MAIDSTONE GRAMMAR SCHOOL FOR GIRLS DEPARTMENT OF MATHEMATICS

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1 MAIDSTONE GRAMMAR SCHOOL FOR GIRLS DEPARTMENT OF MATHEMATICS Introduction to A level Maths

2 INDUCTION BOOKLET INTRODUCTION TO A LEVEL MATHS AT MGGS Thank you for choosing to study Mathematics in the sith form at Maidstone Grammar School for Girls. Over the course, you will study topics in Pure Maths, Mechanics and Statistics. If you have chosen to study Further Mathematics as well as Maths then, you will study etra topics in these three areas. The Mathematics Department is committed to ensuring that you make good progress throughout your A level course. In order that you make the best possible start to the course, we have prepared this booklet. It is vital that you spend time working through the questions in this booklet over the summer. You need to have a good knowledge of these topics before you commence your course in September. You should have met all the topics before at GCSE. Work through what you need to from each chapter, making sure that you understand the eamples. Then tackle the eercise to ensure you understand the topic thoroughly. The answers are at the back of the booklet. You will need to be very organised so keep your work in a folder & mark any queries to ask at the beginning of term. In the first or second week of term you will take a test to check how well you understand these topics, so it is important that you have completed the booklet by then. The pass mark is 75%. If you do not pass this test, you may be asked to leave the course or to complete a programme of additional work in order to bring your basic algebra skills to the required standard. If you have to complete more work you will be re-tested later on. A mock test is provided at the back of this booklet. Use this introduction to give you a good start to your Year 1 work that will help you to enjoy, and benefit from, the course. The more effort you put in, right from the start, the better you will do. Mr C Ansette (Second in Charge of Mathematics) Mrs S Squibb (Head of Mathematics) Sources for further help are indicated throughout the booklet. All topics can be found on the MyMaths website. If you need to access this, our username is maidstone and our password is formula. You may also find the following book useful AS-Level Maths Head Start Published by CGP Workbooks ISBN: Cost: 4.95

3 Contents Contents... Reading List Fractions... 5 Algebra Epanding Brackets Linear Equations Equations Containing Fractions Forming Equations Linear Inequalities Simultaneous Equations Factorising Linear Epressions Factorising Quadratic Epressions Completing the Square Solving Quadratic Equations Changing the Subject Indices Surds Algebraic Fractions Linear and Quadratic Simultaneous Equations Epanding More Than Two Binomials Quadratic Inequalities Using Completing the Square to Find Turning Points Functions Function Notation Composite Functions Inverse Functions Graphs Straight line graphs Basic shapes of curved graphs Factors Trigonometry Trigonometric Equations Practice Booklet Test Solutions to the Eercises Solutions to the Practice Booklet Test... 89

4 Reading List As a student who is choosing to study Mathematics at A Level, it is logical to assume that you have an interest in the subject. With that said, the following books may be of interest to you. Ale s Adventures in Numberland by Ale Bellos Cabinet of Mathematical Curiosities by Ian Stewart The Num8er My5teries by Marcus du Sautoy How Many Socks Make a Pair?: Surprisingly Interesting Maths by Rob Eastway The Curious Incident of the Dog in the Night-time by Mark Haddon The Penguin Dictionary of Curious & Interesting Numbers by David Wells The Calculus Wars by Jason Socrates Bardi The Code Book by Simon Singh 50 Mathematical Ideas You Really Need to Know by Tony Crilly

5 1 Fractions To add or subtract fractions, find the lowest common denominator of the two fractions and then rewrite the fractions accordingly. Ensure that you simplify as far as possible. Eamples + 4 = = 17 1 = = = 9 5 When multiplying fractions, it is far more efficient to cancel first; this avoids trying to simplify fractions with unnecessarily large numerators and/or denominators. To multiply with fractions, simply multiply the numerators and denominators together. Eample = 7 = 4 1 To divide by a fraction, we simply multiply by the reciprocal of the second fraction (i.e. we flip the second fraction over ). Eample 10 9 = 9 10 = = 5

6 For addition and subtraction with mied numbers, add (or subtract) the integer (whole number) parts first and then work with the fractions. Eamples = = = = = = 1 5 = = 7 15 = 7 15 = To multiply and divide with mied, convert the mied numbers to improper fractions and then calculate as normal = = = = = 6 5 = = = 1 10 = 1 10 It should also be noted that in the study of A Level Mathematics, answers are preferred as improper fractions rather than mied numbers. EXERCISE More help is available from MyMaths: Adding/Subtracting Fractions, Multiplying Fractions, Dividing Fractions, Mied Numbers

7 Algebra.1 Epanding Brackets To remove a single bracket multiply every term in the bracket by the number or epression outside: Eamples 1) ( + y) = + 6y ) -( - ) = (-)() + (-)(-) = 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)( + ) = = + + or 1 ( +1)( + ) = = + + ) ( - )( + ) = ( + ) - ( +) = = 6 or ( - )( + ) = = 6 or ( +)( - ) = = - - 6

8 EXERCISE A Multiply out the following brackets and simplify. 1. 7(4 + 5). -(5-7). 5a 4(a - 1) 4. 4y + y( + y) 5. - ( + 4) 6. 5( - 1) ( - 4) 7. ( + )( + ) 8. (t - 5)(t - ) 9. ( + y)( 4y) 10. 4( - )( + ) 11. (y - 1)(y + 1) 1. ( + 5)(4 ) Two Special Cases Perfect Square: Difference of two squares: ( + a) = ( + a)( + a) = + a + a ( + a)( a) = a ( ) = ( )( ) = ( + )( ) = 4 9 EXERCISE B Epand the following 1. ( - 1). ( + 5). (7 - ) 4. ( + )( - ) 5. ( + 1)( - 1) 6. (5y - )(5y + More help is available from MyMaths: Brackets

