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solve each of the following systems of linear inequalities graphically
3x-y<=6 ; x>=1 ; y<=3 if you do 3x - 0 =6 then 3x=6 and x<=2; and 3(1) - y(-3) =6 3+3=6; so you would graph (2,0) and (1,-3) but then on x>=1 you would put that at 1 on the x axis; and for y<=3 you would put that on the 3 of the y axis. is that the correct way to solve this problem??
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you need to draw out all the line , first ignore the < or > , draw 3x-y=6 , x=1 ,y=3 . 3x-y<=6 divides the plane in two sides , only one of them satisfies the need. Notice that (0,0) is above the line, assume x=0. y=0, then the inequality 3x0-0<6 , it satisfies, so it's the upper area , then find the intersection of the three , done .
Last edited by Stanley_Marsh (2007-02-02 08:50:35)
Numbers are the essence of the Universe
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thanks that makes sense now
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