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6 Algebra: Equations and Inequalities > 6.4 Linear Inequalities in One Variable

66. I find the hardest part in solving a word problem is writing the equation that models the verbal conditions.

67. I solved a formula for one of its variables, so now I have a numerical value for that variable.

68. After a 35% reduction, a computer's price is $780, so I determined the original price, x, by solving

69. The price of a dress is reduced by 40%. When the dress still does not sell, it is reduced by 40% of the reduced price. If the price of the dress after both reductionsis $72, what was the original price?

70. In a film, the actor Charles Coburn plays an elderly “uncle” character criticized for marrying a woman when he is 3 times her age. He wittily replies, “Ah, but in 20years time I shall only be twice her age.” How old is the “uncle” and the woman?

71. Suppose that we agree to pay you for every problem in this chapter that you solve correctly and fine you for every problem done incorrectly. If at the end of26 problems we do not owe each other any money, how many problems did you solve correctly?

72. It was wartime when the Ricardos found out Mrs. Ricardo was pregnant. Ricky Ricardo was drafted and made out a will, deciding that $14,000 in a savingsaccount was to be divided between his wife and his child-to-be. Rather strangely, and certainly with gender bias, Ricky stipulated that if the child were a boy, he wouldget twice the amount of the mother's portion. If it were a girl, the mother would get twice the amount the girl was to receive. We'll never know what Ricky was thinkingof, for (as fate would have it) he did not return from the war. Mrs. Ricardo gave birth to twins—a boy and a girl. How was the money divided?

73. A thief steals a number of rare plants from a nursery. On the way out, the thief meets three security guards, one after another. To each security guard, the thief isforced to give one-half the plants that he still has, plus 2 more. Finally, the thief leaves the nursery with 1 lone palm. How many plants were originally stolen?

In Exercises 74–75, solve each proportion for x.

74.

75.

Group Exercise76. One of the best ways to learn how to solve a word problem in algebra is to design word problems of your own. Creating a word problem makes you very aware ofprecisely how much information is needed to solve the problem. You must also focus on the best way to present information to a reader and on how much informationto give. As you write your problem, you gain skills that will help you solve problems created by others.

The group should design five different word problems that can be solved using linear equations. All of the problems should be on different topics. For example, thegroup should not have more than one problem on price reduction. The group should turn in both the problems and their algebraic solutions.

6.4 Linear Inequalities in One Variable

What am I Supposed to Learn?After you have read this section, you should be able to:

1 Graph subsets of real numbers on a number line.

2 Solve linear inequalities.

3 Solve applied problems using linear inequalities.

RENT-A-HEAP, A CAR RENTAL company, charges $125 per week plus $0.20 per mile to rent one of their cars. Suppose you are limited by how much money you canspend for the week: You can spend at most $335. If we let x represent the number of miles you drive the heap in a week, we can write an inequality that models thegiven conditions: Chapter 6 Algebra: Equations and Inequalities

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x − 0.35 = 780.

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a

b+c

c

=ax−b

b

c−d

d

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The weekly

charge of $125

▼

125

plus

▼

+

the charge of

$0.20 per mile

for x miles

▼

0.20x

must be less

than

or equal to

▼

≤

$335.

▼

335.



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6 Algebra: Equations and Inequalities > 6.4 Linear Inequalities in One Variable > Graphing Subsets of Real Numbers on a NumberLine

Notice that the highest exponent on the variable in is 1. Such an inequality is called a linear inequality in one variable. The symbol between thetwo sides of an inequality can be (is less than or equal to), (is less than), (is greater than or equal to), or (is greater than).

In this section, we will study how to solve linear inequalities such as Solving an inequality is the process of finding the set of numbers thatmakes the inequality a true statement. These numbers are called the solutions of the inequality and we say that they satisfy the inequality. The set of all solutions iscalled the solution set of the inequality. We begin by discussing how to represent these solution sets, which are subsets of real numbers, on a number line.

