HfH Homespun for Homeschoolers
Web site: http://www.homespun4homeschoolers.com
"Uncle Dan’s Algebra"
Copyright 2000
"It sure helps to have a teacher!"
"Algebra Workbook"
There is a video and answer CD to this work book contact
Email: uncledan@homespun4homeschoolers.com
Table of Contents – Next Page
Workbook TABLE OF CONTENTS
Part 1  Basic Operations and Solving Equations
Introduction to Integers and the Number Line 1
Addition of Integers 2
Subtraction of Integers 2
QUIZ: (Adding and Subtracting Integers) 4
Packages  "Do Me First" 5
Multiplication of Integers 6
Division of Integers 6
Operations with Integers Using Packages 7
QUIZ: (Operations with Integers Using Packages) 8
TEST Operations with Integers 9
Exponents and Raising a Number to a Power 10
Order of Operations 10
QUIZ: (Exponents and Powers) 13
Evaluating Variable Expressions 14
Evaluating Variable Expressions with Exponents 16
More Evaluating Expressions 18
TEST Evaluating Variable Expressions 20
Solving Equations 21
Order of Operations When Solving Equations 25
Combining Like Terms 26
Combining Terms: First Step in Solving Equations 26
Transposing: Are You Ready to Skip a Step? 27
QUIZ: (Solving Equations) 27
TEST Solving Equations 28
The Distributive Law 30
More Solving Equations 32
TEST More Solving Equations 33
Part 2  Solving Word Problems with Algebra
Writing Variable Expressions 34
Number Problems 36
ConsecutiveInteger Problems 39
Age Problems 41
Geometry Problems: Angles and Degree Measure 43
Coin Problems: Penny, Nickel, Dime, Quarter 44
Uniform Motion Problems: Rate x Time = Distance 46
TEST Solving Word Problems with Algebra 49
Part 3  Coordinate Geometry and Systems of Equations
Graphing Linear Equations on a Coordinate Plane 51
Using the SlopeIntercept Form: y = mx + b 56
More on Slopes 58
Solving a System of Equations Graphically 62
TEST Coordinate Geometry 65
Solving Systems of Equations in Two Variable 68
Problems Using Two Variables  Digit Problems 72
TEST Systems of Equations 74
Part 4  Polynomials
Introduction to Polynomials 75
Adding Polynomials 75
Subtracting Polynomials 75
Multiplying and Powers: The Laws of Exponents for Multipliction 76
QUIZ: (Multiplying and Powers) 77
More Complicated Expressions Involving Multiplying and Powers 77
Multiplying a Monomial by a Polynomial 78
Multiplying a Polynomial by a Polynomial 79
Are You Ready to Skip a Step The "FOIL" Method 80
The Laws of Exponents for Division 81
QUIZ: (The Laws of Exponents for Division) 84
Division of a Polynomial by a Monomial 85
Division of a Polynomial by a Binomial 85
TEST Basic Operations with Polynomials 87
Special Products 88
Factoring Out a Common Monomial Factor 89
Factoring SpecialProduct Polynomials 89
Factoring Quadratic Trinomials 90
Factoring Completely 91
TEST Factoring 92
Part 5  Fractions
Equivalent Fractions 93
Multiplying Fractions 96
Dividing Fractions 97
Adding and Subtracting Like Fractions 98
Adding and Subtracting Unlike Fractions 99
Simplifying Complex Fractions 103
Solving Equations Containing Fractions 106
Solving Quadratic Equations by Factoring 112
Fractional Equations that Transform into Quadratic Equations 113
TEST  Operations and Equations with Fractions 115
Solving Word Problems Involving Quadratic Terms, Factoring, Etc. 118
Solving Word Problems Involving Equations with Fractions 121
TEST  Equations involving Quadratic Equations and Fractions 124
Part 6  Radicals and Roots
Introduction to Rational and Irrational Numbers 125
Square Roots 125
The Laws of Square Roots 126
Adding and Subtracting Radicals 132
Muliplying Radicals 134
Dividing Radicals 136
Solving Radical Equations 139
TEST Radicals and Roots 142
Part 7  More on Quadratic Equations
Solving Quadratic Equations by Completing the Square 144
The Derivation of the Quadratic Formula 147
Using the Quadratic Formula 147
The Discriminate and the Nature of the Roots 148
TEST  Completing the Square and The Quadratic Formula 150
Part 8  Inequalities
Introduction to Inequalities 151
Solving Inequalities 154
Combining Inequalities 156
Absolute Values in Open sentences 161
TEST  Inequalities 164
Graphing Linear Inequalities in Two Variables 166
Solving a System of Inequalities Graphically 168
TEST  Graphing Linear Inequalities in Two Variables 169
Part 9  Supplementary Topics
Calculating Square Roots 170
TEST  Calculating Square Roots 173
Changing Repeating Decimals to Fractions 174
TEST  Changing Repeating Decimals to Fractions 175
Scientific Notation 176
TEST  Scientific Notation 177
The Distance Formula 178
TEST  The Distance Formula 179
Workbook
Part 1  Basic Operations and Solving Equations
Introduction to Integers and the Number Line
{1, 2, 3, 4, . . . } = the "counting numbers"
{0, 1, 2, 3, 4, . . . } = the "natural numbers"
{ . . . 4, 3, 2, 1, 0, 1, 2, 3, 4, . . . } = the "integers"
The Number Line:
l l l l l l l l l l l l l l l l l l l
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9
= means "is equal to"
> means "is greater than" or "is to the right of"
< means "is less than" or "is to the left of"
5 = 5 3 + 2 = 5 5 > 3 3 < 5
3 < 2 4 < 2 3 > 5 2 > 5
Practice: (Put in the correct sign: =, >, or <)
8 8 8 10 12 5
4 7 6 1 5 2
3 6 3 0 0 4
5 + 2 7 3 + 1 5 8  7 4
Note: The value of a number not considering the sign is the "absolute value".
l 5 l = 5 l5 l = 5
Practice:
l 8 l = l8 l =
1
Addition of Integers
5 + 3 = (3) + (+5) =
(5) + (3) = (3) + (5) =
Note: Since numbers can be added in any order, we say that addition is
"commutative".
Rule: When adding two integers with different signs:
1) Find the difference between the absolute values of the two integers.
2) Use the sign of the integer with the greater absolute value.
5 + (3) = (5) + 3 =
Practice: (Show number line for each problem)
3 + 4 = 3 + (4) =
4 + 3 = 4 + (3) =
37 + 45 = 33 + 78 =
53 + (93) = 33 + (47) =
Subtraction of Integers
Rule: To subtract one integer from another, add the "opposite" of the number to
be subtracted.
Note: You may use the number line to help you only after you have changed the
subtraction to addition.
5  (+3) = Note: 5  (+3) is the same as : 5  3
When the double signs are different, it is the
same as if there were one "minus" sign.
5  (+3) =
2
5  (3) = Note: 5  (3) is the same as: 5 + 3
When the double signs are the same, it is the
same as if there were one "plus" sign.
5  (3) = And: "minus a minus" is a "plus".
Practice:
7  (+4) = 8  6 =
6  (+9) = 5  8 =
8  (10) = 4  (7) =
When there's more than two of them:
5  2 + (4)  (+3)  (6) + (+5) = 4  9 + (3)  (6)  8 + 0  (+4)  0 =
Practice:
6 + 3  (7) + (4)  (+5) + 8 + (+2) = 7 + (3)  (9)  8 + 4  (+7) + (5) =
 (12) + (18) + 8  0 19  (+36) = 7 + 0  14 + (2)  (6)  (+4) + 3 =
3
QUIZ: (Adding and Subtracting Integers)
Part I  Show number line for each one
54 + (36) =
28  42 =
19 + (48) =
36 + 94 =
42  (14) =
39 + (86) =
78 + (+17) =
74  (+37) =
21 + 65 =
38  (26) =
Part II
42 + 27  (17)  38 + (14)  0 + (+31)  25 =
4
Packages  "Do Me First"
( ) parentheses Note: "Packages" or "grouping symbols"
are usually arranged in this order:
[ ] brackets parentheses on the inside, then
brackets, then braces:
{ } braces { [ ( ) ] }
Rule: First perform the operations indicated within the innermost packages.
Work from the inside to the outside.
7  (8  9) = [ 4  (7  5) ]  [ 8  (3  7) ] =
{ 12  [ 18 + 7  (3  4) ] } + { [ 5  (8) ]  15 } =
Practice:
8  (9  12) = [ (5  8)  (4) ] + [ 9 + (6  9) ] =
{ [7  (8) ]  (3  4 ) }  { (8 + 9) + [ 6  (9) ] } =
5
Multiplication of Integers
3 x 5 = 3 . 5 = 3 (5) = (3) 5 = (3) (5) =
Rule: When multiplying integers:
If the signs are the same, the product is positive.
If the signs are different, the product is negative.
(3) (5) = (3) 5 = 3 (5) = 3 . 5 =
Practice:
4 (7) = (4) (8) = 9 . 7 = 8 (9) =
Note: (8) (9) is not the same as: (8)  (9)
Division of Integers
Note: The rule for division of integers is the same as the rule for multiplication.
10 ÷ 2 = 10 ÷
2 = 10 ÷
2 = 10 ÷ 2 =
10 = 10 = 10 = 10 =
2 2 2 2
10 + 12 = 5 =
2 4 8
Practice:
63 = 48 = 54 = 72 =
9 6 6 8
8  16 = 3 =
2 8 5
6
Operations with Integers Using Packages
3 (7  9) = 4 [ 7  (3 + 8) ] =
2 ( 3  8) = 7 ( 12 + 20 )
5 (12  9)
4
5 =
5 ( 9  8 )
3
2
Practice:
8 ( 4  3 + 7 ) = 3 [ (4  9)  (2) ] =
6 [ 8 + (3 + 10) ] 3 [ 14 + (6  5) ]
=
2
5 [ 7  (4  5) ]
=
2 [ 3  ( 12 + 12 ) ]
2 3
7
QUIZ: (Operations with Integers Using Packages)
5 [ (3  5)  (5) ] 3 [ 4  ( 12 + 3 ) ]
=
3
3 [ 3  (2  4)] =
2 { 4 + [ 4  (3  7) ] }
2
8
TEST  Operations with Integers
Part I  Show number line for each problem
8 + (4) =
7  (3) =
3  (9) =
7 + (8) =
Part II
5 . 4 = 4 (7) = (3) (8) = (5) 9 =
Part III
12 = 8 = 18 = 3 =
3 2 6 4
Part IV Part V
4 + (3) + 7  (5) + 0  (+8) =
3 [ 4  (8 10 ) ]
=
2 ( 16 + 15 )
8 3
9
Exponents and Raising a Number to a Power
52 = = 5 is the "base". 2 is the "exponent".
52 is read: "5 squared".
25 = = 2 is the "base". 5 is the "exponent".
25 is read: "2 to the fifth (power)".
22 = = 23 = =
23 is read: "2 cubed".
34 = = 2 (32) = =
(2 . 3) 2 = =
[ 3 (4) ] 2 = =
Practice:
42 = = 33 = =
104 = = 5 (22) = =
(5 . 2) 2 = =
[ 4 (2) ] 2 = =
Order of Operations
Remember: Any quantities packaged by grouping symbols (parentheses,
brackets, braces, etc.) are saying, "Do me first!"
Rule: Whenever there are no grouping symbols, operations must be done
in the following order:
1) Quantities must be raised to the indicated power.
2) Do multiplications and divisions in order from left to right
3) Do additions and subtractions in order from left to right.
