
Wendy L. answered 04/08/19
Experienced Math Tutor
Hi Mabry,
Yes, it's a system of 3 equations.
You can solve it by using graphs, substitutions, or eliminations, but for now, I will use eliminations.
Even in eliminations, there are multiple ways to do it too.
To make it easier, I will name each equation (A), (B), and (C).
2x - y + z = -2 ...... (A)
x + 2y + 2z = 3 ...... (B)
2x - 2y + z = 0 ...... (C)
(A) - (C):
2x - y + z = -2
-(2x - 2y + z = 0)
y = -2
Substitute y = -2:
(A): 2x - (-2) + z = -2
2x + z = -4 ........... (A')
(B): x + 2(-2) + 2z = 3
x + 2z = 7 ............. (B')
(C): 2x - 2(-2) + z = 0
2x + z = -4 ........... (C')
(A') - 2(B'):
2x + z = -4
-(2x + 4z = 14)
-3z = - 18
z = 6
Substitute y = -2 and z = 6 into (A) and solve for x:
2x - (-2) + 6 = -2
2x + 8 = -2
2x = - 10
x = -5
Double check:
2x - y + z = 2(-5) - (-2) + 6 = -10 + 2 + 6 = -2 check
x + 2y + 2z = -5 + 2(-2) + 2(6) = -5 - 4 + 12 = 3 check
2x - 2y + z = 2(-5) - 2(-2) + 6 = -10 + 4 + 6 = 0 check
Therefore, x = -5, y = -2, z = 6