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Press the ‘Esc’ key to exit at any time PASS WITH PEARSON Maths Achievement Standard 90147 Use straightforward algebraic methods and solve equations (1.1)

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Presentation on theme: "Press the ‘Esc’ key to exit at any time PASS WITH PEARSON Maths Achievement Standard 90147 Use straightforward algebraic methods and solve equations (1.1)"— Presentation transcript:

1 Press the ‘Esc’ key to exit at any time PASS WITH PEARSON Maths Achievement Standard Use straightforward algebraic methods and solve equations (1.1) Gwenda Hill © Pearson Education New Zealand 2005 Start How to use this CD and Legal Notices

2 Algebra Main Menu Achievement MeritExcellence Press to go to desired section. Exit Title Page This CD consists of: 1.A Main Menu (this page) returned to by pressing: 2.Three Sections within the main menu accessed by: 3.Exercise groups within each section accessed by: 4.Questions within each exercise group accessed by: Back to Main Menu Back to ‘Section’ Menu Exercise ‘1’ Q1.

3 Achievement Menu Press the button to go to the menu page for each exercise group. Exercise A9. Exercise A11. Exercise A13. Exercise A7. Exercise A5. Exercise A3. Exercise A1. Exercise A15. Exercise A10. Exercise A12. Exercise A14. Exercise A8. Exercise A6. Exercise A4. Exercise A2. Exercise A16. Like terms More simplifying Expand and simplify Factorising quadratics Algebraic exponent factors Finding linear rule Solve linear equations Solve factorised quadratics Simplifying Expand and simplify Factorise common factors Algebraic exponent powers Substitution More linear rules More linear equations Achievement review Back to Main Menu

4 Q5. Q6. Q7. Q4. Q3. Q2. Q1. Exercise A1 Menu Like Terms Are the following pairs like or unlike terms?4 a, 3 a Are the following pairs like or unlike terms?3 x, 2 x 2 Are the following pairs like or unlike terms?4 a 2, 3 a 3 Are the following pairs like or unlike terms?5 p, 3 x p Select the like terms to 6 x from:5 x, 7, - 2 x, x 2, 4 xy Select the like terms to 2 a 2 from:4 a, 4 a 4, - 2 a 2, 5 a 2, 45 a 2 y Select the like terms to 7 from: 6 x, - 7, 4, 2 a 5, 1 / 4 Answer each question then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu Next

5 Exercise A1 Question 1 1. Are the following pair like or unlike terms?4 a, 3 a Only variable is a, Both have power 1 so Like Next Back to Exercise A1 Menu

6 Exercise A1 Question 2 2. Are the following pair like or unlike terms? 3 x, 2 x 2 Only variable is x, but One is x and other is x 2 so Unlike NextPrevious Back to Exercise A1 Menu

7 Exercise A1 Question 3 3.Are the following pair like or unlike terms? 4 a 2, 3 a 3 Only variable is a, One has power 2 and the other 3, so Unlike NextPrevious Back to Exercise A1 Menu

8 Exercise A1 Question 4 4. Are the following pair like or unlike terms?5 p, 3  p Only variable is p and 3  p = 3 p Both have power 1, so Like NextPrevious Back to Exercise A1 Menu

9 Exercise A1 Question 5 5.Select the like terms to 6 x from: 5 x, 7, - 2 x, x 2, 4 xy Only variable is x, 5 x, - 2 x, x 2, 4 xy have x x 2 has power 2 not 1, 4 xy also has y so 5 x and - 2 x NextPrevious Back to Exercise A1 Menu

10 Exercise A1 Question 6 6.Select the like terms to 2 a 2 from: 4 a, 4 a 4, - 2 a 2, 5 a 2, 45 a 2 y Only variable is a with power 2, - 2 a 2, 5 a 2 and 45 a 2 y have a 2 45 a 2 y also has y so - 2 a 2 and 5 a 2 NextPrevious Back to Exercise A1 Menu

11 Exercise A1 Question 7 7.Select the like terms to 7 from: 6 x, - 7, 4, 2 a 5, 1 / 4 No variable – just a constant - 7, 4 and 1 / 4 are the constants so like terms are: - 7, 4 and 1 / 4 Previous Back to Exercise A1 Menu Next Menu

12 Exercise A2 Menu Simplifying expressions 4 a + 3 a 4 a a 3 4 a 2 – 2 a a 2 5 x + 6 y + 3 x – 2 y Q1. Q8. 3 x – 2 x 5 p – 3 p + 7 p 6 x – 7 – 4 x x + 2 x 2 – 3 x Q3. Q5. Q7. Q2. Q4. Q6. Back to Achievement Menu Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Next

13 Exercise A2 Question 1 1.Simplify the following expression: 4 a + 3 a Like terms = 7 a Back to Exercise A2 Menu NextPrevious

14 Exercise A2 Question 2 2.Simplify the following expression: 3 x – 2 x Like terms = 1 x = x Back to Exercise 2 Menu NextPrevious

15 Exercise A2 Question 3 3. Simplify the following expression, where possible: 4 a a 3 No like terms as different powers So not possible to simplify Back to Exercise 2 Menu NextPrevious

16 Exercise A2 Question 4 4.Simplify the following expression:5 p – 3 p + 7 p Like terms = 2 p + 7 p = 9 p Back to Exercise 2 Menu NextPrevious

17 Exercise A2 Question 5 5.Simplify the following expression:4 a 2 – 2 a a 2 Like terms = 2 a a 2 = 7 a 2 Back to Exercise 2 Menu NextPrevious

18 Exercise A2 Question 6 6.Simplify the following expression:6 x – 7 – 4 x + 2 Like terms together 6 x – 4 x and = 2 x Back to Exercise 2 Menu NextPrevious

19 Exercise A2 Question 7 7.Simplify the following expression:5 x + 6 y + 3 x – 2 y Like terms together 5 x + 3 x and 6 y – 2 y = 8 x + 4 y Back to Exercise 2 Menu NextPrevious

20 Exercise A2 Question 8 8.Simplify the following expression:4 x + 2 x 2 – 3 x Like terms together 4 x – 3 x = x + 2 x 2 Back to Exercise 2 Menu PreviousNext Menu

21 Exercise A3 Menu More simplifying Q5. Q6. Q7. Q4. Q3. Q2. Q1. Q8. 4 a  3 a 4 a 2  3 a 3 4 a 2  - 2 a 3  a 2 5 x  6 y  3 x  2 y 3 x  - 2 x 5 p  3 p 2  p 6 x  - 7 x  2 4 x  2 x 2  3 xy Back to Achievement Menu Next Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).

22 Exercise A3 Question 1 1.Simplify the following expression:4 a  3 a 4  3 = 12, a  a = a 2 so 12 a 2 Back to Exercise A3 Menu Next

23 Exercise A3 Question 2 2.Simplify the following expression:3 x  - 2 x 3  - 2 = - 6, x  x = x 2 so - 6 x 2 Back to Exercise A3 Menu NextPrevious

24 Exercise A3 Question 3 3.Simplify the following expression:4 a 2  3 a 3 4  3 = 12, a 2  a 3 = a 5 so 12 a 5 Back to Exercise A3 Menu NextPrevious

25 Exercise A3 Question 4 4.Simplify the following expression:5 p  3 p 2  p 5  3  1 = 15, p  p 2  p = p = p 4 so 15 p 4 Back to Exercise A3 Menu NextPrevious

26 Exercise A3 Question 5 5.Simplify the following expression:4 a 2  - 2 a 3  a 2 4  - 2  1 = - 8, a 2  a 3  a 2 = a 7 so - 8 a 7 Back to Exercise A3 Menu NextPrevious

27 Exercise A3 Question 6 6.Simplify the following expression:6 x  - 7 x  2 6  - 7  2 = - 84, x  x = x 2 so - 84 x 2 Back to Exercise A3 Menu NextPrevious

28 Exercise A3 Question 7 7.Simplify the following expression:5 x  6 y  3 x  2 y 5  6  3  2 = 180, x  x = x 2, y  y = y 2 so 180 x 2 y 2 Back to Exercise A3 Menu NextPrevious

29 Exercise A3 Question 8 8.Simplify the following expression:4 x  2 x 2  3 xy 4  2  3 = 24, x  x 2  x = x 4 so 24 x 4 y Back to Exercise A3 Menu PreviousNext Menu

30 Exercise A4 Menu Expand and simplify Q5. Q4. Q3. Q2. Q1. Q10. Q9. Q8. Q7. Q6. 4 ( a + 3) a ( a + 5) 4 a (2 a + 3) 5 x (6 y – 3 x ) 2 x 2 (3 xy x 2 y ) 3 ( x – 2) p ( p – 3) 6 x ( - 7 x + 2) - 4 x ( 2 x – 3 y ) 3 x 2 y ( xy 2 – 2 x 3 y ) Back to Achievement Menu Next Expand and simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).

31 Exercise A4 Question 1 4  a = 4 a, 4  3 = 12 so 4 a Expand and simplify:4 ( a + 3) Back to Exercise A4 Menu Next

32 Exercise A4 Question 2 2.Expand and simplify:3 ( x – 2) 3  x = 3 x, 3  2 = 6 so 3 x – 6 Back to Exercise A4 Menu NextPrevious

33 Exercise A4 Question 3 3.Expand and simplify: a ( a + 5) a  a = a 2, a  5 = 5 a so a a Back to Exercise A4 Menu NextPrevious

34 Exercise A4 Question 4 p  p = p 2, p  3 = 3 p so p 2 – 3 p 4.Expand and simplify: p ( p – 3) Back to Exercise A4 Menu NextPrevious

35 Exercise A4 Question 5 4 a  2 a = 8 a 2, 4 a  3 = 12 a so 8 a a 5.Expand and simplify:4 a (2 a + 3) Back to Exercise A4 Menu NextPrevious

36 Exercise A4 Question 6 6 x  - 7 x = - 42 x 2, 6 x  2 = 12 x so - 42 x x or 12 x – 42 x 2 6.Expand and simplify:6 x ( - 7 x + 2) Back to Exercise A4 Menu NextPrevious

37 Exercise A4 Question 7 7.Expand and simplify:5 x (6 y – 3 x ) 5 x  6 y = 30 xy, 5 x  3 x = 15 x 2 so 30 xy – 15 x 2 Back to Exercise A4 Menu NextPrevious

38 Exercise A4 Question 8 8.Expand and simplify: - 4 x (2 x – 3 y ) - 4 x  2 x = - 8 x 2, - 4 x  3 y = - 12 xy so - 8 x 2 – - 12 xy therefore - 8 x xy or 12 xy – 8 x 2 Back to Exercise A4 Menu NextPrevious

39 Exercise A4 Question 9 9.Expand and simplify:2 x 2 (3 xy x 2 y ) 2 x 2  3 xy 2 = 6 x 3 y 2, 2 x 2  2 x 2 y = 4 x 4 y so 6 x 3 y x 4 y Back to Exercise A4 Menu NextPrevious

40 Exercise A4 Question Expand and simplify:3 x 2 y ( xy 2 – 2 x 3 y ) 3 x 2 y  xy 2 = 3 x 3 y 3, 3 x 2 y  2 x 3 y = 6 x 5 y 2 so 3 x 3 y x 5 y 2 Back to Exercise A4 Menu PreviousNext Menu

41 Exercise A5 Menu Expand and simplify Q5. Q4. Q3. Q2. Q1. Q6. Q11. Q10. Q9. Q8. Q7. Q12. ( a + 3) ( a + 4) ( a – 3) ( a + 5) ( a + 4) (2 a + 3) (5 x + y ) ( x – y ) ( a – b ) ( a + b ) ( a + b )2 ( x + 5) ( x + 2) ( p – 6) ( p – 3) ( x – 3) ( - 7 x + 2) ( x + 4) (2 x – 3) ( x – 3) 2 (2 x – 5) (2 x + 5) Next Expand and simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu

42 Exercise A5 Question 1 1.Expand and simplify:( a + 3) ( a + 4) FirstOutsideInsideLast a  a = a 2 a  4=4 a 3  a =3 a 3  4=12 so a a + 3 a + 12 = a a + 12 Back to Exercise A5 Menu Next

43 Exercise A5 Question 2 2.Expand and simplify:( x + 5) ( x + 2) FirstOutsideInsideLast x  x = x 2 x  2=2 x 5  x =5 x 5  2=10 so x x + 5 x + 10 = x x + 10 Back to Exercise A5 Menu NextPrevious

44 Exercise A5 Question 3 3.Expand and simplify:( a – 3) ( a + 5) FirstOutsideInsideLast a  a = a 2 a  5=5 a - 3  a = - 3 a - 3  5= - 15 so a a – 3 a – 15 = a a – 15 Back to Exercise A5 Menu NextPrevious

