Contributed by:

This pdf includes the following topics:-

Cancelling common factors

Multiplication and division of algebraic fractions

Addition and subtraction of algebraic fractions

Cancelling common factors

Multiplication and division of algebraic fractions

Addition and subtraction of algebraic fractions

1.
Arithmetic of

Algebraic Fractions 1.4

Introduction

Just as one whole number divided by another is called a numerical fraction, so one algebraic expression

divided by another is known as an algebraic fraction. Examples are

x 3x + 2y x2 + 3x + 1

, , and

y x−y x−4

In this Section we explain how algebraic fractions can be simplified, added, subtracted, multiplied

and divided.

Prerequisites • be familiar with the arithmetic of numerical

fractions

Before starting this Section you should . . .

Learning Outcomes • add, subtract, multiply and divide algebraic

fractions

On completion you should be able to . . .

62 HELM (2008):

Workbook 1: Basic Algebra

Algebraic Fractions 1.4

Introduction

Just as one whole number divided by another is called a numerical fraction, so one algebraic expression

divided by another is known as an algebraic fraction. Examples are

x 3x + 2y x2 + 3x + 1

, , and

y x−y x−4

In this Section we explain how algebraic fractions can be simplified, added, subtracted, multiplied

and divided.

Prerequisites • be familiar with the arithmetic of numerical

fractions

Before starting this Section you should . . .

Learning Outcomes • add, subtract, multiply and divide algebraic

fractions

On completion you should be able to . . .

62 HELM (2008):

Workbook 1: Basic Algebra

2.
®

1. Cancelling common factors

10

Consider the fraction . To simplify it we can factorise the numerator and the denominator and then

35

cancel any common factors. Common factors are those factors which occur in both the numerator

and the denominator. Thus

10 65×2 2

= =

35 7× 6 5 7

Note that the common factor 5 has been cancelled. It is important to remember that only common

10 2

factors can be cancelled. The fractions and have identical values - they are equivalent fractions

35 7

2 10

- but is in a simpler form than .

7 35

We apply the same process when simplifying algebraic fractions.

Example 49

Simplify, if possible,

yx x x

(a) , (b) , (c)

2x xy x+y

Solution

yx

(a) In the expression , x is a factor common to both numerator and denominator. This

2x

common factor can be cancelled to give

y6x y

=

26x 2

x 1x

(b) Note that can be written . The common factor of x can be cancelled to give

xy xy

16x 1

=

6 xy y

x

(c) In the expression notice that an x appears in both numerator and denominator.

x+y

However x is not a common factor. Recall that factors of an expression are multi-

plied together whereas in the denominator x is added to y. This expression cannot be

simplified.

HELM (2008): 63

Section 1.4: Arithmetic of Algebraic Fractions

1. Cancelling common factors

10

Consider the fraction . To simplify it we can factorise the numerator and the denominator and then

35

cancel any common factors. Common factors are those factors which occur in both the numerator

and the denominator. Thus

10 65×2 2

= =

35 7× 6 5 7

Note that the common factor 5 has been cancelled. It is important to remember that only common

10 2

factors can be cancelled. The fractions and have identical values - they are equivalent fractions

35 7

2 10

- but is in a simpler form than .

7 35

We apply the same process when simplifying algebraic fractions.

Example 49

Simplify, if possible,

yx x x

(a) , (b) , (c)

2x xy x+y

Solution

yx

(a) In the expression , x is a factor common to both numerator and denominator. This

2x

common factor can be cancelled to give

y6x y

=

26x 2

x 1x

(b) Note that can be written . The common factor of x can be cancelled to give

xy xy

16x 1

=

6 xy y

x

(c) In the expression notice that an x appears in both numerator and denominator.

x+y

However x is not a common factor. Recall that factors of an expression are multi-

plied together whereas in the denominator x is added to y. This expression cannot be

simplified.

HELM (2008): 63

Section 1.4: Arithmetic of Algebraic Fractions

3.
Task

abc 3ab

Simplify, if possible, (a), (b)

3ac b+a

When simplifying remember only common factors can be cancelled.

Your solution

abc 3ab

(a) = (b) =

3ac b+a

b

(a) (b) This cannot be simplified.

3

Task

21x3

Simplify ,

14x

Your solution

Factorising and cancelling common factors gives:

21x3 6 7 × 3× 6 x × x2 3x2

= =

14x 6 7 × 2× 6 x 2

Task

36x

Simplify

12x3

Your solution

Factorising and cancelling common factors gives:

36x 12 × 3 × x 3

3

= 2

= 2

12x 12 × x × x x

64 HELM (2008):

Workbook 1: Basic Algebra

abc 3ab

Simplify, if possible, (a), (b)

3ac b+a

When simplifying remember only common factors can be cancelled.

