So I have to solve this: $2x \equiv 3 [10]$ and $2x \equiv 6 [10]$ in $\mathbb{Z}/10\mathbb{Z}$.
For the first one, as $2x \equiv 3 [10]$, it means $2x = 3 + 10n $ for some $n\in \mathbb{Z}$ thus $2x=3$, and we get that $2$ divides 3, which is absurd if we suppose $x \in \mathbb{Z}$, thus no solution.
Now for the second, I have $2x = 6 + 10n = 6$ thus $x = 3$.
Is my reasoning correct?
For the first one, it is correct.
For the second one, note that $2x=6+10n\iff x=3+5n$. So, $3$ is indeed a soltion, but there is another one, which is $8$. And there ar no others, since each number of the form $3+5n$ ($n\in\mathbb Z$) is congruent (modulo $10$) either to $3$ or to $8$.