4.6 Formulas

Learning outcome

Solve a given formula for a specified variable.

Though mathematical, formulas are the backbone of understanding content from many areas of study. They 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. 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.

A common formula is $\require{color}d=rt$ for calculating distance, $d$ , based on rate, $r$ , and time, $t$ . This formula gives the value of $d$ when you substitute in the values of $r$ and $t$ . But what if you have to find the value of $t$ ? We would need to substitute in values of $d$ and $r$ and then use algebra to solve for $t$ . 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 isolate that variable 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. The formula contains letters that represent variables, also called literals. We will mainly use the term variables in this book.

Let’s try a few examples, starting with the distance, rate, and time formula we used above.

example

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

When $d=520$ and $r=65$
Algebraically (assuming $r>0$ )

Show 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. Algebraically
Write the formula.	$d=rt$	$d=rt$
Substitute any given values.	$520=65t$
Divide to isolate $t$ .	${\color{red}{\dfrac{{\color{black}{520}}}{65}}}={\color{red}{\dfrac{\cancel{\color{black}{65}}{\color{black}{\cdot\: t}}}{\cancel{65} \cdot 1}}}$	${\color{red}{\dfrac{{\color{black}{d}}}{r}}}={\color{red}{\dfrac{\cancel{\color{black}{r}}{\color{black}{\cdot\: t}}}{\cancel{r} \cdot 1}}}$
Simplify.	$t=8$	$t=\dfrac{d}{r}$

We say the formula $t=\dfrac{d}{r}$ 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.

We can use the formula $A=\dfrac{1}{2}bh$ to find the area, $A$ , of a triangle when we are given the base, $b$ , and height, $h$ . In the next example, we will solve this formula for the height.

example

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

When $A=90$ and $b=15$
Algebraically (assuming $b>0$ )

Show Solution

Solution:

	1. When $A = 90$ and $b = 15$	2. Algebraically
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.	${\color{red}{2\cdot\: }}{90}={\color{red}{\dfrac{\cancel{2}\cdot 1}{1}\cdot\: }}\dfrac{1}{{\color{red}\cancel{\color{black}{2}}}\cdot 1}\cdot{15}\cdot h$	${\color{red}{2\cdot\: }}{A}={\color{red}{\dfrac{\cancel{2}\cdot 1}{1}\cdot\: }}\dfrac{1}{{\color{red}\cancel{\color{black}{2}}}\cdot 1}\cdot{b}\cdot h$
Simplify.	${\color{red}{\dfrac{{\color{black}{180}}}{15}}}={\color{red}{\dfrac{\cancel{\color{black}{15}}{\color{black}{\cdot\: h}}}{\cancel{15} \cdot 1}}}$	${\color{red}{\dfrac{{\color{black}{2A}}}{b}}}={\color{red}{\dfrac{\cancel{\color{black}{b}}{\color{black}{\cdot\: h}}}{\cancel{b} \cdot 1}}}$
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 $h={\Large\frac{2A}{b}}$ .

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

example

Solve the formula $I=Prt$ for the principal, $P$ :

When $I=\$5,600$ , $r=4\%$ , and $t=7$ years
Algebraically (assuming $r,t>0$ )

Show Solution

Solution:

	1. $I = \$5600$ , $r = 4\%$ , $t = 7$ years	2. Algebraically
Write the formula.	$I=Prt$	$I=Prt$
Convert percentage to a decimal.	$4\%=\dfrac{4}{100}=0.04$
Substitute any given values.	$5600=P(0.04)(7)$
Multiply $r\cdot t$ .	$5600=P(0.28)$	$I=P(rt)$
Divide to isolate $P$ .	${\color{red}{\dfrac{{\color{black}{5600}}}{0.28}}}={\color{red}{\dfrac{{\color{black}{P\cdot\: }}\cancel{\color{black}{0.28}}}{\cancel{0.28} \cdot 1}}}$	${\color{red}{\dfrac{{\color{black}{I}}}{r\cdot t}}}={\color{red}{\dfrac{{\color{black}{P\cdot\: }}\cancel{\color{black}{r\cdot t}}}{\cancel{r\cdot t} \cdot 1}}}$
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.

