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Thread: Partial Fractions

  1. #1
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    Exclamation Partial Fractions

    Hello, Thanks in advance for any help with this question.

    I am fairly comfortable with these, however the minus on the numerator is throwing me off!

    Question) Express as the sum of partial fractions:

    -3 / (2x+1)(x-3)

    Usually I would say A / (2x+1) + B / (x-3)

    So that -3 = A(x -3) + B(2x+1) and then sub x = 3 so that A=0

    Giving -3 = 0 + 7B

    But that means B is a a complicated decimal!! -3/7 = -0.42857 which is no good!

    Thanks again for any assitance.

    Mike
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by mikewhant View Post
    Hello, Thanks in advance for any help with this question.

    I am fairly comfortable with these, however the minus on the numerator is throwing me off!

    Question) Express as the sum of partial fractions:

    -3 / (2x+1)(x-3)

    Usually I would say A / (2x+1) + B / (x-3)

    So that -3 = A(x -3) + B(2x+1) and then sub x = 3 so that A=0

    Giving -3 = 0 + 7B

    But that means B is a a complicated decimal!! -3/7 = -0.42857 which is no good!

    Thanks again for any assitance.

    Mike
    I've not had chance to look through & see if you calculations are correct, but you are overcomplicating the issue.

    Take the fraction $\displaystyle \frac{a}{3}$ for example. If $\displaystyle a=8$, then the fraction would be $\displaystyle \frac{8}{3}$, simple enough.

    If however, $\displaystyle a=\frac{1}{2}$, then you would get $\displaystyle \frac{\frac{1}{2}}{3}$, which is 1, divided by 2, divided by 3.

    Therefore you can write this as $\displaystyle \frac{1}{6}$.

    Therefore in your above question, you have $\displaystyle \frac{\frac{-3}{7}}{x-3}$, using the rules above this would give you...
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  3. #3
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    you can "throw" the negative sign from numerator into either the (2x + 1) or the (x - 3) of the denominator. for example -(x - 3) = 3 - x

    $\displaystyle -\frac{3}{(2x + 1)(x - 3)} = \frac{3}{(2x + 1)(3 - x)}$

    let $\displaystyle \frac{3}{(2x + 1)(3 - x)} = \frac{A}{2x + 1} + \frac{B}{3 - x}$

    Hence, 3 = A(3 - x) + B(2x + 1)
    0x + 3 = (2B - A)x + (3A + B)
    2B - A = 0
    A = 2B ...(1)
    3A + B = 3 ...(2)
    Solving these equation you get A = $\displaystyle \frac{3}{7}$, B = $\displaystyle \frac{6}{7}$

    so $\displaystyle -\frac{3}{(2x + 1)(x - 3)}$ = $\displaystyle \frac{6}{7(2x + 1)} + \frac{3}{7(3 - x)}$
    or = $\displaystyle \frac{6}{7(2x + 1)} - \frac{3}{7(x - 3)}$
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  4. #4
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    excellent, thanks to both of you, i've got it now!
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