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GCSE

Christine Watson

Maths 4th Edition

Foundation Teacher Pack

06_GCSE Maths Foundation Teacher Pack.indd 5 15/10/14 11:44 AMAQA Foundation Teacher Pack.indd 1 10/22/14 12:47 AM

William Collins’ dream of knowledge for all began with the publication of his first book in 1819. A self- educated mill worker, he not only enriched millions of lives, but also founded a flourishing publishing house. Today, staying true to this spirit, Collins books are packed with inspiration, innovation and practical expertise. They place you at the centre of a world of possibility and give you exactly what you need to explore it.

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206745 GCSE Maths Foundation.indd 2 10/21/14 8:46 PM

AQA Foundation Teacher Pack.indd 2 10/21/14 8:48 PM

Maths4th Edition

GCSEFoundation Teacher Pack

Christine Watson

AQA Foundation Teacher Pack.indd 3 10/22/14 12:47 AM

GCSE Maths 4th edition 4 © HarperCollinsPublishers Ltd 2015 Foundation Teacher Pack

Chapter 7 Ratio, speed and proportion Overview 7.1 Dividing in a given ratio 7.3 Understand and use direct proportion

7.2 Calculating with speed, distance, time 7.4 Calculate in a ‘best buy’ context Prior learning

Can cancel fractions Can convert between measurement units (does not include metric to imperial)

Chapter 7

Ensure students understand and can calculate with ratios, speed and proportion. In the examination, students will be expected to: A cancel ratios to simplest form, to divide quantities using a ratio of two or three parts

including unit ratios. Contexts include recipes and finance. B calculate speed, distance or time including from information in a travel graph. C calculate missing values in direct proportion problems including in a Best Buy context.

Extension

Explore real world applications. Devise a way to compare/order ratios. Explore the golden ratio in architecture and art. Explore proportionality in geometry or in graphs.

Curriculum references

Section KS3 NC Programmes of Study

KS4 NC Programmes of Study GCSE specification

7.1 R1 R2 R1 R3 R4 R5 R6 R7 N11 7.2 R5 A (EF) 7 N14 7.3 R4 R3 R10 7.4 R4 R3 R11 Route mapping

Exercise Accessible Intermediate Challenging AO1 AO2 MR CM

AO3 PS EV

Key questions

7A 1–5 6–8 9–12 1–8 9–10 11–12 2, 6, 7, 9 7B 1–4 5–7 8–10 1–5, 7–8 6 9–10 3, 7, 8g 7C 1–4 5–7 8–10 1–6, 9 7 8, 10 5, 7

7D 1–5 6–13 14–16 1–5, 7–8, 11–13

6, 9, 10, 15 14, 16 3, 4, 9, 14

7E 1–5 6–7 8–10 1–6 7–8 8–10 4, 7, 8 7F 1–4 5–6 7–8 1–4 5, 7, 8 6 3, 5, 7 Key questions are those that demonstrate mastery of the concept or which require a step-up in understanding or application. These could be used to identify the questions that students must tackle, to support differentiation, or to identify the questions that should be teacher-marked rather than student-marked.



About this chapter Making connections: The chapter deals with proportional reasoning, with a focus on using tables to organise information and calculating missing values. The four sections are inter-related, with strong connections between 7.1, 7.3 and 7.4 in methods of representation. It builds on concepts met when working on fractions.

Relevance: Comparison of values is an essential skill, not just for shopping. Students should be encouraged to compare proportions in a range of contexts: financial, test scores, sports records, speed comparisons, betting odds, work rates etc. Working mathematically: What is the most effective way to present this information? What is the most efficient way to solve the problem? What are the key words when reading a ratio problem, and are the key words the same in a direct proportion problem?

Assessment: Give students a pair of values. Ask them to create a ratio question, a speed question, a direct proportion question and a best buy question each using this pair of values. Ask students to describe what is the same and what is different about their 4 questions and their methods of solution.

See the CD for suggested assessment tracking foci, and the section plans for further suggested Assessment tasks.

Worked exemplars from Student Book– suggestions for use A Present students with the same question but different numbers. They use the exemplar

to mirror the working, in full or just the notes. B Copy and cut up the exemplar into cards. Students match the working with the notes.

(You may need to remove the words ‘first, second’ etc.) C Copy and cut up the working into cards but split the label/description from the working.

Students put the working in order then match with the descriptions. Answers to Student Book questions at the end of this book (NB: not included in this sample)



Section 7.1 Working with ratios; Dividing in a ratio Learning objectives Resources and homework • Understand the concept of ratio including

simplest form • Be able to apply methods for dividing

quantities in a ratio

• Student Book 7.1 • Practice Book 7.1

Making mathematical connections Making cross-curricular connections • Cancelling fractions • Describing and comparing quantities and

size

• Geography – map scales • Technology – recipes, concrete mixes • Relevance – betting odds; sharing in

proportion vs fairly Prior learning

• It is essential that students can convert between time measures, between metric units. This may need a prior homework to refresh the skill, or a full lesson, depending on the students. Working mathematically

• Encourage students to apply their understanding of fractions (and decimals, percentages) to ratios. Avoid always presenting the simplest case (unit ratios, ratios with two parts) but challenge students to work with more complex ratios and apply their skills to these.

• Offer students the opportunity to discover their own representations for ratios – pictorial, tabular or other – and to make the links between the various representations. Common misconceptions and remediation

• Students see one part of the ratio as the total, not as a part, or see the ratio 1 : 2 as a half. Ensure that the physical image of a ratio is embedded at the start and make the comparison with fractions explicit. Probing questions

• Describe what is the same and what is different about a ratio and a fraction. • Explain how you would set out a ratio calculation – make up a good example to help you.

Literacy focus

• Key vocabulary: cancel ratio common units simplest form • Be explicit in the use of the vocabulary: fraction – ratio – part – total – sharing – comparing – describing.

Part 1 NB: This section may take more than 1 hour Introduce the concept of ratio as a comparison of proportions and make the link to fractions.

• Use the Top Trumps cards [or a set of photos with characteristics]. Give a set to each group. • Sort the cards in different ways. For each, show and verbalise the ratio, show the link to ratio

in a table (for example F : M : total) and make the link to the associated fractions. • Ask what the ratio tells them about the two groups. Prompt for the idea of relative size of the

groups and bring in the term proportion. • Ask for other suggestions for different groupings, allowing more than two groups. Students

should organise their cards into the new grouping, and write the ratio on their boards. Repeat until it is clear students understand how to decide what the ratio is.



• Ask students to work in pairs to decide what they think a ratio is. Ask for a short definition, a long definition, one with more mathematical words. Ask what the difference is between a ratio and a fraction. Give time for students to write down their ‘final’ definition. Part 2 Draw out equivalent ratios visually and using common factors, and make the link to cancelling fractions.

• Return to a two-part ratio that was not in simplest form – ask the students to quickly organise the photos into these two groups, and expect photos to be bunched. Now ask students to create a ‘tidy’ layout, and look for one that is rectangular. Match this one using the cards on the board [or ask students to suggest a more mathematical way to show the photos].

• Ask students if they can see any other ratios in the grouping (for example from 10 : 5, the new layout should highlight 2 : 1). Make the link between the overall ratio and the simplified ratio – what calculations can you do to simplify or increase the values? Which describes the proportions in the two groups? How does this link to other mathematics you have done?

• Ensure the link to equivalent fractions and cancelling is made. • Students can now do Exercise 7A from Student Book.

A 1–5 I 6–8 C 9–12 CM MR 9–10 PS 11–12 EV Key 2, 6, 7, 9 Part 3 Explore dividing in a given ratio as the reverse of simplifying – multiplying up as the inverse of dividing.

• Return to a visual display of a ratio, for example 2 : 6. Revisit that it can be cancelled, and show that it can be multiplied up.

• Return to the table layout for the ratio, with 2 : 6 (total) 8. Ask what the values would be when the total number of items is 16 – 80 – 40 – 4 – 12, showing these in the table one by one. Ensure ratio equivalence underpins the understanding and make the link to equivalent fractions.

• Students can now do Exercise 7B from Student Book. A 1–4 I 5–7 C 8–10 CM MR 6d PS 9–10 EV Key 3, 7, 8g NB: these involve mixed units – a mini-plenary to ensure students are dealing with the units correctly is appropriate.

• Students could then apply their understanding to Exercise 7C (optional).

Part 4 Explore dividing in a given ratio when the quantity is not the total amount.

• Pose a problem where the given quantity does not match the total and ask students to decide how they could tackle the problem.

• Discuss their suggestions and ensure all have a correct adapted original method. • Students can now do Exercise 7C from Student Book.

A 1–4 I 5–7 C 8–10 CM MR 7 PS 8 EV 10 Key 5, 7 • Plenary: (A) Students could create an easy and a hard ratio question and challenge their

partner to complete one. (B) Students could write the steps needed when working with ratios.

Assessment task

• Create a set of Top Trumps cards of your own to meet set criteria – could be heroes/pop stars/sports personalities/tv or film stars… To be peer-assessed in groups next lesson.

• eg female : male is 2 : 3, under 20 : 20 to 30 : over 30 is 2 : 3 : 1, British : American is 4 : 1.



Section 7.2 Speed, distance and time Learning Objective Resources and homework • Use the relationship between speed,

distance and time to work out unknown values.


Making mathematical connections Making cross-curricular connections • Substituting into formulae • Rearranging formulae • Minutes as a fraction or decimal

• Science/PE – measuring speed • Relevance – road safety; appreciation of

racing Prior learning

• It is essential that students can convert minutes to a decimal of an hour. •

Working mathematically • Encourage students to rearrange the formula or substitute then rearrange their calculation. • Use real life examples of speeds – speed limits, athletics, animal speeds – to bring the topic

to life. Common misconceptions and remediation

• Students divide the given two quantities without thinking about their meaning. Encourage use of the formula or of the units to avoid blind substitution.

• Ensure students recognise that 30 minutes is not 0∙30 of an hour. Use the clock face to link to geometric representations of fractions to consolidate understanding. Probing questions

• Is there a diagram that helps me understand speed questions? • How do I use the units in the question to help me decide what to do? • Which is faster, mph or km/h?

Literacy focus

• Key vocabulary: speed distance time average speed • Prompt for correct units, and ensure students use the units to help decide what to do, for

example km/h means km divided by hours.



Part 1 Explore what students already know about speed.

• Create a mind map: Speed at the centre, ask students to (a) individually add what they know about speed (b) share with a partner (c) share with another pair [1 to 2 to 4].

• Draw out the key points, which may include: speed limits; units of speed; athletics records; land/air/sea records; the formula speed = distance/time; how to graph speed; links to acceleration. Part 2 Demonstrate methods for calculating with speed, distance, time.

• Ensure students know that 60 mph means 60 miles in 1 hour 180 miles in 3 hours 30 miles in half an hour.

• Draw out the formula speed = distance ÷ time and check that students can calculate any value by substituting the other two then calculating.

• Students can now do Exercise 7D from Student Book. A 1–5 I 6–13 C 14–16 CM 6b, 9, 15 MR 10 PS 14 EV 16 Key 3, 4, 9, 14 NB: Students need to use time as a decimal. Include a mini-plenary to pose the thinking point: 30 miles divided by 30 minutes doesn’t give 60 mph – what am I doing wrong? Part 3 Plenary

• Display a distance-time graph and challenge students to apply their understanding to find the speed(s).

Assessment task

• Provide a set of examination questions and ask students to (a) rank them in order of difficulty and assign red – amber – green (b) solve an amber question.



Section 7.3 Direct proportion Learning objective Resources and homework • Understand and calculate with direct

proportion. • Student Book 7.3 • Practice Book 7.3

Making mathematical connections Making cross-curricular connections • Link to ratio and fractions • Link to enlargement

• Science – correlations and connections, creating formulae

• Relevance – personal finance, especially shopping; stretch factors in photography

Prior learning

• This section builds on the ratio work and the links need to be made explicitly. Working mathematically

• Encourage students to see different representations of direct proportion: in a table, as pairs of values, as a graph plot, as a form of ratio, as a pictorial form of ratio etc and to see the links between each form.

• Explore which parts of the body are in direct proportion for a group of people, for example wrist circumference to waist circumference, leg length to height. Link this to media representations, for example the stretching of female legs in advertising – see example photographs on http://mathspig.wordpress.com/2010/05/04/make-me-feel-real-loose-like-a-long-necked-goose/. Common misconceptions and remediation

• Students sometimes assume that the unitary method is an essential step rather than using multipliers to move straight to the solution. Include very simple multipliers to draw this out.

• Students may see the original question and the solution as a set of 4 values, isolated from the concept of direct proportion. Include examples in a table where there are many missing values in both variables that can be found in different ways and emphasise this continuity. Use the graph form to show the infinite pairs of values that satisfy the multiplier, noting where the pairs do not connect with the context (for example, if the context is buying quantities the negative pairs do not have a physical meaning). Probing questions

• What is special about pairs of values that are in direct proportion? • What mathematical tests can you apply to check that values are in direct proportion? • When you plot the values as coordinates, what will you get? • Direct proportion values have a multiplier. How many multipliers can you find?



Literacy focus • Key vocabulary: proportion direct proportion multiplier unitary method unit cost • After completing Exercise 7E, discuss the language of the questions. Some make the direct

proportion aspect obvious and some are phrased to make it more obscure. • Select a straightforward wording – ask students to make it more obscure. Select a complex

wording – ask students to make it more straightforward. This may need modelling first to build student skills. Part 1 Recognise the properties of values in direct proportion.

• Create a set of at least six statements for display. For example: A 2 eggboxes hold 12 eggs B 2 people plaster a room in 4 hours C I travel 2 km in 15 minutes D a square of length 2 m has area 4 m² E 2 bags of potatoes cost £3 F a cube of length 2 m has volume 8 m³ G 2 cold-callers ring a batch of people in 5 hours.

• Ask students to group these into types. Look for a sense of increasing together/increase & decrease, and for a sense of which are direct. If necessary, prompt to change the ‘2’ to 4 (or 10), ask what would happen to the other value.

• Label A, C and E as direct proportion. Ask students to describe in pairs what the criteria might be for direct proportion, and draw this out through questioning. Use the term ‘multiplier’ to describe the relationships. Part 2 Calculate missing values in the context of direct proportion.

• Make the link from ratio to direct proportion by setting it out in a table but without a ‘total’. • Use A, C and E as examples of multiplying up to find missing values, and of the unitary

method. 2 boxes 12 eggs 2 km 15 mins 2 bags £3 ↓ ↓ ↓ ↓ ↓ ↓ (÷2) 1 6 (÷2) (×2∙5) 5 km 37.5 mins (×2∙5) (×4) 8 bags £12 (×4)

• Use the eggbox example to show that there is more than one set of relationships: Downwards, divide by 2; upwards, multiply by 2; to the right, multiply by 6, to the left, divide by 6. Emphasise this property of direct proportion.

• Students can now do Exercise 7E from Student Book. A 1–5 I 6–7 C 8–10 CM 7 MR 8 PS 9 EV 8, 10 Key 4, 7, 8 Part 3 Plenary

• Display a direct proportion problem where a number of values must be found in both variables. Ask students to work in pairs to find at least three different ways to solve it. Assessment task

• See the Literacy section.



Section 7.4 Best buys Learning objectives Resources and homework • Understand the concept of ‘value for

money’ • Calculate price per unit and units per

pound/penny


Making mathematical connections Making cross-curricular connections • Link to direct proportion and to simplifying

ratios • Relevance – shopping, value for money

Prior learning • This section builds on the ratio and direct proportion work and the links need to be made

explicitly. Working mathematically

• This section can be approached as a problem-solving task: Present students (in pairs or groups) with sets of pricing scenarios such as those in the exercise. Ask them to decide which is the best buy in each case and to justify their decisions, presenting solutions as a poster. Ask students to group their responses, showing which ones had been tackled using the same method. Students then present their posters to their peers, with the teacher prompting or rephrasing/clarifying but not leading. Students could then note in general form the most effective methods for deciding a best buy. Common misconceptions and remediation

• Some students will mix up their division calculations. Encourage them to set out the initial information in a table, making the link to direct proportion and ratio, and to then use their established method to find the missing values.

