One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you travel if you drove 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 Interstate.) If you said 120 miles, you already know how to use this formula!
The math to calculate the distance might look like this:
\[\begin{split} distance &= \left(\dfrac{60\; miles}{1\; hour}\right) (2\; hours) \\ distance &= 120\; miles \end{split}\]
In general, the formula relating distance, rate, and time is
\[distance = rate \cdot time\]
Definition: Distance, Rate and Time
For an object moving in 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, and t = time.
Notice that the units we used above for the rate were miles per hour, which we can write as a ratio \(\dfrac{miles}{hour}\). Then when we multiplied by the time, in hours, the common units ‘hour’ divided out. The answer was in miles.
Example \(\PageIndex{1}\):
Jamal rides his bike at a uniform rate of 12 miles per hour for \(3 \dfrac{1}{2}\) hours. How much distance has he traveled?
Solution
Step 1. Read the problem. You may want to create a mini-chart to summarize the information in the problem. |
$$\begin{split} d &=\; ? \\ r &= 12\; mph \\ t &= 3 \dfrac{1}{2}\; hours \end{split}$$ |
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 for the situation. Substitute in the given information. |
$$\begin{split} d &= rt \\ d &= 12 \cdot 3 \dfrac{1}{2} \end{split}$$ |
Step 5. Solve the equation. |
d = 42 miles |
Step 6. Check: Does 42 miles make sense? |
|
Step 7. Answer the question with a complete sentence. |
Jamal rode 42 miles. |
Exercise \(\PageIndex{1}\):
Lindsay drove for \(5 \dfrac{1}{2}\) hours at 60 miles per hour. How much distance did she travel?
- Answer
-
330 mi
Exercise \(\PageIndex{2}\):
Trinh walked for \(2 \dfrac{1}{3}\) hours at 3 miles per hour. How far did she walk?
- Answer
-
7 mi
Example \(\PageIndex{2}\):
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. Summarize the information in the problem. |
$$\begin{split} d &= 520\; miles \\ r &= 65\; mph \\ t &=\; ? \end{split}$$ |
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 for the situation. Substitute in the given information. |
$$\begin{split} d &= rt \\ 520 &= 65t \end{split}$$ |
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. |
$$\begin{split} d &= rt \\ 520 &\stackrel{?}{=} 65 \cdot 8 \\ 520 &= 520\; \checkmark \end{split}$$ |
Step 7. Answer the question with a complete sentence. We know the units of time will be hours because we divided miles by miles per hour. |
Rey's trip will take 8 hours. |
Exercise \(\PageIndex{3}\):
Lee wants to drive from Phoenix to his brother’s apartment in San Francisco, a distance of 770 miles. If he drives at a steady rate of 70 miles per hour, how many hours will the trip take?
- Answer
-
11 hours
Exercise \(\PageIndex{4}\):
Yesenia is 168 miles from Chicago. If she needs to be in Chicago in 3 hours, at what rate does she need to drive?
- Answer
-
56 mph
In this chapter, you became familiar with some formulas used in geometry. Formulas are also very useful in the sciences and social sciences—fields such as chemistry, physics, biology, psychology, sociology, and criminal justice. Healthcare workers use formulas, too, even for something as routine as dispensing medicine. The widely used spreadsheet program Microsoft ExcelTM relies on formulas to do its calculations. Many teachers use spreadsheets to apply formulas to compute student grades. It is important to be familiar with formulas and be able to manipulate them easily.
In Example 9.57 and Example 9.58, we used the formula d = rt. This formula gives the value of d when you substitute in the values of r and t. But in Example 9.58, we had to find the value of t. We substituted in values of d and r and then used algebra to solve to t. If you had to do this often, you might wonder why there isn’t a formula that gives the value of t when you substitute in the values of d and r. We can get a formula like this by solving the formula d = rt for t.
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 the other variables and constants on the other side. We will call this solving an equation for a specific variable in general. This process is also called solving a literal equation. The result is another formula, made up only of variables. The formula contains letters, or literals.
Let’s try a few examples, starting with the distance, rate, and time formula we used above.
Example \(\PageIndex{3}\):
Solve the formula d = rt for t: (a) when d = 520 and r = 65 (b) in general.
