Solve the following system of inequalities graphically:

x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0 , y ≥ 1

Given,

x – 2y 3

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

y = -3/2 = -1.5 and x = 3

The required points are (0, -1.5) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true.

Hence the solution area would be on the left of the line`s graph

3x + 4y 12,

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

y = 3 and x = 4

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

Checking for the origin (0, 0)

0 ≥ 12 , this is not true

So the solution area would of include the origin and the required solution area would be on the right side of the line`s graph.

x 0 ,

For all the values of y, the value of x would be same in the given inequality, which would be the region above the x axis on the graph.

y 1

For all the values of x, the value of y would be same in the given inequality.

The solution area of the line would be not include origin as 0 ≥ 1 is not true.

The solution area would be on the left side of the line`s graph.

The shaded area in the graph is the required solution area which satisfies all the given inequalities at the same time.