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Over [174]
3 years ago
11

Solve the system by elimination. -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5

Mathematics
1 answer:
KengaRu [80]3 years ago
8 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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