Cambridge
Lower Secondary
Complete
Mathematics
Second Edition
Deborah Barton
9
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2 Expressions
2
•
•
3 × 5a = 15a
(multiply numbers together, then write
in front of the letter)
2 p × 3q = 2 × p × 3 × q
(multiply numbers first)
2 × 3 × p × q = 6 pq
y × y = y 2 ( we say ‘y squared’)
p × p × p = p3 ( we say ‘p cubed’)
•
a÷b=
•
Expressions
•
Objectives
In this chapter you will learn about
zz expressions, including variables and
constants
zz collecting like terms
z
a
b
b
c
(write as a fraction)
zz expanding
zz forming
brackets
an algebraic expression
2
2
For example:
−
2+5= 3
4 + − 7 = 4 − 7 = −3
−
6 − −1 = −6 + 1 = −5
What’s the point?
The use of symbols or letters
for numbers helps to describe
relationships among variables.
For example, the speed (v ) of a race
car is related to the time (t ) it takes to
travel a particular distance (d ) by v =d ÷t.
3
The area of a rectangle is length × width.
4
The perimeter of a shape is the distance around it.
Work out:
a
b
c
d
e
f
How to add and subtract with negative numbers
Simplify:
p+ p+ p+ p+ p
i
ii G + G
iii b + b + b − b
iv m + m − m − m
v p× p
vi m × m × m
vii t ÷ p
3 x can be written as 3 × x or in full as
x + x + x. Write in full:
i 4m
ii 5 y
−
8 + 10
4 − 12
3 + −9
−
3− 4
−
7 − −5
−
1 + −8
3
What is the area of a rectangle of length
12 cm and width 8 cm?
4
Find the perimeter of this shape.
3 cm
2 cm
4 cm
5 cm
Before you start
You should know ...
Check in
1
1
The basics of algebra:
•
•
•
•
26
a + a + a = 3 × a or 3a for short. No
need for the multiplication symbol
when letters are used. This is called
simplifying or collecting like terms.
a × 5 = 5a (write the number first)
a × b = ab for short
b × 3 × a = 3ab for short (number
first, then letters in alphabetical order)
a
Write the following in a shorter way.
i 4× p
ii t × 3
iii h × k
iv a × b × c
v 2 × 4m
vi 7 y × 5
vii a × 2 × b
viii 3n × 4u
ix 4t × 6r
2.1 Expressions
Here is some key language about, for example, 3 x + 7.
In maths we try to make things simpler by writing
as few words as possible. For example, these
two sentences can be written in a shorter way:
zz3 x + 7
3 apples and 7 bananas = 3a + 7b
6 apples and 12 bananas = 6a + 12b
We have used the letter a to represent apples, the letter
b to represent bananas and the + symbol to replace the
word ‘and’. The process of using letters to represent
unknown numbers or variables is algebra.
is an expression, in this case involving
numbers letters and a + symbol.
constant is a symbol which always means the
same thing, so 7 is a constant.
term is part of an expression so 3 x and 7 are the
two terms in the expression. 3 x is the algebraic
term and 7 is the number term.
zzAn unknown is part of an expression that you
don’t know the value of, so x is the unknown.
zzThe coefficient is the number part of an algebraic
term so, in 3x, the 3 is the coefficient.
zzA
zzA
27
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Acknowledgements
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First published in 2022
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the Rights Department, Oxford University Press, at the address
above.
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must impose this same condition on any acquirer
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Contents
About this book
5
6.
Area, perimeter and volume
76
Circles
Pythagoras’ theorem
Surface area of prisms and cylinders
Volumes of prisms and cylinders
Small and large units of measurement
77
81
84
87
90
1.
Integers, powers and roots
7
1.1
1.2
1.3
Irrational numbers
Estimating square roots and cube roots
Index laws
8
10
12
6.1
6.2
6.3
6.4
6.5
14
15
Consolidation
Summary
Consolidation
Summary
2.
Expressions and formulae
2.1
2.2
2.3
2.4
2.5
2.6
Simplifying and laws of indices
Algebraic fractions
The product of two linear expressions
Substitution into expressions and formulae
Constructing expressions
Changing the subject of a formula
Consolidation
Summary
3.
3.1
3.2
3.3
3.4
7.
Fractions and decimals
99
17
21
23
25
26
27
7.1
7.2
7.3
Factors
Multiplying and dividing fractions
Order of operations and laws of arithmetic
with fractions
Recurring decimals
100
102
30
33
35
Inscribed polygons
Constructions
Using scales and bearings in maps
Reflective symmetries in 3D shapes
38
40
42
45
4.
Place value
4.1
Multiplying and dividing a number by
a power of 10
Standard form
Lower and upper bounds
4.2
4.3
16
Shapes and mathematical
drawings
Consolidation
Summary
Consolidation
Summary
93
97
47
49
51
52
53
56
58
59
5.
Grouped data and sampling
60
5.1
5.2
Averages and range from grouped data
Sampling
61
64
Consolidation
Summary
68
70
Review A
73
7.4
Consolidation
Summary
8.
Equations and inequalities
8.1
8.2
Constructing and solving linear equations
Solving equations with the unknown
in the denominator
Linear inequalities
Linear functions in two variables
Simultaneous equations
8.3
8.4
8.5
Consolidation
Summary
9.
Geometry
9.1
9.2
9.3
9.4
Angles in polygons
Tessellations
Angle geometry problems
Coordinates on a line segment
Consolidation
Summary
10. Presenting data and
interpreting results
10.1
10.2
10.3
10.4
Frequency polygons
Back-to-back stem-and-leaf diagrams
Scatter graphs and correlation
Choosing the most appropriate graph
105
107
109
111
112
114
116
117
121
124
129
131
134
136
140
143
145
148
150
152
154
157
159
165
Contents
Consolidation
Summary
172
176
Review B
11. Ratio and proportion
179
243
244
247
185
Consolidation
Summary
251
253
186
188
192
Consolidation
Summary
194
196
12.1
12.2
12.3
12.4
12.5
Sequences
Functions
Linear functions
Solving simultaneous equations graphically
Real-life functions
198
199
204
206
210
213
15. Probability
256
262
265
Consolidation
Summary
268
271
Review C
273
16. BONUS CHAPTER: Quadratics
278
279
280
289
291
13.1 Enlargement
13.2 Describing transformations
13.3 Combining transformations
228
231
234
Consolidation
Summary
237
240
Consolidation
Summary
13. Transformations
219
222
226
255
15.1 Successive and combined events
15.2 Tree diagrams
15.3 Experimental probability
16.1 Graphs of quadratics
16.2 Expanding two brackets
16.3 Perfect squares and difference between
two squares
16.4 Factorizing quadratic expressions
16.5 Solving quadratic equations
16.6 Word problems
16.7 Factorizing by grouping
16.8 Trial and improvement
Consolidation
Summary
242
14.1 Estimation
14.2 Multiplying and dividing decimals
14.3 Compound percentages
11.1 Solving ratio problems
11.2 Direct proportion
11.3 Inverse proportion
12. Sequences, functions
and graphs
14. Decimals and percentages
Index
281
284
285
286
288
288
293
About this book
About this book
This book follows the Cambridge Lower Secondary
Mathematics curriculum in preparation for the
assessments at this level, but also for study at IGCSE.
