SlideShare a Scribd company logo
1 of 198
Copyright © 2011 Pearson Education, Inc.
CHAPTER

1

1.1
1.2
1.3
1.4
1.5
1.6
1.7

Foundations of Algebra
Number Sets and the Structure of Algebra
Fractions
Adding and Subtracting Real Numbers;
Properties of Real Numbers
Multiplying and Dividing Real Numbers;
Properties of Real Numbers
Exponents, Roots, and Order of Operations
Translating Word Phrases to Expressions
Evaluating and Rewriting Expressions

Copyright © 2011 Pearson Education, Inc.
1.1
1.
2.
3.
4.
5.

Number Sets and the Structure of
Algebra

Understand the structure of algebra.
Classify number sets.
Graph rational numbers on a number line.
Determine the absolute value of a number.
Compare numbers.

Copyright © 2011 Pearson Education, Inc.
Objective 1
Understand the structure of algebra.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 4
Definitions
Variable: A symbol that can vary in value.
Constant: A symbol that does not vary in value.
Variables are usually letters of the alphabet, like x or y.
Usually constants are symbols for numbers, like 1, 2,
¾, 6.74.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 5
Expression: A constant, variable, or any combination
of constants, variables, and arithmetic operations that
describe a calculation.
Examples of expressions:
2+6

or 4x − 5

or

1 2
πr h
3

Copyright © 2011 Pearson Education, Inc.

Slide 1- 6
Equation: A mathematical relationship that contains
an equal sign.
Examples of equations:
2+6=8

or 4x − 5 = 12

Copyright © 2011 Pearson Education, Inc.

or

1 2
V = πr h
3

Slide 1- 7
Inequality: A mathematical relationship that contains
an inequality symbol (≠, <, >, ≤, or ≥).
Symbolic form

Translation

8≠3

Eight is not equal to three.

5<7

Five is less than seven.

7>5

Seven is greater than five.

x≤3

x is less than or equal to three.

y≥2

y is greater than or equal to two.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 8
Objective 2
Classify number sets.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 9
Set: A collection of objects.
Braces are used to indicate a set. For example, the set
containing the numbers 1, 2, 3, and 4 would be written
{1, 2, 3, 4}.
The numbers 1, 2, 3, and 4 are called the members or
elements of this set.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 10
Writing Sets
To write a set, write the members or elements of the
set separated by commas within braces, { }.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 11
Example 1
Write the set containing the first four days of the week.
Answer
{Sunday, Monday, Tuesday, Wednesday}

Copyright © 2011 Pearson Education, Inc.

Slide 1- 12
Numbers are classified using number sets.
Natural numbers contain the counting numbers 1, 2, 3,
4, …and is written {1, 2, 3, …}. The three dots are
called ellipsis and indicate that the numbers continue
forever in the same pattern.
Whole numbers: natural numbers and 0 {0, 1, 2, 3,…}
Integers: whole numbers and the opposite (or negative)
of every natural number {…, −3, −2, −1, 0, 1, 2, 3…}
Rational: every real number that can be expressed as a
ratio of integers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 13
Rational number: Any real number that can be
a
expressed in the form , where a and b are integers
b
and b ≠ 0.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 14
Example 2
Determine whether the given number is a rational
number.
5
a.
b. 0.8
c. 0.3
6

Answer
5
a.
6

Yes, because 5 and
6 are integers.

b. 0.8

c.

Yes, 0.8 can be
expressed as a
fraction 8 over 10,
and 8 and 10 are
integers.

The bar indicates
that the digit
repeats. This is the
decimal equivalent
of 1 over 3. Yes this
is a rational number.

Copyright © 2011 Pearson Education, Inc.

0.3

Slide 1- 15
Irrational number: Any real number that is not
rational.
Examples:
2,

3,

π

Real numbers: The union of the rational and
irrational numbers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 16
Objective 3
Graph rational numbers on a number
line.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 17
Example 3
4
Graph on a number line. 2
5

Answer
The number is located 4/5 of the way between 2 and 3.
-1

0

1

2

2

4 3
5

Between 2 and 3, we divide the number line
into 5 equally spaced divisions. Place a dot on
the 4th mark.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 18
Objective 4
Determine the absolute value of a
number.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 19
Absolute value: A given number’s distance from 0 on
a number line.
←5 units from 0 → ←5 units from 0 →

The absolute value of a number n is written |n|.
The absolute value
The absolute value
of 5 is 5.
of −5 is 5.
|5| = 5
|−5| = 5

Copyright © 2011 Pearson Education, Inc.

Slide 1- 20
Absolute Value
The absolute value of every real number is either
positive or 0.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 21
Example 4
Simplify.
a. |−9.4|

2
b.
9

Answer
a. |−9.4| = 9.4
2
2
b.
=
9
9

Copyright © 2011 Pearson Education, Inc.

Slide 1- 22
Objective 5
Compare numbers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 23
Comparing Numbers
For any two real numbers a and b, a is greater than
b if a is to the right of b on a number line.
Equivalently, b is less than a if b is to the left of a
on a number line.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 24
Example 5
Use =, <, or > to write a true statement.
a. 3 ___ −3
b. −1.8 ___ −1.6
Answer
a. 3 ___ −3
3 > −3
because 3 is farther
right on a number line.

b. −1.8 ___ −1.6
−1.8 < −1.6
because –1.8 is further
to the left on a number
line.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 25
To which set of numbers does −6
belong?
a) Irrational
b) Natural and whole numbers
c) Natural numbers, whole numbers,
and integers
d) Integers and rational numbers
1.1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 26
To which set of numbers does −6
belong?
a) Irrational
b) Natural and whole numbers
c) Natural numbers, whole numbers,
and integers rational numbers
d) Integers and
1.1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 27
Simplify |7|.

a) 7
b) −7
c) 0
d) 1/7

1.1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 28
Simplify |7|.

a) 7
b) −7
c) 0
d) 1/7

1.1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 29
Which statement is false?

a) 7 > 4
b) −2.4 > −1.4
c) 10 < 22
d) −3.6 > −6.4

1.1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 30
Which statement is false?

a) 7 > 4
b) −2.4 > −1.4
c) 10 < 22
d) −3.6 > −6.4

1.1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 31
1.2
1.
2.
3.
4.

Fractions

Write equivalent fractions.
Write equivalent fractions with the LCD.
Write the prime factorization of a number.
Simplify a fraction to lowest terms.

Copyright © 2011 Pearson Education, Inc.
Fraction: A quotient of two numbers or expressions a
a
and b having the form , where b ≠ 0.
b

3
4

← Numerator
← Denominator

The top number in a fraction is called the numerator.
The bottom number is called the denominator.
Fractions indicated part of a whole.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 33
Objective 1
Write equivalent fractions.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 34
Writing Equivalent Fractions
For any fraction, we can write an equivalent
fraction by multiplying or dividing both its
numerator and denominator by the same nonzero
number.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 35
Example 1
Find the missing number that makes the fractions
equivalent.
a. 9 = ?
b. − 18 = ?
15

45

Solution
a. 9 ?
15

=

45

9×
3
27
=
15 ×
3 45
Multiply the numerator and
denominator by 3.

36

b.

2

18 ?
− =
36 2
18 ÷ 18
1
−
=−
36 ÷ 18
2

Divide the numerator and
denominator by 6.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 36
Objective 2
Write equivalent fractions with the
LCD.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 37
Multiple: A multiple of a given integer n is the
product of n and an integer.
We can generate multiples of a given number by
multiplying the given number by the integers.
Multiples of 2

Multiples of 3

2× = 2
1
2 ×2 = 4
2× = 6
3
2 ×4 = 8
2 × = 10
5
2 ×6 = 12

3× = 3
1
3 ×2 = 6
3× = 9
3
3 ×4 = 12
3 × = 15
5
3 ×6 = 18

Copyright © 2011 Pearson Education, Inc.

Slide 1- 38
Least common multiple (LCM): The smallest number
that is a multiple of each number in a given set of
numbers.
Least common denominator (LCD): The least
common multiple of the denominators of a given set
of fractions.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 39
Example 2
7
5
Write and as equivalent fractions with the LCD.
8
6

Solution
The LCD of 8 and 6 is 24.
7 7×
3 21
=
=
8 8×
3 24

5 5 ×4 20
=
=
6 6 ×4 24

Copyright © 2011 Pearson Education, Inc.

Slide 1- 40
Objective 3
Write the prime factorization of a
number.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 41
Factors: If a ⋅ b = c, then a and b are factors of c.
Example: 6 ⋅ 7 = 42, 6 and 7 are factors of 42
Prime number: A natural number that has exactly two
different factors, 1 and the number itself.
Example: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,…
Prime factorization: A factorization that contains only
prime factors.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 42
Example 3
Find the prime factorization of 420.
Solution
420 Factor 420 to 10 and 42. (Any
Factor 10 to 2 and 5, which are
primes. Then factor 42 to 6
and 7.
7 is prime and then factor 6
into 2 and 3, which are primes.

[
42
[ ]

6
[ ]

2

]
10
[ ]

7 2
3

two factors will work.)

5

Answer 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 7

Copyright © 2011 Pearson Education, Inc.

Slide 1- 43
Objective 4
Simplify a fraction to lowest terms.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 44
a
Lowest terms: Given a fraction and b ≠ 0, if the
b

only factor common to both a and b is 1, then the
fraction is in lowest terms.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 45
Simplifying a Fraction with the Same Nonzero
Numerator and Denominator
n 1
= = 1, when n ≠ 0.
n 1

Eliminating a Common Factor in a Fraction
an a × a
1
=
= , when b ≠ 0 and n ≠ 0.
bn b × b
1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 46
These rules allow us to write fractions in lowest terms
using prime factorizations. The idea is to replace the
numerator and denominator with their prime
factorizations and then eliminate the prime factors that
are common to both the numerator and denominator.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 47
Simplifying a Fraction to Lowest Terms
To simplify a fraction to lowest terms:
1. Replace the numerator and denominator with their
prime factorizations.
2. Eliminate (divide out) all prime factors common
to the numerator and denominator.
3. Multiply the remaining factors.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 48
Example 4a
Simplify to lowest terms.

30
42

Solution
2× ×
3 5
30
5
=
=
42
2 × ×7
3
7
Replace the numerator and denominator with their prime
factorizations; then eliminate the common prime factors.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 49
Example 4b
220
Simplify to lowest terms.
2380

Solution
2 ×2 × ×
5 11
220
11
=
=
2380 2 ×2 × ×7 ×
5 17
119
Replace the numerator and denominator with their prime
factorizations; then eliminate the common prime factors.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 50
Example 5
At a company, 225 of the 1050 employees have
optional eye insurance coverage as part of their
benefits package. What fraction of the employees have
optional eye insurance coverage?
Solution
225
3× × ×
3 5 5
3
=
=
1050 2 × × × ×7 14
3 5 5

Answer 3 out of 14 employees have optional eye
insurance.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 51
What is the prime factorization of 360?

a) 6 ⋅ 6 ⋅ 5
b) 23 ⋅ 32 ⋅ 5
c) 22 ⋅ 32 ⋅ 5
d) 32 ⋅ 5 ⋅ 7

1.2

Copyright © 2011 Pearson Education, Inc.

Slide 1- 52
What is the prime factorization of 360?

a) 6 ⋅ 6 ⋅ 5
b) 23 ⋅ 32 ⋅ 5
c) 22 ⋅ 32 ⋅ 5
d) 32 ⋅ 5 ⋅ 7

1.2

Copyright © 2011 Pearson Education, Inc.

Slide 1- 53
Simplify to lowest terms:

a)
b)
c)
d)

1.2

112
280

14
35
1
4
2
5
21
23
Copyright © 2011 Pearson Education, Inc.

Slide 1- 54
Simplify to lowest terms:

a)
b)
c)
d)

1.2

112
280

14
35
1
4
2
5
21
23
Copyright © 2011 Pearson Education, Inc.

Slide 1- 55
1.3
1.
2.
3.
4.

Adding and Subtracting Real Numbers;
Properties of Real Numbers

Add integers.
Add rational numbers.
Find the additive inverse of a number.
Subtract rational numbers.

Copyright © 2011 Pearson Education, Inc.
Objective 1
Add integers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 57
Parts of an addition statement: The numbers added
are called addends and the answer is called a sum.
2+3=5

Addends

Sum

Copyright © 2011 Pearson Education, Inc.

Slide 1- 58
Properties
of Addition

Symbolic Form

Word Form

Additive
Identity

a+0=a

The sum of a number and 0
is that number.

Commutative
Property of
Addition

a+b=b+a

Changing the order of
addends does not affect the
sum.

Associative
Property of
Addition

a + (b + c) = (a + b) + c

Changing the grouping of
three or more addends does
not affect the sum.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 59
Example 1
Indicate whether each equation illustrates the additive
identity, commutative property of addition, or the
associative property of addition.
a. (5 + 6) + 3 = 5 + (6 + 3)
Answer Associative property of addition
b. 0 + (−9) = −9
Answer Additive identity
c. (−9 + 6) + 4 = 4 + (−9 + 6)
Answer Commutative property of addition
Copyright © 2011 Pearson Education, Inc.

Slide 1- 60
Adding Numbers with the Same Sign
To add two numbers that have the same sign, add their
absolute values and keep the same sign.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 61
Example 2
Add.
a. 27 + 12

b. –16 + (– 22)

Solution
a. 27 + 12 = 39
b. –16 + (–22) = –38

Copyright © 2011 Pearson Education, Inc.

Slide 1- 62
Adding Numbers with Different Signs
To add two numbers that have different signs, subtract
the smaller absolute value from the greater absolute
value and keep the sign of the number with the greater
absolute value.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 63
Example 3
Add.
a. 35 + (–17)

b. –29 + 7

Solution
a. 35 + (–17) = 18
b. –29 + 7 = –22

Copyright © 2011 Pearson Education, Inc.

Slide 1- 64
Example 3 continued
Add.
c. 15 + (–27)

d. –32 + 6

Solution
c. 15 + (–27) = –12
d. –32 + 6 = –26

Copyright © 2011 Pearson Education, Inc.

Slide 1- 65
Objective 2
Add rational numbers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 66
Adding Fractions with the Same Denominator
To add fractions with the same denominator, add the
numerators and keep the same denominator; then
simplify.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 67
Example 4
Add.
a. 2 + 4
9

b. − 4 +  − 5 

÷
12  12 

9

Solution
a. 2 + 4 = 2 + 4 = 6
9 9
9
9
2 g3 2
=
=
3 g3 3
Replace 6 and 9 with their prime
factorizations, divide out the
common factor, 3, then multiply
the remaining factors.