Graphing Subsets of Real Numbers on a Number Line

1 Graph subsets of real numbers on a number line.

Table 6.3 shows how to represent various subsets of real numbers on a number line. Open dots indicate that a number is not included in a set. Closed dots indicate that a number is included in a set.

TABLE 6.3 Graphs of Subsets of Real Numbers

Example 1 Graphing Subsets of Real NumbersGraph each set:

a.

b.

c.

SOLUTION

a.

b.

c.



125 + 0.20x ≤ 335≤ < ≥ >

125 + 0.20x ≤ 335.

(∘) (∙)

{x|x < a} ◄ ►x is a real number

less than a.

{x|x ≤ a} ◄ ►x is a real number less

than or equal to a.

{x|x > b} ◄ ►x is a real number

greater than b.

{x|x ≥ b} ◄ ►x is a real number

greater than or equal to b.

{x|a < x < b} ◄ ►x is a real number greater

than a and less than b.

{x|a ≤ x ≤ b} ◄ ►x is a real number greater than or

equal to a and less than or equal to b.

{x|a ≤ x < b} ◄ ►x is a real number greater than

or equal to a and less than b.

{x|a < x ≤ b} ◄ ►x is a real number greater than

a and less than or equal to b.

{x|x < 3}

{x|x ≥ −1}

{x| − 1 < x ≤ 3} .

{x|x < 3} ◄ ►x is a real number

less than 3.

{x|x ≥ −1} ◄ ►x is a real number

greater than or

equal to − 1.

{x| − 1 < x ≤ 3} ◄ ►x is a real number

greater than − 1 and

less than or equal to 3.








Page 377









6 Algebra: Equations and Inequalities > 6.4 Linear Inequalities in One Variable > Solving Linear Inequalities in One Variable

Check Point 1Graph each set:

a.

b.

c.

Solving Linear Inequalities in One Variable

2 Solve linear inequalities.

We know that a linear equation in x can be expressed as A linear inequality in x can be written in one of the following forms:

In each form,

Back to our question that opened this section: How many miles can you drive your Rent-a-Heap car if you can spend at most $335? We answer the question by solving

for x. The solution procedure is nearly identical to that for solving

Our goal is to get x by itself on the left side. We do this by subtracting 125 from both sides to isolate 0.20x:

Finally, we isolate x from 0.20x by dividing both sides of the inequality by 0.20:

With at most $335 per week to spend, you can travel at most 1050 miles.

Great Question!What are some common English phrases and sentences that I can model with linear inequalities?

English phrases such as “at least” and “at most” can be represented by inequalities.

English Sentence Inequality

x is at least 5.

x is at most 5.

x is no more than 5.

x is no less than 5.

We started with the inequality and obtained the inequality in the final step. Both of these inequalities have the same solution set,namely Inequalities such as these, with the same solution set, are said to be equivalent.

We isolated x from 0.20x by dividing both sides of by 0.20, a positive number. Let's see what happens if we divide both sides of an inequality by anegative number. Consider the inequality Divide 10 and 14 by

Because lies to the right of on the number line, is greater than



{x|x < 4}

{x|x ≥ −2}

{x| − 4 ≤ x < 1} .

ax + b = 0.

ax + b < 0, ax + b ≤ 0, ax + b > 0, ax + b ≥ 0.

a ≠ 0.

0.20x + 125 ≤ 335

0.20x + 125 = 335.

0.20x + 125

0.20x + 125 − 125

0.20x

≤ 335

≤ 335 − 125

≤ 210.

This is the given inequality.

Subtract 125 from both sides.

Simplify.

0.20x

0.20

x

≤ 2100.20

≤ 1050.

Divide both sides by 0.20.

Simplify.

x ≥ 5

x ≤ 5

x ≤ 5

x ≥ 5

0.20x + 125 ≤ 335 x ≤ 1050{x|x ≤ 1050} .

0.20x ≤ 21010 < 14. −2:

= −5 and = −7.10

−2

14

−2

−5 −7 −5 −7:

−5 > −7.








