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Question
Show that the following system of linear inequalities has no solution x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution
Given that: x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Let x + 2y = 3
⇒ x = 3 – 2y
x | 3 | 1 | 5 |
y | 0 | 1 | –1 |
Now putting (0, 0) in x + 2y ≤ 3
0 + 0 ≤ 3
0 ≤ 3 True
Therefore, the shading will be towards (0, 0)
Consider the in equation 3x + 4y ≥ 12
Let 3x + 4y = 12
x | 0 | 4 | 2 |
y | 3 | 0 | 1.5 |
Putting (0, 0) in 3x + 4y ≥ 12
0 + 0 ≥ 12
0 ≥ 12 False
Therefore, shading will be on the opposite side of the graph of line.
x ≥ 0 ⇒ Positive side of y-axis will be shaded.
And y ≥ 1 ⇒ Upperside of y = 1 will be shaded.
Let us now draw the graph.
It is clear from the graph that there is no common shaded region.
Hence, the solution is null set.
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