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Question 993856: x+2y+0z+3w=0
0x+y+z+w=1
0x+y+0z+w=2
x+2y+0z+2w=3
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! x+2y+0z+3w=0
0x+y+z+w=1
0x+y+0z+w=2
x+2y+0z+2w=3
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0x+y+z +w =1 Eqn 2
0x+y+0z+w=2 Eqn 3
===================================== Subtract
z = -1 ******************
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x+2y+3w= 0
0x+y+ w= 2 times 2 --> 0x+2y+2w = 4 Eqn A
x +2y+2w = 3
0x+2y+2w = 4 Eqn A
===================================== Subtract
x = -1 **********************
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0x+y+0z+w=2 Eqn 3
x+2y+0z+2w=3 Eqn 4
-----
y + w = 2 Eqn 3
2y + 2w = 4 Eqn 4 times 2 --> y + w = 2, same as Eqn 3
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--> dependent system, no unique solution.
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