Page 378

Notice that the direction of the inequality symbol is reversed:

In general, when we multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol is reversed. When we reversethe direction of the inequality symbol, we say that we change the sense of the inequality.


10 < 14

↕

−5 > −7.

◄Dividing by − 2 changes

the direction of the

inequality symbol.








Page 379


We can summarize our discussion with the following statement:

Solving Linear InequalitiesThe procedure for solving linear inequalities is the same as the procedure for solving linear equations, with one important exception: When multiplying or dividing bothsides of the inequality by a negative number, reverse the direction of the inequality symbol, changing the sense of the inequality.

Example 2 Solving a Linear InequalitySolve and graph the solution set:

SOLUTION

Our goal is to get x by itself on the left side. We do this by first getting 4x by itself, adding 7 to both sides.

Next, we isolate x from 4x by dividing both sides by 4. The inequality symbol stays the same because we are dividing by a positive number.

The solution set consists of all real numbers that are greater than or equal to 3, expressed in set-builder notation as The graph of the solution set isshown as follows:

We cannot check all members of an inequality's solution set, but we can take a few values to get an indication of whether or not it is correct. In Example 2, we found thatthe solution set of is Show that 3 and 4 satisfy the inequality, whereas 2 does not.

Check Point 2Solve and graph the solution set:

Example 3 Solving Linear InequalitiesSolve and graph the solution set:

a.

b.

SOLUTION

In each case, our goal is to isolate x. In the first inequality, this is accomplished by multiplying both sides by 3. In the second inequality, we can do this by dividing bothsides by




4x − 7 ≥ 5.

4x − 7

4x − 7 + 7

4x

≥ 5

≥ 5 + 7

≥ 12


Add 7 to both sides.

Simplify.

4x

4

x

≥ 124

≥ 3

Divide both sides by 4.

Simplify.

{x|x ≥ 3} .

4x − 7 ≥ 5 {x|x ≥ 3} .

5x − 3 ≤ 17.

x < 513

−3x < 21.

−3.














a.

The solution set is The graph of the solution set is shown as follows:

b.



a.

b.


SOLUTION

We will get x by itself on the left side. We begin by subtracting 8x from both sides so that the variable term appears on the left.

Next, we get by itself, adding 12 to both sides.

In order to solve we isolate x from by dividing both sides by The direction of the inequality symbol must be reversed because we are dividingby a negative number.




x13

3 ⋅ x13

<

<

5

3 ⋅ 5


Isolate x by multiplying by 3 on both sides.

↑⏐

The symbol < stays the same because we are multiplying both

sides by a positive number.

x < 15 Simplify.

{x|x < 15} .

−3x

−3x

−3

<

>

2121−3


Isolate x by multiplying by − 3 on both sides.

↑⏐

The symbol < must be reversed because we are dividing both

sides by a negative number.

x > −7 Simplify.

{x|x > −7} .

x < 214

−6x < 18.

6x − 12 > 8x + 2.

6x − 12

6x − 8x − 12

−2x − 12

> 8x + 2

> 8x − 8x + 2

> 2


Subtract 8x on both sides with

the goal of isolating x on the left.

Simplify.

−2x

−2x − 12 + 12

−2x

> 2 + 12

> 14

Add 12 to both sides.

Simplify.

−2x > 14, −2x −2.

−2x

−2

x

< 14−2

< −7

Divide both sides by − 2 and change the sense of the inequality.

Simplify.

{x|x < −7} .








Page 380












SOLUTION

Begin by simplifying the algebraic expression on each side.

d

We will get x by itself on the left side. Subtract 8x from both sides.

Next, we get by itself, adding 6 to both sides.

To isolate x, we must eliminate the negative sign in front of the x. Because means we can do this by dividing both sides of the inequality by Thisreverses the direction of the inequality symbol.



Great Question!Do I have to solve by isolating the variable on the left?

No. You can solve

by isolating x on the right side. Subtract 7x from both sides and add 8 to both sides:

This last inequality means the same thing as

Solution sets, in this case are expressed with the variable on the left and the constant on the right.

