Is $y=\frac{x^{2}-1}{x-1}\;,\;x\neq1$ belong to two-variables linear equation? I have an equation as below:

$$y=\frac{x^{2}-1}{x-1}\;\;,\;\;x\neq1$$

Question:
Does it belong to two-variables linear equation?

My effort:
The equation can be simplified
$$y=\frac{(x+1)(x-1)}{(x-1)}\;\;,\;\;x\neq1$$
$$y=x+1\;\;,\;\;x\neq1$$
The equation above in the form y = mx + c but I think it is not belong to two-variables linear equation because there is a hole (1,2) in line graph. Is it true?
If you calulate $\lim\limits_{x\rightarrow1}f(x) = 2$
This limit is FINITE!
You can thus create an alternate function $\tilde f$ defined by
$$\tilde f(x)=\left\{
\begin{matrix}
\frac{x^2-1}{x-1}\ \text{ if }\ x\ne1\\
2  \text{ if }\ x=1
\end{matrix}
\right.$$
Which is a continuous function and is equal to $y=x+1$ everywhere
If you calulate $\lim\limits_{x\rightarrow1}f(x) = 2$
This limit is FINITE!
You can thus create an alternate function $\tilde f$ defined by
$$\tilde f(x)=\left\{
\begin{matrix}
\frac{x^2-1}{x-1}\ \text{ if }\ x\ne1\\
2  \text{ if }\ x=1
\end{matrix}
\right.$$
Which is a continuous function and is equal to $y=x+1$ everywhere