9 . 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 or 1 etc) Eample 1: Solve the equation 64 = 5 Solution: There are various ways to solve this equation. One approach is as follows: Step 1: Add to both sides (so that the term is positive): 64 = + 5 Step : Subtract 5 from both sides: Step : Divide both sides by : 9 = 1 = So the solution is = 1. Eample : Solve the equation = 5. Solution: Step 1: Begin by adding to both sides = 5 (to ensure that the terms are together on the same side) Step : Subtract 7 from each side: 8 = Step : Divide each side by 8: = 1 4 EXERCISE A: Solve the following equations, showing each step in your working: 1) + 5 = 19 ) 5 = 1 ) 11 4 = 5 4) 5 7 = -9 5) 11 + = 8 6) 7 + = 4 5

10 Eample : Solve the equation ( ) = 0 ( + ) Step 1: Multiply out the brackets: 6 4 = 0 6 (taking care of the negative signs) Step : Simplify the right hand side: 6 4 = 14 Step : Add to each side: 9 4 = 14 Step 4: Add 4: 9 = 18 Step 5: Divide by 9: = EXERCISE B: Solve the following equations. 1) 5( 4) = 4 ) 4( ) = ( 9) ) 8 ( + ) = 4 4) 14 ( + ) =

11 . 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 4: Solve the equation 5 11 Solution: Step 1: Multiply through by (the denominator in the fraction): y 10 Step : Subtract 10: y = 1 Eample 5: Solve the equation 1 ( 1) 5 Solution: Step 1: Multiply by (to remove the fraction) 1 15 Step : Subtract 1 from each side = 14 Step : 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 Solution: Step 1: Find the lowest common denominator: Step : Multiply both sides by the lowest common denominator Step : Simplify the left hand side: Step 4: Multiply out the brackets: Step 5: Simplify the equation: Step 6: Subtract 1 Step 7: Divide by 9: The smallest number that both 4 and 5 divide into is 0. 0( 1) 0( ) ( 1) 0 ( ) ( + 1) + 4( + ) = = = 40 9 = 7 =

12 5 Eample 7: Solve the equation 4 6 Solution: The lowest number that 4 and 6 go into is 1. So we multiply every term by 1: Simplify Epand brackets Simplify Subtract 10 Add 6 Divide by 5 1( ) 1( 5 ) ( ) 4 ( 5 ) = 4 = 4.8 Eercise: Solve these equations 1) 1 ( ) 5 ) 1 4 ) y y 5 4) ) y 1 y 1 y 5 6) 6 7) 1 5 8)

13 .4 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 n + = 96 n = 9 n = 1 So the numbers are 1, and. 1) Find 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. ) 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. More help is available from MyMaths: Solving Equations

14 .5 Linear Inequalities Linear inequalities can be solved using the same techniques as linear equations (for the most part). We may add and subtract the same numbers on both sides and we can also multiply and divide by positive numbers; multiplying/dividing both sides by a negative needs further eplanation. Eample - < 11 Here we can simply add to both sides: < 14 Net, as with linear equations we divide by : < 7 However, if we were to have > 6, we would need to adopt a different technique. If we wish to divide or multiply by a negative number, we must reverse the direction of the inequality. Eample > 6 As before, we would subtract from both sides: > Divide by - and subsequently reverse the inequality: < We can see this working on a more basic level; it is true to state that < 4 but it is incorrect if we multiply both sides by a negative and keep the sign as it was: -6 < -8 is not true. You may find it easier to rearrange the inequality: Eample > 6 If we add to both sides, we remove the hassle: We then subtract 6: Divide by two as normal: > Remember that you can change this round to say < > 6 + > Both of these techniques are acceptable and is more a matter of preference.

15 Eercise: Solve each inequality 1) + 6 < 10 ) y 7 > 14 ) + 6 < 8 4) y ) ) 8 y > 1 7) 7 < 8 8) 19 4y 40 More help is available from MyMaths: Inequalities, Negative Inequalities

16 .6 Simultaneous Equations Eample + y = 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: + y = y = = To eliminate the y terms, subtract equation from equation. We get: 7 = 14 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 + 5y = 16 4y = 1 Solution: Begin by getting the same number of or y appearing in both equation. Multiply the top equation by 4 and the bottom equation by 5 to get 0y in both equations: 8 + 0y = y = 5 As the SIGNS in front of 0y are DIFFERENT, eliminate the y terms from the equations by ADDING: = 69 + i.e. = Substituting this into equation gives: 6 + 5y = 16 5y = 10 So y = The solution is =, y =.

17 Eercise: Solve the pairs of simultaneous equations in the following questions: 1) + y = 7 ) + y = 0 + y = 9 + y = -7 ) y = 4 4) 9 y = 5 + y = y = 7 5) 4a + b = 6) p + q = 15 5a 4b = 4 p + 5q = 14 More help is available from MyMaths: Simultaneous Equations 1, Simultaneous Equations - Medium, Simultaneous Equations - Hard, Simultaneous Equations - Negatives, Solving Simultaneous Equations Graphically

18 .7 Factorising Linear Epressions Eample 1: Factorise 1 0 Solution: Eample : 6 is a common factor to both 1 and 0. Factorise by taking 6 outside a bracket: 1 0 = 6( 5) 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 = ( y) Eample : Factorise 9 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 y 18 y = 9 y(y ) Eample 4: Factorise ( 1) 4( 1) Solution: There is a common bracket as a factor. So we factorise by taking ( 1) out as a factor. The epression factorises to ( 1)( 4) EXERCISE A Factorise each of the following 1) + y ) 4 y ) pq p q 4) pq - 9q 5) 6 6) 8a 5 b 1a b 4 7) 5y(y 1) + (y 1) More help is available from MyMaths: Factorising Epressions

