A formula is an equation with several letters, and rearranging it uses the same four moves as any other equation. Two things are genuinely new: the answer is an expression rather than a number, and dividing by a letter is legal only when that letter is not zero.
Core ideas
The steps are the ones you already have. What changes is what counts as an answer and what has to be said about it.
Decide which letter you are solving for before you write anything. In A = P(1 + rt) solving for r and solving for t are different problems with different answers, and the algebra only knows which one you mean because you decided.
The target letter must stand by itself on one side and appear nowhere on the other. C = S − rC is not an answer — C is still on the right — even though it is a true statement.
C = S/(1 + r) is correct exactly when 1 + r ≠ 0. Nothing in the algebra will remind you; you have to write it down at the moment you divide.
The material
The ordinary steps with letters, the case where the target appears twice, and the condition on the last division.
A literal equation is an equation containing more than one letter. To solve it for a named letter is to produce an equivalent equation with that letter alone on one side and absent from the other.
Formulas are the usual example, which is why the two words get used interchangeably.
The reason to rearrange rather than substitute is economy. Converting twenty temperatures from Fahrenheit to Celsius by solving F = (9/5)C + 32 twenty times is twenty problems; rearranging once gives a formula you then evaluate twenty times.
Every step is one of the four operations applied to both sides, named as it happens, exactly as in “One- and Two-Step Equations”. The letters change nothing about which moves are available.
Undoing operations one at a time cannot isolate a letter that occurs in two places. Gather every term containing the target on one side, then factor it out:
The factoring step is ab + ac = a(b + c) read from right to left, with a = C, b = 1 and c = r. Note the 1: C on its own is 1·C, and forgetting it produces S = C(r), which is a different formula.
If an equation can be written px = q, where neither p nor q contains x, then x = q/p whenever p ≠ 0.
Collecting and factoring is how you get a formula into that shape. The letter you are solving for plays the part of x; every other letter is swept into p and q. What happens when p = 0 is “Identities and Contradictions”.
Factoring two target terms on one side is not enough when they begin on opposite sides. Collect them first, exactly as in “Variables on Both Sides”, and only then factor the target:
If a − c = 0, division is not available and the original formula must be read as “Identities and Contradictions” reads it: b = d gives every x, while b ≠ d gives none. The condition records a real change in the answer, not a technical annoyance.
A = (1/2)bh solved for h gives h = 2A/b, and that statement is valid when b ≠ 0. At b = 0 the original equation becomes A = 0: if A = 0, every h satisfies it, and if A ≠ 0, none does. A physical triangle model normally excludes b = 0 as well, but that modelling restriction is separate from what the equation says.
Sometimes the condition is invisible until you write it. In C = S/(1 + r) the forbidden value is r = −1, not r = 0, because it is 1 + r that gets divided by. Read the divisor, not the letter.
A = s² solved for s is not a rearrangement of this kind. Adding, subtracting, multiplying and dividing undo a linear relationship, and s² is not linear in s. Undoing a square needs a square root, and for A > 0 it produces two answers, s = ±√A, which is Quadratics and Complex Numbers's subject — and for a length one of the two is then discarded on physical grounds, not algebraic ones.
Interactive
The same steps, with letters for numbers
Method
Decide what the unknown is first; after that every other letter behaves like a number.
Write the target at the top of the page if it helps. Half the errors in this lesson are answers to a different question — solving for t when r was asked for.
Multiply through by any denominator, expand any bracket that hides the target. A = P(1 + rt) becomes A = P + Prt, and only then is it obvious where r lives.
Everything else goes to the other side. If the target appears only once, this is the ordinary collecting of “Variables on Both Sides”.
The coefficient is whatever multiplies the target after factoring. Divide by it, and write down that the answer holds when that expression is not zero.
The first has the target in two places; the second has it inside a bracket.
The spot check is worth doing once for every formula you rearrange. It confirms that the rearrangement returns the right value at those particular numbers, which is enough to expose a dropped sign or an inverted fraction. It is not a proof that the algebra is right for all values — the named steps are the argument for that.
In B, (A − P)/(Pt) cannot be shortened by cancelling the two Ps. The P in the numerator is part of the term A − P, not a factor of it. What you may write instead is (A/P − 1)/t, which is the same expression with the division done first.
For a faded rearrangement, take K = pR + qR and solve for R. The supplied structural move is K = R(p + q). Finish the isolation, state the condition on p + q, and choose exact values for p, q and R that spot-check both the original and rearranged formulas.
Rearrange p(t − r) = q(t + s) for t without the lab. Expand, collect both target terms, factor, divide, and state the condition on the divisor. Use exact values to spot-check the result, then say why that one assignment checks signs but does not prove the formula.
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