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 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!

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.

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

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.		$d = r t$

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.

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, 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.

Step 4. Translate. Write the appropriate formula.		$d = r t$
Substitute in the given information.		$520 = 65 t$
Step 5. Solve the equation.		$t = 8$
Step 6. Check. Substitute the numbers into the formula and make sure the result is a true statement.
	$\begin{matrix} d & = & r t \\ 520 & \overset{?}{=} & 65 \cdot 8 \\ 520 & = & 520 ✓ \end{matrix}$
Step 7. Answer the question with a complete sentence. Rey’s trip will take 8 hours.

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 $d = r t$ . This formula gives the value of $d$ , distance, when you substitute in the values of $r and t$ , the rate and time. But in Example 2, we had to find the value of $t$ . We substituted in values of $d and r$ 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 $d and r$ . We can make a formula like this by solving the formula $d = r t$ 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.

Solve the formula $d = r t$ 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.

ⓐ when $d = 520$ and $r = 65$		ⓑ in general
Write the formula.	$d = r t$	Write the formula.	$d = r t$
Substitute.	$520 = 65 t$
Divide, to isolate $t$ .	$\frac{520}{65} = \frac{65 t}{65}$	Divide, to isolate $t$ .	$\frac{d}{r} = \frac{r t}{r}$
Simplify.	$8 = t$	Simplify.	$\frac{d}{r} = t$

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

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

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

Solution

ⓐ when $A = 90$ and $b = 15$	ⓑ in general
Write the formula.	Write the formula.
Substitute.
Clear the fractions.	Clear the fractions.	$A mathematical equation shows both sides being multiplied by 2, represented in red text, to simplify the expression for the area of a triangle, A = (1/2)bh, by removing the fraction.$
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 $h = \frac{2 A}{b}$ .

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

Solve the formula $I = P r t$ to find the principal, $P$ :

ⓐ when $I = $ 5,600, r = 4 %, t = 7 years$ ⓑ in general

Solution

ⓐ $I = $5,600$ , $r = 4%$ , $t = 7 years$	ⓑ in general
Write the formula.	Write the formula.
Substitute.
Simplify.	Simplify.
Divide, to isolate P.	Divide, to isolate P.
Simplify.	Simplify.
The principal is		$A mathematical formula is shown: P = I / rt. The letter P is equal to a fraction where I is the numerator, and the product of r and t is the denominator. The variables are P, I, r, and t.$

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.

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

ⓐ when $x = 4$ ⓑ in general

Solution

ⓐ when $x = 4$	ⓑ in general

Substitute.
Subtract to isolate the $y$ -term.	Subtract to isolate the $y$ -term.
Divide.	Divide.	$Equation showing two y over two equals eighteen over two minus three x over two. The number two in the denominator of each fraction is highlighted.$
Simplify.	Simplify.

In Examples 1.60 through 1.64 we used the numbers in part ⓐ as a guide to solving in general in part ⓑ. Now we will solve a formula in general without using numbers as a guide.

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.

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

Solution


Subtract $6 x$ from both sides to isolate the term with $y$ .
Simplify.
Divide by 5 to make the coefficient 1.	$A mathematical equation is shown with fractions. The left side is 5y over 5, and the right side is 13 minus 6x, all over 5. The denominator 5 on both sides is highlighted in red.$
Simplify.	$A mathematical equation is displayed, showing y equals the fraction with numerator 13 minus 6x and denominator 5.$

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

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: $d = r t$ 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.

Steve drove for $8 \frac{1}{2}$ hours at 72 miles per hour. How much distance did he travel?

Socorro drove for $4 \frac{5}{6}$ hours at 60 miles per hour. How much distance did she travel?

Solution

290 miles

Yuki walked for $1 \frac{3}{4}$ hours at 4 miles per hour. How far did she walk?

Francie rode her bike for $2 \frac{1}{2}$ hours at 12 miles per hour. How far did she ride?

Solution

30 miles

Connor wants to drive from Tucson to the Grand Canyon, a distance of 338 miles. If he drives at a steady rate of 52 miles per hour, how many hours will the trip take?

Megan is taking the bus from New York City to Montreal. The distance is 380 miles and the bus travels at a steady rate of 76 miles per hour. How long will the bus ride be?

Solution

5 hours

Aurelia is driving from Miami to Orlando at a rate of 65 miles per hour. The distance is 235 miles. To the nearest tenth of an hour, how long will the trip take?

Kareem wants to ride his bike from St. Louis to Champaign, Illinois. The distance is 180 miles. If he rides at a steady rate of 16 miles per hour, how many hours will the trip take?

Solution

11.25 hours

Javier is driving to Bangor, 240 miles away. If he needs to be in Bangor in 4 hours, at what rate does he need to drive?

Alejandra is driving to Cincinnati, 450 miles away. If she wants to be there in 6 hours, at what rate does she need to drive?

Solution

75 mph

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?

Philip got a ride with a friend from Denver to Las Vegas, a distance of 750 miles. If the trip took 10 hours, how fast was the friend driving?

Solution

75 mph

Solve a Formula for a Specific Variable

In the following exercises, use the formula $d = r t$ .

