•

•

•

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm

Find the algebraic rule for finding the height of a stake of n books on the bench top if two stacks of books are placed upon a bench top , and a stack of 3 books on the bench top is 50mm tall while a stack of 5 books on the bench top is 70mm tall. First, we can establish that the height of the bench top is a fixed value. Next, the height of a stack of books is dependent on the number n of books in the stack. So, assuming each book has a fixed thickness of a, and the bench top has a fixed height of b, then the algebraic rule for finding the height h of a stack of n books on the bench top is in the form: h = an + b. In order to find a, we work with the information about two different stacks of books. We firstly find the difference between the total height (includes the stack and the bench top) of two different stacks, and compare this difference to that between the number n of books in both given stacks. In this case, the difference in the total height (in mm) is 70 – 50 = 20, while the difference in the number n of books = 5 – 3 = 2. Therefore, this means that the larger stack has two more books than the smaller stack, and is 20mm taller. Therefore, this means that each book is 20mm/2 = 10mm thick. So a = 10. Then we need to find b by substituting in known values. 1. an + b = h 2. (10 x 3) + b = 50 3. 30 + b = 50 Therefore, b = 20 Hence, the algebraic rule for finding the height (in mm) of a stack of n books on the bench top is 10n + 20

Are you able to find algebraic rules for a given pattern? Take a look at the example

below, then answer the questions by shading in the correct bubble.

Read the example below. !

50cm

80cm

40n - 90 30n + 20 10n + 10 20n - 10

Two stacks of bricks are placed upon a bench top. A stack of four bricks on the bench top

is 50cm tall while a stack of seven bricks on the bench top is 80cm tall. Which rule can be

used to work out the height, in centimetres, of a stack of n bricks on the bench top? Q1

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm

50cm 120cm

50mm 100mm

70cm

50cm

A row of three books with equal dimensions and bookends on either side is 50mm wide,

while a row of eight books with equal dimensions and bookends is 100mm wide. What

rule can be used to work out the width of a row of n books between the bookends? Q3

Two rows of tin cans are placed in between poles. A row of 5 cans with the poles is 80cm

wide, while a row of 9 cans with the pole is 120cm wide. What rule can be used to work

out the width, in centimetres, of a row of n cans within the poles? Q2

A stack of three cups is 50cm tall while a stack of five cups is 70cm tall. Formulate your

own rule and determine the height of both stacks of cups if they were stacked on top of

each other. Q4

10n + 20 20n + 10 10n - 30 30n - 10

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm

Natural disasters have a number of impacts, which can be classified as primary,

secondary or tertiary. They can also affect climate change and alter atmospheric

conditions.

Thermometer 1

a) 4x = 20

b) 43 + x = 60

c) x – 35 = 21

d) 132 ÷ x = 11

e) 143.6 + x = 55

Sum of A to E = ___________

Thermometer 2

a) 8x = 64

b) 98 + x = 105

c) 45 – x = 30

d) 72 ÷ x = 9

e) 40 + x = 7.8

Sum of A to E = ___________

Using algebraic methods, solve each question for . Add the answers to find out the range

by which temperatures are expected to rise by 2100 and draw them onto the

thermometers. Lines have been provided for working out if you need them. Q1

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm Remember that each

expression contains only one IP

address. The first expression

has been decoded for you.

While downloading music off of the internet is free, it is also illegal and means that

the people making the music don’t receive payment for their hard work. Help

Jonah, a policeman working in the piracy division, track down a number of criminals that

are known for downloading this content and selling it for profit.

1. (8a + 9b) (14c + 16d) + 4(ad + bc) IP:

2. 5((2f + 2h) (12l + 11o) - (fl - 4ho)) IP:

3. (56d + 52s) (2a + 3x) - 2x(9d + 17s) IP:

4. (23x + 11a) (10w + 11d) IP:

112ac + 132ad + 130bc + 144bd

112.132.130.144

Jonah has received several algebraic expressions that can be decoded to identify the IP

addresses of the criminals. The coefficients of the expanded expressions are individual

parts of IP addresses. Decode them all below to identify the criminals. Q1

Tip

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm Jack

IP: 112.132.130.144 Jess

IP: 210.129.10.12

Chris IP: 230.100.110.253

Meagan IP: 115.110.120.130

Chandler IP: 129.132.130.125

Joanne IP: 112.150.104.122

! Based on the IP addresses that you identified in Question One, circle the faces of the

people below that are guilty of internet piracy. Q2

SAMPLE

•

•

•

SAMPLE

Firstly, thank you for your support of Mighty Minds and our resources. We endeavour to create high-

quality resources that are both educational and engaging, and results have shown that this approach

works.

