System of $\sqrt{7x+y}+\sqrt{x+y}=6$ and $\sqrt{x+y}-y+x=2$ $$\begin{align*}\sqrt{7x+y}+\sqrt{x+y}=6\\\sqrt{x+y}-y+x=2\end{align*}$$
I have tried various things squaring, summing but nothing really helped, got some weird intermediate results which are probably useless such as:
$$(y-x)(y+x+4)+4-x-y=0$$
or
$$x_{1,2}=\frac{2y-3\pm\sqrt{8y-7}}{2}$$
Could you please give me some hints?

This is the way I would solve the given system in my school days. It is very similar to the hint provided by Ross Millikan. It turns out that one of the equations becomes linear. I omit some intermediate steps. The system
$$\left\{ 
\begin{array}{c}
\sqrt{7x+y}+\sqrt{x+y}=6 \\ 
\sqrt{x+y}-y+x=2
\end{array}
\right. \tag{1}$$
is equivalent to
$$\left\{ 
\begin{array}{c}
\sqrt{7x+y}=4+x-y \\ 
\sqrt{x+y}=2-x+y.
\end{array}
\right. $$
By squaring both sides, we get
$$\left\{ 
\begin{array}{c}
7x+y=\left( 4+x-y\right) ^{2} \\ 
x+y=\left( 2-x+y\right) ^{2}
\end{array}
\right. \tag{2}$$
or
$$\left\{ 
\begin{array}{c}
y=\left( 4+x-y\right) ^{2}-7x \\ 
\left( 4+x-y\right) ^{2}-7x=\left( 2-x+y\right) ^{2}-x,
\end{array}
\right. $$
which is equivalent to
$$\left\{ 
\begin{array}{c}
y=\left( 4+x-y\right) ^{2}-7x \\ 
-12y+6x+12=0
\end{array}
\right. $$
and to
$$\left\{ 
\begin{array}{c}
1+\frac{1}{2}x=9-4x+\frac{1}{4}x^{2} \\ 
y=1+\frac{1}{2}x.
\end{array}
\right. \tag{3}$$
This system has two pairs of solutions $\left( x,y\right) =(2,2)$ and $
\left( x,y\right) =(16,9)$, but only $\left( x,y\right) =(2,2)$ is a
solution of $(1)$.
