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Q.Solve the following system of inequalities graphically :
[!FORMULA] x+2y≤8x + 2y \le 8
[!FORMULA] 2x+y≤82x + y \le 8
[!FORMULA] x≥0x \ge 0
[!FORMULA] y≥0.y \ge 0.

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Subjective· 4mImportance★★★★★
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The common (feasible) region satisfying all four inequalities is bounded by the origin, the xx- and yy-intercepts of the two lines, and their point of intersection.

Boundary lines. x+2y=8x+2y=8 passes through (8,0)(8,0) and (0,4)(0,4). 2x+y=82x+y=8 passes through (4,0)(4,0) and (0,8)(0,8).

Intersection of the two lines. Solve simultaneously: from x+2y=8x+2y=8, x=8−2yx=8-2y. Substitute into 2x+y=82x+y=8:

2(8−2y)+y=8 ⇒ 16−4y+y=8 ⇒ 16−3y=8 ⇒ y=83,x=8−163=83.2(8-2y)+y=8\ \Rightarrow\ 16-4y+y=8\ \Rightarrow\ 16-3y=8\ \Rightarrow\ y=\dfrac83,\quad x=8-\dfrac{16}{3}=\dfrac83.

So the lines meet at (83,83)\left(\dfrac83,\dfrac83\right).

Shading. Since x+2y≤8x+2y\le8 and 2x+y≤82x+y\le8 (both "≤\le"), and x≥0,y≥0x\ge0,y\ge0 restrict to the first quadrant, the required region is the set of points on or below both lines, in the first quadrant. Testing the origin (0,0)(0,0) satisfies both inequalities (0≤80\le8), so the region containing the origin is shaded for each line.

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