19 .8 Factorising Quadratic Epressions 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 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 is that of combining factors. Look at factorising on MyMaths or ask a teacher for help with our preferred method but is difficult to eplain on paper. Another method 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 : Factorise the front and back pair of epressions as fully as possible. Step 4: There should be a common bracket. Take this out as a common factor. Eample : Factorise 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, = = ( 4) + ( 4) (the two brackets must be identical) = ( 4)( + ) Difference of two squares: Factorising quadratics of the form a Remember that a = ( + (a)( (a). Therefore: 9 ( )( ) 16 5 ( ) 5 ( 5)( 5) Also notice that: 8 ( 4) ( 4)( 4) and 48y ( 16 y ) ( 4 y)( 4 y)

20 Factorising by pairing or grouping Factorise epressions like y y using the method of factorising by pairing: y y = ( + y) 1( + y) = ( + y)( 1) (factorise front and back pairs, both brackets identical) EXERCISE B Factorise 1) ) ) 4) 5) ) y 17y 1 7) 7y 10y More help is available from MyMaths: Factorising Quadratics 1, Factorising Quadratics 8) 9) ) y y 11) 1) m 81n 1) 4y 9a y 14) 8( 1) ( 1) 10

21 .9 Completing the Square A related process is to write a quadratic epression such as 6 11 in the form ( a) b. This is called completing the square. It is often useful, because 6 11 is not a very transparent epression it contains in more than one place, and it s not easy either to rearrange or to relate its graph to that of. Completing the square for quadratic epressions in which the coefficient of is 1 (these are called monic quadratics) is very easy. The number a inside the brackets is always half of the coefficient of. Eample 1 Write in the form ( + a) + b. Solution is a monic quadratic, so a is half of 6, namely. When you multiply out ( + ), you get [The -term is always twice a, which is why you have to halve it to get a.] isn t quite right yet; we need 4 at the end, not 9, so we can write = ( + ) = ( + ) 5. This version immediately gives us several useful pieces of information. For instance, we now know a lot about the graph of y = : It is a translation of the graph of y = by units to the left and 5 units down Its line of symmetry is = Its lowest point or verte is at (, 5) We also know that the smallest value of the function is 5 and this occurs when =. And we can solve the equation = 0 eactly without having to use the quadratic equation formula, to locate the roots of the function: = 0 ( + ) 5 = 0 ( + ) = 5 ( + ) = 5 [don t forget that there are two possibilities!] = 5 These are of course the same solutions that would be obtained from the quadratic equation formula not very surprisingly, as the formula itself is obtained by completing the square for the general quadratic equation a + b + c = 0.

22 Non-monic quadratics Everyone knows that non-monic quadratic epressions are hard to deal with. Nobody really likes trying to factorise (although you should certainly be willing and able to do so for A Level, which is why some eamples are included in the eercises here). Eample Write in the form a( + b) + c. Solution First take out the factor of : = ( ) [you can ignore the 11.5 for now] Now we can use the method for monic quadratics to write = ( + ) + (something) So we have, so far Half of = ( + ) + c [so we already have a = and b = ] Now ( + ) = ( ) = We want at the end, not 18, so: = ( + ) 18 + = ( + ) + 5. If the coefficient of is a perfect square you can sometimes get a more useful form. Eample Write in the form (a + b) + c. Solution It should be obvious that a = (the coefficient of a is 4). So = ( + b) + c If you multiply out the bracket now, the middle term will be b = 4b. So 4b must equal 0 and clearly b = 5. And we know that ( + 5) = So = ( + 5) = ( + 5) 6. EXERCISE A 1 Write the following in the form ( + a) + b. (a) (b) 10 + (c) + 4 (d) 4 (e) + (f) 5 6 Write the following in the form a( + b) + c. (a) (b) (c) Write the following in the form (a + b) + c. (a) (b) (c)

23 .10 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 + = 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 = More help is available from MyMaths: Quadratic Equations

24 Method : Using the formula The roots of the quadratic equation a b c 0 are given by the formula: b b 4ac a Eample : Solve the equation 5 7 Solution: First we rearrange so that the right hand side is 0. We get We can then tell that a =, b = and c = -1. Substituting these into the quadratic formula gives: ( 1) 105 (this is the surd form for the solutions) 4 If we have a calculator, we can evaluate these roots to get: = 1.81 or = -.1 Eercise 1) Use factorisation to solve the following equations: (a) + + = 0 (b) 4 = 0 (c) = 15 ) Find the roots of the following equations: (a) + = 0 (b) 4 = 0 (c) 4 = 0 ) Solve the following equations either by factorising or by using the formula: (a) = 0 (b) = 0 4) Use the formula to solve the following equations to significant figures where possible (a) = 0 (b) 6 + = 8 (c) 4 7 = 0 (d) + 18 = 0 (e) = 0 f) = 1 16 More help is available from MyMaths: The Quadratic Formula

25 .11 Changing the Subject 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 = 4 +. Solution: y = 4 + Subtract from both sides: y = 4 Divide both sides by 4; y 4 y So is the same equation but with the subject. 4 Eample : Make the subject of y = 5 Solution: Notice that in this formula the term is negative. y = 5 Add 5 to both sides y + 5 = (the term is now positive) Subtract y from both sides 5 = y Divide both sides by 5 y 5 Eample : 5( F ) The formula C is used to convert between Fahrenheit and Celsius. 9 Rearrange to make F the subject. 5( F ) C 9 Multiply by 9 9C 5( F ) (this removes the fraction) Epand the brackets 9C 5F 160 Add 160 to both sides 9C 160 5F Divide both sides by 5 9C 160 F 5 9C 160 Therefore the required rearrangement is F. 5 EXERCISE A 1) y = 7 1 ) 5 y 4 Make the subject of each of these formulae: ) 4y 4) 4( 5) y 9

