AGM - Worked Solutions

8/10/2019 AGM - Worked Solutions

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Chapter 1 MatricesExercise 1A Solutions

1a Number of rows number of columns

= 2 2

b Number of rows number of columns= 2 3

c Number of rows number of columns= 1 4

d Number of rows number of columns= 4 1

2

a There will be 5 rows and 5 columns tomatch the seating. Every seat of bothdiagonals is occupied, and so thediagonals will all be ones, and the rest ofthe numbers, representing unoccupiedseats, will all be 0.

1

0

0

0

1

0

1

0

1

0

0

0

1

0

0

0

1

0

1

0

1

0

0

0

1

b If all seats are occupied, then everynumber in the matrix will be 1.

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

3 i=jfor the leading diagonal only, so theleading diagonal will be all ones, andthe rest of the numbers 0.

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

4 We can present this as a table with thegirls on the top row, and the boys on thebottom row, in order of year level, i.e.years 7, 8, 9, 10, 11 and 12 going fromleft to right.

200

110

180

117

135

98

110

89

56

53

28

33

Alternatively, girls and boys could bethe two columns, and year levels couldrun down from year 7 to 12, in order.This would give:

200180

135

110

56

28

110117

98

89

53

33

5a Matrices are equal only if they have the

same number of rows and columns, and allpairs of corresponding entries are equal.

The first two matrices have the samedimensions, but the top entries are notequal, so the matrices cannot be equal.The last two matrices have the samedimensions and equal first (left) entries,so they will be equal ifx= 4.Thus, [0 x] = [0 4] if x = 4 .

b The first two matrices cannot be equalbecause corresponding entries are notequal, nor can the second and third forthe same reason.

The last matrix cannot equal any of theothers because it has differentdimensions. The only two that can beequal are the first and third.

4

1

7

2 =

x

1

7

2 if x = 4

Cambridge Essential Advanced General Mathematics 3rd Edition Worked Solutions CD-ROM 1


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c All three matrices have the samedimensions and all correspondingnumerical entries are equal. They couldall be equal.

2

1

x

10

4

3 =

y

1

0

10

4

3

= 2

1

0

10

4

3 ifx = 0, y = 2

6

a The entry corresponding toxis 2, andthe entry corresponding toyis 3, sox= 2 andy= 3.

b The entry corresponding toxis 3, andthe entry corresponding toyis 2, sox= 3 andy= 2.

c The entry corresponding toxis 4, andthe entry corresponding toyis 3,sox= 4 andy= 3.

d The entry corresponding toxis 3, andthe entry corresponding toyis 2,sox= 3 andy= 2.

7 LetA,B, CandDbe the columns androws, in that order.There are no roads fromAtoA, so the topleft entry will be 0. There are 3 roads fromAtoB, so the next entry right will be 3.There is 1 road fromAto Cand no roads

directly fromAtoD, so the next twoentries right will be 1 and 0.Continue to fill in the matrix.

0

3

1

0

3

0

2

1

1

2

0

1

0

1

1

0

8 Write it as set out, with each rowrepresenting playersA,B, C,DandErespectively, and columns showing

points, rebounds and assistsrespectively.

21

8

4

14

0

5

2

1

8

1

5

3

1

60

2



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Exercise 1B Solutions

1 Add the corresponding entries.

X + Y = 1 + 3

2 + 0 =

4

2

Double each entry.

2X =

2 1

22 =

2

4

Multiply each entry in Yby 4 and addthe corresponding entry for X.

4Y + X =

4 3 + 1

4 0 + 2 =

13

2

Subtract corresponding entries.

XY = 1 3

2 0 =

2

2

Multiply each entry by 3.

3A =

3 1

3 2

31

3 3 =

3

6

3

9

Add Bto the previous answer.

3A + B =3

6

3

9 +

4

1

0

2

= 1

7

3

7

2

a Double each entry.

3 2

4 2

6 2

2 2

2 2

1 2

=

6

8

12

4

4

2

b 1

4

0

2

0

3 +

2

8

0

4

0

6 +

2

6

1

1

0

4

= 5

18

1

7

0

13

c The average will be the total divided by 3,so divide each entry by 3.

5 3

18 3

1

3

7 3

0

3

13 3

=

5

3

6

13

73

0

133

3 Multiply each entry by the factor.

2A = 2

0

2

4

3A =3

03

6

6A =6

0

6

12

4

a As the matrices have the samedimensions, corresponding terms can beadded. They will simply be added in theopposite order.Since the commutative law holds true fornumbers, all corresponding entries in the

added matrices terms will be equal, so thematrices will be equal.

b As the matrices have the samedimensions, corresponding terms can beadded. The first matrix will add the firsttwo numbers, then the third, and thesecond matrix will add the second andthird numbers first, then add the result tothe first number.Since the associative law holds true fornumbers, all corresponding entries in the

added matrices terms will be equal, so thematrices will be equal.

5a Multiply each entry by 2.

2A = 6

4

4

4

b Multiply each entry by 3.

3B = 0

12

9

3

c Add answers to aand b.

2A + 3B = 6

4

4

4 +

0

12

9

3

= 6

8

5

1



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d Subtract afrom b.

3 B 2A = 0

12

9

3

6

4

4

4

=6

16

13

7

6a Add corresponding entries.

1

0

0

3 +

1

2

1

0 =

0

2

1

3

b Triple entries in Q, then add tocorresponding entries in P.

1

0

0

3 +

3

6

3

0 =

2

6

3

3

c Double entries in P, then subtract Qand addR.

2

0

0

6

1

2

1

0 +

0

1

4

1 =

3

1

3

7

7

a If 2A 3X= B,then 2A B= 3X3X = 2AB

X = 23

A13B

= 23

3

1

1

4 1

3 0

2

10

17

=

2

3

3 1

3

0

23

1 13

2

2

3

1 1

3

10

23

4 13

17

= 2

0

4

3

b If 3A+ 2Y= 2Bthen 2Y= 2B 3A

Y = B 1 12

A

= 0

2

10

17 1 1

2 3

1

1

4

=

0 32 3

2 32

1

10 32 1

17 32

4

=

9

2

12

232

11

8 X + Y = 150 + 160

100 + 100

90 + 90

0 + 0

100 + 120

75 + 50

50 + 40

0 + 0

=

310

200

180

0

220

125

90

0

The matrix represents the total productionat two factories in two successive weeks.



5/417

Exercise 1C Solutions

1 AX = 1

1

2

3

2

1

=

1 2 + 21

1 2 + 31

= 4

5

BX = 3

1

2

1

2

1

=

3 2 + 21

1 2 + 11

= 4

1

AY =

1

1

2

3

1

3

=

1 1 + 2 3

1 1 + 3 3

=5

8

IX = 1

0

0

1

2

1

=

1 2 + 01

0 2 + 11

=

2

1

AC = 1

1

2

3

2

1

1

1

=

1 2 + 2 1

1 2 + 3 1

1 1 + 2 1

1 1 + 3 1

= 0

1

1

2

CA = 2

1

1

1

1

1

2

3

=

2 1 + 11

1 1 + 11

22 + 1 3

12 + 1 3

= 1

0

1

1

Use AC = 0

1

1

2

(AC)X =

0

1

1

2

2

1

=

0 2 + 11

1 2 + 21

= 1

0

Use BX = 4

1

C(BX) = 2

1

1

1

4

1

=

2 4 + 1 1

1 4 + 1 1

= 9

5

AI = 1

1

2

3

1

0

0

1

=

1 1 + 2 0

1 1 + 3 0

1 0 + 2 1

1 0 + 3 1

= 1

1

2

3

IB = 1

0

0

1

3

1

2

1

=

1 3 + 0 1

0 3 + 1 1

1 2 + 0 1

0 2 + 1 1

= 3

1

2

1

AB = 1

1

2

3

3

1

2

1

=

1 3 + 2 1

1 3 + 3 1

1 2 + 2 1

1 2 + 3 1

= 1

0

0

1

BA = 3

121

11

23

=

3 1 + 21

1 1 + 11

32 + 2 3

12 + 1 3

= 1

0

0

1



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A2

= AA = 1

1

2

3

1

1

2

3

=

1 1 + 21

1 1 + 31

12 + 2 3

12 + 3 3

=

3

4

8

11

B2

= BB = 3

1

2

1

3

1

2

1

=

3 3 + 2 1

1 3 + 1 1

3 2 + 2 1

1 2 + 1 1

= 11

4

8

3

Use CA =1

0

1

1

A(CA) =

1

1

2

3

1

0

1

1

=

1 1 + 2 0

1 1 + 3 0

1 1 + 2 1

1 1 + 3 1

= 1

1

3

4

Use A2

= 3

4

8

11

A2C =

3

4

8

11

2

1

1

1

=

3 2 + 8 1

4 2 + 1 1 1

3 1 + 8 1

4 1 + 1 1 1

=2

3

5

7

2

a A product is defined only if the numberof columns in the first matrix equals thenumber of rows of the second.Ahas 2 columns and Yhas 2 rows, soAYis defined.Yhas 1 column and Ahas 2 rows, soYAis not defined.Xhas 1 column and Yhas 2 rows, soXYis not defined.

Xhas 1 column and 2 rows, so is notdefined.

2X

Chas 2 columns and Ihas 2 rows, so CIis defined.Xhas 1 column and Ihas 2 rows, so XIis not defined.

b AB = 2

0

0

0

0

3

0

2

=

2 0 + 03

0 0 + 03

2 0 + 0 2

0 0 + 0 2

=

0

0

0

0

3 No, because Q.2part bshows that ABcan equal O, and AO, BO.

4 LX = [2 1] 2

3

= [2 2 + 13] = [7]

XL = 23

[2 1]

=

2 2

3 2

21

31

= 4

6

2

3

5 A product is defined only if the numberof columns in the first matrix equals thenumber of rows of the second.This can only happen if m= n, in whichcase both products will be defined.

6 a

c

b

d

d

c

b

a

=

ad+b c

cd+d c

a b +b a

c b +da

= adbc

0

0

adbc =

1

0

0

1

For the equations to be equal, allcorresponding entries must be equal,therefore ad bc= 1.When written in reverse order, we get

d

c

b

a

a

c

b

d

=

d

a + b

cca +ac

d

b + b

dcb +ad

= adbc

0

0

adbc =

1

0

0

1

Since ad bc= 1.



