Patterns and Algebra - In Class, with Miss. Coatescoatesj.weebly.com/uploads/6/1/0/2/61028649/95727447.g... · 2019. 11. 10. · Patterns and Algebra - In Class, with Miss. Coates

Patterns and Algebra

Teacher

Ser

ies

G

Copyright © 2009 3P Learning. All rights reserved.

First edition printed 2009 in Australia.

A catalogue record for this book is available from 3P Learning Ltd.

ISBN 978-1-921861-16-1

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Series G – Patterns and Algebra

Contents

Topic 1 – Pa erns and func ons (pp. 1–17)• recursive number sequences ____________________________

• func on number sequences _____________________________

• func on shape pa erns ________________________________

• func on machines and func on tables ____________________

• real life func ons _____________________________________

• func on tables – apply _________________________________

• the “I Do” venue – solve ________________________________

• fabulous Fibonacci and the bunnies – solve _________________

• triangular numbers – inves gate _________________________

• Pascal’s triangle – inves gate ____________________________

Topic 2 – Algebraic thinking (pp. 18–25)• making connec ons between unknown values ______________

• present puzzle – solve __________________________________

• the candy box – solve __________________________________

Topic 3 – Solving equa ons (pp. 26–33)• introducing variables __________________________________

• using variables in an equa on ___________________________

• simplifying algebraic statements _________________________

• happy birthday – solve _________________________________

• squelch juiceteria – solve _______________________________

Topic 4 – Proper es of arithme c (pp. 34–41)• order of opera ons ____________________________________

• commuta ve rule _____________________________________

• distribu ve rule _______________________________________

• equa on pairs – apply _________________________________

Date completed

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Series Authors:

Rachel Flenley

Nicola Herringer

Copyright ©

Series G – Patterns and Algebra

Contents

Sec on 1 – Answers (pp. 1–41)• pa erns and func ons ________________________________ 1

• algebraic thinking ____________________________________ 18

• solving equa ons ____________________________________ 26

• proper es of arithme c ______________________________ 34

Sec on 2 – Assessment with answers (pp. 42–53)• pa erns and func ons – part 1 _________________________ 42

• pa erns and func ons – part 2 _________________________ 44

• pa erns and func ons – part 3 _________________________ 46

• algebraic thinking ____________________________________ 48

• solving equa ons ____________________________________ 50

• proper es of arithme c ______________________________ 52

Series Authors:

Rachel Flenley

Nicola Herringer

Copyright ©

SERIES TOPIC

1G 1Copyright © 3P Learning


a b c

Complete these grid pa erns. Look closely at the numbers in the grid and follow the pa ern going ver cally and horizontally:

What do you no ce about the pa erns a and b in Ques on 1?

_______________________________________________________________________________________

Figure out the missing numbers in each pa ern and write the rule:

Patterns and functions – recursive number sequences

A number pa ern is a sequence or list of numbers that is formed according to a rule. Number pa erns can use any of the four opera ons ( +, –, ×, ÷) or a combina on of these.There are 2 diff erent types of rules that we can use to con nue a number pa ern:1 A recursive rule – fi nd the next number by doing something to the number before it.2 A func on rule – predict any number by applying the rule to the posi on of the number.

Here is an example of a number sequence with a recursive rule.The rule is add 8 to the previous number, star ng with 5.

a

Rule ____________________

d

Rule ____________________

b

Rule ____________________

e

Rule ____________________

c

Rule ____________________

f

Rule ____________________

9 18 36 45

7 13 25 3149 42 28 21

125 100 50 2510 37 46

3 17 24 31

1

2

3

5 13 21 29 37

+ 8 + 8 + 8 + 8

16 19

26

39

46

10 25

24

33

37 52

45

35

34 52 61

10

19 18

36 28 26

17 15 19

27 27

37

47 42 43

18 20 28

28 29 36

38 38 44

48 47

37

29 34

43 53

49

27 54 19 28

1935

55

3714

75

10

0

38

+ 9

– 7

+ 9

+ 6

– 25

+ 7

Both + 9. Just different starting numbers.

+ 1

+ 10 + 9 + 8

+ 5 + 9

SERIES TOPIC

G 12Copyright © 3P Learning


Use a calculator to work out where each pa ern started to go wrong in these single opera on pa erns and circle them. Hint: The fi rst 2 numbers in both are correct.

a

The rule is ________________

b

The rule is ________________

Complete these decimal number sequences according to the recursive rule:

a Start at 2.5 and add 0.5 2.5

b Start at 25 and subtract 0.5 25

c Start at 30 and add 2.5 30

Complete the following number pa erns and write the rule as 2 opera ons in the diamond shapes and describe it underneath.

a

The rule is ________________

b

The rule is ________________

Patterns and functions – recursive number sequences

Complete these sequences according to the recursive rule:

a Start at 3 and add 7 3

b Start at 125 and subtract 5 125

c Start at 68 and add 20 68

4

5

6

7

2 3 6 15× __ – __ × __ – __ × __ – __

1 3 7 15× __ + __ × __ + __ × __ + __

78 100 122 144 166 188 211 222 233

500 466 432 398 364 330 298 266 230

10

120

88

3

24.5

32.5

17

115

108

3.5

24

35

24

110

128

4

23.5

37.5

31

105

148

4.5

23

40

+ 22

– 34

3 3

× 3– 3

33 3 3

2 2

× 2 + 1

21 1 1

SERIES TOPIC



In each table, fi nd the rule and write it in the middle row. Then apply the rule to posi on 20.

a

b

c

Patterns and functions – function number sequences

There are 2 diff erent types of rules that we can apply to fi nd out more about a sequence:1 A recursive rule – gives the next number by applying a rule to the number before it.2 A func on rule – predicts any number by applying a rule to the posi on of the number.

So far we have prac sed the recursive rule to work out the next number in a sequence.Now we will apply the func on rule to this problem:How can we fi nd out the 20th number in this sequence without wri ng out all of the numbers?To use the func on rule we: • Use a table like this one below.• Write each number of the sequence in posi on.• Work out the rule, which is the rela onship between the posi on of

a number and the number in the pa ern. • Use the rule to work out the 20th number in the sequence.

Posi on of number 1 2 3 4 5 20

Rule × 3 + 1 × 3 + 1 × 3 + 1 × 3 + 1 × 3 + 1 × 3 + 1

Number sequence 4 7 10 13 16 61


Rule

Number sequence 6 11 16 21 26


Rule



Rule


1

HINT: a good way to work out the rule is to see what the sequence is going up by. This tells you what the fi rst opera on is and then you adjust. This sequence is the 3 mes tables moved

up one so it is × 3 + 1.

HINT: All of these func on rules consist of 2 opera ons: × and then + or –.

× 5 + 1

× 2 + 3

× 9– 1

× 5 + 1

× 2 + 3

× 9– 1

× 5 + 1

× 2 + 3

× 9– 1

× 5 + 1

× 2 + 3

× 9– 1

× 5 + 1

× 2 + 3

× 9– 1

× 5 + 1

× 2 + 3

× 9– 1

101

43

179

SERIES TOPIC



Unscramble the sequence according to this func on rule: × 9 – 6.

a Again, use the table below to work out how this sequence should go and cross out numbers that do not belong:

Posi on of number 1 2 3 4 5 6 7 8 9 10

Rule

Number sequence

b What will be the number in posi on 50? ____________

Here is another number sequence but this me 4 of these numbers do not belong. Given the func on rule and the fi rst 2 numbers, use the table below to work out how this sequence should go, then cross out the numbers that do not belong:

15

23

31

36

47

59

63

74

79

97

Posi on of number 1 2 3 4 5 6 7 8 9 10

Rule × 8 + 7 × 8 + 7 × 8 + 7 × 8 + 7 × 8 + 7 × 8 + 7 × 8 + 7 × 8 + 7 × 8 + 7 × 8 + 7

Number sequence

Circle true or false for each of the following:

Patterns and functions – function number sequences

2

3

4

5

Here is part of a number sequence. Write these numbers in the table provided. This will help you to answer the ques ons below:

