Lerner Pck Functionl Mthemtics Level 3 Activity N2: Plying drts Lerner Pck Functionl Mthemtics Level 4 Unit 2: Algebr FÁS 2012 1 September 2012
Acknowledgements Acknowledgements This booklet is prt of pck of resources for Functionl Mthemtics Level 4 which FÁS commissioned for use in their trining progrmmes. A similr set of resources hs been developed for Functionl Mthemtics Level 3. A tem from the Ntionl Adult Litercy Agency (NALA) nd the Ntionl Centre for Excellence in Mthemtics nd Science Teching nd Lerning (NCEMS- TL) developed nd edited the mterils. NALA: Bláthnid Ní Chinnéide Mry Gynor Fergus Doln John Stewrt Dr Terry Mguire (Institute of Technology, Tllght) NCEMS-TL: Prof. John O Donoghue Dr. Mrk Prendergst Dr. Mirim Liston Dr. Nimh O Mer FÁS: John O Neill Louise McAvin We re grteful to Kthleen Crmer nd her tem in Newbridge Youth Trining nd Development Centre who gve feedbck on extrcts from the Level 3 mterils. FÁS 2012 2 September 2012
Activity A1: Keeping the score Activity Keeping the score A1 (Google imges) This ctivity links to unit lerning outcome 2.1. Introduction There re different methods of scoring for different gmes. There re lso different numericl vlues of ech type of score. We could llow different vribles to represent the numericl vlue of different scores. Then we could use such expressions to clculte the score of ny gme. Wht will you lern Lerning Outcomes 1. Discuss the presence of vribles in rnge of rel life situtions. 2. Construct lgebric expressions for rel life situtions using the correct terminology nd including rerrngement of formule. FÁS 2012 3 September 2012
Activity A1: Keeping the score Key Lerning Points 1. Understnding the concept of vrible 2. Recognising the presence of vribles in rel life situtions 3. Replcing vribles with vlues through substitution 4. Constructing lgebric expressions for rel life situtions 5. Solving word problems through rerrngement Mterils you will need for this ctivity Prctice Sheet A1 Solution Sheet A1 Wht do you need to know before you strt? In Level 3 Unit 2 we lerned tht lgebr is n importnt topic within mthemtics nd everydy life. It is often clled the lnguge of mthemtics. Algebr involves writing problems mthemticlly using vribles to represent unknown vlues. A vrible is symbol for number we don t know yet. It cn be ny letter of the lphbet nd its vlue cn chnge depending on the problem or sitution. You should be wre of the different methods of scoring nd lso the different numericl vlues of types of scores in vrious gmes. You cn find these by plying or wtching different gmes. FÁS 2012 4 September 2012
Activity A1: Keeping the score Getting Strted Algebr involves writing problems mthemticlly using vribles to represent unknown vlues. When we do this we get n lgebric expression. Worked Exmple Using vribles to clculte scores In the 2010 All Irelnd Senior hurling finl, Kilkenny were iming to win their fifth consecutive title. On the dy they scored 1 gol nd 18 points. Their opponents Tipperry scored 4 gols nd 17 points. ) Allow different vribles to represent the numericl vlues of gols nd points. b) Using those vribles, write n expression to represent the totl possible score (S) tem cn chieve in hurling. c) Use this expression to clculte both Kilkenny nd Tipperry s totl score in the gme. d) Who won the gme nd by how much? FÁS 2012 5 September 2012
Activity A1: Keeping the score Solution ) We could use the following letters s vribles to represent the scores: - Gol = - Point = p b) S = 3 + p c) Kilkenny s score: S = 3 + p S = 3(1) + 18 S = 3 + 18 S = 21 Kilkenny s score = 21 Tipperry s score: S = 3 + p S = 3(4) + 17 S = 12 + 17 S = 29 Tipperry s score = 29 d) Tipperry won the gme by 8 points. FÁS 2012 6 September 2012
Activity A1: Keeping the score Tsk 1 The Interntionl Rules Series is gme plyed between Irish Gelic footbllers nd Austrlin Footbll Legue plyers. The rules re compromise between the two codes (or sets of rules). In Interntionl Rules gmes there re different methods of scoring, with vrious numericl vlues for ech type of score. For exmple: o A gol is worth 6 points. o An over is worth 3 points. o A behind is worth 1 point. The Series is mde up of two gmes, with the winner being whoever scores the most over the two gmes. In 2010, over the course of two gmes Irelnd scored: Austrli scored: ~ 2 gols ~ 0 gols ~ 19 overs ~ 28 overs ~ 23 behinds ~ 18 behinds ) Write n expression for totl score (S) llowing different vribles to represent the numericl vlue of different scores. FÁS 2012 7 September 2012
Activity A1: Keeping the score b) Use this expression to clculte the totl score of ech tem nd find out who won the Series FÁS 2012 8 September 2012
Activity A1: Keeping the score Tsk 2 Rugby is one of the most populr field gmes plyed in Irelnd. There re four different wys to score in the gme. The scoring system in rugby is s follows: Try Conversion gol Penlty gol Dropped gol 5 points 2 points 3 points 3 points In 2009, Leinster won the Heineken Cup for the first time. (Google Imges) In the mtch Leinster scored o 1 try o 1 try conversion o 2 drop gols o 2 penlties ) Write n expression for totl score (S) llowing different vribles to represent the numericl vlue of different scores b) Use this expression to clculte Leinster s totl score in the gme. FÁS 2012 9 September 2012
