Question 250227: In a 3-by-3 linear system, there should be a tripled pair for the system.
This system being:
3x+3y+6z=9
2x+y+3z=7
x+2y-z=-10
what would be the work in solving this? I figured out the tripled pair, but my work doesn't seem to match up exactly. Would you please help me with this problem? At least start me off, so I can finish it!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In a 3-by-3 linear system, there should be a tripled pair for the system.
This system being:
3x+3y+6z=9
2x+y+3z=7
x+2y-z=-10
--------------------
Let the 3rd equation be 1st:
x+2y-z=-10
3x+3y+6z=9
2x+y+3z=7
----
Multiply 1st by 3 and subtract from the 2nd:
Multiply 1st be 2 and subtract from the 3rd:
x+2y-z=-10
0-3y+9z=39
0-3y+5z=27
-----------------------
Subtract the 2nd from the 3rd:
x+2y-z=-10
0-3y+9z=39
0-0y-4z=-12
-----------------
Solve the 3rd equation for "z":
z = 3
-------------------
Substitute z=3 into the 2nd and solve for "y":
-3y + 27 = 39
-3y = 12
y = -4
----------------------
Substitute z=3 and y = -4 into the 1st and solve for "x":
x + 2(-4)-(3) = -10
x -8 - 3 = -10
x -11 = -10
x = 1
------------------------------
Final answer: x = 1 ; y = -4 ; z = 3
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I checked this with a matrix method.
It is correct.
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Cheers,
Stan H.
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