P-BLTZMC07 727-804-hr 21-11-2008 12:58 Page 778 Section ...teachers.dadeschools.net/lberkson/Documents/Ch7_Section5.pdf · Graphing a Linear Inequality in Two Variables 1. Replace

778 Chapter 7 Systems of Equations and Inequalities

Objectives

� Graph a linear inequality intwo variables.

� Graph a nonlinear inequalityin two variables.

� Use mathematical modelsinvolving linear inequalities.

� Graph a system ofinequalities.

Systems of Inequalities

We opened the chapter noting that the modernemphasis on thinness as the ideal body shape

has been suggested as a major cause of eatingdisorders. In this section, as well as in theexercise set, we use systems of linearinequalities in two variables that willenable you to establish a healthy weightrange for your height and age.

Linear Inequalities in TwoVariables and Their SolutionsWe have seen that equations in the form

are straight lines when graphed.If we change the we obtain a linear inequality in two variables.Some examples of linear inequalities intwo variables are and

A solution of an inequality in two variables, and is an ordered pair of realnumbers with the following property:When the is substituted for andthe is substituted for in the inequality, we obtain a true statement. Forexample, (3, 2) is a solution of the inequality When 3 is substituted for and 2 is substituted for we obtain the true statement or Becausethere are infinitely many pairs of numbers that have a sum greater than 1, theinequality has infinitely many solutions. Each ordered-pair solution issaid to satisfy the inequality. Thus, (3, 2) satisfies the inequality

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

Let’s use Figure 7.15 to get an idea of what thegraph of a linear inequality in two variables lookslike. Part of the figure shows the graph of the linearequation The line divides the points inthe rectangular coordinate system into three sets.First, there is the set of points along the line,satisfying Next, there is the set of pointsin the green region above the line. Points in thegreen region satisfy the linear inequality Finally, there is the set of points in the purple regionbelow the line. Points in the purple region satisfy thelinear inequality

A half-plane is the set of all the points on oneside of a line. In Figure 7.15, the green region is ahalf-plane.The purple region is also a half-plane.A half-plane is the graph of a linearinequality that involves or The graph of an inequality that involves or is ahalf-plane and a line. A solid line is used to show that a line is part of a graph. Adashed line is used to show that a line is not part of a graph.

…Ú6 .7

x + y 6 2.

x + y 7 2.

x + y = 2.

x + y = 2.

x + y 7 1.x + y 7 1

5 7 1.3 + 2 7 1,y,xx + y 7 1.

yy-coordinatexx-coordinate

y,x2x - y 6 4.

x + y 7 2, 3x - 5y … 15,

symbol = to 7 , 6 , Ú , or … ,Ax + By = C

Sec t i on 7.5

� Graph a linear inequality in twovariables.

−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 7.15

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Section 7.5 Systems of Inequalities 779

Graphing a Linear Inequality in Two Variables

1. Replace the inequality symbol with an equal sign and graph the correspondinglinear equation. Draw a solid line if the original inequality contains a or symbol. Draw a dashed line if the original inequality contains a or symbol.

2. Choose a test point from one of the half-planes. (Do not choose a point onthe 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 afalse statement results, shade the half-plane not containing this test point.

76

Ú…


Graph:

SolutionStep 1 Replace the inequality symbol the linear equation. Weneed to graph We can use intercepts to graph this line.2x - 3y = 6.

by � and graph

2x - 3y Ú 6.

EXAMPLE 1

We set to findthe x-intercept.

y � 0 We set to findthe y-intercept.

x � 0

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

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

2x = 6 -3y = 6

x = 3 y = -2−1

12345

−2−3−4−5

1 2 3 4 7 865−1−2

y

x

2x − 3y = 6

Figure 7.16 Preparing to graph2x - 3y Ú 6

The is 3, so the line passes through (3, 0). The is so theline passes through Using the intercepts, the line is shown in Figure 7.16 asa solid line. This is because the inequality contains a symbol, inwhich equality is included.

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 divides the planeinto three parts—the line itself and two half-planes. The points in one half-planesatisfy The points in the other half-plane satisfy We needto find which half-plane belongs to the solution of To do so, we test apoint from either half-plane.The origin, (0, 0), is the easiest point to test.

This is the given inequality.