4  5
b. − +  − ÷
12  12 
−4 + ( −5 )
9
=
=−
12
12
3 g3
3
=−
=−
3 g2 g 2
4
Simplify to lowest terms by dividing
out the common factor, 3.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 68
Example 4 continued
Add.
c. 7 +  − 3 

÷
10  10 
Solution
a. 7 +  − 3  = 7 + (−3) = 4

÷
10
10
10  10 
2 g2 2
=
=
2 g5 5
Simplify to lowest terms by dividing
out the common factor, 2.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 69
Adding Fractions
To add fractions with different denominators:
1. Write each fraction as an equivalent fraction with
the LCD.
2. Add the numerators and keep the LCD.
3. Simplify.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 70
Example 5a
1 1
Add: +
3 4

Solution
1 1
+
3 4

1( 4 ) 1(3)
=
+
3 ( 4 ) 4(3)
4 3
= +
12 12
7
=
12

Write equivalent fractions
with 12 in the denominator.

Add numerators and keep
the common denominator.
Because the addends have
the same sign, we add and
keep the same sign.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 71
Example 5b
5 3
Add: − +
6 4

Solution
5 ( 2 ) 3(3)
5 3
+
− + =−
6 ( 2 ) 4(3)
6 4
10 9
=− +
12 12
−10 + 9
=
12
1
=−
12

Write equivalent fractions
with 12 in the denominator.

Add numerators and keep
the common denominator.
Because the addends have
different signs, we subtract and
keep the sign of the number with
the greater absolute value.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 72
Example 5c
7 9
Add: − +
8 30

Solution
7 9
− +
8 30

7 ( 15 ) 9(4)
=−
+
8 ( 15 ) 30(4)

Write equivalent fractions
with 120 in the denominator.

105 36
Add numerators and keep
=−
+
the common denominator.
120 120
−105 + 36
Reduce to lowest terms.
=
120
−69
3 ×23
23
=
=−
=−
120
2 ×2 ×2 × ×
3 5
40
Copyright © 2011 Pearson Education, Inc.

Slide 1- 73
Example 6
Anna has a balance of $378.45 and incurs a debt of
$85.42. What is Anna’s new balance?
Solution
A debt of $85.42 is −$85.42. Her balance is
378.45 + (– 85.42) = $293.03

Copyright © 2011 Pearson Education, Inc.

Slide 1- 74
Objective 3
Find the additive inverse of a
number.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 75
Additive inverses: Two numbers whose sum is 0.
What happens if we add two numbers that have the
same absolute value but different signs, such as
5 + (–5)? In money terms, this is like making a $5
payment towards a debt of $5. Notice the payment
pays off the debt so that the balance is 0.
5 + (–5) = 0
Because their sum is zero, we say 5 and –5 are
additive inverses, or opposites.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 76
Example 7
Find the additive inverse of the given number.
a. 8
b. –2
c. 0
Answers
a. –8 because 8 + (–8) = 0
b. 2 because – 2 + 2 = 0
c. 0 because 0 + 0 = 0
Copyright © 2011 Pearson Education, Inc.

Slide 1- 77
Example 8
Simplify.
a. – (–5)

b. –|2|

c. –| –9|

Answers
a. – (–5) = 5
b. –|2| = –2
c. –| –9| = –9
Copyright © 2011 Pearson Education, Inc.

Slide 1- 78
Objective 4
Subtract rational numbers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 79
Parts of a subtraction statement:
8–5=3
Difference

Minuend
Subtrahend

Copyright © 2011 Pearson Education, Inc.

Slide 1- 80
Rewriting Subtraction
To write a subtraction statement as an equivalent
addition statement, change the operation symbol
from a minus sign to a plus sign, and change the
subtrahend to its additive inverse.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 81
Example 9a
Subtract
a. –17 – (–5)
Solution
Write the subtraction as an equivalent addition.
–17 – (–5)
Change the operation
from minus to plus.

= –17 + 5
= –12

Change the subtrahend
to its additive inverse.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 82
Example 9b
3 1
Subtract: − −
8 4

Solution
3 1
3 1
− − =− −
8 4
8 4
3  1
= − +− ÷
8  4
3  1(2) 
= − +−
÷
8  4(2) 

Write equivalent fractions with
the common denominator, 8.

3  2
5
= − +− ÷ = −
8  8
8
Copyright © 2011 Pearson Education, Inc.

Slide 1- 83
Example 9c
c. 4.07 – 9.03
Solution
Write the equivalent addition statement.
4.07 – 9.03
= 4.07 + (– 9.03)
= –4.96

Copyright © 2011 Pearson Education, Inc.

Slide 1- 84
Example 10
In an experiment, a mixture begins at a temperature
of 52.6°C. The mixture is then cooled to a
temperature of −29.4°C. Find the difference between
the initial and final temperatures.
Solution
52.6 – (–29.4) = 52.6 + 29.4
= 82
Answer The difference between the initial and final
temperatures is 82°C.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 85
Add –6 + (–9).

a) –15
b) −3
c) 3
d) 15

1.3

Copyright © 2011 Pearson Education, Inc.

Slide 1- 86
Add –6 + (–9).

a) –15
b) −3
c) 3
d) 15

1.3

Copyright © 2011 Pearson Education, Inc.

Slide 1- 87
Subtract 5 – (–8).

a) –13
b) −3
c) 3
d) 13

1.3

Copyright © 2011 Pearson Education, Inc.

Slide 1- 88
Subtract 5 – (–8).

a) –13
b) −3
c) 3
d) 13

1.3

Copyright © 2011 Pearson Education, Inc.

Slide 1- 89
Subtract

a)

16
−
21

b)

1
−
2

c)

−

d)

1.3

3  1
− −  ÷.
7 3

2
21

2
21

Copyright © 2011 Pearson Education, Inc.

Slide 1- 90
Subtract

a)

16
−
21

b)

1
−
2

c)

−

d)

1.3

3  1
− −  ÷.
7 3

2
21

2
21

Copyright © 2011 Pearson Education, Inc.

Slide 1- 91
1.4
1.
2.
3.
4.
5.

Multiplying and Dividing Real Numbers;
Properties of Real Numbers

Multiply integers.
Multiply more than two numbers.
Multiply rational numbers.
Find the multiplicative inverse of a number.
Divide rational numbers.

Copyright © 2011 Pearson Education, Inc.
Objective 1
Multiply integers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 93
In a multiplication statement, factors are
multiplied to equal a product.
2 g 3 =

Factors

6

Product

Copyright © 2011 Pearson Education, Inc.

Slide 1- 94
Properties of
Multiplication
Multiplicative
Property of 0
Multiplicative
Identity

Symbolic Form

Word Form

0 ga = 0

The product of a number
multiplied by 0 is 0.

1 ga = a

The product of a number
multiplied by 1 is the
number.

Commutative
Property of
Multiplication

ab=ba

Changing the order of
factors does not affect the
product.

Associative
Property of
Multiplication

a(bc) = (ab)c

Changing the grouping of
three or more factors does
not affect the product.

Distributive
Property of
Multiplication
over Addition

a(b + c) =ab + ac

A sum multiplied by a
factor is equal to the sum
of that factor multiplied by
each addend.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 95
Example 1
Give the name of the property of multiplication that is
illustrated by each equation.
a. 6(−3) = −3 ⋅ 6
Answer Commutative property of multiplication
b. 3(−9 ⋅ 5) = [3(−9)] ⋅ 5
Answer Associative property of multiplication
c. 4(4 – 2) = 4 ⋅ 4 – 4 ⋅ 2
Answer Distributive property of multiplication over
addition
Copyright © 2011 Pearson Education, Inc.

Slide 1- 96
Multiplying Two Numbers with Different Signs
When multiplying two numbers that have different
signs, the product is negative.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 97
Example 2
Multiply.
a. 7(–4)
Solution
a. 7(–4) = –28
b. (–15)3 = –45

b. (–15)3

Warning: Make sure you see the
difference between 7(–4), which
indicates multiplication, and 7 – 4,
which indicates subtraction.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 98
Multiplying Two Numbers with the Same Sign
When multiplying two numbers that have the same
sign, the product is positive.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 99
Example 3
Multiply.
a. –5(–9)

b. (–6)(–8)

Solution
a. –5(–9) = 45
b. (–6)(–8) = 48

Copyright © 2011 Pearson Education, Inc.

Slide 1- 100
Objective 2
Multiply more than two numbers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 101
Multiplying with Negative Factors
The product of an even number of negative factors
is positive, whereas the product of an odd number
of negative factors is negative.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 102
Example 4
Multiply.
a. (–1)(–3)(–6)(7)
Solution Because there are three negative factors (an
odd number of negative factors), the result is
negative. (–1)(–3)(–6)(7) = –126
b. (–2)(–4)(2)(–5)(–3)
Solution Because there are four negative factors(an
even number of negative factors), the result is
positive. (–2)(–4)(2)(–5)(–3) = 240

Copyright © 2011 Pearson Education, Inc.

Slide 1- 103
Objective 3
Multiply rational numbers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 104
Multiplying Fractions
a c ac
g =
, where b ≠ 0 and d ≠ 0.
b d bd

Copyright © 2011 Pearson Education, Inc.

Slide 1- 105
Example 5a
3 4
Multiply − g  ÷.
5 9

Solution
3 4
3  2 g2 
− g  ÷= − g 
÷
5 9
5  3 g3 
4
=−
15

Divide out the common factor, 3.

Because we are multiplying two
numbers that have different signs,
the product is negative.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 106
Example 5b
6  6  12
Multiply − × − ÷×
15  16  15

Solution
6  6  12
2× 
3
2 ×  2 ×2 ×
3
3
− × − ÷×
=−
× −

÷×
15  16  15
3 ×  2 ×2 ×2 ×2  3 ×
5
5
3
=
25

Copyright © 2011 Pearson Education, Inc.

Divide out the common factors.
Because there are an even
number of negative factors, the
product is positive.

Slide 1- 107
Multiplying Decimal Numbers
To multiply decimal numbers:
1. Multiply as if they were whole numbers.
2. Place the decimal in the product so that it has the
same number of decimal places as the total
number of decimal places in the factors.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 108
Example 6a
Multiply (–7.6)(0.04).
Solution
First, calculate the value and disregard signs for now.
0.04
2 places
7.6
+ 1 place
×
024
+0280
3 places
0.3 0 4
Answer –0.304

When we multiply two numbers with
different signs, the product is negative.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 109
Example 6b
Multiply (−3)(5.2)(1.4)(−6.1).
Solution
First, calculate the value and disregard signs for now.
Multiply from left to right.
(−3)(5.2)(1.4)(−6.1) = (15.6)(1.4)(−6.1)
= 21.84(6.1)
= 133.224
Answer 133.224

15.6 = (3)(5.2)
21.84 = 15.6(1.4)

The product of an even number of negative
factors is positive. The factors have a total of 3
decimal places, so the product has three
decimal places.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 110
Objective 4
Find the multiplicative inverse of a
number.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 111
Multiplicative inverses: Two numbers whose product is 1.
3
2
and 2 are multiplicative inverses
3

because their product is 1.
2 3 6
g = =1
3 2 6

Notice that to write a number’s multiplicative
inverse, we simply invert the numerator and
denominator. Multiplicative inverses are also known
as reciprocals.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 112
Example 7
Find the multiplicative inverse.
a. 2
b. − 1
7

c. −9

8

Answer
7
a. The multiplicative inverse is .
2

b. The multiplicative inverse is −8.
1
c. The multiplicative inverse is − .
9
Copyright © 2011 Pearson Education, Inc.

Slide 1- 113
Objective 5
Divide rational numbers.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 114
Parts of a division statement:

8 ÷ 2 =

Dividend

4

Quotient
Divisor

Copyright © 2011 Pearson Education, Inc.

Slide 1- 115
Dividing Signed Numbers
When dividing two numbers that have the same
sign, the quotient is positive.
When dividing two numbers that have different
signs, the quotient is negative.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 116
Example 8
Divide.
a. 56 ÷ (−8)

b. −72 ÷ ( −6 )

Solution
a. 56 ÷ (−8) = −7

b.−72 ÷ ( −6 ) = 12

Copyright © 2011 Pearson Education, Inc.

Slide 1- 117
Division Involving 0

0 ÷ n = 0 when n ≠ 0.
n ÷ 0 is undefined when n ≠ 0.
0 ÷ 0 is indeterminate.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 118
Dividing Fractions
a c a d
÷ = g , where b ≠ 0, c ≠ 0, and d ≠ 0.
b d b c

Copyright © 2011 Pearson Education, Inc.

Slide 1- 119
Example 9
3 4
Divide − ÷ .
10 5

Solution
3 4
3
5
− ÷ =−
g
10 5
10 4

3
5
=−
g
5 g2 2 g2
3
=−
8

Write an equivalent multiplication.

Divide out the common factor, 5.

Because we are dividing two numbers
that have different signs, the result is
negative.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 120
Dividing Decimal Numbers
To divide decimal numbers, set up a long division
and consider the divisor.
Case 1: If the divisor is an integer, divide as if the
dividend were a whole number and place the
decimal point in the quotient directly above its
position in the dividend.
Case 2: If the divisor is a decimal number,
1. Move the decimal point in the divisor to the
right enough places to make the divisor an
integer.
2. Move the decimal point in the dividend the same
number of places.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 121
Dividing Decimal Numbers continued
3. Divide the divisor into the dividend as if both
numbers were whole numbers. Make sure you
align the digits in the quotient properly.
4. Write the decimal point in the quotient directly
above its new position in the dividend.
In either case, continue the division process until
you get a remainder of 0 or a repeating digit (or
block of digits) in the quotient.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 122
Example 10
Divide −44.64 ÷ (−3.6)
Solution
Because the divisor is a decimal number, we move
the decimal point enough places to the right to create
an integer—in this case, one place. Then we move
the decimal point one place to the right in the
dividend. Because we are dividing two numbers
with the same sign, the result is positive.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 123
Example 10 continued
Divide −44.64 ÷ (−3.6)
Solution

12.4
36 446.4
36
86
72
144
144
0
Copyright © 2011 Pearson Education, Inc.

Slide 1- 124
Example 11
Martha was decorating cookies. She used 2/3 of a
container of frosting that was 3/4 full. What
fractional part of the container did she use?
Solution
To find 2/3 of 3/4, multiply.
2 3
2×
3
1
× =
=
3 4 2 ×2 × 2
3

Copyright © 2011 Pearson Education, Inc.

Slide 1- 125
Multiply (–6)(–3)(7).

a) 126
b) −126
c) –63
d) 63

1.4

Copyright © 2011 Pearson Education, Inc.

Slide 1- 126
Multiply (–6)(–3)(7).

a) 126
b) −126
c) –63
d) 63

1.4

Copyright © 2011 Pearson Education, Inc.

Slide 1- 127
Divide −14.6 ÷ ( −0.03) .

a) −48.6
b) 48.6
c) 486.6
d) −486.6

1.4

Copyright © 2011 Pearson Education, Inc.

Slide 1- 128
Divide −14.6 ÷ ( −0.03) .

a) −48.6
b) 48.6
c) 486.6
d) −486.6

1.4

Copyright © 2011 Pearson Education, Inc.

Slide 1- 129
1.5

Exponents, Roots, and Order of
Operations

1. Evaluate numbers in exponential form.
2. Evaluate square roots.
3. Use the order-of-operations agreement to simplify
numerical expressions.
4. Find the mean of a set of data.

