How does a value not part of the domain satisfy the function?

Question

Consider $$y= \frac{(x^2-9)}{x-3}$$ For y to be meaningful x must not equal 3 , So, 3 is not included within the domain of the function y (Call it f(x) if you will)

It follows that, $$yx-3y = x^2 -9 $$ Plug in 3 to x out of curiosity, I find, $$ 0=0 $$ A perfectly valid relation! How is this possible?

This problem is even more predominant in partial fraction decomposition. Have a look at this example I found on the web, Partial fraction decomposition x cannot equal -1 or 2. How is it giving us the right answers then? What is the math behind this? When I do these kind of problems it feels like something's is happening mathematically, hidden from my view

All help is appreciated

Answer 1

we have $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3$$ only for $$x\ne 3$$

Answer 2

By definition, we have

$$y=\frac{x^2-9}{x-3}$$

and at $x=3$,

$$y=\frac00=\text{undefined}$$

If your definition had instead been

$$y(x-3)=x^2-9$$

which is not the same as before, we would then note that at $x=3$,

$$y(0)=0\implies y=\text{anything}$$

which is completely different.

Answer 3

When you multiply the right hand side by $x-3$, you get: $$\frac{x^2-9}{x-3}\cdot (x-3) \neq x^2-9$$

since those two functions are not equal for ALL values of $x$ (in particular $x=3$).

What we can say is that for all $x \neq 3$, the following equality holds, $$\frac{x^2-9}{x-3}\cdot (x-3) = x^2-9.$$

So it is only with the stipulation $x \neq 3$ that we can make the simplication necessary to continue the problem and that is a stipulation you need to carry out throughout the entire problem.