BEAMS_Unit 4 Linear Equations

8/9/2019 BEAMS_Unit 4 Linear Equations

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Unit 1:Negative Numbers

UNIT 4

LINEAR EQUATIONS

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


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TABLE OF CONTENTS

Module Overview 1

Part A: Linear Equations 2

Part B: Solving Linear Equations in the Forms ofx + a = b and xa = b 6

Part C: Solving Linear Equations in the Forms ofax = b anda

x= b 9

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

Part E: Solving Linear Equations in the Form ofa

x+ b = c 15

Part F: Further Practice on Solving Linear Equations 18

Answers 23


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Basic Essentials Additional Mathematics (BEAMS) Module

UNIT 4: Linear Equations

1Curriculum Development DivisionMinistry of Education Malaysia

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.


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PART A:

LINEAR EQUATIONS

LEARNING OBJECTIVES

Upon completion ofPart A, pupils will be able to:

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 unknownby determining if a numerical value is a solution of a given linear

equation in one unknown.

TEACHING AND LEARNING STRATEGIES

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.


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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)

83 = (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


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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, wherex is an unknown

A linear equation in one unknown is an equation that consists ofonly 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


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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).

xx

x + 2 2 = 5 2

x = 3

x + 2 = 5

(a x + 2 = 5

x = ?

x x

x + 2 = 5 x + 22 = 52

x = 3

x= 3


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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 andx 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 xa =b

LEARNING OBJECTIVES

Upon completion ofPart 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) xa = b

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


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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

x3 + 3 = 5 + 3

x = 5 + 3

x = 8

(i) 52 x

x + 22 = 52

x = 52

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


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1. x + 1 = 6 2. x2 = 4 3. x7 = 2

4. 7 +x = 5 5. 5 +x = 2 6. 9 +x =12

7. 12 +x = 36 8. x9 =54 9. 28 +x =78

10. x + 9 =102 11. 19 +x = 38 12. x5 =92

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

TEST YOURSELF B


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PART C:

SOLVING LINEAR EQUATIONS INTHE FORMS OF

ax = b AND ba

x

LEARNING OBJECTIVES

Upon completion ofPart C, pupils will be able to understand the concept ofsolutions of linear equations in one unknown by solving equations in the

form of:(a)ax = b

ba

xb )(



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

equations in the form of:

(a)ax = b

ba

xb )(


Strategy:





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PART C:

SOLVING LINEAR EQUATION

ax = b AND ba

x


(i) 3m = 12 (ii)

43 m

Solutions:

(i) 3 m = 12

3 12

3 3

m

3

12m

m = 4

(ii) 43

m

3433

m

m = 4 3

m = 12

Divide both sides ofthe 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


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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


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LEARNING OBJECTIVE

Upon completion ofPart D, pupils will be able to understand the concept ofsolutions of linear equations in one unknown by solving equations in the

form ofax + 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 andx is an unknown.

Strategy:

Teacher should emphasise the idea of balancing the linear equations. When pupilshave mastered the skills and concepts involved in solving linear equations, they



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PART D:

SOLVING LINEAR EQUATIONS IN THE FORM OFax + b = c

Solve the equation 2x3 = 11.

Solution:

Method 1

2x3 = 11

2x3 + 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.

Divide both sides of

the equation by 2.

Simplify the LHS.

Divide both sides of

the equation by 2.

Simplify the LHS.

Add2

3to 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


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1. 2m + 3 = 7 2. 3p1 = 11 3. 3k+ 4 = 10

4. 4m3 = 9 5. 4y + 3 = 9 6. 4p + 8 = 11

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

10. 43p = 7 11. 102p = 4 12. 82m = 6

TEST YOURSELF D


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PART E


THE FORM OF

cba

x

LEARNING OBJECTIVES

Upon completion ofPart 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 andx 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 andx is an unknown.

Strategy:





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PART E:

SOLVING LINEAR EQUATIONS IN THE FORM OF cb

a

x

Solve the equation 143

x

.

Solution:

Method 1

143

x

4 43

x= 1 + 4

53

x

33 53x

35x

x = 15

Method 2

33

14

3

x

313433

x

312x

x12 + 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


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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. 32m = 7 11. 72

3 m 12. 12 + 5h = 2

TEST YOURSELF E


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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:





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PART F:

FURTHER PRACTICE

Solve the following equations:

(i) 4x5 = 2x + 7

Solution:

Method 1

2

126

126

756

756

7254

x

x

x

x

x

xx

66

55

Method 2

7254 xx

4x5 + 5 = 2x + 7 + 5

4x = 2x + 12

4x2x = 2x2x + 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 by6.

Add 5 to both sides of the equation.


Subtract 2x from both sides of the equation.



Alternative Method:

2

6

12

126

5724

7254

x

x

x

xx

xx

4x2x5 = 2x2x+ 7


EXAMPLES


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(ii) 3(n2)2(n1) = 2 (n + 5)

3n62n + 2 = 2n + 10

n4 = 2n + 10

n2n4 = 2n2n + 10

n4 = 10

n4 + 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



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3

7

21

7

7

217

318337

1837

183364

18)1(3)32(2

)3(62

16

3

326

)3(62

1

3

32

6

32

1

3

32

x

x

x

x

x

xx

xx

xx

xx

xx


Alternative Method:

37

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 theLCM.

Divide both sides of the equation by 7.


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1. 4x5 + 2x = 8x3x 2. 4(x2)3(x1) = 2 (x + 6)

3. 3(2n5) = 2(4n + 7)

2

9

4

3

.4

x

6

5

3

2

2.5

x

2

53.6

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


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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

35. y

8. k= 2

11. p = 3

15. u = 30

3. k= 2

4

36. 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


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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

BEAMS_Unit 4 Linear Equations

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