Ebook Essential Mathematics for Economics and Business - Part 1

Part 1 of ebook Essential mathematics for economics and business (Fourth edition) provide readers with content about: mathematical preliminaries; the straight line and applications; simultaneous equations; non-linear functions and applications; financial mathematics;... Please refer to the part 1 of ebook for details!

Economics and Business Fourth Edition

Economics and Business

Fourth Edition

Teresa Bradley

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2.3 Applications: Demand, Supply, Cost, Revenue 592.4 More Mathematics on the Straight Line 76

2.6 Elasticity of Demand, Supply and Income 83

CHAPTER 3

3.1 Solving Simultaneous Linear Equations 102

3.4 The National Income Model and the IS-LM Model 1333.5 Excel for Simultaneous Linear Equations 137

CHAPTER 4

4.1 Quadratic, Cubic and Other Polynomial Functions 148

4.4 Hyperbolic (Rational) Functions of the Form a /(bx + c) 197

CHAPTER 6

6.1 Slope of a Curve and Differentiation 2606.2 Applications of Differentiation, Marginal Functions, Average Functions 2706.3 Optimisation for Functions of One Variable 2866.4 Economic Applications of Maximum and Minimum Points 304

6.6 Further Differentiation and Applications 334

8.1 Integration as the Reverse of Differentiation 428

8.3 Integration of the Natural Exponential Function 4358.4 Integration by Algebraic Substitution 4368.5 The Definite Integral and the Area under a Curve 441

8.6 Consumer and Producer Surplus 4488.7 First-order Differential Equations and Applications 4568.8 Differential Equations for Limited and Unlimited Growth 4688.9 Integration by Substitution and Integration by Parts website only

9.5 The Inverse Matrix and Input/Output Analysis 518

Many students who embark on the study of economics and/or business are surprised and apprehensive

to find that mathematics is a core subject on their course Yet, to progress beyond a descriptive level

in most subjects, an understanding and a certain fluency in basic mathematics is essential In this text

a minimal background in mathematics is assumed: the text starts in Chapter 1 with a review of basicmathematical operations such as multiplying brackets, manipulating fractions, percentages, use of thecalculator, evaluating and transposing formulae, the concept of an equation and the solution of simpleequations Throughout the text worked examples demonstrate concepts and mathematical methodswith a simple numerical example followed by further worked examples applied to real-world situations.The worked examples are also useful for practice Start by reading the worked example to make sureyou understand the method; then test yourself by attempting the example with a blank sheet of paper!You can always refer back to the detailed worked example if you get stuck You should then be in aposition to attempt the progress exercises In this new edition, the worked examples in the text arecomplemented by an extensive question bank in WileyPLUS and MapleTA

An Approach to Learning

The presentation of content is designed to encourage a metacognitive approach to manage your ownlearning This requires a clear understanding of the goals or content of each topic; a plan of action tounderstand and become competent in the material covered; and to test that this has been achieved

1 Goals

The learning goals must be clear

Each chapter is introduced with an overview and chapter objectives Topics within chapters aredivided into sections and subsections to maintain an overview of the chapter’s logical development.Key concepts and formulae are highlighted throughout A summary, to review and consolidate themain ideas, is given within chapters where appropriate, with a final overview and summary at the end

2 Plan of Action to Understand and Become Competent in the Material Covered

rUnderstanding the rationale underlying methods is essential To this end, concepts and methods are

reinforced by verbal explanations and then demonstrated in worked examples also interjected withexplanations, comments and reminders on basic algebra More formal mathematical terminologiesare introduced, not only for conciseness and to further enhance understanding, but to enable thereaders to transfer their mathematical skills to related subjects in economics and business

rMathematics is an analytical tool in economics and business Each mathematical method is followed

immediately by one or more applications For example, demand, supply, cost and revenue functionsimmediately follow the introduction of the straight line in Chapter 2

Simultaneous linear equations in Chapters 3 and 9 are followed by equilibrium, taxes and subsides(see Worked Example 3.12 below), break-even analysis, and consumer and producer surplus andinput/output analysis

WO R K E D E X A M P L E 3 1 2

TAXES AND THEIR DISTRIBUTION

Find animated worked examples at www.wiley.com/college/bradley

The demand and supply functions for a good are given as

Demand function: Pd= 100 − 0.5Qd (3.17)

