1.2.pdf
- 3. 2-3
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Copyright Notice
© Copyright. All worldwide rights reserved. No part of this publication may be reproduced,
stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e.
photocopy, electronic, mechanical recording or otherwise without the prior written permission of
ST Aerospace Ltd.
Knowledge Levels — Category A, B1, B2, B3 and C Aircraft
Maintenance Licence
Basic knowledge for categories A, B1, B2 and B3 are indicated by the allocation of knowledge levels indicators (1,
2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2
basic knowledge levels.
The knowledge level indicators are defined as follows:
LEVEL 1
A familiarisation with the principal elements of the subject.
Objectives: The applicant should be familiar with the basic elements of the subject.
The applicant should be able to give a simple description of the whole subject, using common words and
examples.
The applicant should be able to use typical terms.
LEVEL 2
A general knowledge of the theoretical and practical aspects of the subject.
An ability to apply that knowledge.
Objectives: The applicant should be able to understand the theoretical fundamentals of the subject.
The applicant should be able to give a general description of the subject using, as appropriate, typical
examples.
The applicant should be able to use mathematical formulae in conjunction with physical laws describing the
subject.
The applicant should be able to read and understand sketches, drawings and schematics describing the
subject.
The applicant should be able to apply his knowledge in a practical manner using detailed procedures.
LEVEL 3
A detailed knowledge of the theoretical and practical aspects of the subject.
A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive
manner.
Objectives: The applicant should know the theory of the subject and interrelationships with other subjects.
The applicant should be able to give a detailed description of the subject using theoretical fundamentals
and specific examples.
The applicant should understand and be able to use mathematical formulae related to the subject.
The applicant should be able to read, understand and prepare sketches, simple drawings and schematics
describing the subject.
The applicant should be able to apply his knowledge in a practical manner using manufacturer's
instructions.
The applicant should be able to interpret results from various sources and measurements and apply
corrective action where appropriate.
- 5. 2-5
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Table of Contents
Module 1.2 Algebra __________________________________________________________9
Transposition_____________________________________________________________9
Basic Rules _____________________________________________________________9
Common Factors ________________________________________________________15
Powers and Roots _______________________________________________________21
Complex Formulae_______________________________________________________29
Linear Equations _________________________________________________________39
Introduction ____________________________________________________________39
Indices and Powers_______________________________________________________49
Basic Laws of Indices and Powers___________________________________________49
Standard Form___________________________________________________________55
Introduction ____________________________________________________________55
Number Systems _________________________________________________________63
Binary_________________________________________________________________63
Octal _________________________________________________________________69
Hexadecimal ___________________________________________________________75
Binary Coded Decimal ____________________________________________________81
Summary ______________________________________________________________87
Simultaneous Equations___________________________________________________89
Methods of Solving ______________________________________________________89
Quadratic Equations ______________________________________________________95
Introduction ____________________________________________________________95
Solution by Factorisation __________________________________________________97
Logarithms_____________________________________________________________107
Why Logs?____________________________________________________________107
Definition _____________________________________________________________107
Common Logarithms ____________________________________________________109
Natural Logarithms______________________________________________________109
Rules of Logarithms_____________________________________________________111
Further Logarithms______________________________________________________113
Some special properties of logarithms_______________________________________115
Complex Numbers_______________________________________________________121
The Number i__________________________________________________________121
The Complex Plane _____________________________________________________122
Complex Arithmetic _____________________________________________________123
- 7. 2-7
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Module 1.2 Enabling Objectives and Certification Statement
Certification Statement
These Study Notes comply with the syllabus of EASA Regulation (EC) No.2042/2003 Annex III
(Part-66) Appendix I, as amended by Regulation (EC) No.1149/2011, and the associated
Knowledge Levels as specified below:
Objective
Part-66
Reference
Licence Category
B1 B2
Algebra 1.2
(a) 2 2
Evaluating simple algebraic expressions,
addition, subtraction, multiplication and
division, use of brackets, simple algebraic
fractions;
(b) 2 2
Linear equations and their solutions
Indices and powers, negative and fractional
indices
Binary and other applicable numbering systems
Simultaneous equations and second degree
equations with one unknown
Logarithms
- 9. 2-9
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Module 1.2 Algebra
Transposition
Basic Rules
1. Most formulae are remembered in a standard form, but for the purpose of solving a
particular problem, it is often necessary to express a formula differently. This involves
changing the subject of the formula and this process is called transposition.
Note:
(a) In the formula, A = LB, A is the subject
(b) In the formula, C = d, C is the subject
(c) In the formula, S = ut + ½ at2
, S is the subject
2. The basic rules of algebra apply equally to transposition of formulae as to solution of
equations. The most important concept being that whatever we do to the left hand side,
we must also do the right hand side.
Examples:
(a) If A = LB, transpose this formula to make L the subject.
Divide both sides by B:
A
B
= L
Reverse the formula: L =
A
B
(b) If Y =
X
Z
, transpose the formula to make X the subject.
Multiply both sides by Z: YZ = X
X = YZ
(c) If a =
b
c
, transpose this formula to make c the subject:
Multiply both sides by c: ac = b
- 10. 2-10
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Divide both sides by a: c =
b
a
(d) If y = x + c, transpose this formula to make x the subject.
Subtract c from both sides: y - c = x
x = y - c
(e) If p =
r
m
q
, transpose this formula to make q the subject.
Multiply both sides by r: pr = q - m
Add m to both sides: pr + m = q
q = pr + m
- 11. 2-11
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
Transpose the following formulae to make the letter in the brackets the subject.
1. y = x + z (x)
2. a = b - c (b)
3. p = q + s (s)
4. l = m - n (n)
5. y = zx (x)
6. y = mx (m)
7. y =
z
x
(z)
8. a =
b
c
(c)
9. v = u + at (u)
10. y = m + c (c)
11. V = E - IR (I)
12. v = u + at (t)
13. P =
RT
V
(T)
14. s = ut + ½ at2
(u)
15.
1
R
=
1
P
(P)
- 13. 2-13
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. x = y - z
2. b = a + c
3. s = p - q
4. n = m - l
5. x =
y
z
6. m =
y
x
7. z = yx
8. c =
b
a
9. u = v - at
10. c = y - m
11. I =
R
V
E
12. t =
a
u
v
13. T =
PV
R
14. u =
s
t
-
at
2
15. P = R
- 15. 2-15
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Common Factors
1. When the subject exists in 2 or more terms the formulae can only be transposed
correctly when the subject is taken out as a common factor.
Examples:
a) Transpose the formula bc + c = a to make c the subject.
Take out c as a common factor: c(b + 1) = a
c =
a
b 1
b) If 2r = pq + rs, make r the subject.
Subtract rs from both sides: 2r - rs = pq
Take out r as a common factor: r(2 - s) = pq
r =
pq
s
2
c) If x =
ab c
a c
, make c the subject.
Multiply both sides by (a + c): x(a + c) = ab + c
Remove brackets: ax + cx = ab + c
Collect terms containing c onto one side: cx - c = ab - ax
c(x - 1) = ab - ax
c =
ab ax
x
1
c =
a b x
x
( )
1
- 17. 2-17
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
Transpose the following formulae to make the letter in the brackets the subject.
1. If XY + X = 7 (X)
2. If ab - b = c (b)
3. If p = st - pq (p)
4. If x =
y
y
3
(y)
5. If
a b
a
= c (a)
6. If d =
t u
u
(u)
7. If
1
a
=
2
b
+
3
c
(b)
8. If
1
R
=
1
1
R
+
1
2
R
(R) (Hint: Find the Common Multiple of R1 and R2)
(Hint: Take the
c
3
to the left side of the equation
then find the Common Multiple of ‘a’ and ‘c’)
- 19. 2-19
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1 X =
7
1
Y
2 b =
c
a 1
3 p =
q
1
st
4 y =
3
1
x
5 a =
b
c
1
6 u =
t
d 1
7 b =
2
3
ca
c a
8 R =
R R
R R
1 2
1 2
- 21. 2-21
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Powers and Roots
1. The square root of a given number is such that, when the square root is multiplied by
itself, the original number is again obtained.
Examples:
a) The square root of 4 is 2 because 22
= 4
b) The square root of 9 is 3 because 32
= 9
c) The square root of 25 is 5 because 52
= 25
2. Instead of writing or saying 'square root', we write .
Examples:
a) The square root of 16 can be written 16
b) 36 simply means the square root of 36
3. The cube root of a given number is such that, when it is cubed, the original number is
again obtained.