8

11

14

17

20


Rule

Number sequence

a The number in the 6th posi on is 24 true / false

c The number in the 20th posi on is 65 true / false

b 32 is in this sequence true / false

d The number in the 100th posi on is 305 true / false

489

27

78 6023

63 36 33

6684 39

312

57 1821

75

30

× 3 + 5 × 3 + 5 × 3 + 5 × 3 + 5 × 3 + 5 × 3 + 5

65201714118

15 23 31 39 47 55 63 71 79 87

3

× 9– 6 × 9– 6 × 9– 6 × 9– 6 × 9– 6 × 9– 6 × 9– 6 × 9– 6 × 9– 6 × 9– 6

12 21 30 39 48 57 66 75 84

444

SERIES TOPIC



Complete the table for each sequence of matchs ck shapes. Use the func on rule for fi nding the number of matchs cks needed for each shape including the 50th shape:

a

Shape number 1 2 3 4 5 6 7 8 9 10 50

Number of matchs cks 4 7 10 13 16

Func on rule Number of matchs cks = Shape number × ______ + 1

b

Shape number 1 2 3 4 5 6 7 8 9 10 50


Func on rule Number of matchs cks = Shape number × ______ + ______

c

Shape number 1 2 3 4 5 6 7 8 9 10 50



d

Shape number 1 2 3 4 5 6 7 8 9 10 50



Patterns and functions – function shape patterns

When you are inves ga ng geometric pa erns, look closely at the posi on of each shape and think about how it is changing each me.How many matchs cks are needed for the fi rst shape?How many more are needed for the next shape?

1

19

26

13

20

22

30

15

23

25

34

17

26

28

38

19

29

31

42

21

32

151

202

101

152

3

4 2

2 1

3 2

SERIES TOPIC



Tyson also made a sequence out of pa ern blocks but stopped a er the fi rst 3 shapes and decided to con nue inves ga ng by using the table.

Shape number 1 2 3 4 5 6 7 8 9 10

Number of crosses 1 2 3

Number of rectangles 0 2 4

Rule for crosses Number of crosses = (2 + number of rectangles) ÷ 2

Rule for rectangles Number of rectangles = (2 × number of crosses) – 2

a How many rectangles will there be in the 12th shape?

b Josie made this shape following Tyson’s sequence.

What is the posi on of this shape? __________

How do you know?

Patterns and functions – function shape patterns

Gia started to make a sequence out of star and pentagon blocks and recorded her fi ndings in the table as she went. She had to stop when she ran out of pentagons. This is where she got up to:

a Help Gia con nue inves ga ng this sequence by using the table below:

Shape number 1 2 3 4 5 6 7 8 9 10 15

Number of stars 1 2 3

Number of pentagons 0 1 2

Rule for stars Number of stars = Number of pentagons + 1

Rule for pentagons Number of pentagons = Number of stars – 1

b How many stars are in the 10th shape?

c How many pentagons are there in the 15th shape?

2

3

4

3

5

4

4

6

10

22

14

6

5

5

8

7

6

6

10

8

7

7

12

9

8

8

14

10

9

9

16

15

10

14

18

9th

Number of crosses = (2 + 16) ÷ 2

= 9

SERIES TOPIC



Complete these func on tables according to the rule:

a

b

Look carefully at the numbers going in these func on machines and the numbers coming out. What 2 rules are they following each me?

a b

Patterns and functions – function machines and function tables

Remember func on machines? Numbers go in, have the rule applied, and come out again.The rule for this func on machine is mul ply by 6.

RULE:

× 6

12

9

5IN

72

54

30OUT

1

2

RULE:

9

5

8

IN

66

38

59

OUT RULE:

30

25

100

IN

5

4

19

OUT

The func on machines showed us that when a number goes in, it comes out changed by the rule or the func on. Func on tables are the same idea – the number goes in the rule and the number that comes out is wri en in the table. The rule goes at the top:

Rule: ÷ 2 + 6

IN 10 24 50 70 48 90 100 80

OUT 11 18 31 41 30 51 56 46

Rule: × 5 – 4

IN 6 9 3 4 7 11 20 8

OUT 26

Rule: × 8 + 1

IN 8 2 3 5 7 9 4 6

OUT 65

× 7 + 3 ÷ 5– 1

17

41

25

11

41

16

57

31

73

51

33

96

49

36

SERIES TOPIC



Complete the func on tables for these real life scenarios:a A pool which fi lls at a rate of 4 litres every minute.

Rule: Number of minutes × 4 = Amount of litres

Minutes 5 10 15 20 25 30 35 40

Litres 20 40 60 80

How full is it a er 1 hour?

b Maya downloads 5 songs a day onto her MP3 player.

Rule: Number of days × ______ = Amount of songs

Days 1 2 3 4 5 6 7 8

Songs 5 10 15 20

How many songs would she have downloaded a er 30 days?

c A car is travelling at a speed of 50 km/h.

Rule: Number of hours × ______ = Amount of km travelled

Hours 1 2 3 4 5 6 7 8

Km travelled 50 100 150 200

How long would it take to travel 800 km?

Patterns and functions – real life functions

So far we have seen that func ons are rela onships between numbers. These numbers are a ached to real life situa ons everywhere you look. It is possible to create a func on table to show the rela onship between many things, for example:• Your high score Live Mathle cs depends on how o en you prac se mental arithme c.• The distance that you run depends on how long you run. • The amount that you can save depends on how much you earn.• The amount of US dollars you get when you travel to Los Angeles depends on the exchange rate.There are many, many more examples. Can you think of any?

We can show these rela onships on a graph. On the right is a graph of the func on table in ques on c. This is known as a travel graph and shows the rela onship between me and distance. Next, we will look at some examples of graphing func ons.

1

125

100

75

50

25

1 2

Leah’s journey

Time (hours)

Dist

ance

(km

)

100

25

250

240 litres

150 songs

16 hours

120

30

300

140

35

350

160

40

400

5

50

SERIES TOPIC



During the day, Crawly’s friend Creepy, crawls 5 cm up a garden wall. At night when he falls asleep, he slides 2 cm back down the wall.

a Complete the table below to show how far he gets in 8 days.

b Write a rule for working out the distance if you know the number of days. Think about the total distance Creepy covers in 24 hours.

c Plot the points on the graph above (just like the one in Ques on 2), then compare the graphs. How are they diff erent?

___________________________________________________________________________________


3

Crawly the caterpillar crawls 4 cen metres per day.

a Complete the table to show how far he gets in 8 days.

b Write a rule for working out the distance if you know the number of days.

2

Rule:

Days 1 2 3 4 5 6 7 8

Distance

Rule:

Days 1 2 3 4 5 6 7 8

Distance

This is the graph of my journey shown in the func on table. Plot the points and then join the points with a straight line.

Crawly’s journey

Num

ber o

f day

s

Distance in cm

876543210

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

Creepy’s journey

Num

ber o

f day

s

Distance in cm

876543210

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

Number of days × 4 cm

Number of days × 3 cm or (5– 2 cm)

4 cm

3 cm

8 cm

6 cm

12 cm

9 cm

16 cm

12 cm

20 cm

15 cm

24 cm

18 cm

28 cm

21 cm

32 cm

24 cm

Crawly covers more distance over the same amount of time.

SERIES TOPIC



c Graph this data by plo ng the points from the table:

Do not join the points because the data is about whole slices of pizza – not parts of slices. Also you cannot have part of a person, the data is about single people.

d How many slices are needed for 11 people?

e How did you work this out?

f How could the graph help you?

g 10 people confi rmed they were coming to the party. How many pizzas will Julie need to buy if each pizza has 12 slices? Will there be any le overs? Show your working.


Julie is planning her birthday party and is planning how much food and drink she needs for her guests. She has sent out 15 invita ons.

a Complete the table to show how much pizza is needed for diff erent numbers of guests. She has based this table on the es ma on that one guest would eat 3 slices of pizza.

b Write a rule in the table for working out the slices of pizzas needed, if you know the number of guests.

