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Precalculus for Team-Based Inquiry Learning 2025 Edition — Instructor Version

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Section 1.2 Applications of Linear Equations (EQ2)

Subsection 1.2.1 Activities

Observation 1.2.1.

Linear equations can be used to solve many types of real-world applications. We’ll investigate some of those in this section.
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Remark 1.2.2.

Distance, rate, and time problems are a standard example of an application of a linear equation. For these, it’s important to remember that
\begin{equation*} d=rt \end{equation*}
where \(d\) is distance, \(r\) is the rate (or speed), and \(t\) is time.
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Often we will have more than one moving object, so it is helpful to denote which object’s distance, rate, or time we are referring to. One way we can do this is by using a subscript. For example, if we are describing an eastbound train (as we will in the first example), it may be helpful to denote its distance, rate, and time as \(d_E\text{,}\) \(r_E\text{,}\) and \(t_E\) respectively. Notice that the subscript \(E\) is a label reminding us that we are referring to the eastbound train.
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Activity 1.2.3.

Two trains leave a station at the same time. One is heading east at a speed of \(75\) mph, while the other is heading west at a speed of \(85\) mph. After how long will the trains be \(400\) miles apart?
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(a)
How fast is each train traveling?
  1. \(r_E=85\) mph, \(r_W=75\) mph
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  2. \(r_E=75\) mph, \(r_W=85\) mph
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  3. \(r_E=400\) mph, \(r_W=400\) mph
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  4. \(r_E=75\) mph, \(r_W=400\) mph
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  5. \(r_E=400\) mph, \(r_W=85\) mph
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Answer.
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(b)
Which of the statements describes how the times of the eastbound and westbound train are related?
  1. The eastbound train is slower than the westbound train, so \(75+t_E=85+t_W\text{.}\)
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  2. The eastbound train left an hour before the westbound train, so if we let \(t_E=t\text{,}\) then \(t_W=t-1\text{.}\)
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  3. Both trains have been traveling the same amount of time, so \(t_E=t_W\text{.}\) Since they are the same, we can just call them both \(t\text{.}\)
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  4. We don’t know how the times relate to each other, so we must denote them separately as \(t_E\) and \(t_W\text{.}\)
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  5. Since the trains are traveling at different speeds, we need the proportion \(\dfrac{r_E}{r_W}=\dfrac{t_E}{t_W}\text{.}\)
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Answer.
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(c)
Fill in the following table using the information you’ve just determined about the trains’ rates and times since they left the station. Some values are there to help you get started.
rate time distance from station
eastbound train \(75t\)
westbound train \(t\)
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Answer.
rate time distance from station
eastbound train \(75\) \(t\) \(75t\)
westbound train \(85\) \(t\) \(85t\)
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(d)
At the moment in question, the trains are \(400\) miles apart. How does that total distance relate to the distance each train has traveled?
  1. The \(400\) miles is irrelevant. They’ve been traveling the same amount of time so they must be the same distance away from the station. That tells us \(d_E = d_W \text{.}\)
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  2. The \(400\) miles is the difference between the distance each train traveled, so \(d_E - d_W = 400\text{.}\)
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  3. The \(400\) miles represents the sum of the distances that each train has traveled, so \(d_E + d_W = 400\text{.}\)
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  4. The \(400\) miles is the product of the distance each train traveled, so \((d_E)( d_W) = 400\text{.}\)
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Answer.
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(e)
Now plug in the expressions from your table for \(d_E\) and \(d_w\text{.}\) What equation do you get?
  1. \(\displaystyle 75t=85t\)
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  2. \(\displaystyle 75t-85t=400\)
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  3. \(\displaystyle 75t+85t=400\)
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  4. \(\displaystyle (75t)(85t)=400\)
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Answer.
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(f)
Notice that we now have a linear equation in one variable, \(t\text{.}\) Solve for \(t\text{,}\) and put that answer in context of the problem.
  1. The trains are \(400\) miles apart after \(2\) hours.
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  2. The trains are \(400\) miles apart after \(2.5\) hours.
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  3. The trains are \(400\) miles apart after \(3\) hours.
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  4. The trains are \(400\) miles apart after \(3.5\) hours.
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  5. The trains are \(400\) miles apart after \(4\) hours.
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Answer.
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Remark 1.2.4.

