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3 June, 20:15

Solving each system by eliminating

x-y+z = - 1

x+y+3z = - 3

2x-y+2z = 0

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  1. 3 June, 20:22
    0
    There's only one system of equations here.

    Let's eliminate x first,

    x - y + z = - 1

    x = - 1 + y - z

    Substituting into x + y + 3z = - 3 gives

    (-1 + y - z) + y + 3z = - 3

    2y + 2z = - 2

    y + z = - 1

    Substituting for x into 2x-y+2z = 0 gives

    2 (-1 + y - z) - y + 2z = 0

    -2 + 2y - 2z - y + 2z = 0

    y = 2

    Got lucky on that one.

    z = - 1 - y = - 1 - 2 = - 3

    x = - 1 + y - z = - 1 + 2 - - 3 = 4

    Answer: x=4, y=2, z = - 3

    Check:

    x-y+z = 4 - 2 + - 3 = - 1, good

    x+y+3x = 4+2+3 (-3) = - 3, good

    2x-y+2x = 2 (4) - 2+2 (-3) = 0, good
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