Solving a Formula for a Specific Variable
Learning Outcomes
- Solve a formula or equation for a specific variable using the properties of equality
To solve a formula for a specific variable means to get that variable by itself 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, 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 [latex]d=rt[/latex] for [latex]t\text{:}[/latex]- When [latex]d=520[/latex] and [latex]r=65[/latex]
- In general.
1. When d = 520 and r = 65 | 2. In general | |
Write the formula. | [latex]d=rt[/latex] | [latex]d=rt[/latex] |
Substitute any given values. | [latex]520=65t[/latex] | |
Divide to isolate t. | [latex]\frac{520}{65}=\frac{65t}{65}[/latex] | [latex]\frac{d}{r}=\frac{rt}{r}[/latex] |
Simplify. | [latex]8=t[/latex] [latex]t=8[/latex] | [latex]\frac{d}{r}=t[/latex] [latex]t=\frac{d}{r}[/latex] |
example
The formula for area of a triangle is [latex]A=\frac{1}{2}bh[/latex]. Solve this formula for [latex]h\text{:}[/latex]- When [latex]A=90[/latex] and [latex]b=15[/latex]
- In general
Answer:
Solution:1. When A = 90 and b = 15 | 2. In general | |
Write the forumla. | [latex]A=\frac{1}{2}bh[/latex] | [latex]A=\frac{1}{2}bh[/latex] |
Substitute any given values. | [latex]90=\frac{1}{2}\cdot{15}\cdot{h}[/latex] | |
Clear the fractions. | [latex]\color{red}{2}\cdot{90}=\color{red}{2}\cdot\frac{1}{2}\cdot{15}\cdot{h}[/latex] | [latex]\color{red}{2}\cdot{A}=\color{red}{2}\cdot\frac{1}{2}\cdot{b}\cdot{h}[/latex] |
Simplify. | [latex]180=15h[/latex] | [latex]2A=bh[/latex] |
Solve for h. | [latex]12=h[/latex] | [latex]\frac{2A}{b}=h[/latex] |
example
Solve the formula [latex]I=Prt[/latex] to find the principal, [latex]P\text{:}[/latex]- When [latex]I=\text{\$5,600},r=\text{4%},t=7\text{years}[/latex]
- In general
Answer:
Solution:1. I = $5600, r = 4%, t = 7 years | 2. In general | |
Write the forumla. | [latex]I=Prt[/latex] | [latex]I=Prt[/latex] |
Substitute any given values. | [latex]5600=P(0.04)(7)[/latex] | [latex]I=Prt[/latex] |
Multiply r ⋅ t. | [latex]5600=P(0.28)[/latex] | [latex]I=P(rt)[/latex] |
Divide to isolate P. | [latex]\frac{5600}{\color{red}{0.28}}=\frac{P(0.28)}{\color{red}{0.28}}[/latex] | [latex]\frac{I}{\color{red}{rt}}=\frac{P(rt)}{\color{red}{rt}}[/latex] |
Simplify. | [latex]20,000=P[/latex] | [latex]\frac{I}{rt}=P[/latex] |
State the answer. | The principal is $20,000. | [latex]P=\frac{I}{rt}[/latex] |
example
Solve the formula [latex]3x+2y=18[/latex] for [latex]y\text{:}[/latex]- When [latex]x=4[/latex]
- In general
Answer:
Solution:1. When x = 4 | 2. In general | |
Write the equation. | [latex]3x+2y=18[/latex] | [latex]3x+2y=18[/latex] |
Substitute any given values. | [latex]3(4)+2y=18[/latex] | [latex]3x+2y=18[/latex] |
Simplify if possible. | [latex]12+2y=18[/latex] | [latex]3x+2y=18[/latex] |
Subtract to isolate the y-term. | [latex]12\color{red}{-12}+2y=18\color{red}{-12}[/latex] | [latex]3x\color{red}{-3x}+2y=18\color{red}{-3x}[/latex] |
Simplify. | [latex]2y=6[/latex] | [latex]2y=18-3x[/latex] |
Divide. | [latex]\frac{2y}{\color{red}{2}}=\frac{6}{\color{red}{2}}[/latex] | [latex]\frac{2y}{\color{red}{2}}=\frac{18-3x}{\color{red}{2}}[/latex] |
Simplify. | [latex]y=3[/latex] | [latex]y=\frac{18-3x}{2}[/latex] |
example
Solve the formula [latex]P=a+b+c[/latex] for [latex]a[/latex].Answer:
Solution: We will isolate [latex]a[/latex] on one side of the equation.We will isolate a on one side of the equation. | ||
Write the equation. | [latex]P=a+b+c[/latex] | |
Subtract b and c from both sides to isolate a. | [latex]P\color{red}{-b-c}=a+b+c\color{red}{-b-c}[/latex] | |
Simplify. | [latex]P-b-c=a[/latex] |
example
Solve the equation [latex]3x+y=10[/latex] for [latex]y[/latex].Answer:
Solution We will isolate [latex]y[/latex] on one side of the equation.We will isolate y on one side of the equation. | ||
Write the equation. | [latex]3x+y=10[/latex] | |
Subtract 3x from both sides to isolate y. | [latex]3x\color{red}{-3x}+y=10\color{red}{-3x}[/latex] | |
Simplify. | [latex]y=10 - 3x[/latex] |
example
Solve the equation [latex]6x+5y=13[/latex] for [latex]y[/latex].Answer:
Solution: We will isolate [latex]y[/latex] on one side of the equation.We will isolate y on one side of the equation. | |
Write the equation. | [latex]6x+5y=13[/latex] |
Subtract to isolate the term with y. | [latex]6x+5y\color{red}{-6x}=13\color{red}{-6x}[/latex] |
Simplify. | [latex]5y=13-6x[/latex] |
Divide by 5 to make the coefficient 1. | [latex]\frac{5y}{\color{red}{5}}=\frac{13-6x}{\color{red}{5}}[/latex] |
Simplify. | [latex]y=\frac{13-6x}{5}[/latex] |
Licenses & Attributions
CC licensed content, Shared previously
- Find the Base of a Triangle Given Area / Literal Equation. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Literal Equations: Solve ax-by=c for y. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Question ID 142912, 142894, 142892, 145640, 142895, 145635. Authored by: Lumen Learning. License: CC BY: Attribution. License terms: IMathAS Community License, CC-BY + GPL.
- Question ID 142895. Authored by: Sousa James. License: CC BY: Attribution. License terms: IMathAS Community License, CC-BY + GPL.
CC licensed content, Specific attribution
- Prealgebra. Provided by: OpenStax License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].