Integrated Programme/mainstream
Secondary Three Mathematics
Name: _________________________________ (
)
Class: _________ Date: _____________
Term 1 : Unit 1 – Advanced Algebraic Manipulations
Assignment 1
_________________________________________________________________________________
Advanced Algebraic Manipulation
Answer all the questions.
Question 1: Simplify the following:
( x  1)( x  1)
 2
( x  1)( x 2  1)
x2  2 x  1
( x  1)( x  1)
a)
 2
4
x 1
( x  1)( x  1)( x  1)


8  2x2
b) 2
x  5 x  14



( x  1)
( x  1)( x  1)
2
,x 1
2(4  x 2 )
( x  7)( x  2)
2(2  x)( 2  x)
( x  7)( x  2)
 2( x  2)( 2  x)
( x  7)( x  2)
 2(2  x)
, x  7
( x  7)
2( x  3)
2
2 x  6 ( x  3)( x  3x  9)
c) 3
x  27
2
 2
( x  3x  9)

Question 2: Simplify the following:
8x y
1
  3
3
y 6 x xy
8x 6 x 1
a) 2 

4
y
y xy
 3 , x3 y  0
3x y
1
x( x  3)
x2  x  2

( x  3)( x  2) 2 x 2  2
x( x  3)
( x  2)( x  1)


2
2
( x  3)( x  2)
2( x 2  1)
x  3x
2x  2
b) 2
 2
x  x6 x  x2
x( x  3)
( x  2)( x  1)


( x  3)( x  2) 2( x  1)( x  1)
x

, x  1
2( x  1)


x(1  x)  2( x  2) ( x  2) 



( x  2)( x  2)  ( x  2)
5 x 
x  x2  2 x  4 5x 
x(1  x)  2( x  2) 
c) 2


  ( x  2)( x  2)  5 x 
x 4 x2 x2


2(1  x)

, x  2
5( x  2)
Question 3: Simplify the following:
1
2


a  3 (a  3)( a  1)
a 1 2

1
2
(a  3)( a  1)
 2
a)
a  3 a  4a  3
a3

(a  3)( a  1)
1

, a  1
a 1
x
2
3( x  2)


( x  1) ( x  2) ( x  2)( x  1)
x( x  2)  2( x  1)  3 x  6

( x  2)( x  1)

b)
x
2
3x  6

 2
x 1 x  2 x  x  2

x 2  2 x  2 x  2  3x  6
( x  2)( x  1)
x 2  3x  4
( x  2)( x  1)
( x  4)( x  1)

( x  2)( x  1)
x4

, x  2
x2

2

(1  3 x 1 )(1  3 x 1 )
(1  3 x 1 )(1  3 x 1 )
(1  3 x 1 )
1  9 x 2

1  6 x 1  9 x 2 (1  3 x 1 )
x3

,x  3
x3
c)
Question 4:
a) Given that y = 3 x  1 , make x the subject.
y 2  9( x  1)
9x  y2  9
y2  9
x
9
1
b) Given that p = (2 x  1) 3 , make x the subject.
p3  2 x  1
2x  p3  1
x
p3  1
2
2abx 3
c) Given that r =
, make x the subject.
c  x3
r (c  x 3 )  2abx 3
rc  rx 3  2abx 3
rc  2abx 3  rx 3
x 3 (2ab  r )  rc
rc
2ab  r
rc
x3
, r  2ab
2ab  r
x3 
d) Given that g =
x2 1
, make x the subject.
x2  2x  3
x2  1
g  2
x  2x  3
( x  1)( x  1)
g2 
( x  3)( x  1)
x 1
g2 
x3
3g 2  1
x 2
' g  1
g 1
2
3
Question 5:
x
x
y
The diagram shows a rectangular card with shaded squares each of width x cm and the unshaded strips
each of width y cm. Given that the perimeter of the shaded part is equal to that of the unshaded part, find
x
a) the numerical value of , and hence
y
6 y  10 x
x 3

y 5
b)
the numerical value of :
areaofshadedpart
areaofunshadedpart
Area( Shaded )
6x2

Area(Unshaded ) 7 xy  2 y 2
Area( Shaded )
6(0.6 y ) 2

Area(Unshaded ) 7(0.6 y ) y  2 y 2
Area( Shaded )
54 31


Area(Unshaded ) 25 5
Area( Shaded )
54

(3s. f )
Area(Unshaded ) 155
Question 6:
Simplify the following:
5( x 2  4)

( x 2  9)( x 2  4)
5x 2  20
a) 4
x  13 x 2  36
5
 2
x 9
4
b(a  1)  (a )( a  1)
4a 2

12ab
a (a  1)
2
2
1
ab  b  a  a a  a (b  a )( a  1)



2
12ab
4a
3b
(a  1)
ba

,b  0
3b

b)
Question 7:
Given that c 
ab
.Make b
ab
the subject of the above formula.
c(a  b)  ab
ac  bc  ab
ab  bc  ac
b(a  c)  ac
b
ac
, a  c
ac
Question 8:
Express y as the subject of the formula
x 1 2  y
.

x
3y
3 y ( x  1)  x(2  y )
3 xy  3 y  2 x  xy
2 xy  3 y  3 x
y (2 x  3)  3 x
2x
, x  1 . 5
2x  3
Question 9:
x 2  2 x  3 4 x 2  4 xy  y 2
Simplify
. Leave your answer in factorised form.

