3 6 practice compound inequalities form k

C h

a p

t e

r

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’ ’

8.1

8.2

8.3

8.4

Compound Inequalities in One Variable

Absolute Value Equations and Inequalities

Compound Inequalities in Two Variables

Linear Programming

8 More on Inequalities The practice of awarding degrees originated in the universities of medieval

Europe. The first known degree, a degree in civil law, was awarded in Italy at the

University of Bologna in the twelfth century. The University of Paris awarded its

first bachelor’s degree in the thirteenth century. By the time the first colleges were

opened in the American colonies, the process of awarding degrees was firmly

established. At first, American schools offered only a few types of degrees. The

colleges established in the colonies were primarily to train young men for the

ministry. Notable were Harvard (1636; Puritan), William and Mary (1693; Anglican),

Yale (1701; Congregationalist), Princeton (1746; New Lights Presbyterian), Brown

(1765; Baptist), and Rutgers (1766; Dutch Reformed).

The industrial revolution sparked a demand for training in many areas.Today,

approximately 1500 types of degrees are granted by academic institutions in the

United States. Over 1 million Bachelor of Arts (B.A.) or of Science (B.S.) degrees are

granted annually. Over one-quarter of a million Master of Arts (M.A.) or Science

(M.S.) degrees are awarded annually.

The growth of bachelor s and master s degrees is modeled with

linear equations in Exercise 87 of Section 8.1. In that exercise

we also use inequalities and compound inequalities to discuss

the growth of these degrees.

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508 Chapter 8 More on Inequalities 8-2

8.1 Compound Inequalities in One Variable

The inequality a < x < b, from Section 2.8, is one type of compound inequality in one variable. This inequality indicates that x is both greater than a and less than b. That is, x is between a and b. In this section we will study other types of compound inequalities in one variable.

In This Section

U1V Compound Inequalities

U2V Graphing the Solution Set

U3V Applications

U1V Compound Inequalities Inequalities involving a single inequality symbol are simple inequalities. If we join two simple inequalities with the connective “and” or the connective “or,” we get a compound inequality. A compound inequality using the connective “and” is true if and only if both simple inequalities are true.

Now do Exercises 1–3

E X A M P L E 1 Compound inequalities using the connective “and” Determine whether each compound inequality is true.

a) 3 > 2 and 3 < 5 b) 6 > 2 and 6 < 5

Solution a) The compound inequality is true because 3 > 2 is true and 3 < 5 is true.

b) The compound inequality is false because 6 < 5 is false.

A compound inequality using the connective “or” is true if one or the other or both of the simple inequalities are true. It is false only if both simple inequalities are false.

E X A M P L E 2 Compound inequalities using the connective “or” Determine whether each compound inequality is true.

a) 2 < 3 or 2 > 7 b) 4 < 3 or 4 : 7

Solution a) The compound inequality is true because 2 < 3 is true.

b) The compound inequality is false because both 4 < 3 and 4 : 7 are false. U Helpful Hint V

There is a big difference between “and” and “or.” To get money from an automatic teller you must have a bank card and know a secret num­ ber (PIN). There would be a lot of problems if you could get money by having a bank card or knowing a PIN.

Now do Exercises 4–6

If a compound inequality involves a variable, then we are interested in the solu­ tion set to the inequality. The solution set to an “and” inequality consists of all num­ bers that satisfy both simple inequalities, whereas the solution set to an “or” inequality consists of all numbers that satisfy at least one of the simple inequalities.

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8-3 8.1 Compound Inequalities in One Variable 509

E X A M P L E 3 Solutions of compound inequalities Determine whether 5 satisfies each compound inequality.

a) x < 6 and x < 9 b) 2x - 9 - 5 or -4x : -12

Solution a) Because 5 < 6 and 5 < 9 are both true, 5 satisfies the compound inequality.

b) Because 2 · 5 - 9 - 5 is true, it does not matter that -4 · 5 : -12 is false. So 5 satisfies the compound inequality.

A

A n B

Figure 8.1

Now do Exercises 7–14

B U2V Graphing the Solution Set If A and B are sets of numbers, then the intersection of A and B is the set of all num­ bers that are in both A and B. The intersection of A and B is denoted as A n B (read “A intersect B”). This set is illustrated with a Venn diagram in Fig. 8.1. If A = {1, 2, 3} and B = {2, 3, 4, 5}, then A n B = {2, 3} because only 2 and 3 are in both A and B. See Appendix B for more details about sets.

The solution set to a compound inequality using the connective “and” is the inter­ section of the solution sets to each of the simple inequalities. Using graphs, as shown in Example 4, will help you understand compound inequalities.

E X A M P L E 4 Graphing compound inequalities Graph the solution set to the compound inequality x > 2 and x < 5.

Solution First sketch the graph of x > 2 and then the graph of x < 5, as shown in Fig. 8.2. The inter­ section of these two solution sets is the portion of the number line that is shaded on both graphs, just the part between 2 and 5, not including the endpoints. In symbols, (2, o) n (-o, 5) = (2, 5). So the solution set is the interval (2, 5), and its graph is shown in Fig. 8.3. Recall from Section 2.8 that x > 2 and x < 5 is also written as 2 < x < 5.

Now do Exercises 15–18

Figure 8.3

-1 0 1 2 8 93 75 64

(2, 5)

Figure 8.2

-1-2-3 0 1 2 8 93 75 64

(-o, 5) (2, o)

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510 Chapter 8 More on Inequalities 8-4

A B

A U B

Figure 8.4

E X A M P L E 5

E X A M P L E 6

If A and B are sets of numbers, then the union of A and B is the set of all num­ bers that are in either A or B. The union of A and B is denoted as A U B (read “A union B”). This set is illustrated with a Venn diagram in Fig. 8.4. If A = {1, 2, 3} and B = {2, 3, 4, 5}, then A U B = {1, 2, 3, 4, 5} because all of these numbers are in either A or B. Notice that the numbers in A and B are in A U B.

The solution set to a compound inequality using the connective “or” is the union of the solution sets to each of the simple inequalities.

Graphing compound inequalities Graph the solution set to the compound inequality x > 4 or x < -1.

Solution First graph the solution sets to the simple inequalities as shown in Fig. 8.5. The union of these two intervals is shown in Fig. 8.6. Since the union does not simplify to a single inter­ val, the solution set is written using the symbol for union as (-o, -1) U (4, o).

