Proving if $\gcd(c,m)=1$ then $\{x\in \Bbb Z \mid ax\equiv b \pmod m\} =\{x\in \Bbb Z \mid cax\equiv cb \pmod m\}$ Okay so I'm confused on how to approach this question.

If $\gcd(c,m)=1$, then $S=T$ where $S=\{x\in \Bbb Z \mid  ax\equiv b \pmod m\}$ and $T=\{x\in \Bbb Z \mid cax\equiv cb \pmod m\}$.

I know that since $c$ and $m$ are coprime, then there exists two integers $y$ and $z$ such that $cy+mz=1$. Also, I know that to prove $S = T$, I need to show that $S\subseteq T$ and $T\subseteq S$.
But I'm stuck here and don't know how to proceed. Any help would be really appreciated. Thanks!

It might be worth stating and proving Bezut here. Bezuts lemma says that if $(a,b)=1$, then $\exists x,y$ st $ax+by=1$. To prove this, consider the set $S:=\{d>0|\exists x,y,ax+by=d\}$. Let $d_0$ be the minimal element of this set and use the division algorithm on $a$ and $d_0$ to find out that $d_0|a$ (Specifically, the remainder is of the form of elements of $S$ but is smaller than the least element of $S$, so the remainder is zero). Then repeat with $b$ and conclude $d_0=1$