It has been written by a highly experienced teacher,
examiner and author.
This book is part of a series of nine books. There
are three student textbooks covering stages 7,
8 and 9 and three homework books written to match
the textbooks closely, as well as a teacher book for
each stage.
The books are carefully balanced between all the
content areas in the framework: number, algebra,
geometry, measure and handling data. Some of the
questions in the exercises and the investigations within
the book are underpinned by thinking and working
mathematically methods, providing a structure for the
application of mathematical skills.
Features of the book include:
Objectives – showing skills required for the
Cambridge Secondary Stage 9.
What’s the point? – providing rationale for
inclusion of topics in a real-world setting.
Before you start – for each chapter to assess
whether the student has the required prior
knowledge.
Notes and worked examples – in a clear style
using accessible English and culturally appropriate
material.
Exercises – carefully designed to increase
gradually in difficulty, providing plenty of practice
of techniques.
Considerable variation in question style
– encouraging deeper thinking and learning,
including open questions.
Comprehensive practice – plenty of initial
questions for practice followed by varied questions
for stretch, challenge, crossover between topics and
links to the real world with questions set in context.
Extension questions – providing stretch and
challenge for students:
z questions with a box, e.g. 1 , provide
challenge for the average student
z questions with a filled box, e.g. 1 , provide
extra challenge for high-attaining students.
Investigation and activity boxes – providing extra
fun, challenge and interest.
Full colour presentation with modern artwork –
pleasing to the eye, interesting to look at, drawing
the attention of the reader.
Consolidation examples and exercises –
providing review material on each chapter.
Summary and Check out – providing a quick
review of each chapter’s key points, aiding revision
and enabling you to to assess progress.
Review exercises – provided every five chapters
with mixed questions covering all topics.
Bonus chapter – the work from Chapter 16 is not
in the Cambridge Secondary Stage 9 Mathematics
curriculum. It is in the Cambridge IGCSE®
curriculum and is included to stretch and challenge
high-attaining students.
A note from the author
If you don’t already love maths as much as I do, I hope
that after working through this book you will enjoy
it more. Maths is more than just learning concepts
and applying them. It isn’t just about right and wrong
answers. It is a wonderful subject full of challenges,
puzzles and beautiful proofs. Studying maths develops
your analysis and problem-solving skills and improves
your logical thinking – all important skills in the
workplace.
Be a responsible learner – if you don’t understand
something, ask or look it up. Be determined and
courageous. Keep trying without giving up when
things go wrong. No one needs to be ‘bad at maths’.
Anyone can improve with hard work and practice in
just the same way athletes improve their skills through
training. Look for challenges, then maths will never be
boring.
Most of all, enjoy the book. Do the ‘training’, enjoy
the challenges and have fun!
Deborah Barton
5
2
Expressions and formulae
Objectives
In this chapter you will learn about:
z constructing expressions
z finding the product of two linear
expressions
z
z
z
z
laws of indices
substituting into expressions
changing the subject of a formula
simplifying algebraic fractions.
What’s the point?
When you learn about algebra you
learn how to solve real-life problems,
but in an abstract way. Algebra works
in a logical and ordered way with rules
that must be followed. Without rules
like these, computers would not work.
Computer games depend on programs
that follow logical ordered codes similar
to those used in algebra.
Before you start
16
You should know ...
Check in
1
1
How to add and subtract negative
numbers.
For example:
8 + −3 = 8 − 3 = 5
8 − −3 = 8 + 3 = 11
Work out:
−
a 6+ 2
b −2 − 3
c − 3 − −6
d −6 + 5
e − 3 + −6 − −2
2 Expressions and formulae
2
2
How to multiply and divide negative
numbers.
6 × −3 = −18
−
6 × − 3 = 18
6 ÷ −3 = −2
−
6 ÷ −3 = 2
Calculate:
a −3 × 5
b − 2 × −6
c
d
e
3
How to substitute numbers for letters.
For example:
if x = 10 and y = 6
then 2 x + 3 y = 2 × 10 + 3 × 6
−
8
4
−
15 ÷ −3
−3 +
5
2
3
If a = 5 and b = 3, find:
a ab
b 2a + b
c a + 2b
d ab + 2b
= 20 + 18
= 38
4
How to work with indices.
For example:
7 2 × 73 = 7 2 + 3 = 75
85 ÷ 8 2 = 85 − 2 = 83
4
Simplify:
a 95 × 9 4
b 810 ÷ 82
c 56 ÷ 55
d 36 × 32
5
How to construct an algebraic expression
for the perimeter or area of a rectangle.
For example:
5
Write an expression for the area and perimeter
of these rectangles.
a
4
3x
d
b
3d
Perimeter = 3d + d + 3d + d
= 8d
If d = 2 cm, then
6
2y
perimeter = 8 × 2 cm =16 cm.
6
2.1
6
How to factorize.
For example:
Factorize 14x + 21
The HCF (highest common factor) of 14
and 21 is 7.
14x + 21 = 7(2x + 3)
Factorize:
a 9m + 6
c 26 − 13y
b
d
15t − 5
24x + 18y
Simplifying and laws
of indices
In algebra, multiplication signs are often missed out.
2 × 3 × p × r = 6 pr
For example:
2 × 5m × 10 n = 100 mn
12 × x = 12 x
1
3
3 × p × q = 3 pq
× 5 x × (−9 z ) = −15 xz
It may help you to use the longer form when
simplifying fractions.
17
2 Expressions and formulae
Example 1
3 x 3 y2 z 2
x 2 y3 z
=
3× x × x × x × y× y×z ×z
x×x×y×y×y×z
=
3 × x × x/ × x/ × y/ × y/ × z × z/
x/ × x/ × y × y/ × y/ × z/
=
3× x ×z
y
=
3 xz
y
You can multiply expressions that are powers of the
same number or letter by adding the powers.
