Models and Applications

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.

Writing a Linear Equation to Solve an Application

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 $C$.

$C=0.10x+50$

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

Verbal	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ Translation to Math Operations
One number exceeds another by a	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x,\;x+a$
Twice a number	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2x$
One number is a more than another number	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x,\;x+a$
One number is a less than twice another number	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x,\;2x-a$
The product of a number and a, decreased by b	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;ax-b$
The quotient of a number and the number plus a is three times the number	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\frac {x}{x+a}}=3x$
The product of three times a number and the number decreased by b is c	$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3x\left(x-b\right)=c$

How To: Given a real-world problem, model a linear equation to fit it

1. Identify known quantities.
2. Assign a variable to represent the unknown quantity.
3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
4. Write an equation interpreting the words as mathematical operations.
5. Solve the equation. Be sure the solution can be explained in words including the units of measure.

Example: 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?

Show Solution

1. The model for Company A can be written as $A=0.05x+34$. This includes the variable cost of $0.05x$ plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of $40, but a lower variable cost of $0.04x$. Company B’s model can be written as $B=0.04x+40$.
2. If the average number of minutes used each month is 1,160, we have the following:
   ${\begin{array}{l}{\text{Company }}A&=0.05\left(1,160\right)+34\\&=58+34\\&=92\\\\{\text{Company }}B&=0.04\left(1,160\right)+40\\&=46.4+40\\&=86.4\end{array}}$

   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.

3. If the average number of minutes used each month is 420, we have the following:
   ${\begin{array}{l}{\text{Company }}A&=0.05\left(420\right)+34\\&=21+34\\&=55\\\\{\text{Company }}B&=0.04\left(420\right)+40\\&=16.8+40\\&=56.8\end{array}}$

   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.

4. 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 $\left(x,y\right)$ 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.
   ${\begin{array}{l}0.05x+34=0.04x+40\\0.01x=6\\x=600\end{array}}$

   Check the x-value in each equation.

   ${\begin{array}{l}0.05\left(600\right)+34=64\\0.04\left(600\right)+40=64\end{array}}$

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

Licensing

Content obtained and/or adapted from:

• Models and Applications, Lumen Learning College Algebra under a CC BY license