Section 2.1 Solving Multistep Linear Equations
We solved equations in Section 1.5 where only one step was needed to isolate the variable. Now we will work with equations that need more than one step.
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?
You might recognize that this is a ârateâ scenario like we explored in Section 1.8. If you didnât notice that, you can still explore the given information with a table. You may find a pattern that helps you understand the question better. (And even if you do not find a pattern, spending some time thinking about these quantities stills helps to understand the question better.)
| Time Tap Running (in Minutes) |
Water in Tank (in Gallons) |
| \(0\) | \(\phantom{0}5\) |
| \(1\) | \(20\) (\(15\) more than \(5\)) |
| \(2\) | \(35\) (\(30\) more than \(5\)) |
| \(3\) | \(50\) (\(45\) more than \(5\)) |
| \(4\) | \(65\) (\(60\) more than \(5\)) |
| \(\vdots\) | \(\phantom{0}\vdots\) |
Each additional minute of time gives us \(15\) more gallons of water. So after \(t\) minutes, weâve added â\(15\) times \(t\)â gallons of water to the \(5\) gallons that we started with. So after \(t\) minutes, we have \(15t+5\) gallons. To find when the tank will be full with \(140\) gallons, we can write the equation
\begin{equation*}
15t+5=140
\end{equation*}
This is the same equation we would get if we apply the rate model formula.
To solve this, frst we isolate the variable term, \(15t\text{.}\) We need to âremoveâ the \(5\) from the left side of the equation. We can do this in a legal way using the addition property of equality by subtracting \(5\) from each side of the equation. Once the variable term is isolated, we can eliminate its coefficient and solve for \(t\text{.}\)
The full process is:
\begin{align*}
15t+5\amp=140\\
15t+5\subtractright{5}\amp=140\subtractright{5}\\
15t\amp=135\\
\divideunder{15t}{15}\amp=\divideunder{135}{15}\\
t\amp=9
\end{align*}
We should check this possible solution by substituting \(9\) in for \(t\) in the original equation:
\begin{align*}
15t+5\amp=140\\
15(\substitute{9})+5\amp\wonder{=}140\\
135+5\amp\confirm{=}140
\end{align*}
And the solution \(9\) is verified.
This problem had context. It was not simply solving an equation. It came with a story about a tank filling with water. So we should report a conclusion that uses that context. Something like âThe tank will be full after \(9\) minutes.â
In solving the two-step equation in Example 2, we first isolated the variable expression \(15t\) and then eliminated the coefficient \(15\) by dividing each side of the equation by \(15\text{.}\) These two steps are the heart of our approach to solving linear equations. Try these two steps in the following exercise.
Checkpoint 2.1.3.
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 termâs coefficient.
\begin{equation*}
\begin{aligned}
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{aligned}
\end{equation*}
Checking the solution \(y=5\text{:}\)
\begin{equation*}
\begin{aligned}
7-3y\amp=-8\\
7-3(\substitute{5})\amp\wonder{=}-8\\
7-15\amp\confirm{=}-8
\end{aligned}
\end{equation*}
So the solution to the equation \(7-3y=-8\) is \(5\) and the solution set is \(\{5\}\text{.}\)
In Section 1.8, there was Example 1.8.4. In that example, some background information let us set up an equation, but we didnât try to solve it. Now we can try solving it.
Checkpoint 2.1.4.
Your savings account starts with \(\$500\text{.}\) Then each month, there is an automatic deposit of \(\$150\text{.}\) You need \(\$1700\) to afford a deposit on a new apartment. How long will this take?
Explanation.
