3.6 Solve a Formula for a Specific Variable

Learning Objectives

By the end of this section, you will be able to:

  • Use the Distance, Rate, and Time formula
  • Solve a formula for a specific variable

Try It

Before you get started, take this readiness quiz:

1) Solve: [latex]15t=120[/latex]
2) Solve: [latex]6x+24=96[/latex]

Use the Distance, Rate, and Time Formula

One formula you will use often in algebra and in everyday life is the formula for distance travelled by an object moving at a constant rate. Rate is an equivalent word for “speed.” The basic idea of rate may already familiar to you. Do you know what distance you travel if you drive at a steady rate of [latex]60[/latex] miles per hour for [latex]2[/latex] hours? (This might happen if you use your car’s cruise control while driving on the highway.) If you said [latex]120[/latex] miles, you already know how to use this formula!

For an object moving at a uniform (constant) rate, the distance travelled, the elapsed time, and the rate are related by the formula:

[latex]\begin{align*} d&=rt,\;\;\;when\\ d&=distance\\ r&=rate\\ t&=time \end{align*}[/latex]

We will use the Strategy for Solving Applications that we used earlier in this chapter. When our problem requires a formula, we change Step 4: In place of writing a sentence, we write the appropriate formula. We write the revised steps here for reference.

HOW TO

Solve an application (with a formula).

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are looking for.
  3. Name what we are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation. Write the appropriate formula for the situation. Substitute in the given information.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

You may want to create a mini-chart to summarize the information in the problem. See the chart in this first example.

Example 3.6.1

Jamal rides his bike at a uniform rate of [latex]12[/latex] miles per hour for [latex]3\frac{1}{2}[/latex] hours. What distance has he travelled?

Solution

Step 1: Read the problem.

Step 2: Identify what you are looking for.

distance travelled

Step 3: Name. Choose a variable to represent it.

Let [latex]d[/latex] = distance.

Step 4: Translate: Write the appropriate formula.

[latex]\begin{align*} d&=rt\\ d&=?\\ r&=12\;mph\\ t&=3\frac12\;hours \end{align*}[/latex]

Step 5: Substitute in the given information.

[latex]d=12\cdot3\frac{1}{2}[/latex]

Step 6: Solve the equation.

[latex]d=42\;miles[/latex]

Step 7: Check
Does 42 miles make sense?

Jamal rides:

12 miles in 1 hour, 24 miles in 2 hours, 36 miles in 3 hours, 48 miles in 4 hours. 42 miles in 3 and a half hours is reasonable
Figure 3.6.1

Step 8: Answer the question with a complete sentence.

Jamal rode 42 miles.

Try It

3) Lindsay drove for [latex]5\frac{1}{2}[/latex] hours at [latex]60[/latex] miles per hour. How much distance did she travel?

Solution

[latex]330[/latex] miles

4) Trinh walked for [latex]2\frac{1}{3}[/latex] hours at [latex]3[/latex] miles per hour. How far did she walk?

Solution

[latex]7[/latex] miles

Example 3.6.2

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of [latex]520[/latex] miles. If he can drive at a steady rate of [latex]65[/latex] miles per hour, how many hours will the trip take?

Solution

Step 1: Read the problem.

Step 2: Identify what you are looking for.

How many hours (time)

Step 3: Name.
Choose a variable to represent it.

Let [latex]t[/latex] = time.

[latex]\begin{align*} d&=520\;miles\\ r&=65\;mph\\ t&=?\;hours \end{align*}[/latex]

Step 4: Translate.
Write the appropriate formula. Substitute in the given information.

[latex]\begin{align*} d&=rt\\ 520&=65t\\ t&=8 \end{align*}[/latex]

Step 5: Solve the equation.

[latex]t=8[/latex]

Step 6: Check.
Substitute the numbers into the formula and make sure the result is a true statement.

[latex]\begin{align*} d&=rt\\ 520&\overset?=65\times8\\ 520&=520\checkmark \end{align*}[/latex]

Step 7: Answer the question with a complete sentence.

Rey’s trip will take [latex]8[/latex] hours.