26 Eample 4: Make the subject of y w Solution: y w Subtract y from both sides: w y (this isolates the term involving ) Square root both sides: w y Remember the positive & negative square root. Eample 5: Make a the subject of the formula t 1 5a 4 h Solution: Multiply by 4 Square both sides Multiply by h: Divide by 5: EXERCISE B: 16t h 5 1 5a t 4 h 5a 4t h 5a 16t h a 16 5 th a Make t the subject of each of the following 1) P wt r ) V 1 t h 5) Pa w( v t) g ) wt P r 4) P t g 6) r a bt More help is available from MyMaths: Rearranging Equations

27 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 t a b y a b t y Eample 7: Make W the subject of the formula Wa 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 Eercise C Make the subject of these formulae: 1) a b c ) ( a) k( ) ) y 5 4) 1 a b More help is available from MyMaths: Higher Rearranging

28 .1 Indices Basic rules of indices y 4 means y y y y. 4 is called the inde (plural: indices), power or eponent of y. There are basic rules of indices: 1) ) ) ( a ) m n m n a a a e.g. m n m n a a a e.g. m n mn Further eamples a e.g. 5 y 5y 5y 4 7 4a 6a 4a 5 c c 6c 6 8 4d 4d d 8d d (multiply the numbers and multiply the a s) (multiply the numbers and multiply the c s) (divide the numbers and divide the d terms by subtracting the powers) EXERCISE A Simplify the following: Remember that b b 1 1) b 5b 5 5) 8n n 8 ) c c 5 6) d d 11 9 ) b c bc 4) 6 n ( 6 n ) 7) a 4 8) d

29 Zero inde: Remember 0 a 1 For any non-zero number, a. 0 Therefore Negative powers A power of -1 corresponds to the reciprocal of a number, i.e. Therefore This result can be etended to more general negative powers: This means: a 1 1 (Find the reciprocal of a fraction by turning it upside down) a n a 1. n a There is a particularly nice way of understanding the negative powers. Consider the following: Every time you move one step to the right you multiply by. Now consider the sequence continuing, right-to-left: Each time you move one step to the left you divide by. Take particular care when there are numbers as well as negative powers Eample 10 but or (10) 1. The usual rules of powers and brackets tell you that 10 1 is not the same as (10) 1.

30 Fractional powers: Fractional powers correspond to roots: 1/ a a 1/ a a 1/ 4 4 a a 1/n n In general: a a Therefore: 1/ 1/ 1/ m / n 1/ n A more general fractional power can be dealt with in the following way: a a So / / 1/ / / m EXERCISE B: Find the value of: 1) ) 1/ 4 1/ 7 4) 5) ) 11) 8 7 / ) 1 1/ 9 6) 7) 7 1 / 7 9) 8 / 10) 1/ ) 1 16 / Simplify each of the following: 1) 14) a a 1/ 5 / 4 15) y 1/ More help is available from MyMaths: Indices 1, Indices, Indices

31 .1 Surds A surd is a root of a number that cannot be epressed as an integer. Surds are part of the set of irrational numbers. Eample: and 8 are surds but 4 is not as it equals. Simplifying Surds Start to simplify surds by using two rules: ab = a b and a b = a b By using the multiplication rule, simplify surds by finding a factor of the number you are taking a root of which is a square number. Always try to find the largest square number factor, otherwise you will have to simplify further. Eample: 8 = 4 = 1 = 4 = = = 600 = 00 = 100 = 10 EXERCISE A Simplify 1) 50 ) 7 ) 7 4) 80 5) 60 6) 900

32 Multiplying and Dividing with Surds The rules of algebra are true for any numeric value; these include surds. To multiply and divide epressions with surds, deal with any integers together and then deal with any surds. Eamples: = = = = = 8 10 (5 + ) = = 4 (1 + )( ) = + 6 ( + )( ) = ( ) = 1 In this eample, you could epand as usual but this is an eample of the difference of two squares. EXERCISE B Simplify 1) 7 ) ) 6 4) 8 7 5) ) ) ( + 1)( + 5) 8) (5 )( 8)

33 Addition and Subtraction with Surds You can only add or subtract with surds if the surd is the same; sometimes if they are not the same, you may be able to simplify them so that the same surd is present. Eample: = Here add the and 4 as the same surd is present but you cannot add the = = 5 5 By simplifying 45 to 5, you can add the two surds together. These methods also work for subtraction of surds. Eercise C Simplify 1) + 7 ) ) ) ) ) 5 5 7) ) ) )

34 Rationalising the Denominator It is far easier to calculate with a fraction if the surd if the denominator is a rational number (i.e. not a surd). The process of this is known as rationalising the denominator. To do this, multiply by the surd in the denominator, doing so makes use of the fact that a a = ( a) = a Eample: 1 Multiply the denominator by to rationalise it and so multiply the numerator by also: 1 = Eample : Eample : 4 = 4 = 4 = + = ( + ) = = 5 If there is more than just a surd in the denominator, we make use of the difference of two squares by multiplying by its conjugate. Eample: Rationalise 7 We multiply the numerator and denominator by its conjugate: + 7 It s a difference of two squares so epand as usual = ( + 7) ( 7)( + 7) ( + 7) = ( 7) ( + 7) = 9 7 ( + 7) = = + 7

35 Eercise D Rationalise the following: 1 1 a) 5 d) 7 g) 4+ 7 b) 5 e) h) c) 10 5 f) i) a) d) b) e) c) 6 7 More help is available from MyMaths: Surds 1, Surds

36 .14 Algebraic Fractions Algebraic fractions can be treated in eactly the same way as numerical fractions. Eample Solution Eample Multiply by. 7y = 6, so the answer is 6 6. (Not as this is just an equivalent fraction!) 7y 14y Divide y 4 by y. Solution y 4 y = y 4 1 y = y 4y = y (Don t forget to simplify!) 4 Eample Solution Divide PQR 100 PQR 100 by T. T = PQR T = PQR 100T PQR 100, which is a mi (as well as a mess!) Here it would be wrong to say just T Double fractions, or mitures of fractions and decimals, are always wrong. For instance, if you want to divide y z by, you should not say 0.5y z but y z. This sort of thing is etremely important when it comes to rearranging formulae.