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7 We can use any values of a, b, cand das long as ad bc= 1.

(B + C)A =1

0

3

4

1

4

2

3

=

1 1 + 3 4

0 1 + 4 4

1 2 + 3 3

0 2 + 4 3

=

11

16

7

12

For example, a= 5, d= 2, b= 3, c= 3satisfy ad bc= 1 and give

AB = 5

3

3

2

2

3

3

5 =

1

0

0

1

BA = 2

3

3

5

5

3

3

2 =

1

0

0

1

9

5

2.50

12

3.00

1

2 =

5 1 + 1 2 2

2.50 1 + 3.00 2

= 29

850

Other values could be chosen.

8 One possible answer.

A = 1

4

2

3, B =

0

2

1

3 , C =

1

2

2

1

1 5 min plus 2 12 min means 29 minfor one milkshake and two banana splits.The total cost is $8.50.

A + B = 1 + 0

4 + 2

2 + 1

3 + 3 =

1

6

3

6

B + C = 0 + 1

2 + 2

1 + 2

3 + 1 =

1

0

3

4

A(B + C) = 1

4

2

3

1

0

3

4

=

11 + 2 0

41 + 3 0

1 3 + 2 4

4 3 + 3 4

=1

4

11

24

AB = 1

4

2

3

0

2

1

3

=

1 0 + 2 2

4 0 + 3 2

1 1 + 2 3

4 1 + 3 4

=

4

6

7

13

AC = 1

4

2

3

1

2

2

1

=

11 + 22

41 + 32

1 2 + 2 1

4 2 + 3 1

= 5

10

4

11

AB + AC = 4

6

7

13 +

5

10

4

11

=

4 + 56 + 10

7 + 413 + 11

=1

4

11

24

5

2.50

12

3.00

1

2

2

1

0

1

=

5 1 + 1 2 2

2.5 1 + 3 2

5 2 + 1 2 1

2.5 2 + 3 1

5 0 + 1 22.5 0 + 3

= 29

8.50

22

8.00

12

3.00

The matrix shows that John spent 29 minand $8.50, one friend spent 22 min and$8.00 (2 milkshakes and 1 banana split)while the other friend spent 12 min and$3.00 (no milkshakes and 1 banana split).

10

01

1

1

0

0

0

0

1

1

1

1

0

1

0

1

1

0

1

1

2.003.00

2.50

3.50

=

0 2.00 + 0 3.00 + 1 2.50 + 1 3.50

1 2.00 + 0 3.00 + 1 2.50 + 1 3.501 2.00 + 0 3.00 + 0 2.50 + 0 3.501 2.00 + 1 3.00 + 1 2.50 + 1 3.500 2.00 + 1 3.00 + 0 2.50 + 1 3.50

=

6.00

8.002.00

11.00

6.50

This shows the total amount spent onmagazines by each student.Aspent $6.00,Bspent $8.00, Cspent $2.00,Dspent$11.00 andEspent $6.50.



8/417

11

a SC = s11

s21

s12

s22

s13

s23

c1

c2

c3

=

s11c1 +s12c2 +s13c3s21c1 +s22c2 +s23c3

b SCrepresents the income from car sales for each showroom.

c SC =

s11

s21

s12

s22

s13

s23

c1

c2

c3

u1

u2

u3

=

s11c1 +s12c2 +s13c3

s21c1 +s22c2 +s23c3

s11u1 +s12u2 +s13u3

s21u1 +s22u2 +s23u3

SCnow represents the income from each showroom for both new and used car sales.

d CV gives the profit on each new car and each used car for the three models.



9/417


10/417

B1

A1

= 1

3

0

1

12

0

12

1

=

1 1

2+ 0 0

3 12+ 1 0

1 12

+ 01

3 12+ 11

=

1

2

32

12

52

(AB)1

= B1

A1

4a det(A) = 4 1 3 2 = 2

A1

= 12

1

2

3

4

=

1

2

1

32

2

b If AX= , multiply both sides

from the left by

3

1

4

6

1.A

A1

AX = A1

3

1

4

6

IX = X

=

1

2

1

32

2

3

1

4

6

=

1

2 3 + 3

2 1

1 3 + 2 1

12

4 + 32

6

1 4 + 2 6

=0

1

7

8

c If YA= 3

1

4

6 , multiply both sides

from the right by1.A

YAA1

=

3

1

4

6A

1

YI = Y

= 3

1

4

6

1

2

1

32

2

=

3 1

2+ 4 1

1 12

+ 6 1

3 32

+ 42

1 32

+ 62

=

5

2

112

72

212

5

a If AX+ B= Cthen AX= C B

AX = 3

2

4

6

4

2

1

2

=1

0

5

4

det(A) = 3 6 2 1 = 16

A1

= 1

16

6

1

2

3

=

3

8

116

18

316

If AX=1

0

5

4 , multiply both sides

from the left by1.A

A1

AX = A1

1

0

5

4



11/417

IX= X

=

3

8

116

18

316

1

0

5

4

=

38

1+ 18

0

116

1+ 316

0

38

5 + 18

4

116

5 + 316

4

=

3

8

116

118

716

b If YA+ B= Cthen YA= C B

YA =

3

2

4

6

4

2

1

2

=1

0

5

4

From part a, A1

=

3

8

116

18

316

If YA= , multiply both sides

from the right by

1

0

5

4

1.A

YAA1

=

1

0

5

4

A1

YI = Y

=1

0

5

4

3

8

116

18

316

=

1 3

8+ 5 1

16

0 38

+ 4 116

1 18

+ 5 316

0 18

+ 4 316

X=

1116

14

1716

34

6 Amust be a11

0

0

a22 .

det(A) = a11 a22 0 0 = a11a22det(A) 0 since a110 and a220 andthe product of two non-zero numberscannot be zero.

Ais regular.

A1

= 1a11a22

a22

0

0

a11

=

1

a11

0

0

1a22

7 IfA is regular, it will have an inverse,1.A

Multiply both sides of the equationAB = 0

from the left by

1

.

A A

1AB = A

10

IB = 0B = 0

8 Let Abe any matrix a

c

b

d .

If the determinant is n, then the inverse

of Ais given by 1n d

c

b

a .

a

c

b

d

= 1

n

d

c

b

a

a =dnand d=a

n

Substituting for d, a =a nn

=a

n2

This gives2 1n = , or n= 1.

If n= 1, a= dand b= b, which givesb= 0 and similarly c= 0.

det(A) =2 1ad a= =

This leads to two matrices, and 1

0

0

1

1

00

1 .



12/417

If n= 1, a= d; there are no restrictions onband cbut the determinant = ad bc= 1.

a2

+bc = 1 (since a = d)

If b= 0, a= 1, giving , which

can be written

1

c

0

1

1

k0

1 or

1

k01

.

If b0, = 1 gives2

a bc+

c = 1 a2

b, giving

a

1 a2

b

b

a

, which

includes the cases 1

0

k

1 and

1

0

k

1

whena= 1.



13/417

Exercise 1E Solutions

1 First find the inverse of A.det(A) = 3 1 1 4 = 1

A1

= 1

1

1

4

1

3

=

1

4

1

3

a If AX = Kthen A1

AX = A1

K

IX = X = A1

K

X =1

4

1

3

1

2

=

11 + 1 2

41 + 3 2

= 3

10

b If AX = Kthen A1

AX = A1

KIX = X = A

1K

X =1

4

1

3

2

3

=

12 + 1 3

42 + 3 3

= 5

17

2 First find the inverse of A.det(A) = 3 4 1 2 = 14

A1

= 114

4

2

1

3 =

27

17

114

314

a If AX = Kthen A1

AX = A1

K

IX = X = A1

K

X =

2

7

17

114

314

0

1

=

27

0 + 1

14 1

17

0 + 314

1

=

1

14

314

b If AX = Kthen A1

AX = A1

K

IX = X = A1

K

X =

2

7

17

114

314

2

0

=

2

7 2 + 1

14 0

17

2 + 314

0

=

4

7

27

3

a

2

3

4

1

x

y =

6

1

Determinant = 2 1 4 3 = 14

Inverse = 114

1

3

4

2

=

1

14

314

27

17

x

y =

1

14

314

27

17

6

1

=

1

14 6 + 2

7 1

314

6 + 17

1

=

1

7

107

x = 17

, y = 107



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b1

1

2

4

x

y =

1

2


Inverse = 12

4

1

2

1

=

2

12

112

x

y =

2

12

1

12

1

2

=

21 + 1 2

12

1 + 12

2

=

4

3

2

x = 4, y = 32

or 1.5

c Convert the second equation tox+y= 4, giving

2

1

5

1

x

y =

10

4


Inverse = 17 1

1

5

2

xy

= 17 1

152

10

4

= 17

110 + 5 4

110 + 2 4

= 1730

2

x = 307

, y = 27

d Re-write the second equation as3.5x+ 4.6y= 11.4, giving

1.3

3.52.74.6

xy

=1.211.4

Determinant = 1.3 4.6 2.7 3.5= 15.43

Inverse = 115.43

4.6

3.5

2.7

1.3

x

y = 1

15.43 4.6

3.5

2.7

1.3

1.2

11.4

= 1

15.43

4.61.2 + 2.7 11.4

3.51.2 + 1.3 11.4

= 115.43

36.3

10.62

x 2.35, y 0.69

4 Solve the simultaneous equations2x 3y= 73x+y= 5

2

3

3

1

x

y =

7

5


Inverse = 111 13

32

x

y = 1

11 1

3

3

2

7

5

= 111

1 7 + 3 5

3 7 + 2 5

= 111

22

11

x = 2, y = 1

The point of intersection is (2, 1).

5 Ifxis the number of books they arebuying andyis the number of CDs theyare buying, then the following equationsapply.4x+ 4y= 1205x+ 3y= 114

4

5

4

3

x

y =

120

114


Inverse = 18

3

5

4

4 = 1

83

5

4

4

x

y

=18

3

5

4

4

120

114

= 18

3 120 + 4 114

5 120 + 4 114

= 18 96

144

x = 12, y = 18

One book costs $12, a CD costs $18.



