ContentsContents 1 Aims and Objectives1 Aims and Objectives 1 Introduction1 Introduction 1 Linear equations1 Linear equations 1.2 Developing equation of a line1.2 Developing equation of a line 1.2 Special formats1.2 Special formats 1 Application of linear equations1 Application of linear equations 1.3 Linear cost-output relationships1.3 Linear cost-output relationships 1.3 Break-even analysis1.3 Break-even analysis 1 Model examination questions1 Model examination questions
1 AND OBJECTIVES1 AND OBJECTIVES
After reading the chapter students must be able to:After reading the chapter students must be able to: define algebraic expression, equation & linear equation define algebraic expression, equation & linear equation explain the different ways of formulating or developing equations of a explain the different ways of formulating or developing equations of a line line understand the breakeven point and its applicationunderstand the breakeven point and its application define the cost output relation shipdefine the cost output relation ship explain the different cost elementsexplain the different cost elements
1 INTRODUCTION 1 INTRODUCTION
Mathematics, old and newly created, coupled with innovative applicMathematics, old and newly created, coupled with innovative applications of the rapidlyations of the rapidly evolving electronic computer and directed toward management problems, evolving electronic computer and directed toward management problems, resulted in aresulted in a new field of study called quantitative methods, which has become partnew field of study called quantitative methods, which has become part of the curriculum of the curriculum of colleges of business. The importance of quantitative approaches to mof colleges of business. The importance of quantitative approaches to managementanagement problems is now widely accepted and a course in mathematics, wproblems is now widely accepted and a course in mathematics, with managementith management applications is included in the core of subjects studied by almost applications is included in the core of subjects studied by almost all managementall management
students. This manual develops mathematics in the applied context restudents. This manual develops mathematics in the applied context required for anquired for an understanding of the quantitative approach to management problems of the quantitative approach to management problems.
1 Linear Equations1 Linear Equations
EquationEquation :: - A mathematical statement which indicates two algebraic expressions are - A mathematical statement which indicates two algebraic expressions are equal equal Example:Example: Y = 2X + 3 Y = 2X + 3
Algebraic expressionsAlgebraic expressions : - A mathematical statement indicating that numerical quantities : - A mathematical statement indicating that numerical quantities are linked by mathematical operations. are linked by mathematical operations. Example:Example: X + 2 X + 2
Linear equations: - Linear equations: - are equations with a variable & a constant with degree one equations with a variable & a constant with degree one. - - Are equations whose terms (the parts separated by +, -, = signs) Are equations whose terms (the parts separated by +, -, = signs) - - Are a constant, or a constant times one variable to the first power Are a constant, or a constant times one variable to the first power Example:Example: 2X – 3Y = 7 2X – 3Y = 7
However 2X + 3XY = 7 isn’t a linear equation, because 3XY is aHowever 2X + 3XY = 7 isn’t a linear equation, because 3XY is a constant times the constant times the product of 2 variables of 2 variables.
If Y represents Total Cost, the cost is increased by the rate of the amount of the slope m Y represents Total Cost, the cost is increased by the rate of the amount of the slope m.
if X if X 1 1 XX 22
Slope measures the steepness of a line. The larger the slope theSlope measures the steepness of a line. The larger the slope the more steep (steeper) the more steep (steeper) the line is, both in value and in absolute value is, both in value and in absolute value.
