1. Simultaneous Equations Please choose a question to attempt from the following: 1a 1b 2a 3b EXIT 3a 2b4b 4a

2. Simultaneous Equations : Question 1 Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only EXIT The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2, hence solve 3x + 2y = 17 y = 2x – 2 3x + 2y = 17. Get hint

3. Simultaneous Equations : Question 1 Go to full solution Go to CommentsReveal answer only EXIT The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2, hence solve 3x + 2y = 17 y = 2x – 2 3x + 2y = 17. To draw graph: Construct a table of values with at least 2 x- coordinates. Plot and join points. Solution is where lines cross What would you like to do now?

4. Simultaneous Equations : Question 1 EXIT The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2, hence solve 3x + 2y = 17 y = 2x – 2 3x + 2y = 17. Solution is x = 3 & y = 4 What would you like to do now? Go to full solution Go to CommentsReveal answer only

5. Comments Begin Solution Question 1 Back to Home The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2, hence solve 3x + 2y = 17 y = 2x – 2 1. Construct a table of values with at least 2 x coordinates. y = 2x - 2x 0 5 y -2 8

6. Question 1 Back to Home The diagram shows the graph of 3x + 2y = 17. Copy the diagram and on your diagram draw the graph of y = 2x – 2, hence solve 3x + 2y = 17 y = 2x – 2 2. Plot and join points. Solution is where lines cross. Solution is x = 3 & y = 4 What would you like to do now? Comments Begin Solution

7. Markers Comments Back to Home Next Comment 1. Construct a table of values with at least 2 x coordinates. y = 2x - 2x 0 5 y -2 8 There are two ways of drawing the line y = 2x - 1 Method 1 Finding two points on the line: x = 0 y = 2 x 0 – 1 = -1 x = 2 y = 2 x 2 - 1 = 3 First Point (0,-1) Second Point (-1,3) Plot and join (0,-1), and (-1,3). Begin Comment

8. Markers Comments Back to Home 2. Plot and join points. Solution is where lines cross. Solution is x = 3 & y = 4 Method 2 Using y = mx + c form: y = mx + c gradient y - intercept Plot C(0, -1) and draw line with m = 2 y = 2x - 1 gradient m = 2 y - intercept c = -1 Begin Comment

9. Go to full solution Go to CommentsReveal answer only EXIT Simultaneous Equations : Question 1B The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 Get hint

10. EXIT Simultaneous Equations : Question 1B The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 To draw graph: Construct a table of values with at least 2 x- coordinates. Plot and join points. Solution is where lines cross What would you like to do now? Go to full solution Go to CommentsReveal answer onlyGet hint

11. EXIT Simultaneous Equations : Question 1B The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 Solution is x = 2 & y = 4 What would you like to do now? Go to full solution Go to CommentsReveal answer onlyGet hint

12. Comments Begin Solution Continue Solution Question 1B Back to Home The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 1. Construct a table of values with at least 2 x coordinates. x 0 6 y 3 6

13. Begin Solution Question 1B Back to Home The diagram shows the graph of 2x + y = 8. Copy the diagram and on your diagram draw the graph of y = 1 / 2 x + 3, hence solve 2x + y = 8 y = 1 / 2 x + 3 x 0 6 y 3 6 2. Plot and join points. Solution is where lines cross. Solution is x = 2 & y = 4 What would you like to do now? Comments

14. Markers Comments Back to Home Next Comment 1. Construct a table of values with at least 2 x coordinates. y = 1 / 2 x + 3x 0 6 y 3 6 There are two ways of drawing the line y = ½ x + 3 Method 1 Finding two points on the line: x = 0 y = ½ x 0 + 3 = 3 x = 2 y = ½ x 2 +3 = 4 First Point (0, 3) Second Point (2, 4) Plot and join (0, 3), and (2, 4).

