Let $a$ and $b$ be relatively prime: show $a\mid bc\implies a\mid c$

Question

Let $a,b,c$ be integers, with $a$ and $b$ relatively prime. Then $$ a\mid bc\implies a\mid c. $$

The proof starts as follows:

Let $x,y\in\mathbb Z$ with $xa+yb=1$. From $a\mid bc$ we get $a\mid xac+ybc=\dots$

I can follow the rest of the proof. However, I don't understand why $$ a\mid bc\implies a\mid xac+ybc. $$ I can see that $a\mid bc\implies a\mid ybc.$ But How do they get $xac$?

Answer 1

$a\mid xac$ trivially since $a=xc\cdot a$. Hence $a\mid xac$ and $a \mid ybc$. This implies $a \mid xac+ybc$. As simple as that.

Answer 2

First notice $a|xac$ because $xac=a(xc)$. Then notice $a|ybc$ because $a|bc$ so if $bc=ak$ then $ybc=y(ak)=a(yk)$.

Now that we know that $a|xac$ and $a|ybc$ we have $a|xac+ybc$.