Properties of Equality
• Solve equations that require simplification
• Translate to an equation and solve
• Translate and solve applications

### Try It

Before you get started, take this readiness quiz:

1) Simplify: $-7\cdot \frac{1}{-7}$.
2) Evaluate $9x+2$ when $x=-3$.

## Solve Equations Using the Division and Multiplication Properties of Equality

You may have noticed that 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 on the side of the equation with the variable. Now we will see how to solve equations that have a variable multiplied by a constant and so will require the division to isolate the variable.

Let’s look at our puzzle again with the envelopes and counters in Figure 3.2.1.

The illustration shows a model of an equation with one variable multiplied by a constant. On the left side of the workspace are two instances of the unknown (envelope), while on the right side of the workspace are six counters.

In the illustration, 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?

How do we determine the number? We have to separate the counters on the right side into two groups of the same size to correspond with the two envelopes on the left side. The 6 counters divided into 2 equal groups gives 3 counters in each group (since $6\div2=3$).

What equation models the situation shown in Figure 3.2.2? There are two envelopes, and each contains $x$ counters. Together, the two envelopes must contain a total of 6 counters.

The illustration shows a model of the equation $2x=6$.

 $2x=6$ If we divide both sides of the equation by 2, as we did with the envelopes and counters, $\frac{2x}2=\frac62$ we get: $x=3$

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 do equal six!

This example leads to the Division Property of Equality.

### The Division Property of Equality

For any numbers $a$, $b$, and $c$, if $a=b$, and $c\neq 0$,

If $a=b$,
then $\frac{a}{c}=\frac{b}{c}$

When you divide both sides of an equation by any non-zero number, you still have equality.

The goal in solving an equation is to ‘undo’ the operation on the variable. In the next example, the variable is multiplied by 5, so we will divide both sides by 5 to ‘undo’ the multiplication.

### Example 3.2.1

Solve: $5x=-27$.

Solution

Step 1: To isolate $x$, “undo” the multiplication by 5.

$5x=-27$

Step 2: Divide to ‘undo’ the multiplication.

\begin{align*}&\;&\frac{5x}{\color{red}{5}}&=-\frac{27}{\color{red}{5}}\\ &\text{Simplify.}\;&x&=-\frac{27}{5}\\ \end{align*}

Step 3: Check:

$5x=-27$

Step 5: Substitute $-\frac{27}{5}$ for $x$.

$\begin{eqnarray*}5\left({\color{red}{-}}{\color{red}{\frac{27}{5}}}\right)&\overset?=-27\\-27&=-27\checkmark\end{eqnarray*}$

Since this is a true statement, $x=-\frac{27}{5}$ is the solution to $5x=-27$.

### Try It

3) Solve: $3y=-41$.

Solution

$y=\frac{-41}{3}$

4) Solve: $4z=-55$.

Solution

$z=-\frac{55}{4}$

Consider the equation $\frac{x}{4}=3$. We want to know what number divided by 4 gives 3. So to “undo” the division, we will need to multiply by 4. The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.

### The Multiplication Property of Equality

For any numbers $a$, $b$, and $c$,

If        $a=b$,
Then $ac=bc$

If you multiply both sides of an equation by the same number, you still have equality.

### Example 3.2.2

Solve: $\frac{y}{-7}=-14$.

Solution

Here $y$ is divided by $-7$. We must multiply by $-7$ to isolate $y$.

Step 1: Multiply both sides by $-7$.

\begin{align*} &\;&\frac{5x}{\color{red}{5}}&=-\frac{27}{\color{red}{5}}\\ &\text{Simplify.}\;&x&=-\frac{27}{5}\\ \end{align*}

Step 2: Check: $\frac{y}{-7}=-14$

Step 3: Substitute $y=98$.

\begin{align*}\frac{\color{red}{98}}{-7}&\overset?=-14\\-14&=-14\checkmark\end{align*}

### Try It

5) Solve: $\frac{a}{-7}=-42$.

Solution

$a=294$

6) Solve: $\frac{b}{-6}=-24$.

