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24 October, 03:06

Solve the system by elimination.

-2x+2y+3z=0

-2x-y+z=-3

2x+3y+3z=5

+3
Answers (1)
  1. 24 October, 06:52
    0
    -2x + 2y + 3z = 0 → 2x - 2y - 3z = 0 → 2x - 2y - 3z = 0

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

    2x + 3y + 3z = 5 → 2x + 3y + 3z = 5 - 3y - 2z = - 3

    -2x + 2y + 3z = 0 → 2x - 2y - 3z = 0

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

    2x + 3y + 3z = 5 → 2x + 3y + 3z = 5 → 2x + 3y + 3z = 5

    -2y - 4z = - 2

    -3y - 2z = - 3 → - 6y - 4z = - 6

    -2y - 4z = - 2 → - 2y - 4z = - 2

    -4y = - 4

    -4 - 4

    y = 1

    -3y - 2z = - 3

    -3 (1) - 2z = - 3

    -3 - 2z = - 3

    + 3 + 3

    -2z = 0

    -2 - 2

    z = 0

    -2x + 2y + 3z = 0

    -2x + 2 (1) + 3 (0) = 0

    -2x + 2 + 0 = 0

    -2x + 2 = 0

    - 2 - 2

    -2x = - 2

    -2 - 2

    x = 1

    (x, y, z) = (1, 1, 0)
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