(4, o)(-o, -1)

0 1 2 3 54 6 7-1-2-3-4

Figure 8.5

(-o, -1) U (4, o)

0 1 2 3 54 6 7-1-2-3-4

Figure 8.6

Now do Exercises 19–20

CAUTION When graphing the intersection of two simple inequalities, do not draw too much. For the intersection, graph only numbers that satisfy both inequalities. Omit numbers that satisfy one but not the other inequality. Graphing a union is usually easier because we can simply draw both solution sets on the same number line.

It is not always necessary to graph the solution set to each simple inequality before graphing the solution set to the compound inequality. We can save time and work if we learn to think of the two preliminary graphs but draw only the final one.

Overlapping intervals Sketch the graph and write the solution set in interval notation to each compound inequality.

a) x < 3 and x < 5

b) x > 4 or x > 0

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8-5 8.1 Compound Inequalities in One Variable 511

E X A M P L E 7

Solution a) Figure 8.7 shows x < 3 and x < 5 on the same number line. The intersection of these

two intervals consists of the numbers that are less than 3. Numbers between 3 and 5 are not shaded twice and do not satisfy both inequalities. In symbols, (-o, 3) n (-o, 5) = (-o, 3). So x < 3 and x < 5 is equivalent to x < 3. The solution set is (-o, 3) and its graph is shown in Fig. 8.8.

0 1 2 3 84 9 11 5 6 7 10

(-o, 5)(-o, 3)

Figure 8.7

0 1 2 3 84 9 11 5 6 7 10

Figure 8.8

b) Figure 8.9 shows the graph of x > 4 and the graph of x > 0 on the same number line. The union of these two intervals consists of everything that is shaded in Fig. 8.9. In symbols, (4, o) U (0, o) = (0, o). So x > 4 or x > 0 is equivalent to x > 0. The solution set to the compound inequality is (0, o), and its graph is shown in Fig. 8.10.

0 1 2 3 5-4 -3 -2 -1 4 6 7

(0, o) (4, o)

Figure 8.9

0 1 2 3 5-4 -3 -2 -1 4 6 7

Figure 8.10

Now do Exercises 21–22

Example 7 shows a compound inequality that has no solution and one that is satisfied by every real number.

All or nothing Sketch the graph and write the solution set in interval notation to each compound inequality.

a) x < 2 and x > 6 b) x < 3 or x > 1

Solution a) A number satisfies x < 2 and x > 6 if it is both less than 2 and greater than 6.

There are no such numbers. The solution set is the empty set, �. In symbols, (-o, 2) n (6, o) = �.

b) To graph x < 3 or x > 1, we shade both regions on the same number line as shown in Fig. 8.11. Since the two regions cover the entire line, the solution set is the set of all real numbers (-o, o). In symbols, (-o, 3) U (1, o) = (-o, o).

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

Figure 8.11

Now do Exercises 23–28

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512 Chapter 8 More on Inequalities 8-6

If we start with a more complicated compound inequality, we first simplify each part of the compound inequality and then find the union or intersection.

E X A M P L E 8 Intersection Solve x + 2 > 3 and x - 6 < 7. Graph the solution set.

Solution First simplify each simple inequality:

x + 2 - 2 > 3 - 2 and x - 6 + 6 < 7 + 6

x > 1 and x < 13

The intersection of these two solution sets is the set of numbers between (but not including) 1 and 13. Its graph is shown in Fig. 8.12. The solution set is written in interval notation as (1, 13). Recall from Section 2.8 that x > 1 and x < 13 is also written as 1 < x < 13.

Now do Exercises 29–32

Figure 8.12

0 1 2 3 98 11 12 54 6 7 13 14 10

U Calculator Close-Up V

To check Example 8, press Y= and let y1 = x + 2 and y2 = x - 6. Now scroll through a table of values for y1 and y2. From the table you can see that y1 is greater than 3 and y2 is less than 7 precisely when x is between 1 and 13.

E X A M P L E 9 Union Graph the solution set to the inequality

5 - 7x : 12 or 3x - 2 < 7.

Solution First solve each of the simple inequalities:

5 - 7x - 5 : 12 - 5 or 3x - 2 + 2 < 7 + 2

-7x : 7 or 3x < 9

x - -1 or x < 3

The union of the two solution intervals is (-o, 3). The graph is shown in Fig. 8.13.

Figure 8.13

0 1 2 3 5-6 -5 -4 -3 -2 -1 4

U Calculator Close-Up V

To check Example 9, press Y= and let y1 = 5 - 7x and y2 = 3x - 2. Now scroll through a table of values for y1 and y2. From the table you can see that either y1 : 12 or y2 < 7 is true for x < 3. Note also that for x : 3 both y1 : 12 and y2 < 7 are incorrect. The table supports the conclusion of Example 9.

Now do Exercises 33–40

If x is between a and b and a < b, then we can use the “between” notation, a < x < b, rather than writing x > a and x < b. We solved compound inequalities of this type in Section 2.9. For completeness, we review that method in Examples 10 and 11.

E X A M P L E 10 Using “between” notation Solve the inequality and graph the solution set:

-2 - 2x - 3 < 7

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8-7 8.1 Compound Inequalities in One Variable 513

U Calculator Close-Up V

Do not use a table on your calculator as a method of solving an inequality. Use a table to check your algebraic solution and you will get a better understanding of inequalities.

E X A M P L E 11

U Calculator Close-Up V

Let y1 = 3 - 2x and make a table. Scroll through the table to see that y1 is between -1 and 9 when x is between -3 and 2. The table sup­ ports the conclusion of Example 11.

Solution This inequality could be written as the compound inequality

2x - 3 : -2 and 2x - 3 < 7.

However, there is no need to rewrite the inequality because we can solve it in its original form.

-2 + 3 - 2x - 3 + 3 < 7 + 3 Add 3 to each part.

1 - 2x < 10

1 2x 10 -- - -- < - Divide each part by 2.- 2 2 2 1 -- - x < 5 2

The solution set is [ - , 5)1 - , and its graph is shown in Fig. 8.14.2 1 — 2

-1 0 1 2 3 4 5 6 7

Figure 8.14

Now do Exercises 41–44

Solving a compound inequality Solve the inequality -1 < 3 - 2x < 9 and graph the solution set.

