Math Core

The formula $d = rt$ tells you distance when you know rate and time. But what if you know the distance and the rate and need the time? You could plug in numbers and solve every single time, or you could solve the formula once for $t$ and have a new formula ready to use. That's what this lesson is about.

Equations with more than one letter

Definition

Literal equation

A literal equation is an equation with two or more variables. A formula is a literal equation that describes a real relationship, such as $A = \ell w$ or $d = rt$ . To solve for a variable means to rewrite the equation so that variable is alone on one side.

Here's the key fact: solving for a letter uses exactly the same steps as solving for $x$ in a number equation. The other letters simply stand in for numbers you haven't been told yet. Treat them the way you'd treat a $5$ or a $-3$ .

A good way to see this is to solve a number version and a letter version side by side.

number equation	literal equation
$3x + 7 = 19$	$ax + b = c$
$3x = 19 - 7$	$ax = c - b$
$x = \dfrac{19 - 7}{3}$	$x = \dfrac{c - b}{a}$

Each step matches. The only difference is that on the right you can't finish the arithmetic, so the answer stays as an expression.

Solving for a variable

Circle or highlight the variable you're solving for. Every other letter acts like a number.
Undo the operations on that variable in reverse order, doing the same thing to both sides.
Stop when the chosen variable is alone. The answer is an expression in the other letters.

Solving formulas

Worked example: Distance, rate and time

Solve $d = rt$ for $t$ .

The $t$ is multiplied by $r$ . Undo that by dividing both sides by $r$ :

\frac{d}{r} = \frac{rt}{r} \quad\Longrightarrow\quad t = \frac{d}{r}.

Now a trip of $d = 150$ miles at $r = 60$ miles per hour takes $t = \dfrac{150}{60} = 2.5$ hours.

Worked example: Perimeter of a rectangle

Solve $P = 2\ell + 2w$ for $w$ .

Treat $2\ell$ like a constant term. It's added to $2w$ , so subtract it first, then divide by $2$ .

\begin{aligned} P &= 2\ell + 2w \\ P - 2\ell &= 2w && \text{subtract } 2\ell \\ \frac{P - 2\ell}{2} &= w && \text{divide by } 2 \end{aligned}

So $w = \dfrac{P - 2\ell}{2}$ . Check with numbers: a rectangle with $\ell = 7$ and $w = 3$ has $P = 20$ . The formula gives $w = \dfrac{20 - 14}{2} = 3$ . ✓

Common mistake

When you divide, divide the whole side. From $P - 2\ell = 2w$ , the result is $w = \dfrac{P - 2\ell}{2}$ , not $w = P - \dfrac{2\ell}{2}$ (which would be $P - \ell$ ). Every term on that side gets divided by $2$ . An equivalent correct answer is $w = \dfrac{P}{2} - \ell$ .

Solving for y

In later units you'll often rewrite equations of lines so that $y$ is alone. This is a literal equation too.

Worked example: Getting y by itself

Solve $3x + 2y = 12$ for $y$ .

\begin{aligned} 3x + 2y &= 12 \\ 2y &= 12 - 3x && \text{subtract } 3x \\ y &= \frac{12 - 3x}{2} && \text{divide by } 2 \\ y &= 6 - \frac{3}{2}x && \text{divide each term by } 2 \end{aligned}

Both of the last two lines are correct. The form $y = -\dfrac{3}{2}x + 6$ (the same thing, reordered) is the one you'll use when you graph lines.

Temperature conversion

Worked example: Celsius to Fahrenheit

The formula $C = \dfrac{5}{9}(F - 32)$ converts Fahrenheit to Celsius. Solve it for $F$ to get a formula that goes the other way.

The last thing done to $F$ is multiplying by $\dfrac{5}{9}$ , so undo that first with the reciprocal.

\begin{aligned} C &= \frac{5}{9}(F - 32) \\ \frac{9}{5}C &= F - 32 && \text{multiply by } \tfrac{9}{5} \\ \frac{9}{5}C + 32 &= F && \text{add } 32 \end{aligned}

So $F = \dfrac{9}{5}C + 32$ . Test it: water boils at $100^\circ\text{C}$ , and $\dfrac{9}{5}(100) + 32 = 180 + 32 = 212^\circ\text{F}$ . ✓

Tip

Check a rearranged formula with easy numbers. Pick values that make the original equation true, then see whether your new formula gives back the right value. It's the literal-equation version of substituting to check a solution.

One caution: dividing by a letter is only allowed when that letter isn't zero. Writing $t = \dfrac{d}{r}$ assumes $r \ne 0$ , which makes sense, since a rate of zero would never cover any distance.