Alperton Community School Preparation for A Level Mathematics This induction booklet is for students who wish to start AS Level Maths in Year 1. You are epected to know these topics before your first maths lesson in September as you will be tested on them. You are also epected to complete compulsory summer work on my maths.
Introduction To A-Level Mathematics A level Maths is a highly regarded qualification which prepares students well for all university courses requiring application of mathematical skills. If you are studying Mathematics on its own, you will study two core modules (C1 and C) as well as Statistics (S1) during Year 1. If you have chosen to study Further Mathematics as well, you will take a further 3 modules, Further Pure 1 (FP1), Decision 1 (D1) and Mechanics 1 (M1) in Year 1. In Year 13 you will study Further Pure (FP1), Decision (D) and Mechanics (M). The Mathematics Department is committed to ensuring that you make good progress throughout the course. In order to be fully prepared for the best start possible it is important you spend time working through the topics before the course starts. These topics are grade A/A* GCSE topics in which you must be highly skilled. Work will be set up in my maths that will need to be completed before the start of the autumn term and you should be aiming to get a 100%. This work will be checked to ensure that it has been completed to the required standard. In the first week of term you will take a test to check how well you understand these topics. If you do not pass this test you are likely to be taken off the course. We hope you will find this booklet a useful revision and preparation tool and that you enjoy working through the various problems. The more you practise applying your skills, the better you understand the concepts and the higher your grades are likely to be. H. Gill Mrs Rufo KS Coordinator of Mathematics Head of Mathematics Sources for further help are indicated throughout the booklet. Don t forget you have access to MyMaths (username: alperton password: round and mathswatch (https://www.mathswatchvle.com/); ( centre id: alperton username and password : this is the same as your school log in). There are also a number of Bridging the gap books available of which this is one eample: AS-Level Maths Head Start Published by CGP Workbooks (www.cgpbooks.co.uk) ISBN: 978 1 816 993 Cost:.9
CONTENTS Section1 Removing brackets page 3 Section Linear equations Section 3 Simultaneous equations 9 Section Factors 11 Section Change the subject of the formula 1 Section 6 Solving quadratic equations 17 Section 7 Indices 19 Practice Booklet Test Solutions to Eercises Solutions to Practice Booklet Test Section 1: REMOVING BRACKETS To remove a single bracket multiply every term in the bracket by the number or epression outside: Eamples 1) 3 ( + y) = 3 + 6y ) -( - 3) = (-)() + (-)(-3) = - + 6 To epand two brackets multiply everything in the first bracket by everything in the second bracket. You may have used * the smiley face method * FOIL (First Outside Inside Last) * using a grid. Eamples: 1) ( + 1)( + ) = ( + ) + 1( + ) or ( +1)( + ) = + + + = + 3 + or 1 ( +1)( + ) = + + + = + 3 + ) ( - )( + 3) = ( + 3) - ( +3) = + 3-6 = 6 or ( - )( + 3) = 6 + 3 = 6 3
