The region represented by the system of inequalities $x, y ≥ 0 ;-2 x+y ≤ 4 ; x+y ≥ 3$ and $x-2 y ≤ 2$ is :
Answer & explanation
Correct answer: option 1
Plotting graph for given inequalities
x, y ≥ 0 ⇒ Graph in 1st Quadrant
plotting → -2x + y = 4, x + y = 3, x - 2y = 2
lines first
| -2x+y=4 | x + y = 3 | x - 2y = 2 | ||||||||||||||||||
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Now checking regions
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-2x + y ≤ 4 checking for (0, 0) point -2(0) + 0 ≤ 4 0 ≤ 4 hence side containing (0, 0) contains solution |
x + y ≥ 3 for (0, 0) 0 ≥ 3 side not containing (0, 0) is solution region
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x - 2y ≤ 2 for (0, 0) 0 ≤ 2 side containing (0, 0) is solution region
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Option 1 → region is unbounded in first quadrant