Interpretative ExercisesInterpretative Exercises #2 Suppose the Fixed cost (setup cost) for producing product X be br. 2000.#2 Suppose the Fixed cost (setup cost) for producing product X be br. 2000. After setup After setup it costs br. 10 per X produced. If the total cost is represented by Y:it costs br. 10 per X produced. If the total cost is represented by Y: 1. Write the equation of this relationship in slope-intercept form the equation of this relationship in slope-intercept form. 2. State the slope of the line & interpret the numberState the slope of the line & interpret the number 3. State the Y-intercept of the line & interpret the numberState the Y-intercept of the line & interpret the number
#3. A sales man has a fixed salary of br. 200 a week In addition#3. A sales man has a fixed salary of br. 200 a week In addition; he receives a sales; he receives a sales commission that is 20% of his total volume of sales. State the recommission that is 20% of his total volume of sales. State the relationship between thelationship between the sales man’s total weekly salary & his sales for the week. sales man’s total weekly salary & his sales for the week. Answer Answer Y = 0 + 200 Y = 0 + 200
2. The slope point form2. The slope point form The equation of a non-vertical line, L, of slope, m, that passes throughThe equation of a non-vertical line, L, of slope, m, that passes through the point (X the point (X 11 , Y, Y 11 )) is : defined by the formula Y – Yis : defined by the formula Y – Y 1 1 = m (X – X= m (X – X 11 )) Y – Y Y – Y 11 = m (X – X = m (X – X 11 )) Example #1Example #1 Y – 2 = 4 (X – 1) Y – 2 = 4 (X – 1) Given, slop = 4 and Y – 2 = 4X - 4Given, slop = 4 and Y – 2 = 4X - 4 Point = (1, 2) Y = 4X – 2 Point = (1, 2) Y = 4X – 2
#2 A sales man earns a weekly basic salary plus a sales comm#2 A sales man earns a weekly basic salary plus a sales commission of 20% of his totalission of 20% of his total sales. When his total weekly sales total br. 1000, his total sasales. When his total weekly sales total br. 1000, his total salary for the week is 400 for the week is 400. Derive the formula describing the relationship between total salary and sales the formula describing the relationship between total salary and sales. AnswerAnswer Y = 0 + 200 Y = 0 + 200
#3 If the relationship between total cost and the number of units made #3 If the relationship between total cost and the number of units made is linear, & if costsis linear, & if costs increases by br. 7 for each additional unit made, and if the Total increases by br. 7 for each additional unit made, and if the Total Cost of 10 units is br of 10 units is br. 180. Find the equation of the relationship between Total Cost (Y) & 180. Find the equation of the relationship between Total Cost (Y) & number of unitsnumber of units made (X)made (X) Answer:Answer: Y = 7X + 110 Y = 7X + 110
3. Two-point form3. Two-point form Two points completely determine a straight line & of course, they detTwo points completely determine a straight line & of course, they determine the slope ofermine the slope of the line. Hence we can first compute the slope, then use this vathe line. Hence we can first compute the slope, then use this value of m together withlue of m together with
either point in the point-slope form Y – Yeither point in the point-slope form Y – Y 11 = m (X – X = m (X – X 11 ) to generate the equation of a) to generate the equation of a line. By having two coordinate of a line we can determine the equation of the line. By having two coordinate of a line we can determine the equation of the line.
Example #1 Example #1 given (1, 10) & (6, 0) given (1, 10) & (6, 0)
First slope = First slope = 10016 510 2 , then , then
Y – Y Y – Y 11 = m (X – X = m (X – X 11 ) ) Y – 10 = -2 (X – 1) Y – 10 = -2 (X – 1) Y – 10 = -2X + 2 Y – 10 = -2X + 2 Y = -2X + 12Y = -2X + 12 #2 A salesman has a basic salary &, in addition, receives a c#2 A salesman has a basic salary &, in addition, receives a commission which is a fixedommission which is a fixed percentage of his sales volume. When his weekly sales are Br. 1000, hipercentage of his sales volume. When his weekly sales are Br. 1000, his total salary is br total salary is br. 400. When his weekly sales are 500, his total salary is br. 300. Det400. When his weekly sales are 500, his total salary is br. 300. Determine his basicermine his basic salary & his commission percentage & express the relationship betwsalary & his commission percentage & express the relationship between sales & salary ineen sales & salary in equation form form. Answer:Answer: Y = 0 + 200 Y = 0 + 200
