Stuck on this Algebra 2 question!

what is the solution of the system of equations?
2x+2y+3z=-6
3x+5y+4z=3
2x+3y+4z=-10

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  1. 2x+2y+3z=-6 times 3
    3x+5y+4z=3 times 2

    6 x + 6 y + 9 z = -18
    6 x + 10y + 8 z = 6
    -----------------------subtract
    0 x - 4 y + 1 z = -24

    3x+5y+4z=3 times 2
    2x+3y+4z=-10 times 3

    6 x + 10y + 8 z = 6
    6 x + 9 y + 12z = -30
    ------------------------subtract
    0 x + 1 y - 4 z = 36
    so

    - 4 y + 1 z = -24 times 4
    + 1 y - 4 z = 36

    -16y + 4 z = - 96
    + 1y - 4 z = 36
    ------------------ add
    -15 y = -60
    y = 4
    now go back and get z and x

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    posted by Damon
  2. solve it using matrix inversion method.
    AX=B, where A is the coefficient matrix
    2 2 3
    A= 3 5 4
    2 3 4
    X=[x y z]' and B =[-6 3 -10]'
    then X=(A^-1)B,that is A inverse*B
    answer is x=5, y=4, z=-8

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    posted by gg

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