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by vladmire » Tue Dec 30, 2008 6:46 pm
Can anyone show me how what numbers I can use to show that the first four options are incorrect? I keep coming up with each statement being true.
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by vittalgmat » Wed Dec 31, 2008 4:36 am
This particular problem can be solved faster algebraically as follows.

xy +z = x(y+z)
->xy +z = xy +xz
-> xz -z = 0
-> z(x -1) = 0
So z = 0 OR x -1 =0
ie
z = 0, x = 1


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by Zipper » Wed Dec 31, 2008 4:40 am
xy + z = xy + xz

z = xz

so either

x=1 or z=o
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