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Chapter 13

Solving simultaneous
equations
13
...
However, when an equation contains
two unknown quantities it has an infinite number of
solutions
...
Similarly, for three unknown quantities it is
necessary to have three equations in order to solve for a
particular value of each of the unknown quantities, and
so on
...

Two methods of solving simultaneous equations analytically are:
(a)

by substitution, and

(b) by elimination
...
)

Problem 1
...

Removing the bracket gives
−4 − 8y − 3y = 18
−11y = 18 + 4 = 22
y=

22
= −2
−11

Substituting y = −2 into equation (1) gives
x + 2(−2) = −1
x − 4 = −1
x = −1 + 4 = 3

13
...


DOI: 10
...
00013-2

Thus, x = 3 and y = −2 is the solution to the
simultaneous equations
...
)
Problem 3
...
Thus,
2 × equation (1) gives

6x + 8y = 10

(3)

3 × equation (2) gives

6x − 15y = −36

(4)

Equation (3) – equation (4) gives

(Note: in the above subtraction,
18 − −4 = 18 + 4 = 22
...
The solution
x = 3, y = −2 is the only pair of values that satisfies
both of the original equations
...
e
...
)
Substituting y = 2 in equation (1) gives

Problem 2
...
e
...
e
...

The elimination method is the most common method of
solving simultaneous equations
...
Solve

x + 3(−3) = −7

7x − 2y = 26

(1)

x − 9 = −7

6x + 5y = 29

(2)

x = −7 + 9 = 2

Thus, x = 2, y = −3 is the solution of the simultaneous
equations
...
e
...


92 Basic Engineering Mathematics
5 × equation (1) gives

35x − 10y = 130

(3)

2 × equation (2) gives

12x + 10y = 58

(4)

Adding equations (3)
and (4) gives

47x + 0 = 188

Hence,

x=

188
=4
47

Note that when the signs of common coefficients are
different the two equations are added and when the
signs of common coefficients are the same the two
equations are subtracted (as in Problems 1 and 3)
...
3s + 2t = 12
4s − t = 5

12
...
5m − 3n = 11
3m + n = 8

14
...
5x = 2y
3x + 7y = 41

16
...
3 Further solving of simultaneous
equations
Here are some further worked problems on solving
simultaneous equations
...


Solve

2 = 2y

3 p = 2q

(1)

y=1

4 p + q + 11 = 0

(2)

Checking, by substituting x = 4 and y = 1 in equation
(2), gives
LHS = 6(4) + 5(1) = 24 + 5 = 29 = RHS
Thus, the solution is x = 4, y = 1
...

1
...
2x − y = 2
x − 3y = −9

3
...
3x − 2y = 10
5x + y = 21

5
...
7x + 2y = 11
3x − 5y = −7

7
...
a + 2b = 8
b − 3a = −3

9
...
2x + 5y = 7
x + 3y = 4

11 p + 0 = −22
p=

−22
= −2
11

Substituting p = −2 into equation (1) gives
3(−2) = 2q
−6 = 2q
q=

−6
= −3
2

Checking, by substituting p = −2 and q = −3 into
equation (2), gives
LHS = 4(−2) + (−3) + 11 = −8 − 3 + 11 = 0 = RHS
Hence, the solution is p = −2, q = −3
...
Solve

Problem 7
...
5x + 0
...
6x = 1
...
2y
(2)

Whenever fractions are involved in simultaneous equations it is often easier to firstly remove them
...
e
...
Thus,
multiplying equations (1) and (2) by 100 gives
250x + 75 − 300y = 0

(1)

160x = 108 − 120y

(2)

Rearranging gives

Multiplying equation (2) by 3 gives
39 − y = 9x

(4)

x − 8y = −20

(5)

9x + y = 39

(6)

73x + 0 = 292
292
x=
=4
73
Substituting x = 4 into equation (5) gives
4 − 8y = −20
4 + 20 = 8y
24 = 8y
24
=3
8

Checking, substituting x = 4 and y = 3 in the original
equations, gives

(2):

4 5 1
1
LHS = + = + 2 = 3 = y = RHS
8 2 2
2
3
LHS = 13 − = 13 − 1 = 12
3
RHS = 3x = 3(4) = 12

Hence, the solution is x = 4, y = 3
...
3
1300 130 10

Substituting x = 0
...
3) + 75 − 300y = 0
75 + 75 = 300y
150 = 300y
y=

150
= 0
...
3 and y = 0
...
3) = 48
RHS = 108 − 120(0
...
3, y = 0
...

x
y
1
...

+ =4
2 3
y
x
− =0
−1 = 3q − 5 p
6 9
3
...


a
− 7 = −2b
2
2
12 = 5a + b
3
x 2y
49
+
=
5
3
15
y 5
3x
− + =0
7
2 7

7
...
5x − 2
...
4x + 0
...
e
...


b=1

i
...


LHS = 2 − 4(1) = 2 − 4 = −2 = RHS
Hence, a = 2 and b = 1
...
v − 1 =

8
...
5a = 0
...
6a + 0
...
8

However, since

1
= a,
x

x=

1 1
= or 0
...
5, y = 1
...


Solve
3
1
+
=4
2a 5b
1
4
+
= 10
...
4 Solving more difficult
simultaneous equations
Here are some further worked problems on solving more
difficult simultaneous equations
...


Solve
(1)
(2)

In this type of equation the solution is easier if a
1
1
substitution is initially made
...
5
2

(1)
(2)

(3)
(4)

To remove fractions, equation (3) is multiplied by 10,
giving
 
x

3
y = 10(4)
10
+ 10
2
5
i
...


5x + 6y = 40

(5)

Multiplying equation (4) by 2 gives
8x + y = 21

(6)

Multiplying equation (6) by 6 gives
(5)

48x + 6y = 126

(7)

95

Solving simultaneous equations
Subtracting equation (5) from equation (7) gives

from which
and

43x + 0 = 86
x=

86
=2
43

Hence, x = 5, y = 1
...


Substituting x = 2 into equation (3) gives

Problem 11
...
75
4
4

(2)



Hence, the solution is a = 0
...
2, which may be
checked in the original equations
...
e
...
Solve

5x + 3y = 2 + 5 − 6

4
1
=
x+y
27
4
1
=
2x − y
33

(1)
(2)

1
27(x + y)
x+y



4
= 27(x + y)
27

1−x
6
6
i
...




