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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

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.

INTRODUCTION

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.

iii

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

Unit 6: Coordinates and Graphs of Functions

Unit 7: Linear Inequalities


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

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


PANEL OF CONTRIBUTORS

Abdul Rahim bin Bujang

SM Tun Fatimah, Johor

Ali Akbar bin Asri SM Sains, Labuan

Amrah bin Bahari

SMK Dato’ Sheikh Ahmad, Arau, Perlis

Aziyah binti Paimin SMK Kompleks KLIA, , Negeri Sembilan

Bashirah binti Seleman

SMK Sultan Abdul Halim, Jitra, Kedah

Bibi Kismete binti Kabul Khan SMK Jelapang Jaya, Ipoh, Perak

Che Rokiah binti Md. Isa

SMK Dato’ Wan Mohd. Saman, Kedah

Cheong Nyok Tai SMK Perempuan, Kota Kinabalu, Sabah

Ding Hong Eng

SM Sains Alam Shah, Kuala Lumpur

Esah binti Daud SMK Seri Budiman, Kuala Terengganu

Haspiah binti Basiran

SMK Tun Perak, Jasin, Melaka

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

Hon May Wan

SMK Tasek Damai, Ipoh, Perak

Horsiah binti Ahmad SMK Tun Perak, Jasin, Melaka

Kalaimathi a/p Rajagopal

SMK Sungai Layar, Sungai Petani, Kedah

Kho Choong Quan SMK Ulu Kinta, Ipoh, Perak

Lau Choi Fong

SMK Hulu Klang, Selangor

Loh Peh Choo SMK Bandar Baru Sungai Buloh, Selangor

Mohd. Misbah bin Ramli

SMK Tunku Sulong, Gurun, Kedah

Noor Aida binti Mohd. Zin SMK Tinggi Kajang, Kajang, Selangor

Noor Ishak bin Mohd. Salleh

SMK Laksamana, Kota Tinggi, Johor

Noorliah binti Ahmat SM Teknik, Kuala Lumpur

Nor A’idah binti Johari

SMK Teknik Setapak, Selangor

Writers:

Layout and Illustration:

Aszunarni binti Ayob Mohd. Lufti bin Mahpudz

Assistant Director Assistant Director

Curriculum Development Division Curriculum Development Division

Writers:

Nor Dalina binti Idris

SMK Syed Alwi, Kangar, Perlis

Norizatun binti Abdul Samid

SMK Sultan Badlishah, Kulim, Kedah

Pahimi bin Wan Salleh Maktab Sultan Ismail, Kelantan

Rauziah binti Mohd. Ayob

SMK Bandar Baru Salak Tinggi, Selangor

Rohaya binti Shaari SMK Tinggi Bukit Merajam, Pulau Pinang

Roziah binti Hj. Zakaria

SMK Taman Inderawasih, Pulau Pinang

Shakiroh binti Awang SM Teknik Tuanku Jaafar, Negeri Sembilan

Sharina binti Mohd. Zulkifli

SMK Agama, Arau, Perlis

Sim Kwang Yaw SMK Petra, Kuching, Sarawak

Suhaimi bin Mohd. Tabiee

SMK Datuk Haji Abdul Kadir, Pulau Pinang

Suraiya binti Abdul Halim

SMK Pokok Sena, Pulau Pinang

Tan Lee Fang SMK Perlis, Perlis

Tempawan binti Abdul Aziz

SMK Mahsuri, Langkawi, Kedah

Turasima binti Marjuki SMKA Simpang Lima, Selangor

Wan Azlilah binti Wan Nawi

SMK Putrajaya Presint 9(1), WP Putrajaya

Zainah binti Kebi SMK Pandan, Kuantan, Pahang

Zaleha binti Tomijan

SMK Ayer Puteh Dalam, Pendang, Kedah

Zariah binti Hassan SMK Dato’ Onn, Butterworth, Pulau Pinang

Unit 1:

Negative Numbers

UNIT 1

NEGATIVE NUMBERS

B a s i c E s s e n t i a l

A d d i t i o n a l M a t h e m a t i c s S k i l l s



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


1



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.



2



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.

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.



3



PART A:

ADDITION AND SUBTRACTION OF INTEGERS

USING NUMBER LINES

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.

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

–3 –2 –1 0 1 2 3 4

LESSON NOTES

Rules for Adding and Subtracting Positive Integers

When adding a positive integer, you move to the right on a

number line.

When subtracting a positive integer, you move to the left

on a number line.

–3 –2 –1 0 1 2 3 4

–3 –2 –1 0 1 2 3 4

Positive integers

may have a plus sign

in front of them,

like +3, or no sign in

front, like 3.



4



(i) 2 + 3

Alternative Method:

EXAMPLES

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

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Start

with 2

Add a

positive 3


Start at 2 and move 3 units to the right:

2 + 3 = 5

Make sure you start from

the position of the first

integer.

–5 –4

–3 –2 –1 0 1 2 3 4 5 6



5



(ii) –2 + 5

Alternative Method:


Start by drawing an arrow from 0 to –2, and then,

draw an arrow of 5 units to the right:

–2 + 5 = 3

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Add a

positive 5



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



6



(iii) 2 – 5 = –3

Alternative Method:

–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

Subtract a

positive 5


Start at 2 and move 5 units to the left:

2 – 5 = –3

–5 –4 –3 –2 –1 0 1 2 3 4 5 6



integer.



7



(iv) –3 – 2 = –5

Alternative Method:


Start by drawing an arrow from 0 to –3, and

then, draw an arrow of 2 units to the left:

–3 – 2 = –5

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Subtract a

positive 2

–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



integer.



8



3.0 Addition and Subtraction of Negative Integers

Consider the following operations:

4 – 1 = 3

4 – 2 = 2

4 – 3 = 1

4 – 4 = 0

4 – 5 = –1

4 – 6 = –2

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.

–3 –2 –1 0 1 2 3 4

–3 –2 –1 0 1 2 3 4

–3 –2 –1 0 1 2 3 4

–3 –2 –1 0 1 2 3 4

4 + (–5) = –1

–3 –2 –1 0 1 2 3 4

–3 –2 –1 0 1 2 3 4

4 + (–6) = –2

4 + (–1) = 3

4 + (–2) = 2

4 + (–3) = 1

4 + (–4) = 0



9



Rules for Adding and Subtracting Negative Integers

When adding a negative integer, you move to the left on a

number line.

When subtracting a negative integer, you move to the right

on a number line.

–3 –2 –1 0 1 2 3 4

–3 –2 –1 0 1 2 3 4



10



(i) –2 + (–1) = –3

Alternative Method:

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Adding a negative integer:

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

–2 + (–1) = –3

EXAMPLES

–5 –4 –3 –2 –1 0 1 2 3 4 5 6



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

–2 + (–1) = –3

Add a

negative 1



integer.

This operation of

–2 + (–1) = –3

is the same as

–2 –1 = –3.



11



(ii) 1 + (–3) = –2

Alternative Method:

–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

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



integer.

This operation of

1 + (–3) = –2

is the same as

1 – 3 = –2



12



(iii) 3 – (–3) = 6

Alternative Method:

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Subtracting a negative integer:

Start at 3 and move 3 units to the right:

3 – (–3) = 6

–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 3 units to the right:

3 – (–3) = 6

Subtract a

negative 3

This operation of

3 – (–3) = 6

is the same as

3 + 3 = 6



integer.



13



(iv) –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

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

Subtract a

negative 8

This operation of

–5 – (–8) = 3

is the same as

–5 + 8 = 3

3 + 3 = 6



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

–5 – (–8) = 3



14



Solve the following.

1. –2 + 4

2. 3 + (–6)

3. 2 – (–4)

4. 3 – 5 + (–2)

5. –5 + 8 + (–5)

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

TEST YOURSELF A



15




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.

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.



16



PART B:


USING THE SIGN MODEL

In order to help pupils have a better understanding of positive and negative integers, we have

designed the Sign Model.

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

+

+

+

LESSON NOTES

EXAMPLES

The Sign Model

This model uses the „+‟ and „–‟ signs.

A positive number is represented by „+‟ sign.

A negative number is represented by „–‟ sign.



17



Example 2

What is the value of 53 ?

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 53 ?

NUMBER SIGN

–3 – – –

+5 + + + + +

WORKINGS

i. Pair up the opposite signs.

ii. The number of unpaired signs is the

answer.

Answer 2

_

+ + +

_

+

_

+



18




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

TEST YOURSELF B



19



PART C:

FURTHER PRACTICE ON


OF 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.

LEARNING OBJECTIVE

Upon completion of Part C, pupils will be able to perform computations

involving addition and subtraction of large integers.



20



PART C:

FURTHER PRACTICE ON ADDITION AND SUBTRACTION OF INTEGERS

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.

LESSON NOTES

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.



21



Examples:

i) 34 + 37 =

+ –

34

37

+71

ii) 65 – 20 =

+ –

65 20

+45

iii) –73 + 22 =

+ –

22 73

–51

iv) 228 – 338 =

+ –

228 338

–110

Subtract the smaller number from

the larger number and put the sign

of the larger number in the

answer.

We can just write the answer as

45 instead of +45.




answer.




answer.

Add the numbers and then put the

positive sign in the answer.

We can just write the answer as

71 instead of +71.



22



v) –428 – 316 =

+ –

428

316

–744

vi) –863 – 127 + 225 =

+ –

225

863

127

225 990

–765

vii) 234 – 675 – 567 =

+ –

234

675

567

234 1242

–1008


negative sign in the answer.

Add the two numbers in the „–‟

column and bring down the number

in the „+‟ column.


the larger number in the third row

and put the sign of the larger

number in the answer.










23



viii) –482 + 236 – 718 =

+ –

236

482

718

236 1200

–964

ix) –765 – 984 + 432 =

+ –

432

765

984

432

1749

–1317

x) –1782 + 436 + 652 =

+ –

436

652

1782

1088 1782

–694















Add the two numbers in the „+‟


in the „–‟ column.







24




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

TEST YOURSELF C



25




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.

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.



26



PART D:


INCLUDING THE USE OF BRACKETS

To Accept or To Reject? Answer

+ ( 5 ) Accept +5 +5

– ( 2 ) Reject +2 –2

+ (–4) Accept –4 –4

– (–8) Reject –8 +8

LESSON NOTES

The Accept - Reject Model

„+‟ sign means to accept.

„–‟ sign means to reject.



27



i) 5 + (–1) =

Number To Accept or To Reject? Answer

5

+ (–1)

Accept 5

Accept –1

+5

–1

+ + + + +

–

5 + (–1) = 4

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

5 + (–1) = 5 – 1

+ –

5 1

+4

EXAMPLES

This operation of

5 + (–1) = 4

is the same as

5 – 1 = 4




answer.

We can just write the answer as 4

instead of +4.



28



ii) –6 + (–3) =


–6

+ (–3)

Reject 6

Accept –3

–6

–3

– – – – – –

– – –

–6 + (–3) = –9


–6 + (–3) = –6 – 3 =

+ –

6

3

–9

This operation of

–6 + (–3) = –9

is the same as

–6 –3 = –9





29



iii) –7 – (–4) =


–7

– (–4)

Reject 7

Reject –4

–7

+4

– – – – – – –

+ + + +

–7 – (–4) = –3


–7 – (–4) = –7 + 4 =

+ –

4

7

–3

This operation of

–7 – (–4) = –3

is the same as

–7 + 4 = –3




answer.



30



iv) –5 – (3) =


–5

– (3)

Reject 5

Reject 3

–5

–3

– – – – –

– – –

– 5 – (3) = –8


–5 – (3) = –5 – 3 =

+ –

5

3

–8

This operation of

–5 – (3) = –8

is the same as

–5 – 3 = –8





31



v) –35 + (–57) = –35 – 57 =

Using the Table Model:

+ –

35

57

–92

vi) –123 – (–62) = –123 + 62 =

Using the Table Model:

+ –

62

123

–61

This operation of

–35 + (–57)

is the same as

–35 – 57





of the larger number in the answer.

This operation of

–123 – (–62)

is the same as

–123 + 62



32




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)

TEST YOURSELF D



33



PART E:

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.

LEARNING OBJECTIVE

Upon completion of Part E, pupils will be able to perform computations

involving multiplication of integers.



34



PART E:

MULTIPLICATION OF INTEGERS

Consider the following pattern:

3 × 3 = 9

623

313

003 The result is reduced by 3 in

3)1(3 every step.

6)2(3

9)3(3

93)3(

62)3(

31)3(

00)3( The result is increased by 3 in

3)1()3( every step.

6)2()3(

9)3()3(

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.

positive × positive = positive

(+) × (+) = (+)

positive × negative = negative

(+) × (–) = (–)

negative × positive = negative

(–) × (+) = (–)

negative × negative = positive

(–) × (–) = (+)

LESSON NOTES



35



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

EXAMPLES



36




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)

TEST YOURSELF E



37



PART F:


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.

LEARNING OBJECTIVE

Upon completion of Part F, pupils will be able to perform computations

involving multiplication of integers using the Accept-Reject Model.



38



PART F:


USING THE ACCEPT-REJECT MODEL

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:

To Accept or to Reject Answer

(2) × (3) Accept + 6

(–2) × (–3) Reject – 6

(2) × (–3) Accept – –6

(–2) × (3) Reject + –6

Sign To Accept or To Reject Answer

( + ) × ( + ) Accept +

( – ) × ( – ) Reject –

( + ) × ( – ) Accept – –

( – ) × ( + ) Reject + –

LESSON NOTES

EXAMPLES



39




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) =

TEST YOURSELF F



40




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.

PART G:

DIVISION OF INTEGERS

LEARNING OBJECTIVE

Upon completion of Part G, pupils will be able to perform computations

involving division of integers.



41



PART G:


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

(+) ÷ (–) = (–)

negative ÷ positive = negative

(–) ÷ (+) = (–)

3. Division of any number by zero is undefined.

LESSON NOTES

Undefined means “this

operation does not have a

meaning and is thus not

assigned an interpretation!”

Source:

http://www.sn0wb0ard.com

http://www.sn0wb0ard.com/out/meaning/dictionary.htm

http://www.sn0wb0ard.com/out/interpretation/dictionary.htm



42



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

EXAMPLES



43




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)

TEST YOURSELF G



44



PART H:


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.

LEARNING OBJECTIVE

Upon completion of Part H, pupils will be able to perform computations

involving division of integers using the Accept-Reject Model.