Solution

$b=144$

### Example 3.2.3

Solve: $−n=9$.

Solution

Step 1: Remember $-n$ is equivalent to $-1n$.

$-1n=9$

Step 2: Divide both sides by $-1$.

\begin{align*}&\;&\frac{-1n}{{\color{red}{-}}{\color{red}{1}}}&=\frac9{{\color{red}{-}}{\color{red}{1}}}\\&\text{Divide}\;&n&=-9 \end{align*}

Notice that there are two other ways to solve $-n=9$. We can also solve this equation by multiplying both sides by $-1$ and also by taking the opposite of both sides.

Step 3: Check:

$-n=9$

Step 4: Substitute $n=-9$.

\begin{align*}&\;&-\left(-{\color{red}{9}}\right)&\overset?=9\\&\text{Simplify}\;&9&=9\checkmark\end{align*}

### Try It

7) Solve: $−k=8$.

Solution

$k=-8$

8) Solve: $-g=3$.

Solution

$g=-3$

### Example 3.2.4

Solve: $\frac{3}{4}x=12$.

Solution

Since the product of a number and its reciprocal is $1$, our strategy will be to isolate $x$ by multiplying by the reciprocal of $\frac{3}{4}$.

Step 1: Multiply by the reciprocal of $\frac{3}{4}$.

\begin{align*}&\;&{\color{red}{\frac43}}\cdot\frac34x&={\color{red}{\frac43}}\cdot12\\ &\text{Reciprocals multiply to 1}\;&1x&=\frac{4}{3}\cdot\frac{12}{1}\\ &\text{Mulitply.}\;&x&=16 \end{align*}

Notice that we could have divided both sides of the equation $\frac{3}{4}x=12$ by $\frac{3}{4}$ to isolate $x$. While this would work, most people would find multiplying by the reciprocal easier.

Step 2: Check:

Step 5: Substitute $x=16$.

\begin{align*}\frac{3}{4}x&=12\\\frac34\cdot{\color{red}{16}}&\overset?=12\\12&=12\checkmark\end{align*}

### Try It

9) Solve: $\frac{2}{5}n=14$.

Solution

$n=35$

10) Solve: $\frac{5}{6}y=15$.

Solution

$y=18$

In the next example, all the variable terms are on the right side of the equation. As always, our goal in solving the equation is to isolate the variable.

### Example 3.2.5

Solve: $\frac{8}{15}=-\frac{4}{5}x$.

Solution

Step 1: Multiply by the reciprocal of $-\frac{4}{5}$.

\begin{align*}&\;&\left({\color{red}{-}}{\color{red}{\frac{5}{4}}}\right)\left(\frac8{15}\right)&=\left({\color{red}{-}}{\color{red}{\frac{5}{4}}}\right)\left(-\frac45x\right)\\ &\text{Reciprocals multiply to 1}\;&-\frac{\cancel5\cdot\cancel4\cdot2}{\cancel4\cdot3\cdot\cancel5}&=1x\\ &\text{Mulitply.}\;&-\frac{2}{3}&=x \end{align*}

Step 2: Let $x=-\frac{2}{3}$.

\begin{align*}\frac{8}{15}&=-\frac{4}{5}x\\\frac8{15}&=-\frac45\left({\color{red}{-}}{\color{red}{\frac23}}\right)\\\frac{8}{15}&=\frac{8}{15}\checkmark\end{align*}

### Try It

11) Solve: $\frac{9}{25}=-\frac{4}{5}z$.

Solution

$z=-\frac{9}{5}$

12) Solve: $\frac{5}{6}=-\frac{8}{3}r$

Solution

$r=-\frac{5}{16}$

## Solve Equations That Require Simplification

Many equations start out more complicated than the ones we have been working with.

With these more complicated equations, the first step is to simplify both sides of the equation as much as possible. This usually involves combining like terms or using the distributive property.

### Example 3.2.6

Solve: $14-23=12y-4y-5y$.

Solution

Begin by simplifying each side of the equation.

