MERCHANDISING MATHEMATICS FOR RETAILING FOR ISBN REQUEST

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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
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S Y S T E M
O F
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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
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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
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