i X /XC tA<fW aX-gr; LECTiIC EIDUCATIONAL SRIES. 3I av's a r4 rke h t i&. 5-D B A YoS S LC G-E PS R A~ P1ART FPIST: ON THE ANAA'TIC AND INIDUCTI1VE ET IiO i) S O F N S TIt> T C TtON: N W'G I. T N IR 0 U S P RAC T IC A E R t I SE S DESIGNED FORl COIMMON SCHOOLS AND ACADEMIES., BY'JOSEPHI El";, Y. D. PITO 0 FII 0 A TI IEMIA T LS I I \iOOD DWAIt D OoEI I,{1 L SE VI TIS' D I Ti. Ot P UB L I S I- E I, S: CINCINLNATI, WINTIIROP B3. SMITH & CO., No.{, M'AIAN STREET. TYPE ENLARG-ED-RNE. W STERtEOTYIPE PLA.TES..Each PART' of the Arithmetical Course, as elZl as the Algebraic, is a Compete book in itsslf and is sold separatelyg TIE C'IILD'S ARITHM IET IC. RAY'S.ARITLMETIC, PAR'T FRlST: simple mental Lessons and tables for little learners. MENTAL ARITEUIETIC. RA.Y'S iARITT METIC,. PAR[T SECOND: a thorough course in mental exercises, by induction and analysis; the most comr plete and interesting ntcllectual arithmetic extant. PRACTICALO AR ITH{'[ETIC. RA.Y'S ARITIItMETIC, PART THIRD: for schools and academies; a full and comnplete treatise, on the inductive and aa.lytic mnethods of i:tatrulction. tEY TO RAY'S ARITHMETIC, containing solutions to the quoes tions; also an Appeondi, emhbracing Slate and Blackboard exercises.,LE~JTE TARY AIGEBRA, iAY.'S ALGEBRA, PART FIRST: for comlunon schools and. academies; a simple, progressive, and thorough elementary treatise. HtIGH R ALGEBIA. RATY'S ALGEBRA, PART SECON-D: for advanced students in acadcnmies, and for colleges; a progressive, lucid, and comp0rehensive'work. KEY TO RAY'S ALGEBRA, PARTS FIR ST ANlD SECOND: complete in one volume 12muo. Entered according to Act of Congress, in the yeatr ]iNghteen Hundred and F'ortyEight, by- W~iN'Tr-soP Bi. S.aTIL,. in the Clerk's Ofice of the Disrict, Court of the United States, for the District of Ohio. C. O'. oDISCOLI, TrltSEEOTP-; (CeieLNoATI 0. PREFACE. aTiE object of the study of MIathematics, is two fold —the acquisition of useful knowledge, and the cultivation and discipline of the mental powers. A parent often inquires, "W by should my son study maTthemnatics? I do not expect hlim to be a surveyor, an eQngineer, or an. astronomer."' Yet, the parent is very desirous that his son should be able to reason correctly, and to exercise, in all his relations in life, the energies of a cultivated and disciplineid mind. This is, indeed, of more value'than the mere attilnment of any branch of k now ledge. The science of Algebra, properly taught, stands tlmong the first of those studies essential to both the great object-s of educlation In a. course of inastruction properly arranged, it naturally follows Arihtleti, and should be taught inrmnediaLtely faey it. In the iollowinfg work, the oubject has been, to furnish an elenment.ary treatise, commencing with the first principles, and lesadingt the pupil by gradtal and easy steps, to a know ledge of the elaments of the science. The desilr hbas been, to present these in a brief, cleas~L and scientific man ner, so that the pupil should not be taugoht merrely to perform it certiain routine of exercises unee1.harniCally, but to understand the wchy and the antiejroefsoe of every step, For th.is purpose, every rule is denmonstrated, and every principle analvzed, in order that the mind of the pupil may be disciplined a.nd strengthened so as to prepare him, either for pursuing the study of {ath'emnatics intelligently, or more successfildly attending to any pursuit in liife Some teachers may object, that this work is'too simple, aind too easily understood. A. leading object has been, to miake the pupil feel, that he is not operating on unmieaning symnbols, by means of arbitrary rules: -that Algebra is both a rational. and. a pra(ctical subject, and that he can rely upon his reasoning, and the results iv PRl E F A C EE e of his operations, ith the stanze confidence as in arithnietie. eo r this purpose, ihe is u'rn r hSe, at alm ost every st;ep, -ith t.he means of testing th aciracy of the principl'-es oin which t`he ruesn are founde, and of the resllts htich they piroduce. T hroughout the iwork, the aim has been, to combine the e clear, explanat;ory methods of the French Lmathentaticlanus, Rith thee practical exercises of che.t nglisht a-nd Gerimnln, so that the pupil should acquire both a pyactieal and tcoretical 1knowledge of the subject, IWhile Ce ery pae is the eever-R of othe autors on refecntii.on, and the expesience of imany yeares itn the schoolroonm, it iS also proper t' to state, that a large number of the best treatises on the same sujgeec-, both English and French, httve'been carefully consulted, so t1hat the present work mioght emsbrace the issodern and mnost app-rosved methods of treating the various subjecets presented.'Vit; thes tie. V.i... he o ork is submitted to the judgmen t of. fi:.2llow laborers 7 in t1e field of education.'WVT51IOD7.\) "h I m -to'-' t J 08A34 QtS t'- G E' OT I 0 S TO T E AC E IE. E GS. It i.sin-iended t.l:i the pup)il shall recite' t, e 2 ntsel.ectuae.l BExercises witi tIhe book open beforl hlihn, as in mIlntakl Aritihmetic e. A.Tdvsancd pupils mag. omit t ese exercises. The lo ing sfubil iie' s may b' e omitted by te youtnge r p ei upils., a"I. paissedt oer by those or:l adva.2kncetod un'tilt the book is nreview'ied. Obserevations oi Addition i.land Subatsraetion Articles 60 )-t. ste s'gratecr parx of ChaiO ter!Ii Supplns-e.- t to 2quoea'sm of the irsit Degrees, Aa.tices 1" 1, —l lrolepCrties of the 1'.oot of an Equati lcon of'the etconbead Dogre, rticles 215 —217. I.T revioswin.' tlio boo.k., t>e Puphl t 5 1-ald de monsl.rato t'ihe rules on tale blaobeRo'ad.'T'he weor'wil be f)Its d t? cantai n c rs rgae inuaber of examples for r,(ctie oni. an in si strto d e these too numnerous, a portion of t lhe:t ma.y be omitteld. To tench the subjet sunccesflylSy, t.he principles must lbe first clearly explsaned, atnd t.hou thie pupil exercised in the solhtion of alppropriate examtples s ntil they tre rndterted perfectly fisnnti.ar. C OlNY' iNT SP ntetolleeeuafl Exelc sesxorl, XV'.f..sosn o. 7 — 2o.t Cil'[LPT lERM. L -FU-N]TDAM[En TAR".Pi RULES. Preimirnat:y Definitions'ad'Princ.ipleo s..... 301. 215 — 26 Delinitions of Toerms? ant Ex plu.notion of Sig 1ns6- 2( — 11 toxa.mples to ll.-ustratoe se us of the Signms. 1e — 33 Addition........... 53. 55...- 39 Subtra.ction...............5 — 59 39- 43 Observa-tions n Additioa- ond clbtaeu limon........ 60 64 4.3 46 Iu'ltoiliatill- 11-lo of the Cofifi0ciets 67 4 48:l t1ie of thLe 6lxpon. eni...n e9 48o — 50':leTlrtl 3]1ule for iLhe Signs.. 0. 72 5n Genieral L,~,te..i;,''iptii a tio n....53 TDivision1 of."'['''..olfloost'. Llle of l' e ~e Si..nv.t 5..- - 5:[%ly;eaomi~eals!e.ulo.......... 7 9 B5 6 3 CITfTi:1'I:~1:...I Wi-UgT I.:.:1'mI:S, IFATO' IAi[N c'. balgeric horem<l s.0.... 80- S 63 — 6V8 },c~or~irg'................. IT- 96 68- 7 131 Gre atesfot Conmlor, Divisotr.7 - C -- 80.,ea st Conmon;idultiple'.......' 1.0' 112 80) 82 C:tLAIYTE' l I IT I-ALGEB IC R.. AC'OtIN..C.LTOS. Defillitoions "n31d l01.d:0-e:a3tl':Propos50 iols. o. 1.3. —!.27 83 87 To redule a lb'actL. in to t Loest lerms...2 87 —- 89 a FSritio":. to:m En-tie or Mi.[ixed Qluantit y o.. to ixed. Quantity 1 to a stracti e on t.. 0.;90 Si.u:S of F'ractiolls........... g9( Tfo redtuce Fr'actions to a C ommon If)enolinator-.. 3 9.>- 97 the Loeast Comlnonl Denominaitorns 13t 901w o5 To reduce' (Quont33ty to a Fract;io3n with a givel Denloml.ilnato r 95:To cosnvert a "orac'ttion to 3nother with 3 gia neon Deoiir. 96 Atddit.iou and Stbtra ctio of Fra e ion... 137 nI38 97 - 99'To -oulti31yrply one i'raotionalt uanntitny by aothero.. 139 —10 o0 0-10I T5o divide one rat3 tio-nal lQua&ti;ity by asnother 4, o:.03.107 To0 rel-urce a CorlXlex Fraction to a S imple one 1. 43 07....108 itesolution of Fracetions i~lto Series..... t i S109 CHATlfftll:1X.:r[ —(T,]QUATi N0XS OP THE'T:iI FI[!RST D. Gt RE, Definitio lls n.1 Eleen t ary Principles...45 —. 152 1't0 —-J12 Traonsopoitoll'..... o o.ot53 2 1i —113 To clear'an Equatison of Fractions...... 15 8115 Eojqt0;ir>ns ots ihe oirs.t ODeglee eontaeing oxne'Uinktorosn (u:1 tity. O. o......155 15 —119 Questions producing' ~quations of the First DegroeeX con-th11biin, o3e i QUnknown' uant.ity....... 1..56 e11 —-1 31t ioa'lteions 0o'f tle FirSst Degree containing'two Ukitaoown o Ql as-, tit....ies..7:~ ~ ~ ~v9xvet.~~15' 13 2.. ~V ~~i CON TENTS T. ARTICLtES. P AGES. Elinimination-by Substitution o O a....... v].58 32 by Comparison...... 159 1 33 by Addition and Subtraction... 160 134-136 Questions producing Equattions containing two Unknown Quanltities........ i1..1.316-142.Equations containing three or maore Unknown Quantities 10l 14.3-146 Questions produciug.i:quations containgin three or more U11known (uanaQtoon..............163K 147 —159 CHAPTER V'I-SUPPLEKSTENT TO ]QUATIONS OF TiJTE ]iI.rSTE DGiEE. G-eneralizutionuFor.uma-tion of oRules —ales anpes 16 1-itO 150-11.58.Negative Solutions......... 17 159 Discussion of Problems.... 1o3 - 61 Problems of the Coutriers....... 1 3-165 Cases of Indetermiination and Impossible Problems 7t177 1 65-16 CH IAPTEIt 11t P0OWEI. 1S-I105 —A.RPzDICA LSo nvoliution or 0Formatioun of Powers e.. 178 -6 To raise a Monomial to any given Power.....17'S68 Polynomial to any given Power.... 181 170 Braction to any Power.... 182 J.71 lBinomial Theore em.... 183 —186 1.7 --' 76:Extraction of the Squarea Roolt..... 176I Squarse Root of Nsubers..... 1.87-190 17 — 17 Fractions....... 91 179 Perfect and lmpersect Squares-Theorem n 192 -i80 Approximate Squar e Roots.. 19&-03-19'4 181 —-183 qtluaire lioot of.lo no*ial. s 195 18; —184 Polyhomiilis...190 i$.4 ~S1 Riadietals of the SoeconS.. Pe.'ree —efinitsions. 187s Addition.n.......s90 Subttraction.01 190 STM ultiplcsatio.n. 202, 191 Division... 2 3 92o.To renlder Rationalal, ihe Denominator of a Fractin con co-taning Iadial S...... 20-4 t93 rhmpcple Equthations containing tadieials of the Secoold Degree a 2105 195 —197 CIAPTElR YII —EQUATIONS OF TOII:E SECONPD DEGREI, Delinitions ard. or-ms!.,...0I3 — 28 197 —19 Incomiplete Eqluanti eons of tIhe Second Degree' 09 —210 1.98 —00 Questions produnL in.?; encomplete:Iquations of the Seeond l egree, 211 200 —201 Complete 1E90uatolns of the Second- DeTgee O O 21 2 202 General it.ule. for the Solution of Complete teliuations of the iSeond Derzgee....... 204 —20, i:indoo Method of solving Equatilon s of the Second Degee 213 207I Qucstions prodhucein Complete Etquations of the Secoin Degree, 21'4 209 —213 -Properties of the lloots of a Complete Equation of the Second Degree......... 215.218 213-217 ]lquations containing two Unknown Quantities 219 217 —220 Questions produ.cing'quations of thse Second Degree, containing two Ulnkinown Qutantities.... 219 220-229 CHtAPTrlER VIIIPROGROESSIONS AND IROP'ORTION. Arithmetical Progrression. o.. 2920 —225 2?22-227 Geomuetrical Progression....... 226-230 228 —232 Rlatio o. 233 o.1-239 232- 93i4 Propolltion.o o..... 240-255 2 34- 2 <A L G- E B F A) PART FIRgST. INTELLECTUAL EXERCISES., LESSON I. Nore TO TiAC 1. Ens:. -All t-he exercises in the tIloowing lessonts can be solved in the saene nmanner as hi intellecetual arithlmetic; yet the instruetor should require the pu13pils to perorni t1hem after the m-aniner l"here indicated. in every question let the answer be verified. I. I htve I5 cents, which I wish to divide between'Wi]litam and Daniel, in such a marnner, that Daniel shiall have -twice as many as Williamt; what nuimber must I give to each? If' I ive WXillias ar cea2crca'2 s.vt/Ser, and DlTniel tiwice i thiat nuns.cer, both w ill lave 3 tiules that cettetela snbo' but l:othl. toei-ther are to Ih.,:te I 5 cetts,; -hence, 3 timles a cert:iai nunmber is. 5, Now, if 3 thncets a cortain number is 715, one-third iof 1., or 5, irusit be tthe cucnbero r ience, William recCeived 5 ceits, and Daniel twcice 5, 0 1 0 entcs. If; instead of a celbtizeu vnzmbcr, we repcicresnit the numncber of cents W'illiam is to reci eT, by -, -then the nulber Daniel is to receive,will be representecd by 2z, and cwhat both sei ceive will )be lrepresented. by a added to 2x, or 32. If acx is equal to 15, then Is or c is equal to 51 The learner will see tiat the tcwo methods of solvincg t'his quest ion are theb sane icn iriniciplce 3but -thlat it is snore -oet enirsnt to repcresent thwe ouatity wc cwish to find, by a Single letter, thae bI y one or more -words. In the sance manner, let the learner continue t;o use the letter a to represent ithe imallest of the required nuimbers in the, folloW in questeions, RAY'S ALGEBRA, PA T FIRST. s osr 3E —.x is read x, or one, X, and is the same as Ix. 2: is read'.two T, or 2 times a. 3sx is read throe x, or 3 times a, and so on. 2. What number added to itself will niake 12? Let x represent the number; then z added to a miakes 2x, which is equal to 12; hence if 2x is equal to 12, one x, which is the half. of 2x, is equal to the half of 12, which is 6. Vr rIFICeA lZO-.-6 added to 6 makes 12, 3. What number added. to itself will make 16? If x represents the number, what will represent tshe number added. to itself? What is 2z equal, to? If 2a is equal to 16, what is x equal to? 4. AW-hat numb-er added to itself will mak-e o24? 5. Thonmas and Willianm each have the same nuimber of apples, and. they both together hsave 20; how many apples has each 6. James is as old as John, anti the saun of their ages is 22 years; what is the age of each? 7. Each.of t-wo men is to receive the same sum of money -for a job of Awork, a.nd. t:.ley both together receive 30 doilars; what is the share of each? 80 Daniel had'iS cents after spending a part of them. he fbLound l.e bad1 as m-any left as he had spent; how nmany cents had he spent? 9. A pole 30 beet high was broken by a b!ast of wind; the part broken ot; was equal to the part left standing. what twas the length of each part? Instead of saying x added? to x is equ~al to 20, i~t is moraue convenient to sta:y z prs a is equal to 30. To avoid wariting'the word pluis, we use the sign - w, which mean.s the same, anud. is called the sig. of addtiion. Aflso, insteatd of writing the Word epzi.t.a,'we use the siln -, whbich rmeans the same, and is called thie s.ig. of eqsatlity.:.::0 John, Jmues, and'Thonmas, are each to have equal sha.res of L2 app:::lples; if az represents John's sha're, s at s al tep're sent tme share of JSa:tmes?'What will represent- the share of Thom. as? What expression will represent a2- i a more lbrie f yl If 3ex=1l2, Wth.at is the value of a? Why? II. The sum of four equal numbers is equal to g20 if a represents one of the numbers, what will represent each of the o'thers? Whlat will represent x-t-xz —-a. -z, ~more briefly? If 4::=220, what is x equal to? Why? 12. What is x+-a equal to? Ans. 2a. 13. Whtlt is x+-ax- equal to? 14. Wh-at is - taaa 2 x'+a equal to? TNTIELLECTUAL EXEt RCISEtS ) LES SON It 1. J' Amvs aend Joohn together have 18 cents, anld Jbhn has twiice as many as James8 ho'wr many conts. has each? If s represents theo numlber of cents James has,% what will prepre sent th number aoh h' s? W rhat will represent the Lnumlber th]ey.::both have? t f S a ixs eql to ]8, what iw s z equal to?'Wrhy? NoT, —If -thle pupil does not r eadilly perceivre howf to solve a quresition, let the insrt-ructor ask questions shnilar to the pLrecedhq2.z:1 tO A tra'vels a errtan dshrance one d,'aid tvwice as:fr the next, in th.e two i t 3dys hue raves a3 i how ft does hel trawel cacaJ d ay? 3'i The sum of rae ag es of Sarah ral d? me i 1> J5 yTe;a, raid. the at of Jan'e:is t wice that of -tah w a is "a'' a' "'a? 4. The suan of two inut)be rs iis I, auid thc. large r is 3 ti"nics the,smaller'; what are the nunibers? 5. What nunber addted- to 3 thioes 1t 2 l 11 "'c 2 9? 6. James'[)o'iah aI lem ton aln -a o:'}tce jfr.L 0 cenms o t.Lhe rcm:l""'n cost ionourw- ties as multch as tae' e:; to0tns e -or; t;he prieo of eta *7. I2n af s:i-:c;-r~ ooma conXuta inin g 20 cass,-s t —he m-r o-I.tho'se * la a store roor ri 9 i ii d" C r i i A, that a0 frull is foua r timues n ta xii ier 0 hosce at r epy; how many are there of eat? 8 In a f-lock eo2. in,.28 sheep, therte is one l aCk hleep f)r each six'h.iter sheep; how miany 9e t heref each:iud? 9. Two piveoes of iron together we.gh l28 pounds., and the ]ieavier piece wei-hs three to'imes as miuch s the li'htc, r; ans ti1, weight of etacli' m10. Wirlliam and Thomr.as boulht a o1ot-b111 for 30 cenrts, and Thomuas paid twic a uch as W iltliam; what dii each pa y-? 1. ir-video 35 into t-o pats, sutch that one Shaill be: Iur mes the other. 12. TVie sum o''f t, he ai-:ces of a f,-theti an'd son is AxiturI t. i 25 o years, rnl the a'?te of thie fieathir is six tnimes thAnt of 3his so 7; wIat is tht age of earch? 13. There are -two numbers, the arger sf wiich is equ:ai to uine ti.mes tihe smaller', and their suml is 40 l what- are the nunters' 14, The, stum. of two nui'bers is 5, and thle larger is equal to seven t;ii.es the mal le1r,; mat arit the num1bers? t5.'What is -m+2x equal- to 16. W hat is x+-vx e.ual to? I7. What is -Si-. 4 equel to? 10 RAY9S ALG.EItA., PA.tT FIST. LESSON III.'1 Turan boys areo to share 24 apples between them the second is to have twice a's xm"n' as the first, and the third three times ts inany as the tfirst. If z represents the share of the fi stx, what will represent the share of thef second? What will represent the share of the third? lWhat is the sumn of x-2x-1-3gs If 6x is equal to 2-4, walt is thoe -value of x? What is the Astre of the second? Of the- thijrd VE R-IFI CTIO. —lThe first received 4, {he second ix:wice as.iany, which is 8, and te the third three times, the firs-t, or 12 and 4- added to S and 12, m.ake 24, the whole un-imber to 1)e diidled. 2. There are three numbers whose sunm is 20, the second is equal. to twxice the first, inld the third is equal -to thlre tutnes he" first;'wi1htt are the numbers? 3. There are three nuvm-uners whose sunm i 2,1, the seconld is equal to itnice the -first, and the third is equal to tw ice the seco.a..t If z represents the first, wh'at will represent the seeo<:adP? If [2 represents the second, what will res'present the third? Wht is the stun of xd-x-4-xt? AV-l4 nat aire t.he nlumbersh? 4. A. an tiravels 63 miles in ci days; hle trnavel.sr; t-wice'is fi ar the second day tas the first, anld twivce as far 1the t-hird dty as the second; how m.nuy mniles does lie triel ec-ch day? 5. o ihn had 40 chestnut-s, of w rhich hbe gave to ins hbrothier certain numhberi, and to his sister twice as imany as to his broflther after this h hhad ans many left- as he havd given to his br'othter; how mainya c hestuttits did hle give to ethefi? oA f:uirmer bohtught tt sheep, a cow, atd'a hose, fr (0) doTllars; the cou cost three tiiiies as mucmh as the shee-p, aitid the horse twice as tmuch as the cow:-1 wha t wias the cost of ettch? 7 tgames had. 30 cents; hi losCt a certain unimber afiter'thii he givet v aw'xy aws ivay ts.he hald lost, anld then fiound that lie havd three tiunes's tsuitay' re..-lananiang tas he.' lad given away' howt' mainy did he, lose? 8. Tlhe sumri of -three nunulbersl is 36;'th e second is equalt to twice t,.e fi.rst, a ind -the third is eqviual to threoe timIes the second;'lwhat'trie e theIm.Iberss? 9. Jo:n, Janes, and Wi.llian togiether have 50 cents; John has twtice as ilany as Jamns, antId uJames has three tuimes as mrany as'Villitiam how utr any cents has eachi 1 W What is the suni of x, 2x, and three tiraes 2x? 1 1 WXhat is t.he suni of twice 2x, and three tim.esl 3x? INTELLECTUAL EXERCISES, I LESSON IV. 1. If I lemon costs x:cenmts, w)-hat will tepresent the cost of 2 lemons? Of 3? Of 4? Of 5? OCf 67 Of 7? 2. If I lemon costs 2x cents, whacit will recpresents the cost of 2 lemons? Of 3? Of 4? Of 52 O f 69? 3. James bought a certain number of lemons at n cent.s a piece, and as me. any nlore at 3 cents a piece, all for 25 cente-s; if x represents thie inumner of lemnions at cents w whlat rill 1epresent their cost? What will represent the cost of tile len.ons at 3 cents a piece? iow -ma niy lemons at each price did he buy? 4. Mary bought lemons t atod oranges, of each ana cqual nuiiber ithe lemnons cost anid th e oranges 3 cennts ta piece; the cost of tihe whole'was 30 cents; how mvany were there of each? 5. Daniel bought an equal numnuer of apples, leinons,,and oranges -:or 42 cents; each apple cost 1 cent, eac:h lenmon 2 ceonts, and each orange 3 cents' lhow -ianny of each did lie buy? 6. Thonias bougght a number of oraln es for 30 centsc oee-hall of thenm at 2 and the other haif at 3 cents each; how many oranges did lie buy? Let x- one- half the nunmber 7. Twoe mien are 40 eiles apart; if they travel toward each other at tie rate of 4 miles an hour each, in. ow le. any hours will then meet? S. TWo men are 28 e.illdes asunnder; if they tlret-x l'tonLard eachi other, thie lirst at the rate of 3, and the second at bi rate of 4 iiiles acn. hour, ii howr nmany hours will they iceet? 9. Tiwo.en ntr aiel toward each otier, at' tbhe siaiie rate per hour, from two places h'hose distatice apart is 4S miiles, aied tihrey ineet iun. six houexs hos-w cmany miles peor hour does each travel? 10, iwo men travel tow-,rd each othelr, the ftst goin.g twice as i fast as thIe second, a-tnd they meet in 2 hours; the places are 1I miles apart; howr xmauny imiles per hour does each itratvel 11i James boughti a certain nuciber of lemlons, alld twice tas.11any oranges, fo.r 40 cents; the leTmons cost 2, an.d the oranges 3 cents a piece; hoeiw uiay nwelre there of each? 12.. Tw1ox men travel in opposite directions; the f[irst tr"iTelhs three times s mIanyc miles per hour as the second; it dte end of 3 hours they are 3 c miles apart; how lmany inties pe'r hou.r does each. traxel? 13. A cistetn, containing 100 gallons of water, hbas pipes tio m.pty it; t.'he largner discharges foiur times as in tuy gallons per 12 RAY'S ALGEBRA,'PART FIRST. hour as the sinsaller, and they both enptty it in'2 hours; how r.any ogallIons per hour does each discharge? 14. A. grocer sold I pound of coffee and 2 pounds of tea. for 10) cents, and the price of a pound of tea wans to"ur tuines that of a pound of coffiee what lwas the price of each? If x reprosents th3e price of a pound of coffir, swhat will reipre sent the price of a pound of teoa? ~'What will recpresent the cost of both tthe tea ande coffee 15. A. grocer sold 1 pound of tea, 2 pounds of cofXfe, and 3 pounds of sugar, -for 65 cents; the price of a pountd of coffee wias twice that of a poundct of sugar, and the price of a pound of -tea was three thiumes that of a pound. of' coffee. tequldti the cost of each of the -articles. If2 repese.nts the price of a pound of sugar., what will represent tie price of a pound of coffee? Of a pound of t ea? OWhat will representl t1h0 cost of the wv1o?.,; LESSON V. 1. dames, o it 2.pples anutd 3 peaches, for t1O cents;'tilo price of a peajch wa, s twice th.'l;.-t of Tan apple; wihat, was lthe cost of each? If: reprcsen:-s' the cost of an apple, what iwill re presen:t tilhe cost of at peach)s? hYiat Will reprsent the cost of'" 2 ap'pl,'es s Of 3 peaches 0? Of i0otl apples cnud peacihes? 2. There are twfeo;ru3L1bes, the iarger of w.iche l e is qualt to twice the smallLer, d.d th.e sumu of'the l'arger and twice the smauller is equal to 2S; what are the numbers? 3. Tholmas bought 5 apples and 3 pea ches foe 22 enits —t; ecith peacl- cost -twice as muhc as an apple; w-slt was the cost of each? 4. William bought 2 oranges and s tr0 t 5,0s 5o rtn 2'7 conts; ec ch orange cost Itw'ie as nsuch as a lemion; w7lt.t-, was'thie cost of each? 5. Jeases boughft asn equal nunshmebr of apples and peacies for 21 cents; the apples cost 1 cent, and the peaches 2 cents each; how manny of each did hse buy? 6. i thomas bought -anr equal n.umhber of peaches, lemonrs, and oranges, for 45 cent s; the peaches cost 2, the tlemonso 3, and theL ora, nges 4: cents a piece; how many of each did he buy? 7. Daniel bought twice as niany apples as peatches for 24 cents; e:ach apple cost 2 centss, and each peach 4 ecents; bow m any of ea0 0h did he buy? INT:ELLECwTUAL EXERCISE,. S 8. A.famer bought a -orse, a cow, and a calf for 70 dollarst the cow cost three times as much as the calf, and the horse twice as:ilmuch as the cow; hat: was the cost of each 9. Susan b oug'ht an aptple, a lemon, and an orange, for 16 cents; the lemon cost three times as much as the apple, atld the orarnge as much as both the appleV: and the lemon; wvhat was:v thoe cost of each?:0.:0. Famny bought arT apple, a peach, axnd ain orange, fr I8 cents; the peaci cost twice as rnch as thie apple, and the orange twice as much as both thie apple and the peach; what was the cost of each? L.ES S' O.N VI.. Ja:mes bought a lem-on,anl an orange~; the orantge co0st twicre ats ruclh as the lemton!, and thle difference of their prices i was 2. cents; what iras the cost of each? If c repres;-enA h cost of the laron, h.h a- l ropresen~t ctie cost of the oin:. ge? Wha i is 2$ le8R re rPeientit-L b? 2~. WShat is ls - ia hr Xvit -' e.by; Iat L s3I repo se;ted by? W'hrat is 4t ls r: s,i; renete d ce r' ~t>1ht is 5t x lress 2: X reptieseuteld by? T. he vOrtd,flOo1,8is useds i nstela ofv lSS;:,.nd t-e'sic g -, for the Vsaket ortebr rvity, ia: Used to a oid waitei, the word minus Thius, if we ris to take tl~R e. rt t> n 3it" arnd a, ic we V S less a, m'or Ss 6x.w.. C;:al u e, h x a y be wrtav o:ma ors a; When the sign - is usced, i't i r toie 3 read o3-fiz,~-,8. 3.1hor-ias b.ought a Icrh i r Tonm'id.Tl. ai an.<- thie oranezrre cost tlree times 8as muach as tire hlemon ml d stie'dieh o, c of their pries as8 4 cents; what,was tihe p ice of ealchi?' f X reresents'the cost of the lenmon, whl-ia.t'ilt; repreI 3usent the cst te orng?(3 i atg is S-v- represente'd by? 4. In t school'conta-ining crlcamsse inGrainmr, (- orariphy, and Arit-imetice there ar ee..r.. tiaesa: lst; h-.t g aG sl(eography s Glrrammar, and tw: ice as mai std Ar ithetic as'eor.ph:;l there: are: 10 rirore in the class in Arithretii taalt in hi. at i ra: n mar; how imany more are there in: each c ml lass1?:%r rep..e s Ls n tir e number in tl.e cl.ass in Graimmiar, what wi.ll represeri;tier n mbe 14 R'A.S ALGEBRA, PARlT FI ST in. tihe ciass ii Geography? In the class in Aitrithmetic? WNhat is x —x represented 1by? Wheat is it equal. to? 5 The'toe of Sar'ah is three times the age of Jaem, ant d theo diftlreencie of their aes is 12 years; wheat is the aoe of eacth? 6. The di'ffrence of two numbers is 2i, w S and the reater is equial to eightl t-ilnies the less; what are the nnumberS? 7/ Daniel has iour times as mainy cents ts AWNilliam, and Joseph has twico as many as both of th.eml; but if'twnice the snumbere of Dan.ie1 s cents'be taken from Joseph's, the renrainder is oniy 16G horw many ients has each? S. $usaII bought a lemon, an orange, and a pine-:pple; the oraonge cost twice ass much as the lemoon, and the pine-apple three tims as n.uch as both thie leinon ansd the oranmge -the piie-aipple cost 14 cents more than the orange; what was th e cost of e tch? 9. Janies bought 1 loeon and 2 orianges; a olannge cost ttwice as -xaich as a, leimon, and the diference betw-een ithe cost of the eroalftT(0s olad the sleion was cents what it as th-e cost of eachi 10. Ch(arles btouaht 2 lemons andt 3 orainges an orainge cost twice as much as l ileo, aind thUe diffiere'ce hbetweeni. the cost of tle letmorts tail tefi ortxngoes rwas 8 cen ts; wha-t was the cost of each? I I mn b.oA 1 0 ugit a boor, a eo l a col'f, anld. a ihorse the cow cost t.wEice'as much as -t'e ctal f, asnd the horse twi<ct as inucih as tlhe cow; th1le diXtre.ncoe ortwleewn the price of thhe horse anid that of the calf was 30 doh s tir; lat is thei cost of each? 12. T]hert.re "ar three nel-ssabors, of w-hich t.he second. is th1ree times the firs-',, aind the third is -twice'as much ais both thl first and socolnd, while (the di.firence bet-ween the secoind and third is 10' what are te the nu ers? LESSON Vi~. I. Jantres anud Je4th together hi vYe 11 cients, anrd John has 3 more than James; how alr ti-: lts ea ch? If Jamites has a t en-is, then n. has h +3, aind they bolth hlave x -- t-8, or 2 -— i-3 cents; 3ene, 2nx+-3 are eq to 01 t h ence, if Uc2 a.idl 3 are equal to 1,:2x m.ust be equal to I I. less 3, which is ceqou.ai to 8 Ihel-if 2z is ecqna1 to 8, one x, or x, m.ust be equal to 4. 2. V.:Ailliat It:id IDaniel togecther hl:tae 9 apples, and Da ie l has oe I mtosle than:'lWi i ianm; how issany has ecl ch If z represents INTELLECTUAL EXERCISES. 5 t.he apples William has, what will represent the apples Datniel has?'What will represent th.e nutber they'bot-bh have? 3, in a class containing 13 pupils, there are three more boys than girls; how many are there of each? 4. In a store-room containing 40 barrels, the number of thoese that are empty exceeds the nuaiber filled by 10; how many are there of each? 5, in a flock of fifty sheep, the number of those t/ha"t a.re -whlte exceeds the numaber that are black, by 30; how many are there of each kind? 6. Two men toetacthr can earn 650 dollars in a month9, bvut one of them can earn 10 dollars more thaan the oeher; hlow many dollars c:an each earn?. 7, The sum of two numbers is 25, and the larger exceeds the smaller by 15; what are the n.umbmers? S, Sarah -and Jane boughlt a toy for 25 cents, of'vwhich Jane paid 5 cents more than Sar.ah; howr miucl did catah piay? 9. The difference between two numbers is 4, and their sum is 1 6; what tare the numlbers? If x represents the small er mnbml'er, whatt will repre~s:ent tle uarger? 10. T[he diffirence between two Pnumbers1 is 5, and thlr eir sum1 is 35; what are the numbe4rs? LESCSON' VIII. 1 eJiames and John tog tethor have 15 cents, and Joln has twice as man1y as Jatmes, and 03 more'; bow many has ctti-?7 if z represents the number Jamnes has, then 2x —i —3 ft will r presenlt tilhe numnb(q:.r John has, and a —!-2w —3, or 3Sx —3,?hatl th]ey both:ave, If 3z} — 3 is equ;al to 15, then 3w mmust be eq. ual. to 15 tless 3, or 12; hence x is equal to 4, the number James hetas; ih n.l.l' Jobn I has 1. l. 2. William bought a leimon and an orange for!7 cents; t:he orange cost twice as nmuch ias the lemon and I cent more; l:'Sit1 was tilhe cost of ea.h? 3. There are t.- wo numbers whose sum is 35; the secondi is twice the fi.rst tat-nd 5 more; what are the nulbers? 4. Tn anr. orchard containing apple-trees and cherrl-t; ees, tine number of apple trees is three times tb.hat of tie cherry-trees, and; 7 more; tIhe whole number of trees in the orchard is 51; how many are the-re of each kind? PtA'S ALGEBIi, A PA.T FIT 1ST, 5. A farmuer bought a cow and a calf, for 13 dollars; the cow cost three times as muchli as the c alf, and 1 dollar more; what was the co'st of eaclh 6.'illiam an d Tholmas gave 50 cenrts o a poor womlan; Williem ave twice ris 3nany as lionras1, and 5 c:ts more;, how r i5ny conts did each give? I7 Eliza. and Jane hbo g1ot a doll for 1-4 cents;n Eliza, paid twice ss iuch is 2ane, and 2 cents moroe;' what did each paty? 8. Divride'ie oni, ot r 35 int o two parts, so that one part shall exceed teo otr heor lby 3. 9.Di ide ihe numbr " into two prt so0 theat ie graer p. art shall be 5 more sun w -co the.tioess part. O. The sumni of towr o numbers is 23, and the goreater is equtal to thre times the less, tand 3 more; what aore the numbers? 1.1 Tw o noumber S aIdded I.- eicthe iiake 40; tie great'er is 5 times the less, and 4 more -'wha-t are'the numbers.? A.2. A man litos two flocks of shle; the larger o oncains six tines as mntany c;s ithe smanller, ad (5 ioie, Q c4d the numbter in!bot-h is 82; l3'3ow miany a'ire t'ere o i-a each? tL E S S 0 N I X. 1. J'.ye1 hats ais t an y cOts as John, an. d 2 more, an d Tbho1 as ha, j'RS ii)any as Jo'!0n, and 3 -nore; they all Itie 26 cents howr many hats each? f 1 0. irepresents 1h number or cent-s John hias, what will represent the num oberb James'a" l? Tihe nu'nber T.ihonmasi t'as? The rituolber they all h1c e?,'. Jamens, Thoimas, anid. Jch i llt; 0 uo tio ga-her c iesUnntS> Thormas giniherrct 5 lore ta.h Jan naes, and Johl. 3 miore th',an ionlns..... a nr they aoil g.'Iathere0d 34 ho r many did each gather? 3. A frittor ditri.buted 2ih 5 cents among c his three boys; to:toh rsecond heo o1r, 5 i" moiCre than to 1:i.e firs' and to the thir 3 oe than oto 1the second ) how ma lln y did c he'ive to each? 4. Divide t;he nurmber 19 into'ihre parts, so that the first.may bte 2 mL0ore thlisn the second, and the third twice as niuch as tihe second, aind I uo0rec 5. Divide ]3 apples btecitween thiree boys, so tfiat the second shtht. h]ave I more thsan the first, and the third, 2 muoree ihan ti-he second. 6, A peach, a leron, and an orange, cost 15 cnts; -.the lebton cost I ceunt nore than'twice ns mulch as th.e peach., and the orl.ange0 INTELLECTUAL EXERCISES. 17 2 cents more tbhani three times as. much as the peach; how 7. nlany cents did each cost? 7. Three pieces of lead'together vweigh 47 po-tnuds; the second is twice the weight of the first and the third wveighs 7 pounds more the. the second; what is the weiYght of each pi.ece? 8S The sum, of the ages of Eliza, Jane, and asarah, i. 35, years; Jane is 3 years older than Eliza, and Sarah is 2,years older than Jane; what are -their ages? 9. A. father has three sons, each of wlonm is 2 years older than his next younger brother, and the suns of their age is 2 7 years; what is the age of each?'10,.Th.e suns of three numbers is 29; the second is twbice'the first and 1 mlore. and the third is equal to the second, and 2 snore; wh-at are the numbers? ll, A mnian bouoght 2 pounds of co-ffee and 1 pound o:f tea, for 50 cents' the price of a pound- of tea was 1].0 cents more than twice the price cif ta pound of coffee; what did each cost? Ii),2, a lkrn bough)t 3 pounds of coffee and I pound of tea, for 77 cents; the price of a'pound of tea was eqnal to the price of 2 pounds of cofaie, and 7 centcos more; whiat-i wa.s the price of each? 13, Says.A to B,'" Good morning, master, wiith: your husndred gee.se." Says B, "[ hbav-e not 1 00; but, if I had twice'as imany as I now have, 3and 20 snore, I should have 100." How m:iany ha..dl dx? LESSO N X... If X —01.lrepresent a certainj numbe5r, what il Tl represe'nt twice that number? Sisce twice x is P]x, sntd twice 1 is 2, twice x-t- 1, will be represented by 2 — j-2. PW. Wha.t is 3 tune7s t — Hl? 4 times x-t I? 5 times x —-? 3. If X-k2 representi a certainX number, whst wrill. represen't; twite that nsunmbe? 2 times X is P2i, n d 2 t.intes 2 is 4, Ie'nce, twice: —2 is is 2,-x-H4. 4. What is 3 tilmes xi+2? 4 -—? 5 times X-2? 5. If 2x-i. represent. a pcerrie suher, i in what wir i r epresesnt twice thait number? Twice Px is 4c, astd twice 1. is 2, h:';.oce, twice 2x-j-I is 4n-+2. 6. WAlati is 3 tiunes 2x+t? 4 times 2-tHI? 5 tunes 2X+- -? 7, What 5is 2 timnes 3x+2- 3 tim.es 3-s1-22 4 tines 3S31 —2? S. hat is x,. t, asd xPnd a q2 equal sto 1s RAY'S ALGEBRA, PART FIRST. 9. WIhat is X x —— 1, t and 3x -— 3 equal to? 10. What is x, a-i —. and 2x-: i —>2 equal to? 1 1. A fa-ther divided 15 cents betweenl h4is three boys; giving11 to the second 1 nmore than to the first, and to the third twice as many as to the second; how- many cents did each receive? 12. The sum of 3 numbers is 34; the second is I m-ore than the first, and the third is 3 timnes the second.; w-hat are the nu.).bers 13o Eiza, Jane, and Sar<ah, together have 24 cents; Jain-e has twice as many avs Eliza,, and I rmore, and Sarah has twice as imany as Jane; lhoI many. cents has each? s 14. A man bought I poutl d of coffee and 2 pounds of tea, for 62 cent's; the price of a pound of tea.was eq-ual to that of 2 pounds of coffee, and 1 cent more; 5what- was tihe cost of each? 15.A nman worhked thlree days for 10 dollars; the second clay h le earTned I dollar smore lhman'the first, and the th ird day as much as both thle filrst anLd second howv lA-Mlch did lie earn ea-ch day? 16. Three boys together spent 43 cenits the second spent 5 cents more than thle fir.st, and the t;fird twice as much as the second; how m any cCnts did each spend? 17. Divide then umber 33 into three parts, so t.h at the seco-nd shah1 be 2 lor'e -tha. the first, and thie third equ -to five times -fhe second. 18. T1hree nmess, A, B, and C, ihave 40 dollarst betwreen them; B hles twice as many as A, asid I dollar mnore, and C hiaas 3 tilmes as ma)ny as B; hon-w mnany dollars has eachl 1.9. Divide the nuntber 29 into three part:s, sutch that the second 0shall be equal to thle first, and I more, and the third equal to three tinles the second, 20. A man bought 3 pounds of su.gar- and 2 pounds of coffee, for 41 cesnts; the price of a pound of coffee was 3 cenat's more teaa i that of a pounsd of fsugarg what Twa\s the cost 0of ctLeaCL? 21, J.ames bouh'1t 2 lemnons asnd S oranges, for 27 cents; ain orange cost twice as much as'a lemon, and 1 cent more; what was the cost of each? 22. An apple, a( peahn, and 2 pears, cost 17 centss; the peach cost 1 cent more than the apple, anld each pear t-wice as much as the peach;'whbat vwas the cost of each? 23..An apple, 2 peaches, ancd 3 ipears, cost 14 cents; a peach. cost 1 cent niore than the apple, and a pear I cesnt snore than a peth h; twhat wavs thle cost of each? 24. T wo pears, 3 lemonss and 4 oranges, cost 29 cents; a lemon cost I cent more than a pear, and a1. orange t cent more than. a lemon; whast was mthe cost of each? INTELLECTUAL EERCTISES. 19 L' S S 0 N XI. l. James has 4 cents, atnd John has I c enast less thlan Jam.as; how many cents itas John? YWhat is I less than i?' What is 2 less thian 4? 2. If x represents a certain nulmber, vwhat will reprcsent I less than that n umber? Ans. z —; read Z ninus L. 3. If z represents a cert.ain nunber, what willt represent 2 less than that number?'What will represent 3 less tbn that mnumber? 4. If a certain number less I is equal to 3, whlat i s thei numbier equal, to? 5. 1' z — 1 is equaal to 3, what is equal to? G. If 2 — is e l to 5, is 2xo eqlutal to?: If 2 is ecqual to, what is ze a equal to? 7. f 32 —12 is equal to.0a, what is 3xa equal to 1 If 3x is nqual to Ig, w'at~ is c equal to? 8. If 5x-3 is equo to 1 7, v what is 5x rquj'to? If 5 is equal to 20, what -is.i equal. to? 9 Jtames a.na d John togetloer Itoaae 17 cen as, and J'laes Xslas S cents less than John; how manyS hTas e acth If s represents thle nsnusbero Of cents Jamnes hsa;s, Whati W-11 rLepresent the rnuember John hars? What is x and c — 3 equai t4o? If 2x —-3 is eqnua o 17, what 2z euacl to? f xz is eqill to eq if s to 0, what is x equal to? I0. Divide e m mberl3 t17 in.o two parts, so that one slltiall be 5 less than t.e cother. i1. in o(rangea and a lemoon to oetaner cost S cents, aind t-e leamoen cost two c en s less t han'the olrangct; w rshat t-as the cost of each?'2. The sant of two numtbers is 20~, and the smaller is 4 less than the grate i -ha rt are the nnimbers? 13a. William and. Daniel toge:r hav;Te 20 cents, min]. TDaniel htas twice as anu-'y' as Will iam, -a ntaing t coent; how m s.y cents has each? 14 Tlte sum of tv wo numibers is 24t and the largoer is — wice tlhe smaller, waianting 3; whatj are the numbers a? 15. In a basLket contailnin, 25 apples -aIn peahes, if 5 bte stbtra cted etot twice -the nunmber of a:ppl es, it will givay e i( au35ber of peach.es hc-: m-'any are there of eacha? 16. The suml of ti;wieo number i- s 2is atnd the greater is equal to "3 ~-es -t ~'-I, w,t-p. what are the nuatmbersh? o usnes the — titersn ant in"' 7'w srasos:ane 17. A school cotnstains 37 puptils, the nnml)aer of boys is 3 times the nnsaa.lber o a.n ts, aat:g 3 is the.numbter of et. h 20 FA'XS ALG EBiRA P ARBT FfIRST,. IS. A cow,:a calf, and a sheep, cost 28 dollars; the sheep cost 2 dollars less fthasn the calf;, a.nd the cow cost 4 times as mauch at:s the calf; what was the cost of each? LESE - ON' XtI. 1. W'h-t nu.mber is ~that, to which if 3 be added, the numbeer will be doubled? If I represents the number, what will z+-S be eqnal to? Sia3ce 2x is equal t-o x- -3, it is phain that z is equal to 3. 2. Whbat numhber is that, to which if 5 be added, the tuniumber will be doublet? 3. What insumber is thlat,, to w hiclh if 4 be added, t he Isum aill be 3 times the number? If z represents thce number, x- 4 will be equal to 3x; but if 3c is equal to x-f-4, it is ptlin at-hat 2. is equal to 4, andi tha t z is ecqti ato 2. 4, Whi.t nui;iber is that,'to uwhnch if 9 be addedr, the'tun UwitI be 4 tu-sies the n-umbtr? If Tc rpnleseents tshe lnulmber, whiat. will z — 9 be equal to, If 4st is equal x9, it is plain thsat 3z is equal to 9, an-d that x is equal t o 3. 5. Wha: numborels i that, to which if 15 be added, the stun vwmill Tbe fbour time's the lnumner? 6. There a.re 10 years difference between } tf e alcges of two brothers, and the a's of the elder is 3 times that of the youngeyr; what is tire age of each? 7 Jaie sayss sto mJo hn, 4" I l are 4 tim. 9 es was nb y apples a.s you have; abut if you had 9 apples siiore thaln you now hia;e, ie )e woudt th-en each have an equal imn-ber." Hiow umiany has each? S, The difference of two numberbs is 20, and the grea-ner is 5 times the s malIler; what are the numbers? 9, The age of Eliza exceeds that of dJ1'ane t6 years, while; the acre ot:f the forater is t ive times tha t of the latter; wh'at is re heir t0 Jamesa bloenut a bool and a toy; the book cost six timeis, as Iniuch as the toy, alnd'ilhe diffiertence of their prices wmas 20 centa s; nhow mTuch did. he pay feor eaclha I The diUerenee betweien. the age of a fa-tlher antd tha-t or hi, son, is 30 years, I, nd -the ge of the faltfher is seven times the age of the son s; w'hat are t:heir ag es9 I. Wat mmi7h er is tha t, to whenich. f S32 be added, the suln' will be equal to niine hmces he 0numniber itself? INTELLECTUAL EXERCISES. ]1 13. ViWhat num:ber is that which is 6 less tian 3 tiames the numnber itself? 14. Jatmes is 12 years you.nger tb hatn John; hbut John is only four n imes the age of James; what are their ages? 15. W1ha. number is that, to the double of which, ifS 8'be added, the sum twill be equal to 4 tiines the number? In this case, if x represents -the number, 4r- is equal to 2x-4-j-8; hence 2x munst be equal to 8t, and x equal. to 4. C6. What is the value of z, wmhen 5x is equal to 3x-+6? 17.'What is the value of x, when 5x is equal tIo 2a'4 15? 18. What is the value of X, when 8x is equaLl to 3zf 15? 19. lWhat is the value of x, when I0lx is equal -o; 4xd-24? 209. Whalt number is that, to the double of lhich, if 21 be added, the sumT will be five times the numbe r? 21 If DanX3iel's age e multiplied hby 4,s and 30 added to the product, the sui.m -will be 6 timels hs a aie w~hant is his age? 22. WThatl number added to tvwice o.tsecl nud 32 more, will make a sumn equal to 7 times thle nubiler? 23. What number added t-o itself and 40 more, -ill. ake a sumn equtal to 10 tim:es the nunmber? 24. A 4fither g ae'o lhis Sol') 3 tim)es as man y cents as he then had, his uncle then gave hiinm 40 cents, whehe fond he!ndhl e hvad. 9 times as s rmany a-s at lirst'; how 1any haid lie at tfirst -LESSON X TI I. i. What n-umibe is th.at wh'ich )being m-sinrceat yd b- nl then doubled, thme sun''rill lie equal'to three times the ntlsbe'/'? I-n thisn examlple let zx represent tIhe number, than s —L-5 d eon-bled, will tbe 2,- -10,::O tich is equal to 3x- hence x is eqal d to I0. 23. Sarah is. 2 yoears older than Jane, %a nd tw sice e Sarah's -ta, is equail to three tuimes sthe age of Jane; wr.at is thbLe age of each? 3. William has 8 cents more thamLm Daniel, aind three -imes Witlitin's.money is etqut- to 5'times that of Daniel; how many cen ts has each? 4. Three pounds of coffee cost as miuch as 5 tprounds of sugar, and 1 pound of collbe cost 6 cents move than I poundc of sugar); what is the price of a poiutd of each? 5. A farmer bought 2 hogs and 7 sheepT a hog cost 5 dolltrs more than a sheep, while the hogs and sheep both cost tthie same Smn; wheat -'was thke cost of each?. 22 ~ iRNAY'S ALGEBt-ZA, PARl.T FIRST, 6. William bought 3 oranges and 5 lemnons; an orange cost 2 cents mnore than a lemon, while the oranges and the lemoons each cost the ame stumn; what -was the cost of each? 7,' lVillam has 10 cent-s more than Dianiel; but 7 times Daniel's jmoney is equal o' twice that of William; how tanny centsl has each? 8. The greater of P2 numbers exceeds the less by 14; and 3 times the greater is equal to 10 times the less; what are tlhe nuthbers? 9. Mioses is 16 years youngger thtan his Brot-her Joseph; but 3 times the a.eEgo of Joseph is equal to 5 times that of Moses; what are their ages? 10, Tfhe diffe'rence between the ages of a man and his wife i.s 7 years an::d 6 cimnes the afge of the man is equal' to 8 tm h.es the age of his wife; what: l' [ are thei ages0 LESSON XITV. 1..tf z representis a certalrin il-umber, wha].t will replesetnt one half the ummnber? To divide a. number, vwe dra w a.ine bemea th it. under which we place -the divisor; thus, to divide 1 lby 2, it is written which is read oii. bae/' ort on'ie dtisted by too. In the same mantu1ner, one half In a simnihlas r manner, one third of x is wtri.tten; tw thollirds of 34: is'wrten,-o 2. Itf its eitual to 4, 7what1L is x equal to? a 3: Ifs Ts equa to 5, what.rs z ecxoat to? 4. If is equal to 8, what is x equal to? If Aseo tirds of xu iO s equal to 8u one third ofj x is equal to one half of 8, or 4 (sineo one hal4lf of t/o thirds is oi:tze third); atd if one i/.id' of mu is equal -to 4, r is equal to 6ikree tinmes 4, or 12. Or tlts: if 2x divided'by 3 is equal to 8, Sr must b:e equld to -times ci, or 24; and if 2x is equal to 24, u is equal t.o one half of 24, or J' 2. INTELLE CTUAL EXERCISESS 23 tither of -these methods nmay be used in finding the value of z i milar expressions. 5. If is equal to 9, what is x equal to? 5'a 6. Tf a is equal to 10, what is x equal to? 7. If.-. is e3qua! to 14, wlhat is x equal t;o? 8. If is equal to 9, what is x equal to? 4. 9. If - I equal to 12, what is x equal to? 10. If 5_ is equal to 20, what is x equal to? L. If'7 ais equal to 14, what is x equal to? 94 12. If, iu equal to 18, what is x equal to? x x 13, a3 hat is the sum of x and? or of x- Sinee x is equal to we have a-s equal to -t -+ which S 3z equal to -o. 4-. hat will represent the suns of 2x and or of 2z' - 5, WThat wiltl epresent -the sum of xs-? 16. What will represent the sum of Xz.+- Of 2z+3? x 3x 17. What is the sum of x+-? Of x+- - Of 2a x — -? 183 What is the sum of xa —%? Of x-? Of x+f-? 19o There is a certaiun number, to which if the half of itself'be added, t.he sum will be 15; what is the numbesr? 90. Waillia. has half as mnlty -cents as Danel l nd R Lacthey both tog'ethelr have 2.1; how many cents has eacht.? 21. The age of Mary is one third that of Jane, and the suu of their ages is 24 years; whaat is the age of each? 22. A pasture contains 44 sheep and cows; the nunuber of cows is one third th.e number of sheep; how many are there of each? 23. The sum of the ages of Rutbh and Elizsa is 24'years; while tihe age of the fuormer is thire fifths of that of the latter~ wthat is the age of each? 2,4 WAP "S ALGEBRA, iPART FIRST. 24. Jances and John togetler- have 18 cents, anid Johln has four fifths as lmany as Jamnes; how many has each?] 25. Two places, A and C, are 40 imiles apart; between them is a village which is two thirds as far fiom C as it is from A; what is its distance ftoom each of the places? 26. The sum of two numbers is 21, and the sma, ller number is three fourrths of the laxrgr; what are tle tnumbers? 27. Thomas and Charles have 35 cents, -and Charles has half as many mnore cents as TIomas; how many cents hias each 28. The double of a certain nuimber, increased by one third of itself: is equal to 21.; what is the number-? 29. W.Iallliam Jtnames, lantd lobert, -together, havre 33 cents; Jamles has twice as nany as'William, aind Robert has one third as many as James; how Lmany cents has each? 30.'What numb-er is thea, which being increased by its half and its fourth,: equ.als g 1? 31. What number is th!at, which being increased by its half~ itss fourt h, zand 4 more, equals 259? 32. A. boy, being asked howi much money ie had, replied, t.hat if one half band one third of his money, and 9 cents more, -were added to it, the sumn would be 20 cents; how muchim money had he? 33. Thtere are three numbers, whose suMn is 44; the second is equJ;to one third of the first, and the third is equal to thei second mnid twice the first what are t he numbers? 34.: There are four towns in the order of the letters, A, B, C, and 1); the. distance from B to C is one fifth of tlIe distance from A to B, and the distance from C to D is equal to twice thme distahlnee from) A tLo C; the 3.hole distance from A to D is 72 muiles. ]c:lquired the distance from A to B, from B to C, and from C to P. 35. What number is thL.at, to'which if its; half its fiurth, and. 26 nmore be added, the suml will be equal to 5 times the number? S6. There is a fish whose head is 6 inches loncg, and the tail is fas long as t;ttlhe head anid thalf the body, and the body is as long as t/he head] and tail; whtt is the lengAth of the'whole fish?'37.e A, gentDemnian'beilng askied his age, replied, "If to my age you add. its half, its third, and 28 years, the sunl will be equal to three tim$es my, r no "g. Required his age. 7XY'- The p]recedilg eaxerises will serve to.give the learner soame idea:of -the ntlture of Algebra,, ind of ho nmanner in which it; mllay bo applied to tihe solution. of problems,'i shall now proceed to consider the subject in a regular a.nd suelemwifile manner EL` EiLL 0 lT OI A G EBBA. CHAPTER II No TE To TEA c TI s s-IR gne r1'al, the Introduction, embracing.AArtickles I to 15, nReed not be thoroughly studieid 1nti.1 the pupil reviewrs the book..krU1CJE In..,eb)ra, itjnumbers l and q uantities are re:prcsen e:d hy sgr'yms1 T1les syumols are the le-tters of'the IlphobeLt. JA.T~. 2i, QOuntit' is nlythingt that is capable of ilncrease or decreaseo s, ich as n1u! nn1bs, lin.es, spc"e, time, motion, &o.v ATii.T. ke Q-uQnatity is Oclled manCgni-tiude, wheon presentced or c'cm sidered in an undivided fornm, s uch as a quanatit;y of- wactiler. A:../R l Qi-ifntity is called uditidude, w-hen it ius made tinp of individu4al and. distinct parts, such as th.re" cents, -. which is a qjua.n ti ty comnposed of three single centsA.ArtT. n One of theo single parts of uwhich a qany.tity of multitude is cocnposed, -is caulled the units of qguantity, o0]r mea.a.s rinzgf' tniz; thus, one cent is t1le: i n.ea.sqin{ un22g Of tihe'quantity -three cents. The value or mneasure of every quan.itity, is tohe number of tinies it contains its es-uraing u-nit.An.. G In quantities of iagnittude, where there is zno n aturnal unit, it is necessary to fix, upon in an rtificial unit, tas a stan-Ltdard of ineasuore; and. then -to find the t value of the quantity, w-e nust aseertain honl oten. it contaibnus its unit of imeastre, Thus, to neasuroe the length of a line, we t;ake a certain assumle d distanTce called a foot, and applying it a certaini number of times, say ofive, wi'e ascertain thaft the line is five feet long; in this c.se, one foot is the id/, of,i mestrie,.At.'s7 The nunzericat vhlue of any quantitty, is tihe nutmber ti at expresses how nainny times it contains its unit of mieasn.tre Thus, in:the preceding e.xanple, the line being 5 foet long, its numencrical rit''I2E w'. -To [-Imowi are nnibeeris and quanti-t ies represented in Algebra,? Wii:t arse saymbols?, "What is a quantity? 3. When is iquant.nity called agnitudet 4 -v hen is quantity called multitude 5. Vvldt is the. i ani -o'qt,lnt-[i.? 6 [[o in;'s t.he valrae of a qliainitiy uascelrt-inied, ws there ii eno natural un itt? W u'hat is the uOnimemrical raut e of a -ny qu tn-, y? ii-..? 26 R AYTS ALGEBRA, FART FIR ST. value is 5. The same quantity may have different numleriea values, according to the unit of nmeasure that is assumed. ART,,e A,tunit is a single or whole thing of an order or kind. AarT. 9 AtJ[efmber is an expression denoting a unit, or a collection of units. Numbers are either abstract or concrete. A'T. 1.l@h An abstlrcact number denotes how many times a unit is' to be taiken. A co.ncrete, or a.pplicate number, denotes the units that are taken. Thus, 4 feet is a concrete number; whrhile 4 is an abstract nuambher, which merely shows the number of units that are taken. A concrete number may be defined to be the product of the unit of measure by the corresponding abstract number, Thus, 6 dollars are equal to I dollar multiplied by 6, or I dollar taken 6 times. aRT..ll L In Algebra, quantities are represented by numbers, and the letters used, stand for numbers. ART..1 2. There ae two kinds of questions in Algebra, theorems and problemts. ART. I 3T In a theorem, it is required to demonstrate sonme rela. tion or property of numbers, or abstra'ct quanttieSo AnrT. 14X Ian a problem, it is required to find the value of so3me unknown number or quantity, by means of certain.given rel Tations existing between it and others, which are known,. ART.. It B Alg)ebra is a general method of solving problemis and deson srastting th-eorenms, by meains of.fiygures, t etters', and signs. The -ltters and signs are someliThnes ealiled symbols. igFINI''ION OF TE1MS, AD EX PLANATAIO @OF SIfGN.o ART. B6.??,?io0VC2.t- foqe,.a'litieS are those whiose numnerical values are given, or supposed to be known: stos2nknoiscan yuntitzies are those whbose numerical vtalues are not know-n. ARny. A.7T K nown quantities are generally reprtese-nted by t!he( first letters of th1e alphabet, as i, a, C,, c.; and unknown quantities by the last letters, as z, y, z. A.iT. IliS T'he fo0llowing are the principal signs used in Algebra: _, +, - 9 x,, (), 9 >,. Eacht of these signs is the representative of certain words; H~Bvisw,.So 7What is a unit? 9.'Wh at is nuuber?!.0.'What does an abstract number denote? What does concrete num ber denote? 11'What do the letters used in Algebra represent? 12. Iow tin:Eny kinds of quest, Otions.are there in Algebra? What are they? 13. Whart is a theorer-o? t4. What, is a problem? 1.5. W-7hat is Algebra? 16. What are known quantities? What are unknown quantities? 17. By what are Ikown qua< na tities represented? By what; are ulnknown quantities represented? 