Solve the following system of inequalities graphically: $x \geq 0, y \geq 0, 5x + 3y \leq 15, 4x + 5y \leq 20$.

(N/A) To solve the system of inequalities graphically,we first consider the corresponding equations:
$1$. $5x + 3y = 15$
For $x = 0, y = 5$. For $y = 0, x = 3$. The line passes through $(0, 5)$ and $(3, 0)$.
$2$. $4x + 5y = 20$
For $x = 0, y = 4$. For $y = 0, x = 5$. The line passes through $(0, 4)$ and $(5, 0)$.
$3$. $x \geq 0$ and $y \geq 0$ indicate that the solution lies in the first quadrant.
Testing the origin $(0, 0)$ for both inequalities:
$5(0) + 3(0) = 0 \leq 15$ (True)
$4(0) + 5(0) = 0 \leq 20$ (True)
Since both are true,the shaded region is towards the origin.
The feasible region is the quadrilateral bounded by the vertices $(0, 0), (3, 0), (15/13, 40/13),$ and $(0, 4)$.