SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics. Basic Algebra

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1 SCHOOL OF ENGINEERING & BUILT ENVIRONMENT Mthemtics Bsic Algebr Opertions nd Epressions Common Mistkes Division of Algebric Epressions Eponentil Functions nd Logrithms Opertions nd their Inverses Mnipulting Formule nd Solving Equtions 7 Qudrtic Equtions A Reminder 8 Simultneous Equtions A Reminder Tutoril Eercises Dr Derek Hodson

2 Contents Pge Opertions nd Epressions Common Mistkes Division of Algebric Epressions Eponentil Functions nd Logrithms 8 Opertions nd their Inverses 0 Mnipulting Formule nd Solving Equtions 7 Qudrtic Equtions A Reminder 8 Simultneous Equtions A Reminder Tutoril Eercises 7 Answers to Tutoril Eercises

3 BsAlg / D Hodson Bsic Algebr Opertions nd Epressions You will lred be fmilir with vrious forms of mthemticl epressions It is importnt tht ou re confident in hndling nd mnipulting such epressions, so this section should refresh our memor regrding some bsic lgebric concepts nd techniques The Bsic Arithmetic Opertions In mthemticl epressions, numbers nd / or vribles (ie unknowns re combined using the rithmetic opertions of ddition, subtrction, multipliction, division nd eponentition (ie squring, cubing, etc, long with brckets to group or seprte terms If n epression contins onl sums nd differences, or contins onl multiplictions nd divisions, then the opertions cn be delt with from left to right For emple, 9 0 ; For more involved epressions, we use the BODMAS principle to determine the order of evlution: Brckets ] first priorit Order (ie powers ] second priorit Division Multipliction Addition Subtrction third priorit fourth priorit For emple: (B (O (M (A (

4 Another emple: (B (O (D (S ( 8 b Indices (Powers The simple use of powers or indices to represent repeted multipliction should be fmilir, for emple n [ n terms], but ou must lso be cler on the wider interprettion of indices The lines below collte the importnt spects of indices: 0 n n nd n n nd m m n m n m n m n m n m ( m ( b m n m b m m m m b b These results will be needed in both the epnsion nd simplifiction of lgebric epressions

5 Emples ( ( Epnd ( 7 ( 7 8 ( 7 ( ( 7 ( (b Epnd ( ( 8 ( 8 ( ( ( ( ( (c Simplif 8 ( z z 8 8 ( z z z z z z See the Tutoril Eercises for prctice

6 Common Mistkes Here re just few common mistkes to void when mnipulting lgebric epressions: : ( 8 ( : ( ( d b c d c b : d c b d c d c b c b : c bc ( b bc ( n m : n m n m n m : Division of Algebric Epressions Providing the denomintor (ie the bottom bit of division contins no s or s, the division itself is usull strightforwrd to crr out Emples 0 0 ( ( (b ( 8 8 When the denomintor does contin s nd / or s, we m require polnomil division Note:,, re emples of polnomils

7 Polnomil Division Emple Suppose we wnt to crr out the following polnomil division: First we set it out like long-division: Divisor Dividend Net, we divide the first term of the dividend b the first term of the divisor nd plce the result bove the line, keeping corresponding powers of verticll ligned: First term of quotient Multipl the divisor b the term of the quotient just found Align the result below the dividend nd subtrct: First difference Repet the process from the first difference to give the net term of the quotient Divide first terms:

8 Multipl divisor nd subtrct: 8 Second difference Repet to complete quotient nd determine the reminder: Reminder

9 7 Removing the nnottion we hve 8 9 In generl, Dividend Divisor Reminder Quotient, Divisor so for the bove emple we cn write 9 Importnt note: Cre must be tken when n of the polnomils hve missing terms These should be included with zero coefficients to ensure correct lignment nd to minimise mistkes For emple, if we hve then this should be written s, 0

10 Eponentil Functions nd Logrithms 8 An eponent is nother nme for power or inde Eponents cn be used to crete eponentil functions:, ( > 0 : 0, In this contet, is clled the bse of the eponentil function Commonl used bses re 0 nd the eponentil constnt e 788 : 0 ; e The function with the eponentil constnt e s its bse is so importnt in mthemtics, science nd engineering tht it is referred to s the eponentil function, ll others being subordinte Relted to eponentil functions re logrithms If we epress number N s power of, ie N, then the power is defined to be the (bse logrithm of N : log N ( log of N to the bse Note tht becuse of the importnce of the eponentil constnt, logrithms to bse e re given the specil nme of nturl logrithms nd denoted b ln ( Emples ( ( 00 0 log 0 (00 (b (c log 0 (000 log ( The definition cn be etended to frctionl powers with logrithms to bse 0 nd bse e obtinble from clcultors: ( ( log 0 ( (to deciml plces (b ln ( 779 (to 7 deciml plces