10
3 + 5 . 8 = 12 ÷ 2 + 4 = 12 ÷ 4 . 3 =
7 + 3 . 4  8 + 12 ÷ 2 = 8 . 4  6 ÷ 2 + 10 =
8 + 4 [ 7 + ( 3 + 8 ) ] = 9  3 {  4 [ 2  (6) ] } =
32 . 4 = 2 . 32 = (4)2 . (2)3 = (22 + 42) ÷ 2 =
82  62 = [ ( 52  33 )  6 ] 2 = Note: (8)2 = 64
8  6  82 = 64
2 [ 43  8 . 2 + (8  5) 3 ]
=
 3 ( 36  4 . 22 ) 2
11
5 . 4 + 5 = 20  10 ÷ 5 = 8 . 6 ÷ 4 =
8  12 ÷ 3 + 4  2 . 3 = 7 . 3 + 9 ÷ 3 10 =
 8 + 3 [  4  (8  5) ] =  4 + 2 {  5 [ 2 (6  7) ] } =
23 . 3 = 3 . 52 = (3)3 . (2)2 = (32 + 24) ÷ 5 =
42  52 = [ (34  52 )  50 ] 2 =
4  5
12
Practice:
=  32 = 9
(24 ÷ 6 + 22 ) 2
QUIZ: (Exponents and Powers)
2 { 3 [ 32 (23  42) ] }
=
4 + 3 ( 24  4 . 3  6)
13
[ 32  8 ÷ 4 + (5  8)2] 2 Note: (3)2 = 9
Evaluating Variable Expressions
n + 3 = when n = 2 3 + n = when n = 2 n + 3 = when n = 2
n  5 = when n = 2 5  n = when n = 2 5  n = when n = 2
5n = when n = 2 5n = when n = 2
5n = when n = 2 5n = when n = 2
n = when n = 4 n = when n = 4 2 = when n = 4
2 2 n
5n  4 = when n = 3 2n = when n = 6
3
14
Practice:
n + 4 = when n = 3 4 + n = when n = 3 n + 4 = when n = 3
n  8 = when n = 3 8  n = when n = 3 8  n = when n = 3
4n = when n = 3 4n = when n = 3
4n = when n = 3 4n = when n = 3
n = when n = 6 n = when n = 6 3 = when n = 6
3 3 n
6n  3 = when n = 4 3n = when n = 4
2
15
Evaluating Variable Expressions with Exponents
n2 = when n = 2 n2 = when n = 2  n2 = when n = 2
 n2 = when n = 2 (n)2 = when n = 2 (n)2 = when n = 2
 (n)2 = when n = 2  (n)2 = when n = 2 3n2 = when n = 2
3n2 = when n = 2  3n2 = when n = 2  3n2 = when n = 2
(3n)2 = when n = 2 (3n)2 = when n = 2  (3n)2 = when n = 2
4n3 = when n = 2 4n3 = when n = 2
 (4n)3 = when n = 2  (4n3) = when n = 2
16
n2 = when n = 3 n2 = when n = 3  n2 = when n = 3
 n2 = when n = 3 (n)2 = when n = 3 (n)2 = when n = 3
 (n)2 = when n = 3  (n)2 = when n = 3 4n2 = when n = 3
4n2 = when n = 3  4n2 = when n = 3  4n2 = when n = 3
(4n)2 = when n = 3 (4n)2 = when n = 3  (4n)2 = when n = 3
2n3 = when n = 3 2n3 = when n = 3
 (2n)3 = when n = 3  (2n3) = when n = 3
17
Practice:
(7n  5) + (4n  3) = when n = 3   4n + 4 = when n = 2
3n  8
2n2  n = when n = 5 n2  n  6 = when n = 5
n + 3
4n3  3n2 = when n = 2 (2n)3  4n2 + 8 = when n = 2
4  2n2  (2n)2 +3n  2
18
More Evaluating Expressions
(6n + 2)  (3n + 2) = when n = 2   5n + 5 = when n = 3
3n  8
3n2  n = when n = 4 n2  n  8 = when n = 4
n + 2
2n3  2n2 = when n = 3 (3n)3  6n2 + 45 = when n = 3
9  3n2  (3n)2 +2n + 5
19
Practice:
TEST  Evaluating Variable Expressions
Part 1
3n  6 = when n = 4 10 + 5n = when n = 2
5n  8 = when n = 3 7 + 8n = when n = 1
4n = when n = 6 n  10 = when n = 6
3 3n  14
4n2 = when n = 3  3n2 = when n = 3
(2n)3 = when n = 2  (3n)3 = when n = 2
Part II
(4n3  5)  [ (2n2 + 8) + (8  3n) ] = when n = 3
 2n3  3n2 + n = when n = 2
(2n)2  4n + 6
20
Solving Equations
Using the opposite of addition:
x + 4 = 10 x + 7 = 4 x + 8 = 3 x + 5 = 3
Practice:
x + 5 = 20 x + 5 = 5 x + 9 = 6 x + 3 = 1
Using the opposite of subtraction:
x  4 = 6 x  6 = 2 x  4 = 2 x  3 = 8
Practice:
x  8 = 5 x  9 = 3 x  7 = 5 x  4 = 7
21
x and the "minusone maneuver":
4  x = 3 4  x = 3 6  x = 8 6  x = 8
8  x = 9 8  x = 9 5  x = 7 5  x = 7
Practice: (Complete the solution and do the check)
5  x = 2 5  x = 2 5  x = 10 5  x = 10
+ x + x 5 5 + x + x +5 +5
6  x = 9 6  x = 9 4  x = 8 4  x = 8
+ x + x 6 6 + x + x +4 +4
22
Using the opposite of multiplication:
5x = 25 3x = 12 3x = 9 7x = 35
Practice:
4x = 12 5x = 20 6x = 30 4x = 16
Using the opposite of division:
x = 8 x = 2 x = 4 x = 2
4 3 8 5
Practice:
x = 7 x = 3 x = 6 x = 7
5 4 4 4
23
When the "coefficient" of the "variable" is a fraction:
Remember: The quantity which "can be anything" and is represented by a letter
is called the "variable".
Note: When a variable is multiplied by a number, that number is called the
"coefficient" of that variable. In the expression 5x; 5 is the "coefficient of x".
Rule: In an equation where the coefficient of a variable is a fraction, the equation
can be solved by multiplying both sides of the equation by the "reciprocal" or "flipflop"
of the fraction.
Note: 2x is the same as 2 x
3
3
2x = 24 2 x = 24
3 3
4 x = 20 5 x = 30  7 x = 35  2 x = 30
5 6 8 5
Practice:
3 x = 36 3 x = 30  5 x = 20  2 x = 18
4 5 8 3
24
Note: x = 3 is the same as 1 x = 3
4 4
1 x = 3 Practice: 1 x = 5
4 3
Order of Operations When Solving Equations
Rule: The order of operations when solving equations is the reverse of the order
of operations when evaluating variable expressions:
1) Use additions and subtractions to eliminate terms
2) Use multiplications and divisions to eliminate terms
3x + 2 = 14 5x  7 = 13 4x + 20 = 8 3x  10 = 25
Practice:
7x + 8 = 43 4x  3 = 13 4x + 2 = 10 3x  5 = 20
25
Combining Like Terms
5x + 4x = 3x  2x = 3x + 7x = 5x  10x =
Practice:
3x + 8x = 7x  10x = 2x + x = 3x  x =
More combining like terms:
5x  2  8x + 4 = 3x  2x + 4  5 = 3  9x  4 + 10x  1  2x =
5x  4 + 3x  1 + 3 + 2x = 5x + 4  3x + 2x  5 + 4  4x  3 =
Practice:
7x  1  x + 4 = 8x + 2  3  x = 4x +2  3x + 9x  7  x =
5 + 3x  8  5x + 4  1 = 5 + 6x + 8  3x  4  2x + 1  x =
Combining Terms: First Step in Solving Equations
5x + 3  2x  7 = 10 7x + 5  3x  20 = 27
Practice:
8x + 3  10x + 4 = 9 5x + 3 + 4x  8 = 9 Hint: You might need
the "MinusOne
Maneuver".
26
Transposing: Are You Ready to Skip a Step?
5x  3 = 2x + 9 5x + 4  2x + 3 = 7x  5 + 4x  12
Practice:
7x + 19 = 4x + 4 6x + 3  2x  7 = 5x  8 + 3x + 12
QUIZ: (Solving Equations)
5x  9  3x + 2x + 8  11 = 7  4x  5 + 11x  6  x
27
TEST  Solving Equations
(Please, no "skipping steps" until next page.)
Solve each equation and do the check:
x + 5 = 8 x + 7 = 4 x  6 = 2 x  4 = 2
(Complete) (Complete)
2  x = 3 9  x = 3 3x = 21 6x = 18
+ x + x 9 9
4x = 16 x = 8 x = 4 x = 4
3
6
3
3 x = 21 1 x = 9 3x + 5 = 1 4x  18 = 6
5 3
28
4x + 4  2x  7 = 3 8x  5  2x + 11 = 18
3x  20 = 2x + 5 3x  4 + 2x + 8 = 5x + 1  x
29
The Distributive Law
4(3 + 2) = 4(3) + 4(2)
3(x + 2) = 4(x  3) = 6(3x  5) =
4(6  2x) = 3(5x + 2) = 4(7  3x) =
Practice:
4(x +3) = 3(x  2) = 2(2x  6) =
3(5  3x) = 4(2x + 3) = 2(8  4x) =
Using the distributive law and then combining terms:
5(x  4)  3(2x + 2) = 4(3  2x) + 3(6x  4) =
Practice:
6(x + 2)  4(2x + 3) = 3(4  3x) + 2(5  2x) =
30
Distributing a "minusone" to "subtract" a quantity in parentheses:
Note: A plus sign directly in front of a quantity in parentheses is the same as if
the parentheses were not even there.
4x + (2x  1) = 4x + 2x  1 = 6x  1
But a minus sign directly in front of a quantity in parentheses requires
special attention.
4x  (2x  1) = 4x  1(2x  1) = 4x  2x + 1 = 2x + 1
Rule: Whether or not you think of it as distributing with a "minusone",
to clear a quantity in parentheses which has only a minus sign in front of it,
change the sign of every term within the parentheses.
5x  (2x + 2) = 4x  (3  2x) =
(6x  2)  (3x + 1) =  (x  2)  3(2x  1) =
3(2x  4)  4(x  2)  (6  3x) =
Practice:
6x  (5x + 4) = 3x  (5 + 6x) =
(3x  2)  (4x + 5) =  (3x + 1)  4(5  2x) =
5(3x  2)  2(4x  3)  (4  2x) =
31
More Solving Equations
Using the Distributive Law and Combining Terms as First Steps:
3(4x  7) = 5(3x + 12) 5(x  4)  (x  3) = 3(2x  1) + 2(2x  4)
Practice:
4(3x  5) = 5(2x  8) 7(3x  7)  (3 + 2x) = 3(5x + 1)  (5  2x)
32
TEST  More Solving Equations
Solve and Check:
4(x  5)  3(x + 2) = 6(x  3)  (2x  4)
33
Workbook
Part 2  Solving Word Problems with Algebra
Writing Variable Expressions
the sum of n and 6 _____ 4 more than n _____
n increased by 12 _____ 6 plus n _____
14 minus n _____ the difference when 12 is diminished by n _____
3 subtracted from n _____ the difference when n is subtracted from 15 _____
5 less than n _____ n decreased by 10 _____
5 times n _____ the product when n is multiplied by 8 _____
twice n _____ the product of 9 and n _____
n divided by 6 _____ the quotient when n is divided by 4 _____
18 divided by n _____ the quotient when 10 is divided by n _____
Practice:
the sum of n and 8 _____ 5 more than n _____
n increased by 10 _____ 4 plus n _____
18 minus n _____ the difference when 11 is diminished by n _____
2 subtracted from n _____ the difference when n is subtracted from 12 _____
3 less than n _____ n decreased by 9 _____
4 times n _____ the product when n is multiplied by 6 _____
twice n _____ the product of 3 and n _____
n divided by 2 _____ the quotient when n is divided by 7 _____
15 divided by n _____ the quotient when 8 is divided by n _____
More Writing Variable Expressions:
3 times n increased by 4 __________ 3 times the quantity n increased by 4 _________
the sum of 5 times n and 2 __________ 5 times the sum of n and 2 __________
34
5 more than n divided by 2 __________ 5 more than n all divided by 2 __________
the sum of 6 and n divided by 5 ________ the sum of 6 and n all divided by 5 _______
8 less than n divided by 4 ________ the quantity 8 less than n then divided by 4 _______
17 minus n divided by 3 __________ 17 minus n all divided by 3 __________
6 less than n divided by 5 _______ the quotient when 6 less than n is divided by 5 _______
4 times n divided by 3 __________
the sum of n divided by 3 and 4 __________
Practice:
4 times n increased by 5 __________ 4 times the quantity n increased by 5 __________
the sum of 6 times n and 3 __________ 6 times the sum of n and 3 __________
twice n diminished by 8 _________ twice the quantity n diminished by 8 __________
7 more than n divided by 3 __________ 7 more than n all divided by 3 __________
the sum of 9 and n divided by 4 ________ the sum of 9 and n all divided by 4 ________
6 less than n divided by 2 ________ the quantity 6 less than n then divided by 2 _______
16 minus n divided by 5 __________ 16 minus n all divided by 5 __________
4 less than n divided by 2 _______ the quotient when 4 less than n is divided by 2 _______
7 times n divided by 5 __________
the sum of n divided by 2 and 6 __________
A Useful Concept When Writing Variable Expressions:
The sum of two numbers is 12. If one of them is n, the other can be expressed as _________
Practice:
The sum of two numbers is 36. If one of them is n, the other can be expressed as _________
35
twice n diminished by 9 _________ twice the quantity n diminished by 9 __________
Number Problems
1) The sum of a number and twice that 2) The sum of a number and the same
number is 12. What is the number? number increased by 6 is 48. What is
the number?
3) Three times a number subtracted from 4) Ten more than twice a number is
8 times a number is 25. What is the 24. What is the number?
number?
5) 36 more than 5 times a number is 34. 6) When 8 more than twice a number is
What is the number? subtracted from 3 less than 4 times a
number, the result is 13. What is the
number?
7) Five times a number decreased by 18 is 8) 5 times a number exceeds 2 less than 2
equal to 3 times the number increased by 6. times a number by 14. What is the
What is the number? number?
36
number. Their sum is 31. What are number. Their sum is 17. What is the
the numbers? smaller number?
11) One number is 25 greater than a second 12) The sum of two numbers is 28. 5 times
number. If the lesser number is subtracted the saller diminished by 2 times the
from 3 times the greater number, the differ larger is 14. What are the numbers?
ence is 195. What are the numbers?
Practice:
1) The sum of a number and twice that 2) The sum of a number and the same number
number is 63. What is the number? increased by 9 is 33. What is the number?
3) Four times a number subtracted from 6 times 4) Six more than twice a number is 56.
a number is 24. What is the number? What is the number?
37
9) One number is 5 more than another 10) One number is 3 less than another
5) Eight less than 3 times a number is 62. 6) Two more than a certain number is 15 less
What is the number? than twice the number. What is the
number?
7) Six times a certain number exceeds 8) The sum of 5 more than a certain number
4 times the number by 12. Find the and 10 more than 4 times the number is
number. equal to 6 times the number increased by 3.
Find the number.
9) One number is 8 more than another number. 10) One number is 7 less than another
Their sum is 54. What are the numbers? number. Their sum is 39. What
is the smaller number?
11) One number is 14 greater than a 12) The sum of two numbers is 25. 3 times
second number. If the lesser number the smaller is 5 more than twice the
is subtracted from 4 times the greater larger. Find the numbers.
number, the difference is 23. Find the
numbers.
38
ConsecutiveInteger Problems
Set of consecutive integers: {4, 3, 2, 1, 0, 1, 2, ....} {n, (n + 1), (n + 2), (n + 3), ....}
Set of consecutive odd integers: {5, 3, 1, 1, 3, 5, ....} {n, (n + 2), (n + 4), (n + 6), ....}
Set of consecutive even integers: {4, 2, 0, 2, 4, ....} {n, (n + 2), (n + 4), (n + 6), ....}
1) Find two consecutive integers 2) Find two consecutive even integers
whose sum is 375. whose sum is 446.
3) Find three consecutive odd integers 4) Find three consecutive integers if the
whose sum is 3. sum of the first and third is equal to 246.
Practice:
1) Find three consecutive integers 2) Find three consecutive integers
whose sum is 213. whose sum is 153.
