The inequalities are 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0 (i) The line $l_1$ : 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 $l_2$ : 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 $l_3$ : 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.