How to Change the Subject of a Formula
Changing the subject of a formula means rearranging an equation so a chosen variable is on its own. This is a core GCSE and IGCSE algebra skill, used in maths, science and problem solving whenever you need to make a different letter the subject. In this lesson you will learn a reliable reverse BIDMAS method, see worked examples, and then practise rearranging formulae with instant feedback.
Core ideas
Inverse Operations
Every operation has an inverse. Undo addition with subtraction, multiplication with division, squaring with a square root (write ±), and a fraction with its reciprocal.
Balancing Both Sides
Whatever you do to one side, you must do to the other. This keeps every equation equivalent throughout the rearrangement.
Fractions and Brackets
If the subject is inside a fraction, multiply both sides by the denominator first. If inside brackets, divide out or expand before isolating.
Powers and Roots
If the subject is squared, square-root both sides and write ±. If under a square root, square both sides. Be careful about negative values.
Multiple Occurrences
If the subject appears more than once, collect all such terms on one side, factorise it out, then divide. This is a common higher-tier exam question type.
Case-Sensitive Letters
Capital and lower-case letters represent different quantities. \(V\) and \(v\) are not the same variable. Take care throughout your working.
How to change the subject
The goal is to isolate the target variable on one side of the equals sign. Follow these five steps every time:
- Identify the variable you want to make the subject.
- List every operation attached to it. Read the equation like a flow chart from left to right.
- Undo each operation in reverse BIDMAS order using its inverse.
- Balance: apply every inverse operation to both sides of the equation.
- Simplify until your target variable stands alone on one side.
The "Undo" Order: Reverse BIDMAS
The last thing done to your variable is the first thing to undo.
Operations are applied in BIDMAS order when building an expression. Isolate your variable by undoing them in the exact opposite order, starting from the right.
Basic worked examples
💡 Example 1 (linear): make \(y\) the subject of \(5y + 6 = 4x\)
Step 1. Undo the \(+6\): subtract 6 from both sides.
\[ 5y = 4x - 6 \]Step 2. Undo the \(\times 5\): divide both sides by 5.
\[ y = \frac{4x - 6}{5} \]What's happening?
- Reverse BIDMAS: undo the addition/subtraction first (the \(+6\)), then undo the multiplication (the \(5y\)).
- The entire right-hand side, \(4x - 6\), is divided by 5 in one go. Do not split the fraction.
💡 Example 2 (fraction): make \(x\) the subject of \(\dfrac{x + 5}{3} = C\)
Step 1. Multiply both sides by 3 to clear the denominator.
\[ x + 5 = 3C \]Step 2. Subtract 5 from both sides.
\[ x = 3C - 5 \]What's happening?
- \(x\) is trapped inside a fraction. Multiplying by the denominator (3) frees it in one step.
- After clearing the fraction, it is a standard linear rearrangement: subtract 5.
💡 Example 3 (brackets): make \(L\) the subject of \(P = 2(L + W)\)
Step 1. Divide both sides by 2 to undo the outer multiplier.
\[ \frac{P}{2} = L + W \]Step 2. Subtract \(W\) from both sides.
\[ L = \frac{P}{2} - W \]What's happening?
- Dividing by 2 first (before removing \(W\)) is the cleanest route. Expanding first also works: \(P = 2L + 2W \Rightarrow L = \dfrac{P - 2W}{2}\), which is equivalent.
- Both methods give the same answer. Dividing out the bracket is generally neater when the bracket has a simple multiplier.
💡 Example 4 (power): make \(r\) the subject of \(A = \pi r^2\)
Step 1. Divide both sides by \(\pi\) to isolate \(r^2\).
\[ r^2 = \frac{A}{\pi} \]Step 2. Square-root both sides. Write ±.
\[ r = \pm\sqrt{\frac{A}{\pi}} \]What's happening?
- Squaring and square-rooting are inverse operations. Taking the square root undoes the square.
- Always write ± when square-rooting both sides. In context (\(r\) is a radius), \(r > 0\), but the ± is expected unless told otherwise.
Advanced worked examples
💡 Example 5 (multiple constants): make \(h\) the subject of \(V = \tfrac{1}{3}\pi r^2 h\)
Step 1. Multiply both sides by 3 to clear the \(\tfrac{1}{3}\).
\[ 3V = \pi r^2 h \]Step 2. Divide both sides by \(\pi r^2\).
\[ h = \frac{3V}{\pi r^2} \]What's happening?
- \(\pi\), \(r\), and \(\tfrac{1}{3}\) are all constants in this rearrangement. Treat them as one compound coefficient multiplying \(h\).
- Multiply by 3 first to eliminate the fraction, then divide by \(\pi r^2\) in a single step.
