Summary Unit 1 engineering principles Revision Guide 2
30 views 0 purchase
Course
Unit 1 - Engineering Principles
Institution
PEARSON (PEARSON)
Unit 1 engineering principles Revision Guide 2 includes clear worked examples of engineering principles questions from Pass, Merit, and Distinction Criteria. This document will most likely assist you to achieve your best in the exam and in the weekly assignments.
•
REVISE BTEC NATIONAL I r c 10
• _Which __units___should __ you _ revise_?_
eer1 g This Revision Guide has been designed to support you in preparing for thc;: externally assessed units
of your course. Remember that you won't necessarily he studying ahhie-- unit5 included here - ifwffl __ _
__ t;;le:pe:n_c::I g111:;'1e: q1J_a.lifi9~rtic>n y9u a._re __ ta~i_rig.
BTEC National Qualification Externally assessed units
EVISIO
For each of: 1 ~inee53d ng Principles
Extended Certificate 3 Engineering Product Design and Manufacture
foundation Diploma .
Diploma
Extended Diploma 1 Engineering ___Prfridple5 _____ _
GUIDE Your Revision Guide
_3__fngineering--ProductDesiqnand--Manufacture_
6 Microcontroller Systems for Engineers
Each unit in this Revision Guide contains two types of pages, shown below.
pages help you revise the pages help you prepare for
essential content you need your exam or assessed task.
to know for each unit. Skills pages have a coloured edge and
-are shaded in the table of contents.
Series Consultant: Harry Smith Ba4 a look 0 Nearly UlHa 0 Jlailod :It! Q
DC motors
Authors: Andrew Buckenham, Kevin Medcalf, David Midgley
Bad• look Q H.arly tlwr. 0 Nailed. itl D
and Neil Wooliscroft
------
.I. note from the publisher
----------- ------------
While the publi5her5 have made every attempt to en5ure that advice on the qualification
and it5 a5sessment is accurate, the official specification and associated assessment
guidance materials are the only authoritative source of information and should always
be referred to for definitive guidance.
Thi5 qualification is reviewed on a regular ba5is and may be updated in the f_uture.
Any such updates that affect the content of this Revision Guide will be outlined at
www.pearsonfe.eo.uk/BTECchanges. The eBook ver5ion of this Revision Guide will also
be updated to reflect the late5t guidance a5 soon as possible.
For the lull range ol Pearson revision titles across KSZ,
KS3, GCSE, Functional Skills, .I.SI.I. Level and BTEC visit:
---Usethe Now--tly-this -Glctivities .on--evecy
www.pearsonschools.eo.uk/revise page to help you test your knowledge
and practise the relevant skills. -look· out--for--the sample-response -extracts --
to exam questions or set tasks on the skills
pages. fbst::its -wm-explairrtneir strer-rgtns-and __ _
weaknesses.
