Cambridge International AS & A Level Mathematics Pure Mathematics 1 PDF

Cambridge International A and AS Level Mathematics Pure Mathematics 1 Sophie Goldie Series Editor: Roger Porkess Questions from the Cambridge International Examinations A & AS level Mathematics papers are reproduced by permission of University of Cambridge International Examinations. Questions f...

Cambridge International A and AS Level Mathematics Pure Mathematics 1 Sophie Goldie Series Editor: Roger Porkess Questions from the Cambridge International Examinations A & AS level Mathematics papers are reproduced by permission of University of Cambridge International Examinations. Questions from the MEI A & AS level Mathematics papers are reproduced by permission of OCR. We are grateful to the following companies, institutions and individuals you have given permission to reproduce photographs in this book. C page 106, © Jack Sullivan / Alamy; page 167, © RTimages / Fotolia; page 254, © Hunta / Fotolia; page 258, © Olga Iermolaieva / Fotolia Every effort has been made to trace and acknowledge ownership of copyright. The publishers will be glad to make suitable arrangements with any copyright holders whom it has not been possible to contact. Hachette UK’s policy is to use papers that are natural, renewable and recyclable products and made from wood grown in sustainable forests. The logging and manufacturing processes are expected to conform to the environmental regulations of the country of origin. Orders: please contact Bookpoint Ltd, 130 Milton Park, Abingdon, Oxon OX14 4SB. Telephone: (44) 01235 827720. Fax: (44) 01235 400454. Lines are open 9.00–5.00, Monday to Saturday, with a 24-hour message answering service. Visit our website at www.hoddereducation.co.uk Much of the material in this book was published originally as part of the MEI Structured Mathematics series. It has been carefully adapted for the Cambridge International A & AS level Mathematics syllabus. The original MEI author team for Pure Mathematics comprised Catherine Berry, Bob Francis, C Val Hanrahan, Terry Heard, David Martin, Jean Matthews, Bernard Murphy, Roger Porkess and Peter Secker. © MEI, 2012 First published in 2012 by Hodder Education, a Hachette UK company, 338 Euston Road London NW1 3BH Impression number 54321 Year 2016 2015 2014 2013 2012 All rights reserved. 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 means, electronic or mechanical, including photocopying and recording, or held within any information storage and retrieval system, without permission in writing from the publisher or under licence from the Copyright Licensing Agency Limited. Further details of such licences (for reprographic reproduction) may be obtained from the Copyright Licensing Agency Limited, Saffron House, 6–10 Kirby Street, London EC1N 8TS. C Cover photo by © Joy Fera / Fotolia Illustrations by Pantek Media, Maidstone, Kent Typeset in 10.5pt Minion by Pantek Media, Maidstone, Kent Printed in Dubai A catalogue record for this title is available from the British Library ISBN 978 1444 14644 8 This eBook does not include the ancillary media that was packaged with the printed version of the book.  Contents Key to symbols in this book vi Introduction vii The Cambridge A & AS Level Mathematics 9709 syllabus viii Chapter 1 Algebra 1 Background algebra 1 Linear equations 6 Changing the subject of a formula 10 Quadratic equations 12 Solving quadratic equations 17 Equations that cannot be factorised 20 The graphs of quadratic functions 22 The quadratic formula 25 Simultaneous equations 29 Inequalities 34 Chapter 2 Co-ordinate geometry 38 Co-ordinates 38 Plotting, sketching and drawing 39 The gradient of a line 39 The distance between two points 41 The mid-point of a line joining two points 42 The equation of a straight line 46 Finding the equation of a line 49 The intersection of two lines 56 Drawing curves 63 The intersection of a line and a curve 70 Chapter 3 Sequences and series 75 Definitions and notation 76 Arithmetic progressions 77 Geometric progressions 84 Binomial expansions 95 iii  Chapter 4 Functions 106 C The language of functions 106 Composite functions 112 Inverse functions 115 Chapter 5 Differentiation 123 The gradient of a curve 123 Finding the gradient of a curve 124 Finding the gradient from first principles 126 Differentiating by using standard results 131 Using differentiation 134 Tangents and normals 140 Maximum and minimum points 146 Increasing and decreasing functions 150 Points of inflection 153 The second derivative 154 Applications 160 The chain rule 167 Chapter 6 Integration 173 Reversing differentiation 173 Finding the area under a curve 179 Area as the limit of a sum 182 Areas below the x axis 193 The area between two curves 197 The area between a curve and the y axis 202 The reverse chain rule 203 Improper integrals 206 Finding volumes by integration 208 Chapter 7 Trigonometry 216 Trigonometry background 216 Trigonometrical functions 217 Trigonometrical functions for angles of any size 222 The sine and cosine graphs 226 The tangent graph 228 Solving equations using graphs of trigonometrical functions 229 Circular measure 235 The length of an arc of a circle 239 The area of a sector of a circle 239 iv Other trigonometrical functions 244 Chapter 8 Vectors 254 Vectors in two dimensions 254 Vectors in three dimensions 258 Vector calculations 262 The angle between two vectors 271 Answers 280 Index 310 v  Key to symbols in this book ? This symbol means that you want to discuss a point with your teacher. If you are working on your own there are answers in the back of the book. It is important, however, that you have a go at answering the questions before looking up the answers if you are to understand the mathematics fully. This symbol invites you to join in a discussion about proof. The answers to these questions are given in the back of the book. ! This is a warning sign. It is used where a common mistake, misunderstanding or tricky point is being described. This is the ICT icon. It indicates where you could use a graphic calculator or a computer. Graphical calculators and computers are not permitted in any of the examinations for the Cambridge International A & AS Level Mathematics 9709 syllabus, however, so these activities are optional. This