Solve the following system of inequalities graphically:
$x + 2y \leqslant 8$ ..... $(1)$
$2x + y \leqslant 8$ ..... $(2)$
$x \geqslant 0$ ..... $(3)$
$y \geqslant 0$ ..... $(4)$

(N/A) To solve the system of inequalities graphically,we first draw the lines $x + 2y = 8$ and $2x + y = 8$.
For the line $x + 2y = 8$:
If $x = 0$,$y = 4$. If $y = 0$,$x = 8$. The line passes through $(0, 4)$ and $(8, 0)$.
For the line $2x + y = 8$:
If $x = 0$,$y = 8$. If $y = 0$,$x = 4$. The line passes through $(0, 8)$ and $(4, 0)$.
The inequalities $x + 2y \leqslant 8$ and $2x + y \leqslant 8$ represent the regions below these lines,respectively.
Since $x \geqslant 0$ and $y \geqslant 0$,the solution region is restricted to the first quadrant.
The intersection point of the two lines is found by solving $x + 2y = 8$ and $2x + y = 8$. Multiplying the first by $2$ gives $2x + 4y = 16$. Subtracting the second equation gives $3y = 8$,so $y = 8/3$. Then $x = 8 - 2(8/3) = 8/3$. The intersection point is $(8/3, 8/3)$.
The shaded region in the graph represents the common solution set of the given system of inequalities.