Rearranging Formulas

Undo the operations one at a time until the letter you want is on its own. Only the outermost one can come off: dividing by a before subtracting u would divide u as well.

Worked example — Rearranging Formulas

Moves needed	2
Numbers agree	4 vs 4 ✓

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Rearranging Formulas331 words

Making a letter the subject means producing an equivalent formula with that letter alone on one side. Whether it can be done divides into two questions that look identical from the outside: whether this method can do it, and whether anything can.

What each symbol means

subject: the letter you are isolating. Which one it is changes the difficulty completely; the formula does not care, but the algebra does.
±: what an even root hands back — two candidates and no way to choose between them from the algebra alone.

Assumes: That every operation between you and the subject can be undone one at a time from the outside in. That holds exactly when the letter occurs once.
Breaks when: In two quite different places, and telling them apart is the point. The method runs out when the letter occurs more than once: s = ut + ½at² defeats peeling, and the page says so. But that formula is a quadratic in t and a different method solves it exactly — with u = 3, a = 2, s = 20 the answer is t = 3.216991, and putting it back returns 20.000000. Nothing is unsolvable there; peeling is simply the wrong method for it. The equation itself resists in Kepler’s M = E − e sin E, where E sits both on its own and inside a sine, so no finite run of algebraic operations on M and e will ever isolate it. That one is not stuck: Lagrange and Bessel give it as a series, and Bessel functions were invented for the coefficients. What the series has instead is a boundary. It converges only for eccentricities below the Laplace limit, 0.662743419349 — and Halley’s comet, at e = 0.967, is outside it. So the honest answer to “can this be rearranged?” has three levels: this method cannot, another method can; no method can, but a series can; and the series has a range.

A formula is an equation with better-behaved letters 🖖

Making t the subject of v = u + at is: subtract u from both sides, divide both sides by a. Solving 3x + 4 = 10 is: subtract 4 from both sides, divide both sides by 3. The same two moves, in the same order, for the same reason — you are undoing how the expression was built, outermost operation first. That is the only thing to know, and it is why there is no separate list of rules for formulas. The letters make it feel different because nothing collapses into a number along the way, so the answer stays an expression instead of arriving as a value.

Undoing a square hands back two answers and no way to choose 🖖

Press A = πr² and solve for r. The tool prints ±√(A / π), and it means it: both values, squared, give the same area. The algebra has done everything it can and left you with two candidates. Discarding the negative one is a statement about circles — radii are not negative — and not a step in the working. The same situation in a physics problem often keeps both roots: a projectile at a given height has two times, and the earlier one is usually the answer you did not want but is still real. The algebra cannot tell those two situations apart. You can.

When the letter appears twice, the method runs out 🖖

Press s = ut + ½at² and try to make t the subject. The tool refuses. Peeling a formula works by undoing one operation at a time from the outside, and that only isolates a letter if there is exactly one of it to isolate. Two occurrences and there is nothing to peel toward. This particular formula becomes a quadratic in t and yields to the quadratic formula, but that is a different technique with a different answer shape — two roots rather than one expression. Plenty of formulas have no closed-form rearrangement at all, and the honest thing a tool can do is say which case it is in rather than produce something confident and wrong.

Problems solved in full

The radius when the area of a circle is 78.54 cm² 7 steps
The area of a circle is 78.54 cm². Find the radius — and say precisely where the algebra stops and you start.
1. Start from the formula as given, with A the subject and r buried inside a square.
  
  $A = π r^{2}$
2. Peel from the outside. The outermost operation applied to r is the multiplication by π, so divide both sides by π first.
  
  $\frac{A}{π} = r^{2}$
3. Now the square. Undoing it is the step where one input stops giving one output, which is why the line carries a ± rather than a bare root.
  
  $r = \pm \frac{A}{π}$
4. The ± is doing real work: +5 and −5 both square to 25, so both are correct solutions of the equation r² = 25. The equation does not know what r means.
  
  $(+ 5)^{2} = 25 = (- 5)^{2}$
5. Substituting the number in: r = ±5 cm, and 5 is the answer because a radius is a length. That sentence is not algebra. It is the only step in the whole problem that required knowing what the letters stand for.
  
  $r = \pm \frac{78.54}{π} = \pm 5$
6. Check by going forwards, which is the only check worth doing: put 5 back into the original formula and see whether the area comes back.
  
  $π (5)^{2} = 78.54 ✓$
7. And notice what would change in a different dress. The same equation h² = 25 for a height above and below a line keeps both answers; the same equation t² = 25 for a time keeps −5 only if the clock started somewhere else. The algebra is identical in all three. The discarding is not.
  
