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Homework Help: Algbra II Complex Fractions

  1. Apr 20, 2012 #1
    First of all, thank you, everyone, for all your help with my math conundrums! I've really appreciated the patience and helpful hints - because I do want to learn it myself, so this method is just enough for me to cut the math problems down to size. I'm thankful. And, I have noticed my scores improved because I'm recognizing my trouble areas more often. So, to anyone out there shy to ask for help, I do suggest this forum. It's been very useful!

    1. The problem statement, all variables and given/known data
    Simplify. (a - b)/(a-1 - b-1)

    Answer: -ab

    This lesson is teaching that you can either simplify the numerator and denominator separately and divide; or multiply the whole thing by the LCD and then simpify.

    2. Relevant equations

    3. The attempt at a solution
    (a - b)/(a-1 - b-1)
    Switched to a different format.
    (a-b)/1 ÷ (a-1 - b-1)/1
    Get rid of negative exponents.
    (a-b)/1 ÷ 1/(a - b)
    Invert the second fraction & switch to multiplication.
    (a-b)/1 X (a-b)/1
    Perform multiplication.
    a2 - 2ab - b2
    Last edited: Apr 20, 2012
  2. jcsd
  3. Apr 20, 2012 #2
    Hi velox! This step is wrong! It might be easier to see the two cases you are confusing in fraction form: [itex]a^{-1}-b^{-1}= \displaystyle \frac{1}{a}-\frac{1}{b}[/itex] and [itex](a-b)^{-1} = \displaystyle \frac{1}{a-b}[/itex]

    Whatever is being raised to the negative one gets flipped. All nearby operators stay the same. So in the first case that I showed above, the a and the b, individually, are raised to the negative one. So a and b, individually, get "flipped" and the minus sign stays where it is. In the second case the whole expression is raised to the negative one; so the whole expression gets flipped.

    I hope that helps! Try getting the correct answer now!
  4. Apr 20, 2012 #3
    Hi scurty, it's nice to have your help again!

    That's right, the denominators need to be the same in subtraction with fractions, so I see that the separate 1/a - 1/b and 1/a-b aren't the same. I guess I need to watch out for that.

    So, then...
    (a - b)/(1/a - 1/b)
    Use the LCD of 'ab' to create a common denominator.
    {(a - b)/(b - a)/(ab)}
    Turn the complex fraction into two simple ones.
    (a - b)/1 ÷ (b - a)/(ab)
    Invert the second fraction and change the sign to multiplication (and also change the denominator of the second fraction to the proper form).
    (a - b)/1 X (ab)/(-a + b)
    Factor out the '-1' from the denominator of the second fraction.
    (a - b)/1 X (ab)/-1(a - b)

    Is this correct??
  5. Apr 20, 2012 #4
    Absolutely! I see no problem with your method!
  6. Apr 20, 2012 #5
    Well, that was fast. Thank you so much scurty! Have a good weekend!
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