Activity A1: Keeping the score c) Their opponents in the finl, Leicester Tigers scored 16 points in totl. Their score included 1 try; 1 try conversions nd 0 drop gols. Use the expression to find how mny penlties Leicester Tigers scored. Prctise your skills Use Prctice Sheet A1. FÁS 2012 10 September 2012
Activity A2: Indices Activity Indices A2 This ctivity links to unit lerning outcome 2.2. (Google imges) Introduction In Unit 1, we lerned bout indices when working with scientific nottion. This ctivity will help you to become more fmilir with the lws of indices. Wht will you lern? Lerning Outcomes 1. Demonstrte n understnding of the lws of indices nd the rules of logrithms by using the lws nd rules to simplify expressions, solve equtions, nd trnspose formule. FÁS 2012 11 September 2012
Activity A2: Indices Key Lerning Points 1. Understnding the lws of indices Mterils you will need for this ctivity Clcultor Prctice Sheet A2 Solution Sheet A2 Wht do you need to know before you strt? In Unit 1 you lerned tht writing numbers using index (or power) nottion is the shorthnd for repeted multipliction. For exmple, 3 x 3 is 3 multiplied by itself twice. We cn write this s 3 2, nd we would sy three squred or three to the power of 2. In 3 2, 3 is clled the bse number nd 2 is clled power or index. The plurl of index is indices. A bse number is the number tht is being multiplied. The power (or index) tells us how mny times we multiply the bse number by itself. Getting Strted There re 8 rules of indices in totl. In this section we re going to look t five of these. In the previous ctivity A1, you lerned tht we cn use letters to represent numbers. Such letters re known s vribles. For the rules of indices we will let the vrible be ny number. Tht is, we don t know wht is but it represents ny number. FÁS 2012 12 September 2012
Activity A2: Indices Worked Exmples nd tsks for ech Rule of Indices Rule 1 Exmple 1.1 3 = x x 4 = x x x Let s multiply 3 nd 4 ( x x ) x ( x x x ) We now hve x x x x x x = 7 Exmple 1.2 2 = x 6 = x x x x x Let s multiply 2 nd 6 : ( x ) x ( x x x x x ) We now hve x x x x x x x = 8 Cn you spot ny pttern for multipliction? FÁS 2012 13 September 2012
Activity A2: Indices When we re working with lrger powers such s 27 x 19 it is no longer prcticl to list out ll the ' s nd count them. From Exmple 1.1 we know tht 3 x 4 = 7 So, if we dded the powers we would hve the correct nswer. From Exmple 1.2 we know tht 2 x 6 = 8 This shows tht if we dded the powers we would hve the correct nswer. Therefore for 27 x 19, insted of listing out ll the s we cn dd the powers. 27 x 19 = 27 +19 = 46 Rule 1: m x n = m+n Rule 1 pplies when you re multiplying numbers with the sme bse number. FÁS 2012 14 September 2012
Activity A2: Indices Tsk 1 16 Using the rules of indices, simplify the following: 4 13 4 FÁS 2012 15 September 2012
Activity A2: Indices FÁS 2012 16 September 2012 Rule 2 Exmple 2.1 13 = 4 = Let s divide 13 by 4 : ( ) ( ) = = which is equl to 9
Activity A2: Indices Exmple 2.2 6 = 2 = Let s divide 6 by 2 : ( ) ) ( = = which is equl to 4 Cn you spot ny pttern for division? FÁS 2012 17 September 2012
Activity A2: Indices Like multipliction when we re working with lrger powers it is not prcticl to list out ll the s nd count them. From Exmple 2.1 we know tht 13 4 = 9 It is cler from this exmple tht if we subtrct the powers we would hve the correct nswer. 13 4 = 13-4 = 9 From Exmple 2.2 we know tht 6 2 = 4 It is lso cler from this exmple tht if we subtrct the powers we would hve the correct nswer. Therefore for 27 x 19 insted of listing out ll the s we cn subtrct the powers. 27 x 19 = 27-19 = 8 Rule 2: m n = m-n Rule 2 pplies when you re dividing numbers with the sme bse number. FÁS 2012 18 September 2012
Activity A2: Indices Tsk 2 Using the rules of indices, simplify the following: 115 6 6 48 FÁS 2012 19 September 2012
Activity A2: Indices Rule 3 Exmple 3.1 4 = x x x Therefore ( 4 ) 2 = ( x x x ) ( x x x ) = x x x x x x x = 8 Exmple 3.2 5 = = x x x x ( 5 ) 4 = ( x x x x )( x x x x )( x x x x )( x x x x ) = x x x x x x x x x x x x x x x x x x x = 20 Cn you spot ny pttern? FÁS 2012 20 September 2012
Activity A2: Indices From Exmple 3.1 we know tht ( 4 ) 2 = 8 This exmple shows tht if we multiply the powers we get the correct nswer. ( 4 ) 2 = ( 4x2 ) = 8 From Exmple 3.2 we know tht ( 5 ) 4 = 20 This exmple shows tht if we multiply the powers we get the correct nswer. ( 5 ) 4 = ( 5x4 ) = 20 Therefore for ( 27 ) 13 powers. insted of listing out ll the s we cn multiply the ( 27 ) 13 = ( 27x13 ) = 351 Rule 3: = ( m ) n = ( mxn ) Rule 3 pplies when you hve two powers being pplied to the sme bse number. FÁS 2012 21 September 2012
Activity A2: Indices Tsk 3 Using the rules of indices simplify the following : (3 5 ) 18 FÁS 2012 22 September 2012
Activity A2: Indices Rule 4 Let b be nother unknown number or vrible. b = x b (b) 2 = ( x b) 2 = ( x b) ( x b) = ( x b x x b) (b) 3 = ( x b) 3 = ( x b) ( x b) ( x b) = ( x b x x b x x b) Exmple 4.1 (b) 2 = ( x b) ( x b) = ( x b x x b) = ( x x b x b) = ( 2 x b 2 ) = 2 x b 2 Exmple 4.2 (b) 7 = ( x b) ( x b) ( x b) ( x b) ( x b) ( x b) ( x b) = ( x b x x b x x b x x b x x b x x b) = ( x x x x x x x b x b x b x b x b x b x b) = ( 7 x b 7 ) = 7 x b 7 FÁS 2012 23 September 2012
