# Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42

Question:

Find the linear iAlso the shaded region is in the first quadrant. Therefore, we must have $x \geq 0$ and $y \geq 0$nequations for which the solution set is the shaded region given in Fig. 15.42

Solution:

Considering the line $x+y=4$, we find that the shaded region and the origin $(0,0)$ are on the same side of this line and $(0,0)$ does not satisfy the inequation $x+y \leq 4$ So, the first inequation is $x+y \leq 4$

Considering the line $y=3$, we find that the shaded region and the origin $(0,0)$ are on the same side of this line and $(0,0)$ satisfies the inequation $y \leq 3$ So, the corresponding inequation is $y \leq 3$

Considering the line $x=3$, we find that the shaded region and the origin $(0,0)$ are on the same side of this line and $(0,0)$ satisfies the inequation $x \leq 3$ So, the corresponding inequation is $x \leq 3$

Considering the line $x+5 y=4$, we find that the shaded region and the origin $(0,0)$ are on the opposite side of this line and $(0,0)$ does not satisfy the inequation $x+5 y \geq 4$ So, the corresponding inequation is $x+5 y \geq 4$

Considering the line $6 x+2 y=8$, we find that the shaded region and the origin $(0,0)$ are on the opposite side of this line and $(0,0)$ does not satisfy the inequation $6 x+2 y \geq 8$ So, the corresponding inequation is $6 x+2 y \geq 8$

Thus, the linear inequations comprising the given solution set are given below:

$x+y \leq 4, y \leq 3, x \leq 3, x+5 y \geq 4,6 x+2 y \geq 8, x \geq 0$ and $y \geq 0$