DUM10122 WORKBOOK

DUM10122
ENGINEERING
MATHEMATICS 1

Workbook & Tutorial

Workbook for : NAME :
DUM 10122 ENGINEERING MATHEMATICS 1
1st Published Nov 2020-Apr 2021 _______________________________
2nd Published May-Oct 2021

PROGRAMME:

____________________________________

ASSESSMENT PERCENTAGE

1. THEORY ASSIGNMENT 1 2.5%
Task 1 2.5%
Task 2 5%
Task 3

2. THEORY ASSIGNMENT 2 10%
10%
Report Writing
Affective Domain

3. THEORY TEST 1 15%
4. THEORY TEST 2 15%
5. FINAL EXAMINATION 40%

TOTAL MARKS /100%

Prepared By :
Mathematics Unit
KKTM Kuantan

For Internal Circulation KKTM Kuantan Only

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

UNIT 1 : ALGEBRA
1.1 Operations Of Algebra
1.2 Expansion And Factorization
1.3 Transposition Of Formulae
1.4 Simultaneous Equation
1.5 Operations of Polynomials
1.6 Partial Fractions

UNIT 2 : GEOMETRY
2.1 Phytagoras Theorem
2.2 Perimeter and Area
2.3 Surface area and Volume of Solids
2.4 Circular Measures

UNIT 3 : TRIGONOMETRY
3.1 Introduction Trigonometry
3.2 Trigonometry Ratios
3.3 Graphing Trigonometric Function
3.4 Trigonometric Equation
3.5 Solutions of Triangle

UNIT 4 : INDICES AND LOGARITHMS
4.1 Indices and Laws of Indices
4.2 Logarithms and Laws of Logarithms
4.3 Conversion between Indicial Form and Logarithms Form
4.4 Solutions of Indicial Equations
4.5 Solutions of Logarithmic Equations

UNIT 5 : COMPLEX NUMBERS
5.1 Imaginary Number
5.2 Complex Number in Rectangular Form and Argand Diagram
5.3 Complex Number in Polar Form
5.4 De Moivre's Theorem

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

UNIT 1: ALGEBRA

OVERVIEW

After completing the unit, students should be able to: UNIT LEARNING
OUTCOMES
1. Operate algebraic expressions including:
• BODMAS Rules 4
• Removal Bracket Rules
• Basic Rules of Algebra.

2. Expand algebraic expressions.
3. Factorize algebraic expressions.
4. Solve algebraic and quadratic equations
5. Evaluate a given formula.
6. Transpose a formula to change its subject.

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.1 OPERATIONS OF ALGEBRA

Coefficient Algebraic Term Algebraic Expressions

3m 3m + 2t
Unknown
A combination of two or more
Like Term Unlike Term algebraic terms linked by ‘+’ or ‘-’ or

5m 3m 6k 6b both
m
Different unknown
same unknown

Constant is a quantity or value that remains unchanged (always the same in form or rate)

Variable is a quantity that can vary and has no fixed value. There are dependent variable
and independent variable.

Dependent variable is an unknown which is the subject of a relationship. Its value
depends on the value of another quantity.
Example :
In y = 2x2 , y is the dependent variable. The value of y depends on the value of x.

An independent variable is a variable which is not the subject of a formula and its value
determines the value of the dependent variable.
Example :
In the equation y = 4x − 1 , x is the independent variable.

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.1.1 ADDITION AND SUBTRACTION

Group the like terms → rearrange the expression → add or subtract the coefficients of the like terms.

Example : Addition or subtraction of algebraic expressions Add and subtract
Simplify each of the following: LIKE TERM only

a. 7 + 4 − 2x + 5 b. k + k 2 − 7 + 5 + 5k

c. 3mn − 6 − 5mn + 3 d. 0.7a − 6.35b + 9a − 3.5b

e. k − m f. 2 + 5
k+4 k+4 mn

6

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.1.2 MULTIPLICATION AND DIVISION

Example : Multiplication of algebraic expressions a  b = a(b) = a • b
b. 2ab  5bc 4
a  b = a(b) = a • b
Multiply each of the following expression :

a. 4a  3b = 12ab

c. 2t  1 d. 2.4km  2.5my
7m

Example : Division of algebraic expressions
Solve the following :

a. 6ab2c 3  2bc 2 = 6ab2c 3 b. 10m2n5  2mn 3y
2bc 2 d. 3a2  a

= 3abc 49

c. 4.5pq  2.5qr

7

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.1.3 BODMAS RULES

Example : Operate algebraic expressions using BODMAS Rules.
Simplify.

a. 4s  7t  s2 b. 30a2 15a − 10a
= 28st
s2
= 28t
s

c. (20x  5x − 10x) + 2 d. (2  3a) + (8b  2)

e. 5x  10x2 f. 2m − 6n − (2n + m)
12yz 3z2

8

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.1.4 REMOVAL OF BRACKETS
The rules of removing brackets are as follows :

RULES TO REMEMBER Sign of a and b does
not change in
 + (a + b) = a + b  and .
 + (a – b) = a – b
 – (a + b) = – a – b Sign of a and b
 – (a – b) = – a + b changes in
 and .

