INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Simultaneous Equations You have chosen to study: Please choose a question to attempt from the following:

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Simultaneous Equations You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 2 Menu

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

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. 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?

Simultaneous Equations : Question 1 Go to full solution Go to Comments Go to Sim Eq Menu 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 Try another like this What would you like to do now?

Comments Begin Solution Continue Solution Question 1 Sim Eq Menu 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

Begin Solution Try another like this 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 Comments Sim Eq Menu What would you like to do now?

Markers Comments Sim Eqs Menu 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).

Markers Comments Sim Eqs Menu Back to Home Next Comment 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

Go to full solution Go to Comments Go to Sim Eq Menu Reveal 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

Go to full solution Go to Comments Go to Sim Eq Menu Reveal 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 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 Comments Go to Sim Eq Menu 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?

Comments Begin Solution Continue Solution Question 1B Sim Eq Menu 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

Comments Begin Solution Continue Solution Question 1B Sim Eq Menu 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?

Markers Comments Sim Eqs Menu 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).

Markers Comments Sim Eqs Menu 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

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

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only 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 Go to Sim Eq Menu EXIT Simultaneous Equations : Question 2 Solve 3u - 2v = 4 2u + 5v = 9 Solution is u = 2 & v = 1 Try another like this What would you like to do now?

Comments Begin Solution Continue Solution Question 2 Sim Eq Menu 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 +

Comments Begin Solution Question 2 Sim Eq Menu 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 Try another like this What would you like to do now?

Markers Comments Sim Eqs Menu 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

Markers Comments Sim Eqs Menu 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

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

Go to full solution Go to Comments Go to Sim Eq Menu Reveal answer only 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 Go to Sim Eq Menu 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?

Comments Begin Solution Continue Solution Question 2B Sim Eq Menu 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

Comments Begin Solution Continue Solution Question 2B Sim Eq Menu 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?

Markers Comments Sim Eqs Menu 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

Markers Comments Sim Eqs Menu 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 Add the equations

Go to full solution Go to Comments Go to Sim Eq Menu 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

Go to full solution Go to Comments Go to Sim Eq Menu 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. Form two equations, keeping costs in pence 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 Go to Sim Eq Menu 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. Try another like this Two coffees & five doughnuts = £3.90

Comments Begin Solution Question 3 Sim Eq Menu Back to Home 2c + 3d = 290 3c + 1d = 260 1. Form two equations, keeping costs in pence 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 pence & doughnuts d pence then we have 2. Eliminate either c’s or d’s by making coefficient same. Try another like this

Comments Begin Solution Question 3 Sim Eq Menu 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 pence & doughnuts d pence then we have 2 Two coffees & five doughnuts = (2 x 70p) + (5 x 50p) = £1.40 + £2.50 = £3.90 Try another like this What would you like to do now?

Comments Sim Eqs Menu 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 pence to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - Let coffees cost c pence & doughnuts d pence 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 pence eases working.

Comments Sim Eqs Menu Back to Home Next Comment 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 pence to avoid decimals. (x1) (x3) 3 4 2 1 = (x1) (x3) 1 = 2 43 - Let coffees cost c pence & doughnuts d pence then we have 2. Eliminate either c’s or d’s by making coefficient same.

Comments Sim Eqs Menu Back to Home Next Comment 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 pence & doughnuts d pence 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!!!

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 Comments Go to Sim Eq Menu Reveal answer EXIT Simultaneous Equations : Question 3B Get hint

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 Comments Go to Sim Eq Menu Reveal answer 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?

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 Comments Go to Sim Eq Menu EXIT Simultaneous Equations : Question 3B So this blend is more expensive than the other two.

Comments Begin Solution Continue Solution Question 3B Sim Eq Menu Back to Home 0.70B + 0.30K = 74 0.55B + 0.45K = 71 1. Form two equations, keeping costs in pence 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 pence & one litre of kiwi syrup cost K pence.

Comments Begin Solution Continue Solution Question 3B Sim Eq Menu 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 Question 3B Sim Eq Menu 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?

Markers Comments Sim Eqs Menu Back to Home Next Comment 0.70B + 0.30K = 74 0.55B + 0.45K = 71 1. Form two equations, keeping costs in pence 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 pence & one litre of kiwi syrup cost K pence. 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 pence eases working.

Markers Comments Sim Eqs Menu 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.

Markers Comments Sim Eqs Menu 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.

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

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?

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

Comments Begin Solution Question 4 Sim Eq Menu 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 Try another like this

Comments Begin Solution Question 4 Sim Eq Menu 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 Try another like this

Comments Begin Solution Question 4 Sim Eq Menu 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. Try another like this What would you like to do now?

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, …

Go to full solution Go to Comments Go to Sim Eq Menu Reveal 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

Go to full solution Go to Comments Go to Sim Eq Menu Reveal 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 -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 Comments Go to Sim Eq Menu 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

Comments Begin Solution Continue Solution Question 4B Sim Eq Menu 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

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 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?

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

Comments Sim Eqs Menu 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

Markers Comments Sim Eqs Menu 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 Hence equations: 10w - 2v = -18 9w + 5v = 11 Solve by the method of elimination. End of simultaneous Equations

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Simultaneous Equations You have chosen to study: Please choose a question to attempt from the following:

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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Simultaneous Equations You have chosen to study: Please choose a question to attempt from the following:

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Presentation on theme: "INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Simultaneous Equations You have chosen to study: Please choose a question to attempt from the following:"— Presentation transcript:

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