#3 A printer costs a price of birr 1,400 for printing 100 copies of a repor#3 A printer costs a price of birr 1,400 for printing 100 copies of a report & br. 3000 fort & br. 3000 for printing 500 copies. Assuming a linear relationship what would be theprinting 500 copies. Assuming a linear relationship what would be the price for printing price for printing 300 copies?300 copies? Answer: Answer: Y = 4X + 1000Y = 4X + 1000 Cost = 4 (300) + 1000 = br. Cost = 4 (300) + 1000 = br. 22002200
1.2 Special formats 1.2 Special formats a) Horizontal & vertical linesa) Horizontal & vertical lines When the equation of a line is to be determined from two given points, it is a good idea toWhen the equation of a line is to be determined from two given points, it is a good idea to compare corresponding coordinates because if the Y values are the same the compare corresponding coordinates because if the Y values are the same the line isline is horizontal & if the X values are the same the line is vertical horizontal & if the X values are the same the line is vertical
Example: 1Example: 1 Given the points (3, 6) & (8, 6) the line through them is horizontal because Given the points (3, 6) & (8, 6) the line through them is horizontal because both Y-coordinates are the same i. 6 both Y-coordinates are the same i. 6
1 Application of linear equations1 Application of linear equations
1.3 linear cost output relationships1.3 linear cost output relationships – VC, FC, TC, AC, MC, TR, – VC, FC, TC, AC, MC, TR, ::
TR/TC profit TR TC = TVC + TFC TR/TC profit TR TC = TVC + TFC Or TP region TR = PQ Or TP region TR = PQ TC TP = TR - TC TC TP = TR - TC Loss T Loss T region E BEP region E BEP = PQ – (VC + FC) = PQ – (VC + FC) TVC H TVC H A F G FC = PQ – Q - TFC A F G FC = PQ – Q - TFC TC TC TFC = TFC = Q (P – VC) - FC Q (P – VC) - FC B C D G(No of units) B C D G(No of units) Where Q = units Where Q = units product & units product & units sold in revenue sold in revenue TC = Total Cost TC = Total Cost FC = Fixed Cost FC = Fixed Cost VC = Unit variable VC = Unit variable Cost Cost
Interpretation of the graph:Interpretation of the graph: 1. The vertical distance between AB, FC, GD is the same becauseThe vertical distance between AB, FC, GD is the same because Fixed Cost is the Fixed Cost is the same at any levels of output at any levels of output. 2. There is no revenue without sales (because Total Revenue function passes throughThere is no revenue without sales (because Total Revenue function passes through the origin), but there is cost without production (because of Fixed Cost) &the origin), but there is cost without production (because of Fixed Cost) & the TC the TC function starts from A & doesn’t pass through the origin function starts from A & doesn’t pass through the origin 3. Up to point T, Total Cost is greater than Total Revenue Up to point T, Total Cost is greater than Total Revenue results in loss. While at results in loss. While at point T, (Total Revenue = Total Cost) i. Breakeven. (0 profit), & above point T, (Total Revenue = Total Cost) i. Breakeven. (0 profit), & above point T,point T, TR > TC TR > TC +ve profit. +ve profit. 4. TFC remains constant regardless of the number of units produced. Given thatTFC remains constant regardless of the number of units produced. Given that there is no any difference in scale of production is no any difference in scale of production.
1.2 Breakeven Analysis1.2 Breakeven Analysis Break-even point is the point at which there is no loss or profit to tBreak-even point is the point at which there is no loss or profit to the company. It can behe company. It can be expressed as either in terms of production quantity or revenue level expressed as either in terms of production quantity or revenue level depending on howdepending on how the company states its cost equation company states its cost equation.
Manufacturing companies usually state their cost equation in terms ofManufacturing companies usually state their cost equation in terms of quantity (because quantity (because they produce and sell) where as retail business state their cost they produce and sell) where as retail business state their cost equation in terms ofequation in terms of revenue (because they purchase and sell)revenue (because they purchase and sell)
Case 1: Manufacturing CompaniesCase 1: Manufacturing Companies Consider a Company with equation Consider a Company with equation TC = VC + FC / Total cost = Variable cost + Fixed cost TC = VC + FC / Total cost = Variable cost + Fixed cost TR = PQ/ Total Revenue = Price x Quantity TR = PQ/ Total Revenue = Price x Quantity
At Break-even point, TR = TC i TR – TC = 0 At Break-even point, TR = TC i TR – TC = 0 PQ = VC + FC where Qe = Breakeven Quantity PQ = VC + FC where Qe = Breakeven Quantity PQ – VC = FC FC = Fixed cost PQ – VC = FC FC = Fixed cost Q (P – VC) = FC P = unit selling price Q (P – VC) = FC P = unit selling price
Sales volume = 2000 X 10 = Sales volume = 2000 X 10 = 20,000 br,000 br.
