Study Advice Service
Mathematics
Worksheet
Algebraic
Fractions
This is one of a series of worksheets designed to help you increase your confidence
in handling Mathematics. This worksheet contains both theory and exercises which
cover:1. Simplification
2. Multiplication & Division
3. Addition & Subtraction
There are often different ways of doing things in Mathematics and the methods
suggested in the worksheets may not be the ones you were taught. If you are
successful and happy with the methods you use it may not be necessary for you to
change them. If you have problems or need help in any part of the work then there
are a number of ways you can get help.
For students at the University of Hull
 Ask your lecturers
 Contact the Study Advice Service in the Brynmor Jones Library where you can
access the Mathematics Tutor, or contact us by email
 Come to a Drop-In session organised for your department
 Look at one of the many textbooks in the library.
For others
 Ask your lecturers
 Access your Study Advice or Maths Help Service
 Use any other facilities that may be available.
If you do find anything you may think is incorrect (in the text or answers) or want
further help please contact us by email.
Web: www.hull.ac.uk/studyadvice
Email: studyadvice@hull.ac.uk
Tel: 01482 466199
1. Simplification of Fractions
You can simplify numerical fractions by ‘doing the same thing’ to the top and bottom
of the fraction. You can simplify algebraic fractions in the same way.
a)
12 6  2 2


18 6  3 3
b)
2ab 2  a  b 2b


dividing top and bottom by a .
3a
3 a
3
c)
d)
dividing top and bottom by 6 (or cancelling down).
a
a
1


dividing top and bottom by a .
3ab 3  a  b 3b
2a  b
cannot be simplified as there is no common factor in the numerator.
3a
To check an answer: the number test is a very useful way of checking whether your
statement is incorrect. If you can find a numerical value which makes the statement
incorrect, then the statement itself is incorrect. But be careful, finding numerical
values which make the statement correct, does not prove it is correct.
2a  b 2  b

then putting a  2 , b  1 we
3a
3
2a  b 4  1 5
2  b 2 1
2a  b 2  b

 and

 1 so

get
is incorrect.
3a
6
6
3
3
3a
3
For instance if you are tempted to write
e)
2ab2c 2  a  b  b  c 2bc
dividing top and bottom by ab .


3ab
3 a  b
3
21a3b 2
3  7  a  a  a  b  b 3a 2


f)
dividing top and bottom by 7ab 2 .
2
27 abb
2
14ab
g)
2a  5b
cannot be simplified - no common factor in top and bottom.
2ab
h)
2  4b 2  1  2b  1  2b


dividing top and bottom by 2.
6b
6b
3b
i)
2a  4ab 2a1  2b  2a


dividing top and bottom by (1  2b) .
3  6b
31  2b 
3
Notice here that the term (1  2b) behaves like a single letter. If you write x in place
2a1  2b  2ax 2a


of (1  2b) then
. Any terms in brackets behave as a single
31  2b 
3x
3
term.
page 1
j)
k)
a 2  4ab a a  4b 

which cannot be simplified. Although both top and bottom
3  6b
31  2b 
can be factorised there is no common factor in both top and bottom.
a 2  3a  2
a2  4

a  2a  1  a  1
a  2a  2 a  2
dividing top and bottom by ( a  2)
Exercise 1
Simplify the following, where possible.
24
27
(a)
(b)
(c)
1
32
32
2
2ab
5ab
(a)
(b)
(c)
4ac
4a 2
3
2a  6b
2a  6b
(a)
(b)
(c)
4a  10
4a  12b
4
2a  6b
a2  1
(a)
(c)
(b)
4a  12b
2a  2
25
120
23b
24ac
2ab
4a  6ab
a 2  2a
a2  4
(d)
196
144
24a 2b
20abc
2  4a
(d)
3  2a
8a
(d)
2
4a  4a
(d)
2 Multiplication and division
The rules for multiplying and dividing numerical fractions are covered in the booklet
‘Fractions’. To summarise first: convert any mixed fractions to improper fractions.
then,
 to multiply, you multiply the numerators together and multiply the denominators
1 2 5 2 10
5

together. Hence 1    
.
4 7 4 7 28 14
 to divide you multiply the first fraction by the inverse of the second. Hence
1 2 5 2 5 7 35
3
1      
 4 . (See Fractions booklet for proof)
4 7 4 7 4 2 8
8
Multiplying and dividing algebraic fractions is similar but without the complication of
having mixed fractions.
Hence we have
w y wy
 
,
x z
xz
Examples
3a 4 12a 6a
a)
 

2b b 2b 2 b 2
w y w z wz
   
.
x z x y xy
b)
b
3a 2

5a
5ab
5


4b 12a 2b 12a
c)
3a 4 3a b 3ab 3a
 
 

