We have the following system with $a$ and $b$, real numbers:
$ \begin{cases} ax + y + z = 4 \\ x + 2y + 3z = 6 \\ 3x - y - 2z = b \\ \end{cases} $
Show that $\forall a \in \mathbb{Z}$ there is a $b \in \mathbb{Z}$ such that the system admits a solution composed by integers.
OK, how do you tackle this?
Given $a$, let $x$ be any integer you want. The system, $y+z=4-ax$, $2y+3z=6-x$, has a unique solution in integers $y,z$. Now that you have integers $x,y,z$, the third equation makes $b$ an integer.