Copyright © 2011 Pearson Education, Inc.
Objective 1
Evaluate numbers in exponential
form.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 131
Sometimes problems involve repeatedly
multiplying the same number. In such problems,
we can use an exponent to indicate that a base
number is repeatedly multiplied.
Exponent: A symbol written to the upper
right of a base number that indicates how
many times to use the base as a factor.
Base: The number that is repeatedly
multiplied.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 132
When we write a number with an exponent, we
say the expression is in exponential form. The
24 is in exponential form, where the
expression
base is 2 and the exponent is 4. To evaluate 24 ,
write 2 as a factor 4 times, then multiply.
Four 2s

2 = 2 g 2 g 2 g 2 = 16
4

Base

Exponent

Copyright © 2011 Pearson Education, Inc.

Slide 1- 133
Evaluating an Exponential Form
To evaluate an exponential form raised to a
natural number exponent, write the base as a
factor the number of times indicated by the
exponent; then multiply.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 134
Example 1a
Evaluate. (–9)2
Solution
The exponent 2 indicates we have two factors of –9.
Because we multiply two negative numbers, the result
is positive.
(–9)2 = (–9)(–9) = 81

Copyright © 2011 Pearson Education, Inc.

Slide 1- 135
Example 1b
3

Evaluate.

 3
− ÷
 5

Solution
The exponent 3 means we must multiply the base by
itself three times.
3

 3
 3  3   3 
− ÷ =  − ÷ − ÷ − ÷

 5
 5  5   5 
27
=−
125
Copyright © 2011 Pearson Education, Inc.

Slide 1- 136
Evaluating Exponential Forms with Negative
Bases
If the base of an exponential form is a negative
number and the exponent is even, then the
product is positive.
If the base is a negative number and the exponent is
odd, then the product is negative.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 137
Example 2
Evaluate.
(−3) 4
−34
(−2)3
a.
b.
c.
Solution
a. (−3)4 = (−3)(−3)(−3)(−3) = 81
b.

−34 = −3 ×3 ×3 ×3 = −81

c.

(−2)3 = ( −2)( −2)(−2) = −8

d.

d.

−23

−23 = −2 ×2 ×2 = −8
Copyright © 2011 Pearson Education, Inc.

Slide 1- 138
Objective 2
Evaluate square roots.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 139
Roots are inverses of exponents. More
specifically, a square root is the inverse of a
square, so a square root of a given number is a
number that, when squared, equals the given
number.
Square Roots
Every positive number has two square roots, a
positive root and a negative root.
Negative numbers have no real-number square
roots.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 140
Example 3
Find all square roots of the given number.
Solution
a. 49
Answer ± 7
b. −81
Answer No real-number square roots exist.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 141
The symbol, , called the radical, is used to
indicate finding only the positive (or principal)
square root of a given number. The given number or
expression inside the radical is called the radicand.
Radical

Principal Square Root

25 = 5
Radicand

Copyright © 2011 Pearson Education, Inc.

Slide 1- 142
Square Roots Involving the Radical Sign
The radical symbol denotes only the positive
(principal) square root.
a
=
b

a
b

, where a ≥ 0 and b > 0.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 143
Example 4
Evaluate the square root.
a. 169
b. 64

c.

81

0.64

d.

−25

Solution
a. 169 = 13

b.

c.

d. −25 = not a real number

0.64 = 0.8

64 8
=
81 9

Copyright © 2011 Pearson Education, Inc.

Slide 1- 144
Objective 3
Use the order-of-operations
agreement to simplify numerical
expressions.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 145
Order-of- Operations Agreement
Perform operations in the following order:
1. Within grouping symbols: parentheses ( ),
brackets [ ], braces { }, above/below fraction
bars, absolute value | |, and radicals .
2. Exponents/Roots from left to right, in order as
they occur.
3. Multiplication/Division from left to right, in order
as they occur.
4. Addition/Subtraction from left to right, in order as
they occur.
Copyright © 2011 Pearson Education, Inc.

Slide 1- 146
Example 5a
Simplify.

−26 + 15 ÷ (−5) ×2

Solution
−26 + 15 ÷ (−5) ×2
= −26 + (−3) ×2

Divide 15 ÷ (−5) = –3

= −26 + (−6)

Multiply (–3) ⋅ 2 = –6

= −32

Add –26 + (–6) = –32

Copyright © 2011 Pearson Education, Inc.

Slide 1- 147
Example 5b
Simplify.
Solution

−34 + 2 12 − 20

−34 + 2 12 − 20
= −34 + 2 −8

Subtract inside the absolute value.

= −34 + 2 ×
8

Simplify the absolute value.

= −81 + 2 ×
8

Evaluate the exponent.

= −81 + 16

Multiply.

= −65

Add.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 148
Example 5c
Simplify. ( −3)

2

+ 5 6 − ( 2 + 1)  − 49



Solution ( −3) 2 + 5 6 − ( 2 + 1)  −


= ( −3) + 5 [ 6 − 3] − 49
2

= 9 + 5 ( 3) − 7

49
Calculate within the innermost
parenthesis.
Evaluate the exponential form,
brackets, and square root.

= 9 + 15 − 7

Multiply 5(3).

= 24 − 7

Add 9 + 15.

= 17

Subtract 24 – 7.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 149
Square Root of a Product or Quotient
If a square root contains multiplication or
division, we can multiply or divide first, then find
the square root of the result, or we can find the
square roots of the individual numbers, then
multiply or divide the square roots.
Square Root of a Sum or Difference
When a radical contains addition or subtraction, we
must add or subtract first, then find the root of the
sum or difference.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 150
Example 6a
Simplify.

13.5 ÷ 5 ( −4 ) − 3 142 − 21

Solution

13.5 ÷ 5 ( −4 ) − 3 142 − 21

2

2

= 13.5 ÷ 5 ( −4 ) − 3 121
2

= 13.5 ÷ 5 ( 16 ) − 3(11)

Subtract within the radical.
Evaluate the exponential form and
root.

= 2.7 ( 16 ) − 3 ( 11)

Divide.

= 43.2 − 33

Multiply.

= 10.2

Subtract.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 151
Sometimes fraction lines are used as grouping
symbols. When they are, we simplify the
numerator and denominator separately, then
divide the results.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 152
Example 7a
Simplify.
Solution

8(−5) − 23
4(8) − 8
8( −5) − 23
4(8) − 8
8(−5) − 8
=
4(8) − 8
=

−40 − 8
32 − 8

Evaluate the exponential form in
the numerator and multiply in the
denominator.
Multiply in the numerator and
subtract in the denominator.

−48
=
24

Subtract in the numerator.

= −2

Divide.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 153
Example 7b
Simplify.

9(4) + 12
43 + (8)(−8)
9(4) + 12
43 + (8)(−8)

Solution
=

36 + 12
64 + (8)(−8)

48
=
64 + (−64)
48
=
0

Because the denominator or divisor is 0, the answer is
undefined.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 154
Objective 4
Find the mean of a set of data.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 155
Finding the Arithmetic Mean
To find the arithmetic mean, or average, of n
numbers, divide the sum of the numbers by n.
Arithmetic mean =

x1 + x2 + ... + xn
n

Copyright © 2011 Pearson Education, Inc.

Slide 1- 156
Example 8
Bruce has the following test scores in his biology
class: 92, 96, 81, 89, 95, 93. Find the average of his
test scores.
Solution
92 + 96 + 81 + 89 + 95 + 93
546
=
6
6

Divide the sum of the 6
scores by 6.

= 91

Copyright © 2011 Pearson Education, Inc.

Slide 1- 157
Simplify using order of operations.

( −6 )

2

− 18 ÷ ( 9 − 6 )

a) − 18
b) 6
c) 30
d) 36

1.5

Copyright © 2011 Pearson Education, Inc.

Slide 1- 158
Simplify using order of operations.

( −6 )

2

− 18 ÷ ( 9 − 6 )

a) − 18
b) 6
c) 30
d) 36

1.5

Copyright © 2011 Pearson Education, Inc.

Slide 1- 159
Simplify using order of operations.
2 ( 4 + 23 )

a)

8
300

b)

250
361

c)

6 + 30 − ( 2 + 4 )

2

2
11

d) undefined

1.5

Copyright © 2011 Pearson Education, Inc.

Slide 1- 160
Simplify using order of operations.
2 ( 4 + 23 )

a)

8
300

b)

250
361

c)

6 + 30 − ( 2 + 4 )

2

2
11

d) undefined

1.5

Copyright © 2011 Pearson Education, Inc.

Slide 1- 161
1.6

Translating Word Phrases to Expressions

1. Translate word phrases to expressions.

Copyright © 2011 Pearson Education, Inc.
Objective 1
Translating word phrases to
Expressions

Copyright © 2011 Pearson Education, Inc.

Slide 1- 163
Translating Basic Phrases
Addition

Translation

Subtraction

Translation

The sum of x
and three

x+3

The difference of x
and three

x–3

h plus k

h+k

h minus k

h–k

seven added to t

7+t

seven subtracted
from t

t–7

three more than
a number

n+3

three less than a
number

n–3

y increased by
two

y+2

y decreased by two

y–2

Note: Since addition is a
commutative operation, it does not
matter in what order we write the
translation.

Note: Subtraction is not a
commutative operation; therefore,
the way we write the translation
matters.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 164
Translating Basic Phrases
Multiplication

Translation

The product of
x and three

Translation

The quotient of x
and three

hk

x ÷ 3 or x

h divided by k

3x

h times k

Division

h ÷ k or h

Twice a number

2n

h divided into k

Triple the
number

3n

The ratio of a to b

Two-thirds of a
number

3

k
k ÷ h or k
h
a ÷ b or a
b

2
n
3

Note: Like addition, multiplication
is a commutative operation: it does
not matter in what order we write
the translation.

Note: Division is like subtraction in
that it is not a commutative
operation; therefore, the way we
write the translation matters.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 165
Translating Basic Phrases
Exponents

Translation

c squared

c2

The square of b

k3

The cube of b

b3

n to the fourth
power

n4

y raised to the
fifth power

The square root of
x

Translation

b2

k cubed

Roots

y5

Copyright © 2011 Pearson Education, Inc.

x

Slide 1- 166
The key words sum, difference, product, and quotient
indicate the answer for their respective operations.
sum of x and 3

difference of x and 3

x+3

x–3

product of x and 3

quotient of x and 3
x÷3

x⋅3
Copyright © 2011 Pearson Education, Inc.

Slide 1- 167
Example 1
Translate to an algebraic expression.
a. five more than two times a number
Translation: 5 + 2n or 2n + 5
b. seven less than the cube of a number
Translation: n3 – 7
c. the sum of h raised to the fourth power and twelve
Translation: h4 + 12
Copyright © 2011 Pearson Education, Inc.

Slide 1- 168
Translating Phrases Involving Parentheses
Sometimes the word phrases imply an order of
operations that would require us to use parentheses in
the translation.
These situations arise when the phrase indicates that a
sum or difference is to be calculated before performing
a higher-order operation such as multiplication,
division, exponent, or root.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 169
Example 2
Translate to an algebraic expression.
a. seven times the sum of a and b
Translation: 7(a + b)
b. the product of a and b divided by the sum
of w2 and 4
ab
2
Translation: ab ÷ (w + 4) or
2
w +4

Copyright © 2011 Pearson Education, Inc.

Slide 1- 170
Translate the phrase to an algebraic expression.
Twelve less than three times a number

a) 3n + 12
b) 12 – 3n
c) 3n – 12
d) 3n ⋅ 12

1.6

Copyright © 2011 Pearson Education, Inc.

Slide 1- 171
Translate the phrase to an algebraic expression.
Twelve less than three times a number

a) 3n + 12
b) 12 – 3n
c) 3n – 12
d) 3n ⋅ 12

1.6

Copyright © 2011 Pearson Education, Inc.

Slide 1- 172
Translate the phrase to an algebraic expression.
The difference of a and b decreased by the sum
of w and z

a) (a – b) – (w + z)
b) a – b – (w + z)
c) ab – (w + z)
d) (b – a) – (w + z)

1.6

Copyright © 2011 Pearson Education, Inc.

Slide 1- 173
Translate the phrase to an algebraic expression.
The difference of a and b decreased by the sum
of w and z

a) (a – b) – (w + z)
b) a – b – (w + z)
c) ab – (w + z)
d) (b – a) – (w + z)

1.6

Copyright © 2011 Pearson Education, Inc.

Slide 1- 174
1.7

Evaluating and Rewriting Expressions

1. Evaluate an expression.
2. Determine all values that cause an expression to be
undefined.
3. Rewrite an expression using the distributive property.
4. Rewrite an expression by combining like terms.

Copyright © 2011 Pearson Education, Inc.
Objective 1
Evaluate an expression.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 176
Evaluating an Algebraic Expression
To evaluate an algebraic expression:
1. Replace the variables with their corresponding
given values.
2. Calculate the numerical expression using the order
of operations.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 177
Example 1a
Evaluate 3w – 4(a – 6) when w = 5 and a = 7.
Solution
3w – 4(a − 6)
3(5) – 4(7 – 6)
= 3(5) – 4(1)
= 15 – 4
= 11

Replace w with 5 and a with 7.
Simplify inside the parentheses first.
Multiply.
Subtract.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 178
Example 1b
Evaluate x2 – 0.4xy + 9, when x = 7 and y = –2.
Solution
x2 – 0.4xy + 9
(7)2 – 0.4(7)(–2) + 9
= 49 – 0.4(7)(–2) + 9
= 49 – (–5.6) + 9
= 49 + 5.6 + 9
= 63.6

Replace x with 7 and y with –2.
Begin calculating by simplifying the exponential
form.
Multiply.
Write the subtraction as an equivalent addition.
Add from left to right.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 179
Objective 2
Determine all values that cause an
expression to be undefined.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 180
When evaluating a division expression in which the
divisor or denominator contains a variable or variables,
we must be careful about what values replace the
variable(s).
We often need to know what values could replace the
variable(s) and cause the expression to be undefined or
indeterminate.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 181
Example 2
Determine all values that cause the expression to be
undefined.
−2
a. 8
b.
( x + 2)( x − 9)

x+4

Answer
8
8
a. If x = −4, we have −4 + 4 = 0 , which is undefined
because the denominator is 0.
b. If x = −2 or 9 the fraction will be undefined since
the denominator will = 0.
−2
−2
=
( −2 + 2)(−2 − 9) 0

−2
−2
=
(9 + 2)(9 − 9) 0

Copyright © 2011 Pearson Education, Inc.

Slide 1- 182
Objective 3
Rewrite an expression using the
distributive property.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 183
The Distributive Property of Multiplication over
Addition
a(b + c) = ab + ac

This property gives us an alternative to the order of
operations.
2(5 + 6) = 2(11)

2(5 + 6) = 2⋅5 + 2⋅6

= 22

= 10 + 12
= 22
Copyright © 2011 Pearson Education, Inc.