@ Pearson
, e 127 Product design specification (FDS) 2
w (;9mmerc;ial protection
129 Legislation and standards
193 PICAXE'" and Logicator: program files and error checking
194 FICAXE"' and Logicator: simulation, compiling and
debugging
130 Environmental and safety constraints 195 Microchip, FICkit 3 and MFLAB"': program files and error
131 SecJ rfty constraints checking
Unit!: Engineering Principles GG Reluctance and magnetic screening
132 Mar<eting
133 Form and functionality
19G Microchip, FICkit 4 and MFLAB"': simulation and
debugging
1 Laws of indices G7 Electromagnetic induction 134 Product performance 197 GENIE'": program files
2 Logarithms GB DCmotors 135 Manufacturing processes a1d requirements 198 GENIE'": simulation
3 Exponential function G9 Electric generators 13G Manufacturing needs 199 GENIE"': compiling and debugging
-4 Equations of lines 70 Inductors and self-Inductance 137 Generating design ideas 200 Arduino™ Uno: program files and error checking
5 Simultaneous linear equations 71 Transformers and mutual inductance 138 Development - - - - - - .. - -- 201 Arduino™ Uno: simulation and testing
G Expandirig and fac:;t;oris;>t;ion 72 AC waveforms 139 Des~gn information 202 Flowcode and E-Block: creating and managing program
7 Quadratic equations 1 73 Sinc:Jle phase AC parameters 140 Freehand sketching, diagrams, technical drawings fi les- --- -
8 Quadratic equatlorts 2 74 Analysing AC voltages using phasors 141 Graphical techniques 203 Flowcode and E-Block: simulation
~ R.idlans. arcs and sectors 75 Reactance aiidimpedance 142 Written communication 204 Flowcode and E- Block: simulation and debugging
10 Trigonometric ratios and graphs 7G Rectification 143 Des,gn documentation 205 Coding practice and effident code authorinc:J
11 Cosine rule 77 Your Unit 1 exam 144 Iterative development 20G Coding constructs: inputs and outputs (BASIC a.nd C) 1
12 Sine rule 78 Showing your working 145 Statistical data 1 207 Coding constructs: inputs and outputs (BASIC and CJ 2
13 Vector addition 79 'State' and 'describe' questions 14G Statistical data 2 208 Coding constructs: inputs and outputs (GENIE"' flowchart) 1
14 Surface area and volume 80 'Explain' questions 147 Data handlinc:J and graphs 1 209 Coding constructs: inputs and outputs (GENIE'" flowchart) 2
15 Syst~ms of f9rces 81 'Find' questions 148 Data handling and graphs 2 210 Coding constructs: logic and arithmetic variables and
1G Resolving forces 82 'Calculate' questions 1 149 Frequency distributions arrays
17 Moments and equilibrium 83 'Calculate' questions 2 150 Validating the design 211 Coding constructs: logic and arithmetic
!3 Slmply supported bearrts 84 'Solve' questions 15 1 Benefits and opportunities 212 Coding constructs: program flow and control 1
19 Direct loading 85 'Draw' questions 152 Furber modifications 213 Coding constructs: program flow and control 2
2{) Shear loading BG 'Label' questions 153 Your Unit 3 set task 214 Program flow: iteration
21 Velocity, displacement and acceleration 87 'Identify' questions 154 Reading a brief 215 Control of program sequence: if else
22 Applyinc:J the SUVAT equ;itions 88 Synoptic questions 1 155 Conducting research 21G Control of prOc:Jram sequence: switch 1
23 Force, friction and torque 89 Synoptic questions 2 15G Making notes 217 Control of program sequence: switch 2
24 Work and power 90 Using the formulae booklet 157 Reading further information 218 Structured program design
25 Energy 91 Formulae and constants 158 Interpreting an engineering drawing 219 Number systems: decimal to binary
2G Newton's laws of motion, momentum and enerc:Jy 159 Analysing data 220 Number systems: bina ry to decimal
27 Angular parameter5 1GO Creating a time plan 221 Project analysis
28 Mechanical power transmission
Unit 3: Engineering Product 1G1 Recording changes and action points 222 System design and program planninc:J
29 Submerged surfaces
30 Immersed bodies
Design and·· Manufacture 1G2
1G3
Interpreting a brief
Interpreting numerical data
223
224
System assembly, coding and testing
System testing and operation
31 fluid flow iri taperinq pjpes. 9G .. PesigntriggersJ 1G4 Producing design ideas 1 225 Production of evidence
32 Heat transfer parameters and thermal conductivity 97 Design triggers 2 1G5 Producing design ideas 2 22G Your Unit G set task
33 Heat transfer processes 98 Reducing energy 1GG Modified product proposal 227 Reading a brief
