Rearranging a formula is solving an equation in letters. The rule that makes it manageable: treat the letter you want as the unknown, and undo what has been done to it in reverse order, keeping both sides balanced.
Once you add a two-minute numerical check at the end, you stop relying on hope.
Why does rearranging feel harder than solving?
When you solve 3x + 5 = 20, each step produces a number and you can see progress. In a formula the working stays in letters, so there is no visible “answer”, and it is easy to lose track of which operation applies to which term.
Three habits cause most trouble:
- Changing only part of one side. Dividing one term instead of the whole side.
- Losing brackets. Writing a − b ÷ c when the whole of a − b is divided by c.
- Moving terms in the wrong order. Dividing before removing the added term.
The method, step by step
- Circle the letter you want as the subject.
- Ask what has been done to it, from the outside in (for example: squared, then added to, then divided).
- Undo the last operation first, on both whole sides.
- Repeat until the letter stands alone.
- Check with numbers in the original formula and in your answer.
If the same letter appears in two places, follow the path in the formula lesson with the variable on both sides. You can also practise the working layout in the non-calculator working trainer, which keeps steps visible.
Worked example
The formula v² = u² + 2as links final speed v, starting speed u, acceleration a and distance s. Make a the subject.
Step 1: the letter a sits inside the term 2as, which is added to u².
Step 2, undo the addition: subtract u² from both sides.
v² − u² = 2as
Step 3, undo the multiplication: a is multiplied by 2s, so divide the whole of both sides by 2s.
a = (v² − u²) / (2s)
Step 4, numerical check. Choose u = 3, a = 4, s = 5 in the original formula: v² = 9 + 2 × 4 × 5 = 9 + 40 = 49, so v = 7.
Now put v = 7, u = 3, s = 5 into the new formula: a = (49 − 9) / (2 × 5) = 40 / 10 = 4.
The value 4 is recovered, so the rearrangement is right.
A second, harder example: the subject on both sides
Make x the subject of ax + b = cx + d.
Collect the x terms on the left: ax − cx = d − b.
Factorise: x(a − c) = d − b.
Divide: x = (d − b) / (a − c).
Check with a = 5, b = 1, c = 2, d = 10: 5x + 1 = 2x + 10 gives 3x = 9, so x = 3. The formula gives (10 − 1) / (5 − 2) = 3. Correct.
The mistake to watch for
Mistaken answer: a = v² − u² / 2s
What went wrong: the student divided only u² by 2s. The division applies to the whole side, so v² − u² needed brackets.
How the numerical check catches it: with v = 7, u = 3, s = 5 this gives 49 − 9/10 = 48.1, not 4.
The correction is to write the numerator in brackets and draw the fraction line long enough to cover it: a = (v² − u²) / (2s).
Self-check
1. Make r the subject of C = 2πr.
Show answer
Divide both sides by 2π: r = C / (2π). Check: if r = 3 then C = 6π, and 6π / (2π) = 3.
2. Make h the subject of V = (1/3)πr²h.
Show answer
Multiply both sides by 3, then divide by πr²: h = 3V / (πr²). Check: r = 1, h = 6 gives V = 2π, and 3 × 2π / π = 6.
3. Make t the subject of s = ut + 5t.
Show answer
Factorise: s = t(u + 5), so t = s / (u + 5). Check: u = 1, t = 2 gives s = 12, and 12 / 6 = 2.
Where this leads next
Work through rearranging with the variable on both sides, then check your method on the equations and formulas topic. The percentage-base explorer shows the same “undo the operation” idea with percentages.
When a student can recite steps but still stumbles, the gap is usually in one specific habit. Our teachers look for exactly that in online one-to-one Mathematics tuition.