Question

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

Answer 1

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]}