Algebraic Methods 1

Mathematics 1Level 4

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

Algebraic Methods 1

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 1st 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, 8th Editon, Palgrave 2008.•http://www.mathcentre.ac.uk/•Derive 6

Mathematical SyntaxThe 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 25

And a x a x a x a x a x a is a6

We must also be aware that 2a is not the same as a2

One is a + a the other is a x a

Algebraic Methods 1

IndicesWe can now look at what happens when we combine expressions which have indices:

e.g. a4 x a3 = (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 = a7

When multiplying powers of the same letter we add the indices an x am = an+m

e.g. 34

7

aaaaaaaa

aaaaaaaaa

When dividing powers of the same letter we subtract the indices an am = an-m

Algebraic Methods 1

Examples1. a6 x a2

2. a2 x a2 x a

3. a5 x a x a3

4. a3 a7 what does this mean?

5. a9 a3

6. a10 a5

7. ab4 x a2b2

8. a2b3 x a2b2 ab

Note a = a1 and only like letters can be combined.

Algebraic Methods 1

Common errorWhat does 2a2 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 (a2b)2 simplify these

expressionsAlgebraic Methods 1

Powers Roots and ReciprocalsWhat does (a2)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 = a6

When raising a power of a term to a new power we multiply the indices (an)m = anxm

What does √a8 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 √(an) = anmm

Algebraic Methods 1

Powers Roots and ReciprocalsWhat does a a give us?

a1 a1 = a1-1 = a0 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/a2 give us?

1 a2 = a0 a2 = a0-2 = a-2

When reciprocating a power of a term we change the sign of the index 1/an = a-n

Algebraic Methods 1

Addition and SubtractionConsider 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 7a2

Same with subtraction:

10 apples – 4 apples = 6 applesNote 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

NotesPositive numbers are to the right of zeroNegative 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 signThe only sign that can be omitted between two terms is a multiplication sign.Only similar terms can be added or subtracted

2a2b + 3ab2 cannot be performed.

Subtraction Addition

Simplify

2

2

2

1032

56

.6

1032

435

.5

232

.4

103

56

.3

64

2.2

931

3.1

rq

qq

p

cba

cb

ba

ca

ba

ba

abacb

ac

ba

ab

Algebraic Methods 1

B O D M A SThis indicate the order in which mathematical operations must be carried out in order to generate the correct answer:BracketsOrder (power)DivisionMultiplicationAdditionSubtractionWhat 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 algebraBrackets 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 23 = x answer x = 3

e.g. 4x -3 = x + 18Process 4x -3 = x + 18 add 3 to each side

4x = x + 21 subtract x from each side

3x = 21 divide each side by 3x = 7 answer x = 7

Solution to simple equations

Always aim to make the unknown a positive quantity:e.g. 5 – 2x = x +7Process 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

Examples1. 4a = a + 9 2. 4x – 3 = 2x +33. a – 3 = 2a – 14 4. 7x + 1 = 1 + 6x5. 7(a – 5) = 3(4 – a)

Algebraic Methods 1

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.

Solution to equations involving fractions

14

24

523

xx

42

45

23

xxe.g.

We now multiply through by the lowest common denominator. (5 x 2 x 1 = 10)

1410

2410

52310

xx2 5

6x + 4 - 5x – 20 = 40 combine like termsx - 16 = 40 add 16 to each sidex = 56 Answer

Algebraic Methods 1

Solution to equations involving fractionsLowest 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 = 24But there is a lower number which also is fine and this

is 1212/2 = 6 12/3 = 4 12/4 = 3 12/6 = 2To 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

25

4372

)5(61

1

xx

x

25

43)72(

6)5(

11

xxx

e.g.

LCD = 12

2512

412

3)72(12

6)5(12

1112

xxx

2 4 3 6

12 – 2x - 10 = 8x +28 - 3x + 30 combine like terms2 – 2x = 5x + 58 add 2x to each side2 = 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

53

25

a 24.0

256

a

34

57

21

.5

934

23

.4

52

213

512

.3

32

1)1(21

)1(31

.2

41

85

3.1

aaaxx

xx

yyy

pp

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:

214

34.10

92

31

.9

41

312

.8

24

51

.7

21

43

.6

aax

a

xx

x

Transposition of FormulaeIn the formula: P = I2R 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.

IW

Vso

VIVI

makeIVW

IW

I by sides bothdivide

subject)theV (

e.g. H = I2RT (make R the subject)Algebraic Methods 1

Rl

aRl

RRa

Rbysidesbothdivide

lala

Raabysidesbothmultiply

subjecttheamakeal

R

RIVso

VIVI

RIIbysidebothmultiply

subjecttheVmakeIV

R

)(

)(

Transposition of FormulaeIf we have quotients then we do the following:

e.g.

e.g.

Algebraic Methods 1

Transposition of FormulaeIf 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.

REV

I

IREV

RbysidebothDivide

RIEVsidesbothfromESubtract

subjecttheImakeRIEV

)(

Algebraic Methods 1

Transposition of FormulaeIf we have brackets then we do the following:

e.g.

e.g.

0

0

0

00

000

00

0

by sidesBothDivide

sidesbothfrom Subtract

bracketstheRemove

subject)the(make)(

)(

)()()(by sidesbothMultiply

subject)the(make

R

aRRt

tR

aRRR

tRaRRaR

tRaRR

ttaRR

xRIVxR

xRVxRIxR

VxR

VI

Transposition of Formulaee.g.

e.g.

12

21

21

12

21

1

21

2

21

21

111

)(111

1)1(

)1(

)1()1(

)(1

RRRR

RsoRRRR

RRR

RRR

RRR

subjecttheRmakeRRR

xx

nxbysidesbothDivide

xxnfactorcommonaasoutnTake

xnxnsidesbothfromnSubtract

xnxnsidesbothtoxAdd

nxxnbrackettheoutMultiply

nnxnbysidesbothMultiply

subjectthenmakenn

x

Examples

)(2

5.20)(1

.19

)(12

.18)()(

.17

)(2

.16)(3

.15

)(11

1.14)(

)()(2

.13

)(2

.12)()(.11

)()(.10)(.9

)(2

.8)(13115.7

)(.6)(3259

.5

)(.4)(.3

)(.2)(.1

2

121

uuTTu

Txxq

px

xxx

yCCFCS

P

RrRR

Vrh

rhC

RRR

RalnnanS

d

NpnN

CTtTwsH

hhrrSRrR

EC

rrRR

Vaas

lsqlHCCF

rcrpnuatuv

TSTts

TVRT

P

Transposition of FormulaeIf we have roots and/or powers then we do the

following:

e.g.

e.g.

2

2

2

2

2

22

2

2

44

42

22

)(2

)(

gtlorl

gtgbysidesbothMultiply

gltt

sidesbothSquare

glt

bysidesbothDivide

subjectthelmakegl

t

RtJ

IsoIRtJ

sidesbothRootSquare

IRtJ

RtbysidesbothDivide

subjecttheImakeRtIJ

Examples

)(.16)(.15

)(21

.14)(2

1.13

)(2

4.12)(22.11

)(8

.10)(2

.9

)(.8)(2.1.7

)(.6)(21

.5

)(10.4)(21

.3

)(.2)(2.1

2

2

222

22

22

222

2

2

ppfpf

dD

lrrlV

vmvmghECLC

f

RrR

rsrhhrC

rvlpr

nfEVf

P

yyzxLdLD

ccballg

f

ddwvmvE

rrAhghv

Algebraic Methods 1

29

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.

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Algebraic Methods 1