1

Year 9 set 1 Mathematics notes,

to accompany the 9H book. Part 1:

equations 1.2 (p.12), 1.6 (p. 44), 4.6 (p.196)

sequences 3.2 (p.115)

Pupils use the Elmwood Press “Essential Maths” book by David Raymer (9H for set 1, 9C

for sets 2 & 3). This is essentially a book of practice questions, with very little

explanation – hence these notes. The notes are intended for revision and as a guide to

further study: they do not include as many examples as are discussed in lessons – if

they did, they would be too long to read easily and not really “notes”.

These notes are divided into sections in chronological order. A few items appear earlier

than originally planned in the scheme of work, to match topics coming up in the half-

termly tests.

Pupils should follow the hyperlinks for additional explanation and examples on MyMaths

and other websites (via online pdf version, not the paper copy).

Contents

Scheme of Work. ...................................................................................................................................... 2

Algebra ......................................................................................................................................................... 4

Rules: ......................................................................................................................................................... 4

Multiplying out brackets ..................................................................................................................... 5

Squaring brackets............................................................................................................................. 6

Dividing brackets by something ................................................................................................... 6

Negative numbers ................................................................................................................................. 7

Adding and subtracting negative numbers ................................................................................ 7

Multiplying and dividing by negative numbers .......................................................................... 7

Formulae and function machines ...................................................................................................... 8

Solving simple equations ................................................................................................................... 10

Solving equations using algebra ...................................................................................................... 10

Trial and improvement ........................................................................................................................ 11

Simultaneous Equations ...................................................................................................................... 11

Sequences .................................................................................................................................................. 15

2

1. Reminder about linear sequences ............................................................................................. 15

2. What do “linear” and “quadratic” mean?............................................................................... 16

3. Quadratic sequences - finding the values from the formula: ......................................... 16

Scheme of Work. Modules refer to the National Numeracy Strategy – see my website for descriptions.

Module Core Content Textbook Chapters

A1/2 Linear sequences, nth term; represent problems and

synthesise information in graphical form.

9H Unit 1: 1.2, 1.6

Unit 3:3.2

N1 4 rules of fractions (also in HD2); proportional

reasoning; multiplying and dividing by numbers between 0

and 1. Percentages

9H Unit 1: 1.1

Unit 2: 2.1

Unit 3; 3.3, 3.5

A3 Construct and solve linear equations; trial and

improvement

9H Unit 1; 1.2

Unit2: 2.6

CAME, Functional Skills etc Unit 2:2.4, CAME A5

October Half Term

SSM1 Angles in a polygon; angle properties; loci 9H Unit1: 1.3

Unit 3: 3.1, Unit 6:

6.3

HD1 Design a survey; communicate interpretations of results

using tables, graphs and diagrams.

9H Unit 4: 4.2(not

Ex 2M, 2E and 3M),

4.4

Unit 2: 2.5

CAME, Functional Skills etc Unit 4: 4.5 CAME S4

End of Term, Christmas Break

SSM2 Units of measurement, area and circumference of a

circle.

9H Unit 4: 4.3

Unit 6: 6.2

N2 Powers of 10; 4 rules of decimals; calculator methods. 9H Unit 2: 2.2, 2.3

CAME, Functional Skills etc CAME S5, Exam qu

February Half Term

A4 Gradient and intercept of straight line graphs; real life

graphs; integer powers and roots.

9H Unit4: 4.2 (Ex

2M, 2E, and 3M) 4.6

(graphs),

Unit 5: 5.4

HD2 Probability; 4 rules of fractions 9H Unit 5: 5.3

SSM3 Transformations

Similarity and congruence

9H Unit 4:4.1

Unit 1: 1.4 Unit 6:6.5

CAME, Functional Skills etc CAME D3, Exam Qu

End of Term, Easter Break

A5 Formulae 9H Unit 4: 4.6,

3

Unit 5; 5.2

Unit 6:6.2

SSM3 3D shapes 9H: Unit 3: 3.1, 3.4,

HD Statistical investigation, to include use of Autograph 9H Unit1: 1.5

CAME, Functional Skills etc 9H Unit 5: 5.5

CAME N5, Exam Qu

May Half Term

SSM4 Pythagoras’ theorem; circle theorems 9H Unit 5: 5.1,

A6 Square a linear expression; inequalities. 9H Unit 6:6.1, 6.4,

Unit 4: 4.6

Unit 5:5.1

Fully Functional 3 9H Unit 5: 5.5

CAME D4, Exam Qu

Variations: 4.6 (simultaneous equations) is now covered in October because it is needed

