Does the LCM of $ax$ and $bx$ equal $\operatorname{lcm}(a,b)\cdot x$? Let $a,b,x \in Z^+$. Prove that $\operatorname{lcm}(ax,bx) = \operatorname{lcm}(a,b)\cdot x$.
Here are my thoughts: 
Let $d = \operatorname{lcm}(ax, bx)$. By definition $ax|d$ and $bx|d$. Now it can be seen that $a|d$ and $b|d$. So, let e = lcm(a,b). e is merely the lcm(ax, bx) (which equals d) multiplied by x. So, ex = d, which means that x $\cdot$ lcm(a,b) = d." 
I am not certain of my proof's validity and I feel it is too informal to be considered valid. Any thoughts?

A nice fact that you can use is that for any $a,b\in \mathbb{Z}$, $$\text{lcm}(a,b) = \frac{ab}{\gcd(a,b)}.$$ So then you have that $$\text{lcm}(ax,bx)=\frac{x^2ab}{\gcd(ax,bx)}= \frac{x^2ab}{x \cdot \gcd(a,b)} = x \cdot \left( \frac{ab}{\gcd(a,b)} \right) = x \cdot \text{lcm}(a,b).$$