With any word problem, it is best to clearly define the varaible you will use once you really understand what the question is asking for. In this case, we let \(t\) (for time) be the number of months until you have saved up \(\$1700\text{.}\)
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 savngs account after \(t\) months.
|
Time Passed
(in Months)
|
Account Balance (in Dollars) |
| \(0\) | \(500\) |
| \(1\) | \(650\) (\(150\) more than \(500\)) |
| \(2\) | \(800\) (\(300\) more than \(500\)) |
| \(3\) | \(950\) (\(450\) more than \(500\)) |
| \(4\) | \(1100\) (\(600\) more than \(500\)) |
| \(\vdots\) | \(\vdots\) |
We find the pattern is that after \(t\) months, the total amount saved is \(150t + 500\text{.}\) Using this pattern, we set up the equation
\begin{equation*}
150t + 500 = 1700
\end{equation*}
to represent when we will have saved up \(\$1700\text{.}\) To solve this equation, we start by subtracting \(500\) from each side. Then we can divide each side by \(150\text{.}\)
\begin{equation*}
\begin{aligned}
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{aligned}
\end{equation*}
Checking the solution \(8\text{:}\)
\begin{equation*}
\begin{aligned}
500+150m\amp=1700\\
500+150\amp(\substitute{8})\wonder{=}1700\\
500+1200\amp\confirm{=}1700
\end{aligned}
\end{equation*}
So \(8\) is the solution, and it checks out. This means it will take \(8\) months for the account balance to reach \(\$1700\text{.}\)
Subsection 2.1.2 Solving Multistep Linear Equations
More complicated equations might need a few setup steps before we can do the two important steps of isolating the variable term and eliminating the coefficient. Here is a general guide for what the full process can be like.
Process 2.1.5. 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 terms so that the variable terms are on one side of the equation and the constant terms are on the other side of the equation.
- Clear the Coefficient
- Use multiplication or division to eliminate the variable termâs 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. Itâs important to use the order of operations alone rather than properties like the distributive law. Otherwise you might repeat the same arithmetic errors you (might have) 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 2.1.6.
Ahmed has \(\$2500\) in his savings account and is going to start saving \(\$550\) per month. Julia has \(\$4600\) in her savings account and is going to start saving \(\$250\) per month. If this situation continues, how long will it take for Ahmed to catch up with Julia in savings?
Ahmed saves \(\$550\) per month, so he can save \(550t\) dollars in \(t\) months. With the \(\$2500\) he started with, after \(t\) months he has \(550t + 2500\) dollars. Similarly, after \(t\) months, Julia has \(250t + 4600\) dollars. To find when those two accounts will have the same amount of money, we write the equation
\begin{equation*}
550t + 2500 = 250t + 4600
\end{equation*}
Each side is simplified, but unlike earlier examples, we have the variable \(t\) on both side of the equation. But we can still use properties of equality to have only one \(t\)-term. We can start by subtracting \(250t\) from each side.
\begin{align*}
550t+2500\amp=250t+4600\\
550t+2500\subtractright{250t}\amp=250t+4600\subtractright{250t}\\
300t+2500\amp=4600\\
300t+2500\subtractright{2500}\amp=4600\subtractright{2500}\\
300t\amp=2100\\
\divideunder{300t}{300}\amp=\divideunder{2100}{300}\\
t\amp=7
\end{align*}
Checking the solution \(7\text{:}\)
\begin{align*}
550t+2500\amp=250t+4600\\
550(\substitute{7})+2500\amp\wonder{=}250(\substitute{7})+4600\\
3850+2500\amp\wonder{=}1750+4600\\
6350\amp\confirm{=}6350
\end{align*}
Ahmed will catch up to Julia after \(7\) months.
Checkpoint 2.1.7.
Solve for \(x\) in \(5-2x=5x-9\text{.}\)
Explanation.
\begin{equation*}
\begin{aligned}
5-2x\amp=\nextoperation{5x}-9\\
5-2x\subtractright{5x}\amp=5x-9\subtractright{5x}\\
\nextoperation{5}-7x\amp=-9\\
5-7x\subtractright{5}\amp=-9\subtractright{5}\\
-7x\amp=-14\\
\divideunder{-7x}{-7}\amp=\divideunder{-14}{-7}\\
x\amp=2
\end{aligned}
\end{equation*}
Checking the solution \(2\text{:}\)
\begin{equation*}
\begin{aligned}
5-2x\amp=5x-9\\
5-2(\substitute{2})\amp\wonder{=}5(\substitute{2})-9\\
5-4\amp\wonder{=}10-9\\
1\amp\confirm{=}1
\end{aligned}
\end{equation*}
Therefore the solution is \(2\) and the solution set is \(\{2\}\text{.}\)
In Checkpoint 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, but thereâs no reason why we couldnât have done that. Letâs explore:
\begin{align*}
5-\nextoperation{2x}\amp=5x-9\\
5-2x\addright{2x}\amp=5x-9\addright{2x}\\
5\amp=7x-\nextoperation{9}\\
5\addright{9}\amp=7x-9\addright{9}\\
14\amp=7x\\
\divideunder{14}{7}\amp=\divideunder{7x}{7}\\
2\amp=x
\end{align*}
The solution is the same either way.
