### Let’s Think About It

Mr. Ricky’s Biology class is going to the botanical gardens to study the different kinds of flowers and plants that are there. They 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 single variable multiplication equations.

### Guidance

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*}

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

Since \begin{align*}4 \times 2 = 8\end{align*}

To solve a more complex equation, such as \begin{align*}z\times 7 = 105\end{align*}

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

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 states:

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

This means that if you divide one side of an equation by a nonzero number, \begin{align*}c\end{align*}

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*}

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*}

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

In this equation, -8 is multiplied by \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*}

\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 = -16\end{align*}

### Guided Practice

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

- Write an equation to represent , the number of hours she worked last week.
- Determine how many hours Sarvenaz worked last week.

First, complete part a.

Let \begin{align*}h\end{align*}

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

Next, work on part b.

Solve the equation \begin{align*}8h = 168\end{align*}

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*}

\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.

### Examples

Solve each equation.

#### Example 1

\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 2

\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 3

\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*}.

### Follow Up

Remember Mr. Ricky’s Biology class?

The class has raised 68 dollars for their trip and is 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.

The answer is the students can buy 13 tickets.

### Video Review

https://www.youtube.com/watch?v=zBqIH-E3ero&feature=youtu.be

### Explore More

Solve each single variable multiplication equation for the missing value.

- \begin{align*}4x = 16\end{align*}
- \begin{align*}6x = 72\end{align*}
- \begin{align*}-6x = 72\end{align*}
- \begin{align*}-3y = 24\end{align*}
- \begin{align*}-3y =-24\end{align*}
- \begin{align*}-5x = -45\end{align*}
- \begin{align*}-1.4x = 2.8\end{align*}
- \begin{align*}3.5a = 7\end{align*}
- \begin{align*}7a =-49\end{align*}
- \begin{align*}14b = -42\end{align*}
- \begin{align*}24b = -48\end{align*}
- \begin{align*}-24b = -48\end{align*}
- \begin{align*}34b =-102\end{align*}
- \begin{align*}84x = 252\end{align*}
- \begin{align*}-84x = -252\end{align*}