Section3.1Solving Multistep Linear Equations and Inequalities
¶We have learned how to solve onestep equations and inequalities. In this section, we will learn how to solve multistep equations and inequalities.
Subsection3.1.1Solving TwoStep Equations
Example3.1.2
A water tank can hold \(140\) gallons of water, but it has only \(5\) gallons of water. A tap was 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  Amount of Water in 
Was Turned on  the Tank (in Gallons) 
\(0\)  \(5\) 
\(1\)  \(15\cdot1+5=20\) 
\(2\)  \(15\cdot2+5=35\) 
\(3\)  \(15\cdot3+5=50\) 
\(4\)  \(15\cdot4+5=65\) 
\(\vdots\)  \(\vdots\) 
\(m\)  \(15m+5\) 
We can see the tap can pour \(15m\) gallons of water into the tank in \(m\) minutes. The tank had \(5\) gallons of water in the beginning, so the amount of water in the tank can be modeled by \(15m+5\text{,}\) where \(m\) is the number of minutes since the tap was turned on. To find when the tank will be full (with \(140\) gallons of water), we can write the equation
\begin{equation*} 15m+5=140 \end{equation*}First, we need to isolate the variable term, \(15m\text{,}\) in the equation. In other words, we need to remove \(5\) 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 the coefficient and solve for \(m\text{.}\)
The full process appears as:
\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*}Next, we need to substitute \(m\) with \(9\) in the equation \(15m+5=140\) to check the solution:
\begin{align*} 15m+5\amp=140\\ 15(\substitute{9})+5\amp\stackrel{?}{=}140\\ 135+5\amp\stackrel{?}{=}140\\ 140\amp\stackrel{\checkmark}{=}140 \end{align*}The solution \(9\) is checked.
In summary, the tank will be full after \(9\) minutes.
In solving the twostep equation in Example 3.1.2, 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 will be at the heart of our approach to solving linear equations. For more complicated equations, we may need to simplify some of the expressions first. Below is a general approach to solving linear equations that we will use as we solve more and more complicated equations.
Let's look at some more examples.
Example3.1.5
Solve for \(y\) in the equation \(73y=8\text{.}\)
To solve, we will first separate the variable terms and constant terms into different sides of the equation. Then we will eliminate the variable term's coefficient:
\begin{align*} 73y\amp=8\\ 73y\subtractright{7}\amp=8\subtractright{7}\\ 3y\amp=15\\ \divideunder{3y}{3}\amp=\divideunder{15}{3}\\ y\amp=5 \end{align*}To check our solution, we will replace \(y\) with \(5\) in the original equation:
\begin{align*} 73y\amp=8\\ 73(\substitute{5})\amp\stackrel{?}{=}8\\ 715\amp\stackrel{?}{=}8\\ 8\amp\stackrel{\checkmark}{=}8 \end{align*}Therefore the solution to the equation \(73y=8\) is \(5\) and the solution set is \(\{5\}\text{.}\)
Subsection3.1.2Solving Multistep Linear Equations
Example3.1.6
Shane has saved \(\$2{,}500.00\) in his savings account and is going to start saving \(\$550.00\) per month. Tammy 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 later would Shane catch up with Tammy in savings?
Shane saves \(\$550.00\) per month, so he can save \(550m\) dollars in \(m\) months. Counting \(\$2{,}500.00\) already in his account, the amount of money in his account is \(550m+2500\) dollars. Similarly, the amount of money in Tammy's account is \(250m+4600\) dollars. To find when those two accounts will have the same amount of money, we write the equation
\begin{equation*} 550m+2500=250m+4600 \end{equation*}Here is the full process:
\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 \(7\) in the equation \(550m+2500=250m+4600\text{,}\) we get:
\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*}In summary, Shane will catch up with Tammy in the savings account \(7\) months later.
Let's look at a few more examples.
Example3.1.7
Solve for \(x\) in \(52x=5x9\text{.}\)
Checking the solution \(2\) in the equation \(52x=5x9\text{,}\) we get:
\begin{align*} 52x\amp=5x9\\ 52(\substitute{2})\amp\stackrel{?}{=}5(\substitute{2})9\\ 54\amp\stackrel{?}{=}109\\ 1\amp\stackrel{\checkmark}{=}1 \end{align*}Therefore the solution is \(2\) and the solution set is \(\{2\}\text{.}\)
