SOLUTION: please help!!! I just spent 2 hours trying to figure this out. I have multiplied the second equation by -1 to eliminate two variables but the answer seemed wrong. I tried again and
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Matrices-and-determiminant
-> SOLUTION: please help!!! I just spent 2 hours trying to figure this out. I have multiplied the second equation by -1 to eliminate two variables but the answer seemed wrong. I tried again and
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Question 117744: please help!!! I just spent 2 hours trying to figure this out. I have multiplied the second equation by -1 to eliminate two variables but the answer seemed wrong. I tried again and multiplied the first equation by -3 to eliminate z but this answer also seemed wrong. here is the problem...
solve the linear system without using matrices or detemrinants
0.1x+0.2y+0.2z=0.2
0.3x+0.5y+0.1z+-0.1
0.2x-0.3y-0.5z=0.7 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 0.1x+0.2y+0.2z=0.2
0.3x+0.5y+0.1z+-0.1
0.2x-0.3y-0.5z=0.7
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Multiply every thing by 10 to get rid of the decimals:
x +2y+2z=2
3x+5y+z=-1
2x-3y-5z=7
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Multiply 1st by 3 and subtract from 2nd
Then Multiply 1st be 2 and subtract form 3rd.
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x +2y+2z=2
0 -y -5z=-7
0 -7y-9z=3
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Divide thru 2nd by -1
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x +2y+2z=2
0 y +5z=+7
0 -7y-9z=3
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Multiply 2nd by 7 and add to 3rd:
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x +2y+2z=2
0 y +5z=+7
0 0 26z=52
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Divide thru 3rd by 26 to get:
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x +2y+2z=2
0 y +5z=+7
0 0 z =2
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Now that you know z=2 you can solve 2nd for y; then you
can solve 1st for x.
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Cheers,
Stan H.