25000 25000 TR = 10QTR = 10Q 20000 20000 TC = 5Q + 10000 TVC = 5QTC = 5Q + 10000 TVC = 5Q 15000 15000 TVC TTVC T = 5Q - 10000 = 5Q - 10000 10000 10000 5000 5000 TFCTFC 0 1000 2000 3000 4000 5000 1000 2000 3000 4000 5000 Q Q (no. of units produced (no. of units produced & sold)& sold) -5000 - -5000 - -10000 - -10000 -
Interpretation:Interpretation: When a co. produces & sells 2000 units of output, there will not be any lWhen a co. produces & sells 2000 units of output, there will not be any loss or gain (nooss or gain (no profit, no loss) profit, no loss) The effect of changing one variable keeping other constant The effect of changing one variable keeping other constant
Case 1 - Fixed costCase 1 - Fixed cost Assume for the above problem FC is decreased by Br. 5000, Citrus ParibusAssume for the above problem FC is decreased by Br. 5000, Citrus Paribus (other things (other things being constant)being constant)
TC = 5Q + 5000 TC = 5Q + 5000 Qe = Qe = 50005 = 1000 units = 1000 units
TR = 10Q TR = 10Q Therefore, FC Therefore, FC Qe Qe FC & Qe have direct relationship FC & Qe have direct relationship FC FC Qe Qe
Case 2-Unit variable costCase 2-Unit variable cost Assume for the above problem UVC decreased by 1 br. Citrus Paribus (keeAssume for the above problem UVC decreased by 1 br. Citrus Paribus (keeping otherping other thing constant)thing constant)
TC = 4Q + 10000 TC = 4Q + 10000 Qe = Qe = 100006 667,1 units
TR = 10Q TR = 10Q
Therefore, VC Therefore, VC Qe Qe VC & Qe have direct relationship VC & Qe have direct relationship VC VC Qe Qe
Case 3- Selling priceCase 3- Selling price Assume for the above problem selling price is decreased by br. 1, Citrus Paribus,Assume for the above problem selling price is decreased by br. 1, Citrus Paribus,
TC = 5Q + 10,000 TC = 5Q + 10,000 Qe = Qe = 100004 2500 units
TR = 9Q TR = 9Q Therefore P Therefore P Qe Qe Price and breakeven point have indirect relationship Price and breakeven point have indirect relationship P P Qe Qe
In the above example a company has the following options (to minimize itIn the above example a company has the following options (to minimize its breakevens breakeven point and maximize profit).point and maximize profit).
Markup = Selling price – Variable cost = 150 – 100 = 50Markup = Selling price – Variable cost = 150 – 100 = 50 i. as a function of cost, the markup is 50/100 = 50%i. as a function of cost, the markup is 50/100 = 50% ii. as a function of retail price, the markup is 50/150 = 33 % it is also called margin. as a function of retail price, the markup is 50/150 = 33 % it is also called margin. Margin Cost of goods sold Margin Cost of goods sold
The cost of goods sold = 100% - 33 % = 66% The cost of goods sold = 100% - 33 % = 66% 67% 67% Selling price CGS Selling price CGS Given other selling expense = 1%of the selling price i. 0 Given other selling expense = 1%of the selling price i. 0
So, the TC equation becomes: So, the TC equation becomes: Y = 0 + FC Y = 0 + FC Where: X is sales revenueWhere: X is sales revenue Y is total cost Y is total cost Out of 100% selling price 68% is the variable cost of goods purchased & soldOut of 100% selling price 68% is the variable cost of goods purchased & sold
ExampleExample Suppose a retail business sale its commodities at a margin of Suppose a retail business sale its commodities at a margin of 25% on all items25% on all items purchased & sold. Moreover the company uses 5% commission as sellingpurchased & sold. Moreover the company uses 5% commission as selling expense & br. expense & br. 12000 as a Fixed Cost. 12000 as a Fixed Cost.
Find the Breakeven revenue for the retail business after developing the equationFind the Breakeven revenue for the retail business after developing the equation
Solution Solution Selling price 100% Let X represents selling price Selling price 100% Let X represents selling price Margin Margin 25% 25% Y = total cost Y = total cost CGS 75% FC = 12000 CGS 75% FC = 12000 Comm. Exp. Comm. Exp. 5%5% Xe = Breakeven revenue Xe = Breakeven revenue Total VC 80% Total VC 80% Y = 0 + 12000Y = 0 + 12000
Break even revenue is obtained by making sales revenue & cost equalsBreak even revenue is obtained by making sales revenue & cost equals At breakeven point TC = TR Y = mx + b At breakeven point TC = TR Y = mx + b i. Y = X then, unit variable cost i. Y = X then, unit variable cost
0 + 12000 = X 0 + 12000 = X m mwhere VCP TVCTR or FC m
X = 60,000 br = 60,000 br. When the co. receives br. 60,000 as sales revenue, When the co. receives br. 60,000 as sales revenue, there will be no loss or profit. there will be no loss or profit.
