Linear Equations

Quantitative Methods For Business Management

Quantitative Methods

John Kimani Gichuhi Lecturer Africa Nazarene University Dominic Ngugi Ndungu Africa Nazarene University

THE AUTHORS

SUMMARY CONTENTS CHAPTER 1

SET THEORY AND COUNTING NUMBERS

1

CHAPTER 2

LINEAR EQUATIONS

15

CHAPTER 3

SIMULTANEOUS LINEAR EQUATIONS

48

CHAPTER 4

ALGEBRA

63

CHAPTER 5

BINOMIAL THEOREM

75

CHAPTER 6

PERMUTATIONS & COMBINATIONS

81

CHAPTER 7

MATRICES ALGEBRA

84

CHAPTER 8

DECISION ANALYSIS

155

CHAPTER 9

CALCULUS I: DIFFERENTIATION

167

CHAPTER 10

CALCULUS II: INTEGRATION

218

CHAPTER 11

NON – LINEAR EQUATIONS

228

CHAPTER 12

REVENUE, COST AND PROFIT

241

CHAPTER 13

DECISION TREES

250

CHAPTER 14

MATHEMATICS OF FINANCE

258

CHAPTER 15 CHAPTER 16 CHAPTER 17 CHAPTER 18

LINEAR PROGRAMMING INTRO TO QUANTITATIVE TECHNIQUES L P: SIMPLEX METHOD NETWORK ANALYSIS AND SCHEDULING

274 293 295 313

CHAPTER 19

TRANSPORTATION MODEL

343

CHAPTER 20

ASSIGNMENT MODEL

378

CHAPTER 21

INVENTORY PLANNING & CONTROL

406

CONTENTS CHAPTER 1

SET THEORY AND COUNTING

Introduction 1 Standard Symbols used in set 2 Intersection of Sets 3 Venn diagram 4 Principle of counting 6 Application of Venn 8 Practice Problem 11

CHAPTER 2 Introduction 15 Number line 15 Integers 16 Multiplication and Division 19 Points on x and y Plane 20 Gradients 21 Practice Problem 23 Equation of Straight 25 Practice Problem 30 Sketching a Graph 31 Points of intersection 33 Practice Problem 36

LINEAR EQUATIONS

CHAPTER 3

SIMULTANEOUS LINEAR

Elimination method 38 Practice Problem 39 Solving simultaneous equations with three unknowns 40 Practice Problem 41 Solving Simultaneous Equations more than three unknown 42 Practice Problem 45 Application of Linear Equation to Business 46 Demand Function 47 Supply Function 49 Practice Problem 52 Equilibrium 55 Practice Problem 62

CHAPTER 4 Introduction 63 Practice Problem 63 Simple Linear inequalities 64 Practice Problem 64 Compound Linear inequalities 64 Practice Problem 64 Inequalities and number line 65 Practice Problem 65

ALGEBRA

Solving linear inequalities 66 Practice Problem 66 Solving simultaneous linear 67 Practice Problem 68 Intersecting Regions 69 Application of linear inequalities 70 Practice Problem 72

CHAPTER 5

BINOMIAL THEOREM

Introduction 75 Operations of problem 76 Expanding 78 Practice Problem 79

CHAPTER 6

PERMUTATIONS & COMBINATIONS

Introduction 81 Evaluating permutations 82 Practice Problem 82 Combination 83 Practice Problem 83

CHAPTER 7

Introduction 84 Types of matrices 84

MATRICES ALGEBRA

Square matrix 84 Identity matrix 85 Row vector matrix 85 Column Vector 85 Null vectors 85 Practice Problem 86 Operation of Matrices 87 Additional of matrix 87 Subtraction of matrix 88 Scalar multiplication 88 Matrix transportation 89 Multiplication of matrix 90 Practice Problem 90 Application in Business 92 Practice Problem 94 Inversion of Matrices 95 Practice Problem 99, 109 Application of Matrices 121 Practice Problem 138 Markov Process model141 Characteristics or Assumptions of Markov 141 Practice Problem 151

CHAPTER 8

DECISION ANALYSIS

Introduction 155 Essential elements for decision 155 Maxmin Minimax Approach 158 Minimax Regret Approach 158 Decision making with probabilities 160 Practice Problem 168

CHAPTER 9

CALCULUS I: DIFFERENTIATION

Introduction 167 The gradient of a curve 170 Practice Problem 177 Rules of Differentiation 178 Constant Rule 178 Practice Problem 179 The Sum Rule 180 Practice Problem 179 The Difference rule 181 The Sum Rule 182 The Combination Sum rule and Difference rule 183 Practice Problem 184 The Product rule or Multiplication rule 185 Practice Problem 186 The Quotient Rule or Division Rule 187 The chain rule 190

Practice Problem 192 Maximization and Minimization of Function 193 Stationary Points 193 Application of differentiation 199 Practice Problem 206 Partial Differentiation 208 Practice Problem 212 Constrained Optimization 214 Practice Problem 216

CHAPTER 10

CALCULUS II: INTEGRATION

The anti – derivative 218 Definite Integral 220 Application of Definite Integral 222 Practice Problem 225

CHAPTER 11

NON – LINEAR EQUATIONS

Quadratic functions 228 Discriminant 229 Practice Problem 232 Application of Quadratic equation 235 Practice Problem 239

CHAPTER 12 Introduction 241

REVENUE, COST AND PROFIT

Total Revenue 241 Total Cost 244 Profit 244 Break even analysis 245 Practice Problem 247

CHAPTER 13

DECISION TREES

Introduction 250 Solving decision tree 252 Practice Problem 255

CHAPTER 14

MATHEMATICS OF FINANCE

Simple Interest 258 Compound Interest 259 Investments Analysis 261 Payback Period Method 262 Net Present Value 264 Practice Problem 270

CHAPTER 15 Introduction 274 Steps involved in linear programming 275 Problem Formulation programming 275 Solving linear program using graphical 278 Essential functions in linear 282

LINEAR PROGRAMMING

Formulating a linear programming 282 Application of linear program 283 Practice Problem 288

CHAPTER 16

INTRODUCTION TO QUANTITATIVE

Introduction 293 Stages in Quantitative study 293 Example of Model 294

CHAPTER 17

LINEAR PROGRAMMING:

Solving linear program using simplex 295 Definition of Terms 295 Solving LP using simplex algorithm 296 Shadow Prices 306 Slack variables 306 Practice Problem 306 Dual Program/ Minimizing Simplex Method 311

CHAPTER 18 Introduction 313 Basic Network Terminologies 313 Rules for drawing networks 314 Using the Standard Normal Table 318 Float time 336 Practice Problem 339

NETWORK ANALYSIS AND

CHAPTER 19

TRANSPORTATION MODEL

Introduction 343 Stages of Transportation Algorithm 343 Solving a transportation problem using minimum cost 344 Alternative allocation 350 Vogel method 351 Testing for optimality 357 Testing for optimality using Module 357 Degeracy 369 Practice Problem 374

CHAPTER 20

ASSIGNMENT MODEL

Introduction 378 Hungarian Algorithm 378 Alternative assignment 383 Practice Problem 402

CHAPTER 21 Introduction 406 Basic stock model 407 Assumptions of EOQ 407 Annual cost of ordering 408

INVENTORY PLANNING & CONTROL

Annual cost of holding the stock 408 Total cost in EOQ Model 409 The optimum order quantity 409 Economic Batch Quantity 412 The planned shortage model 417 Practice Problem 420

CHAPTER

1

SET THEORY AND COUNTING NUMBERS

Learning Objectives At the end of this section the students should be able to: • Drawing of the Venn diagram. • Identify elements of sets • Identify the different symbols used in the sets • Apply Venn diagram to the world of business today

A set is a collection of objects or elements. A set of student may contain students who are male and female and also students registered for different degree programs. The set brace { } are used enclose numbers belonging to the set. A set containing of numbers1, 2,3, 4, 5, 6, 7, 8, 9 and 10may be written as follow {1, 2, 3, 4, 5, 6, 7,8, 9,10}

