Question: > 
> Solve the following system of equations:
>  $$\large
> \left\{
> \begin{align\*}
> 3x^2 + xy - 4x + 2y - 2 = 0\\
> x(x + 1) + y(y + 1) = 4
> \end{align\*}
> \right.
> $$
> 
> 
> 

I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.

Answer: 
Solving the first equation for $y$ we get $$y=\frac{-3x^2+4x+2}{2+x}$$ for $$x\neq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$
Substituting for the updated equation yields
$$
x=-\frac{4}{5}, \; y=-\frac{13}{5}
$$
or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).

$$
x=\frac{5y^3 - 26y^2 - 24y + 91}{65},
$$
with 
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$