15. Markers Comments Back to Home Next Comment y = 1 / 2 x + 3x 0 6 y 3 6 2. Plot and join points. Solution is where lines cross. Solution is x = 2 & y = 4 Method 2 Using y = mx + c form: y = mx + c gradient y - intercept Plot C(0, 3) and draw line with m = ½ y = ½ x + 3 gradient m = ½ y - intercept c = +3

16. Go to full solution Go to Comments Reveal answer only EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Get hint

17. EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Eliminate either variable by making coefficient same. Substitute found value into either of original equations. What would you like to do now? Go to full solution Go to Comments Reveal answer only Get hint

18. EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Solution is u = 2 & v = 1 What would you like to do now? Go to full solution Go to Comments Reveal answer only Get hint

19. Comments Begin Solution Continue Solution Question 2 Back to Home Solve 3u - 2v = 4 2u + 5v = 9 3u - 2v = 4 2u + 5v = 9 1. Eliminate either u’s or v’s by making coefficient same. (x5) (x2) Now get: 3 4 2 1 = (x5) (x2) 1 = 2 34 +

20. Comments Begin Solution Question 2 Back to Home Solve 3u - 2v = 4 2u + 5v = 9 3u - 2v = 4 2u + 5v = 9 2. Substitute found value into either of original equations. (x5) (x2) 2 1 Substitute 2 for u in equation 4 + 5v = 9 5v = 5 v = 1 Solution is u = 2 & v = 1 2 What would you like to do now?

21. Markers Comments Back to Home Next Comment 3u - 2v = 4 2u + 5v = 9 1. Eliminate either u’s or v’s by making coefficient same. (x5) (x2) Now get: 3 4 2 1 = (x5) (x2) 1 = 2 34 + Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x + 3y = 4 Subtract the equations

22. Markers Comments Back to Home Next Comment 3u - 2v = 4 2u + 5v = 9 1. Eliminate either u’s or v’s by making coefficient same. (x5) (x2) Now get: 3 4 2 1 = (x5) (x2) 1 = 2 34 + Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x - 3y = 4 Add the equations

23. Go to full solution Go to Comments Reveal answer only EXIT Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 Get hint

24. EXIT Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 Eliminate either variable by making coefficient same. Substitute found value into either of original equations. What would you like to do now? Go to full solution Go to Comments Reveal answer only Get hint

25. EXIT Simultaneous Equations : Question 2B Solve 5p + 3q = 0 4p + 5q = -2.6 Solution is q = -1 & q = 0.6 What would you like to do now? Go to full solution Go to Comments Reveal answer only Get hint

26. Comments Begin Solution Continue Solution Question 2B Back to Home 5p + 3q = 0 4p + 5q = -2.6 1. Eliminate either p’s or q’s by making coefficient same. (x4) (x5) Now get: 3 4 2 1 = (x4) (x5) 1 = 2 43 - Solve 5p + 3q = 0 4p + 5q = -2.6

27. Question 2B Back to Home 2. Substitute found value into either of original equations. Substitute -1 for q in equation 5p + (- 3) = 0 5p = 3 p =3/5 = 0.6 Solution is q = -1 & p = 0.6 1 Solve 5p + 3q = 0 4p + 5q = -2.6 5p + 3q = 0 4p + 5q = -2.6 (x4) (x5) 2 1 What would you like to do now? Comments Begin Solution

28. Markers Comments Back to Home Next Comment 5p + 3q = 0 4p + 5q = -2.6 1. Eliminate either p’s or q’s by making coefficient same. (x4) (x5) Now get: 3 4 2 1 = (x4) (x5) 1 = 2 43 - Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x + 3y = 4 Subtract the equations

29. Markers Comments Back to Home 5p + 3q = 0 4p + 5q = -2.6 1. Eliminate either p’s or q’s by making coefficient same. (x4) (x5) Now get: 3 4 2 1 = (x4) (x5) 1 = 2 43 - Note: When the “signs” are the same subtract to eliminate. When the “signs” are different add to eliminate. e.g.2x + 3y = 4 6x - 3y = 4 Add the equations

30. Go to full solution Go to Comments Reveal answer only EXIT Simultaneous Equations : Question 3 If two coffees & three doughnuts cost $2.90 while three coffees & one doughnut cost $2.60 then find the cost of two coffees & five doughnuts. Get hint

31. EXIT Simultaneous Equations : Question 3 If two coffees & three doughnuts cost $2.90 while three coffees & one doughnut cost $2.60 then find the cost of two coffees & five doughnuts. Form two equations, keeping costs in cents to avoid decimals. Eliminate either c’s or d’s by making coefficient same. Substitute found value into either of original equations. Remember to answer the question!!! What would you like to do now? Go to full solution Go to Comments Reveal answer only Get hint