Slide 1- 184
Example 3
Use the distributive property to write an equivalent
expression and simplify.
a. 3(x + 3)
b. –5(w – 4)
Solution
a. 3(x + 3) = 3 ⋅ x + 3 ⋅ 3
= 3x + 9
b. –5(w – 4) = –5 ⋅ w – (–5) ⋅ 4
= –5w + 20
Copyright © 2011 Pearson Education, Inc.

Slide 1- 185
Objective 4
Rewrite an expression by combining
like terms.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 186
Terms: Expressions that are the addends in an
expression that is a sum.
Coefficient: The numerical factor in a term.
The coefficient of 5x3 is 5.
The coefficient of –8y is –8.
Like terms: Variable terms that have the same
variable(s) raised to the same exponents, or constant
terms.
Like terms
Unlike terms
4x and 7x
2x and 8y different variables
5y2 and 10y2
7t3 and 3t2 different exponents
8xy and 12xy
x2y and xy2 different exponents
7 and 15
13 and 15x different variables
Copyright © 2011 Pearson Education, Inc.

Slide 1- 187
Combining Like Terms
To combine like terms, add or subtract the
coefficients and keep the variables and their
exponents the same.

Copyright © 2011 Pearson Education, Inc.

Slide 1- 188
Example 4
Combine like terms.
a. 10y + 8y
Solution
10y + 8y = 18y
b. 8x – 3x
Solution

8x – 3x = 5x

c. 13y2 – y2
Solution

13y2 – y2 = 12y2
Copyright © 2011 Pearson Education, Inc.

Slide 1- 189
Example 5
Combine like terms in 5y2 + 6 + 4y2 – 7.
Solution
5y2 + 6 + 4y2 – 7
= 5y2 + 4y2 + 6 – 7

Combine like terms.

= 9y2 – 1

Copyright © 2011 Pearson Education, Inc.

Slide 1- 190
Example 6
Combine like terms in 18y + 7x – y – 7x.
Solution
18y + 7x – y – 7x
= 17y + 0
= 17y

Copyright © 2011 Pearson Education, Inc.

Slide 1- 191
Example 7
1
1
Combine like terms in a + 4b + 3 + a − b.
12
6

Solution

1
1
a + 4b + 3 + a − b
12
6
1
1
= a + a + 4b − b + 3
Collect like terms.
12
6
1
1(2)
Write the fraction coefficients as
= a+
a + 4b − b + 3 equivalent fractions with their
12
6(2)
LCD, 12.
1
2
= a + a + 4b − b + 3
12
12
1
3
Combine like terms.
= a + 3b + 3 = a + 3b + 3
4
12
Slide 1- 192
Copyright © 2011 Pearson Education, Inc.
Evaluate the expression 4(a + b) when
a = 3 and b = –2.
a) 4
b) −4
c) 12
d) 20

1.7

Copyright © 2011 Pearson Education, Inc.

Slide 1- 193
Evaluate the expression 4(a + b) when
a = 3 and b = –2.
a) 4
b) −4
c) 12
d) 20

1.7

Copyright © 2011 Pearson Education, Inc.

Slide 1- 194
For which values is the
expression undefined?

8m
(m + 2)(m − 5)

a) 8
b) −2
c) −2 and 5
d) 2 and −5

1.7

Copyright © 2011 Pearson Education, Inc.

Slide 1- 195
For which values is the
expression undefined?

8m
(m + 2)(m − 5)

a) 8
b) −2
c) −2 and 5
d) 2 and −5

1.7

Copyright © 2011 Pearson Education, Inc.

Slide 1- 196
Simplify: 7x + 8 – 2x – 4

a) 9x – 4
b) 9x + 4
c) 5x – 4
d) 5x + 4

1.7

Copyright © 2011 Pearson Education, Inc.

Slide 1- 197
Simplify: 7x + 8 – 2x – 4

a) 9x – 4
b) 9x + 4
c) 5x – 4
d) 5x + 4

1.7

Copyright © 2011 Pearson Education, Inc.

Slide 1- 198

More Related Content

What's hot

Lesson 1.9 grade 8
Lesson 1.9   grade 8Lesson 1.9   grade 8
Lesson 1.9 grade 8wzuri
 
G6 m3-a-lesson 5-t
G6 m3-a-lesson 5-tG6 m3-a-lesson 5-t
G6 m3-a-lesson 5-tmlabuski
 
Solving a linear equation single and multi step
Solving a linear equation   single and multi stepSolving a linear equation   single and multi step
Solving a linear equation single and multi stepjulienorman80065
 
Chapter 1 Review
Chapter 1 ReviewChapter 1 Review
Chapter 1 Reviewwzuri
 
Maths level 1 chapter 1 learner materials
Maths level 1 chapter 1 learner materialsMaths level 1 chapter 1 learner materials
Maths level 1 chapter 1 learner materialscatherinelindsay
 
4. solving inequalities
4. solving inequalities4. solving inequalities
4. solving inequalitiesMedhaKetkar
 
Module 1 solving inequalities notes
Module 1 solving inequalities notesModule 1 solving inequalities notes
Module 1 solving inequalities notesMichelle Barnhill
 
Math v
Math vMath v
Math vDepEd
 
Solving and graphing inequalities lecture
Solving and graphing inequalities lectureSolving and graphing inequalities lecture
Solving and graphing inequalities lectureDaisy Zambrano
 
Problms involved with real numbers
Problms involved with real numbersProblms involved with real numbers
Problms involved with real numbersmekaylagonzales
 
TechMathII - 1.1 - The Set of Real Numbers
TechMathII - 1.1 - The Set of Real NumbersTechMathII - 1.1 - The Set of Real Numbers
TechMathII - 1.1 - The Set of Real Numberslmrhodes
 
Solving Equations
Solving EquationsSolving Equations
Solving Equationsswartzje
 
Pat05 ppt 0106
Pat05 ppt 0106Pat05 ppt 0106
Pat05 ppt 0106wzuri
 
G6 m4-a-lesson 1-t
G6 m4-a-lesson 1-tG6 m4-a-lesson 1-t
G6 m4-a-lesson 1-tmlabuski
 
Rational irrational and_real_number_practice
Rational irrational and_real_number_practiceRational irrational and_real_number_practice
Rational irrational and_real_number_practiceeixarc
 

What's hot (19)

Lesson 1.9 grade 8
Lesson 1.9   grade 8Lesson 1.9   grade 8
Lesson 1.9 grade 8
 
G6 m3-a-lesson 5-t
G6 m3-a-lesson 5-tG6 m3-a-lesson 5-t
G6 m3-a-lesson 5-t
 
Solving a linear equation single and multi step
Solving a linear equation   single and multi stepSolving a linear equation   single and multi step
Solving a linear equation single and multi step
 
Chapter 1 Review
Chapter 1 ReviewChapter 1 Review
Chapter 1 Review
 
Maths level 1 chapter 1 learner materials
Maths level 1 chapter 1 learner materialsMaths level 1 chapter 1 learner materials
Maths level 1 chapter 1 learner materials
 
Real numbers
Real numbersReal numbers
Real numbers
 
4. solving inequalities
4. solving inequalities4. solving inequalities
4. solving inequalities
 
Basic math
Basic mathBasic math
Basic math
 
Lesson 2 even numbers
Lesson 2   even numbersLesson 2   even numbers
Lesson 2 even numbers
 
Module 1 solving inequalities notes
Module 1 solving inequalities notesModule 1 solving inequalities notes
Module 1 solving inequalities notes
 
Math v
Math vMath v
Math v
 
1.3 1.7
1.3 1.71.3 1.7
1.3 1.7
 
Solving and graphing inequalities lecture
Solving and graphing inequalities lectureSolving and graphing inequalities lecture
Solving and graphing inequalities lecture
 
Problms involved with real numbers
Problms involved with real numbersProblms involved with real numbers
Problms involved with real numbers
 
TechMathII - 1.1 - The Set of Real Numbers
TechMathII - 1.1 - The Set of Real NumbersTechMathII - 1.1 - The Set of Real Numbers
TechMathII - 1.1 - The Set of Real Numbers
 
Solving Equations
Solving EquationsSolving Equations
Solving Equations
 
Pat05 ppt 0106
Pat05 ppt 0106Pat05 ppt 0106
Pat05 ppt 0106
 
G6 m4-a-lesson 1-t
G6 m4-a-lesson 1-tG6 m4-a-lesson 1-t
G6 m4-a-lesson 1-t
 
Rational irrational and_real_number_practice
Rational irrational and_real_number_practiceRational irrational and_real_number_practice
Rational irrational and_real_number_practice
 

Similar to Cei03 ppt 01

Review of basic algebraic concept
Review of basic algebraic conceptReview of basic algebraic concept
Review of basic algebraic conceptarvin efriani
 
A Problem Solving Approach To Mathematics For Elementary School Teachers
A Problem Solving Approach To Mathematics For Elementary School TeachersA Problem Solving Approach To Mathematics For Elementary School Teachers
A Problem Solving Approach To Mathematics For Elementary School TeachersKimberly Pulley
 
Addition subtraction-multi-divison(6)
Addition subtraction-multi-divison(6)Addition subtraction-multi-divison(6)
Addition subtraction-multi-divison(6)Taunita Renee
 
1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptx
1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptx1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptx
1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptxJiyoona
 
MATH; discussion_20240304_113651_0000.pdf
MATH; discussion_20240304_113651_0000.pdfMATH; discussion_20240304_113651_0000.pdf
MATH; discussion_20240304_113651_0000.pdfKhrysjellCatie
 
Introduction Combined Number And Dp
Introduction Combined Number And DpIntroduction Combined Number And Dp
Introduction Combined Number And DpAwais Khan
 
GCSE-Surds.pptx
GCSE-Surds.pptxGCSE-Surds.pptx
GCSE-Surds.pptxBimboBbsax
 
mgpia3e_ppt_01_08.ppt
mgpia3e_ppt_01_08.pptmgpia3e_ppt_01_08.ppt
mgpia3e_ppt_01_08.pptelle155401
 
Solving Linear Equations with Notes
Solving Linear Equations with NotesSolving Linear Equations with Notes
Solving Linear Equations with Notesswartzje
 
CBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdf
CBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdfCBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdf
CBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdfShavetaSharma37
 

Similar to Cei03 ppt 01 (20)

P
PP
P
 
Unit 1.1
Unit 1.1Unit 1.1
Unit 1.1
 
Review of basic algebraic concept
Review of basic algebraic conceptReview of basic algebraic concept
Review of basic algebraic concept
 
Ch01 se
Ch01 seCh01 se
Ch01 se
 
A Problem Solving Approach To Mathematics For Elementary School Teachers
A Problem Solving Approach To Mathematics For Elementary School TeachersA Problem Solving Approach To Mathematics For Elementary School Teachers
A Problem Solving Approach To Mathematics For Elementary School Teachers
 
Bzpc5e p 01
Bzpc5e p 01Bzpc5e p 01
Bzpc5e p 01
 
Addition subtraction-multi-divison(6)
Addition subtraction-multi-divison(6)Addition subtraction-multi-divison(6)
Addition subtraction-multi-divison(6)
 
Math Chapter 1 - Integers
Math Chapter 1 - IntegersMath Chapter 1 - Integers
Math Chapter 1 - Integers
 
1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptx
1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptx1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptx
1-1-Slide-Show-Writing-and-Interpreting-Numerical-Expressions.pptx
 
MATH; discussion_20240304_113651_0000.pdf
MATH; discussion_20240304_113651_0000.pdfMATH; discussion_20240304_113651_0000.pdf
MATH; discussion_20240304_113651_0000.pdf
 
Introduction Combined Number And Dp
Introduction Combined Number And DpIntroduction Combined Number And Dp
Introduction Combined Number And Dp
 
Unit .3
Unit .3Unit .3
Unit .3
 
Maths glossary
Maths glossary Maths glossary
Maths glossary
 
GCSE-Surds.pptx
GCSE-Surds.pptxGCSE-Surds.pptx
GCSE-Surds.pptx
 
mgpia3e_ppt_01_08.ppt
mgpia3e_ppt_01_08.pptmgpia3e_ppt_01_08.ppt
mgpia3e_ppt_01_08.ppt
 
SOW for year 7
SOW for year 7SOW for year 7
SOW for year 7
 
SOW for year 7
SOW for year 7SOW for year 7
SOW for year 7
 
Solving Linear Equations with Notes
Solving Linear Equations with NotesSolving Linear Equations with Notes
Solving Linear Equations with Notes
 
1059331.pdf
1059331.pdf1059331.pdf
1059331.pdf
 
CBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdf
CBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdfCBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdf
CBSE-Class-8-NCERT-Maths-Book-Rational-Numbers-chapter-1.pdf
 

Recently uploaded

08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking Men08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking MenDelhi Call girls
 
08448380779 Call Girls In Greater Kailash - I Women Seeking Men
08448380779 Call Girls In Greater Kailash - I Women Seeking Men08448380779 Call Girls In Greater Kailash - I Women Seeking Men
08448380779 Call Girls In Greater Kailash - I Women Seeking MenDelhi Call girls
 
Pigging Solutions Piggable Sweeping Elbows
Pigging Solutions Piggable Sweeping ElbowsPigging Solutions Piggable Sweeping Elbows
Pigging Solutions Piggable Sweeping ElbowsPigging Solutions
 
Maximizing Board Effectiveness 2024 Webinar.pptx
Maximizing Board Effectiveness 2024 Webinar.pptxMaximizing Board Effectiveness 2024 Webinar.pptx
Maximizing Board Effectiveness 2024 Webinar.pptxOnBoard
 
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...Neo4j
 
Snow Chain-Integrated Tire for a Safe Drive on Winter Roads
Snow Chain-Integrated Tire for a Safe Drive on Winter RoadsSnow Chain-Integrated Tire for a Safe Drive on Winter Roads
Snow Chain-Integrated Tire for a Safe Drive on Winter RoadsHyundai Motor Group
 
Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...
Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...
Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...shyamraj55
 
GenCyber Cyber Security Day Presentation
GenCyber Cyber Security Day PresentationGenCyber Cyber Security Day Presentation
GenCyber Cyber Security Day PresentationMichael W. Hawkins
 
Unblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen FramesUnblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen FramesSinan KOZAK
 
Benefits Of Flutter Compared To Other Frameworks
Benefits Of Flutter Compared To Other FrameworksBenefits Of Flutter Compared To Other Frameworks
Benefits Of Flutter Compared To Other FrameworksSoftradix Technologies
 
IAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI SolutionsIAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI SolutionsEnterprise Knowledge
 
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024BookNet Canada
 
FULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | DelhiFULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | Delhisoniya singh
 
Presentation on how to chat with PDF using ChatGPT code interpreter
Presentation on how to chat with PDF using ChatGPT code interpreterPresentation on how to chat with PDF using ChatGPT code interpreter
Presentation on how to chat with PDF using ChatGPT code interpreternaman860154
 
CloudStudio User manual (basic edition):
CloudStudio User manual (basic edition):CloudStudio User manual (basic edition):
CloudStudio User manual (basic edition):comworks
 
AI as an Interface for Commercial Buildings
AI as an Interface for Commercial BuildingsAI as an Interface for Commercial Buildings
AI as an Interface for Commercial BuildingsMemoori
 
Understanding the Laravel MVC Architecture
Understanding the Laravel MVC ArchitectureUnderstanding the Laravel MVC Architecture
Understanding the Laravel MVC ArchitecturePixlogix Infotech
 
Breaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path MountBreaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path MountPuma Security, LLC
 
Hyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your Budget
Hyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your BudgetHyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your Budget
Hyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your BudgetEnjoy Anytime
 
Next-generation AAM aircraft unveiled by Supernal, S-A2
Next-generation AAM aircraft unveiled by Supernal, S-A2Next-generation AAM aircraft unveiled by Supernal, S-A2
Next-generation AAM aircraft unveiled by Supernal, S-A2Hyundai Motor Group
 

Recently uploaded (20)

08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking Men08448380779 Call Girls In Civil Lines Women Seeking Men
08448380779 Call Girls In Civil Lines Women Seeking Men
 
08448380779 Call Girls In Greater Kailash - I Women Seeking Men
08448380779 Call Girls In Greater Kailash - I Women Seeking Men08448380779 Call Girls In Greater Kailash - I Women Seeking Men
08448380779 Call Girls In Greater Kailash - I Women Seeking Men
 
Pigging Solutions Piggable Sweeping Elbows
Pigging Solutions Piggable Sweeping ElbowsPigging Solutions Piggable Sweeping Elbows
Pigging Solutions Piggable Sweeping Elbows
 
Maximizing Board Effectiveness 2024 Webinar.pptx
Maximizing Board Effectiveness 2024 Webinar.pptxMaximizing Board Effectiveness 2024 Webinar.pptx
Maximizing Board Effectiveness 2024 Webinar.pptx
 
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
Neo4j - How KGs are shaping the future of Generative AI at AWS Summit London ...
 
