# 7.10: Single Variable Division Equation

**At Grade**Created by: CK-12

**Practice**Single Variable Division Equations

The incoming class at tennis camp is large this year. This is a special program that will have 6 people in each class. However, there is a maximum number of 42 classes. The incoming people need to be divided into groups of 6, so that the number of groups is 42. What is the maximum capacity, \begin{align*}c\end{align*}, of the incoming class? How can you write an equation for \begin{align*}c\end{align*}, and then solve this equation?

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

### Solving Single Variable Division Equations

To solve an equation in which a variable is divided by a number, you use the inverse of division, multiplication, to isolate the variable and solve the equation.

You can multiply both sides of an equation by a number because of the **Multiplication Property of Equality**, which states:

if \begin{align*}a = b\end{align*}, then \begin{align*}a \times c = b \times c\end{align*}.

This means that if you multiply one side of an equation by a number \begin{align*}c\end{align*}, you must multiply the other side of the equation by that same number \begin{align*}c\end{align*}, to keep the values on both sides of the equation equal.

Here is an example.

Solve the equation for \begin{align*}k\end{align*}.

\begin{align*}\frac{k}{-4} = 12\end{align*}

First, use the multiplication property of equality, and multiply both sides of the equation by -4 to isolate the variable \begin{align*}k\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{k}{-4} &=& 12\\ -4 \times \frac{k}{-4} &=& -4 \times 12\\ \frac{-4k}{-4} &=& -48 \end{array}\end{align*}

Next, separate the fraction and simplify.

\begin{align*}\begin{array}{rcl} \frac{-4}{-4} k &=& -48\\ 1k &=& -48\\ k &=& -48 \end{array}\end{align*}

The answer is \begin{align*}k = -48\end{align*}.

Here is another example.

Solve for \begin{align*}n\end{align*} in the equation \begin{align*}\frac{n}{1.5}=10\end{align*}.

First, use the multiplication property of equality to multiply both sides of the equation by 1.5.

\begin{align*}1.5 \times \frac{n}{1.5} = 10 \times 1.5\end{align*} Next, since \begin{align*}1.5 \frac{n}{1.5} = \frac{1.5}{1} \times \frac{n}{1.5}\end{align*}, you can rewrite that multiplication as one fraction.

\begin{align*}\frac{1.5n}{1.5} = 10 \times 1.5\end{align*}

Next, you separate the fraction and simplify.

\begin{align*}\begin{array}{rcl} \frac{1.5}{1.5} n &=& 15\\ 1n &=& 15\\ n &=& 15 \end{array}\end{align*} The answer is \begin{align*}n=15\end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about the tennis camp.

The incoming students need to be divided into groups of 6, but there can only be 42 classes in total. Can you write a division equation, where \begin{align*}c\end{align*}, is the maximum number of people in the incoming class so that there are 6 people in each group, and then solve it?

First, translate the language into an equation. Let \begin{align*}c\end{align*}, be the maximum number of people in the incoming class. This number, divided by 6, should equal 42 classes.

\begin{align*}\frac{c}{6}=42\end{align*}

Next, use the multiplication property of equality and multiply both sides of the equation by 6.

\begin{align*}6 \times \frac{c}{6}=6 \times 42\end{align*}

Then, re-write the multiplication by a fraction and simplify.

\begin{align*}\begin{array}{rcl} \frac{6}{1} \times \frac{c}{6} &=& 252\\ \frac{6}{6} \times \frac{c}{1} &=& 252\\ 1c &=& 252\\ c &=& 252 \end{array}\end{align*}

The answer is that there can be a maximum number of 252 students in the incoming class.

#### Example 2

Three friends evenly split the total cost of the bill for their lunch. The amount each friend paid was $4.25.

- Write a division equation to represent \begin{align*}c\end{align*}, the total cost, in dollars, of the bill for lunch.
- Solve the equation to solve for the total cost of the bill.

Consider part *a* first.

First, rephrase the question to help you solve the problem: The total cost, \begin{align*}c\end{align*}, divided by three equals 4.25, the amount each person paid.

Then, express this as an equation.

\begin{align*}\frac{c}{3}=4.25\end{align*}

Now consider part *b*.

Solve the equation by using the multiplication property of equality. Multiply both sides of the equation by 3.

\begin{align*}\begin{array}{rcl} \frac{c}{3} &=& 4.25\\ 3 \times \frac{c}{3} &=& 3 \times 4.25 \end{array}\end{align*}

Next, rearrange the multiplication of fractions.

\begin{align*}\frac{3}{3} c=12.75\end{align*}

Now, simplify and solve.

\begin{align*}\begin{array}{rcl} 1c &=& 12.75\\ c &=& 12.75 \end{array}\end{align*}

The answer is that the bill was $12.75.

**Solve each equation.**

#### Example 3

\begin{align*}\frac{x}{-2}=5\end{align*}

First, use the multiplication property of equality and multiply both sides of the equation by -2.

\begin{align*}-2 \frac{x}{-2} = -2 \times 5\end{align*}

Next, simplify and solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{-2}{-2} x &=& -10\\ 1x &=& -10\\ x &=& -10 \end{array}\end{align*}

The answer is \begin{align*}x = -10\end{align*}.

#### Example 4

\begin{align*}\frac{y}{5}=6\end{align*}

First, use the multiplication property of equality and multiply both sides of the equation by 5.

\begin{align*}5 \frac{y}{5}=5 \times 6\end{align*}

Next, simplify and solve for \begin{align*}y\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{5}{5} y &=& 30\\ 1y &=& 30\\ y &=& 30 \end{array}\end{align*}

The answer is \begin{align*}y = 30\end{align*}.

#### Example 5

\begin{align*}\frac{b}{-4}=-3\end{align*}

First, use the multiplication property of equality and multiply both sides of the equation by -4.

\begin{align*}-4 \frac{b}{-4}=-4 \times -3\end{align*}

Next, simplify and solve for \begin{align*}b\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{-4}{-4} b &=& 12\\ 1b &=& 12\\ b &=& 12 \end{array}\end{align*}

The answer is \begin{align*}b = 12\end{align*}.

### Review

Solve each single variable division equation for the missing value.

- \begin{align*}\frac{x}{5} = 2\end{align*}
- \begin{align*}\frac{y}{7} = 3\end{align*}
- \begin{align*}\frac{b}{9} = -4\end{align*}
- \begin{align*}\frac{b}{8} = -10\end{align*}
- \begin{align*}\frac{b}{8} = 20\end{align*}
- \begin{align*}\frac{x}{-3} =10\end{align*}
- \begin{align*}\frac{y}{18} = -20\end{align*}
- \begin{align*}\frac{a}{-9} = -9\end{align*}
- \begin{align*}\frac{x}{11} = -12\end{align*}
- \begin{align*}\frac{x}{3} = -3\end{align*}
- \begin{align*}\frac{x}{5} = -8\end{align*}
- \begin{align*}\frac{x}{1.3} = 3\end{align*}
- \begin{align*}\frac{x}{2.4} = 4\end{align*}
- \begin{align*}\frac{x}{6} = 1.2\end{align*}
- \begin{align*}\frac{y}{1.5} = 3\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.10.

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

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