Elements are numbers contained in a set. The above is a set of number 1 up to 10. When writing down the set of multiples of 4 The listings of the multiples can be done as follows. {4, 8, 12, 16, 20, 24 …} Notice that there has been a pattern established and the last three dots indicate the patterns continue to the last term which could be the nth term. Consider listing the multiples of 4 which are less than 100 {4, 8,12, 16, 20, 24,...,100} Instead of listing all the elements the three dots can be used to show where there is a pattern. When we wish to show that a particular number belong to a particular set we use the following symbol. Thus if we have a set of prime numbers less than ten {2, 3, 5, 7}

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CHAPTER 1 Set theory and counting number Thus to show that an element belong to a set of prime numbers less than ten {2, 3, 5, 7} We can write as follows 5 ∈ {2, 3, 5, 7} 5 is an element of the set containing 2, 3, 5 and 7. Also 3∈{2, 3, 5, 7} If we want to show that an element is not an element of the above set we place slash through the symbol as follows 11 ∉{2, 3, 5, 7} When naming sets we use capital letter for example let the above set be A, written as follows A={2, 3, 5, 7} then 3 ∈ A and 11∉ A Two sets are equal if they contain the same elements for example A= {1, 2, 3} and B= {3, 2, 1}all contain exactly the same elements and are equal. In symbols it is written as follows {1, 2, 3}={3, 2, 1} Sets that do not contain exactly the same elements are not equal. For example set C = {4, 5, 6} and D = {7, 8, 9} are not equal. In symbols it can be written as follows {4, 5, 6} ≠ {7, 8, 9}

Standard Symbols used in set

- used to represent set of natural numbers ,

¥= {0,1, 2, 3, 4, 5,...}

- represent positive and negative integer, = {+4, +3, +2, +1, -1, -2, -3, -4} - represent a set of rational numbers, = {1.2, 1.3, 1.4, 1.5} - represent set of real numbers- ={

CHAPTER 1 Set theory and counting number

Intersection of Sets Intersection of sets is represented by the symbols ∩ We say that the intersection of two set A and B is written as follows A ∩ B which means the elements that are contained in both A and B. For example if set A contain {1, 2, 3, 4, 5} and set B contains the following elements { 2, 4, 6, 8, 10} A = {1, 2, 3, 4, 5} B = {2, 4, 6, 8, 10} Then the intersection is written as follows, A ∩ B = {2, 4}

Example 1.1 Given the following sets write down their intersection G = {4, 5, 6, 7, 8. 9} and H = { 1, 3, 5, 7} the intersection is given by G ∩ H = {5, 7} The set containing all elements in A and B is called Union of set A and B and it is represented by the following symbol A ∪ B given as follows A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10} Consider set C which is a set containing odd numbers {3, 5, 7, 9} and a set D which contain multiple of 2 {2, 4, 6, 8.10} Then there is no intersection C ∩ D Such a set with no members, is called an empty set , it is represented by ∅ Thus the above example is written as C ∩ D = ∅ In most case there are some elements which are contained in a particular question or topic. Such sets are represented by the symbol ∪ it is called the universal set. On the same not all elements of the universal set that are not contained in set A are called compliment of set A which are written as A' (complement set of A) For example given the following universal set = { 3, 4, 5, 6, 7, 8, 9} and that A = {5, 6, 7} the compliment of A is the set of A' = {3, 4, 8, 9}

3

4

CHAPTER 1 Set theory and counting number Note that A ∩ A' = ∅ and A ∪ A' = ∪ Consider set E is also a member of set F, then E is called a subset of F. For example E = {4, 5, 6, 7} is a subset of { 3, 4, 5, 6, 7, 8, 9, 10} the symbol is written as follows E ⊂F In set the notation n(A) is used to represent the number of elements in set A

Venn Diagram A Venn diagram is diagrammatic representation of the relation of relationships of sets A Venn diagram contains of a rectangle which encloses all the elements contained in the universal sets and circles which represent the sets and subsets.

Example 1.2 Set A contains {1, 2, 3, 4, 6, 7} and B = {2, 6, 8, 9, 10} Representing this on Venn diagram is as follows Point of intersection

The intersection will be A ∩ B = {2, 6}

A∩B A

1, 3, 4, 7

2, 6

8, 9, 10

B


Example1.3 Consider the following Set C which is {3, 5, 7, 9} and set C is a subset of set D { 1, 2, 3, 4, 5, 6, 7, 8. 9} then drawing the Venn diagram is draw as follows C ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9} U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

D 2, 4, 6, 8

1, 3, 5

C

7, 8

Example 1.4 Given that E = {1, 2, 3, 4, 5, 6, 7, } and 8, 9 and 10 are not contained are compliments of E that is E' ={8, 9, 10} this can be drawn as shown in the Venn diagram. E E'

Placing the numbers we obtain the following Venn representation.

E 1,2, 3, 4, 5. 6,7 s

E' 8, 9, 10

Two sets which have no elements in common are said to be disjoined sets. For example set K = { 12, 13, 14, 17} and H = {11, 15, 16, 18, 19 }

5

CHAPTER 1 Set theory and counting number This can be drawn as follows in the following Venn diagram.

H 11,15,16,18,19 K 12,13,14,17

Principle of counting Consider set A if we denote the number of elements in set A as discussed earlier we can symbolically write it as n(A) If set A is given by A = {1, 2, 3, 4} the we denote the number of elements contained in set A to be n(A) = 4 and if another set B is given by B = {a, b, c, d, e, f, g, h, h, I, j, k} then n(B) = 11 If set C is an empty set or null set then C = ∅ and can be denoted by n(C) = 0. Principle of Inclusion-Exclusion Let consider two set A and B A = {a, b, c, d, e, f, g, h} B = {b, d, f, k, l, m} Then the universal set will be A = {a, b, c, d, e, f, g, h, k, l, m} then n(A) = 11 A ∩ B = {d, d, f} then A ∩ B = 3 A = {a, b, c, d, e, f, g, h} then n (A) = 8 B = {b, d, f, k, l, m} then n (B) = 6 In general we can use the Inclusion-Exclusion principle which states that given two sets the number of the union is equal to the number of the first set plus the number of the second set minus the intersection of the two sets n(A ∪ B) = n(A) + n(B) – n(A ∩ B) To explain this further lets consider the above example n(A ∪ B) = 11

n(A) = 8 n(B) = 6

n(A ∩ B) = 3

n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 11 = 8 + 6 – 3

6


Example 1.5 In the year 2009, five hundred students graduated in Africa Nazarene University. Of these 500 students, 310 had degrees in business of any sort, 238 had undergraduate degrees in business and 184 had master’s degree in business. Required How many students had both undergraduate and master degree in business?

Solution Let A = {students with an undergraduate degree in business} B = {students with masters degree in business} Then A ∪ B = {students with at least on degree in business} A ∩ B = {students with both undergraduate and masters in business} So n(A) = 238 n(B) = 184 n(A ∪ B) = 310 The question is asking as to check for the n(A ∩ B). By Inclusion-Exclusive principle we have n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 310 = 238 +184 - n(A ∩ B). n(A ∩ B) = 112

Sets Operation A ∪B = B ∪A A ∩B = B∩A A ∪ (B ∪ C) = (A ∪ B) ∪ C A ∪ (B ∪ C) = (A ∪ B) ∪ C A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= Commutative law for Union = Commutative law for intersection = Associative law for union = Associative law for intersection = Distributive law for union = Distributive law for intersection

There are two additional rules called De Morgan Laws ( A ∪ B)C =AC ∩ B C

7


Application of Venn Example 1.7 A survey conducted of 600 students at Africa Nazarene University produced the following results. 190 of the students read Daily Mirror; 180 of the students read Telegraph newspaper; 500 read Magazines; 130 read Daily Mirror and Telegraph newspaper; 110 read the Telegraph newspaper and Magazines; 130 read Daily Mirror and Magazines: 90 read all three. Use the above information to answer the following questions a. How many students read none of the publication? b. How many read magazines c. How many read Daily Mirror and the Telegraph newspaper , but not Magazine.

