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Un daa symboias suivants apparaltra sur la darniira imaga da chaqua microficha, sslon la eas: la symbols — ^ signifia "A SUIVRE", la symbols ▼ signifia "FIN". Mapa. platas, charts, stc, may ba flimad at diffarant raduction ratioa. Thoaa too larga to ba antiraly Included in ona axpoaura ara flimad baginning In tha uppar laft hand comar. iaft to right and top to bottom, aa many framas aa raquirad. Tha following diagrams illustrata tha mathod: Las cartas, pianchas, tablaaux. ate, pauvant Atra filmis A das taux da reduction dlff*rants. Lorsqua la document ast trop grand pour Atra raproduit an un saul cllchA. 11 ast film* A partir da i'angia supAriaur gaucha. da gaucha A droita, at da haut an bas, an pranant la nombra d'imagas nAcsssaira. Las disgrammas suivants iliustrsnt la mAthoda. rrata o lelure. 3 32X 1 2 3 1 2 3 4 5 6 // •i ,; * 'SS I) ALGEBRA FOR BEGINNERS. *,., \\ ALGEBM FOR BEGINNERS WITH NUMEROUS EXAMPLES. BT I TODHUNTER, M.A., F.RS. NEW EDITION. SottDon atili iSTamibYiOse : MACMILLAN AND CO. TORONTO: COPP, CLABK, AND CO. i860. . [All Righti r^erved.l ^#*'"'*. **j«t. *■ f -^mf /(^ dOXo I- // CatiiMifle: 1>IUNTBD BY 0. J. OLAT. M.A. AT THS UNIYBBBITT PRB8B. 220514 \\ PREFACE. Thi present work has been undertaken at the request of many teachers, in order to be placed in the hands of beginners, and to serre as an introduction to the larger treatise published by the author; it is accordingly based on the earlier chapters of that treatise, but is of a mord elementary character. Great pafais have been taken to render the woik intelligible to young students, by the use of simple language and by copious explanations. In determining the subjects to be included and the space to be assigned to each, the author has been guided by the papers given at the yaiious examinations in ele- mentaiy Algebra which are now carried on in this countiy. The book may be said to consist of three parts. The first port contains the elementary operations in intend and fractional expressions ; it occupies eighteen chapters. The second part contains the solution of equkikns and pro- blems; it occupies twelve chapters. The subjects contained in these two parts constitute nearly the whole 6f every ex- amination paper which was consulted, and accordingly they are treated with ample detail of illustration and exercise. The third part forms the remainder of the book; it con- sists of. various subjects which are introduced but rarely into the examination papers, and which are thereforo more briefly discussed. The subjects are arranged in what appears to be the most natural order. But many teachers find it advan- tageous to introduce easy equations and problems at a very early stagey and accordhigly provision has been made for >jf' in accordance with iha «wJ«l »»w been .elected flx)mth.^i'~"* Someofthei» wth referenoe to points which iJ!Z7 •''* "* oonrtmcted ^^^^^^^ / authort expenence u a teacher and an wwrki on the worfc ««!i ^'J^'^^e fluggestions. Anv ^ «»»»«llr receir^ *' ** "^^^ *«! be m«t S» JOHH 'S COILUI, I. TODHUITTBR «on. and abo?3:^^;? 5^"Jf <*«'» *» ^ P««ent edi- «• ««nged in ae^JStll^*"'!' ^^P'« »W<* Th«e addition, have £ ^te^^rth"^*? *«« «»»»Plea eminent teachew, in onler tTf *' '*9»«'t o' "owe woA. ^ "" ""*** to "w*ea«e the utility of the ^^ly 1867. # l| .'^.■' \ \ CONTENTS. rAOB I. The Frinoipal Signs i II. Fftotor. Coeffioi«nt Power. Tenue 5 III. BemainiDg Signe. Brackets 9 IV. Ohmge of ibe order of Terms. Like Terms 19 V. Addition 16 VI. Subtraction 19 , VII. Brackets :,: 22 Vni. Multiplication «5 IX. Division 33 X. General Results in Multiplication 43 XI. Factors 49 XII. Greatest Common Measure 55 XIII. Least Common Multiple r 63 XrV. Fractions , 68 XV. Beduction of Fractious 73 XVI. Addition or Subtraction of Fractions 76 XVIL Multiplication of Fractions 84 XVIII. Division of Fractions 88 XIX. Simple Equations 94 XX. Simple Equations, continued 103 XXI. Problems 113 XXII. Problems, continued ill XXni. Simultaneous Equations of the first degree with two unknown quantities 136 ^XrV. Simultaneous Equations of the first degree with more than two unknown quantities... 145 XXV. Problems which lead to simultaneous equa- tions of the first degree with more than one unknown quantity ^ . 150 Tiu CONTENTS, XXVL Quadraiio Equationf 160 XXYII. EquAtiont whieh maj be lolYed like Qatd- ratios ; 171 XXVin. Problemi wbiob lead to Quadratio EquAtioni 176 XXIX. Simultaaeoua Equatioiii involviiig Quad* ratioi i8a XXX. Problema wbicb lead to Qoadratie EquaUoni witb more than one unknown quantity .<.... 1 90 XXXI. Inyolution 195 XXXII. Erolution 300 XXXIII. Indicei 318 XXXIY. Surdi ;... 125 XXXV. Batio ^30 XXXVI. Proportion 134 XXXVII. Variation , 540 XXXVIII. Arithmetioai Progression «.... 345 XXXIX. G^eometrioal Progression 349 XL. Hannonioal Progression 954 XLI. Permutations and Combinations 356 XLIL Btoomial Theorem 360 XLIII. Scales of Notation ,.... 968 XLIV. Interest 371 MiseellaneouB Examples .........1.... 375 ANSWEBS... :... 305 \\ 160 lad- nt IfMM 176 iftd* • • •• 189 oni • ••• 190 »95 • •••■ 900 • ••• 118 t • • • ««5 n^ it«« «34 • •• 940 • *• 245 • •• 149 • • • «54 • • • 356 • •• a6o • • 968 • • «;« • • 275 305 ALGEBRA FOR BEGINNERS. L The Principal Signi. 1. AioiBAJk is the scienoe in which we reason about nnmben, with the aid of letteito to denote the nnmbenu an4 of certain signs to denote the operations performed on the numbers, and the rohttions of the numbers to each other. 2. Numbers may be either known numbers, or num- bers which have to be found, and which are therefore called unknown numbers. It is usual to represent known numbers by the first letters of the alphaMt^ a, 6, e» &c., and unknown numbers by the last letters at, y, ir; this is however not a necessary rule, and so need not be strictly obeyed. . Numbers may be either whole or fractional The word quantity is often used with the same meaning as number. The word integer is often used instead oftMole number. «, 3. The beginner has to accustom himself to the use of letters for representiiur numbers, and to learn the meaning of the siffns; we shaU begin by explainiiur the most im- "oortant signs and illustrating tneir use. We shall assume ibat the student has a knowledge of the elements of Arith- metic, and thiat he admits the truth of the common notions required in all p^rts of mathematics, such as, if eqtude b^ added to equate the wholes are equal^ and the like. 4. The svn + placed before a number denotes that the number is to be added. Thus a+b denotes that the num- ber r6pre8ented4>y d is to be added to the number repre* T. A. 1 f I 'I ■) 'Is. ..^ ■J* 2 !ra^ PRINCIPAL SIGNS. sented by a. If a represent 9 and b represent 3, then a + & represents 12. The sign 4- is called the plu» sign, and a+& is read thus ''ap/uf b.'' 6. The sign -placed before a number denotes that the number is to be iubttacted. Thus a -& denotes .that the number represented by & is to be subtracted from th6 number represented by a. If a represent 9 and b repre- sent 3, then a—b represents 6. The sign ~ is called the minus $ign, and a-& is read thus "a minut b." 6. Similarly a+(+<; denotes that we are to add b to a, and then add e to the result; a-\-b-c denotes that we are to add b to a, and then subtract c from the result; a'-b-^e denotes that we are to subtract & from a, and tiifin add x^ to the result; a^b—e d^otes that we are to wib- tiact 6 from a/ and'then subtract c frt>m the result 7. ^The sign «= denotes that i^e numbers between which it is placed are equal. Thus a ~ & denotes that the number represented by a is equal to the number repre- sented by 6. And aA-b^c denotes that the smir of "the munbers represented by a and b is equal to the liundber represented by c; so that if a rewesent 9, and & represent 3j then e must represent 12. ^he sigu = is catfed" the 9ign i!(if eqwfiity, and a=^b is read thus '^a equaU b** or *^%iiequ(dtoh,*\ 8. The sign x denoted tiiiai the' nmnbers between which it stands are to be muUiplied together. Thus ax6 denotes that the number represent^ by, a is to J)e multiplied bv the number represented by b. If a remre- sent 9j and o represent 3, then axb represents 27. SSie sign X IS called the sign qf rnuUiplicaiton, and^ ax( i» read thus <*a into b.'' Similarly axbxe denotes the pro- duct of the numbers represented by a, b, and c, 9. The sign of Aiultiplioation is howerer often ondtted for the sake of brevity; thus a& is used inst^ of axb, and has the same meaning; so also abe isiused mst^ of axbx Cf -and has the same meanmg. ^ > The sign of multiplication ,must not be omitt^ when limnbers are expressed in the ordmary way by %ip9A Th|iu 4li cannot be used to represent the product of 4 api ■ M ■Sas.'StSp'^-. M^«9* ^ We moat therefore lef present the product of 4 and 6 m another way, and 4x0 IS the way which is adopted Sometimes, noweyer, a point is used instead of the sign x ; thus 4:5 is vaM. in- stead of 4x5. To prevent any confusion between the point thus used as a sign of mmtiplioation, and thenpcdnt used in the notation for decimal finctaons. it is adyisaUe to place the point in the latter case higher up; thus 4*5 may be kept ^ denote 4 + rr • But in &ct the point is not uM instead pf the sign x except in cases where there can be no ambiguity. Forexample, 1.2.3.4 may be pot for 1x2x3x4 because the ^points here will not be taken for decimal p(^ts. The point is sometimes placed instead of the sign x between two letters; so that a. &. is used instead of a x &. But the point is here superfluous, because, as we have 8aid,^isused inst^idof ax&. Nor is the point, nor the si^ X , necesspkxy between a number expressed in the or- dmary way b;^ a figure and a mimber represented by a letter; so th^t, for example* da is used instead of 3^(0, and his the saSe meaning. " ' ' . ^ , ' ■ ',-/-':^ - . ' ■ >- The aaa -f- denotes that the number which pre- it i^ to be divided by the number which foUows il. lOtes that the number represented by a is to ti^e number represeUted by K If a rem- . represent 4, then a-i-b represefits 2. The ted the sign qf divition, and a-^b m read 10. cedes Thud a-i-b be divid sent 8, sign thus There |»- also another way of denoting that one num- ber is to be divided hf another; the dividend is placed over the divisor with a line between them. Thus j- is used instead ofa-i-by and has the same meaning. 11. The letters of the alphabet^ and the siras which we have already explained, tc^ther with those which may occur hereafter, are c^led mgebraical symbols, because they are Aised to represent the numbers about which W9 may be reasoning, the operations performed on them, and 1~2 f ( 4 JSXAMPLS8. i. ihefar retatibiui to eadi oUier. Any oollectioii of Akobralcal «yinbob is called an al(f^raktU eapreirion, or htieSij an 18. We shall now giro some examples at an ezmsise in the nse of the symbols which have been eilplalned; these examples consist in finding the nnmerical Ttthies of oeirtain algebraical expressions. Suppose asl,d'=8yea3,<^==5,«=6,/-0. !nien Ta+aft-skf+z-T-j-e-io+o-ia-io^a. 8^^^^<8»c~atf+4r«4+48-64L0=52-6=^ I T «l 00 2 * 15 8 *** *" * ^ 4g-t'8g 12+80 42 - . EzAirPLISL I, If o=l,J&f 2, 0=8, «r=4, «=5,/=o, find the nmnerit eal wofis of the foUowu^ expressions: 2. 4o-3a-3$4'5e; 4. B^-'bcd-h9€de''dqf, ^ aM+t00rof a; axaxana iaoijQaithe/ottri0irofa;andsomi, And a itself i§ mmmmMmM^JkrH power ot a. «f f « 6 FACTOR. eOEPriaiSNT POWBB. TSBM8. 16. A power is . more briefly denoted thius : fustead of exprefsing iul the equal &ctors, we express the fiiotor onoe, and place orei* it^-^ linmber whieh indicateii how oftoi it is to be repeated. Thus €? is used to denote axa; o^ is used to denote a xax «; a* is nsed to denote ama x ^u. a; mi. lio on. And a^ mar 1>e nsed to denote t&e first power bif a, iliat is A iti^lf ^ so that <^ has (^ tami meaning at "d * It. A number placed over another bo ideate- how ma^ijr ^mes the latter occurs as a factoir in ajK>wer, 1^ billed an indtf^ qf the po^er yO/raxi eaponentqftAepifwer; ei*, briefly, an »nde^, or tfdriMm^n^. Thus, fpr example^ p <^^9. exponent is 3; in at* the ffl^ponentis^ - 18. The student mudt distinjraish yeijr carefU% between il confident md m exponent, Thus 3^ means three timet c; hei» 3 is a eoeffkient Bat c* means c ^Htthee t ii^eeaf hm 3 is an e^i^pondnf. That is 3e=e+o+c, W 'i^e second po^r of a» that is d(\ is oftei^ taSM the equarij^ a, or a squared ; and the third pow^ of Oy l^at is ^, ii ^Mif called ma cube ei a, or a cubed. There are no fuch words in use for the higher powers; a* il read thus **a 10 the fourth power,* or briefly ** a =6»=12«. a «'• 8 EXAMPLES. IL 1!XAMPL1& It <<-i vmnB^io^ , M^v J III, Remaining Signt, BraekttU. 24. ^e diflerenoe of two nnmbeni is sometimes de- noted hj the sign ~; thus a'^h denotes the cUflforenee of the i^umben represented by a and ft; and is equal to a-^(. or &-a^ aooorduig as a is greater than (, or less than h: hut this symbol -^ is very rareqr required. 2^ ^The ngn >, denotes it ^Mrf^r & denotes that the nnmber represented by a is greater than the number representea by b, and o-i-{rf+«) denotes that the result! ex** pressed by a+5~<; is to be mvided %theleBu^«^^ Dyrf+«. 30. Sometimes instead of usinff l>radieti9 a Hne li drawn oyer the numb ers whic h are t o be trotted as Ibnsd^ i one number. Thus a-b-^cxd+e k used iiii the siWk i meaning M (a-6+ VMning the ligiif •¥• and - ; and hi nraltipUah tion and diviiioii we shall mak of the RuU qfStgm, mean- i^ a role relating to the^a^ins 4- and -« 88. We shall now giro some more examples of finding Ihe numerical values of expressions. Suppose a»ly (■«2, c-i^ - V(*+ 12)- >/(16)-4. y(26+4c)-(2rf-5)4^(4-8x4-8x2-32-16-ie. -y(27+54+36+8)+ ■.•■ i , ■ . ■- ■ • > If a»»l, (">2, €mZ^ if— 6, «— 8, find the nnmerical Values of the litAov^ egressions: % die-^d). KH 12 CHANGE OF THB OBPMB OF TERMS. IV. Change qf th$ order qf Termt, Like Temu, 8Si WlienallthetennsofanexpreMdonareooimected by the sign •f it is indifferent in what order thej are Ded; thus 0+7 and 7-f 6 giye the same retnlt, namely, and 80 also a+b and h-ha give the same result^ namelj; thd som of the nnmben which are represented by a and o. We may express this fiMst algebraicaUy thtu^ a+b'»b-h.^'85-45— 75., ;.: ' ''■'''■■ Wi» oansiot aubtnUst 35 frt>m nothing and then 45 fitm therenulnder, so that the statement lust ^ not here intelligible in itself, separated fr^tnr#eit(^%^. algebraical sentepoe in whidi it mayji6ciur) but it can |^ easily explamed thus: if in the ca|^ of an a]|e)«iilesl operation we have to aubtract 35 frellHk number aip ^pi to subtract 45 from the reataindw, we may subtraefr 75«t once instead.' \ . ■ .•* ' ' ■■'' As the student advances in the sul^Ject he may to coi\jecture that it is possible to gfive some m< the jNfitmosed statement by iikelt, t&t ^ a^iart other ab^bralcal operaUon, and this jemectiire * * ' when a laig^ treatise on JMgelNa 1 1 »■ 0* CO ha tei Bu 01 eqi *': I I* } val \ \- 4V ■'•;,,'» ii f re- ihen lariy I 'I ■ h n- ) ' A -I ^ St IK » u"^"^ ^ it^^^^t^^^ ^ eipknatloii which we A ^^•i.S^ rimpBfvinf oT M^rwwioiis by oonectbff like Gh^l^ ^^'^^ •• we •EOI see in the next two 2Si^-T#>.^t?^^ the JoDowing exprogiioM are aO + a\ +1x0, +^aj terms shall stand in the same column: thus wo hav6 ) th wi efl W€ ho 5b am I 'if flJ»4 i is u 1 can . i volv 'i invo 1 y ■ . :^- ADDITIOJBT. n 4a+Sb-'1e+2d 3a— l>+2c+5d 9a-26- c- d -a+Zb+4c-3d+0 I5a + 6b-2c + 4d+e Here the ietfaa 4a, 3a, 9a, and —a are all like terms; the sum of the pesitiye coefficients is 16; there is one term with a negative coefficient, namely — a, of which the co- ) efficient is 1. The difference of 16 and 1 is 15; so that we obtain + 16a' from these like terms; the sign + may howeyer be omitted by Art. 38. Similarly we haVe 5&--&— 2& + 3&r»5&. And so on. so W« > i47. In the followmg examples the terms are arranged soitably in columns : . 4a^+1a^+ a?-9 -2«3+ ^« 9*+8 -34J»- «2+10«-l 9aj"- a-l 3a"-3a&-76» 4a*+6a6+96* a«-3a6-3J* 9a* — , 7a-ll&, a4-9&. 4«"-3y», 2««-6y», -Jf«+y8, -2««+4y«. 5a+3&+c, '3a+3d+3<;, a+3& + 5c. 3«r+2y-;ir, 24r-2K + 2;ir, ^a + 2y + 3z, 7a-4ft + (?, 6a+3&-6 c+a-5, a+5-«, ^. #-f^+3dT, 12a& + l4^-12«V+12dJ^-4y*, e«V-12ajy»+6yS 4«jf»-4yS yl 18. «'+«y"+j?^-««y-«y^-^4f, la 11. li la 14. 15. 16. ^ v^ "-1 j-46, a? +8. -4,' i f-4irS subtbaCtion. YI. SubiracHon. If 48k Suppose we have to take 7+3 from 12; the repnlt u the same as if we first take 7 from 12, and Uien take 3 from the remamder; that is, the result is denoted by 12-7-3. Thus 12-(7 + 3)-12-7-3. Here we enclose 7 + 3 in brackets in the first expression, because we are. to tsdce the tehole of 7+3 from 12; se^ Art 29. Sunilarly 20-(5+4+2)-20-5-4-2. In tike manner, suppose we have to take h+e from a; the result u the same as if we first take b from a, and then take c frx>m the remainder; that is, the result is denoted by a— d-<;. Thus «-(6+c)=a-6-«, Here we endose b+cia brackets in the first expresrioni because we are to take the whde otb+t from (k Similarly «-(ft+c+rf)-»a-6-c-dL 49. Nelt suppose we have to take 7-3 firom 12. If we lake 7 from 12 we obtain 12-7; but we have thus taken too much from 12, for we had to take, not 7, but 7 diminished by 3. Hence we must increase the result by $| and thus we obtain 12-(7-3)»12-7+3. Sunilariy 12-(7+3-2)-12-7-3+2. In like manner, suppose we have to take 5— « frx>m a. If we take b from a we obtain a-b; but we have thus taken too much frtnn a, for we had to take, not 5, but b diminished by e. Hence we must increase the result by e; and thus we obtain a-(&-c)»a—&+c; Similarly a-^+c-rf)«a-6-c+dl 60. Cr)}. a 2a-(25-rf)-{a-6-(2(j-2^}. 9. d^{2&-(3kJ+25-a)}. 10. 2a-{&-(a-25)}. 11. 3a-{&+(2a-6)-(a-J)}. 12. 7a-[3a-{4a-(6a-.2a)}]. 13. 3a-[6-.{a+(&-3a)}]. 14. 6a-[4&-{4a-(6a-4&)}]. 16. 2a~(3& + 2c)-[6&-(6 ■ 24, f*-[>«'-{^-(4i»-l)}]-(a?*4-4a*+6a!*+4«^l)^ • mULTIPUQATIOK Vi is may ^ the «— 0}. m i^ I Vllh MultipKeation. 67* The student is snppoaed to know that the product of any number of factors is the same in whateyer oraer the factors may be taken; thus 2x3x5=2x5x3=3x5x2; and so on. In like manner a&c»ac&««(c We have Similarly, %a-¥h)»1a-¥lh. In the aaine manner suppose we haye to multiply a4-& bye We have c(a+5)-« rettdt;. and eollect theee resulte to form the complete prodwst, \ 61. III. Suppose we haye to multiply a + & by c + d ^ As in the second case we haye {a-¥h)(c-{-d)-a{C'¥d)-¥h{c-¥d)', also a(c+ h(fi-^d^-bc-¥hd\ therefore (a-&)(+c*~a6-&c-«i by a+&+tf. Amuige acoording to descending powers of a. a + 6 + c if • H • I -3a6<; +6^ +c» This example might also be worked with the aid of brackets, thus, a + (6+c) Then we have 0(6^- Jc + c^ - a(6 + c) (6 + c) =a{6»-J«?+tf»-J*-26(j-c»}= -3a6c; and (&+c)(a»-6ivisic . - ,Thus it will be seen that the EiUe qf Sigm holds in Division as well as in Multiplication. 71. Hence we have the following rule for dividing one simple expression by another: Write the divujkiia oter the divisor with a line^hetween them; if the exprettione have common fcbctorif remove the common faetore ; prefix the tign + if the expreesione have the same sign and the sign - if they have diferent eigne. ? , 72. One power qf any number is divided M another power of the same numiberfhy eubtracting theind^qf the latter power from the index qf theformar, o ^ , s ¥oT example^ 8a|ypdae we have to divide a" by-^i^ By Art 16, af=axaxiixaxa, V^ therefore a» axaxaxaxa axaxa :Ax-a»a'=a""^, ai n 1 ^ -excxexexexexc ^. -^- Similarly -^ = ^ .. ^ .. ^ ,. ^ =c x ^ other case. '^, ! \ Qr we may shew the truth of the rule thus : by. Art 59, .-^^i^ c^xc^=c', therefore =c», 73. If any power of a number occurs in tho dividend and a high^ power of the same number in the divisor, tho quotient can be simplified bv Arts. 71, and 72. SuppoUe^' for example, that 4a^ is to oe divided by 3<^; then, i^e quotient is denoted by ^^ . The fiictor 5^ occurs in botli dividend and divisor; tUs may be removed, and the quo-' tient denoMby g^; thus ^^ = g^. 74. IL The rule for dividing a compound expression by a simple expression will be obtainOd from an exaddna- tion of the corresponding case in Multiplication. For exfunple, we have ^v therefore therefore o (a-6)x -c=— oo+Jc; — c =a-Ji ■■/' ■vj 86 division: ^v; 'n V Hence we hf^ye the foUowing rale for dividing a oom- ponnd expreflsion by a simple expreBsion: divide each term qf the dividend by the divieor, by the ruU in the firH catef and coUeet the reeulte to form the comjjMe quotient. ' J^or example, — i?^=4a'-36c+aft . 75. III. To divide One compound expresrion by another we must proceed as in the operation called Long .Division in Arithmetic The following rule may be given. Arrange both dividend and divieor according to aacend' ing powers cf eome common letter, or both according to descending pomere qf some common tetter. Divide the Jirst term of the dividend by the Jlrst term qf the divisor^ and put the residtfor the first term of the quotient; muh tiply the whole divisor by this term, and subtract t%s product from the dividend, T9 the remainder join di many terms qf the dividend, taken t^ order, as may be required, and repeai the whole operation^ Continue the process until all the terms qf the diHdend have been taken down* ■ : = The reascm for this rule is the same as that for the rule of hon^ Division in Arithmetic, namely, that We may hfesk the dividend Up into parts and find how often the divisor is contained in each part, and then the aggregate of these results is the complete quotient 76. We shall now give some examples of Division arranged in a convenient form. a+bja^+2ab+l^(a+b a^+ah €^+db db+I^ -db-^b^ ^db^b» a'^b)a*''l^(a+b «*+3afJ«*+2a!*-3af^«-'l ab-b* ab-V DIVISION. K -4a^6+13a*6«-22a6»' -4a^ft+ 8a«6«-12aft» 5a«6«-10a6»+16&* Consider the last example. The dividend and divisor are both arranged according to deacendinff powers of ol Hie first term m the dividend is So* and the first term in the divisor is V; dividing the former by the latter we obtain 3a* for the first term of the quotient. We then multiply the whole divisor by 3a*, and place the result so that each term comes b^low the term of the dividend which contains the same power of a; we subtract^ and obtain — 4a'&+13a^&'; and we bring down the next term ot the dividend, namely, — 22aC We divide the first term, — 4a'&, by the first term in the divisor, a*: thus we obtain — 4a& for the next term in tiie quotient. We then multiply the whole divisor by — 4a& and place the result in order under those terms of the dividend with which we are now occupied; we subtract^ and obtain 5a'd*-"l(ki6^; and we brinff down the next term of the dividend, namely, ISft*. We divide fk?l^ by a*, and thus we obtain 5^* for the next term in the quotient We then multiply the whole divisor by 55^ and place the terms as before; we sujiytarac^ and there IS no remainder. As all the terms in the dividend have been brought down, the operation is .completed; and the quotient is 3a* - 4a& + 5d\ It it qf great importance to arrange "both dividend and diviior according to the sama order of some common letter; and to ixttend to this order in every part qf the operation, 77. Ijb may happen, as in Arithmetic, that the division cannot he eaactly performed. Thus, for example* if we divide a*+2db+2¥ by a+b. we shall obtain, as in the fbcat examine oC the preceding Article, a -1-6 in the qaotient, and there nfHihen he a remainder h\ This result is ex- m DlP^tSIOlf. pressed in ways idniikr to those used in Arithmetio; thm we may say that =a+6+ t^ a+5' 6^ that is, there Is a quotient a + 5, and a fhuitional part — r • In general, let A and B d0note two expressions, and suppose that when A is divided by B the quotient is q, and the remaindldr R\ then this result is eipressed algebrai- laUly in the following ways, A^qS^Bi or A--qB=Bf I- I A R A B ^^B^^-^W ^B'^^^'S' \ Hie student will observe that eacli letter hero may re^ g^ resent an expression, simple or compound; it is oftoa onveoient for distinctness and brevity thus to represent an expression by a smgle letter. We shall however consider ak^braical fractions in snb- secmont Chapters, and at present shall confine ourselves to e»unt»les of IXvlnon in wmch the operation can be exactly performed. 78. We give some more examples: IHyide 4^-5«»+7dJ»+2«*-6ay-2 by l+lto-8aj»+«*. Arrange both dividend and divisor according to de^ Bcending powers of ;r. - 2aj* -2«* + 6^ + 2aj»- 6« -2ar« 4-6a^-44^-2a? -2# +6«"-44?-2 - \\ -•— r- DIVISION. 30 Arrange the diyidend according to. descending powers ofo. -J -sH It win be seen that wo arrange these terms according to descending powers of a; then when there are two terms, such as crh and cfc^ whoich inyolve the same power of Oy we select a new letter, as &, and pat the term which contains h before the term which does not; and again, of the terms a^ and qbc^ we put the former firet as involving the h%;her power of 6. ' ^ ' ' This example might also be worked, with the aid of brackets, thus: -a^+c)-a(d*+26c+c*) - — ' '• a(y- ^j+^j^+y+c^ M EXAMPLES. IX. Diyide «*-(a+6+«)«'+(aft+ac+Jc)«-aftc by m-e. — (a + 6)«^ + (aft + flw + 6c)«- ofto abx Erery example of Miilti]^oatiil> 246S-^8a^M+12a&by-da&. %/m X «•* 7^+12 by «-3. 1^ ajS+^-tSi byiilff; 2«»-«*+8«— 9 by 2a?-8. SdJ* + 14;]^ - 49 + 24 by 20^ 4- 6. \^^\ 9«*,+ 3«» +4?- 1 by 3i»- 1. 7«*-24e*+5av-21 by 7«-3. V. ■'--S-vT:-"-. v-, - • i ■ r- /If EXAMPLJBS. 41 >y »-«. 10 miitti^ ioiii» will prodiMl ifaeoUitr Kstion ^ Dhriiio% example risor obliiiu .so ' -Mt} *^,ik:; ■' -U:.:^ ■ ■ ^^tt-:^ ■ '•%%^' ■ < ■ \'«i •■ -.1- '■'. '' ■ , ft •* ■■/- ' A'^f '."■•^ 'vl* ' : :.v -^' ■ 17. 4^-lby«-l. 18. aP-SaS^+i*bya-ei 19. «*-81y«by«-8y, SO. «*-8«V+*^-«y"by«-y. fil. oF-y^hya-y. 82. a"+88&'b7a+2& 53. Stf«4-S7-y". S«. «*-tt«»+ll««-12aT+6by«'-8«+8. 57, V+«»-9a*-l««-4bya»+4«y-4, 58. V-18d^+36by4p'+0Jr+6. 89. «*-f64by4P*+4dr+a Sa «^+lOi^4-85dP*+0Our+24by«*+S#+4. 81. «*+4P*-24«*-3^byaa^-&v+c. 41. «*-?'i^-«y»+y*by4»*+«y+y". j 48. ji»-8«y~y"-lby#-y-l. 48r 00^+ 2l4jy + 12y4f - 16;*' by 1a+3y-^, ^,- #+2a6+ftf— ^ by a+6-c. .•♦«^*-6aftra;. wo may say that Algebra enables us, to prove general ^ihe^et^ .respecting numbers^ and also to eapress those theorems br^y. '^ ' ip^; 5l <* *:»■■ IN MULTIPLICATION. 48 ' For exaiii|)le, the result (a+()(a-&)^=a«Jb2a6+6», where ^ indicates that we may take either the sign 4- or the sign — , keeping throughout tJhe upper tign or the lower eign, a ±0 is read thus^ *^ a plueor minui b/* 82. We shall doTote some Articles to explainhig the use that can be made of the formiUse of Art 79. We shall repeat these formulae, vad number them /or the take <\f eaey and dietinct r^erence to them, (a+bf =a«+2a&+6» (1) . (a-6)« =a«-2a5+6? (2) . (a+6)(a-&)=a^-6« (3) 83. The formulae will sometimes be of use hoi Arith- metical calculations. . For example; required the difference of the squares of 127 and 123. By the formula (3) (127)'^(123)"=(127 + 123)(127-123)=250x4=1000. Thus the required number is obtained more easily than it would be by squaring 127 and 123, and subtracting the second result from the first Agam, by the formula (2) . (29)»=(30-l)»=900-60 + l = 841 ; and thus the square of 29 is found more easily than by multi|^yii|g 29 % 29 directly. Or suppose we have to multiply 53 by 47. By the formula (3) ^ 53x^=e- noteoT-fy by a. Then a-hy+z=a'\-z; and by the use of (1) we have =:a^+2a!y+p*+2xz+2yz+z'f, / Thus(«+y+4f)*=«»+^+;»*+2ajy+2y«+2a?;8. Suppose we require the square of p—q+r-t. Denote p—qij aaadr—ihyb; thenp— g+r-«=a+6. By the use of (1) we have (a+&)"=a»+2a6+6»=(|?-g)*+2(|>-g)(r-#)+(r~t)« Then by the use of (2) we express (p-gf and {r-^if, ThusCp-g'+r-*)* =!>' ~ 2|?g + g* + 2 (pr -p# - ^r + g*) + r* - 2r# -I- «■ =i^ + ^ +**+«* + 2l?r + 2g# - 2pj' - 2p* - 2^^ - 2r#. Suppose we require the product of p^q^-r-i and p-q-r+i. ^ Letp—<2^=a and r~«=&; then p-g-^r-$'^a-\'b,«Ddp'-q-^rA-»=»a-hi, IN MVLTIFLtCATION. 4ft Then by the we of (3) we have (a+?>)(a-&)=• - 2pg + g* - r* + 2r# - J*; or more briefly stilli (p-fl'+r-t)(i?^g-r+#)-(i>-g)»-(r-#)" =*P* - 2pg + g* - f* + 2r^ - #'. But at first the stadent will probably find it pnident to go through the work folly as in the preceding Article. 87. The following example will employ all the three fbrmnlflo. Find the product of the four &ctors a+6+tf, a+6-e, 0-6+c, 6+c-tt» Take the first two &6tord j by (8) and (1) We obtain (a+6+c)(a+6-c)i"(«+^)"-c'««*+2a&+6*~c*. Take the last two fiicton; by (3) and (2) we obtain (a~6+c)(6+c--a)={c+(a-6)}{c--(a-6)} =c«-(a-6)"=c*-a"+2a6-5». We have now to multiply togethei^ (a-&)s=:(a-. J) (a2-2a6 + d«) =*a3-3fl«& + 3a6«- 68, V ,, =a?+3a«(5+c)+3a(6»+26c+c2)+6»+36»c + 3&<^+jC» =a8+&»+c»+3a«(6+c)+3d«(a+c)+3c2(a+&)+6a&«. 89» Useful exercises in Multiplication are formed by^ requiring the student to shew that two expressions agree^fi givinff the same result. For example, shew that If we multiply a-6 by &~c we obtain db—l^—ac+bei then by multiplying this result by c- a we obtak , ca6 1- cJ* - ac* + 5c" - a«6 + a6« + a^ - oftc, thatis a*(c-6)+6«(a-c)+c»(ft-fl^). ^ Agafai; jhewthat(a-6)«+(6-y+^)(«*-«y-y'). 11. ]^^^+3«+l)(a»-2aj*+3a?-l). • llj&i|i)«^^ 13. (a+ft)«((^-2a6-J^ 14. @^ i • I, F I .1 48 EXAMPLES. X ¥5 Shew that the following resnltfl are true; la (a+d+c)"+a^+6«+c"==(a+&)«+(6+tff4.(-c)(«-ft)(4&-e) 34. (a+«*+(a+c)«+(a+rf)«+(6+p)>4^4^+^+||(l 36. {(o^+W+C^y-^^WK^^+W-C^y+W} \ FACTORS. 4» XT, Faetor$. 90. In the precedinff Chapter we have noticed some general resnlts in Multipfication; these results may also be regarded hi connexion with Division, because every ex« I ample in Multiplication furnishes an example or examples in JMyision. ne shall now api>ly some of these results to find what expressions will divide a given expression, or [in other words to resolve easpretsions into theirfactors. 91. For example, by the use of formula (3) of Art 82 I we have Hence we see that a^~5^ is the product of the four factors 0^+6*, a* +6*, a+b, and a— 6. Thus «•-&* is livisible by any of these £Eustors, or by the product of any two of them, or by the product of any three of thenu Again, Thus o'+a^-f 6^ is the ivoduct of the two factors f+ab-hb^ and a*-ab+b% and is therefore divisible by bitherofthem. Besides the results which we have already given, wo " now place a few more before the student. 92. The following examples in ^vision may be easily Ified. 1, sobnu T.A. 50 FACTORS. Also V o^-rf^ '«*-«*y+«y'-y*, x-vy _ZIl = 4^8 - a^ + a%» - a^ + any* - y«, cn-^-y and so on. Also and so on. a?