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= 11
k+y+z =4. |2| 2k+2y+2z=8
2k+3y+2z=10 |1| 2k+3y+2z=10-
y=2
eliminasi persamaan 2 dan 3
2k+3y+2z=10 |3| 6k+9y+6z= 30
k+2y+3z = 8. |2| 2k+4y+6z=16 -
4k+5y = 14
subtitusi nilai y
4k+5y= 14
4k+5(2)=14
4k = 14-10
k = 1
subtitusi nilai y dan k
k + y +z =4
1+2+z= 4
z=1
jadi k=1,y=2,z=1