Form4 AlgebraPart2PaperB
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Algebra Part 2
Section1AlgebraicProducts
Expandingbracketsmeanstoremovethebrackets.
Howwouldweexpandthefollowing?
5(x+2)
ThetermwhichisoutsidethebracketsmustbemultipliedwiththeWHOLEbracket.
(Multiplytermbyterm)
5(x+2)=(5× x)+(5×2)
=5x+10
Consolidation
1. 2(x+1) __________________
2. 4(2+3b) __________________
3. 3(b+2f) __________________
Expandingwithlettersoutsidethebracket
Expand:p(p+4)
p(p+4)=(pxp)+(px4)
=p2+4p
Remember:
Positive(+)xNegative(-)=Negative(-)
Negative(-)xNegative(-)=Positive(+)
Positive(+)xPositive(+)=Positive(+)
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Consolidation
1. r(2r–6) ___________________
2. –s(s+5) ___________________
3. 2y(y2+3y+6) ___________________
Expandingtwobracketsatago
Expand:6q–2(r–2q)=6q-(2xr)–(2x(-2q))
=6q–2r+4q(rememberNegativexNegative=Positive)
=10q–2r(Collectliketerms)
Consolidation
1. 3(y+2z)+5(2y+3z)
2. 6n(n–3)–5(n+2)
3. 3q–2(q–6)
4. t(t–6)–3(t–3)
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Anexpressionsuchas(3y+2)(4y–5)canbeexpandedtogiveaquadraticexpression.
Multiplyingoutsuchpairsofbracketsisusuallycalledquadraticexpansion.
Theruleforexpandingexpressionssuchas(t+5)(3t–4)issimilartothatforexpandingsinglebrackets:multiplyeverythinginonesetofbracketsbyeverythingintheothersetofbrackets.
Example1
Intheexpansionmethod,splitthetermsinthefirstsetofbrackets,makeeachofthemmultiplybothtermsinthesecondsetofbrackets,thensimplifytheoutcome.
Expand(x+3)(x+4)
(x+3)(x+4) = x(x+4)+3(x+4)
= x2+4x+3x+4
= x2+7x+12
Example2
Expand(t+5)(t–2)
(t+5)(t–2) = t(t–2)+5(t–2)
= t2–2t+5t–10
= t2+3t–10
Example3
Expand(k–3)(k–2)
(k–3)(k–2) = k(k–2)–3(k–2)
= k2–2k–3k+6
= k2–5k+6
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Example4
Expand(4x–1)(3x–5)
(4x–1)(3x–5)= 4x(3x–5)–1(3x–5)
= 12x2–20x–3x+5
= 12x2–23x+5
Example5
Expand(3x–2)2
(3x–2)2 = (3x–2)(3x–2)
= 3x(3x–2)–2(3x–2)
= 9x2–6x–6x+4
= 9x2–12x+4
Consolidation
Expandthefollowing:
1. (w+3)(w–1)
2. (m+5)(m+1)
3. (a–1)(a–3)
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4. (x+3)(x–3)
5. (4r–3)(2r–1)
6. (1–3p)(3+2p)
7. (t–5)2
8. (x+6)2–36
SupportExercisePg107Exercise8A
Pg110Exercise8CNos1–4
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Section2Factorisation
Factorisationistheoppositeofexpansion.Itputsanexpressionbackintothebracketsitmayhavecomefrom.
Infactorizationyouhavetolookforthecommonfactorsineverytermoftheexpression.
Example1
6t+9m=3(2t+3m)
3isafactorif6and9
Example2
6my+4py=2y(3m+2p)
2andyareinbothterms.
Example3
5k2–25k=5k(k–5)
Example4
10a2b–15ab2=5ab(2a–3b)
Consolidation
1. 4t2–3t _______________________________
2. 3m2–3mp _______________________________
3. 6ab+9bc+3bd _______________________________
4. 6mt2–3mt+9m2t _______________________________
5. 8ab2+2ab–4a2b _______________________________
SupportExercisePg108Ex8BNos1,2
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Section3SolvingLinearEquations
Someequationscanbesolvedmentally.Tosolvemorecomplicatedequationsthebalancemethodis
used.
Tokeepthebalance,whateveryoudoontheleft-handsideyoumustalsodototherighthandsideof
theequation.
