Skip to main content
permalink

Section 2.1 Solving Multistep Linear Equations

permalink

permalinkWe learned how to solve one-step equations in Section 1.5. In this section, we will solve equations that need more than one step.

permalink
Figure 2.1.1. Alternative Video Lessons
permalink

Subsection 2.1.1 Solving Two-Step Equations

permalink
Example 2.1.2.

A water tank can hold up to 140 gallons of water, but it starts with only 5 gallons. A tap is turned on, pouring 15 gallons of water into the tank every minute. After how many minutes will the tank be full? Let's find a pattern first.

permalink
Minutes Since Tap
Was Turned on
Amount of Water in
the Tank (in Gallons)
0 5
1 151+5=20
2 152+5=35
3 153+5=50
4 154+5=65
m 15m+5
Figure 2.1.3. Amount of Water in the Tank

We can see that after m minutes, the tank has 15m+5 gallons of water. This makes sense since the tap pours 15 gallons into the tank for each of m minutes, and it has 5 gallons to start with. To find when the tank will be full with 140 gallons, we can write the equation

15m+5=140

First, we need to isolate the variable term, 15m. In other words, we need to “remove” the 5 from the left side of the equals sign. We can do this by subtracting 5 from both sides of the equation. Once the variable term is isolated, we can eliminate its coefficient and solve for m.

The full process is:

15m+5=14015m+55=140515m=13515m15=13515m=9

We should check the solution by substituting m with 9 in the original equation:

15m+5=14015(9)+5=?140135+5=140

The solution 9 is checked.

This problem had context. It was not simply solving an equation, rather it comes with the story of the tank filling with water. So we report a conclusion that uses that context: In summary, the tank will be full after 9 minutes.

permalinkIn solving the two-step equation in Example 2.1.2, we first isolated the variable expression 15m and then eliminated the coefficient of 15 by dividing each side of the equation by 15. These two steps are at the heart of our approach to solving linear equations. For more complicated equations, we may need to simplify one or both sides first. Below is a general approach that summarizes all of this.

permalink
Example 2.1.5.

Solve for y in the equation 73y=8.

Explanation

To solve, we first separate the variable terms and constant terms to different sides of the equation. Then we eliminate the variable term's coefficient.

\begin{align*} 7-3y\amp=-8\\ 7-3y\subtractright{7}\amp=-8\subtractright{7}\\ -3y\amp=-15\\ \divideunder{-3y}{-3}\amp=\divideunder{-15}{-3}\\ y\amp=5 \end{align*}

Checking the solution \(y=5\text{:}\)

\begin{align*} 7-3y\amp=-8\\ 7-3(\substitute{5})\amp\stackrel{?}{=}-8\\ 7-15\amp\stackrel{\checkmark}{=}-8 \end{align*}

So the solution to the equation \(7-3y=-8\) is \(5\) and the solution set is \(\{5\}\text{.}\)

permalink

Subsection 2.1.2 Solving Multistep Linear Equations

permalink
Example 2.1.6.

Ahmed has saved $2,500.00 in his savings account and is going to start saving $550.00 per month. Julia has saved $4,600.00 in her savings account and is going to start saving $250.00 per month. If this situation continues, how many months will it take Ahmed catch up with Julia in savings?

Ahmed saves $550.00 per month, so he can save 550m dollars in m months. With the $2,500.00 he started with, after m months he has 550m+2500 dollars. Similarly, after m months, Julia has 250m+4600 dollars. To find when those two accounts will have the same amount of money, we can write the equation

550m+2500=250m+4600.
550m+2500=250m+4600550m+25002500=250m+46002500550m=250m+2100550m250m=250m+2100250m300m=2100300m300=2100300m=7

Checking the solution 7:

550m+2500=250m+4600550(7)+2500=?250(7)+46003850+2500=?1750+46006350=6350

Ahmed will catch up with Julia's savings after 7 months.

permalink
Example 2.1.7.

Solve for x in 52x=5x9.

