Solve the following system of inequalities graphically:
$3x + 2y \leq 150, x + 4y \leq 80, x \leq 15, y \geq 0, x \geq 0$

(N/A) $3x + 2y \leq 150$ .... $(1)$
$x + 4y \leq 80$ .... $(2)$
$x \leq 15$ .... $(3)$
The graph of the lines $3x + 2y = 150$,$x + 4y = 80$,and $x = 15$ are drawn in the figure.
Inequality $(1)$ represents the region below the line $3x + 2y = 150$ (including the line).
Inequality $(2)$ represents the region below the line $x + 4y = 80$ (including the line).
Inequality $(3)$ represents the region on the left side of the line $x = 15$ (including the line).
Since $x \geq 0$ and $y \geq 0$,the solution is the common shaded region in the first quadrant,including the points on the respective lines and the axes.