Example : Removing brackets of algebraic expressions
Simplify :

a. (a + 4b)− (3a + b) b. − (3x − 2y) + (x − 2xy)

= a + 4b − 3a − b
= a − 3a + 4b − b
= −2a + 3b

c. 2x − (2 + x) d. – (7x+2) – (x – 5)

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

TUTORIAL 1.1

1. State the number of terms for each of the following algebraic expression.

a. 2x + 1 b. 9p2 + 7q − 6

c. 5 h d. − 2x5 − x3 + 7x − 9
7

2. Identify the unknown and the coefficient of the following term.

a. 6.5x b. Mn
Unknown = Unknown =
Coefficient = Coefficient =

c. 1 x2 d. 7k
2
Unknown = Unknown =
Coefficient =

Coefficient =

3. Determine whether each of the following pair of terms given below are like or unlike terms.

a. 7k, 2 k b. − 9p, − 9q
3

c. 3.7x, 4.5xy d. m , 4m
3

4. Simplify each of the following algebraic expression.

a. 4k + 3b b. − 2b − 3b

c. 10y − 5 − 3y + 3 d. 10h + h2 − 6h2

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

e. 3x + 2x − x f. f + f
5 55 48

5. Simplify each of the following expression. b. 20  4h
a. 2ab  8a

c. 4p2qr  8pq  7pr d. (12p − 8y)  2

e. 4  5d f. 2p  7q2
b 2b 14pq

6. Simplify each of the following expression. ( )b. 2a2 + 5b − (2b − 3a)

a. (a + 7b) − (a + 7b)

( )c. 3y2 − 4x − (3y + x) d. x − 3 − (m − 5)

e. − (k − m + 2k) + (k + 3m − 4p) f. x − 2p − (b + 3p) − (b − 3p)

11

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.2 EXPANSION AND FACTORIZATION
1.2.1 EXPANSION OF ALGEBRAIC EXPRESSIONS

i. a(b + c) = ab + ac

ii. (a + b)(c + d) = ac + ad + bc + bd

iii. (a + b)(a − b) = a2 − ab + ab − b2

= a2 − b2

iv. (a + b)2 = (a + b)(a + b)

= a2 + ab + ab + b2
= a2 + 2ab + b2

v. (a − b)2 = (a − b)(a − b)

= a2 − ab − ab + b2
= a2 − 2ab + b2

Example : Expansion of a(b + c) Bracket is removed when
a is multiplied with each
Expand each of the following expression.
term in bracket.
a. 4(2x) = 8x
b. − 3x(5y)

c. 3(x + 2) d. 5x(2x − 3)

e. 2 − x(y − x) f. (x − 3)(x − y)

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example : Expansion of (a + b)(c + d) or (a + b)(c - d)

Expand each of the following expression:

a. (x + 2)(x + 7) b. (2a + 3b)(4a −1)

c. (x + 3)(x − 2) x +3
x x2 + 3x
Example : Expansion of (a + b)(a − b)
− 2 − 2x − 6
Expand each of the following expression:
a. (x + 3)(x − 3) (x + 3)(x − 2) = x2 + x − 6

b. (2x −1)(2x + 1)

Example : Expansion of (a + b)2 b. (2x + 1)2

Expand each of the following expression:

a. (x + 4)2

Example : Expansion of (a - b)2 b. (2x − 5)2

Expand each of the following expression:

a. (x − 7)2

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.2.2 FACTORIZATION

Factors And Common Factors

Consider the numbers 4 and 8.
4 can be written as 4 = 1x4

4 = 2x2
The numbers 1 , 2 and 4 are called the factors of 4.

Similarly, 8 can be written as 8 = 1x8

8 = 2x4
The numbers 1 , 2 , 4 and 8 are called the factors of 8.

Since the numbers 1 and 2 appear both in number 4 and 8, they are called the common factors of 4 and 8.

The same method is used to obtain the factors and common factors of algebraic terms.

Consider the algebraic expressions 3ab and 6a.

3ab can be written as 3ab = 1 x 3ab
3ab = 3 x ab

3ab = a x 3b
3ab = b x 3a

Therefore, factors of 3ab are 1 , 3 , a , b, 3a, 3b, ab and 3ab.

Similarly, 6a can be written as 6a = 1 x 6a
= 2 x 3a
= 3 x 2a
= ax6

Therefore, factors of 6a are 1 , 2, 3 , 6, a, 2a, 3a and 6a.

Hence, the common factors of 3ab and 6a are 1, 3, a and 3a.

Factorization

Factorization is the process of writing an expression as a product of two or more factors. It is the reverse operation of the multiplication of factors
(expansion).

