2.4 Solve a Formula for a Specific Variable

Izabela Mazur; Kim Moshenko

2. Solving Linear Equations and Inequalities

2.4 Solve a Formula for a Specific Variable

Learning Objectives

By the end of this section it is expected that you will be able to:

Solve a formula for a specific variable
Use formulas to solve applications

Solve a Formula for Specific Variable

In the last section we have worked with some geometry formulas when solving problems. 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.

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 1

Solve the formula $d=rt$ for $t$ :

when $d=520$ and $r=65$
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 $d=520$ and $r=65$		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.	$\phantom{\rule{1.2em}{0ex}}8=t$	Simplify.	$\frac{d}{r}=t$

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

TRY IT 1

Solve the formula $d=rt$ for $r$ :

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

Show answer

a) $r=45$ b) $r=\frac{d}{t}$

EXAMPLE 2

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.		Write the formula.
Substitute.
Clear the fractions.		Clear the fractions.
Simplify.		Simplify.
Solve for $h$ .		Solve for $h$ .

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 2

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) $h=\frac{2A}{b}$

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

EXAMPLE 3

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.		Write the formula.
Substitute.
Simplify.		Simplify.
Divide, to isolate P.		Divide, to isolate P.
Simplify.		Simplify.
The principal is

TRY IT 3

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) $P=\frac{I}{rt}$

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 4

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

a) when $x=4$ b) in general

Solution

a) when $x=4$		b) in general

Substitute.
Subtract to isolate the $y$ -term.		Subtract to isolate the $y$ -term.
Divide.		Divide.
Simplify.		Simplify.

TRY IT 4

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

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

Show answer

a) $y=1$ b) $y=\frac{10-3x}{4}$

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

EXAMPLE 5

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

Solution

We will isolate $a$ on one side of the equation.
Both $b$ and $c$ are added to $a$ , so we subtract them from both sides of the equation.
Simplify.

TRY IT 5

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

Show answer

$b=P-a-c$

EXAMPLE 6

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

Solution


Subtract $6x$ from both sides to isolate the term with $y$ .
Simplify.
Divide by 5 to make the coefficient 1.
Simplify.

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

TRY IT 6

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

Show answer

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

Geometric formulas often need to be solved for another variable, too. The formula $V=\frac{1}{3}\pi {r}^{2}h$ is used to find the volume of a right circular cone when given the radius of the base and height. In the next example, we will solve this formula for the height.

EXAMPLE 7

Solve the formula $V=\frac{1}{3}\pi {r}^{2}h$ for h.

Write the formula.
Remove the fraction on the right.
Simplify.
Divide both sides by $\pi {r}^{2}.$

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 $h=\frac{3V}{\pi {r}^{2}}.$

TRY IT 7

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

Show answer

$b=\frac{2A}{h}$

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 8

Solve the formula $C=\frac{5}{9}\left(F-32\right)$ for F.

Write the formula.
Remove the fraction on the right.
Simplify.
Add 32 to both sides.

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

TRY IT 8

Solve the formula $F=\frac{9}{5}C+32$ for C.

Show answer

$C=\frac{5}{9}\left(F-32\right)$

Use Formulas to Solve Applications

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 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).

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
Translate into an equation. Write the appropriate formula for the situation. Substitute in the given information.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
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 9

Adam 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.		$d=rt$

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:

Step 7. Answer the question with a complete sentence.		Jamal rode 42 miles.

TRY IT 9

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

Show answer

330 miles

EXAMPLE 10

Rey is planning to drive from his house in Saskatoon to visit his grandmother in Winnipeg, a distance of 520 miles. If he can drive at a steady rate of 65 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 t = time.
	d = 600 km r = 75 km/h t = ? hours
Step 4. Translate. Write the appropriate formula.	$\phantom{\rule{1em}{0ex}}d=rt$
Substitute in the given information.	$520=65t$
Step 5. Solve the equation.	$\phantom{\rule{1.2em}{0ex}}t=8$
Step 6. Check. Substitute the numbers into the formula and make sure the result is a true statement.	$\begin{array}{ccc}\hfill d& =\hfill & rt\hfill \\ \hfill 520& \stackrel{?}{=}\hfill & 65\cdot 8\hfill \\ \hfill 520& =\hfill & 520\hfill \end{array}$
Step 7. Answer the question with a complete sentence. Rey’s trip will take 8 hours.

