1. Slide 1- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Introduction to Real Numbers and Algebraic Expressions 1.1Introduction to Algebra 1.2The Real Numbers 1.3Addition of Real Numbers 1.4Subtraction of Real Numbers 1.5Multiplication of Real Numbers 1.6Division of Real Numbers 1.7Properties of Real Numbers 1.8Simplifying Expressions; Order of Operations 1

3. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley INTRODUCTION TO ALGEBRA Evaluate algebraic expressions by substitution. Translate phrases to algebraic expressions. 1.1a b

4. Slide 1- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Evaluate algebraic expressions by substitution.a

5. Slide 1- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Algebraic Expressions An algebraic expression consists of variables, numerals, and operation signs. x + 38 19 – y When we replace a variable with a number, we say that we are substituting for the variable. This process is called evaluating the expression.

6. Slide 1- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Evaluate x + y for x = 38 and y = 62. Solution We substitute 38 for x and 62 for y. x + y = 38 + 62 = 100

7. Slide 1- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Evaluate and for x = 72 and y = 8. Solution We substitute 72 for x and 8 for y:

8. Slide 1- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Evaluate for C = 30. Solution This expression can be used to find the Fahrenheit temperature that corresponds to 30 degrees Celsius.

9. Slide 1- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Translate phrases to algebraic expressions.b

10. Slide 1- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Translating to Algebraic Expressions per of decreased byincreased by ratio of twice less than more than divided into times minus plus quotient of product of difference of sum of divided bymultiplied bysubtracted from added to DivisionMultiplicationSubtractionAddition

11. Slide 1- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Translate each phrase to an algebraic expression. a) 9 more than y b) 7 less than x c) the product of 3 and twice w Solution PhraseAlgebraic Expression a) 9 more than y y + 9 b) 7 less than xx  7 c) the product of 3 and twice w3 2w or 2w 3

12. Slide 1- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Translate each phrase to an algebraic expression. PhraseAlgebraic Expression Eight more than some number One-fourth of a number Two more than four times some number Eight less than some number Five less than the product of two numbers Twenty-five percent of some number Seven less than three times some number x + 8, or 8 + x 4x + 2, or 2 + 4x n – 8 ab – 5 0.25n 3w – 7

13. Slide 1- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.1 1. Evaluate when a = 13 and b = 5. a) b) c) d) 9

14. Slide 1- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.1 1. Evaluate when a = 13 and b = 5. a) b) c) d) 9

15. Slide 1- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.1 2. Write an algebraic expression: The product of a and b. a) ab b) a + b c) b – a d)

16. Slide 1- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.1 2. Write an algebraic expression: The product of a and b. a) ab b) a + b c) b – a d)

17. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley THE REAL NUMBERS State the integer that corresponds to a real-world situation. Graph rational numbers on a number line. Convert from fraction notation to decimal notation for a rational number. Determine which of two real numbers is greater and indicate which, using ; given an inequality like a > b, write another inequality with the same meaning. Determine whether an inequality like  3  5 is true or false. Find the absolute value of a number. 1.2a d c b e

18. Slide 1- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective State the integer that corresponds to a real-world situation.a

19. Slide 1- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Natural Numbers The set of natural numbers = {1, 2, 3, …}. These are the numbers used for counting. Whole Numbers The set of whole numbers = {0, 1, 2, 3, …}. This is the set of natural numbers with 0 included.

20. Slide 1- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Integers The set of integers = {…,  5,  4,  3,  2,  1, 0, 1, 2, 3, 4, 5, …}.

21. Slide 1- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Integers Integers consist of the whole numbers and their opposites. Integers to the left of zero on the number line are called negative integers and those to the right of zero are called positive integers. Zero is neither positive nor negative and serves as its own opposite. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 0, neither positive nor negative Positive integers Negative integers Opposites

22. Slide 1- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Tell which integer corresponds to each situation. 1. Death Valley is 282 feet below sea level. 2. Margaret owes $312 on her credit card. She has $520 in her checking account. Solution 1. 282 below sea level corresponds to  282. 2. The integers  312 and 520 correspond to the situation.