Step 1: Simplify each side.

$-9=3y$

Step 2: Divide both sides by $3$ to isolate $y$.

\begin{align*}&\;&\frac{-9}{3}&=\frac{3y}{3}\\ &\text{Divide.}\;&-3&=y\end{align*}

Step 3: Check:

$14-23\overset?=12y-4y-5y$

Step 4: Substitute $y=-3$.

\begin{align*}14-23&\overset?=12\left({\color{red}{-}}{\color{red}{3}}\right)-4\left({\color{red}{-}}{\color{red}{3}}\right)-5\left({\color{red}{-}}{\color{red}{3}}\right)\\14-23&\overset?=-36+12+15\\-9&=-9\checkmark \end{align*}

### Try It

13) Solve: $18-27=15c-9c-3c$.

Solution

$c=-3$

14) Solve: $18-22=12x-x-4x$.

Solution

$x=-\frac{4}{7}$

### Example 3.2.7

Solve: $-4(a-3)-7=25$.

Solution

Here we will simplify each side of the equation by using the distributive property first.

Step 1: Distribute.

\begin{align*}&\;&-4a+12-7&=25\\ &\text{Simplify.}\;&-4a+5&=25\\ &\text{Simplify.}\;&-4a&=20 \end{align*}

Step 2: Divide both sides by $-4$ to isolate $a$.

\begin{align*}&\;&\left(\frac{-4a}{{\color{red}{-}}{\color{red}{4}}}\right)&=\left(\frac{20}{{\color{red}{-}}{\color{red}{4}}}\right)\\ &\text{Divide.}\;&a&=-5\end{align*}

Step 3: Check:

$-4(a-3)-7=25$

Step 4: Substitute $a=-5$.

\begin{align*}-4\left({\color{red}{-}}{\color{red}{5}}-3\right)-7&\overset?=25\\-4\left(-8\right)-7&\overset?=25\\32-7&\overset?=25\\25&=25\checkmark\end{align*}

### Try It

15) Solve: $-4(q-2)-8=24$.

Solution

$q=-6$

16) Solve: $-6(r-2)-12=30$.

Solution

$r=-5$

Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.

### Properties of Equality

 Subtraction Property of Equality for any real numbers $a$, $b$ and $c$, If $a=b$, then $a-c=b-c$. Addition Property of Equality for any real numbers $a$, $b$ and $c$, If $a=b$, then $a+c = b+c$. Division Property of Equality for any real numbers $a$, $b$ and $c$, and $c\neq0$ If $a=b$, then $\frac{a}{c}=\frac{b}{c}$. Multiplication Property of Equality for any real numbers $a$, $b$ and $c$, If $a=b$, then $a\times c = b\times c$.

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

## Translate to an Equation and Solve

In the next few examples, we will translate sentences into equations and then solve the equations. You might want to review the translation table in the previous chapter.

### Example 3.2.8

Translate and solve: The number $143$ is the product of $-11$ and $y$.

Solution

Begin by translating the sentence into an equation.

Step 1: Translate.

$\underbrace{The\;number\;143}_{143}\;\underbrace{is}_=\;\underbrace{the\;product\;of\;-11\;and\;y.}_{-11y}$

Step 2: Divide by $-11$.

$\frac{143}{{\color{red}{-}}{\color{red}{11}}}=\frac{-11y}{{\color{red}{-}}{\color{red}{11}}}$

Step 3: Simplify.

$-13=y$

Step 4: Check:

\begin{align*}143&=-11y\\143&\overset?=-11(-13)\\143&=143\checkmark\end{align*}

### Try It

17) Translate and solve: The number $132$ is the product of $−12$ and $y$.

Solution

$132=-12y$;        $y=-11$

18) Translate and solve: The number $117$ is the product of $−13$ and $z$.

Solution

$117=-13z$;        $z=-9$

### Example 3.2.9

Translate and solve: $n$ divided by $8$ is $-32$.

Solution

Step 1: Begin by translating the sentence into an equation.

