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Thursday, April 11, 2013

AIRIL AHMAD


Expand a single pair of brackets
Example 1: Expand w (x + y)

w ( x + y )

= w x x + w x y

= wx + wy
AIRIL AHMAD


Example 2: Expand 3 (a - c)

3 ( a - c )

= 3 x a - 3 x c

= 3a - 3c

AIRIL AHMAD


Example 3: Expand 7g (m + 2n)

7g ( m + 2n )

= 7g x m + 7g x 2n

= 7gm + 14gn

AIRIL AHMAD


Example 4: Expand 4e ( r + 2s – 8t)

4e ( r + 2s - 8t )

= 4e x r + 4e x 2s - 4e x 8t

= 4er + 8es - 32et

AIRIL AHMAD


ALGEBRAIC
EXPRESSIONS
EXPAND DOUBLE PAIRS OF BRACKETS

HJ. AIRIL AHMAD

AIRIL AHMAD


Expand double pairs of brackets
Example 1: Expand (w + v)(x + y)

( w + v ) ( x + y )

= wxx + w x y +vxx + vxy
= wx + wy + vx + vy

AIRIL AHMAD


Example 2: Expand (2e + f)(c + d)

( 2e + f ) ( c + d )

= 2e x c + 2e x d +fxc + fxd
= 2ce + 2de + cf + df

AIRIL AHMAD


Example 3: Expand (2m + n)(p - s)

( 2m + n ) ( p - s )

= 2m x p + 2m x (-s) + n x p + n x (-s)
= 2mp - 2ms + np - ns

AIRIL AHMAD



HJ. AIRIL AHMAD

AIRIL AHMAD


Example 1: Expand (r + 4)(r + 3)

( r + 4 ) ( r + 3 )

= rxr + r x 3 +4xr + 4x3
= r2 + 3r + 4r + 12
= r2 + 7r + 12
AIRIL AHMAD


Example 2: Expand (2e + f)(e + 2f)

( 2e + f ) ( e + 2f )

= 2e x e + 2e x 2f +fxe + f x 2f
= 2e2 + 4ef + ef + 2f2
= 2e2 + 5ef + 2f2
AIRIL AHMAD


EXERCISE

AIRIL AHMAD


Example 3: Expand (w + 4)(w - 3)

( w + 4 ) ( w - 3 )

= wxw + w x (-3) + 4 x w + 4 x (-3)
= w2 - 3w + 4w - 12
= w2 + w - 12
AIRIL AHMAD


In general , (a+b)2= a2 + 2ab + b2 or
(a-b)2= a2 - 2ab + b2

HJ. AIRIL AHMAD

AIRIL AHMAD


Expand same pairs of brackets
Example 1: Expand (r + 4)2

( r + 4 ) ( r + 4 )

= rxr + r x 4 +4xr + 4x4
= r2 + 4r + 4r + 16
= r2 + 8r + 16 AIRIL AHMAD


Example 2: Expand (2e + f)2

( 2e + f ) ( 2e + f )

= 2e x 2e + 2e x f + f x 2e + f x f
= 4e2 + 2ef + 2ef + f2
= 4e2 + 4ef + f2 AIRIL AHMAD


Example 3: Expand (w - 4)2

( w - 4 ) ( w - 4 )

= w x w+ w x (-4) + (-4) x w + -(4) x (-4)
= w2 - 4w - 4w + 16
= w2 -8w + 16 AIRIL AHMAD


State the factors of an algebraic term
Example 1: 2m
1, 2, m, 2m

Example 2: 4g
1, 2, 4, g, 2g, 4g

AIRIL AHMAD


State the factors of an algebraic term
Example 3: 12y
1, 2, 3, 4, 6,12, y, 2y, 3y, 4y, 6y, 12y

Example 4: 15k
1, 3, 5, 15, k, 3k, 5k, 15k

AIRIL AHMAD


Exercise
Write all the factors of the
following algebraic terms.
1. 5j 3. 5 n2
2. 14 p 4. abc

