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 halftermly 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 1
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 A1/2 N1
A3
Core Content
Textbook Chapters
Linear sequences, nth term; represent problems and synthesise information in graphical form. 4 rules of fractions (also in HD2); proportional reasoning; multiplying and dividing by numbers between 0 and 1. Percentages Construct and solve linear equations; trial and improvement CAME, Functional Skills etc
9H Unit 1: 1.2, 1.6 Unit 3:3.2 9H Unit 1: 1.1 Unit 2: 2.1 Unit 3; 3.3, 3.5 9H Unit 1; 1.2 Unit2: 2.6 Unit 2:2.4, CAME A5
October Half Term SSM1
Angles in a polygon; angle properties; loci
HD1
Design a survey; communicate interpretations of results using tables, graphs and diagrams.
CAME, Functional Skills etc
9H Unit1: 1.3 Unit 3: 3.1, Unit 6: 6.3 9H Unit 4: 4.2(not Ex 2M, 2E and 3M), 4.4 Unit 2: 2.5 Unit 4: 4.5 CAME S4
End of Term, Christmas Break SSM2 N2
Units of measurement, area and circumference of a circle. Powers of 10; 4 rules of decimals; calculator methods. CAME, Functional Skills etc
9H Unit 4: 4.3 Unit 6: 6.2 9H Unit 2: 2.2, 2.3 CAME S5, Exam qu
February Half Term A4
Gradient and intercept of straight line graphs; real life graphs; integer powers and roots.
HD2 SSM3
Probability; 4 rules of fractions Transformations Similarity and congruence CAME, Functional Skills etc
9H Unit4: 4.2 (Ex 2M, 2E, and 3M) 4.6 (graphs), Unit 5: 5.4 9H Unit 5: 5.3 9H Unit 4:4.1 Unit 1: 1.4 Unit 6:6.5 CAME D3, Exam Qu
End of Term, Easter Break A5
Formulae
9H Unit 4: 4.6, 2
Unit 5; 5.2 Unit 6:6.2 9H: Unit 3: 3.1, 3.4,
SSM3
3D shapes
HD
Statistical investigation, to include use of Autograph CAME, Functional Skills etc
9H Unit1: 1.5 9H Unit 5: 5.5 CAME N5, Exam Qu
May Half Term SSM4 A6
Pythagoras’ theorem; circle theorems Square a linear expression; inequalities.
Fully Functional 3
9H Unit 5: 5.1, 9H Unit 6:6.1, 6.4, Unit 4: 4.6 Unit 5:5.1 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.
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Algebra Algebra is the use of letters (x, y, z etc.) in place of numbers. complicated problems by splitting them into easier stages:
It lets us solve
(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 x 5 means x x 5 not x 5 x 5 ) – see notes part 2.
“Collecting terms”: we can add similar terms ( 2 x 3x 5x but 2 x 3x cannot be 2
turned into a single term since x and x 2 are not identical – same for 2 x 3 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 x 10 I can also write 2 x 3 10 3 13 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 , start with a complete expression 4
and simplify it in stages:
2 3 10 6 10 16 4 “Look, each equals sign is true – it has the 4 4 4 same value on each side!”
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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):
x 2 x 3 x 2 3x 2 x 6 x 2 5 x 6 (2) split the first bracket into two terms:
x 2 x 3 x x 3 2 x 3 x 2 3x 2 x 6 x 2 5 x 6 (3) multiply using boxes, then sum the terms:
x
2
x
x2
2x
3
3x
6
5
Squaring brackets From BIDMAS, we know that 3p 2 means 3 p p
3 p
2
must mean something different!
“The contents of the brackets are multiplied by themselves”, 3 p 3 p 9 p 2 Examples
ab
2
ab ab a 2b 2 2
5 xy 2 5 xy 2 5 xy 2 25 x 2 y 4 z z z2 z (multiplying fractions: multiply the numerators, multiply the denominators).
a b
2
2 2 a b a b a 2 ab ba b 2 , simplify: a b a 2ab b 2
Remember that (a+b)2 is never equal to a2+b2 For instance (1+2)2 = 32 = 9, NOT 5
Dividing brackets by something When a fraction has a two or more terms in the numerator, they are all divided by the denominator. Example:
6a 4 6 a 4 3a 2 2 2 2 The top of the fraction behaves as if it were written with brackets. It could also be written as 6a 4 2 3a 2
Remember that
Hence
x 2 10 1 (true for all x values except zero: could be , etc). 2 10 x
10a 20a 2 20a a 10 and 4a a 5a 5a
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Negative numbers y x
−7 −6 −5 −4 −3 −2 −1
1
2
3
4
5
6
7
8
9 10
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
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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:
A r2
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
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:
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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.