• It may be appropriate to only look at price per unit as this is an easier concept to understand and leave units per price until revisiting the topic at a later date. Probing questions

• How do you know which is the best buy? Convince me. • Which is better, cost divided by quantity, or quantity divided by cost? Why? • Should you always buy the item that gives you the cheapest costs per unit? Why?

Literacy focus

• Key vocabulary: best buy value for money better value • Ensure students understand the concepts of cheapest/dearest vs best buy/value for money. • This may need some specific scenarios where best buy is dependent on product quality,

money available, or quantity needed. • The language can be obscure and may need decoding. Display sets of statements and ask

students to match them up. For example: cost per unit mass cost of 1 gram (or 1 kg) cost per unit length cost of 1 metre (or 1 cm) mass per unit cost number of g (kg) for £1 (1p) length per unit cost number of m (cm) for £1 (1p)



Part 1 Recognise the properties of values in direct proportion.

• Display price labels from supermarkets.

BASICS MINCE Always Good Value

500g for £1∙20 price per kg £2∙40

BEST VALUE MINCE MEATY GOODNESS

1 kg for £3∙00 price per kg £3∙00

BASICS MINCE 2 kg for £4∙50 TODAY ONLY!

price per kg £2∙25

• Ask students: You need a kilogram of mince. Which would you buy? Why? • Draw out: quality comparisons, size comparisons (you don’t need 2 kg, under what

circumstances would you buy more than you need), the ‘price per…’ element of the label. • Draw out the difference between cheapest and best buy, best value for money.

Part 2 Calculate missing values in the context of direct proportion.

• Introduce price per unit. Ask which is better, a lower price per unit or higher. • Introduce units per price. Ask which is better, lower or higher.

NB: For some classes, ask them to complete some of Exercise 7F using price per unit and check understanding before introducing units per price. Note that for some students meeting both in the same session may not be helpful.

• Ensure students understand the language of the chapter – see Literacy section. • Students can now do Exercise 7F from Student Book.

A 1–4 I 5–6 C 7–8 CM 7, 8 MR 5 PS 6 EV Key 3, 5, 7 Part 3 Plenary and Assessment task

• Use examples of supermarket pricing including ones where the largest is not cheapest, or where a special offer or BOGOF affects the calculation (There are sets of these via, for example, TES). Ask students to identify which of each pair is best value for money.


Kevin EvansKeith Gordon

Brian SpeedMichael Kent

Maths 4th Edition

GCSEHigher Student Book

00_GCSE Maths Higher Student Book.indd 5 10/17/14 6:22 PM206745 GCSE Maths Higher SB_CH24.indd 1 10/17/14 9:10 PM

William Collins’ dream of knowledge for all began with the publication of his �rst book in 1819. A self-educated mill worker, he not only enriched millions of lives, but also founded a �ourishing publishing house. Today, staying true to this spirit, Collins books are packed with inspiration, innovation and practical expertise. They place you at the centre of a world of possibility and give you exactly what you need to explore it.

Collins. Freedom to teach.





Cover design by We are LauraCover images © Procy/Shutterstock, joingate/ShutterstockPrinted by Fuller Davies www.fullerdavies.com

Every effort has been made to contact the copyright holders but if any have been inadvertently overlooked, the publishers will be pleased to make the necessary arrangements at the �rst opportunity.

206745 GCSE Maths Higher SB_CH24.indd 2 10/18/14 11:46 AM

Maths 4th Edition

GCSEHigher Student Book

Kevin EvansKeith Gordon

Brian SpeedMichael Kent

206745 GCSE Maths Higher SB_CH24.indd 3 10/17/14 9:10 PM

Algebra: Algebraic fractions and functions

This chapter is going to show you:

• how to combine fractions algebraically and solve equations with algebraic fractions

• how to rearrange and change the subject of a formula where the subject appears twice, or as a power

• how to fi nd the inverse function and the composite of two functions

• how to fi nd an approximate solution for an equation using the process of iteration.

You should already know:

• how to substitute numbers into an algebraic expression

• how to factorise linear and quadratic expressions

• how to expand a pair of linear brackets to get a quadratic equation.

About this chapter

Without algebra, humans would not have reached the moon and aeroplanes would not fl y. Defi ning numbers with letters allows mathematicians to use formulae and solve the very complicated equations that are needed for today’s technologies. The ability to move from a special case to a generalisation is what makes algebra so useful.

Processes such as manipulating algebraic fractions, rearranging formulae, analysing functions and solving equations by iteration are used in a variety of professions in the areas of science, engineering and computing as well as in the arts. For example, to use a spreadsheet competently, you need to understand how functions work.

Weather forecasting makes use of the iterative process where small changes in the initial conditions can lead to completely different results. This is known as chaos theory. When the forecaster on television or on the internet says that there is a 60% chance of rain, this probability has been determined by running hundreds of simulations through an iterative process.

24Skill focus key

MR mathematical reasoning

CM communicate mathematically

PS problem-solving and making connections

EV evaluate and interpret

4


mktjxs

Sticky Note

Mathematical reasoning - you need to apply your skills and draw conclusions from mathematical information

mktjxs

Sticky Note

Communicating mathematically - you need to show how you have arrived at your answer by using mathematical arguments

mktjxs

Sticky Note

Problem solving - you need to devise a strategy to answer the question, based on the information you are given

mktjxs

Sticky Note

Evaluation - your answer needs to show that you have considered the information you are given and commented upon it

24.1 Algebraic fractionsThis section will show you how to:• simplify algebraic fractions

• solve equations containing algebraic fractions.

Algebraic fractions can be added, subtracted, multiplied or divided using the same rules that apply to numbers.

To add and subtract, fi nd a common denominator and then fi nd equivalent fractions with that denominator.

Addition: For ab

cd

+ , the common denominator is bd, so ab

cd

adbd

bcbd

ad bcbd

+ = + = +

ab

cd

adbd

bcbd

ad bcbd

+ = + = +

Subtraction: For ab

cd

− , the common denominator is bd, so ab

cd

adbd

bcbd

ad bcbd

− = − = −

ab

cd

adbd

bcbd

ad bcbd

− = − = −

This method works for more than two terms.

For example, for ab

cd

ef

+ − , the common denominator is bdf, so

ab

cd

ef

adfbdf

bcfbdf

bdebdf

adf bcf bdebd

+ − = + − = + −ff

ab

cd

ef

adfbdf

bcfbdf

bdebdf

adf bcf bdebd

+ − = + − = + −ff

To multiply, cancel any common factors, then multiply the numerators together and the denominators together.

Multiplication: ab

cd

acbd

× =

To divide, fi nd the reciprocal of the fraction you are dividing by, and then multiply.

Division: ab

cd

ab

dc

adbc

÷ = × =

ab

cd

ab

dc

adbc

÷ = × =

ote that a, b, c and d can be numbers, other letters or algebraic expressions. emember to

• use brackets, if necessary, to avoid problems with signs and help you expand expressions

• factorise if you can

• cancel if you can.

Key wordalgebraic fraction

implify these fractions. a 12xxy

+ b 22bab

−

a ommon denominator is xy

12

1 22x

xy

y x xx y2x y2

+ =+ =2+ =2x+ =x ( )1 2)1 2(1 2(1 2 )y x)y x+y x+y x(y x(y x)( )

( )x y)x y(x y(x y)

= +2

2

2y x+y x+xy

b ommon denominator is b

22

42 2b

ab b2 2b b2 2

ab

− =− = −

= −4

2a

b

Exa

mp

le 1

24.1 lgebraic fractions 5


implify these fractions. a x xx3

22× +

− b x x3

2x x2x x7÷

a ultiplying x xx

x x

x x

322 3 2

2x x2x x3 6x3 6x

2x x2x x

×x x×x x +− = ( )x x( )x x( )x x( )x x 2( )2+( )+

( )3 2( )3 2( )3 2( )3 2x3 2x( )x3 2x3 2−3 2( )3 2−3 2

= +x x+x x3 6−3 6

emember, the line that separates the top and bottom of an algebraic fraction acts as brackets as well as a division sign.

b Dividing x x3

2x x2x x7 3 2

76

÷ =÷ = ( )x( )x ( )7( )7

( )3 2( )3 2( )x( )x3 2( )3 2

=

Exa

mp

le 2

olve this equation. x x+x x+x x− − =1x x1x x3

32 1

ubtract the fractions on the left- hand side 2 1 3 3

2 31

( )2 1)2 12 1+2 1(2 1(2 1) ( )3 3)3 33 3−3 3(3 3(3 3)( )2 3)2 3(2 3(2 3) =

x x2 1x x2 1 3 3x x3 32 1+2 1x x2 1+2 1)x x) −x x− (x x(3 3)3 3x x3 3)3 33 3(3 3x x3 3(3 3

ultiply both sides by 6 x x

se brackets to avoid problems with signs and help you to expand to get a linear equation.

x x 6

x

x

Exa

mp

le 3

a how that the equation 31

21 1

x x1x x1x x−x x− + = can be rewritten as x2 x 6 0.

b ence solve the equation 31

21 1

x x1x x1x x−x x− + = .

a dd the fractions 3 1 2 11 1 1( )3 1( )3 1 ( )2 1( )2 1

( )1 1( )1 1( )1 1( )1 1x x2 1x x2 1( )x x( )3 1( )3 1x x3 1( )3 1 ( )x x( )2 1( )2 1x x2 1( )2 1x x1 1x x1 1( )x x( )1 1( )1 1x x1 1( )1 1( )x x( )1 1( )1 1x x1 1( )1 1+ −( )+ −( )3 1( )3 1+ −3 1( )3 1x x+ −x x( )x x( )+ −( )x x( )3 1( )3 1x x3 1( )3 1+ −3 1( )3 1x x3 1( )3 1 2 1( )2 1−2 1( )2 1

1 1− +1 1( )− +( )1 1( )1 1− +1 1( )1 11 1( )1 1− +1 1( )1 11 1x x1 1− +1 1x x1 1( )x x( )− +( )x x( )1 1( )1 1x x1 1( )1 1− +1 1( )1 1x x1 1( )1 11 1( )1 1x x1 1( )1 1− +1 1( )1 1x x1 1( )1 1 =

ultiply both sides by the denominator x x x x

se brackets to help with expanding and to avoid problems with minus signs.

xpand x x x2

ote that the right- hand side is the difference of two squares.

earrange into the general quadratic form x2 x 6 0

b Factorise and solve x x 0

x or

Exa

mp

le 4

24 lgebra lgebraic fractions and functions6


implify this fraction. x x

xx x

x+x x+x x( )( )

+ −+x x+x x( )( )

+6 2x x6 2x x)6 2)x x)x x6 2x x)x x +6 2+(6 2(x x(x x6 2x x(x x

39 1x x9 1x x9 1x x9 1x x9 1x x9 1x x)9 1)x x)x x9 1x x)x x +9 1+(9 1(x x(x x9 1x x(x x

5

ommon denominator is x x x x x x x

x x+x x x+x x x( )( )x x x)x x x +(x x x(x x x ) ( )x x)x x +(x x(x x )( )

+x x+x x( )( )6 2x x x6 2x x x)6 2)x x x)x x x6 2x x x)x x x+6 2+x x x+x x x6 2x x x+x x x(6 2(x x x(x x x6 2x x x(x x x 5 9x x5 9x x− +5 9− +x x− +x x5 9x x− +x x5 9)5 9) − +5 9− +(5 9(− +(− +5 9− +(− + 1 3x1 3x1 3)1 3) +1 3+(1 3(


xpand the brackets in the numerator and simplify x x x x x x

x x

2 2x x2 2x x x x2 2x x8 1x x8 1x x2 28 12 22 5x x2 5x x2 22 52 2x x2 2x x2 5x x2 2x x2 52 22 52 2 10 9 3x x9 3x x3 5x x3 5x x

x x+ +x x2 2+ +2 2x x2 2x x+ +x x2 2x x8 1+ +8 1x x8 1x x+ +x x8 1x x2 28 12 2+ +2 28 12 2x x2 2x x8 1x x2 2x x+ +x x2 2x x8 1x x2 2x x( )2 2)2 2)2 5)2 52 22 52 2)2 22 52 2x x2 5x x+x x2 5x x2 22 52 2+2 22 52 2x x2 2x x2 5x x2 2x x+x x2 2x x2 5x x2 2x x(2 2(2 22 5(2 52 22 52 2(2 22 52 2)x x)x x2 2)2 2x x2 2x x)x x2 2x x− +x x− +x x2 2− +2 2x x2 2x x− +x x2 2x xx x− +x x2 2− +2 2x x2 2x x− +x x2 2x x +x x+x x(2 2(2 2(x x(x xx x− +x x(x x− +x x2 2− +2 2(2 2− +2 2x x2 2x x− +x x2 2x x(x x2 2x x− +x x2 2x x )9 3)9 3x x9 3x x)x x9 3x x9 3+9 3(9 3(9 3x x9 3x x(x x9 3x x )+x x+x x( )3 5)3 5x x3 5x x)x x3 5x x3 5+3 5(3 5(3 5x x3 5x x(x x3 5x x )

= ( )+ +( )+ +( )x x( )+ +( )+ +x x+ +( )+ +( )x x( )+ +( )+ +x x+ +( )+ + x x( )x x( )+( )+x x+( )+ − +x x− +( )3 2( )+ +( )+ +3 2+ +( )+ +( )x x( )3 2( )x x( )+ +( )+ +x x+ +( )+ +3 2+ +( )+ +x x+ +( )+ + 2 3( )2 3( )+ +( )+ +2 3+ +( )+ + +( )+2 3+( )+ − +2 3− +x x2 3x x( )x x( )2 3( )x x( )+( )+x x+( )+2 3+( )+x x+( )+ − +x x− +2 3− +x x− +( )8 1( )+ +( )+ +8 1+ +( )+ +( )x x( )8 1( )x x( )+ +( )+ +x x+ +( )+ +8 1+ +( )+ +x x+ +( )+ +( )3 2( )8 1( )3 2( )+ +( )+ +3 2+ +( )+ +8 1+ +( )+ +3 2+ +( )+ ++ +( )+ +x x+ +( )+ +3 2+ +( )+ +x x+ +( )+ +8 1+ +( )+ +x x+ +( )+ +3 2+ +( )+ +x x+ +( )+ +( )2 5( )+ +( )+ +2 5+ +( )+ +( )x x( )2 5( )x x( )+ +( )+ +x x+ +( )+ +2 5+ +( )+ +x x+ +( )+ +( )40( )( )2 3( )40( )2 3( )( )60( )( )x x( )60( )x x( )( )2 3( )60( )2 3( )( )x x( )2 3( )x x( )60( )x x( )2 3( )x x( ) ( )− +( )− + +( )+x x( )x x− +x x− +( )− +x x− + x x( )x x+ +x x+ +( )+ +x x+ + x x( )x x+x x+( )+x x+2 3( )2 3− +2 3− +( )− +2 3− +− +x x− +2 3− +x x− +( )− +x x− +2 3− +x x− + 2 2( )2 2+ +2 2+ +( )+ +2 2+ +x x2 2x x( )x x2 2x x+ +x x+ +2 2+ +x x+ +( )+ +x x+ +2 2+ +x x+ + x x2 2x x( )x x2 2x x10( )10 9 3( )9 3+ +9 3+ +( )+ +9 3+ +x x9 3x x( )x x9 3x x+ +x x+ +9 3+ +x x+ +( )+ +x x+ +9 3+ +x x+ +2 29 32 2( )2 29 32 2+ +2 2+ +9 3+ +2 2+ +( )+ +2 2+ +9 3+ +2 2+ ++ +x x+ +2 2+ +x x+ +9 3+ +x x+ +2 2+ +x x+ +( )+ +x x+ +2 2+ +x x+ +9 3+ +x x+ +2 2+ +x x+ + 30( )30x x30x x( )x x30x x +++( )+++( )+( )+ ( )

= + + + − − −

( )27( )3 5( )3 5( ) ( )3 5( )+( )+3 5+( )+

13 52 60+ −60+ −60 13− −13− − 393 2+ +3 2+ +133 213+ +13+ +3 2+ +13+ + 3 2133 213

( )x x( )+( )+x x+( )+ 3 5x x3 5( )3 5( )x x( )3 5( ) ( )3 5( )x x( )3 5( )x x+ +x x+ +13x x13+ +13+ +x x+ +13+ +3 2x x3 2+ +3 2+ +x x+ +3 2+ ++ +13+ +3 2+ +13+ +x x+ +13+ +3 2+ +13+ + x x+ −x x+ −x x+ −x x+ −60x x60+ −60+ −x x+ −60+ − x x− −x x− − 39x x393 2x x3 2 −−−

( )+( )+ ( )= +

( )+( )+ ( )

273 5( )3 5( ) ( )3 5( )+( )+3 5+( )+

13 333 5( )3 5( ) ( )3 5( )+( )+3 5+( )+

( )x x( )+( )+x x+( )+ 3 5x x3 5( )3 5( )x x( )3 5( ) ( )3 5( )x x( )3 5( )x

( )x x( )+( )+x x+( )+ 3 5x x3 5( )3 5( )x x( )3 5( ) ( )3 5( )x x( )3 5( )

Note: t is sometimes simpler to leave an algebraic fraction in a factorised form.