Your solution

abc 3ab

(a) = (b) =

3ac b+a

b

(a) (b) This cannot be simplified.

3

Task

21x3

Simplify ,

14x

Your solution

Factorising and cancelling common factors gives:

21x3 6 7 × 3× 6 x × x2 3x2

= =

14x 6 7 × 2× 6 x 2

Task

36x

Simplify

12x3

Your solution

Factorising and cancelling common factors gives:

36x 12 × 3 × x 3

3

= 2

= 2

12x 12 × x × x x

64 HELM (2008):

Workbook 1: Basic Algebra

4.
®

Example 50

3x + 6

Simplify .

6x + 12

Solution

First we factorise the numerator and the denominator to see if there are any common factors.

3x + 6 3(x + 2) 3 1

= = =

6x + 12 6(x + 2) 6 2

The factors x + 2 and 3 have been cancelled.

Task

12

Simplify .

2x + 8

Your solution

12

=

2x + 8

Answer

6×2 6

Factorise the numerator and denominator, and cancel any common factors. =

2(x + 4) x+4

Example 51

3 3(x + 4)

Show that the algebraic fraction and 2 are equivalent.

x+1 x + 5x + 4

Solution

The denominator, x2 + 5x + 4, can be factorised as (x + 1)(x + 4) so that

3(x + 4) 3(x + 4)

=

x2+ 5x + 4 (x + 1)(x + 4)

Note that (x + 4) is a factor common to both the numerator and the denominator and can be

3 3 3(x + 4)

cancelled to leave . Thus and 2 are equivalent fractions.

x+1 x+1 x + 5x + 4

HELM (2008): 65

Section 1.4: Arithmetic of Algebraic Fractions

Example 50

3x + 6

Simplify .

6x + 12

Solution

First we factorise the numerator and the denominator to see if there are any common factors.

3x + 6 3(x + 2) 3 1

= = =

6x + 12 6(x + 2) 6 2

The factors x + 2 and 3 have been cancelled.

Task

12

Simplify .

2x + 8

Your solution

12

=

2x + 8

Answer

6×2 6

Factorise the numerator and denominator, and cancel any common factors. =

2(x + 4) x+4

Example 51

3 3(x + 4)

Show that the algebraic fraction and 2 are equivalent.

x+1 x + 5x + 4

Solution

The denominator, x2 + 5x + 4, can be factorised as (x + 1)(x + 4) so that

3(x + 4) 3(x + 4)

=

x2+ 5x + 4 (x + 1)(x + 4)

Note that (x + 4) is a factor common to both the numerator and the denominator and can be

3 3 3(x + 4)

cancelled to leave . Thus and 2 are equivalent fractions.

x+1 x+1 x + 5x + 4

HELM (2008): 65

Section 1.4: Arithmetic of Algebraic Fractions

5.
Task

x−1 1

Show that is equivalent to .

x2 − 3x + 2 x−2

First factorise the denominator:

Your solution

x2 − 3x + 2 =

Answer

(x − 1)(x − 2)

Now identify the factor common to both numerator and denominator and cancel this common factor:

Your solution

x−1

=

(x − 1)(x − 2)

Answer

1

. Hence the two given fractions are equivalent.

x−2

Example 52

6(4 − 8x)(x − 2)

Simplify

1 − 2x

Solution

The factor 4 − 8x can be factorised to 4(1 − 2x). Thus

6(4 − 8x)(x − 2) (6)(4)(1 − 2x)(x − 2)

= = 24(x − 2)

1 − 2x (1 − 2x)

Task

x2 + 2x − 15

Simplify

2x2 − 5x − 3

First factorise the numerator and factorise the denominator:

Your solution

x2 + 2x − 15

=

2x2 − 5x − 3

66 HELM (2008):

Workbook 1: Basic Algebra

x−1 1

Show that is equivalent to .

x2 − 3x + 2 x−2

First factorise the denominator:

Your solution

x2 − 3x + 2 =

Answer

(x − 1)(x − 2)

Now identify the factor common to both numerator and denominator and cancel this common factor:

Your solution

x−1

=

(x − 1)(x − 2)

Answer

1

. Hence the two given fractions are equivalent.

x−2

Example 52

6(4 − 8x)(x − 2)

Simplify

1 − 2x

Solution

The factor 4 − 8x can be factorised to 4(1 − 2x). Thus

6(4 − 8x)(x − 2) (6)(4)(1 − 2x)(x − 2)

= = 24(x − 2)

1 − 2x (1 − 2x)