Remark3.1.8
In Example 3.1.7, 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:
Lastly, we could save a step by moving variable terms and number terms in one step:
This textbook will move variable terms and number terms separately throughout this chapter. Check with your instructor for their expectations.
Exercise3.1.9
The next example requires combining like terms.
Example3.1.10
Solve for \(n\) in \(n9+3n=n3n\text{.}\)
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*} n9+3n\amp=n3n\\ 4n9\amp=2n\\ 4n9\subtractright{4n}\amp=2n\subtractright{4n}\\ 9\amp=6n\\ \divideunder{9}{6}\amp=\divideunder{6n}{6}\\ n\amp=\frac{3}{2} \end{align*}Checking the solution \(\frac{3}{2}\) in the equation \(n9+3n=n3n\text{,}\) we get:
\begin{align*} n9+3n\amp=n3n\\ \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}\\ 69\amp\stackrel{?}{=}3\\ 3\amp\stackrel{?}{=}3 \end{align*}The solution to the equation \(n9+3n=n3n\) is \(\frac{3}{2}\) and the solution set is \(\left\{\frac{3}{2}\right\}\text{.}\)
Exercise3.1.11
Example3.1.12
Virginia is designing a rectangular garden. The garden's length must be \(4\) meters less than three times the width, and its perimeter must be \(40\) meters. Find the garden's length and width.
Reminder: A rectangle's perimeter formula is \(P=2(L+W)\text{,}\) where \(P\) stands for perimeter, \(L\) stands for length and \(W\) stands for width.
Assume the rectangle's width is \(W\) meters. We can then represent the length as \(3W4\) meters since we are told that it is \(4\) meters less than three times the width. It's given that the perimeter is \(40\) meters. Substituting those values into the formula, we have:
\begin{align*} P\amp=2(L+W)\\ 40\amp=2(3W4+W)\\ 40\amp=2(4W4)\amp\text{Like terms were combined.} \end{align*}The next step to solve this equation is to remove the parentheses by distribution.
\begin{align*} 40\amp=2(4W4)\\ 40\amp=8W8\\ 40\addright{8}\amp=8W8\addright{8}\\ 48\amp=8W\\ \divideunder{48}{8}\amp=\divideunder{8W}{8}\\ 6\amp=W\text{.} \end{align*}To check this result, we'll want to replace \(6\) in the equation \(40=2(4W4)\text{:}\)
\begin{align*} 40\amp=2(4W4)\\ 40\amp\stackrel{?}{=}2(4(\substitute{6})4)\\ 40\amp\stackrel{?}{=}2(20)\\ 40\amp\stackrel{\checkmark}{=}40\text{.} \end{align*}To determine the length, recall that this was represented by \(3W4\text{,}\) which is:
\begin{align*} 3W4\amp=3(\substitute{6})4\\ \amp=14\text{.} \end{align*}Thus the rectangle's width is \(6\) meters and the length is \(14\) meters.
Exercise3.1.13
We should be careful when we distribute a negative sign into the parentheses, like in the next example.
Example3.1.14
Solve for \(a\) in \(4(3a)=22(2a+1)\text{.}\)
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 \(a\text{:}\)
Checking the solution \(1\) in the original equation, we get:
\begin{align*} 4(3a)\amp=22(2a+1)\\ 4(3(\substitute{1}))\amp\stackrel{?}{=}22(2(\substitute{1})+1)\\ 4(4)\amp\stackrel{?}{=}22(1)\\ 0\amp\stackrel{\checkmark}{=}0 \end{align*}Therefore the solution to the equation is \(1\) and the solution set is \(\{1\}\text{.}\)
Subsection3.1.3Differentiating among Simplifying Expressions, Evaluating Expressions and Solving Equations
Let's look at the following similar, yet different examples.
Example3.1.15
Simplify the expression \(103(x+2)\text{.}\)
An equivalent result is \(43x\text{.}\) Note that our final result is an expression.
Example3.1.16
Evaluate the expression \(103(x+2)\) when \(x=2\) and when \(x=3\text{.}\)
We will substitute \(x=2\) into the expression:
\begin{align*} 103(x+2)\amp=103(\substitute{2}+2)\\ \amp=103(4)\\ \amp=1012\\ \amp=2 \end{align*}When \(x=2\text{,}\) \(103(x+2)=2\text{.}\)
Similarly, we will substitute \(x=3\) into the expression:
\begin{align*} 103(x+2)\amp=103(\substitute{3}+2)\\ \amp=103(5)\\ \amp=1015=5 \end{align*}When \(x=3\text{,}\) \(103(x+2)=5\text{.}\)
Note that the final results here are values of the original expression.
Example3.1.17
Solve the equation \(103(x+2)=x16\text{.}\)
To check whether \(x=5\) is the correct solution of the equation, we substitute \(5\) for \(x\) into the equation, and we have:
\begin{align*} 103(x+2)\amp=x16\\ 103(\substitute{5}+2)\amp\stackrel{?}{=}\substitute{5}16\\ 103(7)\amp\stackrel{?}{=}11\\ 1021\amp\stackrel{?}{=}11\\ 11\amp\stackrel{\checkmark}{=}11 \end{align*}We have checked that \(x=5\) is a solution of the equation \(103(x+2)=x16\text{.}\)
Note that the final results here are solutions to the equations.
Let's summarize 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 3.1.15), but we cannot solve for \(x\) in an expression.
As \(x\) takes different values, an expression has different values. In Example 3.1.16, when \(x=2\text{,}\) \(103(x+2)=2\text{;}\) but when \(x=3\text{,}\) \(103(x+2)=5\text{.}\)
An equation connects two expressions with an equals sign. In Example 3.1.17, \(103(x+2)=x16\) 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 \(103(x+2)=x16\text{,}\) we are looking for a number which makes those two expressions have the same value. In Example 3.1.17, we found the solution to be \(x=5\text{,}\) which makes both \(103(x+2)=11\) and \(x16=11\text{,}\) as shown in the checking part.
Subsection3.1.4Solving Multistep Inequalities
When solving a linear inequality, we follow the same steps in List 3.1.4. The only difference in our steps to solving is that when we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol must switch. We will look at some examples.
Example3.1.19
Solve for \(t\) in the inequality \(3t+5\geq11\text{.}\) Write the solution set in both setbuilder notation and interval notation.
Note that when we divided both sides of the inequality by \(3\text{,}\) we had to switch the direction of the inequality symbol.
The solution set in setbuilder notation is \(\{t\mid t\leq2\}\text{.}\)
The solution set in interval notation is \((\infty,2]\text{.}\)
Remark3.1.20
Since the inequality solved in Example 3.1.19 has infinite solutions, it's difficult to check. We found that all values of \(t\) for which \(t\leq2\) are solutions, so one approach is to check if \(2\) is a solution and additionally if one other number less than \(2\) is a solution.
Here, we'll check that \(2\) satisfies this inequality:
\begin{align*} 3t+5\amp\geq11\\ 3(\substitute{2})+5\amp\stackrel{?}{\geq}11\\ 6+5\amp\stackrel{?}{\geq}11\\ 11\amp\stackrel{\checkmark}{\geq}11 \end{align*}Next, we can check another number smaller than \(2\text{,}\) such as \(5\text{:}\)
\begin{align*} 3t+5\amp\geq11\\ 3(\substitute{5})+5\amp\stackrel{?}{\geq}11\\ 15+5\amp\stackrel{?}{\geq}11\\ 20\amp\stackrel{\checkmark}{\geq}11 \end{align*}Thus both \(2\) and \(5\) are solutions. It's important to note that this doesn't directly verify that all solutions to this inequality check. It's valuable though in that it would likely help us catch an error if we had made one. Consult your instructor to see if you're expected to check your answer in this manner.
Example3.1.21
Solve for \(z\) in the inequality \((6z+5)(2z3)\lt12\text{.}\) Write the solution set in both setbuilder notation and interval notation.
Note that we divided both sides of the inequality by \(4\) and since this is a positive number we did not need to switch the direction of the inequality symbol.
The solution set in setbuilder notation is \(\{z\mid t\lt5\}\text{.}\)
The solution set in interval notation is \((\infty,5)\text{.}\)
Example3.1.22
Solve for \(x\) in \(22(2x+1)\gt4(3x)\text{.}\) Write the solution set in both setbuilder notation and interval notation.
Note that when we divided both sides of the inequality by \(5\text{,}\) we had to switch the direction of the inequality symbol.
The solution set in setbuilder notation is \(\{x\mid x\lt1\}\text{.}\)
The solution set in interval notation is \((\infty,1)\text{.}\)
Example3.1.23
When a stopwatch started, the pressure inside a gas container was \(4.2\) atm (standard atmospheric pressure). As the container was heated, the pressure increased by \(0.7\) atm per minute. The maximum pressure the container can handle was \(21.7\) atm. Heating must be stopped once the pressure reaches \(21.7\) atm. In what time interval was the container safe?
