Worksheet on ALGEBRAIC FRACTIONS (II)

(Multipliction n! "i#ision)
1 What Are Algebraic Fractions?
Algebraic fractions are fractions that include pronumerals (i.e. letters) as
well as numbers.
2 Multiplying and Dividing Algebraic Fractions
The rules used in multiplying and dividing algebraic fractions are exactly
the same as those used for purely numerical fractions. So lets review these
rules with some purely numerical examples.
Well begin with multiplication. ur goal is to find a rule that will allow us
to multiply fractions.
3 O! "eans "ultiply!#
!athematical problems begin with words" ne of the #obs of a
mathematician is to translate the words into mathematical symbols.
$n %nglish the word &of is often used to indicate multiplication.
' When we as( )what is * lots of +,- we mean &what is * multiplied by +,-.
' When we as( )what is a .uarter of /,- we mean )what is 0 multiplied
by /,-
$ Multiplying a raction and a %hole nu"ber
$n maths it is nearly always a good strategy to start with the simplest case.
So lets begin by multiplying a fraction and a whole number. We will then
move on to multiplying two fractions.
We will first wor( out the answers to some simple .uestions by using
&commonsense 1 that is2 by using our (nowledge of what the .uestions
mean. $n this way2 we hope to discover a pattern.
Gary Pocock 12/06/14 1
3emember that a whole number can be thought of as a fraction with a
denominator of 4.
4
5
5 =
2
4
*
* =
2
4
6
6 =
2 etc.
Examples
4. What is a half of 7, (r8 What is 7 lots of a half,)
Solution
Without any calculations2 we can use our (nowledge of what the .uestion
means to wor( out the answer.
$nterpretation $ (e.ual parts)
&What is a half of 7, means &$f 7 is divided into two e.ual parts2 what is the
si9e of each part, :learly2 the answer is *.
$nterpretation $$( repeated addition)
&What is 7 lots of a half, means &What is the total if 7 lots of one'half added
together, Again2 it is clear that the answer is *.
;ets state this conclusion in words8 one-half of 6 equals 3.
<ow lets translate into symbols8
* 7
5
4
=
5. What is a half of =, (r8 What is = lots of a half,)
Solution
Again2 using either interpretation $ or $$2 we can see the answer is * > .
$n words8 one-half of 7 equals 3 .
$n symbols8
5
4
* =
5
4
=
& Finding a pattern
At first glance2 the above multiplications dont seem to display much of a
pattern. ?ut if we re'write the answers as improper fractions a pattern
emerges.
*
5
7
4
7
5
4
= =
5
4
*
5
=
4
=
5
4
= =
Gary Pocock 12/06/14 2
These simple examples suggest a rule for multiplying fractions.
'ule 1( 'ule or "ultiplying t%o ractions
)o "ultiply t%o ractions* "ultiply the nu"erators together +to obtain
the nu"erator o the ans%er,* and "ultiply the deno"inators together
+to obtain the deno"inator o the ans%er,-
.n sy"bols(
bd
ac
d
c
b
a
=
/ )esting 'ule 1
We proposed this rule on the basis of only two examples. We need to test
the rule by seeing if it wor(s for a wider range of cases.
3ule 4 applies to both proper and improper fractions. ?ut it cannot be
applied to mixed numbers. Therefore2 it is important to remember8
When using 'ule 1 al%ays convert "i0ed nu"bers into
i"proper ractions-
Examples
4. What is > of 0 ,
Solution
:ommonsense reasoning tells us that the answer is 4@/. (Araw a number line if
you need to convince yourself.)
Aoes our rule give the correct answer,
/
4
6 5
4 4
6
4
5
4
=

=
Bes"
5. What is
*
4
of
5
4
4
,
Solution
Again2 commonsense reasoning tells us that the answer is > . (Araw a number
line and divide 4 > into three e.ual parts.)
Aoes our rule give the correct answer, (3emember we must convert 4 > into
Gary Pocock 12/06/14 3
an improper fraction.)
5
4
7
*
5 *
* 4
5
*
*
4
5
4
4
*
4
*
*
= =

