Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

Line (geometry) wikipedia, lookup

Rational trigonometry wikipedia, lookup

Pythagorean theorem wikipedia, lookup

Integer triangle wikipedia, lookup

BKL singularity wikipedia, lookup

Euler angles wikipedia, lookup

Trigonometric functions wikipedia, lookup

Euclidean geometry wikipedia, lookup

Transcript

GCSE: Solving Linear Equations Skipton Girlsβ High School Objectives: (a) Solve equations, including with unknowns on both sides and with brackets. (b) Form equations from context (with emphasis on quality of written communication, e.g. "Let x be..."). KEY TERMS This is an example of a: 3 2π₯ Term 3π₯ + 2 Expression 2 5π₯ +1=2 A term is a product of numbers and variables ? (no additions/subtractions) An expression is composed ? of one or more terms, whether added or otherwise. Equation An equation says that the ? expressions on the left and right hand side of the = have the same value. STARTER The perimeter of this work of art is 32. By trial and error (or any other method), find π. 3π + 1 9βπ π=3 ? If we added all four sides of the painting to get the perimeter, weβd have: 3π + 1 + 3π + 1 + 9 β π + 9 β π = 4π + 20 And weβre told the perimeter is 32, so ππ + ππ = ππ. Weβll see today how to βsolveβ equations like this so we can find π. Equations must always be βbalancedβ We already know that the β=β symbol means each side of the equation must have the same value. π=4 +2 2 a 2 4 = π + 2= 6 If we added something to one side of the equation, what do we have to do with the other side? +2 Equations must always be βbalancedβ If we tripled the load on one side of the scales, what do we have to do with the other side? π=4 ×3 a 4 a a 4 4 = ×3 3π = 12 Solving !To solve an equation means that we find the value of the variable(s). 3π + 1 4π + 20 = 32 9βπ Strategy: To get π on its own on one side of the equation, we gradually need to βclaw awayβ the things surrounding it. Solving Bro Note: You can probably see the answer to this in your head because the equation is relatively simple, but this full method is crucial when things become more complicated 4π + 20 = 32 ? -20 ? -20 ? 4π = 12 ? οΈ4 Strategy: Do the opposite operation to βget rid ofβ items surrounding our variable. π₯+4 οΈ4? ? 3 π= Bro Tip: Many students find writing these operations between each equation helpful to remind them what theyβre doing to each side, but youβll eventually want to wean yourself off these. 3π¦ π§ 6 -4? ? οΈ3 ? ×6 π₯ π¦ π§ Test Your Understanding 3π β 5 = 13 ? +5 ? +5 3π = 18 ? ? οΈ3 ? οΈ3 ? π= 6 When the solution is not a whole number 4 + 6π§ = 18 ? -4 -4? 6π§ = 14 ? ? οΈ6 οΈ6? 14 7 π§= ? = 16 3 Your Goβ¦ 3 = 20 + 4π₯ β17 = 4π₯ ? ππ π=β π Bro Note: In algebra, we tend to give our answers as fractions rather than decimals (unless asked). And NEVER EVER EVER recurring decimals. Dealing with Fractions π₯ 3 + = 28 5 ? -3 -3? π₯ ? = 25 5 ? ×5 ? ×5 π₯ = ?125 Exercise 1 Solve the following equations, showing full working. 