Itiseasiertoremember:
CHANGEside…. à CHANGEsign
Example1
4x+3=31
4x=31–3
4x=28
x=28÷4
x=7
Example2
5(a+3)=18
5a+15=18
5a=18–15
a=3/5
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Example3
5(3y+2)=13y+4
15y+10=13y+4
15y–13y=4–10
2y=-6
y=-6÷2
y=-3
Consolidation
1. 3x+2=14
2. 4(2x–4)=8
3. 3x+8=2x–4
4. 2(3x+4)=4(2x–3)
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5. 5(2x+7)=-15
SupportExercisePg144Ex10B1–40
Section4SettingUpEquations
Equationsareusedtorepresentsituations,sothatyoucansolvereal-lifeproblems.Manyreal-lifeproblemscanbesolvedbysettingthemupaslinearequationsandthensolvingtheequation.
Example1
AmanbuysadailynewspaperfromMondaytoSaturdayfordcents.HebuysaSundaypaperfor1.80dollars.Hisweeklypaperbillis7.20dollars.
Whatisthepriceofhisdailypaper?
6d+180=720
6d=720–180
6d=540
d=540÷6
d=90
Thereforethedailypapercosts90cents.
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Consolidation
1. Thediagramshowsarectangle.
a) Whatisthevalueofx?
b) Whatisthevalueofy?
2. Marisahastwobags,eachofwhichcontainsthesamenumberofsweets.Sheeats4sweets.Shethenfindsthatshehas30sweetsleft.Howmanysweetswerethereineachbagtostartwith?
3. Flooringcosts$12.75persquaremeter.Theshopcharges$35forfitting.Thefinalbillwas$137.Howmanysquaremetersofflooringwerefitted?
10x–1
6 4y–2
14
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4. Mariobought8gardenchairs.Whenhegottothetillheuseda$10voucheraspartpayment.Hisfinalbillwas$56.
a. Setthisproblemupasanequation,usingcasthecostofonechair.
b. Solvetheequationtofindthecostofonechair.
SupportExercisePg146Ex10BNos1–15
Section5SolvingFractionalTerms
Inalgebraexpressionssuchas(y+5)÷4areusuallywrittenas54y +
Example1:
Solvetheequation4 8q=
Step1:Removethedenominatorfromtheequation
Multiplybothsidesbyq:4 8q qq× = ×
4=8q
Step2:Equatetheunknown
Dividebothsidesby8:4 88 8
q=
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Simplify:q=12
Wecanreducethestepsandconductsomeofthemmentally.Havealookatthenextexample:
Example2:
Solvetheequation5 25y=
Example3:
Solvetheequation:3( 5) 62q +
=
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Example4:
Solvetheequation16 14x x−= −
Example5:
Solvetheequation2 1 5 52 3 4x x− −
− =
SupportExercisePg150Ex10CNos1–5
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Section6SolvingSimultaneousEquations
Pairofsimultaneousequationsaretwolinearequationsforwhichyouhavetwounknownsandasolutionforeachisrequired.
Elimination Method
Step1
Getthecoefficientsofoneoftheunknownsthesame.
Step2
Eliminatethisunknownbyaddingorsubtractingthetwoequations.(Whenthesignsarethesameyousubtract;whenthesignsaredifferentadduptheequations)
Step3
Solvetheresultingequationwithoneunknown.
Step4
Substitutethevaluefoundbackintoanyoneoftheoriginalequations.
Step5
Solvetheresultingequation.
Step6
Checkthatthetwovaluesfoundsatisfytheoriginalequations.
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Example1
Solvetheequations:6x+y=15and4x+y=11
Labeltheequationssothatthemethodcanbeclearlyexplained.
6x+y=15 (1)
4x+y=11 (2)
Step1: Sincethey-terminbothequationshasthesamecoefficientthereisnoneedtobalanceterm.
Step2: Subtractoneequationfromtheother.(Equation(1)minusequation(2)willgivepositivevalues.)
