Further Mathematics

Further
Mathematics

AQA CERTIFICATE Further Mathematics DAVID PRITCHARD with Val Hanrahan and Roger Porkess i7 HODDER EDUCATION AN H ACHETTE UK COMPANY

AQA CERTIFICATE
Further
Mathematics
DAVID PRITCHARD
with Val Hanrahan
and Roger Porkess
i7
HODDER
EDUCATION
AN H ACHETTE UK COMPANY

Ewrycfforthasbecnmadetotrac,,allcopyrightholders,butifanyhavebttninadvertentlyoverlookedthe Publishers will be pleased to make thr necessary arrangements at the first opportunity, Hachette UK's policy is to use pa pers that ar e natural, renewable and rccydablc products and made from wood grown in sustainable forests, The logging and manufacturing processes are expected toconfonn to thernvironmcntalr c gulationsofthccountryoforigin, Orders: please contact Bookpoint Ltd, 130 Milton Park,Abingdon, Ch:on 0Xl4 4SB, Telephone: {44) 01235 827720.Fu:{44)0l235400454.Lincsareopcn9.00--5,00,MondaytoSaturday,witha24 - hourmcssagr answering 9Crvicc, Visit our website a! www.hoddercducation.com C DavidPritchard,ValHanrahan20l3 Firstpublishedin 2013by Hodd erEducation ,aHach c ttcUKcompany, 338Eu.stonRoad LondonNW13BH All rights reocrYrd, Apart from any use permitted under UK copyright law, no part of this publication may be reproduced or transmitted in any form or by any mean<, electronic or mcchani,al, including photocopyingandrccording,orheldwithinanyinformationstorageandrctrievalsystrm,without permission in writing from the publisher or under liccn c,, from the Copyright Licensing Agency Limited, Funherdetail,ofsuchliccnc,,s(forreprographicrcproduction)maybcobtainedfromtheCopyright Licensing Agency Limited, Saffron Hou .. , 6--10 Kirby Street, London EC l N STS, Covc:rphoto C Miclwel Smith/iSrod:pltoro TypcsctinJ0.5/14inMinionRegufurbyAptara,lnc, Print ed in Great Britain by CPI Group (UK) Ltd, Croydon, CRO 4YY Acataloguerccordforthi,ti tl ei,av ailablefromth c BritishLibrary

Ewrycfforthasbecnmadetotrac,,allcopyrightholders,butifanyhavebttninadvertentlyoverlookedthe
Publishers will be pleased to make thr necessary arrangements at the first opportunity,
Hachette UK's policy is to use pa pers that ar e natural, renewable and rccydablc products and made from
wood grown in sustainable forests, The logging and manufacturing processes are expected toconfonn to
thernvironmcntalr c gulationsofthccountryoforigin,
Orders: please contact Bookpoint Ltd, 130 Milton Park,Abingdon, Ch:on 0Xl4 4SB, Telephone: {44) 01235
827720.Fu:{44)0l235400454.Lincsareopcn9.00--5,00,MondaytoSaturday,witha24 - hourmcssagr
answering 9Crvicc, Visit our website a! www.hoddercducation.com
C DavidPritchard,ValHanrahan20l3
Firstpublishedin 2013by
Hodd erEducation ,aHach c ttcUKcompany,
338Eu.stonRoad
LondonNW13BH
All rights reocrYrd, Apart from any use permitted under UK copyright law, no part of this publication
may be reproduced or transmitted in any form or by any mean<, electronic or mcchani,al, including
photocopyingandrccording,orheldwithinanyinformationstorageandrctrievalsystrm,without permission in writing from the publisher or under liccn c,, from the Copyright Licensing Agency Limited,
Funherdetail,ofsuchliccnc,,s(forreprographicrcproduction)maybcobtainedfromtheCopyright
Licensing Agency Limited, Saffron Hou .. , 6--10 Kirby Street, London EC l N STS,
TypcsctinJ0.5/14inMinionRegufurbyAptara,lnc, Covc:rphoto C Miclwel Smith/iSrod:pltoro
Print ed in Great Britain by CPI Group (UK) Ltd, Croydon, CRO 4YY
Acataloguerccordforthi,ti tl ei,av ailablefromth c BritishLibrary

I Introduction I ~- This book supports the AQA Level 2 Certificate in Further Mathematics. The qualification is designed for high achieving students who have already achieved, or are expected to achieve, grades A or A• in GCSE Mathematics. Many of these students are likely to progress to study Mathematics at A-level and beyond. Higher order mathematical skills will be studied in greater depth with an emphasis on algebraic reasoning, rigorous argument and problem solving skills. Students following this course will be able to achieve their maximum potential at Level 2 and be well prepared to tackle a Level 3 Mathematics qualification. The content is split into Algebra, Geometry, Calculus and Matrices with each section containing work that stretches and challenges, and which goes beyond the Key Stage 4 Programme of Study. The topic areas are frequently linked together as progress is made through the book, highlighting the beauty and inter-connectedness of mathematics. The problem-solving questions will often involve more than one topic area. Two symbols are used throughout the book: IJ This denotes a 'discussion point'. These are prompts to help you to understand the theory that has been, or is about to be, introduced. Answers to these are also included. A This 'warning sign' alerts you either to restrictions that need to be imposed or to possible pitfalls. In addition the book includes a nwnber of activities. These are often used to introduce a new concept, or to reinforce the examples in the text. Throughout the book the emphasis is on understanding the mathematics being used rather than merely being able to perform the calculations but the exercises do, nonetheless, provide plenty of scope for practising basic techniques. It is hoped that students (and teachers) will be inspired and challenged by the rigorous nature of the course and be able to appreciate the power of mathematics for its own sake as well as a problem -solving tool.

I
Introduction
I
This book supports the AQA Level 2 Certificate in Further Mathematics. ~-
The qualification is designed for high achieving students who have already
achieved, or are expected to achieve, grades A or A• in GCSE Mathematics.
Many of these students are likely to progress to study Mathematics at A-level and
beyond.
Higher order mathematical skills will be studied in greater depth with an
emphasis on algebraic reasoning, rigorous argument and problem solving skills.
Students following this course will be able to achieve their maximum potential at
Level 2 and be well prepared to tackle a Level 3 Mathematics qualification.
The content is split into Algebra, Geometry, Calculus and Matrices with each
section containing work that stretches and challenges, and which goes beyond
the Key Stage 4 Programme of Study. The topic areas are frequently linked
together as progress is made through the book, highlighting the beauty and
inter-connectedness of mathematics. The problem-solving questions will often
involve more than one topic area.
Two symbols are used throughout the book:
IJ This denotes a 'discussion point'. These are prompts to help you to understand
the theory that has been, or is about to be, introduced. Answers to these are
also included.
A This 'warning sign' alerts you either to restrictions that need to be imposed or
to possible pitfalls.
In addition the book includes a nwnber of activities. These are often used to
introduce a new concept, or to reinforce the examples in the text. Throughout the
book the emphasis is on understanding the mathematics being used rather than
merely being able to perform the calculations but the exercises do, nonetheless,
provide plenty of scope for practising basic techniques.
It is hoped that students (and teachers) will be inspired and challenged by the
rigorous nature of the course and be able to appreciate the power of mathematics
for its own sake as well as a problem -solving tool.

I Contents ALGEBRA SECTION 1 Number and algebra I Numbersandthenumbersystem Simplifying expressions Solving linear equations Algebra and number 10 Expanding brackets 12 Manipulatingsurds 14 Algebra II 19 Factorising 19 Rearranging formulae 2J Simplifying algebraic fractions 26 Solving equations involving fractions 30 Quadratic identities 32 Algebra Ill 35 Function notation 35 Oomain and range of a function 36 Graphs of functions 39 Graphs of linear functions 40 Findingtheequationofaline 45 Graphs of quadratic functions 50 Graphsoffunctionswithuptothreepartstotheirdomains 54 Algebra IV 59 Quadratic equations 59 Simultaneous equations 64 Factor theorem 71 linear inequalities 75 Quadratic inequalities 78 Index laws 80 Algebraic proof BJ Sequences 85

I
Contents
SECTION 1 ALGEBRA
Number and algebra I
Numbersandthenumbersystem
Simplifying expressions
Solving linear equations
Algebra and number 10
Expanding brackets 12
Manipulatingsurds 14
Algebra II 19
Factorising 19
Rearranging formulae 2J
Simplifying algebraic fractions 26
Solving equations involving fractions 30
Quadratic identities 32
Algebra Ill 35
Function notation 35
Oomain and range of a function 36
Graphs of functions 39
Graphs of linear functions 40
Findingtheequationofaline 45
Graphs of quadratic functions 50
Graphsoffunctionswithuptothreepartstotheirdomains 54
Algebra IV 59
Quadratic equations 59
Simultaneous equations 64
Factor theorem 71
linear inequalities 75
Quadratic inequalities 78
Index laws 80
Algebraic proof BJ
Sequences 85

I SECTION 2 GEOMETRY 93 , n Co-ordinate geometry 95 0 Parallel and perpendicular lines 95 Thedistancebetweentwopoints 96 Themidpointofalinejoiningtwo points 96 Equationofastraightline 101 The intersection of two lines 104 Dividing of a line in a given ratio 106 Equation of a circle 109 Geometry I 119 Mensuration 119 Pythagoras· theorem 119 Angle facts 120 Circle theorems 120 Geometric proof 124 Trigonometry in two dimensions 128 ,,. Angles 0145°, 30° and 60° 133 Problems in three dimensions linesandplanesinthreedimensions 138 Geometry II 149 Trigonometricalfunctionsforanglesofanysize 149 The sine and cosine graphs 150 The tangent graph 152 Theareaofatriangle 153 The sine rule 156 ,., The cosine rule 159 ,.. Using thesineand cosine rules together Solution of trigonometrical equations Trigonometrical identities 168 SECTION 3 CALCULUS 175 Calculus 177 Thegradientofacurve 177 Differentiation 177 Differentiation using standard results 182 Tangents and normals 185 Increasing and decreasing functions 189 Stationary points 190

I
SECTION 2 GEOMETRY 93
Co-ordinate geometry 95 , n 0
Parallel and perpendicular lines 95
Thedistancebetweentwopoints 96
Themidpointofalinejoiningtwo points 96
Equationofastraightline 101
The intersection of two lines 104
Dividing of a line in a given ratio 106
Equation of a circle 109
Geometry I 119
Mensuration 119
Pythagoras· theorem 119
Angle facts 120
Circle theorems 120
Geometric proof 124
Trigonometry in two dimensions ,,. 128
Angles 0145°, 30° and 60° 133
Problems in three dimensions
linesandplanesinthreedimensions 138
Geometry II 149
Trigonometricalfunctionsforanglesofanysize 149
The sine and cosine graphs 150
The tangent graph 152
Theareaofatriangle 153
The sine rule 156
The cosine rule ,., 159 ,..
Using thesineand cosine rules together
Solution of trigonometrical equations
Trigonometrical identities 168
SECTION 3 CALCULUS 175
Calculus 177
Thegradientofacurve 177
Differentiation 177
Differentiation using standard results 182
Tangents and normals 185
Increasing and decreasing functions 189
Stationary points 190

I SECTION 4 MATRICES 199 Matrices 201 Multiplyingmiltrices 201 Transformations 204 Identity matrix 205 Transformationsoftheunitsquare 206 Combining transformations 210 Keywords 213 Exam•style paper 1 215 Exam-style paper 2 219 Answers 224 Index 247 ,;

I
SECTION 4 MATRICES 199
Matrices 201
Multiplyingmiltrices 201
Transformations 204
Identity matrix 205
Transformationsoftheunitsquare 206
Combining transformations 210
Keywords 213
Exam•style paper 1 215
Exam-style paper 2 219
Answers 224
Index 247
,;