11 9 Becuse logs re b definition indices, we cn use the rules for combining indices to determine the so-clled lws of logrithms: log ( R S log R log S [L] R log log R log S S [L] n log ( R n log R [L] We cn use these to epnd or contrct epressions involving logrithms: Emples ( ( Epnd log 0 z log 0 z log 0 ( log 0 ( z [L] log 0 ( log 0 ( log 0 ( z [L] log 0 ( log 0 ( log 0 ( z [L] (b Write ln ( ln ( z ln ( s single logrithm ln ( ln ( z ln ( ln ( ln[( z ] ln ( [L] ln[ ( z ] ln ( [L] ( ln z [L] We shll return to eponentils nd logs lter

12 0 Opertions nd their Inverses Ech bsic rithmetic opertion tht we m encounter (with few eceptions hs ssocited with it corresponding inverse opertion An opertion nd its inverse, when pplied in sequence, effectivel cncel out one nd other For emple, suppose we strt off with If we now dd nd then subtrct, we re bck to gin Tht is, We cn s tht the inverse of dding is subtrcting (nd lso vice vers! Similrl, if we multipl b, then divide b, we re gin bck to : So the inverse of multipling b is dividing b (nd vice vers The tble below shows opertions nd their inverses, including the eponentils nd logs from the previous section, nd some others ou m lred be fmilir with: b b This is specil cse of the entr bove Multipling b b the sme s multipling b nd dividing b b Hence the b inverse opertion is multipling b is or ( ( or ( or ( ( or ( n n n n or ( ( n n or n n ( log 0 0 log 0 (0 ln e ln ( e sin (sin sin (sin cos (cos cos(cos tn ( tn tn ( tn

13 Mnipulting Formule nd Solving Equtions The opertion / inverse opertion effect described in the previous section provides the ke to mnipulting formule nd solving equtions A formul is n eqution tht epresses reltionship between quntities In prticulr, it epresses how to determine the vlue of one quntit from the vlues of one or more other quntities Below re some emples of formule tht ou m hve seen before: C ( F 9 V r π d u t t v u t v i R In ech of these formule, the quntit on the left is clled the subject of the formul Often, when working with formul, we wnt to chnge the subject to one of the other quntities For emple: v i R v i - i is now the subject R This is ver simple emple However, the mnipultion of formule is n spect of lgebr tht cn pose difficulties for students Consider the formul in the bove list tht reltes temperture in degrees Celsius to temperture in degrees Fhrenheit: C ( F 9 It is quite es to mke F the subject of this formul How ou would tckle this would probbl depend on methods brought from school If ou re thinking something like, move terms from one side of the eqution to the other to get F on its own, then plese think gin Although the method of moving terms will probbl get ou the correct nswer in this cse, it is not mthemticll correct w of thinking nd cn cuse problems in more complicted formule Let us now look t the correct w to mnipulte formul or eqution This will let ou see wht is rell hppening when terms pper to move round n eqution Wrning: Ignore the following t our peril!

14 First, let us look t the formul s given nd the mthemticl opertions within it: C ( F 9 Given vlue of F, (following BODMAS we would determine the corresponding vlue of C b: subtrcting multipling b 9 A formul (or, in fct, n eqution is like blnced set of scles: C ( F 9 When working with n eqution, we must not upset the blnce This mens tht if we do something to one side of the eqution, then we must do the ect sme thing to the other side This is the one nd onl rule of lgebric mnipultion [Note: We cn, of course, swp the sides round, but tht is just like turning the scles round; it does not ffect the blnce] To chnge the subject of formul, we ppl crefull chosen opertions to both sides of the eqution whose net effect is to isolte the new subject These opertions re the inverse opertions of the those within the originl formul Tht is: subtrct multipl b 9 re reversed nd inverted to give 9 multipl b dd These opertions re now pplied, in turn, to both sides of the eqution

15 The process in full is s follows: C ( F 9 Multipl both sides b 9 : 9 9 C ( F 9 9 C ( F 9 C F Add to both sides: 9 C F 9 C F 9 Swp sides: F C nd with pictures: Formul: C ( F 9 Multipl both sides b 9 : 9 9 C ( F 9 9 C F Add to both sides: C F 9 C F 9 Swp sides: F 9 C