Snow Chain-Integrated Tire for a Safe Drive on Winter Roads
Snow Chain-Integrated Tire for a Safe Drive on Winter RoadsSnow Chain-Integrated Tire for a Safe Drive on Winter Roads
Snow Chain-Integrated Tire for a Safe Drive on Winter Roads
 
Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...
Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...
Automating Business Process via MuleSoft Composer | Bangalore MuleSoft Meetup...
 
GenCyber Cyber Security Day Presentation
GenCyber Cyber Security Day PresentationGenCyber Cyber Security Day Presentation
GenCyber Cyber Security Day Presentation
 
Unblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen FramesUnblocking The Main Thread Solving ANRs and Frozen Frames
Unblocking The Main Thread Solving ANRs and Frozen Frames
 
Benefits Of Flutter Compared To Other Frameworks
Benefits Of Flutter Compared To Other FrameworksBenefits Of Flutter Compared To Other Frameworks
Benefits Of Flutter Compared To Other Frameworks
 
IAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI SolutionsIAC 2024 - IA Fast Track to Search Focused AI Solutions
IAC 2024 - IA Fast Track to Search Focused AI Solutions
 
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
Transcript: #StandardsGoals for 2024: What’s new for BISAC - Tech Forum 2024
 
FULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | DelhiFULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | Delhi
FULL ENJOY 🔝 8264348440 🔝 Call Girls in Diplomatic Enclave | Delhi
 
Presentation on how to chat with PDF using ChatGPT code interpreter
Presentation on how to chat with PDF using ChatGPT code interpreterPresentation on how to chat with PDF using ChatGPT code interpreter
Presentation on how to chat with PDF using ChatGPT code interpreter
 
CloudStudio User manual (basic edition):
CloudStudio User manual (basic edition):CloudStudio User manual (basic edition):
CloudStudio User manual (basic edition):
 
AI as an Interface for Commercial Buildings
AI as an Interface for Commercial BuildingsAI as an Interface for Commercial Buildings
AI as an Interface for Commercial Buildings
 
Understanding the Laravel MVC Architecture
Understanding the Laravel MVC ArchitectureUnderstanding the Laravel MVC Architecture
Understanding the Laravel MVC Architecture
 
Breaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path MountBreaking the Kubernetes Kill Chain: Host Path Mount
Breaking the Kubernetes Kill Chain: Host Path Mount
 
Hyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your Budget
Hyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your BudgetHyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your Budget
Hyderabad Call Girls Khairatabad ✨ 7001305949 ✨ Cheap Price Your Budget
 
Next-generation AAM aircraft unveiled by Supernal, S-A2
Next-generation AAM aircraft unveiled by Supernal, S-A2Next-generation AAM aircraft unveiled by Supernal, S-A2
Next-generation AAM aircraft unveiled by Supernal, S-A2
 