Explanation
\begin{align*} 5-2x\amp=5x-9\\ 5-2x\subtractright{5}\amp=5x-9\subtractright{5}\\ -2x\amp=5x-14\\ -2x\subtractright{5x}\amp=5x-14\subtractright{5x}\\ -7x\amp=-14\\ \divideunder{-7x}{-7}\amp=\divideunder{-14}{-7}\\ x\amp=2 \end{align*}

Checking the solution \(2\text{:}\)

\begin{align*} 5-2x\amp=5x-9\\ 5-2(\substitute{2})\amp\stackrel{?}{=}5(\substitute{2})-9\\ 5-4\amp\stackrel{?}{=}10-9\\ 1\amp\stackrel{\checkmark}{=}1 \end{align*}

Therefore the solution is \(2\) and the solution set is \(\{2\}\text{.}\)

permalink
Example 2.1.8.

In Example 2.1.7, we could have moved variable terms to the right side of the equals sign, and number terms to the left side. We chose not to. There's no reason we couldn't have moved variable terms to the right side though. Let's compare:

52x=5x952x+9=5x9+9142x=5x142x+2x=5x+2x14=7x147=7x72=x

Lastly, we could save a step by moving variable terms and number terms in one step:

52x=5x952x+2x+9=5x9+2x+914=7x147=7x72=x

For the sake of a slow and careful explanation, the examples in this chapter will move variable terms and number terms separately.

permalink
Checkpoint 2.1.9.

permalinkThe next example requires combining like terms.

permalink
Example 2.1.10.

Solve for n in n9+3n=n3n.

Explanation

To start solving this equation, we'll need to combine like terms. After this, we can put all terms containing \(n\) on one side of the equation and finish solving for \(n\text{.}\)

\begin{align*} n-9+3n\amp=n-3n\\ 4n-9\amp=-2n\\ 4n-9\subtractright{4n}\amp=-2n\subtractright{4n}\\ -9\amp=-6n\\ \divideunder{-9}{-6}\amp=\divideunder{-6n}{-6}\\ \frac{3}{2}\amp=n \end{align*}

Checking the solution \(\frac{3}{2}\text{:}\)

\begin{align*} n-9+3n\amp=n-3n\\ \substitute{\frac{3}{2}}-9+3\left(\substitute{\frac{3}{2}}\right)\amp\stackrel{?}{=}\substitute{\frac{3}{2}}-3\left(\substitute{\frac{3}{2}}\right)\\ \frac{3}{2}-9+\frac{9}{2}\amp\stackrel{?}{=}\frac{3}{2}-\frac{9}{2}\\ \frac{12}{2}-9\amp\stackrel{?}{=}-\frac{6}{2}\\ 6-9\amp\stackrel{\checkmark}{=}-3 \end{align*}

The solution to the equation \(n-9+3n=n-3n\) is \(\frac{3}{2}\) and the solution set is \(\left\{\frac{3}{2}\right\}\text{.}\)

permalink
Checkpoint 2.1.11.

permalinkWe should be careful when we distribute a negative sign into the parentheses, like in the next example.

permalink
Example 2.1.12.

Solve for a in 4(3a)=22(2a+1).

Explanation

To solve this equation, we will simplify each side of the equation, manipulate it so that all variable terms are on one side and all constant terms are on the other, and then solve for \(a\text{:}\)

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-3+a\amp=-2-4a-2\\ 1+a\amp=-4-4a\\ 1+a\addright{4a}\amp=-4-4a\addright{4a}\\ 1+5a\amp=-4\\ 1+5a\subtractright{1}\amp=-4\subtractright{1}\\ 5a\amp=-5\\ \divideunder{5a}{5}\amp=\divideunder{-5}{5}\\ a\amp=-1 \end{align*}

Checking the solution \(-1\text{:}\)

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-(3-(\substitute{-1}))\amp\stackrel{?}{=}-2-2(2(\substitute{-1})+1)\\ 4-(4)\amp\stackrel{?}{=}-2-2(-1)\\ 0\amp\stackrel{?}{=}-2+2\\ 0\amp\stackrel{\checkmark}{=}0 \end{align*}

Therefore the solution to the equation is \(-1\) and the solution set is \(\{-1\}\text{.}\)

permalink

Subsection 2.1.3 Revisiting Applications from Section 1.8

permalinkIn Section 1.8, we set up equations given some background information, but we didn't try to solve the equations yet. Now we can do that.

permalink
Example 2.1.13.

Here we revisit Example 1.8.2.

A savings account starts with $500. Each month, an automatic deposit of $150 is made. Find the number of months it will take for the balance to reach $1,700.

