Subject appearing more than once

If the new subject appears in more than one term, collect all those terms on one side, factorise the subject out, then divide by the bracket. If it is in a denominator, first multiply both sides by that denominator.

Why: \(at+bt\) contains \(t\) twice, but \(at+bt=t(a+b)\) contains it once, so dividing by \(a+b\) leaves \(t\) alone. The division needs \(a+b\ne0\), so the new formula comes with a restriction.

Method

1. Multiply both sides by any denominator that contains the subject.
2. Expand the brackets.
3. Put every term with the subject on one side and every other term on the other side.
4. Factorise out the subject, divide by the bracket and state the value that makes the bracket zero.

Example

Make \(t\) the subject of \(a(t+2)=b(3-t)\).

\(\begin{aligned}at+2a&=3b-bt \\ at+bt&=3b-2a \\ t(a+b)&=3b-2a \\ t&=\dfrac{3b-2a}{a+b},\qquad a+b\ne0\end{aligned}\)

Subject in a denominator: make \(R\) the subject of \(\dfrac{2}{R}=\dfrac{3}{a}-\dfrac{1}{b}\). Combine the right side first: \(\dfrac2R=\dfrac{3b-a}{ab}\). Turning both sides over gives \(\dfrac R2=\dfrac{ab}{3b-a}\), so \(R=\dfrac{2ab}{3b-a}\), where \(3b\ne a\).

Check: you cannot turn over term by term. With numbers, \(\dfrac12+\dfrac12=1\), but \(2+2=4\), not \(1\). Turn over only when each side is a single fraction.