Q.Draw the graph of the solution set of the inequations 2x+y≥2, x-y≤1, x+2y≤8, x≥0 and y≥0, also shade the solution region. (Graph paper not necessary)

Locate each boundary line, determine the feasible side (origin test) for each inequation, and find the corner points of the resulting bounded quadrilateral region.

Boundary lines:

$L_1: 2x+y=2$ (for $2x+y\ge2$) — origin gives $0<2$, so origin is on the infeasible side; feasible region is on the side away from the origin.

$L_2: x-y=1$ (for $x-y\le1$) — origin gives $0\le1$, true, so feasible side contains the origin.

$L_3: x+2y=8$ (for $x+2y\le8$) — origin gives $0\le8$, true, so feasible side contains the origin.

Plus $x\ge0,y\ge0$ restrict to the first quadrant.

Corner points (intersections of boundary lines that satisfy all constraints):

$L_1\cap y\text{-axis}$: $x=0\Rightarrow y=2$. Point $(0,2)$ — check: $x-y=-2\le1$✓, $x+2y=4\le8$✓. Valid.

$L_1\cap L_2$: solving $2x+y=2,\ x-y=1$ gives $x=1,y=0$. Point $(1,0)$ — check: $x+2y=1\le8$✓, $x,y\ge0$✓. Valid (also lies on the $x$-axis).

$L_2\cap L_3$: solving $x-y=1,\ x+2y=8$: from $x=1+y$, $(1+y)+2y=8\Rightarrow y=\dfrac73,\ x=\dfrac{10}3$. Point $\left(\dfrac{10}3,\dfrac73\right)$ — check $2x+y=9\ge2$✓. Valid.

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