Task

x2 + 2x − 15

Simplify

2x2 − 5x − 3

First factorise the numerator and factorise the denominator:

Your solution

x2 + 2x − 15

=

2x2 − 5x − 3

66 HELM (2008):

Workbook 1: Basic Algebra

6.
®

Answer

(x + 5)(x − 3)

(2x + 1)(x − 3)

Then cancel any common factors:

Your solution

(x + 5)(x − 3)

=

(2x + 1)(x − 3)

Answer

x+5

2x + 1

Exercises

1. Simplify, if possible,

19 14 35 7 14

(a) , (b) , (c) , (d) , (e)

38 28 40 11 56

14 36 13 52

2. Simplify, if possible, (a) , (b) , (c) , (d)

21 96 52 13

5z 25z 5 5z

3. Simplify (a) , (b) , (c) 2

, (d)

z 5z 25z 25z 2

4. Simplify

4x 15x 4s 21x4

(a) , (b) , (c) , (d)

3x x2 s3 7x3

5. Simplify, if possible,

x+1 x+1 2(x + 1) 3x + 3 5x − 15 5x − 15

(a) , (b) , (c) , (d) , (e) , (f) .

2(x + 1) 2x + 2 x+1 x+1 5 x−3

6. Simplify, if possible,

5x + 15 5x + 15 5x + 15 5x + 15

(a) , (b) , (c) , (d)

25x + 5 25x 25 25x + 1

x2 + 10x + 9 x2 − 9 2x2 − x − 1

7. Simplify (a) , (b) , (c) ,

x2 + 8x − 9 x2 + 4x − 21 2x2 + 5x + 2

3x2 − 4x + 1 5z 2 − 20z

(d) , (e)

x2 − x 2z − 8

6 2x 3x2

8. Simplify (a) , (b) 2 , (c)

3x + 9 4x + 2x 15x3 + 10x2

x2 − 1 x2 + 5x + 6

9. Simplify (a) , (b) .

x2 + 5x + 4 x2 + x − 6

HELM (2008): 67

Section 1.4: Arithmetic of Algebraic Fractions

Answer

(x + 5)(x − 3)

(2x + 1)(x − 3)

Then cancel any common factors:

Your solution

(x + 5)(x − 3)

=

(2x + 1)(x − 3)

Answer

x+5

2x + 1

Exercises

1. Simplify, if possible,

19 14 35 7 14

(a) , (b) , (c) , (d) , (e)

38 28 40 11 56

14 36 13 52

2. Simplify, if possible, (a) , (b) , (c) , (d)

21 96 52 13

5z 25z 5 5z

3. Simplify (a) , (b) , (c) 2

, (d)

z 5z 25z 25z 2

4. Simplify

4x 15x 4s 21x4

(a) , (b) , (c) , (d)

3x x2 s3 7x3

5. Simplify, if possible,

x+1 x+1 2(x + 1) 3x + 3 5x − 15 5x − 15

(a) , (b) , (c) , (d) , (e) , (f) .

2(x + 1) 2x + 2 x+1 x+1 5 x−3

6. Simplify, if possible,

5x + 15 5x + 15 5x + 15 5x + 15

(a) , (b) , (c) , (d)

25x + 5 25x 25 25x + 1

x2 + 10x + 9 x2 − 9 2x2 − x − 1

7. Simplify (a) , (b) , (c) ,

x2 + 8x − 9 x2 + 4x − 21 2x2 + 5x + 2

3x2 − 4x + 1 5z 2 − 20z

(d) , (e)

x2 − x 2z − 8

6 2x 3x2

8. Simplify (a) , (b) 2 , (c)

3x + 9 4x + 2x 15x3 + 10x2

x2 − 1 x2 + 5x + 6

9. Simplify (a) , (b) .

x2 + 5x + 4 x2 + x − 6

HELM (2008): 67

Section 1.4: Arithmetic of Algebraic Fractions

7.
Answers

1 1 7 7 1

1. (a) , (b) , (c) , (d) , (e) .

2 2 8 11 4

2 3 1

2. (a) , (b) , (c) , (d) 4

3 8 4

1 1

3. (a) 5, (b) 5, (c) 2

, (d) .

