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?

Let 
$x=\tan^2\theta$
and $y=\sec^2\theta$


*

*$\tan\theta+\sec^2\theta=7$

*$\tan^2\theta+\sec\theta=11$


Take 1) $\tan\theta+1/\cos^2\theta=7$
Solve it and you will get 
$$7\sin^2\theta+\sin\theta-6=0$\tag{A}$$
Now similarly take equation (2) solve it 
you will get 
$$12\cos^2\theta-\cos\theta-1=0\tag{B}$$
Take (A). 
Let 
$\sin\theta=t$,  $7t^2+t-6=0$.
Solve it you will get 
$t=-1$  or   $t=6/7$
$\sin\theta=-1$  or $\sin\theta=6/7$.
Now 
Take eq. (B)
Do similarly
As equation (A)
$12\cos^2\theta-\cos\theta-1=0$
Solve it as above 
$\cos\theta=1/3$ or  $\cos\theta=-1/4$
$\cos\theta$ value exists in 3 quadrant not -ve
So $-1/4$ not possible.
$\cos\theta=1/3$
$\sin\theta$ value exist in 3 quadrant not +ve  so 
$\sin\theta=-1$.
Now $\tan\theta=\sin\theta/\cos\theta=-1/(1/3)=-3$. 
$x=\tan^2\theta=(-3)^2=9$
Put in equation 1 $y=4$
Therefore 
$x=9$    and $y=4$ is answer.
