# 2.4.2: Multi-Step Equations with Like Terms and Distribution

- Page ID
- 4359

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

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

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

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

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

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

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

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

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

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

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

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

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

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

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

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

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

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

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

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

You are asked to go to the store to buy 4 cartons of milk. You are given $15, which is enough to cover the cost of the milk with $2 leftover dollars for you as payment for the chore. How much does each carton of milk cost?

In this concept, you will learn to solve equations involving combining **like terms**.

**Combining Like Terms**

When you add or subtract the terms in an expression, you can only combine **like terms**. **Like terms** are terms that contain the same variable, and these terms can be combined.

Let’s look at an example.

At the fruit stand, you buy 3 yellow peaches, 5 white peaches and 2 red apples. Write an equation to represent the number of peaches and apples you bought.

First, notice these are different kinds of fruit you bought at the fruit stand. Therefore, let p represent the number of peaches and a represent the number of apples.

Next, write your equation to show the number of peaches and apples.

3 peaches+5 peaches+2 apples

Or

3p+5p+2a

Then, combine like terms.

3p+5p+2a=8p+2a

Let’s take a look at how you can apply what you know about combining like terms to solving algebraic equations.

Solve for r in the following equation.

5r−r−9=15

First, combine the like terms on the left side of the equation.

5r−r−9=15

(5r−r)−9=15

4r−9=15

Notice that 9 cannot be combined with 4r because they are not like terms.

Then you can solve the equation as we would solve any two-step equation. The next step is to isolate the term with the variable, 4r, on one side of the equation. Since 9 is * subtracted* from 4r, you should add 9 to both sides of the equation to isolate that term.

4r−9=15

4r−9+9=15+9

4r=24

Then, 4r means 4×r, you should divide each side of the equation by 4 to get the r by itself on one side of the equation.

4r=24

4r/4=24/4

r=6

The answer is 6.

Let’s look at another example.

Solve for “n” in the following equation.

6n+3+8n+2=23

First, combine the like terms on the left side of the equation. The terms 6n and 8n are like terms since each has the same variable, n. The numbers 3 and 2 are also like terms, so they can be combined as well.

Use the ** commutative property of addition** to help you reorder the terms being added. This property states that terms can be added in any order. Then use the

**associative property of addition**to group the terms so like terms are being added. The associative property of addition states that the grouping of terms being added does not matter.

6n+3+8n+2=33

6n+(3+8n)+2=33

6n+(8n+3)+2=33

(6n+8n)+(3+2)=33

Next, now that the like terms are grouped together with parentheses, combine them.

(6n+8n)+(3+2)=33

14n+5=33

Then, solve as you would solve any two-step equation. The next step is to isolate the term with the variable, 14n, on one side of the equation. Since 5 is added to 14n, you should subtract 5 from both sides of the equation to isolate the variable.

14n+5=33

14n+5−5=33-5

14n=28

Then, since 14n means 14×n, you should divide each side of the equation by 14 to get the n by itself on one side of the equation.

14n=28

14n/14=28/14

n=2

The answer is 2.

**Examples**

Example 2.4.2.1

Earlier, you were given a problem about the task of buying milk.

You are given $15, which is enough to cover the cost of the milk with $2 leftover dollars for you as payment for the chore.

**Solution**

First, you need to name your variable. Since you are going to buy milk, let m represent the price of the milk.

Next, write your equation. You are buying 4 cartons of milk so 4m. You will have $2 leftover after paying for the milk out of the $15. So your equation is:

4m+2=15

Then, since you are adding 2 to the left side, isolate your variable by subtracting 2.

4m+2=15

4m+2−2=15-2

4m=13

Then, solve for m by dividing by 2.

4m=13

4m/4=13/4

m=3.25

The answer is 3.25.

Each carton of milk costs $3.25.

Example 2.4.2.2

Yesterday, Tanya biked 3 more miles than she biked today. She biked a total of 13 miles on both days.

- Let m stand for the number of miles Tanya biked today. Write an algebraic equation to represent the number of miles Tanya biked on both days.
- Find the number of miles Tanya biked today.
- Find the number of miles Tanya biked yesterday.

**Solution**

Part 1.

First, you know that m represents the number of miles Tanya biked today. Use that variable to write an expression for the number of miles Tanya biked yesterday. You know that she biked 3 more miles yesterday than today and the total for the two days was 13. Therefore:

m+(m+3)=13

Part 2.

Next, combine like terms. Use the commutative property of addition to rearrange the terms being added so it is easier to see how to add the like terms.

m+(m+3)=13

(m+m)+3=13

2m+3=13

Then, solve the equation for m as you would solve any two-step equation. Subtract 3 from both sides of the equation.

2m+3=13

2m+3−3=13-3

2m=10

Then divide both sides by 2 to solve for m.

2m=10

2m/2=10/2

m=5

The answer is 5. Therefore Tanya biked 5 miles today.

Part 3.

First, you know that Tanya biked 3 more miles yesterday than today. You also know that she biked 5 miles today. Therefore you need to substitute 5 in for m.

3+m=3+5

3+m=8

The answer is 8.

Tanya biked 8 miles yesterday.

Example 2.4.2.3

3p+5p+2=18

**Solution**

First, combine like terms.

3p+5p+2=18

8p+2=18

Next, subtract 2 from both sides to isolate the variable.

8p+2=18

8p+2−2=18-2

8p=16

Then, solve for p by dividing both sides by 8.

8p=16

8p/8=16/8

p=2

The answer is 2.

Example 2.4.2.4

3x+6x+14=41

**Solution**

First, combine like terms.

3x+6x+14=41

9x+14=41

Next, subtract 14 from both sides to isolate the variable.

9x+14=41

9x+14−14=41-14

9x=27

Then, solve for x by dividing both sides by 9.

9x=27

9x/9=27/9

p=3

The answer is 3.

Example 2.4.2.5

3x+5x+9x−7=44

**Solution**

First, combine like terms.

3x+5x+9x−7=44

17x−7=44

Next, add 7 from both sides to isolate the variable.

17x−7=44

17x−7+7=44+7

17x=51

Then, solve for x by dividing both sides by 17.

17x=51

17x/17=51/17

x=3

The answer is 3.

### Review

Practice combining like terms as you simplify each expression.

1. 8x+3x+2

2. 5y−3y+8

3. 6x+9x+x−4

4. 9x+4x−8+2x

5. 2y−10y+16

6. 3x+4x+5−6+2x

Combine like terms and solve each equation.

7. 8x+3x+2=24

8. 5y+2y+6=48

9. 4x−6x+3=13

10. 7y−10y+6=9

11. 5x+8x+4=30

12. 9a+3a−4=44

13. 7a+4a+6=83

14. 12x−14x+3=19

15. 10y−16y+5=35

### Review (Answers)

To see the Review answers, open this PDF file and look for section 3.5.

### Vocabulary

Term | Definition |
---|---|

Commutative Property |
The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example a+b=b+a and\,(a)(b)=(b)(a). |

like terms |
Terms are considered like terms if they are composed of the same variables with the same exponents on each variable. |

### Additional Resources

Video:

**Solving Multi-Step Equations by Distributing - Example 1**

PLIX: Play, Learn, Interact, eXplore: **Multi-Step Equations with Like Terms: Shipments**

Activity: **Distributive Property for Multi-Step Equations Discussion Questions**

Study Aid: **Linear Equations Study Guide**

Practice: **Multi-Step Equations with Like Terms and Distribution**

Real World Application: **Car Loan**