18. Write on a slate, or a blackboard, the principal signs used in Algebra. What do the signs represent? For what purpose are they used4f? DEFTIN'TIOTS AND.rNOTATION. 27 they are used for the purpose of expressing the various operations, in the most clear and brief mannner. Ant. 19, The si-gn of equzaldy, -, is read e.quacl to. It denotes that'the quantities between which it is placed are equal to each other. Thus, a=3, denotes that, t he quantity represented by a is equal to 3. ArTr. 20. The sign of addition, d-, is read phtus. t; denotes tihat the quantity to which it is prefixed, is to be added to some other quantity. Thus, a —5b d8aenotes that b is to be added to ac. f a=2 and b —39 then a-&-b= L-3, which are -5. AnT. 21. Thle sign of subtractionr, -, is rear!d mint-i's, tt denotes that the quantity to which it is prefixed is to be subtracted, Thus, a-b denotes that b is to -b subtracterd from a. If — 5 ztand b=,-3, then 5 —3 —2. AnT. 2g< The esiS -4C and - are called /the signs; the irmler is called the,positive, land the latter the ntcyrati.ve sign,; they are said to be conl,,trrzt or opcostle. AnT.r 2 3 Everv qnuantity is supposed to be preceded by one or the other of these s'inrls. Quantities haring the positive sign are called potsitive; and thos e having the negative sign are called.negative. When a quantity has no sign prelixed to it, it is considered positive..ART. Sd4i Quantities having the sanme sign are said to have li/ke signs; those having different signs are said to have utnli/ce signs. Thus, A-a andt — b, or -au and -b have like signs; wrhile — c and — d lhave unlike signs. _Any. S'. The sign of na,2tZlp? iiicoinoo, X(, is read rinto, or vsezlt/ipliUed by. It drenotes that the quantities between witich it is placerd, are to be mTultiplied together. A dot or point is somuetines usered instead of the sidn X. Thus, a;Xb and a.bh, both mean that b is to be, tnultipl;ed by b.o Thle drot is Atot used to deno ti e sunpultiplication of figures, becautse it is used to separate whtole numbers and decimals. The product of two or more letters is general!I denoted by writinog them in close succession, Tius, ab denotes the same as aXb, or a.b; and abe means the samne as aXbXc> or a.b.c. Rvtw.o —l19. 1ltonw is the sign of equality, =, read? aWhat does it denote? 20. How is the sign -- read? What does it denote? o21o, low is the sign. roeadct? Whal does it denote? 22. W/list are the signs plus and minuls called, by way of distinction? Which is positive, and which nega-'tive? 23. When quantities are preceded by thle sign plus, what are ithey said to be? By the sign minus? When a quantitr hasy no sign pre'fixed, what sign is undersltood? 2-,o When do quantities hrve like signs?' sTen unlike signs? 25. I-Iow is the sign X read, dnd wha..t does it denote?'What ot.her methods are there of representinig multiplication, besides ttle sign X N? 28 R AYIS ALGEBRAn PAR-1T FIRST. Ant..36 QuaoAntities that are to be multiplied. together, are called fact/ors. The conztinuzed p3rodutct of several factors, means that them product of the- first and second is to be multiplied by the third, this product by'the fourth, and so on. Thus, the continued product of a, b, and c, is expressed by aXbXc, or abec. If ra —2, b3 —, and c- 5, then abca-2'X3/X(5=-6X5w30. ART. 2P The sig n of ditviion, -e-, is read dieiuted tby. It denotes that the quantity preceding it is to be divided by that followin r' it. The division of two quantifies is more frequently represented, by placing the dividend as the numerator, and the divisor as the 1de.0 nominator of a fraction. Thus, a- b, or /m means, ttht ac is to be divided by b. If az - A:2 ulb 6-3, then a -%b_128_3z_4; or a 12 b 3 Division is also'represeinted thus, alb, where a denotes the dividend, and b the divisor. AnRt, 73 The sign >, is called the sign of ieaeqatciy.o It denotes that oxae of the two quantities betiwceen which it is placed, is greater than tire other, the opcni-,tt, of the sign being turned towards th; e g reC'er' quantity. Thus, oa>b denotes that b iS greater than b. It is read, a. greeter lSitabn I f t I -- a nd —5 b —3, then 5>3o.Also, c<d denotes that c is less than di. It iS read, c ler tsitanit. d i:f c=4,and Cd-i7, tlen 4<7..AT. 29. The sign. mz, denotes a quantity greate-r t. han any thatt:roalln be assigned; that is, a qunmutity indefinitely great, or inflinity. a.:Ts. 2.- @ The nmz..neiralt cocJ/gCicesnt of a quanlT tity is a nuiber pro-.fixe:d to it, to show h ow en the quantity is to be talken. Thus, it':lhe quantity rerepresented by cc is to be added to itself several i:mes, as a-i-) a —ac, w-e write it but once, and place a nu m er b.efore iti, to show how often it is taken. Thus, ad+a-c+a- a — 4ac; and ax L~x+xas —S3ax.Z AnT 241. The literal cobfYicie.nt of a quantity, is a qcua3-tity by which it is multiplied. Thus, in the quantity ay,,a may be considered the coefficient of y, or y may be considered thre eoifeiCen.t of ta. T'he literal coefthcients is generally renatrded as a. known qtua anti.tyo R.sir: w'-.-20. What are faetors? Rowmanyfactors in a? In eb? in abe? In tabe? 27. ot-tw is the sign -- reaid. and, what does it denote? Wha. t other methods are there of reprosenting the division of two qluantities? 28. What does the sign of inequality, >, denote? Whiclh quantity is placed at the opening? 29. What does the sign mo denote? 30. W lhat is dt1e lnr1neral coi0fficient of' a qcuantity? Iow ot often is ar -taken in the- expression 3ao? In 5? In t oa? 31. Wha. t is the literal eoiEfficiet of ta quantity? DEFINITIONrS AND NOTATION. 29.ART. 5% The coficient of a quantity may consist of a number, and also of a literal part,. Thus, in the quantity 5ax, 5ac may be regarded as t.he coi-fficient of x. Iff a —2, then 5ci 0, and 5cax —10x.'Whein no numieral coicient is pre-fixed to a quTntit7y, its coilf ficient is understood to be unity. Thus, ac is the same as la, and bx is the same as bx. AuTr. S. The 2oewter of a quantity is tihe product arisintg from nmultiplying tle quantity by itself one or more times. When tlhe quantity is taken tiwice as a factor, the product is called its squ-are, or Seco1nd power; wrhen th. ree times, the cuzbe, or t&icrd power; when four times, the joizt/7h/ power, and so on. Thius, cc(.Xat —aa, is the second power of0 c; caXaXa:\c a-aa,' is the /third power of ac aXa/ccX ca=caaac, is the fiourt/c power of a. Instead of repeating -the same quantit ty as a ft factor, a sma11 liguare, called an extponzet, is placed to the right, and a little above it, to point out the nunber of times the quantity is taklen as a factor. Thus, caa is written tac; actcc is iwritten a'; tact ct is written ac "; ctabbb is written ca/bi. W'hen a letter has no exponent, it is considered to be e thfi',7St Or sio.p/e power of the quantity, and unity is considered to be its expoi nenit. T'hius, a is the samne as ccla exco pressino tthe first mpower of' cc. ATm. l4o1. To involve or raise a quantity to any given power, is to find thalt power of the quantity. Ar.PT. 355 The')oo0 of any quantity is another quatntity, somne po)wer of which is equal to the given qcluntity. The root is called toe square root, cube root, fourth root, &c., according to the number 0E times it nmust be taken as a factor to produce the given quanitity. T hus, since a/Sa —aci, therefore a is the second root, or square root of ao t In the same manner, x is the third root, or cube root of X 3, since Xj'XSr —X3s ARnT. 9 To ex~tract; any root of a quantity, is to'find th"a't root. ART.'.o The sign V, -is called the radical sign. When placed'before a quantity it indicates that its root is to be extracted. Thus V/a., or 1/, denotes the squnare root of c; o cf, tenotfes the cube root of ac; {Va/ denotes the fourth root of a..Ant. 38"6 The numlber placed over the radical sign. is call:ed the index of the root. Thus, 2 is the index of the squarel root, 3 of the ear. E. r w. —32. When a, quacntity has no coefficient written, what eoLfflcient is understood? 33. hat is the power of it quantity? What is meant )y the second power of a? By the third power of a? What is an exponent;? For what is it used? I-ow many times is a, taken as a f- actor in x,? In Ax In a? WIn? Where no exponent is rwrittem, wha.t exponent is understood?:35. What is tihe root of a quantity? 37,. What is the sign called, anil what does it den1ote 9 30 n.1nRA.YS ALGEBRA, PART FItSU.~ cube root, 4 of the fourth root, and so on W-hen the radicatl ha-s no index over it, 2 is understood. Thus, ~9-3,,/8-=2, t/16 —2. AinT.'9. Every quantity writt;en i algebraic liguag3e, that isf, by means of algebraic symbols, is called an ci,/csbraic jucutdiny?, or an algebraic ex,nretsion. Thus, 3a is the algebraic expression of 3 times the number a; 3a —4b, is the algebraic expression for 3 times the number a, diminimshed by 4 times the number b; 2aS2- a, is the algebraic expression ftor twice -the square of a, increased by 3 times tre product of ahe number ac by the number b. AinRt. 4"~ A n algebraic quantitynot united to many otherby the sign of addition or subtraction, is called a monomia~l, or a quantity of one tern, or simply a. er. A monomial is sometimes called asizmple tuaci'zi?/y. Thus, a a, 3Tc -a 2b, i2sn are mono-ianls, or sin p].e quantities. A'L, 4.1 o An algebraic expression cor.posed of two or lore terms, is called a _polyoiomiual, or a cosyuou9nd oian/u. Thus, cr2dl-b is a polynomial. AT 421. n A poly0nomiat composed of'two tlerms, is called a binzomial. Tihus, at -b, acb, i and cdc toe bi a l nomial s. A binomin, in Tlwhich the second term is rne;ai tve, as -b, is so1etimes call, ed a residtual uanftily. T.Am, 43. A polynonmial consistihn of hllree terzs., is called a trionttial. Tlms, a b+6 c, and a t —-b are trineomials. AiT. r44 The, tcmericas ual le of ea;n aloebraic ecipression is the numlber obtained. by giving pactticular tvaluaes to the letterss and then perforlming the operaftions indicated. In the alprebtai expression 2ca. ~-3b, if a-4, aaind b-5, then 2a — 8, and 3b-1,,anid the numerical va:tlu is 8+-15 — 23. A-IT. r- T0he vtalXuet of ta olynoinmial is not affiected by changing thle order of the terms, provided each teirm retains its respectire sign. Thust, ca- -2a-b is the same as ba 2a. This is self-evldent. ART. z'-6. Eacih of the litera factors of any sinple quantivi or ternm is called a /dineutn.sion of that, termi; uist1 the deriee of any termn depends on the numlnber of its literal Gfactors. Thus, ax consists of two literal factors, a and x, and is of the secontd deiree. The quantity ab contains three literal factors, ca a,.R v -..383 Whlati is the. number placed over the radical sign called? 30. What is an Mgaebrelc quantity? 40. What is a nmonomial? What is a asmple quantity)? 4i1. What is a potlyomial? 42. A hbinomial? A residual quantily 43. A im.tlinomtial? 44. What is meant by the numnericet value of an algebrasie expression? DEFINITIOXTS AND N\OTATIONT, 31 andi. b, and is of the t7/'chd de/greeo. "ea3'2 C contain -.1s five literal farctors, a, a, ct, t, ancd s, sand is of th-e J./i'h degr/e; annd so on..nrT.:4.t 1A polyIno3iial is said to be ovihoC;e.~o'es, wheli each of its ternms is of tlce saiiue degree. Thus, the quantity a-36-b-c is of thie first ree, tand hoeiogeneous; (a -3boc-x, is of the second dearee, andhl. hlomorenlleous; x'3ani/ is of tnhe third degree, and honogene os c — is not homogeneous. AR.T. 4S' A aotenti/es/i ( ), is used to sho-w'~ thatt nll t0h1e ter-,ms of a compound quantity are to )be considered together as a single term. thus, 4l(a -b) means Iatl ai —b is to be multiplied by 4; (a- -x) (tc,-) means that a+cx is to be multiplied by c-a,: i 0 —ja —}-c) means thtat a —-ec is to be subtracted from 10; (ae —-b) means thate a-b is to he 3ra ised to the second power andi so (on, ART. 490.. A vinoU1hC,, ~ —-, is sometimes us]ed. instead of a parenthesis. Te hus,.. —b'z mneans the same as (ca-b)r. Somi e —tiimes -the vineoulm is placed vertica.lly, it is then called a bare Thuis,.1i lits the seie meaning. as (c —a:-4i —4/.A..'0,,gieii",'or hi/ce quannttites are these copr sosd of the same letters, afficted wth. the same exponents.'Thus,'t7b and — 3ab, also 4a6c& a/nd 7'c2bt, are similar ternis. The cquantities ac- band. 2cgb are amt: siLs arJ, fbr,t though they are comiposed of the same letters, yc ethese letters have different expoinenits. ARr. fi.t The,'reci'ocad of a quantity, is unity divided bhy that quantityv Th-us, the tceir'ocal of 2 is and of a isART. 2:2.'i The sasie letter accented, is often used to deni-ote qu antit, ies which occupy similar positions in different equations or investigatilons. Tihus, Ca, ac., ca", aci, retIpresent four different -tantities; of xwh Itich tcd is read, a prilue; cc" is read, a secon.d; ac" is read, a third, and so on. E X A.iM P L E S e The following examples are intended to exercise the learne.s r in the use and rmeaning of the signs, R'Cvirw.-460 What is the dinmension of a term? On what does tie degtee of a ternrm depend? klWhat is the degree of the termni xa? Of x'yz? Of 2ctry? 47' When is a polynomuial said to be homlogeneouns? 48. For -iwhait is a parenttesis used? 49. WhaTt is a vinenelum, anid for what is it used? 50. What ax.re similar, or like qtmantities? 51. Wlat is the reciprocal of a quantity? 52. When a letter, as a', has onet ac.eeint, what doees it.retrsentea and how is it re'ad? tow is a witll tvwo accents read?.? C) 2L-RAY'S ALGEBRA, PART FIRST. Let the pupil, copy eaci exaimple on s.is slate, or on he blacki board, and then expresse it in cimmOn lan. guar,go Also, let the nunmr-erical vallue:of each expression be found, on the supposition that ~ —z4, b —3, c=5, dl-,, x —'2, and y/ —6.. c —d —b... Ans. 12, b-+ -d —4 ~. 4g-z.. eeL o 6......ie 4. 2a. 4a. O Y e Ans. 14. c u 8 3. -8- 3a.... Ans. — 24. cy, 8d I, ac)... Ans. 192o. o xo _, -c S. 3ct+2cri-b.. o Ans. 41. 1 b -b. 9. a(a-b...... Ans. 72. 10. c6 —LXa,- b.................ns. 1'3 l, 1 — T(.-b) (ft-b)............... As. 7 I.: X: ( —3t(:, —( —x(o-x)2+5by...... Ans. 4. 13 6:c, (Gx) 4.x2a,. Anos. 7 In cas.Be f. —',"v' e.ercises should be reqtuilred to teacch the pupils tIe utse of the tsinirs the Iollowing equivalent expressions may be emiployed, in which each letter may have any value whateVer,.provided tlat the s-ame value be attributed to the same letter throubghonL t;he satie et3xpressiton~ 15. 3(a —c) i( —C c) =3 t2 - co2. i 6o 5j(o;-b)S-~tl —— f2 —1 0:b-i-5jbh 16 (- 4 — t( xy-2-.'f 17. — us fio?''f S8. - C-tLL' —: -min us 3) ito -i —- te 0 Exmples, in i.u icsh. w3 ods are to be con'v't. ed into talgcbraic I. Three times a, plus b, mlinus Gotur timrhes c. Or,, three into a, p8t8p lu s, vi.iu e 4 into c c 20 aFiv tlIus., divided by three i itmces bo 30 a s inus b into thre e t. i mes c 4. at, suainus tli hree tim5es b into cs 5. a prlus b, dividedt by thliuee c 6. a, plus b divided by -tit:ree oc 7. 5 into t: minxtus thr'ee hito b, divided by C. mims &, S3. a squaroed, minqus uittree a into b, plus 5 ti-mes c into d squ.ared 9. x cubed minus b cubed, divided by bs squared minaus b squared. 10. Five a squared, in to a plus b, into c:minus cd, minus t hree tiues z fourth lpower. 1. a fifth power minus b fifth power, divided by at muiins b, raised to't;}3e fifth powler, EXAMPLES I NOTATION.. 12. a sqtuaretd pl-s i squatred, divided by a plust b, squuared. 1.3. The s.juare root of ra, minus -the square root of x. 14. The square root of a, minus x, 15. The sLuare root of a nminus x. 16. The soquare root of b squared minus four into a into c. A N S' WE RS. 13e46 _I. _ b, 5 9. xZ23 —b 2 52,. tS3b 10. d5i'2(a- -L) ( — 4)-2 Szrt 3. (a Li 11 tz ktC 4.a 30 —- 3c (ea- b)5. 4, a1.... 1 /-':.. 3,,.,1 S 1'ax at 00 15. / ( — a). cd 16. - ( 2 — e). A DD IT ION AnT. S3 ADm3rIox ini Algebra, is the process of collecting two or more algebraic quantidties into one exupression ca'lled tf1ir sumn. CASE X;'liFe 1/t-oe.'.tatt.l..ics (0 s itatr, ci? n:(d zave 1e satce silg, 1. Jame is s 3 pockets, each containing apples-; hi tie first he hats 3 apisples, in the ssecond 4 tipplets, andl in third 5 apples. In order t- bfind {how many apples he has, suppose hie proceeds to find their smum irtn kl.she fo.llo wi.ng manner. 3 apples, 4 atpplls, 5 c~7p/pe,7i 12 apples. Suppose, oewever, -that, instead of mwritjing the word apples, lhe should mere1lyv use, tite letter a, thus: 3Sc 4a 5a Rg.iv.w.sr, —-.53o'What is alogebraic addition? Whl:en. quanrtn itaies are ss:n-r. lapb a'nd have toe siae sig, how arie, ey addedt' together? 34 RAY-'S'ALGEBRA, FPA'RT P RST It is evident that the sumxi of 3 times a, 4 tim-es a, and 5 times a, would'be 12 timnes a, or 12a, whatever a might represent. 2 In te she same manner the sum of -3a, -4a, and -5a -3 a would. be -12, — 4 -— 5a, Ilence, the - ila FOR ADDING SIMILARt QUANTITIES WITH LIKE SIGNS..di toge/Iter the co&-icieaxts of ife severale qanZtiZies, and to their suvia. annex the cos atiot. letter, or lettemrs, 2Jrf/in# t.he common isign. Nf o0r T 1. -L aet;he pupiil i:e reminded, that when a quantity ha s no coiLffiielntt prefixed, I is understoed; the, a is the same as 1Ta.' ovTS 2. — Let the pupil also be reminded, that the sun. of any ounmuber of quantities is the saue, in whatever order they are ta.-ken. This is self-. evident; butt it-,n:iy be ilustrated by nutbers in the followin'g m.nner. Suppose it is req uilred to find the sumni of the nulhbers 1.6, 25, ).and i4; in addi1ng these nurs togbether,. they masy be vwritten in six different w'aysi in each of whichl the saum is the sae. Tnulr: 10 1 g 25 25 34 34 25 34 16 34 10 25 24 25 34 10 25 16 75 5 7 75 75;i: A ll P L __ S. 3. 4. 5, 6. Pa — fi? 2... —.a-.a'b 2a -sXI Pa2 -4cb a -47'i 57a — 52ah.um =-1 la -- 4a:y J 7a 2 -- 4a-b n. the first exaImple, wve will suppose a-2, theno 3a — 3><2 —-, 2a:= —X2X: — a -2, 5a= —5X2i 1O; theirs um is0 +4- 2-+- 10-22. But the sum, 22. is mtore easily found from the algebraic saun, la, for 1Ha — 11X "222. In the second example, let x-3 i and y?2, and the va,lue of its terms will be 0xy=z-XX2X336 x-y — 3PX2- 6 4xyv-4X3X2_24 3xy —3 X3 X2S-18 the sum of their values is' —84 But this sum is more easily found from the algebraic sum;'for Ir. 1 1EvI- w. -1Wh en several quantities are to be a-dded togetheq, is the result affected by the order in which they are taken? ADDITION. 35 14zy-=14X3>3 —- 84. A.s aSll t.hese terms are negative, their sum is — 84 In the fitfh exam.ple let a, represent 3 feet, then 2a2c —ca. —2X X3- 18 square feet, 3a2S3aa -=3X3X3=-27... 5 a2-=5arSa — X3\ X3-45. T_7J2 ca-c7X3X3 63 " and their sum is 153... Or thfe sum n-17a-t=1?7XX3 —53 square feet. N'T E. — It is recoummnended to the learnerl, thus to exemplify the exam.piles numeri.ally, by assigning certain values to the letters; observing throughout each:: exa.-mple, to adhere to thie same numerical, value for the same letter. What is the sumn 7. Of 3b, 5b,'b, and 9b? Ans. 24b. 8. Of 2ab, 5ab, 8f.b, and lI ab? Ans,. 26cb. 9. Of abc, 3abc, 7abe, and i2tbc? Ans. 23abe. 10. Of 5 dollars, 8a dollars, I Ia dollars, and 13a dollars? A-n. 37a dola-rs. 11. Of -3ac, — 5sex, — 7 —ax, and -— 4ca? A.iss. - — 1..... 12. Of -- -by, — 2 —,-by, and -St y? ss - - ti y, 13. 14. 15. 3ay- i 8S -47 33. —d2ax tay8-j 5 —- o 5a"- a2cy-+-4 7x- 6y 7 a ---- 5.tx 5ay.t 6x -2y7 4ai-'4ca ART. 31". hTV/ee quni.ie/ias aoe /i/e, /B11 h/voe sunliZce sg.yns. 1. If James receives frorm one man 6 ctents, from another 9 cents, and from a third 10 cents; and then spends, for candy 4 cents, and for apples 3 cents, howv much money will he have left? If the quantities he received be considered positive, then t1hose he spent may be considered negative; asnd the question is, to find the sum of +-6c, -1-9tc, -Ie Oc, — 4c and -3c, which may be written thus: 46c +90 IHere, it is evident, the true result will. be founf, by i 10e adding t'he positive quantities into one sum7, and the -4c negative quantities into another, and then taking -3e 3 their difference. ItI its thus ound that h.e received +78jc 25c, and spent 7c, which left 1Sc. 2. Suppose James should receive 5 cents, and then spend 7 cents, what sum would he have left? S36 PAY'S ALGEBR.Ai PXR-UT FIRST. If we denote tihe 5c as positive, the 7c will be negative, and it is required to find the sumn of -5c antd -7c. In its present forn, however, it is evident that the question is impossible. But if'we suppose that James had a certain sume o money.before he received the 5c, we may inquire how rauch less money he had after the operation, than before it; or, in other words, what effect the operation had upon his money. The answer, it is obvious, -would be, that iis moneys was dintinis. ed 2 cents; this would be indicated by the snum of +-S and — 7c, being — c. It is'thus we say, that the sus of a positive aned negativse quantilty is equal to the difrencew between the two; the object being to find what the oLnited eftl l of the two will be upon some third quantitity. This imay be farther illustrated byr the following esxample. 3. A nierchant lias a cortain capital; during the year it is i ncreased by hca andc 8cc dollars, and dimizshniced by Sa aelnd 5a dollars; how mruach wvill his capital be increased or diminished at the close of the year? If -we dernote the goains as positive, the losses will be negative. The sui of- -Soa, S-8na, — 2a, anud — 5' is I It —7, n-hich is equalto -4a. ilcnce, wir say, that the merchant's ca, pital will be inc reasaed by 4cet dolltrs; and whaSteover the capltal rmay l.ave been, the result will. be the same to increase it by 4c dollars, as first to) increase it, by 3a a.-ind 8 a dollars, and then to dimeinish it by ha anittd 5n dollars. had the loss been greater than the gain, the effect would be to din'.inis'2 the capital; and this -wrould be indicated, by the suni of tghe gains and Icosses being neaiNue. If the g-in and loss were equal, it is evident the capital would neither bei increased nor diminished; or. in other words, if the amount of.the positive quaintities was equanl to that of the negative, their sunl w.ould be 0. Thus, a -3 ha 0 —ag. If a 4, -3 ha+-1i2 and... --— a —2, and -12-1. =0. Froem this the pupil will percecive, that tto add a nmegative qualntity is thie saune as to subtract a positive quantity, ine such cases, the process of addition is called calelebrai additOie, o, and the sum is called the algebraic szum, to distingguish them from arithmetical addition, and arithmetical sum. tence, the 0OR THE][ ADIDITION OF QITANTITIES VISICHI ARE A-LIKE) BUT. HAE UN:LIKE SIGNS. Pinde the suztm h f 1/e coiscicl tis of /ie simnetior )positgive qcuanitities; also, /the suztm of 1t/e COfficient-iS f tihe simzilar -'rgeative qutttities. Subtract the tess sum from the greeater; then, to the difimrene prefzi h/ie sign ofj/ te greater, and acnnie th/e coMn.MiM, literal p2art. ADDITION. 37 4. What is the sumt of +-3a, — 5ac, t —a, — 6a, and ---? Here, the sumn of the coifficients of the positLive terms, is. + -i 7 — D The suon of the coifficients of tie negative terms, is....5 —6( —1 t The difference between 19 and 11 is 8, to which, prefixing the sign of the greater, and annexing the literal part, we have for the required sum — ifa, In practice, it is most conv.enient to write the 3a different terns under each other. Thus, -5a 9a -ha Sum-:Sa Beginners, however, will sometimes find it easier Sa-S5 to arrange t ihe positve quantities in one coltlumn, a —6c and the.egae.tive in another. The preceding ex- 7Ta ample may be arranged as in the margin. I-9 1 t-8a 5. What is the suim of Sa and -5aY? Ans. 3a, 6.'What is the sulm of 5c and -a? Ans. -3a, 7. What is the, sum of — ax, hex, 6a, and --- ax? Ans. ax 8. What is the sumn of S5abx, — 7abx, 3abx, — rbx, zand 4abx? Ans. 41abx 9. Add togfether, 4cac, S(ac, -3ac, r7ac, — 6ac, —:2ac, D9ac, and 17'ac. Ans..ac...3cc, 10. Find the sumn of a-4b, 3c-f-b, -a- b, and — { —t-9b.A nso a —-b. I1. Find the sumt of 8ax-2-gy, -— 2ax —-3hby, cax-4,yf, andt -DWax-' y 8......-. / Ans. 5:by. 12, Find th.e sum of 3ab —lOx, -3,ab — zx, 3ab —6x, — cb — -- ancd -2acb —, Ans 0. 130 Find the sumz of 4a2-2fb, _ —6ati —ib, " " 2ahb -o 52..b, and — 3e' 9b. Ans;: - — 82 -— i2b 14P Find t}he sum. of xy —-aec, 3xy —9ac, -7Ty-Sa.c, 4y ghc, and -2cc.Aan s c iori:.- — T.Ihe operatrion of colleeting the similar ternms in any algebralic expression linto one sum, as exeaplified ilt this ease, is somtaetnets calledt thlse Rdesctonetoft o oP?-lyoniaofa. The following are examtpls, I5. RBeduc.e 3ab-5-5c —-:Tab — Sc-Sab- Itd -lc b 2I —'c to its sirsplest forlmA,' is, 2ab. RIu:vsi;'w.. —-5. -li0ow are quanlities added together dtut ate sirtilar, buht have unalke signs? :38 R, AY'S ALGEBRA, PART FIRST, 16. Reduce 5a<tc —3b-,:55~ —Sa2 5 o its sim lest form. Ans. ac t+4b.2 ART., a57 }/c. W he qt7tfCbcatiaiaes a.re'itSudit, or' pai'/fy like.b'nd partl;y untiie. 1. Thonits has a m.arbles in one hand, and b marbles in'the other; whait expression will represent the naunber in both? If a is represented by 3, and b by 4, then the number in both would be represented by 3+4, or 7. In the samne manner the number in both wxould be represented by a — b; but ur:tess the numerical values of a and b are g(iven, it is evidently impossible to represent their sum more concisely, than by a+b. In the same nanner, the sum of the cquntities a0+b and c -i-H is represented by ad —. b —— tc+, f, in any expression, there are two or more like quantit.ics, it is obvious, tt -they mltay be reduced to a single expression by the preceding -rules. hbus, the sum of 2ab-xL aind tca-,i is equal to 2a+-g3atd-x +, which reduces to 5a- q-+- U. It is evident thist t c ase e mbraces the two preceding casese7 hence, the UPENEAL RULE, OR o T'.t.E ADDm.TION OF ALGEBRLnic QUAXINTITrIS. M'5rite the quant-iice to be addoed, p'lacin-g those that aCe s-imila r under each/ other; Iiho, leduce the simnilr' teirms, and annex the otler cerams wit/I their proper sig.-s. L.a. -x au.,-If a reason is asked for p1eing similar ters unlder each other, the Yreply is, that it is not absolttely necessary; but as w'e caL only add 5similar termns toge't, heri it is ct omatteer jf con.veaien7ce, to place them nwder each other~ Add together 2. 6a —4c- 3b, and -2a-3c —5b. 4s 4ac —.c —- 2b. 3. 2ab —-c, 4ax-C —2c-r-14, 12S-2az, and 6abd c — x Vnrs. 8ca b-i —2x-,.2c-t- 26 — x 4. 14aad-x, 1.b —, — 1la --- y, and — 2a —S 2b-/az. Ans. ad-b x+, g-U%~z 5. a —-b, 2b-c, 2c-d, d -e, end S9 b-, d,. — 7b —3c, 4b b —-- 3b-b c, and c2e-2. Alrs. x. 7. 3(a+b), 5(a+b), and 7(al- b). Ans. 15(a-+b), vlr, a ~ ew. —55.'What is the general rule ior the addition of algebraio tuc gmtities? T n writing them, whly aure sizilar quaintities placed under achli other? Ai:.DITIOYh 39 N'orT,.-The learnlr should be reminded, that the quantities in thoe parentheses are to be considered as one, quantity; then it is evident, that 3 times, 5 times, and 7 times any qua1ntity whtatever, will be equal to 15 times than quaintity. Add topgether 83 3a(b6 ex), 5'a(b+x), 7a(b+x —), and-I1la(b+ x). Ais. 4c(b-V —x). 9. 2e(a-b2), -3c(ce2 —'), 6C(a2-_b2), and -4c(t 2 —-b'), Ans. c(at-b2). 10. 3ta —4btt-8, — 2at-i-+5by+, 5az+6-T by — 7, -and — 8az..i i by, Ans. — 2 az —4. IL. Si — 3cz, — S5ax — h5a, a5x+2cz, and -4a -4Hcz. Anrs 0. 12. 8an —b, -2ct a2atIn be, -6b-5e-I-U3d,:Icd -- -?7c- _2P Ans. 2ct-b 5i- c -3d. 13. 7 x —-6y-z —,. U, —.' h y x Kt 2y z —-3 cgl, and xS — 81 z-{-1-9' Ans. 4zx+3y} 2 -5y' 14. 2,a"%-5b —xay, 7a ga3tb h3 y S, — 3a-7ab-5xy, and 9i -..- b-2 0.:y Ans. a — xy, I5'. 5a3b-8an2b~I x3 2'F - 2,r 4Cab 2b3 b7 —ghx" y2 —6 Xy~, 3aWRb2 -k-3a2M32'-}, heIT1 - )X2y2 3,Y3 _ryv n 8 a b<3 +he3b-t-gzx 5xy-1Sx and 2Kb3 at i 2y-xyaty Ans. n'b3+'ty, SUBTIRACT IONr AnT. ~5&t SUBTRACT.tox in Algebra, is the process of finding the simplest expression fir the difference between two algebraic quantetites In Algebra, as in Arithmet.ic, Lthe quantity to be su'btracted is called the saubtrahiend. The quantity from which the subtraction is to be made, is called the tin'tuelnd. The quantity left, after the subtraction is perforlxed, is called the deiJ]rence, or remainleer. Rlxsrrs-.- _The word subtrahend means, to be subtrCoted; the word minuend, to be dim-i..izhted. 1. Thonmas has 5n cents; if he give 2,C cents to his brother, how many will hie have left,? Since 5 ti-mes any quantity, diminished by 2 tises the same quan tity, leaves 3 times th.e quantity, the answer is evi.dently 3tz; that is 5ac —2a —3e. Hence, to find the difference bet-wee-Dp two positive similar quan.titites8, w fied t7e ofijrence between,'thfeir coaficietnts, antd prefix it to th e comnmon l'te"t, or letters. 40 PIRA.Y'S ALGEBRA PART FIRST, Let it be noted, that tthe sign of the quantity to be sub$tracted, is changed from pF s to in,ts.o 2., 4. 5 From 5:,: 7atob 8x 1y (f Take 3X 3a'b Sx Sa Remainder 2x 4dab 3x 6alo2m 6. From.l 9a, take 4ca..............Ans 5a 7. Froma l.tb take l]..b......... An. An 0. 8. From II alExy, take 3axy........... Ans. Sexq. 9. From it12bx, take 5bex.......... Anis. 7bcx. 10. From 13/hrp, take 9/........... 1. Aiis./7op l, From 3ca2, take 2ca................ An1is. ct~. 12. From 7t "xy, take 4boy........... ns. 3.b. AT. -- 7 fitmas hbias a. numhber of apples, r.-rersented by a; if he give away a quantlity, represented by b, wht expression will repre-sent the number of apples he has left;? If a represents 6, and b 4, then the- number lefteR wourld be represented by 6-4, whickh is equal to 2; and whateve;r numnbers ac and b represent, it is evideent that their difference may be expressed in the sane way, that is, by a -b, e-Tence, to find kie difbrence between two q-uantilies itat are not xsimilar, we place i/e sia,,mie'metts before i/ce /u( i n/tit i/.at is to be subtactedi.l, Let the pupil here notice again, that the sign of the quantity to be subtracted, is chf11ged from 2alts to nzimnbss 2. Froim c, talrke......... o Ans. c-cl.:3. From d3o2, take 3m'b............ 2. d2,-3n. 4. From 5b, take 3c............,.., 51o —3c. 5. From ab, take c............ Ans. ab —-cd. 6. From cai, take ax,....,,......ns. a, cte —-ax, 7. From. 22, take X..........., Ast X, -:. S. From xy, take yz........... Ans. ag —-yz. ART. 5-' —1, Let it be required to subtract 5+3 f'rom 9. if we subtract 5 from 9, the remainder will be t: —-5 but we wish to subtract, anot only 5, but also 3; hence, after -v have subtracted. 5, w'e must also sulbtract 3; this gives for thi: remainder, 9 —5 —3, which. is equal to 1, R i'. wi, i -.. — What is Subtraction in Algeebra.?'What's the quantity to be subtracted, called? What is the quantity called, rioma rhe-ie the subtraction is to be made? What does subtrahend mean?.? tat does minuendu meanl? Slow do you find the difference between two p. -)stive sinila.r qua.ntitries? 57. R ow do you find the difference between two qantitioes ntha, are not similart? SUBTRACTION. 41 2. Again, suppose that it is required to subtract 5-3 from 9. If w'e subtract 5 friom 9, the remainder will be 9 —-5; but the quantity to be subtracted is 3 less thanl 5, and we have, there-ibre, subtracted 3 too much; we lust, therefore, add 3 to, 9 —5, which gives for the true relmainder,. 95-5 —3, which is equal to 7. 3. Let it now-I be required to subtract b-c from,a. If'we take b from a, the remainder is a —b; but, in doing this, we have subitracted c too much; hence, to obtain the tru-ee result, we must add c. This gives, -for the true remainder, a —— b —-c. If az —9, bz-5, and c —3, the operation and illustration by figures would stand thus: from a froma 9 -9 take i.... talke 5 —3 -= Rema ind er, ca b —- c Rem. 9 —5 —3 — 7 The same principle may be further illustrated by'the following examples. 4. a-(c —a) z-r c —c a~:=2a —c. a —(a —-C) -=a —-ap — c c, + b —(a —b) ~a b -— a — a b =2b. Let it be noted, that in the result in each of tthe preceding examples, -the signs of the quarntity *to be subtracted lhave been chtang,.ed flroml pn.1s -to )in aS, and -from mam-a'iTmes to phtus; 1hence, in order to subtract a quantity, it is muerel y necessary to change the signs and add it. Hence, the H X E s FOnl FINDING TItE DIFFEI'E-NCE nTETr EEN TWO ALCGE-RAIC QUANTITIES. TFW'le Itfhe quanltitL/y to be subtbracted u?,nder ~that, t from ichick itC is to he tlaken, plakci'n.y sn'iilar termns.under eacck otf her. Csnceive the sigzns of' all the term'?,s of the subtrahend to be chatngied, and thi:en- redutce the resuzt to itS siintplJest jon:,1. N orT a. -— It is a good plan with beginners, to direct themr to wvrite the exanmple a second tilme, tned then aetuat:lly change the signs, and add, as in -the followin: example. They should do this, however, only till they become fa.miliar with the rule. From' 5a: —-3t-b —-c The smtne, with tihe 15a —Sb —-c Take 2a —,-2b —-3 scigns of the subtrav -2a —bL26-Se Remain, 3a —+5b+2c hend changed. Oa,i 5b $C:EXA A - P L E S -. 6. 7.I S. Fromn 3ax-P-g, 4cx-Sby1 Syzlt Saz..- S Take 2Pcax+y 2ex _ —3by2 5xyz-S az+8. Remainder, ax5 4 —-y 4e-c... 3.. — 6i —i 4 4A Y'9S ALGEBl A, PART FIRST. 9. 10. 1. Frorm x e —4 — 3cti —2b Gax —4y21-3 lTake (r — y 5a —3Tb 3ax- 6y-1- 12. 9Fromi take a — 5....... o Ans. 19-at.?'. Them a tio a.......... Ans. b. 13. F rom a-.-b, tl9 a0.1.......... nsa b 14. From a, take a —b............ Ars. b. 15. Fronm x, take X —5............ O ans. 5. 160 Praom T3a takle:at-J 7....... AAns. a-c7. 17. RLom --, take -............ Ans. 2/o 1S. From x —y, take a-Fy..........n. A 2y. 9. Fraons x —, takle j —:.......... Ains. 2x —2y. 20. Prom yz, take -y-al.... y..As. 2y z. 21. From Sx" o s..... T a,., Ais. a-Z. 22. From a, take — t............ As, c2a. T3. PFrom e8, ta ke —3c........,., As. l a. 24. Promi a, t — 4a...,........ Ans. 5a. 2 5. From Sb, t'.ke I b......... Ans.O -— 6b. 26. From Ca, take — b............. AOs., a-g-b. 27. Froes 3Tat, ake b........... ns. 3a- -b., 28. From — lc, take c..4 O.......A — 2a. 29. FPreut -Ta, take T7a............. 5ns o O T30. Prom -G-ee, tak4e — oa......... s. t,. 31. From — 6.a,,::.,I-e — S a......... Ans. -c. 32. From -3a, take — Sa.... As. 3.ac.-5, r 5b-16 82. Froml -a, take 365.. o s. o, 5, A l-e 1. 34. Prom -9, take —........... r.. s. 7. 35. Fromt 12,- take -- S............ Ac 0. 82. Fro — 4, —. 1 C ta T.....Is. Ans, -- 9. 87. Fromi 3 —2bit —, take 2 —7'b —3.... Ans. o a+o b-j- 9. 3S. From 3 -— b- -9c -- d, take 8e —b- 9c-10d-I ge 1i. Ans. 5,:,[.+4b-1-7d Z 2' 32. Prom — 7a- -— S, take -6ea-5, - 2x- 3d. Anrs. — a-t —S sr-3Te 40, Prot32a3b, t Ilake 75a-19b..IM.... A is. 2%-4 aLt. 41. From 6cl 5 —3b, le — a — 9b —8... Ans. Sa +6b- 13. 42. From 3c —-- -5c, take 8l-7-Tc —..... Ans. c --— 61 43. Froml 3axt —2Tf, take — 5aX —8 2...... Ains. S8ex 6e. 44. From 2 —3a'c-8''%9, take a-L-+5xa — 3, A. ns. x' —Sa 1-Sa-f-t2. 45. PFrroma 4-ay — 5cz —i-im, take — c-i-Ts 2" -4tz. Ains, 2 iy'-f —-Sam. 46. From a —-1 ls-fz —3a, talke — 6x-f 7 —a —52,-S. Artns. 2xSa-Sc-7, 4:7. Prim S-),,. takle 2(xa- y)...... A.n s. 3( —-y) SUBTRACTA0IONo 43 48. From. 3ew(x —z), talea(a, -x%........... Ans. a(a-...). itt) rolm l7&(c —— aeb(c —),. tubi Z) (.c-z) —— hob(c-dl. Aii (d. 2a"(c-z' )- 4a.b(c-d) d ARTn,' 9i it is sometimes convenient to i dicatfe the subtraction of a poly-nonmial without actually performiniig tlhe operation. This may be done, if it is a monomial, by placing the siogn minus i )'Ore it; and, if it is a polynomial, by enclosing it in a ptarenthesis, Land then placing the sign minus befbre it. Thus, to sub-tract a-b from 2ac, vwe nay write i;t.a —-(a-b), which reduces to a -+b. By this transformation., the samne polynonial may be written in several differu-nt forms,; thus: - -a —-— (d —CE —— d —-( t —b Qs-t — ( —-c=-j — d) Let the puiiIl, in each of tile following examlples, intlroduce atll the quantities, except tho:e first, into a pmarenthesis, nd precede it by the sign innurs, without altering the value of the expression. 1. ac — b-..................Ans.. —(. —c).'3. t-2xTr!o,.. c...o...... (..( —). 4. a b-c Sc-cd h......... x t s (c —bc —'3 i- >2e 6-+s...v.. Aii.e..... i.(a jCZ-t —S). $ l ~., —,-B-:-is.............Anso. (,, —() It will be ibund a useful exercise for the pupil, to tai.ke ea h of the preeedi:,. polynomials, and witlhout clhanging tiheir vadlues, write the i ni all possible modes, by including either two or more terms in a pa? enthesis. OBSER:t'-VA IONS ON ADDITION AND SUBTRACT trON. ART. 6O00 t has been shown, that Algebrai.c Addition is the process of coll.eting, into one, the quantities contained in two or more express:ons. The pupil has already learned, t-hat these expressions may be all positive, or all n tive, orpartly positive -andl partly negati:-e. If t1hey are either all positive, or all nega ive, the sum'will be greater than either of the individual cluantities; but, if some.f the quantities are positive and others negative, the aggregate an S:.,y be less than either of them, or, it m.ay even be Riv. xE VI.w.- in subtraeting b —c from a, afler t'a.king away, hal-ove we subtracted! too mluch, or too little? What mustt; be added, to obtasin the true result?'hvy? What is the general rule for finding the difference between twio l::g1ebraie quantitiess? 59. R-ow calm th? stil:ed:ractlion of an a.lgebr.s.l.e q..uml..t:y beo indtil. a:ted? 44 RAY'S ALGE BRA, PART FIRST. nothiing. Thus, tehe sumi of -4a aund -3a, is a; while thiat of a —a and. —-— a, is zero, or 0. As the pupil should hsave clear views of thle use and ean-ing of the various exprexssions employed, it nitmay be asked, wTha-t idea is he to a.ttach to th:e operations of algebraic addition. and subtraction. ARt, 6i tn com mon or arithmetical addition, when we say, th latt thhe stu. of 5 and 3 is 8, we mean, that their stlum is 8 greater than 0. In aloebra, Awhen we say that 5 and — 3 are equal to 2, we eana, that tlhe aggregate effect of adding 5 and subtracting 3, is the same as that of adding 2. Whien we say, that the sum of 5 and 3, is -2, i'e mean, tha it the result of subtracting 5, and adding 3', is the sa-me as that of subtracting 2. Some aie( - bsaists say, that numnbers with a positive sign represent quantities greater than 0, while those talh a neg;tive sign, such. li --— 3, represent quanltities less /haan nZothizg. The phrase, less thana notiing. howevere, cant not convey an intelligible., idea, with any sionifi cation th at would be attachedl to it in the ordinary use of language; but, if we are to understand by it, that any negative quantity, owhen added to a positive quantity, will produce a result less teha~n if nodi~h'ang hlazd beet,,, dded fo it; or,, that a'neative quantity, when subtrttCted fi'olu a' positive quarntity, will produ.ce a resuIt greater ha' ir sj nothiet had been tafrken, f.; oR, n i,, then t he phrase ha.s a col rect mnela. ini.n. The idea, howi ever, wiould be properly expressed, by sayingr, that inegative quantities are reiaz tively less than zero. Thus, if we ttrle anly nuimber, for instance 10, and add to it trhe numbers 3, 2, g, -1, -2, and —, we see, thait adding a negatrive number produces a less result t.han adding zero. 10 10 10 0 10 1 0 10 3 2 1 0 -1 — 2 -3 1i 1'll) 11 10 9 8 7 Fro i this, lre also see, that adding a negative number, produces the samre result, as subtracting an equa.1 positive nunmber. Again, if we take any number, for example 10, and subtrac t fron i it the nunm bers 3, 0, -1, -2,, -and -3, we see, that su.btrac e tin a negative nunber produces a greater resu'It than subtraacthing zero: 10 10 10 10 10 10 10 3 2 1 0 -1 -2 - 7 s 9 10 Ii 12 13 From this, we also see, that subtr-1acting a negative nunmber, produces the same result, as adding an equalI positive numther. Rtvrxw. —60. When is the sum of two a lgebraie quainilties less tha. eirier of them? When is the usau etqual to zero? OBSERVATIONS. 4'5 AMrt. 6st' In consequence of the results they produce, it is customary to say, of two negative algebraic quantities, that the leastt is that which contains th.le greaeesd numlber of units. Thus, It is said. to be less thin -2'. But, of two negativte quanltities, that which contains the greatest number of units is said to be -.tonmeri' catlly the greatest; thus, -3 is nusmerically gretater th-an -2. ART. 6S33 A correct idea of the natureo of the addition of positive and negative quantities, may be gained by the consideration of sueh questions ts the following: Suppose the siums of money put into a drarwer to be positive quantities, and those taken out to be negative; how wvvrill the money in the drawer be alffected. if, in one dvay, there are 20 dollars taken out, afterwards 15 dollars put in, after this 8 dollars taken out, and t-hen 10 dollars put in 9? Or, in oether words, vwhat is the sum of — -5, — 8, and +-107? The answ er, evidently, is — 3; that is, the result of the whole operation diminishes the amount of money in tae drawer 3 dollars. Had, the sutm of the (quantities been positive, the result of tihe operation wouald have been, an increase of tho amtou nit of money in tLhe drawer. A gain, suppose litoud naorth of the equator to be reckoned H-,3 and that south, -; and ttha the degrees over which a ship sfails north, are designated by -while those she sail.s over south, are de'signated by —, lnd t3lat have the fo1low-in-g question:.A. ship, in latitudoe 10 degrees'nort',t sails 5 dcgreess south, then 7 degrees north, tbhen 9 deog-rees south, and then degfrees north; 1what is her present latti(:itde? This question is the saie as t find the su:i of the quantities -10, — 5, 0 -9, ld.'' til -o i; his is evidently -H-}; that is, the shi0 is in 6 degtreeS no0l'th l:tLti-i ede'. }-a'd the su-in of the nega4tive numberss beeon the great'r, it follows, that tle ship w oul.d have bean found in south latiItu-de. Other questions of a similar nature may be used by't[he instru'ctor, to illustrate the siubject. ART. 6i4. Subtraction, in arithmetic, shows the method of finding the excess of one qua-ntity over another of the sanie Bkind. I rn this case, the number to be subt ractled must be less than that from which it is to be taen; and, as they are considered.t without refertvR E w.-61. WT hat is menet,'by saying that the sum of -q5 and -3, is equal to 42? What is meant, by saying that the sum of -5 sand'-3' is equtal to -2? Is it correct to say, that any qlua.ntity is less than nlotthing? What is the1 eff.ect of a.dding a positive qutantity? Of adding al negafive quantity? Of subtracting a positive~ quantity? Of su.btra.eting13 a negative q:antity? 62. In comparing two negative alge!braic quantitlies7 which is called the least? Wh1ich is -numerically the greahtest? 46 TRAY'S ~ ALGIEBRAP FIRT FIRtST. ence to sign, it i.s eq(it5.l.eti t.oi: r0;',1;ari.)- t~l of ty11 he a1tll.e s.gn, Alge'braic SubtractioD, n sIn,ews t.im`o.~!.A o fi.i.dineg the di/Tevence between two quit-.titics,:viii; h:we h t r h.e 10ae0 or unlike signs; aind it frequently happens, that hi d e i than either of the quantitities. To understand this properly Lrequires ra knowledge of the -nature of positive and negative quantities. All quanltities are to be reafrded as positive, uniess, for some special reason, they are otcherwise desiglnated. Negative quauti ties emlbrace those that are, in their nature, thle oppLosi'i, of positive quantities. Thus, if a merchant's gains are positive, his losses are negaive, if latitude north of the equator is reckoned 4-, -that south, would be -; if distance to the rioght of a certain line is reckoned -, then distance to the left would be -; if elevation aboYe a certain point, or plane, is regarded as -, then distance below wcould be —; if time after certain hour is —, then time before that hour is —; if motion in one direction is a-, then motion in an opposite direct.ion would be -; and so on. With tehis knowvledge of the neianintg of t.he sig';n minus, it is easy to see how the diAffrence' of twor quantities having the same sign, is equnal to their diffelrence; anud also, how the difference of two quantitties having, different signs, is equal to their smn. 1. One place is situated 10, and alotller 6 degrees north of the equator, w-cr hat is their difference of latitutde? Here'we are required to fintd the difference between -1 0 and +6, which is evidently -4-4; by which we tare to understand that the first place is 4 degrees fiarther north tlhanl the second. 2.o Two places are situated, one in 10, aind the o-ther in 6 degrees south latitude; what is Wthe difference of latitu1de? Here we are -required to find the diifference betweein -6 and -- -10, which is ev}idently — 4, by wijch we learn, th at the first p!ace is 4 degrees farther south than the second. 3. One place is situated in 10 degrees north, and antother in G degrees south latitude what is their driferene of latitude.? Here wre are required to find the difference bet-ween -i0 and 6, or to take — 6 fron -t- I 0, lwhich, by the rul1e for subtraction, leaves + 16; which is evidently the difference of their latitudes, and from whiclh we learn, that the first place is 1.6 deegrees fiarther north than the oitter. It is thus, when properly understood, the results are always carpiable of a satisfactory explanation. R i v e 7i w. —64. In whatt respects does algebraic differ from aritfllmetical Subtraetion? (In what respect do inegativo 1uamntities ditiS'er fr'om positivfe?.llu.ftrnt t:-he l differenee hry cxan.iples. ,)ULTIPLICATIOL, 47 3MULTIPLICATION. A..Tc..5 u a'ULTirPeLCATnO N, in Algebra, is the process of taking one algeb'raie expression, as often as there are units in another. In Algebra, as in Arithmetic, the quantity to be multiplied is called the qnnittipjtcad,; the quantity by w-hich we multiply, tihe s'auttitiTer, and the, result of the operation, the pirodct.. The mul-b tiplicand and multiplier are generally called factoVrs RT. 6 Since the quantity a, taken once, is represented by a, when taken twice, by a.-{-a, or 2ca, when t aken three times, by at-a-+a, or 3a, It is evident, that to 2,dltip23)ly a Hitaerat quat. titiq. by a inumber, it i.s onlyZ.f ecssary'o wirie the i / tic'lu iei r as athe coa. yicient of /ihe l;terial quanit2ity 1, If I le m on costs a cents, how ninny cents wili u lemons cost? I' ofcze lemon costs a cents, five lemons will cost fve tiimes as mnuch, that is 5.a cents. 2. If 1 or'ange costs c cent, how many cents will i oranges cost? 3. A merlbant bou ght a pieces of cloth, each containing b yards, at, c dollars per yard' howar many dollazs did the whole cost? In a pieces, the number of yards would be repr esented by ab, or ba, and the cost of ab yairds at c dollars per yard, would be represented by c taken ab times, that is, by abXc, which is represented by abc, Any. GM It is shown in "tay's Aritlhinetic," Part III, A-rt. 44, tfat the product of two factors is the salie, wrhichever be made the multiplier; we w-ill, however, denmonstrate the principle here. Suppose we have a sash containing a tvetitcal, and b horizontal rows; there will be a panes in each horizontatl row, and b panes in. each veartiCal row; it is retuired to find the number of panes in the window. It is evidentt, that tlhe whole number of panes in i3 e window will be equal to'te number in one:io;w taken as mnany times as there are rows. Then, since tlhere are a vertical rows, and b panes:in: each row, the whole nuniber of panes will be represented by b taken a timnes, that is, by ab. Again, since there are b horizontal rows, and a panes in each row', the w hole number of panes -will be represented by a taken b times, that is, by ba. But., since either of the expressions, ba or r vimw. —65. Wshat is t'Imltiplicatio-n in Algebra l What is dthe miltiplictald? Thle multiplier? The procduct?.'Wlhat ire tih multiplieacl aL.nd multiplier generally called? 66. tIow do you multiply a literal quatttity by a anumber? i8 RtAY'S ALG-EBRA. PART FI-ZST. at, represents tle, wlhole' number of panes in the windowt, -they are equattl to each other, that is, ab is equal to ia. Hence, it follows, that lice p1rot'Ccl ojf wo felctrs s tis 4e same, wt.cdc2tever be imade the vcacitleie.r. By tak:ing az3 and bz:4, the figure in the margin may be used to illustrate the principle in a panrticular case. Int the sa;me manner, the product of three or im-ore quantities is the rsame, in whatever order they are ttaken. Thus, 2X3X4}4 —3X2X4=4X2X3, since the product in each case is 24. L What will 2 boxes, each containing a lemons, cost at b cents per lemon? One box will cost ab cents, and 2 boxes will cost twice as much -as t box, that is, 2Tab cents. 2. What is the product of 2b, multiplied by 3a? The produc~t will be Tepresented by 2bX3a, or by 3a/Xb, or by i2X3/Xab, since the product is the same, in whatever order the f.tactors are placed. But 2X3 is equal to 6, hence the product b.X.'. a is equal to l6ab...Ienteo, w1re see, tihat in multiplying one monomial by anothler,;,he corfl~ci.en.t oj' the 2trooctcZt is, obtainedZ by mutllctlycig tzcgqeti/cer the cotffiicisens f the.mlat'Iptelic.d anzd m'.ttlltlier. This is termed, t/he r? otf fZle cos/scti/cisuo An.r. ~6, SSinee tihe product of two or moIre factors is t. he Same, in what-ever order they are written, if vwe t.ake the product of any two factors, as,2X3, and multiply it by any nurmber, as 5, t he product may be written 5X2X3, or 5/X3X/2, that is, 10/3, or 5X/2, either of which is equal to' 30. From which we see, that a when eiti7er qf thie factors of a PirodZuct'is nSamlttited, e t pre'odct ibsey is muclltoliec& AItTn.9s-t WVhat is the product of a by a? The product of b by a is written ab, hence, the product of a by ai would be written act; but -this, (Art. 33,).for the sake of brevity, is writtein aO 2, What is the product of a' by a? Since a~ may be written thus, ca, the product of a2 by a may bo tRnE v ar, w-OtG Prove cthat 3 times 4:' is the same as 4: t imes 3 Provo that a times b is tlhe sane as b times., Is the produnct of any nuimber of factors changed by altering their:a. rrangement? In multiplying one 0m1onomifta hy another, how is the cotffi cient of the product obt.ained? 68.t If you multiply one Of the factors of a product, hoiw does it affbect the product? 069 Holo may the product of a by a be written? low mal y the product of Oc by a be written? IMULTIPLICATION. 49 expresrsesd thus, aaXa,, or aaa, which, for the sake of brevity, is awri.tten as tHence, Ike eaOxeZIt of a letttelr in he product, is equal to t,-he s.ot Q'f it.s expo-.e2ts.int 7he tiwo actors. TPhis is terined, the ru,.. of Ihe exponents. 3. What is the product of a by a2?... Ans. aaaa, or a, 4. What is the product of a2b by ab? Ans. aaabb, or a'ish 5. What is the product of 2ab2 by 3ab? Ans. 6aabbb, or 6cabi. Hen c,'the R U~LE, Pto0.U'LTIJPLYXING ONE POSITIVE NI, jlldlipZ/ f the coJi0ciet'2ts of the..two terms together, and to their' roduzt can2nex atll tie letters in both quantities, giv9ring to each letter ane exponent eqCual io the su-co of its e3xponents itn Ie two fictors. No T-'. -I is 1ustolmar ly to write the letters in the order of the alphabet. Thuls, abX is generally written ate., 6. Muitiply ab by x.....A..... eiAns. atx. 7. Multiply 2be by sm........ Ans. 2 zbcmn. 8. Multiply 4 ab by 5xy...? A... o o ns. 20abxy. 09 Multiply 7ax by 4cd... D...... ACs. 28acdx. 10. Multiply 6by by 3ax........ Ans. 18abxy. i!. ~LMultiply 3at b by 4 ab........... Ans. 12aL?b 12. Mulatiply 2xy2 by 3x5y... 6s Ox. 13. Multiply 4abtx by 5ax2y....... s. n20a. 2O b%3y. 14. What is the product of 3ahb'c by 5abc i?.. Ans. 5a'b4cc4 1 5. What is t, he product of 7aytz by 8x yz?.. Ans. 56x z4y'yz.,orx. -The learner must be ca.reful to distinguish. between theo coi6ffl cient and the expon.ent. Thus, 2a is different from at' To fix this in his mind, let him answer such questions as the following: W'hat is 2a —a2 equal to, when a is?...... Ans. 1. What is a& -2a equal to, when a is 5.?..... Ans. 15, What is a3 —3a equal to, when a is 4?...... Ans. 52. What is a~ —4a equal- to, when a is 3?...... An8. 69. ART. 7t-l'. Suppose you purchase 5 oranges at 4 cents a pieee, and pay' for them, and then purchase 2 lemons at fthe same price; what will be the cost of the whole? 5 oranges, at 4 cents each, will cost 20 cents; 2 lemnons, at 4 cents each, will cost 8 cents, and'the cost of -the whtole wNill be 2O+18-z2S cents. The wo-rk a ty be written thus: 5~ +2 4 20+82S8 cents. :R 9aYS ALG-EBRA, PART FlIRST. If you purchase cc oranges at c cents a piece, and b) lemons at c cents a piece, whatt will be the cost of the whole? The cost of a. oranges, at c cents each, will e ac cent.s; the cost of.b lemons, at c cents each, will be bc cents, and tl.he whole cost will be ac —i-b cents. The work may be written thuss: a —— b ac 4-bc Hence, when the sign of each term is positive, we have the following rRa IMULTI'PLYING A POLYNOMIAL BY fA MONOlMIAL. jiith7tiply eca2 ie'at of ihe muztilfpicanmd by tlhe multijplier. EXAM PL ES. 2, Multiply a-}-d by b....... Ans, abt+bd 3. Multiply ecbe hby d.........Ans, acdctt+bcd. 4. Multiply 4x-5y by 0 a....... ins. 12aex+-15ayo 5. Multiply 2x+3z by 2b..........Ans. 4bx-k6bz 6, Multiply Ta+2;n b 3b,izo Sc. m A.ns. Sa-62:. 7~ Multiply xrLy by ax........... AAns cc-s2-atxy. 8. j;Mu.ltip}~y.2-,yh by xy............ An s. -xy 9. Multiply 2x+Oay by abx.. As.. 2ai2-+5asxSSby.,0. M ultiply 3xz2-l-2xz by 2xz....... Ans. 6Mz-1L4M22. 11. [fultiply 3a —{-2b+ D — by 4d.....Ans. 2adSbd2 —P28Oced 12. Multiply bc+ +,fl-m-x by 3ax.. Ans. 3abcxr+SjfixS/+3amxn 13, Multiply ab-c-ax+xy by abxya. e Ans, ta 2bty+a 2 2y- — abx2 y o Ant, 11.-I. Let it be required to find the product of x+y by a-+b, HIere the lmultiplieand is to be taken as many times as there are units in a+b, and the whiole productv will. vidently be equal io the stur of the two partial products. Thus, a+b thmas: 5 —-. -0y-Q12h: munmtiplica.nd taken 2 times. b 8-by=:zthe multiplieand taken 3 times............. 10+427+f18 55-the multiplicand taken 5 timues. TLJLTIPLIOATION. 51I -IHenc, when all the teTrms in. each are positive, we have the {fllowing RULE, F'O5, 3/U'LTIPL'YING ONE POLY, NO0MIAL.BY ANOTLEIo, itidf4tipiy echc term of the mutipican ii by each termn of the'm:lti_plier, and add the products together. 2. 39.t s.o1 -V a a2 ab at'3b+bcd ab ~a a I- ab2ed" d; a —2-abtA-b aab2 I-a2bcd2~ abcd -cC2d3 4. ul.tiply a-t-b by V.... AnS. ac-t.c ad- 5. Multiply 2x 3y by~ -!+2 o..