# 3.4: Solve Equations Involving Inverse Properties of Subtraction and Division

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

**Practice**Two-Step Equations with Subtraction and Division

Did you ever solve a dilemma about wrapping paper? Take a look at this one.

Brandon and Felicia sold rolls of wrapping paper for a school fundraiser. Brandon sold 3 less than half the number of rolls that Felicia sold. Brandon sold a total of 9 rolls of wrapping paper.

Write an algebraic equation to represent \begin{align*}f\end{align*}, the number of rolls of wrapping paper that Felicia sold. Then, find the number of rolls of wrapping paper that Felicia sold.

Do you know how to solve this dilemma? Notice that there will be two parts to your answer. Pay attention to this Concept and you will know how to figure out this problem.

### Guidance

You are going to learn how to solve two step equations with subtraction and division. Let’s begin.

To solve a two-step equation, we will need to use more than one inverse operation. Let's take a look at how to solve a two-step equation now. When we perform inverse operations to find the value of a variable, we work to get the variable alone on one side of the equals. This is called isolating the variable. It is one strategy for solving equations. You can use isolating the variable whether you are solving one step or two step equations.

*Solve for \begin{align*}z\end{align*}: \begin{align*}\frac{z}{6} - 7 = 3\end{align*}.*

**Notice that there are two terms on the left side of the equation, \begin{align*}\frac{z}{6}\end{align*} and 7. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{z}{6}\end{align*}, by itself on one side of the equal (=) sign.**

**In the equation, 7 is** *subtracted***from \begin{align*}\frac{z}{6}\end{align*}. So, we can use the inverse of subtraction—addition. We can add 7 to both sides of the equation, like this:**

\begin{align*}\frac{z}{6} - 7 & = 3\\ \frac{z}{6} - 7 + 7 & = 3 + 7\\ \frac{z}{6} + (-7 + 7) & = 10\\ \frac{z}{6} + 0 & = 10\\ \frac{z}{6} &= 10\end{align*}

Now, the term that includes a variable, \begin{align*}\frac{z}{6}\end{align*}, is by itself on one side of the equation.

We can now use inverse operations to get the \begin{align*}z\end{align*} by itself. Since \begin{align*}\frac{z}{6}\end{align*} means \begin{align*}z \div 6\end{align*}, we can use the inverse of division—multiplication. We can multiply both sides of the equation by 6, like this:

\begin{align*}\frac{z}{6} &= 10\\ \frac{z}{6} \times 6 & = 10 \times 6\\ \frac{z}{\bcancel{6}} \times \frac{\bcancel{6}}{1} & = 60\\ \frac{z}{1} & = 60\\ z & = 60\end{align*}

**The value of \begin{align*}z\end{align*} is 60.**

Let’s review our steps to solving this two step equation.

*Take a few minutes to write these steps in your notebook.*

#### Example A

\begin{align*}\frac{x}{3} - 8 = 9\end{align*}

**Solution:\begin{align*}x = 51\end{align*}**

#### Example B

\begin{align*}\frac{y}{7} - 2 = 13\end{align*}

**Solution:\begin{align*}y = 105\end{align*}**

#### Example C

\begin{align*}\frac{a}{7} - 2 = 12\end{align*}

**Solution:\begin{align*}a = 98\end{align*}**

Now let's go back to the dilemma at the beginning of the Concept.

**Consider part** *a***first.**

Use a number, an operation sign, a variable, or an equal sign to represent each part of that problem. Since Brandon sold 9 rolls of wrapping paper, represent the number of rolls Brandon sold as 9. Use the key words from the chart to help you translate the rest of the problem into an equation. For example, you can translate “half the number of rolls...Felicia sold” as \begin{align*}\frac{f}{2}\end{align*}.

\begin{align*}& \underline{Brandon} \ \underline{sold} \ \underline{3} \ \underline{less \ than} \ \underline{half \ the \ number \ldots \ Felicia \ sold.}\\ & \quad \ \downarrow \qquad \ \downarrow \quad \ \searrow \qquad \downarrow \qquad \qquad \quad \swarrow\\ & \quad \ \downarrow \qquad \ \downarrow \quad \ \searrow \qquad \downarrow \qquad \qquad \quad \swarrow\\ & \quad \ \downarrow \qquad \ \downarrow \quad \ \swarrow \qquad \downarrow \qquad \qquad \quad \searrow\\ & \quad \ \ 9 \qquad = \ \ \frac{f}{2} \qquad - \qquad \qquad \quad 3\end{align*}

So, this equation, \begin{align*}9 = \frac{f}{2} - 3\end{align*}, represents \begin{align*}f\end{align*}, the number of rolls of wrapping paper that Felicia sold.