45



PART H:

DIVISION OF INTEGERS USING THE ACCEPT-REJECT MODEL

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 + –

)(

)(

LESSON NOTES



46



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 + –

EXAMPLES



47



To Accept or To Reject Answer

)2(

)8(

Accept + 4

)2(

)8(

Reject – 4

)2(

)8(

Accept – – 4

)2(

)8(

Reject + – 4

EXAMPLES



48




1. 18 ÷ (–6)

2. 2

12

3.

8

24

4. 5

25

5. 3

6

6. – (–35) ÷ 7

7. (–32) ÷ (–4)

8. (–45) ÷ 9 ÷ (–5) 9.

)6(

)30(

10. )5(

80

11. 12 ÷ (–3) ÷ (–2) 12. – (–6) ÷ (3)

TEST YOURSELF H



49




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.

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.



50



PART I:

COMBINED OPERATIONS INVOLVING INTEGERS

1. 10 – (–4) × 3

=10 – (–12)

= 10 + 12

= 22

2. (–4) × (–8 – 3 )

= (–4) × (–11 )

= 44

3. (–6) + (–3 + 8 ) ÷5

= (–6 )+ (5) ÷5

= (–6 )+ 1

= –5

LESSON NOTES

EXAMPLES

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



51




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)

TEST YOURSELF I



52



TEST YOURSELF A:

1. 2

2. –3

3. 6

4. –4

5. –2

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 6

ANSWERS



53



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



54



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

Unit 1:

Negative Numbers

UNIT 2

FRACTIONS





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

1



PART 1

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

objectives which can be taught separately. Teachers may use any parts of the

module as and when it is required.


UNIT 2: Fractions

2



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

3




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

4



numerator

denominator

Fraction is written in the form of:

b

a

Examples:

3

4 ,

3

2

Proper Fraction Improper Fraction Mixed Numbers

The numerator is smaller

than the denominator.

Examples:

20

9 ,

3

2

The numerator is larger

than or equal to the denominator.

Examples:

12

108 ,

4

15

A whole number and

a fraction combined.

Examples:

65

71 8 ,2

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.

LESSON NOTES


UNIT 2: Fractions

5



1.0 Addition And Subtraction of Fractions with the Same Denominator

1.1 Addition of Fractions with the Same Denominators

8

5

8

4

8

1 i)

2

1

8

4

8

3

8

1 ii)

fff

651 iii)

EXAMPLES

Add only the numerators and keep the

denominator same.

Write the fraction in its simplest form.


denominator the same.



8

1

8

4

8

5


UNIT 2: Fractions

6



1.2 Subtraction of Fractions with The Same Denominators

2

1

8

4

8

1

8

5 i)

7

4

7

5

7

1 ii)

nnn

213 iii)


Subtract only the numerators and keep

the denominator the same.





8

5

8

1

2

1

8

4


UNIT 2: Fractions

7



1.3 Addition and Subtraction Involving Whole Numbers and Fractions

.8

11 Calculate i)

7

29

7

1

7

28

7

14

7

14

5

18

5

2

5

20

5

24

5

33

3

12

3

1

3

12

3

14

y

yy

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 8

1

8

11

8

9

+

8

8

+

8

1


UNIT 2: Fractions

8



n

n

nn

n

n

52

5252

k

k

k

k

kk

32

323

2

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



UNIT 2: Fractions

9



1.4 Addition or Subtraction Involving Mixed Numbers and Fractions

.8

4

8

11 Calculate i)

7

5

7

15

7

5

7

12

= 7

20 =

7

62

9

4

9

29

9

4

9

23

= 9

25 =

9

72

88

11

88

31

xx

= 8

11 x

First, convert the mixed number to improper fraction.

Then, add or subtract only the numerators and keep the denominator the same.

8

11

8

4

8

51

8

13

+

8

9

+

8

4


UNIT 2: Fractions

10



2.0 Addition and Subtraction of Fractions with Different Denominators

.2

1

8

1 Calculate i)

To make the denominators the same, multiply both the numerator and the denominator of

the second fraction by 4:

Now, the question can be visualized like this:

?

The denominators are not the same.

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!

8

1

8

4

+

8

5

8

4

2

1

4

4

Now, the denominators

are the same. Therefore,

we can add the fractions

together!

8

1

2

1

+

?


UNIT 2: Fractions

11



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.

(i) 6

5

3

1

6

5

6

2

6

7

= 6

11

(ii) 4

3

12

7

12

9

12

7

12

2

6

1

Change the first fraction to an equivalent

fraction with denominator 6.

(Multiply both the numerator and the denominator of the first fraction by 2):

6

2

3

1

2

2



Change the second fraction to an equivalent fraction with denominator 12.

(Multiply both the numerator and the

denominator of the second fraction by 3):

12

9

4

3

3

3

Subtract only the numerators and keep the



Convert the fraction to a mixed number.


UNIT 2: Fractions

12



(iii) vv 5

91

vv 5

9

5

5

v5

14

Change the first fraction to an equivalent

fraction with denominator 5v.

(Multiply both the numerator and the denominator of the first fraction by 5):

vv 5

51

5

5




UNIT 2: Fractions

13



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

One Another

Method I

4

3

6

1

(i) Find the Least Common Multiple (LCM)

of the denominators.

2) 4 , 6

2) 2 , 3

3) 1 , 3

- , 1

LCM = 2 2 3 = 12

The LCM of 4 and 6 is 12.

(ii) Change each fraction to an equivalent

fraction using the LCM as the

denominator.

(Multiply both the numerator and the

denominator of each fraction by a whole

number that will make their

denominators the same as the LCM

value).

= 4

3

6

1

= 12

9

12

2

= 12

11

Method II

4

3

6

1

(i) Multiply the numerator and the

denominator of the first fraction with

the denominator of the second fraction

and vice versa.

= 4

3

6

1

= 24

18

24

4

= 24

22

= 12

11

Write the fraction in its

simplest form.

This method is preferred but you

must remember to give the

answer in its simplest form. 3

3 2

2

4

4 6

6


UNIT 2: Fractions

14



Multiply the first fraction with the second denominator and

multiply the second fraction with the first denominator.

1. 5

1

3

2

= 5

5

3

2

+

3

3

5

1

15

3

15

10

= 15

13

2. 8

3

6

5

=

8

8

6

5

–

6

6

8

3

= 48

18

48

40

= 48

22

= 24

11


EXAMPLES

Multiply the first fraction by the

denominator of the second fraction and multiply the second fraction by the

denominator of the first fraction.


denominator of the second fraction and

multiply the second fraction by the denominator of the first fraction.






UNIT 2: Fractions

15



3. 7

1

3

2g

= 3

3

7

7

7

1

3

2

g

= 21

3

21

14

g

= 21

314 g

4. 53

2 hg

3

3

55

5

3

2

hg

15

3

15

10 hg

15

310 hg

5. dc

46

= c

c

d

d

dc

46

cd

c

cd

d 46

= cd

cd 46

Multiply the first fraction by the denominator of the second fraction and

multiply the second fraction by the


Write as a single fraction.










UNIT 2: Fractions

16



Convert the mixed numbers to improper fractions.


2.3 Addition or Subtraction of Mixed Numbers with Different Denominators

1. 4

32

2

12

= 4

11

2

5

= 4

11

2

5

2

2

= 4

11

4

10

= 4

21

4

15

2. 4

31

6

53

= 4

7

6

23

= 6

6

4

4

4

7

6

23

= 24

42

24

92

= 24

50

= 12

25

= 12

12

Change the first fraction to an equivalent fraction

with denominator 4. (Multiply both the numerator and the denominator

of the first fraction by 2)

The denominators are not multiples of one another:

Multiply the first fraction by the denominator

of the second fraction.

Multiply the second fraction by the






Change the fraction back to a mixed number.






UNIT 2: Fractions

17



The denominators are not multiples of one another Multiply the first fraction with the second denominator

Multiply the second fraction with the first denominator



2.4 Addition or Subtraction of Algebraic Expression with Different Denominators

1. 22

m

m

m

= )2(

)2(

2

2

22

m

mm

m

m

=

22

2

22

2

m

mm

m

m

= )2(2

)2(2

m

mmm

= )2(2

22 2

m

mmm

= )2(2

2

m

m

2. y

y

y

y 1

1

= )1(

)1(1

1

y

y

y

y

y

y

y

y

= )1(

)1)(1(2

yy

yyy

= )1(

)1( 22

yy

yy

= )1(

122

yy

yy

= )1(

1

yy

Remember to use brackets

Write the above fractions as a single fraction.






Expand:

m (m – 2) = m2 – 2m

Expand:

(y – 1) (y + 1) = y2 + y – y – 1

2

= y2 – 1

Expand:

– (y2 – 1) = –y

2 + 1

Write the fractions as a single fraction.







UNIT 2: Fractions

18





3. 24

5

8

3

n

n

n

= n

n

n

n

n

n

n 8

8

24

4

4

5

8

3

2

2

= )4(8

)5(8

)4(8

1222

2

nn

nn

nn

n

= )4(8

)5(812

2

2

nn

nnn

= )4(8

84012

2

22

nn

nnn

= )4(8

404

2

2

nn

nn

= )8(4

)10(42nn

nn

= 28

10

n

n

Factorise and simplify the fraction by canceling

out the common factors.

Expand:

– 8n (5 + n) = –40n – 8n2

Subtract the like terms.








UNIT 2: Fractions

19



Calculate each of the following.

1. 7

1

7

2

2. 12

5

12

11

3. 14

1

7

2

4. 12

5

3

2

5. 5

4

7

2

6. 7

5

2

1

7. 313

22

8. 9

72

5

24

9. ss

12

10. ww

511

TEST YOURSELF A


UNIT 2: Fractions

20



11. aa 2

12

12. ff 3

52

13. ba

42

14. qp

51

15. nmnm5

3

7

2

5

2

7

5

16.

)2(2

1p

p

17.

5

3

2

32 yxyx

18.

xx

x 5

2

412

19.

x

x

x

x 1

1

20.

2

4

2 x

x

x

x


UNIT 2: Fractions

21



21.

4

84

2

36 yxyx

22.

29

4

3

2

n

n

n

23.

r

rr

15

25

5

2

24.

p

p

p

p

2

232

25.

n

n

n

n

10

34

5

322

26.

n

n

mn

nm 33

27.

mn

nm

m

m

5

5

28.

mn

mn

m

m

3

3

29.

24

5

8

3

n

n

n

30.

m

p

m

p 1

3


UNIT 2: Fractions

22



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

23




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

24



1.0 Multiplication of Fractions

Recall that multiplication is just repeated addition.

Consider the following:

32

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

Therefore, the above multiplication 32 can be represented visually as follows:

This means that 3 units are being repeated twice, or mathematically can be written as:

6

33 32

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

This means that 2 units are being repeated twice, or mathematically can be written as:

4

22 22

LESSON NOTES

3 + 3 = 6

2 + 2 = 4

2 groups of 3 units

2 groups of 2 units


UNIT 2: Fractions

25



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

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

211 12

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?

Let’s consider .2

12

Since represents 1 whole unit, therefore 2

1unit can be represented by the

following shaded area:

Then, we can represent visually the multiplication of 2

12 as follows:

This means that 2

1unit is being repeated twice, or mathematically can be written as:

1

2

2

2

1

2

1

2

12

1 + 1 = 2

2

1 +

2

1 = 1

2

2

2 groups of 1 unit

2 groups of 2

1 unit


UNIT 2: Fractions

26



Let’s consider again .22

1 What does it mean? It means ‘

2

1 out of 2 units’ and the

visualization will be like this:

Notice that the multiplications2

12 and 2

2

1 will give the same answer, that is, 1.

How about ?23

1

Since represents 1 whole unit, therefore 3

1unit can be represented by the

following shaded area:

Then, we can represent visually the multiplication 23

1 as follows:

This means that 3

1unit is being repeated twice, or mathematically can be written as:

3

2

3

1

3

1 2

3

1

3

1 +

3

1 =

3

2

The shaded area is 3

1unit.

2

1 out of 2 units 12

2

1


UNIT 2: Fractions

27



Let’s consider 23

1 . What does it mean? It means ‘

3

1out of 2 units’ and the visualization

will be like this:

Notice that the multiplications3

12 and 2

3

1 will give the same answer, that is,

3

2.

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

2

1

3

1

This means ‘3

1 out of

2

1 units’ and the visualization will be like this:

Consider now this multiplication:

2

1

3

2

This means ‘3

2 out of

2

1 units’ and the visualization will be like this:

2

1unit

3

1 out of 2 units

3

22

3

1

3

1 out of

2

1 units

6

1

2

1

3

1

2

1unit

3

2 out of

2

1 units

6

2

2

1

3

2


UNIT 2: Fractions

28



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:

6

1

2

1

3

1

9

2

3

1

3

2

So, what do you think the answer for 3

1

4

1 ? Do you get

12

1 as the answer?

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

1.1 Multiplication of Simple Fractions

Examples:

a) 35

6

7

3

5

2

b) 35

6

5

3

7

2

c) 35

12

5

2

7

6

d) 35

12

5

2

7

6

Steps to Multiply Fractions:

1) Multiply the numerators together and

multiply the denominators together.

2) Simplify the fraction (if needed).

Remember!!!

(+) (+) = +

(+) (–) = –

(–) (+) = –

(–) (–) = +

Multiply the two numerators together and the two denominators together.


UNIT 2: Fractions

29



1.2 Multiplication of Fractions with Common Factors

6

5

7

12 or

6

5

7

12

1.3 Multiplication of a Whole Number and a Fraction

6

152

=

6

31

1

2

=

6

31

1

2

= 3

31

= 3

110

Second Method:

(i) Simplify the fraction by canceling

out the common factors.

6

5

7

12

(i) Then, multiply the two

numerators together and the two

denominators together, and

convert to a mixed number, if

needed.

6

5

7

12

7

31

7

10

2

1

Convert the mixed number to improper

fraction.

Simplify by canceling out the common

factors.

Remember

2 = 1

2

First Method:

(ii) Multiply the two numerators

together and the two

denominators together:

6

5

7

12 =

42

60

(ii) Then, simplify.

7

31

7

10

42

60

10

7

3 Multiply the two numerators together and

the two denominators together.

Remember: (+) (–) = (–)


1

1

2


UNIT 2: Fractions

30



1. Find 10

15

12

5

Solution: 10

15

12

5

= 8

5

2. Find 5

2

6

21

Solution : 5

2

6

21

= 5

2

6

21

5

7

= 5

21


factors.

Note that 3

21 can be further simplified.

Simplify further by canceling out the

common factors.

3

1

Simplify by canceling out the common factors.

EXAMPLES

Multiply the two numerators together and the

two denominators together.

Remember: (+) (–) = (–)

Multiply the two numerators together and


Remember: (+) (–) = (–)

3

1

1

7

Change the fraction back to a mixed

number.