32. EXIT Simultaneous Equations : Question 3 If two coffees & three doughnuts cost $2.90 while three coffees & one doughnut cost $2.60 then find the cost of two coffees & five doughnuts. Two coffees & five doughnuts = $3.90 Go to full solution Go to Comments Reveal answer only Get hint

33. Comments Begin Solution Question 3 Sim Eq Menu Back to Home 2c + 3d = 290 3c + 1d = 260 1. Form two equations, keeping costs in cents to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - If two coffees & three doughnuts cost $2.90 while three coffees & one doughnut cost $2.60 then find the cost of two coffees & five doughnuts. Let coffees cost c cents & doughnuts d cents then we have 2. Eliminate either c’s or d’s by making coefficient same. Try another like this

34. Comments Begin Solution Question 3 Back to Home 3. Substitute found value into either of original equations. Substitute 70 for c in equation 210 + d = 260 d = 50 If two coffees & three doughnuts cost $2.90 while three coffees & one doughnut cost $2.60 then find the cost of two coffees & five doughnuts. 2c + 3d = 290 3c + 1d = 260 (x1) (x3) 2 1 Let coffees cost c cents & doughnuts d cents then we have 2 Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) = $1.40 + $2.50 = $3.90 What would you like to do now?

35. Comments Back to Home Next Comment Step 1 Form the two simultaneous equations first by introducing a letter to represent the cost of a coffee ( c) and a different letter to represent the cost of a doughnut (d). 2c + 3d = 290 3c + 1d = 260 1. Form two equations, keeping costs in cents to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - Let coffees cost c cents & doughnuts d cents then we have 2. Eliminate either c’s or d’s by making coefficient same. i.e.2c +3d=290 1d +2c=260 Note change to cents eases working.

36. Comments Back to Home Step 2 Solve by elimination. Choose whichever variable it is easier to make have the same coefficient in both equations. 2c + 3d = 290 3c + 1d = 260 1. Form two equations, keeping costs in cents to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - Let coffees cost c cents & doughnuts d cents then we have 2. Eliminate either c’s or d’s by making coefficient same.

37. Comments Back to Home 3. Substitute found value into either of original equations. Substitute 70 for c in equation 210 + d = 260 d = 50 2c + 3d = 290 3c + 1d = 260 Let coffees cost c cents & doughnuts d cents then we have Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) = $1.40 + $2.50 = $3.90 Step 3 Once you have a value for one variable you can substitute this value into any of the equations to find the value of the other variable. It is usually best to choose an equation that you were given in question. Step 4 Remember to answer the question!!!

38. A company plan to introduce a new blend of tropical fruit drink made from bananas & kiwis. They produce three blends of the drink for market research purposes. Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? Go to full solution Go to CommentsReveal answer EXIT Simultaneous Equations : Question 3B Get hint

39. A company plan to introduce a new blend of tropical fruit drink made from bananas & kiwis. They produce three blends of the drink for market research purposes. Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? EXIT Simultaneous Equations : Question 3B Form two equations, eliminating decimals wherever possible. Eliminate either c’s or d’s by making coefficient same. Substitute found value into either of original equations. Remember to answer the question!!! What would you like to do now? Go to full solution Go to Comments Reveal answer

40. A company plan to introduce a new blend of tropical fruit drink made from bananas & kiwis. They produce three blends of the drink for market research purposes. Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? EXIT Simultaneous Equations : Question 3B So this blend is more expensive than the other two. Go to full solution Go to Comments

41. Comments Begin Solution Continue Solution Question 3B Back to Home 0.70B + 0.30K = 74 0.55B + 0.45K = 71 1. Form two equations, keeping costs in cents to avoid decimals. (x10) (x100) 3 4 2 1 = (x10) (x100) 1 = 2 2. Get rid of decimals Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? Let one litre of banana syrup cost B cents & one litre of kiwi syrup cost K cents.

42. Question 3B Back to Home 3 4 = (x15) 3 54 - Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? 3. Eliminate either B’s or K’s by making coefficient same. 5 4 Comments Begin Solution Continue Solution

43. Comments Question 3B Back to Home 3 4 (x15) Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p. Blend 2 has 55% banana syrup, 45% kiwi syrup and costs 71p per litre to produce. A third blend is made using 75% banana syrup and 25% kiwi. How does its cost compare to the other two blends? 4. Substitute found value into an equation without decimals. Substitute 80 for B in equation 560 + 3K = 740 3K = 180 3 K = 60 5. Use these values to answer question. 75%B+25%K =(0.75 x 80p)+(0.25 x 60p) = 60p + 15p = 75p So this blend is more expensive than the other two. What would you like to do now?

