838 Chapter 8 Systems of Equations and Inequalities€¦ · 18. Roses sell for $3 each and carnations for $1.50 each. If a mixed bouquet of 20 ﬂ owers consisting of roses and carnations

838 Chapter 8 Systems of Equations and Inequalities

In Exercises 13–16, write the partial fraction decomposition of each rational expression.

13. x2 - 6x + 3

(x - 2)3 14. 10x2 + 9x - 7

(x + 2)(x2 - 1)

15. x2 + 4x - 23

(x + 3)(x2 + 4) 16.

x3

(x2 + 4)2

17. A company is planning to manufacture PDAs (personal digital assistants). The fi xed cost will be $400,000 and it will cost $20 to produce each PDA. Each PDA will be sold for $100.

a. Write the cost function, C, of producing x PDAs. b. Write the revenue function, R, from the sale of x PDAs. c. Write the profi t function, P, from producing and selling x

PDAs. d. Determine the break-even point. Describe what this

means.

18. Roses sell for $3 each and carnations for $1.50 each. If a mixed bouquet of 20 fl owers consisting of roses and carnations costs $39, how many of each type of fl ower is in the bouquet?

19. Find the measure of each angle whose degree measure is represented with a variable.

y

x3y + 20

20. Find the quadratic function y = ax2 + bx + c whose graph

passes through the points (-1, 0), (1, 4), and (2, 3). 21. Find the length and width of a rectangle whose perimeter is

21 meters and whose area is 20 square meters.


statement 3 + 2 7 1, or 5 7 1. Because there are infi nitely many pairs of numbers that have a sum greater than 1, the inequality x + y 7 1 has infi nitely many solutions. Each ordered-pair solution is said to satisfy the inequality. Thus, (3, 2) satisfi es the inequality x + y 7 1.

Objectives � Graph a linear inequality

in two variables. � Graph a nonlinear

inequality in two variables.

� Use mathematical models involving linear inequalities.

� Graph a system of inequalities.

Systems of Inequalities SECTION 8.5

W e opened the chapter noting that the modern emphasis on thinness as the ideal body shape has been suggested as a major cause of eating disorders. In this section (Example 5), as well as in the Exercise Set (Exercises 77–80), we use systems of linear inequalities in two variables that will enable you to establish a healthy weight range for your height and age.

Linear Inequalities in Two Variables and Their Solutions We have seen that equations in the form Ax + By = C are straight lines when graphed. If we change the symbol = to 7 , 6 , Ú , or … ,we obtain a linear inequality in two variables . Some examples of linear inequalities in two variables are x + y 7 2, 3x - 5y … 15, and 2x - y 6 4.

A solution of an inequality in two variables , x and y, is an ordered pair of real numbers with the following property: When the x@coordinate is substituted for x and the y@coordinate is substituted for y in the inequality, we obtain a true statement. For example, (3, 2) is a solution of the inequality x + y 7 1. When 3 is substituted for xand 2 is substituted for y, we obtain the true

M14_BLIT7240_06_SE_08-hr.indd 838 13/10/12 12:23 PM

Section 8.5 Systems of Inequalities 839

The Graph of a Linear Inequality in Two Variables We know that the graph of an equation in two variables is the set of all points whose coordinates satisfy the equation. Similarly, the graph of an inequality in two variables is the set of all points whose coordinates satisfy the inequality.

Let’s use Figure 8.14 to get an idea of what the graph of a linear inequality in two variables looks like. Part of the fi gure shows the graph of the linear equation x + y = 2. The line divides the points in the rectangular coordinate system into three sets. First, there is the set of points along the line, satisfying x + y = 2. Next, there is the set of points in the green region above the line. Points in the green region satisfy the linear inequality x + y 7 2. Finally, there is the set of points in the purple region below the line. Points in the purple region satisfy the linear inequality x + y 6 2.

A half-plane is the set of all the points on one side of a line. In Figure 8.14 , the green region is a half-plane. The purple region is also a half-plane. A half-plane is the graph of a linear inequality that involves 7 or 6 . The graph of a linear inequality that involves Ú or … is a half-plane and a line. A solid line is used to show that a line is part of a solution set. A dashed line is used to show that a line is not part of a solution set.

� Graph a linear inequality in two variables.

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Half-planex + y > 2

Half-planex + y < 2

Line x + y = 2

FIGURE 8.14

Graphing a Linear Inequality in Two Variables

1. Replace the inequality symbol with an equal sign and graph the corresponding linear equation. Draw a solid line if the original inequality contains a … or Ú symbol. Draw a dashed line if the original inequality contains a 6 or 7 symbol.

