Upload
hoangdan
View
223
Download
2
Embed Size (px)
Citation preview
Algebraic fractions
12Chapter
Contents: A Simplifying algebraic fractions
B Multiplying and dividing algebraicfractions
C Adding and subtracting algebraicfractions
D More complicated fractions
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\295IB_10P-2_12.CDR Monday, 4 February 2008 9:15:33 AM PETERDELL
296 ALGEBRAIC FRACTIONS (Chapter 12)
Fractions which involve unknowns are called algebraic fractions.
Algebraic fractions occur in many areas of mathematics. We have already seen them in
problems involving similar triangles.
CANCELLATION
We have observed previously that number fractions can be simplified by cancelling common
factors.
For example, 1228 = 4£3
4£7 = 37 where the common factor 4 was cancelled.
The same principle can be applied to algebraic fractions:
If the numerator and denominator of an algebraic fraction are both written in factored form
and common factors are found, we can simplify by cancelling the common factors.
For example,4ab
2a=
2 £ 2 £ a£ b
2 £ affully factorisedg
=2b
1fafter cancellationg
= 2b
For algebraic fractions, check both numerator
and denominator to see if they can be expressed
as the product of factors, then look for common
factors which can be cancelled.
ILLEGAL CANCELLATION
Take care with fractions such asa + 3
3:
The expression in the numerator, a + 3, cannot be written as the product of factors other
than 1 £ (a + 3): a and 3 are terms of the expression, not factors.
A typical error in illegal cancellation is:a + 3
3=
a + 1
1= a + 1.
You can check that this cancellation of terms is incorrect by substituting a value for a.
For example, if a = 3, LHS =a + 3
3=
3 + 3
3= 2,
whereas RHS = a + 1 = 4.
SIMPLIFYING ALGEBRAIC FRACTIONSA
When cancelling inalgebraic fractions,only factors can be
cancelled, not terms.
1 1
11
1
1
1
1
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\296IB_10P-2_12.CDR Tuesday, 5 February 2008 9:35:35 AM PETERDELL
ALGEBRAIC FRACTIONS (Chapter 12) 297
Simplify: a2x2
4xb
6xy
3x3c
x + y
x
a2x2
4x
=2 £ x£ x
4 £ x
=x
2
b6xy
3x3
=6 £ x£ y
3 £ x£ x£ x
=2y
x2
cx + y
xcannot be simplified
as x+ y is a sum,
not a product.
Simplify: a(x + 3)(x¡ 2)
4(x + 3)b
2(x + 3)2
x + 3
a(x + 3)(x¡ 2)
4(x + 3)
=x¡ 2
4
b2(x + 3)2
x + 3
=2(x + 3)(x + 3)
(x + 3)
= 2(x + 3)
EXERCISE 12A.1
1 Simplify if possible:
a6a
3b
10b
5c
3
6xd
8t
te
t + 2
t
f8a2
4ag
2b
4b2h
2x2
x2i
4a
12a3j
4x2
8x
kt2 + 8
tl
a2b
ab2m
a + b
a¡ cn
15x2y3
3xy4o
8abc2
4bc
p(2a)2
aq
(2a)2
4a2r
(3a2)2
3as
(3a2)2
9a2t
(3a2)2
18a3
2 Split the following expressions into two parts and simplify if possible.
For example,x + 9
x=
x
x+
9
x= 1 +
9
x:
ax + 3
3b
4a + 1
2c
a + b
cd
a + 2b
b
e2a + 4
2f
3a + 6b
3g
4m + 8n
4h
4m + 8n
2m
3 Which of the expressions in 2 produced a simplified answer and which did not? Explain
why this is so.
