The Free High School Science Texts: Textbooks for High School Students Mathematics

The Free High School Science Texts: Textbooks for High School Students Mathematics
FHSST Authors
The Free High School Science Texts:
Textbooks for High School Students
Studying the Sciences
Mathematics
Grades 10 - 12
Version 0
September 17, 2008
ii
iii
Copyright 2007 “Free High School Science Texts”
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FHSST Core Team
Mark Horner ; Samuel Halliday ; Sarah Blyth ; Rory Adams ; Spencer Wheaton
FHSST Editors
Jaynie Padayachee ; Joanne Boulle ; Diana Mulcahy ; Annette Nell ; René Toerien ; Donovan
Whitfield
FHSST Contributors
Rory Adams ; Prashant Arora ; Richard Baxter ; Dr. Sarah Blyth ; Sebastian Bodenstein ;
Graeme Broster ; Richard Case ; Brett Cocks ; Tim Crombie ; Dr. Anne Dabrowski ; Laura
Daniels ; Sean Dobbs ; Fernando Durrell ; Dr. Dan Dwyer ; Frans van Eeden ; Giovanni
Franzoni ; Ingrid von Glehn ; Tamara von Glehn ; Lindsay Glesener ; Dr. Vanessa Godfrey ; Dr.
Johan Gonzalez ; Hemant Gopal ; Umeshree Govender ; Heather Gray ; Lynn Greeff ; Dr. Tom
Gutierrez ; Brooke Haag ; Kate Hadley ; Dr. Sam Halliday ; Asheena Hanuman ; Neil Hart ;
Nicholas Hatcher ; Dr. Mark Horner ; Mfandaidza Hove ; Robert Hovden ; Jennifer Hsieh ;
Clare Johnson ; Luke Jordan ; Tana Joseph ; Dr. Jennifer Klay ; Lara Kruger ; Sihle Kubheka ;
Andrew Kubik ; Dr. Marco van Leeuwen ; Dr. Anton Machacek ; Dr. Komal Maheshwari ;
Kosma von Maltitz ; Nicole Masureik ; John Mathew ; JoEllen McBride ; Nikolai Meures ;
Riana Meyer ; Jenny Miller ; Abdul Mirza ; Asogan Moodaly ; Jothi Moodley ; Nolene Naidu ;
Tyrone Negus ; Thomas O’Donnell ; Dr. Markus Oldenburg ; Dr. Jaynie Padayachee ;
Nicolette Pekeur ; Sirika Pillay ; Jacques Plaut ; Andrea Prinsloo ; Joseph Raimondo ; Sanya
Rajani ; Prof. Sergey Rakityansky ; Alastair Ramlakan ; Razvan Remsing ; Max Richter ; Sean
Riddle ; Evan Robinson ; Dr. Andrew Rose ; Bianca Ruddy ; Katie Russell ; Duncan Scott ;
Helen Seals ; Ian Sherratt ; Roger Sieloff ; Bradley Smith ; Greg Solomon ; Mike Stringer ;
Shen Tian ; Robert Torregrosa ; Jimmy Tseng ; Helen Waugh ; Dr. Dawn Webber ; Michelle
Wen ; Dr. Alexander Wetzler ; Dr. Spencer Wheaton ; Vivian White ; Dr. Gerald Wigger ;
Harry Wiggins ; Wendy Williams ; Julie Wilson ; Andrew Wood ; Emma Wormauld ; Sahal
Yacoob ; Jean Youssef
Contributors and editors have made a sincere effort to produce an accurate and useful resource.
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v
vi
Contents
I
Basics
1
1 Introduction to Book
1.1
II
3
The Language of Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . .
Grade 10
3
5
2 Review of Past Work
7
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.2
What is a number? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.3
Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.4
Letters and Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.5
Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
2.6
Multiplication and Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
2.7
Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
2.8
Negative Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.9
2.8.1
What is a negative number? . . . . . . . . . . . . . . . . . . . . . . . . 10
2.8.2
Working with Negative Numbers . . . . . . . . . . . . . . . . . . . . . . 11
2.8.3
Living Without the Number Line . . . . . . . . . . . . . . . . . . . . . . 12
Rearranging Equations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.10 Fractions and Decimal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.11 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.12 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.12.1 Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.12.2 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.12.3 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.12.4 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.13 Mathematical Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.14 Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.15 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3 Rational Numbers - Grade 10
23
3.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.2
The Big Picture of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.3
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
vii
CONTENTS
CONTENTS
3.4
Forms of Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.5
Converting Terminating Decimals into Rational Numbers . . . . . . . . . . . . . 25
3.6
Converting Repeating Decimals into Rational Numbers . . . . . . . . . . . . . . 25
3.7
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.8
End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4 Exponentials - Grade 10
29
4.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.3
Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.3.1
Exponential Law 1: a0 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.3.2
Exponential Law 2: am × an = am+n . . . . . . . . . . . . . . . . . . . 30
4.3.3
Exponential Law 3: a−n =
4.3.4
4.4
m
1
an , a
n
6= 0 . . . . . . . . . . . . . . . . . . . . 31
Exponential Law 4: a ÷ a = am−n . . . . . . . . . . . . . . . . . . . 32
4.3.5
Exponential Law 5: (ab)n = an bn . . . . . . . . . . . . . . . . . . . . . 32
4.3.6
Exponential Law 6: (am )n = amn . . . . . . . . . . . . . . . . . . . . . 33
End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5 Estimating Surds - Grade 10
37
5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5.2
Drawing Surds on the Number Line (Optional) . . . . . . . . . . . . . . . . . . 38
5.3
End of Chapter Excercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
6 Irrational Numbers and Rounding Off - Grade 10
41
6.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6.2
Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6.3
Rounding Off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
6.4
End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
7 Number Patterns - Grade 10
7.1
45
Common Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
7.1.1
Special Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
7.2
Make your own Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . 46
7.3
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
7.3.1
7.4
Patterns and Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
8 Finance - Grade 10
53
8.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
8.2
Foreign Exchange Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
8.3
8.2.1
How much is R1 really worth? . . . . . . . . . . . . . . . . . . . . . . . 53
8.2.2
Cross Currency Exchange Rates
8.2.3
Enrichment: Fluctuating exchange rates . . . . . . . . . . . . . . . . . . 57
. . . . . . . . . . . . . . . . . . . . . . 56
Being Interested in Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
viii
CONTENTS
8.4
Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
8.4.1
8.5
8.6
8.7
CONTENTS
Other Applications of the Simple Interest Formula . . . . . . . . . . . . . 61
Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
8.5.1
Fractions add up to the Whole . . . . . . . . . . . . . . . . . . . . . . . 65
8.5.2
The Power of Compound Interest . . . . . . . . . . . . . . . . . . . . . . 65
8.5.3
Other Applications of Compound Growth . . . . . . . . . . . . . . . . . 67
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
8.6.1
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
8.6.2
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
9 Products and Factors - Grade 10
71
9.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
9.2
Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
9.2.1
Parts of an Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
9.2.2
Product of Two Binomials . . . . . . . . . . . . . . . . . . . . . . . . . 71
9.2.3
Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
9.3
More Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
9.4
Factorising a Quadratic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
9.5
Factorisation by Grouping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
9.6
Simplification of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
9.7
End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
10 Equations and Inequalities - Grade 10
83
10.1 Strategy for Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
10.2 Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
10.3 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
10.4 Exponential Equations of the form ka(x+p) = m . . . . . . . . . . . . . . . . . . 93
10.4.1 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
10.5 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
10.6 Linear Simultaneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
10.6.1 Finding solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
10.6.2 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
10.6.3 Solution by Substitution
. . . . . . . . . . . . . . . . . . . . . . . . . . 101
10.7 Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
10.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
10.7.2 Problem Solving Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . 104
10.7.3 Application of Mathematical Modelling
. . . . . . . . . . . . . . . . . . 104
10.7.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 106
10.8 Introduction to Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . 107
10.9 Functions and Graphs in the Real-World . . . . . . . . . . . . . . . . . . . . . . 107
10.10Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
ix
CONTENTS
CONTENTS
10.10.1 Variables and Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
10.10.2 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
10.10.3 The Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
10.10.4 Drawing Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
10.10.5 Notation used for Functions
. . . . . . . . . . . . . . . . . . . . . . . . 110
10.11Characteristics of Functions - All Grades . . . . . . . . . . . . . . . . . . . . . . 112
10.11.1 Dependent and Independent Variables . . . . . . . . . . . . . . . . . . . 112
10.11.2 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
10.11.3 Intercepts with the Axes . . . . . . . . . . . . . . . . . . . . . . . . . . 113
10.11.4 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
10.11.5 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
10.11.6 Lines of Symmetry
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
10.11.7 Intervals on which the Function Increases/Decreases . . . . . . . . . . . 114
10.11.8 Discrete or Continuous Nature of the Graph . . . . . . . . . . . . . . . . 114
10.12Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
10.12.1 Functions of the form y = ax + q . . . . . . . . . . . . . . . . . . . . . 116
10.12.2 Functions of the Form y = ax2 + q . . . . . . . . . . . . . . . . . . . . . 120
10.12.3 Functions of the Form y =
a
x
+ q . . . . . . . . . . . . . . . . . . . . . . 125
10.12.4 Functions of the Form y = ab(x) + q . . . . . . . . . . . . . . . . . . . . 129
10.13End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
11 Average Gradient - Grade 10 Extension
135
11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
11.2 Straight-Line Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
11.3 Parabolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
11.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
12 Geometry Basics
139
12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
12.2 Points and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
12.3 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
12.3.1 Measuring angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
12.3.2 Special Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
12.3.3 Special Angle Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
12.3.4 Parallel Lines intersected by Transversal Lines . . . . . . . . . . . . . . . 143
12.4 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
12.4.1 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
12.4.2 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
12.4.3 Other polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
12.4.4 Extra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
12.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
12.5.1 Challenge Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
x
CONTENTS
13 Geometry - Grade 10
CONTENTS
161
13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
13.2 Right Prisms and Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
13.2.1 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
13.2.2 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
13.3 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
13.3.1 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
13.4 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
13.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
13.4.2 Distance between Two Points . . . . . . . . . . . . . . . . . . . . . . . . 172
13.4.3 Calculation of the Gradient of a Line . . . . . . . . . . . . . . . . . . . . 173
13.4.4 Midpoint of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
13.5 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
13.5.1 Translation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
13.5.2 Reflection of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
13.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
14 Trigonometry - Grade 10
189
14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
14.2 Where Trigonometry is Used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
14.3 Similarity of Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
14.4 Definition of the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 191
14.5 Simple Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . 195
14.5.1 Height and Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
14.5.2 Maps and Plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
14.6 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 199
14.6.1 Graph of sin θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
14.6.2 Functions of the form y = a sin(x) + q . . . . . . . . . . . . . . . . . . . 200
14.6.3 Graph of cos θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
14.6.4 Functions of the form y = a cos(x) + q
. . . . . . . . . . . . . . . . . . 202
14.6.5 Comparison of Graphs of sin θ and cos θ . . . . . . . . . . . . . . . . . . 204
14.6.6 Graph of tan θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
14.6.7 Functions of the form y = a tan(x) + q . . . . . . . . . . . . . . . . . . 205
14.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
15 Statistics - Grade 10
211
15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
15.2 Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
15.2.1 Data and Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . 211
15.2.2 Methods of Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . 212
15.2.3 Samples and Populations . . . . . . . . . . . . . . . . . . . . . . . . . . 213
15.3 Example Data Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
xi
CONTENTS
CONTENTS
15.3.1 Data Set 1: Tossing a Coin . . . . . . . . . . . . . . . . . . . . . . . . . 213
15.3.2 Data Set 2: Casting a die . . . . . . . . . . . . . . . . . . . . . . . . . . 213
15.3.3 Data Set 3: Mass of a Loaf of Bread . . . . . . . . . . . . . . . . . . . . 214
15.3.4 Data Set 4: Global Temperature . . . . . . . . . . . . . . . . . . . . . . 214
15.3.5 Data Set 5: Price of Petrol . . . . . . . . . . . . . . . . . . . . . . . . . 215
15.4 Grouping Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
15.4.1 Exercises - Grouping Data
. . . . . . . . . . . . . . . . . . . . . . . . . 216
15.5 Graphical Representation of Data . . . . . . . . . . . . . . . . . . . . . . . . . . 217
15.5.1 Bar and Compound Bar Graphs . . . . . . . . . . . . . . . . . . . . . . . 217
15.5.2 Histograms and Frequency Polygons . . . . . . . . . . . . . . . . . . . . 217
15.5.3 Pie Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
15.5.4 Line and Broken Line Graphs . . . . . . . . . . . . . . . . . . . . . . . . 220
15.5.5 Exercises - Graphical Representation of Data
. . . . . . . . . . . . . . . 221
15.6 Summarising Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
15.6.1 Measures of Central Tendency . . . . . . . . . . . . . . . . . . . . . . . 222
15.6.2 Measures of Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
15.6.3 Exercises - Summarising Data
. . . . . . . . . . . . . . . . . . . . . . . 228
15.7 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
15.7.1 Exercises - Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . 230
15.8 Summary of Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
15.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
16 Probability - Grade 10
235
16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
16.2 Random Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
16.2.1 Sample Space of a Random Experiment . . . . . . . . . . . . . . . . . . 235
16.3 Probability Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
16.3.1 Classical Theory of Probability . . . . . . . . . . . . . . . . . . . . . . . 239
16.4 Relative Frequency vs. Probability . . . . . . . . . . . . . . . . . . . . . . . . . 240
16.5 Project Idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
16.6 Probability Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
16.7 Mutually Exclusive Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
16.8 Complementary Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
16.9 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
III
Grade 11
17 Exponents - Grade 11
249
251
17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
17.2 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
√
m
17.2.1 Exponential Law 7: a n = n am . . . . . . . . . . . . . . . . . . . . . . 251
17.3 Exponentials in the Real-World . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
17.4 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
xii
CONTENTS
CONTENTS
18 Surds - Grade 11
18.1 Surd Calculations . . . . . . . . . .
√
√ √
18.1.1 Surd Law 1: n a n b = n ab
√
p
n
a
18.1.2 Surd Law 2: n ab = √
. .
n
b
√
m
18.1.3 Surd Law 3: n am = a n . .
255
. . . . . . . . . . . . . . . . . . . . . . . . 255
. . . . . . . . . . . . . . . . . . . . . . . . 255
. . . . . . . . . . . . . . . . . . . . . . . . 255
. . . . . . . . . . . . . . . . . . . . . . . . 256
18.1.4 Like and Unlike Surds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
18.1.5 Simplest Surd form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
18.1.6 Rationalising Denominators . . . . . . . . . . . . . . . . . . . . . . . . . 258
18.2 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259
19 Error Margins - Grade 11
261
20 Quadratic Sequences - Grade 11
265
20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
20.2 What is a quadratic sequence? . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
20.3 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
21 Finance - Grade 11
271
21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
21.2 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
21.3 Simple Depreciation (it really is simple!) . . . . . . . . . . . . . . . . . . . . . . 271
21.4 Compound Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
21.5 Present Values or Future Values of an Investment or Loan . . . . . . . . . . . . 276
21.5.1 Now or Later . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
21.6 Finding i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
21.7 Finding n - Trial and Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
21.8 Nominal and Effective Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . 280
21.8.1 The General Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
21.8.2 De-coding the Terminology . . . . . . . . . . . . . . . . . . . . . . . . . 282
21.9 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
21.9.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
21.9.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
21.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
22 Solving Quadratic Equations - Grade 11
287
22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
22.2 Solution by Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
22.3 Solution by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . 290
22.4 Solution by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . 293
22.5 Finding an equation when you know its roots . . . . . . . . . . . . . . . . . . . 296
22.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
xiii
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CONTENTS
23 Solving Quadratic Inequalities - Grade 11
301
23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
23.2 Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
23.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
24 Solving Simultaneous Equations - Grade 11
307
24.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
24.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
25 Mathematical Models - Grade 11
313
25.1 Real-World Applications: Mathematical Models . . . . . . . . . . . . . . . . . . 313
25.2 End of Chatpter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
26 Quadratic Functions and Graphs - Grade 11
321
26.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
26.2 Functions of the Form y = a(x + p)2 + q
. . . . . . . . . . . . . . . . . . . . . 321
26.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
26.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323
26.2.3 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
26.2.4 Axes of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
26.2.5 Sketching Graphs of the Form f (x) = a(x + p)2 + q . . . . . . . . . . . 325
26.2.6 Writing an equation of a shifted parabola . . . . . . . . . . . . . . . . . 327
26.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
27 Hyperbolic Functions and Graphs - Grade 11
329
27.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
27.2 Functions of the Form y =
a
x+p
+q
. . . . . . . . . . . . . . . . . . . . . . . . 329
27.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
27.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331
27.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332
27.2.4 Sketching Graphs of the Form f (x) =
a
x+p
+ q . . . . . . . . . . . . . . 333
27.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
28 Exponential Functions and Graphs - Grade 11
335
28.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
28.2 Functions of the Form y = ab(x+p) + q . . . . . . . . . . . . . . . . . . . . . . . 335
28.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336
28.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
28.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
28.2.4 Sketching Graphs of the Form f (x) = ab(x+p) + q . . . . . . . . . . . . . 338
28.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
29 Gradient at a Point - Grade 11
341
29.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
29.2 Average Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
29.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
xiv
CONTENTS
30 Linear Programming - Grade 11
CONTENTS
345
30.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
30.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
30.2.1 Decision Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
30.2.2 Objective Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
30.2.3 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
30.2.4 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 346
30.2.5 The Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
30.3 Example of a Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
30.4 Method of Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
30.5 Skills you will need . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
30.5.1 Writing Constraint Equations . . . . . . . . . . . . . . . . . . . . . . . . 347
30.5.2 Writing the Objective Function . . . . . . . . . . . . . . . . . . . . . . . 348
30.5.3 Solving the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350
30.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352
31 Geometry - Grade 11
357
31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
31.2 Right Pyramids, Right Cones and Spheres . . . . . . . . . . . . . . . . . . . . . 357
31.3 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
31.4 Triangle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
31.4.1 Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
31.5 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368
31.5.1 Equation of a Line between Two Points . . . . . . . . . . . . . . . . . . 368
31.5.2 Equation of a Line through One Point and Parallel or Perpendicular to
Another Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
31.5.3 Inclination of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
31.6 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373
31.6.1 Rotation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373
31.6.2 Enlargement of a Polygon 1 . . . . . . . . . . . . . . . . . . . . . . . . . 376
32 Trigonometry - Grade 11
381
32.1 History of Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381
32.2 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 381
32.2.1 Functions of the form y = sin(kθ) . . . . . . . . . . . . . . . . . . . . . 381
32.2.2 Functions of the form y = cos(kθ) . . . . . . . . . . . . . . . . . . . . . 383
32.2.3 Functions of the form y = tan(kθ) . . . . . . . . . . . . . . . . . . . . . 384
32.2.4 Functions of the form y = sin(θ + p) . . . . . . . . . . . . . . . . . . . . 385
32.2.5 Functions of the form y = cos(θ + p) . . . . . . . . . . . . . . . . . . . 386
32.2.6 Functions of the form y = tan(θ + p) . . . . . . . . . . . . . . . . . . . 387
32.3 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
32.3.1 Deriving Values of Trigonometric Functions for 30◦ , 45◦ and 60◦ . . . . . 389
32.3.2 Alternate Definition for tan θ . . . . . . . . . . . . . . . . . . . . . . . . 391
xv
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CONTENTS
32.3.3 A Trigonometric Identity . . . . . . . . . . . . . . . . . . . . . . . . . . 392
32.3.4 Reduction Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
32.4 Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 399
32.4.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399
32.4.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
32.4.3 Solution using CAST diagrams . . . . . . . . . . . . . . . . . . . . . . . 403
32.4.4 General Solution Using Periodicity . . . . . . . . . . . . . . . . . . . . . 405
32.4.5 Linear Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . 406
32.4.6 Quadratic and Higher Order Trigonometric Equations . . . . . . . . . . . 406
32.4.7 More Complex Trigonometric Equations . . . . . . . . . . . . . . . . . . 407
32.5 Sine and Cosine Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409
32.5.1 The Sine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409
32.5.2 The Cosine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412
32.5.3 The Area Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414
32.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416
33 Statistics - Grade 11
419
33.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
33.2 Standard Deviation and Variance . . . . . . . . . . . . . . . . . . . . . . . . . . 419
33.2.1 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
33.2.2 Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421
33.2.3 Interpretation and Application . . . . . . . . . . . . . . . . . . . . . . . 423
33.2.4 Relationship between Standard Deviation and the Mean . . . . . . . . . . 424
33.3 Graphical Representation of Measures of Central Tendency and Dispersion . . . . 424
33.3.1 Five Number Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 424
33.3.2 Box and Whisker Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 425
33.3.3 Cumulative Histograms . . . . . . . . . . . . . . . . . . . . . . . . . . . 426
33.4 Distribution of Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428
33.4.1 Symmetric and Skewed Data . . . . . . . . . . . . . . . . . . . . . . . . 428
33.4.2 Relationship of the Mean, Median, and Mode . . . . . . . . . . . . . . . 428
33.5 Scatter Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429
33.6 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432
33.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435
34 Independent and Dependent Events - Grade 11
437
34.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
34.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
34.2.1 Identification of Independent and Dependent Events
. . . . . . . . . . . 438
34.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441
IV
Grade 12
35 Logarithms - Grade 12
443
445
35.1 Definition of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445
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35.2 Logarithm Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446
35.3 Laws of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447
35.4 Logarithm Law 1: loga 1 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447
35.5 Logarithm Law 2: loga (a) = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 448
35.6 Logarithm Law 3: loga (x · y) = loga (x) + loga (y) . . . . . . . . . . . . . . . . . 448
35.7 Logarithm Law 4: loga xy = loga (x) − loga (y) . . . . . . . . . . . . . . . . . 449
35.8 Logarithm Law 5: loga (xb ) = b loga (x) . . . . . . . . . . . . . . . . . . . . . . . 450
√
35.9 Logarithm Law 6: loga ( b x) = logab(x) . . . . . . . . . . . . . . . . . . . . . . . 450
35.10Solving simple log equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452
35.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454
35.11Logarithmic applications in the Real World . . . . . . . . . . . . . . . . . . . . . 454
35.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455
35.12End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455
36 Sequences and Series - Grade 12
457
36.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
36.2 Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
36.2.1 General Equation for the nth -term of an Arithmetic Sequence . . . . . . 458
36.3 Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
36.3.1 Example - A Flu Epidemic . . . . . . . . . . . . . . . . . . . . . . . . . 459
36.3.2 General Equation for the nth -term of a Geometric Sequence . . . . . . . 461
36.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461
36.4 Recursive Formulae for Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 462
36.5 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463
36.5.1 Some Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463
36.5.2 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463
36.6 Finite Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465
36.6.1 General Formula for a Finite Arithmetic Series . . . . . . . . . . . . . . . 466
36.6.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
36.7 Finite Squared Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468
36.8 Finite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
36.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470
36.9 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471
36.9.1 Infinite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . 471
36.9.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
36.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
37 Finance - Grade 12
477
37.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477
37.2 Finding the Length of the Investment or Loan . . . . . . . . . . . . . . . . . . . 477
37.3 A Series of Payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478
37.3.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479
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37.3.2 Present Values of a series of Payments . . . . . . . . . . . . . . . . . . . 479
37.3.3 Future Value of a series of Payments . . . . . . . . . . . . . . . . . . . . 484
37.3.4 Exercises - Present and Future Values . . . . . . . . . . . . . . . . . . . 485
37.4 Investments and Loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485
37.4.1 Loan Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485
37.4.2 Exercises - Investments and Loans . . . . . . . . . . . . . . . . . . . . . 489
37.4.3 Calculating Capital Outstanding . . . . . . . . . . . . . . . . . . . . . . 489
37.5 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
37.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490
37.5.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490
37.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490
38 Factorising Cubic Polynomials - Grade 12
493
38.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
38.2 The Factor Theorem
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
38.3 Factorisation of Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 494
38.4 Exercises - Using Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 496
38.5 Solving Cubic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496
38.5.1 Exercises - Solving of Cubic Equations . . . . . . . . . . . . . . . . . . . 498
38.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498
39 Functions and Graphs - Grade 12
501
39.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
39.2 Definition of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
39.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
39.3 Notation used for Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
39.4 Graphs of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
39.4.1 Inverse Function of y = ax + q . . . . . . . . . . . . . . . . . . . . . . . 503
39.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504
39.4.3 Inverse Function of y = ax2
. . . . . . . . . . . . . . . . . . . . . . . . 504
39.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504
39.4.5 Inverse Function of y = ax . . . . . . . . . . . . . . . . . . . . . . . . . 506
39.4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506
39.5 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507
40 Differential Calculus - Grade 12
509
40.1 Why do I have to learn this stuff? . . . . . . . . . . . . . . . . . . . . . . . . . 509
40.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510
40.2.1 A Tale of Achilles and the Tortoise . . . . . . . . . . . . . . . . . . . . . 510
40.2.2 Sequences, Series and Functions . . . . . . . . . . . . . . . . . . . . . . 511
40.2.3 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512
40.2.4 Average Gradient and Gradient at a Point . . . . . . . . . . . . . . . . . 516
40.3 Differentiation from First Principles . . . . . . . . . . . . . . . . . . . . . . . . . 519
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40.4 Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521
40.4.1 Summary of Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . 522
40.5 Applying Differentiation to Draw Graphs . . . . . . . . . . . . . . . . . . . . . . 523
40.5.1 Finding Equations of Tangents to Curves
. . . . . . . . . . . . . . . . . 523
40.5.2 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524
40.5.3 Local minimum, Local maximum and Point of Inflextion . . . . . . . . . 529
40.6 Using Differential Calculus to Solve Problems . . . . . . . . . . . . . . . . . . . 530
40.6.1 Rate of Change problems . . . . . . . . . . . . . . . . . . . . . . . . . . 534
40.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535
41 Linear Programming - Grade 12
539
41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539
41.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539
41.2.1 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 539
41.3 Linear Programming and the Feasible Region . . . . . . . . . . . . . . . . . . . 540
41.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546
42 Geometry - Grade 12
549
42.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
42.2 Circle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
42.2.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
42.2.2 Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550
42.2.3 Theorems of the Geometry of Circles . . . . . . . . . . . . . . . . . . . . 550
42.3 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566
42.3.1 Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566
42.3.2 Equation of a Tangent to a Circle at a Point on the Circle . . . . . . . . 569
42.4 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571
42.4.1 Rotation of a Point about an angle θ . . . . . . . . . . . . . . . . . . . . 571
42.4.2 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573
42.4.3 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573
42.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574
43 Trigonometry - Grade 12
577
43.1 Compound Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577
43.1.1 Derivation of sin(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 577
43.1.2 Derivation of sin(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578
43.1.3 Derivation of cos(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578
43.1.4 Derivation of cos(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 579
43.1.5 Derivation of sin 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579
43.1.6 Derivation of cos 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579
43.1.7 Problem-solving Strategy for Identities . . . . . . . . . . . . . . . . . . . 580
43.2 Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . 582
43.2.1 Problems in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 582
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43.2.2 Problems in 3 dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 584
43.3 Other Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
43.3.1 Taxicab Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
43.3.2 Manhattan distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
43.3.3 Spherical Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587
43.3.4 Fractal Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588
43.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589
44 Statistics - Grade 12
591
44.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591
44.2 A Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591
44.3 Extracting a Sample Population . . . . . . . . . . . . . . . . . . . . . . . . . . . 593
44.4 Function Fitting and Regression Analysis . . . . . . . . . . . . . . . . . . . . . . 594
44.4.1 The Method of Least Squares
. . . . . . . . . . . . . . . . . . . . . . . 596
44.4.2 Using a calculator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597
44.4.3 Correlation coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 599
44.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600
45 Combinations and Permutations - Grade 12
603
45.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
45.2 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
45.2.1 Making a List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
45.2.2 Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
45.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
45.3.1 The Factorial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
45.4 The Fundamental Counting Principle . . . . . . . . . . . . . . . . . . . . . . . . 604
45.5 Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605
45.5.1 Counting Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . 605
45.5.2 Combinatorics and Probability . . . . . . . . . . . . . . . . . . . . . . . 606
45.6 Permutations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606
45.6.1 Counting Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 607
45.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608
45.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610
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43.2.2 Problems in 3 dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 584
43.3 Other Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
43.3.1 Taxicab Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
43.3.2 Manhattan distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
43.3.3 Spherical Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587
43.3.4 Fractal Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588
43.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589
44 Statistics - Grade 12
591
44.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591
44.2 A Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591
44.3 Extracting a Sample Population . . . . . . . . . . . . . . . . . . . . . . . . . . . 593
44.4 Function Fitting and Regression Analysis . . . . . . . . . . . . . . . . . . . . . . 594
44.4.1 The Method of Least Squares
. . . . . . . . . . . . . . . . . . . . . . . 596
44.4.2 Using a calculator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597
44.4.3 Correlation coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 599
44.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600
45 Combinations and Permutations - Grade 12
603
45.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
45.2 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
45.2.1 Making a List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
45.2.2 Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
45.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
45.3.1 The Factorial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
45.4 The Fundamental Counting Principle . . . . . . . . . . . . . . . . . . . . . . . . 604
45.5 Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605
45.5.1 Counting Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . 605
45.5.2 Combinatorics and Probability . . . . . . . . . . . . . . . . . . . . . . . 606
45.6 Permutations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606
45.6.1 Counting Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 607
45.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608
45.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610
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Part I
Basics
1
Chapter 1
Introduction to Book
1.1
The Language of Mathematics
The purpose of any language, like English or Zulu, is to make it possible for people to communicate. All languages have an alphabet, which is a group of letters that are used to make up
words. There are also rules of grammar which explain how words are supposed to be used to
build up sentences. This is needed because when a sentence is written, the person reading the
sentence understands exactly what the writer is trying to explain. Punctuation marks (like a full
stop or a comma) are used to further clarify what is written.
Mathematics is a language, specifically it is the language of Science. Like any language, mathematics has letters (known as numbers) that are used to make up words (known as expressions),
and sentences (known as equations). The punctuation marks of mathematics are the different signs and symbols that are used, for example, the plus sign (+), the minus sign (-), the
multiplication sign (×), the equals sign (=) and so on. There are also rules that explain how
the numbers should be used together with the signs to make up equations that express some
meaning.
3
1.1
CHAPTER 1. INTRODUCTION TO BOOK
4
Part II
Grade 10
5
Chapter 2
Review of Past Work
2.1
Introduction
This chapter describes some basic concepts which you have seen in earlier grades, and lays the
foundation for the remainder of this book. You should feel confident with the content in this
chapter, before moving on with the rest of the book.
So try out your skills on the exercises throughout this chapter and ask your teacher for more
questions just like them. You can also try making up your own questions, solve them and try
them out on your classmates to see if you get the same answers.
Practice is the only way to get good at maths!
2.2
What is a number?
A number is a way to represent quantity. Numbers are not something that you can touch or
hold, because they are not physical. But you can touch three apples, three pencils, three books.
You can never just touch three, you can only touch three of something. However, you do not
need to see three apples in front of you to know that if you take one apple away, that there will
be two apples left. You can just think about it. That is your brain representing the apples in
numbers and then performing arithmetic on them.