$\underbrace{n\;divided\;by\;8}_\frac n8\;\underbrace{is}_=\;\underbrace{-32}_{-32}.$

Step 2: Multiple both sides by $8$.

\begin{align*}&\;&{\color{red}{8}}\cdot\frac n8&={\color{red}{8}}\left(-32\right)\\ &\text{Simplify.}\;&n&=-256\\ \end{align*}

Step 3: Check:

Is $n$ divided by $8$ equal to $−32$?

Step 4: Let $n=-256$.

\begin{align*}\frac{-256}{8}&\overset?=-32\\-31&=-32\checkmark\end{align*}

### Try It

19) Translate and solve: $n$ divided by $7$ is equal to $-21$.

Solution

$\frac{n}{7}=-21$;         $n=-147$

20) Translate and solve: $n$ divided by $8$ is equal to $-56$.

Solution

$\frac{n}{8}=-56$;          $n=-448$

### Example 3.2.10

Translate and solve: The quotient of $y$ and $-4$ is $68$.

Solution

Begin by translating the sentence into an equation.

Step 1: Translate.

$\underbrace{The\;quotient\;of\;y\;and\;-4}_\frac y{-4}\;\underbrace{is}_=\;\underbrace{68.}_{68}$

Step 2: Multiply both sides by $-4$.

\begin{align*} &\;&{\color{red}{-}}{\color{red}{4}}\left(\frac y{-4}\right)&={\color{red}{-}}{\color{red}{4}}\left(68\right)\\ &\text{Simplify.}\;&y&=-272\\ \end{align*}

Step 3: Check:

Is the quotient of $y$ and $-4$ equal to $68$?

Step 4: Let $y=-272$.

\begin{align*}\frac{-272}{-4}&\overset?=68\\68&=68\checkmark\end{align*}

### Try It

21) Translate and solve: The quotient of $q$ and $-8$ is $72$.

Solution

$\frac{q}{-8}=72$;          $q=-576$

22) Translate and solve: The quotient of $p$ and $-9$ is $81$.

Solution

$\frac{p}{-9}=81$;         $p=-729$

### Example 3.2.11

Translate and solve: Three-fourths of $p$ is $18$.

Solution

Begin by translating the sentence into an equation. Remember, “of” translates into multiplication.

Step 1: Translate.

$\underbrace{Three-fourths\;of\;p}_{\frac34p}\;\underbrace{is}_=\;\underbrace{18.}_{18}$

Step 2: Multiply both sides by $\frac{4}{3}$.

\begin{align*} &\;&{\color{red}{\frac43}}\cdot\frac34p&={\color{red}{\frac43}}\cdot18\\ &\text{Simplify.}\;&p&=24\\ \end{align*}

Step 3: Check:

Is three-fourths of $p$ equal to $18$?

Step 5: Let $p=24$.

Is three-fourths of $24$ equal to $18$?

\begin{align*}\frac{3}{4}\times24&\overset?=18\\18&=18\checkmark\end{align*}

### Try It

23) Translate and solve: Two-fifths of $f$ is $16$.

Solution

$\frac{2}{5}=16$;  $f=40$

24) Translate and solve: Three-fourths of $f$ is $21$.

Solution

$\frac{3}{4}=21$; $f=28$

### Example 3.2.12

Translate and solve: The sum of three-eighths and $x$ is one-half.

Solution

Begin by translating the sentence into an equation.

Step 1: Translate.

$\underbrace{The\;sum\;of\;three-eighths\;and\;x}_{\frac38+x}\;\underbrace{is}_=\;\underbrace{\frac12}_\frac12$

Step 2: Subtract $\frac{3}{8}$ from each side.

\begin{align*}&\;&\frac38-{\color{red}{\frac38}}+x&=\frac12-{\color{red}{\frac38}}\\ &\text{Simplify and rewrite fractions with common denominators.}\;&x&=\frac{4}{8}-\frac{3}{8}\\ &\text{Simplify.}\;&x&=\frac18 \end{align*}

Step 3: Check:

Is the sum of three-eighths and $x$ equal to one-half?

