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In this section you will:
  • Set up a linear equation to solve a real-world application.
  • Use a formula to solve a real-world application.
Many students studying in a large lecture hall
Credit: Kevin Dooley

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.

Setting up a linear equation to solve a real-world 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 0.10 x . 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 = 0.10 x + 50

When dealing with real-world applications, there are certain expressions that we can translate directly into math. [link] 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 2 x
One number is a more than another number x , x + a
One number is a less than twice another number x , 2 x a
The product of a number and a , decreased by b a x b
The quotient of a number and the number plus a is three times the number x x + a = 3 x
The product of three times a number and the number decreased by b is c 3 x ( x b ) = c

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.

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 17 and their sum is 31. Find the two numbers.

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

x + ( x + 17 ) = 31 2 x + 17 = 31 Simplify and solve . 2 x = 14 x = 7 x + 17 = 7 + 17 = 24

The two numbers are 7 and 24.

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Practice Key Terms 3

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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