In one of the solutions to an AIME question from a few years back (reference at the end), one comes across the following system:
$$ P\sin\theta + Q\cos\theta = \cos\theta - \frac{1}{2}P$$ $$ P\cos\theta + Q\sin\theta = -2(Q-1)$$
The next line in the solution states that this leads to
$$\frac{P}{Q} = \frac{\cos\theta}{2+\sin\theta}$$
No explanation for this is given, which leads me to think that there should be a relatively straightforward way (in the solution's author's mind, at least) to get there, but I've looked at it several times over the past few days and am not seeing the light.
If there's something simple I'm overlooking, then a hint or two to point me in the right direction would be much appreciated; if, on the other hand, this is actually a more complex question, then a solution would be great, but I would be even more grateful for the thought process that led to the solution.
[This is from 2013 AIME I, Problem 14, Solution 1]
P.S. This is the first question I'm posting on math.stackexchange, so any suggestions on how I can improve it would also be welcome.
Re-arrange the first equation:
$$P(2\sin \theta+1)=2(1-Q)\cos \theta$$
Combine with the second equation:
\begin{align} P(2\sin \theta+1) &= (P\cos \theta+Q\sin \theta)\cos \theta \\ P(2\sin \theta+1-\cos^2 \theta) &= Q\sin \theta \cos \theta \\ P\sin \theta(2+\sin \theta) &= Q\sin \theta \cos \theta \\ \frac{P}{Q} &= \frac{\cos \theta}{2+\sin \theta} \end{align}