39
whose sum is 260. the second and fourth is 352.
5) Find two consecutive even integers if the 6) Find two consecutive integers if the lesser
larger is equal to 6 less than twice is equal to one more than twice the greater.
the smaller.
7) Find four consecutive even integers if 8) Find four consecutive odd integers if twice
twice the third is equal to the sum of the the sum of the first and the third is equal to
fourth and 3 times the second. 38 more than three times the sum of the
second and fourth.
40
3) Find four consecutive even integers 4) Find four consecutive integers if the sum of
Age Problems
1) Jack is 20 years older than Jill. Five 2) Jack is now 8 years old. Jill is now 2 years
years ago Jack was 5 times as old as Jill old. In how many years will Jack be twice
was then. How old is each now? as old as Jill will be then?
Now 5 years ago Now In x years .
Jack Jack .
Jill Jill .
Let x = Let x =
3) The sum of Jack's age and Jill's age 4) The sum of Jack's age and Jill's age
is 50. Eight years from now Jack is 32. Jack's age one year from now
will be twice as old as Jill will be then. will be 7 times Jill's age one year ago.
How old is each now? How old is each now?
Now 8 years from now Now 1 year from now 1 year ago
Jack Jack
Jill Jill .
Let x = Let x =
Practice:
1) Jack is 3 times as old as Jill. Ten years 2) Jack is 32 years old and Jill is 14. How
from now Jack will be twice as old as Jill many years ago was Jack 4 times as
will be then. How old is each now? old as Jill was then?
now 10 years from now now x years ago
Jack Jack .
Jill Jill .
Let x = Let x =
41
3) Jack is 4 times as old as Jill. Eight years 4) Jack's age exceeds Jill's by 16 years.
from now, Jack's age will exceed 3 times Four years ago Jack was twice as old
Jill's age by 2 years. How old is each of as Jill was then. How old is each of
them now? them now?
now 8 years from now now 4 years ago
Jack Jack .
Jill Jill .
Let x = Let x =
5) Five years from now, Jack will be 3 times as old as Jill will be then.
Four years ago Jack was 30 years older than Jill was then.
How old is each of them now
4 years ago now 5 years from now
Jack .
Jill .
Hint: Let x = Jill's age 5 years from now
42
Geometry Problems: Angles and Degree Measure
Remember:
1) The sum of the measures of the angles in any triangle is 180o.
2) Two angles are supplementary if their sum is equal to 180o. If the measure of one of
the angles is xo, the measure of the other can be expressed as (180  x)o
3) Two angles are complementary if their sum is equal to 90o. If the measure of one of
the angles is xo, the measure of the other can be expressed as (90  x)o
1) Find the measures of two complementary 2) Find the measure of an angle for which the
angles if the second measures 12o more sum of the measures of its complement and
than twice the first. its supplement is 194o .
3) The first angle of a triangle measures 3 times the second angle. The third angle
measures 4o less than the sum of the measures of the other two angles. Find the
measure of each angle.
Practice:
1) Find the measures of two supplementary 2) Find the measure of an angle whose
angles if the second measures 100o less supplement measures 8o more than
than three times the first. twice its complement.
43
3) The first angle of a triangle measures 20o more than the second angle. The third angle
measures 30o less than twice the sum of the first two angles. Find the measure of each
angle in the triangle.
Coin Problems: Penny, Nickel Dime, Quarter
Note:
In the USA: a penny = 1 cent, a nickel = 5 cents, a dime = 10 cents, a quarter = 25 cents
1) A jar contains 45 coins, all dimes and nickels, with a total value of $3.50. How many are
there of each kind of coin?
number of coins x value of each = total value .
dimes .
nickels . .
2) A jar contains 3 more nickels than dimes, with a total value of $1.95. How many are there
of each kind of coin?
number of coins x value of each = total value .
dimes .
nickels .
44
1) A jar contains 50 coins, all dimes and nickels, with a total value of $4.00. How many are
there of each kind of coin?
number of coins x value of each = total value .
dimes .
nickels .
2) A jar has has twice as many dimes as nickels, and 2 more quarters than nickels. The total
value is $4.00. How many are there of each kind of coin?
number of coins x value of each = total value .
dimes .
nickels .
quarters .
3) A jar has 50 coins; all pennies, nickels and dimes; with a total value of $3.10. There are
twice as many nickels as pennies. How many are there of each kind of coin?
number of coins x value of each = total value .
pennies .
nickels .
dimes .
Hint: Whatever is neither pennies nor nickels, must be dimes.
45
Practice:
Uniform Motion Problems: Rate x Time = Distance
1) In a car race, one car went 12 miles/hr. 2) A helicopter flew at 150 miles/hr. to rescue
faster than the other and got to the finish the boat people who were 330 miles away.
line in 6 hours. The slower car took an The boat continued toward the helicopter at a
hour longer. How fast was each car going? speed of 15 miles/hr. How long did it take
How many miles was the race? the helicopter to reach the boat people?
r x t = d r x t = d .
slow car helicopter .
fast car boat .
3) The slow car leaves town 3 hours before 4) John rides his motorcycle from city "A" to city
the fast car. The slow car travels 40 "B" at 30 miles/hr. He makes the return trip
miles/hr, and the fast car 50 miles/hr. at 24 miles/hr. If he makes the entire trip in
How long before the fast car overtakes 9 hours, how far apart are the cities?
the slow car?
r x t = d r x t = d .
slow car going .
fast car returning
46
different route which was 8 miles longer, but he increased his speed by 12 miles/hr and got
home in one hour less time. Find the rate that he traveled from home to the hotel, the rate
he traveled on the way home, and the total number of miles traveled round trip.
r x t = d .
going .
returning .
Practice:
1) A freight train and a passenger train left 2) Two planes fly toward each other from cities
city "A" at 11 A.M.. The freight train 2050 miles apart. One leaves at 9 A.M. and
got to city "B" at 4 P.M. The passenger and flies at 150 miles/hr. The other leaves
train got there at 2 P.M. If the passenger at 11 A.M. and flies at 200 miles/hr. At
train went 40 miles/hr faster than the freight what time will they pass each other?
train, how fast was each train going?
r x t = d r x t = d .
freight slow plane _
passenger fast plane .
47
5) A man drove his car from his home to the hotel in 4 hours. On the return trip he followed a
3) Jack and Jill start at the same place on their 4) Joe swam the length of the pond using the
motorbikes. If Jack leaves at 12 noon and butterfly stroke at a rate of 1.5 m/sec. He
travels at 22 miles/hr and Jill leaves 2 swam back using the breast stroke at
hours later and travels at 30 miles/hr, at what 1.2 m/sec. The entire swim took 3 minutes
time will Jill be only 4 miles behind Jack? How long is the pond? (3 min. = 180 sec.)
r x t = d r x t = d .
Jack butterfly .
Jill breast .
5) A man drove from his home to the beach at 25 miles/hr and returned home at 30 miles/hr
by a different route 5 miles longer. The return trip home took 1 hour less than the trip to
the beach. How long did each trip take and how many miles were traveled in all?
r x t = d .
going .
returning .
48
TEST  Solving Word Problems with Algebra
1) The VCR costs $90 less than twice the 2) Find three consecutive odd integers such
cost of the CD player. The TV costs that the sum of the second and the third is 9
$10 more than the VCR and the CD less than the first.
player put together. All three together
cost $550. How much does each cost?
3) Today Billy is 3 years older than Bobby. 4) Find the angle whose supplement is 30o
Four years ago Billy's age was 11 years more than twice its complement.
less than twice the age that Bobby was then.
How old is each now? (Make your own chart.)
49
5) A passenger train left Northvalley at 11 A.M. heading towards Southberg at 90 miles per
hour. Two hours later a freight train left Southberg heading towards Northvalley at 80
miles per hour. If the two cities are 1030 miles apart, at what time do the two trains pass
by each other? At that time, how many miles had the freight train traveled? (Make your
own chart.)
6) A jar has 47 coins: pennies, nickels, dimes, and quarters. The total value of the coins is
$3.48. There are 2 more dimes than quarters, and there are 4 more nickels than dimes.
The rest of the coins are pennies. How many are there of each kind of coin? (Make your
own chart.)
50
Workbook
Part 3  Coordinate Geometry
and
Systems of Equations
Graphing Linear Equations on a Coordinate Plane
Vocabulary: coordinate geometry, coordinate plane, Rene Descartes, Cartesian
coordinates, xaxis, yaxis, origin, ordered pair, coordinates, xcoordinate,
absissa, ycoordinate, ordinate, plotting points
y
Plot these points:
A (3, 5)
B (5, 3)
C (6, 4)
D (4, 7)
E (4, 7) x
F (0, 5)
G (6, 0)
H (0, 0)
Practice: y
Plot these points:
A (2, 7)
B (7, 2)
C (5, 3)
D (3, 4)
x
E (3, 4)
F (0, 4)
G (5, 0)
H (0, 0)
51
Vocabulary: table of values, graph, linear equations, dependent variable, independent
variable, constant
1) Graph the equation: y = 2x  4 y
x 2x  4 y
x
2) Graph the equation: y = 1/2x + 3 y
x 1/2 x + 3 y
x
52
x y
x
4) Graph the equation: 2x + 3y = 3 y
x y
x
53
3) Graph the equation: 4x  y = 3 y
Practice:
1) Graph the equation: y = 3x  5 y
x 3x  5 y
x
2) Graph the equation: y = 1/3x + 1 y
x  1/3 x + 1 y
x
54
3) Graph the equation: 3x  y = 4 y
y x
x
4) Graph the equation: 3x + 2y = 6 y
x y
x
55
Using the SlopeIntercept Form: y = mx + b
Vocabulary: slope, yintercept, y = mx + b, positive slope, negative slope
1) Graph the equation: y = 2x  4 2) Graph the equation: y = 1/2 x + 3
y y
x x
3) Graph the equation: 4x  y = 3 4) Graph the equation: 2x + 3y = 3
y y
x x
56
Practice:
1) Graph the equation: y = 3x  5 2) Graph the equation: y = 1/3x + 1
y y
x x
3) Graph the equation: 3x  y = 4 4) Graph the equation: 3x + 2y = 6
y y
x x
57
More on Slopes
Rule: The slope of a line through two given y2  y1
points is equal to the difference in "y" divided m =
by the difference in "x": x2  x1
Find the slopes of the lines that go through
the given pairs of points:
(0, 0) (2, 6) (3, 2) (6, 4) (2, 3) (6, 1) (5, 4) (4, 5)
Practice:
(2, 1) (4, 5) (6, 1) (2, 3) (6, 0) (3, 3) (1, 2) (4, 3)
Note: It is possible to find the equation of a line once you know the slope of the
line and the coordinates of just one point on the line. Since we have already
figured out the slopes of the lines that go through the pairs of points given above,
now let's find the equations of those lines by substituting the coordinates of one
of the given points, and also the slope, into the equation: y = mx + b
y = mx + b y = mx + b
y y
x x
58
y = mx + b y = mx + b
y y
x x
Practice:
y = mx + b y = mx + b
y y
x x
59
y = mx + b y = mx + b
y y
x x
The slope of a line parallel to A line parallel to the "y axis" is said to
the "xaxis" is zero. have no slope. This is because for
( Example: y = 2 ) any difference in y, the difference in x
is zero, and it is "impossible" to divide
by zero. Example: ( x = 2 )
y y
x x
60
If two different lines have the same If two lines are perpendicular to each
slope, then those lines are parallel. other, the slope of one is the "negative
Example: y = 2x + 2 and y = 2x  3 reciprocal" of the other. (The product
of the two slopes is 1) Example:
y = 2x + 2 and y = 1/2 x + 1
y y
x x
Show that these two lines are perpendicular: y
2x + 3y = 3 and 3x + 2y = 8
x
61
Practice: Show that these two lines are perpendicular: y
2x + y = 2 and x + 2y = 2
x
Solving a System of Linear Equations Graphically: y
2x + y = 6 and y  x = 3
x
Practice: Solve the system of equations graphically: y
3x  4y = 4 and y = 2x + 4
x
62
Now let's try these: y
x + y = 3 and y = 5  x
x
2x + y = 1 and 3y = 3  6x y
x
Note: Two equations are said to be "consistent" if their slopes are unequal
and thus their graphs intersect in a common point.
Two equations are said to be "inconsistent" if they have the same
slope but different yintercepts; their graphs will be parallel and will not intersect
in a common point.
Two equations are said to be "dependent" if they are actually
different forms of the same equation and represent the same line.
63
Tell whether the following pairs of equations are consistent (intersecting lines),
inconsistent (parallel lines), or dependent (the same line)
x + y = 5 and x + y = 10 x + y = 12 and x  y = 2
2x = 5y + 5 and 4x  10y = 10 y = x and 3y  3x = 4
Practice:
2x + 3y = 6 and 3x + 2y = 10 x  4y = 7 and x  9 = 4y
x + 2y = 6 and x = 2  2y 3x  2y = 6 and 4y = 6x  12
64
TEST  Coordinate Geometry
Complete the table of values and graph the equation:
3x + 2y = 8 y
x y
x
Convert to y = mx + b form and graph the equations:
2x + y =  4 x  2y = 4
y y
x x
65
Find the slope of the line going through the Show that the graphs of these two equations
given pair of points. Then using the slope are perpendicular by showing that the product
and one of the given points, substitute into of the slopes is 1. Then graph the lines:
y = mx + b to find the yintercept. Finally,
write the equation of the line and do the graph. x + 3y = 6 3x  y = 1
(2, 5) (3, 5) y = mx + b
y y
x x
Solve this system of linear equations graphically. In the spaces provided, write the ordered
pair of the point of intersection of the graphs, and the solutions. Also do the checks by
substituting the solutions into the original equations: y
( , )
x + y = 3 and x  2y = 0 x =
y =
x
66
Tell whether the following pairs of equations are consistent (intersecting lines), inconsistent
(parallel lines), or dependent (the same line).
