Solving these two equations 
Solve for $x$ and $y$
  $$5x(1+\frac{1}{x^2+y^2})=12$$ $$5y(1-\frac{1}{x^2+y^2})=4$$

Combining these two equations we get 
$$5x^3-15x^2y+5xy^2-15y^3+5x+15y=0$$
Which could be factored if it had $-15y$ instead of $+15y$ for the last term.
Generally I do not post questions involving solving simple equations, but I find it really hard!

We have $$1+\frac{1}{x^2+y^2}=\frac{12}{5x}$$ and
$$1-\frac{1}{x^2+y^2}=\frac{4}{5y},$$ which gives
$$\frac{6}{x}+\frac{2}{y}=5$$ or
$$y=\frac{2x}{5x-6},$$
which after substitution to the first equation gives 
$$25x^4-120x^3+209x^2-156x+36=0$$ or
$$(5x^2-12x+6.5)^2-2.5^2=0$$ or
$$(x-2)(5x-2)(5x^2-12x+9)=0.$$
Id est, we got the answer:
$$\{(2,1),(0.4,-0.2)\}$$
