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When we add the two inequalities \(0<2x+3y<50\) and \(-50<3x+2y<0\), we obtain \(-50<5x+5y<50\), or \(-20<-2x-2y< 20\).
Statement I.
Adding the two inequalities \(-50<3x+2y<0\) and \(-20<-2x-2y< 20\) yields \(-70<x<20\).
So x may not be greater than zero.
Statement I may not be true.
Statement II.
By multiplying all sides of \(0<2x+3y<50\) by \(-3\), we have \(-150<-6x-9y< 0\).
By multiplying all sides of \(-50<3x+2y<0\) by \(2\), we have \(-100<6x+4y< 0\).
By adding the above inequalities, we have \(-250<-5y<0\) or \(0<y<50\).
Statement II is true.
Statement III.
Since \(0<2x+3y<50\) is equivalent to \(-50<-2x-3y<0\) and \(-50<3x+2y<0\), adding the two inequalities yields
\(-100<x-y<0\). This implies that \(x < y\).
Statement III must be true.
Therefore, the answer is E.
Answer : E
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