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Gemiola [76]
4 years ago
6

Solve the system using elimination...

Mathematics
1 answer:
Galina-37 [17]4 years ago
3 0

Answer: x = -1, y = 1, z = 0

<u>Step-by-step explanation:</u>

EQ 1: -2x + 2y + 3z = 0   ⇒   1(-2x + 2y + 3z = 0)   ⇒   -2x + 2y + 3z = 0

EQ 2: -2x -  y + 3z = -3   ⇒  -1(-2x -  y + 3z = -3)    ⇒   <u> 2x +   y  - 3z</u> = <u>3 </u>

                                                                                                3y       = 3

                                                                                                   y      = 1

EQ 1: -2x + 2y + 3z = 0   ⇒   -2x + 2(1) + 3z = 0   ⇒   -2x + 2 + 3z = 0

EQ 3:  2x + 3y + 3z = 5   ⇒    2x + 3(1) + 3z = 5   ⇒  <u>  2x + 3 + 3z</u> = <u>5 </u>

                                                                                             5 + 6z = 5

                                                                                                    6z = 0

                                                                                                       z = 0

EQ 1: -2x + 2y + 3z = 0  

        -2x + 2(1) + 3(0) = 0

        -2x  + 2              = 0

         -2x                    = -2

            x                     = -1

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