= =

Bes"
*. What is
+
*
of
5
4
,
Solution :ommonsense reasoning tells us one'fifth of
5
4
e.uals
4C
4
. Dence
three!fifths of
5
4
must e.ual three lots of
4C
4
' that is2
4C
*
.
Aoes our rule give the correct answer,
4C
*
5 +
4 *
5
4
+
*
=

=
Bes"
$n fact2 3ule 4 al"a#s gives the correct answer for multiplying fractions 1
that is2 it always gives the same answer as that obtained by commonsense
reasoning.
This is very useful to (now because it means that we no longer have to
always resort to &commonsense reasoning (which can often ta(e .uite a
long time to wor( out) when multiplying fractions. $nstead2 all we have to
do is apply the rule.
1 20tending 'ule 1 to the "ultiplication o three +or "ore, ractions
3ule 4 is easily extended. Dere is the rule for multiplying three fractions
together. (The extension to four or more fractions follows the same pattern.)
'ule 1
3
( 'ule or "ultiplying three ractions
)o "ultiply three ractions* "ultiply the nu"erators together +to
obtain the nu"erator o the ans%er,* and "ultiply the deno"inators
together +to obtain the deno"inator o the ans%er,-
.n sy"bols(
bdf
ace
f
e
d
c
b
a
=
Gary Pocock 12/06/14 4
Examples
4. :alculate
=
4
+
*
5
4

Solution
=C
*
= + 5
4 * 4
=
4
+
*
5
4
=


=
5. :alculate
+
6
/
*
E
/

Solution
4+
6
*7C
E7
+ / E
6 * /
+
6
/
*
E
/
56
56
= =


=

4 5ancelling do%n! ractions

ften the phrase &cancelling do"n (or #ust &cancelling) is used to describe
this process of converting a fraction to its simplest form.
$ will try to avoid this phrase because it can be misleading. $t tends to ma(e
people thin( that there is some special mathematical operation called
&cancelling2 whereas in reality it is #ust a process involving ordinary
division.
Gary Pocock 12/06/14 5
?ut the phrase is widely used (and2 $ admit2 it can be useful")2 so you need
to be aware of it. ?ut remember8 cancelling down! 6ust "eans converting
to simplest form!-
7 8sing cancellation! to si"pliy the process o "ultiplying ractions
;oo( again at the last example of F=. <otice how the numbers that had to be
multiplied were .uite large. (Bou probably needed a calculator.) Dowever2
after simplifying2 the answer turned out to involve .uite small numbers.
The process used in F= involved two steps.
Step ( !ultiply the fractions.
Step * :onvert the answer to simplest form (&cancel down).
$t turns out that it is often simpler to re)erse these two steps 1 that is2 &cancel
down first2 and then multiply. Aoing things in this order means that the
numbers to be multiplied are as small as possible.
Examples
4. :alculate
+
6
/
*
E
/

Solution
+ / E
6 * /
+
6
/
*
E
/


=


+ / E
6 / *


=


+ / E
6 / *
4 *
4 4


=
4+
6
+ 4 *
6 4 4
=


=
Gary Pocock 12/06/14 6
(i) Combine the numbers into a single fraction.
(ii) Look for any numbers in the numerator that have a
common factor with a number in the denominator. Here,
3 and 9 have a common factor, as do and .
(iii) !earrange the numbers so that those with a common
factor are vertically aligned. ("his ste# is o#tional.)
(iv) $or each #air of numbers with a common factor, divide
through by that factor.
(v) %ulti#ly the remaining numbers using the rule for
fraction multi#lication.
(vi) Check you answer to see if there are any common factors
that you may have missed. (&n this e'am#le there are no
such factors.)
5. :alculate
6C
44
**
57
4*
4C

Solution
7
4
45
5
* 4 6
4 5 4
* * * 4 C 6
4 4 7 5 C 4
6C ** 4*
44 57 4C
6C
44
**
57
4*
4C
* 4 6
4 5 4
= =