A number represents quantity because we can look at the world around us and quantify it using
numbers. How many minutes? How many kilometers? How many apples? How much money?
How much medicine? These are all questions which can only be answered using numbers to tell
us “how much” of something we want to measure.
A number can be written many different ways and it is always best to choose the most appropriate
way of writing the number. For example, “a half” may be spoken aloud or written in words,
but that makes mathematics very difficult and also means that only people who speak the same
language as you can understand what you mean. A better way of writing “a half” is as a fraction
1
2 or as a decimal number 0,5. It is still the same number, no matter which way you write it.
In high school, all the numbers which you will see are called real numbers and mathematicians
use the symbol R to stand for the set of all real numbers, which simply means all of the real
numbers. Some of these real numbers can be written in a particular way and some cannot.
Different types of numbers are described in detail in Section 1.12.
2.3
Sets
A set is a group of objects with a well-defined criterion for membership. For example, the
criterion for belonging to a set of apples, is that it must be an apple. The set of apples can
then be divided into red apples and green apples, but they are all still apples. All the red apples
form another set which is a sub-set of the set of apples. A sub-set is part of a set. All the green
apples form another sub-set.
7
2.4
CHAPTER 2. REVIEW OF PAST WORK
Now we come to the idea of a union, which is used to combine things. The symbol for union
is ∪. Here we use it to combine two or more intervals. For example, if x is a real number such
that 1 < x ≤ 3 or 6 ≤ x < 10, then the set of all the possible x values is
(1,3] ∪ [6,10)
(2.1)
where the ∪ sign means the union (or combination) of the two intervals. We use the set and
interval notation and the symbols described because it is easier than having to write everything
out in words.
2.4
Letters and Arithmetic
The simplest things that can be done with numbers is to add, subtract, multiply or divide them.
When two numbers are added, subtracted, multiplied or divided, you are performing arithmetic 1 .
These four basic operations can be performed on any two real numbers.
Mathematics as a language uses special notation to write things down. So instead of:
one plus one is equal to two
mathematicians write
1+1=2
In earlier grades, place holders were used to indicate missing numbers in an equation.
1+=2
4−=2
+ 3 − 2 = 2
However, place holders only work well for simple equations. For more advanced mathematical
workings, letters are usually used to represent numbers.
1+x=2
4−y =2
z + 3 − 2z = 2
These letters are referred to as variables, since they can take on any value depending on what
is required. For example, x = 1 in Equation 2.2, but x = 26 in 2 + x = 28.
A constant has a fixed value. The number 1 is a constant. The speed of light in a vacuum
is also a constant which has been defined to be exactly 299 792 458 m·s−1 (read metres per
second). The speed of light is a big number and it takes up space to always write down the
entire number. Therefore, letters are also used to represent some constants. In the case of the
speed of light, it is accepted that the letter c represents the speed of light. Such constants
represented by letters occur most often in physics and chemistry.
Additionally, letters can be used to describe a situation, mathematically. For example, the
following equation
x+y =z
(2.2)
can be used to describe the situation of finding how much change can be expected for buying
an item. In this equation, y represents the price of the item you are buying, x represents the
amount of change you should get back and z is the amount of money given to the cashier. So,
if the price is R10 and you gave the cashier R15, then write R15 instead of z and R10 instead
of y and the change is then x.
x + 10 = 15
(2.3)
We will learn how to “solve” this equation towards the end of this chapter.
1 Arithmetic
is derived from the Greek word arithmos meaning number.
8
CHAPTER 2. REVIEW OF PAST WORK
2.5
2.5
Addition and Subtraction
Addition (+) and subtraction (-) are the most basic operations between numbers but they are
very closely related to each other. You can think of subtracting as being the opposite of adding
since adding a number and then subtracting the same number will not change what you started
with. For example, if we start with a and add b, then subtract b, we will just get back to a again
a+b−b=a
5+2−2=5
(2.4)
If we look at a number line, then addition means that we move to the right and subtraction
means that we move to the left.
The order in which numbers are added does not matter, but the order in which numbers are
subtracted does matter. This means that:
a+b
= b+a
a−b
6= b − a if a 6= b
(2.5)
The sign 6= means “is not equal to”. For example, 2 + 3 = 5 and 3 + 2 = 5, but 5 − 3 = 2 and
3 − 5 = −2. −2 is a negative number, which is explained in detail in Section 2.8.
Extension: Commutativity for Addition
The fact that a + b = b + a, is known as the commutative property for addition.
2.6
Multiplication and Division
Just like addition and subtraction, multiplication (×, ·) and division (÷, /) are opposites of each
other. Multiplying by a number and then dividing by the same number gets us back to the start
again:
a×b÷b=a
5×4÷4=5
(2.6)
Sometimes you will see a multiplication of letters as a dot or without any symbol. Don’t worry,
its exactly the same thing. Mathematicians are lazy and like to write things in the shortest,
neatest way possible.
abc =
a·b·c =
a×b×c
a×b×c
(2.7)
It is usually neater to write known numbers to the left, and letters to the right. So although 4x
and x4 are the same thing, it looks better to write 4x. In this case, the “4” is a constant that
is referred to as the coefficient of x.
Extension: Commutativity for Multiplication
The fact that ab = ba is known as the commutative property of multiplication.
Therefore, both addition and multiplication are described as commutative operations.
2.7
Brackets
Brackets2 in mathematics are used to show the order in which you must do things. This is
important as you can get different answers depending on the order in which you do things. For
2 Sometimes
people say “parenthesis” instead of “brackets”.
9
2.8
CHAPTER 2. REVIEW OF PAST WORK
example
(5 × 5) + 20 = 45
(2.8)
5 × (5 + 20) = 125
(2.9)
whereas
If there are no brackets, you should always do multiplications and divisions first and then additions
and subtractions3 . You can always put your own brackets into equations using this rule to make
things easier for yourself, for example:
a×b+c÷d =
5 × 5 + 20 ÷ 4 =
(a × b) + (c ÷ d)
(5 × 5) + (20 ÷ 4)
(2.10)
If you see a multiplication outside a bracket like this
a(b + c)
(2.11)
3(4 − 3)
then it means you have to multiply each part inside the bracket by the number outside
a(b + c)
= ab + ac
(2.12)
3(4 − 3) = 3 × 4 − 3 × 3 = 12 − 9 = 3
unless you can simplify everything inside the bracket into a single term. In fact, in the above
example, it would have been smarter to have done this
3(4 − 3) = 3 × (1) = 3
(2.13)
3(4a − 3a) = 3 × (a) = 3a
(2.14)
It can happen with letters too
Extension: Distributivity
The fact that a(b + c) = ab + ac is known as the distributive property.
If there are two brackets multiplied by each other, then you can do it one step at a time
(a + b)(c + d)
= a(c + d) + b(c + d)
= ac + ad + bc + bd
(a + 3)(4 + d)
= a(4 + d) + 3(4 + d)
(2.15)
= 4a + ad + 12 + 3d
2.8
2.8.1
Negative Numbers
What is a negative number?
Negative numbers can be very confusing to begin with, but there is nothing to be afraid of. The
numbers that are used most often are greater than zero. These numbers are known as positive
numbers.
A negative number is simply a number that is less than zero. So, if we were to take a positive
number a and subtract it from zero, the answer would be the negative of a.
0 − a = −a
3 Multiplying and dividing can be performed in any order as it doesn’t matter. Likewise it doesn’t matter which
order you do addition and subtraction. Just as long as you do any ×÷ before any +−.
10
CHAPTER 2. REVIEW OF PAST WORK
2.8
On a number line, a negative number appears to the left of zero and a positive number appears
to the right of zero.
negative numbers
-3
-2
positive numbers
-1
0
1
2
3
Figure 2.1: On the number line, numbers increase towards the right and decrease towards the
left. Positive numbers appear to the right of zero and negative numbers appear to the left of
zero.
2.8.2
Working with Negative Numbers
When you are adding a negative number, it is the same as subtracting that number if it were
positive. Likewise, if you subtract a negative number, it is the same as adding the number if it
were positive. Numbers are either positive or negative, and we call this their sign. A positive
number has positive sign (+), and a negative number has a negative sign (-).
Subtraction is actually the same as adding a negative number.
In this example, a and b are positive numbers, but −b is a negative number
a − b = a + (−b)
(2.16)
5 − 3 = 5 + (−3)
So, this means that subtraction is simply a short-cut for adding a negative number, and instead
of writing a + (−b), we write a − b. This also means that −b + a is the same as a − b. Now,
which do you find easier to work out?
Most people find that the first way is a bit more difficult to work out than the second way. For
example, most people find 12 − 3 a lot easier to work out than −3 + 12, even though they are
the same thing. So, a − b, which looks neater and requires less writing, is the accepted way of
writing subtractions.
Table 2.1 shows how to calculate the sign of the answer when you multiply two numbers together.
The first column shows the sign of the first number, the second column gives the sign of the
second number, and the third column shows what sign the answer will be. So multiplying or
a
+
+
-
b
+
+
-
a × b or a ÷ b
+
+
Table 2.1: Table of signs for multiplying or dividing two numbers.
dividing a negative number by a positive number always gives you a negative number, whereas
multiplying or dividing numbers which have the same sign always gives a positive number. For
example, 2 × 3 = 6 and −2 × −3 = 6, but −2 × 3 = −6 and 2 × −3 = −6.
Adding numbers works slightly differently, have a look at Table 2.2. The first column shows the
sign of the first number, the second column gives the sign of the second number, and the third
column shows what sign the answer will be.
a
+
+
-
b
+
+
-
a+b
+
?
?
-
Table 2.2: Table of signs for adding two numbers.
11
2.8
CHAPTER 2. REVIEW OF PAST WORK
If you add two positive numbers you will always get a positive number, but if you add two
negative numbers you will always get a negative number. If the numbers have different sign,
then the sign of the answer depends on which one is bigger.
2.8.3
Living Without the Number Line
The number line in Figure 2.1 is a good way to visualise what negative numbers are, but it can
get very inefficient to use it every time you want to add or subtract negative numbers. To keep
things simple, we will write down three tips that you can use to make working with negative
numbers a little bit easier. These tips will let you work out what the answer is when you add or
subtract numbers which may be negative and will also help you keep your work tidy and easier
to understand.
Negative Numbers Tip 1
If you are given an equation like −a + b, then it is easier to move the numbers around so that the
equation looks easier. For this case, we have seen that adding a negative number to a positive
number is the same as subtracting the number from the positive number. So,
−a + b
=
−5 + 10 =
b−a
(2.17)
10 − 5 = 5
This makes equations easier to understand. For example, a question like “What is −7 + 11?”
looks a lot more complicated than “What is 11 − 7?”, even though they are exactly the same
question.
Negative Numbers Tip 2
When you have two negative numbers like −3 − 7, you can calculate the answer by simply adding
together the numbers as if they were positive and then putting a negative sign in front.
−c − d =
−7 − 2 =
−(c + d)
−(7 + 2) = −9
(2.18)
Negative Numbers Tip 3
In Table 2.2 we saw that the sign of two numbers added together depends on which one is bigger.
This tip tells us that all we need to do is take the smaller number away from the larger one,
and remember to put a negative sign before the answer if the bigger number was subtracted to
begin with. In this equation, F is bigger than e.
e−F
=
2 − 11 =
−(F − e)
(2.19)
−(11 − 2) = −9
You can even combine these tips together, so for example you can use Tip 1 on −10 + 3 to get
3 − 10, and then use Tip 3 to get −(10 − 3) = −7.
Exercise: Negative Numbers
1. Calculate:
(a) (−5) − (−3)
(d) 11 − (−9)
(g) (−1) × 24 ÷ 8 × (−3)
(j) 3 − 64 + 1
(m) −9 + 8 − 7 + 6 − 5 + 4 − 3 + 2 − 1
12
(b) (−4) + 2
(e) −16 − (6)
(h) (−2) + (−7)
(k) −5 − 5 − 5
(c) (−10) ÷ (−2)
(f) −9 ÷ 3 × 2
(i) 1 − 12
(l) −6 + 25
CHAPTER 2. REVIEW OF PAST WORK
2.9
2. Say whether the sign of the answer is + or (a) −5 + 6
(d) −5 ÷ 5
(g) −5 × −5
(j) 5 × 5
2.9
(b) −5 + 1
(e) 5 ÷ −5
(h) −5 × 5
(c) −5 ÷ −5
(f) 5 ÷ 5
(i) 5 × −5
Rearranging Equations
Now that we have described the basic rules of negative and positive numbers and what to do
when you add, subtract, multiply and divide them, we are ready to tackle some real mathematics
problems!
Earlier in this chapter, we wrote a general equation for calculating how much change (x) we can
expect if we know how much an item costs (y) and how much we have given the cashier (z).
The equation is:
x+y =z
(2.20)
So, if the price is R10 and you gave the cashier R15, then write R15 instead of z and R10 instead
of y.
x + 10 = 15
(2.21)
Now, that we have written this equation down, how exactly do we go about finding what the
change is? In mathematical terms, this is known as solving an equation for an unknown (x in
this case). We want to re-arrange the terms in the equation, so that only x is on the left hand
side of the = sign and everything else is on the right.
The most important thing to remember is that an equation is like a set of weighing scales. In
order to keep the scales balanced, whatever, is done to one side, must be done to the other.
Method: Rearranging Equations
You can add, subtract, multiply or divide both sides of an equation by any number you want, as
long as you always do it to both sides.
So for our example we could subtract y from both sides
x+y
x+y−y
=
=
x =
x =
=
z
z−y
(2.22)
z−y
15 − 10
5
so now we can find the change is the price subtracted from the amount handed over to the
cashier. In the example, the change should be R5. In real life we can do this in our head, the
human brain is very smart and can do arithmetic without even knowing it.
When you subtract a number from both sides of an equation, it looks just like you moved a
positive number from one side and it became a negative on the other, which is exactly what
happened. Likewise if you move a multiplied number from one side to the other, it looks like it
changed to a divide. This is because you really just divided both sides by that number, and a
13
2.9
CHAPTER 2. REVIEW OF PAST WORK
x+y
z
x+y−y
z−y
divide the other side too.
Figure 2.2: An equation is like a set of weighing scales. In order to keep the scales balanced,
you must do the same thing to both sides. So, if you add, subtract, multiply or divide the one
side, you must add, subtract, multiply or
divide the other side too.
number divided by itself is just 1
a(5 + c)
a(5 + c) ÷ a
a
× (5 + c)
a
1 × (5 + c)
5+c
c
= 3a
= 3a ÷ a
a
= 3×
a
= 3×1
= 3
(2.23)
= 3 − 5 = −2
However you must be careful when doing this, as it is easy to make mistakes.
The following is the wrong thing to do
5a + c
5+c
=
6=
4
3a
(2.24)
3a ÷ a
Can you see why it is wrong? It is wrong because we did not divide the c term by a as well. The
correct thing to do is
5a + c =
5+c÷a =
c÷a =
Exercise: Rearranging Equations
14
3a
3
3 − 5 = −2
(2.25)
CHAPTER 2. REVIEW OF PAST WORK
2.10
1. If 3(2r − 5) = 27, then 2r − 5 = .....
2. Find the value for x if 0,5(x − 8) = 0,2x + 11
3. Solve 9 − 2n = 3(n + 2)
4. Change the formula P = A + Akt to A =
5. Solve for x:
2.10
1
ax
+
1
bx
=1
Fractions and Decimal Numbers
A fraction is one number divided by another number. There are several ways to write a number
divided by another one, such as a ÷ b, a/b and ab . The first way of writing a fraction is very hard
to work with, so we will use only the other two. We call the number on the top, the numerator
and the number on the bottom the denominator. For example,
1
5
numerator = 1
denominator = 5
(2.26)
Extension: Definition - Fraction
The word fraction means part of a whole.
The reciprocal of a fraction is the fraction turned upside down, in other words the numerator
becomes the denominator and the denominator becomes the numerator. So, the reciprocal of 32
is 32 .
A fraction multiplied by its reciprocal is always equal to 1 and can be written
b
a
× =1
b
a
(2.27)
This is because dividing by a number is the same as multiplying by its reciprocal.
Extension: Definition - Multiplicative Inverse
The reciprocal of a number is also known as the multiplicative inverse.
A decimal number is a number which has an integer part and a fractional part. The integer
and the fractional parts are separated by a decimal point, which is written as a comma in South
14
Africa. For example the number 3 100
can be written much more cleanly as 3,14.
All real numbers can be written as a decimal number. However, some numbers would take a
huge amount of paper (and ink) to write out in full! Some decimal numbers will have a number
which will repeat itself, such as 0,33333 . . . where there are an infinite number of 3’s. We can
write this decimal value by using a dot above the repeating number, so 0,3̇ = 0,33333 . . .. If
there are two repeating numbers such as 0,121212 . . . then you can place dots5 on each of the
repeated numbers 0,1̇2̇ = 0,121212 . . .. These kinds of repeating decimals are called recurring
decimals.
Table 2.3 lists some common fractions and their decimal forms.
5 or
a bar, like 0,12
15
2.11
CHAPTER 2. REVIEW OF PAST WORK
Fraction
1
20
Decimal Form
0,05
1
16
0,0625
1
10
0,1
1
8
0,125
1
6
0,166̇
1
5
0,2
1
2
0,5
3
4
0,75
Table 2.3: Some common fractions and their equivalent decimal forms.
2.11
Scientific Notation
In science one often needs to work with very large or very small numbers. These can be written
more easily in scientific notation, which has the general form
a × 10m
(2.28)
where a is a decimal number between 0 and 10 that is rounded off to a few decimal places. The
m is an integer and if it is positive it represents how many zeros should appear to the right of
a. If m is negative then it represents how many times the decimal place in a should be moved
to the left. For example 3,2 × 103 represents 32000 and 3,2 × 10−3 represents 0,0032.
If a number must be converted into scientific notation, we need to work out how many times
the number must be multiplied or divided by 10 to make it into a number between 1 and 10
(i.e. we need to work out the value of the exponent m) and what this number is (the value of
a). We do this by counting the number of decimal places the decimal point must move.
For example, write the speed of light which is 299 792 458 ms−1 in scientific notation, to two
decimal places. First, determine where the decimal point must go for two decimal places (to
find a) and then count how many places there are after the decimal point to determine m.
In this example, the decimal point must go after the first 2, but since the number after the 9 is
a 7, a = 3,00.
So the number is 3,00 × 10m , where m = 8, because there are 8 digits left after the decimal
point. So the speed of light in scientific notation, to two decimal places is 3,00 × 108 ms−1 .
As another example, the size of the HI virus is around 120 × 10−9 m. This is equal to 120 ×
0,000000001 m which is 0,00000012 m.
2.12
Real Numbers
Now that we have learnt about the basics of mathematics, we can look at what real numbers
are in a little more detail. The following are examples of real numbers and it is seen that each
number is written in a different way.
√
3, 1,2557878,
56
, 10, 2,1,
34
− 5,
− 6,35,
−
1
90
(2.29)
Depending on how the real number is written, it can be further labelled as either rational,
irrational, integer or natural. A set diagram of the different number types is shown in Figure 2.3.
16
CHAPTER 2. REVIEW OF PAST WORK
2.12
N
Z
Q
R
Figure 2.3: Set diagram of all the real numbers R, the rational numbers Q, the integers Z and
the natural numbers N. The irrational numbers are the numbers not inside the set of rational
numbers. All of the integers are also rational numbers, but not all rational numbers are integers.
Extension: Non-Real Numbers
All numbers that are not real numbers have imaginary components. We√will not see
imaginary numbers in this book but you will see that they come from −1. Since
we won’t be looking at numbers which are not real, if you see a number you can be
sure it is a real one.
2.12.1
Natural Numbers
The first type of numbers that are learnt about are the numbers that were used for counting.
These numbers are called natural numbers and are the simplest numbers in mathematics.
0, 1, 2, 3, 4 . . .
(2.30)
Mathematicians use the symbol N to mean the set of all natural numbers. The natural numbers
are a subset of the real numbers since every natural number is also a real number.
2.12.2
Integers
The integers are all of the natural numbers and their negatives
. . . − 4, −3, −2, −1, 0, 1, 2, 3, 4 . . .
(2.31)
Mathematicians use the symbol Z to mean the set of all integers. The integers are a subset of
the real numbers, since every integer is a real number.
2.12.3
Rational Numbers
The natural numbers and the integers are only able to describe quantities that are whole or
complete. For example you can have 4 apples, but what happens when you divide one apple
into 4 equal pieces and share it among your friends? Then it is not a whole apple anymore and
a different type of number is needed to describe the apples. This type of number is known as a
rational number.
A rational number is any number which can be written as:
a
b
(2.32)
where a and b are integers and b 6= 0.
The following are examples of rational numbers:
20
,
9
−1
20
,
,
2
10
17
3
15
(2.33)
2.12
CHAPTER 2. REVIEW OF PAST WORK
Extension: Notation Tip
Rational numbers are any number that can be expressed in the form ab ; a, b ∈ Z; b 6= 0
which means “the set of numbers ab when a and b are integers”.
Mathematicians use the symbol Q to mean the set of all rational numbers. The set of rational
numbers contains all numbers which can be written as terminating or repeating decimals.
Extension: Rational Numbers
All integers are rational numbers with denominator 1.
You can add and multiply rational numbers and still get a rational number at the end, which is
very useful. If we have 4 integers, a, b, c and d, then the rules for adding and multiplying rational
numbers are
c
a
+
b
d
a
c
×
b
d
=
=
ad + bc
bd
ac
bd
(2.34)
(2.35)
Extension: Notation Tip
The statement ”4 integers a, b, c and d” can be written formally as {a, b, c, d} ∈ Z
because the ∈ symbol means in and we say that a, b, c and d are in the set of integers.
Two rational numbers ( ab and dc ) represent the same number if ad = bc. It is always best
to simplify any rational number so that the denominator is as small as possible. This can be
achieved by dividing both the numerator and the denominator by the same integer. For example,
the rational number 1000/10000 can be divided by 1000 on the top and the bottom, which gives
8
(Figure 2.4).
1/10. 23 of a pizza is the same as 12
8
12
Figure 2.4:
8
12
2
3
of the pizza is the same as
2
3
of the pizza.
You can also add rational numbers together by finding a lowest common denominator and then
adding the numerators. Finding a lowest common denominator means finding the lowest number
that both denominators are a factor 6 of. A factor of a number is an integer which evenly divides
that number without leaving a remainder. The following numbers all have a factor of 3
3, 6, 9, 12, 15, 18, 21, 24 . . .
and the following all have factors of 4
4, 8, 12, 16, 20, 24, 28 . . .
6 Some
people say divisor instead of factor.
18
CHAPTER 2. REVIEW OF PAST WORK
2.12
The common denominators between 3 and 4 are all the numbers that appear in both of these
lists, like 12 and 24. The lowest common denominator of 3 and 4 is the number that has both
3 and 4 as factors, which is 12.
For example, if we wish to add 43 + 23 , we first need to write both fractions so that their
denominators are the same by finding the lowest common denominator, which we know is 12.
We can do this by multiplying 43 by 33 and 32 by 44 . 33 and 44 are really just complicated ways of
writing 1. Multiplying a number by 1 doesn’t change the number.
3 2
+
4 3
=
=
=
=
=
3 3 2 4
× + ×
4 3 3 4
3×3 2×4
+
4×3 3×4
9
8
+
12 12
9+8
12
17
12
(2.36)
Dividing by a rational number is the same as multiplying by its reciprocal, as long as neither the
numerator nor the denominator is zero:
a
c
a d
ad
÷ = . =
b
d
b c
bc
(2.37)
A rational number may be a proper or improper fraction.
Proper fractions have a numerator that is smaller than the denominator. For example,
−1 3 −5
, ,
2 15 −20
are proper fractions.
Improper fractions have a numerator that is larger than the denominator. For example,
−10 13 −53
, ,
2 15 −20
are improper fractions. Improper fractions can always be written as the sum of an integer and a
proper fraction.
Converting Rationals into Decimal Numbers
Converting rationals into decimal numbers is very easy.
If you use a calculator, you can simply divide the numerator by the denominator.
If you do not have a calculator, then you unfortunately have to use long division.
Since long division, was first taught in primary school, it will not be discussed here. If you have
trouble with long division, then please ask your friends or your teacher to explain it to you.
2.12.4
Irrational Numbers
An irrational number is any real number that is not a rational number. When expressed as
decimals these numbers can never be fully written out as they have √
an infinite number of
decimal places which never fall into a repeating pattern, for example 2 = 1,41421356 . . .,
π = 3,14159265 . . .. π is a Greek letter and is pronounced “pie”.
Exercise: Real Numbers
19
2.13
CHAPTER 2. REVIEW OF PAST WORK
1. Identify the number type (rational, irrational, real, integer) of each of the
following numbers:
(a) dc if c is an integer and if d is irrational.
(b) 23
(c) -25
(d) 1,525
√
(e) 10
2. √
Is the following pair of numbers real and rational or real and irrational? Explain.
4; 18
2.13
Mathematical Symbols
The following is a table of the meanings of some mathematical signs and symbols that you should
have come across in earlier grades.
Sign or Symbol
>
<
≥
≤
Meaning
greater than
less than
greater than or equal to
less than or equal to
So if we write x > 5, we say that x is greater than 5 and if we write x ≥ y, we mean that x
can be greater than or equal to y. Similarly, < means ‘is less than’ and ≤ means ‘is less than
or equal to’. Instead of saying that x is between 6 and 10, we often write 6 < 10. This directly
means ‘six is less than x which in turn is less than ten’.
Exercise: Mathematical Symbols
1. Write the following in symbols:
(a)
(b)
(c)
(d)
2.14
x is greater than 1
y is less than or equal to z
a is greater than or equal to 21
p is greater than or equal to 21 and p is less than or equal to 25
Infinity
Infinity (symbol ∞) is usually thought of as something like “the largest possible number” or “the
furthest possible distance”. In mathematics, infinity is often treated as if it were a number, but
it is clearly a very different type of “number” than the integers or reals.
When talking about recurring decimals and irrational numbers, the term infinite was used to
describe never-ending digits.
20
CHAPTER 2. REVIEW OF PAST WORK
2.15
2.15
End of Chapter Exercises
1. Calculate
(a) 18 − 6 × 2
(b) 10 + 3(2 + 6)
(c) 50 − 10(4 − 2) + 6
(d) 2 × 9 − 3(6 − 1) + 1
(e) 8 + 24 ÷ 4 × 2
(f) 30 − 3 × 4 + 2
(g) 36 ÷ 4(5 − 2) + 6
(h) 20 − 4 × 2 + 3
(i) 4 + 6(8 + 2) − 3
(j) 100 − 10(2 + 3) + 4
2. If p = q + 4r, then r = .....
3. Solve
x−2
3
=x−3
21
2.15
CHAPTER 2. REVIEW OF PAST WORK
22
Chapter 3
Rational Numbers - Grade 10
3.1
Introduction
As described in Chapter 2, a number is a way of representing quantity. The numbers that will
be used in high school are all real numbers, but there are many different ways of writing any
single real number.
This chapter describes rational numbers.
3.2
The Big Picture of Numbers
Real Numbers
Rationals
Integers
All numbers inside
the grey oval are rational numbers.
3.3
Whole
Natural
Irrationals
Definition
The following numbers are all rational numbers.
10
,
1
21
,
7
−1
,
−3
10
,
20
−3
6
(3.1)
You can see that all the denominators and all the numerators are integers1 .
Definition: Rational Number
A rational number is any number which can be written as:
a
b
where a and b are integers and b 6= 0.
1 Integers
are the counting numbers (1, 2, 3, ...), their opposites (-1, -2, -3, ...), and 0.
23
(3.2)
3.4
CHAPTER 3. RATIONAL NUMBERS - GRADE 10
Important: Only fractions which have a numerator and a denominator that are integers
are rational numbers.
This means that all integers are rational numbers, because they can be written with a denominator
of 1.
Therefore, while
√
2
,
7
−1,33
,
−3
π
,
20
−3
6,39
(3.3)
are not examples of rational numbers, because in each case, either the numerator or the
denominator is not an integer.
Exercise: Rational Numbers
1. If a is an integer, b is an integer and c is not an integer, which of the following
are rational numbers:
(a)
2. If
3.4
a
1
5
6
(b)
a
3
(c)
b
2
(d)
1
c
is a rational number, which of the following are valid values for a?
√
(d) 2,1
(a) 1
(b) −10
(c) 2
Forms of Rational Numbers
All integers and fractions with integer numerators and denominators are rational numbers. There
are two more forms of rational numbers.
Activity :: Investigation : Decimal Numbers
You can write the rational number 12 as the decimal number 0,5. Write the
following numbers as decimals:
1.
2.
3.
4.
5.
1
4
1
10
2
5
1
100
2
3
Do the numbers after the decimal comma end or do they continue? If they continue,
is there a repeating pattern to the numbers?
You can write a rational number as a decimal number. Therefore, you should be able to write a
decimal number as a rational number. Two types of decimal numbers can be written as rational
numbers:
1. decimal numbers that end or terminate, for example the fraction
24
4
10
can be written as 0,4.
CHAPTER 3. RATIONAL NUMBERS - GRADE 10
3.5
2. decimal numbers that have a repeating pattern of numbers, for example the fraction
can be written as 0,333333.
1
3
For example, the rational number 65 can be written in decimal notation as 0,83333, and similarly,
the decimal number 0,25 can be written as a rational number as 41 .
Important: Notation for Repeating Decimals
You can use a bar over the repeated numbers to indicate that the decimal is a repeating decimal.
3.5
Converting Terminating Decimals into Rational Numbers
A decimal number has an integer part and a fractional part. For example, 10,589 has an integer
part of 10 and a fractional part of 0,589 because 10 + 0,589 = 10,589. The fractional part can
be written as a rational number, i.e. with a numerator and a denominator that are integers.
Each digit after the decimal point is a fraction with denominator in increasing powers of ten.
For example:
•
•
1
10
is 0,1
1
100
is 0,01
This means that:
0
3
1
+
+
10 100 1000
103
= 2
1000
2103
=
1000
2,103 = 2 +
Exercise: Fractions
1. Write the following as fractions:
(a) 0,1
3.6
(b) 0,12
(c) 0,58
(d) 0,2589
Converting Repeating Decimals into Rational Numbers
When the decimal is a repeating decimal, a bit more work is needed to write the fractional part
of the decimal number as a fraction. We will explain by means of an example.
If we wish to write 0,3 in the form
follows
a
b
(where a and b are integers) then we would proceed as
x =
0,33333 . . .
10x =
9x =
3,33333 . . .
3
3
1
=
9
3
x =
(3.4)
multiply by 10 on both sides
subtracting (3.4) from (3.5)
25
(3.5)
3.7
CHAPTER 3. RATIONAL NUMBERS - GRADE 10
And another example would be to write 5,432 as a rational fraction
x =
1000x =
999x =
x =
5,432432432 . . .