Supply function: Ps= 10 + 0.5Qs (3.18)

(a) Calculate the equilibrium price and quantity

(b) Assume that the government imposes a fixed tax of £6 per unit sold

(i) Write down the equation of the supply function, adjusted for tax

(ii) Find the new equilibrium price and quantity algebraically and graphically

(iii) Outline the distribution of the tax, that is, calculate the tax paid by the consumerand the producer

Further applications such as elasticity, budget and cost constraints, equilibrium in the labour market

and the national income model are available on the website www.wiley.com/college/bradley Financial

mathematics is an important application of arithmetic and geometric series and also requires the use

of rules for indices and logs Applications and analysis based on calculus in Chapters 6, 7, 8 and 10 areessential for students of economics

rGraphs help to reinforce and provide a more comprehensive understanding by visualisation Use

Excel to plot graphs since parameters of equations are easily varied and the effect of changes areseen immediately in the graph Exercises that use Excel are available on the website

rA key feature of this text is the verbal explanations and interpretation of problems and of their

solutions This is designed to encourage the reader to develop both critical thinking and solving techniques

problem-rThe importance of practice, reinforced by visualisation, is summarised succinctly in the following

quote attributed to the ancient Chinese philosopher Lao Tse:

You read and you forget;

You see and you remember;

You do and you learn

3 Test whether goals were achieved

There are several options:

rAttempt the worked examples without looking at the text.

rDo the progress exercises Answers and, in some cases, solutions are given at the back of the text.

rIn this new edition a large question bank, with questions classified as ‘easy’, ‘average’ or ‘hard’, is

provided in WileyPLUS and MapleTA Many of the questions are designed to reinforce solving techniques and so require several inputs from the reader not just a single final numericanswer

problem-rA test exercise is given at the end of each chapter Answers to test exercises are available to lecturers

online

Structure of the Text

Mathematics is a hierarchical subject The core topics are covered in Chapters 1, 2, 3, 4, 6 and 8 Someflexibility is possible by deciding when to introduce further material based on these core chapters, such

as linear algebra, partial differentiation and difference equations, etc., as illustrated in the followingchart

Chapter 1 Preliminaries

Chapter 2 Straight line

Chapter 3 Simultaneous Equations

Chapter 4 Quadratics: Indices

Logs: 1/(ax + b)

Chapter 5 Financial Mathematics

Chapter 6 Differentiation

Chapter 7 Partial Differentiation

Chapter 8 Integration

Chapter 9 Linear Programming Matrices Determinants

Chapter 10 Difference Equations

Note: An introductory course may require the earlier sections from the core chapters and a limitednumber of applications

WileyPLUS is a powerful online tool that provides instructors and students with an integrated suite

of teaching and learning resources, including an online version of the text, in one easy-to-use site To learn more about WileyPLUS, request an instructor test drive or view a demo, please visitwww.wileyplus.com

web-WileyPLUS Tools for Instructors

WileyPLUS enables you to:

rAssign automatically graded homework, practice, and quizzes from the test bank.

rTrack your students’ progress in an instructor’s grade book.

rAccess all teaching and learning resources, including an online version of the text, and student

and instructor supplements, in one easy-to-use website These include an extensive test bank ofalgorithmic questions; full-colour PowerPoint slides; access to the Instructor’s Manual; additionalexercises and quizzes with solutions; Excel problems and solutions; and, answers to all the problems

in the book

rCreate class presentations using Wiley-provided resources, with the ability to customise and add

your own materials

WileyPLUS Resources for Students within WileyPLUS

In WileyPLUS, students will find various helpful tools, such as an e-book, additional Progress Exercises,Problems in Context, and animated worked examples

re-book of the complete text is available in WileyPLUS with learning links to various features and

tools to assist students in their learning

rAdditional Progress Exercises are available affording students the opportunity for further practice

of key concepts

rProblems in Context provide students with further exposition on the mathematics elements of a

key economics or business topic

rAnimated Worked Examples utilising full-colour graphics and audio narration; these animations of

key worked examples from the text provide the students with an added way of tackling more difficultmaterial

Ancillary Teaching and Learning Materials

All materials are housed on the companion website, which you can access at www.wiley.com/ lege/bradley

col-Students’ companion website containing:

rAnimations of key worked examples from the text.

rProblems in Context covering key mathematic problems in a business or economics context.

rExcel exercises, which draw upon data within the text.

rIntroduction to using Maple, which has been updated for the new edition.

rAdditional learning material and exercises, with solutions for practice.

