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MATHEMATICS FOR ECONOMICS AND BUSINESS; NINTH EDITION
MATHEMATICS FOR ECONOMICS AND BUSINESS; NINTH EDITION
IAN JACQUES
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MATHEMATICS FOR ECONOMICS AND BUSINESS IAN JACQUES NINTH EDITION @ChristofVanDerWalt MyLab Math MyLab Math provides the practice, instruction and personalised learning path that students need to master concepts and develop critical numeracy skills. With over 5 million global student registrations, MyLab Math combines proven results and engaging experiences to grow student conﬁdence and improve their understanding of mathematics. Designed for lecturers and students MyLab Math provides lecturers with a rich and ﬂexible set of course materials, along with coursemanagement tools that make it easy to deliver all or a portion of their course online. It provides students with a personalised interactive learning environment, where they can learn at their own pace and measure their progress. Trusted partner MyLab Math lets you teach your course your way. Ideal as a selfstudy resource for students who need extra help, or you can take advantage of the advanced customisation options and create your own course. MyLab Math allows you to quickly and easily create tests to assess your students’ skills from basic numeracy right up to undergraduate calculus. ‘Each student received a study plan based on his/her exam result and practised the required area using MyLab Math. I can now monitor each student’s progress on a weekly basis . . . MyLab Math also made a signiﬁcant saving on the workload for me as the lecturer.’ Dr Shadi Ostavari, University of Greenwich To ﬁnd out more, visit: www.pearson.com/mylab/mathglobal This page intentionally left blank MATHEMATICS FOR ECONOMICS AND BUSINESS At Pearson, we have a simple mission: to help people make more of their lives through learning. We combine innovative learning technology with trusted content and educational expertise to provide engaging and effective learning experiences that serve people wherever and whenever they are learning. From classroom to boardroom, our curriculum materials, digital learning tools and testing programmes help to educate millions of p; eople worldwide – more than any other private enterprise. Every day our work helps learning flourish, and wherever learning flourishes, so do people. To learn more, please visit us at www.pearson.com/uk MATHEMATICS FOR ECONOMICS AND BUSINESS IAN JACQUES NINTH EDITION Harlow, England • London • New York • Boston • San Francisco • Toronto • Sydney • Dubai • Singapore • Hong Kong Tokyo • Seoul • Taipei • New Delhi • Cape Town • São Paulo • Mexico City • Madrid • Amsterdam • Munich • Paris • Milan PEARSON EDUCATION LIMITED Kao Two Kao Park Harlow CM17 9NA United Kingdom Tel: +44 (0)1279 623623 Web: www.pearson.com/uk First published 1991 (print) Second edition published 1994 (print) Third edition published 1999 (print) Fourth edition published 2003 (print) Fifth edition published 2006 (print) Sixth edition published 2009 (print) Seventh edition published 2013 (print and electronic) Eighth edition published 2015 (print and electronic) Ninth edition published 2018 (print and electronic) © AddisonWesley Publishers Ltd 1991, 1994 (print) © Pearson Education Limited 1999, 2009 (print) © Pearson Education Limited 2013, 2015, 2018 (print and electronic) The right of Ian Jacques to be identified as author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. The print publication is protected by copyright. Prior to any prohibited reproduction, storage in a retrieval system, distribution or transmission in any form or by any means, electronic, mechanical, recording or otherwise, permission should be obtained from the publisher or, where applicable, a licence permitting restricted copying in the United Kingdom should be obtained from the Copyright Licensing Agency Ltd, Barnard’s Inn, 86 Fetter Lane, London EC4A 1EN. The ePublication is protected by copyright and must not be copied, reproduced, transferred, distributed, leased, licensed or publicly performed or used in any way except as specifically permitted in writing by the publishers, as allowed under the terms and conditions under which it was purchased, or as strictly permitted by applicable copyright law. Any unauthorised distribution or use of this text may be a direct infringement of the author’s and the publisher’s rights and those responsible may be liable in law accordingly. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. Pearson Education is not responsible for the content of thirdparty internet sites. ISBN: 9781292191669 (print) 9781292191706 (PDF) 9781292191713 (ePub) British Library CataloguinginPublication Data A catalogue record for the print edition is available from the British Library Library of Congress CataloginginPublication Data Names: Jacques, Ian, 1957– author. Title: Mathematics for economics and business / Ian Jacques. Description: Ninth edition.  Harlow, United Kingdom : Pearson Education, [2018]  Includes bibliographical references and index. Identifiers: LCCN 2017049617 ISBN 9781292191669 (print)  ISBN 9781292191706 (pdf)  ISBN 9781292191713 (epub) Subjects: LCSH: Economics, Mathematical.  Business mathematics. Classification: LCC HB135 .J32 2018  DDC 512.024/33—dc23 LC record available at https://lccn.loc.gov/2017049617 10 9 8 7 6 5 4 3 2 1 22 21 20 19 18 Front cover image © Getty Images Print edition typeset in 10/12.5pt Sabon MT Pro by iEnergizer Aptara®, Ltd. Printed in Slovakia by Neografia NOTE THAT ANY PAGE CROSS REFERENCES REFER TO THE PRINT EDITION To Victoria, Lewis and Celia Contents Preface xi INTRODUCTION: Getting Started 1 Notes for students: how to use this text Chapter 1 Linear Equations 1 5 1.1 Introduction to algebra 1.1.1 Negative numbers 1.1.2 Expressions 1.1.3 Brackets Key Terms Exercise 1.1 Exercise 1.1* 6 7 9 12 17 18 20 1.2 Further algebra 1.2.1 Fractions 1.2.2 Equations 1.2.3 Inequalities Key Terms Exercise 1.2 Exercise 1.2* 22 22 29 33 36 36 38 1.3 Graphs of linear equations Key Terms Exercise 1.3 Exercise 1.3* 40 51 52 53 1.4 Algebraic solution of simultaneous linear equations Key Term Exercise 1.4 Exercise 1.4* 55 65 65 66 1.5 Supply and demand analysis Key Terms Exercise 1.5 Exercise 1.5* 67 80 80 82 1.6 Transposition of formulae Key Terms Exercise 1.6 Exercise 1.6* 84 91 91 92 1.7 National income determination Key Terms Exercise 1.7 Exercise 1.7* Formal mathematics Multiple choice questions Examination questions 93 105 105 106 109 112 116 Contents vii Chapter 2 Nonlinear Equations Quadratic functions Key Terms Exercise 2.1 Exercise 2.1* 2.2 Revenue, cost and profit Key Terms Exercise 2.2 Exercise 2.2* 2.3 Indices and logarithms 2.3.1 Index notation 2.3.2 Rules of indices 2.3.3 Logarithms 2.3.4 Summary Key Terms Exercise 2.3 Exercise 2.3* 2.4 The exponential and natural logarithm functions Key Terms Exercise 2.4 Exercise 2.4* Formal mathematics Multiple choice questions Examination questions 2.1 Chapter 3 Mathematics of Finance Percentages 3.1.1 Index numbers 3.1.2 Inflation Key Terms Exercise 3.1 Exercise 3.1* 3.2 Compound interest Key Terms Exercise 3.2 Exercise 3.2* 3.3 Geometric series Key Terms Exercise 3.3 Exercise 3.3* 3.4 Investment appraisal Key Terms Exercise 3.4 Exercise 3.4* Formal mathematics Multiple choice questions Examination questions 3.1 Chapter 4 Differentiation 4.1 The derivative of a function Key Terms Exercise 4.1 Exercise 4.1* 121 122 136 137 138 140 148 148 150 151 151 155 161 167 168 168 170 172 182 182 183 186 189 193 197 198 204 208 210 210 213 216 226 226 228 230 238 238 239 241 253 253 255 257 259 263 267 268 277 277 278 viii Contents 4.2 4.3 4.4 4.5 4.6 4.7 4.8 Rules of differentiation Rule 1 The constant rule Rule 2 The sum rule Rule 3 The difference rule Key Terms Exercise 4.2 Exercise 4.2* Marginal functions 4.3.1 Revenue and cost 4.3.2 Production 4.3.3 Consumption and savings Key Terms Exercise 4.3 Exercise 4.3* Further rules of differentiation Rule 4 The chain rule Rule 5 The product rule Rule 6 The quotient rule Exercise 4.4 Exercise 4.4* Elasticity Key Terms Exercise 4.5 Exercise 4.5* Optimisation of economic functions Key Terms Exercise 4.6 Exercise 4.6* Further optimisation of economic functions Key Term Exercise 4.7* The derivative of the exponential and natural logarithm functions Exercise 4.8 Exercise 4.8* Formal mathematics Multiple choice questions Examination questions Chapter 5 Partial Differentiation 5.1 5.2 5.3 279 279 280 281 286 286 288 290 290 297 299 301 301 302 304 305 307 310 312 313 314 326 326 327 329 345 345 347 348 359 359 361 370 371 373 376 382 389 Functions of several variables Key Terms Exercise 5.1 Exercise 5.1* Partial elasticity and marginal functions 5.2.1 Elasticity of demand 5.2.2 Utility 5.2.3 Production Key Terms Exercise 5.2 Exercise 5.2* 390 400 401 402 404 404 407 413 415 416 418 Comparative statics Key Terms Exercise 5.3* 420 429 429 Contents ix 5.4 Unconstrained optimisation Key Terms Exercise 5.4 Exercise 5.4* 433 444 444 445 5.5 Constrained optimisation Key Terms Exercise 5.5 Exercise 5.5* 447 456 457 458 5.6 Lagrange multipliers Key Terms Exercise 5.6 Exercise 5.6* 460 468 469 470 Formal mathematics Multiple choice questions Examination questions Chapter 6 Integration 472 474 477 483 6.1 Indefinite integration Key Terms Exercise 6.1 Exercise 6.1* 484 495 496 497 6.2 Definite integration 6.2.1 Consumer’s surplus 6.2.2 Producer’s surplus 6.2.3 Investment flow 6.2.4 Discounting Key Terms Exercise 6.2 Exercise 6.2* 499 503 504 506 508 509 509 510 Formal mathematics Multiple choice questions Examination questions Chapter 7 Matrices 513 515 518 523 7.1 Basic matrix operations 7.1.1 Transposition 7.1.2 Addition and subtraction 7.1.3 Scalar multiplication 7.1.4 Matrix multiplication 7.1.5 Summary Key Terms Exercise 7.1 Exercise 7.1* 524 526 527 530 531 539 539 540 542 7.2 Matrix inversion Key Terms Exercise 7.2 Exercise 7.2* 545 560 560 561 7.3 Cramer’s rule Key Term Exercise 7.3 Exercise 7.3* 564 572 572 573 x Contents Formal mathematics Multiple choice questions Examination questions Chapter 8 Linear Programming 576 577 581 585 8.1 Graphical solution of linear programming problems Key Terms Exercise 8.1 Exercise 8.1* 586 600 601 602 8.2 Applications of linear programming Key Terms Exercise 8.2 Exercise 8.2* 604 612 612 614 Formal mathematics Multiple choice questions Examination questions Chapter 9 Dynamics 617 618 623 627 9.1 Difference equations 9.1.1 National income determination 9.1.2 Supply and demand analysis Key Terms Exercise 9.1 Exercise 9.1* 628 634 636 639 639 640 9.2 Differential equations 9.2.1 National income determination 9.2.2 Supply and demand analysis Key Terms Exercise 9.2 Exercise 9.2* 643 649 651 653 654 655 Formal mathematics Multiple choice questions Examination questions Answers to Problems 658 659 662 664 Chapter 1 664 Chapter 2 674 Chapter 3 683 Chapter 4 687 Chapter 5 698 Chapter 6 705 Chapter 7 709 Chapter 8 715 Chapter 9 719 Glossary 723 Index 730 Preface This text is intended primarily for students on economics, business studies and management courses. It assumes very little prerequisite knowledge, so it can be read by students who have not undertaken a mathematics course for some time. The style is informal, and the text contains a large number of worked examples. Students are encouraged to tackle problems for themselves as they read through each section. Detailed solutions are provided so that all answers can be checked. Consequently, it should be possible to work through this text on a selfstudy basis. The material is wide ranging and varies from elementary topics such as percentages and linear equations to more sophisticated topics such as constrained optimisation of multivariate functions. The text should therefore be suitable for use on both low and highlevel quantitative methods courses. This text was first published in 1991. The prime motivation for writing it then was to try to produce a text that students could actually read and understand for themselves. This remains the guiding principle when writing this ninth edition. One of the main improvements is the inclusion of over 200 additional questions. Each chapter now ends with both multiple choice questions and a selection of longer examinationstyle questions. Students usually enjoy tackling multiple choice questions since they provide a quick way of testing recall of the material covered in each chapter. Several universities include multiple choice