Chapter 1 Basic Algebra

Engineering Mathematics 1-DBM10013

1.4.4 IMPROPER FRACTION

What is improper fraction?
The degree of numerator (top) is greater than/equal to denominator (bottom)

How to solve an improper fraction to partial fraction for algebraic terms?
First, an improper fraction will convert to the proper fraction using LONG DIVISION

For example, Let ( ) be improper fraction, then the long division will be ,
( )

quotient ()
remainder
() ()
()

()

Now the improper fraction state as :

() ( )+ ()
( )= ()

Improper Proper

Quotient

Then, express the proper fraction, ( ) as a partial fraction.

()

51

Engineering Mathematics 1-DBM10013

REVIEW OF LONG DIVISION
Numberic Terms

Example 1.22 1528616 remainder

Algebra Terms 1 3√1 9 8 7 2 0 1 6
Example 1.23
13

68

65

37
26
1 12
10 4

80
78

21
13

86
78

8

−1 6 6
(−) 6
+2 +5
−6 remainder

8 +5

52

Engineering Mathematics 1-DBM10013

Example 1.24 5 +8 into partial fraction
Express
( + 3 − 2)
Step 1
Convert improper fraction to proper fraction using long division

5 − 15

+3 −2 5 +8
(−) 5 + 15 − 10
−15 + 10 + 8
(−) − 15 − 45 + 30
55 − 30 + 8

Step 2 Write the proper fraction

5 +8 55 − 30 + 8
( + 3 − 2) = 5 − 15 + + 3 − 2

Step 3 Express remainder over divisor into partial fraction

55 − 30 + 8 = + +
( + 3 − 2) +3 −2

55 − 30 + 8 ( + 3 − 2) ( + ) ( + 3 − 2)
= + +3 −2

55 − 30 + 8 = ( + 3 − 2) + ( + )( )

55 − 30 + 8 = + 3 − 2 + +

Combine like terms gives,
55 − 30 + 8 = ( + ) + (3 + ) − 2

Compare coefficient of the constants,
−2 = 8
= −4

53

Engineering Mathematics 1-DBM10013

Compare coefficient of ,
3 + = −30

3(−4) + = −30
= −18

Compare coefficient of ,
+ = 55

−4 + = 55
= 59

Write the partial fraction

5 +8 4 59 − 18
( + 3 − 2) = 5 − 15 − + + 3 − 2

54

Engineering Mathematics 1-DBM10013

Exercise 1.14

Solve the following fraction into partial fraction

1. − 3 − 6
−4

35
+ 1 + 2 + 2( − 4)

55

Engineering Mathematics 1-DBM10013

2. 3 + 7 + 11 + 5
( − 1)(3 − 8)

26 128 + 223
1 − 5( − 1) + 5(3 − 8)

56

Engineering Mathematics 1-DBM10013

3. 3 − 1
− 7 + 12

26 47
3− −3+ −4

57

Engineering Mathematics 1-DBM10013

4. 2 + 5
( − 1)

77
2 + 2 + 2( − 1) − 2( + 1)

58

Engineering Mathematics 1-DBM10013

PAST YEAR QUESTIONS

JUN 2018

QUESTION 1

a. Express each of the following expressions in the simplest form:

i. 1 (4 + 3 ) − 5(2 − ) 13
2 −8 + 2

2 +1 4 −1 2 −7 +2
ii. 4 − 2 4

iii. 12 − 4 4 3−
6 −7 +2÷2 −1 3 −2

b. Calculate the roots for equations below by using the given method:

i. 2 + 5 − 12 = 0 ( ℎ) 3
= 2 , −4

ii. 3 = 4 + 9 ( ) [ = 2.5226, −1 .1893]

iii. 4 − 12 + 8 = 0 ( ℎ ℎ) [ = 2, 1]

QUESTION 2 −15 7
= 8 , =8
a. Determine the value of A and B for partial fraction below:
− + 10
− 8 + 16 = + 5 + − 3

b. Calculate the partial fraction for the following equations: −11 9
11 − 5 4( − 4) − ( − 4)

i. − 8 + 16 4 −7 + 22
ii. 10 + 5 − 3( + 1) + 3( + 2)

( + 1)( + 2) 59

iii. Engineering Mathematics 1-DBM10013
−5 +6
99
1+ −3+ −2

DECEMBER 2018

QUESTION 1

a. State each of the following expression in the simplest form: [ +2 ]
i. ( + ) − (2 − )
4
4 +2 3−2 5 −1
ii. 10 − 2 − 1 − 5
−( − 3)
iii. +2 −4 +4 3( − 2)
3 −9÷ −6 +9

b. Solve the following quadratic equations by using the given method

i. = 4 + 21 ( ) [ = 7, −3]
7
ii. 3 − 4 = 7
= 3 , −1
iii. + 8 − 2 = 0 [ = 0.243 , −8.243]

QUESTION 2

a. Determine the value of A and B for the partial fraction decomposition below:

5 + 13 [ = 3, = 2]
+4 −5= −1+ +5

b. Decompose the following fractions into partial fraction: 21
3 −5 ( + 3) + ( − 4)

i. − − 12 22 3
+2 − − 1 + ( − 1)

ii. ( − 1)

60

3 −5 Engineering Mathematics 1-DBM10013
iii. ( − 1)( + + 1)
−2 2 + 13
3 −5 3( − 1) + 3( + + 1)
iv.
7
−2 3 +6+ −2

JUN 2019 3 +2
QUESTION 1 −2
a. Complete each of the following expressions:
+2
4
i. − 2 − 1

22
ii. − 4 ÷ − 2

b. Solve the following quadratic equations using the given method:

i. − 6 + 2 = 0 ( ) [ = 5.646 , 0.35]
[ = 3.189 , −0.523]
ii. 3 = 5 + 8 ( ℎ)

c. Construct the partial fraction for the following equations 1 + − 1 +2
2(1 + ) 2 ( 2 + 4)

i. ( + 4) 31
−2 +3 −7+ +4

ii.
+5 +4

61