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Thread: systems of equations

  1. #1
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    systems of equations

    xy + xz = -1
    xy + yz = -9
    yz + xz = -4

    solve for (x,y,z)
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  2. #2
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    Quote Originally Posted by stasis View Post
    xy + xz = -1
    xy + yz = -9
    yz + xz = -4

    solve for (x,y,z)
    I'd suggest substituting xy = a, xz = b and yz = c. Solve the three new linear equations simultaneously for a, b and c. NOw use these values to solve for x, y and z.

    By the way ..... There are several good reasons NOT delete questions once they've been answered. (I refer to your edit at http://www.mathhelpforum.com/math-he...-question.html)
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  3. #3
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    Quote Originally Posted by stasis View Post

    xy + xz = -1 (1)
    xy + yz = -9 (2)
    yz + xz = -4 (3)

    solve for (x,y,z)
    Hello Stasis. This problem can easily be solved by doing some basic elimination and substitution.

    My approach starts with (1) - (2) to give $\displaystyle xz -yz = 8$ call this (4)

    then (3) + (4) give $\displaystyle 2xz = 4 \ \ \Rightarrow \ \ xz = 2$
    substitute $\displaystyle xz = 2$ into (1) and (3) to get $\displaystyle yz=-6$ and $\displaystyle xy = -3$

    so we have:
    $\displaystyle xz = 2 \ \ \ \ \ \ (5)$
    $\displaystyle yz=-6 \ \ \ \ (6)$
    $\displaystyle xy = -3 \ \ \ \ (7)$

    now (7)*(5) gives $\displaystyle x^2yz = -6$ (6) gives $\displaystyle yz=-6$ therefore $\displaystyle x^2 = 1 \ \ \Rightarrow \ \ x = \pm 1$

    putting $\displaystyle x = \pm 1$ in (5) gives $\displaystyle z = \pm 2$
    and into (7) $\displaystyle y = \mp 3$

    so your two solutions are $\displaystyle ( \pm 1 , \mp 3 , \pm 2 )$

    Bobak
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