copyright © 2009 pearson education, inc. publishing as pearson addison-wesley numbers, variables,...
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Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Numbers, Variables, and Expressions
Natural Numbers and Whole Numbers
Prime Numbers and Composite Numbers
Variables, Algebraic Expressions, and Equations
Translating Words to Expressions
1.1
Natural Numbers and Whole Numbers
The set of natural numbers are also known as the counting numbers.
1, 2, 3, 4, 5, 6,…Because there are infinitely many natural numbers, three dots are used to show that the list continues in the same pattern without end.The whole numbers can be expressed as
0, 1, 2, 3, 4, 5, …
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Prime Numbers and Composite Numbers
When two natural numbers are multiplied, the result is another natural number.The product of 6 and 7 is 42.
6 7 = 42The numbers 6 and 7 are factors of 42.A prime number has only itself and 1 as factors.A natural number greater than 1 that is not prime is a composite number.Any composite number can be written as a product of prime numbers.
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Prime Factorization
The prime factorization of 120.
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120 2 2 2 3 5
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EXAMPLE
Solution
Classifying numbers as prime or composite
Classify each number as prime or composite. If a number is composite, write it as a product of prime numbers.a. 37 b. 3 c. 45 d. 300
a. 37 The only factors of 37 are 1 and itself. The number is prime.
b. 3 The only factors of 3 are 1 and itself. The number is prime.
c. 45 Composite because 9 and 5 are factors. Prime factorization: 32 5
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EXAMPLE
Solution
Classifying numbers as prime or composite
Classify each number as prime or composite. If a number is composite, write it as a product of prime numbers.a. 37 b. 3 c. 45 d. 300
d. 300 Prime factorization
300
30 10
6 5 2 5
2 3 300 2 2 3 5 5
Variables, Algebraic Expressions, and Equations
Variables are often used in mathematics when tables of numbers are inadequate. A variable is a symbol, typically an italic letter used to represent an unknown quantity.An algebraic expression consists of numbers, variables, operation symbols, such as +, , , and , and grouping symbols, such as parentheses.An equation is a mathematical statement that two algebraic expressions are equal.A formula is a special type of equation that expresses a relationship between two or more quantities.
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Slide 9Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Evaluating algebraic expressions with one variable
Evaluate each algebraic expression for x = 6.a. x + 4 b. 4x c. 20 – x d.
a. x + 4
( 4)x
x
6 + 4 = 10
b. 4x
4(6) = 24
c. 20 – x
20 – 6 = 14 d.
( 4)x
x 4
6
(6 )
6
32
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EXAMPLE Evaluating algebraic expressions with two variables
Solution
Evaluate each algebraic expression for y = 3 and z = 9a. 5yz b. z – y c.
a. 5yz
5(3)(9) = 135
b. z – y
9 – 3 = 6
z
y
c. z
y3
9
3
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EXAMPLE
Solution
Evaluating formulas
Find the value of y for x = 20 and z = 5. a. y = x + 4 b. y = 9xz
a. y = x + 4
y = 20 + 4
b. y = 9xz
y = 9(20)(5)
= 24 = 900
Translating Words to Expressions
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Slide 13Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Translating words to expressions
Translate each phrase to an algebraic expression.a. Twice the cost of a bookb. Ten less than a numberc. The product of 8 and a number
a. Twice the cost of a book
b. Ten less than a number
c. The product of 8 and a number
2c where c is the cost of the book
n – 10 where n is the number
8n where n is the number
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EXAMPLE
Solution
Finding the area of a rectangle
The area A of a rectangle equals its length L times its width W.a. Write a formula that shows the relationship between these three quantities.b. Find the area of a yard that is 100 feet long and 75 feet wide.
a. The word times indicates the length and width should be multiplied. The formula is A = LW.
b. A = LW
= (100)(75)
= 7500 square feet
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Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Fractions
Basic Concepts
Simplifying Fractions to Lowest Terms
Multiplication and Division of Fractions
Addition and Subtraction of Fractions
Applications
1.2
Basic Concepts
The parts of a fraction are named as follows.
