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# Single Variable Multiplication Equations

## Solve one - step equations using multiplication.

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Practice Single Variable Multiplication Equations
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Single Variable Multiplication Equations
Credit: Eagle Brook School
Source: https://www.flickr.com/photos/eaglebrook/7296634422/in/photolist-c7MacL-c7M8D3-c7Mbhu-c7M9A1-c7M95Y-c7MazQ-c7M9aE-c7MbjA-c7M9Uh-c7Maty-c7M9N3-c7M89W-c7Mbmo-c7M9sL-c7M925-c7M8C7-c7M7Ch-c7M8Mf-c7M9kW-c7MaHd-c7M9Es-c7M9vh-c7Mb6w-c7MafY-c7Mar7-c7M8gw-c7M8sd-c7M97J-c7Ma9u-c7Ma77-c7M8yJ-c7MawG-c7M8PC-c7ManG-c7M7uo-c7MaYY-c7Mbd3-c7Ma5G-c7M8Zj-c7Mbbq-c7M9K1-c7M8Uw-c7MaKh-c7M8xA-c7MaRj-c7M9XA-c7M9RA-c7M8qh-c7M9xq-c7Maeb

Mrs. Jackson’s class has done well during the first half of the school’s field day events. One of her students won the 100-meter dash and another placed second in the high jump. Now, the students are hot and thirsty. Twelve of the students give Mrs. Jackson money to buy them a drink from the concession stand. The total for the 12 drinks is 21. When she returns, she passes out the drinks and, because she teaches math, asks the students to use a single variable multiplication equation to figure out how much each drink cost. That way, she will know how much change to give each of them. Can you figure out how much each cost? In this concept, you will learn to solve single-variable multiplication equations. ### Solving Single-Variable Multiplication Equations Here is a multiplication equation with a single variable \begin{align*}x\end{align*}. \begin{align*}5x=30\end{align*} You can figure out the value of \begin{align*}x\end{align*} in two ways. 1. Use mental math 2. Use the inverse operation To use mental math, ask yourself, “What times five is equal to thirty?” Then, using the times table, you can figure out that 5 times 6 is equal to thirty. So, the value of \begin{align*}x\end{align*} is 6. To use the inverse operation, use the opposite operation of multiplication, which is division. First, get the variable alone on one side of the equation by dividing both sides by the number next to the variable. In this example, divide both sides by 5. \begin{align*}\frac{5x}{5}=\frac{30}{5}\end{align*} The fives cancel each other out because five divided by five is one, and “\begin{align*}x\end{align*}” times 1 is “\begin{align*}x\end{align*}”. On the right side, thirty divided by five is six. \begin{align*}\begin{array}{rcl} \frac{\bcancel{5}x}{\bcancel{5}} &=& \frac{30}{5}\\ x &=& 6 \end{array}\end{align*} You can check your work by substituting the value of \begin{align*}x\end{align*} back into the original equation. If both sides are equal, then your work is accurate and correct. \begin{align*}\begin{array}{rcl} 5x &=& 30\\ 5(6) &=& 30 \\ 30 &=& 30 \end{array}\end{align*} The answer is correct. Here is another example. \begin{align*}7y=49\end{align*} First, use the inverse operation of multiplication and divide both sides by 7 to get the variable alone. \begin{align*}\frac{7y}{7}=\frac{49}{7}\end{align*} Next, the 7’s cancel each other out, leaving \begin{align*}y\end{align*} on one side of the equation. On the other side, divide 49 by 7. You are left with \begin{align*}y = 7\end{align*}. \begin{align*}\begin{array}{rcl} \frac{\bcancel{7}y}{\bcancel{7}} &=& \frac{49}{7} \\ y &=& 7 \end{array}\end{align*} Then, check your work by substituting 7 back into the original problem for \begin{align*}y\end{align*}. \begin{align*}\begin{array}{rcl} 7y &=& 49 \\ 7(7) &=& 49 \\ 49 &=& 49 \end{array}\end{align*} The answer is correct. ### Examples #### Example 1 Earlier, you were given a problem about Mrs. Jackson and her students looking for relief from the heat. Mrs. Jackson bought 12 drinks priced the same and the total cost was21. How can she use a single variable multiplication equation to figure out how much each drink cost?

First, write an equation to represent the situation.

