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Equations, Expressions and InequalitiesPages
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2023
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QA #3 Equations, Expressions and Inequalities
QuestionsΒ andΒ AnswersΒ SheetΒ 3
Β
Equations,Β ExpressionsΒ and InequalitiesΒ Β
QuestionΒ #1Β
(a)Β WhatΒ isΒ theΒ logicalΒ firstΒ stepΒ inΒ solvingΒ theΒ equationΒ
Β
|2π₯Β βΒ 1|Β =Β 5?
Β Β Β
(b)Β WhatΒ isΒ theΒ logicalΒ firstΒ stepΒ inΒ solvingΒ theΒ inequalityΒ
Β
Β
|3π₯Β +Β 2|Β β€Β 8?
Β Β
Answer:Β
StepΒ 1Β
(a)Β ConsiderΒ theΒ givenΒ equationΒ Β
|2π₯Β βΒ 1|Β =Β 5
Β .Β
InΒ this equation,Β onlyΒ absoluteΒ valueΒ termΒ is onΒ oneΒ sideΒ ofΒ theΒ equation.Β
Hence,Β weΒ canΒ useΒ theΒ propertyΒ ofΒ absoluteΒ valueΒ toΒ obtainΒ twoΒ equations.Β
Thus,Β Β
|2π₯Β βΒ 1|Β =Β 5
Β Β isΒ equivalentΒ toΒ Β
2π₯Β βΒ 1Β =Β 5Β πππΒ 2π₯Β βΒ 1Β =Β β5
Β .Β
ThatΒ is,Β theΒ logicalΒ stepΒ inΒ solvingΒ theΒ equationΒ Β
|2π₯Β βΒ 1|Β =Β 5
Β Β willΒ beΒ toΒ useΒ theΒ propertyΒ ofΒ absoluteΒ valueΒ
whichΒ says Β
|π₯|Β =Β π
Β Β is equivalentΒ toΒ Β
|π₯|Β =Β Β±π
Β .Β
Hence,Β theΒ givenΒ equationΒ Β
|2π₯Β βΒ 1|Β =Β 5
Β Β isΒ equivalentΒ toΒ Β
2π₯Β βΒ 1Β =Β 5Β πππΒ 2π₯Β βΒ 1Β =Β β5
Β .Β
StepΒ 2Β
(b)Β ConsiderΒ theΒ givenΒ inequalityΒ Β
|3π₯Β +Β 2|Β β€Β 8
Β .Β
As itΒ is clearΒ fromΒ theΒ aboveΒ table,Β solutionΒ setΒ ofΒ inequalityΒ canΒ beΒ writtenΒ asΒ Β
β8Β Β β€Β 3π₯Β +Β 2Β Β β€Β 8
Β .Β
OnceΒ weΒ haveΒ writtenΒ theΒ inequalityΒ inΒ this form,Β weΒ canΒ splitΒ andΒ writeΒ itΒ inΒ terms ofΒ twoΒ inequalitiesΒ Β
3π₯Β +
2Β Β β₯Β β8Β πππΒ 3π₯Β +Β 2Β Β β€Β 8
Β .Β
TheΒ firstΒ logicalΒ stepΒ inΒ solvingΒ theΒ inequalityΒ Β
|3π₯Β +Β 2|Β β€Β 8
Β Β wouldΒ beΒ toΒ useΒ theΒ propertiesΒ ofΒ absoluteΒ
valueΒ inequalitiesΒ andΒ obtainΒ theΒ followingΒ inequalities.Β
Β
3π₯Β +Β 2Β Β β₯Β β8Β πππΒ 3π₯Β +Β 2Β Β β€Β 8
Β .
Β
QuestionΒ #2Β
MinΒ Β
πΒ =Β 4π₯Β +Β 10π¦
Β Β
SubjectΒ to:Β
Β
2π₯Β +Β 6π¦Β Β β₯Β 24Β
Β
Β
4π₯Β +Β 2π¦Β Β β₯Β 18Β
Β
Answer:Β
StepΒ 1Β
Let's solveΒ theΒ twoΒ inequalities treatingΒ themΒ as equationsΒ andΒ thenΒ seeΒ ifΒ theΒ solutionΒ ofΒ theΒ systemΒ ofΒ
equations violateΒ anyΒ ofΒ theΒ inequalities.Β Β
Β
2π₯Β +Β 6π¦Β =Β 24
Β Β ----------------EqnΒ (1)Β Β
Β
4π₯Β +Β 2π¦Β =Β 18
Β Β ---------------EqnΒ (2)Β Β
EqnΒ Β
(1)Β βΒ 3
Β Β eqnΒ (2)Β gives:Β Β
Β
(2π₯Β βΒ 12π₯)Β =Β 24Β βΒ 3βΒ ΓΒ β18
Β Β Β
Or,Β Β
β10π₯Β =Β β30
Β Β
Hence,Β Β
π₯Β =Β 3
Β Β Β
FromΒ eqnΒ (1):Β Β
Β
π¦Β =Β (24Β βΒ 2Β ΓΒ 3)/6Β =Β 3
Β Β Β
StepΒ 2Β Β
Hence,Β theΒ minimumΒ Β
πΒ =Β 4π₯Β +Β 10π¦Β =Β 4Β Β ΓΒ 3Β +Β 10Β Β ΓΒ 3Β =Β 42
Β Β Β
StepΒ 3Β
Hence,Β theΒ finalΒ answer:Β Β
π₯Β =Β 3,Β π¦Β =Β 3,Β πΒ =Β 42
Β Β
QuestionΒ #3Β
DescribeΒ theΒ strategy youΒ wouldΒ useΒ toΒ solveΒ Β
logΒ 6Β π₯Β βΒ βΒ logΒ 6Β 4Β +Β βΒ logΒ 6Β 8
Β .Β
a.Β UseΒ theΒ productΒ ruleΒ toΒ turnΒ theΒ rightΒ sideΒ ofΒ theΒ equationΒ intoΒ aΒ singleΒ logarithm.Β RecognizeΒ thatΒ theΒ
resultingΒ valueΒ is equalΒ toΒ x.Β
b.Β Express theΒ equationΒ inΒ exponentialΒ form,Β setΒ theΒ exponentsΒ equalΒ toΒ eachΒ otherΒ andΒ solve.Β
c.Β UseΒ theΒ factΒ thatΒ theΒ logs haveΒ theΒ sameΒ baseΒ toΒ addΒ theΒ expressionsΒ onΒ theΒ rightΒ sideΒ ofΒ theΒ equationΒ
together.Β Express theΒ resultsΒ inΒ exponentialΒ form,Β setΒ theΒ exponentsΒ equalΒ toΒ eachΒ otherΒ andΒ solve.Β
d.Β UseΒ theΒ factΒ thatΒ sinceΒ bothΒ sides ofΒ theΒ equations haveΒ logarithmsΒ withΒ theΒ sameΒ baseΒ toΒ setΒ theΒ
expressionsΒ equalΒ toΒ eachΒ otherΒ andΒ solve.