“FACTORIZATION” is the REVERSE of “EXPANSION”

For example,

2(x +3) EXPANDING 2x + 6

FACTORING
The factorized form of an algebraic expression is considered to be the simplest form of the expression. If the algebraic expression cannot be
factorized, then the expression is already in its simplest form, e.g. 3x + 1, x – 1.

i. ab  ac = a(b  c)

ii. a2 − b2 = (a + b)(a − b)

iii. a2 + 2ab + b2 = (a + b)2

iv. a2 − 2ab + b2 = (a − b)2

v. x2 − (a + b)x + ab = (x − a)(x − b)

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example : Factorization of (ab + ac) This method of factorization
is called TAKING OUT
Factorize completely : COMMON FACTORS.

a. 2a + 4 Take out the b. .

= 2(a + 2) common factor 2. 3x2 − 12x

c. 2ab2 − 4a2b d. 2m + 3n + 4m2 + n2

( )Example : Factorization of a2 - b2 The method is

Factorize the following. FACTORIZATION OF DIFFERENCE
OF TWO SQUARES.

.

a. x2 − 16 b. 4x2 − 49

= x2 − 42

= (x + 4)(x − 4)

c. m2 − 4n2 d. 9a2 − 25b2

15

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

( )Example : Factorization of a2  2ab + b2 The method is
FACTORIZATION OF
Factorize completely each of the following.
PERFECT SQUARE.
a. x2 + 6x + 9 a = x and b = 3 b. .

4m2 − 4mn + n2

= x2 + 2(x)(3) + 32

= (x + 3)2

Example : Factorization of x2 − (a + b)x + ab

Factorize completely each of the following.
a. x 2 + 4x + 3
b. x 2 − 2x − 8
c. 3x 2 − 2x − 5
d. 6b2 + 7b + 2

Solution : x2 + 4x + 3
a.
= (x + 1)(x + 3)
a = 1 and b = 3

b. x 2 − 2x − 8

= (x − 4)(x + 2) a = 2 and b = -4

c. 3x 2 − 2x − 5

= (3x − 5)(x + 1)

d. 6b2 + 7b + 2

= (3b + 2)(2b + 1)

16

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

TUTORIAL 1.2(a)

1. Expand each of the following expression:

a. − 6(7p) b. − 7t(− 2t) c. 3(y − 3)

d. 5(3x + 4) ( )e. p2 − 2 11 f. 4t(− 3p − 2t)

2. Expand each of the following algebraic expressions:

a. (2 − 3x)(1− 2x) b. (2m − n)(3m − n)

c. (7 − t)(1+ t) d. (2x + 3)(2x − 3)

e. (a + 3)(a + 2) f. (3 + 4y)(2y − 3)

g. (3p − 4)2 h. (4 − 3h)2

17

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

3. Factorize each of the following expression.

a. x 2 + 6x b. 2x2 + x

c. 3x2 − 12x d. a − ab
e. 6x2y − xy f. 2y2x + 4yx2

g. x2 − 64 h. a2 − 16b2
i. 4x2 − y2 j. x2 + 2x + 1

k. x2 − 8x + 16 l. 4x2 + 4x + 1

m. 9x 2 − 12x + 4 n. 4t 2 − 9t + 2
o. m2 − m − 2 p. x2 − 2xy − 15y2

18

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.2.3 SOLUTIONS OF ALGEBRAIC EQUATIONS

An equation is a statement to show that two values are equal. The symbol = represents equality
in an equation.
A linear equation is an equation consisting of linear algebraic terms and numbers, or linear
algebraic expressions and numbers.
Solutions of linear equations are actually finding the value of the unknown in the equation.
“ = ” is read as “is equal to”
“ ≠ “ is read as ‘”is not equal to”

4x on the left side of the equal 4x = 5 5 on the right side of the
sign is called the equal sign is call right-hand

left-hand side (LHS) side (RHS)

Example :Solving algebraic equations Or solve directly
Solve each of the following. x + 11 = 9
x = 9 −11
Using subtraction x = −2
x + 11 = 9
When + 11 is moved to the right of the equation,
x + 11 −11 = 9 −11 it becomes −11
x = −2

Subtract 11 from both sides of the equation

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Using subtraction Or solve directly
x − 7 = −5 x − 7 = −5
x = −5 + 7
x − 7 + 7 = −5 + 7 x=2
x=2
When − 7 is moved to the
Add 7 to both sides of the equation right of the equation, it becomes + 7

Using division Or solve directly
4x = 18 4x = 18
4x = 18 x = 18
44 4
x=9 x=9
2 2

Divide both sides of the equation When the multiplier 4 is moved to the right of the
by 4 equation, it becomes the divisor 4

Using multiplication Or solve directly
x = −6 x = −6
3 3
x = −6  3
x  3 = −6  3 x = −18
3
When the divisor 3 is moved to the right of the
x = −18 equation, it becomes the multiplier

