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

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Please choose a question to attempt from the following:

1 2 3 4

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Simultaneous Equations : Question 1

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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.

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3x + 2y = 17.



y = 2x – 2 ,


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

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3x + 2y = 17.



y = 2x – 2 ,


y = 2x – 2

3x + 2y = 17.Solution is x = 3 & y = 4

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Question 1

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The diagram shows the graph

of 3x + 2y = 17.



y = 2x – 2 ,


y = 2x – 2

1. Construct a table of values with at least 2 x coordinates.

y = 2x - 2 x 0 5

y -2 8

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of 3x + 2y = 17.



y = 2x – 2 ,


y = 2x – 2

2. Plot and join points. Solution is where lines cross.

Solution is x = 3 & y = 4 Comments

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y = 2x - 2 x 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).

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Solution is x = 3 & y = 4

Method 2Using y = mx + c form:

y = mx + c

gradient y - intercept

Plot C(0, -1) and draw line with m = 2

y = 2x - 1

gradientm = 2

y - interceptc = -1

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Simultaneous Equations : Question 1B


2x + y = 8.



y = 1/2x + 3 ,

hence solve 2x + y = 8

y = 1/2x + 3

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2x + y = 8.



y = 1/2x + 3 ,


y = 1/2x + 3

To draw graph:

Construct a table of

values with at least 2 x-

coordinates.

Plot and join points. Solution is where lines

cross


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2x + y = 8.



y = 1/2x + 3 ,


y = 1/2x + 3

Solution is x = 2 & y = 4What would you like to do now?

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Question 1B

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2x + y = 8.



y = 1/2x + 3 ,


y = 1/2x + 3


y = 1/2x + 3 x 0 6

y 3 6

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Question 1B

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of 2x + y = 8.



y = 1/2x + 3 ,


y = 1/2x + 3

y = 1/2x + 3 x 0 6

y 3 6




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y = 1/2x + 3 x 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).

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y = 1/2x + 3 x 0 6

y 3 6



Method 2Using y = mx + c form:

y = mx + c

gradient y - intercept

Plot C(0, 3) and draw line with m = ½

y = ½ x + 3

gradientm = ½

y - interceptc = +3

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Solve 3u - 2v = 4

2u + 5v = 9

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Solve 3u - 2v = 4

2u + 5v = 9

Eliminate either

variable by making

coefficient same.

Substitute found value into either of

original equations.


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Solve 3u - 2v = 4

2u + 5v = 9

Solution is u = 2 & v = 1



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Question 2

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Solve 3u - 2v = 4

2u + 5v = 9 3u - 2v = 42u + 5v = 9

1. Eliminate either u’s or v’s by making coefficient same.

(x5)(x2)

15 10 20

4 10 18

u v

u v

Now get:

19 38u

3

4

2

1

= (x5)

(x2)

1

= 2

3 4+

2u

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Question 2

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Solve 3u - 2v = 4

2u + 5v = 9 3u - 2v = 42u + 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

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3u - 2v = 42u + 5v = 9


(x5)(x2)

15 10 20

4 10 18

u v

u v

Now get:

19 38u

3

4

2

1

= (x5)

(x2)

1

= 2

3 4+

2u

Note:

When the “signs” are the same subtract to eliminate.

When the “signs” are different add to eliminate.

e.g. 2x + 3y = 46x + 3y = 4

Subtract the equations

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3u - 2v = 42u + 5v = 9


(x5)(x2)

15 10 20

4 10 18

u v

u v

Now get:

19 38u

3

4

2

1

= (x5)

(x2)

1

= 2

3 4+

2u

Note:



e.g. 2x + 3y = 46x - 3y = 4

Add the equations

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Solve 5p + 3q = 0

4p + 5q = -2.6

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Solve 5p + 3q = 0

4p + 5q = -2.6

Eliminate either

variable by making

coefficient same.


original equations.


