maanvco08unit+three+algebraic+methodsch6 7-8-11

8/3/2019 MaANVCO08Unit+Three+Algebraic+MethodsCh6 7-8-11

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Math A/B, Unit Three: Algebraic Methods; Unit 5: Graphs and Transformations Ch 6, 7, 8, 11NV-College

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Selected Solved Problems:

G.MaA.6.2.2-6.2.8

G.MaA.6.2.9-6.2.19

G.MaA.6.2.19-6.2.26

V.MaA.6.2.27 G.MaA.6.3.1-6.3.3

G.MaA.6.3.4-6.3.15

G.MaA.6.3.16-

6.3.17

V.MaA.6.4.22-

6.4.27

G.MaA.6.5.3-

6.5.7

G.MaA.6.5.8-

6.5.10

V.MaA.6.5.11-

6.5.14

V.MaA.6.5.15

V.MaA.6.5.18-6.5.20

V.MaA.6.5.20 V.MaA.6.6.1-2 V.MaA.6.6.3-4 V.MaA.6.6.5-6 M.MaA.6.6.7-8

M.MaA.6.6.9-

6.6.10

V.MaA.6.6.11 M.MaA.6.6.12.

G7.5.1 G.MaA.7.5.3 G.MaA.7.5.4 V.MaA.7.5.5 V.MaA.7.5.6 M.MaA.7.5.7

M.MaA.7.5.8 M.MaA.7.5.9 V.MaA.7.5.10 V.MaA.7.5.11 V.MaA.7.5.12 V.MaA.7.5.13

V.MaA.7.5.14

V.MaA.8.1.26 G.MaA.8.2.17 G.MaA.8.2.10 G.MaA.8.2.15

M.MaA.11.1.14 G.MaA.11.2.8 M.MaA.11.2.13 M.MaA.11.2.14

G.MaA.11.3.1-3 G.MaA.11.3.4 G.MaA.11.3.5 G.MaA.11.3.6

G.MaA.11.3.7 G.MaA.11.3.8 G.MaA.11.3.9 M.MaA.11.3.10 M.MaA.11.3.11 M.MaA.11.3.12

G.MaA.11.4.1-4

M.MaA.11.4.18 M.MaA.11.4.19 V.MaA.11.4.20


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Summery: Unit Three: Algebraic Methods:

Ch6 Polynomial Arithmetic and Factoring

Conjugate Binomials ( )( ) 22 BABABA =+

Multiplication Monomials:To multiply two monomials, group like

factors together. Then multiply the like

factors

( ) ( )

=

2

345

2

435

6

512

6

512

y

yxx

y

xyx

( ) ( )( )( ) yxyxy

xyx 99

2

435 1010

6

512 ==

Dividing Monomials:To divide a monomial by another

monomial, group the quotients of like

factors together. Then divide the like

factors:

=

7

3

2

4

72

34

3

15

3

15

y

y

x

x

yx

yx

( )( )4

2

4

2

72

34 515

3

15

y

x

yx

yx

yx =

=

Multiplying a Polynomial by a Monomial:

To multiply a polynomial by a monomial,

multiply each term of the polynomial by

the monomial.

( )( )=+ 4352 2435 yxyx 355539 8610 yxyxyx +=

( ) xxxx 1015235 2 =

Dividing a Polynomial by a Monomial:

To Divide a polynomial by a monomial,

divide each term of the polynomial by the

monomial.

=+

22

4334

3

639

yx

yyxx

22

4

22

33

22

4

3

6

3

3

3

9

yx

y

yx

yx

yx

x+=

2

2

2

2 23

x

yxy

y

x+=

Multiplying a Polynomial by a Polynomial: ( )( ) ( ) ( ) == 2372352375 xxxxx 14211015 2 += xxx

143115 2 += xx

Factoring, Factorization ( )2351015 2 = xxxx Factoring the difference between two

squares

( )( )BABABA += 22

Quadratic Binomial ( ) 222 2 BABABA ++=+

( ) 222 2 BABABA += Factoring ( )222 2 BABABA +=++

( )222 2 BABABA =+ Finding the greatest Common Factor: GCF:

The greatest common factor of a

polynomial is the greatest monomials that

all terms of the polynomial share.

Factoring a Polynomial by Removing the GCF:

123551015 223 +=+ yxxyxyxyyx

xy5 is the GCF of xyxyyx 51015 23 + .

Factorizing qpxx ++2 ( )( )bxaxqpxx ++=++2


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Think of two numbers a and b that:

when multiplied together gives q . i.e.:

qba =.

when added together gives p . i.e.:

pba=+

qba =.

pba =+

Ex.: Four different cases:

( )( )34122 += xxxx

( )( )34122

+=+ xxxx ( )( )341272 ++=++ xxxx ( )( )341272 =+ xxxx

Factor Completely:

A polynomial is factored completely when

each of its factors cannot be factored

further.

Ex.:

( )( )11818728 23

+=

=

yyxy

yxyxyxy

Quadratic Equations: 02 =++ qpxx

1x and 2x are roots or solutions of the

quadratic equation:

pxx =+ 21

qxx =21.

02 =++ qpxx

qpp

x

+=

2

1

22

qpp

x

=

2

222

Ch 7: Operating With Algebraic Fractions

Simplifying Algebraic fractions:7.1 Writing Algebraic Fractions in Lowest

Terms:

Ex. Simplify 6144

6242

3

++

xx

xyyx

1.( )

( )3722146

6144

6242

2

2

3

++

=

++

xx

xxy

xx

xyyx

( )( )( )( )3122

121232

6144

6242

3

++/

+/=

++

xx

xxxy

xx

xyyx

2.( )

( )3123

6144

6242

3

+

=

++

x

xxy

xx

xyyx

3.

3

36

6144

624 2

2

3

+

=

++

x

xyyx

xx

xyyx

1. Factor the numerator and denominator.2. Cancel out any common factors that are

found in the numerator and denominator.

3. Multiply the remaining factors in the

numerator and denominator.

7.2 Multiplying and Dividing Algebraic

Fractions:

Express672

4

32

442

22

+

++

xx

x

x

xxin the

simplest possible terms.

672

4

32

442

22

+

++

xx

x

x

xx

4

672

32

442

22

+

++

x

xx

x

xx

1.Factor where possible.2.Cancel out where possible.3.Multiply the remaining factors.


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( )( )

( )( )( )( )22

232

32

22

+

+

xx

xx

x

x

( )

21

1

1

2+=

+x

x

2672

432

442

22

+=+

++ xxx

xx

xx

7.3 Converting Units of Measurement:

Multiplication Property of ConversionFactors.

Finding a Conversion Factor.

Multiplication Property of ConversionFactors:

( ) sh

shh 00018

360055 =

//=

Finding a Conversion Factor:

////

/

//

//

/

///=

s

h

mk

m

h

mkhkm

0063

1001009/90

4

10

sms

mhkm /25

4

100/90 ==

7.4 Adding and Subtracting Algebraic

Fractions :

Combining Algebraic FractionsHaving the Same Denominator: Just

combine the numerators over the

common denominator.

Combining Algebraic FractionsHaving Different Denominators: First

determine LCD (Lowest Common

Denominator) of the fractions. Then

multiply and divide each fractions by

LCD. Cancel out the common terms

wherever possible. Finally combine (add

or subtract) the numerators over LCD.

Combining Algebraic Fractions Havingthe Same Denominator:

( )yxy

xx

yxy

x

yxy

x

25

5237

25

52

25

37

++=

+

+

yxy

xx

yxy

x

yxy

x

25

5237

25

52

25

37

+=

+

( )

( ) yxy

x

yxy

x

yxy

x 1

25

25

25

52

25

37=

=

+

Combining Algebraic Fractions HavingDifferent Denominators:

( )243

4

7

84

3

4

7

+=

+

xxxx

( )24 += xxLCD

+

+

+=

+

x

x

xx

x

xxx 84

3

2

2

4

7

84

3

4

7

( )

( )24

327

84

3

4

7

+

+=

+

xx

xx

xx

( )243147

84

3

4

7

+

+=

+

xx

xx

xx

( )( )

( )2472

24

144

84

3

4

7

+

+=

+

+=

+

xx

x

xx

x

xx

( )( ) xx

x

xx

x

xx 22

7

22

7

84

3

4

72 +

+=

+

+=

+

7.5 Solving Equations With Fractions: Change the equation into an equivalent

equation that does not have any fraction. To eliminate the fractions, multiply each

M7.5.9 Solvey

yy

y 4

1

8

11 =

+ .

LCD is y8 . Multiply both sides by LCD,


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side of the equation by the LCD of all

fractional terms.

Simplify the equation. Solve the equation

G7.5.1 Solve 532

nn=

.