The Breakeven revenue (BER = The Breakeven revenue (BER = FC 1 m ) method is useful, because we can use a single) method is useful, because we can use a single
formula for different goods so far as the company uses the same amount of profiformula for different goods so far as the company uses the same amount of profit margint margin
for all goods. However, in Breakeven quantity method or BEQ = for all goods. However, in Breakeven quantity method or BEQ = FC VP it is not it is not
possible and hence we have to use different formula for different items and hence we have to use different formula for different items.
Example #1Example #1 It is estimated that sales in the coming period will be It is estimated that sales in the coming period will be br. 6000 & that FCbr. 6000 & that FC will be br. 1000 & variable costs br. 3600, develop the total cost ewill be br. 1000 & variable costs br. 3600, develop the total cost equation & thequation & the breakeven revenue revenue.
Answer:Answer: Y = Y = 60003600 X + 1000 = 0 + 1000X + 1000 = 0 + 1000
Where Y = Total Cost Where Y = Total Cost X = Total revenue X = Total revenue
BER = Xe = BER = Xe = 1000 6 1000 4 br.
At the sales volume of br. 2500, the company breaks even the sales volume of br. 2500, the company breaks even.
1 Model examination questions1 Model examination questions 1. XYZ company’s cost function for the next four months is XYZ company’s cost function for the next four months is
v. If you have any one item at a price of Br. 15/unit how do you convertIf you have any one item at a price of Br. 15/unit how do you convert the cost equation in terms of revenue in to a cost equation in termthe cost equation in terms of revenue in to a cost equation in terms ofs of quantity?quantity?
UNIT 2 MATRIX ALGEBRA AND ITS APPLICATION
Contents 2 Aims and Objectives2 Aims and Objectives 2 Introduction2 Introduction 2 Matrix algebra2 Matrix algebra 2.2 Types of matrices2.2 Types of matrices 2.2 Matrix operation2.2 Matrix operation 2.2 The multiplicative inverse of a matrix2.2 The multiplicative inverse of a matrix 2 Matrix Application2 Matrix Application 2.3 Solving systems of linear equations2.3 Solving systems of linear equations 2.3 Word problems2.3 Word problems 2.3 Markov Chains2.3 Markov Chains
2 AIMS AND OBJECTIVES2 AIMS AND OBJECTIVES
After reading this chapter students will be able to:After reading this chapter students will be able to: explain what a matrix isexplain what a matrix is define the different types of matricesdefine the different types of matrices perform matrix operationsperform matrix operations find inverse of a square matrixfind inverse of a square matrix explain how to solve a systems of linear equationsexplain how to solve a systems of linear equations solve word problems applying matricessolve word problems applying matrices understand the concept of Markov chainunderstand the concept of Markov chain
2 INTRODUCTION 2 INTRODUCTION
Brevity in mathematical statements is achieved through the use Brevity in mathematical statements is achieved through the use of symbols. The priceof symbols. The price paid for brevity, of course, is the effort spent in learning the meaning of the symbol for brevity, of course, is the effort spent in learning the meaning of the symbol.
In this unit we shall learn the symbols for matrices, and apply themIn this unit we shall learn the symbols for matrices, and apply them in the statement and in the statement and solution of input-output problems and other problem involving linear systemssolution of input-output problems and other problem involving linear systems
2 MATRIX ALGEBRA2 MATRIX ALGEBRA
Algebra is a part of mathematics, which deals with operations (+, -, x, Algebra is a part of mathematics, which deals with operations (+, -, x, ).).
A matrix is a rectangular array of real numbers arranged in m rows & A matrix is a rectangular array of real numbers arranged in m rows & n columns. It isn columns. It is symbolized by a bold face capital letter enclosed by a bracket or parentheses by a bold face capital letter enclosed by a bracket or parentheses.
