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.
 A: \begin{align}
5x\left(1+\frac{1}{x^2+y^2}\right)&=12\\
5y\left(1-\frac{1}{x^2+y^2}\right)&=4\\
\end{align}
Obviously, $x\neq0$ and $y\neq0$.
\begin{align}
\left(1+\frac{1}{x^2+y^2}\right)&=\frac{12}{5x}\\
\left(1-\frac{1}{x^2+y^2}\right)&=\frac{4}{5y}\\
1&=\frac{6}{5x}+\frac{2}{5y}&=\frac{6y+2x}{5xy}\\
\frac{1}{x^2+y^2}&=\frac{6}{5x}-\frac{2}{5y}&=\frac{6y-2x}{5xy}\\
x^2+y^2&=\frac{6y+2x}{6y-2x}\\
5xy&=6y+2x&=(x^2+y^2)(6y-2x)\\
\end{align}
Now, consider $x=y.\alpha$. Remember that $y\neq 0$
\begin{align}
y(5\alpha) &=6+2\alpha&=y^2(1+\alpha^2)(6-2\alpha)
\end{align}
$$y=\frac{6+2\alpha}{5\alpha}$$
$$25\alpha^2=(1+\alpha^2)(36-4\alpha^2)$$
$$4\alpha^4-7\alpha^2-36=0$$
$$\alpha=\pm 2$$ (There are two other imaginary roots for $\alpha$ that gives two more solutions in $\mathbb C$)
Hence two solutions : $(x,y)=(2,1)$ or $(x,y)=(\frac{2}{5},-\frac{1}{5})$
A: $$\left\{ {\begin{array}{*{20}{c}}{5x\left( {1 + \frac{1}{{{x^2} + {y^2}}}} \right) = 12}\\{5y\left( {1 - \frac{1}{{{x^2} + {y^2}}}} \right) = 4}\end{array}} \right.$$
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 $(1,-2)$.
A: Hint: Use the transformation $x=r\cos \theta,\ y=r\sin \theta$ and solve for $r$ first, by eliminating $\theta$ and then solve for $\theta$, once $r$ is solved for.