23. Slide 1- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Graph rational numbers on a number line.b

24. Slide 1- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Fractions such as ½ are not integers. A larger system called rational numbers contains integers and fractions. The rational numbers consist of quotients of integers with nonzero divisors. The following are rational numbers:

25. Slide 1- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Numbers The rational numbers consist of all numbers that can be named in the form, where a and b are integers and b is not equal to 0 (b  0).

26. Slide 1- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example To graph a number means to find and mark its point on the number line. Graph: Solution The number can be named, or 3.5. Its graph is halfway between 3 and 4. 10-9-7-5-313579-10-8-4048-10-26-6102 3.5

27. Slide 1- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Graph:  2.8 Solution The graph of  2.8 is 8/10 of the way from  2 to  3. 10-9-7-5-313579-10-8-4048-10-26-6102 -2.8

28. Slide 1- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Convert from fraction notation to decimal notation for a rational number.c

29. Slide 1- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Each rational number can be named using fraction notation or decimal notation.

30. Slide 1- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Find decimal notation for Solution Because means 7  40, we divide. We are finished when the remainder is 0.

31. Slide 1- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Find decimal notation for Solution Divide 1  12 Since 4 keeps reappearing as a remainder, the digits repeat and will continue to do so; therefore,

32. Slide 1- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine which of two real numbers is greater and indicate which, using ; given an inequality like a > b, write another inequality with the same meaning. Determine whether an inequality like  3  5 is true or false.d

33. Slide 1- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Real-Number System The set of real numbers = The set of all numbers corresponding to points on the number line. Decimal notation for rational numbers either terminates or repeats. Decimal notation for irrational numbers neither terminates nor repeats.

34. Slide 1- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Positive Integers: 1, 2, 3, … IntegersZero: 0 Rational numbers Negative integers: -1, -2, - 3, … Real numbers Rational numbers that are not integers: 2/3, - 4/5, 19/-5, -7/8, 8.2, Irrational numbers: pi, square roots, 5.363663666…

35. Slide 1- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Numbers are written in order on the number line, increasing as we move to the right. For any two numbers on the line, the one to the left is less than the one to the right. The symbol < means “is less than,”  4 < 8 is read “  4 is less than 8.” The symbol > means “is greater than,”  6 >  9 is read “  6 is greater than  9.”

36. Slide 1- 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Use either for to form a true sentence. 1.  7 32. 8  33.  21  9 Solution 10-9-7-5-313579-10-8-4048-10-26-6102 1.  7 3 3.  21  9 2. 8  3 Since  7 is to the left of 3, we have  7 < 3. Since 8 is to the right of  3, we have 8 >  3. Since  21 is to the left of  9, we have  21 <  9.  7 < 3.8 >  3.  21 <  9

37. Slide 1- 37 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Use either for to form a true sentence. 1.  7.2 2. Solution 1.  7.2 2. Convert to decimal notation: 10-9-7-5-313579-10-8-4048-10-26-6102 < >

38. Slide 1- 38 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Order; >, < a a.

39. Slide 1- 39 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H Write another inequality with the same meaning. a.  4 >  10 b. c <  7 Solution a. The inequality  10 <  4 has the same meaning. b. The inequality  7 > c has the same meaning.

40. Slide 1- 40 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Positive numbers bNegative numbers a a < 0b > 0 If b is a positive real number, then b > 0. If a is a negative real number, then a < 0.

41. Slide 1- 41 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the absolute value of a number.e

42. Slide 1- 42 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Absolute Value The absolute value of a number is its distance from zero on a number line. We use the symbol |x| to represent the absolute value of a number x.  5 units from 0 

43. Slide 1- 43 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding Absolute Value a) If a number is negative, its absolute value is positive. b) If a number is positive or zero, its absolute value is the same as the number.

44. Slide 1- 44 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example I Find the absolute value of each number. a. |  5|b. |36| c. |0|d. |  52| Solution a. |  5| The distance of  5 from 0 is 5, so |  5| = 5. b. |36| The distance of 36 from 0 is 36, so |36| = 36. c. |0| The distance of 0 from 0 is 0, so |0| = 0. d. |  52| The distance of  52 from 0 is 52, so |  52| = 52.

