3 years ago

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

1 answer:

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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