=


=


=
19 Multiplying Algebraic Fractions With :ronu"erals
The method for multiplying fractions with pronumerals is exactl# the same as
that for numerical fractions.
Examples
4. Simplify
+
5
*

x x

Solution

4+
) 5 (
+ *
) 5 (
+
5
*

=

x x x x x x
5. Simplify
4 5
=
6 +

x
x
Solution
) 4 5 ( 6
=
4 5
=
6 +
=
+

x
x
x
x

The multiplication of algebraic fractions gets a bit more complicated (but not
that much") when common factors are involved. ?efore doing some examples2
lets revise the rules for multiplying and dividing with powers 1 i.e. the index
rules.
Gary Pocock 12/06/14 7
"here is no need to e'#and the
brackets. $actorised form is nearly
always best.
11 Revision: The Index Rules (aka Power Rules or Exponent Rules)
;For "ore detail see "y Wor<sheet on .ndices!-=
$ndices were invented to save time when writing certain arithmetic and
algebraic expressions. Gor example2 instead of writing + H + H + H +2 it is much
more convenient to +
6
. (We say )five to the power of 6-.)
<ote that +
6
does not e.ual + H 6 I 5C. 3ather it e.uals + H + H + H + I 75+.
The lower number (or letter) is referred as the base. The upper number is
called the po"er (or exponent or index).
+
6

There are two simple rules for multiplying and dividing numbers (or
expressions) written in index form.
)%o .nde0 'ules
'ule .1 +"ultiplication,( )o "ultiply t%o nu"bers +or e0pressions,
%ith the sa"e base* add the po%ers-
'ule .2 +division,( )o divide t%o nu"bers +or e0pressions, %ith the
sa"e base* subtract the po%ers-
Gary Pocock 12/06/14 8
base
Jower@exponent@index
Examples
4. Kse index rules to simplify (where possible) the following expressions.
(a) 5
5
H 5
+
(b) =
6
H =
*
H =
5
(c) +
5
H 5
*
(d) x
7
H x
6
(e) x
*
H y
+
Solutions
(a) 5
5
H 5
+
I 5
=
(b) =
6
H =
*
H =
5
I =
E
(c) +
5
H 5
*

cannot be simplified +base numbers aren,t the same-
(d) x
7
H x
6
I x
4C
(e) x
*
H y
+
I x
*
y
+
(cannot be simplified further as the bases aren,t the same-

When applying the index rule for division2 the resulting power will sometimes
be 9ero2 or a negative number. There are two more index rules that tell us what
such powers mean.
)%o More .nde0 'ules
'ule .3 +>ero po%er rule,( A nu"ber raised to po%er o >ero is e?ual to 1-
)hat is* a
9
@ 1- +20ception( 9
9
is undeined-,
'ule .$ +negative po%er rule,( a
An
@ 1Ba
n
+a C 9,
Gary Pocock 12/06/14 9
Examples
5. Kse index rules to simplify (where possible) the following expressions.
(a) 5
=
5
6
(b) +
6
H +
*
+
7
(c) 7
5
7
+
(d) x
5
H x
6
x
7
(e) x
*
y
+
Solutions
(a) 5
=
5
6
I 5
*
(b) +
6
H +
*
+
7
I +
4
I +
(c) 7
5
7
+

I 7
'*
I 4@7
*
(d) x
5
H x
6
x
7
I x
C
I 4
(e) x
*
y
+
I x
*
@

y
+
(cant be simplified further as the bases arent the same)

12 A Mathe"atical 5onvention For Writing Fractions With :ronu"erals
$n the next section we will apply the index rules to the multiplication of
fractions. ?ut first its important to explain a certain mathematical convention
for writing fractions with pronumerals.
A Mathe"atical 5onvention
When %riting a raction such as t%oAthirds o 0!* the 0! "ay be placed
either in the nu"erator

*
5x
* or on the side

x
*
5
- )hese are 6ust t%o
dierent %ays o e0pressing the sa"e thing-
Gary Pocock 12/06/14 10
This convention can cause confusion when the number in the numerator is 4.
Gor example2 the convention implies that
6
4
6
4 x
x =
Dowever2 when the coefficient of x is 4 we normally #ust write &x rather than
&4x. Thus we usually write

6 6
4 x
x =
Examples
4. Where possible2 write the following fractions in a different way using the
above convention.
(a)
x
=
4
(b)
+
*x
(c)
E
x

(d)
x
+
Solutions
(a)
= =
4 x
x =
(b)
x
x
+
*
+
*
=
(c)
x
x
E
4
E
=

(d)
x
+
(Cannot be written
differently as ' is in the
denominator.)
13 Applying .nde0 'ules to the Multiplication o Fractions
Gary Pocock 12/06/14 11
Examples
4. Simplify
4+
+ 5
*
x
x x

