2.4: Solving Formulas for a Specific Variable

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Learning outcome

• Solve any given formula for a specific variable

Though mathematical, formulas are the backbone of understanding content form 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 $d=rt$ for calculating distance based on rate and time. 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 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

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

1. When $d=520$ and $r=65$
2. Algebraically

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. ${\Large\frac{520}{65}}={\Large\frac{65t}{65}}$ ${\Large\frac{d}{r}}={\Large\frac{rt}{r}}$ 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

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We can use the formula $A=\Large\frac{1}{2}\normalsize bh$ to find the area of a triangle when we are 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 $h\text{:}$

1. When $A=90$ and $b=15$
2. Algebraically

Solution:

 1. When A = 90 and b = 15 2. Algebraically 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

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Previously, 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

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. Algebraically

Solution:

 1.  I = $5600, r = 4%, t = 7 years 2. Algebraically 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

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

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

1. When $x=4$
2. Algebraically

Solution:

 1. When x = 4 2. Algebraically 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. $12\color{red}{-12}+2y=18\color{red}{-12}$ $3x\color{red}{-3x}+2y=18\color{red}{-3x}$ Simplify. $2y=6$ $2y=18-3x$ Divide. $\Large\frac{2y}{\color{red}{2}}\normalsize =\Large\frac{6}{\color{red}{2}}$ $\Large\frac{2y}{\color{red}{2}}\normalsize =\Large\frac{18-3x}{\color{red}{2}}$ Simplify. $y=3$ $y=\Large\frac{18-3x}{2}$

In the previous examples, we used the numbers in part (a) as a guide to solving algebraically in part (b). Do you think you’re ready to solve a formula in general without using numbers as a guide?

example

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

So, $a=P-b-c$

try it

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example

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. $3x\color{red}{-3x}+y=10\color{red}{-3x}$ Simplify. $y=10 - 3x$

try it

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example

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. $6x+5y\color{red}{-6x}=13\color{red}{-6x}$ Simplify. $5y=13-6x$ Divide by 5 to make the coefficient 1. $\Large\frac{5y}{\color{red}{5}}\normalsize =\Large\frac{13-6x}{\color{red}{5}}$ Simplify. $y=\Large\frac{13-6x}{5}$

try it

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

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