1 BASIC ALGEBRA

1.1 Simplifying Algebraic Fractions

Algebraic fraction has the same properties as numerical fraction. The only difference being that

the numerator (top) and denominator (bottom) are both algebraic expressions. Fractional algebra

is a rational number usually stated in the form of q

p , where p and q are integers. Integer 'p' is

known as the numerator and integer 'q' as denominator.

Meanwhile, an ordinary fraction is usually use to represent a part of an object or figure. For

example: a cake is cut into 6 equal parts. One part can be represented with fractional expression

of 1/6, similarly if expressed in the fractional algebra. There are some important terms to be

familiarize with before solving fractional algebra.

i. Equivalent fractions are fractions having same value.

2

1 =

4

2 =

8

4

ii. Single fraction is any fractional phrase.

iii. Lowest fractional form is a fraction that cannot be simplified further, or, its numerator

and denominator does not have a common factor.

q

p

Example 1

Complete the equivalent fractions below :

a) 20

6

105

2

Numerator

Denominator

2 BASIC ALGEBRA

b) 20

312

c) AB

CABBC

A

2

d)

XYZ

XYZ

XZ

XYZ

2

Solution:

a) 20

8

15

6

10

4

5

2

b) 10

20

3

6

1

22

c) CB

AB

CAB

BA

BC

A22

2

d)

XYZ

ZXY

YZX

XYZ

XZ

XYZ 2

22

2

Example 2

Simplify each of the following fractions:

a) 27

2

b

b b)

2

2

6

3

x

xx c) mmm 4)2(6

Solution:

(a) bbb

b

b

b

7

2

7

2

7

22

(b) x

x

xx

xx

x

xx

6

3

)6(

)3(

6

32

2

NOTE: The cancellation in (b) is allowed since x is a common factor of the numerator and the

denominator. Sometimes extra work is necessary before an algebraic fraction can be reduced to a

simpler form.

(c) mmmmmm 4264)2(6

3 BASIC ALGEBRA

m12

Example 3

Simplify the algebraic fraction: 32

122

2

xx

xx

Solution

In this case, the numerator and denominator can be factorised into two terms. Thus,

22 )1(12 xxx and )3)(1(322 xxxx

So,

)1)(3(

)1)(1(

32

122

2

xx

xx

xx

xx

3

1

x

x

Exercise 1

Simplify each of the following algebraic fractions.

(a) 45

36

14

7

ba

ba

(b) 25

1072

2

y

yy

Solution

(a) The fraction is 45

36

14

7

ba

ba. Instead of expanding the factors, it is easier to use the rule of

indices (powers) :

nm

n

m

aa

a , Thus,

4

3

5

6

45

36

14

7

14

7

b

b

a

a

ba

ba

4 BASIC ALGEBRA

4356

2

1 ba

4356

2

1 ba

b

a

2

(b) In this case, initial factorization is needed. So,

)2)(5(1072 yyyy and

)5)(5(252 yyy

Thus,

)5)(5(

)2)(5(

25

1072

2

yy

yy

y

yy

5

2

y

y

Exercise 2

Which of the following is a simplified version of 23

432

2

tt

tt?

2

4)(

t

ta

2

4)(

t

tb

2

4)(

t

tc

2

4)(

t

td

Solution

Factorize the numerator and denominator is respectively

)4)(1(432 tttt and )2)(1(232 tttt

so that,

)2)(1(

)4)(1(

23

432

2

tt

tt

tt

tt

5 BASIC ALGEBRA

2

4

t

t

So far, simplification has been achieved by cancelling common factors from the numerator and

denominator. There are fractions which can be simplified by multiplying the numerator and

denominator by an appropriate common factor, thus obtaining an equivalent, simpler expression.

Exercise 3

Simplify the following fractions:

a)

2

14

1y

b)

2

13

xx

Solution

(a) In this case, multiplying both the numerator and the denominator by 4 gives:

2

41

)2

1(4

)4

1(4

2

14

1

yyy

(b) To simplify this expression, multiply the numerator and denominator

by 2 x. Thus

x

x

x

xxx

xx

2

13

2

)1

3(

2

13 2

Exercise 4

6 BASIC ALGEBRA

Simplify each of the following algebra fractions.

(a) 2

2

34 y

(b)

2

13

1

z

z

Solution

(a) The fraction is simplified by multiplying both the numerator and the denominator by 2.

4

38

22

)2

34(2

2

2

34

y

yy

(b) In this case, since the numerator contains the fraction 1/3 and the denominator contains the fraction 1/2, the common factor needed is 2 3 = 6. Thus

36

26

)2

1(6

)3

1(6

2

13

1

z

z

z

z

z

z

Exercise 5:

Which of the following is a simplified version of1

1

1

x

xx

?