Multiply both sides of the equation
by 3

We noticed that, when we changed side:

 A positive term becomes a negative (+  −)
 A negative term becomes a positive term (−  +)
 A term that is multiplied becomes a term that is divided (  )
 A term that is divided becomes a term that is multiplied (    )

20

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example : Solving linear equations. b. 2p − 4 = 8p − 13
Solve each of the following.

a. 5y − 12 = 3y − 4

c. 2x + 5(x − 4) = 6 + 3(2x + 3) d. x − 8 = 3

• Define the Variable
• Form the Equation
• Solve the equation

1.2.4 SOLVING APPLIED PROBLEMS LINEAR EQUATIONS.

ADD SUBTRACT MULTIPLY DIVIDE

Altogether Difference Per Quotient
Increase Decrease Times Divided by
More Less Product Divided into
Plus Fewer Double (2x) Per
Sum Reduce Triple (3x) Share
Total Minus Quadruple (4x) Split
Combine

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example :Solving Applied Problems Linear Equations.
Solve each of the following.

a. A customer pays 50 dollars for a coffee maker after a discount of 20 dollars.
What is the original price of the coffee maker?

b. Half a number plus 5 is 11.What is the number?

Solution: Define the Variable
Form the Equation
a) Let x be the original price. Solve the equation
x - 20 = 50
x - 20 + 20 = 50 + 20 Define the Variable
x = 70
Form the Equation
b) Let x be the number. Solve the equation
Always replace "is" with an equal sign

(1/2)x + 5 = 11
(1/2)x + 5 - 5 = 11 – 5
(1/2)x = 6
2 × (1/2)x = 6 × 2

x = 12

22

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Natasha was going shopping with RM 250. When she checked her wallet she found
that six notes, all of the same value of RM x were missing. She had RM 190 left.
Write this as an equation. Solve it to find the value of the missing note.

TUTORIAL 1.2(b) b. 4x + 2 = 18
1. Find the value of x in the following equations.

a. x − 5 = 28

c. 8 − 3(x − 1) = 7x d. 3x − 2 = x
e. 4x = 3x + 2 5

5 f. x + 4 = 2x − 3
2

23

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

2. Solve each of the following equations. b. 3(x − 6) = 2(x + 1)

a. 7(x + 2) = 2x

c. 4(3y + 2) − 4y = 5(2 − 3y) − 7 d. 6s − 5(s − 2) = 14 − 3(s + 4)

3. Solve each of the following equations. b. 1 = 5
4s
a. s = 12
2

c. x = 3x − 25 d. 5v − 2v = 3
5 5

24

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

4. Solve each of the following equations. b. x + 12 = 11

a. x − 4 = 5

c. x + 49 = 2x + 7

25

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.2.5 SOLUTIONS OF QUADRATIC EQUATIONS
The general form of a quadratic equation can be written as

+ + =

where a, b, c are constants, a ≠ 0, x is a variable / unknown. We also can use the other letters.

The property of the a. . one variable only
quadratic equation b. The highest power of x is 2

Example : Determine quadratic equations.

1. (/) the quadratic equations (b) 2x = − 7 (c) 2x2 = 3

(a) x2 − 7x = 1 (e) 6x + 5 = 3x2 (f) x(1− x) = 4
(d) 2x3 − 5 = 0

2. Rewrite each of the following quadratic equation in the general form, then determine the values of a, b
and c.

a. x2 − 6x = 7 b. y2 = 3y − 4

x2 − 6x − 7 = 0
a = 1, b = −6 , c = −7

c. x(x + 1) = 4(x − 5) d. y2 = 16

ROOT OF THE QUADRATIC EQUATIONS

General form of the quadratic equation ax 2 + bx + c = 0

i. x is called the root or solution of the quadratic equation.
ii. Roots of the quadratic equation are the values of unknown that satisfies the equations.
iii. Quadratic equation has at most 2 roots only.

26

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

SOLUTIONS OF QUADRATIC EQUATION

Solving the quadratic equation is to find the roots of the equation. Quadratic equation
ax2 + bx + c = 0 can be solved by factorization and quadratic formula.

Method 1 FACTORIZATION The solutions depend on
the property that pq = 0
If p x q = 0, therefore p = 0 @ q = 0 or if (x - a) (x – b) = 0 →p=0@q=0
Therefore (x – a) = 0 @ x = a
(x – b) = 0 @ x = b

Where a & b are the roots of the quadratic equations.

We use the factorization method when ax2 + bx + c can be factorized.

Example : Solutions of quadratic equations by factorization.