Example 2 Self Tutor
Example 1 Self Tutor
In these examplesis the
common factor.( + 3)x
1
2
1
1
2 1
11
1
11
1
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\297IB_10P-2_12.CDR Monday, 4 February 2008 9:26:40 AM PETERDELL
298 ALGEBRAIC FRACTIONS (Chapter 12)
4 Simplify:
a3(x + 2)
3b
4(x¡ 1)
2c
7(b + 2)
14
d2(n + 5)
12e
10
5(x + 2)f
15
5(3 ¡ a)
g6(x + 2)
(x + 2)h
x¡ 4
2(x¡ 4)i
2(x + 2)
x(x + 2)
jx(x¡ 5)2
3(x¡ 5)k
(x + 2)(x + 3)
2(x + 2)2l
(x + 2)(x + 5)
5(x + 5)
m(x + 2)(x¡ 1)
(x¡ 1)(x + 3)n
(x + 5)(2x¡ 1)
3(2x¡ 1)o
(x + 6)2
3(x + 6)
px2(x + 2)
x(x + 2)(x¡ 1)q
(x + 2)2(x + 1)
4(x + 2)r
(x + 2)2(x¡ 1)2
(x¡ 1)2x2
FACTORISATION AND SIMPLIFICATION
It is often necessary to factorise either the numerator or denominator before simplification
can be done.
Simplify: a4a + 8
4b
3
3a¡ 6b
a4a + 8
4
=4(a + 2)
4
=(a + 2)
1
= a + 2
b3
3a¡ 6b
=3
3(a¡ 2b)
=1
a¡ 2b
Simplify: aab¡ ac
b¡ cb
2x2 ¡ 4x
4x¡ 8
aab¡ ac
b¡ c
=a(b¡ c)
b¡ c
=a
1
= a
b2x2 ¡ 4x
4x¡ 8
=2x(x¡ 2)
4(x¡ 2)
=x
2
Example 4 Self Tutor
Example 3 Self Tutor
1
1
1
1
1
1
1
1
1
2
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\298IB_10P-2_12.CDR Monday, 4 February 2008 9:29:05 AM PETERDELL
ALGEBRAIC FRACTIONS (Chapter 12) 299
Simplify: a3a¡ 3b
b¡ ab
ab2 ¡ ab
1 ¡ b
a3a¡ 3b
b¡ a
=3(a¡ b)
¡1(a¡ b)
= ¡3
bab2 ¡ ab
1 ¡ b
=ab(b¡ 1)
¡1(b¡ 1)
= ¡ab
Simplify:x2 ¡ x¡ 6
x2 ¡ 4x + 3
x2 ¡ x¡ 6
x2 ¡ 4x + 3
=(x + 2)(x¡ 3)
(x¡ 1)(x¡ 3)
=x + 2
x¡ 1
EXERCISE 12A.2
1 Simplify by cancelling common factors:
a6
2(x + 2)b
2x + 6
2c
3x + 12
3d
3x + 6
6
e5x + 20
10f
3a + 12
9g
xy + xz
xh
xy + xz
z + y
iab + bc
ab¡ bcj
3x¡ 12
6(x¡ 4)2k
(x + 3)2
6x + 18l
2(x¡ y)2
6(x¡ y)
2 Simplify:
a4x + 8
2x + 4b
mx + nx
2xc
mx + nx
m + nd
x + y
mx + my
e2x + 4
2f
x2 + 2x
xg
x2 + 2x
x + 2h
x
bx + cx
i3x2 + 6x
x + 2j
2x2 + 6x
2xk
2x2 + 6x
x + 3l
ax2 + bx
ax + b
Example 6 Self Tutor
Example 5 Self Tutor
1
Don’t forget to expandyour factorisations to
check them.