37 Eample Simplify Solution Use a common denominator. [You must treat ( 1) and ( + 1) as separate epressions with no common factor.] 1 ( 1) ( 1) 1 1 ( 1)( 1) 1 4 ( 1)( 1) ( 1)( 1). Do use brackets, particularly on top otherwise you are likely to forget the minus at the end of the numerator (in this eample subtracting -1 gives +1). Don t multiply out the brackets on the bottom. You will need to see if there is a factor, which cancels out (although there isn t one in this case). Eample Simplify Solution A common denominator may not be obvious, you should look to see if the denominator factorises first is a common factor, so the 1 ( 1) ( 1)( 1) common denominator ( 1) 5 ( 1)( 1) is ( 1)( 1). 15 ( 1)( 1) 17 ( 1)( 1) If one of the terms is not a fraction already, the best plan is to make it one. Eample Write +1 + as a single fraction. Solution ( 1) This method often produces big simplifications when roots are involved.

38 Eample Solution Write + as a single fraction. 1 ( ) ( ) It is also often useful to reverse this process that is, to rewrite epressions such as. The problem with this epression is that appears in more than one place and it is not very easy to manipulate such epressions (for eample, in finding the inverse function, or sketching a curve). Here is a very useful trick. Eercise 1 Write as single fractions (a) 1 (b) (c) 1 (d) 1 1 (e) 1 (f) 1 (g) 4( 1) 4 1 Write as single fractions. 1 (a) (b) (c) ( ) More help is available from MyMaths: Cancelling Algebraic Fractions, Multiplying Algebraic Fractions, Adding Algebraic Fractions

39 .15 Linear and Quadratic Simultaneous Equations I am sure that you will be very familiar with the standard methods of solving simultaneous equations (elimination and substitution). You will probably have met the method for solving simultaneous equations when one equation is linear and one is quadratic. Here you have no choice; you must use substitution. Eample 1 Solve the simultaneous equations + y = 6 + y = 10 Solution Make one letter the subject of the linear equation: = 6 y Substitute into the quadratic equation (6 y) + y = 10 Solve 10y 6y + 6 = 0 (y 1)(5y 1) = 0 to get two solutions: y = 1 or.6 Substitute both back into the linear equation = 6 y = or 1.8 Write answers in pairs: (, y) = (, 1) or ( 1.8,.6) You can t just square root the quadratic equation. You could have substituted for y instead of (though in this case that would have taken longer try to avoid fractions if you can). It is very easy to make mistakes here. Take great care over accuracy. It is remarkably difficult to set questions of this sort in such a way that both pairs of answers are nice numbers. Don t worry if, as in this eample, only one pair of answers are nice numbers. Questions like this appear in many GCSE papers. They are often, however, rather simple (sometimes the quadratic equations are restricted to those of the form + y = a) and it is important to practice less convenient eamples.

40 Eercise Solve the following simultaneous equations. 1 + y = y = 1 + y = 10 y = 1 + y + y = 1 4 y + y = 1 + y = 1 y = 5 c + d = y = 15 c + 4d = y = y + y = y + 6y = 4 + 4y = 1 + y = y = y + y = 0 + y = 5 + y = y + 5y = 15 1 y + + y = 7 y = 1 y = y + 5y = y y = 0 y = 1 y = y 1 1 = y y = 11 + y = 7 More help is available from MyMaths: Quadratic Simultaneous Equations

41 .16 Epanding More Than Two Binomials You should already be able to epand algebraic epressions of the form (a + b)(c + d). e.g. ( + 1)( ) = = 6 e.g. (5 + 4)(5 4) = = 5 16 We are now going to algebraic epressions of the form (a + b)(c + d)(e + f). To simplify the product of three binomials, first epand any two of the brackets and then multiply this answer by each term in the third bracket Eample 1: Epand and simplify ( )( + )( + 7) ( )( + ) = First epand two of the brackets (You may prefer to use the grid method) = 6 Simplify Now ( )( + )( + 7) = ( + 7)( 6) = ( 6) + 7( 6) Multiply your epansion by each term in the rd bracket = = Simplify Eample : Show that ( + 5)( 1)(4 ) = for all values of. ( + 5)( 1) = First epand any two of the brackets. = + 5 Simplify Now ( + 5)( 1)(4 ) = (4 )( + 5) = 4( + 5) ( + 5) Multiply your epansion by each term in the rd bracket = Remember the minus outside the nd bracket changes each sign inside the nd bracket = Simplify To simplify the product of four binomials, first epand any two of the brackets and then epand the other two brackets and then multiply these answers.

42 Eample : Epand and simplify ( + )( )( + 1)(5 6) ( + )( ) = 9 Epand two of the brackets ( + 1)(5 6) = Epand the other two brackets ( + )( )( + 1)(5 6) = ( 9)(10 7 6) Use the two epansions above = (10 7 6) 9(10 7 6) Multiply the nd bracket by each term in the 1st bracket = = Simplify EXERCISE: 1. Epand and simplify (a) ( + 1)( + 4)( + 5) (b) ( + 7)( + 1)( + 8) (c) ( )( 1)( ) (d) ( + 8)( )( 5) (e) (5 1)( + 5)( ) (f) (4 + 1)( + 7)(4 1) (g) ( + 4) ( 7) (h) (6 + 5)( 1) (i) ( 1)( + 1)(4 1)( 5) (j) ( + 5) ( ). Show that ( + ) = for all values of.. Show that ( 4) ( + ) simplifies to + a + b + c where a, b and c are integers. 4. Epress ( 1) 4 in the form a 4 + b + c + d + e where a, b, c, d and e are integers. 5. ( + 5)( 4)( ) = 9 + A + B + 40 Work out the value of A and the value of B. 6. ( )( + 1)(A + 1) = 8 + B + C Work out the value of A, the value of B and the value of C. 7. Here is a cuboid All measurements are in centimetres. Show that the volume of the cuboid is ( ) cm.

