For $D$ a GCD domain , let $a,b,x \in D \setminus \{0\}$ , then is it true that $\gcd (ax,bx)=x \cdot \gcd (a,b)$? [duplicate]

Let $$D$$ be a GCD domain, and let $$a,b,x \in D \setminus \{0\}$$. Then is it true that $$\gcd (ax,bx)=x \cdot \gcd (a,b)$$ ?

Let $$c=\gcd (a,b)$$ and $$d=\gcd(ax,bx)$$, then as $$cx|ax$$ and $$cx|bx$$ so $$cx|d$$. We would be done if we could show $$d|cx$$, but I am unable to show that. Please help me to solve this problem.

Hint $$\$$ By below $$\,(a,b)\,$$ and $$\,(ax,bx)/x\,$$ divide each other. See here for a few more proofs.
$$\ c\mid(a,b)\!\iff\! c\mid a,b \!\iff\! cx\mid ax,bx \!\iff\! cx\mid (ax,bx) \!\iff\! c\mid (ax,bx)/x,\$$ for any $$\,c.\,$$
Since $$x \mid ax$$ and $$x \mid bx$$ it follows that $$x \mid d$$. The codivisor $$dx^{-1}$$ of $$x$$ in $$d$$, divides $$(ax)x^{-1}=a$$ and $$(bx)x^{-1}=b$$, thus $$dx^{-1}$$ divides $$\gcd(a,b)=c$$. Thus $$d=(dx^{-1})x$$ divides $$cx$$.
• $x$ is not necessarily a unit ... – user228168 Sep 19 '16 at 9:05
• Yes I know that $x$ is not necessarily a unit in $D$. This is why I made a point of saying "[t]he codivisor $dx^{−1}$ of $x$ in $d$." Since $x \mid d$, there is some element $m \in D$ such that $mx = d$. This $m$ is $dx^{-1}$. You can take this as notation, or you can think of $x^{-1}$ being an element of the quotient field of $K$. And you know that $dx^{-1}$ is an element of $D$. It is like writing $\frac{12}{2}$, which is an integer, while $2$ is not a unit in the integers. – quid Sep 19 '16 at 12:41