4iii•M~II ALGEBRA

4iii•M~II
ALGEBRA

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Number and algebra I The only thing that separates successful people from the ones who aren't is the willingness to work very, very hard. Helen Gurley Brown (American Businesswoman) Numbers and the number system Number will be tested implicitly throughout the course. Apart from manipulation of surds (pages 14-18), examination questions will always test number with at least one of algebra, calculus or geometry. These questions will appear at appropriate times in the book. The following exercise provides practice of some number skills that may be needed. 1 ABCD isastraightline.AB = 4cm , AC = 10cm,AD =22c m. A C D Work out these ratios, giving your answers in their simplest form. (ii AC:AB ,,, (iii A B :BC [iii) AD:AB ., [i v ) BC:CD BD:BC Workout [ii 60%of£115 [iii 33f% of780 (iii] 17.5%of64an. J Workout ., [ii 95%of7540 [iii 12 ! % of53.76 [iii) 4.2%of£150. [ii I ncrease80by5%. [iii Increase £240 by 75%. [iii] Decrease£20by40%. [iv ] Decrease 36 by 66f%. 5 [ii I ncrease650by 14%. [iii Decrease3250b y3.5%. t [iii] Decrease £3650 by 64%. [iv ] Increase £46 by 5 %. •6 Workout, giving your answers as fractions in their simplest form (± J 3 2 5 3~-I [ii - + - x - [iii + 4 liiil 5 3 6 5 4

Number and algebra I
The only thing that separates successful people from the ones who
aren't is the willingness to work very, very hard.
Helen Gurley Brown (American Businesswoman)
Numbers and the number system
Number will be tested implicitly throughout the course. Apart from
manipulation of surds (pages 14-18), examination questions will always test
number with at least one of algebra, calculus or geometry. These questions will
appear at appropriate times in the book. The following exercise provides practice
of some number skills that may be needed.
1 ABCD isastraightline.AB = 4cm , AC = 10cm,AD =22c m.
A C D
Work out these ratios, giving your answers in their simplest form.
(ii AC:AB ,,, (iii A B :BC [iii) AD:AB
., [i v ) BC:CD BD:BC
Workout
[ii 60%of£115 [iii 33f% of780 (iii] 17.5%of64an.
J Workout
., [ii 95%of7540 [iii 12 ! % of53.76 [iii) 4.2%of£150.
[ii I ncrease80by5%. [iii Increase £240 by 75%.
[iii] Decrease£20by40%. [iv ] Decrease 36 by 66f%.
5 [ii I ncrease650by 14%. [iii Decrease3250b y3.5%.
[iii] Decrease £3650 by 64%. [iv ] Increase £46 by 5 t %.
•6 Workout, giving your answers as fractions in their simplest form
(± J
[ii - + - x - 3 2 5 [iii + 4 liiil 3~-I
5 3 6 5 4

I 7 A bag contains blue, green and white beads. The ratio of blue beads to green beadsis4 :3. The ratio of green beads to white beads is 2: 7. Work out the smallest possible number of beads in the bag. 8 [ii Work out, giving your answer to 3 significant figures 52.7 + 4.93 [iii Work out, giving your answer to 2 significant figures 5.9 - 0.53 X 1.8 [iii] Work out, giving your answer to I significant figure 0.23 x 0.14 + 0.09 2 [i v) Work out, giving your answerto 2 decimal places 19 + 36 144 - 52 9 55% of teachers in a school are female. 36 teachers are male. Work out the number of teachers in the school. D A large ice cream costs 40p more than a small one. Two large ice creams cost the same as three small ones. What is the cost of each size of ice cream? This is an example of the type of question that you might find in a puzzle book or the puzzle section of a newspaper or magazine. H ow would you set about tackling it? 0 You may think that the following question appears to be very similar. What happens when you try to solve it? A large ice cream costs 40p more than a small one. Five small ice creams plus three large ones cost 80p less than three small ice creams plus five large ones. What is the cost of each size of ice cream? Simplifying expressions When you are asked to simplify an algebraic expression you need to write it in its most compact form. This will involve techniques such as collecting like terms, removing brackets, factorising and finding a common denominator (if the expression includes fractions). EXAMPLE1 .1 Simplify this expression. 3a + 4b-2c + a - Jb - c SO LUTION Expression = 3a + a + 4b - 3b - 2c - c collecting like terms = 4a + b - 3c

I
7 A bag contains blue, green and white beads. The ratio of blue beads to green
beadsis4 :3.
The ratio of green beads to white beads is 2: 7.
Work out the smallest possible number of beads in the bag.
8 [ii Work out, giving your answer to 3 significant figures 52.7 + 4.93
[iii Work out, giving your answer to 2 significant figures 5.9 - 0.53 X 1.8
[iii] Work out, giving your answer to I significant figure 0.23 x 0.14 + 0.09 2
[i v) Work out, giving your answerto 2 decimal places 19 + 36
144 - 52
9 55% of teachers in a school are female. 36 teachers are male.
Work out the number of teachers in the school.
D A large ice cream costs 40p more than a small one. Two large ice creams cost the
same as three small ones. What is the cost of each size of ice cream?
This is an example of the type of question that you might find in a puzzle book
or the puzzle section of a newspaper or magazine. H ow would you set about
tackling it?
0 You may think that the following question appears to be very similar. What
happens when you try to solve it?
A large ice cream costs 40p more than a small one. Five small ice creams plus
three large ones cost 80p less than three small ice creams plus five large ones.
What is the cost of each size of ice cream?
Simplifying expressions
When you are asked to simplify an algebraic expression you need to write it in its
most compact form. This will involve techniques such as collecting like terms,
removing brackets, factorising and finding a common denominator (if the
expression includes fractions).
EXAMPLE1 .1 Simplify this expression.
3a + 4b-2c + a - Jb - c
SO LUTION
Expression = 3a + a + 4b - 3b - 2c - c collecting like terms
= 4a + b - 3c

I EXAMPLE 1.2 Simplify this expression. 2(3x- 4y) - 3(x+ 2y) I z . SO LUTION . , Expression = 6x- Sy- 3x- y removing the brackets 1 = 3x- 14y == •M =P ~ L~ E = ' ·' -., Simplify this expression. 3x 2 yz x 2xy 3 SO LUTION Expression = (3 x 2) x (x 2 x x) x (yx y3) x z collecting like terms =6x3y'z EXAMPLE 1.4 Simplify this expression. 6a 2 b 3 c 3ab 4 c 3 SO LUTION Look where the higher powers of a, band c occur. You may find that it helps if you split them up like this. 6a 2 b 3 c _ $x 2 x tfx ax~ 3 xc' 3ab 4 c 3 - $xtfxlflx bxc'x c 2 You can then cancel as indicated to give '" b2 EX AMPLE 1.5 Factorise this expression. SO LUTION First you need to look for the highest common factor of the two terms, which is 3abhere. 3a 2 b+6ab 2 = 3ab(a+2b) Since3a'b = 3<>bX<> and6<>b' = 3<>bx2b. D Explain what the word factorise means.

I
EXAMPLE 1.2 Simplify this expression.
2(3x- 4y) - 3(x+ 2y) . I z
SO LUTION . ,
Expression = 6x- Sy- 3x- y removing the brackets
= 3x- 14y 1
== •M =P ~ L~ E = ' ·' -., Simplify this expression.
3x 2 yz x 2xy 3
SO LUTION
Expression = (3 x 2) x (x 2 x x) x (yx y3) x z collecting like terms
=6x3y'z
EXAMPLE 1.4 Simplify this expression.
6a 2 b 3 c
3ab 4 c 3
SO LUTION
Look where the higher powers of a, band c occur. You may find that it helps if
you split them up like this.
6a 2 b 3 c _ $x 2 x tfx ax~ 3 xc'
3ab 4 c 3 - $xtfxlflx bxc'x c 2
You can then cancel as indicated to give
'"
b2
EX AMPLE 1.5 Factorise this expression.
SO LUTION
First you need to look for the highest common factor of the two terms, which is
3abhere.
3a 2 b+6ab 2 = 3ab(a+2b) Since3a'b = 3<>bX<>
and6<>b' = 3<>bx2b.
D Explain what the word factorise means.

I EXAMPLE 1.6 Simplify this expression. 2x 2 i!!'.:_ Jyz-:- . Sz 2 SO LUTION . zxi 5z 2 Express10n = 3yz x xy 4 2 Sx, ~ EXAMPLE 1.7 Simplify this expression. SO LUTION Expression = ~-~+~ _ 5x-8y+ !Oz - 20t G!Dilmllillll 1 Simplify the following expressions. [ii 12a+3b -7c-2a- 4b+5 c [iii 4x - Sy+ 3z+ 2x+ 2y- 7z [iii] J(Sx- y) + 4 (x + 2y) [i v ) 2(p+Sq)-(p-4q) (vl x(x +3 )-x(x-2) [v ii a(2a+3)+3(3a-4) [v iii 3p(q-p)-3q(p-q) (v iii) 5f{g+2h)-5g(,h-f) 2 Factorise the following expressions by taking out the highest common factor. [ii 8 - 10x 2 [iii 6ab+ Sbc [iii] 2a 2 +4ab [i v ] pq3 - plq [vl Jx 2 y+ 6xy 4 [v ii 6p3q-4p2q2 + 2pq3 [v iii 15/m 2 - 9/lm 3 + 12/2m 4 [v iii) 84a 5 b 4 -96a 4 b 5 3 Simplify the following expressions and factorise the answers. (ii 4 (3x+2y) +B (x-3y) [iii x(x- 2) -x(x- 8) + 6 [iii] x(y + z )-y(x + z)

I
EXAMPLE 1.6 Simplify this expression.
Jyz-:- 2x 2 . i!!'.:_ Sz 2
SO LUTION
Express10n . = 3yz zxi x 4 5z xy 2 2
Sx,
~
EXAMPLE 1.7 Simplify this expression.
SO LUTION
Expression = ~-~+~
_ 5x-8y+ !Oz
- 20t
G!Dilmllillll
1 Simplify the following expressions.
[ii 12a+3b -7c-2a- 4b+5 c
[iii 4x - Sy+ 3z+ 2x+ 2y- 7z
[iii] J(Sx- y) + 4 (x + 2y)
[i v ) 2(p+Sq)-(p-4q)
(vl x(x +3 )-x(x-2)
[v ii a(2a+3)+3(3a-4)
[v iii 3p(q-p)-3q(p-q)
(v iii) 5f{g+2h)-5g(,h-f)
2 Factorise the following expressions by taking out the highest common factor.
[ii 8 - 10x 2
[iii 6ab+ Sbc
[iii] 2a 2 +4ab
[i v ] pq3 - plq
[vl Jx 2 y+ 6xy 4
[v ii 6p3q-4p2q2 + 2pq3
[v iii 15/m 2 - 9/lm 3 + 12/2m 4
[v iii) 84a 5 b 4 -96a 4 b 5
3 Simplify the following expressions and factorise the answers.
(ii 4 (3x+2y) +B (x-3y)
[iii x(x- 2) -x(x- 8) + 6
[iii] x(y + z )-y(x + z)

I [iv) p()q- r) + r(p- 2q) [vl k(l+m+n)-km I z [vii a(a-2)-a(a+4)+2(a-4) . [viii Jx(x+ y)-Jy(x-2y) [viii) a(a - 2) - a(a - 4) + 8 . , 4 Simplify the following expressions as much as possible. 1 [ii 2a 2 bx5ab 3 [iii 6p 3 qx2q 3 r [iii] lmx mnx np [iv] Jr 3 x6s 2 x2rs "' abx2bcx4cdx8de [vii Jxy 2 x 4yz 2 x 5x 2 z [viii 2ab 3 x 6a 4 x 7b 6 5 Simplify the following fractions as much as possible. 4a 2 b p!_ 1:_ [ii 2ab [iii q p X [iii] .§E_x§.t [iv) Mx:!£4 ,,, 3b 2 4a 2 2c 2 6a 2 ~ [vi) ~ _,_ 2a 4 12y, 9b 3 "Ts1i ?f.:9. -,- ~ [viii "" 8rs 2 6 Simplify the following expressions as single fractions. ¥-i+?f ±f-+~ [ii [iii ¥-}+~ it_ _~ !iii] 3 4 [iv) ,,, ~-t+* 2._ _ _l_ ~-E: [viii 2p lq [viii) 3y 2y Solving linear equations D What is an equation? What does solving an equation mean?

I
[iv) p()q- r) + r(p- 2q)
[vl k(l+m+n)-km
[vii a(a-2)-a(a+4)+2(a-4) . I z
[viii Jx(x+ y)-Jy(x-2y)
[viii) a(a - 2) - a(a - 4) + 8 . ,
4 Simplify the following expressions as much as possible.
1
[ii 2a 2 bx5ab 3
[iii 6p 3 qx2q 3 r
[iii] lmx mnx np
"' [iv] Jr 3 x6s 2 x2rs
abx2bcx4cdx8de
[vii Jxy 2 x 4yz 2 x 5x 2 z
[viii 2ab 3 x 6a 4 x 7b 6
5 Simplify the following fractions as much as possible.
[ii 2ab 4a 2 b [iii p!_ q 1:_ p
X
[iii] .§E_x§.t [iv) Mx:!£4
,,, 3b 2 4a 2 2c 2 6a 2
~ [vi) ~ _,_ 2a 4
12y, 9b 3 "Ts1i
[viii ?f.:9. -,- ~ ""
8rs 2
6 Simplify the following expressions as single fractions.
±f-+~ ¥-i+?f
[ii [iii
it_ _~ ¥-}+~
!iii] 3 4 [iv)
,,,
~-t+*
[viii 2._ _ _l_ ~-E:
2p lq [viii) 3y 2y
Solving linear equations
D What is an equation?
What does solving an equation mean?

I Since both sides of an equation are equal, you may do what you wish to the equation, provided that you do exactly the same thing to both sides. The examples that follow illustrate this in great detail. In practice you would expect to omit some of the working. E XA LE .8 Solve this equation. 3(3x- 17 ) = 2(x- I ) SO LUTION Open the brackets 9x- 51 2 =) = lx- Subtract 2x from both sides 9x- 51 - 2x 2x =) = 2x- 2- Tidy up => 7x-51 = - 2 Add 51 to both sides => ?x-51 +51 = -2 +51 Tidy up ~, => 7x = 49 Divide both sides by 7 EXA LE . Solve this equation. Jex + 8) = lx+ {( 4x - 5) SO LUTION Start by clearing the fractions by multiplying by 6 (the least common multiple of2and3). Multiply both sides by 6 => 6xf(x+8 ) = 6x2x+6x{(4x-S) Tidy up => 3(x+ 8) = 12x+ 2( 4x - 5) Open the brackets => 3x + 24 = llx+ Bx- 10 Tidy up 3x+24 = 20x-10 =) Subtract3xfrombothsides 24 = 17x - 10 =) Add 10 to both sides 34 ~, = 17 x =) Divide both sides by 17 0 Wh y have the variable a nd the number changed sides on the last line? Sometimes you will need to set up the equation as well as solve it. When you are doing this, make sure that you define any variables that you introduce. EX AM LE . 10 I na triangle, the largest angle is nine times as big as the smallest. The third angle is 60°. [ii Write this information in the form of an equation for a, the size in degrees of the smallest angle. (iii Solve the equation to find the sizes of the three angles.

I
Since both sides of an equation are equal, you may do what you wish to the
equation, provided that you do exactly the same thing to both sides. The
examples that follow illustrate this in great detail. In practice you would expect to
omit some of the working.
E XA LE .8 Solve this equation.
3(3x- 17 ) = 2(x- I )
SO LUTION
Open the brackets =) 9x- 51 = lx- 2
Subtract 2x from both sides =) 9x- 51 - 2x = 2x- 2- 2x
Tidy up => 7x-51 = - 2
Add 51 to both sides => ?x-51 +51 = -2 +51
Tidy up => 7x ~, = 49
Divide both sides by 7
EXA LE . Solve this equation.
Jex +
8) = lx+ {( 4x - 5)
SO LUTION
Start by clearing the fractions by multiplying by 6 (the least common multiple
of2and3).
Multiply both sides by 6 => 6xf(x+8 ) = 6x2x+6x{(4x-S)
Tidy up => 3(x+ 8) = 12x+ 2( 4x - 5)
Open the brackets => 3x + 24 = llx+ Bx- 10
Tidy up =) 3x+24 = 20x-10
Subtract3xfrombothsides =) 24 = 17x - 10
Add 10 to both sides =) 34 ~, = 17 x
Divide both sides by 17
0 Wh y have the variable a nd the number changed sides on the last line?
Sometimes you will need to set up the equation as well as solve it. When you are
doing this, make sure that you define any variables that you introduce.
EX AM LE . 10 I na triangle, the largest angle is nine times as big as the smallest. The third angle is 60°.
[ii Write this information in the form of an equation for a, the size in degrees of
the smallest angle.
(iii Solve the equation to find the sizes of the three angles.

I SOLUTION Let the smallest angle = a". I z . So the largest angle is 9a 0 • The sum of all three angles is 180° . , ~ a+9a+60 =1 80 ~ IOa =1 20 1 = 12 This gives 9a = 108, so the angles are 12°, 60° and 108°. , Solve the following equations. [ii 2x-3=x+4 (iii 5a+3 = 2a-3 ,,, [iii] 2(x+S) = 14 (iv) 7(2y-5) = -7 5(2c-8) = 2(3c-10) [vii 3(p+2) ~4 (p-1) [viii J(2x- I) = 6(x+2) + 3x [viii] i+7 = 5 ,,, [ix) ~ 11 - J - ~+~ = 35 [xii = 7 ¥-¥=4 [xiii 5p;4_2p2+3 [xiii] p+j ( p+l ) +i(p+2 ) = ~ 2 The length, I metres, of a field is 80 m greater than the width. The perimeter is600m. [ii Write the information in the form of an equation for I. [iii Solve the equation and so find the area of the field. 3 Louise and Molly are twins and their brother Jonathan is four years younger. The total of their three ages is 17 years. [ii Write this information in the form of an equation inj, Jonathan's age in years. [iii Whatarealltheirages? 4 In a multiple-choice examination of 20 questions, four marks are given for each correct answer and one mark is deducted for each wrong answer. There is no penalty for not attempting a question. A candidate attempts a questions andgetsccorrect. [ii Write down, and simplify, an expression for the candidate's total mark in termsofaandc . [iii A candidate attempts three-quarters of the questions and scores 40. Write down, and solve, an equation for the number of correct questions.

I
SOLUTION
Let the smallest angle = a". . I z
So the largest angle is 9a 0 •
The sum of all three angles is 180° .
,
~ a+9a+60 =1 80
~ IOa =1 20 1
= 12
This gives 9a = 108, so the angles are 12°, 60° and 108°.
,
Solve the following equations.
[ii 2x-3=x+4 (iii 5a+3 = 2a-3
,,, [iii] 2(x+S) = 14 (iv) 7(2y-5) = -7
5(2c-8) = 2(3c-10) [vii 3(p+2) ~4 (p-1)
[viii J(2x- I) = 6(x+2) + 3x ,,, [viii] i+7 = 5
[ix) ~ 11 - J - ~+~ = 35
[xii ¥-¥=4 [xiii 5p;4_2p2+3 = 7
[xiii] p+j ( p+l ) +i(p+2 ) = ~
2 The length, I metres, of a field is 80 m greater than the width. The perimeter
is600m.
[ii Write the information in the form of an equation for I.
[iii Solve the equation and so find the area of the field.
3 Louise and Molly are twins and their brother Jonathan is four years younger.
The total of their three ages is 17 years.
[ii Write this information in the form of an equation inj, Jonathan's age in
years.
[iii Whatarealltheirages?
4 In a multiple-choice examination of 20 questions, four marks are given for
each correct answer and one mark is deducted for each wrong answer. There
is no penalty for not attempting a question. A candidate attempts a questions
andgetsccorrect.
[ii Write down, and simplify, an expression for the candidate's total mark in
termsofaandc .
[iii A candidate attempts three-quarters of the questions and scores 40.
Write down, and solve, an equation for the number of correct
questions.

I 5 John is three times as old as his son, Michael, and in xyears' time he will be twice as old as him. (ii Write down expressions for John's and Michael's age in xyears' time. [iii Write down, and solve, an equation in x. 6 A square has sides of length la metres, and a rectangle has length 3a metres and breadth 3 metres. [ii Find, in terms of a, the perimeter of the square. [iii Find, in terms of a, the perimeter of the rectangle. !iii] The perimeters of the square and the rectangle are equal. Find a. 7 The sum of five consecutive numbers is equal to 105. Let m represent the middle number. [ii Write down the five numbers in terms of m. [iii Form an equation in m and solve it. [iii] What are the five consecutive numbers? 8 One rectangle has a length of ( x+ 2) cm and a breadth of2cm, and another, of equal area, has a length of 5cm and a breadth of (x - 3 )cm. [ii Write down an equation in x and solve it. [iii What is the area of each of the rectangles? Algebra and number Some algebra questions will involve using number skills. SO LUTION ais75%ofb a = _!1_b 100 a = .!__b ID 4 b b:c = 3:2 3 ' 2 b = I, "' 2 10

I
5 John is three times as old as his son, Michael, and in xyears' time he will be
twice as old as him.
(ii Write down expressions for John's and Michael's age in xyears' time.
[iii Write down, and solve, an equation in x.
6 A square has sides of length la metres, and a rectangle has length 3a metres
and breadth 3 metres.
[ii Find, in terms of a, the perimeter of the square.
[iii Find, in terms of a, the perimeter of the rectangle.
!iii] The perimeters of the square and the rectangle are equal. Find a.
7 The sum of five consecutive numbers is equal to 105. Let m represent the
middle number.
[ii Write down the five numbers in terms of m.
[iii Form an equation in m and solve it.
[iii] What are the five consecutive numbers?
8 One rectangle has a length of ( x+ 2) cm and a breadth of2cm, and another, of
equal area, has a length of 5cm and a breadth of (x - 3 )cm.
[ii Write down an equation in x and solve it.
[iii What is the area of each of the rectangles?
Algebra and number
Some algebra questions will involve using number skills.
SO LUTION
ais75%ofb a = _!1_b
100
a = .!__b ID
4
b:c = 3:2 b 3
' 2
b = I, "'
2
10

