# 7.10: Single Variable Division Equations

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

**Practice**Single Variable Division Equations

Have you ever been to a theater with limited seating? Well, Marc and Kara had this happen with their grandparents.

Marc and Kara went to see a play with their grandparents. When they arrived at the theater, the manager divided their group and several other people into six smaller groups. Each of these groups was lead to a section of the theater where there were empty seats. Each group had six people in it too.

If this was the division, how many people did the manager divide up to start with?

Write a division equation and solve it to complete this dilemma.

**This Concept is all about single variable division equations. You will know how to do this by the end of the Concept.**

### Guidance

Sometimes, you will see equations that have division in them. Remember, that we can use a fraction bar to show division.

**To solve an equation in which a variable is divided by a number, we can use the inverse of division––multiplication.** We can multiply *both* sides of the equation by that number to solve it.

We must multiply *both* sides of the equation by that number because of the ** Multiplication Property of Equality**, which states:

if \begin{align*}a=b\end{align*} and \begin{align*}c \neq 0\end{align*}, then \begin{align*}a \times c=b \times c\end{align*}.

So, if you multiply one side of an equation by a nonzero 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 equal.

Now let's apply this information.

\begin{align*}k \div (-4)=12\end{align*}.

In the equation, \begin{align*}k\end{align*} is *divided* by -4. So, we can *multiply* both sides of the equation by -4 to solve for \begin{align*}k\end{align*}. You will need to use what you know about multiplying integers to help you solve this problem. It may help to rewrite \begin{align*}k \div (-4)\end{align*} as \begin{align*}\frac{k}{-4}\end{align*}.

\begin{align*}k \div (-4) &= 12\\ \frac{k}{-4} &= 12\\ \frac{k}{-4} \times (-4) &= 12 \times (-4)\\ \frac{k}{-4} \times \frac{-4}{1} &= -48\\ \frac{k}{\cancel{-4}} \times \frac{\cancel{-4}}{1} &= -48\\ \frac{k}{1} &= -48\\ k &= -48\end{align*}

The –4's will cancel each other out when they are divided. Then we multiply.

**The value of \begin{align*}k\end{align*} is –48.**

*Remember the rules for multiplying integers will apply when working with these equations!! Think back and use them as you work.*

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

In the equation, \begin{align*}n\end{align*} is *divided* by 1.5. So, we can *multiply* both sides of the equation by 1.5 to solve for \begin{align*}n\end{align*}.

\begin{align*}\frac{n}{1.5} &= 10\\ \frac{n}{1.5} \times 1.5 &= 10 \times 1.5\\ \frac{n}{1.5} \times \frac{1.5}{1} &= 15\\ \frac{n}{\cancel{1.5}} \times \frac{\cancel{1.5}}{1} &= 15\\ \frac{n}{1} &= 15\\ n &= 15\end{align*}

**The value of \begin{align*}n\end{align*} is 15.**

When an equation has division in it, you can use the ** Multiplication Property of Equality** to solve it, and then the property has been useful. Always remember to think about the

**and associate them with the different properties. This will help you to keep it all straight and not get mixed up.**

*inverse operations*Now it is time for you to practice solving a few of these equations.

Solve each equation for the missing variable.

#### Example A

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

**Solution:\begin{align*}x = -10\end{align*}**

#### Example B

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

**Solution:\begin{align*}y = 30\end{align*}**

#### Example C

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

**Solution:\begin{align*}b = 12\end{align*}**

Here is the original problem once again.

Marc and Kara went to see a play with their grandparents. When they arrived at the theater, the manager divided their group and several other people into six smaller groups. Each of these groups was lead to a section of the theater where there were empty seats. Each group had six people in it too.

If this was the division, how many people did the manager divide up to start with?

Write a division equation and solve it to complete this dilemma.

First, we can write an equation.

Some number of people divided by six is six.

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

Now we can solve it by multiplying.

\begin{align*}6 \times 6 = 36\end{align*}

\begin{align*}x = 36\end{align*}

**This is our answer.**

### Vocabulary

Here are the vocabulary words in this Concept.

- Isolate the Variable
- this means that we want to work to get the variable alone on one side of the equals.

- Inverse Operation
- Opposite operation

- Division Property of Equality
- states that we can solve an equation by multiplying the same value to both sides of the equation.

- Multiplication Property of Equality
- states that we can solve an equation by dividing both sides of the equation by the same value.

### Guided Practice

Here is one for you to try on your own.

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

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

b. Determine the total cost of the bill.

**Answer**

Consider part \begin{align*}a\end{align*} first.

Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. Because the friends split the bill evenly, write a division equation to represent the problem.

\begin{align*}& \underline{Three \ friends} \ \underline{evenly \ split} \ the \ \underline{total \ cost} \ldots amount \ each \ friend \ paid \ \underline{was} \ \underline{\$4.25} \ldots\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ c \qquad \qquad \qquad \ \div \qquad \qquad \quad 3 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ = \quad 4.25\end{align*}

This equation, \begin{align*}c \div 3=4.25\end{align*}, represents \begin{align*}c\end{align*}, the total cost of the lunch bill.

Next, consider part \begin{align*}b\end{align*}.

Solve the equation to find the total cost, in dollars, of the lunch bill.

\begin{align*}c \div 3 &= 4.25\\ \frac{c}{3} &= 4.25\\ \frac{c}{3} \times 3 &= 4.25 \times 3\\ \frac{c}{\cancel{3}} \times \frac{\cancel{3}}{1} &= 12.75\\ c &= 12.75\end{align*}

**The total cost of the lunch bill was $12.75.**

### Video Review

Here is a video for review.

- This is a James Sousa video on solving single variable division equations.

### Practice

Directions: Solve each single-variable division equation for the missing value.

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

2. \begin{align*}\frac{y}{7}=3\end{align*}

3. \begin{align*}\frac{b}{9}=-4\end{align*}

4. \begin{align*}\frac{b}{8}=-10\end{align*}

5. \begin{align*}\frac{b}{8}=-10\end{align*}

6. \begin{align*}\frac{x}{-3}=-10\end{align*}

7. \begin{align*}\frac{y}{18}=-20\end{align*}

8. \begin{align*}\frac{a}{-9}=-9\end{align*}

9. \begin{align*}\frac{x}{11}=-12\end{align*}

10. \begin{align*}\frac{x}{3}=-3\end{align*}

11. \begin{align*}\frac{x}{5}=-8\end{align*}

12. \begin{align*}\frac{x}{1.3}=3\end{align*}

13. \begin{align*}\frac{x}{2.4}=4\end{align*}

14. \begin{align*}\frac{x}{6}=1.2\end{align*}

15. \begin{align*}\frac{y}{1.5}=3\end{align*}

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Inverse Operation

Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.Product

The product is the result after two amounts have been multiplied.Quotient

The quotient is the result after two amounts have been divided.### Image Attributions

Here you'll learn to solve single variable division equations.