Solve the following system of inequalities graphically:
$5x + 4y \leq 40$ ..... $(1)$
$x \geq 2$ ..... $(2)$
$y \geq 3$ ..... $(3)$

(N/A) First,we draw the graphs of the lines $5x + 4y = 40$,$x = 2$,and $y = 3$.
For $5x + 4y = 40$,the intercepts are $(8, 0)$ and $(0, 10)$. The inequality $5x + 4y \leq 40$ represents the region on or below this line.
The inequality $x \geq 2$ represents the region on or to the right of the vertical line $x = 2$.
The inequality $y \geq 3$ represents the region on or above the horizontal line $y = 3$.
The solution to the system is the common shaded region bounded by these three lines,forming a triangle with vertices at $(2, 3)$,$(5.6, 3)$,and $(2, 7.5)$.