math

Solve the following system of equations:
4x+y–2z=10
2x–y+2z=–9
x+y+z=6

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  1. I noticed that the coefficients of y are ± 1 , so ...
    add the first and the second:
    6x = 1
    x = 1/6 , well that was lucky

    add the second and third:
    3x + 3z = -3
    x + z = -1
    1/6 + z = -1
    z = -7/6

    sub into the last:
    1/6 + y - 7/6 = 6
    times 6
    1 + 6y - 7 = 36
    y = 42/6 = 7

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  2. 4x+y–2z=10
    2x–y+2z=–9
    -----------add
    6 x = 1
    x = 1/6

    1/6 + y + z = 6
    is
    6 y + 6 z = 35

    2(1/6)-y+2z = -9 --> -3 y + 6 z = -28
    so use those two
    6 y + 6 z = 35
    -3y + 6 z = -28
    --------------------subtract
    9 y = 63
    y = 7

    1/6 + 7 + z = 6
    1 + 42 + 6 z = 36
    6 z = 36-43 = -7
    z - -7/6

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