systems of equations and inequalities...systems of inequalities we now consider systems of...

Systems of Equations

and Inequalities

Copyright © Cengage Learning. All rights reserved.

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Copyright © Cengage Learning. All rights reserved.

5.5 Systems of Inequalities

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Objectives

■ Graphing an Inequality

■ Systems of Inequalities

■ Systems of Linear Inequalities

■ Application: Feasible Regions

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Graphing an Inequality

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We begin by considering the graph of a single inequality.

We already know that the graph of y = x2, for example, is

the parabola in Figure 1.

If we replace the equal sign by the

symbol , we obtain the inequality

y x2

Figure 1

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Its graph consists of not just the parabola in Figure 1, but

also every point whose y-coordinate is larger than x2.

We indicate the solution in Figure 2(a) by shading the

points above the parabola.

y x2

Figure 2(a)

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Similarly, the graph of y x2 in Figure 2(b) consists of all

points on and below the parabola.

y x2

Figure 2(b)

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However, the graphs of y > x2 and y < x2 do not include the

points on the parabola itself, as indicated by the dashed

curves in Figures 2(c) and 2(d).

y < x2

Figure 2(d)

y > x2

Figure 2(c)

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The graph of an inequality, in general, consists of a region

in the plane whose boundary is the graph of the equation

obtained by replacing the inequality sign ( , , >, or < )

with an equal sign.

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To determine which side of the graph gives the solution set

of the inequality, we need only check test points.

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Example 1 – Graphs of Inequalities

Graph each inequality.

(a) x2 + y2 < 25 (b) x + 2y 5

Solution:

(a) Graph the equation. The graph of the equation

x2 + y2 = 25 is a circle of radius 5 centered at the origin.

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Example 1 – Solution

The points on the circle itself do not satisfy the inequality

because it is of the form <, so we graph the circle with a

dashed curve, as shown in Figure 3.

Figure 3

cont’d

Graph of x2 + y2 < 25

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Graph the inequality. To determine whether the inside

or the outside of the circle satisfies the inequality, we

use the test points (0, 0) on the inside and (6, 0) on the

outside.

To do this, we substitute the coordinates of each point

into the inequality and check whether the result satisfies

the inequality.

cont’d

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Note that any point inside or outside the circle can serve

as a test point. We have chosen these points for

simplicity.

Our check shows that the points

inside the circle satisfy the inequality.

A graph of the inequality is shown in

Figure 3.

cont’d

Figure 3

Graph of x2 + y2 < 25

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(b) Graph the equation. We first graph the equation of

x + 2y = 5. The graph is the line shown in Figure 4.

cont’d

Figure 4

Graph of x + 2y 5

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Graph the inequality. Let’s use the test points (0, 0)

and (5, 5) on either sides of the line.

Our check shows that the points above the line satisfy

the inequality.

cont’d

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A graph of the inequality is shown in Figure 4.

cont’d

Figure 4

Graph of x + 2y 5

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IMPORTANT!!

We can write the inequality in Example 1 as

From this form of the inequality we see that the solution

consists of the points with y-values on or above the line

.

So the graph of the inequality is the region above the line.

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Systems of Inequalities

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We now consider systems of inequalities.

The solution set of a system of inequalities in two

variables is the set of all points in the coordinate plane that

satisfy every inequality in the system.

The graph of a system of inequalities is the graph of the

solution set.

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To solve a system of inequalities, we use the following

guidelines.

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Example 2 – A System of Two Inequalities

Graph the solution of the system of inequalities, and label

its vertices (vertices are the intersection points).

x2 + y2 < 25

x + 2y 5

Solution:

These are the two inequalities of Example 1. Here we want

to graph only those points that simultaneously satisfy both

inequalities.

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Graph each inequality. In Figure 5(a) we graph the

solutions of the two inequalities on the same axes

(in different colors).


x2 + y2 < 25

x + 2y 5

Figure 5(a)

cont’d

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Graph the solution of the system. The solution of the

system of inequalities is the intersection of the two graphs.

