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Q.Solve the following inequalities with graphical method: 2x+y≥82x + y \geq 8, x+2y≥10x + 2y \geq 10.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2017Subjective· 5mImportance★★★★★
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Both boundary lines are drawn from their intercepts; since the origin fails both inequalities, the shaded solution region lies on the far side of each line from the origin, meeting at (2,4)(2,4).

Line 1: 2x+y=82x+y=8. Intercepts: setting x=0x=0 gives y=8y=8, i.e. (0,8)(0,8); setting y=0y=0 gives x=4x=4, i.e. (4,0)(4,0).

Line 2: x+2y=10x+2y=10. Intercepts: setting x=0x=0 gives y=5y=5, i.e. (0,5)(0,5); setting y=0y=0 gives x=10x=10, i.e. (10,0)(10,0).

Testing the origin (0,0)(0,0) in each inequality:

  • 2(0)+0=0≥82(0)+0=0 \geq 8? False. So the region for 2x+y≥82x+y\geq8 is the half-plane on the side AWAY from the origin.
  • 0+2(0)=0≥100+2(0)=0\geq10? False. So the region for x+2y≥10x+2y\geq10 is also the half-plane away from the origin.

Intersection point of the two lines: solving simultaneously, from Line 1: y=8−2xy=8-2x. Substitute into Line 2: x+2(8−2x)=10⇒x+16−4x=10⇒−3x=−6⇒x=2x+2(8-2x)=10 \Rightarrow x+16-4x=10 \Rightarrow -3x=-6 \Rightarrow x=2, and y=8−4=4y=8-4=4. So the lines cross at (2,4)(2,4).

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