Question

OKAY this one is really confusing to me...
Subtract, simplify by removing a factor of 1 when possible
11fc/f^2-c^2 - f-c/f+c

Answer 1

always start by finding a common denominator.

Answer 2

11 and 2 are not common.. so 1??

Answer 3

any factorization will be done after you balance the denominators. So, you might not see it immediately. Math problems will often look crazy until you make some adjustments.

Answer 4

\[\Large \frac{11fc}{f^2-c^2} - \frac{f-c}{f+c}\]



\[\Large \frac{11fc}{(f-c)(f+c)} - \frac{f-c}{f+c}\]



\[\Large \frac{11fc}{(f-c)(f+c)} - \frac{(f-c)(f-c)}{(f-c)(f+c)}\]



\[\Large \frac{11fc-(f-c)(f-c)}{(f-c)(f+c)}\]



\[\Large \frac{11fc-(f^2-2fc+c^2)}{(f-c)(f+c)}\]



\[\Large \frac{11fc-f^2+2fc-c^2}{(f-c)(f+c)}\]



\[\Large \frac{-f^2+13fc-c^2}{(f-c)(f+c)}\]



\[\Large \frac{-(f^2-13fc+c^2)}{(f-c)(f+c)}\]



\[\Large -\frac{f^2-13fc+c^2}{(f-c)(f+c)}\]



\[\Large -\frac{f^2-13fc+c^2}{f^2-c^2}\]



So \[\Large \frac{11fc}{f^2-c^2} - \frac{f-c}{f+c}=-\frac{f^2-13fc+c^2}{f^2-c^2}\]

Answer 5

there you go, that dude is CRAZY!

Answer 6

The key here is to get all denominators equal to (f-c)(f+c). From there, you can combine the fractions and simplify.

Answer 7

yeah Jim is a beast I'm in love with him <3 lol

Answer 8

it says its not rifght :(

Answer 9

then try \[\Large \frac{-f^2+13fc-c^2}{f^2-c^2}\]


if that doesn't work, then try \[\Large \frac{-f^2+13fc-c^2}{(f-c)(f+c)}\]


the computer is very picky and will only accept one answer even though there are multiple ways of saying the same thing

Answer 10

nvm i typed it in wrong!!

Answer 11

ah gotcha

Answer 12

well nvm its still not wright lol

Answer 13

it's okay though :D

Answer 14

well it's one of forms listed above

Answer 15

i didnt put parenthesis.. thats why

Answer 16

often, even the order matters (even though it shouldn't)

Answer 17

i see