Question 57957: Solve each system in three variables form
x+2y-z=5
-3x-2y-3z=11
4x+4y+5z=-18
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! x + 2y - z = 5
-3x - 2y- 3z = 11
4x + 4y + 5z = -18
:
Looking at this we can see that if we mult the 1st equation by -1, something
wonderful will happen: (x & y are eliminated!)
-x - 2y + z = -5
-3x -2y - 3z = 11
+4x +4y + 5z = -18
-------------------- add all three equations:
0x + 0y + 3z = -12
z = -12/3
z = -4
:
Now take the 1st two equations, substitute -4 for z:
x + 2y -(-4) = 5
3x - 2y -3(-4) = 11
:
we have:
x + 2y + 4 = 5
3x -2y + 12 = 11
----------------- add, eliminates y, leaves us with x
4x + 0y + 16 = 16
4x = 16 - 16
4x = 0
x = 0
:
Find y using the 1st equation:
0x + 2y + 4 = 5
2y = 5 - 4
2y = 1
y = 1/2
:
Check our solutions x=0, y=.5, z=-4 in the 2nd and 3rd equations
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