3x-y+2z=15 2x+y+z=13 3x+2y+2z=24 dengan mengguna metode eliminasi substitusi

(1)
3x-y+2z=15
2x+y+z=13
(+)=5x+3z=28

(2)
3x+2y+2z=24[*1] 3x+2y+2z = 24
2x+y+z=13. [*2]4x+2y+2z = 26
(-)= -x=-2 --> x=2

(3)
5x+3z=28 (x=2)
10+3z=28
3z=18 >> z=6

(4)
2x+y+z=13 (x=2,z=6)
4+y+6=13
y= 3

(x,y,z) (2,6,3)