Solving a Formula For a Specific Variable

Learning Outcomes

Solve a formula for a specific variable using the properties of equality
Evaluate a formula for given values of the variables

Formulas are 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. Financial tools and calculators such as those in spreadsheets and applets offered by banks and financial advisors online also rely on formulas. 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.

Here’s an example that uses a formula you may have seen before: $d=rt$ , or distance = rate times time. This formula gives the value of the distance $d$ when you substitute in the values of a rate $r$ , and a time $t$ . We encounter this formula every day in an alternate form: $r=\dfrac{d}{t}$ , or the rate = distance per time. You may recognize it in the more familiar phrase describing a rate in miles per hour. We are able to solve the original formula for the variable $r$ by dividing $t$ away on both sides. See the example below for a demonstration.

Example

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

Solution:
$d=rt$

$\dfrac{d}{t}=\dfrac{r \cancel{t}}{\cancel{t}}$ divide by $t$ on both sides

$\dfrac{d}{t}=r$

We can also solve for $t$ . And we can find the value of one of the variables by substituting in particular values for the others. For example, to find the value of $t$ for particular values of $d$ and $r$ , we can first solve the formula for $t$ , then substitute in the particular values of $d$ and $r$ . Equations that are formulas for real-world relationships are often called literal equations, since the letters in the equation (the literals) each stand for a real value. See more examples below of solving a formula for a specific variable.

To solve a formula for a specific variable means to get that variable by itself with a coefficient of

1

$1$ 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

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

When $d=520$ and $r=65$
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.

	1. When $d = 520$ and $r = 65$	2. In general
Write the formula.	$d=rt$	$d=rt$
Substitute any given values.	$520=65t$
Divide to isolate t.	${\Large\frac{520}{65}}={\Large\frac{65t}{65}}$	${\Large\frac{d}{r}}={\Large\frac{rt}{r}}$
Simplify.	$8=t$ $t=8$	${\Large\frac{d}{r}}=t$ $t={\Large\frac{d}{r}}$

We say the formula $t={\Large\frac{d}{r}}$ is solved for $t$ . We can use this version of the formula any time we are given the distance and rate and need to find the time.

Try it

The formula $A=\Large\frac{1}{2}\normalsize bh$ can be used to find the area of a triangle when given the base and height. In the next example, we will solve this formula for the height.

example

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

When $A=90$ and $b=15$
In general

Show Solution

Solution:

	1. When A = 90 and b = 15	2. In general
Write the forumla.	$A=\Large\frac{1}{2}\normalsize bh$	$A=\Large\frac{1}{2}\normalsize bh$
Substitute any given values.	$90=\Large\frac{1}{2}\normalsize\cdot{15}\cdot{h}$
Clear the fractions.	$\color{red}{2}\cdot{90}=\color{red}{2}\cdot\Large\frac{1}{2}\normalsize\cdot{15}\cdot{h}$	$\color{red}{2}\cdot{A}=\color{red}{2}\cdot\Large\frac{1}{2}\normalsize\cdot{b}\cdot{h}$
Simplify.	$180=15h$	$2A=bh$
Solve for h.	$12=h$	${\Large\frac{2A}{b}}=h$

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

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

try it

The formula $I=Prt$ is used to calculate simple interest, where $I$ is interest, $P$ is principal, $r$ is rate as a decimal, and $t$ is time in years.

example

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

When $I=\text{\$5,600},r=\text{4%},t=7\text{years}$
In general

Show Solution

Solution:

	1. I = $5600, r = 4%, t = 7 years	2. 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.	$\Large\frac{5600}{\color{red}{0.28}}\normalsize =\Large\frac{P(0.28)}{\color{red}{0.28}}$	$\Large\frac{I}{\color{red}{rt}}\normalsize =\Large\frac{P(rt)}{\color{red}{rt}}$
Simplify.	$20,000=P$	$\Large\frac{I}{rt}\normalsize =P$
State the answer.	The principal is $20,000.	$P=\Large\frac{I}{rt}$

try it

Watch the following video to see another example of how to solve an equation for a specific variable.

The following examples just ask you to solve a formula in general, without finding particular values.

example

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

Show Solution

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.		$P\color{red}{-b-c}=a+b+c\color{red}{-b-c}$
Simplify.		$P-b-c=a$

So, $a=P-b-c$

Module 9QR: Financial Literacy

Learning Outcomes

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