The inequalities are 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
(i) The line l1 : 4x + 3y = 60 passes through (15, 0), (0, 20) and it is represented by AB.
Consider the inequality 4x + 3y ≤ 60
Putting x = 0, y = 0.
0 + 0 = 0 ≤ 60 which is true, therefore, origin lies in this region.
Thus, region is below the line AB and the points lying on the line AB represents the inequality 4x + 3y ≤ 60.
(ii) The line l2 : y = 2x passes through (0, 0). It is represented by CD.
Consider the inequality y ≥ 2x.
Putting x = 0, y = 5 in y – 2x ≥ 0
5 > 0 is true.
∴ (0, 5) lies in this region.

Region lying above the line CD and including the points on the line CD represents y ≥ 2x
(iii) x ≥ 3 is the region lying on the right of line l3 : x = 3 and points lying on x = 3 represents the inequality x ≥ 3.
∴ The shaded area ΔPQR in which x ≥ 0 and y ≥ 0 is true for each point, is the solution of given inequalities.