Also, we could save a step by moving variable terms and constant 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 in separate steps.
The next example requires combining like terms.
Example 2.1.8.
Solve for \(n\) in \(n-9+3n=n-3n\text{.}\)
Explanation.
We start by combining like terms. After this, we can separate the \(n\)-terms and constant terms, proceeding as before.
\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\wonder{=}\substitute{\frac{3}{2}}-3\left(\substitute{\frac{3}{2}}\right)\\
\frac{3}{2}-9+\frac{9}{2}\amp\wonder{=}\frac{3}{2}-\frac{9}{2}\\
\frac{12}{2}-9\amp\wonder{=}-\frac{6}{2}\\
6-9\amp\confirm{=}-3
\end{align*}
The solution to the equation \(n-9+3n=n-3n\) is \(\frac{3}{2}\) and the solution set is \(\left\{\frac{3}{2}\right\}\text{.}\)
Example 2.1.9.
Solve for \(a\) in \(4 - (3 - a)=-2 - 2(2a + 1)\text{.}\)
Explanation.
This time we start by simplifying each side of the equation, which involves distributing as well as combining likee terms. Here is a reminder to be careful when distributing a negative sign over a group of terms.
\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\wonder{=}-2-2(2(\substitute{-1})+1)\\
4-(4)\amp\wonder{=}-2-2(-1)\\
0\amp\wonder{=}-2+2\\
0\amp\confirm{=}0
\end{align*}
Therefore the solution to the equation is \(-1\) and the solution set is \(\{-1\}\text{.}\)
Checkpoint 2.1.10.
Solve for \(x\) in \(4 + 3(12 - x) = 12x - 5(2x - 2)\text{.}\)
Explanation.
\begin{equation*}
\begin{aligned}
4 + \nextoperation{3(12 - x)} \amp= 12x\nextoperation{{}-5(2x - 2)}\\
4 + 36 - 3x \amp= 12x - 10x + 10\\
40 - 3x \amp= 2x + 10\\
40 - 3x\addright{3x} \amp= 2x + 10\addright{3x}\\
40 \amp= 5x + 10\\
40\subtractright{10} \amp= 5x + 10\subtractright{10}\\
30 \amp= 5x\\
\divideunder{30}{5}\amp=\divideunder{5x}{5}\\
6\amp=x
\end{aligned}
\end{equation*}
Checking the solution \(6\text{:}\)
\begin{equation*}
\begin{aligned}
4 + 3(12 - x) \amp= 12x - 5(2x - 2)\\
4 + 3(12 - \substitute{6}) \amp\wonder{=} 12\substitute{(6)} - 5(2\substitute{(6)} - 2)\\
4 + 3(6) \amp\wonder{=} 72 - 5(12 - 2)\\
4 + 18 \amp\wonder{=} 72 - 5(10)\\
22 \amp\confirm{=} 72 - 50
\end{aligned}
\end{equation*}
Therefore the solution is \(6\) and the solution set is \(\{6\}\text{.}\)
Subsection 2.1.3 Revisiting Applications
In Section 1.8, we explored several âword problemâ scenarios that led to equations, but we did not try to solve those equations. Letâs revisit some of those applications and try to solve them.
Here we revisit Example 1.8.6.
Example 2.1.11.
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.
We have an initial amount of water (2.5 ft3), a rate at which the amount of water is changing (1.75 ft3), and a final amount of water we are going to reach (6.25 ft3). So we can use the pattern for rate modeling from (1.8.1).