Explanation

To set up an equation, we might start by making a table in order to identify a general pattern for the total amount in the account after \(m\) months. In Figure 2.1.14, we find the pattern is that after \(m\) months, the total amount saved is \(500+150m\text{.}\)

Using this pattern, we determine that an equation representing when the total savings equals \(\$1700\) is:

\begin{equation*} 500+150m=1700 \end{equation*}
permalink
Months Since
Saving Started
Total Amount Saved
(in Dollars)
\(0\) \(500\)
\(1\) \(500+150=650\)
\(2\) \(500+150(2)=800\)
\(3\) \(500+150(3)=950\)
\(4\) \(500+150(4)=1100\)
\(\vdots\) \(\vdots\)
\(m\) \(500+150m\)
Figure 2.1.14. Amount in Savings Account

To solve this equation, we can start by subtracting \(500\) from each side. Then we cna divide each side by \(150\text{.}\)

\begin{align*} 500+150m\amp=1700\\ 500+150m\subtractright{500}\amp=1700\subtractright{500}\\ 150m\amp=1200\\ \divideunder{150m}{150}\amp=\divideunder{1200}{150}\\ m\amp=8 \end{align*}

Checking the solution \(8\text{:}\)

\begin{align*} 500+150m\amp=1700\\ 500+150\amp(\substitute{8})\stackrel{?}{=}1700\\ 500+1200\amp\stackrel{\checkmark}{=}1700 \end{align*}

So \(8\) is the solution, and it checks out. This means it will take \(8\) months for the account balance to reach \(\$1700\text{.}\)

permalink
Example 2.1.15.

Here we revisit Example 1.8.4.

A bathtub contains 2.5 ft3 of water. More water is being poured in at a rate of 1.75 ft3 per minute. How long will it be until the amount of water in the bathtub reaches 6.25 ft3?

Explanation

Since this problem refers to when the amount of water will reach a certain amount, we immediately know that the unknown quantity is time. As the volume of water in the tub is measured in ft3 per minute, we know that time needs to be measured in minutes. We'll define \(t\) to be the number of minutes that water is poured into the tub. To determine this equation, we'll start by making a table of values:

permalink
Minutes Water
Has Been Poured
Total Amount
of Water (in ft3)
\(0\) \(2.5\)
\(1\) \(2.5+1.75=4.25\)
\(2\) \(2.5+1.75(2)=6\)
\(3\) \(2.5+1.75(3)=7.75\)
\(\vdots\) \(\vdots\)
\(t\) \(2.5+1.75t\)
Figure 2.1.16. Amount of Water in the Bathtub

Using this pattern, we determine that the equation representing when the amount will be 6.25 ft3 is:

\begin{equation*} 2.5+1.75t=6.25 \end{equation*}

To solve this equation, we can start by subtracting \(2.5\) from each side. Then we cna divide each side by \(1.75\text{.}\)

\begin{align*} 2.5+1.75t\amp=6.25\\ 2.5+1.75t\subtractright{2.5}\amp=6.25\subtractright{2.5}\\ 1.75t\amp=3.75\\ \divideunder{1.75t}{1.75}\amp=\divideunder{3.75}{1.75}\\ t\amp=\frac{375}{175}=\frac{15}{7}\approx2.143 \end{align*}

Checking the solution \(\frac{15}{7}\text{:}\)

\begin{align*} 2.5+1.75t\amp=6.25\\ 2.5+1.75\left(\substitute{\frac{15}{7}}\right)\amp\stackrel{?}{=}6.25\\ 2.5+\frac{7}{4}\cdot\frac{15}{7}\amp\stackrel{?}{=}6.25\\ 2.5+\frac{15}{4}\amp\stackrel{?}{=}6.25\\ 2.5+3.75\amp\stackrel{\checkmark}{=}6.25 \end{align*}

So \(\frac{15}{7}\) (about \(2.143\)) is the solution, and it checks out. This means it will take about \(2.143\) minutes for the tub to fill to 6.25 ft3.

permalink
Example 2.1.17.

Here we revisit Example 1.8.6.

Jakobi's annual salary as a nurse in Portland, Oregon, is $73,290. His salary increased by 4% from last year. What was Jakobi's salary last year?

