(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