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Question
Show that the following system of linear equations has no solution:
\[x + 2y \leq 3, 3x + 4y \geq 12, x \geq 0, y \geq 1\]
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Solution
\[\text{ We have }, \]
\[x + 2y \leq 3 . . . . . \left( i \right)\]
\[3x + 4y \geq 12 . . . . . \left( ii \right)\]
\[x \geq 0 . . . . . \left( iii \right)\]
\[y \geq 1 . . . . . \left( iv \right)\]
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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