Solve the following system of inequalities graphically:
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
Given,
3x + 2y ≤ 150
Putting value of x = 0 and y = 0 in equation one by one, we get value of
y = 75 and x = 50
The required points are (0, 75) and (50, 0)
Checking for the origin (0, 0)
0 ≤ 150, this is true
Hence the solution area for the line would be on the left side of the line`s graph which would be including the origin too.
x + 4y 
80,
Putting value of x = 0 and y = 0 in equation one by one, we get value of
y = 20 and x = 80
The required points are (0, 20) and (80, 0)
Checking for the origin (0, 0)
0 ≤ 80, this is also true so the origin lies in the solution area.
The required solution area would be toward the left of the line`s graph.
x 
15,
For all the values of y, x would be same
Checking for the origin (0, 0)
0 ≤ 15, this is true so the origin would be included in the solution area. The required solution area would be towards the left of the line`s graph.
y 
0, x 
0
Since x and y are greater than 0, the solution would lie in the 1st quadrant.
The shaded area in the graph satisfies all the given inequalities and hence is the solution area for given inequalities.