Β
Answer:Β
Β
logΒ 6Β π₯Β =Β logΒ 6Β 4Β +Β logΒ 6Β 8
Β Β
log
6
π₯Β =Β log
6
(32)Β β{log
π
πΒ +Β log
π
πΒ =Β log
π
(ππ)}
Β Β ProductΒ rule.Β
Β π₯Β =Β 32
Β Β Β
OptionΒ AΒ isΒ correct.Β
Β
QuestionΒ #4Β
1)Β Β
4π
3
π
3
4π
β3
π
2
Β Β
2)Β Β
3ππ
4
4π
0
Β
3)Β Β
3π’
β2
3π’
β1
π£
0
Β
4)Β Β
βπ
β6
π
β5
β2π
2
ββ β(π
3
π
β3
)
6
Β
Answer:Β
InΒ algebra,Β toΒ useΒ numerals,Β symbols,Β andΒ letters calledΒ variables,Β orΒ pro-numerals,Β andΒ combinationsΒ ofΒ
both.Β They standΒ forΒ theΒ unknownΒ values.Β Algebraic expressions orΒ equations,Β theΒ unknownΒ values areΒ
representedΒ byΒ pro-numerals orΒ variables.Β
TheΒ variableΒ is aΒ symbolΒ thatΒ standsΒ forΒ aΒ number.Β AΒ symbolΒ orΒ letterΒ representingΒ anΒ unknownΒ memberΒ
ofΒ aΒ set.Β InΒ algebraic expressions,Β aΒ variableΒ stands forΒ aΒ value.Β Sometimes itΒ is calledΒ anΒ unknown.Β TheΒ
sameΒ variableΒ may haveΒ differentΒ values underΒ differentΒ conditions.Β Β
''SinceΒ youΒ haveΒ postedΒ aΒ questionΒ withΒ multipleΒ sub-parts,Β weΒ willΒ solveΒ theΒ firstΒ fourΒ sub-partsΒ forΒ you.Β
ToΒ getΒ theΒ remainingΒ sub-partΒ solved,Β pleaseΒ repostΒ theΒ completeΒ questionΒ andΒ mentionΒ theΒ sub-partsΒ toΒ
beΒ solved.''Β
1)Β CancelΒ theΒ commonΒ factorΒ 4Β andΒ simplifyingΒ Β
π
3
/π
β3
Β πππ£ππ
Β
π
6
Β Β andΒ Β
π
3
/π
2
Β is nΒ usingΒ theΒ ruleΒ Β
π/πΒ =
ππ
β1
Β Β
Β
4π
3
π
3
4π
β3
π
2
=
π
3
π
3
π
β3
π
2
Β Β
Β
=Β π
3β(β3)
β Β π
3β2
Β Β
Β
=Β π
6
π
Β Β
2)Β Apply theΒ ruleΒ Β
π
0
=Β 1
Β whereΒ Β
πΒ Β β Β 0
Β Β
3ππ
4
4π
0
=
3ππ
4
4β 1
Β Β
Β
=
3ππ
4
4
Β Β
3)Β Apply theΒ exponentΒ ruleΒ Β
π
0
=Β 1
Β Β whereΒ Β
πΒ Β β Β 0
Β ;Β cancelΒ outΒ theΒ commonΒ factorΒ 3Β andΒ Β
π’
β2
/π’
β1
Β Β isΒ
1/π’
.Β
Β
3π’
β2
3π’
β1
π£
0
=
3π’
β2
3π’
β1
β 1
Β Β
Β
=
3π’
β2
3π’
β1
Β Β
Β
π’
β2
π’
β1
Β Β
=Β π’
β2β(β1)
Β Β
Β
=Β π’
β2+1
Β Β
Β
=Β π’
β1
Β
Β
3π’
β2
3π’
β1
π£
0
=
1
π’
Β Β
4)Β Apply theΒ fractionΒ ruleΒ Β
βπ/βπΒ =Β π/π
Β Β
Β
βπ
β6
π
β5
β2π
2
β (π
3
π
β3
)
6
=
π
β6
π
β5
2π
2
(π
3
π
β3
)
6
Β Β
Β
=
π
β6
2π
2+5
(π
3
π
β3
)
6
Β Β
Β
=
π
β6
2π
7
(π
3
π
β3
)
6
Β Β
Β
=
π
β6
2π
7
(π
6.3
π
β18
)
Β Β
Β
=
π
β6
2β
1
π18
π
18
π
7
Β Β
Β
=
1
2β
1
π18
π
24
π
7
Β Β
Β
=
1
2π24
π11
Β Β
Β
=
π
11
2π
24
Β Β
Β
QuestionΒ #5Β
SimplifyΒ theΒ followingΒ expressionsΒ usingΒ theΒ imaginary numberΒ i:Β Β
1.Β Β
5ββ12
Β Β
2.Β Β
ββ8
Β Β
3.Β Β
3ββ7
Β Β
4.Β Β
β3ββ200
Β Β
WriteΒ theΒ followingΒ numbers usingΒ theΒ imaginary numberΒ i,Β andΒ thenΒ performΒ theΒ operationsΒ necessaryΒ
andΒ simplifyΒ yourΒ answer.Β
1.Β Β
3ββ9Β β Β 2ββ49
Β Β
2.Β Β
ββ12Β β Β ββ36
Β
3.Β
πββ36Β βΒ π2
Β Β
SimplifyΒ theΒ followingΒ expressionsΒ usingΒ theΒ imaginary numberΒ iΒ whenΒ needed:Β
8.Β
π8Β
Β
9.Β
π75
Β
10.Β
(π2)(π6)
Β
SimplifyΒ andΒ rationalizeΒ theΒ denominators.Β
11.Β Β
4ββ3/2
Β Β
12.Β Β
β3ββ2/27
Β Β Β
13.Β Β
ββ4/27
Β Β
PerformΒ theΒ operationΒ indicatedΒ inΒ eachΒ problemΒ below:Β
14.Β Β
(8Β +Β 6π)Β +Β (2Β +Β π)
Β Β
15.Β Β
β9Β +Β (4Β βΒ 2π)
Β Β
16.Β Β
(1Β +Β π)Β βΒ (1Β βΒ π)
Β Β
17.Β Β
3πΒ βΒ (4Β +Β 15π)
Β Β
PerformΒ theΒ operationΒ indicatedΒ inΒ eachΒ problemΒ below:Β
18.Β
(β2Β +Β 3π)(β2Β βΒ 3π)
Β Β
19.Β Β
(β27Β βΒ 3π)(β3Β +Β π)
Β Β
20.Β Β
(β2Β βΒ πβ3)(β2Β +Β πβ3)
Β Β Β
21.Β Β
2π(3πΒ βΒ 2)
Β Β
22.Β Β
(3Β βΒ 6π)2
Β Β
DetermineΒ theΒ sumΒ orΒ differenceΒ inΒ geometricΒ terms,Β andΒ thenΒ checkΒ itΒ algebraically.Β RememberΒ thatΒ Β
πΒ β
πΒ =Β πΒ +Β (π)
Β .Β
23.Β Β
(8Β βΒ 5π)Β βΒ (2Β +Β 3π)
Β Β
24.Β Β
(β6Β +Β π)Β βΒ (β2Β βΒ 3π)
Β Β
25.Β Β
(3Β +Β π)Β +Β (β2Β +Β 2π)Β +Β (8Β βΒ 2π)
Β Β
SolveΒ theΒ followingΒ equationsΒ usingΒ theΒ quadratic formula.Β
26.Β Β
πΒ 2Β βΒ 4πΒ +Β 13Β =Β 0
Β Β
27.Β Β
πΒ 2Β +Β 49Β =Β 0Β
Β
28.Β Β
πΒ 2Β +Β 5πΒ +Β 8Β =Β 0
Β Β
29.Β Β
π2Β βΒ 3πΒ +Β 4Β =Β 0
Β Β
Answer:Β
1)Β TheΒ givenΒ expressionΒ isΒ Β
5ββ12
Β Β
SimplifyΒ theΒ expressionΒ
Β
5ββ12Β =Β 5ββ1β12
Β Β
Β
=Β 5πβ4β3
Β Β
Β
=Β 5π(2)β3
Β
Β
=Β 10πβ3
Β Β
2)Β TheΒ givenΒ expressionΒ isΒ Β
ββ8
Β
SimplifyΒ theΒ expressionΒ
Β
ββ8Β =Β ββ1β8
Β Β
Β
=Β πβ4β2
Β Β
Β
=Β 2πβ2
Β Β
3)Β TheΒ givenΒ expressionΒ isΒ Β
3ββ7
Β
SimplifyΒ theΒ expressionΒ
Β
3ββ7Β =Β 3ββ1β7
Β Β
Β
=Β 3πβ7
Β Β
4)Β TheΒ givenΒ expressionΒ isΒ Β
β3ββ200
Β .Β
SimplifyΒ theΒ expressionΒ
Β
β3ββ200Β =Β β3ββ1β200
Β Β
Β
=Β β3πβ2β100
Β Β
=Β β3π(10)β2
Β Β
Β
