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Show that xz < yz whenever x > y and z > 0.
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that is never true.
There is a way out
z should be negative and not z > 0
so..what's the question?
Show that xz < yz whenever x > y and z < 0. ??
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but u wrote z>0
Sorry for the inconvenience. The question is Show that xz < yz whenever x > y and z < 0.
x > y so multiply z to both sides xz > yz since z is negative... -xz > -yz divide both sides by -1 xz < yz right?
I need an analytical solution
wasn't that analytical?
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it seems algebraic to me
sure
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