Cei03 ppt 01

  • 1. Copyright © 2011 Pearson Education, Inc.
  • 2. CHAPTER 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Foundations of Algebra Number Sets and the Structure of Algebra Fractions Adding and Subtracting Real Numbers; Properties of Real Numbers Multiplying and Dividing Real Numbers; Properties of Real Numbers Exponents, Roots, and Order of Operations Translating Word Phrases to Expressions Evaluating and Rewriting Expressions Copyright © 2011 Pearson Education, Inc.
  • 3. 1.1 1. 2. 3. 4. 5. Number Sets and the Structure of Algebra Understand the structure of algebra. Classify number sets. Graph rational numbers on a number line. Determine the absolute value of a number. Compare numbers. Copyright © 2011 Pearson Education, Inc.
  • 4. Objective 1 Understand the structure of algebra. Copyright © 2011 Pearson Education, Inc. Slide 1- 4
  • 5. Definitions Variable: A symbol that can vary in value. Constant: A symbol that does not vary in value. Variables are usually letters of the alphabet, like x or y. Usually constants are symbols for numbers, like 1, 2, ¾, 6.74. Copyright © 2011 Pearson Education, Inc. Slide 1- 5
  • 6. Expression: A constant, variable, or any combination of constants, variables, and arithmetic operations that describe a calculation. Examples of expressions: 2+6 or 4x − 5 or 1 2 πr h 3 Copyright © 2011 Pearson Education, Inc. Slide 1- 6
  • 7. Equation: A mathematical relationship that contains an equal sign. Examples of equations: 2+6=8 or 4x − 5 = 12 Copyright © 2011 Pearson Education, Inc. or 1 2 V = πr h 3 Slide 1- 7
  • 8. Inequality: A mathematical relationship that contains an inequality symbol (≠, <, >, ≤, or ≥). Symbolic form Translation 8≠3 Eight is not equal to three. 5<7 Five is less than seven. 7>5 Seven is greater than five. x≤3 x is less than or equal to three. y≥2 y is greater than or equal to two. Copyright © 2011 Pearson Education, Inc. Slide 1- 8
  • 9. Objective 2 Classify number sets. Copyright © 2011 Pearson Education, Inc. Slide 1- 9
  • 10. Set: A collection of objects. Braces are used to indicate a set. For example, the set containing the numbers 1, 2, 3, and 4 would be written {1, 2, 3, 4}. The numbers 1, 2, 3, and 4 are called the members or elements of this set. Copyright © 2011 Pearson Education, Inc. Slide 1- 10
  • 11. Writing Sets To write a set, write the members or elements of the set separated by commas within braces, { }. Copyright © 2011 Pearson Education, Inc. Slide 1- 11
  • 12. Example 1 Write the set containing the first four days of the week. Answer {Sunday, Monday, Tuesday, Wednesday} Copyright © 2011 Pearson Education, Inc. Slide 1- 12
  • 13. Numbers are classified using number sets. Natural numbers contain the counting numbers 1, 2, 3, 4, …and is written {1, 2, 3, …}. The three dots are called ellipsis and indicate that the numbers continue forever in the same pattern. Whole numbers: natural numbers and 0 {0, 1, 2, 3,…} Integers: whole numbers and the opposite (or negative) of every natural number {…, −3, −2, −1, 0, 1, 2, 3…} Rational: every real number that can be expressed as a ratio of integers. Copyright © 2011 Pearson Education, Inc. Slide 1- 13
  • 14. Rational number: Any real number that can be a expressed in the form , where a and b are integers b and b ≠ 0. Copyright © 2011 Pearson Education, Inc. Slide 1- 14
  • 15. Example 2 Determine whether the given number is a rational number. 5 a. b. 0.8 c. 0.3 6 Answer 5 a. 6 Yes, because 5 and 6 are integers. b. 0.8 c. Yes, 0.8 can be expressed as a fraction 8 over 10, and 8 and 10 are integers. The bar indicates that the digit repeats. This is the decimal equivalent of 1 over 3. Yes this is a rational number. Copyright © 2011 Pearson Education, Inc. 0.3 Slide 1- 15
  • 16. Irrational number: Any real number that is not rational. Examples: 2, 3, π Real numbers: The union of the rational and irrational numbers. Copyright © 2011 Pearson Education, Inc. Slide 1- 16
  • 17. Objective 3 Graph rational numbers on a number line. Copyright © 2011 Pearson Education, Inc. Slide 1- 17
  • 18. Example 3 4 Graph on a number line. 2 5 Answer The number is located 4/5 of the way between 2 and 3. -1 0 1 2 2 4 3 5 Between 2 and 3, we divide the number line into 5 equally spaced divisions. Place a dot on the 4th mark. Copyright © 2011 Pearson Education, Inc. Slide 1- 18
  • 19. Objective 4 Determine the absolute value of a number. Copyright © 2011 Pearson Education, Inc. Slide 1- 19
  • 20. Absolute value: A given number’s distance from 0 on a number line. ←5 units from 0 → ←5 units from 0 → The absolute value of a number n is written |n|. The absolute value The absolute value of 5 is 5. of −5 is 5. |5| = 5 |−5| = 5 Copyright © 2011 Pearson Education, Inc. Slide 1- 20
  • 21. Absolute Value The absolute value of every real number is either positive or 0. Copyright © 2011 Pearson Education, Inc. Slide 1- 21
  • 22. Example 4 Simplify. a. |−9.4| 2 b. 9 Answer a. |−9.4| = 9.4 2 2 b. = 9 9 Copyright © 2011 Pearson Education, Inc. Slide 1- 22
  • 23. Objective 5 Compare numbers. Copyright © 2011 Pearson Education, Inc. Slide 1- 23
  • 24. Comparing Numbers For any two real numbers a and b, a is greater than b if a is to the right of b on a number line. Equivalently, b is less than a if b is to the left of a on a number line. Copyright © 2011 Pearson Education, Inc. Slide 1- 24
  • 25. Example 5 Use =, <, or > to write a true statement. a. 3 ___ −3 b. −1.8 ___ −1.6 Answer a. 3 ___ −3 3 > −3 because 3 is farther right on a number line. b. −1.8 ___ −1.6 −1.8 < −1.6 because –1.8 is further to the left on a number line. Copyright © 2011 Pearson Education, Inc. Slide 1- 25
  • 26. To which set of numbers does −6 belong? a) Irrational b) Natural and whole numbers c) Natural numbers, whole numbers, and integers d) Integers and rational numbers 1.1 Copyright © 2011 Pearson Education, Inc. Slide 1- 26
  • 27. To which set of numbers does −6 belong? a) Irrational b) Natural and whole numbers c) Natural numbers, whole numbers, and integers rational numbers d) Integers and 1.1 Copyright © 2011 Pearson Education, Inc. Slide 1- 27
  • 28. Simplify |7|. a) 7 b) −7 c) 0 d) 1/7 1.1 Copyright © 2011 Pearson Education, Inc. Slide 1- 28
  • 29. Simplify |7|. a) 7 b) −7 c) 0 d) 1/7 1.1 Copyright © 2011 Pearson Education, Inc. Slide 1- 29
  • 30. Which statement is false? a) 7 > 4 b) −2.4 > −1.4 c) 10 < 22 d) −3.6 > −6.4 1.1 Copyright © 2011 Pearson Education, Inc. Slide 1- 30
  • 31. Which statement is false? a) 7 > 4 b) −2.4 > −1.4 c) 10 < 22 d) −3.6 > −6.4 1.1 Copyright © 2011 Pearson Education, Inc. Slide 1- 31
  • 32. 1.2 1. 2. 3. 4. Fractions Write equivalent fractions. Write equivalent fractions with the LCD. Write the prime factorization of a number. Simplify a fraction to lowest terms. Copyright © 2011 Pearson Education, Inc.
  • 33. Fraction: A quotient of two numbers or expressions a a and b having the form , where b ≠ 0. b 3 4 ← Numerator ← Denominator The top number in a fraction is called the numerator. The bottom number is called the denominator. Fractions indicated part of a whole. Copyright © 2011 Pearson Education, Inc. Slide 1- 33
  • 34. Objective 1 Write equivalent fractions. Copyright © 2011 Pearson Education, Inc. Slide 1- 34
  • 35. Writing Equivalent Fractions For any fraction, we can write an equivalent fraction by multiplying or dividing both its numerator and denominator by the same nonzero number. Copyright © 2011 Pearson Education, Inc. Slide 1- 35
  • 36. Example 1 Find the missing number that makes the fractions equivalent. a. 9 = ? b. − 18 = ? 15 45 Solution a. 9 ? 15 = 45 9× 3 27 = 15 × 3 45 Multiply the numerator and denominator by 3. 36 b. 2 18 ? − = 36 2 18 ÷ 18 1 − =− 36 ÷ 18 2 Divide the numerator and denominator by 6. Copyright © 2011 Pearson Education, Inc. Slide 1- 36
  • 37. Objective 2 Write equivalent fractions with the LCD. Copyright © 2011 Pearson Education, Inc. Slide 1- 37
  • 38. Multiple: A multiple of a given integer n is the product of n and an integer. We can generate multiples of a given number by multiplying the given number by the integers. Multiples of 2 Multiples of 3 2× = 2 1 2 ×2 = 4 2× = 6 3 2 ×4 = 8 2 × = 10 5 2 ×6 = 12 3× = 3 1 3 ×2 = 6 3× = 9 3 3 ×4 = 12 3 × = 15 5 3 ×6 = 18 Copyright © 2011 Pearson Education, Inc. Slide 1- 38
  • 39. Least common multiple (LCM): The smallest number that is a multiple of each number in a given set of numbers. Least common denominator (LCD): The least common multiple of the denominators of a given set of fractions. Copyright © 2011 Pearson Education, Inc. Slide 1- 39
  • 40. Example 2 7 5 Write and as equivalent fractions with the LCD. 8 6 Solution The LCD of 8 and 6 is 24. 7 7× 3 21 = = 8 8× 3 24 5 5 ×4 20 = = 6 6 ×4 24 Copyright © 2011 Pearson Education, Inc. Slide 1- 40
  • 41. Objective 3 Write the prime factorization of a number. Copyright © 2011 Pearson Education, Inc. Slide 1- 41
  • 42. Factors: If a ⋅ b = c, then a and b are factors of c. Example: 6 ⋅ 7 = 42, 6 and 7 are factors of 42 Prime number: A natural number that has exactly two different factors, 1 and the number itself. Example: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,… Prime factorization: A factorization that contains only prime factors. Copyright © 2011 Pearson Education, Inc. Slide 1- 42
  • 43. Example 3 Find the prime factorization of 420. Solution 420 Factor 420 to 10 and 42. (Any Factor 10 to 2 and 5, which are primes. Then factor 42 to 6 and 7. 7 is prime and then factor 6 into 2 and 3, which are primes. [ 42 [ ] 6 [ ] 2 ] 10 [ ] 7 2 3 two factors will work.) 5 Answer 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 7 Copyright © 2011 Pearson Education, Inc. Slide 1- 43
  • 44. Objective 4 Simplify a fraction to lowest terms. Copyright © 2011 Pearson Education, Inc. Slide 1- 44
  • 45. a Lowest terms: Given a fraction and b ≠ 0, if the b only factor common to both a and b is 1, then the fraction is in lowest terms. Copyright © 2011 Pearson Education, Inc. Slide 1- 45
  • 46. Simplifying a Fraction with the Same Nonzero Numerator and Denominator n 1 = = 1, when n ≠ 0. n 1 Eliminating a Common Factor in a Fraction an a × a 1 = = , when b ≠ 0 and n ≠ 0. bn b × b 1 Copyright © 2011 Pearson Education, Inc. Slide 1- 46
  • 47. These rules allow us to write fractions in lowest terms using prime factorizations. The idea is to replace the numerator and denominator with their prime factorizations and then eliminate the prime factors that are common to both the numerator and denominator. Copyright © 2011 Pearson Education, Inc. Slide 1- 47
  • 48. Simplifying a Fraction to Lowest Terms To simplify a fraction to lowest terms: 1. Replace the numerator and denominator with their prime factorizations. 2. Eliminate (divide out) all prime factors common to the numerator and denominator. 3. Multiply the remaining factors. Copyright © 2011 Pearson Education, Inc. Slide 1- 48
  • 49. Example 4a Simplify to lowest terms. 30 42 Solution 2× × 3 5 30 5 = = 42 2 × ×7 3 7 Replace the numerator and denominator with their prime factorizations; then eliminate the common prime factors. Copyright © 2011 Pearson Education, Inc. Slide 1- 49
  • 50. Example 4b 220 Simplify to lowest terms. 2380 Solution 2 ×2 × × 5 11 220 11 = = 2380 2 ×2 × ×7 × 5 17 119 Replace the numerator and denominator with their prime factorizations; then eliminate the common prime factors. Copyright © 2011 Pearson Education, Inc. Slide 1- 50
  • 51. Example 5 At a company, 225 of the 1050 employees have optional eye insurance coverage as part of their benefits package. What fraction of the employees have optional eye insurance coverage? Solution 225 3× × × 3 5 5 3 = = 1050 2 × × × ×7 14 3 5 5 Answer 3 out of 14 employees have optional eye insurance. Copyright © 2011 Pearson Education, Inc. Slide 1- 51
  • 52. What is the prime factorization of 360? a) 6 ⋅ 6 ⋅ 5 b) 23 ⋅ 32 ⋅ 5 c) 22 ⋅ 32 ⋅ 5 d) 32 ⋅ 5 ⋅ 7 1.2 Copyright © 2011 Pearson Education, Inc. Slide 1- 52
  • 53. What is the prime factorization of 360? a) 6 ⋅ 6 ⋅ 5 b) 23 ⋅ 32 ⋅ 5 c) 22 ⋅ 32 ⋅ 5 d) 32 ⋅ 5 ⋅ 7 1.2 Copyright © 2011 Pearson Education, Inc. Slide 1- 53
  • 54. Simplify to lowest terms: a) b) c) d) 1.2 112 280 14 35 1 4 2 5 21 23 Copyright © 2011 Pearson Education, Inc. Slide 1- 54
  • 55. Simplify to lowest terms: a) b) c) d) 1.2 112 280 14 35 1 4 2 5 21 23 Copyright © 2011 Pearson Education, Inc. Slide 1- 55
  • 56. 1.3 1. 2. 3. 4. Adding and Subtracting Real Numbers; Properties of Real Numbers Add integers. Add rational numbers. Find the additive inverse of a number. Subtract rational numbers. Copyright © 2011 Pearson Education, Inc.
  • 57. Objective 1 Add integers. Copyright © 2011 Pearson Education, Inc. Slide 1- 57
  • 58. Parts of an addition statement: The numbers added are called addends and the answer is called a sum. 2+3=5 Addends Sum Copyright © 2011 Pearson Education, Inc. Slide 1- 58
  • 59. Properties of Addition Symbolic Form Word Form Additive Identity a+0=a The sum of a number and 0 is that number. Commutative Property of Addition a+b=b+a Changing the order of addends does not affect the sum. Associative Property of Addition a + (b + c) = (a + b) + c Changing the grouping of three or more addends does not affect the sum. Copyright © 2011 Pearson Education, Inc. Slide 1- 59
  • 60. Example 1 Indicate whether each equation illustrates the additive identity, commutative property of addition, or the associative property of addition. a. (5 + 6) + 3 = 5 + (6 + 3) Answer Associative property of addition b. 0 + (−9) = −9 Answer Additive identity c. (−9 + 6) + 4 = 4 + (−9 + 6) Answer Commutative property of addition Copyright © 2011 Pearson Education, Inc. Slide 1- 60
  • 61. Adding Numbers with the Same Sign To add two numbers that have the same sign, add their absolute values and keep the same sign. Copyright © 2011 Pearson Education, Inc. Slide 1- 61
  • 62. Example 2 Add. a. 27 + 12 b. –16 + (– 22) Solution a. 27 + 12 = 39 b. –16 + (–22) = –38 Copyright © 2011 Pearson Education, Inc. Slide 1- 62
  • 63. Adding Numbers with Different Signs To add two numbers that have different signs, subtract the smaller absolute value from the greater absolute value and keep the sign of the number with the greater absolute value. Copyright © 2011 Pearson Education, Inc. Slide 1- 63
  • 64. Example 3 Add. a. 35 + (–17) b. –29 + 7 Solution a. 35 + (–17) = 18 b. –29 + 7 = –22 Copyright © 2011 Pearson Education, Inc. Slide 1- 64
  • 65. Example 3 continued Add. c. 15 + (–27) d. –32 + 6 Solution c. 15 + (–27) = –12 d. –32 + 6 = –26 Copyright © 2011 Pearson Education, Inc. Slide 1- 65
  • 66. Objective 2 Add rational numbers. Copyright © 2011 Pearson Education, Inc. Slide 1- 66
  • 67. Adding Fractions with the Same Denominator To add fractions with the same denominator, add the numerators and keep the same denominator; then simplify. Copyright © 2011 Pearson Education, Inc. Slide 1- 67
  • 68. Example 4 Add. a. 2 + 4 9 b. − 4 +  − 5   ÷ 12  12  9 Solution a. 2 + 4 = 2 + 4 = 6 9 9 9 9 2 g3 2 = = 3 g3 3 Replace 6 and 9 with their prime factorizations, divide out the common factor, 3, then multiply the remaining factors. 4  5 b. − +  − ÷ 12  12  −4 + ( −5 ) 9 = =− 12 12 3 g3 3 =− =− 3 g2 g 2 4 Simplify to lowest terms by dividing out the common factor, 3. Copyright © 2011 Pearson Education, Inc. Slide 1- 68
  • 69. Example 4 continued Add. c. 7 +  − 3   ÷ 10  10  Solution a. 7 +  − 3  = 7 + (−3) = 4  ÷ 10 10 10  10  2 g2 2 = = 2 g5 5 Simplify to lowest terms by dividing out the common factor, 2. Copyright © 2011 Pearson Education, Inc. Slide 1- 69
  • 70. Adding Fractions To add fractions with different denominators: 1. Write each fraction as an equivalent fraction with the LCD. 2. Add the numerators and keep the LCD. 3. Simplify. Copyright © 2011 Pearson Education, Inc. Slide 1- 70
