<s>[INST] 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.
 [/INST] for $$x,y\ne 0$$ we obtain
$$1+\frac{1}{x^2+y^2}=\frac{12}{5x}$$
$$1-\frac{1}{x^2+y^2}=\frac{4}{5y}$$
adding both we get
$$5=\frac{6}{x}+\frac{2}{y}$$
from here we obtain
$$y=\frac{2x}{5x-6}$$
can you proceed?
after substitution and factorizing we get this equation:
$$- \left( x-2 \right)  \left( 5\,x-2 \right)  \left( 5\,{x}^{2}-12\,x+9
 \right)=0 
$$
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