In Activity 1.2.3, we examined the motion of two objects moving at the same time in opposite directions. In Activity 1.2.5, we will examine a different perspective, but still apply \(d=rt\) to solve.
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Activity 1.2.5.

Jalen needs groceries, so decides to ride his bike to the store. It takes him half an hour to get there. After finishing his shopping, he sees his friend Alex who offers him a ride home. He takes the same route home as he did to the store, but this time it only takes one-fifth of an hour. If his average speed was \(18\) mph faster on the way home, how far away does Jalen live from the grocery store?
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We’ll use the subscript \(b\) to refer to variables relating to Jalen’s trip to the store while riding his \(b\)ike and the subscript \(c\) to refer to variables relating to Jalen’s trip home while riding in his friend’s \(c\)ar.
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(a)
How long does his bike trip from home to the store and his car trip from the store back home take?
  1. \(t_b=18\) hours, \(t_c=18\) hours
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  2. \(t_b=\dfrac{1}{5}\) of an hour, \(t_c=\dfrac{1}{2}\) of an hour
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  3. \(t_b=\dfrac{1}{2}\) of an hour, \(t_c=\dfrac{1}{5}\) of an hour
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  4. \(t_b=2\) hours, \(t_c=5\) hours
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  5. \(t_b=5\) hours, \(t_c=2\) hours
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Answer.
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(b)
Which of the statements describes how the speed (rate) of the bike trip and the car trip are related?
  1. Both the trip to the store and the trip home covered the same distance, so \(r_b=r_c\text{.}\) Since they are the same, we can just call them both \(r\text{.}\)
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  2. We don’t know how the two rates relate to each other, so cannot write an equation comparing them and must leave them as separate variables \(r_b\) and \(r_c\text{.}\)
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  3. Jalen’s rate on the trip home in the car was \(18\) mph faster than his trip to the store on his bike, so if we let \(r_b=r\text{,}\) then \(r_c=r-18\text{.}\)
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  4. Jalen’s rate on the trip home in the car was \(18\) mph faster than his trip to the store on his bike, so if we let \(r_b=r\text{,}\) then \(r_c=r+18\text{.}\)
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Answer.
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(c)
Fill in the following table using the information you’ve just determined about the Jalen’s rates and times on each leg of his grocery store trip. Then fill in the distance column based on how distance relates to rate and time in each case.
rate time distance covered
bike trip (to the store)
car trip (going back home)
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Answer.
rate time distance covered
bike trip (to the store) \(r\) \(\dfrac{1}{2}\) \(\dfrac{1}{2}r\)
car trip (going back home) \(r+18\) \(\dfrac{1}{5}\) \(\dfrac{1}{5}(r+18)\)
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(d)
Our goal is to figure out how far away Jalen lives from the store. To help us get there, write an equation relating \(d_b\) and \(d_c\text{.}\)
  1. The distance he traveled by bike is the same as the distance he traveled by car, so \(d_b = d_c\text{.}\)
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  2. The distance he traveled by bike took longer than the distance he traveled by car, so \(d_b + \dfrac{1}{2} = d_c + \dfrac{1}{5}\text{.}\)
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  3. The distance, \(d\text{,}\) between his house and the grocery store is sum of the distance he traveled on his bike and the distance he traveled in the car, so \(d_b + d_c = d\text{.}\)
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  4. The distance, \(d\text{,}\) between his house and the grocery store is sum of the difference he traveled on his bike and the distance he traveled in the car, so \(d_b - d_c = d\text{.}\)
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Answer.
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(e)
Now plug in the expressions from your table for \(d_b\) and \(d_c\) into the equation you just found. Notice that it is a linear equation in one variable, \(r\text{.}\) Solve for \(r\text{.}\)
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Answer.
\(r=12\)
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(f)
Our goal was to determine the distance between Jalen’s house and the grocery store. Solving for \(r\) did not tell us that distance, but it did get us one step closer. Use that value to help you determine the distance between his house and the store, and write your answer using the context of the problem.
  1. The grocery store is \(6\) miles away from Jalen’s house.
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  2. The grocery store is \(8\) miles away from Jalen’s house.
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  3. The grocery store is \(10\) miles away from Jalen’s house.
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  4. The grocery store is \(12\) miles away from Jalen’s house.
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  5. The grocery store is \(14\) miles away from Jalen’s house.
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Hint.
Can you find an expression involving \(r\) that we made that represents that distance?
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Answer.
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Remark 1.2.6.

