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Section 2.1 Solving Multistep Linear Equations

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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.
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Figure 2.1.1. Alternative Video Lessons
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Subsection 2.1.1 Solving Two-Step Equations

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Example 2.1.2.

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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?
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You might recognize that this is a “rate” scenario like we explored in Section 1.8. If you didn’t 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 05
1 20 (15 more than 5)
2 35 (30 more than 5)
3 50 (45 more than 5)
4 65 (60 more than 5)
⋮ 0⋮
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Each additional minute of time gives us 15 more gallons of water. So after t minutes, we’ve added “15 times t” 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
15t+5=140
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This is the same equation we would get if we apply the rate model formula.
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To solve this, first we isolate the variable term, 15t. We need to “remove” 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.
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The full process is:
15t+5=14015t+5−5=140−515t=13515t15=13515t=9
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We should check this possible solution by substituting 9 in for t in the original equation:
15t+5=14015(9)+5=?140135+5=✓140
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And the solution 9 is verified.
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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 “The tank will be full after 9 minutes.”
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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. These two steps are the heart of our approach to solving linear equations. Try these two steps in the following exercise.
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Checkpoint 2.1.3.

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Solve for y in the equation 7−3y=−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.
7−3y=−87−3y−7=−8−7−3y=−15−3y−3=−15−3y=5
Checking the solution y=5:
7−3y=−87−3(5)=?−87−15=✓−8
So the solution to the equation 7−3y=−8 is 5 and the solution set is {5}.
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In Section 1.8, there was Example 1.8.4. In that example, some background information let us set up an equation, but we didn’t try to solve it. Now we can try solving it.
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Checkpoint 2.1.4.

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Your savings account starts with $500. Then each month, there is an automatic deposit of $150. 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 variable 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.
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 savings 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)
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We find the pattern is that after t months, the total amount saved is 150t+500. Using this pattern, we set up the equation
150t+500=1700
to represent when we will have saved up $1700. To solve this equation, we start by subtracting 500 from each side. Then we can divide each side by 150.
500+150m=1700500+150m−500=1700−500150m=1200150m150=1200150m=8
Checking the solution 8:
500+150m=1700500+150(8)=?1700500+1200=✓1700
So 8 is the solution, and it checks out. This means it will take 8 months for the account balance to reach $1700.
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Subsection 2.1.2 Solving Multistep Linear Equations

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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.
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Example 2.1.6.

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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?
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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
550t+2500=250t+4600
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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.
550t+2500=250t+4600550t+2500−250t=250t+4600−250t300t+2500=4600300t+2500−2500=4600−2500300t=2100300t300=2100300t=7
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Checking the solution 7:
550t+2500=250t+4600550(7)+2500=?250(7)+46003850+2500=?1750+46006350=✓6350
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Ahmed will catch up to Julia after 7 months.
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Checkpoint 2.1.7.

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Solve for x in 5−2x=5x−9.
Explanation.
5−2x=5x−95−2x−5x=5x−9−5x5−7x=−95−7x−5=−9−5−7x=−14−7x−7=−14−7x=2
Checking the solution 2:
5−2x=5x−95−2(2)=?5(2)−95−4=?10−91=✓1
Therefore the solution is 2 and the solution set is {2}.
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In Checkpoint 7, we could have moved variable terms to the right side of the equal sign and number terms to the left side. We chose not to, but there’s no reason why we couldn’t have done that. Let’s explore:
5−2x=5x−95−2x+2x=5x−9+2x5=7x−95+9=7x−9+914=7x147=7x72=x
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The solution is the same either way.
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Also, we could save a step by moving variable terms and constant terms in one step:
5−2x=5x−95−2x+2x+9=5x−9+2x+914=7x147=7x72=x
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For the sake of a slow and careful explanation, the examples in this chapter will move variable terms and number terms in separate steps.
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The next example requires combining like terms.
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Example 2.1.8.

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Solve for n in n−9+3n=n−3n.
Explanation.
We start by combining like terms. After this, we can separate the n-terms and constant terms, proceeding as before.
n−9+3n=n−3n4n−9=−2n4n−9−4n=−2n−4n−9=−6n−9−6=−6n−632=n
Checking the solution 32:
n−9+3n=n−3n32−9+3(32)=?32−3(32)32−9+92=?32−92122−9=?−626−9=✓−3
The solution to the equation n−9+3n=n−3n is 32 and the solution set is {32}.
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Example 2.1.9.

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Solve for a in 4−(3−a)=−2−2(2a+1).
Explanation.
This time we start by simplifying each side of the equation, which involves distributing as well as combining like terms. Here is a reminder to be careful when distributing a negative sign over a group of terms.
4−(3−a)=−2−2(2a+1)4−3+a=−2−4a−21+a=−4−4a1+a+4a=−4−4a+4a1+5a=−41+5a−1=−4−15a=−55a5=−55a=−1
Checking the solution −1:
4−(3−a)=−2−2(2a+1)4−(3−(−1))=?−2−2(2(−1)+1)4−(4)=?−2−2(−1)0=?−2+20=✓0
Therefore the solution to the equation is −1 and the solution set is {−1}.
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Checkpoint 2.1.10.

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Solve for x in 4+3(12−x)=12x−5(2x−2).
Explanation.
4+3(12−x)=12x−5(2x−2)4+36−3x=12x−10x+1040−3x=2x+1040−3x+3x=2x+10+3x40=5x+1040−10=5x+10−1030=5x305=5x56=x
Checking the solution 6:
4+3(12−x)=12x−5(2x−2)4+3(12−6)=?12(6)−5(2(6)−2)4+3(6)=?72−5(12−2)4+18=?72−5(10)22=✓72−50
Therefore the solution is 6 and the solution set is {6}.
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Subsection 2.1.3 Revisiting Applications

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In Section 1.8, we explored several “word problem” scenarios that led to equations, but we did not try to solve those equations. Let’s revisit some of those applications and try to solve them.
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Here we revisit Example 1.8.6.
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Example 2.1.11.

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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. 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:
2.5+1.75t=6.25
Now we have the skills to solve this equation.
2.5+1.75t=6.252.5+1.75t−2.5=6.25−2.51.75t=3.751.75t1.75=3.751.75t≈2.14
So it will take about 2.14 minutes for the tub to have 6.25 ft3 of water.
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Here we revisit Example 1.8.9.
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Example 2.1.12.

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Jakobi’s annual salary as a nurse this year is $73,290. That’s following a 4% raise over last year’s salary. What was his salary the previous year?
Explanation.
As soon as you understand that the solution will be Jakobi’s 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 variable. Let S represent Jakobi’s salary from last year.
To set up the equation, we need to think about how he arrived at this year’s salary. His employer took last year’s salary and added 4% to that. In words, we have:
(last year's salary)+(4% of last year's salary)=(this year's salary)
We represent “4% of last year’s salary” with 0.04S since 0.04 is the decimal equivalent to 4%. So out equation is:
S+0.04S=73290
Now we have the skills to solve this equation.
S+0.04S=732901.04S=732901.04S1.04=732901.04S≈70471
So last year, Jakobi’s salary was about $70,471.
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Here we revisit Checkpoint 1.8.12.
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Checkpoint 2.1.13.

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A shirt is on sale for 20% off. The sale price is $51.00. What was the shirt’s original price?
Explanation.
Let x represent the original price of the shirt. Since 20% is removed to bring the cost down to $51, we can set up the equation:
x| original −| minus 0.20| 20% ⋅| of x| original =| is 51| $51 
Now we have the skills to solve this equation.
x−0.20x=510.80x=510.80x0.80=510.80x=63.75
The shirt’s original price was $63.75.
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Subsection 2.1.4 Differentiating between Simplifying Expressions, Evaluating Expressions and Solving Equations

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Consider the following three examples, which have similarities on the surface, but are fundamentally different from each other.
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Example 2.1.14.

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Simplify the expression 10−3(x+2).
Explanation.
Distribute the −3 and then combine like terms.
10−3(x+2)=10−3x−6=−3x+4
Note that our final result is an expression.
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Example 2.1.15.

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Evaluate the expression 10−3(x+2) when x=2.
Explanation.
Substitute 2 in for x in the expression:
10−3(x+2)=10−3(2+2)=10−3(4)=10−12=−2
Note that our final result here is a number.
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Example 2.1.16.

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Solve the equation 10−3(x+2)=x−16.
Explanation.
Follow the process we have been using in this section to solve a linear equation.
10−3(x+2)=x−1610−3x−6=x−16−3x+4=x−16−3x+4−x=x−16−x−4x+4=−16−4x+4−4=−16−4−4x=−20−4x−4=−20−4x=5
So the solution set is {5}.
Note that our final result here is a solution set.
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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.
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List 2.1.17. A summary the differences between simplifying expressions, evaluating expressions, and solving equations.
  • An expression like 10−3(x+2) can be simplified to −3x+4 (as in Example 14). However we cannot “solve” an expression like this. It is incorrect to say you will “solve 10−3(x+2)”.
  • 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, and the expression evaluates to −2.
  • An equation connects two expressions with an equal sign. In Example 16, 10−3(x+2)=x−16 has one expression on either side of equal 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).
  • When we solve the equation 10−3(x+2)=x−16, 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. 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.)
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Checkpoint 2.1.18. Matching Vocabulary.

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Reading Questions 2.1.5 Reading Questions

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1.

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Describe the five steps you might need to take when solving a linear equation.
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2.

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In this section there is a reminder to take care with negative numbers when doing what?
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3.

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Explain what is wrong with saying “I need to solve 3x+x−8.”
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Exercises 2.1.6 Exercises

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Vocabulary

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7.
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Quotes are given from a student following a quiz. Fill in the blanks with the appropriate vocabulary terms.
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(a)
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The algebra quiz was fun because you had to
  • evaluate
  • simplify
  • solve
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all these equations.
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(b)
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The last exercise seemed hard at first, but it became more clear once you
  • evaluated
  • simplified
  • solved
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each side.
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(c)
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At first your answer to #4 of the quiz might be pretty messy looking, but it looks more reasonable after you
  • evaluate
  • simplify
  • solve
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it.
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(d)
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One of the word problems was asking you to take a given expression for the height of a ball and
  • evaluate
  • simplify
  • solve
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it for t=7.
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8.
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This exercise walks you through an economics problem. Fill in the blanks with the appropriate vocabulary terms.
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(a)
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If the revenue from selling x items is 2x dollars but the cost of production is 0.5x, then the profit is 2x−0.5x. But that can be
  • evaluated
  • simplified
  • solved
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.
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(b)
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We need to know how much profit there was after 1000 items, so we can
  • evaluate
  • simplify
  • solve
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the profit expression.
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(c)
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If we want to work out how many sales would give us a profit of $3000, we can
  • evaluate
  • simplify
  • solve
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the equation 1.5x=3000.
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(d)
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At first we find that it would take 30001.5 sales, but we can
  • evaluate
  • simplify
  • solve
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that result.
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Skills Practice

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Two-Step Equations.
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Solve the equation.
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25.
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−3.6h=−7.6h−2.4
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More Steps.
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Solve the equation.
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29.
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−9E−4=−130+9E
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30.
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−2K−4=−10+4K
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31.
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7Q−4−3Q=−8−5Q−50
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32.
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−5W+6−7W=−6−3W−42
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34.
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−8(h−5)=3h+117
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35.
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2(n−6)−10=−5(n−4)
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36.
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8(t−6)+11=−9(t+6)
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37.
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−4(z−4)+6z=−7(z+3)−35
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38.
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6(E+6)−3E=−7(E+4)+114
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39.
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−2J−6=−6J−7
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41.
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4W−7−9W=−4+3W+9
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42.
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−7c−7+6c=2−8c+9
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43.
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−9(2h−7)=7h−4
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45.
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3(t−6)+4t=−6(4t+8)+4
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46.
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9(y+5)−8y=−5(2y−2)+6
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47.
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−28−(E−8)=−9E−4
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48.
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−24−(8J−9)=−3J+5
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49.
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−Q−(−3Q−9)=29−(Q−4)
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50.
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−7W−(5W−9)=−8−(W+5)
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51.
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−7(b+4)−(−3b+3)=−6
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Applications

