How can I solve this system? Having this system:
$$\begin{cases}
2 x(x^2 + y - 11) + x + y^2 - 7 = 0,\\
x^2 + y - 11 + 2 y (x + y^2 - 7) = 0 .
\end{cases}$$
How can I solve it without using a computer?. 
Thank you!
 A: Note first that a substitution of $x=0$ yields the system $$\begin{cases}y^2-7=0\\y-11+2y(y^2-7)=0,\end{cases}$$ which you can show has no solutions in $y$. Thus, we needn't worry about any problems that might arise from having $x=0$.

Now, multiplying the second equation in our original system by $2x$ gives us the equivalent system $$\begin{cases}2x(x^2+y-11)+x+y^2-7=0\\2x(x^2+y-11)+4xy(x+y^2-7)=0,\end{cases}$$ from which it follows that $$(4xy-1)(x+y^2-7)=0,$$ meaning $$y=\frac1{4x}\quad\text{or}\quad x=7-y^2.$$ Now, by $x=7-y^2$ and $x^2+y-11+2y(x+y^2-7)=0,$ we find $$0=(7-y^2)^2+y-11=y^4-14y^2+y+38.$$ Using the rational root test and a little trial and error, we find that $2$ is a root of that equation, and we can factor it as $$0=(y-2)(y^3+2y^2-10y-19).$$ Thus, $(3,2)$ is one solution to the system. We can explicitly calculate the roots of that cubic factor using Cardano's Method--be warned, they are real, even though they won't look like it, and they aren't very nice in any case--and from substituting those $y$-values back into $x=7-y^2$ find the corresponding $x$-values.
From $y=\frac1{4x}$ and $2x(x^2+y-11)+4xy(x+y^2-7)=0$ we find $$0=2x^3+\frac12-22x+x+\left(\frac1{4x}\right)^2-7=2x^3-21x-\frac{13}2+\frac1{16x^2},$$ whence multiplication by $16x^2$ yields $$0=32x^5-336x^3-104x^2+1.$$ This is messier, still. No rational roots exist, and quintics don't have a nice general method of solution. I wouldn't hold out much hope....
A: Let $a = x^2+y-11$ and $b = x+y^2-7$.
The equations we get are as follows:
$$2ax+b=0$$
$$a+2by=0$$
Replacing $a$ in the first equation yields:
$$-4bxy+b = 0$$
which gives us two options: $1-4xy = 0$ ($x = 1/4y$) or $b = 0$ ($x = 7-y^2)$. In both cases, we have expressed $x$ explicitly, we can insert it back the second original equation and solve it.
1.) $(1/4y)^2 + y - 11 + 2y(1/4y + y^2 -7) = 0$
that is (hopefully) equivalent to
$$ 32y^5 -14y^4+16y^3-21\cdot 18y^2+1 = 0$$ 
which looks painful.
2.) $(7-y)^2 + y - 11 = 0$ which is easily solvable.
