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Complex fractions

  1. Jan 23, 2012 #1
    step 1. [(3 - (3 + h))/(3(3+h))]/h

    step 2. -h/(3(3+h)h

    My book says that those steps are valid. I don't see how.

    If you want my opinion, I times the numerator (3 - (3 + h))/(3(3+h)) by the inverse of the denominator h/1 which makes:

    step 2. -h/(9+3h) * h

    step 3. -h/(9+3)

    step 4. -h/12
     
  2. jcsd
  3. Jan 23, 2012 #2
    Never mind. I figured out what I was doing wrong.

    In a simpler format

    (a/(2 + 3a)) * a

    where a = 3

    is not the same as

    a/2 + 3
     
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