Explanation

We need to find Jakobi's salary last year. So we'll introduce \(s\text{,}\) defined to be Jakobi's salary last year (in dollars). To set up the equation, we need to think about how he arrived at this year's salary. To get to this year's salary, his employer took last year's salary and added \(4\%\) to it. Conceptually, this means we have:

\begin{equation*} (\text{last year's salary})+(4\%\text{ of last year's salary}) = (\text{this year's salary}) \end{equation*}

We'll represent \(4\%\) of last year's salary with \(0.04s\) since \(0.04\) is the decimal representation of \(4\%\text{.}\) This means that the equation we set up is:

\begin{equation*} s+0.04s=73290 \end{equation*}

To solve this equation, we can start by simplifing the left side, and proceed from there.

\begin{align*} s+0.04s\amp=73290\\ 1s+0.04s\amp=73290\\ 1.04s\amp=73290\\ \divideunder{1.04s}{1.04}\amp=\divideunder{73290}{1.04}\\ s\amp\approx70471.15 \end{align*}

Checking the solution \(70471.15\text{:}\)

\begin{align*} s+0.04s\amp=73290\\ \substitute{70471.15}+0.04(\substitute{70471.15})\amp\stackrel{?}{=}73290\\ 70471.15+2818.846\amp\stackrel{?}{=}73290\\ 73289.996\amp\stackrel{?}{=}73290 \end{align*}

In the check, those values are not equal, but they are very close. And it is reasonable to believe the only reason they are at all different comes from when we rounded the real solution to \(70471.15\text{.}\) So Jakobi's salary was \(\$70471.15\) last year.

permalink
Checkpoint 2.1.18.

Here we revisit Exercise 1.8.7.

permalink
Example 2.1.19.

Here we revisit Example 1.8.8.

The price of a refrigerator after a 15% discount is $612. What was the price before the discount?

Explanation

We'll let \(c\) be the original price of the refrigerator. To obtain the discounted price, we take the original price and subtract \(15\%\) of that amount. Conceptually, this looks like:

\begin{equation*} (\text{original price})-(15\%\text{ of the original price})= (\text{discounted price}) \end{equation*}

Since the amount of the discount is \(15\%\) of the original price, we'll represent this with \(0.15c\text{.}\) The equation we set up is then:

\begin{equation*} c-0.15c=612 \end{equation*}

To solve this equation, we can start by simplifing the left side, and proceed from there.

\begin{align*} c-0.15c\amp=612\\ 1.00c-0.15c\amp=612\\ 0.85c\amp=612\\ \divideunder{0.85c}{0.85}\amp=\divideunder{612}{0.85}\\ c\amp=720 \end{align*}

Checking the solution \(720\text{:}\)

\begin{align*} c-0.15c\amp=612\\ \substitute{720}-0.15(\substitute{720})\amp\stackrel{?}{=}612\\ 720-108\amp\stackrel{\checkmark}{=}612 \end{align*}

The solution \(720\) checks out. So the refrigerator cost \(\$720\) before the discount was applied.

permalink
Checkpoint 2.1.20.

Here we revisit Exercise 1.8.9.

permalink

Subsection 2.1.4 Differentiating between Simplifying Expressions, Evaluating Expressions and Solving Equations

permalinkLet's look at the following similar, yet different examples.

permalink
Example 2.1.21.

Simplify the expression 103(x+2).

Explanation
\begin{align*} 10-3(x+2)\amp=10-3x-6\\ \amp=-3x+4 \end{align*}

An equivalent result is \(4-3x\text{.}\) Note that our final result is an expression.

permalink
Example 2.1.22.

Evaluate the expression 103(x+2) when x=2.

Explanation

We will substitute \(x=2\) into the expression:

\begin{align*} 10-3(x+2)\amp=10-3(\substitute{2}+2)\\ \amp=10-3(4)\\ \amp=10-12\\ \amp=-2 \end{align*}

So when \(x=2\text{,}\) \(10-3(x+2)=-2\text{.}\)

Note that our final result here is a numerical value.

permalink
Example 2.1.23.

Solve the equation 103(x+2)=x16.

Explanation
\begin{align*} 10-3(x+2)\amp=x-16\\ 10-3x-6\amp=x-16\\ -3x+4\amp=x-16\\ -3x+4\subtractright{4}\amp=x-16\subtractright{4}\\ -3x\amp=x-20\\ -3x\subtractright{x}\amp=x-20\subtractright{x}\\ -4x\amp=-20\\ \divideunder{-4x}{-4}\amp=\divideunder{-20}{-4}\\ x\amp=5 \end{align*}

So the solution set is \(\{5\}\text{.}\) (We should probably check that.)

Note that our final result here is a solution set.

permalinkHere is a summary collection of the distinctions that you should understand between simplifying expressions, evaluating expressions and solving equations.

permalink
List 2.1.24. A summary the differences among simplifying expressions, evaluating expressions and solving equations.
  • An expression like 103(x+2) can be simplified to 3x+4 (as in Example 2.1.21). However we cannot “solve” an expression like this.

  • As variables take different values, an expression can be evaluated to different values. In Example 2.1.22, when x=2, 103(x+2)=2; but when x=3, 103(x+2)=5.

  • An equation connects two expressions with an equals sign. In Example 2.1.23, 103(x+2)=x16 has the expression 103(x+2) on the left side of equals sign, and the expression x16 on the right side.

  • When we solve the equation 103(x+2)=x16, we are looking for a number which makes those two expressions have the same value. In Example 2.1.23, we found the solution to be 5. It makes both 103(x+2) and x16= equal to 11.

permalink

Reading Questions 2.1.5 Reading Questions

permalink
1.

Describe the five steps that you might need to go through when solving a general linear equation in one variable.

permalink
2.

In percent questions like tipping at a restaurant when you know the total bill, or purchasing something where sales tax is applied and you know the total charge, describe a common misunderstanding of what number to apply the percentage to.

permalink
3.

Explain what is wrong with saying “I need to solve 3x+x8.

permalink

Exercises 2.1.6 Exercises

Warmup and Review
permalink
1.

Solve the equation.

r+4=1

permalink
2.

Solve the equation.

r+1=7

permalink
3.

Solve the equation.

t7=1

permalink
4.

Solve the equation.

t4=2

permalink
5.

Solve the equation.

21=3t

permalink
6.

Solve the equation.

10=5x

permalink
7.

Solve the equation.

54t=3

permalink
8.

Solve the equation.

45a=9

Solving Two-Step Equations
permalink
9.

Solve the equation.

7c+4=25

permalink
10.

Solve the equation.

4A+2=30

permalink
11.

Solve the equation.

10C1=101

permalink
12.

Solve the equation.

6m5=11

permalink
13.

Solve the equation.

6=3p+3

permalink
14.

Solve the equation.

70=9q+2

permalink
15.

Solve the equation.

18=6y6

permalink
16.

Solve the equation.

28=3t4

permalink
17.

Solve the equation.

3a+4=22

permalink
18.

Solve the equation.

6c+1=13

permalink
19.

Solve the equation.

9A9=18

permalink
20.

Solve the equation.

4C6=30

permalink
21.

Solve the equation.

9=m+2

permalink
22.

Solve the equation.

17=p+7

permalink
23.

Solve the equation.

8q+80=0

permalink
24.

Solve the equation.

5y+35=0

Application Problems for Solving Two-Step Equations
permalink
25.

A gym charges members $25 for a registration fee, and then $20 per month. You became a member some time ago, and now you have paid a total of $405 to the gym. How many months have passed since you joined the gym?

months have passed since you joined the gym.

permalink
26.

Your cell phone company charges a $16 monthly fee, plus $0.17 per minute of talk time. One month your cell phone bill was $75.50. How many minutes did you spend talking on the phone that month?

You spent talking on the phone that month.

permalink
27.

A school purchased a batch of T-shirts from a company. The company charged $6 per T-shirt, and gave the school a $65 rebate. If the school had a net expense of $2,575 from the purchase, how many T-shirts did the school buy?

The school purchased T-shirts.

permalink
28.

James hired a face-painter for a birthday party. The painter charged a flat fee of $75, and then charged $5.50 per person. In the end, James paid a total of $141. How many people used the face-painter’s service?

people used the face-painter’s service.

permalink
29.

A certain country has 465.5 million acres of forest. Every year, the country loses 6.65 million acres of forest mainly due to deforestation for farming purposes. If this situation continues at this pace, how many years later will the country have only 206.15 million acres of forest left? (Use an equation to solve this problem.)

