▪ Writing a Linear Equation to Solve an Application*

Learning Outcomes

Write a linear equation to express the relationship between unknown quantities.
Write a linear equation to model a real-world situation.
Write a linear equation in one variable to solve problems with two unknowns.

Recall expressions and equations

Recall that a mathematical expression consists of terms connected by addition or subtraction, each term of which consists of variables and numbers connected by multiplication or division. A number that multiplies a variable, such as the [latex]2[/latex] in the term [latex]2x[/latex] is called the coefficient of the variable.

An equation is a mathematical statement of the equivalency of two expressions, one on either side of an equal sign.

Expressions may be combined or simplified by rearranging or doing operations on the terms. Equations may be solved for the value or values of the variable that make the statement of equivalency true.

Consider a car rental agency that charges $150 per week plus $0.10 per mile driven to rent a compact car. We can use these quantities to write an equation that models the cost of renting the car for a week [latex]C[/latex] given a certain number of miles [latex]x[/latex] driven.

[latex]C=150+0.10x[/latex]

To set up a linear equation that models a real-world situation, we must first determine the known quantities and define the unknown quantity as a variable. Then, we 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 per mile, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write [latex]0.10x[/latex] to model the portion of the weekly cost generated by miles driven. 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. In the car rental example, there is a flat fee of $150 to rent the car, independent of the number of miles driven. In applications involving costs, amounts such as this flat fee that do not change are often called fixed costs.

When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table below 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]\Large\frac{x}{x+a}\normalsize =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]

In the car-rental example above, we identified the unknown quantity (number of miles driven), and assigned it the variable [latex]x[/latex]. Then we identified known quantities, and translated the words in the situation into relationships between the known and unknown quantities to write an equation that models the situation. We could use the model to answer questions about the situation such as finding the total cost for a week if 500 miles were driven or how many miles could be driven in a week on a budget of $375.

How To: Given a real-world situation, write a linear equation to model it

Identify known and unknown 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 in the problem as mathematical operations.
Solve the equation, check to be sure your answer is reasonable, and give the answer using the language and units of the original situation.

Example

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.

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Try It

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.

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Recall the forms of equations of lines

Slope-intercept form: [latex]y=mx+b[/latex], where [latex]x \text{and } y[/latex] represent the coordinates of any point on the line, [latex]m[/latex] represents the slope of the line, and [latex]b[/latex] represents the initial value, or the y-intercept. You can solve for [latex]b[/latex] by substituting a known point for [latex]x \text{ and } y[/latex] and the slope of the line for [latex]m[/latex].

The point-slope form, [latex]y-{y}_{1}=m\left(x-{x}_{1}\right)[/latex] simplifies to slope-intercept form when solved for [latex]y[/latex]. Substitute the coordinates of a point for [latex]{y}_1 \text{ and } {x}_1[/latex] and the slope for [latex]m[/latex] then simplify.

Example: write a Linear Equation in the form of 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?

Show 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.

The following video shows another example of using linear equations to model and compare two cell phone plans.

try it

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $0.04/min talk-time. Company B charges a monthly service fee of $45 plus $0.03/min talk-time. Use [latex]x[/latex] for your variable.

a) Write an equation that models the monthly cost for company A.

b) Write an equation that models the monthly cost for company B.

C) If the average number of minutes used each month is 1,162, how much is the monthly cost for each company?

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Try It

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?

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We can also use a linear equation in one variable to solve a problem with two unknowns by writing an expression for one unknown in terms of the other.

example: write a linear equation in one variable to model and solve an application

A bag is filled with green and blue marbles. There are 77 marbles in the bag. If there are 17 more green marbles than blue marbles, find the number of green marbles and the number of blue marbles in the bag.

How many marbles of each color are in the bag?

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Watch the following video for another example of using a linear equation in one variable to solve an application.

Try it

A cash register contains only five dollar and ten dollar bills. It contains twice as many five’s as ten’s and the total amount of money in the cash register is 560 dollars. How many ten’s are in the cash register?

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Module 1: Equations and Inequalities