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 lines graph which would be including the origin too.

We have 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 lines graph.

Given 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 lines graph.

Consider y ≥ 0, x ≥ 0

Since x and y are greater than 0, the solution would lie in the 1^{st} quadrant.

The shaded area in the graph satisfies all the given inequalities and hence is the solution area for given inequalities.

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