Activity A2: Indices Cn you spot ny pttern from the exmples bove? Discuss this with your group. See if together you cn spot the pttern or rule. Then red the explntion below. From Exmple 4.1 we know tht (b) 2 = 2 x b 2. It is cler from this exmple tht if we pply power to product ech fctor cn be rised to the power. (b) 2 = 2 x b 2. From Exmple 4.2 we know tht (b) 7 = 7 x b 7 If we pply power to product ech fctor cn be rised to the power. ( Fctor mens the prt tht is being multiplied tht is, ech number) (b) 7 = 7 x b 7 Rule 4: (b) m = m x b m Rule 4 pplies when you hve one power being pplied to two or more numbers which re being multiplied. FÁS 2012 24 September 2012
Activity A2: Indices Tsk 4 Using the rules of indices simplify the following: (3 ) 18 FÁS 2012 25 September 2012
Activity A2: Indices Rule 5 Wht hppens when you hve 0? Let s explore this ide using Rule 2. Exmple 5.1 4 4 = We know tht ny number divided by itself is 1. So here we re left with () 4 () 4 = 1 From Rule 2 we lso know tht () 4 () 4 = () 4-4 = () 0 Therefore, () 4 () 4 = = () 4-4 = () 0 = 1. FÁS 2012 26 September 2012
Activity A2: Indices Exmple 5.2 7 7 = From lgebr we know tht ny number divided by itself is 1 so here we re = 1 left with 7 7 From rule 2 we lso know tht 7 7 = 7 7 0 Therefore, 7 7 = = 7 7 0 = 1 Cn you spot ny pttern? From Exmples 5.1 nd 5.2 we know tht 0 = 1. Therefore ny number to the power of 0 is equl to 1. Rule 5: () 0 = 1 FÁS 2012 27 September 2012
Activity A2: Indices Tsk 5 Evlute the following: ) 16 0 = b) 6392 0 = You cn use your clcultor to confirm this rule: (i) Press 16 to the power of button 0 = This will give you n nswer of 1 (i) Press 6392 to the power of button 0 = This will give you n nswer of 1 Prctise your skills Use Prctice Sheet A 2. Which is bigger: ) Three cubed or five squred? b) Fifteen squred or six cubed? FÁS 2012 28 September 2012
Activity A3: The Richter scle Activity The Richter scle A3 This ctivity links to unit lerning outcomes 2.1 nd 2.2. (Google imges) Introduction In the previous ctivity you lerned tht writing numbers using index (or power) nottion is the shorthnd for repeted multipliction. Here we will look t nother kind of shorthnd for repeted multipliction. Before clcultors nd computers, people used their knowledge of logrithms (logs) in order to do long, complicted multiplictions. John Npier invented logs in the 1600 s. Logs sved lot of time, prticulrly for scientists nd stronomers when working with very lrge numbers. It mde their work more productive. FÁS 2012 29 September 2012
Activity A3: The Richter scle Wht will you lern? Lerning Outcomes 1. Demonstrte n understnding of the lws of indices nd the rules of logrithms by using the lws nd rules to simplify expressions, solve equtions, nd trnspose formule. 2. Discuss the presence of vribles in rnge of rel life situtions. Key Lerning Points 1. Understnding the lws of indices nd the rules of logs 2. Recognising the links between both 3. Applying such knowledge to number of bsic exmples where you convert from log to index form nd vice vers 4. Using the knowledge of indices nd logs to simply expressions, solve equtions nd trnspose formul Mterils you will need for this ctivity Prctice Sheet A3 Solution Sheet A3 FÁS 2012 30 September 2012
Activity A3: The Richter scle Wht do you need to know before you strt? The Richter scle is used to mesure the intensity of n erthquke. In the mths world, the eqution used to mesure the intensity of n erthquke is log 10 I R R is the mgnitude of the erthquke on the Richter scle. I is the intensity of the erthquke mesured reltive to the smllest erthquke tht cn be mesured, tht is, 10 0 = 1. Every increse of 1 in the Richter scle mens the mgnitude of the erthquke is 10 times greter. FÁS 2012 31 September 2012
Activity A3: The Richter scle Getting Strted Logs Logs re nother method of writing indices. There is connection between logs nd indices. For exmple 5 to the power of 2 is 25 tht is, 5 2 =25 It follows tht the log of 25 to the bse 5 is 2. We cn write this s log525 = 2 In generl if b L =N then logbn=l For Exmple: 10 2 = 100 is equivlent to log10100 = 2 where 10 is the bse, 2 is the logrithm (tht is, the power) nd 100 is the number. We sy this log number like this: log 100 to the bse 10 equls 2. FÁS 2012 32 September 2012
Activity A3: The Richter scle Uses of Logs The Richter scle The Richter scle mesures the severity of erthqukes. It ws developed by Chrles Richter in 1935. The Richter scle is bsed on subtrcting two logs from ech other. The ph Scle The ph scle mesures how cidic or bsic substnce is. It is lso log scle. The ph scle is determined s the negtive log of hydrogen ion concentrtion. The Decibel Scle The decibel scle mesures sound levels. It is widely used in electronics, signls nd communiction. Chnges in sound pressure ffect volume. This is mesured by log rtio. FÁS 2012 33 September 2012
Activity A3: The Richter scle Worked Exmple - Rel life sitution in which Logs re used In Jnury 2012, Donegl ws hit by mini erthquke which mesured 2.2 on the Richter scle. How mny times more intense ws this erthquke when it is compred with the smllest erthquke tht cn be mesured? Remember: The eqution used to mesure the intensity of n erthquke is log 10 I R, where R is the mgnitude of the erthquke on the Richter scle nd I is the intensity of the erthquke mesured reltive to the smllest erthquke tht cn be mesured, tht is, 10 0 = 1. (Hint: Clculte log 10 I 2. 2 ). FÁS 2012 34 September 2012
