Question

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


pls solve in ur note book and show pls dont scribble

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Answer 1

Answer:

We have x + 2y ≤ 3, 3x + 4y > 12, x > 0, y ≥ 1

Now let’s plot lines x + 2y = 3, 3x + 4y = 12, x = 0 and y = 1 in coordinate plane.

Line x + 2y = 3 passes through the points (0, 3/2) and (3, 0).

Line 3jc + 4y = 12 passes through points (4, 0) and (0, 3).

For (0, 0), 0 + 2(0) – 3 < 0.

Therefore, the region satisfying the inequality x + 2y ≤ 3 and (0,0) lie on the same side of the line x + 2y = 3.

For (0, 0), 3(0) + 4(0)- 12 ≤0.

Therefore, the region satisfying the inequality 3x + 4y ≥ 12 and (0, 0) lie on the opposite side of the line 3x + 4y = 12.

The region satisfying x > 0 lies to the right hand side of the y-axis.

The region satisfying y > 1 lies above the line y = 1.

These regions are plotted as shown in the following figure.

It is clear from the graph that the Shaded portions do not have common region. So, solution set is null set.