Solution -1 - 3 < 3 - 2x - 3 < 9 - 3 Subtract 3 from each part of the inequality.

-4 < -2x < 6

2 > x > -3 Divide each part by -2 and reverse both inequality symbols.

-3 < x < 2 Rewrite the inequality with the smallest number on the left.

The solution set is (-3, 2), and its graph is shown in Fig. 8.15.

-4 -3 -2 -1 0 1 2 3

Figure 8.15

Now do Exercises 45–52

U3V Applications When final exams are approaching, students are often interested in finding the final exam score that would give them a certain grade for a course.

E X A M P L E 12 Final exam scores Fiana made a score of 76 on her midterm exam. For her to get a B in the course, the average of her midterm exam and final exam must be between 80 and 89 inclusive. What possible scores on the final exam would give Fiana a B in the course?

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514 Chapter 8 More on Inequalities 8-8

Warm-Ups ▼

Fill in the blank. 1. A inequality consists of two simple inequali­

ties connected by “and” or “or.” 2. The compound inequality x > 5 x > 8 is true only

if both simple inequalities are true. 3. The compound inequality x > 5 x > 8 is true if

either simple inequality is true.

4. If a < b < c, then a < b b < c.

5. The solution set to x > 3 and x < 9 is the of the intervals (3, o) and (-o, 9).

6. The solution set to x > 3 or x < 9 is the of the intervals (3, o) and (-o, 9).

True or false? 7. 3 < 5 and 3 - 10

8. 3 < 5 or 3 < 10

9. 3 > 5 and 3 < 10

10. 3 : 5 or 3 - 10

11. 4 < 8 and 4 > 2

12. 4 < 8 or 4 > 2

13. -3 < 0 < -2

14. (3, o) U [8, o) = [8, o)

15. (3, o) n [8, o) = [8, o)

16. (-2, o) n (-o, 9) = (-2, 9)

8 .1

Solution Let x represent her final exam score. Between 80 and 89 inclusive means that an average between 80 and 89 as well as an average of exactly 80 or 89 will get a B. So the average of the two scores must be greater than or equal to 80 and less than or equal to 89.

80 - 7 x +

2 76

7 - 89

160 - x + 76 - 178 Multiply by 2.

160 - 76 - x - 178 - 76 Subtract 76.

84 - x - 102 If Fiana scores between 84 and 102 inclusive, she will get a B in the course.

Now do Exercises 77–78

U Helpful Hint V

When you use two inequality sym­ bols as in Example 12, they must both point in the same direction. In fact, we usually have them both point to the left so that the numbers increase in size from left to right.

Exercises

U Study Tips V • When studying for a midterm or final, review the material in the order it was originally presented. This strategy will help you to see

connections between the ideas. • Studying the oldest material first will give top priority to material that you might have forgotten.

U1V Compound Inequalities 3. 1 < 5 and 1 > -3

Determine whether each compound inequality is true. See 4. 3 < 5 or 0 < -3 Examples 1 and 2.

5. 6 < 5 or -4 > -31. -6 < 5 and -6 > -3

2. 4 - 4 and -4 - 0 6. 4 - -4 or 0 - 0

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8-9

Determine whether -4 satisfies each compound inequality. See Example 3.

7. x < 5 and x > -3

8. x > -5 and x < 0

9. x < 5 or x > -3

10. x < -9 or x > 0

11. x - 3 : -7 or x + 1 > 1

12. 2x - -8 and 5x - 0

13. 2x - 1 < -7 or -2x > 18

14. -3x > 0 and 3x - 4 < 11

U2V Graphing the Solution Set Graph the solution set to each compound inequality. See Examples 4–7.

15. x > -1 and x < 4

16. x - 5 and x : 4

17. x - 3 and x - 0

18. x > 2 and x > 0

19. x : 2 or x : 5

20. x < -1 or x < 3

21. x - 6 or x > -2

22. x > -2 and x - 4

23. x - 6 and x > 9

24. x < 7 or x > 0

25. x - 6 or x > 9

26. x : 4 and x - -4

27. x : 6 and x - 1

28. x > 3 or x < -3

8.1 Compound Inequalities in One Variable 515

Solve each compound inequality. Write the solution set using interval notation and graph it. See Examples 8 and 9.

29. x - 3 > 7 or 3 - x > 2

30. x - 5 > 6 or 2 - x > 4

31. 3 < x and 1 + x > 10

32. -0.3x < 9 and 0.2x > 2

1 1 33. -- x > 5 or --- x < 2

2 3

1 34. 5 < x or 3 - -- x < 7

2

35. 2x - 3 - 5 and x - 1 > 0

3 1 36. -- x < 9 and --- x - -15

4 3

1 1 1 2 1 37. -- x - -- : --- or -- x - --

2 3 6 7 10

1 1 1 1 38. -- x - -- > --- and -- x < 2

4 3 5 2

39. 0.5x < 2 and -0.6x < -3

40. 0.3x < 0.6 or 0.05x > -4

Solve each compound inequality. Write the solution set in interval notation and graph it. See Examples 10 and 11.

41. -3 < x + 1 < 3

42. -4 - x - 4 - 1

43. 5 < 2x - 3 < 11

44. -2 < 3x + 1 < 10

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8-10 516 Chapter 8 More on Inequalities

45. -1 < 5 - 3x - 14

46. -1 - 3 - 2x < 11

3m + 1 47. -3 < -- - 5

2

3 - 2x 48. 0 - -- < 5

2

1 - 3x 49. -2 < -- < 7

-2

2x - 1 50. -3 < -- < 7

3

51. 3 - 3 - 5(x - 3) - 8

1 52. 2 - 4 - --(x - 8) - 10

2

Write each union or intersection of intervals as a single interval if possible.

53. (2, o) U (4, o) 54. (-3, o) U (-6, o)

55. (-o, 5) n (-o, 9) 56. (-o, -2) n (-o, 1)

57. (-o, 4] n [2, o) 58. (-o, 8) n [3, o)

59. (-o, 5) U [-3, o) 60. (-o, -2] U (2, o)

61. (3, o) n (-o, 3] 62. [-4, o) n (-o, -6]

63. (3, 5) n [4, 8) 64. [-2, 4] n (0, 9]

65. [1, 4) U (2, 6] 66. [1, 3) U (0, 5)

Write either a simple or a compound inequality that has the given graph as its solution set.