-Ans. C6;-i —ay/)f.I-/ - by 6. Multiply 2a+bt by 3c-d.. Ains 6Oac-t-9bc-+2ad- 3tbd. 7. filiply,min,-t-'by xA-z o.... o Ans.o nmx-VfZx+mz+i}nz. 8. Multiply 4a-i-3b by 2ca+t... An. s. ct a.l- 10ab +3 9. Multiply 4xz-5y by 2a-V3x. Ais. Sar lay 1c- - q5 Iq/ 10o lultipy 3x-}-2y1'biy 2x+3y..... ans. x6Z-)t13Siy -yf 12. Multiply Oai+2bi by 2a V-bt... Ars_ _a' 1'5-'t'bl-t4 1 3o. Mlulti ply a -at bV by a-Vb... Ans. a2 a' 2at- a.2,-cbt214. Multiply 3a+2-d2 by 2cd -.1.. hTi I j 3a- d j. 1 6 Multiply aiV2x'yyabi by r-$-y.. h.:s, A ty-t-Oy b 2tfr-ya'. A T, In be prcdingex.,a, it w.. as dt n.'r. CT2 II n the P'reedling exalples, it was assuamed. Vliat, the produ.ct of two positive quantities, is also positive. It may, however, be shown as tfollows: lst. Let it be required to finld the product of -b by a,. The quantity b, taken once, is -t —; taken. tFwice, is evidently, -+2b; t, aken 3 tinmes, is +3b, and so ons There-fore, taken. a tiilegs, it is +ab. IHience, the product of two positive. quantities is posi.-T: f;ive; or, as it mnay be mlore briefly expressed, p.us rml1tiphled'by gpts, gives )pluts, 2d. Let it be required. to find the product of t. — by a, IixvTr:: w. —To whatf is the exponent of a letter in the produet equal,? What is the rule for mul-tiplying one positive monosmina by;no'ther! " 0. What, is the productt of a plus b, by c? When all the terms in: eacTl are positive, how do you mdultiply a polynomidal by a amnomuoixal?'iA. W. hen all t-,he terms in each tare positive, how do you find the produict of two polynor3ilus? 52 RY'S ALGEBRA., PART Pl-ST. The quantity — b, taken once, is -b-B taken twice, is -2b; taken 3 times, is -.; and hence, taken a times, is -ab; that is., a negative quantity m ultiplied by alo;ositive quantity, gives a.negagive product. This is generally expressed, by saying, that.ein-t.us multiplied by pFzus, gives 2nimets. 3d. Let it be required to nultiply b by -a, Since, when. two quantities are to be'multiplied together, either rmay be made the multiplier (Art. 67), this is the same as to multiply -a by b, which gives -ab. Tlhat is, a positive quanLtity mul tiplied by a negative quantity, gives a negative product; or, more:briefly, ius multiplied by iainnus, gives mz-zus 4the Let it be required to multiply -3 by -2. The negative nultiplier signifies, that the aultiplieshnd izs to be taken. positively, s8 many times as there are lunits in -the -multiplier, and then subtracted. The product, of -3 by -+2 is -6, then, chaneing the sign to subtract, the -6 becomes — 6; 2and, in the same manner, the product of — b by -a is -Pab. ience, the product of two negative quantities is positive; or, more briefly,,cinu6s multiplied by mnzi;zs, gives plusNi' oE TE.- -The following proof of the last prineiple, that the product of two negative quantities is positive, is geenerally regarded by mathiemati'ci.ans as rore satisfactory than the preceding, thorg i it is not quite so simple. The instructor car.n use either method. 5th. To find the productl of two neaatirve quantitles. To do this, let us find the product of c —t by a —b. Here it is requlired to take c —d as umany times as there are units in a.c-b. It is obvlious that this will be done by taking c-ct as iany tirtls as there. are runits in c, and then subtractig frtom this produte ca-c taLoen as many t;imes aIs there are units in b. Since plns multiplied by pi US gives plu s, and minus 3.3r1ultipli od by plus gives minLus, the Prtoduct of c —d by a, is ac-atd. In the s8to;e maanner, the product of c —d by b, is bc —b-d chtLaging the signs of the last product to subtract it, it becomes -— bc-i-bcd; hence te(3 product of cS-d by a6B, is acr —a-ctabc-+d. rBut the last term, — bd, is the product of -— d by — b, hence ho0 product of two negative quantitries is positive; or, more briefly,irS.is nutiprlied by mtibn'tes produces pluts. The msultiplication of c —d by a-b may be written thuis: c —d -ccae.-c'd==c-t taken a times. -- c-I —bc=: —c-d taken b times, arind then subtracted. -a-cct.-bc-f-bd A:KULrTIPLICA-TIO.~ 53 The operation may be illustrated by figures; thus, let it he required to find the product of 7-4 by 5 —3 7-4 We first ttalke 5 times 7-4-; this gives a product too 5 —3 great, by 3 times 7 —4, or 21-12, whicho being subtracted 35 —20 from the first producti gives for the true result, 3 5 —-.'-12, -21 —12 which reduces to — 6. This is evidently eorrect, for 7 —4 - 1.-1.-2 3, and 5 —3-2, and the product of 3 by 2 is 60 Fromn tT opreceding illustratioin-s, we derive the following GENERAtL RULE, FOR THE SIGNS, P.)Zls iinlltsZiecd by pb. zls, m si nti eus ml.iZ.'ied by aintust, give.s pinS,.P.u7s nr ttitcpeecl b,y mainis, o sniuss neud, giptie l by p2ls, gives i nenuSo:.w pSroduclc t Ltof iier si'ntis gives plus, and ojuf'like signs gives zinnzs. I'roni all tlie pr'ceding, wive derive tahe GENERAL It ULE, FOR TE MlUTI t'LICATION OF ALGEBR AIC QUANTITiES...i't,,,?pI y every.ere' o/f ihe tnTullci2.'cacnd, by eacAh [temi of g7e zncnieil;,iier. Observt —i2?n, ist, Tf-hat ie coujcient of cay u term is equal to lthe procl.l of..'he cocjj]ice'zfes Of eit fct.C-lors c2d 2T hat t ihe exaoneant of atiy lehtCr i- n the i pioclc-ac is eql.ual tio tie sun. oJf its exroznent.' ic [l ti7e two ficlors. ci T/, hat t'e prodhtf l of ti6ke sig9ns, gives pitS in', tl.he prodtuct, aend Ut7ikle su.ns, gives'tulZs. fIhen, add tthe seveeral Pjaretal prodiutls togetheel TO VERIFY THE RULE OF THE SIGNSo 1. MuIltiply 8 —3 by 5... o 40-1i 5=255X5o 2. Muiltiply 20- 13 by 4. A... Ans. 80 —52=-28-|7X4, 3 Miultiply 13 —7 by 11 —8 An.ns, 143-81+- t56 —18 —6X3, -4, Minultiply u10-3 by 3 —5., Ans 30 —41-15= —26.=13X-2. 5. uliply 9 5 by 8y A8ns. 72 —58-5 - 10 -24A —-4X6 6. Mlultiply 8 —7 by 5 —3. Ans. 40-5 9 -2 - _ 2.'ine't wer. —r2o What is the product of -h by -',-a? Why? What is the product of -- by h a? Why? What is thle product of -~-b by -— a? Why? What is the prod-act of -3 by — 2? What does a negatidve mnultfpier sige nify? Whiat does manius m-ultiplied by minus produce a'What: is the geunl oralt rule for ite signs? What is the general rule for the multiplication of ttgebraee qn-latities? RA'S ALGE3BRA, PAR.T FI RST. G EN l E AL.i P L ES., Multiply 3a'2xy by 7ay'..........s. 21 a %Yyt 2, Multiply -5a20 by 3atb3'........ n s - 15o 7. 3, Multiply — 5x2yy by —5xy'........ ns 251xa/1, 4. Mullrlty 3Ca-2b by 4c.....u As. 12act -St. 5. Multiply 3xd-2 by -2x....... Ans. -x —-- -6 w4xy 6. Multiply a+- byr x-y........ us. ax-af- tbx by. 7 Irul'tiply a-b by a-..... Ans. a — 2abt —b 8. Jlulitiply f2 —acc-c' by a-c........., i Ans. a. 9 YMultuiply m-1-' by.-o....... ns m-) 0o:.ulplv aS-2ab+-t by a+b..Ais. a'-a b —abt-" t. i 1. Multiply 3xy —2 y3 +yi by 2xy+y" Ans. 6Xz y2+3 1xx3F - -4xyf-U!2. Multiplv a- +2acbq-1b by a' —2t-+-b2,. Ans. a' —.2cat2b-b,.3 i- ultiply p2,-,- I Iby y —+-. Ius....y.lS.1, U. M,4. Mfultiply x3-1-) by x: —y'........ Ans. x2 —"o i5o M)ultiply as 3c'8 by cca —3..... ou oa' — a S 4 I I6, Multiply 2": —3:y-py t)7 2 —Sti, 17V Mu2tiply 3ad-5b) by 3ac —5b..... s- 9 o e A8,ns: 2.:, MT-ltiply c2a" —4co-"-2' by a —3x.r Ans. Cc'I —P1a c —! oczxc s O'C Muitip-is.p3 13) 9t 2', M.ttipty 3a-t -i-2 axi -3X 2y 2 ax.. cns. Ct ot1axc.t 9. Miultiply a Si —cxi2 2 by c2a~ 22. Multiply 3o-?+-5ax-2x by axJ. 24, Multiply xIS 0 by I2 —..... u.. x... Ans. p25, Multiply cca —t batc-j-6:3 by cabt Ans. a4t —b.' tn the followinguf examples, let the pupil prrin 1e a-multipl i,cationus indicated by mnultiplying together the cjuant-t ies conttained in the parenttheses. 2o (-. x-3(x —3)(23xY) O....... x'ns. X-9X 272 —-- 27 27. (3 —4)(x — ) ((t)-4)(;+-5)......s, t /' it 4L4010 2S. ('.c(c) (a, —c) (c-c-)(a —— c....... A, As. -2: -~+C. 29. (a-2 AI-'s —c — (a —tt7bc c). o A.. A'.c —-ti-"-p c -— 3 saBc. 30. (I,'-j-nj )-(n'AJ-a-blc).... 1 )(nI) (iI.. u. Ans -2r'-I-:1 DIVISION. 55 AT. r OT, -3 )IV-Islo:I in Algebra, is the process of finding how - 0ten One alge'1raic quantity is contained in another. Or, it mayt be dcluned thus H aing thoe product o tWo ftctors, <tnd one of then given, Division teaches tihe mu'ethaod odf finding t'ite other. The number by which we diide, is called th-e dincisor; the nunihiier to be divided, is called. the dtividenzdt the number of times the divisor is contained in the dividend, is called tuhe quaodiczt. AnT. 74- Since Division is the reverse of M~aultiplication, the quotient, mnuliplied by the divisor, must produce the dividend. The tsnual inethod ofI indicating Division, is to Write the divisor ni:der the dividend in the fnrm of a frac'tio. Thus, to indicate ab that ab is to be divided' by., we wriAte, lu.bAlgebraic Division, however, is somletiines indicated, like that of whole nunmbers, thus, a)ab; lhere o is thea divscr, and ab the dividend, NoE TO xci;.e u,.. — o solvin g ide following exsamples, let the pupil cgive tco reason for the answeri as in the solutionl to the finst question. Althoaghi the examples cts be solved mentally, it swill be found umst adaininGageons, to work them on the slate or blackboard; s te learner, byr this means, will be preparing for the performanue of moor- dffitcult operations. ]. I:Iowr often is x contained in 4Jx?........ns. 4 —.4,?This solution is to be given by the pupilu thus - 4x divided by x, 5s equal to 4, bcc.use the product of 4 by z is 4x. 2 pow often is a ccontained in. ia?....... -A. 5, S I-ow often is a contained in ab?.......Anes. b. 4. Iow cote is b contained in 3ab?.....,. A, ns. A a. 5. How- often is a, contained in adbx?...... ns bx. G. Ho oftsen is a contained in 5abxi?..,....s. 5dbx 7. How often is 2 contained. iri 4a?..... Ans. 2.t, 8. blow often is 2a contained in 4cb??.......Ans, 2b, 1. IVowr often is a contained in 3as?........ A ns. 3a. 1 I. How often is a ccontained in a2o.?...... Aris. 23a. 12. II-ow often is ab contained in 5a?2d..... Ans. 5a. R.n v Ir w.r —73. What is Algebraic Division? What is the divisor? The dividend? The quotieint? 74. To what is the prodLct of tihe quotien;t antd thle cdiviso equal? Why? What hs the usua l nmetihoi of in.ldicating divi.i sion? 6S \RY'S ALGEBRA, PART FIRST. 13o. H-ow often is 2c.c contained i lOnta3? o D,.luAns. 5a" 14, How often is 3a, contained in. 2ctb?. o. Ans. 4ab 15. -low often is 4dab contained in 12a'bc?. I APus. Si Cb 16. -lTow often is 2a' contained in 6aab? 2a2 2 In obtaining this quotient, we readily see, Ist, The co~fficient of the quotient, lust be such a.number, that when multiplied by 2, it shall produce 6; hence, to obtain it, we divide 6 bv 2. 2d. The exponent of a mninst be such a num'ber, that when 2, the exponent of a in the divisor, is added to it, the snmn sihall be 5, hence,'to obtain it, we must subtract 2 fro 5; that is, 5 —2 is equal to 3, the exponenit of a in the quotient. 3d. The letter b, which is a fact-eor of the dividend, but not of the divisor, must be found in the quotient, in order that the product of the divisor nid quotient imay equal the dividend. APR.'-'tM It reima ms to ascertain the rule for the signs. Since -+g mltiplited bay -b- -aCb, therieore, -— a; henea plus divided b3y 1hts, gives pltas. -— ad Since -— a nu-ltiplled. by h-e-=-c —b, therefore, -- hence, mints div2 ided by pis, give Rs iinust. -ab Since -— a multiplied by — d-ab, therefo'e, — f-a; hene,,cmies divided bly minu2-s, gi0ves p.as. +ab_ Since -— a multiplied by —b ab, therefore, e — e —/e -; hen-ce,.plus divided by sinuits, gives minnuts. From this, we see, that in Division, iike signs give 3la s, and utnlike signs givoe inustt, Htence the on DIVIDIT eNGc ON'E INOLBi IUTKO n uovssn, Divide th.e coij//cient o fe di/ct cidend, by tihat of t/e div /isor; observing, h/at like sai2ns gite p/ats, and c naike,7 signs give mi'tius. 4fier tlhe eocIieit w.rite the letters co e- tmon to bot/c divisor and dlnvidend, cij7en to each asz exponet, enqtal to t he exces's of he exponet of thce srote let/eri ih- tice cividend, over tat in ie dtivisore.. i tife qucotient, write tice letters witUc their respective exponentts, that are oinzd is /ce c/civ/ideci, bult scot /.,/he divisco. DIVISION. 57 N ovaE..T -. Te ppil mus t recollect, that when a letter has no, exponent eAxp'tressed 1.i t.11ndenrt. ood1; thus, i the Same as ca.'X EX AMiPLES8 1., DNvide I 5alJ c'b)y ~, o 6. 6.. 9.iaI.17N Diiid I c w,,, AresSa. "S lb cc 2 x V3 y -o pa. o " o o o A o -'bsjo 191). Divide Sat1x by -ax........... Ans. Sat.,0. vids 2,abx by 5ay............ Ans. 5b. 21, Divide toa bvy c.............. As, oa"'0'9' 4c 5-A b-2y 2c o O8,, A 2 x. 22 t ] — Ive ly t4s y....... An...At. 23. 1 I t]. - by 4caxyt/....,Are. 3xe-t, 2i C cvid":.cvcf by 4ayc....... A ns. -- c 2 D11'; Ii"i 3,,,, c y a......Ans. 2e / 10 ( l( O O * e o YdiA a c ta o e o o t. iy});, 8.. Di5vide — 1} athy-....... Ans. r9 ~ 3c% 29 v, C2Xace:yv by I!4ax..P....... Arns.- cY'-ce Wi ItirA 6tae' iy l2aex...... s. N'r i" t:...houg h meti off operlon in each of te folloi x mnliec is t heo:acrn sa in ttcc crecetding, hey mn bth poised over, until the;31. /l)do (a-b.by (a/)..... An. (a-"). 3do, D hivijde Qa.c-) Toy 2(c -).... Ans ti').. 35, I)iividte a'jl c"f`i- a bya).... Anos. -(2 c:3:-'. IL'Ditide 6:'l(s'-+?/ by 2cbHs- 2)ns.. Acm., c(a; —;o q 31 ~scvcic: ("" y- ) ( l: liy (a. —c, inrs. (a:-) —6)o tby another Ai the follo w integ cases. w let. WorT the coi'6@ eiren of thr e d iv idendi is not exactly divisiblec,by the eoeif"'ernt ofh t.he divisor.gl f tl d la r x 2.d When the same lteral fa ofr bp s a great er exponent ii thgen;d irvisor r tiiah'tmXin hictW-1i tth,o the tdivit ldhend. i3d theo rh divisor cntains one or mor e literal fa vctors, no foundk i the dividend. Io each eo these cases, the division is to be indicated by writing It vim. v —-. W/hen the signs of the dividend acid divisor are (a-J What C tie rule fr dividing one. oni by another? 8 AYS ALGEBRA, PART FIRST. fraction thus fBund, may often be reduced to lower terms. For the method of doing this, see Art. 129. ARn. 7S It has been show\n, in Art. 68, that any product is multiplied, by miultiplying either of its factors; hence, conwersely, any div idee-d will be divided, by dividngg either of its fvttores. 4X6 Or, -— X4XS —l12, by dividing the factor 6. ARTi. 8Y, Siwce, in multiplying a polynomial by a monomial, we multiplr each. term of the multiplicand by he nmultiplier; therefore, wre seave hlhe following FOR. DoIVIDING A POiLYN03oiAL nBY A AONOIALo Dividcp enacit terios of thf.e divvide.nd, by the divitsor, accord'i-my [to ihe f'.0/e,/%?' zie disvi(s.ona nzo iexomicasnt E X AX I M L E S. 1. O 6x 12e-] by 8............ o Ar i 2:-4y.o 2. Dividce L's —-20 by 5,........ A. s3n -4b, 3. Divide 21 a-C}S b by -—......... Ans. — 5b 4. -Divide 6ctz-t —9ay by 3t.........O e e2 -- 5. Divide abc ac by a. Ai o s s b+ sS-c, 6. Divide ac ssJ byac........... bf.. 7, Divide l2ay —Sac byey -4a......s. - As 3y? Sc. S. D:ivide I0a —15ayt by -- Sa.... Ans. 9, Divide 12,bx —-18 y AOm........ Ans 2b —3x. BO Divid.e abll-2abial bvy.b........0nsA. 20. Ans. -4ab —- 3-2 b, 12o Divide 1~cSabic —21a Scb by 3aCb... Ami. 5bis -7b to. 1. 3.Divide 6a'bc +2a2c -4amc hb acc. A nus. 3acbbC —2:. Nor —The following examplass rmay be oimitted until the book is r'viewed. 14. Diviide 6(a —c)-F9(a+l-) by3. o Ans, 2(ac c) — 3(a+ix)'5. Divide 5cL) —I Oay()x-. y) y 5Cq. Ans. (z —y)..,a(g —a g). 16. Divide a'b(c+-d) —ab(tC2d) by ab. Ans. a(c+d) I b(c'o_ —d) "RL v x a w -70, In what case is the eixct division of one monoomial by a othXer imxpossible? 8 hat is the rule for dividing a polynomial by mean aomnil? DIVISION. 59 1., Divide c,( y m+~Z)-bc(.mj) by"' r-i-n 1.. Anit ac -be. li8. Divide l2(a,-b)2-j-)6c(a — b)3 by 2(2a —b). Ans. 6(a;b)L -i3c(ab) 2, 19. iivde 2aY(x+y)3 —2ac2(.F —y)t by 2ac(x+:y)2 Ans, a(x-1-y)-Hc(xJy)2 20. Divide ('eb+)(xq-y)%>l(,-l x-) i by rn —., Ans. (+?,i)V(:-tx —): DIVISON fOF 6 OTNE POLY?'HAIS-L fL Y AN'OT.ItEt..I AgRT. l'90 To explain the mnethod of dividing one polynomial by another, we may regard the dividend as a product, of which the divisor and the quotient are -the two factors. We shall. exanine the method of forming this product, and then, by a reverse operma tion, explain the process of division. 3nultipliaMono, or foryarmtion of a p'roduct. Division, or decomposition of a product. 2a-..b 2ca-3'2 -I-a2' —--- __ —-b___ 2cat —2a 2b 2o — ab 2Sciv-cr. t st. rein. -— %-ab.b2 Qs219b-i-c:bS2 I 0)1).Oh9 04 1-3c~i~i,-i-} 3d. re i n, If we mnultip ly.2ac —ab by ca-b, and arrange the teriis according to the powers of a, awie shall find the product to be 2ca3 —-3c2b+-abt In this multiplication we remark, 1st. Since each term in the mulaiplicand is mulliplied by each term in the multiplier, if no reduction tIakes place in adding t'the several partial. products together, the nuZlmtber f. terms in the final product will be equal to the number produced tby multiplying together the um1ber of terms in the tw-o'factors,. Thus, if one factor have 3 terms, and the other 2, the number of terms in. the product'will be six., Frequently, however, a: reduction takes pla3ce, by wvhich the nuntber of terms is lessened. Thus, in the above example, two terms'being add ed together, there are only 3 ternms in the producet. 2d. In every case of ultipliecation, there are two te.trms which can never be united with any other. These are, first: that termlwhich is the product of the two terms in'the factors, wvhicl conta:in the highest power of -the same letter-; and second: the terlm which is the product of the trwo terms in the factors, which contain t.he lowest power of the same letter. From the last principle it follows, that if the term conta;inisng tihe highest power of any letter in the dividend, be divided by the termn containing the highest power of the same letter in the divisor, 60 RA.YS ALGEBtRA, PA TT FIRST. the rosult will be the term of the quotient containing the highest power of that letter, Hence, if 2a' be dclivided by a the result, 2aa, will be the -term containing the hiUgest power of a, in the quotient~ The dividend expresses the s-u of the partial producLts of the dtivisor, by the different t;erms of the quotient. If, then, we forn the product of the divisor by the first term, 2c,, of the quotient, and subtract it fron the dividena tie renainde r -ci c:I, wil be the suni of the other partial products of the divisor, by the renmaining terms of the quotient. Ntow, since this reema.inder is produced, by multplyi nug n the divisor by the remaining terms of the quotient, it -follows, as in the method of obtamning the first term of the quo tient,, t hat if the tern contanainig the 2iysest power of a: particular letter in this remtainder, be divided by the termx contlining the highest power of the sa Axe Ietlt er i.a the divisor, the quotient will be the term containing thie highest power of that letter in the remaining t e rms of the cquotienl Ience, if --- tb be divided by a., the quotient!, -alb, will be alother term of the quotient. flu!ftiplying tlhe divisor by this second terin, mnd substractig, e ~ind tlthe second re minder is 0 hence, the exact quotitent is 2 d —cs filad tihee u een a second remainn der, the thlird term of'.ote quotient would have been obtained from it in the same er as the second terni was obtained froni the flirt remacindel; Since each terra of the uotient is found, by (cividivng that term. of the dividend conrctaining the highest power of a particular letter, by the terma of the divisor containing the hi-hlest power of ~bthe stmne letter, it is more convnient to place the terms of dthie dividend and the divisor, so that the exponents of the samce letter shall either increase regrular ly, or diminiish regularly, iom the left to the ri ght. This is terni.ed, circ,ngsi/g/ei cti d cviclend ands d/visoiti,?/s reorense so at cerai, c / he letter who letter itsh tefteence to whicih a quantity is arrnneld, i.s called lhe lttelr 0Ja c o~ierrczayesie. The divicsor is piaced on lthe right of the dividend, because it is n. tOre easily muRltplied by the respective term s of the quotient, as they are fbund. Fromt tl- precedi.;g,,ei de1rive tIlIle e8" v Tu liE DVISiO.N O1':r O"'E POLYNaOm tAL xiY ANOTIr:ER, At? ranue lice thicc2cend clcxc> d it. fos,- i'7 ttr/ c7rezce to a cce-rt cci.n letot er, ar pd. aceC tihe divsor on tie rig/hc' i/ce cr;tidezi2. DIVISION. 61 Divide t7ze first terIm, of I7he dividend b1y ihe J-first lersz - of tze divisor,.7the result wvii bee e first term fqf i/he quotienLt.filtigdpy Zike d'ivt-isor by tis term, tied sab acsub ct te product/ jv em Ifa devidend..Divide thIe frst tern" of thie remcna'der by thef.irast ie2s of tie div-i sor, the result z ill be Ie second term, of the qsiotoiesntf ifliufiiy Ishe divisor by this Ierm,'mand subtract t.he product from/t tie last r&e mainirder. Proceed -i' tlhe same mantaner, vne i0f you oblain 0 for a reimaineirder sthe division is sraid to be exact. ti -An i: K.-I St It is not absolteiy neeeessary to a'ranoge the dividend and divisor with referenee to a certa ian letter; it sheoulcl al ys be donle, howervet as a matter of convenience. 2d. The divisor may be placec on the left of the dividend, instead of the riht, as directed in the rule When tihe divaisor is a monomial, it is more convenicent to -lace it on the leat; but, when it is a polynomial, to placett it on the right. 3d. If there're more thln two terms in the quotient it is not necess'ary to bring down aly more terms of the remainder, at each successive subotraetion, than htvre corresponding terms in the quantity to be snbtra cted. 4th. It is a useful exercise for the learnesr to perforsin the samn example in two different ways. First, by arranging' the dividend an-d divisor, so that the powers of -he same letter shall cl-bimninoh from left to right,; and, secondll5y so that the powers of the samn letter esbha ic ree ae frome left to,right., 5th, It is evident-, that the exact division of one polynoenial by another will be i-mpossible, whesn the first term of the arr angeht dividenad is not exactly d isible by the first term of the arraungeo divisor; or when the first term of any of the remainders is not divisibl e by the first term of the di visor. Lo D]ivide G6atl 3azx —0ts5 by 2a —3:, 6e2-13azI-ber+fla —gz OG-e-9az 3a —2x Quotients -4esx-z — u5z' — 4az —gaX2 2. Divide Xt.-.y by zy —y —. 3. Divide - 4 -7a by a3-t x y — Quotieent. na -ai X a — cz-{-, mQuot,',y-__..! -ytj —s a xzz x?/.-..y..-c,?z-$-cz ~~~~~~~~~~~~~~~.v ~,232 ~..3 |-t,! _i a:, 2S ~i RAY'S A LNGEBIPA, PART FII RS T. 4. Divide o5ax — 5a'+-a3x3 by 4ax — c —+x'". % -0a2x'+c5acx2+x31a2+I-4ax-X' a'-4-L-2x-t-a - 2- ad-x Quotient. a2x-I-4ax2+x3 In this example, neither divisor nor dividend being arranged with reference to either ac or a, we arruange them with reference to a, and then proceed to perform the division. 5. Divide a"2-c4 —.5a+3a5 by at —a2 Division performed, by arranging Dbivision peforrmed, by 13r ranging both quantities according to the as- both quantities accordmog to the der lendcing powers of a. scending powers of a. cc' -1-a cc 5-4" 3a2a —-a 3a -5a e-',-a3-+a" a-4 a a' a-~-C13 aS- 21~~" —a a a3 h a I -3, a a —3 - 2 c-+- a q2-c3 — 5a Quotient. — a --- a' Quotient. 2a, >2a -2a04 +-2,c hbo pupil will perceive- that the two quotients are the sanue2 but didffrently as rr':nge d. EIXAMPL. 7 t 6, Dsivide 4a t —-Saxz-+4x by 2a —- 2x.... Ans. 2a —2x. ~ Divide 2xV'+7xyu 6/- by x-Il —2y.. o Aun 2aot'py8{ 8. Divide 2223mx1+34- 2t10m 52 sns+12 nby xS-5, s. zn —s-, 9,'Divide e.:i/, f? —y by x- y...... n x+y i0, Divide 8,a S —8x'2: by 2a 2 —x...... Ans. 4ta+4X2 11, Divide actc —bc-ad- bd by a —' b.. c - 1 2 D1ivide xa-3i-y+s-5x<y?-a-5x> by A' —4xy ——. ee.. A ns x]-y 13, Divide ac — 9a-t+27a-27.by a,.... Ans. a — 6a 9. 1,4. ]Divid.e 4 —4.5a2+xz by 2a, — -3ax-x Ans. 2a'- -3acc —:x2 15. Divide f-y hbby y. An.as y —-a.y a4 p. s v:ts w.-f 79on multiplying one polynomia b. ~ another1'wh-at t:.lrms in. the prod1ct 1t annot he at ed to gethei? I-ow is to ti of tlhe qnotienot fonuld, which contains -the highest power of i:Ly partictular letter? Aftc obtai ning the first rematinder, how is the second term of the quotient fo: und What, is unders:tood by a rralging the dividend andu d -ivisor with reterenct to a, certain letter? What is the letter of arrangement2? Why is the divisor plheeod on t;he right of the quotient? Wihat is the rule:for thl division of one polynomial by another? When is the exact division of one polynomial by anot.her impossible? ALGEBRAIC THEOREMiSS. 13 11.{ Divide c — ba by a2+cab+......... Ans. a —-b, 17. DivideP A —y Sxy — 3xy3 by x-y..o Ans. 2 -2.Ty- -y; t8, Divide 4x4-64 by 2x —4... 2. 2: —4x2+8x-i f16 19. Divide cdt4 —5t.- x+ c 2-i 10 a:.iP —10a-3+5as -— Pz by a'-c2acr+;. Ainsm. cl-33a2c +3Sax-.s 0. Divide 4a0 — 2Sas 2xt+20axs-4x by 2c3 —5acS —-2 Aims. 2-V-5ea —2x3. 21. Divide y3+ - by y......... An ns y2 —-- 1. 22, Divide 6 -.:t- 4ac-9cc-3x3 3-22c by, 2c -'-a,,'2. Ans. 3ct.. ax —2x. 23. Diide 34 d -e 2t3VbS-3ac2 — 5bt —3bc'by c -— b a Ansm 32a2 —5bt — 3c 24. Divide x-3ic41l/2 —3y -'y6 by?" t-3a32.y —-3 3.y 3 Ans.? V- 3:.v —3xya -. /ly3 NUa \8CRLLANEOL E xgltiutc I. 3a 5x-bz9c-V'-7 d 5-c z-3d —oc —(4c-Vt2x 8c-4d)-) -wh.att? A.ns. 4c —c,, 2. 6tab —.3cx —5dab-t-5Scx-8 —-(3c'-i-cx-3d)-:whatt A-ns. 2cabt —cx. 3. a- -b(2a —-3b)- (5 cc-7b) (-13a+2b)what? At.S. 7a-Sb, 4, (ac tb) (c'b — (a b)(a- ) what?.... Ans. 2ac2 —-[-2b' 5. (+-z) (x- (z)- - ) (x-a) — wh;at?......... 4xz2 6. (a3-l-a+ag) (a2 —1)- -( c.a)(ce-a)=wht?... Ans. 0. 7. ( -— ) (c-az+-)- (a+z) (a-)-z)tzvhat? A. az-i-2s"z. 3o ( — 1+ coW) + ( - j-aa) -+(1 -+a2.)( 1 —cc,) — nhat? A, 2 —+cn, 9. c - - --. a )-)(-b)what? Am —-, 4abc. CHAPTER HIL ALGER B At0 C T HEOREMU'So IETi'E9 F0D rcc MULTIPILICATION AND IIVYSION'. Aiz@, @O. —-Iff we squae ac-b, that is, multiply a —b'b-y itselfthe product will be caa3&-2+ b+b-; tlhus: a-+b +a-2b+-t a2.J,+2ab+b'.i 04 RAY'S ALGEBRA, PART FIRST. Buit ad-b is the stun of the quantities, a and b; hence i/ce squcare of'.,e Sutn of tf o qguantities, is eqzal. to ice squarce of the'fist, pflus twzice tie prodzuct of lhc.e fi'rst by -the second, ptlus ih e sqecrwe of Ozthe second.t EX AM'P LE o v N TTo, -ThS instructor should road each of the followin lg lexa,1miles aloud, and r-equire the pupil, by applying the theorem, to write at once the rlesult on a slate, or blackboard. The examples may be cnu;ciated thus: What is the square of 2-+-b? 8. (2x-t-3y)mtx- Wy F g;7 y'l 5. (+c')-Vc<2- 2y — -925. 6. (2c, —+3aj:-=4c: t — tW2c c-Pc-9ak. ART. l-t -fJ we square b —, ithat is, nmultiply ca —..b by itseif, the product will be 2i —-2abj+bb2. Thus: a-b a-b aC —-ab g —oh-f-b o3 —2ab. —b_ a2-2 a VbBut ca —-b is the difference o the quantities a and b; he nce T }I IO E'' EM II, ~Te squactre of the disfferen.ce of -zo quantities, ie, is equal to the square of lihe first, minzs ttice the 2roductt of the first by t he seconzd, ptls the square of the second. EX AM PLE E S. 10 (5 —44) 2v5 -— 40-0+16zlo_ 2. (2a —b) — 4c _ —-4b+bS, (S —2y)j2=h9" x-12xy+-4yJ 2 e 2 (~ 2t ) t~tv 2 4:_ 4. (s' —y')'zzzs xy'+y Lbe (5a — b'.-r25ac — 1i_ n b-+bt. Amr, t...-.f'we'multRiply a-J-b c)y a —b, the product will be a2-b'. Thus- a-+b,ab..Wa-, ALGEBRAIC THEORE S. 65 But ad-b represents the sum of two quantities, and a-b, their difference; hence, TH EOPRET1 IIL The produte of the suta, and di ffcrence of o two tantities, is equal to the dijThrene of their spquares. E A E SXLE 1. (5+3)(5-3) =25-9=16=8XX2. 2. (2a+-b) (2a-b) 4a2 — b-'a (2x+3y) (2x-ty;) =4,-9,y2, 4. (-a+-4b) (5a-4b)-=25a-2 6Wbo 5, (ai+bi) (ai(-b2) -aL4 b 6. (2aA+3bin) (2am) — 3 b-t)=4aj2r"mt-9b-2o1.A2. T. -..If w-e divide a3 by a5, since the rule ir the exponents requires that the exponent of the divisor should be subtracted from that of the dlvidend, we have -~=ct-5=ac2But, since the value of a fraction is not altered by dividing both terms by the same quantity, (Art. 127), if we divide both numerator and denominator by a3, we ha ve - a'ene aC —C 1since each equals a3 In the same manner by subtracting the exponents — aOr,'by dividing both ternms by a, a..o.nce, a...:". -'" —- -I. Tbherefore, an-m THEl 0 REMI IV. The recipocaZ of a quantity is equal to the same quantity with te sign of its ezponent changed. Thus, since is the reciprocal of am' (Art. 51); affl= —: and a-z~6e r oa -ma3 ->am C- -atrt ----— _ a Isa~blO. bm... b S ~~b G1~abi From this we see, that any fcgtor may be tran.sferred /from'o one term of frvaction to the other, if; at thAe sa 2 time, the ht sign of its exponent be chatnged. 66 RAYPS ALGEBRA, PAiRT FIRST. 2 b.2 -Let it be requde a' by a'. Bythe rule AIRT. B49 —Let i-; be reuitrled ~o divide t2 by a", [By ~ the rule for the exponents, (Art. 73), a — ac- b — ut since any quantity is contained in itself once, --— ]I 1Similarly, ya"7 3 —........~,CtO b I c-~A - therefore t0~' —i since each a at? is equal to -,. Xt1enele, TSSS@OPIEEM V. iAny (.i'ot.'ity /OS etos;s<peori 0is 0 is egq.u.al t o 2tun.ity. Tlhis notation is useed, when wre iwish to preserve the trace of a etter, whlich has disappeaic red in t the operation of division. Thus, if it is required to divide nt-12 by nzt', the quotient will be -~rS.&r zli oi Bsiin-ce nis. Iow, Lte quo-tient is correetly expressed eith4ler by i-it, or ni, since bot h have the same vall. he. The first h-orn is used, wvhen it is necessary to show that n oriuinadlly enitered as a fctor into the dividend and divisor. AR.: —t. If we divide a —.b by a-b, the quvotient will be a b. 2. If we divide ca-ba by a-b, trhe quotient will be ac2+ab-[-b. In the s3ame mantner, we wrvould find, by trial, -that t.he differentce of thle sanie powrs of lwo quantities is always divisible by'the difference of the quant-ities, Thedirect proof of this theorem is as follows. Let us divide a"'n1-b by a-b. a" —b0 (6 —b o,?_- __, e~-~6 b(a -7-.....) _. — b m.,Quotient,,;SI — B- 6 a-b — b("''-b...) Rema. indero In performing this division, we see that the first:~ term of the quo.tient is a —-, tand thagt the first remainder is b(a"".l-b"-n.).. The -remainder consists of two fatetorst, b and a. t..:' —O-b"-l: Bow, it is evident, that if the second of these rfators is divisible by a —b, then will the t quantity ea"-b,1' be divisible by a -b. Thus, if a-b is contained c times in al''-b l'', tl he whole quiotient of a.'-,_r divided by ca —b, would be at^"'-d-bc, ALGEBRAIC T'.IEORtEM.S 6(!7 Prom thi, we see that Is u" —-T>b1 is divisible by a —b, titin.'will a -' —b" be also divisibleo by i-t. That 1is, IF i Ile d:ibe'retce ott/th sa'mie power.:~s odf two ce..nit:-dfties'is cdivisible by fte dtofc'7i"e —c e 1( /he i i ianet li ieI.i e c/a'etce / e, et h eir k:/xez}.e?'X i. Ite sanme!lUeniliies, be dievb'isib.e by hke d'flfb eece of c1/'?e pq'eti e..i But we havs seen, alreardy, thatt CT'-b:2 is divisible ay a —; h1elce, it foXllows,., th..ata t —bs isX alSO divisible by a-b. sice ai: divisible y a —b, it again follows, that c- is divisibl by it;',:sd so o n, without limit, 0-Ienle, we hate:'te'cifl"enee o~f Ike sit:ze powers3 of fIwo qa.unt. iezlots, is acioeys o-sibie by I/t diteSblene of the g.afiZ.tieo. The qfuoti- nts obitained 1by divid in the di-fference of the soune ~V; iwers of tw;o quantifies, by: tihe diifforence of thlose quantiti. es, s'.:-'nllo " t s pl law. nThus: (Li b1)c (a-b) a-a- bt ( b" —0) (a —b)=ca —+-a bLb (: —zb ):. (a o-l ). f:-c4+~.,b+-' Fb_'%-U. The exponent of the first letter decre lases by u.nit.y, Swhile that..f the second increases by unity. A.RT. S-G —Since a.. ——.". is always divisible by a-,b, if w'e lu-t. — e ifor b, thhen a-b will becoiie a-i-+c, and, since 0=" will beco.me c-, when iot is even, )s a, i6, &c., and -oC', w<hen srn is odd, as 3, 5, 7, &c., therefore, a" — b w(ill becomie an —c", when tn is evein, tneil a"+od.c. is -heol i dd, becauase ac.-.b'" -a'"-( —c"~)..a.c —,,l "; t'herefore,'F-c" is alwoays divisible by ct-Fe, wlhen in is vuen, and,.1 "& is ai way s divvsib e b -y m.-o when a is odld, These trauths re CXpresCse iI ste follomwing theorems, THE IIexpRess V I I Ltoe d:/i.-ff'ci o0t' 7ke eveoz mpoioei's' Ji7e saatie dei'rce of I'wo qmaanti t'/ies, is gtha/ts dpivittsible by the sam, of' ie qgaaiiti.ies. Thus: (a — ba ) (a+ b)=6c — b. (at — b —-) (a - b)=z at'ab-a ctb-, —TAb. ra vie:w.-.8.0. To what is the square of the same of two quantit'ies eqtl? 81. To what is the square of'tde difference of txi-o quantiies equal? 82. To what i. the product of th1e sur andi difference of two quanltiies equal? S3I.`V'low may the reciprocal of any quantity be expressed? 1t(ow may any facto: i )e tra3nsferre from ono ter3n O' of a fraetiol to -the other? In what other tform may ton he written? c-c —? 8t. What Is the value of any q.antity whoe 2'exponeni t is zero? 68 RRAYS ALGEBRA, PART FIRST. T HE oRE31 V. -XI o rgse suum op t7ie odd power s of tie same d(egriee of two guantitiet, s' always ditvisible by the sumo qf the quantities. It~ls: (+-V-Ib') -,(a Lb)=-a- a- b+. (a5;) ( f$(a+b)=za4 abi- ab-a7-ab'- -b.( a't-+ b) (a + b)f a-ao b-+a4'-a'h+ ab4- a b% — b". FACTOR!ING. FACTORS, AND DIVISORS OF ALGEBRATC QUANTITIES..A'rT. rS-A divisor or mneasure of a quantit3' is any quantity ithat divides it wiithout, remainder, or thait is exactly contained.ia it. Thus, 2 is a divisor of 6; and a is a divisor or Ieasure of at. ARn. g~. A p2rime etndber, is one which hasp no divisors except itself and unity. A. composit nuzber, is one which has one or more divisors besides itself and unity. ifence 1ll nunmbers are either prime or composite; and every composite number is the product of two or more priume numbers. The following is a list ofL the prime numbers under 100: 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, s59 61, 67, 71 73,7 79, 83, 89, 97. The composite numbers are,, 6, 8, 9, 10, 12, &c. RULE, FOnI BESOLYING ANY COMlrOSITE NUIaBERI INTTO ITS PRIEiE FACTOeRS Divide b7y azy prime nuzmber iha' wilt exactly divide it; divide the qutZient again int the samne matnner; and so cozntiu.zte io divide, uustil a quotlient is obtained, whiclh is a 7?prime ren u.? mbert /e', the last qutotient. and the seeralt divisoir,, wi constitte t/he pricmne jci ori o-t if the givem, numnber. R A. r l.~The reason of this rule is evident from the nature of primo and conpoSite numbers. O t will e fotund nost convenient to divide first by the smallest prime number that is a fIctor.-See Ra y's Arith metie Ptt III., FcsTOrIG. R: vi E. w. —85. 3By what is the difference of the same powers of two quantities always divisible? 86. By what is the difference of the even powers of the s.me degree of two quantities always divisible? By what is. ithe sum of the odd powers of the same degree of two quantities a lways divisible? FACTORING. 69 1. The composite numbers under 100, that is, 4, 6, 8, &e., may be given as exanmples. Every pupil should learn to give the factors of these quantities readily. 2.'What are the prime factors of 1 05?..... Ans. 3, 5, 7. 3. What are the prime factors of 210... Ans. 2, q3, 5, 7. 4. Resolve 4290 into its prime factors. Anus. 2, S3 5, j, 3,. ART. ~9- A prinme qtsantiFy, in Algebra, is one which is exactly divisible only by' itself and by unity-.'Thus, a., b, iand b —-c are prime quantities; while ab and ab+ac are not prime. AxT. 9@.-Two quantities, like two numbers, are said to be prime to each other, or relativezlFy ptinme, when no quantity except unity will exa.etly divide them both. Thus, ab and ad are prime to each other. Asr..-~ A compf770osite,numZber, or a comVposite gftatt ity, is one which is the product of two or more factors, neither of w hich is unity. Thus, ax is a composite quantity, of which the factors are a and x. tR x oa. x — -A mono-ial mtay be a composite quantity, as ax; and a polynomial may not be a composite quantity, as a2 —2. AR. 92.-To sleparate a mS ono1T al into its prime factors. RULiE. Resotlve the coil/ gicient into its prime actors; ten ithese, USwith thte i-teralfactors of the mnonomias, will formw the;.rime fazct.ors of the given, qguanicZity The reason of this rule is self-evident, Find the prime factors of the following n ominalsa. 15abe............... Ans. 3X5,.a.ab.c. 2. 21ab2d............... IAns. 7,a,b.b.d. 3. 35'abc2x%........... Ans. 5X7.a.bc.czx. 4. 39c2mn,......... Ans. 3X,.a.a.m.mn.2 ART, 9,.-To separate a polynomial into its factors, when one of them is a mnonomial and the other a polynonmial. P RFvitcw. —8?. What is the divisor of a quantity? 88. What is a prime number? What is a composite number? Namel setveral of the prime numier8s, begotinni wit;l unity. NIame several of the composite numibers, beginning with 4. What is the rule for resolving any compositei ntumber into its prime factors? 89. What is a prime quantity? Give an examnplo 90, When are two quantities prime to each other? Give an example. 9i, What is a composito qwuantity? Give an example. 92. Wbht is the rule for sepasating a monominal into its prime faetor.s? 70 IRqAY'S.ALGEBRA, P ARt FIRST.o i it ~ 5t.iD;vide the given quantz i? by t'he 9at ufre. is ~onomnv ithat wilt exacttq divdle cach7 of /its terms. Then the monomial divisor -w-ill be one jamtor, antd',the. quotfien-'t thfe ot/iher- The reasion of this rule is selfevident. Separatc thue following e.xpressions into factolr's: 1 X. A.az...............&n~,x(1-i-a). 2 on cc.- -c............. Anvs. ak(n+c). 3, bc2'- bad........... Aris. bc(ceLd). 4. 4K:H-G:ay........... A s. 2xa.nSi(2x+3y). 5.' oax —9,?/- -;y2- 2cx2y.. e. Ains. i3xy(2ax-z 3by —4cx)o 6, 5azu-3 fa Jx + 5a: y..... Avs. 5a*z( l-7xycty: ye). 7. I4a'2y —21 cti y-i -35 y' 5.tan. cs Txy(2cax-+38y —5t y). 8. 6bc: -i5bc -3b....... Ans. 3Sbc(2x —5 b). 96. ttCe~Aacm+.C;1hm'~-Jc2aT........ Ains. a'cnt(aA-l-cn)) At. r. 94I.. —To separtite a quaintity w\hiclih is the product-7 of two or nmore polyinoomialss, into its pruio factors. No gen.ea:tll rule can )be glven, for this case. WIhen the given quarn-t:ity does jnot consist of minfre than 4three term-s, the pupil will generally be abl1 to acconmplish it., if lie is ifmiliar vieit.h the t;hle'oems in the ptreceding X setiotno l.st. Any trinomrial can be sep:;arated into two bino-mtial factors, when the extrenmes are squiares and positive, and the middle termn is'twice the product olf the square 00roots of the.extSremes. See Articles 79 and 80o Thus: a;-}-2ab+b`- =(a-b)(aL b) at2 —2ab b-'V —( (a- 6b) (a —b) 2d, Any binonmial, wvhich is the difibrence of two squares, ctn be separiated into two factors, one of wfhiich is the sttn, arni the other the diff'rence of the roots. See Arbt. S1. Thmus:'b —b='-a-=b,%.)(c — ). 3d. When any expression consists of th di' ere e of the sane powers of htwo quauntities, it can be separated into at least two factors, one of.-which is the difference of the qu-1antities. See Art. 84. This eeai> — b -b) (ce"'5-i-en"l 2b., t b7 —-+ where a, b, and iieI, imays be any quantitities wvhlatever. In this case, one of the fietors being thle difference of the qjuantities, the other l will be fosund by dividing the givesn esxpressioi by this dilffrenee. Thus, to find the other factor of aJ-b5, divide by a —b, tg e quotient wvill bhe found to be a - — abj — b;'; ]scne, P3 —b'.z.(a- I) (ac tni' —,$.-bT',)l. FACTOrI0.L:. O JI. a simi.ltr mnanner, c.- -5=(ac b) (.tc+c- a c3,-i-2t bl_.bv4). 4th.'When any espression consist~s of the difference of the even powers of'two quantities, higher than the second degree, it can be separated into at least three factors, one of which. is the suni, ~a4nd another t11he difference of the quantities. See Articles A5 and 86. Thus, ac4-b is exactly divisible by a+b, according to Article 86; and, according to Article 85, it is exactly divisible by a —b; hence, it is exactly divisible by both ac- b and c-b; and the other factor will be found by cividing by tbheir product. Or, it may be separatedi into factors, according t6 paragraph 2d, above, thus:? b k (oi1bi) (ai 62b)(Ci+t2Z) (cc-t,) (a —b). 5tlh. When any expression consists of the sum of the odd powers of two oqunuitmies, it may be separated into t1v least two fictors, one of w.ich is the sumn of the. quaMntities (See Art:. 86). The other factor rrill b fzouncd, by dividing the given expression by this sumn. Thues, we knuow that cd-b, is exactly divisible by a+-b, andti b3 division, we find the other f ctor to be a' —a —b; hence, cci'V —bi=(cc b-)(a a —b17-bi). Separate the bfollowing expressions into their simplest faEct.ors,'2. 9ta" u12ab44b, t130 "X, o39 4-J-_I 2x- l.4 9X2 I O..x- 34 _to.-. 2rigH >: l'n11. b.- " a2,2c&:cdbab 13 xI-I., 4x2P-.20c-2'-5z, 14. 8a3_27b, j7 T2_Y 2 1f5. c5- i1b5. 1. (r.-y)(xT- y), 16 cx(+ cx)(c.-M slt~ ( W8+~)(t +s8)oi z(x+a +)(a —-)o 2. ( 3c-+2b) (3 a- ib). i. (yZ+ —b2") (xi2,_B). (x2 2) (2x —5z)(2 4 (224 3 (2+(43b (+) ( ) 4. (in,-tx)(Tn. —4-n- ). 12 - ),) r6. r xt) (2x or). 14. (a 3 o)(I4-', 6ab-b. 9r (a. itti) (Ct 4 c ) ~ 1 (cc -b) ( cc4 cc S. ( ab tc) a - ) (U'4b3) J\JW-WTJ) (a+ b) e 9. (cabi bi-cd)(ab-cd). -_(ac- b) (j-c b-i-b5) (a —b) (a + — _b-?-b2) - -(cc -4-() (c' a- — b2) (ac2_btb —b)C _ 6g 26 >)(t~t,,}~,2a',,, rScS W~A.~i-{RA-.Y S ALGEIt;LRA,.PTA.RT FlItShIT. ART. 35. —To separate a quadratic lainomial into its fact.ors. A quadratic trinomial is of the form, x2+ca.-j-b, in which the signs of the second and third terms may be3 either plus or minus. When this operation is practicable, the method of doing it, mnay be learned by observing the relation that exists betweenr two binormial factors and their product. 1. (I+a) (X+b)rX2+(a+b)xz —tab. 2. (x-aT) U(x-b)- z x-(Ca-Hb)x-+a-b. 3. (x+a) (-b7)-X2+(a-b)x-ab. 4. (x-a) (x+b)=x2+ (b-a)x —ab. From the preceding, we see, t hat when the first term of a quad& ratio trinomial is a square, with the coficient of its second term equal to the sum of any two quantities, which, being multiplied together, will produce the third term, it may be resolved into two binomrnial factors by inspection. Decompose each of the following trinomials into two binomial factors., 1. x+5x+6........... Ans. (x+2)(x+3). 2. a2+-7a+12.....,......Ans. (c+3) (a-H-4). 3. x2-5z+x6............. Ans. (x-2) (x-3). 4 x2 x2f0 -i.os. (x-4) (x-5). 5. X2-z-,............ As. (x 3)(x2- ) 6. z2 —x —................As. (. —.~)~+~) 7. 2-+x-2......,.... Ans. (x-+-)(c-1). 8. x2-13x+40.,......... n s. (X SL ). 9. x-7x —.......... Ans. ( -8)(xt+ t ). 10. x2+7x-s1....... o... ALs., (x+9)(-s2). I. oxs-x-30...............Ans. (x — 6)(x- 5). In the same manner, we may often separate other trinomials into factors, by first taking out the monomial factor common to each term. Thus, 5az2 — -4a= —d - cz5zfa(aX2-2x-S)= —a(x-4)(x-L-2) 12, 3x2-12 —15..... A.ns. 3(x+5)( -). 13. a2e —9a x+ 14a6....... Ans. a2(x-7) (x-2). 14. 2accbx-4abx —OOb....,..Q Ans. 2 ab(.c —10)(x —+3), 15. 2xt-4x2 —30x......, Ans. 2x(x-5) (x+3). e v i.a w.-93. WYrhat is the rule for separating a polynomial into its primne factors, when one of them is a monomial, and the other a polynomial? 94. When can a irinomial be separated into two binomial factors?. What are the factors of ma-2+2na-n:2? Of c2-2-cd+1d2? When can a tinomial be separated into two binomial factors? What are the factors of 5..-.. ya 2? Of 9a2-IP6b? PWhat is one of the factors of ao2b? Of c3 —b? Of XP —/? What. are two of the factors of et,-'? Of d-_/?6 GREATEST CO3.BtON DIVISO0R.o 73 A.e.T.. 96ro —-The principal use of factoring, is to shorten the sw orki, and simplify the results of algebraic operations. Thus, when it is required to mnultiply and divide by algebraic expressions, if the multiplie- a.nd divisor contLain a common factor, t rimay be canceled, or left out in both, without affecting the value of the result. Thus, if it is required to multiply any quantity by c-b2, and then to divide the product by a-j.-b, the result will be the same, as to multiply at once by a-b. Whentever there is an opportunity oe can'eling co.-mmon factors, the operations to be perfoirnmed sh.ould bei merely indicated, as the common factors will then be more easily discovered. The pupil will see the application of this principle, by solvingr the following exanples.. Muitulply c — b by i -2x Yi-2 y. and dvide the produ.ct by x+y. x-i-'?t/.Y —~7 t -aeam+ay -bs- y 2. *{u4mtiply xt —3 by x2 1, a.nd dieide the perodu b l by, b acto. ring. A ns e —2x-3, 3 ]Dividea -V: by z-q-j, and aultiply'the quotient by z,1 by factori3ng Ans. i - -t}z-. 4. DivideC Ge c 12c -t1-o2Te by 2icc -f, by eto htino Arns. 3 (a —b). 5. Multiply Get -y -ay.:,an —9 a d divide l the product by 4X2+ti 2x-y-9'/, by factoring, Ans. 3a(2:. —I.g). 6, iMultiply -5x+6 bc-G -y j.- n; 1 di ide'the quotient by - A t d by f9 by ringo.eas. ( -2f)(r-4). Other examples in which w the; c't iple may te applied, will be iburnd in tghe multiplication. "an d::ion t, of fractions. tEATEST T I C OMON DIV I SOR. A-rn.t' 9-t-AcN quoantety tha t will exactly' divide tawo oUir sore quantities, is called a. comreimeon dtivisor, or common meassree, of those quantities. Thuas 2 is ena comon divisor of S and 122; and a is a corn.luen divisor of ab and cOX. ErM A u i. —Two quantiftes may sonmetines have more than one omi meon divisor. Thus, 8 and 12 have two commone divisors, 2 a.d 4.. tl vi: w. — 94. Whtis one of the fiactors of t:i-?3? WhSat Vis 0e 0o the fat.ors,, of -!a 95.'Whsat is a quadratic trinleomi? 74 RAY'S ALG-BRA, PART FIRST. ART. 9 --— That common divisor of two quantities, which is the greatest, both'wNith regard to the coifficients and exponents, is called their gr-eattest comrmons, divisor, or greafes-t cornmozn mea~s'ure. Thus, the greatest common divisor of 4a'xr y and 6ax2't is 2a'x.2-, ART. ID9- -Quantities that have a comml-on divisor, are said to be com2mens6arable; and those that have no commuon divisor, are said to t be inco mmttetle Ineomninsu-rable quantities are also said to be pritme to each other, or relsatively )priosc. ATr. 10[.I —— To find the greatest comamon divisor of;wo or more nonomials. 1. Let it be required to find the greatest common divisor of the two monomlials, 6ab and I5 a2c. By separating each quantity into its prime factors, we have 6ab=z>2X3aab, 5a'e — f3X5a.ac. Here'we see, that 3 and a are t.he only factors common to both terms; hence, bothl the quantities catn be exactly divided, either by 3 or a., or by their product 3sa, and by no other quantity whatever; consequently, 3a is their greatest common divisor. Hence, the Fo rI TDING TIHE GREATEST C0ossON DXIVISOR OF TWO o1t:ORa-4 MOINOMIALS& Resoilve ithe q. uantities into their primte jtctors; ihent, the.prodtuc of those fictors Z7hs are coa-mmonz to eac/s of the t'erms,:wilz jfbt the greest cost co;itn, divisort. No TE. —-The great.est coimnmon divisor of the literal parts of the quantities, may generally be more easily found by inspeetion, by tiBdng each letter with the highest power, that; J t-.omanmon to all the quantities. 2. Find the greatest cou:m -:divisor of 4a's-, aG3a', and 10.ask. 4a"zx I. 2X22a'' Here Nwe see, that 2, a2, antsd s are the only 6Gcx2:-2X3cx's freattors common to all the quantities; hence, 10a%=x: 2X5a4t 2Qa'. is the greatest commnon divisor. Find the greatest common divisor of the folloowing qua.ntitieso 3. 4a'.:', and I Oax'........... Ans. 2a 4, 9abct, and 12bc............ o Ans. 38be. 5, 4a3b'%xcy3', and Ss%'t 2 y.. /...........os. 4scsx g2 6. 3a4'/, 6ca z yl, and 9a yg..... Ans. 3a yl. 7 8ax' gaT, 1xz5:f, a1 24.32sz o a. o: 0.. eAn s. 4x'2:.,S 6ca/Cy,, 1.2a3ya'/5, 9ca'xy4, and 24a'yz... Ans. 3ay. A-T. @l0T —To: find the greatest common divisor of two polyno1mials. First. Let AD and BD be either two monoimials, or polynomials, of which D is a common divisor; and let A) be greeater than BD. GRPLIkATEST COMMNtON DIVI SOR. 75 Divide eAD by BD; then, if it gives an exact quotientR BDn:lust be the greatest comnron divisor, since no BR)ARQ, quarntity can have a. divisor greater than BRQ itself. But, if BD is not contained an exact v - number of times in AD, uppose it is con- BD-IB R ttained Q tines writh af rnemninder, which rmay be called I.t: Then, since the remaiander is found, by' ubtracting the product of the divisor by the quotient froun the dividend, we have R=A.D —BBDQ Dividing both sides by D, we get D-zA —BQ; But A arti BQ are each e.utire quantities hence R Thich is equal to their dilbrence, must be an nt ire quan'tyi. Hence, it follows, that any ctmmon etdi;isor of two quonttidtcs, 10wi. asttays exactly divide their remainder after d'ivision. l ntld, sinceic the greatest common divisor is a common divisor, it folliows th;.at the greatest commonn ctivisor of two quantities, will alwaiys exactly dcivide their reimainder after division, tR a r n i. --- A n the above artictle we have used two axioms, which mau-y be new to some pupils. They a re, first: f/ tlico eq',ual qact.tlies be dtriided by thie sace ern.tclity, their qeotlienlts witl he eq-cual Aid, second: 2hte c:if'ere.ce of tieo clv eire quanemtiles is also ace entc-ire cjluactcty. The pupil can easily see, that teho sum, or dlifferuee of two whole inumbers must also be,w whole nunber; amid that the same is likewise true of two entiro quantities. This, and tthe next article will both be better uiderlstood by the pupil, after he haes steudied simple equations. aRT. hl1 c. @ Second. Suppose, now, th t it is retuired to find the greatest coummonm diivisor of twio) poln01 nomiaSi, A, and B, of which A is the greatcr. If we divide A by B, and cholrd is no remaindner B is, evidently, the greatest BQ coarmmon divisor, since It can hav-e ro di A —BQ=rII, tst Ram~ visor greater thtan iRsel.I Dividing A by B, and calling the quo- R)B(Q' tiient Q, if there is a remalinder I, it is evidently less than either of the quantd i- --. ties A and B; aid, by the preceding theorems, it is also exactly, divisibl by tle by tSip co i the greatestelommon divisor; hence, the great- r B — EQ' —-! divcidvend is est comimion divisor rnust divide A, B, and equal to the B, and cian not be greater thatn R. But product of the divisor by if R will exacatly divide B, it will also c c the quotient plus the reaetly divide A, since A —BQ —R, et and will dc be the greatest conmnmon divisor souglit 76 RAgY'S ALGEBRA, PART FIRST Suppose, however, that when we divide -I into B, to ascerftain if it Awlill exactly divide it, we find that the quotient is 0Q, with a remainder, iI'. No w, it has beenl shown, that whlratever exactly divides two quan.tities, will divide their remainder after division; then, since the greatest common divisor of A and, Bd, h.as been shown to divide B and I, it will also divide their remainder t', and can not be greater than W. Alnd, if B'_ exactly divides R, it will also divide B, since B RQ'+I -; and whatever exactly divides B aud R,;will also exactly divide A,,since A-BQ R-1-; theerefore, if it' exactly divides B, it will exactly divide both. A and 3, and will be their greatest common divisor. In thl.e same manner, by continuing to divide the last divisor by th!e last remainder, it may always be shown, that the greatest common divisor of A and B will exactl;y divide every new remainder, and, of course, can not be greater than either of them. It may, also, always be shown, as above, in the case o, f l', that any -~remainder, which exactly divides the. precteding divisor, will also exactly d.i.ide A. and B. Thlen, since the greatest common dixvisor of A and B can not be greater than this remainder, and, as this remainder is a conimon divisor of 2A. an:d B, it will beO theirl great est commnon divisor sought. To illustrate the sa1me principle by numbers, let it be required to find the greatest common divisor of 14 -a'nd 20. If we divide 20 by 14, and there is no remain- 14)20(1 der, 14 is, evidently, the greatest common divisor, 3 4 since it can have no divisor greater than itself 6)14(2 Dividilng 20 by 1.4, e efind the quotient is 1, and 12, the rema.inder 6, which is, necessarily, less than either of the quantities, 20 and 14; and by the 6 theoreum, Article 98, it is exactly divisible by their greatesst commnon divisor; hence, the greatest commnlon divisor nmust divide 20, 14, and 6, and cannot be greater -than 6. Now, if 6'will exactly divide 14, it will v also exactly divide 2{), since 20=14+6, and will be the gre atest comunol n divisor souh'glt. But when. we divide 6 into 14, to ascertain if it will e exactly divide it, we find that the quotiejnt is 2, with a remainder, 2; theln, RlE v IJ wv. —-95.'When can a quadratic trZiomiat1 bhe sepanrated. int[o')i:nonial factors? 96. What is the principal emse of f0aetoring? 97T. Whtt is a common divisor of two or more quantities? Gxive an examnple. 28. W8 hat is the greoatest cornmmon divisor of two ciuantities? Give aun exa. mple. 99. WVlhein are quantities commensurable? When are quantities incommonsurable? 100., How do you fin(d the greatest common divisor of two or more 01oneomials? 101. Prove tllat any comimon divisor of two quantifieis Will alwaysY exactly divide their remainder, after, division. G lEATEST COMAbONT DIIISOR,. 77 ly the preceding theorem, the grneatest common divisor of 14.nd 6 will halso divide 2, and therefore, can not be greater than 2. Now, if 2 will exactly divide 6, it will, also, exactly divide 14, since 14 —6>(2X -2; and whatiever will exactly divide 6 and 14, ivill Mso divide 20. But 2 exactly divides 6; hence it is the greatest common divisor of 14 and'0O ARTn. @ en3 —Whe'n the remaimdnders decrease to unity, or when ire3 arrive at a remainder which does not contain the letter of arrangelmentR, we conclude that there is no common divisor to -the quant itries. APrTM r 4,-If one of the quantities contains a fiactor not found in the other, it t may be ca.nceled without affecting the con1maon divisor (see exa.ple 3); and if both quantities contain a common fact.or, it may t.be set aside t a factor of the coimnnmon divisor- and we may p roced to find the greatest coinmnon divisor of the other factors of the" givena quantities. This is self-evidento See Example 29 ArT.:,-I-~e may imumltiply either quantity, by a factor not fil.nd in the otler, without affecting the greatest conmmon divisor. hrks, in th- - action 2alx- the grea"test connmon divisor of the' 3abc twmo terms, is evidently ab. ITler e n may cancel tie factors 2 and. sn the un mnlerator, or 3 and c in the denominator, without alfectiin: thie common divisor; for the common divisor of }D c or of i S Still a.bo If we mnultiply t he dividend by 4, a fector not found in -the divisot, e hae a —e of wihich the common divisor is still cb. In'the saime Iimanner we imay mnultiply the divisor by anr factor not fo und in the dividend, andthe conmmron divisor will still renmain the same. If, howeverr, we mu-ltiply tihe numnerator byv 3, iwhich is a factor of the denominator, the result is mdc of mhich: the greatest comu mon divisvor is 3ab, mnd not ab as before. Hene, we see, that the greatest comimon divisor wvill lie changed, by nultiplPying one of the quantities by a factor of the other. 