Download GCSE Mathematics - STEM CPD Module

Analytical Toolbox
Introduction to arithmetic & algebra
By
Dr J. Whitty
Why study mathematics
2
The Laws of Mathematics
 Asscociative



3+(4+5)=(3+4)+5
a+(b+c)=(a+b)+c
a(bc)=(ab)c
 Commutative



4+5=5+4
a+b=b+a
ab+=ba
 Distributive



3(4+5)=3x4+4x5
a(b+c)=ab+ac
(a+b)/c=a/c+b/c
3
Module Leaning Objectives

1.
2.
3.
4.
To achieve this unit a learner must:
Determine the fundamental algebraic laws and apply algebraic
manipulation techniques to the solution of problems involving algebraic
functions, formulae and graphs
Use trigonometric ratios, trigonometric techniques and graphical
methods to solve simple problems involving areas, volumes and
sinusoidal functions
Use statistical methods to gather, manipulate and display scientific and
engineering data
Use the elementary rules of calculus arithmetic to solve problems that
involve differentiation and integration of simple algebraic and
trigonometric functions.
4
Assessment Methods
 Assignment 1:
Introductory mathematical statistics
Milestone test 1:
 Mock examination mathematics questions
Assignment 2:
 Introductory mathematical engineering systems modelling
Milestone test 2:
 Three mock-exam science questions
Milestone test 3:
 Computer modelling methods tutorial
Examination (informal)






5
Recommended Reading
 Engineering
mathematics KE
Stroud & DJ Booth
Palgrave
macmillian,
Fifth Edition,
London, 2001
6
Session Learning objectives
 After this session you should be able to:
 Compute numeric expressions using BIDMAS
 Evaluate numeric expressions in standard form
 Transpose simple formulae
 Solve simple linear equations inc those involving
fractions
 Solve elementary non-linear equations inc. squares
and square roots
7
BIDMAS
 Can you remember
what this stands for?






Brackets
Indices
Division
Multiplication
Additon
Subtraction
Let a=3, b=-1, c=2
1. 2a+5b+3c

2x3 + 5x(-1) +3x2

=7
2. 6a-7b

6x3 - 7x(-1) =

18 - -7 =25
3. 2a2 - 3b2

2xaxa - 3xbxb

2x3x3 -3x(-1)x(-1)

=15
8
Remove the brackets
 3(a+b)
 =3a + 3b
 5(2a-5c)
 =10a-25c
 4(2a-3b)-5(a-6b)
 =8a-12b-5a+30b
 =3a+18b
 6c-(3a+2b-5c)
 =6c-3a-3b+5c
 =11c-3a-3b
 4(5a-b)-2(3a-b-c)
 =20a- 4b-6a+2b+2c
 =14a-2b+2c
9
Try some yourselves
(a)( x  2)( x  2); x  4
l
1
(b)2   ;   3 , l  98, g  10
7
g
(c)( 2 x  1)(3x  2); x  3,1.5
(d )( p  q) ; p  3, q  1
2