45. Slide 1- 45 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.2 1. Use, or = for to write a true sentence:  0.53  0.39 a)  0.53 >  0.39 b)  0.53 <  0.39 c)  0.53 =  0.39 d) cannot be determined

46. Slide 1- 46 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.2 1. Use, or = for to write a true sentence:  0.53  0.39 a)  0.53 >  0.39 b)  0.53 <  0.39 c)  0.53 =  0.39 d) cannot be determined

47. Slide 1- 47 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.2 2. Find the absolute value of |  13|. a) 13 b)  13 c) 169 d)

48. Slide 1- 48 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 10.1 2. Find the absolute value of |  13|. a) 13 b)  13 c) 169 d)

49. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley ADDITION of REAL NUMBERS Adding real numbers without using a number line. Find the opposite, or additive inverse, of a real number. Solve applied problems involving addition of real numbers. 1.3a b c

50. Slide 1- 50 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Adding real numbers without using a number line.a

51. Slide 1- 51 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Addition on a Number Line To do the addition a + b, we start at 0. Then we move to a, and then move according to b. a) If b is positive, we move b to the right. b) If b is negative, we move b to the left. c) If b is 0, we stay at a.

52. Slide 1- 52 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Add: 3 + (  6). Solution 3 + (  6) =  3 Start at 3. Move 6 units to the left.

53. Slide 1- 53 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Add:  5 + 8. Solution  5 + 8 = 3 Start at  5. Move 8 units to the right.

54. Slide 1- 54 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rules for Addition of Real Numbers 1. Positive numbers: Add the same as arithmetic numbers. The answer is positive. 2. Negative numbers: Add absolute values. The answer is negative. 3. A positive and a negative number: Subtract the smaller absolute value from the larger. Then: a) If the positive number has the greater absolute value, the answer is positive. b) If the negative number has the greater absolute value, the answer is negative. c) If the numbers have the same absolute value, the answer is 0. 4. One number is zero: The sum is the other number.

55. Slide 1- 55 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Add. 1.  5 + (  8) = 2.  9 + (  7) = Solution 1.  5 + (  8) =  13 2.  9 + (  7) =  16 Add the absolute values: 5 + 8 = 13. Make the answer negative.

56. Slide 1- 56 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Add. 1. 4 + (  6) = 2. 12 + (  9) = 3.  8 + 5 = 4.  7 + 5 = Solution 1. 4 + (  6) = 2. 12 + (  9) = 3.  8 + 5 = 4.  7 + 5 = Think: The absolute values are 4 and 6. The difference is 2. Since the negative number has the larger absolute value, the answer is negative,  2. 22 3 33 22

57. Slide 1- 57 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Add: 16 + (  2) + 8 + 15 + (  6) + (  14). Solution Because of the commutative and associate laws for addition, we can group the positive numbers together and the negative numbers together and add them separately. Then we add the two results. 16 + (  2) + 8 + 15 + (  6) + (  14) = 16 + 8 + 15 + (  2) + (  6) + (  14) = 39 + (  22) = 17

58. Slide 1- 58 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the opposite, or additive inverse, of a real number.b

59. Slide 1- 59 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Opposites, or Additive Inverses Two numbers whose sum is 0 are called opposites, or additive inverses, of each other.

60. Slide 1- 60 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Find the opposite, or additive inverse, of each number. 1. 522.  123. 04. Solution 1. 52 The opposite of 52 is  52 because 52 + (  52) = 0 2.  12 The opposite of  12 is 12 because  12 + 12 = 0 3. 0 The opposite of 0 is 0 because 0 + 0 = 0 4. The opposite of is because

61. Slide 1- 61 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Symbolizing Opposites The opposite, or additive inverse, of a number a can be named  a (read “the opposite of a,” or “the additive inverse of a”). The Opposite of an Opposite The opposite of the opposite of a number is the number itself. (The additive inverse of the additive inverse of a number is the number itself.) That is, for any number a  (  a) = a.