(1) –(2) 2x=4
Step3: x=4÷2
x=2
Step4: Substitutex=2intooneoftheoriginalequations.(Usuallytheonewiththesmallestvaluesistheeasiest)
Sosubstituteinto: 4x+y=11
Whichgives: 4(2)+y=11
Step5: Solvethisequation: 8+y=11
y=11–8
y=3
Step6: Testthesolutionintheoriginalequations.Sosubstitutex=2andy=3into6x+y,whichgives12+3=15andinto4x+y,whichgives8+3=11.Thesearecorrect,soyoucanconfidentlysaythatthesolutionisx=2andy=3.
Example2
Solvetheseequations. 3x+2y=18 (1)
2x–y=5 (2)
Step1: Multiplyequation(2)by2.Thereareotherwaystobalancethecoefficientsbutthisistheeasiestandleadstolessworklater.Withpractice,youwillgetusedtowhichwillbethebestwaytobalancethecoefficients.
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2×(2) 4x–2y=10 (3)
Labelthisequationasequation(3)
Becarefultomultiplyeverytermandnotjustthey-term.Youcouldwrite:
2×(2x–y=5)→4x–2y=10 (3)
Step2: Asthesignsofthey-termsareopposite,addtheequations.
(1) +(3) 7x=28
Becarefultoaddthecorrectequations.Thisiswhylabelingthemisuseful.
Step3: Solvethisequation: x=28÷7
x=4
Step4: Substitutex=4intoanyequation,say2x–y=5→8–y=5
Step5: Solvetheequation: 8–5=y
y=3
Step6:Check:(1),3×4+2×3=18and(2),2×4–3=5,whicharecorrectsothesolutionisx=4andy=3.
Example3
Solvetheseequations: 4x+3y=27 (1)
5x–2y=5 (2)
Bothequationshavetobechangedtoobtainidenticaltermsineitherxory.
However,youcanseethatifyoumakethey-coefficientsthesame,youwilladdtheequations.Thisisalwayssaferthansubtraction,sothisisobviouslythebetterchoice.Wedothisbymultiplyingthefirstequationby2(they-coefficientoftheotherequation)andthesecondequationby3(they-coefficientoftheotherequation).
Step1: (1)×2or2×(4x+3y=27)→8x+6y=54 (3)
(2)×3or3×(5x–2y=5)→15x–6y=15 (4)
Labelthenewequations(3)and(4)
Step2: Eliminateoneofthevariables:(3)+(4) 23x=69
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Step3: Solvetheequation: x=69÷23
x=3
Step4: Substituteintoequation(1) 4(3)+3y=27
Step5: Solvetheequation: 12+3y=27
3y=27–12
3y=15
y=15÷3
y=5
Step6:Check:(1),4×3+3×5=12+15=27,and(2),5×3–2×5=15–10=5,whicharecorrectsothesolutionisx=3andy=5.
Consolidation
1. 4x+y=17and2x+y=9
2. 2x+y=7and5x–y=14
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3. 5x+2y=4and4x–y=11
4. 3x+4y=7and4x+2y=1
5. 2x–3y=15and5x+7y=52
6. 2x+3y=30and5x+7y=71
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7. 2x+5y=37andy=11–2x
8. 4x–3y=7andx=13–3y
SupportExercisePg153Exercise10DNos1–20
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Section7Settingupsimultaneousequations
Example1
Oneangleinatriangleis90°andthedifferencebetweentheothertwoanglesis36°.Findthelargerofthetwounknownangles.
Letxbethelargerangle.
Thesumofthethreeanglesis180°
Therefore x+y=90 (1)
Thedifferencebetweenxandyis36°,
Therefore x–y=36° (2)
Thetwoequationsare:
x+y=90 (1)
x–y=36 (2)
2x=126
x=126÷2
x=63°
x+y=90
63+y=90
y=90–63
y=27°
Thelargerangleis63°.
y
x
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Consolidation
1. Thelengthsofthesidesofanequilateraltriangleare(3a+2)cm,(2b–a)cm,and(b+
3)cm.
a) Findaandb.b) Findtheperimeterofthetriangle.
2. Findtwonumberssuchthattwicethefirstaddedtothesecondis26andthefirstaddedtothesecondis28.
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3. Acupandsaucertogethercost€2.05.Acupandtwosaucerscost€2.70.Findthecostofacupandsaucer.
4. Arectangleisacmlongandbcmwide.Theperimeteroftherectangleis48cmandthelengthis5cmmorethanthewidth.Findthelengthoftherectangle.
SupportExercisePg154Exercise10ENos1–5