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Just Enough Algebra 2026 Edition

Section 3.1 Solving linear equations

Back in Chapter 2, we’ve mostly been learning about evaluating functions: given a particular value of the independent variable, figuring out the resulting value of the dependent variable. You might call this process “plugging in.”
Here in Chapter 3, we will instead be interested in solving equations: given a particular value of the dependent variable, figuring out what value of the independent variable it came from.

Solutions, solving, and evaluating

A solution to an equation is the value of the independent variable we plug in to get the desired value of the dependent variable.
When we are solving an equation, this means we know a value of the dependent variable (an output) and we want to figure out the corresponding value of the independent variable (the input) that produced it.
When we are evaluating a function, this means we know a value of the independent variable (an input) and we want to figure out the corresponding value of the dependent variable (the output).
Another way to think about the difference between evaluating and solving is that it is the difference between doing and undoing. When we evaluate a function, we are doing the steps in the equation to the input, so that we can figure out the corresponding output. On the other hand, when we solve an equation, we already have the output, so we want to undo all the things that the equation said to do. That way we can get back to the original input.

Opening Activity 3.1.1.

A very useful idea whenever we talk about solving is the shoes and socks principle, which we’ll explore in this activity.
Bleepblorp the friendly alien has arrived on our planet, and is interested in human footwear.
  1. Write Bleepblorp a set of careful, step-by-step instructions about how to put on shoes and socks.
  2. Oh no! Bleepblorp absolutely hates wearing shoes and socks! Write Bleepblorp a set of careful, step-by-step instructions about how to remove the footwear.
  3. What do you notice about the steps in your putting-on instructions vs. the steps in your removing instructions?
Here’s the moral of this activity:

Shoes and Socks Principle

In order to undo a set of steps, we have to do the opposite steps in the opposite order.

Subsection Figuring out the plumber’s time

Your kitchen sink keeps getting clogged. Very annoying. Last time the plumber was able to fix it pretty quickly. But now the sink is clogged again. This time when the plumber comes and unclogs the sink, he suggests redoing the trap and a few other things that were causing the problem. You are pretty tired of it clogging up and tell him to “go ahead.” While you’re glad that the sink works when he’s done, you’re a bit shocked when his bill arrives a few days later for parts plus $278.75 in labor. Does that seem right?
Remember our plumber charged $100 for just showing up and then $75 per hour for the service. Using the variables
\begin{align*} T \amp= \text{ time plumber takes (hours) } \sim \text{ indep}\\ P \amp= \text{ total plumber's charge (\$) } \sim \text{ dep} \end{align*}
we found that the equation was
\begin{equation*} P=100+75T \end{equation*}
Let’s figure out how many hours of work would add up to a bill of $278.75. Our first approach might be to look at a table. From earlier we had
\(T\) 0 1/2 1 2 3
\(P\) 100.00 137.50 175.00 250.00 325.00
Since $278.75 is between $250.00 and $325.00, we see that the time must be between 2 and 3 hours. You remember the plumber being there over 2 hours, so this is certainly a reasonable answer. Well, a lot of money, but mathematically it makes sense.

Activity 3.1.2.

Let’s revisit the truck hauling bags of grass seed. Remember, the truck weighs 3,900 pounds when it is empty. Each bag of grass seed it carries weighs 4.2 pounds. The equation for the total weight is
\begin{equation*} W = 3{,}900 + 4.2B \end{equation*}
  1. Name the variables in this story, including units and dependence.
    letter = everyday words (units) ~ dep or indep
    = \(\qquad\qquad\qquad\qquad\qquad\) \(\displaystyle\Big(\qquad\Big)\) ~
    = \(\displaystyle\Big(\qquad\Big)\) ~
  2. Choose a few reasonable values of \(B\) to evaluate this equation. Record your answers in the table below.
    \(B\) 0                                                
    \(W\) 3,900
  3. Use your table of values to estimate how many bags of grass seed the truck is carrying if its total weight is 14,500 pounds.