2

1

4

5


UNIT 2: Fractions

31



1.4 Multiplication of Algebraic Fractions

1. Simplify 4

52 x

x

Solution : 4

52 x

x

= 2

5

= 2

12

2. Simplify

m

n

n4

9

2

Solution:

m

n

n4

9

2

=

1

4

2

9

2

mn

n

n

= 1

)2(

2

9 mn

= nm22

9

1 2

1 1 Simplify the fraction by canceling out the x’s.



Simplify the fraction by canceling the

common factor and the n.

Multiply the two numerators together

and the two denominators together.


Change the fraction back to a mixed

number.

2

1

1

1


UNIT 2: Fractions

32



1. Calculate 27

25

5

9

2. Calculate – 20

14

7

3

12

45

3. Calculate

4

112

4. Calculate

5

14

3

1

5. Simplify

k

m3

6. Simplify )5(2

mn

7. Simplify

14

3

6

11

x

8. Simplify )32(2

dan

9. Simplify

yx

10

95

3

2

10. Simplify

x

x 120

4

TEST YOURSELF B1


UNIT 2: Fractions

33



2.0 Division of Fractions

Consider the following:

36

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

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

This means that 6 units are being divided into a group of 3 units, or mathematically

can be written as:

2 36

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:

.8

1

2

1

LESSON NOTES

How many 8

1 is in

?2

1

6 units are being divided into a group of 3

units:

2 36


UNIT 2: Fractions

34



This means ‘How many is in ?

8

1

2

1

The answer is 4:

Consider now this division:

.4

1

4

3

This means ‘How many is in ?

4

1

4

3

The answer is 3:

But, how do you

calculate the answer?

How many 4

1 is in ?

4

3


UNIT 2: Fractions

35



Consider again .236

Actually, the above division can be written as follows:

3

16

3

636

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:

4

2

8

1

8

2

1

8

1

2

1 (i)

The reciprocal of a

fraction is found by

inverting the

fraction.

Change the second fraction to its

reciprocal and change the sign to .

The reciprocal

of 8

1 is .

1

8

These operations are the same!

The reciprocal

of 3 is .3

1


UNIT 2: Fractions

36



3

1

4

4

3

4

1

4

3 (ii)

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

2.1 Division of Simple Fractions

Example:

7

3

5

2

= 3

7

5

2

= 15

14

Change the second fraction to its reciprocal

and change the sign to .



Steps to Divide Fractions:

1. Change the second fraction to its


2. Multiply the numerators together and

multiply the denominators together.

3. Simplify the fraction (if needed).

Tips:

(+) (+) = +

(+) (–) = –

(–) (+) = –

(–) (–) = +

Change the second fraction to its


The reciprocal

of 4

1 is .

1

4


UNIT 2: Fractions

37



2.2 Division of Fractions With Common Factors

Examples:

9

2

21

10

= 2

9

21

10

= 2

9

21

10

= 7

15

= 7

12

7

6

5

3

6

7

5

3

10

7

7

65

3

1

5 3

7

1

2

Express the fraction in division form.

Change the second fraction to its reciprocal and

change the sign to .

Simplify by canceling out the common factors.




Then, simplify by canceling out the common

factors.



Remember: (+) (–) = (–)




UNIT 2: Fractions

38



1. Find 6

25

12

35

Solution : 6

25

12

35

= 25

6

12

35

= 10

7

2. Simplify –4

52 x

x

Solution : –xx 5

42

= –25

8

x

3. Simplify 2

x

y

Solution :

2x

y

2

1

x

y

x

y

2

5

7


and change the sign to . Then, simplify by canceling out the common

factors.

Method I

EXAMPLES



Multiply the two numerators together and the two

denominators together.

Express the fraction in division form.


and change to .

Multiply the two numerators together and the two

denominators together.

Remember: (+) (–) = (–)



2

1


UNIT 2: Fractions

39



Multiply the numerator and the denominator of

the given fraction with x

2

x

y

= 2

x

y

x

x

= x

xx

y

2

= x

y

2

4. Simplify 5

)1( 1r

Solution:

5

)1( 1r

= 5

)1

1(r

r

r

= r

r

5

1

The given fraction.

r is the denominator of r

1.

Multiply the given fraction with r

r.

Note that:

1)1

1( rrr

Method II

The numerator is also

a fraction with

denominator x

Multiply the numerator and the denominator of the

given fraction by x.


UNIT 2: Fractions

40



1. Calculate 2

21

7

3

2. Calculate 16

5

8

7

9

5

3. Simplify 3

48 y

y

4. Simplify

k

2

16

5. Simplify

3

5

2

x

6. Simplify n

m

n

m

3

24 2

7. Simplify 8

1

4

y

8. Simplify

x

x

11

TEST YOURSELF B2


UNIT 2: Fractions

41



9. Calculate 5

)1(341

10. Simplify y

x15

11. Simplify

32

941 x

12. Simplify

15

1

1

p


UNIT 2: Fractions

42



TEST YOURSELF A:

1. 7

3

2. 2

1

3. 14

5

4. 4

1

5. 35

38 or

35

31

6. 14

3

7. 13

67 or

13

25

8. 45

73or

45

281

9. s

3

10. w

6

11. a2

5

12. f3

1

13. ab

ab 42

14. pq

pq 5

15. nm

16. 2

33 p

17. 10

1716 yx

18. x

x 12

19. )1(

1

xx

20. 2

21. 2

8 yx

22. 29

47

n

n

23. r

r

3

12

24. 2

2

2

6

p

p

25. 2

2

10

647

n

nn

26. m

m1

27. n

n

5

5

28. n

n

3

3

29. 28

10

n

n

30. m

p

3

34

ANSWERS


UNIT 2: Fractions

43



TEST YOURSELF B1:

1. 3

21

3

5or 2.

8

11

8

9 or 3.

2

15

2

11or

4. 5

21

5

7 or 5.

k

m3 6.

2

5mn

7. 4

x 8. ndna

2

3 9. yx

5

3

3

10

10. 4

15 x

TEST YOURSELF B2:

1. 49

2 2.

9

51

9

14 or 3.

2

6

y

4. 8k

5. x5

6 6.

m

6

7. )1(2

1

y 8.

1

2

x

x

9. 20

9

10. xy

x 15 11.

6

13x 12.

p4

5

Unit 1:

Negative Numbers

UNIT 3

ALGEBRAIC EXPRESSIONS

AND

ALGEBRAIC FORMULAE





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


1 Curriculum Development Division


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.






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.

PART A:

PERFORMING OPERATIONS ON

ALGEBRAIC EXPRESSIONS

LEARNING OBJECTIVES

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

expressions.





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 – n

2,

g

3

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.

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

3

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

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

2

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

Examples: 3ab, –5ab are like terms.

3x2,

2

5

2x are like terms.

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

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

LESSON NOTES

6 xy Coefficient Unknowns





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.

Examples: .2

,2

4,

6

2,

15

322

22

2

2

yxyx

yx

grg

gr

h

m

10. To simplify an algebraic fraction, identify the common factor of both the numerator and the

denominator. Then, simplify it by elimination.





Simplify the following algebraic expressions and algebraic fractions:

(a) 5x – (3x – 4x) 64

)e(ts

(b) –3r –9s + 6r + 7s z

yx

2

3

6

5)f(

(c) 2

2

2

4

grg

gr

g

f

e2)g(

qp

43)d(

(h) x

x

3

2

13

Solutions:

(a) 5x – (3x – 4x)

= 5x – (– x)

= 5x + x

= 6x

(b) –3r –9s + 6r + 7s

= –3r + 6r –9s + 7s

= 3r – 2s

2

2

2

4)c(

grg

gr

gr

r

grg

gr

2

4

)2(

4

2

2

Perform the operation in the bracket.

Arrange the algebraic terms according to the like terms.

.

Unlike terms cannot be simplified.

Leave the answer in the simplest form as shown.

Algebraic expression with like terms can be simplified by

adding or subtracting the coefficients of the unknown.

Simplify by canceling out the common factor and the

same unknowns in both the numerator and the

denominator.

1

1

EXAMPLES





pq

pq

pq

p

pq

q

qp

43

43

43)d(

12

23

26

2

34

3

64)e(

ts

ts

ts

z

xy

z

yx

z

yx

4

5

22

5

2

3

6

5)f(

fg

e

gf

eg

f

e

2

2

12)g(

x

x

x

x

x

x

x

x

x

x

6

16

3

1

2

16

3

2

16

3

2

1

2

)2(3

3

2

13

)h(

The LCM of p and q is pq.

The LCM of 4 and 6 is 12.


factor, then multiply the numerators

together and followed by the

denominators.

Change division to multiplication of the

reciprocal of 2g.

Equate the denominator.

2

1





ALTERNATIVE METHOD

Simplify the following algebraic fractions:

(a) x

x

3

2

13

= x

x

3

2

13

2

2

= )2(3

)2(2

1)2(3

x

x

= x

x

6

16

(b) 5

23

x = 5

23

x

x

x

x

x

x

xxx

5

23

)(5

)(2)(3

x

x

x

xx

x

x

xxx

4

316

)2(2

)2(2

3)2(8

2

2

2

2

38

2

2

38

)c(

The denominator of x2

3 is 2x. Therefore,

multiply the algebraic fraction byx

x

2

2.

Each of the terms in the numerator and

denominator is multiplied by 2x.

.

The denominator of 2is2

1. Therefore,

multiply the algebraic fraction by2

2.


denominator of the algebraic fraction is

multiplied by 2.

The denominator of x

3 is x. Therefore,

multiply the algebraic fraction byx

x.


denominator is multiplied by x.





x

x

x

xx

36

21

288

21

)7(4)7(7

8

)7(3

7

7

47

8

3

47

8

3)d(

The denominator of 7

8 x is 7.

Therefore, multiply the algebraic

fraction by7

7.

Each of the terms in the numerator

and denominator is multiplied by 7.

Simplify the denominator.





Simplify the following algebraic expressions:

1. 2a –3b + 7a – 2b

2. − 4m + 5n + 2m – 9n

3. 8k – ( 4k – 2k )

4. 6p – ( 8p – 4p )

xy 5

13.5

5

2

3

4.6

kh

c

ba

2

3

7

4.7

dc

dc

3

8

2

4.8

yzz

xy.9

w

uv

vw

u

2.10

65

2.11

x

54

24

.12

x

x

TEST YOURSELF A






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.

PART B:

EXPANSION OF ALGEBRAIC

EXPRESSIONS

LEARNING OBJECTIVE

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

expressions.





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 – 15b

2

= 12a

2 + 8ab – 15b

2

4. Useful expansion tips:

(i) (a + b)2 = a

2 + 2ab + b

2

(ii) (a – b)2 = a

2 – 2ab + b

2

(iii) (a – b)(a + b) = (a + b)(a – b)

= a2 – b

2

LESSON NOTES





Expand each of the following algebraic expressions:

(a) 2(x + 3y)

(b) – 3a (6b + 5 – 4c)

Solutions:

(a) 2 (x + 3y)

= 2x + 6y

(b) –3a (6b + 5 – 4c)

= –18ab – 15a + 12ac

1293

2)c( y

= 123

29

3

2 y

= 6y + 8

= (a + 3) (a + 3)

= a2 + 3a + 3a + 9

= a2 + 6a + 9

When expanding two brackets, each term

within the first bracket is multiplied by

every term within the second bracket.

1293

2)c( y

2523)e( k

2)3()d( a

)5)(2()f( pp

2)3()d( a

When expanding a bracket, each term

within the bracket is multiplied by the term

outside the bracket.

When expanding a bracket, each term

within the bracket is multiplied by the term

outside the bracket.

1

3

1

4

EXAMPLES


factor, then multiply the numerators

together and followed by the denominators.





(c) (4x – 3y)(6x – 5y)

– 18 xy

– 20 xy

– 38 xy

= 24x2 – 38 xy + 15y

2

2523)e( k

= –3(2k + 5) (2k + 5)

= –3(4k2 + 20k + 25)

= –12k2 – 60k – 75

)5( )2( )f( qp

= pq – 5p + 2q – 10

ALTERNATIVE METHOD

Expanding two brackets

(a) (a + 3) (a + 3)

= a2 + 3a + 3a + 9

= a2 + 6a + 9

(b) (2p + 3q) (6p – 5q)

= 12p2 – 10 pq + 18 pq – 15q

2

= 12p2 + 8 pq – 15q

2




When expanding two

brackets, write down the

product of expansion and

then, simplify the like

terms.








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

4

324.1 n

162

1.2 q

yxx 326.3

)(22.4 baba

)6()3(2.5 pp

3

26

3

1.6

yxyx

121.72

ee

nmmnm 2.82

gfggfgf 2.9

ihiihih 32.10

TEST YOURSELF B






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:

2

2

2

32

3

32

2433

43

)27(

)154()3(

)27(

)2()3(])3(3)[27(

27

)3(.2

)1549()57(2

)6()57(])57(28[2

)57(2.1

x

xx

x

xxx

dx

dy

x

xy

xxx

xxxxdx

dy

xxy

Strategy

1. Pupils revise the techniques of factorisation.

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.





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 + b

2 = (a + b)

2

a2 – 2ab + b

2 = (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 – b

2 = (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.

LESSON NOTES





(a) Factorising the Common Factors

i) mn + m = m (n +1)

ii) 3mp + pq = p (3m + q)

iii) 2mn – 6n = 2n (m – 3)

(b) Factorising Algebraic Expressions with Four Terms

i) vy + wy + vz + wz

= y (v + w) + z (v + w)

= (v + w)(y + z)

ii) 21bm – 7bs + 6cm – 2cs

= 7b(3m – s) + 2c(3m – s)

= (3m – s)(7b + 2c)

Factorise the first and the second terms

with the common factor y, then factorise

the third and fourth terms with the

common factor z.

.

(v + w) is the common factor.

Factorise the first and the second terms with

common factor 7b, then factorise the third

and fourth terms with common factor 2c.

(3m – s) is the common factor.

EXAMPLES

Factorise the common factor m.

.

Factorise the common factor p.

.

Factorise the common factor 2n.

.





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

i) x2 – 16 = x

2 – 4

2

= (x + 4)(x – 4)

ii) 4x2

– 25 = (2x)2 – 5

2

= (2x + 5)(2x – 5)

(d) Factorising the Expressions by Using the Cross Method

i) x2

– 5x + 6

xxx

x

x

523

2

3

x2

– 5x + 6 = (x – 3) (x – 2)

ii) 3x2

+ 4x – 4

xxx

x

x

462

2

23

3x2 + 4x – 4 = (3x – 2) (x + 2)

The summation of the cross

multiplication products should

equal to the middle term of the

quadratic expression in the

general form.

