$5x\big(1+\frac{1}{x^2 +y^2}\big)=12$ ; $5y\big(1-\frac{1}{x^2 +y^2}\big)=4$ find $x$ and $y$ I already tried to solve using substitution and cross multiplication method .
I got the first simplified 
(1)$$\frac{12}{5x}=1+ \frac{1}{x^2} +y^2$$
(2)
$$\frac{4}{5y}=1- \frac{1}{x^2+y^2}$$
Adding (1) and (2)
$$12y +4x= 50xy$$
I am pretty sure that it is irrelevant as there has to be at be $2$ solutions for $x$ and $y$. Please help
Thanks in advance

Let $x=r\cos\varphi,\, y=r\sin\varphi$
$$\begin{cases}
\frac{12}{5r\cos\varphi}=\frac{r^2+1}{r^2}\\
\frac{4}{5r\sin\varphi}=\frac{r^2-1}{r^2}
\end{cases}$$
$$\begin{cases}
\frac{5\cos\varphi}{12}=\frac{r}{r^2+1}\\
\frac{5\sin\varphi}{4}=\frac{r}{r^2-1}
\end{cases}$$
$$\begin{cases}
\cos\varphi=\frac{12}{5}\cdot\frac{r}{r^2+1}\\
\sin\varphi=\frac{4}{5}\cdot\frac{r}{r^2-1}
\end{cases}$$
$$\left(\frac{12}{5}\cdot\frac{r}{r^2+1}\right)^2+\left(\frac{4}{5}\cdot\frac{r}{r^2-1}\right)^2=1,$$
which yields $\frac{5 r^4-26 r^2+5}{r^2-1} = 0$
$r^2=5$ or $r^2=\frac{1}{5}$, substituting $r^2$ back, we obtain $x$ and $y$
