BEAMS_Unit 4 Linear Equations

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Unit 1:

Negative Numbers

UNIT 4

LINEAR EQUATIONS

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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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

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1Curriculum Development Division Ministry 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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2Curriculum Development Division Ministry of Education Malaysia

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 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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3Curriculum Development Division Ministry of Education Malaysia 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

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4Curriculum Development Division Ministry of Education Malaysia 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

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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).

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

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6Curriculum Development Division Ministry of Education Malaysia 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.

PART B:

SOLVING LINEAR EQUATIONS IN

THE FORMS OF

x + a = b AND x – a = b

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

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7Curriculum Development Division Ministry of Education Malaysia PART B:

SOLVING LINEAR EQUATIONS IN THE FORM OF

x + a = b OR x – a = b

Solve the following equations.

(i) x25 (ii) x 3 5 Solutions: (ii) x 3 5 x – 3 + 3 = 5 + 3 x = 5 + 3 x = 8 (i) x25 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 2 5 5 2      x x x Alternative Method: 8 3 5 5 3      x x x Simplify the LHS. Simplify the RHS. Simplify the RHS. EXAMPLES

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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

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PART C:

SOLVING LINEAR EQUATIONS IN

THE FORMS OF

ax = b AND

_a

x



b

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 b a x b)  (

Pupils face difficulty when solving linear equations in one unknown by solving equations in the form of:

(a) ax = b b a x b)  (

Strategy:

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10Curriculum Development Division Ministry of Education Malaysia PART C:

SOLVING LINEAR EQUATION

ax = b AND b a

x _

(i) 3m = 12 (ii) 4 3  m Solutions: (i) 3m = 12 3 12 3 3 m   3 12  m m = 4 (ii) 4 3  m 3 4 3 3   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 12 3    m m m Alternative Method: 12 4 3 4 3     m m m Simplify the RHS. EXAMPLES

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1. 2p = 6 2. 5k = – 20 3. – 4h = 24 4. 7l56 5. 8j72 6. 5n60 7. 6v72 8. 7y42 9. 12z96 10. 4 2  m 11. 4 r = 5 12. 8 w = –7 13. 8 8   t 14. 9 12 s 15. 6 5  u TEST YOURSELF C

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12Curriculum Development Division Ministry of Education Malaysia 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.

PART D:

SOLVING LINEAR EQUATIONS IN

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:

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13Curriculum Development Division Ministry of Education Malaysia 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 2 2 14 2x _ 2 14  x x = 7 Method 2 2x311 2 2 2 11 3 2   x 2 11 2 3   x 2 3 2 3     2 11 2 3 x 2 14  x x7

Add 3 to both sides of 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. Add 2 3 to both sides of the equation.

Simplify both sides of the equation. Alternative Method: 2 2 14 14 2 3 11 2 11 3 2        x x x x x Alternative Method: 7 2 14 2 3 2 11 2 11 2 3 2 2 11 3 2         x x x x x Simplify the RHS. EXAMPLES

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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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PART E

SOLVING LINEAR EQUATIONS IN

THE FORM OF

c

b

a

x





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 b

a 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 b

a x

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

Strategy:

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16Curriculum Development Division Ministry of Education Malaysia PART E:

SOLVING LINEAR EQUATIONS IN THE FORM OF b c

a x _ _

Solve the equation 4 1

3  x . Solution: Method 1 4 1 3   x 44 3 x = 1 + 4 5 3  x 353 3 x 3 5  x x = 15 Method 2 3 3      4 1 3 x 3 4 3 1 3 3     x x123 x – 12 + 12 = 3 + 12 x312 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. Multiply both sides of

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

the equation. Add 12 to both sides of

the equation. Simplify both sides of

the equation. Alternative Method: 15 5 3 5 3 4 1 3 1 4 3         x x x x x EXAMPLES

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1. 3 5 2   m 2. 2 1 3  b 3. 2 7 3  k 4. 3 + 2 h = 5 5. 4 + 5 h = 6 6. 1 2 4   m 7. 5 4 2h  8. 6 k + 3 = 1 9. 2 5 3h  10. 3 – 2m = 7 11. 7 2 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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19Curriculum Development Division Ministry of Education Malaysia PART F:

FURTHER PRACTICE

Solve the following equations:

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

Subtract 2x from both sides of the equation. Simplify 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. Simplify both sides of the equation.

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

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

Add 5 to both sides of the equation. EXAMPLES

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20Curriculum Development Division Ministry of Education Malaysia (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 1 1

Expand both sides of the equation. Simplify the LHS.

Subtract 2n from both sides of the equation.

Add 4 to both sides of the equation.

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

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21Curriculum Development Division Ministry of Education Malaysia 3 7 21 7 7 21 7 3 18 3 3 7 18 3 7 18 3 3 6 4 18 ) 1 ( 3 ) 3 2 ( 2 ) 3 ( 6 2 1 6 3 3 2 6 ) 3 ( 6 2 1 3 3 2 6 3 2 1 3 3 2                                          _      x x x x x x x x x x x x x x x

Add 3 to both sides of the equation.

Alternative Method: 3 7 21 21 7 3 18 7 18 3 7 18 3 3 6 4 18 ) 1 ( 3 ) 3 2 ( 2 6 3 2 1 3 3 2 6 3 2 1 3 3 2                         _      x x x x x x x x x x x x x (iii) Simplify LHS. Expand the brackets.

Multiply both sides of the equation by the LCM.

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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  2 5 3 . 6 x x  6 13 5 2 . 7 y  y 2 9 4 1 3 2 . 8 x  x  0 8 4 3 6 5 2 . 9 x  x  12 7 4 9 7 2 . 10 x   x TEST YOURSELF F

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23Curriculum Development Division Ministry of Education Malaysia 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

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24Curriculum Development Division Ministry of Education Malaysia TEST YOURSELF F: 1. x = − 2 2. x = − 17 3. 14 1  n 4. x = 6 5. x = 3 6. x = 15 7. y = 3 8. x = 7 9. x = −8 10. x = 19