# Single Variable Multiplication Equations

## Solve one - step equations using multiplication.

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Single Variable Multiplication Equations

Mr. Ricky’s Biology class is going to the botanical gardens to study the different kinds of flowers and plants that are there. The students have raised 68 dollars for tickets so far. The student rate for each ticket is 5 dollars. Mr. Ricky asked the students to figure out how many tickets they can buy. How many tickets can they afford with 68 dollars?

In this concept, you will learn to solve equations with one variable and involve multiplication.

### Solving Single Variable Multiplication Equations

In the algebraic equation below, the variable \begin{align*}z\end{align*} represents a number.

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

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

You can find out by asking yourself, “What number, when multiplied by 2, equals 8?”

Since \begin{align*}4 \times 2 = 8\end{align*}\begin{align*}z\end{align*} must be equal to 4.

Sometimes, particularly with larger numbers, it can be more difficult to just 'know' the answer. Consider \begin{align*}z\times 7 = 105\end{align*}, since you may not know right away what number, multiplied by 7, equals 105, you should use another strategy for solving the equation.

To solve a more challenging equation in which a variable is multiplied by a number, you can use the inverse operation of multiplication-division. To isolate the variable, divide both sides of the equation by the number the variable is multiplied by.

You can divide both sides of the equation by the same number and not change the equality because of the Division Property of Equality, which says that two things that are the same will still be the same if they are both divided by an equal amount. In math language, this looks like:

If \begin{align*}a=b\end{align*} and \begin{align*}c \neq 0\end{align*}, then \begin{align*}\frac{a}{c} = \frac{b}{c}\end{align*}.

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

Let’s look at an example.

Solve for \begin{align*}z\end{align*}.

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

In this equation, \begin{align*}z\end{align*} is multiplied by 7. So, to isolate the variable \begin{align*}z\end{align*}, you can divide both sides of the equation by 7.

First, using the division property of equality, divide both sides of the equation by 7.

\begin{align*}\begin{array}{rcl} z \times 7 &=& 105 \\ \frac{z \times 7}{7} &=& \frac{105}{7} \end{array}\end{align*}

Next, separate the fraction \begin{align*}\frac{z \times 7}{7}\end{align*}, and simplify.

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

The answer is \begin{align*}z = 15\end{align*}.

Here is another example.

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

In this equation, -8 is multiplied by \begin{align*}r\end{align*}. So, using the division property of equality, you can divide both sides of the equation by -8 to solve for \begin{align*}r\end{align*}.

First, divide both sides of the equation by -8.

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

Next, separate the fraction \begin{align*}\frac{-8r}{-8}\end{align*}, and simplify.

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

The answer is \begin{align*}r = \text{-}16\end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about Mr. Ricky’s Biology class.

The students have raised 68 dollars for their trip and are wondering how many 5 dollar tickets they can buy.

First, write an equation to represent this information. Let \begin{align*}d\end{align*}, be the number of tickets they can buy. You can say that \begin{align*}d\end{align*} times the price of each ticket, 5 dollars, is 68 dollars.

\begin{align*}5d=68\end{align*}

Next, use the division property of equality and divide both sides of the equation by 5.

\begin{align*}\frac{5d}{5}=\frac{68}{5}\end{align*}

Then, separate the fraction \begin{align*}\frac{5d}{5}\end{align*} and simplify.

\begin{align*}\begin{array}{rcl} \frac{5}{5} d &=& 13.6 \\ 1d &=& 13.6 \\ d &=& 13.6 \end{array}\end{align*}

Next, interpret the result.

The students can buy 13.6 tickets. You can’t buy .6 of a ticket. So, the students can only buy 13 tickets with 3 dollars left over.

#### Example 2

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

1. Write an equation to represent \begin{align*}h\end{align*}, the number of hours she worked last week.
2. Determine how many hours Sarvenaz worked last week.

First, complete part a.

Let \begin{align*}h\end{align*} be the number of hours Sarvenaz worked. She earns $8 for each hour she works, so you multiply the number of hours she worked by$8 to find the total amount she earned. Write a multiplication equation.

\begin{align*}8h = 168\end{align*}

Next, work on part b.

Solve the equation \begin{align*}8h = 168\end{align*} to find \begin{align*}h\end{align*}, the number of hours she worked last week.

First, use the division property of equality to divide both sides of the equations by 8.

\begin{align*}\begin{array}{rcl} 8h &=& 168 \\ \frac{8h}{8} &=& \frac{168}{8} \end{array}\end{align*}

Next, separate the fraction \begin{align*}\frac{8h}{8}\end{align*} and simplify.

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

The answer is Sarvenaz works 21 hours last week.

Solve each equation.

#### Example 3

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

First, use the division property of equality, and divide both sides of the equation by -4.

\begin{align*}\begin{array}{rcl} -4x &=& 12 \\ \frac{-4x}{-4} &=& \frac{12}{-4} \end{array}\end{align*}

Next, separate the fraction \begin{align*}\frac{-4x}{-4}\end{align*} and simplify.

\begin{align*}\begin{array}{rcl} \frac{-4}{-4} x &=& -3 \\ 1x &=& -3 \\ x &=& -3 \end{array}\end{align*}

The answer is \begin{align*}x = -3\end{align*}.

#### Example 4

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

First, use the division property of equality and divide both sides of the equation by 8.

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

Next, separate the fraction \begin{align*}\frac{8a}{8}\end{align*} and simplify.

\begin{align*}\begin{array}{rcl} \frac{8}{8} a &=& 8 \\ 1a &=& 8 \\ a &=& 8 \end{array}\end{align*}

The answer is \begin{align*}a = 8\end{align*}.

#### Example 5

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

First, use the division property of equality and divide both sides of the equation by 9.

\begin{align*}\begin{array}{rcl} 9b &=& 81 \\ \frac{9b}{9} &=& \frac{81}{9} \end{array} \end{align*}

Next, separate the fraction \begin{align*}\frac{9b}{9}\end{align*} and simplify.

\begin{align*}\begin{array}{rcl} \frac{9}{9} b &=& 9 \\ 1b &=& 9 \\ b &=& 9 \end{array}\end{align*}

The answer is \begin{align*}b = 9\end{align*}.

### Review

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 =-102\end{align*}
14. \begin{align*}84x = 252\end{align*}
15. \begin{align*}-84x = -252\end{align*}

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### Vocabulary Language: English

TermDefinition
Division Property of Equality The division property of equality states that two equal values remain equal if both are divided by the same number. For example: If $2x = 8$, then $1x = 4$.
Inverse Operation Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Isolate the variable To isolate the variable means to manipulate an equation so that the indicated variable is by itself on one side of the equals sign.