💡 Example 6 (subject in denominator): make \(T\) the subject of \(\dfrac{1}{T} = 2R + 5\)
Step 1. \(T\) is in the denominator, so take reciprocals of both sides.
\[ T = \frac{1}{2R + 5} \]What's happening?
- When the subject appears only in the denominator of one side, taking reciprocals of both sides solves it in a single step.
- The right-hand side becomes \(\dfrac{1}{2R+5}\): the whole expression \(2R + 5\) moves to the denominator.
- Alternatively: multiply both sides by \(T\) to get \(1 = T(2R+5)\), then divide by \((2R+5)\). Same result.
💡 Example 7 (general formula): make \(y\) the subject of \(Ax^2 + By = C\)
Step 1. Subtract \(Ax^2\) from both sides.
\[ By = C - Ax^2 \]Step 2. Divide both sides by \(B\).
\[ y = \frac{C - Ax^2}{B} \]What's happening?
- Although \(x\) is in the equation, we are making \(y\) the subject, so treat \(Ax^2\) as a single block to move.
- The whole numerator \(C - Ax^2\) is divided by \(B\). Do not split the fraction: \(\frac{C}{B} - Ax^2\) is incorrect.
💡 Example 8 (quadratic in the subject): make \(t\) the subject of \(s = ut + \tfrac{1}{2}at^2\)
Step 1. Move all terms to one side to form a quadratic in \(t\).
\[ \tfrac{1}{2}at^2 + ut - s = 0 \]Step 2. Apply the quadratic formula with \(A = \tfrac{1}{2}a\), \(B = u\), \(C = -s\).
\[ t = \frac{-u \pm \sqrt{u^2 + 2as}}{a} \]What's happening?
- \(t\) appears in two separate terms with different powers (\(t^1\) and \(t^2\)). This makes the equation a quadratic in \(t\), not a simple linear rearrangement.
- Collect all terms on one side to give the standard form \(At^2 + Bt + C = 0\), then apply the quadratic formula.
- Both signs (+ and −) are usually kept. Context from the physical problem may later determine which is relevant.
The hardest type: subject appears in two places
When the target variable appears in more than one term, you cannot isolate it by simple inverse operations. Instead: cross-multiply to remove any fraction, expand, collect all subject terms on one side, factorise the subject out, then divide.
💡 Example 9 (collect and factorise): make \(b\) the subject of \(ab + 2b = 3a - 1\)
Step 1. Both terms on the left contain \(b\), so factorise \(b\) out.
\[ b(a + 2) = 3a - 1 \]Step 2. Divide both sides by \((a + 2)\).
\[ b = \frac{3a - 1}{a + 2} \]What's happening?
- \(b\) appears in two separate terms (\(ab\) and \(2b\)), so a single inverse operation cannot reach it.
- Factorising converts \(ab + 2b\) into a single term \(b(a+2)\) so that \(b\) appears only once so one division isolates it.
- This is the pattern: collect subject terms, factorise, divide.
💡 Example 10 (cross-multiply then factorise): make \(x\) the subject of \(y = \dfrac{x + 2}{x - 3}\)
Step 1. Multiply both sides by \((x - 3)\) to remove the fraction.
\[ y(x - 3) = x + 2 \]Step 2. Expand the left side.
\[ xy - 3y = x + 2 \]Step 3. Collect all \(x\) terms on the left; move all non-\(x\) terms to the right.
\[ xy - x = 2 + 3y \]Step 4. Factorise \(x\) out of the left side.
\[ x(y - 1) = 2 + 3y \]Step 5. Divide both sides by \((y - 1)\).
\[ x = \frac{2 + 3y}{y - 1} \]What's happening?
- \(x\) is in both the numerator and denominator of the fraction. This is the hardest type.
- Multiplying by the denominator \((x-3)\) brings \(x\) out of the fraction and into two separate terms.
- Once \(x\) appears in separate terms (one on each side), move them all to the same side and factorise.
- The factorised form \(x(y-1)\) has \(x\) appearing exactly once. One division gives the answer.
- Note: \(y \neq 1\) (division by zero). If asked, state this restriction.
🔑 Key points
- Undo operations in reverse BIDMAS: + and − are undone first.
- Keep equations balanced: every inverse operation applies to both sides.
- Write one equals sign per line in your working.
- When square-rooting both sides, always write ±.
- If the subject appears twice, factorise it out before dividing.
⚠️ Common pitfalls
- Forgetting ± when taking a square root.
- Applying an operation to only one term on a side rather than the whole side.
- Splitting a fraction incorrectly: \(\dfrac{C - Ax^2}{B} \neq \dfrac{C}{B} - Ax^2\).
- Confusing "change the subject" with "solve": leave letters as letters in the answer.
- Mixing up capital and lower-case letters. \(V\) and \(v\) are different quantities.