Solve the following system of inequalities graphically:

2x + y ≥ 6, 3x + 4y ≤ 12

Given 2x + y 6……………1

3x + 4y 12 …………….2

2x + y ≥ 6

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

y = 6 and x = 3

So the point for the (0,6) and (3,0)

Now checking for (0,0)

0 ≥ 6 which is not true, hence the origin does not lies in the solution of the equality. The required region is on the right side of the graph.

Checking for 3x + 4y 12

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

We get y = 3, x = 4

The points are (0, 3) , (4, 0)

Now checking for origin (0, 0)

0 ≤ 12 which is true,

so the origin lies in solution of the equation.

The region on the right of the equation is the region required.

The solution is the region which is common to the graphs of both the inequalities.

The shaded region is the required region.