For example:
a 5 × a 3 = (a × a × a × a × a) × (a × a × a)
= a 5 + 3 = a8
Similarly, you can divide expressions that are powers
of the same number or letter by subtracting the powers.
a 5 = a × a × a/ × a/ × a/ a 5 − 3 a 2
=
=
a3
a/ × a/ × a/
You can raise a power to another power.
(a 2 )3 = a 2 × a 2 × a 2 = a × a × a × a × a × a
The rules for indices are
am × an = am + n
c
3 x 3 × 4 x 2 = 3 × 4 × x 3 × x 2 = 12 × x 3 + 2
= 12 x 5
d
y 2 ÷ y10 = y 2 −10 = y −8 =
e
( x 5 )3 = x 5× 3 = x 15
Example 3
2
3
3 2
Simplify a 5bc2 × ab c4
ab
bc
Multiply the two numerators together and the two
denominators together:
a 2 bc3ab 3c 2
a 5b 2 bc 4
a/ × a/ × b/ × c/ × c/ × c/ × a/ × b/ × b/ × b × c/ × c
=
a/ × a/ × a/ × a × a × b/ × b/ × b/ × c/ × c/ × c/ × c/
= bc2
a
Alternatively, using the rules for indices:
a 2 bc3 × ab 3c 2 = a 2 + 1 × b1 + 3 × c3 + 2
a5b 2
bc 4
a5 × b 2 + 1 × c4
a3b 4 c5
= 5 3 4
abc
= a3 − 5 × b 4 − 3 × c5 − 4
bc
= a −2 × b × c = 2
a
am ÷ an = am − n
(a m )n = a mn
a0 = 1
a–n =
1
n
a
Note that negative powers go beyond what you
are expected to do and are included here as
extension work.
Exercise 2A
1
Example 2
Simplify:
a p × p4
b m 7 ÷ m3
3
2
c 3x × 4 x
2
10
d y ÷y
e ( x 5 )3
a
b
18
p × p4 = p1 + 4 = p5
m 7 ÷ m3 = m 7 − 3 = m 4
1
y8
2
Simplify:
a a 2 × a3
c r4 × r6
e p × p 2 × p5
g am × an
i ma × m b × mc
b
d
f
h
j
q5 × q3
s10 × s 5
j3 × j7 × j 9
pa × pb
xa × xm × xc
Simplify, leaving your answer in index form.
2
b b8 ÷ b 5
a q ÷q
8
4
c y ÷y
d 8 p 7 ÷ 2 p3
10
5
e 6x6 ÷ 2x 2
f 14 y ÷ 7 y
g
10 n6
2 n3
h
50 x 3
5x5
2 Expressions and formulae
3
Simplify:
a
c
4
2a × 4 a
12 p3 ÷ 2 p2
3
4
3c × c × 2c
10 q 5 ÷ 5q
2
b
d
3
4
(a 3 ) 4 = a 3 × a 3 × a 3 × a 3
= a 3 + 3 + 3 + 3 = a12
Use the method above to simplify:
2 3
a (x )
b ( x 3 )2
3 3
c (x )
d (2 x 2 )4
−2 4
e (x )
f (3 x 2 )4
5
m n
Write (a ) in another way.
6 Simplify:
a
c
−3
x ×x ×x
y 4 × y −3 × y −5
2
4
p ×p ÷p
q 5 × q −4 ÷ q −2
7
b
d
5
3
7 Write with a single positive index.
a
c
e
8
b x −2 × x −5 × x 4
d y 2 × y −5 ÷ y −3
( k 2 × k −3 ) ÷ ( k 4 ÷ k −1 )
Give possible numbers and letters for the
blank boxes. You are not allowed to use the
number 1.
a
c
e
9
p−5
q −5 ÷ q 4


 ×  = 12x16

36y8 ÷  = 
2 x = 2

b

d
(3p)3 = 9p9
4t 4 × 4t 4 = 16t 16

3× =3
b
d
8m8 ÷ 2m2 = 4m4
8y 0 = 1
b
x 5 y 2 z 4 ÷ y3 z 3
b
x 4 yz 3
10 Simplify:
a
lm 3 n 2 ÷ mn 2
11 Simplify:
a
c
p3 q
pq
4
4 2
2 4
xy z
5
k l m
kl 2m 3
12 Simplify:
a
c
2
2( p q)
(3 p2q )2
3 pq
2
x2 × x3
x10
x3
x4
x2 × x
x 11 × x 2 ÷ x 10
× x2
×
x2
x×x
x 4 × x2 × x
x7 ÷ x2
x5 ÷ x8
x 4 ÷ (x
4 2
)
x
14 Put these expressions in order, starting with
the lowest index.
( p5 × p3 ) 2
( p × p3 )
( p 4 × p 2 )3
p30
2
8 2
c
e
p5
(p × p )
15 Simplify:
a ( p3 ) 2
A5 ÷ A3 × A 4
x2 × x7
x4
g
(3T 2 + T 2 ) ÷ T
i
g3 + (g 4 + g 4 ) ÷ 2g
 
( ) = 16m12
Here is Jamil’s homework. Correct any
answers that are wrong.
a
c
13 Copy out the expressions below. Draw lines to
show which are equivalent. One is done for
you.
b
x2 × x3 ÷ x4
d
(m × m 4 ) 2
f
h
( y 2 × y3 )2
( y × y 2 )3
2
 h8  × h 3
 h2 
An expression is a collection of terms separated by
plus or minus signs.
Like terms are terms with the same combination of
letters, raised to the same power.
For example, in
3ab + 5ab + 6
3ab and 5ab are like terms.
Similarly,
2a and 5a
ab2 and 10ab2
xy3 and 6xy3
are like terms.
3
b
d
( pq )3
p2 q3
4 p 2 ( q 2 r )3
3 pr × 2 qr 2
19
2 Expressions and formulae
You can simplify expressions with several terms by
collecting like terms together.
2
Here is Hannah’s homework on simplifying
expressions. Correct any answers that are
wrong.
a 3ab2 + 5ba2 + b2a = 9ab2
b 7p2 − 6p2 = 1p2
c (3x)2 − 3x2 = 0
3
Collect like terms together and simplify:
a 5 xy + 2 z − 3 xy − 5z
b 3ab + pq − ab + 5 pq
c a 2 + b 2 + 6a 2 − 3b 2
d 4 a 3 − a − a 3 + 5a
4
Simplify:
a 4 p + 9q − 2 p − 3q
b 6z 3 − z − 7z 3 + 4 z
c a 2 + b 2 + 6a 2 − 3b 2
2
2
d p + 2 pq + 6 pq − 4 p
5
Simplify:
b
a −3 × 7 y
c (−3q) × 4 p
d
e 3a × 2a × a
f (−2 x ) × 5 x × (−4 x )
Example 4
Simplify:
a 3a + 4 b − 2a + 5b
b x + 3 y + 7 x − 12 x
a
3a + 4 b − 2a + 5b
= (3a − 2a) + (4 b + 5b)
= a + 9b
b
x + 3 y + 7 x − 12 x
= 3 y + ( x + 7 x − 12 x )
= 3y − 4 x
You can simplify more complicated expressions in
the same way.