symbol and a dotted line down the right-hand side of the page indicates material that you are likely to have met before. You need to be familiar with the material before you move on to develop it further. This symbol and a dotted line down the right-hand side of the page indicates material which is beyond the syllabus for the unit but which is included for completeness. vi Introduction This is the first of a series of books for the University of Cambridge International Examinations syllabus for Cambridge International A & AS Level Mathematics 9709. The eight chapters of this book cover the pure mathematics in AS level. The series also contains a more advanced book for pure mathematics and one each for mechanics and statistics. These books are based on the highly successful series for the Mathematics in Education and Industry (MEI) syllabus in the UK but they have been redesigned for Cambridge users; where appropriate new material has been written and the exercises contain many past Cambridge examination questions. An overview of the units making up the Cambridge International A & AS Level Mathematics 9709 syllabus is given in the diagram on the next page. Throughout the series the emphasis is on understanding the mathematics as well as routine calculations. The various exercises provide plenty of scope for practising basic techniques; they also contain many typical examination questions. An important feature of this series is the electronic support. There is an accompanying disc containing two types of Personal Tutor presentation: examination-style questions, in which the solutions are written out, step by step, with an accompanying verbal explanation, and test yourself questions; these are multiple-choice with explanations of the mistakes that lead to the wrong answers as well as full solutions for the correct ones. In addition, extensive online support is available via the MEI website, www.mei.org.uk. The books are written on the assumption that students have covered and understood the work in the Cambridge IGCSE syllabus. However, some of the early material is designed to provide an overlap and this is designated ‘Background’. There are also places where the books show how the ideas can be taken further or where fundamental underpinning work is explored and such work is marked as ‘Extension’. The original MEI author team would like to thank Sophie Goldie who has carried out the extensive task of presenting their work in a suitable form for Cambridge International students and for her many original contributions. They would also like to thank Cambridge International Examinations for their detailed advice in preparing the books and for permission to use many past examination questions. Roger Porkess Series Editor vii  The Cambridge A & AS Level Mathematics syllabus P2 Cambridge AS Level IGCSE P1 S1 Mathematics Mathematics M1 M1 S1 S2 A Level P3 Mathematics S1 M1 M2 viii 1 Algebra P1 1 Background algebra Sherlock Holmes: ‘Now the skillful workman is very careful indeed … He will have nothing but the tools which may help him in doing his work, but of these he has a large assortment, and all in the most perfect order.’ A. Conan Doyle Background algebra Manipulating algebraic expressions You will often wish to tidy up an expression, or to rearrange it so that it is easier to read its meaning. The following examples show you how to do this. You should practise the techniques for yourself on the questions in Exercise 1A. Collecting terms Very often you just need to collect like terms together, in this example those in x, those in y and those in z. ? What are ‘like’ and ‘unlike’ terms? EXAMPLE 1.1 Simplify the expression 2x + 4y − 5z − 5x − 9y + 2z + 4x − 7y + 8z. SOLUTION Collect like Expression = 2x + 4x − 5x + 4y – 9y − 7y + 2z + 8z − 5z terms    = 6x − 5x + 4y − 16y + 10z − 5z Tidy up    = x − 12y + 5z This cannot be simplified further and so it is the answer. Removing brackets Sometimes you need to remove brackets before collecting like terms together. 1  EXAMPLE 1.2 Simplify the expression 3(2x − 4y) − 4(x − 5y). P1 SOLUTION Open the brackets 1 Expression = 6x − 12y − 4x + 20y Notice (–4) × (–5y) = +20y Algebra    = 6x − 4x + 20y − 12y Collect like terms    = 2x + 8y Answer EXAMPLE 1.3 Simplify x(x + 2) − (x − 4). SOLUTION Expression = x 2 + 2x − x + 4 Open the brackets = x2 + x + 4 Answer EXAMPLE 1.4 Simplify a(b + c) − ac. SOLUTION Expression = ab + ac − ac Open the brackets    = ab Answer Factorisation It is often possible to rewrite an expression as the product of two or more numbers or expressions, its factors. This usually involves using brackets and is called factorisation. Factorisation may make an expression easier to use and neater to write, or it may help you to interpret its meaning. EXAMPLE 1.5 Factorise 12x − 18y. 6 is a factor of both 12 and 18. SOLUTION Expression = 6(2x − 3y) EXAMPLE 1.6 Factorise x 2 − 2xy + 3xz. SOLUTION x is a factor of all three terms. Expression = x(x − 2y + 3z) 2  Multiplication Several of the previous examples have involved multiplication of variables: cases like P1 a × b = ab   and   x × x = x 2. 1 Background algebra In the next example the principles are the same but the expressions are not quite so simple. EXAMPLE 1.7 Multiply   3p 2qr × 4pq 3 × 5qr 2. You might well do this line in your head. SOLUTION Expression = 3 × 4 × 5 × p 2 × p × q × q 3 × q × r × r 2 = 60 × p 3 × q 5 × r 3 = 60p 3q 5r 3 Fractions The rules for working with fractions in algebra are exactly the same as those used in arithmetic. x – 2y + z EXAMPLE 1.8 Simplify. 2 10 4 SOLUTION As in arithmetic you start by finding the common denominator. For 2, 10 and 4 this is 20. Then you write each part as the equivalent fraction with 20 as its denominator, as follows. 10x 4y 5z This line would often Expression = 20 – 20 + 20 be left out. 10x – 4y + 5z = 20 x2 y2 EXAMPLE 1.9 Simplify –. y x SOLUTION x3 y3 The common Expression = – denominator is xy. xy xy x3 – y3 = xy 3  3x 2 5yz P1 EXAMPLE 1.10 Simplify × 5y 6x. 1 SOLUTION Since the two parts of the expression are multiplied, terms may be cancelled top Algebra and bottom as in arithmetic. In this case 3, 5, x and y may all be cancelled. 2 5yz Expression = 3x × 5y 62x = xz 2 (x – 1)3 EXAMPLE 1.11 Simplify. 