Show that $ax \equiv b (mod\ m) $ has solution iff $gcd(a,m)$ divides $b$

Question

Here's what I have:

$ax \equiv b (mod\ m)$ has answer if there are $x$ and $y$ such that

$b = ax + my$

Let $d = gcd(a,m)$. Then:

$d|a$ and $d|m \Leftrightarrow d|ax$ and $d|my \Leftrightarrow d|(ax+my)$

Since $m$ divides the right part of the equation, it also has to divide the left part.

Is this a valid proof for what I want?

Answer 1

If $\;d=gcd(a,m)\;$, then there exist $\;r,s\in\Bbb Z\;$ s.t. $\;ra+sm=d\;$ , so

$$b=cd\implies b=c(ra+sm)=a(cr)+m(cs)$$

and we have a solution.