1 2 3 4 5 6 π β 4 = 10 2π₯ + 3 = 9 5π₯ β 4 = 36 9π₯ β 2 = 61 9 = 1 + 4π¦ 8π + 3 = 75 7 3π₯ = 7 8 5π₯ + 2 = 11 9 8π₯ β 2 = 3 10 3 + 10π = 7 11 5 + 3π = 4 12 7π + 23 = 11 13 14 + 9π = 3 π = ππ ? π=π ? π=π ? π=π ? π=π ? π=π ? π π= ? π π π= ? π π π= ? π π π= ? π π π=β ? π ππ π=β ? π ππ π=β ? π 14 15 16 17 18 19 20 21 22 23 24 N π₯ =5 π = ππ ? 7 π +3=8 π = ππ ? 4 π β1=5 π = ππ ? 2 π 1+ =7 π = ππ ? 3 π₯ +3= 4 π=π ? 5 π¦ +8=5 π = βππ? 4 π 5= +9 π = βππ? 6 2π₯ 3+ =7 π = ππ ? 5 5π ππ ? β 3 = 10 π= 6 π 3π₯ π 5+ =3 π=β ? 4 π 6π₯ π 11 = +9 π=β ? 7 π 6 π₯+3 ππ 5 3= +9 π=β ? 8 π What happens if variable appears on both sides? 5π + 3 = 2π + 9 What might our strategy be? Collect the variable terms (i.e. The terms involving a) on one side of the equation, and the βconstantsβ (i.e. The ? individual numbers) on the other side. What happens if variable appears on both sides? Strategy? Collect the variable terms on the side of the equation where thereβs more of them (and move?constant terms to other side). Letβs move the βπβ terms to the left (as 5 > 2) and the constants to the right. 5π + 3 = 2π + 9 This is to get rid of the constant term on the left. 5π = 2π + 6 -3 -3 ? ? -2a ? -2a 3π = 6 ? ? οΈ3 ? οΈ3 π =? 2 We could have done these two steps in either order. More Examples 11π₯ β 4 = 2π₯ β 13 3π¦ + 4 = 8π¦ β 5 9 π¦= 5 π₯ = β1? Way weβd have previously done itβ¦ 2 = β3π₯ 2? 2 π₯= =β β3 3 5 = 3 β 3π₯ Both methods are valid, but I prefer the second β itβs best to avoid dividing by negative numbers, and is less likely to lead to error. ? Or where we put π₯ term on side where itβs positive: 5 + 3π₯ = 3 3π₯ = β2 ?2 π₯=β 3 Test Your Understanding 3π₯ β 3 = π₯ + 5 ππ = π ? π=π 3 β 5π₯ = 5 + 2π₯ βπ = ππ π ? π=β π Dealing with Brackets If thereβs any brackets, simply expand them first! 2 2π₯ + 3 = 9 ππ + π = π ππ = π π? π= π 3 β 4 2π₯ β 3 = 7π₯ π β ππ + ππ = ππ ππ β ππ = ππ ? ππ = πππ π=π Test Your Understanding 7 3π₯ β 1 = 21 + 14π₯ πππ β π = ππ + πππ ππ = ππ ? π=π 5β2 π₯+2 =4β3 2βπ₯ π β ππ β π = π β π + ππ π β ππ = βπ + ππ π = ππ ? π π= π Exercise 2 Solve the following. 1 2 3 4 5 6 7 8 9 10 11 12 13 3π₯ = π₯ + 4 6π¦ = 4π¦ β 4 5π₯ + 3 = 3π₯ + 7 8π¦ β 3 = 6π¦ + 7 10π₯ + 3 = 7π₯ β 3 π₯ =2βπ₯ 2π§ = 9 β π§ 9π‘ = 99 β 2π‘ π₯ + 6 = 2π₯ β 6 7π¦ + 1 = 9π¦ + 5 5 π₯ β 2 = 30 2 π₯+1 =9+π₯ 4 π₯ β 1 = 3π₯ β 1 14 3 π₯β2 =3+π₯ 15 5 2π₯ + 3 = 20 π =?π π =?βπ π =?π π =?π π =? βπ π =?π π =?π π =?π π =?ππ π =?βπ π =?π π =?π π =?π π π =? π π π =? π 16 17 18 19 20 21 22 N ππ 10 β 5 π₯ β 2 = 3 + 4π₯ π =? π 9 β 4π₯ = 4 β 9π₯ π =?βπ π 2 3π₯ + 1 = 3 + 2 2π₯ β 1 π = ?β π π 3 + 2 3π₯ + 1 = 7 + π₯ π =? π 2β π₯β2 =π₯β 2βπ₯ π =?π 6+3 π¦β2 =5 π¦+4 π =?βππ π₯ β 2 2 β π₯ = 3 π₯ β 2 β π₯ π =?βπ 1 3 1 π π₯ + π₯ β 2 = 2π₯ β 3 π =? 2 4 3 π GCSE: Forming and Solving Equations Skipton Girlsβ High School RECAP :: Forming/Solving Process Worded problem [JMC 2008 Q18] Granny swears that she is getting younger. She has calculated that she is four times as old as I am now, but remember that 5 years ago she was five times as old as I was at that time. What is the sum of our ages now? Stage 1: Represent problem algebraically Stage 2: βSolveβ equation(s) to find value of variables. Let π be my age and π be Grannyβs age. π β 5 = 5π β 25 π = 5π β 20 5π β 20 = 4π π = 20 β΄ π = 80 π = 4π π β 5 = 5(π β 5) We previously learnt how to form expressions given a worded context. Weβll learn how to actually solve these equations formed now! Weβll first focus on problems where the expressions have already been specified. Example The angles of a triangle are as pictured. Determine π₯. π + ππ ππ Step 1: Think of a sentence which would have the word βisβ in it. Write each part of sentence as an algebraic expression, with βisβ giving =. π β ππ Step 2: Solve! Sentence with isβisβ in it? βSum of angles 180°β Expr π₯ + 10 + 2π₯1? + π₯ β 50 = 180 Expr 2? 4π₯ β 40 = 180 4π₯ = 220 Solve! π₯ = 55 Another Example π₯+4 3 π₯ 10 The rectangle and triangle have the same area. Determine the width of the rectangle. βArea βisβ of triangle is area of rectβ sentence? Expr Expr 2? 3 π₯ 1? + 4 = 5π₯ 3π₯ + 12 = 5π₯ 12 = 2π₯ π₯ =Solve! 6 Therefore width of rectangle is 6 + 4 = 10 Check Your Understanding 1 The following diagram shows the angles of a quadrilateral. Determine π₯. 3π₯ 2π₯ + 10 2π₯ + 20 120 βTotal angle is 360β Expr 2? Expr 1? ππ + ππ + ππ + ππ + ππ + πππ = πππ ππ + πππ = πππ ππ = Solve! πππ π = ππ° 2 The area of the triangle is 1 more than the π₯+4 5 N π₯ 12 area of the parallelogram. Determine π₯. Area1? of π« Expr is 1 more than area of par ππ = π + ππ + ππ + π = ππ π = ππSolve! Exprπ 2? π+π [JMO 1999 A9] Skimmed milk contains 0.1% fat and pasteurised whole milk contains 4% fat. When 6 litres of skimmed milk are mixed with π litres of pasteurised whole milk, the fat content of the resulting mixture is 1.66%. What is the value of π? Fat content of skimmed + Fat content of pasteurised = fat content of mixture ? 6 × 0.001 + 0.04π = 0.0166 π + 6 β π=4 Exercise 3 (See printed sheet) 1 The ages of three cats are π, π + 4 and 2π β 3. 7 [JMC 1998 Q18] The three angles of a Their total age is 21. Determine π. ππ + π = ππ π=π triangle are π₯ + 10 °, 2π₯ β 40 °, 3π₯ β 90 °. Which statement about the triangles is correct? It is: A right-angled isosceles B right-angled, but not isosceles C equilateral D obtuse-angled and isosceles E none of A-D Solution: C ? 2 Three angles in a triangle are π₯, 2π₯ + 20° and 7π₯ β 10°. What is π₯? ? π = ππ° 3 Two angles on a straight line are π₯ + 30° and 2π₯ β 60°. What is π₯? ? π = ππ° ? 4 The perimeter of this rectangle is 44. What is π₯? Solution: 7 2π₯ + 3 ? 8 The following triangle is isosceles. Determine its perimeter (by first determining π₯). π₯β2 5 The area of this rectangle is 48. Determine π₯. ? π=π 3π₯ + 10 π₯+5 π₯+5 4 6 An equilateral triangle has lengths 3π₯ + 6, 5π₯, 5π₯. What is π₯? π=π ? 5π₯ β 50 ? ππ + ππ = ππ β ππ β π = ππ° Perimeter = ππ + πππ + πππ = πππ Exercise 3 9 (See printed sheet) [JMO 2013 A7] Calculate the value of π₯ in the diagram shown? Solution: 36 ? 