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Get ready for different kinds of
information
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Math is a language…a foreign language
So you need to just plain memorize some of the
terms and concepts
I suggest keeping a stack of note cards next to you
and write down definition words so you can flash
test yourself as the weeks go on!
For this first night, use the supplied blank sheets of
paper to note what you will put to cards this week.
It’s not easy but…
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It IS very systematic. You can do the same thing
step by step over and over and get the right answer
every time.
Expect to stretch your brain!
The learning will go well IF you erase your learned
fears of math. Kids can get it --- so can YOU!!!
Practice!!!
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So my suggestion is you do EVERY problem in
the book (well Chapter 1,2,3,5,8.1 and 8.2 for this
part, part I).
I’ll supply the worked answers for ALL nonassigned problems
If you just don’t have time for all the problems…at
least copy the worked problems onto a fresh piece
of paper! Pass it through your brain!
Due for this week…
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Homework 1 (on MyMathLab – via the Materials
Link)  Sunday night at 6pm.
Read Chapter 2
Do the MyMathLab introductions and Self-Check
for week 1.
Learning team toolkit and team charter.
Read about the Week 5 presentation and begin
discussion how your team want’s to do this (what
topics and math points).
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 5
1.1
Numbers, Variables, and
Expressions
Natural Numbers and Whole Numbers
Prime Numbers and Composite Numbers
Variables, Algebraic Expressions, and
Equations
Translating Words to Expressions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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, …
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7
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 = 42
The 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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8
Prime Factorization
The prime factorization of 120.
120  2  2  2  3  5
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9
EXAMPLE
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
Solution
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
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 10
EXAMPLE
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
Solution
d. 300
Prime factorization
300
30
6
2
10
5
3
2
5
300  2  2  3  5  5
** Try exercises 15-46
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 11
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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 12
EXAMPLE
Evaluating algebraic expressions with
one variable
Evaluate each algebraic expression for x = 6.
x
a. x + 4
b. 4x
c. 20 – x
d.
( x  4)
Solution
a. x + 4
b. 4x
6 + 4 = 10
c. 20 – x
20 – 6 = 14
4(6) = 24
d.
x
6
6

 3
( x  4) (6  4) 2
** Try exercises 47-56
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 13
EXAMPLE
Evaluating algebraic expressions with
two variables
Evaluate each algebraic expression for y = 3 and z = 9
z
a. 5yz
b. z – y
c.
y
Solution
b. z – y
a. 5yz
5(3)(9) = 135
9–3=6
z
9
c.
 3
y
3
** Try exercises 57-62
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 14
EXAMPLE
Evaluating formulas
Find the value of y for x = 20 and z = 5.
a. y = x + 4
b. y = 9xz
Solution
a. y = x + 4
b. y = 9xz
y = 20 + 4
= 24
y = 9(20)(5)
= 900
** Try exercises 63-74
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 15
Translating Words to Expressions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 16
EXAMPLE
Translating words to expressions
Translate each phrase to an algebraic expression.
a. Twice the cost of a book
b. Ten less than a number
c. The product of 8 and a number
Solution
a. Twice the cost of a book
2c where c is the cost of the book
b. Ten less than a number
n – 10 where n is the number
c. The product of 8 and a number
8n where n is the number
** Try exercises 75-90
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 17
EXAMPLE
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.
Solution
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
** Try exercises 91-106
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 18
1.2
Fractions
Basic Concepts
Simplifying Fractions to Lowest Terms
Multiplication and Division of Fractions
Addition and Subtraction of Fractions
Applications
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Basic Concepts
The parts of a fraction are named as follows.
Numerator
Denominator
7
8
Fraction bar
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 20
EXAMPLE
Identifying numerators and denominators
Give the numerator and denominator of each fraction.
mn
b.
p
8
a.
19
cd
c.
f 7
Solution
a. The numerator is 8 and the denominator is 19.
b. The numerator is mn, and the denominator is p.
c. The numerator is c + d, and the denominator is f – 7.
** Try exercises 1-20
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 21
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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 22
EXAMPLE
Finding the greatest common factor
Find the greatest common factor (GCF) for each pair
of numbers.
a. 14, 21
b. 42, 90
Solution
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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 23
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.
** Try exercises 21-30
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 24
EXAMPLE
Simplifying fractions to lowest terms
Simplify each fraction to lowest terms.
a. 9
b. 20
c. 45
15
28
135
Solution
a. The GCF of 9 and 15 is 3.
9
33
3