Activity A3: The Richter scle Solution R = 2.2 log 10 I 2.2 2.2 10 I (Use your clcultor to clculte this). 10 2. 2 158.4893192 Intensity = 158.489 times more intense thn the smllest erthquke tht cn be mesured. FÁS 2012 35 September 2012
Activity A3: The Richter scle Tsk 1 Convert the following from index to log form: ) 5 3 = 125 b) 8 2 = 64 c) 2 5 = 32 FÁS 2012 36 September 2012
Activity A3: The Richter scle Tsk 2 Convert the following from log to index form: ) log5625 = 4 b) log2128 = 7 c) log3243 = 5 FÁS 2012 37 September 2012
Activity A3: The Richter scle Tsk 3 The strongest erthquke ever recorded in the UK took plce in Colchester on April 22 nd 1884. It mesured 9.5 on the Richter scle. How mny times more intense ws this erthquke when it is compred with the smllest erthquke tht cn be mesured? FÁS 2012 38 September 2012
Activity A3: The Richter scle Tsk 4 How much more intense is n erthquke tht mesures 6 on the Richter scle when you compre it with n erthquke tht mesures 3 on the Richter scle? Prctise your skills Use Prctice Sheet A3. FÁS 2012 39 September 2012
Activity A4: Cr rentl Activity Cr rentl A4 This ctivity links to unit lerning outcomes 2.4 nd 2.5. (Google imges) Introduction An eqution is mthemticl sentence with numbers, letters nd n equls sign (=). In this ctivity, you will be ble construct nd solve liner equtions. Wht will you lern? Lerning Outcomes 1. Construct lgebric expression nd formule for rel life situtions using the correct terminology nd including rerrngement of formule. 2. Solve liner equtions. 3. Discuss the presence of vribles in rel life situtions. FÁS 2012 40 September 2012
Activity A4: Cr rentl Key Lerning Points 1. Understnding the concept of n eqution 2. Constructing lgebric expressions nd equtions for rel life situtions 3. Replcing vribles with vlues through substitution 4. Solving liner equtions Mterils you will need for this ctivity Prctice Sheet A4 Solution Sheet A4 Wht do you need to know before you strt? An eqution is like blnce scle. Everything must be equl on both sides. FÁS 2012 41 September 2012
Activity A4: Cr rentl Getting Strted In order to construct or build n lgebric eqution for rel life sitution this is wht we do: Allow vrible to stnd for our unknown vlue. Construct the eqution round this vrible. We then solve the eqution by isolting the vrible on one side of the equls sign. We solve equtions by blncing. The Golden Rule: Whtever we do to one side of n eqution, we must do the sme to the other side See Activity A3 in the Level 3 Lerner Pck if you would like to revise this before moving on. FÁS 2012 42 September 2012
Activity A4: Cr rentl Worked Exmple Solving liner equtions You cn rent cr for 25 per dy plus 0.20 per kilometre. How mny kilometres were driven if the totl rentl fee for dy ws 65? Solution Step 1: Let vrible, x, represent the number of kilometres. Step 2: Build the eqution: 0.20x + 25 = 65 Step 3: Solve. In order to solve this eqution nd find out the number of kilometres we need to isolte the vrible x on the left hnd side of the eqution by itself. The opposite of dding 25 is subtrcting 25. If we subtrct 25 from both sides, we will remove the + 25 on the left. 0.20x + 25 = 65 0.20x + 25-25 = 65 25 0.20x = 40 0.20x mens 0.20 multiplied by x. The opposite of multiplying by 0.20 is dividing by 0.20. If we divide by 0.20 on both sides, we will remove the 0.20 on the left. x = 200 200 kilometres were driven. FÁS 2012 43 September 2012
Activity A4: Cr rentl Tsk 1 You cn rent cr for n initil chrge of 30, plus 15 per dy plus 0.15 per kilometre. ) If the rentl fee for the cr over period of time ws 84, construct n eqution representing the bove informtion. b) How mny kilometres were driven if the cr ws rented for 2 dys? FÁS 2012 44 September 2012
Activity A4: Cr rentl Tsk 2 Eoin is considering his cr rentl options. From Rentlcrs.com (Option A) he cn rent n Opel Insigni for 90 per dy. A rivl compny, Wheels4U (Option B) offers the sme cr for 30 per dy plus 0.30 per kilometre. (Google imges) ) Construct two expressions representing the bove informtion. b) Eoin wnts to rent the cr for three dys. During this time he intends to drive round trip from Belfst Airport to Wterford city (680 kilometres). Which cr compny should Eoin rent the cr from? Prctise your skills Use Prctice Sheet A4. FÁS 2012 45 September 2012
Activity A5: Using formule Activity Using formule A5 This ctivity links to unit lerning outcomes 2.1 nd 2.5. (Google imges) Introduction A formul is n eqution tht indictes how number of vribles re relted to one nother. Wht will you lern? Lerning Outcomes 1. Construct lgebric expressions nd formule for rel life situtions using the correct terminology nd including rerrngement of formule. 2. Discuss the presence of vribles in rnge of rel life situtions. FÁS 2012 46 September 2012