2b b 2b 4 8b
8
d)
e)
3a
3a 2b 6ab
 2b 


 6a
b
b
1
b
(writing 2b as
3a 4 9a 3a 4 2b
24ab
4
 

 


2
2b b 2b 2b b 9a 18ab
3b
2b
)
1
page 2
f) 5a 
g)
h)
3
3
5a
3
15a 3
5a
: writing 5a as




this gives 5a 
10a
1
10a
1 10a 10a 2
2a
2a ab 2a 1
2a
2
 ab 





3b
3b 1
3b ab 3ab2 3b2
2a  3 6b2

: factorising 4a  6 this gives
3b
4a  6

2a  3 6b 2
2a  3  6b 2 6b 2



b
3b
4a  6 3b  22a  3
6b
i)
2a  4 2b 2  3b  9 2a  2 2b  3b  3 a  2b  3




6b  9
4a  2
32b  3
22a  1
32a  1
j)
2ab  3a 6ab  9a 2ab  3a
4b  1
a2b  3  4b  1 4b  1





3b
4b  1
3b
6ab  9a
3b  3a2b  3
9b
k)
2a  4
2a  2
1
2
note 2  a  ( a  2)
 2  a  


3b
3b
 a  2
3b
Exercise 2
Simplify the following, cancelling down as far as possible, where possible.
22 5
2 12
21 25 20
16
 
  11
(b)
(c)
(d) 5   32
1 (a) 
3 14
5 32 49
3 8
25
a  4 5b  5
a  2b
1 a  2b

(b) 
(c) 3 
2 (a)
2b  2
4a
3 ab
ab
2
2
b
2a  3b
(c)

 ab  b2
3 (a) a  1  4b  20 (b) 4b  8  2b  4b
2
3a 2ab  3b
3b  15
a 1
4a  8
6


3. Addition & Subtraction
(a) Equivalent fractions
As you know when you add or subtract two numerical fractions you need to change
the fractions so that they have the same denominator; these are called equivalent
fractions. This is covered in the Fractions booklet.
Examples
(a)
12 6  2 2


18 6  3 3
so
12
18
(b)
3 3  4 12


5 5  4 20
so
3
5
(c)
a a  c ac
a
ac


so
and
are equivalent fractions
b
b b  c bc
bc
and
and
2
3
12
20
are equivalent fractions
are equivalent fractions
page 3
(d)
2 2  c 2c
2c
2


so and
are equivalent fractions
b
b b  c bc
bc
(e)
2  a c are equivalent fractions
2  a 2  a   c 2  a c
2a


so
and
b
bc
bc
b
bc
(f)
c
c
c  1  c 
c1  c 
c1  c 
so
and
are equivalent fractions


1 c
1  c 1  c 1  c  1  c 2
1  c2
Exercise 3
Complete the following
3 ?
?
30 3a
?




1. 
5 10 25
?
? 15b
c
?
?
cc  b  3cd
?





2.
2
b bd b
?
?
4ba  2b 
(b) Adding and subtracting
As you know when you add or subtract two numerical fractions you need to change
the fractions so that they have the same denominator. This is covered in the
Fractions booklet.
Examples
1 1 3 2 5
   
(i)
2 3 6 6 6
3
5
9 10
19

3
(ii) 1  2  3 
8
12
24 24
24
5
2
3
10 8 9
9
3
6 6
(iii) 3  5  2  3  5  2    
6
3
4
12 12 12
12
4
In the same way to add algebraic fractions they must have the same denominator.
1 1
1
 
.
a b ab
This can be checked, using the number test mentioned earlier, by putting, say, a  1
and b  1:
1 1 1 1
1
1
1
    2 but


a b 1 1
a  b 11 2
1 1
2
 
note that
is also wrong (try a  1 and b  1).
a b ab
A common error is to write
Writing the two fractions with a common denominator ab gives
1 1
b
a
ba
 


.
a b ab ab
ab
The lowest denominator is usually called the Lowest Common Multiple (LCM) or
Lowest Common Denominator (LCD)
page 4
Algebraic fractions are dealt with in the same way as numerical fractions; you need
to find the LCM of the fractions, the simplest expression that the denominators of the
fractions will go into. This is often quite easy to see. It is also possible to use larger,
or more complicated, common denominators. The first example above could have
1 1 30 20 50 5
 



been written as:
2 3 60 60 60 6
50
5
cancelling down the fraction
to by dividing top and bottom by 10 (see the
6
60
Fractions booklet if you have any problems with this). Both 6 and 60 are common
multiples of 2 and 3 but 6 is (obviously) the Lowest Common Multiple (LCM). If the
LCM of, say, a, b, c is d then a, b, c are all factors of d (ie go into d without
leaving a remainder).
1
1

