Algebraic Methods 1

Algebraic Methods 1 Mathematics 1 Level 4 © University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License .

The following presentation is an introduction to the Algebraic Methods – part one for level 4 Mathematics. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st year undergraduate programme.
The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.
Contents
Indices
BODMAS
Simple Equations
Transposition
In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see:
KA Stroud & DJ Booth, Engineering Mathematics, 8 th Editon, Palgrave 2008.
http://www.mathcentre.ac.uk/
Derive 6
Algebraic Methods 1

Mathematical Syntax The area of a rectangle is given by: Area = Length x Breath Area = L x B or L.B or LB When two symbols are written next to each other then this implies that they are multiplied. This is not true for numbers 23 is not 2 x 3. Also a shorthand for 2 x 2 x 2 x 2 x 2 is 2 5 And a x a x a x a x a x a is a 6 We must also be aware that 2a is not the same as a 2 One is a + a the other is a x a Algebraic Methods 1

Indices We can now look at what happens when we combine expressions which have indices: e.g. a 4 x a 3 = (a x a x a x a) x (a x a x a) (a x a x a x a) x (a x a x a) = a x a x a x a x a x a x a = a 7 When multiplying powers of the same letter we add the indices a n x a m = a n+m e.g. When dividing powers of the same letter we subtract the indices a n  a m = a n-m Algebraic Methods 1

Examples
a 6 x a 2
a 2 x a 2 x a
a 5 x a x a 3
a 3  a 7 what does this mean?
a 9  a 3
a 10  a 5
ab 4 x a 2 b 2
a 2 b 3 x a 2 b 2  ab
Note a = a 1 and only like letters can be combined.
Algebraic Methods 1

Common error What does 2a 2 mean? Does it mean 2 x a x a or 2a x 2a It means the first one! If we wanted the second one we would write it as (2a) 2 So the indices only effect the adjacent letter unless a bracket is used. e.g. 1. (2a) 3 2. (3ab) 2 3. (ab) 3 x (a 2 b) 2 simplify these expressions Algebraic Methods 1

Powers Roots and Reciprocals What does (a 2 ) 3 give us? (a x a) x (a x a) x (a x a) = a x a x a x a x a x a = a 6 When raising a power of a term to a new power we multiply the indices (a n ) m = a nxm What does √a 8 give us? a x a x a x a x a x a x a x a = (a x a x a x a) x (a x a x a x a) = (a x a x a x a) 2 When rooting a power of a term we divide the indices √ (a n ) = a n  m m Algebraic Methods 1

Powers Roots and Reciprocals What does a  a give us? a 1  a 1 = a 1-1 = a 0 but we also know that this equals 1 Any term raised to the power 0 equals 1 and conversely 1 can be thought as any term to the power 0 What does 1/a 2 give us? 1  a 2 = a 0  a 2 = a 0-2 = a -2 When reciprocating a power of a term we change the sign of the index 1/a n = a -n Algebraic Methods 1

Addition and Subtraction Consider the equation: 6 + 8 = 14 We know that 3 x 2 4 x 2 7 x 2 i.e. three lots of something plus four lots of something equals seven lots of something or 3a + 4a = 7a Take care – it does not equal 7a 2 Same with subtraction: 10 a pples – 4 a pples = 6 a pples Note addition and subtraction only work if the letter is the same. 5 apples plus 6 oranges does not equal 11 ? Algebraic Methods 1

Number lines -8 -7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7 +8 Notes Positive numbers are to the right of zero Negative numbers are to the left of zero So 5a – 7a means start at 5 and move 7 to the left. The only sign that can be omitted in front of a term is a positive sign The only sign that can be omitted between two terms is a multiplication sign. Only similar terms can be added or subtracted 2a 2 b + 3ab 2 cannot be performed. Subtraction Addition

Simplify Algebraic Methods 1

B O D M A S This indicate the order in which mathematical operations must be carried out in order to generate the correct answer: B rackets O rder (power) D ivision M ultiplication A ddition S ubtraction What is the answer to the equation: 7 + 2 x 4 Most calculators have BODMAS built in so if we want 7 added onto 2 then multiplied by 4 we must use brackets. is it 36 or 15?

Brackets in algebra Brackets allow us to treat whatever is inside as a single quantity. i.e. 4a – (2a + 7b) means that 2a + 7b as a whole is subtracted from 4a. 5(5x – 2y) means that 5x – 2y is all multiplied by 5 Consider 3(4 + 1) – 2(2 + 4) 3 x 5 – 2 x 6 = 15 – 12 = 3  12 + 3 – 4 + 8 = 19  12 + 3 – 4 – 8 = 3  Algebraic Methods 1

Rules for the removal of simple brackets Every term inside the bracket must be multiplied by the quantity outside the bracket. If the sign in front of the bracket is positive the signs inside the bracket remain unchanged. (no sign is assumed to be positive. If the sign in front of the bracket is negative the signs inside the bracket are changed. 1. 3(a+b) + 4(5a+b) 2. 3(a-b) + 4(2a+b) 3. 6(a+2b-c) + (a-b+c) 4. 2(3a-4b) – (a+b) + 2(a+b) 5. 3(a-2b+3c) – 2(b+4c) 6. -4(a-3b) – 3(-3a-b) 7. 2p(q+r) – p(3q-2r) 8. 11a(2b+c) – 3a(3b-2c) 9. a(a+b-c) – b(a-b+c) 10. a(b+c) – b(c+a) + c(a+b) Algebraic Methods 1

Solution to simple equations If we have an equation with one unknown quantity then it is possible to determine the unknown quantity. e.g. 9 = 3 + 2x x is the unknown quantity. To determine its value we need to rearrange the equation so that we have x = Process: 9 = 3 + 2x subtract 3 from each side. 6 = 2x divide each sides by 2 3 = x answer x = 3 e.g. 4x -3 = x + 18 Process 4x -3 = x + 18 add 3 to each side 4x = x + 21 subtract x from each side 3x = 21 divide each side by 3 x = 7 answer x = 7

Solution to simple equations Always aim to make the unknown a positive quantity: e.g. 5 – 2x = x +7 Process 5 – 2x = x +7 add 2x to each side 5 = 3x + 7 subtract 7 from each side -2 = 3x divide each side by 3 -2/3 = x Answer x = -2/3 Examples 1. 4a = a + 9 2. 4x – 3 = 2x +3 3. a – 3 = 2a – 14 4. 7x + 1 = 1 + 6x 5. 7(a – 5) = 3(4 – a) Algebraic Methods 1

Solution to equations involving fractions Useful steps – any compound numerator or denominator should be placed in brackets and any number not written as a fraction should be made into a fraction. e.g. We now multiply through by the lowest common denominator. (5 x 2 x 1 = 10) 2 5 6x + 4 - 5x – 20 = 40 combine like terms x - 16 = 40 add 16 to each side x = 56 Answer Algebraic Methods 1

Solution to equations involving fractions Lowest Common Denominator (LCD) is the lowest number that a series of numbers can be divided by and still give a whole number. e.g. the numbers 2, 3, 4, 6 when multiplied gives 144 144/2 = 72 144/3 = 48 144/4 = 36 144/6 = 24 But there is a lower number which also is fine and this is 12 12/2 = 6 12/3 = 4 12/4 = 3 12/6 = 2 To determine this lowest value we write down the factors of each number. 2 = 2 3 = 3 4 = 2 x 2 6 = 3 x 2 Leaving us 2 x 2 x 3 = 12 Algebraic Methods 1

Solution to equations involving fractions e.g. LCD = 12 2 4 3 6 12 – 2x - 10 = 8x +28 - 3x + 30 combine like terms 2 – 2x = 5x + 58 add 2x to each side 2 = 7x + 58 subtract 58 from each side -56 = 7x divide each side by 7 -8 = x Answer x = -8 Algebraic Methods 1

Solution to equations involving fractions If we have a single fraction equal to a another single fraction then we can cross multiply: 5 x 5a = 3 x 2 25a = 6 Examples:

Transposition of Formulae In the formula: P = I 2 R P is said to be the subject of the formula. We may know I and P and need to work out R – to do this we must make R the subject of the formula. The process of altering the formula is called transposition. The best way of seeing who this is achieved is by looking at examples. e.g. e.g. H = I 2 RT (make R the subject) Algebraic Methods 1

Transposition of Formulae If we have quotients then we do the following: e.g. e.g. Algebraic Methods 1

Transposition of Formulae If we have plus or minus signs then we do the following: e.g. T = t + 273 (make t the subject) subtract 273 from both sides T – 273 = t t = T - 273 e.g. Algebraic Methods 1

Transposition of Formulae If we have brackets then we do the following: e.g. e.g.

Transposition of Formulae e.g. e.g.

Examples

Transposition of Formulae If we have roots and/or powers then we do the following: e.g. e.g.

Examples Algebraic Methods 1

This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme.
© 2009 University of Wales Newport
This work is licensed under a Creative Commons Attribution 2.0 License .
The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence.
The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher.
The name and logo of University of Wales Newport is a trade mark and all rights in it are reserved. The name and logo should not be reproduced without the express authorisation of the University.
Algebraic Methods 1

http://www.slideshare.net	13
file://	1
http://moodle.bcu.ac.uk	1

Algebraic Methods 1

by School of Design Engineering Fashion & Technology (DEFT), University of Wales, Newport on Jan 21, 2010

Accessibility

Categories

Upload Details

Usage Rights

3 Embeds 15

Statistics

Algebraic Methods 1 Presentation Transcript