We should clearly identify the variable first though. The solution is supposed to represent an amount of time. So a reasonable variable to use is \(t\text{.}\) Let \(t\) be the amount of time, in minutes, that it takes for the tub to reach 6.25 ft3. And we have the equation:
\begin{equation*}
2.5 + 1.75t = 6.25
\end{equation*}
Now we have the skills to solve this equation.
\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\approx2.14
\end{align*}
So it will take about \(2.14\) minutes for the tub to have 6.25 ft3 of water.
Here we revisit Example 1.8.9.
Example 2.1.12.
Jakobiâs annual salary as a nurse this year is \(\$73{,}290\text{.}\) Thatâs following a \(4\%\) raise over last yearâs salary. What was his salary the previous year?
Explanation.
As soon as you understand that the solution will be Jakobiâs salary from last year, that is when you should clearly define a variable. Since it will represent a salary, we choose to use \(S\) as the variablbe. Let \(S\) represent Jakobiâs salary from last year.
To set up the equation, we need to think about how he arrived at this yearâs salary. His employer took last yearâs salary and added \(4\%\) to that. In words, we have:
\begin{equation*}
(\text{last year's salary})+(4\%\text{ of last year's salary}) = (\text{this year's salary})
\end{equation*}
We represent â\(4\%\) of last yearâs salaryâ with \(0.04S\) since \(0.04\) is the decimal equivalent to \(4\%\text{.}\) So out equation is:
\begin{equation*}
S + 0.04S = 73290
\end{equation*}
Now we have the skills to solve this equation.
\begin{align*}
S + 0.04S \amp= 73290\\
1.04S \amp= 73290\\
\divideunder{1.04S}{1.04}\amp=\divideunder{73290}{1.04}\\
S\amp\approx70471
\end{align*}
So last year, Jakobiâs salary was about \(\$70{,}471\text{.}\)
Here we revisit Checkpoint 1.8.12.
Checkpoint 2.1.13.
A shirt is on sale for \(20\%\) off. The sale price is \(\$51.00\text{.}\) What was the shirtâs original price?
Explanation.
Let \(x\) represent the original price of the shirt. Since \(20\%\) is removed to bring the cost down to \(\$51\text{,}\) we can set up the equation:
\begin{equation*}
\begin{aligned}
\pinover{x}{original}\pinover{-}{minus}\pinover{0.20}{20\%}\pinover{\cdot}{of}\pinover{x}{original}\amp\pinover{=}{is}\pinover{51}{\$51}
\end{aligned}
\end{equation*}
Now we have the skills to solve this equation.
\begin{equation*}
\begin{aligned}
x - 0.20x \amp= 51\\
0.80x \amp= 51\\
\divideunder{0.80x}{0.80}\amp=\divideunder{51}{0.80}\\
x\amp=63.75
\end{aligned}
\end{equation*}
The shirtâs original price was \(\$63.75\text{.}\)
Subsection 2.1.4 Differentiating between Simplifying Expressions, Evaluating Expressions and Solving Equations
Consider the following three examples, which have similarities on the surface, but are fundamentally different from each other.
Example 2.1.14.
Simplify the expression \(10-3(x+2)\text{.}\)
Explanation.
Distribute the \(-3\) and then combine like terms.
\begin{align*}
10-3(x+2)\amp=10-3x-6\\
\amp=-3x+4
\end{align*}
Note that our final result is an expression.
Example 2.1.15.
Evaluate the expression \(10-3(x+2)\) when \(x=2\text{.}\)
Explanation.
Substitute \(2\) in for \(x\) in 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*}
Note that our final result here is a number.
Example 2.1.16.
Solve the equation \(10-3(x+2)=x-16\text{.}\)
Explanation.