Instructor’s companion website containing:

rInstructor’s Manual containing comments and tips on areas of difficulty; solutions to test exercises;

additional exercises and quizzes, with solutions; sample test papers for introductory courses andadvanced courses

rPowerPoint presentation slides containing full-colour graphics to help instructors create stimulating

lectures

rTest bank including algorithmic questions, which is provided in Microsoft Word format and can

be accessed via Maple T.A

rAdditional exercises and solutions that are not available in the text.

At the end of this chapter you should be able to:

rPerform basic arithmetic operations and simplify algebraic expressions

rPerform basic arithmetic operations with fractions

rSolve equations in one unknown, including equations involving fractions

rUnderstand the meaning of no solution and infinitely many solutions

1.1 Some Mathematical Preliminaries

Brackets in mathematics are used for grouping and clarity.

Brackets may also be used to indicate multiplication

Brackets are used in functions to declare the independent variable (see later)

Powers: positive whole numbers such as 23, which means 2× 2 × 2 = 8:

(anything)3= (anything) × (anything) × (anything)

(x)3= x × x × x (x+ 4)5= (x + 4)(x + 4)(x + 4)(x + 4)(x + 4)

Note:

Brackets: (A)(B) or A × B or AB

all indicate A multiplied by B.

Variables and letters: When we don’t know the value of a quantity, we give that quantity a symbol,

such as x We may then make general statements about the unknown quantity or variable, x For example, ‘For the next 15 weeks, if I save x per week I shall have $4500 to spend on a holiday’ This

statement may be expressed as a mathematical equation:

15× weekly savings = 4500

15× x = 4500

Now that the statement has been reduced to a mathematical equation (see Section 1.4 for more on

equations), we may solve the equation for the unknown, x:

Algebra: The branch of mathematics that deals with the manipulation of symbols (letters) is called

algebra An algebraic term consists of letters/symbols: therefore, in the above example, x is referred to

as an algebraic term (or simply a term) An algebraic expression (or simply an expression) is a formula

that consists of several algebraic terms (constants may also be included), e.g., x + 5x + 8.

Square roots: The square root of a number is the reverse of squaring:

(2)2= 4 →√4= 2(2.5)2= 6.25 →√6.25 = 2.5

Accuracy: rounding numbers correct to x decimal places

When you use a calculator you will frequently end up with a string of numbers after the decimalpoint For example, 15/7 = 2.142 857 1 For most purposes you do not require all these numbers.

However, if some of the numbers are dropped, subsequent calculations are less accurate To minimisethis loss of accuracy, there are rules for ‘rounding’ numbers correct to a specified number of decimalplaces, as illustrated by the following example.

Consider: (a) 15/7 = 2.142 857 1; (b) 6/7 = 0.857 142 8 Assume that three numbers after the

deci-mal point are required To round correct to three decideci-mal places, denoted as 3D, inspect the number

in the fourth decimal place:

rIf the number in the fourth decimal place is less than 5, simply retain the first three numbers after

the decimal place: (b) 6/7 = 0.857 142 8: use 0.857, when rounded correct to 3D.

rIf the number in the fourth decimal place is 5 or greater, then increase the number in the third

decimal place by 1, before dropping the remaining numbers: (a) 15/7 = 2.142 857 1, use 2.143,

when rounded correct to 3D

To get some idea of the greater loss of accuracy incurred by truncating (chopping off after a specifiednumber of decimal places) rather than rounding to the same number of decimal places, consider(a) the exact value of 15/7, using the calculator, (b) 15/7 truncated after 3D and (c) 15/7 rounded to3D While these errors appear small they can become alarmingly large when propagated by furthercalculations A simple example follows in which truncated and rounded values of 15/7 are each raised

to the power of 20

Note: Error= exact value − approximate value

(a) Exact value (b) Truncated to 3D (c) Rounded to 3D

20

= (2.142 857 143 )20

= 4 167 392(integer part of result)