Subsection Finding an approximate solution

Still curious, you’d like to know exactly how many hours and minutes the plumber worked. We could use successive approximation. For example, for 2.5 hours the bill would have been
\begin{equation*} P=100 + 75 \ast 2.5 = 100+75\times \underline{2.5}=\$287.50 \end{equation*}
which is more than the charge. Continuing to guess and check we get
\(T\) 2 3 2.5 2.3 2.4 2.35 2.37 2.38
\(P\) 250.00 325.00 287.50 272.50 280.00 276.25 277.75 278.50
vs. \(278.75\) low high high low high low low close enough
The plumber worked approximately 2.38 hours. Converting units we calculate
\begin{equation*} 0.38 \cancel{\text{ hours}} \ast \frac{60 \text{ minutes}}{1 \cancel{\text{ hour}}} = 0.38 \times 60 = 22.8 \approx 23 \text{ minutes} \end{equation*}
The plumber took about 2 hours, 23 minutes. Thinking back, the plumber had arrived around 10:30 in the morning and stayed past lunch, probably until around 1:00 p.m. That’s about right.

Activity 3.1.3.

The equation for the total weight of the truck carrying bags of grass seed is
\begin{equation*} W = 3{,}900 + 4.2B \end{equation*}
Use successive approximation to find approximately how many bags the truck is carrying if its total weight is 14,500 pounds. Give your answer correct to the nearest bag (there’s no such thing as a decimal number of bags, anyway).
(indep)                                                                                                            
(dep)
hi / lo

Subsection Solving symbolically

Wait a minute! We could have figured the plumber’s cost out much more quickly. If the labor cost was $278.75, we know the first $100 was the trip charge. That leaves
\begin{equation*} \$278.75-\$100.00 = \$178.75 \end{equation*}
in hourly charges. At $75 per hour that comes to
\begin{equation*} \$178.75\ast \frac{1\text{ hour}}{\$75}=178.75 \div 75 = 2.3883333\ldots \approx 2.388 \text{ hours} \end{equation*}
which comes to around 2 hours, 23 minutes as before. See how we used the $75/hour as a unit conversion here? Very clever.
That worked well. But, can we figure out this sort of calculation in other problems? What is the general method we’re using? Can we write down our method in an organized fashion so that someone else could follow our thinking here? Turns out there is a formal way to show this calculation, called (symbolically) solving the equation. Officially any method of getting a solution to an equation is considered solving the equation, but in the rest of this book, and in most places that use algebra, when we refer to “solving the equation” or give the instruction to solve, we mean symbolically.
Here’s how it works. We want to figure out when \(P=278.75\text{.}\) We know from our equation that \(P = 100 + 75T\text{.}\) So we want to find the time \(T\) where
\begin{equation*} 100 + 75T=278.75 \end{equation*}

Aside

Remember that the equal sign means that the two sides are the same number. On the left-hand side we have \(100 + 75T\text{.}\) On the right-hand side we have \(278.75\text{.}\) Looks different, but same thing: both of them are \(P\text{.}\) We are looking for the value of \(T\) that makes these two sides equal.
The first thing we did was subtract the $100 trip charge. In this formal method, we subtract 100 from each side of our equation. If the left-hand side and the right-hand side are the same number, then they will still be equal when we take away 100 from each side. We get
\begin{alignat*}{2} \cancel{100}\amp + 75T \amp =\amp 278.75 \\ -\cancel{100}\amp \amp \amp -100 \end{alignat*}
which simplifies to
\begin{equation*} 75T=278.75-100=178.75 \end{equation*}
because the \(+100\) and \(-100\) cancelled.
The next thing we did to figure out the answer was divide by the $75/hour charge. In this formal method, we can divide each side of our equation by 75. Again, if the left-hand side and right-hand side are the same number, then they will still be equal when we divide by 75. Here goes.
\begin{equation*} \frac{\cancel{75}T}{\cancel{75}}=\frac{178.75}{75} \end{equation*}
Notice that we wrote the division in fraction form (instead of using \(\div\)). To understand why the 75’s cancelled, remember that \(75T\) is short for \(75\ast T\) and so
\begin{equation*} \frac{75T}{75} = \frac{75\ast T}{75} = 75 \times T \div 75=T \end{equation*}
because the \(\times 75\) and \(\div 75\) cancelled. So we have
\begin{equation*} T = \frac{178.75}{75} = 178.75 \div 75 = 2.3883333\ldots \end{equation*}
as before. Yet again, our answer is around 2 hours, 23 minutes.
This calculation is a good example of the Shoes and Socks Principle. To evaluate the function \(P = 100 + 75T\text{,}\) we would do the following steps:
  1. Multiply by 75
  2. Add 100
And so, when we solved, we did the opposite steps in the opposite order. The opposite of adding 100 is subtracting 100, and the opposite of multiplying by 75 is dividing by 75:
  1. Subtract 100
  2. Divide by 75