62. Slide 1- 62 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Evaluate  x and  (  x) when x = 12. Solution We replace x in each case with 12. a) If x = 12, then  x =  12 =  12 b) If x = 12, then  (  x) =  (  12) = 12

63. Slide 1- 63 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H Evaluate  (  x) for x =  7. Solution We replace x with  7. If x =  7, then  (  x) =  (  (  7)) =  7

64. Slide 1- 64 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Sum of Opposites For any real number a, the opposite, or additive inverse, of a, expressed as  a, is such that a + (  a) =  a + a = 0.

65. Slide 1- 65 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example I Change the sign (Find the opposite.) a)  9b) 8 Solution a)  9  (  9) = 9 b) 8  (8) =  8

66. Slide 1- 66 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example J On a recent day, the price of a stock opened at a value of $72.37. During the day, it rose $3.25, dropped $6.32, and rose $4.12. Find the value of the stock at the end of the day. Solution We let t = the stock price at the end of the day t = starting price + 1 st rise – drop + 2 nd rise = $72.37 + $3.25 – $6.32 + $4.12 = $73.42 The value of the stock at the end of the day was $73.42

67. Slide 1- 67 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.3 1. Add:  6 + 12 a)  6 b) 0 c) 6 d) 72

68. Slide 1- 68 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.3 1. Add:  6 + 12 a)  6 b) 0 c) 6 d) 72

69. Slide 1- 69 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.3 2. Add:  6 + 12 + (  5) + 3 a)  8 b) 4 c) 14 d) 26

70. Slide 1- 70 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.3 2. Add:  6 + 12 + (  5) + 3 a)  8 b) 4 c) 14 d) 26

71. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SUBTRACTION of REAL NUMBERS Subtract real numbers and simplify combinations of additions and subtractions. Solve applied problems involving addition and subtraction of integers. 1.4a b

72. Slide 1- 72 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Subtract real numbers and simplify combinations of additions and subtractions.a

73. Slide 1- 73 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtraction a  b The difference a  b is the number c for which a = b + c.

74. Slide 1- 74 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Subtract 4  9. Solution Think: 4  9 is the number that when added to 9 gives 4. What number can we add to 9 to get 4? The number must be negative. The number is  5: 4 – 9 = –5. That is, 4  9 =  5 because 9 + (  5) = 4.

75. Slide 1- 75 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtracting by Adding the Opposite For any real numbers a and b, a – b = a + (– b). (To subtract, add the opposite, or additive inverse, of the number being subtracted.)

76. Slide 1- 76 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Subtract. 1.  15  (  25)2.  13  40 Solution 1.  15  (  25) =  15 + 25 Adding the opposite of  25 = 10 2.  13  40 =  13 + (  40) Adding the opposite of 40 =  53

77. Slide 1- 77 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Subtract. 1. 3 – 7 = 2. –5 – 93. –4 – (–10) Solution 1. 3 – 7 = 3 + (–7) = –4 2. –5 – 9 = –5 + (– 9) = –14 3. –4 – (–10) = –4 + 10 = 6 The opposite of 7 is –7. We change the subtraction to addition and add the opposite. Instead of subtracting 7, we add –7.

78. Slide 1- 78 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Simplify:  4  (  6)  10 + 5  (  7). Solution  4  (  6)  10 + 5  (  7) =  4 + 6 + (  10) + 5 + 7 =  4 + (  10) + 6 + 5 + 7 =  14 + 18 = 4 Adding opposites Using a commutative law

79. Slide 1- 79 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve applied problems involving addition and subtraction of integers.b

80. Slide 1- 80 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E The Johnson’s were taking a vacation and one day they drove from mile marker 54 to mile marker 376. How far did they drive? Solution 376 – 54 = 376 + (  54) = 322 miles Adding the opposite of 54.

81. Slide 1- 81 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.4 1. Subtract: 4 – (  11) a)  15 b)  7 c) 7 d) 15

82. Slide 1- 82 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.4 1. Subtract: 4 – (  11) a)  15 b)  7 c) 7 d) 15

83. Slide 1- 83 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.4 2. Subtract: 3.1 – 9.6 a)  6.5 b)  5.5 c) 12.7 d) 29.76

84. Slide 1- 84 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.4 2. Subtract: 3.1 – 9.6 a)  6.5 b)  5.5 c) 12.7 d) 29.76

85. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley MULTIPLICATION of REAL NUMBERS Multiply real numbers. Solve applied problems involving multiplication of real numbers. 1.5a b

86. Slide 1- 86 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Multiply real numbers.a

87. Slide 1- 87 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiplication of real numbers is like multiplication of arithmetic numbers. The difference is that we must determine whether the answer is positive or negative. The Product of a Positive and a Negative Number To multiply a positive number and a negative number, multiply their absolute values. The answer is negative.