43 8. f() = + 4 Epress f( + ) in the form a + b + c + d. 9. The smallest of three consecutive positive odd numbers is ( 1). Work out the product of the three numbers. Give your answer in the form a + b + c + d.

44 .17 Quadratic Inequalities You should be able to solve quadratic equations of the form a + b + c = 0 e.g. 4 = 0 ( 4)( + 1) = 0 = 4 or = 1 e.g = 0 ( )( 4) = 0 = or = 4 e.g. = = 0 ( + 5)( ) = 0 = 5 or = You should also know the shape of a quadratic curve. If the coefficient of is positive, the curve is smiling. If the coefficient of is negative, the curve is frowning. If f() > 0 or f() 0 we want the values of where f() is above the -ais. If f() < 0 or f() 0 we want the values of where f() is below the -ais. Eample 1: Solve ( + 8)( ) 0 First factorise your quadratic epression Critical values are = 8 and = Solve ( + 8)( ) = 0 8 Always draw a sketch of your curve Show where the curve cuts the -ais by solving ( + 8)( ) = 0 8 and We want the area where y 0 If you are asked to write the solution set of the inequality then the answer would be: { : 8, } NOTE: There are TWO regions so we write the answer as TWO inequalities.

45 Eample : Find the solution set of the inequality 6( + ) < < 17 First epand the bracket < 0 Rearrange to the form a + b + c < 0 ( 4)( ) < 0 Factorise in order to find where it cuts the -ais Critical values are = 4/ and / Solve ( 4)( ) = < < Sketch the curve and shade below the ais We want the region where f() is below the -ais Solution set = { : 4 < < } Make sure your answer is given in the correct form NOTE: There is ONE region so we write the answer as ONE inequality. Eample : Solve ( + 9) 0 ( + 9) 0 Critical values are = 0 and = 9 This is already factorised with 0 on one side so there is no need to epand the brackets 9 0 Sketch the curve and shade below the ais 9 0 We want the region where f() is below the -ais There is only one region so write as one inequality

46 Eample 4: Solve the inequality < < 0 Rearrange to the form a + b + c < 0 ( + )(7 ) < 0 Factorise in order to find where it cuts the -ais 7 The curve is 'frowning' as we have < and > 7 We want the region where f() is below the -ais OR < 0 Rearrange to the form a + b + c < > 0 Multiply each term by 1 which changes < to > ( + )( 7) > 0 Factorise in order to find where it cuts the -ais 7 The curve is 'smiling' as we have + < and > 7 This method gives the same answer as the 1st method EXERCISE 1. Solve each of these inequalities. (a) (b) 0 < 0 (c) ( )( + 7) > 0 (d) 5 0 (e) < 0 (f) (5 + )( 1 ) 0 (g) (h) 1 > 0 (i) (5 ) > 0 (j) > 5. Find the solution set for each of these inequalities.. Solve (a) (b) + 4 < 0 (c) ( + ) > 48 (d) (e) > 11 1 (f) 16 6 (g) 7 + (4 15) 0 (h) 4( + 6) > 8 (i) (5 ) > 0 (j) ( + 5) Find the solution set for which Find the set of values for which 6 + and + < 6. Find the solution set for ( )( + ) < (1 ) 5 More help is available from MyMaths: Quadratic Inequalities

47 .18 Using Completing the Square to Find Turning Points You should already be able to epress a quadratic equation in the form a( + b) + c by completing the square. e.g. 6 + = ( ) 9 + = ( ) 6 e.g = [ + ] + 5 = [( + 1) 1] + 5 = ( + 1) + We are now going to deduce the turning points of a quadratic function after completing the square. Eample 1: Given y = + 6 5, by writing it in the form y =( + a) + b, where a and b are integers, write down the coordinates of the turning point of the curve. Hence sketch the curve. y = = ( + ) 9 5 Remember to halve the coefficient of = ( + ) 14 and subtract ( ) to compensate The turning point occurs when ( + ) = 0, i.e. when = When =, y = ( + ) 14 = 0 14 = 14 So the coordinates of the turning point is (, 14) The graph y = cuts the y-ais when = 0, i.e. y = 5 Sketch: y (, 14) (0, 5) When y = ( + a) + b then the coordinates of the turning point is ( a, b). The minimum or maimum value of y is b.

48 Eample : Given that the minimum turning point of a quadratic curve is (1, 6), find an equation of the curve in the form y = + a + b. Hence sketch the curve. y = ( 1) 6 If the minimum is when = 1, we know we have ( 1) = ( + 1) 6 If the minimum is when y = 6, we know we have (...) 6 = 5 An equation of the curve is y = 5 The graph cuts the y-ais when = 0, i.e. at y = 5 Sketch: y It is a minimum turning point so the shape is (0, 5) (1, 6) NOTE: There are other possible equations as, for eample y = 4( 1) 6 also has a turning point of (1, 6). If it was a maimum turning point then the coefficient of would be negative. Eample : Find the maimum value of and sketch the curve = ( 4 + 7) First take out the minus sign = [( ) 4 + 7] Remember to use square brackets = [( ) + ] = ( ) Multiply ( ) and + by 1 The maimum value is y (, ) (0, 7) It is a maimum value so the shape is

49 Eercise: 1. By writing the following in the form y = ( + a) + b, where a and b are integers, write down the coordinates of the turning point of the curve. Hence sketch the curve. (a) y = (b) y = 10 1 (c) y = (d) y = (e) y = (f) y = 5. Given the following minimum turning points of quadratic curves, find an equation of the curve in the form y = + a + b. Hence sketch each curve. (a) (, ) (b) ( 4, 1) (c) ( 1, 5) (d) (, 1) (e) (1, 7) (f) ( 4, 1). Find the maimum or minimum value of the following curves and sketch each curve. (a) y = (b) y = 1 6 (c) y = + (d) y = (e) y = 1 (f) y = The epression + 8 can be written in the form ( a) + b for all values of. (i) Find the value of a and the value of b. The equation of a curve is y = f() where f() = + 8 The diagram shows part of a sketch of the graph of y = f(). The minimum point of the curve is M. (ii) Write down the coordinates of M.