or - - 3 3-6 ( +3)( - ) = + 3 - - 6 = - - 6 EXERCISE A 1. 7( + ). -3( - 7) 3. a (3a - 1). y + y( + 3y). -3 ( + ) Multiply out the following brackets and simplify. 7. ( + )( + 3) 8. (t - )(t - ) 9. ( + 3y)(3 y) 10. ( - )( + 3) 11. (y - 1)(y + 1) 6. ( - 1) (3 - ) 1. (3 + )( ) Two Special Cases Perfect Square: Difference of two squares: ( + a) = ( + a)( + a) = + a + a ( - a)( + a) = a ( - 3) = ( 3)( 3) = 1 + 9 ( - 3)( + 3) = 3 = 9 EXERCISE B Epand the following 1. ( - 1). (3 + ) 3. (7 - ). ( + )( - ). (3 + 1)(3-1) 6. (y - 3)(y + 3 Section : LINEAR EQUATIONS When solving an equation whatever you do to one side must also be done to the other. You may add the same amount to both side subtract the same amount from each side multiply the whole of each side by the same amount divide the whole of each side by the same amount. If the equation has unknowns on both sides, collect all the letters onto the same side of the equation. If the equation contains brackets, you often start by epanding the brackets. A linear equation contains only numbers and terms in. (Not or 3 or 1/ etc) More help on solving equations can be obtained by downloading the leaflet available at this website: http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-simplelinear.pdf
Eample 1: Solve the equation 6 3 = Solution: There are various ways to solve this equation. One approach is as follows: Step 1: Add 3 to both sides (so that the term is positive): 6 = 3 + Step : Subtract from both sides: Step 3: Divide both sides by 3: 39 = 3 13 = So the solution is = 13. Eample : Solve the equation 6 + 7 =. Solution: Step 1: Begin by adding to both sides 8 + 7 = (to ensure that the terms are together on the same side) Step : Subtract 7 from each side: 8 = - Step 3: Divide each side by 8: = -¼ Eercise A: Solve the following equations, showing each step in your working: 1) + = 19 ) = 13 3) 11 = ) 7 = -9 ) 11 + 3 = 8 6) 7 + = Eample 3: Solve the equation (3 ) = 0 3( + ) Step 1: Multiply out the brackets: 6 = 0 3 6 (taking care of the negative signs) Step : Simplify the right hand side: 6 = 1 3 Step 3: Add 3 to each side: 9 = 1 Step : Add : 9 = 18 Step : Divide by 9: = Eercise B: Solve the following equations. 1) ( ) = ) ( ) = 3( 9) 3) 8 ( + 3) = ) 1 3( + 3) =
EQUATIONS CONTAINING FRACTIONS When an equation contains a fraction, the first step is usually to multiply through by the denominator of the fraction. This ensures that there are no fractions in the equation. y Eample : Solve the equation 11 Solution: Step 1: Multiply through by (the denominator in the fraction): y 10 Step : Subtract 10: y = 1 Eample : Solve the equation 1 ( 1) 3 Solution: Step 1: Multiply by 3 (to remove the fraction) 1 1 Step : Subtract 1 from each side = 1 Step 3: Divide by = 7 When an equation contains two fractions, you need to multiply by the lowest common denominator. This will then remove both fractions. Eample 6: Solve the equation 1 Solution: Step 1: Find the lowest common denominator: The smallest number that both and divide into is 0. Step : Multiply both sides by the lowest common denominator Step 3: Simplify the left hand side: 0( 1) 0( ) 0 0 ( 1) 0 ( ) 0 ( + 1) + ( + ) = 0 Step : Multiply out the brackets: + + + 8 = 0 Step : Simplify the equation: 9 + 13 = 0 Step 6: Subtract 13 9 = 7 Step 7: Divide by 9: = 3 6
Eample 7: Solve the equation 3 6 Solution: The lowest number that and 6 go into is 1. So we multiply every term by 1: 1( ) 1(3 ) 1 6 Simplify 1 3( ) (3 ) Epand brackets 1 3 6 6 10 Simplify 1 6 18 10 Subtract 10 6 18 Add 6 = Divide by =.8 Eercise C: Solve these equations 1) 1 ( 3) ) 1 3 3 3) y y 3 3 ) 3 7 1 ) 7 1 13 y 1 y 1 y 6) 3 6 7) 1 3 8) 3 10 1 FORMING EQUATIONS Eample 8: Find three consecutive numbers so that their sum is 96. Solution: Let the first number be n, then the second is n + 1 and the third is n +. Therefore n + (n + 1) + (n + ) = 96 3n + 3 = 96 3n = 93 n = 31 So the numbers are 31, 3 and 33. Eercise D: 1) Find 3 consecutive even numbers so that their sum is 108. ) The perimeter of a rectangle is 79 cm. One side is three times the length of the other. Form an equation and hence find the length of each side. 3) Two girls have 7 photographs of celebrities between them. One gives 11 to the other and finds that she now has half the number her friend has. Form an equation, letting n be the number of photographs one girl had at the beginning. Hence find how many each has now. 7