Solution
We’ll write the solutions side-by-side so you can see 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. |
d = rt |
d = rt |
Substitute any given values. |
520 = 65t |
|
Divide to isolate t. |
$$\dfrac{520}{65} = \dfrac{65t}{65}$$ |
$$\dfrac{d}{r} = \dfrac{rt}{r}$$ |
Simplify. |
$$\begin{split} 8 &= t \\ t &= 8 \end{split}$$ |
$$\begin{split} \dfrac{d}{r} &= t \\ t &= \dfrac{d}{r} \end{split}$$ |
Notice that the solution for (a) is the same as that in Example 9.58. We say the formula t = \(\dfrac{d}{r}\) is solved for t. We can use this version of the formula anytime we are given the distance and rate and need to find the time.
Exercise \(\PageIndex{5}\):
Solve the formula d = rt for r: (a) when d = 180 and t = 4 (b) in general
- Answer a
-
\(r = 45\)
- Answer b
-
\(r = \frac{d}{t}\)
Exercise \(\PageIndex{6}\):
Solve the formula d = rt for r: (a) when d = 780 and t = 12 (b) in general
- Answer a
-
\(r = 65\)
- Answer b
-
\(r = \frac{d}{t}\)
We used the formula A = \(\dfrac{1}{2}\)bh in Use Properties of Rectangles, Triangles, and Trapezoids to find the area of a triangle when we were given the base and height. In the next example, we will solve this formula for the height.
Example \(\PageIndex{4}\):
The formula for area of a triangle is A = \(\dfrac{1}{2}\)bh. Solve this formula 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 = \(\dfrac{1}{2}\)bh |
A = \(\dfrac{1}{2}\)bh |
Substitute any given values. |
$$90 = \dfrac{1}{2} \cdot 15 \cdot h$$ |
|
Clear the fractions. |
$$\textcolor{red}{2} \cdot 90 = \textcolor{red}{2} \cdot \dfrac{1}{2} \cdot 15 \cdot h$$ |
$$\textcolor{red}{2} \cdot A = \textcolor{red}{2} \cdot \dfrac{1}{2} \cdot b \cdot h$$ |
Simplify. |
180 = 15h |
2A = bh |
Solve for h. |
12 = h |
\(\dfrac{2A}{b}\) = h |
We can now find the height of a triangle, if we know the area and the base, by using the formula
\[h = \dfrac{2A}{b}\]
Exercise \(\PageIndex{7}\):
Use the formula A = \(\dfrac{1}{2}\)bh to solve for h: (a) when A = 170 and b = 17 (b) in general
- Answer a
-
\(h = 20\)
- Answer b
-
\(h = \frac{2A}{b}\)
Exercise \(\PageIndex{8}\):
Use the formula A = \(\dfrac{1}{2}\)bh to solve for b: (a) when A = 62 and h = 31 (b) in general
- Answer a
-
\(b = 4\)
- Answer b
-
\(b = \frac{2A}{h}\)
In Solve Simple Interest Applications, we used the formula I = Prt to calculate simple interest, where I is interest, P is principal, r is rate as a decimal, and t is time in years.
Example \(\PageIndex{5}\):
Solve the formula I = Prt to find the principal, P: (a) when I = $5,600, r = 4%, t = 7 years (b) in general
Solution
|
(a) when I = $5,600, r = 4%, t = 7 years |
(b) in general |
Write the forumla. |
I = Prt |
I = Prt |
Substitute any given values. |
5600 = P(0.04)(7) |
I = Prt |
Multiply r • t. |
5600 = P(0.28) |
I = P(rt) |
Divide to isolate P. |
$$\dfrac{5600}{\textcolor{red}{0.28}} = \dfrac{P(0.28)}{\textcolor{red}{0.28}}$$ |
$$\dfrac{I}{\textcolor{red}{rt}} = \dfrac{P(rt)}{\textcolor{red}{rt}}$$ |
Simplify. |
20,000 = P |
\(\dfrac{I}{rt}\) = P |
State the answer. |
The principal is $20,000. |
$$P = \dfrac{I}{rt}$$ |
Exercise \(\PageIndex{9}\):
Use the formula I = Prt. Find t: (a) when I = $2,160, r = 6%, P = $12,000; (b) in general
- Answer a
-
\(t = 3\) years
- Answer b
-
\(t = \frac{I}{Pr}\)
Exercise \(\PageIndex{10}\):
Use the formula I = Prt. Find r: (a) when I = $5,400, P = $9,000, t = 5 years; (b) in general
- Answer a
-
\(r = 0.12 = 12\%\)
- Answer b
-
\(t = \frac{I}{Pt}\)
Later in this course, 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. To do this, we will follow the same steps that we used to solve a formula for a specific variable.
