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Cei03 ppt 01
- 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.
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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.
- 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
- 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
- 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
- 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.
- 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
- 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
- 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
- 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
- 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.
- 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
- 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
- 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.
- 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.
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- 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