Activity A5: Using formule Key Lerning Points 1. Constructing lgebric expressions nd equtions for rel life situtions 2. Using correct terminology 3. Solving word problems using formuls through rerrngement 4. Replcing vribles with vlues through substitution Mterils you will need for this ctivity Prctice Sheet A5 Solution Sheet A Wht do you need to know before you strt? You should be fmilir with constructing expressions nd equtions. You prctised this in the previous ctivities. Getting Strted In formul, we use vribles nd mthemticl symbols (such s +, -, =) to represent words. We often use formul s kind of shorthnd for expressing stted rule or reltionship. Mths words: The plurl of formul is formule or formuls. In lgebr, we use formule. FÁS 2012 47 September 2012
Activity A5: Using formule Worked Exmple Using formule To convert temperture expressed in Celsius degrees, C, to Fhrenheit degrees, F, we multiply the Celsius temperture by nd then dd 32. ) Construct formul for this reltionship. b) When the temperture is 30 degrees Celsius wht is the equivlent temperture in degrees Fhrenheit? (Google imges) FÁS 2012 48 September 2012
Activity A5: Using formule Solution ) Stte the rule simply in words: Fhrenheit equls nine fifths times Celsius plus 32. Now write tht in mthemticl symbols: b) FÁS 2012 49 September 2012
Activity A5: Using formule Tsk 1 A formul for the norml brking distnce, d, in metres of cr is ) 2, where s is the speed in kilometres (km) per hour. (Google imges) How much greter is the norml breking distnce for cr trvelling t speed of 100 km per hour thn cr trvelling t speed of 60km per hour? FÁS 2012 50 September 2012
Activity A5: Using formule Tsk 2 In bsebll, btting verge,, is clculted by dividing the number of hits mde, h, by the number of times up t the bt, n. (Google imges) ) Trnslte this reltionship to formul. b) In modern times, seson btting verge higher thn.300 is considered to be excellent, nd n verge higher thn.400 is nerly n unchievble gol. Ty Cobb holds the record for the highest creer btting verge. In his creer he ws up t the bt 11,429 nd mde 4189 hits. Clculte his btting verge. Prctise your skills Use Prctice Sheet A5. FÁS 2012 51 September 2012
Activity A6: Txi fre Activity Txi fre A6 This ctivity links to unit lerning outcomes 2.4 nd 2.6. (Google Imges) Introduction You hve lredy lerned tht in n eqution everything must be equl on both sides. However, it is different for inequlities. In inequlities, one side is not equl to the other side. Hence we get set of solutions for n inequlity, s opposed to just one solution. Wht will you lern? Lerning Outcomes 1. Solve liner inequlities of one vrible. 2. Solve problems for rel life situtions by mthemtising the sitution nd mking n initil model, pply mthemticl techniques nd discussing nd mking conclusions. FÁS 2012 52 September 2012
Activity A6: Txi fre Key Lerning Points 1. Understnding the concept of n inequlity 2. Solving liner inequlities of one vrible 3. Formulte rel life situtions into mthemticl models mking ssumptions if necessry 4. Solving using pproprite mthemticl techniques 5. Discussing nd mking conclusions if necessry Mterils you will need for this ctivity Prctice Sheet A6 Solution Sheet A6 FÁS 2012 53 September 2012
Activity A6: Txi fre Wht do you need to know before you strt? These re the four different inequlity signs: o x > 2 mens x is greter thn 2 o x < 2 mens x is less thn 2 o x 2 mens x is greter thn or equl to 2 o x 2 mens x is less thn or equl to 2 All txis re leglly obliged to hve tximeter correctly instlled nd clibrted in ccordnce with the Ntionl Mximum Txi Fre. The driver must turn on the tximeter when the txi is hired nd turn it off t the end of the journey for which it is hired. The tble below shows the Ntionl Mximum Txi Fre rte. FÁS 2012 54 September 2012
Activity A6: Txi fre Getting Strted In order to solve liner inequlity we use similr methods s solving equtions, except tht there re extr rules when using multipliction nd division. You my multiply or divide both sides by positive number. However, when you multiply or divide both sides by negtive number, you must turn the inequlity sign round. Worked Exmple Inequlities Look t the Txi Fre tble. You will see there tht the stndrd rte of the ntionl mximum txi fre includes n initil chrge 4.10 nd 1.03 per kilometre. Srh hs no more thn 12 to spend on fre. She will be trvelling by txi sometime between 8.00h nd 20.00h so she will be chrged the stndrd rte. ) Construct n inequlity tht represents Srh s sitution. b) How mny kilometres cn Srh trvel without exceeding her budget? FÁS 2012 55 September 2012
Activity A6: Txi fre Solution ) Let k = number of kilometres We ssume tht Srh is trvelling between 8.00 nd 20.00h nd will chrge the stndrd rte. 1.03k + 4.10 12 b) 1.03k + 4.10 12 1.03k + 4.10 4.10 12 4.10 1.03k 12 4.10 1.03k 7.9 k 7.669 Srh cn trvel 7 kilometres or less before exceeding her budget. Tsk 1 John finishes working in nightclub t 1.30m nd decides to get txi home. He only hs 16 with him to spend on fre. ) Use the informtion in the Txi Fre tble to determine the mximum initil chrge nd mount per kilometre tht txi cn chrge t this time. FÁS 2012 56 September 2012
Activity A6: Txi fre b) Construct n inequlity tht represents John s sitution. c) How mny kilometres cn John trvel without exceeding his budget? Tsk 2 Mrtin nd Nicky decide to get txi home on Bnk Holidy Mondy fternoon. Between them they only hve 21 to spend on fre. ) Use the informtion from the tble to determine the mximum initil chrge nd mount per kilometre nd ny extr tht txi cn chrge. FÁS 2012 57 September 2012
