calculation – How can a function with a hole (removable discontinuity) be equivalent to a function without a hole?

I have done some research and hope that someone will be able to control me. My question was:

Suppose I have the function $ f (x) = frac {(x-3) (x + 2)} {(x-3)} $, so he has removable discontinuity to $ x = $ 3. We remove this discontinuity with algebra: $ f (x) = frac {(x-3) (x + 2)} {(x-3)} = (x + 2) $. BUT, the graph of the first function has a hole to $ x = $ 3, and the graph of the second function is continuous everywhere. How can they be "equal" if one has a hole and the other does not?

I thought that's the answer:

Because the original function is undefined at the point $ x = $ 3we have to limit the domain to $ mathbb {R} setminus $ 3. And when we manipulate this function with algebra, the final result, $ f (x) = (x + 2) $ always use this restricted domain. So, even if the function $ f (x) = (x + 2) $ would not have a hole if the domain were all $ mathbb {R} $, we're sort of "imposing" a hole in $ x = $ 3 by continuing to throw this point out of the domain.

And then just to close the loop: Removing the removable discontinuity is useful because it allows us to "pretend" that we are working with a function that is everywhere continuous, which helps us to easily find the limit. But the reality is that the function $ f (x) = (x +2) $ is in fact NOT continuous everywhere when we restrict the domain by rejecting point 3. Or am I pushing things too far?

Thanks in advance!