4x (x – 1) SOLUTION (x − 1) is a common factor of both top and bottom, so may be cancelled. However, x is not a factor of the top (the numerator), so may not be cancelled. 2 Expression = (x – 1) 4x 24x + 6 EXAMPLE 1.12 Simplify. 3(4x + 1) SOLUTION When the numerator (top) and/or the denominator (bottom) are not factorised, first factorise them as much as possible. Then you can see whether there are any common factors which can be cancelled. 6(4x + 1) Expression = 3(4x + 1) =2 EXERCISE 1A 1 Simplify the following expressions by collecting like terms. (i) 8x + 3x + 4x − 6x (ii) 3p + 3 + 5p − 7 − 7p − 9 (iii) 2k + 3m + 8n − 3k − 6m − 5n + 2k − m + n (iv) 2a + 3b − 4c + 4a − 5b − 8c − 6a + 2b + 12c (v) r − 2s − t + 2r − 5t − 6r − 7t − s + 5s − 2t + 4r 4 2 Factorise the following expressions. (i) 4x + 8y (ii) 12a + 15b – 18c P1 (iii) 72f − 36g − 48h (iv) p 2 − pq + pr 1 (v) 12k 2 + 144km − 72kn Exercise 1A 3 Simplify the following expressions, factorising the answers where possible. (i) 8(3x + 2y) + 4(x + 3y) (ii) 2(3a − 4b + 5c) − 3(2a − 5b − c) (iii) 6(2p − 3q + 4r) − 5(2p − 6q − 3r) − 3(p − 4q + 2r) (iv) 4(l + w + h) + 3(2l − w − 2h) + 5w (v) 5u − 6(w − v) + 2(3u + 4w − v) − 11u 4 Simplify the following expressions, factorising the answers where possible. (i) a(b + c) + a(b − c) (ii) k(m + n) − m(k + n) (iii) p(2q + r + 3s) − pr − s(3p + q) (iv) x(x − 2) − x(x − 6) + 8 (v) x(x − 1) + 2(x − 1) − x(x + 1) 5 Perform the following multiplications, simplifying your answers. (i) 2xy × 3x 2y (ii) 5a 2bc 3 × 2ab 2 × 3c (iii) km × mn × nk (iv) 3pq 2r × 6p 2qr × 9pqr 2 (v) rs × 2st × 3tu × 4ur 6 Simplify the following fractions as much as possible. (i) ab (ii) 2e (iii) x2 ac 4f 5x 4a 2b 6p 2q 3r (iv) (v) 2ab 3p 3q3r 2 7 Simplify the following as much as possible. a ×b ×c 3x 8y 5z p2 q2 (i) (ii) × × (iii) × b c a 2y 3z 4x q p 2fg 4gh 2 32f h3 kmn × 6k 2m3 (iv) × × (v) 16h 4f h 12f 3 3n3 2k 3m 8 Write the following as single fractions. (i) x+x (ii) 2x – x + 3x (iii) 3z + 2z − 5z 2 3 5 3 4 8 12 24 2x − x y 5y 4y (iv) (v) – + 3 4 2 8 5 9 Write the following as single fractions. 3+5 1+1 4+x (i) (ii) (iii) x x x y x y p q 1 –1+1 (iv) + (v) q p a b c 5  10 Write the following as single fractions. P1 (i) x +1+ x –1 (ii) 2x – x – 1 (iii) 3x – 5 + x – 7 4 2 3 5 4 6 1 3(2x + 1) – 7(x – 2) 4x + 1 + 7x – 3 (iv) (v) 5 2 8 12 Algebra 11 Simplify the following expressions. x+3 6(2x + 1)2 2x(y – 3)4 (i) (ii) (iii) 2x + 6 3(2x + 1)5 8x 2(y – 3) 6x – 12 (3x + 2)2 × x 4 (iv) (v) x–2 6x 6x + 4 Linear equations ? What is a variable? You will often need to find the value of the variable in an expression in a particular case, as in the following example. EXAMPLE 1.13 A polygon is a closed figure whose sides are straight lines. Figure 1.1 shows a seven-sided polygon (a heptagon). Figure 1.1 An expression for S °, the sum of the angles of a polygon with n sides, is S = 180(n − 2). ? How is this expression obtained? Try dividing a polygon into triangles, starting from one vertex. 6 Find the number of sides in a polygon with an angle sum of (i) 180° (ii) 1080°.  SOLUTION (i) Substituting 180 for S gives 180 = 180(n − 2) This is an equation P1 which can be Dividing both sides by 180 ⇒ 1 = n−2 solved to find n. 1 Adding 2 to both sides ⇒ 3=n Linear equations The polygon has three sides: it is a triangle. (ii) Substituting 1080 for S gives 1080 = 180(n − 2) Dividing both sides by 180 ⇒ 6 = n−2 Adding 2 to both sides ⇒ 8=n The polygon has eight sides: it is an octagon. Example 1.13 illustrates the process of solving an equation. An equation is formed when an expression, in this case 180(n − 2), is set equal to a value, in this case 180 or 1080, or to another expression. Solving means finding the value(s) of the variable(s) in the equation. Since both sides of an equation are equal, you may do what you wish to an equation provided that you do exactly the same thing to both sides. If there is only one variable involved (like n in the above examples), you aim to get that on one side of the equation, and everything else on the other. The two examples which follow illustrate this. In both of these examples the working is given in full, step by step. In practice you would expect to omit some of these lines by tidying up as you went along. ? ! Look at the statement 5(x – 1) = 5x – 5. What happens when you try to solve it as an equation? This is an identity and not an equation. It is true for all values of x. For example, try x = 11: 5(x − 1) = 5 × (11 − 1) = 50; 5x − 5 = 55 − 5 = 50 ✓, or try x = 46: 5(x − 1) = 5 × (46 − 1) = 225; 5x − 5 = 230 − 5 = 225 ✓, or try x = anything else and it will still be true. To distinguish an identity from an equation, the symbol ≡ is sometimes used. Thus 5(x − 1) ≡ 5x − 5. 7  EXAMPLE 1.14 Solve the equation 5(x − 3) = 2(x + 6). P1 SOLUTION 1 Open the brackets ⇒ 5x − 15 = 2x + 12 ⇒ 5x – 2x − 15 = 2x − 2x + 12 Algebra Subtract 2x from both sides Tidy up ⇒ 3x − 15 = 12 Add 15 to both sides ⇒ 3x − 15 + 15 = 12 + 15 Tidy up ⇒ 3x = 27 Divide both sides by 3 ⇒ 3x = 27 3 3 ⇒ x=9 CHECK When the answer is substituted in the original equation both sides should come out to be equal. If they are different, you have made a mistake. Left-hand side Right-hand side 5(x − 3) 2(x + 6) 5(9 − 3) 2(9 + 6) 5×6 2 × 15 30 30 (as required). EXAMPLE 1.15 Solve the equation 12 (x + 6) = x + 13 (2x − 5). SOLUTION Start by clearing the fractions. Since the numbers 2 and 3 appear on the bottom line, multiply through by 6 which cancels both of them. Multiply both sides by 6 ⇒ 6 × 12 (x + 6) = 6 × x + 6 × 13 (2x − 5) Tidy up ⇒ 3(x + 6) = 6x + 2(2x − 5) Open the brackets ⇒ 3x + 18 = 6x + 4x − 10 Subtract 6x, 4x, and 18 from both sides ⇒ 3x − 6x − 4x = − 10 − 18 Tidy up ⇒ −7x = −28 Divide both sides by (–7) ⇒ –7x = –28 –7 –7 ⇒ x =4 CHECK Substituting x = 4 in 12 (x + 6) = x + 13 (2x – 5) gives: Left-hand side Right-hand side 1 2 (4 + 6) 4 + 13 (8 – 5) 10 2 4 + 33 8 5 5 (as required). EXERCISE 1B 1 Solve the following equations. (i) 5a − 32 = 68 P1 (ii) 4b − 6 = 3b + 2 1 (iii) 2c + 12 = 5c + 12 Exercise 1B (iv) 5(2d + 8) = 2(3d + 24) (v) 3(2e − 1) = 6(e + 2) + 3e (vi) 7(2 − f ) – 3(f − 4) = 10f − 4 (vii) 5g + 2(g − 9) = 3(2g − 5) + 11 (viii) 3(2h − 6) − 6(h + 5) = 2(4h − 4) − 10(h + 4) 1 1 (ix) 2 k + 4 k = 36 1 (x) 2 (l − 5) + l = 11 1 1 1 (xi) 2 (3m + 5) + 1 2 (2m − 1) = 5 2 1 1 5 (xii) n + 3 (n + 1) + 4 (n + 2) = 6 2 The largest angle of a triangle is six times as big as the smallest. The third angle is 75°. (i) Write this information in the form of an equation for a, the size in degrees of the smallest angle. (ii) Solve the equation and so find the sizes of the three angles. 