15/417

7 Enter the 4 4 matrix Aand the 4 1matrix Binto the graphics calculator.

6

a 2

4

3

6

x

y =

3

6

Fill in missing coefficients with zeros,so that r+ s= 1 becomes0p+ 0q+ 1r+ 1s= 1b det(A) = 2 6 3 4 = 0, so the

A =

1

0

2

0

1

0

1

1

1

1

2

0

1

1

0

1 ; B =

5

1

2

0

A1

B =

2

4

1

2

matrix is singular.

c Yes. For examplex= 0,y= 1 is anobvious solution.

d You should notice that the secondequation is simply the first with bothsides multiplied by 2.There is an infinite number ofsolutions to these equations, just asthere is an infinite number of orderedpairs that make 2x 3y= 3 a trueequation.

p= 2, q= 4, r= 1, s= 2



16/417

Solutions to Multiple-choice Questions

1 The dimension is number of rows bynumber of columns, i.e. 4 2. B

2 The matrices cannot be added as theyhave different dimensions. E

3 DC = 1

2

3

3

1

1

2

1

3

0

1

2

= 1 2

2 1

3 3

3 0

1 1

1 2

= C1

1

0

3

0

1

4 Multiply every entry by 1.

M = 4

2

0

6

= E 4

2

0

6

5 2M 2N = 2 0

3

2

1 2

0

3

4

0

= 0

6

4

2

0

6

8

0

C= 0

12

4

2

6 A+ Bwill have the same dimension asAand B, i.e. m n. A

7 The number of columns of Qis not thesame as the number of rows of P, sothey cannot be multiplied. E

8 Determinant = 2 1 2 1= 4 A

9 Determinant = 1 2 1 1= 1

Inverse = 11

2

1

1

1

=

2

1

1

1 E

10 NM = 0

3

2

1

0

3

2

1

=

0 0 + 2 3

3 0 + 1 3

02 + 2 1

32 + 1 1

=6

3

2

5 D



17/417

Solutions to Short-answer Questions

1

a A + B = 1

2

0

3 +

1

0

0

1

= 0

204

AB = 1

2

0

3

1

0

0

1

= 2

2

0

2

(A + B)(AB) = 0

2

0

4

2

2

0

2

= 0

12

0

8

b A2 = AA =

12

03

12

03

= 1

8

0

9

B2

= BB =1

0

0

1

1

0

0

1

= 1

0

0

1

A2B

2=

1

8

0

9

1

0

0

1

= 0

8

0

8

2 Find the inverse of 3

6

4

8 .

Determinant = 3 8 4 6 = 0This is a singular matrix.

If A then this corresponds to the

simultaneous equations:

=xy

,

3x+ 4y= 86x+ 8y= 16

The second equation is equivalent to thefirst, as it is obtained by multiplyingboth sides of the first by 2.

Thus ifx= a,3a + 4y = 8

4y = 8 3a

y = 2 3a4

The matrices may be expressed as

a

2 3a4

.

3

a For a product to exist, the number ofcolumns of the first matrix must equalthe number of rows of the second.This is true only for AC, CDand BE, so

these products exist.

b DA = [2 4] 1

3

2

1

=

2 1 + 4 3

2 2 + 41

= 14

0

det(A) = 1 1 2 3 = 7

A1

= 17

1

3

2

1

= 17 1

3

2

1 or

17

37

27

17

4 AB = 1

5

2

1

1

2

1

1

3

4

6

8

=

1 1 + 2 1 + 1 3

5 1 + 1 1 + 2 3

1 4 + 2 6 + 1

5 4 + 1 6 + 2

= 2

2

0

2

det(C) = 1 4 2 3 = 2

C1

= 12

4

3

2

1

=

2

32

1

12



18/417

5 Find the inverse of 1

3

2

4 .


Inverse = 12

4

3

2

1

= 1243

21

Multiply by the inverse on the right:

A = 5

12

6

14 1

24

3

2

1

=1

3

2

5

6 A2

=

2

0

0

0

0

2

0

2

0

2

0

0

0

0

2

0

2

0

=

40

0

04

0

00

4

A1

=

1

2

0

0

0

0

12

0

12

0

7 The determinant must be zero.

1x 2 4 = 0x 8 = 0

x = 88

a i MM =

2

1

1

3

2

1

1

3

= 3

5

5

8

ii MMM = MM(M)

= 3

5

5

8

2

1

1

3

= 1

18

18

19

iii Determinant = 2 3 1 1 = 7

M

1

=

1

7 3

1

1

2

b M1

Mx

y = M

1

3

5

x

y = 1

7 3

1

1

2

3

5

= 17 14

7

= 2

1

x = 2, y = 1



19/417

Chapter 2 Algebra IExercise 2A Solutions

1a Add indices:

x3

x4

=x3 + 4

=x7

b Add indices:

a5

a3

=a5 + 3

=a2

c Add indices:

x2

x1

x2

=x2 + 1 + 2

=x3

d Subtract indices:

y3

y7 =y

3 7

=y

4

e Subtract indices:

x8

x4

=x8 4

=x12

f Subtract indices:

p5

p2 =p

5 2=p

7

g Subtract indices:

a12

a23

=a3646

=a16

h Multiply indices:

(a2

)4

=a2 4

=a8

i Multiply indices:

(y2

)7

=y27

=y14

j Multiply indices:

(x5)

3=x

5 3=x

15

k Multiply indices:

(a20

)

35

=a20 3

5=a

12

l Multiply indices:

x

12

4

=x1

24

=x2

m Multiply indices:

(n10

)

15

=n10 1

5=n

2

n Multiply the coefficients and add theindices:

2x

12

4x3

= (2 4)x

12

+ 3

= 8x

72

o Multiply the first two indices and addthe third:

(a

2

)

52

a

4

=a

2 52

a

4

=a5 + 4

=a1

=a

p1

x4

=x1

14

=x4

q

2n2

5

5

(43n

4) = 2

5n

25

5

((22)

3n

4)

= 25n

2

(26n

4)

= 25 6

n2 4

= 21n6 = 12n

6

r Multiply the coefficients and add theindices.

x3

2x

12

4x3

2= (1 24)x

3 + 12

+ 32

= 8x2

s (ab3)

2a

2b4

1

a2b3

=a2b

6a

2b4

a2

b3

=a2 + 2 + 2

b6 + 4 + 3

=a2

b5

t (22p

3 4

3p

5

((6p3

))0

= 1

Anything to the power zero is 1.



20/417

2

a 25

12

= 25 = 5

b 64

13

= 3 64 = 4

c16

9

12 =

16

12

9

12

=16

9= 4

3

d 161

2=

1

16

12

=

1

16 =14

e49

36

12 =

1

49

36

12

=1

49

36

=36

49= 6

7

f 27

13

= 3 27 = 3

g 144

12

= 144 = 12

h 64

23

=

64

13

2

= 42

= 16

i 9

32

=

9

12

3

= 33

= 27

j81

16

14 =

81

14

16

14

= 32

k23

5

0

= 1

l 128

37

=

128

17

3

= 23 = 8

3

a 4.352

= 18.9225

18.92

b 2.45

= 79.62624

79.63

c 34.6921 = 5.89

d 0.023

= 125 000

e 3 0.729 = 0.9

f 4 2.3045 = 1.23209.. .

1.23

g (345.64)1

3= 0.14249. . .

0.14

h (4.558)

25

= 1.83607.. .

1.84

i1

(0.064)1

3

= (0.064)

1

3 = 0.4

4

aa

2b

3

a2

b4

=a2 2

b3 4

=a4b

7

b2a

2(2b)

3

(2a)2

b4 =

2a2

23b

3

22

a2

b4

= 24

a2

b3

22

a2

b4

= 24 2

a2 2

b3 4

= 26a

4b

7= 64a

4b

7



21/417

ca2

b3

a2

b4

=a2 2

b3 4

=a0b

1=b

d

a2b

3

a2b4 ab

a1b1 =

a2 + 1

b3 + 1

a2 + 1b4 + 1

= a

3b

4

a3

b5

=a3 3

b4 5

=a6b

9

e(2a)

2 8b

3

16a2

b4

=4a

2 8b

3

16a2

b4

=32a

2b

3

16a2

b4

= 32

16a

2 2b

3 4

= 2a4b

7

f2a

2b

3

8a2

b4

16ab

(2a)1

b1 =

2a2b

3

8a2

b4

(2a)1

b1

16ab

=2a

2b

3

8a2

b4

21

a1

b1

16ab

=2

1 + 1a

2 + 1b

3 + 1

8 16a2 + 1

b4 + 1

=2

0a

1b

2

128a1

b3

= 1128

a1 1

b2 3

=a2b

5

128

52

n 8

n

22n

16=

2n

(23)

n

22n

24

=2

n 2

3n

22n

24

=2

n + 3n 2n

24 = 2

2n 2

4

6 2x

3x

62x

32x

22x

= (2 3)x

62x

(2 3)2x

= 6x

62x

62x

= 6x + 2x + 2x

= 63x

7 In each case, add the fractional indices.

a 2

13

2

16

22

3= 2

26

+ 16

+ 46

= 21

6=

12

16

b a

14

a

25

a 1

10=a

520

+ 820

+ 220

=a

1120

c 2

23

2

56

22

3= 2

46

+ 56

+ 46

= 2

56

d

2

13

2

2

12

5

= 2

23

2

52

= 2

46

+ 156

= 2

196

e 2

13

2

2

13

22

5= 2

23

2

13

22

5

= 2

23

+ 13

+ 25

= 2

35

8

a 3 a3b

2

3 a2b1

= (a3b

2)

13

(a2b1

)

13

=a1b

23

a

23b

13

=a1 2

3b

23

13

=a

13b

b a3b

2 a

2b1

= (a3b

2)

12

(a2b1

)

12

=a

32b

1a

1b

12

=a

32

+ 1

b1 + 1

2=a

52b

12

c 5 a3b

2 5 a

2b1

= (a3b

2)

15

(a2b1

)

15

=a

35b

25

a

25b

15

=a

35

+ 25b

25

+ 15

=ab

15



22/417

d a4

b2

a3b1

= (a4

b2)

12

(a3b1

)

12

=a2

b1

a

32b

12

=a2 + 3

2b

1 + 12

=a1

2b

12

=b

12

a

12

=b

a

12

f 5 a3b

2

5 a2b1

= (a3b

2)

15

(a2b1

)

15

=a

35b

25

a

25b

15

=a

35

25b

25 1

5=a

15b

35

ga

3b

2

a2b1

c5

a4

b2

a3b1 a

3b1

=(a

3b

2)