Another type of application of linear equations is called a mixture problem. In these we will mix together two things, like two types of candy in a candy store or two solutions of different concentrations of alcohol.
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Activity 1.2.7.

Ammie’s favorite snack to share with friends is candy salad, which is a mixture of different types of candy. Today she chooses to mix Nerds Gummy Clusters, which cost \(\$8.38\) per pound, and Starburst Jelly Beans, which cost \(\$7.16\) per pound. If she makes seven pounds of candy salad and spends a total of \(\$55.61\text{,}\) how many pounds of each candy did she buy?
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(a)
There are two "totals" in this situation: the total weight (in pounds) of candy Ammie bought and the total amount of money (in dollars) Ammie spent. Let’s begin with the total weight. If we let \(N\) represent the pounds of Nerds Gummy Clusters and \(S\) represent the pounds of Starburst Jelly Beans, which of the following equations can represent the total weight?
  1. \(\displaystyle N-S=7\)
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  2. \(\displaystyle NS=7\)
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  3. \(\displaystyle N+S=7\)
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  4. \(\displaystyle \dfrac{N}{S}=7\)
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Answer.
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(b)
Which expressions represent the amount she spent on each candy? Again, we will let \(N\) represent the pounds of Nerds Gummy Clusters and \(S\) represent the pounds of Starburst Jelly Beans.
  1. \(N\) spent on Nerds Gummy Clusters; \(S\) spent on Starburst Jelly Beans
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  2. \(8.38N\) spent on Nerds Gummy Clusters; \(7.16S\) spent on Starburst Jelly Beans
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  3. \(8.38 + N\) spent on Nerds Gummy Clusters; \(7.16 + S\) spent on Starburst Jelly Beans
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  4. \(8.38 - N\) spent on Nerds Gummy Clusters; \(7.16 - S\) spent on Starburst Jelly Beans
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Answer.
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(c)
Now we focus on the total cost. Which of the following equations can represent the total amount she spent?
  1. \(\displaystyle N+S=55.61\)
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  2. \(\displaystyle 8.38N+7.16S=55.61\)
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  3. \(\displaystyle 8.38+ N + 7.16 + S=55.61\)
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  4. \(\displaystyle 8.38- N + 7.16 - S=55.61\)
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Answer.
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(d)
We are almost ready to solve, but we have two variables in our weight equation and our cost equation. We will get the cost equation to one variable by using the weight equation as a substitution. Which of the following is a way to express one variable in terms of the other?
  1. If \(N\) is the total weight of the Nerds Gummy Clusters, then \(7-N\) could represent the weight of the Starburst Jelly Beans.
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  2. If \(N\) is the total weight of the Nerds Gummy Clusters, then \(7+N\) could represent the weight of the Starburst Jelly Beans.
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  3. If \(S\) is the total weight of the Starburst Jelly Beans, then \(7-S\) could represent the weight of the Nerds Gummy Clusters.
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  4. If \(S\) is the total weight of the Starburst Jelly Beans, then \(7+S\) could represent the weight of the Nerds Gummy Clusters.
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Hint.
More than one answer may be correct here!
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Answer.
A or C
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(e)
Plug your expressions in to the total cost equation.
  1. \(\displaystyle 8.38N+7.16(7-N)=55.61\)
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  2. \(\displaystyle 8.38S+7.16(7-S)=55.61\)
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  3. \(\displaystyle 8.38(7-N)+7.16N=55.61\)
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  4. \(\displaystyle 8.38(7-S)+7.16S=55.61\)
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Hint.
More than one of these may be correct!
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Answer.
A or D, depending on the substitution chosen.
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(f)
Now solve for \(N\) and \(S\text{,}\) and put your answer in the context of the problem.
  1. Ammie bought \(2.5\) lbs of Nerds Gummy Clusters and \(4.5\) lbs of Starburst Jelly Beans.
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  2. Ammie bought \(3.5\) lbs of Nerds Gummy Clusters and \(3.5\) lbs of Starburst Jelly Beans.
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  3. Ammie bought \(4.5\) lbs of Nerds Gummy Clusters and \(2.5\) lbs of Starburst Jelly Beans.
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  4. Ammie bought \(5.5\) lbs of Nerds Gummy Clusters and \(1.5\) lbs of Starburst Jelly Beans.
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Answer.
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Activity 1.2.8.