Section 3: SIMULTANEOUS EQUATIONS Eample 3 + y = 8 + y = 11 and y stand for two numbers. Solve these equations in order to find the values of and y by eliminating one of the letters from the equations. In these equations it is simplest to eliminate y. Make the coefficients of y the same in both equations. To do this multiply equation by, so that both equations contain y: 3 + y = 8 10 + y = = To eliminate the y terms, subtract equation from equation. We get: 7 = 1 i.e. = To find y substitute = into one of the original equations. For eample put it into : 10 + y = 11 y = 1 Therefore the solution is =, y = 1. Remember: Check your solutions by substituting both and y into the original equations. Eample: Solve + y = 16 3 y = 1 Solution: Begin by getting the same number of or y appearing in both equation. For eample, multiply the top equation by and the bottom equation by to get 0y in both equations: 8 + 0y = 6 1 0y = As the SIGNS in front of 0y are DIFFERENT, eliminate the y terms from the equations by ADDING: 3 = 69 + i.e. = 3 Substituting this into equation gives: 6 + y = 16 y = 10 So y = The solution is = 3, y =. If you need more help on solving simultaneous equations, you can download a booklet from the following website: http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-simultaneous1.pdf 8
Eercise: Solve the pairs of simultaneous equations in the following questions: 1) + y = 7 ) + 3y = 0 3 + y = 9 3 + y = -7 3) 3 y = ) 9 y = + 3y = -6 y = 7 ) a + 3b = 6) 3p + 3q = 1 a b = 3 p + q = 1 Section : FACTORISING Taking out a common factor Eample 1: Factorise 1 30 Solution: Eample : 6 is a common factor to both 1 and 30. Factorise by taking 6 outside a bracket: 1 30 = 6( ) Factorise 6 y Solution: is a common factor to both 6 and. Both terms also contain an. Factorise by taking outside a bracket. 6 y = (3 y) Eample 3: Factorise 9 3 y 18 y Solution: 9 is a common factor to both 9 and 18. The highest power of that is present in both epressions is. There is also a y present in both parts. So we factorise by taking 9 y outside a bracket: 9 3 y 18 y = 9 y(y ) Eample : Factorise 3( 1) ( 1) Solution: Eercise A There is a common bracket as a factor. So we factorise by taking ( 1) out as a factor. The epression factorises to ( 1)(3 ) Factorise each of the following 1) 3 + y ) y 3) pq p q ) 3pq - 9q ) 3 6 6) 8a b 1a 3 b 7) y(y 1) + 3(y 1) 9
Factorising quadratics Simple quadratics: Factorising quadratics of the form b c The method is: Step 1: Form two brackets ( )( ) Step : Find two numbers that multiply to give c and add to make b. Write these two numbers at the end of the brackets. Eample 1: Factorise 9 10. Solution: Find two numbers that multiply to make -10 and add to make -9. These numbers are -10 and 1. Therefore 9 10 = ( 10)( + 1). General quadratics: Factorising quadratics of the form a b c One method (you may not have seen) is: Step 1: Find two numbers that multiply together to make ac and add to make b. Step : Split up the b term using the numbers found in step 1. Step 3: Factorise the front and back pair of epressions as fully as possible. Step : There should be a common bracket. Take this out as a common factor. Eample : Factorise 6 + 1. Solution: We need to find two numbers that multiply to make 6-1 = -7 and add to make 1. These two numbers are -8 and 9. Therefore, 6 + 1 = 6-8 + 9 1 = (3 ) + 3(3 ) (the two brackets must be identical) = (3 )( + 3) Difference of two squares: Factorising quadratics of the form a Remember that a = ( + a)( a). Therefore: 9 3 ( 3)( 3) 16 ( ) ( )( ) Also notice that: and 8 ( ) ( )( ) 3 3 8y 3 ( 16 y ) 3 ( y)( y) Factorising by pairing or grouping Factorise epressions like y y using the method of factorising by pairing: y y = ( + y) 1( + y) (factorise front and back pairs, both brackets identical) = ( + y)( 1) If you need more help with factorising, you can download a booklet from this website: http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-factorisingquadratics.pdf 10
