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Study Guides > Intermediate Algebra

Read: Solve an Application Using a Formula

Learning Objectives

  • Solve distance, rate, and time problems
  • Solve area, volume, and perimeter problems
  • Solve temperature conversion problems
  • Rearrange formulas to isolate specific variables
  • Identify an unknown given a formula
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.

Distance, Rate, and Time

Example

It takes Andrew [latex]30[/latex] min to drive to work in the morning. He drives home using the same route, but it takes [latex]10[/latex] min longer, and he averages [latex]10 mi/h[/latex] 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 [latex]30[/latex] min, or [latex]\Large\frac{1}{2}[/latex] h at rate [latex]r[/latex]. His drive home takes [latex]40[/latex] min, or [latex]\Large\frac{2}{3}[/latex] h, and his speed averages [latex]10 mi/h[/latex] 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]\Large\frac{1}{2}[/latex]
To Home [latex]d[/latex] [latex]r - 10[/latex] [latex]\Large\frac{2}{3}[/latex]
Write two equations, one for each trip.
[latex]\begin{array}{ll}d=r\left(\Large\frac{1}{2}\normalsize\right)\hfill & \text{To work}\hfill \\ d=\left(r - 10\right)\left(\Large\frac{2}{3}\normalsize\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(\Large\frac{1}{2}\right)\hfill&=\left(r - 10\right)\left(\Large\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, [latex]40 mph[/latex]. Substituting [latex]40[/latex] into the rate on the return trip yields [latex]30 mi/h[/latex]. 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(\Large\frac{1}{2}\normalsize\right)\hfill \\ \hfill&=20\hfill \end{array}[/latex]
The distance between home and work is [latex]20 mi[/latex].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-display]\begin{array}{l}r\left(\Large\frac{1}{2}\normalsize\right)\hfill&=\left(r - 10\right)\left(\Large\frac{2}{3}\normalsize\right)\hfill \\ 6\times r\left(\Large\frac{1}{2}\normalsize\right)\hfill& =6\times \left(r - 10\right)\left(\Large\frac{2}{3}\normalsize\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-display]

In the next example, we will find the lengths of a rectangular field given it's perimeter and a relationship between it's side lengths.

Example

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?

Answer: 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.

The following video shows an example of finding the dimensions of a rectangular field given it's perimeter. https://youtu.be/VyK-HQr02iQ

Example

The perimeter of a tablet of graph paper is [latex]48[/latex] in. 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 [latex]6[/latex] 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.

The following video shows another example of finding the area of a rectangle given it's perimeter and the relationship between it's side lengths. https://youtu.be/zUlU64Umnq4

Volume

Example

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 [latex]1,600[/latex] 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.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.

Think About It

Express the formula for the surface area of a cylinder, [latex]s=2\pi rh+2\pi r^{2}[/latex], in terms of the height, h. In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h. [practice-area rows="1"][/practice-area]

Answer: Isolate the term containing the variable, h, by subtracting [latex]2\pi r^{2}[/latex]from both sides.

[latex]\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Next, isolate the variable h by dividing both sides of the equation by [latex]2\pi r[/latex].

[latex]\begin{array}{r}\Large\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\\Large\frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]h=\Large\frac{S-2\pi r^{2}}{2\pi r}[/latex]

In the following video we show how to find the volume of a right circular cylinder formed from a given rectangle. https://youtu.be/KS1pGO_g3vM

Isolate Variables in Formulas

Sometimes, it is easier to isolate the variable you you are solving for when you are using a formula. This is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly. In the next examples, we will use algebraic properties to isolate a variable in a formula.

Example

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle:  

[latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex].

Answer: First, isolate the term with w by subtracting 2l from both sides of the equation.

[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by [latex]2[/latex].

[latex]\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\Large\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\Large\frac{p-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]w=\Large\frac{p-2l}{2}[/latex]

Example

Use the multiplication and division properties of equality to isolate the variable b given [latex]A=\Large\frac{1}{2}\normalsize bh[/latex]

Answer:

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\Large\frac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\Large\frac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\Large\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\Large\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

[latex]b=\Large\frac{2A}{h}[/latex]

Use the multiplication and division properties of equality to isolate the variable h given [latex]A=\Large\frac{1}{2}\normalsize bh[/latex]

Answer:

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\Large\frac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\Large\frac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\Large\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\Large\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

[latex]h=\Large\frac{2A}{b}[/latex]

 

Temperature

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

[latex]C=\left(F--32\right)\cdot\Large\frac{5}{9}[/latex]

Example

Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].

Answer: Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot\Large\frac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot\Large\frac{5}{9}\\\\\left(\Large\frac{9}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\Large\frac{108}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Example

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F. [latex-display]C=\left(F--32\right)\cdot\Large\frac{5}{9}[/latex-display]

Answer: To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].

[latex]\begin{array}{l}\\\,\,\,\,\left(\Large\frac{9}{5}\normalsize\right)C=\left(F-32\right)\left(\Large\frac{5}{9}\normalsize\right)\left(\Large\frac{9}{5}\normalsize\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\Large\frac{9}{5}\normalsize C=F-32\end{array}[/latex]

Add 32 to both sides.

[latex]\begin{array}{l}\Large\frac{9}{5}\normalsize\,C+32=F-32+32\\\\\Large\frac{9}{5}\normalsize\,C+32=F\end{array}[/latex]

Answer

[latex]F=\Large\frac{9}{5}\normalsize C+32[/latex]

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  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay, et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at : http://cnx.org/contents/[email protected]:1/Preface.
  • Ex: Find the Dimensions and Area of a Field Given the Perimeter Mathispower4u . Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex: Find the Area of a Rectangle Given the Perimeter. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Find the Volume of a Right Circular Cylinder Formed from a Given Rectangle. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.