4

Rule:

Number of guests 1 2 3 4 5 6 7 8

Slices of pizza 3

Pizza catering

Slic

es o

f pizz

a

Number of guests

30

28

26

24

22

20

18

16

14

12

10

8

6

4

2

0 1 2 3 4 5 6 7 8 9 10 11 12

Number of guests × 3 slices

11 × 3 = 33 slices

10 × 3 = 30 slices

3 pizzas = 3 × 12 = 36 slices

6 leftover slices

Used the rule.

Continue the plotted points.

6 9 12 15 18 21 24

SERIES TOPIC



What to do next

What to do

Make up your own scoring table where extra points are given for certain answers.

You could also decide on a ‘killer number’. This number means you wipe out all your points.

1 Each player writes their ini als at the top of each column in the scoring tables.2 For each round, roll the dice for ¬ and J. 3 Use the value for ¬ and J in the rule.4 Each player writes the answer in the scoring table which becomes their

running score.5 Players add their scores to the previous score.6 The winner is the player with the highest score at the end of the round.

The overall winner is the player who wins the most points a er 3 rounds.For example If I roll the dice and get 4 for ¬ and 6 for J and I am working with

(2 × ¬) + J, I would calculate (2 × 4) + 6 and my answer would be 14. So I would write 14 in the fi rst row of the table. The next answer I get I add to 14 and so on un l the end of the table.

Function tables apply

You and your partner need 2 dice, a pencil and this page.

Total Total Total

Round 1(2 × ¬) + J

Round 2(3 × ¬) + J

Round 3(6 × ¬) – (2 × J)

Ge ng ready

Answers will vary.

Answers will vary.

SERIES TOPIC



What to do

The “I Do” venue solve

Table and Chair Arrangement 1 2 3 4 5 6 7 8 9 10

Tables 8 12 16

Chairs 12 16 20 24 28 32 36 40 44 48

Rule for tables

Rule for chairs

A very popular wedding recep on venue has a strict policy in the way they put the tables and chairs together. Below is a bird’s eye view of this arrangement. They must only be arranged in this sequence to allow room for their famous ice sculptures in the centre of each table arrangement.

Look carefully at the diagram of the fl oor plan above.a Complete the table below.b Write the rule in the table for the number of tables needed if you know the table

and chair arrangement number.c Write the rule in the table for the number of chairs needed if you know the table

and chair arrangement number.d Draw what Table and Chair Arrangement 4 would look like in the grid at the

bo om of this page.

Table and Chair Arrangement 4

Table and ChairArrangement 1



Ge ng ready

20

Number of tables = Table arrangement × 4 + 4

Number of chairs = Table arrangement × 4 + 8

24 28 32 36 40 44

SERIES TOPIC



The “I Do” venue solve

The latest Bridezilla to hire out the “I Do” venue, wants to know how many guests can fi t into the space at this venue.

Bridezilla wants to be head of the largest table, which seats 36 guests. This is shown on the fl oor plan. Work out how many guests she can invite to her wedding by seeing how many will fi t in the venue space. The table and chair arrangements must follow the sequence described on the previous page (page 12). So, each table arrangement will be a diff erent size.

Hint: Try to get 5 more tables in this fl oor plan. Each table should seat fewer than 36 guests. There should be space between the chairs from all the tables so that guests do not bump against each other when ge ng up from the table.

What to do next

Number of guests: ______

36 guests

136

Answers will vary.

24 guests20 guests

12 guests

28 guests

16 guests

36 guests

SERIES TOPIC



Look carefully at the table above to understand the problem. If we kept going, the table would get very wide indeed and quite confusing! So it is up to you to fi gure out the pa ern. Here is a closer look. Can you see what is happening? What are the next 2 numbers?

1 1 2 3 5 8

Now, back to the bunnies. Use the table below to answer Mr Fibonacci.“How many pairs of rabbits will there be a year from now?”

What to do next

What to do

Fabulous Fibonacci and the bunnies solve

A famous mathema cian by the name of Leonardi di Pisa became known as Fibonacci a er the number sequence he discovered. He lived in 13th century Italy, about 200 years before another very famous Italian, Leonardo da Vinci.

His number sequence can be demonstrated by this maths problem about rabbits:

“How many pairs of rabbits will there be a year from now, if …?”

1 You begin with one male rabbit and one female rabbit. These rabbits have just been born.

2 A er 1 month, the rabbits are ready to mate.3 A er another month, a pair of babies is born – one male and one female.4 From now on, a female rabbit will give birth every month.5 A female rabbit will always give birth to one male rabbit and one female rabbit.6 Rabbits never die.

MonthBabies from

1st PairBabies from

2nd PairBabies from

3rd PairTotal

Pairs of Rabbits

1

2

3

4

5

6

Fibonacci now wants to know:“How many pairs of rabbits will there be 2 years from now?” Use a calculator. Hint: The table below should just con nue from the previous one.

Key

= 1 pair of rabbits

Months 1 2 3 4 5 6 7 8 9 10 11 12

Pairs of bunnies 1 1 2 3 5

Months 13 14 15 16 17 18 19 20 21 22 23 24

Pairs of bunnies

Ge ng ready

8

2 584233

13

4 181377

21

6 765610

34

10 946987

55

17 7111 597

89

28 657

144

13 21

46 368

SERIES TOPIC



What to do next

What to do Let’s inves gate a faster way to fi nd the 10th number:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

Work from the outside in, un l you reach the halfway point adding the numbers.

What is the answer each me? __________

Half of 10 is __________ so that means we have 5 lots of 11, so the 10th triangular

number is __________.

What is the 20th number?

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20

Work from the outside in, un l you reach the halfway point adding the numbers.

What is the answer each me? __________

Half of 20 is __________ so that means we have __________ lots of __________,

so the 20th triangular number is __________.

Find the 30th triangular number without wri ng down the numbers. __________

Hint ques ons:What are the fi rst and the last numbers? ________ What do they add to? ______

Triangular numbers investigate

Write a number sentence for each part of the triangular number pa ern and con nue to complete this list:1st 1 = 1 2nd 3 = 1 + 23rd 6 = 1 + 2 + 34th 10 = 1 + 2 + 3 + 4

5th 15 = _______________________________________________

6th 21 = _______________________________________________

7th 28 = _______________________________________________

8th ____ = _______________________________________________

Ge ng ready

1st 2nd 3rd 4th 5th

1 + 2 + 3 + 4 + 5

1 + 2 + 3 + 4 + 5 + 6

1 + 2 + 3 + 4 + 5 + 6 + 7

1 + 2 + 3 + 4 + 5 + 6 + 7 + 836

11

5

55

21

10

210

10 21

465

1 + 30 31

15 × 31 = 465 So the 30th triangular number is 465.

SERIES TOPIC



What to do Look carefully at the numbers in

the triangle. Can you see how you might go about comple ng it? Once you have worked this out, complete the rest of what you see of Pascal’s triangle:

Pascal’s triangle investigate

Pascal’s triangle is named a er Blaise Pascal and is fascina ng to inves gate because of all its hidden pa erns. Blaise Pascal was born in France in 1623 and displayed a remarkable talent for maths at a very young age. His father, a tax collector, was having trouble keeping track of his tax collec ons, so he built his father a mechanical adding machine! (And you thought washing up a er dinner was helpful!)

Pascal was actually lucky that this triangle was named a er him as it was known about at least 5 centuries earlier in China.

HINT: Start with the 1s at the top of the triangle and add them, you get 2.

Ge ng ready

1

1

1

1

1

4

3

2

1

6

3

1

4

1

1

Complete the missing sec ons of Pascal’s triangle below.

a b c

10

9 84 210

45 210

120

252

165 330 792

165

1 1269 8436 3684 9126 1

1 568 2828 856 170

1 217 721 135 35

1 66 115 20 15

1 15 10 10 5

36 126

120 120

462 495

330

55

462

SERIES TOPIC



Pascal’s triangle investigate

Can you see any other pa erns in Pascal’s triangle?

Look along the diagonals and describe as many pa erns as you can.

See if you can fi nd Fibonacci’s sequence.

What to do next Check that the Pascal’s triangle on page 16 is correct. Then copy the numbers into

the triangle below and colour in all the mul ples of 3 – red; hexagons with 1 less than a mul ple of 3 – green; and all hexagons with 2 less than a mul ple of 3 – blue.