To assist you in using this resource, we have compiled some brief tips and reminders below.

About this resource

This Mighty Minds ‘Fundamentals’ Lesson focusses on one subtopic from the NAPLAN Tests and

presents this skill through a theme from the Australian Curriculum (History, Science or Geography).

This lesson is also targeted at a certain skill level, to ensure that your students are completing work

that is suited to them.

How to use this resource

Our ‘Fundamentals’ Lessons are split into two main sections, each of which contain different types of

resources.

The student workbook contains

• The main title page; and

• The blank student worksheets for students to complete.

The teacher resources section contains

• This set of instructions;

• The Teacher’s Guide, which offers information that may be needed to teach the lesson;

• The Item Description, which gives a brief overview of the lesson and its aims, as well as extension

ideas;

• The student answer sheets, which show model responses on the student worksheets to ensure

that answers to the questions are clear;

• The teacher’s answer sheets, which provide a more detailed explanation of the model responses

or answers; and

• Finally, the ‘end of lesson’ marker.

We suggest that you print the student workbook (the first set of pages) for the students. If students

are completing this lesson for homework, you may also like to provide them with the student answer

pages.

Feedback and contacting us

We love feedback. Our policy is that if you email us with suggested changes to any lesson, we will

complete those changes and send you the revised lesson – free of charge.

Just send your feedback to resources@mightyminds.com.au and we’ll get back to you as soon as we

can.

SAMPLE

Like Terms

A term in algebra is the simplest unit in an expression or equation. It can contain numbers,

pronumerals (representing an unknown value and usually expressed as a letter of the alphabet) and

indices. Like terms are terms that have the same pronumeral or pronumerals and powers, but can

have differing coefficients (a number that comes before a pronumeral – e.g. with 3x, 3 is the

coefficient and x is the pronumeral). Therefore, 3xy3 and 8xy3 are like terms, but 9x2z and 2xy5 are

not. Like terms can be added and subtracted from each other, but unlike terms cannot. However, both

like and unlike terms can be multiplied and divided.

Example: 4de + 4d2+ 54de – d2 + 3d = 48de + 3d2 + 3d

Distributive Law The distributive law states that

a (b + c) = ab + ac

The law is applied to expand brackets in order to solve equations. The opposite process is called

factorisation, i.e. ab + ac → a (b+c)

Example: 4(g2 + 2h) = 4g2 + 8h

Index Laws

There are six index laws which dictate how to perform calculations with numbers/pronumerals in

index form (i.e. numbers with a power)

1. Multiplication: when multiplying powers with the same base, add the indices.

Rule: ax x ay = abx + y

Example: d5 x d8 = d13

2. Division: when dividing powers with the same base, subtract the indices.

Rule: ax ÷ ay = ax – y

Example: t 17 ÷ t4 = t13

Note: division sums are often written as a fraction. Thus, is the same as t 17 ÷ t4 .

3. Power of zero: any number, except 0, raised to the power of 0 is equal to 1.

Rule: a0 = 1, when a ≠ 0

4. Index of an index: when an index form is raised to another power, multiply the indices.

Rule: (ax)y = axy

Example: (k4)3 = k12

5. Powers of products: when a product is raised to a power, every factor of the product is raised to

the power.

Rule: (ab)x = ax x bx

6. Powers of quotients: when a quotient is raised to a power, both the numerator and denominator

are raised to the power.

Rule: =

ax bx ( )

x a b

t 17

t4 SAMPLE

Please note: any activity that is not completed during class time may be set for homework or

undertaken at a later date.

‘Amazing Algebra’, ‘Impacts of Disasters’ and ‘Tracker

Takedown’

• Activity Description: • The first activity, ‘Amazing Algebra’, reminds students of some key concepts of algebra,

giving them a chance to practise their numeracy skills before diving into the rest of the

lesson.

• In the second activity, ‘Impacts of Disasters’, students have to calculate sums using

algebraic methods.

