MISSION · Algebra
Rearranging Formulas
Solving for a letter when every other letter stays a letter
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The idea in one sentence
Rearranging a formula is solving an equation whose answer happens to be made of letters instead of numbers, and every step is the same as it always was.
Nothing new is happening
To solve \(3x + 7 = 22\) you subtract 7 and divide by 3. To make \(x\) the subject of \(ax + b = c\) you subtract \(b\) and divide by \(a\):
\[x = \frac{c - b}{a}\]Identical steps. The only thing that changed is that the answer cannot be simplified to a number, and that feels like being stuck when it isn’t. A formula with letters in the answer is the answer.
Two habits make it easier. Circle the letter you are solving for before you start, so you always know what you are aiming at. And treat every other letter as if it were a number you happen not to know the value of.
Undo in reverse order
Take \(F = \frac{9C}{5} + 32\) and make \(C\) the subject.
Read what has been done to \(C\): multiplied by 9, divided by 5, then 32 added. Undo them backwards.
\(F - 32 = \frac{9C}{5}\) \(5(F - 32) = 9C\) \(C = \frac{5(F - 32)}{9}\)
This is exactly the linear equation method, and the ordering rule is the same: peel off the outermost operation first.
When the letter appears more than once
If your target letter shows up twice, you cannot isolate it by peeling. You have to gather it first.
Make \(x\) the subject of \(ax + b = cx + d\):
\(ax - cx = d - b\) \(x(a - c) = d - b\) \(x = \frac{d - b}{a - c}\)
The factorising step is what does the work. Every term with \(x\) goes to one side, everything else to the other, then take \(x\) out as a common factor and divide by the bracket.
Letters under a root or in a denominator
Same principle, different undoing.
For \(T = 2\pi\sqrt{\frac{l}{g}}\), make \(l\) the subject. Divide by \(2\pi\), square both sides, multiply by \(g\):
\[l = \frac{gT^2}{4\pi^2}\]Squaring squares the whole side, which is where the \(4\pi^2\) comes from. Squaring term by term is the standard error here.
For \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), make \(f\) the subject. Combine the right side over a common denominator to get \(\frac{1}{f} = \frac{u + v}{uv}\), then flip both sides: \(f = \frac{uv}{u + v}\).
Flipping is legal only when each side is a single fraction. \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) does not give \(f = u + v\), which you can check with \(u = v = 2\): the real answer is 1, not 4.
Why do it at all
You could substitute the numbers first and solve each time. Rearranging first wins whenever you are going to use the formula more than once — a spreadsheet column, a graph, a program, a physics experiment with thirty readings.
It is also how a formula reveals what depends on what. Rearranged as \(l = \frac{gT^2}{4\pi^2}\), the pendulum formula shows at a glance that quadrupling the length doubles the period.
Common mix-ups
- Doing something to one term instead of one side. Whatever you do, do it to everything on both sides.
- Squaring or rooting term by term. \(\sqrt{a + b} \ne \sqrt a + \sqrt b\).
- Cancelling a letter that is added rather than multiplied. In \(\frac{a + b}{a}\) the \(a\)s do not cancel.
- Stopping too early, leaving the target letter still on both sides.
- Losing a sign when dividing by a negative, which also flips an inequality if there is one.
Check yourself
Make \(r\) the subject of \(C = 2\pi r\), finish \(C = \frac{5(F-32)}{?}\) from \(F = \frac{9C}{5} + 32\), and make \(h\) the subject of \(V = lwh\).
Divide by \(2\pi\): \(r = \frac{C}{2\pi}\). The denominator is 9. And dividing by \(lw\) gives \(h = \frac{V}{lw}\).
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