Show that the following system of linear inequalities has no solution x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Let’s plot the region of each inequality and then find the common region of all
x + 2y ≤ 3
Line: x + 2y = 3
x | 3 | 1 |
y | 0 | 1 |
Also, (0, 0) satisfies the x + 2y ≤ 3, hence region is towards the origin
3x + 4y ≤ 12
Line: 3x + 4y = 12
x | 0 | 4 |
y | 3 | 0 |
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
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