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 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.

A common formula is [latex]\require{color}d=rt[/latex] for calculating distance, [latex] d [/latex], based on rate, [latex] r [/latex], and time, [latex] t [/latex]. This formula gives the value of [latex]d[/latex] when you substitute in the values of [latex]r[/latex] and [latex]t[/latex]. But what if you have to find the value of [latex]t[/latex]? We would need to substitute in values of [latex]d[/latex] and [latex]r[/latex] and then use algebra to solve for [latex]t[/latex]. If you had to do this often, you might wonder why there isn’t a formula that gives the value of [latex]t[/latex] when you substitute in the values of [latex]d[/latex] and [latex]r[/latex]. We can get a formula like this by solving the formula [latex]d=rt[/latex] for [latex]t[/latex].

To solve a formula for a specific variable means to isolate that variable with a coefficient of [latex]1[/latex] 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 [latex]d=rt[/latex] for [latex]t[/latex]:

  1. When [latex]d=520[/latex] and [latex]r=65[/latex]
  2. Algebraically (assuming [latex] r>0 [/latex])

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

example

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

  1. When [latex]A=90[/latex] and [latex]b=15[/latex]
  2. Algebraically (assuming [latex] b>0 [/latex])

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

example

Solve the formula [latex]I=Prt[/latex] for the principal, [latex]P[/latex]:

  1. When [latex] I=\$5,600 [/latex], [latex] r=4\% [/latex], and [latex] t=7 [/latex] years
  2. Algebraically (assuming [latex] r,t>0 [/latex])

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 [latex]x[/latex] and [latex]y[/latex]. You might be given an equation that is solved for [latex]y[/latex] and you need to solve it for [latex]x[/latex], or vice versa. In the following example, we’re given an equation with both [latex]x[/latex] and [latex]y[/latex] on the same side and we’ll solve it for [latex]y[/latex]. To do this, we will follow the same steps that we used to solve a formula for a specific variable.

example

Solve the equation [latex]3x+2y=18[/latex] for [latex]y[/latex]:

  1. When [latex]x=4[/latex]
  2. Algebraically

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 [latex]P=a+b+c[/latex] for [latex]a[/latex].

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 [latex]x=\dfrac{y+1}{2y-3}[/latex] for [latex]y[/latex]. Assume [latex] x\ne \dfrac{1}{2} [/latex] and [latex] y\ne \dfrac{3}{2} [/latex].

Example

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

[latex] \displaystyle\frac{f’}{f} = \frac{v + v_o}{v} [/latex]

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

  1. [latex] f [/latex]
  2. [latex] v_o [/latex]
  3. [latex] v [/latex]

Example

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