(2.142)20= 4 134 180(integer part of result)

(2.143)20 = 4 172 952(integer part of result)

Addition and subtraction

Adding: If all the signs are the same, simply add all the numbers or terms of the same type and give

the answer with the common overall sign

Subtracting: When subtracting any two numbers or two similar terms, give the answer with the sign

of the largest number or term

If terms are of the same type, e.g., all x-terms, all xy-terms, all x2-terms, then they may be added orsubtracted as shown in the following examples:

Add/subtract with numbers, mostly Add/subtract with variable terms

The x-term is different, so it cannot be

subtracted from the others

The x-term is different from the x2-terms, so

it cannot be subtracted from the x2-terms

WO R K E D E X A M P L E 1 1

ADDITION AND SUBTRACTION

For each of the following, illustrate the rules for addition and subtraction:

(a) 2+ 3 + 2.5 = (2 + 3 + 2.5) = 7.5

(b) 2x + 3x + 2.5x = (2 + 3 + 2.5)x = 7.5x

(c) −3xy − 2.2xy − 6xy = (−3 − 2.2 − 6)xy = −11.2xy

(d) 8x + 6xy − 12x + 6 + 2xy = 8x − 12x + 6xy + 2xy + 6 = −4x + 8xy + 6

(e) 3x2+ 4x + 7 − 2x2− 8x + 2 = 3x2− 2x2+ 4x − 8x + 7 + 2 = x2− 4x + 9

Multiplication and division

Multiplying or dividing two quantities with like signs gives an answer with a positive sign Multiplying

or dividing two quantities with different signs gives an answer with a negative sign

WO R K E D E X A M P L E 1 2 a

MULTIPLICATION AND DIVISION

Each of the following examples illustrates the rules for multiplication

(a) 5× 7 = 35 (d)−5 × 7 = −35

(b)−5 × −7 = 35 (e) 7/5= 1.4

(c) 5× −7 = −35 (f) (−7)/(−5) = 1.4

0× (any real number) = 0

0÷ (any real number) = 0

But you cannot divide by 0

(m) 2(x + 2) = 2x + 4 multiply each term inside the bracket by the term outside

the bracket

(n) (x + 4)(x + 2) multiply the second bracket by x, then multiply the second

bracket by (+4) and add

= x(x + 2) + 4(x + 2)

= x2+ 2x + 4x + 8

= x2+ 6x + 8

multiply each bracket by the term outside it; add or

subtract similar terms, such as 2x + 4x = 6x

Indicate the order in which arithmetic operations should be carried out

The precedence of arithmetic operators is summarised as follows:

(i) Simplify or evaluate the terms within brackets first

(ii) When multiplying/dividing two terms: (1) determine the overall sign first; (2)multiply/divide the numbers; (3) finally, multiply/divide the variables (letters).(iii) Finally, add and/or subtract as appropriate

WO R K E D E X A M P L E 1 2 b

The precedence of arithmetic operators is illustrated by the following examples:

(a) 2x − y + 4(x − 2y + 2x) = 2x − y + 4(3x − 2y) simplify within brackets first

= 2x − y + 12x − 8y multiply the bracket by 4

= 2x + 12x − y − 8y add/subtract similar terms

= 14x − 9y

(b) 15− 5(3y − 2(y + 2)) = 15 − 5(3y − 2y − 4) simplify the inner bracket first

= 15 − 5(y − 4) multiply the bracket by− 5

= 15 − 5(y) − 5(−4) be careful with the negatives!

3 ← 3 is called the numerator

7 ← 7 is called the denominator

The method for adding or subtracting fractions is:

Step 1: Take a common denominator, that is, a number or term which is divisible by the denominator

of each fraction to be added or subtracted A safe bet is to use the product of all the individualdenominators as the common denominator

Step 2: For each fraction, divide its denominator into the common denominator, then multiply the

result by its numerator

Step 3: Simplify your answer if possible.

WO R K E D E X A M P L E 1 3

ADD AND SUBTRACT FRACTIONS

Each of the following illustrates the rules for addition and subtraction of fractions

= (2)(5)(3)(7) = 10



= (−2)(7)(3)(5) = −14

15

(c) 3×2

5 =

31

 25



= (3)(2)(1)(5) = 6

5 = 115

The same rules apply for fractions involving variables, x, y, etc.

