# 3.5: Solve Equations Involving Combining Like Terms

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

**Practice**Multi-Step Equations with Like Terms

Have you ever had a long rehearsal for something? Take a look at what is happening at band practice.

“Wow, that was quite a day rehearsal,” Jake said as he put his things away in the band room.

“I agree. I’m beat,” Anica said.

“We rehearsed for longer today than we did yesterday,” Jake said.

“Yes, 45 minutes longer. So we rehearsed for a total of five hours counting yesterday and today,” Anica said.

“Wait a minute you left me in the dust. How many minutes did we rehearse yesterday, and how many did we rehearse today?” Jake asked sitting down in a band chair.

“Okay, let me show you. You need an equation,” Anica said taking out a piece of paper and a pencil.

**Do you know how Anica can figure this out? You will once you know how to work with multi-step equations. Pay attention to this Concept and you will be able to figure out how long the rehearsals were at the end of it.**

### Guidance

**Consider this simple problem.**

**Suppose you bought 3 yellow peaches, 5 white peaches and 2 red apples at a fruit stand.**

You could also say that you bought 8 peaches and 2 apples because:

\begin{align*}3 \ peaches + 5 \ peaches + 2 \ apples = 8 \ peaches + 2 \ apples\end{align*}.

However, you couldn't say that bought 10 peaches. When you determine how many peaches you bought, you can add 3 yellow peaches to 5 white peaches, but you cannot add the 2 red apples to that total. That is because apples and peaches are *different* kinds of fruit.

**When you add or subtract the terms in an expression, you can only combine** *like terms***.**

Consider this expression:

\begin{align*}3p + 5 p + 2a\end{align*}

This expression represents the problem above. The variable \begin{align*}p\end{align*} stands for peaches. The variable \begin{align*}a\end{align*} stands for apples.

Just as you can combine the yellow peaches with the white peaches because they are both peaches, you can combine \begin{align*}3p\end{align*} and \begin{align*}5p\end{align*} because they are like terms. Each of those terms includes the same variable, \begin{align*}p\end{align*}. However, you could not combine \begin{align*}5p\end{align*} with \begin{align*}2a\end{align*}, because they are not like terms. Each of those terms has a different variable.

*Like terms***are terms that contain the same variable, and these terms can be combined.**

This shows how you could simplify the expression above by combining like terms:

\begin{align*}3p + 5p + 2a = 8p + 2a\end{align*}.

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

*Solve for \begin{align*}r\end{align*}: \begin{align*}5r - r - 9 = 15\end{align*}.*

**First, combine the like terms—\begin{align*}5r\end{align*} and \begin{align*}r\end{align*} on the left side of the equation. It may help to remember that \begin{align*}r = 1r\end{align*}.**

\begin{align*}5r - r - 9 & = 15\\ (5r - 1r) - 9 & = 15\\ 4r - 9 & = 15\end{align*}

**Notice that 9 cannot be combined with \begin{align*}4r\end{align*} because they are** *not***like terms.**

**Now that we have combined like terms, we can solve the equation as we would solve any two-step equation.**

**Our next step is to isolate the term with the variable, \begin{align*}4r\end{align*}, on one side of the equation. Since 9 is** *subtracted***from \begin{align*}4r\end{align*}, we should add 9 to both sides of the equation to isolate that term.**

\begin{align*}4r - 9 & = 15\\ 4r - 9 + 9 & = 15 + 9\\ 4r +(-9 + 9) & = 24\\ 4r + 0 & = 24\\ 4r & = 24\end{align*}

Since \begin{align*}4r\end{align*} means \begin{align*}4 \times r\end{align*}, we should divide each side of the equation by 4 to get the \begin{align*}r\end{align*} by itself on one side of the equation.

\begin{align*}4r & = 24\\ \frac{4r}{4} & = \frac{24}{4}\\ 1r & = 6\\ r & = 6\end{align*}

**The value of \begin{align*}r\end{align*} is 6.**

*Solve for \begin{align*}n\end{align*}: \begin{align*}6n + 3 + 8n + 2 = 33\end{align*}.*

**First, combine the like terms on the left side of the equation. The terms \begin{align*}6n\end{align*} and \begin{align*}8n\end{align*} are like terms since each has the same variable, \begin{align*}n\end{align*}. 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

**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.**

*associative property of addition*\begin{align*}6n + 3 + 8n + 2 & = 33\\ 6n + (3 + 8n) + 2 & = 33\\ 6n + (8n + 3) + 2 & = 33\\ (6n + 8n) + (3 + 2) & = 33\end{align*}

**Now, that the like terms are grouped together with parentheses, combine them.**

\begin{align*}(6n + 8n) + (3 + 2) & = 33\\ 14n + 5 & = 33\end{align*}

**Now, we can solve as we would solve any two-step equation.**

**The next step is to isolate the term with the variable, \begin{align*}14n\end{align*}, on one side of the equation. Since 5 is** *added***to \begin{align*}14n\end{align*}, we should subtract 5 from both sides of the equation to do this.**

\begin{align*}14n + 5 & = 33\\ 14n + 5 - 5 & = 33 - 5\\ 14n + 0 & = 28\\ 14n & = 28\end{align*}

**Since \begin{align*}14n\end{align*} means \begin{align*}14 \times n\end{align*}, we should divide each side of the equation by 14 to get the \begin{align*}n\end{align*} by itself on one side of the equation.**

\begin{align*}14 & = 28\\ \frac{14n}{14} & = \frac{28}{14}\\ 1n & = 2\\ n & = 2\end{align*}