=Β β30πβ2
Β Β
Β
QuestionΒ #6Β
FindΒ theΒ standardΒ equationΒ ofΒ any parabolaΒ thatΒ Β
has vertexΒ
π
.Β
Β
π(β3,1)
Β
Β
Answer:
Β
StepΒ 1Β
TheΒ equationΒ is aΒ statementΒ thatΒ consistsΒ ofΒ equal symbolΒ betweenΒ twoΒ algebraic expressions.Β TheΒ
solutionΒ forΒ theΒ variableΒ ofΒ theΒ equationΒ mustΒ satisfyΒ theΒ equationΒ whenΒ weΒ resubstituteΒ theΒ solutionΒ inΒ theΒ
equation.Β
StepΒ 2Β
QuadraticΒ equationsΒ areΒ theΒ polynomialΒ equations ofΒ degreeΒ 2Β inΒ oneΒ variableΒ ofΒ typeΒ Β
π(π₯)Β =Β ππ₯
2
+Β ππ₯Β +
π
Β Β whereΒ
π,Β π,Β π,
Β areΒ realΒ valuesΒ andΒ
Β πΒ Β β Β 0Β
.Β TheΒ equationΒ ofΒ aΒ parabolaΒ is aΒ quadraticΒ equation.Β LetΒ itΒ beΒ
as Β
π¦Β =Β ππ₯
2
+Β ππ₯Β +Β π
.Β ToΒ getΒ theΒ standardΒ formΒ ofΒ parabolaΒ equationΒ withΒ vertexΒ at4Β substituteΒ this pointΒ inΒ
theΒ vertexΒ equationΒ ofΒ parabolaΒ andΒ simplifyΒ itΒ toΒ writeΒ itΒ inΒ standardΒ formΒ as follows;Β
TheΒ parabolaΒ equationΒ withΒ vertexΒ atΒ
(β,Β π)
Β is Β
π¦Β =Β π(π₯Β βΒ β)
2
+Β π
Β .Β
So,Β substituteΒ theΒ givenΒ vertexΒ pointΒ
(β3,1)
Β inΒ theΒ vertexΒ equationΒ ofΒ parabolaΒ withΒ Β
(β,Β π)Β =Β (β3,1)
Β Β asΒ
follows;Β
Β
π¦Β =Β π(π₯Β βΒ β)
2
+Β π
Β Β
Β
π¦Β =Β π(π₯Β βΒ (β3))
2
+Β 1
Β Β ............. SubstituteΒ theΒ vertexΒ Β
(β,Β π)Β =Β (β3,1)
Β .Β
Β
π¦Β =Β π(π₯Β +Β 3)
2
+Β 1
Β
Β
=Β π(π₯
2
+Β 6π₯Β +Β 9)Β +Β 1
Β Β
Β
=Β ππ₯
2
+Β 6ππ₯Β +Β 9πΒ +Β 1
Β Β
Hence,Β theΒ standardΒ equationΒ ofΒ theΒ parabolaΒ withΒ theΒ vertexΒ atΒ
(β3,1)
Β is Β
π¦Β =Β ππ₯
2
+Β 6ππ₯Β +Β 9πΒ +Β 1
Β ,Β whereΒ
aΒ is non-zeroΒ realΒ number.Β
Β
QuestionΒ #7Β
ExplainΒ why theΒ solutions ofΒ theΒ simultaneousΒ equationsΒ
Β
ππ₯Β +Β ππ¦Β +Β πΒ =Β 0Β ππ₯Β +Β ππ¦Β +Β πΒ =Β 0
Β Β
haveΒ rationalΒ expressionΒ inΒ Β
{π,Β π,Β π,Β π,Β π,Β π}Β
Β wheneverΒ Β
ππΒ Β β Β ππΒ
.Β
Answer:Β
StepΒ 1Β
givenΒ simultaneous equationsΒ are,Β
Β
ππ₯Β +Β ππ¦Β +Β πΒ =Β 0
Β Β andΒ
Β
ππ₯Β +Β ππ¦Β +Β πΒ =Β 0
Β Β
StepΒ 2Β
givenΒ systemΒ canΒ beΒ writtenΒ as,Β
Β
ππ₯Β +Β ππ¦Β =Β βπ
Β Β andΒ
Β
ππ₯Β +Β ππ¦Β Β±Β π
Β Β
aboveΒ systemΒ is nonΒ homogeneous systemΒ ofΒ equation,Β
compaireΒ itΒ withΒ Β
ππ₯Β =Β π
Β Β
Β
a
b
x
c
d
e
y
f
β
ο©
οΉΒ ο©Β οΉΒ ο©
οΉ
=
οͺ
οΊΒ οͺΒ οΊΒ οͺ
οΊ
β
ο«
ο»Β ο«Β ο»Β ο«
ο»
Β
whenΒ Β
ππΒ =Β ππ
Β Β thenΒ weΒ getΒ aΒ solutionΒ inΒ integer.Β
andΒ ifΒ Β
ππΒ Β β Β ππ
Β Β thenΒ weΒ haveΒ toΒ solveΒ systemΒ ofΒ simultaneous equationΒ by usingΒ rowΒ transformation.Β
Β
3
3
2
(
9Β )
r
r
r
οΒ =Β β
a
b
x
c
d
e
y
f
β
ο©
οΉΒ ο©Β οΉΒ ο©
οΉ
=
οͺ
οΊΒ οͺΒ οΊΒ οͺ
οΊ
β
ο«
ο»Β ο«Β ο»Β ο«
ο»
Β
Β
1
1
1
b
c
x
R
a
a
y
a
d
e
f
ο©
οΉ
ο©
οΉ
β
ο©Β οΉ
οͺ
οΊ
οͺ
οΊ
ο
=
οͺΒ οΊ
οͺ
οΊ
οͺ
οΊ
ο«Β ο»
β
ο«
ο»
ο«
ο»
Β Β
Β
π
2
=Β π
2
βΒ ππ
1
Β Β gives,Β
Β
1
0
(
)
b
c
x
a
a
bd
y
f
cd
e
a
ο©
οΉ
ο©
οΉ
οͺ
οΊ
β
ο©Β οΉΒ οͺ
οΊ
=
οͺ
οΊΒ οͺΒ οΊΒ οͺ
οΊ
ο«Β ο»
οͺ
οΊ
βΒ +
β
ο«
ο»
οͺ
οΊ
ο«
ο»
Β Β
by solvingΒ xΒ andΒ yΒ weΒ getΒ rational expressions,Β
soΒ weΒ haveΒ aΒ rational expressionsΒ inΒ Β
{π,Β π,Β π,Β π,Β π,Β π}
Β Β wheneverΒ Β
ππΒ Β β Β ππ
Β .Β
Β Β
QuestionΒ #8Β
SolveΒ theΒ followingΒ quadratic expressions by factoring.Β FirstΒ writeΒ theΒ expressionsΒ inΒ completely factoredΒ
form.Β ThenΒ writeΒ theΒ realΒ numerΒ solutions.Β
[Hint:Β RememberΒ toΒ useΒ properΒ notationΒ whenΒ writingΒ theΒ realΒ numberΒ solutions.ForΒ example:Β ifΒ theΒ
solutions areΒ Β
π₯Β =Β 1
Β Β andΒ Β
π₯Β =Β 4
Β ,Β writeΒ
Β π₯Β =Β 1,4
Β .Β
1)Β Β
π₯
2
+Β 9π₯Β +Β 20Β =Β 0
Β Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?Β
2)Β Β
π₯
2
βΒ π₯Β βΒ 72Β =Β 0
Β Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?Β
3)Β Β
π₯
2
βΒ 36Β =Β 0
Β Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?Β
4)Β Β
π₯
2
βΒ 10π₯Β +Β 25Β =Β 0
Β Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?Β
Answer:Β
1)Β Given:Β Β
π₯
2
+Β 9π₯Β +Β 20Β =Β 0
Β Β
Β
β
Β
π₯
2
+Β 4π₯Β +Β 5π₯Β +Β 20Β =Β 0
Β Β
Β
βΒ π₯(π₯Β +Β 4)Β +Β 5(π₯Β +Β 4)Β =Β 0
Β Β
Β
βΒ (π₯Β +Β 4)(π₯Β +Β 5)Β =Β 0
Β Β
FactoredΒ formΒ
Β
(π₯Β +Β 4)(π₯Β +Β 5)
Β
RealΒ numberΒ solutionΒ
Β
βΒ (π₯Β +Β 4)(π₯Β +Β 5)Β =Β 0
Β Β
Β βΒ π₯Β =Β β4,Β π₯Β =Β β5
Β Β
Β
π₯Β =Β β4,Β β5
Β Β
2)Β Β
βΒ π₯
2
βΒ π₯Β βΒ 72Β =Β 0
Β Β
Β
βΒ π₯
2
βΒ 9π₯Β +Β 8π₯Β βΒ 72Β =Β 0
Β Β
Β
βΒ π₯(π₯Β βΒ 9)Β +Β 8(π₯Β βΒ 9)Β =Β 0
Β Β
Β
βΒ (π₯Β βΒ 9)(π₯Β +Β 8)
Β Β
FactoredΒ formΒ
Β
(π₯Β βΒ 9)(π₯Β +Β 8)
Β Β
RealΒ numberΒ solutionΒ
Β
βΒ (π₯Β βΒ 9)(π₯Β +Β 8)Β =Β 0
Β Β
Β
βΒ π₯Β =Β 9,Β π₯Β =Β β8
Β Β
Β
π₯Β =Β 9,Β β8
Β Β
3)Β Β
π₯
2
βΒ 36Β =Β 0
Β
Β
βΒ π₯
2
=Β 36
Β Β
Β
βΒ π₯Β =Β β36
Β Β
Β
βΒ π₯Β =Β 6
Β Β
FactoredΒ form.Β
Β
π₯Β =Β 6
Β Β
RealΒ numberΒ solution.Β
Β
π₯Β =Β 0.6
Β Β
4)Β Β
π₯
2
βΒ 10π₯Β +Β 25Β =Β 0
Β Β
Β
βΒ π₯
2
βΒ 5π₯Β βΒ 5π₯Β +Β 25Β =Β 0
Β Β
Β
βΒ π₯
2
βΒ 5π₯Β βΒ 5π₯Β +Β 25Β =Β 0
Β Β
Β
βΒ π₯(π₯Β βΒ 5)Β βΒ 5(π₯Β βΒ 5)Β =Β 0
Β Β
Β
βΒ (π₯Β βΒ 5)(π₯Β βΒ 5)Β =Β 0
Β Β
FactoredΒ formΒ
Β
(π₯Β βΒ 5)(π₯Β βΒ 5)
Β Β
RealΒ numberΒ solutionΒ
Β
(π₯Β βΒ 5)(π₯Β βΒ 5)Β =Β 0
Β Β
Β
π₯Β =Β 5,5
Β
Β
QuestionΒ #9Β
WriteΒ
π,Β π,Β π
,Β andΒ
πΒ
inΒ leastΒ toΒ greatest.Β
GivenΒ theΒ inequalityΒ equation:Β
Β
πΒ >Β πΒ +
1
3
;Β πΒ +Β 1Β <Β πΒ βΒ 4;Β πΒ +
5
8
>Β πΒ +Β 2
Β
Β
Answer:Β
ConceptΒ used:Β
InΒ mathematics,Β anΒ inequalityΒ isΒ aΒ relationΒ whichΒ makes aΒ non-equalΒ comparisonΒ betweenΒ twoΒ numbers orΒ
otherΒ mathematicalΒ expression.Β ItΒ is usedΒ mostΒ oftenΒ toΒ compareΒ twoΒ numbers onΒ theΒ numberΒ lineΒ by theirΒ
size.Β
AnΒ inequalityΒ compares twoΒ values,Β showingΒ ifΒ oneΒ isΒ less than,Β greaterΒ than,Β orΒ simply notΒ equal toΒ
anotherΒ value.Β Β
πΒ Β β Β π
Β Β says thatΒ aΒ is notΒ equalΒ toΒ b.Β Β
πΒ <Β π
Β Β says that
Β π
Β isΒ lessΒ than
Β π
. Β
πΒ >Β π
Β Β says thatΒ
π
Β isΒ
greaterΒ thanΒ
π
.Β
ThereΒ areΒ fourΒ differentΒ typesΒ ofΒ inequalities:Β
GreaterΒ thanΒ -(>);Β Less thanΒ -(<);Β GreaterΒ thanΒ orΒ equal toΒ Β
β(β₯)
Β ;Β Less thanΒ orΒ equalΒ toΒ Β -
(β€)
Β Β
ForΒ inequalityΒ equation:Β IfΒ Β
πΒ >Β πΒ Β βΒ πΒ <Β πΒ ππΒ Β πΒ <Β πΒ Β βΒ πΒ >Β π
Β Β
Rules forΒ solvingΒ inequalityΒ equations:Β
TheseΒ things doΒ notΒ affectΒ theΒ directionΒ ofΒ theΒ inequality:Β
-AddΒ (orΒ subtract)Β aΒ numberΒ fromΒ bothΒ sidesΒ
-Multiply (orΒ divide)Β bothΒ sides by aΒ positiveΒ numberΒ
-SimplifyΒ aΒ sideΒ
ButΒ theseΒ thingsΒ doΒ changeΒ theΒ directionΒ ofΒ theΒ inequalityΒ (''<''becomes''>''Β forΒ example):Β
-Multiply (orΒ divide)Β bothΒ sides by aΒ negativeΒ numberΒ
-SwappingΒ leftΒ andΒ rightΒ handΒ sidesΒ
Calculation:Β
Check eachΒ inequalityΒ equation:Β
Β
πΒ +Β 1Β <Β πΒ βΒ 4
Β Β
Β
πΒ +Β 1Β βΒ 1Β <Β πΒ βΒ 4Β βΒ 1
Β Β
Β
πΒ <Β πΒ βΒ 5
Β Β
Β
πΒ +
5
8
>Β πΒ +Β 2
Β Β
Β
πΒ +
5
8
β
5
8
>Β πΒ +Β 2Β β
5
8
Β Β
πΒ >Β πΒ +
16
8
β
5
8
Β
Β
πΒ >Β πΒ +
11
8
Β Β
πΒ >Β πΒ +
1
3
Β Β
LetΒ Β
πΒ =Β 10
Β ;Β
Β
πΒ <Β πΒ βΒ 5Β Β βΒ πΒ <Β 10Β βΒ 5Β Β βΒ πΒ <Β 5
Β Β
Β
πΒ >Β πΒ +
11
8
βΒ πΒ >Β 10Β +Β 1.375Β βΒ πΒ >Β 11.375
Β Β
Β
πΒ >Β πΒ +
1
3
βΒ πΒ >Β 11.375Β +Β 0.33Β βΒ πΒ >Β 11.705
Β Β
ifΒ
Β πΒ =Β 10Β ,Β π‘βππΒ Β πΒ >Β 11.705;Β Β πΒ <Β 5πΒ >Β 11.375
Β Β
OrderΒ fromΒ leastΒ toΒ greatest:Β Β
πΒ <Β πΒ <Β πΒ <Β π
Β Β
Thus,Β theΒ orderΒ fromΒ leastΒ toΒ greatest:Β Β
πΒ <Β πΒ <Β πΒ <Β π
Β Β
Β
QuestionΒ #10Β
Graphs ofΒ theΒ functions
Β π
Β andΒ
π
Β areΒ given.Β
Β
a)Β WhichΒ isΒ larger,Β
π(0)
Β orΒ
π(0)
?Β
Β
b)Β WhichΒ isΒ larger,
π(β3)
Β orΒ
π(β3)
?Β
Β
c)Β ForΒ whichΒ valuesΒ ofΒ
π₯
Β is Β
π(π₯)Β =Β π(π₯)?
Β Β
Β
Β
Answer:Β
AΒ singleΒ variableΒ functionΒ canΒ beΒ definedΒ asΒ aΒ relationΒ betweenΒ theΒ functionΒ variableΒ andΒ theΒ functionΒ
valueΒ whichΒ variesΒ withΒ theΒ functionΒ variable.Β AndΒ ifΒ theΒ functionΒ hasΒ oneΒ uniqueΒ valueΒ forΒ eachΒ valueΒ ofΒ x,Β
itΒ isΒ calledΒ aΒ oneΒ toΒ oneΒ functionΒ Β
StepΒ 2Β
a)Β
π(0)
andΒ
π(0)
Β implies theΒ valueΒ ofΒ theΒ functionΒ yΒ whenΒ theΒ valueΒ ofΒ xΒ is zero.Β FromΒ theΒ graphΒ atΒ Β
π₯Β =Β 0
Β ,Β
theΒ valueΒ ofΒ theΒ functionΒ
π(π₯)
Β is
Β 3
Β andΒ theΒ valueΒ ofΒ
π(π₯)
Β is aroundΒ
0.5.
Β ThereforeΒ atΒ Β
π₯Β =Β 0,Β π(π₯)Β >Β π(π₯)
Β Β
b)Β As inΒ theΒ aboveΒ question
π(β3)
Β andΒ
π(β3)
Β impliesΒ theΒ valueΒ ofΒ theΒ functionΒ y,Β whenΒ theΒ valueΒ ofΒ xΒ isΒ -3.Β
FromΒ theΒ graphΒ ofΒ theΒ functions,Β theΒ valueΒ ofΒ f(x)Β atΒ Β
π₯Β =Β β3
Β Β isΒ -1Β andΒ theΒ valueΒ ofΒ
π(π₯)
Β atΒ Β
π₯Β =Β β3Β
Β is 2.Β
ThereforeΒ
π(β3)
Β isΒ greaterΒ than
π(β3)
Β Β
c)Β TheΒ pointΒ whereΒ Β
π(π₯)Β =Β π(π₯)
Β Β is theΒ pointΒ whereΒ theΒ graphs ofΒ theΒ twoΒ functionsΒ intersects.Β FromΒ theΒ
graph,Β theΒ twoΒ functionsΒ intersectsΒ atΒ theΒ pointsΒ
Β π₯Β =Β β2,2
Β .Β ThereforeΒ forΒ Β
π₯Β =Β β2Β
Β andΒ
2
,Β
π(π₯)
willΒ beΒ equalΒ
toΒ
π(π₯).