Example 5
Simplify:
a 3 pq − 8 pq + 4 pq
b xy + 2ab − 5 xy + 10 ab
c 2n 2 + m 2 − n 2 + 2m 2
a
b
3 pq − 8 pq + 4 pq = − pq
xy + 2ab − 5 xy + 10 ab
= ( xy − 5 xy) + (2ab + 10 ab)
= −4 xy + 12ab
c
2n 2 + m 2 − n 2 + 2m 2
= ( 2n 2 − n 2 ) + (m 2 + 2m 2 )
= n 2 + 3m 2
Exercise 2B
1
20
Simplify:
a 3x + 5x
c 14 b − 2b
e 8a + 3a − 2a
g 4 p − 7p − 5 p
i 2a 2 + 7a 2
2
2
2
k x + 3x + 7 x
b
d
f
h
j
l
8a + 2a
−
7y + 4y
7b − 3b − b
3ab + 5ab + 9ab
9b 2 − 2b 2
8 y3 − 2 y3 − 4 y3
6
Simplify:
a 6a ÷ 2
c 5a ÷ 2a
e 6ab ÷ 2ab
g 10 ÷ 2 x
i a 2 b ÷ ab
k 7ab4 ÷ 3a3b
b
d
f
h
j
l
( − 6 p) × ( − 5)
(− 5 y) × (−3 y)
8b ÷ 4 b
4 pq ÷ q
8 pq ÷ 2q
3 ÷ 6y
a 2 b 2 ÷ ab 2
4 x 2 y 2 z 2 ÷ xy3 z 2
7 Simplify:
a mn + 4 xy + 17 xy − mn + xy
b 3 p2 q + lm + 14 p2 q + 6lm
c pqr + abx + mny + 5abx
d 15ab − 29ab + 4 pq + ab
e x 2 y 2 − 4 xy + 13 xy + 3 x 2 y 2
8 Simplify:
a 2mn + 5mn − 3mn
b 4l 2 m 2 − 7l 2 m 2
c 8 pqr − 9 pqr − 15 pqr
d −3 × 5m
e (−4 pq) × (−2 p2 q 3 )
f 7a2b3 ÷ 2ab
g 2lm 2 n 3 ÷ 3l 3 m 2 n
2 Expressions and formulae
2.2 Algebraic fractions
Adding and subtracting algebraic
fractions
When adding fractions the numerators can be added
when the denominators are the same:
2
7
3
7
+ =
+
p
6
=
Exercise 2C
1
w+p
6
c
2
The same method is used for subtracting:
3 m
−
t
t
=
3−m
t
If the denominators are not the same then you must
first make them the same, using equivalent fractions
to find a common denominator:
1 2
+
3 7
=
7
+ 6
21 21
3
= 13
21
21 is the common denominator here. The same method
is used in algebra.
Example 6
a
b
2
b
a
Simplify:
+b
7
b
x
5
+4
y
First, give the fractions a common
denominator, 7b:
2
b
+b=
x
5
+4=
7
y
14
7b
xy
5y
2
+b =
7b
+ 20 =
5y
14 + b 2
7b
xy + 20
5y
You can work out more complex problems in the
same way.
Example 7
Simplify
2
( x − 1)
+
Work out:
a
4
7
The same applies when the numerators or
denominators include letters:
w
6
Note: Example 7 goes beyond what you are expected
to do and is included here as extension work.
4
3
4
+1
8
5 1
−
6 3
b
2
3
+1
d
5
7
−2+1
b
2a
9
+a
−a+a
4
3
4
Simplify:
a
a
5
+a
c
3a + 2a
d
3a
4
6
3
e
a
7
− 2a
f
5a a
+
21 7
+a
b
3y
11
d
x
4
f
x2
9
3
3
3
5
3
Simplify:
a
2x
3
c
y
2
e
3x
2
y
6
−
+ 2x + z
5
2
5 y2
2
−
+
y
4
+
+
x
3
y2
2
+
2y
7
Simplify:
a
2
7b
b
2
y
c
6
pq
+
−
3
14 b
1
8y
+
e
x + 2x
5
3p
p−
8
f
3 + 2x
g
5 − 4y
d
Hint: x can be
r
2 pq
written as 5x .
5
Hint: 3 can be
written as 3x .
x
5 Simplify:
2
( x − 1)
+
3
( x – 2)
3
( x – 2)
=
2( x − 2)
( x − 1)( x − 2)
=
2( x − 2)+3( x − 1)
( x − 1)( x − 2)
=
2 x − 4 +3 x − 3
( x − 1)( x − 2)
+
3( x − 1)
( x − 1)( x − 2)
=
a
p
3q
c
5
4
e
xy
z
+
2q
p
+ 3xz
+ az
y
b
5
m
d
1
pq
+3
f
ab
5c
+ bc
+
4l
3m
r
4a
5x − 7
( x − 1)( x − 2)
21
2 Expressions and formulae
6 The mean of p numbers is 48. The mean of
another q numbers is 51. Express, as a single
fraction, the mean of p + q numbers.
7 Avril cycles 10 km at x km/h and then 14 km at
y km/h. Express the time for the journey as a
single fraction.
8 b bananas cost $1.45 and a apples cost $2.35.
a How much does one banana cost?
b How much does one apple cost?
c Write as a single fraction the cost of one
banana and one apple.
Example 8
Simplify 35 x + 10
15
First factorize the numerator
5(7 x + 2)
15
Then cancel the 5 on the numerator with the 15 in
the denominator
5
=1
15 3
5(7 x + 2) 7 x + 2
=
15
3
as
9 Simplify:
a
c
e
( x + 3) ( x − 1)
+
4
5
4
3
+
( x + 1) ( x + 2)
1
( x − 1)
+
1
( x + 1)
b
d
f
( x − 2) (2 x − 1)
–
5
7
5
3
+
( x + 3) ( x − 1)
6
( x + 1)
−
Exercise 2D
1
4
( x − 3)
Complete the work to simplify these fractions.
a
b
Investigation
c
x and y are positive integers and x > y. Which of
d
these two fractions is closer to 1?
x or y
y
x
e
as
10 x + 5
,
5
2
you can think of it in the same way as
1
(10 x + 5). Every term inside
5
by 1 so 1 (10 x + 5) = 2 x + 1 .