Activity A6: Txi fre b) Construct n inequlity tht represents Mrtin nd Nicky s sitution. c) How mny kilometres cn they trvel without exceeding their budget? d) They live 14 kilometres wy. If it wsn t Bnk Holidy, would they hve hd enough money for the txi fre to get ll the wy home? Discuss. Prctise your skills Use Prctice Sheet A6 to pply wht you hve lerned. FÁS 2012 58 September 2012
Activity A7: Exm time Activity Exm time A7 This ctivity links to wrd lerning outcomes 2.4 nd 2.5. (Google Imges) Introduction Different prices re often chrged for different tickets t events. We cn use simultneous equtions to clculte the price of tickets if we know how mny different type tickets re sold nd the tkings for ech. Wht will you lern? Lerning Outcomes 1. Solve lgebric equtions including simultneous liner equtions of two unknowns. 2. Construct lgebric expressions nd formul for rel life situtions using the correct terminology. FÁS 2012 59 September 2012
Activity A7: Exm time Key Lerning Points 1. Solving simultneous liner equtions of two unknowns 2. Constructing lgebric equtions for rel life situtions Mterils you will need for this ctivity Prctice Sheet A7 Solution Sheet A7 Wht do you need to know before you strt? Simultneous equtions re two equtions with t lest two unknowns, for exmple, x nd y. In order to solve such equtions they must be simultneously stisfied by prticulr vlues of x nd y. Simultneously mens t the sme time. FÁS 2012 60 September 2012
Getting Strted Activity A7: Exm time Worked Exmple - Solving simultneous equtions An exm pper hd two sections, Section A nd Section B. Questions from ech section were worth different mrks. Brry nswered 6 questions from section A correctly nd 3 questions form section B correctly, nd he got totl of 82.5 mrks. Mebh nswered 4 questions from ech section correctly resulting in 90 mrks. Write two equtions to represent the bove informtion. How much were the questions from ech section worth? FÁS 2012 61 September 2012
Activity A7: Exm time Solution Step 1: Construct the equtions Allow x to represent the worth of questions from Section A. Allow y to represent the worth of questions from Section B. Brry: 6x + 3y = 82.5 Mebh: 4x + 4y = 90 Step 2: Lbel the equtions A nd B. 6x + 3y = 82.5 4x + 4y = 90 [A] [B] Step 3: Get the sme coefficients for either x or y. If we multiply eqution [A] by 4 nd [B] by 3 then we will hve the sme y coefficients. 24x + 12y = 330 12x + 12y = 270 [A] [B] Step 4: Mke sure the chosen coefficients hve opposite signs (i.e. + nd -). If we multiply eqution [B] by -1 then we will hve opposite signs. 24x + 12y = 330 [A] -12x - 12y = -270 [B] FÁS 2012 62 September 2012
Activity A7: Exm time Step 5: Add the two equtions together. 24x + 12y = 330 [A] -12x - 12y = -270 [B] 12x = 60 Step 6: Solve for y. 12x = 60 Divide both sides by 12 x = 5 Step 7: Replce x in either eqution to solve for y. 12(5) + 12y = 270 60 + 12y = 270 60-60 + 12y = 270-60 12y = 210 Divide both sides by 12 y = 17.5 Therefore: o Questions from Section A were worth 5 mrks. o Questions from Section B were worth 17.5 mrks. FÁS 2012 63 September 2012
Activity A7: Exm time Tsk 1 An exm worth 145 points hs 50 questions. Some of the questions re worth two points nd some re worth five points. ) Write two equtions to represent the bove informtion. b) How mny two point questions re on the test? c) How mny five point questions re on the test? FÁS 2012 64 September 2012
Activity A7: Exm time Tsk 2 An exmintion pper consists of 40 questions. There re 5 mrks for ech correct nswer. 3 mrks re deducted for ech incorrect nswer. Kenny nswered ll 40 questions, getting x correct nd getting y incorrect. His totl score for the exmintion ws 56 mrks. ) Write two equtions to represent the bove informtion. b) Solve these equtions to find how mny questions Kenny nswered correctly. FÁS 2012 65 September 2012
Prctise your skills Activity A7: Exm time Use Prctice Sheet A7. Robbie Kene entered penlty competition during the Irish soccer tem s trining. He hd to tke 10 penlties on Shy Given. Robbie erned 3 points for every penlty he scored nd he lost 2 points for every penlty he missed. When he ws finished, he hd scored x mount of his penlties but hd missed y mount of his penlties. His finl score ws 15 points. ) Write two equtions to represent the bove informtion. b) Solve these equtions to find how mny penlties Robbie scored nd how mny he did not score. FÁS 2012 66 September 2012
Activity A8: Qudrtic equtions Activity Qudrtic equtions A8 This ctivity links to unit lerning outcome 2.3. (Google imges) Introduction An eqution of the form x 2 + bx + c = 0 is clled qudrtic eqution. Remember is the x 2 coefficient, b is the x coefficient nd c is constnt. In this section, you will lern how to fctorise qudrtic equtions using the guide number method. FÁS 2012 67 September 2012
Activity A8: Qudrtic equtions Wht will you lern? Lerning Outcomes 1. Solve qudrtic equtions using fctors. Key Lerning Points 1. Understnding the concept of qudrtic eqution 2. Solving such equtions using fctors Mterils you will need for this ctivity Prctice Sheet A8 Solution Sheet A8 FÁS 2012 68 September 2012
