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

Use the Distance, Rate, and Time Formula

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

Distance, Rate, and Time

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

\begin{array}{ccccccccc}d=rt\hfill & & & \hfill \text{where}\hfill & & & \hfill d& =\hfill & \text{distance}\hfill \\ & & & & & & \hfill r& =\hfill & \text{rate}\hfill \\ & & & & & & \hfill t& =\hfill & \text{time}\hfill \end{array}

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 1

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

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. distance traveled
Step 3. Name. Choose a variable to represent it. Let d = distance.
Step 4. Translate: Write the appropriate formula. d=rt

d=?

r=12 mph

t=3\frac{1}{2} hours
Substitute in the given information. d=12\cdot 3\frac{1}{2}
Step 5. Solve the equation. d=42 miles
Step 6. Check

Does 42 miles make sense?

Jamal rides:

 \\12 \text{ miles in } 1 \text{ hour}\\ 24 \text{miles in } 2 \text{ hours}
\begin{array} \quad&\left\ \begin{tabular}{@{}l@{}} 36 miles in 3 hours \\ 48 miles in 4 hours \end{tabular} \right\} \end{array} \text{42 miles in } 3\frac{1}{2} \text{hours seems reasonable}}
Step 7. Answer the question with a complete sentence. Jamal rode 42 miles.

TRY IT 1.1

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

Show answer

330 miles

TRY IT 1.2

Trinh walked for 2\frac{1}{3} hours at 3 miles per hour. How far did she walk?

Show answer

7 miles

EXAMPLE 2

Rey is planning to drive from his house in Saskatoon to visit his grandmother in Winnipeg, a distance of 798 km. If he can drive at a steady rate of 76 km 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 t = time.
d = 798 km

r = 76 km/h

t = ? hours
Step 4. Translate.
Write the appropriate formula.		 \phantom{\rule{1em}{0ex}}d=rt
Substitute in the given information. 798=76t
Step 5. Solve the equation. \phantom{\rule{1.2em}{0ex}}t=10.5 hours
Step 6. Check. Substitute the numbers into
the formula and make sure the result is a
true statement.		 \begin{array}{ccc}\hfill d& =\hfill & rt\hfill \\ \hfill 798& \stackrel{?}{=}\hfill & 76\cdot 10.5\hfill \\ \hfill 798& =\hfill & 798\hfill \end{array}
Step 7. Answer the question with a complete sentence. Rey’s trip will take 10.5 hours.

TRY IT 2.1

Lee wants to drive from Kamloops to his brother’s apartment in Banff, a distance of 495 km. If he drives at a steady rate of 90 km/h, how many hours will the trip take?

Show answer

5 1/2 hours

TRY IT 2.2

Yesenia is 168 km from Toronto. If she needs to be in Toronto in 2 hours, at what rate does she need to drive?

Show answer

84 km/h

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 1) and (Example 2), we used the formula d=rt. This formula gives the value of d, distance, when you substitute in the values of r\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t, the rate and time. But in (Example 2), we had to find the value of t. We substituted in values of d\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r and then used algebra to solve for t. If you had to do this often, you might wonder why there is not a formula that gives the value of t when you substitute in the values of d\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r. We can make a formula like this by solving the formula d=rt for t.

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. 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

Solve the formula d=rt for t:

  1. when d=520 km and r=65 km/hr
  2. 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 d=520 km and r=65 km/hr b) in general
Write the formula. \phantom{\rule{1em}{0ex}}d=rt Write the formula. d=rt
Substitute. 520=65t
Divide, to isolate t. \frac{520}{65}=\frac{65t}{65} Divide, to isolate t. \frac{d}{r}=\frac{rt}{r}
Simplify. \phantom{\rule{1.2em}{0ex}}8=t

t=8 hours

 Simplify. \frac{d}{r}=t

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

TRY IT 3.1

Solve the formula d=rt for r:

a) when d=180\phantom{\rule{0.2em}{0ex}}\text{km and}\phantom{\rule{0.2em}{0ex}}t=4 hr  b) in general

Show answer

a) r=45 km/hr b) r=\frac{d}{t}

TRY IT 3.2

Solve the formula d=rt for r:

a) when d=780\phantom{\rule{0.2em}{0ex}}\text{km and}\phantom{\rule{0.2em}{0ex}}t=12 hr  b) in general