1 1
1
1
1
b¡ a = ¡1(a¡ b)is a useful rule for converting into .It can sometimes allow us to cancel common factors.
b a a b� � � �¡ ¡
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\299IB_10P-2_12.CDR Tuesday, 5 February 2008 9:38:42 AM PETERDELL
300 ALGEBRAIC FRACTIONS (Chapter 12)
3 Simplify, if possible:
a2a¡ 2b
b¡ ab
3a¡ 3b
6b¡ 6ac
a¡ b
b¡ ad
a + b
a¡ b
ex¡ 2y
4y ¡ 2xf
3m¡ 6n
2n¡mg
3x¡ 3
x¡ x2h
xy2 ¡ xy
3 ¡ 3y
i6x2 ¡ 3x
1 ¡ 2xj
4x + 6
4k
12x¡ 6
2x¡ x2l
x2 ¡ 4
x¡ 2
mx2 ¡ 4
x + 2n
x2 ¡ 4
2 ¡ xo
x + 3
x2 ¡ 9p
m2 ¡ n2
m + n
qm2 ¡ n2
n¡mr
3x + 6
4 ¡ x2s
16 ¡ x2
x2 ¡ 4xt
x2 ¡ 4
4 ¡ x2
u5x2 ¡ 5y2
10xy ¡ 10y2v
2d2 ¡ 2a2
a2 ¡ adw
4x2 ¡ 8x
x2 ¡ 4x
3x2 ¡ 6x
4 ¡ x2
4 Simplify:
ax2 ¡ x
x2 ¡ 1b
x2 + 2x + 1
x2 + 3x + 2c
x2 ¡ 4x + 4
2x2 ¡ 4x
dx2 + 4x + 3
x2 + 5x + 4e
x2 ¡ 4
x2 ¡ 3x¡ 10f
x2 + 7x + 12
2x2 + 6x
gx2 + 4x¡ 5
2x2 + 6x¡ 20h
x2 + 6x + 9
x2 + 3xi
2x2 ¡ 7x¡ 4
x2 ¡ 2x¡ 8
j3x2 + 5x¡ 2
3x2 ¡ 4x + 1k
2x2 ¡ 3x¡ 20
x2 ¡ x¡ 12l
8x2 + 14x + 3
2x2 ¡ x¡ 6
The rules for multiplying and dividing algebraic fractions are identical to those used with
numerical fractions. These are:
To multiply two or more fractions, we multiply
the numerators to form the new numerator, and
we multiply the denominators to form the new
denominator.
To divide by a fraction we multiply by its reciprocal.
a
b£ c
d=
a£ c
b £ d=
ac
bd
a
b¥ c
d=
a
b£ d
c=
ad
bc
MULTIPLICATION
Step 1: Multiply numerators and multiply denominators.
Step 2: Separate the factors.
Step 3: Cancel any common factors.
Step 4: Write in simplest form.
MULTIPLYING AND DIVIDINGALGEBRAIC FRACTIONS
B
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\300IB_10P-2_12.CDR Monday, 4 February 2008 9:31:33 AM PETERDELL
ALGEBRAIC FRACTIONS (Chapter 12) 301
For example,n2
3£ 6
n=
n2 £ 6
3 £ n
=n£ n£ 2 £ 3
3 £ n
=2n
1
= 2n
Simplify: a4
d£ d
8b
5
g£ g3
a4
d£ d
8=
4 £ d
d£ 8
= 12
b5
g£ g3 =
5
g£ g3
1
=5 £ g £ g £ g
g
=5g2
1
= 5g2
DIVISIONFor example,m
2¥ n
6=
m
2£ 6
n
=m£ 6
2 £ n
=m£ 6
2 £ n
=3m
n
Step 1: To divide by a fraction, multiply by
its reciprocal.
Step 2: Multiply numerators and multiply
denominators.
Step 3: Cancel any common factors.
Step 4: Write in simplest form.