11211 in which a in which ajjjj are real numbers are real numbers
Each number appearing in the array is said to be an element or cEach number appearing in the array is said to be an element or component of the matrix of the matrix. Element of a matrix are designated using a lower case form of theElement of a matrix are designated using a lower case form of the same letter used to same letter used to symbolize the matrix itself. These letters are subscripted as asymbolize the matrix itself. These letters are subscripted as aijij, to give the row & column, to give the row & column location of the element with in the array. The first subscript alocation of the element with in the array. The first subscript always refers to the rawlways refers to the raw location of the element; the second subscript always refers to itslocation of the element; the second subscript always refers to its column location. Thus, column location. Thus, component acomponent aijij is the component located at the intersection of the i is the component located at the intersection of the ithth raw and j raw and jthth column. column.
The number of rows (m) & the number of columns (n) of the array give its orderThe number of rows (m) & the number of columns (n) of the array give its order or its or its dimension, M x n (reads “M” by “n”)dimension, M x n (reads “M” by “n”)
Eg. The following are examples of matricesEg. The following are examples of matrices
1 7 element a 1 7 element a 1212 = 7 = 7 A = 5 3 this is 3 x 2 matrix a A = 5 3 this is 3 x 2 matrix a 2121 = 5 = 5 4 2 a 4 2 a 3232 = 2 = 2
X = 1 5 9 15 This is a 4 x 4 matrixX = 1 5 9 15 This is a 4 x 4 matrix
N.BN. Each identity matrix is a square matrix. Each identity matrix is a square matrix
2.2 Matrix operations (Addition, Subtraction, Multiplication)2.2 Matrix operations (Addition, Subtraction, Multiplication) Matrix Addition/ SubtractionMatrix Addition/ Subtraction
Two matrices of the same dimension are said to be CONFORMABTwo matrices of the same dimension are said to be CONFORMABLE FOR ADDITION FOR ADDITION. Adding corresponding elements from the two matrices & entering the result Adding corresponding elements from the two matrices & entering the result in the samein the same raw-column position of a new matrix perform the addition-column position of a new matrix perform the addition.
If A & B are two matrices, each of site m x n, then the sumIf A & B are two matrices, each of site m x n, then the sum of A & B is the m x n matrix of A & B is the m x n matrix C whose elements are:C whose elements are:
C Cijij = a = aijij +b +bijij for i = 1, 2 --------m C for i = 1, 2 --------m C 11 11 = a= a 11 11 + b+ b 1111 j = 1, 2 ---------n C j = 1, 2 ---------n C 22 22 = a= a 22 22 + b+ b 2222 C C 1212 = a = a 1212 + b + b 1212 etc etc
eg. 1 3 7 9 8 12 eg. 1 3 7 9 8 12 2 4 + 8 -10 = 10 –6 2 4 + 8 -10 = 10 –
eg. –2 7 2 8 7 These two matrices aren’t eg. –2 7 2 8 7 These two matrices aren’t 4 6 9 + 6 4 = conformable for addition 4 6 9 + 6 4 = conformable for addition because they aren’t of the because they aren’t of the same dimension. same dimension.
Laws of matrix additionLaws of matrix addition
The operation of adding two matrices that are conformable for addition hasThe operation of adding two matrices that are conformable for addition has these two these two basic properties properties.
The laws of matrix addition are applicable to laws of matrix subtraction, given The laws of matrix addition are applicable to laws of matrix subtraction, given that the two matrices are conformable for subtraction A – B = A + (-B) that the two matrices are conformable for subtraction A – B = A + (-B)
eg= 1 2 B = 0 1 eg= 1 2 B = 0 1 3 4 2 5 3 4 2 5
A – B = 1 1 A – B = 1 1
1 -1 1 - Matrix MultiplicationMatrix Multiplication
a)a) By a constant (scalar multiplication) By a constant (scalar multiplication)
A matrix can be multiplied by a constant by multiplying each component in the A matrix can be multiplied by a constant by multiplying each component in the matrix by a constant. The result is a new matrix of the same dimension as the matrix by a constant. The result is a new matrix of the same dimension as the original matrix. original matrix.
KA = K KA = Kaij aij (m x n)(m x n)
eg. If X = 6 5 7 , then 2X = (2 x 6) (2 x 5) (2 x 7) eg. If X = 6 5 7 , then 2X = (2 x 6) (2 x 5) (2 x 7)
2X = 12 10 14 2X = 12 10 14
Laws of scalar multiplication Laws of scalar multiplication
