Select the point which is in the feasible region
of the system of inequalities. 

2x + 3y < 8
5x + 2y < 7
x > 0, y > 0 


A. (1,2)
B. (1,1)
C. (0,3)
D. (3,2)

Substitute each of these points in each of the inequalities
to fnd out which one(s) satisies them all:

Try A:
Substituting (x,y) = (1,2) in the first inequality:

2x + 3y < 8

2(1) + 3(2) < 8
     2 + 6 < 8
         8 < 8
That is true.  So we substitute it in the second inequality:

5x + 2y < 7
5(1) + 2(2) < 7
 5 + 4 < 7
     9 < 7

That is false, so we know (1,2) is not in the feasible region,
and (A) is not a correct choice.

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Try B:
Substituting (x,y) = (1,1) in the first inequality:

2x + 3y < 8

2(1) + 3(1) < 8
     2 + 3 < 8
         5 < 8
That is true.  So we substitute it in the second inequality:

5x + 2y < 7
5(1) + 2(1) < 7
 5 + 2 < 7
     7 < 7

That is also true. So we substitute it in the third inequality:

x > 0

1 > 0

That is also true. So we substitute it in the fourth inequality:

y > 0

1 > 0

That is also true, so (1,1) is in the feasible region. So
(B) is a correct choice.

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Let's see if it could be either of the other two:

Try C:
Substituting (x,y) = (0,3) in the first inequality:

2x + 3y < 8

2(0) + 3(3) < 8
     0 + 9 < 8
         9 < 8

That is false.  So we have eliminated C.

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Try D:
Substituting (x,y) = (3,2) in the first inequality:

2x + 3y < 8

2(3) + 3(2) < 8
     6 + 6 < 8
        12 < 8

That is false.  So we have eliminated D.

So the only correct choice is (B).

Edwin