12

a2b1

c5

(a4

b2)

12

a3b1 (a

3b1

)

12

= a

32b

1

a2b1

c5

a2

b1

a3b1a

32b

12

=a

32

2

b

1 1

c

0 5

a

2 3

b

1 1

a

32

b

12

=a1

2b

2c

5a

5b

2a

32b

12

=a1

2+ 5 + 3

2b

2 + 2 + 12c

5

=a4

b

72c

5

e a3b

2c3

a2b1

c5

= (a3b

2c3

)

12

(a2b1

c5

)

12

=a

32

b1

c

32

a1

b

12

c

52

=a

32

+ 1

b1 + 1

2c

32

+ 52

=a

52b

12c4



23/417


1

a 47.8 = 4.78 101

= 4.78 10

b 6728 = 6.728 103

c 79.23 = 7.923 101

= 7.923 10

d 43 580 = 4.358 104

e 0.0023 = 2.3 103

f 0.000 000 56 = 5.6 107

g 12.000 34 = 1.200 034 101

= 1.200 034

10h Fifty million = 50 000 000

= 5.0 107

i 23 000 000 000 = 2.3 1010

j 0.000 000 0013 = 1.3 109

k 165 thousand = 165 000

= 1.65 105

l 0.000 014 567 = 1.4567 105

2

a The decimal point moves 8 places to the

right = 1.0 108

b The decimal point moves 23 places to

the right = 1.66 1023

c The decimal point moves 5 places to the

right = 5.0 105

d The decimal point moves 3 places to the

left = 1.853 18 103

e The decimal point moves 12 places to

the left = 9.463 1012

f The decimal point moves 10 places to

the right = 2.998 1010

3

a The decimal point move 13 places to theright = 75 684 000 000 000

b The decimal point move 8 places to theright = 270 000 000

c The decimal point move 13 places to theleft = 0.000 000 000 000 19

4

a 324 000 0.000 000 74000

=3.24 10

5 7 10

7

4 103

= 3.24 74

105 + 7 3

= 5.67 105

= 0.000 056 7

b 5 240 000 0.842 000 000

=5.24 10

6 8 10

1

4.2 107

=41.92 10

5

4.2 107

=4192 10

3

42 000 103

= 419242 000

= 2622625

5

a3 a

b4

=3 2 10

9

3.2154

=3 2 3 10

9

106.8375. . .

= 1.2599. . . 103

106.8375. . .

= 0.011 792.. . 103

11.8

b4

a4b

4 =4

2 1012

4 0.054

=4 2 4 10

12

4 0.000 006 25

=1.189 2. . . 10

3

4 6.25 106

= 0.047 568.. . 109

4.76 107



24/417

Exercise 2C Solutions

1

a 3x + 7 = 153x = 15 7

= 8

x = 83

b 8 x2

= 15

x2

= 1 5 8

= 7

x2

2 = 72

x = 14

c 4 2 + 3x = 223x = 22 42

= 20

x = 203

d 2x3

15 = 27

2x3

= 27 + 15

= 422x

3

3

2

= 42 3

2x = 63

e 5(2x + 4) = 1310x + 20 = 13

10x = 13 20 = 7

x = 710

= 0.7

f 3(4 5x) = 2412 + 15x = 24

15x = 24 + 12 = 36

x = 3615

= 125

= 2.4

g 3x + 5 = 8 7x3x + 7x = 8 5

10x = 3

x =310 = 0.3

h 2 + 3 (x 4) = 4(2x + 5)2 + 3x 12 = 8x + 20

3x 10 = 8x + 203x 8x = 20 + 10

5x = 30

x = 305

= 6

i 2x5

34

= 5x

2x5

20 34

20 = 5x 20

8x 15 = 100x8x 100x = 15

92x = 15

x = 1592

j 6x + 4 =x3

3

6x 3 + 4 3 =x3

3 3 3

18x + 12 =x 918xx = 9 1217x = 21

x = 2117

2

a x2

+ 2x5

= 16

x2

10 + 2x5

10 = 16 10

5x + 4x = 1609x = 160

x = 1609



25/417

b 3x4

x3

= 8

3x4

12 x3

1 2 = 8 12

9x 4x = 965x = 96

x = 965

= 19.2

c 3x 22

+x4

= 18

3x 22

4 +x4

4 = 18 4

2(3x 2) +x = 726x 4 +x = 72

7x = 72 + 4 = 68

x = 687

d 5x4

43

= 2x5

5x4

60 43

60 = 2x5

60

75x 80 = 24x75x 24x = 80

51x = 80

x = 8051

e x 4

2+ 2x + 5

4= 6

x 42

4 + 2x + 54

4 = 6 4

2(x 4) + (2x + 5) = 242x 8 + 2x + 5 = 24

4x = 24 + 8 5 = 27

x = 274

= 6.75

f 3 3x10

2(x + 5)6

= 120

3 3x10 60

2(x + 5)6 60 =

120 60

6(3 3x) 20(x + 5) = 318 18x 20x 100 = 3

38x = 3 18 + 100 = 85

x = 8538

g 3 x4

2(x + 1)5

= 24

3 x4

20 2(x + 1)5

20 = 24 20

5(3 x) 8 (x + 1) = 4801 5 5x 8x 8 = 480

13x = 480 15 + 8 = 487

x = 48713

h 2(5 x)8

+ 67

= 4(x 2)3

2(5 x)8

168 + 67

168 = 4(x 2)3

168

42(5 x) + 144 = 224(x 2)210 + 42x + 144 = 224x 448

42x 224x = 448 + 210 144182x = 382

x = 382182

= 19191

3

a 3x+ 2y= 2; 2x 3y= 6Use elimination. Multiply the first equationby 3 and the second equation by 2.9x + 6y = 6 4x 6y = 12 + :

13x = 18

x = 18

13

Substitute into the first equation:

3 1813

+ 2y = 2

5413

+ 2y = 2

2y = 2 5413

= 2813

y = 1413



26/417

b 5x+ 2y= 4; 3xy= 6 e 7x 3y= 6;x+ 5y= 10Use elimination. Multiply the secondequation by 2.

Use substitution. Makexthe subject ofthe second equation.

5x + 2y = 4 6x 2y = 12 + :

11x = 16x = 16

11

x=10 5ySubstitute into the first equation:7(10 5y) 3y = 6

70 35y 3y = 638y = 6 70

= 76

y =7638

= 2

Substitute into the second, simplerequation:

3 1611

y = 6

4811

y = 6

y = 6 4811

y = 18

11

Substitute into the second equation:

x + 5 2 = 1 0x + 1 0 = 1 0

x = 0

f 15x+ 2y= 27; 3x+ 7y= 45Use elimination. Multiply the second

equation by 5.15x + 2y = 27

15x + 35y = 225 :

33y = 198

y =19833

= 6

c 2xy= 7; 3x 2y= 2

Use substitution. Makeythe subject ofthe first equation.y= 2x 7Substitute into the second equation:3x 2(2x 7) = 2

3x 4x + 14 = 2x = 2 1 4

x = 12

Substitute into the second equation:

3x + 7 6 = 453x + 42 = 45

3x = 45 42 = 3

x = 1

Substitute into the equation in whichyisthe subject:

y = 2 1 2 7= 17

d x+ 2y= 12;x 3y= 2Use substitution. Makexthe subject ofthe first equation.

x= 12 2ySubstitute into the second equation:12 2y 3y = 2

5y = 2 12 = 10

y = 2

Substitute into the first equation:x+ 2 2 = 12

x+ 4 = 12x= 8



27/417

Exercise 2D Solutions

1

a 4(x 2) = 604x 8 = 60

4x= 60 + 8= 68

x= 17

b The length of the square is 2x + 74

.

2x + 7

4

2

= 49

2x + 74

= 7

2x + 7 = 7 4 = 282x = 2 8 7 = 2 1x = 10.5

c The equation is length = twice width.x 5 = 2(12 x)x 5 = 24 2x

x + 2x = 24 + 53x = 29

x = 293

d y= 2((2x+ 1) + (x 3))= 2(2x+ 1 +x 3)= 2(3x 2)

= 6x 4

e Q= np

f If a 10% service charge is added, the totalprice will be multiplied by 110%, or 1.1.R= 1.1pS

g Using the fact that there are 12 lots of5 min in an hour (60 12 = 5),60n5

= 2400

h a = circumference 60360

= 2(x + 3) 60360

= 2(x + 3) 16

=3

(x + 3)

2 Let the value of Bronwyns sales in thefirst week be $s.s + (s + 500) + (s + 1000) + (s + 1500) + (s + 2000) = 1

5s + 5000= 175s = 12s = 2

The value of her first weeks sales is $2500.

3 Let dbe the number of dresses boughtand hthe number of handbags bought.65d+ 26h = 598

d+h = 11

Multiply the second equation by 26 (thesmaller number).65d+ 26h = 598 26d+ 26h = 286

:39d= 312

d= 31239

= 8

h + 8 = 11h = 3

Eight dresses and three handbags.

4 Let the courtyards width be wmetres.3w+ w+ 3w+ w= 67

8w= 67w= 8.375

The width is 8.375 m.The length = 3 8.375 = 25.125 m.

5 Letpbe the full price of a case of wine.The merchant will pay 60% (0.6) on the25 discounted cases.25p+ 25 0.6p= 2260

25p+ 15p= 226040p= 2260

p= 56.5The full price of a case is $56.50.



28/417

6 Letxbe the number of houses with an$11 500 commission andybe thenumber of houses with a $13 000commission.

9 Let rkm/h be the speed Kim can run.Her cycling speed will be (r+ 30) km/h.Her time cycling will be 48 + 48 3 = 64 min.Converting the times to hours ( 60) andusing distance = speed time gives thefollowing equation:

We only need to findx.x+y= 22

r 4860

+ (r+ 30) 6460

= 60

48r+ 64(r+ 30) = 60 6048r+ 64r+ 1920 = 3600

112r+ 1920 = 3600112r= 1680

r= 1680112

= 15

11 500x+ 13 000y= 272 500To simplify the second equation, divideboth sides by 500.23x+ 26y= 545Using the substitution method:

23x + 26y = 545y = 2 2 x

23x + 26(22 x) = 54523x + 572 26x = 545

3x = 545 572 = 27

x = 9

10 Let cg be the mass of a carbon atom

andxg be the mass of an oxygen atom.(ois too confusing a symbol to use)

He sells nine houses with an $11 500commission.

