System of Equation. (Complicated) Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt, 
I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.

First of all get rid of $x^2+y^2$ 
$$ \frac{12}{4x} +\frac{4}{5y} = 2,\quad 2x+6y = 5xy $$
It is  a rectangular hyperbola passing through origin with asymptotes parallel to axes reminding one of $ x= t, y= 1/t $ when axes are asymptotes. So solve for $x,y$
$$ x= \frac{6y}{5y-2},\, y= \frac{2x}{5x-6}$$
The asymptotes are 
$$ x= \frac65 ,\, y= \frac{2}{5} $$
We take hyperbola in the form 
$$  ( x-\frac65 ) (y- \frac25) = k $$
Since it goes through the origin, $k$ must be the product $k=\dfrac65 \cdot \dfrac25 = \dfrac{12}{25} $ so as not to leave behind any constant term.
We can now find $x,y$ upto a parameter $t$
$$  ( x-\frac65 ) (y- \frac25) = \frac{12}{25}$$
$$\rightarrow x= \frac65 + t \sqrt{12}/5,\, y= \frac25 + \sqrt{12}/{5t},\, $$