Simplify: a6
x¥ 2
x2b
8
p¥ 2
a6
x¥ 2
x2=
6
x£ x2
2
=3 £ 2 £ x£ x
x£ 2
= 3x
b8
p¥ 2 =
8
p£ 1
2
=8 £ 1
p£ 2
=4
p
Example 8 Self Tutor
Example 7 Self Tutor
1
2
1
1
1
1
1 1
11
1
4
1 1
11
fStep 1g
fSteps 2 and 3g
fStep 4g
1
3
fStep 1g
fStep 2g
fStep 3g
fStep 4g
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\301IB_10P-2_12.CDR Tuesday, 5 February 2008 10:05:12 AM PETERDELL
302 ALGEBRAIC FRACTIONS (Chapter 12)
EXERCISE 12B
1 Simplify:
aa
2£ b
3b
x
4£ 2
xc
c
4£ 2
cd
a
2£ a
3
ea
b£ x
yf
x
y£ y
xg
x
3£ x h
x
4£ 8
y
in
2£ 1
n2j
6
p£ p
2k
m
x£ x
nl x£ 2
x
m5
t£ t2 n
µx
y
¶2
o
µ4
d
¶2
pa
b£ b
c£ c
a
2 Simplify:
ax
3¥ x
2b
3
y¥ 6
yc 3 ¥ 1
xd 6 ¥ 2
y
e3
p¥ 1
pf
c
n¥ n g d¥ 5
dh x¥ x
3
i 1 ¥ a
bj
3
d¥ 2 k
4
x¥ x2
2l
4
x¥ 8
x2
m a¥ a2
3n
x
y¥ x2
yo
5
a¥ a
2p
a2
5¥ a
3
The rules for addition and subtraction of algebraic fractions are identical to those used with
numerical fractions.
To add two or more fractions we obtain the
lowest common denominator and then add the
resulting numerators.
a
c+
b
c=
a + b
c
To subtract two or more fractions we obtain
the lowest common denominator and then
subtract the resulting numerators.
a
c¡ d
c=
a ¡ d
c
To find the lowest common denominator, we look for the lowest common multiple of the
denominators.
For example, when adding 34 + 2
3 , the lowest common denominator is 12,
when adding 23 + 1
6 , the lowest common denominator is 6.
The same method is used when there are variables in the denominator.
ADDING AND SUBTRACTINGALGEBRAIC FRACTIONS
C
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\302IB_10P-2_12.CDR Monday, 4 February 2008 9:41:34 AM PETERDELL
ALGEBRAIC FRACTIONS (Chapter 12) 303
For example, when adding4
x+
5
y, the lowest common denominator is xy,
when adding4
x+
3
2x, the lowest common denominator is 2x,
when adding1
3a+
2
5b, the lowest common denominator is 15ab.
To findx
2+
3x
5we find the LCD and then proceed in the same manner as for ordinary
fractions.
The LCM of 2 and 5 is 10, so the LCD is 10.
)x
2+
3x
5=
x£ 5
2 £ 5+
3x£ 2
5 £ 2
=5x
10+
6x
10
=11x
10
Simplify: ax
3+
5x
6b
3b
4¡ 2b
3
ax
+5x
6
=7x
6
b3b ¡ 2b
=12
Simplify: a2
a+
3
cb
7
x¡ 5
2x
a2
+3
=2c
ac+
3a
ac
=2c + 3a
ac
b7 ¡ 5
2
=14
2x¡ 5
2x
=9
2x
Example 10 Self Tutor
Example 9 Self Tutor
3 6
=x£ 2
3 £ 2+
5x
6fLCD = 6g
=2x
6+
