B1. Find all integer solutions $x$ and $y$ for the equation

$$
3 x y+2 x+y=12
$$
B1. We rearrange the equation to $x(3 y+2)=12-y$. Clearly, $3 y+2 \neq 0$ and divides $12-y$. Therefore, the number $3 y+2$ divides $3(12-y)+(3 y+2)=38$. Since $3 y+2$ leaves a remainder of 2 when divided by 3, we have four possibilities. The number $3 y+2$ is equal to $-19, -1, 2$ or 38. Consequently, the number $y$ is equal to $-7, -1, 0$ or 12, and then the number $x$ is equal to $-1, -13, 6$ or 0. The solutions for $(x, y)$ are $(-1, -7), (-13, -1), (6, 0)$ and $(0, 12)$.

The expression of one of the equations $x(3 y+2)=12-y$ or $y(3 x+1)=12-2 x$ or the statement

![](https://cdn.mathpix.com/cropped/2024_06_07_4050f14f84cf4f327f67g-10.jpg?height=66&width=1579&top_left_y=1829&top_left_x=273)
The conclusion that $3 y+2$ divides $12-y$ (or $3 x+1$ divides $12-2 x$ or the estimate $|3 y+2| \leq|12-y|$ or the estimate $|3 x+1| \leq|12-2 x|$) .................................................................. Justification that only a finite number of possibilities remain and listing all these possibilities . . 2 points All solutions $(-1, -7), (-13, -1), (6, 0)$ and $(0, 12) \ldots \ldots \ldots \ldots \ldots \ldots . .$. (If the contestant writes 2 or 3 solutions, award 1 point.)