Solve the following system of inequalities graphically:

3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

3x + 4y ≤ 60,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 15 and x = 20

The required points are (0, 15) and (20, 0)

Checking if the origin lies in the required solution area (0,0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area of the line`s graph.

The required solution area would be on the left of the line`s graph.

x +3y ≤ 30,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 10 and x = 30

The required points are (0, 10) and (30, 0)

Checking for the origin (0, 0)

0 ≤ 30, this is true.

Hence the origin lies in the solution area which is given by the left side of the line`s graph.

x ≥ 0,

y ≥ 0,

The given inequalities imply the solution lies in the first quadrant only.

Hence the solution of the inequalities is given by the shaded region in the graph.