2. Choose a test point from one of the half-planes. (Do not choose a point on the line.) Substitute the coordinates of the test point into the inequality.

3. If a true statement results, shade the half-plane containing this test point. If a false statement results, shade the half-plane not containing this test point.

EXAMPLE 1 Graphing a Linear Inequality in Two Variables

Graph: 2x - 3y Ú 6.

SOLUTION Step 1 Replace the inequality symbol by � and graph the linear equation. We need to graph 2x - 3y = 6. We can use intercepts to graph this line.

We set y � 0 to fi nd the x@intercept.

We set x � 0 to fi ndthe y@intercept.

2x - 3y = 6 2x - 3y = 6

2x - 3 # 0 = 6 2 # 0 - 3y = 6

2x = 6 -3y = 6

x = 3 y = -2

The x@intercept is 3, so the line passes through (3, 0). The y@intercept is -2, so the line passes through (0, -2). Using the intercepts, the line is shown in Figure 8.15 as a solid line. This is because the inequality 2x - 3y Ú 6 contains a Ú symbol, in which equality is included.

−1

12345

−2−3−4−5

1 2 3 4 7 865−1−2

y

x

2x − 3y = 6

FIGURE 8.15 Preparing to graph 2x - 3y Ú 6



Step 2 Choose a test point from one of the half-planes and not from the line. Substitute its coordinates into the inequality. The line 2x - 3y = 6 divides the plane into three parts—the line itself and two half-planes. The points in one half-plane satisfy 2x - 3y 7 6. The points in the other half-plane satisfy 2x - 3y 6 6. We need to fi nd which half-plane belongs to the solution of 2x - 3y Ú 6. To do so, we test a point from either half-plane. The origin, (0, 0), is the easiest point to test.

2x - 3y Ú 6 This is the given inequality.

2 # 0 - 3 # 0 Ú? 6 Test (0, 0) by substituting 0 for x and 0 for y.

0 - 0 Ú 6?

Multiply.

0 Ú 6 This statement is false.

Step 3 If a false statement results, shade the half-plane not containing the test point. Because 0 is not greater than or equal to 6, the test point, (0, 0), is not part of the solution set. Thus, the half-plane below the solid line 2x - 3y = 6 is part of the solution set. The solution set is the line and the half-plane that does not contain the point (0, 0), indicated by shading this half-plane. The graph is shown using green shading and a blue line in Figure 8.16 . ● ● ●

Check Point 1 Graph: 4x - 2y Ú 8.

When graphing a linear inequality, choose a test point that lies in one of the half-planes and not on the line dividing the half-planes . The test point (0, 0) is convenient because it is easy to calculate when 0 is substituted for each variable. However, if (0, 0) lies on the dividing line and not in a half-plane, a different test point must be selected.

EXAMPLE 2 Graphing a Linear Inequality in Two Variables

Graph: y 7 - 23

x.

SOLUTION Step 1 Replace the inequality symbol by � and graph the linear equation. Because we are interested in graphing y 7 - 23 x, we begin by graphing y = - 23 x. We can use the slope and the y@intercept to graph this linear function.

y-intercept = 0Slope = =

y= x+0–2

3−23

riserun

The y@intercept is 0, so the line passes through (0, 0). Using the y@intercept and the slope, the line is shown in Figure 8.17 as a dashed line. This is because the inequality y 7 - 23 x contains a 7 symbol, in which equality is not included.

Step 2 Choose a test point from one of the half-planes and not from the line. Substitute its coordinates into the inequality. We cannot use (0, 0) as a test point because it lies on the line and not in a half-plane. Let’s use (1, 1), which lies in the half-plane above the line.

y 7 - 23

x This is the given inequality.

1 7? -

23# 1 Test (1, 1) by substituting 1 for x and 1 for y.

1 7 - 23

This statement is true.

−1

12345

−2−3−4−5

1 2 3 4 7 865−1−2

y

x

FIGURE 8.16 The graph of 2x - 3y Ú 6

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Test point: (1, 1)

23y = − x

Rise = −2

Run = 3

FIGURE 8.17 The graph of y 7 - 23 x



Step 3 If a true statement results, shade the half-plane containing the test point. Because 1 is greater than - 23 , the test point (1, 1) is part of the solution set. All the points on the same side of the line y = - 23 x as the point (1, 1) are members of the solution set. The solution set is the half-plane that contains the point (1, 1), indicated by shading this half-plane. The graph is shown using green shading and a dashed blue line in Figure 8.17 . ● ● ●

TECHNOLOGY Most graphing utilities can graph inequalities in two variables with the � SHADE � feature. The procedure varies by model, so consult your manual. For most graphing utilities, you must fi rst solve for y if it is not already isolated. The fi gure shows the graph of y 7 - 23 x. Most displays do not distinguish between dashed and solid boundary lines.