-¥y^ a-k-y w+y =iiA-a!y+f/^f <^a^-iifly+a^i^^ay^+y^. The student can carry on these operations as far aa lie pleases, and he will thus gain confidence in the truth of ^e statements which we awXL now make, and which are strictly demonstrated in the higher parts of lai^r works onAlgebitt. The following are the stotements: iif — y" is diyisible by a?— y if n be any whole number ; af^—y is diyisible by ar+y if n be any even whole number; a^+y*vi divisible by a; + y if n be any odd whole number. We might also put into words a statement of the forms of the quotient in the three cases; but the student will most readily learn these, forms by looking at Ihe above examples and, if necessary, carrying the operations still &rther. "We may add that x*+y^ is never divisible by a+y or ;r~y, when ra is an 0r«9» whole number. 93. The student' will be asldsted in vemembering the results of the preceding Article by noticing the simplest FACTORS. 51 case in eadi of the four resiiltB, and referrluff other eases to it ?or ezanvple^ suppoie we wish to eonnder whether \t^"-y' 18 diyisible oy x—y or by x-\-y\ the hidex 7 is an \odd whde number, and the simplest case of this Und is \x-y, which is divisible by x-y^ but not by x-^y; so we infer that sff-y' Sa di?isible hj x-y and not by x-¥-ff. lAg^. take sfi—f^; the index 8 is an even whole nnmbcn*, land the simplest case of this kind is «*— ^. whiVh is ^visible both by x-y and x+y; so we infer that u, -|^ diyisible both by x-y and x+y^ 94. The following are additional examples of resohiny expressions into &ctors. afi-j^={a^-hy')(a^-ffy ={x+y)(a}*-xy-\-y^(x"y)(x*+xy+y'^; 86'-27c>=(26)S-(3c)8=t(2&-3c){(2&)«+-26x3+6&c+9c«); :{2a&+2crf+a«+&«-<^-fiK}{2a&+2crf--a«-6«+»+«-2i V 34. (a*+8-256. 47. a*+9a6+206« . 49. (a+6)"-llc(a+&)+30c". 50. 2(«+y)«-7(«+y)(a+ft)+3(a+ft)« Shew that the following results are true; 51. (a+26)a»-(6+2a)ft»=(a-ft)(a+6)». 62.- a(a--26)»-&(6-2a)»=(a-6)(a+6)». 38. aj'+ll^+SOi 40. ^-20^+100t» 42. a!«-7a?-44. 44. 0^+125. 46. ^•-64. 48. «*-13a?y+4?y*. \i GBBATSST COMMON MEASURE. 55 XIL Chreatat Common Meature* 97. In Artthmetio a whole nmnber which dtvldefl another whole number exnoUy is said to be a meaaurB of it, or to iMOiure it; a whole number which divif^es two or more whole numben enetly is Mid to be a eommm fiMOffUfv of thenu In Algebra an ennreBaion which diyides another ex- prenion ezacUj is saia to be a meoiura of it, or to memmon measure of simple expressions. The following is the Rule for finding the a.aic. of gmple expressions. Find by Arithmetw the acic. qf numerieal coefficients; qfter this number put every tter which is ^pmmon to all the expressions, and give each letter respectively the lea^t tndex which it has the expressions^ . 66 OREA TEST COMMON MEASURE. 100. For example; reauired the 0.0.11. of 16a^ and 20cfi9^. Here the nnmencal coeffidentiare 16 and 20, And their g.o.m. is 4. The letters common to both the expressions are a and hi the least index of a is 8, and the least index of 5 is 2. Thus we obtain 4a^ as the re- quired O.O.M. Again; required the o.o.m. of 8aVter on Factors, will enable the student to determine in many cases the o.o.m. of com- pound expressions. For example; required the, O.C.M. of 4a"(a+6)« and 606 (a«- 6"), Here 2a Is the o.o.if. of the factors 4cfi and 6a&; and a+b is a factor of (a+t^f and of a* -6', and is the only common factor. Tne product 2a(a + b) is then the o.o.m. of the given expressions. But this method cannot be applied to complex ex- amples, because the general theorv of the resolution of expressions into factors is beyond the present stage of the iitudenVs knowledge; it is therefore necessary te adopt GREATEST COMMON MEASURE. 07 toother method, and we shall now glye the ufoal definition and role. IDS. The following may be giren as the definition of the greatest common measure of compound exprasfions. LH two or more compound e»pre$Hont contain power$ qf iome common letter; then the /actor Let A and B denote the two ex- B) A ( jr^' ' V Sressions. Divide A hj B\ let /i enote the quotient, and C the re- mainder. Divide B by C\ let q de- note the quotient, and 2> the remam- dcr. Divide G by i>, and suppose that Uiere is no remainder, and let r JO) C [r denote the quotient rD Thus we have the following results: -4=i>-B+(7, B=qC+D, C=rD. We shall first shew that /> is a common measure of A and B. Because C=^rDy therefore D measures C\ therefore, by Art. 106, D measures ^C7, and also 0^(7+ J9; that is, D measures B, Again, since D measures B and (7, it measures pB+C; that is, D measures A, Thus D measures A and B. We have thus shewn that J) ia a common measure of A and B; we shaU now shew that it is their grecUett common measure. By Art 106 every common measure of A and B mea* surest ~p^, that is C\ thus every common measure of A and ^ is i^ common measure of B and (7. Similtely, every common measure of B and (7 is a conmioii In^^Miff GREATEST COMMON MEASURE. of C and D. Therefore eyery oommon measure of A and ^ is a measure of 2>. But no exfiression of higher dimen- sions than 2> can divide D. Therefore Z> is the greakii common measure of ^ and ,0. 108. It If obyiouB that, evefjf measure qf a common meoiureqftwo or more ej^etsiont is a common measure qf those expressions, 109. It is shewn in Art 107 that every common measure of ^ and B measures D; that is, every common measure of two eaipressions measures their greatest com* mon measure. 110. We shall now state and exemplify a rule whick is adopted in order to avoid fractioni) in the quotient; by I the use of the rule the woik is ^mpMed. We refer to the Chapter on the Greatest Common Measure in the laiiper I Algebra, for the demonstration of the rule. . Before placing a firesh term in any quotient, we mait [divide the divisor, or the dividend, by, any expression ''ifhieh has no factor which is common to the expressions Dhose greatest common measure is required; or, we my multiply the dividend at^such a stage by any eX" session which has no factor VuU occurs in the divisor, 111. For ezam^e; required the ao.H. of 2^-r74T+6. id 3a^-7«+4. Here we take 2aj*-7«+6 as divisor; rat if we divide Za^ by 2a^ the quotient is a fraction; ta ivoid this we miQtiply the dividend by 2, and then divide. 2««-7«+6j6aj*-14«+ 8^3 6«*-21a?+16 7«- 7 If we now make *lx-l a divisor and 2d!*- 7^+ 5 the ividend, the first term of the quotient will be fractional; ut the fiictor 7 occurs in every term of the pr(^K>Bed Ivisori and wo remove this, and ihen divide. I 60 GREATEST COMMON MEASURK Thus WQ obti&i «-l as the O.O.M. required. Here it will be seen that we used the second pait of the rule of Art. HO, at the beginning of the process, and the first part of the rule later. The first part of the rule should be used if ]{K>ssible; and if not, the second part We have used the word easpression in stating the rule, but in the examples which the student will have to solye, the factors introduced or removed will be almost always rk^ mertMl /acton, as they are in the preceding example. ^ We will now give another example; required the q,cm, of 2«*-7«'-4«'+«-4 and 3a?*-ll«*~2^-4«-16. Multiply the latter expression by 2 and then take ii for dividend. 2jjP*--7«'-4«*+»-4^ 6ar*~22a;»- 4aj"-8ar-32 (,3 6a?*-21«»-12ir8+3d?-12 - «»+ aiJ*-ll«-20 We may multiply every term of this remainder by -^-l before using it as a new diyisor; that is, we may chauge the sign of every term. «»-8«*+lla?+20j2«*-^ TdJ*- 4«*+a?-4 ^24r+9 2«*-16a!8 + 22aj«+40a^ 9aj8-264J*- 39«- 4 9a?'-72aj»+ 994?+ 180 46a»--138i»-18^ ^ere 46 is a &ctor of every term of the remainder; we remove it before usmg the remainder a^ a new div&or. EXAMPLES. XIL 61 -6a"+l&»+20 Thus 0^-30^-4 is the O.O.M. required. 112. Suppose the origmal expressions to contain a common fieictor F, which is obvious on inspection ; let A=aF and B=^bP. Then, by Art. 109, -PwiU be a factor of the O.O.H. Find the g.oIm. of a and 5, and multiply it by F'f tibie product will be the o.o.m. of A and B, 113. We now proceed to the o.o.u. of more tlian two compound expressions. Suppose we requi^ the O.O.M. of three expressions Ay B, C. Find the q.om. of any two of them, say of ^ and B; let 2> denote this O.O.H.; then the O.O.M. of 2> and C will be the required o.o.ic. of J, B, and (7. For, by Art 108, every common measure of 2> and Oh a common measure of A, By and C; and by Art 109 every common measure of A, B, and C7 is a common measure of D and (7. Therefore the a.o.M. of /> and (7 is the o.ojc. of Ay By tixid, G, 114. In a similar manner we may find the o.o.m. of \four expressions. Or we may find the o.o.m. of two of the given expressions, and also the o.o.if. of the other two; then the o.o.m. of the two results thus obtained will bo the aax. of the four given expressions. Examples. XIL . Fmd the greatest common measure in the following Examples: 1. 15«*, 18«>. 5. 4(«+l)^ 6(4»"-l). 2. 16aV, 20a^bK 4. ZSan^aVy 49a«6*«V. 6. 6(«4-l)», 9(«»-l). 62 13. 14 15. 16. 17. 18. 19. m 21. 22. 23. ,1/24. 25. 26. 27. .^ 29. 30. 31. 32. 33. 34. 35.; EXAMPLES, XIL 12(fl?+JV, 8(a^-&*). a aJ»-y», a*-y*. aj'+Sar+lS, iii>+9a;+20. «»-9«+14, «»-lldr+2a a^+2»~120, aj»-2ii?-80, * a^-15d? + 36, «"--9j?-36. oj* + 6** + 13a? + 12, 4;» + 7aj» + 16d? + 16. «»-9;B*+23a?-12, «'-l(to»+2ai?-16. a^-29a?+42, «3+aj»-36a?+49. ;88-41«-30, «*-ll«»+26a?+26. «*+7«*+l7« + 16, ;c»+8«'+19«+12. ;ij'-10a?*+26«-8, «'-9«' + 23a?-12. i 4(aj*-«+l), 3(a?*+^+l). ^^ ^ 6(««-a?+l), 4(aj«-l). ^ 6«'+ii[;-2, 9«'+48«*+62d?+16. *— ^ «^-4aj»+2«+3, 2aj*-9«' + 12^-7* «*+«'-6, a?*-3a?'+2. M «^- 2a?* +30? -6, a?*-aj8-«*-2a?. «*--l, 3a?*+2a?*+4aj'+2aj'+ar. aJ*-9aj*-30a?-25, aj"+a?*-7aj*+6a?. 36a?»+47a;«+13a?+l, 42ii?*+4l«3_9;B2-.9;,,«i. «• - Sa?* + 6a?* - 7*8 + 6«* - 3a? + 1, a?*-aj*+2a?*-a!»+2a?»-;t+l, 2a?*-6a;'+3a?«-3aj+l, a?-3a;*+a?'-4a;»+12a?-4. a?*-l, a?^®+af4^+.2a!^ + 2a;* + 2j;*+aj*+4?-l^l, ir»-3a?-70, a;S_39;,.+7o, ^^s. 43,^+7^ a?*— ajy-12y«, a?*+5ajy+6y'. ^ • ?a?«+3aa?+a«, 3af'+2aa?-^al ; ' " aj*-3a'a?— 2a', aj*-aaj*— 4^. 3dj'-3a?V+^'-"^> 4a?V-"5^+y''. /\^. LEAST COMMON MULTIPLE. 63 Xllt Least Common MtUtipU, 116. In Aiithmetic a whole number which is measored by another whole nnmber is said to be a multiple of it ; a whole number Which is measored by two or mor^ whole nmnbem is said to be a common multiple of them. . ^^ 116. In Arithmetic the least common multiple of two or more whole numbers is the least whole number which is measured by than all The term least common multipk is also used in Algebra, but here it is not very apinropriate; see Art 98. The letters L.ojf. will often be used for shortness instead of this term. We have now to explain in what sense the term is used in Algebra. 117. It is usual to say, that by the least common mul- tiple of two or more simple expressions, is meant the leaet eo^esnon which i> measured op them all; but this defi- nition will not be fully understood until we have giyen and exemphfied the nile for finding the least common multiple of smiple expressions. The following is the Rule for finding the L.o.if. of simple expresswns. Find by Arithmetic the L.o.if. qf the numerical eoeffidente; after this nun^ber put every letter which occurs in the expressions^ and give to each letter respectively the greatest index which it has in the ex- pressions, 118. For example; re<|uired the I1.0.M. of 16a^ and ZOd^lfld, Here the numencal coefficients are 16 and 20, [and their L.O.M. is 80. The letters which occur in the ex- Ipressions are a, h, c, and d; and their greatest indices are ~ ipecUyely 4, 3, 1, and 1. Thus We obtain SOa*¥cd as the luiredLbCM. Ag^ ; required the L.O.M. of Sa^c^a^yj^, 12a*6cj^^, id Wmfiifiy^. Here the l.o.m. of the numerical coefficients 48. 1%e letters wluch occur in the expressions are s, hf Cf^fty^ and z; and theur greatest indices are respec- Ively 4^ 3, 3, fi, 4, and 3. Thus we obtain 48a*WajV^ « requhed X1.0.M. „ 64 tEAST COMMON MULTIPLE. 119. The following statement gives the best prnetical notion of what is meant by the tenn least common multiple in Algebia,- as it shews the sense of the word koit here^ When the least common multiple qf two or more ewpree^ ifione ii divided by thote expreetione the quotients have no dommon measure, » Take the first example of Art 118, and divide the L.0.1C. by the expressions; the quotients are SUM and 4ac^ and meae quotients haye no common measure. Again; take the second example of Art. 118, and ditide the L.aM. by the expressions; the quotients are Qa^Ci^f 4i^!i?ysi?^ and ^at^a^afi^ and these quotients haye no com* mon measure. iT ; 120. The notion which is supplied by the i>receding« Article, willi the aid of the Chapter on Factors, will enable the student to determine in many cases the l.o.m. of com- pound eamressions. For example, reqmred the l.o.m. of 4€fi(a+bf and edbia^-h'). The l.o.m. of Atfi and edb is I2€^b, Also (a -f by And a* ~'&* have the common factor a+b, so that (a+b){a+b)(a-b) is a multiple of {a+bf and of a^—b^; and on diyiding this by (a+ 5)* and a'-d* we obbiin Uie quotientB a—b and a+ (, which haye no common pleasure. Thus we obtain 12a%(a+d)'(a~&][ as the re* quired ii.aM. 121. The following may be giyen as the definition qf the L.O.M. qf two or more compound expressions. Let two or more oompopkd expressions contain powers of some common letter; then the expression of lowest dimeni|ions in that letter which is measured by each of th^ expres- sions is called their least common multiple. 122. We shall now shew how to find the itSijL i^f two compound, expressions. The demonstration howeyer will not be fiilly understood at the present stage of the 8tm!9.^s knowledge. ;.; L^ A and B denote the two expressions, ffreatest common measure. Suppose A^al>^ . iXh^ from the nature of tiie greatest common \ LEAST COMMON MULTIPLE^ ^ and h have no common fiietor, and therefore their least common multiple is ah* Hence the expression of lowest dimensions ^hich is measured by oD ana hD is ad/>. And dbD=Ah^Ba='^ . , Hence we have the following Rule fbr finding the L.o.if. of two compound expressions. Divide the product qf the expreuUma hy their o.o.m. Or we may give the rule thus :^ Divide one qf the expressions by their G.O.M., and mid' tiply the quotient by the other expression, 123. For example; required the L.C.V. of a^— 4^+3 and 4a^ - 9a^-l6x + 18. The G.O.M. is a?— 3; see Art. 105. Divide o^— 4^7+3 by xS; the quotient is x—l. Therefore the l.o.m. is (a?-l)(4i»-9«*-16a?+18); and this gives, by multiplying out, 4d?*-13a:»-6aj2+33d?-18. It is however often convenient to havo^the L.o.ic.. expressed in factors, rather than multiplied out. We kuow that the O.O.H., which is a;— 3, will measure the ex- pression 4a^-da^-l5x+lS; by division we obtain the quotient Hence the L.O.H. is (a?-3)(a?-l)(4aj*+3«-6). For another example, suppose we require the L.O.M. of 2«* - 7« + 6 and 3dJ» - 7a? + 4. The O.O.M. is a?- 1 : see Art. 111. Also (2a!«-7a?+5)-J-(«~l)=2a?-6, and (3aJ*--7«+4)-r(a:-l)=3;i?-4. Hence the luOM. is (a?-l)(2a?-6)(3«-4). > Again; required the L.O.M. of 2a?*-7«'— 4«*+d?^4, id 3«*-.ll«»-2a:*-4a?-16. The o.o.]f. is a;*— da;-4: see Art. 111. Also i2a^-1afi- 4a^+ii?-4)4-<«'- 3«-4) = 2«"- x+l. Id ^-ll«»-2«■-4«-K)■^(«'-3a?-4)=3«■-2i^?+4. ! T.A. 5 06 LEAST COMMON MVlTIPtlB. Henoe thd L.O.X. is («»- 8«-4) (2««-ar+ 1) (8««-2i+4). 124. It is obvious that, werjf rnidtipU ^f a multiple qftwo or more ewpressiam ie a eommof^ ^thoeeexpreeeU/M, 125. Every common mtdtiple of two Mpr9$iion94ia multiple qf their least common multiple* Xet A and B denote the two expressions M their Zi.o.if.; and let N denote any other common multipla Sop- pose, if possible, that when N is divided by M there is a remainder iS ; let g denote the quotient Thus B^N—qM. Now A and B measure JIf and N, and therefore they mea- ■iire B (Art 106). Bilt bj the nature of divbioii K is of Uwer dimensions than Jf ; and thus there is a oommon multiple of A and B which is of lower dimensions than tiieir L.O.H. This is absurd. Therefore there can be no remainder iS; that is, iVis a multiple of M. 126. Suppose now that we require the L.ojf. of three compound expressions, A^ B%0, Find the l.o«m. of any two of them, say of A and J7; let itf denote this ii.o.kc.: then the l.o.x. of M and C will be the required L,p Ji. of ^^^^anda For every common multiple of M and is a common multiple of A, B, and C> by Art 124. And every oonunon multiple of A and J9 is a multiple of Jf, by Art 125 ; hence evenr common mul^le of M and C is a common multiple of Ay Bj and O, l%erefore the l.o.v» of M and C7 is the I..O.X. of ^, J9, and C ' 127. In a similar manner we may ^d the tu^M, ct lour Expressions. 128. The theories of the greatest oOnunonmeasaHie and of the least common multiple are not neeessaiy for the subsequent Qhapters of the present work, and any diffi- culties which the student may find in them tttiiQr bo post- poned until he has read the Theory of Equations* /Hie examples however attached to the preceding (3hai^ and to the present Ohapter should be carefollv woHceo^ on a«Q- count qf the exercise which they afford In all tbe ln|^ mental'processes of Algebra. 1 v.- EXAMPLBS. XUl «r BZAMPUBB. Jtlll. Bind Ad leiit common multiple in the following ex- ampltii; . - 1. 4afh, edf^. ' 2. 12a«6«c, 18a5V. -5. 4 «*-6«»+2a»-ia - 14. /p*+a?«*+a*, «*-a«*-a'«+a*. '16. 4a'Wc, 6a6»c«, I8a«6c». la B(€?"V), 12(«+&)", 20(a-&)". 17. 4(a+d)/ 6(a^~(>), 8(a''+&'). SHHl la 16(a*6-afi^), 21(0^-06*), 35(aJ»+6^* la «^^1, ii^+1, «»-l, 2a fl^^l, ^+1, aj*+l, aj8-l. 21. ««-l, «»+l, i»»-l, «»+l, '22, #+t»+2, «>+44?+3, flj^+^p+e. 2a 4^+2ir-3, «»+3«"-»-3, a^-^4afl+a-e, 24. #+6«+I0, «»-19iJ?-30, «»-16;»-60. 6—2 ea FRACTIONS. XIY. FraeHoM, 129* In this Chapter and the foHowing four Oha^len we shall treat of Fractions; and the student wfll finathat the roles and demonstrations closely resemble thoie with which he is already familiar in Arithmetic 130. ' By the expression 7 we indicate that a nnit is to be cUvided into b equal parts, imd that a of snch parts i^« to be taken. Here r is called a /^tustion; a is called the numerator, and h is called the denominator. Thus the denominator indicates into how many equal parts the unit is to be divided, and the numerator mdicates how many of those parts are to be taken. Every integer or integral expression may be considered as a fraction with unity for its denominator; that is, for eza^lde, a I «=-, 131. In Alffebra. as in Arithmetic, it is usual to ffive the following Rule lor expressing a fraction as a m&ed quantity: Divide the numerator by the denomiwttor, at far at poetible, and annex to the quotient a fraction hating the remainderfor numerator, and the divieor for denominator^ Ezamples. 24a ^ , 9a 7 7 a'+3a& 2db =/? + «»-6ay+14 a+y =«+3+ -df+2 «"'-3«+4 or s«+8— «*-3«4-4 «"-3«t4* \- 'J-'^^^ m FRACTIONS. 6» The rtodent is reoommended to pay parHetdar aitsn- Hon to the last step; it is reallT an example of the use of brackets, namely, +(-*+25=-(a7-2)L 132. Rule for multiplying a fraction by an integer. Either multiply the numerator by that integer, or divide the denominator by that integer. Let r denote any fraction, and o any integer; then will 7 X e=s -r- . For in each of *hi fractions ^ and -g- the nnit is divided into b equal parts» and e times as many parts are taken in -T- as in ^; hence -r* is c times r . ! This demonstrates the first form of the Rule. Again; It ^ denote an, f^<». «.d « «^ mteg.; then will ^^^=5* ^^ hi each of the i^?actions ^ and ? the same number of parts is taken, but each part . fai ? is tunes as laige as each part in r- , because in r the unit is divided into e times as noany parts as in g ; hence £ is 6 times T-« This demonstrates the second form of the Bule. 133. Rule for dividing A fraction by an integer. Either multiply the denominator by that integer, or divide the numerator by that integer. a Let r denote any fraction, and cany integer; then will y^e-Ti* For r is c tunes ^, by Art 132; and therefore^ is -th of f . oe e. This demonstrates the first form of the Rule. FSdOnONS. ae Again; let y denote any fraotion, and any Inti0|(fr; ae ae then wiU y -{- 134. J[f the numerator and denominator ^fanyfroi^ Hon be mvltipUed by th^ eame integer, the value qf ih$ fratOionie not altered. 'Vtft if the nnmerator of a fraction be mnltiptied by any integer^ the firaction win be mtdtiplied by that integer; and the result will be divided by that integer if its de- nominator be multiplied by that faiteger. But u we multiply any number by an integer, and then diyide the resultl>y the same integer, the number is not altered. The result may also be stated thus: if the numerator and denominator of any fraction be divided by the same integer, the yalue of the fraction is not altered. Both these verbal statements are included in the alge- This lesult is of very great importance ; many of the operations in Fractions dej^nd on it, as we shall see in the next 4^0 Chapters. 135. The demonstrations given hi this Chapter are satisfactory only when every letter denotes some poeitive whole number; but the results are ateumed ioloe true whatever the letters denote. For the grounds of ^s assumption the student may hereafter consult tiie lai]ger Algebra. The result contained in Art. 134 is the most important; the student will therefore observe that heiice- forth we assume that it is ahcaye^^ana!^ in. Algebra that r = T-, whatever a, &, and e may denote. ' ■ \ I Foi^ezampley if we put - 1 for c we have 7 = — ^ ^ : .. P SXAMPLE& XIV. 71 Soalfo £1 like maimer, by amainfag that t >< ^ ii afwayt equal 00 to y we obtain ...h re«ilt. M tb. following : a - -2a EZAMPUfl. XIV. Express the Mowing fractions as mixed quantities : 25* ^ 3eae+4e 8a*-f36 lag'-Sy 6. «* + * - 3a'#— 3a' 2^-6df-l *-3 a— 2a a 9. Multiply a(a-&) 10. 12. 8(a«+ft^ 9i?::p)^y^(*-^>- la «^ by 4(a«-a&+y). 14, . >,_^ bydy+I IMvide 15. gby2ar. ld(a^-y) 3(a+&) 16. 9a'~4y a+& («*-!)• by3a-2&. 17. 18. by6(a'+a6+J^. bya^-«+l. .72 REDUCTION OF FRACTIONS. XV. RedudHon of FractioM. 136. The result contained in Art. 134 will now be applied to two important operations, the reduction of a mtctiop to its lowest terms, and the reduction of fractions to a common denominator. 137. Rule for reducing a fhtction to its lowest terms. Divide the numerator and denominator of the fraction hy their greatest common measure. For example; reduce , ,^ , to its lowest terms. The o.o.M. of the numerator and the denominator is ia^*; dividing both numerator and denominator by 4a'&', we obtain for the required result -r^. That is, r^ is equal to ^^ . , but it is expressed in a more simple form; and it is said to be in the lowest terms, because it cannot be further simplified by the aid of Art. 134. ■ij^"~"4^H"3 ' Again ; reduce 4^,9^^15^4.18 *^ ^^ ^^^^^ *®™*- ' The G.aif. of the numerator and the denominator is a—S; dividing both numerator and denominator by «— 3 a? — 1 we obtain for the required result j^ — _ _ , \ In some examples we may perceive that the numerator and denominator nave a common factor, without using the rule for finding the g.o.m. Thus, for example^ ' \ (a~&)*— c* _ {a-J>-rc)(a-h—c) _ a—b+o BEDUCTION OF FB ACTIONS. 78 ttow be )n of a ractiona t terms. s. [nator is by 4a»6«, simple )caase it t terms. 1 138. Rule for reducing fractions to a common denomi- nator. Multiply th€ numerator qf each fraction hy j i &nd ^ to a common de- nominator. a_a^ c _^ e _ebd > b^biif* d^dlff* f'fbd' Thus a^f clff and ehd are fractions of the same e AC value respectively ^ Xf 2* *^^ ?' ^^ ^^^ ^^® ^^ common denominator b^f. The Rule given in this Article will always reduce frac- tions to a common denominator, but not always to the lowest common denominator; it is therefore often con- venient tb employ another Rule which we shall now give. 139. Rule for reducing fi«ctions to their lowest com- mon denominator. Find the least common multiple qf the denominator»f and take this for ,the common denomi- nator; then for the new numerator corresponding to any qf the proposed fractions, multij^^ the numerator qfthat fraction hy the quotient which is obtained by dividing the least common multiple by the denominator qf that fraction, l^fst example; reduce — , — , — to the lowest com- *^ yz^ zx^ say mon denominator. The least common multiple of the de- nominatoFB is xyz\ and an ayz* zx xyz e scy ez ocyz' 74 MXAIBPLBS. Xr. N.,., EZAMPLBS. XV, Rodace iho following fractions to their lowest terms: 1. 4. 7. 9. II. 13. 15. 17. 19. 21. 23. 25. 2. 5. 12a*W« 18a«6«y' 10a*a? 5a'«— 16ay2* ;i^+3a?+2 .'c2+6«+5* 2a^+a?-15 2«2-i9aj + 35' dg'-(a+&)j?-t-a6 a?'»-10a?+21 ;»»-46a?-2l' «3~10«=*+21a? + ia' 20a^+a?-12 12«»-6«« + 6a?-6' 2iB»-6«*-8a?-16 2a6 • 4(a+&)> 3. 6. aj"+10a?4-21 8. 10. 12. )4, 18. Y.20. af2-2a?-16* /g*-i-(a+5)d?+« + 33a?-27' d?^+5;i? + 6 ;«.•* + a? +10' a?* + 9a?+20 aj3 + 7a!*+14;r+8' e^j-11^+5 3d?»-2i5*-r d;^-2g ;g-t - 79 \uantity or Bub- ih •60? + 14 -3a?+4 * addition ^(aH6^, The bednner should pay particular attention to this example. He is very liable to take the product of the denominators for the common denominator, and thus to render the operations extremely laborious. The second fraction contains the factor h—a in its de- nominator, and this factor differs from the &ctor a-h, which occurs in the denominator of the first fraction, only in the sign of each term ; and by Art. 135, & h {h-'C){h-ar (b-e){a-by Also the denominator of the third fiuction can be put in a form which is more convenient for our object ; for by the Mule of Signs we have (c-a)(c-&)=(a-* 6» <^ (a-6)(a-c) "^ (6-a)(6-c) "^ (c-a)(c-^ _ 1 , 1 «*— (&+c)ii?+fcc' 55. '«+« «+& flj'-(a+ft)«+a6 «*-(a+c)«+a6) _ (a--6)» ft«(a+ft)« "6(a+6)' i I 8 C C DIVISION OF FRACTIONS. 89 Invert of the a 151. Complex fractional ezpreaaions may be simplified by the aid of some or all of the roles respeoting fractions wnich hare now been given. The following are examples. Simplify /^+^U(^-^|. '^ ^ \a-h a-^h) {a-h a+b) a+b ' a~&_ (a4-ft)«+(a-6y _ 2a'4-2y a-b^a+b" {a-b)(a+b) ~ 4fi-V ' a+b a-b (g+&)«-(g~6)' 4a& «-6 a+b ia^b)(fl+b) "a«--6»* 2a»4-2y . 4db 2a«+26« a«-5» _ a»4-y a2_/^ • a«-j»~ a»-6» ^ 4ab " 2ab ' In this example the factors a—b and a+& are mtdU- lied together, and the result a*-&> is used instead of a+6)(a— &); m general however the student will &ftd it advisable not to mtdtiply the factors together in the course of the operation, because an opportunity may occur of striking out a common factor from the numerator and denominator of his result f. ^-v. Simplify a+ 14 a+l 3-a - a4-l _3-<8 g+l _ 3~q+g+l 3-a""3--a'*"3-a"' 3-a 1^ 4 _1 3-:a_3--a 3-a 14 4 * 3-a 4a . 3-a 3 + 3a , 3+3a 14 4 1-r — :; — = r X 4 3-a* 1 3+3a 3+3a* 90 DIVISION OF FRAOTIONa. . 2ab a 2a&~a(a+5) a&— a' 2^— a=s — _— _s i 1 ss r-: 2ar-6« "■ a+ft • a+6 1 a+b Therefore 2x-b a+b a+b ^ a+b ^ ab-l^ tlierefore ab-a* ajp—a) '^db-l^^bia-by / 2g~a Y_ / a\*_ ^ \2af'-bJ''\b)'''b** a b' . , a db a(a+b)-ab a^ Again, a-4r«j-— j=-^^^j3 ^^5 . _b db b(a+b)-iib ^ V l^a+b" a+b a+h' a— « a* a+b Therefore ^— ^= — x-«--^x= -tt * / b'-a a+b a+b a+b Therefore ^^ _ __ = t _ _ =o. 6«' agam 152. The results given in Art. 147 most be given '- ^ re in connexion with Division of Fractious. ae ae Since ?x *-5=-£j, and -iX^=-?^ bd ae b d bd ae we have -rj-^ — j=rj a»d — rj-T- j=— i- »(]{*<;& bd' d ae Also dnce -¥ x - js= £3, we have ^ M' ae id' \ BXAMPZSa. XVttL 91 V -6^ 1 jfiyen BxiMFLu. xyni. TA, ' 1 1 8. 3 — 3 by . *• a(a+ft)«'^?a(a^-t^' 6. 8«* by 4«' 9. 10. 11. 12. .1 ^+»ff±y^ by «^-!^ jy*— 8j?+2 - «"-5»+6 ('*3('-J) by y 13. ««»-^Jby«+^ 14. oF-^bya-- 15. ** er9 of the eqnation. The ezprawion to the left of the sign of eanality la called the JUrH aide^ laid the ezpreasion to the rignt ia called the 9eGond aide. 1C4. An idenHeal equation ia one in which the two aides are equal whateyer numbers the letters represent; for ezamplei the followhig are identical equations, t (ir+a)(«-a)=:«*-a^, («+a)(«'-«a+a«)=fl^+a^ ; that is, these algebraical statements are true whatever numbers a and a may represent The student will see that up to the present point he has been almost ezchisiTely occupied with results of this Idnd, tbskt is, with identical equations. An identical equation is called briefly an identity. 155. An equation qf condition is one which is not true whateyer nunibers the letters represent, but only when the letters represent some particular number or numbers. For example, a +1 — 7 cannot be true unless 47=6. An equation of condition is called briefly an equation, 156. A leifter to which a particular yalue or values must be giyen in order that the statement contained in an equation may be true, is called an unknown quantity, Buch particular yalue of the unlpiown quantity is said to eoil^ the equation, and is called a root qf the equation. To eitlve an equation is to find the root or roots. 157. An equation inyolying one unknown quantify la said to be of as many dimensions as the index (^ tlie faighest power of the unknown quantity. Thus, if m dehote i \ lation. if the Ion to Belaid e two Mont; \ liever ill see flively ntiou true when iben. An flues in an Bach fthe SIMPLE BQUATION& M the unknown quantity, the equation is said to be of oim dimension when a oocnrs only in the ^rH power; snob an equation is also called a $impU equaitonf or an equation of the /irtt degree. If «* occurs, and no higher power of «, the equation is said to be of two dimensions; such an eouation is also called a guadroHo equationy or an equation of the $eeond degree. If «* occurs, and no higher power of Wf the equation is said to be of three dimensions ; sui^ an equation is also called a eubie equation^ or an equation of the third degree. And so on. It must be obsenred that these definitions suppose both members of the equation to be integral expreenone eojar a$ r^atee to x. 158. In the present Ohapter we shaU shew how to soke simple equations. We have first to indicate some opera- tions which may be performed on an equation without destroying the equality which it expresses. 159. If every term on each side qf an equoHon he multiplied ly the tame number the reeuUe are equal. The truth of this statement follows from the obyious principle, that if equals be multiplied by the same number the results are equal ; and the uee of this statement will be seen immediately. Likewise if wery term on each tide qf an equation^ be divided by the tame number the retultt are equal, 160. The principal use of Art 159 is to dear an eqwh turn qf fractumt; this is effected hj multiplyinff every term by the product of all the denominators of t£e frac- tions, or, if we please, by the least common multiple of those denominators. Suppose, for example, that ^846 Multiply every term by 3 x 4 x 6 ; thus 4 X 6 X 07 + 3 X 6 X iT + 3 X 4 X « = 3 X 4 X 6 )$ 9, that is, 24«+1&i;+12«b648; divide evcv^ term by 6 ; thus ^ .4;p+3a;+2j9=108. 