The summation of the cross

multiplication products should

equal to the middle term of the

quadratic expression in the

general form.

a2 – b

2 = (a + b)(a – b)





ALTERNATIVE METHOD

Factorise the following quadratic expressions:

i) x 2 – 5x + 6

ac b

+ 6 – 5

–2 –3

(x – 2) (x – 3)

)3)(2(65 2 xxxx

ii) x 2 – 5x – 6

ac b

– 6 – 5

+1 – 6

(x + 1) (x– 6)

)6)(1(65 2 xxxx

+1 (–6) = –6

+1 (–6) = –6

+1 – 6 = –5

a=+1 b= –5 c = –6

REMEMBER!!!

An algebraic expression can

be represented in the general

form of ax2 + bx + c, where

a, b, c are constants and

a ≠ 0, b ≠ 0, c ≠ 0.

+1 (+ 6) = + 6 –2 (–3) = +6

–2 + (–3) = –5

a=+1 b= –5 c =+6





TEST YOURSELF C

(iii) 2x2 – 11x + 5

ac b

+ 10 –11

–1 – 10

2

10

2

1

52

1

(2x – 1) (x – 5)

)5)(12(5112 2 xxxx

(iv) 3x2 + 4x – 4

ac b

– 12 + 4

– 2 +6

23

2

3

6

3

2

The coefficient of x2 is 2,

divide each number by 2.

(+2) (+5) = +10

–1 (–10) = +10

–1 + (–10) = –11

–2 + 6 = 4

The coefficient of x2 is 3, divide each

number by 3.

3 (– 4) = –12

a=+2 b = –11 c =+5

a =+ 3 b=+ 4 c = –4

(3x – 2) (x + 2)

The coefficient of x2 is 2,

multiply by 2:

5)(12

52

5

21

21

xx

xx

xx

The coefficient of x2 is 3, multiply by 3:

2)(23

23

2

32

32

xx

xx

xx

)2)(23(443 2 xxxx





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

TEST YOURSELF C





11. 2x 2 + x – 10

12. 3x 2 + 2x – 8

13. 3p 2 – 5p – 12

14. 4p2 – 3p – 1

15. 2x2

– 3x – 5

16. 4x 2 – 12x + 5

17. 5p 2 + p – 6

18. 2x2

– 11x + 12

19. 3p + k + 9pr + 3kr

20. 4c2 – 2ct – 6cw + 3tw






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 2

2

1rA .

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

hrV

bhA

2

2

1

y, A and V are the

subjects of the

formulae.

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.





PART D:

CHANGING THE SUBJECT OF A FORMULA

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

sign.

Examples:

hrV

bhA

2

2

1

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.

Examples: bhA2

1

a2 = b

2 + c

2

hTrT 2

2

1

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.

LESSON NOTES

A is the subject of the formula because it is

expressed in terms of other unknowns.

a

2 is not the subject of the formula

because the power ≠ 1

T is not the subject of the formula

because it is found on both sides of the

equation.





1. Given that 2x + y = 2, express x in terms of y.

Solution:

2x + y = 2

2x = 2 – y

x = 2

2 y

2. Given that yyx

52

3

, express x in terms of y.

Solution:

yyx

52

3

3x + y = 10y

3x = 10y – y

3x = 9y

x = 3

9y

x = 3y

No fraction and brackets.

Group:

Retain the x term on the left hand side of the

equation by grouping all the y term to the

right hand side of the equation.

Fraction:

Multiply both sides of the equation by 2.

Group:

Retain the x term on the left hand side of the

equation by grouping all the y term to the

right hand side of the equation.

Solve:

Divide both sides of the equation by 2 to

make the coefficient of x equal to 1.

Solve:



EXAMPLES

Steps to Change the Subject of a Formula

(i) Fraction

(ii) Brackets

(iii) Group

(iv) Factorise

(v) Solve





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

Solution:

yx 2

x = (2y)2

x = 4y2

4. Given that px

3, express x in terms of p.

Solution:

px

3

2

2

9

)3(

3

px

px

px

5. Given that yxx 23 , express x in terms of y.

Solution:

2

2

2

2

2

22

23

23

yx

yx

yx

yxx

yxx

Solve:

Square both sides of the equation to make the

power of x equal to 1.

Fraction:

Multiply both sides of the equation by 3.

Solve:

Square both sides of the equation to make

the power of x equal to1.

Group:

Group the like terms

Solve:



Solve:

Square both sides of equation to make the

power of x equal to 1.

Simplify the terms.





6. Given that 4

11x – 2(1 – y) = xp2 , express x in terms of y and p.

Solution:

4

11x – 2 (1 – y) = xp2

11x – 8(1 – y) = xp8

11x – 8 + 8y = 8xp

11x – 8xp = 8 – 8y

x(11 – 8p) = 8 – 8y

x = p

y

811

88

7. Given that n

xp

5

32 = 1 – p , express p in terms of x and n.

Solution:

n

xp

5

32 = 1 – p

2p – 3x = 5n – 5pn

2p + 5pn = 5n + 3x

p(2 + 5n) = 5n + 3x

p = n

xn

52

35

Fraction:

Multiply both sides of the equation

by 4.

Bracket:

Expand the bracket.

Group:

Group the like terms.

Factorise:

Factorise the x term.

Solve:

Divide both sides by (11 – 8p) to


Fraction:

Multiply both sides of the equation by

5n.

Solve:

Divide both sides of the equation by

(2 + 5n) to make the coefficient of p

equal to 1.

Group:

Group the like p terms.

Factorise:

Factorise the p terms.





1. Express x in terms of y.

a) 02 yx

b) 032 yx

c) 12 xy

d) 22

1 yx

e) 53 yx

f) 43 xy

TEST YOURSELF D





2. Express x in terms of y.

a) xy

b) xy 2

c) 3

2x

y

d) xy 31

e) 13 xyx

f) yx 1





3. Change the subject of the following formulae:

a) Given that 2

ax

ax, express x in terms

of a .

b) Given that x

xy

1

1, express x in terms

of y .

c) Given that vuf

111 , express u in

terms of v and f .

d) Given that 4

3

2

2

qp

qp, express p in

terms of .q

e) Given that mnmp 23 , express m in

terms of n and p .

f) Given that

C

CBA

1, express C in

terms of A and B .

g) Given that yx

xy2

2

, express y in

terms of x.

h) Given that g

lT 2 , express g in

terms of T and l.





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) hrV 2 , then V is the of the formula.

7) An can be represented by a letter.

10) 21232 xxxx .

ACTIVITIES





VERTICAL

2) An algebraic consists of two or more algebraic terms combined by

addition or subtraction or both.

6) 252212 2 xxxx .

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. Calculate

.3

5

12

D) 5

1 O) 1

W) 3

11 N)

15

11

2. Simplify yxyx 7693 .

F) yx 23 W) yx 169

E) yx 23 X) yx 29

3. Simplify 23

qp .

L) 6

32 qp A)

6

32 qp

N) 6

23 pq R)

6

23 qp





4. Expand )7()4(2 xx .

A) 1x D) 15x

U) 13 x C) 153 x

5. Expand )52(3 cba .

S ) acab 156 C) acab 156

T) acab 156 R) acab 156

6. Factorise 252 x .

E) )5)(5( xx T) )5)(5( xx

I) )5)(5( xx C) )25)(25( xx

7. Factorise qpq 4 .

D) )41( qpq E) )4( pq

T) )4( qp S) )4( pq

8. Factorise 1282 xx .

I ) )6)(2( xx W) )6)(2( xx

F) )3)(4( xx C) )3)(4( xx

9. Given that 42

3

x

yx, express x in terms of y.

L) 5

yx C)

5

yx

T) 11

yx N)

3

8 yx





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

1. Calculate

.3

15

x

A) 3

5 x O)

x

x

3

5

I ) 5

3

x

x N)

5

3

x

2. Simplify r

qp

54

3 .

F) q

pr

4

15 R)

pr

q

15

4

W) r

pq

20

3 B)

r

pq

5

3

3. Simplifyz

xy

yz

x

2 .

N)2

2

y D)

2

2

2z

x

L) 22z

x I)

2

2

z

x

4. Solve ).3(2

yxxyx

E) xyyx 222 D) xyyx 222

I ) xyxyx 222 3 N) xyyx 222





5. Expand 25p .

I) 252 p N) 252 p

D) 25102 pp L) 25102 pp

6. Factorise 1572 2 yy .

F) )5)(32( yy D) )5)(32( yy

W) )5)(32( yy L) )52)(3( yy

7. Factorise 5112 2 pp .

R) )5)(12( pp B) )5)(12( pp

F) )5)(1( pp W) )52)(1( pp

8. Given that ACC

B )1( , express C in terms of A and B.

L) AB

BC

R)

ABC

1

C) AB

ABC

N)

AB

ABC

9. Given that 25 xyx , express x in terms of y.

O) 16

42

yx B)

24

42

yx

I )

2

2

1

yx U)

2

4

2

yx





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

FURTHER

EXPLORATION

http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut11_simp.htm

http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut11_simp.htm

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

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

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





TEST YOURSELF A:

1. 9a – 5b

2. – 2m – 4n

3. 6k

4. 2p

5. xy

yx

5

15

6.

15

620 kh

7. c

ab

7

6

8. dc

dc

3

)4(4

9. 2z

x

10. 2

2

v

11. x

x

65

2

12. x

x

54

24

TEST YOURSELF B:

1. – 8n + 3 6. x + y

2. 3q + 2

1

7. 2e

3. – 12x2 + 18xy 8. mnmn 22

4. – 3b 9. fgf 22

5. p 10. 22 52 iihh

ANSWERS





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:

1. (a) x = 2 – y (b)

2

3 yx

(c) x = 2y – 1

(d) x = 4 – y (e) 3

5 yx

(f) x = 3y – 4

2. (a) x = y2

(b) 24yx

(c) 236 yx

(d)

2

3

1

yx

2

2

1)e(

yx (f) 12 yx

3. (a) ax 3

(b) 1

1

y

yx

(c) fv

fvu

(d) 2

7qp

(e) 32

n

pm

(f) AB

BC

(g)

)1(2

x

xy (h)

2

24

T

lg





ACTIVITIES

CROSSWORD PUZZLE

RIDDLES

RIDDLE 1

2 F

3

A

1

N

5

T

4

A

7

S

6

T

8

I

9

C

RIDDLE 2

2

W 1

O

3

N

5

D

4

E

7

R

6

F

9

U

8

L

Unit 1:

Negative Numbers

UNIT 4

LINEAR EQUATIONS





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

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

x= b 9

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

Part E: Solving Linear Equations in the Form of a

x+ b = c 15

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 numerical value is a solution of a given linear equation

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.





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

(b) Addition

(c) Division

(d) Multiplication

Arithmetic

8 = (4) (2)

8 – 3 = (4) (2) – 3

Algebra

a = b

a – c = b – c

;

Arithmetic

8 = (4) (2)

8 + 3 = (4) (2) + 3

Algebra

a = b

a + c = b + c

Arithmetic

8 = 6 + 2

8 6 2

3 3

Algebra

a = b

a b

c c c ≠ 0

Arithmetic

8 = (6 +2)

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

Algebra

a = b

ac = bc

OVERALL LESSON NOTES





PART A:

LINEAR EQUATIONS

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?

LESSON NOTES

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

http://www.free-graphics.com/clipart/People/boy-spike-hair.sht





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.

2.

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

x x

x + 2 – 2 = 5 – 2

x = 3

x + 2 = 5

(a) x + 2 = 5

x = ?

x x

x + 2 = 5 x + 2 – 2 = 5 – 2

x = 3

x = 3






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

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

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 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






PART B:

SOLVING LINEAR EQUATIONS IN THE FORM OF

x + a = b OR x – a = b

Solve the following equations.

(i) 52 x (ii) 3 5x

Solutions:

(ii) 3 5x

x – 3 + 3 = 5 + 3

x = 5 + 3

x = 8

(i) 52 x

x + 2 – 2 = 5 – 2

x = 5 – 2

x = 3

Subtract 2 from both

sides of the equation.

Simplify the LHS.

Add 3 to both sides of

the equation.

Alternative Method:

3

25

52

x

x

x

Alternative Method:

8

35

53

x

x

x

Simplify the LHS.

Simplify the RHS.

Simplify the RHS.

EXAMPLES






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

TEST YOURSELF B





PART C:


THE FORMS OF

ax = b AND ba

x

LEARNING OBJECTIVES

Upon completion of Part C, pupils will be able to understand the concept of


form of:

(a) ax = b

b

a

xb )(



Pupils face difficulty when solving linear equations in one unknown by solving

equations in the form of:

(a) ax = b

b

a

xb )(


Strategy:








PART C:

SOLVING LINEAR EQUATION

ax = b AND ba

x


(i) 3m = 12 (ii)

43

m

Solutions:

(i) 3m = 12

3 12

3 3

m

3

12m

m = 4

(ii) 43

m

3433

m

m = 4 3

m = 12

Divide both sides of

the equation by 3.

Multiply both sides of

the equation by 3.

Simplify the LHS.

Simplify the LHS.

Simplify the RHS.

Alternative Method:

4

3

12

123

m

m

m

Alternative Method:

12

43

43

m

m

m

Simplify the RHS.

EXAMPLES






1. 2p = 6

2. 5k = – 20

3. – 4h = 24

4. 567 l

5. 728 j

6. 605 n

7. 726 v

8. 427 y

9. 9612 z

10. 42

m

11. 4

r = 5

12. 8

w= –7

13. 88

t

14. 912

s

15. 65

u

TEST YOURSELF 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.

PART D:


THE FORM OF

ax + b = c


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

Solve the equation 2x – 3 = 11.

Solution:

Method 1

2x – 3 = 11

2x – 3 + 3 = 11 + 3

2x = 14

22

142

x

2

14x

x = 7

Method 2

1132 x

222

1132

x

2

11

2

3x

2

3

2

3

2

11

2

3x

2

14x

7x


the equation.

Simplify both sides of

the equation.


the equation by 2.

Simplify the LHS.


the equation by 2.

Simplify the LHS.

Add 2

3 to both sides

of the equation.


the equation.

Alternative Method:

2

2

14

142

3112

1132

x

x

x

x

x

Alternative Method:

7

2

14

2

3

2

11

2

11

2

3

2

2

1132

x

x

x

x

x

Simplify the RHS.

EXAMPLES






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

TEST YOURSELF D





PART E


THE FORM OF

cba

x

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

of ba

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


Pupils face difficulty when solving linear equations in one unknown by solving

equations in the form of ba

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

Strategy:








PART E:

SOLVING LINEAR EQUATIONS IN THE FORM OF cba

x

Solve the equation 143

x

.