67.

-3 -2 -1 0 1 2 3 4 5 6 7 8

68.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

69.

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5

70.

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5

71.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

72.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

73.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

74.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

75.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

76.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

U3V Applications

Solve each problem by using a compound inequality. See Example 12.

77. Aiming for a C. Professor Johnson gives only a midterm exam and a final exam. The semester average is computed

1 2by taking -- of the midterm exam score plus -- of the final 3 3

exam score. To get a C, Beth must have a semester average between 70 and 79 inclusive. If Beth scored only 64 on the midterm, then for what range of scores on the final exam would Beth get a C?

78. Two tests only. Professor Davis counts his midterm as 2 1 -- of the grade, and his final as -- of the grade. Jason 3 3 scored only 64 on the midterm. What range of scores on the final exam would put Jason’s average between 70 and 79 inclusive?

79. Car costs. A company uses the expression 0.0004x + 20 to estimate the cost in cents per mile for operating a company car and the expression 20,000 - 0.2x to estimate the value of the car in dollars, where x is the number of miles on the

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8-11 8.1 Compound Inequalities in One Variable 517

odometer. If the company plans to replace any car for which the operating cost is greater than 40 cents per mile and the value is less than $12,000, then for what values of x is a car replaced? Use interval notation.

80. Changing plans. The company in Exercise 79 has changed its policy and has decided to replace any car for which the operating cost is greater than 40 cents per mile or the value is less than $12,000. For what values of x is a car replaced? Use interval notation.

81. Supply and demand. An energy minister in a small country uses the expression 20 + 0.1x to estimate the amount of oil in millions of barrels per day that will be supplied to his country and the expression 30 - 0.5x to estimate the demand for oil in millions of barrels per day, where x is the price of oil in dollars per barrel. The govern­ ment must get involved if the supply is less than 22 million barrels per day or if the demand is less than 15 million barrels per day. For what values of x must the government get involved? Use interval notation.

82. Predicting recession. The country of Exercise 81 will be in recession if the supply of oil is greater than 23 million barrels per day and the demand is less than 14 million barrels per day. For what values of x will the country be in recession? Use interval notation.

D eg

re es

( in

th ou

sa nd

s) 1500

1000

500

Bachelors

Masters

83. Keep on truckin’. Abdul is shopping for a new truck in a city with an 8% sales tax. There is also an $84 title and license fee to pay. He wants to get a good truck and plans to spend at least $12,000 but no more than $15,000. What is the price range for the truck?

84. Selling-price range. Renee wants to sell her car through a broker who charges a commission of 10% of the selling price. The book value of the car is $14,900, but Renee still owes $13,104 on it. Although the car is in only fair condi­ tion and will not sell for more than the book value, Renee must get enough to at least pay off the loan. What is the range of the selling price?

85. Hazardous to her health. Trying to break her smoking habit, Jane calculates that she smokes only three full cigarettes a day, one after each meal. The rest of the time she smokes on the run and smokes only half of the cigarette. She estimates that she smokes the equivalent of 5 to 12 cigarettes per day. How many times a day does she light up on the run?

86. Possible width. The length of a rectangle is 20 meters longer than the width. The perimeter must be between 80 and 100 meters. What are the possible values for the width of the rectangle?

87. Higher education. The formulas

B = 16.45n + 1062.45 and M = 7.79n + 326.82

can be used to approximate the number of bachelor’s and master’s degrees in thousands, respectively, awarded per year, n years after 1990 (National Center for Educational Statistics, www.nces.ed.gov).

a) How many bachelor’s degrees were awarded in 2000?

b) In what year will the number of bachelor’s degrees that are awarded reach 1.4 million?

c) What is the first year in which both B is greater than 1.4 million and M is greater than 0.55 million?

d) What is the first year in which either B is greater than 1.4 million or M is greater than 0.55 million?

10 20 30 Years since 1990

Figure for Exercise 87

88. Senior citizens. The number of senior citizens (65 and over) in the United States in millions n years after 1990 can be estimated by using the formula

s = 0.38n + 31.2

(U.S. Bureau of the Census, www.census.gov). See the figure on the next page. The percentage of senior citizens living below the poverty level n years after 1990 can be estimated by using the formula

p = -0.25n + 12.2.

a) How many senior citizens were there in 2000?

b) In what year will the percentage of seniors living below the poverty level reach 7%?

c) What is the first year in which we can expect both the number of seniors to be greater than 40 million and fewer than 7% living below the poverty level?

http:www.census.gov
http:www.nces.ed.gov
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518 Chapter 8 More on Inequalities 8-12

Se ni

or s

(i n

m ill

io ns

) a) -2 - x < 3 b) -4 : x < 7 c) -1 - x > 0 d) 6 < x - -8 e) 5 : x : -9

60

40

91. Discussion20 In each case, write the resulting set of numbers in interval notation. Explain your answers.

a) Every number in (3, 8) is multiplied by 4. b) Every number in [-2, 4) is multiplied by -5.

10 20 30 Years since 1990

Figure for Exercise 88

Getting More Involved

89. Discussion

If -x is between a and b, then what can you say about x?

90. Discussion

For which of the inequalities is the notation used correctly?

c) Three is added to every number in (-3, 6). d) Every number in [3, 9] is divided by -3.

92. Discussion

Write the solution set using interval notation for each of these inequalities in terms of s and t. State any restric­ tions on s and t. For what values of s and t is the solution set empty?

a) x > s and x < t b) x > s and x > t

Math at Work Pediatric Dosing Rules

A drug is generally tested on adults, and an appropriate adult dose (AD) of the drug is deter­ mined. When a drug is given to a child, a doctor determines an appropriate child’s dose (CD) using pediatric dosing rules. However, no single rule works for all children. Determining a child’s dose also involves common sense and experience.

Clark’s rule is based on the ratio of the child’s body weight to the mean weight of an

adult, 150 pounds. By Clark’s rule, CD = · AD. A dose determined by body

weight alone might be too little to be effective in a small child. Young’s rule is based on the assumption that age approximates body weight for patients

over 2 years old. Of course there is a great variability of body weight of children of any given

age. By Young’s rule, CD = · AD.

