Lesson 7: Solve a Formula for a Specific Variable

New Concept

This lesson teaches you how to rearrange formulas, like the distance, rate, and time equation ($d=rt$). By isolating any variable, such as solving for time ($t = d/r$), you can efficiently find missing values in many real-world applications.

What’s next

You've got the big picture! Next, you'll tackle interactive examples that show how to isolate variables, followed by a series of practice problems.

Property

For an object moving in 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.

Examples

• A cyclist travels at a steady rate of 15 miles per hour for 4 hours. The total distance is $d = 15 \cdot 4 = 60$ miles.

• A bus covers a distance of 120 miles in 2 hours. Its average rate of speed is $r = \frac{120}{2} = 60$ miles per hour.

Property

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. This process is also called solving a literal equation.

Examples

• To solve the perimeter formula for a rectangle, $P = 2L + 2W$, for $L$, first subtract $2W$ to get $P - 2W = 2L$. Then divide by 2: $L = \frac{P - 2W}{2}$.

• To solve the simple interest formula $I = Prt$ for the principal $P$, you divide both sides by the product of rate and time: $P = \frac{I}{rt}$.

Property

The formula for the area of a triangle is $A = \frac{1}{2}bh$. To solve this for one of its variables, such as the height $h$, you can apply inverse operations.

Examples

• To solve $A = \frac{1}{2}bh$ for $h$, first multiply both sides by 2 to get $2A = bh$. Then, divide by $b$ to isolate $h$: $h = \frac{2A}{b}$.

• If a triangle has an area of 90 square units and a base of 15 units, its height is $h = \frac{2 \cdot 90}{15} = \frac{180}{15} = 12$ units.

Property

To solve a linear equation like $3x + 2y = 18$ for the variable $y$, you must isolate the $y$ term on one side of the equation and then make its coefficient equal to 1.

Examples

• To solve $3x + y = 10$ for $y$, subtract $3x$ from both sides, which gives you $y = 10 - 3x$.

• To solve $6x + 5y = 13$ for $y$, first subtract $6x$ to get $5y = 13 - 6x$. Then divide everything by 5: $y = \frac{13 - 6x}{5}$.