Question

Please Help Me ASAP!
Which is equal to

11f f-6
--------- + ----
(2f+3)(f-4) 2f+3

A) f^2 + f + 24
------------
(2f+3) (f-4)

B) 22f^2 + 34f - 6
-------------
(2f+3) (f-4)

C) 2f^2 - 3f - 18
-------------
(2f+3) (f-4)

D) f^2 + 13f - 24
-------------
(2f+3) (f-4)

Answer 1

What do all of the answers have in common?

Answer 2

They have the same denominator

Answer 3

Good, now how would you change the original to have the same denominator?

Answer 4

2f+3 of the second equation is combined with the first equation to create (2f+3) (f-4)

Answer 5

I mean how would you get the original to be a single fraction with denominator (2f+3)(f-4)?

Answer 6

(Hint, multiply the top and bottom of the second fraction by (f-4))

Answer 7

hmmm so the denominator is (2f+3(f-4) and the top of the whole fraction would have to be the first top fraction plus the second top equation with the f-4 added to the second equation?

Answer 8

Not just added, multiplied. It should look like this: (11f)/(2f+3)(f-4) + (f-6)(f-4)/(2f+3)(f-4)

Answer 9

\(\Large \frac{11f}{(2f~+~3)(f~-~4)}~+~\frac{f~-~6}{2f~+~3}\)

Is that your problem?

It is all REALLY jumbled up for me

Answer 10

yes :)

Answer 11

Okay thanks

Answer 12

I think a possible answer would be either A or B but I think it may be A

Answer 13

You are correct with A

Answer 14

Did you follow what I wrote then? Just needed to match denominators

Answer 15

Yes, thanks I didn't knew I had to add the (f-4) to the top of the second fraction

Answer 16

Thank you guys for helping me out so much! :)

Answer 17

Certainly, no problem. Sometimes we just don't see the solution right in front of us, I'm pretty guilty of it >.>