Solve the system of inequalities graphically 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, x ≥ 0.

x + 2y ≤ 80 ………………. (2)

x ≤ 15 ……………… (3)

x ≥ 0 ……………… (4)

Graph of inequality (i). Let us draw the graph of the line 3x + 2y = 150.

At y = 0 ⇒ 3x + 2(0) = 150 ⇒ x = 50

and at x = 0 ⇒ 3(0) + 2y = 150 ⇒ y = 75

(50, 0) and (0, 75) are the points on the line 3x + 2y = 150.

Putting x = y = 0 in (i) we have 0 ≤ 150 which is true.

Hence half-plane region containing the origin is the solution region of this inequalities.

Graph of inequality (ii). Let us draw the graph of the line x + 4y = 80.

At y = 0 ⇒ x + 4(0) = 80 ⇒ x = 80

x = 0 ⇒ 0 + 4y = 80 ⇒ y = 20

∴ (80, 0) and (0, 20) are the points on the line x + 4y = 80.

Putting x = y = 0 in (ii) we have 0 ≤ 80 which is true.

Hence half-plane region containing the origin is the solution region of this inequality.

Graph of inequality (iii). Let us draw the graph of the line x = 15 which is parallel to the y – axis and is at a distance of 15 units from it.

Putting x = 0 in (iii), we have 0 ≤ 15 which is true, the solution region of the given this inequality.

Graph of inequality (iv). Clearly x ≥ 0 represents the region lying on the right side of y axis.

The common region of the above four regions represent the solution set of the given linear system.