Β
Β
QuestionΒ #11Β
UseΒ theΒ Gauss-JordanΒ methodΒ toΒ solveΒ theΒ followingΒ systemΒ ofΒ equations.Β
Β
3π₯Β βΒ 4π¦Β +Β 4π§Β =Β 10
Β Β
Β
3π₯Β +Β 5π¦Β βΒ π§Β =Β 15
Β Β
Β
12π₯Β βΒ 7π¦Β +Β 11π§Β =Β 45
Β Β
SelectΒ theΒ correctΒ choiceΒ belowΒ and,Β ifΒ necessary,Β fillΒ inΒ theΒ answerΒ boxΒ toΒ completeΒ yourΒ choice.Β
A.Β TheΒ solutionΒ isΒ (Β _Β ,Β _Β ,_Β )Β inΒ theΒ orderΒ x,Β y,Β z.Β
(SimplifyΒ yourΒ answers.)Β
B.Β ThereΒ isΒ anΒ infiniteΒ numberΒ ofΒ solutions.Β
TheΒ solutionΒ is (Β _Β ,Β _,Β _Β )Β whereΒ zΒ is anyΒ realΒ number.Β
(SimplifyΒ yourΒ answers.Β UseΒ integers orΒ fractions forΒ anyΒ numbers inΒ theΒ expressions.)Β
C. ThereΒ isΒ noΒ solution.
Β
Answer:Β
StepΒ 1Β
Β
3π₯Β βΒ 4π¦Β +Β 4π§Β =Β 10
Β Β
Β
3π₯Β +Β 5π¦Β βΒ π§Β =Β 15
Β Β
Β
12π₯Β βΒ 7π¦Β +Β 11π§Β =Β 45
Β Β
StepΒ 2Β
By usingΒ Gauss-JordanΒ methodΒ solveΒ theΒ systemΒ ofΒ equationsΒ as follows.Β
1
1
4
4
10
1
3
4
4
10
3
3
3
(Β 3
5
1Β 15)
(Β 3
5
1Β 15)(
3
12
7
11
45
12
7
11
45
r
r
β
β
β
=
β
οΒ =
β
β
Β Β
Β
2
2
1
4
4
10
1
3
3
3
(Β 0
9
5
5Β )(
3Β )
12
7
11
45
r
r
r
β
=
β
οΒ =Β β
β
Β
Β
3
3
1
4
4
10
1
3
3
3
(0
9
5
5Β )(
12Β )
0
9
5
5
r
r
r
β
=
β
οΒ =Β β
β
Β
Β
2
2
4
4
10
1
3
3
3
5
5
(0
1
)(
)
9
9
9
0
9
5
5
r
r
β
β
=
οΒ =
β
Β Β
StepΒ 3Β
OnΒ furtherΒ simplification,Β
2
1
1
16
110
1
0
27
27
3
4
4
10
4
5
5
(Β 3
5
1Β 15)
(0
1
)(
9
9
3
12
7
11
45
0
9
5
5
r
r
r
β
β
β
=
οΒ =Β +
β
β
Β
Β
16
110
1
0
27
27
5
5
(0
1
)
9
9
0
0
0
0
β
=
Β Β
3
3
2
(
9Β )
r
r
r
οΒ =Β β
Β
StepΒ 4
a
b
x
c
d
e
y
f
β
ο©
οΉΒ ο©Β οΉΒ ο©
οΉ
=
οͺ
οΊΒ οͺΒ οΊΒ οͺ
οΊ
β
ο«
ο»Β ο«Β ο»Β ο«
ο»
Β
FromΒ theΒ aboveΒ calculationΒ thereΒ areΒ infinitelyΒ many solutions.Β
LetΒ zΒ is aΒ realΒ numberΒ so,Β Β
π₯Β =
110
27
β
16
27
π§Β πππΒ π¦Β =
5
9
(1Β +Β π§)
Β Β
Therefore,Β theΒ solutionΒ ofΒ theΒ givenΒ systemΒ ofΒ equationΒ is Β
(
110
27
β
16
27
π§,
5
9
(1Β +Β π§),Β π§)
Β .
Β
QuestionΒ #12Β
Graphs ofΒ theΒ functions fΒ andΒ gΒ areΒ given.Β
a)Β WhichΒ isΒ larger,Β f(6)Β orΒ g(6)?Β
b)Β WhichΒ isΒ larger,Β f(3)Β orΒ g(3)?Β
c)Β FindΒ theΒ valuesΒ ofΒ xΒ forΒ whichΒ Β
π(π₯)Β =Β π(π₯)
Β .Β
Β
Answer:Β
a)Β Β
π(6)Β =Β 3.5
Β Β andΒ Β
π(6)Β =Β 5.5
Β Β
ClearlyΒ
π(6)
Β isΒ larger.Β
b)Β Β
π(3)Β =Β 6
Β andΒ Β
π(3)Β =Β 3
Β Β
ClearlyΒ
π(3)
is larger.Β
c)Β FromΒ theΒ graphΒ weΒ canΒ seeΒ thatΒ theΒ twoΒ curves meetsΒ atΒ Β
π₯Β =Β 2Β πππΎ
Β andΒ
πππΎΒ π₯Β =Β 7
Β Β
Hence,Β Β
π(π₯)Β =Β π(π₯)
Β Β whenΒ Β
π₯Β =Β 2,Β 7
.Β Β
Β
QuestionΒ #13Β
SolveΒ forΒ theΒ unknownΒ variableΒ fromΒ theΒ inequalityΒ equation.Β
Given:Β
TheΒ inequalityΒ equation:Β
πππΎΒ β
1
2
πΒ β€Β 6
Β
Β
Answer:Β
ConceptΒ used:Β
InΒ mathematics,Β anΒ inequalityΒ isΒ aΒ relationΒ whichΒ makes aΒ non-equalΒ comparisonΒ betweenΒ twoΒ numbers orΒ
otherΒ mathematicalΒ expression.Β ItΒ is usedΒ mostΒ oftenΒ toΒ compareΒ twoΒ numbers onΒ theΒ numberΒ lineΒ by theirΒ
size.Β
AnΒ inequalityΒ compares twoΒ values,Β showingΒ ifΒ oneΒ isΒ less than,Β greaterΒ than,Β orΒ simply notΒ equal toΒ
anotherΒ value.Β Β
πΒ Β β Β π
Β Β saysΒ thatΒ aΒ is notΒ equalΒ toΒ b.Β Β
πΒ <Β π
Β Β says thatΒ aΒ isΒ less thanΒ b.Β Β
πΒ >Β π
Β Β says thatΒ aΒ isΒ
greaterΒ thanΒ b.Β
ThereΒ areΒ fourΒ differentΒ typesΒ ofΒ inequalities:Β
GreaterΒ thanΒ -(>);Β Less thanΒ -(<);Β GreaterΒ thanΒ orΒ equal toΒ Β
β(β₯)
Β ;Β Less thanΒ orΒ equalΒ toΒ Β -
(β€)
Β Β
ForΒ inequalityΒ equation:Β IfΒ Β
πΒ >Β πΒ Β βΒ πΒ <Β πΒ ππΒ πΒ <Β πΒ Β βΒ πΒ >Β πΒ
Β
Rules forΒ solvingΒ inequalityΒ equations:Β
TheseΒ things doΒ notΒ affectΒ theΒ directionΒ ofΒ theΒ inequality:Β
-AddΒ (orΒ subtract)Β aΒ numberΒ fromΒ bothΒ sidesΒ
-Multiply (orΒ divide)Β bothΒ sides by aΒ positiveΒ numberΒ
-SimplifyΒ aΒ sideΒ
ButΒ theseΒ thingsΒ doΒ changeΒ theΒ directionΒ ofΒ theΒ inequalityΒ (''<''becomes''>''Β forΒ example):Β
-Multiply (orΒ divide)Β bothΒ sides by aΒ negativeΒ numberΒ
-SwappingΒ leftΒ andΒ rightΒ handΒ sidesΒ
Calculation:Β
TheΒ inequalityΒ equation:Β Β
β
1
2
πΒ β€Β 6
Β Β
Solve:Β
Β
β
1
2
πΒ β€Β 6
Β
Β
β2Β β Β β
1
2
πΒ β₯Β β2Β β Β 6
Β Β [Multiply bothΒ sides byΒ -2Β andΒ reverseΒ theΒ inequality]Β
Β
πΒ Β β₯Β β12
Β Β
SolutionΒ ofΒ theΒ inequalityΒ equationΒ isΒ Β
πΒ Β β₯Β β12
Β Β
SolutionΒ set:Β Β
[β12,Β β)
Β Β
AΒ closed,Β orΒ shaded,Β circleΒ is usedΒ toΒ representΒ theΒ inequalities greaterΒ thanΒ orΒ equalΒ toΒ Β (\geq)Β Β orΒ lessΒ