5
5
the brackets is multiplied
It is very common for students to show this incorrect
working:
10 x + 5/
= 10 x
5/
3
7
49 p − 77
=
7
1
 − ) =  − 
16 − 72 y (2 − ) (2 − ) 4 − 
=
=
12
12
 = 
16 − 72 y (4 − ) 4 − 
=
=
12
12

7(
Simplify these algebraic fractions.
a
27 x + 3
9
b
18 x + 24 y
3
c
4
12 − 2 y
d
25 x + 15 y − 5 z
10
e
42p − 63
28
Sort these fractions into those that are
equivalent to 2x + 3 and those that are not.
2 x + 9 4 x + 12 2 x + 15 + 8 x
,
,
, 1 , 1 (8 x + 12) ,
2x + 3 4
3
5
2

You can use the correct method above to deal with
algebraic fractions or you can use factorizing and
cancelling as in Example 8.
=
f The method used in part e is quicker than
the method used in part d. Explain how
you can do the work in fewer steps.
Simplifying algebraic fractions
When you expand brackets, e.g. 3(2x + 5), every term
inside the brackets is multiplied by 3 so 3(2x + 5)
= 6x + 15. When you have an algebraic fraction such
(6 x + ) = 6 x + 
10

12 y − 16 4( − )  − 
=
=
40
40

30 x + 15
10
2 x + 3, x + 1.5 , 20 x − ( −80 x + 150)
0.5
1
50
4
Franco says
4(25 p + 10)
simplifies
80
to
25 p + 10
20
His teacher says this is not finished. Explain
what his teacher means.
22
2 Expressions and formulae
5
Notice that each term in one bracket is multiplied by each
term in the other bracket. The box below shows this.
Simplify:
5(3 x + 12)
30
a
b
28
2(14 m − 35)
(x + 3) × (x + 2)
6 Here is Andrea’s working to simplify
= x2 + 2x + 3x + 6
= x2 + 5x + 6
24 m 2 + 18 m
9m
24m 2 + 18m
9m
=
6( 4m 2 + 3m )
9m
2
= 4m + 3m
3m
Explain how she can improve her work.
7 Simplify:
24 x 2 + 20 x
64 x
a
b
15 x 3 − 18 x 2
24 x 2
8 Simplify:
a
3( y − 2)
4( y − 2)
b
c
3 y2 − 9 y
2y − 6
d
4x + 4
x +1
8 x 3 + 16 x 2
4x2 +8x
9 Show that these are equivalent fractions.
10 x 2 + 15
2
16 x + 24
2.3
and
When multiplying two linear expressions like this, you
need to remember to collect the x terms together to
simplify the expression formed.
The expression formed, here x2 + 5x + 6, has x2 in it.
An expression like this, where the highest power of x
is 2, is called a quadratic expression. You can find out
more about these in Chapters 8 and 16.
Exercise 2E
In this exercise, simplify all answers.
1
25 x − 5 x 2
40 x − 8 x 2
x
The product of two linear
expressions
b
3x
6
x+3
x2
2x
x
x
2
The area of the big rectangle
= sum of the four smaller rectangles
= x 2 + 2 x + 3x + 6
= x 2 + 5x + 6
Also, the area of the big rectangle = ( x + 2) × ( x + 3)
So, ( x + 2) × ( x + 3) = x 2 + 5 x + 6.
2
= x × ( x + 2) + 3 × ( x + 2)
= x 2 + 2 x + 3x + 6
= x 2 + 5x + 6
(x + 5) × (x + 1)
x2
5
Use the distributive law to work out:
a ( x + 3) × ( x + 8)
b ( x + 6) × ( x + 2)
c ( x + 7) × ( x + 2)
d ( x + 1) × ( x + 2)
Note:
( x − 3) × ( x − 4) can be written as (x − 3)(x − 4),
with the × sign left out.
3
Expand and simplify:
a ( x + 7)( x − 2)
b ( x + 8)( x − 7)
c ( x + 3)( x − 7)
d ( x − 3)( x + 2)
4
Expand and simplify:
This shows how the product of two expressions in
brackets can be found.
Another way is to use the distributive law:
( x + 3) × ( x + 2)
3
1
x
x
(x + 3) × (x + 4)
x2
x
Look at this rectangle.
3
By considering the areas of the rectangles,
simplify the products:
a 4
a
c
(x − 7)(x − 3)
(x − 2)(x − 2)
b
d
(x − 9)(x − 7)
(x − 1)(x − 1)
23
2 Expressions and formulae
5
a
c
6
Difference of two squares
Expand and simplify:
( x − 6)( x + 7)
(x − 4)(x − 11)
b
d
( x − 4)( x − 7)
( x + 3)( x − 10)
Fatima and Aisha were both working out the
answer to ( x + 3)2
Fatima wrote: (x + 3)2 = x2 + 9
Aisha wrote: (x + 3)2 = (x + 3)(x + 3)
= x2 + 3x + 3x + 9 = x2 + 6x + 9
Whose working is correct?
7
8
Expand and simplify:
b ( x + 4)2
a ( x + 5)2
2
d (7 − x )2
c ( x − 1)
2
f (4 + t )2
e ( p + 9)
Write down an expression for the area of each
shape in the form x2 + ax + b.
a
x+7
For example, (x + 4)(x − 4) = x2 + 4x − 4x − 16 = x2 − 16
x2 – 16 can be written as x2 − 42
In general (x + a)(x − a) = x2 − a2.
Perfect squares
In Exercise 2E question 7, these questions are called
perfect squares.
For example, (x + 4)2 = (x + 4)(x + 4)
= x2 + 4x + 4x + 16 = x2 + 8x + 16
x2 + 8x + 16 can be written as x2 + (2 × 4)x + 42
In general, (x + a)2 = x2 + 2ax + a2.
Exercise 2F
x−4
b
Question 10 in Exercise 2E gives answers that are the
difference of two squares.