1

1)(

2

2

xx

xxa

1

1)(

2

2

x

xxb

1

1)(

2

2

xx

xc

1

1)(

2

2

x

xd

Solution:

7 BASIC ALGEBRA

For 1

1

1

x

xx

, the common multiplier is ( x + 1). Multiplying the numerator and the denominator

by this gives:

)1)(1(

)1

1)(1(

1

1

1

xx

xxx

x

xx

)1(

)1

1)(1()1(

2

x

xxxx

)1(

)1(2

2

x

xx

Exercise 7:

1. Simplified the following.

a) tsrrs 3232

b) )12

5(4 232 xqpqp

c) abcab 624 2

d) hgf

hgf22

23

9

36

e) npq

npqqp

10

153 22

f) xzxyzzxy 181226 32

g) 22222 3)4(7 rssrsr

h) 45326 xymnxymn

i) )32(5 byab

j) 3)12156( 2 cabmk

k) )42()35( 22 zxyzxy

l) )3()23( 22 kpmnrsmnkp

m) )48(2

19 nm

n) 2

1210)35(2

xx

Answer for Exercise 7:

8 BASIC ALGEBRA

1. a) tsr436

b)

xqp 35

3

5

c) bc4

d) f4

e) 22

2

9qp

f) 339xyz

g) 222 311 rssr

h) 473 xymn

i) xyab 1510

j) 2452 cabmk

k) 27 zxy

l) rsmnkp 24

m) nm 249 n) 1215 x

9 BASIC ALGEBRA

1.2 Solving Algebraic Fractions Using Addition, Subtraction, Multiplication and Division

It is make revenue process of finding increase and revenue push of two or more fraction algebra.

In settling these operations there were 3 moves that must be followed.

ADDITION

Example 1: Calculate

5

1

5

3

.

Solution: Since the denominators are the same, the denominator of the answer will be 5. Adding

the numerators , 3 + 1 = 4. The result will be:

5

4

5

13

5

1

5

3

Example 2:

Calculate

3

1

2

1

Solution: The denominators are different, so you must build each fraction to a form where both

have the same denominator. Since 6 is the common factor for 2 and 3, build both

fractions to a denominator of 6.

6

3

3

3

2

11

2

1

and,

6

2

2

2

3

11

3

1

Then

10 BASIC ALGEBRA

3

1

2

1

6

5

6

23

6

2

6

3

SUBTRACTION

Example 1:

Calculate 5

1

5

3

Solution: The denominators are the same, so you can skip step 1. The denominator of the answer

will be 5. Subtract the numerators for the numerator in the answer. 3 - 1 = 2. The answer

is

5

2

5

13

5

1

5

3

Example 2: Calculate

3

1

2

1

Solution:

Since the denominators are not the same, you must build each fraction to a form where

both have the same denominator. As 6 is the common factor for 2 and 3, use 6 as the

denominator.

6

3

3

3

2

11

2

1

and,

6

2

2

2

3

11

3

1

Then,

3

1

2

1

6

1

6

23

6

2

6

3

11 BASIC ALGEBRA

MULTIPLICATION

Example 1:

9

8

4

3

Solution:

Multiply the numerators and the denominators, and simplify the result. Thus,

3

2

36

24

94

83

Example 2:

15

6

7

2

Solution:

Multiply the numerators and the denominators, and simplify the result. Thus,

35

4

105

12

157

62

DIVISION

Example 1:

16

9

4

3

Solution:

Change the division sign to multiplication and invert the fraction to the right of the sign.

9

16

4

3

Multiply the numerators and the denominators, and simplify the result.

3

4

36

48

94

163

12 BASIC ALGEBRA

Example 2:

8

9

40

3

Solution:

Change the division sign to multiplication and invert the fraction to the right of the sign.

9

8

40

3

Multiply the numerators and the denominators, and simplify the result.

15

1

360

24

940

83

9

8

40

3

Exercise 1: Simplify 23

32

10

)8)(5(

ts

tsst

Solution: 223

43

23

32

410

40

10

)8)(5(t

ts

ts

ts

tsst

Exercise 2: Simplify 22

2

)3(

12

ab

ba

Solution: 342

2

22

2

3

4

9

12

)3(

12

bba

ba

ab

ba

Exercise 3: Simplify m

mnnm

2

62 2

Solution: Form the given fraction into 2 fractions,

both with denominator 2m. Thus,

nmnm

mn

m

nm

m

mnnm3

2

6

2

2

2

62 22

Exercise 4: Simplify

45

13

x

x

13 BASIC ALGEBRA

Solution 1 - Multiplying by the reciprocal

Change the top expression into a single fraction with denominator x.

x

x

x

1313

Change the below expression into a single fraction with denominator x.

x

x

x

454

5

Thus, the question has become:

x

xx

x

x

x45

13

45

13

Look at the right hand side expression as a division process. Thus,

x

x

x

x 4513

Instead of division, we can carry out the multiplication by a reciprocal method.

x

x

x

x

x

x

45

13

45

13

The x's cancelled out, and we have our final answer, which is in its simplest form.