5z 5z

4 15 4

4. (a) , (b) , (c) 2 , (d) 3x

3 x s

1 1

5. (a) , (b) , (c) 2, (d) 3, (e) x − 3, (f) 5

2 2

x+3 x+3 x+3 5(x + 3)

6. (a) , (b) , (c) , (d)

5x + 1 5x 5 25x + 1

x+1 x+3 x−1 3x − 1 5z

7. (a) , (b) , (c) , (d) , (e)

x−1 x+7 x+2 x 2

2 1 3

8. (a) , (b) , (c) .

x+3 2x + 1 5(3x + 2)

x−1 x+2

9. (a) , (b) .

x+4 x−2

2. Multiplication and division of algebraic fractions

To multiply together two fractions (numerical or algebraic) we multiply their numerators together

and then multiply their denominators together. That is

Key Point 19

Multiplication of fractions

a c ac

× =

b d bd

Any factors common to both numerator and denominator can be cancelled. This cancellation can be

performed before or after the multiplication.

To divide one fraction by another (numerical or algebraic) we invert the second fraction and then

68 HELM (2008):

Workbook 1: Basic Algebra

1 1 7 7 1

1. (a) , (b) , (c) , (d) , (e) .

2 2 8 11 4

2 3 1

2. (a) , (b) , (c) , (d) 4

3 8 4

1 1

3. (a) 5, (b) 5, (c) 2

, (d) .

5z 5z

4 15 4

4. (a) , (b) , (c) 2 , (d) 3x

3 x s

1 1

5. (a) , (b) , (c) 2, (d) 3, (e) x − 3, (f) 5

2 2

x+3 x+3 x+3 5(x + 3)

6. (a) , (b) , (c) , (d)

5x + 1 5x 5 25x + 1

x+1 x+3 x−1 3x − 1 5z

7. (a) , (b) , (c) , (d) , (e)

x−1 x+7 x+2 x 2

2 1 3

8. (a) , (b) , (c) .

x+3 2x + 1 5(3x + 2)

x−1 x+2

9. (a) , (b) .

x+4 x−2

2. Multiplication and division of algebraic fractions

To multiply together two fractions (numerical or algebraic) we multiply their numerators together

and then multiply their denominators together. That is

Key Point 19

Multiplication of fractions

a c ac

× =

b d bd

Any factors common to both numerator and denominator can be cancelled. This cancellation can be

performed before or after the multiplication.

To divide one fraction by another (numerical or algebraic) we invert the second fraction and then

68 HELM (2008):

Workbook 1: Basic Algebra

8.
®

Key Point 20

Division of fractions

a c a d ad

÷ = × = b 6= 0, c 6= 0, d 6= 0

b d b c bc

Example 53

2a 4 2a c 2a 4

Simplify (a) × , (b) × , (c) ÷

c c c 4 c c

Solution

2a 4 8a

(a) × = 2

c c c

2a c 2ac 2a a

(b) × = = =

c 4 4c 4 2

(c) Division is performed by inverting the second fraction and then multiplying.

2a 4 2a c a

÷ = × = (from the result in (b))

c c c 4 2

Example 54

1 1

Simplify (a) × 3x, (b) × x.

5x x

Solution

3x 1 1 3x 3x 3

(a) Note that 3x = . Then × 3x = × = =

1 5x 5x 1 5x 5

x 1 1 x x

(b) x can be written as . Then × x = × = = 1

1 x x 1 x

HELM (2008): 69

Section 1.4: Arithmetic of Algebraic Fractions

Key Point 20

Division of fractions

a c a d ad

÷ = × = b 6= 0, c 6= 0, d 6= 0

b d b c bc

Example 53

2a 4 2a c 2a 4

Simplify (a) × , (b) × , (c) ÷

c c c 4 c c

Solution

2a 4 8a

(a) × = 2

c c c

2a c 2ac 2a a

(b) × = = =

c 4 4c 4 2

(c) Division is performed by inverting the second fraction and then multiplying.

2a 4 2a c a

÷ = × = (from the result in (b))

c c c 4 2

Example 54

1 1

Simplify (a) × 3x, (b) × x.

5x x

Solution

3x 1 1 3x 3x 3

(a) Note that 3x = . Then × 3x = × = =

1 5x 5x 1 5x 5

x 1 1 x x

(b) x can be written as . Then × x = × = = 1

1 x x 1 x

HELM (2008): 69

Section 1.4: Arithmetic of Algebraic Fractions

9.
Task

1 y

Simplify (a) × x, (b) × x.

y x

Your solution

1 1 x x

(a) ×x= × =

y y 1 y

y y x yx

(b) ×x= × = =y

x x 1 x

Example 55

2x

y

Simplify

3x

2y

Solution

2x 3x

We can write the fraction as ÷ .

y 2y

Inverting the second fraction and multiplying we find

2x 2y 4xy 4

× = =

y 3x 3xy 3

70 HELM (2008):

Workbook 1: Basic Algebra

1 y

Simplify (a) × x, (b) × x.