Solve each of the following quadratic equations by factorization.

a. x2 − 3x − 10 = 0 b. x2 − 9 = 0

(x + 2)(x − 5) = 0 x = 3, x = -3

x + 2 = 0,x − 5 = 0 d. (x + 1)(2x − 3) = 12
x = −2 , x = 5

c. x2 − 3x = 18

27

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Method 2 QUADRATIC FORMULA

We can also solve the quadratic equation by using the quadratic formula.

x = − b  b2 − 4ac
2a

Example : Solutions of quadratic equations by formula.
Solve each of the following quadratic equations by using formula. Give your answer correct to 2
decimal places.

a. x2 + 2x − 3 = 0 b. 4x2 − 8x − 1 = 0
a = 1, b = 2 , c = −3

x = − 2  22 − 4(1)(−3)

2(1)

= − 2  4 + 12
2

= −2 4 =−6,2
2 22

 x = −3.00 , 1.00

28

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

c. 2x2 = 10x + 5 d. 3x2 = 6x − 2

29

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

TUTORIAL 1.2(c)

1. Determine whether the given values are the roots of the following quadratic equation or not

(a) x2 − x − 6 = 0 ; x = 2 , −1, −1, −2
(b) 2x2 + 3x − 2 = 0 ;
; x =1, 1 , −1, −2
(c) 2x2 − 3x − 5 = 0 2

x = 3 , 5 ,1, −1
2

2. Rearrange each of the following quadratic equations in the general form
ax2 + bx + c = 0 then express the values of a, b, c

a. x2 − 6x = 7 b. 2 − 3 = 4 c. x(3 − 2x) = −4

d. (3p − 2)2 = 8 e. (2z − 5)2 = 0 f. x(x − 2) = 2x(3 − 2x)

g. 4 ( + 3) = 11m h. (3 + x)(3x − 4) = 0

3. Solve the equations below by factoring. b. x 2 − 9x = 0

( )a. 2 x2 −1 = 3x

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

c. 4(x − 2)2 = 25 d. x(2x + 1) = 15

Solve each of the following quadratic equations, using the factorization method.

(a) x2 + 5x + 4 = 0
(b) x2 + 2x + 1 = 0
(c) 2x2 + 14x + 20 = 0
(d) x(x − 7) = 18
(e) x2 + 4x = 21
(f ) (x − 1)(x − 5) = 21
(g) x2 − 8x = 0
(h) x(2x − 5) = 3

Solve each quadratic equation using factoring:

1) x2 – 3x + 2 = 0 2) z2 – 5z + 4 = 0 3) x2 – 8x + 16 = 0

4) r2 – 12r + 35 = 0 5) c2 + 6c + 5 = 0 6) m2 + 10m + 9 = 0

7) x2 – 49 = 0 8) z2 – 4 = 0 9) m2 – 64 = 0

10) 3x2 – 12 = 0 11) d2 – 2d = 0 12) s2 – s = 0

13) 2x2 – 5x + 2 = 0 14) 3x2 – 10x + 3 = 0 15) 3x2 – 8x + 4 = 0

16) 5x2 + 11x + 2 = 0 17) y2 = 8y + 20 18) x2 = 9x – 20

19) x2 = 30 + x 20) 2x2 – x = 15 21) x2 + 3x – 4 = 50

22) 2x2 + 7 = 5 – 5x 23) x(x – 2) = 35 24) y(y – 3) = 4

25) 10x2 – 5x + 11 = 9x2 + x + 83 26) 4x2 + 3x – 12 = 6x2 – 7x – 60

31

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

4. Solve the quadratic equation below by using quadratic formula

a. 2x 2 − 10 = 0 b. (x + 3)2 = 4

c. x2 − 4x + 3 = 0 d. x 2 − 3x − 4 = 0

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UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1
f. 5 x2 + 3x + 1 = 0
e. x + 1 = x − 2
3
3x + 2 2x − 3

Solve each of the following quadratic equations, using the formula quadratic.

(a) x2 − 4x − 3 = 0
(b) 2x2 + 6x + 3 = 0
(c) 3x2 + 7x = 5
(d) 3x2 = 6x − 2
(e) x(4x + 1) = 1
(f ) (5x − 3)(x + 1) = 0
(g) 2x2 = 7x + 4
(h) 4x2 = 3x + 2

Solve each equation using the quadratic formula:

1) x2 - 7x + 6 = 0 2) x2 + 4x – 5 = 0 3) x2 + 3x + 2 = 0
6) 3x2 + 5x + 2 = 0
4) 2x2 + x – 1 = 0 5) 3x2 + 5x + 2 = 0 9) x2 + 10x = -25
12) x2 = x + 2
7) x2 + 6x + 9 = 0 8) 4x2 – 4x + 1 = 0 15) x2 – 9 = 0
18) x2 = 5x
10) x2 + x = 12 11) x2 + 2x = 24 21) x2 + 2x – 4 = 0
24) 4x2 = 2x + 1
13) x2 + 8 = 6x 14) 2x2 – 10 = x

16) 5x2 = 20 17) x2 – 3x + 1 = 1

19) x2 – 2x – 2 = 0 20) x2 – 10x + 4 = 0

22) x2 – 2 = 4x 23) 2x2 – 8x + 7 = 0

33

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.3 TRANSPOSITION OF FORMULAE

A formula is an equation relating two or more variables.