3 Miriam and Saloma are twins and their sister Rohana is 2 years older than them. The total of their ages is 32 years. (i) Write this information in the form of an equation for r, Rohana’s age in years. (ii) What are the ages of the three girls? 4 The length, d m, of a rectangular field is 40 m greater than the width. The perimeter of the field is 400 m. (i) Write this information in the form of an equation for d. (ii) Solve the equation and so find the area of the field. 5 Yash can buy three pencils and have 49c change, or he can buy five pencils and have 15c change. (i) Write this information as an equation for x, the cost in cents of one pencil. (ii) How much money did Yash have to start with? 9  6 I n a multiple-choice examination of 25 questions, four marks are given for P1 each correct answer and two marks are deducted for each wrong answer. One mark is deducted for any question which is not attempted. 1 A candidate attempts q questions and gets c correct. Algebra (i) Write down an expression for the candidate’s total mark in terms of q and c. (ii) James attempts 22 questions and scores 55 marks. Write down and solve an equation for the number of questions which James gets right. 7 Joe buys 18 kg of potatoes. Some of these are old potatoes at 22c per kilogram, the rest are new ones at 36c per kilogram. (i) Denoting the mass of old potatoes he buys by m kg, write down an expression for the total cost of Joe’s potatoes. (ii) Joe pays with a $5 note and receives 20c change. What mass of new potatoes does he buy? 8 In 18 years’ time Hussein will be five times as old as he was 2 years ago. (i) Write this information in the form of an equation involving Hussein’s present age, a years. (ii) How old is Hussein now? Changing the subject of a formula The area of a trapezium is given by A = 12 (a + b)h where a and b are the lengths of the parallel sides and h is the distance between them (see figure 1.2). An equation like this is often called a formula. b h a Figure 1.2 The variable A is called the subject of this formula because it only appears once on its own on the left-hand side. You often need to make one of the other variables the subject of a formula. In that case, the steps involved are just the same as those in solving an equation, as the following examples show. 10 EXAMPLE 1.16 Make a the subject in A = 12 (a + b)h. P1 SOLUTION 1 It is usually easiest if you start by arranging the equation so that the variable you Changing the subject of a formula want to be its subject is on the left-hand side. 1 2 (a + b)h = A Multiply both sides by 2 ⇒ (a + b)h = 2A Divide both sides by h ⇒ a + b = 2A h Subtract b from both sides ⇒ a = 2A − b h EXAMPLE 1.17 Make T the subject in the simple interest formula I = PRT. 100 SOLUTION Arrange with T on the left-hand side PRT = I 100 Multiply both sides by 100 ⇒ PRT = 100I Divide both sides by P and R ⇒ T = 100I PR EXAMPLE 1.18 Make x the subject in the formula v = ω a 2 – x 2. (This formula gives the speed of an oscillating point.) SOLUTION Square both sides ⇒ v 2 = ω2(a 2 − x 2) Divide both sides by ω 2 ⇒ v 2 = a2 − x2 ω2 Add x 2 to both sides ⇒ v 2 + x2 = a2 2 ω2 2 Subtract v 2 from both sides ⇒ x2 = a2 − v 2 ω ω v2 Take the square root of both sides ⇒ x = ± a2 – 2 ω EXAMPLE 1.19 Make m the subject of the formula mv = I + mu. (This formula gives the momentum after an impulse.) SOLUTION Collect terms in m on the left-hand side and terms without m on the other. ⇒ mv − mu = I Factorise the left-hand side ⇒ m(v − u) = I Divide both sides by (v − u) ⇒ m= I 11 v –u  EXERCISE 1C 1 Make (i) a (ii) t the subject in v = u + at. P1 2 Make h the subject in V = l wh. 1 3 Make r the subject in A = πr 2. the subject in v 2 − u 2 = 2as. Algebra 4 Make (i) s (ii) u 5 Make h the subject in A = 2πrh + 2πr 2. 6 Make a the subject in s = ut + 12 at 2. 7 Make b the subject in h = a 2 + b 2. l 8 Make g the subject in T = 2π g. 9 Make m the subject in E = mgh + 12 mv 2. 1 = 1 + 1 10 Make R the subject in. R R1 R2 11 Make h the subject in bh = 2A − ah. 12 Make u the subject in f = uv. u+v 13 Make d the subject in u 2 − du + fd = 0. 14 Make V the subject in p 1VM = mRT + p 2VM. ? All the formulae in Exercise 1C refer to real situations. Can you recognise them? Quadratic equations EXAMPLE 1.20 The length of a rectangular field is 40 m greater than its width, and its area is 6000 m2. Form an equation involving the length, x m, of the field. SOLUTION Since the length of the field is 40 m greater than the width, the width in m must be x − 40 and the area in m2 is x(x − 40). x – 40 So the required equation is x(x − 40) = 6000 x or x 2 − 40x − 6000 = 0. Figure 1.3 12  This equation, involving terms in x 2 and x as well as a constant term (i.e. a number, in this case 6000), is an example of a quadratic equation. This is in P1 contrast to a linear equation. A linear equation in the variable x involves only terms in x and constant terms. 1 Quadratic equations It is usual to write a quadratic equation with the right-hand side equal to zero. To solve it, you first factorise the left-hand side if possible, and this requires a particular technique. Quadratic factorisation EXAMPLE 1.21 Factorise xa + xb + ya + yb. Notice (a + b) is a SOLUTION common factor. xa + xb + ya + yb = x (a + b) + y (a + b) = (x + y)(a + b) The expression is now in the form of two factors, (x + y) and (a + b), so this is the answer. You can see this result in terms of the area of the rectangle in figure 1.4. This can be written as the product of its length (x + y) and its width (a + b), or as the sum of the areas of the four smaller rectangles, xa, xb, ya and yb. x y a xa ya b xb yb Figure 1.4 The same pattern is used for quadratic factorisation, but first you need to split the middle term into two parts. This gives you four terms, which correspond to the areas of the four regions in a diagram like figure 1.4. 