Solution
x or
x
x
x x
x x *
5
*
5
4+
4C
4+
+ 5
5
* *
= =
5. Simplify
.
x#
x
#.
#
x 4C
+
5
+
*
5
5


Solution

. x or
x
.
. # x
. # x
.
x#
x
#.
#
x
4
5 *
5 5 5
*
5
5
+
6
+
6
5+
5C 4C
+
5
+

= =
*. Simplify8
5 5
*
7
5
) 5 (
6
5
*
x
x
x
x
x
x

Solution
5
5
* 5
6
5 5
*
) 5 (
5
) 5 ( 7
) 5 ( 45
7
5
) 5 (
6
5
*

x
x
x x
x x
x
x
x
x
x
x
1$ )he :artial 5ancellation! 2rror
Gary Pocock 12/06/14 12
ne very common error made when wor(ing with fractions is what $ call the
&partial cancellation error. Dere is an example.
x
#
# x
5
5
=
+
TD$S $S W3<L""
Three .uestions can be as(ed about this error.
/uestion (8 What is the correct answer,
/uestion *8 What is wrong with the above,
/uestion 38 Why do people ma(e this error,

/uestion (8 What is the correct answer,
The correct answer is that the expression cannot be simplified any further"
M M
5
#
# x +
is as simple as the expression can get.
/uestion *8 What is wrong with the above,
o $n F/ it was emphasised that &cancellation really means &dividing
through by a common factor. <ow a factor is only &common if it
appears in e)er# term in the expression. This means that &y is not a
common factor of the above expression because it does not appear in
the &5x term.
o Deres another way of putting it. $f you are going to divide an
expression through by &y you must divide e)er# term by &y 1
including the &5x term. ;i(e this8
4
5
4
4
5
5
+ =
+
=
+

#
x #
x
#
# x
#
# #
TD$S $S :33%:T"
So
M M
5
#
# x +
could be written as
M M
4
5
+
#
x
. (This is not a simpler way
of writing the expression2 but it is a valid alternative.)
o ne final way of showing what is wrong with &partial cancellation is
to try the procedure on a purely numerical expression. (This is a very2
very useful techni.ue which should always be used whenever you are
in doubt about an algebra step.) :onsider the expression
5
5 6 +
. This
expression can obviously be simplified as follows8
*
5
7
5
5 6
= =
+
Gary Pocock 12/06/14 13
?ut now suppose we tried to simplify the expression using the &partial
cancellation procedure. We would get8
6
5
5 6
=
+
,,,
$t is clear that the procedure gives the wrong answer2 and so cannot be
valid.
/uestion 38 Why do people ma(e this error,
$n algebra2 there are rules for addition and rules for multiplication. The two
sets of rules are .uite similar2 but not identical. !any mista(es2 including this
one2 are made through getting the rules for addition and multiplication
confused.
:onsider the following8
x
#
# x
5
5
=
TD$S $S :33%:T"
$n this example there are only two terms2 and &y is common to both terms. So
dividing by &y gives &full cancellation rather than &partial cancellation.
Algebra always re.uires &full cancellation.
Examples
4. Simplify (if possible)8
+ 5
+ 6

x
x

Solution
There is no common factor. The expression cannot be simplified any further.
5. Simplify (if possible)8
5
/ 6 x
Solution
There is a common factor of 5. The expression can be simplified by dividing
every term by 5.
6 5
4
6 5
5
/ 6
5
/ 6
5
5 5
=

