It’s Teacher’s Planning Day at Jerry’s school, so Jerry and four of his friends have decided to spend the day at the local amusement park. Jerry and his friends rarely ride any of the attractions. They mostly spend their time playing video games in the game center. Each of the boys needs 15 tokens to play all of their favorite games. They nominate Jerry to stand in line at the token booth to purchase their tokens. As he waits in line, Jerry tries to figure out how many tokens he needs to get so he and his four friends will each have 15 tokens. How can Jerry use a single variable division equation to solve this problem?

In this concept, you will learn how to solve single-variable division equations.

### Solving Single-Variable Division Equations

Division equations require you to use multiplication to solve for the variable. Remember that multiplication is the inverse operation for division.

There are two different types of division equations that you will be solving. Let’s look at a problem that is the first type of division equation.

\begin{align*}\frac{x}{3}=12\end{align*}

This type of division problem has a missing numerator. You don’t know the value of the numerator so you use a variable in place of the unknown number.

To figure out the numerator, multiply the denominator with the value on the right side of the equals.

\begin{align*}\begin{array}{rcl}
\frac{x}{3}&=&12\\
x&=&3(12)\\
x&=&36
\end{array}\end{align*}

To check the answer, substitute it back into the original equation for the variable. If one side equals the other side, then your work is accurate and correct.

\begin{align*}\begin{array}{rcl}
\frac{36}{3}&=&12\\
12&=&12
\end{array}\end{align*}

The answer is correct.

Now let’s look at the second type of division equation – one where the denominator is an unknown variable.

\begin{align*}\frac{4}{x}=2\end{align*}

To solve this equation you need to multiply the denominator with the value on the right side of the equals. In this case, the denominator is a variable. Multiply it by two and rewrite the problem.

\begin{align*}\begin{array}{rcl}
\frac{4}{x}&=&2\\
2x&=&4
\end{array}\end{align*}

Now you have a multiplication problem to solve. Solve it by using division. To get the variable alone, divide both sides of the equation by 2.

\begin{align*}\begin{array}{rcl}
\frac{2x}{2}&=&\frac{4}{2}\\
x&=&2
\end{array}
\end{align*}

To check your work, substitute the value of

back into the original problem. \begin{align*}\begin{array}{rcl}
\frac{4}{x}&=&2\\
\frac{4}{2}&=&2\\
2&=&2
\end{array}\end{align*}

It is accurate.

### Examples

#### Example 1

Earlier, you were given a problem about Jerry and his friends at the arcade.

Jerry needs to get enough tokens so that he and his four friends will each have 15 tokens. How can Jerry use a single variable division equation to figure out how many tokens he needs to purchase?

First, mentally write an equation to represent the situation.

This equation means, an unknown number of tokens \begin{align*}(x)\end{align*}

Next, use the inverse of division, which is multiplication, and multiply the denominator (5) and the number (15) on the other side of the equal sign.

That leaves

alone on one side of the equal signs.

Then, check the answer by substituting it back into the original equation for the variable. If one side equals the other side, then your work is accurate and correct.

\begin{align*}\begin{array}{rcl}
\frac{x}{5}&=&15\\
\frac{75}{5}&=&15\\
15&=&15
\end{array}\end{align*}

The answer is correct.

**Solve the equations and write your answer in the form \begin{align*}\text{Variable}=\underline{\;\;\;\;\;\;\;}.\end{align*} Variable=−−−−.**

#### Example 2

\begin{align*}\frac{x}{4}=6\end{align*}

First, use the inverse of division, which is multiplication, and multiply the denominator (4) and the number on the other side of the equal sign, 6.

\begin{align*}4 \times 6=24\end{align*}

That leaves

alone on one side of the equal signs and 24 on the other side.

Next, check the answer by substituting it back into the original equation for the variable. If one side equals the other side, then your work is accurate and correct.

\begin{align*}\begin{array}{rcl}
\frac{x}{4}&=&6\\
\frac{24}{4}&=&6\\
6&=&6
\end{array}\end{align*}

The answer is correct.

#### Example 3

\begin{align*}\frac{x}{5}=7\end{align*}

First, use the inverse of division, which is multiplication, and multiply the denominator (5) and 7.

That leaves

alone on one side of the equal signs and 35 on the other side.

Next, check the answer by substituting it back into the original equation for the variable.

\begin{align*}\begin{array}{rcl}
\frac{x}{5}&=&7\\
\frac{35}{5}&=&7\\
7&=&7
\end{array}\end{align*}

The answer is correct.

#### Example 4

\begin{align*}\frac{x}{2}=3\end{align*}

First, use the inverse of division, which is multiplication, and multiply the denominator (2) and the number (3) on the other side of the equal sign.

That leaves

alone on one side of the equal signs and 6 on the other side.

Next, check the answer by substituting it back into the original equation for the variable. If one side equals the other side, then your work is accurate and correct.

\begin{align*}\begin{array}{rcl}
\frac{x}{2}&=&3\\
\frac{6}{2}&=&3\\
3&=&3
\end{array}\end{align*}

The answer is correct.

#### Example 5

\begin{align*}\frac{12}{x}=6\end{align*}

First, use the inverse of division, which is multiplication, and multiply the denominator

and the number (6) on the other side of the equal sign.

That leaves 12 alone on one side of the equal signs and

on the other side.\begin{align*}12 = 6x\end{align*}

Now you have a multiplication problem to solve. Solve it by using division.

Next, to get the variable alone, divide both sides of the equation by 6.

\begin{align*}\frac{12}{6}=\frac{6x}{6}\end{align*}

The 6’s cancel each other, leaving

.Then, substitute 2 for

in the original problem to check your work.\begin{align*}\begin{array}{rcl}
\frac{12}{x}&=&6\\
\frac{12}{6}&=&6\\
6&=&6
\end{array}\end{align*}

It is accurate.

### Review

Solve the following equations. Write your answer in the form: \begin{align*}\text{Variable}=\underline{\;\;\;\;\;\;\;}.\end{align*}

- \begin{align*}\frac{6}{x}=3\end{align*}
6x=3 - \begin{align*}\frac{x}{9}=9\end{align*}
x9=9 - \begin{align*}\frac{x}{5}=3\end{align*}
x5=3 - \begin{align*}\frac{20}{y}=4\end{align*}
20y=4 - \begin{align*}\frac{x}{2}=18\end{align*}
x2=18 - \begin{align*}\frac{x}{3}=2\end{align*}
x3=2 - \begin{align*}\frac{x}{12}=9\end{align*}
- \begin{align*}\frac{x}{14}=3\end{align*}
- \begin{align*}\frac{x}{11}=9\end{align*}
- \begin{align*}\frac{x}{13}=4\end{align*}
- \begin{align*}\frac{12}{x}=3\end{align*}
- \begin{align*}\frac{44}{x}=4\end{align*}
- \begin{align*}\frac{32}{x}=8\end{align*}
- \begin{align*}\frac{90}{x}=30\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 12.9.