Question

simplify the complex fraction..
n-6
____

n^2+11n+24
_____________

n+1
____

n+3

Answer 1

@NNL1991

Answer 2

@Dido525

Answer 3

Recall that, when you divide fractions, you can rewrite the problem as multiplication by the reciprocal. So \(\frac{\frac{a}{b}}{\frac{c}{d}}\) becomes \(\frac{a}{b}*\frac{d}{c}\)

Answer 4

You can then factor and you will find that there will be terms to cancel.

Answer 5

first step is multiply the bottom fraction by (n+3)/(n+1) (the inverse or "flipped")
this makes the bottom 1

you must also multiply the top fraction by (n+3)/(n+1)
see eSpex post

Answer 6

next, you should factor n^2+11n+24

can you do that ?

Answer 7

n-6 n+1 what i got so far.
------------ * ----
n^2+11n+24 n+3

Answer 8

almost, but it should be (n+3)/(n+1)

Answer 9

factoring i got( n+8 )(n+3)

Answer 10

\[\frac{ \frac{n-6}{n^2+11n+24} \cdot \frac{n+3}{n+1} } { \frac{\cancel{n+1}}{n+3} \cdot \frac{n+3}{\cancel{n+1}}} \]

Answer 11

the n+3 in the bottom also cancel, so the complicated bottom becomes 1 and you are left with the top

you now have
\[ \frac{n-6}{(n+8)(n+3) } \cdot \frac{n+3}{n+1} \]

it looks like you can cancel (n+3) from top and bottom

Answer 12

n-6
---- n \[\neq-8,-3 or -1\]
n^2+9n+8

Answer 13

yes, looks good

Answer 14

whewww :) thanks!

Answer 15

the only part that confused you was "flipping" the bottom fraction -- that is the "trick" you should learn . if you see a/b / c/d change it to
a/b * d/c

Answer 16

alright! will do(: thank you!

Answer 17

phi has done a good job with the answer, so nothing left for me to do. :)