Finding x in an Olympiad simultaneous equation

Consider the second equation and define $z=\frac xy$. So, you have $$z-\frac 1z=7.875$$ which is a quadratic in $z$; its roots are $z=-\frac 18$ (to be discarded) and $z=8$ (to keep). So $xy=2048$ and $\frac xy=8$.

I am sure that you can take from here.

When solving quiz problems (and to some extent even real-world problems), always ask yourself if the problem is easier than it seems. In this case, 2048 is easily recognizable as a power of 2: $2^{11} = 2048$. Because $x$ and $y$ are integers, they have to be powers of two as well since $xy = 2048$. Moreover if $x = 2^n$, then $y = 2^{11-n}$. That's only a few combinations to try (the equation is symmetric on $x$ and $y$ except for the sign of the result), so you immediately get the solution $x = 2^7, y = 2^4$.

In this case the problem specifies that x and y are positive integers, so this approach was guaranteed to work. But it would have been worth exploring even if it was just a hunch.

The best way to solve math olympiad questions is the fast way - A long solution is just as bad as none, given the time constraints. Therefore, I would recommend the following:

First, notice that $2048 = 2^{11}$. With the stipulation that $x$ and $y$ are integers, this means that they must both be integer powers of two, and so are $\frac xy$ and $\frac yx$.

Next, notice that $\frac yx$ must be less than one, since $\frac xy$ is greater than one, since the difference is greater than one.

Thus, the only candidate for $\frac xy$ is 8. Thus, since there is guaranteed to be a solution, $\frac xy = 8 = 2^3$. Thus, $x = 2^7$ and $y = 2^4$.