1
Complete these expansions using the
generalization (x + a) (x – a) = x2 – a2
a (x + 2)(x − 2) = x2 –
b (x + 9)(x − 9) = x2 –
c (x − 11)(x + 11) = –
d (6 + x)(6 − x) =
–
2
Complete these expansions using the
generalization (x + a)2 = x2 + 2ax + a2
a (x + 2)2 = x2 + 4x +
b (x + 5)2 = x2 + x + 25
c (x − 3)2 = x2 – x +
d (x + 12)2 = x2 – x +
3
Expand and simplify:
a (x + 10)(x − 10)
b (x + 12)(x − 12)
c (x − 11)2
d (x + 13)2
x−8
x−4
c
x+2
x+4
9
Check that your answers for question 5 are
correct, by letting x = 2.
10 Expand and simplify:
a (x + 5)(x − 5)
b (x + 8)(x − 8)
c (x − 7)(x + 7)
d (2 + x)(2 − x)
11 Use the distributive law to explain why
( x + a)( x + b) = x 2 + (a + b) x + ab
12 Write the product (a + b)(c + d ) without
brackets.
24
4 Complete these statements.
a x2 + 20x + 100 = (x + )2
b x2 − 16 + 64 = (x – )2
c x2 − 36 = (x + ) (x – )
d x2 − 169 = ( + ) ( – )
e x2 − y2 = ( + ) ( – )
5 Expand (x − 1)(x + 1) and simplify your
answer. Show how you can use this result to
help you work out 49 × 51.
6 Expand and simplify (x + 1)2
Without using a calculator, use this result to
find a 912 b 4012
2 Expressions and formulae
Challenge
3
Find the value of:
a x 2 + 5 when x = 0.2
b 1000(T + 1)3 + T when T = 0.1
c 9 + m 2 when m = − 3
d 10r 2 + 2r − 4 when r = 0.2
e 4W 2 + 1 + W when W = 1
2
f 10 − 200 v 2 − 10 v when v = 0.4
g 5 p4 + 2 p2 + 20 when p = −10
4
Copy these expressions into your exercise book.
6xy
4x + y2
(xy)2
48x2 − y2 − y
Tick () which expressions have the same
value when x = 0.5 and y = −6.
5
A = 2m4 − m2 − 190m and B =
When m = 10 find:
Show that (x + 2)(x + 5)(x + 3)
= x3 + 10x2 + 31x + 30
2.4
Substitution into
expressions and
formulae
When substituting into expressions and formulae,
letters are replaced by numbers, multiplication signs
need to be put back in, and the order of operations
(BIDMAS) needs to be followed.
Example 9
Using the formula A = 100t − 5p2, find A when
p = −3 and t = 0.5
Replace the p
with (-3) and
the t with (0.5).
A = 100t − 5 p2
= 100(0.5) − 5(− 3)2
= 100 × 0.5 − 5 × (− 3)2
BIDMAS tells
us to work out
indices first:
(-3)2 = -3 x -3
= 9
= 100 × 0.5 − 5 × 9
= 50 − 45
=5
a
A+B
b
AB
3m 2 ( m − 4)
110 − m
c
A
B
6
Using s = ut + 1 at2, find s when:
2
a u = 25, a = 0.7 and t = 10
b u = 80, a = −3 and t = 5
7
Using A = πx2 − πy2, find A when x = 0.3 and
y = 0.2. (Use π = 3.142)
8 V = πr2h gives the volume of a cylinder with
radius r and height h.
3
V = 4π r gives the volume of a sphere with
3
radius r.
Which of these two solids has the smaller
volume?
BIDMAS tells us to
work out multiplication
before subtraction.
r = 6cm
h = 7cm
Exercise 2G
1
2
If m = −4, x = 2, y = −5 and r = 3, find the value
of:
a 10(mx + y)
b −3x − y + r
c
5my − 6r
e
5x − 2 y
( my)2
d
5 xmr
y
f
x (m + r )
4
Repeat question 1, this time using m = 0.5,
x = 0.3, y = 2 and r = 0.8.
r = 6cm
9 The distance, d, between two sets of
coordinates, (x1,y1) and (x2,y2), can be
found using the formula
d = ( x − x )2 + ( y − y )2 . What sort of
triangle joins the points (0, 3), (3, 9) and (6, 0)?
2
1
2
1
25
2 Expressions and formulae
Investigation
Heron of Alexandria (also known as Hero) was an
ancient Greek mathematician and engineer who lived
around 2000 years ago. He worked out a way of
finding the area of a triangle from its side lengths.
Heron’s (or Hero’s) formula states that the area, A,
of a triangle with side lengths a, b and c is
A = s(s − a)(s − b)(s − c) , where s is half the
perimeter (the ‘semiperimeter’) of the triangle:
a+ b+ c
s=
2
Using the right-angled triangle with side lengths 3, 4
and 5, work out the area of the triangle using 1 bh,
2
Then subtract the area of the small rectangle.
x
x+3
(x + 2)(x + 6) − x(x + 3)
Expand the brackets.
x2 + 2x + 6x + 12 − x2 − 3x
Simplify.
5x + 12
then work it out using Heron’s formula. Check that
the two answers agree. Investigate this further.
Exercise 2H
2.5
In this exercise, expand all brackets and simplify
where possible.
Constructing
expressions
1
In Student Books 7 and 8 you have learned how to
construct algebraic expressions. You are now going to
construct more complex expressions and use your
skills from earlier in this chapter to expand and
simplify the expressions.
Find an expression for the area of these shapes
made using rectangles.
a
x–1
x+3
Example 10
Find an expression for the area of this compound
shape made from rectangles.
Expand and simplify your expression.
b
x+8
x
x+3
x–8
x+2
c
x+5
x+6
x+5
Find the area of the big outer rectangle.
x+2
x+2
d
x+6
26
x+5
x+4
x+8
2 Expressions and formulae
2
A rectangle has width n + 2. The length of the
rectangle is 5 more than the width. Find an
expression for the area of the rectangle.
3
A rectangle has width w. The length of the
rectangle is 7 more than double the width.
Find an expression of the area of the rectangle.
4
5
x
Let’s say the rectangle in question 3 has a
width of 3 cm. Use substitution to check your
answer to question 3.
Amy works out the area of this square as
x2 + 36. Use the substitution of x = 5 to show
she is not correct.
x+6
x+6
6
9 Find an expression for the length of the
diagonal of this square.
I think of a number, n, add 3 then square it,
then multiply by 2. Find an expression for the
number I end up with.
x
You may want to look
ahead to Chapter 6
to find out about
Pythagoras’ theorem.
10 A rectangle has area x2 + nx − 24 where n is an
integer. The width of the rectangle is x – 3.
Find an expression for the length and the
value of n.