44. Markers Comments Back to Home Next Comment 0.70B + 0.30K = 74 0.55B + 0.45K = 71 1. Form two equations, keeping costs in cents to avoid decimals. (x10) (x100) 3 4 2 1 = (x10) (x100) 1 = 2 2. Get rid of decimals Let one litre of banana syrup cost B cents & one litre of kiwi syrup cost K cents. Step 1 Form the two simultaneous equations first by introducing a letter to represent the cost per litre of banana syrup ( B) and a different letter to represent the cost per litre of kiwis fruit (K). i.e.0.70 B + 0.30K =74 0.55B+ 0.45K =71 Multiply all terms by 100 to remove decimals. Note change to cents eases working.

45. Markers Comments Back to Home Next Comment 3 4 = (x15) 3 54 3. Eliminate either B’s or K’s by making coefficient same. 5 4 - Step 2 Solve by elimination. Choose whichever variable it is easier to make have the same coefficient in both equations.

46. Markers Comments Back to Home Next Comment Step 3 Once you have a value for one variable you can substitute this value into any of the equations to find the value of the other variable. It is usually best to choose an equation that you were given in question. Step 4 Remember to answer the question!!! 3 4 (x15) 4. Substitute found value into an equation without decimals. Substitute 80 for B in equation 560 + 3K = 740 3K = 180 3 K = 60 5. Use these values to answer question. 75%B+25%K =(0.75 x 80p)+(0.25 x 60p) = 60p + 15p = 75p So this blend is more expensive than the other two.

47. Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two termsadd first two to obtain 3 rd add 2 nd & 3 rd to obtain 4 th add 3 rd & 4 th to obtain 5 th etc eg starting with 3 & 7, next four terms are 10, 17, 27 & 44. (a)If the first two terms of such a sequence are P and Q then find expressions for the next 4 terms in their simplest form. (b)If the 5 th & 6 th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. Get hint

48. Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two termsadd first two to obtain 3 rd add 2 nd & 3 rd to obtain 4 th add 3 rd & 4 th to obtain 5 th etc eg starting with 3 & 7, next four terms are 10, 17, 27 & 44. (a)If the first two terms of such a sequence are P and Q then find expressions for the next 4 terms in their simplest form. (b)If the 5 th & 6 th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. Write down expressions using previous two to form next. To find values of P & Q : Match term from (a) with values given in question. Establish two equations. Eliminate either P’s or Q’s by making coefficient same. Solve and substitute found value into either of original equations. What would you like to do now?

49. Go to full solution Go to Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows……. Start with two termsadd first two to obtain 3 rd add 2 nd & 3 rd to obtain 4 th add 3 rd & 4 th to obtain 5 th etc eg starting with 3 & 7, next four terms are 10, 17, 27 & 44. (a)If the first two terms of such a sequence are P and Q then find expressions for the next 4 terms in their simplest form. (b)If the 5 th & 6 th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. P + Q,P + 2Q 2P + 3Q 3P + 5Q P = 7 Try another like this

50. Comments Begin Solution Question 4 Back to Home 1. Write down expressions using previous two to form next. (a)If the first two terms of such a sequence are P and Q then find expressions for the next 4 terms in their simplest form. (a) First term = P & second term = Q 3 rd term = P + Q 4 th term = Q + (P + Q) = P + 2Q 5 th term = (P + Q) + (P + 2Q) = 2P + 3Q 6 th term = (P + 2Q) + (2P + 3Q) = 3P + 5Q

51. Comments Begin Solution Question 4 Back to Home 1 2 = (x3) 1 43 - 1. Match term from (a) with values given in question. 3 4 (b) If the 5th & 6th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. (x2) 2. Eliminate either P’s or Q’s by making coefficient same. = 2

52. Begin Solution Question 4 Back to Home 3. Substitute found value into either of original equations. Substitute -2 for Q in equation 2P + (-6) = 8 2P = 14 2 P = 7 (b) If the 5th & 6th terms are 8 and 11 respectively then write down two equations in P and Q and hence find the values of P and Q. 1 2 (x3) (x2) First two terms are 7 and –2 respectively. What would you like to do now?