AIRIL AHMAD


CONCEPT OF
FACTORISATION OF
ALGEBRAIC EXPRESSIONS

AIRIL AHMAD

AIRIL AHMAD


Highest Common Factor (HCF)
Find the HCF of the following algebraic terms.
Example 1 : Example 2 :
6h & 4gh 2pq & 4q
2 6h , 4gh 2 2pq , 4q
h 3h , 2gh q pq , 2q
3 , 2g p , 2
HCF = 2 x h HCF = 2 x q
= 2h = 2q
AIRIL AHMAD


Highest Common Factor (HCF)
Find the HCF of the following algebraic terms.
Example 3 : Example 4 :
14v2 & 8v 5cd2 & 10cd
2 14v2 , 8v 5 5cd2 , 10cd
v 7v2 , 4v cd cd2 , 2cd
7v , 4 d , 2
HCF = 2 x v HCF = 5 x cd
= 2v = 5cd
AIRIL AHMAD


FACTORISE
ALGEBRAIC
EXPRESSIONS

AIRIL AHMAD

AIRIL AHMAD


Factorise the following algebraic term
Example 1: 2pq + 4qr

2 2pq + 4qr
q pq + 2qr
(p +
p 2r
2r)

AIRIL AHMAD


Example 2: mn – mn2

m mn - mn2
n n - n2
(1 - n
1 n)

AIRIL AHMAD


Example 3: 3ab + 6bc - 9b2

3 3ab + 6bc - 9b2
b ab + 2bc - 3b2
( a + 2c - 3b3b)

AIRIL AHMAD


Factorise ab + ac +bd + cd
Example 9: ag + 3a + 2g + 6
(ag + 3) )+ (2g + 6 )
a
(g 3a a 2
(g 3)2

(a (g + 3) + 2 (g + 3)
)

AIRIL AHMAD


Factorise ab + ac +bd + cd
Example 10: mn + 4n + 2m + 8
( mn + 4n) +( 2(m +2 )
(m + 4) n
n 2m 8 4)

(n (m + 4) + 2 (m + 4)
)

AIRIL AHMAD


Exercise
Factorise the following algebraic expressions:
1. ac + ad + bc + bd
2. ax + ay + bx + by
3. gh + 2h + 3g + 6
4. a2 + ab + 2a + 2b
5. de + d2 + ef + df

AIRIL AHMAD


Factorise a2- b2
Example 11: a 2 – b2
( a 2 – - b2 ) )
(a + b
a

AIRIL AHMAD


Factorise a2- b2
Example 12: m 2 – n2
((m2 –- n2 ) )
m + n n

AIRIL AHMAD


Factorise a2- b2
Example 13: 16m2 – 25n2
16m2 – 25n2
( 42m2 – -5n ) )
(4m + 55n
4m n
2 2

AIRIL AHMAD


Factorise a2 + 2ab + b2
Example 1: y2 + 8y + 16
4
1st Step:
16 = 1 x 16 (y 4)
+
+4y
16 = 2 x 8
16 = 4 x 4
2nd Step:
+
+4y
8= 4+ 4 (y + 4)
AIRIL AHMAD


Example 2: 7
2
y2 + 9y + 14
1st Step:
14 = 1 x 14 (y 2)
+
+2y
14 = 2 x 7

2nd Step:
+
+7y
9= 2+ 7
(y + 7)
AIRIL AHMAD


Exercise:
• Page 135
• Ex.6.2C
•1 - 8
AIRIL AHMAD


Factorise completely (Factorise a2 + 2ab + b2)
1. m2 + 12m + 36 2. a2 + 4a +4
3. j2 + 4j +3 4. k2 + 7k +12
5. b2 + 8b +15 6. f2 + 6f +8
7. g2 + 9g +14 8. s2 + 10s +16
9. h2 + 5h +6 10. r2 + 7r +10

