A reader asks:

I need to solve $\frac ac \frac {NP}{N_0 + N} = mP$ for $N$, and I don’t know where to start. Help!

I had a maths teacher in the early 90s who loved nothing more than making the class groan with bad jokes. If she showed up late, it was because she’d caught the rhom bus. She was afraid of 7, because 7 8 9. And she made a point, when rearranging algebra stuff with more than one instance of an unknown, to ‘get all of your $x$s in one basket.’

Two things: one, that’s not especially helpful if you don’t have an $x$ (although the idea isn’t any different – there’s nothing special about $x$ as a variable); and two, it’s not especially helpful if you don’t know what you have to do. So let’s put that straight.

The steps you need to follow in a question like this are:

  • Rearrange to get all of the unknowns on the ‘top line’, on the same side, and everything else on the other side
  • Factorise the unknown out
  • Divide by the resulting bracket

Let’s go through that here.

Step 1: rearrange

When rearranging, I like to ask ‘what’s ugly?’ – what’s stopping this from being a nice, simple thing to solve? Here, it’s the fractions. To get rid of the fractions, multiply both sides by $c(N_0 + N)$ to get

$aNP = cmP(N_0 + N)$

Since we’re solving for $N$, we need all of the $N$s on one side (I pick the left) and everything else on the other, so I’ll expand the bracket:

$aNP = cmPN_0 + cmPN$

… and subtract $mPN$:

$aNP - cmPN = cmPN_0$

Step 2: Factorise

That’s easy:

$NP(a-cm) = cmPN_0$

Step 3: Divide and tidy up

Divide by everything except the variable you’re solving for:

$N = \frac{cmPN_0}{P(a-cm)}$

Lastly, there’s a factor of $P$ top and bottom that you can remove, giving a final answer:

$N= \frac{cmN_0}{a-cm}$

Awesome, job done. This will be the way to approach it almost all of the time, especially at GCSE, but there’s one potential thing to watch out for: if you end up with $N^2$s or similar knocking around, you’ll probably need to employ some quadratic skills to get the answer(s)!