Try It

5) Lee wants to drive from Phoenix to his brother’s apartment in San Francisco, a distance of [latex]770[/latex] miles. If he drives at a steady rate of [latex]70[/latex] miles per hour, how many hours will the trip take?

Solution

[latex]11[/latex] hours

6) Yesenia is [latex]168[/latex] miles from Chicago. If she needs to be in Chicago in [latex]3[/latex] hours, at what rate does she need to drive?

Solution

[latex]56[/latex] mph

Solve a Formula for a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine, they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

In Example 3.6.1 and Example 3.6.2, we used the formula [latex]d = rt[/latex]. This formula gives the value of [latex]d[/latex], distance, when you substitute in the values of [latex]r[/latex] and [latex]t[/latex], the rate and time. But in Example 3.6.2 we had to find the value of [latex]t[/latex]. We substituted in values of [latex]d[/latex] and [latex]r[/latex] and then used algebra to solve for [latex]t[/latex]. If you had to do this often, you might wonder why there is not a formula that gives the value of [latex]t[/latex] when you substitute in the values of [latex]d[/latex] and [latex]r[/latex]. We can make a formula like this by solving the formula [latex]d=rt[/latex] for [latex]t[/latex].

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of [latex]1[/latex]. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

Example 3.6.3

Solve the formula [latex]d=rt[/latex] for [latex]t[/latex]:

a. when [latex]d=520[/latex] and [latex]r=65[/latex]
b. in general

Solution

We will write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.

a. when [latex]d=520[/latex] and [latex]r = 65[/latex] b. in general
Step 1: Write the formula. [latex]d = rt[/latex] Step 1: Write the formula. [latex]d=rt[/latex]
Step 2: Substitute. [latex]520 = 65t[/latex]
Step 3: Divide, to isolate [latex]t[/latex]. [latex]\frac{520}{65}=\frac{65t}{65}[/latex] Step 2: Divide, to isolate [latex]t[/latex]. [latex]\frac{d}{r}=\frac{rt}{r}[/latex]
Step 4: Simplify. [latex]8 = t[/latex] Step 3: Simplify. [latex]\frac{d}{r}=t[/latex]

We say the formula [latex]t=\frac{d}{r}[/latex] is solved for [latex]t[/latex].

Try It

7) Solve the formula [latex]d=rt[/latex] for [latex]r[/latex]

a. when [latex]d = 180[/latex] and [latex]t = 4[/latex]
b. in general

Solution

a.[latex]r=45[/latex]
b.[latex]r=\frac{d}{t}[/latex]

8) Solve the formula [latex]d=rt[/latex] for [latex]r[/latex]

a. when [latex]d = 780[/latex] and [latex]t = 12[/latex]
b. in general

Solution

a.[latex]r=65[/latex]
b. [latex]r=\frac{d}{t}[/latex]

 

Example 3.6.4

Solve the formula [latex]A=\frac{1}{2}bh[/latex] for [latex]h[/latex]:

a. when [latex]A=90[/latex] and [latex]b=15[/latex]
b. in general

Solution
a. when [latex]A=90[/latex]         and          [latex]b=15[/latex] b. in general
Step 1: Write the formula. [latex]A=\frac12bh[/latex] Step 1: Write the formula. [latex]A=\frac12bh[/latex]
Step 2: Substitute. [latex]{90=\frac12\cdot}{\color{red}{15}}{\cdot h}[/latex]
Step 3: Clear the fractions. [latex]{\color{red}{2}}{\cdot90=}{\color{red}{2}}{\cdot\frac1215h}[/latex] Step 2: Clear the fractions. [latex]{\color{red}{2}}{\cdot A=}{\color{red}{2}}{\cdot\frac12bh}[/latex]
Step 4: Simplify. [latex]180=15h[/latex] Step 3: Simplify. [latex]2A=bh[/latex]
Step 5: Solve for [latex]h[/latex]. [latex]12=h[/latex] Step 4: Solve for [latex]h[/latex]. [latex]\frac{2A}{b}=h[/latex]

We can now find the height of a triangle, if we know the area and the base, by using the formula [latex]h=\frac{2A}{b}[/latex]