(3.6)
5432,432432432 . . .
5427
5427
201
=
999
37
multiply by 1000 on both sides
(3.7)
subtracting (3.6) from (3.7)
For the first example, the decimal number was multiplied by 10 and for the second example, the
decimal number was multiplied by 1000. This is because for the first example there was only
one number (i.e. 3) that recurred, while for the second example there were three numbers (i.e.
432) that recurred.
In general, if you have one number recurring, then multiply by 10, if you have two numbers
recurring, then multiply by 100, if you have three numbers recurring, then multiply by 1000. Can
you spot the pattern yet?
The number of zeros after the 1 is the same as the number of recurring numbers.
But √
not all decimal numbers can be written as rational numbers, because some decimal numbers
like 2 = 1,4142135... is an irrational number and cannot be written with an integer numerator
and an integer denominator. However, when possible, you should always use rational numbers
or fractions instead of decimals.
Exercise: Repeated Decimal Notation
1. Write the following using the repeated decimal notation:
(a) 0,11111111 . . .
(b) 0,1212121212 . . .
(c) 0,123123123123 . . .
(d) 0,11414541454145 . . .
2. Write the following in decimal form, using the repeated decimal notation:
(a) 32
3
(b) 1 11
(c) 4 65
(d) 2 91 . . .
3. Write the following decimals in fractional form:
(a) 0,6333 . . .
(b) 5,313131
(c) 11,570571 . . .
(d) 0,999999 . . .
3.7
Summary
The following are rational numbers:
• Fractions with both denominator and numerator as integers.
• Integers.
• Decimal numbers that end.
• Decimal numbers that repeat.
26
CHAPTER 3. RATIONAL NUMBERS - GRADE 10
3.8
3.8
End of Chapter Exercises
1. If a is an integer, b is an integer and c is not an integer, which of the following are rational
numbers:
(a)
(b)
(c)
(d)
5
6
a
3
b
2
1
c
2. Write each decimal as a simple fraction:
(a) 0,5
(b) 0,12
(c) 0,6
(d) 1,59
(e) 12,277
3. Show that the decimal 3,21̇8̇ is a rational number.
4. Showing all working, express 0,78̇ as a fraction
27
a
b
where a, b ∈ Z.
3.8
CHAPTER 3. RATIONAL NUMBERS - GRADE 10
28
Chapter 4
Exponentials - Grade 10
4.1
Introduction
In this chapter, you will learn about the short-cuts to writing 2 × 2 × 2 × 2. This is known as
writing a number in exponential notation.
4.2
Definition
Exponential notation is a short way of writing the same number multiplied by itself many times.
For example, instead of 5 × 5 × 5, we write 53 to show that the number 5 is multiplied by itself
3 times and we say “5 to the power of 3”. Likewise 52 is 5 × 5 and 35 is 3 × 3 × 3 × 3 × 3. We
will now have a closer look at writing numbers using exponential notation.
Definition: Exponential Notation
Exponential notation means a number written like
an
when n is an integer and a can be any real number. a is called the base and n is called the
exponent.
The nth power of a is defined as:
an = 1 × a × a × . . . × a
(n times)
(4.1)
with a multiplied by itself n times.
We can also define what it means if we have a negative index, −n. Then,
a−n = 1 ÷ a ÷ a ÷ . . . ÷ a
(n times)
(4.2)
Important: Exponentials
If n is an even integer, then an will always be positive for any non-zero real number a. For
example, although −2 is negative, (−2)2 = 1 × −2 × −2 = 4 is positive and so is (−2)−2 =
1 ÷ −2 ÷ −2 = 14 .
29
4.3
4.3
CHAPTER 4. EXPONENTIALS - GRADE 10
Laws of Exponents
There are several laws we can use to make working with exponential numbers easier. Some of
these laws might have been seen in earlier grades, but we will list all the laws here for easy
reference, but we will explain each law in detail, so that you can understand them, and not only
remember them.
a0
m
a × an
a−n
am ÷ an
n
(ab)
(am )n
4.3.1
= 1
(4.3)
= am+n
1
=
an
= am−n
(4.4)
(4.5)
(4.6)
n n
(4.7)
(4.8)
1, (a 6= 0)
(4.9)
= a b
= amn
Exponential Law 1: a0 = 1
Our definition of exponential notation shows that
a0
=
For example, x0 = 1 and (1 000 000)0 = 1.
Exercise: Application using Exponential Law 1: a0 = 1, (a 6= 0)
1. 160 = 1
2. 16a0 = 16
3. (16 + a)0 = 1
4. (−16)0 = 1
5. −160 = −1
4.3.2
Exponential Law 2: am × an = am+n
Our definition of exponential notation shows that
am × an
=
1 × a× ...× a
×1 × a × . . . × a
= 1 × a× ...× a
=
(m times)
(n times)
(m + n times)
am+n
For example,
27 × 23
= (2 × 2 × 2 × 2 × 2 × 2 × 2) × (2 × 2 × 2)
= 210
= 27+3
30
(4.10)
CHAPTER 4. EXPONENTIALS - GRADE 10
4.3
teresting This simple law is the reason why exponentials were originally invented. In the
Interesting
Fact
Fact
days before calculators, all multiplication had to be done by hand with a pencil
and a pad of paper. Multiplication takes a very long time to do and is very
tedious. Adding numbers however, is very easy and quick to do. If you look at
what this law is saying you will realise that it means that adding the exponents
of two exponential numbers (of the same base) is the same as multiplying the two
numbers together. This meant that for certain numbers, there was no need to
actually multiply the numbers together in order to find out what their multiple
was. This saved mathematicians a lot of time, which they could use to do
something more productive.
Exercise: Application using Exponential Law 2: am × an = am+n
1. x2 .x5 = x7
2. 2x3 y × 5x2 y 7 = 10x5 y 8
3. 23 .24 = 27 [Take note that the base (2) stays the same.]
4. 3 × 32a × 32 = 32a+3
4.3.3
Exponential Law 3: a−n =
1
,a
an
6= 0
Our definition of exponential notation for a negative exponent shows that
a−n
=
=
=
1 ÷ a÷ ...÷ a
1
1 × a× ···× a
1
an
(n times)
(4.11)
(n times)
This means that a minus sign in the exponent is just another way of writing that the whole
exponential number is to be divided instead of multiplied.
For example,
2−7
=
=
1
2×2×2×2×2×2×2
1
27
Exercise: Application using Exponential Law 3: a−n =
1. 2−2 =
2.
3.
1
22
=
1
4
1
1
2−2
32 = 22 .32 = 36
( 23 )−3 = ( 32 )3 = 27
8
31
1
an , a
6= 0
4.3
4.3.4
CHAPTER 4. EXPONENTIALS - GRADE 10
= mn4
4.
m
n−4
5.
a−3 .x4
a5 .x−2
=
x4 .x2
a3 .a5
x6
a8
=
Exponential Law 4: am ÷ an = am−n
We already realised with law 3 that a minus sign is another way of saying that the exponential
number is to be divided instead of multiplied. Law 4 is just a more general way of saying the
same thing. We can get this law by just multiplying law 3 by am on both sides and using law 2.
am
an
=
am a−n
=
am−n
(4.12)
For example,
27 ÷ 23
2×2×2×2×2×2×2
2×2×2
= 2×2×2×2
=
= 24
= 27−3
Exercise: Exponential Law 4: am ÷ an = am−n
1.
a6
a2
= a6−2 = a4
2.
32
36
= 32−6 = 3−4 =
3.
4.
4.3.5
2
32a
4a8
3x
a
a4
= 8a
−6
=
1
34 [Always
give final answer with positive index]
8
a6
= a3x−4
Exponential Law 5: (ab)n = an bn
The order in which two real numbers are multiplied together does not matter. Therefore,
(ab)n
=
=
a× b × a × b × ...× a× b
(n times)
a× a × ...× a
(n times)
×b × b × . . . × b
= an b n
(n times)
For example,
(2 · 3)4
= (2 · 3) × (2 · 3) × (2 · 3) × (2 · 3)
= (2 × 2 × 2 × 2) × (3 × 3 × 3 × 3)
= (24 ) × (34 )
= 24 34
32
(4.13)
CHAPTER 4. EXPONENTIALS - GRADE 10
4.3
Exercise: Exponential Law 5: (ab)n = an bn
1. (2x2 y)3 = 23 x2×3 y 5 = 8x6 y 5
49a2
b6
n−4 3
2
2. ( 7a
b3 ) =
3. (5a
4.3.6
) = 125a3n−12
Exponential Law 6: (am )n = amn
We can find the exponential of an exponential just as well as we can for a number. After all, an
exponential number is a real number.
(am )n
=
=
=
am × am × . . . × am
(n times)
a× a× ... × a
(m × n times)
(4.14)
amn
For example,
(22 )3
=
(22 ) × (22 ) × (22 )
=
=
(2 × 2) × (2 × 2) × (2 × 2)
(26 )
=
2(2×3)
Exercise: Exponential Law 6: (am )n = amn
1. (x3 )4 = x12
2. [(a4 )3 ]2 = a24
3. (3n+3 )2 = 32n+6
Worked Example 1: Simlifying indices
2x−1
x−2
.9
Question: Simplify: 5 152x−3
Answer
Step 1 : Factorise all bases into prime factors:
=
=
52x−1 .(32 )x−2
(5.3)2x−3
52x−1 .32x−4
52x−3 .32x−3
Step 2 : Add and subtract the indices of the sam bases as per laws 2 and 4:
33
4.4
CHAPTER 4. EXPONENTIALS - GRADE 10
= 52x−1−2x−3 .32x−4−2x+3
= 52 .3−1
Step 3 : Write simplified answer with positive indices:
=
25
3
Activity :: Investigation : Exponential Numbers
Match the answers to the questions, by filling in the correct answer into the
Answer column. Possible answers are: 32 , 1, -1, -3, 8.
Question
23
73−3
( 23 )−1
87−6
(−3)−1
(−1)23
4.4
Answer
End of Chapter Exercises
1. Simplify as far as possible:
(a) 3020
(e) (2x)3
(b) 10
(f) (−2x)3
(c) (xyz)0
(g) (2x)4
(d) [(3x4 y 7 z 12 )5 (−5x9 y 3 z 4 )2 ]0
(h) (−2x)4
2. Simplify without using a calculator. Leave your answers with positive exponents.
(a)
3x−3
(3x)2
(b) 5x0 + 8−2 − ( 12 )−2 .1x
(c)
5b−3
5b+1
3. Simplify, showing all steps:
(a)
2a−2 .3a+3
6a
(b)
a2m+n+p
am+n+p .am
(c)
3n .9n−3
27n−1
2a
3
(d) ( 2x
y −b )
34
CHAPTER 4. EXPONENTIALS - GRADE 10
(e)
23x−1 .8x+1
42x−2
(f)
62x .112x
222x−1 .32x
4. Simplify, without using a calculator:
(a)
(−3)−3 .(−3)2
.
(−3)−4
(b) (3−1 + 2−1 )−1
(c)
9n−1 .273−2n
812−n
(d)
23n+2 .8n−3
43n−2
35
4.4
4.4
CHAPTER 4. EXPONENTIALS - GRADE 10
36
Chapter 5
Estimating Surds - Grade 10
5.1
Introduction
You should known by now what the nth root of a number means. If the √
nth root√of a number
cannot
be
simplified
to
a
rational
number,
we
call
it
a
surd.
For
example,
2 and 3 6 are surds,
√
but 4 is not a surd because it can be simplified to the rational number 2.
√
In this chapter
we
will only look at surds that look like n a, where a is any positive
√
√
√ number, for
example 7 or 3 5. It√
is very common for n to be 2, so we usually do not write 2 a. Instead we
write the surd as just a, which is much easier to read.
It is sometimes useful to know the approximate value of a surd√without having to use a calculator.
For example, we want to be able to guess where a surd like 3 is on the number√line. So how
do we know where surds lie on the
√ number line? From a calculator we know that 3 is equal to
1,73205....
It
is
easy
to
see
that
3 is above 1 and below 2. But to see this for other surds like
√
18 without using a calculator, you must first understand the following fact:
teresting If a and b are positive whole numbers, and a < b, then √n a < √n b. (Challenge:
Interesting
Fact
Fact
Can you explain why?)
If you don’t believe this fact, check it for a few numbers to convince yourself it is true.
√
How do we use this fact to help us √
guess what
18 is? Well, you can easily see that √
18 < 25?
√
2
18
<
25.
But
we
know
that
5
=
25
so
that
25 = 5.
Using our rule, we also know that
√
√
Now it is easy to simplify to get 18 < 5. Now we have a better idea of what 18 is.
√
Now we know that 18 is less than 5, but this is only half the story. We can use the same trick
again, but this time with 18 on
side. You will agree that 16 < 18. Using our
√
√ the right-hand
rule again,
we
also
know
that
16
<
18.
But
we
know that 16 is a perfect square, so we can
√
√
simplify 16 to 4, and so we get 4 < 18!
√
Can you see now that we now
√ have shown that 18 is between 4 and 5? If we check on our
calculator, we can see that 18 = 4,1231..., and we see that our idea was right! You will notice
that our idea used perfect squares that were close to the number 18. We found the closest
perfect square underneath 18, which was 42 = 16, and the closest perfect square above 18,
which was 52 = 25. Here is a quick summary of what a perfect square or cube is:
teresting A perfect square is the number obtained when an integer is squared. For example,
Interesting
Fact
Fact
9 is a perfect square since 32 = 9. Similarly, a perfect cube is a number which is
the cube of an integer. For example, 27 is a perfect cube, because 33 = 27.
37
5.2
CHAPTER 5. ESTIMATING SURDS - GRADE 10
To make it easier to use our idea, we will create a list of some of the perfect squares and perfect
cubes. The list is shown in Table 5.1.
Table 5.1: Some perfect squares and perfect cubes
Integer Perfect Square Perfect Cube
0
0
0
1
1
1
2
4
8
3
9
27
4
16
64
5
25
125
6
36
216
7
49
343
8
64
512
9
81
729
10
100
1000
√
3
Similarly, when given
the surd √
52 you should be able to tell that it lies somewhere
√
√ between 3
3
3
and 4, because 27 = 3 and 64 = 4 and 52 is between 27 and 64. In fact 3 52 = 3,73 . . .
which is indeed between 3 and 4.
5.2
Drawing Surds on the Number Line (Optional)
√
How can√we accurately draw a surd like 5 on the number line? Well, we could use a calculator
to find 5 = 2,2360... and measure the distance along the number line using a ruler. But for
some surds, there is a much easier way.
√
Let us call the surd we are working with a. Sometimes, we can write a as
sum of two
√ the √
perfect squares, so a = b2 + c2 . We know from Pythagoras’ theorem that a = b2 + c2 is
the length of the hypotenuse of a triangle that has sides that have lengths of b and c. So if we
draw a triangle on the number line with sides of length b and c, we can use a compass to draw
a circle
√ from the top of the hypotenuse down to the number line. The intersection will be the
point b on the number line!
teresting Not all numbers can be written as the sum of two squares. See if you can find a
Interesting
Fact
Fact
pattern of the numbers that can.
Worked Example 2: Estimating Surds
√
Question: Find the two consecutive integers such that 26 lies between them.
(Remember that consecutive numbers that are two numbers one after the other, like
5 and 6 or 8 and 9.)
Answer
Step 1 : From the table find √
the largest perfect square below 26
This is 52 = 25. Therefore 5 < 26.
38
CHAPTER 5. ESTIMATING SURDS - GRADE 10
5.3
Step 2 : From the table find
√ smallest perfect square above 26
This is 62 = 36. Therefore 26 < 6.
Step 3 : Put the √
inequalities together
Our answer is 5 < 26 < 6.
Worked Example 3: Estimating Surds
√
Question: 3 49 lies between: (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 (d) 4 and 5
Answer
Step 1 :√Consider (a) as the solution
If 1 < 3 49 < 2 then cubing√ all terms gives 1 < 49 < 23 . Simplifying gives
1 < 49 < 8 which is false. So 3 49 does not lie between 1 and 2.
Step 2 :√Consider (b) as the solution
If 2 < 3 49 < 3 then cubing √
all terms gives 23 < 49 < 33 . Simplifying gives
3
8 < 49 < 27 which is false. So 49 does not lie between 2 and 3.
Step 3 :√Consider (c) as the solution
If 3 < 3 49 < 4 then cubing all
terms gives 33 < 49 < 43 . Simplifying gives
√
3
27 < 49 < 64 which is true. So 49 lies between 3 and 4.
5.3
1.
2.
3.
4.
5.
6.
7.
8.
End of Chapter Excercises
√
√5 lies between
√10 lies between
√20 lies between
30 lies between
√
3
5 lies between
√
3
10 lies between
√
3
20 lies between
√
3
30 lies between
(a)
(a)
(a)
(a)
(a)
(a)
(a)
(a)
1
1
2
3
1
1
2
3
and
and
and
and
and
and
and
and
2
2
3
4
2
2
3
4
(b)
(b)
(b)
(b)
(b)
(b)
(b)
(b)
2
2
3
4
2
2
3
4
and
and
and
and
and
and
and
and
39
3
3
4
5
3
3
4
5
(c)
(c)
(c)
(c)
(c)
(c)
(c)
(c)
3
3
4
5
3
3
4
5
and
and
and
and
and
and
and
and
4
4
5
6
4
4
5
6
(d)
(d)
(d)
(d)
(d)
(d)
(d)
(d)
4
4
5
6
4
4
5
6
and
and
and
and
and
and
and
and
5
5
6
7
5
5
6
7
5.3
CHAPTER 5. ESTIMATING SURDS - GRADE 10
40
Chapter 6
Irrational Numbers and Rounding
Off - Grade 10
6.1
Introduction
You have seen that repeating decimals may take a lot of paper and ink to write out. Not only
is that impossible, but writing numbers out to many decimal places or a high accuracy is very
inconvenient and rarely gives better answers. For this reason we often estimate the number to
a certain number of decimal places or to a given number of significant figures, which is even
better.
6.2
Irrational Numbers
Activity :: Investigation : Irrational Numbers
Which of the following cannot be written as a rational number?
Remember: A rational number is a fraction with numerator and denominator as
integers. Terminating decimal numbers or repeating decimal numbers are rational.
1. π = 3,14159265358979323846264338327950288419716939937510 . . .
2. 1,4
3. 1,618 033 989 . . .
4. 100
Irrational numbers are numbers that cannot be written as a rational number. You should know
that a rational number can be written as a fraction with the numerator and denominator as
integers. This means that any number that is not a terminating decimal number or a repeating
decimal number are irrational. Examples of irrational numbers are:
√
√
√
3
2,
3,
4, π,
√
1+ 5
≈ 1,618 033 989
2
Important: When irrational numbers are written in decimal form, they go on forever and
there is no repeated pattern of digits.
41
6.3
CHAPTER 6. IRRATIONAL NUMBERS AND ROUNDING OFF - GRADE 10
Important: Irrational Numbers
If you are asked to identify whether a number is rational or irrational, first write the number in
decimal form. If the number is terminated then it is rational. If it goes on forever, then look for
a repeated pattern of digits. If there is no repeated pattern, then the number is irrational.
When you write irrational numbers in decimal form, you may (if you have a lot of time and
paper!) continue writing them for many, many decimal places. However, this is not convenient
and it is often necessary to round off.
6.3
Rounding Off
Rounding off or approximating a decimal number to a given number of decimal places is the
quickest way to approximate a number. For example, if you wanted to round-off 2,6525272 to
three decimal places then you would first count three places after the decimal.
2,652|5272
All numbers to the right of | are ignored after you determine whether the number in the third
decimal place must be rounded up or rounded down. You round up the final digit if the first
digit after the | was greater or equal to 5 and round down (leave the digit alone) otherwise.
So, since the first digit after the | is a 5, we must round up the digit in the third decimal place
to a 3 and the final answer of 2,6525272 rounded to three decimal places is
2,653
Worked Example 4: Rounding-Off
Question: Round-off the following numbers to the indicated number of decimal
places:
1.
120
99
= 1,2121212121̇2̇ to 3 decimal places
2. π = 3,141592654 . . . to 4 decimal places
√
3. 3 = 1,7320508 . . . to 4 decimal places
Answer
Step 1 : Determine the last digit that is kept and mark the cut-off point with
|.
1.
120
99
= 1,212|1212121̇2̇
2. π = 3,1415|92654 . . .
√
3. 3 = 1,7320|508 . . .
Step 2 : Determine whether the last digit is rounded up or down.
1. The last digit of
120
99
= 1,212|1212121̇2̇ must be rounded-down.
2. The last digit of π = 3,1415|92654 . . . must be rounded-up.
√
3. The last digit of 3 = 1,7320|508 . . . must be rounded-up.
Step 3 : Write the final answer.
1.
120
99
= 1,212 rounded to 3 decimal places
2. π = 3,1416 rounded to 4 decimal places
√
3. 3 = 1,7321 rounded to 4 decimal places
42
CHAPTER 6. IRRATIONAL NUMBERS AND ROUNDING OFF - GRADE 10
6.4
6.4
End of Chapter Exercises
1. Write the following rational numbers to 2 decimal places:
(a)
1
2
(b) 1
(c) 0,111111
(d) 0,999991
2. Write the following irrational numbers to 2 decimal places:
(a) 3,141592654 . . .
(b) 1,618 033 989 . . .
(c) 1,41421356 . . .
(d) 2,71828182845904523536 . . .
3. Use your calculator and write the following irrational numbers to 3 decimal places:
√
(a) 2
√
(b) 3
√
(c) 5
√
(d) 6
4. Use your calculator (where necessary) and write the following irrational numbers to 5
decimal places:
√
(a) 8
√
(b) 768
√
(c) 100
√
(d) 0,49
√
(e) 0,0016
√
(f) 0,25
√
(g) 36
√
(h) 1960
√
(i) 0,0036
√
(j) −8 0,04
√
(k) 5 80
5. Write the following irrational numbers to 3 decimal places and then write √
them as a rational
number to get an approximation
to
the
irrational
number.
For
example,
√ 3 = 1,73205 . . ..
√
732
183
= 1 250
. Therefore, 3 is approximately
To 3 decimal places, 3 = 1,732. 1,732 = 1 1000
183
.
1 250
(a) 3,141592654 . . .
(b) 1,618 033 989 . . .
(c) 1,41421356 . . .
(d) 2,71828182845904523536 . . .
43
6.4
CHAPTER 6. IRRATIONAL NUMBERS AND ROUNDING OFF - GRADE 10
44
Chapter 7
Number Patterns - Grade 10
In earlier grades you saw patterns in the form of pictures and numbers. In this chapter we learn
more about the mathematics of patterns.Patterns are recognisable regularities in situations such
as in nature, shapes, events, sets of numbers. For example, spirals on a pineapple, snowflakes,
geometric designs on quilts or tiles, the number sequence 0, 4, 8, 12, 16,....
Activity :: Investigation : Patterns
Can you spot any patterns in the following lists of numbers?
1.
2.
3.
4.
7.1
2;
1;
1;
5;
4; 6; 8; 10; . . .
2; 4; 7; 11; . . .
4; 9; 16; 25; . . .
10; 20; 40; 80; . . .
Common Number Patterns
Numbers can have interesting patterns. Here we list the most common patterns and how they
are made.
Examples:
1. 1, 4, 7, 10, 13, 16, 19, 22, 25, ...
This sequence has a difference of 3 between each number. The pattern is continued by
adding 3 to the last number each time.
2. 3, 8, 13, 18, 23, 28, 33, 38, ...
This sequence has a difference of 5 between each number. The pattern is continued by
adding 5 to the last number each time.
3. 2, 4, 8, 16, 32, 64, 128, 256, ...
This sequence has a factor of 2 between each number. The pattern is continued by
multiplying the last number by 2 each time.
4. 3, 9, 27, 81, 243, 729, 2187, ...
This sequence has a factor of 3 between each number. The pattern is continued by
multiplying the last number by 3 each time.
45
7.2
7.1.1
CHAPTER 7. NUMBER PATTERNS - GRADE 10
Special Sequences
Triangular Numbers
1, 3, 6, 10, 15, 21, 28, 36, 45, ...
This sequence is generated from a pattern of dots which form a triangle. By adding another row
of dots and counting all the dots we can find the next number of the sequence.
Square Numbers
1, 4, 9, 16, 25, 36, 49, 64, 81, ...
The next number is made by squaring where it is in the pattern. The second number is 2 squared
(22 or 2 × 2) The seventh number is 7 squared (72 or 7 × 7) etc
Cube Numbers
1, 8, 27, 64, 125, 216, 343, 512, 729, ...
The next number is made by cubing where it is in the pattern. The second number is 2 cubed
(23 or 2 × 2 × 2) The seventh number is 7 cubed (73 or 7 × 7 × 7) etc
Fibonacci Numbers
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
The next number is found by adding the two numbers before it together. The 2 is found by
adding the two numbers in front of it (1 + 1) The 21 is found by adding the two numbers in
front of it (8 + 13) The next number in the sequence above would be 55 (21 + 34)
Can you figure out the next few numbers?
7.2
Make your own Number Patterns
You can make your own number patterns using coins or matchsticks. Here is an example using
dots:
3
Pattern 1
4
2
b
1 dot
b
b b
b b
6 dots
b
b
b
b
b b
b
b
b
b
b
b
b
b
b
b
b
15 dots
b
b
b
b
b
b b b
b
b b
b
b
b b
b b
b b
b
b
b
b
b
b
b
b
28 dots
b
46
CHAPTER 7. NUMBER PATTERNS - GRADE 10
7.3
How many dots would you need for pattern 5 ? Can you make a formula that will tell you how
many coins are needed for any size pattern? For example if the pattern 20? The formula may
look something like
dots = pattern × pattern + ...
Worked Example 5: Study Table
Question: Say you and 3 friends decide to study for Maths, and you are seated at
a square table. A few minutes later, 2 other friends join you and would like to sit at
your table and help you study. Naturally, you move another table and add it to the
existing one. Now six of you sit at the table. Another two of your friends join your
table, and you take a third table and add it to the existing tables. Now 8 of you can
sit comfortably.
Figure 7.1: Two more people can be seated for each table added.
Examine how the number of people sitting is related to the number of tables.
Answer
Step 1 : Tabulate a few terms to see if there is a pattern
Number of Tables, n
1
2
3
4
..
.
n
Number of people seated
4=4
4+2=6
4+2+2=8
4 + 2 + 2 + 2 = 10
..
.
4 + 2 + 2 + 2 + ... + 2
Step 2 : Describe the pattern
We can see for 3 tables we can seat 8 people, for 4 tables we can seat 10 people and
so on. We started out with 4 people and added two the whole time. Thus, for each
table added, the number of persons increases by two.
7.3
Notation
A sequence does not have to follow a pattern but when it does we can often write down a formula
to calculate the nth -term, an . In the sequence
1; 4; 9; 16; 25; . . .
where the sequence consists of the squares of integers, the formula for the nth -term is
an = n 2
47
(7.1)
7.3
CHAPTER 7. NUMBER PATTERNS - GRADE 10
You can check this by looking at:
a1
=
12 = 1
a2
a3
=
=
22 = 4
32 = 9
a4
a5
=
=
42 = 16
52 = 25
...
Therefore, using (7.1), we can generate a pattern, namely squares of integers.
Worked Example 6: Study Table continued ....
Question: As before, you and 3 friends are studying for Maths, and you are seated
at a square table. A few minutes later, 2 other friends join you move another table
and add it to the existing one. Now six of you sit at the table. Another two of your
friends join your table, and you take a third table and add it to the existing tables.
Now 8 of you sit comfortably as illustrated:
Figure 7.2: Two more people can be seated for each table added.
Find the expression for the number of people seated at n tables. Then, use the
general formula to determine how many people can sit around 12 tables and how
many tables are needed for 20 people.
Answer
Step 1 : Tabulate a few terms to see if there is a pattern
Number of Tables, n
1
2
3
4
..
.
n
Number of people seated
4=4
4+2=6
4+2+2=8
4 + 2 + 2 + 2 = 10
..
.
Formula
= 4 + 2 · (0)
= 4 + 2 · (1)
= 4 + 2 · (2)
= 4 + 2 · (3)
..
.
4 + 2 + 2 + 2 + ...+ 2
= 4 + 2 · (n − 1)
Step 2 : Describe the pattern
The number of people seated at n tables is:
an = 4 + 2 · (n − 1)
Step 3 : Calculate the 12th term
Using the general formula (36.1) and considering the example from the previous
section, how many people can sit around, say, 12 tables? We are looking for a12 ,
that is, where n = 12:
an
=
a12
=
=
=
=
a1 + d · (n − 1)
4 + 2 · (12 − 1)
4 + 2(11)
4 + 22
26
48
CHAPTER 7. NUMBER PATTERNS - GRADE 10
7.3
Step 4 : Calculate the number of terms if an = 20
an
=
a1 + d · (n − 1)
20 =
20 − 4 =
4 + 2 · (n − 1)
2 · (n − 1)
16 ÷ 2 =
8+1 =
n
n−1
n
=
9
Step 5 : Final Answer
26 people can be seated at 12 tables and 9 tables are needed to seat 20 people.
It is also important to note the difference between n and an . n can be compared to a place
holder, while an is the value at the place “held” by n. Like our “Study Table”-example above,
the first table (Table 1) holds 4 people. Thus, at place n = 1, the value of a1 = 4, and so on:
n
an
1
4
2
6
3
8
4
10
...
...
Activity :: Investigation : General Formula
1. Find the general formula for the following sequences and then find a10 , a50 and
a100 :
(a) 2, 5, 8, 11, 14, . . .
(b) 0, 4, 8, 12, 16, . . .
(c) 2, −1, −4, −7, −10, . . .
2. The general term has been given for each sequence below. Work out the missing
terms.
(a) 0; 3; ...; 15; 24
(b) 3; 2; 1; 0; ...; 2
(c) 11; ...; 7; ...; 3
7.3.1
n2 − 1
−n + 4
−13 + 2n
Patterns and Conjecture
In mathematics, a conjecture is a mathematical statement which appears to be true, but has
not been formally proven to be true under the rules of mathematics. Other words that have a
similar in meaning to conjecture are: hypothesis, theory, assumption and premise.
For example: Make a conjecture about the next number based on the pattern 2; 6; 11; 17 : ...
The numbers increase by 4, 5, and 6.
Conjecture: The next number will increase by 7. So, it will be 17 + 7 or 24.
49
7.4
CHAPTER 7. NUMBER PATTERNS - GRADE 10
Worked Example 7: Number patterns
Question: Consider the following pattern.
12 + 1
=
22 + 2
32 + 3
=
=
42 + 4
=
22 − 2
32 − 3
42 − 4
52 − 5
1. Add another two rows to the end of the pattern.
2. Make a conjecture about this pattern. Write your conjecture in words.
3. Generalise your conjecture for this pattern (in other words, write your conjecture
algebraically).