as part of their assessment. The final section in each chapter entitled “Examination Questions” contains longer problems which require knowledge and understanding of more than one topic. Although these have been conveniently placed at the end of each chapter it may be best to leave these until the end of the academic year so that they can be used during the revision period just before the examinations. Ian Jacques This page intentionally left blank This page intentionally left blank INTRODUCTION Getting Started NOTES FOR STUDENTS: HOW TO USE THIS TEXT I am always amazed by the mix of students on firstyear economics courses. Some have not acquired any mathematical knowledge beyond elementary algebra (and even that can be of a rather dubious nature), some have never studied economics before in their lives, while others have passed preliminary courses in both. Whatever category you are in, I hope that you will find this text of value. The chapters covering algebraic manipulation, simple calculus, finance, matrices and linear programming should also benefit students on business studies and manage ment courses. The first few chapters are aimed at complete beginners and students who have not taken mathematics courses for some time. I would like to think that these students once enjoyed mathematics and had every intention of continuing their studies in this area, but somehow never found the time to fit it into an already overcrowded academic timetable. However, I suspect that the reality is rather different. Possibly they hated the subject, could not understand it and dropped it at the earliest opportunity. If you find yourself in this position, you are probably horrified to discover that you must embark on a quantitative methods course with an examination looming on the horizon. However, there is no need to worry. My experience is that every student is capable of passing a mathematics examination. All that is required is a commitment to study and a willingness to suspend any prejudices about the subject gained at school. The fact that you have bothered to buy this text at all suggests that you are prepared to do both. To help you get the most out of this text, let me compare the working practices of economics and engineering students. The former rarely read individual books in any great depth. They tend to skim through a selection of books in the university library and perform a large number of Internet searches, picking out relevant information. Indeed, the ability to read selectively and to compare various sources of information is an important skill that all arts and social science students must acquire. Engineering students, on the other hand, are more likely to read just a few books in any one year. They read each of these from cover to cover and attempt virtually every problem en route. Even though you are most definitely not an engineer, it is the engineering approach that you need to adopt while studying mathematics. There are several reasons for this. First, a mathematics text can never be described, even by its most ardent admirers, as a good bedtime read. It can take an hour or two of concentrated effort to understand just a few pages of a mathematics text. You are therefore recommended to work through this text systematically in short bursts rather than to attempt to read whole chapters. Each section is designed to take between one and two hours to complete, and this is quite sufficient for a single session. Secondly, mathematics is a hierarchical subject in which one topic follows on from the next. A construction firm building an office block is hardly likely to erect the fiftieth storey without making sure that the intermediate floors and foundations are securely in place. Likewise, you cannot ‘dip’ into the middle of a mathematics text and expect to follow it unless you have satisfied the prerequisites for that topic. Finally, you actually need to do mathematics yourself before you can understand it. No matter how wonderful your lecturer is, and no matter how many problems are discussed in class, it is only 2 INTRODUCTION Getting Started by solving problems yourself that you are ever going to become confident in using and applying mathematical techniques. For this reason, several problems are interspersed within the text, and you are encouraged to tackle these as you go along. You will require writing paper, graph paper, pens and a calculator for this. There is no need to buy an expensive calculator unless you are feeling particularly wealthy at the moment. A bottomoftherange scientific calculator should be good enough. Answers to every question are printed at the back of this text so that you can check your own answers quickly as you go along. However, please avoid the temptation to look at them until you have made an honest attempt at each one. Remember that in the future you may well have to sit down in an uncomfortable chair, in front of a blank sheet of paper, and be expected to produce solutions to examination questions of a similar type. At the end of each section there are two parallel exercises. The nonstarred exercises are intended for students who are meeting these topics for the first time and the questions are designed to consolidate basic principles. The starred exercises are more challenging but still cover the full range so that students with greater experience will be able to concentrate their efforts on these questions without having to pickandmix from both exercises. The chapter dependence is shown in Figure I.1. If you have studied some advanced mathematics before, you will discover that parts of Chapters 1, 2 and 4 are familiar. However, you may find that the sections on economics applications contain new material. You are best advised to test yourself by attempting a selection of problems from the starred exercise in each section to see if you need to read through it as part of a refresher course. Economics students in a desperate hurry to experience the delights of calculus can miss out Chapter 3 without any loss of continuity and move straight on to Chapter 4. The mathematics of finance is probably more relevant to business and accountancy students, although you can always read it later if it is part of your economics syllabus. At the end of every chapter you will find a multiple choice test and some examination questions. These cover the work of the whole chapter. We recommend that you try the multiple choice questions when you have completed the relevant chapter. As usual, answers 1 Linear equations 7 Matrices 8 Linear programming 2 Nonlinear equations 3 Mathematics of finance 4 Differentiation 5 Partial differentiation 6 Integration 9 Dynamics Figure I.1 INTRODUCTION Getting Started 3 are provided at the back of the book so that you can check to see how well you have done. If you do get any of the questions wrong, it would be worth redoing that question perhaps writing down full working so that you can spot your mistake more easily. The final section contains several examinationstyle problems which are more challenging. They tend to be longer than the questions encountered so far in the exercises and require more confidence and experience. We recommend that you leave these until the end of the course and use them in your buildup to the final exams. I hope that this text helps you to succeed in your mathematics course. You never know, you might even enjoy it. Remember to wear your engineer’s hat while reading the text. I have done my best to make the material as accessible as possible. The rest is up to you! Chapter 1 Linear Equations The main aim of this chapter is to introduce the mathematics of linear equations. This is an obvious first choice in an introductory text, since it is an easy topic which has many applications. There are seven sections, which are intended to be read in the order that they appear. Sections 1.1, 1.2, 1.3, 1.4 and 1.6 are devoted to mathematical methods. They serve to revise the rules of arithmetic and algebra, which you probably met at school but may have forgotten. In particular, the properties of negative numbers and fractions are considered. A reminder is given on how to multiply out brackets and how to manipulate mathematical expressions. You are also shown how to solve simultaneous linear equations. Systems of two equations in two unknowns can be solved using graphs, which are described in Section 1.3. However, the preferred method uses elimination, which is considered in Section 1.4. This algebraic approach has the advantage that it always gives an exact solution and it extends readily to larger systems of equations. The remaining two sections are reserved for applications in microeconomics and macroeconomics. You may be pleasantly surprised by how much economic theory you can analyse using just the basic mathematical tools considered here. Section 1.5 introduces the fundamental concept of an economic function and describes how to calculate equilibrium prices and quantities in supply and demand theory. Section 1.7 deals with national income determination in simple macroeconomic models. The first six sections underpin the rest of the text and are essential reading. The final section is not quite as important and can be omitted at this stage. Section 1.1 Introduction to algebra Objectives At the end of this section you should be able to: ● Add, subtract, multiply and divide negative numbers. ● Understand what is meant by an algebraic expression. ● Evaluate algebraic expressions numerically. ● Simplify algebraic expressions by collecting like terms. ● Multiply out brackets. ● Factorise algebraic expressions. Algebra is boring There is no getting away from the fact that algebra is boring. Doubtless there are a few enthusiasts who get a kick out of algebraic manipulation, but economics and business students are rarely to be found in this category. Indeed, the mere mention of the word ‘algebra’ is enough to strike fear into the heart of many a firstyear student. Unfortunately, you cannot get very far with mathematics unless you have completely mastered this topic. An apposite analogy is the game of chess. Before you can begin to play a game of chess, it is necessary to go through the tedium of learning the moves of individual pieces. In the same way it is essential that you learn the rules of algebra before you can enjoy the ‘game’ of mathematics. Of course, just because you know the rules does not mean that you are going to excel at the game, and no one is expecting you to become a grandmaster of mathematics. However, you should at least be able to follow the mathematics presented in economics books and journals as well as to solve simple problems for yourself. Advice If you have studied mathematics recently, then you will find the material in the first few sections of the text fairly straightforward. You may prefer just to try the questions in the starred exercise at the end of each section to get yourself back up to speed. However, if it has been some time since you have studied this subject, our advice is very different. Please work through the material thoroughly even if it is vaguely familiar. Make sure that you do the problems as they arise, checking your answers with those provided at the back of this text. The material has been broken down into three subsections: ● ● ● negative numbers; expressions; brackets. You might like to work through these subsections on separate occasions to enable the ideas to sink in. To rush this topic now is likely to give you only a halfbaked understanding, which will result in hours of frustration when you study the later chapters of this text. Section 1.1 Introduction to algebra 7 1.1.1 Negative numbers In mathematics numbers are classified into one of three types: positive, negative or zero. At school you were probably introduced to the idea of a negative number via the temperature on a thermometer scale measured in degrees centigrade. A number such as −5 would then be interpreted as a temperature of 5 degrees below freezing. In personal finance a negative bank balance would indicate that an account is ‘in the red’ or ‘in debit’. Similarly, a firm’s profit of −500 000 signifies a loss of half a million. The rules for the multiplication of negative numbers are negative 3 negative 5 positive negative 3 positive 5 negative It does not matter in which order two numbers are multiplied, so positive 3 negative 5 negative These rules produce, respectively, (−2) × (−3) = 6 (−4) × 5 = −20 7 × (−5) = −35 Also, because division is the same sort of operation as multiplication (it just undoes the result of multiplication and takes you back to where you started), exactly the same rules apply when one number is divided by another. For example, (−15) ÷ (−3) = 5 (−16) ÷ 2 = −8 2 ÷ (−4) = −1/ 2 In general, to multiply or divide lots of numbers it is probably simplest to ignore the signs to begin with and just to work the answer out. The final result is negative if the total number of minus signs is odd and positive if the total number is even. Example Evaluate (a) (−2) × (−4) × (−1) × 2 × (−1) × (−3) (b) 5 3 (–4) 3 (–1) 3 (–3) (–6) 3 2 Solution (a) Ignoring the signs gives 2 × 4 × 1 × 2 × 1 × 3 = 48 There are an odd number of minus signs (in fact, five), so the answer is −48. (b) Ignoring the signs gives 60 5343133 5 55 12 632 There are an even number of minus signs (in fact, four), so the answer is 5. 