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: 
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 
$$
5y^4 + 9y^3 - 11y^2 - 12y - 13=0.
$$
The resultant of $3\,{x}^{2}-xy-4\,x+2\,y-2$ and $
x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is
$$ 10\,{x}^{4}-24\,{x}^{3}-10\,{x}^{2}+42\,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots, 
$x \approx -1.287147510$ and $x \approx 0.2049816008$, which correspond to 
$y \approx -2.469872787$ and $y \approx 1.500750095$ respectively.

EDIT: For the corrected question, the resultant of $3\,{x}^{2}+xy-4\,x+2\,y-2$ and $x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is
$$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$
Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.