Year 9 set 1 Mathematics notes, to accompany the 9H book.


 Hubert Parsons
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1 Part 1: Year 9 set 1 Mathematics notes, to accompany the 9H book. equations 1. (p.1), 1.6 (p. 44), 4.6 (p.196) sequences 3. (p.115) Pupils use the Elmwood Press Essential Maths book by David Raymer (9H for set 1, 9C for sets & 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.... 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 Solving equations using algebra Trial and improvement Simultaneous Equations Sequences
2 1. Reminder about linear sequences What do linear and quadratic mean? Quadratic sequences  finding the values from the formula: Scheme of Work. Modules refer to the National Numeracy Strategy see my website for descriptions. Module Core Content Textbook Chapters A1/ Linear sequences, nth term; represent problems and synthesise information in graphical form. N1 4 rules of fractions (also in HD); proportional reasoning; multiplying and dividing by numbers between 0 and 1. Percentages A3 Construct and solve linear equations; trial and improvement CAME, Functional Skills etc 9H Unit 1: 1., 1.6 Unit 3:3. 9H Unit 1: 1.1 Unit :.1 Unit 3; 3.3, 3.5 9H Unit 1; 1. Unit:.6 Unit :.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 SSM Design a survey; communicate interpretations of results using tables, graphs and diagrams. CAME, Functional Skills etc End of Term, Christmas Break Units of measurement, area and circumference of a circle. 9H Unit 4: 4.(not Ex M, E and 3M), 4.4 Unit :.5 Unit 4: 4.5 CAME S4 9H Unit 4: 4.3 Unit 6: 6. N Powers of 10; 4 rules of decimals; calculator methods. 9H Unit :.,.3 CAME, Functional Skills etc CAME S5, Exam qu A4 February Half Term Gradient and intercept of straight line graphs; real life graphs; integer powers and roots. 9H Unit4: 4. (Ex M, E, and 3M) 4.6 (graphs), Unit 5: 5.4 HD 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. Unit 6:6. 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. 3
4 Algebra Algebra is the use of letters (x, y, z etc.) in place of numbers. complicated problems by splitting them into easier stages: (1) Write an equation (must include an = sign) () Use algebra to manipulate the equation and solve it Without algebra we would have no equations! It lets us solve Rules: We leave out the (times) sign e.g. x becomes x. The meaning is defined by BIDMAS (e.g x 5 means xx 5 not x 5 x 5) see notes part. Collecting terms : we can add similar terms ( x 3x 5x but x 3x cannot be turned into a single term since x and x are not identical same for x 3y) 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 x 10 I can also write x so that I have done the same thing to both sides o This is why it is so horribly wrong to write calculations like 3 = = 16/4 = 4: the first two = signs are completely untrue. If you want to show working for 3 10, start with a complete expression 4 and simplify it in stages: Look, each equals sign is true it has the same value on each side! 4
5 Multiplying out brackets The outside term () 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 x 3 x 3x x 6 x 5x 6 () split the first bracket into two terms: x x 3 x x 3 x 3 x 3x x 6 x 5x 6 (3) multiply using boxes, then sum the terms: x x x x 3 3x 6 5
6 Squaring brackets From BIDMAS, we know that 3p means 3p p 3p must mean something different! The contents of the brackets are multiplied by themselves, 3p3p 9p Examples ab ab ab a b 4 5xy 5xy 5xy 5x y z z z z (multiplying fractions: multiply the numerators, multiply the denominators). a b a ba b a ab ba b, simplify: a b a ab b Remember that (a+b) is never equal to a +b For instance (1+) = 3 = 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 6a 4 3a The top of the fraction behaves as if it were written with brackets. It could also be written as 6a 4 3a x Remember that 1 x (true for all x values except zero: could be, etc). Hence 10 a 10 a and 0 0 a a a 4a 5a 5a 6
7 Negative numbers y x 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+() is the same as 5=3 ] o To subtract a negative number, move to the right on the number line [Think, 5() is the same as 5+=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:
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 righthand side and get a value out for the lefthand 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 r 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 = B1B is a formula that will calculate the profit as the number in cell B1 minus the number in cell B. When I finish entering the formula it automatically does the calculation: 8
9 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 could be shown as: r Square A We can go backwards through a function machine to reverse the process: r Square root A e.g. A = 50 m, find r (this is r ) r 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) would be n 54 P 9
10 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=3 x x 5 3x 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 3 Add 5 to each side to cancel out the 5 on the left: 3x , simplify: 3x 18 Divide each side by 3 to turn 3x into x: 18 x
11 (ii) Solve 9x 7 7x 15 Take 7 from each side to cancel out the +7 on the left: 9x 7 7 7x15 7, simplify: 9x7x 8 Take 7x from each side, so there are no x terms on the right: 9x 7x 7x 7x 8 Hence x 8, x 4 Always finish by checking the answer works in the original equation. Putting x 4 into the equation gives LHS = 9x , RHS = 7x both the same, so the equation is satisfied. (iii) Solve x1 7 x 3 Multiply out the brackets: x 4 7x 1 Most of the x terms are on the right, so swap sides: 7x 1 x 4 Add 1 each side: 7x x 4 1 x 45 Take x 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