y x

Your solution

1 1 x x

(a) ×x= × =

y y 1 y

y y x yx

(b) ×x= × = =y

x x 1 x

Example 55

2x

y

Simplify

3x

2y

Solution

2x 3x

We can write the fraction as ÷ .

y 2y

Inverting the second fraction and multiplying we find

2x 2y 4xy 4

× = =

y 3x 3xy 3

70 HELM (2008):

Workbook 1: Basic Algebra

10.
®

Example 56

4x + 2 x+3

Simplify ×

x2 + 4x + 3 7x + 5

Solution

Factorising the numerator and denominator we find

4x + 2 x+3 2(2x + 1) x+3 2(2x + 1)(x + 3)

× = × =

x2 + 4x + 3 7x + 5 (x + 1)(x + 3) 7x + 5 (x + 1)(x + 3)(7x + 5)

2(2x + 1)

=

(x + 1)(7x + 5)

It is usually better to factorise first and cancel any common factors before multiplying. Don’t remove

any brackets unnecessarily otherwise common factors will be difficult to spot.

Task

Simplify

15 3

÷

3x − 1 2x + 1

Your solution

Answer

To divide we invert the second fraction and multiply:

15 3 15 2x + 1 (5)(3)(2x + 1) 5(2x + 1)

÷ = × = =

3x − 1 2x + 1 3x − 1 3 3(3x − 1) 3x − 1

HELM (2008): 71

Section 1.4: Arithmetic of Algebraic Fractions

Example 56

4x + 2 x+3

Simplify ×

x2 + 4x + 3 7x + 5

Solution

Factorising the numerator and denominator we find

4x + 2 x+3 2(2x + 1) x+3 2(2x + 1)(x + 3)

× = × =

x2 + 4x + 3 7x + 5 (x + 1)(x + 3) 7x + 5 (x + 1)(x + 3)(7x + 5)

2(2x + 1)

=

(x + 1)(7x + 5)

It is usually better to factorise first and cancel any common factors before multiplying. Don’t remove

any brackets unnecessarily otherwise common factors will be difficult to spot.

Task

Simplify

15 3

÷

3x − 1 2x + 1

Your solution

Answer

To divide we invert the second fraction and multiply:

15 3 15 2x + 1 (5)(3)(2x + 1) 5(2x + 1)

÷ = × = =

3x − 1 2x + 1 3x − 1 3 3(3x − 1) 3x − 1

HELM (2008): 71

Section 1.4: Arithmetic of Algebraic Fractions

11.
Exercises

5 3 14 3 6 3 4 28

1. Simplify (a) × , (b) × , (c) × , (d) ×

9 2 3 9 11 4 7 3

5 3 14 3 6 3 4 28

2. Simplify (a) ÷ , (b) ÷ , (c) ÷ , (d) ÷

9 2 3 9 11 4 7 3

3. Simplify

x+y 1 2

(a) 2 × , (b) × 2(x + y), (c) × (x + y)

3 3 3

4. Simplify

x+4 1 3 x x+1 1 x2 + x

(a) 3 × , (b) × 3(x + 4), (c) × (x + 4), (d) × , (e) × ,

7 7 7 y y+1 y y+1

πd2 Q Q

(f) × 2, (g)

4 πd πd2 /4

6/7

5. Simplify

s+3

3 x

6. Simplify ÷

x + 2 2x + 4

5 x

7. Simplify ÷

2x + 1 3x − 1

5 14 9 16

1. (a) , (b) , (c) , (d)

6 9 22 3

10 8 3

2. (a) , (b) 14, (c) , (d)

27 11 49

2(x + y) 2(x + y) 2(x + y)

3. (a) , (b) , (c)

3 3 3

3(x + 4) 3(x + 4) 3(x + 4) x(x + 1) x(x + 1)

4. (a) , (b) , (c) , (d) , (e) , (f) Q/4,

7 7 7 y(y + 1) y(y + 1)

4Q

(g)

πd2

6

5.

7(s + 3)

6

6.

x

5(3x − 1)

7.

x(2x + 1)

72 HELM (2008):

Workbook 1: Basic Algebra

5 3 14 3 6 3 4 28

1. Simplify (a) × , (b) × , (c) × , (d) ×

9 2 3 9 11 4 7 3

5 3 14 3 6 3 4 28

2. Simplify (a) ÷ , (b) ÷ , (c) ÷ , (d) ÷

9 2 3 9 11 4 7 3

3. Simplify

x+y 1 2

(a) 2 × , (b) × 2(x + y), (c) × (x + y)

3 3 3

4. Simplify

x+4 1 3 x x+1 1 x2 + x

(a) 3 × , (b) × 3(x + 4), (c) × (x + 4), (d) × , (e) × ,

7 7 7 y y+1 y y+1

πd2 Q Q

(f) × 2, (g)

