# 7.9: Single Variable Multiplication Equations

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

**Practice**Single Variable Multiplication Equations

While visiting their grandparents, Kara has been handling the money for their rides on the subway. For the past two days, Marc has been watching her calculating and handing out money. Then when they were coming back from the mall, Marc noticed something on a sign about a day pass and a month long pass.

“How much money do we have set aside for the subway?” Marc asked Kara the day after he saw the sign.

“We have $45.00 each allotted for the subway,” she said.

“Well, a day pass is $9.00. But a month long pass is $20.00. I think that is a better deal,” he told her.

“Let’s figure out how many days we can ride for $9.00 per day.”

Kara wrote the following equation on a piece of paper.

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

Solving equations is essentially for Kara and Marc at this point in time. Then they will know whether day passes or a month long pass is the way to go.

**You can figure this out too. Solving equations is what you will learn in this Concept. Take what you learn and then apply it back to this problem. At the end of the Concept, you will know which train option makes the most sense.**

### Guidance

As you saw in the last few Concepts, we can have equations with addition and subtraction in them. We can also have equations with multiplication and division in them. These single-variable equations can be solved for the value of the variable. We want to solve the equation and figure out what the variable is equal to.

**How does this work?**

In the algebraic equation below, the variable \begin{align*}z\end{align*}

\begin{align*}z \times 2=8\end{align*}

What number does \begin{align*}z\end{align*}

Since \begin{align*}4 \times 2=8, z\end{align*}**We solved this equation using mental math.**

**When we determine the value for a variable in an equation, we are solving that equation**. If the equation involves a simple number fact, such as \begin{align*}4 \times 2=8\end{align*}

**What happens when we have a more challenging equation to solve?**

Some equations are more complex and mental math will not get the job done. To solve a more complex equation, such as \begin{align*}z \times 7=105\end{align*}

**One strategy for solving an equation is to use inverse operations to** ** isolate the variable.** Remember, isolating the variable means getting the variable by itself on one side of the equal (=) sign.

**To solve an equation in which a variable is** *multiplied***by a number, we can use the** *inverse operation***of multiplication –– division.** We can divide *both* sides of the equation by that number to find the value of the variable.

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

if \begin{align*}a=b\end{align*}

**That is very simple. If you divide one side of an equation by a nonzero number, \begin{align*}c\end{align*} c, you must divide the other side of the equation by that same number, \begin{align*}c\end{align*}c, to keep the values on both sides equal.**

Let's apply this information.

\begin{align*}z \times 7=105\end{align*}

In the equation, \begin{align*}z\end{align*}*multiplied* by 7. So, we can *divide* both sides of the equation by 7 to solve for \begin{align*}z\end{align*}

\begin{align*}z \times 7 &= 105\\
\frac{z \times 7}{7} &= \frac{105}{7}\\
z \times \frac{7}{7} &= 15\\
z \times 1 &= 15\\
z &= 15\end{align*}

It may help to separate out the factors like we did above.

**Having divided both sides by 7, we can see that the value of \begin{align*}z\end{align*} z is 15.**

Here is another one.

\begin{align*}-8r=128\end{align*}

In the equation, -8 is *multiplied* by \begin{align*}r\end{align*}. So, we can *divide* both sides of the equation by -8 to solve for \begin{align*}r\end{align*}.

\begin{align*}-8r &= 128\\ \frac{-8r}{-8} &= \frac{128}{-8}\end{align*}

We need to use what we know about dividing integers to help us solve this problem.

\begin{align*}\frac{-8r}{-8} &= \frac{128}{-8}\\ 1r &= -16\\ r &= -16\end{align*}

**The value of \begin{align*}r\end{align*} is -16 .**

**Remember the integer rules for dividing positive and negative numbers? Let’s review.**

Remember that a property is a rule for performing operations in mathematics that makes our lives simpler. These properties are no exception. If you can look at an equation, see that it is a multiplication equation and know that division is the way to solve it, then the property of the ** Division Property of Equality** is being utilized.

Now you can practice a few of these on your own. Solve each problem for the missing variable.

#### Example A

\begin{align*}-4x=12\end{align*}

**Solution:\begin{align*}x = -3\end{align*}**

#### Example B

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

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

#### Example C

\begin{align*}9b=81\end{align*}

**Solution:\begin{align*}b = 9\end{align*}**

Here is the original problem once again.

While visiting their grandparents, Kara has been handling the money for their rides on the subway. For the past two days, Marc has been watching her calculating and handing out money. Then when they were coming back from the mall, Marc noticed something on a sign about a day pass and a month long pass.

“How much money do we have set aside for the subway?” Marc asked Kara the day after he saw the sign.

“We have $45.00 each allotted for the subway,” she said.

“Well, a day pass is $9.00. But a month long pass is $20.00. I think that is a better deal,” he told her.

“Let’s figure out how many days we can ride for $9.00 per day.”

Kara wrote the following equation on a piece of paper.

\begin{align*}\underline{9x=45}\end{align*}

Solving equations is essentially for Kara and Marc at this point in time. Then they will know whether day passes or a month long pass is the way to go.

**First, let’s solve the equation and figure out how many days Kara and Marc can purchase day passes given their budget.**

\begin{align*}9x &= 45\\ \frac{9x}{9} &= \frac{45}{9}\\ x &= 5\end{align*}

**According to this, they can purchase day passes for only five days.**

**A month long pass is $20.00 each. This is a much better deal for them.**

\begin{align*}20(2) = \$40.00\end{align*}

**The total subway fares will cost them $40.00 instead of much more than $45.00.**

**Kara and Marc purchase month long T passes the very next morning.**

### 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.

### Guided Practice

Here is one for you to try on your own.

Sarvenaz earns $8 for each hour she works. She earned a total of $168 last week.

a. Write an equation to represent \begin{align*}h\end{align*}, the number of hours she worked last week.

b. Determine how many hours Sarvenaz worked last week.

**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 that problem. She earns $8 for each hour she works, so you could multiply the number of hours she worked by $8 to find the total amount she earned. Write a multiplication equation.

\begin{align*}& \underline{\$8 \ for \ each \ hour}\ldots She \ \underline{earned} \ a \ \underline{total \ of \ \$168} \ last \ week.\\ & \qquad \downarrow \qquad \qquad \qquad \qquad \qquad \downarrow \qquad \qquad \ \downarrow\\ & \qquad 8h \qquad \qquad \qquad \qquad \quad \ = \qquad \quad \ 168\end{align*}

So, this equation, \begin{align*}8h=168\end{align*}, represents \begin{align*}h\end{align*}, the number of hours she worked last week.

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

Solve the equation to find \begin{align*}h\end{align*}, the number of hours she worked last week.

\begin{align*}8h &= 168\\ \frac{8h}{8} &= \frac{168}{8}\\ 1h &= 21\\ h &= 21\end{align*}

**Sarvenaz worked 21 hours last week.**

### Video Review

Here is a video for review.

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

### Practice

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

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

2. \begin{align*}6x=72\end{align*}

3. \begin{align*}-6x=72\end{align*}

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

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

6. \begin{align*}-5x=-45\end{align*}

7. \begin{align*}-1.4x=2.8\end{align*}

8. \begin{align*}3.5a=7\end{align*}

9. \begin{align*}7a=-49\end{align*}

10. \begin{align*}14b=-42\end{align*}

11. \begin{align*}24b=-48\end{align*}

12. \begin{align*}-24b=-48\end{align*}

13. \begin{align*}34b=-72\end{align*}

14. \begin{align*}84x=252\end{align*}

15. \begin{align*}-84x=-252\end{align*}

### Image Attributions

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