Equation

EQUATION

This e-book is a lecture note
for first topic for Business
Mathematics.
It is suitable for all
Polytechnic students who
take this course.

Azmah bt Salleh
Norimah bt Annuar @ Mohd Tahir

i

Contents

Page

Introduction of equations 2

First Degree Equation 3

Second Degree Equation 8

** Factoring Method 9

** Quadratic Formula 13

** Simultaneous Equation 16

The Use Equantion In

Business 21

Management Mathematical

Formulas 24

ii

EQUATIONS

At the end of this topic, you
should be able to :

Understand the methods used to solve
linear, quadratic and simultaneous
equations.

Apply methods used in solving linear,
quadratic and simultaneous equations
to common business problems such as
cost budgeting, renenue and profit.

Understand the relationship beyween
voliume, cost and profit using
algebra.

1

INTRODUCTION

2

First Degree Equation

A linear
equation is form
in two variables,

x and y that is
y= mx+c

For example: The highest degree
5 = 3x + 8 of linear equation
10 = 4x - 5
9 = -4x + 7 is 1.
8 = 2x +2
The value of x that
satisfied the

equation is called
the solution.

3

Solve the following linear equation:

Example 1: Example 2:

5= 3x + 8 10=4x-5

Solution: Solution:
3x+8 = 5 4x – 5 =10

3x = 5-8 4x = 10 +5
3x = -3 4x = 15
x = -1 x = 15/4

4

Example 3:

9= -4x+ 7

Solution:
-4x + 7 = 9

-4x = 9 - 7
-4x = 2

x = -1/2

Example 4:

8=2x-2

Solution:
2x – 2 =8

2x = 8 +2
2x = 10
x=5

5

Find The Value Of X In The
Following Equations

Do the cross
multiplication

Example 1:=

2= 5x Solution:
4 10-2x 2(10-2x) = 4(5x)
20-4x = 20 x
-4x-20x = -20

-24x = -20
x = -20/-24

= 5/6

6

Example 2 Multiply x-3
with 4

2x = 4
x-3

Solution:
2x = 4(x-3)
2x = 4x – 12
2x-4x = -12
-2x = -12
x = -12 / -2
=6
7

Second Degree Equation

• Second • The • Quadratic
degree highest equation
equation is degree of can be
called quadratic form by
quadratic equation is using the
equation. 2 factoring
The general method or
form of the
quadratic quadratic
equation is formula
ax² + bx + c

8

Factoring Method

Solve the equation below :

Example 1: Student can
x² + x- 6 = 0 use

calculator to
get the value

Solution:

(x+3) (x-2) = 0

x+3 = 0 or x-2= 0

x = -3 x=2

9

Step using the calculator the find
the values

of quadratic equation

1. Press Mode 3
times

2. Select 1 : EQN

3. Press right
arrow

4. Select 2 Degrees

5. Key in the values
of a,b and c

6. Write the values
as the answers.

10

Example 2:

4z² + 8z = 0

Solution:
z(4z +8) = 0

z =0
or

4z+8 = 0
4z = -8
z = -2

11

Example 3: Expand the
(2x-3)(x+2) = -3 equation
through
Solution:
multiplication

(2x-3)(x+2) = -3
2x² + 4x -3x -6 = -3
2x² + x -3 = 0

(2x+3)(x-1)=0

2x+3=0 or x-1= 0

2x =-3 x =1

x= -3/2

12

Quadratic Formula

The quadratic
equation can also
be factorized by
using quadratic

formula

x = -b ± √b2 -4ac
2a

13

Solve the following quadratic equations

Example 1:

2x2 + 6x+2 = 0 The equation must
be equal to 0
Solution:
a= 2 List out the value
b= 6 of a,b and c
c= 2

x = -b ± √b2 -4ac
2a

14

x = -b ± √b2 -4ac
2a

x= -b ± √b2 -4ac

2a
= -6 ± √ 62 -4(2)(2)

2(2)
= -6 ± √36-16

4
= -6+√20 or = -6-√20

44

= -0.4 or = -2.6

15

Simultaneous Equation

Solving Two Equation with Two
Unknown

Example 1: Solution: 1
x+y = 8 2
x+y=8 2x + y = 10
2x + y = 10

x = 2 -2 1
Substitute x=2 into
equation 1
x+y = 8
2+y = 8

y = 8-2
y= 6

x = 2 and y = 6

16

Simultaneous Equation

Solving Three Equation with
Three Unknown

Example 2:
2x + y –z = 3
x – 2y-z = -5
5x +y +2z = 6

Solution:

17

Simultaneous Equation

Solving Three Equation with
Three Unknown

Solution:
2x + y –z = 3 1
x – 2y - z = -5 2

x + 3y = 8 1 - 2 4

Eliminate z from equations 1 and 3

by addition

1 x 2; 4x +2y -2z = 6 5

5x +y + 2z = 6

3

5 + 3 9x + 3y = 12 6

18

Simultaneous Equation

Then, solve equation 4 and 6
for x and y.

