Rearranging Formulae

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1. 1

   Identify what is being done to the subject

   v = u + at, make a the subject → a is multiplied by t, then added to u

   List the operations acting on the target letter before touching anything. Rearranging fails far more often from starting in the wrong place than from a wrong step, and thirty seconds of reading prevents that.

2. 2

   Undo in reverse order

   v − u = at

   The last operation applied is the first one undone. Since u was added last, subtracting it comes first — the same order-of-operations logic that governs solving equations, applied to letters instead of numbers.

3. 3

   Keep undoing until the subject is alone

   (v − u)/t = a, so a = (v − u)/t

   Dividing both sides by t isolates a. Note the bracket: the whole of v − u is divided, not just the v. Losing that bracket is the most frequent error in the topic and it changes the formula entirely.

4. 4

   When the subject appears twice, collect it first

   ax + b = cx + d → ax − cx = d − b → x(a − c) = d − b → x = (d − b)/(a − c)

   Gather every term containing the subject on one side, everything else on the other, then factorise out the subject. The factorising step is what makes this case solvable at all, and it is the step students most often do not think to attempt.

5. 5

   Check by substituting numbers

   u=2, a=3, t=4 → v=14. Then (14−2)/4 = 3 = a ✓

   Pick easy values, compute with the original formula, then feed them into the rearranged one. If both agree, the rearrangement is right. This is the only reliable check, because a rearranged formula gives no clue on its own about whether it is correct.

This is solving equations with letters instead of numbers

Students who can solve 3x + 5 = 20 in seconds often freeze at v = u + at, because the absence of numbers makes it feel like a different subject. It is not. The operations are identical and only the notation has changed, and saying so directly removes a surprising amount of the difficulty.

The practical bridge is to solve the numerical version first and then repeat the identical steps with letters. Once a student has seen the two sets of working side by side, the letters stop being an obstacle.

• Do the same to both sides, every time
• Undo operations in reverse order
• Bracket anything that moves as a whole
• Subject appearing twice → collect, then factorise
• Check by substituting easy numbers

Squares and roots

To free a letter from inside a square root, square both sides; to free it from a square, take the root of both sides. In an exam context the positive root is normally the one wanted, because the quantities are lengths, times or speeds — but the question decides, not habit.

The common error is squaring only part of a side. Squaring both sides of √(x + 3) = 5 gives x + 3 = 25, not x + 9 = 25. The square applies to the entire side, which is another argument for bracketing generously.

Where the marks go

Almost every lost mark in this topic is a missing bracket. Dividing v − u by t gives (v − u)/t, and writing v − u/t means something entirely different — only the u is divided. On paper with a horizontal fraction bar the grouping is visible; typed in a line it is not.

The habit worth building is to bracket any expression of more than one term the moment it moves. Redundant brackets cost nothing and the missing ones cost the question.

How we teach rearranging

We pair every rearrangement with its numerical twin for the first few lessons. Solving 2x + 6 = 14 and then making x the subject of ax + b = c in the same breath makes the correspondence impossible to miss.

We also require the substitution check on anything that will be used later, particularly in physics-style formulae. A rearranged formula that is wrong produces confidently wrong answers for the rest of the question, so it is worth thirty seconds to confirm.