11, r v iri w. —102. Show, that by diviiding the last divisor by ithe last remainder, the greatest common divisor of two polynomiams will exact.ly divide both the firsl; ald second remainlers after dfivislon, 78 Tr.AY'S ALGEBRA, PART FIRST. A2nT, ItC — -In the general demnonst, ation, Art. I01, i:t has been shown, that the greatest common divisor of two- quantities, also exactly divides each of the successive remainders; hence, the preceding principles apply to the successive remainders that arise, in the course of the operations necessary to find the greatest comnmon divisor. The preceding principles will be illustrated by some exasmples. 1, Find the greatest common divisor of x3-y' and x1i —x 2y2 Here the second quantity contains a2 as a factor, but it is not a factor of the first; wNe nay, therefore, cancel it, and th.e second quantity becomes z-2y". Divide the first by it. After dividing, we findl that:, is a factor of the _, pf fot p remainder, but not of s —:Y2,% the dividend,. -Inc, by caneeling it, the divisor becomes zny; then, di- / viding by this, we find there is no remainder: there- X'2Y — fore x. —y- is the greatest comnmon divisor. 0or (x-y) —)?/ XY -- y2 2. Find the greatest common. divisor oft x-:a1x amlud U —L —a 2 The factor,aZ2 ais ceomMnon to both these quantits; te a 2 2 it; therefore forms parkt of' the greatest common dvli-. or, and ma.y be ttaken out anli reserved. LDoin g "X3-_'t X (x thLs, -the quantiies'beconme a-4+Af —o anntd cc<.c~.. af,-t —t. The first quantity still contains a common factor, x, Or which the latter does not; canceling this, it be- a — xa comRes a;3 —a_,3$ Then, proceedilng as in the first 2 1 example, we find the greatest common divisor is - 8. Find the greatest common divisor of a56 lGC.a and 2 at x —-.f x -I-,. it,:Zx -{-.'x H-erei 5i is a factor of the first 0m 2 dus2 I a"1 -- t qnantity only, ard: za., of the. second I _ 2i only. S-ppressd ing7 these fa.cts, and _ ( proceed:ig its in the previous ex-am- ax 2-:3 ples, we find cc. —x is the greatest or, (a-+-x)xv co:ammon:,livior, C2+2axf-+xa' a la+xc a-f —az (a-t-x ax-f-c2 tacc mi GREATEST COLMMON DIV. SOP. 79 4. Find the greatest common divisor of O -#' 6x and. In solvin tils et ample, 4e F-a6 I2 —i- 1-a-.-1 &/-'i-x 2' there are two instances 1i1 4ac 03~. 2a (o... f which i. is necessaryv to multiply the dividend, in 80:sz ~M-' } 2(:a'- 3:d orde rthat ith coilttiexlito f Or, ( 8aa3 —a2$- c iO-2 2ax2~.....)x'',\ the first term may )be, exactly divisible by the di- 2a__-aX2_t Xm visor. See Art. 105. The 4greatest cosmmon divisor 8a4 4 j S24'I 8l 2 -X 2 oa1 2 is found to be 2,a,-2'3r. 8X. 4ilO- 3os 9I i9 S c ct -—.s<rt~ — --.. —— x' (a 2 cC — 6a 2+3ax-' I24x1 __..2a________............9 Sar- 2 etv....3x'i~,%-j2 3;5:" or, -3 1 xW(2} —3 Sca -— 2a x 22x' (4 —32 a 1.2t — — 2acx _-_ -3a — 2ax 3xteons From the preceding demonstrations and examples, we derive the soR FINDINX TIE- GREATEST COMMON IVit SOR Or TW'O:.wPOLYNOMi:TAL',S., st., Divide the grdeater. poiynoonsia.l by/ the less, cac'd if there i.,' to Temaintjder, the less quant.iy wtill be the divisor souyc7to 2d.'If /there is a renmainider, divide thejfirst divisor by it, and con3ticnte to divide the ast dtivisor by the aStd reaneinde-r, until a divisorl is ob wained, scwhich oleaves nto remaincder;td. this will bise tie great.est coimmoct, divisor of thfe Lwo given. 7olymon2tia iZt it aE I X.: a s s.-102. Explain tihe principles used, in finding the greatest comAlmVon divisor, by finding it fior the nuambers 14 acnid 20. 103. When do we conclude that there is no conMMon divisor to two quantities? 104. l.ow is the common divisor of two quantities affected, by cancelsing a factor in one of them, not found in the other? When both quasnitities contain a commoAn factr, how may it be treated? 105. I-ow is the greatest common. divisor of two quantities affected, by multiplying either of them by a fa:ctor not found in the other? What, is the rule for finding the greatest comamon divisor of two polynomials? Iow do youx find the greatest common divisor of three or score quantities? 8O8 Ri&$RAY'S ALGEBRA, PART FIRSl. NfOT'r s. —L When the highest power of the ea.tdi;yg letter is tlhe smeor in both, it is.immaterial whiieh of the qua ntities is lade the dividfenild. 2. If both quantities contain a, common factor, let it 1)e set aside, as tirmling a tfactor of the commosl divisor, aend proceed ito find th-e grea.test com. ]no1 dtiis"orT of the rellaining fa ctors, as in Example 2. 3. If either quantit contains a factor not found in tthe other, it may be canceled, befire comme-ncing the operation, as in Example J' See Art. I104. 4. Whenever it becomes necessary, the dividend lmay be nmultiplied by any quantity which'will render the first term exactly divisible by the divi.. soro See Art. 105, 5. If, in any case, the renaincder does not contain the leading letter,' thati is, if it is is idepndende of that letter, there is no comlmon divisor. 6. To find the g'rea2test common divisor of three or more quan1tities, first find the greatest common divisor of two of theim; then., of that divisor and one of the othler quauntities, and so on. The last divisor ti.hss fo'und, will b3 thie greoatest coimmonn divisor soughtr.'7 Sifonc e greatest cormmaon divisor of two or m3ore quantities containsl all the factors comnnon to these qualtititfes, it may be founld mnost easily by separating lthe quanti tiies into factors, where this can be done, by m _eans of the rslies in the.1 preceding artirle. ]?ind the oreatcest conmmon divisor of the following quantitises 5. 4a tA5 —-aC,:Sn a,-3 t 4......,, Ains. - Va' 7 s0 c:.dx an a ac oh1..............,. tc. 6. ifaKxS 3 and..5+6.....,... Ais. -a-. -(... ci [-Sa —.' amd Ce6e2+Te-a., Avis oX Se+3ox 9. A-.n5 l e. e —3 a ad -.adV'.....3... A. s C25-.3a l 0 tt1.-2 itd.!,,, _Xt,_' z.sLB R 2 z,exactly, hcs, i a u3dt-i of 2, or of 3; ttnI 24 is a snlti - 1 0:. & n.r-:S, 6 is -a.1mltip le of 2. o.. 3; and.4.s.a. m. l tia pie of 2 3 4e S roc. a s 8a"b is a mutiple of 2ah, of 2ec~, of 2a," b te;'nd 4(ca —) ivs a ausutiple of (a- ), tof a24t, of 4Ay, &e, 2T.m @,I-8- - i quantety thati contains two or mdore tunstxitisexactly, is a co,'aeos snu'tis;ie of thenT. Thus, 12 is a comnvni)o mnltleci o 2 and 3;'end Lax is a commonn multiple of 2, 3, o c, and rn, LEAST C3OMIMON iM3ULTiPLE. 81 -AirT. 1 9,-The least ieottu b -um,.?d1h, ninth3,o./e of mtwo or m-tle quanttidntes, is tihe least qnantitvy ihat will conttain them exactly. Thtus, 6 is tie least common multiple of 2 and 3; and 1Oxy is the leasts comzon e uitn1iiple of Sn and 5y. Rnt r h~ a i sI. two or more quantities can have but ocne least osutmmont muitiple. whrih they tmay heare an unlimited numnber of cemmon multples. Thus, bhile 6 -is flhe least comon tti multiple of 2 atnd 3, any multiple of 6, ifur inst, nee. 2, 18, 24-, e.,'will beS a common amultiple of these ll1umbers. A_:. t 1@ T —To find the least commtton multiple of twvo or inore quatitles,. tt is eyvi dnt, that one quantity willy11i not cotaitn anothler exactlty, unless -it contains the satme prim' e f'ett1rs. Thus, 30 does not eo-w a ctily centt in 14, becrause 30 X:- ix3Xa and 14- 2'7; the -prime factor 7, / not beilg one of the prime i'cirs of 30. A:RT.'1.l-.0 e..'ly-s-y quanltity wvill. contain another exactly, if it contains al the prime factors iof tt quanitity. T is, 30 01 tains 6 exactly, because 0 30_-2X3X5, Lanid 6-X-S3; the pxrisme factors 2 and 3 of the dlivxisor, beingl, aliso iLCetors, of the dividell dl. h1ence, in or'ier thie i; on0e quasntlitl sballi conlta i anxotit.r etxact]y, it is otlty ncc, i stary tltdat it sholxuld conttt:.in att the'}io::ne. fctors of that quanti;-lt. T3loreolver, in olrdier that aoy q-uailnt-i-y sihall i e —y con.t in two or more quatiti'is, it msnttt coitain l'It.tlhe d.ifi-'Oer'i; prhine afeotsr's of thtliose qni.tlities. Aittind, to ie the leoat; y.:,n-i'tit.t that shall e0'actuly contitlain themln, it shouldti conttain tie's dtil'r -et prime fcto'i rs only once, sand no other i'actors bestides. once, tIe least cam'.'o-iem to opl' tao or meoe qotitatilcs io' co lts i/Zcc h d4i/feea' ett~icic/tne ":'~s ci~t'S' y i/ee qajtton/i/ies oxnce, stisd doies sto, cmn',io,. 3?/other, f/-oras Thus, tihe least commlno n smultiple of ac2te actxld C, is acbexs, since it- Contains all t:he factors in each of these quan'tities, end does not contai n any other factor. With this principle, let us find tle least commonlo niuiltiple of eCx, bx, and (tebc. a e bz a.s b Arranging'tlhe quantiticies as in the sltargtin, iwe see, ti cat a is a fect:or cotmm1on to two of the terms; hercee it must be a factor of' the itea.st, b b be t b 1 1 __ colmm ton mnultiple, ailed we p it on'lie left I I cl of the quantities. We then cancoel this factor in each of t-le quantities in which it is found, whiesch is dolne byv dividiug by it. By examining the remnainiug factors, it is seen that xi ais a ccm1on factor in the first and second terutls, We then place it on the left, v and cancel it in thxose t.erums in whichf t ii t is 8~ B5RAYI'S ALGEBRA, PpARRT FiT.', found.'We next see, that b is a factor common to two of the cquantities; hence, as before, wo place it on the left, and canel it in those terms in which it is found. We thus find,'that ca x, t, and e, are all the prime factors in the given quantities; therefore, ltheir product, abcx, will be the least common multiple of these quoa ntities. lTence, the RU IIE, eFOR FINDIeNG THIE LEAST COM?3.ON ~MULTIPLE OF TWg.O OnR'.ORT QUANTITIES. Ist. A rrange the quant'ities in a hori zontal t ite, atznd diide t zern by any prizme f oet/tz that will divide twoo ort sw.ore of them,'wi?,t/hout c remainder, and se-t the q'uot'ieLntst, togci hee sij tlhe tziv'doided iatttitities, in ct lizne bezieatth. Sclt. Contiuau e dividttfyf aor J bo t/te, utttil.vo /ri) e;ctor', excep6 scit, wti/i d.ivide tio o0r no0 J te / 1/ctatitces, wzihou.lt a. renza indtte' 3do tidtZipty 1t/ ie diciscort and lite qzitcnt-ilies i, the Iast in due toqetlher, catnd the js)prodet.ct tvdilJ be 1the leest. cosisio ntt.nnll/tlipe i.epired. Or, separatte thi.e q' n 9 tii ti'ies, ito their, ptnime /o:e' c tl.r,, cur t/hen mu.ifttii// itolaeihet, s't/c f/ th/tore ji.ctots as ctte 9 tnecestoay t /ltr'tt a, ptrod'.rc tha7., {i,7, co'ttdtc at, lhe. p'riccm/itce ors i. each.q'tantiiy t /his pi-fodilt -idtt. he theA least Cttti til reittitatcn ectt f-~.6I 2','od}>13/:u~.2, w4iit 9 tiZe ~. g' com.} f'mto~,mfIM. I' ple refq'uire& AxT."' Since the greatest common divisor of tiwo qua. L —m dtities, contains a1ll the factore s commono to them, it follows, that if e d diidd e 1i rod'/ttet' qftwvo qiuactitiies, by thei'.r g~/reaest cotmneocz divisori, he jquotietlt twill, be t/ceir tecast commotnt, iic/tlipctle Find tvhe least; co1mmon multiple in each of the following examples.. 4 ca, 3ca r, a.nd" G(I i'.ya.........Ari s, io s 2a'y 2. t2aczO, 6a', mid 8scy'T Z. A ris o, a,.4',,' 5 Z 3. 2,tz, an'rd, aSnd t:c z.. o...... 94t4nso 4'-z'. 4.t a6:(a-, ) eot"l (cj- c........n. 2.. c i) 6. 4S (a —x), t' and ax( r-c)...... Ans. 124a x"(,. -. —x) o 7. O8%"2'(y —y ), 3 t", tt. and 12a 1Xy.... Ans. 24a Xy'(X 7-)..a ~lxt — y)7 1xI[x y) fa....1)2, l.2('_y2) A.Jf\ 6..Ot,22_..(.2__..2'sv-r s'wi- l0i7 WPehat is', mlltiple of a quaantity? Give- aIn exsample. 108. What;1 is a colmmon multiple of two or moe qu-antitiios? Give 1_an, xample. 1009, What is the least comnmon multiple of two or more quanntities? Give an ex;aple. Iow matly conmmon mulltiples may a quantity liavs? 11.0, When is one quanlity not containei extactly in another? (-ive an ex — ample. 1 11. When is one quan t ity contained in another exactly? Give an example~. What is neesseary, in order that one quantityi mnay exactly:ontl..in uo or lmore qutattitles? ALGEBRAIC FRACT7ONS. 8 3 D:i'F N'NITIONS AND FUNIDS)AENTAL'x P roros5:.TCONS. ART. Liti.j.... — ri a unit, or whole thing, is divided into any nnm. - Obeu' of equaLt parts, one of the parts, or lan'y nl:ubter of them, is ctalled a fracti-on. Thus, if the line A B be supposed c d, e to represent one foot, and be divided -d I'W..'!..-i.!s;...Ai B into four equal parts, one of those parts, as Ac, is called one fourth (-) -two of them, as Ad, are called two f)ur ths (<); and. three of theam, as Ae, are called three fourths (). 1 1 Tn the algebraic fraction -, if c —4 and I denotes 1 foot, then denotes one fonrth of a ftot. In the fraction a if aC3 and 1 c c 4 of a foot,; then- reprepsents three fourths (o) of a fot...ART. 11t. —Ervery quat.ntity not expressed under the forin of a fraction, is ctalled an entire algebraic quantity. Thus, c' —1-.) is aon entire quantity. AnT. 11St —Every quantity conmposed partly of an entire qualtm.. tity and partly of a(. fraction, is called a,m~ixed cqtan.tity. Thus, ab —T is $a mnixed quantity. ART. 11t —.VAn i.yproper algebraic /ractiao;, is one whose, nsu:rlerator ean be divided. by the denominator, either -with or without ab ax2 A a remainder. Thus, -—. and -, are lproper frac-tions. Ia c ArT. Ill t —.A single expression, as.:, t, or - is calledi sirple ficc-tion. It may be either proper or improper. t-t'xsir w. —-1ll. What, is necesesary, in order that any quantit.y may be the least,- t.hat sha.ll rontain two or more quantities exactly? What factors does the east common mu31 ltiple of two or more quatltities coneain? What, is the ruile for finding the least common multiple of two or:t0ore quantities? Ilow may the least common multiple of two or smore quantilties be foLunld. by sepat1iting themr into fictors? 112. If the product of t~wo quantities obe divided'by their greatoest colmmnon divisor) wha, m will the, quotielnt; e? I30. What is a fraction? 114. What is an entire algebraic quantity? G-ive an example. 115. What is a mixed quantity? Give a~n example.'116, lWhat is a$n improper algebraic fraefilon? Give an example. 84 R A.TS ALGEBRA, PART FTRST. 2 2 m ma Ar. t', t —..-A fraction of a fraction, as I of, or - of is cailed a conmpoum0 Jfractio'n.2. Ant. " g $69 —When a. f'iaction ha, s a fraction, either in its nuoecrator, or in its denomincaltor, or in both of themli it is called a com, tlex I l b b 2 UO ac c ao 2fl'6/ractionwi. Thus,' -2, and — a re conu-mplex fiactions, e, AR T.. 2 20 —AlgeL brai fri ations r1 le represented in tihe saine umanne ias a commnian ftasCtionr s in Arith etic.'he int a ilber or qulantity below the lince, is called thte deotnl/ont, nc ecanso it td.-,o-.it't'etS, ot )/ l0'hows thle ii. hcer of pa1rts into iwhich the unit is (.nvtn-.I(:. and thO nunber or (quanitity above the line, is called th1e aoti c oazo, because it z... berst, or o sh a tows t any p)urts arce taken.'flts, in. the firaetio:a-, the leenoninetort, 4I, shows, that the unlit (fi'[ insitance,o 1 toot, i t.ivided4 nto 4 oeqtuail ptarts, antl tie inumea tort 3, Showns t nl.t, i of these parts -are takein. A n, the f fttii.I o-n' trhe e.e m':Ti1ltr c, 5s1otwsi, thaet a unit'is ditlvded into c iequ.a. ]):c pit, ando ei, shows, ti-hat: a ilf these parts are taken. The.uut.ne Jattor an -l, n.- called t 1-h1e lri i of tit fietaction1. A...e. G —T n. e the lr1feidi deul:.0 to. s 0Of numerator a.d doeont inatocl, rueinc is!i.d to a t:2? o;ly. This is'th-e s.lnaplc'ast mot. lod ottf con.tsideriil ati fraction; but there is a, nothler pont of cviewv, in w ic it i is pror i t, exainine it. If' it be otnuaired to divide 3 a plpes cu. selly w 4 boys, it can'be et f tetied, by lv iding each of the 3 apples into 4 equal parts, axnd theil g-3vin- to each boy 3 of tios parts cxpressed by I Now, the pa:,i's being eq ial o each o-t.her in size, it will )be the staote, for an i n dividlual to receiveo 3 parts fro:I apple, or 1 p:t fron catch of the 3 apple s; that is, of one apple, is the so as - of 3 apples; or, 3 of I unit, is the se s of 3 units. hus, may blt regarded. as expressing wto ni tis of one thing, or oit filh of two things~ Rlu.:w...t17. Wts-1 t iiL s hiple fraetion Gine gn exanple. 118o Wha1 t is a compountld fraetion? (ive au exasmple. 119. iWhlatis a complex friction? ive an example..12 20. In Al-ebraic Fiact ion, whl t is the quantity bielow t;he line callied? Why?. Above -ti line? Wihly? G'iva an exampl e. W11hat do you u.nderstand by the terms of a africt3ion? ALGEBI.AIC FRACTIONS. 85 So, i- is eitherl the, fraction of of e unit talken as timnes, or3 it is the nzth of in units. Hence, the numerator may be regarded, as showing the number of units to be divided; and the denominator, ais showing the divisor, or liat.paert is tak'en friom each. N O T v o T o r A., c:s E p s. —Although it is impotrant tht t;e pupil. should 1)oe prfetctly fam:iliar with the principles contained in the following propositions, the denlonslrations ma:ly be omitted, especially by the younger clases of pupils, until the booi is reviewed. PRIOPOSITiON I. changing Ithe deanom inaor, the vaute q/ the,/3"acfioi is in i'ctreased as,snnbty times as there are nun, s in the.nitdtipttier. If we nul-tiply the umeeator or of the fraction'y 3, without changing the denominator, re get-. Thue 2X3 6 Now, an - have the same denominator, and, therefore expr:ess parts of tIhe soame size; but thlse second fract'ion, -$j, has three tinmes as large a num- rator uas the first,:; it therefore expresses tihree times as nmany of those equal parts as the -first, alnd is, consiequently, three ti:es cas large. And the same lsay be shown of asny fraction Al-satever. A.MTs. 1 2 -7" see dCir:hfl;e the -anmertcdor tf a., fi'iretionZ, 2wihout ciaangizng the thctajo'>ila:o, o' 1 rva~t,.c f ~he /}oh'acgiob is vdineintishala.nei, as tmain-,'l tuijnes as thefre arfe,n'icts i.'t' l-,1 divisor-. If wie take tOe f'acction'm and divide the niumerator by -, w-ithout changing the deno1minator, we get T..ns 4-+-2 2 - 7 4 Nowr -, tntd a hrve the sa.e denominator, anad, th, erefore, exproess' Parts, of the saime size'; but thte numlerator of the second fr actito, -, is o nly one half as large as the numlerator of -the first, it therefore expresses only one half asi many of those equal, parts ais the first, tand.is, consejquently, only one half as large. And'the, same may be shown of other frac1tions. lisav s:w..-121. In what two ditferent pointts of view luay very frac-tiorl be regtarded? Give exanples. 122,. Tow is the vinuo a a f:t'racti.il:on niaeticeR byr multahiplying'the 1nm11crator only? how is this proposition provcI? 1.23 HR.ow is the vatlte of a fractiont a:it'cted by dividing the num.eraltor only? fIow is this proposition proved?..? 86 RA.Y'S ALGEBRA, PART FB: ST. PR.OPOSITIOIN 1i.. AuRT Q124, If we san//gly 1/ic denoniuna4%r v/ a />ue/ioa, wi//a out cha nging7 te nunaerator,te te (/i i va/c of,i:./;-ract'ion'> is d-im' inish5 ed' as maa t ny imes ae th.eire are un.-its in?I/.c,mtt?$liecr. If we take( the fira ct, ion:[, and mL ultiply the denominator' by 2, without changing ~the numerator, we get -'. Thus: 3 3 Now, each of the fractions, $ and.', have the same )numerator, and, therefore, express the same number of parts; but, in -the second, the parts are only one 1half the size of those in the:first; consequently, the whole value of the second fraction, is only one hal that of the first. And the samie mlay be shown of any fraction whatever. Ant..f5.i -f w iv ide edth de2no'in ti'cor qf a f jiror/acli, wLi//ootit c/?'ieigni/ th.ce n.)1erat..or, the/ va/te oif the [il dacion- is is'creased as mztuany haites aIs Iter e.are uia/s ii'n the td iviso'r.o If we'take the ifraction!, and divide the denominator by 3, without changing the numlerator, we get'-2-. Thus: 2:2 9~0 3 3 Now, each of the firactions, -? and %, have the same nuinera tor, and, therefore, express the same number of parts; but, in the second, the ptsarrts ae three times the size of tho'se of the first; consequently, the whole value of the second fraction is three timnes that of the first. And the sanie may be shon of othiler fractions. P O.POITION V. ART i-:t. 2tpzyi.:ig tbeo/i tfe-rih'Ms Jif ja fcciton j, ti/ he sate zinZber or quat Z ity, chzanges the Jfin- of the j'taction, b'ut does not alter its va/ce. If we mutRltiply the nturerator of a fraction by any number, its value (by PrIop,.) is increatsed, as many times as there are units in the multiplier-p anld, if we tmultiply the denominator, the value (by' [Prop. IIL.) is dc/reasend, as many times as there are units in the m.-iultiplior Hence, if both terms of a fraction are multiplied by tlhe same number, the increase from multiplying the numerator, Rt v -E w. 124. How is the value of a fraction affected by lmiltiplying only the denominrnator? tiow is this proposition proved? 125. RIow is the value of t. fraction affected by' dividing the denominator only? iHown is- this proposition proved? 120. I-eow is the value of a fraction afflected by mali tiplying both.'terass by the same qua.ntity? Why? ALGIEBRAIC FRACTIO0NS. 847 is equal to the decrease friom m sultiplying th-to denominatore consequently, the value remrains unchanged. P lO P OSI g tI Oi''., AnT. 12t -Dicvi'din bot7 teres of 2a/eo tez ty I/ie salmie tnumbee olr quantit?/, chancces t/,efb'ma of:/he jf'action, but does not atdter its vacte. If we divide the numerator of t fraction by any nurmber, its value (by Prop. II.) is decrcascd, as many times as atheree are units.zn the divisor; and if we divide the denominator, the value (by Prop. IV.) is zincrecased, as many timres as there are units in -the divisor, HI-ence, if both terms of a fraction are divided by the same munber, the decrease fronm dividing the numerator is equal. to the increase from dividing the denominatorl consequently,'the;a:lue teomines unchmanged CsAS IoI T}O >->i'-.:j.: -.. t iFA<K CT.[ON TO ITS LOWEST TERAS,.;.V.r. 1lS~, — CSninee t heo value of a fraction is not changed by dividing both terns by the same quantity (See Art. 127), we have the following R U LE, liTvide bot/h termts by tmeir greautest cotmmon cdiais'or. Or, Resolve tihe numterator cad denno oinator i'Lao thei/r' pribme fe, tors, oatd thea cam-ncel those fictoJrs comm to boeth term, __ -t a A Rae. —The last rule i't i be found tmost convenien twhen olls or both terms are monomials. 1, tduce -'b 0 to its lowest terms. 4ab= 2abX'2b 2ab __- -tD Ans. 6bx — 3x X 2b 3Xx' Reduce the following fractions to their lowest terrms,. 4la,'', 2zh u 1 2,Ii2f'r1 3T/s ha4' T a &2e z. 2 S a)X3 4x' 12ab2 d-4~abc' 3be/ 3224, — Ans. 8 xa ax 4. t;4gz 3- S. r Ans. 8. Allso b-{-c 82ab4 ee 4?y4 1 o0Lxc (e O9rb3pzo 3~ Giab _o.-5abU' + 5 a..a,.n..... —..12,4 _' y 2' b 0cn,,d, c't Rvw, -127. 1eowv is the value of a fraction aec.teed by divisdin both -terms by the smne quantity? Whyhy 128., eHow do you reduce n firactioin to its lowest ternms? 8.RAY'S ALGEBRA, PART FIRBST 10. e D.. Ans. 3- - 24z y-40sUf. Sac 1 1 1~ 2 x >.......... —. — 1o 1t. A. 2An dc it 1 i2v/ — >8syf 2~ 23 7/ N or at. -1n the preceding examples, the greatest conmmon divisor in ea-ch is a monomial; in those which follow, it is a polynormia-l; but,; by separati in the quantities into factors, or by the rule (Art. I06,) the grtatestX commnoon divisor is readily found. 3a2 -— 3aB 13. 5bSP'This is eqnal to 3a(a c — b) _3cta-i-b) (a-6b) 3a(a —b) s5(a-t-iq) b~(a-1J-b) -,.,, Ax.'' ns. 2 —-- y' 14"] niy 92 1. ~ob _ ni Sx 9.:t_2 ti4a"~ Vab) n X.a 0 9x" o a1 6is.f 2 5e Ans.7 <3a P —x —6'L: o c2'r Ari 1i!E — xerc ises ix Dii ison ( aSee a5t 76 x -t. iix 3atc 2Ans7. Dvd 7i by 25 c.',, s 1implde for.mA. XB. 8 + Z rr X+, L. Divide ~' n by l i li...... e.. Arts. 5b 3. Divide 25abc by 5ac........ A 55a 4. Divide am —ni by Am.a, In a similar'manner. when one polynomi1al can not be (-,exactly divided Iby another, the division maay be indica, ted, and the result reduced. to its most si.mple form. 5.. Divide 25a:x by 5axP —5aSxy....A. AIRS. 6. Divide 3',~O"~ by 15m2+'-l'5n~......... A n s, D, Ilivide xa~-bxya by azy-azy%.. Ans, ~, I a~~~~~~~~( REDUCTO ION OF FRACTIONS. 89 8, Divide 4a-K-4:b by 2a —-2b2...An s a —b. 9. Divide n2 —2n, by'nt —4n-n+4.... Ans.. n —2 10. Divide +-2x —3 by x2-5x-. -6r....... zi —o 1 CASE 11. TO REDUCE A FRACTION TO AN ENTIRE OR 1MIXEQD QUANTITY ART, l g-,Sinc Oe the numerator of the fraction may be regarded as a dividend, and the denominator as a divisor, this is merely ta case of division. Hence, the RULE. Divide the nmzlerator by the deomisnator, for the entiA e part, and, i~K there be a remainder, place it over the denomincator for lthe, fra tional part. NS oT.-The fractionat part should be reduced to its lowest terms. 1. Reduce ax-b to a xed quantity. 3axbz —-b2 6 Izz3a+ — ns. Reduce ihe following fractions to entire or mixed quantities. 2 +.. a........ Ans. b a'tivb a 2a A —-n. 2x. a. —-c - 5..ax......I. Ans. 2ax-EZ. a2'_+3 3 —x 6. oo.-o 9 *Qo.,.. Ans. -z+ a. a+Xo.a-, xa -{e2a —e.Dga-x c.0 —2axc ace a.. a 9- ------- Ans.'".a -.... 10. 4 _z.. Ans. 3 M —-3 S4 -a...... @_S _ 90O RAY'S ALGFBRA8, ~I~J~TPART FIRST TO REDUCE A i-f11X1ED QUANTITY' TO THIE FORM-i OF 1A FIRACTION. ART,,q..SI —I. I 21 how many thirds? In 1 unit there are 3 thirds; hence, in 2 units, there are twice as many, that is,'5; then, 6 thirds plus. ithird, are equal to 7 thirds; that is, 2 —1 are equal to. In the same mzanner, a-p is ecual toacm whicv h is equalo to - Hence, the FOR REDUCING A BMIXETD QUANTITY TO THE FORI:31 OF A FRACTION,.ZltujigQy the entldire )ya.r, bo y rt e b tyien.otin'iincto'r of ti.c /"racctoa'; theint' facdd ti/e nptumertor wit/., its proper yu, to the product, a.nd p lace the 1'eslbtt over t.ile dnt. oiaOi.lzz or, RIt.r EM A I o —Caitses 1Io alnd III., are the reverse of, and mutual-ly provo ceac.h o t. her. Before proceedoing further, it is important for the learner to consider TH-IE S'IGNS OFI FTRAUCTI@ONS. At ir.l. g.3. —It has been already statedt (Sec Art. 121,) cthat in every fraction the numerator is a dividend, the denominaItor a divisor, and the value of the fraction the quotient. The signs prefixed. to the terms of a fracti0on, affect only those terms; and the sigu placed befbre a fraction, affects its whole value. Thus, in the fraction -— + d' the sign of a'., the first term of -the nmerat.or, is plus; of the second, 6b, minus; wh lile the sign of each ternm of the denomninator, is'lus. But the sign of the fraction, tahen us a whole, is uminus. By the rule for the s$igns'ln Division, Art. 75, we have +ab__,) b;q or, changingt the signs of both terms, — tZj b. But, if we change the sign of thle numerator, we have --.b. And, if we change'the sign. of the denoninator, we hav-te _- -.. -5 -a Hence, the signts of botch serm.s o If a fr.actionit mnay be changed,'without alt'ering its vatiue, or ch.angig its sigZn. but, if the signl ofvJ either ter,, qf' a,;fot'ction be chy:tanged, anzd not tha.t qf' tZhe other, tile si'gn, of the fa/vct.- gion will be ch/tangeid. REDUCTION OP Fli CT-iONS. 91 FroDn this, it Ralso follows, that ihe stijs of eit/ier te)rms of s /j;aclion az.y be c/t, 7;anfe,; i' ihou.;t tt-i,.7zg its'.vastc, if Iic sI;i,,0' of I/C f:is'tion' be c/ian getd at the scame tne. Thus.... _ EX AMPLE Sg 1. Reduce 3aa ~- ~ to a fractional formi 3a ax —, 3cx-l-aox —a 4anx —a Sa - and —- _ 4 i- Ans, 2 geduce 4a -- to a fractional form. 12atc 12,cc at-b 2ac - (a-b) 12ac -a b 4a- 3c, rd S 3s 3c 3c o Ans. In' ARtIE.-I-n sol.ving this example, the learner shouldt observe, tha't a —b is to be subtracted from t4a. We reduce 4s to a yqualnstity- whose lenominator is 3e; then nmake the subtracetion, and write 1lhe result over the comtmon dellnominator, 3e, Reduce the following quantities to inlmproper ifractions. 3. 4 cib5' 5 is b a-b 1c:,~- bba- 1 4. 5 c-............. S - -— d As 3! c — c 5I.o, -- o o, o o,, o o..in. i o. - 7. 8'+S.f' -?ns. c5 Az 5 —'S. Lxv as c 110 hzow do y&ou reduce a fretion ta e ntires n edll Ra n-v i}w.-130, -low do you reduce a fraction to aln entlire or miXed quanlut.y t? 131 howr do you rieduce a mixed quanti.ty to the fi3orm of a fractios? 132o What do the sigj;ns prefixed to fthe terms of a cfraeti on affect? 7Wh1iat does the sign placed before the whole fraction, afc-e t? Whatt effect does it have upon the value of a T'raction, or upon its sign, ito ethsnge the signs of both terms? To change the sign or signs of one tlrm, and not of the other? To chmange the sign of the frection, and one of its terms? s9 RLA.Y'S ALGEBRA, PALRE FIRST. 10. 4y -... As. 4,?/-I0s-5z 7W,+z 2 _ 3 —5e Ans. c-45 8 ~ e9a8 b2{gO-a3 2asx2+ab 12, 30oFx.......... AOns. 2a%0 x x 1F2 2' 2a2 e13. a-x ~ +2.............. Ans.-22. aa —x a,~-X 4. lcy2___ 1ns. C X x 2ts!'- 22 m. c6 —x~+tX Ans. 2 c-ax....ax a —t-x a-%x' TO0 RETI)UCE'RAC'TiTONS OF' o)]o'imfLcnoi't' )E?'i:fNYAT'O.0S't)0 E-QUIVALt't FR ACTIOlNS, HAtVING, COMtO.ON DTNOMIrNATORo, ART. 133 —1 Reduce - and - to a commaon denominator. b d If we nmultiply both terms of the first fraction, % by d, the deac aXd ad nominator of the second, we sha.ll have _-nbXd d, if we miultiply both terms of the second fraction, b, by b, the denominator of the first, -we shall have d dcIXb 64d In. this solution ie observe; first, the values of the fractions are not changed, since, in each firaction, both terms are multiplied by the same quantity; and, secoind, the denominiators in each must be the same, sRince they consist of the product of the sa-me quantities. 2 Rtedce -, -, a -nd, to a common denominator. Here, we are at liberty to multiply both terms of each ffraction, by the same quantity, since this (See Art. 126) will not change its value. Now, if we multiply both terms of each fraction, by the deno-minatons of the other two fraetions, the new denominators in each will. be the same, since, in each case, they will onsist of the product of the same factors, that is, of all the denoiminatorq. REDUCTION 0'T iA FRACT IONS. 93 This...... a. XnT. Xr annr JmX't>KI4' at.. bXmXr/ bmr n>,<xmXr,m,'-n.' 2X 2X - -n Cnr It is evident, that the vnalue of each fraction is not chtag d, and that they have t he sae adenominat.orso. H-ence, the 0o REDUCING FRACTIOI's TO A COM PON DrXNOINATOlRo.Jiculipty botli teerms qf eaci fractio ti by the product qf ac/: 1,e denomrtinators, except its own. t EMA r. — Since eachl denominator of the new fractions, will consist; of the product of all the denouminators of the given fractions, it is unne% essary to perform the sa.me mltiplication more th an:once. Reduce the following fractions, in each example, to others, iaving a common denominator. ta 1 2ad 2cbc bd $, /, ~, and e,.o. o. o, Ans.-.,6 2, and Phd' /-c cx Ny+ay 4...... A s. -, n 2 3a ---- 8b 9ab 12 —2y 5.,, anid b.. s.An 2b — -2 and 3' 4 b...... An 1' 1 1 2b I 2. 3x.sI Oxz 9xy 5ay 0. 5,, d a..p....Ai-is. ips 1 3y' 5z' and a.O O.' -'. O 95yz' a O nt5 Od 15 I ax an' y az x2z xy/j2 7. and' ^An s.',,atnd N. xyz pyz x./z IP e+22 3z~x 3z"-f —2x~,~: *n'-9 + o &8. g'yand. Ans. w'. - nd 2 -L11'c- y 24Oxc'+ 2 &:Y 3b'A - 3an ed 10. a, -, td, and. 5.6An. ancd. i, Lr - w.,-1.33. }low do yon reduce fractions of different denominators to eqZivalent fractions having the same, denominator? Why is the v-Mun of each fraction not chl:nged by this process? Why does thin process give to each fraetion t.he sam denoi minator? 94 PAY'S ALGEBRA, PART FIRST. ai3 ia ot a n. i-1a'nd 3ta+3 3.' 3aK 2+3ava % 2t2, 3a~)2~-i-3aSzo AnrT..8340 —It frequently happens, that the denominators of the fractions to be reduced, contain one or more commnaon factors. In such cases, the preceding rule does not give the least comnlon denominator. From the preceding Article we see, that the Comin meon denominator is a rultiple of all the denominators; and, -thiat, each numerator is multiplied by a quantity which is equal'to the quotient obtained, by dividing this multiple by its deinonrinator. Thus, in the second example, *,ra, nr, and laln, thile quantities by which each numerator is respectively irultiplied, nay be regarded as the quotients obtained, by dividing 9271r successively, by hr. i,2, and r. Now, if we obtain the least conon mul-tiple of -the det romninators, by the rule, Case [It., and then divide it by each dernoiniator respec'tivelv, and multiply the quotients by the nuinerators respecti'vely wie shall obt ain a now class of friactions, equivalent to -the former, and havinrg for a common denominatorto -the least common nrmultiple of the given denominators. It is easily tseen, that botl 0terms of each fraiction are multiplied by the sam.eo quantity, aind heence, tha-t the resulting fractions are equivalent to'the given ones. I. Reduce 7-b-,'b- and d to equivalent fractions, haviing the least conmmon denominator. Th1e l east commnon multiple of the denuominators. is easily found to be bed; dividing this by b, the denomninator of the first fraction, the quiotient is c;d their multliplying both ternms of -- by ed, tfhI 3oct result is.d Then bed —h cd, and h.. f h' v~ Xb 7V Also, bcd- cd=b, a nl ud d The process of mult liplyinr the dentomiators by the quotien'ts may be omitted, as the product in each case will be equval to the least comanon multriple. tIence,'tLe FOR, xREDiUCINc FRACTIONS OF DIFFERENT r EN0'INTiNAO tl0S, TO EQ'UI'V`'NT FRAC'tIcrNoS, HAVrING THE LEAST COOtM'ioOz 0ENOINPt AToOR,. $st. 1i 7td i the ea st contmmo~n mltiple of dalt tife dno'n'tinator".s; this rogt1 be the cooZmonsnor Zn ol i'nzattor. REDUCTION OF FRACTIONS. 95 2d. Divide the least coC on rZ nultipe, by they jirjst of the given denominators, and muttdipty the quotient by the fir/st of the give nuaz erators; the prodct twill be thejfirst of the required nunmerator.a 3d. Proceed, in a similar manner, to t find eac/c of the otheir nuaioerators. No T' -Each fraction should be in its lowest terms, beforo comsmencing the operation~ Reduce the following fractions in each example, to equivalent fractions, having -the least commonn denominator. 2a 3x 55y 4ad 18bx 5ey 2. 4- c' annd.... Ans, ed' ibcad 2 >l 2 oa T 1 S, ~, ad - - @ d3bcg7 cd 6bbd ~~bcd5 4bcd' 6bcd' in ci rc bcdac acccZ Wb.r 3, ac' and... abc2d ab ct and.4 abod x ~ ~_~ (A+?/)y (Xiry)i xLy: 4 -,and - ~ i ~ As c, and X-t xt - 2 22x2+yr X x,4y),.V)_2 2- 2'- y Other exercises will be found in the addition of ifreactions. N eorr. -Tho two followeing Articles depend on the sname principle as the two preeelding, anti are, therefore, introduced here. They -will bot be tbeud.ndl of frequent use, pa'ticualrnly in completing the square, in the solution of equl-a. tions of the second degree. AnT. 13;35.-To rediuce anf entire quantity to the form of a fractiJ.on having a given denominator. 1. Let it be required to reduce a to a fraction having b for its denominator. Since any quantity mayz be reduced to the form of a fraction, by writing I beneath it, a is th)e same as; if we sultiply both terms by b, whichi will not change its value (See Art. 1 26), we hvte i — b for the required fraction. Hence, the RULE, FOR REDUCINcG A ENTIRE Q QUANTITY TO TE FORMi O A FRTACTION THAVING A GIA EN DENiOMINATOR. J.f.glt 7y i./e eltire quanti.ty by /Ae givent de,.onc[ine.tzor, aczcd write lhle product over it, $EX AMUPLES. 4x 20 Reducne x to a fraction, whose denominator is 4. A.ns, -4 4* 3, Re1duce it to a fraction, whose denominator is Oa.t Ansoa.' Rv BTrw. — t 343 t!hOW dlO yo rednee fractions of different denomhinatbors to equivalentl fr.actionis,, hb.aving the least comnon delnominal;or? 96 I.I.Y'S ALG-EBRA P)ART Fl IRST. 4. Reduce 3c-V 5 to a fraction whose denominator is 6c2. Ans. 48;c+0c5. i educe c- b to a fraction, whose denominmator is a —2abd-7b. a3 —3gab+3abc2_b2 (a-b)3 c2 —2ab-[b2 (a b) * ART. 3g@. —To convert a fraction to an equivalent one, havting a denominator equal to stome multiple of the denominator of the given fraction. 1. Reduce to a fraction, whose denominator is bc. It is evident, that the terms must be multiplied by the same quantity, so as n6t to change the value of the fraction. It is then required to find, what the denominator, b, must be multiplied by, that the product shIall becoime be; but, it is evident, this multiple will be found, by dividing be by b, which gives the quotient, e. Then, multiplying both terms of the fraction 6 by c, the result is ae a which is equal to the given fraction b' and has, for its denominator be. Hence, the RULE, FOR CONVERTING A FRACTION TO AN EQUIVrALENT ONE;, ]TAVING A GIVEN DENOMIINATOR, DDivide the yiven denomincator by the denominator of the giuven fraction, and mulZti2y both terms by ithe quotient. R E I A R x.-This rule is perfectly general, but it is never applied, except where the required denominator is a multiple of the given one. In other cases, it would produce a complex friaction. Thus, if it is required to colnvert ~- into an equivMeent fraction, whose denominator is 5, the numerat-or of the new fraction would be 2%. 2. Convert 3 to an equivalent fraction, hbaving the de'nomtina4 1g tor 1 6. Ans. 3. Convert to an equivalent fraction, having the denomina 3a tor. 9. Ans. 4. Convert b to an equivalent fraction, having the denonmina r eh is t i tsAsowest ters bre 1E Vni. tw.134. If each fractiosn is not in its l 1 es3 terms, beoreu com-t mencing the f o eration, what is to e doe? 135, fl dono you reduce an entire qua.ntity to the forum of a fraction having' a given denominator? ADDITIO N AND SUBTRACTION' OF FRACTJIONS. 97 5. Convert~: to an equiv alent fraction, 1having the denomi-,. z -l. 22 Ans. i...onlA;ert tO an equivalent fraction, having the denzomiats'or a(b- )i Ans a(bc) ADDIiTION A-ND SCUT- TSCTIONT. OF FRACTIONS.', =t..B — I. Let it be required to find the sun of a rd - J bre, both parts being of the same kind, th.at is, fifts, we may add. them together, and the sum is c fifths, (i). a fb 2. Let it be requi.red. to find the sum of _and liere, the parts being of'the same kind, that is, /iths, we may, as in the first case, add the ttnlerators, ancl write the result over the colmnuon denomina-tor. a rb a +b Thus....... — t-..... i 3. Again, let it be required to find the sum of a nd c IeTre, thle parts not being of the same kind, that is, the denominator's being different, we can not add the numerators together, and call thiem by the sam nca e. e ma y, however, reduce tihen to a common denonminator, and then add themn together. 42 c cA 5%5 Cm anc e ra i't 21i itt MU11 iZT6 Mt W20M Ttence, th.e R U L E2, FOR THE ADDITION OP FRACTIONS. Reduce the dfracttion.s, if necessaXry, to a commona denominator; add the nomicraotrys toget7her, anmd pl.ace their su'm onver' the common' IA r. 1 4,o —It is obvious, that the same principles would apply, if it were required to find the difference between two fractions; that is, if their denomiinators were the ame,'the numerators nmight be subt1racte~d; but, if'their denominators were different, it would be necessary to reduce -them to the same denominator, before perforaning the subtractiont Hence, the ORt THIE SUBTR.ACTION OF FR ACTIONS Reduce the fractiotTzs, if ny necessary, to a commoa deneomtnator; cahen subteract.he nwumerator of' thefraction to be subtracted jfront the nvtumerator of tthe other, an.d placte thi'- remaintder over tZze comin,1o.r denomit i aety, 1 or. 98 RAY'S ALGEBRA, PART FIRST. ET-,,A~olPLES IN.&DDITION OA FRtACTI ONS a, a a 4. Add, and, t, ogether.. An s a.. 3 6x x x dxtogethe....... 5, Add,,~ and togeth Anns 7 6. Add I 1nd together........ a' An d s ab, x y G 6x-+4y7+3z 7. Add 2 and - together. ~. e o. Ans.o 11 3:' 4 12 3x 4 rAn 143r 23 8 Q Add 4 -- and -- together. o S -23x-. -4P "' 6 60 x- - -',-,?/? /,1 n 1- 2a 10. Add nd togethcer. e...an s. a-f-b aa-b —i'. Add nd-Y together....... s.f X12. Add 5:c3-ad a d 5 —x and b- together.' eAns/ - 13. Add and toge'the. Ans. 0. ab t e ace 14o. Add and together... Ans. 1-x' 1ic- -n VWhen entire quantities and fractions are to be added togerther, they may'be connected by the sign of addition, or the entire quantities a-d the fractions may be reduced to a commnon denominator, and the -addition then performed. Sn 2, 37S 15. Add 2x, 3x+-, and x-2 1 together. A.ns. 6x+y-7z') ~ 9 45' -2. r — S 16. Add a5xr — anC d 4r, t-. together. 17. Add 83 d Sa-2 anl.d 7p~L together7. Ans. 195 — ax a% d b a —o 4a-a, 8 Add ab a-o and 2 together. A..sns a —— b' a-ib' a ib trvavin, W-I3O. Roly do yen con Vert a fraction to an eqaival(ent one, tP.eiog a given dellomflinlator? Explain the operation by ac examplef. 137, Whson fractions havte the same denominator, how dco yon add themli t3ogether When fractions have different deanominators, how do yot add thein. togetherl/ UBTIRAGCTIION OF FRACTIONS, 99 ^ii> APKMI.< tXU 1/ 1/ @U 1/NBT C tUIO~' )F U A CTSX No. 1. From w take o o Ans.... 2 sr 3 2 40. From - take A...ns-. 9a 1 Gax 23' 56 l 3 5 M3x1 —I — Oa 5 rom ae............ 1~. ti an 3 a 4x We_ 1 64xy 7. From. - take ti- s. 4x 3ai.. a..0...... aen-2-aY 8. From a?- / tt;ao e a...... ii..... x —-y x-by X... y 8. F of tak........ Anns 4r 2a-Fb 3ab 126-a 9, F rm2a- b'take -—... An. 2b — 5'c......7 e e50 35c -b bxi-cx-B2 19. Fronm 5xxb take 2x -— ob. b 3x b6-tiS a ~ b ee - 22b i i. From -- take a,. r.....n. a.. a i+11 1 a2bL a- - 12. From ab take Ans. 13. From xa -ake2 Am- A 2ys. + XY'x2f 14. Them 3 takc,,,,,Ans-. 1 —._ 2(a- a-b-) a, a-b 15, lFron a.n. —-.. An a —-b' 1 2 ~-2x1 13 17. From x+-y] tiake A-.i.An.n..... 18, Froin. 2a —--— + A. a,+p-+,Z; 19. From a+x+,-; take a —x. Anns, 2x — --.tirvrxW. —438, If two fractions have the same denominator, how do you find their difference? When two fractions have diffe rent denominators, howa do you'nd their difference? I00 t'RA S ALGE1BRA PART FIRST. CASE Vi; TO0 N'IJLTIPLY~ OlNEt FtNACTIONAL QUANTITY BY ATNOT}IEiR ART. at 9 — To multiply a fraction by an entire quantity, or an entire quanyt1y by a fraction. It is evident froni Prop. I., Art,. 122, tlat in mnultiplyig the ium-erator of a fraction by an entire quantity, the fraction is inc reased. as many t hes as there are units in the nultiplier. Thus, f,,takekn tice,'is b; and taken NI times, is A/1ntin, when two quantities are Zto be nultiplied together, either niay bet mnade the mlultiplier (Art. 67); to multiply 4 iby ib i'the eame as to muRliply 5 by 4. Or, to rutipl y by' tie same ut t as t yo b laitpy b y m. Hence,, tiie'FOI, nTE:M MrULTIPLtCTIOC X Or tl 0A A iC Nrrt s3 N lNTIRE QUANTITSY ORS OF AN ENTIRE QUAINTITY Bi A FRACT'ON.,itf//i/gJi.y I/te,'rroneia/or by t~he ent/ire guarcnli,' cand wIt'ril the pro —,edtC over the deoZ.omiiTnaeor. Since (See Art. 125,) dividing the denominator of a fracttion cfiereases th.e value of the fraction, as -many tiimes as there are units iln the divisor, it is evidenit, tiat nay fraction'will. be ilaltit plied by an entire quantity, if the denominator of the fraction be divided by the entire quantity. Thus, in multiplying: by 2, we m.r a divide the denomrinator bIy 2, and the result will be, irhic is the, same as to multiply by 2, and redtuce the resultingi fiateion tLo its lowest terms. Ilence, in mn/trUp/ying a fl-ac/ion anal cr, e/ire ta.l'i/,gy togeh0er, we sioTt/dau cnts cdividc t. e dt. c;ite niiag22or of tie't"ae tiorlt by e1 entire ua gact6t, n Wt rei', it cCr;rZ be donze Znv,ith/iout a'"ecainzode'r N.ir X A Im K, —-The expression, "1 What is two thirds of 6? has the s'a m mean1 ain'g as "W Iht is the product of 6 multiplied by 2?" The reaion of thae rutle ier thl multitliesation of an eintire qua ntirty y a frctioni ay beo shoewn othrwise, th ius: oe third of a's;a two thirds is twices a' much as one t.ird, that is, two thirds of a is 2 Also, - of is -t and tfhe - part of is _ B r v i ra... 1399. Howe do you multiply a fraction by an entire qcanttuatity, or tn entirt,e qrantity by a fraction?' When tihe denominator of the fraction is a multiple of the exntire quantity, -what is the shortest inmethod of findhin thlesir prod.cet? iMfLTIPlIttCATlOY OF FRA1CTION S. 101 be AXAM P-. 1, Multiply Pj o e e d b s n a 6-cA axui —-bxy. Mtuliply - by oyO s. s. 43 Mdultiply — 7- by y...ns.s.'' 4. ul ipl0y 1- eby by..A - a a 9aa bl-c _b2 a. i ultipl'y —' by a-b..... s..AcI.. MJ/ultiply 2 a+ c alq B2 A ons. 2_14XV 51 7+a93B tO Muitlyll IO-2 1 br Cby- oy a r G Ans. -i-_ flje 0. B-ult-iply X~__.y l. Anso2 ab ab l0. B[Mult iply () y'x.... Ans. t(ae- b)( b) - b) i5 a — 4d by P ta -b) 1/ ISI. m'iedti ply 4 b by a: z7b b).-( L. Ans. l Y bly eli&) a 1y 44,- tuluply by a........... Ai-ius. --— a., or, I4 82[ultipy aby ca- (is...b 10 A niupty~r~ mtbh~rni by 0x-F-5y. ~ ~~ Aens -c —a, or Seince, we, saee, that i2a micion i n mu tplicc.qbuy a quantit y eqtal to its d enomina-efr, the roct ull be eaa io ts -lc e n.u tera or 15. g'rlt;ipl y — by c+d........... AnsAi-is...-b 16. fultiply Iby 2-Or52y........ A ns. A;'z.A —n=0.Aiv., 14a —To multiplyT a fraction by a fractionms 1. Let it be required to find'the product of multiplied by o Sinc 4} is t'he same ais 2 multiplied by -a, it is required to.m-uliply'l by 2, and t.-,ke. of the product. Nlow, aulreplied by 2, is equa -to m —,and ~ of is equaM to -1% (since, to'take is to divide by 3, and any 5:action is divided, by multiplying is denorinator, by.Art. 124.) }IIcne, the product of and is -.. (02 RAY'S ALGEBRA PAR FIRST, In the same maniner, if it were required to multiply[ by — - a sinc —-:nX-,.r e would multiply by % and tae of the!o since,l imo. i_ I c iitcc product. Thus, — Xm= -—, and - of I-Hence, te Cf C A2 C Ile3 RULE, FOR I-tE ULTIPLICATION OF A FRACTION rY A FRACTION..e..iZtly the 3 nucrator together, fors.b new c numerator; cod the dertomi.ctaooras together, for a new denwominatior. t, Macs sA ltCS. - f either of the faetors is a mixed quantiy, it is best to reduce it to an improper fraction, before comamencirng the operation. 2d. The expression, "What is one third of one fourth," hbas the same meaning as "What is the product of I multiplied by -I". Also the expres sion, "What is two thirds of thine -fourths/, has the same meanin as "What is the product of a multipliedi y h. 3d. When the numerators and denominators have- comtmion factors, it is best to indicate the multip.ication, and then cancel the factors common to both termns, after which, the renma.ining termnls may be multiplied toge ther. 1 4ac 5e 2X'$ac 2ac Thus, j-X XX S3t 51d $ X3X3Xjtb- d 9 ~bd_ 5oa a b 5a(a+b) _5 to,~ -1 —X'2 a a(ctc+b)( ja- b)V92(a —b) 2a.X An S 1 Multi;ply 4by.. sO t32 4a 3x 12 2. Nfultiply by.......... ins. 2c 4ca 8a5 3. Mfu.ltiply +- by -K.... as. ~ 3s" 5y x 4. fuiltiply by 9 Q Ans. -_ 3(a+S-) 4x ill. -ultiply... 4 ao s x r, v j ~ w.-,140. I{ow do yoot mult;iply one fraction by another? Explain the reason of the rule, by analyzing,an exanmple. When one of tC'a factors is a mixed quantity: what ought to be done? WThat is the smeaning of the sexpression, "Whitt is one thi'rd of one fo-?rth? " ow may the wiork be short. end, when tihe numerator and denominator have common:fia-ctors? MULTIPLICATION' OF FRACTIONS. I 03 0b x+.y b 8s. M]~utiply D Dl by Fr...Ans. 1. 3 9. Multiply a —2__b 6y Ans.a b b - a —x ay a1 10. Mhultiply p by a2....... Ans, a2 —-X2 aa S. Multiply and ---—, ogether.... Ans.. +flp y x y= A —Y2 12 MultiplyXy' 2 2 Wand a together.. Ans. a(x+y2)o a.... b zo. 3tultiply, 6 2.b and agb toet.er~.. nso l, 14. Muf'ltiplyi by 2'... Ans., +-~ I "i. multipl t c X a ). Ans a 283. llt —-o d.xide, -ctiond by a entire qtuantit..y i bu:st been shuwn, nin lI A~ito 123 and4 Art. ll24L, that a f~rtactionz is diy ided by an ent ire ouantih; byV dividin' its numerator, or.multh. plyintg its denonain vt;or, ihus, 4 divided by 2, or 4 of is 4 3Pa o Pa an m t 7sa,. ] -b 5i;i]{id(d by 3, or - -- drxd y us, or -' of 26 uup~y 2I -----— b 2 )r'. b-)y mdultipl g he deoinaded by U2, is2 eual to Jo''t si3lt he:number of parts in the numerat3or is the y same, but oly h'tal has larhge a's bot 1J2 bein g the half f. o ance, t s detiv ded'; if ne otir q uatil, by l denomidina it by tiee et iore qutliy ld wri.-e ti-7e sdenoineat or Toers t'ie soessdt.1 No, x — If the num erato of rte fiaction ani tihe enlire qant.ty, ct>, tia coemon fa3ctoes, it is bes t o in iceat tre operatio, atnd cancet i: cmmon factotrs; the o result n thus will he in its lowestor terms, :.:0 34: )RAY'S ALGEBRA, PART FIRST, Thie preceding rule ma, y be derived in another manner, thus: To divide a'number by 2, is to take 1 of it, or to multiply it by'*; to' divide by 3, is to take - of it, or to multiply it by:. In the samne mananer, to divide a quantity by Xn, is to take - of it, or to multiply it by - Ilenee, to divide a fraction by an entire qtanliy, wie wriSe the diivsor fin the form of a fr'action (thus, nu=antd itinvert it, and the'n p2roceed as in lts mdiplication of firactions. EX AMiPLE. S. 1. Divide =77-21 by 3ab.............. Ans. 2. Divide -17 by 5dcO A 7bd' 176'4 26_dt i 4aC5'hs 3. Divide - by 7ac.......... As 35 o "a b (- 7d 4, Divide - b by 5......... Ans.. 5. Divide - by.c+d. A. - t s'+2x y —y,, —,Aus 7, Divide b'y.x.. i —-, a c+d a'~a+1~ 2 a) &S Divide -by a —b. ~ e Anso, -2' b+b W2-I-7a'-5a 2a-, —a9. Divide 6a-ra —- by ~.. a.-...Ans.:,-' —xtr-+3 2r 1. Divide -by 4x' —2x —35 Ans. 2,> —d2 abqtcd?,a b, ( 11, Divide 3 byb............ Ans. 9' O 12, Divide a6 — by b.......... ab3d. bc at-b R vs si, w.-i 41, ow (c o you. divide a fraction by n entire quentity? Explain the reason of the rule, by analyzing an example. tRow may the:r' r'b-a:revia. t ed, wheien thfe numerators of the fraction andt the entire K.....: -s " i:lna,-):commonA factors? DV.'S:ION 01' FiiACTl-IONS. 1905 q 4. 2dz+3y 2z+3?y'!4. Divide d- by'y..... Axns. 3a+Oe 3a{-5c 1B. Divide by 22 —Sy. Ans, _ - et L "a bo s'3 <2x-n3y,us —9y 2 c~ 18. Divide s c — - o c+-, a.. _ e Axis.a fi t yt -Lt;9 1 2at2 K b 19' no 21 t' D at ab S 7.v D e, d c - b-y 3-b............... Axns. b+ 20, Divide - by 32,,. Ans., 21.dDivie id bb.y:,..'..... Ans. -,-, 1g7J auli-i~-t.7a ac_ 4. is equal t —o - 2.th;ir. i e n 23. Dividofe;'" a is equal t.?o dn asx i ( Zm );? t]J. Dfiyidsct3 by xa- l:Y. Ans.' " L hido i —':l b:" y a (i-ab-',-b-,... I. Htow oten is contai ned in 4, or -what ics nti quotieint of 4 di'vidcd by by ~is cqmi to 4, anid 2 thirds ()o is co-taied i 12 thirds 2. Ho o ften is' cont aineO in a,? a is eql ton, and - is containedn in na- inas ma R tes as its contain'd inl za:, tha, t is -3 times. O r, 1 is contained in a, na:es; hencer e, twe s is co ntained i-a 1 t3. (ow often is as 8 s contained d n 9, that is, s and ~ —-T~ and 8 twl~h (?s} is contatinedlj;v < 4.: 2(.2 RAYTS ALGEBRA, PART FIRST. 4:e, t'w ofte.n. is contained in -? * 5 C Rteducing these fractiionsto a cominon denominator, -.. -.... andl..;G n oZC n2t a e a snow, is contained in- as often as nc is contained in za, that is,'- times. This is the sause result as that produced by a 92?a a', n-na ult'ipyi n by s inverted, that is /X -- na.An exsaminiation of each of these examples, willi show that the process consists in riedncein/ t.he quantities to a. comnitoyt dCenominator, antd thea divtiditng the n1.um2erator of the dividend, by the niiitoeratour of the divisors, But, as lthe common denominator of the aeaction is not used in perfornsing the division, the result wvill be lthe samue as if we invert the divisor, and proceed as in nusltiplica — t'ion. Hence, the PU LE, rOn. DLVIDING A'h INTEGRiAL OR FltXCTIONAL QUANTI'ri~TY BY A rFICTfION. A'?edtuce both dividteid a-nd divisor to the jbom oJf -a fi action; these invertg the terms qj' the divisor, and n.ultl)lyd the tnmertatosrs to/gelter fog a ew evumneraZtor, and the denominaZlors loether iJ'br a new denoitntator. NoET — After invertin the divisor, the wiork may be bb3reviated, by canceling all the factors colmmon to botJi terms of;he result. IE- X A L UE S. 1. Divide 4 by.. Ans -12 4a 2, Divide 4 by....Ans.~ a 3. Divide a'by -.. 0 ns. 4a. a o 2 ( t. Divide by cOO As Q.n K1.t sv'i. --- 142. iow do yo u divide a na integrlM or fraetion a quantity iy' a. fiaction? Explaini the reason of this rule, by analyzing an exsample.'W teV.V: and.,( lieo, ca.nn thIie work be abbreviated? DIrVIStON OF FRACTIONS'. 1 07 O s... 9a. 70 ]Dvide a - by zo 4 4 0 6 G 0 9.-o a, q ~ u {-U2 C12 ab 89 tv140 -a- \ a Bo v2 &Zv Ze Ans. a D. Divide -- by o A2 -2 a. cnd by...ns 9 e 3 b,2 i* Pd6X 4 9a 10. Divde a6x by A A s,a 5,, Arts 1, I1o Divide,-y- b y *':....... Anes.:2: 12. Divide by Ai 1... 5 a3- b 1 by....r...... oArs. a.a 6z+4 3z+' 2 13. Divide b y y...... Ansrts- 9 ~,-' b a.-bb) i4. Div~ide — by o AnS.'a:,7 —4 z.s z2 e 2 158. Div ide -asb a+1..... A... da + a12 aca-j-9 (32 19. Divide S b b..... a i_.g._.i-r 79. Divide 2's by...t As a, +o 4a+12 3a+9 21. Divide by 20 livide -- b by,..o A.. (e.A,a-b a 221. Divide --- by -,2 ~. (2 ) (.o. Ans. I2 —) 2( —2 — ded Dthedividorare either factios mixed q.... ities. Ths. Di., 3i he (- ) by Ans. 34 (a. -2axx-:) 3 ~Z. Also,, is the samne as to divide a b- by m.-: —-' ln-t- r.-? — 108 tR.AY'S ALGEBRA, PART FIRST. 7 7 7 22 ba ~ ( m acb'-t a c-,b.; aer —br C / I C r' - C 12121v2+n cMrZ+!Cn, In the same manner, let the following examples be solved. a i tad 1. Reduce -to a simple fraction. A~s 2G 3~ Redu.ce t - to a sinple fraction.... Arnse —s5. Reduce - to a simple fraction,. Ans. --— o Re.t~edute t;o at ilzpie ~ractt-'io..t K. Re-edce -- - to a simnple fraction,, A.... i. Reduace to a AimleX fix tion.. s. I A conijplex fraction liay also be reduccd to a sihmple one, by multip'lying both terms by the least comionl multiple of t'.he denominators of the fractional parts of eachl termn Thus, we may reduce to a si.mpe f. tetion, by mult lyinym both toerins by 6, t.he 1tcast commoluil mlult.iple of 2 and 3; tile result is'2s In some cases this is a shoter' method, ti an!by divisioo Efiti er methrod may bc usedd APrT. R 4 R4 —tesolution of &ftrictions intfo series, An inficnbite series consists' of an unlimit. ed unlmber of termns, which observe -the same law. The lawt of a series is a relation existinc betw"eena its ter'ms, so that when soim e o-f. e th.emare known, the succeeding terms i may be easily derived, I i vi,I W.. —-. t3 OI-ow do you reduce a complex fraction to a simple one, by divisiin? -iw't, by inultiplicati on? RESOLtTION OF FRACTIONS INTO SERPES. 