Example \(\PageIndex{6}\):
Solve the formula 3x + 2y = 18 for y: (a) when x = 4 (b) in general
Solution
|
(a) when x = 4 |
(b) in general |
Write the equation. |
3x + 2y = 18 |
3x + 2y = 18 |
Substitute any given values. |
3(4) + 2y = 18 |
3x + 2y = 18 |
Simplify if possible. |
12 + 2y = 18 |
3x + 2y = 18 |
Subtract to isolate the y-term. |
$$12 \textcolor{red}{-12} + 2y = 18 \textcolor{red}{-12}$$ |
$$3x \textcolor{red}{-3x}+ 2y = 18 \textcolor{red}{-3x}$$ |
Simplify. |
2y = 6 |
2y = 18 - 3x |
Divide. |
$$\dfrac{2y}{\textcolor{red}{2}} = \dfrac{6}{\textcolor{red}{2}}$$ |
$$\dfrac{2y}{\textcolor{red}{2}} = \dfrac{18 - 3x}{\textcolor{red}{2}}$$ |
Simplify. |
y = 3 |
$$y = \dfrac{18 - 3x}{2}$$ |
Exercise \(\PageIndex{11}\):
Solve the formula 3x + 4y = 10 for y: (a) when x = 2 (b) in general
- Answer a
-
\(y = 1\)
- Answer b
-
\(y = \frac{10-3x}{4}\)
Exercise \(\PageIndex{12}\):
Solve the formula 5x + 2y = 18 for y: (a) when x = 4 (b) in general
- Answer a
-
\(y = -1\)
- Answer b
-
\(y = \frac{18-5x}{2}\)
In the previous examples, we used the numbers in part (a) as a guide to solving in general in part (b). Do you think you’re ready to solve a formula in general without using numbers as a guide?
Example \(\PageIndex{7}\):
Solve the formula P = a + b + c for a.
Solution
We will isolate a on one side of the equation.
We will isolate a on one side of the equation. |
|
Write the equation. |
P = a + b + c |
Subtract b and c from both sides to isolate a. |
$$P \textcolor{red}{-b -c} = a + b + c \textcolor{red}{-b -c}$$ |
Simplify. |
P − b − c = a |
So, a = P − b − c.
Exercise \(\PageIndex{13}\):
Solve the formula P = a + b + c for b.
- Answer
-
b = P - a - c
Exercise \(\PageIndex{14}\):
Solve the formula P = a + b + c for c.
- Answer
- c = P - a - b
Example \(\PageIndex{8}\):
Solve the equation 3x + y = 10 for y.
Solution
We will isolate y on one side of the equation.
We will isolate y on one side of the equation. |
|
Write the equation. |
3x + y = 10 |
Subtract 3x from both sides to isolate y. |
$$3x \textcolor{red}{-3x} + y= 10 \textcolor{red}{-3x}$$ |
Simplify. |
y = 10 - 3x |
Exercise \(\PageIndex{15}\):
Solve the formula 7x + y = 11 for y
- Answer
-
y = 11 - 7x
Exercise \(\PageIndex{16}\):
Solve the formula 11x + y = 8 for y.
- Answer
-
y = 8 - 11x
Example \(\PageIndex{9}\):
Solve the equation 6x + 5y = 13 for y.
Solution
We will isolate y on one side of the equation.
We will isolate y on one side of the equation. |
|
Write the equation. |
6x + 5y = 13 |
Subtract to isolate the term with y. |
$$6x + 5y \textcolor{red}{-6x} = 13 \textcolor{red}{-6x}$$ |
Simplify. |
5y = 13 - 6x |
Divide 5 to make the coefficient 1. |
$$\dfrac{5y}{\textcolor{red}{5}} = \dfrac{13 - 6x}{\textcolor{red}{5}}$$ |
Simplify. |
$$y = \dfrac{13 - 6x}{5}$$ |
Exercise \(\PageIndex{17}\):
Solve the formula 4x + 7y = 9 for y.
- Answer
-
\(y = \frac{9-4x}{7}\)
Exercise \(\PageIndex{18}\):
Solve the formula 5x + 8y = 1 for y.
- Answer
-
\(y = \frac{1-5x}{8}\)
ACCESS ADDITIONAL ONLINE RESOURCES
Distance = Rate x Time
Distance, Rate, Time
Simple Interest
Solving a Formula for a Specific Variable
Solving a Formula for a Specific Variable
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?