The inequalities are x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1 (i) The line $l_1$ : x – 2y = 3 passes through (3, 0) and $\left(0,-\frac{3}{2}\right)$ This is represented by AB. Consider the inequality x – 2y ≤ 3, putting x = 0, y = 0 we get 0 < 3, which is true. ⇒ Origin lies in the region of x – 2y ≤ 3. ∴ Region on the above of this line and including its points represents x – 2y ≤ 3 (ii) The line $l_2$ : 3x + 4y = 12 passes through (4, 0) and (0, 3). CD represents this line. Consider the inequality 3x + 4y ≥ 12 putting x = 0, y = 0, we get 0 ≥ 12 which is false. ∴ Origin does not lie in the region of 3x + 4y ≥ 12. The region above the line CD and including points of the line CD represents 3x + 4y ≥ 12. (iii) x ≥ 0 is the region on the right of Y-axis and all the points lying on it. (iv) The line l3 : y = 1 is the line parallel to X-axis at a distance 1 from it. Consider y ≥ 1 or y – 1 ≥ 0, putting y = 0 in y – 1 ≥ 0 We get –1 ≥| 0, origin does not lie in the region. ∴ y ≥ 1 is the region above y = 1 and the points lying on it. ∴ The shaded region shown in figure represents the solution of the given inequalities.