The pressure increases by \(0.7\) atm per minute, so it increases by \(0.7m\) after \(m\) minutes. Counting in the original pressure of \(4.2\) atm, pressure in the container can be modeled by \(0.7m+4.2\text{,}\) where \(m\) is the number of minutes since the stop watch started.
The container is safe when the pressure is \(21.7\) atm or lower. We can write and solve this inequality:
\begin{align*} 0.7m+4.2\amp\leq21.7\\ 0.7m+4.2\subtractright{4.2}\amp\leq21.7\subtractright{4.2}\\ 0.7m\amp\leq17.5\\ \divideunder{0.7m}{0.7}\amp\leq\divideunder{17.5}{0.7}\\ m\amp\leq25 \end{align*}In summary, the container was safe as long as \(m\leq25\text{.}\) Assuming that \(m\) also must be greater than or equal to zero, this means \(0\leq m\leq 25\text{.}\) We can write this as the time interval as \([0,25]\text{.}\) Thus the container was safe between 0 minutes and 25 minutes.
Subsection3.1.5Exercises
Solving TwoStep Equations
1
Solve the equation.
\(\displaystyle{ {8m+1}={25} }\)
2
Solve the equation.
\(\displaystyle{ {5p+5}={40} }\)
3
Solve the equation.
\(\displaystyle{ {2q3}={17} }\)
4
Solve the equation.
\(\displaystyle{ {8y2}={18} }\)
5
Solve the equation.
\(\displaystyle{ {9} = {5r+6} }\)
6
Solve the equation.
\(\displaystyle{ {12} = {2a+4} }\)
7
Solve the equation.
\(\displaystyle{ {18} = {7b3} }\)
8
Solve the equation.
\(\displaystyle{ {37} = {4A1} }\)
9
Solve the equation.
\(\displaystyle{ {2C+9}={21} }\)
10
Solve the equation.
\(\displaystyle{ {5m+6}={16} }\)
11
Solve the equation.
\(\displaystyle{ {8p3}={29} }\)
12
Solve the equation.
\(\displaystyle{ {2q10}={20} }\)
13
Solve the equation.
\(\displaystyle{ {7} = {y+2} }\)
14
Solve the equation.
\(\displaystyle{ {15} = {r+6} }\)
15
Solve the equation.
\(\displaystyle{ {9a+18}={0} }\)
16
Solve the equation.
\(\displaystyle{ {6b+42}={0} }\)
Application Problems for Solving TwoStep Equations
17
A gym charges members \({\$30}\) for a registration fee, and then \({\$22}\) per month. You became a member some time ago, and now you have paid a total of \({\$426}\) to the gym. How many months have passed since you joined the gym?
months have passed since you joined the gym.
18
Your cell phone company charges a \({\$23}\) monthly fee, plus \({\$0.18}\) per minute of talk time. One month your cell phone bill was \({\$84.20}\text{.}\) How many minutes did you spend talking on the phone that month?
You spent talking on the phone that month.
19
A school purchased a batch of Tshirts from a company. The company charged \({\$9}\) per Tshirt, and gave the school a \({\$75}\) rebate. If the school had a net expense of \({\$3{,}795}\) from the purchase, how many Tshirts did the school buy?
The school purchased Tshirts.
20
Rita hired a facepainter for a birthday party. The painter charged a flat fee of \({\$95}\text{,}\) and then charged \({\$2.50}\) per person. In the end, Rita paid a total of \({\$122.50}\text{.}\) How many people used the facepainter’s service?
people used the facepainter’s service.
21
A certain country has \(788\) million acres of forest. Every year, the country loses \(9.85\) 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 \(334.9\) million acres of forest left? (Use an equation to solve this problem.)
After years, this country would have \(334.9\) million acres of forest left.
22
Priscilla has \({\$71}\) in her piggy bank. She plans to purchase some Pokemon cards, which costs \({\$1.85}\) each. She plans to save \({\$56.20}\) to purchase another toy. At most how many Pokemon cards can he purchase?
Write an equation to solve this problem.
Priscilla can purchase at most Pokemon cards.
Solving Equations with Variable Terms on Both Sides
23
Solve the equation.
\({6r+4} = {r+54}\)
24
Solve the equation.
\({9a+8} = {a+64}\)
25
Solve the equation.
\(\displaystyle{ {5b+3} = {b13} }\)
26
Solve the equation.
\(\displaystyle{ {8A+8} = {A34} }\)
27
Solve the equation.
\(\displaystyle{ {810B} = {8B+44} }\)
28
Solve the equation.
\(\displaystyle{ {55m} = {8m+70} }\)
29
Solve the equation.
\(\displaystyle{ {2p+9}={4p+8} }\)
30
Solve the equation.
\(\displaystyle{ {7q+7}={3q+6} }\)
31
Solve the equation.

