Changing the Subject of Formulas
Rearranging Equations
Essential Understanding: The subject of a formula is the variable that is expressed in terms of all the others. Changing the subject means rearranging the formula to isolate a different variable. This skill is essential for solving real-world problems where you need to find different quantities.
Key Principle: Use inverse operations
Goal: Isolate the new subject
The Golden Rule
Whatever you do to one side, you must do to the other side
To change the subject, we use inverse (opposite) operations to "undo" what has been done to our target variable.
| Operation | Inverse Operation | |
|---|---|---|
| + (add) | ⟷ | − (subtract) |
| × (multiply) | ⟷ | ÷ (divide) |
| ² (square) | ⟷ | √ (square root) |
| ³ (cube) | ⟷ | ∛ (cube root) |
Step-by-Step Strategy
1
Identify the new subject (the variable you want to isolate)
2
Locate where the new subject appears in the formula
3
Remove operations one at a time using inverse operations
4
Work from the outside in (undo the last operation first)
Worked Examples
Easy
Example 1: Make \(r\) the subject of \(A = \pi r^2\)
Current subject: \(A\) New subject: \(r\)
\(A = \pi r^2\)
Start
\(\frac{A}{\pi} = r^2\)
Divide both sides by \(\pi\)
\(\sqrt{\frac{A}{\pi}} = r\)
Square root both sides
\[ r = \sqrt{\frac{A}{\pi}} \]
Medium
Example 2: Make \(r\) the subject of \(V = \frac{4}{3}\pi r^3\)
Current subject: \(V\) New subject: \(r\)
\(V = \frac{4}{3}\pi r^3\)
Start
\(\frac{3V}{4\pi} = r^3\)
Multiply by 3, divide by \(4\pi\)
\(\sqrt[3]{\frac{3V}{4\pi}} = r\)
Cube root both sides
\[ r = \sqrt[3]{\frac{3V}{4\pi}} \]
CSEC Level
Example 3: Make \(l\) the subject of \(T = 2\pi\sqrt{\frac{l}{g}}\)
Current subject: \(T\) New subject: \(l\)
This is the formula for the period of a pendulum.
\(T = 2\pi\sqrt{\frac{l}{g}}\)
Start
\(\frac{T}{2\pi} = \sqrt{\frac{l}{g}}\)
Divide both sides by \(2\pi\)
\(\left(\frac{T}{2\pi}\right)^2 = \frac{l}{g}\)
Square both sides
\(\frac{T^2}{4\pi^2} = \frac{l}{g}\)
Simplify left side
\(\frac{gT^2}{4\pi^2} = l\)
Multiply both sides by \(g\)
\[ l = \frac{gT^2}{4\pi^2} \]
CSEC Level
Example 4: Make \(M\) the subject of \(M = \sqrt{P + 2M}\)
Note: The subject appears on both sides — we need to collect terms!
\(M = \sqrt{P + 2M}\)
Start
\(M^2 = P + 2M\)
Square both sides
\(M^2 - 2M = P\)
Subtract \(2M\) from both sides
\(M^2 - 2M - P = 0\)
Rearrange to standard form
This is now a quadratic in \(M\). Use the quadratic formula:
\[ M = \frac{2 \pm \sqrt{4 + 4P}}{2} = 1 \pm \sqrt{1 + P} \]
Common Formula Types
| Formula Type | Example | Key Strategy |
|---|---|---|
| Linear | \(y = mx + c\) | Add/subtract, then multiply/divide |
| With Powers | \(E = mc^2\) | Isolate the power, then take root |
| With Roots | \(v = \sqrt{2gh}\) | Square both sides to remove root |
| With Fractions | \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) | Find common denominator or cross-multiply |
| Subject on Both Sides | \(ax + b = cx + d\) | Collect subject terms on one side, factor out |
CSEC Practice Questions
Test Your Understanding
1
Make \(x\) the subject of \(y = 3x + 7\)
Solution:
\(y = 3x + 7\)
\(y - 7 = 3x\) (subtract 7)
\(\frac{y - 7}{3} = x\) (divide by 3)
\(y = 3x + 7\)
\(y - 7 = 3x\) (subtract 7)
\(\frac{y - 7}{3} = x\) (divide by 3)
2
Make \(h\) the subject of \(V = \frac{1}{3}\pi r^2 h\)
Solution:
\(V = \frac{1}{3}\pi r^2 h\)
\(3V = \pi r^2 h\) (multiply by 3)
\(\frac{3V}{\pi r^2} = h\) (divide by \(\pi r^2\))
\(V = \frac{1}{3}\pi r^2 h\)
\(3V = \pi r^2 h\) (multiply by 3)
\(\frac{3V}{\pi r^2} = h\) (divide by \(\pi r^2\))
3
Make \(u\) the subject of \(v^2 = u^2 + 2as\)
Solution:
\(v^2 = u^2 + 2as\)
\(v^2 - 2as = u^2\) (subtract 2as)
\(\sqrt{v^2 - 2as} = u\) (square root)
\(v^2 = u^2 + 2as\)
\(v^2 - 2as = u^2\) (subtract 2as)
\(\sqrt{v^2 - 2as} = u\) (square root)
4
Make \(x\) the subject of \(y = \frac{x + 3}{x - 2}\)
Solution:
\(y = \frac{x + 3}{x - 2}\)
\(y(x - 2) = x + 3\) (cross multiply)
\(xy - 2y = x + 3\) (expand)
\(xy - x = 2y + 3\) (collect x terms)
\(x(y - 1) = 2y + 3\) (factor out x)
\(x = \frac{2y + 3}{y - 1}\) (divide)
\(y = \frac{x + 3}{x - 2}\)
\(y(x - 2) = x + 3\) (cross multiply)
\(xy - 2y = x + 3\) (expand)
\(xy - x = 2y + 3\) (collect x terms)
\(x(y - 1) = 2y + 3\) (factor out x)
\(x = \frac{2y + 3}{y - 1}\) (divide)
Key Steps to Remember
- Identify the new subject
- Use inverse operations to isolate it
- Work from outside in (undo operations in reverse order)
- If subject appears on both sides: collect and factor
- Check your answer by substituting values
CSEC Examination Tips
- Show all steps clearly — each operation earns marks
- When squaring: \((a + b)^2 \neq a^2 + b^2\) — remember to use FOIL or formula
- Watch for negative signs when subtracting from both sides
- If the subject is under a root, square both sides
- If the subject is in a denominator, flip the equation or cross-multiply