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MERCHANDISING MATHEMATICS FOR RETAILING SITI FATIMAH BT. AHMAD HISAM TABLE OF CONTENTS 04 Systems of Linear Equations 14 Cost Volume Profit 23 Simple Interest & Discount 42 Markup & Markdown 55 Business & Consumer Credit 67 83 Transportation Problem 98 Assignment Problem 115 References 114 Glossary Purchase Discounts & Records The only good hero is the kind who survives to talk about mathematics merchandising. S.F.A.H Acknowledgement A YEAR OF MEMORIES A million thanks to God for giving us life, health and knowledge without all those Gifts, we believe we could not finally manage to write of this book after going through so many constraints. First of all, special thanks and appreciation to my respected and understanding Head of Department of Commerce Department En. Salehan b. Marsim and also Head of Diploma in Retail Management Pn. Noor Yanti bt. Supian for the ideas, comment, advice and guidance that has been given to me in order to accomplish this workbook. All my love and appreciation also dedicated to my beloved parent and family who always give me all the support that we want throughout these difficult times. And to all my friends, thanks for all the support that all of you have given to us. Once again, thank you, everybody. Sincere from SITI FATIMAH BT AHMAD HISAM CHAPTER 1 LINEAR EQUATIONS C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S 1 Chapter 1.0 System of Linear Equations Course learning outcomes (CLO): Understand the method used to solve linear and simultaneous equations Differentiate and form linear equations functions 1.1 Introduction An equation is a mathematical statement that asserts the equality of two expressions. It states the equality of two algebraic equations. The algebraic expressions may be stated in terms of one or more variables. Equations are widely used by managers and economists to evaluate and estimate costs, sales, demand, supply profit and etc. The purpose of this chapter is to look at the solution of elementary simultaneous linear equations. Before we do that, let’s just have a look at a relatively straightforward single equation. The equation we are going to look at is 2x − y = 3 4 C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S This is a linear equation. It is a linear equation because there are no terms involving x2, y2 or x*y, or indeed any higher powers of X and Y. The only terms we have got are terms in x, terms in y and some numbers. So this is a linear equation. 1.2 A system with One Variable A common form of the linear equation in two variables, x and y is y = mx+c, where m and c are constants. For example, 8 = 3x + 2 The solution of an equation is the value for the variable x that satisfies the equation. Example 1 Solve the following linear equations: a) b) 6x – 4 = 20 12x + 8 = 32 Solution a) 6x – 4 = 20 6x = 20 + 4 x = 24 6 =4 5 C H A P T E R b) 12x + 8 12x x 1 : S Y S T E M O F L I N E A R E Q U A T I O N S = 32 = 32 - 8 = 24 12 =2 Practice Questions (1) Solve the following linear equations: a) b) c) d) e) 5x + 7 3x – 4 4x + 10 3(8 –h) 10x – 7 = 52 =x–8 = 7x – 17 = 5h + 16 = 5(x +1) 1.3 A system with Two Variable Also known as simultaneous equation because containing a few variables. The general form of simultaneous equations with two unknown is ax + by = c px + qy = r Where a, b, p, and q are constants; x and y are variables. Solving simultaneous equations means finding the values of x and y that satisfy the system of equations simultaneously. Two common methods used to solve these equations which are a method of addition/subtraction and method of substitution. 6 C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S 1.3.1 Method of addition / subtraction In the addition /subtraction method, the two equations in the system are added or subtracted to create a new equation with only one variable. In order for the new equation to have only one variable, the other variable must cancel out. In other words, we must first perform operations on each equation until one term has an equal and opposite coefficient as the corresponding term in the other equation. Example 2 Solve the following simultaneous equation for x and y using the method of subtraction. a) x+y =8 2x + y = 10 x+y 2x + y x =8 = 10 =2 [1] [2] [2] – [1] 7 C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S Substitute x = 2 into equation x+y =8 2+y =8 y =8–2 y =6 Example 3 The procedure to solve simultaneous equations using the method of substitution is as follows: 1. 2. 3. Solve one of the equations for one unknowns in terms of the other. Substitute that into the other equation. Solve the equation in one unknown. Solve the following simultaneous equation for x and y using the method of substitution. 2x – y = 9 x - y = 3 Solution Equation [2]: [1] [2] x–y=3 x=3+y Substitute this into equation [1]: 2x – y = 9 2(3 + y) – y = 9 This equation has only one unknown, y. 6 + 2y – y =9 6+y =9 y =3 To find x, substitute y = 3 into one of the equations [1]: 2x – y =9 2x – 3 =9 2x = 12 x =6 8 C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S Source: stufiles.sanjac.edu 1.4 A system with Three Variable The general form of simultaneous equations with three unknowns is ax + by + cz = d ex + fy + gz = h px + qy + rz = s Where a, b, c, d, e, f, g, h, p, q, and s are constants; x, y and z are variables. 9 C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S To solve this system, we need to identify two pairs of equations in order to reduce the three equations in three unknowns to two equations in two unknowns. This can be done by eliminating one of the unknowns from two pairs of equations, either from equations [1] and [2] or [1] and [3] or [2] and [3]. Example 4 Solve this system of equations: x+y+z 2x – y + z x + y – 3z = 10 =4 =6 Solution Eliminate y from equations [1] and [2], and then from equations [2] and [3]. [1]: [2]: [1] + [2]: x+y+z 2x – y + z 3x + 2z = 10 =4 = 14 [4] [2]: [3]: [2] + [3]: 2x – y + z x + y – 3z 3x – 2z =4 =6 = 10 [5] Solve z using equations [4] and [5] [4]: 3x + 2z = 14 [5]: 3x – 2z = 10 [4] – [5]: 4z =4 z =1 To solve for x, substitute z = 1 into equation [5]. [5]: 3x – 2z 3x – 2(1) 3x – 2 3x x = 10 = 10 = 10 = 12 =4 10 C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S Finally, to solve for y, substitute the values of x and z into one of the original equations. [1]: x+y+z 4+y+1 y y = 10 = 10 = 10 – 5 =5 The solution is x = 4, y = 5 and z = 1. Practice Questions (2) 1. Find the values of X and Y for the following systems of equations: (a) 2p – 3q =7 3p + 2q =4 (b) 2. 2x – 4y 2x – y =3 =5 Solve the following systems of equations. (a) x+y+z =2 4x + 2y + z = 1 9x + 3y + z = 8 (b) x + 2y + z 2x – 3y – z x – 2y - 2z =4 =4 =3 (c) 2x + y – 2z x+y 3x + 2y – z =4 =1 =2 11 C H A P T E R 1 : S Y S T E M O F L I N E A R E Q U A T I O N S Methods for Solving Systems of Linear Equations SUBSTITUTION Find the solution to 1) Choose an equation and solve for a variable: -4x + 2y = 8 -3x = 6 - y 2) Using the other equation, substitute in the expression that you’ve just found for the isolated variable in the Step 1: ELIMINATION 1) Rewrite the equations so that the x’s and y’s are on the same side of the equation, and the constants are on the other. 2) Stack the equations so like terms are lined up. 3) Modify the equations so that the coefficients of one of the variables. 3) Solve the equation from Step 2 for the variable that was not isolated in the Step 1: 4) Add the equations so that one variable is eliminated. 4) Substitute the value of the variable from Step 3 into the equation with the isolated variable from Step 1: 5) Solve the equation from Step 4: 6) Substitute the value of the variable just solved for in the Step 5 into one of the original equations. 5) Solve the equation from Step 4 for the final, remaining variable: Solution 7) Solve for the final, remaining variable: 12