Verifying my proof: "the equation $2x - 6y = 3$ has no integer solution to $x$ and $y$" Question:

Prove the equation $2x - 6y = 3$ has no integer solution to $x$ and $y$.

I need to verify my proof I think I did it correctly, but am not fully sure since I don't have solutions in my book. I basically proved by contradiction and assumed there was an integer solution for x or y. I then solved for $x $ and $y$ in $2x - 6y = 3$ getting $x = 3y + 3/2$ and $y = x/3 - 1/2$ .since both $x,y$ are not integers I said it contradicts that $x$ or $y$ had an integer solution, meaning the original statement was correct. Did I prove this right, or should I redo?

Rewind to the point where you say $x=3y+3/2$. We rearrange this to $x-3y=3/2$, then note that since we have taken $x$ and $y$ to be integers, $x-3y$ is also an integer. But $3/2$ is not an integer, a contradiction.
