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Math Help - Find x and y for (y/2)-x=2 and 6x-(3y/2)=3 using gaussian elimination

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    Find x and y for (y/2)-x=2 and 6x-(3y/2)=3 using gaussian elimination

    Find x and y for (y/2)-x=2 and 6x-(3y/2)=3 using gaussian elimination
    Last edited by mr fantastic; March 26th 2010 at 11:29 PM. Reason: Posted question from title into main body of post
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    Quote Originally Posted by brumby_3 View Post
    As above!
    \frac{y}{2} - x = 2

    6x - \frac{3y}{2} = 3.



    3R_1 \to R_1



    -3x + \frac{3y}{2} = 6

    6x - \frac{3y}{2} = 3



    R_2 + R_1 \to R_2



    -3x + \frac{3y}{2} = 6

    3x = 9.



    Solve for x:

    x = 3.


    Back substitute and solve for y:

    -3(3) + \frac{3y}{2} = 6

    -9 + \frac{3y}{2} = 6

    \frac{3y}{2} = 15

    3y = 30

    y = 10.


    So the solution is (x, y) = (3, 10).
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    Hi Prove it,
    I believe that I'm supposed to show the answer by using the gaussian elimination by using matrixes?
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    Quote Originally Posted by brumby_3 View Post
    Hi Prove it,
    I believe that I'm supposed to show the answer by using the gaussian elimination by using matrixes?
    Then you should have said that.

    Rewrite the system as

    \left[\begin{matrix}-1 & \frac{1}{2} \\ 6 & -\frac{3}{2}\end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}2 \\ 3\end{matrix}\right].

    Then follow the exact same process I said above.
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    Hello, brumby_3!

    Solve by Gaussian elimination: . \begin{array}{cccc}\dfrac{y}{2}-x &=& 2 & [1] \\ \\[-3mm] 6x-\dfrac{3y}{2} &=& 3 & [2]\end{array}

    We have: . \left|\begin{array}{cc|c} \text{-}1 & \frac{1}{2} & 2 \\ \\[-4mm] 6 & \text{-}\frac{3}{2} & 3 \end{array}\right|


    . \begin{array}{c} \\ R_2+6R_1\end{array}\left|\begin{array}{cc|c}<br />
\text{-}1 & \frac{1}{2} & 2 \\ \\[-4mm] 0 & \frac{3}{2} & 15 \end{array}\right|


    . . . . \begin{array}{c} \\ \frac{2}{3}R_2\end{array} \left|\begin{array}{cc|c}<br />
\text{-}1 & \frac{1}{2} & 2 \\ 0 & 1 & 10 \end{array}\right|


    \begin{array}{c}R_1-\frac{1}{2}R_2 \\ \\ \end{array}<br />
\left|\begin{array}{cc|c} \text{-}1 & 0 & \text{-}3 \\ 0 & 1 & 10 \end{array}\right|


    . . . \begin{array}{c}\text{-}1\!\cdot\! R_1 \\ \\ \end{array}<br />
\left| \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 10 \end{array}\right|


    Therefore: . x \:=\:3,\;y\:=\:10

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