Solution Let letter D represent the Daily Mirror Let letter T represent Telegraph Newspaper Let letter M represent the Magazines Thus The number of sets in Daily mirror are n(D) = 190 The number of sets in Telegraph Newspaper are n(s) = 180 The number of sets in Magazines are n(M) = 500 The number of elements of intersection of Daily Mirror and Telegraph Newspaper n(D ∩ S) = 130 The number of elements of intersection of telegraph Newspaper and Magazines n(S ∩ M) = 110

8

9

CHAPTER 1 Set theory and counting number The number of elements of intersection of Daily mirror and Magazines n(D ∩ M) = 130 Intersection of the three n(D ∩ S ∩ M) = 90 Those who read Daily Mirror and Magazines only are 130- 90 = 40 Those who read Daily nation and telegraph Newspaper 130 – 90 = 40 Those that read Telegraph Newspaper and Magazines only = 110 – 90 =20 Those who read Daily Mirror only are 190 – 90 – 40 – 40 = 20 Those students who read telegraph Newspaper are 180 – 90 – 40 – 20 = 30 Those students that read Magazines only 500 – 90 – 40 – 2- = 350 Total number of students who read the three publications are 90 + 40 + 40 +20 +30 + 20 + 350 = 590 Those students who didn’t read any publication 600 – 590 = 10 students

D ∪S∪ M = 600

M 350 40

D

20

90 40

Answers a. None of the publication 600 – 590 =10 b. Magazine only 350 c. Daily mirror and Telegraph but not Magazine 40

20 30

S


Example1.8 Infotell a research firm collected the following information about particular buying behavior of certain families and gave the following information; 150 families bought food, clothes and luxuries; 250 families bought food and luxuries; 420 families bought food; 200 families bought clothes and food; 60 families bought luxuries and clothes, but not food; 40 families bought luxuries, but neither food nor clothes; 100 families bought none of the three; 180 families bought clothes but not luxuries. Using the Venn diagram answer the following question; How many families were surveyed? How man families bought clothes? How many families bought luxuries but not clothes? How many families did not bring luxuries? How many families bought food or costumes?

Solution Let letter F represent food let letter C represent clothes Let letter L represent luxuries Thus The intersection of the three n(F ∩ C ∩ L) = 150 The number of set in food is n(F) = 420 The intersection of Food and Luxuries n(F ∩ L) = 250 The intersection of food and clothes n(F ∩ C) = 200 We have no intersection of clothes and luxuries n(L ∩ C)

10

11

CHAPTER 1 Set theory and counting number U = 750

Clothes 130 50 Food

120

150 100

60 Luxuries 40

Family Surveyed = 120 + 100 + 150 + 50 + 40 + 60 + 130 + 100 = 750 50 + 150 + 60 + 130 = 390 100 + 40 = 140 750 – 100 – 120 – 130 = 400 750 – 100 – 40 = 610

Practice Problem 1. In certain college club of Abel, Ang, Ban, Charles and Debal are doing a accounting majors with Ang and Debal being student representative. Gor, Batt, Hesh, Ila and Joy are choir members with Ila and Joy doing degree courses. Use set A to represent club members, set B choir members, set C student Representatives, set D accounting major students. a) Specify element of the element of the four sets A, B, C and D b) Draw a Venn diagram representing sets A, B, C and D c) Identify members of the following sets and state in words the implication of d) A ∩ C, D ∪ C and D ∩ C e) Establish a universal set of the student’s membership. 2. In a study on the pass rate of CPA sec IV class, 60 candidates passed QT, 80 passed FA III and 50 passed law II. 20 passed QT and FAIII, 15 passed Law II and FAIII, 25 passed QT and Law II while 10 passed all the three subjects. a) What is the total number of candidates in the class? b) If a candidate who fails one subject is referred, how many candidates were referred?

CHAPTER 1 Set theory and counting number 3. The main daily newspapers in Town K are: A, B and C. The management of one dailies is concerned about low sales volume of their paper. In a recent survey of 1000 families in the town K the numbers that the various newspapers were found to be as follows:Newspaper No. of Readers The people 280 The people and East Africa Standard 80 The East Africa Standard 300 The People and Daily Nation 100 The Daily Nation 420 The East Africa Standard and Daily Nation 50 All the three newspaper 30 a) Present this information in a Venn Diagram b) Determine the number of families who did not read any of the three c) Advice the manager of a car dealing business which newspaper to advertise in order to reach the largest audience. 4. The manager of Karibu Café kept a record of his customers’ breakfast orders for one week. He discovered the following: 200 ordered Samosa 180 ordered Hot dog 150 ordered Hamburger 100 ordered Samosa and Hot dog 80 ordered Samosa and Hamburger 60 ordered Hot dog and Hamburger 30 ordered Samosa , Hot dog and Hamburger. Determine the customers; a) Who ordered Samosa only, b) Who ordered Hot dog only, c) Who ordered Hamburger only, d) Who ordered Samosa and Hot dog only, e) Who ordered Samosa and Hamburger only, f) Who ordered Hot dog and Hamburger only. 5. Some enrolment totals Africa Nazarene University for the 1st trimester 2011/2012 are the following; 150 enrolled in Business Statistics 640 enrolled in Management Mathematics I 310 enrolled in Christian Beliefs 90 enrolled in both Business Statistics and Management Mathematics I

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CHAPTER 1 Set theory and counting number 60 enrolled in both Business Statistics and Christian Beliefs 200 enrolled in both Management Mathematics I and Christian Beliefs 50 enrolled in all three courses Determine the number of students enrolled in; a) Business Statistics only b) Management mathematics I only, c) Business Statistics and Management Mathematics I and not Christian Beliefs, d) Christian Beliefs only, e) Christian Beliefs and Management Mathematics I but not Business Statistics, f) Christian Beliefs and Business Statistics, but not Management Mathematics I. 6. Suppose set A has 8 elements, set B has 12 elements and set C has 10 elements. Furthermore, suppose that the numbers of elements in A n B, A n C, B n C and A n B n C are 4, 2, 3, and 1 respectively. Determine the number of elements in A n B’ n C’, B n A’ n C’ and C n A’ n B’. 7. Explain the purpose of a Venn diagram in modern business world. 8. A study by Strategic Research Consultants on the cause of business failure of 1500 micro enterprise firms gave the following results: 600 failed due to lack of capital only 110 failed due to lack of capital and inexperienced management 140 failed due to lack of capital and a poor location 230 failed due to inexperienced management only 150 failed due to inexperienced management and a poor location 60 failed due to all the three 80 failed due to other causes but not any of the three. a) Present the information above in a Venn diagram and determine the number of micro enterprise firms that failed due to: b) A poor location only c) At least two of the causes mentioned above d) Exactly one of the causes 9. A milling company ran an advertising campaign to popularize its products. After a month the marketing department conducted a survey to gauge the success of the campaign. They interviewed 250 people. Out of the 250, 70 had not seen the adverts. 130 of those who saw the adverts had bought their products. In total 160 people bought the

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CHAPTER 1 Set theory and counting number company’s products. How many of the interviewees neither saw the adverts nor bought the company’s products? 10. A certain publisher is interested in the reading habits of the people in a certain small city. A survey indicates that the number of subscribers ( in thousands) to the Time magazine is 25, the number of subscribers Economist is 18 and the number of subscribers to the East African is 12. The number of subscribers to both Time and Economist is 10, the number subscribing to both Time and the East African is 1, and the number subscribing to both Economist and the East African is 2. The number of people with subscriptions to all three magazines is 1. Determine the number subscribing to exactly one of the magazines. 11. The following represent 450 workers and the mode of transport that the use regularly Mode of transport Workers Train 290 Public transport 350 Boda Boda 290 All three 200 Both train and Boda Boda 210 Both train and public transport 250 Boda boda and Public transport 280 Determine: a) Represent the information in Venn diagram b) How many used at least two type of transport c) How many used boda boda only 12. In Mombasa city, three newspapers A, B and the C. A survey shows that 30% of the population of the city read A. B is read by355 read C and 155 read both the A and B. 5% read both A and the C, 6% read both the A and the C and 2% read all the three. a) Draw a Venn diagram to represent the information b) What percentage of the city dwellers read none of the papers c) How many read at least two of the papers d) Africa Nazarene wants to advertise in two newspapers. Which two newspaper would give them the best coverage

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CHAPTER

2

LINEAR EQUATIONS

Learning Objectives At the end of this chapter the student should be able to do the following: • Simple addition, subtraction, multiplication and division of positive and negative numbers; • Perform simple operation on a number line; • Identifying and plotting points on Cartesian plane; • Getting gradient of a line; • Drawing a line on Cartesian plane using gradient and its intercept; • Solving simultaneous linear equations.

