Solve the following system: $(x^2+1)(y^2+1) = 12xy$ and $(x+1)^2(y+1)^2=30xy$ I have been given the following system to solve:
$$(x^2+1)(y^2+1) = 12xy$$
$$(x+1)^2(y+1)^2=30xy$$
I noticed that the system is symmetric, but any of the methods I know to solve symmetric systems doesn't seem right. I have also tried writing the second equation as $$((x^2+1)+2x)((y^2+1)+2y) = (x^2+1)(y^2+1)+2y(x^2+1)+2x(y^2+1)+4xy = 30xy$$,
but this didn't get me anywhere. I would appreciate any hints towards the right solution.

$$
\left\{ 
\begin{array}{c}
\left(x^2+1 \right)\left(y^2+1 \right)=12xy \\ 
\left(x+1 \right)^2\left(y+1 \right)^2=30xy
\end{array}
\right. 
$$
$$
\iff \left\{ 
\begin{array}{c}
\left(x^2+1 \right)\left(y^2+1 \right)=12xy \\ 
\left(x^2+1+2x \right)\left(y^2+1+2y \right)=30xy
\end{array}
\right. 
$$
$$
\iff \left\{ 
\begin{array}{c}
\left(x+\frac{1}{x} \right)\left(y+\frac{1}{y} \right)=12 \\ 
\left(x+\frac{1}{x}+2 \right)\left(y+\frac{1}{y}+2 \right)=30
\end{array}
\right. 
$$
Put $A=x+\frac{1}{x}$ and $B=y+\frac{1}{y}$, we have
$$
\left\{ 
\begin{array}{c}
AB=12 \\ 
\left(A+2 \right)\left(B+2 \right)=30
\end{array}
\right. 
$$
Solving the equations, we have $A=3, B=4$ or $A=4, B=3$.
When $A=3, B=4$ , we have
$$
\left\{ 
\begin{array}{c}
x+\frac{1}{x}=3 \\ 
y+\frac{1}{y}=4
\end{array}
\right. 
$$
$$
\left\{ 
\begin{array}{c}
x=\frac{3\pm\sqrt{5}}{2}\\ 
y=2\pm\sqrt{3}
\end{array}
\right. 
$$
When $A=4, B=3$ , we have
$$
\left\{ 
\begin{array}{c}
x+\frac{1}{x}=4 \\ 
y+\frac{1}{y}=3
\end{array}
\right. 
$$
$$
\left\{ 
\begin{array}{c}
x=2\pm\sqrt{3}\\ 
y=\frac{3\pm\sqrt{5}}{2}
\end{array}
\right. 
$$