Try It

9) Use the formula [latex]A=\frac{1}{2}bh[/latex] to solve for [latex]h[/latex]

a. when [latex]A=170[/latex] and [latex]b=17[/latex]
b. in general

Solution

a. [latex]h=20[/latex]
b.[latex]h=\frac{2A}{b}[/latex]

10) Use the formula [latex]A=\frac{1}{2}bh[/latex] to solve for [latex]b[/latex]

a. when [latex]A=62[/latex] and [latex]h=31[/latex]
b. in general

Solution

a.[latex]b=4[/latex]
b.[latex]b=\frac{2A}{h}[/latex]

The formula [latex]I=Prt[/latex] is used to calculate simple interest, [latex]I[/latex], for a principal, [latex]P[/latex], invested at rate, [latex]r[/latex], for [latex]t[/latex] years.

Example 3.6.5

Solve the formula [latex]I=Prt[/latex] to find the principal, [latex]P[/latex]:

a. when [latex]I=\$5,600[/latex], [latex]r=4\%[/latex], [latex]t=7\;years[/latex]
b. in general

Solution
a. [latex]\$5,600, r=4\%, t=7 years[/latex] b. in general
Step 1: Write the formula. [latex]I=Prt[/latex] Step 1: Write the formula. [latex]I=Prt[/latex]
Step 2: Substitute. [latex]5600=P(0.04)(7)[/latex]
Step 3: Simplify. [latex]5600=P(0.28)[/latex] Step 2: Simplify. [latex]I=P(rt)[/latex]
Step 4: Divide, to isolate [latex]P[/latex]. [latex]\frac{560}{{\color{red}{0}}{\color{red}{.}}{\color{red}{28}}}=\frac{P\left(0.28\right)}{{\color{red}{0}}{\color{red}{.}}{\color{red}{28}}}[/latex] Step 3: Divide, to isolate [latex]P[/latex]. [latex]\frac I{{\color{red}{r}}{\color{red}{t}}}=\frac{P\left(rt\right)}{{\color{red}{r}}{\color{red}{t}}}[/latex]
Step 5: Simplify. [latex]20,000=P[/latex] Step 4: Simplify. [latex]\frac{I}{rt}=P[/latex]
The principal is [latex]$20,000[/latex] [latex]P=\frac{I}{rt}[/latex]

Try It

11) Use the formula [latex]I=Prt[/latex] to find the principal, [latex]P[/latex]:

a. when [latex]I=\$2,160[/latex], [latex]r=6\%[/latex], [latex]t=3 years[/latex]
b. in general

Solution

a.[latex]\$12,000[/latex]
b. [latex]P=\frac{I}{rt}[/latex]

12) Use the formula [latex]I=Prt[/latex] to find the principal, [latex]P[/latex]:

a. when [latex]I=\$5,400[/latex], [latex]r=12\%[/latex], [latex]t=5\;years[/latex]
b. in general

Solution

a. [latex]\$9,000[/latex]
b. [latex]P=\frac{I}{rt}[/latex]

Later in this class, and in future algebra classes, you’ll encounter equations that relate two variables, usually [latex]x[/latex] and [latex]y[/latex]. You might be given an equation that is solved for [latex]y[/latex] and need to solve it for [latex]x[/latex], or vice versa. In the following example, we’re given an equation with both [latex]x[/latex] and [latex]y[/latex] on the same side and we’ll solve it for [latex]y[/latex].

Example 3.6.6

Solve the formula [latex]3x+2y=18[/latex] for [latex]y[/latex]:

a. when [latex]x=4[/latex]
b. in general

Solution
a. when [latex]x=4[/latex] b. in general
Step 1: Substitute. [latex]3(4)+2y=18[/latex]
Step 2: Subtract to isolate the
[latex]y[/latex]-term.		 [latex]{12}{\color{red}{-}}{{\color{red}{12}}+2y=18{\color{red}{-}}}{\color{red}{12}}{}[/latex] Step 1: Subtract to isolate the
[latex]y[/latex]-term.		 [latex]{3x}{\color{red}{-}}{{\color{red}{3}}{\color{red}{x}}+2y=18{\color{red}{-}}{\color{red}{3}}}{\color{red}{x}}{}[/latex]
Step 3: Divide. [latex]\frac{2y}{\color{red}{2}}=\frac6{\color{red}{2}}[/latex] Step 2: Divide. [latex]\frac{2y}{\color{red}{2}}=\frac{18}{\color{red}{2}}-\frac{3x}{\color{red}{2}}[/latex]
Step 4: Simplify. [latex]y=3[/latex] Step 3: Simplify. [latex]y=-\frac{3x}{2}+9[/latex]