for test 1; Pythagoras in November for test 2.

4

Algebra Algebra is the use of letters (x, y, z etc.) in place of numbers. It lets us solve

complicated problems by splitting them into easier stages:

(1) Write an equation (must include an = sign)

(2) Use algebra to manipulate the equation and solve it

Without algebra we would have no equations !

Rules:

We leave out the × (times) sign e.g. 2 x becomes 2x .

The meaning is defined by BIDMAS (e.g 2 5x means 5x x not 5 5x x )

– see notes part 2.

“Collecting terms”: we can add similar terms ( 2 3 5x x x but 22 3x x cannot be

turned into a single term since x and 2x are not identical – same for 2 3x y )

The = sign means that the left side and the right side have the same value

o We can only do things that preserve this truth. E.g. if 2 10x I can also

write 2 3 10 3 13x so that I have “done the same thing to both

sides”

o This is why it is so horribly wrong to write calculations like

2×3 = 6 + 10 = 16/4 = 4: the first two = signs are completely untrue. If

you want to show working for 2 3 10

4

, start with a complete expression

and simplify it in stages:

2 3 10 6 10 164

4 4 4

“Look, each equals sign is true – it has the

same value on each side!”

5

Multiplying out brackets

The “outside” term (2) multiplies both the inside terms.

( )

( )

( )

For “brackets brackets”, each term in the first bracket multiples every term in the

second bracket.

3 ways to think of this:

(1) draw lines reminding you which terms to multiply, think FOIL (First in

each bracket, Outer, Inner, Last):

2 22 3 3 2 6 5 6x x x x x x x

(2) split the first bracket into two terms:

2 22 3 3 2 3 3 2 6 5 6x x x x x x x x x x

(3) multiply using boxes, then sum the terms:

x 2

x x2 2x

3 3x 6

6

Squaring brackets

From BIDMAS, we know that 23p means 3 p p

2

3p must mean something different!

“The contents of the brackets are multiplied by themselves”, 23 3 9p p p

Examples

2 2 2ab ab ab a b

22 2 2 2 4

2

5 5 5 25xy xy xy x y

z z z z

(multiplying fractions: multiply the numerators, multiply the denominators).

2 2 2a b a b a b a ab ba b , simplify:

2 2 22a b a ab b

Dividing brackets by something

When a fraction has a two or more terms in the numerator, they are all divided by the

denominator.

Example:

6 4 6 43 2

2 2 2

a aa

The top of the fraction behaves as if it were written with brackets.

It could also be written as 6 4 2 3 2a a

Remember that 1x

x (true for all x values except zero: could be

2

2,

10

10 etc).

Hence 10

10a

a and

220 204

5 5

a a aa

a a

Remember that (a+b)2 is never

equal to a2+b2

For instance (1+2)2 = 32 = 9,

NOT 5

7

Negative numbers

Adding and subtracting negative numbers

To add a positive number, move to the right → on the number line

o To subtract a positive number, move to the left ← on the number line

To add a negative number, move to the left ← on the number line

[ think, 5+(-2) is the same as 5-2=3 ]

o To subtract a negative number, move to the right → on the number line

[Think, 5-(-2) is the same as 5+2=7 ]

Multiplying and dividing by negative numbers

Every time we multiply or divide by a negative number, we change the sign (positive or

negative) of our answer.

if your multiplication contains an even number of –signs, your answer is positive

if your multiplication contains an odd number of –signs, your answer is negative

Examples:

2 3 6

2 3 4 24

2 1

4 2

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10

x

y

8

Formulae and function machines

A formula is just a mathematical description of how to do a calculation.