TRY IT 10

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

When we solve geometry applications, we adapt our problem solving strategy, use some common geometry formulas, and draw a figure and label it with given information.

The next example involves the area of a triangle. The area of a triangle is one-half the base times the height. We can write this as $A=\frac{1}{2}bh,$ where b = length of the base and h = height.

EXAMPLE 11

The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	height of a triangle
Step 3. Name.
Choose a variable to represent it.	Let $h=$ the height.
Draw the figure and label it with the given information.	Area = 126 sq. in.

Step 4. Translate.
Write the appropriate formula.	$\phantom{\rule{1.8em}{0ex}}A=\frac{1}{2}bh$
Substitute in the given information.	$\phantom{\rule{1em}{0ex}}126=\frac{1}{2}\cdot 18\cdot h$
Step 5. Solve the equation.	$\phantom{\rule{1em}{0ex}}126=9h$
Divide both sides by 9.	$\phantom{\rule{1.5em}{0ex}}14=h$
Step 6. Check. $\begin{array}{ccc}\hfill A& =\hfill & \frac{1}{2}bh\hfill \\ \hfill 126& \stackrel{?}{=}\hfill & \frac{1}{2}\cdot 18 \cdot 14\hfill \\ \hfill 126& =\hfill & 126✓\hfill \end{array}$
Step 7. Answer the question.	The height of the triangle is 14 inches.

TRY IT 11

The area of a triangular church window is 90 square metres. The base of the window is 15 metres. What is the window’s height?

Show answer

The window’s height is 12 metres.

The next example is about the perimeter of a triangle. Since the perimeter is just the distance around the triangle, we find the sum of the lengths of its three sides. We can write this as $P=a+b+c,$ where a, b, and c are the lengths of the sides.

EXAMPLE 12

One side of a triangle is three inches more than the first side. The third side is two inches more than twice the first. The perimeter is 29 inches. Find the length of the three sides of the triangle.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the lengths of the three sides of a triangle
Step 3. Name. Choose a variable to represent the length of the first side.	$\begin{array}{ccc}\hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x& =\hfill & \text{length of}\phantom{\rule{0.2em}{0ex}}{1}^{\text{st}}\phantom{\rule{0.2em}{0ex}}\text{side.}\hfill \\ \hfill x+3& =\hfill & \text{length of}\phantom{\rule{0.2em}{0ex}}{2}^{\text{nd}}\phantom{\rule{0.2em}{0ex}}\text{side}\hfill \\ \hfill 2x+2& =\hfill & \text{length of}\phantom{\rule{0.2em}{0ex}}{3}^{\text{rd}}\phantom{\rule{0.2em}{0ex}}\text{side}\hfill \end{array}$
Step 4. Translate. Write the appropriate formula. Substitute in the given information.
Step 5. Solve the equation.
Step 6. Check. $\phantom{\rule{2em}{0ex}}29\stackrel{?}{=}6+9+14$ $\phantom{\rule{2em}{0ex}}29=29✓$
Step 7. Answer the question.	The lengths of the sides of the triangle are 6, 9, and 14 inches.

TRY IT 12

One side of a triangle is seven inches more than the first side. The third side is four inches less than three times the first. The perimeter is 28 inches. Find the length of the three sides of the triangle.

Show answer

The lengths of the sides of the triangle are 5, 11 and 12 inches.

EXAMPLE 13

The perimeter of a rectangular soccer field is 360 feet. The length is 40 feet more than the width. Find the length and width.

