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Math Help - Partial Fraction concept

  1. #1
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    [SOLVED]Partial Fraction concept

    if a equation is 3x/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)

    will the next step is

    = A(x-3) + B(x-1)(x-2) + C(x-1)

    or

    = A(x-3) + B(x-1)(x-3) + C(x-1)

    thank you.
    Last edited by gilagila; December 18th 2011 at 12:12 AM.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Partial Fraction concept

    Hi gilagila!

    I'm afraid neither is correct.

    To add all fractions together, you need to make the denominators equal.
    For the first term with A this means:
    A / (x-1) = A(x-2)(x-3) / [(x-1)(x-2)(x-3)].
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  3. #3
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    Re: Partial Fraction concept

    Quote Originally Posted by ILikeSerena View Post
    Hi gilagila!

    I'm afraid neither is correct.

    To add all fractions together, you need to make the denominators equal.
    For the first term with A this means:
    A / (x-1) = A(x-2)(x-3) / [(x-1)(x-2)(x-3)].
    oops...just noticed that need to multiple all the demoninator
    if a equation is 3x/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)

    it should look like this, correct me if worng

    3x = A(x-2)(x-3)/(x-1) + B(x-1)(x-3)/(x-2) + C(x-1)(x-2)/(x-3)
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Partial Fraction concept

    You have the nominators correct now, but you should leave out the denominators.

    You have multiplied the left hand side with (x-1)(x-2)(x-3) to give you 3x.
    The same should be done on the right hand side, meaning there would be no denominators left.
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  5. #5
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    Re: Partial Fraction concept

    Quote Originally Posted by ILikeSerena View Post
    You have the nominators correct now, but you should leave out the denominators.

    You have multiplied the left hand side with (x-1)(x-2)(x-3) to give you 3x.
    The same should be done on the right hand side, meaning there would be no denominators left.
    oops, forgot to put in the denominator at left

    3x/(x-1)(x-2)(x-3) = A(x-2)(x-3)/(x-1) + B(x-1)(x-3)/(x-2) + C(x-1)(x-2)/(x-3)

    then only look like this

    3x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: Partial Fraction concept

    Quote Originally Posted by gilagila View Post
    oops, forgot to put in the denominator at left

    3x/(x-1)(x-2)(x-3) = A(x-2)(x-3)/(x-1) + B(x-1)(x-3)/(x-2) + C(x-1)(x-2)/(x-3)

    then only look like this

    3x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)
    Ah well, it should be:
    3x/[(x-1)(x-2)(x-3)] = A(x-2)(x-3)/[(x-1)(x-2)(x-3)] + B(x-1)(x-3)/[(x-1)(x-2)(x-3)] + C(x-1)(x-2)/[(x-1)(x-2)(x-3)]

    but nevermind, you're result is correct.
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