Solution:

Method 1

143

x

4 43

x = 1 + 4

53

x

33 53

x

35x

x = 15

Method 2

33

14

3

x

313433

x

312x

x – 12 + 12 = 3 + 12

123x

15x


the equation.


the equation.


the equation by 3.

Simplify both sides of the

equation.


the equation by 3.

Expand the LHS.


the equation.


the equation.


the equation.

Alternative

Method:

15

53

53

413

143

x

x

x

x

x

EXAMPLES






1. 532

m

2. 123

b

3. 723

k

4. 3 + 2

h= 5

5. 4 +5

h = 6 6. 21

4

m

7. 54

2 h

8. 6

k+ 3 = 1 9. 2

53

h

10. 3 – 2m = 7

11. 72

3 m

12. 12 + 5h = 2

TEST YOURSELF E





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

Solve the following equations:

(i) – 4x – 5 = 2x + 7

Solution:

Method 1

2

126

126

756

756

7254

x

x

x

x

x

xx

66

55

Method 2

7254 xx

– 4x – 5 + 5 = 2x + 7 + 5

– 4x = 2x + 12

– 4x – 2x = 2x – 2x + 12

– 6x = 12

2

126

x

x

66

Subtract 2x from both sides of the equation.

Simplify both sides of the equation.


Divide both sides of the equation by –6.

Add 5 to both sides of the equation.


Subtract 2x from both sides of the equation.


Divide both sides of the equation by – 6.

Alternative Method:

2

6

12

126

5724

7254

x

x

x

xx

xx

–4x – 2x – 5 = 2x – 2x + 7


EXAMPLES





(ii) 3(n – 2) – 2(n – 1) = 2 (n + 5)

3n – 6 – 2n + 2 = 2n + 10

n – 4 = 2n + 10

n – 2n – 4 = 2n – 2n + 10

– n – 4 = 10

– n – 4 + 4 = 10 + 4

– n = 14

14

14

n

n

11

Expand both sides of the equation.

Simplify the LHS.

Subtract 2n from both sides of the equation.


Alternative Method:

14

14

1024

1022263

)5(2)1(2)2(3

n

n

nn

nnn

nnn

Divide both sides of the equation by – 1.





3

7

21

7

7

217

318337

1837

183364

18)1(3)32(2

)3(62

16

3

32 6

)3(62

1

3

32 6

32

1

3

32

x

x

x

x

x

xx

xx

xx

xx

xx


Alternative Method:

3

7

21

217

3187

1837

183364

18)1(3)32(2

632

1

3

326

32

1

3

32

x

x

x

x

x

xx

xx

xx

xx

(iii)

Simplify LHS.

Expand the brackets.

Multiply both sides of the equation by the

LCM.

Divide both sides of the equation by 7.






1. 4x – 5 + 2x = 8x – 3 – x

2. 4(x – 2) – 3(x – 1) = 2 (x + 6)

3. –3(2n – 5) = 2(4n + 7)

2

9

4

3 .4

x

6

5

3

2

2 .5

x

253

.6 xx

6

135

2 .7

yy

2

9

4

1

3

2 .8

xx

08

43

6

52 .9

xx

12

74

9

72 .10

xx

TEST YOURSELF F





TEST YOURSELF B:

1. x = 5

4. x = –2

7. x = 48

10. x = –111

13. x = –107

2. x = 6

5. x = –7

8. x = –45

11. x = 57

14. x = 247

3. x = 9

6. x = –3

9. x = –50

12. x = –87

15. x = –115

TEST YOURSELF C:

1. p = 3

4. l = 8

7. v = 12

10. m = 8

2. k = – 4

5. j = – 9

8. y = – 6

11. r = 20

3. h = –6

6. n = 12

9. z = 8

12. w = – 56

13. t = – 64

TEST YOURSELF D:

1. m = 2

4. m = 3

7. p = 2

10. p = −1

14. s = 108

2. p = 4

2

3 5. y

8. k = 2

11. p = 3

15. u = 30

3. k = 2

4

3 6. p

9. x = –1

12. m = 1

TEST YOURSELF E:

1. m = 4

4. h = 4

7. h = 12

10. m = −2

10. b = 9

5. h = 10

8. k = −12

11. m = −8

11. k = 15

6. m = 12

9. h = 5

12. h = −2

ANSWERS





TEST YOURSELF F:

1. x = − 2 2. x = − 17 3. 14

1n 4. x = 6

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

9. x = −8 10. x = 19

Unit 1:

Negative Numbers

UNIT 5

INDICES





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

3.0 Verifying nmnm aaa

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


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

1.0 Verifying mnnm aa )(

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

4.0 Verifying n

n

aa

1

15

5.0 Verifying nn aa

1

16

Activity 20

Answers 22

Basic Essential Additional Mathematics Skills (BEAMS) Module UNIT 5: Indices

1



PART 1

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:

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.


2



LEARNING OBJECTIVES


1. express repeated multiplication as an and vice versa;

2. find the value of an;

3. verify nmnm aaa ;

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.

PART A:

INDICES I


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.


3



1.0 Expressing Repeated Multiplication As an

and Vice Versa

(i) 3332

(ii) )4)(4)(4(3)4(

(iii) rrrr 3

(iv) )6)(6()6( 2 mmm

2.0 Finding the Value of an

2 factors of 3

3 factors of (4)

3 factors of r

2 factors of (6+m)

32

is read as

‘three to the power of 2’

or

‘three to the second power’.

32

base

81

16

3333

2222

3

2

3

2iii)(

125

)5)(5)(5()5()ii(

32

222222)i(

4

44

3

5

LESSON NOTES A

index

(a) What is 24?

(b) What is (−1)3?

(c) What is an?


4




325

32

213

2

437

43

)1()1(

)]1)(1)(1[()]1)(1[()1()1()iii(

77

)77(777)ii(

22

)2222()222(22)i(

yy

yyyyyyy

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

Base

nmnm aaa

6

515

11

8383

8

14343

3

1

3

1

3

1

3

1)iii(

)5(

)5()5()5()ii(

6

6666)i(


5



3433133

236236

10755255525

15

4

15

4

2

1

5

4

3

2)iii(

30523)ii(

)i(

qpqpqpp

rstrst

nmnmnnmm

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

Same Base

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

Bases

7.0 Simplifying Multiplication of Algebraic Terms Expressed In Index Notation with

Different Bases

4133

52323

402011920119

64242

)iv(

)()()()iii(

6632)ii(

)i(

t

s

t

s

t

s

t

s

abababab

wwwww

pppp

4

4

4

4

44

,Conversely

t

s

t

s

t

s

t

s

45423423

17103147331473

312384384

5

3

2

1

5

3

2

1

5

3

2

1

2

1)iii(

75757755)ii(

2323233(i)

Note: Sum up the indices

with the same base.

numbers with

different bases

cannot be simplified.

555

555

)(

,Conversely

)(

abba

baab


6



1. Find the value of each of the following.

(a)

243

3333335

(b) 36

(c) 4)4(

(d)

5

5

1

(e)

3

4

3

(f)

2

5

12

(g) 47

(h)

5

3

2

2. Simplify the following.

(a)

5

2323

12

1243

m

mmm

(b) bbb 42 35

(c) 342 3)3(2 xxx

(d) 323 )()2(7 ppp

EXAMPLES & TEST YOURSELF A


7




(a)

576

96434 23

(b) 232 22)3(

(c) 343 )7()7()1(

(d)

232

5

4

3

1

3

1

(e) 423 5522

(f)

7

2

3

2

7

2

3

2223


(a) 2424 1234 gfgf

(b) 232 32)3( srr

(c) 343 )3()7()( vww

(d)

232

5

4

5

1

7

3kkh


8



PART B:

INDICES II

LEARNING OBJECTIVES


1. verifynmnm aaa ;

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.


9




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

3

5412

54

12

7

310

3

10

4

239239

6

2828

3

333

3 (iv)

5

55

5(iii)

7

7777(ii)

4

444 (i)

LESSON NOTES B

Note:

1

1

0

0

a

a

aaa

aaaa

m

mmm

mmmm

(a) What is 2

5 ÷ 2

5?

(b) What is 20?

(c) What is a0?

nmnm aaa

1

1 1

1

1

/

1

/

1

/

1 1 1

23

23

297

29

352

35

)2()2(

)2)(2(

)2)(2)(2()2()2()iii(

55

55

55555555555)ii(

22

222

2222222)i(

pp

pp

ppppp

1

/

1

/

1 1


10



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

Base

3

8

3

8

3

8 (iii)

445

20(ii)

(i)

23

2

3

437

3

7

24646

hhh

h

kkk

k

nnnn

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

Bases

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

Different Bases

REMEMBER!!!

Numbers with

different bases cannot

be simplified.

45

2638

23

68

6

11

6

11

6

415

64

156415

5

4

5

4

60

48)ii(

333

3

939 (i)

qp

qpqp

qp

k

h

k

h

k

h

kh

hkhh


11




(a)

144

12

121212

2

3535

(b) 999 310

(c)

3

9

8

8

(d)

1218

3

2

3

2

(e)

18

20

)5(

)5(

(f)

24

1018

3

33


(a)

7

512512

q

qqq

(b) 79 84 yy

(c)

8

10

15

35

m

m

(d)

88

1114

2

2

b

b


(a)

45

1549

4

59

2

9

2

9

8

36

nm

nmnm

nm

(b)

76

1316

12

64

dc

dc

(c)

34

96

12

64

gf

fgf

(d)

56

489

12

378

vu

uvu

EXAMPLES & TEST YOURSELF B


12



PART C:

INDICES III

LEARNING OBJECTIVES

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

1. derivemnnm aa )(

;

2. simplify

(a) numbers;

(b) algebraic terms, expressed in index notation raised to a power;

3. verify n

n

aa

1

; and

4. verify nn aa

1

.


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.


13



1.0 Verifying mnnm aa )(

24

23

8

6

44

33

4

3

4

32

4

3

35391527

555999

595959359

236

33

3323

15

11

15

11

15

11

15

11

15

11

15

11)iii(

2323

23

)23)(23)(23()23()ii(

22

2

22)2()i(

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

245

396

385

31363

85

136(iv)

20715421075342)10(75

34(iii)

1593525395725)397(2(ii)

121062106)2(10(i)

mnnm aa )(

LESSON NOTES C


14



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

518

518

273615

23

76535

23

7653

15

20

15

20

15

205

53

5455

3

4

412

412

4

412

4143

44

3

201510

5453525432

105

52552

3

32

3

32

3

32

12

42

12

4)2()v(

32

32

)2(

)2(2)iv(

625

1

625

5

5

1

5

1)iii(

)()ii(

3

3)3((i)

qp

qp

qp

qp

qpp

qp

qpp

n

m

n

m

n

m

n

m

n

m

ba

ba

ba

baba

gfe

gfegfe

x

xx

Note:

A negative number raised to an even power is positive.

A negative number raised to an odd power is negative.


15



4. 0 Verifying n

n

aa

1

Alternative Method

n

n

10

110

10

1

100

110

10

1

10

110

110

1010

10010

100010

0001010

2

2

1

1

0

1

2

3

4

n

n

aa

1

352

3

52

2

2

264

2

64

777

1

77777

7777)ii(

3

13

333

1

333333

333333)i(

Hint: 100

?

1000


16



5.0 Verifying nn aa

1

pp

pp

p

p

pp

p

p

mm

mm

mmm

1

1

1

11

55

1

55 5

1

5

1

5

1

5

1

5

1

55

5

5

1

5

5

1

15

5

15

5

1

2

1

2

1

2

1

2

2

1

2

2

1

12

2

12

2

1

)iii(

22

222222

22

22

222(ii)

33

333

33

33

333(i)

Take square root on both sides

of the equation.

Note:

mnn

m

nn

aa

aa

1

(a) What is 2

1

4 ?

(b) What is 2

3

4 ?

(c) What is n

m

a ?

nn aa

1


17




(a) (b)

32 ])1[(

(c)

2

2

3

7

2

(d)

32

5

3

(e)

32

5

3

(f)

2. (a) Simplify the following.

(i) 824

4246426

32

3232

(ii) 2346 52

(iii) 5132 44

(iv)

32

5

2

4

3

(v)

23

7

3

4

7

(vi)

4

422

5

43

12

5

327682

22

15

3535

4

232

EXAMPLES & TEST YOURSELF C


18



2. (b) Simplify the following.

(i)

15

155

535153

32

2

))(2(2

x

x

xx

(ii) 674 yx

(iii) 3122 ww

(iv) 779 84 yy

(v)

2

68

59

9

36

qp

qp

(vi)

3. Simplify the following expressions:

(a)

32

1

2

12

5

5

(b)

1

4

3

(c)

4

23y

x

(d)

51

4

6

2

ts

st

(e)

3

23

12

2 km

nm

(f)

2

63

32

2

8

ba

cab

4423 3 2 mnnm


19




(a)

4

6464 33

1

(b) 2

5

100

(c)

4

3

81

(d) 2

1

2

1

273

(e) m

1

m235

110 )()( aaa

(f)

3

4

27

1


20



1. 52

10

44

4

P 04 O

34 R 174 T

134

2. 2327551010

T 514510 O 65510 N 55510 B 614510

3.

2

22

4

32

D 4

22

E 2

2

2

3 N 2

2

4

3 O

3

42

4. xyxy 239 82

M 4

27xy A 4

114

x

y L

4

21xy K 2

74

x

y

5. 425 32

A 820 32 N 69 32 T

620 32 S 89 32

6. 4225 nnmm

T 87nm U

810nm L 67nm E

610nm

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!

ACTIVITY


21



7.

3243

5

2

5

2

5

2

5

2

F 12

5

2

A 2

5

2

V 6

5

2

E 5

5

2

8.

5

3

2

4

7

Y

15

10

4

7 R

8

7

4

7 M

8

10

4

7 A

15

7

4

7

9. 36

59

5

25

ba

ba

L 81515 ba I

835 ba S 235 ba T

5615 ba

10.

5232

5

2

5

2

3

1

3

1

P 105

5

2

3

1

E 76

5

2

3

1

I 75

5

2

3

1

R 106

5

2

3

1

11. 23

76

3

12

qp

qp

Y 3

53qp A

534 qp R 993

1

qp D

993 qp

Congratulations! You have completed this activity.

1 2 3 4 5 6 7 8 9 10 11

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


22



TEST YOURSELF A:

1.

(a) 243

(b) 216

(c) 256 (d)

3125

1

(e)

64

27

(f)

25

214

(g) 2401 (h)

243

32

2.