15 3  5
5
20 4  5 5


b. The GCF of 20 and 28 is 4.
28 4  7 7
45
45 1 1


c.The GCF of 45 and 135 is 45.
135
45  3 3
** Try exercises 31-42
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 25
Multiplication of Fractions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 26
EXAMPLE
Multiplying fractions
Multiply each expression and simplify the result when
appropriate.
3 4
a. 
7 9
3
b. 16 
4
m 5
c. 
n r
Solution
a. 3  4  3  4  12  4  3  4
7 9
b. 16 
7 9
63
21  3
21
16 3
16  3
48 12  4
3
 



 12
1 4
1 4
4
4
4
m 5
5m
m 5


c. 
nr
nr
n r
** Try exercises 43-58
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 27
EXAMPLE
Finding fractional parts
Find each fractional part.
a. One-third of one-fourth
b. One half of three-fourths
Solution
a. 1  1  1 1  1
3 4
b.
1 3

2 4

3 4
12
1 3
24

3
8
** Try exercises 59-64
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 28
Division of Fractions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 29
EXAMPLE
Dividing fractions
Divide each expression.
1 2
a. 
5 3
d f
c. 
6 g
9
b. 27 
2
Solution
a. 1  2  1  3  1  3  3
5
3
5 2
52
10
9 27 2
27  2
54
69


b. 27    
6
2
1 9
1 9
9
1 9
d f
c. 
6 g

d g
dg


6 f
6 f

dg
6f
** Try exercises 69-86
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 30
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 31
Fractions with Like Denominators
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 32
EXAMPLE
Adding and subtracting fractions with common
denominators
Add or subtract as indicated. Simplify your answer to
lowest terms when appropriate.
a.
7 2

11 11
b.
17 11

18 18
Solution
a.
7 2
72


11 11
11
b.
17 11
17  11


18 18
18
9

11
6

18
6
1
The fraction
can be simplified to .
18
3
** Try exercises 87-92
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 33
Fractions with Unlike Denominators
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 34
EXAMPLE
Rewriting fractions with the LCD
Rewrite each set of fractions using the LCD.
a. 2 , 3
b. 1 , 4 , 9
8 5 10
3 8
Solution
a. The LCD is 24
2 8 2  8 16
 

3 8 3  8 24
3 3 33 9
 

8 3 8  3 24
1 5 1 5 5

b. The LCD is 40.  
8 5 8  5 40
4 8 4  8 32
 

5 8 5  8 40
9 4 9  4 36
 

10 4 10  4 40
** Try exercises 105-112
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 35
EXAMPLE
Adding and subtracting fractions with unlike
denominators
Add or subtract as indicated. Simplify your answer to
lowest terms when appropriate.
a.
5 1

6 9
b.
4 1

5 2
Solution
a.
17
5 3 1 2
15 2

  
 
18
6 3 9 2
18 18
b.
4 2 1 5
  
5 2 2 5
8 5
3
 

10 10 10
** Try exercises 113-128
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 36
EXAMPLE
Applying fractions to carpentry
A pipe measures 36
3
inches long and needs to be
8
cut into three equal pieces. Find the length of each
piece.
Solution
Begin by writing 36
291
3
as the improper fraction
.
8
8
1
291
291 1 291
, or 12 inches
3 
 