The area rule is often used for drugs required in radioactive imaging. It is based on the idea that (body mass)213 is approximately the body surface area. For radioactive imaging the adult’s body mass (MA) often

determines AD. By the area rule, CD = -( ( M M

C A)

) 2

2

1

1

3

3

- · AD, where MC is the child’s body mass.

Webster’s rule uses age to approximate the ratio in the area rule and agrees well with the area rule until age 11 or 12. By Webster’s rule CD = -a

a g g e e

+

+

1 7

- · AD.

Fried’s rule is generally used for patients less than one year old. By Fried’s rule CD = -age in

15 m 0 onths

- · AD.

age in years -- age in years + 12

child’s weight in lbs ---

150 lbs

C hi

ld ’s

d os

e (%

o f

ad ul

t d os

e) 100

40

80

60

20

0 20 64 10 8 16 1412 Child’s age (years)

Young’s rule

CD a

a 12 100

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8-13 8.2 Absolute Value Equations and Inequalities 519

8.2 Absolute Value Equations and Inequalities

In Chapter 1 we learned that absolute value measures the distance of a number from 0 on the number line. In this section we will learn to solve equations and inequalities involving absolute value.

In This Section

U1V Absolute Value Equations

U2V Absolute Value Inequalities

U3V All or Nothing

U4V Applications

U Helpful Hint V

Some students grow up believing that the only way to solve an equa­ tion is to “do the same thing to each side.” Then along come absolute value equations. For an absolute value equation we write an equiva­ lent compound equation that is not obtained by “doing the same thing to each side.”

U1V Absolute Value Equations Solving equations involving absolute value requires some techniques that are different from those studied in previous sections. For example, the solution set to the equation

I x I = 5

is {-5, 5} because both 5 and -5 are five units from 0 on the number line, as shown in Fig. 8.16. So I x I = 5 is equivalent to the compound equation

x = 5 or x = -5.

5 units 5 units

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

Figure 8.16

The equation I x I = 0 is equivalent to the equation x = 0 because 0 is the only number whose distance from 0 is zero. The solution set to I x I = 0 is {0}.

The equation I x I = -7 is inconsistent because absolute value measures distance, and distance is never negative. So the solution set is empty. These ideas are summa­ rized as follows.

Summary of Basic Absolute Value Equations

Absolute Value Equation Equivalent Equation Solution Set

I x I = k (k > 0) x = k or x = -k {k, -k} I x I = 0 x = 0 {0} I x I = k (k < 0) �

We can use these ideas to solve more complicated absolute value equations.

E X A M P L E 1 Absolute value equal to a positive number Solve each equation.

a) I x - 7 I = 2 b) I 3x - 5 I = 7

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520 Chapter 8 More on Inequalities 8-14

Solution a) First rewrite I x - 7 I = 2 without absolute value:

x - 7 = 2 or x - 7 = -2 Equivalent equation

x = 9 or x = 5

The solution set is {5, 9}. The distance from 5 to 7 or from 9 to 7 is 2 units. b) First rewrite I 3x - 5 I = 7 without absolute value:

3x - 5 = 7 or 3x - 5 = -7 Equivalent equation

3x = 12 or 3x = -2

x = 4 or x = -- 2 3

-

The solution set is �--2 3 -, 4�.

U Calculator Close-Up V

Use Y= to set y1 = abs(x - 7). Make a table to see that y1 has value 2 when x = 5 or x = 9. The table supports the conclusion of Example 1(a).

Now do Exercises 1–6

E X A M P L E 2

U Helpful Hint V

Examples 1, 2, and 3 show the three basic types of absolute value equations—absolute value equal to a positive number, zero, or a negative number. These equations have 2, 1, and no solutions, respectively.

Absolute value equal to zero Solve I 2(x - 6) + 7 I = 0.

Solution Since 0 is the only number whose absolute value is 0, the expression within the absolute value bars must be 0.

2(x - 6) + 7 = 0 Equivalent equation

2x - 12 + 7 = 0

2x - 5 = 0

2x = 5

x = - 5 2

-

The solution set is �-5 2 -�. Now do Exercises 7–12

E X A M P L E 3 Absolute value equal to a negative number Solve each equation.

a) I x - 9 I = -6 b) -5 I 3x - 7 I + 4 = 14

Solution a) The equation indicates that I x - 9 I = -6. However, the absolute value of any

quantity is greater than or equal to zero. So there is no solution to the equation.

b) First subtract 4 from each side to isolate the absolute value expression:

-5 I 3x - 7 I + 4 = 14 Original equation -5 I 3x - 7 I = 10 Subtract 4 from each side.

I 3x - 7 I = -2 Divide each side by -5.

There is no solution because no quantity has a negative absolute value.

Now do Exercises 13–24

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8-15 8.2 Absolute Value Equations and Inequalities 521

The equation in Example 4 has an absolute value on both sides.

E X A M P L E 4 Absolute value on both sides Solve I 2x - 1 I = I x + 3 I.

Solution Two quantities have the same absolute value only if they are equal or opposites. So we can write an equivalent compound equation:

2x - 1 = x + 3 or 2x - 1 = -(x + 3)

x - 1 = 3 or 2x - 1 = -x - 3

x = 4 or 3x = -2

x = 4 or x = -- 2

3 -

Check 4 and --2 3 - in the original equation. The solution set is �-- 2 3

-, 4�. Now do Exercises 25–30

U2V Absolute Value Inequalities Since absolute value measures distance from 0 on the number line, I x I > 5 indicates that x is more than five units from 0. Any number on the number line to the right of 5 or to the left of -5 is more than five units from 0. So I x I > 5 is equivalent to

x > 5 or x < -5.

The solution set to this inequality is the union of the solution sets to the two simple inequalities. The solution set is (-o, -5) U (5, o). The graph of I x I > 5 is shown in Fig. 8.17.

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Figure 8.17

The inequality I x I - 3 indicates that x is less than or equal to three units from 0. Any number between -3 and 3 inclusive satisfies that condition. So I x I - 3 is equivalent to

-3 - x - 3.

The graph of I x I - 3 is shown in Fig. 8.18. These examples illustrate the basic types of absolute value inequalities.