2x  y = 5 and 2x  y = 10 2x + y = 8 and x  y = 7
3x = 2y + 7 and 6x  4y = 14 x = y and 2y  2x = 6
67
Solving Systems of Equations in Two Variables
Solving by Substitution:
x + y = 12 4x + 3y = 12
3x  y = 4 x  4y = 3
5x  3y = 1 3x  4y = 5
x + y = 3 x + 7y = 10
Practice:
x  y = 1 2x  y = 2
x + y = 5 3x  2y = 3
68
3x  5y = 8 4x  3y = 15
x + 2y = 1 x  2y = 0
Solving by Addition:
x + y = 7 x  y = 4
x  y = 3 4x  y = 2
3y  5x = 19 (watch the order) 3x  5y = 36
2x + 3y = 5 3x + 5y = 12
69
x  y = 9 3x  y = 10
x + y = 17 2x  y = 7
2y = 21  5x (watch the order) 8 = 2y  3x (watch the order)
7x = 2y + 39 16 = 6y  3x
Solving by Multiplication and Addition:
x + 2y = 1 3x + 5y = 9
3x + y = 8 9x + 2y = 12
70
Practice:
2x + 3y = 0 5x + 3y = 25
5x  2y = 19 4x + 2 y = 2
Practice:
x + 11y = 6 5 x + 12 y = 24
2x + y = 9 9x + 4y = 8
5x + 6y = 16 4 x + 20 y = 8
6x  5y = 7 5x + 8y = 27
71
Problems Using Two Variables  Digit Problems
Note: If u represents the units digit, and t represents the tens digit,
then the value of a twodigit number can be expressed as: 10t + u
For example: if u = 7 and t = 6, then 10t + u = 10(6) + 7 = 67
The value of the number with the digits reversed can be expressed as: 10u + t
In the case of this example: 10u + t = 10(7) + 6 = 76
1) The sum of the digits of a twodigit 2) The sum of the digits of a twodigit
number is 11. The value of the number number is 9. If the order of the digits
is 13 times the units digit. Find the is reversed, the result is a number
number. exceeding the original number by 9.
Find the original number.
3) The units digit of a twodigit number 4) Extra for Experts Find a three
is 3 times the tens digit. The sum digit number whose units digit is 3 times
of the digits is 12. Find the number. its hundreds digit and 2 times its tens
digit, and the sum of whose digits is 11.
72
1) The sum of the digits of a twodigit 2) The sum of the digits of a twodigit
number is 9. The number with the number is 14. If 18 is subtracted from
digits reversed is 9 times the original the number, the result is the number
tens digit. Find the original number. with the digits reversed. Find the
original number.
3) The units digit of a twodigit number 4) Extra for Experts A threedigit
exceeds 2 times the tens digit by 2. The number is 198 more than itself with the
sum of the digits is 11. Find the number. digits reversed. The sum of the digits is
19. The hundreds digit is 3 times the
tens digit. Find the original number.
73
Practice:
TEST  Solving Systems of Equations
1) Solve by Substitution: 2) Solve by addition: (Change order as necessary.)
3x + 4y = 6 2x  3y = 9
x  5y = 17 3y = 9  4x
3) Solve by multiplication and then addition: 4) When the digits of a twodigit number
(One of the answers is a fraction) are reversed, the new number is 10
more than twice the original number.
8x  2y = 10 The units digit of the original number
20x  3y = 21 is 3 times its tens digit. Find the
original number.
74
Workbook
Part 4  Polynomials
Introduction to Polynomials
Expressions with only one term are called "monomials": 2, 3x, 2x2, (2x2)3, [4x3y2]2
Expressions with more than one term are called "polynomials": 5x3  2x2 + 3x  4
If a polynomial has two terms it is called a "binomial": 5x + 2, 3x + 2y, x2 + 3y2
If a polynomial has three terms it is called a "trinomial": 3x2  2x + 4, x2 + 2xy + y2
Adding Polynomials
(5x + 3) + (2x  4) = 5x + 3 (3x2  4x + 1) + (2x2  5) = 3x2  4x + 1
2x  4 2x2  5
+ +
Practice:
(2x  3) + (5x + 5) = 2x  3 (5x2  2x + 5) + (3x2 + 5x) = 5x2  2x + 5
5x + 5 3x2 + 5x
+ +
Subtracting Polynomials
(5x  2)  (3x + 4) = 5x  2 (4x2  3x + 1)  (2x2  3x + 5) = 4x2  3x + 1
3x + 4 2x2  3x + 5
 
Practice:
(3x  4)  (7x  3) = 3x  4 (3x2 + 2x  5)  (3x2 + 5x  7) = 3x2 + 2x  5
7x  3 3x2 + 5x  7
 
75
Multiplying and Powers: The Laws of Exponents for Multiplication
The Multiplication Law:
22 . 23 = Note: 25 = 2 . 2 . 2 . 2 . 2 = 32
24 = 2 . 2 . 2 . 2 = 16
23 = 2 . 2 . 2 = 8
x2 . x3 = (x2)(x) = x2 . x2 . x . x4 = 22 = 2 . 2 = 4
21 = 2
xa . xb = xa+b Rule: The Multiplication Law  When multiplying like
quantities raised to powers, add the exponents.
(2x2)(3x) = (3x)(x4) =
(x3)(4x) = 2(x2)(4x3) = (3x3)(x2)= (2x2)(x4)(x) =
Practice:
x4 . x = (x2)(x5 ) = x . x2 . x3 . x4 = x2 . x4 =
(x2)(2x) = (3x3)(x2) = (2x2)(x3) = (5x4)(4x5) =
ThePowerofaPower Law:
(22)3 = (x2)3 =
(xa)b = xab Rule: The PowerofaPower Law  When raising a power of a
quantity to a power, multiply the exponents
(x3)2 = (x3)3 = (x3)4 = (x4)3 =
Practice:
(x5)6 = (x4)4 = (x2)5 = (x5)2 =
The PowerofaProduct Law:
(2 . 3)2 = (xy)2 = (3x)2 =
(xy)a = xaya Rule: The Power of a Product Law:  When the product of two
quantities is raised to a power, the result is equal to the
product of each of those quantities raised to that power.
76
(4x)2 = (2x)3 = (5xy)3 =
(2x2)2 = (2x2)3 =  (3x2y)2=
Practice:
(5x)2 = (3x)3= (4xy)4 =
(3x3)2 = (3x3)3 =  (2x3y2)2 =
Remember: The Laws of Exponents for Multiplication
xa . xb = xa+b The Multiplication Law
(xa)b = xab The Power of a Power Law
(xy)a = xaya The Power of a Product Law
QUIZ: (Multiplying and Powers)
x(x3) = (x2)(x6) = (x3)4 =
(xy)2 = (xy2)2 = (4x)2 =
(4x4)2 = (3x2)3 =  (4x3)3=
(5x3y)3 = (3x5)4 =
More Complicated Expressions Involving Multiplying and Powers
Simplify:
5x3  x(3x2) = (3x2y)(2x2y3) + (xy)4 =
3(xy)2 + 2xy(xy)  4x(2xy2) = (6x2yz3)(3xy2z) + (2z2)(9x3y2)(yz2) =
77
(3x2y)3 + (2x2y)3 = (2xy2)4 (3x2y4)2  (4x4y8)2 =
Practice:
(x2)(2x) + (x)(x2) = (2x3)(2x)4  (4x)2(2x5) =
3xy(xy) + 8x(xy2)  2(xy)2 = (7y2)(2x2)  (4y)(3y)(5x2) + (5y2)(2x)(3x) =
x(2x3) + 3x2(x2)  (2x2)2 = (2x3y3)2  (3x2y2)3 =
(2xy)(3x2y3)3 + (xy6)(2x3y2)2 =
Multiplying a Monomial by a Polynomial
Rule: To multiply a monomial by a polynomial, use the distributive law and
multiply the monomial by each term of the polynomial.
4(x  3) = 3x(2x + 5) =
4x2(x  2) = 2x3(3x2  2x + 5) =
Practice:
3(2x + 5) = 5x(3x  2) =
3x2(4x  8) = 3x3(x2  2x + 1) =
78
Multiplying a Polynomial by a Polynomial
Rule: To multiply two polynomials, multiply each term of one by each term of the
other.
(x + 2)(x  3) = x + 2 (5x + 2)(3x  1) = 5x + 2
x  3 3x  1
(2x  3y)(5x  2y) = 2x  3y (x  3)(x2 + 3x + 9) = x2 + 3x + 9
5x 2y x  3
(x + 3)(x2  4) = x2  4 Note: A space is left for the
x + 3 missing term, 0x
.
Practice:
(x + 1)(x  4) = x + 1 (3x  2)(2x  3) = 3x  2
x  4 2x  3
(5x  4y)(2x + 3y) = 5x  4y (x  y)(x2 + xy + y2) = x2 + xy + y2
2x + 3y x  y
79
(x  5)(2x2 + 3) = 2x2 + 3
x  5
Are You Ready to Skip a Step? The "FOIL" Method
Note: You can multiply two binomials by "inspection", or by doing it "mentally" or
"in your head", by using the "FOIL" method: First, OutsideInside, Last
(2x + 1) (3x + 2) =
First OutsideInside Last
(x + 3)(x + 2) = (x + 3)(x  2) = (x  3)(x  2) =
(2x + 3)(3x  4) = (4x + y)(x  2y) =
Practice:
(x + 1)(x + 4) = (x + 1)(x  4) = (x  1)(x  4) =
(3x  2)(5x + 3) = (7x  3y)(2x  4y) =
80
The Laws of Exponents for Division
The Division Law:
25
= (2)(2)(2)(2)(2) = Note: 23 = (2)(2)(2) = 8
23
(2)(2)(2) 22 = (2)(2) = 4
21 = 2
23
= (2)(2)(2) = 20 = 1
25 (2)(2)(2)(2)(2) 21
= 1 = 1
21 2
x4 = x2 = x2 = 22
= 1 = 1
x2 x2 x4 22 4
xa
= xa  b Rule: The Division Law  When dividing like quantities raised to
xb powers, subtract the exponent of the denominator from the
exponent of the numerator.
Simplify:
x3
= x6
= 4x4
= 8x2
= 6x2
=
x x9 2x3 2x2 8x2
Practice:
x5
= x7 = 9x3
= 10x2
= 6x2
=
x2 x11 3x 5x2 10x2
Simplify:
12x2y = 18x6y4
= 10x3y4
=
4x2 3x3y 5x4y3
6x2y3 =  42x3y3 = 14x6y10
=
12xy4 6x4y3 28x7y11
Practice:
15x5y3
= 8xy2
= 8x2y =
5x2y 2x 24x4
21x5y2 =  27x7y4 = 12x4y =
7x4y4 54x6y 48xy
81
Simplify:
(4xy2)3 = (3xy2)2(2x2y)3 =
(2xy2)4 (9x2y)(4x2y)2
Practice:
(4xy2)2 = (4x2y3)2(xy2)3 =
8x2y4  (2x2y2)2(x3y)2
The PowerofaQuotient Law:
.2. 2
= (2)2
=
.x.a
= (x)a Rule: The PowerofaQuotient Law  The power of
.3ÿ (3)2 .yÿ (y)a a quotient (a division expressed as a fraction)
is equal to the power of the numerator over
the power of the denominator.
Simplify:
.x.3
= .2x.2
= .3x.3
= .2x
2y. 3
=
.yÿ . y2ÿ . x3 ÿ . 3y
3 ÿ
Practice:
.x.2
= . x .3
= .2x.3
= .3x
3y
2. 2
=
.yÿ .3yÿ . x2 ÿ . 2y
3 ÿ
Negative Exponents:
2
2 = 1 = 1 x
a = 1 Rule: Negative exponents  The negative
22 4 xa power of any number is equal to the
reciprocal of its positive power.
Note: By changing the sign of its exponent, any factor of the numerator may be
made a factor of the denominator, or any factor of the denominator may be
made a factor of the numerator.
82
2
3 = x
2 = 1 = 1 =
2x
2 (2x)
3
Practice:
32
= x3
= 1 = 1 =
2x
3 (3x)2
Simplify: (Express the result using only positive exponents.)
x
2 = x2y
3 = 5x
1y 2 = 2x
2y =
y
3
5 = 2 = 2x
2 = (2x)
2 =
x
2 3x
3
(x2y)
3 = (xy
2)3 = .1 . 1
= .x.2 =
.x ÿ .yÿ
(3x
2)4 = x3 = 2x
2 =
x
4 x
4
Practice:
x
3 = x3y
4 = 4x
2y
3 = 4x
3y
2 =
y
2
3 = 3 = 3x
3 = (3x)
3 =
y
4 4x
2
(x3y)
2 = (xy
3)2 = .1.2
= .x.1 =
.xÿ .yÿ
83
(2x
3)3 = x2 = 3x
3 =
x
3 x
5
Simplify: (Express the results without any denominators.)
x
2 = x2 = 5xy
3 = 8x
3y
3 =
y
3 y
1 z4 (2x
4)2
Practice:
x
3 = x3 = 3x2y = 12x
2y

3 =
y
2 y
2 z3 (2y
3)2
QUIZ: (Laws of Exponents for Division)
Simplify:
x4 = 2x3 = 14x4y2 = 15x5y2 =
x3 x4 7x2y3 3x6y2
5x4y3 = (3x2y)3 =
15x5y5 (2xy3)3
(xy3)3(2x2y2)2 =
(3x2y)2(2x3y2)2
.2x2. 3 = . xy . 3 =
. x ÿ .2x2yÿ
Simplify: (Express results using only positive exponents.)