LCD is 10 . Multiply both sides by LCD,i.e. 10:

510310

210

nn=

Simplify the equation: nn 2305 = 02305 = nn 0303 =n

Solve the equation: 303 =n

3

30=n Answer: 10=n

Check:5

32

nn=

5

103

2

10=

235 = Checks!

i.e. y8 . Simplify it

y

yy

yy

yy

//

//=

/

+/

//

4

18

8

18

18

2

( ) ( )1218 =+ yyy

2282

= yyy 0822 2 =++ yyy

01032 =+ yy

Find two numbers a and b that

when multiplied together gives 10 . i.e.:10. =ba

when added together gives 3 : 3=+ ba Due to the fact that 10. =ba , a and b

must have different signs. Due to the fact

that their sum is positive 3=+ ba , thelarger number should be positive.Therefore: 5=a and 2=b

( )( ) 0251032 =+=+ yyyy 05 =+y 51 =y ; 02 =y 22 =y

Ch 8: Radicals and Right Triangles

A square root of a nonnegative numberN

is one of two equal numbers whose productis N:

NNN =

) ) NNN = Ex: Nx =2 Nx =

Ex: 642 =x 864 ==x

81 =x and 82 =x

64 is called radicand

82 =x is called principal square root

Rational Numbers:A number that can be

expressed as a fraction with an integer

numerator and a nonzerinteger numeratoris called a rational number.

m

nx = is rational if both n and m are integer

numbers.

Perfect Square:Any number whose square

root is rational is called a perfect square.121 is a perfect square because 11121 =

169

625is a perfect square because

13

25

169

625=

Irrational Numbers:A real number with

nonrepeating decimal number that do not

end re called irrational number.

097801688724223730950481.414213562 =

059527446341575688772931.732050803 =

705119221267454639892933.6055512713 =

176746211414917961047219.8488578097 =


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Simplifying Radicals: Simplify by factoring and cancelling. Remove any perfect squares from the

radical.

Factor radicand, if possible, so that oneof the factors is a perfect square.

Write the radical over each factor. Evaluate the square root of the perfect

squares(s)

V.MaA.8.1.26. Simplify:63

25

3

54

yx

yx

291818

3

54224

2

63

25

===y

x

y

x

y

x

yx

yx

263

25

233

54

y

x

yx

yx=

Solving quadratic equations with no

second term, no x-term:

Solve for 2x Take the square root of both sides of the

question.

Do not forget the negative term:

Solve: 16992 =x .

Solve for 2x :9

1692 =x

Take the square root of both sides of the

question.

3

13

9

169==x

Do not forget the negative term:

3

131 =x and

3

132 =x

8.2 Arithmatics operations with

Radicals

Square root radicals such as 57 and

53 are called like radicals.

Adding and Subtracting Radicals: A radical may be first simplified before

it is combined with another radical

Like radicals may be added orsubtracted:

( ) bcabcba +=+

( ) bcabcba =

5105357 =+

545357 =

G.MaA.8.2.15: Perform the operation and give

the answer in the simplest possible radical

form: 4812275 + .Suggested solution:

3163423254812275 +=+

34322354812275 +=+

354812275 =+

Multiplying Radicals:

( ) xyabybxa =.G.MaA.8.2.17: Perform the operation and give


form: )823.2 .Suggested solution:) 164328223823.2 ==

) 246423823.2 ===

Dividing Radicals:

y

x

b

a

yb

xa=

G.MaA.8.2.10: Perform the operation and give


form: ( )253 .Suggested solution:

( ) ( ) 45595353 222 ===


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Squaring Radicals eliminates the

radical. ( ) NN =2

Rationalizing Denominator:If a radical appears in the denominator

of a fraction,cb

a, the denominator is

irrational. To rationalize the

denominator incb

awe may multiply

the fraction byc

c:

bc

ca

c

c

cb

a

cb

a==

Unit 5: Graphs and Transformations:

(K4/4000 Grafer och Funktioner, K2/3000)

Ch 11: Linear Equations and Their Graphs

Locating Points in the coordinate system: A point is located by writing an ordered pair

of numbers of the form )yx aa , where xa is

the x-coordinate of the point, andya is the y-

coordinate of the point

xa : x-coordinate of the point gives the

horizontal distance and direction from the

origin.

When xa is positive, the point lies to the

right of the origin When xa is negative, the point lies to the

left of the origin

ya : y-coordinate of the point gives the

vertical distance and direction from the

origin.

Whenya is positive, the point lies above

the origin. When ya is negative, the point lies below

the origin.

Finding Area Using Coordinates


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11.2 Midpoint and Distance FormulasThe length of the line segment that connects

points )yx aaA , and )yx bbB , is:

( ) ( )22

yyxx ababAB += The coordinates of midpoint of the segment

AB is :

++

2,

2

yyxxbaba

M

G.MaA.11.2.8: Find the area and

circumference of a circle if the end points of

a diameter are ( )7,1A and ( )17,9 B .Find the coordinate of the circle. The unit

length is cm00.1

Suggested solution:

The length of the diameter of the circle is

( ) ( )22 yyxx ababABd +==

( )( ) ( )22 71719 +== ABd

cmABd 0.26241022 =+==

cmcmd

r 0.132

0.26

2===

( ) 222 16913 cmrArea === 22

531169 cmcmArea =

cmcmdC 7.8126 ==

cmcmC 7.8126 =

We may use

++

2,

2

yyxxbaba

M to find

the coordinates of the centre of the circle:

+

2

177

,2

91

O or ( )5,4

O

Ch 11: Linear Equations and Their Graphs

11.3 Slope of a lineIn the linear equation mxky +=

k is the slope of the line:

12

12

xx

yyk

=

and m is the y-intercept, i.e. the point the

graph of the function crosses the y-axes.

Positive slope: If as x-increases, y-increases:

52 += xy

52 = xy

Negative slope: If as x-increases, y-decreases

52 += xy

52 = xy

The gradient or slope of a line is a measure

of how fast the line is rising or falling. It is

defined as the rate at which y increases

compared with x between any two points on

the line.

Example.Find the equation of the line joining the

points ( )3,2A and ( )9,5B .The slope (gradient) of the line is

23

6

25

39==

=

=

AB

ABAB

xx

yyk

Note that the order of the points of no

importance, as long as the coordinates of a

given point are both either first or second.

i.e.:


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Slope of parallel lines:Parallel lines share identical slope:

Ex. 52 = xy and 52 += xy are parallel.

If CDAB thenCDAB

kk =

Slope of perpendicular lines:Slope of perpendicular lines are negative

reciprocals:

Ex. 52 = xy and 52

1+= xy are.

perpendicular.

If CDAB thenCD

AB kk

1=

Collinear Points:Collinear Points are points that lie on the

same line.

23

6

52

93=

=

=

=

BA

BAAB

xx

yyk

The equation of the line is:

mxky += mxy += 2

The y-intercept of the line may be foundusing the fact that the line passes through

both ( )3,2A and ( )9,5B and therefore,their coordinates must satisfy the equation of

the line, i.e.:

mxy += 2 passes through ( )3,2A :143223 ==+= mm :

The equation of the line is: 12 = xy

The coordinates of ( )9,5B mustsatisfy the equation of the line

12 = xy : 91101529 ===


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G.MaA.6.2.2 Find the quotient ofxy

yx

13

26 24.

Suggested Solution: yxy

y

x

x

xy

yx 32424

213

26

13

26==

G.MaA.6.2.3 Find the product of 23xy and 325 yx

Suggested Solution: ( )( ) ( )( )( ) 53322322 155353 yxyyxxyxxy ==

G.MaA.6.2.4 Find the product of ( )3245.1 xy and 36.2 xy ?Suggested Solution: ( ) 6332 45.145.1 yxxy =

( )( ( ) ( )( ) ( )( )( ) 94363363332 77.36.245.16.245.16.245.1 yxyyxxxyyxxyxy ===

G.MaA.6.2.5 Find the quotient of xxx 73514 23 + divided by x7 where 0x .

Suggested Solution: 1527

7

7

35

7

14

7

73514 22323

+=+=+

xxx

x

x

x

x

x

x

xxx

G.MaA.6.2.6. Martin had a garden that was in the shape of a rectangle. Its length was twice its

width. He decided to make a new garden that was 2 meters longer and 2 meters wider than the

first garden. Ifx represents the original width of the garden, find the difference between the

area of the new garden and the area of the original garden?

Suggested Solution: Answer: 246 mxAA originalnew +=

The original width of the garden: mx

The original length of the garden: mx2 The original area of the garden: 2222 mxxxAoriginal ==

The new width of the garden: mx 2+

The new length of the garden: mx 22 +

The new area of the garden: ( ) ( ) 2222 mxxAnew ++=

( ) ( ) ( ) ( ) 46242422222222 2222 ++=+++=+++=++= xxxxxmxxxmxxAnew The difference between the area of the new garden and the area of theoriginal garden:

( ) 2222 462462 mxmxxxAA originalnew +=++=

G.MaA.6.2.7 Find the product of

ba2

2

1and 38ab

Suggested Solution: ( ) ( ) ( )( ) 433232 482

18

2

1babbaaabba =

=

G.MaA.6.2.8 Find the product of x2 , 23x , and 34x .

Suggested Solution: ( )( )( ) 63232 2424432 xxxxxxx ==


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G.MaA.6.2.9 Find the product of 34.0 y , and215.0 xy .

Suggested Solution: ( )( ) ( )( )( ) ( ) 52323 6.015.04.015.04.0 xyyyxxyy ==

G.MaA.6.2.10 Find the quotient of 2xy divided by yx3 where 0x and 0y .

Suggested Solution:22

2

33

2

1xyy

xyy

xx

yxxy ===

G.MaA.6.2.11 Find the quotient of 53x divided by 427x where 0x .