The subject of a formula is the variable that is expressed in terms of other variables. It always
appears on the left hand side of the equal sign, “ = ”.

For example, for the expression F = ma,

F is the subject of formula

F is the subject and F is expressed in terms of m and a.

We are usually required to find the value of a variable if the values of other variables in the formula
are known.

Example : Evaluate a given formula

Given y = mx 2 , find the value of y if m = 3 Given V = 0.3t3 − 4t , calculate V if t = 5
and x = −2 .
5

Example : Transpose a formula to change its subject

a. If F = ma , express m in b. Make v the subject of the c. Given that
Ft − M = W − F(3 − x) ,
terms of F and a. formula F = mv 2 .
r make F the subject of the
F = ma formula.
Interchange the sides.

ma = F

Divide both sides by a. ma = F
aa
m is now the subject of m=F
the formula.
a

34

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example : Applied problems involving formula
Ohm’s Law states that V = IR. Find R when I = 0.5 A and V = 200 V.

Solution :

TUTORIAL 1.3

1. The area A of a rectangle is given by the 2. The power P watts of an electrical circuit
formula A = bl. Evaluate the area when
l = 14.21 cm and b = 7.46 cm. is given by the formula P = v 2 . Find
R

the power when V is 14.8 volts and R is
19.5 ohms.

3. Given the formula v2 = u² + 2as, find the 4. Distance is given by s = ut + 1 at2 .
value of u if v = 3, a = 1 and s = 3. 2

Given that s = 500, t = 4 , a = - 5, find u.

35

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

5. Change the subject of each of the following formulae to the variable stated in the brackets

(a) z = x + 3y , [x] (b) y = x – 6z , [z]

(c) y = mx2 , [x] (d) 3 z = w , [z]

(e) z = x , [y] (f) z2 + 5 = 3x , [ z]
2y [x]
(h) 5x + 5a = 2x + b , [x]
(g) y = y2 + 5xz ,

(i) = − [ ] (j) = [ ]

(k) − (4 + ) = [ ] (l) p - q = 6p , [x]
x [ ]
(m) = [ ]
(n) = 4 −1
1−
3 √

36

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

6. The area of circle is A = r 2 , 7. The volume of a cube is
where r is the radius. Make r as V = a3 . Make a as the
the subject in term of A and  . subject.

8. The volume of cuboid is given by 9. Given that T = a + (n − 1)d , make
V = lwh . Express h as the subject.
n as the subject.

10. The volume of cone is V = 1 r 2h . Make r 11. Given S = n (a + l). Make a as
3
2
as the subject. the subject in terms of S, n and l.

37

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

12. The total surface area of a sphere 13. Given that C = W 2L2 , make L
S = 4r 2 . Make r as the subject.
24P 2

the subject of the formula.

14. Given that T = 2 l , make l the 15. Given the formula V = IR ,
g
a. Make I the subject of the formula
subject of the formula.
b. Find the value of I if V = 120, and R =

200.

16. A drill draws a current, , and the resistance, R. The power, P, in watts is given as P = I 2R
a. Make R as the subject of the formula.
b. Given the current is 4.5 A and the power is 324 watts. Find the value of resistance, R.

38

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.4 SIMULTANEOUS EQUATION

Sometimes we need to find the solutions of a two linear equation or three linear equations. In this
unit we only focus to two methods, namely:
a) Elimination Method
b) Substitution Method

1.4.1 TWO LINEAR EQUATIONS

Method 1 Elimination Method

Make the coefficients Eliminate one variable Solve the equation for
for one of the by addition or the value of one
subtraction variable, then the
variables numerically other.
equal.

Example : Solve simultaneously two linear equations using elimination method.

Solve the following simultaneous equation using elimination method.

a. 3x + y = 1 and x + y = 3 . 3 − 5 = 12
− 2 = 6

3x + y = 1 .......... ......... (1)
x+y=3 .......... ......... (2)

39

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

. 2 + 15 = 16 d. Rina buys 3 apples and 2 lemons for RM2.80.
4 + 5 = 7 Amira buys 1 apple and 4 lemons for RM
2.60. How much is the price of the apple and
lemon respectively?

40

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Method 2 Substitution Method

Express one Substitute the Solve the equation
variable in term of equation into the for the value of
other equation.
the other in one one variable, then
equation. the other.

Example 1.25 : Solve simultaneously two linear equations using substitution method.