5x
6
=2x + 5x
4 3
=3b£ 3
4 £ 3¡ 2b£ 4
3 £ 4fLCD = 12g
=9b
12¡ 8b
12
b
a c
=2 £ c
a£ c+
3 £ a
c£ afLCD = acg
x x
=7 £ 2
x£ 2¡ 5
2xfLCD = 2xg
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\303IB_10P-2_12.CDR Tuesday, 5 February 2008 9:40:58 AM PETERDELL
304 ALGEBRAIC FRACTIONS (Chapter 12)
Simplify: ab
3+ 1 b
a
4¡ a
=3
=4¡
4
=¡3a
4or ¡ 3a
4
EXERCISE 12C
1 Simplify by writing as a single fraction:
ax
2+
x
5b
x
3¡ x
6c
x
4+
3x
5d
x
2¡ x
5
e2t
3¡ 7t
12f
11n
21¡ n
7g
a
2+
a
3h
a
2+
b
4
in
3+
2n
15j
5g
6¡ g
3k
4s
5¡ 2s
3l a¡ 3a
5
mx
3+
x
2+
x
6n
y
2+
y
4¡ y
3o
z
4+
z
6¡ z
3p 2q ¡ q
3+
2q
7
2 Simplify:
a3
a+
2
bb
4
a+
3
dc
5
a¡ 3
bd
3a
m¡ 2a
m
ea
y+
b
3yf
4
a¡ 5
2ag
3
a¡ 2
abh
c
a+
b
d
i4
b+
a
bj
2
a¡ c
dk
5
x+
x
3l
p
6¡ 2
d
mm
3+
n
mn
2m
p¡ m
no
3b
5+
b
4p
5b
3¡ 3b
5
3 Simplify:
ax
3+ 2 b
m
2¡ 1 c
a
3+ a d
b
5¡ 2
ex
6¡ 3 f 3 +
x
4g 5 ¡ x
6h 2 +
3
x
i 6 ¡ 3
xj b +
3
bk
5
x+ x l
y
6¡ 2y
4 Simplify:
ax
3+
3x
5b
3x
5¡ 2x
7c
5
a+
1
2ad
6
y¡ 3
4y
e3
b+
4
cf
5
4a¡ 6
bg
x
10+ 3 h 4 ¡ x
3
Example 11 Self Tutor
ab
3+ 1 =
b
3+
3
3b + 3
ba
4¡ a =
a
4¡ a£ 4
1 £ 4a 4a
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\304IB_10P-2_12.CDR Tuesday, 12 February 2008 10:30:20 AM PETERDELL
ALGEBRAIC FRACTIONS (Chapter 12) 305
Addition and subtraction of more complicated algebraic fractions can be made relatively
For example:x + 2
3+
5 ¡ 2x
2=
2
2
µx + 2
3
¶+
3
3
µ5 ¡ 2x
2
¶fachieves LCD of 6g
=2(x + 2)
6+
3(5 ¡ 2x)
6fsimplify each fractiong
We can then write the expression as a single fraction and simplify the numerator.
Write as a single fraction: ax
12+
x¡ 1
4b
x¡ 1
3¡ x + 2
7
ax
12+
x¡ 1
4
=x
12+
3
3
µx¡ 1
4
¶=
x + 3(x¡ 1)
12
=x + 3x¡ 3
12
=4x¡ 3
12
bx¡ 1
3¡ x + 2
7
=7
7
µx¡ 1
3
¶¡ 3
3
µx + 2
7
¶=
7(x¡ 1)
21¡ 3(x + 2)
21
=7(x¡ 1) ¡ 3(x + 2)
21
=7x¡ 7 ¡ 3x¡ 6
21
=4x¡ 13
21
Write as a single fraction: a2
x+
1
x + 2b
5
x + 2¡ 1
x¡ 1
a2
x+
1
x + 2
=2
x
µx + 2
x + 2
¶+
µ1
x + 2
¶x
x
fLCD = x(x + 2)g
=2(x + 2) + x
x(x + 2)
=2x + 4 + x
x(x + 2)
=3x + 4
x(x + 2)
b5
x + 2¡ 1
x¡ 1
=
µ5
x + 2
¶µx¡ 1
x¡ 1
¶¡µ
1
x¡ 1
¶µx + 2
x + 2
¶fLCD = (x + 2)(x¡ 1)g
=5(x¡ 1) ¡ 1(x + 2)
(x + 2)(x¡ 1)
=5x¡ 5 ¡ x¡ 2
(x + 2)(x¡ 1)
=4x¡ 7
(x + 2)(x¡ 1)
MORE COMPLICATED FRACTIONSD
Example 13 Self Tutor
Example 12 Self Tutor
straightforward if we adopt a consistent approach.