x
x x x

Gary Pocock 12/06/14 14
Examples
*. Simplify (if possible)8
#
x
x
x# 4 *
4 6
6 +

+

Solution
4 6
) 4 * ( 6
) 4 6 (
) 4 * ( 6 4 *
4 6
6
+
+
=
+
+
=
+

+ x
x x
x #
x # x
#
x
x
x#

1& Division .nvolving :urely Du"erical Fractions
<ow lets turn to the operation of division. Again2 we start with purely
numerical fractions.
Some terminology8
The goal is to find a rule for dividing fractions.
ur strategy is to calculate a few simple divisions by using our
commonsense understanding of what division means. We will then
examine our results in the hope of finding a pattern.
Dowever2 our &commonsense understanding of division is actually rather
complex. There are in fact three different ways of loo(ing at division.
They all give the same answer2 of course2 but sometimes it is easier to
view things one way rather than another. ;ets consider these three
interpretations in relation to division with whole numbers.
o .nterpretation . +Division is 6ust an alternative %ay o e0pressing
"ultiplication-,
%very statement involving division can be re'written as a statement
involving multiplication" (This is a very important point"")
e.g. 4C + I 5 is 0ust another "a# of sa#ing 4C I 5 H +.
Gary Pocock 12/06/14 15
0 y
dividend divisor
This means that the following two .uestions are equi)alent (i.e. they
are #ust two different ways of as(ing the same thing).
Aivision .uestion8 )What does 4C divided by + e.ual,-
!ultiplication .uestion8 )What number2 when multiplied by +2 gives 4C,-
o .nterpretation .. +e?ual parts,
Deres another way of thin(ing about 4C +8
)$f 4C is divided into + equal parts2 what is the si9e of each part,-
o .nterpretation ... +repeated subtraction,
Ginally2 4C + can viewed as as(ing the following8
)Dow many lots of + can be subtracted from 4C,-
We will now loo( at some examples of division involving fractions. We
will wor( out the answer using one of more of the above interpretations.
%ventually2 we hope to find a pattern"
As always2 we start with the simplest examples 1 i.e. cases involving
fractions divided by whole numbers.
Gary Pocock 12/06/14 16
Examples
4. What is a half divided by *,
Solution
$nterpretation $8
The e.uivalent multiplication .uestion is8
)What number2 when multiplied by *2 gives one'half.-
$nterpretation $$8
)$f one'half is divided into * e.ual parts2 what is the si9e of each part,-
($magine half a ca(e divided into * e.ually si9ed pieces.)
$nterpretation $$$8
(Aoesnt ma(e sense in this case because the dividend (>) is smaller than the
divisor (*).)
n the basis of interpretations $ and $$ it is clear that the answer is one'sixth.
$n words8 one-half di)ided b# three equals one-sixth.
$n symbols8
7
4
*
5
4
=
5. What is two'thirds divided by 5,
Solution
$nterpretation $8
The e.uivalent multiplication .uestion is8
)What number2 when multiplied by 52 gives two'thirds.-
$nterpretation $$8
)$f two'thirds is divided into 5 e.ual parts2 what is the si9e of each part,-
($magine two'thirds a ca(e divided into 5 e.ually si9ed pieces.)
$nterpretation $$$8
(Aoesnt ma(e sense in this case because the dividend (>) is smaller than the
divisor (*).)
n the basis of interpretations $ and $$ it is clear that the answer is one'third.
$n words8 t"o-thirds di)ided b# t"o equals one-third.
$n symbols8
*
4
5
*
5
=
<ow lets do some examples in which the di)isor is a fraction.
Gary Pocock 12/06/14 17
Examples
*. What is * divided by 4@5,
Solution
$nterpretation $8
The e.uivalent multiplication .uestion is8
)What number2 when multiplied by >2 gives *.-
$nterpretation $$8
(Aoesnt ma(e sense in this case because the divisor is a fraction.)
$nterpretation $$$8
Dow many lots of one'half can be subtracted from *,
n the basis of interpretations $ and $$$ it is clear that the answer is 7.
$n words8 1hree di)ided b# one-half equals six.
$n symbols8
7
5
4
* =
6. What is 4@* divided by >,
Solution
$nterpretation $8
The e.uivalent multiplication .uestion is8
)What number2 when multiplied by >2 gives 4@*,-
$nterpretation $$8
(Aoesnt ma(e sense in this case because the divisor is a fraction.)
$nterpretation $$$8
(Aoesnt ma(e sense because the dividend is smaller than the divisor.)
n the basis of interpretation $ it is clear that the answer is 5@*.
(Ao the multiplication yourself")
$n words8 2ne-third di)ided b# one-half equals t"o-thirds.
$n symbols8
*
5
5
4
*
4
=
1/ Finding a pattern
Gary Pocock 12/06/14 18
:an you see a pattern, ;ets write out the above divisions again2 but this
time well write the whole numbers as fractions with a denominator of 4. $f
you focus on the diagonals you should notice a pattern.
7
4
4
*
5
4
=
5
4

