Finding x in an Olympiad simultaneous equation I have been practicing for a an upcoming intermediate math olympiad and I came across the following question:

Let $x$ and $y$ be positive integers that satisfy the equations $$\begin{cases} xy = 2048 \\ \frac{x}{y}-\frac{y}{x}= 7.875.\end{cases}$$ Find x.    

My approach
$$xy=2048$$
$$x=\frac{2048}{y}$$
$$\frac{\frac{2048}{y}}{y} - \frac{y}{\frac{2048}{y}} = 7.875$$
$$\frac{2048}{y^2} -\frac{y^2}{2048} = 7.875$$
$$2048^2 -y^4 =7.875\cdot 2048 y^2$$
If we compare the the first equation (x/y) - (y/x) = 7.875 to the euqtion that we just found ((sqrt. 2048)/y) - (y/(sqrt. 2048)) = 7.875 , x is in the same place so, x must be equal to the sqrt. of 2048. 
Is this solution correct or are there any better solutions? 
Thank you :) 

Your solution is incorrect because it was stated that x is a positive integer, and $\sqrt{2048}$ is not an integer, therefore it can't be a solution.
Hints to solve the problem:
$$2048 = 2^{11}$$
$$ 7.875 = 8 - \frac{1}{8} $$
(Spoiler alert) My solution:
Because $2048 = xy = 2^{11}$, we can say that $x = 2^a$ and $y = 2^b$, where $a+b=11$ (equation 1).
Now, we see that: 
$$\frac{2^a}{2^b} - \frac{2^b}{2^a} = 7.875$$
$$\frac{2^a}{2^b} - \frac{2^b}{2^a} = 8 - 1/8$$
Therefore:
$$\frac{2^a}{2^b} = 8$$ 
Which can be rewritten as:
$$2^{a-b} = 2^3$$
$$a-b = 3 (Equation 2)$$
Now, adding equation 1 and 2:
$$2a = 14$$
$$a = 7$$
Therefore, $x = 2^{7} = 128$
