x and y are positive integers such that x + 2y > 20 and 3x – 30 < -y

Given conditions: x + 2y > 20 and 3x - 30 < -y; x and y are positive integers
Objective: To find the difference between the minimum possible value of x and the minimum value of y

3x - 30 < -y can be written as 3x + y < 30

Please refer to the following 3 graphs.

The area that satisfies the first inequality x + 2y > 20 is the region above the red line.


The area that satisfies the second inequality 3x + y < 30 is the region below the green line.


Because x and y are positive integers, the area of interest is restricted to I quadrant and the area enclosed in the triangle ABC are values of x and y that satisfy both the inequalities.


Let us find the coordinates of point C
Solve the two equations x + 2y = 20 and 3x + y = 30
3x + 6y = 60
3x + y = 30
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5y = 30
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Or y = 6. Substituting y = 6 in equation (1), we get x = 8.
So, coordinates of point C are (8, 6)

It is evident from the graph that lowest point among the three from y coordinates is C, So, the minimum value of y enclosed in the triangle is an integer greater than 6. So, it has to be 7.
From the graph, we can also deduce that the minimum value of x enclosed in the triangle is an integer greater than 0. So, it has to be 1.

The positive difference between the minimum value of x and minimum value of y is 6.