1

1

1

1

1

4

3

2

1

6

3

1

4

1

1

B = Blue

G = Green

R = Red

Diagonals on the left and right edges are ones.

Next row of diagonals is counting numbers in order.

Next row of diagonals is triangular numbers in order.

1 1269 8436 3684 9126 1

1 568 2828 856 170

1 217 721 135 35

1 66 115 20 15

1 15 10 10 5

B

BB

BGB

BRRB

BBRBB

BGBBGB

BRRGRRB

BBRGGRBB

BGBGBGBGB

BRRRRRRRRB

SERIES TOPIC

G18Copyright © 3P Learning

Patterns and Algebra2

Find out the value of both symbols: ««

Algebraic thinking – making connections between unknown values

The balance strategy is what we use when we need to fi nd the value of one symbol. Once we know the value of the fi rst symbol, we can fi nd out the value of the second symbol.

Clue 1 «« + 40 = 60Clue 2 «« × = 100

Use the balance strategy to fi nd the value of «« «« + 40 = 60 «« + 40 = 60 – 40 «« = 20

Now we know the value of «« we can work out the value of «« × = 100 20 × = 100 20 × = 100 ÷ 20 = 5

Using the balance strategy we do the same to both sides which gives us the answer.

1

a Clue 1 «« – 15 = 45 Clue 2 «« × = 120

«« – 15 = 45

«« – 15 = 45 + 15

«« =

«« × = 120

× = 120

= 120 ÷ 60

=

b Clue 1 «« × 9 = 81 Clue 2 – «« = 96

«« × 9 = 81

«« × 9 = 81

«« =

– «« = 96

– = 96

= 96

=

Doing the inverse to the other side of the equa on cancels out a number and makes things easier to solve. This is called the balance strategy.

60

60

2

÷ 9

9

9

+ 9

105

SERIES TOPIC

19GCopyright © 3P Learning

Patterns and Algebra 2


Now that you have had prac ce following the clues and using the step prompts, try these on your own. Set your work out carefully and always use a pencil so that you can erase mistakes and try again.

Find out the value of both symbols: ««

a Clue 1 6 × «« + 12 = 84

Clue 2 «« × = 96

Steps for fi nding «« 6 × «« = 84 – 6 × «« =

« « × 6 =

« « = ÷

« « =

Now you can fi nd

b Clue 1 9 × «« – 42 = 21 Clue 2 «« + = 100

Steps for fi nding ««

« « =

Now you can fi nd

a Clue 1 «« × 8 = 64 Clue 2 – «« = 75

b Clue 1 «« × 7 = 49 Clue 2 «« + = 100

3

2

It is easier if we put the star on the le hand side. We can swap numbers around with addi on and mul plica on.

«« × 7 = 49

« « = 49 ÷ 7

« « = 7

7 + = 100

= 100 – 7

= 93

«« × 8 = 64

«« × 8 = 64 ÷ 8

« « = 8

– 8 = 75

= 75 + 8

= 83

9 × ««– 42 = 21 «« × 9 = 21 + 42 «« × 9 = 63 « « = 63 ÷ 9

«« + = 100 7 + = 100 = 100 – 7 = 93

«« × = 96 12 × = 96 × 12 = 96 = 96 ÷ 12 = 8

12

72

72

72 6

712

SERIES TOPIC



This me you have 3 clues to work through. There are no step prompts, you are on your own except for one hint: start with the clue where you can fi nd the value of one symbol. Set your working out clearly. Use each box to work out the value of each symbol.

a

b


4

Clue 1 2 × «« = 3 × Clue 2 3 × = 4 × Clue 3 6 × «« = 72

«« =

=

=

Clue 1 5 × «« = 3 × Clue 2 4 × = 60 Clue 3 45 ÷ «« = ÷ 4

«« =

=

=

Find the value of these 3 symbols. You must look closely at each clue. There are hints along the way.

Clue 1 Clue 2 Clue 3


2 × « « = 3 × 2 × 12 = 3 × 24 = 3 × 3 × = 24 = 24 ÷ 3 = 8

5 × « « = 3 × 5 × « « = 3 × 15

5 × « « = 45

« « = 45 ÷ 5

« « = 9

3 × = 4 × 3 × 8 = 4 × 24 = 4 × 4 × = 24 = 24 ÷ 4

= 6

4 × = 60

= 60 ÷ 4

= 15

6 × « « = 72 « « = 72 ÷ 6 « « = 12

45 ÷ « « = ÷ 4

45 ÷ 9 = ÷ 4

÷ 4 = 5

= 5 × 4

= 20

12

9

8

15

6

20

SERIES TOPIC



If you were able to complete the last few pages, then you are ready for the next level of algebraic thinking. This me you have to work a bit harder to fi nd the value of the fi rst unknown. However it is easy if you follow these steps and look very closely at the clues. There are clues within the clues!This page is a worked example. Each step is worked through to help you do this on your own on the next few pages.

Find the value of: ««

Clue 2 tells us that: «« + = + Looking at Clue 1, we can swap the star and triangle for 2 circles:

+ + + = 20 So, = 5

Clue 3 tells us that: «« = + 4

Looking at Clue 2, we can swap the star for the triangle plus 4, so:

+ + 4 = + We know is 5, so: + + 4 = 10

Use the balance strategy + + 4 = 10 – 4

+ = 6So, = 3

Now that we know the value of the triangle, we can fi nd out the value of the star with Clue 3:

Clue 3 «« = + 4 « « = 7

By looking closely at the clues, we have found out the value of all 3 symbols:

= 5 = 3

« « = 7


Clue 1 «« + + + = 20 Clue 2 «« + = + Clue 3 «« = + 4

SERIES TOPIC




This page is very similar to the last page. There are step prompts to help you along the way.

Find the value of these 3 symbols: =

You must look closely at each clue. =

There are hints along the way. «« =

Clue 2 tells us that: «« + = Looking at Clue 1, we can swap the star and triangle for a circle. Now we have + = 50.

+ = So, =

Clue 3 tells us that: «« = + 15

Looking at Clue 2, we can swap the star for the triangle plus 15 so we have:

+ + 15 =

We know is ______ , so: + + 15 =

Use the balance strategy + =

+ =

So, =

Now that we know the value of the triangle, we can fi nd out the value of the star with Clue 3:

Clue 3 «« = + 15

«« =

5

Clue 1 «« + + = 50 Clue 2 «« + = Clue 3 «« = + 15

50 25

25

25

25 – 15

10

5

20

5

20

25

SERIES TOPIC




This me, there are 2 ac vi es where you must use the clues. One has step prompts, the other does not:

a Find the value of these 2 symbols: =

You must look closely at each clue. «« =

There are hints along the way.

Clue 1 tells us that:

+ =

Looking at Clue 2, this means that:

+ =

So, =

We know the value of , so we can put this into Clue 3:

Clue 3 =

–

So, «« =

Clue 1 «« + «« = Clue 2 «« + «« + = 200 Clue 3 «« = – 50

Clue 1 «« + + = 100 Clue 2 «« + = Clue 3 = + 40

6

b Find the value of these 3 symbols: =

You must look closely at each clue. =

«« =


100

50

««

««

««

200

100

100

50

50

10

40

50

«« + + = 100

+ = 100

= 50

«« + = «« + 10 = 50

« « = 50 – 10

« « = 40

= + 40

50 = + 40

+ 40 = 50

= 50 – 40

= 10

SERIES TOPIC



What to do next

What to do Can you work out how much was spent on each present?

Label each present with the amount it is worth.

Present puzzle solve

Three students each brought in some presents for the charity drive. Each student spent $36.

Ge ng ready

$9 $6 $5

SERIES TOPIC



What to do next

What to do Read the clues for Compe on 1 and look carefully

at how to solve the problem. This will help you win Compe on 2.

Compe on 1 clues1 There are 36 candies in the box which are a mixture

of choc drops, mallow swirls and caramel dreams.2 The number of mallow swirls equals four mes the

number of choc drops.3 The number of caramel dreams is equal to the

number of mallow swirls.

Compe on 2 clues

See if you could win this box of candy by using these 3 clues to work out the exact contents in the box. Follow the same steps as shown to you in Compe on 1.

