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Study Guides > MATH 1314: College Algebra

Using a Formula to Solve a Real-World Application

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 3: 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?


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? Solution

Example 4: 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?


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.
A rectangle with the length labeled as: L = W + 3 and the width labeled as: W. Figure 3
Now we can solve for the width and then calculate the length.
[latex]\begin{array}{l}P=2L+2W\hfill \\ 54=2\left(W+3\right)+2W\hfill \\ 54=2W+6+2W\hfill \\ 54=4W+6\hfill \\ 48=4W\hfill \\ 12=W\hfill \\ \left(12+3\right)=L\hfill \\ 15=L\hfill \end{array}[/latex]
The dimensions are [latex]L=15[/latex] ft and [latex]W=12[/latex] ft.

Try It 4

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. Solution

Example 5: 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.


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? Solution

Example 6: 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.


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.

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