Dividing by a fraction is the same as multiplying by the fraction inverted

To demonstrate this general rule, consider an example such as dividing 5 by a half; we know theanswer is 10 because there are 10 halves in 5: 5

12

 = 10

This answer may also be obtained by applying the rule stated above:

‘Dividing by a fraction is the same as multiplying

by that fraction inverted’

↓5

12

 = 5 ×

21



=5

1 ×

21

 115



=2215



81

=7

3 ×1

8 = 724

= 4x(x − y)

3x(x + y) =

4(x − y) 3(x + y)

Note: The same rules apply to all fractions, whether the fractions consist of numbers or

variables

Reducing a fraction to its simplest form and equivalent fractions

A fraction that has common factors in the numerator and denominator may be simplified by dividingthe common factors into each other giving unity This is described as the cancellation of commonfactors and is illustrated in the following examples:

3 = 221

5 +2

5 −65



=73

For example, the equation x + 4 = 10 has the solution x = 6 We say x = 6 ‘satisfies’ the equation.

We say this equation has a unique solution.

Methods for solving equations

To solve an equation for an unknown, x, rearrange the equation in order to isolate the x on one side.

This may be achieved by a variety of techniques such as:

radding/subtracting a number or term to each side

rmultiplying/dividing each side by a number or term (except division by 0!).

It is vitally important to remember that whatever operation is performed on the LHS of the equationmust also be performed on the RHS, otherwise the equation is changed

WO R K E D E X A M P L E 1 6

SOLVING EQUATIONS

(a) Given the equation x + 3 = 2x − 6 + 5x, solve for x.

(b) Given the equation (x + 3)(x − 6) = 0, solve for x.

(c) Given the equation (x + 3)(x − 3) = 0, solve for x.

(a) x + 3 = 2x − 6 + 5x

x + 3 = 7x − 6 adding the x-terms on the RHS

x + 3 + 6 = 7x − 6 + 6 to cancel the −6 on the RHS, add +6 to both sides

x + 3 + 6 = 7x bringing over−6 to the other side

x + 9 = 7x

9= 7x − x bringing x over to the other side

Hence the frequently quoted rule: ‘bring the

−6 to the other side and change the sign

Each of these solutions can be confirmed by checking that they satisfy the original equation.Check the solutions:

Substitute x = −3 into (x + 3)(x − 6) = 0: (−3 + 3)(−3 − 6) = 0: (0)(−9) = 0: 0 = 0.

True

Substitute x = 6 into (x + 3)(x − 6) = 0: (6 + 3)(6 − 6) = 0: (9)(0) = 0: 0 = 0 True.

(c) This is similar to part (b), a product on the LHS, zero on the right, hence the equation

(x – 3)(x + 3) = 0 has two solutions: x = 3 from the first bracket and x = –3 from the

If x = 3 → x2= 9, so satisfying the equation

If x = −3 → x2= (−3)2= 9, also satisfying the equation

Therefore, when solving equations of the form x2= number, there are always two solutions:

x= +√number and x= −√number

Not all equations have solutions In fact, equations may have no solutions at all or infinitely manysolutions Each of these situations is demonstrated in the following examples.

Case 1: Unique solutions An example of this is given above: x+ 4 = 10 etc

Case 2: Infinitely many solutions The equation x + y = 10 has solutions (x = 5, y = 5), (x =

4, y = 6), (x = 3, y = 7), etc In fact, this equation has infinitely many solutions or pairs of values

(x , y) which satisfy the formula x + y = 10.

Case 3: No solution The equation 0(x) = 5 has no solution There is simply no value of x which

can be multiplied by 0 to give 5

WO R K E D E X A M P L E 1 7

SOLVING A VARIETY OF SIMPLE ALGEBRAIC EQUATIONS

Find animated worked examples at www.wiley.com/college/bradley

In this worked example, try solving the following equations yourself The answers are givenbelow, followed by the detailed solutions

(d) Infinitely many solutions for which x = y + 4 (e) x = 0, x =√2, x = −√2

Suggested solutions to Worked Example 1.7

3 = x 3= 5x multiply each side by x

5x= 3 swap sides

x= 3

5 = 0.6 dividing each side by 5 (c) x2+ 4x − 6 = 2(2x + 5)

This time, simplify first by multiplying out the brackets and collecting similar terms:

x2+ 4x − 6 = 4x + 10

x2+ 4x − 4x = 10 + 6 bring all x-terms to one side and numbers to the other side

x2= 16

x= ±4

(d) (x − y) = 4: Here we have one equation in two unknowns, so it is not possible to find a

unique solution The equation may be rearranged as

x = y + 4 This equation now states that for any given value of y (there are infinitely many values),

x is equal to that value plus 4 So, there are infinitely many solutions.

