Solve the following system of inequalities graphically: 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

Given, 4x + 3y ≤ 60,

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

y = 20 and x = 15

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

Checking for the origin (0, 0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area. The required area would include be on the left of the lines graph.

We have y ≥ 2x,

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

y = 0 and x = 0

Hence the line would pass through origin.

To check which side would be included in the lines graph solution area, we would check for point (15, 0)

⇒ 0 ≥ 15, this is not true so the required solution area would be to the left of the line’s graph.

Consider, x ≥ 3,

For any value of y, the value of x would be same.

Also the origin (0, 0) doesn’t satisfies the inequality as 0 ≥ 3

So the origin doesn’t lies in the solution area, hence the required solution area would be the right of the lines graph.

We have x, y ≥ 0

Since given both x and y are greater than 0 the solution area would be in the first I^st quadrant only.

The shaded area in the graph shows the solution area for the given inequalities