53. Comments Sim Eqs Menu Back to Home Next Comment 1. Write down expressions using previous two to form next. (a) First term = P & second term = Q 3 rd term = P + Q 4 th term = Q + (P + Q) = P + 2Q 5 th term = (P + Q) + (P + 2Q) = 2P + 3Q 6 th term = (P + 2Q) + (2P + 3Q) = 3P + 5Q For problems in context it is often useful to do a simple numerical example before attempting the algebraic problem. Fibonacci Sequence: 3, 7,10, 17, 27,44, …… P Q P + Q P + 2Q 4, 6,10, 16, 26,42, …… Then introduce the variables: P, Q,P + Q, P + 2Q, 2P + 3Q, …

54. Go to full solution Go to CommentsReveal answer EXIT Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 5 3 -4 -1 8 -1 -5 7 -6 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. 3v 2w w – 2v v + w (A) (B) v + w v – 3w 6w v - w -18 11 Get hint

55. EXIT Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 5 3 -4 -1 8 -1 -5 7 -6 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. 3v 2w w – 2v v + w (A) (B) v + w v – 3w 6w v - w -1811 Work your way to an expression for the top row by filling in the middle rows. Form 2 equations and eliminate either v’s or w’s by making coefficient same. Substitute found value into either of original equations. Go to full solution Go to CommentsReveal answer

56. EXIT Simultaneous Equations : Question 4B In a “Number Pyramid” the value in each block is the sum of the values in the two blocks underneath ….. 5 3 -4 -1 8 -1 -5 7 -6 1 The two number pyramids below have the middle two rows missing. Find the values of v and w. 3v 2w w – 2v v + w (A) (B) v + w v – 3w 6w v - w -1811 V = 4 Go to full solution Go to Comments Reveal answer

57. Comments Begin Solution Question 4B Back to Home 1. Work your way to an expression for the top row by filling in the middle rows. 3v 2w w – 2v v + w (A) -18 (B) v + w v – 3w 6w v - w 11 Pyramid (A) 2 nd row 3v + 2w, -2v + 3w, -v + 2w 3 rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 Pyramid (B) 2 nd row 2v - 2w, v + 3w, v + 5w 3 rd row 3v + w, 2v + 8w Top row 5v + 9w= 11

58. Comments Begin Solution Question 4B Back to Home 3v 2w w – 2v v + w (A) -18 (B) v + w v – 3w 6w v - w 11 2. Form 2 equations and eliminate either v’s or w’s by making coefficient same. 1 2 (x5) (x2) = (x5) 1 43 + 3 4 = 2 (x2) Now get: What would you like to do now?

59. Comments Begin Solution Continue Solution Question 4B Sim Eq Menu Back to Home 3v 2w w – 2v v + w (A) -18 (B) v + w v – 3w 6w v - w 11 1 2 (x5) (x2) 3. Substitute found value into either of original equations. Substitute -1 for W in equation 5V + (-9) = 11 5V = 20 V = 4 2

60. Comments Back to Home Next Comment 1. Work your way to an expression for the top row by filling in the middle rows. Pyramid (A) 2 nd row 3v + 2w, -2v + 3w, -v + 2w 3 rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 Pyramid (B) 2 nd row 2v - 2w, v + 3w, v + 5w 3 rd row 3v + w, 2v + 8w Top row 5v + 9w= 11 Use diagrams given to organise working: 3v + 2w 3w - 2v 2w - v -18 (A) 3v 2w w – 2v v + w 5w + v 5w - 3v (B) v + w v – 3w 6w v - w 11 2v - 2w 3w + v 5w + v 3v + w 8w + 2v

61. Markers Comments Back to Home 1. Work your way to an expression for the top row by filling in the middle rows. Pyramid (A) 2 nd row 3v + 2w, -2v + 3w, -v + 2w 3 rd row v + 5w, -3v + 5w Top row -2v + 10w = - 18 Pyramid (B) 2 nd row 2v - 2w, v + 3w, v + 5w 3 rd row 3v + w, 2v + 8w Top row 5v + 9w= 11 Hence equations: 10w - 2v = -18 9w + 5v = 11 Solve by the method of elimination. End of simultaneous Equations