45 Exercise A5 Question 4 4.Expand and simplify:( p – 6) ( p – 3) FirstOutsideInsideLast p  p = p 2 p  - 3= - 3 p - 6  p = - 6 p - 6  –3=18 so p 2 – 3 p – 6 p + 18 = p 2 – 9 p + 18 Back to Exercise A5 Menu NextPrevious

46 Exercise A5 Question 5 FirstOutsideInsideLast a  2 a =2 a 2 a  3=3 a 4  2 a =8 a 4  3=12 so 2 a a + 8 a + 12 = 2 a a Expand and simplify:( a + 4) (2 a + 3) Back to Exercise A5 Menu NextPrevious

47 Exercise A5 Question 6 6.Expand and simplify:( x – 3) ( - 7 x + 2) Back to Exercise A5 Menu NextPrevious FirstOutsideInsideLast x  - 7 x = - 7 x 2 x  2=2 x - 3 x –7 x =21 x - 3  2= - 6 so - 7 x x + 21 x – 6 = - 7 x x – 6

48 Exercise A5 Question 7 7. Expand and simplify: (5 x + y ) ( x – y ) First OutsideInsideLast 5 x  x =5 x 2 5 x  - y = - 5 xy y  x = xy y  - y = - y 2 so 5 x 2 – 5 xy + xy + - y 2 = 5 x 2 – 4 xy – y 2 Back to Exercise A5 Menu NextPrevious

49 Exercise A5 Question 8 8. Expand and simplify:( x + 4) (2 x – 3) Back to Exercise A5 Menu NextPrevious FirstOutsideInsideLast x  2 x =2 x 2 x  - 3= - 3 x 4  2 x =8 x 4  - 3= - 12 so 2 x 2 – 3 x + 8 x – 12 = 2 x x – 12

50 Exercise A5 Question 9 9. Expand and simplify:( a – b ) ( a + b ) Back to Exercise A5 Menu NextPrevious FirstOutsideInsideLast a  a = a 2 a  b = ab - b  a = - ab - b  b = - b 2 so a 2 + ab – ab – b 2 = a 2 – b 2

51 Exercise A5 Question Expand and simplify:( x – 3) 2 Back to Exercise A5 Menu NextPrevious = ( x –3) ( x –3) FirstOutside InsideLast x  x = x 2 x  –3= - 3 x - 3  x = - 3 x - 3  - 3=9 so x 2 – 3 x – 3 x + 9 = x 2 – 6 x + 9

52 Exercise A5 Question Expand and simplify:( a + b ) 2 Back to Exercise A5 Menu NextPrevious = ( a + b ) ( a + b ) FirstOutsideInsideLast a  a = a 2 a  b = ab b  a = ab b  b = b 2 so a 2 + ab + ab + b 2 = a ab + b 2

53 Exercise A5 Question Expand and simplify:(2 x – 5)(2 x + 5) Back to Exercise A5 Menu Previous FirstOutside Inside Last 2 x  2 x =4 x 2 2 x  5=10 x - 5  2 x = - 10 x - 5  5= - 25 so 4 x x – 10 x – 25 = 4 x 2 – 25 Next Menu

54 Exercise A6 Menu Factorising common factors Q1. 3 a – 15 a a 2 xy + 2 xt 12 x + 3 p 15 a 2 b ab 2 4 x + 24 p – 6 pq 3 q + 15 q 2 9 x 2 y + 30 xy 2 30 x 2 y – 21 x 2 y 3 Q10. Q5. Q4.Q3. Q2. Q9. Q8.Q7. Q6. Next Fully factorise the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu

55 Exercise A6 Question 1 1. Factorise fully:3 a – 15 Back to Exercise A6 Menu Next = 3  a – 3  5 = 3 ( a – 5)

56 Exercise A6 Question 2 2. Factorise fully:4 x + 2 Back to Exercise A6 Menu NextPrevious = 4  x + 4  6 = 4 ( x + 6)

57 Exercise A6 Question 3 3. Factorise fully: a a Back to Exercise A6 Menu NextPrevious = a  a + a  4 = a ( a + 4)

58 Exercise A6 Question 4 4. Factorise fully: p – 6 pq Back to Exercise A6 Menu NextPrevious = p  1 – p  6 q = p (1 – 6 q )

59 Exercise A6 Question 5 5. Factorise fully:2 xy + 2 xt Back to Exercise A6 Menu NextPrevious = 2  x  y + 2  x  t = 2 x ( y + t )

60 Exercise A6 Question 6 6. Factorise fully:3 q + 15 q 2 Back to Exercise A6 Menu NextPrevious = 3  q + 3  5  q  q = 3 q (1 + 5 q ) Note: Must have the 1

61 Exercise A6 Question 7 7. Factorise fully:12 x + 3 p Back to Exercise A6 Menu NextPrevious = 3  4 x + 3  p = 3 (4 x + p )

62 Exercise A6 Question 8 8. Factorise fully:9 x 2 y + 30 xy 2 Back to Exercise A6 Menu NextPrevious = 3  3  x  x  y + 3  10  x  y  y Common factors are 3, x, y = 3 xy (3 x + 10 y )

63 Exercise A6 Question 9 9. Factorise fully:15 a 2 b ab 2 Back to Exercise A6 Menu NextPrevious = 3  5  a  a  b  b  b + 5  5  a  b  b Common factors are 5, a, b and b = 5 ab 2 (3 ab + 5)

64 Exercise A6 Question Factorise fully:30 x 2 y – 21 x 2 y 3 Back to Exercise A6 Menu Previous = 3  10  x  x  y – 3  7  x  x  y  y  y Common factors are 3, x 2 and y = 3 x 2 y (10 – 7 y 2 ) Next Menu

65 Exercise A7 Menu Factorising quadratics Q1. a a + 3 a a – 5 a a – 16 x 2 + x – 20 a 2 – 9 x x + 4 x 2 – 4 x – 5 p 2 – 6 p + 8 k 2 – 6 k – 7 a a + 9 Q10. Q5. Q4. Q3. Q2. Q9. Q8. Q7. Q6. Next Fully factorise the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu

66 Exercise A7 Question 1 1. Factorise fully: a a + 3 Back to Exercise A7 Menu Next No common factors Pairs that multiply to give 3 are 1, 3 and - 1, = 4 so ( a + 1)( a + 3) or ( a + 3)( a + 1)

67 Exercise A7 Question 2 2. Factorise fully: x x + 4 Back to Exercise A7 Menu NextPrevious No common factors Pairs that multiply to give 4 are: 1, 4; 2, 2; - 1, - 4 and - 2, = 4 so ( x + 2)( x + 2) = ( x + 2) 2

68 Exercise A7 Question 3 3. Factorise fully: a a – 5 Back to Exercise A7 Menu NextPrevious No common factors Pairs that multiply to give - 5 are: 1, - 5 and 5, = 4 so ( a + 5)( a – 1) or ( a – 1)( a + 5)

69 Exercise A7 Question 4 4. Factorise fully: x 2 – 4 x – 5 Back to Exercise A7 Menu NextPrevious No common factors Pairs that multiply to give - 5 are: 1, - 5 and 5, = - 4 so ( x + 1)( x – 5) or ( x – 5)( x + 1)

70 Exercise A7 Question 5 5. Factorise fully: a a – 16 Back to Exercise A7 Menu NextPrevious No common factors Pairs that multiply to give - 16 are: 1, - 16; 2, - 8; 4, - 4; 8, - 2 and 16, = 6 so ( a + 8)( a – 2) or ( a – 2)( a + 8)

71 Exercise A7 Question 6 6. Factorise fully: p 2 – 6 p + 8 Back to Exercise A7 Menu NextPrevious No common factors Pairs that multiply to give 8 are: 1, 8; 2, 4; - 1, - 8 and - 2, = - 6 so ( p – 2)( p – 4) or ( p – 4)( p – 2)

72 Exercise A7 Question 7 7. Factorise fully: x 2 + x – 20 Back to Exercise A7 Menu NextPrevious No common factors Pairs that multiply to give - 20 are: 1, - 20; 2, - 10; 4, - 5; 5, - 4; 10, - 2 and 20, - 1 x means 1 x and = 1 so ( x + 5)( x – 4) or( x – 4)( x + 5)

73 Exercise A7 Question 8 8. Factorise fully: k 2 – 6 k – 7 Back to Exercise A7 Menu NextPrevious No common factors Pairs that multiply to give - 7 are: 1, - 7 and 7, = - 6 so ( k + 1)( k – 7) or ( k – 7)( k + 1)

74 Exercise A7 Question 9 9. Factorise fully: a 2 – 9 Back to Exercise A7 Menu NextPrevious No common factors Can be written as: a p – 9 Pairs that multiply to give - 9 are: 1, - 9; 3, - 3 and 9, = 0 so ( a + 3)( a – 3) or ( a – 3)( a + 3)

75 Exercise A7 Question Factorise fully: a a + 9 Back to Exercise A7 Menu Previous No common factors Pairs that multiply to give 9 are: 1, 9; 3, 3; - 1, - 9 and 3, = 6 so ( a + 3)( a + 3) = ( a + 3) 2 Next Menu

76 Exercise A8 Menu Algebraic exponent powers Q3. Q1. (3 x ) 2 Q4. Q2. (4 y ) 3 Next Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu (5 a 2 ) 2 (2 x 2 y ) 4

77 Exercise A8 Question 1 1. Simplify:(3 x ) 2 Back to Exercise A8 Menu Next = 3 x  3 x = 9 x 2

78 Exercise A8 Question 2 2. Simplify:(4 y ) 3 Back to Exercise A8 Menu NextPrevious = 4 y  4 y  4 y = 4 3 y 3 = 64 y 3

79 Exercise A8 Question 3 3.Simplify:(5 a 2 ) 2 Back to Exercise A8 Menu NextPrevious = 5 a 2  5 a 2 = 52( a 2 ) 2 = 25 a 4

80 Exercise A8 Question 4 4. Simplify:(2 x 2 y ) 4 Back to Exercise A8 Menu Previous = 2 4  ( x 2 ) 4  ( y ) 4 = 16  x 8  y 4 = 16 x 8 y 4 Next Menu

81 Exercise A9 Menu Algebraic exponent factors Q4.Q3.Q2.Q1. Q5.Q8.Q7.Q6. Next Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu

82 Exercise A9 Question 1 1. Simplify: Back to Exercise A9 Menu Next Common factors on top and bottom are: 4 and x 4 xy = 4 x  y 8 x = 4 x  2

83 Exercise A9 Question 2 2. Simplify: Back to Exercise A9 Menu NextPrevious Common factors on top and bottom are: 3 and x 12 x = 3 x  415 xy = 3 x  5 y

84 Exercise A9 Question 3 3. Simplify: Back to Exercise A9 Menu NextPrevious Common factors on top and bottom are: 3 and x 3 x 2 = 3 x  x 6 xy = 3 x  2 y

85 Exercise A9 Question 4 4. Simplify: Back to Exercise A9 Menu NextPrevious Common factors on top and bottom are: 3, x 2 and y 2 12 x 3 y 2 = 3 x 2 y 2  4 x 15 x 2 y 3 = 3 x 2 y 2  5 y

86 Exercise A9 Question 5 5. Simplify: Back to Exercise A9 Menu NextPrevious Common factors on top and bottom are: 12 and a (Note: It does not matter if you only notice one or a smaller factor – this just means you will take more steps as you find other common factors) 36 a 2 = 12 a  3 a 60 ab 2 = 12 a  5 b 2

87 Exercise A9 Question 6 6. Simplify: Back to Exercise A9 Menu NextPrevious Common factors on top and bottom are: 2 and x 2 2 x 2 = 2 x 2  1, 4 x 2 y = 2 x 2  2 y

88 Exercise A9 Question 7 7. Simplify: Back to Exercise A9 Menu NextPrevious Common factors on top and bottom are: 12, p and q 36 p 2 q 4 = 12 pq  3 pq 3, 60 pq = 12 pq  5

89 Exercise A9 Question 8 8. Simplify: Back to Exercise A9 Menu Previous Top is 25 x 2 Common factors on top and bottom are: 5 and x 25 x 2 = 5 x  5 x, 60 xy 3 = 5 x  12 y 3 Next Menu

90 Exercise A10 Menu Substitution If A = 0.5( a + b ) h, find A when a = 3, b = 7 and h = 6 If k = 0.6 b 2, find k when b = 4 S =. What is S when a = 3, c = 5 and d = 2? The volume of a sphere is found by where r is radius. Find the volume of a sphere with radius 5. A new symbol  has the meaning that x  y = 3 x 2 – y. Find the value of -2  4. To convert Fahrenheit temperature, f, to Celsius, C, use the formula. Convert 360 o Fahrenheit to o Celsius. Q5. Q6. Q4. Q3. Q2. Q1. Next Substitute within the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu

91 Exercise A10 Question 1 1.If A = 0.5( a + b ) h, find A when a = 3, b = 7 and h = 6 Next Back to Exercise A10 Menu A = 0.5  (3 + 7)  6 = 30

