Answer :

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.



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