96 SIMPLE EQUATIONS, Instead of multiplying every term by 3 x 4 x 6, we may multiply every term oy 12, which is the l.o.m. of the deno- mimitors 3, 4, and 6 ; we shoiild then obtain at once 4^+3^+2^=108; , \ that is, 9^=108; divide both sides by 9 ; therefore ^ 108 ,o ^=— = 12. Thus 12 is the root of the proposed equation. We may renfy this bv putting 12 for x in the original equation. The first side oecomes i IQ IQ |0 V + ^ + ^, thati84+3+2, thatis9; o 4 o which agrees with the second side. 161. Any term may he transposed from one side qf an equation to the other side by changing its sign. Suppose, for example, that x—a^h-y. Add a to each side ; then x—a+a^h-y+a^ that is x^h—y-ha. Subtract h from each side ; thus Here we see that —a has been removed from one , side of the equation, and appears as + a on the other side ; and +& has been removed from one side-and appears as — & on the other side. 162. If the sign of every term qf an equation he changed the equality still holds, \ This follows from Art. 161, by transposing every t^nxi. Thus suppose, for example, that ;r—a~o—2f« SIMPLE EQUATIONS. 97 9 may deno- emay latioii. ide qf ,side and n he By tra]U^K)Bitioxi that 18, a-~w=y-'b; and this result is what we shall obtain if we change the sign of eyery term in the original equation. 163. We can now give a Rule for the solution of any simple equation with one unknown quantity, dear the equation of jf'ractionSj if necessary; transpose all the terms which involve the unknown quantity to one side cf the eqwUion^ and the knovon quantities to the other side; divide both sides by the coefficient^ or the sum qf the co- ejfficients, qfthe unknown quantity^ and the root required u obtained, t 164. We shall now give some examples. ' Solye 7«+26=»35 + 5a?. > Here there are no fractions ; by transposing we have 7a?-6a?=35-265 that is, 2a?=10; divide by 2; therefore w=-^=5. We may verify this result by putting 6 for « in the original equation; then each side is equal to 60. 166. Solve 4(3^-2)-2(4a?-3)-3(4-a?)=0. Perform the multiplications indicated; thus 12d? - a- (Sa? - 6)- (12 - 3a?) = 0. Bemove the brackets; thus 12a?-8-8ii?-4-6-12 + 3i»=0; collect the terms, 7«?-14=0; transpose, *Jaf=l4; 14 divide by 7, a- = 2. - The student will find it a useful exercise to verify the correctness of bis solutions. Thus in the above example^ 98 SIMPLE EQUATIONS. / if we put 2 for x in the original equation we shall obtun 16-10-6, that is 0, as it should be. 166. Solye «-2-(2a?-3)=^-^. Remove the brackets; thus fl?-2-2a?+3=— ~, 3a? + l 2-2a?==3a?+l, 2-l = 2a?+3a?; l=5a?, or6ii?=l; t 1 o tliatisy multiply by 2, transpose, lihatis, « therefore 167. Solve 6a?+4 = 6}-- 10 28 5}= — ; the L.O.M. of the denominators is 10; multiply by 10; thus that is, transpose^ that is, therefore 6(6a?+4)-(7a?+6)=28x2-6(aj-l); 26ar+20-7d?-6 = 66-6a? + 6; 25a?-7^+6^=66 + 5-20 + 6; 23^=46; 46 « X— — =2. 23 The begmner is recommended to put down all the work at fuU, as in this example, in order to ensure ar^curacy. Mistakes with respect to the si^ns are often made u\ clear- ing an equation of fractions. In the above equation the faction -1^ ha. to be multipHed by 10. «d i^ I. ad- visable to put the result first in the form —(7^+5), and afterwaids in the form — 7«-5, in order to secoie Iktten- tioa to the signs. y> tl ti tl J th tr cl tl tl EXAMPLES. XfX 99 obtain lultiply 168. Solve |(&»+3)-^(16-5«)=37-4«. By Art 146 this is the same as Multiply by 21; thus 7(6iP+3)-3(l6-64?)=21(37-4p), that is, 364?+21-48 + 16««777-84«; > transpose, 35;i;+ 15a; +84^=777 -21 +48; that is, 134a;=804; therefore 169. Solve 804 ^ ^=134=^- , 607+15 a»-10 Ax-l 11 7 5 • Multiply by the product of 11, 7, and 5 ; thus 36(6a?+15)-55(a»-10)=77(4a?-7), that is, 210a7+525-440a;+550=30&v-539; transpose, 210a? - 440a? -30a»=- 639 -525 -560; change the signs, 440a?+308a?-210a;=539 +525+ 550, that is, 5380?= 1614; 1614 therefore 0?= 638 =3, 1^ > work iuracy. dear- \xi the is ad- ), and Itten- KXAMPJJES. XIX. 1. 5a;+50=4a?+56. 2. 16a?-ll=7a7+70. 3. 24a?-49=19a?-14. 4. 3a? +23 =78- 2a?. {fc y{a?-18)=3(a?-14). 6. 16a? =38 -3 (4 -a?). 1 7(a?-3)=9(a?+l)-38. 8. 5 (a? -7) +63= 9a?. »t 59(a?-7)=61(9-a?)-2. 10. 72(a?- 6)=^ 63(5-0?). X . X 11. '28(o?+9)=27(46-a?). 12. «+o+3-ll 7-2 100 EXAMPLES. XIX. Of X 5 3 21. ~ + 12=^+6. 3 o 83. '|-6=^*-a 4 O 14. ~+24=2rf?+.6. 4« 16. 36-~=a 3^ 5dT ^ 18. ^+fi=T + 2. .4 6 2a? 176 --4a ? "3" fi~' 24. f-8=74-g. 26. f-.^=*-2. 27. 4(«-3)-7(«-4)=6-a?. 3""3 4'*"4'"6~6"6^6' ' X 2a? So? 30. 2a?- 19-2a? 2a?-ll 2 2 „ a?+l 3a?-l' 31. — = = — =a?-2. 3 92. 0?+ 3a?~9 =4- ffa?-12 S3. 34. 10^+3 6ar-7 3 2 3 • . = 100?- 10, 5a?-7 2a?+7 3 =30?- 14. 36. 4?-l o?-2 . a?-3 '.A>'-.-;.'^ 2 3 =a >,-'v 6. 2. *» w*' n -;=■- »"«* J >i i '*.- V . I EXAMPLES. XIX. »+3 . «+4 . «+5 101 37. 88. 89. 40. ,«• 48. 44. 45. 46. 47. 7«+9 8 =7+«- 4 r=l«. 8a?— 4 ftp--5 _ 8a?-l 2 """i 16 • «-8_«--5 «— 1 4 """T""*'T"* m tliatb, By t»n«podath i < SIMPLE MQUATION& 174 Sohe-j^-jj^ + gj^j. 105 Here it ie oonyenient to multiply by ix-¥i, that ii by 4(«+l); therefore 8»+12-4»-5=:i^^l .V thatiis M;^tiply by 8«+ 1 ; thus (SdT-l- l)(4«-f 7)== I2(«+ 1)*; thatis, ' 12Ai^+2&v+7»12d^+24p+12. Subtnust 12«* from each glde^ and transpose; thus 2&9-24cv=12-7, that is, 47=5.^ 175. Bdye Wehaye And d^-2 "" «-3 ~ «— 6 ~ «— 6* J?-l 4?--2 ^ (jp-l)(jp~3)-(a?-2y »-r2~«-3 («-2)(i-8) («-2)(«-3) "■(aT-2)(«-3)* «-5 ar^e fl^~lftg-f24-(jg>-l(to4.25) ^ (;»-e)(«-6) (4T-6)(«-6) 1 ■(«-5)(«-6)* Thus tibe proposed equation becomes " • • - 1 L 1 "'(«-2)(«-3) («-5)(«-6)' 106 BIMFUt SqUAnONS. OhMge the «ign.,.tlia. ^j_^j_„^_jj^. Clear of fraotioiiB; thus («-5)(«-6)=:(tf-S)(«-8}f ■» f ttiatifl, therefore that is, therefore therefore a^-lld;+S0=«'-&»+6 ; 6a7=:24; n^ Soke *Q«4 •4&r-'75 1-2 -a^-'B I •« •2 •9 To ensure aoenraey it is advisable to express all the dedmahi as oommon finictions ; thus -I 10 6 \100 100/ " 2 ^ 10 9 \10 10/' ««p«^. M(i-i)=«-(M)' - thatis, ^^??-5=6_£ + ?, ^ 2 4 4 3 3 Multiply by 12, 6^-f 9a;-15-72-4a7+a; 1907=72+8+16=95; ^ 96' ^=i9=«- therefore 177. Equations may be proposed in which letUir$ are used to represent known quantities ; we shall oontmiie to represent the unknown quantity by x, and any other tetter wul^ be supposed to represent a known quantity. We ^ will sdye three such equations. i .Vv«, SIMPLE MQlTATTOim 107 \ 17a CMto f+f«a lAiH^by a&; thus (dr+AVaoBt diyidebya+d; thiu 179. Sohro (a+«)(d+«)-a(6+c)+^+^. Here a5+aar+ddr+«««a6+ac+^+««; 1 ! therefore diyidebya+ft;thn8 «=2. 18a. Solve ^ = (f ;) ; . Clear^ of fractions : thus (ar-a)(ai?-&)«=(af-ft)'(ai?-a)«j thatia, («-a)(4i?«-4^+ft2)=(a.-ft)(4aj»-4^+<^ Multiplying out we obtain 4aj»-4««(a+6)+«(4a&+5*)-ai!?« s=4iJ»-4aj«(a+&)+«(4a&+«^-a«&; therefore af^-dt^z^axfi^aHf; therefore a?(a«-d^=a«d-a6«=a&(a-&); .a&(a~ft) db f s X- a«-^ ~a+d" v» MXAMFLBSk XX. 181. AUhoiigh the fbOowiM MpMilta bekms toJOie proNiit Ohnter w^ ghrt it no duBoaltj in ibllowing tbe ileiM aT «tM win www M a model for ifanikr Kfomblee those ab«edy solred, hi the we obtehi onlj a Wf^i;^ nhie of the Mve iy«+V(«-l«)«a Sy tnmspodtioii, ^(a-lB)^S'-^a; ■qnarebothiideB; thiie «-16=(8-V«?=e4-l6,/«4-#; therefore -16~64-16V«; *«n»poii^ 16^a?r=64+16»80| i therafora ijxsilii Qi»nivK*'v <• ^mh^RWf J * ^PijivlM 11. 109 #-1 7#-21 IB. #-3-(8-#)(«+l)»#(«-8)+a le. JI-«-«(«-l)(«-l-2)-(«-8)(5-2l «4-12 4y~l «--5 15-2» 9-» 7 4 "* 82 ■*"^'^"' 2~"8' /.J 28.. ^+*(«-2)«(«-l)«. 24. ^+(«-l)(«-2)^«»-2i»-4U ^ 8g'-2jp~8 (7ir~2)(ag-6) ■1 EXAMPLES. XX. 29. «-5 «-6 a?-8 *«— 9' 30. i» «-9 a?+l a?-8 31. 3-2a? 2ii?-6 4aj>-l l-2a? ^-7~ 7-16a?+4«8* 32. 3+a? 2+a? 1+a? - 3-iP 2-a? l-o?"" • 33. d?-6 as^+6 «2-2 aj'-ar+l 7 3^2 6 34. Oc+l)(x+2)(a!+Z) f3. 35. 36. // 37. 38. 39. 40. 41. « 43. 44. 45. = (a;-l)(a?-a)(a?-3)+3(4«-2)(a?+l). («-9)(a?-7)(«-6)(4?-l) =(a?-2)(a?-4)(aj-6)(a?-10). (8a?-3)2(a?-l)=(4a?-f')2(4a?-6). as^-a+l . aj2+a?+l „ — ^ _ — =2«'. •6a? - 2 = •25a? + '20? - 1. •54? + '60? - '8 = '75^ + •25. •135a?-^225 '36 •09;»-^18 •15a? + •6 •2 '9 • .- a?-a ,a?45 42. a— r^ + 6 — ^ = 0?. a a— a ,6+0? a —J- o =0?. a x{pD''a)-{-x{x-'b)=2{x-a){ps-h), \ ' (a?-a) (« - J) (« + 2« + 26) «(a?+2a) (a?+26) (4T-, a ft a— 6 111 48. 49. 50. 51. tB-^a x-^b W'^c* a h _ a-\-h I 1 a-b x—b «*— oft' 1 1 x—a 1 fnx—a—b mx—a-^c n nx—c—d" nx—b^d* 52. {a-b){x-c)-(p-c)(x-a)-'{C'-thatis56. 184. I^ivide £2, 10s. among A, B, and C, so that B mav have 5s. more than A, and C may have aa much as A and B together. Let X denote. the number of shillings In A*b share, then a+5 will denote the number of shiUings in B*m share, and 2^+5 will denote the number of shillings in 0*b share. PROBLEMS. ilZ The whole number of shillings is 50 ; therefore «+« + 6+2aT+6=605 that is, 4d;-l- 10=50; therefore 4a;=50-10=40; therefore a; =10. Thus A*% share is 10 shillings, ZTs share is 15 shillings^ and (7's share is 25 shillings. 185. A certain sum of money was divided between Ay B, and C; A and B together receiyed ;£17. 15«. ; A an :PROBLBMS. 100- a; lbs. is ^(100-^) shillings.. And the whole value is to b^3 ^ X 100 sinkings; therefore 5x100= 2a? + 5(100-0?); multiply by 2, thus 600 = 4a; + 700 - 7^ ; therefore 7d?-4iP= 700-600; that is, 3a?=200; 200 therefore «=• a \ Thus there must be 66|lbs. of the first sort, and 33ilbs. of the second sort. < ■ • ■ , " 187. A line is 2 feet 4 inches long; it is required to divide it into two parts, such that one part may be three- fourths of the other part Let X denote the number of inches in the larger part ; then -7- will denote the number of inches in the other part. 4 The number of inches in the whole line is 28 ; therefore 3^ «o «?+-j=2S; therefore 4a?+3a?=112; that is, 7a?=;=112; therefore a?=16. Thus one part is 16 inches long, and the other part 12 inches long. 188, A person had £1000, part of wlHch he Jent at 4 per cent., and the rest at 6 per cent.; the whole annual interest received was j£44 : how much was l^t at 4 per cetot. I Mm. PROBLEMS. lis per Let X denote the number of pounds lent at 4 per cent ; then 1000— d; will denote the number of pounas lent at 5 per cent The annual interest obtained from the former is 7^ , and from the latter mn^ * 100 therefore therefore that isy therefore 100 4^ 5(1000~d?) , 100"^ 100 ' 4400 = 4a? + 6(1000-a?); 4400 = 4^ + 5000 - 5aT ; a?=6000-"4400=600. Thus ;£600 was lent at 4 per cent 1 I 189. The student will find that the'only difficulty in solving a problem consists in translating statements ex- pressMl in ordinary language into Algebraical lan^^uage; and he should not be cuscouraged, if he is sometmies a little perplexed, since nothing but practice can give him readiness and certainty in this process. One remark may be made, which is very importont for beginners; what is called the unknown piantity is really an unknown number^ and this should be distincUy noticed in forming the equa- tion. Thus, for example, in the second problem which we have solved, we begin by sayii^, let x denote the number of shillings in A^% share; beginners often say, let a?=^'s money, which is not definite, because A*% money may be expressed in various ways, in pounds, or in shillings, or as a nuction of the whole sum. Again, in the fifth problem which we have solved,^ we begin oy saying, let x denote the number of inches in the longer part; oeginners often say, let a;= the longer part, or, let a; = a part, and to these phrases the same objection applies as to that already noticed. 190. Beginners often find a difficulty in translating a problem from ordinary language into Algebraical language, because they do not understand what is meant by uie ordinary language. If- no consistent meaning can be as- signed lo the words, it is of course impossible to translate them; but it often happens that the words are not ab- 8—2 EXAMPLES. XXL solutoly nnintelligiblejVut appear to be sosoeptible of more than one meaning. The student should then select one meaning, express that meaning in Algebraical symbok. and deduce from it the result to which it will lead, jlf the result bo inadmissible, or absurd, the student should try another meaning of the words. But if the .result is satis- factory he may mfer that he has probably understood the words corrector ; though it may still be interesting to try the other possible meanings, in order to see if the enun- ciation really is susceptible of more than one meaning. 191. A student in solving the problems which are eiven for exercise, may find some which he can readily solve by Arithmetic, or by a process of g^ess and trial ; and he may be thus inclined to undervalue the power of Algebra^ and look on its aid as unnecessary. But we may remark, that by Algebra the student is enabled to solve all these problems, without any uncertaintv ; and moreover, he will find as he proceeds, that by Algebra he can solve pro- blems which would be extremely- difficult or altogether impracticable, if he relied on Arithmetic alone. / Examples. XXI. 1. Find the number which exceeds its fifth part by 24. % A father is 30 years old, and his son is 2 yeak old : In how many years will the fiEither be eight times as old as the son? 3. Tho differonco of two numbers is 7; and their sum is 33 : find the numbers. 4. The sum of .£165 was raised by -4, B, and tog^ ther ; B contributed ^15 more than A, and C ^20 more than B : how much did each contribute i 5. The difference of two numbers is 14, and thefa* sum is 48 : find the numbers. \ , S. AiA twice as old as B^ and seven years ago thoir united, ages amounted to as many years as now t epCM ii it the age of .4: find the ages of ^ and A . ^liv > . • .■VS',!^;^;"^.-;,; EXAMPLES. XXL 117 moxo P8IIIII thoir 7. If 5d be added to a certain number, the result is treble &at number: find the number. 8. A child is bom in November, and on the tenth day of December he is as many days old as the month was on the day of his birth : when was he bom ? 9. Find that number the double of which increased by 24 exceeds 80 as much as the number itself is below 100. 10. There is a certain fish, the head of which is 9 inches long; the tail is as long as the head and half the back ; and the back is as long as the head and tail toge- ther : what is the length of the back and of the tail ? 11. Divide the number 84 into two ^arts such that three times one part may be equal to four times the other. . 12. The sum of £76 was raised by Ay By and C toge- ther; B contributed as much as A and j£10 more, and (7 as much as A and B together : how much did each con- tribute? 13. Divide the number 60 into two parts such that a seventh of one part may be equal to an eighth of the other part. 14. After 34 gallons had been drawn out of one of two equal casks, and 80 gallons out of the other, there remained just three times as much in one cask as in the other: what did each cask contain when full ? 15. Divide the number 75 into two parts such that 3 times the greater may exceed 7 times the less by 15. 16. A person distributes 20 shilling among 20 per- sons, giving sixpence each to some, and sixteen pence each to the rest : how many persons received sixpence each ? 17. Divide the number 20 into two parts such that the sum of three times one part, and five times the other pari^ may be 84. I8«^> The price of a work which comes out in parts is £% ld». 8^. ; out if the price of each part were 13 pence more than it is, the price of the work would be £Z, 78. 6d, : how niany parts were there 1 19. Divide 45 into two parts such that the first divided by It fdiaU be equal to the second multiplied by 2. 118 EXAMPLES. XXL 20. A father is three times as old as his son; four years affo the father was four times as old as his son then* was : what is the age of each ? 21. Divide 188 into two parts such that the fourth of one part may exceed the eighth of the other by 14. 22. A person meeting a company of beggars gave four Eence to each, and had sixteen pence left ; ne found that e should have required a shilling more to enable him to give the beggars sixpence each : how many beggars were there? 23. Divide 100 into two parts such that if a third of one part be subtracted from a fourth of the other the re- mainder may be 11. 24. Two persons, A and B. engage at play; A haS; £*J2 and B has j£d2 when they begin, and after a cei*t»in • number of games have been won and lost between them, A has three times as much money as B : how much did A wini 25. Divide 60 into two parts such that the difference between the greater and 64 may be equal to twice the difference between the less and 38. 26. The sum of ;£276 was raised by A, B, and C^ toge- ther; B contributed twice as much as A and £,\2 more; and C three times as much as B and £12 more: how much did each contribute ? 27. Find a number such that the sum of its fifth and its seventh shall exceed the sum of its eighth and its twelfth by 113. 28. An anny in a defeat loses one-sixth of its number in killed and wounded, and 4000 prisoners ; it is reinforced by 3000 men, but retreats, losing one-fourth of its number in doing so ; there remain 18000 men : what was the ori- ginid force 1 29. Find a number such that the sum of its fifth and its seyenth shall exceed the difference of its foiu*th and its seventh by 99. 30. One-half of a certain number of persons receiye^ eighteen-pence each, one-third received two shilUngs each^' and the rest received half a crown each ; the whole sum distributed was £2. 4#. : how many persons were ^ere t . EXAMPLES, XXL 119 SI. A peraon had ^^00 ; part of it he lent u Ihe rate of 4 per cent^ and part at the rate of per cent, and he received equal sums as interest from the two parts : how much did he lend at 4 per cent. ? 32. A father has six sons, each of whom is four years older than his next younger brother; and the eldest i» three times as old as the youngest: find their respectiTO ages. 33. Divide the number 92 into four such parts thai the first may exceed the second by 10, the third by 18, and the fourth by 24 34. A gentleman left ;£560 to be divided among four servants A,BfC,D; of whom B was to have twice as much OS A, V M much as A and B together, and D as much as C and B together : how much had each ? 35. Find two consecutive numbers such that the half and the fifth of the first ieiken together shall be equal to the third and the fourth of the second taken together. 36. A sum of money is to be distributed among three persons A, B^ and C; the shares of A and B together amount to £60 ; those of A and C to £80 ; and those of B and C to £92 : find the share of each person. 37. Two persons A and B are travelling together ; A has £100, and B has £48; they are met by robbers who take twice as much from A as from Bf and leave to A three times as much as to ^ : how much was taken from each ? 38. The sum of £600 was divided among four persons^ so that the first and second together received £280, the first and third together £260, and the first and fourth together £220 : find the share of each. 39. After A has received £10 from B he has as much money as B and £6 more ; and between them they have £40: what money had each at first ? 40. A wine merchant has two sorts of wines, one sort worth 2 shillings a quart, and the other worth Ss. Ad. a quart; from these he wants to make a mixture of 100 quarts worth 28, id, a quart: how many quarts must he take from each sort f 120 EXAMPLES. XXI. 41. In a mixture of wine and water the wine composed 25 gallons more than half of the mixture, and the water gallons less than a third of the mixture : how many gal- lons were there of each ? 42. In a lottery consisting of 10000 tickets, half the number of prizes added to one-third the number of blanks was 8600 : how many prizes were there in the lottery ? 43. In a certain weight of gunpowder the saltpetre composed 6 lbs. more than a half of the weight, the sulphur fflbs. less than a third, and the charcoal 3 lbs. less than a fourth : how many lbs. were there of each of the three ingredients ? 44. A general, after having lost a battle, found that he had left fit for action 3600 men more than half of bis ; army ; 600 men more than one-eighth of his army were wounded ; and the remainder, forming one-fifth of the army, were slain, taken prisoners, or missing : what was the number of the army ? 46. How many sheep must a person buy at £7 each that after paying one shilling a score for folding them at mght he may gain £19, I6s, by selling the^::^ at £S each ? 46. A certain sum of money was shared among five persons A, B^ C, D, and E; B received £l(i less than A ; t7 received £\^ more than B ; 2> received £6 less than C\ and E received ;£16 more than D ; and it was found that E received as much as A and B together : how much did each receive ? 47. A tradesman starts with a certain sum of money ; at the end of the first year he had doubled his original stodc. all but ;£100 ; also at the end of the second year he had doubled the stock at the beginning of the second year, all but j^lOO; also in like manner at we end of the third year ; and at the end of the third year he was three times as rich as at first : find his original stock. 48. A person went to a taveni with a certain sum of money ; there he borrowed as much as he had about him, and spent a shilling out of the whole ; with the remainder he went to a second tavern, where he borrowed as nluch as he had left, and also spent a shilling ; and he then went to a third tavem,- borrowing and spending as before, after which he had nothing left : how much had he at first ! c o 81 n ii tl m m a- th is th th mi th th PROBLEMS. 121 XXII. Prciblemif continued. 192. 'We shall now giye some examples In which the process of translation from ordinary language to algebrai- cal language is rather more difficult than in the examples of the preceding Chapter. 193. It is required to diyide the number 80 into four such parts, that the first increased by 3, the second dimi- nishea by 3, the third multiplied by 3, and the fourth divided by 3 may all be equaL Let the number x denote the first part; then if it be increased by 3 we obtain •;»+ 3, and this is to be equal to the second part diminished by 3, so that the second part must heof+S; again, d7+3 is to be equal to the third part multiplied by 3, so that the third part must be — ^— ; and «+ 3 is to be equal to the fourth part dirided by 3, so that the fourth foxt must be 3(a: + 3). And the sum of the parts is to be equal to 80. Therefore that Is, that Is, fl?+a?+6-l-~^+3(j?+3)=80, 2a?+6+^|^+3ii?+9=80, 5«+^=80-16=66; multiply by 3; thus 16a?+a5+3=196, that Is, 16a?=192; a— — =12. 16 therefore Thus the parts are 12, 18, 5, 45. I2i PROBLEMS. I94k A alone can perform a pieoe of work in 9 dam and B alone can perform it in 12 days : In wh«t time wUi they perform it if they work together 1 Let w denote the reqnhred number of dayi. In one.day A can perform ^ th of the work ; therefore in w daya he can perform ^ths of the work. In one day B can perform l^th of the work; therefore in x days he can perform gether perform the whole work, the ram of the fradioM of the work must be equal to «mroblems of a more difficult character than those hitherto given. 200. It is required to divide the number 56 into two parts such that one may be to the other as 3 to 4. Let the number a denote the first part; then the other part n^ust be 56 -x; and since a;i8tobcto56-«iras3to4 we have a 8 66-^"" 4' Clear of fractions; thus 4^=3 (56 -a?); that is, 4a?=168-3a?; therefore 7^=168; therefore «=-^=24. Thus the first part is 24 and the other part is 66-24, that is 32. The i>recedinff method of solution is the most natural for a beghmer; w.e following however is much shorter. Lot the number 3x denote the first part; then the second put must be 4d;, because the first part is to the second as 3 to 4. Then the sum of the two parts is equal to 56: thus that is, therefore 3a?+4d;=56, 7a?=56j Thus the &nt part is 3 x 8, that is 24; and the second partis4x8,thatis32. 126 I'ROBLEMS. 201. A cask, A, oontainB 12 gallons of wine and 18 gallons of water; and another cask, B^ contains 9 gallons of wine and 3 gallons of water: how many gp&Ilons must be (kawn from each cask so as to produce oy their nqisture 7 gallons of wine and 7 gallons of water ? '^ Let X denote the number of gallons to be drawn from A\ then. since the mixture is to consist of 14 gallons, 14-^ will denote the number of gallons to be drawn from J?. Now the nimiber of gallons in ^ is 30, of which 12 are 12 wine; that is, the wine i<9 35 of the whole. Therefore the X gallons drawn from A contain -r— gallons of wine. vO Similarly the 14 - a? gallons drawn from B contain ^ \Z \ gallons of wine. And the mixture is to contain 7 gallons of wine; therefore m 9(14-0?) 30 ^12 :' thatis, 2x Z(U-x) 6^ 4. "-'* therefore 8a?+16(14-a?)=xl40, thatis, 8a:+210-15d?=140; therefore 7^=70; therefore «?=10. Thus 10 gallons must be drawn from Ay and 4 from B, 202. At what time between 2 o'clock and 3 o'clock is one hand of a watch exactly over the other 9 Let X denote the required number of minutes. after 2 o'clock. In X mimttes the long hand will move over X divisirTis of the watch 'feuse; and as the long han4 o^oyes twelve times as &st as the short hand, the short hwd wUl move over r^ divisions in « nunutes. At 2 o'clock the PROBLEMS. 127 Bhort hand is 10 divisions in advance of the long hand; so that in the x minntes the long hand must pass over 10 more divisions than the short hand; therefore therefore therefore therefore ^=^ + 10; 12d;=«+120; lla?=120; 120 ,^,o the 203. A hare takes four leaps to a g^honnd's three, but two of the sreyhound's leaps are equivalent to three of the hare's; the hare has a start of fifty leaps: Yam many * ^leaps must the greyhound take to catch the hare? Suppose that 307 denote the number of leaps taken by the greyhound; then Ax ^U denote the number of leaps taken by the hare in the same time. Let a d^iote the num- ber of inches in one leap 4^ Jlie hare; then 3a denotes the . number of inches in three leaps of the hare, and therefore also the number of inches in two lei^ of the greyhound; therefore ~ denotes the number of inches in one leap of the greyhound. Then Zx leapa of the greyhoiypd will con- 3(1 ^ • tain 307 x~ inches. And Hd-^Ax leaps of the .hare will contain (50+ 4or}a inches; therefore Divide by a; thus -^a60+4o7; .therefore 9oj=100+8o?; ... therefore o;=100. Thus the greyhound must take 300 leaps. The student will see tbat we have introduced an auzl- liaiy symbol a, to enable us t6 form the equaUon e^ily; and tmit we eaa- remove it by division when the equation is formed. 128 EX4MPLE8. XXII. 204. Four gamestera, A, B, C, D, each with a ^^Ifiimt stock of money, sit down to plav; A wiiis half of I^wt stock, B wins a third part of C\ C wins a fourth part of jD's, and D wins a fifth part of ^'s ; and then each of thio gamesters has £2^. Fmd the stock of each at first ' Let X denote the number of pounds which D won from A\ then 6x will denote the number in A'% first stodk. Thus 4ar, together with what A won from By mi^e up 23; therefore 23— 4a; denotes the number of pounds which' ^ won from B, And, since A won half of JErs stock, 23—4^0 also denotes what was left with B after his loss to A, Again, 23-4ar, together with what B won from C, make up 23 ; therefore Ax denotes the number of pounds which B won from C, And, since B won a third of C*f^ first stock, \2x denotes (7's first stock; and therefore 8« denotes what was left with C after his loss to B, Again, ^x, together with what C won from D, make up 23; therefore 23— 8a; denotes the number of pounds which O won from 2>. And, since C won a fourth of D's first stodi, 4(23—80;) denotes i>*s first stock; and therefore 3(23— 8a;) denotes what was left with D after his loss to (7. Finally, 3 (23 -8a;), together with or, which D won from ^1 make up 23; thus 23=3 (23 -8a?) + a?; therefore therefore 23a;=46; a;=2. Thus the stocks at first were 10, 30, 24, 28* Examples. XXII. > 1. A privateer runidng at the rate of 10 miles an hour discovers a ship 18 miles off, running at the rate of 8 miles an hour: how many miles can the ship run before jt is overtaken 1 . 2. Divide the number 60 into two parts such that if three-fourths Qf one part be added to five^tbs of the other part the sum may be 40. EXAMPLES. XXIL 129 from hotv [miles It is 3« SuppoM the distance between London uid'Bdin- liiirghvis 360 miles, and that one traveller starts from Edinboigh and travels at the rate of 10 miles an hour, while another starts at the same time from London and travels at the rate of 8 miles an hour : it is required to know where tiiey will meet. 4. Find two numbers whose difference is 4^ and the difference of their squares 112. 5. A sum of 24 shillings is received from 24 people ; some contribute M. each, and some IS^dL each : how many contributors were there of each kind ? 6. Divide the number 48 into two parts such that the excess of one part over 20 may be three times the excess of 20 over the other part 7. A person has ;£98 ; part of it he lent at the rate of 6 per cent simple interest, and the rest at the rate of 6 per cent simple interest; and the interest of the whole in 15 years amounted to j£81 : how much was lent at ^ per cent? 8. A person lent a certain sum of money at 6 per cent simple interest ; in 10 years the interest amountea to £Vl less than the sum lent : what was the sum lent ? 9. A person rents 25 acres of land for ^^7. 12«. ; the land consists of two sorts, the better sort he rents at 8t. per acre, and the worse at 5«. per acre : how mapy acres are there of each sort % 10. A cistern could be filled in 12 minutes bj two pipes which run into it ; and it would be filled in 20 minutes bv one alone : in what time could it be filled by the other alone ? 11. Divide the number 90 into four parts such that the first increased by % the second diminished by 2, the third multiplied by 2, and the fourth divided by 2 may all be equfd. . 12. A person bought 30 lbs. of sugar of two different sorts, and paid for the whole 19f. \ the better sort cost 10r him- Bed bT which of that 4 per ual in- as the loser half quan- le and how ( 66. A clodc has two hands turning on the same centre ; the swifter makes a revolution every twelve hours, and the slower every sixteen hours: in what time will the swifter £^ just one complete revolution on the slower? 67. At what time between 3 o'clock and 4 o'clock is one hand of a watch exactly in the direction of the other hand produced? 68. The hands of a watch are at right angles to each other at 3 o'clock: when are they next at right angles? 69. A certain sum of money lent at simple interest amounted to ;£297. 