Solve the following system of inequalities graphically: 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

Asked by Abhisek | 1 year ago |  113

##### Solution :-

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 1st 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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Answered by Pragya Singh | 1 year ago

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