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Stat1 7x-2y>0
<=> y <7>0 and y>0), in cadran III (x<0 and y<0>0 and y<0> 0, we can be either in cadran I or IV. Thus, y <0> 0 (Fig1).

INSUFF.

Stat2 -y> x
<=> y < -x

Thus, we are below the line y = -x in the XY plan. We can be either in cadran II (x<0>0), in cadran III (x<0 and y<0>0 and y<0> 0, we are necesserary in the cadran IV with y < 0 (Fig2).

SUFF.

Attachments

Fig1_Below 3.5 x X and X positive.GIF [ 3.07 KiB | Viewed 867 times ]

Fig2_Below X and X positive.GIF [ 2.53 KiB | Viewed 863 times ]

Last edited by Fig on 05 Feb 2007, 23:13, edited 1 time in total.

-y< x ; y can also be negative where |y|<x or simply be positive

and from both 1 & 2 does not constrain y hence E. Simply draw the lines y = 3.5 x and y= -x and the area between them is the solution and in both cases Y is not bounded to the first quadrant. hence E.