Test (0, 0) by substituting 0 for and 0 for

Multiply.

This statement is false.

Step 3 If a false statement results, shade the half-plane not containing the testpoint. Because 0 is not greater than or equal to 6, the test point, (0, 0), is not part ofthe solution set. Thus, the half-plane below the solid line is part of thesolution set. The solution set is the line and the half-plane that does not contain thepoint (0, 0), indicated by shading this half-plane. The graph is shown using greenshading and a blue line in Figure 7.17.

Check Point 1 Graph: 4x - 2y Ú 8.

2x - 3y = 6

0 Ú 6

0 - 0 Ú

?6

y.x 2 # 0 - 3 # 0 Ú

?6

2x - 3y Ú 6

2x - 3y Ú 6.2x - 3y 6 6.2x - 3y 7 6.

2x - 3y = 6

Ú2x - 3y Ú 610, -22.

-2,y-interceptx-intercept

−1

12345

−2−3−4−5

1 2 3 4 7 865−1−2

y

x

Figure 7.17 The graph of2x - 3y Ú 6

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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 convenientbecause 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.


Graph:

SolutionStep 1 Replace the inequality symbol the linear equation. Becausewe are interested in graphing we begin by graphing We can usethe slope and the to graph this linear function.

The is 0, so the line passes through (0, 0). Using the and theslope, the line is shown in Figure 7.18 as a dashed line. This is because the inequality

contains a 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 pointbecause it lies on the line and not in a half-plane. Let’s use (1, 1), which lies in thehalf-plane above the line.



This statement is true.

Step 3 If a true statement results, shade the half-plane containing the testpoint. Because 1 is greater than the test point (1, 1) is part of the solution set.All the points on the same side of the line as the point (1, 1) are membersof 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 adashed blue line in Figure 7.18.

y = - 23 x

- 23 ,

1 7 - 23

y.x 1 7

?-

23

# 1

y 7 - 23

x

7y 7 - 23 x

y-intercepty-intercept

y-intercept = 0Slope = =

y= x+0–23

−23

riserun

y-intercepty = -

23 x.y 7 -

23 x,

by � and graph

y 7 - 23

x.

EXAMPLE 2

−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 7.18 The graph ofy 7 -

23 x

TechnologyMost graphing utilities can graph inequalities in twovariables with the feature.The procedure varies by model, so consult your manual. For most graphing utili-ties, you must first solve for if it is not already isolated.The figure shows the graph of Most displays donot distinguish between dashed and solid boundary lines.

y 7 - 23 x.

y

� SHADE �

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Check Point 2 Graph:

Graphing Linear Inequalities without Using Test PointsYou can graph inequalities in the form or without usingtest points. The inequality symbol indicates which half-plane to shade.

• If shade the half-plane above the line

• If shade the half-plane below the line

Observe how this is illustrated in Figure 7.18 in the margin on the previous page.Thegraph of is the half-plane above the line

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

y = - 23 x.y 7 -

23 x

y = mx + b.y 6 mx + b,

y = mx + b.y 7 mx + b,

y 6 mx + by 7 mx + b

y 7 - 34

x.

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

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

x 7 a, • If shade the half-plane abovey = b.

y 7 b,

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

x 6 a, • If shade the half-plane belowy = b.

y 6 b,

Graphing Inequalities without Using Test Points

Graph each inequality in a rectangular coordinate system:

a. b.

Solution

a.

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.

y … -3

x 7 2.y … -3

EXAMPLE 3

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

Study TipContinue using test points to graphinequalities in the form or The graph of

can lie above or belowthe line given by depending on the value of The samecomment applies to the graph ofAx + By 6 C.

B.Ax + By = C,

Ax + By 7 CAx + By 6 C.

Ax + By 7 C

b.

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.

x 7 2

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Graphing a Nonlinear Inequality in Two Variables

Graph:

SolutionStep 1 Replace the inequality symbol graph the nonlinear equation. Weneed to graph The graph is a circle of radius 3 with its center at the origin.The graph is shown in Figure 7.19 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. Substituteits coordinates into the inequality. The circle divides the plane into three parts—thecircle itself, the region inside the circle, and the region outside the circle. We need todetermine whether the region inside or outside the circle is included in the solution.Todo so, we will use the test point (0, 0) from inside the circle.



Square 0:

Add. This statement is true.

Step 3 If a true statement results, shade the region containing the test point. Thetrue statement tells us that all the points inside the circle satisfy Thegraph is shown using green shading and a solid blue circle in Figure 7.20.

x2+ y2

… 9.

0 … 9

02= 0. 0 + 0 …

?9

y.x 02+ 02

…

?9

x2+ y2

… 9

…

x2+ y2

= 9.with � and

x2+ y2

… 9.

EXAMPLE 4

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

a. b.

Graphing a Nonlinear Inequality in Two VariablesExample 4 illustrates that a nonlinear inequality in two variables is graphed in thesame way that we graph a linear inequality.

x … -2.y 7 1

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Figure 7.19 Preparing to graphx2

+ y2… 9

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Figure 7.20 The graph ofx2

+ y2… 9

� Graph a nonlinear inequality intwo variables.

Check Point 4 Graph: x2+ y2

Ú 16.

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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 7.21Source: U.S. Department of Health and Human Services

� Use mathematical modelsinvolving linear inequalities.

Modeling with Systems of Linear InequalitiesJust as two or more linear equations make up a system of linear equations, two ormore linear inequalities make up a system of linear inequalities. A solution of asystem of linear inequalities in two variables is an ordered pair that satisfies eachinequality in the system.

Does Your Weight Fit You?

The latest guidelines, which apply to both men and women, give healthy weightranges, rather than specific weights, for your height. Figure 7.21 shows the healthyweight region for various heights for people between the ages of 19 and 34,inclusive.

EXAMPLE 5

If represents height, in inches, and represents weight, in pounds, the healthyweight region in Figure 7.21 can be modeled by the following system of linearinequalities:

Show that point in Figure 7.21 is a solution of the system of inequalities thatdescribes healthy weight.

Solution Point has coordinates (70, 170).This means that if a person is 70 inchestall, or 5 feet 10 inches, and weighs 170 pounds, then that person’s weight is within thehealthy weight region. We can show that (70, 170) satisfies the system of inequalitiesby substituting 70 for and 170 for in each inequality in the system.yx

A

A

b4.9x - y Ú 1653.7x - y … 125.

yx

true 173 Ú 165, 343 - 170 Ú

?165

4.91702 - 170 Ú

?165

4.9x - y Ú 165

true 89 … 125, 259 - 170 …

?125

3.71702 - 170 …

?125

3.7x - y … 125

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

Check Point 5 Show that point in Figure 7.21 is a solution of the system ofinequalities that describes healthy weight.

B

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� Graph a system of inequalities. Graphing Systems of Linear InequalitiesThe solution set of a system of linear inequalities in two variables is the set of allordered pairs that satisfy each inequality in the system. Thus, to graph a system ofinequalities in two variables, begin by graphing each individual inequality in thesame rectangular coordinate system. Then find the region, if there is one, that iscommon to every graph in the system. This region of intersection gives a picture ofthe system’s solution set.

Graphing a System of Linear Inequalities

Graph the solution set of the system:

Solution Replacing each inequality symbol with an equal sign indicates thatwe need to graph and We can use intercepts to graphthese lines.

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

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.

2x + 3y = 12.x - y = 1

b x - y 6 12x + 3y Ú 12.

EXAMPLE 6

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.

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Graphing a System of Inequalities


Solution We begin by graphing Because equality is included in we graph as a solid parabola. Because (0, 0) makes the inequality

true (we obtain ), we shade the interior portion of the parabolacontaining (0, 0), shown in yellow in Figure 7.22.

0 Ú -4y Ú x2- 4

y = x2- 4

Ú ,y Ú x2- 4.

by Ú x2- 4

x - y Ú 2.

EXAMPLE 7

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = x2 − 4

Figure 7.22 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 7.23 Adding the graph of x - y Ú 2

Now we graph in the same rectangular coordinate system. Firstwe graph the line using its 2, and its Because (0, 0) makes the inequality false (we obtain ), we shadethe half-plane below the line. This is shown in Figure 7.23 using green verticalshading.