34 Linear expansivity and phases of matter 99 Hybrids and energy recovery systems 1G7 Justifying a design 228 Creating a task plan
35 Specific heat capacity, latent and sensible heat 100 Sustainability and cost over product life cycle 1G8 Justifying materials and processes 229 Monitoring progress
3G Heat pllmp performance ratios 101 t1igh~value manufacturing and designing out risk 1G9 Developing sustainability 230 Monitoring changes 1
37 Enthalpy and entropy 102 Systems, equipment and interfaces 170 Validating a design proposal 231 Monitoring changes 2
3S· Thermal efficiency of.heat erigines··· ..103 .Systemcompromk;es 171 Evaluating with tools and techniques 232 Analysing a brief for product requirements
39 Thermodynamic process parameters 104 Equipment specifications and cost effectiveness 233 Completing a test plan
40 Gas laws 105 .Me:<::h;;inicalpr:()p~r:tie:s 234 Formulating a system design
41 Current flow 10G Physical and thermal properties Unit 6: Microcontroller Syste ms 235 Proposing system connections
42 Coulomb's law and electrostatic force
43 Resistance, condl1Ctarlce arid temperature
107 Electrical and magnetic properties
f08 Advanced materials
for Engineers 23G
237
Planning a program structure
System assembly and programming
44 Types of resistor 109 Surface treatments and coatings 172 Microcontrollers and Unit G 238 System testing
45 Field strength 110 Lubrication 173 Comparing different microcontrollers 239 Results analysis
4G Capacitance 111 Modes of failure of materials 174 Project boards 240 Recording a system in operation
47 Cap.acitors - non-polarised. . 11.2 Mechanical motion 175 Using flowcharts 241 Recording a commentary
48 Capacitors - polarised 113 Mechanical linkages 17G BA::IC as a programming language
49 Ohm's law, powerand effici~ncy 1 114 Fower sources 177 Casa programming lanc:Juage 242 ANSWERS
50 Ohm's law, power and efficiency 2 115 Controlling power transmission 178 Input and output devices
51 Kirchoff's voltage and current laws 11 G Processing metals 179 Switches
52 Capaeitors - charging and enerc:JY 117 Powder metallurgy and additive manufacturing 180- Visible and infrared light-sensing devices A small bit of small print
53 Capacitors - networks 118 Joining and assembly 181 Temperature and humidity sensors Pearson publishes Sample Assessment Material and the
54 Capac\tors fn··circuits - the time constant 119 Processing pofymers 182 Input interfacing requirements Specification on its website. This is the official content
55 Capacitors in circuits - RC transients 120 Processing ceramics 183 Ultrasonic and control potentiometers and this book should be used in conjunction with it.
The questions in Now try this have been written to help
SG Diodes. .,,. b.iLJs and;ippllcations 121 f'rocessingflbrecreinforced ..comp.osites 184 Optoelectronic output devices
you test your knowledge and skills. Remember: the real
57 DC power sources 122 Effects of processing 185 Electromechanical output devices assessment may not look like this.
58 Resistors in series or parallel 123 Scales of production 18G Audio output devices
59 Resistors in series and parallel combinations 124 Customers 187 Transistor output stages
GO Resistors and diodes in series 125 Product and service requirements 188 Selecting hardware: input devices
Gi Capaeito61n 5erie5 or para11e1 ·· 12G f'roctuct de5igrrspeciflcatton (PDS) 1· 189 Selecting hardware: output devices
G2 Capacitors in series and parallel combinations 190 Generating a system desig1
G3 Magnetism and magnetic fields 191 Safe use of typical electronic tools
G4 Permeability 192 Assembling and operating a microcontroller system
GS B/H curves, loops and hysteresis
iii iv
, Bad a look 0
Laws o
Nearly there
•
1n
0 Nailed it!
•
ces
0
-- ·-- Bad a look 0
Logarithms
Nearly there 0 Nailed it! D
The laws of indices make it possible__to simplify_and solve equations that contain i ndices~ . Logarithms (or logs) are a way of wdting ..facts.about .powers.
_ °Q$i.~g __tl\e __laws of indices Logs
Numbers 1 to 3 below are the main laws of indices. Numbers 4 to 8 are useful applications and special
Remembering the order···
These two statements mean the same thing:
cases of the main laws. .. The key .t.o bei11gco11fi.d .e .11tinloq .q.ue.stio11s is
logab =xH ii' = b remembering the basic definition. Start at the
Multiplication a"' x a" = a Cm + zi) Index =0 a0 =1 -ba5e, and work. in a drcle~ -
/ Y ou say, ~013 to the base a of b equals x.'
