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2x+3z=12
y+2z=3
eliminasi:
2(x-2y)=2(5)-2x+3z=12
2x-4y=10-(2x+3z=12)
-4y-3z=-2...(1)
4(y+2z=3)=4y+8z=12
4y+8z=12+(-4y-3z=-2)
5z=10
z=2....(2)
y+2z=3
y=3-2(2)=-1.......(3)
2x+3z=12
2x=12-3(2)
x=3....(4)
x+y+z=3+(-1)+2=4