Manipulating this $\frac{x-y}{z-y}$ to $1+\frac{x-z}{z-y}$

Question

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.

Answer 1

Suppose that we have the expression $\dfrac{x-y}{z-y}$ and we want the numerator to be independent of $y$. Then the only way to do this is to make part of the numerator like the denominator, so that both cancel: $$\frac{x-y}{z-y}=\frac{z-y+x-z}{z-y}=\frac{z-y}{z-y}+\frac{x-z}{z-y}=1+\frac{x-z}{z-y}$$ This is sometimes useful in integration, factorisation and other related topics.

Answer 2

$ \frac{x-y}{z-y} = \frac{z - y + x - z}{z-y} = 1 + \frac{x - z}{z-y} $

Answer 3

Hint: Write $$\frac{x-y}{z-y}=\frac{z-y+x-z}{z-y}$$ I think it is a nulladdition

Answer 4

You are certainly right to say that $$\frac{y}{z-y}=\frac{z}{z-y}-\frac{y}{y}$$ Since $$\frac{z}{z-y}-\frac{y}{y}=\frac{z}{z-y}-1=\frac{z}{z-y}-\frac{z-y}{z-y}=\frac{y}{z-y}$$

Answer 5

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.. $$\begin{align} \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} &= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - (2\sin x - 1) + (2\sin x - 1)\\ &= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - \frac{(2\sin x - 1)(\sin x - 2)}{\sin x - 2} + (2\sin x - 1)\\ &= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - \frac{2\sin^2 x -5\sin x + 2}{\sin x - 2} + (2\sin x - 1)\\ &= 0+ (2\sin x - 1)\\ &= 2\sin x - 1\\ \end{align}$$