8 Chapter 1 Linear Equations Advice Attempt the following problem yourself both with and without a calculator. On most machines a negative number such as −6 is entered by pressing the button labelled ( ]) followed by 6. Practice Problem 1. (1) Without using a calculator, evaluate (a) 5 × (−6) (b) (−1) × (−2) (c) (−50) ÷ 10 (d) (−5) ÷ (−1) (e) 2 × (−1) × (−3) × 6 (f) 2 3 (–1) 3 (–3) 3 6 (–2) 3 3 3 6 (2) Confirm your answer to part (1) using a calculator. To add or subtract negative numbers it helps to think in terms of a number line: ]4 ]3 ]2 ]1 0 1 2 3 4 If b is a positive number, then a−b can be thought of as an instruction to start at a and to move b units to the left. For example, 1 − 3 = −2 because if you start at 1 and move 3 units to the left, you end up at −2: ]4 ]3 ]2 ]1 0 1 2 3 4 ]1 0 1 2 3 4 Similarly, −2 − 1 = −3 because 1 unit to the left of −2 is −3. ]4 ]3 ]2 On the other hand, a − (−b) is taken to be a + b. This follows from the rule for multiplying two negative numbers, since −(−b) = (−1) × (−b) = b Consequently, to evaluate a − (−b) Section 1.1 Introduction to algebra 9 you start at a and move b units to the right (that is, in the positive direction). For example, −2 − (−5) = −2 + 5 = 3 because if you start at −2 and move 5 units to the right, you end up at 3. ]4 ]3 ]2 0 ]1 1 2 3 4 Practice Problem 2. (1) Without using a calculator, evaluate (a) 1 − 2 (b) −3 − 4 (c) 1 − (−4) (d) −1 − (−1) (e) −72 − 19 (f ) −53 − (−48) (2) Confirm your answer to part (1) using a calculator. 1.1.2 Expressions In algebra, letters are used to represent numbers. In pure mathematics the most common letters used are x and y. However, in applications it is helpful to choose letters that are more meaningful, so we might use Q for quantity and I for investment. An algebraic expression is then simply a combination of these letters, brackets and other mathematical symbols such as + or −. For example, the expression P 1+ r 100 n can be used to work out how money in a savings account grows over a period of time. The letters P, r and n represent the original sum invested (called the principal – hence the use of the letter P), the rate of interest and the number of years, respectively. To work it all out, you not only need to replace these letters by actual numbers, but you also need to understand the various conventions that go with algebraic expressions such as this. In algebra, when we multiply two numbers represented by letters, we usually suppress the multiplication sign between them. The product of a and b would simply be written as ab without bothering to put the multiplication sign between the symbols. Likewise, when a number represented by the letter Y is doubled, we write 2Y. In this case we not only suppress the multiplication sign but adopt the convention of writing the number in front of the letter. Here are some further examples: P × Q is written as PQ d × 8 is written as 8d n × 6 × t is written as 6nt z × z is written as z2 (using the index 2 to indicate squaring a number) 1 × t is written as t (since multiplying by 1 does not change a number) 10 Chapter 1 Linear Equations In order to evaluate these expressions it is necessary to be given the numerical value of each letter. Once this has been done, you can work out the final value by performing the operations in the following order: Brackets first (B) Indices second (I) Division and Multiplication third (DM) Addition and Subtraction fourth (AS) This is sometimes remembered using the acronym BIDMAS, and it is essential to use this ordering for working out all mathematical calculations. For example, suppose you wish to evaluate each of the following expressions when n = 3: 2n2 and (2n)2 Substituting n = 3 into the first expression gives 2n2 = 2 × 32 (the multiplication sign is revealed when we switch from algebra to numbers) =2×9 (according to BIDMAS, indices are worked out before multiplication) = 18 whereas in the second expression we get (2n)2 = (2 × 3)2 =6 (again, the multiplication sign is revealed) (according to BIDMAS, we evaluate the inside of the brackets first) 2 = 36 The two answers are not the same, so the order indicated by BIDMAS really does matter. Looking at the previous list, notice that there is a tie between multiplication and division for third place, and another tie between addition and subtraction for fourth place. These pairs of operations have equal priority, and under these circumstances you work from left to right when evaluating expressions. For example, substituting x = 5 and y = 4 in the expression, x − y + 2, gives x−y+2=5−4+2 =1+2 (reading from left to right, subtraction comes first) =3 Example (a) Find the value of 2x − 3y when x = 9 and y = 4. (b) Find the value of 2Q2 + 4Q + 150 when Q = 10. (c) Find the value of 5a − 2b + c when a = 4, b = 6 and c = 1. (d) Find the value of (12 − t) − (t − 1) when t = 4. Solution (a) 2x − 3y = 2 × 9 − 3 × 4 = 18 − 12 =6 (substituting numbers) (multiplication has priority over subtraction) Section 1.1 Introduction to algebra 11 (b) 2Q2 + 4Q + 150 = 2 × 102 + 4 × 10 + 150 (substituting numbers) = 2 × 100 + 4 × 10 + 150 (indices have priority over multiplication and addition) (multiplication has priority over addition) = 200 + 40 + 150 = 390 (c) 5a − 2b + c = 5 × 4 − 2 × 6 + 1 (substituting numbers) = 20 − 12 + 1 (multiplication has priority over addition and subtraction) = 8 + 1 (addition and subtraction have equal priority, so work from left to right) =9 (d) (12 − t) − (t − 1) = (12 − 4) − (4 − 1) =8−3 (substituting numbers) (brackets first) =5 Practice Problem 3. Evaluate each of the following by replacing the letters by the given numbers: (a) 2Q + 5 when Q = 7. (b) 5x2y when x = 10 and y = 3. (c) 4d − 3f + 2g when d = 7, f = 2 and g = 5. (d) a(b + 2c) when a = 5, b = 1 and c = 3. Like terms are multiples of the same letter (or letters). For example, 2P, −34P and 0.3P are all multiples of P and so are like terms. In the same way, xy, 4xy and 69xy are all multiples of xy and so are like terms. If an algebraic expression contains like terms which are added or subtracted together, then it can be simplified to produce an equivalent shorter expression. Example Simplify each of the following expressions (where possible): (a) 2a + 5a − 3a (b) 4P − 2Q (c) 3w + 9w2 + 2w (d) 3xy + 2y2 + 9x + 4xy − 8x Solution (a) All three are like terms since they are all multiples of a, so the expression can be simplified: 2a + 5a − 3a = 4a ➜ 12 Chapter 1 Linear Equations (b) The terms 4P and 2Q are unlike because one is a multiple of P and the other is a multiple of Q, so the expression cannot be simplified. (c) The first and last are like terms since they are both multiples of w, so we can collect these together and write 3w + 9w2 + 2w = 5w + 9w2 This cannot be simpified any further because 5w and 9w2 are unlike terms. (d) The terms 3xy and 4xy are like terms, and 9x and 8x are also like terms. These pairs can therefore be collected together to give 3xy + 2y2 + 9x + 4xy − 8x = 7xy + 2y2 + x Notice that we write just x instead of 1x and also that no further simplication is possible since the final answer involves three unlike terms. Practice Problem 4. Simplify each of the following expressions, where possible: (a) 2x + 6y − x + 3y (b) 5x + 2y − 5x + 4z (c) 4Y2 + 3Y − 43 (d) 8r2 + 4s − 6rs − 3s − 3s2 + 7rs (e) 2e2 + 5f − 2e2 − 9f (f) 3w + 6W (g) ab − ba 1.1.3 Brackets It is useful to be able to take an expression containing brackets and rewrite it as an equivalent expression without brackets, and vice versa. The process of removing brackets is called ‘expanding brackets’ or ‘multiplying out brackets’. This is based on the distributive law, which states that for any three numbers a, b and c a(b 1 c) 5 ab 1 ac It is easy to verify this law in simple cases. For example, if a = 2, b = 3 and c = 4, then the lefthand side is 2(3 + 4) = 2 × 7 = 14 However, ab = 2 × 3 = 6 and ac = 2 × 4 = 8 and so the righthand side is 6 + 8, which is also 14. This law can be used when there are any number of terms inside the brackets. We have a(b + c + d) = ab + ac + ad a(b + c + d + e) = ab + ac + ad + ae and so on. Section 1.1 Introduction to algebra 13 It does not matter in which order two numbers are multiplied, so we also have (b + c)a = ba + ca (b + c + d)a = ba + ca + da (b + c + d + e)a = ba + ca + da + ea Example Multiply out the brackets in (a) x(x − 2) (b) 2(x + y − z) + 3(z + y) (c) x + 3y − (2y + x) Solution (a) The use of the distributive law to multiply out x(x − 2) is straightforward. The x outside the bracket multiplies the x inside to give x2. The x outside the bracket also multiplies the −2 inside to give −2x. Hence x(x − 2) = x2 − 2x (b) To expand 2(x + y − z) + 3(z + y) we need to apply the distributive law twice. We have 2(x + y − z) = 2x + 2y − 2z 3(z + y) = 3z + 3y Adding together gives 2(x + y − z) + 3(z + y) = 2x + 2y − 2z + 3z + 3y = 2x + 5y + z (collecting like terms) (c) It may not be immediately apparent how to expand x + 3y − (2y + x) However, note that −(2y + x) is the same as (−1)(2y + x) which expands to give (−1)(2y) + (−1)x = −2y − x Hence x + 3y − (2y + x) = x + 3y − 2y − x = y after collecting like terms. 14 Chapter 1 Linear Equations Advice In this example the solutions are written out in painstaking detail. This is done to show you precisely how the distributive law is applied. The solutions to all three parts could have been written down in only one or two steps of working. You are, of course, at liberty to compress the working in your own solutions, but please do not be tempted to overdo this. You might want to check your answers at a later date and may find it difficult if you have tried to be too clever. Practice Problem 5. Multiply out the brackets, simplifying your answer as far as possible. (a) (5 − 2z)z (b) 6(x − y) + 3(y − 2x) (c) x − y + z − (x2 + x − y) Mathematical formulae provide a precise way of representing calculations that need to be worked out in many business models. However, it is important to realise that these formulae may be valid only for a restricted range of values. Most large companies have a policy to reimburse employees for use of their cars for travel: for the first 50 miles they may be able to claim 90 cents a mile, but this could fall to 60 cents a mile thereafter. If the distance, x miles, is no more than 50 miles, then travel expenses, E (in dollars), could be worked out using the formula E = 0.9x. If x exceeds 50 miles, the employee can claim $0.90 a mile for the first 50 miles but only $0.60 a mile for the last (x − 50) miles. The total amount is then E = 0.9 × 50 + 0.6(x − 50) = 45 + 0.6x − 30 = 15 + 0.6x Travel expenses can therefore be worked out using two separate formulae: ● E = 0.9x when x is no more than 50 miles ● E = 15 + 0.6x when x exceeds 50 miles. Before we leave this topic, a word of warning is in order. Be careful when removing brackets from very simple expressions such as those considered in part (c) in the previous worked example and practice problem. A common mistake is to write (a + b) − (c + d) = a + b − c + d This is NOT true The distributive law tells us that the −1 multiplying the second bracket applies to the d as well as the c, so the correct answer has to be (a + b) − (c + d) = a + b − c − d In algebra, it is sometimes useful to reverse the procedure and put the brackets back in. This is called factorisation. Consider the expression 12a + 8b. There are many numbers which divide into both 8 and 12. However, we always choose the biggest number, which is 4 in this case, so we attempt to take the factor of 4 outside the brackets: 12a + 8b = 4(? + ?) Section 1.1 Introduction to algebra 15 where each ? indicates a mystery term inside the brackets. We would like 4 multiplied by the first term in the brackets to be 12a, so we are missing 3a. Likewise, if we are to generate an 8b, the second term in the brackets will have to be 2b. Hence 12a + 8b = 4(3a + 2b) As a check, notice that when you expand the brackets on the righthand side, you really do get the expression on the lefthand side. Example Factorise (a) 6L − 3L2 (b) 5a − 10b + 20c Solution (a) Both terms have a common factor of 3. Also, because L2 = L × L, both 6L and −3L2 have a factor of L. Hence we can take out a common factor of 3L altogether. 