We have the inequalities :
2x + y ≥ 4 ... (1)
x + y ≤ 3 ... (2)
2x – 3y ≤ 6 ... (3)

We first draw the graph of lines
l1 : 2x + y = 4, l2 : x + y = 3 and l3 : 2x – 3y = 6
(i) 2x + y = 4, passes through (2, 0) and (0, 4)
which is represented by AB.
Consider the inequality 2x + y ≥ 4
Putting x = 0, y = 0 in 2x + y ≥ 4, we get 0 ≥ 4 is false.
∴ Origin does not lie in the region of 2x + y ≥ 4
This inequality represents the region above the line AB and all the points on the line AB.
(ii) Again, x + y = 3 is represented by the line CD, passes through (3, 0) and (0, 3). Consider the inequality x + y ≤ 3 , putting x = 0, y = 0 in x + y ≤ 3, we get 0 ≤ 3 is true.
∴ Origin lies in the region of x + y ≤ 3
∴ x + y ≤ 3 represents the region below the line CD and all the points on the line CD.
(iii) Further, 2x – 3y = 6 is represented by EF passes through (0, –2) and (3, 0).
Consider the inequality 2x – 3y ≤ 6, putting x = 0, y = 0 in 2x – 3y ≤ 6, we get 0 < 6, which is true.
∴ Origin lies in it.
∴ 2x – 3y ≤ 6 represents the region above the line EF and all the points on the line EF.
∴ Shaded triangular area in the figure is the solution of given inequalities.