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7
8Fraction bar
Numerator
Denominator
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EXAMPLE
Solution
Identifying numerators and denominators
Give the numerator and denominator of each fraction.
a. b. c.
a. The numerator is 8 and the denominator is 19.
b. The numerator is mn, and the denominator is p.
8
19
mn
p 7
c d
f
c. The numerator is c + d, and the denominator is f – 7.
Simplifying Fractions to Lowest Terms
When simplifying fractions, we usually factor out the greatest common factor (GCF) for the numerator and the denominator. The greatest common factor is the largest factor common to both the numerator and the denominator.
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Slide 20Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Finding the greatest common factor
Find the greatest common factor (GCF) for each pair of numbers.
a. 14, 21 b. 42, 90
a. Because 14 = 7 ∙ 2 and 21 = 7 ∙ 3, the number 7 is the largest factor that is common to both 14 and 21. Thus the GCF of 14 and 21 is 7.
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EXAMPLE continued
b. When working with larger numbers, one way to determine the greatest common factor is to find the prime factorization of each number.
42 = 6 ∙ 7 = 2 ∙ 3 ∙ 7 and
90 = 6 ∙ 15 = 2 ∙ 3 ∙ 3 ∙ 5
The prime factorizations have one 2 and one 3 in common. Thus the GCF for 42 and 90 is 6.
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EXAMPLE
Solution
Simplifying fractions to lowest terms
Simplify each fraction to lowest terms.a. b. c.
a. The GCF of 9 and 15 is 3.
9
15
20
2845
135
9
15 5
3
3
3
3
5
b. 20
28 7
4
4
5
The GCF of 20 and 28 is 4.
5
7
c. The GCF of 45 and 135 is 45. 45
135
145
5 34
1
3
Multiplication of Fractions
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Slide 24Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Multiplying fractions
Multiply each expression and simplify the result when appropriate.
a. b. c.
a.
3 4
7 9
316
4
5m
n r
b.
c.
3 4
7 9
3 4
7 9
12
63
3
4
1
3
2
4
21
316
4
16 3
1 4
16 3
1 4
48
4
4
12 4 12
5m
n r
5m
n r
5m
nr
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EXAMPLE
Solution
Finding fractional parts
Find each fractional part.
a. One-third of one-fourthb. One half of three-fourths
a.
b.
1 1
3 4
1 1
3 4
1
12
1 3
2 4 1 3
2 4
3
8
Division of Fractions
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Slide 27Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Dividing fractions
Divide each expression.
a. b. c.
a.
1 2
5 3
927
2
6
d f
g
b.
c.
27 2
1 9
27 2
1 9
54
9
9
6
1
9
6
6
d g
f
1 2
5 3
1 3
5 2
1 3
5 2
3
10
927
2
6
d f
g
6
d g
f
6
dg
f
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Fractions with Like Denominators
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EXAMPLE
Solution
Adding and subtracting fractions with common denominators
Add or subtract as indicated. Simplify your answer to lowest terms when appropriate.
a. b.
a.
7 2
11 11
17 11
18 18
b.
7
11 1
2
1
1
7
1
2
1
9
1
8
7
18
1 11
1
1
17
8
11
8
6
1
6 1The fraction can be simplified to .
18 3
Fractions with Unlike Denominators
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Slide 32Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Rewriting fractions with the LCD
Rewrite each set of fractions using the LCD.a. b.
a. The LCD is 24
2 3,
3 8
1 4 9, ,
8 5 10
82 8 2 16
3 8 3 248
b. The LCD is 40.
33 3 3 9
8 3 8 3 24
51 5 1 5
8 5 8 5 40
84 8 4 32
5 8 5 408
49 4 9 36
10 4 10 404
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EXAMPLE
Solution
Adding and subtracting fractions with unlike denominators
Add or subtract as indicated. Simplify your answer to lowest terms when appropriate.
a. b.
a.
5 1
6 9
4 1
5 2
b.
