1. Formulas Where the Subject Appears Once
What actually is a "formula", and what's a "subject"?
Think of a formula like a recipe written in code. Instead of saying "add the two parallel sides, multiply by the height, then halve it" to find the area of a trapezium, we compress that whole sentence into:
Every formula has an equals sign, and one letter usually sits completely by itself on one side — that's the subject. In y = mx + c, the subject is y, because it's alone on the left and everything else is bundled up on the right.
Plain words: the subject is "the letter the formula is currently solved for." Changing the subject just means solving the same equation for a different letter instead.
The Method — Step by Step
Rearranging a formula uses exactly the same logic as solving an equation: whatever you do to one side, you must do to the other, to keep the "balance" true. Here's the general order of attack:
- Remove any fractions first. Multiply both sides by the lowest common denominator — this clears the messiest part of the problem immediately.
- Expand any brackets that contain the letter you're solving for (more on this below).
- Use inverse operations to peel away everything that isn't the target letter — undo addition with subtraction, undo multiplication with division, undo squaring with square-rooting, and so on. Undo the operations in the reverse order to how they were "built up" (this is the exact same logic used when solving normal equations).
Worked Example
Make x the subject of (5x + 6)/2 = y
Step 1 — clear the fraction: multiply both sides by 2 → 5x + 6 = 2y
Step 2 — undo the +6: subtract 6 from both sides → 5x = 2y − 6
Step 3 — undo the ×5: divide both sides by 5 → x = (2y − 6) / 5
Note: (2y−6)/5, 2y/5 − 6/5, 2(y−3)/5, 0.4(y−3), and 0.4y − 1.2 are all the same answer written differently. Examiners accept any fully simplified equivalent form.
Should I expand brackets, or divide instead?
This is a decision point students often get wrong on autopilot — so stop and actually look at where the target letter sits.
The Key Question
Is the letter you want
trapped inside the bracket? If yes, you generally need to expand it out to "release" the letter. If the letter is
outside the bracket (the whole bracket is just multiplying it), you can often skip expanding entirely and just divide by the bracket.
Worked Example — letter inside the bracket
Make x the subject of 3(1 + x) = y
Route A (expand first): 3 + 3x = y → 3x = y − 3 → x = (y − 3)/3
Route B (divide first): 1 + x = y/3 → x = y/3 − 1
Both routes are 100% valid and give equivalent answers — pick whichever feels more natural to you.
Worked Example — letter outside the bracket (no expansion needed!)
Make x the subject of (1 + k)x = y
Here, x is not inside the bracket — the whole bracket (1 + k) is just a multiplier sitting next to x. So simply divide both sides by (1 + k):
x = y / (1 + k)
What if I end up with a fraction inside a fraction?
Sometimes the rearranging leaves you with something ugly like:
Don't panic — this is just "3/t divided by 2." There are two clean ways to tidy it up:
Method 1 — Rewrite with a ÷ sign
x = 3/t ÷ 2 = 3/t ÷ 2/1 = 3/t × 1/2 = 3/(2t)
Remember: dividing by a fraction = multiplying by its flipped (reciprocal) version.
Method 2 — Multiply top and bottom by the LCD
For x = (5/y) / (t/8), multiply the top and bottom of the big fraction by 8y (the LCD of the two inner fractions):
x = [(5/y) × 8y] / [(t/8) × 8y] = 40 / (ty)
What if I end up dividing by a negative?
This is one of the most common places marks get lost — not because the maths is hard, but because students panic about where to put the minus sign. Here's the rule that unlocks it:
In words: a minus sign can legally live underneath (attached to the denominator), on top (attached to the numerator), or outside the whole fraction — they're all the exact same value.
Worked Example
−2x = y − 3, make x the subject.
Divide both sides by −2 (minus below): x = (y − 3) / (−2)
Same as (minus above, brackets required): x = −(y − 3) / 2
Same as (minus outside, brackets assumed): x = −(y − 3)/2
Expand for the tidiest form: x = (3 − y) / 2
Careful!
(−a)/(−b) is
not the same as −(a/b). Two negatives cancel out, so (−a)/(−b) = a/b (positive). Only
one negative sign flips the overall sign — don't accidentally drop or duplicate a minus when you move it around.
Practice Questions
Question 1
Make x the subject of: 4m + 5x = 3
Question 2
Make x the subject of: A = 9(1 − 4x) / (2g)
2. Formulas Where the Subject Appears Twice
Why does this need a different approach?
So far, every example had the target letter appearing just once. But what happens when it shows up in two different places in the same formula, like:
You can't "collect" x and xy the way you'd collect 3x and 5x, because xy is a completely different term (it also contains y). The trick that unlocks this is factorising — pulling x out the front as a common factor so it only appears once, multiplying a bracket.
