Solve the following systems of inequalities graphically 3x + 4y ≤ 60, x + 3y ≥ 0, y ≥ 0

y ≥ 0. ….(iv)

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

3x + 4y = 60 ...(v)

At y = 0 ⇒ 3x + 4(0) = 60

⇒ x = 20

x = 0 ⇒ 3(0) + 4y = 60

⇒ y = 15

(20, 0) and (0, 15) are on the line (v)

Putting x = y = 0 in (i) we get 0 ≤ 60 which is true.

Hence, half plane region containing the origin is the solution region of the given inequality.

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

x + 3y = 30 ...(vi)

At y = 0 ⇒ x + 3(0) = 30

⇒ x = 30

x = 0⇒ 0 + 3y = 30

⇒ y = 30

(30, 0) and (0, 30) are on the line (vi)

Putting x = y = 0 in (ii) we get 0 < 30 which is true. Hence, half plane region containing the origin is the solution region of the given inequality

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

The common region of the above four regions is the solution set.