2.6 Changing the subject of
a formula
7 Ranjit has a rectangular piece of card. He cuts
a rectangle out of this card.
a
area, A
us, r
radi
a
b
A function machine can be used to find the radius, r, of
a circle given its area, A.
x
Remember, the area of a circle,
A = πr2
y
He writes down two different ways of finding
the shaded area of card that remains. Show
that these two expressions are equal. Has
Ranjit found the correct area?
Area method 1: 2ab + y(x − b)
Area method 2: 2ax + (x – b)(y – 2a)
π r2
r2
so,
r
and
r
That is, r =
×π
Square
A
π
A
π
÷π
A (area)
A
A
π
8 n is an even number. Use algebra to show that
the product of the next two odd numbers after
n is odd.
27
2 Expressions and formulae
Exercise 2I
1
2
3
7
Draw a function machine to show how to find:
a V using V = π r 2 h, starting with r
b S using S = π r (2 h + r ) starting with h
c s using s = t (u + 12 at ) starting with a
Use the reverse machine for each part of
question 1 to rewrite the formula. Make the
letter you started with the subject.
The time, T, for each
complete swing of a
pendulum of length l is
given by the formula
T = 2π
l
g
d
Taking π as 3.142, use your machine to
find T when l = 20 and g = 980.
a
Draw the reverse machine for question 3a
and show that the formula can be
rearranged as
l=
b
5
b
28
( ) ×g
T
2π
×m
9 a
b
Draw a function machine to show how to
find y, starting with x:
i y = r (x − s) + t
ii y = m(nx + l )
Use the reverse machine to rearrange the
formula, making x the subject.
Using the balance method
2
If T = 1.571 and g = 32, use the formula in
part a to find l. (Take π as 3.142)
x
a
6
T
b
Rearrange the formula to make x the
subject.
Use this rearranged formula to find x
when the area, A, is 63
What is the length of the rectangle in part c?
8 V = π h( R 2 − r 2 )
a Find V when h = 2 cm, R = 7.5 cm and
r = 2.5 cm, taking π = 3.142.
b Draw a function machine to show how to
find V when starting with R.
c Use the reverse machine to rewrite the
formula to give R in terms of V, h, r and π.
where g is the
÷g
l
7
c
l
Derive the formula for the area, A, of this
rectangle.
2x + 4
b
acceleration due to gravity.
T is the subject of the formula.
a This function machine is for finding T
when you start with l.
Copy and complete the machine.
4
a
÷c
y
Use this machine to write down a formula
with y as its subject.
Use the reverse machine to rewrite the
formula with x as its subject.
V = 1 π r 2h
3
a
Draw a function machine to show how to
find:
i
V, starting with r
ii r, starting with V.
b
Find r if h = 2.1 cm and v = 100 cm3.
(Take π as 3.142)
As you learned in Student Book 8, it is not always
possible to use a function machine to change the
subject of a formula, particularly when the chosen
letter appears more than once (you will learn about this
after Stage 9).
Remember that a formula stays balanced like an
equation. By keeping it balanced you can rearrange the
formula to change the subject.
2 Expressions and formulae
Example 11
Rearrange m =
formula.
6t − 1
5
to make t the subject of the
5m = 6t − 1
5m + 1 = 6t
[divide by 6]
5m + 1
6
or
It does not matter
if t is on the
left-hand side or
right-hand side of
the equals sign,
as long as it is on
its own it is the
subject.
t=
P = x−e
a
Don’t forget
the brackets.
c
f
g− y
Notice that when
(g – y) is in
the denominator
brackets are no
longer necessary
because of the long
fraction line.
P = 2 x−e
5
Step 1: Add e to both sides.
P+e=2x
5
Step 2: Divide by 2 . This is the same as
5
5
multiplying by 2 .
5
( P + e) =
2
x
Don’t forget
the brackets.
Step 3: You can leave your answer like this
or you can write the x first.
x=
M = 4(t − x)
Divide both sides by 4.
M
4
x(g − y) = f
x=
+y
Step 1:
f
x
Step 3: Divide both sides by (g − y).
Make x the subject of these formulae.
a M = 4(t − x)
c
+y
Step 2: Multiply both sides by x so it is no
longer in the denominator.
5m + 1
6
Example 12
f
x
2
5
f
x
g−y=
=t
Take care when the letter you want to make the subject is
negative or in the denominator of a fraction. Example 14
shows you how to avoid mistakes. It is easier if the letter
you are trying to make the subject is positive. It is also
easier if the letter is in the numerator of a fraction.
g=
g=
Step 1: Subtract y from both sides.
[multiply by 5]
[add 1]
b
b
=t−x
5( P + e)
2
This means
exactly the same
thing as
5(P + e )
=x
2
Step 2: Add x to both sides to make it positive.
M
4
+x=t
Step 3: Subtract
x=t−
M
4
from both sides.
M
4
29
2 Expressions and formulae
Exercise 2J
1
5 Make t subject of each formula.
Rearrange the formulae to make each of these
the subject.
m( v − u)
t
a
m in P =
b
c
d
r in V = πr2h
u in v2 = u2 + 2as
x in r2 = (x – a)2 + y2
e
g in T = 2π
c
6 P=
α
a
a
3
b
t
a
S = 4πr2
b
V = 3πrh
c
A = 1 h(R + r)
2
d
V = πh(R2 – r2)
Angelique
Maisie
P = 2(l + w)
P = 2(l + w)
P = 2l + 2w
P
2
P
2
P − 2w
2
b
k
m
a
b
c
d
e
3y + 6x = 27
x = 5y + 10
5(y – 3) = 2x
4x – 2y = 0
3(4 – 2y) = 12x
= l + w
− w = l
= l
Who is correct? Explain your answer.
Consolidation
Example 1
Simplify these expressions.
a
Example 2
6ab 2 − 4 a 2 b − 2ab 2 + a 2 b
= 4 ab 2 − 3a 2 b
b
3
a
2x y
6 x yz
4
=
2× x× x× x× y× y
6× x× x× x× x× y×z×z×z×z
=
2 × x × x × x × y ×y
6 × x × x × x ×x× y ×z×z×z×z
3
=
y
3× x × z× z× z× z
3
a
−4
b
Common denominator is ab:
1
30
Simplify:
3 2
4
=
y
d
T = k(αt + b)
+ mv
Angelique and Maisie each rearrange
P = 2(l + w) to make l the subject. Here is
their work:
P – 2w = 2l
u = v – ft
c
v
d
d
7 By rearranging each of these equations to the
form y = mx + c, find the gradient and the
y-intercept for each line.
Make r the subject of each formula.