60 = 12x and x =

60
=5
12

Substituting x = 5 in equation (3) gives
27 = 4(5) + 4y



5+ y
+6
2



 
5
=6
6

(1 − x) + 3(5 + y) = 5
−x + 3y = 5 − 1 − 15

(3)

Similarly, in equation (2) 33 = 4(2x − y)

Equation (3) + equation (4) gives



1 − x + 15 + 3y = 5

27(1) = 4(x + y)

33 = 8x − 4y

(3)

Multiplying equation (2) by 6 gives




27 = 4x + 4y

5x + 3y = 1

Hence,

To eliminate fractions, both sides of equation (1) are
multiplied by 27(x + y), giving

i
...


5
1−x 5+ y
+
=
6
2
6

Multiplying equation (1) by 15 gives

1
1 1
and since
= y, b = = or 0
...
e
...


1 1
1
= x, a = = or 0
...
5 Practical problems involving
simultaneous equations

5(2) + 3y = 1

There are a number of situations in engineering and
science in which the solution of simultaneous equations
is required
...


10 + 3y = 1
3y = 1 − 10 = −9
y=

−9
= −3
3

Checking, substituting x = 2, y = −3 in equation (4)
gives
LHS = −2 + 3(−3) = −2 − 9 = −11 = RHS
Hence, the solution is x = 2, y = −3
...
The law connecting friction F and
load L for an experiment is of the form F = a L + b
where a and b are constants
...
6 N,
L = 8
...
4N , L = 2
...
Find the
values of a and b and the value of F when
L = 6
...
6 and L = 8
...
6 = 8
...

+ = 14
x
y
5 3
− = −2
x
y
3
...

−
=
5
4
5
3 + 2r 5 − s
15
+
=
4
3
4

4
2
...


3
= 18
b
5
= −4
b

Substituting F = 4
...
0 into F = a L + b
gives
4
...
0a + b

(2)

Subtracting equation (2) from equation (1) gives
1
...
0a
a=
Substituting a =

5 3
+ = 1
...
1
x
y

1
...
2
6
...
6 = 8
...
6 = 1
...
6 − 1
...


7
...
If 5x −

(1)

i
...


b=4

1
Checking, substituting a = and b = 4 in equation (2),
5
gives
 
1
+ 4 = 0
...
4 = LHS
RHS = 2
...
5, F = a L + b = (6
...
3 + 4,
5
i
...
F = 5
...

Hence, a =

97

Solving simultaneous equations
Removing the brackets from equation (1) gives

Problem 13
...
If a straight line passes through the
point where x = 1 and y = −2, and also through
the point where x = 3
...
5, find the
values of the gradient and the y-axis intercept

27 = 1
...
5I1 − 8I2 = 27

Substituting x = 1 and y = −2 into y = mx + c gives
−2 = m + c

Removing the brackets from equation (2) gives

(1)

Substituting x = 3
...
5 into y = mx + c
gives
10
...
5m + c

−26 = 2I2 − 8I1 + 8I2
Rearranging gives
−8I1 + 10I2 = −26

(2)

Subtracting equation (1) from equation (2) gives

47
...


15
...
5(2) − 8I1 = 27
19 − 27 = 8I2
−8 = 8I2

(2)
and

l2

26 V

8V
1
...
1

Solve the equations to find the values of currents I1
and I2

31
=2
15
...
When Kirchhoff’s laws are applied
to the electrical circuit shown in Figure 13
...
5I1 + 8(I1 − I2 )

(6)

Adding equations (5) and (6) gives

RHS = (3
...
5 − 7 = 10
...
5
=5
12
...
5m, from which, m =
2
...

Problem 15
...
Determine the initial
velocity and the acceleration given that s = 42 m
when t = 2 s, and s = 144 m when t = 4 s
...
e
...
e
...
Thus, dividing equation (1) by equation
(2) gives

Multiplying equation (1) by 2 gives
84 = 4u + 4a

(1)

12
=6
2

Substituting a = 15 and u = 6 in equation (2) gives
RHS = 4(6) + 8(15) = 24 + 120 = 144 = LHS
Hence, the initial velocity u = 6 m/s and the acceleration a = 15 m/s2
...

1
Hence, s = (6)(3) + (15)(3)2 = 18 + 67
...
e
...
5 m
...
The resistance R  of a length of
wire at t ◦C is given by R = R0 (1 + αt ), where R0
is the resistance at 0◦ C and α is the temperature
coefficient of resistance in /◦C
...
e
...
004
α=
1250 250
1
into equation (1) gives
250



1
30 = R0 1 + (50)
250

Substituting α =

30 = R0 (1
...
2

1
and R0 = 25 in equaChecking, substituting α =
250
tion (2), gives



1
RHS = 25 1 + (100)
250
= 25(1
...
004/◦ C and R0 = 25 
...
The molar heat capacity of a solid
compound is given by the equation c = a + bT ,
where a and b are constants
...
Determine the values
of a and b

Solving simultaneous equations
When c = 52,

T = 100, hence
52 = a + 100b

(1)

When c = 172, T = 400, hence
172 = a + 400b

(2)

Equation (2) – equation (1) gives
120 = 300b
from which,

b=

120
= 0
...
4 in equation (1) gives
52 = a + 100(0
...
4
Now try the following Practice Exercise
Practice Exercise 52 Practical problems
involving simultaneous equations (answers
on page 345)
1
...
If W = 40 when P = 12
and W = 90 when P = 22, find the values of
a and b
...
Applying Kirchhoff’s laws to an electrical
circuit produces the following equations:
5 = 0
...
4I2 − 2(I1 − I2 )
Determine the values of currents I1 and I2
3
...

If v = 20 when t = 2 and v = 40 when t = 7,
find the values of u and a
...
5
4
...
Find the respective costs of a car
and a van
...


y = mx + c is the equation of a straight line
of slope m and y-axis intercept c
...
5, find the slope and y-axis intercept
of the straight line
...
The resistance R ohms of copper wire at t ◦C
is given by R = R0 (1 + αt ), where R0 is the
resistance at 0◦ C and α is the temperature coefficient of resistance
...
44  at 30◦C
and R = 32
...
The molar heat capacity of a solid compound
is given by the equation c = a + bT
...

Find the values of a and b
...
In an engineering process, two variables p and
q are related by q = ap + b/ p, where a and b
are constants
...

9
...
6 Solving simultaneous equations
in three unknowns
Equations containing three unknowns may be solved
using exactly the same procedures as those used with
two equations and two unknowns, providing that there
are three equations to work with
...