Number line A number line is a horizontal line with both negative and positive numbers. Generally as student of mathematics if you cant understand the number line you miss the point of mathematics. A number line consist of integers. { 1, 2, 3, 4, 5, 6, 7, } These numbers are called counting numbers or natural numbers. In general between two integers are rational numbers. For instance between 1 and 2 there are 1.2, 1.5, 1.8, .1.9. These numbers can be written in ratios as follows respectively. 6 3 9 19 , , , 5 2 5 10 Thus rational numbers are numbers that can be expressed in ratio of integers. Numbers that are not rational are called irrational numbers. For instance irrational numbers are.

π , 2, 6, 7, 10, 12

CHAPTER 2 Linear equations

Integers When you think of integer it better to think of zero, whole numbers which are positive and negative. Always when you think of a number line consider it as made of integers. For instance 23, 1 ,457 -128, -2367 are all integers, but numbers like3/5,0.0001, 2.23004, and -4.3 are not. In general we that an integer is a set of {-5,-4,-3,-2,-1, 0, 1, 2, 3, 4, 5, } the three dots means you keep on going both directions. Consider two integers x and y The relationships between these two numbers can be expressed as follows. x< y In general this means x is on the left of y in the number line. We read this as x is less than y Sometime x and y is written as x≤ y Which means x is less than or equal to y .It coincides on the number line. Again x and y can be written as follows x> y This means that x is greater than y and that x is on right of y on the number line. This means that x is greater than y or equal to y and that it coincides on the number line x≥ y Addition and Subtraction. To add or subtract numbers in a number line we think of walking forward and backwards f h h k f d

Example 2.1 Evaluate i) -3 – 4 ii) -7 + 10 iii) -5 + 4 Solution To evaluate -3 - 4 It will be x (+y) You at point (-3) you move 4 units or steps on your left or backwards

16

17


Figure 1.1 Count 4 units or steps on left -8

-7

-6 -5

-4 -3

-2

-1 0

1

2

3

Thus -3 - 4 = -7

To evaluate -7 + 10 In terms of x and y it x +y At point (-7) you move 1steps forward. Picture this that you own your friend 7 sweets and now you have just brought 10 sweets if you give your friend 7 sweets you are left with 3 sweets.

Figure 1.2

Count 10 units or steps toward right -8

-7

-6 -5

Thus -7 + 10 = 3

-4 -3

-2

-1 0

1

2

3

5 6

18


To evaluate -5 + 4 In general x + y At point (-5) you move 4 steps forward.

Figure 1.3

Count 10 units or steps toward right -8

-7

-6 -5

-4 -3

Thus -5 + 4 = -1

Practice Problems Use a number line to evaluate the following. -1+ 6 -2 - 3 4-8 –9+3

-2

-1 0

1

2

3

5 6


Multiplication and Division Multiplication of numbers involving negatives and positives one must note following order of multiplication. ( − × − = + ) negative × negative = positive

( − × + = − ) negative × positive = negative ( + × − = − ) positive × negative = negative ( + × + = + ) positive x positive = positive From the above rule we evaluate the following respectively. 3 × -5=-15 -4 × 4=-16 -5 × -6=30 7 × 9=63 In division these rules also apply.

( − ÷ − = + ) negative ÷ negative = positive ( − ÷ + = − ) negative ÷ positive = negative ( + ÷ − = − ) positive ÷ negative = negative ( + ÷ + = + ) positive ÷ positive = positive

We can solve the following using the above rules

Gradients Evaluate −3 × 6 × −9 × 2 −18

19

20


Solution We multiply the first part called the numerator. −3 × 6 × −9 × 2 = −324

324 = −18 −18

Points on x and y Plane Any given line has two of the points which one can use to plot a line. These are called the coordinates. Coordinates are inform of x and y. Consider line PQ, the line has two points P (x, y) and Q (x, y). This line can plotted in graph on x on the horizontal axis and y on the y axis this is called as the Cartesian plane. O is the intersection of the x-axis and y- axis and is called the origin that is the point with the coordinates (0, 0)

Figure 1.4

Vertical axis or y- axis

O

Horizontal axis or x- axis

21


Gradients The change on the y-axis over the change of the x-axis is called gradient. We can think of the gradient of a line is the measure of its steepness or angle of slope. If we have two points the gradient of the line is defined as the vertical change divided by the horizontal change. In general if we have point A ( x1, y1 ) and B ( x2 , y2 ) the gradient of the line segment AB is given as follows. From the line AB we can express this as: Change in y (y distance) Change in x (x distance)

=

∆y ∆x

∆( y2 − y1 ) (Where ∆ is the Greek capital D ‘Delta’ which is used to mean ∆( x2 − x1 )

change) Diagrammatically the line AB can be represented as bellow on Cartesian plane.

Figure 1.5

Change in y (∆y )

y-axis A( x1 y1 )

Change in x (∆x)

x-axis

22


Example 2.2 Consider three points A, B, and C .A(0,3), B ( 2, 4) and C (10, 8) Find the gradient of AB and BC. What can you say about the points.

Solution ∆y Change in y (vertical distance) = Change in x (horizontal distance) ∆x

Figure 1.6 9 8

C (10, 8)

7 6 B (2, 4)

5 4 3

A (0, 3)

2 1 1

2

3

4

5

6

7

8

9

10

Taking point AB. A (0, 3) and B (2, 4 ) we get the slope as by

∆y 4 − 3 1 = = ∆x 2 − 0 2 Taking point AB, A(0,3),and B (2,4) we can also obtain the gradient

∆y 8 − 4 4 1 = = = ∆x 10 − 2 8 2

23

CHAPTER 2 Linear equations Taking point BC, B (2,4) and C (10,8) We can get the slope by

∆y 8 − 4 4 1 = = = ∆x 10 − 2 8 2 What we can see is that the gradient is the same .Thus A, B, and C are points on a straight line. In general any two pairs of a point selected they give the same gradient

Figure 1.7 9 8 7 6

C (10, 8) B (2, 4)

5 4 3

A (0, 3)

2 1 1

2

3

4

5

6

7

8

9

10

Practice Problem 1 Find the gradient of the following coordinates (2, 3) and (7, 6) (-2, 7 ) and (-1, -1 ) (4, 4)and (5, 7 ) 2. If point (1, 1 ) and (4, 7) and (5, y) are point on a straight line find the value of y.

CHAPTER 2 Linear equations 3. The line from (1, 1) to (x, 3x) has a gradient 2 .Find x. 4. The four corners of quadrilateral are (1, 1), (2, 2), (4, 3), and (5, 3). Show that the quadrilateral is trapezium. 5.Let points A, B and C have coordinates respectively .If A ( x1, y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) are on straight line show that y1 x2 − y2 x1 + y2 x3 + y3 x2 + y3 x1 − y1 x3 = 0

24


Equation of Straight The gradient is the steepness or the slope of line, curve, or arc. When we have the gradient of a straight line we can be able to find the equation given any particular point that it passes through. The standard form of the equation of a straight line is give by

= y mx + c Where m is the gradient and c is constant which represents the y intercept. In general given point ( x, y ) and the gradient is m then taking another point of the line ( x1, y1 ) we can express m in terms of y ∆( y − y1 ) =m ∆( x − x1 ) Multiplying all through by ( x − x1 )

( x − x1 )

y − y1 =− m( x x1 ) x − x1

This leads to the standard equation of a straight line. The equation can be expressed as follows using the points (x, y) and (4, 2) y−4 = −2 x−2

y − y1 = m( x − x1 ) = y mx + c = y mx + c Where c is a constant and m the gradient

25


Example 1 Find the gradient of the line through the points (4, 7), (5, 9)

Solution ∆y 9 − 7 2 = = = 2 ∆x 5 − 4 1 The equation must be inform of = y mx + c Taking any point for instance along the line (4, 7) y−7 =2 x−4 Multiplying by x − 4 on both side ( x − 4)

y−7 2( x 4) =− x−4

y − 7 = 2x − 8 y = 2x − 8 + 7 = y 2x −1 This is an equation of a positive gradient when plotting the graph the line moves from left to right upwards.

