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x+2y = -3
x+y+2z = 5
_________ _
y-2z = -8 .... (4)
eliminasi 4 dan 2
y-2z = -8
y+2z = 4
_________ _
-4z = -12
z = 3
substitusi y
y+2z = 4
y+2(3) = 4
y+6 = 4
y = 4-6
y = -2
substitusi x
x+2y = -3
x+2(-2) = -3
x+(-4) = -3
x = -3+4
x = 1
(x,y,z)
(1,-2,3)
nilai 3(x+z) = 3(1+3) = 3(4) = 12