Activity A8: Qudrtic equtions Wht do you need to know before you strt? If we multiply 7 nd 8 together we get 56. We cn sy tht 7 nd 8 re fctors of 56. They divide exctly into the number with no reminder. Other fctors of 56 would include: 1 nd 56 2 nd 28 4 nd 14 x + 4 nd x + 5 re two lgebric expressions. If we wnt to multiply these expressions together, tht is, (x + 4) (x + 5), we sy tht the x nd 4 in the first set of brckets must be multiplied by the x nd the 5 in the second set of brckets. We cn only multiply two numbers together t time so we use the following method: x(x + 5) + 4(x + 5) Continue s follows: X(x) + x (5) + 4(x) + 4(5) x 2 + 5x + 4x + 20 Add similr or like terms: x 2 + 9x + 20 The result x 2 + 9x + 20 is lso clled n lgebric expression. We cn sy tht (x + 4) nd (x + 5) re fctors of x 2 + 9x + 20. FÁS 2012 69 September 2012
Getting Strted Activity A8: Qudrtic equtions Tking out Common Fctor Consider the lgebric expression 3x + 3y. In order to fctorise this expression, we need to tke wht is common. We cn see tht 3 divides into both terms. Therefore we cn tke 3 outside brcket nd divide every term by 3. Therefore we cn sy tht both 3 nd (x +y) re fctors of 3x + 3y Exmples: Fctorise ech of the following expressions ) 5t + 10s b) 15q 30p c) 2x 2 + 6x FÁS 2012 70 September 2012
Activity A8: Qudrtic equtions Fctorising by Grouping Consider the lgebric expression 4x + 4y + x + y. There is no fctor common to ll four terms. However, we cn see tht 4 is common to the first two terms nd is common to the second two terms. Here, we use fctorising by grouping. We group terms together where there is common fctor nd then fctorise like before. (4x + 4y) + (x + y) 4(x + y) + (x + y) The terms inside both brckets re the sme. Therefore my fctors re (4 + ) nd (x + y). Exmples: Fctorise ech of the following expressions ) c + bc + 3 + 3b b) 2b + 2bc + 3d + 3cd c) 4x 4y + x - y FÁS 2012 71 September 2012
Activity A8: Qudrtic equtions Fctorising qudrtic equtions We hve lredy looked t the qudrtic expression x 2 + 9x + 20 with fctors (x+4) nd (x+5). We will look t how to get such fctors for ny qudrtic expression in detil in the Worked Exmples. Firstly, consider x 2 + 9x + 20 = 0 This expression hs now become n eqution. However, its fctors remin the sme. x 2 + 9x + 20 = 0 (x+4)(x+5) = 0 In order to solve this eqution for x we must relise tht the product of the two fctors is zero. Therefore, t lest one of these fctors is zero. So we hve x + 4 = 0 or x + 5 = 0 x = - 4 or x = - 5 We hve now solved the qudrtic eqution nd got two vlues for x: x = - 4 or x = - 5. FÁS 2012 72 September 2012
Activity A8: Qudrtic equtions Worked Exmple 1 Solve the following eqution: x 2 + 6x + 8 = 0 Solution Step 1: Find the Guide Number (GN). We find the guide number by multiplying the x 2 coefficient (in this cse 1) by the constnt (in this cse 8). GN = 1 x 8 = 8 Step 2: Write out ll the fctors of the Guide Number. Fctors of 8 re: 1 x 8-1 x - 8 2 x 4-2 x - 4 Step 3: Pick the fctors of 8 which dd to give you the x coefficient in the originl qudrtic (in this cse 6). 2 x 4 FÁS 2012 73 September 2012
Activity A8: Qudrtic equtions Step 4: Split the x term in the originl qudrtic into two terms using the vlues found in Step 3. x 2 + 6x + 8 = 0 = x 2 + 2x + 4x + 8 = 0 Step 5: Proceed by fctorising by grouping. x 2 + 2x + 4x + 8 = 0 (x 2 + 2x) + (4x + 8) = 0 X(x + 2) + 4(x + 2) = 0 The terms inside both brckets re the sme. Therefore my fctors re (x + 4) nd (x + 2). (x + 4) (x + 2) = 0 Step 6: Solve the eqution for x by llowing both fctors equl to 0. x + 4 = 0 or x + 2 = 0 x = - 4 or x = - 2 FÁS 2012 74 September 2012
Activity A8: Qudrtic equtions Tsk 1 Solve the following equtions: ) x 2 + 3x + 2 = 0 b) x 2 + 7x + 10= 0 c) x 2 + 5x + 6 = 0 FÁS 2012 75 September 2012
Activity A8: Qudrtic equtions Worked Exmple 2 Solve the following eqution: x 2-8x + 12 = 0 Solution Step 1: Find the Guide Number (GN) We find the guide number by multiplying the x 2 coefficient (in this cse 1) by the constnt (in this cse 12). GN = 1 x 12 = 18 Step 2: Write out ll the fctors of the Guide Number Fctors of 12 re: 1 x 12-1 x - 12 2 x 6-2 x - 6 3 x 4-3 x - 4 Step 3: Pick the fctors of 12 which dd to give you the x coefficient in the originl qudrtic (in this cse -8). - 2 x 6 FÁS 2012 76 September 2012
Activity A8: Qudrtic equtions Step 4: Split the x term in the originl qudrtic into two terms using the vlues found in Step 3. x 2-8x + 12 = 0 = x 2-2x - 6x + 12 = 0 Step 5: Proceed by fctorising by grouping x 2-2x - 6x + 12 = 0 (x 2-2x) + (- 6x + 12) = 0 x(x - 2) - 6(x - 2) = 0 We took out 6 from the second brcket so the terms in both brckets would be the sme. Therefore my fctors re (x - 6) nd (x - 2). (x - 6) (x - 2) = 0 Step 6: Solve the eqution for x by llowing both fctors equl to 0. x - 6 = 0 or x - 2 = 0 x = 6 or x = 2 FÁS 2012 77 September 2012
Activity A8: Qudrtic equtions Tsk 2 Solve the following equtions; ) x 2-5x + 4 = 0 b) x 2-10x + 25= 0 c) x 2-5x + 6 = 0 FÁS 2012 78 September 2012
Activity A8: Qudrtic equtions Worked Exmple 3 Solve the following eqution: 3x 2-13x - 10 = 0 Solution Step 1: Find the Guide Number (GN) We find the guide number by multiplying the x 2 coefficient (in this cse 3) by the constnt (in this cse - 10). GN = 3 x - 10 = - 30 Step 2: Write out ll the fctors of the Guide Number Fctors of - 30 re: 1 x - 30-1 x 30 2 x - 15-2 x 15 3 x - 10-3 x 10 5 x - 6-5 x 6 Step 3: Pick the fctors of - 30 which dd to give you the x coefficient in the originl qudrtic (in this cse -13). 2 x 15 FÁS 2012 79 September 2012