I Substitute <Zl in© a = ~ x ~c 4 2 I z . 9 8' a = . , 8a = 9c 1 AMPLE1 ,12 p:q =4 :5 Work out p + 2q: 4q, giving your answer in the simplest form. SOLUTION Think in terms of parts. pis4parts,qis5parts. p+ 2qis4+ 2 x 5 = 14 parts 4qis20parts p+2q:4q= 14:20 =7: 10 (ii JOo/oofb !iii yo/oof450 [iiil co/oofd 2 60%ofp =40 %ofq Work out pas a percentage of q. 3 Write expressions for the following, giving your answers in the simplest form. [ii a increased by 20% (iii b increased by 5% [iii] kdecreased by 35% [iv] m decreased by 2% 4 a increased by 80% is equal to b increased by 50%. Show that !!_ = 1.2. 5 p increased by 25% is equal to q decreased by 25%. Work out pas a percentage of q. 6 x:y = 2:3andy:z =4 :9 Workout x: y: z, giving your answer in the simplest form. 7 a:b = S:2 (ii Writeaintermsofb. [iii Workout 2a + b: b,givingyour answer in the simplest form. [iii] Work out 7a - Sb: 4a, giving your answer in the simplest form. 8 m:n = 3:8andris20%ofn. Workout m: r. 11

I
Substitute <Zl in© a = ~ x ~c
4 2 . I z
a = 8' 9 .
,
8a = 9c
1
AMPLE1 ,12 p:q =4 :5
Work out p + 2q: 4q, giving your answer in the simplest form.
SOLUTION
Think in terms of parts.
pis4parts,qis5parts.
p+ 2qis4+ 2 x 5 = 14 parts
4qis20parts
p+2q:4q= 14:20
=7: 10
(ii JOo/oofb !iii yo/oof450 [iiil co/oofd
2 60%ofp =40 %ofq
Work out pas a percentage of q.
3 Write expressions for the following, giving your answers in the simplest form.
[ii a increased by 20% (iii b increased by 5%
[iii] kdecreased by 35% [iv] m decreased by 2%
4 a increased by 80% is equal to b increased by 50%.
Show that !!_ = 1.2.
5 p increased by 25% is equal to q decreased by 25%.
Work out pas a percentage of q.
6 x:y = 2:3andy:z =4 :9
Workout x: y: z, giving your answer in the simplest form.
7 a:b = S:2
(ii Writeaintermsofb.
[iii Workout 2a + b: b,givingyour answer in the simplest form.
[iii] Work out 7a - Sb: 4a, giving your answer in the simplest form.
8 m:n = 3:8andris20%ofn.
Workout m: r. 11

I Expanding br ackets A quadratic expression is one in which the highest power of its terms is 2. F or example, ,, x 2 +3 2y 2 -3y+5 are all quadratic expressions. IJ Why is (x + 5)(2x-J) a quadratic expression? E XA LE .1 Expand (x + 5)(2x- 3). SO LUTION Expression = x(lx- 3) + 5(2x- 3) = 2x 2 -3x+ IOx - 15 = 2x 2 +7x-15 This method has multiplied everything in the second bracket by each term in the first bracket. An alternative way of setting this out is used in the next example. AMPLE 1.f4 Expand (Jx-5) 2 • SO LUTION Writotho squar.,asth< Expression = (Jx- S)(Jx- 5) productoftwobracketsso)'UU dou'tforgctthemiddlotcrm. Jx-5 Jx-5 9x 2 -15x - ISx+ 25 12

I
Expanding br ackets
A quadratic expression is one in which the highest power of its terms is 2.
F or example,
,, x 2 +3
2y 2 -3y+5
are all quadratic expressions.
IJ Why is (x + 5)(2x-J) a quadratic expression?
E XA LE .1 Expand (x + 5)(2x- 3).
SO LUTION
Expression = x(lx- 3) + 5(2x- 3)
= 2x 2 -3x+ IOx - 15
= 2x 2 +7x-15
This method has multiplied everything in the second bracket by each term in the
first bracket. An alternative way of setting this out is used in the next example.
AMPLE 1.f4 Expand (Jx-5) 2 •
SO LUTION Writotho squar.,asth<
Expression = (Jx- S)(Jx- 5) productoftwobracketsso)'UU dou'tforgctthemiddlotcrm.
Jx-5
Jx-5
9x 2 -15x
- ISx+ 25
12

I EXAMPLE 1.15 Multiply (x 3 + 2x-4) by(x 2 -x+ 3). ~ S OLUTION I z . xl + 2x - 4 . xl - X + 3 , Multiply top line by x 1 x5 +2x 3 -4x 2 Multiply top line by (-x) -x 4 -2x 2 + 4x 1 Multiply top line by 3 3x' + 6x-12 x 5 x4 - +5x 3 -6x 2 + IOx - 12 A Keep a separate column for each power of x. Sometimes it is necessary to leave gaps. In arithmetic zeros are placed in the gaps. For example four thousand and five is written 4005. SOLUTION (a-2) 3 = (a-2)(a-2) 2 Work out (a - 2) 2 (a -2) (a -2) = a(a -2) -2(a -2) = a2-2a-2a+4 = a2-4a +4 Multiply by (a - 2) (a - 2)(a 2 - 4a + 4) = a(a 2 - 4a + 4) -2(a 2 - 4a + 4) = a3-4a 2 +4a -2a 2 + Sa- 8 = a3-6a 2 + 12a -8 [ii (x+S)(x+ 4) [iii (x +3 )(x + I) ,,, [iii) (a +5 )(2a- 1) (i v) (2p +3 )(3p-2) (x+3) 2 [v ii (2x+3)(2x-3) ,,, (v iii (2- 3m)(m - 4) (v ii ii (6 +5t )(2- r) (i x] (4-3x) 2 (m-311) 2 2 [ii Multiply(x 3 -x 2 +x-2)by(x 2 + ! ). [iii Multiply (x4- 2x 2 + 3) by (x2 + 2x- ! ). [iii) Multiply(2x 3 -3x+S)by(x 2 -2x+ ! ). (i v) Multiply (x 5 + x 4 + x 3 + x 2 + x+ I) by (x- 1). 13

I
EXAMPLE 1.15 Multiply (x 3 + 2x-4) by(x 2 -x+ 3). ~
S OLUTION . I z
xl + 2x - 4 .
xl - X + 3 ,
Multiply top line by x 1 x5 +2x 3 -4x 2
Multiply top line by (-x) -x 4 -2x 2 + 4x 1
Multiply top line by 3 3x' + 6x-12
x 5 - x4 +5x 3 -6x 2 + IOx - 12
A Keep a separate column for each power of x. Sometimes it is necessary to leave
gaps. In arithmetic zeros are placed in the gaps. For example four thousand and
five is written 4005.
SOLUTION
(a-2) 3 = (a-2)(a-2) 2
Work out (a - 2) 2
(a -2) (a -2) = a(a -2) -2(a -2)
= a2-2a-2a+4
= a2-4a +4
Multiply by (a - 2)
(a - 2)(a 2 - 4a + 4) = a(a 2 - 4a + 4) -2(a 2 - 4a + 4)
= a3-4a 2 +4a -2a 2 + Sa- 8
= a3-6a 2 + 12a -8
[ii (x+S)(x+ 4) [iii (x +3 )(x + I)
,,, [iii) (a +5 )(2a- 1) (i v) (2p +3 )(3p-2)
(x+3) 2 [v ii (2x+3)(2x-3)
(v iii (2- 3m)(m - 4) ,,, (v ii ii (6 +5t )(2- r)
(i x] (4-3x) 2 (m-311) 2
2 [ii Multiply(x 3 -x 2 +x-2)by(x 2 + ! ).
[iii Multiply (x4- 2x 2 + 3) by (x2 + 2x- ! ).
[iii) Multiply(2x 3 -3x+S)by(x 2 -2x+ ! ).
(i v) Multiply (x 5 + x 4 + x 3 + x 2 + x+ I) by (x- 1).
13

I (vl Expand(x+2)(x-l)(x+3). ~ (v ii Expand ( 2x+ l)(x-2)(x+4). [v iii Expand and simplify (x+ 1 )3. [v iii) Expand and simplify ( p- 5)3. [ix] Expand and simplify (la + 3) 3 • (xl Simplify(2x 2 - i )(x +2 )-4(x +2 )2. (xii Simplify (x 2 - l )(x+ l )-(x 2 + l )( x-1). Manipulating surds Simplifying expressions containing square roots In mathematics there are times when it is helpful to be able to manipulate square roots, rather than just find their values from your calculator. This ensures that you are working with the exact value, not just a rounded version. AMPLE 1. 17 Simplify the following. Jg Ji (ii (ii] Jj X .Jii - ,JIS (iii] [; , ) ( 4+ Jj)(4 - Jj) SOLUTION Js = fi7w [ii = .fi .fi .fi X X = (.fi)' .fi X = 2.fi "6x../3 =~ [iii =~ = (/j)' .fi X = 3.fi ./ii - Fis =~ - ~ (iii] Startbylookingforth c = ,.fi _ ,.fi largestsquarcnumb<r factorsof32andl8 . = .fi J;,J (4 + Jj)(4 - Jj) = 16 - 4Jj + ,J, -( Jj)' 14 = 16 - 3 = 13

I
(vl Expand(x+2)(x-l)(x+3). ~
(v ii Expand ( 2x+ l)(x-2)(x+4).
[v iii Expand and simplify (x+ 1 )3.
[v iii) Expand and simplify ( p- 5)3.
[ix] Expand and simplify (la + 3) 3 •
(xl Simplify(2x 2 - i )(x +2 )-4(x +2 )2.
(xii Simplify (x 2 - l )(x+ l )-(x 2 + l )( x-1).
Manipulating surds
Simplifying expressions containing square roots
In mathematics there are times when it is helpful to be able to manipulate square
roots, rather than just find their values from your calculator. This ensures that
you are working with the exact value, not just a rounded version.
AMPLE 1. 17 Simplify the following.
(ii Jg (ii] Ji X Jj
(iii] .Jii - ,JIS [; , ) ( 4+ Jj)(4 - Jj)
SOLUTION
[ii Js = fi7w
= .fi X .fi X .fi
= (.fi)' X .fi
= 2.fi
[iii "6x../3 =~
=~
= (/j)' X .fi
= 3.fi
(iii] ./ii - Fis =~ - ~ Startbylookingforth c
= ,.fi _ ,.fi largestsquarcnumb<r factorsof32andl8 .
= .fi
J;,J (4 + Jj)(4 - Jj) = 16 - 4Jj + ,J, -( Jj)'
14 = 16 - 3
= 13

I Notice that in this last example there is no square root in the answer. In the next example, all the numbers involve fractions with a square root on the I z bottom line. It is easier to work with numbers if any square roots are only on the top . line. Manipulating a number to that form is called mtionalising the denominator. . , D What is a rational number? 1 AMP LE 1. 18 Simplify the following by rationalising their denominators. i Ji 2 [ii T, [iii ( iii] SO LUTION Jj r,= 2 2 [ii T,XT, ,Jj = (,{3)' 2,/j Ji - 3 - r, Jj [ii i Jj .F, = Ts XTs Jj -JS (-JS)' X = ./15 5 Ji - ll (ii i - ..fs =w Jj .r, Jj = 2./, XT,_ .r, Jj X = 2(./,)' -J6 15 4

I
Notice that in this last example there is no square root in the answer.
In the next example, all the numbers involve fractions with a square root on the I z
bottom line. It is easier to work with numbers if any square roots are only on the top .
line. Manipulating a number to that form is called mtionalising the denominator. .
,
D What is a rational number?
1
AMP LE 1. 18 Simplify the following by rationalising their denominators.
2 i Ji
[ii T, [iii ( iii]
SO LUTION
[ii r,= 2 T,XT, 2 Jj
,Jj
= (,{3)'
2,/j
Ji - 3
[ii i - r, Jj
Jj .F,
= Ts XTs
Jj -JS
= (-JS)' X
./15
5
Ji - ll
(ii i - ..fs
=w Jj
Jj .r,
= 2./, XT,_
Jj .r,
X
= 2(./,)'
-J6
15
4

