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 [latex]0.10x[/latex]. 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 [latex]C[/latex].

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 |
[latex]x,\text{ }x+a[/latex] |

Twice a number | [latex]2x[/latex] |

One number is a more than another number |
[latex]x,\text{ }x+a[/latex] |

One number is a less than twice another number |
[latex]x,2x-a[/latex] |

The product of a number and a, decreased by b |
[latex]ax-b[/latex] |

The quotient of a number and the number plus a is three times the number |
[latex]\frac{x}{x+a}=3x[/latex] |

The product of three times a number and the number decreased by b is c |
[latex]3x\left(x-b\right)=c[/latex] |

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

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

### Example 1: Modeling a Linear Equation to Solve an Unknown Number Problem

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[/latex] and their sum is [latex]31[/latex]. Find the two numbers.

### Solution

Let [latex]x[/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[/latex]. The sum of the two numbers is 31. We usually interpret the word *is* as an equal sign.

The two numbers are [latex]7[/latex] and [latex]24[/latex].

### Try It 1

Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[/latex], find the numbers.

### 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.

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

### Solution

- The model for Company
*A*can be written as [latex]A=0.05x+34[/latex]. This includes the variable cost of [latex]0.05x[/latex] plus the monthly service charge of $34. Company*B*’s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[/latex]. Company*B*’s model can be written as [latex]B=0.04x+\$40[/latex]. - If the average number of minutes used each month is 1,160, we have the following:
[latex]\begin{array}{l}\text{Company }A\hfill&=0.05\left(1,160\right)+34\hfill \\ \hfill&=58+34\hfill \\ \hfill&=92\hfill \\ \hfill \\ \text{Company }B\hfill&=0.04\left(1,160\right)+40\hfill \\ \hfill&=46.4+40\hfill \\ \hfill&=86.4\hfill \end{array}[/latex]
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. - If the average number of minutes used each month is 420, we have the following:
[latex]\begin{array}{l}\text{Company }A\hfill&=0.05\left(420\right)+34\hfill \\ \hfill&=21+34\hfill \\ \hfill&=55\hfill \\ \hfill \\ \text{Company }B\hfill&=0.04\left(420\right)+40\hfill \\ \hfill&=16.8+40\hfill \\ \hfill&=56.8\hfill \end{array}[/latex]
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. - 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 [latex]\left(x,y\right)[/latex] 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.*[latex]\begin{array}{l}0.05x+34=0.04x+40\hfill \\ 0.01x=6\hfill \\ x=600\hfill \end{array}[/latex]Check the

*x-*value in each equation.[latex]\begin{array}{l}0.05\left(600\right)+34=64\hfill \\ 0.04\left(600\right)+40=64\hfill \end{array}[/latex]Therefore, a monthly average of 600 talk-time minutes renders the plans equal.

### 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?