92 Exercise A10 Question 2 2. If k = 0.6 b 2, find k when b = 4 NextPrevious Back to Exercise A10 Menu k = 0.6  (4) 2 = 9.6

93 Exercise A10 Question 3 3. S =. What is S when a = 3, c = 5 and d = 2? NextPrevious Back to Exercise A10 Menu = (15  3)  (5 – 2) = 15

94 Exercise A10 Question 4 4.The volume of a sphere is found by where r is radius. Find the volume of a sphere with radius 5. NextPrevious Back to Exercise A10 Menu Note: There was only 1 sig fig in the radius, so it might be acceptable to write 500 to 1 sig fig. However, you must show working steps and it is a good idea to show the unrounded answer. V =V = = 4  3    5 3 = = 524 to 3 sig fig

95 Exercise A10 Question 5 5.A new symbol  has the meaning that x  y = 3 x 2 – y. Find the value of - 2  4 NextPrevious Back to Exercise A10 Menu - 2 takes the place of x and 4 takes the place of y - 2  4 = 3( - 2) 2 – 4 Note: Use brackets to square a negative = 8

96 Exercise A10 Question 6 6. To convert Fahrenheit temperature, f, to Celsius, C, use the formula. Convert 360 o Fahrenheit to o Celsius. Previous Back to Exercise A10 Menu = so Celsius to 1 d.p. Note: Other rounding acceptable. Next Menu

97 Exercise A11 Menu Finding linear rule 8, 12, 16, 20, … 10, 17, 24, 31, … 24, , 34.5, … 35, 32, 29, 26, … Q5. Q7. Q3. Q1. Next 22, 24, 26, 28, … 15, 15.5, 16, 16.5, … 104, 102, 100, 98, … 12.5, 12.1, 11.7, 11.3,... Q6. Q4. Q2. Q8. Find the rule for finding the n th term for the following sequences then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Back to Achievement Menu

98 Exercise A11 Question 1 1. Find the rule for finding the n th term for the sequence: 8, 12, 16, 20, … Next Back to Exercise A11 Menu Difference between terms is 4 Rule is from the 4 x table 4, 8, 12, 16, … Need 4 more to give the sequence Therefore rule is 4 n + 4

99 Exercise A11 Question 2 2. Find the rule for finding the n th term for the sequence: 22, 24, 26, 28, … NextPrevious Back to Exercise A11 Menu Difference between terms is 2 Rule is from the 2 x table 2, 4, 6, 8, … Need 20 more to give the sequence Therefore rule is 2 n + 20

100 Exercise A11 Question 3 3. Find the rule for finding the n th term for the sequence: 10, 17, 24, 31, … NextPrevious Back to Exercise A11 Menu Difference between terms is 7 Rule is from the 7 x table 7, 14, 21, 28, … Need 3 more to give the sequence Therefore rule is 7 n + 3

101 Exercise A11 Question 4 4. Find the rule for finding the n th term for the sequence: 15, 15.5, 16, 16.5, … NextPrevious Back to Exercise A11 Menu Difference between terms is 0.5 Multiplying 1, 2, 3, 4, … by 0.5 gives: 0.5, 1, 1.5, 2, … Need 14.5 more to give the sequence Therefore rule is 0.5 n

102 Exercise A11 Question 5 5. Find the rule for finding the n th term for the sequence: 24, , 34.5, … NextPrevious Back to Exercise A11 Menu Difference between terms is 3.5 Multiplying 1, 2, 3, 4, … by 3.5 gives: 3.5, 7, 10.5, 14, … Need 20.5 more to give the sequence Therefore rule is 3.5 n

103 Exercise A11 Question 6 6. Find the rule for finding the n th term for the sequence: 104, 102, 100, 98, … NextPrevious Back to Exercise A11 Menu Difference between terms is - 2 Multiplying 1, 2, 3, 4, … by - 2 gives: - 2, - 4, - 6, - 8, … Need 106 more to give the sequence Therefore rule is - 2 n or 106 – 2 n

104 Exercise A11 Question 7 7. Find the rule for finding the n th term for the sequence: 35, 32, 29, 26, … NextPrevious Back to Exercise A11 Menu Difference between terms is - 3 Multiplying 1, 2, 3, 4, … by - 3 gives: - 3, - 6, - 9, - 12, … Need 38 more to give the sequence Therefore rule is - 3 n + 38 or 38 – 3 n

105 Exercise A11 Question 8 8. Find the rule for finding the n th term for the sequence: 12.5, 12.1, 11.7, 11.3,... Previous Back to Exercise A11 Menu Difference between terms is – 0.4 Multiplying 1, 2, 3, 4, … by – 0.4 gives: – 0.4, – 0.8, – 1.2, – 1.6, … Need 12.9 more to give the sequence Therefore rule is – 0.4 n or 12.9 – 0.4 n Next Menu

106 Exercise A12 Menu Linear rules diagram or table Q1. A matches design. Find a rule and work out how many matches would be needed. Q2. A rectangular shape. Find a rule to find the perimeter of the 10th rectangle that would be drawn in this sequence. Q3. The telephone account. Complete the table to show the monthly costs. Write a rule to show how much is paid for n calls. Q4. A rectangular courtyard is being tiled. Find a rule to find the number of tiles after n edges have been added. Q5. The savouries table. Find an expression for the number of savouries allowed for n guests. Q6. The amount of sewing pattern material. Write a general rule to give the amount of material for width ( w ) Q7. The number of doors needed. If a block of flats is to have 17 storeys, how many doors are needed? Next Find the rule for finding the n th term for the following sequences then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Back to Achievement Menu

107 Answer Exercise A12 Question 1 1.The following design is being created with matches. Find a rule and work out how many matches would be needed to make a design with 20 of the basic shapes. Sequence is 7, 12, 17, … Difference is 5 5 x table would give: 5, 10, 15, … Need 2 more each time so 5 n basic shapes, n = 20, Need 5  = 102 matches Next Back to Exercise A12 Menu

108 Answer Exercise A12 Question 2 2.A rectangular shape is being increased by adding 1 metre to each side at the corner as shown. It starts as size 5 m by 4 m so has perimeter of 18 m. Find a rule to find the perimeter of the 10th rectangle that would be drawn in this sequence. Sequence is 18, 26, 34, Difference is 8 8 x table is 8, 16, 24, … Need 10 more each time. Rule is 8 n th rectangle would have n = 10 Therefore 8  = 90 NextPrevious Back to Exercise A12 Menu

109 Answer Exercise A12 Question 3 3.Steven pays $20 a month for the telephone account. Phone calls to his girlfriend in another town cost $2.75 each. Complete the table to show the monthly cost for different numbers of phone calls. Write a rule to show how much he would pay for n calls. Calls Cost Complete table with 28.25, 31, Difference = 2.75 This gives 2.75, 5.5, 8.25, … Needs 20 added Rule is 2.75 n + 20 NextPrevious Back to Exercise A12 Menu

110 Answer Exercise A12 Question 4 4.A rectangular courtyard is being tiled. Fifteen tiles form the rectangular centre as shown and the tiles are being placed on each side. Find a rule to find the number of tiles after n edges have been added. ANSWER NextPrevious Back to Exercise A12 Menu Number of tiles is 21, 27, 33, … Difference is 6 6 x table gives: 6, 12, 18, … Need to add 15 to each Therefore rule is 6 n + 15

111 Answer Exercise A12 Question 5 5.The table shows the number of savouries allowed by a catering company for different numbers of guests. Find an expression for the number of savouries allowed for n guests. Guests Savouries NextPrevious Back to Exercise A12 Menu Guests go up by 10 Savouries go up by 20, so 2 savouries per person. But for 10, 20, 30, 40, this would give: 20, 40, 60, 80, so need to add 5 Therefore rule is 2 n + 5

112 Exercise A12 Question 6 6.The amount of material needed for a sewing pattern is given for different sizes as shown in the table. Write a general rule to give the amount of material for the width ( w ). Waist (cm) Material (m) Answer NextPrevious Back to Exercise A12 Menu Waist goes up by 5. Material goes up by 0.2, so 0.04 metres per cm. But for 70, 75, 80, 85, this would give: 2.8, 3, 3.2, 3.4, so need to take 0.1 Therefore rule is 0.04 w – 0.1

113 Answer Exercise A12 Question 7 7.The number of doors needed to be supplied for a block of flats depends on the number of storeys high the flats are. For one storey 27 doors are needed. If the building is two-storied there are 51 doors needed; three- storied, 75 doors; four-storied, 99 doors and so on. If a block of flats is to have 17 storeys, how many doors are needed? Previous Back to Exercise A12 Menu Number of doors is 27, 51, 75, 99, … Difference is x table gives 24, 48, 72, 96, … Need to add 3 to each Therefore rule is 24 n + 3 For 17 storeys, n = 17 so 24  = 411 doors. Next Menu

114 Exercise A13 Menu Solve linear equations Q7. Q5. Q3. Q1. Q8. Q6. Q4. Q2. 5 x + 7 = 42 2( x + 7) = = 3 x – 2 3 x – 8 = 82 5 x + 19 = – x = 16 Next Solve the following equations showing working, then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Back to Achievement Menu

115 Exercise A13 Question 1 5 x + 7= 42 – 7 – 7 5 x = 35  5  5 x = 7 1. Solve the following equation showing working: 5 x + 7= 42 Next Back to Exercise A13 Menu

116 Exercise A13 Question 2 2. Solve the following equation showing working: 3 x – 8 = x = 90  3  3 x = 30 NextPrevious Back to Exercise A13 Menu

117 Exercise A13 Question 3 3. Solve the following equation showing working: 2( x + 7) = 28 Note: Can divide by 2 or expand brackets as first step. This example shows expanding brackets. 2 x + 14 = 28 – 14 – 14 2 x = 14  2 x = 7 NextPrevious Back to Exercise A13 Menu

118 Exercise A13 Question 4 4. Solve the following equation showing working: 5 x + 19 = 32.4 – 19 – 19 5 x = 13.4  5 x = 2.68 NextPrevious Back to Exercise A13 Menu

119 Exercise A13 Question 5 5. Solve the following equation showing working: 26 = 3 x – = 3 x 3 x = 28 Usual to have x term on left side so swap.  3  3 Can swap sides at any time since sides equal x = so 9.33 (to 2 d.p.) NextPrevious Back to Exercise A13 Menu

120 Exercise A13 Question 6 6. Solve the following equation showing working: 4 – x = 16 – 4 – 4 - x = 12  - 1  - 1 x = - 12 NextPrevious Back to Exercise A13 Menu

121 Exercise A13 Question 7 7. Solve the following equation showing working:  8  8 x = 192 – 3 – 3 NextPrevious Back to Exercise A13 Menu

122 Exercise A13 Question 8 8. Solve the following equation showing working: Previous Back to Exercise A13 Menu x 8 x 8 x = 24 Next Menu

123 Exercise A14 Menu Solve more linear equations Q7. Q5. Q3. Q1. Q8. Q6. Q4. Q2. 5 x + 7 x = ( x + 7) = x = 1.8 x x – 8 = 2.4 x 5 x + 19 = 30 x – 2 x = 16 x Next Solve the following equations showing working, then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Back to Achievement Menu

124 Exercise A14 Question 1 1. Solve the following equation showing working: 5 x + 7 x = x =  12 x = 3.5 Next Back to Exercise A14 Menu

125 Exercise A14 Question 2 2. Solve the following equation showing working: 3.6 x – 8 = 2.4 x – 3.6 x – 3.6 x - 8 = x x = - 8 Can swap sides at any time   x = 6.67 (to 2 d.p.) NextPrevious Back to Exercise A14 Menu

126 Exercise A14 Question 3 3. Solve the following equation showing working: 2.5( x + 7) = 27.5 Can be done by expanding bracket or dividing by 2.5 This example is done by dividing by ( x + 7) = 27.5  2.5 x + 7 = 11 – 7 – 7 x = 4 NextPrevious Back to Exercise A14 Menu

127 Exercise A14 Question 4 4. Solve the following equation showing working: 5 x + 19 = 30 x – 19 – 19 Could take any term from both sides as a correct first step. 5 x = 30 x + 85 – 30 x – 30 x - 25 x = 85  - 25  - 25 x = NextPrevious Back to Exercise A14 Menu

128 Exercise A14 Question 5 5. Solve the following equation showing working: 5.6 x = 1.8 x + 7 – 1.8 x – 1.8 x 3.8 x = 7  3.8  3.8 x = 1.84 (to 2 d.p.) NextPrevious Back to Exercise A14 Menu

129 Exercise A14 Question 6 6. Solve the following equation showing working: 14 – 2 x = 16 x + 2 x + 2 x 14= 18 x 18 x = 14  18  18 x = 0.78 (to 2 d.p.) NextPrevious Back to Exercise A14 Menu