• The final activity, ‘Tracker Takedown’, extends on the previous two activities by asking

students to calculate algebraic problems in order to identify the mystery culprit of a crime.

• Purpose of Activity: • To improve students’ abilities to perform algebraic calculations.

• KLAs: • Mathematics, Science

• CCEs: • Recognising letters, words and other symbols (α1)

• Interpreting the meaning of words or other symbols (α4)

• Calculating with or without calculators (Ф16)

• Suggested Time Allocation: • This lesson is designed to take approximately one hour to complete – 20 minutes per

activity.

• Teaching Notes: • Once students have completed each activity discuss the answers as a class. Encourage

debate and justification from students if they have differing answers.

• If students are struggling with the activities, complete the first question with them by

identifying each step necessary to calculate the answer. The teacher’s answer sheet

should help with this, if necessary.

• Follow Up/ Class Discussion Questions: • Which aspects algebra do you find to be the most challenging?

• Why is the order of operations an important aspect of algebra?

Item Description

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm

Find the algebraic rule for finding the height of a stake of n books on the bench top if two stacks of books are placed upon a bench top , and a stack of 3 books on the bench top is 50mm tall while a stack of 5 books on the bench top is 70mm tall. First, we can establish that the height of the bench top is a fixed value. Next, the height of a stack of books is dependent on the number n of books in the stack. So, assuming each book has a fixed thickness of a, and the bench top has a fixed height of b, then the algebraic rule for finding the height h of a stack of n books on the bench top is in the form: h = an + b. In order to find a, we work with the information about two different stacks of books. We firstly find the difference between the total height (includes the stack and the bench top) of two different stacks, and compare this difference to that between the number n of books in both given stacks. In this case, the difference in the total height (in mm) is 70 – 50 = 20, while the difference in the number n of books = 5 – 3 = 2. Therefore, this means that the larger stack has two more books than the smaller stack, and is 20mm taller. Therefore, this means that each book is 20mm/2 = 10mm thick. So a = 10. Then we need to find b by substituting in known values. 1. an + b = h 2. (10 x 3) + b = 50 3. 30 + b = 50 Therefore, b = 20 Hence, the algebraic rule for finding the height (in mm) of a stack of n books on the bench top is 10n + 20

Are you able to find algebraic rules for a given pattern? Take a look at the example

below, then answer the questions by shading in the correct bubble.

Read the example below. !

50cm

80cm

40n - 90 30n + 20 10n + 10 20n - 10

Two stacks of bricks are placed upon a bench top. A stack of four bricks on the bench top

is 50cm tall while a stack of seven bricks on the bench top is 80cm tall. Which rule can be

used to work out the height, in centimetres, of a stack of n bricks on the bench top? Q1

Height difference (in cm): 80 – 50 = 30

Difference in number of bricks: 7 – 4 = 3

So, a = 30 ÷ 3 = 10

an + b = h

(10 x 4) + b = 50

40 + b = 50

Therefore, b = 10. The rule is 10n+10.

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm

50cm 120cm

50mm 100mm

70cm

50cm

A row of three books with equal dimensions and bookends on either side is 50mm wide,

while a row of eight books with equal dimensions and bookends is 100mm wide. What

rule can be used to work out the width of a row of n books between the bookends? Q3

Two rows of tin cans are placed in between poles. A row of 5 cans with the poles is 80cm

wide, while a row of 9 cans with the pole is 120cm wide. What rule can be used to work

out the width, in centimetres, of a row of n cans within the poles? Q2

A stack of three cups is 50cm tall while a stack of five cups is 70cm tall. Formulate your

own rule and determine the height of both stacks of cups if they were stacked on top of

each other. Q4

10n + 20 20n + 10 10n - 30 30n - 10

10n + 30

500cm ÷ 5m

Width difference = 120 – 80 = 40

Difference in number of cans = 9 – 5 = 4

So, a = 40 ÷ 4 = 10

an + b = w

(10 x 5) + b = 80

50 + b = 80. Hence, b = 30.

Width difference = 100 – 50 = 50

Difference in the number of books = 8 – 3 = 5

So, a = 50/5 = 10

(10 x 3) + b = 50

30 + b = 50. Therefore, b = 20

Hence, the rule is 10n + 20.

