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Mathematics Number & Algebra Revision Notes for Higher Tier 3/29/2011 Thomas Whitham Sixth Form S J Cooper
Factors, Primes & Prime Factors
Approximations
Fractions: Equivalent, expressing as, fractions of a quantity.
Percentages: Expressing as, Percentage of a quantity, finding the
original amount, compound measures.
Proportion: Ratio, Direct proportion and Inverse proportion.
Standard from.
Surds
Index Notation
Algebra: Collection of like terms, Solving equations
Factorisation
Graphs: Linear.
Solving simultaneous equations.
Sequences
Inequalities
Algebraic Fractions
Thomas Whitham Sixth Form Page 2
Number Revision
Factors
A factor is any number which will divide into a given number an exact
number of times.
Example 3 is a factor of 12 since 3 divides into 12 exactly 4 times.
Example List all the factors of 18.
Factors of 18 = {1, 2, 3, 6, 9, 18}
Prime Factors
A prime number is a number who’s factors are itself and 1.
A prime factor is a factor which is a prime number. For example the prime
factors of 12 are found in the diagram below.
The prime factors of 12 are 2 x 2 x 3 = 22 x 3
Example
Express each of the following as products of their prime factors
(i) 135 (ii) 54
24
6
2
4
3 2 2
Thomas Whitham Sixth Form Page 3
(i) (ii)
135 = 5 x 3 x 3 x 3 54 = 3 x 3 x 3 x 2
= 5 x 33 = 33 x 2
NB Here the highest common factor (HCF) = 33 = 27
Approximation
Example
Find an approximate value of 02.0
8.43.54
02.0
8.43.54
135
5 27
3 9
3 3
54
9 6
3 2 3 3
Nearer 50 than 60 Nearer 5 than 4
Nearer 0 than 1. So
leave as 0.02
Thomas Whitham Sixth Form Page 4
02.0
250
02.0
550
02.0
8.43.54
2
25000
= 12500
Example
(a) Estimate 96.1
25.360.2 giving your answer to 2 significant figures
(b) Evaluate 96.1
25.360.2 giving your answer to 2 significant figures
(a) 5.42
9
2
33
96.1
25.360.2
(b) 3.496.1
45.8
96.1
25.360.2
Example Estimate the value of 02.0
83.7986.4 giving your answer to
one significant figure.
20002
4000
02.0
40
02.0
85
02.0
83.7986.4
Clearly here if we round 0.02 off we
have 0 which won’t do!
Simplify numerator
Get rid of decimal by multiplying
top and bottom by 100
Thomas Whitham Sixth Form Page 5
Example
Lewis uses his calculator to calculate 54.1 x 0.036 and gets
the answer 19.476.
Use estimation to work out whether his answer is reasonable.
54.1 x 0.036 50 x 0.04 = 2
Answer is unreasonable, as he is approximately a factor of 10 out.
Fractions
1. Equivalent fractions
Example
Express each of the following fractions in their simplest form.
(a) 6
10 (b)
24
42 (c)
60
84
(a) Here we notice that we have a common factor of 2. Since 2 will divide into
both 6 and 10.
6
10
2 3
2 5
3
5
(b) 24
42
6 4
6 7
4
7
Thomas Whitham Sixth Form Page 6
(c) 60
84
6 10
6 12
10
12
10
12
2 5
2 6
5
6
With practice you will not require the intermediate step but move from the
given fraction to the final answer.
Example
Eight pupils out of ninety two pupils failed to turn up for their Mathematics
examination.
What fraction of the group (a) failed to turn up? (b) did turn up for the exam?
(a) Fraction who did not turn up = 8
92
2
23 {no marks are awarded for “8 out of 92”}
(b) fraction who did turn up = 21
23 {i.e. the rest}
Example
On Saturday 9000 people won ten pounds on the National Lottery draw.
However 360 people failed to claim their £10 prize. What fraction failed to
claim their prize?
Fraction failed to claim prize = 360
9000
36
900
4
100
1
25
Notice here we have not completely
cancelled down the fraction so we must
repeat the process.
Thomas Whitham Sixth Form Page 7
Example
What fraction has been shaded in for each of the following shapes?
(a)
Here there is a total of 15
squares, of which 3 are shaded
fraction shaded = 3
15
1
5
(b)
Fraction shaded = 15
24
5
6
2. Fraction of a calculator (without calculator)
Example Find 1
448 of
1
448 of means
1
448 {i.e. 48 divided by 4}
Answer = 12
However we have a technique for showing our working as follows:
1
448 12
1
12
Thomas Whitham Sixth Form Page 8
Example Find 3
570 of
3
570 of means
3
570
{again we could find 1
5
of 70 by dividing 70 by 5}
{so we find 3
5 of 70 by dividing 70 by 5 and then multiplying the answer by 3}
3
570
3
570 42
1
14 of
Example Find 2
7238 of
2
7238
2
7238 68
1
34 of 7 23 8
3 42
Example
In a school with 720 pupils, 9
10 stay in school at lunch time, and
3
8 of these
pupils bring a packed lunch. How many pupils bring a packed lunch?
Number staying at school = 9
10720 648
1
72
7 2
9
2 4 3
More complicated divisions may require
some additional calculations at the side
of your page
Thomas Whitham Sixth Form Page 9
Number with packed lunch = 3
8648 243
1
81 8 64881
4. Fractions to percentages/decimals
Example
Express each of the following as fractions in their simplest form.
(a) 45% (b) 0.34 (c) 2.6 (d) 12%
(a) 45% = 45
100
9
20 (b) 0.34 =
34
100
17
50
(c) 2.6 = 26
102
3
5 (d) 12% =
25
3
100
12
5. Addition and subtraction of fractions
Example Work out 4
7
5
7
4
7
5
7
9
71
2
7
Which is easily done provided the denominators are the same.