In algebra, we often encounter equations that relate two variables, usually $x$ and $y$ . You might be given an equation that is solved for $y$ and you 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

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

When $x=4$
Algebraically

Show Solution

Solution:

	1. When $x = 4$	2. Algebraically
Write the equation.	$3x+2y=18$	$3x+2y=18$
Substitute any given values.	$3(4)+2y=18$
Simplify if possible.	$12+2y=18$
Subtract to isolate the $y$ -term.	$\underset{\large{\color{red}{-12}}}{12}+2y=\underset{\large{\color{red}{-12}}}{18}$	$\underset{\large{\color{red}{-3x}}}{3x}+2y=\underset{\large{\color{red}{-3x}}}{18}$
Simplify.	$2y=6$	$2y=18-3x$
Divide.	${\color{red}{\dfrac{\cancel{\color{black}{2}}{\color{black}{\cdot\: y}}}{\cancel{2} \cdot 1}}}={\color{red}{\dfrac{{\color{black}{6}}}{2}}}$	${\color{red}{\dfrac{\cancel{\color{black}{2}}{\color{black}{\cdot\: y}}}{\cancel{2} \cdot 1}}}={\color{red}{\dfrac{{\color{black}{18-3x}}}{2}}}$
Simplify.	$y=3$	$y=\dfrac{18-3x}{2}$

In the previous examples, we used the numbers in part 1 as a guide to solving algebraically in part 2. From now on we will only solve a formula in general.

example

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

Show Solution

We will isolate $a$ on one side of the equation.

$\begin{align} \underset{\large{\color{red}{-b-c}}}{P}&=a+\underset{\large{\color{red}{-b}}}{b}+\underset{\large{\color{red}{-c\mathstrut}}}{c}\\ P-b-c&=a \end{align}$

So, $a=P-b-c$ .

In the following video we show another example of how to solve an equation for a specific variable.

In the following examples, the variable that we will be solving for will appear more than once in the equation and/or it will be in a denominator of a rational expression.

example

Solve the equation $x=\dfrac{y+1}{2y-3}$ for $y$ . Assume $x\ne \dfrac{1}{2}$ and $y\ne \dfrac{3}{2}$ .

Show Solution

We will isolate $y$ on one side of the equation. Note that this time the variable $y$ that we ought to solve for is present in two places in the equation. Thus, isolating it on one side of the equation is not straightforward. We will follow the intuition in terms of “what we see” and “what seems to be the most natural next step” until only one instance of $y$ is left in the equation.

$\begin{align} x &= \frac{y+1}{2y-3}\\[5pt] x {\color{red}\: \cdot\: (2y-3)} &= \frac{y+1}{{\color{red}\cancel{\color{black}{(2y-3)}}}\cdot 1}{\color{red}\: \cdot\: \dfrac{\cancel{(2y-3)}\cdot 1}{1}} && \color{blue}{\textsf{clear the fraction}}\\[5pt] \underset{\large{\color{red}{-y}}}{2xy}-\underset{\large{\color{red}{+3x\mathstrut}}}{3x} &= \underset{\large{\color{red}{-y}}}{y}+\underset{\large{\color{red}{+3x\mathstrut}}}{1} && \color{blue}{\textsf{distribute $x$ and isolate $y$-terms}}\\[5pt] 2xy-1y &= 3x+1\\[5pt] y(2x-1) &= 3x+1 && \color{blue}{\textsf{factor out $y$}}\\[5pt] {\color{red}{\dfrac{{\color{black}{y\cdot\: }}\cancel{\color{black}{(2x-1)}}}{\cancel{(2x-1)} \cdot 1}}} &= {\color{red}{\dfrac{{\color{black}{3x+1}}}{2x-1}}} && \color{blue}{\textsf{divide to isolate $y$}}\\[5pt] y &= \frac{3x+1}{2x-1} \end{align}$

Example

When an observer moves relative to a wave source, the frequency observed changes. The Doppler effect equation accounts for this:

$\displaystyle\frac{f’}{f} = \frac{v + v_o}{v}$

where $f’$ is the observed frequency, $f$ is the emitted frequency, $v$ is the wave velocity, and $v_o$ is the observer’s velocity. Assuming all variables are positive and $f'\ne f$ , solve the Doppler effect equation for the indicated variable.

$f$
$v_o$
$v$

Show Solution

1. Since $f$ appears only once, in the denominator of the LHS, once we clear the fractions by multiplying both sides by the LCD, isolating $f$ should be straightforward. (Note that since we have an equality of two single rational expressions, we can refer to clear the fractions by cross-multiplying. Please refer to this description to review cross-multiplying.)

$\begin{align} \frac{f’}{f}{\color{red}\: \cdot\: f\cdot v} &= \frac{v + v_o}{v}{\color{red}\: \cdot\: f\cdot v}\\[5pt] \frac{f’}{{\color{red}\cancel{\color{black}{f}}}}\cdot \frac{{\color{red}\cancel{\color{black}{f}}}\cdot v}{1} &= \frac{v + v_o}{{\color{red}\cancel{\color{black}{v}}}}\cdot \frac{f\cdot {\color{red}\cancel{\color{black}{v}}}}{1} && \color{blue}{\textsf{multiply by the LCD}}\\[5pt] {\color{red}{\dfrac{{\color{black}{f'v}}}{v+v_o}}} &= {\color{red}{\dfrac{\cancel{\color{black}{\left(v+v_o\right)}}{\color{black}{\: \cdot\: f}}}{\cancel{\left(v+v_o\right)} \cdot 1}}} && \color{blue}{\textsf{divide to isolate $f$}}\\[5pt] \frac{f’v}{v+v_o} &= f \end{align}$

So, $f=\dfrac{f’v}{v+v_o}$ .