88. Slide 1- 88 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Multiply. 1. (7)(  9)2. 40(  1)3.  3  7 Solution 1. (7)(  9) = 2. 40(  1) = 3.  3  7 =  63  40  21

89. Slide 1- 89 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Product of Two Negative Numbers To multiply two negative numbers, multiply their absolute values. The answer is positive.

90. Slide 1- 90 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Multiply. 1. (  3)(  4)2. (  11)(  5)3. (  2)(  1) Solution 1. (  3)(  4) = 2. (  11)(  5)= 3. (  2)(  1) = 12 55 2

91. Slide 1- 91 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To multiply two nonzero real numbers: a) Multiply the absolute values. b) If the signs are the same, the answer is positive. c) If the signs are different, the answer is negative.

92. Slide 1- 92 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Multiplication Property of Zero For any real number a, a  0 = 0  a. (The product of 0 and any real number is 0.)

93. Slide 1- 93 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Multiply. 1.  9  3(  4) 2.  6  (  3)  (  4)  (  7) Solution 1.  9  3(  4) =  27(  4) = 108 2.  6  (  3)  (  4)  (  7) = 18  28 = 504 Multiplying the first two numbers Multiplying the results Each pair of negatives gives a positive product.

94. Slide 1- 94 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The product of an even number of negative numbers is positive. The product of an odd number of negative numbers is negative.

95. Slide 1- 95 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Evaluate 3x 2 when x = 4 and x =  4. Solution 3x 2 = 3(4) 2 = 3(16) = 48 3x 2 = 3(  4) 2 = 3(16) = 48

96. Slide 1- 96 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Evaluate (  x) 2 and  x 2 when x = 6. Solution (  x) 2 = (  6) 2 = (  6)(  6) = 36  x 2 =  (6) 2 =  36

97. Slide 1- 97 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve applied problems involving multiplication of real numbers.b

98. Slide 1- 98 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F The temperature in a chemical compound was 4  at 1:00. During a reaction, it increased in temperature 2  per minute until 1:12. What was the temperature at 1:12? Solution Number of minutes temp. rises is 12. (1:12 – 1:00) Rises 2  per minute = 2(12) = 24  4  + 24  = 28  The temperature was 28  at 1:12.

99. Slide 1- 99 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.5 1. Multiply: 6  (  12) a)  72 b)  2 c) 2 d)

100. Slide 1- 100 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.5 1. Multiply: 6  (  12) a)  72 b)  2 c) 2 d)

101. Slide 1- 101 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.5 2. Multiply: a) b) c) d)

102. Slide 1- 102 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.5 2. Multiply: a) b) c) d)

103. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley DIVISION of REAL NUMBERS Divide integers. Find the reciprocal of a real number. Divide real numbers. Solve applied problems involving multiplication and division of real numbers. 1.6a b c d

104. Slide 1- 104 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Divide integers.a

105. Slide 1- 105 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Division The quotient a  b or, where b  0, is that unique real number c for which a = b  c.

106. Slide 1- 106 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Divide, if possible. Check each answer. 1. 15  (  3)2. Solution 1. 15  (  3) =  5 2. Think: What number multiplied by –3 gives 15? The number is –5. Check: (–3)(–5) = 15. Think: What number multiplied by –5 gives 45? The number is –9. Check: (–5)(–9) = 45.

107. Slide 1- 107 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To multiply or divide two real numbers (where the divisor is nonzero): a) Multiply or divide the absolute values. b) If the signs are the same, the answer is positive. c) If the signs are different, the answer is negative. Excluding Division by 0 Division by zero is not defined: a  0, or is not defined for all real numbers a.

108. Slide 1- 108 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dividends of 0 Zero divided by any nonzero real number is 0:

109. Slide 1- 109 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Divide, if possible:  72  0. Solution is undefined. Think: What number multiplied by 0 gives  72? There is no such number because anything times 0 is 0.

110. Slide 1- 110 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the reciprocal of a real number.b

111. Slide 1- 111 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Reciprocals Two numbers whose product is 1 are called reciprocals, or multiplicative inverses, of each other.

112. Slide 1- 112 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Find the reciprocal. 1. 2.  63.4. Solution 1. The reciprocal of is 2. The reciprocal of  6 is 3. The reciprocal of is 4. The reciprocal of is

113. Slide 1- 113 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Properties of Reciprocals For a  0, the reciprocal of a can be named and the reciprocal of is a. The reciprocal of any nonzero real number can be named The number 0 has no reciprocal. The Sign of a Reciprocal The reciprocal of a number has the same sign as the number itself.

114. Slide 1- 114 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Number Opposite (Change the sign.) Reciprocal (Invert but do not change the sign.) 25  25  8.5 8.5 00Undefined

115. Slide 1- 115 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Divide real numbers.c

116. Slide 1- 116 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Reciprocals and Division For any real numbers a and b, b  0, (To divide, multiply by the reciprocal of the divisor.)

117. Slide 1- 117 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Rewrite the division as multiplication. 1.  9  52. Solution 1.  9  5 is the same as 2.

118. Slide 1- 118 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Divide by multiplying by the reciprocal of the divisor. Solution Multiply by the reciprocal of the divisor Factoring and identifying a common factor Removing a factor equal to 1

119. Slide 1- 119 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Divide.  18.6  (3) Solution  18.6  (3) = Do the long division. The answer is negative.

120. Slide 1- 120 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve applied problems involving division of real numbers.d

121. Slide 1- 121 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G After diving 110 m below sea level, a diver rises at a rate of 8 meters per minutes for 6 minutes. Where is the diver in relation to the surface? Solution We first determine by how many meters the diver rose altogether. 8 meters  6 = 48 meters The diver was 110 below sea level and rose 48 meters.  110 + 48 =  62 meters 110 m

122. Slide 1- 122 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.6 1. Divide: a) b) c) d)

123. Slide 1- 123 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.6 1. Divide: a) b) c) d)

124. Slide 1- 124 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.6 2. Divide:  4.985  (  0.5) a) 2.4925 b)  2.4925 c) 9.97 d)  9.97

125. Slide 1- 125 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.6 2. Divide:  4.985  (  0.5) a) 2.4925 b)  2.4925 c) 9.97 d)  9.97

126. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROPERTIES OF REAL NUMBERS Find equivalent fraction expressions and simplify fraction expressions. Use the commutative and associative laws to find equivalent expressions. Use the distributive laws to multiply expressions like 8 and x – y. Use the distributive law to factor expressions like 4 x – 12 + 24 y. Collect like terms. 1.7a b c d e

127. Slide 1- 127 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find equivalent fraction expressions and simplify fraction expressions. a

128. Slide 1- 128 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equivalent Expressions Two expressions that have the same value for all allowable replacements are called equivalent. The Identity Property of 0 For any real number a a + 0 = 0 + a = a (The number 0 is the additive identity.)

129. Slide 1- 129 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Identity Property of 1 For any real number a a  1 = 1  a = a (The number 1 is the multiplicative identity.)

130. Slide 1- 130 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Simplify: Solution Look for the largest factor common to both the numerator and the denominator and factor each. Factoring the fraction expression. 8x/8x = 1 Removing a factor of 1 using the identity property of 1

131. Slide 1- 131 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Use the commutative and associative laws to find equivalent expressions.b

132. Slide 1- 132 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Evaluate x + y and y + x when x = 7 and y = 8. Solution We substitute 7 for x and 8 for y. x + y = 7 + 8 = 15 y + x = 8 + 7 = 15

133. Slide 1- 133 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Evaluate xy and yx when x = 7 and y = 8. Solution We substitute 7 for x and 8 for y. xy = 7(8) = 56 yx = 8(7) = 56

134. Slide 1- 134 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Commutative Laws Addition: For any numbers a, and b, a + b = b + a. (We can change the order when adding without affecting the answer.) Multiplication. For any numbers a and b, ab = ba (We can change the order when multiplying without affecting the answer.)

135. Slide 1- 135 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Calculate and compare: 4 + (9 + 6) and (4 + 9) + 6. Solution 4 + (9 + 6) = 4 + 15 = 19 (4 + 9) + 6 = 13 + 6 = 19

136. Slide 1- 136 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Associative Laws Addition: For any numbers a, b, and c, a + (b + c) = (a + b) + c. (Numbers can be grouped in any manner for addition.) Multiplication. For any numbers a, b, and c, a  (b  c) = (a  b)  c (Numbers can be grouped in any manner for multiplication.)

137. Slide 1- 137 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Use the distributive law to multiply expressions like 8 and x – y.c

138. Slide 1- 138 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Distributive Law of Multiplication over Addition For any numbers a, b, and c, a(b + c) = ab + ac. The Distributive Law of Multiplication over Subtraction For any numbers a, b, and c, a(b  c) = ab  ac.

139. Slide 1- 139 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Multiply. 4(a + b). Solution 4(a + b) = 4  a + 4  b = 4a + 4b Using the distributive law of multiplication over addition.

140. Slide 1- 140 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Use the distributive law to write an expression equivalent to each of the following: 1. 8(a – b) 2. (b – 7)c 3. –5(x – 3y + 2z) Solution 1. 8(a – b) = 8a – 8b 2. (b – 7)c = c(b – 7) = c  b – c  7 = cb – 7c

141. Slide 1- 141 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 3. –5(x – 3y + 2z) = –5  x – (–5  3)y + (–5  2)z = –5x – (–15)y + (–10)z = –5x + 15y – 10z

142. Slide 1- 142 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Use the distributive laws to factor expressions like 4x – 12 + 24y.d

143. Slide 1- 143 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factoring is the reverse of multiplying. To factor, we can use the distributive laws in reverse: ab + ac = a(b + c) and ab – ac = a(b – c). Factor To factor an expression is to find an equivalent expression that is a product.

144. Slide 1- 144 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Factor. a.6x – 12 b. 8x + 32y – 8 Solution a. 6x – 12 = 6  x – 6  2 = 6(x – 2) b. 8x + 32y – 8 = 8  x + 8  4y – 8  1 = 8(x + 4y – 1)

145. Slide 1- 145 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H Factor. Try to write just the answer, if you can. a. 7x – 7y b. 14z – 12x – 20 Solution a. 7x – 7y = b. 14z – 12x – 20 = 7(x – y) 2(7z – 6x – 10)

146. Slide 1- 146 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Collect like terms.e

147. Slide 1- 147 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A term is a number, a variable, a product of numbers and/or variables, or a quotient of two numbers and/or variables. Terms are separated by addition signs. If there are subtraction signs, we can find an equivalent expression that uses addition signs. The process of collecting like terms is based on the distributive laws.

148. Slide 1- 148 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Like Terms Terms in which the variable factors are exactly the same, such as 9x and –5x, are called like, or similar terms. Like TermsUnlike Terms 7x and 8x8y and 9y 2 3xy and 9xy5ab and 4ab 2

149. Slide 1- 149 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example I Combine like terms. Try to write just the answer. 1. 8x + 2x 2. 3x – 6x 3. 3a + 5b + 2 + a – 8 – 5b Solution 1. 8x + 2x = (8 + 2) x = 10x 3. 3a + 5b + 2 + a – 8 – 5b = 3a + 5b + 2 + a + (–8) + (–5b) = 3a + a + 5b + (–5b) + 2 + (–8) = 4a + (–6) = 4a – 6 2.