8
8
8 3 24
** Try exercises 129-140
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 37
1.3
Exponents and Order of Operations
Natural Number Exponents
Order of Operations
Translating Words to Expressions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Exponent
5  5 = 52 = 25 square inches
Base
The expression 52 is an exponential expression with
base 5 and exponent 2.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 39
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 40
EXAMPLE
Writing products in exponential notation
Write each product as an exponential expression.
a.
88888  8
b.
2 2 2 2  2 4
    
3 3 3 3 3
c.
y  y  y  y  y  y  y6
5
** Try exercises 17-26
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 41
EXAMPLE
Evaluating exponential notation
Evaluate each expression.
a.
54  5  5  5  5  625
3
b.
2 2 2 2 8
   3  3  3  27
3
** Try exercises 27-36
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 42
Order of Operations
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 43
How to remember them!
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
Please excuse my dear Aunt Sally
Parenthesis Exponents
Multiplication Division
Addition Subtraction
EXAMPLE
Evaluating arithmetic expressions
Evaluate each expression by hand.
a. 12 – 6 – 2
b. 12 – (6 – 2)
8
c.
33
Solution
a. 12 – 6 – 2
b. 12 – (6 – 2)
6–2
12 – 4
4
8
8
c.
33
8 4

6 3
** Try exercises 49-64
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 45
EXAMPLE
Evaluating arithmetic expressions
Evaluate each expression.
a. 15  2  3
b. 3  4  2  (8  1)
4  32
c.
82
Solution
b. 3  4  2  (8  1)
3 4 2  7
387
a. 15  2  3
15 – 6
9
11  7
4
2
4

3
c.
82
49
82
13
6
** Try exercises 65-72
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 46
EXAMPLE
Writing and evaluating expressions
Write each expression and then evaluate it.
a. Two to the fifth power plus three
b. Twenty-four less two times four
Solution
a. Two to the fifth power plus three
25  3  2  2  2  2  2  3  32  3  35
b. Twenty-four less two times four
24  2  4  24  8  16
** Try exercises 73-82
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 47
1.4
Real Numbers and the Number Line
Signed Numbers
Integers and Rational Numbers
Square Roots
Real and Irrational Numbers
The Number Line
Absolute Value
Inequality
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Signed Numbers
The opposite, or additive inverse, of a number a is
−a.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 49
EXAMPLE
Finding opposites (or additive inverses)
Find the opposite of each expression.
a. 29
9
b. 
11
6
c. 8 
2
d. −(−13)
Solution
a. The opposite of 29 is −29.
9
9
b. The opposite of 
is .
11
11
6
6
c. 8   8  3  5, so the opposite of 8  is  5.
2
2
d. −(−13) = 13, so the opposite of −(−13) is −13.
** Try exercises 27-34
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 50
EXAMPLE
Finding an additive inverse (or opposite)
4
Find the additive inverse of –x, if x =  .
9
Solution
4

The additive inverse of −x is x =
because −(−x) = x
9
by the double negative rule.
** Try exercises 35-38
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 51
Integers and Rational Numbers
The integers include the natural numbers, zero, and
the opposite of the natural numbers.
…,−2, −1, 0, 1, 2,…
A rational number is any number
that can be expressed
p
as the ratio of two integers, q where q ≠ 0.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 52
EXAMPLE
Classifying numbers
Classify each number as one or more of the following:
natural number, whole number, integer, or rational
number.
a.
21
3
b. −9
c. 
15
7
Solution
21
21

7
a. Because
, the number 3 is a natural
3
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.
15

c. The fraction 7 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.
** Try exercises 51-56
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 53
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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 54
EXAMPLE
Calculating principal square roots
Evaluate each square root. Approximate to three
decimal places when appropriate.
a.
64
b.
169
c.
23
Solution
a. 64  8 because 8 ∙ 8 = 64 and 8 is nonnegative.
b.
169  13 because 13 ∙ 13 = 169 and 13 is
nonnegative.
c. 23 is a number between 4 and 5. We can estimate
the value of 23 with a calculator. 23  4.796
** Try exercises 7
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 55
Real and Irrational Numbers
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 56
A special note on pi –
a famous irrational number

pi = 3.141592653589…
pi and other special numbers are ALSO found on the
number line, but they are called irrational. Why?
You can’t make them using two integers divided by
one another. They don’t repeat or end…
Irrational Numbers

