How do you solve for x, y, z in #x+3y-z=4#, #-x-2y+3z=4#, #2x-y-2z=-6#?
1 Answer
Apr 30, 2015
x + 3y - z = 4 (1)
-x - 2y + 3z = 4 (2)
2x - y - 2z = -6 (3)
(1) + (2) --> y + 2z = 8 --> y = 8 - 2z
2.(2) + (3) --> -5y + 4z = 2 --> -5.(8 - 2z) + 4z = -40 + 10z + 4z = 2
--> 14z = 42 --> z = 3 --> y = 8 - 6 = 2.
(1) x = 4 + Z - 3Y = 4 + 3 - 6 = 1
CHECK:
(1) x + 3y - z = 1 + 6 - 3 = 4 Correct.