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Thread: Solve the system

  1. #1
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    Solve the system

    4x1 + 3x2 = 4
    (2/3)x1 + 4x2 = 3


    I can't seem to use back substitution
    How do I solve this system?
    Last edited by rhizome; Sep 9th 2019 at 06:06 AM.
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  2. #2
    Member Cervesa's Avatar
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    Re: Solve the system

    Quote Originally Posted by rhizome View Post
    4x1 + 3x2 = 4
    (2/3)x1 + 4x2 = 3


    I can't seem to use back substitution
    How do I solve this system?
    $4x_1 + 3x_2 = 4$
    $\dfrac{2}{3} x_1 + 4x_2 = 3$

    from the first equation, $x_1 = 1 - \dfrac{3}{4}x_2$

    substitute for $x_1$ in the second equation

    $\dfrac{2}{3}\left(1 - \dfrac{3}{4}x_2\right) + 4x_2 = 3$

    $\dfrac{2}{3} - \dfrac{1}{2} x_2 + 4x_2 = 3$

    $\dfrac{7}{2} x_2 = \dfrac{7}{3}$

    $x_2 = \dfrac{2}{3}$

    back substitute to find $x_1$
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  3. #3
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    Re: Solve the system

    I don't understand your result after
    (2/3)(1 - (3/4)x2) + 4x2 = 3
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  4. #4
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    Re: Solve the system

    Ah I understand now.
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