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
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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.
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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.
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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.
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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
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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 .
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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.
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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
Determinant = 1 4 2 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
Determinant = 2 1 5 1 = 7
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
Determinant = 2 1 3 3 = 11
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
Determinant = 4 3 4 5 = 8
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.
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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
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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
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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
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5 Find the inverse of 1
3
2
4 .
Determinant = 1 4 2 3 = 2
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
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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.
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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
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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
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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
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Exercise 2B Solutions
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
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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
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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
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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
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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.
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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
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Exercise 2E Solutions
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.
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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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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
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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
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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}
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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}
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Exercise 3B Solutions
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
expressed as a terminating or recurring
decimal, so long as the denominator 0.
For instance, 110
120
= 110
20 = 2.
2
a No, because the sum cannot be
expressed as a terminating or recurring
decimal. For instance, 2 + 3 .
a No, because the sum cannot be
expressed as a terminating or recurring
decimal. For instance, 2 3 = 6 .
c No, because the sum cannot be
expressed as a terminating or recurring
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
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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.
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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
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Exercise 3E Solutions
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.
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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)
Cambridge Essential Advanced General Mathematics 3rd Edition Worked Solutions CD-ROM 60
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8/10/2019 AGM - Worked Solutions
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