109 Thus, in the infinite series, 1-a z-a c.2 —a. 3 ax —fxt- &e., any ter.m m.ay be found, by multiplying the preceding'terz by -— ax. Any proper algebraic fraction, whose denominator is a polynomial, can, by division, be resolved into ean infinite series; for, the numerator is a dividend, and the denominator a divisor, so related to each other, lthat t.he proesse of division never can teraminate, and the quotient will, therefore, be an infinite series. After a f-v of the terms of the quotient are fouind, the law of the series will, in general, be easily seen, so that the succeeding term-s may be found without continuing the division. EXAMPLE.. Con.vert the fraction - into an infinite series. 1 7 _ * o 11 —z I a x,2z~- -—,, — &Sco'1 The law of this series evidently ~X is, that: each term is equal -to the +4__za preceding termu, nultiplied by +x. 12.z Fron tins, is appears, tlat the f1action 1 —- _T is equa.l to the inlite seriesl, +_ _Lzn2i-xa +x,1 &C. In a simislar man ner, let each of the followin fr:actions be resolved into an infinite series, by division. 2. - g,...... —x2 —--..- a —, & de, to infinity. a _rz' ii 8. [-~ + —bt-b-7-+, &e. 4O s. a al 2 a c I~+: 2x-z-2 A~ —+2x1 —'v, &c. s —41 -di o -1-2 l 1. JP R a isw.-t4. Wint is an infinite series? What is the lai of series? Give an example.'Why can any proper algebraic firaction, wihbse deaconr inator is a polynomnial, be resolved into an infiito series, by division? 110 RIY'~S ALGEBR A, PART FIRST. CH PTER IV, EQUATIONS OP THE FIRST DEGREE:a~ DEFlTiON? AIND EiiLEMlETA'AT PRWNCWLPES. AT. A, 4.. -SThe most useful part of Algebra, is that which relahtes to the sol'ution of problems, This is s perfornmed by means of equations~ An equation is an lgebraic expression, stating the equality between two quantities. Thus, x — 3=-4, is an equation, stating that if 3 be subtracted from x, athe remainder will be equal to 4. Atr. tr. i4 —Every question is composed of two parts, separated fromu each other by the sign of equality. The quantity on the left of the sign of equality, is called the jfes5t w.tenmber, or side of the equation~ Thie quaintity on the riglht, is called t;he seconld member, or sideo The memubeirs or quantities are each copmposed of one or more termso A.T., _l o —-There are generally two classeS of quant-itles in at ecuation, the kncown acnd the tZ.7ro/own. The known quantities are represented either'by numbers, or the first letters of the alphabet, as a, b, c, &e.; and the iunknown quantities by the Last letters of the al1phlabet, as x, y, z, &c.'ARTr.gi4xOSS-Equations are divided into degrees, called first, second, third, and so on. The degree of an equatron, depends on the highest power of the unknown quantity which it contains. An equation which contains no power of the ul.nknow. quantity higher than the first, is calied anz egeatio- oqf thefjirs de.egrc Thus, 2x+55 —9, and ax+-b=c, are eqnuations of the first degcree Equations of the first degree are usually called Simnple Equations~ An equation in which the highest power of the unknownn quan tity is of the second degree, that is, a square, is c alled ac, equation of the seco.ned degree, or a quadratic eqtuation.. wr. w. —t-l5. What is an eqiuation? G ive an example. 1 46. Of how many parts is every equation compIosed)? low are they separated? Wha it is the quan;ity on th left of the sign of equality called? On the righit? Of what is each mexber composed? 147to -Tow many classes of qua.i.).it.es aro there in an equationa? [-low a ire the knownu qtLur ities ropresented? leow are the unknown. qrtantities r ep esented 148 lt-Iow are e guations dividsed d On what does the degree of an equation depend? What is an eqoation1 of the first dgreoe-? G-ive an example. What are equations of the first dogroe usually alled a What is an equation of the secold decrec? Givse an exams pleI'What are eqiuations of the second degree usually called, SIiPPLE EQUATIONS. I Thus 4x- 27.=9, a-d. ax2-Lbx -c, are equtitons of 1he second degree. In a similar m anner, we have equations of the third degree, jburth degree, &e,; the degree of the equation being,alway s the same as the highest power of the unknown quantity which it contains. When any equation contains more thau one unknown quantity,'its degree is equal to the greatest sunm of the expo-nents of the unknown quantities in any of its terms. Thus, iy;+axz-by-ca, is an equation of -tFhe seconid degree. xi-d —-z a - rcz sa, is an equation of the third degree. ArT. 1t4-9 —An idenical t equation, is one in which the two meombers are identical; or, one in which one of the menimbers is the result of the operations indicated in the other. Thus, 2r — =2z — I 5z+Sx3 Sx, and (x —-2 )('-2)-=x —— 4, are identicalt equations. Equataions are als distinguished as numearicatl and literal. A numerical equation is one in rwhich all the known quantities are expressed by-' numbers. hus, 2 -i-2x=3x+7, is a numerieal equation. A literal equation is 1one in which the known quanttiies are represented by letters, or by letters and numbers. Thus, Cax-b=wc —c-d, and a-x2+ ——.bx 2x5, are literal equations. ART.o 1B — Every equation is to be regarded as the statem'ent, in algebraic language, of a particular question. Thus, x —3 —.n.. amay be regarded as tl e statement of the -foliow ing question: To find a number, fromn which, if S be sublracted, the renmabinder will be equal to 4. If we add 3 to each member, we shall have x —— g-3j=4-+3- or An equation is said to be vertified, when the value of the unkrown quantity being substituted for it, the twvo enem bers are rendered equal to each other. Thus, in the equation — 3=4, if 7, the rvalue of x, be substiItuted instead of it, we havee 7-3=4, or 4 4. To sol e an equation, is to find tIhe value of the -unlzcowa qetcalilfy; or, to find a number, which being subshtitut.ed for dthe unknown. (qu.antity, will render the two members identi.tailo R t vi. E w —148. When.an eq ctiteon contains more thban one inklnon -n quantity, to what is its degree equal? Give an example. 149.'VWhat is an identical. equation Give examples. What is a umleriecil quatiis? Give an example. What is a literal equationL Give at example. 150,.flow is every equation to be regamrded? Give an example. When is ialtn aequationa seid to t Tified? What do you understand, by solving an equation? I h2 RAY2S ALGEBRA~ iPARIT TFIRST. A,rT, I..i,-The'value of the unknown quantity in any equation, is called the?root of that equation. C.1AU P L I gE Q U r AI g 0 N SO C C N'TA I NAI Il S D U r 0 C1T N I t 0 W N QUAN~'TkY. ARA. 52,o —-- The operations that we employ,'to fincd'the value of the unknoen. qcuantlity in any equation, are founded on this evident principle: Jf wte peVformn exactly the sanite operaitiont on two equal quantities, tthe resdls will be equaL.c. This principle, or axioam, tray be otherwise stated, as follows:.. tJ; to two e qual cqut estizie, the,sane quantity be added, the stsutsm will be equal. 2. If, front tiwo eqgual u,,tantzities, e thi acrze quantity be subtracted, the remaiinders witv i be equitalo 3. ify woe equal quanztiies be',?t'd.splied by tte samne quattity', tite.prod,'ucts wilt be egclytlI 4. If two equalt quartilies be divided by tihe camne quanti'yi, the quotients w,-ilt be equatl 5. If two eqa!t. l qutant'ies be raised o thze saezme power, th7e resut:1s will be equal. 6. If the scame root of thvo equatl cf.antmties be extgracte, the r'es-ztls will be equal. B utmAPrc. —AT axiom iS n self-evident truth. The preediing axiomns are the foundtItion of a large portion of the reasoning' in.mnlathe.tiest.AnT. 5o2I; —There are two operations of frqequent use in the solution of equations. These are, first, to c/e.ar an equattaio' off J~'actions; and, second, to transit$oe tite t er'ms, in order to fiad /te vatue of,the vntknown quanitt. These e namend iin the order in. mwhitch they are generally used, in the solution of an equation;, we sthall, however, first coanider the subject of Suppose we he'v'e the equatieon -x -3S4..5. Since, by theo precedinog principle, the equality will not be affctedt, by adding the same quantity to both amnembers; or, by subtractin g the same quantity -:rom both mermbers; if'e add 3 to each nmember, we have 2xs-3~3 x+ 53. If we subtract x from each memnber, we hlave 2 x- -.-3g+g$=x —<'x+- 5+!S~ Itsvi.w-. -- t5'1. What is the root of an eqatio n? 1.52'Upon w1hat principle nar tihe operations founded, that are used in solving ant equation? Wihat ar the axiom s whiol this principle embraces? 153, What two operia ticns axe frequently used, in the solution of equations? SIMPLE EQUATIONSo 13 Bu-t, — 3 -3 cancel each. other; so, also, do x —-; om1itting'these, we have h2x —— 5 —3. Now, the result is the same as if we had removed the terms -3 and -— x, to the opposite nmembers of the equation, and, at the saie time, chn-tged their signs. Again, -take the equation ax+-bc — 7x. If we subtract b fron each side, and add dx to each side, we have ax-+d-d c —-b. But, this result is also the samne as if'wte had removed the telrmns +- and -dx t o the opposite memubers of the equation, an'd, at the same tunime, chaneed their signs, THence, A;.oy qtuan tity n)"iey be troansptsosed fir-l' oit. S.ide of aeit equation to tie other,2it at /tIe sam/e'z tite, its sign.' be can:,iged. Ts o CL_ i' A AN EtQUATIN O.F 0 RtA(lCTION%8o ART. 1 1L5 —- Let it be required to clear the following equat tion of fra-ction s~ Since the fIsI; term is, divided by 2, if we multiply it by 2J9, tlhe divisor will be rcov>red; biut if we multiply the firslt ter rm by 2, we munst muvltiplyv all the other terms by 2, in order to preserve'the eqluality of the niembers BMultiplying both sides biy 2, we have <-1 —0. Aigailn, snce the second term is divided by 3, if we multiply iit by 3, the divisor will. be removed' but, if we multiply the second term. by 3 w, Ie ust s ultiply all the terims by 3, in iorde to preserve the equ0ai. ty of the members. Miluttip!ying' both sides'by 3, we have 3x+-2x 30. insteatd of o uit. p nying i rst by 2, and: then by 3, it is plain that y.ve might b1have multipi'ed at once, by XS3, that is, by the product of'the denontinatorsi 2. Again, let it be required to clear the clltowing equtition of fractions. ab 5be Since the first term is divided by aSt, if we multiply it by ab, t2he divisor will be re movled; but, if awe multiply the first terim by ab, we must mualtiplV ll t-he othler termls by ab, in order to preserveo the equalityr of the me'lmbers. PR v t_: i o - 15. l low m n y. qleant'ity be trainsposed fi' um one.member of an -equattion to the ot ieri? sxplain the principle of transposition by a n 10 1 14 RAY'S ALG EBlRA PART FIRST. Again, since tihe second term is divided by be, if wre multi.ply it by be, the divisor will be removed; but, if we multiply the second termn) by be, we must mnultiply all the other terms by be, in order to preserve the equality of tihe members. Hence, if we rmultiply all the terms on both sides, by abXbc, the equation will be cleared of fraction.s. Instead, however, of multiplying every term by abXbc, i.t is evident, that if each term be multiplied by such a quantity as will contaiin the denominators without a remainder, -that 2ll tae denoninators will be removed. This quantity is, evidently, the least Wommon multiple of the denominators, which, in this case, is abc; then, mi.ulItiplying both sides of the equation by abe, we have c-x-atx-abCed I-fence,'the F Lo A CLEhrING AN EQUATrION OF FACTIONs, i tezd clhe least common, tle o ao l nz/'e of ae denominaors, and tmidtiply eac/ tecw' oqf the euat-ionz by i-. Clear the following equations of fiactions. 1....5............. Ans. Cns- 8x-t2xz=30. 4 6 2 — o An. 4x 93x=24. $6 34- us 10x-0-1v I Ans, 20X 15xd-12x:- 60. 2 3 8~~ b....1 4 Ans. 4 0 z a.2.Caz2a —0 3 z.- z.... 4 —_ -_ --....s. 14 -2.4z=1.24 —2 10- 3 t ——'-I-]-~....... 2 3Ans s. t x3 —301 —2x-1-1-.-. 2 6.. 4 Pa 8. +,3 x_. An. 40x —840 —S2= 3 z- n-.- 207. 3 5 10 2R --- 4. do you lear a equation of f'aetions? Eplan t9l p -ri tie.ns. — 21 amp e. th~e prisnciple, by aklu example.p~a SI1MPLE EQUATIONS. 115 xs+1 3 —c 13. x35,2- j —~ X-1S 3 --;,'.'- -( 2 A s. ax-bx-j1 —a —b+S9x —c-c 9-c- -3zaix-abxa3oap —3A,. gL$ - -.! I x: Ans. ax-bxeaxd~ bx:c, o C b -- -- 7c oe /o Ans. adf+bcf+ bdez-df/y x. SOLUT-ON OF EQUATONS 0O HE FIRST DEGI3E, CON.N IN.G ONLY @0. 2_ UNKNOWVN QUA1 0< T'TYW AMRTo A55 T.-he unknown quantity in an equ ation may be combined with the known quantities, either by Addition, Subtraction, 115ultiplication, or Division; or, by two or more of these different met hods. 1. Let it bo rcquireod to find the'valuo of i, in the equation X-+3=5, where -the unknown quamntity is connected by addition. By subtraCt'ing 3 from each side, we have x -5 —3 — 2 2. Let it be required to find the value of x, in the equation xz —3 —5, where the unknown quantity is connected by subtractioa. By a.ddi.p: 3 to each side, we have x- 5+ 3 — 8. 3. Let it be required to find the value of ix, in the equattion where the unknown qua-ntity is connected by mult.iplication. 15 By dividiny each side by 3, we have xc — — =5o 4. Let it be required to find the valaue of il, -the ecuation where the unknown qu antity is connected by division. B~y muttli2:yin each side tby 3, we have x-2X3= 6 Vrom the solution of these examuples, wve see, that iwhen the unkn2own.quantity' is corcCIted by additioa,, i t is to be searat.ed, by subtraction.' Yh/en it is connected bt subtraction it i s to tbe separaled by addition, Wkhenk it is connectcd by inuidtlpication, it is to be separated by divtisiom And, when, it is connecited by division't, it, is to be separated by multinpitication 5. Find. the value of, x n the equat ion 3x-3==-x+5. By trausposing terms — 3 and ix, we haxe reducing, 2x-S8, dividing by 2, ex-.-84, 116 RAY'S ALGEBAA,, PAtRT FIR ST, Let this value of x be sr'bstitutecl instead of x, in the original equation, nd, if it is the true value, it will render the two mneombers equal to each other. Original equai;ion,... 3x, —3- z-5. Substituting 4 in the place of x, it becoines 3X4 —3=-4 —5, or 9=9, The operation of substirtuting the vazue of the unknown qu-antity inistead of itself, in the original equation, to see if it will render the two mebeirs equal to each othl- r, is called uvericoation.m The preoedincr equation may be tolved thius: 3X — 2 — -- -5. By addino 3 to each memlber, we have a. 3. -S.. 3 By subtracting x fronm each iemmber, wye haye 3x —av —-— 3'3-x — l M I~ But -3 e —-3 coaleol eah Other; so, also, do 9 Uand.-. x; omitting these, ad thern reducing, we have 2x —-, Drividin each meiber by, -...-, = 4. 7R E M A I x. —The pupil will p!rceiove that the two methods of solution are the same i'n principle, In the first, we use tra-nsposition, to remove thI known qctuantity from te leift member to the niglsht d the t.nl-known uanctliv from the ri-hri member to the left. In the second the same tDhing is. done, bry adding equals to each mtember; and surbttracting equals ftomu each member-this hbein.g the princtiple on which transposition is founded. It is reconmmended to t.he teacher, to use toe latter mniethod lutil the princople is well understood by the p-apil, alfter which the flrsut me;hod maiy be sed exclusivrel y, 6. Find the value osf x in the equation x —— 5 ~ du1tipying tboth sides 1i y a5, the leas commnon muRltiple of the denominators, ie have 15x.- (5 — 0) g0+-3X-+6 or, 15x- 5:+ - 0 I 60 — x-i-i.'by transposition lixt a, w - 5 606 -109 reducing, 7. -,;56d divding, x=8. 7. Find the 7tvalue of x in the equation c -c -+c. multiplying both. sides by ib, ctx a —c.b bd-=bx +-abc, transposing, ax — b =-abc — abd separating into aectors, (a. b)x- ab(c — d) ab(c. —d) di.vidin by (a t ), a I-= ( Rt n v ~ wn.-55. Whha -are -the m3ethlods by which the mrknown qucntity in san eqiastion Lmay be comibhiied with cknown quantities? Give examples, WX hen the tlnknown quantity is conneited by aLddirtion hiow can it be steparated? W-hn, by subtractioxn? By Ivi ultiplieation? 3By divishiS.? W Vht is werif{ tmtion? SI~MPLE EQUATIONS. 117 Fromn the preceding examples and illustrat ions, wve derive the FOR TIHE`OtL'TON OF AN EQUATION OF T-E PFIRST D3EGREE.. If n ecessary, clear the equation,~ of J'factions IeJ trm all tihe operations indicated; and transpose all the terms co.ntaining the unknozwn quegatity to one side, antd the known quanc6ities to tohe othUer. 2. Reduce eacib minember to ifts simptlest form, and divide both sides by the coeticien of thfe unknoiw'n, guantity. ESXA1'1PLES FOR PRACTIC.o AoTEv * Let the1 pupil yerl tooe value of the tunknown quantityy in each examp>le 1 h 2x-5- 2x' -........n.X...2.', 0xl-3.-32-~-16.,......u.. D b~n~rAs. x =tgo 2. 3 —8-:,.......... Ans. x —l. S o SS X StS B l; 3y o S,' j s5o ~. 5 o o o =.5 1 7........... Ans. x -* 1 4. 2x —= —x-............ns. x. —4. 10: c 10 -— so60:;). n = v~x ~-~ )-~.~ ~x.-o ).... Ans. x=5. 6. 52 ( 14+ x ) -i-')'...... Ans. x='. 7 c )-.e-: t x —4) —96. Ans. x=:6, 4. X 5(x- s )+.21 Ans6 |.. x =8. ko -O / 7I. 14 4 3... Ans. jx — to] i-lL- 15-~ Xj. 2.... o.. B, a. k-i 2 6 513. -f-' " --.~,,&u....... 1I. B:Cl-~ 8 1 O- - -".......... ] 2. - - -- __ — -4 c. Anrs. c —24 44 4 2 x x. 2 x —- 7X-2 fit degras, containin one -{nknown qua.ntity? ;! 8 ]_RAYsS'ALGEBRA, PA iT FIRST. 4 5 1 is. tx-4x+is=:z(4xx+1). A us. 2. 5x 19. }-. -...... -. o AnsO x — x+3. —.... 20.4 6 I... Ans. V= tV 23... I 12 z5 AXs. x.+11. 2..- -.'.3. _. oA xoz 2 —-- 3 _x —1 4 - 23. +6LS — r -- 2 -r3~. A v Zns. x 1R 4 2 24s 2x-',1 - -—: —- -no —, I XC_ 25, bz- t? h 4 s e o, o a A8 * x-1)'7x- 4 x-! 29. 7xI-bx5d —c.. -.......... A-s.,= r 4 3. a —b ooe= —-dx —c o,,,,,, e An-s = —31. 7- xt-'ee — 5x 6 e 5 o Q. x. D ab — i 2 4b(a-bx) (ax —)=b..... s x. b be., a 34.......... A. X e o -'abY aidi —b' 6..a..-. —-....-. —... As x — — c) Sb 3~~'id''2- -i-IIy a~~ 86 6 Icm.......-ir' ~~~~~~~~~~~~~~~~~~~~~~~~~~...... $S-iPIPLE EQUATIONS. l1.9 s.?Xfk-d-c-ca..... A ns. X a.bed. a b a t btl'9 >ab X ao bbe-ab 41o ab Ltvo b..e Ano. xj-:-(abac+bc). 38,.... As. X — 3Cd.... 2 4C e CCt —be __ b x —-b.b. 43 1. Ans. X-al. bts a 1 44. 1+; —--- +. 1 Ans. x= —-. 4 a..b....... 4 a? o.. Q-. ab a-b 3 —a be z-C 2o —. b OIW14Y O NE UK NOWN QrUANTITY. ART. l&5o-The solution of a problem, by Algebra., consists of two distinct parts. 1st. To express the conditions of the problem in Algebraic language; that is, to feormn h,e equalion. 2d. 1b sole the equation; that is, to find the value of the wueknown quantity. With pupils, the most difficult part of the operation of solving a question, is to form the equation, by the solution. of which the value of the unknown quantity is to be found. Sometimes, the $tatement of the question furnishes the equation directly; and, soni etimes, it i;s necessary, from the conditions given, to deduce others, from which to form the equatione. When the condition-s furnish the equation. directly, they are called explicit conditions. When'the conditions are. deduced from those given in the question, they are called.implied conditions. It is impossible to give a precise rule, by meanS of'vw.ich vcry question may be readily stated in the forn of an. eq-1aatiQon. Thei first point, is,. to understand fully'the nature of t.he question, so as to be able to prove whether any proposed answer is cotrect. RI.'vi w. —-r t56 Of what two parts does the solution of as- problemc by Alge'bra, consist? What are explicit conditions? What sari imrpied conditions? 12}0 RtAY'T"S ALGEBRA9, PART FIRST, After this, the equation, by the solution of which the value of the unknown quantity is to be Lfund, may generally be formed by the followting Denote /lhe equired quanztity, by one of tie final letters qf' it e at~plhabetd; then, by zaeans ofj sigmns, i-ndicale th.e sam'?e opecraltions thiat it.itouel be necessary to.nzake o1 tZhce ansiwer, to vert:efj it. R a, sn,-n - s;lst. In solving a qtueston, it is necessary to understand the principles of the science which it involves, at least so far as they rela, to to the questi-on unnder consideration. Thus, when a problem esmbraces the coensideration of Rat-io or Prioportion, in order to solve itk the pupil i1must be famniliar with these suijects, Tn the foillowihg examnples, the learner is supposed to be acqua;inted withlt Ratio and Proportio)n, as fa as theoy are tauighl; in Arithmetic. (See BRatio and Proportion, Ray's Arithmetic, Part III.) 2Pd. The operations concerned in the solution of an equation, involve the removal of copffitcients, the removal of denominators, and t}he tran'sposition of quantities. The first six of the following examples, anid a.l.so those from the 10th to theo 44ith inclusive, are arrangedwith reference to these oS perations. E XA-3IPLE S. 1. There are two nuubers, the second of which is three times the first, and their sum is 48; what are the numbers? Let x.z the first number. Then, by the first condition, 3z-x the second. And, by -te second condition, zx3x —48. Reducing, 4zx48. Dividing by 4, x —,2, the smialler number. Then, 32-36, the larger number. Proof or verifcation. 1.2 — 3d48. 2. A father said to his son, " The difference of our ages,is 48 years, and I alm 5 -tnimes as old as you.'" Vhat were their ages? Let x s- the son's age. Then 5x — the fitther's arge. And 5xz —4S. Pteducing, 4sxzz48, Dividing, x —.2, the son's age. Then 5x —60, the father's age. fierflcatxio. 60 —12 48, the difference of their ages. 3.'What number is that, to whieh, if its thh'd part be added, the Sumi will, be 16? tet x- tIhe required number. RuT v m c w.-Ii560 By w11at general rulen mnay the equawtion of a problem Be found? SIMPLE EQUATIONSo I. Then the lrd part of it will be represented by rAnd, by the conditions of the question, we have the equation x+x416. lMltuiplying it by 3, to clear it of fractions, 3 —x —48. Ieducing, 4 —48. Dividing, x —12. 7elftcations 12+ Y —12+4=16: which shows tlhat the value found is correc!:, since it satisfies the conditions of the question. N* o T.-.he pupil slon11t uver/fy the asnswer in every example. 4. What numiber is that, which being increased by its halt- and then diminished by its two thirds, the remnainder iwill be equal to 105. Let ox — tiet number. TShen the one half will be represented by x, and the two thirds ~s 3 /4A2;.And, by the question x- +- =100 Multiplying by 6, 6Gx+3x-4xz —630. Redueing, 5x=630. Dividing, x-1i6. Arns. ~:When the numbers contained in a solution. are large, it is some.times bettfer to indicat.e the nzultiplication, than to perform. it. The precoding solution may be given thus: z 2 3 0x-3x-4xz z105XG 5z=105X6 _= 21X 6=-&;126. 5. Tt is required to divide a line 25 inches long, into twro parts, so tlhat the greater shall be 3 inches longer than the less.:Let zx the length of the smaller part. Then x+3-z thte greater part. And by the question, xzx+3=-25, Reducing, O2 —x 3-O=25. tra'-nsposing o, 2x=25-3z222. DiRvidine, xa' 1, the smaller part. And -x+3=14, the greater part. 6. It s reqtired to divide 68 dollars between A, B, and C, so that $B shall have 5 dollars.more than A, and 0 7 dollars more t:han B. tRAYS ALGEBRA PART FIRST. Let x== A's share. Then x —5 B's share. And xz-12=: C's share. Then, by the terms of the question, we have x+(x-i5)+(x +2)=68. Reducing, 3x+ /1768. Trasposing, 3x=68 —175T, Dividing, x= 17,.Ass share. x+15 -=22, B's share. xq-1-S=-29, Cs share. 7. What number is'that, which being added to,ts third part, the sum will be equal to its half added to 10. Let z represent the number. Then, the number, with its third part, is represented ly x+>; and its half, added to I0, is expressed by,+10. By the condi4 X fC. tions of the question, these are equal; that is, xw j+ 1 0. Multiplying by 6, 6x+-2x31: +60 ieducing and transposing, 8x'-3.-=60 5x=60. Dividing, x=:l. VFerficltation. 12- = 12+ -10. Or =S16, according to the conditionas. Hereafter, we shall, in general, omit the terms, transposing, dividing, &c., as:the various steps of the solution will be evident by inspection. 8. A cistern was found to be one third full of water, and after emptying into it 17 barrels more, it was found to be half full; what number of barrels will it contain when full? Let x= the number of barrels the cistern will contain. Then - l 7=. 2x+ 102=83x 109x —: Or, by first transposing 3x and 102, we have -x= — 02; and multiplying both sides by -1, we have x:102 The unknown quantity, when its value is found, is geonerally made to stand on the left side of the sign of equality; it is -not material, however, which side it occupies, since, by transposition, it can be readily removed to the other. In effecting the transposition of 102t, so as to bring the x on the left side,:we have made it to consist of two steps; it is, however, geneo ally made in one; the transposition, and multiplying by -1, being both made in one line at the same time. SIMPLE EQUATIONS. 123 No T. -Multiplying by -1 is the same as changing all the signs of both members of the equation. 9. A cistern is supplied with water, by two pipes; the less alone can fill it in 40 minutes, and the greater in 30 minutes; in what time will they fill it, both running at once? Let x — the number of min. in which both together can fill it. Then = the part which both can fill in 1 minute, Since the less can fill it in 40 minutes, it fills I- of it in I min ute. Since the greater can fill it in 30 minutes, it fills 9 of it in 1 minute. Hence, the part of the cistern which both can fill in 1 minute, is represented by +-, and also, by 1, 1 1 1 iHence, 1 40 30-x' Multiply both sides by 120x, and we have 3x-4x —120. 7x= 20. x —12 0=L17W m.ra 10. A laborer, A, can perform a piece of work in 5 days, B can do the same in 6 days, and C in 8 days; in what time can the three together perform the sane work? Let x-= the number of days in which all three can do it. Then I-U the part which all can do in 1 day. If A can do it in 5 days, he does of it in I day. f B 6 6 C 4 Ie Hence, the part of the work done by A, B, and C in 1 day, is represented by - ~+5, + and also, by - lie cc, -t1 5. Hence, 5+8 =-o Or, 24x+20x+ 15x=120. 59x-120 1I. Iow many pounds of sugar at 5 cents, and at 9 cents per pound, must be mix-ed, to make a box of 100 pounds, at 6 cents per pound. Let x= the number of pounds at 5 cents. Then 100-x the number of pounds at 9 cents. Also, 5x= the value of the former. And 9(100 —)= the value of the latter. And 600= the'alue of the mixture. 124 RAY'S ALGEBRA PART FIRST. But the value of the two kinds must be equal to the value of the.mixture. Therefore, 5x-+-9(100 —x)=600 Sx4 900 —9x —600 — 4x — 300 x —75, the numnber of pounds at 5 cents. 100 —x-25,.6 C.. 9 cents. 12. A laborer wa s engaged for 30 days. For each day th-t lie worked, he received 25 cents and his boarding; and, for each day that he was idle, he paid 20 cents for his botrding. At the ex pi ration of the tine, he receiv ed 3 dollars; how nxmany days did he work, and how many days was he idle? Let x= the number of days he worked. Then 30 —x — the number of days he was idle. Also 2 —=- wages due for'ork. And 20(30-tr) —- the amount to be deducted for boarding. Therefore, 25x-20(30 x)'-300 25 -600- 20x -300 45x —900 -— 20 - the number of days he worked. 30-x —0- the number of days he was idle. Proo/f 25X20-500 cents, wages. g02X I0-200 cents, _ boarding. 300 cents, the renmainder. In solvinlg othis eXamiple, we reduce the 3 dollars to cents, in order that the quantities on both sides of the equation -t may be of the snam denonination. For, as wecan only add or subt~ract numibers of the same denominations, it is evident, tlhat ve can only comlpare qua.ntities of the samie iname. Hence, all tie qtcoztiies, in both miem bers of an equation', mUtst be of t/he same cienomniza-ion. 13. A hare is 50 leaps before a greyhound, and ta kes 4 leaps to the greyhound's 3; but 2 of the greyhound's leaps are equal to 3 of the hare's; how many leaps must the greyhound taPke, to catch the hare? Let x be the number of leaps 9taken by the hound. Then, since the hare takes 4 leaps while the hound takes 3, the numxber of leaps taken by the hare, after the starting of the hound, will be 4x r; andl the whole number of leaps taken by the lhare, will be — ~+50, which is equal, in extent, to t-he x leaps run by the hound. Now if the length of the leaps taken by enach were eqaua wee 4 e ut, might put. x equal to V!A~-50; but, by the question, 2 leaips of the SIMPLE' EQUATIONS. 125 hound are equal to 2 of the hare's, that is, I leap of the hound is equal to I2 leaps of thle hare; hence, x leaps of the hound are equal to j- leaps of the hare; and we have the the quation 9x28x- 300 =300, leaps taken by thle greyhound. 14. The hour and intute hands of a watch are exactly together hbetween 8 and 9 o'clock; required the time. Let the number of min-ut;es more than 40, be denoted by a; thiat is, let x== the minutes from VIII to the point of coincidence, P; then, the hour hza nd moves from VIII to the point P, while the minute hand moves from nXII to the same point; or, the former moves over x minutes, while the latter moves over 40d- minutes; but the minute hand m3noves 12 times as tfast as the hour hand. Therefore, 1 2xz-40+x 1I x 40.j — minutes -3 minutes, 383'~T seconds. Hience, the required time is 43 minutes, 382- seconds after 8 o clock. 15. A person spent one fourth of his money, and then received 5 do llars. lIe next spent one half of what lhe then had, end found that lie had only 7 dollars remaining i-wIat su mi hacfd he at first? Let x — the number of dollars he had at first. Then, after spending one ifourth of that, and receiveing dollars, he had c —-— +5, which being reduced, is equal to -+5. Ile now spent the half of this sturn, or e ) - -; Therefore -— ( I- ) =% 7; 3x: 32 5 4 S 2 33x 3mx 5 tor, 4 —--— +2; or, 6x —3x —6 +20; 3x —36 x=12. Ans. 16. Divide 42 cents between A and B, giving to B twice as mlany a.s to A. Ans. A 14, B 28. 17. Divide the number 48 into three parts, so that the second 1may be twice, and the third three timnes the first. Ans. 8, 16, and 24. 126 RAY'S ALGEBRA, PART FIRST. 8.o D3vide the number 60 into 3 parts, so that the second may be three times the first, and the third double the second. Ans. 6, 18, and 36. 19. A boy bought an equal number of apples, lemons, and oranges, for 56 cents; for the apples he gave 1 cent a piece, for thile lemons 2 cents a piece, and for the oranges 5 cents a piece; how many of each did he purchase? Ans. 7. 20. A boy bought 5 apples and 3 lemons, for 22 cents; he gave as much for 1 lemon as for 2 apples; what did he give for each? Atns. 2 cents for an apple, and 4 cents for a lemnon. 21. The age of A is double that of B, the age of B is tw ice that of C0, and the stun of all their ages is 98 yefrs; wh at is the age of each? Ans. A 56 years, B 28 years, and 0 14 years. 22. Four boys, A, B, C, and D, have, between them, 44 cents; of which A has a certain number, B has three times as matny as A, C as many as A and one third as many as B, and PD as many as B and C together;' how many has each? Ans, A 4, B 12, C 8, and D 20. 23, A man has 4 children, the sum of whose ages is 48 years, and the common difference of their ges is equal to twice that of the youngest; required their ages. Ans. 3, 9, 1.5, and 21 years. 24. Divide the number 55 into two parts, in proportion to each other as 2 to 3. Let 2x= one part; then 3x- the other, since 2x is to 3x asi 2 is to 3. 2x.+-3x —55 5x —55 2-22 Ans. 3x=33 Or thus: Let x= one part; then 55 -x= the other. By the question, x: 55-x::2: 3. Then, since, in every proportion, the product of the means is equal to the product of the extremes, we have 3x —2(55-x)-l 110.2 —2x 5x — 11 0 x —22, and 55-xo —33, as before. Or thus: Let x= one part, then 3- the other. And x~ — =55. 3x -2x+3xzl 1O, from which x==22, and — 33. The first method avoids fractions, and is of such fere qent appli cation, that we maygive this general direction: SIMIPLE EQUJKATIO NS. 12 7 tWhesn two or tno-re unknown qzua.ntities in any problemn, are to each other in a given ratio, it is best to assunme each of them a imutuliple of some other Lnknown, q2uantity, so that tlhey shall hare to eacIh otgter the give's ratio. 25. The snu; of two numbers is 60, and the less is to the greater as 5 to 7; what are the nunmbers? Ans. 25 and 35. 26. D)ivide the number 60 into 3 parts, which shall be in proportion to each other as 2, 3, and 5. Ans. I2, 18, and 30. 27. Divide the number 9' into 4 parts, in proportion to each other as the numbLers 3, 5, 7, and S. Ans. 12, 20, 28, and h32 28. Divide the number 60 into 3 such parts, that -4 of the first,: of the second, and 4 of the third, shall 11 be equal to each other. Ans. 12, 18, and 30. Thius question is most conveniently solved, by putting 2z, Sx, and Sr for the parts, since the:, d -4 of these are, respe tively equal to each other. 29. What nnmber is that whose half, third, tand fourrth part e Jogether equtal to i 5? Ans. 60. 30. Wh-at nurber is that., -} of whichie is greaLter than -- by 4? Ans. 70. 31. The age of B is two and four fifth times the a.e of A, and the um of their ages is 76 years; what is the age of each 9 Ans. A 20, B 56 years. 32. Divide 88 dollars between A~, B, and C, giving to B >, and t 0 as mucw as to A, As. f(I A $42, B %28, and C $I8. 83. Divide 440 dollars between three persons, A, B, and C, so that the share of A may be 4 that of B, and the share of B'-3 that of. iAns..'s share 9 0, B's 1 5 0, a.nd C's 20 0. 34. Four towns are situated in the order of the letters A, B1, ~C D. The distatnce from A to D is 120 miles; the distance from A. to B is to the distance from B to G as 3 to 5; and one third of thoe distance front A to B, added to tlhe distance friom B to C, is three, times -the distance from U to D; how far are the towns apart Ans. A to B, 36 miles; B to C, 60 miles; C to D, 24 mniles. 35. A merchant having engaged in trade with a certain capital, lost -$ of it the st year; the 2d year he gained a sum equal to "r of what remained at the close of the 1st year; the 3d year he lost - of what he had a the close of the c e 2d year, when he was worth $1236. What was his original capital? Ans. $1545. 36. The rent of a house this year, is greater, by 5 per cent., tian it was l ast year; this year the rent is 168 dollatrs; what was tl Ila year? Ans. $ 160. 28' RLAY IS ALG-EBlRWA PART FIRST. 37. Divide the number 32 into 2 parts, so that the greater shdall exceed the less by 6. A ans. 13 and 19. 38. At an election, the number of votes given for two eandi dates, was 256; the successful candid ate had a majority of 5 0 votes; how many votes had each? Anso 153 and 103. 39. Divide 1520 dollars between three persons, A, B, C, so that B nmay receive 100 dollars more than A, and 0 270 dollars nmore than 3; twhat is the share of each? Ans. a,350, B $450, and C 1720. 40. A company of 90 persons colnsists of men, women, and children; the men arc 4 more than the women, anIMd the children are 10 more thean both nmen and womenl; what is the number of eachtl? Ans. 18 women, 22 men, and 50 children. 41. AAter cutting off a certain quantity of cloth from a piece con'taining 45 ya:rds, it was found that there rema:ined, 9, ards less than had been cut off; how many yards, had been cut off? Ans. 27. 42. What nurm-lber is that, which, being multitpied by 7, girfves a product as much. greater than 2 0, as the number itself is less th.:an 20? lAns. 5, 43. A persoa dying, left an estate of 6500 dollars, -to be divided between his widow, 2 sons, and 3 daughters, so tha teach. son shall receive twice as much: as a dautghter, end Lthe widovw 500 dollars less th-an all her children together; required the shiae of the widow-, and of each son and daughter. Ans. Widow $3000(, each son $1000, and eaheb da,1ughter $50(0. 44. Two me n set 0out a-t the same time, one from London., elnd the ofte-r front Edinburgl; 0cne goes 20, and the otbher SO miles,a day; in bow many days will they meet, t.he distance being 400 miles? Ans. S days. 45. Two persons, A and. B, depart frotm the sate plmace, to go in the samte directio-n B travels at th.e rate of 3, and A. at the rate of 5 miles an hour, but B has t.he start of A I0 h1ours; in how uanny hours will A o'vertalke B? Ans. i 5. 46. What unrmber is that, of which one half and one third of it dialilnished by 44, is e quald to one fifth of it diminiashed by 6 Ans.: 60. 47. A person beingt asked the time of day, replied,'If, to t4he time past noon, there be added its,and the sum till be equal to - of the time to mindnigtt; required the hour..Ans.i 50 mi.u P;.i" SIMifPLE EQU.ATIONS. S129 48. Divide the number 120 into two such par[t.s, that the smaller may be contained in the greater I~ times. Ans. 48 and 72. 49. "t have a certain number in my mind," said A to B; "if I mnultiplyY it by 7, ad(r 3 to the product, divide this by 2. and subtract 4 fron the q0otienit, the remainder is 5. at is the nnm. 11er.? I.rtAns. 5, 50. What number is th1at, which, if youn m-ltiply it by 5, subtract 24 fromn the prodnuct, divide the remainder': by 0 nd add i.3 to the quotien;it, will give the number itself? Ans. 54, 51. Two persons, A and B, engsged in trade,: the cl apittal of B being -4Ihat of A; B garned, aind A lost, 100 dollars- aft'er w].ich7 if - of wlhat LA head left, be snubtracted frotm lha B now has, the remlnldlider will be 134 dollarS; with what capital did eachl. coLmenee ~? Ans. A $78ti, B. 5:,. 52. A man having1 s.pent 3 dollars more than P of his h niuey, had 7 dollars more than,- of' it left;how rmany dollars ha:d heL at first?.As $.. 53, Two nmei, A and B, have tle same annual incomne A. saves of Ihis, but B spends 2 riollars perl annum more than i, tt A ld t thie end of 5 ye-ars fm'inds h'ie'as saved 200 dollars; what is tie"t annuial income( of' A, 5., 54 In thre comlp.ositif of a qua,ntity of gunosf r;t p ierl, of Athier wh,~ole, pliu i0 pounds:, w'a's: nitr e.of tie whole, pltus r pinu1 a,d was sulhuRir;; and:. of the whol., minuess 17 piounds, as c.. llt.(tre..t'i.houWrll ral,,CO poudi 05 ",-sr1 iSe-te- tell, 0u? cs 695 f.ii 55. A peirsoben boiug.-.,t a chatise, horse, l d r es itau:, ir 245 (.. e. lars; tihe hiorse cot s $ ti:es cs luc h as tme a hatness and it e cli'is cost 19, doll:as less tan'2 tines as -iuch as b. l ot hose alld bIr'tness- wh.ma-t'-l s the cost of' eacI-?h kngs Htarns $les i18, 1irtSt 04 chaise I 1t,' 56. Wlhat tw6o rfi'n ers axre as 3 to 4, to achcl of which, ii 4 be added, the sunis will be to each other as 5 to 6? A.is, C:'.antr S 57. WVha'tR two numberis a're as 2 to 5, from each of iwhrichi, -if be subtracted,:the rermainders will be to each other as S to 8? Ans, 20 a'd 50. 58 The ages of trwo brothers are no01w 25 and Ot years, so t l'at tbeir ages iare as 5 to 6; in how many years will their aigces be ats 8 to 9? A..V IHow many years since'tbheir ages were as I to 2? A. 2 20 yrs. 59. A cistern has 3 pipes to fill it; by the first, it ncan Dbe filled in:1i; hours, by the second, in 34 hours, and by the third, in 5 hours; iln what tine can it be filled, by all three running atA onrce? Ans. 48 min. 130 RAY'S ALGEBRA, PART FIRST. G0. Find the time in which A, B, and C together, can perform a piece of work, which requires 7, 6, and 9 d'ays respectively, when done singly,. Ans. 2U 0 days. 61. From a certain sum I took one third part, and put in its stead 50 dollars; next, from this sum I took the tenth part, and put in its stead 37 dollars; I then counted the money, and found I had 100 dollars; what was the original sum? Ans. $30. 62. A teacher spent 42 of his yearly salary for board and lodging, of the remainder for clothes, and -- of what remained, for books, amd still saved 120 d0ollars per annum; what was his salary? Ans~, $375. 63. A laborer was engaged for a year, at SO dollars and a suit of clothes;.a1fter he had served 7 months, he left and received for his wages, the clothes and 35 dollars; wh at was the value of the suit of clothes? Ans. $28. 64. A n an and his wife ca drink a casik of wine in 6 days, and the rn an alone can drink it in 10 days; how many days fwill it last the woran n? Ans. 15. 65. A steamlboat, that can run 15 miles per hour wvith the current, and.10 miles per hour against it, requires 25 hours to go friom Cincinnati to Louisville, and return:rhat is the distance between those cities? Ans. 150 niles. 66. A and B engaged in a speculation; A with 240 dollars, and B with 96 dollars; A lost twice as muc>h as B, and, upon set. tling their accounts, it appeared, that A had 3 times as much remaining as B; what did each, lose? tAns. A 96, and B $48, 67. In a mixture of wine and water, 4. the9 whole, plus 2-5 gab Ions, was wine, and - of the whole, minus 5 gallons, was water; r'equired th.e quantity of each in the mixture. Ans. 85 galls. of wine, and 3" galls. of w-uater. 68. It ls required to divide tile numtber 91 into 2 osLch parts, that the greater, being divided by their difference, the quotient will be 7. Ans. 49 and 413-. 69. t is required to divide the number 72 into 4 such parts, that if the first be increased'by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the sium, the difference, the product, and the quotient shall all be equal. Ans. 14, 18, and 32. Let the four parts be represented by x-2, x+2, x,: and 2x. 70. A merchant haLving cut 19 yards firom each of 3 equal pieces of ilk, and 17 frot another of the same length, found, that the remunants taken together, mneasured 142 yards; what was the length of each piece? Ans. 54 yds. SIMPLE EQUATIONS. 131 71. Suppose, that for e very 10 sheep a farmer keeps, he should plow an acre of land, and allow 1 acre of pasture for every 4 sheep; how many sheep can the person keep, who farms 161 acres? Ans. 460. 72. It is required to divide the number 34 into 2 such parts, that if 18 be subtracted from the greater, and the less be subtracted from 18, the first remainder shall be to the second as 2 to 3. As. 22 and I2. 73. A person was desirous of giving 3 cents a piece to some beggars, but found that he had not money enough in his pocket by 8 cents; he therefore gave each of them 2 cents, and then had 3 cents remaining; required the number of beggars. Ans. 1. 74. A father distributed a number of apples among his chil dren, as follows: to the first he gave - the whole number, less 8; to the second A the remainder, diminished by 8; ard in the same manner, with the third and fourth; after which, he had 20 apples remaining for the fifth; how many apples did he distribute? Ans. 80. 75, A could reap a field in 20 days, but if B assisted him for 6 days, he could reap it in 16 days in how many days could B reap it alone? Ans. 30 days. 76. There are two numbers in the proportion of - to w, which, being increased respectively, by 6 and 5, are in the proportion of;- to -; required the numbers. Ans. 30 and 40. 77.'When the price of a bushel of barley twanted beut 3 cents to be to the price of a bushel of oats as 8 to 5, nine bushels of oats weoe received as an equivalent for 4 bushels of barley and 90 coe::.I; in amoney what was the price of a bushel of each? Ans. Oats 30 cts., and barley 45 ets. 78. Four places are situated in the order of the 4 letters, A), B, C, and. D; the distance from A to D is 34 miles; the dista.nce from A to B is to the distance front C to D, as 2 to 3; and ~- the distance from A to B, added to - the distance from C to D, is 3 times the distance from B to C. Required the respective distances. Ans. A to B 12, B to C 4, and C to D 18 miles. 79 The ingredients of a loaf of bread are rice, flour, and water, and the weight of the whole is 15 pounds; the weight of the rice increased by 5 pounds, is the weight of the four; and the weight of the water is I the weight of the four and rice together; what' is the weight of each? Ans, Rice 2eb, flour 110lb, and water 2jlTb. 132 RIA.Y'... sALG-EBRA, PAtRT 1FIRST. SiMrLtE EQUATIORSNIS CONTAIMNI: G TfVO UNKNOWNT;-0 QUA iiNTIEWS,.AurT. l.. — Itn order to find the value of any unknown quantity, it is evidenrt, that we nmust obtain a single equation containinlg Ui, and known terms, IHeneo, when we have two or more equations, containing t;w'o or more unknown quantities, we.must obtain friomr them a single equation contnaiing only one unknown quantity. The method~ of doing -this, is tenmned elimiictioen, which may be briefly defined thus: Eligmnlzation is the process of deducing from two or more equations, conttiining tro or more unknown quantities, a less number of equatlions containing one less unknown quantity. There are three mit04hods of elimination.: st, Elimination Mr substitution. 2d. Elhnina'tusion by comparison. 3d. Eiinia:unti n lV- addle-ion and sub-ractdion. EMNSM PWAThJ11ki B3Y SSUBJS NTTUfTlOGN.rr An. 1,-.'.-tinR..tnon by sulbstltuttion, lonsists in finding the value of one of the unkn-own quantities in one of the equations, in terms of the other unkno-wlvn qtiuantilty and known term-s, and. substituting this, instead,1 of the, ua.ntitv, in the other equation. To explain this, suppose we have the followilg equatilons, in which it is required to find thle v-ilue of x and'y. NtoT.veI.-The figures in thoe parentleses, are intended to number the equations for re3ferenee. x,-2y-t7 (1.) 2x3-~3y=28 (2.) By transposhing 2y in the, equation (1 ), we have z t17-2-y. Substituting /i'is value of x, instead of z in equation (2), we have 2(17 *~~y)+:y_ Ss or, 34 —dy- 8/y —28 or, -y —28 34 and x —17 —-gy-17 —12=5. Tence, wlhen wNre have two equations, containing two unknown quantities, we have the followi-no FO ELIsttINrATION BY SUBSTITUTION. Find an expression for tihe vale qf one of' the,znicnozn quan-, tities it either e lquation, and substit2,tde this value innplace of ice scame untci nonl02 quadti/ty in the oltiesl equation; there wilgl tthus be fJoee8E a ew equationici, containing ony onle tcnknown quantily. SIMTPLE EQUATIONMS. 133 NoTE. — i finding an expression fior the valto of on0 of the unknown quantities, let that be taken which is tile least involved. Find the ~values of the unknown quantities in each of the foblowing equations. 1. x+5y-38. Aiis. x3. 6. — y- tI0. Ans. x-'25a 3z-.+4o-, —'37 = I. wc,_/ —p 2~ 2xq-4y =22. Ans. x=:5. 5 - 5x-{-7y=46.o "y3, v xx 2 7. -— I. Ans. x=20. 3. 3z+ 5y- w:5.7 AMIs.; = 4 7 5 4 5x:t-3y-47. Y=9. s: z 3qIy-l O. y —30. 4. 4x —3y-26. Ans. xzz S7 3xe-4gy — 0 n. 7 8a 5. 2x-3yz — 4. Ans. 6:x-t+6. 3,71y ELtSIINA'TION IBYS 40 ACOMIPAI IS 0 ON. AnRTy. 459t-Elin-ination: by comparison, cons0istS in finlding the value of the same unknown quantity in two different equations, and then placing these,values ecqual to each other. To illustrate this method, we will take thie same equations which were used to explain elimination by substitntion. x —2y=17 (:L) 2x-~L3y?28 (2.) By transposing gy in equation (1), we have x —17 —2y, By transposing 3sy in equation (2), and dividing by 2,'we have 28 —3y.2' Placing,th.ese values of x equal to each other, 28-3y 17 2 or, 28 3y-3y 4-4y or, y=6, The value of x ma3ty be found in a similar manner, by first finding the values of y, and placing them equal to each other:., _0 But, after having found the value of one of the unknoiwn quantlties, the value of tie other may be found miost readily by substitution, as in the preceding article. Thus, ax —17 2y1 7 —12z5t. ItB r vIE. —-l5Zo What is necessary in order to find the value of any unknown qufa:ntity? When we have two equations, containing two unknown quantities, what is necelssa.ry, in order to find the value of one of them? What is elimination?.How many methods of eliminatinm are there? 158. In what. does elimination by substitution consist? Wh.a:t is the rule for elimination by substitution? 159. II what does elimination by comparison consist? 134:RAY'S ALGEBRA, PART FIRST. ]Hence, when we have two equations, containing two unknown quantities, we have the following RUJLE FO~R LI.tMIATION BY COMPARISO,. Find an ex.ress-ion for the value of the same unaknown quantity zn each of the given equatiotns and place these values equal to each other; theare will thus be formed a new equation, containing only one unknow~n quantity. Find the value of each of the unknown quantities in the follow> ing equations, by the preceding rule. 1. x+3yz16. Ansx= 7.x yA 6. 3y y A=12. X+-5y —22. y- o3. Ans.O x 2. 3x+5y=1s29. Ans. x —. x y 3x-Sy=l9. y-l= yzz2=8. 3. 5x-2y —4 Ans. xz-2. 2x-?y. -3. 7, _ m-=4. Ans. x —45. 4. - =2. Ans. x —6. ~ —y —— i.'y —3. Any. xz36, 2 ~+~g27 i Ans. x —21, 5. -$ 10. Ans. x36. 7 D 6 =4 y-z24. 9 7 y3 3xx 3~ 2 +2y —x-N... B..... g Ans. x=-20, 3 - 2y, x — 2- t-= 4 A E. la. —Elimn atAon by addITI on and subtration, cons.sts ART. 1.0 —E lmination by addition and subtraction, consists in multiplying or dividing two equations, so as to render the coWficient of one of the unknown quantities, the sane in both; and then, by adding or subtracting, to cause the teru containing it to disappear. To explain this method, we will take the same equations used to illustrate elimination by substitution and comparison. x+2y-=17 (1.) 2x+3y=28S (2.) SIMPLE EQUATIONS. 135 If we multiply equation (1) by 2, so as to make the coefflcient of x the same as in the second equation, we have 2x+4y=34 (3.) 2x-13y-28, equation (2) brought down. Since the cofficient of x has thle samte sign in these equations, if we subtract, the terms containing z will cancel each o:ther, and -the resulting equation will contain only y, the value of which may then readily be found. After this, by substituting the value of y, as before, the value of x is easily obtained. To illustrate the method of eliminating, when the co.fici ents of the unknown quantity to be eliminated, have contrary signs in the two equations, suppose we have the following, in which it is required to eliminate y. 3x-5y=6 (I.) 4x+3yy-37 (2.) It is obvious, that if we multiply equation (1) by 3 and (2) by 5, that the coiefticients of y will be the same. Thus, 9x-15y — 15 20x+ 15yS185 adding, 29z 203 - 7, Substituting this value of z in equation (2), we have 28+3y — 37 3y= 9 y_ 3 From this we see, that after making the coefficients of the quam tity to be eliminated, the same in both equations, if the signs are alike, we must subtract; but if they are unlike, we must add them. Hence, when we have two equations, containing two unknown quantities, we have the following.RULE, FOR LLMINATION BY APDITION AND SUBTRsACTION. Miltiplty, or divide the equations, if necessary, so that one of the,unknown quantities will have thie same coiffiient in both. Theni take the difference, or the sum of the equations, according as the signs of'the equal terms are alike or unlike, and the resulting equation till contain only one uenknown quantity. IR E M A R.-When the coeffiients of the unknown quantities to be climinated are prime to each other, they ma.y be equalized, by multiplying each IE, r iw. —159. Whatis the rule for elimination by comlparison? 160. In what does elimination by addition and subtraction consist? What is the rule for elimination by addition and subtraction? 136 RAY'S ALGEBRA, PAtRT FIRST. equation by the coisfficient of the unknown quantity in the other, When the coefficients are not prime, find their least common multiple, and multiply each equation by the quotient obtained by dividing the least c omnmon multiple by the coefficient of the unknown quantity to be eliminated in the other equation. If -the equations have fractional coefficients, they ought to be clearedt beibre iapplying the rule, Find -the value of the unknown quantities in each of the folXowing equations, by the preceding rule, x t 3x+2y=2 1. Anys a-:5 a y 5. O-t= o 3x-' - 2y= 5 FAnsx —2O, 2. 3x-2 — 79 Ans. i-5 y 5x-32,-410, y-4. xJfy X —y 1. 4-? 7. 2-ay3. Ans.- Aus. x aa x —]-y y O..s a., r a 9 - 3v —= 2 l o AnT. 1.IL Th-The questions contained in Art,. 56, were a-,ll capIa ble of being solved by using 9qe unknown quantity; although, several of the examples contained two, and in some canses more, unknown quantities, In those questions, however, there was such a connection existing between the several quantities, that it was easy to express catch. one in terms of the other, BAut it frequently happens, that in a problem containing two unknown quantities, there nmay be no direct relation existing between t1-ismt, by mleans of which either of them nlmay be found in ter.mxs of the other. In such a case, it becomes necessary to use a separate Sy>mbol for eabch unknoan quantity, and then to find -the equa-tions containinf these symbol.s on the same principle as where there rwas butt one unknownn quantity; that is, in brief,'regCard the sy-mbols as the answer to thte qaestion, amnd then proceed in ther saee sa ae,?nne'' as it zwould be ndeeessaty to do, to p9rovee athe cswier. After the equations are obtained, the vzalues of the unkno0wn quantities may: be found, by either of the three different modes of eliminiation. We shall first 1give two exGamples, which can be solve d by usin:g either one or two unknown quantities, SIMPLE EQUATIONS~ 13 7 In general, no more symbols should be used, than are really necessary; unless, by using them, the solution is rendered more imple. 1. Given, the sum of two numbers equal to 25, and their difference equal to 9, to find the numbers. Solution, by using one unkinown quantity. Let x=- the less number; then x~ 9= the greater, And x+-x-z-9. 25o 2x=l6 x=8, the less numnber; and x+9=1T 7 the greater. Solution, by using two unknown quantities. Let x- the greater, and y= the less. Then xi-y=25 (I.) And x-y- 9 (2.) 2x-34, by adding the two equations together, x=17, the g;reater number. 2y=G6, by subtracting equation (2') front equation (1) y- 8, the less number. 2. The sum of two numbers is 44. and they aire to each other as 5 to 6; required the numbers. Solution, by using one unknown quantity. Let 5xc the less number; then 6x — the gieater. And 5x+6z-z44. x-4 5xs- 20, the less nunber. 6x=24, the greater numberr Solution, by using two unknown. quantities. Let x= the less nunmber, and y —= the greater.'Then x —y — 44 (1.) And x * y: 5 6 - or, 6xSty (2.) by multiplying means and extrem8es 6.x+6y/'64 (3.) by multiplying equation (t) by 6. 6y-264 —5 y, by subtiacting equation (2) fromn (3). 1y= —264 y,=-24 and x=44-y=:20. Several of the f&ilowing questions may also be solved by usinhg only one unknown quantity. 3. There is a certain. number consisting of two places of figures; the sum of the figures s equal to 6, and, if frotm the double of the V E a -..161. In solvin questions, when does it become necessary to use a separate symbnol for eacth uknown quattity? HIow are the quations formed., from which the valaes of the unknown queantities are to be obtained? 12 138 RAY'S ALGEBRA7 PART FIRST. number 6 be subtracted, the remainder is a number whose digits are those of the former in an inverted order; required the number? In solving questions of this kind, the pupil must be reminded, that flny number consisting of two places of figures, is equal to 1 0 times the figure in the ten's place, plus the figure in the unit's place. Thu., 23 is equal to 10 X2+3. In a similar manner, 325 is equal to 100>X3+10X2+5. Let z- = the digit. in the place of tens, and y= that in the place of units. Then I0xzf-y- the number. And 10y+-x= the number, with the digits inverted. Then x-F-yz-6 (I.) And 2(10x+y)-6= Oy-l-+ (2.) or, 20z+S2y-6=1 Oy+-x. 19xz=Sy+Sxz —-Sy48, fron equation (1), by multiplying by 8, and transposing. 27x=54, by adding. x —2 yF-6-2-4.t Ans. 24. 4. What two numbers are those, to which if 5 be added, the suns will be to each other as5 to 6; but, if 5 be subtracted from each, the remainders will be to eachi other as 3 to 4? By the conditions of the question, we have the following proportions x+-5: y+-5: 5: 6 x —5 o y-5:- 3: 4. Since, in every proportion, the product of the means is equal to the product of the extremes, we have the two equtations (x+5s)=5(y —+5) 4(x —5f)3(y-5) From these equations, the values of x and y are readily found to be 20 and 25. R l A K. u —Instead of saying, that the two sums'ill be to each other as 5 to 6, it will be the same to say, that the quotient of the second divided by the first, is equal to -, since 6 divided by 5, expresses the ratio of 5 to 6. This would give the following equations — +E;; -and -= x —5 5 x-S 3 which man be readily obtained from those given above. o otrE. -In solving the following questions, after finding the equations, the values of -the unknown quantities may be found by either of the three methods of elimluation.a SIMTPLE EQUATIONS. 139 5. A grocer sold to one person 5 pounds of coffee and 3 pounds of sugar, fir 79 cents; and to another, at the sa.ne prices, 3 pounds of coffee and 5 pounds of sugar, for 73 cents; what was the price of a pound of each? Ans. Coffee 11 ets., sugar 8 ets. 6. A farmer sold to one person 9 horses and 7 cows, for 300 dollars; and to another, at the same prices, 6 horses and 13 cows, for the sanme sum; what was the price of each? Ans. Horses $24, and cows $12 each. 7. A vintner sold at one time, 20 dozen of port wine and 30 of sherry, and for the whole received 120 dollars; and, at another, 30 dozen of port and 25 of sherry, at the same prices as before, for 140 dollars; what was the price of a dozen of each sort of wine? Ans. Port $3, and sherry $S per doz. S. It is required to find two numbers, such that, of the first and - of the second shall be 22, and 4 of the first and - of the second shall be 12. Aus, 24 and 30. 9. If the g;reater of two numbers be added to ~ of the less, the sum will be 37; but if the less be diminished by 4 of the greater, the difference will be 20; what are the numrbers? Ans. 28 and 27. 10. What two numbers are those, such that 4 of the first diminished by -; (,f the second, shall be 5, and -{ of the first diminished by I of the second, shall be 2? Ans'20 and?15 11. A farmer has 2 horses, and a saddle worth 25 dollars; nouw, if the saddle be put on the first horse, his value will be double that of the second; but, if the saddle be put on the second horse, his value wirt be three timnes that of' the first. Repquired the value of each horsIe. Ans. First $1.5. second $20. 12. A and B are in trade together with different sums; if 50 dollars be acdded to A's property, and 20 dollars taken from B's, thev will have the same sum; and if A's property was 3 times, und B's 5 times as great as each really is, they would together have 2350 dollars; how nuch has each? Ans. A $250, B $320. 13. A has twQo vessels contaiinig wine, and finds, that,: of the first contains 9 gallons less than -t of the second; and that 4 of the second co3ntains as L uch as of the first; how much does each vessel hold? Ans. 720 and512 galls. 14. There is a number consisting of two digits, which, divided by their suam, gives a quotient, 7; but if the digits be written in an inverse order, and the number so arising, be divided by their sum increased by 4, the quotient will be 3. Required the number..Ans. 84, 15. If we add 8 to the numnerator of a certain fraction, its value becomes 2; and if we subtract 5 from the denominator, its value becomes S; required the fraction. A.s, 3. 140 RAY'S ALGEBRA, PART FTRST. 160 If to the ages of A. and B IS be added,'the result will be douhble the tage of A; but, if from their difference 6 be subtracted, the result will be the age of B; required their ages. Ans. A 30, Bl2 yrs. 17. There are two numbers whose sum is 37, and if 3 times the less be subtracted firom 4 times the greater, and the difference divided by 6, the quotient will be 6; what are the numbers? Ans. 1t6 3and I21. 18. It is required to find a fraction, such that if 3 be subtracted from the numerator and denominator, the value wi.ll be; and if be added to the numer'ator and denominator, the value will be 4.A.XAns 1Xs. 9. A father gave his two sons, A and B, together 2400 dollars, to engage in trade: at the close of the year, A has lost - of his capital, -while Bh having gained a sum equal to -4 of his capital, finds that his money is just equal to that of his brother; what w.as the sumn given by thee father to each? Ans. A $1500, B $900. 20. If from the greater of two numbers I be subtracted, the remainder Nwill be equal to 4 times the less; but, if to the less 3 be added, the sum will be -k iof the greater; required the numbers..Alns. S and 33. 21 A said to B, Giv me 10e 0 dollars, and then I shall have as much as you." B said to A, " Give me 100 dollars, and then I shall havte twice as lmuch3 as you." How many dollars h.ad each? Ans. A $ 500, B $700. 22. If the greater of twso numbers be multiplied by 5, a.und the less by 7, the sui of their products is 198; but if the greater be divided by 5, and the less by 7, the sum of their quotients is 6; what are the numbers? Ans. 20 and 14. 23. Seven years ago the age of A was just three times that of B; and seven years hence, A's age will be just double that of B; what are their agtes? Ans. A's 49, B's 21 yrs. 24. There is a certain number consisting of tswo places of figures, which being divided by the sum of its digits, the quotient is 4, and if 27 be added.' to it, the digits will be invert ed; required the nunmber. Ans. 36. 25. A grocer has two kinds of sugar, of such quality that one pound of each are together worth 20 cents; but if 3 pounds of the first, and 5 pounds of the second kind be mixed, a pound of the mixture will be worth II cents; what is the value of a pound of each sort uns. 6 cts., and 14 cts. 26. A boy lays out 84 cents for lemons and oranges, giving 3 cents a piece for the lemons, and 5 cents a piece for the oran-ges; he tafterward sold 4 of the lemons a.nd. of the oranges, for 40 SIMP.LE EqQUATIONSK 141 Pents, and by so doing cleared 8 cents on what he sold; what number of each did he purchase? Ans. 8 lemons and 12 orangesc 27. A person spends 30 cents for peaches and apples, buying his peaches ta 4, and his apples at 5 for a cent; bie afterward sells i- of his peaches, and -, of his apples, at th sae sae rate he bought them, for 13 cents; how many of each did he buy? Ans. 72 peaches and 60 apples. 28. A owes 500 dollars, and B owes 600 dollars, but neither has sufficient money to pay his debts. A said to B, Lend Im-le of your noney, and I shall have enough to discharge my debts." B3 said to A, "~ Len d we of your money, and I can pay mline," H-ow much money has each? Ans. sA. 400, B $500. 29. A nmerchant, bought two pieces of cloth for 23t6 dollars, the first piece at 4, and the second at 7 dollars per yard; bu-t the cloth getting damaged, he sold i of the first piece, and' - of the second, for 60 dollars, by whichli he lost 8 dollars on what iee sold; what was the inmiber of ynrds in each piece? Ans. 24 yards in the -first, and 20 yards in the second. 30. A son said to his ftaier, " iowi old are we? " The fatber replied, " Six years aeo my c:ge wats 3-' times yours, b ut 3 years hence, ny age will be only 2 — times yours." equired the kage of each. Ans. Father's age 36, son's 15 yrs, 3 1. A person has two horses, and two saddles, one of which cost nl50, n the other 3 dLollaurs. If he places the best stddlte upon the first horse, and the other on the second, then the latter is worth. 8 dollars less than the briner; ibut if he puts the worst saddle upon the first, and the best upon the second horse,'then the value of the latter is to that of the former as 15 to 4. Rlequired the value of each horse. A ns, First 830, second 9$70. 32. A farnier having ni'xed a certain number of bushels of oats and rye, found, that if he had mixed 6 bushels more of each, he would have mixed 7 bushels of oats for every C of rye; but if he had mixed 6 bushels less of each, he would have put in 6 bushels of oats for every 5 of rye. H-ow many bushels of each did he mix? ans.l Oats 78, rye 66 bu. 33. A person having laid out a rectangular vard, observed, that if each side had been 4 yards longer, the length would have been to the breadth, as 5 to 4; but, if each had been 4 yards shforter, the length would have been to the breadth, as 4 to 3; requtired the length of the sides. Ans. Length, C3, breadth 28 yards. 34. A fatrmer rents farm fa or 245 dollars per annum;'the ti.ila ble Ianda being ralued at 2 dollars an acre, and. the pasture at l dollar and 40 cents an acre; now the number of acres tillabIe, i 142'8RAYS ALGEBRA, PART FIRST. to the excess of the tillable above the pasture, as 14 to 9; how many were there of each? A. Tillable 98, pastn'e 35 acres. 35. Two shepherds, A and B], are intrusted with the Char';e of two flocks of sheep; at the end of the first year, it is found, that A's flock has increased 10, and B's diminished 20, when their numbers are to each other, as 4 to 3; during the second year, A's flock loses 20, and B's gains 10, when their numbers are to each other as 6 to 7. Required the number in each flock at first. Ans. A's had 70, and B's 80 sheep. 36. After drawing 15 gallons from each of 2 casks of wine, the quantity remaining in the first, is j of that in the second; -aitter drawing 25 guallons more from each, the quantity left in the first, is only half that in the second. Required the number of gallons in each before the first drawing. Ans. 65 and 90 galls. 37. There is a fraction, such that if 1 be added to the numera, for, ancd the nunmerator to the denominator, its value will be 4-; but if the denominator be increasecd by unity, and the numerator by the denominator, its value will be ~; find it. Anis, fr 38. Find two nuimbers in the ratio of 5 to 7, to whlich t.wo other r:equired numbers, in the ratio of 3 to 5, being respectively added, the sunms shall be in the ratio of 9 to 13, and the difference of their sums equal to 16. Ans. 30 and 42, 6 and 10. Let the first two numbers be represented by 5% and 7x, and the other two by 3y and 5y. 39. A farmer, with 28 bushels of barley, worth 28 cents per bushel, would mix rye at 36 cents, and wheat at 48 cents per bushel, so that the whole mllture unay consist of 100 bushels, and be worth 40 cents a bushel; how many bushels of rye, and how many of wheat must be mixed wsith the barley? iAns. Rye 2s0, and wheat 52 bu. 40. Two loaded wargons were weighed, and their weights were found to be in the ratio of 4 to 5; part of their loads, which were in the ratio of 6 to 7, being taken out, their weights wrere then found to be in the ratio of, 2 -to 3, and the sumn of their weights was then 10 tons; what were their weights at first? Ans. 16 and 20 tons. 41. A person had two casks and a certain quantity of wrine in each; in order to have the same quantity in each cask, he poured as much out of the first cask into the second as it already contalned; he next poured as much out of the second, into the first, as it then contained; and lastly, he poured out as much from the irst into the second, as there was remaining in it; after this, he had 16 gallons in each cask; how. n any gallons did each contain ~a. fir st s Aas. First 22. an;d iIsecond 10 galls. SIMPLE EQUATIONS. 143: SIMPLE EQUATFiO,% CONTAINING THEE OR MOR EIUNiNOWN QUATIT1]ES. ART. 162 — Equations involving three or more unknown quaan tities may be solved, by either of the three methods of elimination explained in the preceding Article, as we shall now proceed to show, by solving an example by each of these methods. Suppose we have the three following equations, in which it is required to finad the values of x,y, and z. x+2y+ z=-20 (1.) 2xd- y+3z=31 (2.) 3x+4yl-2z=44 (3.) Solution by subsfitution. From equation (1), x-20-2y —z. Substituting this in equation (2), we have 2(20-2y —z)+y+3z-31. or, 40-4y-2z-+-yt-3z-31. 3y-z=9 (4.) Substituting the same value of z in equation (3), we ha ve 3 (202Gy — ) +4y+2z-=44. or, 60 —6y-3zd-+4y-+.z —44. 2y+z-z=16 (5.) 3y-z=9 (4.) Here the values of y and z are readily found by the rule, Art. 158, to be 5 and 6; then substituting these values in equation (1), we find x=4. Solution by comparison. Fromr equation (1), x=20O-2y-z (2 x- 31-y —3z 44-4y-2z ( 3), x= a Comparing the first and second values of x, we have 20-2yz31 —— 3z or, 40 —4y —2z=3 -y-3z or, 3y-z-=9 (4.) Comparing the first and third values of x, we have 20__ 44- y —2z 3 or, 60-6y-3z — -44 —4y —52z 2y+z=6 I 6 (5.) From equations (4) and (5), the values of y and, an'd then x my be found bythe rule,.At. 1.59. 144Xg R IrS T ALGE BRA, PART F IRST. Solution by addition and subtraction. IlIultiplying equation (1) by 2, to render the eoiffcient of x the same aS in equation (2), we have 2x +4y+2 —=40 equation (2) is 2x- y+3z=3 1 by subtracting, 3y — z- 9 (4.) Next, multiplying equation (1) by 3, to render the coIEfciient of x the same as in equation (3), we have 3x+6y+3z=60 equation (3) is 3x~+4y+2z-44 by subtracting, 2y+ z-16 (5.) 3y- z= 9 (4,) by adding, 5 y -- 2 y = 5 Then 1O+zOl6, and zG6. And x+-0-4-6=20, and x=4. RIt r AR rn.-Tlhe methods of elimination by substitution and comparison, when there are more thian two unknown quantities, are merely an extension of the rules alrealdy presented, in Articles 158 and 159; therefore, it is unnecessary to repeat them here. When the nutber of unknown quantities is three or more, and particularly when each of the unknown quantities is found in all the equations, the method of elimination by addition and subtraction is generally preferred; we stall, therefore, ilustrate it by another example. Let it be required to find the value of each of the unknown tSuantities in the following equations. v-V 2x+3y+4z —30 (1.) 2v+3et y+ zz15 (2.) 3v+ x —2y+3z-23 (3.) 4v+2'x-y+14z=61 (4.) Let us first eliminate v: this may be done by making the cohN-fii cient of v, in one of the equations, the same as in the other three, and then subtracting. 2v+ —4x+6y+Sz:60; by multiplying equation (1) by 2. 2v+3a+y+-z-l 5 (2.) x+35y+ 7z45 (5.), by subtracting 3v- z+9y+l2z-90, by multiplying equation (1) by 3. 3v+x+2y+3oz-23 (3.) 5x+7y-7- 9z-67 (6.), by subtracting. 4v+Sx —+12y-+16z -120, by multiplying equation (1) by 4. 4v-+[2x —y+14z = 61 (4.) d6x+ L3y+2z= 59 (7.), by subtracting. SIMPLE EQUATIONS~ 145 Colleeting into one place, the newr equations (5), (6), nd (7), we find, that the number of unknown quantitie, as well as'the number of equations, is one less. x+-5y +i-7-45 (5.) 5xz+7y+9z-z67 (6.) 6z — 1 3y+2z-s59 (7.) The nex t step is to olimintte x, by making the coefficient of x, in one of the equations, the same a.s in each of the others, and then subtracting. 5z+25y+35z=225, by multiplying equation (5) by 5. 5:c+7y+9z: 67 18y+26z158 (8.) 6x-F3Oy+42z=270, by mnultiplying equation (5) by 6. 6xz13yV+2y= 59 - ft40z J -211 (9.) -rintgng togethernc eoquations (8) and (9), we find, that the numher of equationa, as well as of unknown qtuantites, is now two 1ss. 8lyL26 z=158 (8.) 17y —430z=,2I1 (9.) 30J6y+442z.-2686, by multiplying equation (8) by 17. 306yji720z: — 3798, by multiplying equation (9) by 18. 278z-= 1112 z= 4 Sunbstituting t.he value of z, in equation (9), we get 17y+60 —211 17y= 51 y-~ 3. Substitutiun the values of y and z, in equation (5), we get; x-15t28=-45 x=2 And lastly, substituting the values of x, y, and z, in equation (1), we got v — 4+9+l16-30 or, v —1. From the preceding example, we derive the GEP ERAL RULE, FOR ELTIINATION BY ADDITION. AND SUBTRACTION st. Combiz e an one of the equations wi.th each of the othesr, so as to elimzinate tfhe sa'e,aunknown quantity there will thus ariset a new class of eqpuations, containing one less unknown quantity. 2d. Combine any one o'f these new equations with- each f the other's so as to e~iminate another,unknown quantity; th ere Zwil ihus aritse another class of equations, containing twor 4,'s tunknown quantit e,. 12 146 LAY'S ALGEBIR A-, PA.IRT Fil:rST. 3St. C:/.:i:tisie U&is sue'es qj opvei~tite s tuntrl,, a sti e/c equation' is, obtained, containing but one,unknowan o.uctitt,i, ftinm w. ich its A vaue mZay be easily found; l/tom, by going bt/cc, ald sustilttay:' f Ui/t value in the deri/c ed qcCCIt'aitioacs, Ihe vsCril es of f/he otihc'r,ti/k.o.w~n.t qtt.nttiiies utay be readily fbund. B A n s. K.s —— WI'in. the number of unknown quanrtities in each eb quat.ion, is leses thun the whole number of unknown qouantities i.nvolhed, t:he mothod of substitution will. generally be found the shortest. By solving seovera of'the following exa-mples, by each of the three different mothods, tho pupil! will be able to approciate their rlative excele.nce in different ceases. TO' -1 SOLVrE.D) Y fITt1Rt OF THiE DIrItt iENTttu )IYATODS ciip L~~~t)I5N~~ATisOe-18 1o x-J-y-=50,..-zx y,............... — 3 u+,=4z J' J...........1 2.3x y..........., ro Sx+5y- 1-......... A n s.,.=1 2 Sr~~................p y -8 /-y 106. o' I....... t. x-ky-m z.'2.......Att.. sx x —y+ez-'2. 2......ts 6. (Aris, a;x-:64, 4. x,-0; y+j,-!OO' z+ __-_0 p072 t4> x+ 10 0 _ 0.0 3~~~~~~ at —:y=; — ) — 1.... "~~~~~~~~~~~~~~~~~~ —=,4o 34 rt.19...... Cx t 02.....4. 2.4 An-n xe, I - st-. -Y~-~, Za 8a;.,~-4y~ —2z-~~~~ ~~ ~ ~~~~~~~~~~~~-:% ) z:=5.....t.-.-.:......... x=-, 5 3.l.......:-4. x yz -t —5, 542'3................... 3 x - - k - 4 1 5~~~~~~~~~~~S.0:-, 3x,~"' —'y z. x yiJ1 S!,IaPLE EQUNATIONSl 147 Art s'-a3, fhe n a question contlw t ils t,7ie'e o0n o' nkn n -to wn.quantities0, quations n avolvin them, can be found on the, same principle as in quesletions contarini ng3 o2e or io unknTown quant.t ties. (See Articles ].5G and 16i.) The valaes of the ulnknown quatntities ~may tlh:n be found by either of the ta.ree methods of elimination Pr, E NAP I: t. —tTho rihede" o hf olai nintion io be pr. f'or,',e, will dCf.end oin the manner in twhich tt'l e unkn.own quantl'g..s are co ialed, and 1n ust eo left to the l judgmentof - o pt it lnt.'WheaI suach. a rei'aion exists b etveen ithee different aunnownx quarniti.nes, tat one or 1 tore of theat can toe exp' esscLd direetly in te;rms o hf m.mtc:her, it slhould be done ta thi s generally rendetss ohi solatioh core:siA. dleO 1. A persoi hts 3 ingots, ecr(posmed of 3 dircenat m etals ein d3if 1rcit proportions a'pouind of ther first contaians I ounaes of si -e: 3 of coppr, 6and G of tin; a tpound of the second co rsists of i2 ounces of silver, 3 of copper, and I of -tin tad a pouncd of lthe thi'rd, om f 4 ounces of ie, 7 of coppeCr, atd 5 of tin. How o.uch of each of tho inoaots ltust be -tfaaken, to f).l anthertI inoet of I pound cve tgb a, consisting of S ounr~ of - ulve t of copper,,and 4 — of tin.? Let X, y, o, be,le tnam t.ber of onaces to be ltaki.n of -th- 3 ingoath' reslpeci" veta Then, sinoce! 6 unces of ~he -fir'st cont. in 7 ounces of si].ver, I oriact cilt Ic. ix 3 l T a o f oan eounce of silver~ ad 1aenc. o, z oaunces wiell toeutain ounces of si lver. 12y in the sate1C tXiimaero,' ounces of the second. will contain otunices of silver; attd * ouances of the thaird witt conuttain - ounaces of silver..cut, by the question, the nut ber of ounces of silver in a pound of the new iiigo', is to be 8, hence 7$ t1i,, 4. Or, by clearing iPt of fiactions, tar e0 W —-2.'hait: is the general. aule ler elimainration by f' addition nd subtractioil? When is the metthod of eliminatieon t}y subsIt.tai tion1 to be prtoferred to this "I 13i Upon Wlhat pritnciple are equa ions forited, -whlen a ctuestion contaa ins three, or morte auntknowan quantities? WIthen shltelr ae %tse.o la's aumboer of vyshboles thl5a5n there are lknoe-swon quanteteios? d48 RAY'S ALGEsBRA, P"ART FIRST. Reasoning ifn a similiar manner with reference to the copper and the tin, we have the two following equations: 3x+3y+7z —60 (2.) 6xo+ y+5z=6S (3.) The com eient of y being the simplest, will be most easily elimninated. If we multiply the second equation by 4, and take the first equation rotm the product, the result is 5x+24z= 12 (4.) If we multiply the third equation by 3, and take the second from the p'roduct, the result is 15Sx+Sz —144 (5.) if we tultiply the last equation by 3, and t'ke the preceding equation friom it, the result is 40xz=:320 ta.bst'tduting this valtue of in equation (5), -we h" aT'c 120 —Sz=-=144 z-3 And substituting these values of x and z, in equation (3), 48+y+ 15=68 Hence, the new ingot will contain 8 ounces of the first, 5 of the second, and 3 of the third. 2. The sumns of three numbers, taken two and two, are 27, 32, and 35; required the numbers. Ans. 12, 1 5, and 20. 3. The sum of three numbers is 59;, the difference of the first and second is 5, and -4 the difference of the firs t and third is 9-; required the numbers. Ans 29, 1 9, anad 11. 4. There are three numbers, such that the,:first,, with.;-'the second, is equal to 14; the second, withX part of the third, is equal to IS; and the third, with - part of the first, is equal to 20; required the numbers. Ans. 8, i2, and 18. 5. A person bought tahree silver wIatches; the price of the first, with the price of the other two, was 25 dollars;'the price of the second, with - of the price of the other two, was 26 dollars; and the price of the third, with 4 the price of the other two, was 29 dollars; required the price of each. A $8, $1]8, axld $16. 6. Find thre numbers, such that the first with 4- of tie other t'wo, the second with i of the other two, and the third w.ith'~ of the other two, shall each be equal to 25. An. 1 3, 17, and 19, 7. A boy bought at one time 2 apples and 5 pears, -for 12 cent s at -nothera, 3 petars and 4 peatches, for 18 cents; a't. nother, 4 pears SIMPLE EQUATIONS. 149 tnid 5 oranges, for 28 cents; and at another, 5 peaches and 6 oranges, for 39 cents; required the cost of each kind of fruit. Ans, Apples 1 cent, pears 2, peaches, 3, oranges 4 cts., each. 8.A. and B together possess only ~ as much money as C; B and C together, have G times as much as A; and B has 680 dollars less than A and C together; how much has each? Ans. A $200, B $360, and C $840. 9. A, B, and C together, have 1820 dollars; if B give A 200 dollars, then A will have 160 dollars more than B; but if B receive 70 dollars from C, they w ill both have the same suim; how nmiuch has each? Ans. A $400, B $640, and C $780o 10. Three persons, A, B, and C, compare their -money; A says to B, " Give rme 700 dollars, and I shall have twice as much as you will have left." B says to C, "Give me 1400 dollars, and I shall have three times as much as you will have left,' And C says to A, "Give me 420 dollars, and then I shall have five times as much as you will have lerft" H-ow much has eaclh? Afns. A 980, B 15 40, and C d 2830. I. A certainz numlber is expressed by three figures, and the sum of the figures is 11 the figure in the place of units, is double that in the place of hundreds; and if 297 be added to the number, its figures will be inverted; required the number. Ans. 326, 12. Three persons, A, B, and C, together, have 2000 dollars; if A gives B 200 dollars, then B will htave 1 00 dollars more t.han C; but, if B gives A 100 dollars, then B will have only a's much tas C; requird the smun possessed by each. Ans. A $500, B $700, and C $800. 13. There are three numibers whose sum is 83; if, frof the first and second you subtract 7, the remainders are as 5 to 3; but if from. the second and adthird, you subtract 3, the remainders are to each otlier as 11 to 9; required the numbers. A. 37, 25, 21. 14. Divide 180 dollars between three persons, A, B, and C, so that twice A's share plus 80 dollars, three times B's share, plus 40 dollars, and four times C's share plus 20 dollars, may be all equal tto each other. Ans. A $70, B $60, and C $5&0. 15. There are three numbers w-hose sum is 78; 4 of the first is to 4 of the second, as 1 to g; also, 4 of the second is to - of -the third, as 2 to 3; what are the numbers? Ans. 9, 24, and 45. 16. A, B, and C, have a sum of money; A's share exceeds 4 of the shares of B and 0, by 30 dollars; B's share exceeds - of the shares of A and C, by 30 dollars; and C's share exceeds - of the shares of A and B, by 30 dollars; what is the share of each? A-ns. A'~s $ 150, B's $ 120, an d C's 890. 15C1 ~ RAY'S AL.GEBRA, PA.RT ~113ST, 17. If A and B ctan perofbrin a, certaixn w7ork iln i day,s,,AL andi C in n 5 doavs, and B agnd C in 30 d.ayT, in'nhat time could efach do it alone? Ahns A. 20, ] 30, and C: 60 days. I8. -A sumber, expressed ]}y tl -ree igut-e'-', wren divlIed by i tle suma of the figuroes plus 9, give-s qtlotie te. al so,-,ciddTe figure is equal to half ti.r sun orf the fi's and tr -:rd; and, if 9 "'"S ]be added to the n.lum:ber, we lbbtasi s naumber ih td the s gm'afi" r:o'es:n an inverted order; -what is the nuinber? Ans. 450} 19. A fatrmer.i- nixes barley at 258 cents, iuth rye at 36, nmld -%wheat at' 48 cents per lbushel, so tlhat thle wihole is sa00 )buslw ts, ured -vwo:r th,.. cents per bushel. Itlad he put tsrtwice " nitr h t rye, and 10 bu els msore of vIheat, the rhole 7woslld hatve been wor t eu'`f'tet the sanm.l pe'* bushel; how much of each kinid v a1is fliet Ans. iarley'28 rye 20, 1and wiheat 5:' bushelS. A0. t, nd C, sir, a hunrtirng excurusion, klled 96 bIhdr, wvhic:a thiey m.esn to shaere is equily; in o rosder to do thUis A, A-1ho i-aas the.sest, o.Tes to ]3 and C as ('an y a is r l sd t teal; n serxt, B Sives tir Ai. and C as ma.ny ai s they htad afti the fir.st divisics; anatl sy, 0 gives to A. t'rd B as iny asitt' th" 1tiyv b1il had ater thLe seconsd division; it wtats lhen sfiound, {:Rat each iCe l tlre iO sme nunt.ocr Ihow imanry h l"" ea:tch h at Irt l An s.a. 5s, I f, aral (2 i t: C11 P T E ~ V;S TPPIIEJ`_iL8'::NT. TO EiQUiAT ONS OFi't'le P. T B iR. GT B iNER AL riZ -tiAL I 2 155 Az.r f:lt' --—:Et.:U'iT!1'e te f irmet1i —ed -:, A Ahetn tha kr,:towu':_ qutrnrtatites are ra'epresonted, eit her entirel ort p:atl, i y letters. Qu tam iti.ls re'lpesenrted by letter s, are terued "t, en',al es/ rai.ucs -be cause, iby'gi' insg partioirar Values to tihe lettetrs, tie ssFolutitio l otf one prohlesii, r'irntl.is ta ec.'e " l-al soluition to ail oternrs ol tire salme ltid, h e ansiwer to a. p"a oble:us,'iher'trle'inaowsn suanlitri.es atire refi'e — senafed'by letters, is termeaed /sta.fiv':t.; an1d aa' fot'la, expressed. in ordtinaiy a'-a' -.$ furnitis.es a s/,,'?e. ]y tile applia. l ion of Algebra to fthe solution of oj a eral questen's, a grea t l.ttntiber of usefiil and intreetstig tiruths an.d rules mlay e est abished~i We shall now proceed to illustra'te this sibs jeet, 1by a feinw examples. AzT. — I,.. — Le ot it be required'to find a nuimber, wiarsh haeis d-vided by 8, i, by 5, the sunm of'the qtuotiets -wile l' be 5, GENLE.ALIZATIO:N.' 151 Lot == the siumber: then,+' i:o. 3 5 8x-= 1 -/q15 x —-,' 2. Agai-n,'et i;t oE roquilred to find an-otisxr numb-er, Anlich being divided by 4, and by 7, the sum of the quotiunAs Arill be II. 3 procootl.lt, as in'the preceding qu'estion, wi:find thel. num bor to be 2,. Instead, llowever, of solving every example of the sa-ne kind separately, oTe may give a general solution, that' will embrh-ace all the patrticular quoesstions. Thlus 3. Leot it be required to find a unumber, witich besina divided by two givetln nu3bilrs, a sand b, the sius-) of (te quotien mts mas lbe equal to amnothber given number, c. 1% t;w the nube d- 30 l~ ~;l. -t-' Co a o bx-3-ax=-abc (i b -o)x=r-c' The onseor to this que.stion is termed a, fobrula; it shows, that'the required n7Lumber is equal to t~he continued product of at, b, and a, divided by ta he, s-u of (a anid b. Or, it many be expreseas it1 ordinary lan.guage, t hsus: m1 dlilply togeth er I/e i/re five/ 121-m, boers, s:o di.veide l.1,', prod-luc'ly lbe s'. /s' Lthe divisors; the rose/iull w'ill be itoe r~g:iarecad womm~bu. The pupil mays -est the accuracy of this rulo, by sotliing the 5, Find a numbcer, which boi,';g di.ided by —, as sd- the snrs~ of wlloVvin les, tni ve i lhig he sm:-hs0 Tu. i:l:GTi —-l. - The sum. i of 500 dollars is 3to be divided between two person, A o and B, so that aA may ihave 5 dollars ltess tl. an B. Thna. A, $225'", B $275 eo to mnake thsis question general, let it be stated as:Bilor s: It. v'E a W.-t —-t6'V, When are equations ternmed literal? WhOen a re qoan. - t,i.es ter'med general? When is tie anlswer to a protble termled a o rmtla What is ea hifnussa called, when expressed in ordinary language I? 1tD~ Exrinple 3. Wha,;t is the answer to this qg.uestion, expressed in ordinary, aa.ng. uage? RAY'S ALGEBR A PART FIRST. 2. To divide a given numtber, a, into two such parts, that their difference shall be b. Or thus: The sum of'tw o nuimbers is a-, and their differencea; required the numbers. Let zx- the greater numnbe r and y- t~he less. Then x2- =-a And x —-y-b By addition, 2x=at —b ai-j-b a b By subtraction, 2y=a —b y-b a b This formula, when expressed in ordinary language, Igves the FOR FINJDING TWO QUANTIT IES, 11EN Tn5ITEIR SUM AND DtFFR.ENC:. ARE GIVEN. Tobficd thie twealte?, adrd Zhalf ie dZi/js-en6ce to half'.the salnt. Db fnid te less, stuibhtac-t half the di terence J'aoes, af ti e sum. Let the learxner test the accuracy of the r-ul, by finding two numbers, such that their sun shall be ecual to the first nunber in each of the following examples, and their diference equal to the second. 3e Suns 200, diflference 50......... Ans. 125, 75, 4.o um 100, difrence 25....... Ans. 62- 37. 50 Sum 15, differenc e 10....... n. 2-P, 2) 6. Suams 5, difference........ Ans. 3, 2 ART. e -L1..A can perforsn a certain piece of -work in I days, and B in 4 day; in hat time can they both together do it? Ars. 1. days. To make this question general, let it be stated thus 2 A can perform a certain piece of work in.as days, sand B can do it in it d lays; in how many clays can they both together do it? Let x- the manmlber of days in which they can, both do it. Then -= the part of the work wMhich both can do in one dayL Also, if A can do the work in m, days, he can do - part of it in one day. And, if B can do the work in iA days, he can do paLrt of it in one day. Hence, the part of the mwork which both can do in one day, is represented by -+ -, ud eas by m -21 i GE:NRALIZATION, 53, Therefore,.i n a This result, expressed in ordinarylnn guage, gives'the following RULEo.Divide the prooc4t of the numtbers expressing the time in. which each can p e'1:rm thr e worak by their surna; the quotitzZ ezill be the lime'ir w hi t teCy can jointly pefirm it. The qutes;ion can be made inmore general, by expressing it tihus An agent, ic', can produce a certain effect, e, in a tiue, if; another agent,, ecn. produce the same effect, in a timn, e; in iwhat time can they both do it.? Both the result and the rule would be the sa me as thadt already wivenm The folle-ving examples lill illustrate the rule 3. A cistl'ern is filled by one pipe in 6, and by another in 9 hours; in what timn will it be filled by both togetherv A. 3h his 4. One rnan can drink a keg of cider in 5 days, and another i"n 7dai; in hat tine ami both together drink it? AC). d li 2A}T,'- 6,i0.L-Lte it be rtequired to find a rule for dividinrg the gair or loss in a ptnrtnership, or, a(s it is generally termed, fBllows\hia First, taLke a particultr question. 1. IA, B, and C, engatge in: tradec and put in stock in thle following proportions: A put in 3 dollars, as often as B put in 4, and tv often as C i-ut in 5 dollars. Their gainas l amounted to 60 doSlars; required th;. < share of eacih, the.gains being divided in proportion to trhe stoc3k put in Let 3x-.'-. Ms shlare of the gain, then 4x= B's, and 5'-. O' s (See Exam'u te 24, page 126.) Then 34- tx —- x i... --— GO or, 12=60 x: 5 3x n15, A's sha re. 4x —20, B's "s 5x-25, C's " 2. To rmake this question general, suppose A p'ats in'm dollars, as oft'en as -:o puts in nz dolla-rs, and as often as C puts in ~r dollars; and t1hat tihy gainm c dollars. To find the share of each. R Evw ---— 6. what rule do you fond two quantities, -when their sur tnid di i frnce are given? 167. When the times are given, in which each of two a:':ents can produce a certain effect how is the time feutnd in which they c16 n jointvly producet it? 1 54. 2 gRA.Y''S ALG.-EBRA. P.ARt'tT: FIBST. Let tt-he share, of A be denoted by nix, t]:houn lx —- B's, and rax=. -., i —----,,, 0 C L -x /'. CTawh: "' -—:/ 9/,,, ~.g...,_> t::.o.e: -orm T a-83 ie seew._s' lv esanning these forn m ic), iins s"'a hof g-tah hole gaitn, C, 15 antd r, eaczh one's es3ec tiv e priorion, to o'ut l ai0 M e share. of the gain~ f'i c h.s: iepres e efld lossJ ins3ead 9of O:'i-, if:, same'1 solution 3vonn ieC 0i -:P-r e a Tp e, tiSso tioond each ~ a.: c~ tn7 s sit' r e'of the gai,;n oor t:oes,'we ht. t e i t: 0 A l30ow-. _~,~:~.<>..e.9,s,,:,i o''.' osOj'y,/ sto /'''1,/.o UOpttt to, Ahee, a 4 -a.e 3 doears in:; bt:/d 4 ico e a.Cd Bi d to'r'n&rs,,/ obnth, obs /ee,,ett;:e 3 doliarse 4 piont is,.Cit t,. e e,. 1f. i doll, -for 3,e01 onii S'ahcPl 3vt nni-sccs, eve t'ne o" o3.-,sn tO t doost 4l'thi.0 3 - i 9t 1. hroTfore in t.o oit b 39oA1s o bedivide-d si to"e pirolprt in of 12 to IO that- is, lil pTOptti lion t.. th..le pnrodttt of tichV stBrs $y Li e tipces in wt-ch tLh.io W Tere, aineployed, fit ilen "e wi toin.t- i1 in toml nosl11-ip is ronsi edu.ced, se t -e oave the,'oaset'wht polriownw'hy b'ea -a e o or' S'''108tt'o/}itI- 30 1J 6..-tCLL.9 iCtti, titni3303 6130 stei,) 3 A iB, anhd, C nolnha"fid i.n trade A put in 200 dollarBs, d 30r0, anti C 0; they lo ht _ i ) dollars r on at was caoh sa e's' t?. l. ~ n A's loss tij1 B'I tont, ant C's an:a Btv I, 11d<)J. isS, s 1e gi osl loo e i n:-ilorrstip 1. 33,3t1,,n e, the tusins In edivdct the:stto'o Lsi. s.nmovd Co f 1t 3o t i io t? h an' it i i iitno pophierto toi e t 3ire diithme.-t " GENEIIAJLZATION. 155 Since Ute, sums enLgafaed, evir dently ar to eat ch otherl as, 2 3, and 7, we may either use, t hese numberis, or'tho3e rjer]tesentin.g the. stock. 4. In a trading e petd tion Au pt in 200 dollars 3 mo.un.s -, 150 dollars for 5. monulhs, andC 0 (00 dollars feor 8 mlioth- th cyI gained 215 dollars a at was elach man's share of the gain? Aims. A's share Bi0, 3 s $75, "d arse`80' AnT. I.-B.; —1 o T mn A and A, can perform a ce-i tai piece Oef wiork in a d(ays, a. ne C a.8 days, and. B and C in C days in wh.att time could each olne, alone, perform it? and, in what time co ld tihey peri)-lrm it, a11. workino togethe-? Li~, x, Ay, ad a rep rescnt Ithe days in. which. A, dBald Ca ehan re)Oectiow elh do e J.;T7,,k a? /-, e. ifresn t tihe patrts of the wo'rk Ac.ih., B, ann. C eon eac.to in I day. ink ad B can do it in a days, they do' pseri f eit in I dey. l rep 5f:iesen-[so talO r 1arct of tile 0wor5k'which.L and.l -.3 can do:n CJ A..'ne,!. I1 (LR) andl r,0soling i;n a. sllmrar v'sre. hav! ii 2 1........'dcur ml'retn (o:tifron ( 4er Y z — a b C2 1, I'. aOr,, b b — (.)9y ata, iceagin o) fror(eion)s hob:oIx -'~n l Simitar manner, by sebtirrtg i equation Iso, in. the sa mi- 3.'an1er, is foband = by' j l 1f, t I tC DIMS*'nlo"; }x w,+, re esent, ts the, pa2ri adl can dvo h (2) ~ ~ ~ 0 Poo hetd~ —~-ig tc ~d y-,_ _ 056 bRAL'S ALGEBRPA, PAiT FIRST. one day; if w divde I by (&) 2'theb q otient, whie! is 2c will represent the nunmber of days in which all can ab Lac+ be' perform it..ARTT. 1@%-In the solution of questions, it is someetlhmes neoessary to ise general values for particular quantities, to ascertain the relation which they bear to each other; as in the Iollowing problem, If 4 acres pasture, 40 sheep 4 weeks, and 8 acres pasture 56 slheep 0 weeks, how many sheep will 20 acres pasture 50'weeks, the grass growsun unniformly all the time? The chief difficulty in solving this question, consists in ascertining the relation that exists between the originld quantity of grass on an acre, and the growth on each acre in one week. Let m= th quantity on an acre when the past'urrage began, and = — the gros th on I acre in I week; nl and t repr esenting pounds, or any other measure of the quantity of grass. Then 4n= the growth on 1 acre in 4 weeks. And 1te — the growth on 4 acres in 4 weeks. Also, 4isz-I1Ci. the whole amount of grass on 4 acres in 4 If 40 shtep east 4m-O16n in 4 weeks, then 40 sheep eat 4 a O~mT~. n in one week. + 4 n in it.And I sheep zeats -4 in one week. Again, 8z-S i80n= the whole amount of grass on 8 acres inx 1 0 weeks. If 56 sheep eat S2 r*80n in 10 weeks,,Thon 56 sheep ea't -~-io n in I week. Sit 8 n m, its And I sheep eats 00 _ — - in I Aweek. 400 ir'-= -'2 Or, 7m+28=2-=4', +140n E.. I. 12n or i, —jim; hence, the growth on one acre in I week, is equal to of the original quantity on an acre. Then, I sheep, in I week, eats 4Fm "= i — lq — GENERbALIZATIOONT. I 57 And I sheep, in 50 weeks, eats ~X50=- —. 20 acres have an original quantity of grass, denoted by 2'0to The growth of 1 acre in I week being z, in 50 weeksa it will 50n be -- -oa And the gr~owth of 20 ares, in 50 weeks, will be -20=250m. Then 20m —l-250i?, —270m, the whole amount of grass on 20 acres in 8 weeks. >Thn 540in Then 27mnz -- — 08, the number of sheep reqnired. GENE-.A PROtBEMS,, 1. Divide -the mnber a into two parts, so that one of them shall!be n, times the ofther.. A In., a Als,- and 2. Divide the number a into two parts, so that m?, times one part shal be equal to 2t times the other. r and m "- z —w- and 9 3. Divide the nunber a into two parts, so that when fbe -first is multiplied by nz, and the second by,, the sum of the products may be equal to b Ans. b-' and mza- -b anS,. aad - 4. Find a number, which being divided by ma, and by n, the sum of the quotients shall be equal to a.. nsiz 5. Div ide, a into three such parts, that the second shall bte nt., -and the. third t times the first. a, VWtf?, Ant ~s.' am -n-a i. Dividec a into two such parts, that one of them being divided by b, and the other by c, the sum of the quotients shall be equal to d. b(a —cd) c(bd —— a) b-c b ——. What iL; n-lher mustt be added to a and b, so that the sums shall be to each other as s' to t? Ans b it-.b-n 8 What nmL nber L ust be subtracted from c and b, so that theo differences shall be to each o tiher tas nt to na? r-. —nb 9 tL W ha number must be added to a, and subtra ted from b 4at th sum u ay be to the difr-ence as ni to af? Ans...mb —na i/t-nfu JO t LAY'S ALL fiL s1 PAL FiR~t le0t; Aster p avin a w and -- of mny monCey, I. ad a d.c l tars left;; hog Irlany dollars htlo d I at first i?t 3a 1 a1 rWhoat tputisatty is td_~-,at of wrhichl tuhe - par', dim naished btr 1to pa rt, is equal to.? sa S 12.. f cerion.tum ober of persons p' ad p or the use, of a boat., fs-r a pleasuree,seeroson, a ctI5 each; b t, if teare hatd been b persons 0ess, each wa oulid ha. had to pay c cen-ts; how suinn r y persons Were there?9 c C { —OJ i'".Z~'~0i0 1 A: -% si.ie p07a1' ]{}Oi'8011 Cf {]O~S -]'.;O(;O~ a~3.d -.b(' centsl-S I-:. t t-':].ifg <ie u,'', ve. bu i s Col t' n tn p' co Ls a "rn t.o olid h.ave had d cetl, left,; how masny sepersons were th reA cf — o 14, A.:: reuser mix.es it3 att a ceints per bulshel, with tryet a1t 8 cents per buschel, so that h a bus3'el Of tOe ixts'ret:u is W0oriJe. c eTlts; how ainy b usl s of eah o-wi ll bush'els " he mt'ti.l;t:i-' cata'in o? s'Q'2' —- n t' j (c —-') a P c —u 1'5 iA peson ba } owed as ut mo'e " sy hs, ha hnd in Ti ds pIMurse,'and themn spent o cents; hann,!' oorrorweoo as ans auc -s he hd in his purseo, sf'or which he spent ceOni:s, i borro i t ed aid s penCt, in the sitt-le n.oa'uaer, a s-tird anid foutti tie, after wahihi, he tad nothin o.eft; how mu'i m hdl e no0 at nfirst?~ f ns IC' 6. A pe'su' ~so'ts 2 kinds of con,; it takses p ices5 of tie isi', ani d b 1 pie e sI ot th second, to ctke oe de o llat; t.ow is s pieces of each kin musit be taketn, o that c p ieces ay'Piue t t e.o a dolht' s a? ( c):n s. -L-2 Banned o a.. b- ia a.aRIo.:. 4.17 —-T. it soinsetmlnes hatppens in the s'uOttio n om o n sEit cAtion of t'he frist do'reeo, tlhat tse second or'""e hits'her' poe r o the unknow, n qin rottity occurs; butt, in suelt a matin.er, it' t i, is eonasiy rei.-novedi, or sta e to disappeoris, so tthat ~t e. uantion ca.'ihbe solved in. tie osu'al s~~imu. n The sfoilossg aree e.ss'l s a~tions and t problemns belonging to athis clau ss. C] (liv en L2,L+S I xlr —-- 1 O x. to fti t:le vahls:e o-f'. lBy...vdlt:ing etih sidte by i,'ove holve 2C) —' 1- ~'t O 1 fro niwhi ch11, =2., 2~ Goven 3tofs(r3)( *- ( —-t fld \NiEG&A' tlIE SOLIUIONS'. 150 Porfoming tle operatons indic ated, we have PC, ffie, p1t+2 r 2> 6Dti - htix-z'r -lr2?. xA"T- r; 1 — 6-' f6 " x — -C -- Omitting the n. unts ities on each side wrinch are equal,'we have _,Qa........& 07d from which x=-:8o 4. 34 0 — 2 —2- 52. - o, O 0 tns x —i 4, 4Qjza 6>>;) -t;r/2- X,,;N)0;89-4st;3e X o o o o X,, -— S,.-'' ~tx 1 9 {) -a.: -a')- oAri, O. e se t =. -.... e-9 O 2aA z1 5s —25-~13 x 5 Ans. 9. ax.z:'ms. m-.......... 22')~ I2i. 2 ~. 1.0. s -'b-' -,-Y;...... A i -, e a --—....L I_, The e bo.l.win. w lteoin two numbors is 2, and their product is 8 greater ellni she 4scyquare of the ]ess what are the numbers?.A-ns. > 4 and 6o 12. It is reon: edi to dvi' " the number a i.to two such parts, that thle difference of tRheir squares mruay be co ns 2c — - d 13, If a certain b ook contained 5 more pagt, f?,' ri t i 0 more lines on ol;a page,, the numiber of line s would bet- ifn'lreustid 450; hut if it contained 10 pan'es less,'wit]hi 5 lines less on -a.u,'pe, tl -1e whole. nanmber of line-s wmvould be diminished 450. l'bna Guire d the n-timirber of pages, ain the nuimbeer of lines on a paIno. Ans, I2'0 spa<,s, and 4: 0 lives on a p?J EsG A. AT V R 3 0,A U T I 0 N 8 Aier. 1. fl —, tinas been slated a0re -dy (n.rT, 2 ), thet Ihen a quantity has no sign prefixed, the sign pI.s is undierstood aid also (Art. 64), that naI1l numnbers or quatntities are regtarded'as positive, unless they are otherwise dlesiggnated.o.: e.nse, in aI'probtlemso, it is understood, that the irlesults e atre quiled in posiv.ed-w num)bers, I t somtetnimes happens, h-owever, thAD tbe wrime f th e unk.lown q'uant;tyr in the solution of a problemnl, is R'und to be inuistt. u. S.j3ch a result is ternred a nycgalice aolu.iiotn, e shalt.nowt eslmines a questi.on of tOhis kinrdi lo Wh at number must'bhe added to the nur fmber 5, t;.hait th' suri.1 slim. bo equal to 3? e:: t X::-: thie nua' ber.: Thoen bx-I'..... —5 And:S.....5...o 160 RI~AY"S ALGEBRA PAR1T FIRST Now, -2 added to 5, according to the rule for Algebraic Addition, gives a sum equal to 3; thus, 5-1-( —2)=3. The result, -— 2, is said to satisfy the question in an algebraic sense; b. t the problten is evidently impossible in an arizthmeticcd sense, sir ce any positive number added to 5, must increase, instead of div.:inZishing it;.nd this impo8ssibility is shown, by the result being r.egative, instead of positive. Since adding -2, is the same as subtracting +2 (Art. 61), the result is the answer to the followin.3 question: What number must be subtr'acted from 5, that the reai ainder may be equal to 3? Let the question now be made general, thus: WYhat nunber nmust be added to the number a, tht the suml shall be equal to b? Let x= the number. Then anx=-b, AAnd xz —b- a. Now, since a-L-(b — )- - this v-alue of W w-ill alwa, s satisft the question in an algebraic sense. While b is greater than a, the value of x will be r usitive, and, whatever values are given to b and a, the question wil be consistent, and can- be answ-ered in an artthrecticcai sense. Ths, if b —I 0, and a-S8t, then x-=2o But if b becomes less thlan a, the value of x will tb 9negative; and whatever value7 s are given to b and a, the result ol. tained, will satisfy the question in its algebraic, but not in its arithtcgical sense. Thus, if b=z5, a-nd a-8, then x. —-3. Now 8 — ( —3)=5; that is, if we subtract 3 from 8, the remainder is 5. We t l'.us see, thatt when a becomnes greater than b, the question, to be consistent should read, What number must be subtracted from the number a, that the r"emaainder shall be equal to b? eFrom this we see, Ist. 7t'at a negative soluttion indicates some in consi:/t.ency or abs-urdtity, in the question fjrov which ithe equation was derived. 2d. hfeen a negative solution is obtained, -the question, to which it is~ the answer, tmay be so znocified as to be consistent. Let the pupil now read, carefully, the I" OSEv.rxVoTrss oNr AnDnTION AND SUBTRACTION,'" page 43, and then mondify thi.e following questions, so that they shall be consistent and the results true in an arithmetical sense. 2. What number must be subtracted from 20, that the retaindecr shall be 25? (x=-5.) R n v x z wv. —172. What is a negative solution? When is a r.esult said to satisfy a question in a.n algebraic sense? In an arithmetical sense? What, does a nega'tive solution indicato? .DISCCUoSSJoCI OF POBLMGI. 161 3'Whatl num3ber must be ad'1e1 to I1, ftIiit the s.eSItz 1)0i.ng sin.tiplied by 5,,ohe prodct shall ho 1 40? (-.- ) 4, WfJhat numberC is that, 0ofc which 0I0 tohe f.. 3 5, A. fkather, whose ag.e is 45 years, hlas a son, aed; inh.t/i nicany yeatrs' will'the son be a 1s old as his Ifather ('1 -..) D,j CU Sr O ti 0N 0 a3 Ji( R 0 U L E P L I. AxRT. A'34/2 —AVhen a question has been solved in.. -.newd e. atmanner, that is, by represenuti'ng the lKino-en qurntities 102/' IOIelers,'vwe mraty inquire wha;t values the risc.its wilIl hIsLe, l we orh a p)rtieular suppositions,;are made w ith1 regard( to the kIoos. quan,~tities. The determinatioa of theso values, nd the exadlnation of the variousL versuits wnliell'e -'obttei, constitute what is toerromd. the diisustsioa of the probllel,. Tire various forms whichb t'lhe value of the uniknoiwvn quantity may assume, a.re, shown. in the discussion of the followi ng question. 1. After subtraletimng b front. a, what number, multiplied by t]he remaied: er, - ivye a productl eequal to c? Let, xc~:- ~ —-t. et. th; e I.nuI re ri,. C a-b' No t s resbu. inn'V h'ave five dhilErentf formi s, depen.din.:g oni the values 0of, 1), aInd c. N.o 7. —.1 _[ t olowit ors, A denotes nereAy some qulnitity, l3st..t nn o s le{ss thn. a a. This gives p osi voe -values, of t.he 2d.,.Vheebn bi i res g:ater t hanar a. rahtls gives negs'tive v alues, of the form. 3d.'When b is equal to a. TP:his gives values of tire form Ao 4th. WhIere c is 0, and b either grena-ter or less then a. This gives valutes o' tiie firu r. 5th. Whena b is equal to a, ttnd c is equal. to 0. This gives values of the forn -(. We shall exaemine each of these in sucession, L. When b is less than a. i n thi' s case<, a l —-b is positive, and'b' valune of x is. positi lvel To lialbSt.rate'.;.rs e rm, le te ab 8, b=, a rnd c...0, thei x —::4, J:i4t.-vY' wo —..tV When aI noeg..tive s olut. i 18is obt':ieted, lTow temay tih es troi;tl, to mhichnl, tOe'nl'vesre ba lodified l 173. Wisat ttdo you iunlrrstr nd by the dis(mussion of't p )blem? Tit:5 extpression dividcd by a t.. —b, may have 1131VO - h 1)W 2any 17 s I? i." La ll:m t, iA dif Ietren crmtls:t4 162 IAY'S ALGEBRAl PART FIRST. IL, When b is greater than a. In this case, a-b i a negative quantity, and the valtt ue of x will be negative. This evidently should be so, since minus multiplied by minus produces plus; that is, if a — is iz,/nus, x lmust be ninmus, in order that their product shall be equal to c, a positive quantity. To illustrate this case by numbers, let a-2, b —5 and c=12; then, a-b:-3, x —-4, and -3X-4=-l.2 III. When b is equMa to a. In this case z becomes equal to Nr Wc must now inquire, rhat is the value of a fraction when the denominator is zero. Ist. Suppose the denominator 1, then!=c. 2d. Suppose the denominator, then Oc 3d. Suppose the deuooninattor ri then.t I 0 0c; 3d. Suppose the denominator i 0 then c=100C. 4h. Suppose the denomintaor t hen -lOG0c, While the numerator remains the saute, we see, tiat as the denominator decreases, the value of the fraction increases. Hence, if the denominator be less than any assignable quantity, that is 0, the value of the fraction will be greater than any assig-nable quantty, that is, infinitely great. This is designated by the sign m, that is c — cc This is interpreted by saying, that no finite value of z will satisfy the equation; that is, there is no number, which being multiplied by 0, will give a product equal to c. I.V. When c is 0, and b is either greater or less than a,. If we put a — equal to d, then x= —-O, since dXO0=O; that is, when the product is zero, one of the factors must b)e zero.'V. When b=a, a d c:0. c 0 In this case, we have x= — = or xX0O0. Since any quantity multiplied by 0, gives a product; equal to 0, anyfi/nite vagte of x whatever, will satisfy this equation; hence, x is indeterminate. On thlis account, we say that -8 is the symbol of indetermination; that is, the quantity which it represents, has no particular value. R z vyr w.-lt3. When is x of the form -A? When is x of the form -A? < When is x of the r form co Show how the value of a sfiaction increases, as its denominator decreases. What, is the value of a fraetion whose denomilna, for is Jero? Of x wlen i O, e ndt grea ter or les than a? PROBLEM OF TIlE COURIERS. 163 The form 8 somnetimes arises from a part icular supposition, when the terms of a fraction contain a common factor. Thus, if z —: —- and we make b-a, it reduces to a - - 0; bu~t if we a-b a-a cancel the commnon factor, a-b, and then make b-a, we have x=-2a. This shows, that before deciding the value of the unknown quantity to be indeterminate, we must see that this apparent indetermination has not arisen from the existence of a factor, which, by a particular supposition, becomes equal to zero. The discussion of the following problem, which was originally proposed by Clairaut, will serve to illustrate furthcr the preceding principles, and show, that the results of every correct solution, correspond -to the circumstances of the problem. P 0 BLEM O) T E C URI R 8. Two couriers depart at the same time, from two places,.A and 3, distant a miles from each other; the former travels va miles an hour, and the latter,'n miles; where will they mneet? There are two cases of this question. I. When the couriers travel toward each other.,Let P be the point where they meet, A B;.- and a —AB, the distance between the P two places. Let x-A.tP, the distance lwhich the first travels. Then a —— =B P, the distance which the second travels. Then, the distance each travels, divided by the number of miltes traveled in aun hour, will give the number of hours he was traveline Therefore, - the number of hours the first travels And -- the number of hours the second travels. ft But they both travel the same number of hours, therefore a —x am a —x am a a Ist. Suppose ml-n, then x —= —-, and a-.= t; Ihat is, if the couriers travel at the same rate, each travels precisely Is'aif the distance., 164 RAY'S ALG-EBRA&, PART FIRST.,2d. Suppose- -0, tthen z....= f- t; that is, if the second courier rematins at rest, the first travels the vwholle cdistance from A. to B. Both these results are evidently true, and correspond to the ciwr curnistances of the problem.. IL. Whben the coutriers -' r'V-el in theli sames direction. A.s before, let P 1be tie point of A. s -.. P imeetingn, eacth trvetj'ing in thnt, direc- B'til)n, an-d let a=. —SB the distance between the places. - tanee th_- d8ts: the first, t iravels. a —-c B i thle i astce tce t second travels. Then, reasoni Ing a' tile first cans,,wt e have an'i tt?t,:,se.d u —e-.. Sl —— th Si 9t-' lisst. i-f wVe suppose' greater t, ha'n v', the valnue of x will be positive; thiat i.s, the couriers u-il! meet on the rig ht, of B. This evidcletly correspona: ta th cia- ercunstanes of tle probicns. 2d. If w-e suppoeose'5 greater' hjqan W%, ~.6 v:alue of z, aned also.th't of zi-a, t,v netit. Teis eiat a iv e -v, a o' X shows th-t there is some ilncensistency in thie question Xrt. 172). Int deed, whemn 9it is l3ess tlan' S, i1 is evident that tte'icouriers etyan not; iseet, since t rhe f[-aiyd coanrir is s,, travelin -fa ster-h wu n the hinec-h aost. Let ucs nowv isnqItire how the question n be modified, so that [h'e aluei obtained for az shall be co nsistenll f we supe pose tol e d'irect-ion c1 angred in which tie couri ers tr'avel; tlint is, tht tihe. first traveols F' -;cicrwt2s.3 [e ic -— ia'. g"-?c —~-~BP, we hlave, reaswoning' a's'bhsfore, si- ax 05 7;, i I x a — d a'it -, The disbances travreled are non botfi. positive, and the question willt be consistent, if wie retga, d thei couriers, instead ofi traveling itoa ri ass travi-nelg in the oppos- e direct.ion toward P'. The eihnge of sign,'[tlhus indicatin-l a cha' n-ge of direction (Art. 64). 3d. If wre supjpose sm equal to sz, In tias case; as is equa.l to "-and'.i -— a.. 0 f, POSSIrBLE PROBLEKMS. 165 tAsg has been alroady shown (Art. 173), when the( unklnown. qua-tntity takes this fierso, it is not, sa'tJisfies d bry a -ny finite vstue'; or, it is in6finitely -grea,:. Tihs evidnten v corresponds cto the circumnstances of the probslemi foir, if the coui5iers travel at the same rate, the one canz. neer ovorsaske the oth'er. This is some0times otherwise expressed, by staying, they only meet at aan i., t/site dista-nce fronm t1he point of starting. 4th. If we suppose oa-, then _i —=, ) n 3.-a -- 5'1a —.nd~ -, —-,.'When the u nknown quan'ity takes this oirn, it has been showrt alretady, that; its vaste is., This corresponds to the circlnn:s-tances of the piroblem; Lfor, if tiOhe couriers, are n.to distance apart, then will have to travel 0o (i) distance to be toaetlihet. 5th, If we suppose'ai n a n m:;=-e -,.: In this easeL, t a —=, a- nd - — o,' i-.'tl e unknown qutantity takes thlis form, it l- 1s -beel shosn (Arts, 1i3), that it may have ta.y.?f/i/c sc'ltue whast ever. i This- also, evid ently corresponds to -the circumnastan ces of thle pr loblem; for, i the ocouriers are str,o distaise epart38,...... ~ d >,v at thane tfre'-i ta, they will be cics-,'' together; thlat is, at cs.i dihta'ce - o-' nat rCev or nS t he p)oint (Fi S'artilng Lastly''s, if we su) s a — e -ten —'- ta-. t. s, the first, courier tra-vel frisom-. A -) A v), Sort'iYi5i4T tl. seonl at B. I'-1? BAUvo So'i'3 X.2 —— 2o -J>1 5 —--— 2a, and tihe firs travels twice the distance fron..A to- r,')before ovesl(5ai,- 0 thea second. IBo-th res-ults evide.tly (o0 crspnd to thre cisr,st neos of t.he probl em. CAStES OF15 INSSvi51A e 5qUATtVi5 01i' 1 IR.A.a'rT. YL —An etuatnion is-tIers-sid fitssul.natcd t,. when the relation of the quantities whic h it costains, can not be obtained directly from others with awhich it is eompaxred. Thus, the equation are independent of each other, since the one canit not ble obtained fromn the other in a direct ina:ner. R a -v' I w.-.- 1i73. WhsI at is the value of wrhean ib=-r-cs nd c-=O? What is the valiue of a frlaction whose terms sare both Zer? Show, at that is form sometimes arises ifron the existence of a cormlmon- factior, which, by a, particula.r hbypothesis, reduces tso zero, Discuss the pro3llem of tlie Cosuriers/' asnd shlow, that in e-very Ihypoitesis th me solution correspondsts Ito the cireusnsta.snise of the problem. 166 RPAYS ALGEBRA, P ART FlIRST The eOquastions, x —2yl- I 2x+4y= — 22, are not independent of each other, the second being derived directly fron the first, by mutiplyi ng both sides by 2. ART. r " 5....An equation is said to be indeter'mintae, when it can be verified'by different values of the same unknown quantity. Thus, in the equation xz-y=5, by transposing y, we have x=:5+y. If we make y=l, X=6o If we make y =2, x:7, and so on; from which it is evident, that an untlimited number of vnlues may be given to x and y, that will verify the equation. If we have two equations containing three unknown quantities, we may eliminate one of them; this will leave a single e quation, containing two unknown quantities, which, as in the preceding example, wvill be indeterminate. Thus, if we have z-$ 3yn+ zl 0 and x+2y-z- 6, if we eliminate x we have!y+2z= 4, from which y —4-2z. if we ke make z-, y=2, and x=0 —3y —z=3o If we mnake z —,1, y — I, and x:51-. In the same manner, an unlimited number of values of the three unknown quantities nxay be found, tat will verify both equations. Other examples might be given, but these are sufficient to show, that when t/he n tutber of unkznowz n quantliies exceeds the nitumber of sidependent equation., tiie j.roblten is indeterntihmlae. A question is sometimes indeterminate that involves on ly one unnknown quantity; the equation deduced fromn the conditions, being of that class denominated identictal The following is ain examlple. What number is that, of wvhich the: -, diminished by the - is equal to the n:a increased by the eo Let x — the number. 3u 2u. Then -, - -- -. 4 3 )0 30''lear.ing of fractions, 45x —40x —3x-+2x or, 5x=5x, which will be terified by any value of z w vhatever. ARTo 1 6. —The reverse of the preceding case requires to be considered; that is, when the number of equations is greater than the number of unknown quanttiies. Thus, we may have X+ Y=lo (1.) y- z= 4 (2.o) -2x-3y= 5 (3.) Each of these equations being independent of the other two, one of them is unnecessary, since the values of x and y,'which are 7 an 3, may be determined fronm any two of them. When a IMPOSSIBLE PROBLEMS. 167 problen contains more conditions than are necessary for determining the values of the unknown quantities, those that are unnecessary, are termed rned;undant cond.itions. The number of equations may exceed the number of unknown quantities, so -that the vwalues of the unknown quantties shall be incompatible with each other. Thus, if we have X0+ Y= 9 (1.) x+2Qy-=3 (2.) 2x+-3y-21 (3.) The values of z and y, found from equations (1) and (2), are xw5, y- =4; from equations (1) and (3), are x —6, Sy=3; and from equations (2) and (3), are x=3, y=5. From this it is marnifest, that only two of these equations can be true at the same timne. A question that contains only one unknown quantity, is sometimnes impossible. The followinfg is an exasmple What number is that, of which the 4 and - diminishled by 4, is equal to the - isncreased by S? Let x- the nuz mber, then -+ 4 — 8. Olearing of fractions, 3x+2x-24 —5x+4S. by subtracting equals from -each side, 0=72; which shows, tfhat the question is absurd. R EMA r It P.-Problems from which contradictory equations are deduced, are termed srrantonal or'iapossihle. The pupil should bhable able to detect the character of such questions when they occur, in order that his efforts may not be wasted, in an attempt to perform.n an impossibility. A careftul study of the preceding principles, will enable him to do this, so far as equations of the first degree are concerned. AT'r..l oTake the equation ax —cx-b-d, in which a represents the sunX of the positive, and — e the surn of the negative coeifficients of x; b the sunm of the positive, and -d the sum of the inegative kinown quantities. This will evidently express a simple equation involving one unknown quantity, in its mnost general form. This gives (a —-c)xzb-d. Lot a —c=2=, and b-dt-n, we then have snx —n, or x —-. Now, since n divided by s can give but one quotient, we infor that an equagtion of the frst deyree has but one root; that is, i. a sitmple equation involving but one unknown quantity, there is but one value that will verify the equation. iEriEr, w. —4d.'When is an equation termed independent? Give an example. 17i. When is an equation said -to be indeterminate? Give an example. 176. WhnT t arte redutndant conditiom 168~ lI.MA.S AL.,_'C:EBJIA, PAR[iT FIRST, GO it~bIATION' 0 F POTWERS"E XTRACTION OF THiE SQUARE L OOT - R ICALS,' OF TIRftllEJ SEC ONTD DEGR-:EE. Y iVOLUITUT OI R OH FGRM0WAT5OH MI 0OF 1t'0I'R1iE4I.U A-iS. iiS'-T- -ie term p)oowsi is used to denote 1the product' arising' eino multlIying a quantity by itsel4 a certain.:-llul-rber of s:'naes, and the etuanltity -wlh b is m-ultipled by itself, is catll. d the0'00oo of tile poweo Thus a is ca.l d tl~.-. second. power> y of a, B)caus e a is t:e.kTon. titice as a f. actor; and a is talled the seconid root of ao t So, also, a3 is caleRd Pth e I/i.di. p0ower of ea, because a'Xa-\ci=a3, thea quantlity a bei-ng taklen t/rzee timnes as a factor; and at js ct;!ed the ti/rd rioot iif o. The seconld power is g5"enerally al the d the nd the scond root, the squar'e root:..In like.ma eincr, the third power is cealled the cube, -and tihe third root, thi-e cctube root. The fig.ure indileatin the power tio which th e quaentity is to be raised, is called tuhe index,ar or exponenat; it is tbo be writ't-en on thie right, and a ititle hio'kg.e- than the qa'uintyi (See ( br:ticles "33 and 35,).R.f a P, ii:. —-A poe0'sr imay }e otherwise defined thlus: T/ie,.tl poae:' o/'j a quiaitit, is ft/.e 2 1'od'c ( /' / iSzeJrto-'s, Ce/cA e'lrjl.o thi e t cfo/iat'ity; where w, ulla, be aiy nuimber, as 2 3, L-, and so on.. Therefore, fe qn.iy oSrta/t / i 2/y /Jpoffwri Oft qr ait.tf;/iiiti/ b! tLaAtinij i (a1s afiCtiCi' as8,itia.iiy t a.ifm s, taifere acie iiliit,'di' t]he ezxooeint of i/epfoiier' to wAich it is to be r'aised, T'Zhis rlie alone, is s'afifei.jent for every cquestion in the ifolrmation of Ipowers; brt, for the more easy conprlehllision of pup.ls, it is generally preseiteld i. detail, as in the followilng cases. to RAISE A..ONOxXoAL$ To ANY GCi' R N:POWER, a..RT ]...', iei:. it be required t;o raise 2a:b to thee lthird power. Aee.ccording to tte dSe'mition, tihe third power of'/':,b2, will -be the proluct arising fronm taking it {rice time as sa factor. Thus, (02ai2)s-b ('' X2ab/X,'2W1'2X>2X, )r, &20)2a (ba'.2', 0 t,./ }-/ -'e-?aN? e —1-',- a dtbs.dd 2d 8a 6 2 fia-'+Xb", -' c o.-Ii-,s 1''',o In thils example, we see, th.at the codicicnt of [tb.e,_ow is fbund FORMfATION OF POWERS. 169 by raising tile coiffcient,'2, of the root, to the given power; and, that the exponent of each letter is obtained, by multiplying the exponent of the letter in the root, by 3, the index of the required power. ART. 8 @ —With regard to the signs of the difteret powers, there are two cases. First, when the root is posiive; -and second, when the oot is negatdire. Ist. When the root is positive. Since the product of any number of positive factors is always positive, it is evidenlt, that if the root is positive, all the powers will be positive. Thus, +aX-Ht-+a2 +aXq-+aX+ —-t -a3, and so on. 2d. When the root is negative. Let us examine them different powers of a neogative quantity, as -a. — a:= first power, ntega.tive. — X a —=-K a az second power, positive. -aX-aX -a —aS3 — third power, negative. — aX —aX-aX-a —a+a- fourth power, positive. -aX —— aX-aaX X- a —— a= 5 fifth power, negative, From this, we see, that the product of an even unumber of negative factors is positive, and that the product of an odd number of negative factors is negative. Therefore, the even powers of a negative quantity are all positive, and the odd powers are all negative. Hence we have the following FOR RAISING A MONON3'IAL TO ANY GIVEN POWER.L Raise the numeral cofl tcient to the required powver, and,udltiptlly the expconent of each of the letters, by the ex oneat of the power.if th.e nononminal is positive, all the powers ivill be positive; but, if it is negative, altl the even powers will be positive, and all the odd powers negative. E X A3]P L E S. 1. Find the square of 3ac2y3........ A s. 9 a2y6. 2. Find the square of 5bc...... Ans. 25nbic O 3. Find the cube of 2c23......... Ans. 8 xy. 4. Find the square of -ab2.. e.... Ans. alb 4c2. 5. Find the cube of — abe, a...... Ans. —c -svIEw. -177. Show, that in an equation of the first degree, the nn~ known quantity can have but one valueo 178. What does the term power denote? The terum root?, What is the second power of a? WThy? The third power of a.? Why? What is the second power generally called? The eecond root? That is the index or exponent? Where should itbo written? 15 170 ATY'S ALGEBRA P ARIT FIRST. 6. Find the fourth powver of 3ab63C2.....ls 81 a4b 1c 7. Find the fBurth power of —3ab3c'.. Ans. 8la'i1bcs, 8. Find the fifth power of abcd'c..... Ans. a%'5%61cct 9. Find the fifth power of -ab5cd2.. Ans. -— ab5SdWI. 10. Find the sixth power of ab2C3d..... Ans. al2b6csd 11. Find the seventh power of — mn Q.... Ans.n.. —m'L 12. Find the eighth power of — rtn.. e.. o Ans. nn',t* 13. Find the cube of -3at...... Ans. -27a1' 14. Find the cube of — 3r-..... Ans. -27xs4yY% 15. Find the fourth power of 5adxl A.... j 6Ans, 0625t%". 16. Find the cube of -4a. Q. Ans. -64ax't. 17. Find the cube of — 83x'3,.......Ans. — 512.Py i8. Find the seventh power of -2cyz'... Ans — 128z3x)Zle 19. Find the fourth power of 7c',3.... Ans. 2401al(f12 20. Find the fifth power of $3a'x2z'. Ans. -243a1'1y"OD 1f z ART. ~1. CAS IL TO RAISE A POLYNOM-IAL TO ANY' rOWER. Pind tMe producl of the quantity, taken ats a fctor as n-uzy times as there are units ui the extponet of the power. NoT Yer. - This rule, and t.hat in the succeeding articlet follow directly from the definition of a power. BXAMYPL S. 1. Find the square of ax-+y. (avx - cy) (axc+) ya'X2'-2acxy+ey'. Ans, 2, Find the square of i —...... Ans..-2c+xc,'. o Find the square of x-+i,. r.. o. <. Ans. x2+2x Ia, 4. Find the square of ax-cy. Ans. a' xV-2aaccy-c2'y'2 5. Find the square of x2-3y~'.. Ans. 4X 4 —1 -2x'y-+9y% 6. Bind the cube of a+J-x.... Ans., a'+-3a2x —C3aS +x. 7, Find the cube of z —y...... Ans. A-...3x' y 3y-ey+ 8 -yL 8, Find the cube of 2x —1.... Ans. 8'-12 2xt +6x —l. 9. Find the fourth power of c —.v Anse c.'4c'cGt.'-f4ccc'+c 10. Find the sqiuare of a Ib+c. Ans. a' -2ab-+b' +2ac+2bc-{c' 11, Find the square of a-b+c-d. Ans. ad-2ab+b2+2ac-2ad —c'.-2bc+2sbd —2cdA —d 12. Find the cube of 2'-3x+I1. Aris. 8, —36x5 +66x4.63cL' 33x- 9x+ Ic. FORMATION OF POWERS 17.1 ART. 1~6 V.A SE Ii. TO RAISE A FRACTION TO ANY POWER. RULE. Raise both numerator and denominator to Ike'required power by actual multiplication. 1. Find the square of a+b d a+bx +b a2+2abc-bb c —d c-d c2 —2cdP-d-''~2x 4X 2. Find the square of. Ans 4. 3h3 3. Find the scube of.S4-....... An~ 9 - - 4 Find the sqcubare of - -- An 4 O3...... (25. Find the scuare of....... A ns 5. Find the suae oof —4~3 6'4x+4 ~x-2 2"-41tAns. 6. Find the square of.A......ns. 7. Find the cube of 2a(xy) S. Saa 3 —x.y+3xy"t-)1 S. Find the square of (m —n).. Ans. 4(7nc-2 mn 1n) 3 (m-~- Pn" 9(62 (m —I- 2 m -fa+n')' _O I M'I AL THEOREM. ART. 13. —The Binomial Theorem (discovered by Sir Isaac Newton), explains the method of raising the sum or difference of,any two quantities to any given power, by means of certain rela, tions, that are always found to exist between the exponent of the poower and the different parts of the required result. To discover what these relations are, we sha.ll first, by means of multiplication, find the different powers of a binomial, when both terms are positive; and next, when one term is positive, and the other negative. x vi R w.lt9. hin nraising 22abto to the third power, how is the coaficieint of thepower found? H-owis the exponentofeachletter found? 180. When ther root is positive, what is the sign of the different powers? When it is negative? What is the rule for raising a monomial to any given power? 181. What'is the rule for raising a polynomial to.any given power? 182, What is the frule for raising a fraction to anypower? 183. What does the Binomial Theorem explain? 172'RAY'G ALGEBRA, PART FIRST. 1. We ill first raise a+b to the fifth power. a+ b a -+- b + ab+ b a2+-2a b b... second power of a —b, or (as —b)". a - b at-+-2abt- a 6b ab-+ 2acb-2+ b3 a'+3a2b 3a 6b- b3-= A third power of a4-b, or (a6-b)3. a +b a I+3a3b+ 3a262+- a 6b3 + ab+ 3a262-1- 3a bS-+b' a+b s4a4b6+ 6(c-b"+ 4a6- I- ab + a'b+t 4ea'2-I- 6a2b3.+4a6b4b5 a-5a+'b+-I Oa' b "-I Oa b-+5ab6+b -.....'.... (a+b)'. The first letter, as a, is a.lled the leadinzy quantity; a.nd the second letter, as b, thefollowing quantity. We will next raise a —b to the fifth power. a- b a2- ab'-. a b+ b a' —2a 6+ b-...... (a-b)..'..... a- b a' —2a2b —6 a b2 -- ab+ 2a 6- bs a3 —g3ab-+ 3- o 6'...... ( a —b) -3a3aab+.2b.'b a. a ab-f- 3a'b2 — 3a b6+ b6 a —4ab+ 6a- 2b2_ 4a 6+ 64_ -....... (a-b).o a —- b a'-4a'b+ Gta6'- 4a2b'+ abl -- ab+ 4a - 6a2- Gab+4a" —6' a-Wta6 $ Gl6b'_ IOa'J'+ 5 -B =.D o.......(a-b)' FORMATION OF POWhERS. 173 AT. 14~ —-In exanmining eth different parts of which these results consist, there are evidently four things to'be considered. 1st. The numnber of terms of the power. 2d. The signs of the terms. 3d. The exponents of the letters. 4th. The cotcie enls of the terms.'We shall examine these separately. lst. Of the number of terms. By examining either of these examples, we see, that the second power has three terms, the thired power has four terms, the Jburth power has five terims, the yfth power hlas six terms; hence, we infer, that the nunmber of terms in, atny power of a binomial, is 0one greater than the expozent of the power. 2d. Of the signs of the terms. From an examination of the, examples, it is evident, that when bothf termis of the binomial are positive, all the terms?will be positive. When the first term is piositive, and the second negative, all tIe DDnn termis will be POSITIVE, and the EVEN terlms NEGATIVE. No Tri.-By the odd terms are meant the lst 3d, 5th, anci so on; and, by the evest terms, the 2d, 4th, 6th, anld so on. 3d. Of the exponents of the letters. If we omit the coEffieients, the remaining parts of he fifth powers of a-b and a-b, are (a+b)sf3 a..... aeo eo oe -a4b+aibi+aibi+abi+b5. (a-b)5............ a5-a-b+&b5-a —b: % -ab4-b5. An ex'amination of these and the other different powers of aq-b and a-b, shows, that the exponents of the letters are governed by the following laws st. The exponent q the leadida.g letter in the first term, is the same as that of the.power of the binomial; and the exponents of this letter in the other lerm s, decrease by unity. froneJ ei to right, until tfe last term2, whZ7ik does not contain the leading letter. 2d. The exponent of the second, letter in the second tern is onie; and the olher expoSnents of this letter increase, by uenity, jfromi left to right, ufntil te last termi i which lthe exp)onent is the same as that of the powoer of the binomial. 3d. The sum of the exponents of the two letters in any tesw is always the same, and is equal to the power of te binomial,' xBEVi E.- 184. In examining the different powers of a binomial, what four things are to be considered? XWhat is the number of terms in any power of a binomial? Give examples. ~ When both terms of a binomial are positives what are the signs of the terms? When one term is positive, and the other negative, what are the signs of the odd terms? Of thie even terms? W hat is the exponent of the leading letter in the first term? 174 RAY'S ALGEBRA.,.PA.T FIRST. The pupil mnay now employ these principles, in wri ting; the different powers of binomials without the codicients as in the foI lowing examples (x-yj)t. o x tx —3y+ xy2-xy3+-y (x +y),. o x -xy-f y2+xy~ 3+y4+y. (X -j-y) Z` x 5 -V-x'y y2+x f-}-+y53 + ~x3 ~y5 2 ( X-y)6 o o 6 —XSy-+XI:/2__X3yrs-_ X2yXys-.l- yj, (t y) o.s-~Xl- _ Fxy2-XGyU1t3 X x2y'. x-Ly3 — y —.cyl ry Of the coefficients. An inspection of the different powers of (a-l-b) and (a -— b), plainly shows, PIthat /hIe cotftcieent of the first te rm is allways 1; and the coj/{i:cieat of tihesecond term is t/he sam e as that of the power of hec binomial. The law of the succeeding coeficents is not so readily seen; it is, however, as follows: ff 1h/e co6ffcient of any term be mutlttipied byt the exponent of the leading letter, and the product be divided by thee nmus ber of that term f'0o tm e left, ife quotie-nt will be the couij/icienft of the next term. Omitting the cobfficients, the terms of a-kb raised to the sixth power, are a+ b -a -a b2+ ab3~% ab"a+'b5+bt6. The ecohlcients, according to the above principles, are, 655 15X4 20X3 15X2 6,1 2' 3' 4' 5' or, 1, 6 15, 20, 5, 6, 1. H-ence, (a —Hb)6-at+6a5b+l a2b —- a 203bq3+. Satb'- +6ab5+bt_. From this, we see, that the eiofficients of the followinog terms are equal: the first and the la:st; the second firom the firs3t, and the second front the latest; the third friom the first and the third firom the last;, and so on. Hence, it is only necessary to find the coii.icients of half t4he terms, when their number is even, or one more than half, whenl their numiber is odd; the relnxining eodfficients being equal to those already found. EXAMPLES. 1. Raise x-d-y to the third power. Ans.:c?'-3x2y+3xy2-y+ o 2. Raise (x-y) to the fourth power. Ans. x — 4xy-i 6x2y~2-4:xy y4 8 3. Raise m+-n to the fifth power. Ans, mn5++522n q-10 I. O~s2 7n 1 01m"t4 +Son +n RB.vs'w —1384. flow do the exponents of the leading letter decrea.tso from left to right'? What is the exponent of the second letter in the first term? In the second term? ltow do the exponents of the second letter increase from left to right?? To what is the coefficient of the first term equal? FORMAT ION O POWERS. 175 4. Raise x —z to t1he s.ixth power' Ans. x 6xAi-+ 5rG5x%2 —203"2-j 1l5-.... 6x x5y+-o 5. Wthat is the seventh power of at-? Ans.' c6rg'17bal cb2qt3 boab- ]ba So ib5-+TIa+KV 6. Whlat is the eighth power of t — ns. ranm —$8mt ~+- 228eAir-5 3a:.o.nI- +70Omrln- % -5,3o'_m i n28-?'S —8 o17W -Vfa. 7. Find the inlth power of x —. Ans.:. —9xs-{y-36x7;tyIL — 84x0y'3 - 1 26xyT -126xy,5 +' -- 6x'i t- Hx y —-!/ 8. Find the tenth power of a-.-b. Ans. a,?"' l 0Ocab-345aSb2 —. 120atbi'j-2O Oa%4+-252a5b 2. O10a56 +- 1O0b67+-45ai l2bS- 10a.j —bM 0 ART. FS. —The Binomial Theorem inay be used to find the different powers of a binomial, when one or both terms consist of two or more quantities. I. Find the cube of 2x —ac. Let z2x-,n, and ace22n; then 2z —ac2e~tz —no, at, — z2x?-a c'2 I?2.-:dX2' /'2 —6~."4' Substituting these values of tile different powerCs of us and ia, in the equation above, a,nd we have (2x-ac2) 3-Sa —3X4x- Xac2 41-,X2xXac' —.t3( -8x — i 2ac2x26a2cxi G -ac3. g2 Find the cube of S-t-3b. Ans. 8a-36a%+-54ab 2-27tA. 3. Find the fouarth power of mA-2n. Ans. an — 8-Sm-,4Z-24an412-1-322+35nu2+ I O6n 4, Find the third power of 4az-t +3cy Ans. 64ax-t 144a /y- I08 2S.ctc0x2f 3+27c'Lsf 5. Find the fourth power of 2x — 5X Ans. 106z4- J60xaz_-600xkz2-_-000XV d- 25zt. ART. 16~6 —The Binomial Theorem may likewise be used to raise a trinomial or quadrinomial to any power, by reducing it to a binomial by substitution, and then, after this has been raised to the required power, restoring the vidlues of the letters 1. Find the second power of a+b-{-c. Let b+-c-x; then a+b-+c-=a - x (a x)I — a +2+ax+x2~ 2ax =2a(b+c) xA-(b c)2-b2+2-bc-+c' Then (a+b+c)2a.2+2ab+Sac+b2' 2bc-c'. I.- v x w. —184. Of the second term? flow is the cofllicient of -lny other term found? Of what terms asre the cocfficients equal?? 176 RAY'S ALGEBRAB, iTR FIRST. 2. Find the third power of x+y-z. Ans. x+3Sxy -3z Sx-3+Sxy+6yz+3Sxz +y+3yz+S yz2 +z-. 3. Find the seconld power of at-k+c+-d Ans. a'+2ab+b'+2ac+2bc+c-2+2ad-+2bd —2gcd-+d SEXTRACTION OF TtHE SQUARE ROOT.'ETRACTIO OF TE SQUARE ROT OF NUMBtES ART. 8I 7 —TnrE second root, or square root of a number, is that number, which being multiplied by itself will produce the given number~ Thus, 2 is the square root of 4, because 2X2 —4, The process of finding the second root of a given number, is called the extraction of the square rooto ART. IS S.-The first ten numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and their squares are 1, 4, 9, t 6, 25, 36, 49, 64, 81, 100. The numrbers in the first line, are also the square roots of the numbers in the second. W'e see, fsrom this, that the square root of a number between I and 4, is a nnulmber between 1- and 2; the square root of a numnber between 4 and 9, is a number between 2 and 3; the. square root of a number between 9 and 16, is a nmtuber between 3 and 4, and so on. Since the square root of I is 1, and of any number less than 100, is either one figure, or one figure and a fraction, therefore, when7a e 2nu tber of places of figures in a number is not more tha:n T'Wo, the nuzbe.r of )laces of fiyures:.,in the square root will be oE. Again, take the numbers 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, their squares are 100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000. From this we see, that the square root of 100 is ten; and of any nunmber greater than 100, and less than 10000, the square root will be less than 1.00; that is, when thze nnumber q?ilaces qp7 figures iS more than rTwo, and not nmore thaen a'eoum, the ntonber of pla Ces ojfieignres iln the square root. will be Trwo. In the sane manner, it may be shown, that when the number of places of figures in a given ntmber are more than for, and not more than six, the number of places in the square root will be t.chre, and so ol,. Or thus', when the numiber of places of figures EXTrACTION OF THE SQUARE ROOT. 177 in the number is either one or two, there -will be one figure' in the root; when the number of places is either three orj/bulr, there -will be ttwo figures in the root; when the number of places is either.five or six, there will be three figures in the root, and so on. AnPt. 19. - ---— Every numbe3r may be regarded as being composed of tens and unaits~. Thus, 23 consists of 2 tens and 3 units; 256 consists of 25 tens and 6 units. Therefore, if we represent the tens by t, and the units by u, any nunmber will be represented by t+ut, and its square, by the square of t-ut, or (+-Z)2. (t+u) n i2~Qts2 +uir2-t(2+tu)ut h-Ience, thte square of any number is conmposed of the square of /the tens, pluts a quantity, consisting of twice the tens plus the unils, sseLu titied by qthe ivits. Thus, the square of 23, which is equal to 2 tensa and 3 units, i 2 tens squared- =(20)=-400 (2 tens - 3 units) multiplied by 3=(40-Jr-3)X3 1299 529 1. Let it now be required to extract the square root of 529. Since the number consists of three places?' 3 of figures, its root will consist of two places, 40() aceording to the principles in Art. 188; we 209<_ — 40 1213 therefore separate it into two periods, as in 3 the margin. _ 129 Since the square of 2 tens is 400, and of 3 tens, 900, it is evident, that tihe greatest square contained in 500, is tihe square of 2 tens (20)' the square of two tens (20) is 4009; subtracting this from 529, tihe remainder is 129. Now, according to the preceding theorem, this nuamber 129 consists of twice the tens plus the units, mnultiplied by tihe unit s; that is, by thle formula, it is (2t+V-)uo NoWv, the product of the tens by the units can not give a product less than tens; therefore, the u1.it's -figure (9) forms no part of the double product of the tens by the units.' Then, if we divide the remaining figures (12) by the double of the tens, the quotient iltl be the unit's figure, or a figure greater thanl it. pEsIvir w. - 187. What is the square root of a numnber? Give aln exanmple. 188. When a~ number consists of only one figurle, what is the gre-atest number of figures in its square? Give examples. Wlien a number consists of ti-wo places of figures, what is the greatest number of figures in its square? (-ivTe examples. What relation exists between the nulmber of places of figures in anly number, and the number of places in its square? 189. Of what may every number be regarded as being composed? Prove this, and then illustrate it. R178 AY'S ALGEBRA, PART FIRST.'We then double the tens, which maLes 4 (21), and dividing this into 12, get 3 (n) foir a quotient; this is the unit's figure of the root, This uniit's figure (3) is to be added to the double of the tens (40), and the sumu multiplied by the unit's figure. The double of the tens plus the units, is 40+3==43 (2t-it-n); multiplying this ]by 3 (u), the product is 129, which is the double oof the tens plus the units, multiplied by the units. As there is nothing left afer subtracting this from the first remainder, we conclude that 23 is the exact square root of 529..529 23 In squaring the tens, and also in doubling them, it 4 is customary to omit the ciphers, though they are un- 1derstood. Also,the unit's figure is added to the double 1129 of the tens, by mierely writing it in the unit's place. The actual operation is usually performed as in the margin. 2. Let it be required to extracet the square root of 55225. Since this numnber consists of five places of figures, its rootrwill consist of three places, according to the principles in Art. 188; we therefore separa-te it into th/ree periods. 5a22: In performing this operation, we find the squ-are. root of the number 552, on the same princitle s -- in the preceding ex ample. NWe next consider tt dhe i5 23 as so mun:uty tens, and proceed to find the unit's 12. figure (5) in the same manner as in the precedilng 465j2325 example. If oence the 2325 Fron TIrE EXTRACTION OF THr.E SQUARIE sROOT OF nrUOLaE NUMBl3ERS. 1st. sSeparate the tizven number into periods o0 two p laces each, begizning at /the runit's pelce. (The left period will often contain but one figure.) 2d, Find /the greales square in th./e left per'iod, and place its r"oot on, the'rig/ht, after" tthe ~manner of a quotient in divisioni. Subtract the square (' th/e root je'o.t 1 the left i period, ancd to t1/e remainders brint g dowz, the newt period fot a dividend. 3d. Double t/he,'oo t alb'ead/q found, and place it on the n lefZ,/h cc divisor,.Find how masny times the divieor is contained in the divcidend, excelusive of the rtight haeczdfigc.re, and place the jfigure in the root, and aclso on. 1he ig/ht o1 t/he divisor. 4th. 4,iult(ply the divisor thus increased, biy the last figure of the root; subtract the p3rodtcit jifore the dividee6ld, and to the rem ainCder btring down, the'cnext periodfior a new dividend. 5th. Dosuble the whole root already fjund, for a new diciso.r, and contintce the operation as bebre, until all the periods are broughit down, R v i E. w. —-189. Extract the square root of 529, and show the reason for each step, by referring to the formula. EXTRACTI N OF TIHE SQUARE 0OOTL 179 N o rT E — If, in any case, the dividend will not contain the divisor, the right hand figure of the former being omitted, place a zero in t.he root, and a.lso at the right of the divisor, atnd bringo down the next period. AnRT. 9@.-In Division, when the remainder is greater than the divisor, the last quotient figure may be increased by at least 1; but in extracting the square root, the remainder nay sometimzes be greater than the last divisor, while the last figure of the root can not be increased.' To know when any figure may be increased, the pupil must be acquainted with the relation that exists between the squares of two consecutive numnbers. Let, a and act 1 be two consecutive numbers. Then (at l)2=a2 2at 1, is the square of the greater. (a)2-a2 is the square of the less. Their difference is 2a- 1. Ience, the diqerence of tlhe squares of two consecuZive numbers, is equal to twice tte less number, increased by unity. Consequently, when the rema.inder is less than twice the part of the root already found, plus unity, the last figure can not be increased. Extract the square root of the following nunmbers. 1. 4225..... ns, C5. 7. 6 78978j.. Ans. 824. 2. 9409.... Ans. 97. 8, 950625... Ans. 975. 3. 15129..., Ans. 123. 9 363609... Ans. 603. 4. 120409.. Ans. 347. 10. 1525225... Ans. 1235. 5. 28 444.,. Ans. 538. 11. 12099962250 A. 34785. 6. 498436... Ans. 706. 12. 4t12 t5242$16. Ans, 20304. EX.TAsGLCTION OF T'E qfUARIE ROO OFP FPRAC'I ONS. ArT. V9'OL — Since —, therefore, the square root of ~ is A~T.na X9 —Since, —, that is, H/ I ence, when both ter ms q' a frtactio a're pejreet squares, its square rooet will be found, by extractin-q t/,e squcwe root of both te rms. Before extracting the square root of a fraction, it should be reduced to its lowest terms, unless both numerator and denominator are perfect squares. The reason for this, will be seen by the following example. Find the square root of 42. 12 4X3.H]ere, Now, neither 12 nor 27 are perfect squares; 11 v i E W. —. —89 What is the rule for extraeting the square root of numbers? 190, What is the difference between the squares of two consecutive nuambers? When may any figure of the quotient be increased? 80 RAY'S ALGEBRA, PART FIRST. but, by canceling the common factor 3, the fraction becomes 4, of which thie square root is.o When both terms are perfect squares, and contain. a ommnion factor, the reduction may be made either before, or after the square root is extriacted. Thus, - - or, /and - Find the square root of each of the following fractions. 1. o,.... Ans. Y 4. 6 O7.'..... An...... 9 0 000,' 2. An., -;Ans. S- 5. Ans. f-G. 3. t-3 o.- Q 9 e o Ans. vmrmom -.. Ans- 1000 ART. 192* —A number wrhose square root can be exactly ascertained, is termed a pert ec square. Thus, 4, 9, 16, &e., are perfect squares. Comparatively, these numbers are few. A number whose square root can not be exactly ascertained, is termed an >pz22pef/rct sqguare. Thus, 2, 3, 5, 6, &c., are imperfect squares. Since the difference of twIo consecutive square numbers, a2 and 2 —2+-1k-, is 2aH-l-1 therefore, there are always 2ca imperfect squares between them. Tr1hus, between the square of 4(16), and the square of 5((25), there are 8(2a-2X4) imperfect squares. A root which can not be exactly expressed, is called a surd, or irrational. root. Thus /'2 is an irr tional root; it is 1.414+o The sign —, is sometimes placed after an approximate root, to denote that it is less, and the sign -, that it is greater than the true root. It might be supposed, that when the square root of a whole number can not be expressed by a whole number, that it might be found ex-actly equal to somne fraction. We will, therefore, show, that the square root Q/ an, impeJ: ct square, cane no be a /iact'ion. Let c be an iuperfect square, stuch as 2, and if possible, let its square root be equal to a fraction b' which is supposed to be in its lowest terms. Then 4 — and c- by squaring both sides. Now, by supposition, a and1 b have no common factor, therefore, their squares, 2 a nd U2, can have no conn)mon factor, since to square a number, we merely repeat its factors. Consequently, 2r tust be in its lowest ternms, and can not be equal. to a, whole number, Therefore, the eciuation c -,$ is not true; and hence, the suppow sition is fallse upon which it isfoutnded; that is, that 1/c-; therefobe, thfe square root of an inaerfect square can not be a firaction. EXTASCTION OF THE SQUARE ROOT, 081 APPSROIM,0 ATE SQUARE ROOTS. ART. 19. —To illustrate the method of finding the approximate square root of an imperfect square, let it be required to find th e square root of 2 to within 3-.Reducing 2 to a fi'action whose denominator is 9 (-the squnre of 3, the denominator of the fraction -), we have 2 —=.8 Now, the square root of 18 is greater than 4, and ]ess'than 5; therefore, the square root of 9 is greater than -, and less than; therefore, 4 is the square root of 2 to within less than -3 Hence the FOR EXTRACTING TIE SQUARE ROOT OF A W-THOLE NUMBER TO WITtIN A GIVEN FRACTION. ittdlti2ply the given nmber bqy tle sq arce of the denominator of the wi:action whctich determinzes the de-gree of ap.proxizmation; extract the square root of' this producrt to the nearest tnit, and divide the result by the denominator of thef faction. EXF A iPLE 1. Find the square root of 5 to within Ans. 2' 2. Find the square root of 7 to within - Ans. 2 I 3. Find the square root of 15 to within 1 Ans.'3 4. Find the square root of 27 to within 109... s. 5~ 5. Find the square root of 14 to within... ns. 3.7. 6. Find the square root of 15 to within 00 V.... Auns. 3.87. Since the square of 10 is 100, the square of 100, I0000, andso on, the numlbe of ciphers in the square of the denominator of a decimal fraction is equal to twice the nuboer in the denominator itself. Thetrefoe, wu hen the friaction which deterzmines the eee f approi zmation is a decimal, it is merely necessary to add two ciphers jbr cacT decimal place 1equireed; and, ajer extracting the root, to poeint off fronm, the righlt, one place of decimals for each two ciphers added. 7. Find the square root of 2 to six places of decimals. Ans. 1.414213. 8. Find the square root of 5 to five places of decimals. Ans. 2.23606. t'rvi w.-4191.flow is the square root of a fraction found, when both terms are perfect squares?2 1 192. When is a number a perfect square? Give examples. When is a number an imperfect squarea? flow can you determine the number of imperfect squares between any two consecutive,perfect squares? What is a root called, which can not be exactlyy expressed? Prove that the square root of an imperfect square can not be a fraction. 193 Hflow do you find the approximate square root of an imperfect square to witbhin any given fraction? What is the rule, when the fraction whicl determines the degree of approximation, is a decimal? RAt'S ALGEBRA, PART FIRST. 9. Find the square root of IO. 0. Ans. 3.162277+{-. 10. Find the square root of 101. Ans. 10.049875+. 11. Find the square root of 60. o.7.. AnlS 74596+. ART.. 194 —To find the approximate square root of a fraction 1. Let it be required to find the square root of - to within -. 3 3 \/=7__2 1 Now, since the square root of 21 is greater than 4, and less than 5, therefore,, the square root of 2- is greater than r and less than h; hence is the square root of 3 to within less than -A Hence, i we multittpiy the nuvmerator of a fraction by it8s deootsinator, then extract the square root of t/he product to the nearest znit, nand divide the.result by the denominator, the quotient tcill be the square root of the dfraction to within one of its equaltparts. 2, Find the square root of -y4 to within yA.. c. Ans T 3. Find the square root of:7 to within -.., o. us. 4. Find the square root of to within A i A Since any decimal may be written in the form of a fraction having a denoLm inator a perfect square, by adding ciphers to both terbms (thus8,'4-fo%0-4(-o -0 o0 o &c.), therefore, the squ are root may be found, as in the method of approxim ating to the square root of a whole number, by annexing cmiphers to the gtiven decinalt until the n 6um'ber of decimal places sZalt be equal to double the nzum.ber retquired ia2 the'root. Then, aj2ter exgtractig tue iroot, 0poin ting iff fijom the right, the required number of decinal placeso Fi nd the square root 5. Of.6 to six places of decimals8...... Ans. 774596. 6. Of.29 to six places of decimals...... Ans..538516 The square r'oot of a whole number and a decimal, mnay be found in the sane manner. Thus, the square root of 2,5 is the same as the square root of which, carr'tied out to 6 places of decinmals, is 1.581138%, 7. Find the square root of 10.76 to six places of decimals. Ans. 3.280243. 8, Find the square root of 1.1025...... ns. Ans,.05 When the denominator of a fraction is a perfcit square, its square root may be found by extracting the square root of the umerator to as f dec as manay placeof decinls I8 are required, and diriding the result by the square root of the denominator. Or, by reducing the fraction to a decimal, and then extraceting its square: ~ x w w..-194 fl How do you find the approximate:squaxe root of a f.L a tion to Within one of tse equatl parts of the denominator? 1How do you extsact the squore root of a decimal? otow do you extract the squa-re root of a fraction, when both ternms are not perfect squtres? EXTRACTION OF THIE SQUIRKE ROOT. 18 oot, Y hen t he denoominiator of the fraction is not a perfect square, the latter method should be used. 9. Find the square root of to five places of decimals. V~34.73205+, l/4/-2, z/z --- 1 +3`' r5 o86602N-1i Or,> t.75' and -Vg75=-86602+.!0o Find the square root of 3'.... Ans. 1.795054.+ 11. Find the scuar e root of o..... Ans..661437+o 12. Find the square root of 3,.,... Ans. 1.802'775 13. Find the square root of 5-,.. Ans. 2.426703-t. 14. Find the square root of....n.. A 37796415, Find the square root of - Ans..935414+.!6. Find the square root of'2.. A... ns. 1.527525+i EXT RACt ION OF THEO SETUREU ROOT OFf NOM L. ART. 1 t5.-Fromn' the principles in Art. 179, it is evident, that in order to square a monomial, wNe must square its cohfiicient, and, multiply the exponent of each letter by 2. Thus, ( 3abt)2z 3-tab Kla3abb"., Therefore,'H/9ab' —c:3ab I-ence, the RULE FO. EXTRlACTING TIEM SQUARE ROOT OF A O/'NOiIALo.xtract the square root of 1'the co#cient, and divide the exponent of each letter by 2..Since +a>X+-a=+&a', and caX-az..-a2, Therefore /a2-.+a, or -— a. Hence, the square root of any positive quantity is either plts or rminus.'this is generally expressed, by writing the double sign before the square root. Thus, I/4tai2s~a vwhich is read, plus or tni'nzs 2a. If a monomial is negative, the extraction of the square root is impossible, since the square of any quantity, either positive or negative, is necessarily positive. Th-us, -/V-9, -— 4a', -V -b, are algebraic symbols, which indicate impossible operations. Such expressions are termed imainzaWr quantities. Tey occur, in at tempting to find the value of the unknown quantity in an equation of the second degree, where some absurdity or impossibility exist. in the equation, or in the problem from which it was derived. See Art. 218. R1t I,: w-V195. Itfow do we find the square of a monotmial How, then, do wve fid the square root of af monomial? What is tie Sign of the square root of any positive quantity? Why is the extractioe of; the square,root of a. negati've monomial impossible? Give examples f algebraic symbOl that indicate impossible operations, What arte t;hey tr'.n:n.: d? Under what e ircumstanees do they occur? 184 RAY'S ALGEBRA, PART FIRST. Find the square root of each of the following monomiials. 1. 4ax 2.... Ans. ~2ax. 5. 1G6mnt2,y6.. Ans.:4,mn2y. 2. 9x2y... Ans.:I3xy. 6. 49a2bc8.s A ius.:~:,7ab2ct 3. 2f5a2bcw-. Ans. i5abc. 7. 625x2. z. Ans. i~25.Z4. 36a-bt x. o Ans. ~ia'2bx 8. 11 56a'2xt6. A us. is 34cax t. S( )'a a a2 la' aa 8iince ( ) -y',/, therefore, q i; that is, to find the square root of a tonom iat fractiot, extr.act the sqt are root qf both terms. 4a2 2a 9. Find the square root of t.. Ans.:ly_ 4xy'2 1l0. Find the square root of Ans. i+ — EX'.ACTIONi OF THE iQUA-1E 5 A OOT OF PL0 0 ONM'ILtLS. ART. 196. —In order to deduce a rule, for extractimng the square root of a polynonmial, let us first examine the relation that exists between the several terms of any quantity and its square. ( 2_t___b)' t2 2 2+(2ad-h (a+ - b — )'-a L 2a+ -2 t2b (a+ ) ( (a-t-4 i a at Fa-b r2ac-j —7 tctc-o+ aVt -tA (2a+2t i-c>)c (ab et c 2c )' at' t4t+25ac —2tc+c' 2ad+2'td — 2cd-J-dli,ai= ( — b2a ) +(a+2+b+c) e+(2as 2b+2c+d)d. Or, by. calling rthe successive terms of a polynominal v, i, r'", ",' we shi,1II have (r+r' +t"+ /i)2=,r2- (2tr+)r -1f (2'r -r-2f+/')?' ~- (2s' + r'/+ 2r/-2" ")'/0 I-n this formulIa, -r, r', r /t, may represent any algebraic quantites whatever, either whole or fractional, positive or negattive. HIence, we see, that the square of any polynomial is formed according to the following law: The square of any v polynomial is eqtal to the square of the first term-pihts twice the first term, plus the second, multipitied by the second-plus twice the first and secondci terms, plus thethird, Inultipliect by the third — plus twice the first, second, and thirzd terms, ptlus the fourth, nultiplied by the fourth, and so onm. Ienee, by reversing the operation, we h.lve the RULE, roR ExTRACTING THE SQUAR.E ROOT OF A POLYOMIAL. I st. Arr.ange the polynomial with reference to a certain letter; then ofind the first term of the root, by extracting tbhe square root of the.first term qf the polynomial; place the result otb tfe eight, aed ub.tract its stquarefi'om, the given quantity. EXTRACTION OF THE SQUtARE ROOT. 185 2d. Divide the first te>rm of the r-emain.der, by double the part ofthe root already jbucnd, atnd annex the.result both. to the root hiand the divisoro. Atletiply thze divi.Sor thius iZncreased, by the second teirm of the root, antd subtrac the produczt4from the remainder~ 3d. Double the terms qi' the root already fbund, for a partial divisor, and divide the fi'rst term of the remainder, by ]the first term of the dievior, and annex the restdlt bot to fthe root and t2he partial divisonr, Mi7fuPlytp the divisor thzs creased, by the third tervz ofq the root, and subtract the productfrom fthe last remiainder. Then. proceed in a similar manner, to fitd the other tern'is. R a nr x K.-'In th'e course of the operations on any example, when we find a. remainder, of which the first term is not exactly divisible by double the first term of the root, we ma-y conclude that the polynomial is not a perfect square. EXAMPLE S. 1, Find the square root of r2-2r+ +/tr 2-{+2rr' -f 2irts -'r"+r"2t r,'+2rr'~/r+2rr'+2/r"+s:'21r++r", root. t2dr- i-r.' rr'-/.' 2r+2;r'+/' ~ 2r r"it+2r'r/'+r"2 The square root of the first terL is r, which we write as the first term of the root. We next subtract the square of r frotn the given polynomial, and dividing the first term of the remainder 2rfr, by 2fr, the double of the first term of the root, the quotient is r, the second term of the root. We next place r' in the root, and also in the divisor, and multiply the divisor thus increased, by r', and subtract the product from the first renmainder. We then double the terms sr-F.', of the root already found, fbr a partial divisor, and find that the quotient of 2ir, the first term of the remainder, divided by 2fr, the first term of the divisor, is s', the third term of the root. Completing the divisor, multiplying by H', and subtracting, we find there is nothing left. Ni o T a -The first remainder consists of all the terms after s'2, and the second, of all after r2". It is useless to bring down more terms than have corresponding terms in -the quantity to be subtracted. nRs, vn: w.-1-96. What is the square of acc+? Of a-+1-bj-c? Of a —b -J+c-'d O? O 2fr+r'+s"- +-r"'? What nmy s,', &e., represent? According to w-hat law is the square of any polynomial formed? By reversing this latw, what rule do we obtain, for extracting the square root of a poly-, nomial? When may we conclude t;hat a polynomial is not a perfect squre? 16{ E18 KRRA"TY'S ALGEBRA, - PART FIRSTE 2. Find the square root of the polynomial 25x'2y2 —24x?/3. — I2x3y ~+4xa- 1-6yth Arrangingl the polynomial with reference to x, we have 4x-4-12x yy-fL25x2'i-24xt'1- -y' [2x2-gxy +4fy/, root. -4:Z ag 9:, x2y It is easily seen, that the operation is analogous to th-at of cxtrsaetigng the square root of wsnole numibers. Finid the sqnave root of the, felioivrug polynomials. 3. xlq0xS4 l.............B A my 4+. 6 4x -12x-$1'-9 y...........Alins. 2x —,y. x —3y -l2.. 9 s c x —4. 41 4x-4x 1 6.19 9.x........As..- T. ) 9y42/_ t1293 Ft2 + 2%+., ns y Tq-la..11 1' H-4- is2z Ox 4t 224&2-34-ibx is easily- seen. tha~ th opA —Tati o a 3 n t i' - trathin g th:e Csquareot Cofi oe axbr. a 4, 14. x"?2/ -l 6x I y-'- g...........Ans. Iyx —4. Any. 4 h 4 lge foltwin ars ill.. e....Ans. exsl. b, it becomes 0 2the squarex of adb; a oi subftratoS fronit ab, Ait beeoes A the sq 1uare of au-bS12. i order that a t5rinoi, l r;- be 0 erfec -O as r tOhe tS o extlem, terms mustll b perin t sark r es, aiend the iwddle tousel ti soublt p-droic iof te squa be Poots of the etre;me squas of ance, extomracl t s th e sq arre boots of the to exee terms, an t/s bi it becomes s the squoiare os, ac —brd.i as tect n iliS us 2d In order that a trino binomialy be a perfect squi2 ithe two extreme terms must be perfect squares, and the middle term the sqiuare? W ~hen a trinomial is a perfect squat, how may; it sq:u.:e oct le fotn dl Give an x oa iple~. LADICALS 1 OF T1fHE SECON1) D)EG:tCEEE] 18, Thuas, 1c -- 2- c —4-o v is a perfect square, silnce i/'k t 2 V9I8/c-3O, and -2aX. —-3ScX- I _12ac:. But 9x2-i- rUxyJ 6 iyt`, is not a, perfect square; s ince 1/9x2'-3v, V1fy2- 4?y, and 3; XI4-)2>.2/, xyo whtih is not ecqual to the mniddle t.erm l 1Ixy. RADICALS OF THE SECOND DEGRE E. AnrT. 19V E-Fa.o the rule Art. 195, it is evident, that.,heta a onto.0'miar is ta perfect squa.re, its v3,numeralZ coj'{ficient is a pe-jhct sqnare, a.n'd the exwonent' of eac/, letter is exactly div;isible by 2 Thus, 4c, is a perfect square, while 5a3 is not a perfect square, because the coefficient, 5, is not a perfect squiare, and the exponent, 3, is not exactly divisible by 2. When the exact division of the exponent can not be performed, it may be indicated, by writing the divisor under it, in the forms of a fraction,. Thus, 1/a3 may be written at:' Since a is the same as a' the square root of a lma:y be exTpressted thus, a, 17For this reason, the fractional exponent, -, is used to indicate the extraction of the square roo30 Thus, aJ /-2 atx — j., and (a2 I-2ax-l —2';-, also 1/4 -and 4-, indicate the same opieration: the radical sign, }/, and the fractional exponent,, being reg&arded as equivalent. Quantities of w'hich the square root can not be exact.ly X scertained, are term1ed radicaCls of. 7he second der/oee. They are also called, irratioatccl qutantities, or surcls. Such are the qu.antities,//a, /2, at/b, and 51/3. Or, as they may be otherwise written, a-'N, 2t, abe:t and 5(3g)- The quantity which stands before the radical sign, is called the co;eflientg of the radical. Thus, in the expressions' a-/b, and 3S/5, the quantities a and 3 are called coifficients. Two radicals are said to be simtilar, whean the quantities under the radical sign are the same in both. Thus, 31/2 and 7V1/2 are similar radicals; so, also, are bl/a and 2ct/a. Two Iradicals that are not sinilar, may frequently become so, by simplification. This gives rise to:tI v -.2 w — 19S8. When is a, monomial a perfect square? Give an example, fHow may the square root of a quantity be expressed, witheot the radical sign? What acre radicals of the second degree? What are radicals otherwise called? What is the cofnficient of a radical? When are iwo radiea, ls suinilTr? 188 RAYS ALGETBRA~ PART FIRST. REDUCTION OF RADICAL OF iSE SECOND DE REE. AmRT. 199,-Reductimon of radicals of the second deree, cone sists in changing the form of the quantities without altering their value. It is founded on the following principle. The square root of the p odutc qf hco or more factors, is equal to the produtct of the square oots of those jactors. That is, j/ab —/aX,/b; wrhich is thus proved: (V.b)2=ab and ( /a) /b)7.-V/X/bXV/ aXt/b/VaX/Xa/'bX b —ab. Hence, /oa b and i/caX, /b are equal to each other, since the sqnuare of each is equal to ab. From this principle, we have /36=i/4X9- >X3u, -1/ 44 -— 9X 16=3X4. Any radical of the second degree, can be reduced to a simpler.form when it can be separated into factors, one of which is a perfeet square. Thus, /12=~/4X3= /4XYv3-./3=/3 V/QToac j/9aieX'- 3a ai-r/ ai Xi/3 ac2/v3a From the preceding illustrations, we derive the It U L E8, FOR REDUCING A RADICACL OF TIE SECOND DEnGuREEr TO ITS Sit CLEST FORIm Ist. Separate the quantity tO be reduced, into hvo parts, one of t;,which shalt contain all the Jhetors that are perifect squares, anid the other the remqzini.g Jfitors. 2d. Extract the square suoot of Ute pati that is a pe:fect sctuare, anzd prefix it. as a coijcienzt, to the other part placed under the radical sign. To determine if anuy quantity contains a numeral fator that is a perfect square, ascertain if it is exactly divisible by either of the perfect squares, 4, 9, 16, 25, 36, 49 64, S, 8100, 121, 144, &c. If not thus divisibte, it contains no factor that is a perfect square, and the numerical factor can not be reduced. IReduce each of the following ra dicals to its simplest formn 1. 1/8a. Ans. al,2. 9. "/32abi<c Ans. 4a'3bc/2. 2. l/ a( Ans. 2as 3a. 1. 0/, 404aibit. AA. 2abc/1 / Obc. 3. /1a6db. Ans. 4al/ab. 11. i/44aibe, A. 2abV/I labc. 4. _15a417~. A. 3_abc/2Qbc.- 12. -/4Sa4b%4. Ans. 3a'bic'i5. 5. /'206%c. A. 2abc VI/5abc. 13. -/48ab6cL A. 4a'bici2/33. 6. 3-2f4aV27 Ans. 6aie/6B 14. /75aa3ce. A. 5abtbc/3abc. 7. 4/27a%, A. 12aci/8ac. 15. I/128ta"i". Ao Sa 3b8c1/2. 8. 7/28Saa62. A. 14atc-/a/7a-. 16, i/243ab2ic,.An 9ab,' 3aS RADICALS OF THE SECOND DEGREE, 189 In a similar manner, polynomials may sometimes be simplified. Thus, V {2t&-Q 4b+2ab2t)zv /2a(a2-2abb2)(a....-b) t2a. A fractional radical of the second degree may be reduced to its simplest form, by -the same rule, by first multiplying both terms by any quantity thait will render the denominator a perfect square; separating the ftaction into two factors, one of which is a perfect square, then extracting the squ.are root of the square factor, n4d placing it before the other frotor placed under the radical sign. 17. Reduce I/T to its simplest form. V>V 2X V-/6- VtX-GwXV/6t1t d~ Ans. 3'2, Reduce the followring fractional radictals to their simplest forms 1S. V A 1ns. - /. 22. 9 / I fi Anso 4 / 3o I9. u ij,14. 23. 5t/ o Ans. jV/. 2I o f 3o nse 4 0tv 23 / _6,/ 1 0: I- -- - I/ 21. Vjrs Ans. -v2 25 7A / %3. Ans.8 V. Since a=l/at, and 92/3- /4XV3 — 1/4 X3-V 12, it is obvious, that any quantity rmay be reduced to the form of a radical of the second degree, by squa:ring it, and placing it under the radical sign. By the saure principle, the codfficient of a rtadical may be passed under the radical sign. 26. Reduce 5 to the form of a radical of the second degree. Ans. V25. 27, Reduce 2c to thfe form of a radical of the second degree. 28o. Expess the quantity 3/5, entirely under the radi,. l1 Ans~ 1/45. 29. Pass the coMffiie nt of the quantity 3c2c', un der the rtaJdical. Ans./ l 8c3. 30. Pass the coi~ficient of the quant;ty 5y/3 under t.he r-adical. An s, // DA DTION OF ADOAS O: F li THE 8 COlN D DEGREEi raR. 20@0-1. What is the sunm of 3-t/2 nd a5-/27 It is evident, that 3 times and 5 times any certain quantity must make 8 times th at quantity, therefore 3V2+5V28 5=V2. In the s:ame manner, VA V - V —,/s /2m2 /23X —3/2 Wa. What, is the sumn- of 2/3 and 51/7i Since dissimilar quantities can not be collected into one sut, we ean only add these expressions by placing th e sign of add ition betw een theem. that is, the suam of V2/3 and 5V/77= 23v4.'V/7,1. 10O RkiY'S ALGEBRA, PAR T FIRST..feni.e, the R,ULLE, FOR. TRE ADDITION OF RADICALS, OF THIE SECOND:DEGR'Ee 1 st. Redfuce th.e radicalZs to their simplest Jbrnm. 2d. T2en, if the radicals are similar, reUfix the smn of their co;if cieznts: tho e common ra ical; but, jf they are'not similar, connect them by the3ir proper. signs. Find the sum of the radicals in each of the following examples. 3. V/8 and- 1/. 18........... Ans. 5 2r~ 4. t/12 and V,27. o o An 51/3. 5. 1/20 nd 1/8.......o.... o0. Ans. 6/5. 76 V V/8,- a/3 ad, a........... Ans. 11./6. 8. +P, /490, and -t/250.......... Ans. 101/0. X _,. V/ and j /...s, 9.j/ I/-!~&',ns. -i.. 1/ 14. i/ and 18.I...........a - 1.3 2 /' and I/ 1 8. A Yi 2. 5 OO 14. 2'-/- an3d 3 2./i.,,......... ns. 7 16. 3/.' a"nd 7/' Ans. -','U/62. ]8. Find the sum of /'( 2a —4ac6c+2ac') and,/(2a. ~4cdc- ac'). A ns. a,, //2a. 19. Find ta.e sum of,/aKi x+jvax2-?'