# 12.9: Single Variable Division Equations

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

**Practice**Single Variable Division Equations

Have you ever been on a coach bus?

Carl has been working hard trying to find a coach bus that his sixth grade class can afford. It is more challenging than he expected, so after some frustration he asks his friend Tabitha to help him.

“Here is what I know. The total cost of the bus can’t exceed $143.00 or we can’t afford it. I know this because $143.00 divided by 26 is the cost of the bus per person. Here is the equation I wrote,” Carl explains.

\begin{align*}\frac{143}{26}=y\end{align*}

\begin{align*}y=\$5.50\end{align*}

“The cost per person can’t exceed $5.50 or the cost of the trip becomes too expensive. I have called three different places and the cheapest bus is $1300,” Carl sighed, looking at Tabitha for any new ideas.

“Wow, that is expensive,” Tabitha said.

“I guess its back to the school bus,” Carl frowned.

At that moment Mrs. Hawk came in with a piece of paper in her hand.

“Carl, Buses are Us just called and they will offer us a deal on a bus. It will be only $998.00,” Mrs. Hawk said.

“I think that is still too much money. Let me see how much that is per person.”

Carl wrote this equation down.

\begin{align*}\frac{988}{26}=y\end{align*}

“We have some money in the class account,” Mrs. Hawk said to Carl. “We can take $600 and apply it to the cost of the bus.”

Carl took a pencil to do some quick figuring.

988 – 600 = 388

Now that Carl has this new amount, he is back to the drawing board.

**Can you write and solve this new equation with Carl? This Concept is all about solving multiplication and division single-variable equations. Once you have done this work, Carl and Tabitha will be able to figure out the cost of the bus per person.**

### Guidance

Division equations are a bit tricky because you have to multiply to solve them. We often think in terms of multiplication, but we don’t think in terms of division. When you have a division equation, you have 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 we 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. We don’t know the value of the numerator so we use a variable in place of the unknown number.**

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

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

**Our work is correct.**

Now let's look at one that is the second type of division equation.

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

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

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

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

**Is this accurate?** Let’s substitute it back into the original problem to check.

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

**Our work is accurate and correct.**

Practice solving these equations. Write your answer in the form variable = _____.

#### Example A

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

**Solution: \begin{align*}x = 35\end{align*} x=35**

#### Example B

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

**Solution:\begin{align*}x = 6\end{align*} x=6**

#### Example C

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

**Solution:\begin{align*}x = 2\end{align*} x=2**

Now back to the bus.

Here is the original problem once again.

Carl has been working hard trying to find a coach bus that his sixth grade class can afford. It is more challenging than he expected, and after some frustration he asks his friend Tabitha to help him.

“Here is what I know. The total cost of the bus can’t exceed $143.00 or we can’t afford it. I know this because $143.00 divided by 26 is the cost of the bus per person. Here is the equation I wrote,” Carl explains.

\begin{align*}\frac{143}{26}=y\end{align*}

\begin{align*}y=\$5.50\end{align*}

“The cost per person can’t exceed $5.50 or the cost of the trip becomes too expensive. I have called three different places and the cheapest bus is $1300,” Carl sighed, looking at Tabitha for any new ideas.

“Wow, that is expensive,” Tabitha said.

“I guess its back to the school bus,” Carl frowned.

At that moment Mrs. Hawk came in with a piece of paper in her hand.

“Carl, Buses are Us just called and they will offer us a deal on a bus. It will be only $998.00,” Mrs. Hawk said.

“I think that is still too much money. Let me see how much that is per person.”

Carl wrote this equation down.

\begin{align*}\frac{988}{26}=y\end{align*}

“We have some money in the class account,” Mrs. Hawk said to Carl. “We can take $600 and apply it to the cost of the bus.”

Carl took a pencil to do some quick figuring.

988 – 600 = 388

Now that Carl has this new amount, he is back to the drawing board.

Can you write and solve this new equation with Carl?

Mrs. Hawk has said that the students can apply class funds to the cost of the bus. Carl needs to write an equation and solve for the bus cost per person with these new figures.

The cost of the bus – class funds = new cost

988 – 600 = 388

There are 26 students in the class.

\begin{align*}y\end{align*}

Let’s write the equation. This is a division problem.

\begin{align*}\frac{388}{26} & = y\\
y & = \$14.92 \ \text{per person}\end{align*}

Carl knows that his original figure was $5.50 per person for the bus. To find the difference he subtracts $5.50 from $14.92.

**14.92 – 5.50 = $9.42**

**There is a difference of $9.42 that each person would have to pay to take the coach bus.**

**Carl and Tabitha show their work to Mrs. Hawk. Mrs. Hawk suggests that they ask the class to have a car wash to raise the additional funds. The students think this is a terrific idea!!**

### Vocabulary

Here are the vocabulary words in this Concept.

- Product
- the answer to a multiplication problem

- Quotient
- the answer to a division problem

- Inverse Operation
- the opposite operation

### Guided Practice

Here is one for you to try on your own.

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

**Answer**

To find the answer to this problem, we have to use the inverse of division.

Some number divided by four is six.

We multiply four and six.

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

\begin{align*}x = 24\end{align*}

**This is our answer.**

### Video Review

Here are videos for review.

Khan Academy, Simple Equations

James Sousa, Solving One Step Equation by Multiplication and Division

Math Problem Generator: Solve Single-Step Equation by Division

### Practice

Directions: Solve each single-variable division equation.

1. \begin{align*}\frac{x}{4}=8\end{align*}

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

3. \begin{align*}\frac{x}{9}=9\end{align*}

4. \begin{align*}\frac{x}{5}=3\end{align*}

5. \begin{align*}\frac{20}{y}=4\end{align*}

6. \begin{align*}\frac{x}{2}=18\end{align*}

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

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

9. \begin{align*}\frac{x}{14}=3\end{align*}

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

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

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

13. \begin{align*}\frac{44}{x}=4\end{align*}

14. \begin{align*}\frac{32}{x}=8\end{align*}

15. \begin{align*}\frac{90}{x}=30\end{align*}

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.