Problem 18
...

x +y+z =4

(1)

2x − 3y + 4z = 33

(2)

3x − 2y − 2z = 2

(3)

There are a number of ways of solving these equations
...

The initial object is to produce two equations with two
unknowns
...


100 Basic Engineering Mathematics
Multiplying equation (1) by 4 gives

Now try the following Practice Exercise

4x + 4y + 4z = 16

(4)

Equation (2) – equation (4) gives
−2x − 7y = 17

(5)

Similarly, multiplying equation (3) by 2 and then
adding this new equation to equation (2) will produce
another equation with only x and y involved
...
2x + 3y + 4z = 36 6
...
5x + 5y − 4z = 37 8
...

Equation (7) – equation (5) gives

10x = 20
x =2

from which,
(Note that 8x − −2x = 8x + 2x = 10x)
Substituting x = 2 into equation (5) gives

and

−7y = 17 + 4 = 21
y = −3

Substituting x = 2 and y = −3 into equation (1) gives
2−3+z = 4
from which,

4
...
3x + 2y + z = 14
7x + 3y + z = 22
...
5
10
...
x + 2y + 4z = 16 2
...
3x + 5y + 2z = 6
x − y + 3z = 0
2 + 7y + 3z = −3

Rewriting equation (5) gives
−2x − 7y = 17

In problems 1 to 9, solve the simultaneous equations in 3 unknowns
...


2i1 − 3i2 + 2i3 = 6
Determine the values of i1 , i2 and i3
11
...
They are related by
following simultaneous equations
...
4F1 + 2
...
8F3 = 5
...
2F1 − 1
...
6F3 = 35
...
2F1 + 2
...
4F3 = −5
...
The marks available are shown in brackets at
the end of each question
...


Transpose p − q + r = a − b for b
...


Make π the subject of the formula r =

3
...

5
...

3
E −e
Transpose I =
for E
...

ad − 1

6
...


Transpose A =

8
...


Make

πR 2 θ
for R
...


(2)

13
...
A formula for the focal length f of a convex
1 1
1
lens is = +
...

(4)

L
g
(3)
(2)

the

formula
(5)

the

formula
(5)

10
...
Evaluate b when A = 11 750 mm2 , h = 25 mm and
l = 75 mm
...
The velocity v of water in a pipe appears in
0
...
Evaluate v when
the formula h =
2 dg
h = 0
...
20, L = 80 and g = 10
...
In an engineering process two variables x and y
b
are related by the equation y = ax + , where a
x
and b are constants
...

(5)
16
...


(a) 3a − 8 +

i1 + 8 i2 + 3 i3 = −31
3 i1 − 2 i2 + i3 = −5
2 i1 − 3 i2 + 2 i3 = 6
Determine the values of i1 , i2 and i3

(10)

List of formulae
Laws of indices:

Areas of plane figures:

a m × a n = a m+n
a m/n =

√
n m
a

am
an

= a m−n (a m )n = a mn

a −n =

Quadratic formula:
If ax 2 + bx + c = 0

1
an

Area = l × b

(i) Rectangle

a0 = 1
b

√
−b ± b2 − 4ac
x=
2a

then

Equation of a straight line:

l

(ii) Parallelogram Area = b × h

y = mx + c

Definition of a logarithm:
If y = a x

then

h

x = loga y

Laws of logarithms:
log(A × B) = log A + log B
 
A
= log A − log B
log
B

b

(iii) Trapezium

log An = n × log A

1
Area = (a + b)h
2
a

Exponential series:
ex = 1 + x +

x2 x3
+
+···
2! 3!

h

(valid for all values of x)
b

Theorem of Pythagoras:
b 2 = a 2 + c2

(iv) Triangle

Area =

1
×b×h
2

A

c

B

b
h

a

C
b

List of formulae
Area = πr 2 Circumference = 2πr

(v) Circle

(iii) Pyramid
If area of base = A and
perpendicular height = h then:

r

s

␪

Volume =

1
× A×h
3

r

2π radians = 360 degrees

Radian measure:

h

For a sector of circle:
θ◦
(2πr) = rθ
360

(θ in rad)

1
θ◦
(πr 2 ) = r 2 θ
360
2

(θ in rad)

s=

arc length,

shaded area =

Equation of a circle, centre at origin, radius r:
x 2 + y2 = r 2
Equation of a circle, centre at (a, b), radius r:

Total surface area = sum of areas of triangles
forming sides + area of base
(iv) Cone
1
Volume = πr2 h
3
Curved Surface area = πrl

(x − a)2 + (y − b)2 = r 2

Total Surface area = πrl + πr2

Volumes and surface areas of regular
solids:
l

(i) Rectangular prism (or cuboid)

h

Volume = l × b × h
Surface area = 2(bh + hl + lb)

r

(v) Sphere

l

h

4
Volume = πr3
3
Surface area = 4πr2

b

(ii) Cylinder
Volume = πr2 h
Total surface area = 2πrh + 2πr2
r

r

h

337

338 Basic Engineering Mathematics
Areas of irregular figures by approximate
methods:
Trapezoidal rule


 
width of 1 first + last
Area ≈
interval
2 ordinate

+ sum of remaining ordinates
Mid-ordinate rule
Area ≈ (width of interval)(sum of mid-ordinates)
Simpson’s rule



1 width of
first + last
Area ≈
ordinate
3 interval




sum of even
sum of remaining
+4
+2
ordinates
odd ordinates

For a general sinusoidal function y = A sin (ωt ± α),
then
A = amplitude
ω = angular velocity = 2π f rad/s
ω
= frequency, f hertz
2π
2π
= periodic time T seconds
ω
α = angle of lead or lag (compared with
y = A sin ωt )

Cartesian and polar co-ordinates:
If co-ordinate (x, y) = (r, θ) then

y
r = x 2 + y 2 and θ = tan−1
x
If co-ordinate (r, θ) = (x, y) then
x = r cosθ and y = r sin θ

Mean or average value of a waveform:
area under curve
length of base
sum of mid-ordinates
=
number of mid-ordinates

mean value, y =

Triangle formulae:
Sine rule:
Cosine rule:

b
c
a
=
=
sin A sin B
sin C
a 2 = b2 + c2 − 2bc cos A

B

b

a

If a = first term and d = common difference, then the
arithmetic progression is: a, a + d, a + 2d,
...