Try It

13) Solve the formula [latex]3x+4y=10[/latex] for [latex]y[/latex]:

a. when [latex]x=\frac{14}{3}[/latex]
b. in general

Solution

a. [latex]y=1[/latex]
b.[latex]y=\frac{10-3x}{4}[/latex]

14) Solve the formula [latex]5x+2y=18[/latex] for [latex]y[/latex]:

a. when [latex]x=4[/latex]
b. in general

Solution

a.[latex]y=-1[/latex]
b.[latex]y=\frac{18-5x}{2}[/latex]

In Examples 3.6.3 through 3.6.6 we used the numbers in part a. as a guide to solving in general in part b. Now we will solve a formula in general without using numbers as a guide.

Example 3.6.7

Solve the formula [latex]P=a+b+c[/latex] for [latex]a[/latex]

Solution

Step 1: We will isolate [latex]a[/latex] on one side of the equation.

[latex]P=a+b+c[/latex]

Step 2: Both [latex]b[/latex] and [latex]c[/latex] are added to [latex]a[/latex], so we subtract them from both sides of the equation.

[latex]{P}{\color{red}{-}}{{\color{red}{b}}{\color{red}{-}}{\color{red}{c}}{\color{red}{}}=a+b+c{\color{red}{-}}{\color{red}{b}}{\color{red}{-}}}{\color{red}{c}}{}[/latex]

Step 3: Simplify.

[latex]\begin{align*} P-b-c&=a\\ a&=P-b-b \end{align*}[/latex]

Try It

15) Solve the formula [latex]P=a+b+c[/latex] for [latex]b[/latex]

Solution

[latex]b=P-a-c[/latex]

16) Solve the formula [latex]P=a+b+c[/latex] for [latex]c[/latex]

Solution

[latex]c=P-a-b[/latex]

 

Example 3.6.8

Solve the formula [latex]6x+5y=13[/latex] for [latex]y[/latex]

Solution

Step 1: Subtract [latex]6x[/latex] from both sides to isolate the term with [latex]y[/latex]

[latex]\begin{align*}&\;&6x{\color{red}{-}}{\color{red}{6}}{\color{red}{x}}+5y&=13{\color{red}{-}}{\color{red}{6}}{\color{red}{x}}\\ &\text{Simplify.}\;&5y&=13-6x \end{align*}[/latex]

Step 2: Divide by [latex]5[/latex] to make the coefficient [latex]1[/latex].

[latex]\begin{align*}&\;&\frac{5y}{\color{red}{5}}&=\frac{13-6x}{\color{red}{5}}\\ &\text{Simplify.}\;&y&=\frac{13-6x}{5} \end{align*}[/latex]

The fraction is simplified. We cannot divide [latex]13-6x[/latex] by [latex]5[/latex].

Try It

17) Solve the formula [latex]4x+7y=9[/latex] for [latex]y[/latex].

Solution

[latex]y=\frac{9-4x}{7}[/latex]

18) Solve the formula [latex]5x+8y=1[/latex] for [latex]y[/latex].

Solution

[latex]y=\frac{1-5x}{8}[/latex]

In the next example, we will solve this formula for the height.

Example 3.6.9

Solve the formula [latex]V=\frac{1}{3}\pi r^2h[/latex] for [latex]h[/latex].

Solution

Step 1: Write the formula.

[latex]V=\frac{1}{3}\pi r^2h[/latex]

Step 2: Remove the fraction on the right.

[latex]\begin{align*}&\;&{\color{red}{3}}{\color{red}{\cdot}}V&={\color{red}{3}}{\color{red}{\cdot}}\frac13\pi r^2h\\ &\text{Simplify.}\;&3V&=\pi r^2h \end{align*}[/latex]

Step 4: Divide both sides by [latex]\pi r^2[/latex]

[latex]\frac{3V}{\pi r^2}=h[/latex]

We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula [latex]h=\frac{3V}{\pi r^2}[/latex].