\(\displaystyle{ {9y+8} = {4y+28} }\)

\(\displaystyle{ {4a+8} = {9a42} }\)
32
Solve the equation.

\(\displaystyle{ {9r+7} = {3r+61} }\)

\(\displaystyle{ {3q+7} = {9q5} }\)
Application Problems for Solving Equations with Variable Terms on Both Sides
33
Use a linear equation to solve the word problem.
Two trees are \(6.5\) feet and \(15.5\) feet tall. The shorter tree grows \(2.5\) feet per year; the taller tree grows \(2\) 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.
34
Use a linear equation to solve the word problem.
Massage Heaven and Massage You are competitors. Massage Heaven has \(5300\) registered customers, and it gets approximately \(600\) newly registered customers every month. Massage You has \(7200\) registered customers, and it gets approximately \(500\) 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.
35
Use a linear equation to solve the word problem.
Two truck rental companies have different rates. VHaul has a base charge of \({\$65.00}\text{,}\) plus \({\$0.75}\) per mile. WHaul has a base charge of \({\$56.20}\text{,}\) plus \({\$0.85}\) 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.
36
Use a linear equation to solve the word problem.
Massage Heaven and Massage You are competitors. Massage Heaven has \(6500\) registered customers, but it is losing approximately \(200\) registered customers every month. Massage You has \(1500\) registered customers, and it gets approximately \(300\) 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.
37
Use a linear equation to solve the word problem.
Diane has \({\$85.00}\) in her piggy bank, and she spends \({\$3.00}\) every day.
Alejandro has \({\$13.50}\) in his piggy bank, and he 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, Diane and Alejandro will have the same amount of money in their piggy banks.
38
Use a linear equation to solve the word problem.
Sherial has \({\$90.00}\) in her piggy bank, and she spends \({\$1.50}\) every day.
Rebecca has \({\$26.00}\) 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, Sherial and Rebecca will have the same amount of money in their piggy banks.
Solving Linear Equations with Like Terms
39
Solve the equation.
\(\displaystyle{ {6q+10q+4}={100} }\)
40
Solve the equation.
\(\displaystyle{ {3y+5y+4}={28} }\)
41
Solve the equation.
\(\displaystyle{ {9r+9+6}={105} }\)
42
Solve the equation.
\(\displaystyle{ {6a+4+6}={52} }\)
43
Solve the equation.
\(\displaystyle{ {9+7}={7bb50} }\)
44
Solve the equation.
\(\displaystyle{ {3+2}={10AA67} }\)
45
Solve the equation.
\(\displaystyle{ {3r+68r}={16} }\)
46
Solve the equation.
\(\displaystyle{ {2r+108r}={40} }\)
47
Solve the equation.
\(\displaystyle{ {8t+4+t}={39} }\)
48
Solve the equation.
\(\displaystyle{ {6t+8+t}={42} }\)
49
Solve the equation.
\({18}={2y3y}\)
50
Solve the equation.
\({2}={8r7r}\)
51
Solve the equation.
\(\displaystyle{ {5aa}={10+29} }\)
52
Solve the equation.
\(\displaystyle{ {2bb}={5+\left(9\right)} }\)
53
Solve the equation.
\(\displaystyle{ {47A10}={6} }\)
54
Solve the equation.