The n’th term is: arn−1
Sum of n terms, Sn =

A
c

Arithmetic progression:

a (1 − r n )
a (r n − 1)
or
(1 − r )
(r − 1)

If − 1 < r < 1, S∞ =

a
(1 − r )

C

Area of any triangle
1
= × base × perpendicular height
2
1
1
1
= ab sin C or
ac sin B or
bc sin A
2
2
2

a +b+c
= [s (s − a) (s − b) (s − c)] where s =
2

Statistics:
Discrete data:

#
mean, x¯ =

x

n
)
(
 #

(x − x¯ )2

standard deviation, σ =
n

List of formulae
Grouped data:

#
fx
mean, x¯ = #
f

)
 #$
%(

f (x − x)
¯ 2

#
standard deviation, σ =
f

Standard integrals
y
axn
cos ax

Standard derivatives
sin ax

y or f(x)

dy
= or f (x)
dx

axn

anxn−1

sin ax

a cos ax

cos ax

−a sin ax

eax

aeax

ln ax

1
x

eax
1
x

&

a

y dx
x n+1
+ c (except when n = −1)
n +1

1
sin ax + c
a
1
− cos ax + c
a
1 ax
e +c
a
ln x + c

339

Answers

Answers to practice exercises
Chapter 1

Chapter 2

Exercise 1 (page 2)
1
...

7
...

13
...

17
...
16 m
3
...
£565
6
...
−36 121
9
...
1487
12
...
−70872
15
...
25 cm
d = 64 mm, A = 136 mm, B = 10 mm

1
...

4
...

7
...


5
...

13
...


Exercise 2 (page 5)
(a) 468 (b) 868
2
...

(a) 259 (b) 56
8
...


2
...

6
...

10
...


(a) 12 (b) 360
(a) 90 (b) 2700
(a) 3 (b) 180
(a) 15 (b) 6300
(a) 14 (b) 53 900

Exercise 4 (page 8)
2
...
68

3
...
5

DOI: 10
...
00040-5

3
...

10
1
14
...
1
21
6
...


16
...


2
6
8
2
...

35
9
11
3
1
7
...

5
13
5
3
2
12
...
3
2
5
12
4
3
1
17
...
13
4
9
(a) £60, P£36, Q£16

7
...

15
...


22
9
8
25
3
16
4
27
17
60

4
...

12
...

20
...
5
9
...
−33
10
...
1
2
12
14
...


71
8
11
15
3
16
51
8
52
17
20

19
...

15
5
...
4
20
...
2880 litres

Exercise 7 (page 14)
1
18
7
6
...
2
1
...
22
11
...

(a) 4 (b) 24
(a) 10 (b) 350
(a) 2 (b) 210
(a) 5 (b) 210
(a) 14 (b) 420 420

2
...

3
...

7
...


1
7
4
11
17
30
43
77
9
1
40

1
...
4

1
9
19
7
...
2
20
2
...
1
8
...
7
3

4
...

15
5
...
0
...
14
...
1
...


13
20

7
40
21
1
141
(b)
(c)
(d)
(e)
25
80
500
11
3
7
...
10
25
200
11
1
7
(b) 4
(c) 14
(d) 15
40
8
20
2
...
4
40
41
10
...
(a)

11
...
625

9
250

3
...


6
125

9
...

4
...

10
...


2
...
11
...
185
...
8307
5
...
1581
6
...
571
5
...
1
...
0
...
068
11
...
5 ×10
12
...
5 ×103
−6
−3
4
...
202
...
18
...
6
...
0
...
11
...
0
...
14
...
1
...
2
...
65
...
0
...
329
...
18
...
43
...
72
...
12
...
−124
...
4
...
0
...
−

9
10

4
...
732
8
...
0
...
0
...
0
...
−0
...
0
...
0
...
5
...
2
...
0
...
0
...
0
...
998 2
...
544
3
...
02 4
...
42
456
...
434
...
626
...
1591
...
444 10
...
62963
11
...
563 12
...
455
13
...
8
...
(a) 24
...
812
(a) 0
...
0064 17
...
4˙ (b) 62
...
4
...
0
...
3
...
13
...
50
...
53
...
36
...
12
...
0
...
46
...
1
...
2
...
2
...
30
...
0
...
219
...
5
...
5
...
52
...
0
...
25
...
591
...
69
...
17
...

4
...

10
...
11927
6
...
0944
10
...
325

2
...
30
...
84
...
10
...
2
...


Exercise 10 (page 19)

1
...

9
...

16
...


2
...
0
...
137
...
19
...
515
...
15
...
52
...
0
...
80
...
295
...
59 cm2
159 m/s
0
...

5
...

11
...
78 mm
0
...
8 m2
281
...

6
...

12
...
5
5
...
5
2
...

4
...

10
...


£589
...
508
...
V = 2
...
5
5
...
81 A 6
...
79 s
E = 3
...
I = 12
...
s = 17
...
184 cm2 11
...
327
(a) 12
...
p
...

(c) 13
...
15 h

342 Basic Engineering Mathematics
Exercise 26 (page 43)

Chapter 5

1
...
£66 3
...
450 g 5
...
56 kg
6
...
00025 (b) 48 MPa 7
...
76 litre

Exercise 21 (page 34)
1
...
32%
2
...
4% 3
...
7% 4
...
4%
5
...
5%
6
...
20
7
...
0125 8
...
6875
9
...
462% 10
...
2% (b) 79
...
(b), (d), (c), (a) 12
...

14
...
A = , B = 50%, C = 0
...
30,
2
17
3
F = , G = 0
...
85, J =
10
20

Exercise 27 (page 45)
1
...
170 fr
3
...
8 mm
4
...
(a) 159
...
5 gallons
6
...
4 MPa
7
...
2 mm 8
...

3
...

7
...

14
...
8 kg 2
...
72 m
(a) 496
...
657 g
(a) 14%
(b) 15
...
49% 11
...
2%
2
...
5
...
73 s 4
...
36% 6
...
76 g
9
...
17% 13
...
3
...
25%
37
...
7%

1
...
5 weeks
2
...
(a) 9
...
12 (c) 0
...
50 minutes
5
...
375 m2 (c) 24 × 103 Pa

Chapter 7
Exercise 29 (page 48)

Exercise 23 (page 38)
1
...

9
...

14
...