16 In prctice, we needn t put in s much detil For such simple formul we might simpl set down the following lines: C ( F 9 Multipl both sides b 9 : 9 C F Add to both sides: 9 C F 9 Swp sides: F C This bbrevited version m give the impression tht terms re moving round the eqution, but the re not You must lws keep in mind wht is trul hppening in the bckground Alws think BALANCE Further Emples Note: In the following emples, full detils re shown In prctice, the level of detil ou show in ou own working will depend on our own confidence nd bilit; s these grow, ou will nturll displ less ( For the formul mke the subject [Note: This is the sme s sing, solve for ] Anlse opertions Given, how is clculted? squre multipl b dd Reverse nd invert opertions: subtrct divide b squre root

17 Appl inverse opertions to formul: Formul: Subtrct : Divide b : At this point it is useful to swp sides Squre root: Shortened version (for the more confident: Formul: Subtrct : Swp sides nd divide b : Squre root:

18 (7 For the formul V, mke r the subject r π Anlse opertions Given r, how is V clculted? cube (or rise to the power π multipl b Reverse nd invert opertions: multipl b π cube root (or rise to the power Appl inverse opertions to formul: Formul: V r π V π r Multipl b π : π π V π r V π r Swp sides: r V π Rise to the power r V π : ( ( r V ( π Shortened version: Formul: V π r Swp sides nd multipl b π r V : π Rise to the power r V π : (

19 7 (8 For the eqution e, solve for Note: For BODMAS purposes, ou hve to imgine brckets grouping the power ( terms together, ie e Anlse opertions Given, how is clculted? multipl b e to the power ( ie multipl b ( e Reverse nd invert opertions: divide b tke nturl log ( ie ln ( divide b Appl inverse opertions to formul: Eqution: e Divide b : e e Nturl log: ln ln ( e ln Swp sides: ln Divide b : ln ln

20 8 Shortened version: Eqution: e Divide b : e Swp sides nd tke nturl log: ln Divide b : ln Emples (with less detil: (9 Determine t when 0 t 8 0 This is not formul with subject, so we cnnot write down sequence of opertions It is, however, n eqution (with n unknown nd the concept of blnce still pplies In mnipulting this eqution, we wnt to end up with the form t ( Clerl we must mnipulte t down to the level of the equls sign One of the lws of logrithms will help here: Eqution: 0 t 8 0 0t Tke log (bse 0 of both sides: ( 0 log (8 log 0 0 0t log0 (8 Multipl both sides b : 0t log (8 0 t t log0 80 (8

21 9 t (0 Determine t when e 0 t Eqution: e 0 t Tke nturl log of both sides: ln ( ln (0 e t ln (0 Divide both sides b : t ln (0 (sme s multipling b t 07 ( Determine when Eqution: Tke log (bse 0 of both sides: ( log ( log 0 0 log0 ( log0 ( Divide both sides b log ( : 0 log log 0 0 ( ( 0 Note: You could use nturl logs nd get the sme nswer

22 0 Applied Emples ( The current (i in the brnch of n electronic circuit chnges with time (t in line with the formul i i 0 e k t The initil current (ie when t 0 is ma It tkes 7 s for the current to drop to 7 ma (ie hlf its initil vlue (i From the numericl informtion given, determine the vlues of the prmeters nd k i 0 (ii Determine the current when t s (iii Determine t when the current is % of its initil vlue (i When t 0, i (Note: As long s we re consistent, we don t need to convert ma to A Input these vlues into the formul: i 0 e k 0 i0 i 0 i e k t When t 7, i 7 Input these vlues into the updted formul nd solve for k : 7 e 7 k e 7 k 0 ln 7 k ( e ln (0 7 k ln (0 k ln ( Completed formul is: i e t 078 ; we use this to nswer the remining prts

23 (ii Set t in formul nd evlute i : i e ma (iii % of 7 ; set i 7 in formul nd solve for t : 7 e 078 t e 078 t 0 ln 078 t ( e ln ( t ln (0 ln (0 t 90 s 078 ( The dec of the rdioctive element rdium is modelled b the formul A A t 0 0 (0 where A 0 is the initil mount of rdium nd A is the mount remining fter t ers (i (ii How much rdium remins in kg smple fter 000 ers? How long would it tke for kg smple to dec to 00 kg? (i Set A0, t 000 in formul nd evlute: A ( 0 09 kg (ii Set A0, A 0 0 in formul nd solve for t : 00 (0 t 0 t 0 ln (0 ln (00 t 0 ln (0 ln (00 t 0 ln (00 ln (0 07 ers