(a) 512m

(b) 715b

(c) 918x

(d) 814p

3.

(a) 576

(b) 288

(c) 823543

(d)

6075

16

(e) 000250

(f)

34983

256

ANSWERS


23



4.

(a) 2412 gf

(b) 2554 sr

(c) 3782764 vw (d) 52

125153

144kh

TEST YOURSELF B:

1.

(a) 144

(b) 441531

(c) 144262 (d)

729

64

(e) 25

(f) 81

2.

(a) 7q (b) 2

2

1y

(c) 2

3

7m

(d) 364b

3.

(a) 45

2

9nm

(b) 610

3

16dc

(c) 632 gf

(d) 3714 vu


24



TEST YOURSELF C:

1.

(a) (b) 1

(c)

2401

64

(d)

15625

729

5

36

(e)

125

729

5

33

6

(f)

2. (a)

(i) 8

3224

(ii) 624 52

(iii)

(iv)

)5(2

33

2

(v) 3

2

4

)3(7

(vi)

2

146

5

)4(3

2. (b)

(i) 1532x (ii) 4224 yx

(iii) 3 0

1

w

(iv)

7

14

2

y

(v) 2

16

q

p

(vi) 187162 nm

32768

21677716224

114


25



3.

(a)

32

1

2

15

(b)

3

4

(c) 4

8

81x

y

(d)

9

2

3

1

t

s

(e) 3368 nmk

(f)

16

64

16

1

b

ca

4.

(a) 4

(b) 000100

(c)

27

1

(d) 9

(e) 5a

(f)

81

1

ACTIVITY:

The WONDERWORD is ONEMALAYSIA

Unit 1:

Negative Numbers

UNIT 6

COORDINATES

AND

GRAPHS OF FUNCTIONS





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

Basic Essential Additional Mathematics Skills (BEAMS) Module Unit 6: Coordinates and Graphs of Functions

1



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.


2



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.

PART A:

COORDINATES


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.


3



PART A:

COORDINATES

1.

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

h left –

right +

k up +

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).

Coordinates of P = (h, k)

Start from the

origin.

x

y

O

● P

h units

k units

LESSON NOTES


4



PART A1: State the coordinates of the given points.

1.

Coordinates of A = (2, 3)

1.

Coordinates of A =

2.

Coordinates of B = (–3, 1)

2.

Coordinates of B =

3.

Coordinates of C = (–2, –2)

3.

Coordinates of C =

EXAMPLES TEST YOURSELF

• A

• 4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

Start from

the origin,

move 2 units

to the right.

Next, move

3 units up. • A

•

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

4

3

2

1

-1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

Next, move

1 unit up.

• B

Start from the

origin, move 3 units

to the left.

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• B

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• C

Start from

the origin,

move 2 units

to the left.

Next, move 2

units down.

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• C

EXAMPLES

TEST YOURSELF


5




4.

Coordinates of D = (4, –3)

4.

Coordinates of D =

5.

Coordinates of E = (3, 0)

5.

Coordinates of E =

6.

Coordinates of F = (0, 3)

6.

Coordinates of F =


4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

Start from

the origin,

move 4 units

to the right.

Next, move

3 units

down.

• D

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x • E

Start from the


to the right.

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

Start from

the origin,

move 3 units

up.

• F

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• • D

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x • E

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• F

EXAMPLES

TEST YOURSELF

Do not move

along the y-axis

since y = 0.

Do not move

along the x-axis

since x = 0.


6




7.

Coordinates of G = (–2, 0)

7.

Coordinates of G =

8.

Coordinates of H = (0, –2)

8.

Coordinates of H =

9.

Coordinates of J = (6, 8)

9.

Coordinates of J =


4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

Start from

the origin,

move 2 units

to the left.

• G

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x • G

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• H

Start from the


down.

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• H

8

6

4

2

–2

–4

–6

–8

y

–8 –6 –4 –2 0 2 4 6 8 x

• J

Start from

the origin,

move 6 units

to the right.

Next, move

8units up.

8

6

4

2

–2

–4

–6

–8

y

–8 –6 –4 –2 0 2 4 6 8 x

• J

EXAMPLES

TEST YOURSELF


7




10.

Coordinates of K = (– 6 , 6)

10.

Coordinates of K =

11.

Coordinates of L = (–15, –20)

11.

Coordinates of L =

12.

Coordinates of M = (3, – 4)

12.

Coordinates of M =

8

6

4

2

–2

–4

–6

–8

y

–8 –6 –4 –2 0 2 4 6 8 x

Start from

the origin,

move 6 units

to the left.

• K

Next, move

6 units up.

8

6

4

2

–2

–4

–6

–8

y

–8 –6 –4 –2 0 2 4 6 8 x

• K

20

15

10

5

–5

–10

–15

–20

y

–20 –15 –10 –5 0 5 10 15 20 x

• L

Next, move

20 units

down.

Start from the


to the left.

20

15

10

5

–5

–10

–15

–20

y

–20 –15 –10 –5 0 5 10 15 20 x

• L

M

4

2

–2

–4

y

–4 –2 0 2 4 x

•

Start from

the origin,

move 3 units

to the right.

Next, move 4

units down.

4

2

–2

–4

y

–4 –2 0 2 4 x

• M


EXAMPLES

TEST YOURSELF


8



Write the step by step directions involving integer coordinates that

will get the mouse through the maze to the cheese.

–6 –5 –4 –3 –2 –1

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

0

y

1 2 3 4 5 6 7

x

ACTIVITY A1


9



PART A2: Plot the point on the Cartesian plane given its coordinates.

.

1. Plot point A (3, 4)

1. Plot point A (2, 3)

2. Plot point B (–2, 3)

2. Plot point B (–3, 4)

3. Plot point C (–1, –3)

3. Plot point C (–1, –2)

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 -1 0 1 2 3 4 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x


4

3

2

1

–1

–2

–3

–4

• A

• y

–4 –3 –2 –1 0 1 2 3 4 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• B

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• C

EXAMPLES

TEST YOURSELF


10



PART A2: Plot the point on the Cartesian plane given the coordinates.

.

4. Plot point D (2, – 4)

4. Plot point D (1, –3)

5. Plot point E (1, 0)

5. Plot point E (2, 0)

6. Plot point F (0, 4)

6. Plot point F (0, 3)

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x


4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• D

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x • E

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• F

EXAMPLES

TEST YOURSELF


11




.

7. Plot point G (–2, 0)

7. Plot point G (– 4,0)

8. Plot point H (0, – 4)

8. Plot point H (0, –2)

9. Plot point J (6, 4)

9. Plot point J (8, 6)


4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x • G

8

6

4

2

–2

–4

–6

–8

y

–8 –6 –4 –2 0 2 4 6 8 x

• J

8

6

4

2

–2

–4

–6

–8

y

–8 –6 –4 –2 0 2 4 6 8 x

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• H

EXAMPLES

TEST YOURSELF


12




.

10. Plot point K (– 4, 6)

10. Plot point K (– 6, 2)

11. Plot point L (–15, –10)

11. Plot point L (–20, –5)

12. Plot point M (30, –15)

12. Plot point M (10, –25)

29

10

–10

–20

y

–20 –10 0 10 20 x

• L


8

4

–4

–8

y

–8 –4 0 4 8 x

• K

8

4

–4

–8

y

-8 -4 0 4 8 x

–20 –10 0 10 20

20

10

–10

–20

y

x

20

10

–10

–20

y

–40 –20 0 20 40 x

20

10

–10

–20

y

–40 –20 0 20 40 x

• M

EXAMPLES

TEST YOURSELF


13



1. 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

2 4 –2 –4 x

2

4

–2

y

0

–4

,

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.

Enjoy yourself !

ACTIVITY A2


14



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.

PART B:

GRAPHS OF FUNCTIONS


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.


15



PART B:

GRAPHS OF FUNCTIONS

1. For a standard graph paper, 2 cm is represented by 10 small squares.

2. Some common scales used are as follows:

Scale Note

2 cm to 10 units

10 small squares represent 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



2 cm to 1 unit

10 small squares represent 1 unit


2 cm to 0.1 unit

10 small squares represent 0.1 unit


2 cm

2 cm

LESSON NOTES


16



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.

Scale: 2 cm to 10 units.

[ 1 small square represents 1 unit ]

1. Mark – 6 4, 15 and 26 on the x-axis.



2. Mark –7, –2, 3 and 8on the x-axis.


[ 1 small square represents 0.5 unit ]

2. Mark –8, –3, 2 and 6, on the x-axis.



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.



4. Mark –1.3, – 0.6, 0.5 and 1.6 on the x-axis.

Scale: 2 cm to 1 unit.


4. Mark –1.7, – 0.7, 0.7 and 1.5 on the x-axis.



0 –10 10 20 30

x

7 16 27 –4

x

x

–5 –10 0 5 10

x

–2 3 8 –7

x

–2 –4 2 4

x

1 –3.4 0 –0.8 2.6

x

– 1 –2 1 2

x

0.5 –1.3 0 –0.6 1.6

EXAMPLES

TEST YOURSELF


17




5. Mark – 0.15, – 0.04, 0.03 and 0.17 on the

x-axis.

Scale: 2 cm to 0.1 unit


5. Mark – 0.17, – 0.06, 0.04 and 0.13 on the

x-axis.

Scale: 2 cm to 0.1 unit


6. Mark –13, –8, 2 and 14 on the y-axis.

Scale: 2 cm to 10 units


6. Mark –16, – 4, 5 and 15 on the y-axis.

Scale: 2 cm to 10 units


x

0 –0.1 –0.2 0.1 0.2 0.03 0.17 –0.04 –0.15

x

y y

0

10

–20

20

–10

–13

–8

2

14

EXAMPLES

TEST YOURSELF


18




7. Mark –9, –3, 1 and 7 on the y-axis.



7. Mark –7, – 4, 2 and 6 on the y-axis.



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

y

0

5

–10

10

–9

–3

1

7

–5

y

0

–4

4

–2

–3.2

–0.6

2

1.4

2.4

EXAMPLES

TEST YOURSELF


19










10. Mark – 0.17, – 0.06, 0.08 and 0.16 on the

y-axis.

Scale: 2 cm to 0.1 unit.


10. Mark – 0.18, – 0.03, 0.05 and 0.14 on the

y-axis.

Scale: 2 cm to 0.1 units.


y

y

y

0

1

–2

2

–1

0.4

1.5

– 0.4

–1.6

y

0

0.2

– 0.17

–0.1

– 0.06

0.1

0.08

0.16

–0.2

EXAMPLES

TEST YOURSELF


20



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,

of a function.

x –2 –1 0 1 2

y –2 0 2 4 6

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

function.


of a function.

x –3 –2 –1 0 1

y –2 0 2 4 6



function.


of a function.

x –2 –1 0 1 2

y 5 3 1 –1 –3



function.


of a function.

x –2 –1 0 1 2

y 7 5 3 1 –1



function.

–1 1 x –2 2

–2

6

4

2

y

0

–1 1 x –2 2

–2

6

4

2

y

0

EXAMPLES

TEST YOURSELF


21





of a function.

x – 4 –3 –2 –1 0 1 2

y 15 5 –1 –3 –1 5 15



function.


of a function.

x –1 0 1 2 3 4 5

y 19 4 –5 –8 –5 4 19



function.


of a function.

x –2 –1 0 1 2 3 4

y –7 –2 1 2 1 –2 –7

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

function.


of a function.

x –2 –1 0 1 2 3

y –8 –4 –2 –2 – 4 –8


function.

y

10

5

15

–5

x –3 1 –4 2 0 –1 –2

0

y

2

–6

–2

–4

x 3 4 2 1 –1 –2

EXAMPLES

TEST YOURSELF


22





of a function.

x –2 –1 0 1 2

y –7 –1 1 3 11



function.


of a function.

x –2 –1 0 1 2

y –6 2 4 6 16



function.


of a function.

x –3 –2 –1 0 1 2 3

y 22 5 0 1 2 –3 –20


function.


of a function.

x –3 –2 –1 0 1 2 3

y 21 4 –1 0 1 – 4 –21


function.

x 2 3 1 –2 –3 –1 0

y

20

–20

–10

10

y

10

5

15

–5

x

–2 1 2 –1

0

EXAMPLES

TEST YOURSELF


23



Each table below shows the values of x and y for a certain function.

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.

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

x

y

0

ACTIVITY B1


24



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

below.



a = 7, b = 13, c = – 4, d = –14


below.


below.



a = 2, b = 7.5, c = –3, d = –8.5


below.


below.



a = 0.6, b = 3.4, c = –1.2, d = –2.6


below.

–20 10 20

x

c d 0 –10 a b –20 10 20

x

c d 0 –10 a b

–5 –10 0 5 10

x

c a b d –5 –10 0 5 10

x

c a b d

c –2 – 4 2 4

x

a d 0 b c –2 – 4 2 4

x

a d 0 b

EXAMPLES

TEST YOURSELF


25





below.



a = 0.8, b = 1.4, c = – 0.3, d = –1.6


below.


below.



a = 0.04, b = 0.14, c = – 0.03, d = – 0.16


below.

6. State the values of a, b, c and d on the y-axis

below.


[ 1 small square

represents 1 unit ]

a = 3, b = 17

c = – 6, d = –15


below.

–1 –2 1 2

x

a d 0 c b –1 –2 1 2

x

a d 0 c b

c

x

0 –0.1 –0.2 0.1 0.2 a b d c

x

0 –0.1 – 0.2 0.1 0.2 a b d

y

0

10

–20

20

–10

d

c

a

b

y

0

10

–20

20

–10

d

c

a

b

EXAMPLES

TEST YOURSELF


26





below.


[ 1 small square

represents 0.5 unit ]

a = 4, b = 9.5

c = –2, d = –7.5


below.


below.


[ 1 small square


a = 0.8, b = 3.2

c = –1.2, d = –2.6


below.

y

0

5

–10

10

d

c

a

b

–5

y

0

5

–10

10

d

c

a

b

–5

y

0

–4

4

–2

d

c

2

a

b

y

0

–4

4

–2

d

c

2

a

b

EXAMPLES

TEST YOURSELF


27





below.


[ 1 small square


a = 0.7, b = 1.2

c = – 0.6, d = –1.4


below.


below.