12«. in eight months; and in seven more months it amounted to £306 : what was the sum ? 60. A watch sains as much as a clock loses; and 1799 hours by the dock are equivalent to 1801 hours by the watch: find how much the watch gains and the clock loses per hour. 61. It is between 11 and 12 o'clock, and it is observed that the number of minute spaces between the hands is two-thirds of what it was ten minutes previously: find the tune. 62. A and B made a joint stock of j£600 by which they ^ined £160, of which A had for his share £32 more than n\ what did each contribute to the stock? 63. A distiller has 61 gallons of French brandy, which cost ;iim 8 shillings a gallon; he wishes to buy some £n- gli * brandy at 3 shillings a gallon to mix with the French, and sell the whole at 9 shillings a gallon. How many gal- lons of the English must he take, so that he mav cain 30 per cent on what he gave for the brandy of both kinds? 64. An officer can form his men into a hollow square 4 deep, and also into a hollow square 8 deep; the front in the latter formation contains 16 men fewer than in the former formation: find the mmiber of men, ' 186 SIMULTANEOUS SIMPLE EQUATIONS XXIII. Simultaneout equatiom df thejint degree with two unknown quantitiet, 206. Suppose we have an equation containing two un- known quantities a aiid y^ for example 30^-7^=8. For every value which we please to assign to one of the unknown quantities we can determine the corresponding value of the other ; and thus we can find as many pairs of values as we please which satisfy the given equation. Thus, fbr example, if y=l yre find 3^=15, and therefore «=5; if y=2 we find 8^=22, and therefore a=H; and soon. Also, suppose that there is another equation of the same kind, as for example 2a; +6^ =44; then we can alsb find as many pairs of values as we please which satisfy this equation. But sui)pose wo ask for values of a and y which satisQr both equations; we shall find that there is only one value of a and one value of y. For multiply the first equation by 5; thus 16a?-36y=40; and multiply the second equation by 7 ; thus 1407+35^=308. Thereforei by addition, 150^-352^+1407+35^=40+308; that is, 2907=348; therefore 348 ,„ ^=29=^2. Thus if both equations are to be satisfied a mutt equal 12. Put this value of w in either of the two given equations, for example in the second; thus we obtain \ . . - 24 + 5^=44; therefore 5y=20; theretbre y=4. ONS. SIMULTANEOUS SIMPLE EQUATIONS. 187 9e foith two un- B. For of the }ondiDg ly pain luation. lerefore rj; and of tlie ian al86 sfythis i satisfy e yalue ^uation >?■■ iiall2. ations, 806. Two or more equations which are to be satisfied bj the same values of the unknown quantities are called itmultaneotu equatiofui. In the present Chapter we treat of simultaneous equations inyolving two unknown quanti- ties, where each unknown Quantity occurs only in the first degree, and the product of the unknown quantities does not occur. 207* There are three methods which are usually giyen for solvinff these equations. There is one principle com- mon to allthe methods; namely, from two giren equations containing two unknown quantities a single equation is de- duced containing only one of the unknown quantities. By this process we are said to eliminate the unknown quan- tity which does not appear in the single equation. The single equation contaimng only one unknown quantity can be solved hj the method of Chapter XIX ; and when the value of one of the unknown quantities has thus been determined, we can substitute this value in either of the given equations^ and then determine the value of the other unknown quantity. 208. First method. Multiply the equatione hy $ueh numbers as will make the confident qf one qf the unr known quantities the tame in the resulting equations; then hjf addition or subtraction we can form an equation containing only the other unknown quantity. This method we used in Art 205 ; for another example^ suppose 8«+7y=100, M 124;-6y=88. If we wish to eliminate v we multiply the first equation by 5, which is the coefficient of y in the second equation, and we multiply the second eq|uation by 7, which is the coefficient of y in the first equation. Thus we obtain 40.r+35^=500, '84ar-35^=6I6; therefore, by addition, 40a?+84a?=500 + 616; that is, 124^=1116; therefore «=9. 138 SIMULTANEOUS SIMPLE EQUATIONS. Then put this yaloe of iP in either of the given equatiom^ for example In the second ; thus 108-6ys=88; therefore 20= 5y; ^ therefore y-4. Supposo, however, that in solving these equations we wish to begin by eliminatmg x. If we multiply the first equa- tion by 12, and the second by 8, we obtam 964^+84^=1200, 96a?-40y=704. Therefore^ ly itibtraeticn^ 84y+40y=1200-704; that is, 124y=496; therefore y=4. Or we may render the process more simple ; for we may multiply the first equation by 3, and the second by 2; thus 24a;+21y=300, ' 24a?-10y=176. Therefore, by subtraction, 21y+l()y=300-176; that is, 31j(=124; therefore y=4. 209. Second method. Expreti one qf the unknown guantitiei in termt (^ the other from either equation, and eubetitute this value in the other equation. Thus, taking the example given in the preceding Arti- cle^ we have firom the first equation therefore a»=100-7y; ,_100-7y \ X- 8 .J^^i^ BaB^aa. SIMULTANEOUS SIMPLE EQUATIONS. 139 irti- Substitate this yaliie of ^in the second equation, and we obtain 12(100-7y) g wy-oo, that is, 8Ji00z!LL5y=88; therefore 8(100-7y)-10y=l76; that is, 300-21y-10y=176; therefore 300-l76=21y + lQy; that is, 31^=124; therefore y=4. Then substitute this value of y in either of the given equa- tions, and we shall obtain or «: 9. Or thus : from the first equation we have 7y=100-8a?; 1OO-&0 therefore y = — = — . Substitute this value of y in the second equation, and we obtoin therefore that is, therefore therefore ,2^_«000^)^885 84a?-5(100-8ar)=616; 84^-500+40^=616; 124^=500 + 616=1116; fl?=9. 210. Third method. Eapren the tame unknown quantity in termt qf the other from each equation, and equate the expremom thut obtained. Thus, taking again the same example, fix)m the first 100—72/ equation «= — o~^} • , ., Multiply the second equation by h^ and subtract it from the first; thus that is, therefore Substitute this yalue of or in the second equation; thus ■ . i a— 6 a— a— " 6(a— o) 6(6-a) Or the yalue of y might be found in the samd way as that of » was found. EZAMPLBS. XXIII. 1. Za—4y-2f 2. 7«-6y=24, 3. 847+2y=:32, 4. ll«-7y=37, 5. 7«+5y=60, 6. 6«-7y=42, 7. 10»+9y=290, 8. 3«-4y=:18, 7a>-9y=7. 4a;-3ysll. 20ay-3y=l. av+9y=41. 13ar-lly=10. 7a?-6y=76. 12«~ny=130. 3i»+2y=0. 9. 4.v-|=ll, 2«-3y=0. \ S^AMPLBS. XXIIL m 10. 11. it 18. 14 15. 16. 17. 18. 20. 21. 22. 28. 24. 8 +ay-7, 4dH-2 -4. e«-6yal, 7«-4y=8j. 2142 (6). T.A. 10 Cj I* f Ik- 146 SIMULTANEOUS SIMPLE EQUATIONS. We haye now to find the Talnes of w and y from (4) and (5). Miiltipiy (4) by 9, and multiply (6) by ; thus 226d?+l71y=1184, 2250?+ 66y= 710; therefore, by snbtractiony 106y=4245 therefore y=4. Substitute the Talne of y in (4); thus 26a?+76 = 126; i therefore 26«= 126-76=60; ^ therefore a=2. Substitute the values of47 and y in (1); thus 14+12-2af=16; therefore 10=24?; tiierefora «=6. 217. Sobo i + |-|=l ...(1), Multiply (1) by 2, and add the result to (2); thus 2 4 6 6 4 . 6 „.„; -+--:: + - +- + -=2+24: X yz X y z ' ^ that is, - + -=26 .........(4). ^ X y ^ ^ . I. {SIMULTANEOUS SIMPLE EQUATIONS 147 Multiply (1) by 8, and add the result to (3) ; thus a y z a y » that is, ?-i-' <» Multiply (5) by 4| and add the resnlt to • f.!H.U?.M+«, X y X y that is, X therefore 47=94ar; therefore Substitnte the Talne of x in (6) ; thus 20-?=l7; therefore ?=20-l7=3; therefore 2 y=3- Substitute the values of x and ^ in (1) ; - 2+3-1=1; z therefore 1-'^ therefore ; •-!• 10—2 : ; 148 SIMULTANEOUS SIMPLE EQUATIONS. 218. Solve f-H ('^ 'I « M=» (*^ 5 + 1=4 (8). Subtract (1) from (2) ; thus h e a * that is, J -^=2 (4). By subtracting (4) from (3) we obtain —=2; iiherefore -=1; therefore or = a. By adding (4) to (3) we obtain therefore -=3: therefore z=^e. c ♦ By substituting the value of a; in (1) we find thai y-^. S19. In a similar manner we may proceed if the nun^- ber of equations and unknown quantities should exceed three. ^ EXAMPLES, XXIV. 149 Examples. XXIV. 1 fi«-6y+4jf=15, 7«+4y-air=:19, 2;r+y 4- 6:2^=46. 8. 4i»-5y+4f=6, 7A'-lly4-2«=9, dr+y+3^=:12. 4. 7«-3y=30, 9y-6;»=34, «+y+;2r=33. 6. 3«-y+;»=17, C«+3y~2;»=10, 7«+4y-0;r=d. 6. x+y+z=5, 3«-5y+7jif=75, 9;v-ll;8r+lO=0. 7. «+2y+34f=6, air +4^+2;? =8, 3ay+2y+air=101. 63f— a? 8. S^=l, = 1, 9. 3;»-7 *' 2y-3z a-h2y _ 3j/+4z 6a}+Qi6 7 ■" 8 " 9 y z *' y~2^ _i 3y-2« • t a+y—z=l26» aye* X y z 11. y+;»=a, z-ha^b, a?+y=c. 12. af+y+z=a'hh+e, flj+a=y+6=«++«+y+;y=14, 2«+a7=:2y-f ;7~2, 8»-«+2y+24f=19, 8*4^6 + 2 *' * .■t«.4«r^; '.■ ■>\''.**;Jt-- 150 PROBLEMS. XXV. PrdbUmt which lead to iimultaneom equaiiont f^thejlrtt degree with more than one unknown quantity, 220. We shall now solye some problems which lead to simultaneous equations of the first degree with more than one unknown quantity. Find the fraction which becomes equal to = when the 3 4 numerator is increased by 2, and equal to = when the de- nominator is increased by 4. i Let a denote the numerator, and y the denommator of the required fraction ; then, by supposition, y ""3' y+4"7' Clear the equations of fractions ; thus we obtain av-2^= -6 (1), 7a?-4y= IG (2). Multiply (l) by 2, and subtract it from (2) ; thus 7a?-4y-6a?+4y=16 + 12 ; that is, ^=28. Substitute the value of x in (1) ; thus 84-2y=s-6; therefore 2^=90 ; therefore ^=45. • 28 ' Hence the required fraction is 7^. 221. A sum of money was divided eaually among ia certain number of persons ; if there had oeen six more, each would have received two shillings less than he did ; and if there had been three fewer, each jv^ould have re- ceived two shillings more than he did : finclnthe number of persons, and what each received. (f,V:'', , --i^ya*. PROBLEMS, 151 Let w denote the number of personi, and y the number of shillings which each received. Then xy is the number of shillings In the sum of money which is divided ; an(^ by supposition, (a?+6)(y-2)=ajy (1), (4?-3)(y+2)=«y (2). From (1) we obtain flfy + 6y - 2ii? - 12 = «y ; therefore 6y-2a7=:12 (3). From (2) we obtain «y + 2aT-3y-6 -ary ; therefore 2x~Zy=^ (4). From (3) and (4), by addition, 3y = 18 ; therefort y^^ Substitute the value oty in (4) ; thus 2^-18=6; therefore 2x^2^ ; therefore ;i;=12. Thus there were 12 persons, and each recdved 6 shillings. 222. A certain number of two' digits is equal to five tunes the sum of its digits ; and if nine be added to the number the digits are reversed : find the niunber. Let X denote the digit in the tevi :' riace, and y the digit In the units' place. Then the numuci is 104? +tf; and,Dy supposition, the number is equal to five times the sum of its digits; therefore 10a?+y=6(aj '-:/).., (1). If nine bo added to the number its digits are reversed, that is, we obtain the number \(^ + x ; therefore 10a?+y+9=10y+i» (2). From (1) we obtain 6x=Ay (3). From (2) we Obtain 9^+9 = 9y ; therefore « + 1 =y. 152 PROBLEMS. Babstitate for ^ in (3) ; thns therefore ^=4. Then from (3) we obtain y- 5. Hence the required number is 45. 223. A railway train after travelling an hour is detained 24 minutes, after which it proceeds at six-fifths of its former rate, and arrives 16 mmutes late. If the detention had talcen ^lace 5 miles further on, the train would have arrived 2 mmutes later than it did. Find tiie original iftte of the train, and the distance travelled. \ Let 6^ denote the number of miles per hour at which the train originally travelled, and let y denote the number of miles in the whole distance travelled. Then y—bx will denote the number of miles which remain to be travelled after the detention. At the original rate of the train this distance would be travelled in ^ — hours; at the in-> / bx ' creased rate it will be travelled in \.^ y-6x eof hours. Since the train is detained 24 minutes, and yet is only 16 minutes late at its arrival, it follows that the remainder of the joumev is performed in 9 minutes less than it would have been if the rate had not been increased. And 9 minutes 9 is --of an hour; therefore y—Sx _ y—5x 9^ ex " bx "60 a). If the detention had taken place 6 miles further on, there would hbve been y-bx—b miles left to be travelled. Thus we shall find that \ y—bx—by—bx—b 2_ ex ■" bx ~60 ^(2). PROBLEMS. 153 Sabtaract (2) from (1) ; thus 6«""6a? 60' therefore 50 =80 -2a;; Uierefore 2a; = 10 ; therefore a; = 5. Snbstitate this yalue of a; in (i), and it will be found by Bolying tiie equation that ^=47^. 224. Af Bf and O can together perform a piece of work in 30 days; A and B can together perform it in 32 days; and B and O can together perform it in 120 days: find the tune in which each alone could perform the work. Let a denote the number of days in which A alone could perform it, y the number of days in which B alone could perform it, z the number of days in which alone could perform it. Then we haye a^y z 30 ^*^' 111 . . 5'*'y""32 ^^^* y"^i"l20 (^^• Subtract (2) from (1); thus z 30 32 ""480* Subtract (3) from (1) ; thus 1^2 1^1 5 30"l20 40' Therefore a; =40, and ;v=480; and by substitution in any of the giyen equations we shall find that y= 160. 225. We may obsenre that a problem may often be Bolyed in yarious ways, and with the aid of more or fewer letters to represent the unknown quantities. Thus, to take a yery simple example, suppose we haye to find two 154 EXAMPLES. XXV, numbers snch that one is two^hirds of the other, and ^ehr sum is 100. Wo may proceed thus. Let x denote the greater number, and y the less number; then we have 2a? y=g-, «+y=100. Or we may proceed thus. Let x denote the greater number, then 100— a? will denote the less number; there- fore 100-a?=j. Or we may proceed thus. Let Zx denote the greater number, then 2x will denote the less number; therefore > 2.1?+ 3a? =100. By completing any of these processes we shall find that the required numbers are 60 and 40. The student may accordingly find that he can soke some of the examples at the end of the present Chapter, with the aid of only one letter to denote an unknown quan- tity; and, on the other hand, some of the examples at the end of Chapter xxn. may appear to him most naturally solyed with the aid of two letters. As a general rule it may be stated that the employment of a larger number of unknown quantities renders the work longer, but at the same time allows the successive steps to m more readily followed; and thus is more suitable for beginners. The beginner will find it a good exercise to solve the example given in Art. 204 with the aid of four letters to represent the four unknown quantities Which are required. Examples. XXY, 1. If A*% money were increased by 36 shillings he would have ^hree times as much as ^; and if ^s money were diminished by 5 shillings he would have half as much as A : find the sum possessed by each. 2. Find two numbers such that the first with half the second may make 20, and also that the second with a third of the first may make 20. ^ EXAMPLES. XXV. 155 3. If If were to gite ;£25 to A they would have equal toma of 9(ioney; if A were to give £22 to ^ the monev of B womd be double that of A : find the money whicn each actoally has. 4 Find two numbers such that half the first with a third of the second may make 32, and that a fourth of the first with a fifth of the second may make 18. 5. A person buys 8 lbs. of tea and 3 lbs. of susptr for £1, 29, ; and at another time he buys 5 lbs. of tea and 4 lbs. of sugar for 15«. 2d, : find the price of tea and sugar per lb. 6. Seyen years ago A was three times as old as ^ was; and seven years hence A will be twice as old as J? will be : find their present ages. 7. Find the fraction which becomos equal to i when the numerator is increased by 1, and equal to i when the denominator is increased by 1. 8. A certain fishing rod consists of two parts; the length of the upper part is to the length of the lower as 6 to 7 ; and 9 times the upper part together with 13 times the lower part exceed 11 times the whole rod by 36 inches: find the lengths of the two parts. 9. A person spends half-a-crown in apples and pears, buying the apples at 4 a penny, and the pears at 5. a penny; he sells half his apples and one-third of his pears for 13 pence, which was the price at which he bought them: find how many apples and how many pears he bought. 10. A wine merchant has two sorts of wine, a better and a worse; if he mixes them in the proportion of two quarts of the better sort with three of the worse, the mixture will be worth U. 9d, a quart ; but if he mixes them in the proportion of seven quarts of the better sort with eight of the worse, the mixture will be worth Is. lOd. a quart : find the price of a quart of each sort 11. A farmer sold to one person 30 bushels of wheat, and 40 bushels of barlev for ;£13. 10«. ; to another person he sold 60 bushels of wheat and 30 bushels of Imrley. for ;£17 : find the price of wheat and barley per bushel. .rf,?'-fla^ ..* 156 EXAMPLES. XX K 12. A fEurmer has 28 bushels of barley afc Sir. 4di i bushel: with these he wishes to mix ire at 8t. a bui^V and wheat at 49. a budiel, so that the muture may consist of 100 bushels, and be worth Ss, 4d. a bushel: find hxm many bushels of rye and wheat he must take. " 13. A and B lay a wager of 10 shillmp; if ^ loses he will have as much as B will then haye; if B loses he will have half of what A will then have: find the money of each. 14. If the numerator of a certain fraction be increased b;^ 1, and the denominator be diminished by 1, the Talue will be 1 ; if the numerator be increased by the denomi- nator, and the denominator diminished by toe numerator, the yalue will be 4: find the fraction. 15. A number of posts are placed at equal distanora in a straight line. If to twice the number of them we add the distance between two consecutiye posts, expressed in fedt, thQ sum is 68. If from four times the distance be- tween two consecutiye posts, expressed in feet, we subtract half the number of posts, the remamder is 68. Find the distance between the extreme posts. 16. A gentleman distributing money among some poor men found that he wanted 10 shillings, in order to be able to giye 5 shillings to each man ; therefore he giyes to each man 4 shillings only, and finds that he has 5 shillings left: find the number of poor men and of shillings. 17. A certain company in a tayem found, when they came to pay their bill, that if there had been three more persons to pay the same biU, they would haye paid one shilling each less than they did ; and if there had been two fewer persons thOj would haye paid one shilling each more than they did : find the number of persons and the number of shillings each paid. 18. There is a certain rectangnilar floor, such thai if it hadi^een two feet broader, and three feet longer, it would tS^ been sixty-four square feet larger; but if it had]>e^J||Me feet broader, and two feet longer), it would haye been slxtj-eight square feet larger : find tfie length and breadth of tlie floor. 1 1 v i llJl ^^iertain number of two digits is equal %^ EXAMPLES. XXV. 167 timeB the sum of its digits ; and if 18 be added to the mnnber tiie digits are reversed : find the number. 20. Two digits which form a number change places * on the addition of 9; and the siim of the two numMrs is 33 : find the digits. *-- 21. When a certain number of two digits is doubled, and increased by 36, the result is the same as if the numberl had been reyersed, and doubled, and then diminished by^- 36 ; also the number itself exceeds four times the sum of its d^ts by 3 : find the number. 22. Two passengers have together 5 cwi of liuw;age, and are chained for the excess above the weight imowed ffff. 2 25. A and B can together perform a certain work in 30 days; at the end of 18 days however B is called off and A finishes it alone in 20 more days : find the time in which each could perform the work alone. ^ 26. A, B, and Ccan drink a cask of beer in 15 days; A and B together drink four-thirds of what C does ; and C drinks twice as much as A : find the time in which each alone could drink the cask of beer. f. 27. A cistern holding 1200 gallons is filled by three pipes A, By O together in 24 minutes. The pipe A requires 30 minutes more than (7 to fill the cistern; and 10 gallons less run througii C per minute than through A and B tcffiether. Find the time in which each pipe alone would iilTthe ci !«m. mmmm mni^^mmm mmmmmmm 158 EXAMPLES. XXV. 28. A and B nm a mile. At the first heat A gives B a stu^ of 20 yards, and beats him by 30 seconds. At the second heat A gives B a start of 32 seconds, and beats him by 9^ yards. Find the rate per hour at which A rnna 29. A and B are two towns situated 24 miles apart, oiw the same bank of a river. A man goes from 4 to ^ in 7 hours, by rowing the first half of the distance, and walking the second half. In returning he walks the first half at three-fourths of his former rate, but the stream being with him he rows at double his rate in going ; and he accomplishes the whole distance in 6 hours. Find his rates of walking and rowing. 30. A railway train after travelling an hour is detftined 16 minutes, after which it proceeds at three-fomrths of itp former rate, and arrives 24 minutes late. If the detentio^ had taken place 5 miles further on, the train would have arrived 3 minutes sooner than it did. Find the original rate of the train and the distance travelled. 31. The time which an express train takes to travel a journey of 120 miles is to that taken by an ordinarv train as 9 is to 14. The ordinary train loses as much time in stoppages as it would take to travel 20 miles without stop- ping. The express train oulj loses half as much time m stoppages as the ordinary train, and it also travels 16 miles an nour quicker. Find the rate of each train. 32. Two trains, 92 feet long and 84 feet long respec- tively, are moving with uniform velocities on parallel rails; when they move in opposite directions they are observed te pass each other in one second and a half; but when they move in the same direction the faster train is observed to pass the other in six seconds: find the rate at which each train moves. 33. A railroad runs from A \^ C. A goods' train starts from 4 at 12 o'clock, and a passenger train at 1 o'clock. After going two-thirds of the distance the goods* , train breaks down, and can only travel at three-fourths of its former rate. At 40 minutes past 2 o'clock a collision occurs, 10 miles from C, The rate of the passenger train is double the diminished rate of the goods' frain. Find the distance from A to C7, and tiie rates of the trains. EXAMPLES. XXV, 159 34. A certain sum of money was divided between A, E, and C, so that A^a share exceeded four-sevenths of the shiures of B and C by ^£30 ; also ^'s share exceeded three- eighths of the shares of A and C by £30; and C'b share exceeded two-ninths of the shares of A and E by ;£30. Find the share of each person. 35. A and E working together can earn 40 shillings in 6 days; A and C together can earn 54 shillings in 9 da3rs; and E and C together can earn 80 shillings in 15 days: find what each man can earn alone per day. 36. A certain number of sovereigns, shillings, and six- pences amount to £8. 6s. 6d. The amount of vthe shillings IS a guinea less than that of the sovereigns, and a guinea and a half more than that of the sixpences. Find the number of each coin. <^ 37. A and E can perform a piece of work together in 48 days ; A and C in 30 days ; and E and C in 26j days: find the time in which each could perform the work alone. 38. Thiere is a certain number of three digits which is equal to 48 times the sum of its digits, and if 198 be sub- tracted from the number the digits will be reversed; also the sum of the extreme digits is equal to twice the middle digit: find the number. 39. A man bought 10 bullocks, 120 sheep, and 46 lambs. The price of 3 sheep is equal to that of 5 lambs. A bullock, a sheep, and a lamb together cost a number of shillings greater by 300 than the whole number of animals bought; and the whole sum spent was £468. 6«. Find the price of a bullock, a sheep, and a lamb respectively. 40. A farmer sold at a market 100 head of stock con- sisting of horses, oxen; and sheep, so that the whole realised £2, Is. per head ; while a horse, an ox, and a sheep were sold for £22, £12. 10«., and £1. 10«. respectively. Had he sold one-fourth the number of oxen, and 25 more sheep than he did, the amount received would have been still the same. Find the number of horses^ oxen, and sheep, respec- tifely which were sold. ti 160 QUADRATIC EQUATIONS. XXVL Quadratic Equationt. 226. A quadratic equation is an equation whicll con- tains the tquare of the unknown quantity, but no hi^er power. 227. A pun quadratic equation is one which contains only the square ot the unknown quantity. An affected quadratic equation is one which contains the first power of the unknown quantity as well as its square. Thus, for example, 2^*= 50 is a pure quadratic equation; and 8^- 7^+3=0 is an a^eeted quadratic equation. . 228. The following lis the Rule for solving a pure quadratic equation. Find the value o/* the equare qf the unknown quantity by the Mule /or sotving a simple equO' Hon; Mm, by extracting the square root, the values (/the quantity are found. =6. For example, solve — r— + -^jr- • 1 1 ' 3 10 Clear of fractions by multiplying by 30; thus 10(ii^-13)+3(««-6)=180; therefore 13:1^=180+130+15=325; therefore ^=—=25; extract the square root, thus 4?= :k5. In this example, we find by tiie Rule for solving a ample equation, that a^ is equal to 25; therefore x must be such It number, that if multiplied into itself the pro- duct is 25. That is to say, x must be a square root of 25. In Arithmetic 5 is the square root of 25; in A^ebra we may consider either 5 or -5 as a squareXroot of 25, since, by the Rule cf Signs — 5x — 5a5x5. Hence x majr nave either of the values 5 or —5, and the equation will be satisfied. This we denote thus, d;« dbfi. QUADRATIC EQUATIONS. 161 ties. 289. We proceed to the solution of adfected quadnir If we multiply ^+ r by itself we obtain SB («+|)(;r+|) ax r««+2 2- + ~=««+a»+j.; thus d^+av+^ is a perfect square, for it is the square of « + f. Hence a^+a« is rendered a perfect square by the addition of 7-, that is, hy the addition qfthe square (fha\f the co^cient qf ao. This fact is the essential part of the solution of an adfected quadbratic equation, and we shall now give some examples of it. oi^-¥Qx\ here half the coefficient of or is 3; add 3', and we obtain ^ + 6;v + 3^ that is {jxs-\-2if. a^-Sa; here half the coefficient of 4; is ~o; &dd (-rj, that is (5), and we obtain x^^Sx+y^, that to (*-g'. 4x 2 ' /2\s «a+y ; here half the coefficient of a? is g; add m , and we obtain ^+ x + (5) » *^** ^ (^'*'ft J • a^—-j-; here half the coefficient of a is —3; add ( - |Y, that is f lY, and we obtain x^-^ + (l)\ that A A is (a.~iy. The process here exemplified is called completing the square, T.A. 11 162 QUADRATIC EQUATIONS. i 230. The following is the Rule for solving ftn adfeoted quadratic equation. By tran$po$ition and reduction arrange the equation eo that the terme which involve the unknown quantity are alone on one tide, and the coifficieni qf a^i$ + 1 ; €uid to each tide cf the equation the equare of hd^f the coefficient qf a, and then extract the equare root cfeach tide. It will be aeen from the examples which we shall now solye that the above rule leads us to a point from which we can immediately obtain the values of the unknown quantity. 231. Solve ^~10ar+24=0. By transposition, ii^-^l0x=-24', i add(^Y, «2«iQ^+5a=:«24+25=lj extract the square root^ a-'5= Jbi • transpose^ ^=5akl=5 + l or5-l; hence a-B or 4. It is easv to verify that either of these values satisfies the proposea equation; and it will be useful for the stu- dent thus to verify his results. 232. Solve ai!*-4»-65=0. By transposition, 3^-4^=55; divide by 3, . 4a? 66 3 3 •**(IX 3*\3/ 399' 2 13 extract the square root, «- ; = ±-5- ; .WW \ trmupose, 2 13 . 11 k. QUADRATIC EQUATIONS 233. Scire 2^4-3«-35»0. By transpoBitioii, 2j*+ 3drs 35; divide by 2, 2 2 ' -"©■ ^ 2 W 2 16 16 * 3" )'7 extract the square root» ^"^1^*^* transpose^ 8^17 7 4 4 2 163 234. Solre «»-4«-l = 0. By transpositioii, ai^-4a=l; add 2*, «'-4^+2'=l + 4=5; extract the square root, a-2=^ ^ J5; transpose, a^2^ J5. Here the sanare root of 5 cannot be found exactljr; but we can find dy Arithmetic an approximate value of it to any assigned degree of accuracy, and thus obtain the values of ;i; to any assigned degree of accuracy. 235. In the examples hitherto solved we have found two different roots of a quadratic equation; in some cases however we shall find really only one root. Take, for ex- ample, the equation a^-> 14^+49=0; by extracting the square root we have a? - 7 = 0, therefore a? = 7. It is how- ever found convenient in such a case to say that the quad- ratic equation has two equal rooti, 11—2 164 QUADRATIC EQUATR r:!^ 236. SolTe «>-r6df-l-13-6. By tramposiUon, d^-6^*-13; adds*, «»-6ar+3'--13+9--4. * If we try t6 extract the square root we hare But —4 can have no square root, exact or approximate, because any number, whetner positiye or negatire, if mul- tiplied by itself, giyes a positiye result In this case the quadratic equation has no real root; and this is sometimes expressed by saying that the roots are imoffinary or imposiible. 237. Solve 2(i»-l) ««-l 1 i' \ Here we first clear of fractions by multiplying by 4(^'-l), which is the least common multiple of the de- nominators. Thus 2(a?+l)+ 12=^-1. By transposition, al* —Zof ^15; addl« «2-.2a?+l = 16+l = lC; extract the square root, «—l= :b 4; therefore . ^asldB4=5or -3. 238. Solve ??+3^-«<> 12a?+70 190 16 3(10+a?) Multiply by 670, which is the least common multiple of 16 a^d 190; thus therefore therefore 190(3a?-60) =210-400?; 10+w 190(.^»-60)-(210-40«)(10+i)J QUADRA TIC EQ UA TIONS. that if, 070^-9500 =2100- 190^-40^; therefore AOafl + 760a; » 1 1600 ; therefore «* + 19« = 290 ; 19 S9 extract the square root> a+^^^^^i 2 2 165 therefore ^=:-^-.f=10or-29. 239. Solye /g+3 fl?-3 _ 2^-3 a+2 «-2"" «-l ' Clear of fractions; thus , (aT+3)(i»-2)(«-l)+(i»-3)(«+2)(«-l) «=(2«-3)(a;+2)(aT-2); that is, a^-7a?+6+«'--2«*-5«+6=2«'-3a'-a»+12; that is, 2aj»-2iij3-12i»+12=2«"-3aj«-&»+12; therefore «*— 4iir=0j add2«, «8-4a?