2 = 3 = (x2y2)3 =
4x
3 (2x)3
3x = 8x2y

4 =
x
2 (2y
2)3
84
Division of a Polynomial by a Monomial
25x  15 = 12x2 + 4x = 16x3 + 12x2  40x =
5 2x 4x
9x2y3 + 27xy2  72xy = 18x2y + 30xy2  12x2y2 =
9x2y2 6x2y2
Practice:
24x + 16 = 48x2 + 36x = 24x4 16x3 + 32x2 =
8 6x 8x2
21x2y + 63xy  7xy2 = 40xy2z2  24x2yz2 + 36x2y2z =
7xy2 8x2y2z2
Division of a Polynomial by a Binomial
x2  3x 10 = 3x2 + 12 + 13x = 2x2 + 11x  18 =
x + 2 3x + 4 2x  3
85
3x3  4x  1 = 4x3  2x2 + 18 =
x +1 2x + 3
Practice:
x2  5x + 6 = 2x2 + 4  6x = 12x2 + 4x  18 =
x  2 2x  2 2x + 3
5x3 + 2x2  48 = 6x3 + 4x  56 =
x  2 2x  4
86
TEST  Basic Operations with Polynomials
Perform the indicated operations:
(3x2  2x + 5) + (7x2 + 4x  5) = (7x2  2x  3)  (3x2 + 2x  5) =
2x2(x4) = (3xy2)2 = (2x2)3 = (4x)2(x2)3 =
2x2(x)2  (3x2)2 = 4x2(2x2 + 3x  2) = (3x + 2)(2x + 1) =
(3x  2)(2x2 + 3x + 2) = (x  4)(x2 + 3) =
3x3 = 8x5y = (2xy4)2 = . x2y .3 =
x5 2x4y3 (4xy3)3 .2xy2 ÿ
Express results using only positive exponents:
3x2 = (4x3y
2)2 =
6x
3 (2y
3)2
Divide and check:
4x2y  8xy2 = 6x2  6  5x = 6x3  x2  18 =
2xy 2x  3 2x  3
87
Special Products
Squaring a Binomial:
(x + 1)2 = (x  1)2 = (2x + 3)2 = (2x  3)2 =
(x + y)2 = (x  y)2 = (2x + 3y)2 = (2x  3y)2 =
(a + b)2 = Rule: To square a binomial:
(a  b)2 = Square the first term of the binomial
Square the second term of the binomial
Double the product of the two terms
The result is called a "trinomial square",
or a "perfectsquare trinomial".
Practice:
(Use the "FOIL method".)
(x + 2)2 = (x  2)2 = (3x + 2)2 = (3x  2)2 =
(Use the "shortcut".)
(y + x)2 = (y  x)2 = (3x + 2y)2 = (3x  2y)2 =
The Difference of Two Squares:
(x + 1)(x  1) = (2x + 3)(2x  3) = (x + y)(x  y) = (2x + 3y)(2x  3y) =
(a + b)(a  b) = Rule: When the first terms of the binomials are the
same, but the second terms are the opposites
of each other, the "middle term" drops out, and
the result is called "The Difference of Two Squares".
Practice:
(x + 2)(x  2) = (3x + 2)(3x  2) = (y + x)(y  x) = (3x + 2y)(3x  2y) =
88
Factoring Out a Common Monomial Factor
4x  8 = 21x2 + 14x = 15x2y  10xy + 20xy2 =
4x3  8x2 + 12x = 9x3y2 + 12x2y +12xy2 + 9x2y3 =
Practice:
9x  12 = 20x2 + 15x = 16x2y  8xy + 12xy2 =
5x3  10x2 + 15x = 14x3y2 + 21x2y + 21xy2 + 14x2y3 =
Factoring SpecialProduct Polynomials
Factoring Trinomial Squares:
x2 + 2x + 1 = x2  6x + 9 =
9x2  24x + 16 = 4x2 + 4xy + y2 =
Practice
x2 + 4x + 4 = x2  8x + 16 =
25x2  60 x + 36 = 9x2 + 6xy + y2 =
Factoring the Difference of Two Squares:
x2  4 = 4x2  9 =
x2  y2 = 9x2  16y2 =
89
x2  9 = 9x2  16 =
y2  x2 = 25x2  36y2 =
Factoring Quadratic Trinomials
x2 + 5x + 6 = x2  9x + 8 = x2  2x  8 =
x2 + 2x  35 = x2 + 3x + 10 = x2  x + 6 =
Practice:
x2 + 6x + 8 = x2  10x + 16 = x2  4x  21 =
x2 + 3x  4 = x2 + 10x  24 = x2 + 5x + 24 =
When the coefficient of the quadratic term is other than 1 or 1:
3x2 + 4x + 1 = 2x2  3x + 1 = 4x2 + 3x  1 =
6x2 + 5x + 1 = 3x2  2x  5 = 4x2 + 25x  21 =
16x2  50 x + 25 = 12x2  7x  12 = 6x2  25x  9 =
Practice:
2x2 + 3x + 1 = 3x2  4x + 1 = 3x2  2x  1 =
8x2 + 6x + 1 = 3x2 + 4x  7 = 4x2 + 23x  6 =
8x2 + 3x  5 = 18x2  9x  14 = 6x2  13x + 6 =
90
Practice:
Note: To factor a polynomial completely means to keep factoring as long
as is possible. Sometimes you can factor out a common monomial factor
and then continue to factor what remains.
5x2  45 = x3  25x = 2x2  8x + 8 =
3x2  6x  24 = 3x3  12x2 + 9x = 2x3 + 16x2  40x =
3x2  15x + 18 = 36x3  21x2  30x =
Practice:
5x2  5 = 9x3  x = 12x2  36x + 27 =
6x2  24x 72 = x3 + 7x2 + 12x = 3x2  21x  24 =
18x2 + 9x + 9 = 24x3 + 22x2  10x =
91
Factoring Completely
TEST  Factoring
Factor as completely as possible:
4x2 + 12x + 9 = 4x2  9 =
x2  2x  3 = 6x2  5x  4 =
6x2 + 13x + 5 = 16x3  x =
6x2 + 9x  6 = 8x3 + 46x2 + 12x =
92
Workbook
Part 5  Fractions
Equivalent Fractions
Note: Fractions which have the same value but different forms are called
"equivalent fractions".
Rule: The value of a fraction does not change when both the numerator and
denominator are multiplied by the same quantity. The result is equivalent
to the original, and it is said that the fraction has been raised to "higher terms".
(2) (x – 2)
4 = 4 =
5 5
(x + 2)
(x + 2) =
(x  2)
Practice: Raise each fraction to higher terms by multiplying both numerator
and denominator by the given factor:
(x + 2) (x  2) (2x  3)
3 = x  2 = 2x  3 =
5 x + 2 2x + 3
Rule: The value of a fraction does not change when both the numerator and
denominator are divided by the same quantity. The result is equivalent to the
original, and it is said that the fraction has been "reduced to lower terms".
8 =
10
Note: To reduce a fraction to lowest terms:
8 = Factor completely both numerator and denominator.
10 Cancel out the common factors.
4x  8 = x2 + 4x + 4 =
5x  10 x2  4
Simplify: Reduce to lowest terms:
4  8x = x2  1 =
1  4x2 2x + 2
Practice:
2x  2y = x2  3x =
4x2  4xy x2  4x + 3
93
1) The sign before the fraction
2) The sign of the numerator
3) The sign of the denominator
Rule: The value of a fraction is not changed when any two of the three signs
associated with it are changed.
Note: Often it is useful to change the sign of either the numerator or the
denominator of a fraction, which you may do if you change the sign in front of
the fraction as well.
4 =  4 Changing the sign before the fraction and the sign of the numerator.
2 2
4 =  4 Changing the sign before the fraction and the sign of the denominator.
2 2
Remember: The most useful application of the above 2 = 2 =  2
rule is that if only one of the three signs is negative, it 3 3 3
may be placed in any one of the three positions.
Rule: If the numerator or the denominator of a fraction is a polynomial, when you
change the sign of that polynomial, you must change the sign of each of its terms.
Change the sign of each expression from positive to negative:
(x  y) 1  x
2
Simplify: Reduce to lowest terms.
2x  3 =  y  x =
9  4x2 x – y
(x  5)2 = 4(x  y)2 =
5  x 8(y  x)
94
Note: There are three signs which are associated with a fraction:
x2  2x + 1 x2  9
3x2  10x + 3 =
12  x  x2
Practice:
2  5x = 9x2  1 =
4  25x2 3x  1
(x  6)2 = 8x3  2x =
6  x 2x2  5x + 2
x2  y2 = 4  x =
x2  3xy + 2y2 2x2  9x + 4
3  5x  2x2 = 4y  8x =
4x2  1 2x2 + 3xy  2y2
Rule: When the numerator or the denominator is expressed in factored form,
the value of the fraction is not changed when the signs of an even number of
factors in the numerator or denominator are changed.
(2  x)(5  x) =
(x  2)
Practice:
(x  1)(x  2) = 5(y  x)2
=
(3  x)(2  x)(1  x) 10(x  y)
95
3  3x = 6  x  x2 =
Multiplying Fractions
Rule: To multiply fractions:
Factor completely numerator and denominator.
Cancel out any factors common to any numerator and any denominator.
Multiply across the remaining factors.
9x2  25 . 3x  3 = x2  5x + 6 . x + 2 =
x2  1 9x  15 2x + 4 2x  6
2x + 1 . x2  4x = 3x  6 . 3x2  12 =
x2  16 4x2  1 6x + 12 x2  5x + 6
Practice:
3x  6 . x2  x  6 = 4x  4y . xy =
5x x2  4 xy 2x  2y
5x . 6x + 18 = 4x + 8 . 6x + 15 =
x2  9 15x3 4x2  25 2x2 + 4x
3x2 + 2x  1 . 10x2  13x  3 = x2 + 9x + 14 . x2 + 2x  35 =
5x2  9x  2 2x2  x  3 x2  3x  10 x2 + 4x  21
96
Dividing Fractions
Rule: To divide by a fraction, multiply by the reciprocal of the fraction.
5x + 15 ÷ 10x2 + 10x = x2  9y2 ÷ 3x  9y =
x2  9 4x  12 8x + 4y 12x + 6y
x2  x  20 ÷ x2  7x + 10 = 2x  8 ÷ 4  x =
x2 + 7x + 12 x2 + 9x + 18 x2  4 x2 + 3x + 2
Practice:
x  5 ÷ x2  25 = 12 + 6x ÷ 4  x2 =
x x2 15  3x 25  x2
x  7 ÷ 3x  21 = x2  3x ÷ x2  x  6 =
3x2  8x + 4 6x2  24 x2 + 3x  10 x2  4
x2  5x + 6 ÷ x2  2x  3 = 4x2  1 ÷ 2x2 + x  1 =
x2  4 x2 + 3x + 2 x2 + 2x  3 1  x2
97
Adding and Subtracting Like Fractions
Rule: To Add or Subtract Like Fractions:
Add or subtract the numerators as indicated by their signs.
Write the result over their common denominator.
Reduce to lowest terms.
2 + 3 = 2 + 3 = 5 7  2 = 7  2 = 5 7  3 + 4 = 7  3 + 4 = 8 = 4
8 8 8 8 9 9 9 9 10 10 10 10 10 5
Simplify: (Add or subtract as indicated and reduce the answer)
3 + 2  4 = 5x  2x + 3x = 2x  x  3y =
5x 5x 5x 8 8 8 x  y x  y
2x + 2y = 3x + y + x  4y + 2x  3y = x  y =
x + y x + y 4(x  y) 4(x  y) 4(x  y) x2  y2 x2  y2
x2 + 3x  2x + 12 = 3 + 1 = Note: Sometimes you can make
x2 + 2x  15 x2 + 2x  15 x  4 4  x denominators alike by changing
the sign of one of them.
Practice:
5  2 + 4 = 3x  7x + 9x = 2y  3x  5y =
9x 9x 9x 10 10 10 x + y x + y
98
3x  3y = 6x + 3x  10y  x  6y = 2x  3y =
x  y x  y 2x  y 2x  y 2x  y 4x2  9y2 4x2  9y2
x + 6  x = 7  x + 2x  3 = 5 + 1 =
2(3  x) 2(3  x) x2 + 3x  4 x2 + 3x  4 2x  2 2  2x
x + 1 = x + 3 = 3x  y  x + y =
x + 1 x + 1 x  3 3  x 2x2  3xy + y2 2x2  3xy + y2
Adding and Subtracting Unlike Fractions
Note: The LCD, Least Common Denominator, of two or more given fractions,
is the product of all the factors in the denominators, each being taken the
greatest number of times it occurs in any one of the given denominators.
LCD (20, 50) = LCD (x2 + 2x + 1, x2  1) =
Practice:
LCD (18, 24) = LCD (x2  2x + 1, x2 + x  2) =
99
Rule: To add and subtract unlike fractions,
Factor each denominator and find the LCD.
Replace each fraction with an equivalent one having the LCD as denominator
Find the sum or difference and simplify.
Simplify:
5x + 2x = x + 3 + x + 4 =
6 3 x 3x
1  x = 3  2x  4  3x =
x  y 3x  3y 6 8
x + 3 + x + 1 = 3 + 4 =
4x 3x x  4 x + 4
6  3 = 2 + 3 =
x  2 x + 2 x2  4 x2 + 5x + 6
100
3x  3  3 = 2 + 3 =
5x  15 2x  6 x2  2x + 1 x2 + x  2
3 + 5  6 = 9 + 7  3x  1 =
x  2 x2  4 x + 2 2 + x x  2 4  x2
Practice:
3x  5x = x + 1 + x + 3 =
4 2 3x x
3 + x  2x = 9  4x  3  5x =
x  y 2x  2y 12 9
101
x + 1 + x  3 = 4  3 =
2x 3x x + 5 x  5
4  6 = 7 + 8 =
x  1 x + 1 x2  9 x2 + x  12
3  4 = 3  2 =
x2  4 2  x x2  x x2 + x  2
5  6  1 = 3  2 =
x  3 x2  3x x x2  3x + 2 x2  1
102
Simplifying Complex Fractions
Note: A complex fraction contains a fraction within the numerator or
denominator or both.
Rule: To simplify a complex fraction:
Combine the terms in the numerator and/or denominator into single
fractions.
Set the fraction up as a division problem dividing the numerator by the
denominator.
Invert the divisor and multiply.
Simplify and reduce the result.
x  2 x  4
2
=
x
=
x  2 1 + 2
x
x
Note: If the denominators of both the "top" and "bottom" of the main fraction
are the same, they can be eliminated by multiplying both the "top" and "bottom"
of the main fraction by that factor.
x2  y2 1  1
x + 2y
=
x y
=
x  y 1 + 1
x + 2y x y
Simplify:
x + 1 x .
