Math Detours · Grade 9 · Sharpen · 12

Change the subject

A formula is an equation with several letters. Solve it for the one you want, once, and you never have to repeat the algebra.

Name ____________________Date ____________

Tonight's idea

The formula \(F = \tfrac95 C + 32\) turns a Celsius temperature into Fahrenheit. \(F\) stands alone on one side, so \(F\) is the subject. That suits you if you know \(C\).

If you know \(F\) and want \(C\), you could put the number in and solve an equation, every single time. It is better to solve for \(C\) once, with the letters still in place. That is changing the subject.

The method is the one you use to solve an equation. Do the same thing to both sides, and undo the operations in reverse order. Treat every other letter as if it were a number.

CFCF × 9/5+ 32× 5/9− 32
Forwards: multiply, then add. Backwards: undo the add first, then undo the multiply.

Two things to watch. To undo a square, take a square root (for a length, keep the positive one). And if the letter you want appears twice, collect its terms on one side and factor it out: \(ax + bx = c\) becomes \(x(a + b) = c\), so \(x = \dfrac{c}{a + b}\).

Watch one: Celsius from Fahrenheit

Start with \(F = \tfrac95 C + 32\) and follow the bottom row of the figure.

Subtract 32 from both sides: \(F - 32 = \tfrac95 C\).

Multiply both sides by \(\tfrac59\): \(\tfrac59 (F - 32) = C\).

\(C = \tfrac59 (F - 32)\)

The brackets matter: the subtraction happens first. Check with boiling water, \(F = 212\): \(\tfrac59 \times 180 = 100\). Now any conversion is one line. Body temperature, \(F = 98.6\): \(\tfrac59 \times 66.6 = 37\).

Your turn

  1. Make \(x\) the subject. (a) \(y = 3x - 7\). (b) \(y = \dfrac{x}{4} + 2\). (c) \(ax + b = c\).
  2. (a) Speed is distance over time, \(v = \dfrac{d}{t}\). Make \(t\) the subject. (b) A rectangle has perimeter \(P = 2l + 2w\). Make \(w\) the subject. (c) A circle has area \(A = \pi r^2\). Make \(r\) the subject.
  3. Here the letter you want appears twice. (a) Make \(x\) the subject of \(ax + b = cx + d\). (b) Money \(P\) earning simple interest at rate \(r\) for \(t\) years grows to \(A = P + Prt\). Make \(P\) the subject.
  4. A pendulum of length \(L\) metres takes \(T = 2\pi\sqrt{\dfrac{L}{g}}\) seconds to swing over and back, where \(g = 9.8\). (a) Make \(L\) the subject. (b) How long is a pendulum with \(T = 2\) seconds? (c) You want \(T\) twice as big. Use your formula to say what must happen to \(L\).

Stretch: a formula that undoes itself

Make \(x\) the subject of \(y = \dfrac{x + 1}{x - 1}\). Multiply both sides by \(x - 1\) first; \(x\) then appears twice. What do you notice about your answer? Test it: start with \(x = 3\), find \(y\), then feed that \(y\) back in.

Answers

  1. (a) \(x = \dfrac{y + 7}{3}\). (b) Subtract 2, then multiply by 4: \(x = 4(y - 2)\). (c) \(x = \dfrac{c - b}{a}\).
  2. (a) Multiply by \(t\), divide by \(v\): \(t = \dfrac{d}{v}\). (b) \(2w = P - 2l\), so \(w = \dfrac{P - 2l}{2}\), or \(\dfrac{P}{2} - l\). (c) \(r^2 = \dfrac{A}{\pi}\), so \(r = \sqrt{\dfrac{A}{\pi}}\).
  3. (a) \(ax - cx = d - b\), so \(x(a - c) = d - b\) and \(x = \dfrac{d - b}{a - c}\). (b) \(A = P(1 + rt)\), so \(P = \dfrac{A}{1 + rt}\).
  4. (a) Divide by \(2\pi\), then square: \(\dfrac{T^2}{4\pi^2} = \dfrac{L}{g}\), so \(L = \dfrac{gT^2}{4\pi^2}\). (b) \(L = \dfrac{9.8 \times 4}{4\pi^2} = \dfrac{9.8}{\pi^2} \approx 0.99\) m, very nearly a metre. (c) \(L\) depends on \(T^2\), so doubling \(T\) needs \(L\) to be 4 times as long.

Stretch. \(y(x - 1) = x + 1\), so \(xy - y = x + 1\). Collect: \(xy - x = y + 1\), so \(x(y - 1) = y + 1\) and \(x = \dfrac{y + 1}{y - 1}\). It is the same formula with the letters swapped: doing it twice gets you back. \(x = 3\) gives \(y = \tfrac42 = 2\), and 2 gives \(\tfrac31 = 3\).

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