Exa

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implify this expression. xx x

xx x

+x x+ +x x

− −+x x+x x

35 4x x5 4x x+ +5 4+ +x x+ +x x5 4x x+ +x x

14x x4x x2 2x x2 2x x x x2 2x x+ +2 2+ +x x+ +x x2 2x x+ +x x5 42 25 4+ +5 4+ +2 2+ +5 4+ +x x+ +x x5 4x x+ +x x2 2x x+ +x x5 4x x+ +x x

Factorise the denominators xx x

xx x

+ − −34 1x x4 1x x+ +4 1+ +x x+ +x x4 1x x+ +x x

1( )x x( )x x+ +( )+ +x x+ +x x( )x x+ +x x4 1( )4 1x x4 1x x( )x x4 1x x+ +4 1+ +( )+ +4 1+ +x x+ +x x4 1x x+ +x x( )x x+ +x x4 1x x+ +x x( )x x( )x x4 1( )4 1x x4 1x x( )x x4 1x x+ +4 1+ +( )+ +4 1+ +x x+ +x x4 1x x+ +x x( )x x+ +x x4 1x x+ +x x ( )x x( )x x +( )+ 4( )4

ommon denominator is x x x x x x x

x x x+(x x(x x ) ( )x x)x x +(x x(x x )

+(x x(x x )( )3 1x x3 1x x)3 1) − −3 1− −x x− −x x3 1x x− −x x− −3 1− −(3 1(− −(− −3 1− −(− − 1

1 4x1 4x)1 4) +1 4+(1 4(

xpand and simplify x x

x x

x xx x

2 2x x2 2x x

2 2x x2 2x x

3 1x x3 1x x2 23 12 2

1 4

3 1x x3 1x x x3 1x2 23 12 2x2 2x3 1x2 2x

x x+ −x x2 2+ −2 2x x2 2x x+ −x x2 2x x3 1+ −3 1x x3 1x x+ −x x3 1x x2 23 12 2+ −2 23 12 2x x2 2x x3 1x x2 2x x+ −x x2 2x x3 1x x2 2x x ( )2 2( )2 23 1( )3 1x3 1x( )x3 1x2 23 12 2( )2 23 12 2x2 2x3 1x2 2x( )x2 2x3 1x2 2x3 1−3 1( )3 1−3 1

( )x x( )x x 1 4( )1 4+( )+ ( )x( )x1 4( )1 4x1 4x( )x1 4x1 4+1 4( )1 4+1 4

= x x+ −x x2 2+ −2 2x x2 2x x+ −x x2 2x x3 1+ −3 1x x3 1x x+ −x x3 1x x2 23 12 2+ −2 23 12 2x x2 2x x3 1x x2 2x x+ −x x2 2x x3 1x x2 2x x3 1+3 1( )x x( )x x 1( )1+( )+ ( )x( )x +++( )+++

= ( )+( )+ ( )

( )4( )3 1+3 1+

1 4( )1 4( ) ( )1 4( )+( )+1 4+( )+3 1x3 1

x x( )x x( ) ( )x( )( )1 4( )x( )1 4( )

Exa

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implify this expression. 2 34 9

22 322 324 924 9

2 3x x2 32 322 3x x2 322 34 9x4 9

2 3+ −2 32 3x x2 3+ −2 3x x2 34 9−4 9

Factorise the numerator and denominator 2 3 1

2 3 2 3x x2 3x x2 3

x x2 3x x2 3 2 3x x2 32 3+2 32 3x x2 3+2 3x x2 3( )x x)x x −(x x(x x )

2 3+2 32 3x x2 3+2 3x x2 3( )x x)x x2 3−2 3(x x(x x ) Denominator is the difference of two squares.

ancel common factors 2 32 3 1

2 32 3 2 3

x x2 3x x2 32 3x x2 3

x x2 3x x2 32 3x x2 3 2 3x x2 3

2 3+2 32 3+2 32 3x x2 3+2 3x x2 32 3x x2 3+2 3x x2 3(( )))x x)x x −(x x(x x )2 3+2 32 3+2 32 3x x2 3+2 3x x2 32 3x x2 3+2 3x x2 3(( )))x x)x x2 3−2 3(x x(x x )

f at this stage there isn’t a common factor on the top and bottom, you should check your factorisations.

The remaining fraction is the answer x −( )

( )1

2 3x2 3x −2 3−

Exa

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Exercise 24A 1 implify each of these.

a x x2 3+ b 3

425

x x+ c xyx42+

d x x+ + +12

23 e x x

52 1

3+ + f x x− + −44

2 32

2 implify each of these.

a 34 5x x− b x y

2 3− c xyy42−

d 2 12

3 14

x x+ − + e x x− − −22

34

f x x− − −44

2 32

3 olve the following equations.

a x x+ + + =12

25 3 b 4 1

32

4 2x x+ − + =

c 2 12

17 1x x+ − + = d 3 1

55 1

7 0x x+ − − =


a x x2 3× b 4

32x

yyx

× c x x2

25× −

d x x5

2 13× + e x

x x− × −

510

552


a 27

414

x y÷ b 49

23

2

2

yx

yx

÷ c xx

− ÷ −3

155

2 6

d 2 12

4 24

x x+ ÷ + e x x x6

23

2÷ + f x

x− ÷ −

212

43

g x x x− ÷ −510

55

2

implify this expression. xx

x xx

+−

× x x− −x x+

53

7 1x x7 1x x− −7 1− −x x− −x x7 1x x− −x x 83

2x x2x x

Factorise the quadratic expression xx

x x

x+−

×x x−x x( )( )

+53


3

ultiply x x x

x x

+x x+x x( )( ) +( )x x−x x( )( )5 9x x5 9x x −5 9−5 9x x5 9x x)5 9)x x)x x5 9x x)x x(5 9(x x(x x5 9x x(x x 2

3 33 3x x3 3x xx x3 3x x3 3x x3 3x x3 3x x3 3x x)3 3)x x)x x3 3x x)x x +3 3+(3 3(x x(x x3 3x x(x x

x x x

x x x

+x x+x x( )( ) +( )− +x x− +x x −

5 9x x5 9x x5 9x x5 9x x)5 9)x x)x x5 9x x)x x −5 9−(5 9(x x(x x5 9x x(x x 2

3 33 3x x3 3x xx x3 3x x− +3 3− +− +3 3− +x x− +x x3 3x x− +x xx x− +x x3 3x x− +x x 9

x x x

x+x x+x x( )( ) +( )

−5 9x x5 9x x5 9x x5 9x x)5 9)x x)x x5 9x x)x x −5 9−(5 9(x x(x x5 9x x(x x 2

9 x x

x2 x 0

Exa

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2

3

4

5



6 implify each of these. Factorise and cancel where appropriate.

a 34 4x x+ b 3

4 4x x− c 3

4 4x x×

d 34 4x x÷ e 3 1

22

5x x+ + − f 3 1

22

5x x+ − −

g 3 12

25

x x+ × − h xx

2 910

53

− × − i 2 35

6 910

x x+ ÷ +

j 29

23

2 2x y−

7 how that each algebraic fraction simplifies to the given expression.

a 21

52 3

x x+ + + = simplifies to x2 x 0

b 34 1

42 2

x x+ − + = simplifies to x2 x 0

c 22 1

61 11

x x− − + = simplifies to x2 x 0

8 a implify this expression. xx

xx

++

× −−

2 22

2 22

b se your answer to find the value of xx

xx

++

× −−

2 22

2 22

when x .

9 Write 23

34

45x x x+ + + − +

as a single fraction with an expanded denominator.

10 implify this expression. x xx

x xx

2 25 62

305

+ ++ ÷ − −

+

11 For homework a teacher asks her class to simplify the expression x xx x

2

226

− −+ − .

This is Tom’s answer x xx x

2 1

23

26

− −+ −

−

+

= − −+ = +

+x

xxx

13

13

= − −+ = +

+x

xxx

13

13

Tom has made several mistakes. What are they

12 n expression of the form ax bx cdx

2

2 9+ −

− simplifies to x

x−−

12 3

.

What was the original expression

13 olve the following equations.

a 41

52 2

x x+ + + = b 184 1

11 1

x x− − + = c 2 12

61 1x

x− − + =

d 32 1

43 1 1

x x− − − =

14 implify the following expressions.

a x xx x

2

22 3

2 7 3+ −+ +

b 4 12 5 3

2

2x

x x−

+ − c 6 29 4

2

2x x

x+ −

−

d 4 34 7 3

2

2x xx x

+ −− +

e 4 258 22 5

2

2x

x x−

− +

CM

EV

MR

6

7

9

10

11

12

13

14



15 a Prove that the equation 71

44 3

x x+ + + = simplifies to x2 x 0 0.

b ence solve the equation 71

44 3

x x+ + + = .

16 mma swam 00 metres at a speed of x m s, then a further 00 metres at x m s.

The total swim took mma 00 seconds.

a how that x2 x 0.

b Find the speed at which mma swam the first 00 metres.

Hints and tips emember how speed, distance and time are related.

17 a xpand and simplify x +( )23.

b ence show that 1 23

+( ) 2 .

c Find the value of 1 26

+( ) .

18 implify these expressions.

a x

x

x

x

++( )

− −−( )

43

432 2

b 41

32

23

14x x x x+ + + − + − +

19 implify this expression. x xx x

2

2

3 63 18

+ −− −

24.2 Changing the subject of a formulaThis section will show you how to:• change the subject of a formula where the subject occurs more than once.

When studying algebraic manipulation, you considered how to change the subject of a formula where the subject only appears once. To rearrange formulae where the subject appears more than once, the principle is the same as rearranging a formula where the subject only appears once or solving an equation where the unknown appears on both sides.

ollect all the subject terms on the same side and everything else on the other side. ost often, you then need to factorise the subject out of the resulting expression.

ake x the subject of this formula. ax b cx d

First, rearrange the formula to get all the x-terms on the left- hand side and all the other terms on the right- hand side The rule change sides change signs’ still applies.

ax cx d b

Factorise x out of the left- hand side x a c d b

Divide by the expression in brackets x d ba cd b−d ba c−a c

Exa

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CM 15

16

17

18

19

EV

CM

PS



ake p the subject of this formula. ap bcp d

++

First, multiply both sides by the denominator of the algebraic fraction cp d ap b

xpand the brackets cp d ap b

ow continue as in xample cp ap b d

p c a b d

p b dc a

b d−b dc a−c a

5b d5b d5

Exa

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0

Exercise 24B 1 ake c the subject of each formula.

a c p b c cp

2 ake the subject of each formula.

a F RG + 3 b F

G RG

++ 3

3 ake the letter in brackets the subject of the formula.

a p a b q a b a b p a b q a b b

c a ba c

+−

a d A 2πrh rk r

e v2 u2 av2 v f R xx

−−

32

x

4 a The perimeter of a shape is given by the formula P r kr. ake r the subject of this formula.

b The area of the same shape is given by A 12 [πr 2 r 2 k2 1−( ) . ake r the subject

of this formula.

5 When P is invested for Y years at a simple interest rate of R, the following formula gives the amount, A, at any time.

A P PRY100

ake P the subject of this formula.

6 When two resistors with values a and b are connected in parallel, the total resistance is given by

R aba b+

a ake b the subject of the formula.

b Write the formula when a is the subject.

7 a ake x the subject of this formula. y xx

+−

22

b how that the formula y 42x − can be rearranged to give x 4

1y − .

c ombine the right- hand sides of each formula in part b into single fractions and simplify as much as possible.

d What do you notice

EV

2

3

4

5

6

7

24.2 hanging the subject of a formula 11


8 The volume of the solid shown is given by this formulae. V 2

3πr3 r2h

a xplain why it is not possible to make r the subject of this formula.

b ake the subject.

c f h r, can the formula be rearranged to make r the subject f so, rearrange it to make r the subject.

9 ake x the subject of this formula. W 12 z x y 1

2 y x z

10 The following formulae in x can be rearranged to give the formulae in terms of y as shown.

y xx

++

12

gives x 1 21

−−

yy

y 2 12

xx

++ gives x

1 22

−−

yy

y 3 24 1xx

++ gives x 2

4 3−−

yy

y xx

++

53 2

gives x 5 23 1

−−

yy

Without rearranging the formula, write down y 5 12 3

xx

++ as x . . . and explain how you

can do this without any algebra.

11 lice and rian have been asked to make u the subject of the formula 1 1 1f u v

= + .

lice’s answer is u fv

v f− .

rian’s answer is u 11 1f v− .

a valuate whether either or both of these answers are correct.

b nto which answer is it easier to substitute values

24.3 FunctionsThis section will show you how to:• fi nd the output of a function

• fi nd the inverse function.

This is a function machine.

When you input a number into the function machine, it doubles the number and then adds to produce an output. For example, if the input is , then the output is .

f you are told the output, then you can also fi nd the input by applying the inverse operations in reverse order. For example, if the output is 0, then you subtract and divide by .

0

o the input was .

EV

h

r

EV

EV

Key wordinverse function

� 2I � 8 O

9

10

11



se this function machine to fi nd the output for each the input.

a 6 b

c Find both possible inputs if the output is 0.

SQUAREI � 5 O

a 62 6 6

b 2 6 6

c 0 25 or

Exa

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1

nother way of writing a function is to use the function notation f x . The value of x that you substitute into a function f x is the input and the value of f x is the output of the function. For example, for the function f x x 0, when the input is , the output f is 0 .

The function f x is defi ned as f x x.

Find a f b f . c olve f x 0.

a f 6

b f 0

c When f x 0, x 0, so x .

Exa

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2

Exercise 24C 1 a se this function machine to find the output for each input.

i ii 6 iii iv

� 9I � 2.5 O

b Find the input for each of these outputs.

i ii 0 iii 00 iv

2 a se the function machine to find the output when the input is .