After years, this country would have 206.15 million acres of forest left.

permalink
30.

Randi has $86 in his piggy bank. He plans to purchase some Pokemon cards, which costs $1.95 each. He plans to save $52.85 to purchase another toy. At most how many Pokemon cards can he purchase?

Write an equation to solve this problem.

Randi can purchase at most Pokemon cards.

Solving Equations with Variable Terms on Both Sides
permalink
31.

Solve the equation.

10p+5=p+59

permalink
32.

Solve the equation.

8q+8=q+29

permalink
33.

Solve the equation.

4y+3=y27

permalink
34.

Solve the equation.

10r+7=r56

permalink
35.

Solve the equation.

62a=5a+55

permalink
36.

Solve the equation.

35c=5c+103

permalink
37.

Solve the equation.

8A+6=5A+8

permalink
38.

Solve the equation.

2C+6=10C+5

permalink
39.

Solve the equation.

  1. 9m+4=2m+32

  2. 2r+4=9r17

permalink
40.

Solve the equation.

  1. 6p+10=2p+42

  2. 2n+10=6n6

Application Problems for Solving Equations with Variable Terms on Both Sides
permalink
41.

Use a linear equation to solve the word problem.

Two trees are 8 feet and 13.5 feet tall. The shorter tree grows 3 feet per year; the taller tree grows 2.5 feet per year. How many years later would the shorter tree catch up with the taller tree?

It would take the shorter tree years to catch up with the taller tree.

permalink
42.

Use a linear equation to solve the word problem.

Massage Heaven and Massage You are competitors. Massage Heaven has 2700 registered customers, and it gets approximately 550 newly registered customers every month. Massage You has 6700 registered customers, and it gets approximately 350 newly registered customers every month. How many months would it take Massage Heaven to catch up with Massage You in the number of registered customers?

These two companies would have approximately the same number of registered customers months later.

permalink
43.

Use a linear equation to solve the word problem.

Two truck rental companies have different rates. V-Haul has a base charge of $55.00, plus $0.65 per mile. W-Haul has a base charge of $50.40, plus $0.70 per mile. For how many miles would these two companies charge the same amount?

If a driver drives miles, those two companies would charge the same amount of money.

permalink
44.

Use a linear equation to solve the word problem.

Massage Heaven and Massage You are competitors. Massage Heaven has 9800 registered customers, but it is losing approximately 300 registered customers every month. Massage You has 2650 registered customers, and it gets approximately 350 newly registered customers every month. How many months would it take Massage Heaven to catch up with Massage You in the number of registered customers?

These two companies would have approximately the same number of registered customers months later.

permalink
45.

Use a linear equation to solve the word problem.

Anthony has $75.00 in his piggy bank, and he spends $2.00 every day. Ken has $10.00 in his piggy bank, and he saves $4.50 every day.

If they continue to spend and save money this way, how many days later would they have the same amount of money in their piggy banks?

days later, Anthony and Ken will have the same amount of money in their piggy banks.

permalink
46.

Use a linear equation to solve the word problem.

Penelope has $100.00 in her piggy bank, and she spends $1.50 every day. Haley has $8.00 in her piggy bank, and she saves $2.50 every day.

If they continue to spend and save money this way, how many days later would they have the same amount of money in their piggy banks?

days later, Penelope and Haley will have the same amount of money in their piggy banks.

Solving Linear Equations with Like Terms
permalink
47.

Solve the equation.

5C+2C+2=16

permalink
48.

Solve the equation.

2m+5m+2=58

permalink
49.

Solve the equation.

8p+9+3=60

permalink
50.

Solve the equation.

5q+4+2=21

permalink
51.

Solve the equation.

1+8=2yy20

permalink
52.

Solve the equation.

4+2=5rr56

permalink
53.

Solve the equation.

3y+66y=27

permalink
54.

Solve the equation.

5y+106y=19

permalink
55.

Solve the equation.

7r+5+r=37

permalink
56.

Solve the equation.

5r+8+r=4

permalink
57.

Solve the equation.

52=10m3m

permalink
58.

Solve the equation.

87=7p7p

permalink
59.

Solve the equation.

3qq=2+(5)

permalink
60.