13  EXAMPLE 1.22 Factorise x 2 + 7x + 12. P1 SOLUTION 1 Splitting the middle term, 7x, as 4x + 3x you have Algebra x 2 + 7x + 12 = x 2 + 4x + 3x + 12 = x(x + 4) + 3(x + 4) = (x + 3)(x + 4). How do you know to split the middle term, 7x, into 4x + 3x, rather than say 5x + 2x or 9x − 2x? x 3 x x2 3x 4 4x 12 Figure 1.5 The numbers 4 and 3 can be added to give 7 (the middle coefficient) and multiplied to give 12 (the constant term), so these are the numbers chosen. The coefficient of x is 7. The constant term is 12. x 2 + 7x + 12 4+3=7 4 × 3 = 12 EXAMPLE 1.23 Factorise x 2 − 2x − 24. SOLUTION First you look for two numbers that can be added to give −2 and multiplied to give –24: −6 + 4 = −2    −6 × (+4) = −24. The numbers are –6 and +4 and so the middle term, –2x, is split into –6x + 4x. x 2 – 2x – 24 = x 2 − 6x + 4x − 24 = x(x − 6) + 4(x − 6) = (x + 4)(x − 6). 14  This example raises a number of important points. 1 It makes no difference if you write + 4x − 6x instead of − 6x + 4x. In that case P1 the factorisation reads: 1 x 2 − 2x − 24 = x 2 + 4x − 6x − 24 Quadratic equations = x(x + 4) − 6(x + 4) = (x − 6)(x + 4)  (clearly the same answer). 2 There are other methods of quadratic factorisation. If you have already learned another way, and consistently get your answers right, then continue to use it. This method has one major advantage: it is self-checking. In the last line but one of the solution to the example, you will see that (x + 4) appears twice. If at this point the contents of the two brackets are different, for example (x + 4) and (x − 4), then something is wrong. You may have chosen the wrong numbers, or made a careless mistake, or perhaps the expression cannot be factorised. There is no point in proceeding until you have sorted out why they are different. 3 You may check your final answer by multiplying it out to get back to the original expression. There are two common ways of setting this out. (i) Long multiplication x 2 column x column Numbers column x+4 This is x−6 x (x + 4). x 2 + 4x −6x − 24 This is –6(x + 4). x 2 −2x − 24 (as required) (ii) Multiplying term by term (x + 4)(x – 6) = x 2 – 6x + 4x – 24 = x 2 − 2x − 24   (as required) You would not expect to draw the lines and arrows in your answers. They have been put in to help you understand where the terms have come from. EXAMPLE 1.24 Factorise x 2 − 20x + 100. SOLUTION x 2 − 20x + 100 = x 2 − 10x − 10x + 100 Notice: = x(x − 10) − 10(x − 10) (–10) + (–10) = –20 = (x − 10)(x − 10) (–10) × (–10) = +100 = (x − 10)2 15  Note P1 The expression in Example 1.24 was a perfect square. It is helpful to be able to rec- 1 ognise the form of such expressions. (x + a) 2 = x 2 + 2ax + a 2    (in this case a = 10) Algebra (x − a) 2 = x 2 − 2ax + a 2 EXAMPLE 1.25 Factorise x 2 − 49. SOLUTION Notice this is x 2 – 7 2. x 2 − 49 can be written as x 2 + 0x − 49. x 2 + 0x − 49 = x 2 − 7x + 7x − 49 = x(x − 7) + 7(x − 7) –7 + 7 = 0 (–7) × 7 = –49 = (x + 7)(x − 7) Note The expression in Example 1.25 was an example of the difference of two squares which may be written in more general form as a2 − b2 = (a + b)(a − b). ? What would help you to remember the general results from Examples 1.24 and 1.25? The previous examples have all started with the term x 2, that is the coefficient of x 2 has been 1. This is not the case in the next example. EXAMPLE 1.26 Factorise 6x 2 + x − 12. SOLUTION The technique for finding how to split the middle term is now adjusted. Start by multiplying the two outside numbers together: 6 × (−12) = −72. Now look for two numbers which add to give +1 (the coefficient of x) and multiply to give −72 (the number found above). (+9) + (−8) = +1                     (+9) × (−8) = –72 Splitting the middle term gives 3x is a factor of both 6x 2 and 9x. 6x 2 + 9x − 8x − 12 = 3x(2x + 3) − 4(2x + 3) 16 = (3x − 4)(2x + 3) –4 is a factor of both –8x and –12.  Note The method used in the earlier examples is really the same as this. It is just that in P1 those cases the coefficient of x 2 was 1 and so multiplying the constant term by it had 1 no effect. Solving quadratic equations ! Before starting the procedure for factorising a quadratic, you should always check that the terms do not have a common factor as for example in 2x 2 − 8x + 6. This can be written as 2(x 2 − 4x + 3) and factorised to give 2(x − 3)(x − 1). Solving quadratic equations It is a simple matter to solve a quadratic equation once the quadratic expression has been factorised. Since the product of the two factors is zero, it follows that one or other of them must equal zero, and this gives the solution. EXAMPLE 1.27 Solve x 2 − 40x − 6000 = 0. SOLUTION x 2 − 40x − 6000 = x 2 − 100x + 60x − 6000 = x(x − 100) + 60(x − 100) = (x + 60)(x − 100) ⇒ (x + 60)(x − 100) = 0 ⇒ either x + 60 = 0   ⇒   x = −60 ⇒ or x − 100 = 0   ⇒   x = 100 The solution is x = −60 or 100. ? Look back to page 12. What is the length of the field? Note The solution of the equation in the example is x = –60 or 100. The roots of the equation are the values of x which satisfy the equation, in this case one root is x = –60 and the other root is x = 100. Sometimes an equation can be rewritten as a quadratic and then solved. EXAMPLE 1.28 Solve x 4 – 13x 2 + 36 = 0 SOLUTION This is a quartic equation (its highest power of x is 4) and it isn’t easy to factorise 17 this directly. However, you can rewrite the equation as a quadratic in x2.  Let y = x2 P1 x 4 − 13x 2 + 36 = 0 You can replace 1 ⇒ (x 2)2 − 13x 2 + 36 = 0 x2 with y to get a quadratic equation. ⇒ y 2 − 13y + 36 = 0 Algebra Now you have a quadratic equation which you can factorise. (y − 4)(y − 9) = 0 Don’t stop here. You are asked to find x, not y. So y = 4 or y = 9 Since y = x2 then x2 = 4 ⇒ x = ±2 Remember the or x2 = 9 ⇒ x = ±3 negative square root. You may have to do some work rearranging the equation before you can solve it. EXAMPLE 1.29 Find the real roots of the equation x 2 − 2 = 82. x SOLUTION You need to rearrange the equation before you can solve it. x 2 − 2 = 82 x Multiply by x2: x 4 − 2x 2 = 8 Rearrange: x4 − 2x 2 − 8 = 0 This is a quadratic in x 2. You can factorise it directly, without substituting in for x 2. ⇒ (x 2 + 2)(x 2 − 4) = 0 So this quartic equation So x 2 = −2 which has no real solutions. only has two real roots. You can find out more about roots or x 2 = 4 ⇒ x = ±2 which are not real in P3. EXERCISE 1D 1 