·----~-:'3· ---- -~ --~--x::2 = x4 -···· - - t::·-:3· -- ·-- ... ?~- -===_ 1 _
= a0 -5 = ra
1
Division am + a" = a<m- n> In d ex = z-or 0 . 5 at logab =x 11111 • ii' = b
e.g. x5 + x3 = x2 e.g. 16-t = 16°· 5 = fl6 = 4 ~a is the base of the logarithm.
Powers (am)n = amn Index = _!_ a~ = n.fi
n
e.g. (x3)5 = x15 27-:§- = 3 /27 = 3
e.g.
1
Reciprocals -;;;;; = a-m
Index =1 a1 =a
1 = ··- ~5
e.g. -;f - x e.g. 3.5 1 = 3.5 Laws of logarithms
= -- Learn these four key laws for-manipulating· expressions invohting logs; These laws · work for all logarithms
with the same base.
~l~x-keil e~a:.~~p~e
Start at the beginning of the expression
and look for terms that can be expressed
as decimals to make simpl.1fication eas·1er.
(a) Simplify the expression a-2 b- 1 a~.
log,, X
log 4 8
+ loga y =
+ log4 2 =
loga(xy)
log 4 16 = 2
log. m = -loga x
e.g. a~ = a-1 s.
a- 2 b- 1at = a- 2 b-1a1· 5 .,..~
= a-2a1.sb-1 log a X - lo'da y= IO'da (f)1
log8 ( ; ) . ::: -,,-loga 2 ;;;
10<::19 18- - log9 6 = log9 3 = 2 lo'da(.x") ::: nl0'2ja x
= a-O.Sb-1 10<::1 5 (25 3 ) = 3 lo'ds 25 = 3 X 2 = 6
Add the indices for like terms.
1
= b/a -
Consider alternative ways to
a-1b- 1bt Wor·~a :G ~rna 1:..p~e Changing the base
express the simplified expression. (b) Simplify the expression~. You can change the base of a logarithm usin'd
= a-1b-1bo.sa2b-1 Find: this formula:
Make all the indices decimals_ (a) the positive value of x such that logx49 = 2. 10<::11>-x - · ··· 1°':b 27 3
loga x = log!> a - 10139 27 = logs 9 = 2
x2 = 49
Group together like terms.
x = 7
(b) the value of y such that log5 y = -2. Common logs and natural logs
Add the indices for like terms.
5-2 =y {0 Lo<::1arithms to base 10 are called common
1 Joq.s and the .notation . lq n is sometimes used
y= 25
Consider alternative ways to instead of log10 n; e.g . log10 100 = 2 (note
express the simplfied expression. 100 = 10-?). --
(0 Logarithms to base e are called natural
logs with the notation foqen.
Information Booklet of
Formulae and Constants
Evaluate
In your Unit 1 exam, you will be given Write down the corresponding power fact.
(a) 19° Work in a circle, starting at the base.
an Information Booklet of formulae and
Constants and this includes the multiplication, 2 Express J(x'1 x .0) as a power of x Use law 4 to
-.division.and--powers.-laws o f indices. Ideally, 3 Evaluate 3 ./9 x ./9
6
Find: 2 Express as a single logarithm to base a.
write 3 log. 4
you should be confident in their use without as log. (4 3 ), then
(a) the value of y such that log 3 y = -1 (a) 2 logaS use law 1 to
reference to them.
(b) the value of p such that logP 8 = 3 (b) loga 2 + loga 9 combine the two
The booklet is included in this Revision Guide
(c) the value of log 4 8. (c) 3 loga 4 - loga 8 logarithms.
on pages 91 to 95.
