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2x+2y+3z=-8 (2)
2x-3y+2z=8 (3)
(1) & (2) kita eliminasi y
4x-y+z=4 |x2| 8x-2y+2z=8
2x+2y+3z=-8 |x1| 2x+2y+3z=-8
__________+
10x+5z=0 (4)
(1) & (3) kita eliminasi y
4x-y+z=4 |x3| 12x-3y+3z=12
2x-3y+2z=8 |x1| 2x-3y+2z=8
___________-
10x-z=4 (5)
(4) & (5) kita eliminasi x
10x+5z=0
10x-z=4
________-
6z=-4 ⇒ z=-4/6=-2/3
z kita substitusikan (4)
10x+5z=0 ⇒ 10x=-5z ⇒ 10x=-5(-2/3) ⇒ 10x=10/3 ⇒ x=1/3
x dan z kita substitusikan (1)
4x-y+z=4 ⇒ 4(1/3)-y+(-2/3)=4 ⇒ y=4/3-2/3-4=y ⇒ 2/3-12/3=y ⇒ -10/3=y
Jadi,
(2x+y):z
=(2(1/3)+(-10/3)):(-2/3)
=2/3-10/3:(-2/3)
=(-8/3):(-2/3)
=(-8/3)x(-3/2)
=4
Semangat!