Solve the following system of inequalities graphically:

2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6

Given 2x + y ≥ 4,

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

y = 4 and x =2

The required points are (0, 4) and (2, 0)

Checking for origin (0, 0)

0 ≥ 4, this is not true

Hence the origin doesn’t lies in the solution area of the line`s graph. The solution area would be given by the right side of the line`s graph.

x + y 3,

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

y = 3 and x = 3

The required points are (0, 3) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true

Hence the solution area would include the origin and hence would be on the left side of the line`s graph.

2x – 3y 6

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

y = -2 and x = 3

The required points are (0, -2), (3, 0)

Checking for the origin (0, 0)

0 ≤ 6 this is true

So the origin lies in the solution area and the area would be on the left of the line`s graph.

Hence the shaded area in the graph is the required solution area for the given inequalities.