1 2
, ••
- · ... -
Bad a look 0 Nearly there 0 Nailed it! 0 ' , "
Bad a look 0 Nearly there 0 Nailed it! 0
,
Exponen ial function E uations o •
nes
__ An exponentiaLfonction (a'? is one where the .variab.Jeis the power, not the__base. !he Eulerc~nstant.Jorm
.The equation .of a straight l ine_.c an be.written _in the .form .y = mx + .c, where m. is the_gradient of the
of this expression (e") is found in many engineering disciplines such as aerodynamics, mechanics and
line, and c is the point where it crosses the y-axis.
----electrical principles.
= -- - Point and gradient ~- - -
Exponential growth Wc:rke<t exa:.?.-:p:e
If a population doubles or trebles every year, then Wo-;ke~ examp!e ---If you-are given the gradient m--ot-a straiqht -
line that passes through a point (x1, y 1),
.. itis -5aicfto -pe growfr1g 'exponentialT/_:_
- --tfien you cifrf write its equation ·as: The line L passes through the points (1, 1) and (2, 4).
_____ We representapopulatfonytbat startsat_jQOO - Predict the number of people (y) after 4 years for a
Y -:-Y1 :::: m(>< -:- x1l f;p o~f;ai11 ..an e~p_ressipn c:>f _ Find an equation for L in the form y = mx +c.
and which doubles every year as y = 1000 x (2"), population of 500 that is trebling every year.
y in terms of x. 4-1
-where x represents -the--nlJfllber--of years. y = 500 x (3 4 } = 500 x 81 = 40500 Gradient (m) = 2_1 = 3
--
y=mx+c
~ilorke<.. e;~a:.:r~p:e
Jtt;i~g_ ~ calculator y = 3x +c
.,
0'm
You may need to use the laws
of indices and lo'3arithms when
Most calculators have the xY button; for
example, to find 34, press:
A straight line passes through the point (-3, 2) and
has a gradient -2. Find an equation for this line in the
1 = 3(1} + c (from point (1, 1}}
c = -2 so y = 3x - 2
solvin'3 problems based on exponential '3rowth form ax +by+ c = 0, where a, band care integers.
or decay, see pa'3es 1 and 2.
GJ0GJ8 y - y1 = m(x - x1)
Check using point (2, 4):
4 = 3(2)- 2
and the calculator should show--81-; - -- ------1 y - 2 = -2(x -(-3}}
y - 2 = -2x - 6
Once you"ve evaluated rhe \alue of c. you can
Exponential function e" y+2x+4=0 5Ubstit Jte this ;., the general equation of the
t~~
Graph of y = 2x, Y = 3x
:;tralqht lrne y = mx + c
The Euler notation is a special form of and y = e
If you are given two points on a line, (x,, y 1}
-~----.-
--1-+L-
~--,,_-2_x_~
__ the exponentiaLf.unction~ written e'. The for x = 0, 2x , 3x and
exponential function e' more than doubles, and (x2 , y 2 ), you can calculate the gradient
e are all equal to 1. For usinia:
--- ----but does-not -quite--treble-over--a--period; The
e the gradient at any Y2-Y1
graph opposite shows y = 2X, 3x and e'. -1 0 1 x
m= - --
point on the curve x2-x1
--• A-decaying popuiafiOri, sucli as radiated y = e is equal to e.
y
heat - ~n~~'.3Y or radioactive particles, has
the form y = e-x. -
Intercepts
¥l~:rke{. exa:.i.~f(~e
• Recognise where the laws of indices can You can find-where the line y = mx + c
be used: y = e-x m~ also be written Graph of y = e-x intercepts both the x- and the y-axes.
1 Determine the intercepts for the x- and y-axes of the line
The y- intercept is given by the value of c, 3y-x=6.
y= ef'_
and th~ x-Jnter~pt cc:io. l:?_
e ~v.C1!J..JC1te_(:;Uzy
-1 0
setting the value of y to 0. Rearranging the expression in the form
~-- -
y = mx i
+ c gives y = x + 2, therefore the
w~rke{. ejt.a::-~;p~e
Usiftg a calculator 4yj -
y-axis intercept is +2 (when x
Setting y = 0:
0}. =
~~ .
