Question with Divisibility proof

Question

I have a simple proof question:

Suppose $a,b \in \Bbb Z$ where $a|b$. If $a|(b-c)$, then $a|c$.

I have solved it below, but is my way a valid answer? Is there a better clear way of proving this?

Suppose $a|(b-c)$ is true given $a|b$.

Then $\frac{b}{a} = k, k \in \Bbb Z. $ Re-arrange for b: $b = ka$

Then $\frac{b-c}{a} = \frac{ka-c}{a} = \frac{ka}{a}-\frac{c}{a}$

Since $\frac{b-c}{a} = \frac{ka}{a}-\frac{c}{a}$, it shows that a divides c, $a|c$

QED

Answer 1

If $a,b,c\in\mathbb Z$ and $a\mid b$ and $a\mid b-c$, then $b=ak$ and $b-c=am$ for some $k,m\in\Bbb Z$, so $c=b-am=ak-am=a(k-m)$, where $k-m\in\mathbb Z$, so $a\mid c$.