2. Since $v_o$ appears only once, in the numerator of the RHS, once we clear the fractions, isolating $v_o$ should be straightforward.

$\begin{align} \frac{f’}{f} &= \frac{v + v_o}{v}\\[5pt] f'\cdot v &= f\cdot \left(v+v_o\right) && \color{blue}{\textsf{clear the fractions}}\\[5pt] {\color{red}{\dfrac{{\color{black}{f'v}}}{f}}} &= {\color{red}{\dfrac{\cancel{\color{black}{f}}{\color{black}{\: \cdot\: \left(v+v_o\right)}}}{\cancel{f} \cdot 1}}} && \color{blue}{\textsf{divide to isolate $v + v_o$}}\\[5pt] \frac{f’v}{f} {\color{red}{\: -\: v}} &= v + v_o {\color{red}{\: -\: v}} && \color{blue}{\textsf{subtract $v$ to isolate $v_o$}}\\[5pt] \frac{f’v}{f} - v &= v_o \end{align}$

So, $v_o=\dfrac{f’v}{f} - v$ .

Note, that after clearing fractions, instead of dividing both sides by $f$ , we could have distributed $f$ on the RHS and proceeded from there:

$\begin{align} f'\cdot v &= f\cdot \left(v+v_o\right)\\[5pt] f'\cdot v &= f\cdot v + f\cdot v_o && \color{blue}{\textsf{distribute $f$}}\\[5pt] \underset{\large{\color{red}{-fv}}}{f'v} &= \underset{\large{\color{red}{-fv}}}{fv} + fv_o && \color{blue}{\textsf{subtract $fv$ to isolate $fv_o$}}\\[5pt] {\color{red}{\frac{{\color{black}{f'v - fv}}}{f}}} &= {\color{red}{\frac{\cancel{\color{black}{f}}{\color{black}{\: \cdot\: v_o}}}{\cancel{f} \cdot 1}}} && \color{blue}{\textsf{divide to isolate $v_o$}}\\[5pt] \frac{f'v - fv}{f} &= v_o \end{align}$

So, $v_o=\dfrac{f'v - fv}{f}$ . Note, that $\dfrac{f'v - fv}{f}=\dfrac{f’v}{f} - v$ . (Can you transform the LHS to match the RHS or vice versa?)

3. Using the equation $f'v-fv = fv_o$ from the alternative solution in part 2, we can factor out $v$ on the LHS to obtain $v\left(f'-f\right) = fv_o$ , and divide both sides by $f'-f$ to obtain $v=\dfrac{fv_o}{f'-f}$

Example

Solve the equation $3=\dfrac{2}{t}-\dfrac{5}{z}$ for $z$ . Assume $t,z\ne 0$ and $t\ne\dfrac{2}{3}$ .

Show Solution

If we start with clearing the fractions by multiplying both sides by the LCD, isolating $z$ on one side of the equation should be straightforward.

$\begin{align} 3 {\color{red}{\: \cdot\: t\cdot z}} &= \left(\frac{2}{t}-\frac{5}{z}\right){\color{red}{\: \cdot\: t\cdot z}}\\[5 pt] 3tz &= \frac{2}{{\color{red}\cancel{\color{black}{t}}}\cdot 1}\cdot \frac{{\color{red}\cancel{\color{black}{t}}}\cdot z}{1}-\frac{5}{{\color{red}\cancel{\color{black}{z}}}\cdot 1}\cdot \frac{t\cdot {\color{red}\cancel{\color{black}{z}}}}{1}\\[5 pt] \underset{\large{\color{red}{-2z}}}{3tz} &= \underset{\large{\color{red}{-2z}}}{2z} - 5t\\[5pt] 3tz-2z &= -5t\\[5pt] {\color{red}{\dfrac{{\color{black}{z\: \cdot\: }}\cancel{\color{black}{(3t-2)}}}{\cancel{(3t-2)} \cdot 1}}} &= {\color{red}{\dfrac{{\color{black}{-5t}}}{3t-2}}}\\[5pt] z &= \frac{-5t}{3t-2} \end{align}$

Module 4: Rational Expressions, Functions, and Equations

Learning outcome

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