anonymous
  • anonymous
simplify the complex fraction.. n-6 ____ n^2+11n+24 _____________ n+1 ____ n+3
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@NNL1991
anonymous
  • anonymous
@Dido525
eSpeX
  • eSpeX
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 \(\frac{a}{b}*\frac{d}{c}\)

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eSpeX
  • eSpeX
You can then factor and you will find that there will be terms to cancel.
phi
  • phi
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
phi
  • phi
next, you should factor n^2+11n+24 can you do that ?
anonymous
  • anonymous
n-6 n+1 what i got so far. ------------ * ---- n^2+11n+24 n+3
phi
  • phi
almost, but it should be (n+3)/(n+1)
anonymous
  • anonymous
factoring i got( n+8 )(n+3)
phi
  • phi
\[\frac{ \frac{n-6}{n^2+11n+24} \cdot \frac{n+3}{n+1} } { \frac{\cancel{n+1}}{n+3} \cdot \frac{n+3}{\cancel{n+1}}} \]
phi
  • phi
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
anonymous
  • anonymous
n-6 ---- n \[\neq-8,-3 or -1\] n^2+9n+8
phi
  • phi
yes, looks good
anonymous
  • anonymous
whewww :) thanks!
phi
  • phi
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
anonymous
  • anonymous
alright! will do(: thank you!
anonymous
  • anonymous
phi has done a good job with the answer, so nothing left for me to do. :)

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