Solve the following simultaneous equation using substitution method.

a. b. 2x − 3y = 10 and 4x − 2y = 12
x − 4y = 2
and 3y + 5x = 79

x − 4y = 2 − − − − −(1)

3y + 5x = 79 − − − − − (2)

from(1) x = 2 + 4y − − − − − (3)

substitute (3)int o (2)
3y + 5(2 + 4y) = 79
3y + 10 + 20y = 79

23y = 79 − 10
y = 69
23
y=3

substitute y = 3 int o (3)  x = 2, and y = −2
x = 2 + 4(3)
x = 14

 x = 14 and y = 3

Arina buys two ballpoint pens and four pencils for RM 3. Hakim buys five ballpoint pens and four pencils
for RM 5.70.
a) Write down these two equations in terms of b and p.
b) Solve the two equation to find the cost of one ball point and pen and the cost of one pencil

41

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.4.2 LINEAR EQUATION AND NON-LINEAR EQUATIONS

LINEAR EQUATION NON-LINEAR EQUATION

• equation consisting of linear algebraic • is an equation consisting of non-linear

terms and numbers, or linear algebraic algebraic terms
expressions and numbers.

• The degree of any variable involved does • terms where the unknown is not 1.

not exceed 1. • Example:

• Example: x2 + 3x + 2 = 0 ,

3x + 2 = 7 is a linear equation with one 1 + 1 = 4,

unknown. xy

2x − y = 3 is a linear equation with two x2 + y2 = 1

unknowns.

Example : Solving simultaneous equation of linear and non-linear equation using substitution
method.

Solve the equations simultaneously. Find the value of x and y for the following
y − x = 4 and x2 + y2 = 16 equations

Solution: x2 − 2xy + y2 = 1

x+y =5.

y − x = 4 -----(1) Make one
of the
x2 + y2 = 16 -----(2)
unknown
y = 4 + x -----(3)
→ as the
Substitute (3) into (2).
subject of
the linear
equation.

x2 + (4 + x)2 = 16

( )x2 + 16 + 8x + x2 = 16

2x2 + 8x = 0

2x(x + 4) = 0

2x = 0 x + 4 = 0

x=0 x = −4

Substitute x = 0 x = 4 and into (3)

y =4+0 y = 4 + (− 4) y = 2, y = 3, x = 3, x = 2
y=4
y=0

42

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example : Solving two non-linear equation using substitution method.
Solve x2 = 4 and 1 + 1 = 10 simultaneously.

xy
Solution :

Steps1 ..........(1) Steps2 Steps3
x2 = 4 ..........(2)
1 + 1 = 10 Substitute x = −2 int o (2) Substitute x = 2 int o (2)
xy
1 + 1 = 10 1 + 1 = 10
From (1) x=2 −2 y 2y

x2 −4 = 0 1 = 10 + 1 1 = 10 − 1
y2 y2
(x + 2)(x − 2) = 0 1 = 10.5 1 = 9.5
y y
x = −2 , y= 1 y= 1

10.5 9.5
y= 2 y= 2

21 19

TUTORIAL 1.4

1. Solve the following simultaneous 2. Solve each of the following pairs of
equations using elimination method simultaneous equations.
and substitution method. a. 3x2 − 5y = 10
x + 2y = 1
a) 5x + y = 14 and
3x + y = 10 b. x − 3y = 4
2x2 − 9y2 = 17
b) 7x − 9y = 1 and
y = 5x − 17 c. x2 + 3x + 2 = 0
1+ 1 =4
c) 6x + 7y = 33 and xy
3x + 4y = 18
d.. 3y − 6x = −x − 5
d) 4x − 8 + y = 0 and x2 + 3x + 5y = 20
2x − y = 7
e. x + y = 6
2+5 =3
xy

43

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

1.5 OPERATIONS OF POLYNOMIALS

1.5.1 INTRODUCTION TO POLYNOMIALS

Coefficients and Variable

A polynomial of degree n is written as

( ) = + −1 − +. . . + 1 + 0

Where n is a positive integer
a0,a1,a2,..., an are coefficients
x is the variable

Example : Coefficients and Variable of polynomials

Given a function 5x4 − x3 + x determine the
3

variable and the coefficient of x4,x3 and x

Solutions:

Degree and Class

The highest power of x that occurs in a polynomial defines the degree of the polynomial.
Here are the degree and class of polynomials.

xn Degree Class

x n=1 1 Linear
Quadratic
x2 n = 2 2
Cubic
x3 n = 3 3
Quartic
x4 n = 4 4

44

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example : Class and degree of polynomials

Determine the degree and class of the polynomials

P(x) 2x + 5 9x2 − 5x4 − x3 + x 5x2 − 7x 2x3 + 9x2 − 3

Degree 1
Class linear

1.5.2 OPERATIONS OF POLYNOMIALS

ADDITION AND SUBTRACTION
Example :

1. Given P(x) = x3 + x2 + x + 1 and 2. Given P(x) = 9x4 + 6x3 + 3x2 −12 and
Q(x) = 7x3 − 9x2 − 25 . Calculate Q(x) = 5x3 − x2 − x − 8 . Calculate

a) P(x) + Q(x) a) P(x) + Q(x)
b) Q(x) − P(x) b) 1 P(x) − Q(x)
c) 3P(x) − Q(x)
3

Solutions: Solutions:
a)