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Solve 5p + 3q = 0

4p + 5q = -2.6

Solution is q = -1 & q = 0.6


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Question 2B

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5p + 3q = 04p + 5q = -2.6

1. Eliminate either p’s or q’s by making coefficient same.

(x4)(x5)

20 12 0

20 25 13

p q

p q

Now get:

13 13q

3

4

2

1

= (x4)

(x5)

1

= 2

4 3-

1q

Solve 5p + 3q = 0

4p + 5q = -2.6

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Question 2B

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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 = 04p + 5q = -2.6

(x4)(x5) 2

1


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5p + 3q = 04p + 5q = -2.6


(x4)(x5)

20 12 0

20 25 13

p q

p q

Now get:

13 13q

3

4

2

1

= (x4)

(x5)

1

= 2

4 3-

1q

Note:



e.g. 2x + 3y = 46x + 3y = 4

Subtract the equations

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5p + 3q = 04p + 5q = -2.6


(x4)(x5)

20 12 0

20 25 13

p q

p q

Now get:

13 13q

3

4

2

1

= (x4)

(x5)

1

= 2

4 3-

1q

Note:



e.g. 2x + 3y = 46x - 3y = 4

Add the equations

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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.

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Form two equations,

keeping costs in pence to

avoid decimals.

Eliminate either c’s or

d’s by making coefficient

same.


original equations.

Remember to answer the question!!!


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Two coffees & five doughnuts = £3.90

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Question 3

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2c + 3d = 2903c + 1d = 260

1. Form two equations, keeping costs in pence to avoid decimals.

(x1)(x3)

2 3 290

9 3 780

c d

c d

7 490c

3

4

2

1

= (x1)

(x3)

1

= 2

4 3-

70c

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.


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Question 3

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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 = 2903c + 1d = 260

(x1)(x3) 2

1



2

Two coffees & five doughnuts= (2 x 70p) + (5 x 50p)

= £1.40 + £2.50

= £3.90



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Step 1Form the two simultaneous equations first by introducing aletter to represent the cost of a coffee ( c) and a different letter to represent the cost of a doughnut (d).

2c + 3d = 2903c + 1d = 260


(x1)(x3)

2 3 290

9 3 780

c d

c d

7 490c

3

4

2

1

= (x1)

(x3)

1

= 2

4 3-

70c




i.e. 2c + 3d = 2901d + 2c = 260

Note change to pence eases working.

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Step 2Solve by elimination.

Choose whichever variable it is easier to make have the same coefficient in both equations.

2c + 3d = 2903c + 1d = 260


(x1)(x3)

2 3 290

9 3 780

c d

c d

7 490c

3

4

2

1

= (x1)

(x3)

1

= 2

4 3-

70c




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Substitute 70 for c in equation

210 + d = 260

d = 50

2c + 3d = 2903c + 1d = 260



Two coffees & five doughnuts= (2 x 70p) + (5 x 50p)

= £1.40 + £2.50

= £3.90

Step 3Once you have a value forone variable you can substitutethis value into any of theequations to find the valueof the other variable.

It is usually best to choose anequation that you were givenin question.

Step 4Remember 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?

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Form two equations, eliminating decimals wherever possible.

Eliminate either c’s or

d’s by making coefficient

same.


original equations.

Remember to answer the question!!!






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So this blend is more expensive than the other two.

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Question 3B

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0.70B + 0.30K = 740.55B + 0.45K = 71


(x10)(x100)

7 3 740

55 45 7100

B K

B K

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.

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Question 3B

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7 3 740

55 45 7100

B K

B K

50 4000B

3

4

=

(x15)

(x15)3

5 4-

80B




3. Eliminate either B’s or K’s by making coefficient same.

105 45 11100

55 45 7100

B K

B K

5

4

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Question 3B

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7 3 740

55 45 7100

B K

B K

3

4

(x15)Blend 1 uses 70% banana syrup, 30% kiwi syrup and the cost per litre is 74p.



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


80B


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0.70B + 0.30K = 740.55B + 0.45K = 71


(x10)(x100)

7 3 740

55 45 7100

B K

B K

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 1Form the two simultaneous equations first by introducing a letter to represent the cost per litreof banana syrup ( B) and a different letter to represent the cost per litre of kiwis fruit (K).

i.e. 0.70 B + 0.30K = 740.55B + 0.45K = 71

Multiply all terms by 100 to remove decimals.