-4 -3 -2 -1 0 1 2 3 4

Figure 8.18

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522 Chapter 8 More on Inequalities 8-16

Summary of Basic Absolute Value Inequalities (k > 0)

Absolute Value Equivalent Solution Graph of Inequality Inequality Set Solution Set

I x I > k x > k or x < -k (-o, -k) U (k, o)

I x I : k x : k or x - -k (-o, -k] U [k, o)

I x I < k -k < x < k (-k, k)

I x I - k -k - x - k [-k, k]

-k k

-k k

-k k

-k k

We can solve more complicated inequalities in the same manner as simple ones.

E X A M P L E 5 Absolute value inequality Solve I x - 9 I < 2 and graph the solution set.

Solution Because I x I < k is equivalent to -k < x < k, we can rewrite I x - 9 I < 2 as follows:

-2 < x - 9 < 2

-2 + 9 < x - 9 + 9 < 2 + 9 Add 9 to each part of the inequality.

7 < x < 11

The graph of the solution set (7, 11) is shown in Fig. 8.19. Note that the graph consists of all real numbers that are within two units of 9.

Figure 8.19

1311 12108 975 6

U Calculator Close-Up V

Use Y= to set y1 = abs(x - 9). Make a table to see that y1 < 2 when x is between 7 and 11.

Now do Exercises 31–32

E X A M P L E 6 Absolute value inequality Solve I 3x + 5 I > 2 and graph the solution set.

Solution 3x + 5 > 2 or 3x + 5 < -2 Equivalent compound inequality

3x > -3 or 3x < -7

x > -1 or x < -- 7 3

-

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8-17 8.2 Absolute Value Equations and Inequalities 523

The solution set is �-o, --7 3 -) U (-1, o), and its graph is shown in Fig. 8.20.

Figure 8.20

-5 31 20-4 -3 -2 -1

7 - —

3

E X A M P L E 7 Absolute value inequality Solve I 5 - 3x I - 6 and graph the solution set.

Solution -6 - 5 - 3x - 6 Equivalent inequality

-11 - -3x - 1 Subtract 5 from each part.

- 1 3 1 - : x : --

1 3

- Divide by -3 and reverse each inequality symbol.

-- 1 3

- - x - - 1 3 1 - Write --

1 3

- on the left because it is smaller than - 1 3 1 -.

The solution set is [--1 3 -, - 1 3 1 -� and its graph is shown in Fig. 8.21.

Figure 8.21

-2 4 53-1 1 20

1 - —

3 11 — 3

U Calculator Close-Up V

Use Y = to set y1 = abs(5 - 3x). The table supports the conclusion that y - 6 when x is between --1

3 - and -1

3 1 -

even though --1 3

- and -1 3 1 - do not appear

in the table. For more accuracy, make

a table in which the change in x is -1 3

-.

Now do Exercises 33–34

Now do Exercises 35–38

U3V All or Nothing The solution to an absolute value inequality can be all real numbers or no real num­ bers. To solve such inequalities you must remember that the absolute value of any real number is greater than or equal to zero.

E X A M P L E 8 All real numbers Solve 3 + I 7 - 2x I : 3.

Solution Subtract 3 from each side to isolate the absolute value expression.

I 7 - 2x I : 0

Because the absolute value of any real number is greater than or equal to 0, the solution set is R, the set of all real numbers.

Now do Exercises 59–64

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524 Chapter 8 More on Inequalities 8-18

E X A M P L E 9 No real numbers Solve I 5x - 12 I < -2.

Solution We write an equivalent inequality only when the value of k is positive. With -2 on the right-hand side, we do not write an equivalent inequality. Since the absolute value of any quantity is greater than or equal to 0, no value for x can make this absolute value less than -2. The solution set is �, the empty set.

Now do Exercises 65–68

U4V Applications A simple example will show how absolute value inequalities can be used in applications.

E X A M P L E 10 Controlling water temperature The water temperature in a certain manufacturing process must be kept at 143°F. The com­ puter is programmed to shut down the process if the water temperature is more than 7° away from what it is supposed to be. For what temperature readings is the process shut down?

Solution If we let x represent the water temperature, then x - 143 represents the difference between the actual temperature and the desired temperature. The quantity x - 143 could be posi­ tive or negative. The process is shut down if the absolute value of x - 143 is greater than 7.

1 x - 143 1 > 7 x - 143 > 7 or x - 143 < -7

x > 150 or x < 136

The process is shut down for temperatures greater than 150°F or less than 136°F.

Warm-Ups ▼

Fill in the blank. 1. The of x is the distance from x to 0 on

the number line.

2. The equation I x I = 4 has solutions. 3. The equation I x I = -4 has solutions. 4. The equation I x I = 0 has solution. 5. real numbers satisfy I x I : 0. 6. real numbers satisfy I x I < 0. 7. The solution set to I x I < 3 is . 8. The inequality I x I > 3 is to x > 3

or x < -3.

True or false? 9. If I x I = 2, then x = 2 or x = -2.

10. If I x - 1 I = 7, then x - 1 = 7 or x + 1 = 7. 11. If I x I > 5, then x > 5 or x < -5. 12. If I x I = -4, then x = -4. 13. If I 2x - 8 I = 0, then x = 4. 14. If -3 < x < 3, then I x I < 3. 15. If 5 < x < 9, then x > 5 and x < 9.

16. If x is any real number, then I x I : 0. 17. If I x I + 1 = 5, then x + 1 = 5 or x + 1 = -5. 18. If 1 3x - 99 1 - 0, then x = 33.

Now do Exercises 77–84

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Exercises 8. 2

U Study Tips V • The last couple of weeks of the semester is not the time to slack off. This is the time to double your efforts. • Make a schedule and plan every hour of your time.

U1V Absolute Value Equations U2V Absolute Value Inequalities

Solve each absolute value equation. See Examples 1–3 and the Write an absolute value inequality whose solution set is shown Summary of Basic Absolute Value Equations on page 519. by the graph. See Examples 5–7 and the Summary of Basic

Absolute Value Inequalities on page 522. 1. I a I = 5 2. I x I = 2 3. I x - 3 I = 1

4. I x - 5 I = 2 5. I 3 - x I = 6 6. I 7 - x I = 6 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

7. I 3x - 4 I = 12 8. I 5x + 2 I = -3 32. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

2 3 1 33.9. x - 8 = 0 10. 3 - x = 3 4 4 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 11. I 6 - 0.2x I = 10 12. I 5 - 0.1x I = 0 13. I 7(x - 6) I = -3

34.