Example Work out 3
4
5
12
Thomas Whitham Sixth Form Page 10
First change the first fraction with its equivalent in twelfths.
3
4
3 3
3 4
9
12
3
4
5
12
9
12
5
12
14
121
2
121
1
6
Example Work out 51
32
2
5
51
32
2
57
5
15
6
15
711
15
Example Work out 31
71
2
3
31
71
2
32
3
21
14
21
124
21
14
21
110
21
First add the whole numbers together.
Next the common denominator is 15.
hence replace each fraction with its
equivalent in terms of fifteenths.
Subtract the whole numbers and
place each fraction with common
denominator 21.
Since we cannot subtract 14 from 3 w
must use one of the whole numbers.
hence 321
becomes 2421
and we now
have only 1 whole one.
Thomas Whitham Sixth Form Page 11
Example
A piece of wood is 72
5 metres long. If 1
7
15 metres is cut off, what length of
wood is left?
72
51
7
156
6
15
7
15
521
15
7
15
514
15
6. Multiplication and Division of fractions
Example Work out 2
3
5
7
2
3
5
7
2 5
3 7
10
21
Example Work out 4
9
3
7
4
9
3
7
4
213
1
Length of wood left =
To multiply any two fractions
together simply multiply
numerators together and then
multiply denominators together.
In the event that a number on the
numerator has a common factor to a
number on the denominator, cancel
to start with. i.e. 3 will divide exactly
into both 3 and 9.
Thomas Whitham Sixth Form Page 12
Example Work out 11
22
2
5
11
22
2
9
3
2
20
9
10
3
31
3
1
1
10
3
Example Work out 10
13
5
6
10
13
5
6
10
13
6
5
12
13
2
1
Example Work out 35
91
1
2
35
91
1
2
32
9
3
2
9
32
2
3
18
96
1. Invert the second fraction and
change the ‘‘ to a ‘x’ sign.
2. Cancel where possible
3. multiply across.
change fractions to top heavy fractions
Invert second fraction and cancel where possible.
Finally multiply across
Change fractions to top heavy fractions.
Next cancel where possible
Thomas Whitham Sixth Form Page 13
Example
Find the exact value of ba
11 when
3
2a and
5
4b
2
3
2
31
3
2
11
a That is the reciprocal of
3
2 is
2
3
Similarly the reciprocal of 5
4 is
4
5
4
32
4
11
4
5
4
6
4
5
2
311
ba
Example
Louise has 71
4m of ribbon.
She makes 9 skirts and uses 2
5 of a metre for each one. How much ribbon
does she have left?
5
33
5
18
5
29
Amount left = 5
33
4
17
Thomas Whitham Sixth Form Page 14
= 5
3
4
14
m20
133
20
12
20
253
20
12
20
54
Example
Express each of the following as fractions in their simplest form
a) 0.237
b) 0.2373737.....
c) 0.1373737....
a) 1000
237237.0
b) Let ...237237237.0x
...237237.2371000 x
Subtract from previous expression gives 237999 x
Hence 333
79
999
237x and since ...237237237.0x
333
79...237237237.0
Place over common
denominators
Subtract whole quantities
Since we cannot subtract 12
from 5 here , use one of the
whole ones as 20
20
Thomas Whitham Sixth Form Page 15
c) Again Let ...1373737.0x
...73737.13100 x
Subtract from previous expression gives 6.1399 x
Hence 495
68
990
136
99
6.13x
Percentages
1. Expressing as a percentage
Example Express each of the following percentages as
(a) Decimals (b) Fractions in their simplest form.
(i) 30% (ii) 46% (iii) 2% (iv) 154%
(a)(i) 30% = 3.0100
30 (ii) 46% = 46.0
100
46
(iii) 2% = 02.0100
2 (iv) 154% = 54.1
100
154
(b)(i) 30% = 10
3
100
30 (ii) 46% =
50
23
100
46
(iii) 2% = 50
1
100
2 (iv) 154% =
50
271
100
541
100
154
Thomas Whitham Sixth Form Page 16
Example Mark scored 32 out of 40 in a recent mathematics test.
Express his score as a percentage.
Test result = %8010020
16
40
32 5
1
Example The height of a tree increased from 2.73m to 2.98m in one
year. What percentage increase is this?
Increase = 2.98 – 2.73 = 0.25
Percentage increase = %16.910073.2
25.0
2. Percentage of a quantity
a) Without a calculator
Example What is 20% of 40 kg
Method 1: Using fractions
20% = 5
1
100
20 as a fraction
hence 20% of 40kg = 5
1 of 40 = kg840
5
1 8
1
Thomas Whitham Sixth Form Page 17
Method 2 Using unity
10% of 40kg = 4 kg {i.e. divide by 10}
20% of 40 kg = 2 x 4 = 8kg
Example What is 35% of £60?
Method 1: Using fractions
35% = 20
7
100
35 as a fraction
hence 35% of £60 = 20
7 of 60 = 21£60
20
7 3
1
Method 2 Using unity
10% of £60 = £6 {i.e. divide by 10}
5% of £60 = £3 {i.e. half of 10%}
30% of £60 = 3 x 6 = £18
35% of £60 = 18 + 3 = £21
Example What is 25% of 144cm?
Method 1: Using fractions
25% = 4
1
100
25 as a fraction
63
4144 2
Thomas Whitham Sixth Form Page 18
hence 25% of 144cm = 4
1 of 144 = cm36144
4
1 36
1
63
4144 2
Method 2 Using unity
10% of 144cm = 14.4cm {i.e. divide by 10}
5% of 144cm = 7.2cm {i.e. half of 10%}
20% of 144cm = 2 x 14.4 = 28.8cm
25% of 144cm = 28.8 + 7.2 = 36cm
Example
The price of a new television is £176 plus VAT at 17½%.