thanΒ orΒ equalΒ toΒ Β
(β€)
Β .Β TheΒ pointΒ is partΒ ofΒ theΒ solution.Β
SolutionΒ setΒ onΒ theΒ numberΒ line.Β
Β
Thus,Β theΒ solutionΒ ofΒ theΒ inequalityΒ equationΒ Β
β
1
2
πΒ β€Β 6πππΎππ πππΎπΒ β₯Β β12
Β .Β
Β
QuestionΒ #14Β
SolveΒ forΒ theΒ unknownΒ variableΒ fromΒ theΒ inequalityΒ equation.Β
Given:Β
TheΒ inequalityΒ equation:Β Β
πΒ +Β 2Β Β β₯Β 4
Β Β
Answer:Β
ConceptΒ used:Β
InΒ mathematics,Β anΒ inequalityΒ isΒ aΒ relationΒ whichΒ makes aΒ non-equalΒ comparisonΒ betweenΒ twoΒ numbers orΒ
otherΒ mathematicalΒ expression.Β ItΒ is usedΒ mostΒ oftenΒ toΒ compareΒ twoΒ numbers onΒ theΒ numberΒ lineΒ by theirΒ
size.Β
AnΒ inequalityΒ compares twoΒ values,Β showingΒ ifΒ oneΒ isΒ less than,Β greaterΒ than,Β orΒ simply notΒ equal toΒ
anotherΒ value.Β Β
πΒ Β β Β π
Β Β says thatΒ aΒ is notΒ equalΒ toΒ b.Β Β
πΒ <Β π
Β Β says thatΒ aΒ isΒ less thanΒ b.Β Β
πΒ >Β π
Β Β says thatΒ aΒ isΒ
greaterΒ thanΒ b.Β
ThereΒ areΒ fourΒ differentΒ typesΒ ofΒ inequalities:Β
GreaterΒ thanΒ -(>);Β Less thanΒ -(<);Β GreaterΒ thanΒ orΒ equal toΒ Β
β(β₯)
Β ;Β Less thanΒ orΒ equalΒ toΒ Β
β(β€)
Β Β
ForΒ inequalityΒ equation:Β IfΒ Β
πΒ >Β πΒ Β βΒ πΒ <Β πΒ ππΒ πΒ <Β πΒ Β βΒ πΒ >Β π
Β Β
Rules forΒ solvingΒ inequalityΒ equations:Β
TheseΒ things doΒ notΒ affectΒ theΒ directionΒ ofΒ theΒ inequality:Β
-AddΒ (orΒ subtract)Β aΒ numberΒ fromΒ bothΒ sidesΒ
-Multiply (orΒ divide)Β bothΒ sides by aΒ positiveΒ numberΒ
-SimplifyΒ aΒ sideΒ
ButΒ theseΒ thingsΒ doΒ changeΒ theΒ directionΒ ofΒ theΒ inequalityΒ (''<''becomes''>''Β forΒ example):Β
-Multiply (orΒ divide)Β bothΒ sides by aΒ negativeΒ numberΒ
-SwappingΒ leftΒ andΒ rightΒ handΒ sidesΒ
Calculation:Β
TheΒ inequalityΒ equation:Β Β
πΒ +Β 2Β Β β₯Β 4
Β Β
Solve:Β
Β πΒ +Β 2Β Β β₯Β 4
Β Β
Β
πΒ Β±Β 2Β Β β₯Β 4Β βΒ 2Β
Β [SubtractΒ 2Β fromΒ bothΒ sides]Β
Β
πΒ Β β₯Β 2
Β Β
AΒ closed,Β orΒ shaded,Β circleΒ is usedΒ toΒ representΒ theΒ inequalities greaterΒ thanΒ orΒ equalΒ toΒ Β
(β₯)
Β Β orΒ less thanΒ
orΒ equalΒ toΒ Β
(β€)
.Β TheΒ pointΒ is partΒ ofΒ theΒ solution.Β
SolutionΒ setΒ onΒ theΒ numberΒ lineΒ Β
Β
SolutionΒ ofΒ theΒ inequalityΒ equationΒ isΒ Β
πΒ Β β₯Β 2
Β Β
SolutionΒ set:Β Β
[2,Β β)
Β
Thus,Β theΒ solutionΒ ofΒ theΒ inequalityΒ equationΒ Β
πΒ +Β 2Β Β β₯Β 4Β ππ Β πΒ Β β₯Β 2Β
.
Β
QuestionΒ #15Β
SolveΒ forΒ theΒ unknownΒ variableΒ fromΒ theΒ inequalityΒ equation.Β
Given:Β
TheΒ inequalityΒ equation:Β Β
πΒ βΒ 8Β Β β€Β 7Β
Β
Answer:Β
ConceptΒ used:Β
InΒ mathematics,Β anΒ inequalityΒ isΒ aΒ relationΒ whichΒ makes aΒ non-equalΒ comparisonΒ betweenΒ twoΒ numbers orΒ
otherΒ mathematicalΒ expression.Β ItΒ is usedΒ mostΒ oftenΒ toΒ compareΒ twoΒ numbers onΒ theΒ numberΒ lineΒ by theirΒ
size.Β
AnΒ inequalityΒ compares twoΒ values,Β showingΒ ifΒ oneΒ isΒ less than,Β greaterΒ than,Β orΒ simply notΒ equal toΒ
anotherΒ value.Β Β
πΒ Β β Β π
Β Β saysΒ thatΒ aΒ is notΒ equalΒ toΒ b.Β Β
πΒ <Β π
Β Β says thatΒ aΒ isΒ less thanΒ b.Β Β
πΒ >Β π
Β Β says thatΒ aΒ isΒ
greaterΒ thanΒ b.Β
ThereΒ areΒ fourΒ differentΒ typesΒ ofΒ inequalities:Β
GreaterΒ thanΒ -(>);Β Less thanΒ -(<);Β GreaterΒ thanΒ orΒ equal toΒ Β
β(β₯)
Β ;Β Less thanΒ orΒ equalΒ toΒ Β
β(β€)
Β
ForΒ inequalityΒ equation:Β IfΒ Β
πΒ >Β πΒ Β βΒ πΒ <Β πΒ ππΒ πΒ <Β πΒ Β βΒ πΒ >Β π
Β Β
Rules forΒ solvingΒ inequalityΒ equations:Β
TheseΒ things doΒ notΒ affectΒ theΒ directionΒ ofΒ theΒ inequality:Β
-AddΒ (orΒ subtract)Β aΒ numberΒ fromΒ bothΒ sidesΒ
-Multiply (orΒ divide)Β bothΒ sides by aΒ positiveΒ numberΒ
-SimplifyΒ aΒ sideΒ
ButΒ theseΒ thingsΒ doΒ changeΒ theΒ directionΒ ofΒ theΒ inequalityΒ (''<''becomes''>''Β forΒ example):Β
-Multiply (orΒ divide)Β bothΒ sides by aΒ negativeΒ numberΒ
-SwappingΒ leftΒ andΒ rightΒ handΒ sidesΒ
Calculation:Β
TheΒ inequalityΒ equation:Β Β
πΒ βΒ 8Β Β β€Β 7
Β Β
Solve:Β
Β
πΒ βΒ 8Β Β β€Β 7
Β Β
Β
πΒ βΒ 8Β +Β 8Β Β β€Β 7Β +Β 8
Β Β [AddΒ 6Β toΒ bothΒ sides]Β
Β
πΒ Β β€Β 15
Β Β
AΒ closed,Β orΒ shaded,Β circleΒ is usedΒ toΒ representΒ theΒ inequalities greaterΒ thanΒ orΒ equalΒ toΒ Β
(β₯)
Β orΒ less thanΒ
orΒ equalΒ toΒ Β
(β€)
Β .Β TheΒ pointΒ is partΒ ofΒ theΒ solution.Β
SolutionΒ setΒ onΒ theΒ numberΒ lineΒ
Β
SolutionΒ ofΒ theΒ inequalityΒ equationΒ isΒ Β
πΒ Β β€