.
y
=
x + y
=
x  1
y
.
y
x + y
103
x2  4y2 3  1
x2 =
x
=
x + 2y 3 + 1
x x
Practice:
x  4 x  9
4
=
x
=
x  4 1 + 3
x
x
x2  y2 2 + 1
x  2y
=
x y
=
x + y
2  1
x  2y x y
1  3 x + y
x
=
y x
=
1  9 1 .
x2
xy
104
1 + 1 x .
x2 y2 =
x + 3
=
2 1  x .
xy x + 3
x  y
y x
=
y
 1
x
2  2 .
x + y x  y
=
4 .
x2  y2
105
Solving Equations Containing Fractions
Rule: To solve an equation containing fractions:
Clear the equation of fractions by multiplying both members of the
equation by the LCD of the denominators.
Solve the resulting equation.
Check to see that the "root", or answer, "satisifies" the given equation
Solve and Check:
x + 2 = x x + x = 3
3 2 2 4
3x + 4x + 1 = 4  2x 3x + 5 + 3x  4 = 3x + 2
4 6 3 3 6
106
x  4 + x + 3 = 5 2 (4x  1)  3 (x + 1) = 7
3 2 6 3 5
Practice:
5x + 2 = 7 5  31x = 2  7x 5x  29 = 5  3x  4x
6 6 9 3 2 5 2 15
x + 5  x  3 = 7 4x + 1  2x  1 = 3 4 + x  3 = 3x  1
2 5 10 7 6 2 3 4 6
107
x + 2  x  3 = 1 7x + 5  3x + 15 = 2
4 3 2 8 10
x  3 + x + 2 =  7 3 (2x + 1)  1 (4x  1) =  19
2 4 16 2 3 6
108
Rule: When a variable appears in the denominator of one or more terms of an
equation, you must eliminate as a "root" or "solution" to the equation, any value for the
variable that would result in a division by zero. Any solution that causes a division by zero
in the original equation is called an "extraneous root". If the only solution to the equation
is an "extraneous root", the equation is said to have "no solution".
Solve and Check:
5 = 6 x  1 = 2 3  2 = x .
x 18 x + 3 3 x + 2 3 x + 2
5 + 3 = 6 x + 3 = x + 2
x + 1 x  1 x2  1 x + 1 x + 4
5 + 2 = 3 6x  x  6 = 5x2  12
x  2 2  x 2 x + 3 x  3 x2  9
109
Practice:
x = 3 1 = 6  5 2 = 5 .
x + 2 5 2 x 2x x x  3
11 = 2 + 5 3 + 6 = 14 x + 3 + x  7 = 5
.
x x 3x  6 x  2 3 x x 3
2 + x = 3 + 1 3  1 = 7  9
.
6x 5x 30 x 5x 5
110
4  6 = 24 4x + x  3 = 5x2  33
.
x  3 x + 3 x2  9 x + 2 x  2 x2  4
Note: If the equation consists of only one fraction on each side, and is thus in
the form of a "proportion", then you can apply the rule "the product of the means
equals the product of the extremes":
If a = c , then ad = bc
b d
5 = 6 5 = 6 x  1 = 2 x  1 = 2
x 18 x 18 x + 3 3 x + 3 3
Practice: (Just do them as proportions)
x = 3 2 = 5 .
x + 2 5 x x  3
111
Solving Quadratic Equations by Factoring
Rule: If the product of two factors is zero, then either one or both of the factors is
zero. So; if ab = 0, then a = 0 or b = 0 or both.
Note: An equation in one unknown in which the highest power of the variable
is the second power, is called a "quadratic equation".
Rule: To solve a quadratic equation:
Write all the terms on one side, so that the other side equals zero.
Factor the nonzero side of the equation.
Set each factor equal to zero and solve.
Check each root by substituting into the original equation:
x2  2x  15 = 0 x2 + 10 = 7x 2x2 + x = 15
Practice:
x2 + 6x + 8 = 0 x2 + 8x = 15 10 = 13x + 3x2
23x  6 = 4x2 x + 4 = 3x2 2x2  3x = 1
112
Fractional Equations that Transform into Quadratic Equations
11x  12 = 2x 4  3 = 5 .
x x 2x + 3
3 + 10 = 5 .
x2  1 x  1
113
Practice:
8x + 3 = x 4  7 = 2 .
3x x  2 x  3 15
5 + 12 = 3 .
x2  4 x  2
114
TEST  Operations and Equations with Fractions
Simplify:
x2  4x = 3  x = 3x  3y =
x2  x  12 3x2  5x  12 6y  6x
2x + 8 . 3x2  12x =
2x2  11x + 12 2x2 + 5x  12
x2  9 ÷ x2  3x =
3x  6 x2 + 2x  8
2x + 3x  4 = 1 + x  1 =
x  2 2  x x  1 x + 1 x2  1
115
x + 3 2  2 .
2x x y
= =
4x2  9 x2  y2 .
2x y
Solve and Check: (Show any values of x that cause division by zero
and declare any "extraneous roots")
x  2 + x + 5 = 29 5 + 4 = 11 .
3 5 15 x  2 x + 2 x2  4
116
10x2 = 4x + 6 2 + 3x = 18x .
x + 3 x  3 x2  9
117
Solving Word Problems Involving Quadratic Terms, Factoring, Etc.
1) The side of one square is 2 cm longer than the 2) The difference of the squares of two
side of another. The area of the smaller square consecutive positive integers is 55.
is 36 cm2 less than the area of the larger square. Find both integers.
Find the length of the side of each square.
3) Find two consecutive positive integers 4) The square of a positive integer exceeds
whose product is 42. 3 times the integer by 54. Find the
integer.
5) The dimensions of a rectangular garden are 6) The sum of two numbers is 17 and their
12 m by 8 m. The garden is surrounded by product is 42. Find the numbers.
a walkway of uniform width. The combined
area of the garden and the walkway is 192 m2.
Find the width of the walkway.
118
perimeter is 60 cm and whose area is 161 cm2. is 4 cm less than twice the length of the
other leg. The area of the triangle is 24
cm2. Find the length of each leg. (Area
of right triangle is equal to one half the
product of the legs.)
Practice:
1) A rectangular picture is 3 cm longer than it 2) The difference of the squares of two
is wide. It is surrounded by a frame that is consecutive odd integers is 72. Find
2 cm wide The area of the frame and picture both integers.
together is 100 cm2 greater than that of the picture
alone. Find the dimensions of the picture.
3) Find two consecutive positive integers 4) Find two consecutive negative integers
whose product is 72. such that the sum of their squares is 61.
119
7) Find the dimensions of a rectangle whose 8) The length of one leg of a right triangle
5) The length of a rectangle is 3 cm less than 6) The sum of two integers is 4. The sum
twice its width. The area of the rectangle is of the squares of the two integers is 16
77 m2. Find the dimensions of the rectangle. greater than the product of the two
integers. Find the integers.
7) Find the dimensions of a rectangle whose 8) The length of one leg of a right triangle
perimeter 48 cm and whose area is 95 cm2. is 2 cm more than twice the length of the
other leg. The area of the triangle is 30
cm2. Find the length of each leg. (Area
of right triangle is equal to one half the
product of the legs.)
120
Solving Word Problems Involving Equations with Fractions
1) Five sixths of a number is 6 more than 2) What number added to both the numerator
half the number. What is the number? and the denominator of the fraction 4/7
results in a fraction equivalent to 4/5?
3) The sum of the reciprocals of two positive 4) Find two consecutive even integers such
consecutive integers is 11/30. Find the that 5 times the reciprocal of the lesser is
integers. equal to 6 times the reciprocal of the
greater.
5) One number is 16 more than another 6) The denominator of a fraction is 4 less
number. Four ninths of the greater number than 5 times the numerator. The fraction
is equal to four fifths of the lesser number. is equivalent to 1/4. Find the fraction.
Find the numbers.
121
7) When a number is added to the numerator 8) Find two consecutive integers such that
of the fraction 4/7, and twice the same number the reciprocal of the lesser increased by
is subtracted from the denominator, the result 3 times the reciprocal of the greater is
is equal to 7. Find the number. equal to 9 times the reciprocal of the
product of those integers.
Practice:
1) Three fourths of a number is 1 more than 2) What number added to both the numerator
two thirds of the number. What is the number? and the denominator of the fraction 5/9
results in a fraction equivalent to 3/4?
3) The sum of the reciprocals of two positive 4) Find two consecutive even integers such
consecutive integers is 9/20. Find the that 9 times the reciprocal of the lesser is
integers. equal to 10 times the reciprocal of the
greater.
122
5) One number is 4 less than another 6) The numerator of a fraction is 10 less
number. Three fifths of the greater number than 2 times the denominator. The fraction
is equal to two thirds of the lesser number. is equivalent to 3/4. Find the fraction.
Find the numbers.
7) When a number is subtracted from the 8) Find two consecutive integers such that
denominator of the fraction 5/9, and one the reciprocal of the lesser increased by
more than twice the same number is added 2 times the reciprocal of the greater is
to the numerator, the result is equal to 2. equal to 25 times the reciprocal of the
Find the number. product of those integers.
123
TEST  Problems Involving Quadratic Equations and Fractions
1) Find the dimensions of a rectangle whose perimeter is 20 cm and whose area is 21 cm2.
2) Find two consecutive odd integers such that 3/5 the reciprocal of the lesser plus 5/8 the
reciprocal of the greater is equal to 13/40.
124
Workbook
Part 6  Radicals and Roots
Introduction to Rational and Irrational Numbers
Remember:
{1, 2, 3, 4, . . . } = the "counting numbers"
{0, 1, 2, 3, 4, . . . } = the "natural numbers"
{ . . . 4, 3, 2, 1, 0, 1, 2, 3, 4, . . . } = the "integers"
Note: The counting numbers are a subset of the natural numbers.
The natural numbers are a subset of the integers.
The integers are a subset of the rational numbers.
A "rational number" is a number that can be expressed in the form of a quotient
or ratio or as one integer divided by another. All rational numbers can also
be expressed as either terminating or repeating decimals.
Examples of rational numbers:
1 3 5 1 5  2  4 6 7
2 4 8 3 3 5 3 1
An "irrational number" is a number that cannot be expressed as a ratio or as one
integer divided by another.
___
Example: \/ 2
Neither the rational nor the irrational numbers are subsets of each other.
Together they make up the set called the "real numbers".
Square Roots
Note: Just as the inverse operation of addition is subtraction, and the inverse
operation of multiplication is division, so the inverse operation of squaring a
number is called finding a "square root".
___
Example: 22 = 4 \/ 4 = 2
The square root of a number is one of its two equal factors.
A positive real number has two square roots: a positive or "principal" square root,
and a negative square root.
____ ____
\/ 4 = 2 because (2)(2) = 4, but (2)(2) = 4 also, so \/ 4 = 2 or 2
__ __
The principal square root of a number is represented by: \/ x , so: \/ 4 = 2
____ ____
The negative square root of a number is represented by:  \/ x , so:  \/ 4 = 2
125
Unless otherwise stated, when we speak of the square root of a number as \/ x ,
we will be talking only about the positive or principal square root.
____
\/ 4 is not a real number. It is called an "imaginary" number. The square root of
any negative quantity is "imaginary".
__ __ __ __ __
\/ 22 = \/ 4 = 2 \/ 32 = \/ 42 = \/ x2 =
Simplify:
___ ___ __ ____
\/ 49 =  \/ 16 = \/ 82 = \/ (3)2 =
____ ___ ___ ___
 \/ (5)2 = \/ 36 = \/ y2 =  \/ x2 =
Practice:
___ ___ ___ ____
\/ 64 =  \/ 25 = \/ 72 = \/ (4)2 =
____ ___ ___ ___
 \/ (9)2 = \/ 9 = \/ x2 =  \/ y2 =
The Radical Sign
____
Note: The symbol \/ is called the "radical sign". The quantiy placed beneath
the radical sign is called the "radicand", and the whole quantity is called called
a "radical expression" or simply a "radical".
Rule: The first requirement for a radical to be in "simplified form" is that the
radicand has no factor that is a perfect square.
Note: When we simplify a radical we speak of removing a quantity that is a
perfect square and "taking it out from under the radical."
The Laws of Square Roots
___
\/ 36 = Rule: The Product Property of Square Roots 
The square root of a product is equal to the
___ __ __ the product of the square roots of its factors.
\/ ab = \/ a \/ b
126
Simplify:
__ ___
\/ 8 = \/ 48 =
___ ___
3 \/ 75 = \/ 40 =
2
Practice:
___ ___
\/ 12 = \/ 45 =
___ ____
\/ 72 = \/ 250 =
___ ___
2 \/ 28 = 2 \/ 18 =
___ ___
\/ 54 = \/ 32 =
3 6
Simplify:
___ ___ ___ ___
\/ 4x2 = \/ 3x2 = \/ 16x = \/ x2y =
_____ ____ ____ ____
\/ 9x2 y2 = \/ 4xy2 = \/ 7x2y = 3 \/ 8xy2 =
127
___ ___ ____ ___
\/ 9x2 = \/ 5x2 = \/ 25x = \/ xy2 =
______ _____ _____ _____
\/ 16x2y2 = \/ 36xy2 = \/ 10x2y = 2 \/ 12x2y =
Simplify:
___ ___ ___
\/ x4 = \/ x3 = \/ x
8 =
___ ___ ____
\/ x7 = \/ 9x8 = \/ 16x5 =
___ ___
\/ 5x3 = \/ 8x9 =
Practice:
___ ___ ___
\/ x6 = \/ x5 = \/ x10 =
___ ___ ____
\/ x9 = \/ 4x6 = \/ 25x7 =
____ ____
\/ 11x3 = \/ 20x9 =
Simplify:
________ _______ ___________
\/ (x2 + y2)2 = \/ 3(x + y)3 = \/ x2 + 2xy + y2 =
128
Practice:
_______ _______ ____________
\/ (x2  y2)2 = \/ 5(x + y)5 = \/ 4x2 + 4xy + y2 =
Rule: The Quotient Property of Square Roots 
The square root of a fraction or quotient is equal to the quotient of the
square roots of the numerator and the denominator.
