Rewrite the equation as $x(3y+2) = 12 - y$. Obviously, $3y+2 \neq 0$ divides $12 - y$, so $3y+2$ divides $3(12 - y) + (3y+2) = 38$. Since $3y+2$ gives the remainder of $2$ when divided by $3$, there are four possibilities. The number $3y+2$ is equal to $-19, -1, 2$ or $38$, so $y$ is equal to $-7, -1, 0$ or $12$ and $x$ is equal to $-1, -13, 6$ or $0$ respectively. The solutions are $(-1, -7), (-13, -1), (6, 0)$ and $(0, 12)$.