4
*

7
4
*
4
7
5
4
5
*
5

= =
7
4
7
5
4
4
*
= =
*
5
5
4
*
4
=
These examples suggest that weve found a rule for dividing fractions. All
one has to do is &multiply diagonally.
Dowever2 most maths boo(s express the rule in a slightly different way.
($ts still the same ruleN its #ust put differently.) Deres how the rule is
usually expressed.
'ule 2( 'ule or dividing ractions +invert and "ultiply! rule,
)o divide one raction by another raction do the ollo%ing(
Etep 1( .nvert the divisor +that is* turn the 2
nd
raction upside do%n,
Etep 2( multiply#
.n sy"bols(
bc
ad
c
d
b
a
d
c
b
a
= =
11 )esting 'ule 2
We proposed this rule on the basis of only four examples. We need to test
the rule by seeing if it wor(s for a wider range of cases.
Gary Pocock 12/06/14 19
3ule 5 applies to both proper and improper fractions. ?ut it cannot be
applied to mixed numbers. Therefore2 it is important to remember8
When using 'ule 2 al%ays convert "i0ed nu"bers into
i"proper ractions-
Examples
4. What is 5@+ divided by 4@4C ,
Solution
$nterpretation $8
The e.uivalent multiplication .uestion is8
)What number2 when multiplied by 4@4C2 gives 5@+,-
$nterpretation $$8
(Aoesnt ma(e sense in this case because the divisor is a fraction.)
$nterpretation $$$8
Dow many lots of 4@4C can be subtracted from 5@+,
$f you thin( about these two .uestions carefully you will see that the answer
is 6. (Dint8 5@+ I 6@4C)
Aoes our rule give the correct answer,
6
+
5C
4 +
4C 5
4
4C
+
5
4C
4
+
5
= =

= =
Bes"
5. What is
*
4
divided
5
4
4
,
Solution
$nterpretation $8
The e.uivalent multiplication .uestion is8
)What number2 when multiplied by
5
4
4
2 gives
*
4
,-
(The other two interpretations dont ma(e sense.)
This is a harder .uestion to answer. ?ut after some trial and error you should
find that the answer is
E
5
.
Aoes our rule give the correct answer, (3emember we must convert 4 > into
an improper fraction.)
E
5
* *
5 4
*
5
*
4
5
*
*
4
5
4
4
*
4
=

= = =
Bes"
Gary Pocock 12/06/14 20
$n fact2 3ule 5 al"a#s gives the correct answer for dividing fractions 1 that
is2 it always gives the same answer as that obtained by commonsense
reasoning.
This is very useful to (now because it means that we no longer have to
always resort to &commonsense reasoning (which can often ta(e .uite a
long time to wor( out) when dividing fractions. $nstead2 all we have to do is
apply the rule.
14 Dividing Algebraic Fractions With :ronu"erals
The method for dividing fractions with pronumerals is exactl# the same as that
for numerical fractions.
As was the case with multiplying fractions2 it is important to remember the
index rules and factorisation rules.
Examples
4. Simplify
+
5
*

x x

Solution

) 5 ( *
+
5
+
* +
5
*
=

x
x
x
x x x
5. Simplify
4 5
=
6 +

x
x
Solution
5/
) 4 5 (
=
4 5
6 4 5
=
6
+
=
+
=
+

x x x x
x
x

17 Applying .nde0 'ules to the Division o Fractions
Gary Pocock 12/06/14 21
"here is no need to e'#and the
brackets. $actorised form is nearly
always best.
Examples
4. Simplify
4+ +
5
*
x
x

Solution

6
6 6 *
*
7
7
+
*C 4+
+
5
4+ +
5

= = = x or
x x x x
x
x
5. Simplify

.
x#
.
# x
x#
. 6
+
/
+
*
5
5
5

Solution
(3emember8 always do the operation inside the brac(ets first")

5
5 5
5
5 5
5
*
5
5
5
*
5
5
5
*
5
5
5
5+
5
5+
5
+
5
+
5C
/
+
6 +
/
+
6
+
/
+
#
. x#
x.
.
x
x#
.
x#.
#. x
x#
.
x#
.
.
# x
x#
.
.
x#
.
# x
x#
.
=
=
=
=

