IF you were given that "x= 5a" then, yes, multiplying on both sides by 3, "3x= 15a". But in your first post you **didn't** say "x= 5a". You said "I have variable x which it gives me volume 5a" which I would interpret as saying that you have some function that, applied to x, gives 5a. But there are infinitely many different functions, f, such that f(x)= 5a. One such is \(\displaystyle f(u)= \frac{5au}{x}\). When u= x, \(\displaystyle f(x)= \frac{5ax}{x}= 5a\). And for that function, \(\displaystyle f(3x)= \frac{5a(3x)}{x}= 15a\). But another such function is \(\displaystyle f(u)= \frac{5ax}{u}\). Again, \(\displaystyle f(x)= \frac{5ax}{x}= 5a\). But now \(\displaystyle f(3x)= \frac{5ax}{3x}= \frac{5}{3}a\).