1

27

3

27

34

5

4

5x

xx

x

x

x==

=

G.MaA.6.2.12 Find the quotient of 328 ba divided by5212 ba where 0a and 0b .


3

2

2

52

32

3

211

3

2

12

8

12

8

bbb

b

a

a

ba

ba===

G.MaA.6.2.13 Find 4235 35.005.1 yxyx where 0x and 0y .

Suggested Solution:y

x

yx

y

y

x

x

yx

yxyxyx

33

4

3

2

5

42

354235 313

35.0

05.1

35.0

05.135.005.1 ====

G.MaA.6.2.14 Find

236

1

2

1

yxxywhere 0x and 0y .

Suggested Solution: yx

y

y

x

x

xy

yx

yxxy

22323

233

2

6

2

6

6

1

2

1===

G.MaA.6.2.15 Find the product of y5 and 483 yy

Suggested Solution: ( )( ) yyyyyyyyyyy 2040545855485 2433 ==

G.MaA.6.2.16 Find the product of 73 +x , and 92 x .

Suggested Solution: Answer: ( )( ) 631369273 2 =+ xxxx ( )( ) ( ) ( ) 63142769279239273 2 +=+=+ xxxxxxxx

G.MaA.6.2.17 Find the product of 85 w , and 85 +w .

Suggested Solution: ( )( ) ( ) 6425858585 222 ==+ wwww

G.MaA.6.2.18 Find the product of 34 b , and 2+b .

Suggested Solution:

( )( ) ( ) ( ) 65463842324234 22 +=+=++=+ bbbbbbbbbb

G.MaA.6.2.19 Find the product of 13.0 2 +y , and 13.0 2 y .

Suggested Solution:

( )( ) ( ) 109.013.013.013.0 42222 ==++ yyyy


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G.MaA.6.2.19 Find the product of 13.0 2 +y , and 13.0 2 y .

Suggested Solution:

( )( ) ( ) 109.013.013.013.0 42222 ==++ yyyy

G.MaA.6.2.20. Find the product of x31 , and x+2 .Suggested Solution:

( )( ) ( ) ( ) 253362232231 22 +=+=++=+ xxxxxxxxxx

G.MaA.6.2.21. Find ( )232 x .

Suggested Solution: ( ) ( ) ( )( ) 91243232232 2222 +=+= xxxxx

G.MaA.6.2.22. Find the quotientrs

srr

9

2718 234 .

Suggested Solution: srsr

rssr

rsr

rssrr 2

3234234

329

279

189

2718 ==

G.MaA.6.2.23. Find the quotientc

ccc

3

31221 23

+.


3

3

12

3

21

3

31221 22323

+=

+

+

=

+cc

c

c

c

c

c

c

c

ccc

G.MaA.6.2.24. Find the quotienta

baa

7.0

05.14.0 23 .


3

7

4

7.0

05.1

7.0

4.0

7.0

05.14.0 22323 aba

a

ba

a

a

a

baa==

G.MaA.6.2.25. How does the square of a number,x , compare to the product obtained by

multiplying together the two numbers obtained by increasing x by 6 and decreasing x by 6.

Suggested Solution: ( )( ) 3666 2 =+ xxx

V.MaA.6.2.26. Martina claims that the sum of any five consecutive integers is always evenly

divisible by 5. Explain why you agree with Martina.

Suggested Solution:If the number is the middle numberx , the other four consecutive numbers

in the neighbourhood ofx are 2x , 1x , 1+x , 2+x , and the sum of five

consecutive numbers may be calculated as:( ) ( ) ( ) ( ) xxxxxxS 52112 =++++++= xS 5= is always divisible by 5.


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V.MaA.6.2.27. In the accompanying diagram, the width of the inner rectangle is represented

by x and the length by 12 x . The widthof the outer rectangle is represented by

3+x and the length by 5+x .a) Express the area of the shaded region

as a trinomial in terms ofx . b) If the perimeter of the outer rectangle

is 24, what is the value ofx .

c) What is the area, in square units, of

the shaded region?

d) The area of the inner rectangle is what

percent, correct to the nearest tenth of

a percent, of the area of the outer

rectangle?

Suggested Solution:

a) Answer: The area of the shaded region 1592 ++= xxAshaded

( )( ) ( ) 15921581253 222 ++=+++=++= xxxxxxxxxxAshaded b) Answer: 2=x .

( ) ( ) 22

88162442410262245232 =====+++=+++= xxxxxxP

c) Answer: auAshaded 29=

( ) ( ) auxxAshaded 29151841529215922 =++=++=++=

d) Answer: The area of the inner rectangle is %17 of the area of the

outer rectangle.

( ) ( ) ( ) auxxxxAinner 62822221222 =====

( )( ) ( ) ( ) auxxxxAouter 3515164152821585322 =++=++=++=++=

%1717.035

6===

outer

inner

A

A

G.MaA.6.3.1. If prtpA += , find p in terms of A, r and t.

Suggested Solution:

prtpA += ( )rtpA += 1 ( ) Artp =+1

( )rt

Ap

+=

1

rt

Ap

+=

1

G.MaA.6.3.2. If bydcay += , find y in terms of a, b, c, and d.

Suggested Solution:

bydcay += cdbyay += ( ) dcbay += ( )ba

dcy

+=

ba

dcy

+=

G.MaA.6.3.3. Factor xx 6152

Suggested Solution: ( )253615 2 = xxxx

12 x

x3+x

5+x


14/56



G.MaA.6.3.4. Factor 22 77 qp +

Suggested Solution: ( )2222 777 qpqp +=+

G.MaA.6.3.5. Factor xxx ++ 23


++=++ xxxxxx

G.MaA.6.3.6. Factor 357 1263 yyy +

Suggested Solution: ( )4231263 243357 +=+ yyyyyy

G.MaA.6.3.7. Factor ba 44

Suggested Solution: ( )baba += 444

G.MaA.6.3.8. Factor 5225 408 wuwu

Suggested Solution: 33225225 58408 wuwuwuwu =

G.MaA.6.3.9. Factor 1442 y

Suggested Solution: ( )( )12121442 += yyy

G.MaA.6.3.10. Factor xyx 36.024.0 2

Suggested Solution: ( )yxxxyx 6.046.036.024.0 2 =

G.MaA.6.3.11. Factor

32

4

3

4

1

abba

Suggested Solution: ( )2324

1

4

3

4

1baababba =

G.MaA.6.3.12. Factor 281 x

Suggested Solution: ( )( )xxxx +== 99981 222

G.MaA.6.3.13. Factor9

12 p

Suggested Solution:

+

=

=

31

31

31

91

2

22 pppp

G.MaA.6.3.14. Factor 36.02 b

Suggested Solution: ( )( )60.060.036.02 += bbb

G.MaA.6.3.15. Factor 19

4 2 c

Suggested Solution:

+

=

= 1

3

21

3

21

3

21

9

42

2cccc


15/56



G.MaA.6.3.16. Factor 22 49100 ba

Suggested Solution: ( )( )bababa 71071049100 22 +=

V.MaA.6.3.17. Factor ( ) ( )1714 2 xxx

Suggested Solution: ( ) ( ) ( ) 741171422

= xxxxx


16/56



V.MaA.6.4.22 Factor completely xyxy 7283 .

Suggested Solution: ( ) ( )( )33898728 23 +== yyxyyxyxyxy

V.MaA.6.4.23 Factor completely 20142

2 yy .

Suggested Solution: ( )( )522107220142 22 ++=++= yyyyyy Think of two numbers a and b that:

when multiplied together gives 10 . i.e.: 10. =ba , both numbers have the same sign. when added together gives 1. i.e.: 7=+ ba : Both numbers are positive. 2=a , 5=b

V.MaA.6.4.24 Factor completely xxx 11222 23 + .

Suggested Solution: ( )( )78256211222 223 +=+=+ xxxxxxxxx

Think of two numbers a and b that: when multiplied together gives -56. i.e.: 56. =ba , one is positive the other negative when added together gives 1. i.e.: 1=+ ba : The larger one is positive. 8=a , 7=b

V.MaA.6.4.25 Factor completely 234 5005010 yyy + .

Suggested Solution: ( ) ( )( )51010505105005010 222234 +=+=+ yyyyyyyyy Think of two numbers a and b that:

when multiplied together gives 50 . i.e.: 50. =ba , one is positive the other negative when added together gives 1. i.e.: 5=+ ba : The larger one is positive. 10=a , 5=b

V.MaA.6.4.26 Factor completely 23 5018 xyx

Suggested Solution: ( )( )yxyxxyxxxyx 5353225925018 2223 +==

V.MaA.6.4.27 Factor completely xx 182

1 3

Suggested Solution:

+

=

= 3

213

2129

41218

21 23 xxxxxxx


17/56



6.5 Solving Quadratic Equations by Factoring

G.MaA.6.5.3 If one of the roots of the equation 02 =+ qxx is 3, what is the other root?

(a) -2 (b) 2 (c) -1 (d) -4

Suggested Solution: Answer: alternative (a): 22 =x

If we name the other root 2x , it should satisfy the condition that:( )132 =+x 132 =+x 312 =x 22 =x

G.MaA.6.5.4 For what value(s) x is the fraction32

42

+

xx

xnot defined?