4. Prove that your conjecture is true.
Answer
Step 1 : The next two rows
52 + 5 = 62 − 6
62 + 6 = 72 − 7
Step 2 : Conjecture
Squaring a number and adding the same number gives the same result as squaring
the next number and subtracting that number.
Step 3 : Generalise
We have chosen to use x here. You could choose any letter to generalise the pattern.
x2 + x = (x + 1)2 − (x + 1)
Step 4 : Proof
Lef t side : x2 + x
Right side : (x + 1)2 − (x + 1)
Right side
= x2 + 2x + 1 − x − 1
= x2 + x
= lef t side
T heref ore x2 + x = (x + 1)2 − (x + 1)
7.4
Exercises
1. Find the nth term for: 3, 7, 11, 15, . . .
2. Find the general term of the following sequences:
50
CHAPTER 7. NUMBER PATTERNS - GRADE 10
(a)
(b)
(c)
(d)
(e)
7.4
−2,1,4,7, . . .
11, 15, 19, 23, . . .
x − 1,2x + 5,5x + 1, . . .
sequence with a3 = 7 and a8 = 15
sequence with a4 = −8 and a10 = 10
3. The seating in a section of a sports stadium can be arranged so the first row has 15 seats,
the second row has 19 seats, the third row has 23 seats and so on. Calculate how many
seats are in the row 25.
4. Consider the following pattern:
22 + 2
2
3 +3
42 + 4
= 32 − 3
= 42 − 4
= 52 − 5
(a) Add at least two more rows to the pattern and check whether or not the pattern
continues to work.
(b) Describe in words any patterns that you have noticed.
(c) Try to generalise a rule using algebra i.e. find the general term for the pattern.
(d) Prove or disprove that this rule works for all values.
5. The profits of a small company for the last four years has been: R10 000, R15 000, R19 000
and R23 000. If the pattern continues, what is the expected profit in the 10 years (i.e. in
the 14th year of the company being in business)?
6. A single square is made from 4 matchsticks. Two squares in a row needs 7 matchsticks
and 3 squares in a row needs 10 matchsticks. Determine:
(a)
(b)
(c)
(d)
the first term
the common difference
the formula for the general term
how many matchsticks are in a row of 25 squares
7. You would like to start saving some money, but because you have never tried to save money
before, you have decided to start slowly. At the end of the first week you deposit R5 into
your bank account. Then at the end of the second week you deposit R10 into your bank
account. At the end of the third week you deposit R15. After how many weeks, do you
deposit R50 into your bank account?
8. A horizontal line intersects a piece of string at four points and divides it into five parts, as
shown below.
1
3
b
b
2
5
b
b
4
51
7.4
CHAPTER 7. NUMBER PATTERNS - GRADE 10
If the piece of string is intersected in this way by 19 parallel lines, each of which intersects
it at four points, find the number of parts into which the string will be divided.
52
Chapter 8
Finance - Grade 10
8.1
Introduction
Should you ever find yourself stuck with a mathematics question on a television quiz show, you
will probably wish you had remembered the how many even prime numbers there are between 1
and 100 for the sake of R1 000 000. And who does not want to be a millionaire, right?
Welcome to the Grade 10 Finance Chapter, where we apply maths skills to everyday financial
situations that you are likely to face both now and along your journey to purchasing your first
private jet.
If you master the techniques in this chapter, you will grasp the concept of compound interest,
and how it can ruin your fortunes if you have credit card debt, or make you millions if you
successfully invest your hard-earned money. You will also understand the effects of fluctuating
exchange rates, and its impact on your spending power during your overseas holidays!
8.2
Foreign Exchange Rates
Is $500 (”500 US dollars”) per person per night a good deal on a hotel in New York City? The
first question you will ask is “How much is that worth in Rands?”. A quick call to the local bank
or a search on the Internet (for example on http://www.x-rates.com/) for the Dollar/Rand
exchange rate will give you a basis for assessing the price.
A foreign exchange rate is nothing more than the price of one currency in terms of another.
For example, the exchange rate of 6,18 Rands/US Dollars means that $1 costs R6,18. In other
words, if you have $1 you could sell it for R6,18 - or if you wanted $1 you would have to pay
R6,18 for it.
But what drives exchange rates, and what causes exchange rates to change? And how does this
affect you anyway? This section looks at answering these questions.
8.2.1
How much is R1 really worth?
We can quote the price of a currency in terms of any other currency, but the US Dollar, British
Pounds Sterling or even the Euro are often used as a market standard. You will notice that the
financial news will report the South African Rand exchange rate in terms of these three major
currencies.
So the South African Rand could be quoted on a certain date as 6,7040 ZAR per USD (i.e.
$1,00 costs R6,7040), or 12,2374 ZAR per GBP. So if I wanted to spend $1 000 on a holiday
in the United States of America, this would cost me R6 704,00; and if I wanted £1 000 for a
weekend in London it would cost me R12 237,40.
This seems obvious, but let us see how we calculated that: The rate is given as ZAR per USD,
or ZAR/USD such that $1,00 buys R6,7040. Therefore, we need to multiply by 1 000 to get the
53
8.2
CHAPTER 8. FINANCE - GRADE 10
Table 8.1: Abbreviations and symbols for some common currencies.
Currency
Abbreviation Symbol
South African Rand
ZAR
R
United States Dollar
USD
$
British Pounds Sterling
GBP
£
Euro
EUR
e
number of Rands per $1 000.
Mathematically,
$1,00 =
∴
1 000 × $1,00 =
=
R6,0740
1 000 × R6,0740
R6 074,00
as expected.
What if you have saved R10 000 for spending money for the same trip and you wanted to use
this to buy USD? How much USD could you get for this? Our rate is in ZAR/USD but we want
to know how many USD we can get for our ZAR. This is easy. We know how much $1,00 costs
in terms of Rands.
$1,00 =
$1,00
=
∴
6,0740
1,00
$
=
6,0740
R1,00 =
=
R6,0740
R6,0740
6,0740
R1,00
1,00
6,0740
$0,164636
$
As we can see, the final answer is simply the reciprocal of the ZAR/USD rate. Therefore, R10 000
will get:
R1,00 =
∴
10 000 × R1,00 =
=
$
1,00
6,0740
10 000 × $
1,00
6,0740
$1 646,36
We can check the answer as follows:
∴
$1,00 =
1 646,36 × $1,00 =
=
R6,0740
1 646,36 × R6,0740
R10 000,00
Six of one and half a dozen of the other
So we have two different ways of expressing the same exchange rate: Rands per Dollar (ZAR/USD)
and Dollar per Rands (USD/ZAR). Both exchange rates mean the same thing and express the
value of one currency in terms of another. You can easily work out one from the other - they
are just the reciprocals of the other.
If the South African Rand is our Domestic (or home) Currency, we call the ZAR/USD rate a
“direct” rate, and we call a USD/ZAR rate an “indirect” rate.
In general, a direct rate is an exchange rate that is expressed as units of Home Currency per
54
CHAPTER 8. FINANCE - GRADE 10
8.2
units of Foreign Currency, i.e., Domestic Currency / Foreign Currency.
The Rand exchange rates that we see on the news are usually expressed as Direct Rates, for
example you might see:
Table 8.2: Examples of
Currency Abbreviation
1 USD
1 GBP
1 EUR
exchange rates
Exchange Rates
R6,9556
R13,6628
R9,1954
The exchange rate is just the price of each of the Foreign Currencies (USD, GBP and EUR) in
terms of our Domestic Currency, Rands.
An indirect rate is an exchange rate expressed as units of Foreign Currency per units of Home
Currency, i.e. Foreign Currency / Domestic Currency
Defining exchange rates as direct or indirect depends on which currency is defined as the Domestic
Currency. The Domestic Currency for an American investor would be USD which is the South
African investor’s Foreign Currency. So direct rates from the perspective of the American investor
(USD/ZAR) would be the same as the indirect rate from the perspective of the South Africa
investor.
Terminology
Since exchange rates are simple prices of currencies, movements in exchange rates means that
the price or value of the currency changed. The price of petrol changes all the time, so does the
price of gold, and currency prices also move up and down all the time.
If the Rand exchange rate moved from say R6,71 per USD to R6,50 per USD, what does this
mean? Well, it means that $1 would now cost only R6,50 instead of R6,71. The Dollar is now
cheaper to buy, and we say that the Dollar has depreciated (or weakened) against the Rand.
Alternatively we could say that the Rand has appreciated (or strengthened) against the Dollar.
What if we were looking at indirect exchange rates, and the exchange rate moved from $0,149
1
1
per ZAR (= 6,71
) to $0,1538 per ZAR (= 6,50
).
Well now we can see that the R1,00 cost $0,149 at the start, and then cost $0,1538 at the end.
The Rand has become more expensive (in terms of Dollars), and again we can say that the Rand
has appreciated.
Regardless of which exchange rate is used, we still come to the same conclusions.
In general,
• for direct exchange rates, the home currency will appreciate (depreciate) if the exchange
rate falls (rises)
• For indirect exchange rates, the home currency will appreciate (depreciate) if the exchange
rate rises (falls)
As with just about everything in this chapter, do not get caught up in memorising these formulae
- that is only going to get confusing. Think about what you have and what you want - and it
should be quite clear how to get the correct answer.
Activity :: Discussion : Foreign Exchange Rates
In groups of 5, discuss:
1. Why might we need to know exchange rates?
2. What happens if one countries currency falls drastically vs another countries
currency?
55
8.2
CHAPTER 8. FINANCE - GRADE 10
3. When might you use exchange rates?
8.2.2
Cross Currency Exchange Rates
We know that the exchange rates are the value of one currency expressed in terms of another
currency, and we can quote exchange rates against any other currency. The Rand exchange rates
we see on the news are usually expressed against the major currency, USD, GBP and EUR.
So if for example, the Rand exchange rates were given as 6,71 ZAR/USD and 12,71 ZAR/GBP,
does this tell us anything about the exchange rate between USD and GBP?
Well I know that if $1 will buy me R6,71, and if £1.00 will buy me R12,71, then surely the GBP
is stronger than the USD because you will get more Rands for one unit of the currency, and we
can work out the USD/GBP exchange rate as follows:
Before we plug in any numbers, how can we get a USD/GBP exchange rate from the ZAR/USD
and ZAR/GBP exchange rates?
Well,
USD/GBP = USD/ZAR × ZAR/GBP.
Note that the ZAR in the numerator will cancel out with the ZAR in the denominator, and we
are left with the USD/GBP exchange rate.
Although we do not have the USD/ZAR exchange rate, we know that this is just the reciprocal
of the ZAR/USD exchange rate.
USD/ZAR =
1
ZAR/USD
Now plugging in the numbers, we get:
USD/GBP =
=
=
=
USD/ZAR × ZAR/GBP
1
× ZAR/GBP
ZAR/USD
1
× 12,71
6,71
1,894
Important: Sometimes you will see exchange rates in the real world that do not appear to
work exactly like this. This is usually because some financial institutions add other costs
to the exchange rates, which alter the results. However, if you could remove the effect of
those extra costs, the numbers would balance again.
Worked Example 8: Cross Exchange Rates
Question: If $1 = R 6,40, and £1 = R11,58 what is the $/£ exchange rate (i.e.
the number of US$ per £)?
Answer
Step 1 : Determine what is given and what is required
The following are given:
• ZAR/USD rate = R6,40
• ZAR/GBP rate = R11,58
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CHAPTER 8. FINANCE - GRADE 10
8.2
The following is required:
• USD/GBP rate
Step 2 : Determine how to approach the problem
We know that:
USD/GBP = USD/ZAR × ZAR/GBP.
Step 3 : Solve the problem
USD/GBP =
=
=
=
USD/ZAR × ZAR/GBP
1
× ZAR/GBP
ZAR/USD
1
× 11,58
6,40
1,8094
Step 4 : Write the final answer
$1,8094 can be bought for £1.
Activity :: Investigation : Cross Exchange Rates - Alternate Method
If $1 = R 6,40, and £1 = R11,58 what is the $/£ exchange rate (i.e. the number
of US$ per £)?
Overview of problem
You need the $/£ exchange rate, in other words how many dollars must you pay
for a pound. So you need £1. From the given information we know that it would
cost you R11,58 to buy £1 and that $ 1 = R6,40.
Use this information to:
1. calculate how much R1 is worth in $.
2. calculate how much R11,58 is worth in $.
Do you get the same answer as in the worked example?
8.2.3
Enrichment: Fluctuating exchange rates
If everyone wants to buy houses in a certain suburb, then house prices are going to go up - because
the buyers will be competing to buy those houses. If there is a suburb where all residents want
to move out, then there are lots of sellers and this will cause house prices in the area to fall because the buyers would not have to struggle as much to find an eager seller.
This is all about supply and demand, which is a very important section in the study of Economics.
You can think about this is many different contexts, like stamp-collecting for example. If there
is a stamp that lots of people want (high demand) and few people own (low supply) then that
stamp is going to be expensive.
And if you are starting to wonder why this is relevant - think about currencies. If you are going
to visit London, then you have Rands but you need to “buy” Pounds. The exchange rate is the
price you have to pay to buy those Pounds.
Think about a time where lots of South Africans are visiting the United Kingdom, and other
South Africans are importing goods from the United Kingdom. That means there are lots of
Rands (high supply) trying to buy Pounds. Pounds will start to become more expensive (compare
this to the house price example at the start of this section if you are not convinced), and the
57
8.3
CHAPTER 8. FINANCE - GRADE 10
exchange rate will change. In other words, for R1 000 you will get fewer Pounds than you would
have before the exchange rate moved.
Another context which might be useful for you to understand this: consider what would happen
if people in other countries felt that South Africa was becoming a great place to live, and that
more people were wanting to invest in South Africa - whether in properties, businesses - or just
buying more goods from South Africa. There would be a greater demand for Rands - and the
“price of the Rand” would go up. In other words, people would need to use more Dollars, or
Pounds, or Euros ... to buy the same amount of Rands. This is seen as a movement in exchange
rates.
Although it really does come down to supply and demand, it is interesting to think about what
factors might affect the supply (people wanting to “sell” a particular currency) and the demand
(people trying to “buy” another currency). This is covered in detail in the study of Economics,
but let us look at some of the basic issues here.
There are various factors affect exchange rates, some of which have more economic rationale
than others:
• economic factors (such as inflation figures, interest rates, trade deficit information, monetary policy and fiscal policy)
• political factors (such as uncertain political environment, or political unrest)
• market sentiments and market behaviour (for example if foreign exchange markets perceived a currency to be overvalued and starting selling the currency, this would cause the
currency to fall in value - a self fulfilling expectation).
Exercise: Foreign Exchange
1. I want to buy an IPOD that costs £100, with the exchange rate currently at
£1 = R14. I believe the exchange rate will reach R12 in a month.
(a) How much will the MP3 player cost in Rands, if I buy it now?
(b) How much will I save if the exchange rate drops to R12?
(c) How much will I lose if the exchange rate moves to R15?
2. Study the following exchange rate table:
Country
Currency
United Kingdom (UK) Pounds(£)
United States (USA)
Dollars ($)
Exchange Rate
R14,13
R7,04
(a) In South Africa the cost of a new Honda Civic is R173 400. In England the
same vehicle costs £12 200 and in the USA $ 21 900. In which country is
the car the cheapest if you compare it to the South African Rand ?
(b) Sollie and Arinda are waiters in a South African Restaurant attracting many
tourists from abroad. Sollie gets a £6 tip from a tourist and Arinda gets
$ 12. How many South African Rand did each one get ?
8.3
Being Interested in Interest
If you had R1 000, you could either keep it in your wallet, or deposit it in a bank account. If it
stayed in your wallet, you could spend it any time you wanted. If the bank looked after it for
you, then they could spend it, with the plan of making profit off it. The bank usually “pays” you
to deposit it into an account, as a way of encouraging you to bank it with them, This payment
is like a reward, which provides you with a reason to leave it with the bank for a while, rather
than keeping the money in your wallet.
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CHAPTER 8. FINANCE - GRADE 10
8.4
We call this reward ”interest”.
If you deposit money into a bank account, you are effectively lending money to the bank - and
you can expect to receive interest in return. Similarly, if you borrow money from a bank (or from
a department store, or a car dealership, for example) then you can expect to have to pay interest
on the loan. That is the price of borrowing money.
The concept is simple, yet it is core to the world of finance. Accountants, actuaries and bankers,
for example, could spend their entire working career dealing with the effects of interest on
financial matters.
In this chapter you will be introduced to the concept of financial mathematics - and given the
tools to cope with even advanced concepts and problems.
Important: Interest
The concepts in this chapter are simple - we are just looking at the same idea, but from many
different angles. The best way to learn from this chapter is to do the examples yourself, as you
work your way through. Do not just take our word for it!
8.4
Simple Interest
Definition: Simple Interest
Simple interest is where you earn interest on the initial amount that you invested, but not
interest on interest.
As an easy example of simple interest, consider how much you will get by investing R1 000 for
1 year with a bank that pays you 5% simple interest. At the end of the year, you will get an
interest of:
Interest = R1 000 × 5%
5
= R1 000 ×
100
= R1 000 × 0,05
= R50
So, with an “opening balance” of R1 000 at the start of the year, your “closing balance” at the
end of the year will therefore be:
Closing Balance =
=
=
Opening Balance + Interest
R1 000 + R50
R1 050
We sometimes call the opening balance in financial calculations Principal, which is abbreviated
as P (R1 000 in the example). The interest rate is usually labelled i (5% in the example), and
the interest amount (in Rand terms) is labelled I (R50 in the example).
So we can see that:
I =P ×i
and
Closing Balance =
=
=
=
Opening Balance + Interest
P +I
P + (P × i)
P (1 + i)
59
(8.1)
8.4
CHAPTER 8. FINANCE - GRADE 10
This is how you calculate simple interest. It is not a complicated formula, which is just as well
because you are going to see a lot of it!
Not Just One
You might be wondering to yourself:
1. how much interest will you be paid if you only leave the money in the account for 3 months,
or
2. what if you leave it there for 3 years?
It is actually quite simple - which is why they call it Simple Interest.
1. Three months is 1/4 of a year, so you would only get 1/4 of a full year’s interest, which
is: 1/4 × (P × i). The closing balance would therefore be:
Closing Balance = P + 1/4 × (P × i)
= P (1 + (1/4)i)
2. For 3 years, you would get three years’ worth of interest, being: 3 × (P × i). The closing
balance at the end of the three year period would be:
Closing Balance =
=
P + 3 × (P × i)
P × (1 + (3)i)
If you look carefully at the similarities between the two answers above, we can generalise the
result. In other words, if you invest your money (P ) in an account which pays a rate of interest
(i) for a period of time (n years), then, using the symbol (A) for the Closing Balance:
Closing Balance,(A) = P (1 + i · n)
(8.2)
As we have seen, this works when n is a fraction of a year and also when n covers several years.
Important: Interest Calculation
Annual Rates means Yearly rates. and p.a.(per annum) = per year
Worked Example 9: Simple Interest
Question: If I deposit R1 000 into a special bank account which pays a Simple
Interest of 7% for 3 years, how much will I get back at the end?
Answer
Step 1 : Determine what is given and what is required
• opening balance, P = R1 000
• interest rate, i = 7%
• period of time, n = 3 years
We are required to find the closing balance (A).
Step 2 : Determine how to approach the problem
We know from (8.2) that:
Closing Balance,(A) = P (1 + i · n)
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CHAPTER 8. FINANCE - GRADE 10
8.4
Step 3 : Solve the problem
A =
=
=
P (1 + i · n)
R1 000(1 + 3 × 7%)
R1 210
Step 4 : Write the final answer
The closing balance after 3 years of saving R1 000 at an interest rate of 7% is R1 210.
Worked Example 10: Calculating n
Question: If I deposit R30 000 into a special bank account which pays a Simple
Interest of 7.5% ,for how many years must I invest this amount to generate R45 000
Answer
Step 1 : Determine what is given and what is required
• opening balance, P = R30 000
• interest rate, i = 7,5%
• closing balance, A = R45 000
We are required to find the number of years.
Step 2 : Determine how to approach the problem
We know from (8.2) that:
Closing Balance (A) = P (1 + i · n)
Step 3 : Solve the problem
Closing Balance (A) =
R45 000 =
(1 + 0,075 × n) =
0,075 × n
=
n
=
n
=
P (1 + i · n)
R30 000(1 + n × 7,5%)
45000
30000
1,5 − 1
0,5
0,075
6,6666667
Step 4 : Write the final answer
n has to be a whole number, therefore n = 7.
The period is 7 years for R30 000 to generate R45 000 at a simple interest rate of
7,5%.
8.4.1
Other Applications of the Simple Interest Formula
61
8.4
CHAPTER 8. FINANCE - GRADE 10
Worked Example 11: Hire-Purchase
Question: Troy is keen to buy an addisional hard drive for his laptop advertised for
R 2 500 on the internet. There is an option of paying a 10% deposit then making
24 monthly payments using a hire-purchase agreement where interest is calculated
at 7,5% p.a. simple interest. Calculate what Troy’s monthly payments will be.
Answer
Step 1 : Determine what is given and what is required
A new opening balance is required, as the 10% deposit is paid in cash.
•
•
•
•
10% of R 2 500 = R250
new opening balance, P = R2 500 − R250 = R2 250
interest rate, i = 7,5% = 0,075pa
period of time, n = 2 years
We are required to find the closing balance (A) and then the montly payments.
Step 2 : Determine how to approach the problem
We know from (8.2) that:
Closing Balance,(A) = P (1 + i · n)
Step 3 : Solve the problem
A =
=
=
Monthly payment =
=
P (1 + i · n)
R2 250(1 + 2 × 7,5%)
R2 587,50
2587,50 ÷ 24
R107,81
Step 4 : Write the final answer
Troy’s monthly payments = R 107,81
Worked Example 12: Depreciation
Question: Seven years ago, Tjad’s drum kit cost him R12 500. It has now been
valued at R2 300. What rate of simple depreciation does this represent ?
Answer
Step 1 : Determine what is given and what is required
• opening balance, P = R12 500
• period of time, n = 7 years
• closing balance, A = R2 300
We are required to find the rate(i).
Step 2 : Determine how to approach the problem
We know from (8.2) that:
Closing Balance,(A) = P (1 + i · n)
Therefore, for depreciation the formula will change to:
Closing Balance,(A) = P (1 − i · n)
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CHAPTER 8. FINANCE - GRADE 10
8.5
Step 3 : Solve the problem
A = P (1 − i · n)
R2 300 = R12 500(1 − 7 × i)
i
= 0,11657...
Step 4 : Write the final answer
Therefore the rate of depreciation is 11,66%
Exercise: Simple Interest
1. An amount of R3 500 is invested in a savings account which pays simple interest
at a rate of 7,5% per annum. Calculate the balance accumulated by the end
of 2 years.
2. Calculate the simple interest for the following problems.
(a) A loan of R300 at a rate of 8% for l year.
(b) An investment of R225 at a rate of 12,5% for 6 years.
3. I made a deposit of R5 000 in the bank for my 5 year old son’s 21st birthday.
I have given him the amount of R 18 000 on his birtday. At what rate was the
money invested, if simple interest was calculated ?
4. Bongani buys a dining room table costing R 8 500 on Hire Purchase. He is
charged simple interest at 17,5% per annum over 3 years.
(a) How much will Bongani pay in total ?
(b) How much interest does he pay ?
(c) What is his montly installment ?
8.5
Compound Interest
To explain the concept of compound interest, the following example is discussed:
Worked Example 13:
Using Simple Interest to lead to the concept Com-
pound Interest
Question: If I deposit R1 000 into a special bank account which pays a Simple
Interest of 7%. What if I empty the bank account after a year, and then take the
principal and the interest and invest it back into the same account again. Then I
take it all out at the end of the second year, and then put it all back in again? And
then I take it all out at the end of 3 years?
Answer
Step 1 : Determine what is given and what is required
• opening balance, P = R1 000
• interest rate, i = 7%
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8.5
CHAPTER 8. FINANCE - GRADE 10
• period of time, 1 year at a time, for 3 years
We are required to find the closing balance at the end of three years.
Step 2 : Determine how to approach the problem
We know that:
Closing Balance = P (1 + i · n)
Step 3 : Determine the closing balance at the end of the first year
Closing Balance = P (1 + i · n)
= R1 000(1 + 1 × 7%)
= R1 070
Step 4 : Determine the closing balance at the end of the second year
After the first year, we withdraw all the money and re-deposit it. The opening
balance for the second year is therefore R1 070, because this is the balance after the
first year.
Closing Balance = P (1 + i · n)
= R1 070(1 + 1 × 7%)
= R1 144,90
Step 5 : Determine the closing balance at the end of the third year
After the second year, we withdraw all the money and re-deposit it. The opening
balance for the third year is therefore R1 144,90, because this is the balance after
the first year.
Closing Balance = P (1 + i · n)
= R1 144,90(1 + 1 × 7%)
= R1 225,04
Step 6 : Write the final answer
The closing balance after withdrawing the all the money and re-depositing each year
for 3 years of saving R1 000 at an interest rate of 7% is R1 225,04.
In the two worked examples using simple interest, we have basically the same problem because
P =R1 000, i=7% and n=3 years for both problems. Except in the second situation, we end up
with R1 225,04 which is more than R1 210 from the first example. What has changed?
In the first example I earned R70 interest each year - the same in the first, second and third year.
But in the second situation, when I took the money out and then re-invested it, I was actually
earning interest in the second year on my interest (R70) from the first year. (And interest on
the interest on my interest in the third year!)
This more realistically reflects what happens in the real world, and is known as Compound
Interest. It is this concept which underlies just about everything we do - so we will look at more
closely next.
Definition: Compound Interest
Compound interest is the interest payable on the principal and its accumulated interest.
Compound interest is a double edged sword, though - great if you are earning interest on cash
you have invested, but crippling if you are stuck having to pay interest on money you have
borrowed!
In the same way that we developed a formula for Simple Interest, let us find one for Compound
Interest.
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CHAPTER 8. FINANCE - GRADE 10
8.5
If our opening balance is P and we have an interest rate of i then, the closing balance at the
end of the first year is:
Closing Balance after 1 year = P (1 + i)
This is the same as Simple Interest because it only covers a single year. Then, if we take that
out and re-invest it for another year - just as you saw us doing in the worked example above then the balance after the second year will be:
Closing Balance after 2 years =
=
[P (1 + i)] × (1 + i)
P (1 + i)2
And if we take that money out, then invest it for another year, the balance becomes:
= [P (1 + i)2 ] × (1 + i)
= P (1 + i)3
Closing Balance after 3 years
We can see that the power of the term (1 + i) is the same as the number of years. Therefore,
Closing Balance after n years = P (1 + i)n
8.5.1
(8.3)
Fractions add up to the Whole
It is easy to show that this formula works even when n is a fraction of a year. For example, let
us invest the money for 1 month, then for 4 months, then for 7 months.
Closing Balance after 1 month =
Closing Balance after 5 months =
=
=
=
Closing Balance after 12 months =
1
P (1 + i) 12
Closing Balance after 1 month invested for 4 months more
4
1
[P (1 + i) 12 ] 12
1
4
P (1 + i) 12 + 12
5
P (1 + i) 12
Closing Balance after 5 month invested for 7 months more
5
7
=
[P (1 + i) 12 ] 12
=
P (1 + i) 12 + 12
=
=
P (1 + i) 12
P (1 + i)1
5
7
12
which is the same as investing the money for a year.
Look carefully at the long equation above. It is not as complicated as it looks! All we are doing
is taking the opening amount (P ), then adding interest for just 1 month. Then we are taking
that new balance and adding interest for a further 4 months, and then finally we are taking the
new balance after a total of 5 months, and adding interest for 7 more months. Take a look
again, and check how easy it really is.
Does the final formula look familiar? Correct - it is the same result as you would get for simply
investing P for one full year. This is exactly what we would expect, because:
1 month + 4 months + 7 months = 12 months,
which is a year. Can you see that? Do not move on until you have understood this point.
8.5.2
The Power of Compound Interest
To see how important this “interest on interest” is, we shall compare the difference in closing
balances for money earning simple interest and money earning compound interest. Consider an
amount of R10 000 that you have to invest for 10 years, and assume we can earn interest of 9%.
How much would that be worth after 10 years?
65
8.5
CHAPTER 8. FINANCE - GRADE 10
The closing balance for the money earning simple interest is:
Closing Balance =
=
=
P (1 + i · n)
R10 000(1 + 9% × 10)
R19 000
The closing balance for the money earning compound interest is:
Closing Balance =
=
=
P (1 + i)n )
R10 000(1 + 9%)10
R23 673,64
So next time someone talks about the “magic of compound interest”, not only will you know
what they mean - but you will be able to prove it mathematically yourself!
Again, keep in mind that this is good news and bad news. When you are earning interest on
money you have invested, compound interest helps that amount to increase exponentially. But
if you have borrowed money, the build up of the amount you owe will grow exponentially too.
Worked Example 14: Taking out a Loan
Question: Mr Lowe wants to take out a loan of R 350 000. He does not want
to pay back more than R625 000 altogether on the loan. If the interest rate he is
offered is 13%, over what period should he take the loan
Answer
Step 1 : Determine what has been provided and what is required
• opening balance, P = R350 000
• closing balance, A = R625 000
• interest rate, i = 13% peryear
We are required to find the time period(n).
Step 2 : Determine how to approach the problem
We know from (8.3) that:
Closing Balance,(A) = P (1 + i)n
We need to find n.
Therefore we covert the formula to:
A
= (1 + i)n
P
and then find n by trial and error.
Step 3 : Solve the problem
A
= (1 + i)n
P
625000
= (1 + 0,13)n
350000
1,785... = (1,13)n
Try n = 3 :
(1,13)3 = 1,44...
Try n = 4 :
Try n = 5 :
(1,13)4 = 1,63...
(1,13)5 = 1,84...
Step 4 : Write the final answer
Mr Lowe should take the loan over four years
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CHAPTER 8. FINANCE - GRADE 10
8.5.3
8.5
Other Applications of Compound Growth
Worked Example 15: Population Growth
Question: South Africa’s population is increasing by 2,5% per year. If the current
population is 43 million, how many more people will there be in South Africa in two
year’s time ?
Answer
Step 1 : Determine what has been provided and what is required
• opening balance, P = 43 000 000
• period of time, n = 2 year
• interest rate, i = 2,5% peryear
We are required to find the closing balance(A).
Step 2 : Determine how to approach the problem
We know from (8.3) that:
Closing Balance,(A) = P (1 + i)n
Step 3 : Solve the problem
A =
=
=
P (1 + i)n
43 000 000(1 + 0,025)2
45 176 875
Step 4 : Write the final answer
There are 45 176 875 − 43 000 000 = 2 176 875 more people in 2 year’s time
Worked Example 16: Compound Decrease
Question: A swimming pool is being treated for a build-up of algae. Initially, 50m2
of the pool is covered by algae. With each day of treatment, the algae reduces by
5%. What area is covered by algae after 30 days of treatment ?
Answer
Step 1 : Determine what has been provided and what is required
• opening balance, P = 50m2
• period of time, n = 30 days
• interest rate, i = 5% perday
We are required to find the closing balance(A).
Step 2 : Determine how to approach the problem
We know from (8.3) that:
Closing Balance,(A) = P (1 + i)n
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8.6
CHAPTER 8. FINANCE - GRADE 10
But this is compound decrease so we can use the formula:
Closing Balance,(A) = P (1 − i)n
Step 3 : Solve the problem
A =
=
=
P (1 − i)n
50(1 − 0,05)30
10,73m2
Step 4 : Write the final answer
Therefore the area still covered with algae is 10,73m2
Exercise: Compound Interest
1. An amount of R3 500 is invested in a savings account which pays compound
interest at a rate of 7,5% per annum. Calculate the balance accumulated by
the end of 2 years.
2. If the average rate of inflation for the past few years was 7,3% and your water
and electricity account is R 1 425 on average, what would you expect to pay
in 6 years time ?
3. Shrek wants to invest some money at 11% per annum compound interest. How
much money (to the nearest rand) should he invest if he wants to reach a sum
of R 100 000 in five year’s time ?
8.6
Summary
As an easy reference, here are the key formulae that we derived and used during this chapter.
While memorising them is nice (there are not many), it is the application that is useful. Financial
experts are not paid a salary in order to recite formulae, they are paid a salary to use the right
methods to solve financial problems.
8.6.1
P
i
n
Definitions
Principal (the amount of money at the starting point of the calculation)
interest rate, normally the effective rate per annum
period for which the investment is made
8.6.2
Equations

Closing Balance - simple interest 
Solve for i
= P (1 + i · n)

Solve for n
68
CHAPTER 8. FINANCE - GRADE 10
8.7

Closing Balance - compound interest 
Solve for i
= P (1 + i)n

Solve for n
Important: Always keep the interest and the time period in the same units of time (e.g.
both in years, or both in months etc.).
8.7
End of Chapter Exercises
1. You are going on holiday to Europe. Your hotel will cost e200 per night. How much will
you need in Rands to cover your hotel bill, if the exchange rate is e1 = R9,20.
2. Calculate how much you will earn if you invested R500 for 1 year at the following interest
rates:
(a) 6,85% simple interest
(b) 4,00% compound interest
3. Bianca has R1 450 to invest for 3 years. Bank A offers a savings account which pays
simple interest at a rate of 11% per annum, whereas Bank B offers a savings account
paying compound interest at a rate of 10,5% per annum. Which account would leave
Bianca with the highest accumulated balance at the end of the 3 year period?
4. How much simple interest is payable on a loan of R2 000 for a year, if the interest rate is
10%?
5. How much compound interest is payable on a loan of R2 000 for a year, if the interest rate
is 10%?
6. Discuss:
(a) Which type of interest would you like to use if you are the borrower?
(b) Which type of interest would you like to use if you were the banker?
7. Calculate the compound interest for the following problems.
(a) A R2 000 loan for 2 years at 5%.
(b) A R1 500 investment for 3 years at 6%.
(c) An R800 loan for l year at 16%.
8. If the exchange rate 100 Yen = R 6,2287 and 1 AUD = R 5,1094 , determine the exchange
rate between the Australian Dollar and the Japanese Yen.
9. Bonnie bought a stove for R 3 750. After 3 years she paid for it and the R 956,25 interest
that was charged for hire-purchase. Determine the simple rate of interest that was charged.
69
8.7
CHAPTER 8. FINANCE - GRADE 10
70
Chapter 9
Products and Factors - Grade 10
9.1
Introduction
In this chapter you will learn how to work with algebraic expressions. You will recap some of
the work on factorisation and multiplying out expressions that you learnt in earlier grades. This
work will then be extended upon for Grade 10.
9.2
Recap of Earlier Work
The following should be familiar. Examples are given as reminders.
9.2.1
Parts of an Expression
Mathematical expressions are just like sentences and their parts have special names. You should
be familiar with the following names used to describe the parts of an mathematical expression.
a · xk + b · x + cm = 0
p
d·y +e·y+f ≤0
Name
term
expression
coefficient
exponent (or index)
base
constant
variable
equation
inequality
binomial
trinomial
9.2.2
(9.1)
(9.2)
Examples (separated by commas)
a · xk ,b · x, cm , d · y p , e · y, f
a · xk + b · x + cm , d · y p + e · y + f
a, b, d, e
k, p
x, y, c
a, b, c, d, e, f
x, y
a · xk + b · x + cm = 0
d · yp + e · y + f ≤ 0
expression with two terms
expression with three terms
Product of Two Binomials
A binomial is a mathematical expression with two terms, e.g. (ax + b) and (cx + d). If these
two binomials are multiplied, the following is the result:
71
9.2
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
(a · x + b)(c · x + d)
= (ax)(c · x + d) + b(c · x + d)
= (ax)(cx) + (ax)d + b(cx) + b · d
Worked Example 17: Product of two Binomials
Question: Find the product of (3x − 2)(5x + 8)
Answer
(3x − 2)(5x + 8) =
=
(3x)(5x) + (3x)(8) + (−2)(5x) + (−2)(8)
15x2 + 24x − 10x − 16
15x2 + 14x − 16
=
The product of two identical binomials is known as the square of the binomials and is written
as:
(ax + b)2 = a2 x2 + 2abx + b2
If the two terms are ax + b and ax − b then their product is:
(ax + b)(ax − b) = a2 x2 − b2
This is known as the difference of squares.
9.2.3
Factorisation
Factorisation is the opposite of expanding brackets. For example expanding brackets would
require 2(x + 1) to be written as 2x + 2. Factorisation would be to start with 2x + 2 and to end
up with 2(x + 1). In previous grades you factorised based on common factors and on difference
of squares.
Common Factors
Factorising based on common factors relies on there being common factors between your terms.
For example, 2x − 6x2 can be factorised as follows:
2x − 6x2 = 2x(1 − 3x)
Activity :: Investigation : Common Factors
Find the highest common factors of the following pairs of terms:
(a) 6y; 18x
(f) 2xy; 4xyz
(b) 12mn; 8n
(g) 3uv; 6u
(c) 3st; 4su
(h) 9xy; 15xz
72
(d) 18kl; 9kp
(i) 24xyz; 16yz
(e) abc; ac
(j) 3m; 45n
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
9.2
Difference of Squares
We have seen that:
(ax + b)(ax − b) = a2 x2 − b2
(9.3)
Since 9.3 is an equation, both sides are always equal. This means that an expression of the form:
a2 x2 − b2
can be factorised to
(ax + b)(ax − b)
Therefore,
a2 x2 − b2 = (ax + b)(ax − b)
For example, x2 − 16 can be written as (x2 − 42 ) which is a difference of squares. Therefore the
factors of x2 − 16 are (x − 4) and (x + 4).
Worked Example 18: Factorisation
Question: Factorise completely: b2 y 5 − 3aby 3
Answer
b2 y 5 − 3aby 3
=
by 3 (by 2 − 3a)
Worked Example 19: Factorising binomials with a common bracket
Question: Factorise completely: 3a(a − 4) − 7(a − 4)
Answer
Step 1 : bracket (a − 4) is the common factor
3a(a − 4) − 7(a − 4) =
(a − 4)(3a − 7)
Worked Example 20: Factorising using a switch around in brackets
Question: Factorise 5(a − 2) − b(2 − a)
Answer
Step 1 : Note that (2 − a) = −(a − 2)
5(a − 2) − b(2 − a)
= 5(a − 2) − [−b(a − 2)]
= 5(a − 2) + b(a − 2)
= (a − 2)(5 + b)
73
9.3
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
Exercise: Recap
1. Find the products of:
(a) 2y(y + 4)
(d) (y + 8)(y + 4)
(b) (y + 5)(y + 2)
(e) (2y + 9)(3y + 1)
(c) (y + 2)(2y + 1)
(f) (3y − 2)(y + 6)
(b) 20a − 10
(e) 16k 2 − 4k
(h) −2ab − 8a
(k) 12k 2 j + 24k 2 j 2
(n) a(a − 1) − 5(a − 1)
(q) 3b(b − 4) − 7(4 − b)
(c) 18ab − 3bc
(f) 3a2 + 6a − 18
(i) 24kj − 16k 2 j
(l) 72b2q − 18b3 q 2
(o) bm(b+4)−6m(b+4)
(r) a2 b2 c2 − 1
2. Factorise:
(a)
(b)
(c)
(d)
(e)
2l + 2w
12x + 32y
6x2 + 2x + 10x3
2xy 2 + xy 2 z + 3xy
−2ab2 − 4a2 b
3. Factorise completely:
(a) 7a + 4
(d) 12kj + 18kq
(g) −6a − 24
(j) −a2 b − b2 a
(m) 4(y − 3) + k(3 − y)
(p) a2 (a + 7) + a(a + 7)
9.3
More Products
We have seen how to multiply two binomials in section 9.2.2. In this section we learn how to
multiply a binomial (expression with two terms) by a trinomial (expression with three terms).
Fortunately, we use the same methods we used to multiply two binomials to multiply a binomial
and a trinomial.
For example, multiply 2x + 1 by x2 + 2x + 1.
=
(2x + 1)(x2 + 2x + 1)
2x(x2 + 2x + 1) + 1(x2 + 2x + 1) (apply distributive law)
=
=
[2x(x2 ) + 2x(2x) + 2x(1)] + [1(x2 ) + 1(2x) + 1(1)]
4x3 + 4x2 + 2x + x2 + 2x + 1 (expand the brackets)
=
=
4x3 + (4x2 + x2 ) + (2x + 2x) + 1 (group like terms to simplify)
4x3 + 5x2 + 4x + 1 (simplify to get final answer)
Important: Multiplication of Binomial with Trinomial
If the binomial is A + B and the trinomial is C + D + E, then the very first step is to apply the
distributive law:
(A + B)(C + D + E) = A(C + D + E) + B(C + D + E)
If you remember this, you will never go wrong!
74
(9.4)
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
Worked Example 21: Multiplication of Binomial with Trinomial
Question: Multiply x − 1 with x2 − 2x + 1.
Answer
Step 1 : Determine what is given and what is required
We are given two expressions: a binomial, x − 1, and a trinomial, x2 − 2x + 1. We
need to multiply them together.
Step 2 : Determine how to approach the problem
Apply the distributive law and then simplify the resulting expression.
Step 3 : Solve the problem
(x − 1)(x2 − 2x + 1)
x(x2 − 2x + 1) − 1(x2 − 2x + 1) (apply distributive law)
[x(x2 ) + x(−2x) + x(1)] + [−1(x2 ) − 1(−2x) − 1(1)]
=
=
x3 − 2x2 + x − x2 + 2x − 1 (expand the brackets)
x3 + (−2x2 − x2 ) + (x + 2x) − 1 (group like terms to simplify)
=
=
x3 − 3x2 + 3x − 1
=
(simplify to get final answer)
Step 4 : Write the final answer
The product of x − 1 and x2 − 2x + 1 is x3 − 3x2 + 3x − 1.
Worked Example 22: Sum of Cubes
Question: Find the product of x + y and x2 − xy + y 2 .
Answer
Step 1 : Determine what is given and what is required
We are given two expressions: a binomial, x + y, and a trinomial, x2 − xy + y 2 . We
need to multiply them together.
Step 2 : Determine how to approach the problem
Apply the distributive law and then simplify the resulting expression.
Step 3 : Solve the problem
(x + y)(x2 − xy + y 2 )
=
=
x(x2 − xy + y 2 ) + y(x2 − xy + y 2 ) (apply distributive law)
[x(x2 ) + x(−xy) + x(y 2 )] + [y(x2 ) + y(−xy) + y(y 2 )]
=
x3 − x2 y + xy 2 + yx2 − xy 2 + y 3
=
=
3
2
2
2
(expand the brackets)
2
x + (−x y + yx ) + (xy − xy ) + y 3
x3 + y 3 (simplify to get final answer)
Step 4 : Write the final answer
The product of x + y and x2 − xy + y 2 is x3 + y 3 .
75
(group like terms to simplify)
9.3
9.4
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
Important: We have seen that:
(x + y)(x2 − xy + y 2 ) = x3 + y 3
This is known as a sum of cubes.
Activity :: Investigation : Difference of Cubes
Show that the difference of cubes (x3 − y 3 ) is given by the product of x − y and
2
x + xy + y 2 .
Exercise: Products
1. Find the products of:
(a) (−2y 2 − 4y + 11)(5y − 12)
(c) (4y 2 + 12y + 10)(−9y 2 + 8y + 2)
(e) (10y 5 + 3)(−2y 2 − 11y + 2)
(g) (−10)(2y 2 + 8y + 3)
(i) (6y 7 − 8y 2 + 7)(−4y − 3)(−6y 2 − 7y − 11)
(k) (8y 5 + 3y 4 + 2y 3 )(5y + 10)(12y 2 + 6y + 6)
(m) (4y 3 + 5y 2 − 12y)(−12y − 2)(7y 2 − 9y + 12)
(o) (9)(8y 2 − 2y + 3)
(q) (−6y 4 + 11y 2 + 3y)(10y + 4)(4y − 4)
(s) (−11y 5 + 11y 4 + 11)(9y 3 − 7y 2 − 4y + 6)
(b) (−11y + 3)(−10y 2 − 7y − 9)
(d) (7y 2 − 6y − 8)(−2y + 2)
(f) (−12y − 3)(12y 2 − 11y + 3)
(h) (2y 6 + 3y 5 )(−5y − 12)
(j) (−9y 2 + 11y + 2)(8y 2 + 6y − 7)
(l) (−7y + 11)(−12y + 3)
(n) (7y + 3)(7y 2 + 3y + 10)
(p) (−12y + 12)(4y 2 − 11y + 11)
(r) (−3y 6 − 6y 3 )(11y − 6)(10y − 10)
(t) (−3y + 8)(−4y 3 + 8y 2 − 2y + 12)
2. Remove the brackets and simplify:(2h + 3)(4h2 − 6h + 9)
9.4
Factorising a Quadratic
Finding the factors of a quadratic is quite easy, and some are easier than others.
The simplest quadratic has the form ax2 , which factorises to (x)(ax). For example, 25x2
factorises to (5x)(5x) and 2x2 factorises to (2x)(x).
The second simplest quadratic is of the form ax2 + bx. We can see here that x is a common
factor of both terms. Therefore, ax2 + bx factorises to x(ax + b). For example, 8y 2 + 4y
factorises to 4y(2y + 1).
The third simplest quadratic is made up of the difference of squares. We know that:
(a + b)(a − b) = a2 − b2 .
This is true for any values of a and b, and more importantly since it is an equality, we can also
write:
a2 − b2 = (a + b)(a − b).
This means that if we ever come across a quadratic that is made up of a difference of squares,
we can immediately write down what the factors are.
76
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
9.4
Worked Example 23: Difference of Squares
Question: Find the factors of 9x2 − 25.
Answer
Step 1 : Examine the quadratic
We see that the quadratic is a difference of squares because:
(3x)2 = 9x2
and
52 = 25.
Step 2 : Write the quadratic as the difference of squares
9x2 − 25 = (3x)2 − 52
Step 3 : Write the factors
(3x)2 − 52 = (3x − 5)(3x + 5)
Step 4 : Write the final answer
The factors of 9x2 − 25 are (3x − 5)(3x + 5).
The three types of quadratic that we have seen are very simple to factorise. However, many
quadratics do not fall into these categories, and we need a more general method to factorise
quadratics like x2 − x − 2?
We can learn about how to factorise quadratics by looking at how two binomials are multiplied
to get a quadratic. For example, (x + 2)(x + 3) is multiplied out as:
(x + 2)(x + 3) = x(x + 3) + 2(x + 3)
= (x)(x) + 3x + 2x + (2)(3)
= x2 + 5x + 6.
We see that the x2 term in the quadratic is the product of the x-terms in each bracket. Similarly,
the 6 in the quadratic is the product of the 2 and 3 the brackets. Finally, the middle term is the
sum of two terms.
So, how do we use this information to factorise the quadratic?
Let us start with factorising x2 + 5x+ 6 and see if we can decide upon some general rules. Firstly,
write down two brackets with an x in each bracket and space for the remaining terms.
( x
)(
x
)
Next decide upon the factors of 6. Since the 6 is positive, these are:
Factors of 6
1
6
2
3
-1
-6
-2
-3
Therefore, we have four possibilities:
Option 1
(x + 1)(x + 6)
Option 2
(x − 1)(x − 6)
Option 3
(x + 2)(x + 3)
77
Option 4
(x − 2)(x − 3)
9.4
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
Next we expand each set of brackets to see which option gives us the correct middle term.
Option 1
(x + 1)(x + 6)
x2 + 7x + 6
Option 2
(x − 1)(x − 6)
x2 − 7x + 6
Option 3
(x + 2)(x + 3)
x2 + 5x + 6
Option 4
(x − 2)(x − 3)
x2 − 5x + 6
We see that Option 3 (x+2)(x+3) is the correct solution. As you have seen that the process
of factorising a quadratic is mostly trial and error, however the is some information that can be
used to simplify the process.
Method: Factorising a Quadratic
1. First divide the entire equation by any common factor of the coefficients, so as to obtain
an equation of the form ax2 + bx + c = 0 where a, b and c have no common factors and
a is positive.
2. Write down two brackets with an x in each bracket and space for the remaining terms.
( x
)( x
)
(9.5)
3. Write down a set of factors for a and c.
4. Write down a set of options for the possible factors for the quadratic using the factors of
a and c.
5. Expand all options to see which one gives you the correct answer.
There are some tips that you can keep in mind:
• If c is positive, then the factors of c must be either both positive or both negative. The
factors are both negative if b is negative, and are both positive if b is positive. If c is
negative, it means only one of the factors of c is negative, the other one being positive.
• Once you get an answer, multiply out your brackets again just to make sure it really works.
Worked Example 24: Factorising a Quadratic
Question: Find the factors of 3x2 + 2x − 1.
Answer
Step 1 : Check whether the quadratic is in the form ax2 + bx + c = 0 with a
positive.
The quadratic is in the required form.
Step 2 : Write down two brackets with an x in each bracket and space for
the remaining terms.
( x
)(
x
)
(9.6)
Write down a set of factors for a and c. The possible factors for a are: (1,3).
The possible factors for c are: (-1,1) or (1,-1).
Write down a set of options for the possible factors for the quadratic using the
factors of a and c. Therefore, there are two possible options.
Option 1
(x − 1)(3x + 1)
3x2 − 2x − 1
78
Option 2
(x + 1)(3x − 1)
3x2 + 2x − 1
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
9.5
Step 3 : Check your answer
(x + 1)(3x − 1) =
=
=
=
x(3x − 1) + 1(3x − 1)
(x)(3x) + (x)(−1) + (1)(3x) + (1)(−1)
3x2 − x + 3x − 1
x2 + 2x − 1.
Step 4 : Write the final answer
The factors of 3x2 + 2x − 1 are (x + 1) and (3x − 1).
Exercise: Factorising a Trinomial
1. Factorise the following:
(a) x2 + 8x + 15
(d) x2 + 9x + 14
(b) x2 + 10x + 24
(e) x2 + 15x + 36
(c) x2 + 9x + 8
(f) x2 + 13x + 36
2. Factorise the following:
(a) x2 − 2x − 15
(b) x2 + 2x − 3
(c) x2 + 2x − 8
(d) x2 + x − 20
(e) x2 − x − 20
3. Find the factors for the following quadratic expressions:
(a) 2x2 + 11x + 5
(b) 3x2 + 19x + 6
(c) 6x2 + 7x + 2
(d) 12x2 + 7x + 1
(e) 8x2 + 6x + 1
4. Find the factors for the following trinomials:
(a) 3x2 + 17x − 6
(b) 7x2 − 6x − 1
(c) 8x2 − 6x + 1
(d) 2x2 − 5x − 3
9.5
Factorisation by Grouping
One other method of factorisation involves the use of common factors. We know that the factors
of 3x + 3 are 3 and (x + 1). Similarly, the factors of 2x2 + 2x are 2x and (x + 1). Therefore, if
we have an expression:
2x2 + 2x + 3x + 3
then we can factorise as:
2x(x + 1) + 3(x + 1).
You can see that there is another common factor: x + 1. Therefore, we can now write:
(x + 1)(2x + 3).
We get this by taking out the x + 1 and see what is left over. We have a +2x from the first
term and a +3 from the second term. This is called factorisation by grouping.
79
9.6
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
Worked Example 25: Factorisation by Grouping
Question: Find the factors of 7x + 14y + bx + 2by by grouping
Answer
Step 1 : Determine if there are common factors to all terms
There are no factors that are common to all terms.
Step 2 : Determine if there are factors in common between some terms
7 is a common factor of the first two terms and b is a common factor of the second
two terms.
Step 3 : Re-write expression taking the factors into account
7x + 14y + bx + 2by = 7(x + 2y) + b(x + 2y)
Step 4 : Determine if there are more common factors
x + 2y is a common factor.
Step 5 : Re-write expression taking the factors into account
7(x + 2y) + b(x + 2y) = (x + 2y)(7 + b)
Step 6 : Write the final answer
The factors of 7x + 14y + bx + 2by are (7 + b) and (x + 2y).
Exercise: Factorisation by Grouping
1. Factorise by grouping: 6x + 9 + 2ax + 3
2. Factorise by grouping: x2 − 6x + 5x − 30
3. Factorise by grouping: 5x + 10y − ax − 2ay
4. Factorise by grouping: a2 − 2a − ax + 2x
5. Factorise by grouping: 5xy − 3y + 10x − 6
9.6
Simplification of Fractions
In some cases of simplifying an algebraic expression, the expression will be a fraction. For
example,
x2 + 3x
x+3
has a quadratic in the numerator and a binomial in the denominator. You can apply the different
factorisation methods to simplify the expression.
=
=
x2 + 3x
x+3
x(x + 3)
x+3
x provided x 6= −3
80
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
Worked Example 26: Simplification of Fractions
Question: Simplify: 2x−b+x−ab
ax2 −abx
Answer
Step 1 : Factorise numerator and denominator
Use grouping for numerator and common factor for denominator in this example.
(ax − ab) + (x − b)
ax2 − abx
a(x − b) + (x − b)
ax(x − b)
(x − b)(a + 1)
ax(x − b)
=
=
=
Step 2 : Cancel out same factors
The simplified answer is:
=
a+1
ax
Worked Example 27: Simplification of Fractions
2
2
+x
Question: Simplify: x x−x−2
÷ xx2 +2x
2 −4
Answer
Step 1 : Factorise numerators and denominators
=
(x + 1)(x − 2) x(x + 1)
÷
(x + 2)(x − 2) x(x + 2)
Step 2 : Multiply by factorised reciprocal
=
(x + 1)(x − 2) x(x + 2)
×
(x + 2)(x − 2) x(x + 1)
Step 3 : Cancel out same factors
The simplified answer is
=
Exercise: Simplification of Fractions
1. Simplify:
81
1
9.6
9.7
CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10
2a+10
4
a2 −4a
a−4
9a+27
9a+18
2
y−8xy
(h) 16x12x−6
3a+9
(j) 14 ÷ 7a+21
a+3
12p2
(l) 3xp+4p
÷
8p
3x+4
(y) 24a−8
÷ 9a−3
12
6
2
+pq
(p) p 7p
÷ 8p+8q
21q
2
a2
(r) f fa−f
−a
3a
15
5a+20
a+4
3a2 −9a
2a−6
(g) 6ab+2a
2b
(i) 4xyp−8xp
12xy
2
−5a
(k) a2a+10
÷ 3a+15
4a
2
+8x
16
÷ 6x 12
(x) 2xp+4x
2
÷ 2a+4
(o) a +2a
5
20
6b2
÷
(q) 5ab−15b
4a−12
a+b
(b)
(d)
(f)
(a)
(c)
(e)
2. Simplify:
9.7
x2 −1
3
×
1
x−1
−
1
2
End of Chapter Exercises
1. Factorise:
(a) a2 − 9
(d) 16b6 − 25a2
(g) 16ba4 − 81b
(j) 2a2 − 12ab + 18b2
(m) 125a3 + b3
(p) 64b3 + 1
(b) m2 − 36
(e) m2 − (1/9)
(h) a2 − 10a + 25
(k) −4b2 − 144b8 + 48b5
(n) 128b7 − 250ba6
(q) 5a3 − 40c3
(c) 9b2 − 81
(f) 5 − 5a2 b6
(i) 16b2 + 56b + 49
(l) a3 − 27
(o) c3 + 27
(r) 2b4 − 128b
2. Show that (2x − 1)2 − (x − 3)2 can be simplified to (x + 2)(3x − 4)
3. What must be added to x2 − x + 4 to make it equal to (x + 2)2
82
Chapter 10
Equations and Inequalities - Grade
10
10.1
Strategy for Solving Equations
This chapter is all about solving different types of equations for one or two variables. In general,
we want to get the unknown variable alone on the left hand side of the equation with all the
constants on the right hand side of the equation. For example, in the equation x − 1 = 0, we
want to be able to write the equation as x = 1.
As we saw in section 2.9 (page 13), an equation is like a set of weighing scales, that must always
be balanced. When we solve equations, we need to keep in mind that what is done to one side
must be done to the other.
Method: Rearranging Equations
You can add, subtract, multiply or divide both sides of an equation by any number you want, as
long as you always do it to both sides.
For example, in the equation x + 5 − 1 = −6, we want to get x alone on the left hand side of the
equation. This means we need to subtract 5 and add 1 on the left hand side. However, because
we need to keep the equation balanced, we also need to subtract 5 and add 1 on the right hand
side.
x+5−1
= −6
x+0+0
x
= −11 + 1
= −10
x+5−5−1+1
= −6 − 5 + 1
In another example, 23 x = 8, we must divide by 2 and multiply by 3 on the left hand side in
order to get x alone. However, in order to keep the equation balanced, we must also divide by
2 and multiply by 3 on the right hand side.
2
x
3
2
x÷2×3
3
2 3
× ×x
2 3
1×1×x
x
= 8
= 8÷2×3
8×3
2
= 12
= 12
=
These are the basic rules to apply when simplifying any equation. In most cases, these rules
have to be applied more than once, before we have the unknown variable on the left hand side
83
10.2
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
of the equation.
We are now ready to solve some equations!
Important: The following must also be kept in mind:
1. Division by 0 is undefined.
2. If
x
y
= 0, then x = 0 and y 6= 0, because division by 0 is undefined.
Activity :: Investigation : Strategy for Solving Equations
In the following, identify what is wrong.
4x − 8
4(x − 2)
4(x − 2)
(x − 2)
4
10.2
=
=
=
=
3(x − 2)
3(x − 2)
3(x − 2)
(x − 2)
3
Solving Linear Equations
The simplest equation to solve is a linear equation. A linear equation is an equation where the
power on the variable(letter, e.g. x) is 1(one). The following are examples of linear equations.
2x + 2
2−x
3x + 1
4
x−6
3
= 1
= 2
= 7x + 2
In this section, we will learn how to find the value of the variable that makes both sides of the
linear equation true. For example, what value of x makes both sides of the very simple equation,
x + 1 = 1 true.
Since the highest power on the variable is one(1) in a linear equation, there is at most one
solution or root for the equation.
This section relies on all the methods we have already discussed: multiplying out expressions,
grouping terms and factorisation. Make sure that you are comfortable with these methods,
before trying out the work in the rest of this chapter.
2x + 2 =
2x =
2x =
1
1 − 2 (like terms together)
−1 (simplified as much a possible)
Now we see that 2x = −1. This means if we divide both sides by 2, we will get:
x=−
84
1
2
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.2
If we substitute x = − 21 , back into the original equation, we get:
=
=
=
2x + 2
1
2(− ) + 2
2
−1 + 2
1
That is all that there is to solving linear equations.
Important: Solving Equations
When you have found the solution to an equation, substitute the solution into the original
equation, to check your answer.
Method: Solving Equations
The general steps to solve equations are:
1. Expand(Remove) all brackets.
2. ”Move” all terms with the variable to the left hand side of equation, and all constant terms
(the numbers) to the right hand side of the equal to-sign. Bearing in mind that the sign
of the terms will chance(from (+) to (-) or vice versa, as they ”cross over” the equal to
sign.
3. Group all like terms together and simplify as much as possible.
4. Factorise if necessary.
5. Find the solution.
6. Substitute solution into original equation to check answer.
Worked Example 28: Solving Linear Equations
Question: Solve for x: 4 − x = 4
Answer
Step 1 : Determine what is given and what is required
We are given 4 − x = 4 and are required to solve for x.
Step 2 : Determine how to approach the problem
Since there are no brackets, we can start with grouping like terms and then simplifying.
Step 3 : Solve the problem
4−x =
−x =
−x =
−x =
−x =
∴ x =
4
4 − 4 (move all constant terms (numbers) to the RHS (right hand side))
0 (group like terms together)
0 (simplify grouped terms)
0
0
Step 4 : Check the answer
85
10.2
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Substitute solution into original equation:
4−0=4
4=4
Since both sides are equal, the answer is correct.
Step 5 : Write the Final Answer
The solution of 4 − x = 4 is x = 0.
Worked Example 29: Solving Linear Equations
Question: Solve for x: 4(2x − 9) − 4x = 4 − 6x
Answer
Step 1 : Determine what is given and what is required
We are given 4(2x − 9) − 4x = 4 − 6x and are required to solve for x.
Step 2 : Determine how to approach the problem
We start with expanding the brackets, then grouping like terms and then simplifying.
Step 3 : Solve the problem
4(2x − 9) − 4x = 4 − 6x
8x − 36 − 4x = 4 − 6x
(expand the brackets)
8x − 4x + 6x = 4 + 36 (move all terms with x to the LHS and all constant terms to the RHS of the =)
(8x − 4x + 6x) = (4 + 36) (group like terms together)
10x = 40
10
40
x =
10
10
x = 4
(simplify grouped terms)
(divide both sides by 10)
Step 4 : Check the answer
Substitute solution into original equation:
4(2(4) − 9) − 4(4) =
4(8 − 9) − 16 =
4(−1) − 16 =
−4 − 16 =
−20 =
4 − 6(4)
4 − 24
−20
−20
−20
Since both sides are equal to −20, the answer is correct.
Step 5 : Write the Final Answer
The solution of 4(2x − 9) − 4x = 4 − 6x is x = 4.
Worked Example 30: Solving Linear Equations
Question: Solve for x:
Answer
2−x
3x+1
=2
86
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.2
Step 1 : Determine what is given and what is required
2−x
We are given 3x+1
= 2 and are required to solve for x.
Step 2 : Determine how to approach the problem
Since there is a denominator of (3x+1), we can start by multiplying both sides of
the equation by (3x+1). But because division by 0 is not permissible, there is a
restriction on a value for x. (x 6= −1
3 )
Step 3 : Solve the problem
2−x
3x + 1
(2 − x)
2−x
= 2
= 2(3x + 1)
= 6x + 2 (remove/expand brackets)
−x − 6x = 2 − 2
−7x = 0
theref ore
x
x
(move all terms containing x to the LHS and all constant terms (numbers) to the RHS.)
(simplify grouped terms)
= 0 ÷ (−7)
= 0
zero divide by any number is 0
Step 4 : Check the answer
Substitute solution into original equation:
2 − (0)
3(0) + 1
2
1
=
2
=
2
Since both sides are equal to 2, the answer is correct.
Step 5 : Write the Final Answer
2−x
= 2 is x = 0.
The solution of 3x+1
Worked Example 31: Solving Linear Equations
Question: Solve for x: 34 x − 6 = 7x + 2
Answer
Step 1 : Determine what is given and what is required
We are given 34 x − 6 = 7x + 2 and are required to solve for x.
Step 2 : Determine how to approach the problem
We start with multiplying each of the terms in the equation by 3, then grouping like
terms and then simplifying.
Step 3 : Solve the problem
4
x−6 =
3
4x − 18 =
4x − 21x =
−17x =
−17
x =
−17
x =
7x + 2
21x + 6 (each term is multiplied by 3
6 + 18 (move all terms with x to the LHS and all constant terms to the RHS of the =)
24
24
−17
−24
17
(simplify grouped terms)
(divide both sides by -17)
87
10.2
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Step 4 : Check the answer
Substitute solution into original equation:
4 −24
×
−6
3
17
4 × (−8)
−6
(17)
(−32)
−6
17
−32 − 102
17
−134
17
Since both sides are equal to
−134
17 ,
=
=
=
=
=
−24
+2
17
7 × (−24)
+2
17
−168
+2
17
(−168) + 34
17
−134
17
7×
the answer is correct.
Step 5 : Write the Final Answer
The solution of 34 x − 6 = 7x + 2 is, x =
−24
17 .
Exercise: Solving Linear Equations
1. Solve for y: 2y − 3 = 7
2. Solve for w: −3w = 0
3. Solve for z: 4z = 16
4. Solve for t: 12t + 0 = 144
5. Solve for x: 7 + 5x = 62
6. Solve for y: 55 = 5y +
3
4
7. Solve for z: 5z = 3z + 45
8. Solve for a: 23a − 12 = 6 + 2a
9. Solve for b: 12 − 6b + 34b = 2b − 24 − 64
10. Solve for c: 6c + 3c = 4 − 5(2c − 3).
11. Solve for p: 18 − 2p = p + 9
12. Solve for q:
4
q
=
16
24
13. Solve for q:
4
1
=
q
2
14. Solve for r: −(−16 − r) = 13r − 1
15. Solve for d: 6d − 2 + 2d = −2 + 4d + 8
16. Solve for f : 3f − 10 = 10
17. Solve for v: 3v + 16 = 4v − 10
18. Solve for k: 10k + 5 + 0 = −2k + −3k + 80
19. Solve for j: 8(j − 4) = 5(j − 4)
20. Solve for m: 6 = 6(m + 7) + 5m
88
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.3
10.3
Solving Quadratic Equations
A quadratic equation is an equation where the power on the variable is at most 2. The following
are examples of quadratic equations.
2x2 + 2x = 1
2−x
= 2x
3x + 1
4
x − 6 = 7x2 + 2
3
Quadratic equations differ from linear equations by the fact that a linear equation only has one
solution, while a quadratic equation has at most two solutions. There are some special situations
when a quadratic equation only has one solution.
We solve quadratic equations by factorisation, that is writing the quadratic as a product of two
expressions in brackets. For example, we know that:
(x + 1)(2x − 3) = 2x2 − x − 3.
In order to solve:
2x2 − x − 3 = 0
we need to be able to write 2x2 − x − 3 as (x + 1)(2x − 3), which we already know how to do.
Activity :: Investigation : Factorising a Quadratic
Factorise the following quadratic expressions:
1. x + x2
2. x2 + 1 + 2x
3. x2 − 4x + 5
4. 16x2 − 9
5. 4x2 + 4x + 1
Being able to factorise a quadratic means that you are one step away from solving a quadratic
equation. For example, x2 − 3x − 2 = 0 can be written as (x − 1)(x − 2) = 0. This means
that both x − 1 = 0 and x − 2 = 0, which gives x = 1 and x = 2 as the two solutions to the
quadratic equation x2 − 3x − 2 = 0.
Method: Solving Quadratic Equations
1. First divide the entire equation by any common factor of the coefficients, so as to obtain
an equation of the form ax2 + bx + c = 0 where a, b and c have no common factors. For
example, 2x2 + 4x + 2 = 0 can be written as x2 + 2x + 1 = 0 by dividing by 2.
2. Write ax2 + bx + c in terms of its factors (rx + s)(ux + v).
This means (rx + s)(ux + v) = 0.
3. Once writing the equation in the form (rx + s)(ux + v) = 0, it then follows that the two
solutions are x = − rs or x = − uv .
Extension: Solutions of Quadratic Equations
There are two solutions to a quadratic equation, because any one of the values can
solve the equation.
89
10.3
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Worked Example 32: Solving Quadratic Equations
Question: Solve for x: 3x2 + 2x − 1 = 0
Answer
Step 1 : Find the factors of 3x2 + 2x − 1
As we have seen the factors of 3x2 + 2x − 1 are (x + 1) and (3x − 1).
Step 2 : Write the equation with the factors
(x + 1)(3x − 1) = 0
Step 3 : Determine the two solutions
We have
x+1=0
or
3x − 1 = 0
1
3.
Therefore, x = −1 or x =
Step 4 : Write the final answer
3x2 + 2x − 1 = 0 for x = −1 or x = 13 .
Worked Example 33: Solving Quadratic Equations
√
Question: Solve for x: x + 2 = x
Answer
Step 1 : Square both sides of the equation
Both sides of the equation should be squared to remove the square root sign.
x + 2 = x2
Step 2 : Write equation in the form ax2 + bx + c = 0
x2
(subtract x2 to both sides)
=
=
0
0
(divide both sides by -1)
x2 − x + 2 =
0
x+2 =
x + 2 − x2
−x − 2 + x2
Step 3 : Factorise the quadratic
x2 − x + 2
The factors of x2 − x + 2 are (x − 2)(x + 1).
Step 4 : Write the equation with the factors
(x − 2)(x + 1) = 0
Step 5 : Determine the two solutions
We have
x+1=0
or
x−2=0
90
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.3
Therefore, x = −1 or x = 2.
Step 6 : Check whether solutions are valid
√
Substitute x = −1into the original equation x + 2 = x:
LHS
p
(−1) + 2
√
1
1
=
=
=
but
RHS =
(−1)
Therefore LHS6=RHS
Therefore x 6= −1
√
Now substitute x = 2 into original equation x + 2 = x:
LHS
√
2+2
=
√
4
=
= 2
and
RHS
= 2
Therefore LHS = RHS
Therefore x = 2 is the only valid solution
Step
7 : Write the final answer
√
x + 2 = x for x = 2 only.
Worked Example 34: Solving Quadratic Equations
Question: Solve the equation: x2 + 3x − 4 = 0.
Answer
Step 1 : Check if the equation is in the form ax2 + bx + c = 0
The equation is in the required form, with a = 1.
Step 2 : Factorise the quadratic
You need the factors of 1 and 4 so that the middle term is +3 So the factors are:
(x − 1)(x + 4)
Step 3 : Solve the quadratic equation
x2 + 3x − 4 = (x − 1)(x + 4) = 0
Therefore x = 1 or x = −4.
Step 4 : Write the final solution
Therefore the solutions are x = 1 or x = −4.
91
(10.1)
10.3
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Worked Example 35: Solving Quadratic Equations
Question: Find the roots of the quadratic equation 0 = −2x2 + 4x − 2.
Answer
Step 1 : Determine whether the equation is in the form ax2 + bx + c = 0,
with no common factors.
There is a common factor: -2. Therefore, divide both sides of the equation by -2.
−2x2 + 4x − 2 =
x2 − 2x + 1 =
0
0
Step 2 : Factorise x2 − 2x + 1
The middle term is negative. Therefore, the factors are (x − 1)(x − 1)
If we multiply out (x − 1)(x − 1), we get x2 − 2x + 1.
Step 3 : Solve the quadratic equation
x2 − 2x + 1 = (x − 1)(x − 1) = 0
In this case, the quadratic is a perfect square, so there is only one solution for x:
x = 1.
Step 4 : Write the final solution
The root of 0 = −2x2 + 4x − 2 is x = 1.
Exercise: Solving Quadratic Equations
1. Solve for x: (3x + 2)(3x − 4) = 0
2. Solve for a: (5a − 9)(a + 6) = 0
3. Solve for x: (2x + 3)(2x − 3) = 0
4. Solve for x: (2x + 1)(2x − 9) = 0
5. Solve for x: (2x − 3)(2x − 3) = 0
6. Solve for x: 20x + 25x2 = 0
7. Solve for a: 4a2 − 17a − 77 = 0
8. Solve for x: 2x2 − 5x − 12 = 0
9. Solve for b: −75b2 + 290b − 240 = 0
10. Solve for y: 2y = 13 y 2 − 3y + 14 32
11. Solve for θ: θ2 − 4θ = −4
12. Solve for q: −q 2 + 4q − 6 = 4q 2 − 5q + 3
13. Solve for t: t2 = 3t
14. Solve for w: 3w2 + 10w − 25 = 0
15. Solve for v: v 2 − v + 3
16. Solve for x: x2 − 4x + 4 = 0
17. Solve for t: t2 − 6t = 7
18. Solve for x: 14x2 + 5x = 6
19. Solve for t: 2t2 − 2t = 12
20. Solve for y: 3y 2 + 2y − 6 = y 2 − y + 2
92
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.4
10.4
Exponential Equations of the form ka(x+p) = m
examples solved by trial and error)
Exponential equations generally have the unknown variable as the power. The following are
examples of exponential equations:
2x
2
3x+1
4
−6
3
−x
=
1
=
2
=
7x + 2
You should already be familiar with exponential notation. Solving exponential equations are
simple, if we remember how to apply the laws of exponentials.
Activity :: Investigation : Solving Exponential Equations
Solve the following equations by completing the table:
2x = 2
-3
-2
-1
x
0
1
2
3
-3
-2
-1
x
0
1
2
3
2x
3x = 9
3x
2x+1 = 8
-3
-2
-1
x
0
1
2
3
2x+1
10.4.1
Algebraic Solution
Definition: Equality for Exponential Functions
If a is a positive number such that a > 0, then:
ax = ay
if and only if:
x=y
.
This means that if we can write all terms in an equation with the same base, we can solve the
exponential equations by equating the indices. For example take the equation 3x+1 = 9. This
can be written as:
3x+1 = 32 .
Since the bases are equal (to 3), we know that the exponents must also be equal. Therefore we
can write:
x + 1 = 2.
This gives:
x = 1.
93
10.4
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Method: Solving Exponential Equations
Try to write all terms with the same base.
Activity :: Investigation : Exponential Numbers
Write the following with the same base. The base is the first in the list. For
example, in the list 2, 4, 8, the base is two and we can write 4 as 22 .
1. 2,4,8,16,32,64,128,512,1024
2. 3,9,27,81,243
3. 5,25,125,625
4. 13,169
5. 2x, 4x2 , 8x3 , 49x8
Worked Example 36: Solving Exponential Equations
Question: Solve for x: 2x = 2
Answer
Step 1 : Try to write all terms with the same base.
All terms are written with the same base.
2x = 21
Step 2 : Equate the indices
x=1
Step 3 : Check your answer
2x
= 2(1)
= 21
Since both sides are equal, the answer is correct.
Step 4 : Write the final answer
x=1
x
is the solution to 2 = 2.
Worked Example 37: Solving Exponential Equations
Question: Solve:
2x+4 = 42x
94
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.4
Answer
Step 1 : Try to write all terms with the same base.
2x+4
=
42x
2x+4
=
22(2x)
2x+4
=
24x
Step 2 : Equate the indices
x + 4 = 4x
Step 3 : Solve for x
x+4
= 4x
x − 4x = −4
−3x = −4
−4
x =
−3
4
x =
3
Step 4 : Check your answer
LHS =
=
RHS
2x+4
4
2( 3 +4)
16
=
23
=
=
(216 ) 3
42x
=
=
42( 3 )
8
43
=
(48 ) 3
=
((22 )8 ) 3
=
(216 ) 3
=
LHS
1
4
1
1
1
Since both sides are equal, the answer is correct.
Step 5 : Write the final answer
x=
4
3
is the solution to 2x+4 = 42x .
Exercise: Solving Exponential Equations
1. Solve the following exponential equations.
a. 2x+5 = 25
d. 65−x = 612
b. 32x+1 = 33
e. 64x+1 = 162x+5
2. Solve: 39x−2 = 27
95
c. 52x+2 = 53
f. 125x = 5
10.5
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
3. Solve for k: 81k+2 = 27k+4
4. The growth of an algae in a pond is can be modeled by the function f (t) = 2t .
Find the value of t such that f (t) = 128?
5. Solve for x: 25(1−2x) = 54
6. Solve for x: 27x × 9x−2 = 1
10.5
Linear Inequalities
graphically;
Activity :: Investigation : Inequalities on a Number Line
Represent the following on number lines:
1. x = 4
2. x < 4
3. x ≤ 4
4. x ≥ 4
5. x > 4
A linear inequality is similar to a linear equation and has the power on the variable is equal to 1.
The following are examples of linear inequalities.
2x + 2
2−x
3x + 1
4
x−6
3
≤ 1
≥ 2
< 7x + 2
The methods used to solve linear inequalities are identical to those used to solve linear equations.
The only difference occurs when there is a multiplication or a division that involves a minus sign.
For example, we know that 8 > 6. If both sides of the inequality are divided by −2, −4 is not
greater than −3. Therefore, the inequality must switch around, making −4 < −3.
Important: When you divide or multiply both sides of an inequality by any number with a
minus sign, the direction of the inequality changes.
For example, if x < 1, then −x > −1.
In order to compare am inequality to a normal equation, we shall solve an equation first. Solve
2x + 2 = 1.
2x + 2 = 1
2x = 1 − 2
2x = −1
1
x = −
2
If we represent this answer on a number line, we get
96
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
-3
x = − 12
b
-1
0
-2
1
10.5
2
3
2
3
Now let us solve the inequality 2x + 2 ≤ 1.
2x + 2
≤ 1
2x ≤ 1 − 2
2x ≤ −1
1
x ≤ −
2
If we represent this answer on a number line, we get
x ≤ − 21
-3
-2
b
-1
0
1
As you can see, for the equation, there is only a single value of x for which the equation is true.
However, for the inequality, there is a range of values for which the inequality is true. This is
the main difference between an equation and an inequality.
Worked Example 38: Linear Inequalities
Question: Solve for r: 6 − r > 2
Answer
Step 1 : Move all constants to the RHS
−r > 2 − 6
−r > −4
Step 2 : Multiply both sides by -1
When you multiply by a minus sign, the direction of the inequality changes.
r<4
Step 3 : Represent answer graphically
r<4
0
1
2
3
bc
4
5
Worked Example 39: Linear Inequalities
Question: Solve for q: 4q + 3 < 2(q + 3) and represent solution on a number line.
Answer
Step 1 : Expand all brackets
4q + 3
4q + 3
< 2(q + 3)
< 2q + 6
97
10.5
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Step 2 : Move all constants to the RHS and all unknowns to the LHS
4q + 3
4q − 2q
2q
<
<
<
2q + 6
6−3
3
Step 3 : Solve inequality
2q
q
< 3 Divide both sides by 2
3
<
2
Step 4 : Represent answer graphically
q<
0
3
2
1
c
b
2
3
4
5
Worked Example 40: Compound Linear Inequalities
Question: Solve for x: 5 ≤ x + 3 < 8 and represent solution on a number line.
Answer
Step 1 : Subtract 3 from Left, middle and right of inequalities
5−3≤ x+3−3
2≤
x
<8−3
<5
Step 2 : Represent answer graphically
0
b
2
1
2≤x<5
3
4
bc
5
Exercise: Linear Inequalities
1. Solve for x and represent the solution graphically:
(a) 3x + 4 > 5x + 8
(b) 3(x − 1) − 2 ≤ 6x + 4
2x−3
(c) x−7
3 >
2
(d) −4(x − 1) < x + 2
(e) 21 x + 13 (x − 1) ≥ 65 x −
1
3
2. Solve the following inequalities. Illustrate your answer on a number line if x is
a real number.
(a) −2 ≤ x − 1 < 3
98
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.6
(b) −5 < 2x − 3 ≤ 7
3. Solve for x: 7(3x + 2) − 5(2x − 3) > 7.
Illustrate this answer on a number line.
10.6
Linear Simultaneous Equations
Thus far, all equations that have been encountered have one unknown variable, that must be
solved for. When two unknown variables need to be solved for, two equations are required
and these equations are known as simultaneous equations. The solutions to the system of
simultaneous equations, are the values of the unknown variables which satisfy the system of
equations simultaneously, that means at the same time. In general, if there are n unknown
variables, then n equations are required to obtain a solution for each of the n variables.
An example of a system of simultaneous equations is:
2x + 2y = 1
2−x
=2
3y + 1
10.6.1
(10.2)
Finding solutions
In order to find a numerical value for an unknown variable, one must have at least as many independent equations as variables. We solve simultaneous equations graphically and algebraically/
10.6.2
Graphical Solution
Simultaneous equations can also be solved graphically. If the graphs corresponding to each
equation is drawn, then the solution to the system of simultaneous equations is the co-ordinate
of the point at which both graphs intersect.
x = 2y
(10.3)
y = 2x − 3
1
1
−1
2
y=
1 x
2
3
2x
−
−1
b
y=
−2
(2,1)
3
Draw the graphs of the two equations in (10.3).
The intersection of the two graphs is (2,1). So the solution to the system of simultaneous
equations in (10.3) is y = 1 and x = 2.
99
10.6
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
This can be shown algebraically as:
x =
∴ y
y − 4y
−3y
y
=
=
=
=
Substitute into the first equation: x =
=
2y
2(2y) − 3
−3
−3
1
2(1)
2
Worked Example 41: Simultaneous Equations
Question: Solve the following system of simultaneous equations graphically.
4y + 3x =
4y − 19x =
100
12
Answer
Step 1 : Draw the graphs corresponding to each equation.
For the first equation:
4y + 3x =
4y =
y
=
100
100 − 3x
3
25 − x
4
and for the second equation:
4y − 19x =
4y
=
y
=
12
19x + 12
19
x+3
4
=
x
19
40
4y
4y + 3x
=
100
12
−
30
20
10
−8
−6
−4
−2
2
4
6
8
Step 2 : Find the intersection of the graphs.
The graphs intersect at (4,22).
Step 3 : Write the solution of the system of simultaneous equations as given
by the intersection of the graphs.
x =
y =
100
4
22
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.6.3
10.6
Solution by Substitution
A common algebraic technique is the substitution method: try to solve one of the equations
for one of the variables and substitute the result into the other equations, thereby reducing the
number of equations and the number of variables by 1. Continue until you reach a single equation
with a single variable, which (hopefully) can be solved; back substitution then yields the values
for the other variables.
In the example (??), we first solve the first equation for x:
x=
1
−y
2
and substitute this result into the second equation:
2−x
3y + 1
2 − ( 21 − y)
3y + 1
1
2 − ( − y)
2
1
2− +y
2
= 2
= 2
= 2(3y + 1)
= 6y + 2
y − 6y
= −2 +
−5y
∴
x
1
2
=
y
= −
=
=
=
=
1
+2
2
1
10
1
−y
2
1
1
− (− )
2
10
6
10
3
5
The solution for the system of simultaneous equations (??) is:
3
5
x
=
y
= −
1
10
Worked Example 42: Simultaneous Equations
Question: Solve the following system of simultaneous equations:
4y + 3x =
100
4y − 19x =
12
Answer
Step 1 : If the question, does not explicitly ask for a graphical solution, then
the system of equations should be solved algebraically.
101
10.6
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Step 2 : Make x the subject of the first equation.
4y + 3x =
3x =
x =
100
100 − 4y
100 − 4y
3
Step 3 : Substitute the value obtained for x into the second equation.
100 − 4y
) =
3
12y − 19(100 − 4y) =
4y − 19(
12
36
12y − 1900 + 76y
88y
=
=
36
1936
y
=
22
Step 4 : Substitute into the equation for x.
x =
=
=
=
100 − 4(22)
3
100 − 88
3
12
3
4
Step 5 : Substitute the values for x and y into both equations to check the
solution.
4(22) + 3(4) = 88 + 12 =
100 X
4(22) − 19(4) = 88 − 76 =
12 X
Worked Example 43: Bicycles and Tricycles
Question: A shop sells bicycles and tricycles. In total there are 7 cycles and 19
wheels. Determine how many of each there are, if a bicycle has two wheels and a
tricycle has three wheels.
Answer
Step 1 : Identify what is required
The number of bicycles and the number of tricycles are required.
Step 2 : Set up the necessary equations
If b is the number of bicycles and t is the number of tricycles, then:
b+t =
2b + 3t =
7
19
Step 3 : Solve the system of simultaneous equations using substitution.
102
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
b
=
Into second equation: 2(7 − t) + 3t
14 − 2t + 3t
=
=
t
Into first equation: b
=
=
=
10.7
7−t
19
19
5
7−5
2
Step 4 : Check solution by substituting into original system of equations.
2+5
=
2(2) + 3(5) = 4 + 15 =
7 X
19 X
Exercise: Simultaneous Equations
1. Solve graphically and confirm your answer algebraically: 3a − 2b7 = 0 , a −
4b + 1 = 0
2. Solve algebraically: 15c + 11d − 132 = 0, 2c + 3d − 59 = 0
3. Solve algebraically: −18e − 18 + 3f = 0, e − 4f + 47 = 0
4. Solve graphically: x + 2y = 7, x + y = 0
10.7
Mathematical Models
10.7.1
Introduction
Tom and Jane are friends. Tom picked up Jane’s Physics test paper, but will not tell Jane what
her marks are. He knows that Jane hates maths so he decided to tease her. Tom says: “I have
2 marks more than you do and the sum of both our marks is equal to 14. How much did we
get?”
Let’s help Jane find out what her marks are. We have two unknowns, Tom’s mark (which we shall
call t) and Jane’s mark (which we shall call j). Tom has 2 more marks than Jane. Therefore,
t=j+2
Also, both marks add up to 14. Therefore,
t + j = 14
The two equations make up a set of linear (because the highest power is one) simultaneous
103
10.7
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
equations, which we know how to solve! Substitute for t in the second equation to get:
t+j
= 14
j+2+j
2j + 2
= 14
= 14
2(j + 1) = 14
j+1 = 7
j
= 7−1
= 6
Then,
t
= j+2
= 6+2
= 8
So, we see that Tom scored 8 on his test and Jane scored 6.
This problem is an example of a simple mathematical model. We took a problem and we able
to write a set of equations that represented the problem, mathematically. The solution of the
equations then gave the solution to the problem.
10.7.2
Problem Solving Strategy
The purpose of this section is to teach you the skills that you need to be able to take a problem
and formulate it mathematically, in order to solve it. The general steps to follow are:
1. Read ALL of it !
2. Find out what is requested.
3. Let the requested be a variable e.g. x.
4. Rewrite the information given in terms of x. That is, translate the words into algebraic
language. This is the reponse
5. Set up an equation (i.e. a mathematical sentence or model) to solve the required variable.
6. Solve the equation algebraically to find the result.
Important: Follow the three R’s and solve the problem... Request - Response - Result
10.7.3
Application of Mathematical Modelling
Worked Example 44: Mathematical Modelling: One variable
Question: A fruit shake costs R2,00 more than a chocolate milkshake. If three fruit
shakes and 5 chocolate milkshakes cost R78,00, determine the individual prices.
Answer
Step 1 : Summarise the information in a table
Fruit
Chocolate
Price number
x+2
3
x
5
104
Total
3(x + 2)
5x
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.7
Step 2 : Set up an algebraic equation
3(x + 2) + 5x = 78
Step 3 : Solve the equation
3x + 6 + 5x = 78
8x = 72
x
= 9
Step 4 : Present the final answer
Chocolate milkshake costs R 9,00 and the Fruitshake costs R 11,00
Worked Example 45: Mathematical Modelling: Two variables
Question: Three rulers and two pens cost R 21,00. One ruler and one pen cost R
8,00. Find the cost of one ruler and one pen
Answer
Step 1 : Translate the problem using variables
Let the cost of one ruler be x rand and the cost of one pen be y rand.
Step 2 : Rewrite the information in terms of the variables
3x + 2y
x+y
=
=
21
8
Step 3 : Solve the equations simultaneously
First solve the second equation for y:
y =8−x
and substitute the result into the first equation:
3x + 2(8 − x) = 21
3x + 16 − 2x = 21
x
= 5
therefore
y
y
= 8−5
= 3
Step 4 : Present the final answers
one Ruler costs R 5,00 and one Pen costs R 3,00
105
(10.4)
(10.5)
10.7
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Exercise: Mathematical Models
1. Stephen has 1 l of a mixture containing 69% of salt. How much water must
Stephen add to make the mixture 50% salt? Write your answer as a fraction.
2. The diagonal of a rectangle is 25 cm more than its width. The length of
the rectangle is 17 cm more than its width. What are the dimensions of the
rectangle?
3. The sum of 27 and 12 is 73 more than an unknown number. Find the unknown
number.
4. The two smaller angles in a right-angled triangle are in the ratio of 1:2. What
are the sizes of the two angles?
5. George owns a bakery that specialises in wedding cakes. For each wedding cake,
it costs George R150 for ingredients, R50 for overhead, and R5 for advertising.
George’s wedding cakes cost R400 each. As a percentage of George’s costs,
how much profit does he make for each cake sold?
6. If 4 times a number is increased by 7, the result is 15 less than the square of
the number. Find the numbers that satisfy this statement, by formulating an
equation and then solving it.
7. The length of a rectangle is 2 cm more than the width of the rectangle. The
perimeter of the rectangle is 20 cm. Find the length and the width of the
rectangle.
10.7.4
End of Chapter Exercises
1. What are the roots of the quadratic equation x2 − 3x + 2 = 0?
2. What are the solutions to the equation x2 + x = 6?
3. In the equation y = 2x2 − 5x − 18, which is a value of x when y = 0?
4. Manuel has 5 more CDs than Pedro has. Bob has twice as many CDs as Manuel has.
Altogether the boys have 63 CDs. Find how many CDs each person has.
5. Seven-eighths of a certain number is 5 more than one-third of the number. Find the
number.
6. A man runs to a telephone and back in 15 minutes. His speed on the way to the telephone
is 5 m/s and his speed on the way back is 4 m/s. Find the distance to the telephone.
7. Solve the inequality and then answer the questions:
x
x
3 − 14 > 14 − 4
(a) If xǫR, write the solution in interval notation.
(b) if xǫZ and x < 51, write the solution as a set of integers.
8. Solve for a:
1−a
2
−
2−a
3
9. Solve for x: x − 1 =
>1
42
x
10. Solve for x and y: 7x + 3y = 13 and 2x − 3y = −4
chapterFunctions and Graphs - Grade 10
106
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.8
10.8
Introduction to Functions and Graphs
Functions are mathematical building blocks for designing machines, predicting natural disasters,
curing diseases, understanding world economies and for keeping aeroplanes in the air. Functions
can take input from many variables, but always give the same answer, unique to that function.
It is the fact that you always get the same answer from a set of inputs, which is what makes
functions special.
A major advantage of functions is that they allow us to visualise equations in terms of a graph.
A graph is an accurate drawing of a function and is much easier to read than lists of numbers.
In this chapter we will learn how to understand and create real valued functions, how to read
graphs and how to draw them.
Despite their use in the problems facing humanity, functions also appear on a day-to-day level,
so they are worth learning about. A function is always dependent on one or more variables, like
time, distance or a more abstract quantity.
10.9
Functions and Graphs in the Real-World
Some typical examples of functions you may already have met include:• how much money you have, as a function of time. You never have more than one amount
of money at any time because you can always add everything to give one number. By
understanding how your money changes over time, you can plan to spend your money
sensibly. Businesses find it very useful to plot the graph of their money over time so that
they can see when they are spending too much. Such observations are not always obvious
from looking at the numbers alone.
• the temperature is a very complicated function because it has so many inputs, including;
the time of day, the season, the amount of clouds in the sky, the strength of the wind, where
you are and many more. But the important thing is that there is only one temperature
when you measure it. By understanding how the temperature is effected by these things,
you can plan for the day.
• where you are is a function of time, because you cannot be in two places at once! If you
were to plot the graphs of where two people are as a function of time, if the lines cross it
means that the two people meet each other at that time. This idea is used in logistics, an
area of mathematics that tries to plan where people and items are for businesses.
• your weight is a function of how much you eat and how much exercise you do, but everybody
has a different function so that is why people are all different sizes.
10.10
Recap
The following should be familiar.
10.10.1
Variables and Constants
In section 2.4 (page 8), we were introduced to variables and constants. To recap, a variable
can take any value in some set of numbers, so long is the equation is consistent. Most often, a
variable will be written as a letter.
A constant has a fixed value. The number 1 is a constant. Sometimes letters are used to
represent constants, as its easier to work with.
Activity :: Investigation : Variables and Constants
In the following expressions, identify the variables and the constants:
107
10.10
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
1. 2x2 = 1
2. 3x + 4y = 7
3. y =
−5
x
x
4. y = 7 − 2
10.10.2
Relations and Functions
In earlier grades, you saw that variables can be related to each other. For example, Alan is two
years older than Nathan. Therefore the relationship between the ages of Alan and Nathan can
be written as A = N + 2, where A is Alan’s age and N is Nathan’s age.
In general, a relation is an equation which relates two variables. For example, y = 5x and
y 2 + x2 = 5 are relations. In both examples x and y are variables and 5 is a constant, but for a
given value of x the value of y will be very different in each relation.
Besides writing relations as equations, they can also be represented as words, tables and graphs.
Instead of writing y = 5x, we could also say “y is always five times as big as x”. We could also
give the following table:
x
2
6
8
13
15
y = 5x
10
30
40
65
75
Activity :: Investigation : Relations and Functions
Complete the following table for the given functions:
x
y = x y = 2x y = x + 2
1
2
3
50
100
10.10.3
The Cartesian Plane
When working with real valued functions, our major tool is drawing graphs. In the first place, if
we have two real variables, x and y, then we can assign values to them simultaneously. That is,
we can say “let x be 5 and y be 3”. Just as we write “let x = 5” for “let x be 5”, we have the
shorthand notation “let (x, y) = (5, 3)” for “let x be 5 and y be 3”. We usually think of the
real numbers as an infinitely long line, and picking a number as putting a dot on that line. If
we want to pick two numbers at the same time, we can do something similar, but now we must
use two dimensions. What we do is use two lines, one for x and one for y, and rotate the one
for y, as in Figure 10.1. We call this the Cartesian plane.