(e)(2 x  3 y )  2 x  3 y ; x  3 y
2
2
2
2
10
Indices
 Consider 26
 The 2 is called the base, the 6 is the power or
the index. The above is said “2 to the power
of 6” and is calculated 2 x 2 x 2 x 2 x 2 x 2 =
64
 On the calculator press 2 x y 6 =
 26 DOES NOT EQUAL 12
11
Negative Indices
 Consider 2
-6
 The rule to eliminate the negative power is :
write 1 divided by the old base, to the positive
power
 This is equal to
12
Fractional Indices
 40.5 is the same as the square root of 4
4
1/8 is
the same as the eighth root of 4 and is
again calculated using the x y button on the
calculator (= 1.1892)
13
Rules of Indices
 If two powers are multiplied, then if the bases
are the same, the powers are ADDED
 25 x 27 = 212
 If two powers are divided, then if the bases
are the same, the powers are SUBTRACTED
 38  35 = 3 3
14
Examples on Indices
47  48 =
4-1 = 1/4
5-2 x 5- 4 =
5 -6 =
1/15625
3-3  3-1 =
3-2=1/9
54 x 52  5-3 =
59 = 1953125
100=
1
90=1 a0 = 1
3-7  3-5 x 27=
3
15
Standard Form
 Standard Form is used to write very large or
very small numbers in a more convenient way
 A number in standard form is written:
a x 10 n
 where a is a number between 1 and 10, and
n is an integer, positive or negative
16
Standard & Engineering Form examples
2,000,000
= 2x10 6
35,800
= 3.58x10 4 (35.8 x103)
5,927,000,000
= 5.927x109 (5.9x109)
43
= 4.3 x 10 1 (43)
0.000 004
= 4x10-6 (4x10-6)
0.000 458
= 4.58x10- 4 (458x10-6)
0.000 000 000 021
= 2.1x10-11 (21x10-12)
0.35
= 3.5x10-1 (350x10-3)
17
Using a calculator
To calculate 3.98 x 10 12 x 4.2 x 10 11
press the following keys:
3.98 exp 12 x 4.2 exp 11 =
Do NOT type x10, the exp button does this
automatically.
The answer should appear as 1.6716 24 on
screen.
The correct answer is then 1.6716 x 10 24
18
Evaluate, in standard form
3.2 x 10 6 x 4 x 10 4
= 1.28x10 11
4.5 x 10 16 + 4 x 10 15
=4.9x10 16
2.8 x10 27 x 3.5x 10-25
= 980
= 9.8x10 2
4.4 x10 -10 x 5.2x10 -4
= 2.288x10 -13
5 x 10 -8 2 x 10 -7
= 0.25
2.5x10 -1
19
Rules of Indices
x x x
n
m
nm
n
x
n
m
nm
x /x x
m
x
x 
n m
x
nm
20
Corollaries
n
x
nn
0
 n 1 x  x 1
x
0
x
1
n
0n
x  x  n  n
x
x
x 1
0
x
m
n
1
 n
x
x x
n
n/m

m
x   x 
n
n
1
m
x
n
m
n 1
x
n 1 n
1
NB : x  x  x  n  x
x
x
1
21
Let’s use them:
3 3 3
(a)
34
2
3
(c)(36)
 12
(b)(6)(2 x 0 )
(d )(16)
 34

2   3 
(e)
3 2
2 3
4
3
22
They still work algebraically!
3
12 x y
(a)
2
4x y
2
 a 3b 2c 4  a 2 
 2 
(b) 4
 a bc  c 

(c) b c ab c a
3 2
3 2
0

2
23
Recap: Make x the subject
Last lecture we examined the differences between
equations and formulae and their subsequent
solution protocols:
Formula:
Equation:
gx + h = k
3x+2 = 23
gx + h - h = k - h
3x+2 - 2= 23 -2
gx = k- h
3x = 23 - 2
x = (k - h)/g
x = (23-2)/3
x=7
24
Transposition Of Formulae
 The rules are exactly the same as for
algebra, except the final result is an algebraic
expression instead of a numerical answer.
25
Simple Transposition
 In the Science units you
will come across very
simple formulae for
instance

Density
Newton’s second law
(mechanics)
Electrical charge

Ohms Law


m

V
F  ma
Q
I
t
V  IR
26
Simple Transposition
 Here the same
rules apply as the
letters in the
formulae are just
numbers in
disguise
F  ma
F
F
a
m
m
a
F
ma
27
Activity
 In groups use the systematic transposition (or
otherwise) approach to develop calculation
transposition triangles for the formulae, describing:
Density
 Electrical Charge
 Voltage
In each case define each of the variables you have
used