(a) -1, 3 (b) 1, -3 (c) -3, -1 (d) -4

Suggested Solution: Answer: alternative (a): 3x and 1x The fraction is not defined if its denominator is zero. Therefore, first wefactorize the denominator. Then cancel out all terms that could be so.Finally do not allow the denominator to be zero. i.e.:

324

2 +xx

x ( )( )13

4+

+xx

x 3x and 1x . Therefore, alternative (a)

G.MaA.6.5.5 Solve ( ) 012 =xx .

Suggested Solution: Answer: 01 =x and2

12 =x

( ) 012 =xx

====

==

2

1

2

112012

00

2

1

xxxx

xx

G.MaA.6.5.6 Solve 0232 =++ yy .

Suggested Solution: Answer: 11 =y and 22 =y

0232 =++ yy ( )( ) 021 =++ yy

===+

==+

2202

101

2

1

yyy

yy

First we looked for two numbers banda such that 2=ab

Then we chose those numbers that satisfied 3=+ ba These two numbers are 1=a and 2=b


18/56



G.MaA.6.5.8 Solve 122 += xx .

Suggested Solution: Answer: 41 =x and 32 =x

122 += xx 0122 = xx ( )( ) 034 =+ xx

==+

==

303

404

2

1

xx

xx

First we moved both terms x and 12to the left hand side of the

equation Then we looked for two numbers banda such that 12=ab

Then we chose those numbers that satisfied 1=+ ba These two numbers are 4=a and 3=b

G.MaA.6.5.9 Solve 013 2 = nn .

Suggested Solution: Answer: 01 =n and 132 =n

013 2 = nn ( ) 013 = nn

==

==

13013

00

2

1

nn

nn

G.MaA.6.5.10 Solve 2235 rr =+ .

Suggested Solution: Answer:2

11 =r and 32 =r

2235 rr =+ 0352 2 = rr ( )( ) 0312 =+ rr

===

===+

3303

2

112012

2

1

rrr

rrr

First we moved both terms r5 and 3 to the right hand side of the

equation (where the coefficient of the quadratic term is a positivenumber.)

Then we looked for two numbers banda such that 3=ab

Then we chose those numbers that satisfied 52 =+ ba These two numbers are 1=a and 3=b


19/56



V.MaA.6.5.11 Solve 01825 2 =+ bb .

Suggested Solution: Answer: 5.41 =b and 22 =b

01825 2 =+ bb 01852 2 = bb ( )( ) 0292 =+ bb

===+

====

2202

5.42992092

2

1

bbb

bbb

First we multiplied both sides of the equation by 1 . This wasnecessary because we wanted the coefficient of the quadratic termto be positive.

Then we looked for two numbers dandc such that 3=cd

Then we chose those numbers that satisfied 52 =+ dc These two numbers are 9=c and 2=d

V.MaA.6.5.12 Solve 04129 2 =+ xx .


221 == xx

04129 2 =+ xx ( ) ( ) ( )( ) 023223 22 =+ xx ( )223 x

3

223023 21 ==== xxxx

Using Quadratic Binomial: ( ) 222 2 BABABA ++=+

V.MaA.6.5.13 Solve2

32

xx = .


23

2x

x = 26 xx = 062 = xx ( ) 06 =xx

==

==

606

00

2

1

xx

xx

V.MaA.6.5.13 Solve 376 2 += tt .


1 =t and31

2 =t

376 2 += tt 0376 2 = tt ( )( ) 01332 =+ tt

===+

===

3

113013

2

332032

2

1

ttt

ttt


Then we chose those numbers that satisfied 723 =+ ba These two numbers are 3=a and 1+=b


20/56



V.MaA.6.5.15 Solve 5143 2 =+ pp .


11 =p and 52 =t

51432

=+ pp 051432

=+ pp ( )( ) 0513 =+ pp

===+

===

55053

113013

2

1

ppp

ppp


Then we chose those numbers that satisfied 143 =+ ba These two numbers are 1=a and 5=b


21/56



V.MaA.6.5.18 Solve the equationx

x

x

x

2

3

2

3 +=

.


LCD is ( )22 xx

Multiply both sides of the equation by LCD, i.e. by ( )22 xx :( ) ( )22

2

3

2

322 //

//

+=

xx

x

x

x

xxx

( ) ( ) ( )2332 += xxxx

( ) ( ) 63223262 22 +=+= xxxxxxxx 662

22 += xxxx

0662 22 =+ xxxx

0672 =+ xx

Solve the equation: 0672 =+ xx

067

2

=+ xx ( )( ) 016 = xx 61 =x and 12 =x To solve the quadratic equation 0672 =+ xx think of two numbers a and b that: when multiplied together gives 6 . i.e.: 6. =ba , both a and b have the signwhen added

together gives 1. i.e.: 7=+ ba : Both a and b are negative numbers: .

6=a , 1=b

Substitute 61 =x into the original equation: 62

36

26

36

+=

12

9

4

3=

Checks!

Substitute 12

=x into the original equation: 12

31

21

31

+=

2

4

1

2=

Checks!

V.MaA.6.5.19. Solve the equationx

x

x

x

3

42 +=

.

Suggested Solution: Answer: 5=x LCD is x3

Multiply both sides of the equation by LCD: xx

x

x

xx //

//

+=

/

/ 3

3

423

Simplify and solve the equation:( ) 423 += xx 463 += xx 643 += xx

102 =x 52

10==x

Substitute 5=x into the original equation:53

45

5

25

+=

15

9

5

3= Checks!


22/56



V.MaA.6.5.20. The perimeter of a certain rectangle is cm24 . When the length of the

rectangle is doubled and the width is tripled, the area of the rectangle is increased by2160 cm : Find the dimensions of the original rectangle.

Suggested Solution: Answer: 81 =w and cm41 =l , or 42 =w and cm82 =l

Lets cml : length of the rectangle

cmw : width of the rectangle

cmcmp 240 = : perimeter of the original rectangle2

0 cmA : area of the original rectangle

( )wcmcmp +== l2240 cmcm

w 122

24==+l

cmw 12=+l and wcmA = l20

( ) ( ) wwcmcmAcmAnew ll 63216022

0

2 ==+=

220 1606 cmcmAw +=l

222 1606 cmcmwcmw += ll 21605 cmw =l

22 325

160cmcmw ==l

232 cmw =l

Therefore, we must solve the two simultaneous equations: 232 cmw =l and

cmw 12=+l .

cmw 12=+l cmw= 12l . Substitute cmw= 12l in232 cmw =l

( ) 3212 = ww 3212 2 = ww 032122 =+ ww

Solve the quadratic equation: 032122 =+ ww ( )( ) 048 = ww To solve the quadratic equation 032122 =+ ww think of two numbers a

and b that:

when multiplied together gives 32 . i.e.: 32. =ba , both a and b have thesign when added together gives 1. i.e.: 12=+ ba : Both a and b are

negative numbers: the numbers are 4=a and 8=b

( )( ) 048 = ww 81 =w and 42 =w

Substitute it into cmw= 12l 81 =w cmcmw 481212 11 ===l cm41 =l

42 =w cmcmw 841212 22 ===l cm82 =l

Note that even though seemingly there are two sets of answers, in realitythey are identical rectangles, they are just rotated by 90 degrees.


23/56



V.MaA.6.6.1

Suggested Solution: Answer: ftp 54=

Data: 2180 ftwA == l , ftw 3= l ; ( ) ?2min =+= wp l

=

=

ftw

ftw

3

180 2

l

l ( ) 21803 ft= ll 018032 = ll ( )( ) 01215 =+ ll

To factorize 018032 = ll , and therefore to find the length of the triangle,we must look for two numbers a and b in such a way that their product is

180= ba and their sum is 3=+ ba . Due to the fact that 180= ba is

negative, one of the numbers, a is negative and the other number, b is

positive. On the other hand 3=+ ba indicates that the negative number

should have larger magnitude. Simultaneously due to the fact that3=+ ba is a relatively small number, the magnitude of the numbers

should be close to each other. The search results in the prediction that15=a and 12=b .

Checking: ( )( ) OKba 1801215 == , and ( ) OKba 31215 =+=+ .Therefore: 018032 = ll ( )( ) 01215 =+ ll

==+

=====

rejected

ftwft

12012

12315315015

ll

lll

( ) ( ) ( ) ( ) ( ) ftwp 54272330231522322322 =====+=+= llll Answer: ftp 54=

V.MaA.6.6.2.

Suggested Solution: Answer: inw 8= ; in13=l

Data: inh 2= , inw 5+=l ; 3208 inV = ; ?=l , ?=w

( ) 320825 inwwhwV =+== l ( ) 20825 =+ ww ( ) 1042

2085 ==+ ww

010452 =+ ww

To factorize 010452 =+ ww , and therefore to find the width of the base ofthe box, we must look for two numbers a and b in such a way that their

product is 180= ba and their sum is 3=+ ba . Due to the fact that104=ba is negative, one of the numbers,a is positive and the other

number, b is negative. On the other hand 5=+ ba indicates that thepositive number should have larger magnitude. Simultaneously due to thefact that 5=+ ba is a relatively small number, the magnitude of the

numbers should be close to each other. The search results in theprediction that 13=a and 8=b .Checking: ( )( ) OKba 104813 == , and ( ) OKba 5813813 ==+=+ .Therefore: 010452 =+ ww ( )( ) 0813 =+ ww

=+=+===

==+

inwinww

rejectedww

13585808

13013

l

Answer: inw 8=

; in13=l


24/56



V.MaA.6.6.3.

