Find the linear inequalities for which the shaded region in the given figure is the solution set.

(N/A) First,we consider the line $3x + 2y = 48$,which intersects the $X$-axis at $(16, 0)$ and the $Y$-axis at $(0, 24)$.
Since the origin $(0, 0)$ lies within the shaded region,it must satisfy the inequality $3x + 2y \leq 48$.
Next,we consider the line $x + y = 20$,which intersects the $X$-axis at $(20, 0)$ and the $Y$-axis at $(0, 20)$.
Since the origin $(0, 0)$ lies within the shaded region,it must satisfy the inequality $x + y \leq 20$.
From the figure,it is clear that the shaded region is in the first quadrant,which implies $x \geq 0$ and $y \geq 0$.
Thus,the system of linear inequalities representing the shaded region is $3x + 2y \leq 48$,$x + y \leq 20$,$x \geq 0$,and $y \geq 0$.