29 )he Elash! Ey"bol or division
Sometimes you will find the slash symbol (@)2 or a fraction line2
4
being
used to indicate division. The safest way to deal with this is to first
convert the slash or fraction line to the familiar division symbol ()
and than apply 3ule 5.
4
Deres a bit of trivia for you. The correct term for the fraction line is the )inculum.
Gary Pocock 12/06/14 22
$f there is more than one fraction line2 it is the longer line that indicates
division.
Examples
4. %valuate
*
+
5
Solution

4+
5
*
4
+
5
4
*
+
5
*
+
5
= = =
5. %valuate
*
5
7
Solution

E
4
E
5 4
* 7
5
*
4
7
*
5
4
7
*
5
7
4
*
= =

= = =

*. %valuate
5
4
4
6
*
Solution
5
4
45
7
*
5
6
*
5
*
6
*
5
4
4
6
*
5
4
4
6
*
= = = = =

21 Fuestions
4. Simplify the following fractions.
(a)
6
*
4
=
7
*
5

Gary Pocock 12/06/14 23
(b)

7
4
4C
E
+
*
(c)
=
*
*
*
4
5
(d)
/
+
*
+

(e)
5
5
/
*
6 #
x.
#
x

(f)
x
x
+
5
*

(g)
5
=
6#
(h)
#
x
+
+
(i)
x#
# x
6
5 4

(#)
x
x
*
6
4
(()
6 +
5 5
5
x
x x
x

+
(l)
a
bc
c ab
*
5
5
5

(m)

/
4
5
4
5
# #
(n)
*
+ 7 5
+

+
x x
x
(o)
b a
b a
5 5
4
*
5 5
+

22 Ans%ers
1-
(a) 4
5
5
4 4 5
4 5 4
= * 6
= 7 5
6 = *
= 7 5
6
=
=
7
*
5
6
*
4
=
7
*
5
4 4 5
4 5 4
= =


=


=


= =
Gary Pocock 12/06/14 24
(b)
6
*
45
+ E
C 7 *
E
7C
+
*
7C
E
+
*
7 4C
4 E
+
*
7
4
4C
E
+
*
4 *
45 4
= =

= = =

(c) /
* =
6 5 =
=
56
*
=
=
*
*
*
4
5
4 4
/ 4
=

= =
(d)
*
5
5
*
/
* +
/ +
+
/
*
+
/
+
*
+
= =

= =
(e)
.
x#
x#.
# x
x.
#
#
x
#
x.
#
x
*
5
45
/
*
/
6 /
*
6
5 5 5 5
5
5
= = =
(f)
4+
5
+
5
* +
5
*
5
x x x
x
x
= =
(g)
=
5
46
6
5
4
=
6
5
=
6
5
=
6
# # # #
#
= = = =
(h)
x
#
x
# #
x # x
#
x
= = = =
+
+
+
+ + +
+
+
(i)
5 5 5 5
5
4
6
5
6
4 5
6
5
6
5 4
6
5 4
# x # x x# x#
x#
x#
x#
# x
x#
# x
= = = =

(#)
45
4
45 *
4
6
*
6
*
6
4
= = = =
x
x
x
x
x
x
x
x
(()
) + ( 5
) 5 (
) + ( 6
) 5 ( 5
6 +
5 5
5 5
x
x x
x x
x x x
x x
x

+
=

+
=

+
(l)
c
b a
bc
c b a
bc
a c ab
a
bc
c ab
5
5
5 5
5
5 5
5
7 7 *
4
5
*
5 = = =
(m)
#
#
#
# #
# # # # # # #
6
4
6
5
/
4
/
5
4
/
4
5
4
/
4 4
5
4
/
4
5
4
5 5
5 5 5
= = = = =

Gary Pocock 12/06/14 25
(n)
x x x
x
x x
x
x x
x 4C
) * (
) * ( 4C
) * (
+ ) * ( 5
*
+ 7 5
=
+
+
=
+
+
=
+

+
(o)
7 ) ( 7
) )( (
) ( 5 *
) )( (
5 5
4
*
5 5
b a
b a
b a b a
b a
b a b a
b a
b a
=
+
+
=
+
+
=
+

Gary Pocock 12/06/14 26