3 2
3 2
5 1
6 9
15 2
18 18
17
18
2 5
2 5
4 1
5 2
8 5
10 10
3
10
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EXAMPLE
Solution
Applying fractions to carpentry
A pipe measures inches long and needs to be cut into three equal pieces. Find the length of each piece.
336
8
Begin by writing as the improper fraction 3
368
291.
8
2913
8
291 1
8 3
291 1, or 12 inches
24 8
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Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Exponents and Order of Operations
Natural Number Exponents
Order of Operations
Translating Words to Expressions
1.3
Natural Number Exponents
The area of a square equals the length of one of its sides times itself. If the square is 5 inches on a side, then its area is
5 5 = 52 = 25 square inches
The expression 52 is an exponential expression with base 5 and exponent 2.
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Base
Exponent
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EXAMPLE Writing products in exponential notation
Write each product as an exponential expression.
a.
b.
c.
8 8 8 8 8 58
2 2 2 2
3 3 3 3
y y y y y y
42
3
6y
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EXAMPLE Evaluating exponential notation
Evaluate each expression.
a.
b.
45 5 5 5 5 625
32
3
2 2 2 8
3 3 3 27
Order of Operations
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Slide 42Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Evaluating arithmetic expressions
Evaluate each expression by hand.a. 12 – 6 – 2 b. 12 – (6 – 2) c.
a. 12 – 6 – 2
8
3 3
6 – 2
4
b. 12 – (6 – 2)
12 – 4
8
c. 8
3 38 4
6 3
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EXAMPLE
Solution
Evaluating arithmetic expressions
Evaluate each expression.a. b. c.
a.
24 3
8 2
15 – 6
9
b. c.
15 2 3 3 4 2 (8 1)
15 2 3 3 4 2 )8 1( 3 4 2 7
3 8 7
11 7
4
24
2
3
8
4 9
8 2
13
6
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EXAMPLE
Solution
Writing and evaluating expressions
Write each expression and then evaluate it. a. Two to the fifth power plus threeb. Twenty-four less two times four
a. Two to the fifth power plus three
b. Twenty-four less two times four
52 3 2 2 2 2 2 3 32 3 35
24 2 4 24 8 16
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Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Real Numbers and the Number Line
Signed Numbers
Integers and Rational Numbers
Square Roots
Real and Irrational Numbers
The Number Line
Absolute Value
Inequality
1.4
Signed Numbers
The opposite, or additive inverse, of a number a is −a.
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Slide 48Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Finding opposites (or additive inverses)
Find the opposite of each expression.
a. 29 b. c. d. −(−13)
a. The opposite of 29 is −29.
b. The opposite of is
c.
9
11
68
2
9
11
9.
116 6
8 8 3 5, so the opposite of 8 is 5.2 2
d. −(−13) = 13, so the opposite of −(−13) is −13.
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EXAMPLE
Solution
Finding an additive inverse (or opposite)
Find the additive inverse of –x, if x = .
The additive inverse of −x is x = because −(−x) = x by the double negative rule.
4
9
4
9
Integers and Rational Numbers
The integers include the natural numbers, zero, and the opposite of the natural numbers.
…,−2, −1, 0, 1, 2,…
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A rational number is any number that can be expressed as the ratio of two integers, where q ≠ 0.
p
q
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EXAMPLE
Solution
Classifying numbers
Classify each number as one or more of the following: natural number, whole number, integer, or rational number.
a. b. −9 c.
a. Because , the number is a natural number, whole number, integer, and rational number.
b. The number −9 is an integer and rational number, but not a natural number or a whole number.
c. The fraction is a rational number because it is the ratio of two integers. However it is not a natural number, a whole number, or an integer.
21
3
15
7
217
3
21
3
15
7
Square Roots
Square roots are frequently used in algebra. The number b is a square root of a number a if b ∙ b = a. Every positive number has one positive square root and one negative square root.
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Slide 53Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Calculating principal square roots
Evaluate each square root. Approximate to three decimal places when appropriate.
a. b. c.
a. because 8 ∙ 8 = 64 and 8 is nonnegative.
b. because 13 ∙ 13 = 169 and 13 is nonnegative.
c. is a number between 4 and 5. We can estimate the value of with a calculator.
