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Math Help - Elimination method with 3 equations

  1. #1
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    Elimination method with 3 equations

    http://i1209.photobucket.com/albums/...g?t=1333868898

    I've tried this problem about 6 times and can't get the right answer in the book, can someone please show me how to do this...
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  2. #2
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    Re: Elimination method with 3 equations

    Quote Originally Posted by batosaims View Post
    http://i1209.photobucket.com/albums/...g?t=1333868898

    I've tried this problem about 6 times and can't get the right answer in the book, can someone please show me how to do this...
    \begin{cases}3a+5b-3c=-4 \\a-3b+c=6 \\-4a+6b+2c=-6\end{cases}

    Multiply 2nd. by -3 and add to the 1st. ,also..multiply 2nd. by 4 and add to the 3rd. :

    \begin{cases}a-3b+c=6 \\14b-6c=-22 \\-6b+6c=18\end{cases}

    Add 2nd to the 3rd. :

    \begin{cases}a-3b+c=6 \\14b-6c=-22\\8b=-4\end{cases}

    So , solution of the system is :

    (a,b,c)=\left(2,\frac{-1}{2},\frac{5}{2}\right)
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  3. #3
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    Re: Elimination method with 3 equations

    Quote Originally Posted by princeps View Post
    \begin{cases}3a+5b-3c=-4 \\a-3b+c=6 \\-4a+6b+2c=-6\end{cases}
    Short cut; add 1st 2 equations to get:
    4a + 2b - 2c = 2 ;add to 3rd equation:
    -4a+ 6b +2c = -6
    =============
    8b = -4
    b = -1/2 (as per Princeps)
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