(e) x is common to both terms, therefore we can separate or factor x from each term as follows:

Table 1.1 Euro exchange rates

The calculations (correct to four decimal places) are carried out as follows:

Step 1: State the appropriate rates from Table 1.1.

Step 2: Set up the identity: 1 unit of given currency = y units of required currency.

Step 3: Multiply both sides by x: x units of given currency = x × (y units of required currency) (a) (i) The price is given in US dollars, the price is required (currency) in Euros, hence

Step 1: From Table 1.1 write down the exchange rates for 1 Euro in both British pounds and

$1.3003= £0.8346 since they are each equivalent to 1 Euro

The given price is in US dollars, the price is required in British pounds

Step 2: US$1= £1 ×0.8346

1.3003 dividing both sides of the previous equation

by 1.3003 to get rate for $1 in £

Step 3: US$20= £20 ×0.8346

1.3003 multiplying both sides by 20

US$20= £12.8370

US$1.3003= A$1.2429 since they are each equivalent to 1 Euro

The given price is US$, the required price (currency) is A$

Step 2: US$1= A$1 × 1.2429

1.3003 dividing both sides by 1.3003

Step 3: US$20= A$20 ×1.2429

1.3003 multiplying both side by 20

Here we are given A$500 and we require its equivalent in £ sterling

A$1.2429= £0.8346 since they are each equivalent to 1 Euro

P RO G R E S S E X E R C I S E S 1 2

Use Basics to Solve Equations

Solve the following equations Remember that equations may have no solution or infinitely manysolutions

to two decimal places

34 An iPad is priced at 400 Canadian dollars Calculate the price of the iPad in (a) Euros,

(b) rupees and (c) ringgits

35 How many Euros are equivalent to (a) 800 US dollars, (b) 800 zlotys?

36 A flight is priced at 240 South African rand What is the equivalent price in Brazilian

reals?

37 Calculate a table of exchange rates for Hong Kong dollars for the first five currencies in

the third column of Table 1.1

38 A new textbook is priced at £58 sterling Calculate the equivalent price in Hungarian

forints

39 Convert 500 US dollars to (a) Swiss francs, (b) Australian dollars, (c) Singaporean dollars.

40 You have 400 Polish zlotys How much would you have left (in Euros) if you bought a book

for £35 sterling and a T-shirt for 420 Hong Kong dollars?

1.6 Simple Inequalities

An equation is an equality It states that the expression on the LHS of the ‘=’ sign is equal to theexpression on the RHS An inequality is a statement in which the expression on the LHS is eithergreater than (denoted by the symbol>) or less than (denoted by the symbol <) the expression on the

RHS For example, 5= 5 or 5x = 5x are equations, while

5> 3 5x > 3x

are inequalities which read, ‘5 is greater than 3’, ‘5x is greater than 3x’ (for any positive value of x).

Note: Inequalities may be read from left to right, as above, or the inequality may be read from right to

left, in which case the above inequalities are

5> 3 (5 is greater than 3) is the same as 3 < 5 (‘3 is less than 5’)

5x > 3x (‘5x is greater than 3x’) is the same as 3x < 5x (‘3x is less than 5x’)

Inequality symbols

> greater than < less than

≥ greater than or equal to ≥ less than or equal to

The number line

The number line is a horizontal line on which every point represents a real number The centralpoint is zero, the numbers on the left are negative, numbers on the right are positive, as illustrated forselected numbers in Figure 1.1

Figure 1.1 Number line, numbers increasing from left to right

Look carefully at the negative numbers; as the numbers increase in value they decrease in magnitude,for example:−1 is a larger number than −2; −0.3 is a larger number than −0.5 Another way of looking

at it is to say the numbers become less negative as they increase in value (Like a bank account, youare better off when you owe £10 (−10) than when you owe £1000 (−1000).)

