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?
 A: First rearrange the equations like this:
$$y-4=3-\sqrt{x} \\x-9=2-\sqrt{y}$$ 
then multiply two sides of these equations to each other;
$$(y-4)(x-9)=(3-\sqrt{x})(2-\sqrt{y})$$
Since $x,y\ge 0$ we can factor LHS :
$$(\sqrt{y}-2)(\sqrt{y}+2)(\sqrt{x}-3)(\sqrt{x}+3)=(3-\sqrt{x})(2-\sqrt{y})$$
Implies that$$ (\sqrt{y}-2)(\sqrt{x}-3)((\sqrt{y}+2)(\sqrt{x}+3)-1)=0 $$
at least one of the factors must be zero.
1. $$(\sqrt{y}-2)=0 \Rightarrow y=4$$ 
by substitution $y=4$ in the first eq. we have $x=9$.
and by checking in the second eq. implies that  $$x=9\qquad and \qquad y=4$$ is a solution.
or
2. $$(\sqrt{x}-3)=0 \Rightarrow x=9$$
by substitution in the first eq. we have $y=4$  and by checking in the second eq. implies that this is not a new solution.
or
3. $$(\sqrt{y}+2)(\sqrt{x}+3)-1=0 \Rightarrow (\sqrt{y}+2)(\sqrt{x}+3)=1$$
but it has no solution for $x$ and $y$ because 
 $$ (\sqrt{y}+2)\ge 2 ; \quad (\sqrt{x}+3) \ge 3 \Rightarrow (\sqrt{y}+2)(\sqrt{x}+3)\ge 6 $$
as a result $(\sqrt{y}+2)(\sqrt{x}+3)=1$ has no solution for $x$ and $y$.
A: 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 .
A: I'll solve a related system of equations: $$A + B^2 = 7$$ and $$B +A^2=11$$
Solving for $A$ in the first gives $$A=7-B^2$$
Substituting this into the second equation gives $$B+(7-B^2)^2 = 11$$ from which we arrive at $$B^4 -14B^2+B+38=0$$

Added
Here, my goal is to find some solution(s) to this quartic (i.e. fourth degree) equation. There is a theoretical result that tells us what the rational (this includes integers) solutions will be, if there are any. 
Basically, the rational root theorem says that in this case, any rational root will divide $38$.
We (the royal we ... I) then employ a technique known as synthetic (polynomial) division. I will not go into much detail, except to mention that I usually test smaller potential roots first... so testing $\pm 1, \pm 2 $ gives us that $B = 2$ is a solution. When we have exhausted our potential rational roots, we see that this is the only rational solution.
It then follows that $A=3$, since if $B = 2$, the first equation can be rewritten as $$A + 2^2 = 7$$
A: The OP asked for the "easiest steps" to solve this system of equations. Here is what I would do:
Given that the occurring expressions are so simple I would try to get a global overview by drawing a figure. This means that we have to intersect the two graphs
$$y=7-\sqrt{x}\quad(x\geq0)\ ,\qquad y=(11-x)^2\quad(x\leq11)$$
(note restriction of $x$ for the second graph!). Looking at the figure one immediately sees that there is exactly one intersection point; one guesses that this is the point $(9,4)$. Plugging this point into the original system of equations one verifies that it is indeed a solution.
In order to prove that we were not cheated by our eyes we have to prove that the auxiliary function
$$f(x)\ :=\ (11-x)^2-(7-\sqrt{x})=114-22x+x^2+\sqrt{x}$$
has exactly one zero in the interval $[0,11]$. Unfortunately the derivative
$$f'(x)=-22+2x+{1\over 2\sqrt{x}}$$
is not strictly negative in this interval; so we have to take some extra measures:
When $0\leq x\leq5$ then $f(x)\geq 114-22x\geq4$, and when $10\leq x\leq11$ then $f(x)\leq 1-7+\sqrt{x} < -2$. For the intermediate interval $[5,10]$ we now can show that $f$ is strictly decreasing:
$$f'(x)\leq -22+20+{1\over 2\sqrt{5}} < -1\ .$$
A: we can rewrite equations as $(x-4)+\sqrt{y}-3=0$,$(y-9)+\sqrt{x}-2=0$ so we have 
$(\sqrt{x}-2)(\sqrt{x}+2)+\sqrt{y}-3=0$ ,$(\sqrt{y}-3)(\sqrt{y}+3)+\sqrt{x}-2=0$ substitute $\sqrt y$ from first equation and factor we have $(\sqrt{x}-2)((\sqrt{x}+2)(\sqrt{y}+3)-1)=0$ so we have $x=4,y=9$
A: 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.
