Solve the following systems of inequalities graphically x - 2y ≤ 3, 3x +4y ≥ 12, x ≥ 0, y ≥ 1.

3x +4y ≥ 12 …………… (2)

x ≥ 0 …………… (3)

y ≥ 1 …………… (4)

Graph of inequality (i): Let us draw the graph of the line

x - 2y = 3

y = 0 ⇒ x – 2(0) = 3 ⇒ x = 3.

x = 0 ⇒ 0 – 2y = 3 ⇒ y = .
(3, 0) and (0, ) are on the line x - 2y = 3.

Putting x = y = 0 in (i) we get 0 ≤ 3 which is true.

Hence, half-plane region containing the origin is the solution region.

Graph of inequality (ii): Let us draw the graph of the line.

3x +4y = 12

y = 0 ⇒ 3x + 4(0) = 12 ⇒ x = 4.

x = 0 ⇒ 3(0) + 4y = 12 ⇒ y = 3.

∴ (4, 0) and (0, 3) are the points on the line 3x + 4y = 12.

Putting x = y = 0 in (ii) we get 0 ≥ 12 which is false.

Hence, half-plane region not containing the origin is the solution region.

Graph of inequality (iii): Clearly, x ≥ 0 represents the region lying on the right side of y-axis.

Graph of inequality (iv): Let us draw the graph of the line y = 1 which is parallel to x – axis and is at a distance of 1 unit from it.

Putting y = 0 in (iv), we have 0 ≥ 1 which is false.

Hence, half-plane region not containing the origin is the solution region of the given inequality.

The common region of the above four regions represents the solution set of the given linear constraints.