Solving Literal Equations

Most of the equations you've solved so far have had numbers and one variable. A literal equation is different—it's an equation with multiple variables, usually representing a formula. When you solve a literal equation, you're not finding a numerical answer. Instead, you're isolating one specific variable in terms of the others.

Why does this matter? Because formulas are literal equations, and you often need to rearrange them to solve for different variables depending on what information you have.

What's a Literal Equation?

Here are some examples of literal equations you might recognize:

The distance formula: $d = rt$ (distance equals rate times time)

The area of a rectangle: $A = lw$ (area equals length times width)

The perimeter of a rectangle: $P = 2l + 2w$

The formula for Celsius to Fahrenheit: $F = \frac{9}{5}C + 32$

Each of these formulas has multiple variables. Solving a literal equation means isolating one variable on one side while leaving the other variables on the other side.

A Simple Example

Let's start with the distance formula.

Example: Solve $d = rt$ for $t$.

We want to isolate $t$, which means getting $t$ by itself on one side. Right now, $t$ is being multiplied by $r$. To undo multiplication, we divide.

Divide both sides by $r$: $$\frac{d}{r} = \frac{rt}{r}$$

The $r$ cancels on the right: $$\frac{d}{r} = t$$

Or, written more conventionally: $$t = \frac{d}{r}$$

Now if you know the distance and the rate, you can find the time by dividing distance by rate. This is the same formula, just rearranged.

Example: Solve $d = rt$ for $r$.

This time we want to isolate $r$. It's being multiplied by $t$, so we divide both sides by $t$: $$\frac{d}{t} = r$$

Or: $r = \frac{d}{t}$

If you know distance and time, you can find rate by dividing distance by time.

Formulas with Addition or Subtraction

Example: Solve $P = 2l + 2w$ for $l$.

We want $l$ by itself. First, let's get rid of the $2w$ term by subtracting $2w$ from both sides: $$P - 2w = 2l$$

Now $l$ is being multiplied by 2, so divide both sides by 2: $$\frac{P - 2w}{2} = l$$

Written conventionally: $$l = \frac{P - 2w}{2}$$

Notice that the entire left side got divided by 2. You can't just divide $P$ by 2 and leave $-2w$ alone.

Example: Solve $P = 2l + 2w$ for $w$.

Subtract $2l$ from both sides: $$P - 2l = 2w$$

Divide by 2: $$w = \frac{P - 2l}{2}$$

The Area and Perimeter of Basic Shapes

Understanding how to manipulate these formulas is useful for geometry problems.

Example: The area of a triangle is $A = \frac{1}{2}bh$. Solve for $h$.

Multiply both sides by 2 to get rid of the fraction: $$2A = bh$$

Now divide by $b$: $$h = \frac{2A}{b}$$

So if you know the area and the base of a triangle, you can find the height.

Example: The area of a trapezoid is $A = \frac{1}{2}(b_1 + b_2)h$. Solve for $h$.

Multiply by 2: $$2A = (b_1 + b_2)h$$

Divide by $(b_1 + b_2)$: $$h = \frac{2A}{b_1 + b_2}$$

Notice we divided by the entire sum $(b_1 + b_2)$, not just by one of the bases.

Temperature Conversion

The formula to convert from Celsius to Fahrenheit is $F = \frac{9}{5}C + 32$. What if you want to convert from Fahrenheit to Celsius?

Example: Solve $F = \frac{9}{5}C + 32$ for $C$.

First, subtract 32 from both sides: $$F - 32 = \frac{9}{5}C$$

Now $C$ is being multiplied by $\frac{9}{5}$. To undo this, multiply by the reciprocal, which is $\frac{5}{9}$: $$\frac{5}{9}(F - 32) = C$$

Or: $$C = \frac{5}{9}(F - 32)$$

This is the Fahrenheit-to-Celsius conversion formula. If you know the temperature is 68°F, you can plug it in: $C = \frac{5}{9}(68 - 32) = \frac{5}{9}(36) = 20$°C.

Working with Fractions

When a variable is in the denominator of a fraction, you need to be careful.

Example: Solve $d = \frac{m}{V}$ for $V$.

Here, $V$ is in the denominator. To get it out, multiply both sides by $V$: $$dV = m$$

Now divide by $d$: $$V = \frac{m}{d}$$

Example: The formula for average speed is $s = \frac{d}{t}$. Solve for $t$.

Multiply both sides by $t$: $$st = d$$

Divide by $s$: $$t = \frac{d}{s}$$

A More Complex Example

Example: Solve $A = P(1 + rt)$ for $r$.

This is a formula from finance (simple interest). We want to isolate $r$.

First, divide both sides by $P$: $$\frac{A}{P} = 1 + rt$$

Subtract 1 from both sides: $$\frac{A}{P} - 1 = rt$$

Divide by $t$: $$r = \frac{\frac{A}{P} - 1}{t}$$

You could also write this as: $$r = \frac{A - P}{Pt}$$

Both forms are correct, though the second is probably cleaner.

Practice Problems

Solve $y = mx + b$ for $x$
Solve $C = 2\pi r$ for $r$
Solve $A = \frac{1}{2}h(b_1 + b_2)$ for $b_1$
Solve $V = lwh$ for $h$
Solve $ax + b = c$ for $x$
Solve $I = Prt$ for $P$

Solutions:

Subtract $b$: $y - b = mx$, then divide by $m$: $x = \frac{y - b}{m}$
Divide by $2\pi$: $r = \frac{C}{2\pi}$
Multiply by 2: $2A = h(b_1 + b_2)$, divide by $h$: $\frac{2A}{h} = b_1 + b_2$, subtract $b_2$: $b_1 = \frac{2A}{h} - b_2$
Divide by $lw$: $h = \frac{V}{lw}$
Subtract $b$: $ax = c - b$, divide by $a$: $x = \frac{c - b}{a}$
Divide by $rt$: $P = \frac{I}{rt}$

Common Mistakes

Trying to isolate two variables at once doesn't work. If asked to solve for $x$, you'll have $x$ alone on one side and everything else (including other variables) on the other side.

When dividing by an expression like $(a + b)$, you must divide the entire other side by that expression. You can't split it up and divide different parts by $a$ and $b$ separately.

Forgetting to use the reciprocal when dividing by a fraction leads to errors. If $x$ is multiplied by $\frac{3}{4}$, you multiply by $\frac{4}{3}$ to isolate it.

Not checking your work is risky. Pick some simple numbers for the other variables and make sure your rearranged formula gives the same result as the original.

Why This Matters

Being able to rearrange formulas is incredibly useful in science, engineering, finance, and everyday life. You might memorize $d = rt$, but what if you're trying to figure out how long a trip will take? You need $t = \frac{d}{r}$. What if you want to know what temperature Fahrenheit corresponds to 25°C? You need the rearranged formula.

Learning to manipulate literal equations gives you flexibility. Instead of memorizing dozens of variations of formulas, you can derive what you need from the basic form.