Check Point 2 Graph: y 7 - 34

x.

Graphing Linear Inequalities without Using Test Points You can graph inequalities in the form y 7 mx + b or y 6 mx + b without using test points. The inequality symbol indicates which half-plane to shade.

• If y 7 mx + b, shade the half-plane above the line y = mx + b. • If y 6 mx + b, shade the half-plane below the line y = mx + b.

Observe how this is illustrated in Figure 8.17 in the margin on the previous page. The graph of y 7 - 23 x is the half-plane above the line y = - 23 x.

It is also not necessary to use test points when graphing inequalities involving half-planes on one side of a vertical or a horizontal line.

For the Vertical Line x � a: For the Horizontal Line y � b:

• If x 7 a, shade the half-plane to the right of x = a.

• If y 7 b, shade the half-plane above y = b.

• If x 6 a, shade the half-plane to the left of x = a.

• If y 6 b, shade the half-plane below y = b.

x < a in theyellow region.

x > a in thegreen region.

x

x = a

y

y < b in theyellow region.

y > b in thegreen region.

x

y = b

y

GREAT QUESTION! When is it important to use test points to graph linear inequalities?

Continue using test points to graph inequalities in the form Ax + By 7 C or Ax + By 6 C. The graph of Ax + By 7 C can lie above or below the line given by Ax + By = C, depending on the values of A and B. The same comment applies to the graph of Ax + By 6 C.



SOLUTION a. y … -3

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

Graph y = −3, a horizontal line with y-intercept−3. The line is solid because equality is includedin y � −3. Because of the less than part of �,shade the half-plane below the horizontal line.

b. x 7 2

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

Graph x = 2, a vertical line with x-intercept 2.The line is dashed because equality is notincluded in x > 2. Because of >, the greaterthan symbol, shade the half-plane to the rightof the vertical line.

Check Point 3 Graph each inequality in a rectangular coordinate system:

a. y 7 1 b. x … -2.

Graphing a Nonlinear Inequality in Two Variables Example 4 illustrates that a nonlinear inequality in two variables is graphed in the same way that we graph a linear inequality.

EXAMPLE 4 Graphing a Nonlinear Inequality in Two Variables

Graph: x2 + y2 … 9.

SOLUTION Step 1 Replace the inequality symbol with � and graph the nonlinear equation. We need to graph x2 + y2 = 9. The graph is a circle of radius 3 with its center at the origin. The graph is shown in Figure 8.18 as a solid circle because equality is included in the … symbol.

Step 2 Choose a test point from one of the regions and not from the circle. Substitute its coordinates into the inequality. The circle divides the plane into three parts—the circle itself, the region inside the circle, and the region outside the circle. We need to determine whether the region inside or outside the circle is included in the solution. To do so, we will use the test point (0, 0) from inside the circle.

x2 + y2 … 9 This is the given inequality.

02 + 02 …? 9 Test (0, 0) by substituting 0 for x and 0 for y.

0 + 0 …? 9 Square 0: 02 = 0.

0 … 9 Add. This statement is true.

Step 3 If a true statement results, shade the region containing the test point. The true statement tells us that all the points inside the circle satisfy x2 + y2 … 9. The graph is shown using green shading and a solid blue circle in Figure 8.19 .

● ● ●

� Graph a nonlinear inequality in two variables.

EXAMPLE 3 Graphing Inequalities without Using Test Points

Graph each inequality in a rectangular coordinate system:

a. y … -3 b. x 7 2.



Check Point 4 Graph: x2 + y2 Ú 16.

Modeling with Systems of Linear Inequalities Just as two or more linear equations make up a system of linear equations, two or more linear inequalities make up a system of linear inequalities . A solution of a system of linear inequalities in two variables is an ordered pair that satisfi es each inequality in the system.

EXAMPLE 5 Does Your Weight Fit You?

The latest guidelines, which apply to both men and women, give healthy weight ranges, rather than specifi c weights, for your height. Figure 8.20 shows the healthy weight region for various heights for people between the ages of 19 and 34, inclusive.

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

FIGURE 8.18 Preparing to graph x2 + y2 … 9

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

FIGURE 8.19 The graph of x2 + y2 … 9

� Use mathematical models involving linear inequalities.