150. Slide 1- 150 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.7 1. Multiply: –3(  x – 7y + 2). a) –3x – 21y + 6 b) 3x + 21y – 6 c) 3x – 21y + 6 d) 3x + 21y + 6

151. Slide 1- 151 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.7 1. Multiply: –3(  x – 7y + 2). a) –3x – 21y + 6 b) 3x + 21y – 6 c) 3x – 21y + 6 d) 3x + 21y + 6

152. Slide 1- 152 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.7 2. Collect like terms: 14a – b – 30a – 7b. a) –16a – 8b b) 44a – 6b c) –16a + 6b d) 44a + 8b

153. Slide 1- 153 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.7 2. Collect like terms: 14a – b – 30a – 7b. a) –16a – 8b b) 44a – 6b c) –16a + 6b d) 44a + 8b

154. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SIMPLIFYING EXPRESSION; ORDER OF OPERATIONS Find an equivalent expression for an opposite without parentheses, where an expression has several terms. Simplify expressions by removing parentheses and collecting like terms. Simplify expressions with parentheses inside parentheses. Simplify expressions using rules from order of operations. 1.8a b c d

155. Slide 1- 155 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find an equivalent expression for an opposite without parentheses, where an expression has several terms.a

156. Slide 1- 156 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Property of  1 For any real number a,  1  a =  a (Negative one times a is the opposite, or additive inverse, of a.)

157. Slide 1- 157 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Find an equivalent expression without parentheses.  (4x + 5y + 2) Solution  (4x + 5y + 2) =  1(4x + 5y + 2) =  1(4x) +  1(5y) +  1(2) =  4x – 5y – 2 Using the property of  1 Using a distributive law Using the property of  1

158. Slide 1- 158 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Find an equivalent expression without parentheses.  (  2x + 7y  6) Solution  (  2x + 7y  6) = 2x – 7y + 6 Changing the sign of each term

159. Slide 1- 159 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Simplify expressions by removing parentheses and collecting like terms.b

160. Slide 1- 160 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Remove parentheses and simplify. (8x + 5y – 3)  (4x – 2y  6) Solution (8x + 5y – 3)  (4x – 2y  6) = 8x + 5y – 3 – 4x + 2y + 6 = 4x + 7y + 3

161. Slide 1- 161 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Remove parentheses and simplify. (3a + 4b – 8) – 3(–6a – 7b + 14) Solution (3a + 4b – 8) – 3(–6a – 7b + 14) = 3a + 4b – 8 + 18a + 21b – 42 = 21a + 25b – 50

162. Slide 1- 162 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Simplify expressions with parentheses inside parentheses.c

163. Slide 1- 163 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When more than one kind of grouping symbol occurs, do the computations in the innermost ones first. Then work from the inside out.

164. Slide 1- 164 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Simplify. a. 5(3 + 4) – {8 – [5 – (9 + 6)]} b. [6(x + 3) – 4x] – [4(y + 3) – 8(y – 4)] Solution a. 5(3 + 4) – {8 – [5 – (9 + 6)]} = 5(7) – {8 – [5 – 15]} = 35 – {8 – [ –10]} Computing 5(7) and 5 – 15 = 35 – 18 Computing 8 – [–10] = 17

165. Slide 1- 165 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E continued b. [6(x + 3) – 4x] – [4(y + 3) – 8(y – 4)] = [6x + 18 – 4x] – [4y + 12 – 8y + 32] = [2x + 18] – [  4y + 44] Collecting like terms within brackets = 2x + 18 + 4y – 44 Removing brackets = 2x + 4y – 26 Collecting like terms

166. Slide 1- 166 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Simplify expressions using the rules for order of operationsd

167. Slide 1- 167 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rules for Order of Operations 1. Do all calculations within grouping symbols before operations outside. 2. Evaluate all exponential expressions. 3. Do all multiplications and divisions in order from left to right. 4. Do all additions and subtractions in order from left to right.

168. Slide 1- 168 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Simplify. 1. 2. Solution 1. 2.

169. Slide 1- 169 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Simplify: Solution

170. Slide 1- 170 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.8 1. Simplify: 3(2a + 4b) – a – 6 a) 5a + 4b – 6 b) 5a + 12b – 6 c) 9ab – 6 d) 12b

171. Slide 1- 171 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.8 1. Simplify: 3(2a + 4b) – a – 6 a) 5a + 4b – 6 b) 5a + 12b – 6 c) 9ab – 6 d) 12b

172. Slide 1- 172 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.8 2. Simplify:  3(12) – [5(  4) – 2 3 ] a) 8 b)  8 c) –24 d)  64

173. Slide 1- 173 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 1.8 2. Simplify:  3(12) – [5(  4) – 2 3 ] a) 8 b)  8 c) –24 d)  64