130 Exercise A14 Question 7 7. Solve the following equation showing working: Multiply all terms by 5 3 x – 15 = 10 x Note that each term has been  5 – 3 x – 3 x - 15= 7 x 7 x = -15 Can swap sides at any time  7  7 x = (to 2 d.p.) NextPrevious Back to Exercise A14 Menu

131 Exercise A14 Question 8 8.Solve the following equation showing working: Multiply all terms by 6 x + 18 = - 30 x Note that each term has been  6 – x 18 = - 31 x - 31 x = 18 Can swap sides at any time  - 31  - 31 x = (to 2 d.p.) Previous Back to Exercise A14 Menu Next Menu

132 Exercise A15 Menu Solve factorised quadratic equations ( x – 3)( x – 2) = 0 ( x + 1)( x + 4) = 0 (3 x + 2)( x – 2) = 0 x (2 x + 7) = 0 Q5. Q7. Q3. Q1. ( x – 5)( x + 3) = 0 (2 x – 5)( x + 4) = 0 (3 x + 2)(2 x – 5) = 0 5 x (3 x – 7) = 0 Q6. Q4. Q2. Q8. Next Solve the following equations showing working, then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Back to Achievement Menu

133 Exercise A15 Question 1 1. Solve the following equation showing working: ( x – 3)( x – 2) = 0 x – 3= 0 or x – 2 = x = 3 x = 2 Next Back to Exercise A15 Menu

134 Exercise A15 Question 2 2. Solve the following equation showing working: ( x – 5)( x + 3) = 0 x – 5 = 0 or x + 3 = – 3 – 3 x = 5 x = -3 NextPrevious Back to Exercise A15 Menu

135 Exercise A15 Question 3 3. Solve the following equation showing working: ( x + 1)( x + 4) = 0 x + 1 = 0 or x + 4 = 0 – 1 – 1 – 4 – 4 x = -1 x = - 4 NextPrevious Back to Exercise A15 Menu

136 Exercise A15 Question 4 4. Solve the following equation showing working: (2 x – 5)( x + 4) = 0 2 x – 5 = 0 or x + 4 = – 4 – 4 2 x = 5 x = - 4  2  2 x = 2.5 NextPrevious Back to Exercise A15 Menu

137 Exercise A15 Question 5 5. Solve the following equation showing working: (3 x + 2)( x – 2) = 0 3 x + 2 = 0 or x – 2 = 0 – 2 – x = - 2 x = 2  3  3 x = NextPrevious Back to Exercise A15 Menu

138 Exercise A15 Question 6 6. Solve the following equation showing working: (3 x + 2)(2 x – 5) = 0 3 x + 2 = 0 or 2 x – 5 = 0 – 2 – x = - 22 x = 5  3  3  2  2 x = (2 d.p.) or x = 2.5 NextPrevious Back to Exercise A15 Menu

139 Exercise A15 Question 7 7. Solve the following equation showing working: x (2 x + 7) = 0 x = 0 or 2 x + 7 = 0 – 7 – 7 2 x = - 7  2  2 x = NextPrevious Back to Exercise A15 Menu

140 Exercise A15 Question 8 8. Solve the following equation showing working: 5 x (3 x – 7) = 0 5 x = 0 or 3 x – 7= 0  5  x = 0 3 x = 7  3  3 x = 2.33 (2 dp) Previous Back to Exercise A15 Menu Next Menu

141 Click on the buttons to go to the review questions, solve each one, then press the answer button. Or scroll to each question using the next arrow (below). Click to move from step to step in the solution. Exercise A16 Menu Achievement review Expand the equations Find the rule Solve the equation Q5. Q7. Q3. Q1. Factorise the equations Find the surface area Solve the equations Q6. Q4. Q2. Next Back to Achievement Menu

142 Answer b Exercise A16 Question 1 1.Expand and simplify: (a) 3 xy (2 x – 7)(b) (2 x + 3)( x – 6) Answer a Next Back to Exercise A16 Menu (a) 3 xy  2 x – 3 xy  7 = 6 x 2 y – 21 xy (b) First Outside Inside Last 2 x  x – 2 x   x – 3  6 = 2 x 2 – 12 x + 3 x – 18 = 2 x 2 – 9 x – 18

143 Answer a Exercise A16 Question 2 2. Factorise fully: (a) 5 x 2 y – 20 x 3 y 4 (b) x 2 – 8 x + 15 Answer b (b) Pairs that multiply to give 15 are: 1, 15; 3, 5; - 1, - 15 and - 3, and - 5 add to give - 8 x 2 – 8 x + 15 = ( x – 3)( x – 5) or ( x – 5)( x – 3) NextPrevious Back to Exercise A16 Menu (a)= 5  x  x  y – 5  4  x  x  x  y  y  y  y = 5 x 2 y (1 – 4 xy 3 )

144 Exercise A16 Question 3 3. Simplify: (a) (b) Answer a Answer b NextPrevious Back to Exercise A16 Menu (b) Top = 9  p  p  p  p  q  q Bottom = 9  p Common factor is 9 p (a) = 5 3  ( x 2 ) 3  2 2  x 2  y 2 = 125  4  x 6  x 2  y 2 = 500 x 8 y 2

145 Exercise A16 Question 4 4.S.A. = gives the surface area of a sphere for given radius r. What is the surface area of a sphere with radius 8 cm? Answer NextPrevious Back to Exercise A16 Menu S. A. = 4    8  8 = so 804 (3 sig fig)

146 Exercise A16 Question 5 5. Hedge plants are to be placed around the edge of rectangular fields. The diagram shows the spacing for different 2 m wide rectangular fields. Find a rule to give the number of plants needed for a 2 m rectangular field that is n metres long. Answer For n values 0.5, 1, 1.5,… the number of plants is 10, 12, 14 so that the values 1, 2, 3, … would give 12, 16, 20 With difference 4, the 4 x table gives 4, 8, 12, … Therefore need to add 8. The rule for length n is 4 n + 8 NextPrevious Back to Exercise A16 Menu

147 Answer bAnswer a Exercise A16 Question 6 6. Solve: (a) 5 x – 7 = 32(b) 6 x – 19 = 3.6 x + 21 NextPrevious Back to Exercise A16 Menu (a) 5 x – 7 = x = 39  5  5 x = 7.8 (b) 6 x – 19 = 3.6 x x = 3.6 x + 40 – 3.6 x – 3.6 x 2.4 x = 40  2.4 x = (2 d.p.)

148 Exercise A16 Question 7 7.Solve: (3 x + 5)(6 – x ) = 0 3 x + 5 = 0 or 6 – x = 0 – 5 – 5 x = 6 3x=-53x=-5  3  3 x = (2 d.p.) PreviousMerit Menu Back to Exercise A16 Menu Back to Main Menu

149 Merit Menu Press the button to go to the menu page for each exercise group. Exercise M9. Exercise M11. Exercise M7. Exercise M5. Exercise M3. Exercise M1. Exercise M10. Exercise M8. Exercise M6. Exercise M4. Exercise M2. Simplifying Change the subject Equations and inequations Harder quadratic equations Substitution in equations Context equation solving Rational expressions Quadratic patterning Solving inequations Simple quadratic equations Elimination in equations Back to Main Menu

150 Q5. Q6. Q7. Q4. Q3. Q2.Q1. Exercise M1 Menu Adding and subtracting rational expressions Answer each question as a single expression. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Next Q8. Q9. Back to Merit Menu

151 Exercise M1 Question 1 3 and 5 have common multiple of 15 Rewrite x / 3 as 5 x / 15 by multiplying top and bottom by 5 Rewrite x / 5 as 3 x / 15 by multiplying top and bottom by 3 Write common bottom line and combine the top lines. 1. Write as a single expression: Next Back to Exercise M1 Menu

152 Exercise M1 Question 2 4 and 7 have common multiple of 28 Rewrite x / 4 as 7 x / 28 by multiplying top and bottom by 7 Rewrite x / 7 as 4 x / 28 by multiplying top and bottom by 4 Write common bottom line and combine the top lines. 2. Write as a single expression: NextPrevious Back to Exercise M1 Menu

153 Exercise M1 Question 3 4 and 5 have common multiple of 20 Rewrite 2 x / 5 as 4 x / 10 by multiplying top and bottom by 2 Write common bottom line and combine the top lines. 3. Write as a single expression: NextPrevious Back to Exercise M1 Menu

154 Exercise M1 Question 4 4 and 5 have common multiple of 20 Rewrite 3 x / 4 as 15 x / 20 by multiplying top and bottom by 5 Rewrite 2 x / 5 as 8 x / 20 by multiplying top and bottom by 4 Write common bottom line and combine the top lines. 4. Write as a single expression: NextPrevious Back to Exercise M1 Menu

155 Exercise M1 Question 5 x and 5 multiply to give 5 x Rewrite 4 / x as 20 / 5 x by multiplying top and bottom by 5 Rewrite x / 5 as by multiplying top and bottom by x Write common bottom line and combine the top lines. 5. Write as a single expression: NextPrevious Back to Exercise M1 Menu

156 Exercise M1 Question 6 x 2 and x have common multiple of x 2 Rewrite as by multiplying top and bottom by x Write common bottom line and combine the top lines. 6. Write as a single expression: NextPrevious Back to Exercise M1 Menu

157 Exercise M1 Question 7 3 and x multiply to give 3 x Rewrite x / 3 as by multiplying top and bottom by x Rewrite 5 / x as 15 / x by multiplying top and bottom by 3 Write common bottom line and combine the top lines. 7. Write as a single expression: NextPrevious Back to Exercise M1 Menu

158 Exercise M1 Question 8 4 and 8 x both divide into 8 x Rewrite x / 4 as by multiplying top and bottom by 2 x Write common bottom line and combine the top lines. 8. Write as a single expression: NextPrevious Back to Exercise M1 Menu

159 Exercise M1 Question 9 Common denominator 10 p 2 First term multiply top and bottom by 2 Second term multiply top and bottom by p 2 Write common bottom line and combine the top lines. 9. Write as a single expression: Next MenuPrevious Back to Exercise M1 Menu

160 Exercise M2 Menu Simplifying rational expressions Answer each question as a single expression. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Next Q5. Q6. Q7. Q4. Q3. Q2.Q1. Q8. Q9. Back to Merit Menu

161 Exercise M2 Question 1 x 2 – 3 x = x ( x – 3) 3 x = x  3 Cancel out the x 1. Simplify fully: Next Back to Exercise M2 Menu

162 Exercise M2 Question 2 3 x + 6 = 3( x + 2) 3 = 3  1 Cancel out the 3 2. Simplify fully: NextPrevious Back to Exercise M2 Menu

163 Exercise M2 Question 3 2 x 2 – 4 x = 2 x ( x – 2) 10 x = 2 x  5 Cancel out the 2 x 3. Simplify fully: NextPrevious Back to Exercise M2 Menu

164 Exercise M2 Question 4 x x + 2 = ( x + 1)( x + 2) x + 2 = ( x + 2)  1 Cancel out the ( x + 2) 4. Simplify fully: NextPrevious Back to Exercise M2 Menu

165 Exercise M2 Question 5 x – 4 = ( x – 4)  1 x 2 – 2 x – 8 = ( x – 4)( x + 2) Cancel out the ( x – 4) 5. Simplify fully: NextPrevious Back to Exercise M2 Menu

166 Exercise M2 Question 6 x – 5 = ( x – 5)  1 x x – 40 = ( x + 8)( x – 5) Cancel out the ( x – 5) 6. Simplify fully: NextPrevious Back to Exercise M2 Menu

167 Exercise M2 Question 7 x 2 – 16 = ( x – 4)( x + 4) 2 x – 8 = 2( x – 4) Cancel out the ( x – 4) 7. Simplify fully: NextPrevious Back to Exercise M2 Menu

168 Exercise M2 Question 8 x 2 + 5x = x ( x + 5) 5 x x 2 = 5 x ( x )( x + 5) Cancel out the x ( x + 5) 8. Simplify fully: NextPrevious Back to Exercise M2 Menu

169 Exercise M2 Question 9 p p – 15 = ( p + 5)( p – 3) p 2 – 9 = ( p + 3)( p – 3) Cancel out the ( p – 3) 9. Simplify fully: Previous Back to Exercise M2 Menu Next Menu

170 Q3. Q1. Exercise M3 Menu Quadratic patterning Solve the quadratic pattern in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Next Q4. Q2. Complete the table and write a rule for when n metres has been added. Write a rule for the number of triangles needed for Design n. Find a rule to give the number of plants needed if n rounds are to be added. How many tins would there be in x layers? Back to Merit Menu