This is the region where the two regions overlap, which is

the purple region graphed in Figure 5(b).


x2 + y2 < 25

x + 2y 5

Figure 5(b)

cont’d

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Find the Vertices. The points (–3, 4) and (5, 0) in

Figure 5(b) are the vertices of the solution set. They are

obtained by solving the system of equations

x2 + y2 = 25

x + 2y = 5

We solve this system of equations by substitution.

cont’d

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Solving for x in the second equation gives x = 5 – 2y, and

substituting this into the first equation gives

(5 – 2y)2 + y2 = 25

(25 – 20y + 4y2) + y2 = 25

–20y + 5y2 = 0

–5y(4 – y) = 0

Thus y = 0 or y = 4.

Substitute x = 5 – 2y

Expand

Factor

Simplify

cont’d

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When y = 0, we have x = 5 – 2(0) = 5, and when y = 4, we

have x = 5 – 2(4) = –3. So the points of intersection of

these curves are (5, 0) and (–3, 4).

Note that in this case the vertices are not part of the

solution set, since they don’t satisfy the inequality

x2 + y2 < 25 (so they are graphed as open circles in the

figure). They simply show where the “corners” of the

solution set lie.

cont’d

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Systems of Linear Inequalities

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An inequality is linear if it can be put into one of the

following forms:

ax + by c ax + by c ax + by > c ax + by < c

In the next example we graph the solution set of a system

of linear inequalities.

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Example 3 – A System of Four Linear Inequalities

Graph the solution set of the system, and label its vertices.

x + 3y 12

x + y 8

x 0

y 0

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Graph each inequality. In Figure 6 we first graph the lines

given by the equations that correspond to each inequality.

To determine the graphs of the first two inequalities, we

need to check only one test point.

Figure 6(b)Figure 6(a)

Answer is the entire

shaded area.

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For simplicity let’s use the point (0, 0).

Since (0, 0) is below the line x + 3y = 12, our check shows

that the region on or below the line must satisfy the

inequality.

cont’d

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Likewise, since (0, 0) is below the line x + y = 8, our check

shows that the region on or below this line must satisfy the

inequality.

The inequalities x 0 and y 0 say that x and y are

nonnegative.

cont’d

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These regions are sketched in Figure 6(a).

Figure 6(a)

cont’d

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Graph the solution of the system. The solution of the

system of inequalities is the intersection of the graphs. This

is the purple region graphed in Figure 6(b).

cont’d

Figure 6(b)

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Find the Vertices. The coordinates of each vertex are

obtained by simultaneously solving the equations of the

lines that intersect at that vertex. From the system

x + 3y = 12

x + y = 8

we get the vertex (6, 2). The origin (0, 0) is also clearly a

vertex. The other two vertices are at the x- and y-intercepts

of the corresponding lines: (8, 0) and (0, 4). In this case all

the vertices are part of the solution set.

cont’d

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A region in the plane is called bounded if it can be

enclosed in a (sufficiently large) circle. A region that is not

bounded is called unbounded.

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For example, the region graphed in Figure 8 is bounded

because it can be enclosed in a circle, as illustrated in

Figure 10(a).

Figure 8A bounded region can

be enclosed in a circle.

Figure 10(a)

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But the regions graphed in Figure 4 and 10b are unbounded,

because we cannot enclose either of them in a circle as

illustrated.

The shaded areas go on forever, so you can’t enclose them

in a circle.

Figure 4

Graph of x + 2y 5

An unbounded region cannot

be enclosed in a circle.

Figure 10(b)

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Example:


its vertices, and determine whether the solution set is

bounded.

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Example:

This is the same problem we just did, but there is one small change on it. Can you see which inequality changed? How does this affect the shading on the graph?

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Example (like #63 on assignment)


its vertices, and determine whether the solution set is

bounded. (This example is just like #63.)

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Assignment:

Section 5.5:

problems 1-21 odd, 27-63 odd

systems of equations and inequalities...systems of inequalities we now consider systems of...

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