2c + 6x = 2.45 1022

x =c3

7 It is easiest to let the third boy have Use substitution.mmarbles, in which case the second boywill have 2mmarbles and the first boywill have 2m14.

2c + 6c3

= 2.45 1022

2c + 2c = 2.45 1022

4c = 2.45 1022

c = 2.45 1022

4

= 6.125 1023

x =c3

= 6.125 1023

3

2.04 1023

(2m 14) + 2m+ m= 715m 14 = 71

5m= 85m = 17

The first boy has 20 marbles, the second

boy has 34 and the third boy has 17marbles, for a total of 71.

8 Let Belindas score be b.Annies score will be 110% of Belindasor 1.1b.

The mass of an oxygen atom isCassies will be 60% of their combinedscores: 0.6(1.1b+ b) = 0.6 2.1b 2.04 10

23g.

= 1.26b1.1b +b + 1.26b = 504

3.36b = 504

b = 5043.36

= 150

Belinda scores 150Annie scores 1.1 150 = 165Cassie scores 0.6 (150 + 165) = 189



29/417


1 Let kbe the number of kilometrestravelled in a day.The unlimitedkilometre alternative will become moreattractive when 0.32k+ 63 > 108.

Solve for 0.32k+ 63 = 108:0.32k= 108 63

= 45

k= 450.32

= 140.625

The unlimited kilometre alternative willbecome more attractive when you travelmore than 140.625 km.

2 Let gbe the number of guests. Solve forthe equality.300 + 43g = 450 + 40g

43g 40g = 450 3003g = 150g = 50

Company A is cheaper when there aremore than 50 guests.

3 Let abe the number of adults and cthenumber of children.45a+ 15c= 525 000

a+ c= 15 000Multiply the second equation by 15.45a + 15c = 525 000

15a + 15c = 225 000

:30a = 300 000

a = 10 000

10 000 adults bought tickets.

4 Let $mbe the amount the contractor paid aman and $bthe amount he paid a boy.

8m+ 3b= 22406m+ 18b= 4200Multiply the first equation by 6.48 m + 18b = 13 440 6m + 18b = 4200 :

42m = 9240m = 220

Substitute in the first equation:

8 220 + 3b = 22401760 + 3b = 2240

3b = 2240 1760

= 480b = 160

He paid the men $220 each and the boys $160.

5 Let the numbers bexandy.x +y = 212 xy = 42

+ :2x = 254x = 127

127 +y = 212y = 85

The numbers are 127 and 85.

6 LetxL be the amount of 40% solutionandyL be the amount of 15% solution.Equate the actual substance.

0.4x + 0.15y = 0.24 700 = 168

x +y = 700 Multiply the second equation by 0.15.

0.4x + 0.15y = 168 0.15x + 0.15y = 105 :

0.25x = 63

x = 63 4 = 252

252 +y = 700y = 448

Use 252 L of 40% solution and 448 L of15% solution.

7 Form two simultaneous equations.x +y = 220

xx2

=y 40

x

2

y = 40

+ :3x2

= 180

x = 120120 +y = 220

y = 100 They started with 120 and 100 marblesand ended with 60 each.



30/417

8 Let $xbe the amount initially invested at10% and $ythe amount initiallyinvested at 7%. This earns $31 000.

9 Let abe the number of adults and sthenumber of students who attended.30a + 20s = 37 000

a +s = 1600

20a + 20s = 1600 20 = 32 000

:10a = 5000

a = 500500 +s = 1600

s = 1100

0.1x+ 0.07y= 31 000When the amounts are interchanged, sheearns $1000 more, i.e. $32 000.

0.07x+ 0.1y= 32 000Multiply the first equation by 100 andthe second equation by 70.10x + 7y = 3 100 000

4.9x + 7y = 2 240 000 :

5.1x = 860 000

x = 860 0005.1

168 627.451

500 adults and 1100 students attendedthe concert.

10 168 627.451 + 7y = 3 100 0001 686 274.51 + 7y = 3 100 000

7y = 1 413 725.49

y = 201 960.78 The total amount invested isx+y= 168 627.45 + 201 960.78

= $370 588.23



31/417

Exercise 2F Solutions

1

a v =u +at

= 15 + 2 5

= 25

b I=PrT100

= 600 5.5 10100

= 330

c V=r2h

= 4.252

6

340.47

d S= 2r(r+h)= 2 10.2 (10.2 + 15.6) 1653.48

e V= 43r

2h

= 4 3.582

11.43

612.01

f s =ut+ 12at

2

= 25.6 3.3 + 12

1.2 3.32

77.95

g T= 2 lg

= 2 1.459.8

= 2 0.3846. . .

2.42

h 1 = 1v

+ 1u

= 13

+ 17

= 1021

f= 2110

= 2.1

i c2

=a2

+b2

= 8.82

+ 3.42

= 89

c = 89

9.43

j v2

=u2

+ 2as

= 4.82

+ 2 2.25 13.6 = 91.04

v = 91.04

9.54

2

a v =u +at

vu =at a =vu

t

b S=n2(a +l)

2S=n(a +l)

a +l = 2Sn

l = 2Sn

a

c A =12bh

2A =bh

b = 2Ah

d P =I2R

PR

=I2

I= PR

e s =ut+ 12at2

sut= 12at

2

2(sut) =at2

a =2(sut)

t2



32/417

f E= 12

mv2

2E=mv2

v2

= 2Em

v = 2Em

g Q = 2gh

Q2

= 2gh

h =Q2

2g

h xyz =xy +zxyxy =z +z

2xy = 2z

x = 2z

2y

= zy

i ax +byc

=xb

ax +by =c(xb)ax +by =cxbcaxcx = bcby

x(ac) = b(c +y)

x = b(c +y)ac

=b(c +y)ca

j mx +bxb

=c

mx +b =c(xb)mx +b =cxbc

mxcx = bcbx(mc) = b(c + 1)

x = b(c + 1)mc

3

a F= 9C5

+ 32

= 9 285

+ 32

= 82.4

b F= 9C5

+ 32

F 32 = 9C5

9C= 5(F 32)

C= 5(F 32)9

Substitute F= 135.

C= 5(135 32)9

= 515

9

57.22

4

a S= 180(n 2)= 180(8 2)

= 1080

b S= 180(n 2)S

180=n 2

n = S180

+ 2

= 1260180

+ 2

= 7 + 2 = 9

Polygon has 9 sides (a nonagon).

5

a V= 13

r2h

= 13

3.52

9

115.45 cm3

b V= 13

2h

3V=r2h

h =3V

r2

=3 210

42

12.53 cm



33/417

c V= 13

r2h

3V=r2h

r2

= 3V

h

r= 3Vh

=3 262 10

5.00 cm

6

a S=n2(a +l)

= 72(3 + 22)

= 66.5

b S=n2

(a +l)

2S=n(a +l)2Sn

=a +l

a = 2Sn

l

= 2 104013

156

= 4

c S=n2

(a +l)

2S=n(a +l)

n = 2Sa +l

= 2 11025 + 5

= 11

There are 11 terms.



34/417

Exercise 2G Solutions

1

a 2x3

+ 3x2

= 4x + 9x6

=13x6

b 3a2

a4

= 6aa4

= 5a4

c 3h4

+ 5h8

3h2

= 6h + 5h 12h8

= h8

d 3x4

y6x

3= 9x 2y 4x

12

= 5x 2y12

e 3x

+ 2y

= 3y + 2xxy

f 5x 1

+ 2x

= 5x + 2(x 1)x(x 1)

= 5x + 2x 2x(x 1)

= 7x 2x(x 1)

g 3x 2

+ 2x + 1

= 3(x + 1) + 2(x 2)(x 2)(x + 1)

= 3x + 3 + 2x 4(x 2)(x + 1)

= 5x 1(x 2)(x + 1)

h2x

x + 3 4x

x 33

2

= 4x(x 3) 8x(x + 3) 3(x + 3)(x 3)2(x + 3)(x 3)

= 4x2 12x 8x

2 24x 3(x

2 9)

2(x + 3)(x 3)

= 4x2 12x 8x

2 24x 3x

2+ 27

2(x + 3)(x 3)

=7x2 36x + 27

2(x + 3)(x 3)

i 4x + 1

+3

(x + 1)2 =

4(x + 1) + 3

(x + 1)2

=4x + 4 + 3

(x + 1)2

=4x + 7

(x + 1)2

j a 2a

+a4

+ 3a8

= 8(a 2) + 2a2

+ 3a2

8a

= 5a2

+ 8a 168a

k 2x6x2 4

5x= 10x

2 (6x

2 4)

5x

=10x

2 6x

2+ 4

5x

= 4x2

+ 45x

= 4(x2

+ 1)5x

l2

x + 4

3

x2

+ 8x + 16= 2

x + 4

3

(x + 4)2

=2(x + 4) 3

(x + 4)2

=

2x + 8 3

(x + 4)2

=2x + 5

(x + 4)2



35/417

m 3x 1

+ 2(x 1)(x + 4)

= 3(x + 4) + 2(x 1)(x + 4)

= 3x + 12 + 2(x 1)(x + 4)

= 3x + 14(x 1)(x + 4)

n 3x 2

2x + 2

+4

x2 4

= 3x 2

2x + 2

+ 4(x 2)(x + 2)

= 3(x + 2) 2(x 2) + 4(x 2)(x + 2)

= 3x + 6 2x + 4 + 4(x 2)(x + 2)

= x + 14(x 2)(x + 2)

o 5x 2

3x

2+ 5x + 6

+ 2x + 3

= 5x 2

3(x + 2)(x + 3)

+ 2x + 3

= 5(x + 3)(x + 2) 3(x 2) + 2(x 2)(x + 2)(x 2)(x + 2)(x + 3)

= 5(x2

+ 5x + 6) 3x + 6 + 2 (x2 4)

(x 2)(x + 2)(x + 3)

= 5x2

+ 25x +3 0 3x + 6 + 2x2 8

(x 2)(x + 2)(x + 3)