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\305IB_10P-2_12.CDR Tuesday, 12 February 2008 10:30:45 AM PETERDELL
306 ALGEBRAIC FRACTIONS (Chapter 12)
EXERCISE 12D.1
1 Write as a single fraction:
ax
4+
x¡ 1
5b
2x + 5
3+
x
6c
x
7+
2x¡ 1
6
da + b
2+
b¡ a
3e
x¡ 1
4+
2x¡ 1
5f
x + 1
2+
2 ¡ x
7
gx
5¡ x¡ 3
6h
x¡ 1
6¡ x
7i
x
10¡ 2x¡ 1
5
jx
6¡ 1 ¡ x
12k
x¡ 1
3¡ x¡ 2
5l
2x + 1
3¡ 1 ¡ 3x
8
2 Write as a single fraction:
a2
x + 1+
3
x¡ 2b
5
x + 1+
7
x + 2c
5
x¡ 1¡ 4
x + 2
d2
x + 2¡ 4
2x + 1e
3
x¡ 1+
4
x + 4f
7
1 ¡ x¡ 8
x + 2
g1
x + 1+
3
xh
5
x¡ 2
x + 3i
x
x + 2+
3
x¡ 4
j 2 +4
x¡ 3k
3x
x + 2¡ 1 l
x
x + 3+
x¡ 1
x + 2
m2
x(x + 1)+
1
x + 1n
1
x¡ 1¡ 1
x+
1
x + 1
o2
x + 1¡ 1
x¡ 1+
3
x + 2p
x
x¡ 1¡ 1
x+
x
x + 1
PROPERTIES OF ALGEBRAIC FRACTIONS
Writing expressions as a single fraction can help us to find when the expression is zero.
However, we need to be careful when we cancel common factors, as we can sometimes lose
values when an expression is undefined.
Write as a single fraction: a3
(x + 2)(x¡ 1)+
x
x¡ 1b
¡3
(x + 2)(x¡ 1)+
x
x¡ 1
a3
(x + 2)(x¡ 1)+
x
x¡ 1
=3
(x + 2)(x¡ 1)+
µx
x¡ 1
¶µx + 2
x + 2
¶fLCD = (x + 2)(x¡ 1)g
=3 + x(x + 2)
(x + 2)(x¡ 1)
=x2 + 2x + 3
(x + 2)(x¡ 1)which we cannot simplify further.
Example 14 Self Tutor
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\306IB_10P-2_12.CDR Monday, 4 February 2008 9:50:00 AM PETERDELL
REVIEW SET 12A
ALGEBRAIC FRACTIONS (Chapter 12) 307
b¡3
(x + 2)(x¡ 1)+
x
x¡ 1
=¡3
(x + 2)(x¡ 1)+
µx
x¡ 1
¶µx + 2
x + 2
¶fLCD = (x + 2)(x¡ 1)g
=¡3 + x(x + 2)
(x + 2)(x¡ 1)
=x2 + 2x¡ 3
(x + 2)(x¡ 1)
=(x + 3)(x¡ 1)
(x + 2)(x¡ 1)
=x + 3
x + 2
EXERCISE 12D.2
1 Write as a single fraction:
a2
x(x + 1)+
1
x + 1b
2
x(x + 1)+
x
x + 1c
2x
x¡ 3+
4
(x + 2)(x¡ 3)
d2x
x¡ 3¡ 30
(x + 2)(x¡ 3)e
3
(x¡ 2)(x + 3)+
x
x + 3f
x
x + 3¡ 15
(x¡ 2)(x + 3)
g2x
x + 4¡ 40
(x¡ 1)(x + 4)h
x + 5
x¡ 2¡ 63
(x¡ 2)(x + 7)
2 a Write2
(x + 2)(x¡ 3)+
2x
x¡ 3as a single fraction.
b Hence, find x when2
(x + 2)(x¡ 3)+
2x
x¡ 3is i undefined ii zero.