\(\displaystyle{ {34B7}={4} }\)
55
Solve the equation.
\(\displaystyle{ {2+5}={10m67m+24m} }\)
56
Solve the equation.
\(\displaystyle{ {3+\left(7\right)}={7n92n+26n} }\)
57
Solve the equation.
\(\displaystyle{ {q46q} = {63q+10} }\)
58
Solve the equation.
\(\displaystyle{ {x1010x} = {44x+4} }\)
59
Solve the equation.
\(\displaystyle{ 8r+3r = 104r15 }\)
60
Solve the equation.
\(\displaystyle{ 7a+4a = 84a16 }\)
61
Solve the equation.
\({9b+3} = {5b+34b}\)
62
Solve the equation.
\({6A+7} = {3A+74A}\)
Application Problems for Solving Linear Equations with Like Terms
63
A \(162\)meter rope is cut into two segments. The longer segment is \(14\) 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.
64
In a doctor’s office, the receptionist’s annual salary is \({\$149{,}000}\) less than that of the doctor. Together, the doctor and the receptionist make \({\$225{,}000}\) per year. Find each person’s annual income.
The receptionist’s annual income is .
The doctor’s annual income is .
65
Emily and Brent went picking strawberries. Emily picked \(122\) fewer strawberries than Brent did. Together, they picked \(250\) strawberries. How many strawberries did Brent pick?
Brent picked strawberries.
66
Connor and Bobbi collect stamps. Bobbi collected \(13\) fewer than three times the number of Connor’s stamps. Altogether, they collected \(767\) stamps. How many stamps did Connor and Bobbi collect?
Connor collected stamps.
Bobbi collected stamps.
67
Penelope and Charity sold girl scout cookies. Penelope’s sales were \({\$21}\) more than four times of Charity’s. Altogether, their sales were \({\$671}\text{.}\) How much did each girl sell?
Penelope’s sales were .
Charity’s sales were .
68
A hockey team played a total of \(98\) games last season. The number of games they won was \(14\) 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.
Solving Linear Equations with Parentheses
69
Solve the equation.
\(\displaystyle{ {2\!\left(a+2\right)}={16} }\)
70
Solve the equation.
\(\displaystyle{ {8\!\left(b+9\right)}={152} }\)
71
Solve the equation.
\(\displaystyle{ {5\!\left(A7\right)}={55} }\)
72
Solve the equation.
\(\displaystyle{ {2\!\left(B4\right)}={2} }\)
73
Solve the equation.
\(\displaystyle{ {56}={8\!\left(m+2\right)} }\)
74
Solve the equation.
\(\displaystyle{ {60}={5\!\left(n+6\right)} }\)
75
Solve the equation.
\(\displaystyle{ {38}={2\!\left(q10\right)} }\)
76
Solve the equation.
\(\displaystyle{ {77}={7\!\left(x5\right)} }\)
77
Solve the equation.
\(\displaystyle{ {\left(r9\right)}={13} }\)
78
Solve the equation.
\(\displaystyle{ {\left(t3\right)}={7} }\)
79
Solve the equation.
\(\displaystyle{ {4}={\left(7b\right)} }\)
80
Solve the equation.
\(\displaystyle{ {12}={\left(3A\right)} }\)
81
Solve the equation.
\(\displaystyle{ {10\!\left(6B10\right)}={320} }\)
82
Solve the equation.
\(\displaystyle{ {7\!\left(10m10\right)}={70} }\)
83
Solve the equation.
\(\displaystyle{ {52}={2\!\left(43n\right)} }\)
84
Solve the equation.
\(\displaystyle{ {60}={5\!\left(82q\right)} }\)
85
Solve the equation.