(a) Work out the VAT to be added to the price of the television.
(b) What is the total cost of the television?
(a) 10% of £176 = £17.60
5% of £176 = £ 8.80 {i.e. half of 10%}
2½% of £176 = £ 4.40 {i.e. half of 5%}
17½% of £176 = £30.80 =VAT {i.e. add the previous answers together}
(b) Total cost = 176+30.80 = £206.80
Division more complicated so
done at the side of our page
Thomas Whitham Sixth Form Page 19
Example The number of girls attending football matches is expected to
increased by 3% this year. If there were 12500 girls attending
matches last year, how many girls are expected to attend this
year?
1% of 12500girls = 125 girls {i.e. divide by 100}
3% of 12500girls = 3 x 125 = 375girls
Number of girls = 12500 + 375 = 12875
3. Finding the original percentage
Here we make use of the formula:
Example The population of Villanova has increased by 63% during the
last five years and is now 124 000. What was its population
five years ago?
Here we have been told the answer. That is the new value is 124 000.
after an increase of 63%
163% of original value = 124 000 {i.e. increase means 63+100%}
1% of original value = ...73.760163
124000
{i.e divide by 163 to find 1%}
New value = Percentage of Original value
Or
New value = Percentage x Original value
Thomas Whitham Sixth Form Page 20
100% of original value = original value = 76074...73.760100
NB original value is always 100% and we have either increased it or
decreased it to find the new value
Example The price of houses in Villanova has increased by 20% during
the last year. If the house costs $36 000 now, what would it
have cost a year ago?
Here we have been told the answer. That is the new value is $36 000.
after an increase of 18%
120% of original value = 36 000 {i.e. increase means 20+100%}
1% of original value = 300120
36000
{i.e divide by 120 to find 1%}
100% of original value = original value = 30000$300100
Example The attendance of Burnley football club fell by 7% in 2001. If
2030 fewer people went to matches in 2001, how many went
in 2000?
Here we have been told the answer. That is the new value is 2030. which
represents the 7%!
7% of original value = 2030
1% of original value = 2907
2030 {i.e divide by 7 to find 1%}
100% of original value = original value = 29000.290100 people
Thomas Whitham Sixth Form Page 21
Example During a Grand Prix race, the tyres on a car are reduced in
weight by 10%. If they weigh 360 kg at the end of the race,
how much did they weigh at the start?
Here we have been told the answer. That is the new value is 360kg. after
an decrease of 10%
90% of original value = 360 {i.e. increase means 100 – 10%}
1% of original value = 490
360 {i.e divide by 90 to find 1%}
100% of original value = original value = kg4004100
Example A car, which failed its MOT test, was sold for £456, thereby
making a loss of 35% on the cost price. What was the cost
price?
Here we have been told the answer. That is the new value is £456. after
an decrease of 35%
65% of original value = 456 {i.e. increase means 100 – 35%}
1% of original value = ...015.765
456 {i.e divide by 65 to find 1%}
100% of original value = original value = 54.701£...015.7100
Thomas Whitham Sixth Form Page 22
Ratio & Proportion
Example Simplify each of the following ratios
a) 4 : 28
b) 18 : 27
c) £3 : £1.80
d) 300m : 5.1km
e) 1250 cm2 : 3 litres
a) 4 : 28 = 1 : 7 {Divide both sides by 4}
b) 18 : 27 = 2 : 3 {Divide both sides by 9}
c) £3 : £1.80 = 300 : 180 {Change both into pence}
= 30 : 18 { Divide by 10}
= 5 : 3 { Divide by 6}
d) 300m : 5.1km = 300 : 5100 { Change both into metres}
= 3 : 51 { Divide by 100}
= 1 : 17 { Divide by 3}
e) 1250 cm2 : 3 litres = 1250 : 3000 {Change both into cm2}
= 25 : 60 {divide both sides by 50}
= 5 : 12 {divide both sides by 5}
Example A school decides to give 20% of the proceeds of a jumble sale
to charity and the rest to the school fund. In what ratio are
the proceeds to be divided?
Charity to school fund = 20 : 80 = 1 : 4
Thomas Whitham Sixth Form Page 23
Example One bottle of wine holds 750 cm2 whereas another holds 1.2
litres. Give the simplest ratio of their capacities.
Ratio = 750cm2 : 1.2 litres = 750 : 1200 {change units into cm2}
= 15 : 24
= 5 : 8
Example Jane took 45 minutes to do her homework, but her sister
Lucy took 1¼ hours. What is the simplest ratio of their times
taken?
Jane to Lucy = 45 : 1¼ = 45 : 75 {change units into minutes}
= 9 : 15 = 3 : 5
Example The standard gauge of railway track is 1.43m. A model is to
be made with gauge 11mm. Calculate the scale in the form 1
: n, where n>1.
Model to Standard gauge = 11mm : 1.43m
= 11 : 1430 {change into mm}
= 1 : 130
Thomas Whitham Sixth Form Page 24
Example £420 is divided between two people in the ratio 2 : 5. Work
out what each person will receive.
There are two ways of looking at this problem.
Method 1
There are a total of 7 parts 7 parts represents £420
1 part represents 60£7
420
2 parts = £60 x 2 = £120 and 5 parts = £60 x 5 = £300
Method 2
The fraction for the first person is 2 out of 7 parts, that is 7
2 and the fraction
for the second person will be 7
5
First person = 120£4207
2 and
second person = 420 - 120 = £300
Both methods work for most cases.