171 Answer Exercise M3 Question 1 1.A garden area is a rectangle 4 m by 3 m. It is being added to by extending 1 metre at a time along 2 adjacent sides. After 1 metre is added it has area 5 m by 4 m = 20 m 2. After a second metre is added it has area 6 m by 5 m = 30 m 2. Complete the table and write a rule for when n metres has been added. Length added Area5x4 =20 6x5 =30 7x6 =42 Next Question Back to Exercise M3 Menu 3 gives 7  6 = 42 4 gives 8  7 = 56 5 gives 9  8 = 72 6 gives 10  9= 90 n gives ( n + 4)( n + 3) Alternative Method

172 Exercise M3 Q1 (Answer b) Length added, n AreaDiff2 nd diff Question 1: Alternative method Half second difference = 1, so rule is to do with (1) n 2. n 2 gives 1, 4, 9, 16, 25, … But Area is 20, 30, 42, 56, 72, … Need an extra 19, 26, 33, 40, 47,… This is a linear pattern with difference 7, so can be found by 7 n + 12 Therefore rule is n n + 12 Back to the question Next Question Back to Exercise M3 Menu

173 Answer Exercise M3 Question 2 2. A pattern is made from the basic design as shown. Write a rule for the number of triangles needed for Design n. Next QuestionPrevious One way of seeing this pattern is: Design 1 = 6 triangles (2 by 2 square doubled to get triangles – 2 corner triangles) = 2 x (2 x 2) – 2 x 1 Design 2 = 2 x (3 x 4) – 2 x (2 x 2) doubled rectangle – 2 triangles which form a square Design 3 = 2 x (4 x 6) – 2 x (3 x 3) Design 4 would have rectangle 5 by 8 = 2 x (5 x 8) – 2 x (4 x 4) Design n would be = 2( n + 1)(2 n ) – 2( n x n ) = 4 n n – 2 n 2 = 2 n n Back to Exercise M3 Menu Alternative Method

174 Exercise M3 Q2 (Answer b) Design, n TrianglesDiff2 nd diff Question 2: Alternative method There are several different ways of seeing the design which lead to the same answer. This approach is the mathematical one: Half second difference = 2, so rule is to do with 2 n 2. n 2 gives 1, 4, 9, 16, 25, … 2 n 2 gives 2, 8, 18, 32, 50 Triangles is 6, 16, 30, 48, 70, … Need an extra 4, 8, 12, 16, 20,… This is a linear pattern with difference 4, so can be found by 4 n Therefore rule is: 2 n n Next QuestionPrevious Back to the question Back to Exercise M3 Menu

175 Answer Exercise M3 Question 3 3.Plants are to be spaced out equally in a rectangular field. From a centre rectangular area with 8 plants, the plants are set out in rounds as shown in the diagram. Find a rule to give the number of plants needed if n rounds are to be added. Next QuestionPrevious 1 round 4 by 6 = 24 2 rounds 6 by 8 = 48 3 rounds 8 by 10 = 80 Dimensions are going up in 2s so linear patterns. n rounds (2 n + 2) by (2 n + 4) So (2 n + 2)(2 n + 4) plants Back to Exercise M3 Menu Alternative Method

176 Exercise M3 Q3 (Answer b) Rounds added, n PlantsDiff2 nd diff Question 3: Alternative method There are several different ways of seeing the design which lead to the same answer. This approach is the mathematical one: Half second difference = 4, so rule is to do with 4 n 2. n 2 gives 1, 4, 9, 16, 25, … 4 n 2 gives 4, 16, 36, 64, 100 Triangles is 24, 48, 80, 120, 168, … Need an extra 20, 32, 44, 56, 68,… This is a linear pattern with difference 12, so can be found by 12 n + 8 Therefore rule is: 4 n n + 8 Next QuestionPrevious Back to the question Back to Exercise M3 Menu

177 Answer Exercise M3 Question 4 4.Cans are stacked on a shelf as shown in the diagrams. There are 4 cans in 1 layer, 11 cans if there are 2 layers, 21 cans if there are 3 layers and so on. How many cans would there be in x layers? Previous 1 layer = 4 cans 2 layers = 0.5 x (2 x 11) 3 layers = 0.5 x (3 x 14) 4 layers = 0.5 x (4 x 17) Linear pattern for values in brackets, so x layers has 0.5 x x x (3 x +5) =0.5 x (3 x +5) Back to Exercise M3 Menu Next Menu Alternative Method

178 Exercise M3 Q4 (Answer b) Layers, x CansDiff2 nd diff Question 4: Alternative method There are several different ways of seeing the design which lead to the same answer. This approach is the mathematical one: Half second difference = 3, so rule is to do with 1.5 x 2. x 2 gives 1, 4, 9, 16, 25, … 1.5 x 2 gives 1.5, 6, 13.5, 24, 37.5 Cans is 4, 11, 21, 34, 50, … Need an extra 2.5, 5, 7.5, 10, 12.5,… This is a linear pattern with difference 2.5, so can be found by 2.5 x Therefore rule is: 1.5 x x Previous Back to the question Back to Exercise M3 Menu Next Menu

179 Q5. Q3. Q1. Exercise M4 Menu Change the subject Make k the subject in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Next x = 15 k + 7 m = k 2 – 5 z = Q6. Q4. Q2. P = 3( k – 2) B = g = 4 a 2 k – 2 p Back to Merit Menu

180 Exercise M4 Question 1 1. Make k the subject in: x = 15 k k + 7 = x – 7 – 7 15 k = x – 7  15  15 k = ( x – 7)  15 or NextPrevious Back to Exercise M4 Menu

181 Exercise M4 Question 2 2. Make k the subject in: P = 3( k – 2) 3( k – 2)= P  3  3 k – 2= P / k = P / or NextPrevious Back to Exercise M4 Menu

182 Exercise M4 Question 3 3. Make k the subject in: m = k 2 – 5 k 2 – 5 = m k 2 = m + 5 take square root k = NextPrevious Back to Exercise M4 Menu

183 Exercise M4 Question 4 4. Make k the subject in B = NextPrevious Back to Exercise M4 Menu x 3 x 3 3 B = k – 5 k – 5 = 3 B k = 3 B + 5

184 Exercise M4 Question 5 5. Make k the subject in z = + q NextPrevious 4 z – q 4 k Back to Exercise M4 Menu – q – q z – q = 4 / k x k x k k ( z – q )= 4  ( z – q )  ( z – q ) k =

185 Exercise M4 Question 6 6. Make k the subject in g = 4 a 2 k – 2 p Previous g + 2 p 4a24a2 Back to Exercise M4 Menu 4 a 2 k – 2 p = g + 2 p + 2 p 4 a 2 k = g + 2 p  4a2 4a2 4a2 4a2 k = Next Menu

186 Exercise M5 Menu Solving Inequations Next Q5. Q3. Q1. Q6. Q4. Q2. 4 x – 8 < 20 4(2 – x ) > 16 3 x + 8 > 5 x Solve the inequations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Back to Merit Menu

187 Exercise M5 Question 1 1.Solve:4 x – 8 < x < 28  4  4 x < 7 Next Back to Exercise M5 Menu

188 Exercise M5 Question 2 2. Solve:3 x + 8 > 5 x Swap sides, reversing inequation 5 x < 3x + 8 – 3 x – 3 x 2 x < 8  2  2 x < 4 NextPrevious Back to Exercise M5 Menu

189 Exercise M5 Question 3 3. Solve: 4(2 – x ) > 16  4  4 2 – x > 4 – 2 – 2 - x > 2 - x  - 1 < 2  - 1 Note: Change of sign direction x < - 2 NextPrevious Back to Exercise M5 Menu

190 Exercise M5 Question 4 4. Solve: x 3 x 3 k – 5 < k < 29 NextPrevious Back to Exercise M5 Menu

191 Exercise M5 Question 5 5. Solve: k / 5 > 14 x 5 x 5 k > 70 NextPrevious Back to Exercise M5 Menu

192 Exercise M5 Question 6 6. Solve: – 6 - k / 6 > 5 x 6 x 6 - k > 30 - k  - 1 < 30  - 1 Note: Change of sign direction k < - 30 Previous Back to Exercise M5 Menu Next Menu

193 Exercise M6 Menu Form and solve equations and inequations Five more than three times a number is forty seven. What is the number? How many items of jewellery were sold on commission? How many weeks will it be before the account is overdrawn? What day will Marama be jogging more than 5 km? Click on the buttons to go to each problem or scroll to each question/answer using the ‘next’ arrow (below). Write an equation or inequation to represent each problem, then solve it. Press the answer button for the correct answer. Click to move from step to step in the solution. After how many weeks will Tyrone have more than $500? What is the mechanic charging per hour for labour? How many calls can be made, without spending more than $60 in a week? How many glasses of drink can be poured? Q1. Q5. Q3. Q7. Q2. Q6. Q4. Q8. Next Back to Merit Menu

194 Answer Exercise M6 Question 1 1. Write an equation or inequation to represent: Five more than three times a number is forty seven. What is the number? 3 x + 5 = 47 – 5 – 5 3 x = 42  3  3 x = 14 Next Back to Exercise M6 Menu

195 Answer Exercise M6 Question 2 2. Write an equation or inequation to represent: Tyrone saves $16 a week for a holiday. If he starts with $83, after how many weeks will he have more than $500? 16 w + 83 > 500 w = number of weeks – 83 – w > 417  16  16 w > Therefore 27 weeks NextPrevious Back to Exercise M6 Menu

196 Answer Exercise M6 Question 3 3. Write an equation or inequation to represent: Trish sells jewellery on commission at a jewellery evening. She gets $8 for each item she sells, but it costs her $45 to cover the costs of the jewellery evening. If she makes $203 one evening, how many items of jewellery did she sell? 8 j - 45 = 203 j = number of jewellery items j = 248  8  8 j = 31 Therefore 31 items NextPrevious Back to Exercise M6 Menu

197 Answer Exercise M6 Question 4 4. Write an equation or inequation to represent: A mechanic charges $56 for parts and has worked for 3.5 hours on the car. If the total account was $248.50, what is the mechanic charging per hour for labour? 3.5 h + 56 = h = hourly rate – 56 – h =  3.5  3.5 h = 55 Therefore $55 per hr NextPrevious Back to Exercise M6 Menu

198 Answer Exercise M6 Question 5 5. Write an equation or inequation to represent: Jenny has $1364 in an account. If she pays her rent from this account and it is $75 per week, how many weeks will it be before her account is overdrawn? 1364 – 75 w < 0 w = number of weeks – 1364 – w < w  - 75 >  - 75 w > Therefore 19 weeks NextPrevious Back to Exercise M6 Menu

199 Answer Exercise M6 Question 6 6. Write an equation or inequation to represent: Steven pays $20 a month for the telephone account. Phone calls to his girlfriend in another town cost $2.75 each. He wants to call his girlfriend as much as possible. How many calls can he make but not spend more than $60 in a week? c < 60 c = number of calls – 20 – c < 40  2.75  2.75 c < Therefore 14 calls NextPrevious Back to Exercise M6 Menu

200 Answer Exercise M6 Question 7 7. Write an equation or inequation to represent: Marama is increasing the distance she jogs by one more telephone pole length, which means she adds 24 m each day she jogs. If on day 1 she jogged 2 km (2000 m), on what day’s jogging will she be jogging more than 5 km? ( d – 1) > 5000 d = number of days d > 5000 – 1976 – d > 3024  24  24 d > 126 Therefore Day 127 NextPrevious Back to Exercise M6 Menu

201 Answer Exercise M6 Question g + 30 = 9500 g = number of glasses Note: Could be < instead of = – 30 – g = 9470  150  150 g = Therefore 63 glasses 8. Write an equation or inequation to represent: From a 9.5 L (9500 ml) drink container, 150 ml glasses are being poured. If 30 ml was spilt from the container, how many glasses of drink could be poured? Back to Exercise M6 Menu Next MenuPrevious

202 Exercise M7 Menu Solving simple quadratic equations x x + 5 = 0 x x – 12 = 0 x x = 16 x 2 – 3 x = 10 Q5. Q3. Q1. Q7. x x + 18 = 0 x 2 – 5 x – 24 = 0 x x = -24 x 2 – 2 x = 8 Q6. Q4. Q2. Q8. Next Solve the simple quadratic equations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Back to Merit Menu

203 Exercise M7 Question 1 1. Solve: x x + 5 = 0 Pairs of factors of 5 are 1, 5; - 1, - 5 Pair that add to give 6 are 1, 5 Therefore ( x + 1)( x + 5) = 0 x + 1 = 0 or x + 5 = 0 – 1 – 1 – 5 – 5 x = - 1 or x = - 5 Next Back to Exercise M7 Menu

204 Exercise M7 Question 2 2. Solve: x x + 18 = 0 Pairs of factors of 18 are: 1, 18; 2, 9; 3, 6; - 1, - 18; - 2, - 9; - 3, - 6 Pair that add to 9 are 3, 6 Therefore ( x + 3)( x + 6) = 0 x + 3 = 0 or x + 6 = 0 – 3 – 3 – 6 – 6 x = - 3 or x = - 6 NextPrevious Back to Exercise M7 Menu

