Solve a Formula for a Specific Variable

Use the distance, rate, and time formula

One formula you’ll use often in algebra and in everyday life is the relationship between distance, rate, and time.

Example. Jamal rides his bike at a steady rate of $12$ miles per hour for $3\tfrac{1}{2}$ hours. How much distance has he traveled?

Substituting $r = 12$ and $t = 3\tfrac{1}{2}$ into $d = rt$:

$d = 12 \cdot 3\tfrac{1}{2} \qquad\Rightarrow\qquad d = 12 \cdot \tfrac{7}{2} \qquad\Rightarrow\qquad d = 42 \text{ miles}$

Jamal rode $42$ miles.

Example. Rey drives $520$ miles, at a steady rate of $65$ miles per hour. How many hours did his trip take?

Substituting $d = 520$ and $r = 65$ into $d = rt$, and then solving for $t$:

$520 = 65t \qquad\Rightarrow\qquad \frac{520}{65} = \frac{65t}{65} \qquad\Rightarrow\qquad 8 = t$

Rey’s trip took $8$ hours.

Solve a formula for a specific variable

In the distance, rate, and time examples, we substituted numbers for some of the variables in the formula before we solved for the unknown. Sometimes, though, it’s more useful to solve a formula for one specific variable in general, without substituting any numbers at all. This is called solving a formula for a specific variable, and it works the same way as solving any other equation — the object is still to isolate the variable we’re solving for, treating the other variables as if they were numbers.

Example. Solve the formula $d = rt$ for $t$: (a) when $d = 520$ and $r = 65$, and (b) in general.

(a) Substituting the given numbers and dividing both sides by $65$:

$520 = 65t \qquad\Rightarrow\qquad \frac{520}{65} = \frac{65t}{65} \qquad\Rightarrow\qquad 8 = t$

(b) This time, there are no numbers to substitute for $d$ and $r$. We divide both sides by $r$ to isolate $t$, exactly as we divided by $65$ above:

$d = rt \qquad\Rightarrow\qquad \frac{d}{r} = \frac{rt}{r} \qquad\Rightarrow\qquad \frac{d}{r} = t$

We now have an equation that solves $d = rt$ for $t$ in general: $t = \tfrac{d}{r}$. Whenever we know values for $d$ and $r$, we can substitute them into this equation to find $t$.

Example. Solve the formula $A = \tfrac{1}{2}bh$ for $h$: (a) when $A = 90$ and $b = 15$, and (b) in general.

(a) Substituting the given numbers:

$90 = \tfrac{1}{2}(15)h \qquad\Rightarrow\qquad 90 = 7.5h \qquad\Rightarrow\qquad 12 = h$

(b) Following the same steps, without any substitutions:

$A = \tfrac{1}{2}bh \qquad\Rightarrow\qquad 2A = 2 \cdot \tfrac{1}{2}bh \qquad\Rightarrow\qquad 2A = bh \qquad\Rightarrow\qquad \frac{2A}{b} = \frac{bh}{b} \qquad\Rightarrow\qquad \frac{2A}{b} = h$

So $h = \tfrac{2A}{b}$.

Example. Solve the formula $I = Prt$ for $P$: (a) when $I = \text{\textdollar}5{,}600$, $r = 4\%$, $t = 7$ years, and (b) in general.

(a) Substituting $r = 0.04$:

$5600 = P(0.04)(7) \qquad\Rightarrow\qquad 5600 = 0.28P \qquad\Rightarrow\qquad 20{,}000 = P$

(b) Following the same steps, without any substitutions, dividing both sides by $rt$:

$I = Prt \qquad\Rightarrow\qquad \frac{I}{rt} = \frac{Prt}{rt} \qquad\Rightarrow\qquad \frac{I}{rt} = P$

So $P = \tfrac{I}{rt}$.

Formulas can also involve two variables that play a similar role, like $x$ and $y$. We solve for one of them the same way — by isolating it on one side of the equation, treating the other variable like a number.

Example. Solve the formula $3x + 2y = 18$ for $y$: (a) when $x = 4$, and (b) in general.

(a) Substituting $x = 4$:

$3(4) + 2y = 18 \qquad\Rightarrow\qquad 12 + 2y = 18 \qquad\Rightarrow\qquad 2y = 6 \qquad\Rightarrow\qquad y = 3$

(b) Following the same steps, without substituting a number for $x$: we subtract $3x$ from both sides, then divide by $2$.

$3x + 2y = 18 \qquad\Rightarrow\qquad 2y = 18 - 3x \qquad\Rightarrow\qquad y = \frac{18 - 3x}{2}$

So $y = \tfrac{18 - 3x}{2}$.

Example. Solve the formula $P = a + b + c$ for $a$.

We want to isolate $a$, so we subtract $b$ and $c$ from both sides:

$P = a + b + c \qquad\Rightarrow\qquad P - b - c = a$

So $a = P - b - c$.

Example. Solve the equation $3x + y = 10$ for $y$.

We isolate $y$ on one side of the equation by subtracting $3x$ from both sides:

$3x + y = 10 \qquad\Rightarrow\qquad 3x - 3x + y = 10 - 3x \qquad\Rightarrow\qquad y = 10 - 3x$

Example. Solve the equation $6x + 5y = 13$ for $y$.

This time, the coefficient of $y$ isn’t $1$, so after isolating the term with $y$, we also need to divide by that coefficient:

$6x + 5y = 13 \qquad\Rightarrow\qquad 5y = 13 - 6x \qquad\Rightarrow\qquad \frac{5y}{5} = \frac{13 - 6x}{5} \qquad\Rightarrow\qquad y = \frac{13 - 6x}{5}$

Key terms

distance, rate, time formula — the relationship $d = rt$, where $d$ is distance traveled, $r$ is the rate (speed), and $t$ is time. solve a formula for a specific variable — to isolate that variable on one side of the equation, treating the other variables as if they were numbers; also called solving a literal equation.

This section is adapted from Prealgebra 2e, Section 9.7: Solve a Formula for a Specific Variable by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the Be Prepared quiz, Links to Literacy and Media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.