# How many solution sets do systems of linear inequalities have? Do solutions to systems of linear inequalities need to satisfy both inequalities? In what case might they not?

embizze | High School Teacher | (Level 2) Educator Emeritus

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There is only one solution set to a system of linear inequalities.

Assuming that you have two inequalities in two variables, any solution must satisfy both inequalities except in the following case: If one or both of the inequalities is not strict (e.g `>=,<=` ) then a solution might satisfy one of the inequalities but not the other.

The solution set to a system of 2 linear inequalities in two unknowns is 1/4 of the plane. The underlying "lines" divide the plane into 4 parts and all of the solutions lie in one of these parts. If the inequlaities are not strict, then some solutions lie on the "lines". If both inequalities are not strict and the lines intersect, there is a solution point that satisfies neither inequlaity.

pramodpandey | College Teacher | (Level 3) Valedictorian

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An infinite no. of solution to system of linear inequality.

Both inequality must be satisfied, otherwise no region in common so no solution.

Let's say we have a system of inequalities

x>0 and y<0 ,

these are two inequalities.

X>0 ,this satisfies in first and fourth quadrant of xy-plane.

y<0  ,this satfies  in third and fourth quadrant of xy -plane.

Both inequalities have common fourth quadrant of xy plane.

Fourth quadrant has an infinite no. of points that satisfies these inequality.