Is the equation $3x+3=0$ solvable in Z6? I start off by adding 3 to each side of the equation because 3 is the additive inverse of itself in Z6.
$3x+3+3=0$
Then, because 3+3=0 in Z6, I have:
$3x=3$
Next I need to find the multiplicative inverse of 3 in Z6 to go further, but there is no number which can multiply by 3 in order to get 1 in Z6. Does this mean that the equation is unsolvable?
 A: Notice that $6=2\times 3$, hence for $3x+3$ to be equal to $0$ in $\mathbb Z_6$ it needs to be zero in both $\mathbb Z_2$ and $\mathbb Z_3$. It's always zero in $\mathbb Z_3$ (can you see why?) hence we only need $3x+3=x+1=0$ in $\mathbb Z_2$. So $x=1$ is the unique solution in $\mathbb Z_2$, which translates to $x=1,3,5$ in $\mathbb Z_6$.
Note that you don't actually need to find the inverse of $3$ for example to find the solution. If for some $n$, $m$ is a unit (meaning it has an inverse) in $\mathbb Z_n$, then the equation $m(x+k)=0$ has a unique root regardless of $k$, because $x+k=m^{-1}0=0$, so the unique solution is $x=-k$. Note that this however does not imply that if $m$ is not a unit, then the equation has no solution. It simply means it might not be unique. This is not to say that surely it must have one or many solutions, but the simple fact that $m$ is not a unit is not enough to draw any conclusions whatsoever.
A: Hint $\,\ 3x = -3 + 6n\!\!\overset{\rm cancel\ 3\!\!\!}\iff x = -1+ 2n\iff x\, $ is odd.
Expressed in (nonstandard) fractional language it is
$\qquad\quad x\equiv \dfrac{-3}3\pmod{\! 6}\,\equiv\,\dfrac{-1}1\pmod{\!2} $
where we need to cancel $3$ from the numerator, denominator and modulus, corresponding to all $3$ summands in the original equation. See this answer for a rigorous presentation of the fractional viewpoint