10 In the following diagram, π΄π· = π΅π·, π΄π΅ = π΄πΆ, β πΆπ΄π· = 7π₯ and β π΄π΅π· = 9π₯. Determine π₯. π΄ 7π₯ π· 9π₯ π΅ πΆ ππ + ππ + ππ = πππ ? π=π 11 A very large jug of π₯ litres of orange squash is 10% concentrate (the rest water). When 5 litres of concentrate is added the jug is now 12% concentrate. a. Form an equation in terms of π₯ (by considering the amount of concentrate we have). π. ππ + π = ? π. ππ(π + π) b. Hence determine π₯. π. ππ + π = π. πππ + π. π ? π. πππ = π. π π = πππ ππππππ Tougher Examples [JMC 2013 Q7] After tennis training, Andy collects twice as many balls as Roger and five more than Maria. They collect 35 balls in total. How many balls does Andy collect? Let π be the number of balls Roger collects. Then Andy collects ππ balls and Maria collects ππ β π. Total balls collected: π + ππ + ππ ? β π = ππ ππ β π = ππ π=π So Andy collected π × π = ππ balls. [TMC Regional 2014 Q9] In a list of seven consecutive numbers a quarter of the smallest number is five less than a third of the largest number. What is the value of the smallest number in the list? Let smallest number be π. Then numbers are π, π + π, π + π, π + π, π + π, π + π, π + π π π Therefore π π = π π + π β π π π π = π?+ π β π π π π π = π β π = ππ ππ So smallest number is 36. Test Your Understanding 1 The length of the rectangle is three times the width. The total perimeter is 56m. Determine its width. Let π be the width of the rectangle. Then the length is ππ. Then the perimeter is: π + ππ +?π + ππ = ππ ππ = ππ π=π β΄ The width is 7m. Bro Reminder: You should usually start with βLet β¦β 2 In 4 years time I will be 3 times as old as I was 10 years ago. How old am I? Let π be my age. Then π + π = π π β ππ π+π= ? ππ β ππ ππ = ππ π = ππ Exercise 4 (Maths Challenge Questions) 1 The sum of 5 consecutive numbers is 200. What is the smallest number? Solution: 38 (a quick non-algebraic method is to realise the middle ? of the five number is the average numbers, i.e. 40) 2 3 In 5 years time I will be 5 times as old as I was 11 years ago. Form a suitable equation, and hence determine my age. π + π = π π β ππ π = ππ ? In 6 years time I will be twice as old as I was 8 years ago. Determine my age. π+π=π πβπ ? π = ππ 4 I have three times as many cats as Alice but Bob has 7 less cats than me. In total we have 56 cats. How many cats do I have? Solution: 27 ? 5 Bob is twice as old as Alice at the moment. In 4 years time their total age will be 71. What is Aliceβs age now? 4 years time: Alice π + π Bob ππ + π ππ + π = ππ? β π = ππ 6 [TMC Final 2012 Q1] A Triple Jump consists of a hop, step and jump. The length of Keithβs step was three-quarters of the length of his hop and the length of his jump was half the length of his step. If the total length of Keithβs triple jump was 17m, what was the length of his hop, in metres? Solution: 8 metres ? Exercise 4 7 [JMO 2003 A6] Given a βstartingβ number, you double it and add 1, then divide the answer by 1 less than the starting number to get the βfinalβ number. If you start with 2, your final number is 5. If you start with 4, your final number is 3. What starting number gives the final number 4? ππ + π =π πβπ ππ + π = π π β π ? π π= π Exercise 4 8 [JMO 2012 A3] In triangle π΄π΅πΆ, β πΆπ΄π΅ = 84°; π· is a point on π΄π΅ such that β πΆπ·π΅ = 3 × β π΄πΆπ· and π·πΆ = π·π΅. What is the size of β π΅πΆπ·? 