� 3 O� 7I � 4

b Find an algebraic expression for the output when the input is x.

c se your answer for b to find the input when the output is 0.

3 For this function machine, find the input for each of these outputs.

a 0 b 60 c

� 10 O� 2I cube root

1

2

3

24.3 Functions 13


4 For which input do these function machines also have the same output

� 3I � 4 O

� 6I � 2 O

5 f x x2 Find the value of each of these for f x .

a f b f c f

d f 0 e f 0 f f 3

g iven that f k , find both values of k.

6 g x x2 Find the value of each of these for g x .

a g b g c g

d g e g 13 f g 0.

g olve g x 0.

7 a iven that f x x2 x, find the value of

i f 6 and ii f .

g x x2 x 6

b Find the inputs for which f x and g x have the same outputs.

Inverse functionsn inverse function is a function that performs the opposite process of the original function, such as

adding instead of subtracting or multiplying instead of dividing. f the original function turns an input of into an output of , then the inverse function turns the output of back into the input of , and it will do this for all inputs and outputs. The notation used for an inverse function is f– 1 x .

For example, the inverse function of f x x is f– 1 x x − 12

.

To fi nd the inverse function, write f x as y make x the subject of the function replace x with f– 1 x and then replace y with x.

onsider f x x .

Writing x as y gives y x .

aking x the subject of the function gives x y − 1

2 .

eplacing x with f– 1 x and then y with x gives f– 1 x x − 12 .

Find the inverse of the function f x x3 0.

Write f x as y y x3 0

ubtract 0 from both sides y 0 x3

ube root each side y − 103 x

everse the sides x y − 103

eplace x with f– 1 x and then y with x f– 1 x x − 103

Exa

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3MR

MR

5

6

7

4



Exercise 24D 1 Find an expression for f– 1 x for

a f x x b f x x3 c f x 101x +

d f x 0 x e f x x − 76

f f x 3 5x

+

2 a iven that f x xx

+−

23 5 , find an expression for f– 1 x .

b Find the value of f .

c ubstitute f into f– 1 x to verify that the answer is .

3 a Find the inverse functions of

i f x x and ii g x 12x

.

What do you notice

b Find the inverse function of f x 3 84 3xx

+− . What do you notice

c Prove that if f x ax bcx a

+−

, then f– 1 x ax bcx a

+− .

24.4 Composite functionsThis section will show you how to:• fi nd the composite of two functions.

composite function is a combination of two functions to create a third function. For two functions f x and g x , the function created by substituting g x into f x is called fg x . ou work this out by fi nding g x fi rst and then substituting your answer into f x .

2

3 MR

Key wordcomposite

The functions f x and g x are defi ned as f x x and g x 12 x . Find each of

the following.

a f b fg c ff

Exa

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4

a f 0

b g 12

f

c f

f

The functions f x and g x are defi ned as f x x and g x 12x . Find each of

the following.

a fg x b gf x c ff x

Exa

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5

a ubstitute g x into f xf x 1

2 x 1

2 x 1

2 x

b ubstitute f x into g xg x 1

2 x

12 x 1

2

12x 1

2

c ubstitute f x into f x f x x x x

24.4 omposite functions 15


Exercise 24E 1 iven that f x 0 x and g x x − 3

2 , find the value of each of the following.

a fg b gf c ff d gg 0 e fgfgfg .

Hints and tips For fg a , calculate g a and substitute this into f x .

2 iven that f x 241x − and g x x , find the value of each of the following.

a fg . b gf . c fgf 0 d gfg e ff f gg

3 f x x3 6 g x x h x x

a Find simplified expressions for each of the following.

i gf x ii hg x iii fh x iv gg x v ff x

b Prove that gh x can never equal hg x for any value of x.

Hints and tips For gf x , substitute f x into g x .

4 f x x2 a g x x b

f fg x gf x , find x in terms of b.

24.5 IterationThis section will show you how to:• fi nd an approximate solution for an equation using the

process of iteration.

any equations cannot be solved exactly using any of the techniques you have met already. ou could use trial and improvement to solve an equation like this but there is a process called iteration which is more effi cient and does not require you to make a new decision after each attempt. This involves solving the equation many times, using your result from the previous version each time to make the answer more accurate.

To perform iteration, fi rst rearrange the equation so that x is the subject, although there will be x terms on the other side the right hand- side as well. The x that is the subject is called x

n + 1 and any x term on the right hand side is called x

n.

For example, xn + 1 2 6x

n+ can be used to solve the quadratic equation x2 x 6 0.

ubstitute an initial value, called x1, into the right- hand side, and call the value obtained from this substitution x2.

2

3

4

CM

PS

Key worditeration

Find the fi rst four iterations of the iterative formula xn + 1 x

n with x1 .

x2 x1 x3 x2 x4 x3 x5 x4

0

0 Exa

mp

le 1

6



n approximate solution for the equation x3 6x 0 can be found using the iterative formula x

n + 1 16 93 xn

− and an initial value of x1 .

a Find the fi rst six iterations, correct to decimal places.

b erify that .6 is a solution of the equation, correct to decimal places.

a nter the initial value on the calculator 4

nter the iterative formula

This substitutes x1 into the formula x2 . 0

Press to substitute the value of x2 into the formula x3 .

Pressing four more times gives x4 . 00 x5 .6 0 x6 .6 6 x7 .6 0

b oth x6 and x7 round to .6 , correct to decimal places.

Exa

mp

le 1

7

These steps can be used to fi nd an approximate value for x3 6x .

Step 1: tart with x .

Step 2: Find the value of 6 83 6 8x6 86 8+6 8 , correct to decimal places.

Step 3: ompare your answer with the value of x you substituted. f it is the same, you have found the answer. f it is not the same, go back to step .

Find the solution of x3 6x given by this process.

nter the initial value on the calculator

nter the iterative formula

First iteration 6 3 83 × +6 3× +6 3 . 6 dp ot the same as so return to step

econd iteration . dp ot the same as . 6 so return to step

Third iteration . 0 dp ot the same as . so return to step

Fourth iteration . dp ot the same as . 0 so return to step

Fifth iteration . dp ot the same as . so return to step

ixth iteration . dp The same as . from the fi fth iteration so this is the answer.

Exa

mp

le 1

8sing a calculator makes iteration much easier.

To solve the quadratic equation x2 x 6 0 using xn + 1 2 6x

n+

et x1 .

n your calculator, type 4 . This records the number as the fi rst answer’.

ext type

ote your calculator may work in a slightly different way, and require a different key press order. heck how this works on your calculator before starting any iteration questions.

x2 2 61x + 14 . dp

ou then substitute this value back into the right hand side to generate the term x3, and so on.

x3 2 62x + 2 3 7417 6× +. .6 0 dp

x4 2 63x + 2 3 6720 6× +. .6 dp

ou can just keep pressing on your calculator to generate further iterations.

24.5 teration 17


Exercise 24F 1 Find the first four iterations using the iterative formula x

n + 1 xn with x1 .

2 For the iterative formula xn + 1 x

n , find x5 for each initial value.

a x1 6 b x1 c x1 . d x1

3 These steps can be used to find an approximate value for x3 x .

Step 1: tart with x .

Step 2: Find the value of 19 343 x + , correct to decimal places.

Step 3: ompare your answer with the value of x you substituted. f it is the same, you have found the answer. f it is not the same, go back to step .

Find the solution to x3 x given by this process.

Hints and tips Type in the value of x1, then use the button to type in the iterative formula.

4 n approximate solution for the equation x3 x 0 0 can be found using the iterative formula x

n + 1 7 103 xn

+ and an initial value of x1 .

Find the first five iterations, correct to decimal places.

5 The iterative formula xn + 1 x

n23 5+ can be used to find an approximate solution for

the equation x3 x2 0.

a sing x1 , show that x5 . , correct to decimal places.

b erify that . is a solution, correct to decimal places.

6 se the iterative formula xn + 1 5 23 x

n+ to show that . is an approximate

solution for the equation x3 x 0.

CM

CM

For the iterative formula xn + 1 3

3 − xn

, fi nd the value of x200 when x1 .

x2 33 53 5−3 5

32

x3 33 3

2− −(− −(− − ) 392

23

x4 33 2

3− ( ) 373

97

x5 33 9

7− ( ) 3127

74

x6 33 7

4− ( ) 354

125

x7 33 12

5− ( ) 335

ince you started with , this sequence will now cycle round , 32 , 2

3 , 97 , 7

4 , 125 , returning to 12

5 for every multiple of 6 x6, x12, x18, etc. .

The largest multiple of 6 below 00 is , which is less than 00, so x200 will be the same as x2.

ence x200 32

.

Exa

mp

le 1

9

2

3

4

5

6



7 a olve, by factorisation, the equation x2 0x 0.

b se the iterative formula xn + 1 10 21x

n− with x1 to determine which of the

two answers is generated by the formula.

c nvestigate what happens with each of these initial values.

i x1 .00 ii x1 . iii x1 00 iv x1

d eneralise what happens for all values of x.

8 The iterative formula xn + 1 x

n23 5+ can be used to find an approximate solution for

the equation x3 x2 0.

a how that the equation x xx

+−

41

can be rearranged as x2 x 0 and hence

xn + 1

xx

n

n

+−

41

is an iterative formula for the equation x2 x 0.

b When x1 , find the values of x2, x3 and x4, writing the answers as fractions, where appropriate.

c When x1 . , find the values of x2, x3 and x4,writing the answers as fractions.

d When x1 . , find the values of x2, x3 and x4,writing the answers as fractions.

e When x1 5, find the values of x2, x3 and x4, writing the answers as surds.

f se the results from parts b to e to deduce a solution for the equation x2 x 0.

9 square has sides of x2 cm and x cm.

a how that x2 x 0.

b se the iterative formula xn + 1 x

n+ 7 and an initial input of x1 to find the area

of the square, correct to the nearest integer.

c ow reliable is your answer for the area of the square

10 a For the iterative formula xn + 1 x

n2 , find the value of x36 when x1 .

b For the iterative formula xn + 1 1

1 − xn

, find the value of x100 when x1 .

Hints and tips enerate x2, x3, x4 and so on, and look for a pattern.

11 The equation x3 x is to be solved by an iterative formula with x1 .

a nvestigate what happens when the iterative formula used is given by

xn + 1 9

42xn

−.

b nvestigate what happens when the iterative formula used is given by

xn + 1

xn

3 94−

c nvestigate what happens when the iterative formula used is given by

xn + 1 4 9+

xn

.

EV

CM

EV

PS

EV

7

8

9

10

11

24.5 teration 19


Worked exemplars 1 The equation x3 x 0 can be written as the iterative formula x

n + 1 19 93 xn

− .

a sing x1 , find the first two iterations, correct to decimal places.

b how that . 0 is a solution to the equation, correct to decimal places.

This is a question on communicating mathematics so you need to state your method clearly.

a First iteration

x 19 913 x − 19 4 93 × − 673

.06 dp

tart by finding the first iteration.

lthough the method shown is correct, it is more efficient to register as an answer in the calculator’s memory and then to type in the iterative formula using

ns for xn, especially if you need more

than two iterations.

econd iteration

x 19 923 x − 19 67 933 × −

.0 dp

Then move on to the second iteration.

ote that you should be using the exact answer from the first iteration rather than the rounded .06 . This would definitely happen if you were using the efficient calculator method. n this example, .06 also gives a second iteration of .0 , but this is not guaranteed.

b x4 .0 dp

x .0 dp

x6 .0 dp

x .0 dp

ince you are asked to show that . is a solution, you are being asked to

communicate information accurately.

ontinue to find iterations until two of them repeat.

ince x6 and x both equal .0 correct to decimal places, . 0 is a solution of the

equation correct to decimal places.

nclude a conclusion.

CM



2 iven that f x x xx x

2

23 10

2 9 10+ −− + , prove that f – 1 .

This is a problem solving question so you need to plan a strategy to solve it and, most importantly, communicate your method clearly. ou need to show each step clearly.

There are two different methods shown here.

Method 1

f x x xx x

2

23 10

2 9 10+ −− + ( )( )

( )( )x xx x+ −− −

5 22 5 2

f x xx+−

52 5

Factorise and cancel the numerator and denominator.

y xx+−

52 5

y x x

xy y x

xy x y

x y y

x 5 52 1yy

+−

f x 5 52 1xx

+−

Find the inverse function.

f 5 3 52 3 1

× +× − 20

5

ubstitute x .

Method 2

x xx x

2

23 10

2 9 10+ −− +

Put the function equal to .

x x 0 x x 0 ultiply by the denominator.

6x x 0 x x 0

x 0x 0 0

x 6x 0

implify.

x x 0

x or

Factorise.

ot x because f x is undefined for x in its original form since both the numerator and denominator would equal ero, and division by ero is forbidden.

x , so f

xplain why x would not be allowed.

PS

2124 Worked exemplars


Ready to progress? can fi nd the output of a function given an input.

can rearrange more complicated formulae where the subject may appear twice or as a power.

can fi nd an inverse function by rearranging. can fi nd a composite function by combining two functions together. can combine and simplify algebraic fractions. can use iteration to fi nd a solution to an equation to an appropriate degree of accuracy.

Review questions 1 f x 0 x2. Find the value of f .

2 a ake x the subject of the formula 6x K a Cx.

b ence find the value of x when a , K and C .

3 a Write f x xx x x− −

−( )39

3 as a single fraction in its simplest form.

b ence find the inverse function f – 1 x .

4 implify fully 21 79 1

2

2x xx

−− .

5 The iterative formula xn + 1 6 1335 x

n+ can be used to solve the equation x5 6x3 .

a tarting with x1 . , find the first four iterations, all correct to decimal places.

b Find x5 correct to decimal places and compare it with x4.

6 f x x g x x3

a Find a simplified expression for fg x .

b sing the expression from part a, verify that fg .

7 Find the inverse of each function.

a f x px q b f x a x3 c f x a

x c+ 8 f x 2

2+( )x .

a Find each of the following.

i f 0 ii ff 0 iii fff 0 iv ffff 0 v fffff 0

b Find the nth term of the sequence given by the answers to part a.

9 how, by iteration, that a solution of the equation x3 x is given by . , correct to decimal places.

10 a implify f x 2 3 145 6

2

2x xx x

+ −− + .

g x 12 2− xx

b olve gf x .

2

3

4

5

6

7

8

9

10

EV

CM

MR

CM



11 lex was working out f – 1 x for the function f x 42 3

−−

xx

.

Find the mistakes in lex’s solution and write the correct solution.

y 42 3

−−

xx

y x x

xy x

xy x

x xy

x y

x 1 32

− y

ence f – 1 x 1 32

− x

12 xn + 1 2

2 − xn

f x1 , find each of the following.

i x219 ii x238 iii x257 iv x276

13 implify fully x x x xx x

+( ) +( ) −( ) −+ +

2 4 67 19 12

3

2

14 f x x2 g x x

a olve f x g x , giving both answers correct to significant figures.

b olve fg x � 0.

c olve gf x 0.

15 right- angled triangle has a base of x cm and a perpendicular height of x cm. The area of the triangle is cm2.

a how that x2 6x 60 0.

The equation x2 6x 60 0 can be rewritten as the iterative formula x

n + 1 60 6− xn

.

b Find the value of x, correct to significant figures.

16 f x x2 x g x x

a Find fg .

b how that if fg x g f x , then the answer can be written in the form a b 3.

17 argaret has n beads in a bag, of which are green. he removes two beads at random from the bag at the same time.

The probability that neither bead is green is 722 .

a how that n2 n 0.

b ow many beads were in the bag originally

EV

PS

PS

PS

PS

CM

PS

2324 eview questions


Rob Ellis

Maths 4th Edition

GCSEHigher Practice BookUse and apply standard techniques

01_GCSE Maths Higher Practice Book.indd 5 10/17/14 6:59 PM208105 GCSE Higher Practice Book CH24.indd 1 10/17/14 8:27 PM

Rob Ellis

Maths4th Edition








Cover design by We are LauraCover images © Procy/Shutterstock, joingate/ShutterstockIllustrations by Ann PaganuzziPrinted by Fuller Davies www.fullerdavies.com


208105 GCSE Higher Practice Book CH24.indd 2 10/17/14 8:27 PM

Rob Ellis

Maths4th Edition



24 Algebraic fractions and functions

24.1 Algebraic fractions 1 Simplify each of these.

a 23

45

x x+ b x x+ + +13

32

c 2 32

5 13

x x– –+

2 Simplify each of these.

a 34

25

x x– b x x+ +22

15– c 4 1

22 4

3x x– – –

3 Solve the following equations.

a 23

45 11x x+ = b x x+ + + =1

33

2 10 c 2 52

13 1x x– – – =


a 32

45

x x× b xx

+ × +1

43

2 2 c 2 1

24

3 1x

x–

–×


a x x4

25÷ b x x+ ÷ +3

22 6

5 c 4 2

32 1

4x x– –÷

6 Show that 32

52 1 2

x x+ + =– simplifies to 4x2 − 5x − 11 = 0.

7 Solve the following equations.

a 31

22 3 5

x x– + + = b 53 2

32 3 4

x x+ =– – c 53

22 6 4

x x+ + + =

8 Simplify the expression x xx x

2

22 3

2 10 12– –

– + .

24.2 Changing the subject of a formula 1 Make the stated term the subject of each formula.

a 4(x − 2y) = 3(2x − y) (x) b p(a − b) = q(a + b) (a) c A = 2ab2 + ac (a)

d s(t + 1) = 2r + 3 (r) e st − r = 2r − 3t (t)

1

2

3

4

5

6

7

8

1

24 Algebraic fractions and functions4

The questions below are differentiated with colours to show progression. Green is the most accessible, moving through blue to the pink questions, which are the most challenging.


2 Make x the subject of these equations.

a ax = b − cx b x(a − b) = x + b

c a − bx = dx − a d x(c − d) = c(d − x)

3 a The perimeter of the shape on the right is given by the formula P = 2πr + 4r.

Make r the subject of the formula.

b The area of the same shape is given by A = πr2 + 4r2.

Make r the subject of this formula.

4 a Make x the subject of y = xx

++

23

.

b Make x the subject of y = 2 31

––

xx

.

5 Make b the subject of a = 2 32

+ ab –

.

6 The resistance when two resistors with values a and b are connected in parallel is given by:

R = aba b+

a Make b the subject of the formula.

b Write down the formula when a is the subject.

7 a Make x the subject of the formula y = xx

++

12

.

b Show that the formula y = 1 − 12x + can be rearranged to give x = −2 − 1

1y – .

8 a Rearrange the formula y = xx + 1 to make x the subject.

b Rearrange the formula z = xx

+ 11–

to make x the subject.

c Equate the answers to part a and b, cross multiply and expand each side.

Make y the subject of the resultant formula.

2 4.3 Functions 1 Find the input for this function machine when the output is:

a 12 b 16 c 8

� 2 3�� 4I O

2 For which input do these function machines also have the same output?

a � 2 � 5I O b � 3 � 4I O

2

3

4

5

6

7

8

r

2r

24.3 Functions 5


3 f(x) = 3x2 − 2a Find the value of:

i f(2) ii f(5) iii f(−1) iv f(−4) v f( 3 )

b Given that f(k) = 25, find both values of k.

4 g(x) = 8 − x2

a Find the value of:

i g(2) ii g(−3) iii g(6)

iv g(−4) v g( 7 ) vi g(−0.5)

b Solve g(x) = −1.

5 f(x) = 2x2 − 6x + 4

a Find the value of:

i f(−1) ii f(5) iii f(−2)

b Solve f(x) = 0.

6 Find an expression for f−1(x) for:

a f(x) = 10x − 1 b f(x) = 13

x + 4 c f(x) = −5x − 10

d f(x) = (x − 3)2 e f(x) = x − 4

7 Given that f(x) = xx+−

32 1

, find an expression for f−1(x).

8 Given that f(x) = 3 52 3xx

+− , find an expression for f−1(x).

9 Given that f(x) = 4 34

xx

+− , find an expression for f−1(x).

10 What do you notice about your answers to questions 8 and 9? Make up a question of this type and find the inverse.

24.4 Composite functions 1 Evaluate each composite value

a f(x) = 3x − 5 g(x) = x2 Find fg(3).

b f(x) = −9x − 9 g(x) = ( )x − 9 Find fg(10).

c f(x) = −4x + 2 g(x) = ( )x − 8 Find fg(12).

d f(x) = −3x + 4 g(x) = x2 Find gf(−2).

e f(x) = −2x + 1 g(x) = x2 5− Find gf(2).

2 Find each composite function.

a f(x)= −9x + 3 g(x) = x4 Find fg(x).

b f(x)= 2x − 5 g(x) = x + 2 Find fg(x).

c f(x)= x2 + 7 g(x) = x − 3 Find fg(x).

d f(x)= 4x + 3 g(x) = x2 Find gf(x).

e f(x)= x − 1 g(x) = x2 + 2x − 8 Find gf(x).

3

4

5

6

7

8

9

10

1

2



3 For homework a teacher asks his class to evaluate the composite function f(x) = x2 − 3 g(x) = 5x.

Find fg(−3)

This is Wayne’s answer:

f(−3) = (−3)2 − 3 g(x) = 5x fg(−3) = 6(−15)

f(−3) = 9 − 3 g(−3) = 5(−3) = −90

f(−3) = 6 g(−3) = −15

What has he done wrong? Solve the problem correctly.

24.5 Iteration 1 Find the first five iterations of the following iterative formulae:

a xx

nn

+ =+

1

26

b xx

nn

+ = +1 5 4 c x

xnn

+ = −1

25

Start each one with x1 3= .

2 Find a root of the quadratic equation 2 3 4 02x x+ − = using the iterative formula:

xx

nn

+ =−

1

4 32

.

Start with x1 2= giving your answer to 2 decimal places.

3 Show that x x2 1 0+ − = can be rearranged into the iterative formula x xn n+ = −1 1

Use the iterative formula together with a starting value of x1 0 5= . to obtain a root of the equation correct to 2 decimal places.

4 Show that x x2 9 2 0− + = can be re–arranged into the iterative formula

x xn n+ = −1 9 2 .

Use the iterative formula together with a starting value of xn

= 8 to obtain a root of the equation correct to 2 decimal places.

5 A rectangle has sides of x – 3( ) cm and x +( )4 cm and an area of 26 2cm .

a Show that x x2 38 0+ =– .

b Use the iteration formula x xn n+ = −1 38 and an initial input of x1 3= to find the

length of each side of the rectangle, correct to 2 decimal places.

6 Show that xx

= − 5 3 can be rearranged into the equation x x2 3 5 0+ =– .

Use the iterative formula xxn

n+ = −1

5 3 to find a root of the equation giving your

answer to 2 decimal places.

7 Solve the equation x x3 2 3 5– + = using an iterative formula.

3

1

2

3

4

5

6

7 7

24.5 Iteration 7


Sandra Wharton

Maths 4th Edition

GCSEHigher Skills BookReason, interpret and communicate mathematically and solve problems

03_GCSE Maths Higher Skills Book.indd 5 15/10/14 11:42 AM208105 HB_Higher_SW_P001_003.indd 1 10/20/14 9:30 PM







Cover design by We are LauraCover images © Procy/Shutterstock, joingate/ShutterstockIllustrations by Ann PaganuzziPrinted by Fuller Davies www.fullerdavies.com


208105 HB_Higher_SW_P001_003.indd 2 10/20/14 9:30 PM

Sandra Wharton

Maths 4th Edition

GCSEHigher Skills BookReason, interpret and communicate mathematically and solve problems


How to use this bookFocused on the new assessment objectives AO2 and AO3, this book is full of expertly written practice questions to help students succeed in mathematical reasoning and problem-solving.

This book is ideal to be used alongside the Practice Book or Student Book. It is structured by strand to encourage students to tackle questions without already knowing the mathematical context in which they sit.

The longer questions can be tried in class to generate discussions, and new questions require students to think and analyse their work, encouraging independence. Students should take their time with the longer and multi-step questions.

Some questions have a hint at the back of the book to get you started, but students should be encouraged to try to answer the question �rst, before looking at the hint.

Full mark schemes will be provided online – a brief sample is included in this Evaluation Pack.

The questions in this book are differentiated with colours to show progression. Green is the most accessible, moving through blue to the pink questions, which are the most challenging.

4 How to use this book


1 Algebraic fractions and functions 1 Explain the difference between each of these expressions.

a 2n and n + 2 b 3(c + 5) and 3c + 5 c n² and 2n d 2n² and (2n)²

2 Explain how you know if a letter symbol represents an unknown or a variable.

3 Look at each of the examples below. Identify the errors in each one and explain how they should be corrected. Provide some written feedback to the person who did them that does not give them the answer but would help them to see their mistake.

a 5(c + 4) = 5c + 4 b 6(t – 2) = 6t – 4 c –3(4 – s) = –12 – 2s

d 15 – (n – 4) = 11 – n

4 Look at the list of formulae below.

i V = lwh ii s ut at= + 12

2 iii x = 3y – 2 iv u vt

t+( )v V 2 = u2 + 2ad vi A = 2πr2 + 2πrh

a Which of these examples might be easy/difficult to substitute in to? Explain what makes them easy/difficult and identify what the classic mistakes might be.

b Now imagine you are going to change the subject of the formula in each case. Again explain which of the examples might be easy/difficult to rearrange. Explain what makes them easy/difficult and identify what the classic mistakes might be.

5 Look at the list of formulae below.

i y = 2x + 3 ii z = 2(x + 3) iii t = −2(3 – x) iv − +( )2 2zx

If you are substituting a negative value for the variable in each case, explain which of these might be difficult and why. What typical mistakes might people make? Do you have any suggestions to help them avoid making these mistakes?

6 Explain the similarities and differences between rearranging a formula and solving an equation.

7 When you substitute s = 6 and t = 2 into the formula: zs t

=+( )3 2

12 you get 2.5.

Can you make up two more formulae that also give z = 2.5 when s = 6 and t = 2 are substituted? Explain what you are doing and try to use two different methods as well.

An identity can be described as two expressions that mean exactly the same thing. Alternatively they could be described as an equation that is TRUE no matter the value of the variable. For example:

a a4 4= ÷ or 2n(n + 4) = 2n2 + 8n or 2a × 2b = 2 a + bExa

mp

le 1

8 Could this expression be written in a different way, but still be the same? 2 62

n +

9 The height of an isosceles triangle is three times its base. The area is 6 cm2. What is its height?

7

1

2

3

4

5

6

1 Algebraic fractions and functions 5


10 a Write an expression that is equivalent to z q q2 1 4− +( ) −

b Write an expression that would simplify to:

i 3 25

x y+ ii 5 45 2x x +( )

11 a When working with a quadratic expression explain what it means to factorise.

b Explain what information you need to complete this.

c Explain what it tells you about the factors if the constant term of the quadratic is negative.

d Explain what difference it makes if the constant term is zero.

12 a Write an expression which can be written as the difference of two squares. Explain how you know it can be written as the difference of two squares.

b Why must 1000 × 998 give the same result as 9992 – 1?

13 Give two examples of algebraic fractions that can be cancelled and two that cannot be cancelled. Explain how you decided on your examples.

14 An expression of the form ax bcx dx e

2

2 2−

+ − simplifies to 3 42

xx

++ .

What was the original expression?

15 For homework, a teacher asks her class to simplify the expression xx x

2

29

2 3−

+ −This is Phillips’ answer:

xx x

2

29

2 3−

+ −

/ − // +( ) − /

xx x

2 3

1

92 3

= −+ − = −

−x

xxx

32 1

31

= −+ − = −

−x

xxx

32 1

31

Phillip checked the answer and it was correct. However, when the teacher marked the homework, she found that Phillip had in fact made several mistakes.

Explain the mistakes that Phillip made.

16 Explain how the product of two linear expressions of the form (2a ± b) are different from the product of two linear expressions of the form (a ± b)?

17 Six friends agreed to buy each other a chocolate Easter egg.

Four of the friends are girls and two of them are boys.

Each girl gives each boy a red egg.

Each boy gives each girl a blue egg.

Each girl gives each of the other girls a yellow egg.

And each boy gives each of the other boys a green egg.

How many eggs of each colour do the friends buy between them? Make sure you show your working.

10

11

12

13

14

16

15



18 a Using the sine rule and the general formula for the area of a triangle show that the area of any triangle is equal to 1

2absinC.

b Show that if x = 4 the area of this triangle is equal to 3 2.

x + 2

x – 2

45°

19 a Explain why:

i ‘I think of a number and double it’ is different from

ii ‘I think of a number and double it. The answer is 12.’

b 6 = 2p – 8

i How many solutions does this equation have?

ii Write another equation with the same solution.

iii Why do they have the same solution? How do you know?

iv Can you write a rule for generating an infinite number of solutions?

20 a Imagine you are teaching a short session on expressions, equations, formulae and identities. Explain carefully the similarities and differences between them.

b Produce a short activity that will help your class to become confident and fluent when using expressions, equations, formulae and identities.

21 The functions f and g are such that

f(x) = 3 – 7x g(x) = 7x + 3

Prove that f−1(x) + g−1(x) = 0 for all values of x.

22 Without drawing the graphs, compare and contrast features of the following pairs of graphs.

i y = 2x ii y = x + 5 iii y = 4x – 5 iv y = 2x y = 2x + 6 y = x – 6 y = −4x + 6 y x= 1

2Make sure your explanations use as much mathematical vocabulary as possible.

23 For real-life problems that generate linear functions:

a Explain what the gradient means in terms of the original problem.

b Do the intermediate points have any practical meaning?

c What does the intercept mean in terms of the original problem?

19

20

22

23

21



24

Do you know the story of the race between the hare and the tortoise? This graph shows information about the race between these two creatures. The hare is red and the tortoise is grey.

a Who was ahead after five minutes?

b What happened at nine minutes?

c After how long did the tortoise draw level with the hare?

d Who won the race and by how much?

e If they both continued at the same speed they were travelling at the end of the race how much further would the race need to be run over for the hare to retake the lead?

f Can you write your own question similar to this one that uses the graph of a linear function?

25 a What is special about the two linear expressions that, when expanded, have:

i a positive x coefficient?

ii a negative x coefficient?

iii no x coefficient?

b Give an example of an expression in the form (x + a)(x + b) which when expanded has:

i the x coefficient equal to the constant term

ii the x coefficient greater than the constant term.

iii What does the sign of the constant term tell you about the original expression?

0Time (minutes)

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

2

0

4

6

8

10

12

14

16

Dist

ance

(km

)

The tortoise and the hare

25

24



This example will remind you about completing the square. You will then try a couple of questions on completing the square before using this technique to deduce the turning points for quadratic equations.

What number should you add to x2 + 3x to complete the square?

The general formula for completing the square is x bx b22

2+ + ( ) .

So in this case you will need to add 32

94

2( ) = .Exa

mp

le 2

26 a What number should be added to x2 + 5x to complete the square?

b Solve the quadratic equation 2x2 + 10x – 5 = 0 by completing the square.

c Adam tried to solve a quadratic by completing the square but he made a number of mistakes. Look at his working below. What mistakes did he make?

x2 + 5x – 5 = 0

x +( ) = ( )52

52

2 2

x +( ) =52

252

2

x = +52 5 2

x + =52

252

d Copy his working and write some comments which show Adam where he has gone wrong. Write them in a way that does not correct his answer for him but will help make sure he does not make the same mistake again. Your answer should include guidance on why he should have known he had made a mistake this time as well as how to make sure he doesn’t make the same mistake again.

27 For each part of this question think of a quadratic equation of the type y = (x + a)2 + b. Give an example of a quadratic that fits the description. Remember to justify your example in each case, which means giving an explanation as well as an example.

a The turning point has a positive x-value.

b The turning point has a positive y-value.

c The y-intercept is positive.

28 a Sketch the graph of f( )x x x= + +2 94

b Hence determine whether f(x +3) – 2 = 0 has any real roots. Give a clear justification for your answer.

26

27

28



Hints and tips1, 2, 6 You may �nd a couple of examples will help you explain this.

9 You may �nd it useful to draw a diagram and make sure you read the question carefully.

11 Explain how you work out the two linear factors.

18 First use what you did in part 1 to show that the area of this triangle can be written

as 2

4 42x −( ). Also sin 45 is equal to 22

20 You may want to use some examples to support your explanation.

23 It might help your explanation if you use some real life examples.

25 In each case consider what it means when you expand two linear functions of the type (ax + b)(bx + c).

10 Hints and tips


SolutionsGuidance on the use of codes for this mark scheme

M1 Method mark E1 Explanation or justi�cation mark

A1 Accuracy mark oe Or equivalent

B1 Working mark ft Follow through

C1 Communication mark

Question Working Answer Mark Notes

1(a)

(b)

(c)

(d)

Example

2 × 3 = 6

2 + 3 = 5

3(c + 5) = 3c + 15

Example

32 = 3 × 3 = 9

2 × 3 = 6

Example

2n2 = 2 × (32 ) = 2 × 9 = 18

(2 × 3)2 = 6 × 6 = 36

C1

C1

M1

C1

C1

C1

C1 for explanation that 2n means 2 × n which is different from n + 2. An example could be given to support the argument.

An additional mark can be given for identifying the exception which is when n = 2.

M1 for multiplying out the brackets to show that the two expressions are not equivalent.

C1 for explanation that n2 means n × n which is different from 2n. An example could be given to support the argument.

An additional mark can be given for identifying the exception which is when n = 2.

C1 for an explanation that using BIDMAS for 2n2 tells you to calculate the power �rst. BIDMAS tells you that (2n)2 tells you that you do the calculation inside the bracket �rst. An example could be given to support the argument.

6

2 C1

C1

C1

C1

An explanation that a variable is a category of number that has a range of values.

With an example such as

F C= +95

32

An unknown is a discrete quantity that you can solve for.

With an example such as 3x + 2 = 8

4

Solutions 11



3 (a)

(b)

(c)

(d)

5(c + 4) = 5c + 20

6(t – 2) = 6t – 12

–3(4 – s) = –12 + 3s

15 – (n – 4) = 15 – n + 4 = 15 + 4 – n = 19 – n

M1

C1

M1

C1

M1

C1

M1

C1

C1

M1 for correctly multiplying out the brackets.

Comment such as “Don’t forget to multiply out both terms in the brackets.


Comment such as “Don’t forget 6(……..) means MULTIPLY both terms by 6.


Comment such as “Don’t forget –3(……..) means MULTIPLY both terms by 6 and a minus × minus = ……...


Comment such as “Don’t forget –1(n – 4) means ………

Extra communication mark for giving the advice in the form of a question.

9

4 (a)

(b)

M1

E1

M1

E1

E1

5

M1 for identifying at least 2 examples

E1 for justi�cations such as:

• Examples with brackets are more dif�cult with negative numbers

• Divisions and powers can be more dif�cult when substituting non integer values.

M1 for identifying at least 2 examples

E1 for justi�cations such as:

• Multi step problems are more dif�cult.

• Examples that include all the possible variations of BIDMAS are more dif�cult

Identi�cation of a classic mistake such as 22 as 2 × 2.

Incorrect use of BIDMAS.

Solutions12



5 M1

E1

C2

Identify that examples iii or iv might be more dif�cult because they have negative numbers outside the brackets.

The mistake would be especially common if there is a negative number inside the bracket as people would either forget or not understand that a negative × negative will give a positive.

C2 for a comment such as –3(n – ……) Will the second term be positive or negative when you multiply out the brackets? 1 mark if the answer is given in the comment.

5

6

C1

C1

Explanation such as:

SAME - When rearranging a formula or solving an equation you will be moving variables or numbers across the equals using inverse operations.

DIFFERENT when rearranging a formula you are not looking for an exact value for the variable. Whilst when solving an equation you are looking to identify the value of the unknown in the equation. 2

7 M1

C1

M1

M1 for correct example.

C1 for accurate explanation of how the examples were found.

Extra method mark for having a systematic approach to �nding examples. 3

8 n + 3 Yes M2 M2 for simplest form.

M1 for other equivalent example 12

2 6( )n +2

9

x

3x

12

3 42

( )x x x× =

42

6x =

4x = 12

x = 3 so height is 6 cm C1

M1

A1

3

C1 for drawing a suitable diagram

A1 cao

Solutions 13



10 (a)

b (i)

(ii)

M1

C1

M1

C1

M1

C1

M1







Extra method mark for having a systematic approach to �nding examples.

5

11 (a)

(b)

(c)

(d)

M2

E2

E2

E1

C1

M1 for suitable explanation, for example “it is �nding what we can multiply to get the quadratic.”

M2 for also identifying that the factors identify the roots or the x intercepts of the quadratic.

Identifying that you need to know the coef�cients of x2 and x and the constant term of the quadratic. E1 for identify 2 of these.

Identifying in words that it tells you that the brackets are of the form (x + a)(x – b)

Identifying in words that it tells you that the brackets are of the form x(x ± a)

Extra C1 for generalising using algebra to support explanation.

8

Solutions14



12 (a)

(b) 9992 – 1

Completing the square gives:

(999 + 1)(999 – 1)

This is the same as

1000 × 998

As required.

M1

E2

C1

M2

M1 for a correct example.

E1 for explaining that the sign of the constant term is negative and it is a square number. Need both for E2

x2 – b2 C1 for generalising using algebra to support explanation.

M1 with no explanation

6

13 M1

C2

M1 for correct examples.


Extra method mark for having a systematic approach to �nding examples.

3

14 Using the difference of two squares for the numerator gives (3x + 4)(3x – 4). Assuming the new term in the numerator is also in the denominator gives:

( )( )( )( )3 4 3 4

2 3 4x xx x

+ −+ −

If you then multiply out the numerator you will get:

9x2 – 12x + 12x – 16

= 9x2 – 16

Expand the brackets in the denominator to give:

3x2 – 4x + 6x – 8

= 3x2 + 2x – 8

The original expression:

9 163 2 8

2

2x

x x−

+ −

B2

C1

A1

M1

B2 for correct working B1 for ft

C1 for explanation of process

A1 cao

M1 extra method mark for using a value to check answer. For example substituting x = 1 into both the simpli�ed and the

original equation gives 73

5

Solutions 15



15 In line 2 Phillip has initially rearranged x2 + 2x – 3 to x(x + 2) – 3 and then he has incorrectly simpli�ed in line 3

/ − // + − /

xx x

2

129

( ) 33

M2

C1

M1 for each mistake

C1 for clear explanation

3

16 The coef�cient of a2 for the product of (2a ± b) is four times the coef�cient of a2

for the product of (a ± b).

The coef�cient of a for the product of (2a ± b) is twice the coef�cient of a for the product of (a ± b).

The constant term is the same for both.

M3

C1

M1 for each coef�cient and one for the constant.

C1 for effect use of algebraic notation to support explanation.

3

17 4 × 2 = 8 blue eggs

4 × 2 = 8 red eggs

4 × 3 = 12 yellow eggs

2 × 1 = 2 green eggs

B3

A1

B3 for correct working

B1 if don’t realise that when calculating the totals for girls to girls and boys to boys it is n × (n × 1)

A1 cao

4

18 (a)

Using trigonometric function

sin A = hc

sin C = ha

Therefore:h = c sin A A = a sin C

Then using the basic formula for area of a triangle

Area ∆ = 12 hb

Substituting for h gives:

Area ∆ = 12 a sin C × b

Area ∆ = 12 ab sin C as

required

M4

C1

M1 for drawing diagram

M3 for correct method

C1 extra mark for good explanation possibly using connectives effectively.

h

A

B

C

a

b

c

Solutions16



(b)

x + 2

x – 2

45°

Area = 12

ab sin45

= 12

ab 22

= 22 (x + 2)(x – 2)

= 22 (x2 – 4)

Therefore if x = 4

= 22 (42 – 4)

= 22 =× 12

=3 2 as required

M3

C1

M3 for correct method

C1 extra mark for good explanation possibly using connectives effectively.

5

19 (a)

(b) (i) One

(ii) Any suitable example

(iii) Something similar to – Both sides have been doubled.

(iv) 6n = n(2p – 8) or equivalent

E1

A1

A1

E1

A2

An explanation such as: Number (i) is an expression whilst number (ii) is an equation and therefore can be solved for n.

6

20 (a)

(b)

C2

C3

At least 3 differences and 2 similarities well explained.

Suitable activity deduct a marks for any inaccuracies.

5

Solutions 17



21f x x− = − −( )1 3

7( )

g x x− = −( )1 37

( )

Therefore:

f x−1( ) + g x−1( )

= − −( ) + −( )x x37

37

= 0 for all values of x as required.

M2

C1 C1 for clear layout and use of connectives.

3

22 (i)

(ii)

(iii)

(iv)

E1

E1

E2

E1

C1

Both have a gradient of 2 but the �rst crosses the y axis at 0 the other at 6

Both have a gradient of 1 but the �rst crosses the y axis at 5 the other at –6

The magnitude of the gradient is the same but the �rst is positive whilst the second is negative. The �rst crosses the y axis at –5 the other at 6

The �rst has a gradient of 2 the second has a gradient of 1

2 . They

both cross the y axis at the origin.

C1 for good use of mathematical vocabulary.

7

23 (a)

(b)

(c)

E1

E2

E1

C1

The gradient describes how much the dependent variable changes for a given change in the independent variable, oe

A straight line graph should describe a functional relationship that is de�ned for all values of x within the stated domain, oe

The intercept is the value of the dependent variable when the independent variable is equal to 0, oe

C1 extra mark for good use of mathematical vocabulary.

5

Solutions18



24 (a)

(b)

(c)

(d)

(e)

(f)

Hare

The Hare stops

12 minutes

The Tortoise won the race by 2 km

About 4 km

A1

A1

A1

A2

M1

A1

B2

Calculate approximate gradient for each line and use this to work out when the 2 lines are equivalent.

B1 for suitable question B2 for quality of question

9

25 (a)

(i) (ii)

(iii)

b (i)

(ii)

(iii)

(x + a)(x + b) As the coef�cient for x needs to be 1

(x + 1)2

ab < 1

If the constant term is positive the signs are the same. If negative the signs are different.

E1

E1

E2

C1

E1

E1

E1

The constant terms in the linear expressions are either both negative OR both positive, oe

One of the constant terms in the linear expressions is negative the other is positive, oe.

The constant terms in the linear expressions are either both 0. When multiplied out this will give ax2, oe.


5

26 (a) 52

2( ) A1

(b) To solve 2 10 5 02x x+ − =Divide through by 2

x x2 5 52

0+ − =Rearrange

x x2 5 52

+ +

x +( ) − ( ) =52

52

52

2 2

M2

B4

C1

Award working marks for clarity of presentation. Deduct marks if not aware of ± or if not simpli�ed.

C1 extra mark for good use of mathematical vocabulary and/or commentary

Solutions 19



(c)

(d)

Completing the square

x +( ) − =52

254

52

2

Rearrange

x +( ) = +52

52

254

2

x +( ) =52

354

2

Square root both sides

x + = ±52

354

x = ± −354

52

x = ± −( )12

35 5

A2

C2

Award marks for identifying error

Award marks for suitable formative comments

10

27 (a)

(b)

(c)

A1

E1

A1

E1

A1

E1

C1

Suitable quadratic of the form ( )x a b− ±2

Explanation to refer to a positive translation of x2 in the x axis

Suitable quadratic of the form ( )x a b± +2

Explanation to refer to a positive translation of x2 in the y axis

Any suitable example for which y > 1 when x = 0

E1 for explanation of why example chosen.


6

28 (a)

(b) One real root at

(–7/2,0)

y

x

654321

–1–2

–2–3 –1 0 1 2 3

A2

E2

A1

Explanation that f(x + 3) – 2 is a translation 3 left and 2 down of f(x). Therefore as the turning point moves 2 down it will turn on the x axis giving one real root. (–7/2,0), actual value not necessary as long as accurate explanation of why one repeated root.

5

Solutions20


Maths 4th Edition

GCSE

Christine Watson

Higher Teacher Pack

05_GCSE Maths Higher Teacher Pack.indd 5 15/10/14 11:35 AMAQA Higher Teacher Pack.indd 1 10/22/14 12:49 AM







Cover design by We are LauraCover images © Procy/Shutterstock, joingate/ShutterstockPrinted by Fuller Davies www.fullerdavies.com



AQA Higher Teacher Pack.indd 2 10/21/14 9:51 PM

Maths 4th Edition

GCSEHigher Teacher Pack

Christine Watson

AQA Higher Teacher Pack.indd 3 10/22/14 12:49 AM

GCSE Maths 4th edition 4 © HarperCollinsPublishers Ltd 2015 Higher Teacher Pack

Chapter 24 Algebraic fractions and functions Overview 24.1 Algebraic fractions 24.4 Composite functions

24.2 Changing the subject of a formula 24.5 Iteration 24.3 Functions Prior learning

Manipulate algebraic expressions and solve equations. Use rules to generate sequences. Use trial and improvement to solve equations.

Chapter 24

Ensure students can solve equations involving algebraic fractions by manipulation, find inverse and composite functions, and use iteration to find solutions. In the examination, students will be expected to A manipulate, simplify and rearrange algebraic expressions involving fractions B substitute into functions, find inverse and composite functions C rearrange polynomials to the iterative form; solve iterations to a given number of

iterations or number of decimal places.

Extension

Explore more complex algebraic fractions, for example partial fractions. Use a wider range of mathematics in functions including trigonometric functions. Explore iteration in coding, in particular the generation of fractals.

Curriculum references

Section KS4 NC Programmes of Study GCSE specification

24.1 A (ER) 10 A4, A6 24.2 A (ER) 3 A5, A6 24.3 A (ER) 6 A6, A7 24.4 A (EF) 7 A6, A7 24.5 A (EF) 2 A6, A20



Route mapping

Exercise Accessible Intermediate Challenging AO1 AO2 MR CM

AO3 PS EV

Key questions

24A 1–7 8–19 1–6, 8–10, 13–14, 18

7, 12, 15, 17

11, 16, 19

6, 8, 10, 12, 16

24B 1–6 7–11 1–6, 9 7–8, 10–11 4, 7, 9

24C 1–4 5–7 1–3, 5–7 4, 7 2, 7 24D 1–2 3 1–2 3 1e, 1f, 2

24E 1–2 3–4 1–2 3 4 2a, 2f, 3i, 3v

24F 1–7 8–11 1–5 5–6, 8 7, 9–11 6, 9 Key questions are those that demonstrate mastery of the concept or which require a step-up in understanding or application. These could be used to identify the questions that students must tackle, to support differentiation, or to identify the questions that should be teacher-marked rather than student-marked. About this chapter Making connections: The chapter brings together algebraic manipulation, solving equations, rearranging formulae. The work on iteration builds on understanding of sequences and graphical representation of sequences is a clear way of demonstrating convergence/divergence. For much of the chapter, the next step is A level content.

