3x - 2y + 2z = - 2

x+6y-2z=-2

x+2y = 0

(x, y, z) → ( - 2, 1, 3)

given the 3 equations

3x - 2y + 2z = - 2 → (1)

x + 6y - 2z = - 2 → (2)

x + 2y = 0 → (3)

By a process of elimination we can solve for x, y and z

(1) + (2) term by term will eliminate z

4x + 4y = - 4 → (4)

from (3) x = - 2y which we can substitute into (4)

4 ( - 2y) + 4y = - 4

- 8y + 4y = - 4

- 4y = - 4 (divide both sides by - 4)

y = 1

substitute y = 1 into (3)

x + 2 = 0 ⇒ x = 0 - 2 = - 2

substitute x = - 2, y = 1 into (1) or (2) and solve for z

(1) : - 6 - 2 + 2z = - 2

- 8 + 2z = - 2 (add 8 to both sides)

2z = 6 (divide both sides by 2)

z = 3