how to solve 2x=4 in $Z_{12}$

Question

How do I solve 2x=4 in $Z_{12}$

I know the $gcd(2,12) = 2$ and $2|4$ therefore there are 2 solutions, but I'm not sure how to solve this. I tried using the euclidean algorithm but it doesn't seem to work with numbers this small.

Answer 1

$\begin{align}{\bf Hint}\qquad\ \ \ 2x&\equiv 2a\!\pmod{\!2n}\\[.2em] \iff \color{#c00}2x &= \color{#c00}2a + \color{#c00}2nk \ \ \ {\rm for\ some}\ \ k\in\Bbb Z\\[.2em] \iff\ \ x &= \ \ a +\ \ nk \ \ \ {\rm for\ some}\ \ k\in\Bbb Z\\[.2em] \iff\ \ x&\equiv\ \ a\!\pmod{\!n} \end{align}$

Answer 2

That's very simple: since $d$ divides all coefficients and the modulus, you can simplify, since $\mathbf Z$ is an integral domain: $$2x\equiv 4\mod 12\iff x\equiv 2\mod 6$$