4x2  y 2
2x2  2x
y
( x  3)( x  1)
(2 x  y )( 2 x  y )

(2 x  y )( 2 x  y )
2 x( x  1)
( x  3) (2 x  y )


(2 x  y )
2x
( x  3)( 2 x  y )

, x  0, y  2 x
2 x(2 x  y )

Question 10:
Express each of the following as a fraction in its simplest form:
5
3a  b  3a  4b

5
10
6(a  b)  (3a  4b)

10
a)
6a  6b  3a  4b

10
3a  10b

10
b)
3
1

2(2 x  1) 1  2 x
3
1


2( 2 x  1) (2 x  1)
3 2

2( 2 x  1)
1

, x  0 .5
2( 2 x  1)
Question 11:
a)
If
x
=
2
y7
, express y in terms of x.
3 y
x2 y  7

4 3 y
3 x 2  x 2 y  4 y  28
x 2 y  4 y  28  3 x 2
y
b)
28  3 x 2
, x  2
x2  4
Simplify
y=
x 2  6 xy  9 y 2
1
. Hence, find the value of this expression when x = and
2
3
x  3xy
1
.
12
x 2  6 xy  9 y 2
x 2  3 xy
( x  3 y)2

x( x  3 y )
( x  3 y)

x
1

4
6
Question 12:
a) Simplify the following expression giving your answer as a single fraction.
3
1
2


2
x  4 2x  4 x
3
1
2


( x  2)( x  2) 2(x  2) x
6 x  x( x  2)  4( x  2)( x  2)

2( x  2)( x  2) x

 3x 2  8 x  16
, x  0, x  2
2( x  2)( x  2) x
b) Express the following expression, giving your answer as a single fraction in its lowest terms.
7
4

( x  4)( x  3) ( x  3)( x  1)
7
4

( x  4)( x  3) ( x  3)( x  1)
7( x  1)  4( x  4)

( x  1)( x  3)( x  4)
7 x  7  4 x  16

( x  1)( x  3)( x  4)
3x  9

( x  1)( x  3)( x  4)
3

, x  1, x  4, x  3
( x  1)( x  4)
Question 13: (EOY 2007)
a) Simplify
b)
6
5
4


.
3x  2 x  1 2  3x
Make P the subject of the formula t 
P2 w
.
4P2  h
Solution :
6
5
4
12  13x


a)

3 x  2 x  1 2  3 x  3x  2  x  1
t
P2w
4  P2  h
P
th
4t  w
b)
Question 14 : (EOY 2008)
a) Factorise completely
7
3a 3  27 a ,
3ab  2bc  15ad  10cd
i)
ii)
ab
, find the value(s) of x such that
ba
2 Ф ( x Ф 1 ) = ( 3x Ф 1 ).
b) If a Ф b means
Solution :
3a 3  27a

a) i)  3a a 2  9

 3aa  3a  3
a) ii) 3ab-2bc-15ad+10cd
= b(3a-2c)-5d(3a-2c)
=(3a-2c)(b-5d)
 x  1  3x  1

 1  x  1  3x
x 1
2
1  x  3x  1
x 1
1  3x
2
1 x
3  x 3x  1

3x  1 1  3x
 3  x   3x  1
b) 2Ф 
x  2
Question 15 : (EOY 2010)
1
3x  5
a) Solve for x when
.
2x  3 
2
2
b) Make h the subject of the formula
c) Simplify
g 1

2r y
h2  1
.
2  h2
x 2  3x  2 2 x 2  7 x  6 x 2  x  2

 2
.
2 x2  x  3
8 x 2  18
x  2x  1
Solution:
1
3x  5
2 x  3 3x  5
(a)
2x  3 


2
2
4
2
 4x  6  12x  20
x
13
4
8
g 1 h2  1
g2
1  h2  1 





2r y 2  h 2
4r 2 y 2  2  h 2 
(b)
 g 2 y 2 (2  h2 )  4r 2 (h2  1)
 4r 2 h2  h2 g 2 y 2  2 g 2 y 2  4r 2
 h2 
2 g 2 y 2  4r 2
4r 2  g 2 y 2
h
(c )
2 g 2 y 2  4r 2
4r 2  g 2 y 2
x 2  3x  2 2 x 2  7 x  6 x 2  x  2

 2
2 x2  x  3
8 x 2  18
x  2x  1

( x  1)( x  2) 2(2 x  3)(2 x  3) ( x  2)( x  1)


(2 x  3)( x  1) (2 x  3)( x  2)
( x  1) 2

2( x  2)
2x  4
or
x 1
x 1
9