Try It

19) Use the formula [latex]A=\frac{1}{2}bh[/latex] to solve for [latex]b[/latex].

Solution

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

20) Use the formula [latex]A=\frac{1}{2}bh[/latex] to solve for [latex]h[/latex].

Solution

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

In the sciences, we often need to change temperature from Fahrenheit to Celsius or vice versa. If you travel in a foreign country, you may want to change the Celsius temperature to the more familiar Fahrenheit temperature.

Example 3.6.10

Solve the formula [latex]C=\frac{5}{9}(F-32)[/latex] for [latex]F[/latex].

Solution

Step 1: Write the formula.

[latex]C=\frac59(F-32)[/latex]

Step 2: Remove the fraction on the right.

[latex]\begin{align*}&\;&{\color{red}{\frac95}}C&={\color{red}{\frac95}}\cdot\frac59\left(F-32\right)\\ &\text{Simplify.}\;&\frac95C&=(F-32) \end{align*}[/latex]

Step 4: Add 32 to both sides.

[latex]\frac95C+32=F[/latex]

We can now use the formula [latex]F=\frac{9}{5}C+32[/latex] to find the Fahrenheit temperature when we know the Celsius temperature.

Try It

21) Solve the formula [latex]F=\frac{9}{5}C+32[/latex] for [latex]C[/latex].

Solution

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

22) Solve the formula [latex]A=\frac{1}{2}h(b+B)[/latex] for [latex]b[/latex].

Solution

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

The next example uses the formula for the surface area of a right cylinder.

Example 3.6.11

Solve the formula [latex]S=2\pi r^2+2\pi rh[/latex] for [latex]h[/latex].

Solution

Step 1: Write the formula.

[latex]S=2\pi r^2+2\pi r h[/latex]

Step 2: Isolate the [latex]h[/latex] term by subtracting [latex]2\pi r^2[/latex] from each side.

[latex]\begin{align*}&\;&S{\color{red}{-}}{\color{red}{2}}{\color{red}{π}}{\color{red}{r}}^{\color{red}{2}}&=2\pi r^2{\color{red}{-}}{\color{red}{2}}{\color{red}{π}}{\color{red}{r}}^{\color{red}{2}}+2\pi rh\\ &\text{Simplify.}\;&S-2\pi r^2&=2\pi r h \end{align*}[/latex]

Step 3: Solve for [latex]h[/latex] by dividing both sides by [latex]2\pi r[/latex]

[latex]\begin{align*}&\;&\frac{S-2\pi r^2}{2\pi r}&=\frac{2\pi r h}{2\pi r}\\ &\text{Simplify.}\;&\frac{S-2\pi r^2}{2\pi r}&=h \end{align*}[/latex]

Try It

23) Solve the formula [latex]A=P + Prt[/latex] for [latex]t[/latex].

Solution

[latex]t=\frac{A-P}{Pr}[/latex]

24) Solve the formula [latex]A=P + Prt[/latex] for [latex]r[/latex].

Solution

[latex]r=\frac{A-P}{Pt}[/latex]

Key Concepts

  • To Solve an Application (with a formula)
    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what we are looking for.
    3. Name what we are looking for. Choose a variable to represent that quantity.
    4. Translate into an equation. Write the appropriate formula for the situation. Substitute in the given information.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.
  • Distance, Rate and Time
     For an object moving at a uniform (constant) rate, the distance travelled, the elapsed time, and the rate are related by the formula: [latex]d=rt[/latex] where [latex]d = distance[/latex], [latex]r = rate[/latex], [latex]t = time[/latex].
  • To solve a formula for a specific variable means to get that variable by itself with a coefficient of [latex]1[/latex] on one side of the equation and all other variables and constants on the other side.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

What does this checklist tell you about your mastery of this section? What steps will you take to improve?

 

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Fanshawe Pre-Health Sciences Mathematics 1 Copyright © 2022 by Domenic Spilotro, MSc is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted.

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