Follow the process we have been using in this section to solve a linear equation.
\begin{align*}
10-3(x+2)\amp=x-16\\
10-3x-6\amp=x-16\\
-3x+4\amp=x-16\\
-3x+4\subtractright{x}\amp=x-16\subtractright{x}\\
-4x+4\amp=-16\\
-4x+4\subtractright{4}\amp=-16\subtractright{4}\\
-4x\amp=-20\\
\divideunder{-4x}{-4}\amp=\divideunder{-20}{-4}\\
x\amp=5
\end{align*}
So the solution set is \(\{5\}\text{.}\)
Note that our final result here is a solution set.
Simplifying, evaluating, and solving are three different algebra tasks. Students often use these vocabulary terms incorrectly, using one when they meant another. Here is a summary collection of the differences that you should understand between these algebra tasks.
- An expression like \(10-3(x+2)\) can be simplified to \(-3x+4\) (as in Example 14). However we cannot âsolveâ an expression like this. It is incorrect to say you will âsolve \(10-3(x+2)\)â.
- An expression like \(10-3(x+2)\) can be evaluated, but only once you have a number to use in place of the variable. This is what happens in Example 15, where \(x=2\text{,}\) and the expression evaluates to \(-2\text{.}\)
- An equation connects two expressions with an equals sign. In Example 16, \(10-3(x+2)=x-16\) has one expression on either side of equals sign. You can solve this equation, because it is an equation. You can also solve inequalities. You just cannot solve an expression like \(10-3(x+2)\text{.}\)
- When we solve the equation \(10-3(x+2)=x-16\text{,}\) we are looking for a number which makes those two expressions evaluate to the same value. In Example 16, we found the solution was \(5\text{.}\) That number \(5\) makes both \(10-3(x+2)\) and \(x-16\) evaluate to the same number (and that number is \(-11\) if you were curious.)
Checkpoint 2.1.18. Matching Vocabulary.
Reading Questions 2.1.5 Reading Questions
1.
Describe the five steps you might need to take when solving a linear equation.
2.
In this section there is a reminder to take care with negative numbers when doing what?
3.
Explain what is wrong with saying âI need to solve \(3x+x-8\text{.}\)â
Exercises 2.1.6 Exercises
Review and Warmup
One-Step Equations.
Solve the equation.
1.
\({d+3}={-19}\)
2.
\({-8}={\frac{i}{17}}\)
3.
\({-p}={-3}\)
4.
\({{\frac{1}{7}}+u}={-{\frac{1}{12}}}\)
5.
\({-z}={{\frac{7}{10}}}\)
6.
\({9.93+F}={-1.76}\)
Vocabulary
7.
Quotes are given from a student following a quiz. Fill in the blanks with the appropriate vocabulary terms.
(a)
The algebra quiz was fun because you had to
- evaluate
- simplify
- solve
all these equations.
(b)
The last exercise seemed hard at first, but it became more clear once you
- evaluated
- simplified
- solved
each side.
(c)
At first your answer to #4 might be pretty messy looking, but it looks more reasonable after you
- evaluate
- simplify
- solve
it.
(d)
One of the word problems was asking you to take a given expression for the height of a ball and
- evaluate
- simplify
- solve
it for \(t=7\text{.}\)
8.
This exercise walks you through an economics problem. Fill in the blanks with the appropriate vocabulary terms.
(a)
If the revenue from selling \(x\) items is \(2x\) dollars but the cost of production is \(0.5x\text{,}\) then the profit is \(2x-0.5x\text{.}\) But that can be
- evaluated
- simplified
- solved
.
(b)
We need to know how much profit there was after \(1000\) items, so we can
- evaluate
- simplify
- solve
the profit expression.
(c)
If we want to work out how many sales would give us a profit of \(\$3000\text{,}\) we can
- evaluate
- simplify
- solve
the equation \(1.5x=3000\text{.}\)
(d)
At first we find that it would take \(\frac{3000}{1.5}\) sales, but we can
- evaluate
- simplify
- solve
that result.
Skills Practice
Two-Step Equations.
Solve the equation.
9.
\({-5X-8}={7}\)
10.
\({4c+6}={-30}\)
11.
\({-23}={-8i+1}\)
12.
\({-12}={2n-4}\)
13.
\({8u+9}={9u}\)
14.
\({-4z+18}={5z}\)
15.
