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
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 [email protected] 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