6L − 3L2 = 3L(2) − 3L(L) = 3L(2 − L) (b) All three terms have a common factor of 5, so we write 5a − 10b + 20c = 5(a) − 5(2b) + 5(4c) = 5(a − 2b + 4c) Practice Problem 6. Factorise (a) 7d + 21 (b) 16w − 20q (c) 6x − 3y + 9z (d) 5Q − 10Q2 We conclude our discussion of brackets by describing how to multiply two brackets together. In the expression (a + b)(c + d) the two terms a and b must each multiply the single bracket (c + d), so (a + b)(c + d) = a(c + d) + b(c + d) The first term a(c + d) can itself be expanded as ac + ad. Likewise, b(c + d) = bc + bd. Hence (a + b)(c + d) = ac + ad + bc + bd This procedure then extends to brackets with more than two terms: (a + b)(c + d + e) = a(c + d + e) + b(c + d + e) = ac + ad + ae + bc + bd + be 16 Chapter 1 Linear Equations Example Multiply out the brackets (a) (x + 1)(x + 2) (b) (x + 5)(x − 5) (c) (2x − y)(x + y − 6) simplifying your answer as far as possible. Solution (a) (x + 1)(x + 2) = x(x + 2) + (1)(x + 2) = x2 + 2x + x + 2 = x2 + 3x + 2 (b) (x + 5)(x − 5) = x(x − 5) + 5(x − 5) = x2 − 5x + 5x − 25 = x2 − 25 the xs cancel (c) (2x − y)(x + y − 6) = 2x(x + y − 6) − y(x + y − 6) = 2x2 + 2xy − 12x − yx − y2 + 6y = 2x2 + xy − 12x − y2 + 6y Practice Problem 7. Multiply out the brackets. (a) (x + 3)(x − 2) (b) (x + y)(x − y) (c) (x + y)(x + y) (d) (5x + 2y)(x − y + 1) Looking back at part (b) of the previous worked example, notice that (x + 5)(x − 5) = x2 − 25 = x2 − 52 Quite generally (a + b)(a − b) = a(a − b) + b(a − b) = a2 − ab + ba − b2 = a2 − b2 The result a2 ] b2 5 (a 1 b)(a ] b) is called the difference of two squares formula. It provides a quick way of factorising certain expressions. Section 1.1 Introduction to algebra 17 Example Factorise the following expressions: (a) x2 − 16 (b) 9x2 − 100 Solution (a) Noting that x2 − 16 = x2 − 42 we can use the difference of two squares formula to deduce that x2 − 16 = (x + 4)(x − 4) (b) Noting that 9x2 − 100 = (3x)2 − (10)2 (3x)2 5 3x 3 3x 5 9x 2 we can use the difference of two squares formula to deduce that 9x2 − 100 = (3x + 10)(3x − 10) Practice Problem 8. Factorise the following expressions: (a) x2 − 64 (b) 4x2 − 81 Advice This completes your first piece of mathematics. We hope that you have not found it quite as bad as you first thought. There now follow a few extra problems to give you more practice. Not only will they help to strengthen your mathematical skills, but also they should improve your overall confidence. Two alternative exercises are available. Exercise 1.1 is suitable for students whose mathematics may be rusty and who need to consolidate their understanding. Exercise 1.1* contains more challenging problems and so is more suitable for those students who have found this section very easy. Key Terms Difference of two squares The algebraic result which states that a2 − b2 = (a + b)(a − b). Distributive law The law of arithmetic which states that a(b + c) = ab + ac for any numbers, a, b, c. Factorisation brackets. Like terms The process of writing an expression as a product of simpler expressions using Multiples of the same combination of algebraic symbols. 18 Chapter 1 Linear Equations Exercise 1.1 1. Without using a calculator, evaluate (a) 10 × (−2) (e) 24 ÷ (−2) (b) (−1) × (−3) (c) (−8) ÷ 2 (d) (−5) ÷ (−5) (f) (−10) × (−5) 20 (g) –4 (h) (i) (−6) × 5 × (−1) ( j) –27 –9 2 3 (–6) 3 3 (–9) 2. Without using a calculator, evaluate (a) 5 − 6 (b) −1 − 2 (c) 6 − 17 (d) −7 + 23 (e) −7 − (−6) (f) −4 − 9 (g) 7 − (−4) (h) −9 − (−9) (i) 12 − 43 ( j) 2 + 6 − 10 3. Without using a calculator, evaluate –30 – 6 –18 (a) 5 × 2 − 13 (b) (e) 1 − 6 × 7 (f) −5 + 6 ÷ 3 (i) (−2)2 − 5 × 6 + 1 (c) (–3) 3 (–6) 3 (–1) (d) 5 × (1 − 4) 2–3 (g) 2 × (−3)2 ( j) (h) −10 + 22 (–4)2 3 (–3) 3 (–1) (–2)3 4. Simplify each of the following algebraic expressions: (a) 2 × P × Q (b) I × 8 (c) 3 × x × y (d) 4 × q × w × z (e) b × b (f) k × 3 × k 5. Simplify the following algebraic expressions by collecting like terms: (a) 6w − 3w + 12w + 4w (b) 6x + 5y − 2x − 12y (c) 3a − 2b + 6a − c + 4b − c (d) 2x2 + 4x − x2 − 2x (e) 2cd + 4c − 5dc (f) 5st + s2 − 3ts + t2 + 9 6. Without using a calculator, find the value of the following: (a) 2x − y when x = 7 and y = 4. (b) x2 − 5x + 12 when x = 6. (c) 2m3 when m = 10. (d) 5fg2 + 2g when f = 2 and g = 3. (e) 2v + 4w − (4v − 7w) when v = 20 and w = 10. 7. If x = 2 and y = −3, evaluate (a) 2x + y (b) x − y (c) 3x + 4y (d) xy (e) 5xy (f) 4x − 6xy Section 1.1 Introduction to algebra 19 8. (a) Without using a calculator, work out the value of (−4)2. (b) Press the following key sequence on your calculator: (]) 4 x2 Explain carefully why this does not give the same result as part (a) and give an alternative key sequence that does give the correct answer. 9. Without using a calculator, work out (a) (5 − 2)2 (b) 52 − 22 Is it true in general that (a − b)2 = a2 − b2? 10. Use your calculator to work out the following. Round your answer, if necessary, to two decimal places. (a) 5.31 × 8.47 − 1.012 (b) (8.34 + 2.27)/9.41 (c) 9.53 − 3.21 + 4.02 (d) 2.41 × 0.09 − 1.67 × 0.03 (e) 45.76 − (2.55 + 15.83) (f) (3.45 − 5.38)2 (g) 4.56(9.02 + 4.73) (h) 6.85/(2.59 + 0.28) 11. Multiply out the brackets: (a) 7(x − y) (b) 3(5x − 2y) (c) 4(x + 3) (d) 7(3x − 1) (e) 3(x + y + z) (f) x(3x − 4) (g) y + 2z − 2(x + 3y − z) (a) 25c + 30 (b) 9x − 18 (c) x 2 + 2x (d) 16x − 12y (e) 4x2 − 6xy (f) 10d − 15e + 50 12. Factorise 13. Multiply out the brackets: (a) (x + 2)(x + 5) (b) (a + 4)(a − 1) (c) (d + 3)(d − 8) (d) (2s + 3)(3s + 7) (e) (2y + 3)(y + 1) (f) (5t + 2)(2t − 7) (g) (3n + 2)(3n − 2) (h) (a − b)(a − b) 14. Simplify the following expressions by collecting together like terms: (a) 2x + 3y + 4x − y (b) 2x2 − 5x + 9x2 + 2x − 3 (c) 5xy + 2x + 9yx (d) 7xyz + 3yx − 2zyx + yzx − xy (e) 2(5a + b) − 4b (f) 5(x − 4y) + 6(2x + 7y) (g) 5 − 3(p − 2) (h) x(x − y + 7) + xy + 3x 15. Use the formula for the difference of two squares to factorise (a) x2 − 4 (b) Q2 − 49 (c) x2 − y2 (d) 9x2 − 100y2 16. Simplify the following algebraic expressions: (a) 3x − 4x2 − 2 + 5x + 8x2 (b) x(3x + 2) − 3x(x + 5) ➜ 20 Chapter 1 Linear Equations 17. A law firm seeks to recruit topquality experienced lawyers. The total package offered is the sum of three separate components: a basic salary which is 1.2 times the candidate’s current salary together with an additional $3000 for each year worked as a qualified lawyer and an extra $1000 for every year that they are over the age of 21. Work out a formula that could be used to calculate the total salary, S, offered to someone who is A years of age, has E years of relevant experience and who currently earns $N. Hence work out the salary offered to someone who is 30 years old with five years’ experience and who currently earns $150 000. 18. Write down a formula for each situation: (a) A plumber has a fixed callout charge of $80 and has an hourly rate of $60. Work out the total charge, C, for a job that takes L hours in which the cost of materials and parts is $K. (b) An airport currency exchange booth charges a fixed fee of $10 on all transactions and offers an exchange rate of 1 dollar to 0.8 euros. Work out the total charge, C, (in $) for buying x euros. (c) A firm provides 5 hours of inhouse training for each of its semiskilled workers and 10 hours of training for each of its skilled workers. Work out the total number of hours, H, if the firm employs a semiskilled and b skilled workers. (d) A car hire company charges $C a day together with an additional $c per mile. Work out the total charge, $X, for hiring a car for d days and travelling m miles during that time. Exercise 1.1* 1. Without using a calculator, evaluate (a) (12 − 8) − (6 − 5) (b) 12 − (8 − 6) − 5 (c) 12 − 8 − 6 − 5 2. Put a pair of brackets in the lefthand side of each of the following to give correct statements: (a) 2 − 7 − 9 + 3 = −17 (b) 8 − 2 + 3 − 4 = −1 (c) 7 − 2 − 6 + 10 = 1 3. Without using a calculator, work out the value of each of the following expressions in the case when a = 3, b = −4 and c = −2: (a) a(b − c) (e) c+b 2a (b) 3c(a + b) (f) 2(b2 – c) (c) a2 + 2b + 3c (g) b a – 2c 3b (d) 2abc2 (h) 5a − b3 − 4c2 4. Without using a calculator, evaluate each of the following expressions in the case when x = −1, y = −2 and z = 3: (a) x3 + y2 + z (b) x2 1 y2 1 z x2 1 2xy 2 z 5. Multiply out the brackets and simplify (x − y)(x + y) − (x + 2)(x − y + 3) (c) xyz(x 1 z)(z 2 y) (x 1 y)(x 2 z) Section 1.1 Introduction to algebra 21 6. Simplify (a) x − y − ( y − x) (b) (x − ((y − x) − y)) (c) x + y − (x − y) − (x − (y − x)) (a) (x + 4)(x − 6) (b) (2x − 5)(3x − 7) (c) 2x(3x + y − 2) (d) (3 + g)(4 − 2g + h) (e) (2x + y)(1 − x − y) (f) (a + b + c)(a − b − c) (a) 9x − 12y (b) x2 − 6x (c) 10xy + 15x2 (d) 3xy2 − 6x2y + 12xy (e) x3 − 2x2 (f) 60x4y6 − 15x2y4 + 20xy3 7. Multiply out the brackets: 8. Factorise 9. Use the formula for the difference of two squares to factorise (a) p2 − 25 (b) 9c2 − 64 (c) 32v2 − 50d 2 (d) 16x4 − y4 10. Evaluate the following without using a calculator: (a) 50 5632 − 49 4372 (b) 902 − 89.992 (c) 7592 − 5412 (d) 123 456 7892 − 123 456 7882 11. A specialist paint manufacturer receives $12 for each pot sold. The initial setup cost for the production run is $800 and the cost of making each tin of paint is $3. (a) Write down a formula for the total profit, π, if the firm manufactures x pots of paint and sells y pots. (b) Use your formula to calculate the profit when x = 1000 and y = 800. (c) State any restrictions on the variables in the mathematical formula in part (a). (d) Simplify the formula in the case when the firm sells all that it manufactures. 12. Factorise (a) 2KL2 + 4KL (b) L2 − 0.04K2 (c) K2 + 2LK + L2 Section 1.2 Further algebra Objectives At the end of this section you should be able to: ● Simplify fractions by cancelling common factors. ● Add, subtract, multiply and divide fractions. ● Solve equations by doing the same thing to both sides. ● Recognise ● the symbols <, >, ≤ and ≥. Solve linear inequalities. This section is broken down into three manageable subsections: ● fractions; ● equations; ● inequalities. The advice offered in Section 1.1 applies equally well here. Please try to study these topics on separate occasions and be prepared to work through the practice problems as they arise in the text. 1.2.1 Fractions For a numerical fraction such as 7 8 the number 7, on the top, is called the numerator and the number 8, on the bottom, is called the denominator. In this text we are also interested in the case when the numerator and denominator involve letters as well as numbers. These are referred to as algebraic fractions. For example, 1 2 x 22 and 2x 2 2 1 y1z are both algebraic fractions. The letters x, y and z are used to represent numbers, so the rules for the manipulation of algebraic fractions are the same as those for ordinary numerical fractions. It is therefore essential that you are happy manipulating numerical fractions without a calculator so that you can extend this skill to fractions with letters. Two fractions are said to be equivalent if they represent the same numerical value. We know that 3/4 is equivalent to 6/8 since they are both equal to the decimal number 0.75. It is also intuitively obvious. Imagine breaking a bar of chocolate into four equal pieces and eating three Section 1.2 Further algebra 23 of them. You eat the same amount of chocolate as someone who breaks the bar into eight equal pieces and eats six of them. Each piece is only half the size so you need to compensate by eating twice as many. Formally, we say that when the numerator and denominator are both multiplied by the same number, the value of the fraction remains unchanged. In this example we have 3 332 6 5 5 4 432 8 This process can be reversed so equivalent fractions are produced when the numerator and denominator are both divided by the same number. For example, 16 16/8 2 5 5 24 24/8 3 so the fractions 16/24 and 2/3 are equivalent. A fraction is said to be in its simplest form or reduced to its lowest terms when there are no factors common to both the numerator and denominator. To express any given fraction in its simplest form, you need to find the highest common factor of the numerator and denominator and then divide the top and bottom of the fraction by this. Example Reduce each of the following fractions to its lowest terms: (a) 14 48 2x 3a x22 (b) (c) (d) (e) 21 60 3xy 6a 1 3b (x 2 2)(x 1 1) Solution (a) The largest number which divides into both 14 and 21 is 7, so we choose to divide top and bottom by 7: 14 14/7 2 5 5 21 21/7 3 An alternative way of writing this (which will be helpful when we tackle algebraic fractions) is: 14 2 3 7 2 5 5 21 3 3 7 3 (b) The highest common factor of 48 and 60 is 12, so we write: 48 4 3 12 4 5 5 60 5 3 12 5 (c) The factor x is common to both 2x and 3xy, so we need to divide top and bottom by x; that is, we cancel the xs: 2x 23x 2 5 5 3xy 3 3 x 3 y 3y (d) Factorising the denominator gives 6a + 3b = 3(2a + b) ➜ 24 chApter 1 LInEAr EqUATIonS which shows that there is a common factor of 3 in the top and bottom which can be cancelled: 3a 3a a 5 5 6a 1 3b 3(2a 1 b) 2a 1 b (e) We see immediately that there is a common factor of (x − 2) in the top and bottom, so this can be cancelled: x22 (x 2 2) (x 1 1) 5 1 x 11 Before we leave this topic, a word of warning is in order. Notice that you can only cancel by dividing by a factor of the numerator or denominator. In part (d) of the above example you must not get carried away and attempt to cancel the as, and write something daft like: a 1 5 2a 1 b 2 1 b This is NOT true To see that this is totally wrong, let us try substituting numbers, a = 3, b = 4, say, into both sides. The lefthand side gives a 3 3 whereas the righthand side gives 5 5 2a 1 b 23314 10 1 1 1 5 5 , which is not the same value. 