[ 1 small square


a = 0.03, b = 0.07

c = – 0.04, d = – 0.18


below.

y

0

1

–2

2

–1

a

b

c

d

y

0

1

–2

2

–1

a

b

c

d

y

0

d

–0.1

c

a

–0.2

0.2

b

0.1

y

0

d

c

a

–0.2

0.2

b

0.1

–0.1

EXAMPLES

TEST YOURSELF


28



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

when (a) x = 1.5

(b) x = –2.8

(a) 7 (b) –1.6


when (a) x = 0.6

(b) x = –1.7

(a) (b)


when ( a ) x = 0.14

( b ) x = – 0.26

(a) 1.5 (b) 11.5


when ( a ) x = 0.07

( b ) x = – 0.18

(a) (b)

–1 1 x –2 2

–2

6

4

2

y

0

– 2.8

1.5

7

– 1.6

–1 1 x –2 2

–2

6

4

2

y

0

– 0.26

1.5

0.14

11.5

x –0.1 – 0. 2 0.1 0.2

y

10

–10

5

–5

0 x –0.1 –0. 2 0.1 0.2

y

10

–10

5

– 5

0

EXAMPLES

TEST YOURSELF


29





when ( a ) x = 0.6

( b ) x = –2.7

( a ) 11 ( b ) –3.5


when ( a ) x = 1.2

( b ) x = –1.8

( a ) ( b )


when (a) x = 1.4

(b) x = –1.5

(a) 3 (b) –5.8


when (a) x = 2.7

(b) x = –2.1

(a) (b)

y

10

5

15

–5

x –3 1 – 4 2 0 –1 –2

11

0.6

– 2.7

– 3.5

y

10

5

15

–5

x –3 1 – 4 2 0 –1 –2

x 3 4 2 1 –1 –2 0

y

2

– 6

– 2

– 4

1.4

3

– 1.5

– 5.8

x 3 4 2 1 –1 –2 0

y

2

– 6

– 2

– 4

EXAMPLES

TEST YOURSELF


30





when (a) x = 1.7

(b) x = –1.3

(a) 5.5 (b) –3.5


when (a) x = 1.2

(b) x = –1.9

(a) (b)


when (a) x = 1.6

(b) x = –2.3

(a) –9 (b) 25


when (a) x = 2.8

(b) x = –2.6

(a) (b)

y

10

5

15

–5

–2 x 1 2 –1 0

5.5

1.7

– 1.3

– 3.5

y

10

5

15

–5

–2 x 1 2 –1 0

x 2 3 1 –2 –3 –1 0

y

20

–20

–10

10

1.6

– 9

– 2.3

25

x 2 3 1 –2 –3 –1 0

y

20

–20

–10

10

EXAMPLES

TEST YOURSELF


31




7. Based on the graph below, find the value of x

when (a) y = 5.4

(b) y = –1.6

(a) 1.4 (b) –2.8


when (a) y = 2.8

(b) y = –2.4

(a) (b)


when ( a ) y = 4

( b ) y = –7.5

(a) – 0.07 (b) 0.08


when ( a ) y = 6.5

( b ) y = –7

(a) (b)

–1 1 x –2 2

–2

6

4

2

y

0

x –0.1 –0. 2 0.1 0.2

y

10

–10

5

– 5

0

–1 1 x –2 2

–2

6

4

2

y

0

– 2.8

1.4

5.4

– 1.6

– 0.07

4

0.08

– 7.5

x –0.1 –0. 2 0.1 0.2

y

10

–10

5

–5

0

EXAMPLES

TEST YOURSELF


32




9. Based on the graph below, find the values of x

when (a) y = 8.5

(b) y = 0

(a) –3.1 , 2.1 (b) –2 , 1


when (a) y = 3.5

(b) y = 0

(a) (b)


when (a) y = 2.6

(b) y = – 4.8

(a) 0.6 , 2.1 (b) –1.2 , 3.9


when (a) y = 1.2

(b) y = – 4.4

(a) (b)

x 3 4 2 1 –1 –2 0

y

2

– 6

– 2

– 4

x –3 1 – 4 2 –1 –2 2.1 – 3.1

8.5

0

y

10

5

15

–5

x 3 4 2 1 –1 –2 0

y

2

– 6

– 2

– 4

0.6 2.1

– 1.2 3.9

2.6

– 4.8

x –3 1 – 4 2 –1 –2 0

y

10

5

15

–5

EXAMPLES

TEST YOURSELF


33





when (a) y = 14

(b) y = –17

(a) 2.6 (b) –2.3


when (a) y = 11

(b) y = –23

(a) (b)


when (a) y = 6.5

(b) y = 0

(c) y = –6

(a) – 0.8 (b) 1.3 (c) 2.3


when (a) y = 7.5

(b ) y = 0

(c) y = –9

(a) (b) (c)

x 2 3 1 –2 –3 –1 0

y

20

–20

–10

10

2.6

– 2.3

– 17

14

x 2 3 1 –2 –3 –1 0

y

20

–20

–10

10

y

10

5

15

–5

–2 x 1 2 –1 0

y

10

5

15

–5

–2 x 1 2 –1 0

6.5

– 6

1.3 – 0.8 2.3

EXAMPLES

TEST YOURSELF


34



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.

0

5

10

15

20

25

30

35

40

45

50

55

60

y

1 2 3 4 5 6 7 8 9 x

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.

ACTIVITY B2


35



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).

ANSWERS


36



PART A2:

1.

4.

2.

5.

3.

6.

4

3

2

1

–1

–2

–3

-–4

–4 –3 –2 –1 0 1 2 3 4

y

x

• B

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• A

4

3

2

1

–1

–2

–3

–4

–4 –3 –2 –1 0 1 2 3 4

y

x

• D

4

3

2

1

–1

–2

–3

–4

–4 –3 –2 –1 0 1 2 3 4

y

x • E

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• C

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

• F


37



7.

10.

8.

11.

9.

12.

4

3

2

1

–1

–2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x • G •

K

8

4

–4

–8

y

–8 –4 0 4 8 x

4

3

2

1

–1

-2

–3

–4

y

–4 –3 –2 –1 0 1 2 3 4 x

–

• H

–20 –10 0 10 20

20

10

–10

–20

y

x

• L

8

6

4

2

–2

–4

–6

–8

y

–8 –6 –4 –2 0 2 4 6 8 x

• J

• M

20

10

–10

–20

y

–40 –20 0 20 40 x


38



ACTIVITY A2:

YAKOMI ISLANDS

2

4

–2

y

O

–4 RM 1 million

U

A

B C

D

E F

P Q

R S

T

2 4 –2 –4 x

,


39



PART B1:

1

2.

3.

4.

5.

6.

7.

8.

9.

10.

0 –10 10 20 30

x

4 15 26 –6 –5 –10 0 5 10

x

–3 2 6 –8

–2 –4 2 4

x

–3.2 0 –1 2.8 1.2 –1 –2 1 2

x

0.7 –1.7 0 –0.7 1.5

x

0 –0.1 –0.2 0.1 0.2 0.04 0.13 –0.06 –0.16

y

0

10

–20

20

–10

–16

–4

5

15

y

0

5

–10

10

–7

–4

2

6

–5

y

0

1

–2

2

–1

0.3

1.7

–0.8

–1.5

y

0

0.2

– 0.18

– 0.1

– 0.03

0.1

0.05

0.14

– 0.2

y

0

–4

4

–2

–3.4

–1.4

2

0.8

2.8


40



PART B2:

1.

2.

3.

4.

5.

6.

–2

6

4

2

y

0 x –3 1 –1 –2 –1 1 x –2 2

–2

6

4

2

y

0

x 4 –1 5 1 0

y

10

5

15

–5

2 3

y

–4

–8

–2

–6

0 x 3 2 1 –1 –2

y

10

5

15

–5

–2 x 1 2 –1 0

x 2 3 1 –2 –3 – 1 0

y

20

–20

–10

10


41



ACTIVITY B1:

–4 –3 –2 –1 0 1 2 3 4 x

2

4

6

8

10

12

14

16

18

20

y


42



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)

Unit 1:

Negative Numbers

UNIT 7

LINEAR INEQUALITIES



Curriculum Development Division Ministry of Education Malaysia

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

Basic Essential Additional Mathematics Skills (BEAMS) Module Unit 7: Linear Inequalities

1 Curriculum Development Division Ministry of Education Malaysia

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

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

−3 < − 1

−3 is less than − 1

and

−1 > − 3

−1 is greater than − 3

1 < 3

1 is less than 3

and

3 > 1

3 is greater than 1

OVERALL LESSON NOTES

−1 − 2 − 3 x

0 1 2 3


______________________________________________________________________________


3.0 Properties of Inequalities

(a) Addition Involving Inequalities

Arithmetic Form Algebraic Form

812 so 48412

92 so 6962

If a > b, then cbca

If a < b, then cbca

(b) Subtraction Involving Inequalities


7 > 3 so 5357

2 < 9 so 6962

If a > b, then cbca

If a < b, then cbca

(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)

12 > 9 so 12 9

3 3

If a > b and c > 0 , then ac > bc

If a > b and c > 0, then a b

c c

52 so )3(5)3(2

128 so 2

12

2

8

If ba and 0c , then bcac

If ba and 0c , then c

b

c

a


______________________________________________________________________________


(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)

6 < 7 so 6(−3) > 7(−3)

16 > 8 so 16 8

4 4

10 <15 so 10 15

5 5

If a > b and c < 0, then ac < bc

If a < b and c < 0, then ac > bc

If a > b and c < 0, then a b

c c

If a < b and c < 0, then a b

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

m4

(b) A solution of an inequality is any value of the variable that satisfies the inequality.

Examples:

(i) Consider the inequality 3x

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 3x can be written as

,...8 ,7 ,6 ,5 ,4x


______________________________________________________________________________


A number line is normally used to represent all the solutions of an inequality.

(ii) x > 2

(iii) 3x

The solid dot

means the value

3 is included.

The open dot

means the value

2 is not

included.

3 − 2 − 1 1 0 2 x

4

o

0 − 1 − 2 x

1 2 3 4

To draw a number line representing 3x , 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.


______________________________________________________________________________


PART B:

POSSIBLE SOLUTIONS FOR A

GIVEN LINEAR INEQUALITY IN

ONE UNKNOWN


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.

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) hx

(d) x h

(ii) representing a linear inequality:

(a) x h

(b) x h

(c) hx

(d) x h on a number line and vice versa.


______________________________________________________________________________


PART B:

POSSIBLE SOLUTIONS FOR

A GIVEN LINEAR INEQUALITY IN ONE UNKNOWN

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:

The possible integers are: 5, 6, 7, …

(2) 3x

Solution:

The possible integers are: – 4, − 5, −6, …

(3) 13 x

Solution:

The possible integers are: −2, −1, 0, and 1.

−2 −5 −8 x

−1 0 2 1 −7 −4 −6 −3 3 4

EXAMPLES

4 1 −2 x

5 6 8 7 −1 2 0 3 9 10

−2 −5 −8 x

−1 0 2 1 −7 −4 −6 −3 3 4


______________________________________________________________________________


Draw a number line to represent the following inequalities:

(a) x > 1

(b) 2x

(c) 2x

(d) 3x

TEST YOURSELF B


______________________________________________________________________________



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.

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.

PART C:

COMPUTATIONS INVOLVING

ADDITION AND SUBTRACTION ON

LINEAR INEQUALITIES


______________________________________________________________________________


PART C:

COMPUTATIONS INVOLVING ADDITION AND SUBTRACTION

ON LINEAR INEQUALITIES

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

Adding 1 to both sides of the inequality:

The inequality

sign is

unchanged.

LESSON NOTES

1 x

2 3 4

2 < 4

4 x

2 3 5

2 + 1 < 4 + 1

3 < 5


______________________________________________________________________________


(ii) 4 > 2

Subtracting 3 from both sides of the inequality:

(1) Solve 145x .

Solution:

9

51455

145

x

x

x

(2) Solve 3 2.p

Solution:

3 2

3 3 2 3

5

p

p

p

Subtract 5 from both sides

of the inequality.

Simplify.

Add 3 to both sides of the

inequality.

Simplify.

The inequality

sign is

unchanged.

EXAMPLES

x −1 0 1 2

1 x

2 3 4

4 > 2

4 − 3 > 2 − 3

1 > − 1


______________________________________________________________________________


Solve the following inequalities:

(1) 24 m

(2) 3.4 2.6x

(3) 613 x

(4) 65.4 d

(5) 1723 m

(6) 78 54y

(7) 9 5d

(8) 2 1p

(9) 1

32

m

(10) 3 8x

TEST YOURSELF C


______________________________________________________________________________



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)

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.

PART D:


DIVISION AND MULTIPLICATION

ON LINEAR INEQUALITIES


______________________________________________________________________________


PART D1:


MULTIPLICATION AND DIVISION ON LINEAR INEQUALITIES

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

Multiplying both sides of the inequality by 3:

LESSON NOTES

x

The inequality

sign is

unchanged.

1 x

2 3 4

2 < 4

2 3 < 4 3

6 < 12

x 6 8 10 12 14


______________________________________________________________________________


(ii) − 4 < 2

Dividing both sides of the inequality by 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

Dividing both sides of the inequality by −1:

The inequality

sign is reversed.

x −6 −5 −4 −3

3 x

4 5 6

The inequality

sign is

unchanged.

−4 x

− 2 0 2

4 < 6

4 (−1) > 6 (−1)

− 4 > − 6

− 4 < 2

− 4 2 < 2 2

− 2 < 1

−2 − 1 0 1 2

x


______________________________________________________________________________


(ii) 1 > −3

Multiply both sides of the inequality by −1:

Solve the inequality 3 12q .

Solution:

(i) 3 12q

3

12

3

3

q

4q

Divide each side of the

inequality by −3.

Simplify.

The inequality

sign is reversed.

EXAMPLES

The inequality

sign is reversed.

1 > −3

x −3 −2 −1 0 1

(− 1) (1) < (−1) (−3)

31

x −1 0 1 2 3


______________________________________________________________________________



(1) 7 49p

(2) 6 18x

(3) −5c > 15

(4) 200 < −40p

(5) 243 d

(6) 82 x

(7) x312

(8) y525

(9) 162 m

(10) 276 b

TEST YOURSELF D1


______________________________________________________________________________


PART D2:

PERFORM COMPUTATIONS INVOLVING

MULTIPLICATION OF LINEAR INEQUALITIES

Solve the inequality 32

x .

Solution:

32

x .

3)2()2

(2 x

6x

Multiply both sides of the

inequality by −2.

Simplify.

The inequality

sign is reversed.

EXAMPLES


______________________________________________________________________________


1. Solve the following inequalities:

(1) − 38

d

(2) 82

n

(3) 5

10y

(4) 67

b

(5) 0 128

x

(6) 8 06

x

TEST YOURSELF D2


______________________________________________________________________________



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.

LEARNING OBJECTIVES

Upon completion of Part E, pupils will be able perform computations

involving linear inequalities.

