# 3.2: One-Step Equations Transformed by Multiplication-Division

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

**Practice**One-Step Equations Transformed by Multiplication/Division

What if you had an algebraic equation involving multiplication or division like \begin{align*}-5x = 3\end{align*}? How could you solve it for the unknown variable *x*? After completing this Concept, you'll be able to solve equations like this one.

### Watch This

CK-12 Foundation: 0302S Solving Equations with Multiplication and Division (H264)

### Guidance

Suppose you are selling pizza for $1.50 a slice and you can get eight slices out of a single pizza. How much money do you get for a single pizza? It shouldn’t take you long to figure out that you get \begin{align*}8 \times \$1.50 = \$12.00\end{align*}. You solved this problem by multiplying. Here’s how to do the same thing algebraically, using \begin{align*}x\end{align*} to stand for the cost in dollars of the whole pizza.

#### Example A

*Solve* \begin{align*} \frac{1}{8} \cdot x = 1.5\end{align*}.

Our \begin{align*}x\end{align*} is being multiplied by one-eighth. To cancel that out and get \begin{align*}x\end{align*} by itself, we have to multiply by the reciprocal, 8. Don’t forget to multiply **both sides** of the equation.

\begin{align*}8 \left ( \frac{1}{8} \cdot x \right ) &= 8(1.5)\\ x &= 12\end{align*}

#### Example B

*Solve \begin{align*}0.25x = 5.25\end{align*}.*

0.25 is the decimal equivalent of one fourth, so to cancel out the 0.25 factor we would multiply by 4.

\begin{align*}4(0.25x) &= 4(5.25)\\ x &= 21\end{align*}

Solving by division is another way to isolate \begin{align*}x\end{align*}. Suppose you buy five identical candy bars, and you are charged $3.25. How much did each candy bar cost? You might just divide $3.25 by 5, but let’s see how this problem looks in algebra.

#### Example C

*Solve \begin{align*}5x = 3.25\end{align*}.*

To cancel the 5, we divide both sides by 5.

\begin{align*} \frac{5x}{5} &= \frac{3.25}{5}\\ x &= 0.65\end{align*}

#### Example D

*Solve \begin{align*}1.375x = 1.2\end{align*}.*

Divide by 1.375

\begin{align*} x &= \frac{1.2}{1.375}\\ x &= 0.8 \overline{72}\end{align*}

Notice the bar above the final two decimals; it means that those digits recur, or repeat. The full answer is 0.872727272727272....

To see more examples of one - and two-step equation solving, watch the Khan Academy video series starting at http://www.youtube.com/watch?v=bAerID24QJ0.

Watch this video for help with the Examples above.

CK-12 Foundation: Solving Equations with Multiplication and Division

### Vocabulary

- An equation in which each term is either a constant or the product of a constant and a single variable is a
**linear equation**. - We can add, subtract, multiply, or divide both sides of an equation by the same value and still have an
**equivalent equation**. - To solve an equation,
**isolate**the unknown variable on one side of the equation by applying one or more arithmetic operations to both sides.

### Guided Practice

*Solve:*

a) \begin{align*} \frac{9x}{5} = 5\end{align*}.

b) \begin{align*}7x = \frac{5}{11}\end{align*}.

**Solutions:**

a) \begin{align*}\frac{9x}{5}\end{align*} is equivalent to \begin{align*} \frac{9}{5} \cdot x\end{align*}, so to cancel out that \begin{align*} \frac{9}{5}\end{align*}, we multiply by the reciprocal, \begin{align*} \frac{5}{9}\end{align*}.

\begin{align*} \frac{5}{9} \left ( \frac{9x}{5} \right ) &= \frac{5}{9}(5)\\ x &= \frac{25}{9}\end{align*}

b) Divide both sides by 7.

\begin{align*}x &= \frac{5}{11.7}\\ x &= \frac{5}{77} \end{align*}

### Practice

For 1-5, solve the following equations for \begin{align*}x\end{align*}.

- \begin{align*}7x = 21 \end{align*}
- \begin{align*}4x = 1 \end{align*}
- \begin{align*}\frac{5x}{12} = \frac{2}{3}\end{align*}
- \begin{align*}0.01x = 11\end{align*}
- \begin{align*}\frac{-2x}{9} = \frac{10}{3}\end{align*}

For 6-10, solve the following equations for the unknown variable.

- \begin{align*}21s = 3 \end{align*}
- \begin{align*}-7a = -5 \end{align*}
- \begin{align*}\frac{7f}{11} = \frac{7}{11} \end{align*}
- \begin{align*}6r = \frac{3}{8} \end{align*}
- \begin{align*}\frac{9b}{16} = \frac{3}{8} \end{align*}

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### Image Attributions

Here you'll learn how to use multiplication and division to solve algebraic equations for their unknown variable.