26

27


Figure 1.8 9 8

Positive gradient

7 6 5 4 3

= y 2x −1

2 1 -1

1

2

3

4

5

6

7

8

9

10

Find the equation of line passing through the following points (2, 4) and (1, 6) The gradient is given as

Solution Change in y (vertical distance) ∆y = Change in x (horizontal distance) ∆x

Example 2

∆y 6 − 4 2 = = = −2 ∆x 1 − 2 −1 Multiply by on both sides x − 2

28


y − 4 =−2( x − 2) y − 4 =−2 x + 4 y =−2 x + 4 + 4 y= −2 x + 8 Plotting the graph the line moves (slopes) from top left to right downwards. This kind of line is called negative gradient. 9

Figure 1.9 8 7 6 Negative gradient

5 4 3 2

y= −2 x + 8

1 1

2

3

4

5

6

7

8

9

10

Example 2.3 Consider points A(4, 2) and B(7, 2) The gradient of the line passing through these points The equation we obtain is supposed to be inform of = y mx + c ∆y 2 − 2 0 = = = 0 ∆x 7 − 4 3 The gradient is zero Thus taking any point along the line we obtain the following equation y−2 = 0 taking point (4, 2) x−4

29

CHAPTER 2 Linear equations Thus taking any point on the line (x, y) and taking point (4, 2) We get the following y − 2= 0( x − 4) y−2= 0 y=2 When you plot the graph it is a horizontal line.

Figure 2.0 9 8 Zero or horizontal gradient

7 6 5 4 3 2

y=2

1 1

2

3

4

5

6

7

8

9

Suppose you have two points (4, 2) and (4, 6) the gradient is

∆y 6 − 2 4 = = = ∞ ∆x 4 − 4 0 The equation is either inform of x = 1, 2,3, 4...........

The graph can be plotted as follows it is a vertical line.

10

30


Figure 2.1 Vertical gradient

98 7 6 5 4

x = 1, 2,3, 4,5...n

3 2 1 1

2

3

4

5

6

7

8

9

10

Practice Problem Find the equation of following straight lines. Gradient 2 through (5, 6) Gradient -3 through (1, 2) Gradient 0 through (2, 5) Gradient ∞ through (1, 1) 1 Gradient through (3, 3) 2 Through (2, 4) and (6, 8) Through (1, 1) and (0, 9) Through (1, 3) and (2, 3) Through (1, 2) and (1, 5) Find the gradient of the following lines −2 x + 4 a) y = y 15 + 10 b) 5 = c) 3 x + 6 y + 3 = 0 d) 3 y − x − 1 =4 x 6y + 7 e) 2= 3. A straight line goes through the points (a, 0) and (0, b) show that the equation x y written as + = 1 a b

31


Sketching a Graph When sketching a graph of a straight line there are two ways which can be used.

Method 1 Consider a straight line 2 x + y = 8 Given a range of −2 ≤ x ≤ 4 In general you consider −2 ≤ x ≤ 4 as the distance of the x-axis. First make sure that the line is in the format = y mx + c So we get y= −2 x + 8

With the points of x and y we plot these points into a graph For instance point (-2, 12), (-1, 10) and so on the Cartesian plane

Figure 2.2 8 7 6 5 4 3 2

y= −2 x + 8

1 1

2

3

4

5

6

7

8

9

10

32


Method 2 This an alternative method which student consider to be the simple Using the same equation of 2 x + y = 8 plot the graph using method 2 Solution Setting x = 0 you get 2(0) + y = 8 0+ y = 8 y =8 Hence the coordinate of the first point of the line is (0, 8) Also setting y = 0 we can obtain the value of x as follows 2x + 0 = 8 2x = 8 x=4 Hence the second point has the following coordinates (4, 0) Using these coordinates (0, 8) which give the y-intercept that is where the line cuts on intersect with the y axis on the Cartesian plane. The coordinates of (4, 0) this is where the line will intersects with the x-axis called the xintercepts.

Figure 2.3 y − Intercept 8 7 6 5 4

y= −2 x + 8

3 2

x − Intercept

1 1

2

3

4

5

6

7

8

9

10


This is the easiest method in algebra but remember if you are looking for accurate point it is necessary to use a graph paper or solve the simultaneous equation.

Practice Problem Find the coordinate of the two points of the following lines and hence sketch their graphs. i)

8 x − 4y =

ii)

3x − 4 y = 7

iii)

6 − x + 2y =

Points of intersection A point of intersection is where two or more straight line cut each other or meets. We will be dealing with simultaneous equation with two or more unknowns. In general a point of intersection gives the solution of any given simultaneous equations.

Example 2.4 Consider the following 3x + 4 y = 18.............................(i ) 6x − 3y = 3...............................(ii ) Required

33

34

CHAPTER 2 Linear equations Find the point of intersection of the two linear equations. Plot the point on a graph.

Solution From the previous topic we are already familiar with how to get the points Take the first equation 3x + 4 y = 18.............................(i ) It passes through the point (0, 4.5) and (6, 0) 6x − 3y = 3...............................(ii ) The second equation passes through points (0, -1) and (0.5, 0) These two line sketched on a graph the point of intersection is (2, 3) Always when you have gotten the solution it is better to verify whether you have gotten the right solution by substituting the values. By doing this you are assured of minimizing common mistakes that students make not that they didn’t know but they overlooked. I call these mathematical remedy.

The point of intersection is correct because it satisfies our equations

Figure 2.4 8 7 6

6x − 3y = 3

5 4

Point of intersection (2,3)

3 2

3x + 4 y = 18

1 -1

1

2

3

4

5

6

7

8

9

10


However some simultaneous equation that has no point of intersection that means that they have no solution.

Example 2.5 Find the point of intersection and comment on the answer of 2x + 4 y = 11................................(i ) 6 x + 12 y = 33..............................(ii )

Solution Using either of the methods discussed earlier for our case let use the second method. Taking equation 2x + 4 y = 11................................(i ) When x = 0 we get

2(0) + 4 y + 11 4 y = 11 y = 2.75 So we have points (0, 2.75) Also when y = 0 2 x + 4(0) = 11 2 x = 11 x = 5.5 The second point on the line is (5.5, 0) Considering the second equation 6 x + 12 y = 33..............................(ii ) When x = 0 6(0) + 12 y = 33 12 y = 33 y = 2.75 The first point on the second line is (0, 2.75) When we set y = 0 6 x + 12(0) = 33 6 x = 33 x = 5.5

35

36

CHAPTER 2 Linear equations The second point is (5.5, 0) With the above points we can draw a graph

Figure 2.5 6 x + 12 y = 633 Coincidental lines

5 4 3 2 1 1

2

3

4

5

6

7

8

9

10

The above lines on the same line and are called coincident line

Practice Problem Find the point of intersection of a)

5 x + 10 y = 25 15 x + 30 y = 75

b)

7x + 9 y = 3 2x − 5 y = 16

b)

5x − 7 y = 9 x− y = 3

2x + 4 y = 11

CHAPTER

3

SIMULTANEOUS LINEAR EQUATIONS

Learning Objectives At the end of this section the student should be able to: • Solving a system or two equation using substitution and elimination method; • Solve a system of 3 unknown s using elimination method; • Solve a system of simultaneous equation of more than 3 unknowns using Gauss method; • Detect when the system of the above equation has no solution; • Detect when the system of the above equation has more solution that is infinite;

Example 3.1 Consider the two equation we had earlier on the other section 3x + 4 y = 18.............................(i ) 6x − 3y = 3...............................(ii ) The coefficient of the first equation (i) of x is 3 and the coefficient for the second equation for the second equation (ii) of x is 6 We can multiply the equation (i) by 2 so that we can eliminate x by subtracting equation (ii) from the result of equation (i) = 3 x + 4 y 18................................(i )..... × 2

= 6 x − 3 y 3..................................(ii ).... ×1 The result of equation (i) we subtract equation (ii) x cancel when subtracted