24 ( In the formul R R R R R mke R the subject Becuse the new subject ppers in more thn one position in the formul, we cnnot deconstruct the formul into sequence of opertions tht revel how it should be mnipulted This mens we hve to tr nd determine the pproprite opertions s we go Ultimtel, we wnt to chieve to form R ( epression with R nd onl R, so we must mnipulte the eqution in such w tht the be wkwrd to hndle, so let us get rid of the frction R 's re combined Frctions cn Multipl both sides of the formul b the denomintor R R : ( R R R R R ( R R R R This simplifies to ( R R R R R Now multipl out the brckets: R R R R R R Subtrct R R from both sides: R R R R R R R R R R R R R R R R R now ppers on one side of the eqution nd t the sme level Swp sides nd tke out common fctor: R ( R R R R Now divide both sides b R R to give the desired result: R R R RR See over for the shortened version with less detil:

25 Formul: R R R R R Mnipultions: ( R R R R R R R R R R R R R R R R R R ( R R R R R R R R R The bove emples illustrte the process of lgebric mnipultion, but ever possible twist nd turn cnnot be covered You must now develop our understnding of the process through prctice nd be prepred to wrestle with problems to chieve result

26 7 Qudrtic Equtions A Reminder One tpe of eqution tht crops up quite frequentl is the qudrtic eqution: b c 0 This tpe of eqution is solved either b fctoristion (which isn t lws possible or b use of the qudrtic formul b ± b c Emples ( ( Solve 8 0 b fctoristion 8 0 ( ( 0 ( ( or 0 0 or (b Solve 8 0 b the qudrtic formul 8 0, b, c 8 b ± b c ± ( 8 8 ± ± or or Note: Qudrtic equtions m hve two rel solutions, one rel solution or no rel solutions, depending on the vlue of the discriminnt b c

27 8 Simultneous Equtions A Reminder Equtions cn contin more thn one unknown For emple, the eqution hs two unknowns A solution of this eqution is mde up of n -vlue nd -vlue tht together stisf the eqution We could hve ( 0, or (, or n one of n infinite number of solutions When plotted on Crtesin es sstem, ll possible solutions of this eqution lie on stright line (see the hnd-out Coordinte Geometr The Bsics If we hve second eqution, s, this lso hs n infinite number of solutions, which lso lie on stright line However, there is one solution tht is common to both equtions Grphicll, this common solution is given b the coordintes of the point of intersection of the two stright-line grphs To find this solution, we could drw the grphs nd red off the vlues of nd This would be fine for the bove equtions which, s it turns out, hve integer vlues in the solution For more generl method of solving equtions simultneousl, we require n lgebric pproch There re vrious ws of setting this out Method Elimintion b Substitution Tke either eqution nd epress one unknown in terms of the other: Substitute this into the other eqution, thereb eliminting one of the unknowns nd leving n eqution with single unknown tht is esil solved: ( 8 This vlue is then substituted into to give Solution:,

28 Method Elimintion b the Addition or Subtrction of Equtions Line up the equtions one bove the other: If necessr, multipl up one or other or both equtions to obtin common coefficients on one of the unknowns: or Net, eliminte the unknown with the common coefficient b dding or subtrcting corresponding sides of the equtions: Add : 8 or Subtrct : Solve : Solve : Solution:, Whichever method ou use, lws check our nswers b substituting the vlues into the originl equtions: st eqution : LHS RHS ( nd eqution : LHS ( RHS

29 7 Tutoril Eercises ( Epnsion of Algebric Epressions Contining Brckets (Revision ( Epnd the following epressions involving products nd powers: (i ( ( (ii (8 ( 9 (iii (v (vii ( (7 8 9 (iv ( ( 7 9 ( ( 7 (vi (b ( b ( ( 8 (viii ( b ( b (i ( ( ( b b ( ( b( b ( b ( Open out the following brcketed epressions: (i ( ( (ii ( ( 7 (iii ( ( (iv ( 8( (v ( (vi ( (vii (i ( (viii ( ( ( ( 7 ( ( ( ( ( Fctorise the following qudrtic epressions: (i (ii (iii (iv (v (vi (vii (viii (i ( ( Show tht ( b( b b nd use the result to (i epnd: ( ( ; ( ( (ii fctorise: ;