EXAMPLE
Classifying numbers
Identify the natural numbers, whole numbers, integers,
rational numbers, and irrational numbers in the
following list.
5.7, 4,
17
, 25,  7, and  23
9
Solution
Natural numbers: 4 and 25  5
Whole numbers: 4 and 25  5
Integers: 4, 25  5, and  23
17
Rational numbers: 5.7, 4, , 25  5, and  23
9
Irrational numbers:  7
** Try exercises 57-68
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 59
EXAMPLE
Plotting numbers on a number line
Plot each real number on a number line.
5
a. 
2
b.
7
7
c.
2
Solution
5
a.   2.5 Plot a dot halfway between −2 and −3.
2
b.
c.
7  2.65 Plot a dot between 2 and 3.
7
 3.5 Plot a dot halfway between 3 and 4.
2
** Try exercises 69-78
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 60
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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 61
EXAMPLE
Finding the absolute value of a real number
Write the expression without the absolute value sign.
a. 9
b. 0
c. 16
d.  y
Solution
a. 9  9 because the distance between the origin
and −9 is 9.
b. 0  0 because the distance is 0 between the origin
and 0.
c. 16  16
d.  y  y
** Try exercises 79-88
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 62
EXAMPLE
Ordering real numbers
List the following numbers from least to greatest. Then
plot these numbers on a number line.
4,  , 3, and 2.4
Solution
4,
3, 2.4, and 
** Try exercises 101-106
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 63
1.5
Addition and Subtraction of Real
Numbers
Addition of Real Numbers
Subtraction of Real Numbers
Applications
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
There are four arithmetic operations: addition,
subtraction, multiplication, and division.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 65
Addition of Real Numbers
In an addition problem the two numbers added are
called addends, and the answer is called the sum.
5 + 8 = 13
5 and 8 are the addends
13 is the sum
The opposite (or additive inverse) of a real number a
is a.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 66
EXAMPLE
Adding Opposites
Find the opposite of each number and calculate the
sum of the number and its opposite.
a. 78
3
b.  4
The opposite is  78.
Sum = 78  ( 78)  0
3
The opposite is .
4
3 3
Sum:    0
4 4
** Try exercises 13-18
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 67
Addition of Real Numbers
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 68
EXAMPLE
Adding real numbers
Evaluate each expression.
a. 3  ( 8)  11
3 9
b.  
4 10
3
15
 
4
20
9 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
** Try 35-52
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 69
Subtraction of Real Numbers
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 70
EXAMPLE
Subtracting real numbers
Find each difference by hand.
a. 12 – 16
b. –6 – 2
Solution
a. 12 – 16
b. –6 – 2
12 + (–16)
4
–6 + (–2)
8
** Try exercises 53-68
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 71
EXAMPLE
Adding and subtracting real numbers
Evaluate each expression.
a. 6  7  (8)  2
b. 6.3  5.8 10.4
Solution
a. 6  7  (8)  2
6  (7)  8  2
1  8  2
9
b. 6.3  5.8 10.4
6.3  5.8  (10.4)
0.5  (10.4)
10.9
** Try exercises 69-82
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 72
EXAMPLE
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
Solution
Find the sum of the five numbers.
326  (20)  (15)  200  (150)
306  200  (15)  (150)
506  ( 165)
341
The final balance is $341.
** Try exercises 93-98
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 73
1.6
Multiplication and Division of Real Numbers
Multiplication of Real Numbers
Division of Real Numbers
Applications
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Multiplication of Real Numbers
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 75
EXAMPLE
Multiplying real numbers
Find each product by hand.
a. −4 ∙ 8
4 3
b. 
9 7
c. 3.4  60 d.  4.5 6 5 3
Solution
a. The resulting product is negative because the
factors have unlike signs. Thus −4 ∙ 8 = −32.
b. The product is positive because both factors are
positive. 4  3  12  4
9 7 63 21
c. Since both factors are negative, the product is
positive. 3.4  60  204
d.  4.5 6 5 3   27  5 3  135 3  405
** Try exercises 19-40
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 76
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 77
EXAMPLE
Evaluating real numbers with exponents
Evaluate each expression by hand.
a. (−6)2
b. −62
Solution
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.
b. This is the negation of an exponential expression with
base 6. Evaluating the exponent before negative results
in −62 = −(6)(6) = −36.
** Try exercises 43-50
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 78
Division of Real Numbers
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 79
EXAMPLE
Dividing real numbers
Evaluate each expression by hand.
1
a. 24 
3
b.
2
5
6
6
c.
52
d. 4  0
Solution
1 24 3 72
 
