The inequalities are
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
(i) The line l1 : x – 2y = 3 passes through
(3, 0) and (0,−

3

2

)
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 l2 : 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.