16
...
5%
2
...
£310
4
...
£20 000 7
...
45 8
...
25
£39
...
£917
...
£185 000 12
...
2%
A 0
...
9 kg, C 0
...
3 t
20 000 kg (or 20 tonnes)
13
...
5 mm 17
...
27
6
...
128
7
...
100 000
8
...
24
10
...
96
9
...
1 6
...
16 4
...
01 10
...
76
7
...
1000 9
...
36 13
...
34 15
...
25
1
1
1
17
...
49 19
...
5
20
...
128

2
...
36 : 1

2
...
5 : 1 or 7 : 2 3
...
96 cm, 240 cm 5
...
£3680, £1840, £920 7
...
£2172

Exercise 31 (page 52)
1
...
9

Exercise 25 (page 42)
1
...
76 ml
3
...
12
...
14
...
25 000 kg

147
148
17
13
...


2
...
±3
10
...
64

19
56

32
25
1
7
...


11
...
4

1
2

4
...
±
13
1
12
...

3
...

7
...

11
...

15
...

19
...

23
...

27
...

4
...

8
...

12
...

16
...

20
...

24
...

28
...

y
20
...
(x − y)(a + b)
1
...

5
...

9
...

13
...


2x(y − 4z)
2x(1 + 2y)
4x(1 + 2x)
x(1 + 3x + 5x 2 )
r(s + p + t )


2 p q 2 2 p2 − 5q
2x y(y + 3x + 4x 2 )
7y(4 + y + 2x)
2r
18
...

t
21
...
(a − 2b)(2x + 3y)

2
...

6
...

10
...

14
...


1
...
2

6
...
2

11
...
2

4
...
6
1
8
...
6
18
...
−10
17
...
0
14
...
12y 2 − 3y
4
...
1

7
...
6a 2 + 5a −
11
...
9x 2 +

1
...
−2

3
...
15

7
...
5

11
...
2

16
...
6

21
...
−3

1
4
1
3

10
...
10a 2 − 3a + 2

15
...
5

1
2

1
3

4
...
12

9
...
13

13
...
−11

15
...
3

19
...
10

23
...
±4

Exercise 44 (page 79)
1
...

5
...


10−7
2
...
3
...
8  (b) 30 
digital camera battery £9, camcorder battery £14
800 
7
...
12 cm, 240 cm2
4
...
30 kg

2
...
004
5
...
12 m, 8 m

3
...
£312, £240
9
...
5 N

Chapter 12
Exercise 46 (page 84)
1
...
3

Exercise 43 (page 76)

Exercise 41 (page 72)
1
...
4b − 15b2
3
5
...
2

c
p
V
5
...
r =
2π
3
...
a =
t
1
6
...
x =
m
2
...
R =
I
5
13
...
T =

10
...
C =

ω ωL −

12
...
f =

13
...
λ =

R2


5

345

+ , 63
...
r =
or 1 −
S
S
2
...
f =

AL
3F − AL
or f = F −
3
3

4
...
t =

R − R0
R0 α

6
...
b =
9
...
R =

t 2g
4π 2


360 A
12
...
080
14
...
L =

11
...
a =
m −n
3(x + y)
3
...
b = √
1 − a2
a( p2 − q 2 )
7
...
t2 = t1 +
11
...
725



√
v 2 − 2as

M
+ r4
π
mrCR
4
...
r =
x+y



2
...
965
10
...
03L
8
...
5
p = 2, q = −1
x = 3, y = 2
a = 5, b = 2
s = 2, t = 3
m = 2
...
5
x = 2, y = 5

2
...

6
...

10
...

14
...


x = 3, y = 4
x = 4, y = 1
x = 1, y = 2
a = 2, b = 3
x = 1, y = 1
x = 3, y = −2
a = 6, b = −1
c = 2, d = −3

Exercise 50 (page 94)

13
...

3
...

7
...

11
...

15
...

3
...

7
...

4
...

8
...
30, b = 0
...
x = , y =
2
4
1
1
3
...
c = 3, d = 4
3
7
...
a = , b = −
3
2
4
...
r = 3, s =
2
8
...

3
...

6
...


a = 0
...
5, c = 3
α = 0
...
56 
a = 4, b = 10

2
...
47, I2 = 4
...
£15 500, £12 800
7
...
40
9
...
5, F2 = −4
...
x = 2, y = 1, z = 3
3
...
x = 2, y = −2, z = 2
4
...

7
...

10
...


x = 2, y = 4, z = 5 6
...
x = −4, y = 3, z = 2
x = 1
...
5, z = 4
...

4
...

8
...


1
...
0
...
905 A
0
...
38 m
1
...
835 m or 18
...

5
...

9
...
84 cm
0
...
78 cm
7m

Chapter 14
Exercise 58 (page 110)

Exercise 54 (page 104)
1
...
−1
...
5
7
...

13
...


4
−2 or −3
4 or −3
2

19
...
4 or −8
4
5
...
−5
11
...
2 or 7
17
...
−1
...
x = 1, y = 3 and x = −3, y = 7

6
...
x = , y = − and −1 , y = −4
5
5
3
3

9
...

15
...


23
...
5

1
4
25
...

5

1
1
26
...

or −
3
2

1
2 or −1
−4
3 or −3

1
8
1
24
...

8
5
30
...


22
...
x 2 − 4x + 3 = 0
33
...
x 2 − 36 = 0

3
...
3
7
...
−2
9
...

2
3
1
11
...
10 000 13
...
9 15
...
0
...

18
...
4

or −2
3
2

32
...
4x 2 − 8x − 5 = 0
36
...
7x − 1
...
4

3
...
−3

5
...
−3
...
268
3
...
468 or −1
...
2
...
307

1
8

3
...
−3
...
637
4
...
290 or 0
...
−2
...
351

1
...
log 12

2
...
log 500

3
...
log 100

4
...
log 6

9
...
log 1 = 0 11
...
log 243 or log 35 or 5 log3
13
...
log 64 or log26 or 6 log2

Exercise 56 (page 107)
1
...

5
...

9
...

13
...
637 or −3
...
781 or 0
...
608 or −1
...
851 or −2
...
481 or −1
...
167
4
...
438

2
...

6
...

10
...


0
...
792
0
...
693
1
...
232
2
...
086
4
...
676
7
...
641

15
...
5
16
...
5
19
...
x = 2

17
...
5 18
...
a = 6 22
...
1
...
3
...
0
...
6
...
2
...
3
...
2
...
−0
...
316
...

2
...

4
...


(a)
0
...
0988
(a) 4
...
04106
2
...