Solve the equation.

9yy=6+17

permalink
61.

Solve the equation.

43r10=6

permalink
62.

Solve the equation.

210a7=5

permalink
63.

Solve the equation.

b98b=72b+28

permalink
64.

Solve the equation.

A63A=58A+5

permalink
65.

Solve the equation.

6B+4B=23B3

permalink
66.

Solve the equation.

9m+5m=910m57

permalink
67.

Solve the equation.

5p+5=6p+510p

permalink
68.

Solve the equation.

2q+9=4q+910q

permalink
69.

Solve the equation.

9+11=8y103y+56y

permalink
70.

Solve the equation.

6+(10)=5r108r+4+2r

Application Problems for Solving Linear Equations with Like Terms
permalink
71.

A 130-meter rope is cut into two segments. The longer segment is 10 meters longer than the shorter segment. Write and solve a linear equation to find the length of each segment. Include units.

The segments are and long.

permalink
72.

In a doctor’s office, the receptionist’s annual salary is $135,000 less than that of the doctor. Together, the doctor and the receptionist make $207,000 per year. Find each person’s annual income.

The receptionist’s annual income is . The doctor’s annual income is .

permalink
73.

Selena and Charity went picking strawberries. Selena picked 100 fewer strawberries than Charity did. Together, they picked 214 strawberries. How many strawberries did Charity pick?

Charity picked strawberries.

permalink
74.

Stephanie and Haley collect stamps. Haley collected 30 fewer than three times the number of Stephanie’s stamps. Altogether, they collected 626 stamps. How many stamps did Stephanie and Haley collect?

Stephanie collected stamps. Haley collected stamps.

permalink
75.

Emily and Sherial sold girl scout cookies. Emily’s sales were $35 more than five times of Sherial’s. Altogether, their sales were $725. How much did each girl sell?

Emily’s sales were . Sherial’s sales were .

permalink
76.

A hockey team played a total of 129 games last season. The number of games they won was 13 more than three times of the number of games they lost.

Write and solve an equation to answer the following questions.

The team lost games. The team won games.

permalink
77.

After a 55% increase, a town has 1395 people. What was the population before the increase?

Before the increase, the town’s population was .

permalink
78.

After a 45% increase, a town has 145 people. What was the population before the increase?

Before the increase, the town’s population was .

Solving Linear Equations Involving Distribution
permalink
79.

Solve the equation.

4(r+2)=28

permalink
80.

Solve the equation.

10(a+10)=190

permalink
81.

Solve the equation.

7(b7)=91

permalink
82.

Solve the equation.

3(A4)=6

permalink
83.

Solve the equation.

72=9(B+1)

permalink
84.

Solve the equation.

66=6(m+5)

permalink
85.

Solve the equation.

12=3(n9)

permalink
86.

Solve the equation.

72=9(q3)

permalink
87.

Solve the equation.

(y8)=9

permalink
88.

Solve the equation.

(r2)=10

permalink
89.

Solve the equation.

7=(8a)

permalink
90.

Solve the equation.

4=(5b)

permalink
91.

Solve the equation.

2(5A7)=56

permalink
92.

Solve the equation.

8(9B7)=88

permalink
93.

Solve the equation.

18=3(33m)

permalink
94.

Solve the equation.

2=2(72n)

permalink
95.

Solve the equation.

4+6(q+7)=76

permalink
96.

Solve the equation.

2+8(y+6)=42

permalink
97.

Solve the equation.

510(r+6)=5

permalink
98.

Solve the equation.

47(a+6)=101

permalink
99.

Solve the equation.

28=48(b6)

permalink
100.

Solve the equation.

37=103(A6)

permalink
101.

Solve the equation.

48(B6)=124

permalink
102.

Solve the equation.

210(m6)=2

permalink
103.

Solve the equation.

9=10(2n)

permalink
104.

Solve the equation.

1=8(4q)

permalink
105.

Solve the equation.

2(x+7)=15

permalink
106.

Solve the equation.

5(r+9)=8

permalink
107.

Solve the equation.

  1. 6+(a+5)=1

  2. 6(a+5)=1

permalink
108.

Solve the equation.

  1. 3+(b+2)=4

  2. 3(b+2)=4

permalink
109.

Solve the equation.