= 7x2

+ 22x + 28(x 2)(x + 2)(x + 3)

p xy 1xy

= (xy)(xy) 1xy

= (xy)2 1

xy

q 3x 1

4x1 x

= 3x 1

+ 4xx 1

= 4x + 3x 1

r3

x 2 +2

2 x =3

x 22x

x 2

= 3 2xx 2

2

a x2

2y 4y

3

x= 4y

3x

2

2yx

= 2xy2

b 3x2

4y y

2

6x= 3x

2

y2

24yx

=xy8

c 4x3

3

12

8x4

=48x

3

24x4

= 2x

d x2

2y

3xy6

=x2

2y 6

3xy

=6x

2

6xy2

= x

y2

e4 x3a

a2

4 x=a

2(4 x)

3a(4 x)

=a3

f

2x + 5

4x2 + 10x =

2x + 5

2x(2x + 5)

= 12x

g(x 1)

2

x2

+ 3x 4= (x 1)

2

(x 1)(x + 4)

=x 1x + 4

h x2x 6x 3

= (x 3)(x + 2)x 3

=x + 2

ix

2 5x + 4

x2 4x

= (x 1)(x 4)x(x 4)

=x 1x



36/417

j5a

2

12b2

10a6b

=5a

2

12b2 6b

10a

=30a

2b

120ab2

= a

4b

k x 2x

x2 4

2x2

=x 2x

2x

2

x2 4

=x 2x

2x2

(x 2)(x + 2)

= 2x2

x(x + 2)

= 2xx + 2

l x + 2x(x 3)

4x + 8x

2 4x + 3

= x + 2x(x 3)

4(x + 2)(x 1)(x 3)

= x + 2x(x 3)

(x 1)(x 3)4(x + 2)

= 1x

x 14

=x 14x

m 2x

x 1

4x2

x2 1= 2x

x 1x

2 1

4x2

= 2xx 1

(x 1)(x + 1)

4x2

=2x(x + 1)

4x2

=x + 12x

n x2 9

x + 2 3x + 6

x 3

9x

= (x 3)(x + 3)

x + 2

3(x + 2)

x 3

x

9 = 3x(x 3)(x + 3)(x + 2)

9(x + 2)(x 3)

=x(x + 3)3

o 3x9x 6

6x2

x 2 2

x + 5

= 3x3(3x 2)

x 2

6x2

2x + 5

=2x(x 2)

6x

2

(3x 2)(x + 5) = x 2

3x(3x 2)(x + 5)

3

a 1x 3

+ 2x 3

= 3x 3

b 2x 4

+ 2x 3

= 2(x 3) + 2(x 4)(x 4)(x 3)

=2x 6 + 2x 8

x2 7x + 12

= 4x 14x

2 7x + 12

c 3x + 4

+ 2x 3

= 3(x 3) + 2(x + 4)(x + 4)(x 3)

=3x 9 + 2x + 8

x2

+x 12

=5x 1

x2

+x + 12

d2x

x 3+ 2

x + 4= 2x(x + 4) + 2(x 3)

(x 3)(x + 4)

=2x2 + 8x + 2x 6

x2

+x 12

=2x

2+ 10x 6

x2

+x + 12

e1

(x 5)2

+ 2x 5

=1 + 2 (x 5)

(x 5)2

=1 + 2x 10

x2 10x + 25

=2x 9

x

2

10x + 25

f3x

(x 4)2 +

2x 4

=3x + 2(x 4)

(x 4)2

=3x + 2x 8

x2 8x + 16

=5x 8

x2 8x + 16



37/417

g 1x 3

2x 3

= 1x 3

= 13 x

h 2

x 3 5

x + 4= 2(x + 4) 5(x 3)

(x 3)(x + 4)

=2x + 8 5x + 15

x2

+x 12

=2 3 3x

x2

+x 12

i2x

x 3+ 3x

x + 3= 2x(x + 3) + 3x(x 3)

(x 3)(x + 3)

=2x

2+ 6x + 3x

2 9x

x2 9

=5x

2 3x

x2 9

j1

(x 5)2

2x 5

=1 2 (x 5)

(x 5)2

=1 2x + 10

x2 10x + 25

=1 1 2x

x2 10x + 25

k2x

(x 6)3

2

(x 6)2

=2x 2(x 6)

(x 6)3

= 2x 2x + 12(x 6)

3

=12

(x 6)3

l 2x + 3x 4

2x 4x 3

= (2x + 3)(x 3) (2x 4)(x 4)(x 4)(x 3)

=(2x

2 3x 9) (2x

2 12x + 16)

x2 7x + 12

= 2x2

3x 9 2x2

+ 12x 16x

2 7x + 12

=9x 25

x2 7x + 12

4

a 1 x +2

1 x=

1 x 1 x + 2

1 x

=1 x + 2

1 x

= 3 x1 x

b2

x 4+ 2

3=

2 x 4 + 6

3 x 4

c3

x + 4+

2

x + 4=

5

x + 4

d3

x + 4+ x + 4 =

3 + x + 4 x + 4

x + 4

= 3 +x + 4x + 4

= x + 7

x + 4

e3x

3

x + 4 3x

2x + 4 =

3x3 3x

2x + 4 x + 4

x + 4

=3x

3 3x

2(x + 4)

x + 4

=3x

3 3x

3 12x

2

x + 4

= 12x

2

x + 4

f3x

3

2 x + 3+ 3x

2x + 3 =

3x3

+ 6x2

x + 3 x + 3

x + 3

=3x

3+ 6x

2(x + 3)

x + 3

=3x

3+ 6x

3+ 18x

2

x + 3

=

9x3

+ 18x2

x + 3

=9x

2(x + 2)

x + 3



38/417


39/417

Exercise 2H Solutions

1

a ax +n =max =mn

x =mn

a

b ax +b =bxaxbx = b

x(ab) = b

x = bab

This answer is correct, but to avoid anegative sign, multiply numerator anddenominator by 1.

x = bab

11

= bba

c axb

+c = 0

axb

= c

ax = bc

x = bca

d px =qx + 5

pxqx = 5x(pq) = 5

x = 5pq

e mx +n =nxmmxnx = mn

x(mn) = mn

x = mnmn

=m +nnm

f 1x +a

=bx

Take reciprocals of both sides:

x +a =xb

xxb

= a

xb

x =a

xxbb

=a

xxbb

b =ab

xxb =abx(1 b) =ab

x = ab

ab

g bxa

= 2bx +a

Take reciprocals of both sides:xa

b=x +a

2bxa

b 2b =x +a

2b 2b

2(xa) =x +a2x 2a =x +a

2xx =a + 2a

x = 3a

h xm

+n =xn

+m

xm

mn +nmn =xn

mn +mmn

nx +mn2

=mx +m2n

nxmx =m2nmn

2

x(nm) =mn(mn)

x =mn(mn)nm

Note that n m = m +n

= 1(mn) x = mn(nm)

nm = mn



40/417

i b(ax +b) =a(bxa)

abxb2

=abxa2

abxabx = a2

+b2

2abx = a2

+b2

x = ( a2

+b2)

2ab

=a2b

2

2ab

j p2(1 x) 2pqx =q

2(1 +x)

p2p

2x 2pqx =q

2+q

2x

p2x 2pqxq

2x =q

2p

2

x(p2

+ 2pq +q2) =q

2p

2

x = (q

2p

2)

p2

+ 2pq +q2

= p2

q2

(p +q)2

=(pq)(p +q)

(p +q)2

=pqp +q

k xa

1 =xb

+ 2

xa

abab =xb

ab + 2ab

bxab =ax + 2ab

bxax = 2ab +abx(ba) = 3ab

x = 3abba

l xab

+ 2xa + b

=1

a2b

2

x(ab)(a + b)ab

+ 2x(a + b)(ab)a + b

=(a + b)(ab)

a2b

2

x(a + b) + 2x(ab) = 1ax + bx + 2ax 2bx = 1

3axbx = 1

x(3ab) = 1

x = 13ab

m pqxt

+p =qx 1p

pt(pqx)t

+ppt=pt(qx 1)p

p(pqx) +p2t=t(qx 1)

p2pqx +p

2t=qtxt

pqxqtx = tp2p

2t

qx(p +t) = (t+p2

+p2t)

x =t+p2

+p2t

q(p +t)or p

2+p

2t+t

q(p +t)

n1

x +a+ 1

x + 2a= 2

x + 3a

Multiply each term by (x+ a)(x+ 2a)(x+ 3a).(x + 2a)(x + 3a) + (x +a)(x + 3a) = 2(x +a)(x + 2a

x2

+ 5ax + 6a2

+x2

+ 4ax + 3a2

= 2x2

+ 6ax + 4a

2x2

+ 9ax + 9a2

= 2x2

+ 6ax + 4a

2x2 9ax 2x

2 6ax = 4a

2 9a

2

3ax = 5a2

x =5a2

3a

= 5a3

2 ax +by =p ;bxay =q

Multiply the first equation by aand thesecond equation by b.

a2x +aby =ap

b2xaby =bp

+ :

x(a2

+b2) =ap +bq

x =ap +bq

a2

+b2

Substitute into ax+ by=p:

aap +bq

a2

+b2 +by =p

a(ap +bq) +by(a2

+b2) =p(a

2+b

2)

a2p +abq +by(a

2+b

2) =a

2p +b

2p

by(a

2

+b

2

) =a

2

p +b

2

pa

2

pabqby(a

2+b

2) =b

2pabq

y =b(bpaq)

b(a2

+b2)

=bpaq

a2

+b2



41/417


42/417

c Add the starting equations:ax +by +axby =t+s

2ax =t+s

x =t+s2a

Subtract the starting equations:

ax +by (axby) =ts2by =ts

y =ts2b

d Multiply the first equation by aand thesecond equation by b.

a2x +aby =a

3+ 2a

2bab

2

b2x +aby =a

2b +b

3

:

x(a2b

2) =a

3+a

2bab

2b

3

x =a

3

+a

2

bab

2

b

3

a2b

2

=a

2(a +b) b

2(a +b)

a2b

2

=(a

2b

2)(a +b)

a2b

2

=a +b

Substitute into the second, simplerequation.

