Question:
Show that the following system of linear equations has no solution: $x+2 y \leq 3,3 x+4 y \geq 12, x \geq 0, y \geq 1$
Solution:
We have:
$x+2 y \leq 3 \quad \ldots \ldots(\mathrm{i})$
$3 x+4 y \geq 12 \quad \ldots$ (ii)
$x \geq 0 \quad \ldots$ (iii)
$y \geq 1 \quad \ldots \ldots$ (iv)
As, the points satisfying x + 2y = 3 are:
Also, the points satisfying 3x + 4y = 12 are:
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.