How do you solve #1/(x-1) + 2/(x+3) + (2x+2)/(3-2x-x^2)#?

1 Answer
Jul 18, 2016

Answer:

Maybe you see that the third fraction can be factorized.

Explanation:

#3-2x-x^2=-(x^2+2x-3)=-(x-1)(x+3)#

Which leaves the whole thing as:
#1/(x-1)+2/(x+3)-(2x+2)/((x-1)(x+3))# (mark the #-# sign)

Next step: make the bottom parts equal:
#1/(x-1)xx(x+3)/(x+3)+2/(x+3)xx(x-1)/(x-1)-(2x+2)/((x-1)(x+3))=#

#(1xx(x+3))/((x-1)(x+3))+(2xx(x-1))/((x-1)(x+3))-(2x+2)/((x-1)(x+3))=#

#((x+3)+(2x-2)-(2x+2))/((x-1)(x+3))=#

Put the numbers and the #x#'s together:
#((x+cancel(2x)-cancel(2x))+(3-2-2))/((x-1)(x+3))=((x-1))/((x-1)(x+3))=#

Now cancel:
#cancel((x-1))/(cancel((x-1))(x+3))=1/(x+3)#

NOTE:
Forbidden solutions are #x=1and x=-3# as the numerator of the fraction would be #=0#
graph{1/(x+3) [-16.02, 16.02, -8, 8.02]}