Solution 2 - Multiplying top and bottom

Multiply "x" to both the numerator and denominator. So, by just multiplying

the top and bottom part by x, everything will be simplified.

x

x

x

x

x

x

45

13

)45

(

)1

3(

14 BASIC ALGEBRA

QUADRATIC EQUATION

A Quadratic equation in a unknown (variable) is one equation have one unknown only and his

unknown's supreme power is 2.

ROOT OF A QUADRATIC EQUATION

Root of a quadratic equation can be found by using three methods:

a) Factorization b) Quadratic formula c) Completing the square

a. Factorization

Example:

Solve the quadratic equations below, using factorization method.

a) 2x2 + 13x + 10 = 0

b) 6x2 20 = 2x

c) 3x = - 6x2

Solution:

a) Given: 010132 2 xx

Then: 0)5)(32( xx

03 x or 05 x

3x or 5x

b) Given: xx 2206 2

Then: 02026 2 xx

0)42)(53( xx

053 x or 042 x

3

5x or 2x

c) Given: 263 xx

Then: 036 2 xx

15 BASIC ALGEBRA

0)36( xx

0x or 3x

b. Quadratic formula

If quadratic equation given by form 02 cbxax , then value x can find through quadratic

formula, a

acbbx

2

42

Example:

Solve quadratic equation using quadratic formula method

a. 018219 2 xx

b. 35172 2 xx

c. 0245 2 xx

Solution:

a. 018219 2 xx

Thus: a = 9, b = 21 and c = -18

18

64844121 x

18

108921x

18

3321x

18

3321x or

18

3321

92

18942121 2 x

16 BASIC ALGEBRA

18

12x or

18

54

3

2x or 3

b. 35172 2 xx then: 0351722 xx

So, a=2, b=-17, and c=35

)2(2

)35)(2(417)17( 2 x

4

28028917 x

4

917 x

4

317 x

4

317 x or

4

317

5x or 2

7

c. 0245 2 xx

52

2544)4( 2 x

10

40164 x

17 BASIC ALGEBRA

10

564 x

10

48.74 x

10

48.74 x or

10

48.74

10

48.11x or

10

48.3

148.1x or 0.348

c) Completing the square

Example

Solve the quadratic equation using completing the square

a) 4x2 + 5 = -9x

b) x2 = 4x + 4

c) 2x2 +4x 8=0

Solution

a) Given 4x2 + 5 = -9x

Step 1: Arrange to the form into cbxax 2

594 2 xx

Step 2: Divide with 4 to make the coefficient of x2 equal to 1

4

5

4

9

4

4 2

xx

4

5

4

92 x

x

18 BASIC ALGEBRA

Step 3: Add 22 )8

9()

2

1

4

9( to both sides

222 )8

9(

4

5)

8

9(

4

9

xx

22 )8

9(

4

5)

8

9( x

64

8180

64

1)

8

9( 2 x

64

1)

8

9( x

8

1

8

1

8

9x or

8

1

8

9x

8

8x or

8

10x

1x or 4

5x

b) Given 4x2 = 4x + 4

Step 1: Arrange to the form into cbxax 2

4x2 4x = 4

Step 2: Divided by 4 to make the coefficient of x2 equal to 1

4

4

4

4

4

4 2

xx

19 BASIC ALGEBRA

x2 x = 1

Step 3: Add 22 )2

1()

2

11( to both sides

222 )2

1(1)

2

1( xx

4

11)

2

1( 2 x

4

5

4

5

2

1x

4

5

2

1x

118.15.0 x

118.15.0 x or 118.15.0 x

618.1x or 618.0x

c) Given 2x2+4x-8 =0

Step 1: Arrange to the form into cbxax 2

2x2 + 4x = 8

Step 2: Divided by 2 to make the coefficient of x2 equal to 1

2

8

2

4

2

2 2

xx

20 BASIC ALGEBRA

422 xx

Step 3: Add 22 )1()2

12( to both sides

222 1412 xx

5)1( 2 x

51 x

51x

236.21x

236.21x or 236.21x

236.1x or 236.3x

Exercise:

1. Solve quadratic equation

i. by factoring

(a) 0862 xx

(b) 05143 2 xx

(c) 04133 2 xx

ii. by quadratic formula

(a) 0162 2 xx

(b) 0935 2 xx

(c) 1092

1 2 xx

iii. by completing the square

(a) 0762 xx

(b) 062 2 xx

(c) 01263 2 xx

Answer:

21 BASIC ALGEBRA

1. i. By factoring

(a) 4,2 xx

(b) 5,3

1 xx

(c) 4,3

1 xx

ii. By quadratic formula

(a) 177.0,823.2 xx

(b) 675.1,075.1 xx

(c) 05.19,05.1 xx

iii. By completing the square

(a) 1,7 xx

(b) 0,3 xx

(c) 236.3,236.1 xx