Show that the following system of linear equations has no solution: $x + 2 y \leq 3, 3 x + 4 y \geq 12, x \geq 0, y \geq 1$

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Solution

$We have, x + 2 y \leq 3 . . . . . (i) 3 x + 4 y \geq 12 . . . . . (ii) x \geq 0 . . . . . (iii) y \geq 1 . . . . . (iv)$

As, the points satisfying x + 2y = 3 are:

x 3 0 5

y 0 1.5 $-$ 1

Also, the points satisfying 3x + 4y = 12 are:

x 0 4 8

y 3 0 $-$ 3

Now, the region representing the given inequalities is as follows:

Since, there is no common region.

So, the given system of inequalities has no solution.

x 3 0 5

y 0 1.5 $-$ 1

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