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rearranging a formula

Also called: rearrange, solve for, solving for

Turning a formula around to solve for a different quantity by doing the same operation to both sides, undoing the operations around the unknown in reverse order. From c·Q − F = 0 you get Q = F/c.

Most business formulas answer one question, but you often need the opposite one. Profit = c·Q − F tells you the profit at a given volume; break-even asks which volume gives zero profit. Rearranging turns one into the other.

The rule: an equation is a balance, so whatever you do to one side, do to the other. Undo the operations around the unknown from the outside in. In $c \cdot Q - F = 0$ , add F to both sides, then divide both by c: $Q = F / c$ .

Roots and squares work the same way. From $SE = \sqrt{p(1-p)/n}$ , square both sides and swap SE and n to get the sample size $n = p(1-p)/SE^2$ . Dividing by a share makes a number bigger: an investment X for a share s gives a post-money value of X/s.

Example

Fixed costs are 870,000 UAH a month and each lesson leaves 174 UAH.

174·Q − 870,000 = 0.
Add 870,000 to both sides: 174·Q = 870,000.
Divide both by 174: Q = 5,000 lessons.

Common mistakes

Doing an operation to one side only. The equation stops being true.
Undoing in the wrong order. Peel off the outermost operation first.
Forgetting to round up when the answer counts whole units you need enough of.

Learn it in the course

The Σ sign: sums, limits and running totals · Read Σ with its limits, see why deviations from the mean add up to zero, use sums of products, running totals and double indices, and two shortcuts: n(n + 1)/2 and k(k − 1)/2.
Rearranging a formula: solve for what you need · Do the same to both sides and undo in reverse order: break-even Q = F/c, runway, the regression intercept, sample size, a fraction inside a denominator, and why a gross-up divides by (1 − f).
Back-of-the-envelope estimates · Check any number in ten seconds: round to one significant figure, count the zeros, and give a range instead of false precision.
e, ln, odds and the sigmoid · ln undoes a power, so it solves for time; odds = p ÷ (1 − p), log-odds and the sigmoid p = 1 ÷ (1 + e⁻ᶻ), eᵇ as an odds multiplier, and payback with churn.