Suggested Solution: Answer: ftnew 12=l ; ftwnew 10=

Data: ftw 6= , ft8=l ; 22 1207248 ftftAnew =+= ; ?8 =+= ftxnewl ,

?6 =+= ftxwnew

( )( ) 212068 ftxxwnewnew =++=l 01206848 2 =+++ xxx 072142 =+ xx To factorize 072142 =+ xx , and therefore to find x , we must look for twonumbers a and b in such a way that their product is 72=ba and theirsum is 14=+ ba . Due to the fact that 72=ba is negative, one of thenumbers,a is positive and the other number, b is negative. On the other

hand 14=+ ba indicates that the positive number should have larger

magnitude. Simultaneously due to the fact that 14=+ ba is a relatively

large number, the magnitude of the numbers should be far from oneanother. The search results in the prediction that 18=a and 4=b .

Checking: ( )( ) OKba 72418 == , and ( ) OKba 14418418 ==+=+ .

Therefore: 072142 =+ xx ( )( ) 0418 =+ xx

=+==+===

==+

ftwftftxx

rejectedxx

newnew 1046;1248404

18018

l

Answer: ftnew 12=l ; ftwnew 10=

V.MaA.6.6.4.

Suggested Solution: Answer: 2_ 81 ftA gardenflower = ;2117 ftAveg = ;

2162 ftApatio =

Data: xgardenflowertheofLength , ftxveg 4+=l ; xpatio 2=l 2360 ftAoriginal = ;

?=gardenA , ?=patioA Due to the fact that the flower garden is an square, its width is also x .

( ) 236024 ftxxxxxx =+++ 36024 222 =+++ xxxx 036044 2 =+ xx

Divide both sides of the equation by 4: 0902 =+ xx

To factorize 0902 =+ xx , and therefore to find x , we must look for twonumbers a and b in such a way that their product is 90=ba and theirsum is 1=+ ba . Due to the fact that 90=ba is negative, one of the

numbers,a is positive and the other number, b is negative. On the other


magnitude. Simultaneously due to the fact that 1=+ ba is a relativelysmall number, the magnitude of the numbers should be close to eachother. The search results in the prediction that 10=a and 9=b .Checking: ( )( ) OKba 90910 == , and ( ) OKba 9110110 ==+=+ .

Therefore: 0902 =+ xx ( )( ) 0910 =+ xx

====+=+===

==+

ftxftxftxx

rejectedxx

patioveg 18922;13494909

10010

ll

222 819 ftxAgarden === ; ( ) ( )21171394994 ftxxAveg ==+=+= ;

222 1628129222 ftxxxApatio =====

Answer: 2_ 81 ftA gardenflower = ;2117 ftAveg = ;

2162 ftApatio =


25/56



V.MaA.6.6.5. Find two consecutive positive integers such that the square of the first

decreased by 25 equals three times the second.

Suggested Solution: Answer: 7=n ; 81 =+n

Data: ?=n , ?1 =+n ; 0>n ; ( )13252 += nn

( )13252 += nn 33252 += nn 032532 = nn 02832 = nn

To factorize 02832 = nn , and therefore to find n , we must look for twonumbers a and b in such a way that their product is 28=ba and their

sum is 3=+ ba . Due to the fact that 28=ba is negative, one of the


hand 3=+ ba indicates that the negative number should have larger


small number, the magnitude of the numbers should be close to eachother. The search results in the prediction that 4=a and 7=b .Checking: ( )( ) OKba 2874 == , and ( ) OKba 37474 ==+=+ .

Therefore: 02832 = nn ( )( ) 074 =+ nn

=+=+==

==+

8171707

404

nnn

rejectednn

Check: ( ) 2424242549832571325??

22 ===+= nn QED

Answer: 7=n ; 81 =+n

V.MaA.6.6.6. If the second of three positive consecutive integers is added to the product of

the first and the third, the result is 71. Find the three integers.

Suggested Solution: Answer: 71 =n 8=n ; 91 =+n

Data: ?1 =n , ?=n , ?1 =+n ; 0>n ; ( )( ) 7111 =++ nnn ( )( ) 7111 =++ nnn 7112 =+ nn 07112 =+ nn 0722 =+ nn

To factorize 0722 =+ nn , and therefore to find n , we must look for twonumbers a and b in such a way that their product is 72=ba and their





small number, the magnitude of the numbers should be close to eachother. The search results in the prediction that 9=a and 8=b .

Checking: ( )( ) OKba 7289 == , and ( ) OKba 18989 ==+=+ .

Therefore: 0722 =+ nn ( )( ) 089 =+ nn

=+=+====

==+

9181;7181808

909

nnnn

rejectednn

Check: ( )( ) 7111 =++ nnn ( )( ) 71717163871978??

==+=+ QED

Answer: 71 =n 8=n ; 91 =+n


26/56



M.MaA.6.6.7. A positive number is one more than twice another number. If the difference of

their square is 40, find the larger of the two numbers.

Suggested Solution: Answer: 3=m ; 7=n

Data: ?=m , ?=n ; 0>n ; 0>m ; 12 += mn ; 4022 = mn

+=

=

12

4022

mn

mn

( ) 401222

=+ mm 04014422

=++ mmm 039432

=+ mm

To factorize 03943 2 =+ mm , as ( )( ) 033943 2 =++=+ bmammm and thereforeto find m , we must look for two numbers a and b in such a way that their

product is 39=ba . Due to the fact that 39=ba is negative, one of the

numbers,a is positive and the other number, b is negative. We may try

13=a and 3=b . i.e.:

( )( ) 031333943 2 =+=+ mmmm Check: ( )( ) ( ) ( ) 3943391393313333133 22 +=+=+=+ mmmmmmmmmm QED

Therefore: 039432

=+ mm ( )( ) 03133 =+ mm

=+=+=+===

==+

71613212303

3

130133

mnmm

rejectedmm

Check: 404040949403740??

2222 ==== mn QEDAnswer: 3=m ; 7=n

M.MaA.6.6.8. The side of a certain square is 3 feet longer than the side of another square. If

the sum of the areas of the squares is2

117 ft , find the length of a side of the smaller square.

Suggested Solution: Answer: 6=x Data: ?squaresmallertheofside = x , ftx 3squarelargertheofside += ; 0>n ;

( ) 222 1173 ftxx =++ 01179622 =+++ xxx 010862 2 =+ xx Divide both sides of the equation by 2: 05432 =+ xx

To factorize 05432 =+ xx , and therefore to find x , we must look for twonumbers a and b in such a way that their product is 54=ba and their




magnitude. Simultaneously due to the fact that 3=+

ba is a relativelysmall number, the magnitude of the numbers should be close to eachother. The search results in the prediction that 9=a and 6=b .Checking: ( )( ) OKba 5469 == , and ( ) OKba 36969 ==+=+ .

Therefore: 05432 =+ xx ( )( ) 069 =+ xx

=+=+==

==+

ftxftxx

rejectedxx

9363606

909

Check: ( ) 1171171178136117961173??

22222 ==+=+=++ ftxx QED

Answer: 6=x ;


27/56



M.MaA.6.6.9. A rectangular piece of cardboard is twice as long as its wide. From each of itsfour corners a square piece of 3 inches on a side is cut out. The flaps at each corner are then

turned up to form an open box. If the volume of the box is3

168 in , what were the original

dimensions of the piece of cardboards?

Suggested Solution: Answer: inw 10= ; in20=l

Data: ?=l , ?=w ; If w2=l , ( ) ( ) 3168663 inwV == l Divide both sides of the equation ( ) ( ) 1686623 = ww by 3:( ) ( ) 56662 = ww ( ) ( ) 566662 = www 056366122 2 =+ www

0201822 = ww

Divide both sides of the equation 020182 2 = ww by 2: 01092 = ww

To factorize 01092 = ww , and therefore to find w , we must look for twonumbers a and b in such a way that their product is 10=ba and their



hand 9=+ ba indicates that the negative number should have largermagnitude. Simultaneously due to the fact that 9=+ ba is a relatively

large number, the magnitude of the numbers should be far from oneanother. The search results in the prediction that 9=a and 6=b .Checking: ( )( ) OKba 10101 == , and ( ) OKba 9101101 ==+=+ .

Therefore: 01092 = ww ( )( ) 0110 =+ ww

=====

==+

inwinww

rejectedww

201022;10010

101

l

Check:

( ) ( ) ( ) ( ) 16816816841431686106203168663??3 ===== inwV l QED

Answer: inw 10= ; in20=l

M.MaA.6.6.10. If the length of one side of a square garden is increased by ft3 , and the

length of an adjacent side is increased by ft2 , the area of the garden increases to272 ft .

What is the length of a side of the original garden?

Suggested Solution: Answer: ftx 2.13=

Data: xgardensquaretheofside , ( )( ) 22 7223 ftxxx +=++ ;

( ) ( )

2272232 ftxxxx +=+++ 222 72632 ftxxxx +/=+++/ 07265 =+x

0665 =x 665 =x ftx 2.135

66==

Check: ( ) fftx 24.1742.13 22 == ( )( ) ( )( ) ( ) ( ) 224.2462.152.1622.1332.1323 ftxxAnew ==++=++=

27224.17424.246 ftAA oldnew == QED


28/56



V.MaA.6.6.11. A rectangular picture cm30 wide and cm50 long is surrounded by a frame

having a uniform width. If the combined area of the picture and the frame is 22016 cmAtot = ,

what is the width of the frame?

