Question

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

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 :- See the attachment for graph

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

Hope it helps you a lot @itzheartcracer

Answer 2

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• These regions are plotted as shown in the following figure. ( REFER TO ATTACHMENT )

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

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