+2«=:4; extract the square root, A;-2=:sfc2, therefore a; = 2 it 2 = 4 or 0. We hare given the last three lines in order to com- plete the solution of the equation in the same manner as m the former examples; but the results may be obtained more simply. For tne e€[uation d^— 4a;= may be written (a;— 4)ar=0; and in this form it is sufficiently obvious that we must have either or— 4=0, or «=0, that is, «=4 or 0.. The student will observe that in this example 2a^ is found on both sides of the equation, after we have cleared of fractions; acoordingly it can be removed by subtraction^ and so the equation remains a quadratic equation. "■■■^ ■' '';.dBti>v.-.^."tf;.,-.j',;,/ 166 Q UAD RA TIC EQ UA TIONS. 240. Everp quadratic equation can be put in the form x^+px+ q«0, lehere p and <| represent tome known numbers, whoie or fraetional, positive or negative. For a quadratic equation, by definition, coiitains no Eswer of the unknown quantity higher than the second, et all the terms be brought to one side, and, if necessary, change the signs of all the terms so that the coefficient of the square of the unknown quantity may be a positive number; then divide every term by this coefficient^ and the equation takes the assigned form. For example, suppose 7x—4a^=5, Here we have 7a?-4a?*— 6=0; therefore 4a^-1x+5=0i i therefore ^ 7a? 6 ^ ^ Ik Thus in this example we have p= -^ ^'^^ ^=i* J. 241. Solve ' By transposition, a^-^-px+q^O, a^-¥px=-q\ add ri^' ^.;>-(fy=-..?=^; extract the square root, « + § = * v^^"* ^f . 2 2 thei^fore ^^^t^tMz^^ZP^^!^^:!^ , 2 2 2 242. We have thus obtained a general formula for the roots of the quadratic equation ;r+j7;r+g=0, namely, that X must be equal to -y-f-N/(j>«-4g) _^^ -j>-^(p«-4g ) . orto ^ . \ We shall UvW deduce from this general formula some ▼er^ important inferences, which wiu hold for any quad- ratic equation, by Art. 240. qUADBATIC EQUATIONS. 167 243. A quadratic equation cannot have more than tteo roots. For we have seen that the root must be one or tho other of two assigned expressions. 244. In a quadratic equation where the terms are all on one side, and the coefficient of the square of the unknown quantity is unity, the sum of the roots is equal to the coefficient qf the second term with its sign changed^ and the product qfthe roots is equal to the last term. For let the equation be «* -{-px + g = ; the sum of the roots is -P*^f-*^) ^ -p-^f-*^) , thati. -p; SI J» the product of the roots is that is p"-(p*-4^ 168 QUADRATIC EQUATIONS. 246. Solve aa^-{l'X^C'mO, By transposition, aa^+hx^^-ci divide by a, a a extract the square root, a; + r- = db ^^ — '- ; therefore 2a 247. The general formulsu given in Arts. 241 and 246 may be employed in solving anv quadratic equation. Take for example the equation S^c'*— 4a?— 55=0; divide by 3, thus we have ■^ 4a? 55 ^ Take the formula in Art. 24 i, which gives the roots of A. f\f\ a^+px+q=0; and put p=-^) and q=—-^; we shall thus obtain the roots of the prciy>«ed equation. But it is more convenient to use the formula in Art. 246, as we thus avoid fractions. The proposed equation being 3aj2-4a?-55=0, we must put a=3, d=— 4, and/«-W=0. 5. 2V(^-2»+l)+«*=23 + ap. 6. »*-2aj»+«^=86. 7. V(^-8«+16)+(«-3)"=ia. 8. 9y(«^-94^+28)+9«=«»+36. 9. 2aj»+6iP=226-V(«'+3aT-8). 10, fl?*-4aj8-2>/(«*-4aj«+4)=31. 11. «+2Ay(^+&»+2)=ia 12.' 3«+Ay(«»+7i»+6)=19. 13. «=7V(2-«a). 14. ^/(«+9)=2^/a?-3. 15. ^/(^+8)- V(«+3)=Vj?. 16. 5>/(l-««)+6a?=7. 17. ^/(3«-3)+^/(6a?-19)=V(2«+8). la V(2ar+l)+^(7i»-27)=N/(3«+4). 19. ^/(ft"+aa?)-^/(aH6«)=a+6. 20. 21. 23. 24. 25. 26. 2a?^/(a + «*) + 2a?» = a' - A £+_5/(12fl^-£) __ a +1 ii?-^/(12a^--ip)""a-l' a?+7 fl?-l «+l a;-7 22. = 0. 34? 1-4? 1+4? 1+4J*' = 4?. 4?-V(«8-l) 4? + V(««-l) ^ =8^/(45*-l). fl?+a 4?-a &+4? &- 4? 4?-a 4?+a d-4? t+4?" 27. «»+3a4J*=4a^. 28. 5««(a-4f)=(a«-4;^(4?+3«). 176 PROBLEMS. XXYIII. Problems which lead to Quadratic EquatioiM, 258. Find two numbers such that their sum 'is 15, and their product is 54. Let X denote one of the numbers, then 15i— a; will denote the oUier number; and by supposition a; (16— a?) =54. By transposition, . «2-.l5;i?=-.54j /15V -. 225 9 therefore «»-15a?+| 54+^^ = 4' therefore therefore 15 3 a?=^*-=9or6. If we take ^=9 we have l^-x^Q, and if we take d;= 6 wo have 1 5 —47= 9. Thus the two numbers are 6 and 9. Here although the quadratic equation gives two values of OB^ yet there is really only one solution of the problem. 259. A person laid out a certain sum of money in goods, which he sold again for ^£24, and lost as much per cent, as he laid out : find how much he laid out. Let X denote the number of pounds which he laid out ; then x^24. will denote the number of pounds which he lost. Now by supposition he lost at the rate of x per cent., X that is the loss was the fraction y^ of the cost ; therefore X therefore ^^100=^^24; a;2-i00a?=-2400. From this quadratic equation we shall obtain ^=40 or 60. Thus all we can infer is that the sum of money laid out was either MO or ;£60; for each of these numbers satisfies all the oonditions of the problem. PROBLEMS. 177 is 15, 'X will . \ 76 take |6 and 9. ktlues of m. mey in uchper d out ; hich he )r cent., ^erei^ore 260. The tnm of £*J. 4m, web divided equally among^ a certain number of persons ; if there had been two fewer persons, each would have received one shilling more : find the number of persons. Let X denote the number of persons; then each person 144 receired — shillings. If there had been x-2 persons each would have receiyed supposition, 144 x-2 144 x-2 144 shillings Therefore, by X Therefore 1444>=144(j?-2)+«(4f-2) ; therefore «'-247=288. From this quadratic equation we shall obtain xts\^ or — 16. Thus the number of persons must be 18, for that is the only number which satisfies the conditions of the problem. The student will naturally ask whether any meaning can be given to the other result, namely —16, and in order to answer this question we shall take another problem dosejy connected with that which we have here solved. 261. The sum of £*J. it, was divided equally among a certain number of persons ; if there had been two more persons, each would have received one shilling let9 : find the number of persons. Let X denote the number of persons. Then proceeding as before we shall obtain the equation 144 144 ^ «2+2d?=288; «=16 or -18. therefore therefore Thus in the former problem we obtained an applicable result, namely 18, and an inapplicable result, namely — 16 ; and in the present problem we obtain an applicable result, namely 16, and an inapplicable result, namely —18. T.A. la IMAGE EVALUATION TEST TARGET (MT-3) 1.0 1.1 J25 m ttt Uii 122 11.25 IHU I 1.6 ^ ^ 7 '/ Fhotogra{^ Sdaioes Corporation 33 WIST MAIN STMIST WnSTM.N.Y. )4SM (71«)t72.4S03 Is ;\ 178 PROBLEMS. 262. In Bolvmg problems it is often fomid, as in Aii 260, that remilti are olraamed which do not api^y to the poUent actaally proposed. The reason appears to be^ that the algebraical mode of expression is more general than oidir* nary language, and thus the equation whidi is a oroper represen&tion of the conditions of the prooiem wul juso apply to oUier conditions. . Experience will cdnvmce the suident that he will always be able to select the result which belongs to the problem he is solving. Jind it will be often possil^, by suitable changes in the enunciation of the original problem, to form a new problem corresponding to any result which was inapplicable to the orighial problem ; this is illustrated m Article 261, and we will now give ano- ther example. 263. Find the price of eggs per score, when ten moiie in half a crown's worth lowers the price threepence p^ score. ■ ^ Let X denote the number of pence in the price of a score of eggs, then each egg costs ^ pence ; and therefore the number of eggs which can be bought for half a crown It AAA is 30 -7-^, that is -xr* ^ ^® P^^^^ '^^^ threepence 20' X x-Z per score less, eadi egg would cost -^ pence^ and the number of eggs which could be t>ought for half a crown would be -^-:l . Therefore, by supposition, «-3* therefore therefore X-Z X +^"' 60;»=60(«-3)+a?(«-3) ; ai*-3a?=180. From this quadratic equation we shall obtain «=15 or ~ 12. Hence the price required is \M, per score. It will be found that l^d. is the result of the following pro- blem; find tiie price of eg^ per score whra tp fewer in half a crown's worth roMti the price threepence per score. . _ /\ <»»- EXAMPLES. XXrilL 179 proUMtt Juit the AH ordir< k proper ince the le result twill be m of the ndingto >roblem; ^veano- »n more ence pd^ rice of a tJierefore a crown reepence and the a crown ■$"> n «=15 core. It inngpro- n fewer ence per /\ Ktamplbr XXYIIL 1. IMyide the nnmber 60 into two parti Buch thi^ their prodnot may be 864. 2. The Bom of two nnmbers is 60, and the sum of their squares is 1872 : find the numbers. 3^ Tbib difference of two numbers is 6, and theur pro- duct is 720 : find the numbers. 4 Find three numbers such that the second shall be two-thirds of the first, and the third half of the first ; and that the sum of the squares of the numbers shall be M9. 5. The difference of two numbers is 2, and the sum of thdr squares is 244 : find the numbers. . 6. IHvide the number 10 into two parts such that their product added to the sum of their squares may maJie 76. 7. Find the number which added to its square root willmidLe2ia. 8. One number is 16 times another; and the product of the numbers is 144: find the numbers. 9. One hundred and ten bushels of coals were divided ainong a Certain number of poor personp; if each jperson had received one bushel more he would have received as many bushels as there were persons: find the number of persons. 10. A company dining together at an inn find their bill amounts to £8. 15f. ; two of them were not allowed to pay, and the rest found that their shares amounted to 10 shillings a man more than if all had paid : find the number of men in the company. 11. A dstem can be supplied with water by two I^pes; by one of them it would be filled 6 hours sooner than by the other, and bj both together in 4 hours: find the time in whidi each pipe alone would fill it 12—2 180 BXAMPLBS. xxrm. 'I ' 12. A jMnoD booffhi * 6«rft«lii nombtr of pfiOii of c^loth for £SB, I6i,, which ho fold afiln st d% $§, por ploooi iiid ho galnod Miniioh in tho wholo ii ft liiii^ piooo ooii} And tho numbor of piocoi of oloth. 18. A ftnd 3 iogoihor cm porform * ploof of work In 14} dayt : and A alono Cftn porform » In 18 dftjo loot than JSr ilono: And tho tlmo m which A alono can par* fomiit* . 14. A man bought a certain quantify of moat to IB fhlUingf. If moat woro to rlfo In jnto ono penny por lb.| ho would got 8 Ibi. loM fior tho Muno fum. find now much moat ho bjought 10. Tho pri«90 of ono kind of fUffar nor atono of 141bf. If 1#. 9d* moro than that of another kina ; and 6 Ibi. leu Of tho flret kii»d can bo got for £1 than of tho aecond: find the price of each kinaper itono. 16. A penon apent a certain fum of money In goodi, which he wud again for £^4, and gained aa much por cent. aa the goods coit him: find what the goode coat. 17. The iide of a sauare la 110 inches long; find the lengtii and breadth of a rectangle which shall hare its penmeter 4 inches longer than thM of tho square, and its area 4 square inches less than that of tho square. 18. Find the price of eggs per dozen, when two less In i^ shilling's worth raises tho price one penny per dozen. 19. Two messengers A and B were despatched at the same time to a pU^ at the distance of 90 milea: the former by riding one mile per hour more than the latter arri?ed at the end of his Journey one hour before him: find at what rate per hour each trareUed. fiO. A person rents a certain number of acres of pas- ture land for £10i he keeps 8 acres In his own possession, and aublets the remainder at 6 shillings per acre more than he gare^ and thus ho corers his rent and has £p orer : find the number of acres. ^ roik in yi Urn la ptr- Ulbf. d: find I goodie nr cent nd the iftTd iti and it* leu In at the I! tllO ktt«r i: find EXAMPLES. XXrill 151 Sl« Vmn two placM «i » diftanoe of 890 nfloi, two poraona A and JS^ aot oat In ordor to moot oaob otbor, A tfiraOod 6 nllaa a day more tban Bi and tho mmibM' of di^a In whkli tlufjr mot waa otjoal to half tiio nnmber of mdoi J9 want In a day. Find how te oach travalM bafofo th^oiot 8S. A poiiondrowaaiiaatltjofwInoilromnftdlToaaal wUch hold 81 gallona, ana thon flllod np tho yoaial with water. He then drew from the mlitore aa mneh aa he before drew of pure wine: and It waa found that 64 aidlona of pure whie remained. Bind how mnch he drew ead time, 28. A certain company of a(ddleri can he formed Into a iolld iqnare; a battalion contliUnff of aeren each eqnal companlea can be formed Into a hoUow agoarOi tho men bemg fonr deep. The hollow aqoaire formed bj the ba^ tallon la alzteen timea aa hu|;e aa the aoUd aouare formed bjoneeompanj* Find the nnmber of men In the company. 24. There are three e^ reaaela A^ P, and {7; the firat contalna water, the aeoond brandy, and the thfad brandy and water. V the contenta al B md O be jmt together, It la found that the fraction obtained by difldfaig the quantity of brandy by the quantity ct water la nfaie thnea aa gfetii aa If the contenta of A and O had been treated in like manner. Find tlie proportion of brandy to water in the reiad 0, 25. AperaonlendaifMOOatacertafairateoflntereat; at the end of a year he reoelTea hla intercat^ ipenda £25 of it and adda the remainder to hla ci^tal; be then lenda hia capital at the aame rate of intereat aa before, and at the end of another year finda that he haa altofether ^6882: determine the rate of intereat ion, than over: iSi SIMULTANEOUS EQUATIONS XXDL Sitntdtaneoiu Eguaiion» inwivinff. QuadraHa, 264. We shall now solre Bome examplM of limnltme- 0118 equations involying quadratics. There are two oases of frequent occurrence for which roles can be giyen; in both iiese cases there are two unknown quantities and two equations. The unknown quantities will always be den6ted by the letters AT and y. * ' - ■ • , 265. First Can, Suppose that one of the equations is of the first degree, and the other of the second degree. Rule. From the equation qfthejlrat degree find ^ value qf either qf the unknown quantitiee in terme kf the other, and eiAetitute thie value in the equoHon qf the eeeond degree. Example. Sdve 3^+4^=18, 5a^-3^=2. From the first equation y- — ^ — ; substitute this Tsilue in the second equation; therefore 4 therefore 204j»-54^+9«*=:8; therefore 29«*-64aT=8. 4 From this quadratic equation we find x^2-63)=36; therefore 3a)'- 33d; +90=0; therefore d^-lldr+30=0. From this quadratic equation we shaH find that«=5 or 6; and then by substituting in the Value of y werfind that y=8 or 12|, 267. Second Case, When the terms involving the un- known quantities in each equation constitute an expressioa which IS homogeneous and of the second degreis; see Art. 23. Rule. .Assume y=vz, and siibstUute in both e^[wi» lions; then by division the value qfyean be found. Example. Solve «'+a!y+2^=44, 2«'-ii?y+y»=16. Assume y=«;p, and substitute for y; thus y(l+»+2»^=44, d?8(2-t>+»>)=16. Therefore^ by division, l+ g-t-2g' _44_ll therefore therefore therefore 2-»+»* 16 4(l+i>+2r!0=ll(2-»+i?^; Sc'-lSc+lS^O; r2-e»+6=0. 184 81MULTANE0(TS EQUATIONS "Ftook this qnadraiio eqiuitioii we ihall obtain 9s8 or ai In the equation a^(l+e+2«*)s44 put 2 for e; tiins wts rft2; uid ilnoe y=v«) we haye y^ ik4. Again, in tiie same equation put 3 for « ; thns «» db j^ ; and atnod ys94r, we have ysASi ' \ Or we might proceed thne: mnltiply the forat of the giyen equations by 2; thna 2«*+2jEy-l-4y*s88; the aecond equation ia 2^-4^ 4- y'== 16. Byfubtraotion 8a)Sf4-8^=:72y therefore ^=24-«y. Again, multiply the aeoond equation by 2 and lubtract the firat equation; thus 3^-30^8-12; therefore «>=«y— 4. Hence, by multiplication «V = (24 - «y ) («y - 4), or 2«y-28«y=-9G. By solying this quadratic we obtain d^=8 or 6. Sub- stitute the former in the given equations ; thus // «'+2y*=36, 2«»+y*=24. Hence we can find a^ and ^. Similariy we may take the other Talue of ^, and then find «^ and ^. 26a Solye 2«»+3ajy+y*«70, 6a*+«y-^=60. Assume yse«, and substitute for y; thus «»(2+3»+fJ«)=70, ii^(6+e-e«)=«60. Therefore by diyision f 2+3g+e* ^70_7. 6+e-e« *60 6' therefore 6(i4-3»+e*)= 7(6 +«-«*); therefore 12ij*+ 89-32=0; \ therefore 39*4-2e~8=0. 2oraL ; tiiiii in the i linQe of tho ibtract \ Biib- kethe INVOLVINO QUADRATICS. 189 Fiom tbift q^iadimtic equation we shall find car W, ? + ?-% «^+y«-a»+ay. la ?+?«! 19. SO. 81. 29. 83. 84. 85. 86. 87'. 8& 89. 80. 81. 82. 83. 34. 8^-«ys56, 80^-^*^48. a*-8«yBl5, «y-8^«i7. «*+3Jlfss88, «y+4/s8. «'+«y-$y*=21, «ry-2y'»4. d^-f8«ys04^ 4V+4y*a:118. i-y «+y"2' ««+y««9a 4^=48. g'+y* _25 4y-fy ig-y ^lO «-y «+y"3' a^-t^^Z. { a^(«+y)-fy(^-y)sl58, 7dr(«+y)s7^(4*-y). fljVC^+y)*^, A:>y(2«-'3y)=80. 2aj'-«y+y*=2y, 2a^+4^=5y. «*+y*=(*. ay-fy fly~y __ a*-f 1 +2(2«+y)»6, 4«y(^+l)s3. 45. «*-l-«yB&p+3, y'+^sSy+C. 46. d^-«y-i2«+ff, «y-y's2y+2. 47. 8«+y-l-6>/(2«+y+4}»23, 4i^-ejrsyi4-9y. 48. 18+9(«+y)«2(«+y)«, 6-(«-y)=(«-y)*. 49. «^-fly-a(«+l)+6+l, a» «y 62. «*=a«+^, y'aay+&;v. ffS. a^xisa, ai/^z^b, »yz^=e, 60. 8y4rH-2«r«->4a^sl6/ 2y;i?~3;ra;+^s5y 4yz-za-Zxy=16, 68. 6(«^+if*+«^«l3(«+y+«^)=:-^, «y««*. «y-y*=ay+6. 6a ;^ + g=18, a* f/* 61. ^-£=12, «• P' = 1. =2. vv^i 4ite^ »t^^i 190 PROBLEMS. XXX Prdtlemi^hich lead to QuadraUe Equatton* with more than one unknown qttantity, 274. There is a certain number x>f two jdigits; the Bum of the squares of i^ cUsitB is e^nal to the nnmbw in- creased by the product of its digits; and if thirty-six be added to the number the digits are retersed: find the number. Let a denote the digit in the tens' place, and if the digit in Uie units' place. Then the number is lOa+y; and if tike digits be reversed we obtain lOy +>. Therefore^ by supposition, we have «"+y»=4;y+lOa?+y ,..(1). 1 10a:-fy+36=10y+a?.... (2). From (2) we obtain 9y=9a?+36; therefore ^=^+4. Substitute in (IX thus thei^fore «*-7«+12=0. From this quadratic equation we obtain a; =3 or 4; and therefore y=7 or 8. Hence the required number must be either 37 or 48 ; each of these numbers satisfies all the conditions of the problem. 275. A man starts from the foot of a mountain to walk to its summit His rate of walking during the second half of the distance is half a mile per hour less than his rate during the first hal^ and he reaches the summit in 6^ hours. He descends in 3| hours by walking at a uni- form rate, which is one mile per hour more than hia rate during the first half of the ascent. Find the distance to the sununit, and his rates of walking. \ Let 2af denote the number of miles to the summit, and suppose that during the first half of the ascent the man fi £t th th Tl th „w-^' PROBLEMS. m X Item ieBam Ksr in- •six be id tbe r tf the y; and ore^by t-4 or 4; amber .Usfies i& to g the B than nut in a nni- rate ice to walked y mfles per hour. Then he took - hours for the w first half of the ascent^ and — ^ hours for the second. y- 2 Therefore - +-£^=6J (1). V" 2 SimUarljr, From (2), iherefore 2a; y+1 T— «fj < (2). Therrfore^ I^ Babstitolioii, ' therefore 15(y+l)(4y-l)=44y(i2y-l),' therefore 28y'-89y+16=0. From this quadratic equation we obtain y=3 or ^ . The value ^ is inapplicable, because by supposition y is 1 15 greater than ^. Therefore y=3; and then ^=-^9 so that the whole distance to the summit is 15 miles. t>and ) man 192 EXAMPLES. XXX. Examples. ZXX. 1. The lam of the squares of two numbers is 170, and th% difference of their squares is 72 : find the numbers. 2. The product of two numbers is 108^ and their sum is t^ce their difference: find the numbers. 3. The product of two numbers is 192, and the sum of their squares is 640 : find the numbers. 4. The product of two numbers is 128, and the differ- i enoe of their squares is 192 : find the numbers. \ 6. The product of two numbers is 6 times their sum, and the sum of their squares is 325 : find the numbers. 6. The product of two numbers is 60 times their differ- ence, and the sum of their squares is 244 : find the numbers. 7. The sum of two numbers is 6 times their difference, and their product exceeds their sum by 23 : find the num- ber& 8. Find two numbers such that twice the first with three times the second may make 60, and twice the square of the first with three times the square of the second may make 840. 9. Find two numbers such that their difference multi- plied into the difference of their squares shall make 32, and their sum multiplied into the sum of their squares shall make 272. 10. Find two numbers such that their difference r.dded to the difference of their squares may make 14, and iheir sum added to the sum of their squares may make 26. 11. Find two numbers such that their product is equal to their sum, and their sum added to the sum oi their squares equal to 12. EXAMPLES. XXX. 193 0, and r sum tarn of differ- > sum, 8. differ- ibers. renoe, nam- with |uare , may nulti- e 32, iiares ided heir qual ;neir 12. Find two numbers sttch that their sum increased by their product is equal to 34, and the sum of their squares diminished by tiieir sum equal to 42. . 13. The difference of two numbers is 3, and the dif- ference of theuf cubes is 279 : find the numbers. 14. The sum of two numbers is 20, end the sum of their cubes is 2240 : find the numbers. 16. A certain rectangle contains 300 square feet; a second rectangle is 8 feet shorter, and 10 feet broader, and also contains 300 square feet: find the length and breadth of the first rectangle. 16. A person bought two pieces of cloth of different sorts; the finer cost 4 shillings a yard more than the cotoer, and he boiufht 10 yards more of the coarseir than of the finer. For we finer piece he paid £18, and for the coarser piece £16. Find the number of yards m each piece. 17. A man has to travel a certain distance ; and when he has travelled 40 miles he increases his speed 2 miles Sor hour. If he had travelled Mth his increased speed uring the whole of his Journey he would have arriv^ 40 minutes earlier; but if he Imd continued at his original speed he would have arrived 20 minutes later. Find the whole distance he had to travel, and his original speed. 18. A number consisting of two digits has one decimal place ; the difference of the squares of the digiiis is 20, and if the digits be reversed, the sum of tiie two numbers is II : find the number. 19. A person buys a quantity of wheat which he sells so as to gain 6 per cent on his outlay, and thus clears £16, If he hf^i sold it at a gain of 6 shillings per quarter, he would have cleared as many pounds as each quarter cost him shillings: find how many quarters he bought, and what each quarter cost. 20. Two workmen, A and ^, were emplo^red by the day at different rates ; A at the end of a certain number of days received £4, 16«., but Bj who was absent six of T. A. 13 194 EXAMPLES, XXX. those days, received onlv £2, 14«. li B had worked the whole time, and A had been absent six days, they would have rei^ived exactly alike. Find the number of days, and what each was paid per day. 21. Two trains start at the same time from two towns, and each proceeds at a miiform rate towards the other town. When they meet it is found that one train has nm 108 miles more than the other, and that if they continue to run at the same rate thev will finish the journey in 9 and l6 hours respectively. Find the distance between the towns and the rates of the trains. 22. A and B are two tdwns situated 18 miles apLtii on the same bank of a river. A man goes from .^ to i? in 4 hours, by rowing the first half of the distance and walkiiffi the second half. In returning he walks the first half at the same rate as before, but the stream being with him, he rows 1^ miles per hour more than in ffoing, and accom- plishes the whole distance in 3^ hours. jPind his rates of walking and rowing. 23. A and B run a race round a two mile course. In the first heat B reaches the winning post 2 minutes before A* I In the second heat A increases nis speed 2 miles per liour, and B diminishes his as much ; and A then aiTives at tne winning post two minutes before B, Find at what rate each man ran in the first heat. 24. Two travellers, A and B^ set out from two places, P and ^, at the same time; A starts from P with the design to pass through Q, and B starts from Q and travels in the same direction as A, When A overtook B it was found that thev had together travelled thirty mil^, that A had passed through Q four hours before, and that By at his rate of travelling, was nine hours' journey distant from P. Find the distance between P and Q, INVOLUTION. 195 2XXL Involutions 276. We have already defined a power to be fhe pro- dact of two or more equal factoraf and we hare explained the notation for denoting powers; see Arts. 15, 16, 17. The frocess of obtaining powers is called Involution ; so that nvolution is only a particular case of Multiplication, but it is a particular case which occurs so often that it is convenient to devote a Chapter to it The student will find that he IS already familiar with some of the results which we ^hall have to notice, and that the whole of the present Chapter follows immediately from the elementary laws of A^ebra. 277. Any even power qf a negative quantity is poii- tive^ and any odd power is negative. This is a simple consequence of the Rttle of Signs. Thus, for example, ~-a x -a—a^^ —a x -a x ^a—a^ x — a= -a'; — ax — ax — ax — a=— >»c^; (2a6»(j»)«=2«a*^>")»=a?+5a*6+10a»^*+lOa26>+6a&*+«>^. (a 4- &)• =^ «• + 6a*6 4- 15a<6* + 20a?ft» + 16a«6* + 6a&« + e^. In like manner the following results may be obtained : : (a-.6)a=««--2a&+&«, (a-ft)*=a*-4a^6+6a26a-4a&8+t*, (a-5)»=a«-6a»Hl5a*6*-20a^6»+15a«6*-6afc*+e*. Thus in the results obtained for the powers of 'a— 5, where any odd power of b occurs, the negative dgp is pre- fixed; and thus any power of a-'b can be immdiaiely deduced from the same power of a+b, by changing tiie 8:|;ns of the terms which involve the odd powers ofW iirrozuTioN. 191 ■•-. f JRaiss 8r, and I6a» 816** Hise of . ScSp h, fifth 2^. ned: •a-6, Bpre- the 282. The stadent will see hereafter that, by the aid of a theorem called the Binomial Theorem, any power of a binomial expression can be obtained without the labour of actual multiplication. 283. The formuloB c^ven in Article 281 may be used in the way we have already explained in Art 84. Sup- pose, for example, we require the fourth power of 2sB—Zy, In the formula for {a-hf put 2x for a, and 3y for h \ thus, (2a;-3jf)*=(2a?)*-4(2a?)«(3y)+6(2a?)«(3y)«-4(2aO(3y)'+(3y)* = 16ar«-96aj»y+216aiy-216a^+81y*. 284. It will be easily seen that we can obtain required results in Inyolution b^ different processes. Suppose, for example, that we require the sixth power of a +6. We may obtain this by repeated multiplication by a+&. Or we may first find the cube of a +6, and then the square of this result; since* the square of {a-^Vf is (a+&)'. Or we may first find the sauare of a +6, and then the cube of this result ; since the cube of (a + 5)^ is (a + h)\ In like manner the eighth power ot a-¥h may be found by taking the square of (o •(-&)*, or by taking the fourth power of (a+&)^. 285. Some examples of Involution in the case of trinomial expresnons have already been given. See Arts. 85 and 88. Thus (a + 6 + c)5 =5 a' + B» + c? + 2aH 26c + 2flk?, (a+6+c)»= • a'+6^+c^+8a«(6+c)+86a(a+c)+3c*(a+6)+6a6& These formulsB may be used in the manner explained in Art 84. Suppose, for example, we require (1— 2d;+3a^'. In the formula for {a+h-\-cf put 1 for a, -207 for 6, and 3«* for c; thus we obtain (l-2a?+3«*)«= (l)»+(-2a>)?+(3ipV+2(lX-2a?) + 2(-2a:)(3a;^+2(l)(3««) = l+4«"+9a;*-4a?-12aj»+6aj» ==l-r4iP+10««-12«»+9;i?*. 198 EXAMPLES. XZXL ■ f Similariy, we hare + 3(I)V-2d?+3a5»)+3(--2a?)«(l+3^+8{«a5«(;i-jtey +6(l)(-2ar)(8ifj^ / = 1- &» + 21dJ*- 44** + 63«* - W«" + 27«^. 286. It is found by obsenratioii that the aonare of any multinomial expression may be obtained by either of two rules. Take, for example, {a-hb+c-k- d^» It inU be found that this - a* + 6* 4- c* + 4^«»+«»jl 83. (l+24y+3;iJ»+4a^\ 36. Xt+8i»+a<»*+«^V 37. (l-6ar+12a?«-8«^». 3a (l+4f«r+6«*+4ij'+aj*)V 8SI- .tatinff three simple consequences of the Ride qf Signe^ we shful then consider in suocessioi^ the extraction of Qie roots of simple expressions, the extraction of the square root of compound expressions ai)d numbers, and the extn^on of the cube root of compound expressions and numbers. ' 288. Any even root qf a positive quantity may he either positive or negative* Thus, for example, axa=aK and —ax —d^a*] there- fore the square root of a' i9 either a or -a, tiiat is, eiUier + aor —a. '289. An^ odd root of a quantity has the same sign as the quantity. Thus, for example^ the cube root of a' Is a, and the cube root of -a» is —a. 290. The^e can he no even root qfa negative quantity. Thus, for example, there can be no square root of -a*; for if any quantity be multiplied by itself the r^ult is a positive quantity. The fact that there can be no eyen root of a negatiye ^uantit^ is sometimes expressed by calling su^ a root an impossible quantity or an imaginary quantity, 291. Eule for obtaining any root of a simple integral expression. Divide the index qf every factor in the expression hy^ the number denoting the root, and give the proper sign to the result. evolution: 201 9ign cube ThuB, for example, J(\Qa^b^) = ^(4^t^ » ^ ial^, y ( - WVc>») = y ( " 2 V W) = - 2aV IF^T'f 202 EVOLUTION. If there were more terms we •honld proceed with a +5 08 we did formerly with a; its square, that is, a*-¥2ah-^h\ hM already been subtracted from the proposed expression, so we should divide, the remainder by 2(a-i-()for a new term in the root Then for a new subtrahend we multiply the sum of 2 (a +5) and the new term, by the new tcirm. The process must be continued until the required root is found. \ . 2S95. Examples. 40^ + 120):^ -f 9^ ^2d? -f 3y 12«y+9y" 4i»*-2ae»+ 37^-3(to+ 9 (^2a«-6aT+ 3 4^ 4«*-5«; -20««+37aj«-3(to+9 -20;i5'+25«* 12^-3(to+9 2a«-4ajy+3y»;6«V'- 12«y»+9y* , 6djV'-12«y'+9y* \ SVOLUTION. 203 4ar»+4ir* 1 1 2;c» + 4«"-4d?-l^ - 2«»-4«»+4«+l - 2«*-4«*+4«+l 296. It has been already observed that all even roots admit of a double sign; see Art. 288. Thus the square root of a'+2a6+^ is either a+6 or —a-h. In fact, in the process of extracting the square root of a*4-2a&-f&*, we begin by extracting the square root of a'; and this may be either a or —a. If we take the latter, and con« tinue the operation as before, we shall arriye at the residt — a-&. A similar remark holds in eyery other case. Take, for example, the last of those worked out in Art 295. Here we begin by extracting the square root of afi\ this may be either a^or —a^. If we take the latter, and con- tinue the operation as before^ we shall arriye ^t the result ~«»-2«*+2«+l. 297. 