In our next example, the inequality has three parts:



7x − 3 > 13x + 33.

2 (x − 3) + 5x ≤ 8 (x − 1) .

7x − 8x − 6

−x − 6

≤ 8x − 8x − 8

≤ −8

−x

−x − 6 + 6

−x

≤ −8 + 6

≤ −2

−x −1x, −1.

−x

−1

x

≥ −2

−1

≥ 2

Divide both sides by − 1 and change the sense of the inequality.

Simplify.

{x|x ≥ 2} .

2 (x − 3) − 1 ≤ 3 (x + 2) − 14.

7x − 6 ≤ 8x − 8

7x − 6 ≤ 8x − 8

7x − 6 − 7x

−6

−6 + 8

2

≤ 8x − 8 − 7x

≤ x − 8

≤ x − 8 + 8

≤ x.

x ≥ 2.

{x|x ≥ 2} ,








Page 381

By performing the same operation on all three parts of the inequality, our goal is to isolate x in the middle.


−3 < 2x + 1 ≤ 3.

▲

2x + 1 is greater than − 3

and less than or equal to 3.









Example 6 Solving a Three-Part InequalitySolve and graph the solution set:

SOLUTION

We would like to isolate x in the middle. We can do this by first subtracting 1 from all three parts of the inequality. Then we isolate x from 2x by dividing all three partsof the inequality by 2.

The solution set consists of all real numbers greater than and less than or equal to 1, represented by The graph is shown as follows:

Check Point 6Solve and graph the solution set on a number line:

As you know, different professors may use different grading systems to determine your final course grade. Some professors require a final examination; others do not. Inour next example, a final exam is required and it counts as two grades.

3 Solve applied problems using linear inequalities.

Example 7 An Application: Final Course GradeTo earn an A in a course, you must have a final average of at least 90%. On the first four examinations, you have grades of 86%, 88%, 92%, and 84%. If the finalexamination counts as two grades, what must you get on the final to earn an A in the course?

SOLUTION

We will use our five-step strategy for solving algebraic word problems.

Steps 1 and 2 Represent unknown quantities in terms of x. Let

Step 3 Write an inequality in x that models the conditions. The average of the six grades is found by adding the grades and dividing the sum by 6.

Because the final counts as two grades, the x (your grade on the final examination) is added twice. This is also why the sum is divided by 6.



−3 < 2x + 1 ≤ 3.

−3

−3 − 1

−4−4

2

−2

< 2x + 1 ≤ 3

< 2x + 1 − 1 ≤ 3 − 1

< 2x ≤ 2

< ≤2x

222

< x ≤ 1


Subtract 1 from all three parts.

Simplify.

Divide each part by 2.

Simplify.

−2 {x| − 2 < x ≤ 1} .

1 ≤ 2x + 3 < 11.

x = your grade on the final examination.

Average =86 + 88 + 92 + 84 + x + x

6








Page 382









Page 383

6 Algebra: Equations and Inequalities > 6.4 Linear Inequalities in One Variable > Concept and Vocabulary Check

To get an A, your average must be at least 90. This means that your average must be greater than or equal to 90.

Step 4 Solve the inequality and answer the problem's question.

You must get at least 95% on the final examination to earn an A in the course.

Step 5 Check. We can perform a partial check by computing the average with any grade that is at least 95. We will use 96. If you get 96% on the final examination,your average is

Because you earn an A in the course.

Check Point 7To earn a B in a course, you must have a final average of at least 80%. On the first three examinations, you have grades of 82%, 74%, and 78%. If the finalexamination counts as two grades, what must you get on the final to earn a B in the course?

Concept and Vocabulary CheckFill in each blank so that the resulting statement is true.

1. On a number line, an open dot indicates that a number _____________________ in a solution set, and a closed dot indicates that a number___________________ in a solution set.

2. If an inequality's solution set consists of all real numbers, x, that are less than a, the solution set is represented in set-builder notation as ___________________.

3. If an inequality's solution set consists of all real numbers, x, that are greater than a and less than or equal to b, the solution set is represented in set-builder notationas ______________________.

4. When both sides of an inequality are multiplied or divided by a/an ___________________ number, the direction of the inequality symbol is reversed.

In Exercises 5–8, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

5. The inequality is equivalent to _____

6. The statement “x is at most 5” is written _____

7. The inequality is equivalent to _____

8. The statement “the sum of x and 6% of x is at least 80” is modeled by _____




Your

average

▼

86+88+92+84+x+x

6

must be greater

than or equal to

▼

≥

90.

▼

90

86+88+92+84+x+x

6

350+2x

6

6 ( )350+2x

6

350 + 2x

350 + 2x − 350

2x

2x

2

x

≥ 90

≥ 90

≥ 6 (90)

≥ 540

≥ 540 − 350

≥ 190

≥ 190

2

≥ 95

This is the inequality that models the given conditions.

Combine like terms in the numerator.

Multiply both sides by 6, clearing the fraction.

Multiply.

Subtract 350 from both sides.

Simplify.

Divide both sides by 2.

Simplify.

= = 90 .86 + 88 + 92 + 84 + 96 + 96

6

542

6

1

3

90 > 90,13

x − 3 > 0 x < 3.

x < 5.

−4x < −20 x > −5.

x + 0.06x ≥ 80.













6 Algebra: Equations and Inequalities > 6.4 Linear Inequalities in One Variable > Exercise Set 6.4

Exercise Set 6.4Practice ExercisesIn Exercises 1–12, graph each set of real numbers on a number line.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

In Exercises 13–66, solve each inequality and graph the solution set on a number line.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.



{x|x > 6}

{x|x > −2}

{x|x < −4}

{x|x < 0}

{x|x ≥ −3}

{x|x ≥ −5}

{x|x ≤ 4}

{x|x ≤ 7}

{x| − 2 < x ≤ 5}

{x| − 3 ≤ x < 7}

{x| − 1 < x < 4}

{x| − 7 ≤ x ≤ 0}

x − 3 > 2

x + 1 < 5

x + 4 ≤ 9

x − 5 ≥ 1

x − 3 < 0

x + 4 ≥ 0

4x < 20

6x ≥ 18

3x ≥ −15

7x < −21

2x − 3 > 7

3x + 2 ≤ 14

3x + 3 < 18

8x − 4 > 12

x < 412

x > 312

> −2x

3

< −1x

4

−3x < 15

−7x > 21

−3x ≥ −15

−7x ≤ −21

3x + 4 ≤ 2x + 7

2x + 9 ≤ x + 2








37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

64.

65.

66.

Practice PlusIn Exercises 67–70, write an inequality with x isolated on the left side that is equivalent to the given inequality.

67. Assume

68. Assume

69. Assume

70. Assume

In Exercises 71–76, use set-builder notation to find all real numbers satisfying the given conditions.

71. A number increased by 5 is at least two times the number.

72. A number increased by 12 is at least four times the number.

73. Twice the sum of four and a number is at most 36.

74. Three times the sum of five and a number is at most 48.

75. If the quotient of three times a number and five is increased by four, the result is no more than 34.

76. If the quotient of three times a number and four is decreased by three, the result is no less than 9.

Application ExercisesThe graphs show that the three components of love, namely passion, intimacy, and commitment, progress differently over time. Passion peaks early in a relationship andthen declines. By contrast, intimacy and commitment build gradually. Use the graphs to solve Exercises 77–84. Assume that x represents years in a relationship.

5x − 9 < 4x + 7

3x − 8 < 2x + 11

−2x − 3 < 3

14 − 3x > 5

3 − 7x ≤ 17

5 − 3x ≥ 20

−x < 4

−x > −3

5 − x ≤ 1

3 − x ≥ −3

2x − 5 > −x + 6

6x − 2 ≥ 4x + 6

2x − 5 < 5x − 11

4x − 7 > 9x − 2

3(x + 1) − 5 < 2x + 1

4(x + 1) + 2 ≥ 3x + 6

8x + 3 > 3(2x + 1) − x + 5

7 − 2(x − 4) < 5(1 − 2x)

− ≤ + 1x

4

3

2x

2

+ 1 ≥ −3x

1015

x

10

1 − > 4x

2

7 − x <45

3

5

6 < x + 3 < 8

7 < x + 5 < 11

−3 ≤ x − 2 < 1

−6 < x − 4 ≤ 1

−11 < 2x − 1 ≤ −5

3 ≤ 4x − 3 < 19

−3 ≤ x − 5 < −123

−6 ≤ x − 4 < −312

Ax + By > C; A > 0.

Ax + By ≤ C; A > 0.

Ax + By > C; A < 0.

Ax + By ≤ C; A < 0.



Page 384

dSource: R. J. Sternberg, A Triangular Theory of Love, Psychological Review, 93, 119–135.

77. Use set-builder notation to write an inequality that expresses for which years in a relationship intimacy is greater than commitment.

78. Use set-builder notation to write an inequality that expresses for which years in a relationship passion is greater than or equal to intimacy.

79. What is the relationship between passion and intimacy for

80. What is the relationship between intimacy and commitment for


{x|5 ≤ x < 7}?

{x|4 ≤ x < 7}?








6 Algebra: Equations and Inequalities > 6.4 Linear Inequalities in One Variable > Exercise Set 6.4

81. What is the relationship between passion and commitment for

82. What is the relationship between passion and commitment for

83. What is the maximum level of intensity for passion? After how many years in a relationship does this occur?

84. After approximately how many years do levels of intensity for commitment exceed the maximum level of intensity for passion?

In more U.S. marriages, spouses have different faiths. The bar graph shows the percentage of households with an interfaith marriage in 1988 and 2008. Also shown isthe percentage of households in which a person of faith is married to someone with no religion.

dSource: General Social Survey, University of Chicago

The formula

models the percentage of U.S. households with an interfaith marriage, I, x years after 1988. The formula

models the percentage of U.S. households in which a person of faith is married to someone with no religion, N, x years after 1988. Use these models to solve Exercises85–86.

85.

a. In which years will more than 33% of U.S. households have an interfaith marriage?

b. In which years will more than 14% of U.S. households have a person of faith married to someone with no religion?

c. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage and more than 14% have a faith/noreligion marriage?

86.

a. In which years will more than 34% of U.S. households have an interfaith marriage?

b. In which years will more than 15% of U.S. households have a person of faith married to someone with no religion?

c. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage and more than 15% have a faith/noreligion marriage?

87. On two examinations, you have grades of 86 and 88. There is an optional final examination, which counts as one grade. You decide to take the final in order to geta course grade of A, meaning a final average of at least 90.

a. What must you get on the final to earn an A in the course?

b. By taking the final, if you do poorly, you might risk the B that you have in the course based on the first two exam grades. If your final average is less than 80, youwill lose your B in the course. Describe the grades on the final that will cause this to happen.

88. On three examinations, you have grades of 88, 78, and 86. There is still a final examination, which counts as one grade.

a. In order to get an A, your average must be at least 90. If you get 100 on the final, compute your average and determine if an A in the course is possible.

b. To earn a B in the course, you must have a final average of at least 80. What must you get on the final to earn a B in the course?



{x|6 < x < 8}?

{x|7 < x < 9}?

I = x + 261

4

N = x + 61

4








Page 385

89. A car can be rented from Continental Rental for $80 per week plus 25 cents for each mile driven. How many miles can you travel if you can spend at most $400for the week?

90. A car can be rented from Basic Rental for $60 per week plus 50 cents for each mile driven. How many miles can you travel if you can spend at most $600 for theweek?

91. An elevator at a construction site has a maximum capacity of 3000 pounds. If the elevator operator weighs 245 pounds and each cement bag weighs 95 pounds,up to how many bags of cement can be safely lifted on the elevator in one trip?