You put numbers instead of the letters on the right-hand side and get a value out

for the left-hand letter.

We call the letters “parameters” or “variables” since they are not a fixed number

– they can take any value.

It may help to think of the letter as a bucket.

It has a name painted on the side (x, y z etc) and there is a number carried

inside it. Often we do not know the number and have to do calculations just using

the bucket’s name. Sometimes we can solve an equation – this tells us the number

inside the bucket.

For instance, to find the area of a circle we would write “area = pi × radius squared” as

a formula:

2A r

If you have used Excel you will have

come across formulae to automatically

calculate cell values

Here the “= B1-B2” is a formula that

will calculate the profit as the number in

cell B1 minus the number in cell B2.

When I finish entering the formula it

automatically does the calculation:

A is the subject, it

show us the purpose

of the formula

The right hand side

show us how to find a

value for A

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We can often represent a formula as a “function machine” that shows the stages in the

calculation that “build up” the final answer.

2A r could be shown as:

We can “go backwards” through a function machine to reverse the process:

e.g. A = 50 m2, find r

5015.92

(this is 2r )

15.92 3.99 mr

The main virtue of function machines is simply to get you thinking about the order of

operations: it is “another way” of visualising the process. The formula P = 4(5n-2) would

be

Square r A

Square root

Ar

×5 -2 ×4 n P

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Solving simple equations

One way of thinking about equations is in terms of a function machine where the output

is known. E.g. to solve 3x+5=23

Solving equations using algebra

Algebra is a better way of solving equations because it can cope with more complex

problems. We use a series of steps that result in all the letters being on the left and

numbers on the right. Possible steps:

Adding or subtracting a term from each side

Multiplying or dividing each side by a number of letter

Multiplying out brackets

Collecting terms and simplifying

Swapping the sides (4 = x is better written as x = 4)

(Year 10) factorising

It is important to show all your working, both to collect “method” marks and so that you

can check it.

Examples:

(i) Solve 3 5 23x

Add 5 to each side to cancel out the -5 on the left:

3 5 5 23 5x , simplify:

3 18x

Divide each side by 3 to turn 3x into x:

186

3x

×3 +5 x 3 5x

÷3 -5 2318

6

3 5 23x

11

(ii) Solve 9 7 7 15x x

Take 7 from each side to cancel out the +7 on the left:

9 7 7 7 15 7x x , simplify:

9 7 8x x

Take 7x from each side, so there are no x terms on the right:

9 7 7 7 8x x x x

Hence 2 8x , 4x

Always finish by checking the answer works in the original equation. Putting

4x into the equation gives LHS = 9 7 36 7 43x , RHS =

7 15 28 15 43x - both the same, so the equation is satisfied.

(iii) Solve 2 12 7 3x x

Multiply out the brackets: 2 24 7 21x x

Most of the x terms are on the right, so swap sides:

7 21 2 24x x

Add 21 each side:

7 2 24 21 2 45x x x

Take 2x from each side;

5 45x hence (÷5) 9x

Trial and improvement

Simultaneous Equations

It is a general rule in algebra that we can have equations with more than one unknown

(e.g. x, y and z) but we cannot solve them unless we have as many equations as we

have unknowns. So we need:

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one equation if our problem only contains x,

two equations to find x and y,

three for x, y and z,

100 equations for 100 unknowns.

To solve such equations we need to combine them into simpler equations until ultimately

we just have one equation in one unknown.

We must always remember the laws of algebra:

“Do the same thing to each side of the equation”

Equation solving always follows a repeating sequence:

Change the values each side (but keeping them equal)

Then simplify

We are only interested in “linear” equations that may contain x, y, z etc but no higher

powers such as 2x . We can solve by elimination or substitution.

Elimination

Example 1

Consider two equations 7x y and 3x y . We want to find a pair of values that

will work in both equations.