1 There are 84 candies in the box which are a mixture of hokey pokies, pep up chews and chomp s x.

2 The number of pep up chews equals twice the number of chomp s x.3 The number of hokey pokies equals double the number of pep up chews.

The candy box solve

Miss Harley, the class teacher of 6H, enjoys ge ng her class to think mathema cally by holding guessing compe ons. Her most famous guessing compe on was when she asked 6H to guess the number of cocoa puff s in a bowl if she used 250 mL of full cream milk.

In her 2 latest compe ons, she has said that the person who correctly guesses the exact contents of the box gets to take home all the candy. This me she has given clues.

Look at the 3 types of candy as 3 diff erent groups. For every 1 choc drop, there are 4 mallow swirls and 4 caramel dreams.

So 9 × = 36

That means = 4

So there are:4 choc drops16 mallow swirls16 caramel dreams

Choc drops

Mallowswirls

Caramel dreams

= =

Chomp s x Pep up chews Hokey pokies

= =

Ge ng ready

7 × = 84 = 12

12 chomp stix, 24 pep up chews and 48 hokey pokies

12 + 24 + 48 = 84

SERIES TOPIC



Using the balance strategy, solve each equa on and then match the le ers to the answers to solve this riddle: What gets we er and we er the more that it dries? The fi rst one has been done for you:

Use the balance strategy to fi nd out the value of y:

Solving equations – introducing variables

Algebra normally uses le ers of the alphabet to stand for unknown parts of an equa on. These le ers are known as variables and are used in the same manner as symbols such as stars, triangles and boxes. Common le ers used in algebra are: x, y, a, b, c, u and v.

– 12 = 38 – 12 = 38 + 12 = 38 + 12 = 50

Same equa on

1

2

x – 12 = 38 x – 12 = 38 + 12 x = 38 + 12 x = 50

a y + 6 = 68 b y – 18 = 42 c y × 8 = 72

O x + 9 = 14

x + 9 = 14 – 9

x = 14 – 9

x = 5

W m + 5 = 19

E y – 5 = 29

T y + 8 = 25

A a + 7 = 15

L 8 + x = 24

8 17 5 14 34 16

y = 68– 6

y = 62

y = 42 + 18

y = 60

y = 72 ÷ 8

y = 9

y – 5 = 29 + 5

y = 34

y + 8 = 25 – 8

y = 17

m + 5 = 19 – 5

m = 14

a + 7 = 15 – 7

a = 8

8 + x = 24

x = 24 – 8

x = 16

A T O W E L

SERIES TOPIC



For each ques on, write an equa on using the variable x for the mystery number, then solve it.

a The sum of 7 and a mystery number is 26.

b A mystery number increased by 15 is 48.

c A mystery number doubled is 64.

d The diff erence between a mystery number and 19 is 42.

Solving equations – using variables in an equation

In algebra, variables are used to represent the unknown number or what we are trying to fi nd out. Look at this example:Amity’s teacher gave the class a mystery number ques on:

“The sum of a mystery number and 18 is 36. What is the number?”

Amity used variable x to stand for the mystery number. She wrote: x + 18 = 36This is really saying, “mystery number plus 18 is 36.”Next, Amity used the balance strategy to solve the equa on:

x + 18 = 36 x + 18 = 36 – 18 x = 18

1

7 + x = 26

x = 26 – 7

x = 19

x + 15 = 48

x = 48 – 15

x = 33

x × 2 = 64

x = 64 ÷ 2

x = 32

x – 19 = 42

x = 42 + 19

x = 61

SERIES TOPIC



Find the length of the side of each of these shapes with algebra. Here you will be using variables to represent the unknown number and the balance strategy to solve the equa on. The fi rst one has been done for you.

a If the perimeter of this square is 28 cm, fi nd the length of one side. Call the side x.

x × 4 = 28

x × 4 = 28 ÷ 4

x = 28 ÷ 4

x = 7 cm

b The perimeter of this pentagon is 40 cm. Find the length of one side. Call the side y.

y × 5 = 40


2

3

Find the value of x and y. First fi nd the value of x by using the balance strategy, then you will be able to fi nd the value of y. Show your working out:

x

y

a x – 15 = 35

x × y = 250

x =

y =

b x × 9 = 72

x × y = 48

x =

y =

c x ÷ 7 = 8

x + y = 60

x =

y =

x – 15 = 35 + 15

x = 50

x ÷ 7 = 8 × 7

x = 56

x × 9 = 72 ÷ 9

x = 8

50 × y = 250

50 × y = 250 ÷ 50

y = 5

8 × y = 48

8 × y = 48 ÷ 8

y = 6

56 + y = 60

56 + y = 60 – 56

y = 4

50 8 565 6 4

y × 5 = 40 ÷ 5

y = 8

y = 8 cm

SERIES TOPIC



Read the mystery number riddle, then solve using algebra. Write the informa on as an equa on, use y to stand for the unknown. Show all your working:


Algebra can help us fi nd out the value of unknowns or mystery numbers. Look at how this perimeter riddle is solved and then solve the rest in the same way. Call each unknown y. The fi rst one has been done for you.

a I am a length between 1 cm and 10 cm. When you add 4 cm to me you get the total length of one side of a square which has a perimeter of 28 cm. What am I?

(y + 4) × 4 = 28 (y + 4) × 4 = 28 ÷ 4 y + 4 = 7 y + 4 = 7 – 4 y = 3 cm

b I am a length between 1 cm and 10 cm. When you add 2 cm to me you get the total length of one side of an octagon which has a perimeter of 40 cm. What am I?

(y + 2) × = 40

(y + 2) × = 40 ÷

y + 2 = –

y = cm

c I am a length between 1 cm and 10 cm. When you add 5 cm to me you get the total length of one side of a pentagon which has a perimeter of 40 cm. What am I?

(y + 5) × 5 = 40

4

5

y + 4

y + 5

y + 2

a I am thinking of a number between 1 and 10. When I add 2, then divide by 3 and mul ply by 5, I get 10. What is the number?

(y + ) ÷ × = 10

(y + ) ÷ × = 10 ÷ 5

y + ÷ =

y + =

y =

b I am thinking of a number between 1 and 10. When I add 5, then divide by 4 and mul ply by 5, I get 10. What is the number?

8

8 8

5 2

3

(y + 5) × 5 = 40 ÷ 5 y + 5 = 8 – 5 y = 3 cm

2

2

3

3

2

5

5

3

2

2

6

4

×

–

3

2

(y + 5) ÷ 4 × 5 = 10

(y + 5) ÷ 4 × 5 = 10 ÷ 5

(y + 5) ÷ 4 = 2 × 4

y + 5 = 8 – 5

y = 3

SERIES TOPIC



Complete the algebraic addi on stacks. Here is a simple example to start you off . The blocks underneath must add to give the block above. In this example, 3a + 4a = 7a.

a b

c d

Simplify these statements. The fi rst one has been done for you:

Match these algebraic statements by connec ng them with a line:

k + k + k + 6 3k + 6

k + k + k + k k + k + k + 2

3k + 2 4k

6k + k + k + 10 8k + 10

Solving equations – simplifying algebraic statements

An algebraic statement is part of an equa on.Some mes algebraic statements can have the same variable many mes.To simplify a + a + a + a + a, we would rewrite it as 5a.5a means 5 × a which is the same as a + a + a + a + a, but is much easier to work with.

1

2

3

a 3k + k + 4k = 8k

c b + b + 5b =

b 6x + 2x + x =

d 8y + 5y – y =

5b

14b

9b 7b

3a

7a

4a

9a

22a

4a

9a

2x

6x 13x

9x 14y 7y

25y

7b

30b

19x 46y

13a

21y

16b

4x

5a

18y

12y

9x

SERIES TOPIC



Solve this riddle in the same way as the ques ons above: What is as light as a feather but impossible to hold for long?