 72
a. 24  
3
1 1
1
2
2  1
2
1
b.
   6          
6 5
5  6
30
15
2
5
6
3
 1  6
 6     

c.
52
 52  52 26
d. 4 ÷ 0 is undefined. The number 0 has no reciprocal.
** Try exercises 51-76
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 80
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 81
EXAMPLE
Converting fractions to decimals
Convert the measurement to a decimal number.
5
2 -inch washer
16
Solution
a. Begin by dividing 5 by 16.
0.3125
16 5.0000
− 48
20
− 16
40
− 32
80
−80
0
5
Thus the mixed number 2
16
is equivalent to 2.3125.
** Try exercises 77-88
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 82
EXAMPLE
Converting decimals to fractions
Convert each decimal number to a fraction in lowest
terms.
a. 0.32
b. 0.875
Solution
a. The decimal 0.32 equals thirty-two hundredths.
32
8 4
8


100 25  4 25
b. The decimal 0.875 equals eight hundred seventy-five
thousandths.
875 7 125 7


1000 8 125 8
** Try exercises 89-96
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 83
EXAMPLE
Application
After surveying 125 pediatricians, 92 stated that they
had admitted a patient to the children’s hospital in the
92
last month for pneumonia. Write the fraction
as a
125
decimal.
Solution
One method for writing the fraction as a decimal is to
divide 92 by 125 using long division. An alternative
8
method is to multiply the fractions by , so the
8
denominator becomes 1000. Then, write the
numerator in the thousandths place in the decimal.
92 8 736
 