\(\displaystyle{ {6+\left(x+8\right)}={7} }\)

\(\displaystyle{ {6\left(x+8\right)}={7} }\)
86
Solve the equation.

\(\displaystyle{ {3+\left(r+5\right)}={10} }\)

\(\displaystyle{ {3\left(r+5\right)}={10} }\)
87
Solve the equation.
\(\displaystyle{ {5+10\!\left(t+10\right)}={35} }\)
88
Solve the equation.
\(\displaystyle{ {3+8\!\left(b+10\right)}={155} }\)
89
Solve the equation.
\(\displaystyle{ {110\!\left(A+9\right)}={119} }\)
90
Solve the equation.
\(\displaystyle{ {47\!\left(B+9\right)}={38} }\)
91
Solve the equation.
\(\displaystyle{ {131}={57\!\left(m9\right)} }\)
92
Solve the equation.
\(\displaystyle{ {8}={22\!\left(n9\right)} }\)
93
Solve the equation.
\(\displaystyle{ 48(q9)=68 }\)
94
Solve the equation.
\(\displaystyle{ 310(x9)=143 }\)
95
Solve the equation.
\(\displaystyle{ {14}={6\left(2r\right)} }\)
96
Solve the equation.
\(\displaystyle{ {9}={9\left(4t\right)} }\)
97
Solve the equation.
\(\displaystyle{ {2\left(b+6\right)}={2} }\)
98
Solve the equation.
\(\displaystyle{ {1\left(c+8\right)}={1} }\)
99
Solve the equation.
\(\displaystyle{ {4\!\left(B+10\right)10\!\left(B1\right)}={20} }\)
100
Solve the equation.
\(\displaystyle{ {3\!\left(m+5\right)8\!\left(m6\right)}={38} }\)
101
Solve the equation.
\(\displaystyle{ {5+10\!\left(n5\right)}={4\left(12n\right)} }\)
102
Solve the equation.
\(\displaystyle{ {4+7\!\left(q3\right)}={9\left(63q\right)} }\)
103
Solve the equation.
\(\displaystyle{ {7\!\left(x7\right)x}={494\!\left(2+3x\right)} }\)
104
Solve the equation.
\(\displaystyle{ {10\!\left(r1\right)r}={144\!\left(6+3r\right)} }\)
105
Solve the equation.
\(\displaystyle{ {7\!\left(14t+6\right)}={14\!\left(38t\right)} }\)
106
Solve the equation.
\(\displaystyle{ {3\!\left(14b+10\right)}={6\!\left(108b\right)} }\)
107
Solve the equation.
\(\displaystyle{ {10+6\!\left(35c\right)}={5\!\left(c4\right)+8} }\)
108
Solve the equation.
\(\displaystyle{ {29+4\!\left(53B\right)}={5\!\left(B8\right)+9} }\)
Application Problems for Solving Linear Equations with Parentheses
109
A rectangle’s perimeter is \({76\ {\rm cm}}\text{.}\) Its base is \({27\ {\rm cm}}\text{.}\)
Its height is .
110
A rectangle’s perimeter is \({58\ {\rm m}}\text{.}\) Its width is \({12\ {\rm m}}\text{.}\) Use an equation to solve for the rectangle’s length.
Its length is .
111
A rectangle’s perimeter is \({114\ {\rm in}}\text{.}\) Its length is \({7\ {\rm in}}\) longer than its width. Use an equation to find the rectangle’s length and width.
Its width is .
Its length is .
112
A rectangle’s perimeter is \({120\ {\rm cm}}\text{.}\) 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 .
113
A rectangle’s perimeter is \({100\ {\rm ft}}\text{.}\) Its length is \({5\ {\rm ft}}\) shorter than four times of its width. Use an equation to find the rectangle’s length and width.
Its width is .
Its length is .
114
A rectangle’s perimeter is \({180\ {\rm ft}}\text{.}\) Its length is \({2\ {\rm ft}}\) longer than three times of its width. Use an equation to find the rectangle’s length and width.
Its width is .
Its length is .
Comparisons
115
Solve the equation.

\(\displaystyle{ {b+6}={6} }\)

\(\displaystyle{ {y+6}={6} }\)

\(\displaystyle{ {C6}={6} }\)

\(\displaystyle{ {c6}={6} }\)
116
Solve the equation.

\(\displaystyle{ {c+6}={6} }\)

\(\displaystyle{ {m+6}={6} }\)

\(\displaystyle{ {y6}={6} }\)

\(\displaystyle{ {r6}={6} }\)
117

Solve the following linear equation:
\(\displaystyle{ {r2}={4} }\)

Evaluate the following expression when \(r=6\text{:}\)
\(\displaystyle{{r2}=}\)
118

Solve the following linear equation:
\(\displaystyle{ {r8}={2} }\)

Evaluate the following expression when \(r=10\text{:}\)
\(\displaystyle{{r8}=}\)
119

Solve the following linear equation:
\(\displaystyle{ {4\!\left(t3\right)3}={9} }\)

Evaluate the following expression when \(t=6\text{:}\)
\(\displaystyle{{4\!\left(t3\right)3}=}\)

Simplify the following expression:
\(\displaystyle{{4\!\left(t3\right)3}=}\)
120

Solve the following linear equation:
\(\displaystyle{ {3\!\left(t+5\right)9}={24} }\)

Evaluate the following expression when \(t=6\text{:}\)
\(\displaystyle{{3\!\left(t+5\right)9}=}\)