4
k
d
b
Rearrange the formula so that the subject is:
l
g
If P = P0(1 + αt), rearrange the formula so that
the subject is:
2
(u + v )t
2
R = PV
mt
s=
a
−4=
b
=
3b 4 a
−
ab ab
3b − 4 a
ab
Example 3
Simplify:
a
3 xz 4
b
x4 × x5
= x4 + 5 = x9
a6 ÷ a 4
= a6 − 4 = a 2
You can look ahead
to Chapter 12 if you
want to find out
more about this topic.
2 Expressions and formulae
c
6a0 = 6 × 1 = 6
d
(2x2)3 = 8x6
Exercise 2
1
Example 4
Expand and simplify:
a
( x + 3)( x − 5)
= x 2 + 3 x − 5 x − 15
2
= x − 2 x − 15
( x − 3)2
2
b
= ( x − 3)( x − 3)
3
b
d
f
y 2 × y6
3x 4 × x 5
y 2 × y3 × y6
Simplify:
5
4
a x ÷x
7
3
c x ÷x
2
7
e x × x ÷ x4
b
d
f
y5 ÷ y5
y8 ÷ y 3
y3 × y 4 ÷ y5
= x 2 − 6x + 9
( x + 3)( x − 3)
Simplify:
a (4 x )0
b 5x 0
3 2
c (2m )
3 2
d (4 xy )
= x 2 + 3x − 3x − 9
e
= x − 3x − 3x + 9
2
c
Simplify:
a x2 × x3
c 2x3 × x 4
e x3 × x4 × x2
= x2 − 9
Example 5
Using the formula M = 10y + 3r4, find M when y = 2.2
and r = –2.
M = 10 y + 3r
f
4
4
= 10(2.2) + 3(−2)4
= 10 × 2.2 + 3 × (− 2)4
= 10 × 2.2 + 3 × 16
= 22 + 48
Example 6
Make x the subject of these formulae.
a
y
3
y
3
y
6
b
h
y = 3(t + 2 x )
i
= t + 2x
[divide both sides by 3]
− t = 2x
[subtract t from both sides]
−t =x
[divide both sides by 2]
2
M=
P
x
−r
M +r =
P
x
x(M + r) = P
x=
P
M +r
[add r to both sides]
[multiply by x to clear the
denominator]
[divide both sides by M + r]
j
5
z 5 x 3 y5
( x 2 y )2 z 3
Simplify:
a y 4 × y6
b a8 ÷ a 4
c x 8 ÷ x 10
d m 7 × m 3 × m −2
4
3
10
e p ×p ÷p
4
5
f 7 x × 3x
g
= 70
3 x 3 y2
6 x 2 y4
8m5
2m 7
q 2 × q10
( q × q9 )2
( p 2 × p3 ) 2
p8
( R 2 × R 4 )3
R10 ÷ R 2
Expand and simplify:
a (x + 6)(x + 2)
b (x − 5)(x + 5)
c (x + 4 )(x − 7)
d (x − 2 )(x + 8)
e (x − 4 )(x − 3)
f (x − 10 )(x − 1)
g ( x + 5)2
h ( x − 9)2
31
2 Expressions and formulae
6
d
Write down a formula for:
a the area, A
b the perimeter, P, of these shapes.
i
x+3
x−5
ii
x−6
9
ab
3n
= j+ y
(n)
f
h=
d + 2s
y
(s)
Make h the subject of these formulae.
c
7
Write as a single fraction.
a
b
c
d
e
f
g
h
8
32
+
)
+h =y
3
A = 2πr2h
1
x
+1=
y
1
h
10 The volume of a sphere, V, is given by the
3
the sphere.
a Make r the subject of this formula.
b What is the radius of a sphere with
volume 200 cm2?
11 Find the value of:
a 3x4 when x = 2
b 10m2 + 5y when m = −0.1 and y = 4
c 4 + y3 when y = −2
d 50h2 + 10h − f when h = 0.4 and f = 3
e 200g4 − 20g when g = 0.3
f 3kw + 20w2 + k when k = −10 and w = 0.1
1
( x + 3)
Rearrange each formula to make the letter in
brackets the subject.
a
b=
(v)
b
T
2v + p
3
y−h
=
4
(h)
c
x=
2m
3
(m)
+y
d
x
2
h
20
formula V = 4 π r 3 , where r is the radius of
d
+d
4
6
3x x
+
5
4
2m
−m
3
10
y 2y y
+ +
3
5
4
3
1
+
m 2m
1− 1
x
3
+ 2
4b
3c
2
( x + 1)
(
R=2 F−
b
Use your formulae to find the areas and
perimeters of the shapes when x = 10.
(r)
e
a
x−2
c
Y = 2(r + P) – t
2 Expressions and formulae
Summary
You should know ...
Check out
1
1
The rules for indices.
am × an = am+n
am ÷ an = am−n
a0 = 1
(a m )n = a mn
Work out:
a b3 × b5
b
c
d
e
2
How to change the subject of a formula.
For example:
2
Make x the subject of y = m x + c
y = m x +c
3
[− c]
y − c=m x
[÷ m]
y−c
m
[square]
x=
3c5 ÷ c3
3a 2 b
2 ab
5a3b 4c2
b 2c 5
(3 x 3 )2
Make x the subject of
these equations.
a y = mx + c
b
y2 = x 2 + c2
c
y=
x+c
p−c
= x
( )
y−c
m
2
How to expand the product of two linear expressions and simplify
the resulting quadratic expression.
For example:
3
Expand and simplify:
a ( x + 5)( x + 1)
b ( x − 3)( x + 4)
c ( x + 2)( x − 6)
d ( x − 3)( x − 8)
2
e ( x + 5)
4
Write as a single
fraction:
Expand and simplify (x + 7)(x − 2)
( x + 7)( x − 2)
= x 2 + 7 x − 2 x − 14
4
= x 2 + 5 x − 14
How to add and subtract simple algebraic fractions by finding
a common denominator.
For example:
d
4
+
3
2y
=
dy + 6
4y
a
e
3
+
2e
5
b
2y
3
c
5
x
+
1
2x
d
1
4
−
+
y
4
d
8p
33
2 Expressions and formulae
5
How to substitute numbers into an expression or formula and find
the result following the rules of BIDMAS
For example:
Using A = π x − π y , find A when x = 5 and y = 2.4.
2
2
Using π = 3.142
A = 3.142 × 52 − 3.142 × 2.4 2
= 3.142 × 25 − 3.142 × 5.76
= 78.55 − 18.09792
= 60.45208
34
5
Using the formula
s = ut + 1 at2, find s
2
when u = 20, a = 2.5
and t = 15.
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Mathematics
Second Edition
9
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