Solve the following system of inequalities graphically:
$x-2y \leq 3, 3x+4y \geq 12, x \geq 0, y \geq 1$

(N/A) The given system of inequalities is:
$x-2y \leq 3$ .....$(1)$
$3x+4y \geq 12$ .....$(2)$
$y \geq 1$ .....$(3)$
$x \geq 0$ .....$(4)$
First,we draw the graphs of the lines $x-2y=3$,$3x+4y=12$,$y=1$,and $x=0$ (the $y$-axis).
For inequality $(1)$,testing the point $(0,0)$: $0-0 \leq 3$ is true,so the region includes the origin. Since the line $x-2y=3$ passes through $(3,0)$ and $(0,-1.5)$,the region is towards the origin.
For inequality $(2)$,testing the point $(0,0)$: $0+0 \geq 12$ is false,so the region is away from the origin. The line $3x+4y=12$ passes through $(4,0)$ and $(0,3)$.
For inequality $(3)$,the region is above the line $y=1$.
For inequality $(4)$,the region is to the right of the $y$-axis.
The common shaded region satisfying all these inequalities is the feasible region shown in the graph.