Is $y-\frac{1}{x} = f(x)-\frac{1}{x}$ equivalent to $y=f(x)$? I really thought I had basic algebra figured out, but I was trying to provide an argument as to why this is true, and I don't know how to justify it.
If $x=0$, then $1/x$ is undefined, but I am also quite sure that $\frac{1}{x} - \frac{1}{x} = 0$.
Is $y-\frac{1}{x} = f(x) - \frac{1}{x}$ defined at $x=0$? If not, then how can it be equivalent to $y = f(x)$, which is defined at $0$?
 A: Note that
$$\frac{1}{x} - \frac{1}{x} = 0$$
is true only for $x\neq 0$ and
$$y-\frac{1}{x} = f(x) - \frac{1}{x}$$
with $y=f(x)$ is true for $x\neq 0$ for all $x$ for which $f(x)$ is defined.
A: Though
$$y-\frac1x=f(x) - \frac1x$$
doesn't make sense when $x=0$, if $f$ and $y$ are continuous and defined at $0$ the fact that they are equal everywhere else implies that they are also equal at $0$.
When software like Mathematica does algebraic manipulation, sometimes it does it without regard to whether the expression is defined at all values, though I have seen it do things like specify $x\neq 0$ in this case. In a certain context, such manipulations are completely correct, such as when we have no intention of substituting anything into $x$ and we are working in the field of rational functions. For your context this is not the case, and we must ensure that $x\neq 0$.
A: It sounds like if $$5+\infty =6+\infty$$ can we say that $5=6?$
The answer is no. 
If $x=0$, then neither  $f(x)-1/x$ nor $y-1/x$ is defined so they can not be equal if they are not defined.
On the other hand if $x\ne 0$ we may cancel $1/x$ and get $ y=f(x)$ provided that $f(x)$ is defined.