Relevance: There is an emphasis on the use of logic and thinking in steps. There are applications in engineering, architecture, manufacturing, project management and many more areas, with STEM careers being a strong focus. Working mathematically: Which is more complex, having a linear algebraic numerator or denominator? Why? What is the connection between fg(x) and gf(x)? When will they be equal? When will the inverse of a function be equal to the function? Find a chain of steps from an equation involving algebraic fractions to the iterative form. How efficient can you make this chain?

Assessment: Give students a set of fractions with linear denominators and simple numerators. Ask them to select two to create an equation of the form y = A + B, where A and B are their chosen fractions. They should then represent this in as many forms as they can: rearrange, simplify, graph, iteration etc.

See the CD for suggested assessment tracking foci, and the section plans for further suggested Assessment tasks.

Worked exemplars from Student Book – suggestions for use A Present students with the same question but different numbers. They use the exemplar

to mirror the working, in full or just the notes. B Copy and cut up the exemplar into cards. Students match the working with the notes.

(You may need to remove the words ‘first, second’ etc.) C Copy and cut up the working into cards but split the label/description from the working.

Students put the working in order then match with the descriptions. Answers to Student Book questions at the end of this book (NB: not included in this sample)



Section 24.1 Algebraic fractions Learning objectives Resources and homework • Apply methods for calculating with

numerical fractions to algebraic fractions • Student Book 24.1 • Practice Book 24.1

Making mathematical connections Making cross-curricular connections • Equivalent fractions • Simplifying linear and quadratic

expressions

• Science – use of formulae • Relevance – developing logical thinking

Prior learning It is essential that students can confidently calculate with numerical fractions; this may need refreshing before the lesson via a homework task. Students should be able to expand, factorise, simplify and solve linear and quadratic expressions. Working mathematically

• Encourage students to articulate their methods for numerical fractions (equivalence including cancelling; addition and subtraction; multiplication and division) and then to apply these to algebraic fractions.

• Structure tasks so students can work out the methods for themselves, either by increasing the difficulty incrementally or through one straightforward and one complex example. Common misconceptions and remediation

• Students can be confused about cancelling terms across two fractions. If this arises – or to force the issue into the open – try this example: 34 − 2

3 → (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 3 𝑐𝑐𝑐𝑐𝑎𝑎 3) 14 − 2

1 → (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑐𝑐𝑐𝑐𝑎𝑎 4) 12 − 1

1 = − 12 Is this a correct final answer?

If not, what has gone wrong? Link to the pictorial representation. Ensure the distinction is made between cancelling within a single fraction 3×2

4×3 or it’s equivalent 34 × 2

3 , and cancelling separate fractions. Link to an algebraic version such as 6𝑥𝑥

𝑥𝑥(𝑥𝑥+1) − 2(𝑥𝑥+1)6𝑥𝑥−4 , asking what can be cancelled and what

cannot. Probing questions

• What is the same and what is different about simplifying these two expressions? 4𝑥𝑥𝑥𝑥+3 + 𝑥𝑥+3

𝑥𝑥 𝑥𝑥𝑥𝑥+3 × 𝑥𝑥+3

4𝑥𝑥 • Which types of algebraic fractions lead to quadratic numerators and which to linear numerators? • How does simplifying algebraic fractions link to the work on irrational fractions?

Literacy focus

• Key vocabulary: cancel algebraic fraction LCM (lowest common multiple) • Be explicit about the language of equivalence and avoid ‘cross-multiplication’.



• Part 1 NB: This section may take more than 1 hour Review calculating with numerical fractions.

• Display fraction calculations on the board. Ask students to complete these then to describe their method(s) as efficiently as possible. For example: 1 − 2

6 316 + 11

12 41 − 1

6 + 112 + 1

3 41 × 1 3

161112

• Draw out the principles: Cancel first for efficiency. For add/subtract, find lowest common multiple (LCM) of denominators & use to find the equivalent fractions then add/subtract. For division, multiply first fraction by the reciprocal of the second fraction.

• It may be useful to show 1 − 26 = 1 − 1

3 = 213 − 1

3 = 21−13 and pause to check the

equivalence of the last two parts. Part 2 Apply the principles of numerical fraction calculations to algebraic fractions, ensuring that cancelling of algebraic terms is correct.

• Present the students with simple algebraic fractions such as 6𝑥𝑥4 + 𝑥𝑥 and ask them to apply

the same principle – can they work out what to do? • Extend the challenge to algebraic denominators such as 4

6𝑥𝑥 + 𝑥𝑥 – can the students extend their method?

• Ask students what difference it makes if this becomes an equation: 6𝑥𝑥4 − 𝑥𝑥

6 = . Draw out the equivalent method of multiplying every term by the lowest common multiple (LCM) then ask which is more efficient: finding the single algebraic fraction then solving, or multiplying all terms by the LCM. NB: there is no right answer; it depends on the starting point!

• Students can now begin Exercise 24A from Student Book. A I 1–7 C CM 7 MR PS EV Key 6 Part 3

• Apply the principles to algebraic fractions with more complex terms. NB: If this extends to a second lesson, clarifying cancelling of algebraic fractions is an appropriate starter for the second lesson.

• Return to 46𝑥𝑥 + 𝑥𝑥 and ask students to suggest more complex terms for the denominators –

but not too complex! Look for linear expressions as a starting point and use these to challenge students to extend their method again.

• Ensure that students are allowed to try methods in pairs, then use the example to build the general method: (A) simplify by cancelling if possible (B) find the LCM of the denominators (C) find the equivalent fractions, using brackets if needed (D) add/subtract (E) simplify.

• Emphasise the importance of the brackets to avoid errors. • If appropriate, pause during this phase to ensure understanding through a more complex

example such as 3𝑥𝑥𝑥𝑥 + 𝑥𝑥 − 2𝑥𝑥+2

𝑥𝑥 +𝑥𝑥 . • Students can now continue Exercise 24A from Student Book.

A I C 8–19 CM 15, 17 MR 12 PS 19 EV 11, 16 Key 8, 10, 12, 16

Assessment task • Use some of these terms to create two algebraic fractions – you can use each more than

once. It is a good idea to think ahead! • 2x (x + 2) x (2 – x) x² 2 (x – 2) 2x² 4 • Add, subtract, multiply and then divide your two fractions – good luck!



Section 24.2 Changing the subject of a formula Learning objectives Resources and homework • Rearrange formulae where the subject

appears more than once • Student Book 24.2 • Practice Book 24.2

Making mathematical connections Making cross-curricular connections • Rearranging simple formulae • Solving equations

• Science – use of formulae • Computing – application of logic • Relevance – programming languages;

business use of flow diagrams Prior learning Students should be able to solve equations and rearrange simple formulae using the concept of ‘balance’. An understanding of inverse operations is an advantage. Working mathematically

• Encourage students to be able to move forwards and backwards within a rearrangement and to be able to articulate how they are ‘balancing’ at each step.

• This activity lends itself to sorting the steps into order (see the Unjumble Assessment task). Ordering the steps and then writing the descriptions of the operations, then doing the same for the reverse order, provides a foundation for the work on inverse functions in section 24.3.

• Extend students by asking them to create formulae that include more complicated elements – powers, roots, trigonometry – and work out how to rearrange these. Common misconceptions and remediation

• Students do not link solving equations with rearranging formulae. Make the parallels explicit by asking students to work in pairs, one solving the equation and the other rearranging the formula, then to swap solutions and discuss. Probing questions

• How many steps will I need to rearrange this 4-term formula? Will it always be the same number of steps? Can you find an exception?

• How do you ‘undo’ (find the inverse of) an add 4? Identify as many types of operations as you can , then list how to ‘undo’ each one – be creative! Literacy focus

• Key vocabulary: formula rearrange balance equivalent inverse • Ask students to describe the steps in their rearrangement, but insist on the language of ‘… to

both sides’. For example: I subtracted 3x from both sides, then added 4 to both sides, then divided by 5 on both sides. This can be turned into a game similar to the children’s game ‘May I…’, where children paid a forfeit if they did not say ‘May I’. Students could have a set of formulae to rearrange then describe their steps in groups, getting points each time they do not say ‘to both sides’; the student with the least points wins.



x k x d

x k x d x x d k

x k x d x 12 k x d

Part 1 Review equivalent expressions/equations.

• Use the Clouding the Picture frame. • Ask students what is happening along each

of the four branches: make explicit the operation and how it is applied.

• Identify the operation that is a step towards x being the subject, and that which is a step towards k being the subject.

• Ask students to identify the most efficient step for each rearrangement. If appropriate, provide additional examples for the discussion.

• Ensure students recognise (A) operations happen to both sides equally (B) multiplication and division happen to every term on each side (C) some operations lead to a simpler form and some lead towards the desired rearrangement. Part 2 Apply the principles of rearranging simple formulae to those where the subject occurs more than once.

• Use the same Clouding the Picture frame but change the central formula to one with a repeated subject (x).

• Repeat the key question from part one: Which operation(s) are the most efficient ways to move towards a formula with only one subject (x)?

• Ask students to find a single step that will lead to each variable only occurring once.

• If appropriate, repeat for other formulae. • Challenge students to extend the principles to equations involving algebraic fractions such

as 𝑥𝑥+2𝑥𝑥− = – what would be the most efficient first move? And the next move?

• Students can now do Exercise 24B from Student Book. A I 1–6 C 7–11 CM MR PS EV 7–8, 10–11 Key 4, 7, 9 Part 3

• Assessing understanding at the end of the session. • Provide students with the Clouding the Picture frame, with the end four boxes completed.

They can be challenged to step back towards the centre to find a formula that can be rearranged into all four outer variations.

Assessment task

• Unjumble: Create a set of cards that show the steps in rearranging a formula. Students should be able to place these in the correct order. For a more challenging task, include incorrect steps.

• Challenge students to create their own formula and matching set of Unjumble cards, including distractors.

x k d

x k d x d k

x k d x x 12 k 12 d

x k d x k d k



Section 24.3 Functions Learning objectives Resources and homework • Know what a function is and use simple

function notation • Use functions including finding the

inverse function


Making mathematical connections Making cross-curricular connections • ‘Doing and undoing’ • Equations, expressions identities and

functions • Mappings, coordinates and graphing

• Science – use of formulae • Computing – logic gates; formulae in

spreadsheets; flow diagrams • Relevance – developing logical thinking

Prior learning Students should be familiar with the concept of ‘flow’ via flowcharts or flow diagrams, and of inverse operations. Previous experience of function machines and mapping diagrams is an advantage. Students should be confident with manipulating expressions. Working mathematically

• Research: find the definition of a function. Write an algebraic statement that is not a function then compare with your class. Can you classify these statements?

• Display a function and ask students to identify the steps in order (see Literacy focus – this could be verbal). Ask “What happens if the steps are not in the same order? Try the steps in different orders. Can you write the function?” Differentiate by providing steps on card to order, or by using matching cards: one set of cards with the steps in various orders, and one set with the functions on. This can include partially complete cards and distractors to extend other students. Common misconceptions and remediation

• Students do not use brackets when converting function machines into algebraic statements – remind them the order of operations applies to algebra as well as number.

• Students forget to reverse the ‘x’ and ‘f(x)’ (or y) when finding inverse functions. When marking/going through questions, award marks for method, for answer, and for remembering to reverse to highlight the issue.

• Students read f(x) as f multiplying an-x-in-brackets. Be consistent in the use of the correct language and insist on ‘f of x’ from students rather than ‘f x’. Probing questions

• What are the pitfalls when working with function notation? How can you avoid them? Literacy focus

• Key vocabulary: function machine function of x function inverse • Ask students to verbalise functions as if they are the variable. If appropriate model this first,

for example for the function f(x) = 6(𝑥𝑥− )11 Say: “I am x. Subtract 5, then square, then multiply

by 6 and divide by 11.” Repeat for other functions with the emphasis on verbal descriptions and the use of ‘then’, ‘and’ to discriminate between steps that must be in sequence and those that can be reversed.



Part 1

Introduce function machine cards to represent mathematical operations. • Display the start (3) and the finish (10) and a set of function machines. Ask students to find

different ways to get from 3 to 10 using these cards only: ×2 ×3 ÷2 square +1 +2 +3 +4 • Ask students for their routes from 3 to 10. List some on the board. Discuss the different ways

found: shortest v longest, most efficient, easiest, complex etc. Refer to order of operations and the need for brackets. Is there an operation that students wish had been in the list? Some routes: +4 +3 ×2 +4 ×3 +1 +1 ×2 +2 +4 +2 +1 +2 ×2 +1 ÷2 +3 ×2 square +1 +1 square ÷2 +2 Part 2 Introducing function notation and the concept of a function.

• Using the ways found in part 1, model conversion to algebraic function notation – replace the 3 with x, then the 10 with y. Then replace the y with f(x) and introduce the language ‘function of x’. For example:

Talk through the steps: x x x + x + 4 = y x + 4 = f(x x) = 2x + 4

x x + 1 (x + 1) ÷ 2 1

2(x + 1) 12(x + 1) + 3 1

2(x + 1) + 3] × x + 1) + 6 = f(x x) =x + 7

• This may be an appropriate point to do the activity in the Literacy focus. • Students can now do Exercise 24C from Student Book.

A 1–4 I 5–7 C CM MR 4, 7 SP EM Key 2, 7 Part 3 Extend to finding inverse functions.

• Use the function machine model to demonstrate the inverse, for example:

• Talk through how to complete the reverse function machine one cell at a time from the ‘end’:

f(x) becomes the starting point x x becomes the end point f(x).

• Ask students to write this algebraically: x = 𝑥𝑥+2

3 .

• Students work in pairs. Create a function and its matching function machine. Student A: from the function, write it as a function machine then find the inverse function machine, then convert to algebraic form. Student B: from the function machine, convert to algebraic form, rearrange to make x the subject (replace f(x) with y for convenience). Students then compare their work.

• Draw out that each method gives the inverse – method A replaces the f(x) with x at the start, method B the x and y also need reversing at the start (or end) of the process.

• Students can now do Exercise 24D from Student Book. A I 1–2 C 3 CM MR 3 PS EV Key 1e, 1f, 2 Assessment task

• Ask students what is different about the function in Exercise 24D Question 2, compared with the functions in Question 1. Challenge them to create a function machine diagram for this function.

x y

x x

x x

x x



Section 24.4 Composite functions Learning objectives Resources and homework • Use function notation • Find composite functions


Making mathematical connections Making cross-curricular connections • Equations, expressions identities and

functions • Mappings , coordinates and graphing

• Science – use of formulae • Computing – logic gates; formulae in

spreadsheets; flow diagrams • Relevance – developing logical thinking

Prior learning Students should be able to use function notation and substitute numerical values into a function. Working mathematically

• Give students a function f(x) and a function h(x) where h(x) = gf(x), and ask students to deduce g(x). This can easily be differentiated: Support – provide a set of g(x) functions to choose from. Challenge – only give h(x) and a set of functions, so students have to identify both f(x) and g(x). Common misconceptions and remediation

• It is easy to forget that fg(x) means ‘do g of x, then substitute the result into f of x’. Be consistent in the use of language. Probing questions

• Which method is best for finding composite functions in simple cases? Does it change if one function has more operations? If the second (last) function has more than one x? Literacy focus

• Key vocabulary: function machine function of x function composite Part 1 Introduce composite function notation.

• Display f(x) = 3x + 1, g(x) = 2(x – 1), fg(x), gf(x). • Ask students to consider what fg(x) and gf(x) might mean, and what fg(5) and gf(5) might

give. This could be done as a Think, Pair, Share. Take initial thoughts, without indicating whether right or wrong.

• Ask students what difference it would make if the notation was f(g(x)) – would this change their ideas?