Eercise B Factorise 1) 6 8) 10 30 ) 6 16 9) 3) 10) 3 y 3y ) 3 11) 1 8 ) 3 1) 16m 81n 6) y 17y 1 13) 3 y 9a y 7) 7y 10y 3 1) 8( 1) ( 1) 10 Section : CHANGING THE SUBJECT OF A FORMULA Rearranging a formula is similar to solving an equation always do the same to both sides. Eample 1: Make the subject of the formula y = + 3. Solution: y = + 3 Subtract 3 from both sides: y 3 = Divide both sides by ; y 3 y 3 So is the same equation but with the subject. Eample : Make the subject of y = Solution: Notice that in this formula the term is negative. y = Add to both sides y + = (the term is now positive) Subtract y from both sides = y Divide both sides by y Eample 3: ( F 3) The formula C is used to convert between Fahrenheit and Celsius. 9 Rearrange to make F the subject. ( F 3) C 9 Multiply by 9 9C ( F 3) (this removes the fraction) Epand the brackets 9C F 160 Add 160 to both sides 9C160 F 11
Divide both sides by Therefore the required rearrangement is F 9C 160 F 9C 160. Eercise A Make the subject of each of these formulae: 1) y = 7 1 ) y 3) y ) 3 y (3 ) 9 Eample : Make the subject of y w Solution: Subtract y from both sides: y w Square root both sides: w y Remember the positive & negative square root. w y (this isolates the term involving ) Eample : Make a the subject of the formula 1 a t h Solution: Multiply by Square both sides Multiply by h: Divide by : 1 a t h a t h a 16t h 16t h a 16th a Eercise B: Make t the subject of each of the following 1) wt P ) 3r P wt 3r 3) ) 1 V t h 3 ) w( v t) Pa g 6) P t g r a bt 1
Harder eamples Sometimes the subject occurs in more than one place in the formula. In these questions collect the terms involving this variable on one side of the equation, and put the other terms on the opposite side. Eample 6: Make t the subject of the formula a t b yt Solution: a t b yt Start by collecting all the t terms on the right hand side: Add t to both sides: a b yt t Now put the terms without a t on the left hand side: Subtract b from both sides: a b yt t Factorise the RHS: a b t( y ) Divide by (y + ): So the required equation is a b t y t a b y Wa Eample 7: Make W the subject of the formula T W b Solution: This formula is complicated by the fractional term. Begin by removing the fraction: Multiply by b: bt bw Wa Add bw to both sides: bt Wa bw (this collects the W s together) Factorise the RHS: bt W( a b) Divide both sides by a + b: W bt a b If you need more help you can download an information booklet on rearranging equations from the following website: http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-formulae-tom.pdf Eercise C Make the subject of these formulae: 1) a 3 b c ) 3( a) k( ) 3) 3 y ) 1 a b 13
Section 6: SOLVING QUADRATIC EQUATIONS A quadratic equation has the form a b c 0. There are two methods that are commonly used for solving quadratic equations: * factorising * the quadratic formula Not all quadratic equations can be solved by factorising. Method 1: Factorising Make sure that the equation is rearranged so that the right hand side is 0. It usually makes it easier if the coefficient of is positive. Eample 1 : Solve 3 + = 0 Factorise ( 1)( ) = 0 Either ( 1) = 0 or ( ) = 0 So the solutions are = 1 or = Note: The individual values = 1 and = are called the roots of the equation. Eample : Solve = 0 Factorise: ( ) = 0 Either = 0 or ( ) = 0 So = 