b(a +b) +ay =a2

+b2

ab +b2

+ay =a2

+b2

ay =a2 +b

2abb2

ay =a2ab

y =a2aba

=ab

e Rewrite the second equation, thenmultiply the first equation by b + candthe second equation by c.(a +b)(b +c)x +c(c +c)y =bc(b +c)

acx +c(b +c)y = abc :

x((a +b)(b +c) ac) =bc(b +c) +abcx(ab +ac +b

2+bcac) =bc(b +c +a)

x(ab +b2

+bc) =bc(a +b +c)xb(a +b +c) =bc(a +b +c)

x =bc(a +b +c)b(a +b +c)

=c

Substitute into the first equation. (It hasthe simpleryterm.)c(a +b) +cy =bcac +bc +cy =bc

cy =bcacbc

cy = ac

y = acc

= a

f First simplify the equations.3x 3a 2y 2a = 5 4a

3x 2y = 5 4a + 3a + 2a3x 2y =a + 5

2x + 2a + 3y 3a = 4a 12x + 3y = 4a 1 2a + 3a2x + 3y = 5a 1

Multiplyby 3 andby 2.9x 6y = 3a + 15 4x + 6y = 10a 2 + :

13x = 13a + 13x =a + 1

Substitute into:

2(a + 1) + 3y = 5a 12a + 2 + 3y = 5a 1

3y = 5a 1 2a 23y = 3a 3y =a 1



43/417

e s =h2

+ah

= (3a2)

2+a(3a

2)

= 9a4

+ 3a3

= 3a3(3a + 1)

5

a s =ah=a(2a + 1)

b Make hthe subject of the second equation.h =a(2 +h)

= 2a +ahhah = 2a

h(1 a) = 2a

h = 2a1 a

f as =a + 2h =a + 2(as) =a + 2a 2s

as + 2s = 3as(a + 2) = 3a

s = 3aa + 2

Substitute into the first equation.s =ah

=a 2a1 a

= 2a2

1 a

g s = 2 +ah +h2

= 2 +aa1

a +

a1

a

2

= 2 +a

2

1 +a

2

2 +

1

a2

= 2a2 1 +

1

a2

c h +ah = 1h(1 +a) = 1

h = 1(1 +a)

= 1a + 1

as =a +h

=a + 1a + 1

=a(a + 1) + 1a + 1

=a2

+a + 1a + 1

s =a2

+a + 1

a(a + 1)

h Make hthe subject of the secondequation.as + 2h = 3a

2h = 3aas

h = 3aas2

Substitute into the first equation.

3sah =a

2

3sa(3aas)2

=a2

6sa(3aas) = 2a2

6s 3a2

+a2s = 2a

2

a2s + 6s = 2a

2+ 3a

2

s(a2

+ 6) = 5a2

s =5a

2

a2

+ 6

d Make hthe subject of the second equation.ah =a +h

ahh =ah(a 1) =a

h = 1a 1

Substitute into the first equation.as =s +h

as =s + aa 1

ass = aa 1

s(a 1) = aa 1

s(a 1)(a 1) =a(a 1)a 1

s(a 1)2

=a

s = a

(a 1)2



44/417

Exercise 2I Solutions

Use your CAS calculator to find thesolutions to these problems. The exactmethod will vary depending on the

calculator used.1a x= a b

b x= 7

c x = a a2

+ 4ab 4b2

2

d x =a +c2

2a (x 1)(x+ 1)(y 1)(y+ 1)

b (x 1)(x+ 1)(x+ 2)

c (a2 12b)(a

2+ 4b)

d (ac)(a 2b +c)

3

a axy +b = (a +c)ybxy +a = (b +c)yDividing byyyields:

ax +by

=a +c

bx +ay

=b +c

let n = 1y

and the equations become:

ax +bn =a +cbx +an =b +c

x =a +b +ca +b

y =a +bc

b x(bc) +byc = 0y(ca) ax +c = 0

(bc)x +by =cax + (ca)y = c

x = (abc)a +bc

y =ab +ca +bc



45/417

Solutions to Multiple-choice Questions

1 5x + 2y = 02y = 5x

yx

= 52

A

2 Multiply both sides of the secondequation by 2.

3x + 2y = 36 6x 2y = 24

+ :9x = 60

x = 203

3 203

y = 12

20 y = 12

y= 8 A

3 t 9 = 3t 17t 3t= 9 17

2t= 8

t= 4 C

4 m =npn +p

m(n +p) =npmn +mp =np

mp +p =nmn

p(m + 1) =n(1 m)

p =n(1 m)1 +m

A

5 3x 3

2x + 3

= 3(x + 3) 2(x 3)(x 3)(x + 3)

=3x + 9 2x + 6

x2 9

=x + 15

x2 9

B

6 9x2y

3

15(xy)3

=9x

2y

3

15(xy)3

=

9x2y

3

15x3y3

= 915x

= 35x

E

7 V= 13

h(l +w)

3V=h(l +w)3V=hl +hwhl = 3Vhw

l = 3Vhwh

= 3Vh

w B

8(3x

2y

3)

2

2x2y

=9x

4y

6

2x2y

= 9x2y

5

2

= 92

x2y

5 B

9 Y= 80%Z= 4

5

Z

X= 150%Y= 32

Y

= 32

4Z5

= 12Z10

= 1.2Z

= 20% greater thanZ B

10 Let the other number be n.x +n

2= 5x + 4

x +n = 2(5x + 4) = 10x + 8n = 10x + 8 x

= 9x+ 8 B



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47/417

5 Let tseconds be the required time.The number of red blood cells to be

replaced is 12

5 1012

= 2.5 1012

2.5 106

t= 2.5 1012

t= 2.5

10

12

2.5 106

= 106

Time = 106seconds

= 106

3600

24 days

11.57 or 11 3154

days

61.5 10

8

3 106

= 0.5 102

= 50 times further

7 Let gbe the number of games the teamlost. They won 2ggames and drew onethird of 54 games, i.e. 18 games.g + 2g + 1 8 = 5 4

3g = 54 18 = 36g = 12

They have lost 12 games.

8 Let bbe the number of blues CDs sold.The store sold 1.1bclassical and

1.5(b + 1.1b) heavy metal CDs, totalling420 CDs.

b + 1.1b + 1.5 2.1b = 4205.25b = 420

b = 4205.25

= 80

1.1b = 1.1 80 = 881.5 2.1b = 1.5 2.1 80 = 252

80 blues, 88 classical and 252 heavymetal (totalling 420)

9

a V= r2h

= 52

12

= 300

942 cm3

b h = Vr

2

=585

52

= 117

5

7.4 cm

c r2

= V

h

r= V

h (use positive root)

= 786 6

=128

40.7 cm

10

a xy +ax =bx(y +a) =b

x = ba +y

b ax +bx =c

axx

+bxx

=cx

a +b =cx

x =a +bc

c xa

=xb

+ 2

xab

a=xab

b+ 2ab

bx =ax + 2ab

bxax = 2abx(ba) = 2ab

x = 2abba



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d adxd

+b =ax +db

bd(adx)d

+bdb =bd(ax +d)b

b(adx) +b2d=d(ax +d)

abbdx +b2d=adx +d

2

bdxadx =d2abb

2d

x(bd+ad) = (ab +b2dd

2)

x = (ab +b2dd

2)

(bd+ad)

=ab +b2dd

2

bd+ad

11

a pp +q

+ qpq

=p(pq) +q(p +q)(p +q)(pq)

=p

2qp +qp +q

2

p2pq +pqq

2

=p

2+q

2

p2q

2

b 1x

2y

xyy2

=(xyy

2) 2xy

x(xyy2)

=xyy

2

x2yxy

2

=y( xy)xy(xy)

= xyx(xy)

= x +yx(yx)

c x2

+x 6x + 1

2x2

+x 1x + 3

= (x 2)(x + 3)x + 1

(x + 1)(2x 1)x + 3

= (x 2)(2x 1)

d

2a

2a +b

2ab +b2

ba2 =

2a

2a +b

b(2a +b)

ba2

=2ab

ba2

= 2a

12 LetAs age be a,Bs age be band Csage be c.

a = 3bb + 3 = 3(c + 3)

a + 15 = 3(c + 15)

Substitute for a and simplify:

b + 3 = 3(c + 3)b + 3 = 3c + 9

b = 3c + 6 3b + 15 = 3(c + 15)3b + 15 = 3c + 45

3b = 3c + 30b = c + 10 = :

3c + 6 = c + 103cc = 10 6

2c = 4c = 2

b = 3

2 + 6 = 12

a = 3 12 = 36

A,Band Care 36, 12 and 2 years oldrespectively.

13

a Simplify the first equation:

a 5 = 17

(b + 3)

7(a 5) =b + 37a 35 =b + 37ab = 38

Simplify the second equation:

b 12 = 15(4a 2)

5(b 12) = 4a 25b 60 = 4a 2

4a + 5b = 58

Multiply the first equation by 5, and addthe second equation.35a 5b = 190

4a + 5b = 58 + :

31a = 248a = 8

Substitute in the first equation:

7 8 b = 3856 b = 38

b = 5 6 3 8 = 1 8



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b Multiply the first equation byp.