● ● ●

Height (inches)

x60 62 64 66 68 70 72 74 76 78

230

Wei

ght (

poun

ds)

y Healthy Weight Region forMen and Women, Ages 19 to 34

90

110

130

150

170

190

210

3.7x − y = 125

4.9x − y = 165

Healthy WeightRegion

B

A

FIGURE 8.20 Source: U.S. Department of Health and Human Services

If x represents height, in inches, and y represents weight, in pounds, the healthy weight region in Figure 8.20 can be modeled by the following system of linear inequalities:

b4.9x - y Ú 1653.7x - y … 125.

Show that point A in Figure 8.20 is a solution of the system of inequalities that describes healthy weight.



SOLUTION Point A has coordinates (70, 170). This means that if a person is 70 inches tall, or 5 feet 10 inches, and weighs 170 pounds, then that person’s weight is within the healthy weight region. We can show that (70, 170) satisfi es the system of inequalities by substituting 70 for x and 170 for y in each inequality in the system.

4.9x - y Ú 165

4.9(70) - 170 Ú? 165

343 - 170 Ú? 165

173 Ú 165, true

3.7x - y … 125

3.7(70) - 170 …? 125

259 - 170 …? 125

89 … 125, true

The coordinates (70, 170) make each inequality true. Thus, (70, 170) satisfi es the system for the healthy weight region and is a solution of the system. ● ● ●

Check Point 5 Show that point B in Figure 8.20 is a solution of the system of inequalities that describes healthy weight.

Graphing Systems of Linear Inequalities The solution set of a system of linear inequalities in two variables is the set of all ordered pairs that satisfy each inequality in the system. Thus, to graph a system of inequalities in two variables, begin by graphing each individual inequality in the same rectangular coordinate system. Then fi nd the region, if there is one, that is common to every graph in the system. This region of intersection gives a picture of the system’s solution set.

EXAMPLE 6 Graphing a System of Linear Inequalities

Graph the solution set of the system:

b x - y 6 12x + 3y Ú 12.

SOLUTION Replacing each inequality symbol with an equal sign indicates that we need to graph x - y = 1 and 2x + 3y = 12. We can use intercepts to graph these lines.

x=1

x-intercept: x-0=1

y=–1

–y=1

y-intercept: 0-y=1

The line passes through (1, 0).

The line passes through (0, –1).

x=6

2x=12

x-intercept: 2x+3 � 0=12

y=4

3y=12

y-intercept: 2 � 0+3y=12



x � y � 1 2x � 3y � 12Set y = 0 ineach equation.

Set x = 0 ineach equation.

Now we are ready to graph the solution set of the system of linear inequalities.

� Graph a system of inequalities.

Height (inches)

x60 62 64 66 68 70 72 74 76 78

230

Wei

ght (

poun

ds)

y Healthy Weight Regionfor Men and Women,

Ages 19 to 34

90

110

130

150

170

190

210

3.7x − y = 125

4.9x − y = 165

HealthyWeightRegion

B

A

FIGURE 8.20 (repeated)



Check Point 6 Graph the solution set of the system:

b x - 3y 6 62x + 3y Ú -6.

EXAMPLE 7 Graphing a System of Inequalities


by Ú x2 - 4x - y Ú 2.

SOLUTION We begin by graphing y Ú x2 - 4. Because equality is included in Ú , we graph y = x2 - 4 as a solid parabola. Because (0, 0) makes the inequality y Ú x2 - 4 true (we obtain 0 Ú -4 ), we shade the interior portion of the parabola containing (0, 0), shown in yellow in Figure 8.21 .

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

x − y = 1: passesthrough (1, 0)and (0, −1)

Graph x − y < 1. The blue line, x − y = 1,is dashed: Equality is not included in x − y < 1.Because (0, 0) makes the inequality true(0 − 0 < 1, or 0 < 1, is true), shade the half-plane containing (0, 0) in yellow.

The graph of x-y<1

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

2x + 3y = 12:passes through

(6, 0) and (0, 4)

Add the graph of 2x + 3y � 12. The red line,2x + 3y = 12, is solid: Equality is included in2x + 3y � 12. Because (0, 0) makes the inequalityfalse (2 � 0 + 3 � 0 � 12, or 0 � 12, is false),shade the half-plane not containing (0, 0) usinggreen vertical shading.

Adding the graph of2x+3y � 12

x − y = 1

x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3−4−5

The solution set of the system is graphedas the intersection (the overlap) of thetwo half-planes. This is the region inwhich the yellow shading and the greenvertical shading overlap.

The graph of x-y<1and 2x+3y � 12

This open dot shows(3, 2) is not inthe solution set.

It does not satisfyx − y < 1.

● ● ●

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = x2 − 4

FIGURE 8.21 The graph of y Ú x2 - 4

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = x2 − 4x − y = 2

FIGURE 8.22 Adding the graph of x - y Ú 2

−1

12345

−2

−4−5

1 3 4 5−1−2−3−4−5

y

x

y = x2 − 4x − y = 2

(−1, −3) (2, 0)

FIGURE 8.23 The graph of y Ú x2 - 4 and x - y Ú 2

Now we graph x - y Ú 2 in the same rectangular coordinate system. First we graph the line x - y = 2 using its x@intercept, 2, and its y@intercept, -2. Because (0, 0) makes the inequality x - y Ú 2 false (we obtain 0 Ú 2 ), we shade the half-plane below the line. This is shown in Figure 8.22 using green vertical shading.

The solution of the system is shown in Figure 8.23 by the intersection (the overlap) of the solid yellow and green vertical shadings. The graph of the system’s solution set consists of the region enclosed by the parabola and the line. To fi nd the points of intersection of the parabola and the line, use the substitution method to solve the nonlinear system containing y = x2 - 4 and x - y = 2. Take a moment to show that the solutions are (-1, -3) and (2, 0), as shown in Figure 8.23 . ● ● ●




b y Ú x2 - 4x + y … 2.

A system of inequalities has no solution if there are no points in the rectangular coordinate system that simultaneously satisfy each inequality in the system. For example, the system

b2x + 3y Ú 62x + 3y … 0,

whose separate graphs are shown in Figure 8.24 , has no overlapping region. Thus, the system has no solution. The solution set is �, the empty set.

EXAMPLE 8 Graphing a System of Inequalities


c x - y 6 2-2 … x 6 4

y 6 3.

SOLUTION We begin by graphing x - y 6 2, the fi rst given inequality. The line x - y = 2 has an x@intercept of 2 and a y@intercept of -2. The test point (0, 0) makes the inequality x - y 6 2 true, and its graph is shown in Figure 8.25 .

Now, let’s consider the second given inequality, -2 … x 6 4. Replacing the inequality symbols by = , we obtain x = -2 and x = 4, graphed as red vertical lines in Figure 8.26 . The line of x = 4 is not included. Because x is between -2 and 4, we shade the region between the vertical lines. We must intersect this region with the yellow region in Figure 8.25 . The resulting region is shown in yellow and green vertical shading in Figure 8.26 .

Finally, let’s consider the third given inequality, y 6 3. Replacing the inequality symbol by = , we obtain y = 3, which graphs as a horizontal line. Because of the less than symbol in y 6 3, the graph consists of the half-plane below the line y = 3. We must intersect this half-plane with the region in Figure 8.26 . The resulting region is shown in yellow and green vertical shading in Figure 8.27 . This region represents the graph of the solution set of the given system.

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

2x + 3y ≥ 6

2x + 3y ≤ 0

FIGURE 8.24 A system of inequalities with no solution

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

x − y = 2

FIGURE 8.25 The graph of x - y 6 2

−1

12345

−2−3−4−5

1 2 3 5−1−3−4−5

y

x

x − y = 2

x = −2x = 4

FIGURE 8.26 The graph of x - y 6 2 and -2 … x 6 4

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

x − y = 2

x = −2x = 4

y = 3

FIGURE 8.27 The graph of x - y 6 2 and -2 … x 6 4 and y 6 3

In Example 7, how did you graph y = x2 - 4?

We used the parabola’s vertex and its x@ intercepts.

a = 1 b = 0 c = −4

y=1x2+0x-4

• a 7 0: opens upward • Vertex:

a- b2a

, f a- b2ab b = (0, f(0)) = (0, -4)

GREAT QUESTION!

• x@ intercepts: x2 - 4 = 0

x2 = 4 x = {2

The graph of y = x2 - 4, shown in blue in Figure 8.21 on the previous page, passes through (-2, 0) and (2, 0) with a vertex at (0, -4).



In Figure 8.27 it may be diffi cult to tell where the graph of x - y = 2 intersects the vertical line x = 4. Using the substitution method, it can be determined that this intersection point is (4, 2). Take a moment to verify that the four intersection points in Figure 8.27 are, clockwise from upper left, (-2, 3), (4, 3), (4, 2), and (-2, -4). These points are shown as open dots because none satisfi es all three of the system’s inequalities. ● ● ●


c x + y 6 2-2 … x 6 1y 7 -3.

Fill in each blank so that the resulting statement is true .

CONCEPT AND VOCABULARY CHECK

1. The ordered pair (3, 2) is a/an of the inequality x + y 7 1 because when 3 is substituted for and 2 is substituted for , the true statement is obtained.

2. The set of all points that satisfy a linear inequality in two variables is called the of the inequality.

3. The set of all points on one side of a line is called a/an .

4. True or false: The graph of 2x - 3y 7 6 includes the line 2x - 3y = 6.

5. True or false: The graph of the linear equation 2x - 3y = 6 is used to graph the linear inequality 2x - 3y 7 6.

6. True or false: When graphing 4x - 2y Ú 8, to determine which side of the line to shade, choose a test point on 4x - 2y = 8.

7. When graphing x2 + y2 7 25, to determine whether to shade the region inside the circle or the region outside the circle, we can use as a test point.

8. The solution set of the system

b x - y 6 12x + 3y Ú 12.

is the set of ordered pairs that satisfy and .

9. True or false: The graph of the solution set of the system

b x - 3y 6 62x + 3y Ú -6.

includes the intersection point of x - 3y = 6 and 2x + 3y = -6.

EXERCISE SET 8.5

Practice Exercises In Exercises 1–26, graph each inequality.

1. x + 2y … 8 2. 3x - 6y … 12 3. x - 2y 7 10 4. 2x - y 7 4

5. y …13

x 6. y …14

x

7. y 7 2x - 1 8. y 7 3x + 2 9. x … 1 10. x … -3 11. y 7 1 12. y 7 -3 13. x2 + y2 … 1 14. x2 + y2 … 4 15. x2 + y2 7 25 16. x2 + y2 7 36 17. (x - 2)2 + (y + 1)2 6 9 18. (x + 2)2 + (y - 1)2 6 16 19. y 6 x2 - 1 20. y 6 x2 - 9 21. y Ú x2 - 9 22. y Ú x2 - 1 23. y 7 2x 24. y … 3x 25. y Ú log2(x + 1) 26. y Ú log3(x - 1)

In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution.

27. b 3x + 6y … 62x + y … 8

28. b x - y Ú 4x + y … 6

29. b 2x - 5y … 103x - 2y 7 6

30. b 2x - y … 43x + 2y 7 -6

31. b y 7 2x - 3y 6 -x + 6

32. b y 6 -2x + 4y 6 x - 4

33. b x + 2y … 4y Ú x - 3

34. b x + y … 4y Ú 2x - 4

35. b x … 2y Ú -1

36. b x … 3y … -1

37. -2 … x 6 5 38. -2 6 y … 5

39. b x - y … 1x Ú 2

40. b 4x - 5y Ú -20x Ú -3

41. b x + y 7 4x + y 6 -1

42. b x + y 7 3x + y 6 -2



43. b x + y 7 4x + y 7 -1

44. b x + y 7 3x + y 7 -2

45. b y Ú x2 - 1x - y Ú -1

46. b y Ú x2 - 4x - y Ú 2

47. b x2 + y2 … 16x + y 7 2

48. b x2 + y2 … 4x + y 7 1

49. b x2 + y2 7 1x2 + y2 6 16

50. b x2 + y2 7 1x2 + y2 6 9

51. b (x - 1)2 + (y + 1)2 6 25(x - 1)2 + (y + 1)2 Ú 16

52. b (x + 1)2 + (y - 1)2 6 16(x + 1)2 + (y - 1)2 Ú 4

53. b x2 + y2 … 1y - x2 7 0

54. b x2 + y2 6 4y - x2 Ú 0

55. b x2 + y2 6 16y Ú 2x 56. b x2 + y2 … 16

y 6 2x

57. c x - y … 2x 7 -2y … 3

58. c 3x + y … 6x 7 -2y … 4

59. d x Ú 0y Ú 02x + 5y 6 103x + 4y … 12

60. d x Ú 0y Ú 02x + y 6 42x - 3y … 6

61. d 3x + y … 62x - y … -1x 7 -2y 6 4

62. d 2x + y … 6x + y 7 2

1 … x … 2y 6 3

Practice Plus In Exercises 63–64, write each sentence as an inequality in two variables. Then graph the inequality.

63. The y@variable is at least 4 more than the product of -2 and the x@variable.

64. The y@variable is at least 2 more than the product of -3 and the x@variable.

In Exercises 65–68, write the given sentences as a system of inequalities in two variables. Then graph the system.

65. The sum of the x@variable and the y@variable is at most 4. The y@variable added to the product of 3 and the x@variable does not exceed 6.

66. The sum of the x@variable and the y@variable is at most 3. The y@variable added to the product of 4 and the x@variable does not exceed 6.

67. The sum of the x@variable and the y@variable is no more than 2. The y@variable is no less than the difference between the square of the x@variable and 4.

68. The sum of the squares of the x@variable and the y@variable is no more than 25. The sum of twice the y@variable and the x@variable is no less than 5.

In Exercises 69–70, rewrite each inequality in the system without absolute value bars. Then graph the rewritten system in rectangular coordinates.

69. b � x � … 2� y � … 3

70. b � x � … 1� y � … 2

The graphs of solution sets of systems of inequalities involve fi nding the intersection of the solution sets of two or more inequalities. By contrast, in Exercises 71–72, you will be graphing the union of the solution sets of two inequalities.

71. Graph the union of y 7 32 x - 2 and y 6 4.

72. Graph the union of x - y Ú -1 and 5x - 2y … 10.

Without graphing, in Exercises 73–76, determine if each system has no solution or infi nitely many solutions.

73. b 3x + y 6 93x + y 7 9

74. b 6x - y … 246x - y 7 24

75. b (x + 4)2 + (y - 3)2 … 9(x + 4)2 + (y - 3)2 Ú 9

76. b (x - 4)2 + (y + 3)2 … 24(x - 4)2 + (y + 3)2 Ú 24

Application Exercises The fi gure shows the healthy weight region for various heights for people ages 35 and older.

Height (inches)

x60 62 64 66 68 70 72 74 76 78

240

Wei

ght (

poun

ds)

y Healthy Weight Region forMen and Women, Ages 35 and Older

100

120

140

160

180

200

220

4.1x − y = 140

5.3x − y = 180

Healthy WeightRegion

A

B

Source: U.S. Department of Health and Human Services

If x represents height, in inches, and y represents weight, in pounds, the healthy weight region can be modeled by the following system of linear inequalities:

b 5.3x - y Ú 1804.1x - y … 140.

Use this information to solve Exercises 77–80.

77. Show that point A is a solution of the system of inequalities that describes healthy weight for this age group.

78. Show that point B is a solution of the system of inequalities that describes healthy weight for this age group.



79. Is a person in this age group who is 6 feet tall weighing 205 pounds within the healthy weight region?

80. Is a person in this age group who is 5 feet 8 inches tall weighing 135 pounds within the healthy weight region?

81. Many elevators have a capacity of 2000 pounds.

a. If a child averages 50 pounds and an adult 150 pounds, write an inequality that describes when x children and y adults will cause the elevator to be overloaded.

b. Graph the inequality. Because x and y must be positive, limit the graph to quadrant I only.

c. Select an ordered pair satisfying the inequality. What are its coordinates and what do they represent in this situation?

82. A patient is not allowed to have more than 330 milligrams of cholesterol per day from a diet of eggs and meat. Each egg provides 165 milligrams of cholesterol. Each ounce of meat provides 110 milligrams.

a. Write an inequality that describes the patient’s dietary restrictions for x eggs and y ounces of meat.

b. Graph the inequality. Because x and y must be positive, limit the graph to quadrant I only.

c. Select an ordered pair satisfying the inequality. What are its coordinates and what do they represent in this situation?

83. On your next vacation, you will divide lodging between large resorts and small inns. Let x represent the number of nights spent in large resorts. Let y represent the number of nights spent in small inns.

a. Write a system of inequalities that models the following conditions:

You want to stay at least 5 nights. At least one night should be spent at a large resort. Large resorts average $200 per night and small inns average $100 per night. Your budget permits no more than $700 for lodging.

b. Graph the solution set of the system of inequalities in part (a).

c. Based on your graph in part (b), what is the greatest number of nights you could spend at a large resort and still stay within your budget?

84. A person with no more than $15,000 to invest plans to place the money in two investments. One investment is high risk, high yield; the other is low risk, low yield. At least $2000 is to be placed in the high-risk investment. Furthermore, the amount invested at low risk should be at least three times the amount invested at high risk. Find and graph a system of inequalities that describes all possibilities for placing the money in the high- and low-risk investments.

The graph of an inequality in two variables is a region in the rectangular coordinate system. Regions in coordinate systems have numerous applications. For example, the regions in the following two graphs indicate whether a person is obese, overweight, borderline overweight, normal weight, or underweight.

30

35

25

20

7060

Bod

y-M

ass

Inde

x (B

MI)

Age

ObeseObese

Overweight

Overweight

Borderline OverweightBorderline Overweight

Normal RangeNormal Range

Underweight Underweight

2010 30 40 50 80

30

35

25

20

15 157060

Bod

y-M

ass

Inde

x (B

MI)

Age2010 30 40 50 80

Females Males

Source: Centers for Disease Control and Prevention

The horizontal axis shows a person’s age. The vertical axis shows that person’s body-mass index (BMI), computed using the following formula:

BMI =703W

H2 .

The variable W represents weight, in pounds. The variable H represents height, in inches. Use this information to solve Exercises 85–86.

85. A man is 20 years old, 72 inches (6 feet) tall, and weighs200 pounds.

a. Compute the man’s BMI. Round to the nearest tenth. b. Use the man’s age and his BMI to locate this information

as a point in the coordinate system for males. Is this person obese, overweight, borderline overweight, normal weight, or underweight?

86. A woman is 25 years old, 66 inches (5 feet, 6 inches) tall, and weighs 105 pounds.

a. Compute the woman’s BMI. Round to the nearest tenth.

b. Use the woman’s age and her BMI to locate this information as a point in the coordinate system for females. Is this person obese, overweight, borderline overweight, normal weight, or underweight?

Writing in Mathematics 87. What is a linear inequality in two variables? Provide an

example with your description. 88. How do you determine if an ordered pair is a solution of an

inequality in two variables, x and y? 89. What is a half-plane? 90. What does a solid line mean in the graph of an inequality?



91. What does a dashed line mean in the graph of an inequality? 92. Compare the graphs of 3x - 2y 7 6 and 3x - 2y … 6.

Discuss similarities and differences between the graphs. 93. What is a system of linear inequalities? 94. What is a solution of a system of linear inequalities? 95. Explain how to graph the solution set of a system of

inequalities. 96. What does it mean if a system of linear inequalities has no

solution?

Technology Exercises Graphing utilities can be used to shade regions in the rectangular coordinate system, thereby graphing an inequality in two variables. Read the section of the user’s manual for your graphing utility that describes how to shade a region. Then use your graphing utility to graph the inequalities in Exercises 97–102.

97. y … 4x + 4 98. y Ú23

x - 2

99. y Ú x2 - 4 100. y Ú12

x2 - 2

101. 2x + y … 6 102. 3x - 2y Ú 6

103. Does your graphing utility have any limitations in terms of graphing inequalities? If so, what are they?

104. Use a graphing utility with a � SHADE � feature to verify any fi ve of the graphs that you drew by hand in Exercises 1–26.

105. Use a graphing utility with a � SHADE � feature to verify any fi ve of the graphs that you drew by hand for the systems in Exercises 27–62.

Critical Thinking Exercises Make Sense? In Exercises 106–109, determine whether each statement makes sense or does not make sense, and explain your reasoning.

106. When graphing a linear inequality, I should always use (0, 0) as a test point because it’s easy to perform the calculations when 0 is substituted for each variable.

107. When graphing 3x - 4y 6 12, it’s not necessary for me to graph the linear equation 3x - 4y = 12 because the inequality contains a 6 symbol, in which equality is not included.

108. The reason that systems of linear inequalities are appropriate for modeling healthy weight is because guidelines give healthy weight ranges, rather than specifi c weights, for various heights.

109. I graphed the solution set of y Ú x + 2 and x Ú 1 without using test points.

In Exercises 110–113, write a system of inequalities for each graph.

110.

x

y

2

4

−2

−4

2 4−3−5

111.

x

y

2

4

−2

−4

2 4−2−4

112.

−1

123

54

−2−3

1 2 3 4−1−2−3−4

y

x

y = x2

113.

78

x

y

1 2 6 7 83 4 5

123

6

45

114. Write a system of inequalities whose solution set includes every point in the rectangular coordinate system.

115. Sketch the graph of the solution set for the following system of inequalities:

b y Ú nx + b (n 6 0, b 7 0)y … mx + b (m 7 0, b 7 0).

Preview Exercises Exercises 116–118 will help you prepare for the material covered in the next section.

116. a. Graph the solution set of the system:

c x + y Ú 6x … 8y Ú 5.

b. List the points that form the corners of the graphed region in part (a).

c. Evaluate 3x + 2y at each of the points obtained in part (b).

117. a. Graph the solution set of the system:

d x Ú 0y Ú 0

3x - 2x … 6y … -x + 7.

b. List the points that form the corners of the graphed region in part (a).

c. Evaluate 2x + 5y at each of the points obtained in part (b).

118. Bottled water and medical supplies are to be shipped to survivors of an earthquake by plane. The bottled water weighs 20 pounds per container and medical kits weigh 10 pounds per kit. Each plane can carry no more than 80,000 pounds. If x represents the number of bottles of water to be shipped per plane and y represents the number of medical kits per plane, write an inequality that models each plane’s 80,000-pound weight restriction.


838 Chapter 8 Systems of Equations and Inequalities€¦ · 18. Roses sell for $3 each and carnations for $1.50 each. If a mixed bouquet of 20 ﬂ owers consisting of roses and carnations

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