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12.8: Single Variable Multiplication Equations

Difficulty Level: At Grade Created by: CK-12
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Remember the amusement park? Look at this dilemma.

Eight of the students bought cotton candy at the amusement park. If the total cost of the cotton candy was $12.00, what was the cost of each cotton candy?

To solve this problem, you will need to write a multiplication equation and solve it. You will learn how to do that in this Concept.


Just like you learned how to solve single-variable equations with addition and subtraction, this Concept will teach you how to solve single-variable multiplication and division equations.

Let’s start with solving multiplication equations.


Here we need to figure out what the value of \begin{align*}x\end{align*}x is. We can do this in two ways.

  1. Use mental math
  2. Use the inverse operation

To use mental math we can think to ourselves, “What times five is equal to thirty?”

Using our times tables, we can figure out that 5 times 6 is equal to thirty. The value of \begin{align*}x\end{align*}x is 6.

To use the inverse operation, we use the opposite operation of multiplication, since this is a multiplication problem. The inverse of multiplication is division.

Once again, we work to get the variable alone on one side of the equation. This time by dividing both sides by the number next to the variable. In this example, we divide both sides by 5.

\begin{align*}\frac{5x}{5}= \frac{30}{5}\end{align*}5x5=305

The fives cancel each other out because five divided by five is one, and "x" times 1 is "x". On the right side, thirty divided by five is six.

\begin{align*}\frac{\bcancel{5}x}{\bcancel{5}}&= \frac{30}{5}\\ x&=6\end{align*}5x5x=305=6

You can check your work by substituting the value of \begin{align*}x\end{align*}x back into the original equation. If both sides are equal, then your work is accurate and correct.

\begin{align*}5(6)&=30\\ 30&=30\end{align*}5(6)30=30=30

Our work is correct.


To do this one, let’s use the inverse operation. We divide both sides by 7 to get the variable alone.


The 7’s cancel each other out, leaving \begin{align*}y\end{align*}y alone. Forty-nine divided by seven is seven.

\begin{align*}\frac{\bcancel{7}y}{\bcancel{7}}&=\frac{49}{7}\\ y&=7\end{align*}7y7y=497=7

Check your work. Substitute 7 back into the original problem for \begin{align*}y\end{align*}y.

\begin{align*}7(7)&=49\\ 49&=49\end{align*}7(7)49=49=49

Our work is accurate.

Practice solving a few equations on your own. Write your answer is the form variable = _____.

Example A


Solution: \begin{align*}x = 8\end{align*}x=8

Example B


Solution:\begin{align*}a = 13\end{align*}a=13

Example C


Solution:\begin{align*}y = 7\end{align*}y=7

Here is the original problem once again.

Eight of the students bought cotton candy at the amusement park. If the total cost of the cotton candy was $12.00, what was the cost of each cotton candy?

First, let's write an equation that illustrates this dilemma.

\begin{align*}8x = 12\end{align*}8x=12

The students bought 8 cotton candies and the total cost was $12.00. We are trying to figure out the cost for one cotton candy.

Now we can solve this by using the inverse of multiplication, division.

\begin{align*}/frac{8x}{8} = \frac{12}{8}\end{align*}/frac8x8=128

\begin{align*}x = 1.50\end{align*}x=1.50

Each cotton candy costs $1.50.


Here are the vocabulary words in this Concept.

the answer to a multiplication problem
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*}12y = 84\end{align*}12y=84


To complete this problem, we can use the inverse of multiplication. This means that we divide 84 by 12 to get the variable "y" by itself.

\begin{align*}y = \frac{84}{12}\end{align*}y=8412

\begin{align*}y = 7\end{align*}y=7

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: Solving Single-Step Equation by Division


Directions: Solve each single-variable multiplication equation.

1. \begin{align*}7y = 14\end{align*}7y=14

2. \begin{align*}3y = 24\end{align*}3y=24

3. \begin{align*}9x = 81\end{align*}9x=81

4. \begin{align*}4x=16\end{align*}4x=16

5. \begin{align*}3y=12\end{align*}3y=12

6. \begin{align*}8a=72\end{align*}8a=72

7. \begin{align*}12v=36\end{align*}12v=36

8. \begin{align*}9x=45\end{align*}9x=45

9. \begin{align*}10y=100\end{align*}10y=100

10. \begin{align*}7x=21\end{align*}7x=21

11. \begin{align*}9a=99\end{align*}9a=99

12. \begin{align*}16x=32\end{align*}16x=32

13. \begin{align*}14y=28\end{align*}14y=28

14. \begin{align*}13y = 39\end{align*}13y=39

15. \begin{align*}7y = 140\end{align*}7y=140


Inverse Operation

Inverse Operation

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


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


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

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Difficulty Level:
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Date Created:
Oct 29, 2012
Last Modified:
Mar 23, 2016
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