Solve for all real values of $x$ and $y$ Solve the system of equations for all real values of $x$ and $y$ 
$$5x(1 + {\frac {1}{x^2 +y^2}})=12$$
$$5y(1 - {\frac {1}{x^2 +y^2}})=4$$
I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5}}\le y \le {\frac 45}$}
I don't know what to do next.

Let $z=x + iy \in \mathbb{C}$, now for $|z|^2 = x^2+y^2 \not = 0$ have that:
$$\left(5x + \frac{5x}{x^2+y^2}\right) + i\left(5y - \frac{5y}{x^2+y^2}\right) = 12 + 4i$$
$$5(x+iy) + \frac{5(x-iy)}{x^2+y^2} = 12 + 4i$$
$$5(x+iy) + \frac{5}{x+iy} = 12 + 4i$$
$$5z + \frac{5}{z} = 12 + 4i$$
$$5z^2 - (12+4i)z + 5 = 0$$
Solving this complex equation we get $z_1=\frac 25 - \frac 15i$ and $z_2 =2+i$, corresponding to the solution $(x,y) = \left\{\left(\frac 25, - \frac 15 \right), (2,1)\right\}$