__ __ __ __
4 = \/ 4 = 2
a = \/ a .
\ 9 \/ 9 3 \ b \/ b
Rule: There are three requirements for a radical to be in "simplified form":
1) The radicand has no factor that is a perfect square.
2) The radicand does not contain a fraction.
3) No radical appears in a denominator.
Note: In order to fulfill the requirement that no radical appears in a denominator
we employ a process called "rationalizing the denominator".
__ __ __ __ __ __
\/ 2 \/ 2 = \/ 3 \/ 3 = \/ 5 \/ 5 =
__
4 =
\ 5
2 =
_
\/ 8
Simplify:
___ ___
3 = 1 =
\ 4 \ 2
1 = 3 =
\/ 3 \/ 2
129
Practice:
9 = 5 =
\ 32 \ 12
12 = 4 =
_
\/ 18 \/12
Practice:
___ ___
7 =
2 =
\ 16 \ 5
3 = 5 =
\/ 5 \/ 8
___ ___
16 =
5 =
\ 27 \ 50
8 = 15 =
\/ 8 \/20
130
Simplify:
___
1 =
1 =
\ x \/ x3
x = 8 =
\/ y \/ x
___ ___
3x = x =
\ y \ y2
__ __
x = 2 =
\ 6 \ 8x
___ ___
2 \/ 3x = x3 =
\/ x3 \ xy
Practice:
___
2 =
1 =
\ x \/ x5
__
y \/ x = 8 =
_
\/ y \/ 2x
131
_____ ___
2x = x2 =
\ y \
y
__ __
12 = 9 =
\
3x \
5x
___ ___
3\/ 5x = x5 =
_
\/ x3 \ 2xy
Adding and Subtracting Radicals
Rule: To add or subtract like radicals, apply the rules for combining terms.
To add or subtract unlike radicals, it must be possible to transform
them into like radicals
Simplify:
__ __ __ __ __ ___
6 \/ 3 + 9 \/ 3 = 8 \/ 2  5 \/ 2 = \/ 3 + \/ 12 =
___ ___ ___ __ __ ___
\/ 45 + \/ 20 = \/ 50  \/ 8 = 3 \/ 5  \/ 20 =
132
__ __ ___ ___ ___ __ ___
3 \/ 2 + 2 \/ 8  3 \/ 32 = 7 \/ 28  4 \/ 63 = 4 \/ 8  \/ 98 =
____ ____ ___ ____ ___ __ _____
\/ 16x + \/ 25x = \/ 5x2 + \/ 20x2 = \/ xy2  y \/ x + \/ 16xy2 =
Practice:
__ __ __ __ __ ___
5 \/ 2 + 6 \/ 2 = 9 \/ 5  7 \/ 5 = \/ 5 + \/ 45 =
___ ___ ___ ___ __ ___
\/ 32 + \/ 50 = \/ 75  \/ 27 = 3 \/ 2 + 2 \/ 18 =
__ ___ ___ ___ ___ ___ __ ___
3 \/ 5  \/ 20 + \/ 45 = 4 \/ 72  2 \/ 50 = \/ 32  5 \/ 8 + 2 \/ 50 =
___ ____ ___ ___ ___ __ _____
\/ 9x  \/ 64x = \/ 9x3  x \/ 4x = y \/ x2y  2x \/ y3 + \/ 9x2y3 =
133
Multiplying Radicals
Simplify:
__ __ ___ __ __ __
\/ 8 \/ 2 = 2 \/ 50 3 \/ 2 = \/ 6 \/ 3 =
__ __ ___ __ __
2 \/ 6 \/ 2 = 4 \/ 18 5 \/ 3 = ( 5 \/ 3 )2 =
__ __ __ __ __
\/ 2 ( 4 + \/ 3 ) = \/ 5 ( \/ 2  \/ 3 ) =
__ __ __ __
( 4 + \/ 2 ) ( 3 + \/ 2 ) = ( 3 + \/ 2 )( 1  \/ 2 ) =
Note: Two binomial factors of the form: __ __
(a + \/ b ) and (a  \/ b )
are called "conjugates" of each other. Their product is a "difference of
two squares", and the "middle term" drops out leaving a rational number
for an answer.
Simplify:
__ __ __ __
( 5 + \/ 2 )( 5  \/ 2 ) = ( 5 + 6 \/ 3 ) ( 5  6 \/ 3 ) =
__ __ __ __
( 4 \/ 3 + \/ 5 ) ( 4 \/ 3  \/ 5 ) =
134
Practice:
___ __ ___ __ ___ __
\/ 27 \/ 3 = 3 \/ 12 4 \/ 3 = \/ 10 \/ 5 =
__ __ ___ __ __
3 \/ 6 \/ 8 = 4 \/ 10 5 \/ 6 = ( 4 \/ 2 )2 =
__ __ __ __ __
\/ 5 ( 3 + \/ 2 ) = \/ 2 ( \/ 5  \/ 3 ) =
__ __ __ __
( 3 + \/ 3 ) ( 2 + \/ 3 ) = ( 2 + \/ 5 ) ( 8  \/ 5 ) =
__ __ __ __
( 4 + \/ 3 )( 4  \/ 3 ) = ( 7 + 2 \/ 5 ) ( 7  2 \/ 5 ) =
__ __ __ __
( 5 \/ 2 + \/ 3 ) (5 \/ 2  \/ 3 ) =
Simplify:
__ __ ___ ___ ____ ____
\/ x \/ x3 = \/ 3x \/ 2x4 = \/ 5x2y \/ 10y =
135
Practice:
___ ___ ___ ___ ___ ____
\/ 2x \/ 8x = \/ 4x \/xy = \/ 3x3y \/ 8xy2 =
Dividing Radicals
Simplify:
___ ___
\/ 50 = 6 \/ 45 =
\/ 2 3 \/ 5
___ ___
\/ 24 = 50 \/ 24 =
\/ 3 5 \/ 2
___ ___
6 \/ 27 = 2 \/ 24 =
12 \/ 3 4 \/ 3
___ ___ ___ ___
\/ 32 + \/ 50 = 10 \/ 72  15 \/ 18 =
\/ 2 5 \/ 3
136
Practice:
___ ___
\/ 48 = 6 \/ 96 =
\/ 3 2 \/ 6
___ ___
\/ 54 = 8 \/ 60 =
\/ 2 4 \/ 5
___ ___
4 \/ 32 = 4 \/ 72 =
12 \/ 2 8 \/ 6
___ ___ ___ ___
\/ 18 + \/ 8 = 8 \/ 60  6 \/ 15 =
\/ 2 2 \/ 3
Simplify:
___ _____ ______
\/ x3 = \/ 12x5 = \/ 75x3y =
\/ x \/ 3x \/ 3xy
137
___ _____ _______
\/ x7 = \/ 18x3 = \/ 125x3y4 =
\/ x3 \/ 2x \/ 5xy2
Simplify: (Rationalize the denominators.)
6 =
2  \/ 2
__
\/ 2 =
\/ 7 + \/ 2
__
\/ 3 + 1 =
\/ 3  1
Practice:
12 =
3  \/ 5
__
\/ 3 =
\/ 5 + \/ 3
__
\/ 2 + 4 =
\/ 2  1
138
Practice:
Solving Radical Equations
Note: A "radical equation" is one in which the unknown appears under a radical sign.
Usually the equation is transformed into one without the radical (the square root)
by squaring both sides. After solving the resulting equation, you must check to
see if the answer satisfies the original equation. If not, then the answer is
considered to be an extraneous root and the equation may have no solution.
When doing the check, you may consider only the principal square root, not the
negative square root.
__ ___ ___ __
\/ x = 7 \/ 5x = 5 \/ 3x = 2 4 \/ x = 8
___ ____ ______
5 \/ 5x = 20 \/ x  5 = 3 \/ 2x + 2 = 6
_____ __ _______
\/ x2 + 3 = x + 1 2x + 6 = 3 \/ x \/ x2 + 15 = 5  x
\/ x
139
____ ____ ______
2 \/ x  1 = 4 3 + \/ 5  x = 2 \/ 12 + x = x
Practice:
__ ___ ___ __
\/ x = 4 \/ 6x = 6 \/ 5x = 3 2 \/ 3x = 6
___ _____ _____
6 \/ 6x = 12 \/ x + 3 = 5 \/ 5x  1 = 3
140
______ ____ _____
\/ x2 + 27 = x + 3 3 \/ x  7 = 6 \/ x2  7 = x  1
\/ x  7
_____ ___ ______
\/ 3x  2 = 1 2  \/ 3x = 5 \/ 2x + 3 = x
141
TEST  Radicals and Roots
Simplify:
___ ___
2 \/ 24 = \/ 63 =
3
___
3 = 4 =
\ 5 \/ 8
___ ___ __ __ ___
4 \/ 27  \/ 48 = \/ 3 (\/ 5 + \/ 20 ) =
___ __ __ __
3 \/ 50 3 \/ 2 = ( 3 + \/ 3 ) ( 3  \/ 3 ) =
___ ___
\/ 48 = 3 \/ 72 =
\/ 3 9 \/ 6
___ ___ __
\/ 40  2 \/ 90 = \/ 3 =
\/ 2 \/ 3 + 1
142
______ ____
2 \/ 12x2y5 = x7 =
\ x2y
___ ____ ____ ____ ______
\/ x2y + \/ 9x2y = \/ 3xy2 \/ 8x3y = \/ 12x2y3 =
\/ 3y
Solve and Check:
____ _____
2 \/ x  2 = 6 \/ 5 + x = 2x
143
Workbook
Part 7  More on Quadratic Equations
Solving Quadratic Equations by Completing the Square
Note: We can solve some quadratic equations by factoring, those that are factorable
and have roots that are rational numbers. By another process called
"completing the square" we can solve any quadratic equation, even those that
are not factorable and have roots that are irrational numbers.
Remember: When we square a binomial, the result is called a "trinomial
square" or a "perfectsquare trinomial".
Rule: In every perfect square trinomial of the form: ax2 + bx + c the constant
term is always equal to the square of onehalf the coefficient of x.
Examples: x2 + 8x + 16 = Onehalf of 8 = 4; then 42 = 16
(x + 4)(x + 4) =
(x + 4)2
x2  10x + 25 = Onehalf of 10 = 5; then (5)2 = 25
(x  5)(x  5) =
(x  5)2
Make each binomial into a perfect square trinomial by completing the square.
Then show each perfect square trinomial as the square of a binomial:
x2 + 6x x2  2x x2 + x
x2  3x x2
 1 x x2 + 3 x
2
4
Practice:
x2 + 10x x2  12x x2 + 7x
x2  5x x2
 x x2  2 x
3
144
Rule: To solve a quadratic equation by completing the square:
Put all terms that contain the unknown on the left side and the constant on the right side.
If the coefficient of x2 is not 1, divide all terms by the coefficient of x2.
Find onehalf the coefficient of x. Square it, and add the result to both sides.
Express the left side as the square of a binomial and simplify the right side.
Take the square root of both sides. Write + before the square root of the right side.
Solve the two separate equations, one for the positive value and one for the negative value.
x2  6x + 8 = 0 x2 + 2x  7 = 0
2x2  3x  2 = 0 x2  x  3 = 0
145
Practice:
x2 + 10x + 16 = 0 x2 + 4x  14 = 0
3x2 + 2x  1 = 0 x2 + 7x + 2 = 0
146
The Derivation of the Quadratic Formula
ax2 + bx + c = 0
Using the Quadratic Formula
Solve and check:
2x2  5x + 2 = 0 3x2  3x  2 = 0
147
Practice: (Solve and check)
2x2  x  3 = 0 3x2 + 5x + 1 = 0
The Discriminate and the Nature of the Roots
Note: When solving equations of the form: ax2 + bx + c
The expression b2  4ac is called the "discriminant" because it can
tell us about the "nature of the roots".
Rule: If the value of the discriminate is a negative number, the roots of the
equation are imaginary.
If the value of the discriminate is zero, the equation will have just
one real root. (Actually there are two roots, but they are equal, so the root
is sometimes called a "double root".)
If the value of the discriminant is a positive number, there are two
unequal real roots. (They will be irrational numbers unless the value of the
discriminate is a "perfect square", in which case the roots will be rational.
148
Find the value of the discriminate for each equation and tell the nature of the roots:
x2 + 3x + 4 = 0 3x2  6x + 3 = 0
2x2  5x  4 = 0 5x2  11x + 2 = 0
Practice:
4x2  3x + 5 = 0 4x2 + 4x + 1 = 0
x2  2x  10 = 0 x2  7x + 10 = 0
149
TEST  Completing the Square and The Quadratic Formula
Solve each equation, first by Completing the Square and then again by the
Quadratic Formula. Do the check for each equation just once. Finally, show
how the discriminant confirms the nature of the roots:
2x2 + 5x + 3 = 0 x2  2x  2 = 0
150
Workbook
Part 8  Inequalities
Introduction to Inequalities
Remember: Example:
= means "is equal to" 4 = 4
> means "is greater than" or "is to the right of" 4 > 3
< means "is less than" or "is to the left of" 3 < 4
Note:
> means "is greater than or equal to" 4 > 3, 4 > 4
< means "is less than or equal to" 3 < 4, 3 < 3
A mathematical sentence containing the = symbol is called an equation.
A mathematical sentence containing any of the symbols >, <, >, or < is called an inequality.
A mathematical sentence containing a variable is called an open sentence.
Examples of open sentences:
x = 4 is read, "x is equal to 4" (an equation).
x < 5 is read, "x is less than 5 (an inequality).
10 > x is read, "10 is greater than or equal to x" (an inequality).
The set of numbers that a variable could possibly stand for is called the
replacement set or the domain of that variable.
The set of numbers that satisfies an equation or an inequality within the
specified domain is called the solution set.