For the following I will use method 1
Thomas Whitham Sixth Form Page 25
Example A sum of money is divided in the ratio 3 : 4 and the smallest
share is equivalent to £27. What is
(a) the amount given to the largest share.
(b) the total amount of money shared?
The smallest share is worth 3 parts so here 3 parts was £27.
1 part = 9£3
27
(a) Largest share = 4 x 9 = £36
(b) Total amount = 36 + 27 = £63
Example Two lines have lengths in the ratio 5 : 2. If the longer line is
15 cm long, find the length of the other line.
Longer line = 5 parts so 5 parts = 15 cm
1 part = cm35
15
Other line = 2 parts = 2 x 3 = 6cm
Example The ratio of my gas bill to my electricity bill was 13 : 5. If my
gas bill was £182, how much was my electricity bill?
Thomas Whitham Sixth Form Page 26
Gas bill = 13 parts so 13 parts = £182
1 part = 14£13
182
Electricity bill = 5 parts = 5 x 14 = £70
Example Three people stake £10 on the national lottery and win £850.
Peter paid £2, John paid £4.50 and Claire paid the rest
towards the stake. The winnings are shared in the ratio of the
contributions. How much does Claire receive?
Ratio of share = Peter to John to Claire = £2 : £4.50 : £3.50
= 200 : 450 : 350
= 4 : 9 : 7
Hence out of 20 parts of the money Claire will receive 7 parts.
20 parts = £850
1 part = 50.42£20
850
Calire = 7 parts = 7 x 42.50 = £297.50
Thomas Whitham Sixth Form Page 27
Example
For every 9 teenagers who like pop music there are 2 that does not.
In a youth club of 187 members, how many do not like pop music?
Ratio = Like pop to not like pop = 9 : 2
11 parts = 187
1 part = 1711
187
Not like pop = 2 parts = 2 x 174 = 34 people
Example A lorry is loaded up with fruit and vegetables for market. The
mass of fruit to vegetables is in the ratio of 7 : 8. If the lorry’s
load is 18.6 tonnes, find the mass of fruit and the mass of
vegetables it is carrying.
15 parts = 18.6 tonnes
1 part = 24.115
6.18 tonnes
Fruit = 7 parts = 7 x 1.24 = 8.68 tonnes
Vegetables = 8 parts = 8 x 1.24 = 9.92 tonnes
Thomas Whitham Sixth Form Page 28
Example When £195 is divided in the ratio 2 : 4 : 7, what is the
difference between the largest share and the smallest?
Difference between the largest and smallest share is 5 parts(7 -2)
13 parts = £195
1 part = 15£13
195
Difference = 5 parts = 5 x 15 = £75
Example
A man and a woman share a bingo prize of £1000 between them in the ratio
1 : 4. The woman shares her part between herself, her mother and her two
daughters in the ratio 2 : 1 : 1
How much does the woman receive?
5 parts = £1000
1 part = 200£5
1000
Woman’s original share = 4 parts = 4 x 200 = £800
Hence 4 parts = £800
1 part = 200£4
800
Woman’s final share = 2 parts = 2 x 200 = £400
Thomas Whitham Sixth Form Page 29
Example £400 is divided between Ann, Brian and Carol so that Ann has
twice as much as Brian and Brian has three times as much as
Carol. How much does Brian receive?
If Carol received one part Brian would have to receive three parts hence Ann
would have to receive six parts. This gives the ratio
Ann : Brian : carol = 6 : 3 : 1
Hence 10 parts = £400
1 part = 40£10
400
Brian = 3 parts = 3 x 40 = £120
Example
Mrs Simms inherits £24 000.
She divides the money between her three children, aged 9, Alice, Brenda,
aged 7 and Charles, aged 8,in the ratio of their ages.
How much does Charles receive?
Ratio = 9 : 7 : 8
24 parts = 24 000
1 part = 1 000
Charles = 8 parts = £8 000
Thomas Whitham Sixth Form Page 30
Direct Proportion
If two quantities are directly proportional to one another then one can be
written as a constant (k) multiplied by the other.
In order to find the constant k, more information needs to be provided.
Example
W and P are both positive quantities.
W is directly proportional to the square of P.
When W = 12, P = 4.
(a) Express W in terms of P.
(b) What is the value of W when P = 6?
(c) What is the value of P when W = 75?
a) Using the definition above
W is directly proportional to the square of P means W = k x P2
Using the information W = 12, P = 4 12 = k x 42 or 12 = k x 16
k = 75.04
3
16
12
W = 4
3P2
b) P = 6 W = 27364
36
4
3 2
= k x
Don’t forget once k is found to
write the equation down
Thomas Whitham Sixth Form Page 31
c) W = 75 2
4
375 P
2
75.0
75P
2100 P
10P
Example
Y are X are both positive quantities.
Y is directly proportional to the square root of X.
When Y = 16, X = 16
a) Express Y in terms of X
b) What is the value of Y when X = 25?
c) What is the value of X when Y = 60?
a) XkY
Y = 16, X = 16 1616 k
4
416
k
k
XY 4
b) X = 25 2054254 Y
c) Y = 60 X460
X15
Thomas Whitham Sixth Form Page 32
225
152
X
Inverse Proportion
If two quantities are inversely proportional to one another then one can be
written as a constant (k) divided by the other.
In order to find the constant k, more information needs to be provided.
Example
Given that M varies inversely to P and that M = 12 when P = 4
a) Obtain an expression for M in terms of P
b) What is the value of M when P = 8?
c) What is the value of P when M = 0.5?
a) P
kM
M = 12, P = 4 4
12k
k412
48k
k =
Thomas Whitham Sixth Form Page 33
PM
48
b) P = 8 68
48M
c) M = 0.5 P
485.0
965.0
48P
Example
y is inversely proportional to the square root of x.