Show answer

a) r=65 km/hr  b) r=\frac{d}{t}

EXAMPLE 4

Solve the formula A=\frac{1}{2}bh for h:

a) when A=90 and b=15 b) in general

Solution
a) when A=90 and b=15 b) in general
Write the formula. A=\frac{1}{2}bh Write the formula. A=\frac{1}{2} bh
Substitute. 90=\frac{1}{2} \cdot \textcolor{red}{15} \cdot h
Clear the fractions.
\textcolor{red}{2} \cdot 90 = \textcolor{red}{2} \cdot \frac{1}{2} 15h		 Clear the fractions. \textcolor{red}{2} \cdot A = \textcolor{red}{2} \cdot \frac{1}{2} bh
Simplify. 180=15h Simplify. 2A=bh
Solve for h. 12=h Solve for h. \frac{2A}{b} = h

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

TRY IT 4.1

Use the formula A=\frac{1}{2}bh to solve for h:

a) when A=170 and b=17 b) in general

Show answer

a) h=20 b) h=\frac{2A}{b}

TRY IT 4.2

Use the formula A=\frac{1}{2}bh to solve for b:

a) when A=62 and h=31 b) in general

Show answer

a) b=4 b) b=\frac{2A}{h}

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

EXAMPLE 5

Solve the formula I=Prt to find the principal, P:

a) when I=\$5,600, r=4\%, t=7\phantom{\rule{0.2em}{0ex}}years\phantom{\rule{0.2em}{0ex}} b) in general

Solution
a)I=\$5,600, r=4\%, t=7 years b) in general
Write the formula. I = Prt Write the formula. I=Prt
Substitute. 5600 = P(0.04)(7)
Simplify. 5600=P(0.28) Simplify. I = P (rt)
Divide, to isolate P. \frac{5600}{\textcolor{red}{0.28}}=\frac{P(0.28)}{\textcolor{red}{0.28}} Divide, to isolate P. \frac{I}{\textcolor{red}{rt}}=\frac{P(rt)}{\textcolor{red}{rt}}
Simplify. 20,000 = P Simplify. \frac{I}{rt}=P
The principal is $20,000 P=\frac{I}{rt}

TRY IT 5.1

Use the formula I=Prt to find the principal, P:

a) when I=\$2,160, r=6\%, t=3\phantom{\rule{0.2em}{0ex}}years\phantom{\rule{0.2em}{0ex}} b) in general

Show answer

a) $12,000 b) P=\frac{I}{rt}

TRY IT 5.2

Use the formula I=Prt to find the principal,P:

a) when I=\$5,400, r=12\%, t=5\phantom{\rule{0.2em}{0ex}}years\phantom{\rule{0.2em}{0ex}} b) in general

Show answer

a) $9,000 b) P=\frac{I}{rt}

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

EXAMPLE 6

Solve the formula 3x+2y=18 for y:

a) when x=4 b) in general

Solution
a) when x=4 b) in general
3x+2y=18 3x+2y=18
Substitute. 3(\textcolor{red}{4})+2y=18
Subtract to isolate the
y-term.		 \begin{gathered} 12\textcolor{red}{-12}+2y=18\textcolor{red}{-12}\\ 2y=6\\ \end{gathered} Subtract to isolate the
y-term.		 \begin{gathered} 3x\textcolor{red}{-3x}+2y=18\textcolor{red}{-3x}\\ 2y=18-3x\\ \end{gathered}
Divide. \frac{2y}{\textcolor{red}{2}}=\frac{6}{\textcolor{red}{2}} Divide. \frac{2y}{\textcolor{red}{2}}=\frac{18}{\textcolor{red}{2}}- \frac{3x}{\textcolor{red}{2}}
Simplify. y=3 Simplify. y=9-\frac{3}{2} x
or
y=-\frac{3}{2} x+9

TRY IT 6.1

Solve the formula 3x+4y=10 for y:

a) when x=\frac{14}{3}     b) in general

Show answer

a)y=-1     b)y=\frac{10-3x}{4}

TY IT 6.2

Solve the formula 5x+2y=18 for y:

a) when x=4     b) in general

Show answer

a)y=-1       b)y=\frac{18-5x}{2}

Now we will solve a formula in general without using numbers as a guide.