(50 x 3) = 150

(70 x 5) = 350

150 + 350 = 500

Hence, the rule is: 500cm ÷ 5cm

SAMPLE

Amazing Algebra

Question One:

In this question, students were asked to shade the bubble that represented the correct rule to work

out the height of a stack of n bricks on a bench top.

Model Response:

10n + 10

Students’ working out does not have to match the working out below exactly, but should be fairly

similar.

Model Response:

Assuming each brick has a fixed thickness of a, and the bench top has a fixed height of b,

then the algebraic rule for finding the height h (in cm) of a stack of n bricks on the bench top is

in the form: h = an + b.

For finding a, Difference in the total height (in cm) = 80 – 50 = 30

Difference in the number n of bricks = 7 – 4 = 3

So, a = 30/3 = 10

Now, solving for b (equation based on smaller stack),

an + b = h

(10 x 4) + b = 50

40 + b = 50

Therefore, b = 10

Hence, the algebraic rule for finding the height (in cm) of a stack of n bricks on the bench top

is 10n+10.

Question Two:

In this question, students were asked to determine which rule could be used to calculate the width (in

cm) of a row of n cans within two poles.

Model Response:

10n + 30

The working out required to reach this answer can be seen on the following page. Students’ working

out did not have to match the model response exactly, but should have been fairly similar in order to

reach the same answer.

This teacher’s answer guide is continued on the next page...

SAMPLE

...This answer guide is continued from the previous page.

Model Response:

Assuming each can has a fixed width of a, and each pair of bookends has a fixed width of b,

then the algebraic rule for finding the width w(in cm) of a row of n cans with the bookends is in

the form: w=an+b.

For finding a,

Difference in the total width (in cm) = 120 – 80 = 40

Difference in the number n of cans = 9 – 5 = 4

So, a = 40/4 = 10

Now, solving for b (equation based on smaller row),

an + b = w

(10 x 5) + b = 80

50 + b = 80

Hence, b = 30

Therefore, the algebraic rule for finding the width (in cm) of a row of n cans with the bookends

is 10n + 30.

Question Three:

In this question, students were required to formulate a rule that can be used to calculate the width of

a row of n books between two bookends.

Model Response:

10n + 20

To find this answer, students’ working out should have been fairly similar to the model response

below.

Model Response:

Assuming each book has a fixed width of a, and each pair of bookends has a fixed width of b,

then the algebraic rule for finding the width w(in mm) of a row of n books with the bookends is

in the form: w=an+b.

For finding a,

Difference in the total width (in mm) = 100 – 50 = 50

Difference in the number n of books= 8 – 3 = 5

So, a = 50/5 = 10

Now, solving for b (equation based on smaller row),

an + b = w

(10 x 3) + b = 50

30 + b = 50

Therefore, b = 20

This teacher’s answer guide is continued on the next page...

SAMPLE

...This answer guide is continued from the previous page.

Hence, the algebraic rule for finding the width w (in mm) of a row of n books with the

bookends is 10n + 20.

Question Four:

In this activity, students were asked to formulate a rule that could be used to calculate the height of

two stacks of cups if they were stacked on top of each other. One stack was 50cm high (three cups)

and one stack was 70cm high (five cups).

Model Response:

(50 x 3) + (70 x 5) = 500

Therefore, the rule that could be used is: 500cm ÷ 5cm

SAMPLE

Natural disasters have a number of impacts, which can be classified as primary,

secondary or tertiary. They can also affect climate change and alter atmospheric

conditions.

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm

Thermometer 1

a) 4x = 20

b) 43 + x = 60

c) x – 35 = 21

d) 132 ÷ x = 11

e) 143.6 + x = 55

Sum of A to E = ___________

Thermometer 2

a) 8x = 64

b) 98 + x = 105

c) 45 – x = 30

d) 72 ÷ x = 9

e) 40 + x = 7.8

Sum of A to E = ___________

5

17

56

12

-88.6

1.4

8

7

15

8

-32.2

5.8

Using algebraic methods, solve each question for . Add the answers to find out the range

by which temperatures are expected to rise by 2100 and draw them onto the

thermometers. Lines have been provided for working out if you need them. Q1

SAMPLE

Impacts of Disasters

Question One:

In this question, students were required to answer a series of algebraic problems and find the sum of

the answers for two different situations. The sums represented two temperatures, 1.4 ̊C and 5.8 ̊C.

