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?

When we see a square root, it is hard to resist the impulse to square. But resist we will, and such a decision is often useful.
The kind of solution we are looking is not specified, but the expressions $\sqrt{7x+y}$ and $\sqrt{x+y}$ suggest we are looking for real solutions.  Let
$$7x+y=p^2 \qquad\text{and}\qquad x+y=q^2,$$
where we can assume that $p$ and $q$ are non-negative. 
Note that $(7x+y)-4(x+y)=3x-3y$, so $3(x-y)=p^2-4q^2$.  So our equations can be rewritten as
$$p+q=6\qquad \text{and}\qquad q+\frac{1}{3}(p^2-4q^2)=2.$$
Now in principle it's all over!  (Substitute $6-q$ for $p$ in the second equation. We get a quadratic in the variable $q$.)

For fun we continue in another way. Observe that the second equation can be written as $p^2-4q^2+3q=6$.  Is that a $6$ on the right? Yes, so we have 
$p^2-4q^2+3q=p+q$, and therefore
$$p^2-4q^2=p-2q.$$
Since $p^2-4q^2=(p-2q)(p+2q)$, we conclude that (a) $p-2q=0$ or (b) $p+2q=1$.
Case (a): We have $p+q=6$ and $p=2q$, so $p=4$, $q=2$. That gives $7x+y=16$, $x+y=4$, so $x=y=2$.  This works. 
Case (b): We have $p+q=6$ and $p+2q=1$, and therefore $p=11$, $q=-5$. This contradicts the fact that $q$ is non-negative.
Comment: It is perhaps worth looking at Case (b) more closely.  We get $7x+y=121$, $x+y=25$, which yields $x=16$, $y=9$. If we interpret $\sqrt{121}$ as being $11$, and $\sqrt{25}$ as being $-5$, then $x=16$, $y=9$ is a solution. But that is a quite peculiar interpretation of $\sqrt{25}$ if we are working in the reals.
However, the manipulations that we did apply equally well to complex numbers. So we have found all complex solutions. But if we are working in the complex numbers, interpreting $\sqrt{121}$ as $11$ and $\sqrt{25}$ as $-5$ is not a problem. Thus we have the slightly paradoxical fact that if we are working in the reals, there is one real solution, while if we are working in the complex numbers, there are two real solutions!  