Exercise 67 (page 134)

0
...
064037
2
...
07482
120
...
446
8
...
08286

Exercise 63 (page 120)
1
...
0601

2
...
389 (b) 0
...

x − 2x 4
3
1
4
...
(a) Horizontal axis: 1 cm = 4 V (or 1 cm = 5 V),
vertical axis: 1 cm = 10 
(b) Horizontal axis: 1 cm = 5 m, vertical axis:
1 cm = 0
...
2 mm
2
...
5 (d) 5
3
...
5
4
...
1 (b) −1
...
The 1010 rev/min reading should be 1070 rev/min;
(a) 1000 rev/min (b) 167 V

1 − 2x 2 −

Exercise 64 (page 122)
1
...
95, 2
...
(a) 28 cm3 (b) 116 min

2
...
65, −1
...
(a) 70◦C (b) 5 minutes

Exercise 65 (page 124)
1
...

3
...

11
...

17
...
55547 (b) 0
...
8941
(a) 2
...
33154 (c) 0
...
4904 4
...
5822 5
...
197
6
...
2
0
...
11
...
1
...
1
...
962 12
...
4
147
...
4
...
3
...
e
...
500 19
...
Missing values: −0
...
25, 0
...
25, 2
...
(a) 4, −2 (b) −1, 0 (c) −3, −4 (d) 0, 4
1
1
1 1
(b) 3, −2 (c) ,
2
2
24 2
4
...
(a) 2,

1
2
2
1
2
5
...
(a) (b) −4 (c) −1
5
6
7
...
(2, 1)

9
...
5, 6)

10
...
(a) 89 cm (b) 11 N (c) 2
...
4 W + 48
12
...
15 W + 3
...
a = −20, b = 412

Exercise 69 (page 144)
1
...
(a) 850 rev/min (b) 77
...
(a) 150◦ C (b) 100
...
(a) 0
...
25L + 12
2
...
21 kPa

3
...
32 volts (b) 71
...
(a) 1
...
293 m 5
...
45 s
6
...
37 N
7
...
04 A (b) 1
...
2
...
£2424

347

9
...
07 A (b) 0
...
5 N (e) 592 N (f) 212 N
4
...
003, 8
...
(a) 22
...
43 s (c) v = 0
...
5
6
...
9L − 0
...
(a) 1
...
89% (c) F = −0
...
21
8
...
00022 (c) 28
...
a = 0
...
3 kPa, 275
...
(a)

Chapter 18

9
...
6, 13
...
6, 0
...
6 or 0
...
x = −1
...
5 (a) −30 (b) 2
...
50
(c) 2
...
8

Exercise 74 (page 161)

Exercise 70 (page 149)
1
...
(a) y (b)

√

x (c) b (d) a

y
1
(c) f (d) e 4
...
(a) (b) 2 (c) a (d) b
x
x
6
...
5, b = 0
...
78 mm2 7
...
15

1
...
5, y = −5
...
(a) x = −1
...
5 (b) x = −1
...
24
(c) x = −1
...
0

3
...
(a) 950 (b) 317 kN
9
...
4, b = 8
...
4 (ii) 11
...
x = −2
...
5 or 1
...
x = −2, 1 or 3, Minimum at (2
...
1),
Maximum at (−0
...
2)
3
...
x = −2
...
4 or 2
...
x = 0
...
5
6
...
3, 1
...
8
7
...
5

Exercise 71 (page 154)
1
...

3
...

5
...

7
...


(a) lg y (b) x (c) lg a (d) lg b
(a) lg y (b) lg x (c) L (d) lg k
(a) ln y (b) x (c) n (d) ln m
I = 0
...
75 candelas
a = 3
...
5
a = 5
...
6, 38
...
0
R0 = 26
...
42
8
...
08e0
...
4 N, μ = 0
...
0 N, 1
...
5, y = 1
...
3, y = −1
...
4, b = 1
...
82◦ 27
2
...
51◦11 4
...
15 44 17 6
...
72
...
27
...
37◦ 57
10
...
reflex 2
...
acute 4
...
(a) 21◦ (b) 62◦ 23 (c) 48◦56 17

Chapter 19

1
...

5
...


Chapter 20

2
...
x = −1, y = 2
6
...
(a) Minimum (0, 0) (b) Minimum (0, −1)
(c) Maximum (0, 3) (d) Maximum (0, −1)
2
...
4 or 0
...
−3
...
9
4
...
1 or 4
...
−1
...
2
6
...
5 or −2, Minimum at (−1
...
1)
7
...
7 or 1
...
(a) ±1
...
3

6
...
(a) 60◦ (b) 110◦ (c) 75◦ (d) 143◦ (e) 140◦
(f ) 20◦ (g) 129
...
Transversal (a) 1 & 3, 2 & 4, 5 & 7, 6 & 8
(b) 1 & 2, 2 & 3, 3 & 4, 4 & 1, 5 & 6, 6 & 7,
7 & 8, 8 & 5, 3 & 8, 1 & 6, 4 & 7 or 2 & 5
(c) 1 & 5, 2 & 6, 4 & 8, 3 & 7 (d) 3 & 5 or 2 & 8
9
...
a = 69◦ , b = 21◦ , c = 82◦ 11
...
1
...
0
...
40◦55

Exercise 78 (page 173)
1
...
a = 40◦ , b = 82◦, c = 66◦,
d = 75◦, e = 30◦ , f = 75◦
3
...
52◦
5
...
5◦
6
...
40◦, 70◦, 70◦, 125◦, isosceles
8
...
a = 103◦, b = 55◦ , c = 77◦ , d = 125◦,
e = 55◦, f = 22◦, g = 103◦, h = 77◦ ,
i = 103◦, j = 77◦, k = 81◦
10
...
A = 37◦, B = 60◦ , E = 83◦

349

4
3
4
3
2
...
sin A = , tan A =
17
15
112
15
, cos X =
4
...
(a)
(b)
(c)
17
17
15
7
24
6
...
(a) 9
...
625

Exercise 79 (page 176)
1
...
proof

Exercise 84 (page 187)
1
...

5
...

