How do you get from one step to another in implicit differentiation?

Question

So trying to understand the step process here,

$$y^2 - 2x = 1 - 2y$$

So after a few simplifications we get:

$$yy' - 1 = -y'$$

But what I am confused on is how that turns into

$$(y+1)y' = 1$$

I know the answer and understand it, but I don't understand how the $-1$ (left side) and $-y'$ (right side) become positive.

Answer 1

\begin{align} y^2-2x &= 1-2y\\[0.5em] {} &\Downarrow {}\\[0.5em] 2yy'-2 &= -2y'\\[0.5em] {} &\Downarrow {}\\[0.5em] yy' - 1 &= -y'\\[0.5em] {} &\Downarrow {}\\[0.5em] yy'+y' &= 1\\[0.5em] {} &\Downarrow {}\\[0.5em] y'(y+1) &= 1\\[0.5em] {} &\Downarrow {}\\[0.5em] y' &= \frac{1}{y+1} \end{align}

Answer 2

Think of $y'$ as a variable, say, $Z$ $$yZ - 1 = -Z$$

Adding $Z+1$ to both side yields $yZ+Z = 1$ which becomes $Z(y+1) = 1$, or $$(y+1)y'=1$$