EXAMPLE 7

Solve the formula P=a+b+c for a.

Solution
We will isolate a on one side of the equation. P=a+b+c
Both b and c are added to a, so we subtract them from both sides of the equation. P \textcolor{red}{-b-c} = a+b+c \textcolor{red}{-b-c}
Simplify. \begin{gathered} P-b-c=a \\ \text{or} \\ a=P-b-c \end{gathered}

TRY IT 7.1

Solve the formula P=a+b+c for b.

Show answer

b=P-a-c

TRY IT 7.2

Solve the formula P=a+b+c for c.

Show answer

c=P-a-b

EXAMPLE 8

Solve the formula 6x+5y=13 for y.

Solution
6x+5y=13
Subtract 6x from both sides to isolate the term with y. 6x \textcolor{red}{-6x} +5y = 13 \textcolor{red}{-6x}
Simplify. 5y=13-6x
Divide by 5 to make the coefficient 1. \frac{5y}{\textcolor{red}{5}}=\frac{13-6x}{\textcolor{red}{5}}
Simplify. y=\frac{13-6x}{5}

The fraction is simplified. We cannot divide 13-6x by 5

TRY IT 8.1

Solve the formula 4x+7y=9 for y.

Show answer

y=\frac{9-4x}{7}

TRY IT 8.2

Solve the formula 5x+8y=1 for y.

Show answer

y=\frac{1-5x}{8}

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 traveled, the elapsed time, and the rate are related by the formula: d=rt where d = distance, r = rate, t = time.
  • To solve a formula for a specific variable means to get that variable by itself with a coefficient of 1 on one side of the equation and all other variables and constants on the other side.

Practice Makes Perfect

Use the Distance, Rate, and Time Formula

In the following exercises, solve.

1. Socorro drove for 4\frac{5}{6} hours at 60 miles per hour. How much distance did she travel? 2. Steve drove for 8\frac{1}{2} hours at 72 miles per hour. How much distance did he travel?
3. Francie rode her bike for 2\frac{1}{2} hours at 12 miles per hour. How far did she ride? 4. Yuki walked for 1\frac{3}{4} hours at 4 miles per hour. How far did she walk?
5. Marta is taking the bus from Abbotsford to Cranbrook. The distance is 774 km and the bus travels at a steady rate of 86 miles per hour. How long will the bus ride be? 6. Connor wants to drive from Vancouver to the Nakusp, a distance of 630 km. If he drives at a steady rate of 90 km/h, how many hours will the trip take?
7. Kareem wants to ride his bike from Golden, BC to Banff, AB. The distance is 140 km. If he rides at a steady rate of 20 km/h, how many hours will the trip take? 8. Aurelia is driving from Calgary to Edmonton at a rate of 85 km/h. The distance is 300 km. To the nearest tenth of an hour, how long will the trip take?
9. Alejandra is driving to Prince George, 450 km away. If she wants to be there in 6 hours, at what rate does she need to drive? 10. Javier is driving to Vernon, 240 km away. If he needs to be in Vernon in 3 hours, at what rate does he need to drive?
11. Philip got a ride with a friend from Calgary to Kelowna, a distance of 890 km. If the trip took 10 hours, how fast was the friend driving? 12. Aisha took the train from Spokane to Seattle. The distance is 280 miles and the trip took 3.5 hours. What was the speed of the train?

Solve a Formula for a Specific Variable

13. Solve for t using d=rt

  1. when d=240\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=60
  2. in general
 14. Solve for t using d=rt

  1. when d=350 and r=70
  2. in general
15. Solve for t using d=rt

  1. when d=175\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=50
  2. in general
 16. Solve for t using d=rt

  1. when d=510\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=60
  2. in general
17. Solve for r using d=rt

  1. when d=420\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t=6
  2. in general
 18. Solve for r using d=rt

  1. when d=204\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t=3
  2. in general

19. Solve for r using d=rt

  1. when d=180\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t=4.5
  2. in general
 20. Solve for r using d=rt

  1. when d=160\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t=2.5
  2. in general
21. Solve for h using A=\frac{1}{2}bh