  • 71. Example 5a 1 1 Add: + 3 4 Solution 1 1 + 3 4 1( 4 ) 1(3) = + 3 ( 4 ) 4(3) 4 3 = + 12 12 7 = 12 Write equivalent fractions with 12 in the denominator. Add numerators and keep the common denominator. Because the addends have the same sign, we add and keep the same sign. Copyright © 2011 Pearson Education, Inc. Slide 1- 71
  • 72. Example 5b 5 3 Add: − + 6 4 Solution 5 ( 2 ) 3(3) 5 3 + − + =− 6 ( 2 ) 4(3) 6 4 10 9 =− + 12 12 −10 + 9 = 12 1 =− 12 Write equivalent fractions with 12 in the denominator. Add numerators and keep the common denominator. Because the addends have different signs, we subtract and keep the sign of the number with the greater absolute value. Copyright © 2011 Pearson Education, Inc. Slide 1- 72
  • 73. Example 5c 7 9 Add: − + 8 30 Solution 7 9 − + 8 30 7 ( 15 ) 9(4) =− + 8 ( 15 ) 30(4) Write equivalent fractions with 120 in the denominator. 105 36 Add numerators and keep =− + the common denominator. 120 120 −105 + 36 Reduce to lowest terms. = 120 −69 3 ×23 23 = =− =− 120 2 ×2 ×2 × × 3 5 40 Copyright © 2011 Pearson Education, Inc. Slide 1- 73
  • 74. Example 6 Anna has a balance of $378.45 and incurs a debt of $85.42. What is Anna’s new balance? Solution A debt of $85.42 is −$85.42. Her balance is 378.45 + (– 85.42) = $293.03 Copyright © 2011 Pearson Education, Inc. Slide 1- 74
  • 75. Objective 3 Find the additive inverse of a number. Copyright © 2011 Pearson Education, Inc. Slide 1- 75
  • 76. Additive inverses: Two numbers whose sum is 0. What happens if we add two numbers that have the same absolute value but different signs, such as 5 + (–5)? In money terms, this is like making a $5 payment towards a debt of $5. Notice the payment pays off the debt so that the balance is 0. 5 + (–5) = 0 Because their sum is zero, we say 5 and –5 are additive inverses, or opposites. Copyright © 2011 Pearson Education, Inc. Slide 1- 76
  • 77. Example 7 Find the additive inverse of the given number. a. 8 b. –2 c. 0 Answers a. –8 because 8 + (–8) = 0 b. 2 because – 2 + 2 = 0 c. 0 because 0 + 0 = 0 Copyright © 2011 Pearson Education, Inc. Slide 1- 77
  • 78. Example 8 Simplify. a. – (–5) b. –|2| c. –| –9| Answers a. – (–5) = 5 b. –|2| = –2 c. –| –9| = –9 Copyright © 2011 Pearson Education, Inc. Slide 1- 78
  • 79. Objective 4 Subtract rational numbers. Copyright © 2011 Pearson Education, Inc. Slide 1- 79
  • 80. Parts of a subtraction statement: 8–5=3 Difference Minuend Subtrahend Copyright © 2011 Pearson Education, Inc. Slide 1- 80
  • 81. Rewriting Subtraction To write a subtraction statement as an equivalent addition statement, change the operation symbol from a minus sign to a plus sign, and change the subtrahend to its additive inverse. Copyright © 2011 Pearson Education, Inc. Slide 1- 81
  • 82. Example 9a Subtract a. –17 – (–5) Solution Write the subtraction as an equivalent addition. –17 – (–5) Change the operation from minus to plus. = –17 + 5 = –12 Change the subtrahend to its additive inverse. Copyright © 2011 Pearson Education, Inc. Slide 1- 82
  • 83. Example 9b 3 1 Subtract: − − 8 4 Solution 3 1 3 1 − − =− − 8 4 8 4 3  1 = − +− ÷ 8  4 3  1(2)  = − +− ÷ 8  4(2)  Write equivalent fractions with the common denominator, 8. 3  2 5 = − +− ÷ = − 8  8 8 Copyright © 2011 Pearson Education, Inc. Slide 1- 83
  • 84. Example 9c c. 4.07 – 9.03 Solution Write the equivalent addition statement. 4.07 – 9.03 = 4.07 + (– 9.03) = –4.96 Copyright © 2011 Pearson Education, Inc. Slide 1- 84
  • 85. Example 10 In an experiment, a mixture begins at a temperature of 52.6°C. The mixture is then cooled to a temperature of −29.4°C. Find the difference between the initial and final temperatures. Solution 52.6 – (–29.4) = 52.6 + 29.4 = 82 Answer The difference between the initial and final temperatures is 82°C. Copyright © 2011 Pearson Education, Inc. Slide 1- 85
  • 86. Add –6 + (–9). a) –15 b) −3 c) 3 d) 15 1.3 Copyright © 2011 Pearson Education, Inc. Slide 1- 86
  • 87. Add –6 + (–9). a) –15 b) −3 c) 3 d) 15 1.3 Copyright © 2011 Pearson Education, Inc. Slide 1- 87
  • 88. Subtract 5 – (–8). a) –13 b) −3 c) 3 d) 13 1.3 Copyright © 2011 Pearson Education, Inc. Slide 1- 88
  • 89. Subtract 5 – (–8). a) –13 b) −3 c) 3 d) 13 1.3 Copyright © 2011 Pearson Education, Inc. Slide 1- 89
  • 90. Subtract a) 16 − 21 b) 1 − 2 c) − d) 1.3 3  1 − −  ÷. 7 3 2 21 2 21 Copyright © 2011 Pearson Education, Inc. Slide 1- 90
  • 91. Subtract a) 16 − 21 b) 1 − 2 c) − d) 1.3 3  1 − −  ÷. 7 3 2 21 2 21 Copyright © 2011 Pearson Education, Inc. Slide 1- 91
  • 92. 1.4 1. 2. 3. 4. 5. Multiplying and Dividing Real Numbers; Properties of Real Numbers Multiply integers. Multiply more than two numbers. Multiply rational numbers. Find the multiplicative inverse of a number. Divide rational numbers. Copyright © 2011 Pearson Education, Inc.
  • 93. Objective 1 Multiply integers. Copyright © 2011 Pearson Education, Inc. Slide 1- 93
  • 94. In a multiplication statement, factors are multiplied to equal a product. 2 g 3 = Factors 6 Product Copyright © 2011 Pearson Education, Inc. Slide 1- 94
  • 95. Properties of Multiplication Multiplicative Property of 0 Multiplicative Identity Symbolic Form Word Form 0 ga = 0 The product of a number multiplied by 0 is 0. 1 ga = a The product of a number multiplied by 1 is the number. Commutative Property of Multiplication ab=ba Changing the order of factors does not affect the product. Associative Property of Multiplication a(bc) = (ab)c Changing the grouping of three or more factors does not affect the product. Distributive Property of Multiplication over Addition a(b + c) =ab + ac A sum multiplied by a factor is equal to the sum of that factor multiplied by each addend. Copyright © 2011 Pearson Education, Inc. Slide 1- 95
  • 96. Example 1 Give the name of the property of multiplication that is illustrated by each equation. a. 6(−3) = −3 ⋅ 6 Answer Commutative property of multiplication b. 3(−9 ⋅ 5) = [3(−9)] ⋅ 5 Answer Associative property of multiplication c. 4(4 – 2) = 4 ⋅ 4 – 4 ⋅ 2 Answer Distributive property of multiplication over addition Copyright © 2011 Pearson Education, Inc. Slide 1- 96
  • 97. Multiplying Two Numbers with Different Signs When multiplying two numbers that have different signs, the product is negative. Copyright © 2011 Pearson Education, Inc. Slide 1- 97
  • 98. Example 2 Multiply. a. 7(–4) Solution a. 7(–4) = –28 b. (–15)3 = –45 b. (–15)3 Warning: Make sure you see the difference between 7(–4), which indicates multiplication, and 7 – 4, which indicates subtraction. Copyright © 2011 Pearson Education, Inc. Slide 1- 98
  • 99. Multiplying Two Numbers with the Same Sign When multiplying two numbers that have the same sign, the product is positive. Copyright © 2011 Pearson Education, Inc. Slide 1- 99
  • 100. Example 3 Multiply. a. –5(–9) b. (–6)(–8) Solution a. –5(–9) = 45 b. (–6)(–8) = 48 Copyright © 2011 Pearson Education, Inc. Slide 1- 100
  • 101. Objective 2 Multiply more than two numbers. Copyright © 2011 Pearson Education, Inc. Slide 1- 101
  • 102. Multiplying with Negative Factors The product of an even number of negative factors is positive, whereas the product of an odd number of negative factors is negative. Copyright © 2011 Pearson Education, Inc. Slide 1- 102
  • 103. Example 4 Multiply. a. (–1)(–3)(–6)(7) Solution Because there are three negative factors (an odd number of negative factors), the result is negative. (–1)(–3)(–6)(7) = –126 b. (–2)(–4)(2)(–5)(–3) Solution Because there are four negative factors(an even number of negative factors), the result is positive. (–2)(–4)(2)(–5)(–3) = 240 Copyright © 2011 Pearson Education, Inc. Slide 1- 103
  • 104. Objective 3 Multiply rational numbers. Copyright © 2011 Pearson Education, Inc. Slide 1- 104
  • 105. Multiplying Fractions a c ac g = , where b ≠ 0 and d ≠ 0. b d bd Copyright © 2011 Pearson Education, Inc. Slide 1- 105
  • 106. Example 5a 3 4 Multiply − g  ÷. 5 9 Solution 3 4 3  2 g2  − g  ÷= − g  ÷ 5 9 5  3 g3  4 =− 15 Divide out the common factor, 3. Because we are multiplying two numbers that have different signs, the product is negative. Copyright © 2011 Pearson Education, Inc. Slide 1- 106
  • 107. Example 5b 6  6  12 Multiply − × − ÷× 15  16  15 Solution 6  6  12 2×  3 2 ×  2 ×2 × 3 3 − × − ÷× =− × −  ÷× 15  16  15 3 ×  2 ×2 ×2 ×2  3 × 5 5 3 = 25 Copyright © 2011 Pearson Education, Inc. Divide out the common factors. Because there are an even number of negative factors, the product is positive. Slide 1- 107
  • 108. Multiplying Decimal Numbers To multiply decimal numbers: 1. Multiply as if they were whole numbers. 2. Place the decimal in the product so that it has the same number of decimal places as the total number of decimal places in the factors. Copyright © 2011 Pearson Education, Inc. Slide 1- 108
  • 109. Example 6a Multiply (–7.6)(0.04). Solution First, calculate the value and disregard signs for now. 0.04 2 places 7.6 + 1 place × 024 +0280 3 places 0.3 0 4 Answer –0.304 When we multiply two numbers with different signs, the product is negative. Copyright © 2011 Pearson Education, Inc. Slide 1- 109
  • 110. Example 6b Multiply (−3)(5.2)(1.4)(−6.1). Solution First, calculate the value and disregard signs for now. Multiply from left to right. (−3)(5.2)(1.4)(−6.1) = (15.6)(1.4)(−6.1) = 21.84(6.1) = 133.224 Answer 133.224 15.6 = (3)(5.2) 21.84 = 15.6(1.4) The product of an even number of negative factors is positive. The factors have a total of 3 decimal places, so the product has three decimal places. Copyright © 2011 Pearson Education, Inc. Slide 1- 110
  • 111. Objective 4 Find the multiplicative inverse of a number. Copyright © 2011 Pearson Education, Inc. Slide 1- 111
  • 112. Multiplicative inverses: Two numbers whose product is 1. 3 2 and 2 are multiplicative inverses 3 because their product is 1. 2 3 6 g = =1 3 2 6 Notice that to write a number’s multiplicative inverse, we simply invert the numerator and denominator. Multiplicative inverses are also known as reciprocals. Copyright © 2011 Pearson Education, Inc. Slide 1- 112
  • 113. Example 7 Find the multiplicative inverse. a. 2 b. − 1 7 c. −9 8 Answer 7 a. The multiplicative inverse is . 2 b. The multiplicative inverse is −8. 1 c. The multiplicative inverse is − . 9 Copyright © 2011 Pearson Education, Inc. Slide 1- 113
  • 114. Objective 5 Divide rational numbers. Copyright © 2011 Pearson Education, Inc. Slide 1- 114
  • 115. Parts of a division statement: 8 ÷ 2 = Dividend 4 Quotient Divisor Copyright © 2011 Pearson Education, Inc. Slide 1- 115
  • 116. Dividing Signed Numbers When dividing two numbers that have the same sign, the quotient is positive. When dividing two numbers that have different signs, the quotient is negative. Copyright © 2011 Pearson Education, Inc. Slide 1- 116
  • 117. Example 8 Divide. a. 56 ÷ (−8) b. −72 ÷ ( −6 ) Solution a. 56 ÷ (−8) = −7 b.−72 ÷ ( −6 ) = 12 Copyright © 2011 Pearson Education, Inc. Slide 1- 117
  • 118. Division Involving 0 0 ÷ n = 0 when n ≠ 0. n ÷ 0 is undefined when n ≠ 0. 0 ÷ 0 is indeterminate. Copyright © 2011 Pearson Education, Inc. Slide 1- 118
  • 119. Dividing Fractions a c a d ÷ = g , where b ≠ 0, c ≠ 0, and d ≠ 0. b d b c Copyright © 2011 Pearson Education, Inc. Slide 1- 119
  • 120. Example 9 3 4 Divide − ÷ . 10 5 Solution 3 4 3 5 − ÷ =− g 10 5 10 4 3 5 =− g 5 g2 2 g2 3 =− 8 Write an equivalent multiplication. Divide out the common factor, 5. Because we are dividing two numbers that have different signs, the result is negative. Copyright © 2011 Pearson Education, Inc. Slide 1- 120
  • 121. Dividing Decimal Numbers To divide decimal numbers, set up a long division and consider the divisor. Case 1: If the divisor is an integer, divide as if the dividend were a whole number and place the decimal point in the quotient directly above its position in the dividend. Case 2: If the divisor is a decimal number, 1. Move the decimal point in the divisor to the right enough places to make the divisor an integer. 2. Move the decimal point in the dividend the same number of places. Copyright © 2011 Pearson Education, Inc. Slide 1- 121
  • 122. Dividing Decimal Numbers continued 3. Divide the divisor into the dividend as if both numbers were whole numbers. Make sure you align the digits in the quotient properly. 4. Write the decimal point in the quotient directly above its new position in the dividend. In either case, continue the division process until you get a remainder of 0 or a repeating digit (or block of digits) in the quotient. Copyright © 2011 Pearson Education, Inc. Slide 1- 122
  • 123. Example 10 Divide −44.64 ÷ (−3.6) Solution Because the divisor is a decimal number, we move the decimal point enough places to the right to create an integer—in this case, one place. Then we move the decimal point one place to the right in the dividend. Because we are dividing two numbers with the same sign, the result is positive. Copyright © 2011 Pearson Education, Inc. Slide 1- 123
  • 124. Example 10 continued Divide −44.64 ÷ (−3.6) Solution 12.4 36 446.4 36 86 72 144 144 0 Copyright © 2011 Pearson Education, Inc. Slide 1- 124
  • 125. Example 11 Martha was decorating cookies. She used 2/3 of a container of frosting that was 3/4 full. What fractional part of the container did she use? Solution To find 2/3 of 3/4, multiply. 2 3 2× 3 1 × = = 3 4 2 ×2 × 2 3 Copyright © 2011 Pearson Education, Inc. Slide 1- 125
  • 126. Multiply (–6)(–3)(7). a) 126 b) −126 c) –63 d) 63 1.4 Copyright © 2011 Pearson Education, Inc. Slide 1- 126
  • 127. Multiply (–6)(–3)(7). a) 126 b) −126 c) –63 d) 63 1.4 Copyright © 2011 Pearson Education, Inc. Slide 1- 127
  • 128. Divide −14.6 ÷ ( −0.03) . a) −48.6 b) 48.6 c) 486.6 d) −486.6 1.4 Copyright © 2011 Pearson Education, Inc. Slide 1- 128
  • 129. Divide −14.6 ÷ ( −0.03) . a) −48.6 b) 48.6 c) 486.6 d) −486.6 1.4 Copyright © 2011 Pearson Education, Inc. Slide 1- 129
  • 130. 1.5 Exponents, Roots, and Order of Operations 1. Evaluate numbers in exponential form. 2. Evaluate square roots. 3. Use the order-of-operations agreement to simplify numerical expressions. 4. Find the mean of a set of data. Copyright © 2011 Pearson Education, Inc.
  • 131. Objective 1 Evaluate numbers in exponential form. Copyright © 2011 Pearson Education, Inc. Slide 1- 131
  • 132. Sometimes problems involve repeatedly multiplying the same number. In such problems, we can use an exponent to indicate that a base number is repeatedly multiplied. Exponent: A symbol written to the upper right of a base number that indicates how many times to use the base as a factor. Base: The number that is repeatedly multiplied. Copyright © 2011 Pearson Education, Inc. Slide 1- 132
  • 133. When we write a number with an exponent, we say the expression is in exponential form. The 24 is in exponential form, where the expression base is 2 and the exponent is 4. To evaluate 24 , write 2 as a factor 4 times, then multiply. Four 2s 2 = 2 g 2 g 2 g 2 = 16 4 Base Exponent Copyright © 2011 Pearson Education, Inc. Slide 1- 133
  • 134. Evaluating an Exponential Form To evaluate an exponential form raised to a natural number exponent, write the base as a factor the number of times indicated by the exponent; then multiply. Copyright © 2011 Pearson Education, Inc. Slide 1- 134
  • 135. Example 1a Evaluate. (–9)2 Solution The exponent 2 indicates we have two factors of –9. Because we multiply two negative numbers, the result is positive. (–9)2 = (–9)(–9) = 81 Copyright © 2011 Pearson Education, Inc. Slide 1- 135
  • 136. Example 1b 3 Evaluate.  3 − ÷  5 Solution The exponent 3 means we must multiply the base by itself three times. 3  3  3  3   3  − ÷ =  − ÷ − ÷ − ÷   5  5  5   5  27 =− 125 Copyright © 2011 Pearson Education, Inc. Slide 1- 136