Part 2 Introducing conventions for composite function notation and applying this to substitution and algebraic notation.

• Display on the board the function machine version:

• Draw out that f(g(x)) would mean putting g(x) where the x is in f(x):

• Model the two methods: • A: combining the two function machines in the correct order

(x – x – x – x – x – 5

• B: substituting g(x) into f(x) and simplifying f(g(x)) = 3[2(x – 1)] + 1 = 3[2x – 2] + 1 = 6x – 5

• Ask students to try both methods for gf(x) and ff(x), then to choose the method that they find most effective. Answers: gf(x) = 2[(3x + 1) – 1] = 6x ff(x) = 3(3x + 1) + 1 = (9x + 3) + 1 = 9x + 4

• Students can now do Exercise 24E from Student Book. A I 1–2 C 3–4 CM 3 MR PS 4 EV Key 2a, 2f, 3i, 3v Part 3 – Assessment task

• Design functions f(x) and g(x

x x x x

x x

x x

x



Section 24.5 Iteration Learning objectives Resources and homework • Use term-to-term sequence notation to

generate sequences • Use iteration to find solutions to

polynomials


Making mathematical connections Making cross-curricular connections • Trial and improvement • Sequences • Solving equations

• Computing – efficient coding using ‘repeat’ or loops

• Art - fractals • Relevance – applications to

manufacturing and project management Prior learning Students should be able to solve polynomials – although not know that name – using trial and improvement. They should be able to generate sequences using simple term-to-term rules, and algebraic position-to-term rules. Working mathematically

• Students use graphing software to identify where the roots of an equation lie: to the nearest whole number, to one decimal place, etc. They then describe how this helps with both trial and improvement and iteration.

• Students use graphing software to generate (for example) = 4 − − 4 and • = + 4 to ensure understanding of equivalence. • Students work in groups of 4, one pair using iteration and one pair using trial and

improvement to solve an equation. They then identify the most efficient method for doing this on their calculator, then compare methods: which is easiest? Quickest? Which is more efficient when the equation is longer? Common misconceptions and remediation

• Students may confuse the work on rearranging formulae – where the subject must appear once – with the rearranging here, where the ‘subject’ must appear on both sides. Probing questions

• What is the same and what is different about ‘trial and improvement’ and ‘iteration’? • Give me an example of a convergent sequence … and another one … and a more general

one… [prompt towards an algebraic form]. Literacy focus

• Key vocabulary: iteration convergence divergence oscillation



Part 1 Introduce the concept of convergence, divergence and oscillation in sequences.

• Display these sequences on the board and ask students to find the first 6 terms of each. 2 1

2 (− ) (−2) −12

• Challenge the students to sketch the graph for each sequence, or to use graph-plotting software to construct the graphs.

• Draw out the 5 sketches. Ask students to classify the 5 sequences and use this to bring in the language convergence, divergence, oscillation. Part 2 Introduce conventions for sequence notation.

• Use the sequence 2 to model the notation +1 = × 2, or +1 = 2 . • Ask students to find the next 3 terms of the sequence if 1 = . • Ask: Does the starting term change the behaviour of the sequence, or does it still diverge? • Students can now begin Exercise 24F from Student Book – questions 1 & 2.

A I 1–2 C CM RM SP EM Key 2b Part 3 Make the link from sequence notation to iteration.

• Use the equation = 4 − − 4. Ask students to find y when x = 2 and when x = 3. x = 2 gives a negative (–18), x = 3 gives a positive (37) therefore there must be a value between 2 and 3 that gives 0. NB: If students are not convinced that this is the case, construct the graph using graph plotting software.

• Explain that this mathematics helps to solve equations such as 4 − − 4 = . • Show the rearrangement to = + 4 then the conversion to sequence notation =

+ 4. • Ask students to find the next 5 terms of the sequence beginning with 1 = 2. What value

does their sequence converge towards? (To 1 • If appropriate, model a second example or ask students to try Exercise 24F question 3, then

go through it as a class. • Students can now continue Exercise 24F from Student Book.

A I 3–7 C 8–11 CM 5, 6, 8 MR PS 10 EV 7, 9, 11 Key 6, 9 Assessment task

• Use the Worked Exemplars – example 1. Cut up the exemplar into separate cards, add some distractors, and ask students to put the cards in order. Students could then choose Exercise 24F Question 11a or 11b or 11c and generate their own complete working with notes and distractors to challenge one another.


MathsSkills Builder

Transition from KS3 to GCSE

Gain confidence in tackling different types of questionswith an introductory worked example and solution

Practise communicating mathematically by givingexplanations and reasons

Challenge yourself with more demanding questionsmarked with an asterisk *

Learn independently and check your work with solutionsand helpful notes at the back of the book

Make connections across different parts of maths in themixed questions section

Chris Pearce

MathsSkills Builder Transition from KS3 to GCSE

Maths Skills B

uilder

Transition from K

S3 to GC

SE

Start preparing for new GCSE Maths by applyingthe concepts and ideas you already know from KS3.Packed with questions that build skills in mathematicalreasoning and problem solving, you can get a head-start from Year 9.

Published

206745_Skill Builder Sample.indd 1 21/10/14 9:15 PM







Cover design by We are LauraCover images © Kiora�lms-JochenTeschke/ Shutterstock Premier, Kachan Eduard/ShutterstockPremierIllustrations by Ann PaganuzziPrinted by Fuller Davies www.fullerdavies.com



Maths Skills BuilderTransition from KS3 to GCSE

Chris Pearce


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ProbabilityWORKED EXAMPLEThere are red, blue and black pens in a drawer.

Tracey takes one pen from the drawer, without looking.

The probability that she takes a red pen is 35 .

The probability that she takes a black pen is 310 .

a. Work out the probability that her pen is:

i. not red ii. blue.

b. Denise says: ‘There are now 15 pens left in the drawer.’

Explain why this is false.

SOLUTION

The question says that Tracey takes the pen without looking. This means that she is equally likely to take a pen of any colour. You could also say that she takes a pen ‘at random’.

a. i. The probability that the pen is not red = 1 − the probability that it is red

= 1 − 35

= 25

ii. The pen must be red, blue or black. The three probabilities must add up to one.

The probability the pen is blue = 1 35

310– +( )

= 1 910

110– =

b. If there are 15 pens left then originally there were 16 pens, since only one has been taken out.

But originally 35 of the pens were red − that is what the probability

tells you.

However, 35 of 16 is not a whole number. Tracey must be wrong.

You could also have used one of the other probabilities ( 310 or 1

10 ) to justify your answer.


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QUESTIONS

1. Graham has seven cards. Each card has a letter and a number on it.

D4

A1

B2

C3

E5

F6

G7

Graham takes a card at random.

Work out the probability that the card has on it:

a. a multiple of 3

b. a letter in the word GRAHAM

c. an even number and a letter in the word CAMERA.

2. Weather each day is put in one of three categories:

sunny

cloudy and dry

wet.

The probability it is sunny today is 0.3.

The probability it is not cloudy and dry is 0.9.

What is the probability it is wet?

3. Lucy has a large jar that contains 80 coloured sweets.

She says: ‘If you take one without looking, the probability that you will not get a red sweet is 4

5.’

Lucy is correct. How many of the sweets in the jar are red? Give a reason for your answer.

4. Gurdeep has a 2p coin, a 10p coin and a 20p coin.

He throws all three coins. Each coin can show a head (H) or a tail (T).

a. Copy and complete this table to show all the possible outcomes. You will need to add more rows.

2p 10p 10p

H H H

b. Work out the probability of his throwing at least one head.

c. Work out the probability of his throwing more heads than tails.


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5. In a game, Sasha throws darts at this target.

The probability that Sasha will miss the target is 10%.

The probability he will hit red is twice the probability he will hit blue.

What is the probability he will hit red?

6. Ewan has a pack of cards.

Each card has on it a two-digit number.

Ewan takes a card at random.

The probability that it is an even number is 34 .

Look at the statements below. Say whether each one is true or false. Give a reason for each answer.

a. A quarter of the cards have odd numbers.

b. The number on Ewan’s card cannot be 98.

c. There could be 50 cards in the pack.

7. This is part of a newspaper report.

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What is the probability that a person chose the number three?


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8. There are 180 raf�e tickets in �ve different colours.

This pie chart shows the proportion of each colour.

Orange

Grey

Pink

Yellow

Green

Leszek chooses one ticket at random.

a. Which colour is most likely?

b. Which two colours have the same probability?

c. The probability of one colour is 112 . Which colour is that?

d. What is the probability that the ticket is not pink?

9. The probability that Hendrick is late for school is one-ninth of the probability that he is not late.

What is the probability that he is late for school?

10. Cassie throws a dice twice and gets a 4 both times.

She is going to throw the dice again.

She says: ‘The probability of a 4 this time is more than 16.’

Is Cassie correct? Give a reason for your answer.

Answers at back of published book in tear-out section.

*

*

*


GCSE Maths 2 year Higher tier route map

For Jeni 23.10.14.indd 1 23/10/2014 15:53

GCSE Maths 4th edition 2 © HarperCollinsPublishers Ltd 2015

2 year Higher tier route map YEAR 10

SEPTEMBER OCTOBER NOVEMBER Weeks 1 – 2

N: Number Skills and

Properties

Weeks 3 – 4

N: Fractions, Ratios and Proportion

Weeks 5 – 6

S: Statistics Statistical diagrams and

averages

Week 7

A: Number and Sequences

Week 8

Holiday

Week 9


NOVEMBER DECEMBER JANUARY Weeks 10 – 12

R: Ratio and Proportion

Weeks 13 – 14

G: Angles

Week 15

G: Length and area

Week 16

Holiday

Week 17

Holiday

Weeks 18 – 19

G: Transformations

Weeks 20 – 21

G: Constructions and Loci

JANUARY FEBRUARY MARCH Week 22

A: Algebraic

Manipulation

Week 23

G: Length, Area and Volume

Week 24

Holiday

Week 25


Weeks 26 – 27

A: Linear Graphs

Weeks 28 – 29

G: Right Angled Triangles Pythagoras and intro to

trigonometry

Week 30

G: Congruency and Similarity

APRIL MAY JUNE Week 31

Holiday

Week 32

Holiday

Week 33


Weeks 34 – 35

P: Probability Exploring and applying

Week 36

N: Powers Laws of indices

Week 37

N: Powers and Standard

Form

Week 38

Holiday

Weeks 39 – 40

A: Algebraic Manipulation

JUNE JULY

Week 41

Summer examinations and revision

Week 42


Week 43


Weeks 44 – 45

N: Equations and Inequalities

Key N: Number S: Statistics P: Probability, proportion and rates of change A: Algebra R: Ratio G: Geometry and measures

For Jeni 23.10.14.indd 2 23/10/2014 15:53



G: Triangles

Trigonometry

Weeks 3 – 4

N: Number Accuracy

Weeks 5 – 6

N: Number Powers and surds

Week 7

A: Quadratic Equations

and graphs

Week 8

Holiday

Weeks 9 – 11

A: Quadratic Equations Graphs, solving, quadratic formula,

difference of 2 squares


S: Statistics

Sampling and more complex diagrams

Week 14

Mock examinations and revision

Week 15


Week 16

Holiday

Week 17

Holiday

Week 18

S: Statistics Sampling and more complex diagrams

Weeks 19 – 20

P: Probability Combined events

Week 21

G: Circle Theorems Properties of

circles


G: Circle Theorems Properties of circles

Week 23

Holiday

Weeks 24 – 25

R: Variation Direct and inverse

Weeks 26 – 27

G: Triangles Trigonometry

Weeks 28 – 29

A: Graphs

Week 30

Holiday


Holiday

Week 32

A: Graphs

Weeks 33 – 35

A: Algebraic Fractions and Functions

Weeks 36 – 37

G: Vectors

Week 38

Holiday

Weeks 39 – 40

Revision

JUNE JULY

Week 41

June examinations

Week 42

June examinations

Week 43

Week 44 Week 45


For Jeni 23.10.14.indd 3 23/10/2014 15:53

For Jeni 23.10.14.indd 4 23/10/2014 15:53

GCSE Maths 3 year Higher tier route map

For Jeni 23.10.14.indd 1 23/10/2014 15:54



N: Number Skills and Properties

Weeks 4 – 6

N: Fractions, Ratios and Proportion

Week 7

Review and revision 1

Week 8

Holiday

Weeks 9 – 10

S: Statistics Statistical diagrams and averages



Weeks 13 – 14

R: Ratio and Proportion Emphasis on core skills

Week 15

Examinations and revision

Week 16

Holiday

Week 17

Holiday

Weeks 18 – 19

R: Ratio and Proportion Emphasis on problem solving and

reasoning

Weeks 20 – 21

G: Angles


G: Length and

area

Week 23

Review and Revision 2

Week 24

Holiday

Weeks 25 – 26

G: Transformations

Weeks 27 – 28

G: Constructions and Loci

Week 29

A: Algebraic manipulation

Week 30



Holiday

Week 32

Holiday

Weeks 33 – 35


Week 36


Week 37


Week 38

Holiday

Weeks 39 – 40

A: Linear Graphs

JUNE JULY

Week 41


Week 42


Weeks 43 – 45

G: Right Angled Triangles Pythagoras and intro to trigonometry

3 year Higher tier route map YEAR 9 Key N: Number S: Statistics P: Probability, proportion and rates of change A: Algebra R: Ratio G: Geometry and measures

For Jeni 23.10.14.indd 2 23/10/2014 15:54


SEPTEMBER OCTOBER NOVEMBER Week 1


Weeks 2 – 3


Weeks 4 – 5

P: Probability Exploring and

applying

Week 6

N: Powers Laws of indices

Week 7


Week 8

Holiday

Weeks 9 – 10

N: Powers and Standard Form



Simultaneous equations and graphs

Week 14


Week 15


Week 16

Holiday

Week 17

Holiday

Weeks 18 – 19

A: Algebraic Manipulation Quadratic equations and graphs

Week 20

Number recap and

review

JANUARY FEBRUARY MARCH Weeks 21 – 22

N: Equations and Inequalities

Week 23

Review and Revision 7

Week 24

Holiday

Weeks 25 – 26


Week 27

Statistics recap and

review

Weeks 28 – 29

N: Number Accuracy

Week 30



Holiday

Week 32

Holiday

Weeks 33 – 35

N: Number Powers and surds

Week 36

N: Number / Review and revision 8

Week 37


Week 38

Holiday

Weeks 39 – 40

A: Quadratic Equations and graphs

JUNE JULY

Week 41


Week 42


Weeks 43 – 45

A: Quadratic Equations Graphs, solving, quadratic formula, difference of 2

squares


For Jeni 23.10.14.indd 3 23/10/2014 15:54


SEPTEMBER OCTOBER NOVEMBER Week 1


Weeks 2 – 4

S: Statistics Sampling and more complex diagrams

Weeks 5 – 6

P: Probability Combined events

Week 7


Week 8

Holiday

Week 9

Algebra recap and

review

Weeks 10 – 11

G: Circle Theorems Properties of circles


R: Variation

Direct and inverse

Week 14


Week 15


Week 16

Holiday

Week 17

Holiday

Weeks 18 – 19


Weeks 20 – 21

A: Graphs



Week 23

Holiday

Weeks 24 – 26

A: Graphs

Weeks 27 – 28

A: Algebraic Fractions and Functions

Week 29


Week 30

Holiday


Holiday

Weeks 32 – 33

A: Algebraic Fractions and Functions Transformation of graphs

Weeks 34 – 35

G: Vectors

Weeks 36 – 37

Revision

Week 38

Holiday

Weeks 39 – 40

Revision

JUNE JULY

Week 41

June examinations

Week 42

June examinations

Week 43

Week 44 Week 45


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