9.7 Solve a Formula for a Specific Variable

Use the Distance, Rate, and Time Formula

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:

In general, the formula relating distance, rate, and time is

Notice that the units we used above for the rate were miles per hour, which we can write as a ratio $\frac{miles}{hour}.$ Then when we multiplied by the time, in hours, the common units ‘hour’ divided out. The answer was in miles.

Example 9.57

Jamal rides his bike at a uniform rate of $12$ miles per hour for $3\frac{1}{2}$ 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.	$d=?$ $r=12\text{mph}$ $t=3\frac{1}{2}\text{hours}$
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.	$d=rt$ $d=12\cdot 3\frac{1}{2}$
Step 5. Solve the equation.	$d=42\text{miles}$
Step 6. Check: Does 42 miles make sense?
Step 7. Answer the question with a complete sentence.	Jamal rode 42 miles.

Solve a Formula for a Specific Variable

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 Excel^TM 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 9.59

Solve the formula $d=rt$ for $t\text{:}$

1. ⓐ when $d=520$ and $r=65$
2. ⓑ 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.

	ⓐ when d = 520 and r = 65	ⓑ in general
Write the forumla.
Substitute any given values.
Divide to isolate t.
Simplify.

Notice that the solution for ⓐ is the same as that in Example 9.58. We say the formula $t=\frac{d}{r}$ 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.

Example 9.60

The formula for area of a triangle is $A=\frac{1}{2}bh.$ Solve this formula for $h\text{:}$

1. ⓐ when $A=90$ and $b=15$
2. ⓑ in general

Solution

	ⓐ when A = 90 and b = 15	ⓑ in general
Write the forumla.
Substitute any given values.
Clear the fractions.
Simplify.
Solve for h.

We can now find the height of a triangle, if we know the area and the base, by using the formula

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

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 9.61

Solve the formula $I=Prt$ to find the principal, $P\text{:}$

1. ⓐ when $I=\text{\$5,600},r=\text{4\%},t=7\text{years}$
2. ⓑ in general

Solution

	I = $5600, r = 4%, t = 7 years	in general
Write the forumla.
Substitute any given values.
Multiply r ⋅ t.
Divide to isolate P.
Simplify.
State the answer.	The principal is $20,000.

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 $y.$ To do this, we will follow the same steps that we used to solve a formula for a specific variable.

Example 9.62

Solve the formula $3x+2y=18$ for $y\text{:}$

1. ⓐ when $x=4$
2. ⓑ in general

Solution

	when x = 4	in general
Write the equation.
Substitute any given values.
Simplify if possible.
Subtract to isolate the y-term.
Simplify.
Divide.
Simplify.

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 9.63

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.
Simplify.		$P-b-c=a$

So, $a=P-b-c$

Example 9.64

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.
Simplify.		$y=10-3x$

Example 9.65

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.
Subtract to isolate the term with y.
Simplify.
Divide 5 to make the coefficient 1.
Simplify.