  • 137. Evaluating Exponential Forms with Negative Bases If the base of an exponential form is a negative number and the exponent is even, then the product is positive. If the base is a negative number and the exponent is odd, then the product is negative. Copyright © 2011 Pearson Education, Inc. Slide 1- 137
  • 138. Example 2 Evaluate. (−3) 4 −34 (−2)3 a. b. c. Solution a. (−3)4 = (−3)(−3)(−3)(−3) = 81 b. −34 = −3 ×3 ×3 ×3 = −81 c. (−2)3 = ( −2)( −2)(−2) = −8 d. d. −23 −23 = −2 ×2 ×2 = −8 Copyright © 2011 Pearson Education, Inc. Slide 1- 138
  • 139. Objective 2 Evaluate square roots. Copyright © 2011 Pearson Education, Inc. Slide 1- 139
  • 140. Roots are inverses of exponents. More specifically, a square root is the inverse of a square, so a square root of a given number is a number that, when squared, equals the given number. Square Roots Every positive number has two square roots, a positive root and a negative root. Negative numbers have no real-number square roots. Copyright © 2011 Pearson Education, Inc. Slide 1- 140
  • 141. Example 3 Find all square roots of the given number. Solution a. 49 Answer ± 7 b. −81 Answer No real-number square roots exist. Copyright © 2011 Pearson Education, Inc. Slide 1- 141
  • 142. The symbol, , called the radical, is used to indicate finding only the positive (or principal) square root of a given number. The given number or expression inside the radical is called the radicand. Radical Principal Square Root 25 = 5 Radicand Copyright © 2011 Pearson Education, Inc. Slide 1- 142
  • 143. Square Roots Involving the Radical Sign The radical symbol denotes only the positive (principal) square root. a = b a b , where a ≥ 0 and b > 0. Copyright © 2011 Pearson Education, Inc. Slide 1- 143
  • 144. Example 4 Evaluate the square root. a. 169 b. 64 c. 81 0.64 d. −25 Solution a. 169 = 13 b. c. d. −25 = not a real number 0.64 = 0.8 64 8 = 81 9 Copyright © 2011 Pearson Education, Inc. Slide 1- 144
  • 145. Objective 3 Use the order-of-operations agreement to simplify numerical expressions. Copyright © 2011 Pearson Education, Inc. Slide 1- 145
  • 146. Order-of- Operations Agreement Perform operations in the following order: 1. Within grouping symbols: parentheses ( ), brackets [ ], braces { }, above/below fraction bars, absolute value | |, and radicals . 2. Exponents/Roots from left to right, in order as they occur. 3. Multiplication/Division from left to right, in order as they occur. 4. Addition/Subtraction from left to right, in order as they occur. Copyright © 2011 Pearson Education, Inc. Slide 1- 146
  • 147. Example 5a Simplify. −26 + 15 ÷ (−5) ×2 Solution −26 + 15 ÷ (−5) ×2 = −26 + (−3) ×2 Divide 15 ÷ (−5) = –3 = −26 + (−6) Multiply (–3) ⋅ 2 = –6 = −32 Add –26 + (–6) = –32 Copyright © 2011 Pearson Education, Inc. Slide 1- 147
  • 148. Example 5b Simplify. Solution −34 + 2 12 − 20 −34 + 2 12 − 20 = −34 + 2 −8 Subtract inside the absolute value. = −34 + 2 × 8 Simplify the absolute value. = −81 + 2 × 8 Evaluate the exponent. = −81 + 16 Multiply. = −65 Add. Copyright © 2011 Pearson Education, Inc. Slide 1- 148
  • 149. Example 5c Simplify. ( −3) 2 + 5 6 − ( 2 + 1)  − 49   Solution ( −3) 2 + 5 6 − ( 2 + 1)  −   = ( −3) + 5 [ 6 − 3] − 49 2 = 9 + 5 ( 3) − 7 49 Calculate within the innermost parenthesis. Evaluate the exponential form, brackets, and square root. = 9 + 15 − 7 Multiply 5(3). = 24 − 7 Add 9 + 15. = 17 Subtract 24 – 7. Copyright © 2011 Pearson Education, Inc. Slide 1- 149
  • 150. Square Root of a Product or Quotient If a square root contains multiplication or division, we can multiply or divide first, then find the square root of the result, or we can find the square roots of the individual numbers, then multiply or divide the square roots. Square Root of a Sum or Difference When a radical contains addition or subtraction, we must add or subtract first, then find the root of the sum or difference. Copyright © 2011 Pearson Education, Inc. Slide 1- 150
  • 151. Example 6a Simplify. 13.5 ÷ 5 ( −4 ) − 3 142 − 21 Solution 13.5 ÷ 5 ( −4 ) − 3 142 − 21 2 2 = 13.5 ÷ 5 ( −4 ) − 3 121 2 = 13.5 ÷ 5 ( 16 ) − 3(11) Subtract within the radical. Evaluate the exponential form and root. = 2.7 ( 16 ) − 3 ( 11) Divide. = 43.2 − 33 Multiply. = 10.2 Subtract. Copyright © 2011 Pearson Education, Inc. Slide 1- 151
  • 152. Sometimes fraction lines are used as grouping symbols. When they are, we simplify the numerator and denominator separately, then divide the results. Copyright © 2011 Pearson Education, Inc. Slide 1- 152
  • 153. Example 7a Simplify. Solution 8(−5) − 23 4(8) − 8 8( −5) − 23 4(8) − 8 8(−5) − 8 = 4(8) − 8 = −40 − 8 32 − 8 Evaluate the exponential form in the numerator and multiply in the denominator. Multiply in the numerator and subtract in the denominator. −48 = 24 Subtract in the numerator. = −2 Divide. Copyright © 2011 Pearson Education, Inc. Slide 1- 153
  • 154. Example 7b Simplify. 9(4) + 12 43 + (8)(−8) 9(4) + 12 43 + (8)(−8) Solution = 36 + 12 64 + (8)(−8) 48 = 64 + (−64) 48 = 0 Because the denominator or divisor is 0, the answer is undefined. Copyright © 2011 Pearson Education, Inc. Slide 1- 154
  • 155. Objective 4 Find the mean of a set of data. Copyright © 2011 Pearson Education, Inc. Slide 1- 155
  • 156. Finding the Arithmetic Mean To find the arithmetic mean, or average, of n numbers, divide the sum of the numbers by n. Arithmetic mean = x1 + x2 + ... + xn n Copyright © 2011 Pearson Education, Inc. Slide 1- 156
  • 157. Example 8 Bruce has the following test scores in his biology class: 92, 96, 81, 89, 95, 93. Find the average of his test scores. Solution 92 + 96 + 81 + 89 + 95 + 93 546 = 6 6 Divide the sum of the 6 scores by 6. = 91 Copyright © 2011 Pearson Education, Inc. Slide 1- 157
  • 158. Simplify using order of operations. ( −6 ) 2 − 18 ÷ ( 9 − 6 ) a) − 18 b) 6 c) 30 d) 36 1.5 Copyright © 2011 Pearson Education, Inc. Slide 1- 158
  • 159. Simplify using order of operations. ( −6 ) 2 − 18 ÷ ( 9 − 6 ) a) − 18 b) 6 c) 30 d) 36 1.5 Copyright © 2011 Pearson Education, Inc. Slide 1- 159
  • 160. Simplify using order of operations. 2 ( 4 + 23 ) a) 8 300 b) 250 361 c) 6 + 30 − ( 2 + 4 ) 2 2 11 d) undefined 1.5 Copyright © 2011 Pearson Education, Inc. Slide 1- 160
  • 161. Simplify using order of operations. 2 ( 4 + 23 ) a) 8 300 b) 250 361 c) 6 + 30 − ( 2 + 4 ) 2 2 11 d) undefined 1.5 Copyright © 2011 Pearson Education, Inc. Slide 1- 161
  • 162. 1.6 Translating Word Phrases to Expressions 1. Translate word phrases to expressions. Copyright © 2011 Pearson Education, Inc.
  • 163. Objective 1 Translating word phrases to Expressions Copyright © 2011 Pearson Education, Inc. Slide 1- 163
  • 164. Translating Basic Phrases Addition Translation Subtraction Translation The sum of x and three x+3 The difference of x and three x–3 h plus k h+k h minus k h–k seven added to t 7+t seven subtracted from t t–7 three more than a number n+3 three less than a number n–3 y increased by two y+2 y decreased by two y–2 Note: Since addition is a commutative operation, it does not matter in what order we write the translation. Note: Subtraction is not a commutative operation; therefore, the way we write the translation matters. Copyright © 2011 Pearson Education, Inc. Slide 1- 164
  • 165. Translating Basic Phrases Multiplication Translation The product of x and three Translation The quotient of x and three hk x ÷ 3 or x h divided by k 3x h times k Division h ÷ k or h Twice a number 2n h divided into k Triple the number 3n The ratio of a to b Two-thirds of a number 3 k k ÷ h or k h a ÷ b or a b 2 n 3 Note: Like addition, multiplication is a commutative operation: it does not matter in what order we write the translation. Note: Division is like subtraction in that it is not a commutative operation; therefore, the way we write the translation matters. Copyright © 2011 Pearson Education, Inc. Slide 1- 165
  • 166. Translating Basic Phrases Exponents Translation c squared c2 The square of b k3 The cube of b b3 n to the fourth power n4 y raised to the fifth power The square root of x Translation b2 k cubed Roots y5 Copyright © 2011 Pearson Education, Inc. x Slide 1- 166
  • 167. The key words sum, difference, product, and quotient indicate the answer for their respective operations. sum of x and 3 difference of x and 3 x+3 x–3 product of x and 3 quotient of x and 3 x÷3 x⋅3 Copyright © 2011 Pearson Education, Inc. Slide 1- 167
  • 168. Example 1 Translate to an algebraic expression. a. five more than two times a number Translation: 5 + 2n or 2n + 5 b. seven less than the cube of a number Translation: n3 – 7 c. the sum of h raised to the fourth power and twelve Translation: h4 + 12 Copyright © 2011 Pearson Education, Inc. Slide 1- 168
  • 169. Translating Phrases Involving Parentheses Sometimes the word phrases imply an order of operations that would require us to use parentheses in the translation. These situations arise when the phrase indicates that a sum or difference is to be calculated before performing a higher-order operation such as multiplication, division, exponent, or root. Copyright © 2011 Pearson Education, Inc. Slide 1- 169
  • 170. Example 2 Translate to an algebraic expression. a. seven times the sum of a and b Translation: 7(a + b) b. the product of a and b divided by the sum of w2 and 4 ab 2 Translation: ab ÷ (w + 4) or 2 w +4 Copyright © 2011 Pearson Education, Inc. Slide 1- 170
  • 171. Translate the phrase to an algebraic expression. Twelve less than three times a number a) 3n + 12 b) 12 – 3n c) 3n – 12 d) 3n ⋅ 12 1.6 Copyright © 2011 Pearson Education, Inc. Slide 1- 171
  • 172. Translate the phrase to an algebraic expression. Twelve less than three times a number a) 3n + 12 b) 12 – 3n c) 3n – 12 d) 3n ⋅ 12 1.6 Copyright © 2011 Pearson Education, Inc. Slide 1- 172
  • 173. Translate the phrase to an algebraic expression. The difference of a and b decreased by the sum of w and z a) (a – b) – (w + z) b) a – b – (w + z) c) ab – (w + z) d) (b – a) – (w + z) 1.6 Copyright © 2011 Pearson Education, Inc. Slide 1- 173
  • 174. Translate the phrase to an algebraic expression. The difference of a and b decreased by the sum of w and z a) (a – b) – (w + z) b) a – b – (w + z) c) ab – (w + z) d) (b – a) – (w + z) 1.6 Copyright © 2011 Pearson Education, Inc. Slide 1- 174
  • 175. 1.7 Evaluating and Rewriting Expressions 1. Evaluate an expression. 2. Determine all values that cause an expression to be undefined. 3. Rewrite an expression using the distributive property. 4. Rewrite an expression by combining like terms. Copyright © 2011 Pearson Education, Inc.
  • 176. Objective 1 Evaluate an expression. Copyright © 2011 Pearson Education, Inc. Slide 1- 176
  • 177. Evaluating an Algebraic Expression To evaluate an algebraic expression: 1. Replace the variables with their corresponding given values. 2. Calculate the numerical expression using the order of operations. Copyright © 2011 Pearson Education, Inc. Slide 1- 177
  • 178. Example 1a Evaluate 3w – 4(a – 6) when w = 5 and a = 7. Solution 3w – 4(a − 6) 3(5) – 4(7 – 6) = 3(5) – 4(1) = 15 – 4 = 11 Replace w with 5 and a with 7. Simplify inside the parentheses first. Multiply. Subtract. Copyright © 2011 Pearson Education, Inc. Slide 1- 178
  • 179. Example 1b Evaluate x2 – 0.4xy + 9, when x = 7 and y = –2. Solution x2 – 0.4xy + 9 (7)2 – 0.4(7)(–2) + 9 = 49 – 0.4(7)(–2) + 9 = 49 – (–5.6) + 9 = 49 + 5.6 + 9 = 63.6 Replace x with 7 and y with –2. Begin calculating by simplifying the exponential form. Multiply. Write the subtraction as an equivalent addition. Add from left to right. Copyright © 2011 Pearson Education, Inc. Slide 1- 179
  • 180. Objective 2 Determine all values that cause an expression to be undefined. Copyright © 2011 Pearson Education, Inc. Slide 1- 180
  • 181. When evaluating a division expression in which the divisor or denominator contains a variable or variables, we must be careful about what values replace the variable(s). We often need to know what values could replace the variable(s) and cause the expression to be undefined or indeterminate. Copyright © 2011 Pearson Education, Inc. Slide 1- 181
  • 182. Example 2 Determine all values that cause the expression to be undefined. −2 a. 8 b. ( x + 2)( x − 9) x+4 Answer 8 8 a. If x = −4, we have −4 + 4 = 0 , which is undefined because the denominator is 0. b. If x = −2 or 9 the fraction will be undefined since the denominator will = 0. −2 −2 = ( −2 + 2)(−2 − 9) 0 −2 −2 = (9 + 2)(9 − 9) 0 Copyright © 2011 Pearson Education, Inc. Slide 1- 182
  • 183. Objective 3 Rewrite an expression using the distributive property. Copyright © 2011 Pearson Education, Inc. Slide 1- 183
  • 184. The Distributive Property of Multiplication over Addition a(b + c) = ab + ac This property gives us an alternative to the order of operations. 2(5 + 6) = 2(11) 2(5 + 6) = 2⋅5 + 2⋅6 = 22 = 10 + 12 = 22 Copyright © 2011 Pearson Education, Inc. Slide 1- 184
  • 185. Example 3 Use the distributive property to write an equivalent expression and simplify. a. 3(x + 3) b. –5(w – 4) Solution a. 3(x + 3) = 3 ⋅ x + 3 ⋅ 3 = 3x + 9 b. –5(w – 4) = –5 ⋅ w – (–5) ⋅ 4 = –5w + 20 Copyright © 2011 Pearson Education, Inc. Slide 1- 185
  • 186. Objective 4 Rewrite an expression by combining like terms. Copyright © 2011 Pearson Education, Inc. Slide 1- 186
  • 187. Terms: Expressions that are the addends in an expression that is a sum. Coefficient: The numerical factor in a term. The coefficient of 5x3 is 5. The coefficient of –8y is –8. Like terms: Variable terms that have the same variable(s) raised to the same exponents, or constant terms. Like terms Unlike terms 4x and 7x 2x and 8y different variables 5y2 and 10y2 7t3 and 3t2 different exponents 8xy and 12xy x2y and xy2 different exponents 7 and 15 13 and 15x different variables Copyright © 2011 Pearson Education, Inc. Slide 1- 187
  • 188. Combining Like Terms To combine like terms, add or subtract the coefficients and keep the variables and their exponents the same. Copyright © 2011 Pearson Education, Inc. Slide 1- 188
  • 189. Example 4 Combine like terms. a. 10y + 8y Solution 10y + 8y = 18y b. 8x – 3x Solution 8x – 3x = 5x c. 13y2 – y2 Solution 13y2 – y2 = 12y2 Copyright © 2011 Pearson Education, Inc. Slide 1- 189
  • 190. Example 5 Combine like terms in 5y2 + 6 + 4y2 – 7. Solution 5y2 + 6 + 4y2 – 7 = 5y2 + 4y2 + 6 – 7 Combine like terms. = 9y2 – 1 Copyright © 2011 Pearson Education, Inc. Slide 1- 190
  • 191. Example 6 Combine like terms in 18y + 7x – y – 7x. Solution 18y + 7x – y – 7x = 17y + 0 = 17y Copyright © 2011 Pearson Education, Inc. Slide 1- 191
  • 192. Example 7 1 1 Combine like terms in a + 4b + 3 + a − b. 12 6 Solution 1 1 a + 4b + 3 + a − b 12 6 1 1 = a + a + 4b − b + 3 Collect like terms. 12 6 1 1(2) Write the fraction coefficients as = a+ a + 4b − b + 3 equivalent fractions with their 12 6(2) LCD, 12. 1 2 = a + a + 4b − b + 3 12 12 1 3 Combine like terms. = a + 3b + 3 = a + 3b + 3 4 12 Slide 1- 192 Copyright © 2011 Pearson Education, Inc.
  • 193. Evaluate the expression 4(a + b) when a = 3 and b = –2. a) 4 b) −4 c) 12 d) 20 1.7 Copyright © 2011 Pearson Education, Inc. Slide 1- 193
  • 194. Evaluate the expression 4(a + b) when a = 3 and b = –2. a) 4 b) −4 c) 12 d) 20 1.7 Copyright © 2011 Pearson Education, Inc. Slide 1- 194
  • 195. For which values is the expression undefined? 8m (m + 2)(m − 5) a) 8 b) −2 c) −2 and 5 d) 2 and −5 1.7 Copyright © 2011 Pearson Education, Inc. Slide 1- 195
  • 196. For which values is the expression undefined? 8m (m + 2)(m − 5) a) 8 b) −2 c) −2 and 5 d) 2 and −5 1.7 Copyright © 2011 Pearson Education, Inc. Slide 1- 196
  • 197. Simplify: 7x + 8 – 2x – 4 a) 9x – 4 b) 9x + 4 c) 5x – 4 d) 5x + 4 1.7 Copyright © 2011 Pearson Education, Inc. Slide 1- 197
  • 198. Simplify: 7x + 8 – 2x – 4 a) 9x – 4 b) 9x + 4 c) 5x – 4 d) 5x + 4 1.7 Copyright © 2011 Pearson Education, Inc. Slide 1- 198