# BEAMS_Unit 4 Linear Equations

August 1, 2017 | Author: rea0079 | Category: Equations, Algebra, Arithmetic, Curriculum, Mathematical Concepts

#### Description

UNIT 4 LINEAR EQUATIONS Unit 1: Negative Numbers

Curriculum Development Division Ministry of Education Malaysia

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

Part D:

Solving Linear Equations in the Form of ax + b = c

12

Part E:

Solving Linear Equations in the Form of

x +b=c a

15

Part F:

Further Practice on Solving Linear Equations

x =b a

9

18 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.

Curriculum Development Division Ministry of Education Malaysia

1

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

PART A: LINEAR EQUATIONS

LEARNING OBJECTIVES Upon completion of Part 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 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 TEACHING AND LEARNING STRATEGIES 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.

Curriculum Development Division Ministry of Education Malaysia

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

OVERALL LESSON NOTES

GUIDELINES: 1.

2.

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. Take note of the following properties of equality:

(a) Subtraction Arithmetic

Algebra

8 = (4) (2)

a=b

8 – 3 = (4) (2) – 3

a–c=b–c

Algebra

8 = (4) (2)

a =; b

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

a+c=b+c

Arithmetic

Algebra

8=6+2

a=b

(c) Division

8 62  3 3

a b  c c

c≠0

(d) Multiplication Arithmetic

Algebra

8 = (6 +2)

a=b

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

ac = bc

Curriculum Development Division Ministry of Education Malaysia

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

PART A: LINEAR EQUATIONS LESSON NOTES

1.

An equation shows the equality of two expressions and is joined by an equal sign. Example:

2.

2  4=7+1

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

x + 1 = 3,

where x is an unknown

A linear equation in one unknown is an equation that consists of only one unknown. 3.

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

4.

When solving equations, (i) always write each step on a new line; (ii) keep the left hand side (LHS) and the right hand side (RHS) balanced by:  adding the same number or term to both sides of the equation;  subtracting the same number or term from both sides of the equations;  multiplying both sides of the equation by the same number or term;  dividing both sides of the equation by the same number or term; and (iii) simplify (whenever possible).

5.

When pupils have mastered the skills and concepts involved in solving linear equations, they can solve the questions by using alternative method. What is solving an equation?

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

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

The puzzle can be visualised by using real life and concrete examples. 1. The equality in an equation can be visualised as the state of equilibrium of a balance. (a) x + 2 = 5

x=3

x=? 2.

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

x

xx

xx + + 22 == 55

x + 2x –+ 2 –=25= –5 2– 2 x =3 3 x=

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

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

AND x – a = b

LEARNING OBJECTIVES Upon completion of Part B, pupils will be able to understand the concept of solutions of linear equations in one unknown by solving equations in the form of: (i) x+a=b (ii) x – a = b where a, b, c are integers and x is an unknown.

TEACHING AND LEARNING STRATEGIES 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.

Curriculum Development Division Ministry of Education Malaysia

6

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

PART B: SOLVING LINEAR EQUATIONS IN THE FORM OF

x+a=b

OR

x–a=b

EXAMPLES

Solve the following equations. (i) x  2  5

(ii) x  3  5

Solutions:

(i)

x25

Subtract 2 from both sides of the equation.

x+2–2=5–2 x=5–2 x=3

(ii)

Simplify the LHS.

Alternative Method:

x25 x 52 x3

Simplify the RHS.

x35

x–3+3=5+3

Add 3 to both sides of the equation.

Alternative Method:

x 35

x=5+3

Simplify the LHS.

x 53

x=8

Simplify the RHS.

x 8

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7

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

TEST YOURSELF B

Solve the following equations. 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

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

PART C: SOLVING LINEAR EQUATIONS IN THE FORMS OF ax = b

AND

x b a

LEARNING OBJECTIVES Upon completion of Part C, pupils will be able to understand the concept of solutions of linear equations in one unknown by solving equations in the form of: (a) ax = b x (b)  b a where a, b, c are integers and x is an unknown.