Activity A8: Qudrtic equtions Step 4: Split the x term in the originl qudrtic into two terms using the vlues found in Step 3. 3x 2-13x - 10 = 0 = 3x 2-15x + 2x - 10 = 0 Step 5: Proceed by fctorising by grouping 3x 2-15x + 2x - 10 = 0 (3x 2-15x) + (2x - 10) = 0 3x(x - 5) + 2(x - 5) = 0 The terms inside both brckets re the sme. Therefore my fctors re (3x + 2) nd (x - 5). (3x + 2) (x - 5) = 0 Step 6: Solve the eqution for x by llowing both fctors equl to 0. 3x + 2 = 0 or x - 5 = 0 3x = - 2 x = - or x = 5 FÁS 2012 80 September 2012
Activity A8: Qudrtic equtions Tsk 3 Solve the following equtions: ) x 2-4x - 5 = 0 b) 6x 2 - x - 2= 0 c) 9x 2-9x - 28= 0 FÁS 2012 81 September 2012
Activity A8: Qudrtic equtions Worked Exmple 4 Solve the following eqution: 4x 2-9 = 0 Solution We cn rewrite this qudrtic eqution s 4x 2 + 0x - 9 = 0 Step 1: Find the Guide Number (GN) We find the guide number by multiplying the x 2 coefficient (in this cse 4) by the constnt (in this cse - 9). GN = 4 x - 9 = - 36 Step 2: Write out ll the fctors of the Guide Number Fctors of - 30 re: 1 x - 36-1 x 36 2 x - 18-2 x 18 3 x - 12-3 x 12 4 x 9-4 x 9 6 x 6-6 x 6 Step 3: Pick the fctors of - 36 which dd to give you the x coefficient in the originl qudrtic (in this cse 0). - 6 x 6 FÁS 2012 82 September 2012
Activity A8: Qudrtic equtions Step 4: Split the x term in the originl qudrtic into two terms using the vlues found in Step 3. 4x 2 + 0x - 9 = 0 = 4x 2-6x + 6x - 9 = 0 Step 5: Proceed by fctorising by grouping 4x 2-6x + 6x - 9 = 0 (4x 2-6x) + (6x - 9) = 0 2x (2x - 3) + 3(2x - 3) = 0 The terms inside both brckets re the sme. Therefore my fctors re (2x + 3) nd (2x - 3). (2x + 3) (2x - 3) = 0 Step 6: Solve the eqution for x by llowing both fctors equl to 0. 2x + 3 = 0 or 2x - 3 = 0 2x = - 3 2x = 3 x = - or x = FÁS 2012 83 September 2012
Activity A8: Qudrtic equtions Tsk 4 Solve the following equtions: ) x 2-4 = 0 b) 4x 2-16= 0 c) 9x 2 25 = 0 Prctise your skills Use Prctice Sheet A8. FÁS 2012 84 September 2012
Activity A9: Grden dimensions Activity Grden dimensions A9 This ctivity links to unit lerning outcome 2.3. (Google imges) Introduction In the previous ctivity you lerned how to fctorise qudrtic equtions using the guide number method. Sometimes, if the fctors re not whole numbers we re unble to use this method. In such instnces we use the qudrtic formul to fctorise. Wht will you lern? Lerning Outcomes 1. Solve qudrtic equtions using the qudrtic formul. FÁS 2012 85 September 2012
Activity A9: Grden dimensions Key Lerning Points 1. Solving equtions using qudrtic formul method 2. Recognising when to use this method Mterils you will need for this ctivity Prctice Sheet A9 Solution Sheet A9 Wht do you need to know before you strt? In the previous ctivity you lerned tht n eqution of the form x 2 + bx + c = 0 is clled qudrtic eqution. is the x 2 coefficient, b is the x coefficient nd c is constnt. Such n eqution cn be solved using the formul method where FÁS 2012 86 September 2012
Activity A9: Grden dimensions Getting Strted Exmple Solve the following eqution: x 2-9x + 12 = 0 Solution using Guide Number Method Step 1: Find the Guide Number (GN) GN = 1 x 12 = 12 Step 2: Write out ll the fctors of the Guide Number Fctors of 12 re: 1 x 12-1 x - 12 2 x 6-2 x - 6 3 x 4-3 x 4 Step 3: Pick the fctors of 12 which dd to give you the x coefficient in the originl qudrtic (in this cse -9). As you cn see, there re no fctors of 12 which dd up to - 9 so we cnnot fctorise the eqution using the guide number method. When this hppens we hve to use the qudrtic formul. FÁS 2012 87 September 2012
Activity A9: Grden dimensions Solution using the Formul Method Step 1: Lbel the vlues, b nd c = 1 (x 2 coefficient) b = - 9 (x coefficient) c = 12 (constnt) Step 2: Substitute the vlues for, b nd c into the formul to solve for x or x = 7.37 or x = 1.63 FÁS 2012 88 September 2012
Activity A9: Grden dimensions Worked Exmple - Rel life sitution where formul is used A rectngulr shped grden is 2 times s long s its width. If its re is 264 m 2, find the length nd width of the lwn. Solution If we llow width = x Then length = 2x 2x x Are of rectngle = length x width Are of lwn = 264 = 2x(x) = 2x 2 2x 2 = 264 2x 2 264 = 0 2x 2 + 0x 264 = 0 FÁS 2012 89 September 2012
Activity A9: Grden dimensions If we try to solve this using the Guide Number method, my Guide Number (GN) will be 528. No fctors of - 528 will dd to give me the x coefficient of 0. Therefore, I hve to solve this eqution using the formul method. Step 1: Lbel the vlues, b nd c = 2 (x 2 coefficient) b = 0 (x coefficient) c = - 264 (constnt) Step 2: Substitute the vlues for, b nd c into the formul to solve for x or x = 11.49 or x = -11.49 Width cnnot be negtive, therefore width = 11.49m If width = 11.49, then length = 2(11.49) = 22.98m FÁS 2012 90 September 2012
Tsk 1 Activity A9: Grden dimensions The length of rectngulr shped grden is 4 times s long s its width. If its re is 323 m 2, find the length nd width of the lwn. FÁS 2012 91 September 2012
Activity A9: Grden dimensions Tsk 2 The length of rectngulr shped grden is 3 times s long s its width. If its re is 282 m 2, find the length nd width of the lwn. Prctise your skills Use Prctice Sheet A9. FÁS 2012 92 September 2012
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