I . Do not use a calculator for this exercise. 1 Simplify the following. ms ~ Jii ;; [ii (iii ,,, .J', 05 [iii] .Js - fi X [iv] 3,m-,.jj .Ji) [vii 4 (3+ fi) - 3(5- ,.Js ,,, [viii 4Fz - [viii] 5(6 - .jjJ + 2(3+ 4Jj J s..[,1 [ix) 2-ms + ,.Js ,c,fi- ,..{,1 - 2( 3.Ji - 2 Simplify the following. [ii (fi- I )' [iii (4 - F](2+fs) ,,, fi)(fi - [iii] (2 - I) (iv] (fs-..{,J(fs+..{,) (3+ fi)(S - 2.Ji) (vii cfi - ,1<2fi + 3J ,,, [viii * ( JJj - 2)(2Jj - 3) [viii] c.Js-JiJ' c2Jz [ix) (5- J,h)(,./2 - I) + 3>2 J Simplify the following by rationalising their denominators. Ts [ii [iii l 8 [iii] To (iv) ,,, ,J, Ts [vii J% 21 5 [viii T, [viii] ,Js ,,, fis 8 [ix) Ti,; Tiis Rationalising denominators with two terms The next examples show how to rationalise denominators that have two terms. EXAMPLE 1. 19 Simplify the following by rationalising the denominator. 16

I
. Do not use a calculator for this exercise.
1 Simplify the following. ms
~ Jii
;; ,,, [ii .J', 05 (iii
[iii] X [iv] .Js - fi
3,m-,.jj [vii 4 (3+ fi) - 3(5- .Ji)
[viii 4Fz - ,.Js ,,, [viii] 5(6 - .jjJ + 2(3+ 4Jj J
[ix) 2-ms + ,.Js ,c,fi- ,..{,1 - 2( 3.Ji - s..[,1
2 Simplify the following.
[ii (fi- I )' [iii (4 - F](2+fs)
,,, [iii] (2 - fi)(fi - I) (iv] (fs-..{,J(fs+..{,)
(3+ fi)(S - 2.Ji) (vii cfi - ,1<2fi + 3J
[viii * ( JJj - 2)(2Jj - 3) ,,, [viii] c.Js-JiJ'
[ix) (5- J,h)(,./2 - I) c2Jz + 3>2
J Simplify the following by rationalising their denominators. Ts
[ii [iii
l
[iii] To 8 (iv)
,,,
,J,
Ts [vii J%
[viii T, 21 [viii] ,Js 5
,,,
fis 8
[ix) Ti,; Tiis
Rationalising denominators with two terms
The next examples show how to rationalise denominators that have two terms.
EXAMPLE 1. 19 Simplify the following by rationalising the denominator.
16

I SO LUTION __J,/i_ _ 3.fi X 4 + ,[s I z 4- ,/s - 4- ,/s 4+ ,/s . 12.fi +3.fi,[s . , = 16 + 4JS - ,J5 -( J5)' 1 = 12.fi + 3,/10 16 - 5 = 12..fi. + 3.fw 11 . 2Fs J3 EXAMPLE 1. 20 + - 4 . b.jj b . Wnte m the form a + 3 , where a and are mtegers. 3 5 SO LUTION ,.jj _ 4 ,.jj _ 4 3.jj _ 5 3J3 + s = ,J3 + s x 313 - s "'.jj )' - 10.jj - 12.jj + 20 = 9( F3 )' - 1s.jj + 1s.jj - 2s = 1s - 22JJ° + 20 27 - 25 = 38 - 22,h 2 = 19 - 11,{3 1 Do not use a calculator for this exercise. 1 Simplify the following by rationalising the denominator. 2.jj J'i 3,{3 [ii [iii (iii) s + fi. 4 - fi. .jj + 1 2+ fi. J'i - 3 10 + ,h [ iv ] 1 , 1 (v ii ~ ~ T,;T, 17

I
SO LUTION
__J,/i_ _ 3.fi X 4 + ,[s I z
4- ,/s - 4- ,/s 4+ ,/s .
12.fi +3.fi,[s . ,
= 16 + 4JS - ,J5 -( J5)'
= 12.fi + 3,/10 1
16 - 5
= 12..fi. + 3.fw
11
EXAMPLE 1. 20 Wnte . 2Fs J3 + - 4 m . the form a + b.jj 3 , where a and b are mtegers. .
3 5
SO LUTION
,.jj _ 4 ,.jj _ 4 3.jj _ 5
3J3 + s = ,J3 + s x 313 - s
"'.jj )' - 10.jj - 12.jj + 20
= 9( F3 )' - 1s.jj + 1s.jj - 2s
= 1s - 22JJ° + 20
27 - 25
= 38 - 22,h
2
= 19 - 11,{3
1
Do not use a calculator for this exercise.
1 Simplify the following by rationalising the denominator.
2.jj J'i 3,{3
[ii s + fi. [iii 4 - fi. (iii) .jj + 1
2+ fi. J'i - 3 10 + ,h
[ iv ] ~ 1 , 1 ~ (v ii T,;T,
17

I . 3..fi. + 6. W, b . 2 Wnte~mtheforma + 2, whereaand aremtegers. . ,,r, . .Js ,{s d d . J Wnte + m the form c 5 + , where c and are mtegers. 4 9 4 Write~ in the formp + :1.-Jj, where p, qand rare integers. 3+2>./3 r Each chapter in this book ends with KEY POINTS, a summary of the essential ideas that you should have understood in the chapter. Chapter I, however, is fundamental to mathematics and you will need to be confident on all of the techniques covered in this chapter if you are to understand the rest of your course. These are: 1 Simplifying algebraic expressions by • collecting like terms • removing brackets • cancelling by common factors • expressing them as a single fraction. 2 When simplifying expressions involving square roots you should • make the number under the square root sign as small as possible • rationalise the denominator. 18

I
2 Wnte~mtheforma . 3..fi. + 6. + W, 2, whereaand b aremtegers. .
. .Js ,,r, . ,{s d d .
J Wnte + m the form c 5 + , where c and are mtegers.
4 9
4 Write~ 3+2>./3 in the formp + :1.-Jj, r where p, qand rare integers.
Each chapter in this book ends with KEY POINTS, a summary of the essential
ideas that you should have understood in the chapter. Chapter I, however, is
fundamental to mathematics and you will need to be confident on all of the
techniques covered in this chapter if you are to understand the rest of your
course. These are:
1 Simplifying algebraic expressions by
• collecting like terms
• removing brackets
• cancelling by common factors
• expressing them as a single fraction.
2 When simplifying expressions involving square roots you should
• make the number under the square root sign as small as possible
• rationalise the denominator.
18

Algebra II If A equals success, then the formula is A equals Xplus Yplus Z, with X being work, Yplay, and Z keeping your mouth shut. Albert Einstein Factor ising Factorising an expression involves writing the expression as a product using brackets. Simple cases of this were seen in Chapter I in Algebra and number. Here, pairs of brackets will be needed. If you have already learnt another method, and use it quickly and accurately, then you should stick with it. With practice, you may be able to factorise some of these expressions by inspection. EXA LE 2. Factorise xa + xb + ya + yb. SOLUTION First take out a common factor of each pair of terms. xa+xb+ ya+ yb=x(a+ b) + y(a+ b) Next notice that (a+ b) is now a common factor. x(a+ b) + = }{a+ b) (a+ b)(x+ y) In practice this relates to areas of rectangles. +: 1 --- :: x+y_____.. , ' I: I Figure2.1 The idea illustrated in figure 2.1 can be used to factorise a quadratic expression containing three terms, but first you must decide how to split up the middle term. "

Algebra II
If A equals success, then the formula is A equals Xplus Yplus Z, with
X being work, Yplay, and Z keeping your mouth shut.
Albert Einstein
Factor ising
Factorising an expression involves writing the expression as a product using
brackets. Simple cases of this were seen in Chapter I in Algebra and number.
Here, pairs of brackets will be needed. If you have already learnt another method,
and use it quickly and accurately, then you should stick with it. With practice, you
may be able to factorise some of these expressions by inspection.
EXA LE 2. Factorise xa + xb + ya + yb.
SOLUTION
First take out a common factor of each pair of terms.
xa+xb+ ya+ yb=x(a+ b) + y(a+ b)
Next notice that (a+ b) is now a common factor.
x(a+ b) + }{a+ b) = (a+ b)(x+ y)
In practice this relates to areas of rectangles.
+: 1 --- :: , x+y_____.. I: ' I
Figure2.1
The idea illustrated in figure 2.1 can be used to factorise a quadratic expression
containing three terms, but first you must decide how to split up the middle term.
"

I '= ·' ~., == •M =P ~L~ E Factorise x2 + 6x + 8. SO LUTION Splitting the 6x as 4x + 2x gives x2 + 6x + B= x2 + 4x + 2x + 8 = x(x + 4) + 2(x + 4) = (x + 4)(x + 2). -<--- .t +4-------... 4 Figure 2.2 The crucial step is knowing how to split up the middle term. D ls the illustration in figure 2.2 the only possibility? To answer this question, notice that • the numbers4 and 2 haveasumof6, which is the coefficientofx(i.e. the number multiplyingx) inx2 + 6x + 8 • the numbers4 and 2 have a product of8 which is the comtant term in x2 + 6.x + 8. There is on ly one pair of numbers that satisfies both of these conditions. EX AM LE 2.3 Factorisexl- - 7x - 18. SOLUTION Pairs of nwnbers with a product of (- 18) are: I and (- 18) 2 and (- 9) 3 and( - 6) 6and (- 3) 9and (- 2) !Sand (- 1) There is only one pair, 2 and (- 9) with a sum of (- 7), so use these. x2 - 7x - 18 = x2 + 2x - 9x - 18 Notiath c signchange = x(x + 2 )- 9(x + 2) du c tothe - signin = front of the 9. (x + 2) (x - 9) 20

I
'= ·' ~.,
== •M =P ~L~ E Factorise x2 + 6x + 8.
SO LUTION
Splitting the 6x as 4x + 2x gives
x2 + 6x + B= x2 + 4x + 2x + 8
= x(x + 4) + 2(x + 4)
= (x + 4)(x + 2).
-<--- .t +4-------...
4
Figure 2.2
The crucial step is knowing how to split up the middle term.
D ls the illustration in figure 2.2 the only possibility?
To answer this question, notice that
• the numbers4 and 2 haveasumof6, which is the coefficientofx(i.e. the
number multiplyingx) inx2 + 6x + 8
• the numbers4 and 2 have a product of8 which is the comtant term in x2 + 6.x + 8.
There is on ly one pair of numbers that satisfies both of these conditions.
EX AM LE 2.3 Factorisexl- - 7x - 18.
SOLUTION
Pairs of nwnbers with a product of (- 18) are:
I and (- 18)
2 and (- 9)
3 and( - 6)
6and (- 3)
9and (- 2)
!Sand (- 1)
There is only one pair, 2 and (- 9) with a sum of (- 7), so use these.
x2 - 7x - 18 = x2 + 2x - 9x - 18
= x(x + 2 )- 9(x + 2) Notiath du c tothe c signchange - signin
= (x + 2) (x - 9) front of the 9.
20