When y = 6, x = 9.
a) What is the value of y when x = 4.
b) What is the value of x when y = 10.
y is inversely proportional to the square root of x means x
ky
When y = 6, x = 9. 9
6k
or 3
6k
k36
18k x
y18
a) X = 4 92
18
4
18y
b) Y =10 x
1810
Using the algebraic knowledge that
the P and 0.5 can be “swapped”
Thomas Whitham Sixth Form Page 34
10
18x
8.1x
24.3x
Standard form
A number expressed in standard form is a number written between 1 and 10
multiplied by 10 to an appropriate power.
The use of standard form is to represent very small numbers or very large
numbers
For example
0.000 000 000 000 32 represented in standard form will be 13102.3
214000 000 000 represented in standard form will be 111014.2
Example Express each of the following in standard form
(i) 0.000 000 462
(ii) 0.004
(iii) 90 000
(iv) 5910 000 000
(i) 0.000 000 462 = 71062.4
(ii) 0.004 = 3104
(iii) 90 000 = 4109
(iv) 5910 000 000 = 91091.5
Thomas Whitham Sixth Form Page 35
Example Express each of the following as ordinary numbers
(i) 8106.1
(ii) 61054.2
(iii) 12107
(iv) 210414.9
(i) 8106.1 = 160 000 000
(ii) 61054.2 = 0.000 00254
(iii) 12107 = 7000 000 000 000
(iv) 210414.9 = 0.09414
Example The surface of the earth is about 509 970 000 km2. Express
this in standard form correct to two significant figures.
509 970 000 = 8101.5 km2
Example
Given that 4106A and 7104B work out, without a calculator
(i) AB (ii) B
A
(i) (ii)
12
11
74
74
104.2
1024
10106
104106
AB
3
7
4
7
4
105.1
10
10
4
6
104
106
B
A
Thomas Whitham Sixth Form Page 36
NB Here we used the laws of indices. That is nmnm aaa and
nm
n
mnm a
a
aaa
Example
Given that 5102.1 X and 9105 Y work out , in standard form
(i) XY (ii) Y
X
(i) (ii)
14
95
106
105102.1
XY
3
9
5
104.2
2400
105
102.1
Y
X
Surds 5,3,2 are irrational numbers expressed in surd form.
A rational number is one which can be written in the form q
p where p and q
are integers. An irrational number cannot be written in this form.
2 for example is an irrational number, and so is the numbers .
Numbers written in the form a are called surds.
Laws
abba
b
a
b
a
Thomas Whitham Sixth Form Page 37
Special case
Example Simplify (i) 12 (ii) 27 (iii) 75
Hence simplify 752712
0
353332752712
= =
= =
= =
aaa
34 32
39 33
325 35
Thomas Whitham Sixth Form Page 38
Example
Express with rational denominator 5
20
Here we multiply by one! However we make one 5
5 as this will help us.
545
520
5
5
5
20
5
20
Example
Given that 3
91227 can be express in the form 𝑎 3. Find the value
a.
333927
323412
333
39
3
3
3
9
3
9
323332333
91227
From the fact that
555
Thomas Whitham Sixth Form Page 39
Example simplify (i) 1898 (ii) 4875
(i) 242327292491898
(ii) 4
5
34
35
316
3254875
Index notation
Laws
Special Cases 10 a
aa 2
1
, n aa n 1
,
nmnm aaa
nmnm aaa
mnnm aa
Thomas Whitham Sixth Form Page 40
n
n
aa
1 , n
na
a
1 ,
nn
a
b
b
a
n maa nm
mn a
Example 288 33
1
Example 3
1
9
1
9
19
2
1
2
1
Example
(a) Write down the value for each of the following
(i) 06 (ii) 32
27 (iii) 43
(i) 160 {anything to the power 0 is 1}
(ii) 93272722
332
(iii) 81
1
3
13
4
4
(b) Simplify 25.0 336
3
2
9
16
3
136336
2
25.0
Power 0.5 means square root
Thomas Whitham Sixth Form Page 41
Example
Simplify each of the following (i) 34 xx (ii) 75 yy (iii) 53t
(i) 734 xxx {Add the powers}
(ii) 275 yyy or 2
1
y {subtract the powers}
(iii) 1553 tt {multiply the powers}
Example
Simplify
24
323
6
23
ba
baa
ba
ba
ba
ba
baa
ba
baa
5
24
39
24
363
24
323
4
6
24
6
83
6
23
Working out 2 cubed
Multiplying powers for a
Multiplying powers for b
Adding powers for a on
numerator Working out
3 x 8
Subtracting powers for b Dividing 24
by 6
Subtracting powers for a
Thomas Whitham Sixth Form Page 42
Algebra Revision
Collection of like terms
When collecting like terms remember we can collect together equivalent
letters i.e. aaa 1037 and we can collect together numerical terms i.e. 4 + 8 – 3 = 9
However we cannot collect together terms that are not alike i.e.
ba 54 cannot be simplified. Nor can 43 a Example Simplify each of the following
a) bababa 855439 {Remember to take note of the sign
in front of each letter}
b) yxyxyx 87626
c) fedfdefed 27647324
d) qpqpqp 553782
Solving simple equations
Example Solve each of the following equations
a) 1743 x
b) 7325 xx
c) 9457 xx
d) 19231 xx
e) 1214 x
Thomas Whitham Sixth Form Page 43
f) 7523 x
g) 53
52
x
h)
95
73
x
Golden rules The equation starts balanced and must remain balanced! So whatever you do to one side you must do to the other side.
i.e. (a) 1743 x
417443 x {Add 4 to both sides of the equation}
213 x {Simplify}
3
21
3
3
x {Divide both sides by 3}
7x {Simplify giving answer}
An alternative way of thinking! When moving a number from one side of the
equal sign to the other we perform the opposite operation.
The opposite of Addition is subtraction and visa versa.
The opposite of Multiplication is Division and visa versa.
Thomas Whitham Sixth Form Page 44
(b) 7325 xx
52
{Simplify} 52
subtract} andover 3 the{take 535
2}subtract andover 2 the{take 2735
7325
.x
x
xxx
xx
xx
(c) 9457 xx
33333.13
4
43
447
5947
9457
x
x
xx
xx
xx
(d) 19231 xx
4
{Simplify} 520
add} andover 3 the{Move 3220
}add! andover 19 the{Move 23191
19231
x
x
xxx
xx
xx
Thomas Whitham Sixth Form Page 45
(e) 1214 x
4
164
before} as rearrangecan we{now 4124
first} brackets {remove 1244
1214
x
x
x
x
x
(f) 7523 x
333333.13
4
6
8
86
1576
7156
7523
x
x
x
x
x
(g) 53
52
x
5
102
5152
times}andover 3 the{take 1552
53
52
x
x
x
x
x
Thomas Whitham Sixth Form Page 46
(h)
95
73
x
8
243
21453
bracket} the{Remove 45213
times}andover 5 the{Take 4573
95
73
x
x
x
x
x
x
Factorisation
Example Factorise each of the following
a) 342 xx
b) 652 xx
c) 1582 xx
d) 542 xx
e) 1522 xx
f) 62 xx
g) 2832 xx
When factorising a quadratic such as 342 xx we express as two
brackets multiplied together. i.e. bxax where a and b are numbers
that
1. Multiply to give, in this case, 3
2. add to give 4.
This means they can be 1 and 3 or –1 and –3.
Since 1 and 3 add to give 4 then 31342 xxxx
Thomas Whitham Sixth Form Page 47
(b) 652 xx
Here the numbers could be 1 and 6, –1 and –6, 2 and 3 or –2 and –3.
Since –2 and –3 add to give –5 the answer is found.
32652 xxxx
(c) 1582 xx
Here the numbers could be 1 and 15, -1 and –15, 3 and 5 or –3 and –5.
Since 3 and 5 add to give 8
531582 xxxx
(d) 542 xx
Here the numbers could be 1 and –5 or –1 and 5 (one positive and one
negative.)
Since 1 and –5 add to give –4.
51542 xxxx
(e) 1522 xx
Here the numbers could be 1 and –15, –1 and 15, 3 and –5 or –3 and 5
Since –3 and 5 add to give 2
531522 xxxx
(f) 62 xx
Thomas Whitham Sixth Form Page 48
Here the numbers could be 2 and –3, –2 and 3, 1 and –6 or –1 and 6
Since 2 and –3 add to give –1.
3262 xxxx
(g) 2832 xx
Here the numbers could be 1 and –28, –1and 28, 2 and –14, 14 and –2, 4 and
–7 or –4 and 7.
Since –4 and 7 add to give 3.
742832 xxxx
Example Factorise each of the following:
a) xyx 32
b) qppq 22 4
c) mnmm 6104 2
d) abcbaba 223 37
Here we cannot place into two brackets since it does not follow the
pattern of 2x followed by x, followed by a constant!
However we have a common factor. So place the common factor outside
a bracket.
a) 332 xyxxyx
Since
yxxyx 2
Since
xx 33
Thomas Whitham Sixth Form Page 49
b) pqpqqppq 44 22
c) nmmmnmm 35226104 2
d) cabaababcbaba 3737 2223
Difference of squares
Example factorise each of the following
(i) 𝑥2 − 36 (ii) 4𝑥2 − 49
(i) Here 𝑥2 − 36 represents the difference of two squares which
are 𝑥2 − 62. When we factorise 𝑥2 − 36 we are looking for two
numbers that have a sum of zero (for the middle term) this is
+6 − 6 = 0
Hence 𝑥2 − 36 = 𝑥 − 6 𝑥 + 6
(ii) 4𝑥2 − 49 = 2𝑥 − 7 2𝑥 + 7
Example (a) factorise the quadratic 1272 xx
(b) Hence solve the equation 01272 xx
(a) 431272 xxxx {since –3 and –4 add
to give –7.
2𝑥 2 72
Thomas Whitham Sixth Form Page 50
(b) 01272 xx is the same as saying 043 xx
Which means 03 x or 04 x
Since 0ba means 0a or 0b
3x or 4x
Example (a) factorise the quadratic 2762 xx
(b) Hence solve the equation 02762 xx
(a) 392762 xxxx
(b) 02762 xx
039 xx
09 x or 03x 9x or 3x
Inequalities
Example Solve each of the following inequalities
a) 932 x
b) 1973 x
c) 8314 xx
d) 7335 xx
e) 9142 x
Thomas Whitham Sixth Form Page 51
f) 73
5
x
g) 252 x
h) 812 x
Solving inequalities can be like solving equations. However don’t
forget to write the correct symbol; and not the equal sign!
a) 932 x
6
{Simplify} 122
side}other the to3 {Add 392
932
x
x
x
x
b) 1973 x
4
123
7193
1973
x
x
x
x
c) 8314 xx
7
734
1834
8314
x
xx
xx
xx
Thomas Whitham Sixth Form Page 52
d) 7335 xx
2
42
435
3735
7335
x
x
xx
xx
xx
e) 9142 x
8
11
118
298
928
9142
x
x
x
x
x
f) 73
5
x
16
521
215
73
5
x
x
x
x
Thomas Whitham Sixth Form Page 53
g) 252 x
5
252
x
x
However when dealing with a quadratic we have two solutions. NB If
6x then 362 x which is also greater than 25. Hence 5x is
a second solution.
h) 812 x
9 and 9
812
xx
x
Example
Given that n is an integer, find the values of n such that −7 ≤ 2𝑛 < 6
−3.5 ≤ 𝑛 < 3 {divide throughout by 2}
Hence n can equal −3,−2,−1 ,0 ,1 ,2
Example List the values of n such that n is an integer value and
a) 23 n
b) 51 n
c) 34 n
Whole number
Thomas Whitham Sixth Form Page 54
Here we are not asked to find the solution by solving an equation but
by listing the possible solutions.
