Question

Show that the following system of linear inequalities has no solution x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1

Answer 1

Let’s plot the region of each inequality and then find the common region of all

x + 2y ≤ 3

Line: x + 2y = 3

Also, (0, 0) satisfies the x + 2y ≤ 3, hence region is towards the origin

3x + 4y ≤ 12

Line: 3x + 4y = 12

Also, (0, 0) satisfies the 3x + 4y ≤ 3, hence region is towards the origin

x ≥ 0 implies that region is right to the y-axis and y ≥ 1 implies that region is up above the line x = 1, Therefore graph is

It is clear from the graph the above system has no common region as solution.