-8-7-6-5-4-3-2-1 0 1 2 3 4 5 6 7 8 14. I 2(a + 3) I = 15

15. I 2(x - 4) + 3 I = 5 35.

16. I 3(x - 2) + 7 I = 6 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

17. I 7.3x - 5.26 I = 4.215 36.

18. I 5.74 - 2.17x I = 10.28 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

Solve each absolute value equation. See Examples 3 and 4.

19. 3 + I x I = 5 37.

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 620. I x I - 10 = -3 21. 2 - I x + 3 I = -6 22. 4 - 3 I x - 2 I = -8 38.

I 3 - 2x I -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

3 1 1

23. 5 - = 4

24. 3 - x - 4 = 2 2 2 Determine whether each absolute value inequality is equivalent

25. I x - 5 I = I 2x + 1 I to the inequality following it. See Examples 5–7. 26. I w - 6 I = I 3 - 2w I I I39. x < 3, x < 3

27. 5 - x x 40. I x I > 3, x > 3 = 2 - 2 2 41. I x - 3 I > 1, x - 3 > 1 or x - 3 < -1

1 1 3 x - I28. x - = 42. x - 3 I - 1, -1 - x - 3 - 14 2 4

29. I x - 3 I = I 3 - x I 43. I x - 3 I : 1, x - 3 : 1 or x - 3 - 1

30. I a - 6 I = I 6 - a I 44. I x - 3 I > 0, x - 3 > 0

31.

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8-20 526 Chapter 8 More on Inequalities

Solve each absolute value inequality and graph the solution set. See Examples 5–7.

45. I x I > 6

46. I w I : 3

47. I t I - 2

48. I b I < 4

49. I 2a I < 6

50. I 3x I < 21

51. I x - 2 I : 3

52. I x - 5 I : 1

1 53. -- I 2x - 4 I < 1

5

1 54. -- I 2x - 1 I < 1

3

55. -2 I 5 - x I : -14

56. -3 I 6 - x I : -3

57. 2 I 3 - 2x I - 6 : 18

58. 2 I 5 - 2x I - 15 : 5

U3V All or Nothing

Solve each absolute value inequality and graph the solution set. See Examples 8 and 9.

59. I x I > 0

60. I x - 2 I > 0

61. I x I - 0

62. I x I < 0 63. I x - 5 I : 0

64. I 3x - 7 I : -3

65. -2 I 3x - 7 I > 6 66. -3 I 7x - 42 I > 18 67. I 2x + 3 I + 6 > 0

68. I 5 - x I + 5 > 5

Solve each inequality. Write the solution set using interval notation.

69. 1 < I x + 2 I 70. 5 : I x - 4 I 71. 5 > I x I + 1 72. 4 - I x I - 6 73. 3 - 5 I x I > -2 74. 1 - 2 I x I < -7 75. I 5.67x - 3.124 I < 1.68

76. I 4.67 - 3.2x I : 1.43

U4V Applications

Solve each problem by using an absolute value equation or inequality. See Example 10.

77. Famous battles. In the Hundred Years’War, Henry V defeated a French army in the battle of Agincourt and Joan of Arc defeated an English army in the battle of Orleans (The Doubleday Almanac). Suppose you know only that these two famous battles were 14 years apart and that the battle of Agincourt occurred in 1415. Use an absolute value equation to find the possibilities for the year in which the battle of Orleans occurred.

78. World records. In July 1985 Steve Cram of Great Britain set a world record of 3 minutes 29.67 seconds for the 1500-meter race and a world record of 3 minutes 46.31 seconds for the 1-mile race (The Doubleday Almanac). Suppose you know only that these two events occurred 11 days apart and that the 1500-meter record was set on July 16. Use an absolute value equation to find the possible dates for the 1-mile record run.

79. Weight difference. Research at a major university has shown that identical twins generally differ by less than 6 pounds in body weight. If Kim weighs 127 pounds, then

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8-21 8.2 Absolute Value Equations and Inequalities 527

in what range is the weight of her identical twin sister Kathy?

80. Intelligence quotient. Jude’s IQ score is more than 15 points away from Sherry’s. If Sherry scored 110, then in what range is Jude’s score?

81. Approval rating. According to a Fox News survey, the presidential approval rating is 39% plus or minus 5 percentage points.

a) In what range is the percentage of people who approve of the president?

b) Let x represent the actual percentage of people who approve of the president. Write an absolute value inequality for x.

82. Time of death. According to the coroner the time of death was 3 A.M. plus or minus 2 hours.

a) In what range is the actual time of death?

b) Let x represent the actual time of death. Write an absolute value inequality for x.

83. Unidentified flying objects. The formula

S = -16t2 + v0t + s0

gives height in feet above the earth at time t seconds for an object projected into the air with an initial velocity of v0 feet per second (ft/sec) from an initial height of s0 feet. Two balls are tossed into the air simultaneously, one from the ground at 50 ft/sec and one from a height of 10 feet at 40 ft/sec. See the accompanying graph.

a) Use the graph to estimate the time at which the balls are at the same height.

H ei

gh t (

fe et

)

0 1 2 3 4

40

30

20

10

0

Time (seconds)

Figure for Exercise 83

b) Find the time from part (a) algebraically. c) For what values of t will their heights above the ground

differ by less than 5 feet (while they are both in the air)?

84. Playing catch. A circus clown at the top of a 60-foot plat­ form is playing catch with another clown on the ground. The clown on the platform drops a ball at the same time as the one on the ground tosses a ball upward at 80 ft/sec. For what length of time is the distance between the balls less than or equal to 10 feet? (Hint: Use the formula given in Exercise 83. The initial velocity of a ball that is dropped is 0 ft/sec.) See the accompanying figure.

Figure for Exercise 84

0 ft/sec

60 ft

80 ft/sec

Getting More Involved

85. Discussion

For which real numbers m and n is each equation satisfied?

a) I m - n I = I n - m I b) I mn I = I m I · I n I

m I m Ic) -- = --� n � I n I

86. Exploration

a) Evaluate I m + n I and I m I + I n I for i) m = 3 and n = 5

ii) m = -3 and n = 5 iii) m = 3 and n = -5 iv) m = -3 and n = -5

b) What can you conclude about the relationship between I m + n I and I m I + I n I?