205 Exercise M7 Question 3 3. Solve: x x – 12 = 0 Pairs of factors of - 12 are: 1, - 12; 2, - 6; 3, - 4; 4, - 3; 6, - 2; 12, - 1 Pair that add to 4 are 6, - 2 Therefore ( x + 6)( x – 2) = 0 x + 6 = 0 or x – 2 = 0 – 6 – x = -6 or x = 2 NextPrevious Back to Exercise M7 Menu

206 Exercise M7 Question 4 4. Solve: x 2 – 5 x – 24 = 0 Pairs of factors of - 24 are: 1, - 24; 2, - 12; 3, - 8; 4, - 6; 6, - 4; 8, - 3; 12, - 2; 24, - 1 Pair that add to - 5 are 3, - 8 Therefore ( x + 3)( x – 8) = 0 x + 3 = 0 or x – 8 = 0 – 3 – x = - 3 or x = 8 NextPrevious Back to Exercise M7 Menu

207 Exercise M7 Question 5 5. Solve: x x = 16 Get all terms on left by – 16 giving x x – 16 = 0 Pairs of factors of - 16 are: 1, - 16; 2, - 8; 4, - 4; 8, - 2; 16, - 1 Pair that add to 6 are 8, - 2 Therefore ( x + 8)( x – 2) = 0 x + 8 = 0 or x – 2 = 0 – 8 – x = - 8 or x = 2 NextPrevious Back to Exercise M7 Menu

208 Exercise M7 Question 6 6. Solve: x x = - 24 Get all terms on left by + 24 giving x x + 24 = 0 Pairs of factors of 24 include: 1, 24; 2, 12; 4, 6 Pair that add to 10 are 4, 6 Therefore ( x + 4)( x + 6) = 0 x + 4 = 0 or x + 6 = 0 – 4 – 4 – 6 – 6 x = - 4 or x = - 6 NextPrevious Back to Exercise M7 Menu

209 Exercise M7 Question 7 7. Solve: x 2 – 3 x = 10 Get all terms on left by – 10 giving x 2 – 3 x – 10 = 0 Pairs of factors of - 10 are: 1, - 10; 2, - 5; 5, - 2; 10, - 1 Pair that add to - 3 are 2, - 5 Therefore ( x + 2)( x – 5) = 0 x + 2 = 0 or x – 5 = 0 – 2 – x = -2 or x = 5 NextPrevious Back to Exercise M7 Menu

210 Exercise M7 Question 8 8. Solve: x 2 – 2 x = 8 Get all terms on left by – 8 giving x 2 – 6 x – 8 = 0 Pairs of factors of - 8 are: 1, - 8; 2, - 4; 4, - 2; 8, - 1 Pair that add to - 2 are 2, - 4 Therefore ( x + 2)( x – 4) = 0 x + 2 = 0 or x – 4 = 0 – 2 – x = - 2 or x = 4 Previous Back to Exercise M7 Menu Next Menu

211 Exercise M8 Menu Solving harder quadratic equations Next 2 x x – 56 = 0 4 x x = p p + 12 = 0 4 x x + 25 = 0 Q5. Q3. Q1. Q7. 3 x x + 9 = 0 3 x 2 – 19 x + 20 = 0 4 x 2 – 7 x + 3 = 0 3 x x = 14 Q6. Q4. Q2. Q8. Solve the quadratic equations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Back to Merit Menu

212 Exercise M8 Question 1 1. Solve: 2 x x – 56 = 0 Common factor of 2, so divide by thisleaving x x – 28 = 0 Factors of - 28 are: 1, - 28; 2, - 14; 4, - 7; 7, - 4; 14, - 2; 28, - 1 Pair that add to give 3 are 7, - 4 Therefore ( x + 7)( x – 4) = 0 x + 7 = 0 or x – 4 = 0 – 7 – x = - 7 or x = 4 Next Back to Exercise M8 Menu

213 Exercise M8 Question 2 2. Solve: 3 x x + 9 = 0 Common factor of 3, so divide by this leaving x x + 3 = 0 Factors of 3 are 1, 3; - 1, - 3; Pair that add to give 4 are 1, 3 Therefore ( x + 1)( x + 3) = 0 x + 1 = 0 or x + 3 = 0 – 1 – 1 – 3 – 3 x = - 1 or x = - 3 NextPrevious Back to Exercise M8 Menu

214 Exercise M8 Question 3 3. Solve: 4 x x = 288 Common factor of 4, so divide by this leaving x 2 + x = 72 Subtract 72 from both sides, so x 2 + x – 72 = 0 Factors of - 72 include 4, - 18; 6, - 12; 8, - 9; 9, - 8; Pair that add to give 1 are: 9, - 8 Therefore ( x + 9)( x – 8) = 0 x + 9 = 0 or x – 8 = 0 – 9 – x = - 9 or x = 8 NextPrevious Back to Exercise M8 Menu

215 Exercise M8 Question 4 4. Solve: 3 x 2 – 19 x + 20 = 0 No common factor; x 2 term and constant multiplied gives 60 x 2 Pairs that multiply to give 60 x 2 include: - x, - 60 x ; - 2 x, - 30 x ; - 3 x, - 10 x ; - 4 x, - 15 x ; - 5 x, - 12 x ; - 6 x, - 10 x Pair that add to give - 19 x are - 4 x and - 15 x 3 x 2 – 4 x – 15 x + 20 = 0 x (3 x – 4) – 5(3 x – 4) = 0 Therefore (3 x – 4)( x – 5) = 0 3 x – 4 = 0 or x – 5 = 0 3 x = x = 4 / 3 or x = 5 NextPrevious Back to Exercise M8 Menu

216 Exercise M8 Question 5 5. Solve:5 p p + 12 = 0 No common factor; x 2 term and constant multiplied gives 60 p 2 Pairs that multiply to give 60 x 2 include: p, 60 p ; 2 p, 30 p ; Pair that add to give to 32 p are 2 p and 30 p 5 p p + 30 p + 12 = 0 p (5 p + 2) + 6(5 p + 2) = 0 Therefore (5 p + 2)( p + 6) = 0 5 p + 2 = 0 or p + 6 = 0 5 p = - 2 – 6 – 6 p = -2 / 5 or x = -6 NextPrevious Back to Exercise M8 Menu

217 Exercise M8 Question 6 6. Solve: 4 x 2 – 7 x + 3 = 0 No common factor; x 2 term and constant multiplied gives 12 x 2 Pairs that multiply to give 12 x 2 include: - x, - 12 x ; - 2 x, - 6 x ; - 3 x, - 4 x ; Pair that add to give to - 7 x are - 3 x and - 4 x 4 x 2 – 3 x – 4 x + 3 = 0 x (4 x – 3) – 1(4 x – - 3) = 0 Therefore (4 x – 3)( x – 1) = 0 4 x – 3 = 0 or x – 1 = 0 4 x = x = 3 / 4 or x = 1 NextPrevious Back to Exercise M8 Menu

218 Exercise M8 Question 7 7. Solve:4 x x + 25 = 0 No common factor; x 2 term and constant multiplied gives 100 x 2 Pairs that multiply to give 100 x 2 include: 2 x, 50 x ; 4 x, 25 x ; 5 x, 20 x ; 10 x, 10 x ; Pair that add to give 20 x are 10 x and 10 x 4 x x + 10 x + 25 = 0 2 x (2 x + 5) + 5(2 x + 5) = 0 Therefore (2 x + 5)(2 x + 5) = 0 2 x + 5 = 0 Note: Other factor gives same result 2 x = -5 x = -5 / 2 or NextPrevious Back to Exercise M8 Menu

219 Exercise M8 Question 8 8. Solve: 3 x x = 14 No common factor; take 14 from both sides 3 x x – 14 = 0 x 2 term and constant multiplied gives - 42 x 2 Pairs that multiply to give - 42 x 2 include: 7 x, - 6 x ; 14 x, - 3 x ; 21 x, - 2 x ; Pair that add to give 19 x are 21 x and - 2 x 3 x x – 2 x – 14 = 0 3 x ( x + 7) – 2( x + 7) = 0 Therefore ( x + 7)(3 x – 2) = 0 x + 7 = 0 or 3 x – 2 = 0 – 7 – 7 3 x = 2 x = - 7 or x = 2 / 3 Previous Back to Exercise M8 Menu Next Menu

220 Exercise M9 Menu Solve simultaneous equations using elimination 2 x + 7 y = x – 7 y = 16 4 x + 9 y = x – 3 y = f + 6 g = 26 f + g = 4 4 k + 7 n = 41 3 k – 2 n = 9 3 t + 4 u = 25 5 t – 6 u = x – 4 y = x + 20 y = 20 7 x + 5 y = 11 4 x – 10 y = d – 20 e = d + 4 e = p + 11 q = 91 4 p – 2 q = 14 2 x + 8 y = 13 5 x – 7 y = - 62 Q5. Q3. Q1. Q7. Q9. Q6. Q4. Q2. Q8. Q10. Next Solve these equations. Press the button for the correct answer or scroll using the ‘next’ arrow (below). Click to move through the solution. Back to Merit Menu

221 Exercise M9 Question 1 1. Solve: 2 x + 7 y = x – 7 y = 16 As 7 y and - 7 y will cancel out, add the two equations to get: 14 x = 84 gives x = 6 Substitute into first equation to find y y = 68 gives 7 y = 56 so y = 8 Check second equation 12  6 – 7  8 = 72 – 56 = 16 x = 6, y = 8 Next Back to Exercise M9 Menu

222 Exercise M9 Question 2 2.Solve: 3 x – 4 y = x + 20 y = 20 As 3 x and - 3 x will cancel out, add the two equations to get: 16 y = 40 gives y = 2.5 Substitute into first equation to find x 3 x – 10 = 20 gives 3 x = 30 so x = 10 Check second equation - 3   2.5 = = 20 x = 10, y = 2.5 NextPrevious Back to Exercise M9 Menu

223 Exercise M9 Question 3 3. Solve: 4 x + 9 y = x – 3 y = - 1 Need to multiply equation 2 by 3 so 9 y and - 9 y will cancel out (Note: could have been by 2 to eliminate the x with 4 x and - 4 x ) - 6 x – 9 y = - 3 Add this new equation to the first equation - 2 x = 8 gives x = - 4 Substitute into first equation to find y y = 11 gives 9 y = 27 so y = 3 Check second equation - 2  - 4 – 3  3 = 8 – 9 = - 1 x = - 4, y = 3 NextPrevious Back to Exercise M9 Menu

224 Exercise M9 Question 4 4. Solve:7 x + 5 y = 11 4 x – 10 y = 32 Need to multiply equation 1 by 2 so 10 y and - 10 y will cancel out 14 x + 10 y = 22 Add this new equation to the second equation 18 x = 54 gives x = 3 Substitute into first equation to find y y = 11 gives 5 y = - 10 so y = - 2 Check second equation 4  3 – 10  - 2 = 12 – - 20 = 32 x = 3, y = - 2 NextPrevious Back to Exercise M9 Menu

225 Exercise M9 Question 5 5. Solve:5 f + 6 g = 26 f + g = 4 Need to multiply equation 2 by - 5 so 5 f and - 5 f will cancel out (Note could have been by - 6 to eliminate using 6 g and - 6 g ) - 5 f g = - 20 Add this new equation to the second equation gives g = 6 Substitute into first equation to find y 5 f + 36 = 26 gives 5 f = - 10 so f = - 2 Check second equation = 4 f = - 2, g = 6 NextPrevious Back to Exercise M9 Menu

226 Exercise M9 Question 6 6. Solve:100 d – 20 e = d + 4 e = 17.2 Multiply equation 2 by 5 so - 20 e and 20 e will cancel out (Note could have been by - 20 to eliminate using 100 d and d ) 25 d + 20 e = 86 Add this new equation to the first equation 125 d = 200 gives d = 1.6 Substitute into first equation to find e 160 – 20 e = 114 gives - 20 e = - 46 so e = 2.3 Check second equation 5   2.3 = = 17.2 d = 1.6, e = 2.3 NextPrevious Back to Exercise M9 Menu

227 Exercise M9 Question 7 7. Solve:4 k + 7 n = 41 3 k – 2 n = 9 Multiply eqn 1 by 2 and eqn 2 by 7 so 14 n and - 14 n will cancel out. (Note could have been eqn 1 by 3 and eqn 2 by - 4 to eliminate using 12 k and - 12 k ) 8 k + 14 n = k – 14 n = 63 Add these 29 k = 145 gives k = 5 Substitute into first equation to find n n = 41 gives 7 n = 21 so n = 3 Check second equation 3  5 – 2  3 = 15 – 6 = 9 k = 5, n = 3 NextPrevious Back to Exercise M9 Menu