In a production process involving heat transfer, Most calculators have the exponential button e'
ix+2=0
the temperature (J °C of a mould, at time t minutes, button; it may be a secondary function of the
is given by (J = 250 + 1soe-0.rn. Determine the 'In-' button. ·rorexampte; ·to· finct-·e2· 5 ~press: -- - x+6=0
temperature of the mould after 5 minutes. -G -5 -4 -3 -2 -1 0
~GJ~8
2 3 4 x
x = -6 Therefore, the x- intercept is -6.:
9 = 250 + 1SOe-{0 ·15 )( 5 > = 250 + 150e-{0 ·75> Graph of y = :ix + 2
You car sketch a "!raph co chec your answer:
= 250 + 70.8549 and the calculator shoulc:t-show ·t2:182-. ; ..
6 = 320°C (to 2 s.f.} ------------ ...
The qrad;enc 15 oosit1ve because tl1e m term (l)
15 oc:, tive and the 'ine passes throui::ili the
x-axis at -G and the -'-axis at + 2.
3
Now try this
A manufacturer quadruples its production of a component from 2000 units per year every year for three years. The line L passes through the point (6, -5) and has gradient _ _!__ Find an equation for Lin the form ax +by+ c = O,
where a, band care integers. 3
Calculate the number of components produced at the end of the third ye'lr.
2 The line L passes through (-4, 2) and (8, 11 ). Find an equation for Lin the form y = mx + c, where m and care
2 The voltage Wcl across a capacitor in a RC circuit is given by Ve= V.(1 - e--7), wher~ (-r) is the time constant and Vs is constants.
the supply voltage. Determine the value of Ve at t = 5-rwhen the supply voltage is 4.SV.
3 4
, ..
-
,_., :CJ. \:1
Bad a look 0 Nearly there 0 Nailed it! 0 ' {Jt:h '.\, ~
Bad a look 0 Nearly there 0 Nailed it! 0
- rr,_,,~~~ ~ ~'Z~~
•
Si taneous 1·near e uati ns Ex pan 1ng an actorisation
Linear equations have the form y = mx + c (i.e. no x2 or y2). Simultaneous equations can be solved using ___ ..Expanding brack.ets._and_factorisinq _expres_sions are tbe basis for manipulation of_engineering.. expressions. _
either the substitution or the elimination method. Whichever method you use, remember to number the and formulae and are therefore essential skills.
equations to keep track of your working. Expanding brackets
y -2x = 2
Substitution Yt -- ---To -expand th-e product of two factors--you--h-ave to-multiply-EVERY-TERM in the -first ·factor by ·EVERY---
TERM in the second factor:
Solve the simultaneous equations:
6 I There are 2 terms in the first factor and 3 terms in the
~w-~d fa-Cto;,·so there will be -2·-x 3 - ~ - 6 te~; -i~ the
y - 2x = 2 Call this equation (1). R.earrange the linear v (0.2,2.4) expanded expressio n BEFORE you collect like terms.
4
-2y + 5 = x Call this equation ( 2 ) . / equation (1) to make (2x + + 4 ) = 10x3 - 2x2 + 8x + 15x2- 3x +
3 )(5x2 - x 12
y = 2x + 2 Call this equation (3). ythe subject.
2 .
~ = 10x3-+13.x2-+5x+--12 --- - - · - - - --- ----- ·
~~~~~.---+~--,--~,1 =-.::::-~1
1~~
- 2(2x + 2) + 5 = x -------- Substitute equatio_n (~) into -6 -4 -2.· 0 2 4 8 x
--------- Simplify your expression by collecting like terms:
_ x _ + = x-------- equation (2) and simplify to find x. .:::2xz + 1sX2 = ·13xz
4 4 5 /-2 x = - 2y + 5
5x = 1 / You have found x. Substitute
I · Factorising· -Assuming-- no resistance·; you can
= k = 0. 2 x = 0.2 into equation (1) to find Y·
-4
calculate the distance travelled by an
I
X
for 5imple expressio ns you__ can_extract .t he
The solutions are x = 0.2, y = 2.4.
