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5 votes
Write a system of linear inequalities with the given characteristic:

-all solutions are in quadrant 3

2 Answers

5 votes

If all solutions are points in Quadrant-3, then the x-coordinates
and y-coordinates of all those points are all negative.

I believe the only possible system of liner inequalities whose
solutions have those characteristics is . . .

x < 0
y < 0 .


User JDong
by
7.8k points
4 votes
In quadrant III x-coordinates are less than 0 and y-coordinates are less than 0.


x<0 \\ y<0

You can modify it to get other examples:

x+1<1 \\ 2y-4<-4 \\ \\ -x>0 \\ y+2<2 \\ \\ (x+y)/(2)<(1)/(2)y \\ -2y+x>x-y
User LRutten
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7.0k points