228 Exercise M9 Question 8 8. Solve:34 p + 11 q = 91 4 p – 2 q = 14 Multiply eqn 1 by 2 and eqn 2 by 11 so 22 q and - 22 q will cancel out. (Note could have been eqn 1 by 2 and eqn 2 by - 17 to eliminate using 68 p and - 68 p ) 68 p + 22 q = p – 22 q = 154 Add these112 p = 336 gives p = 3 Substitute into first equation to find q q = 91 gives 11 q = - 11 so q = - 1 Check second equation 4  3 – 2  - 1 = 12 – - 2 =14 p = 3, q = - 1 NextPrevious Back to Exercise M9 Menu

229 Exercise M9 Question 9 9. Solve:3 t + 4 u = 25 5 t – 6 u = 11.9 Multiply eqn 1 by 3 and eqn 2 by 2 so 12 u and - 12 u will cancel out. (Note could have been eqn 1 by 5 and eqn 2 by - 3 to eliminate using 15 t and - 15 t ) 9 t + 12 u = t – 12 u = 23.8 Add these 19 t = 98.8 gives t = 5.2 Substitute into first equation to find u u = 25 gives 4 u = 9.4 so u = 2.35 Check second equation 5  5.2 – 6  2.35 = 26 – 14.1 = 11.9 t = 5.2, u = 2.35 NextPrevious Back to Exercise M9 Menu

230 Exercise M9 Question Solve:2 x + 8 y = 13 5 x – 7 y = - 62 Multiply eqn 1 by 7 and eqn 2 by 8 so 56 y and - 56 y will cancel out. (Note could have been eqn 1 by 5 and eqn 2 by - 2 to eliminate using 10 x and - 10 x ) 14 x + 56 y = x – 56 y = Add these54 x = gives x = Substitute into first equation to find y y = 13 gives 8 y = 28 so y = 3.5 Check second equation 5  – 7  3.5 = – 24.5 = - 62 x = - 7.5, y = 3.5 Previous Back to Exercise M9 Menu Next Menu

231 Exercise M10 Menu Solve simultaneous equations using substitution x = y – 20 x + 3 y = 40 y = x 4 x – 2 y = 15 d = 5 e – 17 2 d + 6 e = 18 q = 7 – 6 p 4 p – 2 q = 14 y = - 3 x – 7 4 x – 2 y = - 12 Q5. Q3. Q1. Q7. Q9. Q6. Q4. Q2. Q8. Q10. Next Solve these equations. Press the button for the correct answer or scroll using the ‘next’ arrow (below). Click to move from step to step in the solution. y = 2 x x + y = 16 x = y – 11 2 x – 3 y = - 1 f = 2 g + 6 f + g = 4 k = 20 – n 2 n – 3 k = 8 y = 3 x + 17 y = 2 x – 3 Back to Merit Menu

232 Exercise M10 Question 1 1. Solve: y = 2 x x + y = 16 Replace y in equation 2 by what it equals in equation 1: 3 x + (2 x + 8) = 16 Simplify 5 x + 8 = 16 gives 5 x = 8 so x = 1.6 Substitute into first equation to find y y = 2  so y = 11.2 Check second equation 3  = = 16 x = 1.6, y = 11.2 Back to Exercise M10 Menu Next

233 Exercise M10 Question 2 2. Solve: x = y – 20 x + 3 y = 40 Replace x in equation 2 by what it equals in equation 1: ( y – 20) + 3 y = 40 Simplify 4 y – 20 = 40 gives 4 y = 60 so y = 15 Substitute into first equation to find x x = 15 – 20 so x = - 5 Check second equation  15 = = 40 x = - 5, y = 15 NextPrevious Back to Exercise M10 Menu

234 Exercise M10 Question 3 3. Solve: x = y – 11 2 x – 3 y = - 1 Replace x in equation 2 by what it equals in equation 1: 2( y – 11) – 3 y = - 1 Simplify - y – 22 = - 1 gives - y = 21 so y = - 21 Substitute into first equation to find x x = - 21 – 11 so x = - 32 Check second equation 2  - 32 – 3  - 21 = = - 1 x = - 32, y = - 21 NextPrevious Back to Exercise M10 Menu

235 Exercise M10 Question 4 4. Solve: y = x 4 x – 2 y = 15 Replace y in equation 2 by what it equals in equation 1: 4 x – 2(7 + 3 x ) = 15 Simplify - 2 x – 14 = 15 gives - 2 x = 29 so x = Substitute into first equation to find y y = so y = Check second equation 4  – 2  = = 15 x = , y = NextPrevious Back to Exercise M10 Menu

236 Exercise M10 Question 5 5. Solve: f = 2 g + 6 f + g = 24 Replace f in equation 2 by what it equals in equation 1: (2 g + 6) + g = 24 Simplify 3 g + 6 = 24 gives 3 g = 18 so g = 6 Substitute into first equation to find f f = 2  so f = 18 Check second equation = 24 f = 18, g = 6 NextPrevious Back to Exercise M10 Menu

237 Exercise M10 Question 6 6. Solve: d = 5 e – 17 2 d + 6 e = 18 Replace d in equation 2 by what it equals in equation 1: 2(5 e – 17) + 6 e = 18 Simplify 16 e – 34 = 18 gives 16 e = 52 so e = 3.25 Substitute into first equation to find d d = 5  3.25 – 17 so d = Check second equation 2   3.25 = = 18 d = , e = 3.25 NextPrevious Back to Exercise M10 Menu

238 Exercise M10 Question 7 7. Solve: k = 20 – n 2 n – 3 k = 8 Replace k in equation 2 by what it equals in equation 1: 2 n – 3(20 – n ) = 8 Simplify 5 n – 60 = 8 Note: take care with the – (..-..) to get + gives 5 n = 68 so n = 13.6 Substitute into first equation to find k k = 20 – 13.6 so k = 6.4 Check second equation 2  13.6 – 3  6.4 = 27.2 – 19.2 = 8 k = 6.4, n = 13.6 NextPrevious Back to Exercise M10 Menu

239 Exercise M10 Question 8 8. Solve: q = 7 – 6 p 4 p – 2 q = 14 Replace q in equation 2 by what it equals in equation 1: 4 p – 2(7 – 6 p ) = 14 Simplify 16 p – 14 = 14 Note: take care with the – (..-..) to get + gives 16 p = 28 so p = 1.75 Substitute into first equation to find q q = 7 – 6  1.75 so q = 7 – 10.5 = Check second equation 4  1.75 – 2  = = 14 p = 1.75, q = NextPrevious Back to Exercise M10 Menu

240 Exercise M10 Question 9 9. Solve: y = 3 x + 17 y = 2 x – 3 Replace y in equation 2 by what it equals in equation 1: 3 x + 17 = 2 x – 3 Simplify: – 17 from both sides 3 x = 2 x – x gives x = - 20 Substitute into first equation to find y y = 3  so y = = - 43 Check second equation y = 2  - 20 – 3 = - 43 as required x = - 20, y = - 43 NextPrevious Back to Exercise M10 Menu

241 Exercise M10 Question Solve: y = - 3 x – 7 4 x – 2 y = - 12 Replace y in equation 2 by what it equals in equation 1: 4 x – 2( - 3 x – 7) = - 12 Simplify 10 x + 14 = x = - 26 x = Substitute into first equation to find y y = - 3  – 7 so y = 7.8 – 7 = 0.8 Check second equation 4  – 2  0.8 = – 1.6 = - 12 x = - 2.6, y = 0.8 Previous Back to Exercise M10 Menu Next Menu

242 Exercise M11 Menu Solve simultaneous equations in context Q5. Q3. Q1. Next Question Find the price of a chicken pie. Find the price of the drink and chips. Find how many metres of wallpaper. Find the number of 2-legged pets. Q6. Q4. Q2. Find the adult price to swim at the pool. Find the time taken to prepare each sandwich. Find the size of the larger number. Find how many trees Ricky pruned. Q7.Q8. Solve each of the simultaneous equations. Press the button for question details or scroll to each question using the ‘next question’ arrow (below). Click to move from step to step in the solution. Back to Merit Menu

243 1.Chicken pies cost 50c more than mince pies. If 6 mince and 9 chicken pies cost a total of $40.50, solve the pair of simultaneous equations: c = m m + 9 c = 40.5 to find the price of a chicken pie. Answer Exercise M11 Question 1 Next Question Back to the Exercise 11 Menu

244 Question: Find the cost of a chicken pie Answer: c = m m + 9 c = 40.5 Substitute for c in the second equation: 6 m + 9( m + 0.5)= 40.5 Simplifying: 15 m + 4.5= 40.5 Giving: 15 m = 36 m = 2.4 Substitute into the first equation: c = m c = = 2.9 Read the problem and check the words for chicken pies at $2.90 and mince pies $2.40 Chicken pies 50c more than mince, 6 mince and 9 chicken pies = 6  $  $2.90 = $40.50 Chicken pie costs $2.90 Exercise M11 Q1 Answer Back to the question Next Question Back to the Exercise 11 Menu

245 2.A family with 2 adults and 3 children pay $14.80 to go swimming at the pool. Another group with 5 adults and 2 children pay $30.40 in total to swim at the pool. Solve the simultaneous equations 2 a + 3 c = a + 2 c = 30.4 to find the price for an adult to go swimming at the pool. Answer Exercise M11 Question 2 Next QuestionPrevious Back to the Exercise 11 Menu

246 Question: Find the adult price at a swimming pool. Answer:2 a + 3 c = a + 2 c = 30.4 Multiply first equation by 2: 4 a + 6 c = 29.6 Multiply second equation by - 3: - 15 a – 6 c = Adding gives: - 11 a = Dividing by - 11 gives: a = 5.6 This gives adult price $5.60, but best to check out by: Subsitute into the first equation c = 14.8 Giving: 3 c = 3.6 so c = 1.2 Check wording of problem with 2 adults, 3 children $14.80, 5 adults and 2 children (5  $  $1.20) = $30.40 Adults price is $5.60 Exercise M11 Q2 Answer Back to the question Next Question Back to the Exercise 11 Menu

247 3.Six bags of chips and 4 bottles of drink cost $ The money needed for the bottle of drink is 10 cents less than twice the price of the chips. Solve the simultaneous equations 6 c + 4 d = 24.8 d = 2c – 0.1 and use the results to find the price of 1 bottle of drink and 1 bag of chips. Answer Exercise M11 Question 3 Next QuestionPrevious Back to the Exercise 11 Menu

248 Question: Find the price of 1 bottle of drink and 1 bag of chips. Answer:6 c + 4 d = 24.8 d = 2 c – 0.1 Substitute for d in the first equation: 6 c + 4(2 c – 0.1) = 24.8 Simplifying: 14 c – 0.4 = c = 25.2 c = 1.8 Substitute into the second equation: d = 2  1.8 – 0.1 d = 3.5 Read the problem and check the words: 6 chips and 4 drinks = 6  $  $3.50 = $24.80 Drink ($3.50) is 10c less than twice chips Price of 1 drink and 1 chips = d + c = $ $1.80 Price is $5.30 Exercise M11 Q3 Answer Back to the question Next Question Back to the Exercise 11 Menu

249 Exercise M11 Question 4 4.A kitchen worker takes 17.6 minutes to prepare 10 sandwiches and 2 burgers. She takes 14.6 minutes to prepare 4 sandwiches and 5 burgers. Assuming the worker is working at a constant rate, solve the equations: 10 x + 2 y = x + 5 y = 14.6 to find the time taken to prepare each sandwich. Answer Next QuestionPrevious Back to the Exercise 11 Menu

250 Question: Find the time taken to prepare each sandwich. Answer: 10 x + 2 y = x + 5 y = 14.6 x = time for sandwich, y = time for burger Want x, so eliminate y by multiplying first equation by 5: 50 x + 10 y = 88 Multiply second equation by - 2: - 8 x – 10 y = Adding gives: 42 x = 58.8 Dividing by 42: x = 1.4 This gives time for sandwich 1.4 min, but best to check out by: Substitute into the first equation y = 17.6 giving: 2 y = 3.6 so y = 1.8 Check wording of problem with 10 sandwiches, 2 burgers 17.6 min, 4 sandwiches and 5 burgers (4   1.8) = 14.6 Time to prepare sandwich = 1.4 min Exercise M11 Q4 Answer Back to the question Next Question Back to the Exercise 11 Menu

251 Answer Exercise M11 Question 5 5.A formula is used to work out how many metres of wallpaper is needed for a standard room, using the distance around the room and the total width of the windows. The formula gives m for a room with perimeter 28 m and with windows with total width 3.6 m. The formula gives 71.1 m for a room with perimeter 16 m and windows with total length 1.5 m. Solve the equations: 28 x y = x y = 71.1 to find out how many metres of wallpaper is allowed per metre of perimeter and per metre of total window width. Next QuestionPrevious Back to the Exercise 11 Menu

