Question

subtract. simplify, if possible.
x - 2
--- ---
x^2+5x+6 x^2+3x+2

Answer 1

\[\frac{x}{(x+3)(x+2)}-\frac{2}{(x+2)(x+1)}\]

Answer 2

now look at the bottoms
we want the bottoms to be the same
so what do we need to multiply the first bottom by

Answer 3

what happens if i multiply the first bottom by x+1
then i have to multiply the first top by x+1
\[\frac{x*(x+1)}{(x+3)(x+2)*(x+1)}-\frac{2}{(x+2)(x+1)}\]
now what do i need to multiply the second bottom by?

Answer 4

hello?

Answer 5

2

Answer 6

we want the bottoms to be the same
what do we need to multiply the second bottom by for this to happen?

Answer 7

is the answer x-2
(x+3) (x+2)

Answer 8

hello

Answer 9

you never answered my question

Answer 10

by

Answer 11

by 1x

Answer 12

do you see the bottoms?
how can we make them the same?
what factor is the second fraction missing on the bottom that will make them the same?

Answer 13

(x+3)

Answer 14

thats right so if we multiply the second fraction by (x+3)/(x+3)
we will have common denominators so we can combine the fractions like
\[\frac{x*(x+1)}{(x+3)(x+2)(x+1)}-\frac{2*(x+3)}{(x+1)(x+2)*(x+3)}=\frac{x(x+1)-2(x+3)}{(x+1)(x+2)(x+3)}\]

Answer 15

\[\frac{x^2+x-2x-6}{(x+1)(x+2)(x+3)}\]
what do we get on top after combining like terms?

Answer 16

x^2times -6

Answer 17

x^2+x-2x-6 does not equal x^2(-6)

Answer 18

x-2x=-x
so we have on top x^2-x-6

Answer 19

is x^2-x-6 factorable?

Answer 20

yes

Answer 21

what does x^2-x-6 factor to be?

Answer 22

(x-2)(x+3)

Answer 23

close...
it should be (x-3)(x+2)

Answer 24

so we have \[\frac{(x-3)(x+2)}{(x+1)(x+2)(x+3)}\]
does anything cancel?

Answer 25

x+2

Answer 26

right so we have the answer to be
\[\frac{x-3}{(x+1)(x+3)}\]

Answer 27

got it?

Answer 28

yea thanks

Answer 29

:)