Applications of Linear Equations

Learning Objectives

Figure 1

Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce $x$ widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

Source: Rice University, https://openstax.org/books/college-algebra/pages/2-3-models-and-applications
This work is licensed under a Creative Commons Attribution 4.0 License.

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write $0.10x$. This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost $C$.

When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 1 lists some common verbal expressions and their equivalent mathematical expressions.

EXAMPLE 2

Setting Up a Linear Equation to Solve a Real-World Application

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time.

1. ⓐWrite a linear equation that models the packages offered by both companies.
2. ⓑ If the average number of minutes used each month is 1,160, which company offers the better plan?
3. ⓒIf the average number of minutes used each month is 420, which company offers the better plan?
4. ⓓHow many minutes of talk-time would yield equal monthly statements from both companies?

Solution

(a)The model for Company $A$ can be written as $A=0.05 x+34$. This includes the variable cost of $0.05 x$ plus the monthly service charge of $\$ 34$. Company B's package charges a higher monthly fee of $\$ 40$, but a lower variable cost of $0.04 x$. Company B's model can be written as $B=0.04 x+\$ 40$.

(b) If the average number of minutes used each month is $1,160$, we have the following:

$\begin{aligned} \text { Company } A &=0.05(1,160)+34 \\ &=58+34 \\ &=92 \\ \\ \text { Company } B &=0.04(1,160)+40 \\ &=46.4+40 \\ &=86.4 \end{aligned}$

So, Company $B$ offers the lower monthly cost of $\$ 86.40$ as compared with the $\$ 92$ monthly cost offered by Company $A$ when the average number of minutes used each month is 1,160.

(c) If the average number of minutes used each month is 420, we have the following:

$\begin{aligned} \text { Company } A &=0.05(420)+34 \\ &=21+34 \\ &=55 \\ \\ \text { Company } B &=0.04(420)+40 \\ &=16.8+40 \\ &=56.8 \end{aligned}$

If the average number of minutes used each month is $420$, then Company $A$ offers a lower monthly cost of $\$ 55$ compared to Company B's monthly cost of $\$ 56.80$.

(d) To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of $(x, y)$ coordinates: At what point are both the $x$-value and the $y$-value equal? We can find this point by setting the equations equal to each other and solving for $x$.

$\begin{aligned} 0.05 x+34 &=0.04 x+40 \\ 0.01 x &=6 \\ x &=600 \end{aligned}$

Check the $x$-value in each equation.

$\begin{aligned} &0.05(600)+34=64 \\ &0.04(600)+40=64 \end{aligned}$

Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2.

Figure 2

TRY IT #2

Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company's monthly expenses?

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem's question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, $A=L W$; the perimeter of a rectangle, $P=2 L+2 W$; and the volume of a rectangular solid, $V=L W H$. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

EXAMPLE 3

Solving an Application Using a Formula

It takes Andrew $30$ min to drive to work in the morning. He drives home using the same route, but it takes $10$ min longer, and he averages $10 \, \mathrm{mi} / \mathrm{h}$ less than in the morning. How far does Andrew drive to work?

Analysis

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for $r$.

$\begin{aligned} r\left(\frac{1}{2}\right) &=(r-10)\left(\frac{2}{3}\right) \\ 6 \times r\left(\frac{1}{2}\right) &=6 \times(r-10)\left(\frac{2}{3}\right) \\ 3 r &=4(r-10) \\ 3 r &=4 r-40 \\-r &=-40 \\ r &=40 \end{aligned}$

TRY IT #3

On Saturday morning, it took Jennifer $3.6 \, h$ to drive to her mother's house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer $4 \, h$ to return home. Her speed was $5$ mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

EXAMPLE 4

Solving a Perimeter Problem

Solution

Now we can solve for the width and then calculate the length.

The dimensions are $L=15 \, \mathrm{ft}$ and $W=12 \, \mathrm{ft}$.

TRY IT #4

EXAMPLE 5

Solving an Area Problem

Solution

$\begin{aligned} P &=2 L+2 W \\ 48 &=2(W+6)+2 W \\ 48 &=2 W+12+2 W \\ 48 &=4 W+12 \\ 36 &=4 W \\ 9 &=W \\(9+6) &=L \\ 15 &=L \end{aligned}$

Now, we find the area given the dimensions of $L=15$ in. and $W=9$ in.

$\begin{aligned}A &=L W \\A &=15(9) \\&=135 \, \text { in. }^{2}\end{aligned}$

The area is $135$ in. $^{2}$.

TRY IT #5

EXAMPLE 6

Solving a Volume Problem

Solution

The formula for the volume of a box is given as $V=L W H$, the product of length, width, and height. We are given that $L=2 W$, and $H=8$. The volume is 1,600 cubic inches.

$\begin{aligned}V &=L W H \\1,600 &=(2 W) W(8) \\1,600 &=16 W^{2} \\100 &=W^{2} \\10 &=W\end{aligned}$

The dimensions are $L=20$ in., $W=10$ in., and $H=8$ in.

Analysis