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Homework Help: Expanding brackets with fractions

  1. Feb 27, 2010 #1
    I have no problem expanding brackets with fractions generally, unless the fraction contains an unknown variable, such as in the following expression:

    m/4[6m - 8] + m/2[10m - 2]

    I know that the answer is:
    12/2m2 - 3m

    ..but I have no idea how to get to that. Can anyone help?
     
  2. jcsd
  3. Feb 27, 2010 #2

    tiny-tim

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    Hi Gringo123! :smile:

    Don't leave out brackets!! :rolleyes:

    I think you mean:

    [m/4][6m - 8] + [m/2][10m - 2]

    = [13/2]m2 - 3m​

    Does that help? :smile:
     
  4. Feb 28, 2010 #3
    Hello again Tim
    Thanks again for helping out.
    The fractions are definitiely not in brackets. However, I've probably confused everyone with a typo. The answer should be:

    13/2 m2 - 3m
     
  5. Feb 28, 2010 #4

    Char. Limit

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    I guess you just multiply each term in the bracket by the term outside.
     
  6. Feb 28, 2010 #5

    tiny-tim

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    The fractions definitiely are in brackets.

    If they're not, then the [6m - 8] and the [10m - 2] would be on the bottom, and you'd never get the answer given.
     
  7. Feb 28, 2010 #6

    Mentallic

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    I don't believe the brackets matter in this case. m/4(6m-8) is to be read as (m/4)(6m-8). If the second factor were to be in the denominator as you claim then it should be written as m/(4(6m-8)).
    This is how it works in most older calculators that use parenthesis also.

    Yes that's exactly how it's done! :smile:

    [tex]\frac{m}{4}(6m-8)+\frac{m}{2}(10m-2)[/tex]

    [tex]=\frac{m}{4}(6m)-\frac{m}{4}(8)+\frac{m}{2}(10m)-\frac{m}{2}(2)[/tex]
     
  8. Feb 28, 2010 #7

    tiny-tim

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    No it isn't!!

    See eg http://en.wikipedia.org/wiki/BODMAS#The_standard_order_of_operations"
    The vagaries of some computer programs don't affect the standard BODMAS rules for human notation! :smile:
    But I'm not an older calculator! …

    … I'm still fresh! :rolleyes:
     
    Last edited by a moderator: Apr 24, 2017
  9. Feb 28, 2010 #8

    Mentallic

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    Omg you're right!
    I guess I've been misinterpreting 1/2x as (1/2)x this whole time! I always took that if you aren't to use parenthesis, then only the very next symbol is in the denominator, in this case, just the 2.

    and what if we have 1/2xy. Do I interpret this as [tex]\frac{y}{2x}[/tex] or [tex]\frac{1}{2xy}[/tex]? I'm guessing the latter, because of the new rules I have just learnt! :biggrin:
     
  10. Feb 28, 2010 #9

    tiny-tim

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    Yup! 1/2xy = 1/(2xy) :biggrin:
     
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