Examples:
a) The cube root of 8 is 2, because 23
= 8
b) The cube root of 125 is 5, because 53
= 125
4. Instead of writing or saying 'cube root', we write 3 .
Examples:
a) 27
3
simply means the cube root of 27
b) The cube root of 64 can be written 64
3
5. It follows from the above the nth root of a given number is such that when it is raised to
the power n, the original number is obtained.
Examples:
a) If x
n
= y, then x = yn
- 22. 2-22
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
b) If x = yn
, then x
n
= y
6. You should remember that, providing we add or subtract equal numbers or letters to both
sides of equations or formulae, or multiply or divide both sides of equations or formulae
by the same number or letter, the truth of the equation or formula is unaffected. We can
now extend this concept to include powers and roots.
Examples:
a) x2
= 9
Taking square roots of both sides: x2
= 9
x = 3
7. Consider the following examples:
a) If x = y, make x the subject.
Squaring both sides: ( )
x
2
= y2
so x = y2
b) If x2
= y, make x the subject.
Square rooting both sides: x2
= y
so x = y
c) If a = b c , make c the subject.
Squaring both sides: a2
= (b c )2
a2
= b2
( c )2
a2
= b2
c
Divide through by b2
:
a
b
2
2
= c
Reverse: c =
a
b
2
2
- 23. 2-23
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
so: c =
2
b
a
8. You should note that 4 can be either +2 or -2 because both
(+2)2
and (-2)2
= +4.
Similarly, 9 = 3 and, in general, x2
= x
- 25. 2-25
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
Transpose the following formulae to make the letter in brackets the subject:
1. A = d (A)
2. p2
= q (p)
3. x
3
= y (x)
4. a3
= b (a)
5. A = r2
(r)
6. (n - 1)2
= t (n)
7. p = q r (r)
8. z = r x
2 2
(x)
9. a = 2bc (c)
10. x2
= yz3
(z)
- 27. 2-27
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. A = d2
2. p = q
3. x = y3
4. a = b
3
5. r =
A
6. n = t 1
7. r =
p
q
2
8. x = 2
2
r
z
9. c =
a
b
2
2
10. z =
x
y
2
3
- 29. 2-29
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Complex Formulae
1. In previous periods, you were shown some of the basic types of situations that can arise
when changing the subject of a formula. These basic types are often combined into a
single problem. Such problems can be treated in a variety of ways, but, if you are in any
doubt at all, the following sequence should be followed:
a) First: remove root signs These 3 steps may be carried
out in a different order for
b) Second: remove fractions certain problems, but all roots,
fractions and brackets should
c) Third: remove brackets be removed before carrying
out the fourth step.
d) Fourth: rearrange formula, collecting all terms containing the required
letter on one side of the equation and all other terms on the other side.
e) Fifth: take the subject out as a common factor
f) Sixth: divide through by the coefficient of the subject
g) Seventh: take roots (if necessary)
2. It may not always be necessary to use all the above steps, but, nevertheless, the
sequence should be followed.
Examples:
a) Transpose v2
= u2
+ 2fs to make u the subject.
There are no root signs, fractions or brackets, therefore rearrange:
u2
= v2
- 2fs
There are no common factors or coefficients, therefore take roots:
u = v fs
2
2
- 30. 2-30
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
b) Transpose T = 2
L
g
, to make g the subject.
Square both sides: T2
= [2]2
2
g
L
= 42
2
g
L
= 42 L
g
Multiply through by g: T2
g = 42
L
Divide through by coefficient T2
: g =
4 2
2
L
T
- 31. 2-31
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
Transpose the following formulae to make the letter in the brackets the subject.
1. c = d (d)
2. v2
= 2gh (v)
3. I =
E
r R
(R)
4. v = 2gh (h)
5. Q =
1
R
L
C
(C)
6.
1 1 1
u v j
(u)
7. T = 2
L
g
(L)
8. s = ut + ½ ft2
(f)
9. x =
s a
s b
(s)
10. I =
PTR
100
(R)
- 33. 2-33
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. d =
c
2. v = 2gh
3. R = r
I
E
4. h =
v
g
2
2
5. C =
L
QR
( )2
6. u =
j
v
jv
7. L = g
T
2
2
8. f =
2
2
( )
s ut
t
9. s =
x b a
x
2
2
1
10. R =
PT
100I
- 35. 2-35
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
Solve the following equations.
1. n + 8 = 17. 2. n - 5 = 11
3. 3y = 20 4.
x
2
= 9
5. 8x = 0 6. x + 2x = 18
7. 3n - 2 = 10 8.
a 4
3
= 2
9. 4y - y = 21 10.
x 2
4
= 5
11. 3c = c + 5 12. 2p - 8 = p - 3
13. t + 7 = 17 - 4t 14. 2a + 4 = 19 - a
15. 7m - 9 - m = 3m 16. 3(n- 7) = 12
17. 4(2k - 1) = 20 18. 5(2r + 3) = 15
19. 4(t - 5) = 0 20.
x
4
+ 2 = 5
21. m +
m
5
= 12 22.
a
3
+ 1 =
a
4
23.
x
5
-
2
7
= 0 24.
2
3
k
-
k
2
= 1
25.
a
3
+
a
5
= 1 26.
x
2
+
x
3
+
x
4
= 1
27.
5
2
W
+
1
3
=
3
4
28.
4
x
=
2
7
29.
3
p
=
9
10
30.
2
v
+
3
2
=
5
3
- 37. 2-37
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. n = 9 2. n = 16 3. y = 6.666
4. x = 18 5. x = 0 6. x = 6
7. n = 4 8. a = 2 9. y = 7
10. x = 18 11. c = 2.5 12. p = 5
13. t = 2 14. a = 5 15. m = 3
16. n = 11 17. k = 3 18. r = 0
19. t = 5 20. x = 12 21. m = 10
22. a = -12 23. x = 1.4287 24. k = 6
25. a = 1.875 26. x = 0.9230 27. W = 0.1666
28. x = 14 29. P = 3.3333 30. v = 12
- 39. 2-39
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Linear Equations
Introduction
Any statement of equality between two quantities is an equation. This chapter is concerned with
the solution of equations that can be rearranged into the form
ax + b = 0
where x is the unknown (variable), and a and b are constants.
To solve equations of this form (variable) in the equation, we may first need to manipulate the
equation so that all the terms involving the unknown (variable) appear on one side of the
equation, and only constants appear on the other side.
Note. When manipulating the terms of an equation it must remembered that whatever
arithmetic operation is performed to one side of the equation must also be performed
to the other side.
Example:
Solve 2x – 4 = 10
We want to find the numerical value of x that satisfies this equation. By moving -4 to the R.H.S.
of the equation, remembering that we must change the sign (i.e. by adding + 4 to both sides of
the equation) we obtain
2x = 10 + 4
2x = 14
now, by dividing both sides of the equation by 2, we obtain
x = 7
Hence x = 7 is the solution to the equation 2x - 4 = 10.
We can check our answer by substituting it back into the original equation
i.e. 2x – 4 =10
2(7) – 4 = 10
14 – 4 = 10
Since this is true, our solution is correct.
- 40. 2-40
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Example:
Solve 3x – 2 = 2x + 4
We aim to get the terms in x on one side of the equation and the constants on the other.
Now, by moving -2 to the R.H.S. of the equation (i.e. by adding + 2 to both sides of the
equation), we obtain
3x = 2x + 4 + 2
3x = 2x + 6
Now by moving 2x to the L.H.S. of the equation (i.e. by subtracting 2x from both sides of the
equation), we obtain
3x - 2x = 6
x = 6
This is the solution and we can check it by substituting back into the original equation.
i.e. 3x - 2 = 2x + 4
3(6) - 2 = 2(6) + 4
18 - 2 = 12 + 4
16 = 16
Since this is true, our solution is correct.
Example:
Solve
5
4x
= 1
By multiplying both sides of the equation by 5, we obtain
by dividing both sides of the equation by 4, we obtain
5
5
4x
= 5 (1)
4x = 5
x = 1¼
- 41. 2-41
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
We can check this answer by substituting it back into the original equation
i.e.
5
4x
= 1
5
4
(1¼) = 1
5
4
4
5
= 1
Since this is true our solution is correct.
Example:
Solve ½ (3x - 1) = 7
Thisequation can be written as
2
1
3x
= 7
multiplying both sides of the equation by 2, we obtain
2
2
1
3x
= 2(7)
3x - 1 =14
by moving -I to the R.H.S. of the equation (i.e. by adding +1 to both sides of the equation), we
obtain
3x = 14 + 1
3x = 15
by dividing both sides of the equation by 3, we obtain
3
15
3
3x
x = 5
This is the solution.
