Precalc help?

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1 Answer
George C.
May 28, 2018

#(x, y, z) = (1, -1, 2)#

Explanation:

Given:

#{ (2/x-1/y+2/z=4), (1/x+2/y-2/z=-2), (3/x+3/y+4/z=2) :}#

Note that if we define #X=1/x#, #Y=1/y# and #Z=1/z#, then we get linear equations:

#{ (2X-Y+2Z=4), (X+2Y-2Z=-2), (3X+3Y+4Z=2) :}#

We can eliminate #Z# from the first equation by adding the second equation to it. We can eliminate #Z# from the third equation by adding twice the second equation to it. These two combinations give us:

#{ (3X+Y=2), (5X+7Y=-2) :}#

To eliminate #Y#, we can subtract #7# times the first equation from the second to get:

#-16X=-16#

Hence #X=1#.

Then substituting this value of #X# in #3X+Y=2# we find #Y=-1#.

Then substituting the values #X=1# and #Y=-1# in #2X-Y+2Z=4#, we find #Z=1/2#

Hence #x=1/X=1#, #y = 1/Y=-1# and #z=1/Z=2#

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