ORCCA Solving Multistep Linear Equations

Section 2.1 Solving Multistep Linear Equations

Objectives: PCC Course Content and Outcome Guide

MTH 60 CCOG 2.3
MTH 60 CCOG 2.5
MTH 60 CCOG 2.7

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

Figure 2.1.1. Alternative Video Lessons

Subsection 2.1.1 Solving Two-Step Equations

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.

Minutes Since Tap Was Turned on	Amount of Water in the Tank (in Gallons)
$0$	$5$
$1$	$15\cdot1+5=20$
$2$	$15\cdot2+5=35$
$3$	$15\cdot3+5=50$
$4$	$15\cdot4+5=65$
$\vdots$	$\vdots$
$m$	$15m+5$

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

\begin{equation*} 15m+5=140 \end{equation*}

First, we need to isolate the variable term, $15m\text{.}$ 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 $m\text{.}$

The full process is:

\begin{align*} 15m+5\amp=140\\ 15m+5\subtractright{5}\amp=140\subtractright{5}\\ 15m\amp=135\\ \divideunder{15m}{15}\amp=\divideunder{135}{15}\\ m\amp=9 \end{align*}

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

\begin{align*} 15m+5\amp=140\\ 15(\substitute{9})+5\amp\stackrel{?}{=}140\\ 135+5\amp\stackrel{\checkmark}{=}140 \end{align*}

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.

In 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 $15\text{.}$ 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.

Process 2.1.4. Steps to Solve Linear Equations.

Simplify: Simplify the expressions on each side of the equation by distributing and combining like terms.
Separate: Use addition or subtraction to separate the variable terms and constant terms so that they are on different sides of the equation.
Clear Coefficient: Use multiplication or division to eliminate the variable term's coefficient.
Check: Check the solution in the original equation. Substitute values into the original equation and use the order of operations to simplify both sides. It's important to use the order of operations alone rather than properties like the distributive law. Otherwise you might repeat the same arithmetic errors made while solving and fail to catch an incorrect solution.
Summarize: State the solution set. Or in the case of an application problem, summarize the result in a complete sentence using appropriate units.

Example 2.1.5.

Solve for $y$ in the equation $7-3y=-8\text{.}$

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 $\{5\}\text{.}$

Subsection 2.1.2 Solving Multistep Linear Equations

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

\begin{equation*} 550m+2500=250m+4600\text{.} \end{equation*}

\begin{align*} 550m+2500\amp=250m+4600\\ 550m+2500\subtractright{2500}\amp=250m+4600\subtractright{2500}\\ 550m\amp=250m+2100\\ 550m\subtractright{250m}\amp=250m+2100\subtractright{250m}\\ 300m\amp=2100\\ \divideunder{300m}{300}\amp=\divideunder{2100}{300}\\ m\amp=7 \end{align*}

Checking the solution $7\text{:}$

\begin{align*} 550m+2500\amp=250m+4600\\ 550(\substitute{7})+2500\amp\stackrel{?}{=}250(\substitute{7})+4600\\ 3850+2500\amp\stackrel{?}{=}1750+4600\\ 6350\amp\stackrel{\checkmark}{=}6350 \end{align*}

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

Example 2.1.7.

Solve for $x$ in $5-2x=5x-9\text{.}$

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 $\{2\}\text{.}$

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:

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

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

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

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

Checkpoint 2.1.9.

The next example requires combining like terms.

Example 2.1.10.

Solve for $n$ in $n-9+3n=n-3n\text{.}$

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 $\left\{\frac{3}{2}\right\}\text{.}$

Checkpoint 2.1.11.

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

Example 2.1.12.

Solve for $a$ in $4-(3-a)=-2-2(2a+1)\text{.}$

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 $\{-1\}\text{.}$

Subsection 2.1.3 Revisiting Applications from Section 1.8

In 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.

Example 2.1.13.

Here we revisit Example 1.8.2.

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

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*}

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 $\$1700\text{.}$

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:

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.

Example 2.1.17.

Here we revisit Example 1.8.6.

Jakobi's annual salary as a nurse in Portland, Oregon, is $\$73{,}290\text{.}$ 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.

Checkpoint 2.1.18.

Here we revisit Exercise 1.8.7.

Example 2.1.19.

Here we revisit Example 1.8.8.

The price of a refrigerator after a $15\%$ discount is $\$612\text{.}$ 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.

Checkpoint 2.1.20.

Here we revisit Exercise 1.8.9.