6x + 8 y = 36 6x − 3y = 3

Subtract equation (ii) from (i)

CHAPTER 3 Simultaneous Linear equations

11 y = 33 y=3

With y = 3 we can deduce x from either of our two original equation. For instance let’s take equation (i)

3x + 4 y = 18.............................(i ) We substitute y = 3 3 x + 4(3) = 18 3 x + 12 = 18 3= x 18 − 12 3x = 6 x=2 Remember the mathematical remedy check whether the value satisfy the equation

The above method is called the elimination method There are four stages we are supposed to remember when using the elimination method. Stage 1 Add or subtract a multiple of one equation to from the other to eliminate x or y depending on your choice of elimination. Stage 2 Solve the resulting equation for x or y Stage 3 Substitute the value of x or y into one of the original equation to deduce x or y. Stage 4 Mathematical remedy ,check the values whether they satisfy the equations. Lets consider the above system of equation

38


3x + 4 y = 18.............................(i ) 6x − 3y = 3...............................(ii ) We can choose which unknown value to substitute For instance let make x the subject of the equation (i) 3x + 4 y = 18.............................(i )

Use 6 −

4 y as your value of x to substitute from equation (ii) 3 6x − 3y = 3.............................(ii )

4   6 6 − y  − 3y = 3 3   36 − 8 y − 3 y = 3 36 − 11 y = 3 −11 y =− 3 36 −11 y = −33 y=3 With the value of y = 3 our x = 2 The above method is called substitution method.

Practice Problem Using elimination method solve the following system of equation 6x − 4 y = 8 a) 2x − 4 y = 4 b)

c)

6 x + 10 y = 38 4 y − 4 y + 22 = 0 3x + 2 y = 10 −x + 2 y = 8

39


Solving simultaneous equations with three unknowns When solving a simultaneous equation with three unknowns it is important to understand that you must derive other two equations from the three equations to get equation (iv) and (v) Example Consider the following systems 2 x + 6 y − 2 z = 8..................................................(i ) 4 x + 2 y + 2 z = 20................................................(ii ) 6 x − 2 y + 2 z = 8...................................................(iii ) Solve for x, y and z Take the first two equations (i) and (ii)

2 x + 6 y − 2 z = 8......................................................(i ) 4 x + 2 y + 2 z = 20.....................................................(ii ) To eliminate x multiply the first equation with 2 To get 2 x + 6 y − 2 z = 8......................................................(i ).......... × 2

4 x + 2 y + 2 z = 20.....................................................(ii ) Subtract the second equation from the result x cancel out

4 x + 12 y − 4 z = 16 4 x + 2 y + 2 z = 20 10 y − 6 z = −4.......................................................(iv)

Using the (i) and the (iii) equation to obtain the (v) equation

2 x + 6 y − 2 z = 8.........................................................(i ) 6 x − 2 y + 2 z = 20........................................................(iii ) To eliminate x we multiply the first equation by 3 2 x + 6 y − 2 z = 8.........................................................(i )......... × 3 6 x − 2 y + 2 z = 20........................................................(iii )

40

CHAPTER 3 Simultaneous Linear equations The result of equation (i) we subtract equation (iii) Y cancels

6 x + 18 y − 6 z = 24 6 x − 2 y + 2 z = 20 20 y − 8 z = 4.......................................................(v)

Now taking equation (iv) and (v) we can use either of elimination on substitution method to solve for y and z 10 y − 6 z = −4.......................................................(iv) 20 y − 8 z = 4.......................................................(v) Multiply equation (iv) with 2 20 y − 12 z = −8 20 y − 8 z = 4 We get the result and get the difference 20 y − 12 z = −8 20 y − 8 z = 4 − 4 z = −12 z=3 We obtain y through any of equation (iv) or (v) 20 y − 8(3) = 4 20 y = 4 + 24 20 y = 28 28 y= 20 y = 1.4 When you have the values of y and z we can substitute our original three equation the value of x by taking any of the equation. Taking equation (i)

41

CHAPTER 3 Simultaneous Linear equations 2 x + 6(1.4) − 2(3) = 8 2 x + 8.4 − 6 = 8 2 x = 8 − 8.4 + 6 2 x = 5.6 x = 2.8

Practice Problem Find the value of the x, y and z x + y + z =1 a) 2 x + 3 y + 3 z = 4 3x + 2 y + 5 z = 6 8q − 3s + 2t = 20 35 b) 6q + 3s + 12t = 4q + 11 y + 11 = 33 2a − 3b + 10c = 3 c) −a + 4b + 2c =20 5a + 2b + c =−12

Solving Simultaneous Equations more than three unknown Example 3.2 x1 + x2 + 3 x3 − x4 = 2.......................................(i ) 2 x1 + x2 − 3 x3 + x = 0.....................................(ii ) −2...................................(ii 3 x1 + x2 + x3 + x4 =

42

43


Solution Replace equation (ii) with equation (i) time 2 = x1 + x2 + 3 x3 − x4 2.......................................(i )..... × 2 2 x1 + x2 − 3 x3 + x = 0.....................................(ii ) We obtain the following systems 2 x1 + 2 x2 + 6 x3 − 2 x4 = 4.......................................(i ) 2 x1 + x2 − 3 x3 + x = 0.............................................(ii ) Getting the difference we obtain the following result. 2 x1 + 2 x2 + 6 x3 − 2 x4 = 4...........................................(i ) 2 x1 + x2 − 3 x3 + x = 0.................................................(ii ) 3 x2 − 9 x3 + 3 x4 = −4..........................................(iv) Multiply equation (i) by 3 and minus equation (iii) = x1 + x2 + 3 x3 − x4 2...........................................(i )........ × 3 3 x1 + x2 + x3 + x4 = −2.........................................(iii ) The result from the above is as follows 3 x1 + 3 x2 + 9 x3 − 3 x4 = 6...........................................(i ) 3 x1 + x2 + x3 + x4 = −2...............................................(iii ) 4 x2 − 8 x3 + 4 x4 = −8..........................................(v) We can take equation (i), (iv) and (v) to replace equation (ii) and (iii) respectively. Now instead of writing (iv) and (v) we are going to use (ii) and (iii) 3 x1 + 3 x2 + 9 x3 − 3 x4 = 6...........................................(i ) 3 x2 − 9 x3 + 3 x4 = −4..........................................(ii ) 4 x2 − 8 x3 + 4 x4 = −8..........................................(iii ) To eliminate x 2 we divide equation (ii) by 3 or multiply by

1 and equation (iii) by 4 4

44


3 x2 − 9 x3 + 3 x4 = −4..........................................(ii ) ……… × We obtain the following

x 2 − 3 x3 + x 4 = −

4 3

In the (iii) equation when we divide both sides by

1 3

1 4

4 x 2 − 8 x3 + 4 x 4 = −8..........................................(iii )............. ×

1 4

After dividing we get the following result x 2 − 2 x3 + x 4 = −2........................................................(iii ) Replace the (iii) equation with the (ii) by getting the difference

x 4 cancels x 2 cancels

4 − 3 x3 − −2 x3 = − − −2 3 2 x3 = − 3

Using equation (iii) substitute for x3 using equation (ii)

2 4 x 2 − 3(− ) + x 4 = − 3 3 10 x2 = − − x4 3 Where x 4 is an arbitrary figure in other word it is based on subject to individual judgment or preference. Substitute for x3 and x 2 in the first equation

45

CHAPTER 3 Simultaneous Linear equations  10   2 x1 −  − − x 4  + 3 −  − x 4 = 2  3   3 10 x1 + + x 4 − 2 − x 4 = 2 3 10 x1 + − 2 = 2 3 10 x1 + − 2 − 2 = 0 3 10 x1 + − 4 = 0 3 10 − 12 x1 + =0 3 2 2 x1 − = 0 ⇒ x1 = 0 + 3 3 2 x1 = 3 Using equation (i) can get the value of x 4 having the values of x1 =

x3 −

2 3

10 2 , x 2 = − − x 4 and 3 3

x1 − x 2 + 3 x3 − x 4 = 2

With the above figures we substitute. x1 − x 2 + 3 x3 − x 4 = 2 2  10   2 −  − − x 4  − 3 −  − x 4 = 2 3  3   3 2 10 + +x 3 3