Suggested Solution: Answer: cmx 6=

Data: ?frametheofwidthuniform = x , ( )( )2

20165030 cmxxAtot =++= ( )( ) 220165030 cmxxAtot =++= ( ) ( ) 2016505030 =+++ xxx

0201650301500 2 =+++ xxx 0516802 =+ xx

To factorize 0516802 =+ xx , and therefore to find x , we must look for twonumbers a and b in such a way that their product is 516=ba and their





large number, the magnitude of the numbers should be far from oneanother. The search results in the prediction that 86=a and 6=b .Checking: ( )( ) OKba 516686 == , and ( ) OKba 80686 =+=+ .

Therefore: 0516802 =+ xx ( )( ) 0686 =+ xx

==

==+

cmxx

rejectedcmxx

606

86086Check: ( )( ) 220165636650630 cmAtot ==++=

Answer: cmx 6= ;

M.MaA.6.6.12. The art staff at Central High School is determining the dimensions of paper to

be used in the senior yearbook. The area of each sheet is to be2

432 cm . The staff has agreed

on margins of cm3 on each side and cm4 on the top and the bottom. If the printed matter

is to occupy2

192 cm on each page, what must be the overall length and width of the paper?

Suggested Solution: Answer: cmw 18= , and cm24=l

Data: lpapertheoflength , wpapereach theofwidth , ( )( ) 219268 cmw =l ;

2432 cmw =l

w

432=l

( )( ) 219268 cmw =l ( ) ( ) 2192686 cmww =l 01924886 =+ ww ll

028882592014484326432014486 =+== www

www ll

First multiply both sides of the equation 028882592

=+ ww

by w and

then divide both sides of the new equation by 8 and then arrange theterms in decreasing order of the power:

028882592 2 =+ ww 0324362 =+ ww 018182 22 =+ ww ( ) 018 2 =w

cmw 18= , and cmw

2418

432432===l Answer: cmw 18= , and cm24=l

Check:

( ) ( ) ( ) ( ) 1192119211926188241192684322418??

2 ===== wicmw ll


29/56



QED

G.MaA.7.5.3 Solve the equation20

21

5

2

4

3+=

bb.

Suggested Solution: Answer: 3=b

20

21

5

2

4

3

+=

bb

LCD is 20 Multiply both sides by LCD, i.e. 20 and simplify:

02

2102

5

202

4

302

1

1

1

4

1

5

////+

///=

///

bb

21815 += bb 21815 = bb 217 =b

Solve the equation: 217 =b 37

21==b Answer: 3=b

Check: Substitute 3

=b in 20

21

5

2

4

3+=

bb

( ) ( )20

21

5

32

4

33+

=

20

21

5

6

4

9+=

20

21

4

4

5

6

4

9+=

4

9

20

45

20

2124

4

9==

+= checks!

G.MaA.7.5.4 Solve the equation420

12

2

2 yyy=

.


19=y :

420

12

2

2 yyy=

LCD is 20 Multiply both sides by LCD, i.e. 20 and simplify:

1

5

1

1

1

10

402

02

1202

2

202

///=

//

//

/

//

yyy

( ) ( ) yyy 512210 = yyy 5122010 =+ 05122010 =+ yyy Solve the equation: 0193 =y

193 =y 3

19=y Answer:

3

19=y

Check: Substitute3

19=y in

420

12

2

2 yyy=

4

3

19

20

13

192

2

23

19

=

4

3

19

20

3

338

2

3

619

=

12

19

60

35

6

13=

12

19

60

35130=

12

19

60

95=

12

19

125

195=

/

/checks!


30/56



V.MaA.7.5.5 Solve the equationx

x

x 6

1

12

1

3

8=

.

Suggested Solution: Answer: 61 =x , 52 =x :

x

x

x 6

1

12

1

3

8=

LCD is x12 Multiply both sides by LCD, i.e. x12 and simplify:

xx

xx

xx

/////=

//

//

/////

1

2

1

1

1

4

6

121

21

121

3

821

( ) 2132 = xx 232 2 =+ xx 02322 =+ xx 0302 = xx

Solve the equation 0302 = xx ( )( ) 056 =+ xx Using Find twonumbers a and b that when multiplied: 30=ab and when added

1=+

ba . Due to the fact that their product is negative, i.e.: 30=

ab ,a and b must have different signs. Due to the fact that their sum is

negative: 1=+ ba , the larger number should be negative. Therefore:

6=a and 5=b .

06 =x 61 =x

05 =+x 52 =x Answer: 61 =x , 52 =x

Second Method: Using 02 =++ qpxx qpp

x

=

2

22

0302 = xx 5.55.0302

1

2

12

=+

=x

65.55.01 =+=x 61 =x

55.55.02 ==x 52 =x

Check: Substitute 61 =x inx

x

x 6

1

12

1

3

8=

( )( )

( )661

12

16

63

8=

36

1

12

5

18

8=

36

1

3

3

12

5

2

2

18

8=

36

1

36

1516=

checks!

Check: Substitute 52 =x inx

x

x 6

1

12

1

3

8=

( )( )

( )561

12

15

53

8

=

30

1

12

6

15

8

=

30

1

2

1

15

8

=+

30

1

15

15

2

1

2

2

15

8

=+

30

1

30

1516

=+

checks!


31/56



V.MaA.7.5.6 Solve the equation2

41

4 =

x

x.

Suggested Solution: Answer: 41 =x , 22 =x

2

41

4

=

x

x

LCD is x2 Multiply both sides by LCD, i.e. x2 and simplify:

2

4212

42

/

/=

//

xxx

xx

xxx 4282 = 08242 =+ xxx

0822 = xx

Solve the equation 0822 = xx ( )( ) 024 =+ xx Using Find twonumbers a and b that when multiplied: 8=ab and when added

2=+ ba . Due to the fact that their product is negative, i.e.: 8=ab , a and b must have different signs. Due to the fact that their sum is

negative: 2=+ ba , the larger number should be negative. Therefore:

4=a and 2=b

04 =x 41 =x

02 =+x 22 =x


x

=

2

22

082

2 =xx

( ) 31812

2 2=+=

x 4311 =+=x 41 =x

2312 ==x 22 =x

Check: Substitute 41 =x in2

41

4 =

x

x

2

441

4

4 = 00 = checks!


41

4 =

x

x

2

421

2

4 =

312 =

checks!


32/56



M.MaA.7.5.7 Solve the equation3

1

4

1

6

13=

++

x

x

x


11 =x , 22 =x

3

1

4

1

6

13

=

+

+

x

x

x

LCD is x12 . Multiply both sides by LCD, i.e. x12 and simplify:

1

4

1

3

1

2

3

121

4

121

6

1321

///=

/

+//+

//

/// x

xx

x

xx

( ) ( ) ( )1413132 =++ xxxx xxxx 43326 2 =++ 04932 2 =++ xxx 0253 2 =+ xx

Solve the equation 0253 2 =+ xx ( )( ) 0213 =+ xx Using Find twonumbers a and b that when multiplied: 2=ab . Due to the fact that

the sum is positive 2=

a and 1=

b . Note that due to the fact that thecoefficient of the quadratic factor is not one. The sum terms are morecomplicated and is no longer 5=+ ba

013 =x 013 =x 13 =x 3

11 =x and 02 =+x 22 =x


x

=

2

22

0253 2 =+ xx 03

2

3

52 =+ xx 3

2

6

5

6

52

+

=x

3

1

6

7

6

7

6

5

36

49

6

5

36

2425

6

5

12

12

3

2

36

25

6

5

3

2

6

5

6

52

1 ==+=+=+

+=++=+

+=x

3

11 =x

26

12

6

7

6

5

3

2

6

5

6

52

2 ===+

=x 22 =x

Obviously the second method is more complicated and in most casesrequire a calculator. The first method, i.e. factoring method is simplerbut mostly useful when the roots are integer.

Check: Substitute3

11 =x in

3

1

4

1

6

13=

++

x

x

x

3

1

4

13

1

3

16

13

13

=+

+

3

1

4

3

4

2

11=+

//

3

1

12

4= checks!


1

4

1

6

13=

++

x

x

x

( )( )

( )3

1

4

12

26

123=

++

31

41

127 =+

31

33

41

127 =

31

1237 = checks!


33/56



M.MaA.7.5.8 Solve the equation 23

63=

++

xx

Suggested Solution: Answer: 31 =x , 5.12 =x

2

3

63=

++

xx

LCD is )3( + xx

Multiply both sides by LCD, i.e. )3( + xx and simplify:

( ))3(2

3

6)3(

3)3( +=

+++

/+/ xx

xxx

xxx

)3(26)3(3 +=++ xxxx

xxxx //+=//++ 626932

0932 2 = xx

Solve the equation 0932 2 = xx 05.45.12 = xx ( )( ) 05.13 =+ xx

03=

x

31=

x 05.1 =+x 5.12 =x


x

=

2

22

0932 2 = xx 05.45.12 = xx 5.42

5.1

2

5.12

+

=x

( ) 0.325.275.05.475.075.0 21 =+=++=x 31 =x

( ) 5.125.275.05.475.075.0 22 ==+=x 5.12 =x

Check: Substitute 31 =x in 23

63=

++

xx 2

33

6

3

3=

++ checks!