The solution of the system is shown in Figure 7.24 by the intersection (theoverlap) of the solid yellow and green vertical shadings. The graph of the system’ssolution set consists of the region enclosed by the parabola and the line. To find thepoints of intersection of the parabola and the line, use the substitution method tosolve the nonlinear system

Take a moment to show that the solutions are and (2, 0), as shown inFigure 7.24.

Check Point 7 Graph the solution set of the system:

b y Ú x2- 4

x + y … 2.

1-1, -32

b y = x2- 4

x - y = 2.

0 Ú 2x - y Ú 2-2.y-intercept,x-intercept,x - y = 2

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 7.24 The graph of andx - y Ú 2

y Ú x2- 4


b x - 3y 6 62x + 3y Ú -6.

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In Figure 7.28 it may be difficult to tell where the graph of intersects thevertical line Using the substitution method, it can be determined that thisintersection point is (4, 2). Take a moment to verify that the four intersection pointsin Figure 7.28 are, clockwise from upper left, (4, 3), (4, 2), and These points are shown as open dots because none satisfies all three of the system’sinequalities.


cx + y 6 2

-2 … x 6 1y 7 -3.

1-2, -42.1-2, 32,

x = 4.x - y = 2

A system of inequalities has no solution if there are no points in the rectangularcoordinate system that simultaneously satisfy each inequality in the system. Forexample, the system

whose separate graphs are shown in Figure 7.25, has no overlapping region. Thus,the system has no solution. The solution set is the empty set.

Graphing a System of Inequalities


Solution We begin by graphing the first given inequality. The linehas an of 2 and a of The test point (0, 0)

makes the inequality true, and its graph is shown in Figure 7.26.Now, let’s consider the second given inequality, Replacing the

inequality symbols by we obtain and graphed as red vertical linesin Figure 7.27. The line of is not included. Because is between and 4, weshade the region between the vertical lines. We must intersect this region with theyellow region in Figure 7.26. The resulting region is shown in yellow and greenvertical shading in Figure 7.27.

Finally, let’s consider the third given inequality, Replacing the inequalitysymbol by we obtain which graphs as a horizontal line. Because of the lessthan symbol in the graph consists of the half-plane below the line Wemust intersect this half-plane with the region in Figure 7.27. The resulting region isshown in yellow and green vertical shading in Figure 7.28. This region represents thegraph of the solution set of the given system.

y = 3.y 6 3,y = 3,= ,

y 6 3.

-2xx = 4x = 4,x = -2= ,

-2 … x 6 4.x - y 6 2

-2.y-interceptx-interceptx - y = 2x - y 6 2,

cx - y 6 2

-2 … x 6 4y 6 3.

EXAMPLE 8

�,

b2x + 3y Ú 62x + 3y … 0,

−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 7.25 A system ofinequalities 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 7.26 The graph ofx - 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 7.27 The graph ofand -2 … x 6 4x - y 6 2

−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 7.28 The graph of and and y 6 3-2 … x 6 4

x - y 6 2

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Exercise Set 7.5

Practice ExercisesIn Exercises 1–26, graph each inequality.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17.

18.

19. 20.

21. 22.

23. 24.

25. 26.

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

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46. by Ú x2- 4

x - y Ú 2by Ú x2

- 1x - y Ú -1

bx + y 7 3x + y 7 -2

bx + y 7 4x + y 7 -1

bx + y 7 3x + y 6 -2

bx + y 7 4x + y 6 -1

b4x - 5y Ú -20x Ú -3

bx - y … 1x Ú 2

-2 6 y … 5-2 … x 6 5

bx … 3y … -1

bx … 2y Ú -1

bx + y … 4y Ú 2x - 4

bx + 2y … 4y Ú x - 3

by 6 -2x + 4y 6 x - 4

by 7 2x - 3y 6 -x + 6

b2x - y … 43x + 2y 7 -6

b2x - 5y … 103x - 2y 7 6

bx - y Ú 4x + y … 6

b3x + 6y … 62x + y … 8

y Ú log31x - 12y Ú log21x + 12

y … 3xy 7 2x

y Ú x2- 1y Ú x2

- 9

y 6 x2- 9y 6 x2

- 1

1x + 222 + 1y - 122 6 16

1x - 222 + 1y + 122 6 9

x2+ y2

7 36x2+ y2

7 25

x2+ y2

… 4x2+ y2

… 1

y 7 -3y 7 1

x … -3x … 1

y 7 3x + 2y 7 2x - 1

y …

14

xy …

13

x

2x - y 7 4x - 2y 7 10

3x - 6y … 12x + 2y … 8

47. 48.