**The value of \begin{align*}n\end{align*} is 2.**

#### Example A

\begin{align*}3p + 5p + 2 = 18\end{align*}

**Solution: \begin{align*}p = 2\end{align*}**

#### Example B

\begin{align*}13x + 6x + 14a - 9a\end{align*}

**Solution: \begin{align*}19x + 5a\end{align*}**

#### Example C

\begin{align*}3x + 5x + 9x - 7 = 44\end{align*}

**Solution: \begin{align*}x = 3\end{align*}**

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

**First, we need to name the variable. We are looking to figure out times, so we can use \begin{align*}t\end{align*} as our variable.**

\begin{align*}t = time\end{align*}

**Next, we can write an equation. We know that there are two times.**

\begin{align*}t + t\end{align*}

**But also, one day was 45 minutes longer.**

\begin{align*}t + t + 45\end{align*}

**The total sum of time is 5 hours. We need to make sure both our units are the same, so we convert 5 hours into minutes and write 300 minutes.**

\begin{align*}t + t + 45 = 300\end{align*}

**Here is our equation.**

**Next, we solve it for the two times.**

\begin{align*}t + t + 45 &= 300\\ 2t &= 300 - 45\\ 2t &= 255\\ t &= 127.5 \ minutes\end{align*}

**This is the time that the band rehearsed yesterday. They rehearsed 45 more minutes today. We add 45 to the total time from yesterday.**

\begin{align*}127.5 + 45 = 172.5 \ minutes\end{align*}

### Vocabulary

- Like Terms
- terms that include a common variable.

- Commutative Property of Addition
- states that the order that you add different numbers does not change the sum.

### Guided Practice

Here is one for you to try on your own.

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

a. Let \begin{align*}t\end{align*} stand for the number of miles Tanya biked today. Write an algebraic equation to represent the number of miles Tanya biked on both days.

b. Find the number of miles Tanya biked today.

c. Find the number of miles Tanya biked yesterday.

**Solution**

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

You know that \begin{align*}t\end{align*} represents the number of miles Tanya biked today. Use that variable to write an expression for the number of miles Tanya biked yesterday.

\begin{align*}& Yesterday, \ Tanya \ biked \ \underline{3} \ \underline{more \ldots than} \ she \ biked \ \underline{today}.\\ & \qquad \qquad \qquad \qquad \qquad \downarrow \qquad \ \ \downarrow \qquad \qquad \qquad \qquad \downarrow\\ & \qquad \qquad \qquad \qquad \qquad \ 3 \qquad \ + \qquad \qquad \qquad \quad \ \ t\end{align*}

So, you know that Tanya biked \begin{align*}t\end{align*} miles today and \begin{align*}3 + t\end{align*} miles yesterday. You also know that she biked a *total* of 13 miles on both days. Use this information to write an addition equation for this problem.

\begin{align*}&(\text{miles biked today}) + (\text{miles biked yesterday}) = (\text{total miles biked})\\ & \qquad \qquad \downarrow \qquad \qquad \downarrow \ \ \qquad \qquad \quad \downarrow \qquad \qquad \ \ \downarrow \qquad \qquad \ \ \downarrow\\ & \qquad \qquad \ t \qquad \quad \ \ + \qquad \qquad \ \ 3 + t \qquad \quad \ \ = \qquad \qquad 13\end{align*}

So, this problem can be represented by the equation, \begin{align*}t + 3 + t = 13\end{align*}.

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

The variable \begin{align*}t\end{align*} represents the number of miles Tanya biked today. So, solve the equation for \begin{align*}t\end{align*}.

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

\begin{align*}t + (3 + t) &= 13\\ t + (t + 3) &= 13\end{align*}

Now, add the like terms on the left side of the equation.

\begin{align*}t + t + 3 &= 13\\ (t + t) + 3 &= 13\\ 2t + 3 &= 13\end{align*}

Solve the equation for \begin{align*}t\end{align*} as you would solve any two-step equation. Subtract 3 from both sides of the equation.

\begin{align*}2t + 3 &= 13\\ 2t + 3 - 3 &= 13 - 3\\ 2t + 0 &= 10\\ 2t &= 10\end{align*}

Then, divide both sides of the equation by 2.

\begin{align*}2t &= 10\\ \frac{2t}{2} &= \frac{10}{2}\\ 1t &= 5\\ t &= 5\end{align*}

The value of \begin{align*}t\end{align*} is 5, so Tanya biked 5 miles today.

**Consider part** *c***next.**

In part *a*, you determined that Tanya biked \begin{align*}3 + t\end{align*} miles yesterday. Since \begin{align*}t = 5\end{align*}, substitute 5 for \begin{align*}t\end{align*} in the expression to find how many miles she biked yesterday.

\begin{align*}3 + t = 3 + 5 = 8\end{align*}

**Tanya biked 8 miles yesterday.**

### Video Review

### Practice

Directions: Practice combining like terms as you simplify each expression.

- \begin{align*}8x + 3x +2\end{align*}
- \begin{align*}5y-3y+8\end{align*}
- \begin{align*}6x + 9x + x - 4\end{align*}
- \begin{align*}9x+4x-8+2x\end{align*}
- \begin{align*}2y-10y+16\end{align*}
- \begin{align*}3x+4x+5-6+2x\end{align*}

Directions: Combine like terms and solve each equation.

- \begin{align*}8x+3x+2=24\end{align*}
- \begin{align*}5y+2y+6=48\end{align*}
- \begin{align*}4x-6x+3=13\end{align*}
- \begin{align*}7y-10y+6=9\end{align*}
- \begin{align*}5x+8x+4=30\end{align*}
- \begin{align*}9a+3a-4=44\end{align*}
- \begin{align*}7a+4a+6=83\end{align*}
- \begin{align*}12x-14x+3=19\end{align*}
- \begin{align*}10y-16y+5=35\end{align*}

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 .like terms

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

Here you'll learn to solve equations involving combining like terms.

## Concept Nodes:

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 .like terms

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