AIRIL AHMAD


Example 3: y2 - 6y + - 2
84
1st Step:
8= 1x8 (y -2)
-2y
8= 2x4

2nd Step: -4y
-6 = -2 - 4
(y - 4)
AIRIL AHMAD


Example 4: y2 - 10y + - 3
7
21
1st Step:
21 = 1 x 21 (y -3)
-3y
21 = 3 x 7

2nd Step: -7y
-10 = -3 - 7
(y - 7)
AIRIL AHMAD


Exercise
1. y2 – 3y + 2 6. y2 – 8y + 15
2. y2 – 5y + 6 7. y2 – 9y + 18
3. y2 – 7y + 12 8. y2 – 8y + 12
4. y2 – 7y + 10 9. y2 – 12y + 35
5. y2 – 9y + 14 10. y2 – 14y + 45

AIRIL AHMAD


Exercise:
• Page 134
• Ex.6.2B
• 1 - 15
AIRIL AHMAD


Factorise a2 + 2ab - b2
Example 5: y2 + 5y - 36
+9
-4
1st Step:
36 = 1 x 36
(y +9)
36 = 2 x 18 +9y
36 = 3 x 12
36 = 4x 9
36 = 6x 6
-4y
2 Step:
nd

5= 9- 4 (y - 4)
AIRIL AHMAD


Example 6: y2 + 4y - 21
+7
-3
1st Step:
21 = 1 x 21
(y +7)
21 = 3 x 7 +7y

2nd Step:
4= 7- 3
-3y

(y - 3)
AIRIL AHMAD


Exercise
1. y2 + y – 6 6. y2 + 4y – 12
2. y2 + 2y – 15 7. y2 + y – 42
3. y2 + 6y – 27 8. y2 + 4y – 21
4. y2 + 3y – 10 9. y2 + 7y – 30
5. y2 + 3y – 18 10. y2 + 10y – 24

AIRIL AHMAD


Example 7: y2 - 3y - 10
+2
-5
1st Step:
10 = 1 x 10 (y -5)
10 = 2 x 5 -5y

2nd Step:
-3 = - 5 + 2 +2y

(y +2)
AIRIL AHMAD


Example 8: y2 - 3y - 28
+4
-7
1st Step:
28 = 1 x 28 (y -7)
28 = 2 x 14 -7y

28 = 4 x 7
2nd Step:
-3 = - 7 + 4 +4y

(y + 4)
AIRIL AHMAD


Exercise
1. y2 – 2y – 8 6. y2 – 5y – 36
2. y2 – 3y – 18 7. y2 – y – 12
3. y2 – 5y – 14 8. y2 – 2y – 24
4. y2 – 7y – 30 9. y2 – 6y – 7
5. y2 – y – 20 10. y2 – y – 6

AIRIL AHMAD


FACTORISE AND SIMPLIFY
ALGEBRAIC FRACTIONS

AIRIL AHMAD

AIRIL AHMAD


Simplify the following expressions:

Example 1:

4
8 xy 4y
= = 4y
1
2x 1

AIRIL AHMAD


Example 2:

5 a 2
20a b 5a
4ab
=
1
= 5a
1

AIRIL AHMAD


Exercise
Page 138
Ex. 6.2F
No. 1 abcd

AIRIL AHMAD


Simplify the following algebraic
fractions to the lowest terms Working Space

Example 3:
1 2 6g + 4
6g + 4 2(3g + 2)
= 3g + 2
4 42
3g + 2 2 ( 3g + 2 )
=
2
AIRIL AHMAD


Working Space
Example 4:
2 2a - 8b
2a − 8b 2(a − 4b)
= a - 4b
2c 2c
2 ( a – 4b )

a − 4b
=
c AIRIL AHMAD


Exercise
Page 138
Ex.6.2F
No. 2 only

AIRIL AHMAD


ADDITION OF
ALGEBRAIC FRACTIONS

AIRIL AHMAD

AIRIL AHMAD


Simplify the following:

