Solve for y: $\ y = yCe^x -Ce^x + 2 $

Question

How is

$\ y = yCe^x -Ce^x + 2 $

equivalent to

$\ y = [2-Ce^x]/[1-Ce^x] $

On an interval $\ 1 < y <2 $ ? I do not quite undestand how the get to the second form.

Answer 1

$$y=yCe^x-Ce^x+2$$ $$2-Ce^x=y-yCe^x$$ $$2-Ce^x=y(1-Ce^x)$$ $$y=\frac{2-Ce^x}{1-Ce^x}$$

Answer 2

Rearrange as $(1-Ce^x)y=2-Ce^x$, then divide. The range for $y$ is equivalent to $0<1/(1-Ce^x)<1$ i.e. $1<1-Ce^x$ i.e. $Ce^x<0$ , which holds for all real $x$ if $C<0$.