4 πd πd2 /4

6/7

5. Simplify

s+3

3 x

6. Simplify ÷

x + 2 2x + 4

5 x

7. Simplify ÷

2x + 1 3x − 1

5 14 9 16

1. (a) , (b) , (c) , (d)

6 9 22 3

10 8 3

2. (a) , (b) 14, (c) , (d)

27 11 49

2(x + y) 2(x + y) 2(x + y)

3. (a) , (b) , (c)

3 3 3

3(x + 4) 3(x + 4) 3(x + 4) x(x + 1) x(x + 1)

4. (a) , (b) , (c) , (d) , (e) , (f) Q/4,

7 7 7 y(y + 1) y(y + 1)

4Q

(g)

πd2

6

5.

7(s + 3)

6

6.

x

5(3x − 1)

7.

x(2x + 1)

72 HELM (2008):

Workbook 1: Basic Algebra

12.
®

3. Addition and subtraction of algebraic fractions

To add two algebraic fractions the lowest common denominator must be found first. This is the

simplest algebraic expression that has the given denominators as its factors. All fractions must be

written with this lowest common denominator. Their sum is found by adding the numerators and

dividing the result by the lowest common denominator.

To subtract two fractions the process is similar. The fractions are written with the lowest common

denominator. The difference is found by subtracting the numerators and dividing the result by the

lowest common denominator.

Example 57

State the simplest expression which has x + 1 and x + 4 as its factors.

Solution

The simplest expression is (x + 1)(x + 4). Note that both x + 1 and x + 4 are factors.

Example 58

State the simplest expression which has x − 1 and (x − 1)2 as its factors.

Solution

The simplest expression is (x − 1)2 . Clearly (x − 1)2 must be a factor of this expression. Also,

because we can write (x − 1)2 = (x − 1)(x − 1) it follows that x − 1 is a factor too.

HELM (2008): 73

Section 1.4: Arithmetic of Algebraic Fractions

3. Addition and subtraction of algebraic fractions

To add two algebraic fractions the lowest common denominator must be found first. This is the

simplest algebraic expression that has the given denominators as its factors. All fractions must be

written with this lowest common denominator. Their sum is found by adding the numerators and

dividing the result by the lowest common denominator.

To subtract two fractions the process is similar. The fractions are written with the lowest common

denominator. The difference is found by subtracting the numerators and dividing the result by the

lowest common denominator.

Example 57

State the simplest expression which has x + 1 and x + 4 as its factors.

Solution

The simplest expression is (x + 1)(x + 4). Note that both x + 1 and x + 4 are factors.

Example 58

State the simplest expression which has x − 1 and (x − 1)2 as its factors.

Solution

The simplest expression is (x − 1)2 . Clearly (x − 1)2 must be a factor of this expression. Also,

because we can write (x − 1)2 = (x − 1)(x − 1) it follows that x − 1 is a factor too.

HELM (2008): 73

Section 1.4: Arithmetic of Algebraic Fractions

13.
Example 59

3 2

Express as a single fraction +

x+1 x+4

Solution

The simplest expression which has both denominators as its factors is (x + 1)(x + 4). This is the

lowest common denominator. Both fractions must be written using this denominator. Note that

3 3(x + 4) 2 2(x + 1)

is equivalent to and is equivalent to . Thus writing

x+1 (x + 1)(x + 4) x+4 (x + 1)(x + 4)

both fractions with the same denominator we have

3 2 3(x + 4) 2(x + 1)

+ = +

x+1 x+4 (x + 1)(x + 4) (x + 1)(x + 4)

The sum is found by adding the numerators and dividing the result by the lowest common denomi-

nator.

3(x + 4) 2(x + 1) 3(x + 4) + 2(x + 1) 5x + 14

+ = =

(x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4)

Key Point 21

Addition of two algebraic fractions

Step 1: Find the lowest common denominator

Step 2: Express each fraction with this denominator

Step 3: Add the numerators and divide the result by the lowest common denominator

Example 60

1 5

Express + as a single fraction.

x − 1 (x − 1)2

Solution

The simplest expression having both denominators as its factors is (x − 1)2 . We write both fractions

with this denominator.