An integer value means possible whole number answers whether
positive or negative.
a) 23 n means n can be –2 –1, 0 or 1 we write
1,0,1,2 n
b) 51 n Answer: 5,4,3,2,1,0n
c) 34 n Answer: 2,1,0,1,2,3,4 n
Graphs 1. The straight line
Example Draw the graph of 𝑦 = 3𝑥 − 4 for values of 𝑥 from −3 to 3
For any straight line we can get away with plotting three points and then
a line through these three points.
Three points are selected to make sure we have only one straight line.
X – 3 0 3
y – 13 – 4 5
13
49
433
y
4
403
y
11
415
453
y
Thomas Whitham Sixth Form Page 55
Example Draw the graph of 𝑥 + 𝑦 = 7 for values of 𝑥 from −1 to 7
X 4 0 3
y 3 7 4
3
74
y
y 70 y
4
73
y
y
2
4
6
-2
-4
-6
-8
-10
-12
-14
2 -2 0 x
y
Thomas Whitham Sixth Form Page 56
Gradient of a line
With coordinates 11, yxA and 22 , yxB
2
4
6
8
10
-2
2 4 6 8 -2 0 x
y
y
x 0
11, yxA
22 , yxB
Thomas Whitham Sixth Form Page 57
Gradient = 12
12
xx
yym
Or alternatively gradient xinchange
yinchange
Example
Work out the gradient of the line
opposite
Parallel and perpendicular lines
Let two lines have gradients 1m and 2m
Lines parallel 21 mm
Lines perpendicular 121 mm or 2
1
1
mm
2
4
6
8
10
-2
2 4 6 8 -2 0 x
y
Gradient = 29
26
= 7
8
Thomas Whitham Sixth Form Page 58
Equation of a straight line
The equation of a straight line should be expressed in the form
𝑦 = 𝑚𝑥 + 𝑐
Example
Find the equation of the straight line passing through the points (1, 3) and
has gradient 5
Here we know that 𝑚 = 5 so 𝑦 = 5𝑥 + 𝑐
Also we know that when 𝑥 = 1, 𝑦 = 3
Which gives 3 = 5 + 𝑐 i.e. 𝑐 = −2
Hence the equation is 𝑦 = 5𝑥 − 2
Example
Find the equation of the straight line passing through the points (2, 5) and
(4, 11)
The y-intercept
(where it crosses
the y-axis)
The gradient of
the graph
Thomas Whitham Sixth Form Page 59
Here we must first find the gradient.
Gradient = 24
511
= 32
6 so 𝑦 = 3𝑥 + 𝑐
Now we have choice of which point to use (2, 5) or (4, 11)
Using (2, 5) 5 = 6 + 𝑐 so 𝑐 = −1
However using (4, 11) would give the same value
i.e. 11 = 12 + 𝑐
Hence equation is given by 𝑦 = 3𝑥 − 1
Example
Write down the gradient to the line 5𝑥 + 3𝑦 = 6
3𝑦 = 6 − 5𝑥
xy3
52
Hence the gradient is 3
5
Sequences
Here we must rearrange the
equation to be in the form
𝑦 = 𝑚𝑥 + 𝑐
Thomas Whitham Sixth Form Page 60
i) The simplest sequence is the sequence of natural numbers
i.e. 1, 2, 3, 4, 5, 6, .......
ii) Other simple sequences are
1, 3, 5, 7, 9, ....... [Odd numbers]
2, 4, 6, 8, 10, ....... [Even numbers]
2, 3, 5, 7, 11, 13, ...... [Prime numbers]
iii) Continuing a sequence
Example
Find the next two terms in the following sequences:
a) 2, 5, 8, 11, ..... Answer = 14, 17, {add three to the previous term}
b) 1, 4, 9, 16, .... Answer = 25, 36 {square numbers}
c) 1, 1, 2, 3, 5, 8, ... Answer = 13, 21
{sum of the last two gives the next term}
iv) The nth term of a sequence
The nth term of any sequence can be written in many forms
nth term = nn TU etc..
in each case the nth term is a method of calculating directly a given term
for a sequence.
Example
Given the nth term of a sequence is 14 nU n , write down the first
four terms to the sequence.
In this case the first term 1U is found by replacing n with 1 in the formula
3141141 U
Similarly 7181242 U
111121343 U
151161444 U
Thomas Whitham Sixth Form Page 61
Hence first four terms are 3, 7, 11, 15
Example
Using the nth term 1 nnU n , write down the first four terms to the
sequence.
21111 U
61222 U
121333 U
201444 U
2, 6, 12, 20
v) Finding the nth term
There are an infinite number of expressions for the nth terms, but here
are a few pointers to look for.
1. If the sequence goes up by the same amount each time (i.e. its
linear) then the sequence can always be written in the form
cmnU n , where 𝑚 is the common difference and 𝑐 is the
number which must be added to obtain the first value.
