Number & Algebra Revision Notes for Higher Tier -...

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


(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


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


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


(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.


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


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


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


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.


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.


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


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


= 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


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


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


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?


35% = 20

7

100

35 as a fraction

hence 35% of £60 = 20

7 of 60 = 21£60

20

7 3

1


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?


25% = 4

1

100

25 as a fraction

63

4144 2


hence 25% of 144cm = 4

1 of 144 = cm36144

4

1 36

1

63

4144 2


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


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


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%!



2030 {i.e divide by 7 to find 1%}

100% of original value = original value = 29000.290100 people


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%}



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


100% of original value = original value = 54.701£...015.7100


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


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


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


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?


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


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


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


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


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


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


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 =


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”


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


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


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


Special case

Example Simplify (i) 12 (ii) 27 (iii) 75

Hence simplify 752712

0

353332752712

= =

= =

= =

aaa

34 32

39 33

325 35


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


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


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


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


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


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.


(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


(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


(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


(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


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


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


(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


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


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


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


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


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


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


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


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


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


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


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, .


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


2

108

10

y

y

yx

Step3 : Write down your solution

𝑥 = 8, 𝑦 = 2


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



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


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


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


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


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.


(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


Notes


Notes


Notes

Date post:	22-Aug-2018
Category:	Documents
Upload:	dohanh
View:	219 times
Download:	1 times

Number & Algebra Revision Notes for Higher Tier -...

Documents