10 [JMC 2012 Q24] After playing 500 games, my success rate in Spider Solitaire is 49%. Assuming I win every game from now on, how many extra games do I need to play in order that my success rate increases to 50%? A 1 B 2 C 5 D 10 E 50 Let π be the number of extra games played. Then, giving 245 games were won before: πππ + π = π. π πππ + π π = ππ Solution: Let β π¨πͺπ« = π, then β πͺπ«π© = ππ and β π¨π«πͺ = πππ β ππ. Angles in π«π¨πͺπ«: ππ + πππ β ππ + π = πππ π = ππ° πππβπππ Then β π«πͺπ© = = ππ° ? ? (Note: itβs easier to just exploit the fact itβs multiple choice and try the options!) π 9 [JMO 2005 A8] A large container holds 14 litres of a solution which is 25% antifreeze, the remainder being water. How many litres of antifreeze must be added to the container to make a solution which is 30% antifreeze? Let π be the amount of antifreeze added in litres. 25% of 14 is 3.5. Thus: π. π + π = π. π ππ + π π=π ? N1 [JMO 2008 A9] In the diagram, πΆπ· is the bisector of angle π΄πΆπ΅. Also π΅πΆ = πΆπ· and π΄π΅ = π΄πΆ. What is the size of angle πΆπ·π΄? Let β π«πͺπ© = π. Filling in the angles using the information we find angles in π«π«π¨πͺ are π, πππ β ππ and πππ β ππ. This gives π = ππ° . β΄ β πͺπ«π¨ = πππ β π × ππ = πππ° (Note: this is not intended to ? be a full proof!) Exercise 4 N2 [JMO 2010 A10] Inn the diagram, π½πΎ and ππΏ are parallel. π½πΎ = πΎπ = ππ½ = ππ and πΏπ = πΏπ = πΏπΎ. Find the size of angle π½ππ. 60° π₯ π₯ β 60° 60° 300 β 2π₯ π₯ π₯ π₯ β 60° π₯ Since π±π² and π΄π³ are parallel, β π²π±π΄ and β π±π΄π³ are cointerior so add to πππ°. If we let β π²πΆπ³ = π, we can eventually find the angles as pictured. π β ππ + ππ + π β ππ + π = πππ π = ππ° β΄ β π±π΄πΆ = ππ β ππ = ππ° ? N3 [JMO 2013 B2] Pippa thinks of a number. She adds 1 to it to get a second number. She then adds 2 to the second number to get a third number, adds 3 to the third to get a fourth, and finally adds 4 to the fourth to get a fifth number. Pippaβs brother Ben also thinks of a number but he subtracts 1 to get a second. He then subtracts 2 from the second to get a third, and so on until he has five numbers. They discover that the sum of Pippaβs five numbers is the same as the sum of Benβs five numbers. What is the difference between the two numbers of which they first thought? Let π be Pippaβs first number. Then her numbers are π, π + π, π + π, π + π, π + ππ. Sum is ππ + ππ. Let π be Benβs first number. Then his numbers are π, π β π, π β π, π β π, π β ππ. Sum is ππ β ππ Weβre told ππ + ππ = ππ β ππ and the difference between their two starting numbers is π β π. Rearranging: ππ + ππ = ππ β ππ ππ β ππ = ππ β΄πβπ=π ? Exercise 4 (See printed sheet) N4 [JMO 2005 B4] In this figure π΄π·πΆ is a straight line and π΄π΅ = π΅πΆ = πΆπ·. Also, π·π΄ = π·π΅. Find the size of β π΅π΄πΆ. (Full proof needed) Full proof: Let β π«π¨π© = π Then β π«π©π¨ = π (base angles of isosceles π«π«π¨π© are equal) β π¨πͺπ© = π (base angles of isosceles π«π¨π©πͺ are equal) β πͺπ«π© = ππ (exterior angle of triangle π«π«π¨π© is sum of two interior angles) β π«π©πͺ = ππ (base angles of isosceles π«π©πͺπ« are equal) ? the angles in π«π¨π©πͺ: The angles in a triangle sum to πππ°. Using π + π + ππ + π = πππ° ππ = πππ° π = ππ° (We could have also used the angles in π«π©πͺπ«)