Example:
Solve
3
2 (x—1) =
5
4 (2x—3)
- 42. 2-42
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
This equation can he written as
5
3)
4(2x
3
1)
2(x
Using the process of cross-multiplication, we obtain
5[2(x - 1)] = 3[4(2x—3)]
10(x - l) = 12(2x - 3)
eliminating the brackets, we obtain
10x - 10 = 24x - 36
rearranging the equation so that the terms in x are on the R.H.S. of the equation, and the
constants are on the L.H.S., we obtain
-10 + 36 = 24x – 10x
26 = 14x
dividing both sides of the equation by 14, we obtain
14
14x
14
26
x
14
26
7
6
1
x
This is the solution.
Example:
Solve
1)
3(x
2
5x
2
7
10
3
x
9
- 43. 2-43
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
In order to first simplify this equation we can multiply both sides of the equation by 10, thus
eliminating the denominators.
We then have:
1)
3(x
2
5x
10
2
7
10
3
x
9
10
Expanding the brackets we have:
9(x + 3) + 35 = 25x + 30(x - 1)
9x + 27 + 35 = 25x + 30x - 30
9x + 62 = 55x - 30
Rearranging the equation so that the terms in x are on the R.H.S. of the equation, we have:
30 + 62 = 55x - 9x
92 = 46x
x = 2
- 45. 2-45
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1.
2.
3.
4.
5.
i) x – 3 = 4
ii) 3x – 5 = 7
iii) 2 – x = 1
iv) 14 - 5x = -6
v) -3 + 4x = 5
i) 4x + 1 = 3x + 2
ii) 2 – x = 7x - 6
iii) 4 + 2x = 5x - 8
iv) -4 - 3x = -7 - 2x
v) 5x – 2 = -12x - 36
i)
3
8x = 16
ii) 1¾ x = -7
iii) -
3
2x = -6
iv) -3½ x = 4
v)
5
7x = -1
i) 2(3x - 1) = 28
ii) 5(3 - 2x) = 35
iii) 3(2x +1)= -15
iv) -4(1 - x) = 24
v) ¾(5 - 3x) = 15
i) 1)
(x
3)
(x 2
1
3
2
ii) 1)
(6x
2x)
(4
1 14
5
4
1
iii) 2x)
(3
2)
(x 3
4
2
3
iv) 4x
3)
(x
3
2
v)
4
3x
3)
(x
25
2
- 46. 2-46
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
6.
i) 3x = x - 7
ii) 2(3 – 2x) = -2
iii) x – 2 = 7 – 2x
iv)
5
x
= -2
v)
2
1
2
3
4
x
vi)
1
4
3x
4
1
2
x
vii)
6
5
6
x
3x)
(2
3
2
viii) 2(3a -1) = 5(a + 7)
ix)
5
y
1
x)
6
1
3x
2
xi)
5
k
2
3
k
1
xii)
3
x
3
2
xiii)
4x)
2(1
3
4
2x
1
xiv)
1
3x
2
4x
3
xv)
3
4
x
5x)
3( 6
67
xvi) 2(3t + 7) + 4(8 – t) = 8(t + 2)
xvii) ¾ + 2(3 – z) = ½ (2 – 6z)
xviii)
7
3
4x
3)
(x
3
1
2x
2
1
xix)
2
x
7x
3x2
xx)
3
4
x
2)
5(x
2
7
4x
3
- 47. 2-47
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1.
i) x = 7
ii) x = 4
iii) x = 1
iv) x = 4
v) x = 2
2.
i) x = 1
ii) x = 1
iii) x = 4
iv) x = 3
v) x = -2
3.
i) x = 6
ii) x = -4
iii) x = 9
iv) x = - 7
8
v) x = - 7
5
4.
i) x = 5
ii) x = -2
iii) x = -3
iv) x = 7
v) x = -5
5.
i) x = 9
ii) x = 1
iii) x = 25
17
1
iv) x = 7
3
v) x = 21
6
2
- 48. 2-48
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
6.
i) x = -3½
ii) x = 2
iii) x = 3
iv) x = -10
v) x = -2
vi) x = -5
vii) x = 1
viii) a = 37
ix) y = 5
1
x) x = 4
xi) k = 11
xii) x = 2 3
1
xiii) x = 1
xiv) x = -3
xv) x = 2
xvi) t = 5
xvii) z = -5¾
xviii) x = -45
xix) x = 3
xx) x = 4 3
2
- 49. 2-49
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Indices and Powers
Basic Laws of Indices and Powers
The laws are shown below in algebraic form, where a, m and n are any number.
am
= a x a x a x …….to m terms
an
= a x a x a x …….to n terms
Laws:
1. am
an
= am+n
[I]
For example a
3
a
5
= a
3+5
= a
8
am
2. = am-n
[II]
an
a6
For example -- = a6-2
= a4
a2
3. (am
)n
= amn
[III]
For example (a3
)5
= a3x5
= a15
4. am/n
= n
√am
[IV]
For example a7/2
= √a7
1
5. a-n
= [V]
an
1
For example: a-3
=
a3
- 50. 2-50
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
and expressing a4
x a -7
with a positive index gives
a4-7
= a-3
(by law [I])
1
= (by law [V])
a3
a0
=1 [VI]
a2
a3
a2+3
Example: = = (by law [I])
a5
a5
a5
= (by law [II])
a5
= a0
= 1 (by law [VI])
Example:
Simplify a2
b3
c x ab2
c3
and evaluate when a = 1, b = ½ and c = 2
Grouping like terms this becomes a2
x a x b3
x b2
x c x c3
and since a = a1
and
c = c1
, using law [I] this becomes
a(2+1)
x b(3+2)
x c(1+3)
= a3
x b5
x c4
= a3
b5
c4
When a = 1, b = ½ and c = 2, a3
b5
c4
=(1)3
(½)5
(2)4
= ½
- 51. 2-51
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1 Simplify p2
q3
r x pq2
r5
and evaluate when p = 3, q = 2 and r = ½
x½
y2
z3/2
2 Simplify and evaluate when x = 2, y = 2 and z = 9
y3/2
z
3 Simplify a ¼
b2
c-2
x b –3/2
c3
and evaluate when a = 16, b = 1/9 and c = 6
Simplify the following
4 a2
b3
c x a-3
b -5
c2
, expressing the answer with positive indices only
p1/3
q2
r1/5
5
(pq ½
r 3
)1/3
6 (x-2/3
y ½
z3
) x (x 2/3
y 2
z1/3
)1/3
expressing the answer with positive indices only
a2
b ½
c -¼
x (a b) ¼
7
3
√a 2
√b 3
c ½
k3
(√l) m3/2
x l3
√m2
8
k1/3
√l3
- 53. 2-53
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1 p3
q5
r6
13 ½
2 x ½
y ½
z ½
6
3 a ¼
b ½
c 4
c3
4
ab2
5 q 11/6
r – 4/5
y7/6
z3 1/9
6
x4/9
7 a19/12
b-3/4
c–3/4
8 k 2 2/3
l2
m 2 1/2
- 55. 2-55
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Standard Form
Introduction
To multiply a decimal fraction by 10 the decimal point is moved one place to the right, by 100
two places to the right and so on. To divide a decimal fraction by 10, the decimal point is moved
one place to the left and to divide by 100, two places to the left. The value of a number is
unaltered if the number is both multiplied and divided by the same number. For example, the
number 3 is not altered if multiplied by 1000 and divided by 1000, for
3 x 103
÷ 103
= 3.
When solving problems containing decimal or other fractions, the fractions can be expressed in
decimal fraction form with one figure only in front of the decimal point by multiplying or dividing
the number by 10 raised to some power. When this way of writing a number is used it is said to
be written in standard form. Thus a number written in standard form is a number between 1 and
10 multiplied by 10 raised to a power.
To write 43.7 in standard form, for example, it is first divided by 10 by moving the decimal point
one place to the left to give 4.37. But it must now be multiplied by 10 to retain the value of the
original number. So, 43.7 = 4.37 x 10 when written in standard form.
Again, to write 0.0437 in standard form, it is multiplied by 100 or 102
by moving the decimal
point two places to the right and then divided by 100 (or multiplied by 10-2
) to retain its original
value.
Thus 0.0437 = 4.37 x 10-2
when written in standard form.
