Lesson 2.6: Solve a Formula for a Specific Variable

New Concept

Learn to rearrange any formula to solve for a specific variable. We'll start with the familiar distance formula, $d=rt$, and then apply the same algebraic steps to isolate any variable you need in various equations.

What’s next

You'll start with worked examples using the distance formula, $d=rt$. Soon, you will apply this skill in a series of interactive practice cards.

Property

For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula:

where $d$ = distance, $r$ = rate, and $t$ = time. To solve problems with this formula, first identify the knowns and unknowns, substitute the values into the formula, and then solve for the missing variable.

Examples

• Lindsay drove for 4.5 hours at 60 miles per hour. How much distance did she travel? The distance is $d = rt = 60 \cdot 4.5 = 270$ miles.
• Lee wants to drive 600 miles. If he drives at a steady rate of 75 miles per hour, how many hours will the trip take? The time is $t = \frac{d}{r} = \frac{600}{75} = 8$ hours.
• A train trip from Boston to Washington D.C. is 440 miles and takes 5 hours. What was the average speed of the train? The rate is $r = \frac{d}{t} = \frac{440}{5} = 88$ miles per hour.

Explanation

This formula connects speed, time, and distance. Think of it this way: the distance you cover is a product of how fast you travel (rate) and for how long (time). It is a fundamental principle used to describe motion.

Property

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. For example, to solve $d = rt$ for $t$, we divide both sides by $r$ to get $t = \frac{d}{r}$.

Examples

• Solve the formula $d=rt$ for $r$. To isolate $r$, we divide both sides by $t$. The new formula is $r = \frac{d}{t}$.
• Solve the formula for the perimeter of a triangle, $P = a + b + c$, for side $b$. We subtract $a$ and $c$ from both sides, so $b = P - a - c$.
• Solve the formula for volume, $V = LWH$, for the width $W$. We divide both sides by $L$ and $H$. The result is $W = \frac{V}{LH}$.

Explanation

Rearranging a formula lets you find any one piece of information if you know the others. It’s like rewriting a recipe to figure out how much flour you need based on the number of cookies you want to bake.

Property

When solving a formula that includes a fraction, like the area of a triangle $A = \frac{1}{2}bh$, you can first clear the fraction by multiplying both sides of the equation by the denominator. Then, isolate the desired variable. To solve for $h$, we get $2A = bh$, which simplifies to $h = \frac{2A}{b}$.

Examples

• Solve the formula $A = \frac{1}{2}bh$ for $b$. First, multiply by 2 to get $2A = bh$. Then, divide by $h$ to get $b = \frac{2A}{h}$.
• The formula for the average of two numbers is $M = \frac{a+b}{2}$. Solve for $a$. Multiply by 2 to get $2M = a+b$. Then subtract $b$ to get $a = 2M - b$.
• Solve the formula for the area of a trapezoid, $A = \frac{1}{2}h(b_1 + b_2)$, for $h$. Multiply by 2: $2A = h(b_1 + b_2)$. Then divide by $(b_1+b_2)$ to get $h = \frac{2A}{b_1 + b_2}$.

Explanation

To get a variable out of a fraction, first 'free' it by multiplying everything by the denominator. This clears the fraction and makes it much easier to isolate the variable you're looking for using standard algebra steps.

Property

The formula $I = Prt$ calculates simple interest, where $I$ is interest, $P$ is principal, $r$ is rate, and $t$ is time. To solve for a different variable, such as the principal ($P$), you isolate it by dividing both sides by the other variables. The formula for principal becomes $P = \frac{I}{rt}$.

Examples

• Solve $I=Prt$ for $t$. To isolate time, divide the interest by the principal and rate. The formula is $t = \frac{I}{Pr}$.
• Find the principal $P$ if the interest earned $I$ was 300 dollars, the rate $r$ was $5\% (0.05)$, and the time $t$ was 2 years. Using $P = \frac{I}{rt}$, we get $P = \frac{300}{0.05 \cdot 2} = \frac{300}{0.1} = 3000$ dollars.
• Find the rate $r$ if 10000 dollars principal ($P$) earned 2400 dollars interest ($I$) over 3 years ($t$). Using $r = \frac{I}{Pt}$, we get $r = \frac{2400}{10000 \cdot 3} = \frac{2400}{30000} = 0.08$ or $8\%$.

Explanation

By rearranging the interest formula, you can find the initial investment (principal), the interest rate, or the time. This is useful for financial planning when you know your goal but need to figure out one of the starting conditions.

Property

To solve a linear equation like $6x + 5y = 13$ for one variable, use inverse operations to isolate it. To solve for $y$, first subtract $6x$ from both sides to get $5y = 13 - 6x$. Then, divide by the coefficient of $y$, which is 5, to get $y = \frac{13 - 6x}{5}$.

Examples

• Solve the formula $4x + 3y = 9$ for $y$. Subtract $4x$ from both sides to get $3y = 9 - 4x$. Then divide by 3 to get $y = \frac{9 - 4x}{3}$.
• Solve the formula $x - y = -4$ for $y$. First, subtract $x$ to get $-y = -4 - x$. Then multiply the entire equation by $-1$ to get $y = 4 + x$.
• Solve the perimeter formula $P = 2L + 2W$ for $W$. Subtract $2L$ to get $P - 2L = 2W$. Then divide by 2 to get $W = \frac{P - 2L}{2}$.

Explanation

Think of this as getting the variable 'by itself'. First, move any terms added to or subtracted from the variable's term to the other side. Then, divide by the number multiplying your variable to finally isolate it.