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

Curriculum Development Division Ministry of Education Malaysia

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

PART C: SOLVING LINEAR EQUATION

ax = b AND

x b a

EXAMPLES

Solve the following equations. (i) 3m = 12

(ii)

m 4 3

Solutions:

(i)

3  m = 12 3  m 12  3 3 m

12 3

m=4

(ii)

Divide both sides of the equation by 3. Simplify the LHS.

Alternative Method:

3m  12 12 m 3 m4

Simplify the RHS.

m 4 3 m 3  43 3

Multiply both sides of the equation by 3.

m = 4 3

Simplify the LHS.

m = 12

Simplify the RHS.

Alternative Method:

m 4 3 m  3 4 m  12

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

TEST YOURSELF C Solve the following equations. 1.

2p = 6

2.

5k = – 20

3.

– 4h = 24

4.

7l  56

5.

 8 j  72

6.

 5n  60

7.

6v  72

8.

7 y  42

9.

12z  96

10.

m 4 2

11.

r =5 4

12.

13.

14.

s 9 12

15.

t 8 8

Curriculum Development Division Ministry of Education Malaysia

w = –7 8

u  6 5

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

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

LEARNING OBJECTIVE Upon completion of Part D, pupils will be able to understand the concept of solutions of 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.

TEACHING AND LEARNING STRATEGIES 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: 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.

Curriculum Development Division Ministry of Education Malaysia

12

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

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

Solve the equation 2x – 3 = 11. Solution: Method 1 2x – 3 = 11 2x – 3 + 3 = 11 + 3 2x = 14 2 x 14  2 2 x

14 2

x=7

Add 3 to both sides of the equation.

Alternative Method:

2 x  3  11 Simplify both sides of the equation. Divide both sides of the equation by 2.

2 x  11  3 2 x  14 14 2 x2 x

Simplify the LHS. Simplify the RHS.

Method 2 2x  3  11

2 x 3 11   2 2 2

x x

3 11  2 2

3 3 11 3    2 2 2 2

Divide both sides of the equation by 2. Simplify the LHS.

3 to both sides 2

of the equation.

14 x 2 x7

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Alternative Method:

2 x  3  11 2 x 3 11   2 2 2 11 3 x  2 2 14 x 2 x7

Simplify both sides of the equation.

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

TEST YOURSELF D

Solve the following equations.

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

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

PART E SOLVING LINEAR EQUATIONS IN THE FORM OF

x bc a

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

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

Strategy: 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.

Curriculum Development Division Ministry of Education Malaysia

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

PART E: SOLVING LINEAR EQUATIONS IN THE FORM OF

x bc a

EXAMPLES

Solve the equation

x  4  1. 3

Solution: Method 1 x  4 1 3 x 44 = 1 + 4 3 x 5 3

x  3  5 3 3 x  5 3

x = 15

Add 4 to both sides of the equation. Simplify both sides of the equation. Multiply both sides of the equation by 3. Simplify both sides of the equation.

Alternative Method:

x  4 1 3 x 1 4 3 x 5 3 x  3 5 x  15

Method 2 x    4   3  1 3 3 

Multiply both sides of the equation by 3.

x  3  4  3  1 3 3

Expand the LHS.

x  12  3

x – 12 + 12 = 3 + 12 x  3  12 x  15

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Simplify both sides of the equation. Add 12 to both sides of the equation. Simplify both sides of the equation.

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

TEST YOURSELF E

Solve the following equations. 2.

b 2 1 3

h =5 2

5.

4+

h 5 4

1.

m 35 2

4.

3+

7.

2

10.

3 – 2m = 7

3.

k 27 3

h =6 5

6.

m 1  2 4

8.

k +3=1 6

9.

3

11.

3

12.

12 + 5h = 2

Curriculum Development Division Ministry of Education Malaysia

m 7 2

h 2 5

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

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.

