The region represented by the system of inequalities $x, y ≥ 0 ;-2 x+y ≤ 4 ; x+y ≥ 3$ and $x-2 y ≤ 2$ is :

unbounded in first quadrant unbounded in first and second quadrant bounded in first quadrant not feasible

unbounded in first quadrant

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      x   0   -2  y 4 0         x  0  3   y  3   0       x   0   2  y  -1  0    Now checking regions  -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    x - 2y ≤ 2  for (0, 0) 0 ≤ 2  side containing (0, 0) is solution region   Option 1 → region is unbounded in first quadrant