(N/A) The given system of inequalities is:
$x+2y \leq 10$ .....$(1)$
$x+y \geq 1$ .....$(2)$
$x-y \leq 0$ .....$(3)$
$x \geq 0, y \geq 0$ .....$(4)$
To solve this graphically,we first draw the lines $x+2y=10$,$x+y=1$,and $x-y=0$.
$1$. For $x+2y=10$: If $x=0, y=5$; if $y=0, x=10$. The line passes through $(0, 5)$ and $(10, 0)$. Since $(0,0)$ satisfies $x+2y \leq 10$,the region is towards the origin.
$2$. For $x+y=1$: If $x=0, y=1$; if $y=0, x=1$. The line passes through $(0, 1)$ and $(1, 0)$. Since $(0,0)$ does not satisfy $x+y \geq 1$,the region is away from the origin.
$3$. For $x-y=0$: This is a line passing through the origin $(0,0)$ and $(2,2)$. Testing a point like $(0,1)$,we see $0-1 \leq 0$ is true,so the region is above the line $x-y=0$.
The common shaded region in the first quadrant,bounded by these lines,represents the solution set of the given system of linear inequalities.