Solve this system of equation Solve this system of equations for real $x$ and $y$:


*

*$5x\left(1+\dfrac{1}{x^2+y^2}\right)=12$

*$5y\left(1-\dfrac{1}{x^2+y^2}\right)= 4$


I juggled with those equations and got $x-y+\dfrac{x+y}{x^2+y^2}=\dfrac{8}{5}$, from where I guessed a solution $(2,1)$.
But I don't know how to approach mathematically.
Please help me.

$$\left\{ {\begin{array}{*{20}{c}}{5x\left( {1 + \frac{1}{{{x^2} + {y^2}}}} \right) = 12}\\{5y\left( {1 - \frac{1}{{{x^2} + {y^2}}}} \right) = 4}\end{array}} \right.$$
If you sum the above equations, you can find formula of a line all the points on it are solutions to that system of equations;
$$\left\{ {\begin{array}{*{20}{c}}{1 + \frac{1}{{{x^2} + {y^2}}} = \frac{{12}}{{5x}}}\\{1 - \frac{1}{{{x^2} + {y^2}}} = \frac{4}{{5y}}}\end{array}} \right.$$
Here is the formula of the line:
$$1 = \frac{6}{{5x}} + \frac{2}{{5y}}$$
Obviously, zero value for $x$ or $y$ is not in the domain of solutions to the equations. As verification, all the results above can be found by this formula: $(x,y)=(\frac{2}{5},-\frac{1}{5})$, $(2,1)$ and $(1,-2)$.