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53.
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Donavan is driving with an average speed of 58 mph on Interstate-5. He looks out the window and sees mile marker 131 pass by. How long will it take him to reach his destination, which is at mile marker 378?
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54.
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Fox filled the gas tank in their car to 13 gal. When the tank reaches 1 gal, the low gas light will come on. On average, Fox’s car uses 0.044 gal per mile driven. How many miles will Fox’s car be able to drive before the low gas light comes on?
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55.
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You planted a young tree in front of your house, and it was 6 feet tall. Ever since, it has been growing by 12 ft each year. How many years will it take for the tree to grow to be 12 feet tall?
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56.
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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 22 psi. You have a portable device that can pump air into the tire increasing the pressure at a rate of 2.2 (psi(min. How long will it take to fill the tire to the manual’s recommended pressure of 31 psi?
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57.
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To save up for a new phone that costs $640, Matteo sets aside 1% of his pay each week. This works out to $4.60 per week. At present, he has $369 saved up. How long will it be until Matteo has saved up enough?
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58.
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A small town would like to replace its aging water treatment system. This will cost $9,050,000, but the town just needs $1,810,000 up front for downpayment on a loan that will cover the rest. The town treasury has $362,000 in it already for this need, and the town can gather $175,000 per month from taxes. How long will it take to reach enough for downpayment on that loan?
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59.
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Taniya puts a pot of water from the tap onto the stove and turns the burner all the way up. The water temperature starts at 58℉ and climbs steadily up to the boiling point of 212℉, raising at a rate of 25℉min. How long will it take for the pot to boil?
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60.
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Alexa puts a pot of water from the tap onto the stove and turns the burner all the way up. The water temperature starts at 12℃ and climbs steadily up to the boiling point of 100℃, raising at a rate of 15℃min. How long will it take for the pot to boil?
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61.
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Candice baked a pie at 425℉ and just took it out of the oven. It immediately starts to cool at a rate of 28℉min. How long will it take to cool to 225℉?
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62.
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On a cold snowy day, the temperature in your home is a cozy 70℉, but then you lose power and heating. Your home temperature begins to drop at a rate of 11℉hour. How long will it take before your home is 39℉?
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63.
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A restaurant stocks its pantry with 100 lb of onions. When the supply reaches 20 lb, it’s time to order more. Recently, the restaurant has been using 9 lb of onions per day. How long since restocking will it take until another order must be placed?
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64.
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At a recent trip to the casino, Jailyn brought $940 in cash. She knows she needs to hold on to $80 in reserve to pay for dinner later. Unfortunately Jailyn had rough luck and was losing money at the slot machines at an average rate of $230 per hour. How long was Jailyn gambling before she had to stop?
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65.
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For Kaylen’s 8th birthday party, her parents rented a venue that charges a flat fee of $95 plus $12 per guest. Ultimately it cost Kaylen’s parents $371. How many guests were there?
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66.
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Matteo’s annual property tax had been $4,020. But he did some renovation to his house that added more square footage, and now the annual property tax is $4,845. The county assesses property tax at a rate of $1.94 per square foot. How much area did Matteo add to his home?
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67.
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Mariana’s current annual salary as a paralegal is $63,841. This is with a raise of 5.2% over last year’s salary. What was her salary last year?
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68.
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A television is for sale in a state where sales tax applies. The sales tax rate is 6.8% and the total was $355. What was the price before sales tax?
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69.
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An annual wine subscription is on sale with a 5% discount and the final price is $439. What was the original price before the discount is applied?
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70.
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The final bill at a restaurant one night was $97.85, including a 16% tip. What was the bill before the tip was added?
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71.
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One year, the median rent for a one-bedroom apartment in a city was reported to be $1,270. This was reported to be an increase of 1.5% over the previous year. Based on this reporting, what was the median rent for of a one-bedroom apartment the previous year?
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Challenge

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72.
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Write a linear equation whose solution is x=−9. Neither side of your equation may be just “x”.
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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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