<s>[INST] Solve the following system of equations - (3) 
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.
 [/INST] 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 $y = 1$.
</s>