The figure is an illustration of rectangular soccer field.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the length and width of the soccer field
Step 3. Name. Choose a variable to represent it. The length is 40 feet more than the width. Draw the figure and label it with the given information.	Let w = width. $w+40=$ length
Step 4. Translate. Write the appropriate formula and substitute.
Step 5. Solve the equation.
Step 6. Check. $\begin{array}{ccc}\hfill P& =\hfill & 2L+2W\hfill \\ \hfill 360& \stackrel{?}{=}\hfill & 2\left(110\right)+2\left(70\right)\hfill \\ \hfill 360& =\hfill & 360✓\hfill \end{array}$
Step 7. Answer the question.	The length of the soccer field is 110 feet and the width is 70 feet.

TRY IT 13

The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.

Show answer

The length of the swimming pool is 70 feet and the width is 30 feet.

Key Concepts

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.
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.

2.4 Exercise Set

In the following exercises, use the formula $d=rt$ .

Solve for
1. when $d=240\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=60$
2. in general
Solve for
1. when $d=420\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t=6$
2. in general

In the following exercises, solve the formula .

Solve the formula for
1. when $A=176\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b=22$
2. in general
Solve the formula I=Prt for the principal, P for
1. $I=\$5,480$ , $r=4\%$ , $t=7\phantom{\rule{0.2em}{0ex}}\text{years}\phantom{\rule{0.2em}{0ex}}$
2. in general
Solve for the time, t for
1. $I=\$2,376$ , $P=\$9,000$ , $r=4.4\%$
2. in general
Solve the formula for y
1. when $x=3$
2. in general
Solve the formula for y
1. when $x=-2$
2. in general
Solve $180=a+b+c$ for $a$ .
Solve the formula $-4x+y=-6$ for y.
Solve the formula $x-y=-4$ for y.
Solve the formula $P=2L+2W$ for $L$ .
Solve the formula $C=\pi d$ for $d$ .
Solve the formula $V=LWH$ for $L$ .
Solve the formula $A=\frac{1}{2}{d}_{1}{d}_{2}$ for ${d}_{1}.$
Solve the formula $A=\frac{1}{2}{d}_{1}{d}_{2}$ for ${d}_{2}.$
Solve the formula $A=\frac{1}{2}h\left({b}_{1}+{b}_{2}\right)$ for ${b}_{1}.$
Solve the formula $A=\frac{1}{2}h\left({b}_{1}+{b}_{2}\right)$ for ${b}_{2}.$
Solve the formula $h=54t+\frac{1}{2}a{t}^{2}$ for a.
Solve the formula $h=48t+\frac{1}{2}a{t}^{2}$ for a.

In the following exercises, solve using a geometry formula.

A triangular flag has area 0.75 square feet and height 1.5 foot. What is its base?
What is the base of triangular window with area 207 square inches and height 18 inches?
The width of a rectangle is seven metres less than the length. The perimeter is 58 metres. Find the length and width.
The width of the rectangle is 0.7 metres less than the length. The perimeter of a rectangle is 52.6 metres. Find the dimensions of the rectangle.
The perimeter of a rectangle of 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.
The length of the rectangle is three metres less than twice the width. The perimeter of a rectangle is 36 metres. Find the dimensions of the rectangle.
The perimeter of a triangle is 39 feet. One side of the triangle is one foot longer than the second side. The third side is two feet longer than the second side. Find the length of each side.
One side of a triangle is twice the smallest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.
The perimeter of a rectangular field is 560 yards. The length is 40 yards more than the width. Find the length and width of the field.
A rectangular parking lot has perimeter 250 feet. The length is five feet more than twice the width. Find the length and width of the parking lot.

In the following exercises, solve.

Socorro drove for $4\frac{5}{6}$ hours at 60 miles per hour. How much distance did she travel?
Francie rode her bike for $2\frac{1}{2}$ hours at 12 miles per hour. How far did she ride?
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?
Halle 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?
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?
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?
Converting temperature. Yon was visiting the United States and he saw that the temperature in Seattle one day was 50^o Fahrenheit. Solve for C in the formula $F=\frac{9}{5}C+32$ to find the Celsius temperature.