Step 4: Let $x=\frac{1}{8}$

Is the sum of three-eighths and one-eighth equal to one-half?

\begin{align*}&\;&\frac{3}{8}+\frac{1}{8}&\overset?=\frac{1}{2}\\ &\text{Simplify.}\;&\frac{4}{8}&\overset?=\frac{1}{2}\\ &\text{Simplify.}\;&\frac{1}{2}&=\frac{1}{2}\checkmark \end{align*}

### Try It

25) Translate and solve: The sum of five-eighths and $x$ is one-fourth.

Solution

$\frac{5}{8}+x=\frac{1}{4}$;           $x=-\frac{3}{8}$

26) Translate and solve: The sum of three-fourths and $x$ is five-sixths.

Solution

$\frac{3}{4}+x=\frac{5}{6}$;           $x=\frac1{12}$

## Translate and Solve Applications

To solve applications using the Division and Multiplication Properties of Equality, we will follow the same steps we used in the last section. We will restate the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve.

### Example 3.2.13

Denae bought $6$ pounds of grapes for $\10.74$. What was the cost of one pound of grapes?

Solution

Step 1: What are you asked to find?

The cost of 1 pound of grapes

Step 2: Assign a variable.

Let $c$ = the cost of one pound.

Step 3: Write a sentence  that gives the information to find it.

The cost of $6$ pounds is $\10.74$.

Step 4: Translate into an equation.

$6c=10.74$

Step 5: Solve.

\begin{align*} \frac43(12,000) &= \frac43 \times \frac34p \\ 16,000 &= p \end{align*}

The grapes cost $\1.79$ per pound.

Step 6: Check:

If one pound costs $\1.79$, do $6$ pounds cost $\10.74$?

\begin{align*} 6\left(1.79\right)&\overset?=10.74 \\ 10.74 &= 10.74\checkmark \end{align*}

### Try It

27) Translate and solve:

Arianna bought a 24-pack of water bottles for $\9.36$. What was the cost of one water bottle?

Solution

$\0.39$

28) Translate and solve:

At JB’s Bowling Alley, 6 people can play on one lane for $\34.98$. What is the cost for each person?

Solution

$\5.83$

### Example 3.2.14

Andreas bought a used car for $\12,000$. Because the car was 4-years old, its price was $\frac{3}{4}$ of the original price, when the car was new. What was the original price of the car?

Solution

Step 1: What are you asked to find?

The original price of the car

Step 2: Assign a variable.

$p$ = the original price.

Step 3: Write a sentence that gives the information to find it.

$\12,000$ is $\frac{3}{4}$ of the original price.

Step 4: Translate into an equation.

$12,000=\frac{3}{4}p$

Step 5: Solve.

\begin{align*} \frac43\times \left(12,000\right)&=\frac43\times\frac34p \\ 16,000 &= p \end{align*}

The original cost of the car was $\16,000$.

Step 6: Check:

Is $\frac{3}{4}$ of $\16,000$ equal to $\12,000$?

\begin{align*} \frac34\times \left(16,000\right)&\overset?=12,000 \\ 12,000 &= 12,000\checkmark \end{align*}

### Try It

29) Translate and solve:

The annual property tax on the Mehta’s house is $\1,800$, calculated as $\frac{15}{1,000}$ of the assessed value of the house. What is the assessed value of the Mehta’s house?

Solution

$\120,000$

30) Translate and solve:

Stella planted 14 flats of flowers in $\frac{2}{3}$ of her garden. How many flats of flowers would she need to fill the whole garden?

Solution

21 flats

### Key Concepts

• The Division Property of Equality—For any numbers $a$, $b$, and $c$, and $c\neq 0$, if $a=b$, then $\frac{a}{c}=\frac{b}{c}$.
When you divide both sides of an equation by any non-zero number, you still have equality.
• The Multiplication Property of Equality—For any numbers $a$, $b$, and $c$, if $a=b$, then $ac=bc$.
If you multiply both sides of an equation by the same number, you still have equality.

### 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?