7x y would be satisfied by many pairs of (x,y) values, for instance (0,7),

(1,6), (2,5), (3,4), (4,3), (5,2) etc.

3x y would have a different list of possible pairs (6,3), (5,2), (4,1) etc.

We want the pair of x,y values that occurs in both lists i.e. they satisfy both equations.

We can see that (5,2) does this: our answer is x = 5, y = 2.

This shows you what we want to achieve but it is a silly way of doing it. In general the

x, y values will be fractions or decimals and we need a way to calculate them.

First decide which to eliminate (“get rid of”): x or y?

o Looking at the equations, I see +y and –y. If I add them, they will

cancel out (+y-y = 0y)

We write:

x + y = 7

x – y = 3 +

2x + 0y = 10

13

and then ÷2 each side to get x = 5.

The final step is to substitute back into one of the equations to find y:

x + y = 7

5 + y = 7

y = 7-5 = 2

Finally, check it works: x + y = 5 + 2 = 7 OK

x – y = 5 – 2 = 3 OK

If I had chosen to eliminate x, I would have needed to subtract:

x + y = 7

x – y = 3 -

0x + 2y = 4 (remember that y–(-y) = y+y = 2y )

hence y = 2 and then x + y = x + 2 = 7 so x = 5.

“Solve the pair of simultaneous equations 2 3 7x y and 6x y ”

We could choose to eliminate x or eliminate y. I will choose to eliminate y since I can

do it by adding (less likely to go wrong than subtracting!).

2 3 7x y

3 3 18x y (scaling the second equation so it contains a “3y”)

Add the left-hand sides, add the right-hand sides:

2 3 3 3 7 18x y x y

Simplify:

5 25x so 5x

We then put this value back into either of the original equations to find y:

5 6x y y , 6 5 1y

All simultaneous equations in x and y can be represented as a pair of lines on a graph.

There will be one or more (x,y) points on the graph where the pair of x & y values

satisfy both equations at the same time. This point (the solution of the equations)

must lie on both lines so it is where they cross (“intersect”).

14

We could alternatively have used substitution:

6x y so 6y x

Then 2 3 2 3 6 2 18 3 5 18 7x y x x x x x which gives 5 25x as above.

[but please don’t write 5 18 7 5 25x x . This is a horrible mistake - obviously

untrue, 7 and 25 are not the same! ]

2 4 6 8 10

−2

2

4

x

y

5,1

2 3 7x y

6x y

15

Sequences

1. Reminder about linear sequences

1, 3, 5, 7, 9, 11

100, 105, 110, 115, 120, 125

2, 1, 0, -1, -2

A “linear” sequence is a bit like a straight line on a graph. The nth term formula looks

like an+b where a and b are numbers, eg 2n+5.

How to find the formula

Position n 1 2 3 4 5 n

Value t 6 11 16 21 26 ?

Increase: +5 +5 +5 +5

The “+5” increase tells us the formula for the nth term must be t = 5n +

something

Now think how to get the first term t = 6 when n = 1.

, we need to have the “something” = 1.

Hence the sequence 6, 11, 16, 21, 26… has the formula 5n + 1

Examples

(a) Sequence 1, 4, 7, 10, 13, the increase is +3 each time so we need 3n + something.

To make the first term = 1, we think , the formula is 3n – 2

(b) Sequence 5, 3, 1, -1, -3, the increase is -2 each time and the formula is -2n +7

linear sequences (constant difference between terms)

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2. What do “linear” and “quadratic” mean?

Look at these examples.

2 1n is a linear expression since the highest power is just n . The values plot

as a strqight line.

2 3 4n n is a quadratic expression since the highest power is 2n . When

plotted, all quadratics follow a curve known as a parabola.

You won’t get questions on them, so just for reference:

3 2 3 4n n n is a cubic expression (highest power 3n )

4 3 27 5 3 4n n n n is a quartic expression (highest power 4n )

5 4 3 22 7 5 3 4n n n n n is a quintic expression (highest power 5n )

These follow more interesting curves!

