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# 12.8: Single Variable Multiplication Equations

Difficulty Level: At Grade Created by: CK-12
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Practice Single Variable Multiplication Equations

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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. ### Guidance 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. 5x=30\begin{align*}5x=30\end{align*} Here we need to figure out what the value of x\begin{align*}x\end{align*} 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 x\begin{align*}x\end{align*} 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. 5x5=305\begin{align*}\frac{5x}{5}= \frac{30}{5}\end{align*} 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. 5x5x=305=6\begin{align*}\frac{\bcancel{5}x}{\bcancel{5}}&= \frac{30}{5}\\ x&=6\end{align*} You can check your work by substituting the value of x\begin{align*}x\end{align*} back into the original equation. If both sides are equal, then your work is accurate and correct. 5(6)30=30=30\begin{align*}5(6)&=30\\ 30&=30\end{align*} Our work is correct. 7y=49\begin{align*}7y=49\end{align*} To do this one, let’s use the inverse operation. We divide both sides by 7 to get the variable alone. 7y7=497\begin{align*}\frac{7y}{7}=\frac{49}{7}\end{align*} The 7’s cancel each other out, leaving y\begin{align*}y\end{align*} alone. Forty-nine divided by seven is seven. 7y7y=497=7\begin{align*}\frac{\bcancel{7}y}{\bcancel{7}}&=\frac{49}{7}\\ y&=7\end{align*} Check your work. Substitute 7 back into the original problem for y\begin{align*}y\end{align*}. 7(7)49=49=49\begin{align*}7(7)&=49\\ 49&=49\end{align*} Our work is accurate. Practice solving a few equations on your own. Write your answer is the form variable = _____. #### Example A 8x=64\begin{align*}8x=64\end{align*} Solution: x=8\begin{align*}x = 8\end{align*} #### Example B 2a=26\begin{align*}2a=26\end{align*} Solution:a=13\begin{align*}a = 13\end{align*} #### Example C 6y=42\begin{align*}6y=42\end{align*} Solution:y=7\begin{align*}y = 7\end{align*} 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 was12.00, what was the cost of each cotton candy?

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

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

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. /frac8x8=128\begin{align*}/frac{8x}{8} = \frac{12}{8}\end{align*} x=1.50\begin{align*}x = 1.50\end{align*} Each cotton candy costs1.50.

### 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.

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

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.

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

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

### Video Review

Here are videos for review.

### Practice

Directions: Solve each single-variable multiplication equation.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Color Highlighted Text Notes

### Vocabulary Language: English

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.

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