Solve the following system of inequalities graphically: $x \geq 0, y \geq 0, x+y \leq 6, 3x+4y \leq 12$.

(N/A) To solve the system of inequalities graphically,we first consider the corresponding equations to find the boundary lines:
$1$. $x = 0$ (the $y$-axis)
$2$. $y = 0$ (the $x$-axis)
$3$. $x + y = 6$: If $x=0, y=6$; if $y=0, x=6$. The line passes through $(0, 6)$ and $(6, 0)$.
$4$. $3x + 4y = 12$: If $x=0, y=3$; if $y=0, x=4$. The line passes through $(0, 3)$ and $(4, 0)$.
Next,we determine the feasible region:
- $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant.
- For $x + y \leq 6$,the region is towards the origin since $(0,0)$ satisfies $0+0 \leq 6$.
- For $3x + 4y \leq 12$,the region is towards the origin since $(0,0)$ satisfies $0+0 \leq 12$.
The intersection of these regions is the polygon formed by the vertices $(0, 0), (4, 0), (0, 3)$.