Hệ đã cho tương đương với: $\left\{\begin{array}{l}(3x+2y)(x^2+x)=12\\(x^2+x)+(3x+2y)=8\end{array}\right.$
Đặt $u=3x+2y;v=x^2+x$ hệ trở thành:
$\left\{\begin{array}{l}uv=12\\u+v=8\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}u=6\\v=2\end{array}\right.\\\left\{\begin{array}{l}u=2\\v=6\end{array}\right.\end{array}\right.$
Với $\left\{\begin{array}{l}u=6\\v=2\end{array}\right.$ ta có: $\left\{\begin{array}{l}3x+2y=6\\x^2+x=2\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=1\\y=\dfrac{3}{2}\end{array}\right.\\\left\{\begin{array}{l}x=-2\\y=6\end{array}\right.\end{array}\right.$
Với $\left\{\begin{array}{l}u=2\\v=6\end{array}\right.$ ta có: $\left\{\begin{array}{l}3x+2y=2\\x^2+x=6\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=2\\y=-2\end{array}\right.\\\left\{\begin{array}{l}x=-3\\y=\dfrac{11}{2}\end{array}\right.\end{array}\right.$