'She fourth root of an expression may be found by extracting the square root of the square root ; similarly the eighth root may be found, by extracting the square root of the fourth root; and so on. 298. in Arithmetic we Imow that we cannot find the square root of eyery number exactly; for example, we cannot find the square root of 2 exactly. In Algebra we cannot findl^^ square root of eyery^ proposed expression 204 EVOLUTION, eJeactlp, We Bometimes find such an example as the follow^ ing proposed; find foor terms of the square root of 1 -2ar. l-2*(^»-*-2-2 2-a»-^ !-«» 4 a? «* s-s^-^-f;-**-! •-flj'+aj^H 1 — 6*U* sfi iXr 'T""2"" 4 U /pi /n8 /m8 Thus we have a remainder — 4"'" 2*""T' *^' findmg four terms of the squpre root of 1 - 2^; and so we know that (l-*-5-0=l-ar+^ + ^ + ?. » ' 299. The precedm^ investigation of the squar«i root of An Algebraical expression will enable us to demonstrate the nue which is jp^ven in Arithmetic for the extraction o^f the square root of a number. The square root of 100 is 10, the square root of 10000 is 100, the square root of 1000000 is 1000, and so 6a ; hence it follows that, the square root of a number less than 100 must consist of only one figure, the square root of a MrOLUTJO^, 205 iitimber beiwden 100 and 10000 of two plaoes of figdreii, of a number between 10000 and 1000000 of three plaoea of fissures, and so on. If then a point be placed over eyerj second fiffure in any number, be|;inning with the figure in the units' place, the number of points will shew the number of figures in the square root. Thus, for example^ the square root of 4$5ft consists of two figures, and the square root of 611^24 consists of three figures. 30b. Stippose the square root of 324^ required. 2500 100+7J749 749 Point the number according to the rule ; thtis it appears tiiat the root must consist of two places of figuf es. Let a+5 denote the root^ where a is the value of the figure m the tens' place, and h of that in the units' place. Then a must be the greatest multiple of ten, which has its square less than ^200; this is found to be 50. Subtract a\ that is, the square of 50. from the giyen number, and the remainder is 749. Diyiae this re- mainder by 2a, that is, 1^ 100, and the quotient is 7, which ?3 the value of h. Then (2a+&)5, that is, 107 x 7 or 749, is the number to be subtracted ; and as there is now no remainder, we conclude that 50 + 7 or 57 is the required square root. It is stated above that a is the greatest multiple of ten which has its square less than 3200. For a evidently can- not be a greater multiple of ten. If possible, suppose it to be some multiple of ten less than this, say x\ then since X is in the tens' place^ and h in the units' place, «4- 6 is less than a ; therefore the square of a;+& is less thaii a\ and consequently a; + & is less ilum the true square rooi ' If tiie root consisi of three places of figures, let a re^ present the hundreds, and h the tens; then having ob* tained a and h as before, let the hundreds and tens together be considered as a new value of a, and find anew valuo of h for the units* .viiSim** 206 woLunoN. 301. The <^herB may be omitted for the sake^ of brevity, and the following rule may be obtained from the process. Point every second figure^ beginning <^ 3fi4d ^67 voith that in thA unit^ place, and thus 25 divide the whole number into periods, Find the greatest number whose square 107 J 749 is contained in the first period; this y^g is the first figure in the root; subtract its square from tlie first period, and to the remainder bring dozen the next period. Divide this quantity, omitting the last figure, by tvnce the part of the root already found, and annex the result to the root and also to the divisor; then mtdtiply the divisor +3a'&i-3a^4-&' \& a-¥h\ and we shftU be led to a general nile for the extraction of the cube root of any compound exi>re8sipn by observing the manner in which a + & may be derived from c? + 3a^ii> 3aft^ + Jfi^ Arrange the temid ac^ rt*+3a2j+3ai'+ft'(rt+6 cording to the dimensions ^t of one letter a; then the tint term is a', and its cube root is a^ which is the first term o^ the required root. Subtract its cube, that is (j^y from the whole expression, and bring down' the re-^ 3aV3a'6+3rt62+2,^ 3a«6 + 32+&» Find the first term of the root, that is a; put a' under the given expression in the third column and subti'act it Put 3a in the first column, and 3a^ in the second column; divide 3a% by ^\ and thus obtain the quotient &. Add h to the expression in the first column ; multiply the ex- pression now in the first column by 6, and place the pro- duct in the second column, and add it to the expression already there; thus we obtain 3a2+3ai&+&2. Multiply this by by and we obtain 3a%+3a2>2+&^, which is to be placed in the third column and subtracted. We have thus completed the process of subtracting (a+d)' from the original expression. If there were more terms the opera* tion would have to be continued. T. A. 14 210 BVOLUTION. 3a+3& } 308. In oontintdiig the operation we mnst add locfa a tenu to the first column, as to obtain there three timee the part qf the root already found. This is oonyeniently effected thus; we have already in the first column ,3a+&; place 2b below b and itdd; thus we obtain Za+Sby which is three limes a+b, that is, three times the part of the root already found. Moreover, we must add such a term to the second column, as to obtain there three times the eqtiare qf the part of the root already found. This is conyenientlY effected thus; we have already in the second column (Sa + b)b, and below that 3a^+Zdb+b^; place V^ below, and add the expressions in the three lines ; Uius we obtain d(ir+6aib+db^j which is three times (a+bf, that is three times the square of the part of the root already found. (3a+5)& ') da*+3a&+ (3> W 1- 3aS+6a&+35* 309. Example. Extract the cube root of au"-36«' + 102a?*-l7l«»+204««-144a?+64. 6««-3a?) 12aj* Sof) ~&»(6««-3«) 6««-9a?+4 12aj*-18«>+9«« 12a;*-36«»+27aj« 4(6««-9i?4-4) 12a;**-36«»+61«»-36«+ 16 8aj»-36a;"+102a;*-l7l«»+204;««-144»+64(,2i*-3d?+4 - SO;!?" + 102a?« - 171«* + 204aj» - 144a? + 64 -36«*+ 64aj*- 27«" 48a?*- 144a!'+204a^- 144a?+ 64 48.i?*-l44a!»+204iJ«-l444?+64 BVOLVTWN. 211 The ettbe root of d^ is 2a^, which will be the first term of the requhred root; put 8d^ under the given expression in the thfrd column aikd subtract it» Put three times 2a^ in the first column, and three times the square of 2a^ in the second column $ that is, put ^3^ in the firat column, and 12d^ ill the second column^ Divide ^36a^ by 12^. and thus obtain the quotient —Str, which will be the secoUa term of the root; place this term in the first column, and multiply the expression now in the first column, that is 60^—307, by — 3«; place the product under the expression in the second column, and add it to that expression \ thus we obtain 12^ - 18;i^ 4- 9a;' ; multiply this by — ' 3ar, and place the product in the third column and subtract. Thus we have a remainder in the third column, and the purt of the root alroEuly found is 2ix^—^, We must now acMust the first and second columns in the manner explained in Art 308. We put twice - 3ar, that is - 6a;, in the first column, and add the two lines; thus We obtain 6a^->-9ar, which is three times the part of the root already fbund. We put the square of -3a;, that is 9a;', in the second column, and add the last three lines in this column ; thus we obtain 12a;*-36a^+27a;^, which is three times the square of the part of the root already found. Now divide the remainder in the third column by the expression just obtained, and we arrive at 4 for the last term of the rooi^ and with this we proceed as before. Place this term m the first column, and multiply the expression now in the first column, that is 6a:'-^9a;+4, by 4; place the product under the expression in the second column, and add it to that expression; thus we obtain 12a;*~36a;'+51a;'--36a;+16; multiply this by 4 and place the pix)duct in the third column and subtract As there is now no remainder we conclude that 2a^-3a;+ 4 is the required cube root 310. The preceding investigation of the cube root of an Algebraical expression will suggest a method for the extraction of the cube root of any number. The cube root of 1000 is 10, the cube root of 1000000 is 100, and so on; hence it follows that, the cube root of 14—2 212 EVOLUTION. ' a number less than 1000 most condst of onW one figure^ the cube root of a number between 1000 and 1000000 of two places of figures, and so on. If then a point be placed oyer every third figure in any number, begmning with the figure in the units' place, the number of points "will shew the number of figures in the cube root Thus, for example, the cube root of 40&224 consists of two figures, and the cube root of 1^81^^904 consists of throe figures. Suppose the cube root of 274G25 required. 180+5 10800 274625 (,60 + 5 926 1172S 216000 68625 68625 Point the number according to the rule ; thus it appeto that the root must consist of two places of figures. Let a-\-h denote the root, where a is the value of the figure in the tens* place, and h of that in the units' place. Then a must be the greatest multiple of ten which has its cube less than 274000 ; this is found to be 60. Place the cube of 60, that is 216000, in the third column under the given number and subtract Place three times 60, that is 180, ih the first column, and three times the square of 60, that is 10800, in the second colunm. Divide the remainder in the third column by the number in the second column, that is, divide 68625 by 10800; we thus obtain 6, which is the value of 6. Add 6 to the first column, and multiply the sum thus formed by 6, that is, multiply 185 by 6; we thus obtain 925, which we place in the second column and add to the number already there. Thus we obtain 11726; multiply this by 6, place the product in the third column, and subtract The remainder is zero, and therefore 66 is the required cube root. - ' The cyphers may be omitted for brevity, and the pro- cess will stond thus : 185 108 925 11726 27462^(65 21^6 58626 5S625 EVOLUTIOir. 213 811. Bzample. Extract the cube root of 1092. 352. 127 271 14 f 1418 10^21536^^473 64 45215 39823 5392352 5392352 48 889 5689^ 49] 6627 11344 674044 After obtaining the first two figures of the root, namely 47) we adjust the first and second columns in the manner explained in Art 308. We place twice 7 under the first column, and add the two lines, giving 141 ; and we place the square of 7 under the secona column, and add tlie last three lines, giving 6627. Then the operation is continued as before. The cube root is 478. In the course of working this example we might have imagined that the second figure of the root would be 8 or even 9 ; but on trial it will be found that these numbers are too large. As in the <^e of the square root, we are liable occasionally to try too laige a figure, especially at tho early stages of the operation. . 312. Example. Extract the cube root of 8653002877. 605) 1200 g65d00S87t(,2053 10 } 6153 3025 123025 . 25 } 8 653002 615125 37877877 37877877 126075 18459 12625959 In this example the student should notice the occur- rence of the cypher in the root. 814 EVOLUTION. 313. If the root have any nrnnber of decimal itoei^ the cube will have thripe as many ; and therefore the nnm- ber of decimal places in a decimal number, which is a perfect cube, ana in its simplest state, will neoessaHlT be a multiple of ihrte^ and the number of decimal placte m the cube root will necessarily be a third of that number. Hence if the giyen cube number be a dedmal, we pUce a point imr the Jigure in the unitt* place, and o>?er every third fiffure to the right and to the left of it, and proceed as in the extraction of the cube root of an integer; then the number of points in the decimal part of the proposed number will mdicate the number of decimal places In the cube root 814. Example. Extract the cube root of 14102*31)7296. 64 8,' 721 2/ 12 256 n 1456 • 16, Ill02'32}29d(24*l6 8 6108 6824 7236 1728 72r 278327 173521 173521 1. 104806296 104806296 174243 4341 6 17467716 316. If any number, integral or decimal, has no exact cube root, we may annex cyphers, and proceed with the approximation to the cube root to any desired extent. - The following is the extraction of th^ cube i>oo| of *4 to four decimal pUu^s: 913 } 2196 12 } 22088 n EXAMPLES. XXXII. 215 15987 13176| 1611876> 36) 1625088 176704 162685504 •400... (•7368 343 57000 46017 10983000 9671256 1311744000 1301484032 10259968 EzAMPLXS. XXXIt Find the valae of 1. ,J{9€^. 4. ^(16a«6Pc»). //25a^6»\ 6. 8 2. 4^(8a»6»). 3. ^(-64a»Z»«). 3// 216a^6»\ 9. VV326"j* 10. y(^^ Find the sqnare roots of the following expressions: 11. 16a*+40a6+256«. 12. 49a*-84a^+36&« 13. 36a^+124!'+l. 14 64a*+48aJc+96*c'. 15. 25<^-f20a6+4y 250^ rf 2006+40** 16. 9^^-24^Mhl6 44J»-l2»+9 • m 216 EXAMPLES. XXXIl ■ 17. «*+2aj'+34^+24T4.1. 18. l-2»+«««-4«»+4c»+(a»+4<»6 + 62);ij2-2a6(a+6>+a^}* Find the eighth roots of the following expressions : 33. ««+8a^4-28aj«+66a?» + 70ii?*+66a?»+284^+ap+l. 84. {a^-2afip+3a^^-'2a!^-hp*}*, ^ Find the square roots of the following numbers : 35. 1156. 36. 2025. 37. 3721. 3a 5184. 39. 7569. 40. 9801. 41. 15129. 42. 103041. 43. 165649. 44. 3080*25. 45. 41^1^. 46. *835396. 47. 1522756. 48. 29376400. EXAMPLES, XXXIL 217 49. 88492401. 50. 4981*5364. 51. 64128064. 52. -24373969. 53. 144168049. 54. 254076*4836. 55. 3-25513764. 56. 4*54499761. 57. '5687573056. 58. 196540602241. Eztnot the square root of each of the following nam- bers to five places of decimals : 59. -9. 60. 6*21. 61. *43. 62. -00352. 63. 17. 64. 129. 65. 347*259. 66. 14295*387. Find the cube roots of the following expressions: 67. aB»+36a?!^ + 54ary«+27y'. 6a i72aB»+i72apV+fi76ajV+^V. 69. «»-3a:«(a+&) + 3a 319. As ihustional indices and negative i|i|d|^!^ not yet been defined, we are at liberty to give wha| MM* tions we please to them ; and it is found oonveiliJIt lo INDICES. 21» give sudi clefiQiiioDs to them as will mi^e the hnportant relation a* X <»"-'»"•+" -' " * may he. relation a**xa"=a"*'^" tdways true, whatever m and n « For example; required the meaning of aK By supposition we are to have a* x a" = a^ = a. Thus a ^ must be such a number that if it be multipli^ by itself Uie result is a; and the eqiiare root of a is by definition such a number; therefore or must be equivalent to the squarerootof (z, thatiSya^=>/a. . Again; required the meaning of a^, . By supposition we are to have a xa >ia =a a^=a. Hencei as before^ a^ must be equivalent to the cube root of a/that is a^» /J^a. Again; required the meaning of a V faff By supposition, a xVxa xa =a'; therefore a*= ^a\ These examples would enable the student to under- stand what is moant by any fractional exponent; but we will give the definition in general symbols in jttie next two Artides. X 320. Required the meaning qf a" where n U any positive whole number. By supposition, 111 111 ^^ a'xa'xofx ...ton factors = a" «•**•* =a^=ai therefore a" must be equivalent to the n^ root of a, that is, a*^ff^ 220 INDICES. . 321. Required the meaning qf a" where m andn are ^ any positive iohols numbere. By supposition, a"x a"xa" X ... to » factors=a" " • "* =a"»; therefore a* must be equivalent to the n*** root of a*", that is, a^^lja"^. M Hence a" means the n*^ root of the w*^ power of «; that is, in a fractional index the numerator denotes a power and the denominator a root. \ 322. We have thus assigned a meaning to any positfve index, whether whole or fractional; it remains to assign a meaning to negative indices. For example, required the meaning of a"'. By supposition, a' x a"" =«"-'=«*= a, therefore a^ or We will now give the definition in general symbols. < 323. Required the meaning qf a~'/ wh^ren ie any positive number whole orfraetionaL By supposition, whatever m may be, we are to have a"*xa~"=a"'~". Now we may suppose m positive and greater than n^ and then, by what has gone before, we have m > fit rt""" X «• = a"» ; and therefore a""* = -» • Therefore therefore a ,m a* \ INDICES, 221 In orddr to express this in words we will define the Yford reciprocal. One quantity is said to be the recipro' col of another when the product of the two is equal to unity ; thus, for example, a is the reciprocal of - . Hence a~" is the reciprocal of a" ; or we may put this result symbolically in any of the following ways, «--•-„ a "' a xa —II— 1. y 324. It will follow from the meaning which has been given to a negative index that a'"-T-a"=a'"~" when m is less than Uf as well as when m is greater than n. For suppose m loss than n ; we have o a —- = ^-(»-«") = a"*"". n—m Suppose fn=n; then a'^-r-a' is obviously =1; and cr-*=ia\ llie last symbol has not hitherto received a meaning, so tliat we are at liberty to give it the meaning which naturally presents itself; hence we may say that 325. In order to form a complete theory of Indices it would be necessary to give demonstrations of several pro- positions which will be found in the larger Algebra. JBut those propositions follow so naturally from the definitions and the properties of fractions, that the student will not find any difficulty in the simple cases which will come be- fore him. We shall therefore refer for the complete theory to the larger Algebra, and only give hero some examples as specimens. 326. If m and n are positive whole numbers we know that (a'")"=a"^; see Art. 279. Now this result will also hold when m and n are not positive whole numbers. For example, (a*)^=a^. For let (a»)*=d?; then by raising both sides to the fourth power we have a^=^; then by raising both sides f i2i INDICES, to the third t>ower we have a=a;^'; thel«fore x^a^, which was to he diewn. i( 327. If n is a positive whole number' we know that a^xlf^iflbf. This result will also hold when n is not a positive whole number. For example, a^x&^=(a&A For if we raise each side to the third power, we obtain in each case ah\ so that each side is the cube root of ah In like manner we have 111 1 a" X &" X (^ X . . . =i (a&c. . .)". Suppose now that there are m of these quantities a, by c,..., and that all the rest are equal to a\ thus we obtain 1 1^ (a")* = (a*)" j that is, ( ya)- = yo*. Hius the m^ power of the n^ root of is equal to the n* root of the «i* power of a» ' 328. Since a fraction may take different forms without any change in its value, we may expect to be able to give Afferent forms to a quantity with a fractional index, with- out altering the value of the quantity. Thus, for example, 4 9 4 g we may expeet that a^=a^ ; and this is the smoe 3 3 case. Fof if we raise each side to the sixth power, we obtain a*; that is, each side is the sixUi root of a*. \t r 329. We will now give some examples of Algebraical operations involving fractional and negative exponents. Multiply a^^M by ah^fi. 2 17 3"*"2"C' 3 1 13 4 3"'12» thet^fore 3-^5 ?=^L :>-• INDICES. 223 Divide a^jfi by «^y*. 4 2""4» ?«i_l. therefore Multiply 8 6 %* x-^-ar+ar* by a^-i-x'^-aTK Of +«■ +»"' ;»^+2aj^-M -^-t Here in the first line «* x a?=a?^**=i»^, a?^ x a^ »x\ ^'x^'^soj^ssl; andsoon. ; Divide — . ww«w«»i>*»«i < "I 'l 224 EXAMPLES. XtklTL 6. (81)-^. 9. ya-». Examples. XXXIII « Find the value of I. 9"*. 2. 4"*. 3. (100)"*. 4. (lOOOili Simplify 6. {dF)-\ 7. ' (a-")-». 8. Va"*. 10. o^a*xa"i Multiply II. i^+y^ by a?^-y^. 12. /32=V9; 4=4/43=4^64; a=i/a*; a+b= i/{a+h)\ 333. The product of a rational ouantity and a surd may be expressed as an entire surd, by reducing the rational quantity to the form of the surd, and then multi- plymg ; see Art 327. For example, 3 ^/2 = ^9 x ^2 = ^ 18 ; 24/4= ^(8x^4= J/32; a^/&=^/a«x ^b=jj{a% 334. Conversely, an entire surd may be expressed as the prodnct of a rational quantity and a surd, if the root of one mdor can.be extracted. T.A. 15 226 SURDS. For example, ^/32 = ^/(16 x 2) = ^16 x ^2 = 4 ,^2 j 4^48= j|/(8x 6)= ysx 4/6 = 24/65 i/ia*b*)=i/a*xi/b*=ai/b\ 335. A surd fraction can be transformed into an equivalent expression with the surd part integral. ™ 1/3 /3x2 /6 Je jFor example, iJ ~ — U ■■a/ — = 5l_ • ^ *wi cA»uip*o, V g ^8x2 ^16 4 ' 336. Surds which haye not the same index can be transformed into equivalent surds which have ; see Art 327. For example, take ^/5 and 4/11: V6=6*» /yil = (ll)*j 6^=6*= 4/6»= 4/125, (11)*=11^=: 4/(1 1)«= 4/121. 337. We may notice an application of the preceding Article. Suppose we wish to know which is the greater, 1^5 or ^11. When we have reduced them to the same index we see that the former is the greater, because 125 is greater than 121. 338. Surds are said to be Bimilar when they have, or can be reduced to have, the same irrational factors. Thus 4 ^7 and 5 Jl are similar surds ; 5 4/2 and 44/16 are also similar surds, for 4 4/ 16 = 8 4/2. 339. To add or subtract similar surds, add or subtract their coefficients, and affix to the result tbe common irrational factor. 1 For example, ^/12+^/75~^/48=2^/3+6^/3-4^3 = (2 + 5-4)^/3=3^/3. V3^.1 V256 2 Vl2 1 V64xl V2'*"4^T=3'^8"^4^'^'"W _2^12 1 44/1224/12 3 2 4 3 3 • SURDS, 227 340. To multiply simple sards which have the same index, multiply separately the rational factors and the irrational factors. 1/ For example, 3^/2x^3=3^6; 4^5x7^6-28^30; 24^4x3^^2 = 64^8=6x2 = 12. 341. To multiply simple surds which have not the same index, reduce them to equivalent surds which have the same index, and then proceed as before. For example, multiply 4 >/5 by 2 4/11. By Art. 336 ^/6=4^125, 4/11=4^121. Hence the product is 84^(125 x 121), that is, 8^15125. 342. The multiplication of compound surds is per- formed like the multiplication of compound algebraical expressions. For example, (6/^/3-5 ^2) x (2 ^3 + 3 J2) = 36 + 18/s/6-10^/6-30 = 6 + 8^/6. 343. Division by a simple surd is performed by a rule like that for multiplication by a simple surd; the result may bo simplified by Art. 335. w 1 « /« ^ i« 3n/2 3 /2 3 /6 >/6 For example, 3 V2-i-4 ^/3=^ = - V ^ = ^ V ^ = ^- ; 4^5^-^11-2;^- ^121 -^'^121-^'^121x(ll)* 24^1830125 ~ n ^' The student will observe that by the aid of Art. 335 the results are put in forms which are more convenient for nu- merical application; ti^us, if we have to find the approxi- mate numerical value of 3 ,^,/2-r4 ^/3. the easiest method is to extract the square root of 6, and divide the result by 4. 16—2 228 SURDS. • 844. The only case of division by a eomponnd inrd which is of any importance is that in which the divisor is the sum or difference of two quadrcUie surds, that is, surds inyolving square roots. The division is practically effected by an important process which is called rdtionmiting the denominator cf a fraction. For example, take the fraction 4 5 724-2 /a * ^ ^^ multiply both numerator and denomi- nator of this fraction by 6 J2 -2JZ, the value of the frac- tion is not altered, while its denominator i» made rational; 4 4(6^2-2 ^/3) thus 5^2 + 2^/3 (5^/2 + 2^/3){5<>/2-2,ys) ^ 4(5^/2^2^/3) 10^/2-4^/3 60-12 * 19 ' • fiimikrlv N/3 + V2 _ (^/3^^^^2)(2^/3•^-^/2) ^'2 3-^/2"(2^/3-^/2)(2^/3+^/2) ^ 84-3i^6 _8-f3j6 12-2 10 • \ 845. We shall now shew how to find the square root of a binomial expression, one of whose terms is a quadratic nurd. Suppose, for example, that we require the square root of 7+4/^3. Since Ua!'h^v^=^w+y+2j(ap\ it is obvious that if we find values of of and y fi*om x+y=*I, and 2 tJiay) -- 4 ly/S, then the square root of 7 + 4 ^3 will be tjx + Vy. We may arrange the whole process thus : Suppose square, 7+4,/3=A;+y+2^/(a?y). Assume ^+^=7, then 2^(a^)=4^/3; square, and subtract, (w + y)^~'4ay =: 49 - 48 = 1, that is, (;»-y)*=l, therefore a?-y=l. Since a; -I- ^=;: 7 and a; - 2^ = 1, we have or = 4, j^ = 3 ; therefore /s/(7 + 4V3)=>/4+ .^3=2+1/3. Similarly, s/(7-4 J3)=2- ^3. EXAMPLES. XXXIV. 229 EXAHPLES. XXXIY. Simpliiy . 1. 3^/2+4^/8-^/32. % 2^4 + 6^32-4/108. 3. 2^/3 + 3V(li)-^/(6l). 4. ^"i^- • Multiply 6. V5+V(li)-;^by^/3. * ^- 'y^-yr6-^^2^yy^ 7. l+is/3-V2by ^/6-^/2. 8. V3+N/2by-^+^. Rationalise the denominators of the following fractions: 9. 11. 3+V2 2-V2* 2^6+^3 10. n/3+ .^2 >/3-n/2- 12. 2/s/3+^2 3;/6T2;73' '*• 3^/3-2^5• EitrsMst the square ropt of 13. .14 + 6^6. 14. 16-6*77. 15. 8 + 4^/3. 16. 4-^16. Simplify 17. ^/(6+^/24)* 18. n/(7-4V3)- 19: ^^f^^ll 20. V(3+>/5)+V(3-V5). 230 RATIO, XXXV. Ratio. 846. Ratio is the relation which one quantity bears to another with respect to magnitude, the comparison beinff made by considi 4ng what multiple, part^ or parts, the first is of the second. Thus, for example, in comparing 6 with 3, we observe that 6 has a certam magnitude with respect to 3, which it contains twice ; again, in comparing 6 with 2, we see that 6 has now a different relative magnitude, for it contains 2 three times; or 6 is ^eater when compared with 2 than it is when compared with 3. 347. The ratio of a to 5 is usually expressed by two points placed between them, thus, a:b\ and the former is called the antecedent of the ratio, and the latter the conte- qmnt of the ratio. , 348. A ratio is measured! by the fraction which has for its numerator the antecedent of the ratio, and for its denominator the consequent of the ratio. Thus the ratio of a to ( is measured by t ; then for shortness we may say that the ratio of a to & is equal to ^ or is r . 349. Hence we may say that the ratio of a to 5 is equal a to the ratio of c to d, when r = 3 . h d 350. ffthe terms of a ratio he multiplied or divided by the 9am£ quantity the ratio i» not altered. a ma For g = ^ (Art. 136). 361. We compare two or more ratios by reducing the fractions which measure these ratios to a common denominator. Thus, suppose one ratio to be that of a to d, and a 6' RATIO. 231 and another ratio to be that of <; to = !-;,. h ha a ha Hence the first ratio is g^reater than, equal to, or less than the second ratio, according as ad? is greater than, equal to, or less than be, 352. A ratio is called a ratio of greater inequality ^ of leti inequatity^ or of eqtudity^ according as the antecedent is greater than, less than, or equal to the consequent 353. A ratio qf greater inequality it diminiihed^ and a ratio qf lest inequality it increated, by adding any number to both tennt qfthe ratio. Let the ratio be ^ , and let a new ratio be formed by adding x to both terms of the original ratio: then .-— - o-\-x is greater or less than y according as & (a + or) is greater or less than aifi-^-x); that is, according as bx is greater or less than ax, that is, according as 6 is greater or less than a, 354. A ratio qf greater inequality it increased, and a ratio qf lett inequality it diminished, by taking from both termt qf the ratio any number which it lett than each qf thote termt. Let the ratio be v , and let a new ratio be formed by taking x from both terms of the original ratio; then ^^ is greater or less than ^, according as b{a-x) is greater or less than a(b-'X); that is, according as &a; is less or greater than ax, that is^ according as & is less or greater than a. 355. If the antecedents of any ratios be multiplied together, and also the consequents, a new ratio is obtfuned which is said to be compounded of the former ratios. Thus 232 RATIO. the ratio ocxMSa said to be compounded of the two ratios a :&andc : dL When the ratio a : & is compounded with itself the resulting ratio is a' : ^ ; this ratio is som^imes .called the duplicate ratio of a : &. And the ratio a? : 2^ is sometimes called the triplicate ratio of a : &. 356. The following is a very important theorem con- cerning equal ratios. ace Suppose that ^ = ^ = ^, then each of these ratios where jp, q, r, n tre any numbers whatever. \ For lot A;= r = ^ = -^j then Kb=af kd=c, ^=e; J9(*6)"+g(A^"+r(^)"=i?a"+2'6"+r«"; "ph^-^qd'+r/** therefore therefore therefore 1 The same mode of demonstration may be applied, and a similar result obtained when there are more ttum three ratios given equal As a particular example we may suppose n = 1^ then we fE c e see that ^^ i:-^-fy ®<^ ^^ ^^®<3^ ratios is equal to ^f ^^ — >; and then as a special case we may suppose po+qa + rj p=zq=ryEo that each of the given equal ratios m equal to a+c+e b+d+f' EXAMPLES. XXXV. 2da Examples. XXXV. 1. Find the ratio of fourteen shillingB to three guineas. 2. Arrange the following ratios in the order of magni- tude; 3 : 4, 7 : 12, 8 : 9y 2 : 3, 5 : 8. 3. Find the ratio compounded of 4 : 15 and 25 : 36. 4. Two numbers are in the ratio of 2 to 3, and if 7 be added to each the ratio is that of 3 to 4 : find the numbers. 5. Two numbers are in the ratio of 4 to 5, and if 6 be taken from each the ratio is that of 3 to 4 : find the numbers. 6. Two numbers are in the ratio of 5 to 8; if 8 be added to the less number, and 5 taken from the greater number, the ratio is that of 28 to 27 : fii^d the numMrs. 7. Find the number which added to each term of the ratio 5 : 3 makes it three-foiuihs of what it would have be- come if the same number had been taken from each term. 8. Find two numbers in the ratio of 2 to 3, such that their diiSerence has to the dif erence of their squares the ratio of 1 to 25. 9. Find two numbers in the ratio of 3 to 4, such that theu* sum has to the sum of their squares the ratio of 7 to 50. 10. Find two numbers in the ratio of 5 to 6, such that theur sum has to the difference of their squares the ratio of 1 to 7. 11. Find X so that the ratio x : 1 may be the duplicate of the ratio 8 : x, 12. Find x so that the ratio a—xxh-x may be the duplicate of the ratio a : h, 13. A person has 200 coins consisting of guineas, half- sovereigns, and half-crowns; the sums of money in g^mneas, half-sovereigns, and half-crowns are as 14 : 8 : 3; find the numbers of the different coins. 14. If &-a :&+a=4a-& : 6a-&, find a :&. 15. If — £ = i: — = ,then/+m+n=Q. 2S4 PROPORTION. XXXVI. Proportion. 357. Foar numbers are said to be proportional when the first is the same multiple, part, or parts of the second as the third is of the fourth ; that is when 7 = ^ the four o a numbers a^hfC^d are called i>roportional8. This is usually expressed by saying that a is to & as c is to d {p+l)c. Th™ g w less than unity, and ^^ ' is greater than unity. Hence, by hypothesis, ^ is less than unity, and ^^r — is greater than unity. Thus ^ and r are both greater than - , and both less than ~ — , And since this is true however great/? and q may be, we infer that v and -3 cannot be unequal; that is, they must be equal: see Art. 370. Therefore a i b :: c : d. That is, the four quantities a, b, c, d satisfy the alge- braical definition of proportion. 