4 2x + 3y = 8
6 9x + 3y = 12

-6 4 8x = 4 6
x = 1/2

To solve for y, substitute x = ½ into
equation 4

x + 3y = 8
½ + 3y = 8

3y = 8 – 1/2
y = 5/2

19

To solve for z, substitute the
values of x = ½ and y = 5/2 into
one of the original equations.

1 2x + y –z =3

2(1/2) + 5/2 – z = 3

1 + 5/2 – z = 3

7/2 – z =3

z = 3 – 7/2

z =½

The solution is x = ½
y = 5/2
z = 1/2

20

Equations can be form by analyzing
the business and management
problem. From the equation, we can
find the solution because we have
converting them into mathematical
formulae.

21

Some terms that must be understand
by the students:

Fixed cost are business expenses that
are not dependent on the amount of
activities of the business.
For example: Salaries, rent and
copyright

Variable costs are expenses that
changes proportionally to the amount
of activities of a business. All costs
are changes according to the number
of units produced or sold.

22

Variable cost are paid per quantity.

Example:
The total kilogram of flour to bake
100 cakes is 450kg.
So the total kilogram of flour to bake
a cake is
450/100 = 4.5 kg.

23

Management
Mathematical Formulas

Total revenue(TR) = Price x Quantity
= PQ

Total variable cost (TVC) =
Variable cost per unit x quantity
= VC X Q

Total cost = Fixed Cost + Variable Cost
= FC + (VC X Q)

Profit = Total revenue – Total Cost
P = TR - TC
P = pq – (FC + (VC X Q))

24

Management
Mathematical Formulas

Let TP (Target Profit) is given,
We can find the unit that must be
sold to achieve the targeted profit
by using:

Quantity = FC + Targeted Profit
p - VC

25

Solve each of the following equations;
1. 3(p+6) = 8p+8
2. 7x + 7 = 2( x + 1)

3. 6 = 3
x +4 x – 2

4. 4x – 3 = 6 - 2x
52

26

1. 2x2 + 4x =5
2. z2 -8z = 0

3. x2 – 4x = -4
4. x2 -12x = -36

27

1. 3x- 2y = 6
x-y=1

2. 2x – 4y = 3
2x - y = 5

28

1. x + y + z = 2
4x+ 2y+ z= 1
9x+ 3y+ z= 8

2. x + 2y + z = 4
2x - 3y - z = 4
x - 2y - 2z = 3

29

Question 1:

Dini Enterprise launches new
cookies. The variable cost is RM
250 for 50 cookies. The fixed
cost involved is RM50,000. Each
unit will be sold at RM15.
Determine the number of units
that needs to be sold if the
company wishes to gain profit of
RM 60 000.

30

Question 2:
A batik factory started its
operations with a capital of RM 5
000 to buy a sewing machine and
RM 3 000 for a printing machine.
The raw material and labor cost for
10 pieces of fabric is RM 600. The
owner plans to sell the batik fabric
at RM150 per piece. How many
pieces of batik need to be produced
to achieve a profit of RM 5 000?
How much is the profit, if the
number of units produced is
maintained whereas the price is
increased for RM180?

31

Question 3 :
XYZ Enterprise had launched a
new product. The variable cost for
100 units of the product is
RM1350 and the fixed cost of the
product is RM20000. In the first
production, they produced 5 000
units with a selling price of RM 20
per unit. Using the information
given, find:

32

a) The profit gained by XYZ
Enterprise if all of the new
products are sold

b) The total units to be sold in
order to earn a profit of RM 20
000.

c) The total revenue if 6 000
units are sold

d) The profit gained for total
units sold in (c) with the
fixed cost reduced by 10%

33

Question 4:
Company XYZ is in the process of
diversifying the business by
introducing a new product. The
production and finance departments
had provided the data below:

Building RM 15000
Interest RM 10000
Copyright RM 1500
Other Overhead Cost RM 15000
Raw Materials RM 17500
(for 5000 units)
Other Variable Costs RM 1.00
per unit
Price per unit RM 10.50

34

Question 5:

Using the information given,
find
a) The profit if 5 500 units

is sold
b) The profit if 10 000 units

is sold
c) The change of profit

found in (a), if the fixed
cost and the variable cost
are reduced by 10% and
5% respectively.

35

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Rule: Algebra Practice Workbook with Answers. North Charleston SC, United
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36

Thank you