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Talja [164]
3 years ago
14

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

Mathematics
1 answer:
jeyben [28]3 years ago
8 0

Answer:

The answer to your question is:

x = 1; y = 1, z = 0

Step-by-step explanation:

                                            -2x + 2y + 3z = 0        (I)

                                             -2x - y + z = -3           (II)

                                              2x + 3y + 3z = 5      (III)

I and II. Multiply II by 2

                                      -2x + 2y + 3z = 0

                                  (2)(-2x -   y  +   z = -3 )

                                       -2x + 2y + 3z = 0

                                       -4x -  2y  + 2z = -6

                                      -6x           + 5z = - 6       (IV)

II and III. Multiply II by 3

                                      -6x - 3y +  3z = -9     (II)

                                       2x + 3y + 3z = 5      (III)

                                     -4x         + 6z = -4           (V)

IV and V. Multiply IV by 2 and V by -3

                                   -12x  +  10z = -12

                                     12x  -  18z = 12

                                             - 8 z = 0

                                                   z = 0

Substitute z in IV

                              -6x + 5(0) = -6

                             -6x = -6

                                x = 1

Substitute x and z in I

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

                              -2 + 2y = 0

                                      2y = 2

                                        y = 1

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