Example: For the inequality x > 5 to be solved within the domain of the
integers, the solution set is: {6, 7, 8, 9, ...}
Find the solution set given that the domain is the integers:
x > 3 x < 2
x > 2 x < 1
x > 4 x < 7
x > 1 x < 0
151
Practice:
x > 4 x < 6
x > 1 x < 2
x > 2 x < 6
x > 0 x < 1
When the domain of an inequality is the set of real numbers, rather than the
set of just the integers, the solution set includes rational numbers and
irrational numbers as well as integers:
Example: For the inequality x > 2 the solution set would include, among others,
the following members: 2.1, 2 1/2, \/ 5, 3, etc.
Thus, when the domain is the set of real numbers, it is best to show the solution
set as a graph on a number line.
O .
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 2 x < 5
o means that 2 is not a member of the solution set. . means that 5 is a member of the solution set.
The arrowhead on the graph means "and so on" in the indicated direction.
Use graphs to show the solution set of each inequality:(the domain is the real
numbers).
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 3 x < 2
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 2 x <  1
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 4 x < 7
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 2 x < 0
152
Practice:
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 4 x < 6
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 1 x <  2
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 5 x < 6
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x > 0 x < 1
Given the graph, write the inequality underneath the number line:
.
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
o .
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
o o
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
Practice:
o o
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
. .
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
. o
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
153
o
Solving Inequalities
The procedures for solving an inequality are the same as for solving an equation
with one important exception:
Rule: If you multiply or divide each side of an inequality by the same negative
number, you must also reverse the direcion of the inequality symbol.
Solve each inequality and show the solution set as a graph on the number line.
The domain is always the real numbers:
x + 5 > 7 2 > x  4
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
4x < 12 7  4x > 15
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x + 1 < 4 x + 3 < 2
2 4
.
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
154
Note: We will do this problem two ways.
3x + 5 < 4x + 3 3x + 5 < 4x + 3 2(x  4)  9 > 3 (x  6)
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
Practice:
x  4 > 2 3 < x  2
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
3x > 15 8  3x < 2
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
1 < 1 + x x  2 > 2
5 3
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
155
Note: Please do this problem two ways.
3x + 8 < 5x + 4 3x + 8 < 5x + 4 3(x + 5)  4 < 2(2x + 4)
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
Combining Inequalities
We can combine two inequalities by connecting them with either the word "and" or the word "or".
If we put them together with the word "and", the result is called a "conjunction". The solution set of the
conjunction, contains only those elements that are common to both of the solution sets of the separate
parts. This is also known as the "intersection" of those two solution sets.
If we put them together with the word "or", the result is called a "disjunction". The solution set of the
disjunction, contains those elements that are in either one (or both) of the solution sets of the separate
parts. This is also known as the "union" of those two solution sets.
Examples: "x > 3 and x < 4" First, x > 3 can be turned around to be 3 < x
Then 3 < x and x < 4 can be put together into 3 < x < 4
Notice that this can be done because the inequality symbols are in the same direction and
because the last part of the first inequality is identical to the first part of the second inequality.
However, be careful: You can't connect 3 < x and x < 2 to make 3 < x < 2
because 3 is not less than 2.
Now, look at the graphs of each part and of the conjunction: 3 < x < 4 o
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 x >  3
.
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 x < 4
o .
  l l l l l l l l l l l l l l l l l l l 3 < x < 4
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
In words, the solution set for the conjunction, 3 < x < 4
being the intersection of the two parts, is:
"4 and all the real numbers between 3 and 4".
156
o
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 x >  3
.
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 x < 4
  l l l l l l l l l l l l l l l l l l l x > 3 or x < 4
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
In words, the solution set for for the disjunction, x > 3 or x < 4
being the union of the two parts, is:
"all the real numbers".
Solve each inequality and graph the solution sets:
2 < x  1 < 5 3 < 2x + 1 < 13
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
5 > 2  3x > 7 10  2x > 12 and 7x < 4x + 9
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
157
Now, look at the graphs of each part and of the disjunction: x > 3 or x < 4
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
8  2x > 6 or 8  2x < 6 x  1 < 2x + 3 < x + 4
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
2x  1 < 3(x  1) < 2(x  2)
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
158
3 < 2x  1 and 0 > 1  x x + 2 > 6 or x + 2 < 6
1 < x + 3 < 5 14 < 4x  2 < 6
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
8 > 4  2x > 6 6x  12 > 12x and 13 + 4x < 1
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
2 < x + 1 and 3 < 5  2x 4 + x < 3 or 3 + x > 4
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
159
Practice:
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
8  3x < x  4 < 2x + 5
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
3x + 8 < 2 or x + 5 > 1 7  3x > 6  4x > 4  3x
160
Absolute Values in Open Sentences
Remember: The value of a number not considering the sign is the "absolute value".
l5l = 5 l5l = 5
Rule: On a number line, the absolute value of a number is its distance from the
origin not considering the sign. Thus you travel in both the positive and
negative directions to graph the absolute value of a number.
Example: lxl = 3 . .
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
The graph consists of the two points that are three units from the origin, 3 and 3.
This is true because: If x = 3, then l3l is 3. If x = 3, then l3l is 3.
The value of x could be either one, 3 or 3, to produce an absolute value of 3.
Thus, lxl = 3 is equivalent to the disjunction: x = 3 or x = 3.
Example: lxl < 3 o o
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
The graph consists of all the points less than three units from the origin.
This is equivalent to the conjunction: 3 < x < 3.
Example: lxl < 3 . .
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
The graph consists of all the points at most three units from the origin.
This is equivalent to the conjunction: 3 < x < 3.
Example: lxl > 3 o o
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
The graph consists of all the points more than three units from the origin.
This is equivalent to the disjunction: x < 3 or x > 3.
Example: lxl > 3 . .
  l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
The graph consists of all the points at least three units from the origin.
This is equivalent to the disjunction: x < 3 or x > 3.
161
Solve the open sentence and graph the solution set:
lx  4l = 6 l1  4xl = 3
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
lxl < 1 l2x  9l < 1
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
l3  xl > 5 l3  3xl > 7
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
162
lx + 3l = 5 l1  3xl = 10
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
lx  3l < 6 l3x  7l < 2
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
l4  xl > 3 l5  2xl > 3
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
163
Practice:
TEST  Inequalities
Solve each Inequality and graph the solution set:
3 > x  4 4  3x < 13
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
x + 4 < 6 3(x  1)  1 < 4(x + 1)
2
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
4x  5 > 7 or 4x  5 <  9 7  4x > 6  5x > 4(1  x)
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
164
l2x + 3l < 11 l4  2xl > 4
  l l l l l l l l l l l l l l l l l l l   l l l l l l l l l l l l l l l l l l l
10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
165
Graphing Linear Inequalities in Two Variables
1) Graph the inequality: y > 2x  4 2) Graph the inequality: y > 1/2 x + 3
y y
x x
3) Graph the inequality: 4x  y < 3 4) Graph the inequality: 2x + 3y < 3
y y
x x
166
Practice:
1) Graph the inequality: y > 3x  5 2) Graph the inequality: y >  1/3x + 1
y y
x x
3) Graph the inequality: 3x  y < 4 4) Graph the inequality: 3x + 2y < 6
y y
x x
167
Solving a System of Inequalities Graphically: y
2x + y > 6 and y  x < 3
x
Practice: Solve the systems of inequalities graphically: y
3x  4y > 4 and y > 2x + 4
x
y
x + y > 4 and x  y < 2
x
168
TEST  Graphing Linear Inequalities in Two Variables
Solve the system of inequalities graphically:
y
2x + 3y < 6 and 2x  y > 6
x
169
Workbook
Part 9  Supplementary Topics
Calculating Square Roots
Calculate the square roots:
________ _______ _______
\/ 1 2 1 \/ 2 2 5 \/ 7 8 4
________ _________ ___________
\/ 1 3 6 9 \/ 9 2 1 6 \/ 1 3 4 5 6
_____________ _____________
\/ 1 1 9 0 2 5 \/ 2 5 7 0 4 9
170
Practice:
_______ _______ _______
\/ 1 6 9 \/ 3 6 1 \/ 4 4 1
________ _________ ___________
\/ 3 3 6 4 \/ 7 5 6 9 \/ 5 0 6 2 5
______________ _____________
\/ 1 2 0 4 0 9 \/ 1 6 4 8 3 6
171
Rule: When asked to round off a square root to a certain decimal place, always carry
out the operation one place beyond the place you are asked to round the answer to.
Calculate the square roots. Round off the answers to the nearest tenth:
______________ _____________
\/ 2. 0 0 0 0 \/ 7 7. 0 0 0 0
Practice: (Calculate the square roots. Round off the answers to the nearest tenth)
______________ _____________
\/ 3. 0 0 0 0 \/ 9 1. 0 0 0 0
172
TEST  Calculating Square Roots
Calculate the square root:
___________
\/ 7 5 6 2 5
Calculate the square root. Round off the answer to the nearest tenth:
_____________
\/ 4 3. 0 0 0 0
173
Changing Repeating Decimals to Fractions
Remember: All rational numbers can also be expressed as either terminating
or repeating decimals
Examples: 1 = .5 1 = .125 1 = .3 1 = .16 1 = .09 2 1 = 2.1 2 8 3 6 11 9
Rule: To change a repeating decimal into an equivalent fraction:
1) Set up an equation of the form: Let n = "the repeating decimal".
2) Multiply both sides of the equation by a power of 10 whose
exponent equals the number of digits in the repeating block.
3) Subtract the first equation from the second.
4) Solve the resulting equation to obtain the equivalent fraction.
Change these repeating decimals into equivalent fractions:
.3 .16 .09
.123 2.1
174
Practice:
.6 .83 .18
.321 4.2
TEST  Changing Repeating Decimals to Fractions
Change these repeating decimals into equivalent fractions:
.416 3.27
175
Scientific Notation
In "scientific notation", a number is expressed as the product of two numbers,
one of them being a number between 1 and 10, and the other being an integral
power of 10.
Scientific notation is used for two reasons:
1) To provide a more concise notation for very large or very small numbers.
2) To simplify computations by applying the laws of exponents to numbers
expressed in scientific notation.
Examples: 700 = 7 X 102 .75 = 7.5 X 101
750 = 7.5 X 102 .075 = 7.5 X 102
7,500 = 7.5 X 103 .0075 = 7.5 X 103
75,000 = 7.5 X 104 .00075 = 7.5 X 104
Change from decimal notation to scientific notation:
842 = 735,000 =
The approximate distance from our sun to Alpha Centauri, the nearest star
is 25,000,000,000,000 miles.
25,000,000,000,000 =
.01 = .000068 =
The mass of a molecule of water is approximately .00000000000000000000003
gram.
.00000000000000000000003 =
Practice:
7,634 = 9,380,000 =
A light year is the distance light travels in one year, approximately
5,900,000,000,000 miles.
5,900,000,000,000 =
.009 = .0000000056 =
The radius of the orbit of an electron of a hydrogen atom is approximately
.0000000000053 meter.
.0000000000053 =
176
Change from scientific notation to decimal notation:
8 X 103 = 7.4 X 106 =
5 X 104 = 5.12 X 105 =
Practice:
7 X 104 = 2.8 X 108 =
3 X 103 = 6.52 X 107 =
Use scientific notation and the laws of exponents to perform these computations:
(327,000)(200) =
(82,000)(3,000) =
252,000 =
.00126
Practice:
(2,570,000)(3,000) =
(38,000)(400) =
.0096 =
800
TEST  Scientific Notation
Use scientific notation and the laws of exponents to perform these computations:
(432,000,000)(.000007) =
.000072 =
300
177
The Distance Formula
Remember: The Pythagorean Theorm: In any right triangle the square of the
hypotenuse is equal to sum of the squares of other two sides.
y c2 = a2 + b2
It is possible to determine the distance
between any two points that lie in the
coordinate plane.
Example:
x Plot these two points: A (2,2) B (6, 4)
Now plot a third point: C (2,4)
Connecting these three points forms
a right triangle. Label the sides as follows:
Side a is opposite vertex A.
Side b is opposite vertex B.
Side c is opposite vertex C.
It is easy to determine the lengths of the two legs as one is parrallel to the xaxis and the other
to the yaxis.
The length of the leg parallel to the xaxis is found by subtracting one of the xcoordinates from
the other. BC = [2  6] = 8. Taking the absolute value, the number not considering the sign,
the length of the leg is 8 units. You would get the same result if you subtracted the
xcoordinates in the opposite order. [6 (2)] = 8 (This is the side we labeled a ).
The length of the leg parallel to the yaxis is found by subtracting one of the ycoordinates from
the other. AC = [4  2] = 6. Taking the absolute value, the number not considering the sign,
the length of the leg is 6 units. You would get the same result if you subtracted the
ycoordinates in the opposite order. [2  (4)] = 6 (This is the side we labeled b ).
Now, using the Pythagorean Theorem, you can determine the length of the hypotenuse, which is
also the distance between points A and B.
c2 = a2 + b2
= 82 + 62
= 64 + 36
= 100
c = \/ 100
= 10
This process leads to a general formula for the distance between any two points in the
coordinate plane:
d = \/ (x2  x1)2 + (y2  y1)2
178
(2, 1) and (2, 2) (4, 2) and (6, 1) (3,0) and (0, 3)
d = \/ (x2  x1)2 + (y2  y1)2 d = \/ (x2  x1)2 + (y2  y1)2 d = \/ (x2  x1)2 + (y2  y1)2
Practice:
(6, 4) and (6, 1) (3, 2) and (5, 1) (2,3) and (2, 5)
d = \/ (x2  x1)2 + (y2  y1)2 d = \/ (x2  x1)2 + (y2  y1)2 d = \/ (x2  x1)2 + (y2  y1)2
TEST  The Distance Formula:
(1, 2) and (5, 5) (2, 3) and (3, 4) (4,6) and (2, 4)
d = \/ (x2  x1)2 + (y2  y1)2 d = \/ (x2  x1)2 + (y2  y1)2 d = \/ (x2  x1)2 + (y2  y1)2
179
Using the distance formula, find the distance between the given points: