Beams Latest 24 Mac 2010 Edited Version

Basic Essential Additional Mathematics Skills

Curriculum Development Division
Ministry of Education Malaysia
Putrajaya

2010

First published 2010

© Curriculum Development Division,
Aras 4-8, Blok E9
Pusat Pentadbiran Kerajaan Persekutuan
62604 Putrajaya
Tel.: 03-88842000 Fax.: 03-88889917
Website: http://www.moe.gov.my/bpk

Copyright reserved. Except for use in a review, the reproduction or utilization of this
work in any form or by any electronic, mechanical, or other means, now known or
hereafter invented, including photocopying, and recording is forbidden without prior
written permission from the Director of the Curriculum Development Division, Ministry
of Education Malaysia.

TABLE OF CONTENTS

Preface i

Acknowledgement ii

Introduction iii

Objective iii

Module Layout iii

BEAMS Module:

Unit 1: Negative Numbers

Unit 2: Fractions

Unit 3: Algebraic Expressions and Algebraic Formulae

Unit 4: Linear Equations

Unit 5: Indices

Unit 6: Coordinates and Graphs of Functions

Unit 7: Linear Inequalities

Unit 8: Trigonometry

Panel of Contributors

ACKNOWLEDGEMENT

The Curriculum Development Division,

Ministry of Education wishes to express our

deepest gratitude and appreciation to all

panel of contributors for their expert

views and opinions, dedication,

and continuous support in

the development of

this module.

ii

INTRODUCTION
Additional Mathematics is an elective subject taught at the upper secondary level. This
subject demands a higher level of mathematical thinking and skills compared to that required
by the more general Mathematics KBSM. A sound foundation in mathematics is deemed
crucial for pupils not only to be able to grasp important concepts taught in Additional
Mathematics classes, but also in preparing them for tertiary education and life in general.

This Basic Essential Additional Mathematics Skills (BEAMS) Module is one of the
continuous efforts initiated by the Curriculum Development Division, Ministry of Education,
to ensure optimal development of mathematical skills amongst pupils at large. By the
acronym BEAMS itself, it is hoped that this module will serve as a concrete essential
support that will fruitfully diminish mathematics anxiety amongst pupils. Having gone
through the BEAMS Module, it is hoped that fears induced by inadequate basic
mathematical skills will vanish, and pupils will learn mathematics with the due excitement
and enjoyment.

OBJECTIVE
The main objective of this module is to help pupils develop a solid essential mathematics
foundation and hence, be able to apply confidently their mathematical skills, specifically
in school and more significantly in real-life situations.

MODULE LAYOUT
This module encompasses all mathematical skills and knowledge
taught in the lower secondary level and is divided into eight units as
follows:

Unit 2: Fractions
Unit 4: Linear Equations
Unit 5: Indices

iii

Each unit stands alone and can be used as a comprehensive revision of a particular topic.
Most of the units follow as much as possible the following layout:
Module Overview
Objectives
Teaching and Learning Strategies
Lesson Notes
Examples
Test Yourself
Answers

The “Lesson Notes”, “Examples” and “Test Yourself” in each unit can be used as
supplementary or reinforcement handouts to help pupils recall and understand the basic
concepts and skills needed in each topic.

Teachers are advised to study the whole unit prior to classroom teaching so as to familiarize
with its content. By completely examining the unit, teachers should be able to select any part
in the unit that best fit the needs of their pupils. It is reminded that each unit in this module is
by no means a complete lesson, rather as a supporting material that should be ingeniously
integrated into the Additional Mathematics teaching and learning processes.

At the outset, this module is aimed at furnishing pupils with the basic mathematics
foundation prior to the learning of Additional Mathematics, however the usage could be
broadened. This module can also be benefited by all pupils, especially those who are
preparing for the Penilaian Menengah Rendah (PMR) Examination.

iv

PANEL OF CONTRIBUTORS

Advisors:

Haji Ali bin Ab. Ghani AMN
Director

Dr. Lee Boon Hua
Deputy Director (Humanities)

Mohd. Zanal bin Dirin
Deputy Director (Science and Technology)

Editorial Advisor:

Aziz bin Saad
Principal Assistant Director
(Head of Science and Mathematics Sector)

Editors:

Dr. Rusilawati binti Othman
Assistant Director
(Head of Secondary Mathematics Unit)

Aszunarni binti Ayob
Assistant Director

Rosita binti Mat Zain
Assistant Director

Writers:

Abdul Rahim bin Bujang Hon May Wan
SM Tun Fatimah, Johor SMK Tasek Damai, Ipoh, Perak

Ali Akbar bin Asri Horsiah binti Ahmad
SM Sains, Labuan SMK Tun Perak, Jasin, Melaka

Amrah bin Bahari Kalaimathi a/p Rajagopal
SMK Dato’ Sheikh Ahmad, Arau, Perlis SMK Sungai Layar, Sungai Petani, Kedah

Aziyah binti Paimin Kho Choong Quan
SMK Kompleks KLIA, , Negeri Sembilan SMK Ulu Kinta, Ipoh, Perak

Bashirah binti Seleman Lau Choi Fong
SMK Sultan Abdul Halim, Jitra, Kedah SMK Hulu Klang, Selangor

Bibi Kismete binti Kabul Khan Loh Peh Choo
SMK Jelapang Jaya, Ipoh, Perak SMK Bandar Baru Sungai Buloh, Selangor

Che Rokiah binti Md. Isa Mohd. Misbah bin Ramli
SMK Dato’ Wan Mohd. Saman, Kedah SMK Tunku Sulong, Gurun, Kedah

Cheong Nyok Tai Noor Aida binti Mohd. Zin
SMK Perempuan, Kota Kinabalu, Sabah SMK Tinggi Kajang, Kajang, Selangor

Ding Hong Eng Noor Ishak bin Mohd. Salleh
SM Sains Alam Shah, Kuala Lumpur SMK Laksamana, Kota Tinggi, Johor

Esah binti Daud Noorliah binti Ahmat
SMK Seri Budiman, Kuala Terengganu SM Teknik, Kuala Lumpur

Haspiah binti Basiran Nor A’idah binti Johari
SMK Tun Perak, Jasin, Melaka SMK Teknik Setapak, Selangor

Noorliah binti Ahmat
SM Teknik, Kuala Lumpur

Ali Akbar bin Asri Nor A’idah binti Johari

SM Sains, Labuan SMK Teknik Setapak, Selangor

Amrah bin Bahari Nor Dalina binti Idris

SMK Dato’ Sheikh Ahmad, Arau, Perlis SMK Syed Alwi, Kangar, Perlis

Writers:

Nor Dalina binti Idris Suhaimi bin Mohd. Tabiee
SMK Syed Alwi, Kangar, Perlis SMK Datuk Haji Abdul Kadir, Pulau Pinang

Norizatun binti Abdul Samid Suraiya binti Abdul Halim
SMK Sultan Badlishah, Kulim, Kedah SMK Pokok Sena, Pulau Pinang

Pahimi bin Wan Salleh Tan Lee Fang
Maktab Sultan Ismail, Kelantan SMK Perlis, Perlis

Rauziah binti Mohd. Ayob Tempawan binti Abdul Aziz
SMK Bandar Baru Salak Tinggi, Selangor SMK Mahsuri, Langkawi, Kedah

Rohaya binti Shaari Turasima binti Marjuki
SMK Tinggi Bukit Merajam, Pulau Pinang SMKA Simpang Lima, Selangor

Roziah binti Hj. Zakaria Wan Azlilah binti Wan Nawi
SMK Taman Inderawasih, Pulau Pinang SMK Putrajaya Presint 9(1), WP Putrajaya

Shakiroh binti Awang Zainah binti Kebi
SM Teknik Tuanku Jaafar, Negeri Sembilan SMK Pandan, Kuantan, Pahang

Sharina binti Mohd. Zulkifli Zaleha binti Tomijan
SMK Agama, Arau, Perlis SMK Ayer Puteh Dalam, Pendang, Kedah

Sim Kwang Yaw Zariah binti Hassan
SMK Petra, Kuching, Sarawak SMK Dato’ Onn, Butterworth, Pulau Pinang

Layout and Illustration:

Aszunarni binti Ayob Mohd. Lufti bin Mahpudz
Assistant Director Assistant Director
Curriculum Development Division Curriculum Development Division

Basic Essential

Additional Mathematics Skills

UNIT 1
NEGATIVE NUMBERS

Unit 1:
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Addition and Subtraction of Integers Using Number Lines 2

1.0 Representing Integers on a Number Line 3

2.0 Addition and Subtraction of Positive Integers 3

3.0 Addition and Subtraction of Negative Integers 8

Part B: Addition and Subtraction of Integers Using the Sign Model 15

Part C: Further Practice on Addition and Subtraction of Integers 19

Part D: Addition and Subtraction of Integers Including the Use of Brackets 25

Part E: Multiplication of Integers 33

Part F: Multiplication of Integers Using the Accept-Reject Model 37

Part G: Division of Integers 40

Part H: Division of Integers Using the Accept-Reject Model 44

Part I: Combined Operations Involving Integers 49

Answers 52

Basic Essential Additional Mathematics Skills (BEAMS) Module

MODULE OVERVIEW

1. Negative Numbers is the very basic topic which must be mastered by every
pupil.

2. The concept of negative numbers is widely used in many Additional
Mathematics topics, for example:
(a) Functions (b) Quadratic Equations
(c) Quadratic Functions (d) Coordinate Geometry
(e) Differentiation (f) Trigonometry
Thus, pupils must master negative numbers in order to cope with topics in
Additional Mathematics.

3. The aim of this module is to reinforce pupils‟ understanding on the concept of
negative numbers.

4. This module is designed to enhance the pupils‟ skills in

 using the concept of number line;
 using the arithmetic operations involving negative numbers;
 solving problems involving addition, subtraction, multiplication and
division of negative numbers; and
 applying the order of operations to solve problems.

5. It is hoped that this module will enhance pupils‟ understanding on negative
numbers using the Sign Model and the Accept-Reject Model.

6. This module consists of nine parts and each part consists of learning objectives
which can be taught separately. Teachers may use any parts of the module as
and when it is required.

1


PART A:
ADDITION AND SUBTRACTION
OF INTEGERS USING
NUMBER LINES

LEARNING OBJECTIVE

Upon completion of Part A, pupils will be able to perform computations
involving combined operations of addition and subtraction of integers using a
number lines.

TEACHING AND LEARNING STRATEGIES

The concept of negative numbers can be confusing and difficult for pupils to
grasp. Pupils face difficulty when dealing with operations involving positive and
negative integers.

Strategy:

Teacher should ensure that pupils understand the concept of positive and negative
integers using number lines. Pupils are also expected to be able to perform
computations involving addition and subtraction of integers with the use of the
number line.

2


PART A:
ADDITION AND SUBTRACTION OF INTEGERS
USING NUMBER LINES

LESSON NOTES

1.0 Representing Integers on a Number Line

 Positive whole numbers, negative numbers and zero are all integers.

 Integers can be represented on a number line.
Positive integers
may have a plus sign
–3 –2 –1 0 1 2 3 4 in front of them,
like +3, or no sign in
front, like 3.
Note: i) –3 is the opposite of +3

ii) – (–2) becomes the opposite of negative 2, that is, positive 2.

2.0 Addition and Subtraction of Positive Integers

Rules for Adding and Subtracting Positive Integers

 When adding a positive integer, you move to the right on a
number line.

–3 –2 –1 0 1 2 3 4

 When subtracting a positive integer, you move to the left
on a number line.

–3 –2 –1 0 1 2 3 4

3


EXAMPLES

(i) 2 + 3
Start Add a
with 2 positive 3

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Adding a positive integer:

Start by drawing an arrow from 0 to 2, and then,
draw an arrow of 3 units to the right:

2+3=5

Alternative Method:

Make sure you start from
the position of the first
integer.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at 2 and move 3 units to the right:

2+3=5

4


(ii) –2 + 5

Add a
positive 5

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start by drawing an arrow from 0 to –2, and then,
draw an arrow of 5 units to the right:

–2 + 5 = 3

Alternative Method:

integer.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at –2 and move 5 units to the right:

–2 + 5 = 3

5


(iii) 2 – 5 = –3
Subtract a
positive 5

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Subtracting a positive integer:

Start by drawing an arrow from 0 to 2, and then,
draw an arrow of 5 units to the left:

2 – 5 = –3

Alternative Method:

integer.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at 2 and move 5 units to the left:

2 – 5 = –3

6


(iv) –3 – 2 = –5

Subtract a
positive 2

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start by drawing an arrow from 0 to –3, and
then, draw an arrow of 2 units to the left:

–3 – 2 = –5

Alternative Method:

integer.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at –3 and move 2 units to the left:

–3 – 2 = –5

7


3.0 Addition and Subtraction of Negative Integers

Consider the following operations:

4 + (–1) = 3
4–1=3
–3 –2 –1 0 1 2 3 4

4–2=2 4 + (–2) = 2
–3 –2 –1 0 1 2 3 4

4–3=1 4 + (–3) = 1
–3 –2 –1 0 1 2 3 4

4–4=0 4 + (–4) = 0
–3 –2 –1 0 1 2 3 4

4 + (–5) = –1
4 – 5 = –1
–3 –2 –1 0 1 2 3 4

4 – 6 = –2 4 + (–6) = –2
–3 –2 –1 0 1 2 3 4

Note that subtracting an integer gives the same result as adding its opposite. Adding or
subtracting a negative integer goes in the opposite direction to adding or subtracting a positive
integer.

8


Rules for Adding and Subtracting Negative Integers

 When adding a negative integer, you move to the left on a
number line.

–3 –2 –1 0 1 2 3 4

 When subtracting a negative integer, you move to the right
on a number line.

–3 –2 –1 0 1 2 3 4

9


EXAMPLES

(i) –2 + (–1) = –3
This operation of
–2 + (–1) = –3
is the same as
Add a
negative 1 –2 –1 = –3.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Adding a negative integer:

then, draw an arrow of 1 unit to the left:

–2 + (–1) = –3

Alternative Method: Make sure you start from
integer.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at –2 and move 1 unit to the left:

–2 + (–1) = –3

10


(ii) 1 + (–3) = –2
This operation of
1 + (–3) = –2
is the same as
1 – 3 = –2
Add a
negative 3

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Start by drawing an arrow from 0 to 1, then, draw an arrow of
3 units to the left:
1 + (–3) = –2

Alternative Method:
integer.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at 1 and move 3 units to the left:

1 + (–3) = –2

11


(iii) 3 – (–3) = 6

This operation of
3 – (–3) = 6
is the same as
3+3=6
Subtract a
negative 3

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Subtracting a negative integer:

Start by drawing an arrow from 0 to 3, and
then, draw an arrow of 3 units to the right:

3 – (–3) = 6

Alternative Method:

integer.

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at 3 and move 3 units to the right:

3 – (–3) = 6

12


(iv) –5 – (–8) = 3 This operation of
–5 – (–8) = 3
is the same as
–5 + 8 = 3

Subtract a 3+3=6
negative 8

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


then, draw an arrow of 8 units to the right:

–5 – (–8) = 3

Alternative Method:

–5 –4 –3 –2 –1 0 1 2 3 4 5 6


Start at –5 and move 8 units to the right:

–5 – (–8) = 3

13


TEST YOURSELF A

Solve the following.

1. –2 + 4

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

2. 3 + (–6)

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

3. 2 – (–4)

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

4. 3 – 5 + (–2)

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

5. –5 + 8 + (–5)

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

14


PART B:
OF INTEGERS USING
THE SIGN MODEL

LEARNING OBJECTIVE

Upon completion of Part B, pupils will be able to perform computations
involving combined operations of addition and subtraction of integers using
the Sign Model.


This part emphasises the first alternative method which include activities and
mathematical games that can help pupils understand further and master the
operations of positive and negative integers.

Strategy:

Teacher should ensure that pupils are able to perform computations involving
addition and subtraction of integers using the Sign Model.

15


PART B:

USING THE SIGN MODEL

LESSON NOTES

In order to help pupils have a better understanding of positive and negative integers, we have
designed the Sign Model.

The Sign Model

 This model uses the „+‟ and „–‟ signs.
 A positive number is represented by „+‟ sign.
 A negative number is represented by „–‟ sign.

EXAMPLES

Example 1

What is the value of 3 – 5?

NUMBER SIGN

3 + + +
–5 – – – – –

WORKINGS
+ + +
i. Pair up the opposite signs.
    
ii. The number of the unpaired signs is
the answer.

Answer –2

16


Example 2

What is the value of  3  5 ?

NUMBER SIGN

–3 _ _ _

–5 – – – – –

WORKINGS

There is no opposite sign to pair up, so _ _ _ _ _ _ _ _
just count the number of signs.

Answer –8

Example 3

What is the value of  3  5 ?

NUMBER SIGN

–3 – – –
+5 + + + + +

WORKINGS _ _ _
i. Pair up the opposite signs. + + + + +
ii. The number of unpaired signs is the
answer.
Answer 2

17


TEST YOURSELF B


1. –4 + 8 2. –8 – 4 3. 12 – 7

4. –5 – 5 5. 5–7–4 6. –7 + 4 – 3

7. 4+3–7 8. 6–2 +8 9. –3 + 4 + 6

18


PART C:
FURTHER PRACTICE ON
OF INTEGERS

LEARNING OBJECTIVE

Upon completion of Part C, pupils will be able to perform computations
involving addition and subtraction of large integers.


This part emphasises addition and subtraction of large positive and negative integers.

Strategy:

Teacher should ensure the pupils are able to perform computation involving addition
and subtraction of large integers.

19


PART C:
FURTHER PRACTICE ON ADDITION AND SUBTRACTION OF INTEGERS

LESSON NOTES

In Part A and Part B, the method of counting off the answer on a number line and the Sign
Model were used to perform computations involving addition and subtraction of small integers.
However, these methods are not suitable if we are dealing with large integers. We can use the
following Table Model in order to perform computations involving addition and subtraction
of large integers.

Steps for Adding and Subtracting
Integers

1. Draw a table that has a column for + and a column
for –.

2. Write down all the numbers accordingly in the
column.

3. If the operation involves numbers with the same
signs, simply add the numbers and then put the
respective sign in the answer. (Note that we
normally do not put positive sign in front of a
positive number)

4. If the operation involves numbers with different
signs, always subtract the smaller number from
the larger number and then put the sign of the
larger number in the answer.

20


Examples:

i) 34 + 37 =

+ – Add the numbers and then put the
positive sign in the answer.
34
37 We can just write the answer as
71 instead of +71.

+71

ii) 65 – 20 =
Subtract the smaller number from
+ – the larger number and put the sign
of the larger number in the
65 20 answer.

+45 We can just write the answer as
45 instead of +45.

iii) –73 + 22 =

+ – Subtract the smaller number from
the larger number and put the sign
22 73
answer.
–51

iv) 228 – 338 =

+ –
228 338 the larger number and put the sign
–110 answer.

21


v) –428 – 316 =

+ –

428
316
Add the numbers and then put the
negative sign in the answer.
–744

vi) –863 – 127 + 225 =

+ –

225 863 Add the two numbers in the „–‟
column and bring down the number
127 in the „+‟ column.

225 990 Subtract the smaller number from
the larger number in the third row
–765 and put the sign of the larger
number in the answer.

vii) 234 – 675 – 567 =

+ –

567
in the „+‟ column.


22


viii) –482 + 236 – 718 =

+ –

718 in the „+‟ column.


ix) –765 – 984 + 432 =

+ –

432 765
Add the two numbers in the „–‟
984 column and bring down the number
in the „+‟ column.

432 1749
–1317 the larger number in the third row
and put the sign of the larger
x) –1782 + 436 + 652 =

+ –

436 1782 Add the two numbers in the „+‟
652 in the „–‟ column.

1782
1088 Subtract the smaller number from

23


TEST YOURSELF C


1. 47 – 89 2. –54 – 48 3. 33 – 125

4. –352 – 556 5. 345 – 437 – 456 6. –237 + 564 – 318

7. –431 + 366 – 778 8. –652 – 517 + 887 9. –233 + 408 – 689

24


PART D:
OF INTEGERS INCLUDING THE
USE OF BRACKETS

LEARNING OBJECTIVE

Upon completion of Part D, pupils will be able to perform computations
involving combined operations of addition and subtraction of integers, including
the use of brackets, using the Accept-Reject Model.


This part emphasises the second alternative method which include activities to
enhance pupils‟ understanding and mastery of the addition and subtraction of
integers, including the use of brackets.

Strategy:

Teacher should ensure that pupils understand the concept of addition and subtraction
of integers, including the use of brackets, using the Accept-Reject Model.

25


PART D:
INCLUDING THE USE OF BRACKETS

LESSON NOTES

The Accept - Reject Model

 „+‟ sign means to accept.
 „–‟ sign means to reject.

To Accept or To Reject? Answer

+(5) Accept +5 +5

–(2) Reject +2 –2

+ (–4) Accept –4 –4

– (–8) Reject –8 +8

26


EXAMPLES

i) 5 + (–1) =

Number To Accept or To Reject? Answer
5 Accept 5 +5
+ (–1) Accept –1 –1

+ + + + +
–

5 + (–1) = 4

This operation of
5 + (–1) = 4
is the same as
5–1=4

We can also solve this question by using the Table Model as follows:

5 + (–1) = 5 – 1

+ – the larger number and put the sign
5 1
answer.
+4 We can just write the answer as 4
instead of +4.

27


ii) –6 + (–3) =


–6 Reject 6 –6
+ (–3) Accept –3 –3

– – – – – –
– – –
–6 + (–3) = –9

This operation of
–6 + (–3) = –9
is the same as
–6 –3 = –9


–6 + (–3) = –6 – 3 =

+ –

6
3 Add the numbers and then put the

–9

28


iii) –7 – (–4) =


–7 Reject 7 –7
– (–4) Reject –4 +4

– – – – – – –
+ + + +

–7 – (–4) = –3

This operation of
–7 – (–4) = –3
is the same as
–7 + 4 = –3


–7 – (–4) = –7 + 4 =

+ –
4 7 the larger number and put the sign
–3 answer.

29


iv) –5 – (3) =


–5 Reject 5 –5
– (3) Reject 3 –3

– – – – –
– – –
– 5 – (3) = –8

This operation of
–5 – (3) = –8
is the same as
–5 – 3 = –8


–5 – (3) = –5 – 3 =

+ –

5

–8

30


v) –35 + (–57) = –35 – 57 = This operation of
–35 + (–57)
is the same as
–35 – 57

Using the Table Model:

+ –

35
–92

vi) –123 – (–62) = –123 + 62 =
This operation of
–123 – (–62)
is the same as
–123 + 62

Using the Table Model:

+ –

the larger number and put the sign
of the larger number in the answer.
–61

31


TEST YOURSELF D


1. –4 + (–8) 2. 8 – (–4) 3. –12 + (–7)

4. –5 + (–5) 5. 5 – (–7) + (–4) 6. 7 + (–4) – (3)

7. 4 + (–3) – (–7) 8. –6 – (2) + (8) 9. –3 + (–4) + (6)

10. –44 + (–81) 11. 118 – (–43) 12. –125 + (–77)

13. –125 + (–239) 14. 125 – (–347) + (–234) 15. 237 + (–465) – (378)

16. 412 + (–334) – (–712) 17. –612 – (245) + (876) 18. –319 + (–412) + (606)

32


PART E:
MULTIPLICATION OF
INTEGERS

LEARNING OBJECTIVE

Upon completion of Part E, pupils will be able to perform computations
involving multiplication of integers.


This part emphasises the multiplication rules of integers.

Strategy:

Teacher should ensure that pupils understand the multiplication rules to perform
computations involving multiplication of integers.

