SOLUTION: solve the system x+y+z=6 3x+4y-7z=1 2x-y+3z=5

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Question 1799: solve the system
x+y+z=6
3x+4y-7z=1
2x-y+3z=5

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
these are just simultaneous equations...3 unknowns and 3 equations...solvable, if tedious.
You need to start with 2 equations and get rid of 1 variable...looking at the 3 equations, use 1 and 3, as they both have a y, so easy to remove these...
1 is +x+%2B+y+%2B++z+=+6
3 is 2x+-+y+%2B+3z+=+5
the y's have Different signs so aDD (note the DD). Similarly, Same signs, then Subtract!!
so, we get equation 4
3x+%2B+4z+=+11
Now repeat the process with 2 other equations, say 2 and 3
3x+%2B+4y+-+7z+=+1
2x+-++y+%2B+3z+=+5
Again I will choose to get rid of the y term, but i need to increase the lower equation by a factor of 4, to get the 2 equations to match....
3x+%2B+4y+-++7z+=++1
8x+-+4y+%2B+12z+=+20
The y-terms have Different signs therefore we aDD, getting equation 5...
11x+%2B+5z+=+21
Now we use 4 and 5 to get rid of another variable...i choose to remove the z's, by making them both 20z...
equation 4 needs to be multiplied by 5: +3x+%2B+4z+=+11
equation 5 needs to be multiplied by 4: 11x+%2B+5z+=+21
This gives us
15x+%2B+20z+=+55
44x+%2B+20z+=+84
Here the z terms have the Same sign therefore we Subtract the 2 equations, giving us %28-29x%29=%28-29%29, so x=1.
We now substitute this into 4 or 5, to find z (z=2) and then these into equation 1, 2 or 3 to find y (y=3).
THEN you double check each of the original equations (1, 2 and 3) to make sure that your answers are correct.


cheers
Jon.