Your …

Use what you know about algebraic statements to solve these equa ons:

a 2a + 3a = 15 5a = 15 5a ÷ 5 = 15 ÷ 5 a = 3

Solving equations – simplifying algebraic statements

Remember with algebraic statements, a le er next to a number just means mul ply. 6y means 6 × y.You can add and subtract variables that are the same, just like you would for regular numbers 5y + 9y = 14y 20a – 16a = 4a

b 9b – 5b = 24

= 24

÷ = 24 ÷

b =

c 6c – 2c = 36

= 36

÷ = 36 ÷

c =

Use the balance strategy to fi nd out what a is.5a means 5 × a, so with the balance strategy you must do the inverse which is ÷ 5 to both sides.

4

5

H 16r – 4r = 48

A 8i + 5i = 39

B 7m + 2m = 63

R 9p – 2p = 35

T 10f – 3f = 42

E 7x – x = 54

7 5 9 3 6 4

4b 4c

4b 4c4 44 4

6 9

12r = 48

12r ÷ 12 = 48 ÷ 12

r = 4

7p = 35

7p ÷ 7 = 35 ÷ 7

p = 5

13i = 39

13i ÷ 13 = 39 ÷ 13

i = 3

7f = 42

7f ÷ 7 = 42 ÷ 7

f = 6

9m = 63

9m ÷ 9 = 63 ÷ 9

m = 7

6x = 54

6x ÷ 6 = 54 ÷ 6

x = 9

B R E A T H

SERIES TOPIC



Ge ng ready

What to do

What to do next

1 Read the clues.

2 Show your working.

Clue 1 Maya and Josh have 20 candles.

Clue 2 Maya and Lim have 13 candles.

Clue 3 Lim and Josh have 15 candles.

Clue 4 There are 24 candles altogether.

Draw the right amount of candles on each cake:

Happy birthday solve

Three children are having a birthday party.

Can you work out how many candles need to go on each cake?

HINT: If Maya and Josh’s cakes take up 20 candles what is le for Lim?

Happy Birthday Maya Happy Birthday JoshHappy Birthday Lim

24– 20 = 4 (Lim)

x + 4 = 13– 4

x = 9 (Maya)

4 + x = 15– 4

x = 11 (Josh)

11 candles4 candles9 candles

SERIES TOPIC



Use algebra with the clues below to work out the missing prices.

Look carefully at this fi rst example and follow the steps to work out the rest.

Clue 1 A Mango Tango and a Strawberry Squeeze costs $9.

Clue 2 A Mango Tango costs $3 more than a Strawberry Squeeze.

Use algebra to fi nd out the cost of each. Use m for Mango Tango and s for Strawberry Squeeze.

Clue 1 A Cherry Bliss and an Apple Berry cost $12.

Clue 2 A Cherry Bliss costs $1 more than an Apple Berry.

Use algebra to fi nd out the cost of each. Use c for Cherry Bliss and a for Apple Berry.

Squelch juiceteria solve

Ge ng ready

What to do

You work at Squelch Juiceteria, a popular juice bar serving delicious concoc ons to go.

The manager has le you in charge of wri ng the daily specials on the board. She has texted you the 4 diff erent juices she wants you to write up but forgot to text the prices and now her phone is turned off .

Squelch Juiceteria Specials

Mango Tango Strawberry Squeeze

Cherry BlissApple Berry

m + s = $9

m – s = $3

m + s + m – s = $9 + $3

m + s + m – s = 12

m + m = $12 m =

+ s = $9 s =

Step 2 Combine the clues into one statement to cancel out one unknown:

Step 3 Work out the cost of the second juice:

Step 1 Write clues as algebra:

c + a =

c – a =

Step 2 Combine the clues into one statement to cancel out one unknown:

Step 3 Work out the cost of the second juice:

Step 1 Write clues as algebra:

c + a + c – a = $12 + $1

c + c = $13

c = $6.50

a = $5.50

$3

$12

$1

$6

$6

SERIES TOPIC



Check the following sums based on what you know about the order of opera ons. Correct any that are wrong:

Prac se Rule 2, mul plica on and division before addi on and subtrac on:

Prac se Rule 3, working from le to right:

Prac se Rule 1, doing the brackets fi rst:

Properties of arithmetic – order of operations

Mr Gain wrote this equa on on the board: 4 + 5 × 7 = ?Max performed the opera on of addi on fi rst, then mul plica on; Amity performed mul plica on fi rst, then addi on. Now they are confused—they can’t both be right!We need a set of rules so that we can avoid this kind of confusion. This is why for some number sentences we need to remember the 3 rules for the order of opera ons. Rule 1 Solve brackets.Rule 2 Mul plica on and division before addi on and subtrac on.Rule 3 Work from le to right.By following the rules, we can see that Amity was right. Rule 2 says you should always mul ply before you add.

1

2

3

4

4 + 5 × 7 = 63

4 + 5 × 7 = 39

a 7 + (6 × 9) =

c 100 – (25 ÷ 5) =

b 8 + (4 × 7) =

d 30 ÷ (4 + 11) =

a 100 – 4 × 8 =

c 8 + 6 × 9 =

b 60 + 25 ÷ 5 =

d 2 × 7 – 5 =

a 42 ÷ 6 × 4 =

c 32 + 4 – 16 =

b 46 + 10 – 20 =

d 72 ÷ 8 × 3 =

a 50 – (45 ÷ 9) + 8 = 53

b 100 – (7 × 5) + (30 – 6) = 41

c (60 – 8) × 2 + (16 ÷ 4) – 32 = 140

61

68

28

95

62

20

36

65

36

2

9

27

89

76

SERIES TOPIC



Work with a partner to see who can get the biggest number in each round. Roll a die 3 mes and write down the numbers in the equa on frame. Compare your answers. The biggest answer wins ten points. The winner is the player with the highest score at the end of Round 3.

Round 1: × ( + ) =

Round 2: + × =

Round 3: ( + ) × =

In each word problem there is an equa on frame that solves each problem. Use it to solve the problem:

a

b

c

6

7

5

Properties of arithmetic – order of operations

Make these number sentences true by adding an opera on (+, –, ×, ÷):

a 96 3 8 = 40

c 84 12 3 = 21

b 16 4 22 = 26

d 100 5 5 = 15

How much was the total bill if 5 people each had a sandwich worth $8 and 2 people had a drink for $3.

What is the total number of people at a party if 12 invita ons were sent to couples, 7 people could not make it and 5 people turned up unannounced?

30 children went to the water park. 12 went on the water slides fi rst. The rest went in 3 equal groups to the swimming pool. How many were in one of the groups that went to the pool?

My Score: / 30

( × ) + ( × ) =

( × ) – + =

( 30 – 12 ) ÷ =

Don’t forget the order of opera ons!

÷ ÷

÷÷

+ +

–×

5

12

3 6

$8

2

2

7

$3

5

$46

22

Answers will vary.

SERIES TOPIC



Using brackets and changing the order can make it easier to fi nd unknowns. Look at the fi rst ques on as an example, then try the rest.

Use brackets to show which pairs you should mul ply fi rst to make it easier:

Change the order and use brackets to make these equa ons easier:

a 325 + 61 + 75 =

+ + =

b 24 + 12 + 276 =

+ + =

Use brackets to show which pairs you should add fi rst to make it easier:

Properties of arithmetic – commutative rule

Look at 13 + 16 + 4. The rules say we go from le to right but this sum is easier to answer if we add it like this: (16 + 4) + 13. The commuta ve rule lets us do this when it is all addi on or all mul plica on, no ma er which order we do this the answer will be the same. But this is only if the sum is all addi on or all mul plica on. We can use brackets as a signal of what part of the sum to do fi rst. Look at these examples:7 + 34 + 23 = 64 is the same as (7 + 23) + 34 = 642 × 17 × 5 = 170 is the same as (2 × 5) × 17 = 170

1

2

3

4

a 17 + 3 + 8 = b 43 + 18 + 2 = c 62 + 5 + 15 = d 57 + 3 + 16 =

a ( × 12) × 5 = 120

× (12 × 5) = 120

× 60 = 120 ÷ 60

= 2

c 40 + (160 + ) = 300

b ( + 36) + 14 = 100

d 8 × ( × 9) = 144

a 7 × 25 × 4 = b 6 x 8 × 2 = c 50 × 4 × 3 = d 2 × 9 × 8 =

Can you goboth ways with subtrac on and division?

