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Miscellaneous Exercise 7 · Q72

Q.Find graphical solution for each of the following system of linear inequation : 3x+4y≤12, x−2y≥2, y≥−13x + 4y \le 12,\ x - 2y \ge 2,\ y \ge -1

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Draw the three boundary lines: 3x+4y=123x+4y=12 (through (4,0)(4,0) and (0,3)(0,3)), x−2y=2x-2y=2 (through (2,0)(2,0) and (0,−1)(0,-1)), and y=−1y=-1 (horizontal). Shade each half-plane (test the origin for the first two; y≥−1y\ge-1 is immediate) and keep the common overlap. Its three vertices are found by pairing the boundary lines: 3x+4y=123x+4y=12 with y=−1y=-1 gives 3x−4=12⇒x=1633x-4=12\Rightarrow x=\tfrac{16}{3}, point (163,−1)\left(\tfrac{16}{3},-1\right); x−2y=2x-2y=2 with y=−1y=-1 gives x=2−2=0x=2-2=0, point (0,−1)(0,-1); and 3x+4y=123x+4y=12 with x−2y=2x-2y=2 (solve x=2+2yx=2+2y, substitute: 3(2+2y)+4y=12⇒6+6y+4y=12⇒y=0.6, x=3.23(2+2y)+4y=12\Rightarrow6+6y+4y=12\Rightarrow y=0.6,\ x=3.2) gives $\left(\tfrac{1 …

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