\({-5F+5}={-2}\)
16.
\({8K-6}={-7}\)
17.
\({5}={5Q-9}\)
18.
\({-6}={-7X-3}\)
19.
\({7c}={-2c-4}\)
20.
\({i}={7i-9}\)
21.
\({-5.7n+5.6}={7.1}\)
22.
\({6.7u+0.5}={-2.8}\)
23.
\({0.1z+7.3}={-4.5z}\)
24.
\({-6.5E-2.7}={8.5E}\)
25.
\({6.9K}={3.5K+7.4}\)
26.
\({-0.6Q}={-1.5Q-2.5}\)
More Steps.
Solve the equation.
27.
\({5X+7}={-8X-97}\)
28.
\({-6c-3}={7c-68}\)
29.
\({2i-1}={-3+i}\)
30.
\({-9n-1}={39-4n}\)
31.
\({-2t+7-9t}={9-t-42}\)
32.
\({7z-4+4z}={-9-3z-37}\)
33.
\({-5\mathopen{}\left(E-2\right)}={-2E+37}\)
34.
\({4\mathopen{}\left(K-2\right)}={-7K+25}\)
35.
\({-8\mathopen{}\left(Q-3\right)-16}={8\mathopen{}\left(Q+7\right)}\)
36.
\({2\mathopen{}\left(X-3\right)+8}={4\mathopen{}\left(X-4\right)}\)
37.
\({8\mathopen{}\left(c+6\right)-4c}={-3\mathopen{}\left(c-1\right)+59}\)
38.
\({-4\mathopen{}\left(h-4\right)-9h}={-3\mathopen{}\left(h+2\right)+62}\)
39.
\({8n+6}={5n-4}\)
40.
\({2t+1}={-6t-4}\)
41.
\({-5z-4+3z}={6-6z+7}\)
42.
\({7E-4-9E}={-3+9E+8}\)
43.
\({2\mathopen{}\left(4K-4\right)}={-3K+6}\)
44.
\({-6\mathopen{}\left(3Q-5\right)}={7Q-3}\)
45.
\({7\mathopen{}\left(W+6\right)-5W}={-8\mathopen{}\left(2W-6\right)-8}\)
46.
\({-2\mathopen{}\left(c-4\right)+5c}={7\mathopen{}\left(2c-5\right)+2}\)
47.
\({71-\left(-8h-5\right)}={-2h+6}\)
48.
\({20-\left(n-6\right)}={4n-4}\)
49.
\({-9t-\left(8t-6\right)}={28-\left(t+6\right)}\)
50.
\({5z-\left(-3z-6\right)}={-70-\left(z-4\right)}\)
51.
\({5\mathopen{}\left(E-3\right)-\left(2E-7\right)}={8}\)
52.
\({-2\mathopen{}\left(J-7\right)-\left(5J+1\right)}={-9}\)
Applications
53.
Reese is driving with an average speed of \({50\ {\rm mph}}\) on Interstate-65. She looks out the window and sees mile marker 191 pass by. How long will it take her to reach her destination, which is at mile marker 296?
54.
Tommy filled the gas tank in their car to \({16\ {\rm gal}}\text{.}\) When the tank reaches \({1\ {\rm gal}}\text{,}\) the low gas light will come on. On average, Tommyâs car uses \({0.04\ {\rm gal}}\) per mile driven. How many miles will Tommyâs car be able to drive before the low gas light comes on?
55.
You planted a young tree in front of your house, and it was \(5\) feet tall. Ever since, it has been growing by \({{\frac{1}{12}}\ {\rm ft}}\) each year. How many years will it take for the tree to grow to be 11 feet tall?
56.
One of the tires on your car looks a little flat. You measure its air pressure and are alarmed to see it so low at \({20\ {\rm psi}}\text{.}\) You have a portable device that can pump air into the tire increasing the pressure at a rate of \({1.3\ {\textstyle\frac{\rm\mathstrut psi}{\rm\mathstrut min}}}\text{.}\) How long will it take to fill the tire to the manualâs recommended pressure of \({34\ {\rm psi}}\text{?}\)
57.
