# 3.2: Solving Multi-Step Equations

Difficulty Level: At Grade Created by: CK-12

## Introduction

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 lesson and you will be able to figure out how long the rehearsals were at the end of the lesson.

What You Will Learn

In this lesson you will learn how to complete the following skills.

• Solve equations involving the combining of like terms.
• Solve equations involving parentheses and the distributive property
• Solve equations involving both distributing and combining.
• Model and solve real-world problems using multi-step equations.

Teaching Time

I. Solve Equations Involving the Combining of Like Terms

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.

Example

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.

Example

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 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.

\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.

Yes. You can check these by substituting the value of the variable back into the original equation. Just be sure to substitute it anywhere that you see that variable.

Let’s look at the last example.

\begin{align*}6n + 3 + 8n +2 &= 33\\ 6(2) + 3 + 8(2) + 2 &= 33\\ 12 + 3 + 16 + 2 &= 22\\ 33 &= 33\end{align*}

Our work is accurate and checks out.

II. Solving Equations Involving Parentheses and the Distributive Property

The distributive property may also help you solve some equations. This property states that when a factor is multiplied by the sum of two numbers, we can multiply each of the two numbers by that factor and then add them.

Examples: \begin{align*}7 \times (4+k) = (7 \times 4) + (7 \times k) = 28 + 7k\end{align*}

\begin{align*}2(a + 3) = (2 \times a) + (2 \times 3) = 2a + 6\end{align*}

Multiplication can also be distributed over subtraction.

Examples: \begin{align*}7 \times (4 - k) = (7 \times 4) - (7 \times k) = 28 - 7k\end{align*}

\begin{align*}2(a - 3) = (2 \times a) - (2 \times 3) = 2a - 6\end{align*}

Let's see how the distributive property can help us solve some multi-step equations.

Example

Solve for \begin{align*}k\end{align*}: \begin{align*}5(3 + k) = 45\end{align*}

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by 5 and then add those products.

\begin{align*}5(3 + k) &= 45\\ (5 \times 3) + (5 \times k) &= 45\\ 15 + 5k &= 45\end{align*}

Now, solve as you would solve any two-step equation. To get \begin{align*}5k\end{align*} by itself on one side of the equation, subtract 15 from both sides.

\begin{align*}15+5k &= 45\\ 15-15+5k &= 45-15\\ 0+5k &= 30\\ 5k &= 30\end{align*}

To get \begin{align*}k\end{align*} by itself on one side of the equation, divide both sides by 5.

\begin{align*}5k &= 30\\ \frac{5k}{5} &= \frac{30}{5}\\ 1k &= 6\\ k &= 6\end{align*}

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

Example

\begin{align*}2(y - 9) = 40\end{align*}

Now we can distribute the two by multiplying it by both of the terms inside the parentheses. Notice that the second term has a subtraction sign in front of it. Remember to include that sign when we multiply.

\begin{align*}2y - 18 = 40\end{align*}

Next, we solve this for \begin{align*}y\end{align*} as we would with any two step equation.

\begin{align*}2y - 18 &= 40\\ 2y - 18 + 18 &= 40 + 18\\ 2y &= 58\\ y &= 29\end{align*}

The value of \begin{align*}y\end{align*} is 29.

Sometimes, there will be negative numbers that will need to be worked with in an equation. Let’s look at an example like this.

Example

Solve for \begin{align*}x\end{align*}: \begin{align*}3(3 - x) =12\end{align*}

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by 3 and then subtract those products. It may help you to remember that \begin{align*}x = 1x\end{align*}.

\begin{align*}3(3-x) &= 12\\ 3(3 - 1x) &= 12\\ (3 \cdot 3) - (3 \cdot 1x) &= 12\\ 9 - 3x &= 12\end{align*}

Now, solve as you would solve any two-step equation. We need to first get the term that includes a variable, \begin{align*}3x\end{align*}, by itself on one side of the equation. In the equation, \begin{align*}3x\end{align*} is subtracted from 9. Subtracting \begin{align*}9 - 3x\end{align*} is the same as adding \begin{align*}9 + (-3x)\end{align*}. Rewrite the left side of the equation to show that 9 is being added to \begin{align*}-3x\end{align*}, and then subtract 9 from both sides.

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

To get \begin{align*}x\end{align*} by itself on one side of the equation, divide both sides by -3. You will need to use what you know about dividing integers to help you. For example, you know that when you divide two negative integers, the quotient will be positive. Since you know that \begin{align*}-3 \div (-3) = 1\end{align*}, you also know that \begin{align*}-3x \div (-3) = 1x\end{align*}. To review how to compute with integers, look back at Lessons 2.5 and 2.6.

\begin{align*}-3x &= 3\\ \frac{-3x}{-3} &= \frac{3}{-3}\\ 1x &= -1\\ x &= -1\end{align*}

The value of \begin{align*}x\end{align*} is -1.

You can see that we apply the same steps when working with a negative term as we do when working with positive terms. The key is to keep everything straight and follow the steps.

III. Solve Equations Involving Both Distributing and Combining

To solve some multi-step equations you will need to use the distributive property and combine like terms. When this happens, you will see that there is more than one term with the same variable or there is more than one number in the equation. You always want to combine everything that you can before moving on to solving the equation. Let’s start by looking at an example.

