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.

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
$$(5x+4)(x-1)^3=0$$
