Mark M. answered 06/27/18
Tutor
4.9
(954)
Retired math prof. Very extensive Precalculus tutoring experience.
I am assuming that the system is actually:
x + 8y - 2z = 9
3x - 4y + 8z = 27
9x - y - z = 3
Multiply the first equation by -3 and add to the second equation
Multiply the first equation by -9 and add to the third equation:
x + 8y - 2z = 9
-28y+14z = 0
-73y+17z = -78
Divide the second equation by -28:
x + 8y - 2z = 9
y - (½)z = 0
-73y + 17z = -78
Add 73 times the second equation to the third equation:
x + 8y - 2z = 9
y - (½)z = 0
-(39/2)z = -78
From the third equation, z = -78(-2/39) = 4
Plug into equation 2 to get y = 2
So, from the first equation, x + 8(2) - 2(4) = 9
x = 1
SOLUTION: (1, 2, 4)