Mo2men wrote:Let's check the Statement I: x>0
Let x =-1 in first inequality
0< -2+3y<50
2 < 3y < 52
2/3 < y < 17 1/3
Let x =-1 in second inequality
−50 < 3x+2y < 0
−50 < -3+2y < 0
−47 < 2y < 3
−47/2 < y < 3/2
What can we conclude?Can this make statement I invalid? if yes, how?

Here, the blue inequalities yielded by x=-1 are both satisfied by y=1, implying that x=-1 and y=1 constitute a valid solution for 0<2x+3y<50 and -50<3x+2y<0.
Thus, it does NOT have to be true that x>0.

Let's check the Statement II: y>0
Let y = -1 in first inequality
0 < 2+3y < 50
3 < 2x < 53
3/2 < x < 53/2
Let y = - 1 in second inequality
−50 < 3x+2y < 0
−48< 3x < 2
−16 < x < 2/3
What can we conclude?Can this make statement I invalid? if yes, how?

Here, no value for x will satisfy the blue inequalities yielded by y=-1, implying that y<0 is not a valid option for 0<2x+3y<50 and -50<3x+2y<0.
If we follow the same process for y=0, we will see that y=0 is also not a valid option for 0<2x+3y<50 and -50<3x+2y<0.
Since it is not possible that y<0 or that y=0, it must be true that y>0.

Mo2men wrote:If 0<2x+3y<50 and −50<3x+2y<0 , then which of the following must be true?

I. x>0
II. y>0
III. x < y

A. I only
B . II only
C. III only
D. I and III
E. II, and III

To eliminate x from the two inequalities, multiply the first by 3 (to yield 6x) and the second by -2 (to yield -6x) and then ADD the resulting inequalities:

0<2x+3y<50 --> 0<6x+9y<150
-50<3x+2y<0 --> 100>-6x-4y>0 --> 0<-6x-4y<100
Adding together the inequalities in blue, we get:
0<5y<250
0 < y < 50.

To eliminate y from the two inequalities, multiply the first by 2 (to yield 6y) and the second by -3 (to yield -6y) and then ADD the resulting inequalities:

0<2x+3y<50 --> 0<4x+6y<100
-50<3x+2y<0 --> 150>-9x-6y>0 --> 0<-9x-6y<150
Adding together the inequalities in blue, we get:
0<-5x<250
0>x>-50.
-50 < x < 0.

Linking together the two inequalities in green, we get:
-50 < x < 0 < y < 50.
Thus, only Statements II and III must be true.

The correct answer is E.