13
...
7550 2
...
846
3
...
52
(a) 0
...
1010 (c) 0
...
33◦
6
...
25◦
7
...
78◦
8
...
41 54 11
...
05
12
...
3586 14
...
803

Exercise 80 (page 178)

Exercise 85 (page 189)

1
...
54 mm, y = 4
...
9 cm, 7
...
(a) 2
...
3 m

1
...
22 (b) 5
...
87 (d) 8
...
595 (f ) 5
...
(a) AC = 5
...
04◦ , ∠C = 30
...
928 cm, ∠D = 30◦, ∠F = 60◦
(c) ∠J = 62◦, HJ = 5
...
59 cm
(d) ∠L = 63◦ , LM = 6
...
37 cm
(e) ∠N = 26◦, ON = 9
...
201 cm
(f ) ∠S = 49◦, RS = 4
...
625 cm

Exercise 81 (page 180)
1–5
...


3
...
54 m

4
...
40 mm

Chapter 21
Exercise 82 (page 182)

Exercise 86 (page 192)

1
...

7
...

11
...


1
...
15 m
2
...
249
...
110
...
53
...
9
...
107
...
9
...
56 m
9
...
24 m
3
...
54 mm
20
...
7
...
11
...
11 mm 8
...
20 cm each (b) 45◦
10
...
81 km
3
...
132
...
94 mm
14
...
sin Z = , cos Z = , tan X = , cos X =
41
41
9
41

1
...
78◦ and 137
...
53◦ and 351
...
(a) 29
...
92◦ (b) 123
...
14◦
3
...
21◦ and 224
...
12◦ and 293
...
t = 122◦7 and 237◦53
5
...
θ = 39◦44 and 219◦44

Exercise 88 (page 202)
1
...
180◦
3
...
120◦
◦
◦
◦
5
...
2, 144
7
...
5, 720◦
7
9
...
6, 360◦ 11
...
5 ms
2
13
...
100 μs or 0
...
625 Hz 16
...
leading

Exercise 91 (page 209)
1
...
2 cm, Q = 47
...
65◦,
area = 77
...
p = 6
...
83◦, R = 44
...
938 m2
3
...
33◦, Y = 52
...
05◦,
area = 27
...
Z = 29
...
50◦ , Z = 96
...
(a) 40 (b) 25 Hz (c) 0
...
29 rad (or 16
...

3
...

7
...
(a) 122
...
80◦ , 40
...
54◦
(a) 11
...
55◦
4
...
4 m
BF = 3
...
0 m 6
...
35 m, 5
...
48 A, 14
...
(a) 75 cm (b) 6
...
157 s
(d) 0
...
94◦ ) lagging 75 sin 40t
3
...
01 s or 10 ms
(d) 0
...
61◦) lagging 300 sin 200πt
4
...
43) volts

π

A or
5
...
524)A
6
...
2 sin(100πt + 0
...
(a) 5A, 50 Hz, 20 ms, 24
...
093A (c) 4
...
375 ms (e) 3
...
C = 83◦ , a = 14
...
9 mm,
area = 189 mm2
2
...
568 cm, a = 7
...
65 cm2
3
...
0 cm,
area = 134 cm2
4
...
08 mm,
area = 185
...
J = 44◦29 , L = 99◦31 , l = 5
...
132 cm2 , or, J = 135◦ 31 , L = 8◦29 ,
l = 0
...
917 cm2
6
...
2 mm,
area = 820
...
19 mm, area = 174
...
80
...
38◦, 40
...
(a) 15
...
07◦
3
...
25 cm, 126
...
19
...
36
...
x = 69
...
130◦ 8
...
66 mm

Chapter 24
Exercise 94 (page 215)
1
...

3
...

5
...

7
...


(5
...
04◦) or (5
...
03 rad)
(6
...
82◦) or (6
...
36 rad)
(4
...
57◦) or (4
...
03 rad)
(6
...
58◦) or (6
...
54 rad)
(7
...
20◦) or (7
...
55 rad)
(4
...
31◦) or (4
...
12 rad)
(5
...
04◦) or (5
...
74 rad)
(15
...
75◦) or (15
...
37 rad)

Exercise 95 (page 217)
(1
...
830)
2
...
917, 3
...
362, 4
...
(−2
...
154)
(−9
...
400)
6
...
615, −3
...
750, −1
...
(4
...
233)
(a) 40∠18◦, 40∠90◦, 40∠162◦, 40∠234◦, 40∠306◦
(b) (38
...
36), (0, 40), (−38
...
36),
(−23
...
36), (23
...
36)
10
...
0 mm
1
...

5
...

9
...
p = 105◦ , q = 35◦
3
...
r = 142◦, s = 95◦

Exercise 97 (page 225)
1
...
91 cm
2
...
7 cm2 3
...
27
...
18 cm
6
...
(a) 29 cm2 (b) 650 mm2
8
...
3
...
6750 mm
11
...
30 cm2
12
...

4
...

9
...

13
...

18
...


113 cm2
2
...
1790 mm2
2
2
802 mm
5
...
1269 m2
2
1548 m
8
...
0 cm (b) 783
...
46 m 10
...
80 cm, 74
...
86 mm (b) 197
...
26
...
67 cm, 54
...
82
...
748
(a) 0
...
2 m2
17
...
47 m2
2
(a) 396 mm (b) 42
...
701
...
74 mm

Exercise 104 (page 237)
1
...
Centre at (3, −2), radius 4
3
...
Circle, centre (0, 0), radius 6

Chapter 27
Exercise 98 (page 226)
1
...

3
...

7
...
27 cm2 (b) 706
...
(a) 20
...
41 mm
(a) 53
...
9 mm2
6
...
89 m

Exercise 99 (page 228)
1
...
1624 mm2 3
...
918 ha (b) 456 m

Exercise 105 (page 243)
1
...

5
...

10
...

14
...

17
...


1
...
5 cm3
3
...
15 cm3 , 135 g
7
...
(a) 35
...
3 cm2
1
...
37
...
63 cm
13
...
709 cm, 153
...
99 cm
16
...
099 m2
8
...
22 m
18
...
5 min
4 cm
20
...
08 m3

Exercise 100 (page 229)
1
...
80 m2

3
...
14 ha

Chapter 26
Exercise 101 (page 231)
1
...
24 cm 2
...
5 mm 3
...
629 cm 4
...
68 cm
5
...
73 cm 6
...
97
...
201
...
0 cm2 2
...
68 cm3 , 25
...
113
...
1 cm2 4
...
131 cm 5
...
2681 mm3 7
...
083 cm3
(b) 20 106 mm2 or 201
...
8
...
(a) 512 × 106 km2 (b) 1
...
664

Exercise 102 (page 232)

Exercise 107 (page 251)

5π
5π
π
(b)
(c)
6
12
4
2
...
838 (b) 1
...
054
3
...
(a) 0◦ 43 (b) 154◦8 (c) 414◦ 53

1
...