50 5. (i) Sketch the graph of f() = , showing the coordinates of the turning point and the coordinates of any intercepts with the coordinate aes. (ii) Hence, or otherwise, determine whether f() = 0 has any real roots. Give reasons for your answer. *6. The minimum point of a quadratic curve is (1, 4). The curve cuts the y-ais at 1. Show that the equation of the curve is y = 6 1 *7. The maimum point of a quadratic curve is (, 5). The curve cuts the y-ais at 1. Find the equation of the curve. Give your answer in the form a + b + c. * = etension

51 Functions.1 Function Notation In GCSE Mathematics equations are written as shown below: y = + 4 y = ² + 5 Sometimes a different notation is used which is called function notation. We often use the letters f and g and we write the above equations as f() = + 4 g() = ² + 5 Eample 1: Using the equation y = + 4, find the value of y if (a) = 4 (b) = 6 (a) y = (4) + 4 = = 16 Substitute for = 4 in the equation (b) y = ( 6) + 4 = = 14 Substitute for = 6 in the equation Eample : f is a function such that f() = + 4 Find the values of (a) f(4) (b) f( 6) (a) f(4) = (4) + 4 = = 16 Substitute for = 4 in the equation (b) f( 6) = ( 6) + 4 = = 14 Substitute for = 6 in the equation

52 Eample : g is a function such that g() = ² 5 Find the values of (a) g() (b) g( 4) (a) g() = ()² 5 = 18 5 = 1 (b) g( 4) = ( 4)² 5 = 5 = 7 Substitute for = in the equation Substitute for = 4 in the equation Eample 4: The functions f and g are defined for all real values of and are such that f() = ² 4 and g() = Find (a) f( ) (b) g(0.) (c) Find the two values of for which f() = g(). (a) f( ) = ( )² 4 = 9 4 = 5 (b) g(0.) = 4(0.) + 1 = =. (c) ² 4 = Substitute for = in the equation f() Substitute for = 0. in the equation g() Put f() = g() ² = 0 Rearrange the equation as a quadratic = 0 ² 4 5 = 0 Simplify ( 5)( + 1) = 0 Solve the quadratic by factorising 5 = 0 or + 1 = 0 = 5 or = 1

53 Eercise: 1. The function f is such that f() = 5 + Find (a) f() (b) f(7) (c) f( 4) (d) f( ) (e) f( 0.5) (f) f(0.). The function f is such that f() = 4 Find (a) f(4) (b) f(6) (c) f( ) (d) f( 6) (e) f( 0.) (f) f(0.9). The function g is such that g() = + 1 Find (a) g(0) (b) g(1) (c) g() (d) g( 1) (e) g( 0.4) (f) g(1.5) 4. The function f is such that f ( ) 5 Find (a) f(0) (b) f(1) (c) f() (d) f( 1) (e) f( 0.7) (f) f(1.5) 5. f() = 8 Epress f( + ) in the form a + b 6. The functions f and g are such that f() = 5 and g() = (a) Find (i) f( 1) (ii) g() (b) Find the value of for which f() = g(). 7. The functions f and g are such that f() = 1 and g() = 5 + (a) Find f( ) and g( 5) (b) Find the two values of for which f() = g(). More help is available from MyMaths: Functions 1

54 . Composite Functions A composite function is a function consisting of or more functions. The term composition is used when one operation is performed after another operation. For instance: + 5( + ) + 5 This function can be written as f() = 5( + ) Suppose f() = and g() = + What is fg()? Now fg() = f[g()] This means apply g first and then apply f. fg() = f( + ) = ( + )² g() is replaced with + It is then substituted into f() What is gf()? This means apply f first and then apply g. gf() = g(²) = + ² f() is replaced with It is then substituted into g() NOTE: The composite function gf() means apply f first followed by g. NOTE: The composite function fg() means apply g first followed by f. NOTE: fg() can be written as fg and gf() can be written as gf; fg is not the same as gf.

55 Eample 1: f and g are functions such that f() = 1 and g() = Find the composite functions (a) fg (b) gf (a) fg = fg() = f ( ) 1 = 1 (b) gf = gf() = g 1 = = Do g first: Put ( ) instead of g() 1 Substitute ( ) for in 1 Do f first: Put instead of f() 1 Substitute for in ( ) Eample : f() = 7 g() = 4 1 h() = ( 1) Find the following composite functions: (a) gf (b) gg (c) fgh (a) gf = gf() = g(7 ) Do f first: Put (7 ) instead of f() = 4(7 ) 1 Substitute (7 ) for in 4 1 = 7 8 Simplify (b) gg = gg() = g(4 1) Put (4 1) instead of g() = 4(4 1) 1 Substitute (4 1) for in 4 1 = 16 5 Simplify (c) fgh = fgh() = fg[( 1)] Put ( 1) instead of h() = fg ( ) Epand ( 1) = f [4( ) 1] Substitute ( ) for in 4 1 = f (1 1) Simplify = 7 (1 1) Substitute (1 1) for in 7 = 4 Simplify

56 Eample : f() = 7 g() = 4 1 h() = ( 1) Evaluate (a) fg (5) (b) ff ( ) (c) ghf () (a) fg(5) = f(0 1) = f(19) Substitute for = 5 in 4 1 = 7 (19) = 1 Substitute for = 19 in 7 (b) ff( ) = f[7 ( )] = f(11) Substitute for = in 7 and simplify = 7 (11) = 15 Substitute for = 11 in 7 and simplify (c) ghf() = gh(7 6) = gh(1) Substitute for = in 7 and simplify = g[(1 1)] = g(0) Substitute for = 1 in ( 1) and simplify = 4(0) 1 = 1 Substitute for = 0 in 4 1 and simplify Eample 4: f() = + and g() = 7 Solve the equation gf() = gf() = g( + ) Put ( + ) instead of f() = 7 ( + ) g's rule is subtract from 7 = 5 Simplify 7 5 = Put gf() = and solve 5 = 5 Add to both sides = 1

