Write a linear equation to express the relationship between unknown quantities.
Use a linear model to answer questions.
Set up a linear equation involving distance, rate, and time.
Find the dimensions of a rectangle given the area.
Find the dimensions of a box given information about its side lengths.
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.
College students taking an exam. Credit: Kevin Dooley
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 [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].
[latex]C=0.10x+50[/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: 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.
Show 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
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.
Show Solution
11 and 25
Example: Writing an Equation for a Linear Cost Function
Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C where C(x) is the cost for x items produced in a given month.
Show Solution
The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by [latex]C\left(x\right)=1250+37.5x[/latex].
Analysis of the Solution
If Ben produces 100 items in a month, his monthly cost is represented by
Example: Using a Linear Function to Determine the Number of Songs in a Music Collection
Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year?
Show Solution
The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200.
Figure 12
The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line.
We can write the formula [latex]N\left(t\right)=15t+200[/latex].
With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at t = 12.
Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.
Example: Using a Linear Function to Calculate Salary Plus Commission
Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation.
Show Solution
The given information gives us two input-output pairs: (3, 760) and (5, 920). We start by finding the rate of change.
Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.
The value of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.
We can now write the final equation.
[latex]I\left(n\right)=80n+520[/latex]
Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.
Example: Using Tabular Form to Write an Equation for a Linear Function
The table below relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.
w, number of weeks
0
2
4
6
P(w), number of rats
1000
1080
1160
1240
Show Solution
We can see from the table that the initial value for the number of rats is 1000, so b = 1000.
Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.
[latex]P\left(w\right)=40w+1000[/latex]
If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240)
Is the initial value always provided in a table of values like the table in Example above?
No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into [latex]f\left(x\right)=mx+b[/latex], and solve for b.
Try It
A new plant food was introduced to a young tree to test its effect on the height of the tree. The table below shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment.
x
0
2
4
8
12
H(x)
12.5
13.5
14.5
16.5
18.5
Show Solution
[latex]H\left(x\right)=0.5x+12.5[/latex]
Using a Given Input and Output to Build a Model
Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.
How To: Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.
Identify the input and output values.
Convert the data to two coordinate pairs.
Find the slope.
Write the linear model.
Use the model to make a prediction by evaluating the function at a given x value.
Use the model to identify an x value that results in a given y value.
Answer the question posed.
Example: Using a Linear Model to Investigate a Town’s Population
A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues.
Predict the population in 2013.
Identify the year in which the population will reach 15,000.
Solution
The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2000 years ago!
To make computation a little nicer, we will define our input as the number of years since 2004:
Input: t, years since 2004
Output: P(t), the town’s population
To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.
To determine the rate of change, we will use the change in output per change in input.
[latex]m=\frac{\text{change in output}}{\text{change in input}}[/latex]
The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to [latex]t=0[/latex], giving the point [latex]\left(0,\text{6200}\right)[/latex]. Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to [latex]t=\text{5}[/latex], giving the point [latex]\left(5,\text{8100}\right)[/latex].
The two coordinate pairs are [latex]\left(0,\text{6200}\right)[/latex] and [latex]\left(5,\text{8100}\right)[/latex]. Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.
[latex]\begin{cases} m=\frac{8100 - 6200}{5 - 0}\hfill \\ \text{ }=\frac{1900}{5}\hfill \\ \text{ }=380\text{ people per year}\hfill \end{cases}[/latex]
We already know the y-intercept of the line, so we can immediately write the equation:
[latex]P\left(t\right)=380t+6200[/latex]
To predict the population in 2013, we evaluate our function at t = 9.
Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.
Try It
A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.
Write a linear model to represent the cost C of the company as a function of x, the number of doughnuts produced.
Find and interpret the y-intercept.
Solution
C(x)=0.25x+25,000The y-intercept is (0, 25,000). If the company does not produce a single doughnut, they still incur a cost of $25,000.
Try It
A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.
Predict the population in 2014.
Identify the year in which the population will reach 54,000.
Solution
41,100; 2020
Using a Diagram to Model a Problem
It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.
Example: Using a Diagram to Model Distance Walked
Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?
Solution
In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: “How long will it take them to be 2 miles apart?”
In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.
Input: t, time in hours.
Output: [latex]A\left(t\right)[/latex], distance in miles, and [latex]E\left(t\right)[/latex], distance in miles
Because it is not obvious how to define our output variable, we’ll start by drawing a picture.
Figure 3
Initial Value: They both start at the same intersection so when [latex]t=0[/latex], the distance traveled by each person should also be 0. Thus the initial value for each is 0.
Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3.
Using those values, we can write formulas for the distance each person has walked.
For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.
We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful.
Recall that we need to know how long it takes for D, the distance between them, to equal 2 miles. Notice that for any given input t, the outputs A(t), E(t), and D(t) represent distances.
Figure 4
This picture shows us that we can use the Pythagorean Theorem because we have drawn a right angle.
In this scenario we are considering only positive values of [latex]t[/latex], so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t.
They will fall out of radio contact in 0.4 hours, or 24 minutes.
Q & A
Should I draw diagrams when given information based on a geometric shape?
Yes. Sketch the figure and label the quantities and unknowns on the sketch.
Example: Using a Diagram to Model Distance between Cities
There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?
SOLUTION
It might help here to draw a picture of the situation. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0).
Figure 5
Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:
The equation of the road from Westborough to Agritown would be
[latex]W\left(x\right)=\frac{1}{3}x[/latex]
From this, we can determine the perpendicular road to Eastborough will have slope [latex]m=-3[/latex]. Because the town of Eastborough is at the point (20, 0), we can find the equation:
One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.
Try It
There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?
Solution
21.15 miles
Using Formulas to Solve Problems
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, [latex]A=LW[/latex]; the perimeter of a rectangle, [latex]P=2L+2W[/latex]; and the volume of a rectangular solid, [latex]V=LWH[/latex]. 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: Solving an Application Using a Formula
It takes Andrew 30 minutes to drive to work in the morning. He drives home using the same route, but it takes 10 minutes longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?
Show Solution
This is a distance problem, so we can use the formula [latex]d=rt[/latex], where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.
First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or [latex]\frac{1}{2}[/latex] h at rate [latex]r[/latex]. His drive home takes 40 min, or [latex]\frac{2}{3}[/latex] h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance [latex]d[/latex]. A table, such as the one below, is often helpful for keeping track of information in these types of problems.
We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.
On Saturday morning, it took Jennifer 3.6 hours to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 hours to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?
Show Solution
45 [latex]\frac{\text{mi}}{\text{h}}[/latex]
Example: Solving a Perimeter Problem
The perimeter of a rectangular outdoor patio is [latex]54[/latex] ft. The length is [latex]3[/latex] ft. greater than the width. What are the dimensions of the patio?
Show Solution
The perimeter formula is standard: [latex]P=2L+2W[/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as shown below.
Now we can solve for the width and then calculate the length.
The dimensions are [latex]L=15[/latex] ft and [latex]W=12[/latex] ft.
Try It
Find the dimensions of a rectangle given that the perimeter is [latex]110[/latex] cm. and the length is 1 cm. more than twice the width.
Show Solution
L = 37 cm, W = 18 cm
Example: Solving an Area Problem
The perimeter of a tablet of graph paper is 48 in2. The length is [latex]6[/latex] in. more than the width. Find the area of the graph paper.
Show Solution
The standard formula for area is [latex]A=LW[/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.
We know that the length is 6 in. more than the width, so we can write length as [latex]L=W+6[/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.
A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft2 of new carpeting should be ordered?
Show Solution
250 ft2
Example: Solving a Volume Problem
Find the dimensions of a shipping box given that the length is twice the width, the height is [latex]8[/latex] inches, and the volume is 1,600 in.3.
Show Solution
The formula for the volume of a box is given as [latex]V=LWH[/latex], the product of length, width, and height. We are given that [latex]L=2W[/latex], and [latex]H=8[/latex]. The volume is [latex]1,600[/latex] cubic inches.
The dimensions are [latex]L=20[/latex] in., [latex]W=10[/latex] in., and [latex]H=8[/latex] in.
Analysis of the Solution
Note that the square root of [latex]{W}^{2}[/latex] would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.
Key Concepts
A linear equation can be used to solve for an unknown in a number problem.
Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities.
There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the [latex]d=rt[/latex] formula.
Many geometry problems are solved using the perimeter formula [latex]P=2L+2W[/latex], the area formula [latex]A=LW[/latex], or the volume formula [latex]V=LWH[/latex].
Glossary
area
in square units, the area formula used in this section is used to find the area of any two-dimensional rectangular region: [latex]A=LW[/latex]
perimeter
in linear units, the perimeter formula is used to find the linear measurement, or outside length and width, around a two-dimensional regular object; for a rectangle: [latex]P=2L+2W[/latex]
volume
in cubic units, the volume measurement includes length, width, and depth: [latex]V=LWH[/latex]
Licenses and Attributions
CC licensed content, Original
Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
Analysis of the Solution
One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.