Factorise the following expressions. (i) al + am + bl + bm (ii) px + py − qx − qy (iii) ur − vr + us − vs (iv) m 2 + mn + pm + pn (v) x 2 − 3x + 2x − 6 (vi) y 2 + 3y + 7y + 21 (vii) z 2 − 5z + 5z − 25 (viii) q 2 − 3q − 3q + 9 (ix) 2x 2 + 2x + 3x + 3 (x) 6v 2 + 3v − 20v − 10 2 Multiply out the following expressions and collect like terms. (i) (a + 2)(a + 3) (ii) (b + 5)(b + 7) (iii) (c − 4)(c − 2) (iv) (d − 5)(d − 4) (v) (e + 6)(e − 1) (vi) (g − 3)(g + 3) (vii) (h + 5)2 (viii) (2i − 3)2 (ix) (a + b)(c + d) (x) (x + y)(x − y) 18 3 Factorise the following quadratic expressions. (i) x 2 + 6x + 8 (ii) x 2 − 6x + 8 P1 (iii) (v) y 2 + 9y + 20 r 2 − 2r − 15 (iv) (vi) r 2 + 2r − 15 s 2 − 4s + 4 1 x 2 − 5x − 6 x 2 + 2x + 1 Exercise 1D (vii) (viii) (ix) a2 − 9 (x) (x + 3)2 − 9 4 Factorise the following expressions. (i) 2x 2 + 5x + 2 (ii) 2x 2 − 5x + 2 (iii) 5x 2 + 11x + 2 (iv) 5x 2 − 11x + 2 (v) 2x 2 + 14x + 24 (vi) 4x 2 − 49 (vii) 6x 2 − 5x − 6 (viii) 9x 2 − 6x + 1 (ix) t12 − t22 (x) 2x 2 − 11xy + 5y2 5 Solve the following equations. (i) x 2 − 11x + 24 = 0 (ii) x 2 + 11x + 24 = 0 (iii) x 2 − 11x + 18 = 0 (iv) x 2 − 6x + 9 = 0 (v) x 2 − 64 = 0 6 Solve the following equations. (i) 3x 2 − 5x + 2 = 0 (ii) 3x 2 + 5x + 2 = 0 (iii) 3x 2 − 5x − 2 = 0 (iv) 25x 2 − 16 = 0 (v) 9x 2 − 12x + 4 = 0 7 Solve the following equations. (i) x 2 − x = 20 (ii) 3x 2 + 5x = 4 3 (iii) x 2 + 4 = 4x (iv) 2x + 1 = 15 x (v) x − 1 = x6 (vi) 3x + x8 = 14 8 Solve the following equations. (i) x 4 – 5x 2 + 4 = 0 (ii) x 4 – 10x 2 + 9 = 0 (iii) 9x 4 – 13x 2 + 4 = 0 (iv) 4x 4 – 25x 2 + 36 = 0 (v) 25x 4 – 4x 2 = 0 (vi) x −6 x +5 = 0 (vii) x 6 – 9x 3 + 8 = 0 (viii) x− x −6 = 0 9 Find the real roots of the following equations. (i) x 2 + 1 = 22 (ii) x 2 = 1 + 122 x x (iii) x 2 − 6 = 27 (iv) 1 20 1+ 2 − 4 = 0 x2 x x (v) 9 + 4 = 13 (vi) x + 3=3 3 2 x4 x2 x (vii) x + 8 =6 (viii) 2+ = 7 3 19 x x x  Find the real roots of the equation 94 + 82 = 1. P1 10 x x 1 11 The length of a rectangular field is 30 m greater than its width, w metres. (i) Write down an expression for the area A m2 of the field, in terms of w. Algebra (ii) The area of the field is 8800 m2. Find its width and perimeter. 12 A cylindrical tin of height h cm and radius r cm, has surface area, including its top and bottom, A cm2. (i) Write down an expression for A in terms of r, h and π. (ii) A tin of height 6 cm has surface area 54 π cm2. What is the radius of the tin? (iii) Another tin has the same diameter as height. Its surface area is 150 π cm2. What is its radius? 13 When the first n positive integers are added together, their sum is given by 1 2 n(n + 1). (i) Demonstrate that this result holds for the case n = 5. (ii) Find the value of n for which the sum is 105. (iii) What is the smallest value of n for which the sum exceeds 1000? 14 The shortest side AB of a right-angled triangle is x cm long. The side BC is 1 cm longer than AB and the hypotenuse, AC, is 29 cm long. Form an equation for x and solve it to find the lengths of the three sides of the triangle. Equations that cannot be factorised The method of quadratic factorisation is fine so long as the quadratic expression can be factorised, but not all of them can. In the case of x 2 − 6x + 2, for example, it is not possible to find two whole numbers which add to give −6 and multiply to give +2. There are other techniques available for such situations, as you will see in the next few pages. Graphical solution If an equation has a solution, you can always find an approximate value for it by drawing a graph. In the case of x 2 − 6x + 2 = 0 you draw the graph of y = x 2 − 6x + 2 20 and find where it cuts the x axis.  x 0 1 2 3 4 5 6 P1 x2 0 1 4 9 16 25 36 −6x 0 −6 −12 −18 −24 −30 −36 1 Equations that cannot be factorised +2 +2 +2 +2 +2 +2 +2 +2 y +2 −3 −6 −7 −6 −3 +2 y 2 Between Between 0.3 and 0.4 5.6 and 5.7 1 0 1 2 3 4 5 6 x –1 –2 –3 –4 –5 –6 –7 Figure 1.6 From figure 1.6, x is between 0.3 and 0.4 so approximately 0.35, or between 5.6 and 5.7 so approximately 5.65. Clearly the accuracy of the answer is dependent on the scale of the graph but, however large a scale you use, your answer will never be completely accurate. Completing the square If a quadratic equation has a solution, this method will give it accurately. It involves adjusting the left-hand side of the equation to make it a perfect square. The steps involved are shown in the following example. 21  EXAMPLE 1.30 Solve the equation x 2 − 6x + 2 = 0 by completing the square. P1 SOLUTION 1 Subtract the constant term from both sides of the equation: Algebra ⇒   x 2 − 6x        = −2 −6 } Take the coefficient of x : Halve it: −3 Square the answer: +9 ? Explain why this makes the left-hand side a perfect square. Add it to both sides of the equation: ⇒   x 2 − 6x + 9 = −2 + 9 Factorise the left-hand side. It will be found to be a perfect square: ⇒ (x − 3)2 = 7 Take the square root of both sides: ⇒ x−3 =± 7 ⇒   x =3± 7 This is an exact answer. Using your calculator to find the value of 7 This is an approximate answer. ⇒ x = 5.646 or 0.354, to 3 decimal places. The graphs of quadratic functions Look at the curve in figure 1.7. It is the graph of y = x 2 − 4x + 5 and it has the characteristic shape of a quadratic; it is a parabola. y Notice that: it has a minimum point 5 x= 2 (or vertex) at (2, 1) 4 it has a line of symmetry, x = 2. It is possible to find the vertex 3 and the line of symmetry without plotting the points by using the 2 technique of completing the square. 