28
Solution of Linear Equations
 Mathematical
 6x+1=2x+9
 6x+1-1=2x+9-1
 6x=2x+8
 6x-2x =2x-2x+8
 4x=8
 4x/4=8/4 =2
 Systematic
 6x+1=2x+9
 6x-2x= +9-1
 4x=8
 x=8/4=2
Note: The approaches are
exactly the same however in
the systematic approach we
MOVE numbers & variables
29
What about brackets?
 The rule with
 Consider:
brackets is simply
MULTIPLY (that’s all
they mean).
 MULTIPLY
everything inside the
bracket by everything
outside the bracket.
 3(x-2)=9
 3x-6=9
 Solve as before
 4(2r-3)-2(r-4)=3(r-3)-1

8r-12-2r+8=3r-9-1

Solve as before
 Exercise
30
Class Discussion/Exercise
 2(x-1)=4
 2x+5=7
 16=4(t+2)
 2c/3-1=3
 5(f-2)=3(2f+5)-15
 7-4p=2p-3
 2x=4(x-3)
 8-3t=2
 6(2-3y)-42=-2(y-1)
 2x-1=5x+11
 2(g-5)-5=0
 2a+6-5a=0
 4(3x+1)=7(x+4)-2(x+5)
 3x-2-5x=2x-4
 20d-3+3d=11d+5-8
31
Linear Equations with Fractions
 The systematic method
is especially useful with
fractional coefficients:
 e.g.



2x
6
5
Mathematical
Approach
2x
6
5
2x  6  5
65
x
 15
2
2x  5   5 
   6   6  2.5
5  2  2
32
Linear Equations with Fractions
 Consider: (math)
 There are several
methods to attempt
such problems but by
far the best is to
attempt to clear the
fractions (some how)
in order to reduce the
2y
equation to something20
5
simpler which can be
solved
2y 3
1 3y
 5 

5 4
20 2
3
1
3y
 20  20  5  20  20
4
20
2
8 y  15  100  1  30 y
33
Linear Equations with Fractions
 This is the so called
mathematical
2y 3
1 3y
approach
 5 

5 4
20 2
 There is the analogous
systematic approach,
which most engineers 20  2 y  20  3  20  5  1  20  3 y
find a little easier to
5
4
2
apply
8 y  30 y  1  15  100
 Thus:
38 y  114
y  3
34
Class Exercise
1.
d / 3  4  3
2.
3
3
5
2  y  1 y 
4
2
6
3.
1
4 x  1  5  1
3
2
x x 1
 
3 5 2
1
2 x  1  3  1
4
2
x x6 x3


4
5
2
4.
5.
6.
35
Squares and their roots
 Proceed as before but at the end to get rid of
a square root simply square everything

 x  2 2
x 2
Likewise:
4
5
x 2 x 2
x  3  x  35
4
What About:
2 x 8
36
Squares and their roots
 A similar approach can be applied to
squares and powers, thus
x  25
2
What about
15 2

2
4t
3
37
Examples



2 y 5

3 x
 6
1 x

x 
10  5   1
2 

t2
16 
9
6 2a

a 3
11
8
 5 2
2
x
38
Summary
 Have we met our learning objectives,
specifically, are you able to:




Compute numeric expressions using BIDMAS
Evaluate numeric expressions in standard form
Derive the rules of indices from first principles
Evaluate and simplify mathematical expressions
using the rules of indices.
39
Homework
 Evaluate or simplify the following
1
7 7
6 6
2
(
c
)
0
.
5
(a)
(
b
)
7 31
62
3
30
x
 3
0
(d ) 
(e)2876 ( f )
9
7
6
x
 
6
2
6
2
6
2
(i)4 x  3x
( g )36 x / 12 x (h)4 x  3x
7
5x
5 2
6 0
(l )
( n) 5 x
(k ) 3x
7
10 x
2
 
4
2
 
40
More homework
1.
2.
d / 5  3  2
3.
3
4.
5.
6.
3
3
4
y  1 y 
5
4
5
1
1
3x  1  2 
3
4
x x 1
 
5 3 4
1
1
5 x  1  6 
2
4
x x3 x6


5
4
6
41
Further Study
 Foundation topics
 F1 Arithmetic
 F2
Introduction to
algebra
42