The method we have used above is called Gaussian elimination process

Practice Problem Find the values of x1, x 2 , x3 and x 4 from the following linear systems x1 − x 2 − x3 + x 4 = 1 x1 + 2 x 2 + x3 = 1 2 x1 + x 2 − x 4 = 3 − x1 + x 2 − x3 + x 4 = 0

APPLICATION OF LINEAR EQUATION TO BUSINESS Learning Objectives At the end of this section the student should be able to: • Identify and draw a linear demand equation. • Identify and draw a linear supply equation. • Determine the equilibrium price and quantity for and quantity for single product or good • Determine the equilibrium price and quantity for two ,three or more than three products

The application of linear equations in business is also called demand and supply analysis Demand Functions Demand function is an equation that shows the relationships between the quantity demanded and other factors like tastes, presence of substitute and price We are going to concentrate on price as a factor that affect demand. Remember the formulae we had of y = mx + c The value of y depends on the value of x thus y is called the dependent variable. The value of x on the other hand does not depend on y but determines y thus in general it is called independent variable. When we replace y by P which represent the price and x replaced by Q which is the price. We can introduce parameter or constants that is a and b In general the equation will be as follows.

P = aQ + b a and b are counting numbers for instance 1,2,3,4................................n The demand function has a negative gradient a is the gradient and b is the y intercept. Plotting the graph it slopes from bottom left to top right upwards .The demand curve has negative slope coefficient that is − a

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business

47

Figure 2.6

Price

∆y Change in x ∆y = −a ∆x

P= −aQ + b

Change in x (∆x)

Quantity

`

Demand Function The marketing management of Beta shoe company noticed that on a certain month the safari boots cost KES 1000 ,the customers bought 600 pairs and the cost was KES 1600, 300 pairs were bought. Assuming the demand function is a linear function calculate The line that passes through the points Sketch the graph Find the change in a) P when Q = 700 b) Q when P = 2000

Solution We first get the gradient


Taking the points (600, 1000) and (300, 2000) ∆(1600 − 1000) 600 = − = −2 ∆(300 − 600) 300 Let get the line that passes through the points Taking any point lets take point (300, 1600)

P − 1600 = −2 Q − 300 Multiply both sides by Q − 300 (Q − 300)

P − 1600 = −2(Q − 300) Q − 300

P − 1600 =2Q + 600 −2Q + 600 + 1600 P= −2Q + 2200 P= Sketching the graph we are going to use the following coordinate. Remember the previous method we were using to get the coordinates. When P = 0 the value of Q is 1100 so the coordinates are (1100, 0) On the other hand when Q = 0 , P is equal to 2200 hence the coordinates are (0, 2200). We plot the graph as follows.

48


Figure 2.7 2200

P= −2Q + 2200

Price

Quantity

1100

Supply Function The equation of supply consists of the ratio or relationships between the quantity supplied and factors that affect supply such as price. In this course we are going to look at the price as our major factor that affects supply. Consider the following function P = aQ + b

49


Figure 2.8

= P aQ + b Price

Change in y

∆y ∆y =a ∆x

Change in x (∆x)

Quantity

` The supply curve move from bottom left to upwards right as shown by the figure 2.8 above. The supply curve has a positive slope coefficient that is a

Example 3.6 In a Datacom offer the sales manager observed that when the price of a flash disk 1GB was fixed at KES 400 the quantity supplied was 250 and when the price changed to KES 250 the quantity supplied was 200. Assuming that price was the only factors that affected supply and other factors remains constants. Calculate the slope Obtain the equation that passes through these points. Sketch the graph of the equation obtained. Hence find the value of i) P when Q = 400

50

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business ii) Q when P = 650

Solution We get the gradient by

∆(400 − 250) 150 = =3 ∆(250 − 200) 30 We obtain the equation passing through points (400, 250) and (250, 200) by taking (Q, P) and any point as follows. ∆( P − 250) =3 ∆(Q − 200) Multiply both sides by Q − 200 P − 250 = 3(Q − 200) P − 250 =3Q − 200 Collect the like terms together by adding on both sides 250

P =3Q − 200 + 250 = P 3Q + 50 Hence using the equation P = 3Q + 50 we can be able to find the value of P when Q = 400 = P 3Q + 50 Q = 400 Substitute Q by 400

51

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business = P 3(400) + 50 = P 1200 + 50 P = 1250 Also given the value of P = 650 we can obtain the value of Q as follows using the same equation. P = 3Q + 50 Substitute P with the value given of 650 = P 3Q + 50 P = 650 650 = 3Q + 50

Subtract 50 on both sides 650 − 50 = 3Q + 50 − 50 600 = 3Q 200 = Q

200 = Q

can also be written as Q = 200

Practice Problem 1. Suppose a certain commodity has the following linear demand and supply. When the price is 7500 the quantity is 100uniits, when the price is 4625 the quantity is 750 units and when the price is 2525 the quantity is 100 units, when the price is 1525 the quantity is 200 units a) Obtain the linear equation that go through the points given above, clearly explain which is the demand and supply curve. b) Equilibrium price and quantity.

52

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business 2. At a unit price of sh. 16,000, the demand of a product is 300 units and at a unit price of sh.48,000 the demand is 100 units. At a unit price of sh. 30,000 the supply is 550 units and at a unit price of sh. 50,000, the supply is 650 units. Determine the market equilibrium point 3. The quantity demanded per month of product is 250 units when the unit price KES 1400. The quantity demanded per month is 1000 units when the unit price is KES 1100. The suppliers will market 750 units if the price is KES 600 or lower but will supply 2250 units if the price is KES 800. Both the demand and supply functions are known to be linear a) b) c) d)

Find the demand equation Find the supply equation Find the equilibrium price and quantity Illustrate the demand and equation on a graph paper

4. Recently the government liberalized the consumer prices of maize. For the last six months, the Economic Institute have been keeping the average monthly price and quantity of maize consumed and supplied per bag in Kenya, The table shows this data; Price (KES) 15 14 13 12 11 10

Quantity Demanded (Kg) 6 8 10 12 14 16

Quantity supplies (Kg) 20 18 16 14 12 10

a) Obtain the demand and supply equation b) Determine the equilibrium price and quantity

5. The sales manager of Akamba Wood Carvings Ltd finds that he can sell 45 pieces per week at a price of KES 80 per item, but if he changes the price to KES 100 per item he can sell only 40 pieces per week. The production manager determines that the total production cost of 45 items per week is KES 3,000 and that of 40 items per week is KES 2800

53

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business a) Assuming that the unit price is linear function of the quantity sold determine the profit function. b) Assuming the total production cost per week is a linear function of the quantity produced, determine the cost function. c) Calculate the level of sales that will maximize the weekly revenue. d) Calculate the level of sales that will maximize the weekly profit e) Determine the corresponding profit per item sold the weekly profit maximized.

54


Equilibrium This is the state where the supply and demand are equal in other words it occurs when quantity demanded is equal to the quantity supplied. To get the equilibrium price and quantity we just equate the two linear functions of supply and demand. A graphical representation would look like one below.

Figure 2.9

Supply equation

Price

Equilibrium

P

Demand equation Q Quantity

`

Example 3.7 Suppose a commodity A has the following linear functions. Given the price is KES 9500 the quantity is 3000 units also when the price is KES 6625 the quantity is 2750 units. When the price is KES 4525 the quantity is 2100 units. Also when the price KES 2525 the quantity is 2200 units.

55

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business Required a) b) c) d)

56

Obtain the linear functions that passes through the points Identify which of the two equations is demand and supply functions. Calculate the equilibrium price and quantity Sketch the graph of the equations.

Solution We first obtain the change in price and change in quantity.

9500 − 6625 2875 = = 11.5 3000 − 2750 250 Taking any of the point we get P − 9500 = 11.5 Q − 3000 Multiply both sides by Q − 3000 to obtain P − 9500 = 11.5(Q − 3000) P − 9500 = 11.5Q − 34500 Add 9500 both sides to get the following equation P = 11.5Q − 34500 + 9500 = P 11.5Q − 25000 The above is supply function since the Q coefficient 11.5 is positive thus we can write this one as P = 11.5QS − 25000

Taking the other points in part (ii) we can obtain the slope just as we did in the first part.