1 (2, 1) –1 0 1 2 3 4 x 22 Figure 1.7  Rewrite the expression with the constant term moved to the right x 2 − 4x + 5. P1 Take the coefficient of x: −4 1 Divide it by 2: −2 The graphs of quadratic functions Square the answer: +4 Add this to the left-hand part and compensate by subtracting it from the constant term on the right This is the completed x 2 – 4x + 4 + 5 – 4. square form. This can now be written as (x − 2)2 + 1. The minimum value is 1, The line of symmetry is so the vertex is (2, 1). x – 2 = 0 or x = 2. EXAMPLE 1.31 Write x 2 + 5x + 4 in completed square form. Hence state the equation of the line of symmetry and the co-ordinates of the vertex of the curve y = x 2 + 5x + 4. SOLUTION x 2 + 5x +4 5 ÷ 2 = 2.5; 2.52 = 6.25 x 2 + 5x + 6.25 + 4 − 6.25 (x + 2.5)2 − 2.25 (This is the completed square form.) The line of symmetry is x + 2.5 = 0, or x = −2.5. The vertex is (−2.5, −2.25). y x = –2.5 2 1 –5 –4 –3 –2 –1 0 1 2 x –1 –2 Vertex –3 (–2.5, –2.25) Line of symmetry x = –2.5 Figure 1.8 23 P1 ! For this method, the coefficient of x 2 must be 1. To use it on, say, 2x 2 + 6x + 5, you must write it as 2(x 2 + 3x + 2.5) and then work with x 2 + 3x + 2.5. In completed 1 square form, it is 2(x + 1.5)2 + 0.5. Similarly treat −x 2 + 6x + 5 as −1(x 2 − 6x − 5) and work with x 2 − 6x − 5. In completed square form it is −1(x − 3)2 + 14. Algebra Completing the square is an important technique. Knowing the symmetry and least (or greatest) value of a quadratic function will often give you valuable information about the situation it is modelling. EXERCISE 1E 1 For each of the following equations: (a) write it in completed square form (b) hence write down the equation of the line of symmetry and the co-ordinates of the vertex (c) sketch the curve. (i) y = x 2 + 4x + 9 (ii) y = x 2 − 4x + 9 (iii) y = x 2 + 4x + 3 (iv) y = x 2 − 4x + 3 (v) y = x 2 + 6x − 1 (vi) y = x 2 − 10x (vii) y = x 2 + x + 2 (viii) y = x 2 − 3x − 7 1 (ix) y = x 2 − 2x + 1 (x) y = x 2 + 0.1x + 0.03 2 Write the following as quadratic expressions in descending powers of x. (i) (x + 2)2 − 3 (ii) (x + 4)2 − 4 (iii) (x − 1)2 + 2 (iv) (x − 10)2 + 12 ( x − 12 ) 2 (v) + 43 (vi) (x + 0.1)2 + 0.99 3 Write the following in completed square form. (i) 2x 2 + 4x + 6 (ii) 3x 2 − 18x – 27 (iii) −x 2 − 2x + 5 (iv) −2x 2 − 2x − 2 (v) 5x 2 − 10x + 7 (vi) 4x 2 − 4x − 4 (vii) −3x 2 − 12x (viii) 8x 2 + 24x − 2 24  4  he curves below all have equations of the form y = x 2 + bx + c. T In each case find the values of b and c. P1 (i) y (ii) y 1 The quadratic formula (3, 1) x x (–1, –1) (iii) y (iv) y (–3, 2) (4, 0) x x 5 Solve the following equations by completing the square. (i) x 2 − 6x + 3 = 0 (ii) x 2 − 8x – 1 = 0 (iii) x 2 − 3x + 1 = 0 (iv) 2x 2 − 6x + 1 = 0 (v) 5x 2 + 4x − 2 = 0 The quadratic formula Completing the square is a powerful method because it can be used on any quadratic equation. However it is seldom used to solve an equation in practice because it can be generalised to give a formula which is used instead. The derivation of this follows exactly the same steps. To solve a general quadratic equation ax 2 + bx + c = 0 by completing the square: First divide both sides by a: ⇒ x 2 + bx c a + a = 0. Subtract the constant term from both sides of the equation: ⇒ x 2 + bx c a = −a 25  Take the coefficient of x: + b P1 a +b 1 Halve it: 2a 2 Square the answer: +b2 Algebra 4a Add it to both sides of the equation: 2 2 ⇒ x 2 + bx + b 2 = b 2 – c a 4a 4a a Factorise the left-hand side and tidy up the right-hand side: ( ) 2 2 ⇒ x + 2ba = b – 42 ac 4a Take the square root of both sides: 2 ⇒ x + b = ± b – 4ac 2a 2a ⇒ x = –b ± b – 4ac    2 2a This important result, known as the quadratic formula, has significance beyond the solution of awkward quadratic equations, as you will see later. The next two examples, however, demonstrate its use as a tool for solving equations. EXAMPLE 1.32 Use the quadratic formula to solve 3x 2 − 6x + 2 = 0. SOLUTION Comparing this to the form ax 2 + bx + c = 0 gives a = 3, b = –6 and c = 2. Substituting these values in the formula x = –b ± b – 4ac 2 2a gives x = 6 ± 36 – 24 6 = 0.423 or 1.577 (to 3 d.p.). EXAMPLE 1.33 Solve x 2 − 2x + 2 = 0. SOLUTION The first thing to notice is that this cannot be factorised. The only two whole numbers which multiply to give 2 are 2 and 1 (or −2 and −1) and they cannot be added to get −2. Comparing x 2 − 2x + 2 to the form ax 2 + bx + c = 0 26 gives a = 1, b = −2 and c = 2. Substituting these values in x = –b ± b – 4ac 2 2a P1 2 ± 2 ±4 – 48 – 8 gives 2 2 1 2 ± 2 ±–4 –4 The quadratic formula = 2 2 Trying to find the square root of a negative number creates problems. A positive number multiplied by itself is positive: +2 × +2 = +4. A negative number multiplied by itself is also positive: −2 × −2 = +4. Since −4 can be neither positive nor negative, no such number exists, and so you can find no real solution. Note It is not quite true to say that a negative number has no square root. Certainly it has none among the real numbers but mathematicians have invented an imaginary number, denoted by i, with the property that i2 = −1. Numbers like 1 + i and −1 − i (which are in fact the solutions of the equation above) are called complex numbers. Complex numbers are extremely useful in both pure and applied mathematics; they are covered in P3. To return to the problem of solving the equation x 2 − 2x + 2 = 0, look what happens if you draw the graph of y = x 2 − 2x + 2. The table of values is given below and the graph is shown in figure 1.9. As you can see, the graph does not cut the x axis and so there is indeed no real solution to this equation. y x −1 0 1 2 3 5 x2 +1 0 +1 +4 +9 −2x +2 0 –2 −4 −6 4 +2 +2 +2 +2 +2 +2 3 y +5 +2 +1 +2 +5 2 1 –1 0 1 2 3 x –1 Figure 1.9 27  The part of the quadratic formula which determines whether or not there are real P1 roots is the part under the square root sign. This is called the discriminant. 1 x = –b ± b – 4ac 2 2a Algebra The discriminant, b 2 – 4ac If b 2 − 4ac > 0, the equation has two real roots (see figure 1.10). x Figure 1.10 If b 2 − 4ac < 0, the equation has no real roots (see figure 1.11). x Figure 1.11 If b 2 − 4ac = 0, the equation has one repeated root (see figure 1.12). x Figure 1.12 28 EXERCISE 1F 1 Use the quadratic formula to solve the following equations, where possible. (i) x 2 + 8x + 5 = 0 (ii) x 2 + 2x + 4 = 0 P1 (iii) x 2 − 5x − 19 = 0 (iv) 5x 2 − 3x + 4 = 0 1 (v) 3x 2 + 2x − 4 = 0 (vi) x 2 − 12 = 0 Simultaneous equations 2 Find the value of the discriminant and use it to find the number of real roots for each of the following equations. (i) x 2 − 3x + 4 = 0 (ii) x 2 − 3x − 4 = 0 (iii) 4x 2 − 3x = 0 (iv) 3x 2 + 8 = 0 (v) 3x 2 + 4x + 1 = 0 (vi) x 2 + 10x + 25 = 0 3 Show that the equation ax2 + bx − a = 0 has real roots for all values of a and b. 4 Find the value(s) of k for which these equations have one repeated root. (i) x 2 − 2x + k = 0 (ii) 3x 2 − 6x + k = 0 (iii) kx2 + 3x − 4 = 0 (iv) 2x 2 + kx + 8 = 0 (v) 3x 2 + 2kx − 3k = 0 5 The height h metres of a ball at time t seconds after it is thrown up in the air is given by the expression h = 1 + 15t − 5t 2. (i) Find the times at which the height is 11 m. (ii) Use your calculator to find the time at which the ball hits the ground. (iii) What is the greatest height the ball reaches? Simultaneous equations There are many situations which can only be described mathematically in terms of more than one variable. When you need to find the values of the variables in such situations, you need to solve two or more equations simultaneously (i.e. at the same time). Such equations are called simultaneous equations. If you need to find values of two variables, you will need to solve two simultaneous equations; if three variables, then three equations, and so on. The work here is confined to solving two equations to find the values of two variables, but most of the methods can be extended to more variables if required. 