Writing a number in standard form enables a quick check to be made on the approximate value
of a calculation to make sure an error in the position of the decimal point has not occurred. Also
a similar principle is used to denote the size of certain physical quantities. The SI system of
units has adopted the metre as its basic unit of linear measure (length or distance). To measure
the distance between two towns, thousands or tens of thousands of metres would be required,
whereas the length of a small insect such as an ant would be expressed in thousandths of a
metre. Since length and distance can vary so much, large distances are measured in kilometres
or metres x 103
. The Table below gives some of the powers of 10 used to express numbers as
a reasonable size, together with the abbreviations used for these powers of 10 and the name
given to them.
- 56. 2-56
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Powers of ten in common use:
When multiplying The prefix The abbreviation
a number by used is used is
109
Giga G
106
Mega M
103
kilo k
10-1
deci d
10-2
centi c
10-3
milli m
10-6
micro μ
10-9
nano n
10-12
pico p
10-15
femto f
To measure the power output from a large modern alternator in a power station, megawatts
(MW) are used, but the power to drive a small transistor radio would be measured in milliwatts
(mW). The distance between London and Birmingham would be stated in kilometres (km) but
the distance between the ends of a pencil would be measured in centimetres (cm). These units
are selected to keep numbers to a reasonable size. Other units used such as velocity, whose SI
unit is metres per second, will be written as ms-1
or m/s and acceleration, having an SI unit of
metres per second squared, will be written as ms-2
or m/s2
.
When a number is written in standard form, the number is called the mantissa and the factor by
which it is multiplied the exponent.
Thus 4.3 x 105
has a mantissa of 4.3 and an exponent of 105
. Addition and subtraction of
numbers in standard form can be achieved by adding the mantissae provided the exponent is
the same for each of the numbers being added. For example:
4 x 102
+ 5.6 x 102
= 9.6 x 102
This can be verified by writing the numbers as integers, for
4 x 102
+ 5.6 x 102
= 400 + 560 = 960
Also 9.6 x 102
= 960
hence 4 x 102
+ 5.6 x 102
= 9.6 x 102
When the exponents are not the same it is usually better to write the numbers in decimal
fraction form before adding or subtracting.
- 57. 2-57
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
The laws of indices are used when multiplying or dividing numbers given in standard form. For
example:
(3 x 1O3
) x (5 x 102
) = (3 x 5) x (103+2
) = 15 x 105
= 1.5 x 106
8 x 105
8
Similarly, = - x (10 5-3
) = 4 x 102
2 x 103
2
- 59. 2-59
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
Express the following numbers in standard form:
1 (a) 47.44 (b) 83.6 (c) 91.274 (d) 387.7
2 (a) 563 (b) 7210 (c) 63 000 0000 (d) 76271. 85
3 (a) 0.375 (b) 0.14 (c) 0.6 (d) 0.0026 (e) 0.00302
4 (a) 0.0000017 (b) 0.000101 5 (c) 0.10002 (d) 0.07073
5 (a) 63 7/8 (b) 3/20 (c) 468 4/5 (d) 1/500
In the following problems, change the numbers from standard form to integers or decimal
fractions:
6 (a) 3.72 x 102
(b) 6.2174 x 10-2
(c) 1.1004 x 103
(d) 3.27 x 104
(e) 8.27 x 10-1
7. (a) 5.21 x 100
(b) 3 x 10-6
(c) 1.4771 x 10-3
(d) 5.87 x 10
8. (a) 7.176 x 106
(b) 9.98 x 10-4
(c) 4 x 10-5
9. Change the following numbers from standard form to proper or improper fractions:
(a) 9.375 x 10-2
(b) 1.873 5 x 102
(c) 5.625 x 10-1
(d) 3.2475 x 102
- 60. 2-60
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
10. Evaluate and express the answer in standard form:
(a) 3.774 x 10-2
+ 7.28 x 10-2
(b) 6.3 x 103
+ 5.381 x 103
(c) 1.476 x 10 -6
- 1.471 x 10-6
(d) 3.576 x 104
- 4.211 x 104
11. Find the value of the following, giving the answer in standard form:
(a) 1.874 x 10-2
+ 2.227 x 10-3
(b) 5.27 x 10-10
+ 8.371 42 x 10-10
(c) 7.2873 x 10-4
- 3.8771 x 10-4
(d) 9.71 x 102
- 9.998 x 103
12 Rewrite the following statements without using powers of 10:
(a) the freezing temperature of copper is 1.3576 x 103
Kelvin
(b) one kilowatt hour has the same energy as 3.6 x 106
joules
(c) the reciprocal of 1.609 x 103
is 6.214 x 10-4
(d) the volume of one fluid ounce is 2.841 x 10-5
cubic metres
(e) the square root of 4 x 10-4
is 2 x 10-2
13 In these problems, evaluate giving the answers in standard form:
(a) (4.75 x 102
)(8 x 103
) (b) 3 x (4.4 x 103
)
(c) 8 x 10-3
(d) (4.5 x 103
)(3 x 10-2
)
5 x 10-5
2.7 x 104
- 61. 2-61
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1 (a) 4.744 x 10 (b) 8.36 x 10 (c) 9.127 4 x 10 (d) 3.877 x 102
2 (a) 5.63 x 102
(b) 7.21 x 103
(c) 6.3 x 108
(d) 7.627185 x 104
3 (a) 3.75 x 10-1
(b) 1.4 x 10-1
(c) 6 x 10-1
(d) 2.6 x 10-3
(e) 3.02 x 10-3
4 (a) 1.7 x 10-6
(b) 1.015 x 10-4
(c) 1.0002 x 10-1
(d) 7.073 x 10-2
5 (a) 6.3875 x 10 (b) 1.5 x 10-1
(c) 4.688 x 102
(d) 2 x 10-3
6 (a) 372 (b) 0.062174 (c) 1 100.4 (d) 32 700 (e) 0.827
7 (a) 5.21 (b) 0.000003 (c) 0.0014771 (d) 58.7
8 (a) 7 176 000 (b) 0.000998 (c) 0.00004
9 (a) 3/32 (b) 3747/20 (c) 9/16 (d) 1299/4
10 (a) 1.1054 x 10-1
(b) 1.1681 x 104
(c) 5 x 10-9
(d) –6.35 x 103
11 (a) 2.0967 x 10-2
(b) 1.364142 x 10-9
(c) 3.4102 x 10-4
(d) -9.027 x 103
12 (a) 1357.6 K (b) 3 600 000 joules (c) 1609; 0.0006214
(d) 0.00002841 m3
(e) 0.0004; 0.02
13 (a) 3.8 x 106
(b) 1.32 x 104
(c) 1.6 x 102
(d) 5 x 10-3
- 63. 2-63
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Number Systems
Binary
Every number that can be written in decimal can also be written in another system called
Binary. Binary is the main number systems used by computer scientists.
The binary number system is a base 2 number system which uses only the digits 0 and 1. It is
also a place value system which means that each place represents a power of 2, just as the
place represents a power of 10 in the decimal system:
Powers of 2: 25
24
23
22
21
20
. 2-1
2-2
Decimal No.: 32 16 8 4 2 1 . 0.5 0.25
e.g.: 1010.012 0 0 1 0 1 0 . 0 1
The number 1010.012 therefore means:
1 x 8 = 8
+1 x 2 = 2
+ 1 x 0.25 = 0.25
So, 1010.012 = 10.2510
Binary to Decimal Conversions
Example:
Convert 1001001 to Decimal
Write down the powers of 2, and the number to be converted below them, as follows
Then add all the numbers above the 1’s
i.e. 64 + 8 + 1 = 73
64 32 16 8 4 2 1
1 0 0 1 0 0 1
- 64. 2-64
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Decimal to Binary Conversions
Example:
Convert 271 to Binary
Write down the powers of two up to the next higher number (256 in this case)than the number to
be converted
Next write in the first digit 1 under the highest number (256). Subtract the 256 from 271
271 – 256 = 15
Insert 1’s under the numbers which, when added, come to 15.
The Decimal number 27110 is therefore 1000011112
Adding Binary Numbers
Example:
Add 1100010 to 1000111
Line up the numbers as shown, and add each column starting from the left (as you would when
adding decimal numbers). When two 1’s are added, this would normally be 2. But 2 is not
allowed in binary, so write 0 and carry 1 to the next column to the left and include it in the
addition of the next column.