21b 214 6 practice problem 1. Reduce each of the following fractions to its lowest terms: (a) 9 15 (b) 24 30 (c) x 2xy (d) 3x 6x 1 9x2 (e) x(x 1 1) x(x 2 4)(x 1 1) The rules for multiplication and division are as follows: to multiply fractions you multiply their corresponding numerators and denominators In symbols, a c a 3 c ac 3 5 5 b d b 3 d bd to divide by a fraction you turn it upside down and multiply In symbols, a c a d 4 5 3 b d b c 5 ad bc turn the divisor upside down rule of multiplying fractions Section 1.2 FUrTHEr ALgEbrA example Calculate (a) 2 5 3 3 4 (b) 2 3 6 13 (c) 6 4 4 7 21 (d) 1 43 2 Solution (a) The multiplication rule gives 2 5 2 3 5 10 5 3 5 3 4 3 3 4 12 We could leave the answer like this, although it can be simplified by dividing top and bottom by 2 to get 5/6. It is also valid to ‘cancel’ by 2 at the very beginning: that is, 1 2 5 135 5 3 5 5 3 42 332 6 (b) The whole number 2 is equivalent to the fraction 2/1, so 23 12 6 2 6 236 5 3 5 5 13 1 13 1 3 13 13 (c) To calculate 6 4 4 7 21 the divisor is turned upside down to get 21/4 and then multiplied to get 3 3 6 4 6 21 333 9 4 3 5 5 5 7 21 71 42 132 2 (d) We write 3 as 3/1, so 1 1 3 1 1 1 435 4 5 3 5 2 2 1 2 3 6 practice problem 2. (1) Without using a calculator, evaluate (a) 1 3 3 2 4 (b) 7 3 1 4 (c) 2 8 4 3 9 (d) 8 4 16 9 (2) Confirm your answer to part (1) using a calculator. The rules for addition and subtraction are as follows: to add (or subtract) two fractions you write them as equivalent fractions with a common denominator and add (or subtract) their numerators 25 26 Chapter 1 Linear Equations Example Calculate (a) 1 2 1 2 7 5 1 (b) 1 (c) 2 5 5 4 3 12 8 Solution (a) The fractions 1/5 and 2/5 already have the same denominator, so to add them we just add their numerators to get 1 2 112 3 5 1 5 5 5 5 5 (b) The fractions 1/4 and 2/3 have denominators 4 and 3. One number that is divisible by both 3 and 4 is 12, so we choose this as the common denominator. Now 4 goes into 12 exactly 3 times, so 1 133 3 5 5 4 4 3 3 12 multiply top and bottom by 3 and 3 goes into 12 exactly 4 times, so 2 234 8 5 5 3 3 3 4 12 multiply top and bottom by 4 Hence 1 2 3 8 3 1 8 11 5 5 1 5 1 4 3 12 12 12 12 (c) The fractions 7/12 and 5/8 have denominators 12 and 8. One number that is divisible by both 12 and 8 is 24, so we choose this as the common denominator. Now 12 goes into 24 exactly twice, so 7 732 14 5 5 12 24 24 and 8 goes into 24 exactly 3 times, so 5 5 3 3 15 5 5 8 24 24 Hence 7 5 14 15 21 2 5 2 5 12 8 24 24 24 It is not essential that the lowest common denominator is used. Any number will do provided that it is divisible by the two original denominators. If you are stuck, then you could always multiply the original two denominators together. In part (c) the denominators multiply to give 96, so this can be used instead. Now 7 7 3 8 56 5 5 12 96 96 Section 1.2 Further algebra 27 and 5 60 5 3 12 5 5 8 96 96 so 7 5 56 60 56 2 60 24 1 2 5 2 5 5 52 12 8 96 96 96 96 24 as before. Notice how the final answer to part (c) of this example has been written. We have simply used the fact that when a negative number is divided by a positive number, the answer is negative. It is standard practice to write negative fractions like this, so we would write 2 in preference to either 3 4 23 3 23 3 or and, of course, is written as . 4 4 24 24 Before we leave this topic a word of warning is in order. Notice that you can add or subtract fractions only after you have gone to the trouble of finding a common denominator. In particular, the following shortcut does not give the correct answer: a c a1c This is NOT true 1 5 b d b1d As usual you can check for yourself that it is complete rubbish by using actual numbers of your own choosing. Practice Problem 3. (1) Without using a calculator, evaluate (a) 3 1 1 2 7 1 2 (b) 1 (c) 2 7 7 3 5 18 4 (2) Confirm your answer to part (1) using a calculator. Provided that you can manipulate ordinary fractions, there is no reason why you should not be able to manipulate algebraic fractions just as easily since the rules are the same. Example Find expressions for each of the following: (a) x 2 x x26 2 x11 x 1 (b) (c) 2 (d) 3 4 1 2 2 x 2 1 x(x 1 4) x21 x21 x 12 x 12 x12 x11 ➜ 28 Chapter 1 Linear Equations Solution (a) To multiply two fractions we multiply their corresponding numerators and denomin ators, so x 2 2x 2 3 5 5 x 2 1 x(x 1 4) (x 2 1)x(x 1 4) (x 2 1)(x 1 4) the xs cancel top and bottom (b) To divide by x x21 we turn it upside down and multiply, so 2 x 2 x21 2 4 5 3 5 x x x21 x21 x21 the (x ] 1)s cancel top and bottom (c) The fractions x11 x2 1 2 and x26 x2 1 2 already have the same denominator, so to add them we just add their numerators to get x26 x11 x111x26 2x 2 5 1 2 5 5 2 2 2 x 12 x 12 x 12 x 12 (d) The fractions x x12 and 1 x11 have denominators x + 2 and x + 1. An obvious common denominator is given by their product, (x + 2)(x + 1). Now x + 2 goes into (x + 2)(x + 1) exactly x + 1 times, so x x(x 1 1) 5 x 1 2 (x 1 2)(x 1 1) multiply top and bottom by (x 1 1) Also, x + 1 goes into (x + 2)(x + 1) exactly x + 2 times, so 1 (x 1 2) 5 x11 (x 1 2)(x 1 1) multiply top and bottom by (x 1 2) Hence x 1 x(x 1 1) (x 1 2) x(x 1 1) 2 (x 1 2) 2 5 2 5 x 1 2 x 1 1 (x 1 2)(x 1 1) (x 1 2)(x 1 1) (x 1 2)(x 1 1) It is worth multiplying out the brackets on the top to simplify: that is, x2 1 x 2 x 2 2 x2 2 2 5 (x 1 2)(x 1 1) (x 1 2)(x 1 1) Section 1.2 FUrTHEr ALgEbrA 29 practice problem 4. Find expressions for the following algebraic fractions, simplifying your answers as far as possible. (a) 5 x21 3 x21 x12 (b) x2 x 4 x 1 10 x11 (c) 4 1 1 x11 x11 (d) 2 1 2 x11 x12 1.2.2 equations In Section 1.1.2 and again in Section 1.2.1, we have seen how to rewrite an algebraic expression in a simpler but equivalent form. For example, when we write things like x2 + 3x + 3x2 − 10x = 4x2 − 7x (collecting like terms) 1 x x2 2 2 2 5 x 1 2 x 1 1 (x 1 2)(x 1 1) (part (d) of the previous worked example) or we have at the back of our minds the knowledge that the left and righthand sides are identical so that each statement is true for all possible values of x. For this reason the above relations are called identities. Compare these with statements such as: 7x − 1 = 13 or x2 − 5x = 1 These relations are called equations and are only true for particular values of x which need to be found. It turns out that the first equation above has just one solution, whereas the second has two solutions. The latter is called a quadratic equation and will be considered in the next chapter. One naïve approach to the solution of equations such as 7x − 1 = 13 might be to use trial and error: that is, we could just keep guessing values of x until we find the one that works. Can you see what x is in this case? However, a more reliable and systematic approach is to actually solve this equation using the rules of mathematics. In fact, the only rule that we need is: you can apply whatever mathematical operation you like to an equation, provided that you do the same thing to both sides There is only one exception to this rule: you must never divide both sides by zero. This should be obvious because a number such as 11/0 does not exist. (If you do not believe this, try dividing 11 by 0 on your calculator.) The first obstacle that prevents us from writing down the value of x immediately from the equation 7x − 1 = 13 is the presence of the −1 on the lefthand side. This can be removed by adding 1. For this to be legal we must also add 1 to the righthand side to get 7x − 1 + 1 = 13 + 1 7x = 14 30 Chapter 1 Linear Equations The second obstacle is the number 7 which is multiplying the x. This can be removed by dividing the lefthand side by 7. Of course, we must also do the same thing to the righthand side to get 7x 14 5 7 7 x=2 This is no doubt the solution that you spotted earlier by simple trial and error, and you may be wondering why you need to bother with the formal method. The reason is simple: guesswork will not help to solve more complicated equations in which the solution is nonobvious or even simple equations in which the solution is a fraction. In these circumstances we need to follow the approach of ‘balancing the equation’. Example Solve (a) 6x + 1 = 10x − 9 (c) (b) 3(x − 1) + 2(2x + 1) = 4 20 57 3x 2 1 (d) 9 7 5 x 1 2 2x 1 1 (e) 2x 52 x26 Solution (a) To solve 6x + 1 = 10x − 9 the strategy is to collect terms involving x on one side of the equation, and to collect all of the number terms on to the other side. It does not matter which way round this is done. In this particular case, there are more xs on the righthand side than there are on the lefthand side. Consequently, to avoid negative numbers, you may prefer to stack the x terms on the righthand side. The details are as follows: 1 5 4x 2 9 10 5 4x 10 5x 4 (subtract 6x from both sides) (add 9 to both sides) (divide both sides by 4) Hence x = 5/2 = 21/2. (b) The novel feature of the equation 3(x − 1) + 2(2x + 1) = 4 is the presence of brackets. To solve it, we first remove the brackets by multiplying out, and then collect like terms: 3x − 3 + 4x + 2 = 4 (multiply out the brackets) 7x − 1 = 4 (collect like terms) Note that this equation is now of the form that we know how to solve: 7x 5 5 5 x5 7 (add 1 to both sides) (divide both sides by 7) Section 1.2 Further algebra 31 (c) The novel feature of the equation 20 57 3x 2 1 is the fact that it involves an algebraic fraction. This can easily be removed by multi plying both sides by the bottom of the fraction: 20 3 (3x 2 1) 5 7(3x 2 1) 3x 2 1 which cancels down to give 20 = 7(3x − 1) The remaining steps are similar to those in part (b): (multiply out the brackets) (add 7 to both sides) 20 5 21x 2 7 27 5 21x 27 5x 21 (divide both sides by 21) Hence x = 9/7 = 12/7. (d) The next equation, 9 7 5 x 1 2 2x 1 1 looks particularly daunting since there are fractions on both sides. However, these are easily removed by multiplying both sides by the denominators, in turn: 7(x 1 2) 2x 1 1 9(2x + 1) = 7(x + 2) 95 (multiply both sides by x + 2) (multiply both sides by 2x + 1) With practice you can do these two steps simultaneously and write this as the first line of working. The procedure of going straight from 9 7 5 x 1 2 2x 1 1 to 9(2x + 1) = 7(x + 2) is called ‘crossmultiplication’. In general, if a c 5 b d then a b c d ad = bc The remaining steps are similar to those used in the earlier parts of this example: 18x + 9 = 7x + 14 (multiply out the brackets) 11x + 9 = 14 (subtract 7x from both sides) ➜ 32 Chapter 1 Linear Equations (subtract 9 from both sides) 11x = 5 5 x5 11 (divide both sides by 11) (e) The lefthand side of the final equation 2x 52 x26 is surrounded by a square root, which can easily be