  $h^{2} = 25 \Rightarrow h = \pm 5 (both)$
Answer

r = 5 cm. The algebra gives ±5 and stops there, correctly. Choosing 5 is a statement about radii, and it is the one step in the problem that the formula could not have made for you.
Fahrenheit as the subject, and the reading both scales agree on 6 steps
Make F the subject of C = 5(F − 32)/9 and check it on boiling water. Then decide whether a thermometer can ever show the same number on both scales.
1. F is buried three deep: inside a bracket, inside a multiplication by 5, inside a division by 9. Peel from the outside, so the 9 comes off first.
  
  $C = \frac{5 ( F - 32 )}{9}$
2. Multiply both sides by 9, then divide by 5. Two of the three moves, and neither has touched the bracket.
  
  $9 C = 5 (F - 32) ⟹ \frac{9 C}{5} = F - 32$
3. Add 32 and F stands alone. Three moves, which is what the panel counts, and the check card puts 100 °C back through the rearranged formula and gets 212.
  
  $F = \frac{9 C}{5} + 32; C = 100 : \frac{900}{5} + 32 = 180 + 32 = 212$
4. Now the question this method refuses. Asking where the two scales read the same means setting F = C, which puts the letter on both sides. That is the situation the tool stops at, because there is nothing left to peel toward.
  
  $F = C : C = \frac{9 C}{5} + 32$
5. Here both occurrences are linear, so they collect into one and the peeling starts again. The answer is −40, and it checks: 9(−40)/5 + 32 is −72 + 32.
  
  $C - \frac{9 C}{5} = 32 ⟹ - \frac{4 C}{5} = 32 ⟹ C = - 40$
6. The same collection answers the general version. Fahrenheit is double Celsius at 160 °C, which is 320 °F. At k = 1.8 the denominator vanishes and there is no answer, because the two lines are parallel and Fahrenheit runs 32 ahead of 1.8 × Celsius for ever.
  
  $F = k C : C = \frac{32}{k - 1.8}; k = 2 ⟹ C = 160, F = 320$
Answer

F = 9C/5 + 32, and the two scales agree at −40. Three moves turn the formula round, and the panel confirms it by sending 100 °C back to 212 °F. The fixed point needs a fourth move the tool does not offer: collect the two Cs. That works only because both are linear, which is the whole difference between this and the displacement formula the tool declines.

$F = k C ⟹ C = \frac{32}{k - 1.8} : k = 1 \to - 40, k = 2 \to 160$

Learning path

Solving for x, from the minus sign upward

Leads to Two equations at once the same solving, done with letters in place of numbers — and the ability to make any one letter the subject.

References (1)

The series that solves what peeling cannot, and the eccentricity at which it stops: P. Colwell, "Bessel Functions and Kepler's Equation." The American Mathematical Monthly 99:1 (1992), 45–48.

By the 42Math team · Updated 2026-08-26

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

v = u + at - Making t the subject takes two moves: subtract u, divide by a. Exactly the two moves you would make on 3x + 4 = 10, with different letters on them.
A = πr² - Solving for r means undoing a square, and undoing a square has two answers. The tool shows ±, and only the fact that r is a radius rules the negative one out. That is a decision about circles, not about algebra.
s = d / t - Here t is underneath. You cannot peel a denominator by dividing, so the two swap places instead: d / t = s becomes t = d / s.
Celsius from Fahrenheit - Three moves to invert, and the answer is the formula everyone half-remembers. Worth doing once rather than memorising twice.
s = ut + ½at² - Try to make t the subject and the tool stops. t appears in two places, so no sequence of undos can isolate it — this one needs the quadratic formula, and it is a different problem.
T = 2π√(l/g) - The length sits under a square root, so undoing it means squaring both sides — the one move here that is not a subtraction or a division. Ask for g instead and it has to come out of a denominator as well.
E = ½mv² - v is squared, so making it the subject brings a ±. Only the fact that a speed cannot be negative rules one of them out, and that is a fact about the world rather than about the algebra.
V = IR - One move, where every other sample here takes three. Worth doing once: it is what the others would look like if nothing were wrapped around the letter you want.
d = m / V - V is underneath, and a denominator cannot be peeled off by dividing. The two swap places instead: d = m / V becomes V = m / d.
c = √(a² + b²) - Solving for a leg means undoing a square, so the algebra offers two answers and one of them is a negative length. The same ± as the kinetic energy sample, ruled out by the shape rather than by the working.
A = bh / 2 - Written the way it is written at school, A = bh / 2 rather than ½bh — the halved form rearranges to h = A / (½b), which is correct and reads like nothing anyone would write. Two moves either way, and the answer is the formula you would have guessed.