33


PART E:
MULTIPLICATION OF INTEGERS

LESSON NOTES

Consider the following pattern:
3×3=9

3 2  6
positive × positive = positive
3 1  3 (+) × (+) = (+)

3 0  0 The result is reduced by 3 in
positive × negative = negative
3  (1)  3 every step. (+) × (–) = (–)

3  (2)  6

3  (3)  9

(3)  3  9

(3)  2  6
negative × positive = negative
(3)  1  3 (–) × (+) = (–)

(3)  0  0 The result is increased by 3 in
negative × negative = positive
(3)  (1)  3 every step. (–) × (–) = (+)

(3)  (2)  6

(3)  (3)  9

Multiplication Rules of Integers
1. When multiplying two integers of the same signs, the answer is positive integer.
2. When multiplying two integers of different signs, the answer is negative integer.
3. When any integer is multiplied by zero, the answer is always zero.

34


EXAMPLES

1. When multiplying two integers of the same signs, the answer is positive integer.

(a) 4 × 3 = 12

(b) –8 × –6 = 48

2. When multiplying two integers of the different signs, the answer is negative integer.

(a) –4 × (3) = –12

(b) 8 × (–6) = –48

3. When any integer is multiplied by zero, the answer is always zero.

(a) (4) × 0 = 0

(b) (–8) × 0 = 0

(c) 0 × (5) = 0

(d) 0 × (–7) = 0

35


TEST YOURSELF E


1. –4 × (–8) 2. 8 × (–4) 3. –12 × (–7)

4. –5 × (–5) 5. 5 × (–7) × (–4) 6. 7 × (–4) × (3)

7. 4 × (–3) × (–7) 8. (–6) × (2) × (8) 9. (–3) × (–4) × (6)

36


PART F:
USING
THE ACCEPT-REJECT MODEL

LEARNING OBJECTIVE

Upon completion of Part F, pupils will be able to perform computations
involving multiplication of integers using the Accept-Reject Model.


This part emphasises the second alternative method which include activities to
enhance the pupils‟ understanding and mastery of the multiplication of integers.

Strategy:

Teacher should ensure that pupils understand the multiplication rules of integers
using the Accept-Reject Model. Pupils can then perform computations involving
multiplication of integers.

37


PART F:
USING THE ACCEPT-REJECT MODEL

LESSON NOTES

The Accept-Reject Model

 In order to help pupils have a better understanding of multiplication of integers, we have
designed the Accept-Reject Model.

 Notes: (+) × (+) : The first sign in the operation will determine whether to accept
or to reject the second sign.

Multiplication Rules:

Sign To Accept or To Reject Answer

(+) × (+) Accept + 
(–) × (–) Reject – 
(+) × (–) Accept – –
(–) × (+) Reject + –

EXAMPLES

To Accept or to Reject Answer
(2) × (3) Accept + 6

(–2) × (–3) Reject – 6

(2) × (–3) Accept – –6
(–2) × (3) Reject + –6

38


TEST YOURSELF F


1. 3 × (–5) = 2. –4 × (–8) = 3. 6 × (5) =

4. 8 × (–6) = 5. – (–5) × 7 = 6. (–30) × (–4) =

7. 4 × 9 × (–6) = 8. (–3) × 5 × (–6) = 9. (–2) × ( –9) × (–6) =

10. –5× (–3) × (+4) = 11. 7 × (–2) × (+3) = 12. 5 × 8 × (–2) =

39


PART G:
DIVISION OF INTEGERS

LEARNING OBJECTIVE

Upon completion of Part G, pupils will be able to perform computations
involving division of integers.


This part emphasises the division rules of integers.

Strategy:

Teacher should ensure that pupils understand the division rules of integers to
perform computation involving division of integers.

40


PART G:

LESSON NOTES

Consider the following pattern:
3 × 2 = 6, then 6÷2=3 and 6÷3=2

3 × (–2) = –6, then (–6) ÷ 3 = –2 and (–6) ÷ (–2) = 3

(–3) × 2 = –6, then (–6) ÷ 2 = –3 and (–6) ÷ (–3) = 2

(–3) × (–2) = 6, then 6 ÷ (–3) = –2 and 6 ÷ (–2) = –3

Rules of Division

1. Division of two integers of the same signs results in a positive integer.

i.e. positive ÷ positive = positive
(+) ÷ (+) = (+)

negative ÷ negative = positive
(–) ÷ (–) = (+)

2. Division of two integers of different signs results in a negative integer.

i.e. positive ÷ negative = negative
(+) ÷ (–) = (–)

Undefined means “this
negative ÷ positive = negative operation does not have a
(–) ÷ (+) = (–) meaning and is thus not
assigned an interpretation!”

Source:
3. Division of any number by zero is undefined. http://www.sn0wb0ard.com

41


EXAMPLES

1. Division of two integers of the same signs results in a positive integer.

(a) (12) ÷ (3) = 4

(b) (–8) ÷ (–2) = 4

2. Division of two integers of different signs results in a negative integer.

(a) (–12) ÷ (3) = –4

(b) (+8) ÷ (–2) = –4

3. Division of zero by any number will always give zero as an answer.

(a) 0 ÷ (5) = 0

(b) 0 ÷ (–7) = 0

42


TEST YOURSELF G


1. (–24) ÷ (–8) 2. 8 ÷ (–4) 3. (–21) ÷ (–7)

4. (–5) ÷ (–5) 5. 60 ÷ (–5) ÷ (–4) 6. 36 ÷ (–4) ÷ (3)

7. 42 ÷ (–3) ÷ (–7) 8. (–16) ÷ (2) ÷ (8) 9. (–48) ÷ (–4) ÷ (6)

43


PART H:
USING
THE ACCEPT-REJECT MODEL

LEARNING OBJECTIVE

Upon completion of Part H, pupils will be able to perform computations
involving division of integers using the Accept-Reject Model.


This part emphasises the alternative method that include activities to help pupils
further understand and master division of integers.

Strategy:

Teacher should make sure that pupils understand the division rules of integers using
the Accept-Reject Model. Pupils can then perform division of integers, including
the use of brackets.

44


PART H:
DIVISION OF INTEGERS USING THE ACCEPT-REJECT MODEL

LESSON NOTES

 In order to help pupils have a better understanding of division of integers, we have designed
the Accept-Reject Model.

 Notes: (+) ÷ (+) : The first sign in the operation will determine whether to accept
or to reject the second sign.

() : The sign of the numerator will determine whether to accept or
() to reject the sign of the denominator.

Division Rules:


(+) ÷ (+) Accept + +

(–) ÷ (–) Reject – +

(+) ÷ (–) Accept – –

(–) ÷ (+) Reject + –

45


EXAMPLES

To Accept or To Reject Answer

(6) ÷ (3) Accept + 2

(–6) ÷ (–3) Reject – 2

(+6) ÷ (–3) Accept – –2

(–6) ÷ (3) Reject + –2

Division [Fraction Form]:


() Accept + +
()

()
Reject – +
()

()
Accept – –
()

()
Reject + –
()

46


EXAMPLES

To Accept or To Reject Answer

(  8)
Accept + 4
(  2)

(  8)
Reject – 4
(  2)

(  8)
Accept – –4
( 2)

(  8)
Reject + –4
( 2)

47


TEST YOURSELF H


1. 18 ÷ (–6) 12 24
2. 3.
2 8

 25 6 6. – (–35) ÷ 7
4. 5.
5 3

7. (–32) ÷ (–4) 8. (–45) ÷ 9 ÷ (–5) (30 )
9.
(6)

80 11. 12 ÷ (–3) ÷ (–2) 12. – (–6) ÷ (3)
10.
(5)

48


PART I:
COMBINED OPERATIONS
INVOLVING INTEGERS

LEARNING OBJECTIVES

Upon completion of Part I, pupils will be able to:

1. perform computations involving combined operations of addition,
subtraction, multiplication and division of integers to solve problems; and

2. apply the order of operations to solve the given problems.


This part emphasises the order of operations when solving combined operations
involving integers.

Strategy:

Teacher should make sure that pupils are able to understand the order of operations
or also known as the BODMAS rule. Pupils can then perform combined operations
involving integers.

49


PART I:
COMBINED OPERATIONS INVOLVING INTEGERS

LESSON NOTES

 A standard order of operations for calculations involving +, –, ×, ÷ and
brackets:

Step 1: First, perform all calculations inside the brackets.

Step 2: Next, perform all multiplications and divisions,
working from left to right.

Step 3: Lastly, perform all additions and subtractions, working
from left to right.

 The above order of operations is also known as the BODMAS Rule
and can be summarized as:

Brackets
power of
Division
Multiplication
Addition
Subtraction

EXAMPLES

1. 10 – (–4) × 3 2. (–4) × (–8 – 3 ) 3. (–6) + (–3 + 8 ) ÷5
= (–4) × (–11 ) = (–6 )+ (5) ÷5
=10 – (–12)
= 44 = (–6 )+ 1
= 10 + 12 = –5
= 22

50


TEST YOURSELF I


1. 12 + (8 ÷ 2) 2. (–3 – 5) × 2 3. 4 – (16 ÷ 2) × 2

4. (– 4) × 2 + 6 × 3 5. ( –25) ÷ (35 ÷ 7) 6. (–20) – (3 + 4) × 2

7. (–12) + (–4 × –6) ÷ 3 8. 16 ÷ 4 + (–2) 9. (–18 ÷ 2) + 5 – (–4)

51


ANSWERS

TEST YOURSELF A:

1. 2

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

2. –3

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

3. 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

4. –4

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

5. –2

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

52


TEST YOURSELF B:

1) 4 2) –12 3) 5
4) –10 5) –6 6) –6
7) 0 8) 12 9) 7

TEST YOURSELF C:

1) –42 2) –102 3) –92
4) –908 5) –548 6) 9
7) –843 8) –282 9) –514

TEST YOURSELF D:

1) –12 2) 12 3) –19
4) –10 5) 8 6) 0
7) 8 8) 0 9) –1
10) –125 11) 161 12) –202
13) –364 14) 238 15) –606
16) 790 17) 19 18) –125

TEST YOURSELF E:

1) 32 2) –32 3) 84
4) 25 5) 140 6) –84
7) 84 8) –96 9) 72

53


TEST YOURSELF F:

1) –15 2) 32 3) 30
4) –48 5) 35 6) 120
7) –216 8) 90 9) –108
10) 60 11) –42 12) –80

TEST YOURSELF G:

1) 3 2) –2 3) 3
4) 1 5) 3 6) –3
7) 2 8) –1 9) 2

TEST YOURSELF H:

1. –3 2. –6 3. 3
4. 5 5. –2 6. 5
7. 8 8. 1 9. 5
10. –16 11. 2 12. 2

TEST YOURSELF I:

1. 16 2. –16 3. –12
4. 10 5. –5 6. –34
7. –4 8. 2 9. 0

54

Basic Essential


UNIT 2

FRACTIONS

Unit 1:
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Addition and Subtraction of Fractions 2
1.0 Addition and Subtraction of Fractions with the Same Denominator 5
1.1 Addition of Fractions with the Same Denominators 5
1.2 Subtraction of Fractions with The Same Denominators 6
1.3 Addition and Subtraction Involving Whole Numbers and Fractions 7
1.4 Addition or Subtraction Involving Mixed Numbers and Fractions 9
2.0 Addition and Subtraction of Fractions with Different Denominator 10
2.1 Addition and Subtraction of Fractions When the Denominator
of One Fraction is A Multiple of That of the Other Fraction 11
2.2 Addition and Subtraction of Fractions When the Denominators
Are Not Multiple of One Another 13
2.3 Addition or Subtraction of Mixed Numbers with Different
Denominators 16
2.4 Addition or Subtraction of Algebraic Expression with Different
Denominators 17

Part B: Multiplication and Division of Fractions 22
1.0 Multiplication of Fractions 24
1.1 Multiplication of Simple Fractions 28
1.2 Multiplication of Fractions with Common Factors 29
1.3 Multiplication of a Whole Number and a Fraction 29
1.4 Multiplication of Algebraic Fractions 31
2.0 Division of Fractions 33
2.1 Division of Simple Fractions 36
2.2 Division of Fractions with Common Factors 37

Answers 42

UNIT 2: Fractions

MODULE OVERVIEW

1. The aim of this module is to reinforce pupils’ understanding of the concept
of fractions.

2. It serves as a guide for teachers in helping pupils to master the basic
computation skills (addition, subtraction, multiplication and division)
involving integers and fractions.

3. This module consists of two parts, and each part consists of learning

PART 1
objectives which can be taught separately. Teachers may use any parts of the
module as and when it is required.

Ministry of Education Malaysia 1

UNIT 2: Fractions

PART A:
OF FRACTIONS

LEARNING OBJECTIVES

Upon completion of Part A, pupils will be able to:

1. perform computations involving combination of two or more operations
on integers and fractions;

2. pose and solve problems involving integers and fractions;

3. add or subtract two algebraic fractions with the same denominators;

4. add or subtract two algebraic fractions with one denominator as a
multiple of the other denominator; and

5. add or subtract two algebraic fractions with denominators:

(i) not having any common factor;
(ii) having a common factor.


UNIT 2: Fractions


Pupils have difficulties in adding and subtracting fractions with different
denominators.

Strategy:

Teachers should emphasise that pupils have to find the equivalent form of
the fractions with common denominators by finding the lowest common
multiple (LCM) of the denominators.


UNIT 2: Fractions

LESSON NOTES

Fraction is written in the form of:

a numerator
b denominator

Examples:
2 4
,
3 3
Proper Fraction Improper Fraction Mixed Numbers

The numerator is smaller The numerator is larger A whole number and
than the denominator. than or equal to the denominator. a fraction combined.

Examples: Examples: Examples:

2 9 15 108 2 1 ,85
, , 7 6
3 20 4 12

Rules for Adding or Subtracting Fractions

1. When the denominators are the same, add or subtract only the numerators and
keep the denominator the same in the answer.

2. When the denominators are different, find the equivalent fractions that have the
same denominator.

Note: Emphasise that mixed numbers and whole numbers must be converted to improper
fractions before adding or subtracting fractions.


UNIT 2: Fractions

EXAMPLES

1.0 Addition And Subtraction of Fractions with the Same Denominator

1.1 Addition of Fractions with the Same Denominators

Add only the numerators and keep the
1 4 5
i)   denominator same.
8 8 8

1  4  5
8 8 8

1 3 4 denominator the same.
ii)  
8 8 8
1 Write the fraction in its simplest form.

2

1 5 6
iii)   denominator the same.
f f f


UNIT 2: Fractions

1.2 Subtraction of Fractions with The Same Denominators

Subtract only the numerators and keep
5 1 4
i)   the denominator the same.
8 8 8
1
 Write the fraction in its simplest form.
2

4 1
5
 1 
8

2
8 8

1 5 4
ii)   the denominator the same.
7 7 7

3 1 2
iii)   the denominator the same.
n n n


UNIT 2: Fractions

1.3 Addition and Subtraction Involving Whole Numbers and Fractions

1
i) Calculate 1  .
8

1
1 +
8

9
8 1 
 + 8
8 8
1
 1
8

 First, convert the whole number to an improper fraction with the
same denominator as that of the other fraction.
 Then, add or subtract only the numerators and keep the denominator
the same.

1 28 1 2 20 2 1 12 1
4    4    4 y  y
7 7 7 5 5 5 3 3 3

29 18 12  y
  
7 5 3

1 3
 4  3
7 5


UNIT 2: Fractions

 First, convert the whole number to an improper fraction with
the same denominator as that of the other fraction.
 Then, add or subtract only the numerators and keep the
denominator the same.

5 2n 5 2 2 3k
2     3 
n n n k k k

2n  5 2  3k
 
n k


UNIT 2: Fractions

1.4 Addition or Subtraction Involving Mixed Numbers and Fractions
1 4
i) Calculate 1  .
8 8

1 + 4
1
8 8

 9 + 4  13 5
 1
8 8 8 8

 First, convert the mixed number to improper fraction.
 Then, add or subtract only the numerators and keep the

1 5 15 5 2 4 29 4 3 x 11 x
2    3    1   
7 7 7 7 9 9 9 9 8 8 8 8

20 6 25 7 11  x
= = 2 = = 2 =
7 7 9 9 8


UNIT 2: Fractions

2.0 Addition and Subtraction of Fractions with Different Denominators

1 1
i) Calculate  . The denominators are not the same.
8 2 See how the slices are different in
sizes? Before we can add the
fractions, we need to make them the
same, because we can't add them
together like this!

?

1 + 1  ?
8 2
To make the denominators the same, multiply both the numerator and the denominator of
the second fraction by 4:
4

1 4
 Now, the denominators
2 8 are the same. Therefore,
we can add the fractions
4
together!

Now, the question can be visualized like this:

1 + 4  5
8 8 8


UNIT 2: Fractions

Hint: Before adding or subtracting fractions with different denominators, we must
convert each fraction to an equivalent fraction with the same denominator.

2.1 Addition and Subtraction of Fractions When the Denominator of One Fraction is
A Multiple of That of the Other Fraction
Multiply both the numerator and the denominator with an integer that makes the
denominators the same.

Change the first fraction to an equivalent
1 5 fraction with denominator 6.
(i)  (Multiply both the numerator and the
3 6
denominator of the first fraction by 2):

2 5 2
 
6 6 1

2
3 6
7 2

6
1 Add only the numerators and keep the
=1
6 denominator the same.

Convert the fraction to a mixed number.

Change the second fraction to an equivalent
7 3 fraction with denominator 12.
(ii)  (Multiply both the numerator and the
12 4 denominator of the second fraction by 3):

7 9 3
  3 9
12 12 
4 12
2 3
 
12
Subtract only the numerators and keep the
1
  denominator the same.
6

Write the fraction in its simplest form.


UNIT 2: Fractions

Change the first fraction to an equivalent
1 9
(iii)  fraction with denominator 5v.
v 5v (Multiply both the numerator and the
denominator of the first fraction by 5):

5 9 5
  1 5
5v 5v 
v 5v
5

14

5v Add only the numerators and keep the


UNIT 2: Fractions

2.2 Addition and Subtraction of Fractions When the Denominators Are Not Multiple of
One Another

Method I Method II

1 3 1 3
 
6 4 6 4

(i) Find the Least Common Multiple (LCM) (i) Multiply the numerator and the
denominator of the first fraction with
of the denominators.
the denominator of the second fraction
and vice versa.
2) 4 , 6
2) 2 , 3
1 4 3 6
3) 1 , 3 = 
- , 1 6 4 4 6

LCM = 2  2  3 = 12 4 18
= 
24 24
The LCM of 4 and 6 is 12.
22
=
(ii) Change each fraction to an equivalent 24
fraction using the LCM as the
denominator. 11 Write the fraction in its
=
(Multiply both the numerator and the 12 simplest form.
denominator of each fraction by a whole
number that will make their
denominators the same as the LCM
value).
 This method is preferred but you
must remember to give the
1 2 33 answer in its simplest form.
= 
6 2 43

2 9
= 
12 12

11
=
12


UNIT 2: Fractions

EXAMPLES

2 1
1. 
3 5

2 5 1 3
Multiply the first fraction with the second denominator
= + and
3 5 5 3 multiply the second fraction with the first denominator.
Multiply the first fraction by the
denominator of the second fraction and
10 3
  multiply the second fraction by the
15 15 denominator of the first fraction.

13
= Add only the numerators and keep the
15

5 3
2. 
6 8

8 6
5 3
= –
6 8
8 6
40 18
=  multiply the second fraction by the
48 48 denominator of the first fraction.

22 Subtract only the numerators and keep
= the denominator the same.
48

11 Write the fraction in its simplest form.
=
24


UNIT 2: Fractions

2 1
3. g 
3 7

2g  7 1 3
= 
3 7 7 3 Multiply the first fraction by the
multiply the second fraction by the
14 g 3
=  denominator of the first fraction.
21 21
Write as a single fraction.
14 g  3
=
21

2g h
4. 
3 5

5 3
2g h
 
3 5
5 3 Multiply the first fraction by the
10 g 3h multiply the second fraction by the
  denominator of the first fraction.
15 15
Write as a single fraction.
10 g  3h

15

6 4
5. 
c d

6 d 4 c
= 
c d d c
6d 4c multiply the second fraction by the
 
cd cd denominator of the first fraction.

6d  4c Write as a single fraction.
=
cd


UNIT 2: Fractions

2.3 Addition or Subtraction of Mixed Numbers with Different Denominators

1 3
1. 2  2 Convert the mixed numbers to improper fractions.
2 4
Convert the mixed numbers to improper fractions.
5 11
= 
2 4

5 2 11 Change the first fraction to an equivalent fraction
= 
2 2 4 with denominator 4.
(Multiply both the numerator and the denominator
10 11 of the first fraction by 2)
= 
4 4
=

1
5 Change the fraction back to a mixed number.
4

5 3
2. 3  1 Convert the mixed numbers to improper fractions.
6 4
23 7
=  Convert the mixed numbers to improper fractions.
6 4
The denominators are not multiples of one another:
23  4 7 6
= 
6 4 4 6  Multiply the first fraction by the denominator
of the second fraction.
92 42  Multiply the second fraction by the
=  denominator of the first fraction.
24 24

=

25
= Write the fraction in its simplest form.
12

1 Change the fraction back to a mixed number.
= 2
12


UNIT 2: Fractions

2.4 Addition or Subtraction of Algebraic Expression with Different Denominators
m m
1.  The denominators are not multiples of of one another
The denominators are not multiples one another:
m2 2 Multiply the first fraction with the second denominator
Multiply the second fraction with the first denominator
2  ( m2)  Multiply the first fraction by the denominator
m m
=  of the second fraction.
m2 2 2  ( m2)  Multiply the second fraction by the
denominator of the first fraction.

2m mm  2
=  Remember to use
2m  2 2m  2 brackets

2m  m(m  2) Write the above fractions as a single fraction.
=
2(m  2)

2m  m 2  2 m Expand:
=
2(m  2)
m (m – 2) = m2 – 2m

m2
=
2(m  2)

y y 1
2.  The denominators are not multiples of one another:
y 1 y The denominators are not multiples of one another
Multiply the first first fractionthe second denominator
Multiply the fraction with by the denominator
y y y  1 ( y 1) Multiply the second fraction with the first denominator
of the second fraction.
= 
y 1  y y  ( y 1)  Multiply the second fraction by the
denominator of the first fraction.

y 2  ( y  1)( y  1) Write the fractions as a single fraction.
=
y ( y  1)
Expand:
y 2  ( y 2  1) (y – 1) (y + 1) = y2 + y – y – 12
=
y ( y  1)
= y2 – 1

y2  y2  1
= Expand:
y ( y  1) – (y2 – 1) = –y2 + 1

1
=
y ( y  1)


UNIT 2: Fractions

3 5n
3.  The denominators are not multiples of one another:
8n 4n 2
 Multiply the first fraction multiples of one another
The denominators are not by the denominator
3  4n 2 5  n  8n of the second fraction. with the second denominator
Multiply the first fraction
=  Multiply the second fraction with the first denominator
 Multiply the second fraction by the
8n  4n 2 4 n 2  8n denominator of the first fraction.

12n 2 8n (5  n)
= 
8n(4n ) 8n(4n 2 )
2

12 n 2  8n (5  n) Write as a single fraction.
=
8n(4n 2 )
Expand:
12 n 2  40 n  8n 2
= – 8n (5 + n) = –40n – 8n2
8n(4n 2 )
4n 2  40 n Subtract the like terms.
=
8n ( 4 n 2 )

4n (n  10 ) Factorise and simplify the fraction by canceling
= out the common factors.
4n(8n 2 )

n  10
=
8n 2


UNIT 2: Fractions

TEST YOURSELF A

Calculate each of the following.

2 1 11 5
1.   2.  
7 7 12 12

2 1 2 5
3.   4.  
7 14 3 12

2 4
5.   1 5
7 5 6.  
2 7

2
7. 2 3 2 7
13 8. 4  2 
5 9

2 1 11 5
9.   10.  
s s w w


UNIT 2: Fractions

2 5
11.
2 1
  12.  
a 2a f 3f

1 5
13.
2 4
  14.  
a b p q

p 1
5 2 2 3 16.  (2  p) 
15. m  n  m  n  2
7 5 7 5

2 x  3 y 3x  y 12  4 x 5
17.   18.  
2 5 2x x

x x 1
19.   x x4
x 1 x 20.  
x2 x2


UNIT 2: Fractions

6x  3 y 4x  8 y 2 4n
21.   22.  
2 4 3n 9n 2

r 5  2r 2 p3 p2
23.   24.  
5 15 r p2 2p

2n  3 4n  3 3m  n n  3
25.   26.  
5n 2 10n mn n

5m mn m3 nm
27.   28.  
5m mn 3m mn

3 5n
29.   p 1 p
8n 4n 2 30.  
3m m


UNIT 2: Fractions

PART B:
MULTIPLICATION AND DIVISION
OF FRACTIONS

LEARNING OBJECTIVES

Upon completion of Part B, pupils will be able to:

1. multiply:
(i) a whole number by a fraction or mixed number;
(ii) a fraction by a whole number (include mixed numbers); and
(iii) a fraction by a fraction.

2. divide:
(i) a fraction by a whole number;
(ii) a fraction by a fraction;
(iii) a whole number by a fraction; and
(iv) a mixed number by a mixed number.

3. solve problems involving combined operations of addition, subtraction,
multiplication and division of fractions, including the use of brackets.


UNIT 2: Fractions


Pupils face problems in multiplication and division of fractions.

Strategy:

Teacher should emphasise on how to divide fractions correctly. Teacher should
also highlight the changes in the positive (+) and negative (–) signs as follows:

Multiplication Division
(+)  (+) = + (+)  (+) = +
(+)  (–) = – (+)  (–) = –
(–)  (+) = – (–)  (+) = –
(–)  (–) = + (–)  (–) = +


UNIT 2: Fractions

LESSON NOTES

1.0 Multiplication of Fractions

Recall that multiplication is just repeated addition.
Consider the following:
2  3

First, let’s assume this box as 1 whole unit.

Therefore, the above multiplication 2 3 can be represented visually as follows:

2 groups of 3 units

3 + 3 = 6

This means that 3 units are being repeated twice, or mathematically can be written as:
23  3  3
6

Now, let’s calculate 2 x 2. This multiplication can be represented visually as:

2 groups of 2 units

2 + 2 = 4
This means that 2 units are being repeated twice, or mathematically can be written as:
2 2  2  2
4


UNIT 2: Fractions

Now, let’s calculate 2 x 1. This multiplication can be represented visually as:

2 groups of 1 unit

1 + 1 = 2

This means that 1 unit is being repeated twice, or mathematically can be written as:
2 1  1  1  2

It looks simple when we multiply a whole number by a whole number. What if we
have a multiplication of a fraction by a whole number? Can we represent it visually?

1
Let’s consider 2  .
2

1
Since represents 1 whole unit, therefore unit can be represented by the
2
following shaded area:

1
Then, we can represent visually the multiplication of 2 as follows:
2

1
2 groups of unit
2
1 1 2
+ = 1
2 2 2
1
This means that unit is being repeated twice, or mathematically can be written as:
2
1 1 1
2  
2 2 2
2

2
1


UNIT 2: Fractions

1 1
Let’s consider again  2. What does it mean? It means ‘ out of 2 units’ and the
2 2
visualization will be like this:

1 1
out of 2 units 2 1
2 2

1 1
Notice that the multiplications 2 and  2 will give the same answer, that is, 1.
2 2

1
How about 2?
3

1
Since represents 1 whole unit, therefore unit can be represented by the
3
following shaded area:

1
The shaded area is unit.
3

1
Then, we can represent visually the multiplication  2 as follows:
3

1 1 2
+ =
3 3 3
1
This means that unit is being repeated twice, or mathematically can be written as:
3
1 1 1
2 
3 3 3
2

3


UNIT 2: Fractions

1 1
Let’s consider  2 . What does it mean? It means ‘ out of 2 units’ and the visualization
3 3
will be like this:

1 1 2
out of 2 units 2 
3 3 3

1 1 2
Notice that the multiplications 2 and  2 will give the same answer, that is, .
3 3 3

Consider now the multiplication of a fraction by a fraction, like this:

1 1

3 2

1 1
This means ‘ out of units’ and the visualization will be like this:
3 2

1 1 1 1 1
out of units  
1 3 2 3 2 6
unit
2

Consider now this multiplication:

2 1

3 2

2 1
This means ‘ out of units’ and the visualization will be like this:
3 2

1
unit
2

2 1 2 1 2
out of units  
3 2 3 2 6


UNIT 2: Fractions

What do you notice so far?
The answer to the above multiplication of a fraction by a fraction can be obtained by
just multiplying both the numerator together and the denominator together:

1 1 1 2 1 2
   
3 2 6 3 3 9

1 1 1
So, what do you think the answer for  ? Do you get as the answer?
4 3 12

The steps to multiply a fraction by a fraction can therefore be summarized as follows:

Steps to Multiply Fractions: Remember!!!

1) Multiply the numerators together and (+)  (+) = +
multiply the denominators together. (+)  (–) = –
(–)  (+) = –
2) Simplify the fraction (if needed). (–)  (–) = +

1.1 Multiplication of Simple Fractions
Examples:

2 3 6 2 3 6
a)   b)    
5 7 35 7 5 35

6 2 12 6 2 12
c)     d)    
7 5 35 7 5 35

Multiply the two numerators together and the two denominators together.


UNIT 2: Fractions

1.2 Multiplication of Fractions with Common Factors

12 5 12  5 
 or  
7 6 7 6 

First Method: Second Method:

(ii) Multiply the two numerators (i) Simplify the fraction by canceling
together and the two out the common factors.
denominators together:
2 12 5

7 61
12 5 60
 =
7 6 42 (i) Then, multiply the two
numerators together and the two
denominators together, and
(ii) Then, simplify. convert to a mixed number, if
needed.
6010 10 3
 1
42 7 7 2
7 12 5 10 3
  1
7 6 7 7
1

1.3 Multiplication of a Whole Number and a Fraction

2  5 
1
Remember  6
2= 2
 
1
2  31 
=   Convert the mixed number to improper
1  6  fraction.

Simplify by canceling out the common
 31 
12
factors.
=  
1  6 
3
Multiply the two numerators together and
the two denominators together.
31
=  Remember: (+)  (–) = (–)
3
1
=  10 Change the fraction back to a mixed number.
3


UNIT 2: Fractions

EXAMPLES

5 15
1. Find  
12 10
1
5 15 5 Simplify by canceling out the common factors.
Solution:  
12 10 2
4
Multiply the two numerators together and the
two denominators together.
5
= 
8 Remember: (+)  (–) = (–)

21 2
2. Find  Simplify by canceling out the common
6 5
factors.
21 2 1
Solution :  21
3
6 5 Note that can be further simplified.
3
21 2 1
= 7  Simplify further by canceling out the
6 5 common factors.
3
1
7
 Multiply the two numerators together and
5
the two denominators together.
2
= 1
5 Remember: (+)  (–) = (–)

Change the fraction back to a mixed
number.


UNIT 2: Fractions

1.4 Multiplication of Algebraic Fractions

2 5x
1. Simplify 
x 4

2 5x 1
Solution : 1  Simplify the fraction by canceling out the x’s.
x 4
1 2

Multiply the two numerators together and
5 the two denominators together.
=
2
1 Change the fraction back to a mixed
= 2 number.
2

n 9 
2. Simplify   4m 
2 n 

n 9 
Solution:   4m 
2 n 
Simplify the fraction by canceling the
1 2 common factor and the n.
n9 n  4m 
=     
2n 1 1
2 1 
Multiply the two numerators together
9 n ( 2m)
=  and the two denominators together.
2 1

9
=  2nm Write the fraction in its simplest form.
2


UNIT 2: Fractions

TEST YOURSELF B1

9 25 45 3 14
1. Calculate   2. Calculate –     
5 27 12 7 20

 11  1 1 
3. Calculate 2    4. Calculate  4 
4 3 5 

 m n
5. Simplify  3    6. Simplify (5m) 
 k  2

1  3x  n
7. Simplify 1   8. Simplify (2a  3d ) 
6  14  2

2  9  x 1
9. Simplify   5x  y 10. Simplify  20   
3  10  4 x


UNIT 2: Fractions

LESSON NOTES

2.0 Division of Fractions

Consider the following:
6  3

First, let’s assume this circle as 1 whole unit.

Therefore, the above division can be represented visually as follows:

6 units are being divided into a group of 3
units:

6  3  2

This means that 6 units are being divided into a group of 3 units, or mathematically
can be written as:
6  3  2

The above division can also be interpreted as ‘how many 3’s can fit into 6’. The answer is
‘2 groups of 3 units can fit into 6 units’.

Consider now a division of a fraction by a fraction like this:

1
1 1 How many is in
 . 8
2 8 1
?
2


UNIT 2: Fractions

This means ‘How many is in ?

1 1
8 2

The answer is 4:

Consider now this division:
1 3
3 1 How many is in ?
 . 4 4
4 4

This means ‘How many is in ?

1 3
4 4

But, how do you
The answer is 3:
calculate the answer?


UNIT 2: Fractions

Consider again 6  3  2.

Actually, the above division can be written as follows:

6 These operations are the same!
63 
3
1
 6 The reciprocal
3 1
of 3 is .
3

Notice that we can write the division in the multiplication form. But here, we have to
change the second number to its reciprocal.

Therefore, if we have a division of fraction by a fraction, we can do the same, that is,
we have to change the second fraction to its reciprocal and then multiply the
fractions.

Therefore, in our earlier examples, we can have:

1 1
(i)  Change the second fraction to its
2 8
reciprocal and change the sign  to .
1 8
 
2 1
8 The reciprocal

2 1 8
of is .
4 8 1

The reciprocal of a
fraction is found by
inverting the
fraction.


UNIT 2: Fractions

3 1
(ii)  Change the second fraction to its
4 4 reciprocal and change the sign  to .
3 4
 
4 1
3 The reciprocal
1 4
of is .
4 1

The steps to divide fractions can therefore be summarized as follows:

Steps to Divide Fractions: Tips:
1. Change the second fraction to its
reciprocal and change the  sign to .
(+)  (+) = +
2. Multiply the numerators together and (+)  (–) = –
multiply the denominators together. (–)  (+) = –
(–)  (–) = +
3. Simplify the fraction (if needed).

2.1 Division of Simple Fractions

Example:

2 3
 Change the second fraction to its reciprocal
5 7 and change the sign  to  .
2 7
= 
5 3 Multiply the two numerators together and
14 the two denominators together.
=
15


UNIT 2: Fractions

2.2 Division of Fractions With Common Factors

Examples:

10 2
  Change the second fraction to its reciprocal and
21 9
change the  sign to  .
10 9
=  
21 2 Simplify by canceling out the common factors.
10 9
=5   3
7 21 21 Multiply the two numerators together and the
15
=  two denominators together.
7
1 Remember: (+)  (–) = (–)
= 2
7
Change the fraction back to a mixed number.

3
5
6
Express the fraction in division form.
7
3 6
 
5 7 Change the second fraction to its reciprocal
and change the  sign to  .
1
3 7
  Then, simplify by canceling out the common
5 62 factors.

7
 Multiply the two numerators together and the
10


UNIT 2: Fractions

EXAMPLES

35 25
1. Find 
12 6

35 25
Solution : 
12 6
Change the second fraction to its reciprocal
and change the  sign to .
35 61
= 7  Then, simplify by canceling out the common
2 12 25 factors.
7 5
= Multiply the two numerators together and the
10

2 5x
2. Simplify – 
x 4
Change the second fraction to its reciprocal
2 4
Solution : –  and change the  sign to .
x 5x

8 Multiply the two numerators together and the two
= – denominators together.
5x 2

y
3. Simplify x
2
Solution :
Express the fraction in division form.
Method I y
 2
x Change the second fraction to its reciprocal

y
 
1 and change  to  .
x 2
y
  Multiply the two numerators together and the two
2x denominators together.

Remember: (+)  (–) = (–)


UNIT 2: Fractions

Method II
The given fraction.
y
The numerator is also
x
2 a fraction with
denominator x

y
= x 
x Multiply the numerator and the denominator of the
Multiply the numerator and the denominator of
2 x the given fraction withfraction by x.
given x

y
x
= x
2 x

y
= 
2x

(1  1 )
4. Simplify r
5

Solution:
1
(1  1 ) r is the denominator of
r
.
r
5
1 r
(1  ) Multiply the given fraction with
r
.
= r  r
5 r
r 1
= Note that:
5r
1
(1  )  r  r  1
r


UNIT 2: Fractions

TEST YOURSELF B2

3 21 5 7 5
1. Calculate    2. Calculate   
7 2 9 8 16

8 4y 16
3. Simplify   4. Simplify
y 3 2
k

2 4m 2m 2
5. Simplify Simplify   
5 x 6.
n 3n
3

4 x
8. Simplify
y 1 1
1
7. Simplify
8 x


UNIT 2: Fractions

3 (1  1 ) 5 1
9. Calculate 4 x
10. Simplify
5 y


x 1 4
9
 1
p
11. Simplify
2 12. Simplify
1
3 1
5


UNIT 2: Fractions

ANSWERS

TEST YOURSELF A:

3 1 5
1. 2. 3.
7 2 14

1 38 3 3
4. 5. or 1 6. 
4 35 35 14

67 2 73 28 3
7. or 5 8. or 1 9.
13 13 45 45 s

6 5 1
10. 11. 12.
w 2a 3f

2b  4a q  5p 15. m  n
13.
ab 14.
pq

3p  3 16 x  17 y 2x  1
16. 17. 18.
2 10 x

1 8x  y
19. 20. 2 21.
x( x  1) 2

7n  4 r 2 1  p2  6
22. 23. 24.
9n 2 3r 2 p2

7 n  4n 2  6 1 m n5
25. 26. 27.
10 n 2 m 5n

n3 n  10 4p 3
28. 29. 30.
3n 8n 2 3m


UNIT 2: Fractions

TEST YOURSELF B1:

5 2 9 1 11 1
1. or 1 2.  or  1 3. or 5
3 3 8 8 2 2

7 2 3m 5mn
4.  or  1 5. 6.
5 5 k 2

x 3 10 3
7. 8. na  nd 9.  x y
4 2 3 5

1
10. 5x 
4

TEST YOURSELF B2:

2 14 5 6
1. 2.  or  1 3. 
49 9 9 y2

6 6
5. 6.
4. 8k 5 x m

1 x2 9
7. 9.
2( y  1) 8.
x 1
20

5x  1 13x 5
10. 11. 12. 
xy 6 4p


Basic Essential


UNIT 3
ALGEBRAIC EXPRESSIONS
AND
Unit 1:
ALGEBRAIC FORMULAE
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Performing Operations on Algebraic Expressions 2

Part B: Expansion of Algebraic Expressions 10

Part C: Factorisation of Algebraic Expressions and Quadratic Expressions 15

Part D: Changing the Subject of a Formula 23

Activities
Crossword Puzzle 31
Riddles 33

Further Exploration 37

Answers 38

Basic Essential Additional Mathematics Skills (BEAM) Module

MODULE OVERVIEW

1. The aim of this module is to reinforce pupils’ understanding of the concepts and skills
in Algebraic Expressions, Quadratic Expressions and Algebraic Formulae.

2. The concepts and skills in Algebraic Expressions, Quadratic Expressions and
Algebraic Formulae are required in almost every topic in Additional Mathematics,
especially when dealing with solving simultaneous equations, simplifying
expressions, factorising and changing the subject of a formula.

3. It is hoped that this module will provide a solid foundation for studies of Additional
Mathematics topics such as:
 Functions
 Quadratic Equations and Quadratic Functions
 Simultaneous Equations
 Indices and Logarithms
 Progressions
 Differentiation
 Integration

4. This module consists of four parts and each part deals with specific skills. This format
provides the teacher with the freedom to choose any parts that is relevant to the skills
to be reinforced.



PART A:
PERFORMING OPERATIONS ON
ALGEBRAIC EXPRESSIONS

LEARNING OBJECTIVES

Upon completion of Part A, pupils will be able to perform operations on algebraic
expressions.


Pupils who face problem in performing operations on algebraic expressions might have
difficulties learning the following topics:

 Simultaneous Equations - Pupils need to be skilful in simplifying the algebraic
expressions in order to solve two simultaneous equations.
 Functions - Simplifying algebraic expressions is essential in finding composite
functions.
 Coordinate Geometry - When finding the equation of locus which involves
distance formula, the techniques of simplifying algebraic expressions are required.
 Differentiation - While performing differentiation of polynomial functions, skills
in simplifying algebraic expressions are needed.

Strategy:

1. Teacher reinforces the related terminologies such as: unknowns, algebraic terms,
like terms, unlike terms, algebraic expressions, etc.
2. Teacher explains and shows examples of algebraic expressions such as:
8k, 3p + 2, 4x – (2y + 3xy)
3. Referring to the “Lesson Notes” and “Examples” given, teacher explains how to
perform addition, subtraction, multiplication and division on algebraic expressions.
4. Teacher emphasises on the rules of simplifying algebraic expressions.



LESSON NOTES

PART A:
PERFORMING BASIC ARITHMETIC OPERATIONS ON ALGEBRAIC EXPRESSIONS

1. An algebraic expression is a mathematical term or a sum or difference of mathematical
terms that may use numbers, unknowns, or both.

Examples of algebraic expressions: 2r, 3x + 2y, 6x2 +7x + 10, 8c + 3a – n2, 3
g

2. An unknown is a symbol that represents a number. We normally use letters such as n, t, or
x for unknowns.

3. The basic unit of an algebraic expression is a term. In general, a term is either a number
or a product of a number and one or more unknowns. The numerical part of the term, is
known as the coefficient.

Coefficient Unknowns
6 xy

Examples: Algebraic expression with one term: 2r, 3
g

Algebraic expression with two terms: 3x + 2y, 6s – 7t

Algebraic expression with three terms: 6x2 +7x + 10, 8c + 3a – n2

4. Like terms are terms with the same unknowns and the same powers.

Examples: 3ab, –5ab are like terms.

2 2
3x2, x are like terms.
5

5. Unlike terms are terms with different unknowns or different powers.

Examples: 1.5m, 9k, 3xy, 2x2y are all unlike terms.



6. An algebraic expression with like terms can be simplified by adding or subtracting the
coefficients of the unknown in algebraic terms.

7. To simplify an algebraic expression with like terms and unlike terms, group the like terms
first, and then simplify them.

8. An algebraic expression with unlike terms cannot be simplified.

9. Algebraic fractions are fractions involving algebraic terms or expressions.

3m 2 4r 2 g x2  y2
Examples: , , , 2 .
15 6h 2rg  g 2 x  2 xy  y 2

10. To simplify an algebraic fraction, identify the common factor of both the numerator and the
denominator. Then, simplify it by elimination.



EXAMPLES

Simplify the following algebraic expressions and algebraic fractions:

s t
(a) 5x – (3x – 4x) ( e) 
4 6

5x 3 y
(b) –3r –9s + 6r + 7s (f ) 
6 2z

4r 2 g e
(c) (g )  2g
2rg  g 2 f

1
3 4 3x 
(d )  2
p q (h)
3x

Solutions: Algebraic expression with like terms can be simplified by
(a) 5x – (3x – 4x) adding or subtracting the coefficients of the unknown.

= 5x – (– x) Perform the operation in the bracket.

= 5x + x

= 6x

(b) –3r –9s + 6r + 7s
Arrange the algebraic terms according to the like terms.
= –3r + 6r –9s + 7s
.
= 3r – 2s Unlike terms cannot be simplified.
Leave the answer in the simplest form as shown.

4r 2 g
(c)
2rg  g 2

4r 2 g 1 Simplify by canceling out the common factor and the
 same unknowns in both the numerator and the
1
g ( 2r  g )
denominator.
4r 2

2r  g



3 4
(d ) 
p q
3q 4 p The LCM of p and q is pq.
 
pq pq
3q  4 p

pq

s t
(e) 
4 6
3s 2t The LCM of 4 and 6 is 12.
 
43 6 2
3s  2t

12

1 Simplify by canceling out the common
5x 3 y 5x  y
(f )   factor, then multiply the numerators
6 2z 2  2z together and followed by the
2
5 xy
 denominators.
4z

e e 1 Change division to multiplication of the
(g )  2g   reciprocal of 2g.
f f 2g
e

2 fg

Equate the denominator.
1 3 x(2) 1
3x  
(h ) 2 2 2
3x 3x
6x  1
 2
3x
6x  1 1
 
2 3x
6x  1

6x



ALTERNATIVE METHOD

Simplify the following algebraic fractions:

1  1
3x   3x   The denominator of
1
is 2 . Therefore,
2  2 2
(a) =  2
3x 3x 2 2
multiply the algebraic fraction by .
2
1
3 x(2)  (2)
2 Each of the terms in the numerator and
=
3 x(2) denominator of the algebraic fraction is
multiplied by 2.

6x  1
=
6x

3
3 3  The denominator of is x. Therefore,
2   2 x
x x  x
(b) =  x
5 5 x multiply the algebraic fraction by .
x
3
( x )  2( x )
 x Each of the terms in the numerator and
5( x) denominator is multiplied by x.
3  2x

5x

3
The denominator of is 2x. Therefore,
 3    3  2x
8  8   2 x 
 2x     2 x 2x
(c)   multiply the algebraic fraction by .
2 2 2x 2x
Each of the terms in the numerator and
denominator is multiplied by 2x.
 3 
8(2 x)   (2 x)
 2x  .

2( 2 x )

16 x  3

4x



3 3 7 8 x
(d )   The denominator of is 7.
8 x 8 x 7 7
 4  4 Therefore, multiply the algebraic
 7   7 
7
3(7) fraction by .
 7
8 x
  ( 7 )  4( 7 )
 7  Each of the terms in the numerator
21 and denominator is multiplied by 7.

8  x  28
21
 Simplify the denominator.
36  x



TEST YOURSELF A

Simplify the following algebraic expressions:
1. 2a –3b + 7a – 2b 2. − 4m + 5n + 2m – 9n

3. 8k – ( 4k – 2k ) 4. 6p – ( 8p – 4p )

3 1 4h 2k
5.  6. 
y 5x 3 5

4a 3b 4c  d 8
7.  8. 
7 2c 2 3c  d

xy u uv
9.  yz 10. 
z vw 2w

2 4
11 .  2
5
12.  
 6 x
 x 4
 5
 x



PART B:
EXPANSION OF ALGEBRAIC
EXPRESSIONS

LEARNING OBJECTIVE

Upon completion of Part B, pupils will be able to expand algebraic
expressions.


Pupils who face problem in expanding algebraic expressions might have
difficulties in learning of the following topics:

 Simultaneous Equations – pupils need to be skilful in expanding the
algebraic expressions in order to solve two simultaneous equations.
 Functions – Expanding algebraic expressions is essential when finding
composite function.
 Coordinate Geometry – when finding the equation of locus which
involves distance formula, the techniques of expansion are applied.

Strategy:
Pupils must revise the basic skills involving expanding algebraic expressions.



LESSON NOTES

PART B:
EXPANSION OF ALGEBRAIC EXPRESSIONS

1. Expansion is the result of multiplying an algebraic expression by a term or another
algebraic expression.

2. An algebraic expression in a single bracket is expanded by multiplying each term in the
bracket with another term outside the bracket.

3(2b – 6c – 3) = 6b – 18c – 9

3. Algebraic expressions involving two brackets can be expanded by multiplying each term of
algebraic expression in the first bracket with every term in the second bracket.

(2a + 3b)(6a – 5b) = 12a2 – 10ab + 18ab – 15b2

= 12a2 + 8ab – 15b2

4. Useful expansion tips:

(i) (a + b)2 = a2 + 2ab + b2

(ii) (a – b)2 = a2 – 2ab + b2

(iii) (a – b)(a + b) = (a + b)(a – b)

= a2 – b2



EXAMPLES

Expand each of the following algebraic expressions:

(a) 2(x + 3y) (d ) ( a  3) 2

(b) – 3a (6b + 5 – 4c) (e)  32k  5
2

(f ) ( p  2)( p  5)
( c)
2
9 y  12
3

Solutions:

When expanding a bracket, each term
(a) 2 (x + 3y) within the bracket is multiplied by the term
outside the bracket.
= 2x + 6y

When expanding a bracket, each term
(b) –3a (6b + 5 – 4c) within the bracket is multiplied by the term
outside the bracket.
= –18ab – 15a + 12ac

2
(c) 9 y  12
3 Simplify by canceling out the common
2 3 2 4
=  9 y   12 factor, then multiply the numerators
1 3 1 3 together and followed by the denominators.
= 6y + 8

(d ) (a  3) 2

When expanding two brackets, each term
= (a + 3) (a + 3)
within the first bracket is multiplied by
every term within the second bracket.
= a2 + 3a + 3a + 9
= a2 + 6a + 9



(e)  32k  5
2

= –3(2k + 5) (2k + 5) When expanding two brackets, each term
within the first bracket is multiplied by
= –3(4k2 + 20k + 25)

= –12k2 – 60k – 75

(f ) ( p  2) (q  5)
When expanding two brackets, each term
= pq – 5p + 2q – 10 within the first bracket is multiplied by

ALTERNATIVE METHOD

Expanding two brackets

When expanding two
(a) (a + 3) (a + 3) brackets, write down the
product of expansion and
then, simplify the like
= a2 + 3a + 3a + 9 (c) (4x – 3y)(6x – 5y)
terms.
= a2 + 6a + 9

– 18 xy
– 20 xy
– 38 xy
(b) (2p + 3q) (6p – 5q)

= 24x2 – 38 xy + 15y2
= 12p2 – 10 pq + 18 pq – 15q2
= 12p2 + 8 pq – 15q2



TEST YOURSELF B

Simplify the following expressions and give your answers in the simplest form.

 3 1
6q  1
1.  4 2n   2.
 4 2

3.  6 x2 x  3 y  4. 2a  b  2(a  b)

2( p  3)  ( p  6)
6 x  y    x  2 y 
5. 1
6.  
3  3 

7. e  12  2e  1 8. m  n 2  m2m  n 

9. f  g  f  g   g 2 f  g  10 . h  i h  i   2ih  3i 



PART C:
FACTORISATION OF
ALGEBRAIC EXPRESSIONS AND
QUADRATIC EXPRESSIONS

LEARNING OBJECTIVE

Upon completion of Part C, pupils will be able to factorise algebraic expressions
and quadratic expressions.


Some pupils may face problem in factorising the algebraic expressions. For
example, in the Differentiation topic which involves differentiation using the
combination of Product Rule and Chain Rule or the combination of Quotient
Rule and Chain Rule, pupils need to simplify the answers using factorisation.

Examples:

1. y  2 x 3 (7 x  5) 4
dy
  2 x 3 [28(7 x  5) 3 ]  (7 x  5) 4 (6 x 2 )
dx
 2 x 2 (7 x  5) 3 (49 x  15)

(3  x) 3
2. y
7  2x
dy (7  2 x)[3(3  x) 2 ]  (3  x) 3 (2)
 
dx (7  2 x ) 2
(3  x) 2 (4 x  15)

(7  2 x ) 2
Strategy
1. Pupils revise the techniques of factorisation.



LESSON NOTES

PART C:
FACTORISATION OF
ALGEBRAIC EXPRESSIONS AND QUADRATIC EXPRESSIONS

1. Factorisation is the process of finding the factors of the terms in an algebraic expression. It
is the reverse process of expansion.

2. Here are the methods used to factorise algebraic expressions:

(i) Express an algebraic expression as a product of the Highest Common Factor (HCF) of
its terms and another algebraic expression.

ab – bc = b(a – c)

(ii) Express an algebraic expression with three algebraic terms as a complete square of two
algebraic terms.

a2 + 2ab + b2 = (a + b)2

a2 – 2ab + b2 = (a – b)2

(iii) Express an algebraic expression with four algebraic terms as a product of two algebraic
expressions.

ab + ac + bd + cd = a(b + c) + d(b + c)

= (a + d)(b + c)

(iv) Express an algebraic expression in the form of difference of two squares as a product of
two algebraic expressions.

a2 – b2 = (a + b)(a – b)

3. Quadratic expressions are expressions which fulfill the following characteristics:

(i) have only one unknown; and
(ii) the highest power of the unknown is 2.

4. Quadratic expressions can be factorised using the methods in 2(i) and 2(ii).

5. The Cross Method can be used to factorise algebraic expression in the general form of
ax2 + bx + c, where a, b, c are constants and a ≠ 0, b ≠ 0, c ≠ 0.



EXAMPLES

(a) Factorising the Common Factors

Factorise the common factor m.
i) mn + m = m (n +1)
.
Factorise the common factor p.
ii) 3mp + pq = p (3m + q)
.
Factorise the common factor 2n.
iii) 2mn – 6n = 2n (m – 3)
.

(b) Factorising Algebraic Expressions with Four Terms

Factorise the first and the second terms
with the common factor y, then factorise
i) vy + wy + vz + wz
the third and fourth terms with the
= y (v + w) + z (v + w) common factor z.

= (v + w)(y + z) .
(v + w) is the common factor.

ii) 21bm – 7bs + 6cm – 2cs
Factorise the first and the second terms with
= 7b(3m – s) + 2c(3m – s) common factor 7b, then factorise the third
and fourth terms with common factor 2c.
= (3m – s)(7b + 2c)
(3m – s) is the common factor.



(c) Factorising the Algebraic Expressions by Using Difference of Two Squares

a2 – b2 = (a + b)(a – b)

i) x2 – 16 = x2 – 42
= (x + 4)(x – 4)

ii) 4x2 – 25 = (2x)2 – 52

= (2x + 5)(2x – 5)

(d) Factorising the Expressions by Using the Cross Method

i) x2 – 5x + 6
The summation of the cross
multiplication products should
x 3 equal to the middle term of the
x 2 quadratic expression in the
 3 x  2 x  5 x general form.

x2 – 5x + 6 = (x – 3) (x – 2)

ii) 3x2 + 4x – 4
The summation of the cross
multiplication products should
3x 2
equal to the middle term of the
x 2 quadratic expression in the
 2x  6x   4x general form.

3x2 + 4x – 4 = (3x – 2) (x + 2)



ALTERNATIVE METHOD

Factorise the following quadratic expressions: REMEMBER!!!

An algebraic expression can
2
i) x – 5x + 6 be represented in the general
form of ax2 + bx + c, where
a=+1 b= –5 c =+6 a, b, c are constants and
a ≠ 0, b ≠ 0, c ≠ 0.

ac b
+1  (+ 6) = + 6 –2  (–3) = +6
+6 –5
–2 + (–3) = –5
–2 –3

(x – 2) (x – 3)

 x 2  5x  6  ( x  2)(x  3)

ii) x 2 – 5x – 6

a=+1 b= –5 c = –6

+1  (–6) = –6
ac b

–6 –5
+1  (–6) = –6
+1 –6
+1 – 6 = –5

(x + 1) (x– 6)

 x 2  5x  6  ( x  1)(x  6)



(iii) 2x2 – 11x + 5

a=+2 b = –11 c =+5

(+2)  (+5) = +10 ac b

+ 10 –11

–1 – 10 –1  (–10) = +10

1 10 –1 + (–10) = –11
 
2 2

1 The coefficient of x2 is 2,
 5 divide each number by 2.
2

The coefficient of x2 is 2,
multiply by 2:
(2x – 1) (x – 5)
x  12 x  5
 2x  1 x  5
2

 2 x  1)(x  5
 2x 2  11x  5  (2x  1)(x  5)
TEST YOURSELF C

(iv) 3x2 + 4x – 4

a =+ 3 b=+ 4 c = –4

ac b –2 + 6 = 4
3  (– 4) = –12 – 12 +4

–2 +6 The coefficient of x2 is 3, divide each
number by 3.
2 6

3 3
The coefficient of x2 is 3, multiply by 3:


2
2
x  2 x  2
3
3  3x  2 x  2
3

 3x  2)(x  2
(3x – 2) (x + 2)
 3x 2  4x  4  (3x  2)(x  2)



TEST YOURSELF C

Factorise the following quadratic expressions completely.

1. 3p 2 – 15 2. 2x 2 – 6

3. x 2 – 4x 4. 5m 2 + 12m

5. pq – 2p 6. 7m + 14mn

7. k2 –144 8. 4p 2 – 1

9. 2x 2 – 18 10. 9m2 – 169



11. 2x 2 + x – 10 12. 3x 2 + 2x – 8

13. 3p 2 – 5p – 12 14. 4p2 – 3p – 1

15.
2
2x – 3x – 5 16. 4x 2 – 12x + 5

17. 5p 2 + p – 6 18. 2x
2
– 11x + 12

19. 3p + k + 9pr + 3kr 20. 4c2 – 2ct – 6cw + 3tw



PART D:
CHANGING THE SUBJECT
OF A FORMULA

LEARNING OBJECTIVE

Upon completion of this module, pupils will be able to change the subject of
a formula.


If pupils have difficulties in changing the subject of a formula, they probably
face problems in the following topics:
 Functions – Changing the subject of the formula is essential in finding
the inverse function.
 Circular Measure – Changing the subject of the formula is needed to

find the r or  from the formulae s = r  or A  1 r 2 .
2
 Simultaneous Equations – Changing the subject of the formula is the
first step of solving simultaneous equations.

Strategy:
1. Teacher gives examples of formulae and asks pupils to indicate the subject
of each of the formula.
Examples: y=x–2
1 y, A and V are the
A  bh subjects of the
2
formulae.
V  r 2 h



LESSON NOTES

PART D:
CHANGING THE SUBJECT OF A FORMULA

1. An algebraic formula is an equation which connects a few unknowns with an equal
sign.

1
A  bh
Examples: 2
V  r 2 h

2. The subject of a formula is a single unknown with a power of one and a coefficient
of one, expressed in terms of other unknowns.

1 A is the subject of the formula because it is
Examples: A bh
2 expressed in terms of other unknowns.

a2 is not the subject of the formula
a2 = b2 + c2 because the power ≠ 1

T is not the subject of the formula
1 2 because it is found on both sides of the
T Tr h equation.
2

3. A formula can be rearranged to change the subject of the formula. Here are the
suggested steps that can be used to change the subject of the formula:

(i) Fraction : Get rid of fraction by multiplying each term in the formula with
the denominator of the fraction.

(ii) Brackets : Expand the terms in the bracket.
(iii) Group : Group all the like terms on the left or right side of the formula.
(iv) Factorise : Factorise the terms with common factor.
(v) Solve : Make the coefficient and the power of the subject equal to one.



EXAMPLES

Steps to Change the Subject of a Formula
(i) Fraction
(ii) Brackets
(iii) Group
(iv) Factorise
(v) Solve

1. Given that 2x + y = 2, express x in terms of y.
Solution: No fraction and brackets.
2x + y = 2
Group:
2x = 2 – y Retain the x term on the left hand side of the
equation by grouping all the y term to the
2 y
x= right hand side of the equation.
2

Solve:
Divide both sides of the equation by 2 to
make the coefficient of x equal to 1.

3x  y
2. Given that  5 y , express x in terms of y.
2

Solution:

3x  y
 5y Fraction:
2
Multiply both sides of the equation by 2.
3x + y = 10y
Group:
3x = 10y – y
Retain the x term on the left hand side of the
3x = 9y equation by grouping all the y term to the
right hand side of the equation.
9y
x=
3 Solve:
Divide both sides of the equation by 3 to
x = 3y make the coefficient of x equal to 1.



3. Given that x  2 y , express x in terms of y.

Solution:

x  2y Solve:
Square both sides of the equation to make the
2 power of x equal to 1.
x = (2y)
2
x = 4y

x
4. Given that  p , express x in terms of p.
3

Solution:

x
p
3
Fraction:
x  3p Multiply both sides of the equation by 3.

x  (3 p ) 2
x  9 p2 Solve:

Square both sides of the equation to make
the power of x equal to1.

5. Given that 3 x  2  x  y , express x in terms of y.

Solution:
Group:
3 x 2 xy Group the like terms
3 x  x  y2
Simplify the terms.
2 x  y2
y2 Solve:
x
2 Divide both sides of the equation by 2 to
 y 2
2 make the coefficient of x equal to 1.
x 
 2 
Solve:
Square both sides of equation to make the
power of x equal to 1.



11x
6. Given that – 2(1 – y) = 2 xp , express x in terms of y and p.
4

Solution:
Fraction:
11x
– 2 (1 – y) = 2 xp Multiply both sides of the equation
4
by 4.
11x – 8(1 – y) = 8 xp
Bracket:
11x – 8 + 8y = 8xp Expand the bracket.

11x – 8xp = 8 – 8y
Group:
Group the like terms.

x(11 – 8p) = 8 – 8y
Factorise:
8  8y Factorise the x term.
x=
11  8 p
Solve:
Divide both sides by (11 – 8p) to
make the coefficient of x equal to 1.

2 p  3x
7. Given that = 1 – p , express p in terms of x and n.
5n

Solution:

2 p  3x
=1–p Fraction:
5n Multiply both sides of the equation by
2p – 3x = 5n – 5pn 5n.

2p + 5pn = 5n + 3x Group:
Group the like p terms.
p(2 + 5n) = 5n + 3x

5n  3x Factorise:
p= Factorise the p terms.
2  5n

Solve:
Divide both sides of the equation by
(2 + 5n) to make the coefficient of p
equal to 1.



TEST YOURSELF D

1. Express x in terms of y.

a) x  y  2  0 b) 2 x  y  3  0

c) 2 y  x  1
d)
1
x  y   2
2

e) 3x  y  5 f) 3 y  x  4



2. Express x in terms of y.

a) y  x b) 2 y  x

x d) y  1  3 x
c) 2 y 
3

e) 3 x  y  x  1 f) x 1  y



3. Change the subject of the following formulae:
xa 1 x
a) Given that  2 , express x in terms b) Given that y  , express x in terms
xa 1 x
of a . of y .

c) Given that 1  1  1 , express u in d) Given that 2 p  q  3 , express p in
f u v 2p  q 4
terms of v and f . terms of q.

e) Given that p  3m  2mn , express m in f) Given that A  B C  1  , express C in
 
terms of n and p .  C 
terms of A and B .

2y  x l
g) Given that  2 y , express y in h) Given that T  2 , express g in
x g
terms of x.
terms of T and l.



ACTIVITIES

CROSSWORD PUZZLE

HORIZONTAL

1) – 4p, 10q and 7r are called algebraic .

3) An algebraic term is the of unknowns and numbers.

4) 4m and 8m are called terms.

5) V  r 2 h , then V is the of the formula.

7) An can be represented by a letter.

10) x 2  3x  2  x  1x  2 .



VERTICAL

2) An algebraic consists of two or more algebraic terms combined by
addition or subtraction or both.

6) 2 x  1x  2  2 x 2  5 x  2 .

8) terms are terms with different unknowns.

9) The number attached in front of an unknown is called .



RIDDLES

RIDDLE 1

1. You are given 9 multiple-choice questions.
2. For each of the questions, choose the correct answer and fill the alphabet in the box
below.
3. Rearrange the alphabets to form a word.
4. What is the word?

1 2 3 4 5 6 7 8 9

1
2
1. Calculate 5.
3

1
D) O) 1
5

11 11
W) N)
3 15

2. Simplify  3x  9 y  6 x  7 y .

F) 3x  2 y W)  9 x  16 y

E) 3x  2 y X) 9 x  2 y

p q
3. Simplify  .
3 2

2 p  3q 2 p  3q
L) A)
6 6

3q  2 p 3 p  2q
N) R)
6 6



4. Expand 2( x  4)  ( x  7) .

A) x  1 D) x  15

U) 3x  1 C) 3x  15

5. Expand  3a(2b  5c) .

S )  6ab  15ac C) 6ab  15ac

T)  6ab  15ac R) 6ab  15ac

6. Factorise x 2  25 .

E) ( x  5)(x  5) T) ( x  5)(x  5)

I) ( x  5)(x  5) C) ( x  25)(x  25)

7. Factorise pq  4q .

D) pq(1  4q) E) q( p  4)

T) p(q  4) S) q( p  4)

8. Factorise x 2  8x  12 .

I ) ( x  2)(x  6) W) ( x  2)(x  6)

F) ( x  4)(x  3) C) ( x  4)(x  3)

3x  y
9. Given that  4 , express x in terms of y.
2x

y y
L) x   C) x 
5 5

y 8 y
T) x  N) x 
11 3



RIDDLE 2

1. You are given 9 multiple-choice questions.
2. For each of the questions, choose the correct answer and fill the alphabet in the box
below.
3. Rearrange the alphabets to form a word.
4. What is the word?

1 2 3 4 5 6 7 8 9

5
1
1. Calculate x .
3

5 x 5 x
A) O)
3 3x
3x 3
I) N)
x5 x5

3p q
2. Simplify  .
4 5r

15 pr 4q
F) R)
4q 15 pr
3 pq 3 pq
W) B)
20r 5r

x xy
3. Simplify  .
yz 2 z

2 x2
N) D)
y2 2z 2
x x2
L) I) 2
2z 2 z

4. Solve x  y 2  x(3x  y).
E)  2 x 2  y 2  xy D) 2 x 2  y 2  xy

I) x 2  y 2  3x 2  xy N) 2 x 2  y 2  xy



5. Expand  p  5 2 .

I) p 2  25 N) p 2  25

D) p 2  10 p  25 L) p 2  10 p  25

6. Factorise 2 y 2  7 y  15 .

F) (2 y  3)( y  5) D) (2 y  3)( y  5)

W) (2 y  3)( y  5) L) ( y  3)(2 y  5)

7. Factorise 2 p 2  11 p  5 .

R) (2 p  1)( p  5) B) (2 p  1)( p  5)

F) ( p  1)( p  5) W) ( p  1)(2 p  5)

B
8. Given that (C  1)  A , express C in terms of A and B.
C

B 1
L) C  R) C 
BA BA
AB AB
C) C  N) C 
BA BA

9. Given that 5 x  y  x  2 , express x in terms of y.
y2  4 y2  4
O) x  B) x 
16 24

 y 1  y  2
2 2

I) x  U) x   
 2   4 



FURTHER
EXPLORATION

SUGGESTED WEBSITES:

1. http://www.themathpage.com/alg/algebraic-expressions.htm

2. http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut11_si
mp.htm

3. http://www.helpalgebra.com/onlinebook/simplifyingalgebraicexpressions.htm

4. http://www.tutor.com.my/tutor/daily/eharian_06.asp?h=60104&e=PMR&S=MAT&ft=F
TN



ANSWERS

TEST YOURSELF A:

1. 9a – 5b
2. – 2m – 4n
3. 6k
4. 2p

15 x  y 20h  6k
5. 6.
5 xy 15

6ab 4(4c  d )
7. 8.
7c 3c  d

x
9. 2
z2 10.
v2
4  2x
2x 12.
11. 4  5x
5  6x

TEST YOURSELF B:

1. – 8n + 3 6. x + y

1 7. e 2
2. 3q +
2

3. – 12x2 + 18xy 8. n 2  m 2  mn

4. – 3b 9. f 2  2 fg

5. p 10. h 2  2ih  5i 2



TEST YOURSELF C:

1. 3(p 2 – 5) 2. 2(x 2 – 3) 3. x(x – 4)

4. m(5m + 12) 5. p(q – 2) 6. 7m (1 + 2n)

7. (k + 12)(k – 12) 8. (2p – 1)(2p + 1) 9. 2(x – 3)(x + 3)

10. (3m + 13)(3m – 13) 11. (2x + 5)(x – 2) 12. (3x – 4)(x + 2)

13. (3p + 4)(p – 3) 14. (4p + 1)(p – 1) 15. (2x – 5)(x +1)

16. (2x – 5)(2x – 1) 17. (5p + 6)(p – 1) 18. (2x – 3)(x – 4)

19. (1 + 3r)(3p + k) 20. (2c – t)(2c – 3w)

TEST YOURSELF D:

3 y
(b) x 
1. (a) x = 2 – y 2 (c) x = 2y – 1

5 y
(d) x = 4 – y (e) x (f) x = 3y – 4
3

2. (a) x = y2 (b) x  4 y 2 (c) x  36 y 2

1 y 
2
 y  1
2

(d) x    ( e) x    (f) x  y 2  1
 3   2 

y 1 fv
3. (a) x  3a (b) x (c) u 
y 1 v f

p
7q (e) m B
(d) p 2n  3 (f) C 
2 B A

(g) y 
x 4 2 l
(h) g
2( x  1) T2



ACTIVITIES

CROSSWORD PUZZLE

RIDDLES
RIDDLE 1
2 3 1 5 4 7 6 8 9
F A N T A S T I C

RIDDLE 2
2 1 3 5 4 7 6 9 8
W O N D E R F U L


Basic Essential


UNIT 4

LINEAR EQUATIONS

Unit 1:
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Linear Equations 2

Part B: Solving Linear Equations in the Forms of x + a = b and x – a = b 6

x
Part C: Solving Linear Equations in the Forms of ax = b and =b 9
a

Part D: Solving Linear Equations in the Form of ax + b = c 12

x
Part E: Solving Linear Equations in the Form of +b=c 15
a

Part F: Further Practice on Solving Linear Equations 18

Answers 23

Basic Essentials Additional Mathematics (BEAMS) Module
UNIT 4: Linear Equations

MODULE OVERVIEW

1. The aim of this module is to reinforce pupils’ understanding on the concept involved in
solving linear equations.

2. The module is written as a guide for teachers to help pupils master the basic skills
required to solve linear equations.

3. This module consists of six parts and each part deals with a few specific skills.
Teachers may use any parts of the module as and when it is required.

4. Overall lesson notes are given in Part A, to stress on the important facts and concepts
required for this topic.



PART A:
LINEAR EQUATIONS

LEARNING OBJECTIVES


1. understand and use the concept of equality;

2. understand and use the concept of linear equations in one unknown; and

3. understand the concept of solutions of linear equations in one unknown
by determining if a numerical value is a solution of a given linear
equation in one unknown.

a. determine if a numericalLEARNING STRATEGIES linear equation
TEACHING AND value is a solution of a given
in one unknown;
The concepts of can be confusing and difficult for pupils to grasp. Pupils might
face difficulty when dealing with problems involving linear equations.

Strategy:

Teacher should emphasise the importance of checking the solutions obtained.
Teacher should also ensure that pupils understand the concept of equality and
linear equations by emphasising the properties of equality.



OVERALL LESSON NOTES

GUIDELINES:

1. The solution to an equation is the value that makes the equation ‘true’. Therefore,
solutions obtained can be checked by substituting them back into the original
equation, and make sure that you get a true statement.
2. Take note of the following properties of equality:

(a) Subtraction
Arithmetic Algebra

8 = (4) (2) a=b

8 – 3 = (4) (2) – 3 a–c=b–c

(b) Addition
Arithmetic Algebra

8 = (4) (2) ;
a=b

8 + 3 = (4) (2) + 3 a+c=b+c

(c) Division

Arithmetic Algebra

8=6+2 a=b

8 62 a b
  c≠0
3 3 c c

(d) Multiplication
Arithmetic Algebra

8 = (6 +2) a=b

(8)(3) = (6+2) (3) ac = bc



PART A:
LINEAR EQUATIONS

LESSON NOTES

1. An equation shows the equality of two expressions and is joined by an equal sign.
Example: 2  4=7+1

2. An equation can also contain an unknown, which can take the place of a number.

Example: x + 1 = 3, where x is an unknown

A linear equation in one unknown is an equation that consists of only one unknown.

3. To solve an equation is to find the value of the unknown in the linear equation.

4. When solving equations,

(i) always write each step on a new line;

(ii) keep the left hand side (LHS) and the right hand side (RHS) balanced by:

 adding the same number or term to both sides of the equation;

 subtracting the same number or term from both sides of the equations;

 multiplying both sides of the equation by the same number or term;

 dividing both sides of the equation by the same number or term; and

(iii) simplify (whenever possible).

5. When pupils have mastered the skills and concepts involved in solving linear equations,
they can solve the questions by using alternative method.

What is solving
an equation?

Solving an equation is like solving a puzzle to find the value of the unknown.



The puzzle can be visualised by using real life and concrete examples.

1. The equality in an equation can be visualised as the state of equilibrium of a balance.

(a) x + 2 = 5
x=3
x=?
2.

2. The equality in an equation can also be explained by using tiles (preferably coloured tiles).

x x
x

x+2=5
x+2=5 x + 2x – 2 – 25= – 2 2
+ = 5–

x == 3
x3



PART B:
SOLVING LINEAR EQUATIONS IN
THE FORMS OF
x+a=b AND x – a = b

LEARNING OBJECTIVES

Upon completion of Part B, pupils will be able to understand the concept of
solutions of linear equations in one unknown by solving equations in the
form of:
(i) x+a=b
(ii) x – a = b

where a, b, c are integers and x is an unknown.


Some pupils might face difficulty when solving linear equations in one
unknown by solving equations in the form of:
(i) x+a=b
(ii) x–a=b


Strategy:

Teacher should emphasise the idea of balancing the linear equations. When pupils
have mastered the skills and concepts involved in solving linear equations, they
can solve the questions using the alternative method.



PART B:
SOLVING LINEAR EQUATIONS IN THE FORM OF
x+a=b OR x–a=b

EXAMPLES

Solve the following equations.

(i) x  2  5 (ii) x  3  5

Solutions:

(i) x25 Subtract 2 from both Alternative Method:
sides of the equation.
x+2–2=5–2 x25
x 52
x=5–2 Simplify the LHS.
x3
x=3 Simplify the RHS.

(ii) x35
Add 3 to both sides of
Alternative Method:
the equation.
x–3+3=5+3
x 35
x=5+3 Simplify the LHS. x 53
x=8 Simplify the RHS. x 8



TEST YOURSELF B


1. x+1=6 2. x–2 = 4 3. x–7=2

4. 7+x=5 5. 5+x= –2 6. – 9 + x = – 12

7. –12 + x = 36 8. x – 9 = –54 9. – 28 + x = –78

10. x + 9 = –102 11. –19 + x = 38 12. x – 5 = –92

13. –13 + x = –120 14. –35 + x = 212 15. –82 + x = –197



PART C:
THE FORMS OF
x
ax = b AND b
a

LEARNING OBJECTIVES

Upon completion of Part C, pupils will be able to understand the concept of
form of:
(a) ax = b
x
(b)  b
a



Pupils face difficulty when solving linear equations in one unknown by solving
equations in the form of:
(a) ax = b
x
(b)  b
a

Strategy:




PART C:
SOLVING LINEAR EQUATION
x
ax = b AND b
a

EXAMPLES


m
(i) 3m = 12 (ii) 4
3

Solutions:

(i) 3  m = 12
Alternative Method:
3  m 12
 Divide both sides of
3 3 the equation by 3. 3m  12
12
12 m
m Simplify the LHS. 3
3 m4
m=4 Simplify the RHS.

m
(ii) 4
3
Multiply both sides of Alternative Method:
m the equation by 3.
3  43 m
3 4
3
Simplify the LHS. m  3 4
m = 4 3
m  12
m = 12 Simplify the RHS.



TEST YOURSELF C


1. 2p = 6 2. 5k = – 20 3. – 4h = 24

4. 7l  56 5.  8 j  72 6.  5n  60

7. 6v  72 8. 7 y  42 9. 12z  96

m r w
10. 4 11. =5 12. = –7
2 4 8

t s u
13.  8 14. 9 15.   6
8 12 5



PART D:
THE FORM OF
ax + b = c

LEARNING OBJECTIVE

Upon completion of Part D, pupils will be able to understand the concept of
form of ax + b = c where a, b, c are integers and x is an unknown.


Some pupils might face difficulty when solving linear equations in one
unknown by solving equations in the form of ax + b = c where a, b, c are
integers and x is an unknown.

Strategy:




PART D:
SOLVING LINEAR EQUATIONS IN THE FORM OF ax + b = c

EXAMPLES

Solve the equation 2x – 3 = 11.

Solution:

Method 1

2x – 3 = 11 Add 3 to both sides of
Alternative Method:
the equation.
2x – 3 + 3 = 11 + 3
2 x  3  11
2x = 14 Simplify both sides of 2 x  11  3
the equation.
2 x  14
2 x 14
 14
2 2 Divide both sides of x
the equation by 2. 2
14 x2
x
2 Simplify the LHS.

x=7 Simplify the RHS.

Method 2

2x  3  11

2 x 3 11 Divide both sides of
  Alternative Method:
2 2 2 the equation by 2.
2 x  3  11
3 11 2 x 3 11
x  Simplify the LHS.  
2 2 2 2 2
11 3
3 3 11 3 3 x 
x    Add
2
to both sides 2 2
2 2 2 2
14
of the equation. x
14 2
x x7
2
Simplify both sides of
x7 the equation.



TEST YOURSELF D


1. 2m + 3 = 7 2. 3p – 1 = 11 3. 3k + 4 = 10

4. 4m – 3 = 9 5. 4y + 3 = 9 6. 4p + 8 = 11

7. 2 + 3p = 8 8. 4 + 3k = 10 9. 5 + 4x = 1

10. 4 – 3p = 7 11. 10 – 2p = 4 12. 8 – 2m = 6



PART E
THE FORM OF

x
bc
a

LEARNING OBJECTIVES

Upon completion of Part E, pupils will be able to understand the concept of
solutions of linear equations in one unknown by solving equations in the form
x
of  b where a, b, c are integers and x is an unknown.
a


Pupils face difficulty when solving linear equations in one unknown by solving
x
equations in the form of  b where a, b, c are integers and x is an unknown.
a

Strategy:




PART E:
x
SOLVING LINEAR EQUATIONS IN THE FORM OF bc
a
EXAMPLES

x
Solve the equation  4  1.
3

Solution:

Method 1

x
 4 1
3

x
44 = 1 + 4 Add 4 to both sides of Alternative
3 the equation. Method:

x x
5 Simplify both sides of  4 1
3 3
the equation. x
x 1 4
 3  5 3 3
3 Multiply both sides of x
the equation by 3. 5
3
x  5 3
x  3 5
x = 15 Simplify both sides of the x  15
equation.
Method 2
Multiply both sides of
x 
  4   3  1 3 the equation by 3.
3 

x Expand the LHS.
 3  4  3  1 3
3
x  12  3 the equation.

x – 12 + 12 = 3 + 12 Add 12 to both sides of
the equation.
x  3  12
x  15 the equation.



TEST YOURSELF E


m b k
1. 35 2. 2 1 3. 27
2 3 3

h h m
4. 3+ =5 5. 4+ =6 6. 1  2
2 5 4

h k h
7. 2 5 8. +3=1 9. 3 2
4 6 5

10. 3 – 2m = 7 m 12. 12 + 5h = 2
11. 3 7
2



PART F:
FURTHER PRACTICE ON SOLVING
LINEAR EQUATIONS

LEARNING OBJECTIVE

Upon completion of Part F, pupils will be able to apply the concept of
solutions of linear equations in one unknown when solving equations of
various forms.


Pupils face difficulty when solving linear equations of various forms.

Strategy:




PART F:
FURTHER PRACTICE

EXAMPLES

Solve the following equations: Alternative Method:
(i) – 4x – 5 = 2x + 7  4x  5  2x  7
 4x  2x  7  5
 6 x  12
Solution: 12
x
6
x  2
Method 1

 4x  5  2x  7 Subtract 2x from both sides of the equation.
–4x – 2x – 5 = 2x – 2x + 7
 6x  5  7 Simplify both sides of the equation.
 6x  5  5  7  5
Add 5 to both sides of the equation.
 6 x  12
 6 x 12
 Simplify both sides of the equation.
6 6
x  2 Divide both sides of the equation by –6.

Method 2

 4x  5  2x  7

– 4x – 5 + 5 = 2x + 7 + 5 Add 5 to both sides of the equation.

– 4x = 2x + 12 Simplify both sides of the equation.

– 4x – 2x = 2x – 2x + 12
Subtract 2x from both sides of the equation.
– 6x = 12
Simplify both sides of the equation.
 6 x 12

6 6 Divide both sides of the equation by – 6.
x  2



(ii) 3(n – 2) – 2(n – 1) = 2 (n + 5)
Expand both sides of the equation.
3n – 6 – 2n + 2 = 2n + 10
Simplify the LHS.
n – 4 = 2n + 10

n – 2n – 4 = 2n – 2n + 10 Subtract 2n from both sides of the equation.

– n – 4 = 10

– n – 4 + 4 = 10 + 4 Add 4 to both sides of the equation.

– n = 14

 n 14 Divide both sides of the equation by – 1.

1 1
n  14

Alternative Method:

3(n  2)  2(n  1)  2(n  5)
3n  6  2n  2  2n  10
n  4  2n  10
 n  14
n  14



2x  3 x  1
(iii)  3
3 2
 2x  3 x  1 
6    6(3) Multiply both sides of the equation by the
 3 2  LCM.
 2x  3   x  1 
6   6   6(3)
 3   2 
2(2 x  3)  3( x  1)  18 Expand the brackets.
4 x  6  3 x  3  18
7 x  3  18 Simplify LHS.

7 x  3  3  18  3
Add 3 to both sides of the equation.
7 x  21
7 x 21 Divide both sides of the equation by 7.

7 7
x3

Alternative Method:

2x  3 x  1
 3
3 2
 2x  3 x  1 
6    3 6
 3 2 
2(2 x  3)  3( x  1)  18
4 x  6  3 x  3  18
7 x  3  18
7 x  18  3
7 x  21
21
x
7
x3



TEST YOURSELF F


1. 4x – 5 + 2x = 8x – 3 – x 2. 4(x – 2) – 3(x – 1) = 2 (x + 6)

3. –3(2n – 5) = 2(4n + 7) 3x 9
4. 
4 2

x 2 5 x x
5.   6.  2
2 3 6 3 5

y 13 y x  2 x 1 9
7. 5  8.  
2 6 3 4 2

2 x  5 3x  4 2x  7 x7
9.  0 10. 4
6 8 9 12



ANSWERS

TEST YOURSELF B:

1. x=5 2. x=6 3. x=9
4. x = –2 5. x = –7 6. x = –3
7. x = 48 8. x = –45 9. x = –50
10. x = –111 11. x = 57 12. x = –87
13. x = –107 14. x = 247 15. x = –115

TEST YOURSELF C:

1. p=3 2. k=–4 3. h = –6

4. l=8 5. j=–9 6. n = 12

7. v = 12 8. y=–6 9. z=8

10. m=8 11. r = 20 12. w = – 56

13. t = – 64 14. s = 108 15. u = 30

TEST YOURSELF D:

1. m=2 2. p=4 3. k=2

3 3
4. m=3 5. y  6. p 
2 4

7. p=2 8. k = 2 9. x = –1

10. p = −1 11. p = 3 12. m = 1

TEST YOURSELF E:

1. m=4 10. b = 9 11. k = 15

4. h=4 5. h = 10 6. m = 12

7. h = 12 8. k = −12 9. h=5

10. m = −2 11. m = −8 12. h = −2



TEST YOURSELF F:
1
1. x=−2 2. x = − 17 3. n  4. x=6
14

5. x=3 6. x = 15 7. y=3 8. x=7

9. x = −8 10. x = 19


Basic Essential


UNIT 5

INDICES

Unit 1:
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Indices I 2

1.0 Expressing Repeated Multiplication as an and Vice Versa 3

2.0 Finding the Value of an 3
m n
Verifying a  a  a
m n
3.0 4
4.0 Simplifying Multiplication of Numbers, Expressed in Index
Notation with the Same Base 4
5.0 Simplifying Multiplication of Algebraic Terms, Expressed in Index
Notation with Different Bases 5
7.0 Simplifying Multiplication of Algebraic Terms Expressed in Index

Part B: Indices II 8

mn
Verifying a  a  a
m n
1.0 9
2.0 Simplifying Division of Numbers, Expressed In Index Notation
with the Same Base 9

3.0 Simplifying Division of Algebraic Terms, Expressed in Index


5.0 Simplifying Multiplication of Algebraic Terms, Expressed in
Index Notation with Different Bases 10

Part C: Indices III 12

Verifying (a )  a
m n mn
1.0 13
2.0 Simplifying Numbers Expressed in Index Notation Raised
to a Power 13

3.0 Simplifying Algebraic Terms Expressed in Index Notation Raised
to a Power 14
1
a n 
4.0 Verifying an 15
1
5.0 Verifying an na
16

Activity 20

Answers 22

UNIT 5: Indices

MODULE OVERVIEW

1. The aim of this module is to reinforce pupils’ understanding on the
concept of indices.

2. This module aims to provide the basic essential skills for the learning of
Additional Mathematics topics such as:
 PART 1
Indices and Logarithms
 Progressions
 Functions
 Quadratic Functions
 Quadratic Equations
 Simultaneous Equations
 Differentiation
 Linear Law
 Integration
 Motion Along a Straight Line

3. Teachers can use this module as part of the materials for teaching the
sub-topic of Indices in Form 4. Teachers can also use this module after
PMR as preparatory work for Form 4 Mathematics and Additional
Mathematics. Nevertheless, students can also use this module for self-
assessed learning.

4. This module is divided into three parts. Each part consists of a few learning
objectives which can be taught separately. Teachers are advised to use any
sections of the module as and when it is required.


UNIT 5: Indices

PART A:
INDICES I

LEARNING OBJECTIVES


1. express repeated multiplication as an and vice versa;

2. find the value of an;

3. verify a m  a n  a m n ;

4. simplify multiplication of
(a) numbers;
(b) algebraic terms, expressed in index notation with the same base;

5. simplify multiplication of
(a) numbers; and
(b) algebraic terms, expressed in index notation with different bases.


The concept of indices is not easy for some pupils to grasp and hence they
have phobia when dealing with multiplication of indices.

Strategy:

Pupils learn from the pre-requisite of repeated multiplication starting from
squares and cubes of numbers. Through pattern recognition, pupils make
generalisations by using the inductive method.

The multiplication of indices should be introduced by using numbers and
simple fractions first, and then followed by algebraic terms. This is intended
to help pupils build confidence to solve questions involving indices.


UNIT 5: Indices

LESSON NOTES A

1.0 Expressing Repeated Multiplication As an and Vice Versa

(i) 32  3  3 32 is read as
‘three to the power of 2’
2 factors of 3
or
‘three to the second power’.

(ii) (4)3  (4)(4)(4) index
32
3 factors of (4)
base

(iii) r3  r  r  r
3 factors of r
(a) What is 24?
(b) What is (−1)3?
(c) What is an?
(iv) (6  m) 2  (6  m)( 6  m)

2 factors of (6+m)

2.0 Finding the Value of an

(i ) 25  2  2  2  2  2
 32

(ii ) (  5)3  ( 5)(5)(5)
  125

4
2 24
(iii)    4
3 3
 2 2 2 2 
  
 3 3 3 3 
16

81


UNIT 5: Indices

m n
Verifying a  a  a
m n
3.0

(i) 23  24  (2  2  2)  (2  2  2  2)
 27  234

(ii ) 7  7 2  7  (7  7 )
 73  7 12

(iii ) ( y  1) 2 ( y  1)3  [( y  1)( y  1)] [( y  1)( y  1)( y  1)]
 ( y  1)5  ( y  1) 23

am  an  amn

4.0 Simplifying Multiplication of Numbers, Expressed In Index Notation with the Same
Base

(i) 6 3  6 4  6  6 3 41
 68

(ii ) (5) 3  (5) 8  (5) 38
 (5)11

5 15
1 1 1
(iii )     
3 3 3
6
1
  
3


UNIT 5: Indices

5.0 Simplifying Multiplication of Algebraic Terms, Expressed In Index Notation with the
Same Base

(i) p 2  p 4  p 2 4  p 6
(ab) 5  a 5 b 5
Conversely,
(ii ) 2 w9  3w11  w 20  6 w911 20  6 w 40 a 5 b 5  (ab) 5

(iii ) (ab) 3  (ab) 2  ab
3 2
 (ab) 5 4
s
4
s
   4
t t
3 31 4
s s s s Conversely,
(iv )           
t t t t
4
s4  s 
 
t4  t 

6.0 Simplifying Multiplication of Numbers, Expressed In Index Notation with Different
Bases

Note:
(i) 34  38  2 3  348  2 3  312  2 3  Sum up the indices
with the same
(ii ) 53  5 7  714  7 3  537  7143  510  717 base.
 numbers with
different bases
3 2 4 3 2 4 5 4 cannot be
1 1 3 1 3 1 3
(iii )                   simplified.
 2  2 5 2 5  2 5

7.0 Simplifying Multiplication of Algebraic Terms Expressed In Index Notation with
Different Bases

(i) m 5  m 2  n 5  n 5  m 52  n 55  m 7 n10

(ii) 3t 6  2s 3  5r 2  30t 6 s 3 r 2

2 4 1 4 13 3 4 4 3
(iii ) p  p3  q3  p q  p q
3 5 2 15 15


UNIT 5: Indices

EXAMPLES & TEST YOURSELF A

1. Find the value of each of the following.

(a) 35  3  3  3  3  3 (b) 63 
 243

(c) (4) 4  (d) 1
5
  
5

(e)  3
3
(f)  1
2

    2  
 4  5

(g)  74  (h)  2
5

   
 3

2. Simplify the following.

(a) 3m 3  4m 2  12m 3 2 (b) 5b 2  3b 4  b 
 12m 5

(c) 2 x 2  (3x 4 )  3x 3  (d) 7 p 3  (2 p 2 )  ( p)3 


UNIT 5: Indices


(a) 43  32  64  9 (b) (3) 2  23  2 2 
 576

(c) (1)3  (7) 4  (7)3  (d) 2
1 1  4
3 2

      
 3  3  5 

(e) 2  23  52  54  (f) 3 2 2
 2 2  2 2
        
 3 7  3 7


(a) 4 f 4  3g 2  12 f 4 g 2 (b) (3r ) 2  2r 3  3s 2 

(c) (w) 3  (7w) 4  (3v) 3  (d) 2
3  1  4 
3 2

 h  k   k  
7  5  5 


UNIT 5: Indices

PART B:
INDICES II

LEARNING OBJECTIVES


mn
verify a  a  a
m n
1. ;

2. simplify division of
(a) numbers;
(b) algebraic terms, expressed in index notation with the same base;

3. simplify division of
(a) numbers; and
(b) algebraic terms, expressed in index notation with different bases.


Some pupils might have difficulties in when dealing with division of indices.

Strategy:

Pupils should be able to make generalisations by using the inductive method.
The divisions of indices are first introduced by using numbers and simple
fractions, and then followed by algebraic terms. This is intended to help
pupils build confidence to solve questions involving indices.


UNIT 5: Indices

LESSON NOTES B

mn
Verifying a  a  a
m n
1.0
1 1 1
2 2 2 2 2
(i) 2  2 5 3
/ / /
21 21 2 1
 (a) What is 25 ÷ 25?
2 2
 2 53 (b) What is 20?
(c) What is a0?
1 1
555555555
(ii) 5  5  / /
9 2

51 51
5 7
 5 9 2
1 1
(2  p )(2  p )(2  p )
(iii) (2  p ) 3  (2  p ) 2 
1
(2  p )(2  p ) 1
 (2  p)  ( 2  p ) 3 2

Note:
a  a m  a mm  a 0
m

am
am  am  1
am
am  an  amn
 a0  1

2. 0 Simplifying Division of Numbers, Expressed In Index Notation with the Same Base

(i) 48  4 2  48  2
 46
(ii) 79  73  7 2  79  3 2
 74
510
(iii) 3
 510  3
5
 57
312
(iv)  312  4  5
3 3
4 5

 33


UNIT 5: Indices

3.0 Simplifying Division of Algebraic Terms, Expressed In Index Notation with the Same
Base

(i) n 6  n 4  n 6 4  n 2

20k 7
(ii) 3
 4k 73  4k 4
5k

 8h 3 8 8
(iii) 2
  h 32   h
3h 3 3

4.0 Simplifying Multiplication of Numbers, Expressed In Index Notation With Different
Bases

REMEMBER!!!

Numbers with
different bases cannot
be simplified.

5.0 Simplifying Multiplication of Algebraic Terms, Expressed In Index Notation with
Different Bases

9h15
(i) 9h15  3h 4 k 6 
3h 4 k 6
3h15 4 3h11 h11
   3 6
k6 k6 k

48 p 8 q 6 4 83 6  2
(ii ) 3 2
 p q
60 p q 5
4
 p5q 4
5


UNIT 5: Indices

EXAMPLES & TEST YOURSELF B


(a) 12 5  12 3  12 53 (b) 910  93  9 
 12 2

 144
(c) 8 9 (d) 2
18
2
12
     
83 3 3

(e) (5) 20 (f) 318  310
 
(5)18 324


(a) q12  q 5  q125 (b) 4 y9  8 y7 
 q7

(c) 35m10 (d) 214 b11
 
15m8 28 b8


(a) 36m9 n 5 9 94 51 (b) 64c16d 13
 m n 
8m 4 n 2 12c 6 d 7
9
 m5 n 4
2

(c) 4 f 6  6 fg 9 (d) 8u 9  7v8  3u 4
 
12 f 4 g 3 12u 6v5


UNIT 5: Indices

PART C:
INDICES III

LEARNING OBJECTIVES

Upon completion of Part C of the module, pupils will be able to:

derive (a )  a ;
m n mn
1.

2. simplify
(a) numbers;
(b) algebraic terms, expressed in index notation raised to a power;

n 1
3. verify a  ; and
an

1
4. verify a n  n a .


The concept of indices is not easy for some pupils to grasp and hence they
have phobia when dealing with algebraic terms.

Strategy:

Pupils learn from the pre-requisite of repeated multiplication starting from
squares and cubes of numbers. Through pattern recognition, pupils make
generalisations by using the inductive method.

In each part of the module, the indices are first introduced using numbers and
simple fractions, and then followed by algebraic terms. This is intended to
help pupils build confidence to solve questions involving indices.


UNIT 5: Indices

LESSON NOTES C

1.0 Verifying (a m )n  a mn

(i) (23 ) 2  23  23
 23  3
 26  2 3 2

(ii ) (39  2 5 ) 3  (39  2 5 )(39  2 5 )(39  2 5 )
 39  9  9  2 5  5  5
 327  215  39 3  2 5 3

2
 113   113  113 
(iii )  4    4  
 15   15  154 
    
 113  3 
  4 4 
 15 
 
116 113 2
 
158 154 2

(a m ) n  a mn

2. 0 Simplifying Numbers Expressed In Index Notation Raised to a Power

(i) (102 )6  102  6  1012

(ii) (27  93 )5  27  5  93  5  235  915

5
(iii)  43   (710 )2  43  5  710  2  415  720
 
 

3 13  3
 613  639
(iv)    6 
 58  58  3 524
 


UNIT 5: Indices

3.0 Simplifying Algebraic Terms Expressed In Index Notation Raised to a Power

(i) (3 x 2 ) 5  35 x 25
 35 x10

(ii ) (e 2 f 3 g 4 ) 5  e 25 f 35 g 45
 e10 f 15 g 20

4 4
1  1
(iii )  a 3b     a 34 b14
5  5
a12b 4

54
a12b 4

625
1 12 4
 a b
625

5
  2m 4  (2) 5 m 45
(iv ) 
 n3  

  n 35
Note:
(2) 5 m 20
 A negative number raised to
n15 an even power is positive.
 32m 20
 A negative number raised to
n15
an odd power is negative.
m 20
  32 15
n

(2 p 3 ) 5  4 p 6 q 7 2 5  4 p 35  p 6  q 7
( v)  
12 p 3 q 2 12 p 3q 2
32 p1563 q 72

3
18 5
32 p q

3
32 18 5
 p q
3


UNIT 5: Indices

n 1
4. 0 Verifying a 
an
3 3 3 3
(i) 34  36 
3 3 3 3 3 3
1
 2  3 4  6  3 2
3
1
3 2  2
3

77
(ii ) 7 2  75 
77777
1
 3  7 2 5  7 3
7

1
a n 
an

Alternative Method
104  10 000 1000
Hint:  100
10  1000
3 ?

102  100
101  10
100  1
1 1
101   1
10 10
1 1
102   2
100 10

1
10n 
10n


UNIT 5: Indices

1
5.0 Verifying an na

2
 1 1
2 
(i)  32   32  31
 
 
2
 1
 32   3
 
 
Take square root on both sides
2
 1  of the equation.
 32   3
 
 
 1  1 
 3 2  3 2   3
  
  
1
32  3
5
 1 1
5 
(ii)  25   25  21
 
 
5
 1
 25   2
 
 
5
 1
5  25   5
2
 
 
1

 1  1  1  1  1  (a) What is 4 2 ?
5  25  2 5  2 5  2 5  2 5   5
2 3
      (b) What is 4 2 ?
     
1 m

25  5
2 (c) What is a n ?
p
 1 1
 p
(iii ) m p   m p
 m1
 
 
p
 1
p m p  
p
m
 
 
1
p
m p
 m

Note:
1
a n
 n
a
1
 a
m
a n a
n
a n
 n
m


UNIT 5: Indices

EXAMPLES & TEST YOURSELF C


(a) (b)
2  5 3
2 53
[(1) 2 ] 3 
 215  32768

(c) 2 (d) 3
 23   3  2 
 2
7  
     
   5  
 

(e)  32  (f)
3

  
 
4
 5 
   23 2 

 


2. (a) Simplify the following.

(i) 2 6
 32 
4
 2 64  3 24 (ii) 2   5 
6 4 3 2

 2 24  38

(iii)
4   4 
2 3 1 5

(iv) 3 2
2

    
3

4 5

(v)  7 3
3 2 (vi) 2
 32  4 4
4

      5   
   5 
 4 7  12   


UNIT 5: Indices

2. (b) Simplify the following.

(i)
2 x  3 5
 (215 )( x 35 )
(ii) x y 
4 7 6


 25 x15
 32 x15

(iii)
w 2
 w12   3 (iv)
4 y 9
 8y7 7


2m n 3mn 
(v) 2 (vi)
 36 p 9 q 5  4 4

3 2

 9 p8q 6  
 

3. Simplify the following expressions:

(a) (b) 1
2 5 
1 3
  
25 4
1

32

(c)
 x 
4 (d) 2st 4
 2  
 3y  6s 1t 5
 

(e) 3 (f) 2
 m 2 n 1   8ab 2 c 3 
    3 6 
 2a b  
 2m 3 k 2 
   


UNIT 5: Indices


(a) 1 (b) 5
 64 3  3  64 100 2 
 4

(c) 
3 (d) 1 1
81 4
 3  27 
2 2

a  (a
(e) 1 1 (f) 4
10 5 3  2
) (a m ) m   1 
3   
 27 


UNIT 5: Indices

ACTIVITY

Solve the questions to discover the WONDERWORD!
 You are given 11 multiple choice questions.
 Choose the correct answer for each of the question.
 Use the alphabets for each of the answer to form the WONDERWORD!

410
1. 
4 2  45

P 40 O 43 R 417 T 413

2. 107  102  53  5 2 

T 10145 5 O 105 56 N 105 55 B 10145 6

2 2  32
3. 
42
22 32 32 42
D E N O
4 22 42 3

4. 2 y x  8 y x 
9 3 2

y7 x2 4 y 11 y1 x 2 4y7
M A L K
4 x4 4 x2

5. 2 5
 32 
4


A 2 3 2 9  36 2 20  36 2 9  38
20 8
N T S

6. m  m  n  n 
5 2 2 4

T m7 n8 U m10n 8 L m7 n 6 E m10n 6


UNIT 5: Indices

3 4 2 3
2 2 2 2
7.         
 5 5 5 5

12 2 6 5
F 2 A 2 V 2 E 2
       
5 5 5 5

5
 72 
8.  3 
4 
 

 710   77   71 0   77 
Y  15  R  8
4 
 M  8  A  15 
4   4  4 
       

25a 9 b 5
9. 
5a 6 b 3

L 15a15b 8 I 5a 3b 8 S 5a 3b 2 T 15a 6 b 5

2 3 2 5
1 1  2  2
10.         
 3 3  5  5

5 10 6 7 5 7 6 10
1  2 1  2 1  2 1  2
P     E     I     R    
3  5 3  5 3  5 3  5

12 p 6 q 7
11. 
3 p 3q 2

p3q5 1
Y A 4 p3q5 R D 3 p9q9
3 3 p9q9

Congratulations! You have completed this activity.

1 2 3 4 5 6 7 8 9 10 11

The WONDERWORD IS: ........................................................


UNIT 5: Indices

ANSWERS

TEST YOURSELF A:

1.

(a) 243 (b) 216

(c) 256 (d) 1
3125
(e) 27 (f) 21
 4
64 25

(g) 2401 (h) 32
243

2.

(a) 12m5 (b) 15b 7
(c)  18x 9 (d) 14 p 8

3.

(a) 576 (b) 288

(c) 823543 (d) 16
6075

(e) 250 000 (f) 256

83 349


UNIT 5: Indices

4.

(a) 12 f 4 g 2 (b) 54r 5 s 2

(c) 64 827 w7 v 3 (d) 144
h2k 5
153125

TEST YOURSELF B:

1.

(a) 144 (b) 531 441

(c) 262 144 (d) 64
729
(e) 25 (f) 81

2.

(a) q7 (b) 1 2
y
2

(c) 7 2 (d) 64b3
m
3

3.

(a) 9 5 4 (b) 16 1 0 6
m n c d
2 3

(c) 2 f 3g6 (d) 14u 7 v 3


UNIT 5: Indices

TEST YOURSELF C:

1.

(a) 32768 (b) 1

(c) 64 (d) 3
6
729
  
2401 5 15625
(e) 36 729 (f)

5 3

125
2 24  16 777 216

2. (a)

(i) 2 24  3
8
(ii) 224  56

(iii) 411 (iv) 32
2(53 )
(v) 7(32 ) (vi) 36 (414 )

43 52

2. (b)

(i) 32x15 (ii) x 24 y 42

(iii) 1 (iv) y1 4
w30
27
(v)  p
2 (vi) 162m 7 n18
16 
q
 


UNIT 5: Indices

3.

(a) 1 1 (b) 4
5

2 32 3

(c) y8 (d) 1  s2 
81  
x4
3  t9



(e) 8k 6 m 3 n 3 (f) 1  a 4c6 
 
16  b16




4.

(a) 4 (b) 100000

(c) 1 (d) 9
27

(e) (f) 1
a5
81

ACTIVITY:

The WONDERWORD is ONEMALAYSIA


Basic Essential


UNIT 6
COORDINATES
AND
Unit 1:
GRAPHS OF FUNCTIONS
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Coordinates 2

Part A1: State the Coordinates of the Given Points 4

Activity A1 8

Part A2: Plot the Point on the Cartesian Plane Given Its Coordinates 9

Activity A2 13

Part B: Graphs of Functions 14

Part B1: Mark Numbers on the x-Axis and y-Axis Based on the Scales Given 16

Part B2: Draw Graph of a Function Given a Table for Values of x and y 20

Activity B1 23

Part B3: State the Values of x and y on the Axes 24

Part B4: State the Value of y Given the Value x from the Graph and Vice Versa 28

Activity B2 34

Answers 35


MODULE OVERVIEW

1. The aim of this module is to reinforce pupils’ understanding of the concept of
coordinates and graphs.

2. It is hoped that this module will provide a solid foundation for the studies of
Additional Mathematics topics such as:
 Coordinate Geometry
 Linear Law
 Linear Programming
 Trigonometric Functions
 Statistics
 Vectors

3. Basically, this module is designed to enhance the pupils’ skills in:
 stating coordinates of points plotted on a Cartesian plane;
 plotting points on a Cartesian plane given the coordinates of the points;
 drawing graphs of functions on a Cartesian plane; and
 stating the y-coordinate given the x-coordinate of a point on a graph and
vice versa.

4. This module consists of two parts. Part A deals with coordinates in two sections
whereas Part B covers graphs of functions in four sections. Each section deals
with one particular skill. This format provides the teacher with the freedom of
choosing any section that is relevant to the skills to be reinforced.

5. Activities are also included to make the reinforcement of basic essential skills
more enjoyable and meaningful.



PART A:
COORDINATES

LEARNING OBJECTIVES


1. state the coordinates of points plotted on a Cartesian plane; and

2. plot points on the Cartesian plane, given the coordinates of the points.


Some pupils may find difficulty in stating the coordinates of a point. The
concept of negative coordinates is even more difficult for them to grasp.
The reverse process of plotting a point given its coordinates is yet another
problem area for some pupils.

Strategy:

Pupils at Form 4 level know what translation is. Capitalizing on this, the
teacher can use the translation = , where O is the origin and P
is a point on the Cartesian plane, to state the coordinates of P as (h, k).
Likewise, given the coordinates of P as ( h , k ), the pupils can carry out
the translation = to determine the position of P on the Cartesian
plane.

This common approach will definitely make the reinforcement of both the
basic skills mentioned above much easier for the pupils. This approach
of integrating coordinates with vectors will also give the pupils a head start
in the topic of Vectors.



PART A:
COORDINATES

LESSON NOTES

1. y
●P
Start from the
origin.
k units

x
O h units

Coordinates of P = (h, k)

2. The translation must start from the origin O horizontally [left or right] and then vertically
[up or down] to reach the point P.

3. The appropriate sign must be given to the components of the translation, h and k, as shown in the
following table.
Component Movement Sign
left –
h
right +
up +
k
down –

4. If there is no horizontal movement, the x-coordinate is 0.

If there is no vertical movement, the y-coordinate is 0.

5. With this system, the coordinates of the Origin O are (0, 0).



PART A1: State the coordinates of the given points.

EXAMPLES TEST YOURSELF

1. 1.
y y
Start from 4 4
A
the origin, 3
• Next, move
3
A
•
move 2 units
2 3 units up. 2
to the right.
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 –4

Coordinates of A = (2, 3) Coordinates of A =

2. 2.
Start from the y y
origin, move 3 units 4 4
B
to the left. 3
2
• 3
2
B
• 1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
-1 –1
–2 Next, move –2
1 unit up.
–3 –3
–4 –4

Coordinates of B = (–3, 1) Coordinates of B =

3. 3.
y y
Start from 4 4
the origin, 3 3
move 2 units
2 2
to the left.
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1

•
C –2 –2
Next, move 2
units down.
–3
C• –3
–4 –4

Coordinates of C = (–2, –2) Coordinates of C =




TEST YOURSELF
EXAMPLES

4. 4.
y y
Start from 4 4
Next, move
the origin, 3 3
3 units
move 4 units
2 down. 2
to the right.
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 • –3
–4
D
–4 •D
Coordinates of D = (4, –3) Coordinates of D =

5. 5.
Start from the y y
to the right. 3 3
2 2
1 1
E
–4 –3 –2 –1 0 1 2 •3
E
4 x –4 –3 –2 –1 0 1 •
2 3 4 x
–1 –1
Do not move –2 –2
along the y-axis
–3 –3
since y = 0.
–4 –4

Coordinates of E = (3, 0) Coordinates of E =

6. 6.
y y
4 4
Start from
the origin,
•
3 F 3
move 3 units
up.
2
1
2
•F
1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
Do not move
–3 along the x-axis –3
–4 since x = 0.
–4

Coordinates of F = (0, 3) Coordinates of F =




TEST YOURSELF
EXAMPLES

7. 7.
y y
Start from 4 4
the origin, 3 3
move 2 units
2 2
to the left.
1 1
G
•
G
–4 –3 –2 –1 0 1 2 3 4 x •
–4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 –4

Coordinates of G = (–2, 0) Coordinates of G =

8. 8.
Start from the y y
down. 3 3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
•H
–2 •H –2
–3 –3
–4 –4

Coordinates of H = (0, –2) Coordinates of H =

9. 9.
y y
J
Start from
8
• 8
J
the origin,
move 6 units
6
Next, move
6
•
4 4
to the right. 8units up.
2 2

–8 –6 –4 –2 0 2 4 6 8 x –8 –6 –4 –2 0 2 4 6 8 x
–2 –2
–4 –4
–6 –6
–8 –8

Coordinates of J = (6, 8) Coordinates of J =





10. 10.
y y

K
8 Start from
K • 8

• 6
4
the origin,
move 6 units
6
4
to the left.
2 2

–8 –6 –4 –2 0 2 4 6 8 x –8 –6 –4 –2 0 2 4 6 8 x
–2 –2

Next, move –4 –4
6 units up. –6 –6
–8 –8

Coordinates of K = (– 6 , 6) Coordinates of K =

11. 11.
y y
Start from the 20 20
origin, move 15 units
to the left. 15 15
10 10
5 5

–20 –15 –10 –5 0 5 10 15 20 x –20 –15 –10 –5 0 5 10 15 20 x
–5 –5

Next, move –10 –10
20 units –15 •L –15
down.
L • –20 –20

Coordinates of L = (–15, –20) Coordinates of L =

12. 12.
Start from y y
the origin, 4 Next, move 4 4
move 3 units units down.
to the right.
2 2

–4 –2 0 2 4 x –4 –2 0 2 4 x

–2 –2
•M
–4 •M –4

Coordinates of M = (3, – 4) Coordinates of M =



ACTIVITY A1

Write the step by step directions involving integer coordinates that
will get the mouse through the maze to the cheese.

y

7
6
5
4
3
2
1
x
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7
–1
–2
–3
–4
–5
–6



PART A2: Plot the point on the Cartesian plane given its coordinates.

. EXAMPLES TEST YOURSELF

1. Plot point A (3, 4) 1. Plot point A (2, 3)
y A y
4
3
• 4
3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 –4

2. Plot point B (–2, 3) 2. Plot point B (–3, 4)
y y
4 4
B
• 3 3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 -1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 –4

3. Plot point C (–1, –3) 3. Plot point C (–1, –2)
y y
4 4
3 3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2

C • –3 –3
–4 –4



PART A2: Plot the point on the Cartesian plane given the coordinates.


4. Plot point D (2, – 4) 4. Plot point D (1, –3)
y y
4 4
3 3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 •D –4

5. Plot point E (1, 0) 5. Plot point E (2, 0)
y y
4 4
3 3
2 2
1 1
E
–4 –3 –2 –1 0 • 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 –4

6. Plot point F (0, 4) 6. Plot point F (0, 3)
y y
•
4
F 4
3 3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 –4





7. Plot point G (–2, 0) 7. Plot point G (– 4,0)
y y
4 4
3 3
2 2
1 1
G
•
–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 –4

8. Plot point H (0, – 4) 8. Plot point H (0, –2)
y y
4 4
3 3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
–3 –3
–4 •H –4

9. Plot point J (6, 4) 9. Plot point J (8, 6)
y y
8 8
6 6
J
4
• 4
2 2

–8 –6 –4 –2 0 2 4 6 8 x –8 –6 –4 –2 0 2 4 6 8 x
–2 –2
–4 –4
–6 –6
–8 –8

.





10. Plot point K (– 4, 6) 10. Plot point K (– 6, 2)
y y
8 8
K
•
4 4

–8 –4 0 4 8 x -8 -4 0 4 8 x

–4 –4

–8 –8

11. Plot point L (–15, –10) 11. Plot point L (–20, –5)
y y
29 20

10 10

–20 –10 0 10 20 x –20 –10 0 10 20 x

•L –10 –10

–20 –20

12. Plot point M (30, –15) 12. Plot point M (10, –25)
y y
20 20

10 10

–40 –20 0 20 40 x –40 –20 0 20 40 x

–10 –10

•M
–20 –20



ACTIVITY A2

Exclusive News:
A group of robbers stole RM 1 million from a bank. They hid the money
somewhere near the Yakomi Islands. As an expert in treasure hunting, you
are required to locate the money! Carry out the following tasks to get the
clue to the location of the money.

Mark the location with the symbol.

1. Enjoy yourself !
Plot the following points on the Cartesian plane.

P(3, 3) , Q(6, 3) , R(3, 1) , S(6, 1) , T(6, –2) , U(3, –2) ,

A(–3, 3) , B(–5, –1) , C(–2, –1) , D(–3, – 2) , E(1, 1) , F(2, 1).

2. Draw the following line segments:

AB, AD, BC, EF, PQ, PR, RS, UT, ST

YAKOMI ISLANDS
y

4

2

x
–4 –2 0 2 4
,
–2

–4



PART B:
GRAPHS OF FUNCTIONS

LEARNING OBJECTIVES


1. understand and use the concept of scales for the coordinate axes;

2. draw graphs of functions; and

3. state the y-coordinate given the x-coordinate of a point on a graph and
vice versa.


Drawing a graph on the graph paper is a challenge to some pupils. The concept
of scales used on both the x-axis and y-axis is equally difficult. Stating the
coordinates of points lying on a particular graph drawn is yet another
problematic area.

Strategy:

Before a proper graph can be drawn, pupils need to know how to mark numbers
on the number line, specifically both the axes, given the scales to be used.
Practice makes perfect. Thus, basic skill practices in this area are given in Part
B1. Combining this basic skills with the knowledge of plotting points
on the Cartesian plane, the skill of drawing graphs of functions, given the
values of x and y, is then further enhanced in Part B2.

Using a similar strategy, Stating the values of numbers on the axes is
done in Part B3 followed by Stating coordinates of points on a graph in
Part B4.

For both the skills mentioned above, only the common scales used in the
drawing of graphs are considered.



PART B:
GRAPHS OF FUNCTIONS

LESSON NOTES

1. For a standard graph paper, 2 cm is represented by 10 small squares.

2 cm

2 cm

2. Some common scales used are as follows:

Scale Note

10 small squares represent 10 units
2 cm to 10 units
1 small square represents 1 unit

2 cm to 5 units
1 small square represents 0.5 unit

2 cm to 2 units

10 small squares represent 1 unit
2 cm to 1 unit

10 small squares represent 0.1 unit
2 cm to 0.1 unit



PART B1: Mark numbers on the x-axis and y-axis based on the scales given.


1. Mark – 4. 7, 16 and 27on the x-axis. 1. Mark – 6 4, 15 and 26 on the x-axis.
Scale: 2 cm to 10 units. Scale: 2 cm to 10 units.
[ 1 small square represents 1 unit ] [ 1 small square represents 1 unit ]

x x
–10 –4 0 7 10 16 20 27 30

2. Mark –7, –2, 3 and 8on the x-axis. 2. Mark –8, –3, 2 and 6, on the x-axis.
[ 1 small square represents 0.5 unit ] [ 1 small square represents 0.5 unit ]

x x
–10 –7 –5 –2 0 3 5 8 10

3. Mark –3.4, – 0.8, 1 and 2.6, on the x-axis. 3. Mark –3.2, –1, 1.2 and 2.8 on the x-axis.

x x
–4 –3.4 –2 –0.8 0 1 2 2.6 4

4. Mark –1.3, – 0.6, 0.5 and 1.6 on the x-axis. 4. Mark –1.7, – 0.7, 0.7 and 1.5 on the x-axis.
Scale: 2 cm to 1 unit. Scale: 2 cm to 1 unit.

x x
–2 –1.3 – 1 –0.6 0 0.5 1 1.6 2





5. Mark – 0.15, – 0.04, 0.03 and 0.17 on the 5. Mark – 0.17, – 0.06, 0.04 and 0.13 on the
x-axis. x-axis.

Scale: 2 cm to 0.1 unit Scale: 2 cm to 0.1 unit

x x
–0.2 –0.15 –0.1 –0.04 0 0.03 0.1 0.17 0.2

6. Mark –13, –8, 2 and 14 on the y-axis. 6. Mark –16, – 4, 5 and 15 on the y-axis.

Scale: 2 cm to 10 units Scale: 2 cm to 10 units
[ 1 small square represents 1 unit ] [ 1 small square represents 1 unit ]
y y
20

14

10

2
0

–8
–10
–13

–20





7. Mark –9, –3, 1 and 7 on the y-axis. 7. Mark –7, – 4, 2 and 6 on the y-axis.

y y
10

7

5

1
0

–3

–5

–9
–10

8. Mark –3.2, – 0.6, 1.4 and 2.4 on the y-axis. 8. Mark –3.4, –1.4, 0.8 and 2.8 on the y-axis.

y y
4

2.4
2
1.4

0
–0.6

–2

–3.2

–4





9. Mark –1.6, – 0.4, 0.4 and 1.5 on the y-axis. 9. Mark –1.5, – 0.8, 0.3 and 1.7 on the y-axis.

Scale: 2 cm to 1 unit. Scale: 2 cm to 1 unit.
y y
2

1.5

1

0.4

0

– 0.4

–1

–1.6

–2

10. Mark – 0.17, – 0.06, 0.08 and 0.16 on the 10. Mark – 0.18, – 0.03, 0.05 and 0.14 on the
y-axis. y-axis.

Scale: 2 cm to 0.1 unit. Scale: 2 cm to 0.1 units.
y y
0.2

0.16

0.1

0.08

0

– 0.06

–0.1

– 0.17
–0.2



PART B2: Draw graph of a function given a table for values of x and y.


1. The table shows some values of two variables, x and y, 1. The table shows some values of two variables, x and y,
of a function. of a function.

x –2 –1 0 1 2 x –3 –2 –1 0 1
y –2 0 2 4 6 y –2 0 2 4 6
By using a scale of 2 cm to 1 unit on the x-axis and By using a scale of 2 cm to 1 unit on the x-axis and
2 cm to 2 units on the y-axis, draw the graph of the 2 cm to 2 units on the y-axis, draw the graph of the
function. function.
y

6 

4 

2


–2 –1 0 1 2 x
 –2


x –2 –1 0 1 2 x –2 –1 0 1 2
y 5 3 1 –1 –3 y 7 5 3 1 –1
function. function.
y

6

4

2


–2 –1 x
0 
1 2
–2







x –4 –3 –2 –1 0 1 2 x –1 0 1 2 3 4 5
y 15 5 –1 –3 –1 5 15 y 19 4 –5 –8 –5 4 19
function. function.
y
 15 

10

 5 

–4 
–2 –1 0 x
–3  1 2
–5


x –2 –1 0 1 2 3 4 x –2 –1 0 1 2 3
y –7 –2 1 2 1 –2 –7 y –8 –4 –2 –2 –4 –8
function. function.
y
2 
 

–2 –1 0 1 2 3 4 x
 –2 

–4

–6
 






x –2 –1 0 1 2 x –2 –1 0 1 2
y –7 –1 1 3 11 y –6 2 4 6 16
function. function.
y
15

10

5


 x
–2 –1 1 2
0
–5



x –3 –2 –1 0 1 2 3 x –3 –2 –1 0 1 2 3
y 22 5 0 1 2 –3 –20 y 21 4 –1 0 1 –4 –21
function. function.
y

20

10

  
–3 –2 –1 0 1 
2 3 x
–10

–20 



ACTIVITY B1

Each table below shows the values of x and y for a certain function.

FUNCTION 1 FUNCTION 2
x –4 –3 –2 –1 0 x 0 1 2 3 4
y 16 17 18 19 20 y 20 19 18 17 16

FUNCTION 3
x –4 –3 –2 –1 0 1 2 3 4
y 16 9 4 1 0 1 4 9 16

FUNCTION 4
x –3 –2 –1 0 1 2 3
y 9 14 17 18 17 14 9

FUNCTION 5
x –3 –2 –1.5 –1 – 0.5 0
y 9 8 7.9 7 4.6 0

FUNCTION 6
x 0 0.5 1 1.5 2 3
y 0 4.6 7 7.9 8 9

The graphs of all these functions, when drawn on the same axes, form a beautiful logo. Draw the logo on
the graph paper provided by using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis.
y

x
0


PART B3: State the values of x and y on the axes.


1. State the values of a, b, c and d on the x-axis 1. State the values of a, b, c and d on the x-axis
below. below.

x x
–20 d –10 c 0 a 10 b 20 –20 d –10 c 0 a 10 b 20

Scale: 2 cm to 10 units.
[ 1 small square represents 1 unit ]

a = 7, b = 13, c = – 4, d = –14

below. below.

x x
–10 d –5 c 0 a 5 b 10 –10 d –5 c 0 a 5 b 10

[ 1 small square represents 0.5 unit ]

a = 2, b = 7.5, c = –3, d = –8.5

below. below.

x x
–4 d –2 c 0 a 2 b 4 – 4d –2 c 0 a 2 b 4


a = 0.6, b = 3.4, c = –1.2, d = –2.6





below. below.

x x
–2 d –1 c 0 a 1 b 2 –2 d –1 c 0 a 1 b 2
Scale: 2 cm to 1 unit.

a = 0.8, b = 1.4, c = – 0.3, d = –1.6

below. below.

x x
–0.2 d –0.1 c 0 a 0.1 b 0.2 – 0.2 d –0.1 c 0 a 0.1 b 0.2
Scale: 2 cm to 0.1 unit.

a = 0.04, b = 0.14, c = – 0.03, d = – 0.16

6. State the values of a, b, c and d on the y-axis 6. State the values of a, b, c and d on the y-axis
y y
below. below.
Scale: 2 cm to 10 units. 20 20

[ 1 small square b
b
represents 1 unit ]
10 10
a = 3, b = 17
c = – 6, d = –15 a
a
0 0
c
c

–10 –10

d
d
–20 –20





below. y below. y
10 10
Scale: 2 cm to 5 units. b

[ 1 small square b

represents 0.5 unit ]
5 5
a
a = 4, b = 9.5
a
c = –2, d = –7.5
0 0
c
c

–5 –5

d
d
–10 –10

below. y below. y
4 4
Scale: 2 cm to 2 units. b
[ 1 small square b
2 2
a = 0.8, b = 3.2 a
a
c = –1.2, d = –2.6
0 0

c
c

–2 –2
d

d
–4 –4





below. y below. y
2 2
Scale: 2 cm to 1 unit.
b
[ 1 small square
represents 0.1 unit ] b
1 1
a
a = 0.7, b = 1.2 a

c = – 0.6, d = –1.4 0 0

c
c

–1 –1

d
d

–2 –2

below. y below. y
0.2 0.2
Scale: 2 cm to 0.1 unit.
b
[ 1 small square b
0.1 0.1

a
a = 0.03, b = 0.07
a

c = – 0.04, d = – 0.18 0 0

c
c
–0.1 –0.1

d
d
–0.2 –0.2



PART B4: State the value of y given the value x from the graph and vice versa.


1. Based on the graph below, find the value of y 1. Based on the graph below, find the value of y
when (a) x = 1.5 when (a) x = 0.6
(b) x = –2.8 (b) x = –1.7
y y
7
6 6

4 4

2 2
– 2.8

–2 –1 0 1 1.5
2 x –2 –1 0 1 2 x
– 1.6
–2 –2

(a) 7 (b) –1.6 (a) (b)

when ( a ) x = 0.14 when ( a ) x = 0.07
( b ) x = – 0.26 ( b ) x = – 0.18
y y
11.5
10 10

5 5
1.5
– 0.26 0.14 x x
– 0. 2 –0.1 0 0.1 0.2 –0. 2 –0.1 0 0.1 0.2
–5 –5

–10 –10

(a) 1.5 (b) 11.5 (a) (b)





when ( a ) x = 0.6 when ( a ) x = 1.2
( b ) x = –2.7 ( b ) x = –1.8

y y
15 15
11
10 10

5 5
– 2.7

–4 –3 –2 –1 0 0.6
1 2 x –4 –3 –2 –1 0 1 2 x
– 3.5
–5 –5

(a) 11 (b) –3.5 (a) (b)

when (a) x = 1.4 when (a) x = 2.7
(b) x = –1.5 (b) x = –2.1

y y
3
2 2
– 1.5

–2 –1 0 1
1.4
2 3 4 x –2 –1 0 1 2 3 4 x
–2 –2

–4 –4
– 5.8
–6 –6

(a) 3 (b) –5.8 (a) (b)





when (a) x = 1.7 when (a) x = 1.2
(b) x = –1.3 (b) x = –1.9

y y
15 15

10 10
5.5
5 5
– 1.3

–2 –1 0 1 1.7
2 x –2 –1 0 1 2 x
– 3.5
–5 –5

(a) 5.5 (b) –3.5 (a) (b)

when (a) x = 1.6 when (a) x = 2.8
(b) x = –2.3 (b) x = –2.6

y y
25
20 20

10 10
1.6

–3 – 2.3
–2 –1 0 1 2 3 x –3 –2 –1 0 1 2 3 x
–9
–10 –10

–20 –20

(a) –9 (b) 25 (a) (b)





7. Based on the graph below, find the value of x 7. Based on the graph below, find the value of x
when (a) y = 5.4 when (a) y = 2.8
(b) y = –1.6 (b) y = –2.4
y y

6 6
5.4

4 4

2 2
– 2.8

–2 –1 0 1
1.4
2 x –2 –1 0 1 2 x
– 1.6
–2 –2

(a) 1.4 (b) –2.8 (a) (b)

when ( a ) y = 4 when ( a ) y = 6.5
( b ) y = –7.5 ( b ) y = –7
y y
10 10

5 5
4
0.08
– 0.07 x x
–0. 2 –0.1 0 0.1 0.2 –0. 2 –0.1 0 0.1 0.2
–5 –5
– 7.5
–10 –10

(a) – 0.07 (b) 0.08 (a) (b)





9. Based on the graph below, find the values of x 9. Based on the graph below, find the values of x
when (a) y = 8.5 when (a) y = 3.5
(b) y = 0 (b) y = 0

y y
15 15

10 10
8.5
5 5

– 3.1 0 2.1 x 0 x
–4 –3 –2 –1 1 2 –4 –3 –2 –1 1 2
–5 –5

(a) –3.1 , 2.1 (b) –2 , 1 (a) (b)

10. Based on the graph below, find the values of x 10. Based on the graph below, find the values of x
when (a) y = 2.6 when (a) y = 1.2
(b) y = – 4.8 (b) y = – 4.4

y y
2.6
2 2
– 1.2 3.9
0 0.6 2.1 x x
–2 –1 1 2 3 4 –2 –1 0 1 2 3 4
–2 –2

–4 –4
– 4.8
–6 –6

(a) 0.6 , 2.1 (b) –1.2 , 3.9 (a) (b)





when (a) y = 14 when (a) y = 11
(b) y = –17 (b) y = –23

y y
20 20
14

10 10
– 2.3

–3 –2 –1 0 1 2
2.6
3 x –3 –2 –1 0 1 2 3 x
–10 –10
– 17
–20 –20

(a) 2.6 (b) –2.3 (a) (b)

when (a) y = 6.5 when (a) y = 7.5
(b) y = 0 (b ) y = 0
(c) y = –6 (c) y = –9

y y
15 15

10 10
6.5

5 5
– 0.8 1.3 2.3

–2 –1 0 1 2 x –2 –1 0 1 2 x
–5 –5
–6

(a) – 0.8 (b) 1.3 (c) 2.3 (a) (b) (c)



ACTIVITY B2

There is smuggling at sea and you know two possible locations.

As a responsible citizen, you need to report to the marine police these two locations.

Task 1: Two points on the graph given are (6.5, k) and (h, 45).

Find the values of h and k.

Task 2: Smuggling takes place at the locations with coordinates (h, k).

State each location in terms of coordinates.

y

60

55

50

45

40

35

30

25

20

15

10

5

0 x
1 2 3 4 5 6 7 8 9



ANSWERS

PART A:

PART A1:

1. A (4, 2) 2. B (– 4, 3)
2.
3. C (–3, –3) 4. D (3, – 4)

5. E (2, 0) 6. F (0, 2)

7. G (–1, 0) 8. H (0, –1)

9. J (8, 6) 10. K (– 4, 8)

11. L (–10, –15) 12. M (4, –3)

ACTIVITY A1:

Start at (5, 3).

Then, move in order to (4, 3), (4, –3), (3, –3), (3, 2), (1, 2) , (1, –3) , (–3, –3) , (–3, 3),
(– 4, 3), (–
4, 5), (–3, 5) and (–3, 6).



PART A2:

1. 4.
y y
4 4
A
3
2
• 3
2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1
–2 –2
D
–3 –3 •
–4 –4

2. 5.
B y y
• 4
3
4
3
2 2
1 1
E
–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 • 2 3 4 x
–1 –1
–2 –2
–3 –3
-–4 –4

3. 6.
y y
4 4
F
3
•
3
2 2
1 1

–4 –3 –2 –1 0 1 2 3 4 x –4 –3 –2 –1 0 1 2 3 4 x
–1 –1

•
C
–2 –2
–3 –3
–4 –4



7. 10.
y y
4 8
3
2 4
K
G
1
•
•
–4 –3 –2 –1 0 1 2 3 4 x –8 –4 0 4 8 x
–1
–2 –4
–3
–4 –8

8. 11.
y y
4 20
3
2 10
1

–4 –3 –2 –1 0 1 2 3 4 x –20 –10 0 10 20 x
–1
– H •L
-2 –10
–3 •
–4 –20

9. 12.
y y
8 20
J
6
4
• 10
2

–8 –6 –4 –2 0 2 4 6 8 x –40 –20 0 20 40 x
–2
–4 –10
–6
–8 –20
M
•



ACTIVITY A2:

YAKOMI ISLANDS
y

4
A P Q

2
R S
E F
x
–4 –2 O 2 4
B C ,
–2 U
D T

–4
 RM 1 million



PART B1:

1 2.

x x
–10 –6 0 4 10 15 20 26 30 –10 –8 –5 –3 0 2 5 6 10

3. 4.

x x
–4 –3.2 –2 –1 0 1.2 2 2.8 4 –2 –1.7 –1 –0.7 0 0.7 1 1.5 2
y
5. 6. 20

15

x
–0.2 –0.16 –0.1 –0.06 0 0.04 0.1 0.13 0.2 10

5

0

–4

–10

–16

–20

7. y 8. y 9. y 10. y
10 4 2 0.2
1.7
2.8 0.14
6
5 2 1 0.1

0.05
2 0.8
0.3

0 0 0 0
– 0.03

–1.4
–4 –0.8
–5 –2 –1 – 0.1

–7
–1.5
–3.4
– 0.18
–10 –4 –2 – 0.2



PART B2:

1. y 2. y

6  6

4 4

 2 2


–3 –2 –1 x –2 –1 x
0 1 0 1 
2
 –2 –2

3.  y  4. y

15 0 x
–2 –1 1 2 3
10 –2  

5   –4 

–6
–1 0 1 2 3 4 5 x
–5    –8 


5. y 6. y
 
15 20

10 10
 
 
5  x
 –3 –2 –1 0 1 
2 3

0 –10
–2 –1 1 2 x

 –5 –20 



ACTIVITY B1:
y

20
 
 18 
   
 16 

 14 

12

10
 
  8  
 
6
 
 4 

2
 
 x
–4 –3 –2 –1 0 1 2 3 4



PART B3:

1. a = 3, b = 16, c = – 3, d = – 18

2. a = 3.5, b = 7, c = – 2.5, d = – 8

3. a = 1.4, b = 2.4, c = – 1.6, d = – 3.8

4. a = 0.7, b = 1.8, c = – 0.5, d = – 1.4

5. a = 0.08, b = 0.16, c = – 0.02, d = – 0.17

6. a = 6, b = 15, c = – 3, d = – 17

7. a = 2, b = 8, c = – 0.5, d = – 8.5

8. a = 1.4, b = 3.6, c = – 0.8, d = – 3.4

9. a = 0.5, b = 1.7, c = – 0.4, d = – 1.6

10. a = 0.06, b = 0.16, c = – 0.07, d = – 0.15

PART B4:

1. (a) 6.4 (b) – 2.8

2. (a) – 12 (b) 13

3. (a) – 2.5 (b) 9

4. (a) 0.6 (b) – 5.4

5. (a) 8 (b) – 6.5

6. (a) – 16 (b) 22

7. (a) 0.7 (b) – 1.3

8. (a) – 0.08 (b) 0.12

9. (a) – 3.5, 1.5 (b) –3,1

10. (a) – 1.6, 0.6 (b) – 2.7, 1.7

11. (a) 2.2 (b) – 3.5

12. (a) – 2.3 (b) – 0.6 (c) 1.4

ACTIVITY B2:

k =15, h = 1.1, 8.9

Two possible locations: (1.1, 15), (8.9, 15)


Basic Essential


UNIT 7
LINEAR INEQUALITIES

Unit 1:
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Linear Inequalities 2
1.0 Inequality Signs 3
2.0 Inequality and Number Line 3
3.0 Properties of Inequalities 4
4.0 Linear Inequality in One Unknown 5

Part B: Possible Solutions for a Given Linear Inequality in One Unknown 7

Part C: Computations Involving Addition and Subtraction on Linear Inequalities 10

Part D: Computations Involving Division and Multiplication on Linear Inequalities 14
Part D1: Computations Involving Multiplication and Division on
Linear Inequalities 15
Part D2: Perform Computations Involving Multiplication of Linear
Inequalities 19

Part E: Further Practice on Computations Involving Linear Inequalities 21

Activity 27

Answers 29


MODULE OVERVIEW

1. The aim of this module is to reinforce pupils‟ understanding of the concept involved
in performing computations on linear inequalities.

2. This module can be used as a guide for teachers to help pupils master the basic skills
required to learn this topic.

3. This module consists of six parts and each part deals with a few specific skills.
Teachers may use any parts of the module as and when it is required.

4. Overall lesson notes given in Part A stresses on important facts and concepts required
for this topic.



PART A:
LINEAR INEQUALITIES

LEARNING OBJECTIVE

Upon completion of Part A, pupils will be able to understand and use the
concept of inequality.


Some pupils might face problems in understanding the concept of linear
inequalities in one unknown.

Strategy:

Teacher should ensure that pupils are able to understand the concept of inequality
by emphasising the properties of inequalities. Linear inequalities can also be
taught using number lines as it is an effective way to teach and learn inequalities.

______________________________________________________________________________



PART A:
LINEAR INEQUALITY

OVERALL LESSON NOTES

1.0 Inequality Signs

a. The sign “<” means „less than‟.
Example: 3 < 5

b. The sign “>” means „greater than‟.
Example: 5 > 3

c. The sign “  ” means „less than or equal to‟.

d. The sign “  ” means „greater than or equal to‟.

2.0 Inequality and Number Line

x
−3 −2 −1 0 1 2 3

−3 < − 1 1<3
−3 is less than − 1 1 is less than 3

and and

−1 > − 3 3>1
−1 is greater than − 3 3 is greater than 1

______________________________________________________________________________



3.0 Properties of Inequalities

(a) Addition Involving Inequalities

Arithmetic Form Algebraic Form

12  8 so 12  4  8  4 If a > b, then a  c  b  c
29 so 2  6  9  6 If a < b, then a  c  b  c

(b) Subtraction Involving Inequalities


7 > 3 so 7  5  3  5 If a > b, then a  c  b  c
2 < 9 so 2  6  9  6 If a < b, then a  c  b  c

(c) Multiplication and Division by Positive Integers

When multiply or divide each side of an inequality by the same positive number, the
relationship between the sides of the inequality sign remains the same.


5>3 so 5 (7) > 3(7) If a > b and c > 0 , then ac > bc
12 9 a b
12 > 9 so  If a > b and c > 0, then 
3 3 c c

25 so 2(3)  5(3) If a  b and c  0 , then ac  bc
8 12 a b
8  12 so  If a  b and c  0 , then 
2 2 c c

______________________________________________________________________________



(d) Multiplication and Division by Negative Integers

When multiply or divide both sides of an inequality by the same negative number, the
relationship between the sides of the inequality sign is reversed.


8>2 so 8(−5) < 2(−5) If a > b and c < 0, then ac < bc
6<7 so 6(−3) > 7(−3) If a < b and c < 0, then ac > bc
16 8 a b
16 > 8 so  If a > b and c < 0, then 
4 4 c c
10 15 a b
10 <15 so  If a < b and c < 0, then 
5 5 c c

Note: Highlight that an inequality expresses a relationship. To maintain the same
relationship or „balance‟, pupils must perform equal operations on both sides of
the inequality.

4.0 Linear Inequality in One Unknown

(a) A linear inequality in one unknown is a relationship between an unknown and a
number.

Example: x > 12
4m

(b) A solution of an inequality is any value of the variable that satisfies the inequality.

Examples:

(i) Consider the inequality x  3

The solution to this inequality includes every number that is greater than 3.
What numbers are greater than 3? 4 is greater than 3. And so are 5, 6, 7, 8, and
so on. What about 5.5? What about 5.99? And 5.000001? All these numbers are
greater than 3, meaning that there are infinitely many solutions!

But, if the values of x are integers, then x  3 can be written as
x  4, 5, 6, 7, 8,...

______________________________________________________________________________



A number line is normally used to represent all the solutions of an inequality.

To draw a number line representing x  3 , place an
open dot on the number 3. An open dot indicates that
the number is not part of the solution set. Then, to
show that all numbers to the right of 3 are included in
the solution, draw an arrow to the right of 3.

The open dot
means the value
(ii) x>2 2 is not
included.

o
x
−2 −1 0 1 2 3 4

The solid dot
(iii) x3 means the value
3 is included.

x
−2 −1 0 1 2 3 4

______________________________________________________________________________



PART B:
POSSIBLE SOLUTIONS FOR A
GIVEN LINEAR INEQUALITY IN
ONE UNKNOWN

LEARNING OBJECTIVES

Upon completion of Part B, pupils will be able to solve linear
inequalities in one unknown by:

(i) determining the possible solution for a given linear inequality in one
unknown:
(a) x  h
(b) x  h
(c) x  h
(d) x  h

(ii) representing a linear inequality:
(a) x  h
(b) x  h
(c) x  h
(d) x  h
on a number line and vice versa.


Some pupils might have difficulties in finding the possible solution for a given
linear inequality in one unknown and representing a linear inequality on a number
line.

Strategy:

Teacher should emphasise the importance of using a number line in order to solve
linear inequalities and should ensure that pupils are able to draw correctly the
arrow that represents the linear inequalities.

______________________________________________________________________________



PART B:
POSSIBLE SOLUTIONS FOR
A GIVEN LINEAR INEQUALITY IN ONE UNKNOWN

EXAMPLES

List out all the possible integer values for x in the following inequalities: (You can use the
number line to represent the solutions)

(1) x>4

Solution:

x
−2 −1 0 1 2 3 4 5 6 7 8 9 10
The possible integers are: 5, 6, 7, …

(2) x  3

Solution:

x
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4

The possible integers are: – 4, − 5, −6, …

(3)  3  x 1

Solution:

x
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4

The possible integers are: −2, −1, 0, and 1.

______________________________________________________________________________



TEST YOURSELF B

Draw a number line to represent the following inequalities:

(a) x>1

(b) x2

(c) x  2

(d) x3

______________________________________________________________________________



PART C:
COMPUTATIONS INVOLVING
ADDITION AND SUBTRACTION ON
LINEAR INEQUALITIES

LEARNING OBJECTIVES

Upon completion of Part C, pupils will be able perform computations
involving addition and subtraction on inequalities by stating a new
inequality for a given inequality when a number is:
(a) added to; and
(b) subtracted from
both sides of the inequalities.


Some pupils might have difficulties when dealing with problems involving
addition and subtraction on linear inequalities.

Strategy:

Teacher should emphasise the following rule:

1) When a number is added or subtracted from both sides of the inequality,
the inequality sign remains the same.

______________________________________________________________________________



PART C:
COMPUTATIONS INVOLVING ADDITION AND SUBTRACTION
ON LINEAR INEQUALITIES

LESSON NOTES

Operation on Inequalities

1) When a number is added or subtracted from both sides of the inequality, the inequality
sign remains the same.

Examples:

(i) 2 < 4

2<4

x
1 2 3 4

Adding 1 to both sides of the inequality:

The inequality
sign is
2+1<4+1
unchanged.
3<5

x
2 3 4 5

______________________________________________________________________________



(ii) 4>2 4>2

x
1 2 3 4

Subtracting 3 from both sides of the inequality:

4−3>2−3
The inequality
1>−1
sign is
unchanged.

x
−1 0 1 2

EXAMPLES

(1) Solve x  5  14 .

Solution:
Subtract 5 from both sides
x  5  14 of the inequality.
x  5  5  14  5
x9 Simplify.

(2) Solve p  3  2.

Solution:
Add 3 to both sides of the
p3 2
inequality.
p  3 3  2  3
p5 Simplify.

______________________________________________________________________________



TEST YOURSELF C

Solve the following inequalities:

(1) m  4  2 (2) x  3.4  2.6

(3) x  13  6 (4) 4.5  d  6

(5) 23  m  17 (6) y  78  54

(7) 9  d 5 (8) p  2  1

1 (10) 3 x 8
(9) m 3
2

______________________________________________________________________________



PART D:
DIVISION AND MULTIPLICATION
ON LINEAR INEQUALITIES

LEARNING OBJECTIVES

Upon completion of Part D, pupils will be able perform computations
involving division and multiplication on inequalities by stating a new
inequality for a given inequality when both sides of the inequalities are
divided or multiplied by a number.


The computations involving division and multiplication on inequalities can be
confusing and difficult for pupils to grasp.

Strategy:

Teacher should emphasise the following rules:

1) When both sides of the inequality is multiplied or divided by a positive
number, the inequality sign remains the same.
2) When both sides of the inequality is multiplied or divided by a negative
number, the inequality sign is reversed.
3)

______________________________________________________________________________



PART D1:
MULTIPLICATION AND DIVISION ON LINEAR INEQUALITIES

LESSON NOTES

1. When both sides of the inequality is multiplied or divided by a positive number, the
inequality sign remains the same.

Examples:

(i) 2<4

2<4

x
1 2 3 4

Multiplying both sides of the inequality by 3:
The inequality
sign is
unchanged.
2  3<4  3
6 < 12

x
6 8 10 12 14

______________________________________________________________________________



(ii) −4<2

−4<2

x
−4 −2 0 2

Dividing both sides of the inequality by 2:

The inequality
−4  2<2  2 sign is
−2 <1 unchanged.

x
−2 −1 0 1 2

2. When both sides of the inequality is multiplied or divided by a negative number, the
inequality sign is reversed.

Examples:

(i) 4<6

4<6

x
3 4 5 6

Dividing both sides of the inequality by −1:

4  (−1) > 6  The inequality
(−1) sign is reversed.
−4>−6

x
−6 −5 −4 −3
______________________________________________________________________________



(ii) 1 > −3

1 > −3

x
−3 −2 −1 0 1

Multiply both sides of the inequality by −1:

The inequality
(− 1) (1) < (−1) (−3) sign is reversed.
1  3

x
−1 0 1 2 3

EXAMPLES

Solve the inequality 3q  12 .

Solution:

(i) 3q  12
Divide each side of the
 3q 12 inequality by −3. The inequality

3 3 sign is reversed.

q  4 Simplify.

______________________________________________________________________________



TEST YOURSELF D1


(1) 7 p  49 (2) 6 x  18

(3) −5c > 15 (4) 200 < −40p

(5) 3d  24 (6)  2x  8

(7)  12  3x (8) 25  5 y

(9)  2m  16 (10)  6b  27

______________________________________________________________________________



PART D2:
PERFORM COMPUTATIONS INVOLVING
MULTIPLICATION OF LINEAR INEQUALITIES

EXAMPLES

x
Solve the inequality   3.
2
Solution:

x
  3. Multiply both sides of the
2 inequality by −2.
x
 2( )  (2)3
2 Simplify.
x  6

The inequality
sign is reversed.

______________________________________________________________________________



TEST YOURSELF D2

1. Solve the following inequalities:

d n
(1) − 3 (2) 8
8 2

y b
(3) 10   (4) 6 
5 7

x x
(5) 0  12  (6) 8 0
8 6

______________________________________________________________________________



PART E:
FURTHER PRACTICE ON
LINEAR INEQUALITIES

LEARNING OBJECTIVES

Upon completion of Part E, pupils will be able perform computations
involving linear inequalities.


Pupils might face problems when dealing with problems involving linear
inequalities.

Strategy:

Teacher should ensure that pupils are given further practice in order to enhance
their skills in solving problems involving linear inequalities.

______________________________________________________________________________



PART E:
FURTHER PRACTICE ON COMPUTATIONS
INVOLVING LINEAR INEQUALITIES

TEST YOURSELF E1


1. (a) m5 0

(b) x26

(c) 3+m>4

2. (a) 3m < 12

(b) 2m > 42

(c) 4x > 18

______________________________________________________________________________



3. (a) m + 4 > 4m + 1

(b) 14  m  6  m

(c) 3  3m  4  m

4. (a) 4  x  6

(b) 15  3m  12

x
(c) 3 5
4

______________________________________________________________________________



(d) 5x  3  18

(e) 1  3 p  10

x
(f) 3 4
2

x
(g) 3  8
5

p2
(h) 4
3

______________________________________________________________________________



EXAMPLES

What is the smallest integer for x if 5x  3  18 ?

A number line can
be used to obtain the
answer.
Solution:

5x  3  18

5x  18  3
x3
5x  15 O
x 3 x
0 1 2 3 4 5 6
x = 4, 5, 6,…
Therefore, the smallest integer for x is 4.

______________________________________________________________________________



TEST YOURSELF E2

1. If 3x  1  14, what is the smallest integer for x?

2. What is the greatest integer for m if m  7  4m  1 ?

3. x
If  3  4 , find the greatest integer value of x.
2

4. p2
If  4 , what is the greatest integer for p?
3

5. 3 m
What is the smallest integer for m if  9?
2

______________________________________________________________________________



ACTIVITY

1

2 3

4

5 6

7 8

9

10

11 12

______________________________________________________________________________



HORIZONTAL:
4. 1  3 is an ___________.

5. An inequality can be represented on a number __________.

7. 2  6 is read as 2 is __________ than 6.

9. Given 2x  1  9 , x  5 is a _____________ of the inequality.

11.  3x  12

x  4

The inequality sign is reversed when divided by a ____________ integer.

VERTICAL:
x
 1
1. 2
x  2

The inequality sign remains unchanged when multiplied by a ___________ integer.

2. 6 x  24 equals to x  4 when both sides are _____________ by 6.

3. x  5 equals to 3x  15 when both sides are _____________ by 3.

6. ___________ inequalities are inequalities with the same solution(s).

8. x  2 is represented by a ____________ dot on a number line.

10. 3x  6 is an example of ____________ inequality.

12. 5  3 is read as 5 is _____________ than 3.

______________________________________________________________________________



ANSWERS

TEST YOURSELF B:

(a) x
−3 −2 −1 0 1 2 3

(b) x
−3 −2 −1 0 1 2 3

(c) x
−3 −2 −1 0 1 2 3

x
(d) −3 −2 −1 0 1 2 3

TEST YOURSELF C:

(1) m  6 (2) x  6 (3) x  19 (4) d  1.5 (5) m  6
5
(6) y  24 (7) d  4 (8) p  3 (9) m  (10) x  5
2

TEST YOURSELF D1:

(1) p7 (2) x  3 (3) c  3 (4) p  5 (5) d  8

9
(6) x  4 (7) x  4 (8) y  5 (9) m  8 (10) b 
2

TEST YOURSELF D2:

(1) d  24 (2) n  16 (3) y  50 (4) b  42 (5) x  96 (6) x  48

______________________________________________________________________________



TEST YOURSELF E1:

1. (a) m  5 (b) x  8 (c ) m  1
9
2. (a) m  4 (b) m  21 (c ) x 
2
1
3. (a ) m  1 (b) m  4 (c) m 
2
4. (a) x  10 (b) m  1 (c) x  8 (d) x  3 (e) p  3 (f) x  2 (g) x  25 (h) p  10

TEST YOURSELF E2:

(1) x  6 (2) m  1 (3) x  13 (4) p  9 (5) m  14

ACTIVITY:

1. positive
2. divided
3. multiplied
4. inequality
5. line
6. Equivalent
7. less
8. solid
9. solution
10. linear
11. negative
12. greater

______________________________________________________________________________


Basic Essential


UNIT 8

TRIGONOMETRY

Unit 1:
Negative Numbers


TABLE OF CONTENTS

Module Overview 1

Part A: Trigonometry I 2

Part B: Trigonometry II 6

Part C: Trigonometry III 11

Part D: Trigonometry IV 15

Part E: Trigonometry V 19

Part F: Trigonometry VI 21

Part G: Trigonometry VII 25

Part H: Trigonometry VIII 29

Answers 33

Basic Essentials Additional Mathematics Skills (BEAMS) Module

MODULE OVERVIEW

1. The aim of this module is to reinforce pupils’ understanding of the concept
of trigonometry and to provide pupils with a solid foundation for the study
of trigonometric functions.

2. This module is to be used as a guide for teacher on how to help pupils to
master the basic skills required for this topic. Part of the module can be
used as a supplement or handout in the teaching and learning involving
trigonometric functions.

3. This module consists of eight parts and each part deals with one specific
skills. This format provides the teacher with the freedom of choosing any
parts that is relevant to the skills to be reinforced.

4. Note that Part A to D covers the Form Three syllabus whereas Part E to H
covers the Form Four syllabus.



PART A:
TRIGONOMETRY I

LEARNING OBJECTIVE

Upon completion of Part A, pupils will be able to identify opposite,
adjacent and hypotenuse sides of a right-angled triangle with reference
to a given angle.


Some pupils may face difficulties in remembering the definition and
how to identify the correct sides of a right-angled triangle in order to
find the ratio of a trigonometric function.

Strategy:

Teacher should make sure that pupils can identify the side opposite to
the angle, the side adjacent to the angle and the hypotenuse side
through diagrams and drilling.



LESSON NOTES

θ

Opposite side is the side opposite or facing the angle  .

Adjacent side is the side next to the angle  .

Hypotenuse side is the side facing the right angle and is the longest side.



EXAMPLES

Example 1:

θ

AB is the side facing the angle  , thus AB is the opposite side.

BC is the side next to the angle  , thus BC is the adjacent side.

AC is the side facing the right angle and it is the longest side, thus AC is the
hypotenuse side.

Example 2:

θ

QR is the side facing the angle  , thus QR is the opposite side.

PQ is the side next to the angle  , thus PQ is the adjacent side.

PR is the side facing the right angle or is the longest side, thus PR is the
hypotenuse side.



TEST YOURSELF A

Identify the opposite, adjacent and hypotenuse sides of the following right-angled triangles.

1. 2. 3.

Opposite side = Opposite side = Opposite side =
Adjacent side = Adjacent side = Adjacent side =
Hypotenuse side = Hypotenuse side = Hypotenuse side =

4. 5. 6.

Opposite side = Opposite side = Opposite side =
Adjacent side = Adjacent side = Adjacent side =
Hypotenuse side = Hypotenuse side = Hypotenuse side =



PART B:
TRIGONOMETRY II

LEARNING OBJECTIVE

Upon completion of Part B, pupils will be able to state the definition
of the trigonometric functions and use it to write the trigonometric
ratio from a right-angled triangle.


Some pupils may face problem in

(i) defining trigonometric functions; and

(ii) writing the trigonometric ratios from a given right-angled
triangle.

Strategy:

Teacher must reinforce the definition of the trigonometric functions
through diagrams and examples. Acronyms SOH, CAH and TOA can
be used in defining the trigonometric ratios.



LESSON NOTES

Definition of the Three Trigonometric Functions

Acronym:
opposite side
(i) sin  =
hypotenuse side SOH:
Sine – Opposite - Hypotenuse

adjacent side Acronym:
(ii) cos  =
hypotenuse side
CAH:
Cosine – Adjacent - Hypotenuse

opposite side Acronym:
(iii) tan  =
adjacent side
TOA:
Tangent – Opposite - Adjacent

θ

opposite side AB
sin  = =
hypotenuse side AC

adjacent side BC
cos  = =
hypotenuse side AC

opposite side AB
tan  = =
adjacent side BC



EXAMPLES

Example 1:

θ

AB is the side facing the angle  , thus AB is the opposite side.

BC is the side next to the angle  , thus BC is the adjacent side.

AC is the side facing the right angle and is the longest side, thus AC is the hypotenuse
side.

opposite side AB
Thus sin  = =
hypotenuse side AC

adjacent side BC
cos  = =
hypotenuse side AC

opposite side AB
tan  = =
adjacent side BC



Example 2:

θ
You have to identify the
opposite, adjacent and
hypotenuse sides.

WU is the side facing the angle, thus WU is the opposite side.

TU is the side next to the angle, thus TU is the adjacent side.

TW is the side facing the right angle and is the longest side, thus TW is the hypotenuse
side.

opposite side WU
Thus, sin  = =
hypotenuse side TW

adjacent side TU
cos  = =
hypotenuse side TW

opposite side WU
tan  = =
adjacent side TU



TEST YOURSELF B

Write the ratios of the trigonometric functions, sin , cos  and tan  , for each of the diagrams
below:

1. 2. θ 3.

θ
θ

θ

sin  = sin  = sin  =

cos  = cos  = cos  =

tan  = tan  = tan  =

4. 5. 6.

θ θ

θ

sin  = sin  = sin  =

cos  = cos  = cos  =

tan  = tan  = tan  =



PART C:
TRIGONOMETRY III

LEARNING OBJECTIVE

Upon completion of Part C, pupils will be able to find the angle of
a right-angled triangle given the length of any two sides.


Some pupils may face problem in finding the angle when given
two sides of a right-angled triangle and they also lack skills in
using calculator to find the angle.

Strategy:

1. Teacher should train pupils to use the definition of each
trigonometric ratio to write out the correct ratio of the sides
of the right-angle triangle.

2. Teacher should train pupils to use the inverse trigonometric
functions to find the angles and express the angles in degree
and minute.



LESSON NOTES

opposite adjacent opposite
Since sin  = Since cos  = Since tan  =
hypotenuse hypotenuse adjacent

opposite adjacent opposite
then  = sin-1 then  = cos-1 then  = tan-1
hypotenuse hypotenuse adjacent

1 degree = 60 minutes 1 minute = 60 seconds

1o = 60 1 = 60

Use the key D M S or on your calculator to express the angle in degree and minute.

Note that the calculator expresses the angle in degree, minute and second. The angle in
second has to be rounded off. ( 30, add 1 minute and < 30, cancel off.)

EXAMPLES

Find the angle  in degrees and minutes.

Example 1: Example 2:

θ

θ

o 2
sin  =  a 3
h 5 cos  = =
h 5
 = sin-1 2
5  = cos-1 3
5
= 23o 34 4l
= 53o 7 48
= 23o 35
= 53o 8
(Note that 34 41 is rounded off to 35) (Note that 7 48 is rounded off to 8)




θ

θ

tan  = o = 7 cos  = a = 5
a 6 h 7

 = tan-1 7  = cos-1 5
6 7

= 49o 23 55 = 44o 24 55

= 49o 24 = 44o 25


θ

θ

o 5
o 4 tan  = =
sin  = = a 6
h 7

 = sin-1 4  = tan-1 5
6
7
= 39o 48 20
= 34o 50 59
= 39o 48
= 34o 51



TEST YOURSELF C

Find the value of  in degrees and minutes.

1. 2.

θ
θ

3. 4.

θ

θ

5. 6.

θ

θ



PART D:
TRIGONOMETRY IV

LEARNING OBJECTIVE

Upon completion of Part D, pupils will be able to find the
angle of a right-angled triangle given the length of any two
sides.


Pupils may face problem in finding the length of the side of a
right-angled triangle given one angle and any other side.

Strategy:

By referring to the sides given, choose the correct trigonometric
ratio to write the relation between the sides.

1. Find the length of the unknown side with the aid of a
calculator.



LESSON NOTES

Find the length of PR. Find the length of TS.

With reference to the given angle, PR is the With reference to the given angle, TR is the
opposite side and QR is the adjacent side. adjacent side and TS is the hypotenuse
side.
Thus tangent ratio is used to form the
relation of the sides. Thus cosine ratio is used to form the
relation of the sides.
o PR
tan 50 =
5 8
cos 32o =
TS
PR = 5  tan 50 o

TS  cos 32o = 8

8
TS =
cos 32o



EXAMPLES

Find the value of x in each of the following.


3
tan 25o =
x x
sin 41.27o =
5
3
x =
tan 25o x = 5  sin 41.27o
= 6.434 cm = 3.298 cm


x
cos 34o 12 =
6
x
tan 63o =
x = 6  cos 34o 12 9

= 4.962 cm x = 9  tan 63o

= 17.66 cm



TEST YOURSELF D

Find the value of x for each of the following.

1. 2.

3. 4.

10 cm

6 cm

5. 6.
13 cm



PART E:
TRIGONOMETRY V

LEARNING OBJECTIVE

Upon completion of Part E, pupils will be able to state the
definition of trigonometric functions in terms of the
coordinates of a given point on the Cartesian plane and use
the coordinates of the given point to determine the ratio of the
trigonometric functions.


Pupils may face problem in relating the coordinates of a given
point to the definition of the trigonometric functions.

Strategy:

Teacher should use the Cartesian plane to relate the coordinates
of a point to the opposite side, adjacent side and the hypotenuse
side of a right-angled triangle.



LESSON NOTES

θ

In the diagram, with reference to the angle , PR is the opposite side, OP is the adjacent side
and OR is the hypotenuse side.

opposite PR y
sin    
hypotenuse OR r

adjacent OP x
cos   
hypotenuse OR r

opposite PR y
tan    
adjacent OP x



PART F:
TRIGONOMETRY VI

LEARNING OBJECTIVE

Upon completion of Part F, pupils will be able to relate the sign of the
trigonometric functions to the sign of x-coordinate and y-coordinate and to
determine the sign of each trigonometric ratio in each of the four quadrants.


Pupils may face difficulties in determining that the sign of the x-coordinate
and y-coordinate affect the sign of the trigonometric functions.

Strategy:

Teacher should use the Cartesian plane and use the points on the four
quadrants and the values of the x-coordinate and y-coordinate to show how the
sign of the trigonometric ratio is affected by the signs of the x-coordinate and
y-coordinate.

Based on the A – S – T – C, the teacher should guide the pupils to determine
on which quadrant the angle is when given the sign of the trigonometric ratio
is given.

(a) For sin  to be positive, the angle  must be in the first or second
quadrant.

(b) For cos  to be positive, the angle  must be in the first or fourth
quadrant.

(c) For tan  to be positive, the angle  must be in the first or third quadrant.



LESSON NOTES

First Quadrant Second Quadrant

θ
θ

y
y sin  = (Positive)
sin  = (Positive) r
r
x
x cos  = (Negative)
cos  = (Positive) r
r
y
y tan  = (Negative)
tan  = (Positive) x
x
(Only sine is positive in the second
(All trigonometric ratios are positive in the
quadrant)
first quadrant)

Third Quadrant Fourth Quadrant

θ θ

y y
sin  = (Negative) sin  = (Negative)
r r
x
cos  = (Negative) x
cos  = (Positive)
r r
y y y
tan  =  (Positive) tan  = (Negative)
x x x

(Only tangent is positive in the third (Only cosine is positive in the fourth
quadrant) quadrant)



Using acronym: Add Sugar To Coffee (ASTC)

sin  is positive  cos  is positive  tan  is positive 

sin  is negative  cos  is negative  tan  is negative 

S – only sin  is positive A – All positive

T – only tan  is positive C – only cos  is positive



TEST YOURSELF F

State the quadrants the angle is situated and show the position using a sketch.

1. sin  = 0.5 2. tan  = 1.2 3. cos  = −0.16

4. cos  = 0.32 5. sin  = −0.26 6. tan  = −0.362



PART G:
TRIGONOMETRY VII

LEARNING OBJECTIVE

Upon completion of Part G, pupils will be able to calculate the length
of the side of right-angled triangle on a Cartesian plane and write the
value of the trigonometric ratios given a point on the Cartesian plane


Pupils may face problem in calculating the length of the sides of a
right-angled triangle drawn on a Cartesian plane and determining the
value of the trigonometric ratios when a point on the Cartesian plane is
given.

Strategy:

Teacher should revise the Pythagoras Theorem and help pupils to
recall the right-angled triangles commonly used, known as the
Pythagorean Triples.



LESSON NOTES

The Pythagoras Theorem:

The sum of the squares of two sides of
a right-angled triangle is equal to the
square of the hypotenuse side.

PR2 + QR2 = PQ2

(a) 3, 4, 5 or equivalent (b) 5, 12, 13 or equivalent (c) 8, 15, 17 or equivalent



EXAMPLES

1. Write the values of sin , cos  and tan 2. Write the values of sin , cos  and tan 
from the diagram below. from the diagram below.

θ
θ

OB2 = (−12)2 + (−5)2
= 144 + 25
OA2 = (−6)2 + 82 = 169
= 100
OB = 169
OA = 100
= 13
= 10
y 5
y 8 4 sin  = 
sin  =   r 13
r 10 5
x 6 3 cos  = x   12
cos  =    r 13
r 10 5 5
y 8 4 tan  = 
5
tan  =   12 12
x 6 3



TEST YOURSELF G

Write the value of the trigonometric ratios from the diagrams below.

1. 2. 3.
y
B(5,4)

B(5,12)
θ θ
θ θ

x

sin  = sin  = sin  =

cos  = cos  = cos  =

tan  = tan  = tan  =

4. 5. 6.

θ θ
θ

sin  = sin  = sin  =

cos  = cos  = cos  =

tan  = tan  = tan  =



PART H:
TRIGONOMETRY VIII

LEARNING OBJECTIVE

Upon completion of Part H, pupils will be able to sketch the
trigonometric function graphs and know the important features of the
graphs.


Pupils may find difficulties in remembering the shape of the
trigonometric function graphs and the important features of the
graphs.

Strategy:

Teacher should help pupils to recall the trigonometric graphs which
pupils learned in Form 4. Geometer’s Sketchpad can be used to
explore the graphs of the trigonometric functions.



LESSON NOTES

(a) y = sin x

The domain for x can be from 0o to 360o or 0 to 2 in radians.
Important points: (0, 0), (90o, 1), (180o, 0), (270o, −1) and (360o, 0)
Important features: Maximum point (90o, 1), Maximum value = 1
Minimum point (270o, −1), Minimum value = −1
(b) y = cos x

Important points:(0o, 1), (90o, 0), (180o, −1), (270o, 0) and (360o, 1)
Important features: Maximum point (0o, 1) and (360o, 1),
Maximum value = 1 Minimum point (180o, −1)
Minimum value = 1



(c) y = tan x

Important points: (0o, 0), (180o, 0) and (360o, 0)

Is there any
maximum or
minimum point
for the tangent
graph?



TEST YOURSELF H

1. Write the following trigonometric functions to the graphs below:

y = cos x y = sin x y = tan x

2. Write the coordinates of the points below:

(a) (b)

y = cos x y = sin x

A(0,1)



ANSWERS

TEST YOURSELF A:

1. Opposite side = AB 2. Opposite side = PQ 3. Opposite side = YZ

Adjacent side = AC Adjacent side = QR Adjacent side = XZ

Hypotenuse side = BC Hypotenuse side = PR Hypotenuse side = XY

4. Opposite side = LN 5. Opposite side = UV 6. Opposite side = RT

Adjacent side = MN Adjacent side = TU Adjacent side = ST

Hypotenuse side = LM Hypotenuse side = TV Hypotenuse side = RS

TEST YOURSELF B:

AB PQ YZ
1. sin  = 2. sin  = 3. sin  =
BC PR YX
AC QR XZ
cos  = cos  = cos  =
BC PR XY
AB PQ YZ
tan  = tan  = tan  =
AC QR XZ

LN UV RT
4. sin  = 5. sin  = 6. sin  =
LM TV RS
MN UT ST
cos  = cos  = cos  =
LM TV RS
LN UV RT
tan  = tan  = tan  =
MN UT TS



TEST YOURSELF C:

1. sin  = 1 2. cos  = 1
3 2
 = sin-1 1 = 19o 28  = cos-1 1 = 60o
3 2

3. tan  = 5 4. cos  = 5
3 8
 = tan-1 5 = 59o 2  = cos-1 5 = 51o 19
3 8

5. tan  = 7.5 6. sin  = 6.5
9.2 8.4

 = tan-1 7.5 = 39o 11  = sin-1 6.5 = 50o 42
9.2 8.4

TEST YOURSELF D:

4 x
1. tan 32o = 2. sin 53.17o =
x 7

x=
4
= 6.401 cm x = 7  sin 53.17o = 5.603 cm
tan 32o

x o
3. cos 74o 25 = 1 6
10 4. sin 55 =
3 x
x = 10  cos 74o 25 6
x= o
= 7.295 cm
= 2.686 cm sin 55 1
3

x 10
5. tan 47o = 6. cos 61o =
13 x

x = 13  tan 47o = 13.94 cm x=
10
= 20.63 cm
cos 61o



TEST YOURSELF F:

1. 1ST and 2nd 2. 1st and 3rd 3. 2nd and 3rd

4. 1st and 4th 5. 3rd and 4th 6. 2nd and 4th

TEST YOURSELF G:

4 12 4
1. sin  = 2. sin  = 3. sin  =
5 13 5
3 5 3
cos  = cos  = cos  = 
5 13 5
4 12 4
tan  = tan  = tan  = 
3 5 3

4 8 5
4. sin  =  5. sin  =  6. sin  = 
5 17 13
3 15 12
cos  =  cos  =  cos  =
5 17 13
4 8 5
tan  = tan  = tan  = 
3 15 12



TEST YOURSELF H:

1.

y = tan x y = sin x y = cos x

2. (a) A (0, 1), B (90o, 0), C (180o, 1), D (270o, 0)

(b) P (90o, 1), Q (180o, 0), R (270o, 1), S (360o, 0)


Beams Latest 24 Mac 2010 Edited Version

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Beams Latest 24 Mac 2010 Edited Version