12 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. We can solve by elimination or substitution. Elimination Example 1 Consider two equations xy7 and xy 3. We want to find a pair of values that will work in both equations. xy 7 would be satisfied by many pairs of (x,y) values, for instance (0,7), (1,6), (,5), (3,4), (4,3), (5,) etc. xy 3 would have a different list of possible pairs (6,3), (5,), (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,) does this: our answer is x = 5, y =. 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 (+yy = 0y) We write: x + y = 7 x y = 3 + x + 0y = 10 1
13 and then each side to get x = 5. The final step is to substitute back into one of the equations to find y: x + y = y = 7 y = 75 = Finally, check it works: x + y = 5 + = 7 OK x y = 5 = 3 OK If I had chosen to eliminate x, I would have needed to subtract: x + y = 7 x y = 30x + y = 4 (remember that y (y) = y+y = y ) hence y = and then x + y = x + = 7 so x = 5. Solve the pair of simultaneous equations x3y 7 and xy 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!). x3y 7 3x3y 18 (scaling the second equation so it contains a 3y ) Add the lefthand sides, add the righthand sides: x 3y 3x 3y 7 18 Simplify: 5x 5 so x 5 We then put this value back into either of the original equations to find y: x y 5 y 6, y 65 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 ). 13
14 4 y x3y7 5, x xy6 We could alternatively have used substitution: xy 6 so y6 x x 3y x 3 6 x x 18 3x 5x 18 7 which gives 5x 5 as above. Then [but please don t write 5x18 7 5x 5. This is a horrible mistake  obviously untrue, 7 and 5 are not the same! ] 14
15 Sequences 1. Reminder about linear sequences 1, 3, 5, 7, 9, , 105, 110, 115, 10, 15, 1, 0, 1,  A linear sequence is a bit like a straight line on a graph. The n th term formula looks like an+b where a and b are numbers, eg n+5. How to find the formula linear sequences (constant difference between terms) Position n n Value t ? Increase: The +5 increase tells us the formula for the n th 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, 1, 6 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 (b) Sequence 5, 3, 1, 1, 3, the increase is  each time and the formula is n +7 15
16 . What do linear and quadratic mean? Look at these examples. n 1 is a linear expression since the highest power is just n. The values plot as a strqight line. n 3n 4 is a quadratic expression since the highest power is n. When plotted, all quadratics follow a curve known as a parabola. You won t get questions on them, so just for reference: 3 n n n 3 4 is a cubic expression (highest power 4 3 n n n n 3 n ) is a quartic expression (highest power n n n n n 4 n ) is a quintic expression (highest power 5 n ) These follow more interesting curves! 3. Quadratic sequences  finding the values from the formula: A quadratic sequence has n in its formula, e.g. 3n 5n, in general like abc,, are numbers ("coefficients") For instance, n makes a sequence makes a sequence 101, 104, 109, 116, 15, makes a sequence, 6, 1, 0, 30, 1,,3,4,... so the values are 1, 4, 9, 16, 5, 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 ( an values) a linear sequence ( bn c) and then add ing them. Example 1. Find the first 6 terms in the sequence having n 3n 1 as the n th term. Position n n
17 3n Add to get n 3n = 5 4+7= The sequence is 5, 11, 19, 9, 41, 55, Example. Find the first 6 terms in the sequence having 3n 5n as the n th term. Position n n n Add to get 3n 5n 3+(3) = 0 1+(8) = The sequence is 0, 4, 14, 30, 5, 80, 4. How to find the n th term formula from the list of numbers. Any quadratic sequence e.g. a pure n part ( n ) plus a linear part (3n 1). n 3n 1 can be written as two sequences added: We need to find each formula separately. (i) Find how many n you need (ii) subtract the n part from the sequence values. You are then left with a linear sequence that is easy to identify. 17
18 (i) finding the n coefficient. Make a table showing first and second order differences, e.g. for the sequence 8, 13, 0, 9, 40 Position n n Value ? First order difference: Second order difference Not constant, hence not a linear sequence. Increasing by a regular amount each step, hence quadratic Halve the second difference (= ) to find the multiplier for n Here, 1 so the sequence must be 1n 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, 0, 9, 40 from above. We now know the formula must be like 1n bn c. Position n Sequence n bn c n Subtract to find what we need for bn+c 81=7 134=9 09= = = =17 The linear sequence part 7, 9, 11, 13, 15, 17 goes up in steps of and follows the formula n 5. The original sequence adds pairs values from the the complete formula is n n 5 n sequence and the n 5 sequence. 18
19 Another example: find the n th term formula for the sequence 14, 31, 58, 95, 14, 199 First we find whether we need order difference and halving it: 1n, n, 3n or whatever, by finding the second Position n n Value ? First order difference: Second order difference so the sequence is 5n + something Now we take 5n 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 Sequence 5n bn c n 5 1=5 5 4=0 5 9= =80 5 5= =180 Subtract to leave bn+c The bn+c sequence goes in steps of +, with first term 9, so must be n+7 Each number in the original sequence consists of a number from the a number from the n 7 sequence. 5n sequence plus the complete formula is 5n n 7. 19
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