1 5 x−1 5 x−1+5 x+4

+ 2

= 2

+ 2

= 2

=

x − 1 (x − 1) (x − 1) (x − 1) (x − 1) (x − 1)2

74 HELM (2008):

Workbook 1: Basic Algebra

3 2

Express as a single fraction +

x+1 x+4

Solution

The simplest expression which has both denominators as its factors is (x + 1)(x + 4). This is the

lowest common denominator. Both fractions must be written using this denominator. Note that

3 3(x + 4) 2 2(x + 1)

is equivalent to and is equivalent to . Thus writing

x+1 (x + 1)(x + 4) x+4 (x + 1)(x + 4)

both fractions with the same denominator we have

3 2 3(x + 4) 2(x + 1)

+ = +

x+1 x+4 (x + 1)(x + 4) (x + 1)(x + 4)

The sum is found by adding the numerators and dividing the result by the lowest common denomi-

nator.

3(x + 4) 2(x + 1) 3(x + 4) + 2(x + 1) 5x + 14

+ = =

(x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4)

Key Point 21

Addition of two algebraic fractions

Step 1: Find the lowest common denominator

Step 2: Express each fraction with this denominator

Step 3: Add the numerators and divide the result by the lowest common denominator

Example 60

1 5

Express + as a single fraction.

x − 1 (x − 1)2

Solution

The simplest expression having both denominators as its factors is (x − 1)2 . We write both fractions

with this denominator.

1 5 x−1 5 x−1+5 x+4

+ 2

= 2

+ 2

= 2

=

x − 1 (x − 1) (x − 1) (x − 1) (x − 1) (x − 1)2

74 HELM (2008):

Workbook 1: Basic Algebra

14.
®

Task

3 5

Express + as a single fraction.

x+7 x+2

First find the lowest common denominator:

Your solution

Answer

(x + 7)(x + 2)

Re-write both fractions using this lowest common denominator:

Your solution

3 5

+ =

x+7 x+2

Answer

3(x + 2) 5(x + 7)

+

(x + 7)(x + 2) (x + 7)(x + 2)

Finally, add the numerators and simplify:

Your solution

3 5

+ =

x+7 x+2

Answer

8x + 41

(x + 7)(x + 2)

Example 61

5x 3x − 4

Express − as a single fraction.

7 2

Solution

In this example both denominators are simply numbers. The lowest common denominator is 14, and

both fractions are re-written with this denominator. Thus

5x 3x − 4 10x 7(3x − 4) 10x − 7(3x − 4) 28 − 11x

− = − = =

7 2 14 14 14 14

HELM (2008): 75

Section 1.4: Arithmetic of Algebraic Fractions

Task

3 5

Express + as a single fraction.

x+7 x+2

First find the lowest common denominator:

Your solution

Answer

(x + 7)(x + 2)

Re-write both fractions using this lowest common denominator:

Your solution

3 5

+ =

x+7 x+2

Answer

3(x + 2) 5(x + 7)

+

(x + 7)(x + 2) (x + 7)(x + 2)

Finally, add the numerators and simplify:

Your solution

3 5

+ =

x+7 x+2

Answer

8x + 41

(x + 7)(x + 2)

Example 61

5x 3x − 4

Express − as a single fraction.

7 2

Solution

In this example both denominators are simply numbers. The lowest common denominator is 14, and

both fractions are re-written with this denominator. Thus

5x 3x − 4 10x 7(3x − 4) 10x − 7(3x − 4) 28 − 11x

− = − = =

7 2 14 14 14 14

HELM (2008): 75

Section 1.4: Arithmetic of Algebraic Fractions

15.
Task

1 1

Express + as a single fraction.

x y

Your solution

The simplest expression which has x and y as its factors is xy. This is the lowest common denom-

1 y 1 x

inator. Both fractions are written using this denominator. Noting that = and that =

x xy y xy

we find

1 1 y x y+x

+ = + =

x y xy xy xy

No cancellation is now possible because neither x nor y is a factor of the numerator.

Exercises

x x 2x x 2x 3x x 2 x+1 3

1. Simplify (a)+ , (b) + , (c) − , (d) − , (e) + ,

4 7 5 9 3 4 x+1 x+2 x x+2

2x + 1 x x+3 x x x

(f) − , (g) − , (h) −

3 2 2x + 1 3 4 5

2. Find

1 2 2 5 2 3 x+1 x+4

(a) + , (b) + , (c) − , (d) + ,

x+2 x+3 x+3 x+1 2x + 1 3x + 2 x+3 x+2

x−1 x−1

(e) + .

x − 3 (x − 3)2

5 4

3. Find + .

2x + 3 (2x + 3)2

1 11

4. Find s+

7 21

A B

5. Express + as a single fraction.

2x + 3 x + 1

A B C

6 Express + + as a single fraction.

2x + 5 (x − 1) (x − 1)2

A B

7 Express + as a single fraction.

x + 1 (x + 1)2

76 HELM (2008):

Workbook 1: Basic Algebra

1 1

Express + as a single fraction.

x y

Your solution

The simplest expression which has x and y as its factors is xy. This is the lowest common denom-

1 y 1 x

inator. Both fractions are written using this denominator. Noting that = and that =

x xy y xy

we find

1 1 y x y+x

+ = + =

x y xy xy xy

No cancellation is now possible because neither x nor y is a factor of the numerator.

