Manipulating this $\frac{x-y}{z-y}$ to $1+\frac{x-z}{z-y}$ There is probably a very easy explanation for this that is lost on me. Came across a formula that was manipulated into another form and it was presented as a given, so I am trying to figure out how that was done.
Original:
$\frac{x-y}{z-y}=1+\frac{x-z}{z-y}$
So I began to break it down:
$\frac{x-y}{z-y}=\frac{x}{z-y}-\frac{y}{z-y}$
$\frac{x-y}{z-y}=\frac{x}{z-y}-(\frac{z}{z-y}-\frac{y}{y})$ <-- maybe this is wrong, but it works and I don't know why.
Opening up the bracket:
$\frac{x-y}{z-y}=\frac{x}{z-y}-\frac{z}{z-y}+1$
$\frac{x-y}{z-y}=1+\frac{x-z}{z-y}$
As I typed through all this, I see that $\frac{y}{z-y}=\frac{z}{z-y}-1$ is just true, and I get it when I simplify. I don't get how someone could see that to begin and wish to expand a formula like that.
Thank you for your time and patience with my high school level math question.

A super easy way to get what you want is to just add it in (then subtract it so that you don't change the expression's value).
$$\begin{align}
\frac{x-y}{z-y} + (1+\frac{x-z}{z-y}) - (1+\frac{x-z}{z-y}) &= 1+\frac{x-z}{z-y}+\frac{(x-y)-(z-y)-(x-z))}{z-y}\\
&=1+\frac{x-z}{z-y}\\
\end{align}$$
It's especially useful for doing messy trig proofs, e.g..
$$\begin{align}
\frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} &= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - (2\sin x - 1) + (2\sin x - 1)\\
&= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - \frac{(2\sin x - 1)(\sin x - 2)}{\sin x - 2} + (2\sin x - 1)\\
&= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - \frac{2\sin^2 x -5\sin x + 2}{\sin x - 2} + (2\sin x - 1)\\
&= 0+ (2\sin x - 1)\\
&= 2\sin x - 1\\
\end{align}$$
