anonymous
  • anonymous
?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1371150388411:dw| I only know how to graph it sorry
anonymous
  • anonymous
|dw:1371150625304:dw| this is actually closer to the correct graph
anonymous
  • anonymous
|dw:1371150697330:dw|

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anonymous
  • anonymous
|dw:1371171954946:dw| A (-3,3),B (3,3),C(3/2,-3/2) Eq. of line AB is y-3=(3-3)/(3+3) (x+3) y-3=0/6 (x+3) y-3=0 ...(1) Eq. of line BC is y-3=(-3/2-3)/(3/2-3) (x-3) y-3=(-9/2)/(-3/2) (x-3) y-3=-9/2*2/-3 (x-3) y-3=3x-9 3x-y=-3+9=6 3x-y=6 ...(2) Eq. of line AC is y-3=(-3/2 -3)/(3/2 +3) (x+3) y-3=(-9/2)/(9/2) (x+3) y-3=-x-3 x+y=0 ...(3) Put x=1, y=0 in (1),(2) &(3) 0-3<0 ,Hence y-3<0 ...(4) 3*1-0<6,Hence 3x- y<6 ...(5) 1+0 >0 x+y>0 ...(6) (4),(5) &(6) are the required equations.
anonymous
  • anonymous
when you get the equations of three lines to find the enequality put x=0,y=0 in the equations, if the lines do not pass through origin. If any line pass through origin then put x=1,y=0 or any other value.

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