4
...


1
...
90 cm2
(a) 56
...
82 cm2
3
...
57 kg
5
...
4 m2
29
...
5 cm3 (b) 84
...
4 cm3 (b) 32
...
0 cm3

352 Basic Engineering Mathematics

7
...

11
...


(b) 146 cm2 (vi) (a) 86
...
9 cm (b) 38
...
125 cm3
3
2
10
...
5 m
10
...
220
...
1 cm3 , 1027 cm2
2
(a) 1458 litres (b) 9
...
45

147 cm3 , 164 cm2
10480 m3 , 1852 m2
10
...
14 m

14
...
72◦ to the 5 N force
29
...
04◦ to the horizontal
9
...
70◦
9
...
89 m/s at 159
...
62 N at 26
...
07 knots, E 9
...

3
...

7
...

7
...

10
...


2
...
1707 cm2
6
...
(a) 54
...
16◦ (b) 45
...
66◦
2
...
71 m/s at 121
...
55 m/s at 8
...
83
...
6◦ to the vertical
2
...
22
...
78◦ N

Exercise 109 (page 256)
1
...
137
...
4
...
54
...
63
...
4
...
143 m2

Exercise 111 (page 260)
1
...
59 m3

2
...
20
...
(a) 2 A (b) 50 V (c) 2
...
0
...
1 A
5
...
13 cm2 , 368
...

3
...

7
...


i − j − 4k
−i + 7j − k
−3i + 27j − 8k
i + 7
...
6i + 4
...
9k

2
...

6
...

10
...
5j − 10k
2i + 40j − 43k

Chapter 30
Exercise 118 (page 279)

2
...
5 V (b) 3 A
4
...
83 V (b) 0

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Chapter 29
Exercise 119 (page 281)
Exercise 113 (page 266)
1
...

2
...
scalar
4
...
scalar
6
...
vector 8
...
vector

Exercise 114 (page 273)
1
...

3
...

5
...
35 N at 18
...
62◦ to the 12 m/s velocity
16
...
57◦ to the 13 N force
28
...
30◦ to the 18 N force
32
...
80◦ to the 30 m displacement

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Exercise 120 (page 283)
1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395) 4
...
11 sin(ωt + 0
...
8
...
173)

Answers to practice exercises
Exercise 121 (page 284)
11
...
324)
2
...
73 sin(ωt − 0
...
79 sin(ωt − 0
...
695 sin(ωt + 0
...
38 sin(ωt + 1
...
3 sin(314
...
233) V (b) 50 Hz
(a) 10
...
3t + 0
...
(a) 79
...
352)V (b) 150 Hz
(c) 6
...

3
...

5
...

7
...
(a) continuous (b) continuous (c) discrete
(d) continuous
2
...
If one symbol is used to represent 10 vehicles, working correct to the nearest 5 vehicles, gives 3
...
5,
6, 7, 5 and 4 symbols respectively
...
If one symbol represents 200 components, working
correct to the nearest 100 components gives: Mon 8,
Tues 11, Wed 9, Thurs 12 and Fri 6
...

3
...

4
...

5
...

6
...

7
...

P increases by 20% at the expense of Q and R
...
Four rectangles of equal height, subdivided as follows: week 1: 18%, 7%, 35%, 12%, 28%; week 2:
20%, 8%, 32%, 13%, 27%; week 3: 22%, 10%, 29%,
14%, 25%; week 4: 20%, 9%, 27%, 19%, 25%
...

9
...
5◦ , 22
...
5◦, 167
...


353

10
...

11
...
(a) £16 450 (b) 138

Exercise 124 (page 297)
1
...
3–39
...
5–39
...
7–39
...
9–40
...
1–40
...
3–40
...
5–40
...
7–40
...

2
...
35, 39
...
75, 39
...
and
heights of 1, 5, 9, 17,
...
There is no unique solution, but one solution is:
20
...
9 3; 21
...
4 10; 21
...
9 11;
22
...
4 13; 22
...
9 9; 23
...
4 2
...
There is no unique solution, but one solution is:
1–10 3; 11–19 7; 20–22 12; 23–25 11;
26–28 10; 29–38 5; 39–48 2
...
20
...
45 13; 21
...
45 37; 22
...
45 48
6
...
5, 15, 21, 24, 27, 33
...
5
...
3, 0
...
67, 2
...
5 and 0
...

7
...
95 2), (11
...
95 19), (12
...
95
42), (13
...
A graph of cumulative frequency against upper class
boundary having co-ordinates given in the answer
to problem 7
...
(a) There is no unique solution, but one solution is:
2
...
09 3; 2
...
14 10; 2
...
19 11;
2
...
24 13; 2
...
29 9; 2
...
34 2
...
07, 2
...
and heights of 3, 10,
...
095 3; 2
...
195 24; 2
...
295 46; 2
...


(d) A graph of cumulative frequency against upper
class boundary having the co-ordinates given
in part (c)
...
Mean 7
...
Mean 27
...
−2542 A/s
2
...
16 cd/V (b) 312
...
(a) −1000 V/s (b) −367
...
−1
...
(a)
15
2
3
4
...
(a)

5
...


7
...

9
...
(a) 8 x + 8 x3 +
x +c
5

7x 2
+c
2
3
(b) t 8 + c
8
5 4
x +c
(b)
24
3
(b) 2t − t 4 + c
4
(b)

(b) 4θ + 2θ 2 +

5
(a) θ 2 − 2θ + θ 3 + c
2
3
2
3
(b) x 4 − x 3 + x 2 − 2x + c
4
3
2
4
1
(a) − + c
(b) − 3 + c
3x
4x
4√ 5
1√
4 9
(a)
x +c
(b)
x +c
5
9
15 √
10
5
(b)
x +c
(a) √ + c
7
t

7
(b) − cos 3θ + c
3
1
(b) 18 sin x + c
3
−2
(b)
+c
15e5x

13
...
(a) 4x + c

3
sin 2x + c
2
1
11
...
(a) e + c
8
10
...
(a) 1
...
5

2
...
5

3
...
333

4
...
75 (b) 0
...
(a) 10
...
1667

6
...
(a) 1 (b) 4
...
(a) 0
...
638

9
...
09 (b) 2
...
(a) 0
...
099

Exercise 141 (page 334)
1
...
37
...
1
...
proof
6
...
5
9
...
67

3
...
1
10
...
29

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