Check: Substitute 5.12 =x in 23

63=

++

xx 2

35.1

6

5.1

3=

++

242 =+

checks!


34/56



V.MaA.7.5.10. If2

1is added to the reciprocal of a number, the result is 1 less than twice the

reciprocal of the original number. Find the number.


2=x

The task is to solve 121

2

1=+

xxifx is the original number.

LCD is x2 . Multiply both sides by LCD, i.e. x2 and simplify:

122

21

22

12

1

1

/

/=/

/+/

/ xx

xx

xx

xx 242 =+ 242 =+ xx 23 =x

Solve the equation 23 =x 3

2=x Answer:

3

2=x

Substitute

3

2=

xin

1

21

2

1=+

xx

12

3

22

3

2

1=+

12

3

22

//= Checks!

V.MaA.7.5.11. If the reciprocal of a number is multiplied by 3, the result exceeds the

reciprocal of the original number by3

1. Find the number.

Suggested Solution: Answer: 6=x

The task is to solve3

1113 +=

xxor to solve

3

12=

xifx is the original number

3

1113 +=

xx

3

1113 =

xx

3

12=

x

LCD is x3 .Multiply both sides by LCD, i.e. x3 simplify, and solve:

3

13

23

////=

// xx

x x=6 Answer: 6=x

Substitute 6=x in3

1113 +=

xx

3

1

6

1

6

3+=

2

1

6

3

6

21

2

1==

+= Checks!

V.MaA.7.5.12. If the reciprocal of a number is multiplied by 1 less than the original number,

the result exceeds2

1the reciprocal of the original number by

8

5.

Suggested Solution: Answer: 3

10

=x

The task is to solve ( )8

51

2

111 +=

xxx

8

5

2

11+=

xx

x

LCD is x8 .Multiply both sides by LCD, i.e. x8 simplify, and solve:( )

1

1

1

2

8

58

2

18

18

//+

////=

/

/ x

xx

x

xx ( ) xx 5218 += xx 5288 += 8258 += xx

103 =x 3

10=x Answer:

3

10=x

Substitute 3

10

=x in 3

1113 += xx 3

1

6

1

6

3

+= 2

1

6

3

6

21

2

1

==+

= Checks!


35/56



V.MaA.7.5.13. If2222

ba

b

ba

ax

+

= , solve for a in terms of b and x .

Suggested Solution: Answer: bx

a +=1

2222ba

b

ba

ax +=

( )( )( )baba

ba

ba

bax +

+=

+= 22 bax =

1

Multiply both sides ofba

x

=1

by ba

( )( )

( )baba

bax

=1

( ) 1= bax

Expand it: 1= bxax bxax += 1 Divide both sides of the equation by x :

bxax += 1 x

bxa

+=

1or b

xa +=

1Answer: b

xa +=

1

Alternative method

bax

=

1

xba

1= b

xa +=

1

V.MaA.7.5.14. On Tuesday, CityEx delivered 7 less than two times the number of packages ithad delivered on Monday. The number of packages delivered on Wednesday was equal to 10

more than the average number of packages delivered on Monday and Tuesday. If the total

number of packages that were delivered over the 3 days was 661, how many packages weredelivered on Monday?

Suggested Solution: Answer: 147=x was delivered on Monday.If the number of packages delivered on Monday is denoted by x the

problem may be translated to the mathematical equation:

Packages delivered on Monday: x

Packages delivered on Tuesday: 72 x

Packages delivered on Wednesday:( )

102

72+

+ xx

Total number of Packages delivered: ( )( )

661102

7272 =

+

+++

xxxx

The task is to solve the linear equation:

661102

7372 =+

++

xxx

65836612

733 ==

+

xx

Multiply both sides of the equation by 2:

65822

73232 =

/

/+

xx 3161736 =+ xx 1323731619 =+=x 147

9

1323==x

Answer: 147=x


36/56



V.MaA.7.5.15. A student got 75% of the questions on a test correct. If he answered 11 of thefirst 13 questions correctly, and two-third of the remaining questions correctly, how many

questions were on the test?

Suggested Solution:

Answer: Total number of question on the test were 28=x .If the number of questions on the test is taken to be x the problem may

be translated to the mathematical equation:

Total number of questions on test: x

Number of questions that were answered correctly: x75.0

Number of questions that were answered correctly: ( )133

211 + x

Therefore: ( ) xx 75.01332

11=+

( ) xx 43

133

2

11=+

( )4

3

3

132

11

xx=

+

To solve this equation we may, first, multiply both sides of the equationby LCD that is 12:

( )

1

3

1

4

4

321

3

132211112

///=

/

//+

xx

( ) xx 9138132 =+ xx 91048132 =+

xxx == 89104132 28=x Answer: Total number of question on the test were 28=x .


37/56



M.MaA.11.1.14. Find the area, in square units, of the quadrilateral whose vertices are

( )2,4 A , ( )5,0B , ( )3,9C and ( )4,7 D . A B C D E F G H

Suggested Solution: Answer: auAABCD 76=

To solve the problem we may plot the quadrilateral on a coordinatesystem as illustrated below. The area of the quadrilateral may becalculated indirectly by constructing a rectangle EFGH, and calculating

the area of the four rectangles AEH , AFB , BGC and CHD

Area ofEFGH may be calculated using ( ) 134949 =+==HE and( ) 94545 =+==FE . auEFHEAEFGH 117913 === .

Area of AEH may be calculated using ( ) 24242 =+==AE and

( ) 114747 =+==ED . auEDAE

A AEH 112

112

2=

=

= .

Area of AFB may be calculated using ( ) 72525 =+==AF and

( ) 440 ==FB . auFBAFA AFB 14247

2 === .

Area of BGC may be calculated using 9=BG and 235 ==GC .

auGCBG

A BGC 92

29

2=

=

= .

Area of CHD may be calculated using ( ) 74343 =+==CH and

279 ==HD . auHDCH

A CHD 72

27

2=

=

= .

( ) auauAAAtrianglesEFGHABCD

7641117791411117 ==+++==

Answer: auAABCD

76=


38/56



M.MaA.11.2.13: The vertices of right triangle RGA with hypotenuse AG are

( )0.4,0.2R , ( )0.4,0.7A and )yx GGG , . If = 45RGAm , and the unit length iscentimetre.

What are possible coordinates ofG ?

Plot the triangle in a properly scaled coordinate system. What are the possible area of the triangle? What are the possible perimeter of the triangle?Suggested solution: Answer: ( )0.5,0.2 G or ( ).13,0.2G ,

22 415.40 cmcmArea = , cmcm 31290.18P +=

Due to the fact that the triangle is right triangle and one of the angles is

cmRGAm 45= , the triangle is isosceles and RARG = . Therefore:

( ) ( ) ( ) ( ) cmaRaRRA yyxx 0.944722222

=+=+= cmRARG 0.9==

cmRARG 0.9== implies that the x-coordinate of the point )yx GGG , isthe same as that of ( )0.4,0.2R , i.e. cmGx 0.2= , but its y-coordinate is

cmRG 0.9= above or below ( )0.4,0.2R , i.e.: cmGy 130.40.9 =+= or

cmGy 50.40.9 =+= as illustrated in the figure below:

Answer: The possible coordinates of the point )yx GGG , are( )0.5,0.2 G or ( ).13,0.2G

cmRARG 0.9== 222 415.402

0.90.9

22cmcmcm

RARGhbArea =

=

==

( ) ( ) ( )( ) ( ) cmcmGAGAAGyyxx

7.12299941327 222222

==+=+=+=

Instead we could use Pythagoras theorem cmcmAG 7.122999 22 ==+= .

cmcmcm 317.30290.18290.90.9Pperimeter =+=++=

Answer: 241 cmArea ; cm31P


39/56



M.MaA.11.2.14: The coordinates of the vertices of quadrilateral ABCD are ( )0.9,0.2A ,( )0.4,0.11B , ( )0.1,0.6 C , and ( )0.4,0.5D .

Using graph paper, graph the quadrilateral ( )0.9,0.2A on a properly scaled coordinatesystem.

Determine whether diagonals AC and BD bisect each other. Give a reason for youranswer. Calculate the area of the quadrilateral ABCD . Calculate the perimeter of the quadrilateral ABCD .

Suggested solution: Answer: Diagonals AC and BD do not bisect each

other 2.80 cmArea = , cmPABCD 37

-2

-1

0

1

2

3

4

5

6

7

8

9

10

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

x (cm)

y(cm)

( )0.9,0.2A

( )0.1,0.6 C

( )0.4,0.11B

( )0.4,0.5D

E

F

To determine whether diagonals AC and BD bisect each other, we may

find midpoints ofAC and BD . If the midpoints are identical they bisect

each other, otherwise they do not bisect each other:

( )4,42

19,

2

62

2,

2=

+=

++ yyxxAC

cacaM

( )4,3244

,2

115

2,2 =

++

=

++ yyxxDB

bdbdM

Answer: Therefore, due to the fact that ( )4,4ACM and ( )4,3DBM are

not identical, diagonals AC and BD do not bisect each other. This is

evident in the figure above. The area of the quadrilateral ABCD may be calculated directly: It is the

area of two triangles ADB and CDB . They share the same base

( ) cmDB .16511511 =+== . The height ofADB is cmAE 0.549 == , andthe altitude ofCDB is ( ) cmCF 0.514 == . Therefore the area of the

quadrilateral ABCD is:2

.802

516

2

516cmArea =

+

= . Answer:2

.80 cmArea =


40/56



The perimeter of the quadrilateral ABCD is: DACDBCABPABCD +++= :

( )0.9,0.2A , ( )0.4,0.11B , ( )0.1,0.6 C , and ( )0.4,0.5D .