374 It is usually stated that the Algebraical definition of proportion cannot be used in Geometry because there is no method of representing geometrically the result of the operation of division. Straight lines can be represented geometrically, but not the abstract number which expresses ow oftein one straight line is contained in another. But it should be observed that Euclid's definition is rigorous and applicable to incommensurable as well as to commensur- abU quantities ; while the Algebraical definiti<|n is, strictly speaking, confined to the latter. Hence this consideration alone would furnish a sufficient reason for the definition adopted by Euclid, EXAMPLES. XXXVL 239 Examples. XXXVL Find ihd yalue of x in each of the following propor- tiona ]. 3. 6. 6. 7. 7 :: a: : 9 :: 16 ; 42. X, 4 : 7 :: 8 : or. 2. 3 H I X V, X \ 45. 4. X «+4 : a?+2 :: «+8 : «+6. «+4 : 2a?+8 :: 2d?-l : 3a?+2, Sor-f 2 : a;+7 :: 9^-2 : 50^+8. 8. «»+a?+l:62(a?+l)::«*-a?+l :63(ii?-l). 9. 007+^ :&a7+a:: »M;+n :na7+m. 10. If pq=ri, and qt=8Uf then p :r ::t :u, 11. If a : & :: c : ei?, and a' : &' :? ^ : d', then aa^ :W :i ctf : /(; then AC=fnnBI>; therefore AC ac BD, Similarly, if AccB, and CxZ>, and JSccF, then ACE^BDF; and so on. 387. JE/*Ax B, #AenA"x B*. / For let A=^mBy then A^=m*B*\ tl^erefore -4" x J5". VARIATION. 248 38a if A)+ +a. Therefore, by addition, 2«=(^+a)+(^+a) + tow terms =w(/+a); n therefore »=*o(^'*'*) (^^* V /■'' '■■■ 246 ARtTHMEtlCAL PBOQBESSION. Also /=a+(n-l)6 (2), thus «=^{2a+(n-l)6} (3). The equation (3) gives the value of # in terms of the quantities which were supposed known. Equation (1) also gives a convenient expression for «, and furnishes the foUowmg rule: the Bum qf any number qf termt in Arithmetical Progretsion is equal to the product of the number qf the terms into ha{f the sum <^ the first and last terms. We shall now apply the equations in the present Article to solve some examples relating to Arithmetical Pro- gression. \ 394. Find the sum of 20 terms of the series 1, 2, 3, 4,... Here a=l, &=1, n=20; therefore «=^(2 + 19)=10x21=210. 395. Find the sum of 20 terms of the series, 1, 8, 5, 7,... Here a=l, &=2, n=20; therefore, *=|'(2+I9x2)=|^x40=(20)"=400. 396. Find the sum of 12 terms of the series 20, 18, 16,. . . Here o=20, 6=s-2, n=12; therefore «=^(40-2xll)t=6(40-'22)=6xl8=ipa. 397. Find the sum of 8 terms of the series A t ::. 7f L- 12'6 4 3 11 Herea= — , 6=Y2,n=8j therefore 8/2 . 7\ . 9 ^ of the [1) also ^es the in of the Yst and ■Article Pro- I ■ 3, 4,.„ »5,7,... EXAMPLES. XXXVJIL 247 398. How many terms mnst be taken of the Beiies 15, 12, 9,... that the sum may be 42 ? Here #=42, a=15, 5= -3; therefore 42=||30-3(n-l)} =^(33-3n). We have to find n from this quadratic ej[nation ; by solving it we shall obtain n=4 or 7. The series is 15, 1^ 9, 6, C 0,--3, ; and thus it will be found that we ob- tain 42 as the sum of the first 4 terms^ or as the sum of the first 7 terms. 399. Insert fiye Arithmetical means between 11 and 23. Here we have to obtain an Arithmetical Progression consisting of seeen terms, beginning with 11 and ending with 23. Thus a=ll, /=23, n=7 ; therefore by equation (2) of Art 393, 23=11+65, therefore &a2. Thus the whole series is 11, 13, 15, 17, 19, 21, 2a EXAXPLB9. XXXVIII. Sum the following series : 1. 100, 101, 102, to 9 terms. 2. 1, 2|, 4, to 10 terms. 3. 1, 2}, 4^, to 9 terms. 4. 2, 3f, 5^,..., to 12 terms. 6. X, -, 1, tolSterms. 6. 2' "3' ""?'••• tol5terms, 7. Insert 3 Arithmetical means between 12 and 20. 8* Insert 5 Arithmetical means between 14 and 16. us EXAMPLES. XXXVIIL 9. Insert 7 Arithmetical means between 8 and -*4. 10. Insert 8 Arithmetical means between — 1 and 5. 11. The first term of an Arithmetical Pro|^ession is 13, the second term is 11, the sum is 40: find we number of terms. 12. The first term of an Arithmetical Progression is 5, and the fifth term is 11 : find the sum of 8 terms. 13. The sum of four terms in Arithmetical Progression is 44, and the last term is 17 : find the terms. 14. The sum of three numbers in Arithmetical Pro- gression is 21, and the sum of theur squares is 155 : fii^ the numbers. ^ 15. The sum of five numbers in Arithmetical Prcwpres- sion is 15, and the sum of their squares is 55: find the numbers. 16. The seventh term of an Arithmetical Progression is 12, and the twelfth term is 7; the sum of the series is 171 : find the number of terms. 17. A traveller has a journey of 140 miles to perform. He goes 26 miles the first day, 24 the second, 22 the third, and so on. In how many days does he perform the journey? 18. A sets out from a place and travels 2^ miles an hour. B sets out 3 hours after A^ and travels in the same direction^ 3 miles the first hour, 3^ miles the second, 4 miles the third, and so on. In how many hours wUl B overtake A 1 19. The sum of three numbers in Arithmetical Pro- gression is 12 ; and the sum of their squares is 66 : find the numbers. 20. If the sum of n terms of an Arithmetical Pro- gression is always equal to n\ find the first term and the common difference. OEOMETRICAL PROGRESSION. 249 XXXIX. Geometrical Progression, 400. Quantities are said to be in Geometrical Pro- gression when each is equal to the product of the i>recediDff and some constant factor. The constant factor is called the common ratio of the series, or more shortly, the ratio. Thus the following series are in Geometrical Progres- sion. Xf Of 9f 27) ol)..».>t 1 1 1 J_ ^'2*4'8V16' a, ar, ar^, ar^, ar*,. The commcn ratio is found by dividing any term by that which immediately precedes it. In the first example the common ratio is 3, in the second it is - , in the third itisr. ^ 401. Let a denote the first term of a Geometrical Pro- gression, r the common ratio; then the second term is ar, the third term is at^, the fourth term is ar*, and so on. Thus the n^ term is ar^'K 402. To find the sum qfa given number tf terms qfa Geometrical Progression, the first term and the common ratio being supposed knoton. Let a denote the first term, r the common ratio, n the number of terms, s the sum of the terms. Then f=a+ar+ar' + ar' + ...+ar"''*; . therefore sr-ar-k-at^+ai^-^-.^.-^-ar^'^-^-ar^. Therefore, by subtraction, sr—s—ar^—af therefore a(r*-l) ^^^ 25a GEOMETRICAL PROGRESSION. 1(1 denote the last term we have l=ar^'i rl-a therefore i- r-1 .(2X .(3). Ednation (1) gives the value of « in terms of the quantities which were supposed known. Equation (8) is sometimes a convenient form* We shall now apply these eouations to solve some ex- amples relating to Qeometrical rrogression. 403. Find the sum of 6 terms of the series l, 3, 9, 27,. . . Here a=l, r=3, n=6; therefore 3«-l 720-1 #= 3-1 3-1 =364. 404. Find the sum of 6 terms of the series 1, -3, 9, ^2if» ^" Herj a=l, r= -3, n=6; therefore (-3)»-l 729-1 ,^„ -k 405. Find the sum of 8 terms of the series 4, 2, 1, r ,. .. Here a = 4, r = - , n = 8 J therefore ^ V^ " ^ J _ ^ V "¥) 255 2 255 I-' '-I 64 32 406. Find the sum of 7 terms of the series, 8, -4, 2 -1 i Here a =8, r=-=, n=7; therefore -!-. -^' 16 3 8 'Vf of the (8) fa me ex- 9,27,... h "3, OEOMETRICAL PROaRESSIOK 251 407. Insert three Geometrical means between 2 and 32. Here we have to obtain a Qeometrical Progression consisting of Jtte terms, beginning with 2 and ending with 32. Thus a =2, /=32, n=5; therefore, by equation (2) of Art 402, 32=2r*, that is r*^16=2*; therefore r=% Thus the whole series is 2, 4, 8, 1^, 32. 408. We may write the value of », given in Art. 402, thus «(l-r") *= 1-r Now suppose that r is less than unity; then the larger n is, the smaller will r** be, and by taking n large enough r" can be made as small as we please. If we negle<^ r^ we obtain a and we may enunciate the result thus. In a Qeometrical Progression in which the common ratio is numerically less than unity, by taking a sufficient number of terms the sum can be made to difer as lUtle as we please from r-^. i—r 409. For example, take the series 1, = , - , - , . . . ^ 4 o 1 /» Here a=l, **=o; therefore y— =2. Thus by taking a sufficient number of terms the sum can be made to differ as little as we please from 2. In fact if we take four terms the sum is 2—-, if we. take five terms the sum is 8 2- — , If we take six terms the sum is 2^ — , and so on. The result is sometimes expressed thus for shortness, the sum of an irfinite number qf terms qfthis series is 2; or thir, the sum to infinity is 2» * It 252 EXAMPLES. XXXIX. 410. Recurring decimals are examples of what are called infinite Geometrical Progression. Thus for example 3 24 24 24 •3242424... denotes j^ + j^ + ^^ + j^, + ... Here the terms after z-z form a Geometrical Progres- 24 sion, of which the first term is j^, and the common ratio Is rr^. Hence we may say that the sum of an infinite 24 / 1 \ number of terms of this series i^ Tp -^ ( 1 "*To2j » *^** ^^ 24 ^^. Therefore the value of the recurring decipaal is 3^ 2£ ^ 10 9»0* The value of the recurring decimal may be found prac- tically thus: Let 9= •32424...; then 10*= 3-2424..., and 1000«=324'2424... Hence, by subtraction, (1000 - 10) « = 324 - 3 = 321 ; *!. f 321 therefore '"ogo' And any other example may be treated in a similar manner. Examples. XXXIX. Sum the following series : 1. 1,4,16, to 6 terms. 2. 9,3,1, to 5 tends. % 3. 25,10,4, to 4 terms. 4. 1, V2, 2, 2V2, ... tol2terms. leciinal is >und prac- EXAMPLES. XXXIX. 258 5. g> |> g, to 6 terms. 2 3 6. ^» "■1>2» to 7 terms. 7. 1, ""3» 9> to infinity. 8. 1, J, jg, to infinity. 9. 1» ""o* 4* to infinity. 2 10. 6,-2,-, to infinity. o Find the Talue of the following recurring decimals: 11. -ISISIS... 12. -123123123... 13. -4282828... 14. -28131313... 15. Insert 3 Geometrical means between 1 and 256. 16. Insert 4 Geometrical means between 5} and 40j^. 17. Insert 4 Geometrical means between 3 and -729. 18. The sum of three terms in Geometrical Progression is 63, and the difference of the first and third terms is 45: find the terms. 19. The sum of the first four terms of a Geometrical Progression is 40, and the sum of the first eight terms is 3280 : find the Progression. 20. The sum of three terms in ij^eometrical Progres- sion is 21, and the sum of their squares is 189 : find the ierms. 254 HARMONWAl PROGRESSION, XL. Harmonicat Progretnon, 411. Three quantities A^ B, C are said to be in Har- monical Progression when A : :: A- B : B — C, Any number of quantities are said to be in Harmonica! Progression when every three consecutive quantities are in Harmonical Progressiott 412. The reciprocals qf quantities in Harmonical Progression are in Arithmetical Progression, Let Af Bf C he in Harmonical Progression; then A: C::A'-B : B-0. \ Therefore A (B-C)=C{A-'B). Divide by ^5C; thus i -i = -i - 2 • This demonstrates the proposition. * 413. The property established in the preceding Article will enable us to solve some questions relatmg to Har- monical Progression. For example, insert five Harmonical 2 8 means between - and r-. Here we have to insert five 9 lo Arithmetical means between ^ and -3- 3 1 • therefore ^&=-, therefore &= — . o 10 3 25 26 Hence the Arithmetical Progression is - , HencOi by equa- tion (2) of Art 393, 27 28 29 15. 16' 16' 16' 8 ' 2 16 16 16' 16' and therefore the ^a^monical Pro- 16 16 16 8 greasionis-, ^g, -, ^, -, -, -. EXAMPLES. XL. 255 414. Let a and e be any two quantiUes ; let ^ be their Arithmetical mean, G their Geometrical mean, H Uieir Uarmonical mean. Then A-a=e-A \ therefore A = " (a+c). a \ Q \\ G : e\ therefore G= tj{ac). a : e :: a^H : if— c; therefore J7= 2ae a+c" 1 1 Examples. XL. 1. Continue the Harmonical Progression 6, 3, 2 for three terms. 2. Continue the Harmonical Progression 8, 2, 1^ for three terms. 3. Insert 2 Harmonical means between 4 and 2. 4. Insert 3 Harmonical means between - and rr. 5. The Arithmetical mean of two numbers is 9, and the Harmonical mean is 8 : find the numbers. 6. The Geometrical mean of two numbers is 48, and the Harmonical mean is 46^ : find the numbers. 7. Find two numbers such that the sum of their Arith- metical, Geometrical, and Harmonical means is 9f , and the product of these means is 27. 8. Find two numbers such that the product of their Arithmetical and Harmonical means is 27, and the excess of the Arithmetical mean above the Harmonical mean is IJ. 9. If a, &, " ■' 256 PERMUTATIONS AND COMBINATIONS. XLL Permutations and CombinatiQns, 415. The different orders in which a set of things can be arranged are called i\iea permutations. Thus the permutations of the three letters a,&, c, taken two at a time, are a&, &a, aCy ca, be, cb, 416. The combinations of a set of things are the different collections which can be formed out of them, without regarding the order in which the things are placed. Thus the combinations of the three letters a, &, o, taken two at a time, are ab, ac, be ; ab and ba, though different permutations, form the same combination, so also do ac and ca, and be and cb, 417. The number qf permutations of n things taken r at a time is n (n— l)(n— 2) (n-r+ 1). // Let there be n letters a, b, c, d, ; we shall first find the number of permutations of them taken two at a time. Put a before each of the other letters; we thus obtain n-1 permutations in which a stands first. Put b before each of the other letters; we thus obtain n— 1 permuta< tions in which b stands first. Similarly there are n— 1 permutations in which e stands first. And so on. Thus, on the whole, there are n(n — l) permutations of n letters taken two at a time. We shall next find the number of permutations of n letters taken three at a time. It has just been shewn that out of n letters we can form n (n— 1) permutations, each of two letters; hence out of the n—l letters b,c,d, we can form (n-1) (n-2) permutotions, each of two letters : put a before each of these, and we have (n- l)(n— 2) permutations, each of threer letters, in which a stands first Similarly there are (n-1) (n-2) permutations, each of three letters, in whicb b stands first. Similarly there are as many in which e stands first. * And so on. Thus, on the whole, there are n (n — 1) {n—2) per- mutations of n letters taken three at a time. PERMUTATIONS AND COMBINATIONS. 267 From considering these cases, it might be conjootmvd that the number of permutations of n letters talcen r at a time is n(n-l)(n~2)...(n-r+l); and we shall shew that this is the case. For suppose it known that the num- ber of permutations of n letters taken r— 1 at a time is 7t (n-l)(n- 2).. .{n-(r-l)+l}. we shall shew that a similar formula will give the number or permutations of n letters, taken r at a time. For out of the n-\ letters h^ e, d,... we can form (n-l)(n-2) {n-l—(r— 1)4-1} permuta- tions, each of rr>l letters: put a before each of tnese, and we obtain as many permutations, each of r letters, in which a stands first Similarly thr-r* are as many permu- tations, each of r letters, in which b stands first Simi- larly tnere are as many permutations, each of r letters, in which e stands first. And so on. Thus on the whole there are n(n— l)(n-2)....(»-r+l) permutations of n letters taken r at a tima If then the formula holds when the letters are taken r- 1 at a time it will hold when thev are taken r at a time. But it has been shewn to hold when they are taken three at a time, therefore it holds when they are taken four at a time, and therefore it holds when they are taken five at a time, and so on : thus it holds universally. 418. Hence the number of permutations of n things taken all together is n (n — 1) (n — 2) ... 1. 419. For the sake of brevity n(7i- l)(n~2)...l is often denoted by \n; thus \n denotes the product of the natural numbers from 1 to n inclusive. The symbol \n may be resAf factorial w. 420. Any combination of r things wiU produce [r permutations. For by Art. 418 the r things which form the given combination can be arranged in [r different orders. 421. JTie number qf combinations qfn things taken r r,*^4'^^ • n(n-l)(n-2)...(n-r + l) aJt a time %s -^ — r — ^ • T.A. 17 258 PERMUTATIONS AND COMBII^ATfOm For the number of permutatiom of n things takesi r at atimei8n(n-l)(n-2)...(n-r+l)byArt417; and each combination proauces [r permutations by Art 420; hence the number of combinations must be -^ ' „ n(n--l)(n-2)...(n-r4-l) __. ' If we multiply both numerator and denominator of In this expression by |n-r it takes the form -, — -= — , the value of course being unchanged. 422. To find the number qf permutations of n things taken all together which are not all different. Let there be n letters; and suppose p of them to be a, q of them to be &, r of them to be c, and the rest of them to be the letters dy e, ..., each occurring singly: then the number of permutations of them taken all together will be lPL2l£' For suppose N to represent the required number of permutations. If in any^ one of the permutations the p letters a were changed into p new and dififerent letters, then, without changing the situation of an^ of the other letters, we could from the single permutation produce \p different permutations: and thus if the p letters a were changed into p new and different letters the whole number of permutations would be iV x [£. Similarly if the q letters h were also changed into q new and different letters the wliole number of permutations we could now obtain would he Nx\px\qj And if the r letters c were also changed into r now and different letters the whole number of per- mutations would be iV X [p X [£ X [r . But this number must be equal to the number of permutations of n different letters taken all together, that is to [n. \ \n Thus iVx[£x [^x [r = |n; therefore iv= . , . -* And similarly any other case may be treated* EXAMPLES. XLL 259 423. The siadent should notice the peculiar method of demonstration which is employed in Art 417. This is called inathemaiical induction, and may be thus described: Wo shew that if a theorem is true in one case, whateyer that case may be, it is also true in another case so reUted to the former that it may be called the neat case; we also shew in some manner that the theorem t« true in a certain case; hence it is true in the next case, and hence in the next to that, and so on; thus finally the theorem must be true in eveiy case after that with which we began. The method of mathematical induction is frequently used in the higher parts of mathematics* ExAUPLESv XLI. 1 . Find how many parties of 6 men each can be formed from a company of 24 men. 2. Find how many permutations can be formed of the loiters in the word company, taken all together. 3. Find how many combinations can be formed of the letters in the word longitude, taken four at a time. 4. Find how many permutations can be formed of the letters in the word consonant, taken all together. 6. The number of the permutations of a set of things taken ybur at a time is twice as great as the number taken three at a time : find how many things there are in the set. 6. Find how many words each containing two conso- nants and one vowel can be formed from 20 consonants and 6 Towdsi the rowel being the middle letter of the wordk 7. t^te pelhM>nB are to be chosen by lot out of twenty: find in how many ways this ctm be done. Find also how often an assigned person would be chosen. hi A boat^s crew consisting of eight roWei^s and a bVeersman is to be formed out of twelve persons, nine of whom can row but cannot steer, while the other three can steer but cannot rowt find in how many ways the crew can be formed ^ind ite in how many ways the crew could be formed tf one of the three were able both to row and to steer. 17—2 mtm 26a BINOMIAL THEOREM. XLII. Binomiai Theorem^ 424. We liave already seen that («+aJ2=aj8+2aya+a2, and that (ar+a)'=«*+3«*a+3a?a2+a*; the object of the present Chapter is to find an expression for {a+a)* where n is any positive integer. 425. By actual multiplication we obtain (a; + a) (d7 + &) (« + c) = aj« + (a + 6 + c)dj2 + («6 + &<; + ca)^ + a&c, (a+a)(a-{-b)(x+c)(:!B+d)=a!*+(a+h+c+d)a}^ \ + (€ib+ac+ad+bc+bd+cd)a^ \ + {abc+bcd+€da-i-dc^)a}+dbcd, Now in these results we see that the following laws holdt I. The number of terms on the right-hand side is one more than the number of binomial factors which are multi- plied together. II. The exponent of or in the first term is the same as the number of binomial factors, and in the other terms each exponent is less than that of the preceding term by unity. in. The coefficient of the first term is unity; the coefficient of the second term is the sum of the second letters of the binomial factors ; the coefficient of the third term is the sum of the products of the secon^ letters of the binomial factors taken two at a time; the coefficient of the fourth term is the sum of the products of the second letters of the binomial footers taken three at a time; and so on; the last term is the product of all tiie second letters of the binomial factors. We shall shew that these laws always Bold, whatever be the number of binomial factors. Suppose the laws to hold when n-1 factors are multiplied together; that is, BINOMIAL THEOREM. 261 suppose there are n-l factors a;+a, d7+&, d?+c,...«+A, and that {x f «) (a? + 6). . .(a? + /b) = af"^ + jooj""* + ga?"-' + raJ"** +... + «, \v here ^ = the sum of the letters afb,c,...ky ^=the sum of the products of these letters taken two at a tune, r=the sum of the products of these letters taken three at a time, t<=the product of all these letters. Multiply both sides of this identity by another factor a+l, and arrange the product on the nght hand accoixling to powers of a; thus (a? + tf ) (a? + &) (a? + c) . . . (a? + A;) (a? + Q = ;»* + tp + Z)af "* "Now p+l=a+l>-\-c+..,+k+l ' = the sum of all the letters a, h, Cy.^k, I ; q+pl=q+l(a+h+c+,.,+k) =the sum of the products taken two at a time of all the letters a, b,c,.,.k,l; r+ql=r+l{€ib+ac+bc+ ,.. ) =the sum of the products taken three at a time of all the letters a, b,Cf., k,l; td = the product of all the letters. Hence, if the laws hold when n-l factors are multi- plied together, they hold when n factors are multiplied together; but they have been shewn to hold when /our factors are multiplied together, therefore they hold when /vd factors are multiplied together, and so on: thus they hold uniyersally. 262 BINOMIAL THEOREM. We shall write the res* Jt for the multiplication of n factors thus for abbreyiation : (« + a) (« + 6). . .(« 4- Aj) (a? + = af + P«""* + ^«""" Now P is the sum of the letters a, 5, «,...^ t^ which are 91 in number; Q is the sum of the products of these letters two and two, so that there are .^ ^ of these products ; i2 is the sum of 7 » 7 products; and so on. See Art. 421. Suppose &, (;,...A!, I each equal to a. Then P becomes na, Q becomes !t^^:^a%R becomes ^(^-^(^-f) ^. and so on. Thus finally ^ w(n-l)(n-2)(n-3) , „., "*" 1.2.3.4 +a*. 426. The formula just obtained is called the Binomial Theorem; the series on the right-hand side is called the expansion of (x+ay, and when we put this series instead of (w+aY we are said to expand (a+a)\ The theorem was discovered by Newton. It will be seen that we have demonitrated the theorem in the case in which the exponent n is a positive integer; and that we have used in this demonstration the method of mathematical indtiction. 427. Take for example (a}+ a)«. Here w = ^, w(w-l)_6»5_ n(w~l)(n~2) 6.5.4 1.2 1.2 ' 1.2.3 "l.2.3 w(»-l)(n-2)(»-3) 6.6.4.3 20, 1.2.3.4 1.2.3.4 15, w(n-l)(n-2)(»~3)(n-4) __ 6.5.4.8.2 1.2.3.4.5 1.2.3.4.5 -6; BINOMIAL THEOREM. 263 m of n V. lich are theso |f theso and so ecomcs -2) a 3. a»;l?»-3 nomial 9d the Dstead eoreijii 9orem ethod thus (;» + a)* = «• + 6a»+^^-^««+ ... and as this is true for all values of a we may put x for a ; thua /t . \« t. . w(w-l) , n(»-l)(«-2) « 428. We ma^ apply the Binomial Theorem to expand expressions containing more than two terms. For example, required to expand (1 + 2a?— a^)*. Put y for 2a?— a?^. then we have (1 + 2a? - a;^)* = (1 + y)* = 1 + 4y + 6y* + 4^* + 2^ = 1 + 4(20?"" a?») + 6(2a?-a!«)2 + 4(2a?-aj»)8 + (ai?-".'c^)*. Also (2«-«*)2=(2a?)2-2(2a?)ar' + (a?3)2==4iB»-4a?3+a?*, (24? -aPf= {2xf - 3(2a?)«a?3 + 3(2a?) {jx^f- {x^^ =8a?»-12a?* + 6a?»-««, (2a? - aj«)* = (2a?)* - 4 (2a?)»a?" + 6 (2a?)«(aj*)2 - 4 (2a?) (a?2)' + (^«)^ = 16a?* - 32a?« + 24a?« - 8a?7 + a?8. - I l»IFt«lfti«Wi faiwto 264 BINOMIAL THEOREM. Hence, collecting the terms, we obtain (1 + 2a? ~ a^^ 429. In the expansion of (1+x)" the coefficients of terms equally distant from the beginning and the end are the same. The coeflScient of the r*^ term from the beginning is n(n-l)(n-2)...(w-r+2) , i*. i . u lu ^ a '^ ' ^ ^; by multiplying both numerator |r-l In and denominator by [ n — r + 1 this becomes , rT= r . •' L [r-l [ w-r+1 The r* term from the end is the (w-r+2)*term from the beginning, and its coefficient is . n(n-I)..,{w-(w-r+2)+2} thatia ^^^""'^^'"^ '- by multiplying both numerator and denominator by I r- 1 \n ' this also becomes . ^ . _^^^ . 430. Hitherto in speaking of the expansion of {x-\-af we have assumed that n denotes some positive integer. But the Binomial Theorem is also applied to expand (d?+a)" ■vhen w is a positive fraction, or a negative quan- tity whole or fractional. For a discussion of me Binomial Theorem with any exponent the student is referred to the larger Al^bra; it will however be a useful exercise to obtein vanons particular cases from the general formula. Thus the student will asiume for the present that whatever be the values of Xy a, and n, 1.2 1.2.3 n(n-l)(ii~2)(n-3) . ^ 1.2.3.4 ^^ ^-T- If n is not a positive integer the series never ends. rt^0i^ BINOMIAL THEOREM. 265 431. As an example take (1+^)^. Here in the formula of Art. 430 we put 1 for x, y for a, and ^ for n. n{n-\) 2\2 ) 1 1.2 1.2 ""8* n(n~l)(n-2) 2\2"^Jv2" / 1 1.2.3 1.2.3 16' n(w~l)(n-2)(n-3) _ 2(2"v(2"^}(2~V A 1.2.3.4 "■ 1.2.3.4 ""128' and 80 on. Thus i (.i^yf^iy^y-'ly^^t.^'^^y'^ As another example take (1 +^)~^. Here we put 1 for x^ 2^ for a. and —^ for n. n ^_1 M(n~l) _3 n(n-l)(n~2) ^ 5 1.2 8 1.2.3 16 w(n-l)(w-2)(ffl-3) 35 1.2.3.4 {\+yy^=.l^\yAy%^ ^~ , and so on. Thus 16^ 128*^ 8 Again, expand (1 +y)~"». Here we put 1 for ar, y for «, and -wforw n=-m, n(w — 1) w(m+l) 1.2 1.2 n(n~l)(n-2) _ #t(w+l)(m+2) 1.2.3 1.2.3 n(n-l)(n-2)(w~3) m(w-i-lXm+2)(m + 3) 1.2.3.4 1.2.3.4 and so on. WHs&i 266 EXAMPLBS. XML Tho. (1 ^y)-= 1 -my ^.^i^V- '"^^.^^ ^ jf m(m+l)(m4-2)(m4-3) ^_ 1.2.3.4 ^ ••• As a particular case sui^se m=l ; thus (l+y)'"»=l-y+y«-y»+y*-... This may be verified by diyiding 1 by 1 + ^. Again, expand (1 +2dr-d^)^ in powers of x, Pat y for 2.«- «*; thus weJiave (l+2«-a5*)'~(l j-y)^ = 1+2^-8^+162^-128^+ • \ :=l+|(2a;-«»)-|(2»-4jV+^(2«-«y-j~(2«-«»^^ Now expand (2a;-a3*y, (2a?-«2j8^,,, and collect tho terms : thus we shall obtain 1 3 (l+2«-«*)'=l+a?-a;*+«»-s«*+... Examples. XLII. 1. Write down the first three and the last three terms of(«-a?)w 2. "Write down tho expansion of (3- 2;»y, 3. Expand (l-2y)'. 4. Wrif» down the first four terms in tho expansion of(a^^-2y)". • 0. Expand (l+a?-**)*. ^ 6. Expand (!+«+«*/. EXAMPLES. XLir. 267 7. Expand (l-2i» +«?■)*. 8. Find the coefficient of o^ in the expanBion of 9. Find the coefficient ot sfi ixk the expansion of (l-2ii;+3««)». 10. If the second term in the expansion of («+ y)" bo 240, the third term 720, and the fourth term 1080, find Xi y, and n. 11. If the sixth, seventh, and eighth terms in the ex- pansion of {x+yY be respectiyely 112, 7| and 2» find x^ y, andn. 12. Write down the first five terms of the expanrion of (a -2a?)*. 13. Expand to four terms 14. Expand (l-2a?)-\ (-S-)-'- 15. Write down the coefficient of ^ in the expansion of(l-a?)-«. 16. Write down the sixth term in the expansion of {2a-y)'l 17. Expand to five terms (a-SbyV: shew that if a=l and &=- the fourth term is greater than either the third or the fifth. 18. Write down the coefficient of of in the expansion of(l-a?)-*. 19. Expand {lA-a-^a^^ to four terms in powers of a, 20. Expand (1 - « + «*)'* to four terms in powers of x» W/' ■''->,i_>ka^ 268 SCALES OF NOTATION. XLIII. Scales of Notation, 432. The student will of course have learned from Arithmetic that in the ordinary methdd of expressing whole numbers by figures, the number represented by each figure is always some multiple qf some power qf ten, Thng in 523 the 6 represents 6 hundreds, that is 6 times 10'; the 2 represents 2 tens, that is 2 times 10^; and the 3, which represents 3 units, may be said to represent 3 times 10^ see Art. 324. This mode of expressing whole numbers is called the cS^mmon scale qf notation, and ten is said to be the hose or radix of the common scale. a 433. We shall now shew that any positive integer greater than unity may be used instead of 10 for the radix; and then explain how a given whole number may be expressed in any proposed scale. The figures by means of which a number is expressed are called digits. When we speak in future of any radix we shall always mean that this radix is some positive / integer greater than unity. 434. To shew that any whole number may he express- ed in terms of any radix. Let N denote the whole number, r the radix. Suppose that r" is the highest power of r which is not greater than N\ divide iV by r"j let the quotient be a, and the re- mainder P: thus N=ar*+P, Here, b^ supposition, a is less than r, and P is less than r*. Divide P by y"-^j let the quotient be &, and the remainder Q : thus Proceed in this way until the remainder is less than r: thus we find N expressed in the manner < shewn by the following identity, iV=ar"+&r"~^+ and the answer Is seven times with two oven Next we have to find how often eight is con- tained in 25. that is how often ei^ht is contained in twenty^ three) and the answer is twice with seven t^Vet. Next we have to find how often eight is contained tn 70, that is hoW often eight is contained in sis^-three, and tne answer is seven times with seven, over^ Next we have tq find boW often eight is contained in 73> that is how often eight |S contained in sixty-six, and the answer is eight times \dth two oven Next we have to find how often eiffht is con- tained in 22, that is how often eight is contained in twenty> and the answer is tvdce with four oven Thus 4 is the first of the required digits.. ' y We win indicate the )*emiunder of the pfoeess ; the student should carefully work it for himself, and then com>' EXAMPLES. XLIIL pare his result with that which is here obtained. 8 1 72782 81 8210. ..2 271 8 1 1023 .. .a 8|113...6 8|22...5 A • t < 3* Thu8thenumber = 1.8' + 3.8*+6.8*+6.8»+S.8*+2.8+4, 80 that, expressed in the scale of which the radix is eight, it is 1366324. 