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More About This Textbook
Overview
Product Details
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Meet the Author
Until 2012, Teresa Bradley lectured in mathematics and statistics at Limerick Institute of Technology, and has been involved for many years with the University of London on the external Diploma in Economics as well as the BSc in Economics, Business and Management.
Teresa Bradley is also author of Essential Statistics for Economics, Business and Management, published by John Wiley & Sons, Ltd.
WileyPLUS is a researchbased online environment for effective teaching and learning. The course builds studentsâ€™ confidence by taking the guesswork out of studying and provides students with a clear roadmap showing them:
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Read an Excerpt
Essential Mathematics for Economics and Business
By Teresa Bradley Paul Patton
John Wiley & Sons
ISBN: 0470844663Chapter One
Mathematical PreliminariesAt 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 Â£x per week I shall have Â£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 xterms, all xyterms, 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 yterm is different, so it cannot be (ii) 5xy + 8xy + 3xy + added to the others y = 16xy + y
The yterm 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 xterm is different, so it cannot be The xterm 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 + 2Â·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/34x/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 lefthand 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 lefthand side (LHS) of the equation is equal to the righthand 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...)
Table of Contents
Introduction
Ch. 1 Mathematical Preliminaries 1
Ch. 2 The Straight Line and Applications 37
Ch. 3 Simultaneous Equations 101
Ch. 4 Nonlinear 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