Simplify the following expression:
\(\displaystyle{{3\!\left(t+5\right)9}=}\)
121
Choose True or False for the following questions about the difference between expressions and equations.
\(10x10=10x10\text{ is an equation.}\)
?
True
False
\(\text{We can check whether }x=1\text{ is a solution of }10x10=10x10.\)
?
True
False
\(10x10\text{ is an equation.}\)
?
True
False
\(\text{We can check whether }x=1\text{ is a solution of }10x10.\)
?
True
False
\(\text{We can evaluate }10x10=10x10\text{ when }x=1\)
?
True
False
\(10x10\text{ is an expression.}\)
?
True
False
\(\text{We can evaluate }10x10\text{ when }x=1\)
?
True
False
\(10x10=10x10\text{ is an expression.}\)
?
True
False
122
Choose True or False for the following questions about the difference between expressions and equations.
\(\text{We can evaluate }7x+4=4x7\text{ when }x=1\)
?
True
False
\(4x7\text{ is an equation.}\)
?
True
False
\(7x+4\text{ is an expression.}\)
?
True
False
\(7x+4=4x7\text{ is an expression.}\)
?
True
False
\(\text{We can check whether }x=1\text{ is a solution of }7x+4.\)
?
True
False
\(7x+4=4x7\text{ is an equation.}\)
?
True
False
\(\text{We can evaluate }7x+4\text{ when }x=1\)
?
True
False
\(\text{We can check whether }x=1\text{ is a solution of }7x+4=4x7.\)
?
True
False
Solving Linear Inequalities
123
Solve this inequality.
\(\displaystyle{ {4x+4} > {12} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
124
Solve this inequality.
\(\displaystyle{ {5x+10} > {35} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
125
Solve this inequality.
\(\displaystyle{ {50} \geq {6x4} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
126
Solve this inequality.
\(\displaystyle{ {25} \geq {7x3} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
127
Solve this inequality.
\(\displaystyle{ {74} \leq {108x} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
128
Solve this inequality.
\(\displaystyle{ {24} \leq {69x} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
129
Solve this inequality.
\(\displaystyle{ {10x3} \lt {63} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
130
Solve this inequality.
\(\displaystyle{ {2x9} \lt {29} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
131
Solve this inequality.
\(\displaystyle{ {4} \geq {4x+4} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
132
Solve this inequality.
\(\displaystyle{ {2} \geq {5x+2} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
133
Solve this inequality.
\(\displaystyle{ {8} > {2x} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
134
Solve this inequality.
\(\displaystyle{ {5} > {3x} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
135
Solve this inequality.
\(\displaystyle{ {7\!\left(x+1\right)} \geq {28} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
136
Solve this inequality.
\(\displaystyle{ {8\!\left(x+5\right)} \geq {112} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
137
Solve this inequality.
\(\displaystyle{ {10t+6} \lt {5t+31} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
138
Solve this inequality.
\(\displaystyle{ {10t+2} \lt {2t+18} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
139
Solve this inequality.
\(\displaystyle{ {4z+10} \leq {z8} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
140
Solve this inequality.
\(\displaystyle{ {4z+5} \leq {z16} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
141
Solve this inequality.
\(\displaystyle{ {a42a} > {64a+29} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
142
Solve this inequality.
\(\displaystyle{ {a58a} > {1010a+26} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
143
Solve this inequality.
\(\displaystyle{ {5p+58p} \geq {6p+5} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
144
Solve this inequality.
\(\displaystyle{ {8p+25p} \geq {7p+2} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
145
Solve this inequality.
\(\displaystyle{ {112} \lt {8\!\left(p7\right)} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
146
Solve this inequality.
\(\displaystyle{ {18} \lt {9\!\left(p4\right)} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
147
Solve this inequality.
\(\displaystyle{ {\left(x10\right)} \geq {12} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
148
Solve this inequality.
\(\displaystyle{ {\left(x7\right)} \geq {17} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
149
Solve this inequality.
\(\displaystyle{ {22} \leq {24\!\left(z1\right)} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
150
Solve this inequality.
\(\displaystyle{ {133} \leq {310\!\left(z8\right)} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
151
Solve this inequality.
\(\displaystyle{ {2\left(y+8\right)} \lt {10} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
152
Solve this inequality.
\(\displaystyle{ {3\left(y+7\right)} \lt {5} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
153
Solve this inequality.
\(\displaystyle{ {3+10\!\left(x7\right)} \lt {26\left(15x\right)} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
154
Solve this inequality.
\(\displaystyle{ {4+9\!\left(x10\right)} \lt {67\left(13x\right)} }\)
In setbuilder notation, the solution set is .
In interval notation, the solution set is .
Application Problems for Linear Inequalities
155
You are riding in a taxi and can only pay with cash. You have to pay a flat fee of \({\$40}\text{,}\) and then pay \({\$2.70}\) per mile. You have a total of \({\$148}\) in your pocket.
Let \(x\) be the number of miles the taxi will drive you. You want to know how many miles you can afford. Write an inequality to represent this situation in terms of how many miles you can afford:
Solve this inequality. At most how many miles can you afford?
You can afford at most miles.
Use interval notation to express the number of miles you can afford.
156
You are riding in a taxi and can only pay with cash. You have to pay a flat fee of \({\$40}\text{,}\) and then pay \({\$3.40}\) per mile. You have a total of \({\$244}\) in your pocket.
Let \(x\) be the number of miles the taxi will drive you. You want to know how many miles you can afford. Write an inequality to represent this situation in terms of how many miles you can afford:
Solve this inequality. At most how many miles can you afford?
You can afford at most miles.
Use interval notation to express the number of miles you can afford.