5(A+9)10(A1)=60

permalink
110.

Solve the equation.

3(B+3)8(B6)=67

permalink
111.

Solve the equation.

1+9(m5)=53(14m)

permalink
112.

Solve the equation.

4+7(n3)=17(64n)

permalink
113.

Solve the equation.

8(q6)q=683(2+5q)

permalink
114.

Solve the equation.

6(x10)x=393(7+5x)

permalink
115.

Solve the equation.

7(8r+4)=14(35r)

permalink
116.

Solve the equation.

5(8a+8)=10(105a)

permalink
117.

Solve the equation.

5+2(35b)=3(b3)+2

permalink
118.

Solve the equation.

14+5(53A)=3(A12)+3

Application Problems for Solving Linear Equations Involving Distribution
permalink
119.

A rectangle’s perimeter is 78 cm. Its base is 26 cm.

Its height is .

permalink
120.

A rectangle’s perimeter is 50 m. Its width is 8 m. Use an equation to solve for the rectangle’s length.

Its length is .

permalink
121.

A rectangle’s perimeter is 112 in. Its length is 8 in longer than its width. Use an equation to find the rectangle’s length and width.

Its width is .

Its length is .

permalink
122.

A rectangle’s perimeter is 180 cm. Its length is 2 times as long as its width. Use an equation to find the rectangle’s length and width.

It’s width is .

Its length is .

permalink
123.

A rectangle’s perimeter is 92 ft. Its length is 4 ft shorter than four times its width. Use an equation to find the rectangle’s length and width.

Its width is .

Its length is .

permalink
124.

A rectangle’s perimeter is 170 ft. Its length is 1 ft longer than three times its width. Use an equation to find the rectangle’s length and width.

Its width is .

Its length is .

Comparisons
permalink
125.

Solve the equation.

  1. t+9=9

  2. A+9=9

  3. c9=9

  4. x9=9

permalink
126.

Solve the equation.

  1. b+3=3

  2. p+3=3

  3. q3=3

  4. r3=3

permalink
127.
  1. Solve r+1=6.

  2. Evaluate r+1 when r=5.

permalink
128.
  1. Solve r5=4.

  2. Evaluate r5 when r=9.

permalink
129.
  1. Solve 5(r6)3=18.

  2. Evaluate 5(r6)3 when r=3.

  3. Simplify 5(r6)3.

permalink
130.
  1. Solve 4(t+2)9=11.

  2. Evaluate 4(t+2)9 when t=3.

  3. Simplify 4(t+2)9.

permalink
131.

Choose True or False for the following questions about the difference between expressions and equations.

  1. 9x7 is an expression.

    • True

    • False

  2. We can check whether x=1 is a solution of 9x7=7x+9.

    • True

    • False

  3. 9x7=7x+9 is an equation.

    • True

    • False

  4. 7x+9 is an equation.

    • True

    • False

  5. 9x7=7x+9 is an expression.

    • True

    • False

  6. We can evaluate 9x7 when x=1

    • True

    • False

  7. We can evaluate 9x7=7x+9 when x=1

    • True

    • False

  8. We can check whether x=1 is a solution of 9x7.

    • True

    • False

permalink
132.

Choose True or False for the following questions about the difference between expressions and equations.

  1. 7x9 is an equation.

    • True

    • False

  2. 9x+7=7x9 is an equation.

    • True

    • False

  3. We can evaluate 9x+7=7x9 when x=1

    • True

    • False

  4. We can evaluate 9x+7 when x=1

    • True

    • False

  5. We can check whether x=1 is a solution of 9x+7=7x9.

    • True

    • False

  6. 9x+7=7x9 is an expression.

    • True

    • False

  7. We can check whether x=1 is a solution of 9x+7.

    • True

    • False

  8. 9x+7 is an expression.

    • True

    • False

Challenge
permalink
133.

Think of a number. Add four to your number. Now double that. Then add six. Then halve it. Finally, subtract 7. What is the result? Do you always get the same result, regardless of what number you start with? How does this work? Explain using algebra.

permalink
134.

Write a linear equation whose solution is x=9. You may not write an equation whose left side is just “x” or whose right side is just “x.

There are infinitely many correct answers to this problem. Be creative. After finding an equation that works, see if you can come up with a different one that also works.