3 Simplify: a
xx¡2 ¡ 3
x¡ 3b
3xx+4 ¡ 1
x¡ 2c
x2
x+2 ¡ 1
x + 1
d
x2
2¡x+ 9
x¡ 3e
1x2 ¡ 1
4
x¡ 2f
x¡3x2 ¡ 1
16
x¡ 4
1 Simplify:
a6x2
2xb 6 £ n
2c
x
2¥ 3 d
8x
(2x)2
2 Simplify, if possible:
a8
4(c + 3)b
3x + 8
4c
4x + 8
4d
x(x + 1)
3(x + 1)(x + 2)
3 Write as a single fraction:
a2x
3+
3x
5b
2x
3£ 3x
5c
2x
3¥ 3x
5d
2x
3¡ 3x
5
The expression is zerowhen . The
expression is undefinedwhen and alsowhen We can
see this from theoriginal expression.
x
x
x :
� �
� �� �
= 3
= 2=1
¡
¡
1
1
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\307IB_10P-2_12.CDR Tuesday, 12 February 2008 10:31:30 AM PETERDELL
REVIEW SET 12B
308 ALGEBRAIC FRACTIONS (Chapter 12)
4 Simplify by factorisation:
a4x + 8
x + 2b
5 ¡ 10x
2x¡ 1c
4x2 + 6x
2x + 3
5 Write as a single fraction:
ax + 3
4+
2x¡ 2
3b
x¡ 1
7¡ 1 ¡ 2x
2c
2
x + 2+
1
x
6 Simplify by factorisation:
a8 ¡ 2x
x2 ¡ 16b
x2 + 7x + 12
x2 + 4xc
2x2 ¡ 3x¡ 2
3x2 ¡ 4x¡ 4
7 a Write3x
x¡ 4¡ 60
(x + 1)(x¡ 4)as a single fraction.
b Hence, find x when3x
x¡ 4¡ 60
(x + 1)(x¡ 4)is i undefined ii zero.
1 Simplify:
a4a
6ab
x
3£ 6 c 3 ¥ 1
nd
12x2
6x
2 Simplify, if possible:
a3x + 15
5b
3x + 15
3c
2(a + 4)
(a + 4)2d
abc
2ac(b¡ a)
3 Write as a single fraction:
a3x
4+ 2x b
3x
4¡ 2x c
3x
4£ 2x d
3x
4¥ 2x
4 Simplify by factorisation:
a3 ¡ x
x¡ 3b
5x + 10
2x + 4c
3x2 ¡ 9x
ax¡ 3a
5 Write as a single fraction:
ax
5+
2x¡ 1
3b
x
6¡ 1 + 2x
2c
3
2x¡ 1
x + 2
6 Simplify by factorisation:
a2x2 ¡ 8
x + 2b
x2 ¡ 5x¡ 14
x2 ¡ 4c
3x2 ¡ 5x¡ 2
4x2 ¡ 7x¡ 2
7 a Write2x
x + 5¡ 70
(x + 5)(x¡ 2)as a single fraction.
b Hence, find x when2x
x + 5¡ 70
(x + 5)(x¡ 2)is i zero ii undefined.
8 Simplify: a
3x+7x¡1 ¡ 13
x¡ 2b
x2
3¡x¡ 1
2
x¡ 1.
IB10 plus 2nd edmagentacyan yellow black
0 05 5
25
25
75
75
50
50
95
95
100
100 0 05 5
25
25
75
75
50
50
95
95
100
100
Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_12\308IB_10P-2_12.CDR Tuesday, 12 February 2008 10:32:56 AM PETERDELL