1100010 1’s carried to next left column
+1000111
1 1 1
10101001
256 128 64 32 16 8 4 2 1
256 128 64 32 16 8 4 2 1
1 1 1 1 1
- 65. 2-65
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Convert the following Binary numbers to Decimal:
(a) 1101.1 (b) 1001110.11 (c) 100100.1
2. Convert the following Decimal numbers to Binary:
(a) 62 (b) 1,024 (c) 42.25 (d) 51.125
3. Add the following Binary numbers:
(a) 111 and 100 (b) 10010 and 1101 (c) 10110001 and 11100010
- 67. 2-67
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. (a) 13.5 (b) 78.75 (c) 36.5
2. (a) 111110 (b) 10000000000 (c) 101010.01 (d) 110011.001
3. (a) 1011 (b) 11111 (c) 110010011
- 69. 2-69
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Octal
Every number that can be written in decimal can also be written in another system called octal.
Like binary, octal is one of the three main number systems used by computer scientists.
The octal number system is a base 8 number system which uses only the eight digits 0, 1, 2, 3,
4, 5, 6, and 7. It is also a place value system which means that each place represents a power
of 8, just as the place represents a power of 10 in the decimal system:
Powers of 8: 84
83
82
81
80
. 8-1
8-2
Decimal No: 4096 512 64 8 1 . 0.125 0.015625
e.g.: 2378 2 3 7 .
Thus, an octal number such as 2378 = 2 x 82
+ 3 x 81
+ 7 x 80
= 128 + 24 + 7 = 15910.
Converting Binary to Octal and Octal to Binary
To convert a binary number to an octal number, construct a 3-bit binary / octal lookup table like
the one below. Starting at the binary decimal point of the binary number, take the first 3 bits and
find the corresponding octal value from the table. Repeat with next 3 bits and so on. If less than
3 bits remain, pad them with 0's until there are 3 bits. Again use the table.
3-bit binary octal
000
001
010
011
100
101
110
111
0
1
2
3
4
5
6
7
3-bit binary / octal table
- 70. 2-70
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Example:
Convert 11010010 to octal.
1. Take the 3 most right bits, 010 and find the corresponding octal value in the above
lookup table. The octal value is '2'.
2. Take the next 3 bits, 010. The corresponding octal value from the lookup table is '2'
again.
3. Now, only 2 bits, 11 of the binary number remain. Pad the left hand side with a 0 to get
011. The corresponding octal value from the lookup table is '3'.
So , 110100102 = 3228
To convert from octal to binary, write down the binary representation of each octal digit. Note
that each octal digit should take up 3 bits.
Example:
Convert 3228 to binary
3 = 011
2 = 010
2 = 010
So, 3228 = 0110100102
- 71. 2-71
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Convert the following Binary numbers to Octal:
(a) 101010100 (b) 011110100000 (c) 111101001
2. Convert the following Octal numbers to Binary:
(a) 1263 (b) 65217 (c) 426 (d) 5625
- 73. 2-73
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. (a) 524 (b) 3640 (c) 751
2. (a) 1010110011 (b) 110101010001111 (c) 100010110
(d) 101110010101
- 75. 2-75
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Hexadecimal
Every number that can be written in decimal can also be written in another system called
hexadecimal. Hexadecimal is the last of the three main number systems used by computer
scientists.
The hexadecimal number system is a base 16 number system which uses the sixteen digits 0,
1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. Here, we need the extra digits A, B, C, D, E, and F
to represent the numbers 10, 11, 12, 13, 14, and 15, since there are no digits in the decimal
numeral system to do this.
decimal hexadecimal
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
Hexadecimal is also a place value system which means that each place represents a power of
16, just as the place represents a power of 10 in the decimal system:
Powers of 16: 163
162
161
160
. 16-1
Decimal No: 4096 256 16 1 . 0.0625
e.g.: 3AF16 3 A F .
Thus, a hexadecimal number such as 3AF16 = 3 x 162
+ 10 x 161
+ 15 x 160
= 768 + 160 + 15 =
94310.
Note: It is much more difficult to convert from decimal to hexadecimal than it is to convert from
hexadecimal to decimal. If you are asked in the exam to do the latter, take each answer
provided and convert to decimal, until you get the number in the question.
- 76. 2-76
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Converting Hexadecimal to Binary and Binary to Hexadecimal
To convert a binary number to an hexadecimal number, construct a 4-bit binary / hexadecimal
lookup table like the one below. Starting at the binary decimal point of the binary number, take
the first 4 bits and find the corresponding hexadecimal value from the table. Repeat with next 4
bits and so on. If there is less than 4 bits remaining, pad them out to 4 bits.
4-bit binary hexadecimal
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
Example:
Convert 11010010 to hexadecimal.
1. Take the 4 most right bits, 0010 and find the corresponding hexadecimal value in the
above lookup table. The hexadecimal value is '2'.
2. Take the next 4 bits, 1101. Find the corresponding hexadecimal value in the above
lookup table. The hexadecimal value is 'D'.
So, 110100102 = D216
To convert from hexadecimal to binary, write down the binary representation of each
hexadecimal digit. Note that each hexadecimal digit should take up 4 bits.
Example:
Convert 2CF16 to binary
2 = 0010
C = 1100
F = 1111
So, 2CF16 = 0010110011112
- 77. 2-77
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Convert the following Binary bits to Hexadecimal code:
(a) 11100001
(b) 101110001111
(c) 11111100
2. Convert the following Hexadecimal codes to Binary bits:
(a) 4F (b) 1AC (c) 67 (d) 2A8
3. Convert the following Hexadecimal codes to Decimal:
(a) 2D (b) 1AF (c) 21A (d) 1AE
4. Convert the following Decimal numbers to Hexadecimal codes:
(a) 1632 (b) 494 (c) 5174 (d) 67
- 79. 2-79
Module 1.2 Algebra
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ST Aerospace Ltd
© Copyright 2013
Answers
1. (a) E1
(b) B8F
(c) FC
2. (a) 1001111 (b) 110101100 (c) 1100111 (d) 1010101000
3. (a) 45 (b) 431 (c) 538 (d) 430
4. (a) 660 (b) 1EE (c) 1436 (d) 43
- 81. 2-81
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Binary Coded Decimal
The BCD system is a four-bit system representing a decimal character for use with digital
display readouts. It can also be used for addressing to make it more convenient for humans to
use.
BCD number 1001 0010 0011 0000
Decimal equivalent 9 2 3 0
Thus, a BCD number such as 1001001000110000 is 4 sets of 4-bit binary numbers
9, 2, 3 and 0, which, when decoded means decimal 9230.
- 83. 2-83
Module 1.2 Algebra
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ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Convert the following Decimal numbers to BCD:
(a) 94 (b) 429 (c) 2947 (d) 1736
2. Convert the following BCD numbers to Decimal:
(a) 10000101 (b) 011100001001 (c) 001101100100
- 85. 2-85
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. (a) 10010100 (b) 010000101001 (c) 0010100101000111
(d) 0001011100110110
2. (a) 85 (b) 709 (c) 364
- 87. 2-87
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Summary
HEX BINARY DECIMAL BCD OCTAL
0 0 0 0000 0000 0
1 1 1 0000 0001 1
2 10 2 0000 0010 2
3 11 3 0000 0011 3
4 100 4 0000 0100 4
5 101 5 0000 0101 5
6 110 6 0000 0110 6
7 111 7 0000 0111 7
8 1000 8 0000 1000 10
9 1001 9 0000 1001 11
A 1010 10 0001 0000 12
B 1011 11 0001 0001 13
C 1100 12 0001 0010 14
D 1101 13 0001 0011 15
E 1110 14 0001 0100 16
F 1111 15 0001 0101 17
10 1 0000 16 0001 0110 20
11 1 0001 17 0001 0111 21
12 1 0010 18 0001 1000 22
13 1 0011 19 0001 1001 23
14 1 0100 20 0010 0000 24
15 1 0101 21 0010 0001 25
16 1 0110 22 0010 0010 26
17 1 0111 23 0010 0011 27
18 1 1000 24 0010 0100 30
19 1 1001 25 0010 0101 31
1A 1 1010 26 0010 0110 32
1B 1 1011 27 0010 0111 33
1C 1 1100 28 0010 1000 34
1D 1 1101 29 0010 1001 35
1E 1 1110 30 0011 0000 36
1F 1 1111 31 0011 0001 37
- 89. 2-89
Module 1.2 Algebra
For Training Purposes Only
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Simultaneous Equations
Methods of Solving
Method 1: By Substitution
Let 2x + 3y =11 (1)
and 4x + 2y = 10 (2)
11 – 3y
Then from equation (1), x =
2
Let this expression for x be substituted into equation (2). Thus
11 - 3y
4 + 2y = 10
2
This is now a simple equation in y and may be solved. Multiplying both sides of the equation by
2 gives
4 ( 11 - 3y ) + 4y = 20
Removing brackets gives
44 - 12y + 4y = 20
Rearranging gives
44 – 20 = 12y – 4y
24 = 8y
Hence y = 3
This value of y may be substituted into either equation (1) or equation (2). (The result should be
the same in both cases.)