Activity 3.1.4.

The equation for the total weight of the truck carrying bags of grass seed is
\begin{equation*} W = 3{,}900 + 4.2B \end{equation*}
We are interested in finding how many bags the truck is carrying if its total weight is 14,500 pounds.
  1. Is the number 14,500 a value of \(B\) or a value of \(W\text{?}\) (Hint: Think about the units.)
  2. Replace the correct letter in the equation \(W = 3{,}900 + 4.2B\) by 14,500.
  3. In previous activities in this chapter, you evaluated this equation at many different values of \(B\text{.}\) What steps do you do, in what order, to evaluate the equation?
  4. Shoes and Socks Principle: Do the opposite steps in the opposite order to solve this equation.
  5. Check: Is your answer for the number of bags close to the approximate answer you found in Activity 3.1.3?

Subsection Another plumber cost

Let’s practice working with this symbolic way of solving equations. Suppose instead the plumber went to my neighbor’s house and billed her for $160 in labor costs. How long did the plumber work at my neighbor’s? As before, we begin with our equation
\begin{equation*} P = 100 + 75T \end{equation*}
and we are looking for \(P=160\text{.}\) Replace \(P\) in the equation by 160 to get
\begin{equation*} 100+75T =160 \end{equation*}
Remember, opposite steps in opposite order: First, subtract 100 from each side to get
\begin{alignat*}{2} \cancel{100}\amp + 75T \amp =\amp 160 \\ -\cancel{100}\amp \amp \amp -100 \end{alignat*}
which simplifes to
\begin{equation*} 75T=160-100=60 \end{equation*}
Last, divide each side by 75 to get
\begin{equation*} \frac{\cancel{75}T}{\cancel{75}}=\frac{60}{75} \end{equation*}
which simplifies to
\begin{equation*} T = \frac{60}{75} = 60\div75=0.8 \text{ hours} \end{equation*}
We have solved the equation, but it would make more sense to report our answer in minutes so we convert
\begin{equation*} 0.8 \cancel{\text{ hours}} \ast \frac{60 \text{ minutes}}{1 \cancel{\text{ hour}}} = 0.8 \times 60 = 48 \text{ minutes} \end{equation*}
The plumber worked for 48 minutes at my neighbor’s house.
Remember that we can always check our work by evaluating. Since \(T\) is measured in hours we need to go back and use \(T= 0.8\) hours, not 48 which is in minutes. Evaluating in our original function we get
\begin{equation*} P=100 + 75 \ast 0.8 = 100+75\times\underline{0.8}= 160 \quad \checkmark \end{equation*}

Activity 3.1.5.

The equation for the total weight of the truck carrying bags of grass seed is
\begin{equation*} W = 3{,}900 + 4.2B \end{equation*}
Let’s find the number of bags for a different weight.
  1. Very heavy things (like trucks) are often weighed in tons rather than pounds. Use the fact that 1 ton = 2,000 pounds to convert 8 tons to pounds.
  2. Set up and solve an equation to find the number of bags of grass seed being carried by the truck if its total weight is 8 tons. Remember, “set up and solve an equation” means to use the Shoes and Socks Principle.