437. It is easy to form an unl' od number of self- verifying examples. Thus, take two numbers, expressed in the common sade, and obtain their sum, their difference, and their product, and transform these into any proposed scale; next transform the numbers into the proposed scale^ and obtain their sum, their difference, and their pro- duct in this scalo ; the results should of coiurse agree re< spectively with those already obtained. Examples. XLIIL 1. Express 34042 in the scale whose radix is fire. 2. Express 45792 in the scale whose radix is twelvOi 3. Express 1866 in the scale whose radix is two. 4. Express 2745 in the scale whose radix is eleven. 5. Multiply Ht by te\ these being in the scale with radix twelye^ transform them to the common scale and multiply them together. 6. Find in what scale the number 4161 becomes lOlOI. 7. Find in what scale the number 5261 becomes 40205. 8. Express 17161 in the scale whose radix is twelve, and divide it by te in that scale. 9. Find the radix of the scale in which I3> 22, 33 are in geometrical progression. 10. Extract the square root of ^^001> in tha scale whose radix is twelve^ ^. \ \ > ^ \ IMAGE EVALUATION TEST TARGET (MT-3) denote ny given )al,nthe me year, miber jof nepomid Dne year, therefore three of ti ciin.be 444. To find the amount qf a given sum in any given time at compound intereet. Let P be the number of pounds in the prmcipal, n tlie number of years, r the interest of one pound for one year, expressed as a fraction of a pound, M the number of pounds in the amount Let M denote the amount of one pound in one year ; so that E=l+r, Then PE is the amount of P pounds in one year. The amount of PE pounds in one year is PEE, wPE^; which is therefore the amount of P pounds in ^too years. Similarly the amount of PE^ pounds m one year is PE^E, or P/2', which is therefore the amount of P pounds in three years. Proceeding in this way we find that the amount of P pounds in n years is PE^; that is M^PET, The interest gained in n years is / Pi?»-PorP(ir~l). 445. The Present value of an amount due at the end of a given time is that sum which with its interest for the given time will be equal to the amount. That is, the Prin-- cipcd is tiie present value of the Amount; see Art 439. 446. Discount is an allowance made for the payment of a sum of money before it is due. From the definition of present value it follows that a debt is fairly discharged by paying the present value at once: hence the discount is eqim to the amount due diminished by its present value. 447. To find the present value qf a sum qf mofiey due at the end qfa given time, and the discount. Let P be the number of pounds in the present value^ n the number of years, r the interest of one pound for one year expressed as a fraction of a i>ound, M the number of pounds m the sum due, J> the discount. Leti2=l + r. T. A* 18 m EXAMPLES. XLir. At simple interest Jf = P(l + »r), by Art. 442 ; therefore P= 3f l+nr* J)=M"P= Mnr 1+w* At compound interest M^PR", by Art. 444; therefore P=~; 2>=iJf-P--^^^^. 448. In practice it is rery common to allow the interest of a sum of money paid before it is due instead of the discount as here defined. Thus at simple interest in- stead of ,- the payer would be allowed Mnr fbr im- l+nr mediate payment. .": ■• '-. .'V I Examples. XLIY. 1. At what rate per cent, will £a produce the same interest in one year as £b produces when the rate is £c percent? 2. Shew that a sum of money at compound interest becomes greater at a given rate per cent, for a g^yen number of years than it does at twice that rate per cent for half that number of years. 3. Find in how many years a sum of money will double itself at a given rate of simple interest. 4. Shew, by taking the first three terms of the Bi- nomial series for (1 +r)**, that at five per cent compound interest a sum of money will be more than doubled in nfteen years. \ MI8CELLANE0 US EXAMPLES. 27 o ,* Miscellaneous Examples. 1. Find the Tahies when a=5 and & = 4 of ii^+3a«&+3a6a+6», ofa^+lOab + ^, of (a-6)», andof(tf+96)(a-&). 2. Simplify 6a?-3 [2«+9y-2{3ar-4(y-4?)}]. 3. Square 3-5d?+2aj*. 4. Divide 1 by l-or+a;* ^q fQ^j. terms: also divide 1— a? by 1—0? to four terms. 5. Simplify 6aj*-17ay+12* 6j7'-5;»-6, and 6. Find the L.O.M. of 4^-9, -+ 2 -+-+2 7. Simplify —-—-- + -—----. a—a a+a - o , a?— 2 «+6 74?-6 a Solye-5- + -^ = -g-. 9. The first edition of a book had 600 pages and was divided into two parts. In the second edition one quarter of the second part was omitted, and 30 pages were added to the first part; this change made the two parts of the same length. Find the number of pages in each part in the first edition. 10. In paying two bills, one of which exceeded the other by one tinird of the less, the change out of a £5 note was htUi the difference of the bills : find the amount of each bill 11. Add together y+x;8f-g;r, »+2^~'l^*^'^2^''3^' and firom t^e result subtiract g^-V^o^* ^ 18-2 ,^ 27e MISCELLANEOUS EXAMPLES. 12. If a= 1, & a 3, and tf">5, find the yalne of 2a?-6»+c'+a«(6-c)-6«(2a-c)+c»(2a+6)' 13. Simplify(a+6)*-(a+&X«-&)-'{«(2&-2)T(a^-2a)}. 14. Divide 15. Reduce to its lowest termB «*+«»+l 16. Find the UkU. oi ii?«--9a?-l0, a?2-7a?-30, (a?+l)(a?+3)(«-10), and««+4a?+3. 17. Simplify __ 2 3 6 -• 3. I if 5f *»IN < / I *li^ II S4 73. Resolve into simple factors o^ - So; + 2, o^ - 7^ + 10, and«'*6a;+0. t 74. Simplify ^__^^2 + ^-7!^+l6";c«-L+5- 75. Solve i(3;»+y)-^(4^-.2|)=|(6a;-i). 76. Solve 9«»-63aT+68=0. 4 77. A man and a boy being paid for certain days' work, the man received 27 shillings and the boy who had been absent 3 days out of the time received 12 shillings : had the man instead of the boy been absent those 3 days they would both have daimed an equal sum. Find the wages of each per day. 78. Extract the square root of 9a;* - e^e* + 7^— 2;v + 1 ; and shew that the result is true when x=s 10. ' 79. Ifaibiie: d, shew that 80. If a, bfCfdhe in geometrical progression, shew that a'+<^ is greater than J* +c*. \ 81. If n is a whole positive number 7*"'*'* + 1 is divisible by8. MISCELLANEOUS EXAMPLES. 281 81 Find the least common mnltiple of «'-4^, A'+CaV+lJto^'+Sy*, and d^-e«V+12«ir-8y*. 83. Solye | + - 14 3. 2' *'"y'=^*- 84. Solve «■+a» + 2^/(d^+2ar + l)=47. 1, 80. The sum of a certain number consisting of two digits and of the number formed by reversing the digits is 121 ; and the product of the digits is 28 : find the number. 86. Nine ffallons are drawn from a cask Aill of wine^ and it is then filled up with water ; then nine gallons of tho mixture are drawn, and the cask is again filled up witii water. If the quantity of wine now in the cask be to the qiiantity of water in it as 16 is to 9, find |iow much the cairic holds. 87. Extract the square root of 16aj« + 25^- 304?^* - 24a?*y" + 9a?V* + 40«V' 88. In an arithmetical progression the first term is 81, and the fourteenth is 159. In a ^ometrical progression the second term is 81, and the sixth is 16. Fmd the harmonic mean between the fourth terms of the two pro- gressions. 89. If ^/5= 2*23606, find the value to five places of decimals of -7^ — r, , 90. If ^ be greater than 9, shew that kjx is greater than^(a?+18). 91. Divide (a?-y)»-2y(4?«y)«+y«(a?-y) by (j7-2y)«. 92. Find the ao.H. and the l. cm. of 24(«'+«V+*^+^) and 16(«'-a?'y+^-y'). 93. Simplify » . y . _J 1 (I # '.*;>■, 282 MiaCMLLANBOUS EXAMPLES. 4 // 94. Solve -33- + -y-=8 ip. 96. Solve 4jy+20(«-y)=0, y«+30(y-*)=0, 84T-2Jrf>0. 96. Solve 3«*-2ar+V(av>-4ar-6)=18-f2«. 97. ^ rowf at the rate of 8) miles an hour. He leaves Oambridge at the same time that B leaves Ely. A spends 12 minutes in Bly and is back in Oambridge 2 hotkrs and 20 minntes after B gets there. B rows at the rate of 7^ miles an hour; and there is no stream. Find the distance firom Oambridge to Ely. 98. An apple womim finding that apples have this year become so much cheaper that she could sell COimore than she used to do for five shillings, lowered her price and sold them one penny per dozen cheaper. Find the price perdosen. 99. Sum to 8 terms and to infinity 12 + 4 + 1} 4- ... 100. Find three numbers in geometrical progression such tiiat if 1, 3, and 9 be subtracted from them in order they will form an arithmetical progression whose sum is 15. 101. Multiply «?*-«•+«* --«*+«^-«+«*-l by «*+l; and divide 1-0^ by l-ar». 102. Find the L.aM. of a^-a^, ^+a', ^-l-aV+a^ a^-aa^'-a^w-^c^, and a^+oi^-a'a?- a*. 2P"^ 103. Simplify a»-6»+ 1 + a+b a-b 104. Solve 6 5(|.?)-|(3.^)= 4ay-14 . «+10 i — :: — + 10 105. Solve -—-+-—: «-l «-5 18 a+l «+5 MISCELLANEOUS EXAMPLES. 288 loe. Solve afy--y£-zx=47. 1Q7. A and B travel 120 miles together by roil. B intending to oome back again talces a return ticltet for which he pays half as much again as A: and they find that B travels cheaper than A by it, 2d. for every 100 mUes. Find the price of A'§ ticket ' 108. Find a third proportional to the harmonic mean between 3 and z, and the geometric mean between 2 and 18. 109. Extract the square root of y\ yj «\ a yj no. If a :&;:&: c, shew that b*=: .f "V^t'*'^-* * 111. Dhideai-a'i hyJ-a'^. 112. Reduce «»+3aj»-20 ar*-«»»-12 find its value when «=2. to its lowest terms, and 113. Solve af-3 13 3 a+2 a-4 3 3(6-«)' 114. Find the values of m for which the equation mV+(m'+m)aa;+a^=0 will have its roots equal to one another. Solve 8a^+;c*= 10, 5ay-2a^=2. 115. 116. Solve i + i=6, 2 + ^^ Of y * y w 2*. 117. Fhid the fraction such that if you quadruple the numerator and add 3 to the denominator the fraction is doubled ; but if you add 2 to the numerator and quadruple the denominator l^e fraction is halved. ti 284i MISCELLANEOUS EXAMPLES. 118. SimpUfy {-(aj»)*}"*x{~(-a?)-»}*. 119. The third term of an arithmetical progression is IS; and the seventh term is 30: find the sum of 17 terms. 120. If — g-, J, -g- be in harmonical progression, shew that a, &, c are in geometrical progression. 1 121. Simplify a- &+ 6+ ah a—b 122. Extract the square root of 37a?V-30ajV+9^-20aJ2^+4y*. i 123 Besolve 3a^~ 14a^ - 24^ into its simple factors. ,«^ a 1 ^+5 3(5a?+l) 4 ^, 124. Solve s r ^ — — r^ = s T -24. 207-1 6d?+4 2«-l ^ 125. Solve aj»+i = ^. 126. Solve a^-y^=^9, a?+4=3(y-l). 127. Solve y+^/(dJa-l)=2, V(«+l)- V(a?-1)= Vy. 128. If a, &, c, (^ are in Qeometrical Progression^ a : b-^d :: c* : c^+cP. 129. The common difference in an arithmetical pro- gression is equal to 2, and the number of terms is equid to the second term : find what the first term must be that the sum may be 35. / 130. Sum to n terms the series whose m^ term is 2x3'", ni Simnlifv l+N/a~2^) , ^-N/a-2^ )- 131. Smiphfy J— ^^j-^+ ^ ^ . 132. Find the o.CM. of 30aj*+16iB'-50o;*-24a? and 24o?*+14aj»-48aj"-32o?. )88ion is I terms. ression, MISCELLANEOUS EXAMPLES 133. Solve «'-a?-12=0. 285 134. Form a quadratic equation whose roots shall be Sand -2. 135. Solve «*+-=«*+ -\. Or a* 136. Solve a?^ = 1 + V(aj2 + 6) "^^/(dJ2+5)• 137. Having given ,y/3= 1*73205, find the value of to five places of decimals» s/3-1 138. Extract the square root of 61 - 28 ^/3. A* J. «# 139. Find the mean proportional between — - and 140» If a, 6, e be the first, second and last terms of an ^arithmetical progression, find the number of terms. Also find the sum of the terms. /' 141. If d, c, h, a are 2, 3, 4, 5, find the values of a+b+e ah-cd , /a-l 142. In the product of l+4a?+7^+10;»' + l5a:* by 1 + 6a? + 9d;8 + 13^?* 4- 17a:*, find the coefficient of «*. Divide 2ii»»-2a?*-70a;*-23a?«+33a?+27 by 7a:"+4a?*9. «*-&* a-b "^ ^^^^WTM^¥:^k> and fjx-tja'' tjx^r nja" x-\-a 286 MISCELLANEOUS EXAMPLES. 144. Solve the following equations: (1) (2) 60-a? 3^-6 24~3;i; "14 7~=^ — r~" a?+4 5d? + 12 a?+3""43a? ^ ' , . 3a?-t-5y . 5^-3y ^ ^+l_? ^^^ ~20r"^~"8~""^' y + 2~3- 145. Solve the following equations : (3) 3«2-4a?y=7, 3a!y-4y2=6. 146. A bill of J2(} is paid in sovereigns and crowns, and 32 pieces are used: find how many there were of each kind. 147. A herd cost £180, but on 2 oxen being stolen, the rest average £\ a head more than at first : find the number of oxen. 148. Find two numbers when their sum is 40, and the «• sum of their reciprocals is 7t . 48 149. Find a mean proportional to 2^ and 5|; and a third proportional to 100 and 130. 150. If 8 gold coins and 9 silver coins are worth as inuch as 6 gold coins and 19 silver ones, find the ratio of the value of a gold coin to that of a silver coin.' 151. Remove the brackets from (a? - a) (a? - 6) (a? - c) - [&c (oj - a) - {(a + & + c)>a? - a (6 + c)} a?]. 152. Multiplya+24^(a2&)+2^/&bya-2y(a^)+2V6. 153. Fmd the o.cm. of aJ*-16«'4-93«"-234a?+216 and 4«'- 48aj" + 186a? - 234. MISCELLANEOUS EXAMPLES. 287 154. Solve the following equations : 13a?-l (1) (2) 4 • 28-6d? _^ 3a?+l = 17- 3 2a?-8 8 3a?+9 3a?-f3' (3) .-j,=3, 3(14)=llg-i). 155. Solve the following equations: «. (1) ^/(a? + l)+^^(2ar)=7. (2) 7a?- 20^0?= 3. (3) 7djy-5«a=36, 4ajy-3y8=105. 156. A boy spends his monej in oranges ; if he had bought 5 more for his money they would have averaged an half-penny less, if 3 fewer an half-penny more : find how much he spent. 157. Potatoes are sold so as to gain 25 per cent, at 6 lbs. for bd, : find the gain per cent when they are sold at 5 lbs. for 6(^. 158. A horse is sold for £2Ay and the number ex- pressing the profit per cent, expresses also the cost price of the horse : find the cost 159. Simplify V{4a2+V(16a2a;2+8aaj'+ a?*)}. 160. If the sum of two fractions is unitv, shew that the first together with the square of the second is equal to the second together with the square of the first. 161. Simplify the following expressions : a- [&-{«+ (6 -a)}], 25a-19&-[36-{4a-(6&-6c)}]-8a, [{(a-")-}-^-L{(«*")-"}*"]. 288 MISCELLANEOUS EXAMPLES. I 162. Find the o.CM. of 18a«-18a24?+6«a:*-6a^, and 163. Find the L.O.M. of 18(a^-y2), I2{x-yf, and 24(^ + 2/8). 164. Solye the following equations : 2af-4 . 3a?-2 (1) (2) (3) =7. 9a? +20 _ 4jy-12 x 36 "■5a?-4 4' ill 2 r ( (4) 2(d?-y)=3(a?-4y), 14(d?+y) = ll(a?+8). 165. Solve the following equations : (1) 32a?-5a?2=12. (2) ^/(2a?+3)^/(;l?-2)=16. (3) a?2+y2=290, ajy=143. (4) 3;c2-42/8=8, 6«8-6ii?y=32. 4^ 166. ^ and B together complete a work in 3 days which ^would have occupied A sione 4 days: how long would it employ B alone 1 2 167. Find two numbers whose product is - of the sum of their squares, and the difference of their squares is 96 times the quotient of the less number divided by the greater. 168. Find a fraction which becomes ^ on increasii^g its numerator bgr !» and ^ on similarly increasipg its denomi- naton MISCELLANEOUS EXAMPIES. 289 169. U a \h \\ c \ df shew that 1 + 1 .l-l..l + i . 1«1 170. Find a mean proportional between 169 and 266, and a third proportional to 25 and 100. 171. Bemove the brackets from the expression &-2{6~3[a-4(a-&)]}. 172. Simplify the following expressions : y xy a^ arV ' (j'' - fl' — *w)p — (w + ^ -^) 3' + (g + w) wi + m (/? — m) + j'*, 173. Fmd the o.o.m. of ar*+dation in lumber in t lodnngs, ess by 27 Find the )ur terms he sum of eater by 4 203. Add together 1 Take l+x+afl from 1 &P~5 a^-x^-^ 2 + 3a?* (2 + 3a?)«» (2 + 3d?/ * 1 204. Solve the following equations : (1) — g— =39-6a?. (2) (a + &)(a-a?)«a(&-a?). (3) 16 12 2i/=3. I'+ar-a 205. Solve the following equations: (2) 4(a^+3a?)-2^/(4?«+3a?) = 12. (3) aj2+icy=16, ^+ary=10. 206. A person walked out from Cambridge to a village at the rate of 4 miles an hour, fmd on reaching the railway station had to wait ten minutes for the train which was then 4^ miles off. On arriving at his rooms which were a mile trom the Cambridge station he found that he had been out 3^ hours. Find theMistance of the village. 207. The tens digit of a number is less by 2 than the units digit, and if the digits are inverted the new number is to the former as 7 is to 4 : find the number. 208. A sum of money consists of shillings and r^wns, and is such, that the square of the number of cro\. .i^ a. equal to twice the number of shillings; also the sum * . worth as many florins as there are pieces of money: find the sum. 209. Extract the square root of 4il?* + 8aaJ3+4a24^^ + 16&«a;2^.lgaJ2^+1654, . 210. Find the arithmetical progression of which the first term is 7} and the sum of twelve terms is 348. ( f 294 MISCELLANEOUS EXAMPLES ii 211. Divide C 212. Multiply ^ ^ 'l2+41a^+36«* 8 + 60?-- 25;rV+47«V-^5«y+62«y*-46y» by6-2«+ 264r-8««-14 4+7« 213. Reduce to its lowest terms 4aj»-46a^ + 162a?- 185 3-44T «*-16«»+81d?*-186aT+150' 214. Solve the following equations: (1) 3^-2 1-ga? 5 11 =9. (2) a?+-y=l7, y+^«=8. 18' 215. Solve the following equations : I 1 6* (1) i a?+3 (2) lO«y-7^=7,' 5y2-3a?y=20. (3) a+y=6, ;b*+^=272. 216. Divide ;£34. 4s. into two parts such that the num- ber of crowns in the one may be equal to the number of shillings in the other. 217. A number, consisting of three digits whose sum is 9, is equal to 42 times the sum of the middle and left-hand digits; also the right-hand digit is twice the sum of the other two: find the number. 218. A person bought a number of railway shares when they were at a certain price for ;£2626, and afterwards when the price of each share was doubled, sold them all but five for ^£4000: find how many shares he bought 220. 221. mi 222. 223. 2la;8-2 1^24. 225. 226. rate of at the I tanceu he run' 227. to bed daught MISCELLANEOUS EXAMPLES. 295 By*-46y» -14 \. the num- iumber of me sum is left-hand m of the ires when fterwardfl them all ugbt 219. Four ntimbera are in arithmetical proffrewioii; their sum is 60, and the product of the second ana third is 156: find the numbers. 220. Extract the square root of 17 + 12 j^ 221. Divide aj»- 1 by 4b*- 1 ; and tn{qa!^-'ra!)+p(mx*—na^—n(qx~r) hjma-n, 222. Simplify a^bx—b^afi a»-ft2_^.26c 223. Find the L. 0. M. of 7^- 44^-2107+ 12 and 21aj8-26«+8. ^24. Solve the following equations : a»-4 2-a» (1) (2) (3) =7. 7 6 174?-I3y=144, 23a?+19y=890. l^l^l 1 1_1 1_1_8_ ^""y""8* w z~'9^ z y 72' 225. Solve the following equations : (1) -^_ 21^=1 ^ ' 100 25d? 4* (2) •0076a!« + '75d?=160. (3) J{a!+y) + s/ia!-y)==^c, b{x-a)+a{b'-y)=0. 226. A person walked out a certain distance at the rate of 3^ mues an hour, and then ran part of the way back at the rate of 7 miles an hour, walking the remaining dis- tance in 5 minutes. He was out 25 minutes : how Du* did he run? 227. A man leaves his property amounting to ^£7500 to be divided between his wife, his two sons, and his three daughters as follows: a son is to have twice as much as II 296 MISCELLANEOUS EXAMPLES. A tlanghter, and the widow £60^ more than all the five chil- dren together : find how much each person obtained. % 228. A cistern can be filled by two pipes in 1} hours. The larger pipe by itself will fill the cistern sooner than the smiuler oy 2 hours. Find what time each will sepa- rately take to fill it. 229. The third term*of an arithmetical progression is four times the first term \ and the sixth term la 17 : find the series. 230. Sumtonterms3j+2i+l}4-... 231. Simplify the following expressions : _6 fi .»,. .^#- (1) '^4— +— TT=4' ^ ' 4 iC+4 ;, ., vj (2) V(2^+2)^/(4^-3) = 20. ' . (3) V(3a? + 1)-V(2a?-1) = L 297. A siphon would empty a cistern in 48 minutes, a cock would fill it in 36 minutes ; when it is empty both begin to act : find how soon the cistern will be ^led. 298. A waterman rows 30 miles and back in 12 hours, and he finds that he can row 6 miles with the stream in the same time as 3 against it. Find the times of rowing up and down. 299. Insert three Arithmetical means between a-h and a+b, ' 300. Finda?if2«^:2*»::8:l. »♦ that ANSWERS. 1. 22. 2. 26. 0. 274. 10. 89. 6. 10. 11. 6. 7. 6. 12. 5. 3. 89. 8. 6. 13. 9. 4. 564. 9. 34. 14. (^. II. 1. 65. 2. 81. 3. 94. 4. 8. 5. 27. 6. 81. 7. 12. 8. 11. 9. 21. 10. 15. 11. 10. 12. 3. 13. 2. 14. 127. 15. 6. 16. 1. III. 1. 6. 2. 16. 3. 9. 4. 224. 5. 469. 6. 7. 7. 74. 8. 12. 9. 8. 10. 238. 11. 420. 12. 144. 13. 43. 14. 15. 15. 9. 16. 2. IV. 1. 7. 2. 88. 3. 43. 4. 2. 5. 72. 6. 1. 7. 1. 8. 16. 9. 14. 10. 5. 11. 7. 12. 5. 13. 11. 14. 7. 15. 4. 16. 2. V. 1. 16a--96. 2. Sx^-Zy\ 3. 9a+96 + 9c)+abc, 39. afi+a*a^+€fi» 40. «*-6aV+4a*. IX. 1. Soj*. 2. -3a». 3. 3i»y. 4. -8a«W. 6. 4««6V. 6. «"-2a?+4. 7. -a'+4«~6. 8. a^~day+4^, 9. 6aV+a&-4. 10. 16<»V-12aJ'+9a5c"-6c*. 11. «-4. 12. a?~8. 13. a^+x+Z. 14. 3aj"-ap+4. 16. 3aj*+2iP+l. 16. flj"-3a:+7. 17. «*+ic*+aj*+«*+«+l,. 18. fl^+a6-6'. 19. «»+3«V+9*y'+2V» 20. a^-r-a^y+ai^, 21. a*+a^y+a^+ayi^+f^, 22. a^--2a»6+4a*&"-~8a6»+166*. ' 2a 2y. 15. a*««-5V' y 8. 6. 8. 9. 10. 12. 16. 19. 22. 24. 26. 28. 31. XL 1. £^+6«+c». ^ a? + ft' + c* + y. 2a-6&+4», 32. x-l, 33. («~l)(a?-h4}. 20—2 ■4i ^#r-*^'';;^. 87. (»+4)(»+6). 8i^1N^I#ir+6). 89. («-6)(ar-10). 40. («-10)F. 41. (i-ll)(«-|.12). 42. (»+4)(»-ll). 43. («-8)(#+8)frl*+9). 44. («+5)(««-e«+26), ^ 45. («-2)(»+2)(a>"+4)(«*+16). 4e. (;»-2)(4T+2)(«"+2«+4)(«*-2»+4). 47. (a+4&)(a+66). 48. {«-6y)(«-7y> ^ 49. (a+6-5c)(a+6-6c). 50. (2a;+2y-a-6)(a?+y~3a-35). XII. 1. 4. 7aV«V. 7. 4(fl^+ft^. 8. n. 0^-10. 14. d?"-6«+3. 17. «+3. 20. V-«+l. 23. «"-2. '' 26. 09^+3^+5. 28. flr*-2«*+3«"-2a;+l. 80. 47+1. 31. «+7. 2. 4aW 3. 5. 2(^+1). 6. «*-2/". 9. aj+5. 12. «-12. 13. 15. a^-6«+7. 16. 18. a^-4. 19. 21. 3^+2. 22. 24. «-2. 25. 27. 7«"+8«+l. 12a>y^. 3(a7+l).i 10. *-7 ^■+3a?+4. flj"-6dJ-6. «■-»+!. 29. a;*-3*+l. 32. x+Sff, 33. A; + a. 34. a-2a. 36. a?-y. XIIL 1. 12a»6'. 2. 36a»W, 3. 24aVajV« 4. (a+6)(a-&)". 5, 12a6(<^+6»). 6. (a+&)(iiP-n 7. (a»+l)(«+3)(«-4). 8. («+2)(4y+4)(4>«+3»+i> 9. »(2»+l)(3«-l)(4i»+3). / 10. (ii^-5a?+6)(«-l)(a?-4). 11. (4j"+3«+2)(«~3)(ay+5). 12. («i«+i»+I)(«»+l)(^+I)(*"I)« * 13. («»-^-4i»+4)(«-.l)(ar-4). 14. («"- («-l)«(d?+l)* 15. ^. 16. ?^^ 17. |(?rW. 18. ^^:^±1\ Sp a+b Z(a+b) «*+l 3y 26 a-6 oa?- 5(a-6) a-& aj+5 fl?-5 fl?-7 ,0 2±5. 11. *^, 12. 34T-4 14. 17. 20. 23. 26. 29. g-t-3 w—a «*-3ar+l' 3a^+ay+2 2«*+«+3* 1 15. 18. 4a?-3' «"4-7a?+3' 6a?- 5 13. 16. 19. a+b-a-^e' a+5 6se+4 ^^- i»+4- ^^' 2x'+dax+4a*' 24. a;+a «*+••• XVI. 1. 4. - 2eh 6a-66-c a+J+o 2a a+of aa 9. 12. 19. 23. 26. 29. 31. 37. 40. 44. abo a+h 2a-26' a 6. ¥::&' 8. aa-6i ' «-y 10 4a - a+w 11. l-9««; 2a«+9g4 606 16 ^ „ , ^ 2iP— 3 ^::6- ^3. ^^z^' 14. ^(4^«i)- 1^- (^-2)(«+2)»- 17. _^ ' ' ,^ — . 18. aa- (a;»-l)(a?-2) 2a« 6 a*-a^ . 20. 4aj» (at + l)(a;+2)(«+3)* . 2a^ 2^ 24. 2a*+6tf'y 25. 3«2 ..27.^^^±i^. 28. 4^-16 ^ aj«-l' 2a?'-9ig+44 a;*+64 • ■^* 30. 2a ^ a?—2x 31. L 32. ^^rirr^. 33. 0. «(« + l)(il? + 2)* 38. 35. «»+l • 4a?8 48a^ («»-^)(aj»-9aa) 41. 0. (1+«2)(1+^)" 39. , Si6. 2j8» 2y« ^o 2a;»+2 ^^ 4(a*a^~5V ) 45. 246* a(a?-&«)W*-46») 13. 13. 16. 20. 23. 26. * 30. r /^': ^MM 811 At 33. 0. 4ft 49. 47 w(a-¥b)^ab i0^4(m^V)' *^- («-a)(*-i)- ^' (w-a)(m^h) («- 59. £2^8. 60. 200000000 tree, seconds, 61. .40 miniktet past eleyen. 62. £300 and ;£200. 63. 14. 64. 640. XXIII. 1. 10 ; 7. 4. 4 : 1. 5. 5 : 5. 2. 17 ; 19. 6. 21 ; 12. 3. 2; 13. 7. 20: 10. 20. 13; 5. 24. 5 ; 7. 28. 12; 3. 32. 2; 3. 21. 9 ; 7. 22. 10 : 4. 8. 2; -a 9. 3;2, 10. 3; 2. - 11. 3i; 4. 12. 10; 7. 13. 19; 2. 14. 38i; 70. 15. 6; 12. 16. ftf ; W 17. 10; 5. 18. 12; 12. 19. 20; 20. 23. 4 ; 9. 27. 10 ; 8. 31. 3 ; 2. 35. a I &. 25. .2i; 1. 26, -2; '2. 29. 3; 2. /33. 4; 12. 30. 63; 14. 34. a ; b. 36. 39. ab 4h — • a+b* a+b' ae be V 37. 5 ; a. 38. 40. ;0. a-hb* a+b' a+b 42. a+bia-b, 43. (a+b^iia-b)\ 44. 41. a; b, e e a+b* a+b* n XXIV. 1. 2; 1; 3. 2. 3; 4; 6. 3. 2; 1; 3. 4. 9; 11; 13. 5. 4; 0; 5. 6. 5; -5; 5. 7. 45;-2Ul. 8. 10; 7; 3. 9. 51;76;1. lA 2 3 2 ■ 3' 4' 5' 2 12. a?= g (a + 6 + c) -a, &c. abc 14. a:=y = z = .^^^^^^^, 16. «?=3,a?=4,y=5, ;2f=2. 11. a?=-(6+c-a), &C. 13. x=^(p+c\^. 15. a}=a,y=bfZ=e, XXV. 1. 42; 26. 2. 12; 16. 4. 24; 60. 5. ZOd.;Sd. 6. 49; 21.\ 7. 3. 116 ; 166. 4_ 15* 8./45; 63. . 9. 72; 6^ 10. 30<;.; 15fc ^2. 25. 2, dbl. 26. 0, ^J(db). 27. a, -2«i) -2a. 28. a,^, -|. XXYIII. 1. 36,24. 2. 36,24. a 30,24. 4 18, 12, 9. 5. 12, 10. 6. 4, 6. 7. 196. 8. 3,48. 9. 11. 10. 7. 11. 6,12. 12. 15. 13. 24. 14. 27 lbs. 15. S9.9d.,7s. 16, £20, 17. 126,96. 18. Sd, 19. 10, 9 miles. 20. 56. 21. 192,128. 22. 9 gallons. 23. 64. 24. Equal. 25. 4 per cent. XXIX 1. 5, -4; 4, -6. 2 4 -?2. 1 J-1 * 7 ' * 35* 3. a8; ^6. 4. 6, 12; 2, -4. 5. 7, -4; 4, -7. 6.4,-g;3,-g. 7. -24,?; 12,^. 8.6,-i;5,|f. 9. 2, -|^;4, -y. 10. 6,0; 5,0. lUg,0;|,0. 12.3,6;|,|. 13.4,|;8,^. 14. ^,0;^,0. 15.0,5. ANSWERS. 317 4. ,e.«,e»z«l».j,«!i;S?. ,7, a+& a+^ 20. ik5: db4. V2 iaa»0;Oy&. 19. 1^4,^4^; "^3, 21. di7; 1^6. 22. ikl5; 1^7. 23. ^4^ ikH; ikl, >v4^ 23 24. d>9; iii4. 25. ikS, >i>36; >t5, 7 ^-^ 26. ik9; ikS. 27. •kS; di6. 28. ^2; ^\. 29. ±9, :^8^/2; si.7, ^^1% 30. db 4 : db 1. 15 9 31. 0, 1, ;;^ ; 0, 2. ik 22 32. (g-i-l)ft . . {a-Dh 34. :fea. db ^/(2a8+2)'^^(2a>+2)• a+1 .«-! 33. ±a,:fe a-¥h ^/2 ±&,± 22' 7^' 36. 6,-4:4,-6, > ^> 36. 5, 4: 4, 5. 37. 4, 2 : 2, 4. 38. 4, -3; 3, -4. 39. 1|2;2,1. 40. :b4» :fe3; :fe3, ii5: ifa3. 43. 2,1,-1, -2; 1,2,-2, -1. 44. ^ -2*^3 -1*>/13 2'- 2 45. 3, -r ; 6, - 1, -2*^3, -1tn/13 3 2 3 46. 5, -=; 2, -r 3 3" 47. 2: I, 48 4?.l -?.2?^ -? ? «• *>2»4» 4' ^'2* 4'4- 49. «+6+l, - a+5+l a+1 a-fl 60. 3 ^tb* 51. ±2&. 52. 0, a+ft, i(a-&)±5N/{(a+8d)(a-6)}i '^>«+^s(«-^)=Fo'^{(<»+3&){a-&)}. 53.a?=a-s-/y(adc); &c. 2 2 C4. («+yXy+*X*+*)= *<* 20. a^+aM+6*. 21. ^164?"^-12a;~*y"^+9y~\ f V SSO ANSJVSBJS, 22. »+y. 28. J-ahKbK 24 aKb^n-A 25. «*+2ara^ + 3a^*a+2«*a*+a'. 26. w^-ix^y +yK 27. «*-2«"^. 28. a-Z-a'K 29. aj*-5«i+«^. 80i 2a^-8+4«"i XXXIV. 1. 7^/2. 2.9/^4 8. |^/3. 4.^. _ 13^/15 ^- 10 • 9. 4+|^/2. 6. 6J^2 3 7. 2+2^/2-2^3. 8. 2+^V«, 10. 5 + 2^6. 11. K 12. J ( 18+9^6+4 V15+6V10 ) 6 33 K 8-^/7. 15. ^6+1^2. 16. 13. 8+^5. V 2" ^2' ^7« fj^-fj^* 18. 2+^/3. 19. V3. 20. ^/lO. XXXV. 1. |. 2 1 5 2 3 8 8' 3' 4' §• 3 ^ 3- 27- 14,21. 6. 24, 30. 6. 20,32. 7. 1. 15, 10. 9. 6,8. 10. 35,42. 11. 4 12. ^^. la 50, 60, 9a 14. 0, 2 : 5. XXXVI. 1. 14 2. la 3. 15. 4. 12. 5. 4. 6. 4. 7. 2, 2i. 8, 6. 9. 1^ -1. 13. 45, 60, 80. 14. 4, ^9. •II XXXVII. I. 4. 3. 5:2. 4. 2. 5. 4 6. 0. 7. 8. 8. a((?. 9. -p-. ao. j$113}. lU 15.^ 12. £15360. 5. 9. la 16. i^ifiimMii'-^ ANiiWJERS. 821 27' XXXYIII. 1. 936. S. 77i 8. 69. 4. 139}. 6. 37}. 6. -116. 7. H16,18. a 14}, 14J,... 9. 6J,5,... 10. -|, |,... 11. 10,4, 12. 82. 13. 5,9,13,17. 14. 5,7,9. 16. 1,2,3,4,5. 16. 18, 19. 17. 7. 18. 5. 19. 1, 4, 7. 20. 1, 2. XXXIX. 1. 1366. 2.13}. 3.40}. 4.63(^2 + 1). 5. 666 648' 11 1' "• 33' 6. 463 96* 12. 11. ^ 333* 7 ? 7. ^. 212 ^^•496' 9 ? 14. 10. 4}. 667 1980* 15. 4,16,64. 16. 8,12,18,27. 17. -9,27,-81,248. 18. 3, 12, 48; or 36, -64, 81. 19. 1, 3, 9,... 20. 3, 6, 12. XL. 1. g, -,1. 2 1 2 2 ^ A 1 ^' 5' 13' 2* 3 a i? 3. 8, J. 4, jj, jg, ^. 5. 6, 12. 6. 36, 64. 7. 1, 9. 8. 3, 9. XLI. 1. 134696. 2. 6040. 3. 126. 4. 30240. 5. 11. 6. 1900. 7. 16604; 3876. 8. 27; 99. XLII. 1. a«-13a"a?+78a"aj«...-78a«a?»+13a«M-«". 2. 243-810ajVl080«*-726aJ«+240«'-32»*! 3. l-14y+84y«-280y'+560y*-672y»+448y*-128y'. 4. «f +2««->y+2n(«-l);U"-V+ — ^^-*^^«""V. o 5. l+4a?+24:»-8a;^-6««+8a?'+2««-4«'+i»'» 6. l+6« + 16aJ»+ 30«» + 46dJ*+ 61«»4- 45««+ 30a?'+ 1&B»+ 5«»+ «»•. 7. l-8«+28«»-66««+70d?*-66a?»+28««-8«'+«'. 8.5922. 9.1690. 10.af=2,y=3,»=5. ll.«=4,y-2,n=a 1 a"^a? ZaT^o^ la'ha^ *J*Iar^ai^ !. «• 7i 7^ rs 77J5 — • 12. 13. 1 + ^+^ + 2 8 16 128 2^ Z^ 64 • T.A. 14. l+2a?+4«»+8«'+.M 21 322 4NSrrBRS. \. ■ 16. r+1. l«: tIJJjJ»J!(to)-l-y. .U _!• .1* 1A40 t* 4040 ** 18. (r+l)(r+2)(r+3) 1.2.3 1 S^ 8«* 20 1 + 5-.^-^ ^"- ^^2 8 16' XLIII. 1. 2042132. 2. 22600. 3. 11101001010. '4. 2076. 6. H592. 6. Radix a 7. Badix 6. 8. 9e&l\U, 9. Radix 5. 10. eee. XLIV. 1. 5£. a 3. n = -. .. MlSOBLLANVOns. 1. 729, 369, 1, 41. 2. 41^-51^. 3. 9-30*+37«*-20««+4aj*. 4. l+a?-4J»-«*, l_ay+aj«-a:*. 5. ^.t^r^ . 6. (4«?-9)(9a«-4). 8. 3. 3«-4 9. 240,360. 10. £2, £21. n. ^%|^.|,^+1|? + |, 12. 1. 13. 3J«. fl!* + « + l* 59 14. 2«»-d??y-22/» 15. 16. :(^-10)(..+ l)(a,+3). n> (^-10)(a^-H)(ar.f.3) - 18. 5. 19. 7. 20. £40. 21. 2a-26-flf-2y, a+36+4j;+4y. 22. 11. 23. «*-«*. 24. ^ + ^-7. 2 3 4 25. ««-2. . 26. ^^. 27. (16«»-l)(«=»-4). 28. 6. 31. i. \ 62-4a2* 29. 14«, 21*, 52J#. 30. 100. d2. (a^-.a«)(^-&2),' (a?-aK«-ft). ANSWERS. 8S3 a 1 83. 44?*-2«v+«V*-«y'+^. 80. 8(4r-y) 34. ^-2. 36. 1. 37. 4. 38. 2. 39. 30 minutes. 2(3««+j^' 40. £\Sym. 41. 1047+10;2r. 42. 7d^-:2d72^-t-2^*, -«"-6«y+7y*, 12jr*-10^y-jr»y«+20«y8-12/. ;^. 46. (««-4)(4;>-9). 43. a+6-& 44. «•+!. 46. 47. 4P*+jr+2 48. 1. 49. 16 50. 30 lbs. fil. 3a*-6a»&-12a«&»-aft» + 3&*, 3a^-8a«5-4a&«+3&'. 52. 2aJ-5. 53. 2, 64. (^+51*. 55. 1.2. 66. 3; 6. 67. 5; 8. 68. 4; 6; 2. 69. -^ ; ^^. 60. I . 61. 4^+d;a+l+l + i. 62. iij?+2 63. 25: *ii4. 107. 20 Bhiniiigs. 108. 4a a y 109. - + i-^ V a 112. 6 111. «»+l+ar"". SIS 113. 7, J . 114. 1 or -3. llff. iii2; rfil. 117.? lift 1111 "®- 8* 2' 2' 8 3 118. -«-•. 119. 612. 121. 2a'y-ay+y 2aft«-6»+a-6 122. 3a>*-6;»y+2J^" 123. «(3«+4)(«-6). 124. 17 125. 2. ^ lAA M 13 . 5 126. 5, — T ; 4, 7 127. 1, ^; 2: ^ 3 129. 3. // 130. 3(3»-l). 131. - X 132. 2dr(3a;+4). 133. 4, -3. 134. «»-i»-6=0. 136. «*a=a* or -;, 136. *2. 137. 819615. 138. 7-2^3. 139.^^. ay 140. g4-5— gg («+«)(/6. 181. 15. (2) 6; 8. (3) 4, -j. 185. (1) 13, -15. (2) 7. (3) 2, -1. 186. 288,224. 187. 29 miles. 188. On the first day A won 8 games and lost 4 games. 190. -85j^. 18aj*+12«'-43«"+36a?-18 6aj*-20aj«+d?+36 191. 192. 144 4««-15«+13 «"-6«*+lla;-6' (2) 7. (3) 40; 16. 4,2; 196. 56 miles. 199. a«-a6+6«, a«+6». 1 + 9j?-13«» 193. aj*-16y*. 194. (1) 8. 195. (1)|, -|. (2)13. (3)2,4; 201. 197. 24. 198. 23+15. 200. 2,4,8,16. 16«» 3(7-2iif) 202. a^-2ii?+4. 203. (2+3ar)»' 2^ n^fiJ. 204. (1)9. (2)^. (3)6;& .»■'»■ 326 ANSWSJRA 206. (1) 7, |. (2) 1, -4 (3) is J ±2. 206. lO ttdlea. 6 207. 24 , 208. 6 crownB+ 18 shiUings. 209. 2«»+2aa?+46«. 210. 7,11,15,... 211. 3aj»-2ajV+3d^-6y". 213. 4^-26^+37 212. ^ 12 + 6«-28«** ^-lO^^rral^Zao- /^^^ (1)9. (2)16;4 (3) 3; 6; 9. 216. (1) 3, -6. (2) J.7; *6. (3) 2,4; 4,2. 216. 114 of each. 217. 126. 218. 21. 219. 11,12, 13,14 220. 3+2^/2. 221. «•+«*+ 1, paj»+g'aT-r. 2^- 5^' ^S^^ ^'^- (7^-4)(3^-2)(««\.3). 224 (1)9. (2) 23; 19. (3)12; -24; 36. 226. (1)28, -3. ae he (2) 100, -200. (3) 2a+2V(aa-6a)» 2a+2^/(a«-6«)* 226. j^ of a mile. 227. 600; 1000; 4000. 228. 2 hours; ,'/ r 4 hours. 230. f|(9-n). * • 244 64 246. 3,6,8. 242. «r+a. 243. 106 shil- 246. \ 6} per cent 247. 2S^ 248. ^.24^. 6-678, 1234. 251. 2a- d. ■'W^, ?62. a"-a^«, e, ANSWERS. 253. a(2ai'2at). S27 264. (1) 6. (2) 114; 77. (3) 0, |. 256. 112; 96. 256. ^has £6400, J9 has ^200. 257. 7; 13. 258. 80. ' 269. 8; 6. 260. ^080. 261. c>+2&<;. 262. fl?»»-l, -^,(ai?*+3flWJ»-4aV-3a»a?+2a*). 263./-^+l+i+i. 264.gl^3,l. 265,(1)?. (2) 1. (3) 18; 9. 266. (1) 3, -2. (2) 5, ?. (3) 2, 3; 3, 2. a I 268. 45 shillings, 30 shiUings. 270. a^+^-:^» 5-2^/2. 3(a»+^ 271. 0. 272. «»+3a?+8. 2 3' 273. 12a«-8aa?+6a^ ^ 4a?* -.n-^r^ 15aa+ -S. (2) 7, -—. > (3) 1,5. 297. liiraiimtei. 4 298. 4^ hours with the stream, 7^ hoiirs a|;aiiist the stream. 299. a-^h, a^ a^r-h* 300. 3,-1. // THE MD. OAMBBIDOl: PBINTBD AT TBI UNIVIB8ITT niSS. I \ ti I J