30 8 ( Division of Algebric Epressions ( Perform the following divisions: (i b (ii 8b (iii b b 7c d (iv 9c d (v 8 (vi π r h 8π r π r h h ( Use polnomil division to epress the following quotients in the form Reminder Quotient Divisor (i 7 (ii 7 (iii (iv (v 7 (vi

31 9 ( Eponentils nd Logrithms The Bsics ( Use our clcultor functions, nmel log (which is ctull log 0 nd 0, to evlute the following epressions Comment on the results for prts (v (viii (i log 0 (000 (ii log 0 ( (iii 0 (iv 0 (v log (0 (vi log ( 0 log log 0 ( ( 0 (vii (viii ( Use our clcultor functions, nmel ln (which is ctull log e nd e, to evlute the following epressions Comment on the results for prts (v (viii (i ln ( (ii ln ( 7 (iii 0 e (iv e (v 0 ln ( e (vi ln ( e ln ( ln ( 07 (vii e (viii e ( Write ech of the following epressions s sums nd differences of logrithms (where possible, without using powers: (i log 0 (ii ln (iii log 0 00 (iv ln e (v log (vi 0 ln ( ( ( ( Write ech of the following s single logrithm: (i (ii (iii log0 log0 log0 z log0 ( log0 z ln ln (iv ln ( ln z

32 0 ( Chnging the Subject of Formul ( For ech of the following formule, chnge the subject to the quntit indicted in brckets: (i (ii (iii (iv (v v u t ( t s ( t t ( u s u t s u t t ( t v u s ( s (vi (vii v u s ( u b ( (viii t i e ( t (i i t 8e ( t ( (i ( 0 ( ( 0 ( k t (ii c e ( t 0 0 ( For ech of the following formule, chnge the subject to the quntit indicted in brckets: (i ( (ii ( (iii z ( (iv C C C C C ( C

33 ( Solution of Equtions ( Solve the following equtions: (i (iii 8 (ii 0 8 (iv 9 (v (vi 0 0 (vii (i e 07 (viii e 0 e 8 7 ( e (i ln ( (ii ln ( 0 (iii ln ( 0 (iv ln ( 0 (v 0 (vi 0 7 (vii log 0 ( 0 (viii log 0 ( 0 (i ( ( Solve the following qudrtic equtions b fctoristion, then repet using the qudrtic formul: (i 0 (ii (iii 0 0 (iv 7 0

34 ( The following qudrtic equtions do not hve n rel solutions Tr solving them b the qudrtic formul nd see wht hppens: (i (ii 0 0 ( Solve the following sets of simultneous equtions: (i (ii (iii (iv

35 Answers ( (i (ii 8 7 (iii 8 (iv 8 (v (vi b (vii 0 (viii b (i b ( b ( (i (ii (iii 9 (iv 00 0 (v 8 (vi 9 (vii (i 9 (viii ( 7 0 ( (i ( ( (ii ( ( (iii (v ( (iv ( ( ( ( (vi ( ( (vii ( ( (viii ( ( (i ( ( ( ( ( ( (i ( ( ; ( ( 9 (ii ( ( ; ( 8( 8 ( (i (ii b (iii 7 b (iv 8c d (v (vi h r

36 ( (i 0 ( (ii ( 0 (iii (8 (iv [no reminder] (v 0 (vi ( ( (i (ii (iii 77 (iv (v (vi (vii (viii 0 ( (i (ii 0009 (iii (iv 000 (v 0 (vi (vii (viii 07 ( (i log0 log0 log0 (ii ln ln ln (iii log0 ( (iv ln ( (v log ( 0 (vi [ ln ( ln ( ln ( ] ( (i log 0 z (ii log 0 ( z (iii ( ln (iv ln z

37 ( (i t v u (ii t s (iii u s t ( s (iv t t u t (v s v u (vi u v s (vii (i (viii t ln i b t ln ( [ log0 ( ] 8 i (i log ( ] (ii [ 0 t k ln 0 c 0 ( (i (ii (iii z (iv z C C C C C ( (i (ii (iii (iv (v, (vi (vii (i ln ( (viii ln (0 0 0 ln ( 098 ( ln ( (i e 0 0 (ii e (iii e 7 09 (iv ( e 0 (v log 0 ( 0979 (vi log (7 ] [ (vii 0 0 (viii [0 ] (i / Over the pge

38 (i log0 log [could lso use nturl logs] ( log0 log0 8 [could lso use nturl logs] ( (i, (ii, (iii [repeted root] (iv, ( Negtive vlues under the squre root sign indicte no rel solutions Solutions onl possible b moving into comple numbers ( (i, (ii, (iii, (iv,

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