Literal Equations and Formulas

A literal equation is an equation with two or more variables. Formulas you already know — like $d = rt$ for distance, or $A = lw$ for area — are literal equations. Solving a literal equation means rearranging it to isolate one specific variable, treating all other variables as if they were numbers. The algebra is identical to what you use for regular equations; the only difference is that your answer contains variables instead of a single number.

This skill is essential in science, trades, and finance, where you constantly need the same formula rearranged in different ways depending on what information you have.

The Core Idea: Treat Other Variables as Constants

When solving for a specific variable, pretend every other letter is just a number. Then use the same inverse operations you would with any linear equation.

Example 1: Solve $d = rt$ for $t$

You want $t$ alone. Right now $t$ is multiplied by $r$, so divide both sides by $r$:

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

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

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

This tells you: if you know the distance and the rate, divide to find the time.

Example 2: Solve $A = \frac{1}{2}bh$ for $h$

You want $h$ alone. Start by eliminating the fraction — multiply both sides by 2:

$2A = bh$

Now divide both sides by $b$:

$h = \frac{2A}{b}$

Answer: $h = \frac{2A}{b}$

Rearranging Common Formulas

Let’s practice with formulas that appear across many fields.

Perimeter of a Rectangle

The formula is $P = 2l + 2w$. Solve for $w$.

Step 1 — Subtract $2l$ from both sides:

$P - 2l = 2w$

Step 2 — Divide both sides by 2:

$w = \frac{P - 2l}{2}$

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

Simple Interest

The formula is $I = Prt$ where $I$ is interest, $P$ is principal, $r$ is rate, and $t$ is time. Solve for $r$.

Step 1 — Divide both sides by $Pt$:

$r = \frac{I}{Pt}$

Answer: $r = \frac{I}{Pt}$

Temperature Conversion

The formula is $F = \frac{9}{5}C + 32$. Solve for $C$.

Step 1 — Subtract 32 from both sides:

$F - 32 = \frac{9}{5}C$

Step 2 — Multiply both sides by $\frac{5}{9}$:

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

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

Circumference of a Circle

The formula is $C = 2\pi r$. Solve for $r$.

$r = \frac{C}{2\pi}$

Answer: $r = \frac{C}{2\pi}$

Formulas with the Target Variable in Multiple Terms

When the variable you are solving for appears in more than one term, you need to factor it out.

Example 3: Solve $ax + bx = c$ for $x$

The variable $x$ appears in both terms on the left. Factor it out:

$x(a + b) = c$

Divide both sides by $(a + b)$:

$x = \frac{c}{a + b}$

Answer: $x = \frac{c}{a + b}$

Example 4: Solve $3x - bx = 12$ for $x$

Factor out $x$:

$x(3 - b) = 12$

Divide by $(3 - b)$:

$x = \frac{12}{3 - b}$

Answer: $x = \frac{12}{3 - b}$ (where $b \neq 3$)

Example 5: Solve $S = \frac{a}{1 - r}$ for $r$

This is the formula for the sum of an infinite geometric series.

Step 1 — Multiply both sides by $(1 - r)$:

$S(1 - r) = a$

Step 2 — Distribute:

$S - Sr = a$

Step 3 — Subtract $S$ from both sides:

$-Sr = a - S$

Step 4 — Divide by $-S$:

$r = \frac{a - S}{-S} = \frac{S - a}{S}$

Answer: $r = \frac{S - a}{S}$

Real-World Application: Electrician — Ohm’s Law Rearrangements

Ohm’s Law is one of the most important formulas in electrical work:

$V = IR$

where $V$ = voltage (volts), $I$ = current (amps), $R$ = resistance (ohms).

Electricians constantly rearrange this depending on what they need to find.

Solve for $I$ (current):

$I = \frac{V}{R}$

Solve for $R$ (resistance):

$R = \frac{V}{I}$

Applied problem: An electrician is troubleshooting a circuit. The voltage source is 240 V and the circuit has two resistors in series with total resistance $R_1 + R_2$. The measured current is 10 amps. What is the total resistance?

$R = \frac{V}{I} = \frac{240}{10} = 24 \text{ ohms}$

If the electrician knows $R_1 = 14$ ohms, what is $R_2$?

Using $R_1 + R_2 = 24$:

$R_2 = 24 - 14 = 10 \text{ ohms}$

Answer: The second resistor has a resistance of 10 ohms.

Real-World Application: HVAC — BTU Calculations

HVAC technicians use the sensible heat formula:

$Q = 1.08 \times \text{CFM} \times \Delta T$

where $Q$ = heat (BTU/hr), CFM = cubic feet per minute of airflow, and $\Delta T$ = temperature difference (°F).

Solve for CFM:

$\text{CFM} = \frac{Q}{1.08 \times \Delta T}$

Problem: A system must deliver 32,400 BTU/hr of cooling. The desired temperature drop is 25°F. What airflow is needed?

$\text{CFM} = \frac{32{,}400}{1.08 \times 25} = \frac{32{,}400}{27} = 1{,}200$

Answer: The system needs 1,200 CFM of airflow. The technician uses this to select the correct duct size and fan speed.

Real-World Application: Nursing — Drip Rate Formula

Nurses use the IV drip rate formula:

$\text{Rate} = \frac{V \times D}{T}$

where $V$ = total volume (mL), $D$ = drop factor (drops/mL), and $T$ = time (minutes).

Solve for $T$ (infusion time):

$T = \frac{V \times D}{\text{Rate}}$

Problem: A nurse must infuse 500 mL using a drop factor of 15 drops/mL. The prescribed drip rate is 25 drops per minute. How long will the infusion take?

$T = \frac{500 \times 15}{25} = \frac{7{,}500}{25} = 300 \text{ minutes} = 5 \text{ hours}$

Answer: The infusion will take 5 hours. The nurse documents the expected completion time in the patient’s chart.

Common Mistakes to Avoid

1. Treating the target variable like a number. Remember that you are isolating a variable, so your final answer will contain other variables — that is correct.
2. Forgetting to factor when the target variable appears in multiple terms. If $x$ shows up in two places, you must collect those terms and factor before dividing.
3. Dividing by an expression that could be zero. When you write $x = \frac{c}{a + b}$, note that this requires $a + b \neq 0$.
4. Incorrectly handling negative signs during rearrangement. When you move $-Sr$ across the equals sign, it becomes $+Sr$ — be systematic with sign changes.
5. Stopping before fully isolating the variable. The answer to “solve for $h$” must have $h$ completely alone on one side: $h = \ldots$, not $2h = \ldots$.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• A literal equation has two or more variables — formulas like $d = rt$ and $V = IR$ are literal equations
• To solve for a specific variable, treat all other variables as constants and use the same inverse operations as regular equations
• When the target variable appears in multiple terms, collect those terms on one side and factor the variable out before dividing
• Common formulas from geometry, physics, finance, and trades all follow the same rearrangement process
• Always state restrictions — if you divide by an expression, note that it cannot equal zero
• Rearranging formulas is a core skill in every trade and profession that uses math

Return to Algebra for more topics in this section.