Solve the following system of inequalities graphically: $5x + 4y \leq 20, x \geq 1, y \geq 2$.

(N/A) The given system of inequalities is:
$5x + 4y \leq 20$ .....$(1)$
$x \geq 1$ .....$(2)$
$y \geq 2$ .....$(3)$
First,we draw the graphs of the corresponding equations $5x + 4y = 20$,$x = 1$,and $y = 2$.
$1$. For $5x + 4y = 20$,if $x = 0$,$y = 5$; if $y = 0$,$x = 4$. The line passes through $(0, 5)$ and $(4, 0)$. Since the inequality is $\leq$,the region is below the line.
$2$. For $x = 1$,the region is to the right of the vertical line $x = 1$.
$3$. For $y = 2$,the region is above the horizontal line $y = 2$.
The common shaded region represents the solution set of the given system of inequalities,which is a triangle with vertices at $(1, 2)$,$(2.4, 2)$,and $(1, 3.75)$.