removed by squaring both sides to get 2x 54 x26 The remaining steps are ‘standard’: 2x = 4(x − 6) (multiply both sides by x − 6) 2x = 4x − 24 (multiply out the brackets) −2x = −24 (subtract 4x from both sides) (divide both sides by −2) x = 12 Looking back over each part of the previous example, notice that there is a common strategy. In each case, the aim is to convert the given equation into one of the form ax + b = c which is the sort of equation that we can easily solve. If the original equation contains brackets, then remove them by multiplying out. If the equation involves fractions, then remove them by crossmultiplying. Advice If you have the time, it is always worth checking your answer by substituting your solution back into the original equation. For the last part of the above example, putting x = 12 into 2x gives x2 6 2 3 12 24 5 5 4 5 2 ✓ 12 2 6 6 Practice Problem 5. Solve each of the following equations. Leave your answer as a fraction, if necessary. (a) 4x + 1 = 25 (d) 4 55 x21 (b) 4x + 5 = 5x − 7 (e) 3 5 5 x x21 (c) 3(3 − 2x) + 2(x − 1) = 10 Section 1.2 Further algebra 33 1.2.3 Inequalities In Section 1.1.1 we made use of a number line: ]5 ]4 ]3 ]2 0 ]1 1 2 3 4 5 Although only whole numbers are marked on this diagram, it is implicitly assumed that it can also be used to indicate fractions and decimal numbers. Each point on the line corresponds to a particular number. Conversely, every number can be represented by a particular point on the line. For example, −21/2 lies exactly halfway between −3 and −2. Similarly, 47/8 lies 7/8 of the way between 4 and 5. In theory, we can even find a point on the line corresponding to a number such as 2 , although it may be difficult to sketch such a point accurately in practice. My calculator gives the value of 2 to be 1.414 213 56 to eight decimal places. This number therefore lies just less than halfway between 1 and 2. 1 ]5 ]4 ]3 7 2 ]2 2 ]2 ]1 0 1 48 2 3 4 5 A number line can be used to decide whether or not one number is greater or less than another number. We say that a number a is greater than a number b if a lies to the right of b on the line, and we write this as a>b Likewise, we say that a is less than b if a lies to the left of b, and we write this as a<b From the diagram we see that −2 > −4 because −2 lies to the right of −4. This is equivalent to the statement −4 < −2 Similarly, 0 > −1 (or equivalently −1 < 0) 2 > −2 /2 (or equivalently −21/2 < 2) 1 4 7/ 8 . 2 (or equivalently 2 , 4 7/ 8) There are occasions when we would like the letters a and b to stand for mathematical expressions rather than actual numbers. In this situation we sometimes use the symbols ≥ and ≤ to mean ‘greater than or equal to’ and ‘less than or equal to’, respectively. We have already seen that we can manipulate equations in any way we like, provided that we do the same thing to both sides. An obvious question to ask is whether this rule extends to inequalities. 34 chApter 1 LInEAr EqUATIonS Consider the true statement 1<3 (*) ● Adding 4 to both sides gives 5 < 7, which is true. ● Adding −5 to both sides gives −4 < −2, which is true. ● Multiplying both sides by 2 gives 2 < 6, which is true. However, ● Multiplying both sides by −6 gives −6 < −18, which is false. In fact, quite the reverse is true; −6 is actually greater than −18. This indicates that the rule needs modifying before we can extend it to inequalities and that we must be careful when manipulating inequalities. practice problem 6. Starting with the true statement 6>3 decide which of the following are valid operations when performed on both sides: (a) add 6 (b) multiply by 2 (c) subtract 3 (d) add −3 (e) divide by 3 (f) multiply by −4 (g) multiply by −1 (h) divide by −3 (i) add −10 These examples show that the usual rule does apply to inequalities with the important proviso that if both sides are multiplied or divided by a negative number, then the sense of the inequality is reversed By this we mean that ‘>’ changes to ‘<’, ‘≤’ changes to ‘≥’ and so on. To see how this works out in practice, consider the inequality 2x + 3 < 4x + 7 If we try to solve this like we did for equations, the first step would be to subtract 4x from both sides to get −2x + 3 < 7 and then take 3 away from both sides to get −2x < 4 Finally, we divide both sides by −2 to get x > −2 Notice that the sense has been reversed at this stage because we have divided by a negative number. Section 1.2 Further algebra 35 Advice You should check your answer using a couple of test values. Substituting x = 1 (which lies to the right of −2 and so should work) into both sides of the original inequality 2x + 3 < 4x + 7 gives 5 < 11, which is true. On the other hand, substituting x = −3 (which lies to the left of −2 and so should fail) gives −3 < −5, which is false. Of course, just checking a couple of numbers like this does not prove that the final inequality is correct, but it should protect you against gross blunders. Practice Problem 7. Simplify the inequalities (a) 2x < 3x + 7 (b) 21x − 19 ≥ 4x + 15 Inequalities arise in business when there is a budgetary restriction on resource allocation. The following example shows how to set up and solve the relevant inequality. Example A firm’s Human Resources department has a budget of $25 000 to spend on training and laptops. Training courses cost $700 and new laptops are $1200. (a) If the department trains E employees and buys L laptops, write down an inequality for E and L. (b) If 12 employees attend courses, how many laptops could be bought? Solution (a) The cost of training E employees is 700E and the cost of buying L laptops is 1200L. The total amount spent must not exceed $25 000, so 700E + 1200L ≤ 25 000. (b) Substituting E = 12 into the inequality gives 8400 + 1200L ≤ 25 000. 1200L ≤ 16 600 L # 13 5 6 (subtract 8400 from both sides) (divide both sides by 1200) so a maximum of 13 laptops could be bought. 36 Chapter 1 Linear Equations Key Terms Algebraic fraction Ratio of two expressions; p(x)/q(x) where p(x) and q(x) are algebraic expressions such as ax2 + bx + c or dx + e. Denominator The number (or expression) on the bottom of a fraction. Equation Equality of two algebraic expressions which is true only for certain values of the variable. Fractions which may appear different but which have the same numerical Equivalent fractions value. Factor Part of an expression which, when multiplied by all the other factors, gives the complete expression. Identity Equality of two algebraic expressions which is true for all values of the variable. Number line An infinite line on which the points represent real numbers by their (signed) distance from the origin. Numerator The number (or expression) on the top of a fraction. Exercise 1.2 1. Reduce each of the following numerical fractions to their lowest terms: (a) 13 26 (b) 9 12 (c) 18 30 (d) 24 72 (e) 36 27 2. In 2011 in the United States, 35 out of every 100 adults owned a smartphone. By 2013 this figure increased to 56 out of every 100. (a) Express both of these figures as fractions reduced to their lowest terms. (b) By what factor did smartphone ownership increase during this period? Give your answer as a mixed fraction in its lowest terms. 3. Reduce each of the following algebraic fractions to their lowest terms: (a) 6x 9 (b) x 2x 2 (c) b abc (d) 4x 6x 2y (e) 15a 2b 20ab2 4. By factorising the numerators and/or denominators of each of the following fractions, reduce each to its lowest terms: (a) 2p 4q 1 6r (b) x x 2 4x 2 (c) 3ab 6a 1 3a 2 (d) 14d 21d 2 7de (e) x12 x2 2 4 5. Which one of the following algebraic fractions can be simplified? Explain why the other two fractions cannot be simplified. x]1 , 2x ] 2 x22 , x12 5t 10t 2 s Section 1.2 Further algebra 37 6. (1) Without using a calculator, work out the following, giving your answer in its lowest terms: (a) 1 2 1 7 7 (b) 2 5 ] 9 9 (c) 1 1 1 2 3 (d) 3 2 ] 4 5 (e) 1 2 1 6 9 (f) 1 2 1 6 3 (g) 5 3 3 6 4 (h) 4 2 4 15 3 (i) 7 2 3 8 3 ( j) 2 4 4 75 5 (k) 2 43 9 (l) 3 4 2 7 (2) Use your calculator to check your answers to part (1). 7. It takes 11/4 hours to complete an annual service of a car. If a garage has 471/2 hours available, how many cars can it service? 8. Work out each of the following, simplifying your answer as far as possible: (a) 2 1 1 3x 3x (b) 2 x 3 x 5 (c) 3 2 2 2 x x (d) 7 2 1 x y (e) a a 4 2 6 (f) 5c 5d 1 12 18 (g) x12 y25 3 y25 x13 (h) 4gh 2g 4 7 9h (i) t 45 4 ( j) P Q 3 Q P 9. Solve each of the following equations. If necessary give your answer as a mixed fraction reduced to its lowest terms. (a) x + 2 = 7 (b) 3x = 18 (e) 2x − 3 = 17 (f) 3x + 4 = 1 (i) 4 − x = 9 (j) 6x + 2 = 5x − 1 (m) 4x 2 7 52 3 (n) 4 51 x11 x 52 9 x (g) 2 7 5 3 6 (c) (k) 5(3x + 8) = 10 (o) 5 2 (d) x − 4 = −2 (h) 3(x − 1) = 2 (l) 2(x − 3) = 5(x + 1) 1 51 x 10. Which of the following inequalities are true? (a) −2 < 1 (b) −6 > −4 (c) 3 < 3 (d) 3 ≤ 3 (e) −21 ≥ −22 (f) 4 < 25 11. Simplify the following inequalities: (a) 2x > x + 1 (b) 7x + 3 ≤ 9 + 5x (c) x − 5 > 4x + 4 12. Simplify the following algebraic expression: 4 2x 4 x 2y y 13. (a) Solve the equation 6(2 + x) = 5(1 − 4x) (b) Solve the inequality 3x + 6 ≥ 5x − 14 (d) x − 1 < 2x − 3 38 Chapter 1 Linear Equations Exercise 1.2* 1. Simplify each of the following algebraic fractions: (a) 2x 2 6 4 (b) 9x 6x2 2 3x (c) 4x 1 16 x14 (d) x21 12x (e) x16 x 2 2 36 (f) (x 1 3)(2x 2 5) (2x 2 5)(x 1 4) (g) 3x 3 6x 2 15x2 1 9x (h) 4x2 2 25y2 6x 2 15y (d) 2 1 8 3 4 5 12 25 2. (1) Without using your calculator, evaluate (a) 4 25 3 5 28 (b) 2 14 5 3 3 7 25 8 (c) 9 3 4 16 8 (e) 10 12 2 13 13 (f) 5 2 1 9 3 3 3 (g) 2 1 1 7 5 3 3 (i) 3 3 1 5 4 ( j) 3 1 1 3 2 1 5 3 2 (k) (h) 5 5 1 2 3 2 21 6 3 5 (l) 2 9 1 2 11 5 10 2 1 1 5 3 42 4 3 6 13 (2) Confirm your answer to part (1) using a calculator. 3. Find simplified expressions for the following fractions: 2 3 1 xy xy (a) x2 1 6x x22 3 x22 x (b) 1 1 4 x x11 (c) (e) 3 4 1 x x11 (f) 3 5 1 2 x x (g) x 2 2 x11 (d) x x11 1 2 3 (h) 5 2 3 2 1 x(x 1 1) x x11 4. Solve the following equations: (a) 5(2x + 1) = 3(x − 2) (b) 5(x + 2) + 4(2x − 3) = 11 (c) 5(1 − x) = 4(10 + x) (d) 3(3 − 2x) − 7(1 − x) = 10 (e) 9 − 5(2x − 1) = 6 (f) 3 52 2x 1 1 (h) x 1357 2 ( j) 5(x 2 3) 2(x 2 1) 5 2 5 (g) 2 3 5 x 2 1 5x 1 4 (i) 5 2 (k) x 52 3 (2x 2 5) 5 3 (m) (x + 2)2 + (2x − 1)2 = 5x(x + 1) (o) 45 53 2x 2 1 (l) (x + 3)(x − 1) = (x + 4)(x − 3) (n) 2x 1 7 x 2 4 1 5 1 2 3 6 (p) 4 3 1 2 5 x 4 4x 5. Twothirds of Ariadne’s money together with fivesevenths of Brian’s money is equal to threefifths of Catriona’s money. If Ariadne has $2.40 and Catriona has $11.25, write down an equation that you could use to work out how much Brian has. Solve this equation. Section 1.2 Further algebra 39 6. An amount $P is placed in a savings account. The interest rate is r% compounded annually so that after n years the savings, S, will be S 5 P 11 r 100 n (a) Find S when P = 2000, n = 5 and r = 10. (b) Find P when S = 65 563.62, n = 3 and r = 3. (c) Find r when S = 7320.50, P = 5000 and n = 4. 