252 Question: Find out how many metres of wallpaper. Answer:28 x y = x y = 71.1 x = perimeter of room, y = width of window Eliminate y by multiplying first equation by 1.5: 42 x y = Multiply second equation by -3.6: x – 5.4 y = Adding: x = Dividing by : x = 4.5 Substitute into first equation y = giving: 3.6 y = so y = Check wording of problem: with 28 m perimeter, 3.6 m window = , 16 m perimeter, 1.5 m windows (16   - 0.6) = 71.1 Formula allows 4.5 m per metre perimeter and takes 0.6 m per window width metre. Exercise M11 Q5 Answer Back to the question Next Question Back to the Exercise 11 Menu

253 Exercise M11 Question 6 6.The sum of two numbers is 98 and the larger number is five more than twice the smaller number. Solve the equations: x + y = 98 y = 2 x + 5 to find the size of the larger number. Answer Next QuestionPrevious Back to the Exercise 11 Menu

254 Question:Find the size of the larger number. Answer: x + y = 98 y = 2 x + 5 Substitute for y in the first equation: x + (2x + 5) = 98 Simplifying 3 x + 5 = 98 Giving 3 x = 93 x = 31 Substitute into the second equation: y = 2  y = 67 Read the problem and check the words Sum of 31 and 67 is is twice 31 plus 5 Question asks for larger number Larger number is 67 Exercise M11 Q6 Answer Next Question Back to the question Back to the Exercise 11 Menu

255 Exercise M11 Question 7 7.There are 48 pets in a pet parade. The pets either have 4 legs or 2 legs. If there were 150 legs on the pets, solve the equations: x + y = 48 4 x + 2 y = 150 to find how many 2-legged pets are in the parade. Answer Next QuestionPrevious Back to the Exercise 11 Menu

256 Question:Find the number of 2-legged pets. Answer: x + y = 48 4 x + 2 y = 150 y = number of 2-legged pets. Multiply the first equation by x – 4 y = Adding this to the second equation gives - 2 y = - 42 y = 21 As a check, substitute into the first equation, x + 21 = 48 so x = 27 Check legs: 4   21 = = 150 Number of 2-legged pets is 21 Exercise M11 Q7 Answer Back to the question Next Question Back to the Exercise 11 Menu

257 Exercise M11 Question 8 8.If Ricky had pruned 6 more trees he would have pruned twice as many as Steve. Together they pruned 96 trees. Solve the equations: x + 6 = 2 y x + y = 96 to find how many trees Ricky pruned. Answer Previous Back to the Exercise 11 Menu Excellence Menu

258 Question: Find how many trees Ricky pruned. Answer: x + 6 = 2 y can be rearranged to give x – 2 y = - 6 (1) We also know that x + y = 96 (2) x = number of trees Ricky pruned Multiplying equation (2) by 2 gives 2 x + 2 y = 192 (3), Adding equation (3) to equation (1) gives 3 x = 186 Therefore x, the number of trees Ricky pruned, = 62 As a check, substitute into the first equation, = 2 y Therefore y = 34 This also fits the second equation x + y = 96 Exercise M11 Q8 Answer Back to the question Excellence Menu Back to Merit Menu Back to Main Menu

259 Excellence Menu: Exercise E1 Solve equations in context by modelling How many movies were shown each week 5 years ago? Estimate the charge to erect a fence 50 m long. Calculate the worth when the cabinet is full. Calculate is the area of the section. How much do they each earn? Give the dimensions of each of the rectangular regions. Give the maximum length of the garden. Give the side lengths of the original field. Calculate how many tile layers there are. Calculate how many weeks. Q1. Q8. Q3. Q5. Q7. Q2. Q4. Q6. Q9.Q10. Solve each of the equations. Press the button for question details or scroll to each question using the ‘next question’ arrow (below). Click to move from step to step in the solution. Next Question Back to Main Menu

260 Exercise E1 Question 1 1.The average number of movies shown in a movie complex each week this year has increased by 8 since 5 years ago. The average number of viewers per movie was 4 times the number of weekly movies shown 5 years ago, but is now half that number. If there are now 480 viewers per week, how many movies were shown each week 5 years ago? Back to Excellence Menu Answer Next Question

261 Exercise E1 Answer 1 1.How many movies were shown each week 5 years ago? Let number of movies per week 5 years ago = x. The number of movies shown each week has increased by 8 since 5 years ago. Number of movies per week now = x + 8 The average number of viewers per movie was 4 times the number of weekly movies 5 years ago so: Average number of viewers per movie was 4 x but is now half that number so: Average number of viewers per movie is 2 x There are now 480 viewers per week Number of viewers per week now = average number per movie x number of movies per week = 2 x ( x + 8) 2 x ( x + 8) = x x – 480 = 0 Dividing by 2 x x – 240 = 0 Factors of include: - 10, 24; - 12, 20 ( x – 12)( x + 20) = 0 x = - 20 is not realistic. Therefore x = movies shown each week 5 years ago. Back to question Next Question Back to Excellence Menu

262 Exercise E1 Question 2 2.A building floor plan has 2 rectangular regions. The left region has one wall completely in common with the right region. The left region has length 5 m longer than its width. The left region has the shared wall ¾ of the width of the right region and the left region is ½ of the length of the right region. If the total perimeter around the building is 186 m, give the dimensions of each of the rectangular regions. Answer PreviousNext Question Back to Excellence Menu

263 Exercise E1 Answer 2 2.Give the dimensions of each of the rectangular regions. Let the width of the left region = x The left region has length 5m longer than its width. Length of left region = x + 5 The left region has the shared wall ¾ of the width of the right region Width of right region = 4 x  3 and it is ½ of the length of the right region. Length of right region = 2( x + 5) The total perimeter around the building is 186 m 6( x + 5) + 2(4 x  3) = x  = 186 x = 156 x 3  26 = 18 Therefore the smaller region is 18 m by 23 m and the larger region is 46 m by 24 m. Back to question Next Question Back to Excellence Menu

264 Exercise E1 Question 3 3.A fencing firm have a standard fee that they charge for labour and for the number of metres of fencing they build. They charged one customer $ for a 165 m long fence that takes 3 hours to erect. Another customer is charged $ for a fence 42 m long that takes 1.3 hours to erect. What will they estimate as a charge to erect a fence 50 m long that they expect to take 1.5 hours to erect? Next Question Answer Previous Back to Excellence Menu

265 Exercise E1 Answer 3 3.Fencers have a standard fee that they charge for labour and for the number of metres of fencing they build Let charge rate for labour = $ x per hour and cost per metre of fence = y. They charge one customer $ for a 165 m long fence that takes 3 hours to erect. 3 x y = Another customer is charged $ for a fence 42 m long that takes 1.3 hours to erect. 1.3 x + 42 y = To solve these, multiply equation 1 by 1.3: 3.9 x y = Multiply equation 2 by - 3: x – 126 y = Adding gives: 88.5 y = y = 14.5 Substitute into equation 1 3 x = leading to 3 x = 85.9 x = 28.3 Therefore cost for labour is $28.30 per hour and cost per metre of fence is $ What will they estimate as a charge to erect a fence 50 m long that they expect to take 1.5 hours to erect? Charge = (1.5 x $28.30) + (50 x $14.50) = $ Back to question Next Question Back to Excellence Menu

266 Exercise E1 Question 4 4. Concrete is needed for a path that is to surround four sides of a rectangular garden. It is to be 0.85 m wide and 0.05 m thick. The garden is 2 metres longer than it is wide. If 0.8 cubic metres of concrete is needed, what is the maximum length the garden can be? Answer PreviousNext Question Back to Excellence Menu

267 Exercise E1 Answer 4 4. What is the length of the garden? Let the length of the garden = x The garden is 2 metres longer than it is wide.. Width of garden = x – 2 Concrete is to be 0.85 m wide and 0.05 m thick. Area of concrete = 2 x 0.85 x x + 2 x 0.85 x ( x – 2) + 4 x = 1.7 x x – = 3.4 x – 0.51 Volume of concrete = (3.4 x – 0.51) x 0.05 = 0.17 x – cubic metres of concrete is needed 0.17 x – < x < x < Therefore maximum length is 4.85 m Back to question Next Question Back to Excellence Menu

268 Exercise E1 Question 5 5.A jeweller displays watches in a cabinet. The watches are all the same price. When the cabinet is ¾ full, the value of the cabinet and watches is $ When the cabinet is 1 / 3 full, the combined value is $ What is the cabinet and watches worth when it is full? Answer PreviousNext Question Back to Excellence Menu

269 Exercise E1 Answer 5 5.What is the cabinet and watches worth when it is full? Let value of cabinet = $ c and value of watches = $ w. When the cabinet is ¾ full, the value of the cabinet and watches is valued at $ c w = When the cabinet is 1/3 full, the combined value is $ c + w = To solve these, multiply equation 2 by -1 - c – w = Adding gives: w = w = Substitute into equation 1 c = c = 1200 Therefore cost for cabinet full of watches is $ $ = $ Back to question Next Question Back to Excellence Menu

270 Exercise E1 Question 6 6.A square field has a right-angled isosceles triangle area cut from one corner. The sides of the rectangular field where the corner has been cut off are reduced to 2 / 3 of their original length. If the resulting field has a perimeter of 96 m, what was the original side lengths of the field? Answer PreviousNext Question Back to Excellence Menu

271 Exercise E1 Answer 6 6.What was the original side lengths of the field? Let the length of the original sides = x A square field has a right-angled isosceles triangle area cut from one corner The sides off the rectangular field where the corner has been cut off are reduced to 2 / 3 of their original length. The resulting field has a perimeter of 96 m Diagonal length= = = x Perimeter= 2 x + 2( 2 / 3 x ) x = x x = 96 x = m Original side= m Back to question Next Question Back to Excellence Menu

272 Exercise E1 Question 7 7.A right-angled triangle section has the longest side 4 m longer than the shortest side. The other side is 2 m longer than the shortest side. What is the area of the section? Answer PreviousNext Question Back to Excellence Menu

273 Exercise E1 Answer 7 7. A right-angled triangle section has the longest side 4 m longer than the shortest side. Let the shortest side = x Longest side = x + 4 The other side is 2 m longer than the shortest side. Length of other side = x + 2 What is the area of the section? Using Pythagoras’ theorem, ( x + 4) 2 = ( x + 2) 2 + x 2 x x + 16= x x x 2 x x + 16= 2 x x + 4 x 2 – 4 x – 12= 0 ( x – 6)( x + 2)= 0 x = 6 (since x = - 2 not sensible) Area = 0.5  6  8 = 24 m 2 Back to question Next Question Back to Excellence Menu

274 Exercise E1 Question 8 8.A pattern of tiles is shown in the diagram. If 754 tiles are used, how many layers are there? Answer PreviousNext Question Back to Excellence Menu

275 Exercise E1 Answer 8 8. Number of tiles 4, 10, 18, 28, 40, … Difference between terms is 6, 8, 10, 12, … This is increasing at a constant rate of 2 Therefore quadratic pattern involving 1 n 2 Creating table shows difference between 1 n 2 and number of tiles is 3, 6, 9, 12, … Therefore pattern is n n If 754 tiles are used, how many layers are there? n n = 754 n n – 754= 0 Factors of include - 26, 29 ( n – 26)( n + 29)= not sensible so 26 layers. Layers, n Tiles n2n2 diff Back to question Next Question Back to Excellence Menu

276 Exercise E1 Question 9 9.If Susan was paid $30 more a week, she would earn half as much as Sophie. Together they are paid $510. How much do they each earn? Previous Let m = money that Susan earns. Sophie earns 2  ( m + 30) = 2 m + 60 Together they earn m + (2 m + 60) = m + 60= m = 450 (– 60 from both sides) m = 150 (  3 on both sides) Susan earns $150 and Sophie earns $360 Next Question Back to Excellence Menu

277 Exercise E1 Question Nathan decides to exercise each week by jogging. He starts in week 1 by jogging for 20 minutes. In week 2 he jogs for 30 minutes; in week 3 he jogs for 40 minutes, and so on. He keeps a record of the total time he spends jogging. After how many weeks is his total minutes? Answer Previous Back to Excellence Menu EXIT

278 Exercise E1 Answer Total number of minutes is 20, 50, , … Difference between terms is 30, 40, 50, … This is increasing at a constant rate of 10 Therefore quadratic pattern involving 5 n 2 Creating table shows difference between 5 n 2 and number of minutes is 15, 30, 45, 60, … Therefore pattern is 5 n n After how many weeks is his total minutes? 5 n n = n n – 3654= 0 Factors of include - 59, 62 ( n – 59)( n + 62)= not sensible so Week 59. Weeks, n Total5n25n2 diff Back to question EXIT Back to Excellence Menu Back to MAIN Menu


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