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Math Help - partial-fraction-decomp question

  1. #1
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    partial-fraction-decomp question

    I'm having such a hard time solving the problem below

    5-x
    _____
    2xsquared + x -1

    Can you help point me in the right direction?

    Thank you very much.
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  2. #2
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    Quote Originally Posted by MrsPaiva View Post
    I'm having such a hard time solving the problem below

    5-x
    _____
    2xsquared + x -1
    \displaystyle \frac{5-x}{(2x-1)(x+1)} = \frac{A}{2x-1} + \frac{B}{x+1}

    5-x = A(x+1) + B(2x-1)

    choose two "strategic" values for x to help you find A and B
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  3. #3
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    Hi thanks for helping.

    You multiplied by the LCD?
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  4. #4
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    Quote Originally Posted by MrsPaiva View Post
    Hi thanks for helping.

    You multiplied by the LCD?
    I found a common denominator, and set the numerators equal.
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  5. #5
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    So from here applied the distributive property and came up w/

    5-X = AX + 1A +2BX -1B

    Good so far?
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  6. #6
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    Quote Originally Posted by MrsPaiva View Post
    So from here applied the distributive property and came up w/

    5-X = AX + 1A +2BX -1B

    Good so far?
    you can do it this way (equating coefficients and solving a system), although the method I recommended in my initial post is much easier, imho.
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  7. #7
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    My instructor wants us to show our work.

    Thanks again
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  8. #8
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    You can still show work using skeeter's method. This is what he was getting at:
    Quote Originally Posted by skeeter View Post
    5-x = A(x+1) + B(2x-1)

    choose two "strategic" values for x to help you find A and B
    Choose a value of x such that the A is eliminated. So let x = -1:
    \begin{aligned}<br />
5 + 1 &= A(-1 + 1) + B(2(-1) - 1) \\<br />
6 &= A(0) + B(-3) \\<br />
6 &= -3B \\<br />
B &= -2<br />
\end{aligned}
    See what happens? You'll just have one variable B, that you can solve for it easily.

    Now choose a value of x so that B is eliminated.
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  9. #9
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    I see what you mean now. Okay, I misunderstood. I'll work out the problem and post my answer
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