To save up for a new phone that costs \({\$760}\text{,}\) Dorian sets aside 1% of their pay each week. This works out to \({\$6.68}\) per week. At present, they have \({\$327}\) saved up. How long will it be until Dorian has saved up enough?
58.
A small town would like to replace its aging water treatment system. This will cost \({\$13{,}200{,}000}\text{,}\) but the town just needs \({\$1{,}320{,}000}\) up front for downpayment on a loan that will cover the rest. The town treasury has \({\$330{,}000}\) in it already for this need, and the town can gather \({\$110{,}000}\) per month from taxes. How long will it take to reach enough for downpayment on that loan?
59.
Jayda puts a pot of water from the tap onto the stove and turns the burner all the way up. The water temperature starts at \(62\,â\) and climbs steadily up to the boiling point of \(212\,â\text{,}\) raising at a rate of \(21\,\frac{â}{\text{min}}\text{.}\) How long will it take for the pot to boil?
60.
Kelsey puts a pot of water from the tap onto the stove and turns the burner all the way up. The water temperature starts at \(15\,â\) and climbs steadily up to the boiling point of \(100\,â\text{,}\) raising at a rate of \(13\,\frac{â}{\text{min}}\text{.}\) How long will it take for the pot to boil?
61.
Matty baked a pie at \(425\,â\) and just took it out of the oven. It immediately starts to cool at a rate of \(23\,\frac{â}{\text{min}}\text{.}\) How long will it take to cool to \(300\,â\text{?}\)
62.
On a cold snowy day, the temperature in your home is a cozy \(70\,â\text{,}\) but then you lose power and heating. Your home temperature begins to drop at a rate of \(14\,\frac{â}{\text{hour}}\text{.}\) How long will it take before your home is \(45\,â\text{?}\)
63.
A restaurant stocks its pantry with \({100\ {\rm lb}}\) of onions. When the supply reaches \({25\ {\rm lb}}\text{,}\) itâs time to order more. Recently, the restaurant has been using \({12\ {\rm lb}}\) of onions per day. How long since restocking will it take until another order must be placed?
64.
At a recent trip to the casino, Alexa brought \({\$840}\) in cash. She knows she needs to hold on to \({\$100}\) in reserve to pay for dinner later. Unfortunately Alexa had rough luck and was losing money at the slot machines at an average rate of \({\$210}\) per hour. How long was Alexa gambling before she had to stop?
65.
For Carissaâs 8th birthday party, her parents rented a venue that charges a flat fee of \({\$115}\) plus \({\$16}\) per guest. Utimately it cost Carissaâs parents \({\$419}\text{.}\) How many guests were there?
66.
Donavanâs annual property tax had been \({\$4{,}540}\text{.}\) But he did some renovation to his house that added more square footage, and now the annual property tax is \({\$5{,}125}\text{.}\) The county assesses property tax at a rate of \({\$1.52}\) per square foot. How much area did Donavan add to his home?
67.
Coltâs current annual salary as an office manager is \({\$57{,}477}\text{.}\) This is with a raise of \({3.4\%}\) over last yearâs salary. What was his salary last year?
68.
A tablet is for sale in a state where sales tax applies. The sales tax rate is \({5.4\%}\) and the total was \({\$472}\text{.}\) What was the price before sales tax?
69.
A car stereo is on sale with a \({17\%}\) discount and the final price is \({\$354}\text{.}\) What was the original price before the discount is applied?
70.
The final bill at a restaurant one night was \({\$69}\text{,}\) including a \({19\%}\) tip. What was the bill before the tip was added?
71.
One year, the median rent for a one-bedroom apartment in a city was reported to be \({\$1{,}580}\text{.}\) This was reported to be an increase of \({2.6\%}\) over the previous year. Based on this reporting, what was the median rent for of a one-bedroom apartment the previous year?
Challenge
72.
Write a linear equation whose solution is \(x = -9\text{.}\) Neither side of your equation may be just â\(x\)â.
There are infinitely many correct answers to this exercise. After finding an equation that works, see if you can come up with a different one that also works.
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