<s>[INST] 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.
 [/INST] Just to provide another way.
Use polar coordinates $$x=\rho \cos(\theta) \qquad y=\rho \sin(\theta)$$ to make the equations to be $$\frac{5 \left(\rho ^2+1\right) \cos (\theta )}{\rho }=12$$ $$\frac{5 \left(\rho ^2-1\right) \sin (\theta )}{\rho }=4$$ So,$$\cos(\theta)=\frac{12 \rho }{5 \left(\rho ^2+1\right)}\qquad \sin(\theta)=\frac{4 \rho }{5 \left(\rho ^2-1\right)}$$ Now, using $$\sin^2(\theta)+\cos^2(\theta)=1 \implies \frac{16 \rho ^2}{25 \left(\rho ^2-1\right)^2}+\frac{144 \rho ^2}{25 \left(\rho
   ^2+1\right)^2}=1$$ Now, let $t=\rho ^2$ to have $$\frac{16 t}{25 \left(t -1\right)^2}+\frac{144 t}{25 \left(t+1\right)^2}=1$$ Reduce to same denominator to get $$25 t^4-160 t^3+206 t^2-160 t+25=0$$ where you can notice the symmetry of coefficients. 
Factoring gives $$25 t^4-160 t^3+206 t^2-160 t+25=(5-t) (5 t-1) \left(5 t^2-6 t+5\right)=0$$ The quadratic does not show real roots; this makes the solutions to be $$t_1=\frac 15\qquad t_2=5$$ then $$\rho_1=\frac 1 {\sqrt 5}\qquad  \rho_2=\sqrt 5$$ and now compute the corresponding $\theta$'s from $$\tan(\theta)=\frac{\sin(\theta) }{\cos(\theta) }=\frac{t+1}{3 (t-1)}=-\frac 12$$.
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