# 2.3.5: Two-Step Equations with Subtraction and Multiplication

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

## Solving Equations Using Inverse Properties of Addition and Multiplication

You have $58 dollars to spend on a pair of jeans and two T-shirts. The jeans cost$32. How much will the T-shirts have to be?

In this concept, you will solve equations involving inverse properties of subtraction and multiplication.

### Inverse Properties of Adding and Multiplying

To solve a two-step equation, you will need to use more than one inverse operation. When you perform inverse operations to find the value of a variable, you 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.

For example, solve for x in the following equation:

2x−9=17

Notice that there are two terms on the left side of the equation, 2x and 9.

First, use inverse operations to get the term that includes a variable, 2x, by itself on one side of the equal sign. In the equation, 9 is subtracted from 2x. So, you can use the inverse of subtraction — addition.

2x−9=17

2x−9+9=17+9

2x=26

The number +9 in this solution is the additive inverse, or opposite, of -9.

Next, use inverse operations to get the x by itself. Since 2x means 2⋅x, you can use the inverse of multiplication—division. So you can divide both sides of the equation by 2.

2x=26

2x/2=26/2

x=13

Let’s review the steps for solving this two-step equation.

### Examples

Example 2.3.5.1

Earlier, you were given a problem about shopping budget.

You only have $58 to spend on two T-shirts and a pair of jeans that Solution First, walk through the words to write the equation. 2x+32=58 Next, isolate the variable by using the inverse of addition – subtraction. 2x+32=58 2x+32−32=58-32 2x=26 Then, solve for x by using the inverse of multiplication – division. 2x=26 2x/2=26/2 x=13 The answer is 13. Each T-shirt must cost$13.

Example 2.3.5.2

Eight times a number minus four is equal to ninety - two. Write a two-step equation and solve for the missing variable.

Solution

First, walk through the words to write the equation.

8x−4=92

Next, isolate the variable by using the inverse of subtraction – addition.

8x−4=92

8x−4+4=92+4

8x=96

Then, solve for x by using the inverse of multiplication – division.

8x=96

8x8=96/8

x=12

Example 2.3.5.3

9x−5=40

Solution

First, isolate the variable by using the inverse of subtraction – addition.

9x−5=40

9x−5+5=40+5

9x=45

Then, solve for x by using the inverse of multiplication – division.

9x=45

9x9=45/9

x=5

Example 2.3.5.4

9y−6=66

Solution

First, isolate the variable by using the inverse of subtraction – addition.

9y−6=66

9y−6+6=66+6

9y=72

Then, solve for y by using the inverse of multiplication – division.

9y=72

9y/9=72/9

y=8

Example 2.3.5.5

12a−4=44

Solution

First, isolate the variable by using the inverse of subtraction – addition.

12a−4=44

12a−4+4=44+4

12a=48

Then, solve for a by using the inverse of multiplication – division.

12a=48

12a/12=48/12

a=4

### Review

Solve each two-step equation that has multiplication and subtraction in it.

1. 4x−3=13
2. 5y−8=22
3. 7x−11=31
4. 8y−15=25
5. 9x−12=42
6. 12y−9=99
7. 2y−3=23
8. 3x−8=19
9. 5y−2=28
10. 7x−11=38
11. 5y−9=51
12. 6a−12=30
13. 9x−14=13
14. 12x−23=49
15. 13y−3=23
16. 18x−12=42