Substituting in equation (1) gives
2x + 3(3) = 11
Therefore 2x + 9 = 11
2x = 11 - 9
2x = 2
Hence x = 1
- 90. 2-90
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Therefore the solution of the simultaneous equations 2x + 3y = 11 and 4x + 2y = 10 is x = 1
and y = 3. This is the only pair of values that satisfies both equations.
Method 2: By Elimination
let 2x + 3y = 11 (1)
and 4x + 2y = 10 (2)
If equation (1) is multiplied throughout by 2 the resulting equation will be
4x + 6y = 22 (3)
The reason that equation (1) is multiplied by 2 is that the coefficient of x (i.e. the number
multiplying x) in equation (2) and equation (3) is now the same. Sometimes it is necessary to
multiply both equations by constants chosen so that the coefficients of x or y in each equation
become the same.
Equation (2) can now be subtracted from equation (3). Thus
4x + 6y = 22 (3)
4x + 2y = 10 (2)
Subtracting 0 + 4y = 12
Hence 4y = 12
Therefore y = 3
This value of y may now be substituted in equation (1) or equation (2) exactly as in method 1 to
find the value of x.
It will be found from experience that in many cases method 1, that of substitution is
unnecessarily cumbersome, so that method 2, the elimination procedure will be employed.
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Worksheet
Solve the following equations for the unknowns:
1. x + y = 5
x - y = 2
2. 2s + 3t = 5
s + t = 2
3. 3g - 2h = 7
g + 2h = 5
4. 4x – 3y = 18
x + 2y = -1
5. 7a - 4b = 37
6a + 3b = 51
6. 4c = 2 - 5d
3d + c + 3 = 0
7. 3a + 4b – 5 = 0
12 = 5b - 2a
8. a – b = 8
a + b = 12
9. d + e = 3
3d + 2e = 7
10. x + 3y = 11
x + 2y = 8
11. 3m - 2n = -4.5
4m + 3n = 2.5
12. 3x = 2y
4x + y = -11
13. 4x - 3y = 3
3x + 5y = 111
14 6m – 19 = 3n
13 = 5m + 6n
15 4a – 6b + 2.5 = 0
7a – 5b + 0.25 = 0
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16 s + t = 15
s t
- = 1
3 7
17
a
- 11 = -2b
2
3
b = 9 - 3a
5
18 3 1
p - 2q = -
2 2
3
p + q = 6
2
19
1.2a - 1.8b = -21
2.5a + 0.6b = 65
20
2.5x + 0.45 - 3y = 0
1.6x + 0.8y - 0.8 = 0
21
1.2p + q = 1.8
p - 1.2q = 3.94
- 93. 2-93
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Answers
1 x = 3½, y = 1½
2 s = 1, t = 1
3 g = 3, h = 1
4 x = 3, y = -2
5 a = 7, b = 3
6 c = 3, d = -2
7 a= -1, b=2
8 a = 10, b = 2
9 d = 1, e = 2
10 x = 2, y = 3
11 m = - ½, n = 1½
12 x = -2, y = -3
13 x = 12, y = 15
14 m = 3, n = - 1/3
15 a = ½, b = ¾
16 s = 8, t = 7
17 a = 2, b = 5
18 p = 3, q = 2
19 a = 20, b = 25
20 x = 0.30, y = 0.40
21 p = 2.50, q = -1.20
- 95. 2-95
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Quadratic Equations
Introduction
A quadratic function is a function of the form
y = ax2
+ bx + c (1)
where a, b, c are constants, and a * 0
Examples of the graphs of quadratic functions are given below:
A quadratic equation may be written in the form
ax2
+ bx + c = 0 (2)
where a, b, c are constants, and a 0
Note that this corresponds to putting y = 0 in the quadratic function (1)
The solution of the quadratic equation (2) is given by
2a
4ac
b
b
x
2
(3)
N. B. The derivation of formula (3) is not considered here.
It should be noted that if b2
- 4ac < 0 then (3) involves taking the square root of a negative
number. This is not possible in terms of real numbers and such cases are not considered here.
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Example:
Solve x2
- 4x + 3 = 0
Comparing this equation with (2) we see that
a = 1; b = -4; c = 3
(a is the coefficient of x2
; b is the coefficient of x; and c is the constant term)
Using (3) we obtain
2a
4ac
b
b
x
2
=
2(1)
4(1)(3)
(-4)
4 2
2
12
16
4
=
2
4
4
=
2
2
4
=
2
6
and
2
2
x = 3 and 1
The solution of x2
- 4x + 3 = 0 is x = 1 and x = 3
This can be verified by looking at Fig. 1 (i.e. the solution occurs where the curve
y = x2 - 4x + 3 cuts the x axis).
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Solution by Factorisation
An alternative method of solving a quadratic equation is factorisation. However, this method is
only easily applied to some quadratic equations. The method is illustrated as follows:
x2
- 4x + 3 = 0
To factorise the left hand side of this equation we require two numbers the sum of which is -4
(the coefficient of x) and the product of which is 3 (the constant term). These numbers are -3
and -1, and the factors are (x - 3) and (x - 1). We now have:
x2
- 4x + 3 = 0
(x - 3) (x - 1) = 0
The two terms (x - 3) and (x - 1) have a product of zero if either one of the terms equals zero.
Therefore the solution of our quadratic equation is
(x – 3) = 0 or (x - 1) = 0
x = 3 or x = 1
- 98. 2-98
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Example:
Solve -x2
+ 6x - 8 = 0
Comparing this equation with (2), we have
a= -1; b = 6; c = -8
Using (3) we obtain
2a
4ac
b
b
x
2
=
2(-1)
4(-1)(-8)
6
6
- 2
2
-
32
36
6
-
=
2
-
4
6
-
=
2
-
2
6
-
=
2
8
and
2
4
x = 4 and 2
This can be verified by looking at Fig. 2.
Alternatively, using the method of factorisation described in example 1, we have:
-x2
+ 6x - 8 = 0
x2
- 6x + 8 = 0
(x - 4) (x - 2) = 0
x = 4 and 2
Example:
Solve 4x2
- 14x+12=0
The calculations can be made simpler here if the equation is first divided through (on both
sides) by 2.
Hence we obtain
2x2
- 7x + 6 = 0
In this case we have
a = 2; b = -7; c = 6;
- 99. 2-99
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and hence
2a
4ac
b
b
x
2
=
2(2)
4(2)(6)
49
7
4
1
7
=
4
1
7
=
4
8
and
4
6
x = 2 and 1 ½
Example:
Solve 3x2
- 48 = 0
Here we have
a = 3; b = 0; c = -48
and hence
2a
4ac
b
b
x
2
=
2(3)
4(3)(-48)
0
0
=
6
576
=
6
24
x = 4
This example is a special case because b = 0. Whenever this case arises it is not necessary to
employ the quadratic formula given by (3). Instead we could have treated this example as
follows:
3x2
- 48 = 0
Moving -48 to the R.H.S. of the equation (i.e. by adding 48 to both sides of the equation) we
obtain
- 100. 2-100
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3x2
= 48
dividing through by 3 we obtain
x2
= 48/3 = 16
x2
= 16
Hence x = ±4
Example:
Solve x2
- 8x = 0
Here we have
a = 1; b = -8; c = 0
Hence
2a
4ac
b
b
x
2
=
2(1)
4(1)(0)
8)
(
8 2
=
2
0
64
8
=
2
8
8
=
2
16
and 0
x = 8 and 0
This example is a special case because c = 0. Whenever this case arises it is not necessary to
employ the quadratic formula given by (3). Instead we could have treated this example as
follows:
x2
- 8x = 0
Because both terms on the L.H.S. contain x, we can write this equation as
x(x - 8) = 0
That is x = 0 or (x - 8) = 0
Hence the solution is x = 0 and 8
- 101. 2-101
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Example:
Solve (x - 2) (x - 3) = 2x - 6
The first step is to expand the brackets and we have:
(x - 2) (x - 3) = 2x - 6
x2
- 5x + 6 = 2x - 6
Rearranging this equation we have
x2
- 7x + 12=0
We can now solve in the usual way. We have a = 1; b = -7; c = 12;
and hence
2a
4ac
b
b
x
2
=
2
48
49
7
=
2
1
7
=
2
8
and
2
6
x = 4 and 3
- 103. 2-103
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Worksheet
1. Solve the following
i) x2
- 5x + 4 = 0
ii) x2
- 3x = 10
iii) 6x2
= 1 - x
iv) x2
- 4x + 4 = 0
2. Solve the following:
i) x2
– 25 =0
ii) x2
= 49
iii) 4x2
- 576 = 0
iv) x2
- 1=0
v) 16x2
- 1=0
3. Solve the following:
i) x2
- 6x = 0
ii) -3x2
+ 6 = 0
iii) 4x2
-12x = 0
iv) 3x2
– x = 0
v) 7x2
= -x
4. Solve the following:
i) x2
- 8x + 7 = 0
ii) 3x2
- 14x + 8 = 0
iii) -8x2
- 6x - 1=0
iv) 3x2
– 300 = 0
v) l0x2
– x = 0
vi) 42x2
= 13x - 1
vii) (x - 8)(x - 4) + 3 = 0
viii) -2z2
+ 19z - 24 = 0
ix) y2
– l0y + 25 = 0
x) -4x2
= 3x
- 105. 2-105
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Answers
1.
