Non Linear Equations in One Variable and System of Equations in Two Variables Pattern - Rearranging Formulas

Rearranging Formulas

This pattern gives you a formula and asks you to isolate a specific variable. The operations are the same as solving a regular equation — add, subtract, multiply, divide, take a square root — but you're working with letters instead of numbers. The answer is always an expression, not a single number.

The Core Strategy

Treat the target variable like the unknown and everything else like constants. Undo operations in reverse order: if the variable was multiplied then added to, you subtract first, then divide.

Worked Examples

Example 1. $p = q + 5k$. Solve for $k$.

A) $k = 5(p - q)$
B) $k = \dfrac{p}{5} - q$
C) $k = \dfrac{p - q}{5}$
D) $k = \dfrac{p + q}{5}$

Subtract $q$: $p - q = 5k$
Divide by $5$: $k = \dfrac{p - q}{5}$
Gotcha: Option A multiplies by $5$ instead of dividing. Option B only divides the $p$ term by $5$ but doesn't divide $q$ — when you divide both sides, the entire expression $(p - q)$ is divided.
The answer is C.

Example 2. $A = 3P - C$. Solve for $P$.

A) $P = \dfrac{A - C}{3}$
B) $P = \dfrac{A + C}{3}$
C) $P = 3(A + C)$
D) $P = \dfrac{A}{3} + C$

Add $C$: $A + C = 3P$
Divide by $3$: $P = \dfrac{A + C}{3}$
Gotcha: Option A subtracts $C$ instead of adding it. The original has $-C$, so to move it you add $C$. Option D divides only $A$ by $3$ but leaves $C$ undivided.
The answer is B.

Example 3. $P = \dfrac{Q}{25R}$. Solve for $Q$.

A) $Q = \dfrac{25P}{R}$
B) $Q = P - 25R$
C) $Q = 25PR$
D) $Q = \dfrac{R}{25P}$

$Q$ is divided by $25R$, so multiply both sides by $25R$:
$Q = 25PR$
Gotcha: Option A divides by $R$ instead of multiplying. Option B subtracts — but the original operation is division, so the inverse is multiplication. Always match the inverse operation correctly.
The answer is C.

Example 4. $\dfrac{30a}{6b} = 10\sqrt{c + 1}$. Solve for $c$.

A) $c = \dfrac{a}{2b} - 1$
B) $c = \left(\dfrac{a}{2b}\right)^2 - 1$
C) $c = \left(\dfrac{5a}{b}\right)^2 - 1$
D) $c = \left(\dfrac{2a}{b}\right)^2 - 1$

Step 1: Simplify $\dfrac{30a}{6b} = \dfrac{5a}{b}$
Step 2: $\dfrac{5a}{b} = 10\sqrt{c+1}$
Step 3: Divide by $10$: $\dfrac{a}{2b} = \sqrt{c+1}$
Step 4: Square both sides: $\left(\dfrac{a}{2b}\right)^2 = c + 1$
Step 5: Subtract $1$: $c = \left(\dfrac{a}{2b}\right)^2 - 1$
Gotcha: Option A forgets to square. Option C doesn't divide by $10$. When a square root is involved, you must square both sides to eliminate it — but only after isolating the root first.
The answer is B.

Example 5. $V = \dfrac{4}{3}\pi r^3$. Solve for $r$.

Multiply by $\dfrac{3}{4\pi}$: $r^3 = \dfrac{3V}{4\pi}$
Take the cube root: $r = \sqrt[3]{\dfrac{3V}{4\pi}}$
Key principle: Undo operations in reverse order — the last thing done to $r$ was cubing, so the last thing you undo is taking the cube root.

Example 6. $T = 2V - 8W$. Solve for $W$.

Subtract $2V$: $T - 2V = -8W$
Divide by $-8$: $W = \dfrac{T - 2V}{-8} = \dfrac{2V - T}{8}$
Gotcha: Dividing by a negative flips the sign. $\dfrac{T - 2V}{-8}$ is the same as $\dfrac{2V - T}{8}$. Either form is correct, but the answer choices might show only one.

What to Do on Test Day

• Undo in reverse order. If the formula applied operations in the order "multiply, then add," undo by subtracting first, then dividing.
• Divide the whole expression. When dividing both sides by a number, every term gets divided. $\dfrac{A + C}{3}$ is correct; $\dfrac{A}{3} + C$ is wrong.
• Square roots → square both sides. But isolate the root first. Don't square a side that has extra terms outside the root.
• Negative coefficients: When isolating from $-8W$, dividing by $-8$ flips the sign of everything.
• Simplify fractions early. Before solving, simplify any coefficients: $\dfrac{30}{6} = 5$ makes everything cleaner.
• Check your answer. Substitute back into the original formula mentally. If $k = \dfrac{p-q}{5}$, then $5k = p - q$, so $q + 5k = p$ ✓.