Consider the line ` 3x+ 2y=48`, we observe that the shaded and the origin are on the same side of the line ` 3x+2y=48 and (0,0)` satisfy the linear constraint ` 3x+2y le 48`. So, we must have one inequation as ` 3x+2 y le 48`.
Now, condider the line x+ y = 20 . We find that the shaded region and the origin are on the same side of the line x + y = 20 and (0, 0) satisfy the constraints ` x+y le 20`. So, the second inequation is ` x + y le 20`.
We also notice that the shaded region is above X- axis and is on the right side of y-axis, so we must have ` x ge 0, y ge 0`.
Thus, the linear inequations corresponding to the given solution set are `3x + 2 y le 48, x + y le 20 and x ge 0, y ge 0`.