-6
Highest Common Factor (HCF).
= +
object; using s -ut ~at2-:
=~---
-some ·expressions· may provide ·a common
Thinking graphically Graph showing simultaneous factor within the bracketed terms. Worke-l e:xa:.:"'r~p~e
(,0 The solutions to a pair of tinear ·simultaneous equations linear equations
(3x-2) is common to
correspond to the point where the graphs of the equations intersect. y - 2x = 2 and -2y + 5 = x.
~t;;s so ca n be (a) Factorise the RHS of this equation.
(0 The point of intersection has an x value and a y value. ~ ~racted as a factor.
s = t(u+iat)
= - - --
-- - -- ------ 2x(3x - 2) - 7(3x - 2) = (3x - 2)(2x - 7)
Take out the common factor in each
You can substitute ~example term. Don't forget to check your
Worked example for x or y. It is easier answer by multiplying back out. You
to substitute for y should get the original expression.
Solve the simultaneous equations
Solve the simultaneous equations: because there will be
no fractions. 6x+6=5y (1)
y-3x= 8 (1)
3y+ 2x= 7 (2) (b) Factorise 3a(x+4a) + 2(x+4a).
2y + 11 = -9x (2) Remember to number In some cases it may be necessary to group
your equations. Multiply equation (2) by -3 and rearrange the terms to identify a common factor. 3a(x+ 4a) + (x + 4a) = (x+ 4a)(3a + 2)
From (1): y = 3x + 8 (3)
to make the x terms the same.
Substitute (3) into (2) and simplify to find x: Start by group·ing the first two and last two terms. (x + 4a) 1s common to both terms so can
-6x + 21 = 9y (3)
be take n <:>s one of the facto rs. (3a + 2 are
2(3x + 8) + 11 = -9x Then extract the HCF as in part (a) above for each
Add equation (1) to equation (3) to group, which leaves a common factor: (2a- 3). the c:errns o utside the brackets gathered
6x + 16 + 11 = -9x eliminate the term 6x and -6x. togethe r and which form the second fact or:
You can then extract the common factor as in 2 above.
15x = -27, x = -¥s = -1.8 27 = 14y, y = 1.93; now substitute into
(2) to obtain x: (c) Factorise 6a2 - 9a - 4ab + 6b.
Substitute x = 1.8 into equation (1) to find y:
y - 3(-1.8) = 8, y = 8 - 5.4 = 2.6 (3 x 1.93) + 2x = 7, x = 0.61 Special cases (6a 2 - 9a) - (4ab - 6b)
There are some special cases that you should watch out for: 3a(2a - 3 ) - 2b(2a - 3)
Remember that the value of x and y
-~ Completi·ng· the sqoare--(5ee pa;:ae 7 --kw ho w thi s works.)
represent the coordinates of the point
You can check your solution by substituting = (2a - 3)(3a - 2b)
where the simultaneous equations intersect. x = O.G1 into equation (1): x2 + 2bx + c = (x+ b) 2 - b2 + c
(G x O.G1) + G = Sy
~ The difference of two squares
- (a +b)(a- b) = -a2 - b2-
Elimination y = 1.93
Manipulate one of the equations to make eithe~ the x
Now try this
------------- - - - --- -
or the y terms exactly the same in both equations.
-
Hint: to factorise \!j2-V 2 , use the
Expand the brackets to show that (3x - 4)(3x + 4) = 3x2 - 16.
Now try this difference of two squares.
2 Factorise 3x(y- 4) - 2(y- 4).
3
V(~1 - ~ ).
2
Solve the following simultaneous equations: 3 The thrust to speed efficiency of a jet engine is given by the expression
~ -V
2x+l3=-3.5y 2 14=3y+Sx
-3x= -9y 10x=4y+ 7 Factorise and, hence, show t hat the complete expression is equal to ~
~+V
5 6
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller JuliusCesar1. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $20.77. You're not tied to anything after your purchase.
