Simultaneous Equations
Please choose a question to attempt from the following:
1a
1b
2a
3b
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3a
2b 4b
4a
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.
Get hint
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.
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
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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.Solution is x = 3 & y = 4What would you like to do now?
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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
1. Construct a table of values with at least 2 x coordinates.
y = 2x - 2 x 0 5
y -2 8
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
2. Plot and join points. Solution is where lines cross.
Solution is x = 3 & y = 4
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1. Construct a table of values with at least 2 x coordinates.
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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2. Plot and join points. Solution is where lines cross.
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
The diagram shows the graph of
2x + y = 8.
Copy the diagram and on your
diagram draw the graph of
y = 1/2x + 3 ,
hence solve 2x + y = 8
y = 1/2x + 3
Get hint
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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 = 1/2x + 3 ,
hence solve 2x + y = 8
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
What would you like to do now?
Go to full solution
Go to CommentsReveal answer only Get hint
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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 = 1/2x + 3 ,
hence solve 2x + y = 8
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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The diagram shows the graph of
2x + y = 8.
Copy the diagram and on your
diagram draw the graph of
y = 1/2x + 3 ,
hence solve 2x + y = 8
y = 1/2x + 3
1. Construct a table of values with at least 2 x coordinates.
y = 1/2x + 3 x 0 6
y 3 6
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Question 1B
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The diagram shows the graph
of 2x + y = 8.
Copy the diagram and on your
diagram draw the graph of
y = 1/2x + 3 ,
hence solve 2x + y = 8
y = 1/2x + 3
y = 1/2x + 3 x 0 6
y 3 6
2. Plot and join points. Solution is where lines cross.
Solution is x = 2 & y = 4
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1. Construct a table of values with at least 2 x coordinates.
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
2. Plot and join points. Solution is where lines cross.
Solution is x = 2 & y = 4
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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Simultaneous Equations : Question 2
Solve 3u - 2v = 4
2u + 5v = 9
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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.
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Simultaneous Equations : Question 2
Solve 3u - 2v = 4
2u + 5v = 9
Solution is u = 2 & v = 1
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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
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
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
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
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
Add the equations
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Simultaneous Equations : Question 2B
Solve 5p + 3q = 0
4p + 5q = -2.6
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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?
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Get hint
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Simultaneous Equations : Question 2B
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
Question 2B
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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 = 04p + 5q = -2.6
(x4)(x5) 2
1
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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
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
Markers Comments
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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
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
Add the equations
Go to full solution
Go to Comments
Reveal answer only
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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.
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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?
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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
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Question 3
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2c + 3d = 2903c + 1d = 260
1. Form two equations, keeping costs in cents 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 cents
& doughnuts d cents then we have
2. Eliminate either c’s or d’s by making coefficient same.
Try another like this
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Question 3
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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 = 2903c + 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
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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
1. Form two equations, keeping costs in cents 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
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 = 2901d + 2c = 260
Note change to cents 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
1. Form two equations, keeping costs in cents 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
Let coffees cost c cents
& doughnuts d cents then we have
2. Eliminate either c’s or d’s by making coefficient same.
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3. Substitute found value into either of original equations.
Substitute 70 for c in equation
210 + d = 260
d = 50
2c + 3d = 2903c + 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 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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Simultaneous Equations : Question 3B
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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?
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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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Simultaneous Equations : Question 3B
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
1. Form two equations, keeping costs in cents to avoid decimals.
(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
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.
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
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.
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.
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.
80B
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0.70B + 0.30K = 740.55B + 0.45K = 71
1. Form two equations, keeping costs in cents to avoid decimals.
(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 cents & one litre of kiwi syrup cost K cents.
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 cents 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
So this blend is more expensive than the other two.
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.
• 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.
• 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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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.
• 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.
• 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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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.
• 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.
• 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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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.
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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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
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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
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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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
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)
3. Substitute found value into either of original equations.
Substitute -1 for W in equation
5V + (-9) = 11
5V = 20
V = 4
2
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1. Work your way to an expression for the top row by filling in the middle rows.
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
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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1. Work your way to an expression for the top row by filling in the middle rows.
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
Hence equations:10w - 2v = -18 9w + 5v = 11
Solve by the methodof elimination.
End of simultaneous Equations