Note change to pence eases working.

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7 3 740

55 45 7100

B K

B K

50 4000B

3

4

=

(x15)

(x15)3

5 4

80B

3. Eliminate either B’s or K’s by making coefficient same.

105 45 11100

55 45 7100

B K

B K

5

4

-

Step 2Solve by elimination.

Choose whichever variable it is easier to make have the same coefficient in both equations.

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Step 3Once you have a value forone variable you can substitutethis value into any of theequations to find the valueof the other variable.

It is usually best to choose anequation that you were givenin question.

Step 4Remember to answer the question!!!

7 3 740

55 45 7100

B K

B K

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


80B

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Simultaneous Equations : Question 4 A Fibonacci Sequence is formed as follows…….

Start with two terms add first two to obtain 3rd

add 2nd & 3rd to obtain 4th

add 3rd & 4th to obtain 5th 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 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.Get hint

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(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.

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.


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(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.

P + Q, P + 2Q 2P + 3Q 3P + 5Q P = 7 2Q


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Question 4

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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

3rd term = P + Q

4th term = Q + (P + Q) = P + 2Q

5th term = (P + Q) + (P + 2Q)

= 2P + 3Q

6th term = (P + 2Q) + (2P + 3Q)

= 3P + 5Q


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Question 4

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2 3 8

3 5 11

P Q

P Q

1

2

=

(x3)

(x3)1

4 3-

1. Match term from (a) with values given in question.

6 9 24

6 10 22

P Q

P Q

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 (x2)

2Q


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Question 4

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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.

2 3 8

3 5 11

P Q

P Q

1

2

(x3)

(x2)

First two terms are 7 and –2 respectively.



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1. Write down expressions using previous two to form next.

(a)

First term = P & second term = Q

3rd term = P + Q

4th term = Q + (P + Q) = P + 2Q

5th term = (P + Q) + (P + 2Q)

= 2P + 3Q

6th 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, …

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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

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5 3 -4 -1 8 -1 -5

7 -6

1


3v 2w w – 2v v + w

(A) (B)

v + w v – 3w 6w v - w

-18 11

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.

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5 3 -4 -1 8 -1 -5

7 -6

1


3v 2w w – 2v v + w

(A) (B)

v + w v – 3w 6w v - w

-18 11

1W V = 4

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Question 4B

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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)

2nd row 3v + 2w, -2v + 3w, -v + 2w

3rd row v + 5w, -3v + 5w

Top row -2v + 10w = - 18

Pyramid (B)

2nd row 2v - 2w, v + 3w, v + 5w

3rd row 3v + w, 2v + 8w

Top row 5v + 9w = 11

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Question 4B

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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.

2 10 18

5 9 11

V W

V W

1

2

(x5)

(x2)

= (x5)1

4 3+

10 50 90

10 18 22

V W

V W

3

4 = 2 (x2)

68 68W

Now get:

1W


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Question 4B

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3v 2w w – 2v v + w

(A) -18

(B)

v + w v – 3w 6w v - w

11

2 10 18

5 9 11

V W

V W

1

2

(x5)

(x2)


Substitute -1 for W in equation

5V + (-9) = 11

5V = 20

V = 4

2

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Pyramid (A)

2nd row 3v + 2w, -2v + 3w, -v + 2w

3rd row v + 5w, -3v + 5w

Top row -2v + 10w = - 18

Pyramid (B)

2nd row 2v - 2w, v + 3w, v + 5w

3rd row 3v + w, 2v + 8w


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

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Pyramid (A)

2nd row 3v + 2w, -2v + 3w, -v + 2w

3rd row v + 5w, -3v + 5w

Top row -2v + 10w = - 18

Pyramid (B)

2nd row 2v - 2w, v + 3w, v + 5w

3rd row 3v + w, 2v + 8w


Hence equations:10w - 2v = -18 9w + 5v = 11

Solve by the methodof elimination.

End of simultaneous Equations

Documents

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