An inequality statement, such as x > 2, means all numbers greater than, but not including, 2 This

statement is represented graphically as every point on the number line to the right of the number 2, asshown in Figure 1.2, with an open circle at the point 2 to show that it is not included in the interval(a solid circle at 2 would indicate that 2 is included in the interval)

Figure 1.2 The inequality, x > 2

In economics, it is meaningful to talk about positive prices and quantities In this text, we shall

assume that the variable x ≥ 0 when solving inequalities

Intervals defined by inequality statements

When an application uses values within a certain range only, then inequality signs are often used to

define this interval precisely For example, suppose a tax is imposed on all incomes, (£Y), between

£10 000 and £15 000 inclusive; we say the tax is imposed on all salaries within the interval 10 000≤

Y≤ 15 000

A certain bus fare applies to all children of ages (x) 4 but less than 16; we say the fare applies to

those whose ages are in the interval 4≤ x < 16 The age 4 is included, and all ages up to, but not

including, 16

Manipulating inequalities

Inequalities may be treated as equations for many arithmetic operations The inequality remains truewhen constants are added to or subtracted from both sides of the inequality sign, or when both sides ofthe inequality are multiplied or divided by positive numbers or variables For example, the equalitiesabove are still true when 8 or –8 is added to both sides,

5+ 8 > 3 + 8, that is, 13 > 11, similarly 5x + 8 > 3x + 8

5− 8 > 3 − 8, that is −3 > −5 Remember−5 is less than−3 See Figure 1.1.

However, when both sides of the inequality are multiplied or divided by negative numbers or variables,then the direction of the inequality changes:> becomes < and vice versa For example, multiply both

sides of the inequality 5> 3 by –2:

5(−2) > 3(−2), or −10 > −6 is not true

5(−2) < 3(−2), or −10 < −6 is true

Solving inequalities

The solution of an equation is the value(s) for which the equation statement is true For

exam-ple, x + 4 = 10 is true when x = 6 only On the other hand, the solution of an inequality

is a range of values for which the inequality statement is true; for example, x + 4 > 10 is true when x > 6.

WO R K E D E X A M P L E 1 9

SOLVING SIMPLE INEQUALITIES

Find the range of values for which the following inequalities are true, assuming that x > 0.

State the solution in words and indicate the solution on the number line

(a) 10< x − 12 (b) −75

x > 15 (c) 2x − 6 ≤ 12 − 4x

SOLUTION

(a) 10< x − 12 → 10 + 12 < x → 22 < x (or x > 22)

The solution states: 22 is less than x or x is greater than 22 Hence the solution is represented

by all points on the number line to the right of, but not including, 22, as shown inFigure 1.3

Figure 1.3 x > 22

(b) −75

x > 15

Multiply both sides of the inequality by x Since x > 0, the direction of the inequality sign

does not change

(c) 2x − 6 ≤ 12 − 4x

2x + 4x − 6 ≤ 12 add 4x to both sides

6x≤ 12 + 6 add 6 to both sides

6x≤ 18

x≤ 3 dividing both sides by 6

But we assume x > 0, hence the solution is 0 < x ≤ 3.

This is represented by all points on the number line to the right of 0 and up to andincluding 3, as shown in Figure 1.5

rThe increased number

= number + increase = number + x× number

percent-WO R K E D E X A M P L E 1 1 0

CALCULATIONS WITH PERCENTAGES

(a) Calculate (i) 23% of 1534 (ii) 100% of 1534

(b) A salary of £55 240 is to be increased by 12% Calculate (i) the increase, (ii) the newsalary

(c) In 2013, a holiday apartment is valued at £63 600 This is a drop of 40% on the pricepaid for the apartment in 2007 Calculate the price paid in 2007

So the increase in salary is £6628.8

(ii) The new salary is £55 240+ £6628.8 = £61 868.8

Alternatively, the new salary is 112% of the previous salary and may be calculated as

55 240×

112100



= £61 868.8

(c) Let the 2007 price be the basic price The price in 2013 is 60% of the 2007 price, i.e.,

2013 price= 60% × basic price

So £63 400= 60% of the basic price and we want to find 100% of the basic price