Steps to solve this system of equations: $\sqrt{x}+y=7$, $\sqrt{y}+x=11$ I want to solve this system of equations, I have been out of Maths for a long!!
$$\sqrt{x}+y=7$$
$$\sqrt{y}+x=11$$
Just wondering easiest step to find values for $x$ and $y$ from the above equations?

Note that : $x,y \geq 0$
$(7-y)^2=11-\sqrt y$
substitute  $~\sqrt y = t~$ , so :
$(7-t^2)^2=11-t \Rightarrow t^4-14 \cdot t^2+49=11-t \Rightarrow t^4-14 \cdot t^2+t+38=0$
Note that  possible nonnegative integer solutions of equation  $~\sqrt x+y=7~$ are :
$(x,y) \in \{(1,6),(4,5),(9,4),(16,3),(25,2),(36,1),(49,0) \}$
Since pair $(x,y)$ has to be solution of equation $\sqrt y +x =11$ also we have that :
$(x,y)=(9,4) \Rightarrow \sqrt y=2 \Rightarrow t=2 ~$ 
Therefore polynomial $~t^4-14 \cdot t^2+t+38~$ has to be divisible by $(t-2)$
If you divide this polynomial by $(~t-2)~$ using Polynomial long division method you will get polynomial :
$t^3+2t^2-10t-19$
So you should solve equation $~t^3+2t^2-10t-19=0~$ to obtain other solutions .
This can be done using general formula of roots .