(pq)x + (p +q)y = (p +q2)

p(pq)x +p(p +q)y =p(p +q2)

Multiply the second by (p+ q).

qxpy =q2pq

q(p +q)xp(p +q)y = (p +q)(q2pq)

+:

(p(pq) + q(p + q))x =p(p + q)2

+ (p +q)(q2pq)

(p2pq +pq +q

2)x =p(p

2+ 2pq +q

2) +pq

2p

2q + q

3pq

2

(p2

+q2)x =p

3+ 2p

2q +pq

2p

2q + q

3

=p3

+p2q +pq

2+ q

3

=p2(p + q) + q

2(p + q)

= (p + q)(p2

+ q2)

x =p + q

Substitute into the second equation, factorising the right side.q(p +q) py =q

2pq

pq +q2py =q

2pq

py =q2pqpqq

2

py = 2pq

y =2pqp

= 2q

14 Time = distancespeed

Remainder = 50 7 7 = 36 km7x

+ 74x

+ 366x + 3

= 4

7x

+ 74x

+ 122x + 1

= 4

(4x(2x + 1))7x

+ 74x

+ 122x + 1

= 4 4x(2x + 1)

28(2x + 1) + 7(2x + 1) + 48x = 16x(2x + 1)

56x + 28 + 14x + 7 + 48x = 32x2

+ 16x

56x + 2 8 + 1 4x + 7 + 48x 32x2 16x = 0

32x2

+ 102x + 35 = 0

32x2 102x 35 = 0

(2x 7)(16x + 5) = 02x 7 = 0 or 16x + 5 = 0

x > 0, so 2x 7 = 0x = 3.5



50/417

15

a 2n2

6nk2

3n =2n

2 6nk

2

3n

= 12n3k

2

3n

= 4n

2

k

2

b8c

2x

3y

6a2b

3c

3

12xy

15abc2

=8c

2x

3y

6a2b

3c

3xy

30abc2

=8c

2x

3y

6a2b

3c

330abc

2

xy

=240abc

4x

3y

6a2b

3c

3xy

=40cx

2

ab2

16 x+ 515

x 510

= 1 + 2x15

30(x + 5)15

30(x 5)10

= 301 + 2x

15

2(x + 5) 3(x 5) = 30 + 4x2x + 10 3x + 15 = 30 + 4x

2x 3x 4x = 30 10 155x = 5

x = 1



51/417

Chapter 3 Number systems and setsExercise 3A Solutions

1

a A'= {4}

b B' = {1, 3, 5}

c AB = {1, 2, 3, 4, 5}, or

d (A

B)'=

e A'B' =

2

a P' = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16}

b Q'= {1, 3, 5, 7, 9, 11, 13, 15}

c PQ= {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16}

d (PQ)'= {1, 5, 7, 11, 13}

e P' Q' = {1, 5, 7, 11, 13}

3

a A' = {1, 2, 3, 5, 6, 7, 9, 10, 11}

b B'= {1, 3, 5, 7, 9, 11}

c AB= {2, 4, 6, 8, 10, 12}

d (AB)'= {1, 3, 5, 7, 9, 11}

e A' B' = {1, 3, 5, 7, 9, 11}

4

a P' = {10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25}

b Q'= {11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}

c PQ= {10, 12, 15, 16, 20, 24, 25}

d (PQ)'= {11, 13, 14, 17, 18, 19, 21, 22, 23}

e P' Q' = {11, 13, 14, 17, 18, 19, 21, 22, 23}

5

a A' = {R}

b B'= {G, R}

c AB= {L, E, A, N}

d AB= {A, N, G, E, L}

e (AB)'= {R}

f A' B' = {G, R}



52/417

6

a X' = {p, q, u, v}

b Y'= {p, r, w}

c X'Y'= {p}

d X'Y'= {p, q, r, u, v, w}

e XY= {q, r, s, t, u, v, w}

f (X Y)'= {p}

c and fare equal.

7

a X' = {5, 7, 8, 9, 10, 11}

b Y'= {1, 3, 5, 7, 9, 11}

c X'Y'= {1, 3, 5, 7, 8, 9, 10, 11}

d X'Y'= {5, 7, 9, 11}

e XY= {1, 2, 3, 4, 6, 8, 10, 12}

f (X Y)'= {5, 7, 9, 11}

d and fare equal.

8

a

b

c

d

e

f

9

a A' = {E, H, M , S}

b B'= {C, H, I, M}

c AB= {A, T}

d (A B)'= {H, M}

e A'B'= {C, E, H, I, M, S}

f A' B'= {H, M}



53/417


1

a Yes, because the sum can be expressed

as a terminating or recurring decimal.

For instance, 110 + 120 = 320.

b Yes, because the product can be

expressed as a terminating or recurring

decimal. For instance, 110

1

20= 1

200.

c Yes, because the quotient can be


decimal, so long as the denominator 0.

For instance, 110

120

= 110

20 = 2.

2

a No, because the sum cannot be


decimal. For instance, 2 + 3 .

a No, because the sum cannot be


decimal. For instance, 2 3 = 6 .

c No, because the sum cannot be


decimal. For instance,2

3.

3

a 0..2

.7 = 0.272727. . .

0..2

.7 100 = 27.272727. . .

0..2

.7 99 = 27

0..2

.7 = 27

99= 3

11

b 0.12 =12100 =

325

c 0..28571

.4 = 0.285714285714. . .

0..28571

.4 10

6= 285714.285714. . .

0..28571

.4 (10

6 1) = 285714

0. .28571.4 = 285714999999

= 27

d 0..3

.6 = 0.363636. . .

0..3

.6 100 = 36.3636. . .

0..3

.6 99 = 36

0..3

.6 = 36

99= 4

11

e 0..2 = 0.22222. . .

0..2 10 = 2.2222. . .

0.

.

2 9 = 2

0..2 = 2

9

f 0.45 = 45100

= 920

4

a 27

= 7 2.000000. ..

= 0.2857142857. . .

= 0..28571

.4

b511

= 11 5.000000. . .

= 0.454545. . .

= 0..4

.5

c 720

= 20 7.00

= 0.35

d413

= 13 4.000000. . .

= 0.30769230. . .= 0.

.30769

.2

e117

= 17 1.00000000000000000.. .

= 0.0588235294117647058. . .

= 0..058823529411764

.7



54/417

5 Assume 3 is a rational number, so

3 is a fraction in its simplest form, ab

.

3 = ab

3 =a

2

b2

3b2

= a2

ais a multiple of 3.

a = 3k, where kis an integer.

3b2

= (3k)2

3b2

= 9k2

b2

= 3b2

bis a multiple of 3.

But this contradicts the assumption thata

b

is a fraction in its simplest form,

as aand bare both multiples of 3.

Therefore the initial assumption must be

incorrect and 3 is not a rational

number.



55/417


56/417


57/417


58/417

Exercise 3D Solutions

1

a 2 68 640

2 34 320

2 17 160

2 8580

2 4290

3 2145

5 715

11 143

13 13 1

Prime decomposition

= 25 3 5 11 13

b 2 96 096

2 48 048

2 24 024

2 12 012

2 6006

3 3003

7 1001

11 143

13 13 1

Prime decomposition

= 25 3 7 11 13

c 2 32 032

2 16 016

2 8008

2 4004

2 2002

7 1001

11 143

13 13 1

Prime decomposition

= 25 7 11 13

d 2 544 544

2 272 272

2 136 136

2 68 068

2 34 034

7 17 017

11 2431

13 221

17 17 1

Prime decomposition

= 25 7 11 13 17

2 For each part, first find the prime

decomposition of each number.

a 4361 = 72 89

Neither 7 nor 89 are factors of 9281.

HCF = 1

b 999 = 33 37

2160 = 24 3

3 5

HCF = 33

= 27

c 5255 = 5 1051

716 845 is divisible by 5 but not 1051.HCF = 5

d 1271 = 31 41

3875 = 53 31

HCF = 31

e 804 = 22 3 67

2358 = 2 32 131

HCF = 2 3 = 6



59/417


1

a

9 5

H E

x

Since all students do at least one of

these subjects, 9 + 5 +x= 28

x= 14

b i 5 + 14 = 19

ii 9

iii 9 + 14 = 23 or 28 5 = 23

2

a

3 9

7

5

6 42

14

C

A B

b i n(A'C') = 9 + 14 = 23

ii n(AB') = 3 + 6 + 5 + 2 + 7 + 14

= 37

iii n(A'BC') = 9

3

yx

B(60%) G

20%

Since 40% dont speak Greek,y+ 20% = 40%

y= 20%

Since 40% speak Greek,

x+ 20% = 40%

x= 20%

20% speak both languages.

4

yx6

C(25) D(16)

Since 40 25 = 15 dont own a cat,

y+ 6 = 15

y= 9

Since 16 own a dog,x+ 9 = 16

x= 7

Seven students own both.

5

a

a

b cx

E(70) F(50)

J(50)

We must assume every delegate spokeat least one of these languages.

If 70 spoke English, and 25 spoke

English and French, 45 spoke English

but not French.

45 + 50 = 95 spoke either English or

French or both.

105 95 = 10 spoke only Japanese.

If 50 spoke French, and 15 spoke French

and Japanese, 35 spoke French but not

Japanese.

35 + 50 = 85 spoke either French or

Japanese or both.

105 85 = 20 spoke only English.

If 50 spoke Japanese, and 30 spoke

Japanese and English, 20 spoke

Japanese but not English.

20 + 70 = 90 spoke either Japanese or

English or both.

105 90 = 15 spoke only French.



60/417

We can now fill in more of the Venn

diagram.

a

b cx

E(70) F(50)

J(50)

20 15

10

cis the number who dont speak English.105 70 = 10 + c + 15

c + 25 = 35c = 10

x + c = 15x = 5

5 delegates speak all five languages.

b We have already found that 10 spokeonly Japanese.

6 Enter the information into a Venn

diagram.

50 70

60

30

10 4540

P G

F

Number having no dessert

= 350 50 30 70 10 40 45 60

= 45

7 Insert the given information on a Venn

diagram. Placeyas the number taking a

bus only, andzas the number taking a

car only.

z y

x

x

C B ( 33 )

T ( 20 )

4

82

a Using n(T ) = 20, 2x+ 10 = 20

x= 5

b Using n(B) = 33 andx= 5,12 +5 +y= 33

y= 16

c Assume they all used at least one of

these forms of transport.

z+ 4 + 8 + 16 + 2 + 5 + 5 = 40

z= 0

8

a

b i (XYZ) = intersection of all sets

= 36 (from diagram)

ii n(X Y) = number of elements in bothXand Y

= 5 (from diagram)



61/417

9 The following information can be placed

on a Venn diagram.

10

2

5 33

R G

B

x

x

The additional information gives

5 >xandx> 3.

x= 4

Number of students

= 10 + 2 + 4 + 5 + 3 + 3 + 4

= 31

20 bought red pens, 12 bought green

pens and 15 bought black pens.

10 Enter the given information as below.

BMis shaded.

x y

12

z

B M

F

5 5

5

5 + 12 + 5 + 5 + x +y +z = 28

27 +x +y +z = 28x +y +z = 1

This means that exactly one ofx,yandz

must equal 1, and the other two will

equal zero.

Since n(FB) > n(MF), the Venndiagram shows that this meansx>y.

x= 1,y=z= 0

5 5

5

1 0

0

B M

F

a + b = 12

a

b

n(MFB) = n(F')

b= a+ 10

Substitute in a+ b= 12:a + (a + 10) = 12

2a = 12a = 6

b = a + 10 = 16

n(MF) = b+ 0 = 16

11 Enter the given information as below.

p q

r

a

b cx

A(23) S

Documents

AGM - Worked Solutions