Β 15Β Β
SolutionΒ set:Β Β
(ββ,Β 15]
Β Β
Thus,Β theΒ solutionΒ ofΒ theΒ inequalityΒ equationΒ Β
πΒ βΒ 8Β Β β€Β 7Β ππ Β πΒ Β β€Β 15Β
.Β
QuestionΒ #16Β
SolveΒ forΒ theΒ unknownΒ variableΒ fromΒ theΒ inequalityΒ equation.Β
Given:Β
TheΒ inequalityΒ equation:Β Β
πΒ βΒ 6Β Β β₯Β 3
Β
Β
Answer:Β
ConceptΒ used:Β
InΒ mathematics,Β anΒ inequalityΒ isΒ aΒ relationΒ whichΒ makes aΒ non-equalΒ comparisonΒ betweenΒ twoΒ numbers orΒ
otherΒ mathematicalΒ expression.Β ItΒ is usedΒ mostΒ oftenΒ toΒ compareΒ twoΒ numbers onΒ theΒ numberΒ lineΒ by theirΒ
size.Β
AnΒ inequalityΒ compares twoΒ values,Β showingΒ ifΒ oneΒ isΒ less than,Β greaterΒ than,Β orΒ simply notΒ equal toΒ
anotherΒ value.Β Β
πΒ Β β Β π
Β Β says thatΒ aΒ is notΒ equalΒ toΒ b.Β Β
πΒ <Β π
Β Β says thatΒ aΒ isΒ less thanΒ b.Β Β
πΒ >Β π
Β Β says thatΒ aΒ isΒ
greaterΒ thanΒ b.Β
ThereΒ areΒ fourΒ differentΒ typesΒ ofΒ inequalities:Β
GreaterΒ thanΒ -(>);Β Less thanΒ -(<);Β GreaterΒ thanΒ orΒ equal toΒ
β(β₯)
;Β Less thanΒ orΒ equalΒ toΒ Β
β(β€)
Β Β
ForΒ inequalityΒ equation:Β IfΒ Β
πΒ >Β πΒ Β βΒ πΒ <Β πΒ ππΒ πΒ <Β πΒ Β βΒ πΒ >Β π
Β Β
Rules forΒ solvingΒ inequalityΒ equations:Β
TheseΒ things doΒ notΒ affectΒ theΒ directionΒ ofΒ theΒ inequality:Β
-AddΒ (orΒ subtract)Β aΒ numberΒ fromΒ bothΒ sidesΒ
-Multiply (orΒ divide)Β bothΒ sides by aΒ positiveΒ numberΒ
-SimplifyΒ aΒ sideΒ
ButΒ theseΒ thingsΒ doΒ changeΒ theΒ directionΒ ofΒ theΒ inequalityΒ (''<''becomes''>''Β forΒ example):Β
-Multiply (orΒ divide)Β bothΒ sides by aΒ negativeΒ numberΒ
-SwappingΒ leftΒ andΒ rightΒ handΒ sidesΒ
Calculation:Β
TheΒ inequalityΒ equation:Β Β
5
Β Β
Solve:Β
Β
πΒ βΒ 6Β Β β₯Β 3
Β Β
Β
πΒ βΒ 6Β +Β 6Β Β β₯Β 3Β +Β 6
Β Β [AddΒ 6Β toΒ bothΒ sides]Β
Β πΒ Β β₯Β 9Β
Β
AΒ closed,Β orΒ shaded,Β circleΒ is usedΒ toΒ representΒ theΒ inequalities greaterΒ thanΒ orΒ equalΒ toΒ Β
(β₯)
Β Β orΒ less thanΒ
orΒ equalΒ toΒ Β
(β€)
.Β TheΒ pointΒ is partΒ ofΒ theΒ solution.Β
SolutionΒ setΒ onΒ theΒ numberΒ lineΒ
Β
SolutionΒ ofΒ theΒ inequalityΒ equationΒ isΒ Β
πΒ Β β₯Β 9Β
Β
SolutionΒ set:Β Β
[9,Β β)
Β Β
Thus,Β theΒ solutionΒ ofΒ theΒ inequalityΒ equationΒ
Β πΒ βΒ 6Β Β β₯Β 3Β ππ Β πΒ Β β₯Β 9Β
.
Β
Β
QuestionΒ #17Β
SolveΒ forΒ theΒ unknownΒ variableΒ fromΒ theΒ inequalityΒ equation.Β
Given:Β
TheΒ inequalityΒ equation:Β Β
5Β +Β πΒ Β β€Β 1
Β
Β
Answer:Β
ConceptΒ used:Β
InΒ mathematics,Β anΒ inequalityΒ isΒ aΒ relationΒ whichΒ makes aΒ non-equalΒ comparisonΒ betweenΒ twoΒ numbers orΒ
otherΒ mathematicalΒ expression.Β ItΒ is usedΒ mostΒ oftenΒ toΒ compareΒ twoΒ numbers onΒ theΒ numberΒ lineΒ by theirΒ
size.Β
AnΒ inequalityΒ compares twoΒ values,Β showingΒ ifΒ oneΒ isΒ less than,Β greaterΒ than,Β orΒ simply notΒ equal toΒ
anotherΒ value.Β Β
πΒ Β β Β π
Β Β says thatΒ aΒ is notΒ equalΒ toΒ b.Β Β
πΒ <Β π
Β Β says thatΒ aΒ isΒ less thanΒ b.Β Β
πΒ >Β π
Β Β says thatΒ aΒ isΒ
greaterΒ thanΒ b.Β
ThereΒ areΒ fourΒ differentΒ typesΒ ofΒ inequalities:Β
GreaterΒ thanΒ -(>);Β Less thanΒ -(<);Β GreaterΒ thanΒ orΒ equal toΒ Β -(\geq)Β ;Β LessΒ thanΒ orΒ equalΒ toΒ
β(β€)
Β Β
ForΒ inequalityΒ equation:Β IfΒ Β
πΒ >Β πΒ Β βΒ πΒ <Β πβπβΒ <Β πΒ Β βΒ πΒ >Β π
Β Β
Rules forΒ solvingΒ inequalityΒ equations:Β
TheseΒ things doΒ notΒ affectΒ theΒ directionΒ ofΒ theΒ inequality:Β
-AddΒ (orΒ subtract)Β aΒ numberΒ fromΒ bothΒ sidesΒ
-Multiply (orΒ divide)Β bothΒ sides by aΒ positiveΒ numberΒ
-SimplifyΒ aΒ sideΒ
ButΒ theseΒ thingsΒ doΒ changeΒ theΒ directionΒ ofΒ theΒ inequalityΒ (''<''becomes''>''Β forΒ example):Β
-Multiply (orΒ divide)Β bothΒ sides by aΒ negativeΒ numberΒ
-SwappingΒ leftΒ andΒ rightΒ handΒ sidesΒ
Calculation:Β
TheΒ inequalityΒ equation:Β Β
5Β +Β πΒ Β β€Β 1
Β Β
Solve:Β
Β
5Β +Β πΒ Β β€Β 1
Β Β
Β
5Β βΒ 5Β +Β πΒ Β β€Β 1Β βΒ 5
Β Β [SubtractΒ 5Β fromΒ bothΒ sides]Β
Β
πΒ Β β€Β β4
Β Β
AΒ closed,Β orΒ shaded,Β circleΒ is usedΒ toΒ representΒ theΒ inequalities greaterΒ thanΒ orΒ equalΒ toΒ Β (\geq)Β Β orΒ lessΒ
thanΒ orΒ equalΒ toΒ Β
(β€)
Β .Β TheΒ pointΒ is partΒ ofΒ theΒ solution.Β
SolutionΒ setΒ onΒ theΒ numberΒ lineΒ
SolutionΒ ofΒ theΒ inequalityΒ equationΒ isΒ Β
πΒ Β β€Β β4
Β Β
SolutionΒ set:Β Β
(ββ,Β β4]
Β Β
Thus,Β theΒ solutionΒ ofΒ theΒ inequalityΒ equationΒ Β
5Β +Β πΒ Β β€Β 1Β ππ Β πΒ Β β€Β β4
Β .