i) x = 1, 4
ii) x = -2, 5
iii) x = -½, 3
1
iv) x = 2
2.
i) x = ±5
ii) x = ±7
iii) x = ±12
vi) x = ±1
v) x = ±¼
3.
i) x = 0, 6
ii) x = ±1.41
iii) x = 0, 3
iv) x = 0, 3
1
v) x = 0, - 7
1
4.
i) x = 1, 7
ii) x = 3
2 , 4
iii) x = -½, -¼
iv) x = ±10
v) x = 0, 10
1
vi) x = 6
1 , 7
1
vii) x = 5, 7
viii) z = 1½, 8
ix) y = 5
x) x= 0, -¾
- 107. 2-107
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Logarithms
Why Logs?
In the discussion of indices it was noted that whenever a number is "raised" to a power then we
write that in exponential notation and the meaning of it is that the number appearing in the base
is being multiplied by itself the number of times that is indicated by the exponent. The notation
used was such that if we write 53
, what we actually mean is "5 multiplied by itself 3 times".
Logarithms are mathematical inventions in order to answer a slightly different question (notice
the word "invention"; logarithms make certain operations easier to handle and that is all they do,
so you should think of them as a definition). In order to motivate why logarithms are introduced
in the first place, let us invent a scenario. Suppose someone asked you the following question:
What number do I have to raise to the power of 3 in order to get 1,000? Well ... this might seem
pretty simple and obvious. If you multiply 10 × 10 you get 100, and if you multiply 100 × 10 you
get 1,000. So, you would say that 10 multiplied by itself 3 times - or, in our power notation, 103
-
is equal to 1,000.
Now, this is easy to answer by thinking about powers because the above example is simple
powers and simple number, and once can reason it out relatively quickly. However, things can
get more complicated. Suppose now that you were asked "what number do I have to raise 10 to
in order to get 735. All of a sudden the answer is not very obvious. What is so different about
this question?
There is actually nothing different about this question. You still can try doing the same process,
but now the number isn't that pretty and it's not exactly obvious how many times you should
multiply 10 by itself to get 735. If you multiply it by itself 2 times you get 100, but 3 times gives
1,000, and you have already exceeded 735! How do we "get out" this power that we need?
Logarithms are - at the most basic level - invented to answer the more general question of how
does one extract the base or exponent of an algebraic power when one of these is an unknown.
Definition
Continuing on the above reasoning, let us take our simple example again: what number raised
to the power of 3 gives 1,000? If we invent an unknown variable - call it ‘y’ and try to write out
our question in terms of the notation of algebraic powers we have the following situation:
10y
= 1,000
The question is: what is ‘y’ in the above formula? How do we solve for ‘y’? We invent an
operation called the logarithm - abbreviated to Log - and we apply this operation to the above
relation.
- 108. 2-108
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Thus:
If by
= x then logb(x) = y
This is the definition of a log.
How does this help us with anything? It seems like we went in a big loop, and we knew the
answer to begin with anyway! But ... now consider the slightly more complicated question that
we had above: "what number do I raise 10 to, in order to get 735?". Let us apply the logarithmic
process to this situation:
10y
= 735
Log10735 = y
If you take the Log of 735 on your calculator you get, 2.866...! So, 10 raised to the power of
2.866... gives you 735, and the question is answered. Recall that algebraic powers need NOT
be integers, and here we have a clear example of a non-integer power.
A series of logs can be drawn on a graph as shown below
Thus it can be seen that
Log101 = 0
Log1010 = 1
Log10100 = 2
Log101000 = 3
Log1010000 = 4
etc.
x
y
Log10x = y
10 1000
100
1
2
3
1
- 109. 2-109
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And the log of anything between 10 and 100 is between 1 and 2 (but not on a linear
relationship).
It can be said then, that Log10150 = 2.xxx, where the x’s are any numbers after the decimal
point. The ‘2’ is known as the Characteristic of the Log, and the decimals are known as the
Mantissa. Hence, the characteristic of the Log of any number between 100 and 999 is 2. (The
‘Characteristic’ is in fact ‘n-1’ where ‘n’ is the quantity of digits you are taking a log of).
Common Logarithms
There are two basic types of logarithms that are important to know. In the previous section,
where logarithms were defined, you already saw the definition of one kind of logarithms; that
was the so called "log base 10".
The logarithmic operation that we have introduced serves the main purpose of extracting the
exponents in an algebraic power. This is true of the operation of "taking the logarithm".
The logarithm of base 10 is most often useful when powers of 10 are involved, but not
necessarily. It can be used in many other situations. For instance, suppose you were asked the
following question: 3 raised to what power gives 16.8? Again, applying our definition of
logarithm of base 10 - as defined in the previous section - we can answer this question ... but, in
order to do this we need to define some rules of operation for logarithms (this is outlined in the
next few pages).
Logarithms having a base of 10 are called common logarithms and log10 is often abbreviated to
‘lg’.
Natural Logarithms
There is another logarithm that is also useful (and in fact more common in natural processes).
Many natural phenomenon are seen to exhibit changes that are either exponentially decaying
(radioactive decay for instance) or exponentially increasing (population growth for example).
These exponentially changing functions are written as ex, where x represents the rate of the
exponential.
In such cases where exponential changes are involved we usually use another kind of logarithm
called natural logarithm. The natural log can be thought of as Logarithm Base-e. What this
means is that it is a logarithmic operation that when carried out on e raised to some power gives
us the power itself. This logarithm is labelled with Ln (for "natural log") and its definition is:
Ln(ex
) = x
Logarithms having a base of e (where ‘e’ is a mathematical constant approximately equal to
2.7183) are called hyperbolic or natural logarithms, and loge is often abbreviated to ‘ln’.
- 111. 2-111
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Rules of Logarithms
There are three rules of logarithms, which apply to any base.
Rule 1. To multiply two numbers:
log AB = log A + log B
The following may be checked by using a calculator:
lg 10 = 1
Also lg 5 + lg 2 = 0.69897 + 0.301029 = 1
Hence lg(5 x 2) = lg 10 = lg 5 + lg 2
Rule 2. To divide two numbers:
A
log — = log A – log b
B
The following may be checked using a calculator:
5
In — = ln 2.5 = 0.91629...
2
Also In 5 - ln 2 = In 2.5 = 1.60943 - 0.69314 = 0.91629
5
Hence ln — = ln 5 – ln 2
2
- 112. 2-112
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Rule 3. To raise a number to a power
log An
= n log A
The following may be checked using a calculator:
lg52
= lg 25 = 1.39794
Also 2 lg 5 = 2(0.69897) = 1.39794
Hence lg 52
= 2 lg 5
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Further Logarithms
Example
Solve the equation 3 x + 1
= 2 2x – 3
Taking logarithms to base 10 of both sides gives
log10 3 x + 1
= log10 2 2x - 3
(x + 1) log10 3 = (2x – 3) log10 2
x log10 3 + log10 3 = 2x log10 2 - 3 log10 2
A calculator or log tables are required for the next line. It is given here as an example only.