0 or = Method : Using the formula The roots of the quadratic equation a b c 0 are given by the formula: b b ac a Eample 3: Solve the equation 7 3 Solution: First we rearrange so that the right hand side is 0. We get We can then tell that a =, b = 3 and c = -1. Substituting these into the quadratic formula gives: 31 0 3 3 ( 1) 3 10 (this is the surd form for the solutions) If we have a calculator, we can evaluate these roots to get: = 1.81 or = -3.31 If you need more help with the work in this chapter, there is an information booklet downloadable from this web site: http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-quadraticequations.pdf 1
EXERCISE 1) Use factorisation to solve the following equations: a) + 3 + = 0 b) 3 = 0 c) = 1 ) Find the roots of the following equations: a) + 3 = 0 b) = 0 c) = 0 3) Solve the following equations either by factorising or by using the formula: a) 6 - = 0 b) 8 + 10 = 0 ) Use the formula to solve the following equations to 3 significant figures where possible a) +7 +9 = 0 b) 6 + 3 = 8 c) 7 = 0 d) 3 + 18 = 0 e) 3 + + = 0 f) 3 = 13 16 Section 7: INDICES Basic rules of indices y means y y y y. is called the inde (plural: indices) or eponent of y. There are 3 basic rules of indices: m n m n 1) a a a 9 e.g. 3 3 3 m n m n ) a a a e.g. 3 8 / 3 3 = 3 3) ( a m ) n a mn e.g. 10 3 3 Further eamples y y y 3 7 3 a 6a a (multiply the numbers and multiply the a s) c 3c 6 6c 8 (multiply the numbers and multiply the c s) 7 7 d d 3d 8d (divide the numbers and divide the d terms by subtracting the powers) 3d Eercise A Simplify the following: Remember that b b 1 1) ) 3) ) b b 3c c b c bc 3 6 n ( 6 n ) ) 6) 8n n d 8 3 d 11 9 3 7) a 8) d 3 1
Zero inde: Remember 0 a 1 For any non-zero number, a. 0 Therefore 1 1.30 1 Negative powers 0 3 A power of -1 corresponds to the reciprocal of a number, i.e. Therefore 1 1 0. 0. 1 1 1 This result can be etended to more general negative powers: This means: 3 Fractional powers: 1 1 3 9 Fractional powers correspond to roots: In general: a Therefore: 1/ 3 3 8 8 1/n n a 0 a 1 1 (Find the reciprocal of a fraction by turning it upside down) 1 1 16 a n a 1. n a 1 1 1 16 1 1/ 1/ 3 3 1/ a a a a a a 1/ 1/ 10000 10000 10 m / n 1/ n A more general fractional power can be dealt with in the following way: a a 3/ 3 8 So 3 / 3 1/ 3 8 8 7 7 3 9 3/ 3/ 3 3 36 36 6 16 36 1 m Eercise B: Find the value of: 1) ) 1/ 1/ 3 7 3) 1 1/ 9 ) ) 6) 7) 0 18 7 1 / 3 7 8) 9) 3 8 / 3 10) 1/ 0.0 11) 1) 8 7 1 16 / 3 3 / Simplify each of the following: 13) 1/ / a 3a 1) 3 1) 1/ y 16
SOLUTIONS TO THE EXERCISES CHAPTER 1: E A 1) 8 + 3 ) -1 + 1 3) -7a + ) 6y + 3y ) 6) 7 1 7) + + 6 8) t 3t 10 9) 6 + y 1y 10) + 11) y 1 1) 1 + 17 E B 1) + 1 ) 9 + 30 + 3) 9 8 + ) ) 9-1 6) y 9 CHAPTER E A 1) 7 ) 3 3) 1½ ) ) -3/ 6) -7/3 E B 1). ) 3) 1 ) ½ E C 1) 7 ) 1 3) /7 ) 3/3 ) 3 6) 7) 9/ 8) E D 1) 3, 36, 38 ) 9.87, 9.6 3), 8 CHAPTER 3 1) = 1, y = 3 ) = -3, y = 1 3) = 0, y = - ) = 3, y = 1 ) a = 7, b = - 6) p = 11/3, q = /3 CHAPTER E A 1) (3 + y) ) ( y) 3) pq(q p) ) 3q(p 3q) ) ( - 3) 6) a 3 b (a 3b ) 7) (y 1)(y + 3) E B 1) ( 3)( + ) ) ( + 8)( ) 3) ( + 1)( + ) ) ( 3) ) (3-1 )( + ) 6) (y + 3)(y + 7) 7) (7y 3)(y 1) 8) ( 3)( + ) 9) ( + )( ) 10) ( 3)( y) 11) ( )( 1) 1) (m 9n)(m + 9n) 13) y(y 3a)(y + 3a) 1) ( + )( ) CHAPTER E A y 1 9y 0 1) ) y 3) 3( y ) ) 7 1 E B 3rP 3rP 3V Pg Pag 1) t ) t 3) t ) t ) t v 6) w w h w E C c 3 3a k y 3 ab 1) ) 3) ) a b k 3 y b a CHAPTER 6 1) a) -1, - b) -1, c) -, 3 ) a) 0, -3 b) 0, c), - 3) a) -1/, /3 b) 0.,. ) a) -.30, -1.70 b) 1.07, -0.699 c) -1.0, 1. d) no solutions e) no solutions f) no solutions t r a b CHAPTER 7 E A 1) b 6 ) 6c 7 3) b 3 c ) -1n 8 ) n 6) d 7) a 6 8) -d 1 E B 1) ) 3 3) 1/3 ) 1/ ) 1 6) 1/7 7) 9 8) 9/ 9) ¼ 10) 0. 11) /9 1) 6 13) 6a 3 1) 1) y 17