∆(4525 − 2525) 2000 = = −20 ∆(2100 − 2200) − 100 Taking any point we can obtain the line as follows P − 4525 = −20 Q − 2100 Multiply both sides by Q − 2100 to obtain the following equation −20(Q − 2100) P − 4525 = P − 4525 = −20Q + 42000 P − 4525 + 4525 = −20Q + 42000 + 4525 P= −20Q + 46525 The equation is a demand function since the coefficient of Q is negative that means it slopes from top left to bottom right and can be expressed as follows. P = −20QD + 46525

With our equations we can get the equilibrium price and quantity by equating them both. P = −20QD + 46525 = P = 11.5QS − 25000

Collect the like terms together

P − P − 20Q − 11.5Q = −25000 − 46525 −31.5Q = −71525 Q=

−71525 −31.5

Q = 2271

57


By substituting to any of equation of your choice we get the value of P P = 11.5QS − 25000 P =11.5 × 2271 − 25000 P = 1116.5 The price is KES 1116.50 and the quantity is 2271 units, When plot the graph of the two linear equation it appears as follows.

Example 3.8 Suppose a certain commodity has the following linear demand and supply When the price is 7500 the quantity is 100uniits, when the price is 4625 the quantity is 750 units and when the price is 2525 the quantity is 100 units, when the price is 1525 the quantity is 200 units

58

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business Obtain the linear equation that go through the points given above, clearly explain which is the demand and supply curve. Equilibrium price and quantity.

Solution a)

∆P Change in price = Change in quantity ∆Q ∆P 7500 − 4625 2875 = = = 11.5 ∆Q 1000 − 750 280

P − 7500 = 11.5 Q − 1000 P − 7500= 11.5 ( Q − 1000 ) P − 7500 = 11.5Q − 11500 P =11.5Q − 11500 + 7500 P 11.5Q − 4000 = The above is supply function because of the positive co-efficient of Q which is 11.5

Change in price ∆P = Change in quantity ∆Q ∆P 2525 − 1525 1000 = = = −10 ∆Q 100 − 200 −100

P − 1525 = −10 Q − 200 P − 1525 = −10 ( Q − 200 )

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CHAPTER 3 Linear equations: Application Linear equation to business

−10Q + 2000 P − 1525 = −10Q + 2000 + 1525 P= −10Q + 3525 P= This is demand function because of a negative coefficient of -10 b).At equilibrium the demand is equal supply 11.5Q − 4000 = −10Q + 3525 11.5Q + 10Q = 3525 + 4000 21.5Q = 7525 Q = 350 P 11.5Q − 4000 = = P 11.5 ( 350 ) − 4000 P 4025 − 4000 = P = 25 The equilibrium price is 25 and the quantity is 350 units

Example 3.9 At a unit price of KES 16,000, the demand of a product is 300 units and at a unit price of KES 48,000 the demand is 100 units. At a unit price of KES 30,000 the supply is 550 units and at a unit price of KES 50,000, the supply is 650 units. Determine the market equilibrium point

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CHAPTER 4 Linear equations: Application Linear equation to business

Solution Demand function ∆P 48000 − 16000 32000 = = = −160 100 − 300 ∆Q −200 P − 48000 = −160 Q − 100 P − 48000 = −160 ( Q − 100 ) P − 48000 = −160Q + 16000 P= −160Q + 64000 Supply demand ∆P 50,000 − 30,000 20,000 = = = 200 ∆Q 650 − 550 100 P − 30,000 = 200 Q − 550 P − 30,000 =200 ( Q − 550 ) P − 30,000 = 200Q − 110,000 P= 200Q − 110,000 + 30,000 P 200Q − 80,000 = At equilibrium the supply is equal to demand

61

CHAPTER 3 Simultaneous Linear equations: Application Linear equation to business −160Q + 64000 200Q − 80,000 = 200Q + 160Q = 64000 + 80000 360Q = 144000 Q = 400 = P 200 ( 400 ) − 80,000 = P 80000 − 80000 P=0

Practice Problem The demand and supply equations of a particular goods are given by P= −12QD + 360 = P QD + 87 Where P, QS and QD represent the price, quantity supplied and quantity demanded respectively. Calculate the equilibrium price and quantity. Find the value of P when the quantity of supply changes to 300

62

CHAPTER

ALGEBRA

4

Learning Objectives At the end of this section the student should be able to: • Identify the symbols , ≤, and ≥ • Operating of inequalities. • Multiplying of brackets. • Operations of fractions that is multiplying and dividing , adding and subtracting. • Operation of algebraic fractions

Introduction A number line which we met at the beginning of the chapter can be used to decide whether a number is greater than or less than. The symbols < means it is less than > means it is greater than ≤ means it is less than or equal to ≥ it means it is greater than or equal to Consider the number line below

Practice Problem Classify each of the statement as true or false 8 ≥ -3

d) -2 ≥ -5

15 ≤ 13

e) -7 ≤ 6 ≤ 10

g) 7 > 0 > 12 h) -5 ≥ 0

64

CHAPTER 4 Algebra -15 < -21

f) 4 > -4 > -8

i) 0 ≤ -12

Simple Linear inequalities Practice Problem x>5

c) x < −3

y ≤3

d) y > −32

e) y ≥ 16 f)

y ≤8

Compound Linear inequalities Practice Problem

9 < 10 < 15

b) 3 > −2 > −4

c) 7 < x ≤ 12

Note One can convert two simple linear inequalities into a compound inequality

Example 4.1 x < 6 and x ≥ 4 ⇒

x ≤ 3 and x ≥ −7 ⇒

4≤ x 2 and

c)

b)

x
circle the number For ≤ and ≥ shade the circle

Practice Problem Illustrate the following sets on a number line b) − x > −6

a) x < 7 d) − x ≤ −6

e)

x ≥ 2x − 6

c) x ≥ −10 f)

x > −12

Illustrate the following sets on a number line 1≤ x ≤7 d) − 7 < x ≤ −1 g)

−3< y 12, x>4 Again 2x – 4 ≤ 8 ⇒ 2x ≤ 12 ⇒4 < x ≤ 6 Solve -6 < 1 – 2x ≤ 8

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CHAPTER 4 Algebra

Solution The inequality can be reduced to two inequalities that is – 6 < 1 – 2x and 1 – 2x ≤ 8 and then solved separately. – 6 < 1 – 2x -7 < -2x 3.5 > x Again 1 – 2x ≤ 8 -2x ≤ 7 x≥ 3.5 ⇒ -3.5 ≤ x 3.5

Practice Problem Solve the following simultaneous inequalities a)

x + 7 > 53 x − 10 < 3

1 x −3≤1 c) 2 x +7>3

e) 4 x + 2 ≤ x + 5 < 2 x + 9

b)

x+5>3 x −1 ≤ 5

d)

x +6>3 2x − 3 ≤ 5

f)

1 1 x +1< x +2 ≤ x +2 2 8

Representing the following inequalities on a Cartesian Plane Note For < and > use broken line For ≤ and ≥ use continuous line Always shade the unwanted region

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CHAPTER 4 Algebra

Practice Problem Represent the following linear inequalities on a Cartesian plane; a)

x≥4

b) 3 < x < 7

d) 0 ≤ y < 6

c) y > −3 e) x ≥ 12

f) −1 ≤ x ≤ 4

g) x ≥ 4, y < 2

h) y < 0

Represent the following linear inequalities on Cartesian plane a) 2 x + y > 2

b) 5x + 2y ≤ 4

c) x + y ≥ 0

d) 8= x 2y ≥ 6

e) 3= x 4y ≤ 6

f) 3y − x ,5

g) y − x ≤ 2

g) 12= x 3y > 9

Intersecting Regions Represent the following on a Cartesian plane x= y > 0 x 0

y+x 2 x +2 > 0

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CHAPTER 4 Algebra

2x + y < 5 c) y < x + 3 y>0

2x + y < 6 d) x ≤ 2 y≤5

4 x − 3y < 6 e) x < 3 y ≤6

f)

y−x