For expressions with some holes, why are they equal to expressions with less holes.
For instance, (example is in the comments)

Both expressions share a hole of x = 0, but the second one has a hole of x = -1 while the other one does not. I appreciate all insight.

Question

For expressions with some holes, why are they equal to expressions with less holes. 
For instance, (example is in the comments)

Both expressions share a hole of x = 0, but the second one has a hole of x = -1 while the other one does not. I appreciate all insight.

anonymous · Answer

$$\frac{ 1 }{ x } = \frac{ x+1 }{ x(x+1) }$$

anonymous · Answer

The two expressions aren't equal to each other for all values of $x$.

When $x=-1$, you have
$$-1=\frac{0}{0}$$
which is indeterminate, and when $x=0$ you have
$$\frac{1}{0}=\frac{1}{0}$$
which is undefined.

But for any other value of $x$, say $x=4$, the expressions are equivalent.
$$\frac{1}{4}=\frac{5}{20}$$

anonymous · Answer

The reason the expressions are equivalent for all $(-\infty<x<-1)\cup(-1<x<0)\cup(0<x<\infty)$ is because the discontinuity at $x=-1$ can be removed; a factor of $x+1$ is present in both the numerator and denominator, so you can cancel (but only when $x=-1$; otherwise you cannot).

anonymous · Answer

Okay, so effect, we are saying they are equal because the hole can be removed. But if we looked at the hole for any other reason, then it would be not equal.