64 169 23
64 8
169 13
2323 23 4.796
Real and Irrational Numbers
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Slide 55Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Classifying numbers
Identify the natural numbers, whole numbers, integers, rational numbers, and irrational numbers in the following list.
Natural numbers:
175.7,4, , 25, 7, and 23
9
4 and 25 5Whole numbers: 4 and 25 5
Integers:4, 25 5,and 23
Rational numbers:17
5.7,4, , 25 5, and 239
Irrational numbers: 7
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EXAMPLE
Solution
Plotting numbers on a number line
Plot each real number on a number line.
a. b. c.
a. Plot a dot halfway between −2 and −3.
b. Plot a dot between 2 and 3.
c. Plot a dot halfway between 3 and 4.
5
2 7 7
2
52.5
2
7 2.65
73.5
2
Absolute Value
The absolute value of a real number equals its distance on the number line from the origin. Because distance is never negative, the absolute value of a real number is never negative.
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Slide 58Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Finding the absolute value of a real number
Write the expression without the absolute value sign.
a. b. c. d.
a. because the distance between the origin and −9 is 9.
b. because the distance is 0 between the origin and 0.
c.
9 0 16 y
9 9
0 0
16 16
d. y y
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EXAMPLE
Solution
Ordering real numbers
List the following numbers from least to greatest. Then plot these numbers on a number line.
4, , 3, and 2.4
4, 3, 2.4, and
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Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Addition and Subtraction of Real Numbers
Addition of Real Numbers
Subtraction of Real Numbers
Applications
1.5
There are four arithmetic operations: addition, subtraction, multiplication, and division.
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Slide 63Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Addition of Real Numbers
In an addition problem the two numbers added are called addends, and the answer is called the sum. 5 + 8 = 135 and 8 are the addends13 is the sumThe opposite (or additive inverse) of a real number a is a.
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EXAMPLE Adding Opposites
Find the opposite of each number and calculate the sum of the number and its opposite.
a.
b.
78 The opposite is 78.
Sum = 78 ( 78) 0
3
4 3
The opposite is .4
3 3Sum: 0
4 4
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Addition of Real Numbers
Slide 66Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Adding real numbers
Evaluate each expression.
a.
b.
3 ( 8) 11
3 9
4 10
3 15
4 209 18
10 20
The numbers are both negative, add the absolute values. The sign would be negative as well.
15 18 3
20 20 20
Subtraction of Real Numbers
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Slide 68Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Subtracting real numbers
Find each difference by hand.a. 12 – 16 b. –6 – 2
a. 12 – 16
12 + (–16)
4
b. –6 – 2
–6 + (–2)
8
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EXAMPLE
Solution
Adding and subtracting real numbers
Evaluate each expression.a. b.
a. b.
6 7 ( 8) 2 6.3 5.8 10.4
6 7 ( 8) 2 ( 7) 286
1 8 2
9
6.3 5.8 10.4
6.3 5.8 ( 10.4)
0.5 ( 10.4)
10.9
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EXAMPLE
Solution
Balancing a checking account
The initial balance in a checking account is $326. Find the final balance if the following represents a list of withdrawals and deposits:$20, $15, $200, and $150
Find the sum of the five numbers.
326 ( 20) ( 15) 200 ( 150)
306 200 ( 15) ( 150)
506 ( 165) 341
The final balance is $341.
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Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Multiplication and Division of Real Numbers
Multiplication of Real Numbers
Division of Real Numbers
Applications
1.6
Multiplication of Real Numbers
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Slide 74Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Multiplying real numbers
Find each product by hand.
a. −4 ∙ 8 b. c. d.
b. The product is positive because both factors are positive.
c. Since both factors are negative, the product is positive.
4 3
9 7 3.4 60
d.
4.5 6 5 3
a. The resulting product is negative because the factors have unlike signs. Thus −4 ∙ 8 = −32.