 0.736
125 8 1000
** Try exercises 105-108
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 84
1.7
Properties of Real Numbers
Commutative Properties
Associative Properties
Distributive Properties
Identity and Inverse Properties
Mental Calculations
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Commutative Properties
The 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
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 86
EXAMPLE
Applying the commutative properties
Use the commutative properties to rewrite each
expression.
a. 72  56 can be written as 56  72
b. b 4
can be written as 4  b
c. d (e  g )
d (e  g )  (e  g ) d
 ( g  e) d
** Try exercises 15-22
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 87
Associative Properties
The associative property allows us to change how
numbers are grouped.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 88
EXAMPLE
Applying the associative properties
Use the associative property to rewrite each
expression.
a. (7  8)  9  7  (8  9)
b. a (bc )  (ab)c
** Try exercises 23-30
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 89
EXAMPLE
Identifying properties of real numbers
State the property that each equation illustrates.
a. 7  (4 w)  (7  4) w
Associative property of multiplication
because the grouping of the numbers
has been changed.
b. 8  3  3  8
Commutative property for addition
because the order of the numbers has
changed.
** Try exercises 55-58
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 90
Distributive Properties
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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 91
EXAMPLE
Applying the distributive properties
Apply a distributive property to each expression.
a. 4(x + 3)
b. –8(b – 5)
c. 12 – (a + 2)
Solution
c. 12 – (a + 2)
a. 4(x + 3)
=4x+43
= 4x + 12
b. –8(b – 5)
= 8  b  (8)  5
= 8b + 40
= 12 + (1)(a + 2)
= 12 + (1)  a + (–1)  2
= 12  a – 2
= 10  a
** Try exercises 33-44
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 92
EXAMPLE
Inserting parentheses using the distributive
property
Use the distributive property to insert parentheses in
the expression and then simplify the result.
a. 8a  4a
b. 5 y  9 y
Solution
a. 8a  4a
(8  4)a
b. 5 y  9 y
(5  9) y
4y
12a
** Try exercises 47-54
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 93
EXAMPLE
Identifying properties of real numbers
State the property or properties illustrated by each
equation.
Distributive property .
a. 3(8  y )  24  3 y
b. (7  w)  8  w  15
(7  w)  8  w  (7  8)
Commutative and associative
properties for addition.
 w  15
** Try exercises 58-68
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 94
Identity and Inverse Properties
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
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 95
Identity and Inverse Properties
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 96
EXAMPLE
Identifying identity and inverse properties
State the property or properties illustrated by each
equation.
a. 0  ab  ab
Identity property for 0.
b. a  (a)  3  0  3  3
Additive inverse property and the
identity property for 0.
** Try exercises 69-78
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 97
EXAMPLE
Performing calculations mentally
Use the properties of real numbers to calculate each
expression mentally.
a. 32  16  8  4  32  16  8  4
 (32  8)  (16  4)
 40  20  60
b.
1 3
4
 4
4 4
3
1 3
4
  4
4 4
3
1  3 4
   4  
4  4 3
 1 1  1
** Try exercises 79-96
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 98
1.8
Simplifying and Writing Algebraic
Expressions
Terms
Combining Like Terms
Simplifying Expressions
Writing Expressions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Terms
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.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 100
EXAMPLE
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
Solution
a. A number is a term. The coefficient is 97.
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.
** Try exercises 11-22
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 101
EXAMPLE
Identifying like terms
Determine whether the terms are like or unlike.
a. 9x, −15x
b. 16y2, 1
c. 5a3, 5b3
d. 11, −8z
Solution
a. The variable in both terms is x, with the same
power of 1, so they are like terms.
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.
** Try exercises 23-36
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 102
EXAMPLE
Combining like terms
Combine terms in each expression, if possible.
a. −2y + 7y
b. 4x2 – 6x
Solution
a. Combine terms by applying the distributive
property.
−2y + 7y = (−2 + 7)y = 5y
b. They are unlike terms, so they can not be combined.
** Try exercises 39-56
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 103
EXAMPLE
Simplifying expressions
Simplify each expression.
a. 13 + z – 9 + 7z
b. 9x – 2(x – 5)
Solution
a. 13 + z – 9 + 7z
b.
9x – 2(x – 5)
= 13 +(– 9) + z + 7z
= 9x + (– 2)x + (−2)(– 5)
= 13 +(– 9) + (1+ 7)z
= 9x – 2x + 10
= 4 + 8z
= 7x + 10
** Try exercises 57-68
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 104
EXAMPLE
Simplifying expressions
Simplify each expression.
a. 6x2 – y + 9x2 – 3y
b.
Solution
18a  6
3
= (6 + 9)x2 + (–1+ (– 3))y
18a  6
3
18a 6


3
3
= 15x2 –4y
 6a  2
a. 6x2 – y + 9x2 – 3y
= 6x2 + 9x2 + (–1y) + (–3y)
b.
** Try exercises 69-92
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 105
EXAMPLE
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.
11 ft
Solution
a.
b.
11w + 4w + 18w
= (11 + 4 + 18)w
= 33w
33w = 33 ∙ 3 = 99 square feet
w
4 ft
18 ft
** Try exercises 99-102
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 106
End of week 2






You again have the answers to those problems not
assigned
Practice is SOOO important in this course.
Work as much as you can with MyMathLab, the
materials in the text, and on my Webpage.
Do everything you can scrape time up for, first the
hardest topics then the easiest.
You are building a skill like typing, skiing, playing a
game, solving puzzles.
NEXT TIME: Linear Equations and Inequalities