Exercises

x x 2x x 2x 3x x 2 x+1 3

1. Simplify (a)+ , (b) + , (c) − , (d) − , (e) + ,

4 7 5 9 3 4 x+1 x+2 x x+2

2x + 1 x x+3 x x x

(f) − , (g) − , (h) −

3 2 2x + 1 3 4 5

2. Find

1 2 2 5 2 3 x+1 x+4

(a) + , (b) + , (c) − , (d) + ,

x+2 x+3 x+3 x+1 2x + 1 3x + 2 x+3 x+2

x−1 x−1

(e) + .

x − 3 (x − 3)2

5 4

3. Find + .

2x + 3 (2x + 3)2

1 11

4. Find s+

7 21

A B

5. Express + as a single fraction.

2x + 3 x + 1

A B C

6 Express + + as a single fraction.

2x + 5 (x − 1) (x − 1)2

A B

7 Express + as a single fraction.

x + 1 (x + 1)2

76 HELM (2008):

Workbook 1: Basic Algebra

16.
®

Ax + B C

8 Express + as a single fraction.

x2+ x + 10 x − 1

C

9 Express Ax + B + as a single fraction.

x+1

x1 x1 x2 x3

10 Show that is equal to .

1 1 x2 − x3

−

x3 x2

3x x x 3x x x

11 Find (a) − + , (b) − + .

4 5 3 4 5 3

Answers

11x 23x x x2 − 2 x2 + 6x + 2

1. (a) , (b) , (c) − , (d) , (e) ,

28 45 12 (x + 1)(x + 2) x(x + 2)

x+2 9 + 2x − 2x2 x

(f) , (g) , (h)

6 3(2x + 1) 20

3x + 7 7x + 17 1

2. (a) , (b) , (c) ,

(x + 2)(x + 3) (x + 3)(x + 1) (2x + 1)(3x + 2)

2x2 + 10x + 14 x2 − 3x + 2

(d) , (e)

(x + 3)(x + 2) (x − 3)2

10x + 19

3.

(2x + 3)2

3s + 11

4.

21

A(x + 1) + B(2x + 3)

5.

(2x + 3)(x + 1)

A(x − 1)2 + B(x − 1)(2x + 5) + C(2x + 5)

6.

(2x + 5)(x − 1)2

A(x + 1) + B

7.

(x + 1)2

(Ax + B)(x − 1) + C(x2 + x + 10)

8.

(x − 1)(x2 + x + 10)

(Ax + B)(x + 1) + C

9.

x+1

53x 13x

11. (a) , (b)

60 60

HELM (2008): 77

Section 1.4: Arithmetic of Algebraic Fractions

Ax + B C

8 Express + as a single fraction.

x2+ x + 10 x − 1

C

9 Express Ax + B + as a single fraction.

x+1

x1 x1 x2 x3

10 Show that is equal to .

1 1 x2 − x3

−

x3 x2

3x x x 3x x x

11 Find (a) − + , (b) − + .

4 5 3 4 5 3

Answers

11x 23x x x2 − 2 x2 + 6x + 2

1. (a) , (b) , (c) − , (d) , (e) ,

28 45 12 (x + 1)(x + 2) x(x + 2)

x+2 9 + 2x − 2x2 x

(f) , (g) , (h)

6 3(2x + 1) 20

3x + 7 7x + 17 1

2. (a) , (b) , (c) ,

(x + 2)(x + 3) (x + 3)(x + 1) (2x + 1)(3x + 2)

2x2 + 10x + 14 x2 − 3x + 2

(d) , (e)

(x + 3)(x + 2) (x − 3)2

10x + 19

3.

(2x + 3)2

3s + 11

4.

21

A(x + 1) + B(2x + 3)

5.

(2x + 3)(x + 1)

A(x − 1)2 + B(x − 1)(2x + 5) + C(2x + 5)

6.

(2x + 5)(x − 1)2

A(x + 1) + B

7.

(x + 1)2

(Ax + B)(x − 1) + C(x2 + x + 10)

8.

(x − 1)(x2 + x + 10)

(Ax + B)(x + 1) + C

9.

x+1

53x 13x

11. (a) , (b)

60 60

HELM (2008): 77

Section 1.4: Arithmetic of Algebraic Fractions