12 times a number (price of drink) is 21.

\begin{align*}12y=21\end{align*}

Next, use the inverse operation of multiplication and divide both sides by 12 to get the variable alone.

\begin{align*}\frac{12y}{12}=\frac{21}{12}\end{align*}

The 12’s cancel each other out, leaving \begin{align*}y\end{align*} alone on one side of the equal sign. On the other side, divide 21 by 12 to get \begin{align*}y = 1.75\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{\bcancel{12} y}{\bcancel{12}} &=& \frac{21}{12} \\ y &=& 1.75 \end{array}\end{align*}

Then, check your work by substituting 1.75 back into the original problem for \begin{align*}y\end{align*}.

\begin{align*}\begin{array}{rcl} 12y &=& 21 \\ 12(1.75) &=& 21 \\ 21 &=& 21 \end{array}\end{align*}

The price of each drink is \$1.75.

Solve the following equations and write your answer in the form   \begin{align*}\text{Variable} =\underline{\; \; \; \; \; \; \; \; }\end{align*}.

#### Example 2

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

First, use the inverse operation of multiplication and divide both sides by 12 to get the variable alone.

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

Next, the 12’s cancel each other out, leaving \begin{align*}y\end{align*} alone. Dividing 84 by 12 leaves \begin{align*}y = 7\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{\bcancel{12} y}{\bcancel{12}} &=& \frac{84}{12} \\ y &=& 7 \end{array}\end{align*}

Then, check your work by substituting 7 back into the original problem for \begin{align*}y\end{align*}.

\begin{align*}\begin{array}{rcl} 12y &=& 84 \\ 12(7) &=& 84 \\ 84 &=& 84 \end{array}\end{align*}

#### Example 3

\begin{align*}8x=64\end{align*}

First, use the inverse operation of multiplication and divide both sides by 8 to get the variable alone.

\begin{align*}\frac{8x}{8}=\frac{64}{8}\end{align*}

Next, the 8’s cancel each other out, leaving \begin{align*}x\end{align*} alone. Dividing 64 by 8 leaves \begin{align*}x = 8\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{\bcancel{8} x}{\bcancel{8}} &=& \frac{64}{8} \\ x &=& 8 \end{array}\end{align*}

Then, check your work by substituting 8 back into the original problem for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 8x &=& 64 \\ 8(8) &=& 64\\ 64 &=& 64 \end{array}\end{align*}

#### Example 4

\begin{align*}2a=26\end{align*}

First, use the inverse operation of multiplication and divide both sides by 2 to get the variable alone.

\begin{align*}\frac{2a}{2}=\frac{26}{2}\end{align*}

Next, the 2’s cancel each other out, leaving \begin{align*}a\end{align*} alone. So, divide 26 by 2 to get \begin{align*}a = 13\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{\bcancel{2} a}{\bcancel{2}} &=& \frac{26}{2} \\ a &=& 13 \end{array}\end{align*}

Then, check your work by substituting 13 back into the original problem for \begin{align*}a\end{align*}.

\begin{align*}\begin{array}{rcl} 2a &=& 26 \\ 2(13) &=& 26 \\ 26 &=& 26 \end{array}\end{align*}

#### Example 5

\begin{align*}6y=42\end{align*}

First, use the inverse operation of multiplication and divide both sides by 6 to get the variable alone.

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

Next, the 6’s cancel each other out, leaving \begin{align*}y\end{align*} alone. Dividing 42 by 6 leaves \begin{align*}y = 7\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{\bcancel{6} y}{\bcancel{6}} &=& \frac{42}{6} \\ y &=& 7 \end{array}\end{align*}

Then, check your work by substituting 7 back into the original problem for \begin{align*}y\end{align*}.

\begin{align*}\begin{array}{rcl} 6y &=& 42 \\ 6(7) &=& 42 \\ 42 &=& 42 \end{array}\end{align*}

### Review

Solve each single-variable multiplication equation. Write your answer in the form \begin{align*}\text{Variable} =\underline{\; \; \; \; \; \; \; \; }\end{align*}

1. \begin{align*}7y = 14\end{align*}
2. \begin{align*}3y = 24\end{align*}
3. \begin{align*}9x = 81\end{align*}
4. \begin{align*}4x = 16\end{align*}
5. \begin{align*}3y = 12\end{align*}
6. \begin{align*}8a = 72\end{align*}
7. \begin{align*}12v = 36\end{align*}
8. \begin{align*}9x = 45\end{align*}
9. \begin{align*}10y = 100\end{align*}
10. \begin{align*}7x = 21\end{align*}
11. \begin{align*}9a = 99\end{align*}
12. \begin{align*}16x = 32\end{align*}
13. \begin{align*}14y = 28\end{align*}
14. \begin{align*}13y = 39\end{align*}
15. \begin{align*}7y = 140\end{align*}

To see the Review answers, open this PDF file and look for section 12.8.

### 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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