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we have to prove x > 7/3 y Case 1: x> y+4 y = 0 x > 4 means x > (7/3)* 0 , y=1, x> 5 means x >(7/3 ) * 1 etc... its sufficient. Case 2: x > -14/5 y it can be similarly proven using values .... that its sufficient. hence D.

we have to prove x > 7/3 y Case 1: x> y+4 y = 0 x > 4 means x > (7/3)* 0 , y=1, x> 5 means x >(7/3 ) * 1 etc... its sufficient. Case 2: x > -14/5 y it can be similarly proven using values .... that its sufficient. hence D.

how is st. 1 true?

if y = 1, x > 5, then 3x is 15 and 7y is 7. so 3x>7y is true. but if y = 10, x > 14. lets say x = 15 3x is 45, where as 7y is 70.
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1) x>y+4 say 3x>4y+3y => x>y+4/3 y now if 4/3y<4 we are not sure whether 3x>7y holds good !!

INSUFFI

2)-5x<14y => x> -14/5y

now if this is true and y>0 => x>7/3y may o may not be true !! INSUFFI

Consider both : 1) and 2) say x>y+4 since this the positive among the two ,represents overlaping values hence INSUFI to validate the questioin IMO E
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