beams_unit 3 algebraic expressions and algebraic formulae

Unit 1:

Negative Numbers

UNIT 3

ALGEBRAIC EXPRESSIONS

AND

ALGEBRAIC FORMULAE

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

Curriculum Development Division

Ministry of Education Malaysia

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

Unit 3: Algebraic Expressions and Algebraic Formulae

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.





TEACHING AND LEARNING STRATEGIES

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.

Simplify by canceling out the common

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

Simplify by canceling out the common

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

beams_unit 3 algebraic expressions and algebraic formulae

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