Question 400010
2x-y+z=1
x+y-7z=2
3x-y+z=0 
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Move the 2nd equation to the top of the list to get:
x+y-7z=2
2x-y+z=1
3x-y+z=0
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Subtract 2R1 from R2 
Subtract 3R1 from R3
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2x-y+z   =  1
0-3y+15z = -3
0-4y+22z = -6
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Subtract R3 from R2 to get:
2x-y+z   =  1
0+ y-7z =   3
0-4y+22z = -6
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Add 4R2 to R3 to get:
2x-y+z   =  1
0+ y-7z =   3
0 +0-6z =  +6
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Divide R3  by -6 to get:
2x-y  +z   =  1
0+ y -7z  =   3
0 +0 + z =   -1
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Substitute z = -1 into R2 and into R1
2x- y - 1   =  1
0 + y + 7 =  3
0 + 0 + z =  -1
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Solve for "y" in R2 to get:
2x- y -1 =   1
0+ y + 0 =  -4
0 + 0+ z =  -1
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Substitute y = -4 into R1 to get: 
2x + 4 -1 =   1
0+ y + 0 =  -4
0 + 0+ z =  -1
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Solve R1 for "x":
x + 0 + 0 = -1
0 + y + 0 = -4
0 + 0 + z = -1
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cheers,
Stan H.