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 
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Let the 3rd equation be 1st:
x+2y-z=-10 
3x+3y+6z=9
2x+y+3z=7
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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
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Subtract the 2nd from the 3rd:
x+2y-z=-10 
0-3y+9z=39
0-0y-4z=-12
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Solve the 3rd equation for "z":
z = 3
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Substitute z=3 into the 2nd and solve for "y":
-3y + 27 = 39
-3y = 12
y = -4
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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
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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.