How can I solve this hard system of equations? Solve the system below

\begin{align}
&\sqrt {3x} \left( 1+\frac {1}{x+y}  \right) =2\\
&\sqrt {7y} \left( 1-\frac{1}{x+y}  \right) =4\sqrt{2}
\end{align}

Frankly I am disappointed, because I spent around 2 hours in solving this equation, but, finally I didn't do it so, I hope you can help me in approaching this problem.

Squaring 1st equation you have $3x(x+y+1)^2=4(x+y)^2$, squaring second you have $7y(x+y-1)^2=32(x+y)^2$. Put $s=x+y,y=s-x$ and we get $3s^2x+6sx+3x-4s^2=0,7s^3-7s^2x+14sx-46s^2+7s=0$. The first of these gives $x=\frac{4s^2}{3s^2+6s+3}$.
Substituting in the second we get $$s(21s^4-124s^3-178s^2-124s+21)=0$$ or $$s(3s^2-22s+3)(7s^2+10s+7)=0$$ It is clear from the original equations that $s\ne0$ and $7s^2+10s+7=0$ has no real solutions, so we have $s=\frac{1}{3}(11\pm4\sqrt7)$.
Using the equation for $x$ we get $x=\frac{1}{21}(11+4\sqrt7),y=\frac{2}{7}(11+4\sqrt7)$ or $x=\frac{1}{21}(11-4\sqrt7),y=\frac{22}{7}(11-\frac{8}{\sqrt7})$. But checking with the original equations the second solution fails (because $s-1$ is negative).
