$\mathbf{Question}$: Determine the integers $n$ for which $\mathbb{Z}_n$, the set of integers modulo $n$, contains elements $x,y$ so that $x+y=2$, $2x-3y=3$.
$\mathbf{Attempt}$: Let us put $x=n\alpha+p$ and $y=n\beta+q$. Now, $x+y=n(\alpha+\beta)+p+q=2 \implies p+q-2\equiv0 (\mod n)$. Similarly, $2p-3q-3\equiv 0(\mod n)$.
Combining $p+q\equiv 2 \mod(n)$ and $2p-3q\equiv 3 \mod (n)$, we get:
$5p \equiv 9 (\mod n) $ and $5q\equiv 1(\mod n)$. If the solution to the latter linear congruence were to exist, it must be case that $\gcd(5,n)=1$. Therefore, the set of required integers becomes: $\{n \in \mathbb{Z}: \gcd(5,n)=1\}$.
If $\gcd(5,n)=1$, we get the solution to both of the linear congruences. For example, taking $n=6$:
$p \equiv 3 (\mod 6)$ and $q \equiv -1 (\mod 6)$. We get $x+y-2 \equiv0 (\mod 6)$ and $2x-3y-3 \equiv 0 (\mod 6)$.
Is the procedure okay?
Kindly $\mathbf{VERIFY}$
Yes, it is fine. You may simplify the answer using that $5$ is prime: $\gcd(5,n)=1$ simply means that $5$ does not divide $n$.
This being said, the conclusion is obtained in a shorter way considering these conditions as a linear system in $\mathbf Z/n\mathbf Z$; the condition becomes that the determinant $-5$ of the l.h.s. of the linear system is a unit modulo $n$.