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学習ガイド > Prealgebra

Solving a Formula for a Specific Variable

Learning Outcomes

  • Solve a formula or equation for a specific variable using the properties of equality
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 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. In some examples, we used the formula d=rtd=rt. This formula gives the value of dd when you substitute in the values of rr and tt. But in another example, we had to find the value of tt. We substituted in values of dd and rr and then used algebra to solve for tt. If you had to do this often, you might wonder why there isn’t a formula that gives the value of tt when you substitute in the values of dd and rr. We can get a formula like this by solving the formula d=rtd=rt for tt.
To solve a formula for a specific variable means to get that variable by itself with a coefficient of 11 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=rtd=rt for t:t\text{:}
  1. When d=520d=520 and r=65r=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.
1. When d = 520 and r = 65 2. In general
Write the formula. d=rtd=rt d=rtd=rt
Substitute any given values. 520=65t520=65t
Divide to isolate t. 52065=65t65\frac{520}{65}=\frac{65t}{65} dr=rtr\frac{d}{r}=\frac{rt}{r}
Simplify. 8=t8=t t=8t=8 dr=t\frac{d}{r}=t t=drt=\frac{d}{r}
We say the formula t=drt=\frac{d}{r} is solved for tt. We can use this version of the formula any time we are given the distance and rate and need to find the time.   We used the formula A=12bhA=\frac{1}{2}bh to find the area of a triangle when we were 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=12bhA=\frac{1}{2}bh. Solve this formula for h:h\text{:}
  1. When A=90A=90 and b=15b=15
  2. In general

Answer:

Solution:
1. When A = 90 and b = 15 2. In general
Write the forumla. A=12bhA=\frac{1}{2}bh A=12bhA=\frac{1}{2}bh
Substitute any given values. 90=12⋅15⋅h90=\frac{1}{2}\cdot{15}\cdot{h}
Clear the fractions. 2⋅90=2⋅12⋅15⋅h\color{red}{2}\cdot{90}=\color{red}{2}\cdot\frac{1}{2}\cdot{15}\cdot{h} 2⋅A=2⋅12⋅b⋅h\color{red}{2}\cdot{A}=\color{red}{2}\cdot\frac{1}{2}\cdot{b}\cdot{h}
Simplify. 180=15h180=15h 2A=bh2A=bh
Solve for h. 12=h12=h 2Ab=h\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=2Abh=\frac{2A}{b}

  Previously, we used the formula I=PrtI=Prt to calculate simple interest, where II is interest, PP is principal, rr is rate as a decimal, and tt is time in years.

example

Solve the formula I=PrtI=Prt to find the principal, P:P\text{:}
  1. When I=\text{\$5,600},r=\text{4%},t=7\text{years}
  2. In general

Answer:

Solution:
1.  I = $5600, r = 4%, t = 7 years 2. In general
Write the forumla. I=PrtI=Prt I=PrtI=Prt
Substitute any given values. 5600=P(0.04)(7)5600=P(0.04)(7) I=PrtI=Prt
Multiply r ⋅ t. 5600=P(0.28)5600=P(0.28) I=P(rt)I=P(rt)
Divide to isolate P. 56000.28=P(0.28)0.28\frac{5600}{\color{red}{0.28}}=\frac{P(0.28)}{\color{red}{0.28}} Irt=P(rt)rt\frac{I}{\color{red}{rt}}=\frac{P(rt)}{\color{red}{rt}}
Simplify. 20,000=P20,000=P Irt=P\frac{I}{rt}=P
State the answer. The principal is $20,000. P=IrtP=\frac{I}{rt}

  Watch the following video to see another example of how to solve an equation for a specific variable. https://youtu.be/VQZQvJ3rXYg Later in this class, and in future algebra classes, you’ll encounter equations that relate two variables, usually xx and yy. You might be given an equation that is solved for yy and you need to solve it for xx, or vice versa. In the following example, we’re given an equation with both xx and yy on the same side and we’ll solve it for yy. 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=183x+2y=18 for y:y\text{:}
  1. When x=4x=4
  2. In general

Answer:

Solution:
1. When x = 4 2. In general
Write the equation. 3x+2y=183x+2y=18 3x+2y=183x+2y=18
Substitute any given values. 3(4)+2y=183(4)+2y=18 3x+2y=183x+2y=18
Simplify if possible. 12+2y=1812+2y=18 3x+2y=183x+2y=18
Subtract to isolate the y-term. 12−12+2y=18−1212\color{red}{-12}+2y=18\color{red}{-12} 3x−3x+2y=18−3x3x\color{red}{-3x}+2y=18\color{red}{-3x}
Simplify. 2y=62y=6 2y=18−3x2y=18-3x
Divide. 2y2=62\frac{2y}{\color{red}{2}}=\frac{6}{\color{red}{2}} 2y2=18−3x2\frac{2y}{\color{red}{2}}=\frac{18-3x}{\color{red}{2}}
Simplify. y=3y=3 y=18−3x2y=\frac{18-3x}{2}

    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

Solve the formula P=a+b+cP=a+b+c for aa.

Answer:

Solution: We will isolate aa on one side of the equation.
We will isolate a on one side of the equation.
Write the equation. P=a+b+cP=a+b+c
Subtract b and c from both sides to isolate a. P−b−c=a+b+c−b−cP\color{red}{-b-c}=a+b+c\color{red}{-b-c}
Simplify. P−b−c=aP-b-c=a
So, a=P−b−ca=P-b-c

   

example

Solve the equation 3x+y=103x+y=10 for yy.

Answer:

Solution We will isolate yy on one side of the equation.
We will isolate y on one side of the equation.
Write the equation. 3x+y=103x+y=10
Subtract 3x from both sides to isolate y. 3x−3x+y=10−3x3x\color{red}{-3x}+y=10\color{red}{-3x}
Simplify. y=10−3xy=10 - 3x

   

example

Solve the equation 6x+5y=136x+5y=13 for yy.

Answer:

Solution: We will isolate yy on one side of the equation.
We will isolate y on one side of the equation.
Write the equation. 6x+5y=136x+5y=13
Subtract to isolate the term with y. 6x+5y−6x=13−6x6x+5y\color{red}{-6x}=13\color{red}{-6x}
Simplify. 5y=13−6x5y=13-6x
Divide by 5 to make the coefficient 1. 5y5=13−6x5\frac{5y}{\color{red}{5}}=\frac{13-6x}{\color{red}{5}}
Simplify. y=13−6x5y=\frac{13-6x}{5}

  In the following video we show another example of how to solve an equation for a specific variable. https://youtu.be/dG0_i8lN2y0

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