57 Eample 5: (more challenging question) Functions f, g and h are such that f: 4 1 g: 1, h: ( ) Find (a)(i) fg() (ii) hh() (b) Show that hgf() = 1 (a) fg() = f 1 = 4 1 = = 4 1( ) Substitute for g() f s operation is Simplify using a common denominator of + denominator o (b) hh() = h( ) = [ ( ) ] Substitute for h() h s operation is subtract from and then square = [ (4 + )] ( ) = ( )( ) = 4 + = ( + 4 ) (c) hgf() = hg(4 1) 1 = h( ) 4 1 Put (4 1) for f() Substitute (4 1) for in g() = = = 4 + 1, so put for in h() Simplify using a common denominator of = 4 1 =

58 Eercise: 1. Find an epression for fg() for each of these functions: (a) f() = 1 and g() = 5 (b) f() = + 1 and g() = 4 + (c) f() = and g() = 1 (d) f() = and g() = +. Find an epression for gf() for each of these functions: (a) f() = 1 and g() = 5 (b) f() = + 1 and g() = 4 + (c) f() = and g() = 1 (d) f() = and g() = +. The function f is such that f() = Find (i) ff() (ii) Solve the equation ff(a) = a 4. Functions f and g are such that f() = and g() = 5 + Find (a)(i) fg() (ii) gf() (b) Show that there is a single value of for which fg() = gf() and find this value of. 5. Given that f() = 1, g() = + 4 and fg() = gf(), show that 1 = The function f is defined by f() =, 0 Solve ff() = 7. The function g is such that g() = (a) Prove that gg() = (b) Find ggg() for 1 8. Functions f, g and h are such that f() =, g() = 14 and h() = Given that f() = gfh(), find the values of.

59 . Inverse Functions The function f() = +5 can be thought of as a sequence of operations as shown below Now reversing the operations 5 5 The new function, 5, is known as the inverse function. Inverse functions are denoted as f 1 (). Eample 1: Find the inverse function of f() = 4 y = 4 Step 1: Write out the function as y =... = y = y Step : Swap the and y Step : Make y the subject 4 = y f 1 () = 4 Step 4: Instead of y = write f 1 () =

60 Eample : Find the inverse function of f() = 7 y = = 7 y 7 Step 1: Write out the function as y =... Step : Swap the and y 7 = y Step : Make y the subject 7 + = y f 1 () = 7 + Step 4: Instead of y = write f 1 () = Eample : Find the inverse function of f() = 4 y = 4 Step 1: Write out the function as y =... = y 4 Step : Swap the and y = y + 4 Step : Make y the subject 4 = y f 1 () = 4 Step 4: Instead of y= write f 1 () = RULES FOR FINDING THE INVERSE f 1 (): Step 1: Write out the function as y =... Step : Swap the and y Step : Make y the subject Step 4: Instead of y= write f 1 () =

61 Eercise: 1. Find the inverse function, f 1 (), of the following functions: (a) f() = 1 (b) f() = + (c) f() = 1 (d) f() = + 5 (e) f() = 6(4 1) (f) f() = 4 (g) f() = (h) f() = (1 ) (i) f() = 1 (j) f() = 1. The function f is such that f() = 7 (a) Find f 1 (). (b) Solve the equation f 1 () = f().. The function f is such that f() = 8 (a) Find f 1 (). (b) Solve the equation f 1 () = f(). 4. The function f is such that f() = 1, 4. 4 Evaluate f 1 (). [Hint: First find f 1 () and then substitute for = ] 5. f() =, R, (a) If f 1 () = 5, find the value of. (b) Show that ff 1 () = 6. Functions f and g are such that f() = + g() = + 1 Find an epression for (fg) 1 () [Hint: First find fg() ] More help is available from MyMaths: Functions

62 4 Graphs No doubt you will have plotted many graphs of functions such as y = + 4 by working out the coordinates of points and plotting them on graph paper. But it is actually much more useful for A Level mathematics (and beyond) to be able to sketch the graph of a function. It might sound less challenging to be asked to draw a rough sketch than to plot an accurate graph, but in fact the opposite is true. The point is that in order to draw a quick sketch you have to understand the basic shape and some simple features of the graph, whereas to plot a graph you need very little understanding. Many professional mathematicians do much of their basic thinking in terms of shapes of graphs, and you will be more in control of your work, and understand it better, if you can do this too. When you sketch a graph you are not looking for eact coordinates or scales. You are simply conveying the essential features: the basic shape where the graph hits the aes what happens towards the edges of your graph The actual scale of the graph is irrelevant. For instance, it doesn t matter what the y- coordinates are. 4.1 Straight line graphs I am sure that you are very familiar with the equation of a straight line in the form y = m + c, and you have probably practised converting to and from the forms a + by + k = 0 or a + by = k, usually with a, b and k are integers. You need to be fluent in moving from one form to the other. The first step is usually to get rid of fractions by multiplying both sides by a common denominator. Eample 1 Write y in the form a + by + k = 0, where a, b and k are integers. 5 Solution Multiply both sides by 5: 5y = 10 Subtract 5y from both sides: 0 = 5y 10 or 5y 10 = 0 In the first line it is a very common mistake to forget to multiply the by 5. It is a bit easier to get everything on the right instead of on the left of the equals sign, and this reduces the risk of making sign errors. In plotting or sketching lines whose equations are written in the form a + by = k, it is useful to use the cover-up rule: