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Q.Solve the following system of linear inequalities graphically: 2x+y≤6, 3x+4y≤12, x≥0, y≥02x+y\le6,\ 3x+4y\le12,\ x\ge0,\ y\ge0

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Subjective· 4mImportance★★★★★
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The region satisfying 2x+y≤6, 3x+4y≤12, x≥0, y≥02x+y\le6,\ 3x+4y\le12,\ x\ge0,\ y\ge0 is the quadrilateral with vertices (0,0),(3,0),(2.4,1.2),(0,3)(0,0),(3,0),(2.4,1.2),(0,3).

Step 1: Draw the boundary lines.

Line 1: 2x+y=62x+y=6. It meets the axes at (3,0)(3,0) and (0,6)(0,6).

Line 2: 3x+4y=123x+4y=12. It meets the axes at (4,0)(4,0) and (0,3)(0,3).

Step 2: Determine which side of each line satisfies the inequality.

Test the origin (0,0)(0,0) in each:

  • 2(0)+0=0≤62(0)+0=0\le6 ✓, so the region for Line 1 includes the origin (below/left of the line).
  • 3(0)+4(0)=0≤123(0)+4(0)=0\le12 ✓, so the region for Line 2 also includes the origin.

Combined with x≥0, y≥0x\ge0,\ y\ge0, we stay in the first quadrant, on the origin-side of both lines.

Step 3: Find where the two lines intersect.

Solve 2x+y=62x+y=6 and 3x+4y=123x+4y=12 together. From the first, y=6−2xy=6-2x. Substitute:

3x+4(6−2x)=12  ⟹  3x+24−8x=12  ⟹  −5x=−12  ⟹  x=2.43x+4(6-2x)=12 \implies 3x+24-8x=12 \implies -5x=-12 \implies x=2.4

y=6−2(2.4)=1.2y=6-2(2.4)=1.2

So the lines cross at (2.4, 1.2)(2.4,\,1.2).

Step 4: Identify the binding intercepts.

On the xx-axis (y=0y=0): Line 1 gives x=3x=3, Line 2 gives x=4x=4. Since we need BOTH ≤\le conditions, the tighter (smaller) bound x=3x=3 governs — vertex (3,0)(3,0).

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