49. 50.

51.

52.

53. 54.

55. 56.

57. 58.

59. 60.

61. 62.

Practice PlusIn Exercises 63–64, write each sentence as an inequality in twovariables. Then graph the inequality.

63. The is at least 4 more than the product of andthe

64. The is at least 2 more than the product of andthe

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

65. The sum of the and the is at most 4.Theadded to the product of 3 and the does

not exceed 6.

66. The sum of the and the is at most 3.Theadded to the product of 4 and the does

not exceed 6.

67. The sum of the and the is no more than2. The is no less than the difference between thesquare of the and 4.

68. The sum of the squares of the and the isno more than 25. The sum of twice the and the

is no less than 5.x-variabley-variable

y-variablex-variable

x-variabley-variable

y-variablex-variable

x-variabley-variabley-variablex-variable

x-variabley-variabley-variablex-variable

x-variable.-3y-variable

x-variable.-2y-variable

d2x + y … 6

x + y 7 21 … x … 2y 6 3

d3x + y … 62x - y … -1x 7 -2y 6 4

dx Ú 0y Ú 02x + y 6 42x - 3y … 6

dx Ú 0y Ú 02x + 5y 6 103x + 4y … 12

c3x + y … 6x 7 -2y … 4

cx - y … 2x 7 -2y … 3

bx2+ y2

… 16y 6 2xbx2

+ y26 16

y Ú 2x

bx2+ y2

6 4y - x2

Ú 0bx2

+ y2… 1

y - x27 0

b 1x + 122 + 1y - 122 6 161x + 122 + 1y - 122 Ú 4

b 1x - 122 + 1y + 122 6 251x - 122 + 1y + 122 Ú 16

bx2+ y2

7 1x2

+ y26 9

bx2+ y2

7 1x2

+ y26 16

bx2+ y2

… 4x + y 7 1

bx2+ y2

… 16x + y 7 2

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In Exercises 69–70, rewrite each inequality in the system withoutabsolute value bars.Then graph the rewritten system in rectangularcoordinates.

69. 70.

The graphs of solution sets of systems of inequalities involvefinding the intersection of the solution sets of two or moreinequalities. By contrast, in Exercises 71–72, you will be graphingthe union of the solution sets of two inequalities.

71. Graph the union of and

72. Graph the union of and

Without graphing, in Exercises 73–76, determine if each system hasno solution or infinitely many solutions.

73.

74.

75.

76.

Application ExercisesThe figure shows the healthy weight region for various heights forpeople ages 35 and older.

b 1x - 422 + 1y + 322 … 241x - 422 + 1y + 322 Ú 24

b 1x + 422 + 1y - 322 … 91x + 422 + 1y - 322 Ú 9

b6x - y … 246x - y 7 24

b3x + y 6 93x + y 7 9

5x - 2y … 10.x - y Ú -1

y 6 4.y 732 x - 2

b ƒ x ƒ … 1ƒ y ƒ … 2

b ƒ x ƒ … 2ƒ y ƒ … 3

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

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

79. Is a person in this age group who is 6 feet tall weighing205 pounds within the healthy weight region?

80. Is a person in this age group who is 5 feet 8 inches tallweighing 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 children and

adults will cause the elevator to be overloaded.

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

c. Select an ordered pair satisfying the inequality. Whatare its coordinates and what do they represent in thissituation?

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

a. Write an inequality that describes the patient’s dietaryrestrictions for eggs and ounces of meat.