3. Quadratic sequences - finding the values from the formula:

A quadratic sequence has 2n in its formula, e.g. 23 5 2n n , in general like

For instance, n2 makes a sequence 2 2 2 21 ,2 ,3 ,4 ,... so the values are 1, 4, 9, 16, 25,…

makes a sequence 101, 104, 109, 116, 125, …

makes a sequence 2, 6, 12, 20, 30, …

I know you can work out the values in your head or on a calculator but a good way of

doing it (you’ll see why later) is to make a table showing two separate sequences:

the actual quadratic part ( 2an values)

a linear sequence ( )bn c

and then add ing them.

Example 1.

Find the first 6 terms in the sequence having 2 3 1n n as the nth term.

Position

“n”

1 2 3 4 5 6

2n 1 4 9 16 25 36

, , are numbers ("coefficients")a b c

17

3 1n 4 7 10 13 16 19

Add to

get 2 3 1n n

1+4 = 5

4+7=11

19

29

41

55

The sequence is 5, 11, 19, 29, 41, 55,…

Example 2.

Find the first 6 terms in the sequence having 23 5 2n n as the nth term.

Position “n” 1 2 3 4 5 6

23n 23 1 3 23 2 12 27 48 75 108

5 2n -3 -8 -13 -18 -23 -28

Add to get 23 5 2n n

3+(-3) = 0

12+(-8) = 4

14

30

52

80

The sequence is 0, 4, 14, 30, 52, 80,…

4. How to find the nth term formula from the list of numbers.

Any quadratic sequence e.g. 22 3 1n n can be written as two sequences added:

a pure 2n part (

22n )

plus a linear part (3 1n ).

We need to find each formula separately.

(i) Find how many 2n you need

(ii) subtract the 2n part from the sequence values. You are then left with a linear

sequence that is easy to identify.

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(i) finding the 2n coefficient.

Make a table showing first and second order differences, e.g. for the sequence 8, 13,

20, 29, 40

Position n 1 2 3 4 5 n

Value 8 13 20 29 40 ?

First order difference: +5 +7 +9 +11

Second order difference +2 +2 +2

Halve the second difference (= 2) to find the multiplier for 2n

Here, 2

12 so the sequence must be 21n bn c where b and c are numbers.

Our next step is to find what they are.

(ii) finding the linear bn+c part

We make another table starting with sequence values 8, 13, 20, 29, 40 from above. We

now know the formula must be like 21n bn c .

Position “n” 1 2 3 4 5 6

Sequence 2n bn c

8 13 20 29 40 53

2n 1 4 9 16 25 36

Subtract to find

what we “need”

for bn+c

8-1=7 13-4=9 20-9=11 29-16 = 13 40-25=15 53-36=17

The “linear sequence” part 7, 9, 11, 13, 15, 17 goes up in steps of 2 and follows the

formula 2 5n .

The original sequence adds pairs values from the 2n sequence and the 2 5n sequence.

the complete formula is 2 2 5n n

Not constant, hence not a

linear sequence.

Increasing by a regular amount

each step, hence quadratic

19

Another example: find the nth term formula for the sequence 14, 31, 58, 95,

142, 199

First we find whether we need 21n , 22n , 23n or whatever, by finding the second

order difference and halving it:

Position n 1 2 3 4 5 6 n

Value 14 31 58 95 142 199 ?

First order difference: +17 +27 +37 +47

Second order difference +10 +10 +10

105

2 so the sequence is 25n + something

Now we take 25n from each value in the sequence. This is just a way of finding

what extra we would add to the square terms to get the complete sequence.

Position “n” 1 2 3 4 5 6

Sequence 25n bn c

14 31 58 95 142 199

25n 5×1=5 5×4=20 5×9=45 5×16=80 5×25=125 5×36=180

Subtract to

leave bn+c

9 11 13 15 17 19

The bn+c sequence goes in steps of +2, with first term 9, so must be 2n+7

Each number in the original sequence consists of a number from the 25n sequence plus

a number from the 2 7n sequence.

the complete formula is 25 2 7n n .