algebra part 2 2b.pdf · form 4 algebra part 2 paper b 3 an expression such as (3y +2)(4y – 5)...

Form4 AlgebraPart2PaperB

1

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


7

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=


12

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

algebra part 2 2b.pdf · form 4 algebra part 2 paper b 3 an expression such as (3y +2)(4y – 5)...

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