Do you know …

  1. The difference between solving and evaluating?
  2. When you solve an equation, and when you evaluate an equation?
  3. Why we “do the same thing to each side” of an equation when solving?
  4. How to solve a linear equation?
  5. The advantages and disadvantages of solving versus successive approximation?
  6. How to check that a solution is correct using the equation?
If you’re not sure, work the rest of exercises and then return to these questions. Or, ask your instructor or a classmate for help.

Exercises Exercises

Exercises 1-4 are available in a separate workbook format.

1.

A truck hauling bags of grass seed weighs 3,900 pounds when it is empty.
Aside
Each bag of seed it carries weighs 4.2 pounds. The equation for the gross weight \(W\) pounds is
\begin{equation*} W = 3{,}900 + 4.2B \end{equation*}
for \(B\) bags of grass seed.
(a)
Set up and solve an equation to determine the number of bags of grass seed being carried by the truck with gross weight of 14,500 pounds.
(b)
Do the same for a truck with gross weight 8 tons. A ton is 2,000 pounds.

2.

Is laughter really the best medicine? A study examined the impact of comedy on anxiety levels. Subjects’ anxiety levels were rated on a scale of 1 to 5 before and after the study. Levels averaged 4.3 before the study. There was no significant change in subjects in the control group. Subjects who watched the comedy videos showed a noticeable difference, and it depended on the hours of comedy watched. Anxiety levels fell an average of 0.098 (on the 1 to 5 scale) for each hour of comedy watched.
(a)
Name the variables. Anxiety is measured on a unitless scale.
(b)
Make a table showing average anxiety levels for subjects who watched comedy videos for 0 hours (control group), 2 hours, 10 hours, and 20 hours, according to these findings.
(c)
Use successive approximation to guess the approximate number of hours watching comedy needed to lower the average anxiety level below 2.
(d)
Write an equation relating the variables.
(e)
Solve your equation to determine the exact number of hours watching comedy needed to lower the average anxiety level below 2.

3.

Lizbeth wants to send her mom truffles for Mother’s Day. It costs $\(C\) to send a box of \(T\) truffles where
\begin{equation*} C = 1.90T+7.95 \end{equation*}
(a)
Make a table of values showing the charges for a box of 8 truffles, 12 truffles, or 30 truffles.
(b)
What are the units on 1.90 and what does it mean in the story?
(c)
What are the units on 7.95 and what does it mean in the story?
(d)
Draw a graph illustrating the cost of sending truffles. Include \(T=0\text{.}\)
(e)
If Lizbeth was charged $53.55 for the box of truffles she sent her mom, how many truffles were there? Set up and solve an equation to answer the question.

4.

The local burger restaurant had a promotion this summer.
Aside
They reduced the price on a bacon double cheeseburger by 2¢ for each degree in the daily high temperature. The equation is
\begin{equation*} B = 7.16 - 0.02H \end{equation*}
where $\(B\) is the price of the bacon double cheeseburger and \(H\) is the daily high temperature, in °F.
(a)
What is the usual price of a bacon double cheeseburger?
(b)
Ronald paid $5.34 for a bacon double cheeseburger on Tuesday. How hot was the temperature that day? Set up and solve an equation.
(c)
What was the high temperature on Sunday when Wendy bought a bacon double cheeseburger for only $5.70? Set up and solve an equation.
(d)
Leroy is holding out for a $5 burger. What temperature will make Leroy’s wish to come true? Set up and solve an equation.

5.

A charter boat tour costs \(\$C\) for \(P\) passengers, where
\begin{equation*} C = 135.00 + 11.95P \end{equation*}
(a)
Make a table of values showing the charges for no passengers, 4 passengers, 10 passengers, and 20 passengers.
(b)
What does the 135.00 represent and what are its units?
(c)
What does the 11.95 represent and what are its units?
(d)
If Freja was charged \(\$ 326.20\) for use of the boat, how many passengers were there? Set up and solve an equation to answer the question.