. We want to put both fractions over a common denominator.
a 1 a  2
1
1
1
1
1 1


 
Putting x  a  1; y  a  2, then
becomes
a 1 a  2
a 1 a  2 x y
1 1
y
x
y  x a  2  a  1
2a  1




which we can do as above  
a  1a  2 a  1a  2
x y xy xy
xy
Consider
In practice we write this as
a  2  a  1  2a  1
1
1
a2
a 1




a  1a  2 a  1a  2
a  1 a  2 a  1a  2 a  1a  2
with practice step 1, which uses the fact that 1 and
a 1
a2
a 1a  2 are equivalent
fractions, can be left out.
From above we can see that the simplest common denominator (LCM) for terms with
no common factors is the product of those terms.
Denominators ( a  3) and (2a  1) give LCM ( a  3)(2a  1)
Denominators ( a2  1) and ( a  1) give LCM ( a2  1)( a  1)
What about terms which have common factors?
The LCM of a 2 and a is not a3 but a 2 as both a and a 2 will go into a 2 without
leaving a remainder.
Hence the LCM of ( a  1)2 and ( a  1) is ( a  1)2 .
The LCM of 3( a  1)3 and 2(a  1)2 is 6( a  1)3 etc.
Examples In each case step 1 can be left out once you understand how to get to
step 2!
1
2
a 1
2( a  3)
a  1  2a  6
3a  5





(a)
a  3 a  1 a  3a  1 a  3a  1 a  3a  1 a  3a  1
page 5
(b)
1
1
a2
a 3
a  2  a  3
5





a  3 a  2 a  3a  2 a  3a  2 a  3a  2 a  3a  2
(c)
3a
3a
2a  2 3a  2a  2 3a  2a  4 a  4
2 




a2
a2
a2
a2
a2
a2
(d)
(e)
1

a  12
8a
3a  1
3
1
1
a 1
1 a 1
a2




a  1 a  12 a  12 a  12 a  12

5
2a  1
2
Exercise 4
Work out the following
1 1
1. 
a b

28a 
6a  1
3
2.

3a  1  5
6a  1
3
x 2x

5 3

16a  15a  15
6a  1
3
3.
2
1
x
2
3

3x  4 x  6
4.
1
1

x2 x2
5.
3
2

t2 t2
6.
7.
5
1

2y  3 y  2
8.
1
3

z z2
9.
3
s2

s2
s2
12.
10.
3r
r  12
1
13. 4  t
3

5
r 1
11.
14.
1
y

y
 y  2
2

2
x  2
2

a  15
6a  13
1
x2
4
1

3
x 1 x2
15. 3 
2
1

r  2 3r  2
Express as single fractions:
A
B

16.
x  2 x 1
B
C

17. A 
2x 1 2x  1
C
18. Ax  B 
x2
page 6
Answers
Exercise 1
3
27
5
49
b
5b
23b
6a
1(a)
(b)
(c )
(d )
2(a)
(b)
(c )
(d )
4
32
24
36
2c
4a
24ac
5c
a  3b
1
b
2  4a
a  3b
a 1
a
2
3(a)
(b) (c)
(d )
4(a)
(b)
(c )
(d )
2a  5
2
2  3b
3  2a
2a  6b
2
a2
a 1
Exercise 2
4
75
5
1
5a  4
a  2b
3a  2b 
1( a )
(b )
(c )
(d )
2( a )
(b )
(c )
7
56
12
10
8a
3a  b 
ab
4a  1
3
1
3( a )
(b )
(c )
3
ba  2
3aba  b 
Exercise 3
3 6 15 30 3a 9b




1. 
5 10 25 50 5a 15b
Exercise 4
ba
13 x
1.
2.
ab
15
7.
12.
3.
2x
x
3 y  13
4z  2
8.
2 y  3 y  2
z z  2 
3x 3  x 2  x  1
x 2 x  1
13.
12  t
3
2.
2x
2x

2
x  2x  2 x  4
4.
9.
c cd
bc cc  b  3cd 4ca  2b 





b bd b 2 b(c  b ) 3bd 4ba  2b 
x
 x  2 2
14.
4y  4
y  y  2 2
Ax  1  Bx  2  A  B x  2 B  A
16.

x  2x  1
x  2x  1

8r  5
10.
r  12
15.
5.
11.
t  10
2
t 4
6.
11x
3x  4x  6
4s 2  4
s 2 s  2
9r 2  17 r  18
r  23r  2

A 4 x 2  1  B2 x  1  C 2 x  1 4 Ax 2  2 B  2C x  A  B  C
17.

2 x  12 x  1
2 x  12 x  1
18.
 Ax  B x  2  C
x2

Ax 2  B  2 Ax  2 B  C
x2
We would appreciate your comments on this worksheet, especially if
you’ve found any errors, so that we can improve it for future use. Please
contact the Maths tutor by email at studyadvice@hull.ac.uk
Updated 25th November 2004
The information in this leaflet can be made available in an
alternative format on request. Telephone 01482 466199
© 2009
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