i. 2x - 2y - 2z = 9 ii. x - 6y - 3z = -28 iii. 3x + 2y + z = 16 eliminasi kan

2x-2y-2z=9
3x+2y+z=16
------------------- +
5x -z = 25 (a)

2x-2y-2z=9 |3| 6x-6y-6z=27
x-6y-3z=-28 |1| x-6y-3z=-28
------------------------ -
5x - 3z = 55 (b)

5x-z =25
5x-3z=55
-------------- -
2z= -30
z = -15 -> 5x- z =25
5x-(-15)=25
5x =25-15
5x =10
x = 2
substitusi z dan x
2x-2y-2z=9
2(2)-2y-2(-15)=9
4-2y+30 =9
2y =9-4-30
2y = -25
y= -25/2