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Q.Solve the following system of inequalities graphically: 3x+2y≤123x+2y \le 12, x≥1x \ge 1, y≥2y \ge 2 OR Solve the following system of inequalities graphically: x+2y≤8x+2y \le 8, 2x+y≤82x+y \le 8, x≥0x \ge 0, y≥0y \ge 0

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2022Subjective· 5mImportance★★★★★
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Plot each boundary line, shade the half-plane satisfying each inequality, and the feasible (solution) region is where all shaded areas overlap; its corners are found by pairwise intersection of the boundary lines.

Main part: 3x+2y≤12, x≥1, y≥23x+2y\le12,\ x\ge1,\ y\ge2.

  • x≥1x\ge1: the region on or to the right of the vertical line x=1x=1.
  • y≥2y\ge2: the region on or above the horizontal line y=2y=2.
  • 3x+2y≤123x+2y\le12: the region on or below/left of the line 3x+2y=123x+2y=12 (which meets the axes at (4,0)(4,0) and (0,6)(0,6)).

Find the corner points of the overlapping (feasible) region by pairwise intersection:

  • x=1x=1 and y=2y=2: point (1,2)(1,2) — check 3(1)+2(2)=7≤123(1)+2(2)=7\le12 ✓
  • x=1x=1 and 3x+2y=123x+2y=12: 3+2y=12⇒y=4.53+2y=12\Rightarrow y=4.5 — point (1,4.5)(1,4.5)
  • y=2y=2 and 3x+2y=123x+2y=12: 3x+4=12⇒x=8/33x+4=12\Rightarrow x=8/3 — point (83,2)\left(\dfrac83,2\right)

These three points form the vertices of a triangular feasible region: (1,2), (1,4.5), (83,2)(1,2),\ (1,4.5),\ \left(\dfrac83,2\right), bounded on all sides.


OR alternative: x+2y≤8, 2x+y≤8, x≥0, y≥0x+2y\le8,\ 2x+y\le8,\ x\ge0,\ y\ge0.

This restricts to the first quadrant (x≥0,y≥0x\ge0,y\ge0), below both lines x+2y=8x+2y=8 (meets axes at (8,0),(0,4)(8,0),(0,4)) and 2x+y=82x+y=8 (meets axes at (4,0),(0,8)(4,0),(0,8)).

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