However the CAA question may give you the values of log103 and log102
x(0.4771) + 0.4771 = 2x(0.3010) - 3(0.3010)
x = 11.05
Example
Solve the equation x 2.5
= 37.5
Taking logarithms to base 10 of both sides gives
log10 x2.5
= log10 37.5
2.5 log10 x = log10 37.5
log10 x = log10 37.5 2.5
A calculator or log tables are required for the next line. It is given here as an example only.
However the CAA question may give you the values of log1037.5
log10x = 0.6296
x = antilog 0.6296 = 4.262
- 115. 2-115
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Some special properties of logarithms
(i) logb 1 = 0
Let logb 1 = x then bx
= 1 from the definition of a logarithm.
If bx
= 1 then x = 0, from the laws of indices.
Hence logb 1 = 0.
(log10 1 = 0, for example)
(ii) Iogb b = 1
Let logb b = x then bx
= b from the definition of a logarithm.
If bx
= b then x = 1, from the laws of indices
Hence logb b = 1.
(log10 10 = 1, for example)
- 117. 2-117
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Worksheet
In problems 1 to 3, solve the equations for x.
1. (a) log10 x = 4 (b) lg x = 5
2. (a) log3x = 2 (b) lg x = -2
4
3. (a) log8x = - -- (b) ln x = 4
3
In problems 4 to 8, evaluate the given expressions.
4. (a) log10 100 (b) log2 16
5. (a) log7 343 (b) lg 1000
6. (a) log5 125 (b) log2 1/8
7. (a) log4 8 (b) log27 3
8. (a) log10 105
(b) ln e7
In problems 9 and 10, write the given expressions in terms of
log 2, log 3 and log 5 to any base.
9. (a) log 60 (b) log 300
8 x 5 125 x 4
16
10. (a) log (b) log
9 4
813
11. Simplify the following:
(a) log 64 – log 128 + log 32 (b) log 125 + log 25 - log 625
- 119. 2-119
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Answers
1 (a) 10 000 (b) 100 000
2 (a) 9 (b) 0.01
3 (a) 1/16 (b) e 4
4 (a) 2 (b) 4
5 (a) 3 (b) 3
6 (a) 3 (b) -3
7 (a) 1 ½ (b) 1/3
8 (a) 5 (b) 7
9 (a) 2log 2 + log 3 + log 5 (b) 2log 2 + log 3 + 2log 5
10 (a) 3 log 2 + ½ log 5 — 2 log 3 (b) log 2 + 3 log 5 — 3 log 3
11 (a) 4log 2 (b) log 5
- 121. 2-121
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Complex Numbers
The Number i
Consider the Equations 1 and 2 below.
Equation 1 Equation 2
x2
- 1 = 0. x2
+ 1 = 0.
x2
= 1. x2
= -1.
Equation 1 has solutions because the number 1 has two square roots, 1 and -1. Equation 2 has
no solutions because -1 does not have a square root. In other words, there is no number such
that if we multiply it by itself we get -1. If Equation 2 is to be given solutions, then we must
create a square root of -1.
Definition: The imaginary unit i is defined by
The definition of i tells us that i 2
= -1. We can use this fact to find other powers of i.
Example:
i 3
= i 2
x i = -1 x i = - i.
i 4
= i 2
x i 2
= (-1) x (-1) = 1.
We treat i like other numbers in that we can multiply it by numbers, we can add it to other
numbers, etc. The difference is that many of these quantities cannot be simplified to a pure real
number.
For example, 3 i just means 3 times i, but we cannot rewrite this product in a simpler form,
because it is not a real number. The quantity 5 + 3 i also cannot be simplified to a real number.
However, (-i)2
can be simplified. (-i)2
= (-1 x i)2
= (-1)2
x i 2
= 1 x (-1) = -1.
Because i 2
and (-i)2
are both equal to -1, they are both solutions for Equation 2 above.
- 122. 2-122
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The Complex Plane
Definition: A complex number is one of the form a + b i, where a and b are real numbers. a is
called the real part of the complex number, and b is called the imaginary part.
Two complex numbers are equal if and only if their real parts are equal and their imaginary
parts are equal. i.e., a+b i = c+d i if and only if a = c, and b = d.
Examples:
2 - 5 i
6 + 4 i
0 + 2 i = 2 i
4 + 0 i = 4
The last example above illustrates the fact that every real number is a complex number (with
imaginary part 0). Another example: the real number -3.87 is equal to the complex number
-3.87 + 0 i.
It is often useful to think of real numbers as points on a number line. For example, you can
define the order relation c < d, where c and d are real numbers, by saying that it means c is to
the left of d on the number line.
We can visualize complex numbers by associating them with points in the plane. We do this by
letting the number a + b i correspond to the point (a, b).
- 123. 2-123
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Complex Arithmetic
When a number system is extended the arithmetic operations must be defined for the new
numbers, and the important properties of the operations should still hold. For example, addition
of whole numbers is commutative. This means that we can change the order in which two whole
numbers are added and the sum is the same: 3 + 5 = 8 and 5 + 3 = 8.
We need to define the three arithmetic operations on complex numbers.
1. Addition and Subtraction
To add or subtract two complex numbers, you add or subtract the real parts and the imaginary
parts.
(a + b i) + (c + d i) = (a + c) + (b + d) i.
(a + b i) - (c + d i) = (a - c) + (b - d) i.
Example:
(3 - 5 i) + (6 + 7 i) = (3 + 6) + (-5 + 7) i = 9 + 2 i.
(3 - 5 i) - (6 + 7 i) = (3 - 6) + (-5 - 7) i = -3 - 12 i.
Note: These operations are the same as combining similar terms in expressions that have
a variable. For example, if we were to simplify the expression (3 - 5x) + (6 + 7x) by
combining similar terms, then the constants 3 and 6 would be combined, and the terms -5x
and 7x would be combined to yield 9 + 2x.
2. Multiplication
The formula for multiplying two complex numbers is
(a + b i) x (c + d i) = (ac - bd) + (ad + bc) i
You do not have to memorize this formula, because you can arrive at the same result by
treating the complex numbers like expressions with a variable, multiply them as usual, then
simplify. The only difference is that powers of i do simplify, while powers of x do not.
Example
(2 + 3 i)(4 + 7 i) = 2x4 + 2x7 i + 4x3 i + 3x7x i 2
= 8 + 14 i + 12 i + 21x(-1)
= (8 - 21) + (14 + 12) i
= -13 + 26 i.
Notice that in the second line of the example, the i 2
has been replaced by -1.
Using the formula for multiplication, we would have gone directly to the third line.
- 124. 2-124
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
3. Division
Definition: The conjugate (or complex conjugate) of the complex number a + b i is a - b i.
Conjugates are important because of the fact that a complex number times its conjugate is real;
i.e., its imaginary part is zero.
(a + b i)(a - b i) = (a2
+ b2
) + 0 i = a2
+ b2
Example:
Number Conjugate Product
2 + 3 i 2 - 3 i 4 + 9 = 13
3 - 5 i 3 + 5 i 9 + 25 = 34
4 i -4 i 16
Suppose we want to do the division problem (3 + 2 i) ÷ (2 + 5 i). First, we want to rewrite this as
a fractional expression
i
i
5
2
2
3
.
Even though we have not defined division, it must satisfy the properties of ordinary division. So,
a number divided by itself will be 1, where 1 is the multiplicative identity; i.e., 1 times any
number is that number.
So, when we multiply
i
i
5
2
2
3
by
i
i
5
2
5
2
, we are multiplying by 1 and the number is not changed.
Notice that the quotient on the right consists of the conjugate of the denominator over itself. This
choice was made so that when we multiply the two denominators, the result is a real number.
Here is the complete division problem, with the result written in standard form.
i
i
5
2
2
3
=
i
i
5
2
2
3
x
i
i
5
2
5
2
=
)
5
)(2
5
(2
)
5
)(2
2
(3
i
i
i
i
=
29
11
16 i
=
29
16
- i
29
11
- 125. 2-125
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Evaluate the following complex expressions:
(a) (8 - 7 i) + (2 + 3 i)
(b) (3 - 4 i) - (6 + 5 i)
2. Perform the following operations:
(a) (-3 + 4 i) + (2 - 5 i).
(b) 3 i - (2 - 4 i).
(c) (2 - 7 i)(3 + 4 i).
(d) (1 + i)(2 - 3 i).
3. Write (2 - i) ÷ (3 + 2 i) in standard form.
- 127. 2-127
Module 1.2 Algebra
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1.
(a) 10 – 4 i
(b) -3 – 9 i
2.
(a) -1 - i
(b) -2 + 7 i
(c) 34 - 13 i
(d) 5 - i
3. i
13
7
13
4