Example 1:

3x 2 x 3x + 2 x 5x
+ = =
7 7 7 7

AIRIL AHMAD


Example 2: g + 3 2g + 4
+
3 3
g + 3 + 2g + 4
=
3

3g + 7
=
3
AIRIL AHMAD


Exercise

AIRIL AHMAD


SUBTRACTION OF
ALGEBRAIC FRACTIONS

AIRIL AHMAD

AIRIL AHMAD



Example 1:

9p 2p 9p −2p 7p
− = =
10 10 10 10

AIRIL AHMAD


Example 2: 9 f +8 2 f +5
−
hk hk

9 f + 8 − 2f - 5
=
hk
7f +3
=
hk AIRIL AHMAD


8k + 7 7k − 2
Example 3: −
2abc 2abc
8k + 7 − 7k + 2
=
2abc
k+9
=
2abc AIRIL AHMAD


Exercise
Page 139
Ex. 6.3A
No. 1,2,3,5,10,11,12,

13,14,15
AIRIL AHMAD


ADD OR SUBTRACT TWO
ALGEBRAIC FRACTIONS WITH
ONE DENOMINATOR AS A
MULTIPLE OF THE OTHER
DENOMINATOR

AIRIL AHMAD
AIRIL AHMAD



Example 1: 3 xx2 2 x
+
2 x2 4
=6 x+ 2x
4 4
8x
= = 2x
4 AIRIL AHMAD


4h x 3 2h
Example 2: +
5 x 3 15
12h +2h
=
15 15
12h + 2h 14h
= =
15 15
AIRIL AHMAD


5hx 2 2 h
Example 3: −
6 x 2 12
10h −2h
=
12 12
10h − 2h 8h 2h
= = =
12 12 3
AIRIL AHMAD


ADD AND SUBTRACT
ALGEBRAIC FRACTIONS
WITHOUT ANY COMMON
FACTORS

AIRIL AHMAD
AIRIL AHMAD


Example 1:

3 p 5 2r 2
x x 15 p − 4r
− =
2x5 5x 2 10

AIRIL AHMAD


8x n f + x m
5
Example 2: −
m n
x nx m
8n − fm – 5m
=
mn

8n − fm − 5m
=
mn
AIRIL AHMAD


2 − x y 3 xx x
x
Example 3: +
xxy yx x
2y - xy + 3x2
=
xy

2 y − xy + 3x 2
=
xy
AIRIL AHMAD


Exercise
Page
142&143
ALL
Ex.6.3C
AIRIL AHMAD


MULTIPLICATION
ALGEBRAIC FRACTIONS

AIRIL AHMAD

AIRIL AHMAD



Example 1:

3 2 3× 2 6
× = =
5 7 5 × 7 35

AIRIL AHMAD


Example 2:

m p m × p mp
× = =
n q n × q nq
AIRIL AHMAD


5p
Example 3:
1
r
×5p = r ×5p 5 pr
2 =
2 2
AIRIL AHMAD


( x + y) 4g
Example 4: ×
h (e − f )
( x + y) × 4 g
=
h × (e − f )
4 g ( x + y)
=
h (e − f )
AIRIL AHMAD


Exercise
Page 85
Ex.7
No. 1 to 12
AIRIL AHMAD


DIVISION OF TWO
ALGEBRAIC FRACTIONS

AIRIL AHMAD

AIRIL AHMAD



Example 1:

×
3 2 3
÷ =
5 7 5
7 3 × 7 21
= =
2 5 × 2 10
AIRIL AHMAD


5p
Example 2: 1
r
2
×
÷5p =
r
2
1
5p
r ×1 r
= =
2×5 p 10 p
AIRIL AHMAD


Example 3:

÷ = ×
2c 5 2c
3d 4e 3d
4e
5
2c × 4e 8ce
= =
3d × 5 15d
AIRIL AHMAD


Exercise
Page 85
Ex.7
No. 13 to 29
AIRIL AHMAD

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