Look for complements.

28

700

63

96

82

600

461

312

(325

(276

75)

24)

61

12

461

312

76

144

( )

( )

( )

( )

( )

( )

( )

( )

+ (36 + 14) = 100

+ 50 = 100– 50

= 50

(8 × 9) × = 144

72 × = 144 ÷ 72

= 2

(40 + 160) + = 300

200 + = 300 – 200

= 100

SERIES TOPIC



a b c

Write equa ons for these word problems. Once you are sure of which opera on to use, order the numbers in way that suits you.

a Adele loves reading books. One weekend she read 8 pages on Friday night, 17 pages on Saturday night and 12 pages on Sunday a ernoon. How many pages did she read that weekend?

b Two classes competed to see who could raise the most money for charity over 3 days. 6H raised $85 on Monday, $38 on Tuesday and $15 on Wednesday. 6F raised $75 on Monday, $29 on Tuesday and $25 on Wednesday. How much did each class raise?

c Luke has been collec ng aluminium cans for a sculpture he is making. He has been collec ng 5 cans a week for the past 13 weeks but s ll needs double this amount. How many cans does he need in total?

Properties of arithmetic – commutative rule

5 Let’s prac se adding numbers in the order that makes it easier to add. Make a path through each number matrix so that the selected numbers add together to make the total in the shaded box. You can’t go diagonally and not all of the numbers need to be used. Start at the bold number:

325 75 42

61 25 82

12 80 70

250

50 150 42

30 120 75

12 180 25

300

15 85 50

85 70 40

120 80 100

400

6

325 75 42

61 25 82

12 80 70

250

50 150 42

30 120 75

12 180 25

300

15 85 50

85 70 40

120 80 100

400

8 + 17 + 12

= (8 + 12) + 17

= 20 + 17

= 37 pages

6H $85 + $38 + $15

= ($85 + $15) + $38

= $100 + $38

= $138

6F $75 + $29 + $25

= ($75 + $25) + $29

= $100 + $29

= $129

5 × 13 × 2

= (5 × 2) × 13

= 10 × 13

= 130 cans

SERIES TOPIC



Colour match each step of the distribu ve rule. For example, colour the equa on frame labelled ‛1’ in yellow and look for all the parts that match this equa on and colour them yellow too. Then match equa on 2 and so on. By matching all 5 equa ons, you will have the order of the le ers that spell the answer to the ques on below:

1 (30 + 8) × 6 (20 × 2) + (7 × 2) 40 + 14 = 54 A

2 (20 + 7) × 2 (30 × 6) + (8 × 6) 180 + 48 = 228 N

3 (10 + 9) × 3 (10 × 3) + (9 × 3) 30 + 27 = 57 I

4 (40 + 4) × 4 (70 × 3) + (2 × 3) 210 + 6 = 216 S

5 (70 + 2) × 3 (40 × 4) + (4 × 4) 160 + 16 = 176 L

What part of a human is in the Guinness Book of Records for reaching the length of 7.51 metres?

Fill in the missing numbers for the mul plica ons:

The distribu ve rule says that you can split a mul plica on into two smaller mul plica ons and add them.

53 × 4

(50 + 3) × 4

(50 × 4) + (3 × 4)

200 + 12 = 212

1

Properties of arithmetic – distributive rule

a 64 × 5 =

(60 + 4) × 5

( × 5) + ( × 5)

+ =

c 56 × 5 =

( + 6 ) × 5

( × 5) + ( × 5)

+ 30 =

b 73 × 5 =

(70 + 3) × 5

( × 5) + ( × 5)

+ =

d 84 × 6 =

( + ) × 6

( × 6) + ( × 6)

480 + =

2

______1

______2

______3

______4

______5

This comes in handy when the numbers are bigger than normal mes tables ques ons.

320

280 504

365

60

50 80

70

300

50

Y R R

R Y Y

B B B

O G G

G O O

80

250 24

35020

6 4

15

4

4

3

Yellow = Y Red = R Blue = B Orange = O Green = G

N A I L S

320

280 504

365

SERIES TOPIC



Use the distribu ve rule in reverse to solve this problem:

a Over the weekend Blake’s dad made 5 batches of cupcakes on Saturday and 7 batches on Sunday. How many were in a batch if the total amount that he made was 180?

(5 × ) + (7 × ) = 180

(5 + 7) × = 180

b Jenna and Mel made up a game where if you score a goal you get a certain number of points. Jenna scored 6 goals and Mel scored 5 goals. How many points did they each get if the total number of points was 66?

(6 × ) + (5 × ) = 66

3

4

Fill in the missing numbers for the divisions:

Properties of arithmetic – distributive rule

The distribu ve rule can help us fi nd unknowns if we reverse the fi rst 2 steps.

(8 × ) + (3 × ) = 88

(8 + 3) × = 88

11 × = 88

= 8

Both 8 and 3 are to be mul plied by the diamond so we can rewrite this as shown in line 2. Then you can use the balance strategy twice to fi nd the value of the diamond.

a 84 ÷ 4 =

(80 + 4) ÷ 4

( ÷ 4) + ( ÷ 4)

+

=

b 108 ÷ 4 =

(100 + 8) ÷ 4

( ÷ 4) + ( ÷ 4)

+

=

You can also use the distribu ve rule with division.

21

21

27

27

80 100

20 251 2

4 8

12 × = 180

12 × = 180 ÷ 12

= 15

(6 + 5) × = 66

11 × = 66

11 × = 66 ÷ 11

= 6 points

Jenna 6 × 6 = 36 points

Mel 5 × 6 = 30 points

SERIES TOPIC



Ge ng ready

What to do

1 A er shuffl ing the cards, place the 8 ques on cards and 8 answer cards face down in 2 separate arrays like this:

2 Player 1 selects one card from each set and if the ques on and answer match, then the player takes both cards and has another turn. If they don't match then Player 1 must return the cards to the same posi on and then it is Player 2’s turn.

3 Con nue un l there are no cards le .

4 The player with the most pairs wins. Both players check through the winner’s pairs.

Equation pairs apply

Prac se what you have learned in this topic by playing equa on pairs with a friend. You will need to copy both this page and page 41, then cut out the cards.

7 × (5 + 3) 56

42 ÷ 6 × 4 28

27 + 11 – 8 30

copy

SERIES TOPIC



Equation pairs apply

(40 + 6) × 5 230

(20 × 7) + (20 × 4) 220

60 + 25 ÷ 5 65

8 + 6 × 9 62

50 – (45 ÷ 9) + 8 53

42 Series G Topic 1 Assessment Copyright © 3P Learning

Patterns and functions – part 1 Name ____________________

Use the table to complete the number sequence.

6

9

12

15

18

Posi on of number 1 2 3 4 5 6 7 8 20

Rule

Number sequence



b The number in the 20th posi on is 63 true / false

c 32 is in this sequence true / false



a Start at 12 and subtract 0.5

b Start at 20 and add 1.5

Skills Not yet Kind of Got it

• Completes recursive number pa ern and writes the rule

• Creates a number pa ern according to a rule

• Completes and analyses a func on number pa ern with more than one opera on

1

2

3


a

Rule ____________________

b

Rule ____________________

45 21 13 49 42 28 21

43Series G Topic 1 AssessmentCopyright © 3P Learning


Use the table to complete the number sequence.

