Does $\frac{x-2}{3x-6}$ really equal $\frac{1}{3}$?

In my maths lesson today we were simplifying fractions by factorising. One question was something like this: $\frac{x-2}{3x-6}$, which I simplified as $\frac{x-2}{3x-6}=\frac{x-2}{3(x-2)}=\frac{1}{3}$. It got me wondering however, whether these expressions are really equal, specifically in the case $x=2$, where the former expression is undefined but the latter takes the value $\frac{1}{3}$

Since the expressions only differ at a single point are they for all intents and purposes equal, or are they theoretically different? If I wanted to be entirely correct would I have to write $\frac{x-2}{3x-6}=\frac{1}{3}$ where $x\ne 2$?

My maths teacher explained that at $x=2$ the expression evaluates to $\frac{0}{3\times 0}$ and the zeros effectively cancel out. I wasn't altogether satisfied with this explanation because as far as I know $\frac{0}{0}$ is undefined.

Thanks in advance!

In my maths lesson today we were simplifying fractions by factorising. One question was something like this: $\frac{x-2}{3x-6}$, which I simplified as $\frac{x-2}{3x-6}=\frac{x-2}{3(x-2)}=\frac{1}{3}$. It got me wondering however, whether these expressions are really equal, specifically in the case $x=2$, where the former expression is undefined but the latter takes the value $\frac{1}{3}$

Since the expressions only differ at a single point are they for all intents and purposes equal, or are they theoretically different? If I wanted to be entirely correct would I have to write $\frac{x-2}{3x-6}=\frac{1}{3}$ where $x\ne 2$?

My maths teacher explained that at $x=2$ the expression evaluates to $\frac{0}{3\times 0}$ and the zeros effectively cancel out. I wasn't altogether satisfied with this explanation because as far as I know $\frac{0}{0}$ is undefined.

Thanks in advance!