I 0 Do you get the same factors if the order in which you use the 2 and the (- 9) is reversed so that you write it >2 - 9x + lx - 181 NO TE Since the pair of numbers that you are looking for is unique, you can stop listing products when you find one that has the correct sum. EXAMPLE 2.4 Factorise:'2 - 16. SO LUTION First write >2 - 16 = :'2 + 0x - 16. Pairs of numbers with a product of (- 16) are: 1 and (- 16 ) 2and (- 8) 4 and (- 4), .. (stop here) = x(x + 4) - 4(x + 4) = (x + 4)(x - 4) This is an example of a special case called the difference of two sqriares since you have :'2 - 42 = (x + 4)(x - 4) In general a 2 - b2 =( a + b)(a - b) Most people recognise this when it occurs and write down the answer straight away. D Does:'2 + 16 factorise to (x + 4) (x + 4)? EXA E 2.5 Factorise4:'2 - 9f. SO LUTION 4:'2 - 9/ = (2x) 2 - (3y) 2 = (2x + 3y)(2x - 3y) 21

I
0 Do you get the same factors if the order in which you use the 2 and the (- 9) is
reversed so that you write it >2 - 9x + lx - 181
NO TE Since the pair of numbers that you are looking for is unique, you can stop listing
products when you find one that has the correct sum.
EXAMPLE 2.4 Factorise:'2 - 16.
SO LUTION
First write
>2 - 16 = :'2 + 0x - 16.
Pairs of numbers with a product of (- 16) are:
1 and (- 16 )
2and (- 8)
4 and (- 4), .. (stop here)
= x(x + 4) - 4(x + 4)
= (x + 4)(x - 4)
This is an example of a special case called the difference of two sqriares since
you have
:'2 - 42 = (x + 4)(x - 4)
In general
a 2 - b2 =( a + b)(a - b)
Most people recognise this when it occurs and write down the answer straight away.
D Does:'2 + 16 factorise to (x + 4) (x + 4)?
EXA E 2.5 Factorise4:'2 - 9f.
SO LUTION
4:'2 - 9/ = (2x) 2 - (3y) 2
= (2x + 3y)(2x - 3y)
21

I '= ·• -., Y - == •M = P ~L~ E Factorise fully 36f. SO LUTION The instruction to factorise fully tells you that there is likely to be more than one step involved. Take out the highest common factor of the two terms y' - 36/ = r'(l - 36) Use the difference of two squares /Cl - ,,1 = /(y+6)(y - 6J T he technique for finding how to split the middle term needs modifying for examples where the expression starts with a multiple of x?-. T he difference is that you now multiply the two outside numbers together to give the product you want. EXAMPLE 2.7 Factorise2x2 - llx + 15. SO LUTION Here the sum is (- 11) and the product is2 x 15 = 30. Options are: (- !) and (- 30) (- 5) and (- 6) (- 2) and (- 15) (- 6) and (- 5) .. (repeats) (- 3)and( - IO) = 2x?- - I Ix + 15 2x2 - 5x - 6x + 15 = x(2x - 5) - 3(2x - 5) = (2x - S)(x - 3) EXAMPLE 2. 8 FactoriseJx2 - IOxy - Bf. SOLUTION This expression can be factorised using the same method used in the previous example. Here the sum is (- 10) and the product is3 x - 8 =- 24. Option needed is (- 12 ) and 2. l,. n~ tiwproductmran, tha tonenumbcrispositive 3x2 - 10xy - 8y2 = 3x2 - 12xy + 2xy - 8y2 andtheoth a is negati ve . = Jx(x - 4y) + 2y(x - 4y) = (3x + 2y)(x - 4y ) 22

I
'= ·• -.,
== •M = P ~L~ E Factorise fully Y - 36f.
SO LUTION
The instruction to factorise fully tells you that there is likely to be more than one
step involved.
Take out the highest common factor of the two terms
y' - 36/ = r'(l - 36)
Use the difference of two squares
/Cl - ,,1 = /(y+6)(y - 6J
T he technique for finding how to split the middle term needs modifying for
examples where the expression starts with a multiple of x?-. T he difference is that
you now multiply the two outside numbers together to give the product you want.
EXAMPLE 2.7 Factorise2x2 - llx + 15.
SO LUTION
Here the sum is (- 11) and the product is2 x 15 = 30.
Options are:
(- !) and (- 30) (- 5) and (- 6)
(- 2) and (- 15) (- 6) and (- 5) .. (repeats)
(- 3)and( - IO)
2x?- - I Ix + 15 = 2x2 - 5x - 6x + 15
= x(2x - 5) - 3(2x - 5)
= (2x - S)(x - 3)
EXAMPLE 2. 8 FactoriseJx2 - IOxy - Bf.
SOLUTION
This expression can be factorised using the same method used in the previous
example.
Here the sum is (- 10) and the product is3 x - 8 =- 24.
Option needed is (- 12 ) and 2. l,. n~ tiwproductmran,
3x2 - 10xy - 8y2 = 3x2 - 12xy + 2xy - 8y2 tha andtheoth tonenumbcrispositive a is negati ve .
= Jx(x - 4y) + 2y(x - 4y)
= (3x + 2y)(x - 4y )
22

I , Factorise the following expressions. [ii ab - a c+ db - de ,,. (iii 2xy + 2X + "J' + W !iii] 2pq - 8p - 3rq + 12r (iv) 5 - 5m - 2n + 2nm i 2 Factorise the following expressions. [ii x 2 + 5x + 6 [iii y 2 - 5y + 4 ,,, [iii] m 2 - 8m + l6 [iv) m 2 - 8m + l5 x 2 + 3x - lO [vii a 2 + 20a + 96 ,,, [viii x 2 - x - 6 [viii] y 2 - 16y + 48 [ix) k2 + 10k + 24 k2 - 10k - 24 [xii x 2 + 3xy + 2y 2 [xiii x 2 + 4xy - 5y 2 [xiii] a 2 - ab - 12b 2 [xiv) c 2 - llcd + 24d 2 J Each of these is a difference of two squares. Factorise them. [ii x 2 - 4 (iii a 2 - 25 ,,, [iii] 9 - p' (iv) i' - f t 2 - 64 [vii 4x 2 - I ,,, [viii 4x 2 - 9 [viii] 4x 2-y2 [ix) 16x2 - 25 9a 2 - 4b2 [xii (2a + i )2- a 2 [xiii ( 3x + i )2- (x + 4) 2 [xiii] ( 2p - 3 )2-( p + 1) 2 [xiv) 16 -( Sy - 2 )2 4 Factorise the following expressions. [ii 2x 2 + 5x + 2 [iii 2a 2 + lla - 21 ,,, [iii] 15p 2 + 2p - I [iv) 3x 2 + 8x - 3 Sa 2- 9a - 2 (vii 2p 2 + 5p - 3 [viii 8x 2 ,,, + IOx - 3 [viii] 2a 2 - 3a - 27 [ix) 9x 2 - 30x + 25 4x 2 + 4x - 15 (xii zx 2+ Sxy + 2 y2 (xiii 3x 2 + Sxy - 2y 2 [xiii] 5a 2 - 8ab + 3b 2 [xiv) 6c 2 + Scd - 4d 2 5 Factorise fully the following expressions. [ii x3 - 4x (iii a4 - 16a 2 ,,, [iii] 9f - >' [iv) 2x3 - 2x 4p4 - 9JJ (vii lOOx - K [viii 18 ~- 2, [viii] 8x3 - 50xy 2 (i] Workout9 2 and(a 2 ) 2 • ~ Rnn em ~ rthat (al') • = d'I. [iii Show that a4 - 81 is the difference of two squares. [iii) Factorisefullya 4 - 81. Rearranging formulae Th e circumference of a circle is given by C = 21tr where r is the radius. An equation such as this is often called a formula . 23

I
,
Factorise the following expressions.
[ii ab - a c+ db - de (iii 2xy + 2X + "J' + W ,,.
!iii] 2pq - 8p - 3rq + 12r (iv) 5 - 5m - 2n + 2nm i
2 Factorise the following expressions.
,,, [ii x 2 + 5x + 6 [iii y 2 - 5y + 4
[iii] m 2 - 8m + l6 [iv) m 2 - 8m + l5
x 2 + 3x - lO ,,, [vii a 2 + 20a + 96
[viii x 2 - x - 6 [viii] y 2 - 16y + 48
[ix) k2 + 10k + 24 k2 - 10k - 24
[xii x 2 + 3xy + 2y 2 [xiii x 2 + 4xy - 5y 2
[xiii] a 2 - ab - 12b 2 [xiv) c 2 - llcd + 24d 2
J Each of these is a difference of two squares. Factorise them.
,,, [ii x 2 - 4 (iii a 2 - 25
[iii] 9 - p' (iv) i' - f
t 2 - 64 ,,, [vii 4x 2 - I
[viii 4x 2 - 9 [viii] 4x 2-y2
[ix) 16x2 - 25 9a 2 - 4b2
[xii (2a + i )2- a 2 [xiii ( 3x + i )2- (x + 4) 2
[xiii] ( 2p - 3 )2-( p + 1) 2 [xiv) 16 -( Sy - 2 )2
4 Factorise the following expressions.
,,, [ii 2x 2 + 5x + 2 [iii 2a 2 + lla - 21
[iii] 15p 2 + 2p - I [iv) 3x 2 + 8x - 3
Sa 2- 9a - 2 (vii 2p 2 + 5p - 3
[viii 8x 2 + IOx - 3 ,,, [viii] 2a 2 - 3a - 27
[ix) 9x 2 - 30x + 25 4x 2 + 4x - 15
(xii zx 2+ Sxy + 2 y2 (xiii 3x 2 + Sxy - 2y 2
[xiii] 5a 2 - 8ab + 3b 2 [xiv) 6c 2 + Scd - 4d 2
5 Factorise fully the following expressions.
,,, [ii x3 - 4x (iii a4 - 16a 2
[iii] 9f - >' [iv) 2x3 - 2x
4p4 - 9JJ (vii lOOx - K
[viii 18 ~- 2, [viii] 8x3 - 50xy 2
(i] Workout9 2 and(a 2 ) 2 • ~ Rnn em ~ rthat (al') • = d'I.
[iii Show that a4 - 81 is the difference of two squares.
[iii) Factorisefullya 4 - 81.
Rearranging formulae
Th e circumference of a circle is given by
C = 21tr
where r is the radius. An equation such as this is often called a formula . 23

I D C is called the subject of the formula. Explain what this means. In some cases, you want to calculate rdirectlyfrom C. You want rto be the subject of the formula. EXAMPLE 2.9 Make rthe subject ofC = 21tr. SO LUTION ~ = Divide both sides by 21t => r r=£ => 2n A Notice how the new subject should be on its own on the left-hand side of the new formula. EXA LE 2.1 Make x the subject of this formula. h= ~ ( x' + y' ) SO LUTION Square both sides h' = x' + y f Subtract from both sides => h 2- f = x2 Lead with the :.? term x' = h' - y' Take the square root of both sides => x = ±~(h' - y' ) 0 What would you do with the ± sign in the case where h is the hypotenuse of a right-angled triangle with x and y as the other two sides? EX AMPLE 2. 11 Make a the subject of this formula. SO LUTION Subtract u from both sides => v - u Divide both sides by t => - = a t V - 11 Write the answer with a a = - t on the left-hand side "

I
D C is called the subject of the formula. Explain what this means.
In some cases, you want to calculate rdirectlyfrom C. You want rto be the
subject of the formula.
EXAMPLE 2.9 Make rthe subject ofC = 21tr.
SO LUTION
Divide both sides by 21t => ~ = r
=> r=£
2n
A Notice how the new subject should be on its own on the left-hand side of the
new formula.
EXA LE 2.1 Make x the subject of this formula.
h= ~ ( x' + y' )
SO LUTION
Square both sides h' = x' + y
Subtract f from both sides => h 2- f = x2
Lead with the :.? term x' = h' - y'
Take the square root of both sides => x = ±~(h' - y' )
0 What would you do with the ± sign in the case where h is the hypotenuse of a
right-angled triangle with x and y as the other two sides?
EX AMPLE 2. 11 Make a the subject of this formula.
SO LUTION
Subtract u from both sides =>
Divide both sides by t => - v - u = a
t
Write the answer with a a = - V - 11
on the left-hand side t
"