Example
Find the formula for the nth term of the sequence 5, 8, 11, 14, ....
Here the common difference is 3 so the formula must involve nU n 3
-however this would generate the sequence 3, 6, 9, 12, ... which is 2
short each time
23 nU n
Example
Obtain an expression for the nth term of the sequence 4, 11, 18, 25, .
Thomas Whitham Sixth Form Page 62
Here the common difference is 7 so the formula must involve nU n 7
-however this would generate the sequence 7, 14, 21, 28, ... which is 3
over each time
37 nU n
Simultaneous Equations
1. Elimination method
Remember
If the signs are different we add the equations, and
If the signs are the same we subtract.
Example Solve the equations
6
10
yx
yx
Step1: Eliminate y by adding the two equations
8
16 2
6
10
x
x
yx
yx
Step2 : Place the value of 𝑥 = 8 into the first equation
Thomas Whitham Sixth Form Page 63
2
108
10
y
y
yx
Step3 : Write down your solution
𝑥 = 8, 𝑦 = 2
Example Solve the equations
523
82
yx
yx
Step1: Multiply an equation by a number so that we can eliminate x or y.
e.g. multiply the first equation by 2 to give 1624 yx
Step2 : Eliminate the y by adding the two equations
3
21 7
523
1624
x
x
yx
yx
Step3 : Place the value of 𝑥 = 3 into the first equation
2
86
82
y
y
yx
Step4 : Write down your solution
𝑥 = 3, 𝑦 = 2
Thomas Whitham Sixth Form Page 64
Example Solve the equations
634
856
yx
yx
634
856
yx
yx
301520
241518
yx
yx
x2 6
3x
Put this back into one of the original equations
856 yx becomes 8518 y
Hence 105 y
2y
𝑥 = 3,𝑦 = 2
2. Substitution method
Remember
Rearrange one of the equations and substitute into the other
equation and solve!
To eliminate y multiply this
equation by 3
And this one by 5. And then
subtract the two equations
Thomas Whitham Sixth Form Page 65
Example
Example Solve the equations 32,542 xyxxy
Step1 : replace 𝑦 with 2𝑥 − 3 in 𝑦 = 𝑥2 − 4𝑥 + 5
2𝑥 − 3 = 𝑥2 − 4𝑥 + 5
Step2 : tidy up the equation and solve for 𝑥
0 = 𝑥2 − 6𝑥 + 8
0 = 𝑥 − 2 𝑥 − 4
𝑥 = 2, 𝑥 = 4
Step3 : place found values of 𝑥 into either equation to find 𝑦
𝑥 = 2 𝑦 = 2 2 − 3 = 1 𝑥 = 2, 𝑦 = 1
𝑥 = 4 𝑦 = 2 4 − 3 = 5 𝑥 = 2, 𝑦 = 5
Example Solve simultaneously 2 ,832 2 xxyyx
Here we substitute for y from the second equation into the first
832 yx
8232 2 xxx
023 2 xx
Thomas Whitham Sixth Form Page 66
0123 xx 1 ,32x
when 32x ,
928
32
94 2 y
when 1x , 2211 y
The geometrical interpretation here is that the straight line 832 yx
and the parabola 22 xxy intersect at points 928
32 , 2,1
Rearranging equations
Example 𝐶 = 2𝜋𝑟 .......rearrange to make 𝑟 the subject
rC 2
rC
2
divide both sides by 2
Example 𝐴 = 𝜋𝑟2 ............rearrange to make 𝑟 the subject
x
y
(1,2)
0
928
32 ,
} Solutions, 928
32 , yx
2 ,1 yx
Thomas Whitham Sixth Form Page 67
𝐴 = 𝜋𝑟2
2rA
divide both sides by 𝜋
rA
square root both sides
Example 𝑣 = 𝑢 + 𝑎𝑡 .......make a the subject
𝑣 = 𝑢 + 𝑎𝑡
𝑣 − 𝑢 = 𝑎𝑡 take u to the other side
at
uv
divide both sides by t
Example 𝑦 = 𝑥2 − 3 ...... make 𝑥 the subject
𝑦 = 𝑥2 − 3
𝑦2 = 𝑥2 − 3 remove the square root by squaring the other side
𝑦2 + 3 = 𝑥2 take the 3 over to the other side
𝑦2 + 3 = 𝑥 finally square root to find x
Thomas Whitham Sixth Form Page 68
Algebraic Fractions
Example
Simplify each of the following
(i) x
x
2
8 3
(ii)
221
2
xx
x
(iii) 2
322
2
xx
xx
(i) 23
42
8x
x
x
(ii)
221
2
xx
x =
221
2
xx
x
= 21
1
xx
Using laws of indices 𝑥3 ÷ 𝑥 = 𝑥2
8 ÷ 2 = 4
Here we have a
common factor on
the numerator and
denominator. Hence
cancel one on the
top with one of
them on the
bottom.
Thomas Whitham Sixth Form Page 69
(iii) 2
322
2
xx
xx= 21
13
xx
xx
= 21
13
xx
xx
= 2
3
x
x
Example
Solve the equation 72
3
3
12
xx
72
3
3
12
xx
762
36
3
126
xx
5
357
4277
429324
4233122
x
x
x
xx
xx
Factorise numerator and
denominator
Here we have a common
factor of 𝑥 − 1 on top and
bottom
Multipliy throughout by 2 and 3.
i.e. multiply throughout by 6.
Remove the brackets and collect
like terms
Rearrange and solve the now
simple equation
Thomas Whitham Sixth Form Page 70
Notes
Thomas Whitham Sixth Form Page 71
Notes
Thomas Whitham Sixth Form Page 72
Notes