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David Terr
Ph.D. Math, UC Berkeley

 

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7.5. Systems of Inequalities in Two Variables

In Section P.9, we discussed how to solve linear inequalities in one variable, and in Section 2.7, we extended this to polynomial and rational inequalities. In both cases, our solutions consisted of intervals or sets of intervals of real numbers. In this section, we discuss how to solve systems of linear and nonlinear inequalities in two variables. To represent the solution sets of these systems, we need to make graphs of regions of the xy-plane. The idea is to first graph the corresponding equations in the system, which appear as lines or curves in the plane. Then we shade the appropriate side of the regions corresponding to the inequalities. Our solution is then the intersection of the shaded regions.

 

Example 1: Graph the solution set of the following system of linear inequalities:

x ≥ 2

y < 3

Solution: The corresponding equations are x = 2, whose graph is the vertical line intersecting the x-axis at the point (2, 0), and y = 3, whose graph is the horizontal line which intersects the y-axis at the point (0, 3). Now the first inequality tells us the solution set is contained in the region to the right of and including the line x = 2. Thus, we draw this line as a solid line and shade the half-plane lying to the right of this line. Similarly, the second inequality tells us that the solution set is contained in the region lying below and not including the line y = 3. To draw this region, we draw a dashed line corresponding to y = 3 and shade the region below this line. The solution set to this system of inequalities is then equal to the intersection of these two shaded regions, shown in a darker shade as follows:

example 1 solution

 

Often we have more than two inequalities, as the following example shows:

 

Example 2: Solve the following system of linear inequalities:

|x + y| ≤ 1

|x - y| ≤ 1

Solution: This is really a system of four linear inequalities, since the first inequality is equivalent to the system of inequalities x + y ≤ 1 and x + y ≥-1, and similarly, the second inequality is equivalent to the system of inequalities x - y ≤ 1 and x - y ≥-1. The system is thus bounded by the four lines x ± y = ± 1. The solution set is clearly the interior region as shown below:

example 2 solution

 

Example 3: Graph the solution set of the following system of linear inequalities:

(x + 1)2 + y2 < 4

(x - 1)2 + y2 < 4

Solution: The bounding equation (x + 1)2 + y2 = 4 of the first inequality is a circle of radius 2 centered at the point (-1, 0). Similarly, the bounding equation (x - 1)2 + y2 = 4 of the second inequality is a circle of radius 2 centered at the point (1, 0). The regions corresponding to the inequalities are each the interiors of these circles, not including the circles themselves. The solution to the system of inequalities is the intersection of these two regions, the lens-shaped region shown below in dark gray:

example 3 solution

 

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