What does applying Z-transform to an equation mean?

Question

I have read the Wikipedia page on Z-transform and came across one section I couldn't manage to understand.
I am now referring to the section called "Linear constant-coefficient difference equation"
Suppose signal $x$ is mapped to signal $y$ via some linear system. suppose also the system holds$$\sum_{p=0}^Ny(n-p)\alpha_p=\sum_{q=0}^Mx(n-q)\beta_q$$ where $\alpha_p,\beta_q$ are coefficients. It is not stated but I suppose $M,N$ are the support of the signal, and as I understand it, the equation must hold for all $n$.
Now, in the next section the author claims that by applying the Z-transform on each member of the equation we can get a nice expression for the transfer fucntion.
And here is the part I don't understand - the Z-transform is defined for signals, or some discrete serial data, not for plain numbers (as each side of the equation is). What does it mean to apply it to the equation as a whole?

Answer 1

The issue may become clearer if $N=0$ and $\alpha_0=0$. Then $y(n) = \sum_{k=0}^M\beta_k x(n-k)$ for each $n$. Now, write these guys below each other in a huge vector: $$ y = (y(n))_{n\in\mathbb Z} = \left(\sum_{k=0}^M\beta_k x(n-k)\right)_{n\in\mathbb Z} = \sum_{k=0}^M\beta_k(x(n-k))_{n\in\mathbb Z} = \sum_{k=0}^M\beta_kT_kx, $$ where $T_k$ is the time-delay operator. Since $Z(T_kx) = z^{-k}X(z)$, we get $$ Y(z) = X(z)\sum_{k=0}^M\beta_kz^{-k}. $$