I Ol!D:Elf.llllll In this exercise all the equations refer to real sifllations. How many of them can you recognise? 1 Make [i) u liil a thesubjectofv = u + ar. = { 2 Make b the subject of A bh. 3 Make I the subject of P = 2(1 + b). 4 Make rthe subject of A = xr2. 5 MakecthesubjectofA = {(b + c)h. 6 Make h the subject of A = rrr2 + 2xrh. 3r-. = 7 Make I the subject of T 8 Make [i) u [ii) a thesubjectofs = ut + fat 2 • 9 Makexthe subjectofv2 = w2(al - .x2). The following examples show how to rearrange a formula when the letter that is to be the subject appears more than once. "'-" XAo, M ,c PL "' E "2. ,,. 12 ._.. Make t the subject of this formula. at = 3(t + 2) SO LUTION Expand the brackets at = 3t + 6 Collect all the terms in ton one side => at - 3t = 6 Factorise => t (a - 3 )= 6 ~ = - 6 Th e br a chts ar~ notn cc, dcd Dividebothsidesby (a - J ) ~ t inthe denominator. " - 3 '= "' " "' ' "' "" '' '-" "'-"' Make x the subject of this formula. = T+"3x" x + 2 y SO LUTION Multiply both sides by ( I + 3x) -=> y(l + 3x) = x + 2 Expand the brackets y + 3xy = x + 2 Collect all the terms in x on one side with all the other terms on the other side 3xy - x = 2 - y Factorise ~ x ( 3y - 1) = 2 - y = Divide both sides by (3y - I) 2.=..l._ X 3y - 1 25

I
Ol!D:Elf.llllll In this exercise all the equations refer to real sifllations. How many of them can
you recognise?
1 Make [i) u liil a thesubjectofv = u + ar.
2 Make b the subject of A = { bh.
3 Make I the subject of P = 2(1 + b).
4 Make rthe subject of A = xr2.
5 MakecthesubjectofA = {(b + c)h.
6 Make h the subject of A = rrr2 + 2xrh.
7 Make I the subject of T = 3r-.
8 Make [i) u [ii) a thesubjectofs = ut + fat 2 •
9 Makexthe subjectofv2 = w2(al - .x2).
The following examples show how to rearrange a formula when the letter that is
to be the subject appears more than once.
"'-" XAo, M ,c PL "' E "2. ,,. 12 ._.. Make t the subject of this formula.
at = 3(t + 2)
SO LUTION
Expand the brackets at = 3t + 6
Collect all the terms in ton one side => at - 3t = 6
Factorise => t (a - 3 )= 6 ~
Dividebothsidesby (a - J ) = - 6 Th e br a chts ar~ notn cc, dcd
~ t " - 3 inthe denominator.
'= "' " "' ' "' "" '' '-" "'-"'
Make x the subject of this formula.
y = T+"3x" x + 2
SO LUTION
Multiply both sides by ( I + 3x) -=> y(l + 3x) = x + 2
Expand the brackets y + 3xy = x + 2
Collect all the terms in x on one side with
all the other terms on the other side 3xy - x = 2 - y
Factorise ~ x ( 3y - 1) = 2 - y
Divide both sides by (3y - I) X = 2.=..l._
3y - 1 25

I Gl!Dilmf.11. 1 Makemthesubjectof3m = x(m + 2). 2 Makeythesubjectof5y - 2x = xy. 3 Make bthe subject of 4(a + b )= 3(a - b). 4 Make h the subject of S = 2n:r2 + 2rrrh. = ~. x + I 5 Make x the subject of y 6 Make cthe subject of d(2 + c) = I - Jc. 7 [ii Maketthesubjectof x = t~J. (iii Hence, or otherwise, work out the value of twhen x = 3. = 2 8 [ii Make p the subject of r Jp + . 2p + 3 (iii Hence, or otherwise, work out the value of pwhen r= - 1. fMlkliifij ti1 Show that (x + 3) 2 = x2 + 6x + 9. (iii Hence, make xthe subject of y = x2 + 6x + 9. fMlkliifij ti1 Show that (x - 5) 2 + 4 = :l- - IOx + 29. [iii Hence, make xthe subject of p = x2 - I Ox + 29. Simplifying algebraic fractions 0 What is a fraction in arithmetic? What about in algebra? Fractions in algebra obey the same rules as fractions in arithmetic. These cover two pairs of operations: x and +, and + and - . D When can you cancel fractions in arithmetic? What about in algebra? What is a factor in arithmetic? What about in algebra? 26

I
Gl!Dilmf.11. 1 Makemthesubjectof3m = x(m + 2).
2 Makeythesubjectof5y - 2x = xy.
3 Make bthe subject of 4(a + b )= 3(a - b).
4 Make h the subject of S = 2n:r2 + 2rrrh.
5 Make x the subject of y = ~. x + I
6 Make cthe subject of d(2 + c) = I - Jc.
7 [ii Maketthesubjectof x = t~J.
(iii Hence, or otherwise, work out the value of twhen x = 3.
8 [ii Make p the subject of r = Jp + 2 .
2p + 3
(iii Hence, or otherwise, work out the value of pwhen r= - 1.
fMlkliifij ti1 Show that (x + 3) 2 = x2 + 6x + 9.
(iii Hence, make xthe subject of y = x2 + 6x + 9.
fMlkliifij ti1 Show that (x - 5) 2 + 4 = :l- - IOx + 29.
[iii Hence, make xthe subject of p = x2 - I Ox + 29.
Simplifying algebraic fractions
0 What is a fraction in arithmetic?
What about in algebra?
Fractions in algebra obey the same rules as fractions in arithmetic.
These cover two pairs of operations: x and +, and + and - .
D When can you cancel fractions in arithmetic?
What about in algebra?
What is a factor in arithmetic?
What about in algebra?
26

I EXAM LE2,1 Simplify the following. [ii_!! [iii 2x + al - a - 6 2 [iii) 24 3x + 3 al - Sa + lS SOLUTION 1 ,6 [i] 18 X 3 3 M =~ =4 I [iii 2x + y:.-+1} 2 2 2 3x +3= 3J..l:--t'1) =J I A O Look at this calculation for [iii. Why is it wrong? a 2 - a - 6 ~(a + 2) a + 2 [iii) a 2 - 8a + 15 = J9.--f} (a - S )= --;;--=s A O Look at this answer to [iii]. Why is it wrong? EXAMPLE 2.15 Simplify the following. [ii ~x_!_ (iii ~_,__!_ 3 [iii] a'b x~ [iv)~ + ~ 3 14 4 · 16 2c 9ab n + 1 nl - I SOLUTION [i] ~X_!_ = ~ = ~ 3 14 1 X 7 7 (iii ~ + _!_ = = ± ~ x ~ 4 16 4 9 3 (iii) 3albx~ = 2a,2 2c 9ab 3 1 1 [iv) ~ + ~ - ~ ( 2n - 3) x..(P--f1J( n - 1) .(p---1'1) n + l 11 2 - 1 - ~ I I = (2n - 3)(n - 1) 27

I
EXAM LE2,1 Simplify the following.
[ii_!! [iii 2x + 2 [iii) al - a - 6
24 3x + 3 al - Sa + lS
SOLUTION
1 ,6
[i] M =~ 18 X 3 =4 3
I
[iii 3x 2x + +3= 2 2 3J..l:--t'1) y:.-+1} =J 2
I
A O Look at this calculation for [iii.
Why is it wrong?
a 2 - a - 6 ~(a + 2) a + 2
[iii) a 2 - 8a + 15 = J9.--f} (a - S )= --;;--=s
A O Look at this answer to [iii].
Why is it wrong?
EXAMPLE 2.15 Simplify the following.
[ii ~x_!_ (iii ~_,__!_ [iii] 3 a'b x~ [iv)~ + ~
3 14 4 · 16 2c 9ab n + 1 nl - I
SOLUTION
[i] ~X_!_ = ~ = ~
3 14 1 X 7 7
(iii ~ + _!_ = ~ x ~ = ±
4 16 4 9 3
(iii) 3albx~ = 2a,2
2c 9ab 3 1 1
[iv) ~ + ~ - ~ .(p---1'1) ( 2n - 3) x..(P--f1J( n - 1)
n + l 11 2 - 1 - ~
= I I
(2n - 3)(n - 1) 27

I D A Look at this answer to [iv]. ~--- - -----.__, Why is it wrong? D What is a common denominator? To add or subtract fractions it is first necessary to find a common denominator. EXAMPLE 2.16 Simplify the following. 2 3 5x X (ii - + - (iii - + - 3 4 6 4 2 5 " 2 [iii]~ + ~ [iv)~ - ~ SOLUTION til 3- + I = ~ + ~ = !:_ 3 4 12 12 12 [iii ~ + ~ = ~ + ~ = ~ 6 4 12 12 12 A Notice that the common denominator is the lowest common multiple of the original denominators. 2 5 [iii] - - + - - = ~ + ~ (x + I) (x - I) (x + l )(x - I) (x + l)(x - I) 2x - 2 + 5x + 5 (x + l )( x - 1) 7x + 3 (x + l)(x - 1) a 2 a 2 (iv] ~ - ~ = (a - l )( a + l )- ~ a l ( a-1 ) (a - l)( a + I) - (a - !)( a + I) a - 2a + 2 (a - l )( a + I) 2 - , 2B (a - l )( a + l )

I
D A Look at this answer to [iv]. ~--- - -----.__,
Why is it wrong?
D What is a common denominator?
To add or subtract fractions it is first necessary to find a common denominator.
EXAMPLE 2.16 Simplify the following.
(ii - + - 2 3 (iii - 5x + - X
3 4 6 4
[iii]~ 2 + ~ 5 [iv)~ " - ~ 2
SOLUTION
til 3- + I = ~ + ~ = !:_
3 4 12 12 12
[iii ~ + ~ = ~ + ~ = ~
6 4 12 12 12
A Notice that the common denominator is the lowest common multiple of
the original denominators.
[iii] - 2 - + - 5 - = ~ + ~
(x + I) (x - I) (x + l )(x - I) (x + l)(x - I)
2x - 2 + 5x + 5
(x + l )( x - 1)
7x + 3
(x + l)(x - 1)
a 2 a 2
(iv] ~ - ~ = (a - l )( a + l )- ~
a l ( a-1 )
(a - l)( a + I) - (a - !)( a + I)
a - 2a + 2
(a - l )( a + I)
2B 2 - ,
(a - l )( a + l )

I II What is the lowest common multiple of the following? (al 6and4 [bi (x2 - I)and(x2 - 4x + 3) [ii 2(x + 3) 4x - 8 (iii 4x + 12 (x - 2)(x + 8) ;> 6xzy1 !iii] ~~x _+ (iv) ~ 2p 4abi [vl 6p - 2p2 (vii IOa 3 b x 2 + [viii x2 - 4x + 3 xy (viii] x y2 2x - 6 2 _ , + 2 3x 2 + 15x [ix) 1,1 ai - a - 6 !Ox + 2x 2 9x 2 - I Jx 2 + Jxy [xii [xiii xy + y2 9x + 3 6 6 2 Simplify the following. - 3a x - b 1 (ii) xy - y2 X ____!__ [ii bl 6a y x - y [iii] ~ + 4x2 _ 4 [iv) 3a 2 + a - 2 + 6a 2 - a - 2 2x x 2 2 8a + 4 [vl x 2 - 4x + 4 x ~ 2x - 1 2x 2 - x - l [vii x 2 - 2x x 2 - 4 x + 1 x1 + Jx + 2 4 2 [viii P + llx~ (viii] ~7x2 - 6x + 9 p - 3 p 2 + 3 x + l x 2 + x - 2 J Simplify the following. [ii 3a a 5 4 [iii 5 4 3, " 2 1 4 3 [iii]~ - ~ (iv) p - 2 lp + I 2 3 1,1 -- + -- [vii ~ + --2'.'_ a2 + a a 2 - a x - y y - x p/_ [viii - p~I [viii]!!....=!?_ + ~ 1 a + b a - b

