Question 974179: 2x+y+4z=2
3x+2y+5z=4
x+3y+z=6
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! I would use determinants.
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Here's the elimination method:
2x + 1y + 4z = 2
3x + 2y + 5z = 4
1x + 3y + 1z = 6
------------------
4x + 2y + 8z = 4 Eqn 1 times 2
3x + 2y + 5z = 4
--------------------- Subtract
x + 3z = 0 Eqn A
===============================
6x + 3y +12z = 6 Eqn 1 times 3
1x + 3y + 1z = 6 Eqn 3
------------------------- Subtract
5x + 11z = 0 Eqn B
5x + 15z = 0 Eqn A times 5
----------------- Subtract
-4z = 0
z = 0 **************
x + 3z = 0 Eqn A
x = 0 **************
y = 2 **************
--> (0,2,0)
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