A r 2 could be shown as:
Square
r
A
We can “go backwards” through a function machine to reverse the process:
r
A
Square root
e.g. A = 50 m2, find r
50
15.92 (this is r 2 )
r 15.92 3.99 m 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
n
×5
-2
×4
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
x
×3
+5
3x 5 3x 5 23
÷3
6
18
-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 3x 5 23 Add 5 to each side to cancel out the -5 on the left:
3x 5 5 23 5 , simplify: 3x 18 Divide each side by 3 to turn 3x into x:
x
18 6 3
10
(ii)
Solve 9 x 7 7 x 15 Take 7 from each side to cancel out the +7 on the left:
9 x 7 7 7 x 15 7 , simplify: 9x 7 x 8 Take 7x from each side, so there are no x terms on the right: 9x 7 x 7 x 7 x 8 Hence 2 x 8 , x 4 Always finish by checking the answer works in the original equation. Putting
x 4 into the equation gives LHS = 9 x 7 36 7 43 , RHS = 7 x 15 28 15 43 - both the same, so the equation is satisfied. (iii)
Solve 2 x 12 7 x 3 Multiply out the brackets: 2 x 24 7 x 21 Most of the x terms are on the right, so swap sides:
7 x 21 2 x 24 Add 21 each side:
7 x 2 x 24 21 2 x 45 Take 2x from each side;
5x 45 hence (÷5) x 9
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: 11
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 x 2 . We can solve by elimination or substitution. Elimination Example 1 Consider two equations x y 7 and x y 3 . We want to find a pair of values that will work in both equations.
x y 7 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. x y 3 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
12
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 x 3 y 7 and x y 6 ” 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!).
2x 3 y 7 3x 3 y 18 (scaling the second equation so it contains a “3y”) Add the left-hand sides, add the right-hand sides:
2 x 3 y 3x 3 y 7 18 Simplify: 5x 25 so x 5
We then put this value back into either of the original equations to find y:
x y 5 y 6,
y 6 5 1
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”).
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2x 3 y 7
y
4
5,1
2
x
2
4
−2
6
8
10
x y 6
We could alternatively have used substitution:
x y 6 so
y 6 x
Then 2 x 3 y 2 x 3 6 x 2 x 18 3 x 5 x 18 7 which gives 5x 25 as above. [but please don’t write 5x 18 7 5x 25 . This is a horrible mistake - obviously untrue, 7 and 25 are not the same! ]
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Sequences 1. Reminder about linear sequences 1, 3, 5, 7, 9, 11 100, 105, 110, 115, 120, 125
linear sequences (constant difference between terms)
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 th
The “+5” increase tells us the formula for the n 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
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2. What do “linear” and “quadratic” mean? Look at these examples.
2n 1 is a linear expression since the highest power is just n . The values plot as a strqight line.
n2 3n 4 is a quadratic expression since the highest power is n 2 . When plotted, all quadratics follow a curve known as a parabola.
You won’t get questions on them, so just for reference:
n3 n2 3n 4 is a cubic expression (highest power n3 )
n4 7n3 5n2 3n 4 is a quartic expression (highest power n 4 )
n5 2n4 7n3 5n2 3n 4 is a quintic expression (highest power n5 )
These follow more interesting curves!
3. Quadratic sequences - finding the values from the formula: A quadratic sequence has n 2 in its formula, e.g. 3n2 5n 2 , in general like
a, b, c are numbers ("coefficients") For instance, n2 makes a sequence 12 , 22 ,32 , 42 ,... 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 ( an2 values)
a linear sequence (bn c)
and then add ing them. Example 1. Find the first 6 terms in the sequence having n2 3n 1 as the nth term. Position “n”
1
2
3
4
5
6
n2
1
4
9
16
25
36 16
3n 1
4
7
10
13
16
19
Add to get
1+4 = 5
4+7=11
19
29
41
55
n2 3n 1
The sequence is 5, 11, 19, 29, 41, 55,…
Example 2. Find the first 6 terms in the sequence having 3n2 5n 2 as the nth term. Position “n”
1
2
3
4
5
6
3n2
3 12 3
3 22 12
27
48
75
108
5n 2
-3
-8
-13
-18
-23
-28
12+(-8) = 4
14
30
52
80
Add to get
3n2 5n 2 3+(-3) = 0
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. 2n2 3n 1 can be written as two sequences added: a pure n part ( 2n ) plus a linear part ( 3n 1 ). 2
2
We need to find each formula separately. (i) Find how many n 2 you need (ii) subtract the n 2 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 n 2 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
Second order difference
+7 +2
+9 +2
+11 +2
Not constant, hence not a linear sequence. Increasing by a regular amount each step, hence quadratic
Halve the second difference (= 2) to find the multiplier for n 2 Here,
2 1 so the sequence must be 1n2 bn c where b and c are numbers. 2
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 1n2 bn c . Position “n”
1
2
3
4
5
6
Sequence
8
13
20
29
40
53
n2
1
4
9
16
25
36
Subtract to find
8-1=7
13-4=9
20-9=11
29-16 = 13
40-25=15
53-36=17
n bn c 2
what we “need” for bn+c
The “linear sequence” part 7, 9, 11, 13, 15, 17 goes up in steps of 2 and follows the formula 2n 5 . The original sequence adds pairs values from the n 2 sequence and the 2n 5 sequence. 2 the complete formula is n 2n 5
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Another example: 142, 199
find the nth term formula for the sequence 14, 31, 58, 95,
First we find whether we need 1n2 , 2n2 , 3n 2 or whatever, by finding the second order difference and halving it: Position n Value
1
2
3
4
5
6
n
14
31
58
95
142
199
?
First order difference:
+17
Second order difference
+27 +10
+37 +10
+47 +10
10 5 so the sequence is 5n2 + something 2 Now we take 5n2 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
14
31
58
95
142
199
5n2
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
5n bn c 2
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 5n2 sequence plus a number from the 2n 7 sequence. the complete formula is 5n2 2n 7 .
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