**Next, consider part** *b***.**

Solve the equation for \begin{align*}f\end{align*} to find the number of rolls that Felicia sold.

Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{f}{2}\end{align*}, by itself on one side of the equal (=) sign. In the equation, 3 is *subtracted* from \begin{align*}\frac{f}{2}\end{align*}. So, we can use the inverse of subtraction and add 3 to both sides of the equation, like this:

\begin{align*}9 &= \frac{f}{2}-3\\ 9 + 3 & = \frac{f}{2} - 3 + 3\\ 12 & = \frac{f}{2} + (-3 + 3)\\ 12 & = \frac{f}{2} + 0\\ 12 & = \frac{f}{2}\end{align*}

Now, the term that includes a variable, \begin{align*}\frac{f}{2}\end{align*}, is by itself on one side of the equation.

We can now use inverse operations to get the \begin{align*}f\end{align*} by itself. Since \begin{align*}\frac{f}{2}\end{align*} means \begin{align*}f \div 2\end{align*}, we can use the inverse of division and multiply both sides of the equation by 2, like this:

\begin{align*}12 &= \frac{f}{2}\\ 12 \times 2 & = \frac{f}{2} \times 2\\ 24 & = \frac{f}{\bcancel{2}} \times \frac{\bcancel{2}}{1}\\ 24 & = \frac{f}{1}\\ 24 & = f\end{align*}

**The value of \begin{align*}f\end{align*} is 24, so Felicia sold 24 rolls of wrapping paper for the fundraiser.**

### Vocabulary

- Equation
- a mathematical statement with an equal sign where the quantity on one side of the equation is equal to the quantity on the other side.

- Variable
- a letter used to represent an unknown quantity.

- Algebraic Equation
- An equation with at least one variable in it.

- One Step Equation
- An algebraic equation with one operation in it.

- Two Step Equation
- An algebraic equation with two operations in it.

### Guided Practice

Here is one for you to try on your own.

\begin{align*}\frac{x}{6} - 9 = 8\end{align*}

**Solution**

To solve this problem, first we have to add nine to both sides of the equation.

\begin{align*}\frac{x}{6} - 9 + 9 = 8 + 9\end{align*}

\begin{align*}\frac{x}{6} = 17\end{align*}

Next, multiply 6 times 17.

\begin{align*}x = 102\end{align*}

**This is the answer.**

### Video Review

### Practice

Directions: Solve each two step equation that has division and subtraction in it.

- \begin{align*}\frac{x}{5} - 4 = 8\end{align*}
- \begin{align*}\frac{y}{6} - 3 = 8\end{align*}
- \begin{align*}\frac{x}{7} - 7 = 10\end{align*}
- \begin{align*}\frac{x}{8} - 4 = 12\end{align*}
- \begin{align*}\frac{y}{7} - 5 = 11\end{align*}
- \begin{align*}\frac{x}{4} - 10 = 12\end{align*}
- \begin{align*}\frac{y}{4} - 8 = 2\end{align*}
- \begin{align*}\frac{x}{3} - 12 = 9\end{align*}
- \begin{align*}\frac{a}{5} - 3 = 11\end{align*}
- \begin{align*}\frac{b}{4} - 1 = 15\end{align*}
- \begin{align*}\frac{x}{2} - 8 = 4\end{align*}
- \begin{align*}\frac{a}{7} - 4 = 9\end{align*}
- \begin{align*}\frac{b}{4} - 7 = 3\end{align*}
- \begin{align*}\frac{x}{8} - 1 = 12\end{align*}
- \begin{align*}\frac{y}{6} - 8 = 5\end{align*}
- \begin{align*}\frac{x}{2} - 15 = 12\end{align*}

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Algebraic Equation

An algebraic equation contains numbers, variables, operations, and an equals sign.Equation

An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.inverse

Inverse functions are functions that 'undo' each other. Formally: and are inverse functions if .One-Step Equation

A one-step equation is an algebraic equation with one operation in it that requires one step to solve.Two-Step Equation

A two-step equation is an algebraic equation with two operations in it that requires two steps to solve.Variable

A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.### Image Attributions

Here you'll solve equations involving inverse properties of subtraction and division.