92. An elevator at a construction site has a maximum capacity of 2800 pounds. If the elevator operator weighs 265 pounds and each cement bag weighs 65 pounds,up to how many bags of cement can be safely lifted on the elevator in one trip?

93. A basic cellphone plan costs $20 per month for 60 calling minutes. Additional time costs $0.40 per minute. The formula

gives the monthly cost for this plan, C, for x calling minutes, where How many calling minutes are possible for a monthly cost of at least $28 and at most$40?

94. The formula for converting Fahrenheit temperature, F, to Celsius temperature, C, is

If Celsius temperature ranges from to inclusive, what is the range for the Fahrenheit temperature?

Writing in Mathematics95. When graphing the solutions of an inequality, what is the difference between an open dot and a closed dot?

96. When solving an inequality, when is it necessary to change the direction of the inequality symbol? Give an example.


C = 20 + 0.40 (x − 60)

x > 60.

C = (F − 32) .5

9

15° 35°,








6 Algebra: Equations and Inequalities > 6.5 Quadratic Equations

97. Describe ways in which solving a linear inequality is similar to solving a linear equation.

98. Describe ways in which solving a linear inequality is different than solving a linear equation.

Critical Thinking ExercisesMake Sense? In Exercises 99–102, determine whether each statement makes sense or does not make sense, and explain your reasoning.

99. I can check inequalities by substituting 0 for the variable: When 0 belongs to the solution set, I should obtain a true statement, and when 0 does not belong to thesolution set, I should obtain a false statement.

100. In an inequality such as I can avoid division by a negative number depending on which side I collect the variable terms and on which side Icollect the constant terms.

101. I solved and concluded that is the greatest integer in the solution set.

102. I began the solution of by simplifying the left side, obtaining

103. A car can be rented from Basic Rental for $260 per week with no extra charge for mileage. Continental charges $80 per week plus 25 cents for each mile drivento rent the same car. How many miles must be driven in a week to make the rental cost for Basic Rental a better deal than Continental's?

104. A company manufactures and sells personalized stationery. The weekly fixed cost is $3000 and it cost $3.00 to produce each package of stationery. The sellingprice is $5.50 per package. How many packages of stationery must be produced and sold each week for the company to generate a profit?

6.5 Quadratic Equations

What am I Supposed to Learn?After you have read this section, you should be able to:

1 Multiply binomials using the FOIL method.

2 Factor trinomials.

3 Solve quadratic equations by factoring.

4 Solve quadratic equations using the quadratic formula.

5 Solve problems modeled by quadratic equations.

I'm very well acquainted, too, with matters mathematical, I understand equations, both simple and quadratical. About binomial theorem I'm teeming with a lot of news,With many cheerful facts about the square of the hypotenuse.

—Gilbert and Sullivan, The Pirates of Penzance

EQUATIONS QUADRATICAL? CHEERFUL NEWS ABOUT THE SQUARE OF THE hypotenuse? You've come to the right place. In this section, we study two methodsfor solving quadratic equations, equations in which the highest exponent on the variable is 2. (Yes, it's quadratic and not quadratical, despite the latter's rhyme withmathematical.) In Chapter 10 (Section 10.2), we look at an application of quadratic equations, introducing (cheerfully, of course) the Pythagorean Theorem and thesquare of the hypotenuse.

Multiplying Two Binomials Using the FOIL MethodBefore we learn about the first method for solving quadratic equations, factoring, we need to consider the FOIL method for multiplying two binomials. A binomial is asimplified algebraic expression that contains two terms in which each exponent that appears on a variable is a whole number.

1 Multiply binomials using the FOIL method.

Examples of Binomials

Two binomials can be quickly multiplied by using the FOIL method, in which F represents the product of the first terms in each binomial, O represents the product of theoutside terms, I represents the product of the two inside terms, and L represents the product of the last, or second, terms in each binomial.



5x + 4 < 8x − 5,

−2x + 5 ≥ 13 −4

5 − 3 (x + 2) > 10x 2x + 4 > 10x.

x + 3, x + 4, 3x + 4, 5x − 3







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