108
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
y
10.10
4
3
(−3,2)
(2,2)
b
b
2
1
x
−4
−3
−2
−1
1
2
3
4
−1
−2
−3
−4
Figure 10.1: The Cartesian plane is made up of an x−axis (horizontal) and a y−axis (vertical).
10.10.4
Drawing Graphs
In order to draw the graph of a function, we need to calculate a few points. Then we plot the
points on the Cartesian Plane and join the points with a smooth line.
The great beauty of doing this is that it allows us to “draw” functions, in a very abstract way.
Assume that we were investigating the properties of the function f (x) = 2x. We could then
consider all the points (x, y) such that y = f (x), i.e. y = 2x. For example, (1, 2), (2.5, 5), and
(3, 6) would all be such points, whereas (3, 5) would not since 5 6= 2 × 3. If we put a dot at
each of those points, and then at every similar one for all possible values of x, we would obtain
the graph shown in
5
b
4
b
3
b
2
b
b
1
b
1
2
3
4
5
Figure 10.2: Graph of f (x) = 2x
The form of this graph is very pleasing – it is a simple straight line through the middle of
109
10.10
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
the plane. The technique of “plotting”, which we have followed here, is the key element in
understanding functions.
Activity :: Investigation : Drawing Graphs and the Cartesian Plane
Plot the following points and draw a smooth line through them. (-6; -8),(-2; 0),
(2; 8), (6; 16)
10.10.5
Notation used for Functions
Thus far you would have seen that we can use y = 2x to represent a function. This notation
however gets confusing when you are working with more than one function. A more general form
of writing a function is to write the function as f (x), where f is the function name and x is the
independent variable. For example, f (x) = 2x and g(t) = 2t + 1 are two functions.
Both notations will be used in this book.
Worked Example 46: Function notation
Question: If f (n) = n2 − 6n + 9, find f (k − 1) in terms of k.
Answer
Step 1 : Replace n with k − 1
f (n) =
f (k − 1) =
n2 − 6n + 9
(k − 1)2 − 6(k − 1) + 9
Step 2 : Remove brackets on RHS and simplify
=
=
k 2 − 2k + 1 − 6k + 6 + 9
k 2 − 8k + 16
Worked Example 47: Function notation
Question: If f (x) = x2 − 4, calculate b if f (b) = 45.
Answer
Step 1 : Replace x with b
f (b) =
butf (b) =
110
b2 − 4
45
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.10
Step 2 : f(b) = f(b)
b2 − 4 =
2
b − 49 =
b =
45
0
+7or − 7
{ExerciseRecap
1. Guess the function in the form y = . . . that has the values listed in the table.
x
y
1
1
2
2
3
3
40
40
50
50
600
600
700
700
800
800
900
900
1000
1000
2. Guess the function in the form y = . . . that has the values listed in the table.
x
y
1
2
2
4
3
6
40
80
50
100
600
1200
700
1400
800
1600
900
1800
1000
2000
3. Guess the function in the form y = . . . that has the values listed in the table.
x
y
1
10
2
20
3
30
40
400
50
500
600
6000
700
7000
800
8000
900
9000
1000
10000
4. On a Cartesian plane, plot the following points: (1,2), (2,4), (3,6), (4,8), (5,10). Join the
points. Do you get a straight-line?
5. If f (x) = x + x2 , write out:
(a) f (t)
(b) f (a)
(c) f (1)
(d) f (3)
6. If g(x) = x and f (x) = 2x, write out:
(a) f (t) + g(t)
(b) f (a) − g(a)
(c) f (1) + g(2)
(d) f (3) + g(s)
7. A car drives by you on a straight highway. The car is travelling 10 m every second.
Complete the table below by filling in how far the car has travelled away from you after 5,
10 and 20 seconds.
Time (s)
Distance (m)
0
0
1
10
2
20
5
10
20
Use the values in the table and draw a graph of distance on the y-axis and time on the
x-axis.
111
10.11
10.11
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Characteristics of Functions - All Grades
There are many characteristics of graphs that help describe the graph of any function. These
properties are:
1. dependent and independent variables
2. domain and range
3. intercepts with axes
4. turning points
5. asymptotes
6. lines of symmetry
7. intervals on which the function increases/decreases
8. continuous nature of the function
Some of these words may be unfamiliar to you, but each will be clearly described. Examples of
these properties are shown in Figure 10.3.
bE
f (x)
3
2
1
Bb
g(x)
3
2
1
bA
bF
−3 −2 −1
−1
bC
1
−2
2
3
h
−3 −2 −1
−1
2
3
−2
b
D
−3
1
−3
(a)
A
B, C, F
D, E
(b)
y-intercept
x-intercept
turning points
Figure 10.3: (a) Example graphs showing the characteristics of a function. (b) Example graph
showing asymptotes of a function.
10.11.1
Dependent and Independent Variables
Thus far, all the graphs you have drawn have needed two values, an x-value and a y-value. The
y-value is usually determined from some relation based on a given or chosen x-value. These
values are given special names in mathematics. The given or chosen x-value is known as the
independent variable, because its value can be chosen freely. The calculated y-value is known
as the dependent variable, because its value depends on the chosen x-value.
112
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.11.2
10.11
Domain and Range
The domain of a relation is the set of all the x values for which there exists at least one y value
according to that relation. The range is the set of all the y values, which can be obtained using
at least one x value.
If the relation is of height to people, then the domain is all living people, while the range would
be about 0.1 to 3 metres — no living person can have a height of 0m, and while strictly its not
impossible to be taller than 3 metres, no one alive is. An important aspect of this range is that
it does not contain all the numbers between 0.1 and 3, but only six billion of them (as many as
there are people).
As another example, suppose x and y are real valued variables, and we have the relation y = 2x .
Then for any value of x, there is a value of y, so the domain of this relation is the whole set of
real numbers. However, we know that no matter what value of x we choose, 2x can never be
less than or equal to 0. Hence the range of this function is all the real numbers strictly greater
than zero.
These are two ways of writing the domain and range of a function, set notation and interval
notation. Both notations are used in mathematics, so you should be familiar with each.
Set Notation
A set of certain x values has the following form:
{x : conditions, more conditions}
(10.6)
We read this notation as “the set of all x values where all the conditions are satisfied”. For
example, the set of all positive real numbers can be written as {x : x ∈ R, x > 0} which reads
as “the set of all x values where x is a real number and is greater than zero”.
Interval Notation
Here we write an interval in the form ’lower bracket, lower number, comma, upper number,
upper bracket’. We can use two types of brackets, square ones [, ] or round ones (, ). A square
bracket means including the number at the end of the interval whereas a round bracket means
excluding the number at the end of the interval. It is important to note that this notation can
only be used for all real numbers in an interval. It cannot be used to describe integers in an
interval or rational numbers in an interval.
So if x is a real number greater than 2 and less than or equal to 8, then x is any number in the
interval
(2,8]
(10.7)
It is obvious that 2 is the lower number and 8 the upper number. The round bracket means
’excluding 2’, since x is greater than 2, and the square bracket means ’including 8’ as x is less
than or equal to 8.
10.11.3
Intercepts with the Axes
The intercept is the point at which a graph intersects an axis. The x-intercepts are the points
at which the graph cuts the x-axis and the y-intercepts are the points at which the graph cuts
the y-axis.
In Figure 10.3(a), the A is the y-intercept and B, C and F are x-intercepts.
You will usually need to calculate the intercepts. The two most important things to remember
is that at the x-intercept, y = 0 and at the y-intercept, x = 0.
For example, calculate the intercepts of y = 3x + 5. For the y-intercept, x = 0. Therefore the
y-intercept is yint = 3(0) + 5 = 5. For the x-intercept, y = 0. Therefore the x-intercept is
found from 0 = 3xint + 5, giving xint = − 35 .
113
10.11
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.11.4
Turning Points
Turning points only occur for graphs of functions that whose highest power is greater than 1.
For example, graphs of the following functions will have turning points.
f (x)
g(x)
h(x)
= 2x2 − 2
= x3 − 2x2 + x − 2
2 4
x −2
=
3
There are two types of turning points: a minimal turning point and a maximal turning point.
A minimal turning point is a point on the graph where the graph stops decreasing in value and
starts increasing in value and a maximal turning point is a point on the graph where the graph
stops increasing in value and starts decreasing. These are shown in Figure 10.4.
y
y
b
2
y-values decreasing
1
y-values increasing
1
0
−1
minimal turning point
b
2 maximal turning point
0
y-values increasing
−1
−2
(a)
y-values decreasing
−2
(b)
Figure 10.4: (a) Maximal turning point. (b) Minimal turning point.
In Figure 10.3(a), E is a maximal turning point and D is a minimal turning point.
10.11.5
Asymptotes
An asymptote is a straight or curved line, which the graph of a function will approach, but never
touch.
In Figure 10.3(b), the y-axis and line h are both asymptotes as the graph approaches both these
lines, but never touches them.
10.11.6
Lines of Symmetry
Graphs look the same on either side of lines of symmetry. These lines include the x- and yaxes. For example, in Figure 10.5 is symmetric about the y-axis. This is described as the axis
of symmetry.
10.11.7
Intervals on which the Function Increases/Decreases
In the discussion of turning points, we saw that the graph of a function can start or stop
increasing or decreasing at a turning point. If the graph in Figure 10.3(a) is examined, we find
that the values of the graph increase and decrease over different intervals. We see that the
graph increases (i.e. that the y-values increase) from -∞ to point E, then it decreases (i.e. the
y-values decrease) from point E to point D and then it increases from point D to +∞.
10.11.8
Discrete or Continuous Nature of the Graph
A graph is said to be continuous if there are no breaks in the graph. For example, the graph in
Figure 10.3(a) can be described as a continuous graph, while the graph in Figure 10.3(b) has a
114
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.11
1
−2
−1
1
2
−1
Figure 10.5: Demonstration of axis of symmetry. The y-axis is an axis of symmetry, because the
graph looks the same on both sides of the y-axis.
break around the asymptotes. In Figure 10.3(b), it is clear that the graph does have a break in
it around the asymptote.
Exercise: Domain and Range
1. The domain of the function f (x) = 2x + 5 is -3; -3; -3; 0. Determine the range
of f .
2. If g(x) = −x2 + 5 and x is between - 3 and 3, determine:
(a) the domain of g(x)
(b) the range of g(x)
3. Label, on the following graph:
(a)
(b)
(c)
(d)
the x-intercept(s)
the y-intercept(s)
regions where the graph is increasing
regions where the graph is decreasing
4
3
2
1
−3
−2
−1
1
−1
−2
−3
4. Label, on the following graph:
(a) the x-intercept(s)
(b) the y-intercept(s)
115
2
3
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CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
(c) regions where the graph is increasing
(d) regions where the graph is decreasing
3
2
1
−2
−1
1
2
−1
−2
10.12
Graphs of Functions
10.12.1
Functions of the form y = ax + q
Functions with a general form of y = ax + q are called straight line functions. In the equation,
y = ax + q, a and q are constants and have different effects on the graph of the function. The
general shape of the graph of functions of this form is shown in Figure 10.6 for the function
f (x) = 2x + 3.
12
b
b
9
b
6
3
− 32
−5
−4
−3
b
b
b
−2
b
b
b
−1
1
2
3
−3
−6
Figure 10.6: Graph of f (x) = 2x + 3
Activity :: Investigation : Functions of the Form y = ax + q
1. On the same set of axes, plot the following graphs:
(a) a(x) = x − 2
116
4
5
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
(b)
(c)
(d)
(e)
10.12
b(x) = x − 1
c(x) = x
d(x) = x + 1
e(x) = x + 2
Use your results to deduce the effect of q.
2. On the same set of axes, plot the following graphs:
(a)
(b)
(c)
(d)
(e)
f (x) = −2 · x
g(x) = −1 · x
h(x) = 0 · x
j(x) = 1 · x
k(x) = 2 · x
Use your results to deduce the effect of a.
You should have found that the value of a affects the slope of the graph. As a increases, the slope
of the graph increases. If a > 0 then the graph increases from left to right (slopes upwards).
If a < 0 then the graph increases from right to left (slopes downwards). For this reason, a is
referred to as the slope or gradient of a straight-line function.
You should have also found that the value of q affects where the graph passes through the y-axis.
For this reason, q is known as the y-intercept.
These different properties are summarised in Table 10.1.
Table 10.1: Table summarising general shapes and positions of graphs of functions of the form
y = ax + q.
a>0
a<0
q>0
q<0
Domain and Range
For f (x) = ax + q, the domain is {x : x ∈ R} because there is no value of x ∈ R for which
f (x) is undefined.
The range of f (x) = ax + q is also {f (x) : f (x) ∈ R} because there is no value of f (x) ∈ R
for which f (x) is undefined.
For example, the domain of g(x) = x − 1 is {x : x ∈ R} because there is no value of x ∈ R for
which g(x) is undefined. The range of g(x) is {g(x) : g(x) ∈ R}.
Intercepts
For functions of the form, y = ax + q, the details of calculating the intercepts with the x and y
axis is given.
117
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CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
The y-intercept is calculated as follows:
y
=
ax + q
yint
=
=
a(0) + q
q
(10.8)
(10.9)
(10.10)
For example, the y-intercept of g(x) = x − 1 is given by setting x = 0 to get:
g(x)
=
yint
=
=
x−1
0−1
−1
The x-intercepts are calculated as follows:
y
=
0 =
a · xint
=
xint
=
ax + q
(10.11)
a · xint + q
(10.12)
−q
q
−
a
(10.13)
(10.14)
For example, the x-intercepts of g(x) = x − 1 is given by setting y = 0 to get:
g(x) =
0 =
xint =
x−1
xint − 1
1
Turning Points
The graphs of straight line functions do not have any turning points.
Axes of Symmetry
The graphs of straight-line functions do not, generally, have any axes of symmetry.
Sketching Graphs of the Form f (x) = ax + q
In order to sketch graphs of the form, f (x) = ax + q, we need to determine three characteristics:
1. sign of a
2. y-intercept
3. x-intercept
Only two points are needed to plot a straight line graph. The easiest points to use are the
x-intercept (where the line cuts the x-axis) and the y-intercept.
For example, sketch the graph of g(x) = x − 1. Mark the intercepts.
Firstly, we determine that a > 0. This means that the graph will have an upward slope.
The y-intercept is obtained by setting x = 0 and was calculated earlier to be yint = −1. The
x-intercept is obtained by setting y = 0 and was calculated earlier to be xint = 1.
Worked Example 48: Drawing a straight line graph
Question: Draw the graph of y = 2x + 2
118
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.12
3
2
1
(1,0)
−4 −3 −2 −1
−1
b
1
b (0,-1)
2
3
4
−2
−3
−4
Figure 10.7: Graph of the function g(x) = x − 1
Answer
Step 1 : Find the y-intercept
For the intercept on the y-axis, let x = 0
y
=
2(0) + 2
=
2
Step 2 : Find the x-intercept
For the intercept on the x-axis, let y = 0
0 =
2x =
2x + 2
−2
x =
−1
Step 3 : Draw the graph
3
2
1
2x
+2
−2
y=
−3
−1
1
2
−1
−2
119
3
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CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Exercise: Intercepts
1. List the y-intercepts for the following straight-line graphs:
(a)
(b)
(c)
(d)
y=x
y =x−1
y = 2x − 1
y + 1 = 2x
2. Give the equation of the illustrated graph below:
y
(0;3)
(4;0)
x
3. Sketch the following relations on the same set of axes, clearly indicating the
intercepts with the axes as well as the co-ordinates of the point of interception
on the graph: x + 2y − 5 = 0 and 3x − y − 1 = 0
10.12.2
Functions of the Form y = ax2 + q
The general shape and position of the graph of the function of the form f (x) = ax2 + q is shown
in Figure 10.8.
Activity :: Investigation : Functions of the Form y = ax2 + q
1. On the same set of axes, plot the following graphs:
(a)
(b)
(c)
(d)
(e)
a(x) = −2 · x2 + 1
b(x) = −1 · x2 + 1
c(x) = 0 · x2 + 1
d(x) = 1 · x2 + 1
e(x) = 2 · x2 + 1
Use your results to deduce the effect of a.
2. On the same set of axes, plot the following graphs:
(a) f (x) = x2 − 2
(b) g(x) = x2 − 1
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CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.12
9
8
7
6
5
4
3
2
1
−4
−3
−2
−1 −1
1
2
3
4
Figure 10.8: Graph of the f (x) = x2 − 1.
(c) h(x) = x2 + 0
(d) j(x) = x2 + 1
(e) k(x) = x2 + 2
Use your results to deduce the effect of q.
Complete the following table of values for the functions a to k to help with drawing
the required graphs in this activity:
x
−2
−1
0
1
2
a(x)
b(x)
c(x)
d(x)
e(x)
f (x)
g(x)
h(x)
j(x)
k(x)
From your graphs, you should have found that a affects whether the graph makes a smile or a
frown. If a < 0, the graph makes a frown and if a > 0 then the graph makes a smile. This is
shown in Figure 10.9.
b
b
a > 0 (a positive smile)
b
b
a < 0 (a negative frown)
Figure 10.9: Distinctive shape of graphs of a parabola if a > 0 and a < 0.
You should have also found that the value of q affects whether the turning point is to the left
of the y-axis (q > 0) or to the right of the y-axis (q < 0).
These different properties are summarised in Table ??.
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CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
Table 10.2: Table summarising general shapes and positions of functions of the form y = ax2 +q.
a>0
a<0
q>0
q<0
Domain and Range
For f (x) = ax2 + q, the domain is {x : x ∈ R} because there is no value of x ∈ R for which
f (x) is undefined.
The range of f (x) = ax2 + q depends on whether the value for a is positive or negative. We
will consider these two cases separately.
If a > 0 then we have:
x2
ax2
ax2 + q
f (x)
≥ 0
≥ 0
(The square of an expression is always positive)
(Multiplication by a positive number maintains the nature of the inequality)
≥ q
≥ q
This tells us that for all values of x, f (x) is always greater than q. Therefore if a > 0, the range
of f (x) = ax2 + q is {f (x) : f (x) ∈ [q,∞)}.
Similarly, it can be shown that if a < 0 that the range of f (x) = ax2 + q is {f (x) : f (x) ∈
(−∞,q]}. This is left as an exercise.
For example, the domain of g(x) = x2 + 2 is {x : x ∈ R} because there is no value of x ∈ R for
which g(x) is undefined. The range of g(x) can be calculated as follows:
x2
2
x +2
g(x)
≥
0
≥
≥
2
2
Therefore the range is {g(x) : g(x) ∈ [2,∞)}.
Intercepts
For functions of the form, y = ax2 + q, the details of calculating the intercepts with the x and
y axis is given.
The y-intercept is calculated as follows:
y
yint
= ax2 + q
2
= a(0) + q
= q
122
(10.15)
(10.16)
(10.17)
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.12
For example, the y-intercept of g(x) = x2 + 2 is given by setting x = 0 to get:
g(x) =
x2 + 2
yint
02 + 2
2
=
=
The x-intercepts are calculated as follows:
y
=
0 =
ax2int =
xint
=
ax2 + q
(10.18)
ax2int
(10.19)
(10.20)
+q
−q
r
q
± −
a
(10.21)
However, (10.21) is only valid if − aq > 0 which means that either q < 0 or a < 0. This is
consistent with what we expect, since if q > 0 and a > 0 then − aq is negative and in this case
the graph lies above the x-axis and therefore does not intersect the x-axis. If however, q > 0
and a < 0, then − aq is positive and the graph is hat shaped and should have two x-intercepts.
Similarly, if q < 0 and a > 0 then − aq is also positive, and the graph should intersect with the
x-axis.
For example, the x-intercepts of g(x) = x2 + 2 is given by setting y = 0 to get:
g(x) =
0 =
−2 =
x2 + 2
x2int + 2
x2int
which is not real. Therefore, the graph of g(x) = x2 + 2 does not have any x-intercepts.
Turning Points
The turning point of the function of the form f (x) = ax2 + q is given by examining the range of
the function. We know that if a > 0 then the range of f (x) = ax2 + q is {f (x) : f (x) ∈ [q,∞)}
and if a < 0 then the range of f (x) = ax2 + q is {f (x) : f (x) ∈ (−∞,q]}.
So, if a > 0, then the lowest value that f (x) can take on is q. Solving for the value of x at
which f (x) = q gives:
q
=
ax2tp + q
0 =
ax2tp
0 =
x2tp
xtp
=
0
∴ x = 0 at f (x) = q. The co-ordinates of the (minimal) turning point is therefore (0; q).
Similarly, if a < 0, then the highest value that f (x) can take on is q and the co-ordinates of the
(maximal) turning point is (0; q).
Axes of Symmetry
There is one axis of symmetry for the function of the form f (x) = ax2 + q that passes through
the turning point. Since the turning point lies on the y-axis, the axis of symmetry is the y-axis.
Sketching Graphs of the Form f (x) = ax2 + q
In order to sketch graphs of the form, f (x) = ax2 + q, we need to calculate determine four
characteristics:
1. sign of a
123
10.12
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
2. domain and range
3. turning point
4. y-intercept
5. x-intercept
For example, sketch the graph of g(x) = − 21 x2 − 3. Mark the intercepts, turning point and axis
of symmetry.
Firstly, we determine that a < 0. This means that the graph will have a maximal turning point.
The domain of the graph is {x : x ∈ R} because f (x) is defined for all x ∈ R. The range of the
graph is determined as follows:
x2
1
− x2
2
1
− x2 − 3
2
∴ f (x)
≥ 0
≤ 0
≤ −3
≤ −3
Therefore the range of the graph is {f (x) : f (x) ∈ (−∞, − 3]}.
Using the fact that the maximum value that f (x) achieves is -3, then the y-coordinate of the
turning point is -3. The x-coordinate is determined as follows:
1
− x2 − 3 =
2
−3
1
− x2 − 3 + 3 =
2
1
− x2 =
2
Divide both sides by − 21 : x2 =
0
Take square root of both sides: x =
0
x =
0
∴
The coordinates of the turning point are: (0, − 3).
The y-intercept is obtained by setting x = 0. This gives:
yint
=
=
=
1
− (0)2 − 3
2
1
− (0) − 3
2
−3
The x-intercept is obtained by setting y = 0. This gives:
0
3
−3 · 2
−6
1
= − x2int − 3
2
1
= − x2int
2
= x2int
= x2int
which is not real. Therefore, there are no x-intercepts.
We also know that the axis of symmetry is the y-axis.
124
0
0
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
−4 −3 −2 −1
−1
1
2
3
10.12
4
−2
−3
b (0,-3)
−4
−5
−6
Figure 10.10: Graph of the function f (x) = − 12 x2 − 3
Exercise: Parabolas
1. Show that if a < 0 that the range of f (x) = ax2 +q is {f (x) : f (x) ∈ (−∞,q]}.
2. Draw the graph of the function y = −x2 + 4 showing all intercepts with the
axes.
3. Two parabolas are drawn: g : y = ax2 + p and h : y = bx2 + q.
y
g
23
(-4; 7)
(4; 7)
x
3
h
-9
(a)
(b)
(c)
(d)
10.12.3
Find the values of a and p.
Find the values of b and q.
Find the values of x for which ax2 + p ≥ bx2 + q.
For what values of x is g increasing ?
Functions of the Form y =
a
x
+q
Functions of the form y = xa + q are known as hyperbolic functions. The general form of the
graph of this function is shown in Figure 10.11.
Activity :: Investigation : Functions of the Form y =
1. On the same set of axes, plot the following graphs:
(a) a(x) = −2
x +1
(b) b(x) = −1
x +1
125
a
x
+q
10.12
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
5
4
3
2
1
−4 −3 −2 −1
−1
1
2
3
4
−2
Figure 10.11: General shape and position of the graph of a function of the form f (x) =
(c) c(x) =
(d) d(x) =
(e) e(x) =
a
x
+ q.
0
x +1
+1
x +1
+2
x +1
Use your results to deduce the effect of a.
2. On the same set of axes, plot the following graphs:
(a) f (x) = x1 − 2
(b) g(x) = x1 − 1
(c) h(x) = x1 + 0
(d) j(x) = x1 + 1
(e) k(x) = x1 + 2
Use your results to deduce the effect of q.
You should have found that the value of a affects whether the graph is located in the first and
third quadrants of Cartesian plane.
You should have also found that the value of q affects whether the graph lies above the x-axis
(q > 0) or below the x-axis (q < 0).
These different properties are summarised in Table 10.3. The axes of symmetry for each graph
are shown as a dashed line.
Domain and Range
For y =
a
x
+ q, the function is undefined for x = 0. The domain is therefore {x : x ∈ R,x 6= 0}.
We see that y =
a
x
+ q can be re-written as:
y
=
y−q
=
If x 6= 0 then: (y − q)(x)
=
x =
a
+q
x
a
x
a
a
y−q
This shows that the function is undefined at y = q. Therefore the range of f (x) =
{f (x) : f (x) ∈ (−∞,q) ∪ (q,∞)}.
126
a
x
+ q is
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.12
Table 10.3: Table summarising general shapes and positions of functions of the form y =
The axes of symmetry are shown as dashed lines.
a>0
a<0
a
x
+ q.
q>0
q<0
For example, the domain of g(x) =
x = 0.
2
x
+ 2 is {x : x ∈ R, x 6= 0} because g(x) is undefined at
y
=
(y − 2) =
If x 6= 0 then: x(y − 2) =
x =
2
+2
x
2
x
2
2
y−2
We see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}.
Intercepts
For functions of the form, y = xa + q, the intercepts with the x and y axis is calculated by setting
x = 0 for the y-intercept and by setting y = 0 for the x-intercept.
The y-intercept is calculated as follows:
y
=
yint
=
a
+q
x
a
+q
0
which is undefined. Therefore there is no y-intercept.
For example, the y-intercept of g(x) =
2
x
+ 2 is given by setting x = 0 to get:
y
=
yint
=
which is undefined.
127
2
+2
x
2
+2
0
(10.22)
(10.23)
10.12
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
The x-intercepts are calculated by setting y = 0 as follows:
y
a
+q
x
a
+q
xint
=
0 =
a
xint
a
xint
For example, the x-intercept of g(x) =
2
x
(10.24)
(10.25)
=
−q
(10.26)
=
−q(xint )
a
−q
(10.27)
=
(10.28)
(10.29)
+ 2 is given by setting x = 0 to get:
y
=
0
=
−2 =
−2(xint )
=
xint
=
xint
=
2
+2
x
2
+2
xint
2
xint
2
2
−2
−1
Asymptotes
There are two asymptotes for functions of the form y =
the domain and range.
a
x +q.
They are determined by examining
We saw that the function was undefined at x = 0 and for y = q. Therefore the asymptotes are
x = 0 and y = q.
For example, the domain of g(x) = x2 + 2 is {x : x ∈ R, x 6= 0} because g(x) is undefined at
x = 0. We also see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈
(−∞,2) ∪ (2,∞)}.
From this we deduce that the asymptotes are at x = 0 and y = 2.
Sketching Graphs of the Form f (x) =
a
x
+q
In order to sketch graphs of functions of the form, f (x) =
four characteristics:
a
x
+ q, we need to calculate determine
1. domain and range
2. asymptotes
3. y-intercept
4. x-intercept
For example, sketch the graph of g(x) =
2
x
+ 2. Mark the intercepts and asymptotes.
We have determined the domain to be {x : x ∈ R, x 6= 0} and the range to be {g(x) : g(x) ∈
(−∞,2) ∪ (2,∞)}. Therefore the asymptotes are at x = 0 and y = 2.
There is no y-intercept and the x-intercept is xint = −1.
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CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.12
6
5
4
3
2
1
−4 −3 −2 −1
−1
1
2
3
4
−2
−3
Figure 10.12: Graph of g(x) =
2
x
+ 2.
Exercise: Graphs
1. Using grid paper, draw the graph of xy = −6.
(a) Does the point (-2; 3) lie on the graph ? Give a reason for your answer.
(b) Why is the point (-2; -3) not on the graph ?
(c) If the x-value of a point on the drawn graph is 0,25, what is the corresponding y-value ?
(d) What happens to the y-values as the x-values become very large ?
(e) With the line y = −x as line of symmetry, what is the point symmetrical
to (-2; 3) ?
2. Draw the graph of xy = 8.
(a) How would the graph y = 83 + 3 compare with that of xy = 8? Explain
your answer fully.
(b) Draw the graph of y = 38 + 3 on the same set of axes.
10.12.4
Functions of the Form y = ab(x) + q
Functions of the form y = ab(x) + q are known as exponential functions. The general shape of
a graph of a function of this form is shown in Figure 10.13.
Activity :: Investigation : Functions of the Form y = ab(x) + q
1. On the same set of axes, plot the following graphs:
(a) a(x) = −2 · b(x) + 1
(b) b(x) = −1 · b(x) + 1
(c) c(x) = −0 · b(x) + 1
129
10.12
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
4
3
2
1
−4 −3 −2 −1
1
2
3
4
Figure 10.13: General shape and position of the graph of a function of the form f (x) = ab(x) +q.
(d) d(x) = −1 · b(x) + 1
(e) e(x) = −2 · b(x) + 1
Use your results to deduce the effect of a.
2. On the same set of axes, plot the following graphs:
(a)
(b)
(c)
(d)
(e)
f (x) = 1 · b(x) − 2
g(x) = 1 · b(x) − 1
h(x) = 1 · b(x) + 0
j(x) = 1 · b(x) + 1
k(x) = 1 · b(x) + 2
Use your results to deduce the effect of q.
You should have found that the value of a affects whether the graph curves upwards (a > 0) or
curves downwards (a < 0).
You should have also found that the value of q affects the position of the y-intercept.
These different properties are summarised in Table 10.4.
Table 10.4: Table summarising general shapes and positions of functions of the form y =
ab(x) + q.
a>0
a<0
q>0
q<0
Domain and Range
For y = ab(x) + q, the function is defined for all real values of x. Therefore, the domain is
{x : x ∈ R}.
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CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.12
The range of y = ab(x) + q is dependent on the sign of a.
If a > 0 then:
b(x)
a · b(x)
a · b(x) + q
f (x)
≥ 0
≥ 0
≥ q
≥ q
Therefore, if a > 0, then the range is {f (x) : f (x) ∈ [q,∞)}.
If a < 0 then:
b(x)
(x)
a·b
a · b(x) + q
f (x)
≤ 0
≤ 0
≤ q
≤ q
Therefore, if a < 0, then the range is {f (x) : f (x) ∈ (−∞,q]}.
For example, the domain of g(x) = 3 · 2x + 2 is {x : x ∈ R}. For the range,
2x
3 · 2x
≥
≥
3 · 2x + 2 ≥
0
0
2
Therefore the range is {g(x) : g(x) ∈ [2,∞)}.
Intercepts
For functions of the form, y = ab(x) + q, the intercepts with the x and y axis is calculated by
setting x = 0 for the y-intercept and by setting y = 0 for the x-intercept.
