Solving Literal Equations

Most of the equations you've solved so far have had numbers and one variable. A literal equation is different: it has multiple variables and usually represents a formula. When you solve a literal equation, you're not finding a number — 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

Start with the distance formula.

Example: Solve \(d = rt\) for \(t\).

The goal is to isolate \(t\) — get it by itself on one side. Right now, \(t\) is being multiplied by \(r\). What operation undoes that? Show answerDivide both sides by \(r\): \(\frac{d}{r} = \frac{rt}{r}\), so \(t = \frac{d}{r}\). If you know distance and rate, divide to find time.

Example: Solve \(d = rt\) for \(r\).

This time isolate \(r\). It's being multiplied by \(t\), so 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 \(l\).

The goal is \(l\) alone on one side. First, 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 the division was by the entire sum \((b_1 + b_2)\), not by one of the bases on its own.

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 the simple-interest formula from finance. The goal is 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

1. Solve \(y = mx + b\) for \(x\) Show answerSubtract \(b\): \(y - b = mx\), then divide by \(m\): \(x = \frac{y - b}{m}\)
2. Solve \(C = 2\pi r\) for \(r\) Show answerDivide by \(2\pi\): \(r = \frac{C}{2\pi}\)
3. Solve \(A = \frac{1}{2}h(b_1 + b_2)\) for \(b_1\) Show answerMultiply by 2: \(2A = h(b_1 + b_2)\), divide by \(h\): \(\frac{2A}{h} = b_1 + b_2\), subtract \(b_2\): \(b_1 = \frac{2A}{h} - b_2\)
4. Solve \(V = lwh\) for \(h\) Show answerDivide by \(lw\): \(h = \frac{V}{lw}\)
5. Solve \(ax + b = c\) for \(x\) Show answerSubtract \(b\): \(ax = c - b\), divide by \(a\): \(x = \frac{c - b}{a}\)
6. Solve \(I = Prt\) for \(P\) Show answerDivide by \(rt\): \(P = \frac{I}{rt}\)

Why This Matters

Rearranging formulas is constant background work in science, engineering, finance, and everyday life. You might memorize \(d = rt\), but to figure out how long a trip will take, the form you actually want is \(t = \frac{d}{r}\). To convert 25°C to Fahrenheit, you need the rearranged temperature formula. Learning to manipulate literal equations means you don't have to memorize dozens of formula variations — you can derive whatever you need from the base form.

A few patterns to watch out for. When the variable you're isolating is grouped with others in parentheses, like \((a + b)\), you have to divide the entire other side by that group; you can't split it up. When the variable's coefficient is a fraction, multiply by the reciprocal rather than dividing by the fraction directly. And to sanity-check your work, plug simple numbers in for the other variables and confirm the rearranged form gives the same result as the original.