DS inequality problem

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by spanlength » Mon Aug 04, 2008 5:29 am
adding the 2 inequalities we have
m- 3z + 4z - m > 0
hence z > 0 . So now we have to rove that m is either +ve or a -ve no less than z.
from the 1 st inequalitiy we have : Let m is -ve then (-m ) - (3z ) can never be greater than zero.