( ) ( ) cmAB 22.929119112 2222 =+=+=

( ) ( )( ) cmcmBC 07.72555146112222

=+=+= ( )( ) ( ) cmCD 08.125114156 2222 =+=+=

( ) ( ) cmDA 60.8579425 2222 =+=+=

cmcmDACDBCABPABCD

3798.3660.808.1207.722.9 =+++=+++=

Answer: cmPABCD 37


41/56



G.MaA.11.3.1. What is the slope of a line through ( )2,4A and ( )8,6B

(1)5

3 (2)

5

3(3)

3

5(4)

3

5

Suggested solution: Answer: Alternative (2)

( ) 53

10

6

46

28==

=

=

AB

AB

AB xx

yyk

G.MaA.11.3.2. Which pairs of points determine a line that is parallel to the y-axis?

(1) ( )1,1 and ( )3,2 (2) ( )1,1 and ( )3,3 (3) ( )3,2 and ( )5,2 (4) ( )5,2 and ( )5,4

Suggested solution: Answer: Alternative (3)

The points on a line parallel to the y-axis have all the same x-coordinatesbut different y-coordinates. Therefore ( )3,2 and ( )5,2 are pair of pointsthat are on a line parallel to the y-axis: 2=x

G.MaA.11.3.3. Which pairs of points determine a line that is parallel to the x-axis?

(1) ( )3,1 and ( )3,2 (2) ( )1,1 and ( )1,1 (3) ( )3,1 and ( )1,1 (4) ( )1,1 and ( )3,3

Suggested solution: Answer: Alternative (1)The points on a line parallel to the x-axis have all the identical y-

coordinates but different y-coordinates. Therefore ( )3,1 and ( )3,2 arepair of points that are on a line parallel to the x-axis: 3=y


42/56



G.MaA.11.3.4. The points ( )2,3A , and ( )5, xB lie on a line whose slope is

2

7 . Which one of the following alternatives is the correct value ofx .

(1) 5 (2) 6 (3)

7

15(4) 4

Suggested solution: Answer: Alternative (1) 5=x , ( )5,5 B Data: ( )2,3A , ( )5, xB

We may first write the equation of the line AB whose slope is2

7=k and

passes through ( )2,3A . The equation of the line may be written as

mxmkxy +=+=2

7. The y-intercept of the line may be determined using

the fact that coordinates of ( )2,3A must satisfy the equation of the line

mxy += 2

7

. i.e.:

mxy +=2

7 5.12

2

25

2

214

2

21223

2

7==

+=+==+ mm

2

25

2

7+= xy

2

25

2

7+= xy must pass ( )5, xB , therefore the coordinates of ( )5, xB

must satisfy the equation2

25

2

7+= xy

2

25

2

7+= xy 5

7

35357

2

355

2

25

2

7

2

25

2

75 ====+=+= xxxx 5=x

-16

-12

-8

-4

0

4

8

12

16

20

-2 -1 0 1 2 3 4 5 6 7

x

y

2

25

2

7+= xy

( )2,3A

( )5,5 B


43/56



G.MaA.11.3.5. Given points ( )0,0A , ( )2,3B , ( )3,2C , which statement is true?

(1) ACAB (2) ACAB (3) BCAB > (4) CABC

Suggested solution: Answer: Alternative (2) i.e. ACAB

Data: ( )0,0A , ( )2,3B , ( )3,2C

We may first write the equation of the lines AB , AC, BCand calculate the

length ofAB and BC.

AB ( )0,0A ( )2,3B

3

2

03

02=

=

=

AB

AB

AB xx

yyk mxy +=

3

2

The line passes ( )0,0A , therefore 0=m

xy3

2=

AC

( )0,0A ( )3,2C

2

3

02

03=

=

=

AC

AC

AC xx

yyk mxy +=

2

3

The line passes ( )0,0A , therefore 0=m

xy2

3=

BC

( )2,3B ( )3,2C

5

1

32

23=

=

=

BC

BC

BC xx

yyk mxy +=

5

1

The line passes ( )2,3B , therefore

5

13

5

323

5

12 =+=+= mm

5

13

5

1+= xy

5

13

5

1+= xy

Length of

AB ( )0,0A ( )2,3B

( ) ( ) 1332 2222 =+=+=ABAB

xxyyAB ( )0,7E does notlie on the line.

Length ofBC

( )2,3B ( )3,2C

( ) ( ) ( ) ( )( )2222 2332 +=+= CBCB xxyyBC

2651 22 =+=BC

( )0,7E does notlie on the line.

As illustrated in the table above ACAB . This is due to the fact that the

slope of AB and AC are reciprocals:

3

2=

ABk and

2

3=

ACk and therefore:

AC

AB kk

1

2

3

1

3

2===

None of the other alternatives are correct!


44/56



G.MaA.11.3.6. The point whose coordinates are ( )2,4 A , lies on aline whose slope is2

3.

The coordinates of another point on this line could be

(1) ( )0,1B (2) ( )1,2C (3) ( )1,6D (4) ( )0,7E Suggested solution: Answer: Alternative (3), i.e. ( )1,6D lies on the line

82

3= xy

Data: ( )2,4 A ,2

3=k

We may first write the equation of the line that passes ( )2,4 A and

whose slope is2

3=k . The equation of the line may be written as

mxmkxy +=+=2

3. The y-intercept of the line may be determined using the

fact that coordinates of ( )2,4 A must satisfy the equation of the line

mxy +=2

3. i.e.:

mxy +=2

3 86264

2

32 ==+=+= mmm 8

2

3= xy

If any of the points given as the alternative lies on the line 82

3= xy , their

coordinates must satisfy the equation of the line:

( )0,1B 8

2

3= xy 5.681

2

3==y

( )0,1B does not

lie on the line.( )1,2C 8

2

3= xy 58382

2

3===y .

( )1,2C does notlie on the line.

( )1,6D 8

2

3= xy 18986

2

3===y .

( )1,6D lies onthe line.

( )0,7E 8

2

3= xy 05.285.1087

2

3===y .

( )0,7E does notlie on the line.

-12

-10

-8

-6

-4

-2

0

2

4

6

-2 -1 0 1 2 3 4 5 6 7 8

x

y

( )1,6D

82

3= xy

( )2,4 A


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G.MaA.11.3.7. In the accompanying diagram, if the slope of the line l is4

1. What are the

coordinates of the point ( )baP , ?Suggested solution: Answer:

12=a , 5=b , i.e.:( )

5,12P

Data: ( )5,1A , ( )2,0B ( )baP , and ( )5,1A lie on a

horizontal line parallel to the x-axis and therefore, they mustshare the identical y-coordinates,i.e. 5=b . On the other hand the

equation of line l may be written based on the fact that its slope is4

1and

its y-intercept is 2 :

24

1+= xy Equation of the line l .

( )5,aP lies on the line l and therefore, its coordinates may satisfy theequation of the line, i.e.:

24

15 += a 25

4

1=a 3

4

1=a 1243 ==a 12=a

0

1

2

3

4

5

6

-2 0 2 4 6 8 10 12 14 16

x

y

( )5,1A

( )2,0B

l

24

1+= xy

( )5,12P

x

y

( )5,1A

( )2,0B

l

( )baP ,


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G.MaA.11.3.8. If ( )14,xC lies on the same lie as ( )1,6 A and ( )5,2B , what is thevalue ofx ?

Suggested solution: Answer: 82

3+= xy ; 4=x ; ( )14,4C

Data:( )

14,xC ,

( )1,6 A ,

( )5,2B

If ( )14,xC lies on the line AB , its coordinates must satisfy the equation of

the line. Therefore, first we may find the equation of the line AB andrequire the condition of satisfaction:

The slope of the line AB is:( )

2

3

4

6

4

15

62

15==

+=

=

=

AB

AB

AB xx

yyk

mxmxkyAB

+=+=2

3 mxy +=

2

3

To calculate the y-intercept m of the line AB , we may use the fact that

mxy += 2

3passes through ( )1,6 A and therefore the coordinates of the

point must satisfy the equation of the line.

mxy +=2

3 m+= 6

2

31 m+= 91 819 ==m 8

2

3+= xy

If ( )14,xC lies on the line AB , its coordinates must satisfy the equation ofthe line:

82

3+= xy 8

2

314 += x 148

2

3=x 6

2

3=x 123 = x

3

12=x

4=x

-3

-1

1

3

5

7

9

11

13

15

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

x

y

( )1,6 A

( )5,2B

( )14,4C


47/56

Math A/B, Unit Three: Algebraic Methods; Unit 5: Graphs and Transformations Ch

maanvco08unit+three+algebraic+methodsch6 7-8-11

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