Solve for $t$ ⓐ when $d = 350$ and $r = 70$ ⓑ in general

Solve for $t$ ⓐ when $d = 240 and r = 60$ ⓑ in general

Solution

ⓐ $t = 4$ ⓑ $t = \frac{d}{r}$

Solve for $t$ ⓐ when $d = 510 and r = 60$ ⓑ in general

Solve for $t$
ⓐ when $d = 175 and r = 50$
ⓑ in general

Solution

ⓐ $t = 3.5$ ⓑ $t = \frac{d}{r}$

Solve for $r$
ⓐ when $d = 204 and t = 3$ ⓑ in general

Solve for $r$ ⓐ when $d = 420 and t = 6$ ⓑ in general

Solution

ⓐ $r = 70$ ⓑ $r = \frac{d}{t}$

Solve for $r$ ⓐ when $d = 160 and t = 2.5$ ⓑ in general

Solve for $r$ ⓐ when $d = 180 and t = 4.5$ ⓑ in general

Solution

ⓐ $r = 40$ ⓑ $r = \frac{d}{t}$

In the following exercises, use the formula $A = \frac{1}{2} b h$ .

Solve for $b$ ⓐ when $A = 126 and h = 18$ ⓑ in general

Solve for $h$
ⓐ when $A = 176 and b = 22$ ⓑ in general

Solution

ⓐ $h = 16$ ⓑ $h = \frac{2 A}{b}$

Solve for $h$ ⓐ when $A = 375 and b = 25$ ⓑ in general

Solve for $b$ ⓐ when $A = 65 and h = 13$ ⓑ in general

Solution

ⓐ $b = 10$ ⓑ $b = \frac{2 A}{h}$

In the following exercises, use the formula I = Prt.

Solve for the principal, P forⓐ $I = $ 5,480, r = 4 %,$ $t = 7 years$ ⓑ in general

Solve for the principal, P for
ⓐ $I = $ 3,950, r = 6 %,$ $t = 5 years$ ⓑ in general

Solution

ⓐ $P = $ 13, 166.67$ ⓑ $P = \frac{I}{r t}$

Solve for the time, t for ⓐ $I = $ 2,376, P = $ 9,000,$ $r = 4.4 %$ ⓑ in general

Solve for the time, t for
ⓐ $I = $ 624, P = $ 6,000,$ $r = 5.2 %$ ⓑ in general

Solution

ⓐ $t = 2$ years ⓑ $t = \frac{I}{\Pr}$

In the following exercises, solve.

Solve the formula $2 x + 3 y = 12$ for yⓐ when $x = 3$ ⓑ in general

Solve the formula $5 x + 2 y = 10$ for yⓐ when $x = 4$ ⓑ in general

Solution

ⓐ $y = −5$ ⓑ $y = \frac{10 - 5 x}{2}$

Solve the formula $3 x - y = 7$ for yⓐ when $x = −2$ ⓑ in general

Solve the formula $4 x + y = 5$ for yⓐ when $x = −3$ ⓑ in general

Solution

ⓐ $y = 17$ ⓑ $y = 5 - 4 x$

Solve $a + b = 90$ for $b$ .

Solve $a + b = 90$ for $a$ .

Solution

$a = 90 - b$

Solve $180 = a + b + c$ for $a$ .

Solve $180 = a + b + c$ for $c$ .

Solution

$c = 180 - a - b$

Solve the formula $8 x + y = 15$ for y.

Solve the formula $9 x + y = 13$ for y.

Solution

$y = 13 - 9 x$

Solve the formula $−4 x + y = −6$ for y.

Solve the formula $−5 x + y = −1$ for y.

Solution

$y = −1 + 5 x$

Solve the formula $4 x + 3 y = 7$ for y.

Solve the formula $3 x + 2 y = 11$ for y.

Solution

$y = \frac{11 - 3 x}{2}$

Solve the formula $x - y = −4$ for y.

Solve the formula $x - y = −3$ for y.

Solution

$y = 3 + x$

Solve the formula $P = 2 L + 2 W$ for $L$ .

Solve the formula $P = 2 L + 2 W$ for $W$ .

Solution

$W = \frac{P - 2 L}{2}$

Solve the formula $C = π d$ for $d$ .

Solve the formula $C = π d$ for $π$ .

Solution

$π = \frac{C}{d}$

Solve the formula $V = L W H$ for $L$ .

Solve the formula $V = L W H$ for $H$ .

Solution

$H = \frac{V}{L W}$

Everyday Math

Converting temperature While on a tour in Greece, Tatyana saw that the temperature was 40^o Celsius. Solve for F in the formula $C = \frac{5}{9} (F - 32)$ to find the Fahrenheit temperature.

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.

Solution

10°C

Writing Exercises

Solve the equation $2 x + 3 y = 6$ for $y$ ⓐ when $x = −3$ ⓑ in general ⓒ Which solution is easier for you, ⓐ or ⓑ ? Why?

Solve the equation $5 x - 2 y = 10$ for $x$ ⓐ when $y = 10$ ⓑ in general
ⓒ Which solution is easier for you, ⓐ or ⓑ ? Why?

Solution

Answers will vary.

Self Check

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

This is a table that has three rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads “use the distance, rate, and time formula,” and “solve a formula for a specific variable.” The rest of the cells are blank.

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