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Overview

Essential Mathematics for Economics and Business is an introductory textbook on mathematics for students of those subjects. The fundamental mathematical concepts are explained as simply and briefly as possible, using a selection of worked examples, graphs and real-world applications.
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Product Details

  • ISBN-13: 9780471975113
  • Publisher: Wiley, John & Sons, Incorporated
  • Publication date: 7/23/1999
  • Edition number: 1
  • Pages: 740
  • Product dimensions: 7.46 (w) x 9.19 (h) x 1.63 (d)

Meet the Author

Teresa Bradley lectures in mathematics and statistics at Limerick Institute of Technology, and has been involved with the University of London on the external Diploma in Economics as well as the BSc in Economics, Business and Management for many years.
Teresa Bradley is also author of Essential Statistics for Economics, Business and Management, published by John Wiley & Sons, Ltd.

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Essential Mathematics for Economics and Business


By Teresa Bradley Paul Patton

John Wiley & Sons

ISBN: 0-470-84466-3


Chapter One

Mathematical Preliminaries

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

Perform basic arithmetic operations and simplify algebraic expressions

Perform basic arithmetic operations with fractions

Solve equations in one unknown, including equations involving fractions

Understand the meaning of no solution and infinitely many solutions

Currency conversions

Solve simple inequalities

Calculate percentages

In addition, you will be introduced to the calculator and a spreadsheet.

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 [2.sup.3], which means 2 x 2 x 2 = 8:

[(anything).sup.3] = (anything) x (anything) x (anything)

[(x).sup.3] = x x x x x

[(x + 4).sup.5] = (x + 4)(x + 4)(x + 4) (x + 4) (x + 4)

Note

Brackets: (A)(B) or A x 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 unknownquantity, x, for example 'For the next 15 weeks, if I save £x per week I shall have £4000 to spend on a holiday'. This statement may be expressed as a mathematical equation:

15 x weekly savings = 4000 15 x x = 4000 or 15x = 4000

Now that the statement has been reduced to a mathematical equation, we may solve the equation for the unknown, x:

15x = 4000 15x/15 = 4000/15 divide both sides of the equation by 15 x = 266.67

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

[(2).sup.2] = 4 [right arrow] [square root of (4)] = 2

[(2.5).sup.2] = 6:25 [right arrow] [square root of (6.25)] = 2:5

1.1 Arithmetic Operations

Addition and subtraction

Adding: If all the signs are the same, simply add all the terms 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 identical, for example all x-terms, all xy-terms, all [x.sup.2]-terms, then they may be added or subtracted as shown in the following examples:

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

5 + 8 + 3 = 16 similarity [right arrow] 5x + 8x + 3x = 16x

5 + 8 + 3 + y = 16 + y similarity [right arrow] (i) 5x + 8x + 3x + y = 16x + y

The y-term is different, so it cannot be (ii) 5xy + 8xy + 3xy + added to the others y = 16xy + y

The y-term is different, so it cannot be added to the others

7 - 10 = -3 similarity [right arrow] (i) 7x - 10x = -3x

(ii) 7[x.sup.2] - 10[x.sup.2] = -3[x.sup.2]

7 - 10 - 10x = -3 - 10x similarity [right arrow] 7[x.sup.2] - 10[x.sup.2] - 10x = -3[x.sup.2] - 10x

The x-term is different, so it cannot be The x-term is different, so it cannot be subtracted from the others subtracted from the others

Worked example 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 = -11.2 xy

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

(e) 3[x.sup.2] + 4x + 7 - 2[x.sup.2] - 8x + 2 = 3[x.sup.2] - 2[x.sup.2] + 4x - 8[x + 7 + 2 = [x.sup.2] - 4x + 9

Multiplying and dividing

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.

Worked example 1.2 Multiplication and division

Each of the following examples illustrate the rules for multiplication.

(a) 5 x 7 = 35 (b) -5 x -7 = 35 (c) 5 x -7 = -35 (d) -5 x 7 = -35 (e) 7/5 = 1.4 (f) (-7)/(-5) = 1.4 (g) (-7)/5= -1.4 (h) 7/(-5) = -1.4 (i) 5(7) = 35 (j) (-5)(-7) = 35 (k) (-5)y = -5y (l) (-x)(-y) = xy (m) 2(x + 2) = 2x + 4 (n) (x + 4)(x + 2) = x(x + 2) + 4(x + 2) = [x.sup.2] + 2x + 4x + 8 = [x.sup.2] + 6x + 8

(o) [(x + y).sup.2]

= (x + y)(x + y) = x(x + y) + y(x + y) = xx + xy + yx + yy = [x.sup.2] + 2xy + [y.sup.2]

* Remember

It is very useful to remember that a minus sign is a -1, so -5 is the same as -1 x 5

* Remember

0 x (any real number) = 0 0 / (any real number) = 0 But you cannot divide by 0

multiply each term inside the bracket by the term outside the bracket multiply the second bracket by x, then multiply the second bracket by (+4) and add,

multiply each bracket by the term outside it add or subtract similar terms, such as 2x + 4x = 6x

multiply the second bracket by x and then by y add the similar terms: xy + yx = 2xy

The following identities are important:

1. (x + y).sup.2] = [x.sup.2] + 2xy + [y.sup.2] 2. (x - y).sup.2] = [x.sup.2] - 2xy + [y.sup.2] 3. (x + y) = (x - y) = [x.sup.2] -[y.sup.2]

* Remember: Brackets are used for grouping terms together in maths for:

(i) Clarity (ii) Indicating the order in which a series of operations should be carried out

1.2 Fractions

Terminology:

fraction = numerator/denominator

3/7

3 is called the numerator

7 is called the denominator

1.2.1 Add/subtract fractions: method

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 individual denominators as the common denominator.

