Show that if $[m]_7 = [n]_7$, then $[3m]_7 = [3n]_7$ for all $m,n \in \mathbb{Z}.$ 
Show that if $[m]_7 = [n]_7$, then $[3m]_7 = [3n]_7$ for all $m,n \in \mathbb{Z}.$

I remember that we had the property that $a\equiv b \pmod{7}$ iff $[a]_7 = [b]_7$. However, I couldn't find a way to use this here. If I have that $m\equiv n \pmod{7}$ then shouldn't I be able just to multiply the expression by $3$ to get  $3m\equiv 3n \pmod{7}$? I don't have any experience in proving residue classes so apologies if this is poorly expressed.

By definition, we have
$$\begin{align}
[a]_k=[b]_k&\iff a\equiv b\pmod{k}\\
&\iff k\mid a-b\\
&\implies k\mid \ell(a-b)\\
&\iff k\mid \ell a-\ell b\\
&\iff \ell a\equiv \ell b\pmod{k}\\
&\iff [\ell a]_k=[\ell b]_k.
\end{align}$$
Now just let $a=m, b=n, k=7, \ell=3$.