6

9

12

15

18

Posi on of number 1 2 3 4 5 6 7 8 20

Rule

Number sequence



b The number in the 20th posi on is 63 true / false

c 32 is in this sequence true / false



a Start at 12 and subtract 0.5

b Start at 20 and add 1.5


• Completes recursive number pa ern and writes the rule

• Creates a number pa ern according to a rule

• Completes and analyses a func on number pa ern with more than one opera on

1

2

3


a

Rule ____________________

b

Rule ____________________

45 21 13 49 42 28 2137 29 355 14

– 8 – 7

12

20

× 3 +3 × 3 +3 × 3 +3 × 3 +3 × 3 +3 × 3 +3 × 3 +3 × 3 +3 × 3 +3

6 9 12 15 18 21 24 27 63

11.5

21.5

11

23

10.5

24.5

10

26



• Completes the table to describe a growing pa ern

• Completes the rule to describe a growing pa ern for each shape


Complete the table for each sequence of shapes.

a

b Shape 1 Shape 2 Shape 3

Shape number 1 2 3 4 5 6 7 8 9 10 15

Number of squares 1 2 3

Number of circles 3 5 7

Rule for circles Number of circles = Number of squares ____________

Rule for squares Number of squares = Number of circles ____________

1

Shape number 1 2 3 4 5 20

Number of matchs cks 6 10 14


Shape 1 Shape 2 Shape 3 Shape 4



• Completes the table to describe a growing pa ern

• Completes the rule to describe a growing pa ern for each shape


Complete the table for each sequence of shapes.

a

b Shape 1 Shape 2 Shape 3

Shape number 1 2 3 4 5 6 7 8 9 10 15

Number of squares 1 2 3

Number of circles 3 5 7

Rule for circles Number of circles = Number of squares ____________

Rule for squares Number of squares = Number of circles ____________

1

Shape number 1 2 3 4 5 20

Number of matchs cks 6 10 14


Shape 1 Shape 2 Shape 3 Shape 4

18

4

9

5

11

6

13

7

15

8

17

9

19

10

21

15

31

22 82

4

× 2 + 1

– 1 ÷ 2

2


Complete the table and answer the ques ons about these real life func ons.

a A car is travelling at a speed of 80 km/hour.

b A pool fi lls at a rate of 5 litres every minute.


• Works with input and output rela onships and rules

• Can write a rule to describe input and output rela onships

Complete the func on tables.

a

b


1

2

Rule: × 8 + 1

IN 8 2 10 5 9 6 7 11

OUT

Rule: × ______ + 5

IN 3 6 2 11 20 9 4 5

OUT 23 41 17 71 125

Rule: Number of hours × ______ = Number of km travelled (or total km travelled)

Hours 1 2 3 4 5 6 7 8



Rule: Number of minutes × ______ = Number of litres (or total litres)

Minutes 5 10 15 20 25 30 35 40

Litres 25 50 75 100

How many litres a er 1 hour?


Complete the table and answer the ques ons about these real life func ons.

a A car is travelling at a speed of 80 km/hour.

b A pool fi lls at a rate of 5 litres every minute.


• Works with input and output rela onships and rules

• Can write a rule to describe input and output rela onships

Complete the func on tables.

a

b


1

2

Rule: × 8 + 1

IN 8 2 10 5 9 6 7 11

OUT

Rule: × ______ + 5

IN 3 6 2 11 20 9 4 5

OUT 23 41 17 71 125

Rule: Number of hours × ______ = Number of km travelled (or total km travelled)

Hours 1 2 3 4 5 6 7 8



Rule: Number of minutes × ______ = Number of litres (or total litres)

Minutes 5 10 15 20 25 30 35 40

Litres 25 50 75 100

How many litres a er 1 hour?

65 17 81 41 73

400

125

6 hours

300 L

480

150

560

175

640

200

49

59

57

29

89

35

6

80

5



• Finds the value of an unknown represented by a symbol by using the balance strategy

• Subs tutes the value of one symbol to solve both symbols

• Sets out steps correctly

Algebraic thinking Name ____________________

1 Find out the value of both symbols:

a Clue 1 6 × = 36 Clue 2 – = 14

Find out the value of both symbols:

a Clue 1 5 × + 7 = 52

Clue 2 × = 72

Steps for fi nding 5 × = 52 –

5 × =

× 5 =

= ÷

=

Now you can fi nd

b Clue 1 8 × – 12 = 44

Clue 2 + = 30

Steps for fi nding

Now you can fi nd

2


Algebraic thinking Name ____________________


• Finds the value of an unknown represented by a symbol by using the balance strategy

• Subs tutes the value of one symbol to solve both symbols


1 Find out the value of both symbols:

a Clue 1 6 × = 36 Clue 2 – = 14

Find out the value of both symbols:

a Clue 1 5 × + 7 = 52

Clue 2 × = 72

Steps for fi nding 5 × = 52 –

5 × =

× 5 =

= ÷

=

Now you can fi nd

b Clue 1 8 × – 12 = 44

Clue 2 + = 30

Steps for fi nding

Now you can fi nd

2

6 × = 36

= 36 ÷ 6 = 6

8 × = 44 + 12

8 × = 56

× 8 = 56

= 56 ÷ 8

= 7

9 × = 72

= 72 ÷ 9

= 8

7 + = 30

= 30– 7

= 23

– = 14 – 6 = 14

= 14 + 6

= 20

7

45

45

45 5

9


Solving equations Name ____________________

Write an equa on to solve each mystery number ques on. Use m for the mystery number.

a A mystery number doubled is 84.



• Finds the value of an unknown represented by a variable by using the balance strategy

• Writes an equa on using variables to solve an unknown


Using the balance strategy, solve each equa on and then match the le ers to solve this riddle:

What belongs to you but others use it more than you do? Your …

1

2

M x + 5 = 25

E m + 9 = 36

A y – 6 = 46

N y + 8 = 32

24 52 20 27


Solving equations Name ____________________

Write an equa on to solve each mystery number ques on. Use m for the mystery number.

a A mystery number doubled is 84.



• Finds the value of an unknown represented by a variable by using the balance strategy

• Writes an equa on using variables to solve an unknown


Using the balance strategy, solve each equa on and then match the le ers to solve this riddle:

What belongs to you but others use it more than you do? Your …

1

2

M x + 5 = 25

E m + 9 = 36

A y – 6 = 46

N y + 8 = 32

24 52 20 27

x = 25 – 5

x = 20

y = 46 + 6

y = 52

m = 36 – 9

m = 27

m × 2 = 84

m = 84 ÷ 2

m = 42

m + 21 = 94

m = 94 – 21

m = 73

y = 32 – 8

y = 24

N A M E



• Understands and applies rules for order of opera ons

Properties of arithmetic Name ____________________

Check the following sums based on what you know about the order of opera ons. Correct any that are wrong.

a 40 – (25 ÷ 5) + 6 = 41

b 100 – (5 × 6) + (15 – 5) = 60

c (60 – 8) × 2 + (16 ÷ 4) – 32 = 140

Add brackets to make these equa ons true.

a 12 – 6 + 4 × 6 = 8 + 7 + 5 × 3

b 9 + 9 × 9 = 20 + 30 × 2 + 10

c 30 × 2 – 8 + 2 = 7 + 6 × 5 + 13

1

2

3

Show what you know about the order of opera ons.

a 3 + (4 × 4) – 3 – 3 =

c 20 – (36 ÷ 6) + 5 =

e 30 ÷ (3 × 5) + 4 =

b 20 – (25 ÷ 5) × 2 =

d 7 – 2 + (7 × 9) + 8 =

f 36 ÷ (3 × 3) + 5 =



• Understands and applies rules for order of opera ons

Properties of arithmetic Name ____________________

Check the following sums based on what you know about the order of opera ons. Correct any that are wrong.

a 40 – (25 ÷ 5) + 6 = 41

b 100 – (5 × 6) + (15 – 5) = 60

c (60 – 8) × 2 + (16 ÷ 4) – 32 = 140

Add brackets to make these equa ons true.

a 12 – 6 + 4 × 6 = 8 + 7 + 5 × 3

b 9 + 9 × 9 = 20 + 30 × 2 + 10

c 30 × 2 – 8 + 2 = 7 + 6 × 5 + 13

1

2

3

Show what you know about the order of opera ons.

a 3 + (4 × 4) – 3 – 3 =

c 20 – (36 ÷ 6) + 5 =

e 30 ÷ (3 × 5) + 4 =

b 20 – (25 ÷ 5) × 2 =

d 7 – 2 + (7 × 9) + 8 =

f 36 ÷ (3 × 3) + 5 =

13 10

19 76

6

( )

( )

( ) ( ) ( )

( )

( )

9

80

76

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Patterns and Algebra - In Class, with Miss. Coatescoatesj.weebly.com/uploads/6/1/0/2/61028649/95727447.g... · 2019. 11. 10. · Patterns and Algebra - In Class, with Miss. Coates