How is she getting this?

[hopelynnwelch]

This is a solution to one of the problems on the calculus test I just bombed and I for the life of me can't see where she goes from

(2-2x)/[(2x+1)(x-1)]

to

(-2)/(2x+1)

I think I have been looking at math too long...

Can someone just explain this?




The original function was f of x = 3/(2x+1)

 

[Ishuda]

I've had those days too

(2-2x)/[(2x+1)(x-1)]=\(\displaystyle \frac{2-2x}{(2x+1)(x-1)}=\frac{2 (1-x)}{(2x+1)(x-1)}\)

=\(\displaystyle \frac{-2 (x-1)}{(2x+1)(x-1)}=\frac{x-1}{x-1}\frac{-2 }{2x+1}\)\(\displaystyle =\frac{-2}{2x+1}\)

 

[hopelynnwelch]

I've had those days too

(2-2x)/[(2x+1)(x-1)]=\(\displaystyle \frac{2-2x}{(2x+1)(x-1)}=\frac{2 (1-x)}{(2x+1)(x-1)}\)

=\(\displaystyle \frac{-2 (x-1)}{(2x+1)(x-1)}=\frac{x-1}{x-1}\frac{-2 }{2x+1}\)\(\displaystyle =\frac{-2}{2x+1}\)

I swear I factored out the negative like a billion times and still couldn't see that! ugh lol

Thank you!