Example

Solve for \begin{align*}m\end{align*}: \begin{align*}6(1 +2m) -3m = 24\end{align*}

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by 6 and then add those products.

\begin{align*}6(1 + 2m) - 3m &= 24\\ (6 \times 1) + (6 \times 2m) - 3m &=24\\ 6 + 12m - 3m &= 24\end{align*}

Next, subtract the like terms—\begin{align*}12m\end{align*} and \begin{align*}3m\end{align*}—on the left side of the equation.

\begin{align*}6 + 12m - 3m &= 24\\ 6 + (12m - 3m) &= 24\\ 6 + 9m &= 24\end{align*}

Finally, solve as you would solve any two-step equation. Subtract 6 from both sides of the equation.

\begin{align*}6 + 9m &= 24\\ 6-6 + 9m &= 24-6\\ 0 + 9m &= 18\\ 9m &= 18\end{align*}

Now, divide both sides of the equation by 9.

\begin{align*}9m &= 18\\ \frac{9m}{9} &= \frac{18}{9}\\ 1m &= 2\\ m &= 2\\\end{align*}

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

Example

Solve for \begin{align*}b\end{align*}: \begin{align*}-4(2 + 3b) + 5b = 13\end{align*}

Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by -4 and then add those products.

\begin{align*}-4(2 + 3b) + 5b &= 13\\ (-4 \times 2) + (-4 \times 3b) + 5b &= 13\\ -8 + (-12b) + 5b &= 13\end{align*}

Next, add the like terms on the left side of the equation. To add those like terms, \begin{align*}-12b\end{align*} and \begin{align*}5b\end{align*}, you will need to use what you know about adding integers. To review how to add and subtract integers, look back at Lesson 2.5.

\begin{align*}-8 + (-12b) + 5b &= 13\\ -8 + (-12b + 5b) &= 13\\ -8 + (-7b) &= 13\end{align*}

Finally, solve as you would solve any two-step equation. Since -8 is added to \begin{align*}(-7b)\end{align*}, you can subtract -8 from both sides of the equation to solve it.

\begin{align*}-8 + (-7b) &= 13\\ -8 - (-8) + (-7b) &= 13 - (-8)\\ (-8 + 8) + (-7b) &= 13 + 8\\ 0 + (-7b) &= 21\\ -7b &= 21\end{align*}

Now, divide both sides of the equation by -7.

\begin{align*}-7b &= 21\\ \frac{-7b}{-7} &= \frac{21}{-7}\\ 1b &= -3\\ b &= -3\end{align*}

The value of \begin{align*}b\end{align*} is -3.

The skills you have learned in this lesson can also help you solve real-world problems. Let's take a closer look at that now.

IV. Model and Solve Real World Problems Using Multi-Step Equations

Sometimes, you can write a multi-step equation to represent a problem. If so, you can solve the problem by solving that equation. Think back to the key words necessary when writing equations. Then break down the problem one step at a time to write the equation. Then you can use what you have learned to solve the equation.

Example

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.

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.

Now let’s use what we have learned to work on the problem from the introduction.

## Real-Life Example Completed

Band Practice

Here is the original equation once again. First, reread it. Then name the variable that you are looking to figure out and write an equation. Next, solve the equation and write the two rehearsal times down. Your answer will have three parts, an equation and two different rehearsal times.

“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.

Now write an equation and use it to figure out the two rehearsal times. Remember there will be three parts to your answer.

Solution to Real – Life Example

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

Here are the vocabulary words that are found in this lesson.

Like Terms
terms that include a common variable.
states that the order that you add different numbers does not change the sum.
states that you can change the groupings of numbers being added without changing the sum.
Distributive Property
states that you can multiply a term outside of a set of parentheses with the terms inside the parentheses to simplify the set of parentheses.

## Time to Practice

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

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

Directions: Combine like terms and solve each equation.

1. \begin{align*}8x+3x+2=24\end{align*}
2. \begin{align*}5y+2y+6=48\end{align*}
3. \begin{align*}4x-6x+3=13\end{align*}
4. \begin{align*}7y-10y+6=9\end{align*}
5. \begin{align*}5x+8x+4=30\end{align*}
6. \begin{align*}9a+3a-4=44\end{align*}
7. \begin{align*}7a+4a+6=73\end{align*}
8. \begin{align*}12x-14x+3=19\end{align*}
9. \begin{align*}10y-16y+5=35\end{align*}

Directions: Use the distributive property to solve each equation.

1. \begin{align*}2(x+3)=10\end{align*}
2. \begin{align*}5(x+4)=25\end{align*}
3. \begin{align*}9(x-3)=27\end{align*}
4. \begin{align*}7(x+5)=70\end{align*}
5. \begin{align*}5(x-6)=35\end{align*}
6. \begin{align*}8(y-4)=40\end{align*}

Directions: Distribute and combine like terms and then solve each equation.

1. \begin{align*}x+8(x+2)=53\end{align*}
2. \begin{align*}2y+6(y+3)=34\end{align*}
3. \begin{align*}4y+2(y-2)=8\end{align*}
4. \begin{align*}9y+3(y-6)=30\end{align*}

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