7. Solve the following inequalities: (a) 2x − 19 > 7x + 24 (d) 3 1 x , 2(x 1 4) 3 (b) 2(x − 1) < 5(3x + 2) (c) 2x 2 1 x23 $ 5 2 (e) x < 2x + 1 ≤ 7 8. The design costs of an advertisement in a glossy magazine are $9000 and the cost per cm2 of print is $50. (a) Write down an expression for the total cost of publishing an advert which covers x cm2. (b) The advertising budget is between $10 800 and $12 500. Write down and solve an inequality to work out the minimum and maximum area that could be used. 9. List all the whole numbers that satisfy both of the following inequalities simultaneously: −7 ≤ 2x < 6 and 4x + 1 ≤ x + 2 10. (a) Simplify 31x 2 8 14 2 (2x 2 1)(x 1 2) x 1 2 (b) Solve the equation x11 x13 1 2 5 8 4 2 (c) Simplify the inequality (2x + 1)(x − 5) ≤ 2(x + 2)(x − 4) 11. Simplify x2 2x 4 x 1 1 x2 2 1 Section 1.3 Graphs of linear equations Objectives At the end of this section you should be able to: ● Plot points on graph paper given their coordinates. ● Sketch a line by finding the coordinates of two points on the line. ● Solve simultaneous linear equations graphically. ● Sketch a line by using its slope and intercept. Consider the two straight lines shown in Figure 1.1. The horizontal line is referred to as the x axis and the vertical line is referred to as the y axis. The point where these lines intersect is known as the origin and is denoted by the letter O. These lines enable us to identify uniquely any point, P, in terms of its coordinates (x, y). The first number, x, denotes the horizontal distance along the x axis and the second number, y, denotes the vertical distance along the y axis. The arrows on the axes indicate the positive direction in each case. y P (x, y) y x O x Figure 1.1 Figure 1.2 shows the five points A(2, 3), B(−1, 4), C(−3, −1), D(3, −2) and E(5, 0) plotted on coordinate axes. The point A with coordinates (2, 3) is obtained by starting at the origin, moving 2 units to the right and then moving 3 units vertically upwards. Similarly, the point B with coordinates (−1, 4) is located 1 unit to the left of O (because the x coordinate is negative) and 4 units up. Note that the point C lies in the bottom lefthand quadrant since its x and y coordinates are both negative. It is also worth noticing that E actually lies on the x axis since its y coordinate is zero. Likewise, a point with coordinates of the form (0, y) for some number y would lie somewhere on the y axis. Of course, the point with coordinates (0, 0) is the origin, O. Section 1.3 Graphs of linear equations 41 y 5 B 4 A 3 2 1 E ]5 ]4 ]3 ]2 ]1 C O 1 2 3 4 5 ]1 D ]2 ]3 ]4 ]5 Figure 1.2 Practice Problem 1. Plot the following points on graph paper. What do you observe? (2, 5), (1, 3), (0, 1), (−2, −3), (−3, −5) In economics we need to do rather more than just plot individual points on graph paper. We would like to be able to sketch curves represented by equations and to deduce information from such a picture. We restrict our attention in this section to those equations whose graphs are straight lines, deferring consideration of more general curve sketching until Chapter 2. In Practice Problem 1 you will have noticed that the five points (2, 5), (1, 3), (0, 1), (−2, −3) and (−3, −5) all lie on a straight line. In fact, the equation of this line is −2x + y = 1 Any point lies on this line if its x and y coordinates satisfy this equation. For example, (2, 5) lies on the line because when the values x = 2, y = 5 are substituted into the lefthand side of the equation, we obtain −2(2) + 5 = −4 + 5 = 1 which is the righthand side of the equation. The other points can be checked similarly (see Table 1.1). Table 1.1 Point Check (1, 3) (0, 1) (−2, −3) (−3, −5) −2(1) + 3 = −2 + 3 = 1 −2(0) + 1 = 0 + 1 = 1 −2(−2) − 3 = 4 − 3 = 1 −2(−3) − 5 = 6 − 5 = 1 ✓ ✓ ✓ ✓ 42 Chapter 1 Linear Equations The general equation of a straight line takes the form a multiple of x 1 a multiple of y 5 a number that is, dx + ey = f for some given numbers d, e and f. Consequently, such an equation is called a linear equation. The numbers d and e are referred to as the coefficients. The coefficients of the linear equation, −2x + y = 1 are −2 and 1 (the coefficient of y is 1 because y can be thought of as 1 × y). Practice Problem 2. Check that the points (−1, 2), (−4, 4), (5, −2), (2, 0) all lie on the line 2x + 3y = 4 and hence sketch this line on graph paper. Does the point (3, −1) lie on this line? In general, to sketch a line from its mathematical equation, it is sufficient to calculate the coordinates of any two distinct points lying on it. These two points can be plotted on graph paper and a ruler used to draw the line passing through them. One way of finding the coordinates of a point on a line is simply to choose a numerical value for x and to substitute it into the equation. The equation can then be used to deduce the corresponding value of y. The whole process can be repeated to find the coordinates of the second point by choosing another value for x. Example Sketch the line 4x + 3y = 11 Solution For the first point, let us choose x = 5. Substitution of this number into the equation gives 4(5) + 3y = 11 20 + 3y = 11 The problem now is to solve this equation for y: 3y = −9 y = −3 (subtract 20 from both sides) (divide both sides by 3) Consequently, the coordinates of one point on the line are (5, −3). Section 1.3 Graphs of linear equations 43 For the second point, let us choose x = −1. Substitution of this number into the equation gives 4(−1) + 3y = 11 −4 + 3y = 11 This can be solved for y as follows: 3y = 15 y=5 (add 4 to both sides) (divide both sides by 3) Hence (−1, 5) lies on the line, which can now be sketched on graph paper as shown in Figure 1.3. y 5 (]1, 5) 4 3 4x 1 3y 5 11 2 1 ]5 ]4 ]3 ]2 1 ]1 2 3 4 5 x ]1 ]2 ]3 (5, ]3) ]4 ]5 Figure 1.3 Practice Problem 3. Find the coordinates of two points on the line 3x − 2y = 4 by taking x = 2 for the first point and x = −2 for the second point. Hence sketch its graph. In this example we arbitrarily picked two values of x and used the linear equation to work out the corresponding values of y. There is nothing particularly special about the variable x. We could equally well have chosen values for y and solved the resulting equations for x. In fact, the easiest thing to do (in terms of the amount of arithmetic involved) is to put x = 0 and find y and then to put y = 0 and find x. 44 Chapter 1 Linear Equations Example Sketch the line 2x + y = 5 Solution Setting x = 0 gives 2(0) + y = 5 0+y=5 y=5 Hence (0, 5) lies on the line. Setting y = 0 gives 2x + 0 = 5 2x = 5 x = 5/2 (divide both sides by 2) Hence (5/2, 0) lies on the line. The line 2x + y = 5 is sketched in Figure 1.4. Notice how easy the algebra is using this approach. The two points themselves are also slightly more meaningful. They are the points where the line intersects the coordinate axes. y y intercept 5 (0, 5) 4 2x 1 y 5 5 3 2 1 (2.5, 0) ]5 ]4 ]3 ]2 1 ]1 2 ]1 ]2 x intercept ]3 ]4 ]5 Figure 1.4 3 4 5 x Section 1.3 Graphs of linear equations 45 Practice Problem 4. Find the coordinates of the points where the line x − 2y = 2 intersects the axes. Hence sketch its graph. In economics it is sometimes necessary to handle more than one equation at the same time. For example, in supply and demand analysis we are interested in two equations, the supply equation and the demand equation. Both involve the same variables Q and P, so it makes sense to sketch them on the same diagram. This enables the market equilibrium quantity and price to be determined by finding the point of intersection of the two lines. We shall return to the analysis of supply and demand in Section 1.5. There are many other occasions in economics and business studies when it is necessary to determine the coordinates of points of intersection. The following is a straightforward example which illustrates the general principle. Example Find the point of intersection of the two lines 4x + 3y = 11 2x + y = 5 Solution We have already seen how to sketch these lines in the previous two examples. We discovered that 4x + 3y = 11 passes through (5, −3) and (−1, 5), and that 2x + y = 5 passes through (0, 5) and (5/2, 0). These two lines are sketched on the same diagram in Figure 1.5, from which the point of intersection is seen to be (2, 1). It is easy to verify that we have not made any mistakes by checking that (2, 1) lies on both lines. It lies on 4x + 3y = 11 because 4(2) + 3(1) = 8 + 3 = 11 ✓ and lies on 2x + y = 5 because 2(2) + 1 = 4 + 1 = 5. ✓ For this reason, we say that x = 2, y = 1 is the solution of the simultaneous linear equations 4x + 3y = 11 2x + y = 5 ➜ 46 Chapter 1 Linear Equations y 5 4 3 point of intersection 2 1 ]5 ]4 ]3 ]2 (2, 1) 1 ]1 2 3 4 x 5 ]1 4x 1 3y 5 11 ]2 ]3 2x 1 y 5 5 ]4 ]5 Figure 1.5 Practice Problem 5. Find the point of intersection of 3x − 2y = 4 x − 2y = 2 [Hint: you might find your answers to Problems 3 and 4 useful.] Quite often it is not necessary to produce an accurate plot of an equation. All that may be required is an indication of the general shape together with a few key points or features. It can be shown that, provided e is nonzero, any equation given by dx + ey = f can be rearranged into the special form y = ax + b An example showing you how to perform such a rearrangement will be considered in a moment. The coefficients a and b have particular significance, which we now examine. To be specific, consider y = 2x − 3 in which a = 2 and b = −3. Section 1.3 Graphs of linear equations 47 When x is taken to be zero, the value of y is y = 2(0) − 3 = −3 The line passes through (0, −3), so the y intercept is −3. This is just the value of b. In other words, the constant term, b, represents the intercept on the y axis. In the same way it is easy to see that a, the coefficient of x, determines the slope of a line. The slope of a straight line is simply the change in the value of y brought about by a 1 unit increase in the value of x. For the equation y = 2x − 3 let us choose x = 5 and increase this by a single unit to get x = 6. The corresponding values of y are then, respectively, y = 2(5) − 3 = 10 − 3 = 7 y = 2(6) − 3 = 12 − 3 = 9 The value of y increases by 2 units when x rises by 1 unit. The slope of the line is therefore 2, which is the value of a. The slope of a line is fixed throughout its length, so it is immaterial which two points are taken. The particular choice of x = 5 and x = 6 was entirely arbitrary. You might like to convince yourself of this by choosing two other points, such as x = 20 and x = 21, and repeating the previous calculations. A graph of the line y = 2x − 3 is sketched in Figure 1.6. This is sketched using the information that the intercept is −3 and that for every 1 unit along we go 2 units up. In this example the coefficient of x is positive. This does not have to be the case. If a is negative, then for every increase in x there is a y 3 2 2 units 1 ]2 1 unit 1 ]1 2 ]1 y 5 2x ] 3 ]2 intercept Figure 1.6 ]3 3 x 48 Chapter 1 Linear Equations y positive slope b zero slope x negative slope Figure 1.7 corresponding decrease in y, indicating that the line is downhill. If a is zero, then the equation is just y=b indicating that y is fixed at b and the line is horizontal. The three cases are illustrated in Figure 1.7. It is important to appreciate that in order to use the slope–intercept approach, it is necessary for the equation to be written as y = ax + b If a linear equation does not have this form, it is usually possible to perform a preliminary rearrangement to isolate the variable y on the lefthand side. For example, to use the slope–intercept approach to sketch the line 2x + 3y = 12 we begin by removing the x term from the lefthand side. Subtracting 2x from both sides gives 3y = 12 − 2x and dividing both sides by 3 gives y 5 4 2 23 x This is now in the required form with a = −2/3 and b = 4. The line is sketched in Figure 1.8. A slope of −2/3 means that, for every 1 unit along, we go 2/3 units down (or, equivalently, for every 3 units along, we go 2 units down). An intercept of 4 means that it passes through (0, 4). Practice Problem 6. Use the slope–intercept approach to sketch the lines (a) y = x + 2 (b) 4x + 2y = 1 Section 1.3 Graphs of linear equations 49 y 5 4 intercept 3 units 3 2 units 2 1 1 ]1 2 3 4 ]1 5 6 y54] 7 x 2 x 3 Figure 1.8 We conclude this section with two examples showing how linear graphs can be used in business. Example Two new models of a smartphone are launched on 1 January 2018. Predictions of sales are given by: Model 1: S1 = 4 + 0.5n Model 2: S2 = 8 + 0.1n where Si (in tens of thousands) denotes the monthly sales of model i after n months. (a) State the values of the slope and intercept of each line and give an interpretation. (b) Illustrate the sales of both models during the first year by drawing graphs on the same axes. (c) Use the graph to find the month when sales of Model 1 overtake those of Model 2. Solution (a) The intercept for Model 1 is 4. There are 40 000 sales of this phone when the product is launched. The slope is 0.5, so each month, sales increase by 5000. The corresponding figures for Model 2 are 8 and 0.1, respectively. (b) The intercept for Model 1 is 4, so the line passes through (0, 4). For every oneunit increase in n the value of S1 increases by 0.5. So, for example, a twounit increase in n results in a oneunit increase in S1. The line passes through (2, 5), (4, 6) and so on. The line is sketched in Figure 1.9. For Model 2, the line passes through (0, 8), and since the slope is 0.1, it passes through (10, 9). (c) The graphs intersect at (10, 9), so sales of Model 1 overtake sales of Model 2 after 10 months. ➜ 50 Chapter 1 Linear Equations 12 11 10 Model 2 (10, 9) Sales (in 10 000s) 9 8 7 6 Model 1 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 Time (months) Figure 1.9 Example Three companies can supply a university with some mathematical software. Each company has a different pricing structure: Company 1 provides a site licence which costs $130 000 and can be used by anyone at the university; Company 2 charges $1000 per user; Company 3 charges a