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

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

Example 2.1.21.

Simplify the expression $10-3(x+2)\text{.}$

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.

Example 2.1.22.

Evaluate the expression $10-3(x+2)$ 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.

Example 2.1.23.

Solve the equation $10-3(x+2)=x-16\text{.}$

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.

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

List 2.1.24. A summary the differences among simplifying expressions, evaluating expressions and solving equations.

An expression like $10-3(x+2)$ 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 $x=2\text{,}$ $10-3(x+2)=-2\text{;}$ but when $x=3\text{,}$ $10-3(x+2)=-5\text{.}$
An equation connects two expressions with an equals sign. In Example 2.1.23, $10-3(x+2)=x-16$ has the expression $10-3(x+2)$ on the left side of equals sign, and the expression $x-16$ on the right side.
When we solve the equation $10-3(x+2)=x-16\text{,}$ 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 $5\text{.}$ It makes both $10-3(x+2)$ and $x-16=$ equal to $-11\text{.}$

Reading Questions 2.1.5 Reading Questions

1.

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

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.

3.

Explain what is wrong with saying “I need to solve $3x+x-8\text{.}$”

Exercises 2.1.6 Exercises

Warmup and Review

1.

Solve the equation.

$\displaystyle{ {r+4}={1} }$

2.

Solve the equation.

$\displaystyle{ {r+1}={-7} }$

3.

Solve the equation.

$\displaystyle{ {t-7}={1} }$

4.

Solve the equation.

$\displaystyle{ {t-4}={-2} }$

5.

Solve the equation.

$\displaystyle{ {21}={-3t} }$

6.

Solve the equation.

$\displaystyle{ {10}={-5x} }$

7.

Solve the equation.

$\displaystyle{ {{\frac{5}{4}}t} = {3} }$

8.

Solve the equation.

$\displaystyle{ {{\frac{4}{5}}a} = {9} }$

Solving Two-Step Equations

9.

Solve the equation.

$\displaystyle{ {7c+4}={25} }$

10.

Solve the equation.

$\displaystyle{ {4A+2}={30} }$

11.

Solve the equation.

$\displaystyle{ {10C-1}={-101} }$

12.

Solve the equation.

$\displaystyle{ {6m-5}={-11} }$

13.

Solve the equation.

$\displaystyle{ {-6} = {3p+3} }$

14.

Solve the equation.

$\displaystyle{ {-70} = {9q+2} }$

15.

Solve the equation.

$\displaystyle{ {18} = {6y-6} }$

16.

Solve the equation.

$\displaystyle{ {-28} = {3t-4} }$

17.

Solve the equation.

$\displaystyle{ {-3a+4}={22} }$

18.

Solve the equation.

$\displaystyle{ {-6c+1}={13} }$

19.

Solve the equation.

$\displaystyle{ {-9A-9}={18} }$

20.

Solve the equation.

$\displaystyle{ {-4C-6}={-30} }$

21.

Solve the equation.

$\displaystyle{ {9} = {-m+2} }$

22.

Solve the equation.

$\displaystyle{ {17} = {-p+7} }$

23.

Solve the equation.

$\displaystyle{ {8q+80} = {0} }$

24.

Solve the equation.

$\displaystyle{ {5y+35} = {0} }$

Application Problems for Solving Two-Step Equations

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.

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 ${\$75.50}\text{.}$ How many minutes did you spend talking on the phone that month?

You spent talking on the phone that month.

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.

28.

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

people used the face-painter’s service.

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.

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

31.

Solve the equation.

${10p+5} = {p+59}$

32.

Solve the equation.

${8q+8} = {q+29}$

33.

Solve the equation.

$\displaystyle{ {-4y+3} = {-y-27} }$

34.

Solve the equation.

$\displaystyle{ {-10r+7} = {-r-56} }$

35.

Solve the equation.

$\displaystyle{ {6-2a} = {5a+55} }$

36.

Solve the equation.

$\displaystyle{ {3-5c} = {5c+103} }$

37.

Solve the equation.

$\displaystyle{ {8A+6}={5A+8} }$

38.

Solve the equation.

$\displaystyle{ {2C+6}={10C+5} }$

39.

Solve the equation.

$\displaystyle{ {9m+4} = {2m+32} }$
$\displaystyle{ {2r+4} = {9r-17} }$

40.

Solve the equation.

$\displaystyle{ {6p+10} = {2p+42} }$
$\displaystyle{ {2n+10} = {6n-6} }$