Represent the following inequalities graphically in a two dimensional plane. 3x + 4y < 12.

3x + 4y = 12 ...(i)

Putting x - 0, y = 3, therefore the point on y-axis is (0, 3) and the point on x-axis is (4, 0). AB is the graph of (i).

Putting x = 0, y = 0 in the given inequality, we have 3(0) + 4(0) ≤ 12 or 0 ≤ 12, which is true.

Hence, origin lies in the half plane region I.

Clearly, any point on the line satisfy the given inequality.

Hence, the shaded region I including the points on the line is the solution region of the inequality.