  1. when A=176\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b=22
  2. in general
 22. Solve for b using A=\frac{1}{2}bh

  1. when A=126\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}h=18
  2. in general
23. Solve for the principal, P using I = Prt  for

  1. I=\$5,480,r=4\%,t=7\phantom{\rule{0.2em}{0ex}}\text{years}\phantom{\rule{0.2em}{0ex}}
  2. in general
24. Solve for b using A=\frac{1}{2}bh
  1. when A=65\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}h=13
  2. in general.
25. Solve for the time, t  using I = Prt for

  1. I=\$2,376,P=\$9,000 ,r=4.4\%
  2. in general
 26. Solve for the principal, P using I = Prt for

  1. I=\$3,950,r=6\%,t=5\phantom{\rule{0.2em}{0ex}}\text{years}\phantom{\rule{0.2em}{0ex}}
  2. in general
27. Solve the formula 2x+3y=12 for y when

  1. x=3
  2. in general
28. Solve for the time, t for
  1. I=\$624,P=\$6,000,r=5.2\%
  2. in general
29. Solve the formula 3x-y=7 for y when

  1. x=-2
  2. in general
 30. Solve the formula 5x+2y=10 for y when

  1. x=4
  2. in general
31. Solve a+b=90 for

  1. a when b = 50
  2. a in general
 32. Solve the formula 4x+y=5 for y when

  1. x=-3
  2. in general
33. Solve 180=a+b+c for a. 34. Solve a+b=90 for a.
35. Solve the formula 8x+y=15 for y. 36. Solve 180=a+b+c for c.
37. Solve the formula -4x+y=-6 for y. 38. Solve the formula 9x+y=13 for y.
39. Solve the formula 4x+3y=7 for y. 40. Solve the formula -5x+y=-1 for y.
41. Solve the formula x-y=-4 for y. 42. Solve the formula 3x+2y=11 for y.
43. Solve the formula P=2L+2W for L. 44. Solve the formula x-y=-3 for y.
45. Solve the formula C=\pi d for d. 46. Solve the formula P=2L+2W for W.
47. Solve the formula V=LWH for L. 48. Solve the formula C=\pi d for \pi.
49. Solve the formula V=LWH for H.

Everyday Math

50. Converting temperature. Yon was visiting the United States and he saw that the temperature in Seattle one day was 50o Fahrenheit. Solve for C in the formula F=\frac{9}{5}C+32 to find the Celsius temperature. 51. Converting temperature. While on a tour in Greece, Tatyana saw that the temperature was 40o Celsius. Solve for F in the formula C=\frac{5}{9}\left(F-32\right) to find the Fahrenheit temperature.

Writing Exercises

52. Solve the equation 5x-2y=10 for x
a) when y=10
b) in general
c) Which solution is easier for you, a) or b)? Why?		 53. Solve the equation 2x+3y=6 for y
a) when x=-3
b) in general
c) Which solution is easier for you, a) or b)? Why?

Answers

1. 290 miles 3. 30 miles 5. 9 hr
7. 7 hr 9. 75 km/hr 11. 89 km/hr
13.

  1. t=4
  2. t=\frac{d}{r}
 15.

  1. t= 3.5
  2. t=\frac{d}{r}
 17.

  1. r= 70
  2. r=\frac{d}{t}
19.

  1. r= 40
  2. r=\frac{d}{t}
 21.

  1. h= 16
  2. h=\frac{2A}{b}
 23.

  1. P= \$19,571.43
  2. P=\frac{I}{rt}
25.

  1. t= 6
  2. t=\frac{I}{Pr}
 27.

  1. y=2
  2. y=\frac{12-2x}{3}
 29.

  1. y= -13
  2. y=-7+3x or y=3x-7
31. 40; a= 90-b 33. a= 180-b-c 35. y= 15-8x
37. y= 4x-6 39. y= \frac{7-4x}{3} 41. y= 4+x
43. L= \frac{P-2W}{2} 45. d= \frac{C}{\pi} 47. L= \frac{V}{WH}
49. H= \frac{V}{LW} 51. F= 104o 53. Answers will vary

Attributions

This chapter has been adapted from “Solve a Formula for a Specific Variable” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

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Intermediate Algebra II Copyright © 2021 by Pooja Gupta is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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