Β
QuestionΒ #18Β
SolveΒ theΒ followingΒ quadratic expressions by factoring.Β FirstΒ writeΒ theΒ expressionsΒ inΒ completely factoredΒ
form.Β ThenΒ writeΒ theΒ realΒ numerΒ solutions.Β Β
[Hint:Β RememberΒ toΒ useΒ properΒ notationΒ whenΒ writingΒ theΒ realΒ numberΒ solutions.ForΒ example:Β ifΒ theΒ
solutions areΒ Β
π₯Β =Β 1
Β Β andΒ Β
π₯Β =Β 4
Β ,Β writeΒ Β
π₯Β =Β 1,4
Β .Β Β
1)Β Β
π₯
2
βΒ 5π₯Β =Β 0
Β Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?Β
2)Β Β
4π₯
2
+Β 8π₯Β =Β 0
Β Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?Β
3)Β Β
β7π₯
2
βΒ 21π₯Β =Β 0
Β Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?Β
4)Β Β
4π₯
2
βΒ 8π₯Β =Β 0
Β
FactoredΒ formΒ -Β ?Β
RealΒ numberΒ solutionsΒ -Β ?
Β
Answer:Β
1)Β Β
π₯
2
βΒ 5π₯Β =Β 0
Β Β
Β
π₯(π₯Β βΒ 5)Β =Β 0
Β
Β
π₯Β =Β 0Β πππΒ π₯Β =Β 5
Β Β
FactoredΒ formΒ Β
=Β π₯(π₯Β βΒ 5)
Β Β
realΒ numberΒ solutionΒ Β
=Β 0,5
Β Β
2)Β Β
4π₯
2
+Β 8π₯Β =Β 0
Β Β
Β
4π₯(π₯Β +Β 2)Β =Β 0
Β Β
Β
π₯Β =Β 0Β πππΒ π₯Β =Β β2Β
Β
FactoredΒ formΒ Β
=Β 4π₯(π₯Β +Β 2)
Β Β
realΒ numberΒ solutionΒ Β
=Β 0,Β β2
Β Β
3)Β Β
β7π₯
2
βΒ 21π₯Β =Β 0
Β Β
Β
β7π₯(π₯Β +Β 3)Β =Β 0
Β Β
Β
π₯Β =Β 0Β πππΒ π₯Β =Β β3Β
Β
FactoredΒ formΒ Β
=Β β7π₯(π₯Β +Β 3)
Β Β
realΒ numberΒ solutionΒ Β
=Β 0,Β β3
Β Β
4)Β Β 4
π₯
2
βΒ 8π₯Β =Β 0
Β Β
Β
4π₯(π₯Β βΒ 2)Β =Β 0
Β Β
Β
π₯Β =Β 0Β πππΒ π₯Β =Β 2Β
Β
FactoredΒ formΒ Β
=Β 4π₯(π₯Β βΒ 2)
Β
realΒ numberΒ solutionΒ Β
=Β 0,2
Β
Β
QuestionΒ #19Β
TheΒ reducedΒ row-echelonΒ formΒ ofΒ theΒ augmentedΒ matrixΒ forΒ aΒ systemΒ ofΒ linearΒ equations withΒ variables Β
π₯
1
,Β β―Β ,Β π₯
4
Β Β is givenΒ below.Β DetermineΒ theΒ solutions forΒ theΒ systemΒ andΒ enterΒ themΒ below.Β Β
1
0
1
5
|Β 0
0
1
3
3Β |Β 1
β
β
ο©
οΉ
οͺ
οΊ
β
β
ο«
ο»
Β Β IfΒ theΒ systemΒ has infinitely many solutions,Β selectΒ ''TheΒ systemΒ has atΒ leastΒ oneΒ
solution''.Β YourΒ answerΒ may useΒ expressionsΒ involvingΒ theΒ parameters r,Β s,Β andΒ t.Β Β
TheΒ systemΒ has atΒ leastΒ oneΒ solutionΒ Β
π₯
1
=Β 0
Β Β
π₯
2
=Β 0
Β Β
π₯
3
=Β 0
Β Β
π₯
4
=Β 0
Β
Β
Answer:Β
StepΒ 1Β
SinceΒ rank(augmentedΒ matrix)<numberΒ ofΒ column.Β Implies systemΒ has infinitely many solution.Β
StepΒ 2Β
GivenΒ row-reducedΒ formΒ atΒ augmentedΒ matrixΒ
Β
1
0
1
5
|Β 0
0
1
3
3Β |Β 1
β
β
ο©
οΉ
οͺ
οΊ
β
β
ο«
ο»
Β
SoΒ considerΒ systemΒ ofΒ linearΒ equationΒ withΒ variableΒ Β
π₯
1
,Β π₯
2
,Β π₯
3
,Β πππΒ π₯
4
.Β
Β
1
2
3
4
1
0
1
5
0
0
1
3
3
1
x
x
x
x
ο©Β οΉ
οͺΒ οΊ
β
β
ο©
οΉ
ο©Β οΉ
οͺΒ οΊΒ =
οͺ
οΊ
οͺΒ οΊ
οͺΒ οΊ
β
ο«
ο»
ο«Β ο»
οͺΒ οΊ
ο«Β ο»
Β Β
SystemΒ has intinitely many solutionΒ asΒ rank ofΒ augmentedΒ matrixΒ is 2Β whichΒ isΒ strictlyΒ less thanΒ numberΒ ofΒ
(4)Β columnΒ
Β
π₯
1
βΒ π₯
3
βΒ 5π₯
4
=Β 0
Β Β
Β
π₯
2
+Β 3π₯
3
βΒ 3π₯
4
=Β 0
Β Β
SoΒ solutionΒ setΒ Β
1
2
3
4
1
3
4
2
3
(Β ,
,
,
)
5
0
{
}
3
3
0
d
xΒ xΒ xΒ x
x
x
x
S
x
x
x
βΒ β
=
=
+
β
=
Β
Β
1
2
3
4
1
3
4
4
2
3
(Β ,
,
,
)
5
0
{
}
1
(
3Β )
3
xΒ xΒ xΒ x
x
x
x
and
S
x
x
x
=
β
=
=
=
+
Β Β
Β
1
2
3
4
1
3
2
3
5
{(Β ,
,
,
)
5Β }
3
S
xΒ xΒ xΒ x
x
x
x
x
=
=Β β
β
Β
Β
1
2
3
4
1
2
3
{(Β ,
,
,
)
3
5
12Β }
S
xΒ xΒ xΒ x
x
x
x
=
=Β β
β
Β
Β
1
2
3
4
1
2
3
{(Β ,
,
,
)
3
5
12
0}
S
xΒ xΒ xΒ x
x
x
x
=
+
+
=
Β
thenΒ surelyΒ Β
π₯
1
=Β 0,Β π₯
2
=Β 0,Β π₯
3
=Β 0,Β π₯
4
=Β 0
Β Β is oneΒ atΒ theΒ solutionΒ andΒ calledΒ trivialΒ solution.Β
QuestionΒ #20Β
GraphΒ theΒ solutionΒ setΒ forΒ eachΒ compoundΒ inequality,Β andΒ express theΒ solutionΒ setsΒ inΒ intervalΒ notation.Β Β
π₯Β >Β 0
Β Β andΒ Β
π₯Β β»Β 1
Β
Β
Answer:Β
StepΒ 1Β
TheΒ givenΒ compoundΒ inequalityΒ is,Β
Β
π₯Β >Β 0Β β―Β β―Β (π)
Β Β
Β
π₯Β β»Β 1Β β―Β β―Β (π)
Β Β
ToΒ findΒ theΒ solutionΒ ofΒ theΒ compoundΒ inequality,Β weΒ findΒ theΒ commonΒ setΒ ofΒ solutions whichΒ satisfiesΒ bothΒ
theΒ inequality.Β
ForΒ theΒ firstΒ inequality,Β theΒ valueΒ ofΒ
π₯
Β shouldΒ beΒ greaterΒ thanΒ
0
Β andΒ forΒ theΒ secondΒ inequality,Β theΒ valueΒ ofΒ
π₯
Β shouldΒ beΒ
β1
.Β
IfΒ weΒ takeΒ
π
Β valueΒ ofΒ
π₯
Β as greater
Β 0
. thenΒ itΒ willΒ satisfy bothΒ theΒ equations.Β
StepΒ 2Β
Hence,Β theΒ graphΒ ofΒ theΒ twoΒ inequalitiesΒ areΒ as,Β
Β
StepΒ 3Β
Therefore,Β theΒ requiredΒ solutionΒ setΒ is as,Β
Β
π₯Β βΒ (0,Β β)
Β Β Β
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