4 3 12 4
9 7 63 21
3.4 60 204
4.5 6 5 3 27 5 3 135 3 405
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Slide 76Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Evaluating real numbers with exponents
Evaluate each expression by hand.
a. (−6)2 b. −62
b. This is the negation of an exponential expression with base 6. Evaluating the exponent before negative results in −62 = −(6)(6) = −36.
a. Because the exponent is outside of parentheses, the base of the exponential expression is −6. The expression is evaluated as (−6)2 = (−6)(−6) = 36.
Division of Real Numbers
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Slide 78Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Dividing real numbers
Evaluate each expression by hand.
a. b. c. d.
b.
c.
124
3 6
52
d. 4 ÷ 0 is undefined. The number 0 has no reciprocal.
4 0
a.
25 2 2 1 2 1
66 5 5 6 30 15
25
6
1 24 3 7224 72
3 1 1 1
6 1 6 36
52 52 52 26
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Slide 80Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Converting fractions to decimals
Convert the measurement to a decimal number.
a. Begin by dividing 5 by 16.
52 -inch washer
16
16 5.0000− 48
20− 16
40− 32
80−80
0
0.3125
Thus the mixed number
is equivalent to 2.3125.
52
16
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EXAMPLE
Solution
Converting decimals to fractions
Convert each decimal number to a fraction in lowest terms.
a. 0.32 b. 0.875
b. The decimal 0.875 equals eight hundred seventy-five thousandths.
a. The decimal 0.32 equals thirty-two hundredths. 32 8 4 8
100 25 4 25
875 7 125 7
1000 8 125 8
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EXAMPLE
Solution
Application
After surveying 125 pediatricians, 92 stated that they had admitted a patient to the children’s hospital in the last month for pneumonia. Write the fraction as a decimal.
92
125
92 8 7360.736
125 8 1000
One method for writing the fraction as a decimal is to divide 92 by 125 using long division. An alternative method is to multiply the fractions by so the denominator becomes 1000. Then, write the numerator in the thousandths place in the decimal.
8,
8
Slide 83Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Properties of Real Numbers
Commutative Properties
Associative Properties
Distributive Properties
Identity and Inverse Properties
Mental Calculations
1.7
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Commutative PropertiesThe commutative property for addition states that two numbers, a and b, can be added in any order and the result will be the same.
6 + 8 = 8 + 6
The commutative property for multiplication states that two numbers, a and b, can be multiplied in any order and the result will be the same.
9 4 = 4 9
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EXAMPLE Applying the commutative properties
Use the commutative properties to rewrite each expression.
a.
b.
c.
72 56 can be written as 56 72
4b can be written as 4 b
( )d e g ( ) ( )
( )
d d
g d
e g e
e
g
The associative property allows us to change how numbers are grouped.
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Associative Properties
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EXAMPLE Applying the associative properties
Use the associative property to rewrite each expression.
a.
b.
(7 8) 9 7 (8 9)
( )a bc ( )ab c
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EXAMPLE Identifying properties of real numbers
State the property that each equation illustrates.
a.
b.
7 (4 ) (7 4)w w
8 3 3 8
Associative property of multiplication because the grouping of the numbers has been changed.
Commutative property for addition because the order of the numbers has changed.
Distributive Properties
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The distributive properties are used frequently in algebra to simplify expressions.
7(3 + 8) = 7 3 + 7 8
The 7 must be multiplied by both the 3 and the 8.
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EXAMPLE
Solution
Applying the distributive properties
Apply a distributive property to each expression.a. 4(x + 3) b. –8(b – 5) c. 12 – (a + 2)
a. 4(x + 3)
= 4 x + 4 3
= 4x + 12
b. –8(b – 5)
c. 12 – (a + 2)
= 12 + (1)(a + 2)
= 8 b (8) 5
= 8b + 40
= 12 + (1) a + (–1) 2= 12 a – 2
= 10 a
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EXAMPLE
Solution
Inserting parentheses using the distributive property
Use the distributive property to insert parentheses in the expression and then simplify the result.a. b.
a. b.
