SOLUTION: Three system of linear equation: x+3y-z=4 x+14y-5z=11 5x+19y-6z=13

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Question 1016795: Three system of linear equation: x+3y-z=4
x+14y-5z=11
5x+19y-6z=13
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Three system of linear equation:
x+3y-z=4
x+14y-5z=11
5x+19y-6z=13
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Subtract 1st row from 2nd
Subtract 5*1st row from 3rd
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x+3y-z=4
0x+11y-4z = 7
0x+4y-z = -7
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Multiply thru 2nd row by 4
Multiply thru 3rd row by 11
==================
x+3y-z = 4
0x+44y-16z = 28
0x+44y-11z = -77
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Subtract 2nd from 3rd row
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x+3y-z = 4
0x+44y-16z = 28
0x+0y+5z = -105
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Solve 3rd row for "z"::
z = -21
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x+3y-z = 4
0x+44y-16z = 28
0x+0y+z = -21
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Multiply 3rd row by 16 and add to 2nd row::
x+3y-z = 4
0x+44y+0z = -308
0x+0y+z = -21
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Divide thru 2nd by 44 to solve for "y"::
x+3y-z = 4
0x+y+0z = -7
0x+0y+z = -21
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Substitute and solve for x::
x = 4
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Cheers,
Stan H.
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