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Thread: Addition and subtraction with fractions

  1. #1
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    Addition and subtraction with fractions

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  2. #2
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    Quote Originally Posted by klik11 View Post
    $\displaystyle \frac{1}{3(a-1)} + \frac{a+2}{a(a+1)} - \frac{2(2a-1)}{3(a-1)(a+1)}$

    common denominator is $\displaystyle 3a(a-1)(a+1)$

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

    now proceed ...
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  3. #3
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    That exercise doesn't have it's solution on the book, so could you please tell me if I did it right?

    http://img690.imageshack.us/img690/8339/algebra474.png
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  4. #4
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    you have a mistake in the denominator.u didn't factorize well.
    Last edited by Raoh; Nov 8th 2009 at 11:34 AM.
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  5. #5
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    you should end up with ...

    $\displaystyle \frac{2}{a(a+1)}$
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  6. #6
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    Quote Originally Posted by Raoh View Post
    well done "klik11".
    Quote Originally Posted by skeeter View Post
    you should end up with ...

    $\displaystyle \frac{2}{a(a+1)}$
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  7. #7
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    $\displaystyle \frac{a(a+1)}{3a(a-1)(a+1)} + \frac{3(a-1)(a+2)}{3a(a-1)(a+1)} - \frac{2a(2a-1)}{3a(a-1)(a+1)}$$\displaystyle =\frac{a(a+1)+3(a-1)(a+2)-2a(2a-1)}{3a(a-1)(a+1)}$ $\displaystyle = \frac{6(a-1)}{3a(a^{2}-1)}$ $\displaystyle = \frac{2(a-1)}{a(a^{2}-1)}$.
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  8. #8
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    I repeat ...

    you should end up with $\displaystyle \frac{2}{a(a+1)}$
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  9. #9
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    Quote Originally Posted by Raoh View Post
    $\displaystyle \frac{a(a+1)}{3a(a-1)(a+1)} + \frac{3(a-1)(a+2)}{3a(a-1)(a+1)} - \frac{2a(2a-1)}{3a(a-1)(a+1)}$$\displaystyle =\frac{a(a+1)+3(a-1)(a+2)-2a(2a-1)}{3a(a-1)(a+1)}$ $\displaystyle = \frac{6(a-1)}{3a(a^{2}-1)}$ $\displaystyle = \frac{2(a-1)}{a(a^{2}-1)}$.
    which is $\displaystyle \frac{2}{a(a+1)}$
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  10. #10
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    I didn't learn that part yet but okay
    Thanks
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