TEACHING AND LEARNING STRATEGIES Pupils face difficulty when solving linear equations of various forms. 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.

Curriculum Development Division Ministry of Education Malaysia

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

PART F: FURTHER PRACTICE EXAMPLES

Solve the following equations: (i) – 4x – 5 = 2x + 7

Alternative Method:

 4x  5  2x  7  4x  2x  7  5  6 x  12

Solution:

12 6 x  2 x

Method 1  4x  5  2x  7

–4x – 2x – 5 = 2x – 2x + 7  6x  5  7  6x  5  5  7  5  6 x  12  6 x 12  6 6 x  2

Subtract 2x from both sides of the equation. Simplify both sides of the equation. Add 5 to both sides of the equation. Simplify both sides of the equation. Divide both sides of the equation by –6.

Method 2  4x  5  2x  7

– 4x – 5 + 5 = 2x + 7 + 5 – 4x = 2x + 12 – 4x – 2x = 2x – 2x + 12

Add 5 to both sides of the equation. Simplify both sides of the equation. Subtract 2x from both sides of the equation.

– 6x = 12

 6 x 12  6 6 x  2

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Simplify both sides of the equation. Divide both sides of the equation by – 6.

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

(ii) 3(n – 2) – 2(n – 1) = 2 (n + 5) Expand both sides of the equation.

3n – 6 – 2n + 2 = 2n + 10 n – 4 = 2n + 10

Simplify the LHS.

n – 2n – 4 = 2n – 2n + 10

Subtract 2n from both sides of the equation.

– n – 4 = 10 – n – 4 + 4 = 10 + 4

Add 4 to both sides of the equation.

– n = 14

 n 14  1 1 n  14

Divide both sides of the equation by – 1.

Alternative Method:

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

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

(iii)

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

Multiply both sides of the equation by the LCM.

Expand the brackets. Simplify LHS. Add 3 to both sides of the equation. Divide both sides of the equation by 7.

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

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

TEST YOURSELF F Solve the following equations. 1.

4x – 5 + 2x = 8x – 3 – x

2.

4(x – 2) – 3(x – 1) = 2 (x + 6)

3.

–3(2n – 5) = 2(4n + 7)

4.

3x 9  4 2

5.

x 2 5   2 3 6

6.

x x  2 3 5

7.

y 13 y 5  2 6

8.

x  2 x 1 9   3 4 2

9.

2 x  5 3x  4  0 6 8

10.

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2x  7 x7 4 9 12

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

TEST YOURSELF B: 1.

x=5

2.

x=6

3.

x=9

4.

x = –2

5.

x = –7

6.

x = –3

7.

x = 48

8.

x = –45

9.

x = –50

10.

x = –111

11.

x = 57

12.

x = –87

13.

x = –107

14.

x = 247

15.

x = –115

TEST YOURSELF C: 1.

p=3

2.

k=–4

3. h = –6

4.

l=8

5.

j=–9

6. n = 12

7.

v = 12

8.

y=–6

9.

10.

m=8

11. r = 20

12. w = – 56

13.

t = – 64

14. s = 108

15. u = 30

3.

z=8

TEST YOURSELF D: 1.

m=2

2.

4.

m=3

5. y 

7.

p=2

8. k = 2

9.

11. p = 3

12. m = 1

10. p = −1

p=4 3 2

k=2

6. p 

3 4

x = –1

TEST YOURSELF E: 1.

m=4

10. b = 9

11. k = 15

4.

h=4

5.

h = 10

6.

m = 12

7.

h = 12

8.

k = −12

9.

h=5

10. m = −2

11. m = −8

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12. h = −2

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

TEST YOURSELF F: 1.

x=−2

2.

x = − 17

3. n 

5.

x=3

6.

x = 15

7.

9.

x = −8

10.

x = 19

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y=3

4.

x=6

8.

x=7

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