29  Linear simultaneous equations P1 EXAMPLE 1.34 At a poultry farm, six hens and one duck cost $40, while four hens and three 1 ducks cost $36. What is the cost of each type of bird? Algebra SOLUTION Let the cost of one hen be $h and the cost of one duck be $d. Then the information given can be written as: 6h + d = 40  1 4h + 3d = 36.  2 There are several methods of solving this pair of equations. Method 1: Elimination Multiplying equation  1 by 3 ⇒ 18h + 3d = 120 Leaving equation  2 ⇒ 4h + 3d = 36 Subtracting ⇒ 14h = 84 Dividing both sides by 14 ⇒ h = 6 Substituting h = 6 in equation  gives 36 + d 1 = 40 ⇒ d = 4 Therefore a hen costs $6 and a duck $4. Note 1 The first step was to multiply equation  1 by 3 so that there would be a term 3d in both equations. This meant that when equation  2 was subtracted, the variable d was eliminated and so it was possible to find the value of h. 2 The value h = 6 was substituted in equation  1 but it could equally well have been substituted in the other equation. Check for yourself that this too gives the answer d = 4. Before looking at other methods for solving this pair of equations, here is another example. EXAMPLE 1.35 Solve 3x + 5y = 12 1   2x − 6y = −20 2   SOLUTION 1 ×6  ⇒ 18x + 30y = 72 ×5 2 ⇒ 10x − 30y = −100 Adding ⇒ 28x = −28 Giving x = −1 Substituting x = −1 in equation  1 ⇒ −3 + 5y = 12 Adding 3 to each side ⇒ 5y = 15 30 Dividing by 5 ⇒ y = 3 Therefore x = −1, y = 3. Note In this example, both equations were multiplied, the first by 6 to give +30y and the P1 second by 5 to give −30y. Because one of these terms was positive and the other 1 negative, it was necessary to add rather than subtract in order to eliminate y. Simultaneous equations Returning now to the pair of equations giving the prices of hens and ducks, 6h + d = 40  1 4h + 3d = 36  2 here are two alternative methods of solving them. Method 2: Substitution The equation 6h + d = 40 is rearranged to make d its subject: d = 40 − 6h. This expression for d is now substituted in the other equation, 4h + 3d = 36, giving 4h + 3(40 − 6h) = 36 ⇒ 4h + 120 − 18h = 36 ⇒ −14h = −84 ⇒ h = 6 Substituting for h in d = 40 – 6h gives d = 40 − 36 = 4. Therefore a hen costs $6 and a duck $4 (the same answer as before, of course). Method 3: Intersection of the graphs of the equations Figure 1.13 shows the graphs of the two equations, 6h + d = 40 and 4h + 3d = 36. As you can see, they intersect at the solution, h = 6 and d = 4. d 10 9 6h + d = 40 8 7 6 4h + 3d = 36 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 h Figure 1.13 31  Non-linear simultaneous equations P1 The simultaneous equations in the examples so far have all been linear, that 1 is their graphs have been straight lines. A linear equation in, say, x and y contains only terms in x and y and a constant term. So 7x + 2y = 11 is linear but 7x 2 + 2y = 11 is not linear, since it contains a term in x 2. Algebra You can solve a pair of simultaneous equations, one of which is linear and the other not, using the substitution method. This is shown in the next example. EXAMPLE 1.36 Solve x + 2y = 7 1  x 2 + y 2 = 10 2   SOLUTION Rearranging equation  1 gives x = 7 − 2y. Substituting for x in equation 2: (7 − 2y)2 + y 2 = 10 Multiplying out the (7 − 2y) × (7 − 2y) gives 49 − 14y − 14y + 4y 2 = 49 − 28y + 4y 2, so the equation is 49 − 28y + 4y 2 + y 2 = 10. This is rearranged to give 5y 2 − 28y + 39 = 0 A quadratic in y which you ⇒ − 15y − 13y + 39 5y 2 = 0 can now solve using ⇒ 5y(y − 3) − 13(y − 3) = 0 factorisation or the formula. ⇒ (5y − 13)(y − 3) = 0 Either 5y − 13 = 0 ⇒   y = 2.6 Or y − 3 = 0 ⇒   y = 3 1 , x + 2y = 7: Substituting in equation  y = 2.6 ⇒ x = 1.8 y =3 ⇒ x=1 The solution is either x = 1.8, y = 2.6 or x = 1, y = 3. ! Always substitute into the linear equation. Substituting in the quadratic will give you extra answers which are not correct. 32 EXERCISE 1G 1 Solve the following pairs of simultaneous equations. (i) 2x + 3y = 8 (ii) x + 4y = 16 (iii) 7x + y = 15 P1 3x + 2y = 7 3x + 5y = 20 4x + 3y = 11 1 (iv) 5x − 2y = 3 (v) 8x − 3y = 21 (vi) 8x + y = 32 Exercise 1G x + 4y = 5 5x + y = 16 7x − 9y = 28 (vii) 4x + 3y = 5 (viii) 3u − 2v = 17 (ix) 4l − 3m = 2 2x − 6y = −5 5u − 3v = 28 5l − 7m = 9 2 A student wishes to spend exactly $10 at a second-hand bookshop. All the paperbacks are one price, all the hardbacks another. She can buy five paperbacks and eight hardbacks. Alternatively she can buy ten paperbacks and six hardbacks. (i) Write this information as a pair of simultaneous equations. (ii) Solve your equations to find the cost of each type of book. 3 The cost of a pear is 5c greater than that of an apple. Eight apples and nine pears cost $1.64. (i) Write this information as a pair of simultaneous equations. (ii) Solve your equations to find the cost of each type of fruit. 4 A car journey of 380 km lasts 4 hours. Part of this is on a motorway at an average speed of 110 km h−1, the rest on country roads at an average speed of 70 km h−1. (i) Write this information as a pair of simultaneous equations. (ii) Solve your equations to find how many kilometres of the journey is spent on each type of road. 5 Solve the following pairs of simultaneous equations. (i) x 2 + y 2 = 10 (ii) x 2 − 2y 2 = 8 (iii) 2x 2 + 3y = 12 x + y = 4 x + 2y = 8 x − y = –1 (iv) k 2 + km = 8 (v) t12 − t22 = 75 (vi) p+q+5=0 m = k − 6 t1 = 2t 2 p 2 = q2 + 5 (vii) k(k − m) = 12 (viii) p12 − p22 = 0 k(k + m) = 60 p1 + p2 = 2 33  6  he diagram shows the ne