Mathematics, 16.10.2019 21:30 cece4649

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

Answers: 1

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Answer from: wmaingrette1

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

Answer from: Quest

step-by-step explanation:

i dont get

Answer from: Quest

it would open into a square depending on the type of envelope i’m not sure about the second part

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