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

Converting the given inequation,we obtain,x+2y=3,3x+4y=12,x=0,y=1

Region represented by $x + 2 y \leq 3$

The line x+2y=3 meets the coordinate axes at $(0, \frac{3}{2})$ and (3,0).We find that satisfies containing origin represents the solution set of the inequation $x + 2 y \leq 3$ .

Region represented by $3 x + 4 y \geq 12$

The line 3x+4y=12 meets the coordinate axes at (0,3) and (4,0).We find that satisfies containing (0,0) does not satisfy inequation $3 x + 4 y \geq 12$ .

Region represented by $x \geq 0$

Clearly, $x \geq 0$ represents the region lying on the right side of y-axis.

Region represented by $y \geq 1$

The line y=1 is parallel to x-axis (0,0) does not satisfy inequation $y \geq 1$ .So the region lying above the line y=1 is represented by $y \geq 1$ .

From graph we can see that there is no common region bounded by all the four inequalities.So there is no solution set satisfying the given inequalities.

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