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528 Chapter 8 More on Inequalities 8-22

Mid-Chapter Quiz Sections 8.1 through 8.2 Chapter 8

Write the solution set to each inequality in interval notation and graph it.

1. x > 1 and x < 4

2. x : 2 or x : 4

3. x - 3 and x - 5

4. 2x - 4 - 6 or 3x > -6

1 5. -3x + 1 < 7 and --x - -3

2

3x - 2 6. -10 - -- < 5 2

7. 0 < -5x - 3 < 7

8. I a - 3 I > 4

9. I 2w + 6 I - 8

10. I 2x - 7 I + 5 < 3

11. 5 - I 4x I < 9

Solve each equation.

12. I x - 3 I = 4

13. I w - 9 I = 0

14. I x + 3 I = I 2x - 9 I

15. 4 - � -2 1

-x + 1 � = 5

8.3 Compound Inequalities in Two Variables

A simple inequality in two variables involves only one inequality symbol. For example, y > x - 3 is a simple inequality in two variables. We graphed simple inequalities in two variables in Section 3.6. In this section we study compound inequalities in two variables.

In This Section

U1V Satisfying a Compound Inequality

U2V Graphing Compound Inequalities

U3V Absolute Value Inequalities

U4V Inequalities with No Solution

U5V Applications

E X A M P L E 1

U1V Satisfying a Compound Inequality A compound inequality in two variables consists of two simple inequalities joined with “and” or “or.” For example, y > x - 3 and y < 2 - x is a compound inequality in two variables. An ordered pair (or point) satisfies an “and” inequality only if it sat­ isfies both of the simple inequalities. An ordered pair satisfies an “or” inequality if it satisfies one or the other or both inequalities.

Satisfying compound inequalities Determine whether (-2, 3) satisfies each compound inequality.

a) y > x and x - y < -4

b) y > x and x - y > -4

c) y > x or x - y > -4

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8-23 8.3 Compound Inequalities in Two Variables 529

Solution a) Replacing x with -2 and y with 3 in y > x and x - y < -4 yields 3 > -2 and

-2 - 3 < -4. Since both inequalities are correct, (-2, 3) satisfies the compound inequality.

b) Replacing x with -2 and y with 3 in y > x and x - y > -4 yields 3 > -2 and -2 - 3 > -4. Since the second inequality is not correct, (-2, 3) does not satisfy the compound inequality.

c) Replacing x with -2 and y with 3 in y > x or x - y > -4 yields 3 > -2 or -2 - 3 > -4. Since the first inequality is correct and the connecting word is “or,” (-2, 3) satisfies the compound inequality.

Now do Exercises 1–6

U2V Graphing Compound Inequalities The solution set to a compound inequality using “and” is the intersection of the solu­ tion sets to the simple inequalities. Example 2 illustrates two methods for graphing the solution set.

E X A M P L E 2 Graphing a compound inequality with and Graph the compound inequality y > x - 3 and y < --1 2 - x + 2.

Solution The Intersection Method Start by graphing the lines y = x - 3 and y = --1 2 - x + 2. Points that satisfy y > x - 3 lie

above the line y = x - 3, and points that satisfy y < --1 2 - x + 2 lie below the line y = -- 1 2

- x + 2

as shown in Fig. 8.22(a). Since the connective is “and,” only points that are shaded with both colors (the intersection of the two regions) satisfy the compound inequality. The solution set to the compound inequality is shown in Fig. 8.22(b). Dashed lines are used because the inequalities are > and <.

Figure 8.22

y

x

3

6 7

4

3 41 2

(a)

-4

-1

y < - x + 2

-1

-2

1

y

x

3

6 7

4

3 41 2

(b)

-4

-1 -1

-2

-2-3 -2-3

y > x - 3

y > x - 3 and

y < - x + 21 — 2

1 — 2

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530 Chapter 8 More on Inequalities 8-24

Figure 8.23

y

x

3

43

4

2

5

1

1

y � x – 3

(a)

-3-4-5

-3

-2

-4

-5

-2 -1 -1

y

x

3

43

4

2

5

1

(b)

(0, 0)

-4

-5

-1 y � x – 3

and y � - x � 2

(0, 0)

(3, 3)

(4, -5)

(5, 0)

y � - x � 21 — 2

1 — 2

1

-2

-3 -1-5 -4 -2

The Test Point Method

Again graph the lines, but this time select a point in each of the four regions determined by the lines as shown in Fig. 8.23(a). Test each of the four points (3, 3), (0, 0), (4, -5), and (5, 0) to see if it satisfies the compound inequality:

y > x - 3 and y < --1 2 - x + 2

3 > 3 - 3 and 3 < --1 2 - · 3 + 2 Second inequality is incorrect.

0 > 0 - 3 and 0 < --1 2 - · 0 + 2 Both inequalities are correct.

-5 > 4 - 3 and -5 < --1 2 - · 4 + 2 First inequality is incorrect.

0 > 5 - 3 and 0 < --1 2 - · 5 + 2 Both inequalities are incorrect.

The only point that satisfies both inequalities is (0, 0). So the solution set to the compound inequality consists of all points in the region containing (0, 0) as shown in Fig. 8.23(b).

Now do Exercises 7–8

Example 3 involves a compound inequality using “or.” Remember that a compound sentence with “or” is true if one, the other, or both parts of it are true. The solution set to a compound inequality with “or” is the union of the two solution sets.

E X A M P L E 3 Graphing a compound inequality with or Graph the compound inequality 2x - 3y - -6 or x + 2y : 4.

Solution The Union Method Graph the line 2x - 3y = -6 through its intercepts (0, 2) and (-3, 0). Since (0, 0) does not satisfy this inequality, shade the region above this line as shown in Fig. 8.24(a). Graph

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8-25 8.3 Compound Inequalities in Two Variables 531

the line x + 2y = 4 through (0, 2) and (4, 0). Since (0, 0) does not satisfy this inequality, shade the region above the line as shown in Fig. 8.24(a). The union of these two solution sets consists of everything that is shaded as shown in Fig. 8.24(b). The boundary lines are solid because of the inequality symbols - and :.

y

x

3

43

1

4

5

21

y

x

3

43

1

4

2

5

1

2x - 3y - -6 x + 2y : 4