Solve the following system of inequalities graphically: $3x + 4y \leq 60, x + 3y \leq 30, x \geq 0, y \geq 0$

(N/A) $3x + 4y \leq 60$ ...... $(1)$
$x + 3y \leq 30$ ...... $(2)$
The graph of the lines $3x + 4y = 60$ and $x + 3y = 30$ are drawn in the figure below.
Inequality $(1)$ represents the region below the line $3x + 4y = 60$ (including the line $3x + 4y = 60$),and inequality $(2)$ represents the region below the line $x + 3y = 30$ (including the line $x + 3y = 30$).
Since $x \geq 0$ and $y \geq 0$,every point in the common shaded region in the first quadrant,including the points on the respective lines and the axes,represents the solution of the given system of linear inequalities.