The y-intercept is calculated as follows:
y
yint
= ab(x) + q
= ab(0) + q
(10.30)
(10.31)
= a(1) + q
= a+q
(10.32)
(10.33)
For example, the y-intercept of g(x) = 3 · 2x + 2 is given by setting x = 0 to get:
y
yint
=
=
=
=
3 · 2x + 2
3 · 20 + 2
3+2
5
The x-intercepts are calculated by setting y = 0 as follows:
y =
0 =
ab(xint )
=
(xint )
=
b
ab(x) + q
ab(xint ) + q
(10.34)
(10.35)
−q
q
−
a
(10.36)
(10.37)
Which only has a real solution if either a < 0 or q < 0. Otherwise, the graph of the function of
form y = ab(x) + q does not have any x-intercepts.
131
10.12
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
For example, the x-intercept of g(x) = 3 · 2x + 2 is given by setting y = 0 to get:
y
= 3 · 2x + 2
0 = 3 · 2xint + 2
−2 = 3 · 2xint
−2
2xint =
3
which has no real solution. Therefore, the graph of g(x) = 3 · 2x + 2 does not have any
x-intercepts.
Asymptotes
There are two asymptotes for functions of the form y = ab(x) + q. They are determined by
examining the domain and range.
We saw that the function was undefined at x = 0 and for y = q. Therefore the asymptotes are
x = 0 and y = q.
For example, the domain of g(x) = 3 · 2x + 2 is {x : x ∈ R, x 6= 0} because g(x) is undefined
at x = 0. We also see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈
(−∞,2) ∪ (2,∞)}.
From this we deduce that the asymptotes are at x = 0 and y = 2.
Sketching Graphs of the Form f (x) = ab(x) + q
In order to sketch graphs of functions of the form, f (x) = ab(x) + q, we need to calculate
determine four characteristics:
1. domain and range
2. y-intercept
3. x-intercept
For example, sketch the graph of g(x) = 3 · 2x + 2. Mark the intercepts.
We have determined the domain to be {x : x ∈ R} and the range to be {g(x) : g(x) ∈ [2,∞)}.
The y-intercept is yint = 5 and there are no x-intercepts.
6
5
4
3
2
1
−4 −3 −2 −1
1
2
3
4
Figure 10.14: Graph of g(x) = 3 · 2x + 2.
132
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
10.13
Exercise: Exponential Functions and Graphs
1. Draw the graphs of y = 2x and y = ( 12 )x on the same set of axes.
(a) Is the x-axis and asymptote or and axis of symmetry to both graphs ?
Explain your answer.
(b) Which graph is represented by the equation y = 2−x ? Explain your
answer.
(c) Solve the equation 2x = ( 12 )x graphically and check that your answer is
correct by using substitution.
(d) Predict how the graph y = 2.2x will compare to y = 2x and then draw the
graph of y = 2.2x on the same set of axes.
2. The curve of the exponential function f in the accompanying diagram cuts the
f
b
4
B(2,4)
3
2
b
1
y-axis at the point A(0; 1) and B(2; 4) is on f .
A(0,1)
0
1
2
(a) Determine the equation of the function f .
(b) Determine the equation of h, the function of which the curve is the reflection of the curve of f in the x-axis.
(c) Determine the range of h.
10.13
End of Chapter Exercises
1. Given the functions f (x) = −2x2 − 18 and g(x) = −2x + 6
(a) Draw f and g on the same set of axes.
(b) Calculate the points of intersection of f and g.
(c) Hence use your graphs and the points of intersection to solve for x when:
i. f (x) > 0
ii.
f (x)
g(x)
≤0
(d) Give the equation of the reflection of f in the x-axis.
2. After a ball is dropped, the rebound height of each bounce decreases. The equation
y = 5(0.8)x shows the relationship between x, the number of bounces, and y, the height
of the bounce, for a certain ball. What is the approximate height of the fifth bounce of
this ball to the nearest tenth of a unit ?
3. Marc had 15 coins in five rand and two rand pieces. He had 3 more R2-coins than R5coins. He wrote a system of equations to represent this situation, letting x represent the
number of five rand coins and y represent the number of two rand coins. Then he solved
the system by graphing.
133
x
10.13
CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10
(a) Write down the system of equations.
(b) Draw their graphs on the same set of axes.
(c) What is the solution?
134
Chapter 11
Average Gradient - Grade 10
Extension
11.1
Introduction
In chapter 10.7.4, we saw that the gradient of a straight line graph is calculated as:
y2 − y1
x2 − x1
(11.1)
for two points (x1 ,y1 ) and (x2 ,y2 ) on the graph.
We can now define the average gradient between any two points, (x1 ,y1 ) and (x2 ,y2 ) as:
y2 − y1
.
x2 − x1
(11.2)
This is the same as (11.1).
11.2
Straight-Line Functions
Activity :: Investigation : Average Gradient - Straight Line Function
Fill in the table by calculating the average gradient over the indicated intervals
for the function f (x) = 2x − 2:
x1
A-B
A-C
B-C
x2
y1
y2
y2 −y1
x2 −x1
y
2
C(2,2)
b
1
b
−1
1
−1
−2
−3
A(-1,-4)
135
b
−4
B(1,0)
x
11.3
CHAPTER 11. AVERAGE GRADIENT - GRADE 10 EXTENSION
What do you notice about the gradients over each interval?
The average gradient of a straight-line function is the same over any two intervals on the function.
11.3
Parabolic Functions
Activity :: Investigation : Average Gradient - Parabolic Function
Fill in the table by calculating the average gradient over the indicated intervals
for the function f (x) = 2x − 2:
x1
x2
y1
y2
b A(-3,7)
y2 −y1
x2 −x1
b G(3,7)
y
A-B
B-C
C-D
D-E
E-F
F-G
What do you notice about the average
gradient over each interval? What can
you say about the average gradients between A and D compared to the average
gradients between D and G?
B(-2,2)
F(2,2)
b
b
C(-1,-1)
x
b
b E(1,-1)
b
D(0,-2)
The average gradient of a parabolic function depends on the interval and is the gradient of a
straight line that passes through the points on the interval.
For example, in Figure 11.1 the various points have been joined by straight-lines. The average
gradients between the joined points are then the gradients of the straight lines that pass through
the points.
b A(-3,7)
B(-2,2)
b G(3,7)
y
F(2,2)
b
C(-1,-1)
b
b
b E(1,-1)
x
b
D(0,-2)
Figure 11.1: The average gradient between two points on a curve is the gradient of the straight
line that passes through the points.
Method: Average Gradient
Given the equation of a curve and two points (x1 , x2 ):
136
CHAPTER 11. AVERAGE GRADIENT - GRADE 10 EXTENSION
1. Write the equation of the curve in the form y = . . ..
2. Calculate y1 by substituting x1 into the equation for the curve.
3. Calculate y2 by substituting x2 into the equation for the curve.
4. Calculate the average gradient using:
y2 − y1
x2 − x1
Worked Example 49: Average Gradient
Question: Find the average gradient of the curve y = 5x2 − 4 between the points
x = −3 and x = 3
Answer
Step 1 : Label points
Label the points as follows:
x1 = −3
x2 = 3
to make it easier to calculate the gradient.
Step 2 : Calculate the y coordinates
We use the equation for the curve to calculate the y-value at x1 and x2 .
y1
= 5x21 − 4
= 5(−3)2 − 4
= 5(9) − 4
= 41
y2
=
=
=
=
5x22 − 4
5(3)2 − 4
5(9) − 4
41
Step 3 : Calculate the average gradient
y2 − y1
x2 − x1
41 − 41
3 − (−3)
0
=
3+3
0
=
6
= 0
=
Step 4 : Write the final answer
The average gradient between x = −3 and x = 3 on the curve y = 5x2 − 4 is 0.
137
11.3
11.4
11.4
CHAPTER 11. AVERAGE GRADIENT - GRADE 10 EXTENSION
End of Chapter Exercises
1. An object moves according to the function d = 2t2 + 1 , where d is the distance in metres
and t the time in seconds. Calculate the average speed of the object between 2 and 3
seconds.
2. Given: f (x) = x3 − 6x.
Determine the average gradient between the points where x = 1 and x = 4.
138
Chapter 12
Geometry Basics
12.1
Introduction
The purpose of this chapter is to recap some of the ideas that you learned in geometry and
trigonometry in earlier grades. You should feel comfortable with the work covered in this chapter
before attempting to move onto the Grade 10 Geometry Chapter (Chapter 13) or the Grade 10
Trigonometry Chapter (Chapter 14). This chapter revises:
1. Terminology: quadrilaterals, vertices, sides, angles, parallel lines, perpendicular lines,
diagonals, bisectors, transversals
2. Similarities and differences between quadrilaterals
3. Properties of triangles and quadrilaterals
4. Congruence
5. Classification of angles into acute, right, obtuse, straight, reflex or revolution
6. Theorem of Pythagoras which is used to calculate the lengths of sides of a right-angled
triangle
12.2
Points and Lines
The two simplest objects in geometry are points and lines.
A point is something that is not very wide or high and is usually used in geometry as a marker of
a position. Points are usually labelled with a capital letter. Some examples of points are shown
in Figure 12.1.
A line is formed when many points are placed next to each other. Lines can be straight or
curved, but are always continuous. This means that there are never any breaks in the lines. The
endpoints of lines are labelled with capital letters. Examples of two lines are shown in Figure 12.1.
bS
b
b
R
Q
P
Some points
b
B
C
D
E
Some lines
Figure 12.1: Examples of some points (labelled P , Q, R and S) and some lines (labelled BC and DE).
139
12.3
CHAPTER 12. GEOMETRY BASICS
Lines are labelled according to the start point and end point. We call the line that starts at a
point A and ends at a point B, AB. Since the line from point B to point A is the same as the
line from point A to point B, we have that AB=BA.
The length of the line between points A and B is AB. So if we say AB = CD we mean that
the length of the line between A and B is equal to the length of the line between C and D.
In science, we sometimes talk about a vector and this is just a fancy way of saying the we are
referring to the line that starts at one point and moves in the direction of the other point. We
~ referring to the vector from the point A
label a vector in a similar manner to a line, with AB
~ is the line segment
with length AB and in the direction from point A to point B. Similarly, BA
with the same length but direction from point B to point A. Usually, vectors are only equal if
~ 6= BA.
~
they have the same length and same direction. So, usually, AB
A line is measured in units of length. Some common units of length are listed in Table 12.1.
Table 12.1: Some common units of length and their abbreviations.
Unit of Length Abbreviation
kilometre
km
metre
m
centimetre
cm
millimetre
mm
12.3
Angles
An angle is formed when two straight lines meet at a point. The point at which two lines meet
is known as a vertex. Angles are labelled with a ˆ on a letter, for example, in Figure 12.3, the
angle is at B̂. Angles can also be labelled according to the line segments that make up the angle.
For example, in Figure 12.3, the angle is made up when line segments CB and BA meet. So,
the angle can be referred to as ∠CBA or ∠ABC. The ∠ symbol is a short method of writing
angle in geometry.
Angles are measured in degrees which is denoted by ◦ .
C
B
A
Figure 12.2: Angle labelled as B̂, ∠CBA or ∠ABC
C
F
A
(a)
B
E
(b)
G
Figure 12.3: Examples of angles. Â = Ê, even though the lines making up the angles are of
different lengths.
140
CHAPTER 12. GEOMETRY BASICS
12.3.1
12.3
Measuring angles
The size of an angle does not depend on the length of the lines that are joined to make up the
angle, but depends only on how both the lines are placed as can be seen in Figure 12.3. This
means that the idea of length cannot be used to measure angles. An angle is a rotation around
the vertex.
Using a Protractor
A protractor is a simple tool that is used to measure angles. A picture of a protractor is shown
in Figure 12.4.
90◦
◦
120
150◦
60◦
30◦
180◦
0◦
Figure 12.4: Diagram of a protractor.
Method:
Using a protractor
1. Place the bottom line of the protractor along one line of the angle.
2. Move the protractor along the line so that the centre point on the protractor is at the
vertex of the two lines that make up the angle.
3. Follow the second line until it meets the marking on the protractor and read off the angle.
Make sure you start measuring at 0◦ .
Activity :: Measuring Angles : Use a protractor to measure the following
angles:
12.3.2
Special Angles
What is the smallest angle that can be drawn? The figure below shows two lines (CA and AB)
making an angle at a common vertex A. If line CA is rotated around the common vertex A,
down towards line AB, then the smallest angle that can be drawn occurs when the two lines are
pointing in the same direction. This gives an angle of 0◦ .
141
12.3
CHAPTER 12. GEOMETRY BASICS
C
swing point C down
towards AB
0◦
A
C
C
A
B
B
b
180◦
A
B
If line CA is now swung upwards, any other angle can be obtained. If line CA and line AB
point in opposite directions (the third case in the figure) then this forms an angle of 180◦ .
Important: If three points A, B and C lie on a straight line, then the angle between them
is 180◦ . Conversely, if the angle between three points is 180◦ , then the points lie on a
straight line.
An angle of 90◦ is called a right angle. A right angle is half the size of the angle made by a
straight line (180◦ ). We say CA is perpendicular to AB or CA ⊥ AB. An angle twice the size
of a straight line is 360◦ . An angle measuring 360◦ looks identical to an angle of 0◦ , except for
the labelling. We call this a revolution.
C
90◦
A
360◦
B
A
b
C
B
Figure 12.5: An angle of 90◦ is known as a right angle.
Extension: Angles larger than 360◦
All angles larger than 360◦ also look like we have seen them before. If you are given
an angle that is larger than 360◦ , continue subtracting 360◦ from the angle, until
you get an answer that is between 0◦ and 360◦ . Angles that measure more than 360◦
are largely for mathematical convenience.
Important:
• Acute angle: An angle ≥ 0◦ and < 90◦ .
• Right angle: An angle measuring 90◦ .
• Obtuse angle: An angle > 90◦ and < 180◦ .
• Straight angle: An angle measuring 180◦ .
• Reflex angle: An angle > 180◦ and < 360◦ .
• Revolution: An angle measuring 360◦ .
These are simply labels for angles in particular ranges, shown in Figure 12.6.
Once angles can be measured, they can then be compared. For example, all right angles are
90◦ , therefore all right angles are equal, and an obtuse angle will always be larger than an acute
angle.
142
CHAPTER 12. GEOMETRY BASICS
C
12.3
reflex
A
C
b
acute
A
b
obtuse
B
A
B
b
B
C
Figure 12.6: Three types of angles defined according to their ranges.
12.3.3
Special Angle Pairs
In Figure 12.7, straight lines AB and CD intersect at point X, forming four angles: X̂1 , X̂2 ,
X̂3 and X̂4 .
C
3
A
B
2
X
b
1
4
D
Figure 12.7: Two intersecting straight lines with vertical angles X̂1 ,X̂3 and X̂2 ,X̂4 .
The table summarises the special angle pairs that result.
Special Angle
adjacent angles
linear pair
(adjacent angles
on a straight
line)
vertically
opposite angles
supplementary
angles
complementary
angles
Property
share a common
vertex and a
common side
adjacent angles
formed by two
intersecting straight
lines that by
definition add to
180◦
angles formed by
two intersecting
straight lines that
share a vertex but
do not share any
sides
Example
(X̂1 ,X̂2 ), (X̂2 ,X̂3 ),
(X̂3 ,X̂4 ), (X̂4 ,X̂1 )
X̂1 + X̂2
X̂2 + X̂3
X̂3 + X̂4
X̂4 + X̂1
= 180◦
= 180◦
= 180◦
= 180◦
X̂1 = X̂3
X̂2 = X̂4
two angles whose sum is 180◦
two angles whose sum is 90◦
Important: The vertically opposite angles formed by the intersection of two straight lines
are equal. Adjacent angles on a straight line are supplementary.
12.3.4
Parallel Lines intersected by Transversal Lines
Two lines intersect if they cross each other at a point. For example, at a traffic intersection,
two or more streets intersect; the middle of the intersection is the common point between the
streets.
143
12.3
CHAPTER 12. GEOMETRY BASICS
Parallel lines are lines that never intersect. For example the tracks of a railway line are parallel.
We wouldn’t want the tracks to intersect as that would be catastrophic for the train!
All these lines are parallel to each other. Notice the arrow symbol for parallel.
teresting A section of the Australian National Railways Trans-Australian line is perhaps
Interesting
Fact
Fact
one of the longest pairs of man-made parallel lines.
Longest Railroad Straight (Source: www.guinnessworldrecords.com)
The Australian National Railways Trans-Australian line over the
Nullarbor Plain, is 478 km (297 miles) dead straight, from Mile 496,
between Nurina and Loongana, Western Australia, to Mile 793,
between Ooldea and Watson, South Australia.
A transversal of two or more lines is a line that intersects these lines. For example in Figure 12.8,
AB and CD are two parallel lines and EF is a transversal. We say AB k CD. The properties
of the angles formed by these intersecting lines are summarised in the table below.
F
h
C
g
D
a
b
d
A
c
B
e
f
E
Figure 12.8: Parallel lines intersected by a transversal
Extension: Euclid’s Parallel Line Postulate
If a straight line falling on two straight lines makes the two interior angles on the same
side less than two right angles (180◦ ), the two straight lines, if produced indefinitely,
will meet on that side. This postulate can be used to prove many identities about
the angles formed when two parallel lines are cut by a transversal.
144
CHAPTER 12. GEOMETRY BASICS
Name of angle
interior angles
exterior angles
12.3
Definition
the angles that lie
inside the parallel
lines
the angles that lie
outside the parallel
lines
Examples
Notes
a, b, c and d are
interior angles
the word interior
means inside
e, f , g and h are
exterior angles
the word exterior
means outside
alternate interior
angles
the interior angles
that lie on opposite
sides of the
transversal
(a,c) and (b,d) are
pairs of alternate
interior angles,
a = c, b = d
co-interior angles on
the same side
co-interior angles
that lie on the same
side of the
transversal
(a,d) and (b,c) are
interior angles on
the same side.
a + d = 180◦,
b + c = 180◦
corresponding
angles
the angles on the
same side of the
transversal and the
same side of the
parallel lines
(a,e), (b,f ), (c,g)
and (d,h) are pairs
of corresponding
angles. a = e,
b = f , c = g, d = h
Z shape
C shape
F shape
Important:
1. If two parallel lines are intersected by a transversal, the sum of the co-interior angles
on the same side of the transversal is 180◦ .
2. If two parallel lines are intersected by a transversal, the alternate interior angles are
equal.
3. If two parallel lines are intersected by a transversal, the corresponding angles are equal.
4. If two lines are intersected by a transversal such that any pair of co-interior angles on
the same side is supplementary, then the two lines are parallel.
5. If two lines are intersected by a transversal such that a pair of alternate interior angles
are equal, then the lines are parallel.
6. If two lines are intersected by a transversal such that a pair of alternate corresponding
angles are equal, then the lines are parallel.
Exercise: Angles
1. Use adjacent, corresponding, co-interior and
alternate angles to fill in all the angles labeled
with letters in the diagram alongside:
b
g
145
d
f
e
a
c
30◦
12.3
CHAPTER 12. GEOMETRY BASICS
E
A
1
B
2. Find all the unknown angles in the figure
alongside:
1
30◦
2
F
3
1
2
C
2
◦
100
D
A
G
1
3
3
H
1
D
X
4x
3. Find the value of x in the figure alongside:
x
Y
x+20◦
Z
B
C
4. Determine whether there are pairs of parallel lines in the following figures.
M
O
O
Q
S
1
A
115◦
K
Q
2
3
1
3
35◦
1
2
P
N
3
T
a)
R
b)
P
K
T
85◦
U
M
1
2
1
3
V
c)
Y
2
3
N
85
◦
L
C
5. If AB is parallel to CD and AB is parallel to
A
EF, prove that CD is parallel to EF:
E
146
3
L
55◦
B1
45◦
2
2
D
B
F
R
CHAPTER 12. GEOMETRY BASICS
12.4
12.4
Polygons
If you take some lines and join them such that the end point of the first line meets the starting
point of the last line, you will get a polygon. Each line that makes up the polygon is known as
a side. A polygon has interior angles. These are the angles that are inside the polygon. The
number of sides of a polygon equals the number of interior angles. If a polygon has equal length
sides and equal interior angles then the polygon is called a regular polygon. Some examples of
polygons are shown in Figure 12.9.
*
Figure 12.9: Examples of polygons. They are all regular, except for the one marked *
12.4.1
Triangles
A triangle is a three-sided polygon. There are four types of triangles: equilateral, isosceles,
right-angled and scalene. The properties of these triangles are summarised in Table 12.2.
Properties of Triangles
Activity :: Investigation : Sum of the angles in a triangle
1. Draw on a piece of paper a triangle of any size and shape
2. Cut it out and label the angles Â, B̂ and Ĉ on both sides of the paper
3. Draw dotted lines as shown and cut along these lines to get three pieces of
paper
4. Place them along your ruler as shown to see that  + B̂ + Ĉ = 180◦
B
A
C
Important: The sum of the angles in a triangle is 180◦ .
147
C
A
B
12.4
CHAPTER 12. GEOMETRY BASICS
Name
Table 12.2: Types of Triangles
Diagram
Properties
C
b
60◦
equilateral
A
60◦
b
All three sides are equal in length
and all three angles are equal.
60◦
b
B
C
b
Two sides are equal in length. The
angles opposite the equal sides are
equal.
isosceles
b
A
B
b
B
b
This triangle has one right angle.
The side opposite this angle is
called the hypotenuse.
po
hy
te
right-angled
nu
se
b
b
A
C
All sides and angles are different.
scalene (non-syllabus)
148
CHAPTER 12. GEOMETRY BASICS
12.4
B
A
C
Figure 12.10: In any triangle, ∠A + ∠B + ∠C = 180◦
Important: Any exterior angle of a triangle is equal to the sum of the two opposite interior
angles. An exterior angle is formed by extending any one of the sides.
ˆ + BCA
ˆ = CBD
ˆ
BAC
D
B
ˆ + CBA
ˆ = BCD
ˆ
BAC
B
b
b
C
B
b
b
A
ˆ + BAC
ˆ = ACD
ˆ
ABC
A
b
b
b
C
D
A
b
b
C
D
Figure 12.11: In any triangle, any exterior angle is equal to the sum of the two opposite interior
angles.
149
12.4
CHAPTER 12. GEOMETRY BASICS
Congruent Triangles
Label
Description
RHS
If the hypotenuse and one
side of a right-angled
triangle are equal to the
hypotenuse and the
respective side of another
triangle then the triangles
are congruent.
SSS
If three sides of a triangle
are equal in length to the
same sides of another
triangle then the two
triangles are congruent
SAS
If two sides and the
included angle of one
triangle are equal to the
same two sides and
included angle of another
triangle, then the two
triangles are congruent.
AAS
If one side and two angles
of one triangle are equal to
the same one side and two
angles of another triangle,
then the two triangles are
congruent.
Diagram
Similar Triangles
Description
Diagram
a
If all three pairs of corresponding
angles of two triangles are equal,
then the triangles are similar.
If all pairs of corresponding sides
of two triangles are in proportion,
then the triangles are similar.
a
y
x
z
q
p
r
x
p
150
c
b
c
b
=
y
q
=
z
r
CHAPTER 12. GEOMETRY BASICS
12.4
The theorem of Pythagoras
b
If △ABC is right-angled (B̂ = 90◦ ) then
b 2 = a 2 + c2
A
c
b
c
C
a
B
Converse:
If b2 = a2 + c2 , then
△ABC is right-angled (B̂ = 90◦ ).
a
Exercise: Triangles
1. Calculate the unknown variables in each of the following figures. All lengths
are in mm.
a)
N
b)
N
x
c)
x
30o
y
68o
36o
y
68o
x
O
P
O
P
O 68o
P
N
d)
N
N
x
19
P
e)
R
116
O
15
x
76
f)
N
R
12
P
S
g)
14
x
20
O
N
R
x
5
15
9
P
P
O
S
21
T
y
O
2. State whether or not the following pairs of triangles are congruent or not. Give
reasons for your answers. If there is not enough information to make a descision,
say why.
151
12.4
CHAPTER 12. GEOMETRY BASICS
B
E
B
b)
a)
C
A
B
C
A
D
D
E
B
c)
E
d)
C
C
D
A
B
D
A
e)
b
A
C
b
D
12.4.2
Quadrilaterals
A quadrilateral is any polygon with four sides. The basic quadrilaterals are the trapezium,
parallelogram, rectangle, rhombus, square and kite.
Name of quadrilateral
trapezium
parallelogram
rectangle
rhombus
square
kite
Figure
Figure 12.12
Figure 12.13
Figure 12.14
Figure 12.15
Figure 12.16
Figure 12.17
Table 12.3: Examples of quadrilaterals.
Trapezium
A trapezium is a quadrilateral with one pair of parallel opposite sides. It may also be called a
trapezoid. A special type of trapezium is the isosceles trapezium, where one pair of opposite
sides is parallel, the other pair of sides is equal in length and the angles at the ends of each
parallel side are equal. An isosceles trapezium has one line of symmetry and its diagonals are
equal in length.
isosceles trapezium
Figure 12.12: Examples of trapeziums.
152
CHAPTER 12. GEOMETRY BASICS
12.4
Parallelogram
A trapezium with both sets of opposite sides parallel is called a parallelogram. A summary of
the properties of a parallelogram is:
• Both pairs of opposite sides are parallel.
• Both pairs of opposite sides are equal in length.
• Both pairs of opposite angles are equal.
• Both diagonals bisect each other (i.e. they cut each other in half).
D
b
b
/
C
///
b
///
/
A
b
b
B
Figure 12.13: An example of a parallelogram.
Rectangle
A rectangle is a parallelogram that has all four angles equal to 90◦ . A summary of the properties
of a rectangle is:
• Both pairs of opposite sides are parallel.
• Both pairs of opposite sides are of equal length equal.
• Both diagonals bisect each other.
• Diagonals are equal in length.
• All angles are right angles.
D
b
/
b
C
b
B
/
b
/
/
A
b
Figure 12.14: Example of a rectangle.
Rhombus
A rhombus is a parallelogram that has all four side of equal length. A summary of the properties
of a rhombus is:
• Both pairs of opposite sides are parallel.
• All sides are equal in length.
153
12.4
CHAPTER 12. GEOMETRY BASICS
• Both pairs of opposite angles equal.
• Both diagonals bisect each other at 90◦ .
• Diagonals of a rhombus bisect both pairs of opposite angles.
b
x
•
/
•
x
///
b
D
C
b
x
A
b
/
///
•
x
•
b
B
Figure 12.15: An example of a rhombus. A rhombus is a parallelogram with all sides equal.
Square
A square is a rhombus that has all four angles equal to 90◦ .
A summary of the properties of a rhombus is:
• Both pairs of opposite sides are parallel.
• All sides are equal in length.
• All angles are equal to 90◦ .
• Both pairs of opposite angles equal.
• Both diagonals bisect each other at 90◦ .
• Diagonals are equal in length.
• Diagonals bisect both pairs of opposite angles (ie. all 45◦ ).
D
b
•
•
•
b
C
b
B
•
/
/
A
•
b
/
/
b
•
•
•
Figure 12.16: An example of a square. A square is a rhombus with all angles equal to 90◦ .
Kite
A kite is a quadrilateral with two pairs of adjacent sides equal.
A summary of the properties of a kite is:
• Two pairs of adjacent sides are equal in length.
154
CHAPTER 12. GEOMETRY BASICS
12.4
• One pair of opposite angles are equal where the angles must be between unequal sides.
• One diagonal bisects the other diagonal and one diagonal bisects one pair of opposite
angles.
• Diagonals intersect at right-angles.
C
b
x x
b
/
b
b
/
A
B
••
b
D
Figure 12.17: An example of a kite.
12.4.3
Other polygons
There are many other polygons, some of which are given in the table below.
Sides
5
6
7
8
10
15
Name
pentagon
hexagon
heptagon
octagon
decagon
pentadecagon
Table 12.4: Table of some polygons and their number of sides.
octagon
nonagon
decagon
pentagon
hexagon
heptagon
Figure 12.18: Examples of other polygons.
155
12.4
12.4.4
CHAPTER 12. GEOMETRY BASICS
Extra
Angles of regular polygons
You can calculate the size of the interior angle of a regular polygon by using:
 =
n−2
× 180◦
n
(12.1)
where n is the number of sides and  is any angle.
Areas of Polygons
1. Area of triangle:
1
2×
2. Area of trapezium:
h
base × perpendicular height
1
2×
h
(sum of k sides) × perpendicular height
h
3. Area of parallelogram and rhombus: base × perpendicular height
b
4. Area of rectangle: length × breadth
l
s
5. Area of square: length of side × length of side
s
r
6. Area of circle: π x radius2
Exercise: Polygons
1. For each case below, say whether the statement is true or false. For false
statements, give a counter-example to prove it:
(a) All squares are rectangles
(b) All rectangles are squares
(c) All pentagons are similar
(d) All equilateral triangles are similar
(e) All pentagons are congruent
(f) All equilateral triangles are congruent
2. Find the areas of each of the given figures - remember area is measured in
square units (cm2 , m2 , mm2 ).
156
CHAPTER 12. GEOMETRY BASICS
a)
12.5
b)
c)
10cm
10cm
10cm
7cm
d)
b
5cm
5cm
3cm
e)
f)
6cm
10cm
5cm
12cm
g)
h)
10cm
16cm
15cm
9cm
12.5
5cm
8cm
21cm
Exercises
1. Find all the pairs of parallel lines in the following figures, giving reasons in each case.
A
N
B
62
M
◦
137◦
57◦
62◦
(a)
D
C
123◦
G
H
(b)
120◦
60◦
K
60◦
L
(c)
2. Find a, b, c and d in each case, giving reasons.
P
b
a
Q
d c
S
(a)
73◦
R
157
O
P
12.5
CHAPTER 12. GEOMETRY BASICS
W
K
◦
100
a
A
B
L
b
C
c
d N
d
a
U
V
b
F
(c) X
O
(b)
c
50◦
D
M
E
45◦
T
(a) Which of the following claims are true? Give a counter-example for those that are
incorrect.
i. All equilateral triangles are similar.
ii. All regular quadrilaterals are similar.
iii. In any △ABC with ∠ABC = 90◦ we have AB 3 + BC 3 = CA3 .
iv. All right-angled isosceles triangles with perimeter 10 cm are congruent.
v. All rectangles with the same area are similar.
(b) Say which of the following pairs of triangles are congruent with reasons.
D
A
F
C
E
i. B
G
L
J
K
I
ii. H
P
N
iii. M
R
Q
O
U
R
V
S
T
iv. Q
(c) For each pair of figures state whether they are similar or not. Give reasons.
158
CHAPTER 12. GEOMETRY BASICS
(a)
12.5
A
P
√
2 2
2
45◦
3
45◦
C
B
Q
(b)
H
J
7,5
W
5
X
5
L
12.5.1
R
3
Z
K
Y
Challenge Problem
1. Using the figure below, show that the sum of the three angles in a triangle is 180◦ . Line
DE is parallel to BC.
A
D
E
d
e
a
B
c
b
C
159
12.5
CHAPTER 12. GEOMETRY BASICS
160
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