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

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

83. On your next vacation, you will divide lodging between largeresorts and small inns. Let represent the number of nightsspent in large resorts. Let represent the number of nightsspent in small inns.

a. Write a system of inequalities that models the followingconditions:

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

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

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

84. A person with no more than $15,000 to invest plans toplace the money in two investments. One investment ishigh risk, high yield; the other is low risk, low yield. At least$2000 is to be placed in the high-risk investment. Further-more, the amount invested at low risk should be at leastthree times the amount invested at high risk. Find andgraph a system of inequalities that describes allpossibilities for placing the money in the high- and low-riskinvestments.

yx

yx

yx

yx

yx

B

A

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 represents height, in inches, and represents weight, in pounds,the healthy weight region can be modeled by the following systemof linear inequalities:

Use this information to solve Exercises 77–80.

b5.3x - y Ú 1804.1x - y … 140.

yx

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The graph of an inequality in two variables is a region in the rectangular coordinate system. Regions in coordinate systems have numerousapplications. 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 showsthat person’s body-mass index (BMI), computed using the follow-ing formula:

The variable represents weight, in pounds. The variable represents height, in inches. Use this information to solveExercises 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 informationas a point in the coordinate system for males. Is thisperson obese, overweight, borderline overweight, normalweight, or underweight?

86. A woman is 25 years old, 66 inches (5 feet, 6 inches) tall, andweighs 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 infor-mation as a point in the coordinate system for females. Isthis person obese, overweight, borderline overweight,normal weight, or underweight?

Writing in Mathematics87. 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 aninequality in two variables, and

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 and 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?

3x - 2y … 6.3x - 2y 7 6

y?x

HW

BMI =

703W

H2 .

95. Explain how to graph the solution set of a system ofinequalities.

96. What does it mean if a system of linear inequalities has nosolution?

Technology ExercisesGraphing utilities can be used to shade regions in the rectangularcoordinate system, thereby graphing an inequality in two variables.Read the section of the user’s manual for your graphing utility thatdescribes how to shade a region. Then use your graphing utility tograph the inequalities in Exercises 97–102.

97. 98.

99. 100.

101. 102.

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

104. Use a graphing utility with a feature to verifyany five of the graphs that you drew by hand in Exercises 1–26.

105. Use a graphing utility with a feature to verifyany five of the graphs that you drew by hand for the systemsin Exercises 27–62.

Critical Thinking ExercisesMake Sense? In Exercises 106–109, determine whethereach statement makes sense or does not make sense, and explainyour reasoning.

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

107. When graphing it’s not necessary for me tograph the linear equation because theinequality contains a symbol, in which equality is notincluded.

6

3x - 4y = 123x - 4y 6 12,

� SHADE �

� SHADE �

3x - 2y Ú 62x + y … 6

y Ú

12

x2- 2y Ú x2

- 4

y Ú

23

x - 2y … 4x + 4

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Objectives

� Write an objective functiondescribing a quantity thatmust be maximized orminimized.

� Use inequalities to describelimitations in a situation.

� Use linear programming tosolve problems.

Linear ProgrammingSec t i on 7.6


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

109. I graphed the solution set of and withoutusing test points.

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

110. 111.

x

y

2

4

−2

−4

2 4−2−4x

y

2

4

−2

−4

2 4−3−5

x Ú 1y Ú x + 2

Preview ExercisesExercises 116–118 will help you prepare for the material coveredin the next section.

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

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

c. Evaluate at each of the points obtained inpart (b).

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

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

c. Evaluate at each of the points obtained inpart (b).

118. Bottled water and medical supplies are to be shipped tosurvivors of an earthquake by plane.The bottled water weighs20 pounds per container and medical kits weigh 10 poundsper kit. Each plane can carry no more than 80,000 pounds. If represents the number of bottles of water to be shipped perplane and represents the number of medical kits per plane,write an inequality that models each plane’s 80,000-poundweight restriction.

y

x

2x + 5y

dx Ú 0y Ú 0

3x - 2x … 6y … -x + 7.

3x + 2y

dx + y Ú 6

x … 8

y Ú 5.

112. 113.

78

x

y

1 2 6 7 83 4 5

123

6

45

−1

123

54

−2−3

1 2 3 4−1−2−3−4

y

x

y = x2

West Berlin children at Tempelhofairport watch fleets of U.S. airplanesbringing in supplies to circumvent theSoviet blockade. The airlift beganJune 28, 1948 and continued for 15months.

The Berlin Airlift (1948 – 1949) was an operation by the United States and GreatBritain in response to military action by the former Soviet Union: Soviet troops

closed all roads and rail lines between West Germany and Berlin, cutting off supplyroutes to the city. The Allies used a mathematical technique developed during World

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

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

by Ú nx + b 1n 6 0, b 7 02y … mx + b 1m 7 0, b 7 02.

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