fixed amount of $40 000 for the first 60 users and $500 for each additional user. (a) Draw a graph of each cost function on the same set of axes. (b) What advice can you give the university about which company to use? Solution (a) If there are n users, then the cost, C, from each supplier is: Company 1: C = 130 000. The graph is a horizontal line with intercept 130 000 Company 2: C = 1000n. The graph is a line passing through the origin with a slope 1000 Company 3: If n ≤ 60 then C = 40 000 If n > 60 then C = 40 000 + 500(n − 60) = 500n + 10 000 The graph for Company 3 is a horizontal line with intercept 40 000 until n = 60 after which the line bends upwards with a slope 500. The graphs are sketched in Figure 1.10. Section 1.3 Graphs of linear equations 51 Company 2 240 000 Cost Company 3 130 000 Company 1 40 000 40 60 Number of users Figure 1.10 (b) The cheapest supplier depends on the number of users: If n ≤ 40, Company 2 is the cheapest. If 40 ≤ n ≤ 240, Company 3 is the cheapest. If n ≥ 240, Company 1 is the cheapest. Key Terms Coefficient A numerical multiplier of the variables in an algebraic term, such as the numbers 4 and 7 in the expression 4x + 7yz2. Coordinates Intercept A set of numbers that determine the position of a point relative to a set of axes. The point(s) where a graph crosses one of the coordinate axes. Linear equation An equation of the form dx + ey = f. Origin The point where the coordinate axes intersect. Simultaneous linear equations A set of linear equations in which there are (usually) the same number of equations and unknowns. The solution consists of values of the unknowns which satisfy all of the equations at the same time. Slope of a line Also known as the gradient, it is the change in the value of y when x increases by 1 unit. x axis The horizontal coordinate axis pointing from left to right. y axis The vertical coordinate axis pointing upwards. 52 Chapter 1 Linear Equations Exercise 1.3 1. On graph paper draw axes with values of x and y between −3 and 10, and plot the following points: P(4, 0), Q(−2, 9), R(5, 8), S(−1, −2) Hence find the coordinates of the point of intersection of the line passing through P and Q, and the line passing through R and S. 2. An airline charges $300 for a flight of 2000 km and $700 for a flight of 4000 km. (a) Plot these points on graph paper with distance on the horizontal axis and cost on the vertical axis. (b) Assuming a linear model, estimate (i) the cost of a flight of 3200 km; (ii) the distance travelled on a flight costing $400. 3. By substituting values into the equation, decide which of the following points lie on the line, x + 4y = 12: A(12, 0), B(2, 2), C(4, 2), D(−8, 5), E(0, 3) 4. For the line 3x − 5y = 8, (a) find the value of x when y = 2; (b) find the value of y when x = 1. Hence write down the coordinates of two points which lie on this line. 5. If 4x + 3y = 24, complete the following table and hence sketch this line. x y 0 0 3 6. Solve the following pairs of simultaneous linear equations graphically: (a) −2x + y = 2 (b) 3x + 4y = 12 x + 4y = 8 2x + y = −6 (c) 2x + y = 4 (d) x+y=1 4x − 3y = 3 6x + 5y = 15 7. State the value of the slope and y intercept for each of the following lines: (a) y = 5x + 9 (b) y = 3x − 1 (c) y = 13 − x (d) −x + y = 4 (e) 4x + 2y = 5 (f) 5x − y = 6 8. Use the slope–intercept approach to produce a rough sketch of the following lines: (a) y = −x (b) x − 2y = 6 9. A taxi firm charges a fixed cost of $4 plus a charge of $2.50 a mile. (a) Write down a formula for the cost, C, of a journey of x miles. (b) Plot a graph of C against x for 0 ≤ x ≤ 20. (c) Hence, or otherwise, work out the distance of a journey which costs $24. Section 1.3 Graphs of linear equations 53 10. The number of people, N, employed in a chain of cafes is related to the number of cafes, n, by the equation: N = 10n + 120 (a) Illustrate this relation by plotting a graph of N against n for 0 ≤ n ≤ 20. (b) Hence, or otherwise, calculate the number of (i) employees when the company has 14 cafes; (ii) cafes when the company employs 190 people. (c) State the values of the slope and intercept of the graph and give an interpretation. 11. Monthly sales revenue, S (in $), and monthly advertising expenditure, A (in $), are modelled by the linear relation, S = 9000 + 12A. (a) If the firm does not spend any money on advertising, what is the expected sales revenue that month? (b) If the firm spends $800 on advertising one month, what is the expected sales revenue? (c) How much does the firm need to spend on advertising to achieve monthly sales revenue of $15 000? (d) If the firm increases monthly expenditure on advertising by $1, what is the corres ponding increase in sales revenue? Exercise 1.3* 1. Which of the following points lie on the line 3x − 5y = 25? (5, −2), (10, 1), (−5, 0) (5, 10), (−5, 10), (0, −5) 2. Solve the following pairs of simultaneous equations graphically: (a) y = 3x − 1 (b) 2x + y = 6 (c) 2x + 3y = 5 (d) 3x + 4y = −12 y = 2x + 1 x − y = −3 5x − 2y = −16 −2x + 3y = 25 3. State the value of the slope and y intercept for each of the following lines: (a) y = 7x − 34 (b) y = 1 − x (c) 3x − 2y = 6 (e) x − 5y = 0 (f) y = 2 (g) x = 4 (d) −4x + 2y = 5 4. Identify the two lines in the following list which are parallel: (a) 3x + 5y = 2 (b) 5x − 3y = 1 (d) 10x − 6y = 9 (e) y = 0.6x + 2 (c) 5x + 3y = 13 5. (a) The Feelgood Gym has a monthly membership fee of $100 and it charges an addi tional $1 per hour. If I use the gym for x hours in a month, write down an expression for the total cost in terms of x. (b) Repeat part (a) for the FabulousMe Gym, which charges $35 per month and $3.50 per hour. (c) By plotting both graphs on the same axes, or otherwise, find the number of hours which gives the same total cost. ➜ 54 Chapter 1 Linear Equations 6. A bakery discovers that if it decreases the price of its birthday cakes by $1, it sells 12 more cakes each month. (a) Assuming that monthly sales, M, are related to prices, P, by a linear model, M = aP + b, state the value of a. (b) If the bakery sells 240 cakes in a month when the price of the cake is $14, work out the value of b. (c) Use this model to estimate monthly sales when the price is $9. (d) If the bakery can make only 168 cakes in a month, work out the price that it needs to charge to sell them all. 7. (1) Show that the lines ax + by = c and dx + ey = f are parallel whenever ae − bd = 0. (2) Use the result of part (1) to comment on the solution of the following simultaneous equations: 2x − 4y = 1 −3x + 6y = 7 8. Write down the coordinates of the points where the line ax + by = c intercepts the axes. Section 1.4 Algebraic solution of simultaneous linear equations Objectives At the end of this section you should be able to: ● Solve a system of two simultaneous linear equations in two unknowns using elimination. ● Detect when a system of equations does not have a solution. ● Detect when a system of equations has infinitely many solutions. ● Solve a system of three simultaneous linear equations in three unknowns using elimination. In Section 1.3 a graphical method for the solution of simultaneous linear equations was described. Both lines are sketched on the same piece of graph paper and the coordinates of the point of intersection are then simply read off from the diagram. Unfortunately, this approach has several drawbacks. It is not always easy to decide on a suitable scale for the axes. Even if the scale allows all four points (two from each line) to fit on the diagram, there is no guarantee that the point of intersection itself also lies on it. When this happens, you have no alternative but to throw away your graph paper and to start again, choosing a smaller scale in the hope that the solution will now fit. The second drawback concerns the accuracy of the graphical solution. All of the problems in Section 1.3 were deliberately chosen so that the answers had nice numbers in them; whole numbers such as −1, 2 and 5 or, at worst, simple fractions such as 1/2, 21/2 and −1/4. In practice, the coefficients of the equations may well involve decimals and we might expect a decimal solution. Indeed, even if the coefficients are whole numbers, the solution itself could involve nasty fractions such as 7/8 or perhaps something like 231/571. A moment’s thought should convince you that in these circumstances it is virtually impossible to obtain the solution graphically, even if we use a really large scale and our sharpest HB pencil in the process. The final drawback concerns the nature of the problem itself. Quite frequently in economics, we need to solve three equations in three unknowns or maybe four equations in four unknowns. Unfortunately, the graphical method of solution does not extend to these cases. In this section an alternative method of solution is described which relies on algebra. It is called the elimination method, since each stage of the process eliminates one (or more) of the unknowns. This method always produces the exact solution and can be applied to systems of equations larger than just two equations in two unknowns. In order to illustrate the method, we return to the simple example considered in the previous section: 4x + 3y = 11 (1) 2x + y = 5 (2) 56 Chapter 1 Linear Equations The coefficient of x in equation (1) is 4 and the coefficient of x in equation (2) is 2. If these numbers had turned out to be exactly the same, then we could have eliminated the variable x by subtracting one equation from the other. However, we can arrange for this to be the case by multiplying the lefthand side of the second equation by 2. Of course, we must also remember to multiply the righthand side of the second equation by 2 in order for this operation to be valid. The second equation then becomes 4x + 2y = 10 (3) We may now subtract equation (3) from equation (1) to get y=1 You may like to think of this in terms of the usual layout for the subtraction of two ordinary numbers: that is, 4x 1 3y 5 11 4x 1 3y 5 10 2 y5 1 the xs cancel when you subtract This number can now be substituted into one of the original equations to deduce x. From equation (1) 4x + 3(1) = 11 (substitute y = 1) 4x + 3 = 11 4x = 8 x= 2 (subtract 3 from both sides) (divide both sides by 4) Hence the solution is x = 2, y = 1. As a check, substitution of these values into the other original equation (2) gives 2(2) + 1 = 5 ✓ The method of elimination can be summarised as follows. Step 1 Add/subtract a multiple of one equation to/from a multiple of the other to eliminate x. Step 2 Solve the resulting equation for y. Step 3 Substitute the value of y into one of the original equations to deduce x. Step 4 Check that no mistakes have been made by substituting both x and y into the other original equation. Section 1.4 Algebraic solution of simultaneous linear equations 57 Example Solve the system of equations 3x + 2y = 1 (1) −2x + y = 2 (2) Solution Step 1 The coefficients of x in equations (1) and (2) are 3 and −2, respectively. We can arrange for these to be the same size (but of opposite sign) by multiplying equation (1) by 2 and multiplying equation (2) by 3. The new equations will then have x coefficients of 6 and −6, so we can eliminate x this time by adding the equations together. The details are as follows. Doubling the first equation produces 6x + 4y = 2 (3) Tripling the second equation produces −6x + 3y = 6 (4) If equation (4) is added to equation (3), then 6x 1 4y 5 2 26x 1 3y 5 6 1 7y 5 8 the xs cancel when you add (5) Step 2 Equation (5) can be solved by dividing both sides by 7 to get y = 8/7 Step 3 If 8/7 is substituted for y in equation (1), then 3x 1 2 3x 1 8 51 7 16 51 7 16 7 7 2 16 3x 5 7 3x 5 1 2 3x 5 2 x5 (put over a common denominator) 9 7 1 9 3 2 3 7 x 52 (subtract 16/7 from both sides) (divide both sides by 3) 3 7 The solution is therefore x = −3/7, y = 8/7. ➜ 58 Chapter 1 Linear Equations Step 4 As a check, equation (2) gives 22 2 3 8 6 8 6 1 8 14 5 2 ✓ 1 5 1 5 5 7 7 7 7 7 7 Advice In the general description of the method, we suggested that the variable x is eliminated in step 1. There is nothing special about x. We could equally well eliminate y at this stage and then solve the resulting equation in step 2 for x. You might like to solve the above example using this alternative strategy. You need to double equation (2) and then subtract from equation (1). Practice Problem 1. (a) Solve the equations 3x − 2y = 4 x − 2y = 2 by eliminating one of the variables. (b) Solve the equations 3x + 5y = 19 −5x + 2y = −11 by eliminating one of the varia