I
II What is the lowest common multiple of the following?
(al 6and4
[bi (x2 - I)and(x2 - 4x + 3)
[ii 2(x + 3) (iii 4x - 8
4x + 12 (x - 2)(x + 8)
;> 6xzy1
!iii] ~~x _+ (iv) ~
2p 4abi
[vl 6p - 2p2 (vii IOa 3 b
[viii x2 - 4x + 3 (viii] x x 2 + xy y2
2x - 6 2 _
[ix) , + 2 1,1 3x 2 + 15x
ai - a - 6 !Ox + 2x 2
[xii 9x 2 - I Jx 2 + Jxy
9x + 3 [xiii 6 xy + 6 y2
2 Simplify the following.
[ii - 3a x - b 1 (ii) xy - y2 X ____!__
bl 6a y x - y
[iii] ~ + 4x2 _ 4 [iv) 3a 2 + a - 2 + 6a 2 - a - 2
2x x 2 2 8a + 4
[vl x 2 - 4x + 4 x ~ [vii 2x - 1 2x 2 - x - l
x 2 - 2x x 2 - 4 x + 1 x1 + Jx + 2
[viii 4 P 2 + llx~ (viii] ~7x2 - 6x + 9
p - 3 p 2 + 3 x + l x 2 + x - 2
J Simplify the following.
[ii 3a a [iii 5 4
5 4 3, "
2 1 4 3
[iii]~ - ~ (iv) p - 2 lp + I
1,1 -- 2 + -- 3 [vii ~ + --2'.'_
a2 + a a 2 - a x - y y - x
p/_
[viii 1 - p~I [viii]!!....=!?_ + ~ a + b a - b

I D What is the difference between simplifying fractions and 50/vingan equation involving fractions? Solving equations involving fractions When you solved the equations in Chapter I, you used mathematical operations such as +,-, x and 7 to find the value of the variable. The same principle applies for solving equations involving fractions. EX AM LE 2. 1 Solve the following. x + 2 x - 6 15 =~ SO LUTION The LCM of 6 and 2 is 6, so multiply by 6. 1,e'x~ = 3tx (x - 6) 1 { A O When you multiply a fraction, you only multiply its numerator (top line). Why? D H ow does this help? x + 2 = 3x - 18 20 = 2x x = 10 EXAMPLE 2. 18 Solve the following. x + 2 - + 3 = - X 6 5 SO LUTION The LCM of 6 and 5 is 30, so multiply by 30. S (x + 2) 6 )0 X x ~ + 30x3 = )0 xJi "

I
D What is the difference between simplifying fractions and 50/vingan equation
involving fractions?
Solving equations involving fractions
When you solved the equations in Chapter I, you used mathematical operations
such as +,-, x and 7 to find the value of the variable. The same principle applies
for solving equations involving fractions.
EX AM LE 2. 1 Solve the following.
15 x + 2 =~ x - 6
SO LUTION
The LCM of 6 and 2 is 6, so multiply by 6.
1,e'x~ 1 = 3tx (x - { 6)
A O When you multiply a fraction, you only multiply its numerator
(top line). Why?
D H ow does this help?
x + 2 = 3x - 18
20 = 2x
x = 10
EXAMPLE 2. 18 Solve the following.
- x + 2 + 3 = - X
6 5
SO LUTION
The LCM of 6 and 5 is 30, so multiply by 30.
)0 S x ~ (x + 2) + 30x3 = )0 6 xJi X
"

I A O Look at this version of the first stage of the solution. 0 o3 0x;; Why is itwrnngl ,.---- ij , -- -) I ,("'J,)+ s 5x + l0 + 90 = 6x x = 100 EXAMPLE 2. 19 Solve the following. 5 la I ~ - ~ =2 SOLUTION First factorise (a2 - i ) as (a + l )(a - 1). 0 H ow does this help? 5 2a I =2 ~ - (a + l)( a - 1) Multiplyby2 (a + I)(a - 1) I 5 I I 2a 2~ (o - l) x ~ - 2~~ x ~~ 1 = i (a + l )( a - l) x * IO(a - 1 )- 4a =( a + l )( a - 1) 10a - l0 - 4a = a2 - I 0 = a 2 - 6a + 9 0 =( a - 3)( a - 3) a = 3 (repeated root) = (ii x - ~ ~ 2 3 (iii - - - = 2 5 3 0 4o 1 2 3x + 2 x - 1 [iii] - = 3 - - [iv) -- - - = 3 X x + l 2 5 2 1 1 1,1 -- + - = - 2 5 [vii 3x - I x + S 2 - - -- = 0 0 la - I 1 13 [viii - + p + I = - 1 2x [viii) ! + - = - p 3 x - 1 x + l 31

I
A O Look at this version of the first stage of the solution. 0 o3 0x;;
Why is itwrnngl ,.---- ij , ,("'J,)+ -- -) I
s
5x + l0 + 90 = 6x
x = 100
EXAMPLE 2. 19 Solve the following.
~ 5 - ~ la =2 I
SOLUTION
First factorise (a2 - i ) as (a + l )(a - 1).
0 H ow does this help?
5 2a =2 I
~ - (a + l)( a - 1)
Multiplyby2 (a + I)(a - 1)
I 5 I I 2a
2~ (o - l) x ~ - 2~~ x ~~
1
= i (a + l )( a - l) x *
IO(a - 1 )- 4a =( a + l )( a - 1)
10a - l0 - 4a = a2 - I
0 = a 2 - 6a + 9
0 =( a - 3)( a - 3)
a = 3 (repeated root)
(ii x - ~ = ~ (iii - 2 - - 3 = 2
5 3 0 4o
[iii] - 1 = 3 - - 2 [iv) -- 3x + 2 - - x - 1 = 3
X x + l 2 5
1,1 -- 2 + - 1 = - 1 [vii 2 5
3x - I x + S 2 - - -- = 0
0 la - I
1 13
[viii - p + p + I = - 3 [viii) ! + - x - 1 1 = - x + l 2x
31

I 2 A formula used in physics is 7= I I I - + - where f is the focal length of a mirror, u is the distance of the object from the mirror, and vis the distance of the image from the mirror. For a mirror with focal length 20cm, find the distance of the object from the mirror when the image is twice as far away from the mirror as is the object. Quadratic identities When considering a quadratic expression it will sometimes be useful to write it using (x- a)2, where a is a constant. Some uses of this approach will be seen later in sections on quadratic equations and quadratic graphs. NO TE The identity symbol( =) is used when two expressions are equal for all values of x. For example x + x = lx. EXAMPLE 2.2 0 Work out the values of p and qsuch that:? - 6x + 2 = (x- p) 2 + q. SO LUTION Expand the bracket x?--6x+2= x?--2px + p 2 + q ~~~~-~ Equate coefficients of x - 6 =- 2p (x - p)' (x - p )( x - p) = x' - px - px+p' 3 = p = x' - 2 px+p' Equate constants 2 = p2+q 2 = 9 + q EquatecodJkirntsof x rneam makingequalchenurnberofxon - 7 = q eachsideoftheidrnti ty. p = 3 and q =- 7 EXAM PLE 2.21 Work out the values of a, band cs uch that 2:?- + bx + 5 = a(x - 3) 2 + c. SO LUTION Expand the bracket 2x?- + bx + 5 = a(x?- - 6x + 9) + c : ax?- - 6ax + 9a + c Equate coefficients of x?- 2 = a Equate coefficients of x b =- 6a b =- 12 Equate constants 5 = 9a + c 5 = IS + c - 13 =c 32 a= 2, b =- 12 and c =- 13.

I
2 A formula used in physics is
7= I - + - I I
where f is the focal length of a mirror, u is the distance of the object from the
mirror, and vis the distance of the image from the mirror.
For a mirror with focal length 20cm, find the distance of the object from the
mirror when the image is twice as far away from the mirror as is the object.
Quadratic identities
When considering a quadratic expression it will sometimes be useful to write it
using (x- a)2, where a is a constant. Some uses of this approach will be seen later
in sections on quadratic equations and quadratic graphs.
NO TE The identity symbol( =) is used when two expressions are equal for all values of x.
For example x + x = lx.
EXAMPLE 2.2 0 Work out the values of p and qsuch that:? - 6x + 2 = (x- p) 2 + q.
SO LUTION
Expand the bracket x?--6x+2= x?--2px + p 2 + q ~~~~-~
Equate coefficients of x - 6 =- 2p (x - p)' = (x - p x' - px - )( x - px+p' p)
3 = p = x' - 2 px+p'
Equate constants 2 = p2+q
2 = 9 + q makingequalchenurnberofxon EquatecodJkirntsof x rneam
- 7 = q eachsideoftheidrnti ty.
p = 3 and q =- 7
EXAM PLE 2.21 Work out the values of a, band cs uch that 2:?- + bx + 5 = a(x - 3) 2 + c.
SO LUTION
Expand the bracket 2x?- + bx + 5 = a(x?- - 6x + 9) + c
: ax?- - 6ax + 9a + c
Equate coefficients of x?- 2 = a
Equate coefficients of x b =- 6a
b =- 12
Equate constants 5 = 9a + c
5 = IS + c
- 13 =c
32 a= 2, b =- 12 and c =- 13.

I AMPLEZ.22 Work out the values of a, band csuch that 3x2+ 5x - I : a(x + b) 2 + c. SOLUTION Expand the bracket 3x2 + Sx - 1 = a( x2 + 2bx + Er )+ c = nx2 + 2abx + ab2 + c Equate coefficients of x1 3 = a Equate coefficients of x 5 = 2ab 5 = 6b 5 6 = b Equate constants = - J ab 2 + C -1=3 x( i r+ , 25 - 1 = 3x - + c 36 25 - 1 = - + c 12 _ E_ = C 12 a = 3, b ~ = and C =-~ 6 12 1 Workout the values of a and bsuch that:?- + 8x + 10 = (x + a) 2 + b. 2 Workout the values of cand dsuch that x2 - ex + 7 = (x - 1) 2 + d. 3 Work out the values of p and q such that x2 - 12x - 4 = (x - p )2 + q. 4 Work out the values of a and b such that x2 + Sx - 2 = (x + a ) 2 + b. 5 Work out the values of p and q such that 5 + 4x - x2 : p - (x - q )2 . 6 Workout the values of cand dsuch that 2 - x - x2 = c- (x + d )2 . 7 Workout the values of a, band csuch that 2x2 + bx + S = a(x + 2)2+ c. B Workout the values of a, band c such that 5x2 + 30x + 10 = a(x + b)2+ c. 9 Workout thevaluesofp, qand rsuch that 3x2 - 12x + 14 = p(.x + q) 2 + r. 10 Workout the values of a, band c such that 3x2 - bx + I = a(x - 4)2+ c. 11 Workout the values of a, band csuch that 6 + bx - 2x2 = c- a(x - 1 )2 . 12 Workout thevaluesofp, qand rsuch that 5 - 12x - 2x2 = p - q(x + r )2. 13 [ii Workout the values of a and bsuch

AQA Certificate in Further Mathematics