Step 2: For each fraction, divide each denominator into the common denominator, then multiply the answer by the numerator.

Step 3: Simplify your answer if possible.

Worked example 1.3 Add and subtract fractions

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

Numerical example

1/7 + 2/3 - 4/5

Step 1: The common denominator is (7)(3)(5)

Step 2: 1/7 + 2/3 - 4/5

=1(3)(5)+2(7)(5)-4(7)(3)/(7)(3)(5)

Step 3: = 15 + 70 - 84/105 =1/105

1/7 + 2/3

Step 1: The common denominator is (7)(3)

Step 2: 1/7 + 2/3 = 1(3) + 2(7)/(7)(3)

Step 3: = 3 + 14/21 = 17/21

Same example, but with variables

x/7 + 2x/3-4x/5

= x(3)(5) + 2x(7)(5)-4x(7)(3) (7)(3)(5)

= 15x + 70x - 84x/105

= x/105

1/x + 4 + 2/x = 1(x) + 2(x + 4)/(x + 4) (x)

= x + 2x + 8/[x.sup.2]+ 4x

= 3x + 8/[x.sup.2]+ 4x

1.2.2 Multiplying fractions

In multiplication, write out the fractions, multiply the numbers across the top lines and multiply the numbers across the bottom lines.

Note: Write whole numbers as fractions by putting them over 1.

Terminology: Worked example 1.4

Multiplying fractions

(a) (2/3) (5/7) = (2)(5)/(3)(7) = 10/21

(b) (-2/3) (7/5) = (-2)(7)/(3)(5) = -14/15

(c) 3 x 2/5 (3/1)(2/5)=(3)(2)/(1)(5) = 6/5 = 1 1/5

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

(d) (3/x) (x + 3)/(x - 5) = 3(x + 3)/x(x - 5) = 3x + 9/[x.sup.2]- 5x]

1.2.3 Dividing fractions

General rule:

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

Worked example 1.5

Division with fractions

The following examples illustrate how division with fractions operates.

(a) (2/3)/(5/11) = (2/3)(11/5) = 22/15

(b) 5/(3/4) = 5 x 4/3 = 5/1 x 4/3 = 20/3 = 6 2/3

(c) (7/3)/8 = (7/3)/8/1 = 7/3 x 1/8 = 7/24

(d) 2x/x + y/3x/2(x - y) = 2x/x + y 2(x - y)/ 3x

= 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.

Progress Exercises 1.1 Revision on Basics

Show, step by step, how the expression on the left-hand side simplifies to that on the right.

1. 2x + 3x + 5(2x - 3) = 15(x - 1)

2. 4[x.sup.2] + 7x + 2x (4x - 5) = 3x(4x - 1)

3. 2x (y + 2) - 2y(x + 2) = 4(x - y)

4. (x + 2)(x - 4) - 2(x - 4) = x(x - 4)

5. (x + 2)(y - 2) + (x - 3)(y + 2)

= 2xy - y - 10

6. [(x + 2).sup.2] + [(x - 2).sup.2] = 2[(x.sup.2] + 4)

7. [(x + 2).sup.2] - [(x - 2).sup.2] = 8x

8. (x + 2)2 - x(x + 2) = 2(x + 2)

9. 1/3 + 3/5 + 5/7 = 1 68/105

10. x/2 - x/3 = x/6

11. (2/3)/(1/5) = 10/3

12. (2/7)/3 = 2/21

13. 2(2/x - x/2) = 4-[x.sup.2]/x = 4/x - x

14. -12/P (3P/2 + P/2) = -24

15. (3/x)/x + 3 = 3/x(x +3)

16. (5Q/P + 2)/(1/(P + 2)) = 5Q

1.3 Solving Equations

The solution of an equation is simply the value or values of the unknown(s) for which the left-hand side (LHS) of the equation is equal to the right-hand side (RHS).

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.

Not all equations have solutions. In fact, equations may have no solutions at all or may have infinitely many solutions. 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.

* Methods for solving equations

Solving equations can involve a variety of techniques, many of which will be covered later.

(Continues...)



Excerpted from Essential Mathematics for Economics and Business by Teresa Bradley Paul Patton Excerpted by permission.
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
Excerpts are provided by Dial-A-Book Inc. solely for the personal use of visitors to this web site.

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Table of Contents

Introduction

Ch. 1 Mathematical Preliminaries 1

Ch. 2 The Straight Line and Applications 37

Ch. 3 Simultaneous Equations 101

Ch. 4 Non-linear Functions and Applications 147

Ch. 5 Financial Mathematics 207

Ch. 6 Differentiation and Applications 257

Ch. 7 Functions of Several Variables 359

Ch. 8 Integration and Applications 425

Ch. 9 Linear Algebra and Applications 475

Ch. 10 Difference Equations 537

Solutions to Progress Exercises 565

Bibliography 651

List of Worked Examples 653

Index 659

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  • Anonymous

    Posted November 7, 2006

    Good for general business studies, but not for economic studies

    If you don¿t care for mathematical modelling and theory then this is the book for you. It focuses on practical business mathematics, using Excel, and mainly covers economics. I would recommend this book for the general business student who is not majoring in economics but still needs an understanding of practical economic analysis. Each chapter begins with general calculus followed by applications of it. What I liked about this book was the no-nonsense approach, it is very clear to read with plenty of good examples and you are not overwhelmed by extensive theoretical proofs. The trade-off however is that you lack the modelling skills that you need in economic studies, especially at intermediate and advanced economics levels. This book is inadequate as a complement to economic studies, and despite its title, it is also inadequate as an introduction to financial mathematics with only one chapter dedicated to it. Another problem is that the book is simply too big and cluttered with too many exercises, it was difficult keeping consistency and coherency throughout it.

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