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Study Guides > College Algebra

Models and Applications

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 students studying in a large lecture hall Figure 1. 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.

Write 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.

  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: 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.

Answer: 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.

[latex]\begin{array}{l}x+\left(x+17\right)\hfill&=31\hfill \\ 2x+17\hfill&=31\hfill&\text{Simplify and solve}.\hfill \\ 2x\hfill&=14\hfill \\ x\hfill&=7\hfill \\ \hfill \\ x+17\hfill&=7+17\hfill \\ \hfill&=24\hfill \end{array}[/latex]
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.

Answer: 11 and 25

Use 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 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?

Answer: 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.

[latex]d[/latex] [latex]r[/latex] [latex]t[/latex]
To Work [latex]d[/latex] [latex]r[/latex] [latex]\frac{1}{2}[/latex]
To Home [latex]d[/latex] [latex]r - 10[/latex] [latex]\frac{2}{3}[/latex]
Write two equations, one for each trip.
[latex]\begin{array}{ll}d=r\left(\frac{1}{2}\right)\hfill & \text{To work}\hfill \\ d=\left(r - 10\right)\left(\frac{2}{3}\right)\hfill & \text{To home}\hfill \end{array}[/latex]
As both equations equal the same distance, we set them equal to each other and solve for r.
[latex]\begin{array}{l}r\left(\frac{1}{2}\right)\hfill&=\left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ \frac{1}{2}r\hfill&=\frac{2}{3}r-\frac{20}{3}\hfill \\ \frac{1}{2}r-\frac{2}{3}r\hfill&=-\frac{20}{3}\hfill \\ -\frac{1}{6}r\hfill&=-\frac{20}{3}\hfill \\ r\hfill&=-\frac{20}{3}\left(-6\right)\hfill \\ r\hfill&=40\hfill \end{array}[/latex]
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.
[latex]\begin{array}{l}d\hfill&=40\left(\frac{1}{2}\right)\hfill \\ \hfill&=20\hfill \end{array}[/latex]
The distance between home and work is 20 mi.

Analysis of the Solution

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for [latex]r[/latex].
[latex]\begin{array}{l}r\left(\frac{1}{2}\right)\hfill&=\left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ 6\times r\left(\frac{1}{2}\right)\hfill& =6\times \left(r - 10\right)\left(\frac{2}{3}\right)\hfill \\ 3r\hfill& =4\left(r - 10\right)\hfill \\ 3r\hfill& =4r - 40\hfill \\ -r\hfill& =-40\hfill \\ r\hfill& =40\hfill \end{array}[/latex]

Try It 3

On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

Answer: 45 [latex]\frac{text{mi}}{\text{h}}[/latex]

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.

Answer: 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.

[latex]\begin{array}{l}P\hfill&=2L+2W\hfill \\ 48\hfill&=2\left(W+6\right)+2W\hfill \\ 48\hfill&=2W+12+2W\hfill \\ 48\hfill&=4W+12\hfill \\ 36\hfill&=4W\hfill \\ 9\hfill&=W\hfill \\ \left(9+6\right)\hfill&=L\hfill \\ 15\hfill&=L\hfill \end{array}[/latex]
Now, we find the area given the dimensions of [latex]L=15[/latex] in. and [latex]W=9[/latex] in.
[latex]\begin{array}{l}A\hfill&=LW\hfill \\ A\hfill&=15\left(9\right)\hfill \\ \hfill&=135\text{ in}^{2}\hfill \end{array}[/latex]
The area is [latex]135[/latex] in2.

Try It 5

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?

Answer: 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.

Answer: 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.

[latex]\begin{array}{l}V=LWH\hfill \\ 1,600=\left(2W\right)W\left(8\right)\hfill \\ 1,600=16{W}^{2}\hfill \\ 100={W}^{2}\hfill \\ 10=W\hfill \end{array}[/latex]
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]

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