P(x) + Q(x)

( ) ( )= x3 + x2 + x + 1 + 7x3 − 9x2 − 25
( ) ( )= x3 + 7x3 + x2 − 9x2 + x + (1− 25)

= 8x3 − 8x2 + x − 24

b)

Q(x) − P(x)

( ) ( )= 7x3 − 9x2 − 25 − x3 + x2 + x + 1

= 7x3 − 9x2 − 25 − x3 − x2 − x − 1

= 7x3 − x3 − 9x2 − x2 − x − 25 − 1

= 6x3 − 10x2 + x − 26

c)

3P(x) − Q(x)

( ) ( )= 3 x3 + x2 + x + 1 − 7x3 − 9x2 − 25

= 3x3 + 3x2 + 3x + 3 − 7x3 + 9x2 + 25

= 3x3 − 7x3 + 3x2 + 9x2 + 3x + 3 + 25

= −4x3 + 11x2 + 3x + 28

45

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

MULTIPLICATION OF POLYNOMIALS

Example :

1. Given P(x) = x3 + x2 + 1 and Q(x) = x2 − 5 . 2. Given P(x) = 2x3 + 3x2 −12 and

Calculate P(x)Q(x) Q(x) = 5x3 − 8 . Calculate P(x)Q(x)

Solutions: Solutions:

( )( )P(x)Q(x) = x3 + x2 + 1 x2 − 5
( ) ( )= x3 + x2 + 1 x2 + x3 + x2 + 1(− 5)

= x5 + x4 + x2 − 5x3 − 5x2 − 5
= x5 + x4 − 5x3 − 4x2 − 5

DIVISIONS OF POLYNOMIALS

The division of polynomials can be done using two methods which are long divisions and
synthetic divisions

Method 1 Long Divisions

If P(x) is divided by ax + b until a constant remainder R is obtained

P(x) = Q(x) + R
ax + b ax + b

Where

ax + b is the divisor

Q(x) is the quotient

R is the remainder

Gives

P(x) = Q(x)(ax + b) + R

46

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

Example 1.32

By using long division, find the quotient and b) 12x3 + 6x2 − 8x + 2
remainder of 3x + 3

a) 2x3 + 7x2 + 10x + 15 Solutions:
x+2

Solutions:
2x2 + 3x + 4

x + 2 2x3 + 7x2 + 10x + 15

− 2x3 + 4x2
3x2 + 10x

− 3x2 + 6x

4x + 15

− 4x + 8

7

2x3 + 7x2 + 10x + 15 = 2x2 + 3x + 4 + 7
x+2 x+2

Divisor x + 2 Quotient Remainder 7

2x2 + 3x + 4

Method 2 Synthetic Divisions

When dividing a polynomial by a factor of the form − or + a quick form
of divisions is possible and this process is known as synthetic divisions.

Example 1.32

a) 2x3 + 7x2 + 10x + 15 (1)
Solutions: x+2 List the coefficient of

(2) P(x) coefficients P(x)
Start by write 0 under
the first coefficient. 2 7 10 15
0
This is always 0

47

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

(3) 2 7 10 15
Add the coefficient 0 10
and it’s lower value to 7 15
2 −4
get this value 15
−8
(4) 2
From divisor x + 2 −2 0 7
Write -2 as the factor
2
(5)
Multiply factor and the
addition value to get

this value

(6) 2 7 10
Repeat multiply the
factor with addition − 2 0 −4 −6

value until the last 2 3 4 remainder, R
coefficient.

Rewrite the answer Quotients, Q(x)

2x2 + 3x + 4 + 7
x+2

b) c)
4x3 − 8x + 12 30x3 − 3x2 + 12x + 1
x−6
x−5
12
40 −8 460
472
5 0 20 100
R
4 20 92

Quotients, Q(x)

= 4x2 + 20x + 92 + 472
x−5

48

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

TUTORIAL 1.5

1. If P(x) = x3 + 6x2 − 8x + 9 , Q(x) = 3x4 − 9x2 + 6x − 3 and R(x) = x + 4 find

a. P(x) + Q(x) b. P(x) − Q(x)

c. R(x)Q(x) d. Q(x) − 1 P(x)

2

e. Q(x) f. P(x)
R(x) R(x)

49

UNIT 1: ALGEBRA DUM 10122: Engineering Mathematics 1

2. If P(x) = 8x4 + 4x2 −10x + 9 , Q(x) = −4x3 + 6x2 −1and R(x) = x + 8 find

a. P(x) + Q(x) b. P(x) − Q(x)

c. P(x)R(x) d. R(x)+ 1 Q(x)

3

e. Q(x) f. P(x)
R(x) R(x)

50