PART E:

FURTHER PRACTICE ON


LINEAR INEQUALITIES


______________________________________________________________________________


PART E:

FURTHER PRACTICE ON COMPUTATIONS

INVOLVING LINEAR INEQUALITIES


1.

(a) 05 m

(b) 62 x

(c) 3 + m > 4

2.

(a) 3m < 12

(b) 2m > 42

(c) 4x > 18

TEST YOURSELF E1


______________________________________________________________________________


3.

(a) m + 4 > 4m + 1

(b) mm 614

(c) mm 433

4.

(a) 64 x

(b) 12315 m

(c) 54

3 x


______________________________________________________________________________


(d) 1835 x

(e) 1031 p

(f) 432

x

(g) 85

3 x

(h) 43

2

p


______________________________________________________________________________


What is the smallest integer for x if 1835 x ?

Solution:

1835 x

3185 x

155 x O

3x

x = 4, 5, 6,…

Therefore, the smallest integer for x is 4.

3x

A number line can

be used to obtain the

answer.

2 1 0 x

3 4 5 6

EXAMPLES


______________________________________________________________________________


1. If ,1413 x what is the smallest integer for x?

2. What is the greatest integer for m if 147 mm ?

3. If 43

2

x, find the greatest integer value of x.

4. If 4

3

2

p, what is the greatest integer for p?

5. What is the smallest integer for m if 9

2

3

m?

TEST YOURSELF E2


______________________________________________________________________________


1

2

3

4

5

6

7

8

9

10

11

12

ACTIVITY


______________________________________________________________________________


HORIZONTAL:

4. 31 is an ___________.

5. An inequality can be represented on a number __________.

7. 62 is read as 2 is __________ than 6.

9. Given 912 x , 5x is a _____________ of the inequality.

11. 123 x

4x

The inequality sign is reversed when divided by a ____________ integer.

VERTICAL:

1.

2

12

x

x

The inequality sign remains unchanged when multiplied by a ___________ integer.

2. 246 x equals to 4x when both sides are _____________ by 6.

3. 5x equals to 153 x when both sides are _____________ by 3.

6. ___________ inequalities are inequalities with the same solution(s).

8. 2x is represented by a ____________ dot on a number line.

10. 63 x is an example of ____________ inequality.

12. 35 is read as 5 is _____________ than 3.


______________________________________________________________________________


TEST YOURSELF B:

(a)

(b)

(c)

(d)

TEST YOURSELF C:

(1) 6m (2) 6x (3) 19x (4) 5.1d (5) 6m

(6) 24y (7) 4d (8) 3p (9) 2

5m (10) 5x

TEST YOURSELF D1:

(1) 7p (2) 3x (3) 3c (4) 5p (5) 8d

(6) 4x (7) 4x (8) 5y (9) 8m (10) 2

9b

TEST YOURSELF D2:

(1) 24d (2) 16n (3) 50y (4) 42b (5) 96x 48 (6) x

0 − 2 − 3

x

1 2 3 − 1

0 − 2 − 3

x

1 2 3 − 1

0 − 2 − 3

x

1 2 3 − 1

0 − 2 − 3

x

1 2 3 − 1

ANSWERS


______________________________________________________________________________


TEST YOURSELF E1:

1. 5 )( ma 8 )( xb 1 )( mc

2. 4 )( ma 21 )( mb 2

9 )( xc

3. 1

( ) 1 ( ) 4 (c) 2

a m b m m

4. ( ) 10 (b) 1 (c) 8 (d) 3 (e) 3 (f) 2 (g) 25 (h) 10a x m x x p x x p

TEST YOURSELF E2:

(1) 6x (2) 1m (3) 13x (4) 9p (5) 14m

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

Unit 1:

Negative Numbers

UNIT 8

TRIGONOMETRY





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.






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.

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.





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.

LESSON NOTES

θ





Example 1:

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:

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.

EXAMPLES

θ

θ





Identify the opposite, adjacent and hypotenuse sides of the following right-angled triangles.

1.

Opposite side =

Adjacent side =

Hypotenuse side =

2.

Opposite side =

Adjacent side =

Hypotenuse side =

3.

Opposite side =

Adjacent side =

Hypotenuse side =

4.

Opposite side =

Adjacent side =

Hypotenuse side =

5.

Opposite side =

Adjacent side =

Hypotenuse side =

6.

Opposite side =

Adjacent side =

Hypotenuse side =

TEST YOURSELF A





PART B:

TRIGONOMETRY II


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.

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.





Definition of the Three Trigonometric Functions

(i) sin = opposite side

hypotenuse side

(ii) cos = adjacent side

hypotenuse side

(iii) tan = opposite side

adjacent side

sin = opposite side

hypotenuse side

= AB

AC

cos = adjacent side

hypotenuse side =

BC

AC

tan = opposite side

adjacent side=

AB

BC

LESSON NOTES

Acronym:

SOH:

Sine – Opposite - Hypotenuse

Acronym:

CAH:

Cosine – Adjacent - Hypotenuse

Acronym:

TOA:

Tangent – Opposite - Adjacent

θ





Example 1:

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.

Thus sin = opposite side

hypotenuse side =

AB

AC

cos = adjacent side

hypotenuse side =

BC

AC

tan = opposite side

adjacent side =

AB

BC

EXAMPLES

θ





Example 2:

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.

Thus, sin = opposite side

hypotenuse side =

WU

TW

cos = adjacent side

hypotenuse side =

TU

TW

tan = opposite side

adjacent side =

WU

TU

You have to identify the

opposite, adjacent and

hypotenuse sides.

θ





Write the ratios of the trigonometric functions, sin , cos and tan , for each of the diagrams

below:

1.

sin =

cos =

tan =

2.

sin =

cos =

tan =

3.

sin =

cos =

tan =

4.

sin =

cos =

tan =

5.

sin =

cos =

tan =

6.

sin =

cos =

tan =

TEST YOURSELF B

θ

θ

θ

θ

θ

θ

θ





PART C:

TRIGONOMETRY III


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.

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.





Find the angle in degrees and minutes.

Example 1:

sin = 2

5

o

h

= sin-1

2

5

= 23o 34 4l

= 23o 35

(Note that 34 41 is rounded off to 35)

Example 2:

cos = a

h =

3

5

= cos-1

3

5

= 53o 7 48

= 53o 8

(Note that 7 48 is rounded off to 8)

Since sin = opposite

hypotenuse

then = sin-1

opposite

hypotenuse

Since cos = adjacent

hypotenuse

then = cos-1 adjacent

hypotenuse

Since tan = opposite

adjacent

then = tan-1

opposite

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.)

LESSON NOTES

EXAMPLES

θ

θ





Example 3:

tan = o

a =

7

6

= tan-1

7

6

= 49o 23 55

= 49o 24

Example 4:

cos = a

h =

5

7

= cos-1

5

7

= 44o 24 55

= 44o 25

Example 5:

sin = o

h =

4

7

= sin-1

4

7

= 34o 50 59

= 34o 51

Example 6:

tan = o

a =

5

6

= tan-1

5

6

= 39o 48 20

= 39o 48

θ

θ

θ

θ





Find the value of in degrees and minutes.

1.

2.

3.

4.

5.

6.

TEST YOURSELF C

θ θ

θ

θ

θ

θ





PART D:

TRIGONOMETRY IV


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.

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.





Find the length of PR.

With reference to the given angle, PR is the

opposite side and QR is the adjacent side.

Thus tangent ratio is used to form the

relation of the sides.

tan 50o =

5

PR

PR = 5 tan 50o

Find the length of TS.

With reference to the given angle, TR is the

adjacent side and TS is the hypotenuse

side.

Thus cosine ratio is used to form the

relation of the sides.

cos 32o =

8

TS

TS cos 32o = 8

TS = 8

cos32o

LESSON NOTES





Find the value of x in each of the following.

Example 1:

tan 25o =

3

x

x = 3

tan 25o

= 6.434 cm

Example 2:

sin 41.27o =

5

x

x = 5 sin 41.27o

= 3.298 cm

Example 3:

cos 34o 12 =

6

x

x = 6 cos 34o 12

= 4.962 cm

Example 4:

tan 63o =

9

x

x = 9 tan 63o

= 17.66 cm

EXAMPLES





Find the value of x for each of the following.

1.

2.

3.

4.

5.

6.

TEST YOURSELF D

10 cm

6 cm

13 cm





PART E:

TRIGONOMETRY V


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.

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.





In the diagram, with reference to the angle , PR is the opposite side, OP is the adjacent side

and OR is the hypotenuse side.

r

y

OR

PR

hypotenuse

oppositesin

r

x

OR

OP

hypotenuse

adjacentcos

x

y

OP

PR

adjacent

oppositetan

LESSON NOTES

θ





PART F:

TRIGONOMETRY VI


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.

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.





First Quadrant

sin = y

r (Positive)

cos = x

r(Positive)

tan = y

x(Positive)

(All trigonometric ratios are positive in the

first quadrant)

Second Quadrant

sin = y

r (Positive)

cos = x

r

(Negative)

tan = y

x(Negative)

(Only sine is positive in the second

quadrant)

Third Quadrant

sin = y

r

(Negative)

cos = x

r

(Negative)

tan = y y

x x

(Positive)

(Only tangent is positive in the third

quadrant)

Fourth Quadrant

sin = y

r

(Negative)

cos = x

r (Positive)

tan = y

x

(Negative)

(Only cosine is positive in the fourth

quadrant)

LESSON NOTES

θ θ

θ θ





Using acronym: Add Sugar To Coffee (ASTC)

sin is positive

sin is negative

cos is positive

cos is negative

tan is positive

tan is negative

A – All positive

C – only cos is positive T – only tan is positive

S – only sin is positive





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

TEST YOURSELF F





PART G:

TRIGONOMETRY VII


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.

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





The Pythagoras Theorem:

(a) 3, 4, 5 or equivalent (b) 5, 12, 13 or equivalent (c) 8, 15, 17 or equivalent

The sum of the squares of two sides of

a right-angled triangle is equal to the

square of the hypotenuse side.

PR2 + QR

2 = PQ

2

LESSON NOTES





1. Write the values of sin , cos and tan

from the diagram below.

OA2 = (−6)

2 + 8

2

= 100

OA = 100

= 10

sin = 8 4

10 5

y

r

cos = 6 3

10 5

x

r

tan = 8 4

6 3

y

x

2. Write the values of sin , cos and tan

from the diagram below.

OB2 = (−12)

2 + (−5)

2

= 144 + 25

= 169

OB = 169

= 13

sin = 5

13

y

r

cos = 12

13

x

r

tan = 5 5

12 12

EXAMPLES

θ θ





Write the value of the trigonometric ratios from the diagrams below.

1.

sin =

cos =

tan =

2.

sin =

cos =

tan =

3.

sin =

cos =

tan =

4.

sin =

cos =

tan =

5.

sin =

cos =

tan =

6.

sin =

cos =

tan =

TEST YOURSELF G

θ θ θ

θ

θ

θ θ

B(5,4)

B(5,12)

x

y





PART H:

TRIGONOMETRY VIII


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.

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.





(a) y = sin x

The domain for x can be from 0o to 360

o or 0 to 2 in radians.

Important points: (0, 0), (90o, 1), (180

o, 0), (270

o, −1) and (360

o, 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), (90

o, 0), (180

o, −1), (270

o, 0) and (360

o, 1)

Important features: Maximum point (0o, 1) and (360

o, 1),

Maximum value = 1 Minimum point (180o, −1)

Minimum value = 1

LESSON NOTES





(c) y = tan x

Important points: (0o, 0), (180

o, 0) and (360

o, 0)

Is there any

maximum or

minimum point

for the tangent

graph?





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)

A(0,1)

TEST YOURSELF H

y = cos x y = sin x





TEST YOURSELF A:

1. Opposite side = AB

Adjacent side = AC

Hypotenuse side = BC

2. Opposite side = PQ

Adjacent side = QR

Hypotenuse side = PR

3. Opposite side = YZ

Adjacent side = XZ

Hypotenuse side = XY

4. Opposite side = LN

Adjacent side = MN

Hypotenuse side = LM

5. Opposite side = UV

Adjacent side = TU

Hypotenuse side = TV

6. Opposite side = RT

Adjacent side = ST

Hypotenuse side = RS

TEST YOURSELF B:

1. sin = AB

BC

cos = AC

BC

tan = AB

AC

2. sin = PQ

PR

cos = QR

PR

tan = PQ

QR

3. sin = YZ

YX

cos = XZ

XY

tan = YZ

XZ

4. sin = LN

LM

cos = MN

LM

tan = LN

MN

5. sin = UV

TV

cos = UT

TV

tan = UV

UT

6. sin = RT

RS

cos = ST

RS

tan = RT

TS

ANSWERS





TEST YOURSELF C:

1. sin = 1

3

= sin-1

1

3 = 19

o 28

2. cos = 1

2

= cos-1

1

2 = 60

o

3. tan = 5

3

= tan-1

5

3 = 59

o 2

4. cos = 5

8

= cos-1

5

8 = 51

o 19

5. tan = 7.5

9.2

= tan-1

7.5

9.2 = 39

o 11

6. sin = 6.5

8.4

= sin-1

6.5

8.4= 50

o 42

TEST YOURSELF D:

1. tan 32o =

4

x

x = 4

tan 32o = 6.401 cm

2. sin 53.17o =

7

x

x = 7 sin 53.17o = 5.603 cm

3. cos 74o 25 =

10

x

x = 10 cos 74o 25

= 2.686 cm

4. sin 551

3

o

= 6

x

x = 13

6

sin55o

= 7.295 cm

5. tan 47o =

13

x

x = 13 tan 47o = 13.94 cm

6. cos 61o =

10

x

x = 10

cos61o= 20.63 cm





TEST YOURSELF F:

1. 1ST

and 2nd

2. 1st and 3

rd

3. 2nd

and 3rd

4. 1st and 4

th

5. 3rd

and 4th

6. 2nd

and 4th

TEST YOURSELF G:

1. sin = 4

5

cos = 3

5

tan = 4

3

2. sin = 12

13

cos = 5

13

tan = 12

5

3. sin = 4

5

cos = 3

5

tan = 4

3

4. sin = 4

5

cos = 3

5

tan = 4

3

5. sin = 8

17

cos = 15

17

tan = 8

15

6. sin = 5

13

cos = 12

13

tan = 5

12





TEST YOURSELF H:

1.

y = tan x y = sin x y = cos x

2. (a) A (0, 1), B (90o, 0), C (180

o, 1), D (270

o, 0)

(b) P (90o, 1), Q (180

o, 0), R (270

o, 1), S (360

o, 0)

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