This is the expected increase in temperature caused by climate change. Students were also required

to colour this in on a thermometer. Lines were provided for working out if they chose to use them,

however they did not have to use them. Working out has been provided below to assist with

explanations of answers.

Model Response:

Thermometer 1

A) 4x = 20

x = 20 ÷ 4

x = 5

B) 43 + x = 60

x = 60 – 43

x = 17

C) x – 35 = 21

x = 21 + 35

x = 56

D) 132 ÷ x = 11

x = 132/11

x = 12

E) 143.6 + x = 55

x = 55 – 143.6

x = -88.6

Sum of A to E = 1.4

This teacher’s answer guide is continued on the next page...

SAMPLE

...This answer guide is continued from the previous page.

Model Response:

Thermometer 2

A) 8x = 64

x = 64/8

x = 8

B) 98 + x = 105

x = 105 – 98

x = 7

C) 45 – x = 30

x = 45 – 30

x = 15

D) 72 ÷ x = 9

x = 72 ÷ 9

x = 8

E) 40 + x = 7.8

x = 7.8 – 45

x = -32.2

Sum of A to E = 5.8

SAMPLE

Remember that each

expression contains only one IP

address. The first expression

has been decoded for you.

While downloading music off of the internet is free, it is also illegal and means that

the people making the music don’t receive payment for their hard work. Help

Jonah, a policeman working in the piracy division, track down a number of criminals that

are known for downloading this content and selling it for profit.

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm

1. (8a + 9b) (14c + 16d) + 4(ad + bc) IP:

2. 5((2f + 2h) (12l + 11o) - (fl - 4ho)) IP:

3. (56d + 52s) (2a + 3x) - 2x(9d + 17s) IP:

4. (23x + 11a) (10w + 11d) IP:

230xw + 100xd + 110aw + 253ad

230.100.110.253

112ad + 150dx + 104as + 122sx

112.150.104.122

115fl + 110fo + 120hl + 130ho

115.110.120.130

112ac + 132ad + 130bc + 144bd

112.132.130.144

Jonah has received several algebraic expressions that can be decoded to identify the IP

addresses of the criminals. The coefficients of the expanded expressions are individual

parts of IP addresses. Decode them all below to identify the criminals. Q1

Tip

SAMPLE

Height = 0.6cm

Height = 1.1cm

Height = 1.6 cm Jack

IP: 112.132.130.144 Jess

IP: 210.129.10.12

Chris IP: 230.100.110.253

Meagan IP: 115.110.120.130

Chandler IP: 129.132.130.125

Joanne IP: 112.150.104.122

! Based on the IP addresses that you identified in Question One, circle the faces of the

people below that are guilty of internet piracy. Q2

SAMPLE

Tracker Takedown

Question One:

In this question, students were required to expand a number of expressions and gather like terms.

After doing this, they should then have taken the coefficients and used them to determine which of

the suspects was guilty of internet piracy. The coefficients of each expression could be combined to

find the IP address of a guilty pirate.

Model Response:

1. (8a + 9b) (14c + 16d) + 4(ad + bc)

= 112ac + 128ad + 126bc + 144bd + 4ad + 4bc

= 112ac + 132ad + 130bc + 144bd

IP: 112.132.130.144

2. 5((2f + 2h) (12l + 11o) - (fl - 4ho))

= 5 ( 23fl + 22fo + 24hl + 22ho)

= 115fl + 110fo + 120hl + 130ho

IP: 115.110.120.130

3. (56d + 52s) (2a + 3x) - 2x(9d + 17s)

= 112ad + 168dx + 104as +156sx - 18xd - 34xs

= 112ad + 150dx + 104as + 122sx

IP: 112.150.104.122

4. (23x + 11a) (10w + 11d)

= 230xw + 100xd + 110aw + 253ad

IP: 230.100.110.253 :

This teacher’s answer guide is continued on the next page...

SAMPLE

...This answer guide is continued from the previous page.

Question Two:

In this question, students should have used the decoded IP addresses from Question One to identify

the guilty pirates. They should have done so by circling their faces.

Model Response:

Jack IP: 112.132.130.144

Jess IP: 210.129.10.12

Chris IP: 230.100.110.253

Meagan IP: 115.110.120.130

Chandler IP: 129.132.130.125

Joanne IP: 112.150.104.122

SAMPLE

SAMPLE