Worked Example
x + xy = 3 − 2y
Factorise x out of the left-hand side: x(1 + y) = 3 − 2y
Notice — x now appears only once! Divide both sides by (1 + y):
x = (3 − 2y) / (1 + y)
The full strategy when the subject appears twice
- Clear fractions — multiply both sides by the denominator, just like before.
- Expand any brackets that contain the subject letter, so every instance of it is "released" and sitting as a separate term.
- Bring every subject-term to the same side of the equation (and everything else to the other side) by adding/subtracting.
- Factorise — pull the subject letter out the front as a common factor, turning two separate terms into one bracket multiplied by the letter.
- Divide both sides by the bracket to finish isolating the subject.
Worked Example — brackets AND subject on one side
Make x the subject of: c(x + 2) − x = f
Expand the bracket first: cx + 2c − x = f
Keep the x-terms together: cx − x = f − 2c
Factorise x out: x(c − 1) = f − 2c
Divide by (c − 1): x = (f − 2c) / (c − 1)
Worked Example — subject on both sides of the "="
Make x the subject of: 3x = y − px
Bring both x-terms to the same side — add px to both sides: 3x + px = y
Factorise x out: x(3 + p) = y
Divide by (3 + p): x = y / (3 + p)
Full Worked Example — the whole strategy in one go
Rearrange p = (2 − ax) / (x − b) to make x the subject.
Clear the fraction — multiply both sides by (x − b): p(x − b) = 2 − ax
Expand the bracket to release x: px − pb = 2 − ax
Bring x-terms to one side — add ax to both sides: px − pb + ax = 2
Move the non-x term across — add pb to both sides: px + ax = 2 + pb
Factorise x out the front: x(p + a) = 2 + pb
Divide by the whole bracket (p + a): x = (2 + pb) / (p + a)
Why this feels tricky at first
Students often try to "cancel" x's that are on opposite sides without factorising — that doesn't work here because one x is being multiplied by other letters. Factorising is what legally lets you treat "x(p + a)" as a single block you can divide by.
When the subject appears twice with the SAME power
Sometimes both appearances of the subject are squared (or cubed) rather than a mix of x and xy. The idea is identical — collect, then factorise — you just also need to take a root at the very end.
Worked Example
Make x the subject of: x² = −px² + r
Collect the x² terms — add px² to both sides: x² + px² = r
Factorise x² out the front: x²(1 + p) = r
Divide by (1 + p): x² = r / (1 + p)
Finally, square-root both sides: x = ± √[ r / (1 + p) ]
Don't forget the ± !
When you take a square root to undo a square, there are always two possible answers — a positive and a negative one — because both (+n)² and (−n)² give the same positive result. Cube roots, on the other hand, only give
one real answer, so no ± is needed there.
Worked Example — cube roots and a common trap
Make x the subject of: x³ = (t³ + 1) / (t³ + 8)
The entire right-hand side must be cube-rooted together — you cannot root the top and bottom separately and simplify the t's:
x = ∛[ (t³ + 1) / (t³ + 8) ]
Common error to avoid: this does NOT simplify to (t+1)/(t+2). You cannot cancel the cube root against individual terms inside a sum — only against the whole fraction as one block.
Practice Question
Question 3
Make m the subject of: km − 5 = n − 3m
Exam Tips — Common Mistakes & Mark-Scheme Traps
Forgetting to clear fractions first
Trying to move terms around while a fraction is still in place almost always leads to errors. Always multiply through by the denominator (or lowest common denominator) as your very first move.
Expanding brackets that didn't need expanding
Before expanding, check whether the target letter is actually
inside the bracket. If it's outside (the bracket is just a multiplier), you can often skip a step entirely and divide by the whole bracket.
Losing track of minus signs
When you end up dividing by a negative, remember a/(−b) = −(a/b) = (−a)/b. Write the negative
somewhere — don't drop it, and don't accidentally double it up.
Trying to cancel the subject without factorising first
If the subject appears on both sides of the "=" or twice on the same side as different terms, you cannot simply cross it out. Collect the terms, then factorise the subject out as a common factor before dividing.
Forgetting ± when square-rooting
Whenever you undo a square by taking a square root, there are two valid solutions (positive and negative). Mark schemes will often want to see the ± symbol explicitly.
Splitting a root across a fraction incorrectly
∛[(t³+1)/(t³+8)] does
not simplify to (t+1)/(t+2). A root applied to a whole fraction must stay applied to the whole fraction — it cannot be distributed across a sum of terms inside it.
What examiners actually look for
Method marks are usually awarded for each correct
step (clearing fractions, expanding correctly, factorising correctly) — not just the final answer. Show every step clearly, and remember that several algebraically equivalent forms of the final answer are all acceptable, as long as they're fully simplified.