8 4a a 5 9y y
8 4a a(8 4)a
12a
5 9y y
( 5 9)y
4y
Slide 93Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Identifying properties of real numbers
State the property or properties illustrated by each equation.a.
b.
3(8 ) 24 3y y
(7 ) 8 15w w
Distributive property .
Commutative and associative properties for addition. (7 ) 8 (7 8)
15
w w
w
The identity property of 0 states that if 0 is added to any real number a, the result is a. The number 0 is called the additive identity.
3 + 0 = 3
The identity property of 1 states that if any number a is multiplied by 1, the result is a. The number 1 is called the multiplicative identity.
4 1 = 4
Slide 94Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Identity and Inverse Properties
Slide 95Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Identity and Inverse Properties
Slide 96Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Identifying identity and inverse properties
State the property or properties illustrated by each equation.a.
b.
0 ab ab
( ) 3 0 3 3a a
Identity property for 0.
Additive inverse property and the identity property for 0.
Slide 97Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Performing calculations mentally
Use the properties of real numbers to calculate each expression mentally.a.
b.
32 16 8 4
1 3 44
4 4 3
1632 8 4 32 8( )4) 6(1
40 20 60
3 4
4 3
14
4
14
4
3 4
4 3
1 1 1
Slide 98Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Simplifying and Writing Algebraic Expressions
Terms
Combining Like Terms
Simplifying Expressions
Writing Expressions
1.8
Terms
Slide 100Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
A term is a number, a variable, or a product of numbers and variables raised to powers. Examples of terms include
4, z, 5x, and −6xy2.
The coefficient of a term is the number that appears in the term.
Slide 101Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Identifying terms
Determine whether each expression is a term. If it is a term, identify its coefficient.
a. 97 b. 17x c. 4a – 6b d. 9y2
b. The product of a number and a variable is a term. The coefficient is 17.
c. The difference of two terms in not a term.
d. The product of a number and a variable with an exponent is a term. Its coefficient is 9.
a. A number is a term. The coefficient is 97.
Slide 102Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Identifying like terms
Determine whether the terms are like or unlike.
a. 9x, −15x b. 16y2, 1 c. 5a3, 5b3 d. 11, −8z
b. The term 1 has no variable and the 16 has a variable of y2. They are unlike terms.
c. The variables are different, so they are unlike terms.
d. The term 11 has no variable and the −8 has a variable of z. They are unlike terms.
a. The variable in both terms is x, with the same power of 1, so they are like terms.
Slide 103Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Combining like terms
Combine terms in each expression, if possible.
a. −2y + 7y b. 4x2 – 6x
b. They are unlike terms, so they can not be combined.
a. Combine terms by applying the distributive property. −2y + 7y = (−2 + 7)y = 5y
Slide 104Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Simplifying expressions
Simplify each expression.
a. 13 + z – 9 + 7z b. 9x – 2(x – 5)
b. a. 13 + z – 9 + 7z
= 13 +(– 9) + z + 7z
= 13 +(– 9) + (1+ 7)z
= 4 + 8z
9x – 2(x – 5)
= 9x + (– 2)x + (−2)(– 5)
= 9x – 2x + 10
= 7x + 10
Slide 105Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Simplifying expressions
Simplify each expression.
a. 6x2 – y + 9x2 – 3y b.
b. a. 6x2 – y + 9x2 – 3y
= 6x2 + 9x2 + (–1y) + (–3y)
= (6 + 9)x2 + (–1+ (– 3))y
= 15x2 –4y
18 6
3
a
18 6
3
a
18 6
3 3
a
6 2a
Slide 106Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Writing and simplifying an expression
A sidewalk has a constant width w and comprises several short sections with lengths 11, 4, and 18 feet.
a. Write and simplify an expression that gives the number of square feet of sidewalk.
b. Find the area of the sidewalk if its width is 3 feet.
a. 11w + 4w + 18w = (11 + 4 + 18)w = 33w
b. 33w = 33 ∙ 3 = 99 square feet
11 ft
4 ft
18 ft
w