Question

(2x)/(x^2-x-2) - (4x)/(x^2-3x+2)
simplify the difference...

Answer 1

Solve each side first....

Answer 2

\[\frac{2x}{x^2-x-2} - \frac{4x}{x^2-3x+2}\] The first thing I would do is factor the denominators.

Answer 3

so it would be \[2x \div(x+1)(x-2) - 4x \div(x-1)(x-2)\] right?

Answer 4

Yes.

Answer 5

okay and then i have to find the common denominator?

Answer 6

Yes

Answer 7

Yes Ma'am

Answer 8

and that would be x-2?

Answer 9

Yep! (:

Answer 10

so then i multiply each side with x-2?

Answer 11

Yes.

Answer 12

so then it would be (x-2)/(x-2) * (2x)/(x+1)(x-2) for the first side. right?
does any of the (x-2)'s cancel out?

Answer 13

?

Answer 14

No, because it isn't in the correct form for them to cancel out (:

Answer 15

okay so then it would be \[2x ^{2}-4x/x ^{3}-3x ^{2}+4\] on the first side right?

Answer 16

and then on the second side it would be \[4x ^{2}-8x/x ^{3}-5x ^{2}+8x-4\] right?

Answer 17

Correct !! (:

Answer 18

Okay. I don't know what to do after that.

Answer 19

Where exactly are u at with this problem?

Answer 20

What do you mean?

Answer 21

Write out everything you've answered so far..

Answer 22

okay i've got \[2x/(x ^{2}-x-2) - 4x/(x ^{2}-3x+2)\] as the problem. then i multiplied (x-2) to each side after I factored the denominators. so it looked like this: \[(x-2)/(x-2) * 2x/(x+1)(x-2) - 4x/(x-1)(x+2) * (x-2)/(x-2)\] and after i multiplied them i got \[(2x ^{2}-4x)/(x ^{3}-3x+4) - (4x ^{2}-8x)/(x ^{3}-5x ^{2}+8x-4)\]

Answer 23

and thats all i have so far

Answer 24

Shucks, I'm lost!

Answer 25

Haha, me too.

Answer 26

You worked correctly up to this point:\[\frac{2x}{(x+1)(x−2)}−\frac{4x}{(x−1)(x−2)}\]

Answer 27

after this, you correctly noticed the common factor between the two denominators was (x-2).

Answer 28

what you missed was calculating the LCM of the denominators as the next step.

Answer 29

so we have the denominators as:\[(x+1)(x-2)=(x+1)\times(x-2)\]and:\[(x-1)(x-2)=(x-1)\times(x-2)\]so the LCM would be:\[(x+1)(x-1)(x-2)\]

Answer 30

a strategy you can use to find the LCM is as follows:

1. remove all common factor(s) - in this example it was (x+2)
2. multiply all remaining terms together with the removed common factor(s) - in this example it was (x+1)(x-1) multiplied by the common factor of (x+2)

Answer 31

okay so now i multiply each side by all three of those?

Answer 32

no, the next step is to identify what you need to multiply the denominator by in order to get the LCM.
so, for the first term, the denominator is (x+1)(x-2) and the LCM is (x+1)(x-1)(x-2), which means we need to multiply the first fraction by (x-1)/(x-1)

Answer 33

so first fraction becomes:\[\frac{2x}{(x+1)(x−2)}*\frac{x-1}{x-1}=\frac{2x(x-1)}{(x+1)(x-1)(x−2)}\]

Answer 34

now what do you think we need to multiply the second fraction by?

Answer 35

remember the second denominator is (x-1)(x-2) and we know the LCM is (x+1)(x-1)(x-2).
so what do we need to multiply (x-1)(x-2) by in order to get (x+1)(x-1)(x-2)?

Answer 36

(x+1)?

Answer 37

correct, so the second fraction becomes:\[\frac{4x}{(x−1)(x−2)}=\frac{4x}{(x−1)(x−2)}*\frac{x+1}{x+1}=\frac{4x(x+1)}{(x+1)(x−1)(x−2)}\]

Answer 38

do you think you can complete the question from here now?

Answer 39

i think i can try.

Answer 40

good - have a go at it and post your working here - I'll come back and check it for you later - good luck! :)

Answer 41

Thanks.

Answer 42

so would the answer be \[-2x ^{2}+2x/(x+1)(x-1)(x-2)\] right?

Answer 43

not quite, it should be:\[\frac{2x(x-1)}{(x+1)(x-1)(x−2)}-\frac{4x(x+1)}{(x+1)(x−1)(x−2)}=\frac{2x(x-1)-4x(x+1)}{(x+1)(x-1)(x−2)}\]\[\qquad=\frac{2x^2-2x-4x^2-4x}{(x+1)(x-1)(x−2)}\]\[\qquad=\frac{-2x^2-6x}{(x+1)(x-1)(x−2)}\]\[\qquad=\frac{-2x(x+3)}{(x+1)(x-1)(x−2)}\]

Answer 44

alright. I see i forgot the negative on one thing. Well thanks very much! (:

Answer 45

yw