6.

Abduwali has just opened a restaurant.
Aside
He spent $82,500 to get started but hopes to earn back $6,300 each month. Earlier we determined that
\begin{equation*} A = 6{,}300M - 82{,}500 \end{equation*}
describes how Abduwali’s profit $\(A\) is a function of how long he works (\(M\) months).
(a)
Set up and solve an equation to determine how long it will take Abduwali to break even, meaning make a profit of $0?
(b)
Aduwali will consider the restaurant a success once he’s earned $100,000. According to our equation, when will that be?

7.

Between e-mail, automatic bill pay, and online banking, it seems like I hardly ever actually mail something.
Aside
But for those times, I need postage stamps. The corner store sells as many (or few) stamps as I want for 44¢ each but they charge a 75¢ convenience fee for the whole purchase.
(a)
Make a table showing the cost to buy 5 stamps, 10 stamps, or 20 stamps from the corner store.
(b)
Name the variables and write a linear equation showing how the total price depends on the number of stamps I buy.
(c)
My partner bought postage stamps at the corner store and it cost him $7.35. Solve your equation to determine how many stamps she bought.
(d)
How many stamps could I buy for $10? Solve your equation and check your answer.

8.

When Gretchen walks on her treadmill, she burns 125 calories per mile.
Aside
Recall
\begin{equation*} C=125M \end{equation*}
where \(C\) is the number of calories Gretchen burns by walking \(M\) miles.
(a)
Set up and solve an equation to calculate how far Gretchen has to walk to burn 300 calories.
(b)
If Gretchen walks 3.4 miles per hour on her treadmill, how long will it take her to burn those 300 calories? Report your answer to the nearest minute.
(c)
Pecan pie? Yum. Not fitting into your favorite jeans? No fun. How far does Gretchen have to walk to burn off the calories from those two slices of pecan pie she ate last night? Each slice has approximately 456 calories.

9.

The more expensive something is, the less likely we are to buy it. Well, if we have a choice. For example, when strawberries are in the peak of season, they cost about $2.50 per pint at my neighborhood farmer’s market and demand is approximately 180 pints. (That means, people want to buy about 180 pints at that price.) We approximate that the demand, \(D\) pints, depends on the price, $\(P\text{,}\) as described by the equation
\begin{equation*} D = 305 - 50P \end{equation*}
(a)
How many pints of strawberries are in demand when the price is $3.19 per pint?
(b)
Make a table of values showing the demand for strawberries priced at $2.00/pint, $2.25/pint, $2.50/pint, $2.75/pint, $3.00/pint, $3.25/pint, $3.50/pint.
(c)
Draw a graph illustrating the function. Start at $0/pint even though that’s not realistic.
(d)
It’s been a great week for strawberries and there are 240 pints to be sold at my neighborhood farmer’s. What price should the farmer charge for her strawberries in order to sell them all? Estimate your answer from the graph. Then set up and solve an equation to answer the question.

10.

The stretch of interstate highway through downtown averages 1,450 cars per hour during the morning rush hour.
Aside
The economy is improving (finally), but with that the county manager predicts traffic levels will increase around 130 more cars per hour each week for the next couple of weeks. Earlier we found the equation
\begin{equation*} C=1{,}450 + 130W \end{equation*}
where \(C\) is the number of cars per hour during the morning rush \(W\) weeks since the country manager made her projection.
(a)
Significant slowdowns are expected if traffic levels exceed 2,000 cars per hour. When do they expect that will happen? Set up and solve an equation. Don’t forget to check your answer by evaluating.
(b)
If traffic levels exceed 2,500, the county plans to install control lights at the on-ramps. When is that expected to happen? Set up and solve an equation. Don’t forget to check your answer by evaluating.