A chemist needs to mix two solutions to create a mixture consisting of \(30\%\) alcohol. She uses \(20\) liters of the first solution, which has a concentration of \(21\%\) alcohol. How many liters of the second solution (that is \(45\%\) alcohol) should she add to the first solution to create the mixture that is \(30\%\) alcohol?
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Answer.
\(12\) liters
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Remark 1.2.9.

In the next couple of activities, we will explore linear applications in geometry.
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Activity 1.2.10.

A rectangular field is ten times as long as it is wide. If the perimeter of the field is \(1{,}870\) feet, what are the dimensions of the field?
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(a)
Let’s first think about the relationship between the length and width of the rectangular field. How can we express the length of the field (\(l\)) if we let \(w\) represent the width of the field?
  1. \(\displaystyle l=\frac{1}{10}w\)
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  2. \(\displaystyle w=10l\)
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  3. \(\displaystyle l=10w\)
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  4. \(\displaystyle w=\frac{1}{10}l\)
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Answer.
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(b)
Now that we can represent both the length and the width of the field, how can we represent the perimeter of the field in terms of \(w\text{?}\)
  1. \(\displaystyle P=10w+w+10w+w\)
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  2. \(\displaystyle P=20w+2w\)
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  3. \(\displaystyle P=11w\)
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  4. \(\displaystyle P=22w\)
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  5. \(\displaystyle P=10w^2\)
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Answer.
A, B, and D
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(c)
Use your expression from part (b) and the fact that we know the perimeter is \(1{,}870\) feet to create an equation. Then solve for \(w\) to find the width of the field.
  1. \(w=850\) feet
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  2. \(w=14\) feet
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  3. \(w=13\) feet
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  4. \(w=85\) feet
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Answer.
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(d)
Use the value you got in part (c) to find the length of the field.
  1. \(l=850\) feet
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  2. \(l=14\) feet
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  3. \(l=13\) feet
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  4. \(l=85\) feet
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Answer.
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Activity 1.2.11.

Two sides of a triangle are \(5\) and \(7\) inches longer than the third side. If the perimeter measures \(21\) inches, find the length of each side.
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Answer.
The three sides of the triangle measure \(3\) inches, \(8\) inches, and \(10\) inches.
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Remark 1.2.12.

As you can see, there are many applications of linear equations! This section only highlights some of the more common types of applications - have you seen other types of linear applications in your life?
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Subsection 1.2.2 Exercises

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