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Q.Solve the following system of inequalities graphically: 2x+y≥62x + y \geq 6, 3x+4y≤123x + 4y \leq 12, x≥0x \geq 0, y≥0y \geq 0.

Bihar BsebBihar Board Intermediate 1st Year 2025Subjective· 5mImportance★★★★★
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The solution region is the triangle with vertices (125,65)\left(\dfrac{12}{5},\dfrac65\right), (3,0)(3,0), (4,0)(4,0).

Boundary line 1: 2x+y=62x+y=6 passes through (3,0)(3,0) and (0,6)(0,6). Testing the origin: 2(0)+0=0<62(0)+0=0<6, so the origin does NOT satisfy 2x+y≥62x+y\geq6 — shade the side away from the origin.

Boundary line 2: 3x+4y=123x+4y=12 passes through (4,0)(4,0) and (0,3)(0,3). Testing the origin: 3(0)+4(0)=0≤123(0)+4(0)=0\leq12, so the origin DOES satisfy 3x+4y≤123x+4y\leq12 — shade the side containing the origin.

Constraints x≥0, y≥0x\geq0,\,y\geq0: restrict to the first quadrant.

Intersection of the two lines: 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=1253x+4(6-2x)=12\Rightarrow3x+24-8x=12\Rightarrow-5x=-12\Rightarrow x=\dfrac{12}{5}, then y=6−2(125)=65y=6-2\left(\dfrac{12}{5}\right)=\dfrac65.

So the two lines meet at (125,65)=(2.4,1.2)\left(\dfrac{12}{5},\dfrac65\right)=(2.4,1.2).

Checking the first-quadrant intercepts against both inequalities: (3,0)(3,0) satisfies 3x+4y≤123x+4y\leq12 (9≤129\leq12 ✓), and (4,0)(4,0) satisfies 2x+y≥62x+y\geq6 (8≥68\geq6 ✓).

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