+x/'(a- +x)it ) Ans. (1-q-a —2x)/1aq-Lx. ~UTCTP~fUh I -o'Ttr~iicDN)lw/- Lm SUBTSR'IACT f JION @P OFRADI{CALS OF WTHE SECON~D DEGiREE ART. 2@.L —1. Take 3F,/2 from 5%/Q. It is evident that 5 times any quantity minus 3 tim-es the quantity will be equal to 2 times the quantity, therefore 5i 2 —3i/2zi/2, In the sam e manner, /S8- -/2 V/ —2 V/ —0 V/?', qgl:esVlw. — 109. In what does reduction of radicals of the second degree eonsist? On what principle is it founded? Prov-e this principle. What is the rule for the rednetion of a radical of the second degree to its simplest form1? How do you determine if any quanltity contains a numerical factor that is a perfect squaret? }Iow may a fractional radical of the second degree be reduced to its simplest form? 200. What is the rule for the addition of dieottls of' the secon-d egree? RADICALS OF TEE SECOND DEGREE. 191 If the radicals are dissimilar, it is obvious that their difference can only be indicated. Thus, if it be required to take 3-/a from 51/b, the differenee would be expressed by 5V/b —31/ao Froum these illustrations, we derive the RULE, r-o THE SUBTRACTION OF RADICAL OF mcAs THE SECOND DEoGRE. 1 st. Reduce the radicals to th.eir simplestfomr m; then subtract their cotfilcients, and prefix the d'ifetrence to the comnnon raadical. 2d. If the radicas are not similtar, indicate their difference by the proper sin.o EXAM PLE S. 2 i/1 1 8 y.......... ns.1 21/2. 3 /'45a -- 5a a............ Ans. o2ay5 4. 54b —.............. Ans. 2/b. 5. i/ 1 2a'-1/S/28a'c........ ns. 2acO7. O. 1/27b 6 — V12 b1......,.. Ans. bc/3 bc. 7. /36a- 4 — a............. Ans. 4a2Va. 8. V/49abtc-Vdl.abc1..........Ans. 2bc-/ab. 9. V1 6()abc- / 10Oat.3c,....... Ans. 3abt1ab8c 10. 5a!/2i7- 3aV/48....... e.. Ans. 3at/3. 4 3 12 / / -- /...... Ans. /30. 1. Frotm V4a22x take ae/ oe...... s. (a- ax) /x. 17. From V'3Mix-t-6mnx+3On)x take O3m2x- nmnx -- 3 ncb:X.1 Ans. /3 MXULTSQPLICA'II9tiON Or RADICALS OF THE SECN iD DEGRt a AnT. r2, —Sinee Vab-=t/aX, /b, thereforfore -/VaXV=/-ab. See Art. 199. Also, aY bXcd/al-=XcX Vb-X V/d-=acVbd. From which we hayve the'O0 THE MtUTIP LICATIoN OF RADICALS OF THE SECOND DECGEeR It. M1utip/ly the quantities unde7r thfe radical stig togetZiei, and place the result under the radical, 2d. If the radicas have co'fficients,.place their producxt as a cot!f%' ficient efbre the r adical sign. 192 RAY'S ALGEBRA, PART FIRST. EX A M P L ES. 1. Find the product of /6 and 1/8. j/6X /83 V4 v/1/ 6 X3>= 4j/3 o Ans. 2. Find the product of 2/T14 and 3t/2. 2-/14X3/-2= —6'i28-6/4XY6X2f/7l== V2-/7 Ans. 3. Find the product of /8 and -t /2......Ans. 4. 4. Find the product of 2-/Va and 31/a...... Ans. 6a. 5. Find the product of i/27 and 1/3...... Ans. 9. 6. Find the product of 3i/2 and 2i/3...... Ans. 6&V/6G 7. Find the product of 31/3 and 2-/3....... ns. 18. S. Find the product of 1/6 and -/I5,.0.. Ans. 3i/10. 9. Find the product of 2i/15 and 3i/35 An. Ans. 30/21 10. Find the product of i/3abc and 1/abe.... Ans. a2bOe. 11. Find the product of t/s and /..... Ans8; - 12. Find the product of an i/ Ans4 1Ai/5. a 3a 13. Find the productt of 2 and 34ft. Ans. -c -/ When twvo polynomials contain radicals of the second degree, they may hbe multiplied together, in the same mlanner as in multiplication of polynomials, Arto 72, attending, at the same timne, to the directions contained in the preceding rule. 14. Find the product of 2+i1/2 and 2-lf/2-.....Ans. 2 15. Find the product of 1+1/2 and 1-iF/2... Ans.-l. 16o Find'the product of /zx+2 by I/x-2...Ans. i/x2-4. 17. Find the product of i/a+x by l/a+x... Ans. a-+x. 18. Find the product of i/abd-bx by /ab —bx. A. - a bt —bt, 19. Find the product of I /x+2 by /z-+3. Ans. i/x'+5X-i 6. Perform the operations indicated in the following eamnples. 20. (c1/a+d-b/b)X(c/Va-t d Vb). o...Ans. c2a-d-c2b. 21. (7-v+2/6)X (9-5i/6)........ Ans, 3-17i/6, 22. (,i/a-t-tx-Fa-x)(p1/aqf-x —/a-x-)...... rtns. 2x. 23. (x+z-2/aS+a)(x2 —2/ax-a)..... Ans. xa-2ax+a'. 24. (x2 —xi/ +)l (a'x/2 —+1)o.,....., Ans. x+l. DIVISION OF RiADICALS OF TH'E SECOINTD DEGREE. ARnT. 2@*t —Sinee Division is the reverse of TMultiplication, and since l/ b —-Xi /~1c, therefore -Vab-' - -e/a'. Vb] a, -n v i E w.-201. What is the rule for the subtraction of radicais of the second degree? 202. What is tile rule for the multiplication of radicals of the second degree? On what principle does it depend? RADICALS OF THE SECOOND DEGREEC 19E Also, since av/'bXclc/d=acctbd, therefore acj/bT —aV'b-/ ~ _c (bcrelCd.i ence, the RULE, FOR TIIE D IVSSION OF RADICALS OF T'lE SECOND.DGcREEJ, I St. Finzd the quotient, f the parts under the radical, td place it zunder /the common radlical. 24d. If the radicals have cotj'.cients, divide i7he coff/iendt sf Me dividend bgy that of t/e divisor, and prefi the result to the commnts radical. Nor1 Tr..-When a radical quantity has no coefficient prefixed, its co'fflient is understood to be 1. Thus, }/2 is the same as 1t/2. Se Ar't. 32. I. Divide 8-/2 by 2 /o. 81/72 s 2, Divide 1/54 by /6.............. s. 3 3. Divide 6-/54 by 3r/27...........Ansu 2/2. 4. Divide 6d/28 by 2 7 O......... Anus. 6. 5. Divide /160 by -/8-......... A ns. 2-f/K5 6. Divide 15/378 by 5 /6... Ans. 9/7 7. Divide /act by l/aa............ Ans. a. 8. Divide abl/abA by /....a..... Ans, a"b. 9, Divide a by /a........... As. Va., 10. Divide a/b by c1/&. An s or o/ b' [a 62., DivAVtby/.. A, orVi 32, Divide a/ ty ~m... An I,p /6 14. Divide. / y it o.r.... nb. 4, 15. Divide by /.-]A,/,. 6. s Dby V2+ k..... Ans. T ART. -gi4 To reduce a faction -whose denoinao r r s er i.er a monomial or a binomial containing oradicals of the second 4segxeo to an ecqu ivalent fraction having a rational denomnisator. RE. ~vw. -2203. What is the rule for the division of radicals o.f th. second degree On wht principle does it' depend? 194 RAY'S ALGEBRA,. PART FURST. When the frac'tion is of the form A, if we multiply both terms /b by Ti/b, the denominator will become rational. Thus, a aXI/b a_1/b i/b ik/bXi /b b Since the s-um of two quantities, multiplied by their difibrence, is equal to the difference of their squares; if the fraction is of the form a and we mnultiply both terms by b —/c, the denominstor will be made rational, since it will be b2 —c. Thus, a, b-/e ab —a/e b —c b+-l/c b-i/c b For the same reason, if the denominator is b-i/c, the multiplier will be b+i/c- If the denominator is i/b+ic, the multiplier will be /b. —/c; and, if the denominator is /b-/ec, the multiplier will be t/b —/c. These different forms may be embi aced in the folliosn EN&ER AL RULE, If the denominator is a monomial, nmulticpy both terms by th!e radcicalt quantity; but, if iJ t is a binomial, musltiply both terms by the given bitnoial with the sign of faone of its termse changed, and the denominator will be rational. Reduce the following fractions to equivalent; fractions, hatvin~g rational denomninatorso Ass i =1> 2. 4 -3 Ans. yy(b+i/J>. _2/2''2. -Asns. 3 i 5A. 5 / i/+ —, _i 2 l Sg~i3 3 6/3 Rg n, a s,.-The utsstility of these trasforlmations consists in diminis}hing the amount of caternlat.on, necessary to obtimn the nImnerical vrae of a fractionl radical to any required degree of accuraacy, Thuts suppose it is required to obtain the numerical value of the fractiion I1 il, tre-'} to six places of decimals, If' we m ake the alc'ulation Without rendering the denomina-tor rational, it will be found, that we must first extract the square root of 2, to soven R u v x u w.-204. When the den-ominatmor of a fraction is either a. monomial or a bin-omia, containing radicals of the second degree, holws may it be reduced to a fr action having a rational denominator? SIMPLs EQUATIONS CON TAINING RA.DICALS. 195 places of deeimals, alnd then divide 1 by this result.'But if we render the denominator rational, the calculation merely consists in finding the square root of 2 and- then dividing by 2. The work by the lattcr method, requires only about half the labor of that by the fiormero Besides, the operator feels certain, if he has made no mistake, that the last figure of his result is eor rect.. Whereas, by the other mode, as the divisor is too small, the quotient figures soon become too large. Thus in this example, if we use seven decimals for a divisor, the seventh figure of the quotient is too large; if we only use six places of decimals, the sixth figure will be erroneous. 3 7, Find the numerical value of the fraction Ans. 1.3416407j. 8, Find the numerical. value of the fraction - Ans. 3.650281+. 9. Find the numerical valIae of the fraction Ans. 2.805883 —+.R, E a xA R. —It is proper to noties, that the signs 1/ andl 7/, when applied to a smononemial, both have the same meaning. There is a wan-t of uniformity namong the best writers, in the manner of making the radical sign before a mtonomial. SiMPLE EQUATIONS CONTAINING RADICALi OF THE SI.CONID DEGREE. 1N o T o T g As i T e:a a s o.-This ptart of the subject of Equations of the First Degree, could not be treated till after Radicals. It may be omitted entirely by the younger class of pupils. AnT, 9O4 —In -the solution of questions involving radicals, much will depend on the judgment of the pupil; but the etasiest processes can only be learned firom practice, as atmost every question can be solved in several ways. The following - directions will be frequently found useful, 1St. s When the equttion contains one radical expression, transpose it to one side of the equation, and the rational terms to the other side; then involve both sides to a power corresponding to the radical sign. Thus, if we have the equation /(x-1) —l=2, to find z. Transposing, V/ (x —1 )-3 Squaring, x —1 -9, from which x —1 0 2d. When more than one expression is under the radical sign, the operation must be repeated. Z196 WRAhYIS ALGEB RA PART FIRST. Thus, a —'Zv — (aoA+xV/C2+X2"), to find x. Squaring, a'+-2acz+-x-~a2 -al/ +-x~.x2 Reducing and dividing by x, 2 _a_'z /cq-xx t Squaring, 4a 1 4ax+X2-C2+Z2. C2__4a2 whence x= —4a 3d. When thoer are two radical expressions, it is generally bet ter to make one of thenm- stand alone on one side, before squaring. Thus, /(x-5) —3 4 —-/ (x —12), to find xo Transposin ng, / (x- 5)V 7-/ (x — 2). Squaring, x-5 —49 —14/ (x- 2) +x-12. IReduciag and transposing, 141j (x-12)=- 42. Dividing, V-/(rx-2) —3. Squaring, x —12 —9, from which x —21. A XAMJPL]ESE FOR PPRAC T CER. 1, /(x-+3)-+3~7 4..... O...... Ans. xz=13o 2,.A'/(.-I) 1)....... Ans. x-5. 3. -v/ (6+ ) x1)=3 o......... e s o: I o. 4. j/x(ar-tx)=a~ —z............ A s. x —, 5 -'/zx —2=i (/z-8)......... Ans. xz-9 6:x+r-tX__I —7 —:7............... Ans. x-. —-4. 7. 2-f+-V3x —/ 540 e O. -. O G a O. o e.s. -2.. 7. 2+ —_&=j/_5i~x+ -4............Ans. x==l2. _8. /x-j- /x' — 7 5..,,....... Ans. X —-9.. s25a 9. / x-a z / x /a........... A ns. X=T 6. I0. t'x+225 — V./x —-424 —1lzzO.,., as. xz 1000. O1 x+Vax — xa.......... Ans. x'a.~ 12. /xa'/ —a- /..l.. Ar s e44 13, /Vx-1 2+ /........ Ans. x —4t, / s -5x0+6........,., As. 6- /Z-':-c237-l1Ox 16. /~x-'4zz -zz-...,..,.Ans, x==23. -^ —- A -' s. A+-_/x_ -a lax.A X.4 -H/ EQUATIONS OF THIE SECOND DEGREEE. 197 1.9o W(/XTW-tZ~v95v^) Anv t ~ @ ~ ssg Z-b(a+b)2 41A v....... Ans. x= W-b 2 -, CHAPTER VII. EQUATIONS OF THE SECOND DEGRiEE, ARTo. 2@6.-An Equation of the Second Degree (See Art. 148), is one in which the greatest exponent of the unknown qua.ntity is 2. Thus,' x2-9, and 5xz2-3x26 —, are equations of the second degree. An equation containing two or more unknown quantities, in which the greatest exponent, or the greatest sum of the exponents of the unknown quantities, is 2, is also an equation of the second degree. Thus, xy-6, x-2+xy=8, x-y+x-+-t-y 1, are equations of the second degree. Equations of the Second Degree, are frequently denominated Quadratic -Equations. AuRT. 207T.-Equations of the second degree are of two kindsiecoazs ptete and complete. An incomplete equation of the second degree, is of the form Caxz —b, and contains only the second power of the unknown quantity, and known terms. Thus, x2 —9, and 8x2-5x212, a-re incomplete equations of the second degree. An incomplete equation of the second degree, is frequently denominated a pure quadratic equation. A complete equation of the second degree, is of the form ax2-l-bx-c, and contains both the first and second powers of the unknown quantity, and known terms. Thus, 3 x2q+4t. —20, and ax2 —bx2+-drx-exzr-f-g, are complete equations of the second degree. A. complete equation of the second degree, is frequently denominated an q/tclted qeuadratic equation. R:vEr-w.-206. What is an equation of the second degree? Give examples. If an equation contains two unknown quantities, when is it of the second degree? Give examples. 20'7. fow many kinds of equations of the second degree are there? What are they? What is the form of an incomplete equation of the second degree? What does it conbain? Give an example:. What is the form of a complete equation of the second degree? What does it contain? Give an example. What is a pure quadratio equation? What is an affected quadratic equation? 198 nRAY'S ALGEBRA, PART FIRST, ART. t S@ — Every equation of the second degree, nays be reduced to one of the forms ao-b, or a —bx — c. For, in an incomplete equation, all the terms containing x may be collected together, and then, if the co~.fficient of x2 contains more than one term, it may be assumed equal to a single quantity, as a, and the sum of the known quantities, to another quantity, b, and then the equation becomes ax2=b, or axtC-b-O. So a complete equation may be similarly reduced; for all the termns containing X2, may be reduced to one term, as a,.;,and those containing x, to one, as bax; and the known terms to one, as c; then the equation is ax'+bxzc, or a{'2+bx-c-0. Hence, we infer: T'hat every equation of the second degree, mnay be?reduced tgo an8 incomplete equation inevolvizyg two lterms, or to a com?,plaete equation involving t7hree termons. Frequent illustrations of these principles will occur hereafter. INCOMPLETE EQJUATIOS OF THtE i' ECOXHD DBECEt1EEE APT. 209. —-1. Let it be required to find the value of x in the equation x2Il 6 —0. Transposing, xA2-1 6 Extracting the square root of both members, x -— 4, that is, x+-4, or — 4.. Ve}rificalion. (+4) -- 1 6= -- 16=0z or, (-4)2- 16=16 —16z=0. 20 Find the value of x in the equation 5x +4=449. Transposing, 5x2-45 Dividing, x:= 9 Extracting the square root of both sides, x =:-3. 3. Find the value of x in the equation -3-4-s — n. Clearing of f'ractions, S2_- -9X2=GS Reducing, 17x~' —68 Dividing, x2= 4 Extracting the square root, x I:a 4. Given ax'-H_=cx2'+d, to find the value of x. ax. —cx"=d-b or, (a-c)r'd d —b d a-c azz~4%.t EQUATIONS OF TIlE SECOND DEGREE. 99 From the preceding examples, we derive tle FOR TSlE SOLUTION OF AN INCOMPLETE EQUATION OF TiE SECOND 2 DKGRE.. Reduce the equation to the form aox-=zb..D ivide both sides by t ahe coigir ient of X2, and then extract the square root oq botlhe member.o AaR. 2R0.-tf we take the equation ax2-Qb we hav<e X-2 and x -~i7; that is, x aS and b If we assume - m, then x2m2 By tra"nsposing, x2- m2O By separating into factors, (x+m)) (x-7-m-,) 0. Now, this equation can be satisfied in two ways, and in two ontZy; that is, by makinglip- either of the factors equal to 0o By mlaking the second factor equal to 0, we have X —m-:0, or x-=~m. By making the first factor equal to 0, we have x+n-m-O, or x —-m. Since the equation (x+m?) (x-mn )-0, can be satisfied only in these two ways, it follows, that the values of x obtained from these conditions, are the only values of the unknown quantity. H-Tence we conclude st. That every incomplete equation qf the second degrl ee, has two roots, and only two. fd, That these roots are equal, b'ut have contrary siyns. Find the roots of the equation, or the values of x, in each of the following exanples. 2. 3.-.xl. + a.........Ans,=, 3. &a * 2-o-b2 —=0,.......... An x. =~-.t( 4. 7x2 —25-4 —13....... Ans. -z. t, v Era w.-208. To what two forms may every equaiion of the second degree be reduced?- Why? 209. What is the rule for the solution of an incomplete equa.tion of the second degree? 210. Show that every inco01 plete equation of the second degree, has two roots, and only two; and that those roots are equal, but have contrary signs. 200 RAY'S ALGEBRA -PART FRST. 5, 5.X23:=835X52....... AMS x=: z~Y 7. -- -A2:=.3 ns. x=Itz7. 39 iSB. —@ (r b-j) z2. _r rs..........8 3 8. (2x —5)a 2 —20x+73. Ans. x. 4. 10, x~a x-ans. 10d~x==:-4 x-r-a xz —a xloa 10. 1:4 =z~ i 2a.... Ans. xzz:L2c'o a az-a -.2 a2-bz QiESTIONS PRODCING IGNCOMPLETE EQUATION P THE-i SECOND DBEGRF-BE,Amt'l..1 —Itn the solution of a problem producing an equltnion containing the second power of the unknown quantity, the equation is found on the same principle, as in questions producing equationsa f the first degree. See Art. 150. 1 Find. a number, whose - multiplied by its -, will be equal to 600 Let x== -the number; then -— X' - - 5=0 x -.225 x — 15, 2. What number is that, of which the product of its third and fourth parts is equal to 108? Ans. 36. 3. What number is that, whose square diminished by 16, is equal to half its square increased by 16? Ans. 8 4. What number is that, whose square diminished by 54, is equal to tihe square of its half, increased by 54? An's. 12. 5. What number is that, which being divided by 9, gives the same quotient, as 16 divided by tlhe number? An s. 12. 6. What two numbers a-re to each other as 3 to 5, and the difference of whose squares is 64? Let 3x —- the less nrilmutber; then 5xz- the greater. A.nd (5Xz)2(3z)2==4 Or: 25- ax':...x2' 6- =a1664. FSrom which x:=-2; hence, 3x2__-6 aruod 5x==-l O, are the numbers. See general directions, page 127. -] sv n v w.-o21.1. In the solution of a problem producing an e,quation onta.ining the second power of the unknown quantity, upon what pril) ciple!i tho equ.,a..tion found? EQUATIONS OF TiHE SECOND DEGR EE. 2 0 7" What two nunbers are those which are to each other as 3 to 4, and the diffre tne of whose squares is 63?.Ans. 9 and 12. So'What two numbers are those, -which are to each other as 3 to 4, and the sum of whose squares is 100? Ans. 6 and 8. 9. What number is -that, to which if 3 be added, and from which if 3 be subtracted, the product of the sum and difference is 40? Ans. 7. 10. The breadth of a lot of ground is to its length, as 5 to 9, and it contains Iff20 square feet; required the breadth and length. Ans. Breadth 30, length 54 fiet. I]1. A man purchased a farm, giving T- as imany dollars per acre, as there were acres in the farm; the cost of the fia.rm was 1000 dolla.s; required the number of acres and the priee per acre. uns. 100 acres, t10 per acre. 1. Whhat two numbers are those, whose sumn is to the greater, as 10 to 7, and whose sum, multiplied by the less, produces 270? Ans. 21 and 9. Let I; Oxz their sum; then 7x — the greater, and 3x-r the less number. 13. ihant two numnbers are those, whose difference is to the greater as 2 to 0, and the difference of whose squaores is 128? Ans~ 18 and 14o 14. C bouogbht a nunmber of oranges for 48 cents, and the price of an orange was to the number bought, as I to 3; how ianly did he buy, and how much a piece did he pay Ans. 12 oranges, at 4 cents a piece. 15. A person bo ught a piece of muslin for 3 dollatrs and 24 cents, and the number of cents which hte paid -or a yard, was to the nuimber of yards, as 4 to 9; how nmany yards did he by, uand what as the price per yard? Ans. 27 yds., at 12 cents per yd. 16. Find two nuaiers, in the ratio of to -, the sum of whose squares is 225. Ans. 9 and 12. By reducing an-d to a commlon denominator, we find they are to each other as 3 to 4. Then let 3z and 4x represe-nt the numbers. 17. Find three numbers, in the proportion of,:, -,the sumn. of whose squares is 724. Ans. 12, 16, iand 18 ].8 A merchant sold a piece of muslin at such a rate, t.hat the price of a yard was to the number of yards, as 4 to 5; but, if he had'eceived 45 cents more for the same piece, the price of a yarcd rou]ld have been -to the number of yards as 5 to 4; how niany yards were the1'e in the piece, and whhat was the price per yard? A uls. 1 0 ya ds. ad sa cents per yarsd: 202 RAT'S.LGEBRA, PART FIRST. CONMP-LETE EQUATIONCS OF THE SECOND DEUREEE, 1. Let it be required to find the values of x, in. the equation X 2-4x-+4z= 1. It is evident, from Article 197, that the first member of this equation is a perfect square. By extracting the square root of both members, we have z —2= —i Whence x=-2~l-=2+l=3, or 2-1=1. Per' fication. (3)2 —4X3- 4-1, that is, 9-12+ —4=1 0'also, (I)2-4.X14=1, that is, 1- 4 4 —1 Hence, x has two values, +-3 and l1, either of which verifies the equation. 2. Let it be required to find the value of x, in the equation x2+6x=16. If the left member of this equation were a perfect square, we mi'hlt find the value of x, by extracting t;he square root, as in the preceding example. To ascertain what is necessary to be added, to render the first member a perfect square, let us compare it with the square of x+a., which is x-+2axq+at We fin d x-2=x2 2ax -=Gx 2a —,-6 2c =33 h-{ence, by adding 9, whiclh is the square of half the cot.iciena oQf the first.power' of x, to each member, the equation becomes x -F6x+9-=25 Extracting the square root,, x 3 —~-L5 Whence x= — 3=~5- +2, or -8. Eit.her of which values of x will verify the equation. AnRT. 2.1 2 —We will now show the diffirent forms to which every comnplete equation of the second degree may be reduced, and illustrate further, the principle of completing the square. Since every complete equation of the second degree may be reduced to the form ax-+bx=c, if we divide both sides by a,'we have C a a For the ke of simplicity, let b=2 and p The equation then becomes x2+2px-q (1.) If- is negative, and - positive, the equation becomes a a xY-21px=q (2.) EQUATIONS OF T.HE SECOND DEGREE, 203 If is positve, asni negative, the equation becomnes 22+-2pcz-q (3,) 6 c Lastly, if - and are both negative, the equation becomes x2-2px —q (4.) Hence, every conmpplee equation of thle second degree, may be reduced to tfhe %brn x2+Q2xzq, in chick7 2t and q may be e ither ositive or negatiuve, integral orjfactional quantities. We will now proceed to explain the principle, by which the first member of thlis equation nay always be made a perfect square. Since the square of a binomial is equal to the square of the first term, plus twice the product of the first term by the second, plus the square of the second; if we consiAder x,2+2x as the first t.wo terms of the square of a binomial, x2 is the square of thle first term (x), and 2fzvx, the double product of the first term'by the second; therefore, if we divide by 2x (the double t o the first term), or ~v by 2, the quot;ient p (half tic coteffcient of zx), will be the second termn of the binomial, and its square,:p", added to the first nmember,'will render it a perfect square. But,, to preserve the equality, we must add the sam e quantity to both sides. This gives 2 +2X+ +)2=q p__ Extracting the square root, x-J-J ~z q+p2-i' Transposing, --- p —/q It is obvious, that the square may be completed in each of the other forms, on the same principle; tfant is, by taking half tie coefficient of the first power of x, squaring it, and adding it to each member. Thus, in the second form x2 — 2x q ~-p-=q ~/ +1) In the t1ird and fourth forms, the values of x are readily obtained, in the same manner. Collecting the four different forms together, and the values of x in each, we have the following table. (12.) xt-2pxzq_ zzp}/qhpt ($3) x-+2px -q. x= —PE/ —q+P (4,) -2np —A!p x -q... =-i -- qq-p e 204 RPAY'S ALGETBRA PART FIRST. Although the method of finding the salue of x is the sameto in ea.ch of these forms, it is convenient to distinguish between them. See Art. 215. From the preceding we derive the FOR THE SOLUTION OF A COMPLETE EQUATION OrF TE SECOND DEGREE. I st. Reduce the equation, by clearing of fractions and Iransposition (if necessary), to the form ax2 -b-x=c. 2d. Divide each side of the equation by the cofftcizet /' x2, and add to each mnember the square of half the coilticient qf' the first power of X. 3d. itriract f: e square'root of both sides, and Iranspose tJhe known lternm to the second member. 1. Find the roots of the equation x-z-Sxz=33. Completing the square by taking half the coefficient of xt(s), squaring it, and adding the squ.are to each member, we have xz+8-txl 6-33 14C 16=-49 Extracting the root,, x~+4~7 Transposing, x-z-4~7 Whence x —— 4+7=:+3 And x —-4-7- -1 1. erification. ((3)2- 3) —33, that is, 9-24-33. Or ( —1)+S8(-_1 )=33, that is, 121 —8833. In verifying these values of x, it is to be noticed, that the squ re of -1, is 121, and that 8 multiplied by -11, gives -88. 2. Solve the equation Tx2-6x=l16. Completing the square, xt —6+9 0-16+9-25 Extracting the root, x —3::5 Transposing, x=+-3:55 Whence x —=+3+-k-5And x=-3-5=-2. Both of which will be found to verify the equati on. 3. Solve the equation x2+6xz=-5. Completing the square, x 6t x+9-9-5 —4 Extracting the root, x+-3=h 2 Transposing, xZ —3 2 Whence x.z. —3+2: —1 And dx=-3-2- -5 EQUATIONS OF THlE SECOND DEGREE. 205 4. Find the values of x, in the equation x-lOxz — 24. Completing the square, A2 1Ox+25%z25 —24z Extracting the root, x —5 —:I~ Transposing, x= 5~il1 Whence x-c5+1=6 And x-5 —=14. The preceding examples, illustrate the four different forms, when the equation is already reduced. Equations of the second degree, however, generally occur in a more complicated form, and require to be reduced before completing the square. 5. Find the values of x, in the equation 3x —-5= 36 Clearing of fractions, 3x2-5x=7x+36 TVansposing, 3x — 12x36 Dividing, x 2-4x=12 Completing the square, 2 —4xt+4- 16 Extracting the root, x —2= — Transposing, x= —+2i4 W[hence xz —6 or -2. 12x2 13x 6. Find tihe values of x, in the equation +x ~s-5+ —Clearingo of fractions, 1 22+25, =260+ 13x Transposing and reducing, IS 2x-SSa,x260 Dividing, --- x== a Here the coefficient of x is --- the half of which is-.; the square of'this is -, which being added to both sides, we have Extracting the root,.- -_ _ W[hence x -=+3, or - 3'lXAiPLS 0 PRACTICE. NoTE.-The first sixteen.of the following F1xampoles, are trranged to illustrate the for forms, to one of which every complete equation of the second degree may be reduced. 7, *x+X=20.......... Ans. xz2, or -I 0 8. xZ+l 6x- 80..........e Ans. xs —4, or. -20. 9. Xtfr-7x=78....., a e e. e A os r. x1 6 0 13.B:!0. a'+3x-28......... Ans. x ==- 4, or -7... 206 RAY'S ALGEBRA, PART FIRST. 11. x- -Ox — -24....... A.... -i- 12, or — 2. 12.,Y —8x'-20........... AI e, A x1U, or -— 2. 13. x2-5x —6.............. Ans. x —6, or — 1. 1.4. x —21x==100......... A risz-=25, or -4. 18. x~-~4x —' 3...........Ans. as —-: 1~ or —3 15. +6x-z —z —S,.......... Arc. xz= —, or -4, 19. X2+4x- -3.............. Ansr. x —-— i, or -3. 170 xi-S 1..... 5.. Arm. x —---- 35, or 3. 18. x2+7y, -1 —X 2...........Ane. xm —3, or -4.. 19..2-..m —-........... xAs x —4, or 2. 20,'z —Sm-l-15,........, Ans. x=, or 3, 21. x5 —l0xs — 21,.......... Arm. a.' 7 01 3. 25 x2-1x —54........... Ar. x —, 9or 6 23. 3x2 —2x+i23z2-56........ Ars, 5"01 24. 2X2 -5x —=z2..,.......,.. x....-4 o 25'. 2X2'+3xz65............ Ans. x::zo, 0r13 2x'5X2 26. AD 2..,,,... o r. A — x, or - 2 30. x-~-...-......... Arso x=z.3, or 2.;~ 28o 4 -— 4 1... Ans, x~-1, or 1... 6-x 29 a, -a,q-t ~ —x. T.... O,.Ans. x:=4, or....~3. 33. 9x-Ao-=,-..14............Ans. x-2, or 2... 34. z-'.. —3 0.............. Alne. x-2-5, or — 6.-G'4 x2 32. 5x., —~f! 2 xx- x Ans. A -— 4, o r4 35~ 323 4x1t 29'.........,, A m.a, Oi4 36. 2+x.92-3............. Ans. i=5,4 or1 j1. X- 29 -si'U. [ 4...., Ans. x ator 4. 2i. 2 x 36. Aim x c~9,~ Crt —.4, ovP l 37,:x2-Ix......... Ans. x........ -, or. 38 17a'...9 —x 30,. As. x.... i r:e v'T: w. —-212. To what form may every complete eqifatioon of';h-c seh - ood degroee be reduced? What are the four forms thatt this gives, depeand. ig onr twe signs of 2p and q? Explain theo principle, hby maness of which the TrMs mesciber of the equation x2 —2pxq= may be nmade a Peusc t ri ce-(.ree What is the Mle fbor the solution of a complete equation of tbe'eoon.td degre? 'QUATIONS OF THE SECOND DEGREEo 207 x39. 3x #5x=S o,........, Ans. x=-3, o4r- -. 40. 4.r-3x-=8x —-.............Ans. x=,z4 or -2. 41. 2 —-4x —.',. Ans. 2-1/31 —:3 32-, or, 9 684 S2.. 3,sx or4., -6 7,....... 4 65x 0Ix' 13 2x 43. -. As-. x=-.35e, -' xs +3 44. +,. o see Ans. z —12, or -2. x __ 7 41. x-80' Ans. x=t 14, or -10. 46 46. x,- 2'3x-4,........ As x=5, or -2. 22 —xs 15 — 53 0 47. 2.0 e.,, g Ans. x —36, or 12. 48. --'.. —.. o..o 491 -3 25' 3b i, As, x-s 8, or 132~50. 2ax -g —x o.= —ab #b...a.... Ans. x2a-bt-, or — b9 51. ax —a axb2-8,....2.... An s. xx —-a{b4, or 8b. 523 xa 8 3 —4....... as.. a —+b, or - 8-b 53 X —t-i 0 bu-b......... Ans. Ix-, or 48, a54. z x-a. x a,, x-bicz zaoi'i ~'P'-s Axs. xirbzy-ab-8 55. BzXXi2+(=-28> ah =......... Ans. tzE — l, 8or' X2 x 2a-2 2ar2 a a2 ba b........... 57. x (; a1) — 4bar-1 —.0a=O..t.....Ains. x=a, or —-1. 58. x2 (a+b —c)=(a. )a....... Ans. x=ia-l-b, or — c. At. 4.gr -Tif xir)oo As.io O F SIn o Ir o ir9e qrA Isf -...^ Vibex an eu ation is brougha t to the tform axz -'-bx —-c, it may hereduced to a imple equation, wthout divriding by the eotolocioet of x; thu a oiding fractionsr If we moisiply both sides of the equation 1an2~=:, aby a,'of.tient of,. i. becon.es a.x.+.abx... a. ac Now:i ie roegard aXx2-abx, s tJhe first and seond i';' of e squarxe of a binomiaI, a2xb inust be the square of tho or; it erom, ardu c ta e double product of the irst term byt the sco -ind, fie., the sirt termi of the binomial is e/uxai- ax; d e' 4 tb eoid l, wh qoaoient deprivd from d ividing a xi by the double o ax e 208 RAY9S' ALGEBRA, PART FIRST. first term; that is, ~-~- Adding the square of b to each side, 2aCx2' 2 b~ b2 the equation becoLmes a23X2-+abx+-' ac6 0 Now, the left side is a perfect square; but it will still be a perfect square, if we multiply both sides by 4, which will clear it of fractions. Thus, 4a2'x'+44bx+bL-4ac+b2 Extracting the square root, 2arq-b-i/4ac —i-b — bm1/ 4ac+bj Whence == Now, it is evident, that the equation 4ar2+4abx+b2z==4ac'b", may be derived directly frorm the equation ax2'-bxzc, by multi.plying both sides by 4a, the coiefifcient of bx, and theni adding to each member, the square of b, the coefficient of the first power of x. This gives the following FOR THi SOLUTION OF A CO3PLETE EQUATIOi OP TtE SECOND DEGREE. Reduce the equation to the Jbrm aX2 —bx- - and muldtiply both sides, by four times the cofi icient oft x. Add the square f the co/'/t "ficient of z to eaclh side, and thea extract the sq-uare r.oot. [This will give a shimple equationz, fbom which x is easily found. 1. Given 3" —52x-8, to find the values of x. Multiplying both sides by 12, which is 4 times the coefficient of'x2, 362 -60x-=336 Adding to each member 25, tle square of 5, the cogfficient of x, 36x2 —60+25 —361 Extracting the root, 6x- 5 —~l1 9 x —5::19 — 24, or -14 x2' —4, Or --- By the same rule, find the values of the unknown cuantity in each of the following examples. 2t 2x- +5x-33. Ans. x-3, or l3, 5+2- 2xS — 88 Ans. t=4, 1or o 64.o 3tx'-x=zz70O e V 4 < & @a o sAns. xz-5, or - 4 5. x" —-x=42...a.Ae eov ons. x=7, or-'6. 6. x2 +- 8 59,4 9... Ans. x=z6 or- -7I. If further exercisesare desired, the examples in thIe preceding article may be solved by this rule. ItR. v,-X 2, It w 13o axp in the o indoo method of o mple thig, the tstqh e.v EQUATIONS OF THIE SECOND DEGREE, 209 PR O ENl S PiDUCE ~G0 CMPLETE 1I`tKJXA10NS OF rE SECOND DEGREE ART. 214il-1i. What number is that, whose square, diminished by the number itself, is equal to 20? Let z- the number. Then. — x-20 Completing the square, x'-x- 2On I 1 -4 Extracting the, root, x. _. Whence x —z+5, or — 4. Now either of these values of x satisfies the equation; but the negative value — 4, does not fulfill *the conditions of the question in an arituhmetical sense, But, since the subtraction of a negative quantity is equal to the addition of a positive qhantity, the question may be so nodified, that the value -4, will be a correct answer to it, the 4 being considered positive. rThe question thus changed, is: What number is that, whose square increased by the number itself, is equal to 20? 2. A person buys several oranges for 60 cents; had he bought 3 more ibr the samne sum, each orange would have cost himi 1 cent less; how many did he buty? Let x- the numnber he bought. Then -0z the price of each one. X 60 And the price of one, had he bought 3 more for 60 cents. +3 Therefore, 6 x x -j_-3 Clearing of fractions, and reducing, X+3a I-SO180 Completing the square, 2 —3+ 9 4 + I 801S1O 9 Extracting the root, x2 —:: " Whence x-r- 12, or -15. Now either of these values, taken with its proper sign, satisfies the equation froim- which it was derived; but the value 12 is the only one that satisfies the conditions of the question. Since s-j- ~4 and 1 OTj= —5; td since buying and selling are opposite operations, the result, -15, is the answer to this question. A person sells several oranges for 60 cents. Had he sold 3 less for the same sum, he would have recezived one cent m7ore for each. IHosw mtany oranges did he sell? It EAut AR i.FProyn the two preceding examples, we see, thatt the root which is obtained, from giving the plus sign to the radles, satisfies both 19.t 2 0 RA'S'ALGEBRA, PART FIRST. the conditions of the question, and the equation derived fr'om it; while the other root satisfies tile equation only. We see, also, that the root which arises from giving the radical the nega tive sign may be regarded. as the answer to a question differing from the one proposed in this; that certain quantities which were additi~e, ha.lve become subtseractive, and reciprocally. Som.etimes, however, as in the following example, both values of the unknown quantity satisfy the conditions of the question. 3. Find a number, whose square increased by 15, shaMl be 8 times the number. Let x= the numlber; thoen 2+ 15 — Sx Or X2S x —l 5 Whence z-5, or 3. Either of which fulfills the conditions of the question. When there are two unknolwn quantities in a problem, that canr be solved by the use of one symbol, the two values of the symnibol generally give both values of the unknown quantity, as in the following question. 4. Divide the number 24 into two such parts, that their product shall be 95. Let x — one of the parts; then 24 —-_ the other. And x(24-x)-95 O r xs2 2 L4x — 95 W'hence x —i9 and 5 And. 24 —-5, or 19o 5, There are -three numbers, such that the product of the first and third is equal to the square of the second; the sum of the first and second is 10, and the third exceeds the second, by 24; required the numbers. Let x- the first; then 10x=- the second, And 10-x-1-24-34 —- the third. Also ( 10 —x)2=- -(34-x) Or 100-20x —Lxz34x —x2 From which, ax25, or 2. fWhen xs25, 10-x-x: 15, 34-x-9, and the numbers are 25, -15, and 9. When x2, I 0-wS —S, 34 — x-32, and the numbers are 2, 8, and 32. Both these sets of values satisfy the question inm a alngebraic sense; only the last, however, satisfies it in an arithmetical sense. Let us eundeavo:; to ascertain how the question must be modified, so that the first set of numbers shall satisfy it in n ar arthmetical ktrlf EQUATIONS OF THE SECOND DEGREE. 11 The meaning of the negative solution -15, will be understood by considering that the addition of a negative quuantity, is the same as the subtraction of the same quantity taken positively (A rt 61). The first condition of the question theix becomes 25+(-....15) —25 — (+15)-=25 —15 —1 0 and the second is 9 —(-15) — 9- - ( —15) =9+15 —24. This indicates, that -1-5 may be changed to — H 1i5, provided, that instead of the condition of the suma of the first and second numnbers being 10, their differenc be e0; and the second condition may for a similar reason, be changed into this, that the.sum of the second and third is 24. The question, with these modifications, would be: What three numbers are those, such that the product of the first and third, is equal to the square of the second; the d'iefrence of the first and second is 10; and the sutm of the second and third is 24? tR i sx t. I.-in the.following exanmple, the pupil is required to find only that value of the unkinown quantity, hich satisfies the conditions of the question in an arithiextical sense. It forins, however, a good exercise for advanced pupils, to deterti.e the negttive value, and then to modify the question, so th:at this vale'll SatiSsfy the conditions in an arithmetieal sanse. 6. Find a number, su litha t if its esquare be diminished by 6 times the nanber itself, the remainder shall be 7. Ans. 7. 7. Find a number, such that if its square be increased by 8 times the number itself, the slum shall be 9. Ans. 1. S. Find a number, such that twice it s square, plus 3 times the numnber itself, shall be 650 Arns. 5. 9. Find a nunmber, such that if its square be diminished by 1, and - of the remainder be taken, the result shall be equal to 5 times the mn:nber divided by 2. Ans. 4 10. Find a number, such that if 44 be divided by the number diminished by 2, the quotient shall be equal to 4 of the number, diminished by 4. Ans. 24. 11. Find two numbers, whose difference is 8, and product 240. Ans. 12 and 20. 12. A person bought a number of sheep, bor 80 dollars; if he had b ought 4 more for the same money, he would have paid 1 dollar less for each; how many did he buy? A ns. 16. 13. There are two numbers, whose diffrenece is 10, and if 600 be divided by each, the difference of the quotients is also 10; xwhat are te tnumbers? ~Ans. 20 and 30. 14. A pedoestrian, having to walk 45 miles, finds that if he, in-:3rases his spe-ed.f a mile an hour, he will perfobr his t.ask 1 - 212 RAY'S.ALG-EBRA, PART FIRST. hours sooner, than if he walked at his usual rate; what is thnat rate? Ans. 4 miles per hour, 15. Divide the number 14 into two parts, the sum of whose quarses shall be 100. Ans. S and 6. 10. In an orchard contalning 204 trees, there are 5 more trees in a row than there rows; required the number of rows, and the number of trees in a row. A. 12 rows, and 17 trees in a row, 17. A schoolboy, being asked the a.ges of his sidter and himself, replied, that he was 4 years older than his sister, and that twice the square of her age, was 7 less than the square of his own; recquired theiP ages. Ans. 13 and 9 yrs. 8I. A and B start at thle same time to travel 150 msiles; A travels 3 miles an hour if1ster than B, and finishes his jour ey 8 S hours before him; at what ratae per hour did each travel? Ans. 9 and 6 miles per hoeur. 19. A company at a tavern had. 1 dollar and 75 cents to pay; but before the bill was paid tw o of them went away, when those who remained hlad each 10 cents more to paly; how many were in the companly at first? Atns. 7. 20. The product of two numbers it,91o.10, and if I be t-aken f'om the greater, and added to the less, tho'product of the resutl;ing numbers 120; what are the numbers? Ans, 25 and, 4. Let x= the larger number; then.- = the smaller. x 21 If 4: be subtracted'from u father's age, the remainder will be thrice the age of the son; and if 1 be taken from the son's age, half the remainder will be the square root of the father's age. Required the age of each. 4 Ans. 49 and 15 yrs. 2~-4 Let x- the father's age; then — the son's age. 22. A young lady being asked her age, aniswered, " If you add the square root of my age to - of my age,'the sum will be 1n." Required her age. Ans. 16 yrs. 23. What number is that, from rwhich, if - of its square root be taken, the remainider will be 22? Ans. 25. 24. c mnerchant bought a piece of muslin'for 6 dollars; after cutting off 15 yasrds, he sold the remainder ebr 5 dollars 40 cents, by'vwhicl he gained I cent a yard on the amnount sold0; how many yards did lie buy, and at what price? Ans. 75 yds., at 8 cts. per y4, 25. -A man bought a horse, which he afterward sold fore 24 do]' lars, and by so doing, lost as much per cent. upon the price of Sis purchase, a the horse cost hi; wath did her pahy ir the h.orse Aias0 $60. Ara f$4011: EQUATIONS OF THE SECOND DEGREE. 213 PRsI [TIESi OF T-E ROOTS OF A COMiLPLETi J5 1AIUdA'O 0F rTbHE SECOND DEGREEEto TOTE TO To zACi5ERS.-This subject may be omitted entirely, by the younger class of pupils; and passed over, by those more advanced, until the book is reviewed. ART. 2l, —-The pupil may have learned already, by inference, fron the solution of the preceding examples, that an equation of the second degree has two roots, fthat is, that the unknown qua.n* titv has tiwo values. This principle mnay be proved directly, as follows. The general form to which every complete eanuation of the seeond degree may be reduced, is a —+2pxc; in which )p and q may be either both positive or both negative, or one positive and the other negative. Completing the square, we have &x+22X+p= 2_q P2 Now, the first mnaember is equal to (x-p)', and if, for the sake of simplicity, we assume cf-t 2 —-n2', that is, -ff-Pp'2-sn, then Transposing, ('- ".' -0. But, since the left hand ie'fember of this equation, is the diffser ence of two squares, it t.ay be resolved into two factors, Art,. 94, This gives (X-a-i -i n) (x+-p —n) 0. Now, this equation cat be satisfied in two ways, arnd in onaiy twvo; that, is, by making either of the factors equal to 0. If we maoe the second factor equal to zero, we have Or, by transposing, x-= —+pm — p-/qp-2 if we maike the first factor equal to zero, we have x+p- -7=0 Or, by transposing, xz. —~nzw-p-yq/ p.P Hene@, we conclude, 71st That every equcatZion of" the second dertree, has two roots, tnd only tco. 2d. Th'at eery complete equat.aion of the second degfree, reduced8- to the fanm x-~2pxz:q, magy bie decomposed into two binoaaial./heltt, of wthich the first terim in each is x, and the second, the two roots witth their signs changed, Thus, the two roots of the equation xa —5x:z-6, or x — 5a:+i-.0 aTe = —2 and xz3; hence, a-5z-+-6(: --- - 2)(xtr-). s'oro this, it is evident, that the direct mnethod of resolving a quaLdra.tie trinomial into its factors, is to plate it. equal to zero, and. thenl find the roots of the equation, n thnis manner, et the learner solve the questions on page 72, '214 RAY'S ALGEBRA, PR1i FIRST. By reversing the operation, we can readily forn an equation, whose roots shall have any given values. Thus, let it be required to form an equation'whose roots shall be 4 and -6. We mRus. have xz= 4 or ax-4-0 And x=6- or x6 —60 Hence, (x —4) (x+6 x 6x-2x -24=Q Or x2+2x=24, Which is an equation whose roots are +4 and -6. 1. Find an equation whose roots are 7 and 10. Ans. x2-17x=-70. 2. Find an equation whose roots are -3 and -1. Ans. ax2-+4x.= —-.3. Find an equation whose roots are — 2, and — 1. Ans. x x2 —z' ArT. 2at -Resum ing the equation x2-J-+2zt.q. The first vralue of x is q ft The econd value of x is 1p V/Vi{' Their sum is -hich is the co.fflcient of x, taken wvith a contrary sign. Hence; we conclude, [,1Zhat the sutm qt tfe r'"oots of n equattion (f the second degree,; reduced to ie jfbrmn x2-t-2px.2q is equal [to the coe:jicient qf the first power ('x taken, with a contrary sqign. If we take the product of the roots, we have First rootb. -...+ qHj- - Second oot= - — pVq p p 2 —'l/ p2 +pF /q-'-_ (q+p') But -q is the known term of the equation, taken with a contrary sign Hence, we conclude, That the Jproduct of the twio roots of an, equation of the second degree, reduced to the form xH2-l-gvx-q, is eualt to the known te.r taken with a contrary sign. AR: A R.- In the preceding demonstrations, we have regarded 2p and q as both positive; the same course of reasoning, however, will apply when they are both negative, or when one is positive and the other negativ e; so that the conclusions are true in each of the four different forms. AR'r. 2.-In the equation x+ 2pxq, or first form, the two vaues of x are -7+V+p EQUATIONS OF THE SECOND DEGREE. 21.5 If we examine the part /q - p', we see that its value zmust be a quantity greater than p, since the square root of p- alone, is p. Hence, the first root is the difference between p and a positive quantity greater than p; therefore, it is essentially positive. If we take the negative value of the radical part, the second root is equal to the sunm of two negative quantities, one of which is p, and the other a quantity greater -than p; theretfore, it is essenztilly negative. Since the first root is the difference, and the second root the sum7, of the same twno quantities, the second, or negative root, is necessarily greater than the first or positive root See questions 7, 8, 9, 10, page 205. In the equation x2-2px-q, or second form, the two ralues of x are +p l / -p' And p-qp The quantity under the radical being the same as in the pre' ceding form, its square root is greater than p. The first root then, is the tsum of two positivbe quantities, one of which is p, and the other a quantity greater than p; therefore, it is essentiacdypositive. If we ta.ke the negative value of the radical part, the second root is equal to the difference between p, and a negative quantity greater than p; therefore it is essentially nregative. Since the first root is the slum, and the second root the diftsrene of the same two quantities, the first, or positive root, is greater thlan the second, or negative root. See questions 11 12, 138, 14, page' 206. In the equation x2 +2px —-q, or third formi, the two values of x'ire — +-/ —-q__p2 And --— a/ —-p'. If we examine the radical part, V-q+p2, we sce, that its varl.must be a quantity less than p, since the square root of p2 without its being diminished, is p; hence, the first root is the difference between -p, and a positive quantity less tfhan p; therefore, it is essentially n:egative. If we take the negative value of the radical part, the second root is equal to the sum of two negative quantities; therefore, it is essentially negative. R E wi.-W.21i5. To what general form, may every equation of the seer. ond degree, containing one unknown quantity, be reduced? Show th at every equation of the second degree has two roots, and only two. 216. To what.is the sum of the roots of an equation of the second degree equal? To what is the product equal? 217. Show that in the first form one of the roots is positive, and the other negative; and that the negaativv root ih greater than thel positiver 216 RAY'S ALGEBRA, PAR2T FIRST., Hence, in the third form, both roots are negative. See questions 15, 16, 17, 18, page 206. In the equation rz —-2px —q, or fourth form, the two vaxlues of x are + - qf /+p2 And + -— 2 The value of the radiclt- part, being the samne as in the preceding form, it is less than p. The first root, then, is the sunm of two positive quantities, therefore, it is essentially positiveo The second root is the differene betwieni p, and a negative quantity less than p, therefore, it is essentially positive.. Hence, in the fourth form, both roots are positive. See questions 19, 20, 21, 22, page 206. AiT. Si, —In the third and fourth forms, the radical part is @/ -q (-p. Now, if q is greater than pl, this is essentially negafive, and we are required to extract the square root of a negative qutantitr, which is impossible. See Art. 195 Therefore, in the third and fourth forms, when q is greater than p', that is, when the known term is negative, and greater than the squlare of half the coefficient of the first power of x, both values of the unknoiwn quantity are impossible. What is the cause of this impossibilit? To explain this, we must inquire into what two parts, a untmbher must be divided, so thbat. the product of the parts shall be the greatest possible. Let 2p represent any number, and let the parts, into which it is supposed to be divided, be p+-z and p-z. The product of these parts is (p+z) ( —Z)=-)p'-z'. Now, this product is evidently the greatest, when za' is the least; that is, when z2 or z is 0. But, when z is 0, the parts are p and o, that is,;,when a number is dividced into two equal pa-rts, their poduct is g'reater than th.at of anSy other two pacrts iznto whicth ithe numher can, be dividesd Or, s a the same principle niay be otherwise expressed, the protduct qf an.y two.bunequal utmaber is tess than the square qf half their sztmn. As an illustration of this principle, take the nutber 10, and divide it into different parts. I0=9-+,1, and 9Xl'- 9 10 —8+2, and SX2=16 10=7-7- 3, and 7X3= 21 10:6+4, and 6X4==24 }10 %5-5, tand 5X5-=25 gEV iE W.-2lt Slow that in the second form, one rootis positive and the other negative; and that the positive root is greater than the negative. Show that in t;he third form, bothr roots are aegative. Show that in the fourth form, both roots aro positivre EQUATIONS OF TUHE SECOND DEGREE. 217 We thus see, that the product of the patrs becomes greater as they approach to equalit, y, and that it is the.greatest when they,are equal to each other. Now, in Art. 21 5, it h1as been s~hown, t-,ha-t 2p, the cotifitcient of the first power of X, is equal to the s.m't) of the tw'o v'alues of tx, and. that. q is equal to their product. But, when q is greater thant p we have the product of two numbers, greater than the square of half their sum, which, by the preceding theorem, is impossible. It; thein, any problem furnishes an equation in which the known term is negative, anind greater than the square of half the coibfifl cient of the first power of the unknown quantity, ywe i.nfr, that the conditions of the problem are incompatible with ceacth other.. The following is an example. Let it be required to divide the number 12 into two such parts, that their product shall be 40. Leot x and ] 2 —-x reresent the parts. Then: x(12 —-x)-40, or xa......12x.- a: — 40:...4x+3 = — 4 a2 - - G=jV-4, and x-2-6O: l/t 4. The.se expressions for the'values of x, show that the problem is impossible. This we also know, from the preceding theorem, since the number 1 2 can not be divided into any two parts, wthose product will. be greater than 36; thus, the algebraic solution ren — ders manifest the absurdity of an impossible problem. R a A A It. s, —s8t. When the co-ffi:elent of X2 is negative, as in the' equa tion — a'-~:m2eje n,~ the pupil nmay not perceive that it is embraced in the four general forms. This difficult is obviated, by multiplying both sides of the equation by -— L. 24. Since the sign of -the square root of xz, or of (+p-)2, is L:C, it might seem, that when xai2t,: we should have f:r=i~ei that is, — +xt ==mn, and -=~ —-m-.; such is resally the Case, but — a=-nM, is the1 same as +-t'=~ —,z andt. —x== —n- is the same as +-z —- m Ri enTc, i-e~ll..-.'zi embraces all the values of ro. In the same manner, it is necessary to take only the pins sign of the sqiuare root of (c-.z-,)2. EQUTI S OF THES O ECOND SEGOB REE, CONTA IN G TWO* Uka i rKNOWN QUANlTITIES. No0s., —-A:full discussion of equations of this cleass does -not properly belong to aan elementary treatise. Indeed, no directions can be given, tihat will be applicable to all cases. The general method of treating the subject, consists in presenting the solution of a variety of examples, and then furnishing others for the exercise of the student. The following examples a-re intended to emibrace only those capable of solation by siuple Inethlods, See Ray's Algebra, Part Second, I 19.;) 2 I 8 RY'S ALGEBRA7 PETT FIlRST AUr. o l9 —-In solving equations of the second degree, cont-ain-. ing twio, unknown quantities, the first step is to eliminate one of them, so as to ohbtain a single equation involving only one unknown quantity. The elimination may be performed by either of the ~three methods already given. See Articles 158, 159, 160. 7hen a singi6 equation is thus obtained, the value of the unknown quantity is to be found by the rules already girenr. EX A P LES. 1. Civen x —-y2 and x-y2+y21O00 to find x and p. By the first equation, xz-y+2 Substituting this value of x, in the second, (y+2)2+l2= G00 From which we readily find, y-6, or -8 Hence, x-y+2-8, or — 6. 2. Given x+yz-8, and xy=-= 5, to find r and y. Fromn the first equation, x= Substituting this value of x, in the second, y(S y)-= 5 Or Y2 — 8y —I 5 From which y is found to be 5 or 3. ience 3, x-3, r 5. There is a general method of solving questions of this form. without completing the square, with which pupils should be 2tac quainted. To explain it, suppose'we havee the equations x+y —a Squaring the first, zx+2zy+y2 -a2 AMultiplying the second by 4, 4xyI-4b Subtracting, x2 -2xry+y2-=a2 —4b Extracting the square root, x —yp-zi-~/a —4b But xsy-k —-a Adding 2xr-=a-~-v/ 4-r -4 b Or x=zwa~-Vo — 4b Subtracting, 2y=a- — V/a2= 4b Or z=-a V - If we ha the e teqations x —ya and xy-b, wec may find the ivalues of x and /, in a Similar manner, by sequaring each member of the first equation, and adding to each side 4 times the second. Then, extracting the square root, we obtain the value of x-+py:j:-:~/ai+4b; from xwhich, and x —-- a, we find x:=ai-. /aq4b ad yze ==a /a- 2+4k EQUATIONS OF THlE SECOND DEGREE. 219 3. Given x' — ya and x'+y' —b, to find x and y. Squaring the first, x"-'+2-xy-y'a' (3.) iut, Xz2+yb' (2.) Subtracting, 2x'y=a2b' (4.) Take (4) from (2), xt'-2xy-yI"=2b-a' Extracting the root x —'y=-_~/'2b —a x+y-~a Adding and dividiing, x=-a:k.t-+1/2Ab —aSubtraeting and dividing, y-4a: )gj2/2b-a-. 4, Given zI04-y't= and -xy=b, to -find x and gy Adding twice the second to the first, a.2+2-xy+y~__a +2b Extracting the square root, x+-~y= ~' 6-l/a+2b Subtracting twice the second from the first, s. 2fgy+y:-f —2. b Extracting the square root,'x-y=~J/a —b WYhenVe x-~iV/a& 291./,-t~ -26 And g V — ~26+4t i —-2f& 5. Given x.-+y-a- and x-y-b, to find x and y. Dividing the first by the second, - 2.,r+yy2=-' (3.) Squaring the second, x -2 +xy+y,-L=b (4.) b6-a Subtracting (3) from (4), 32xy=.. b~ Or (. ) 4a —3 Take (5) from (3),.x-2:ixy-_-y — 3 Extracting th root, x-Y~Ji ( 3b But'ha"e cx ~4~.i ( 4a —6b ) And y 4N ( 3 4a-b S kn.a similar -manner, if we have x5 —yf=a and x —— ysb we fi14 _, ~ / f( 4a-b 4,... _ [ a-__ 4., ~ 2)20 RA:IYS AL GEBRA, PART FIRSTuT E X A~W PL 2.'c'-y2-$...... Ans.:= S.1 341.'~0'I...... o.... 36f or l?/:=t x —.Y=......As_=?7o 2.-_7X 2+y —1.....~..l..f.A s. a=2, or 7.:y.=:6................y....., ors9 8o~~~~~~.c.......... Ans. x-.:9, or —4. xy........ y —, or 9. x 2. ) Ans. x:=7, or 4O2. x y Si....g........... y=2, or 76 13. x;..y:=5 }.........Ains., xS=8, or — 2, nd%.y..t — j.. Y= —.-=2, or. 1'4..~ —-y~ — 2....:....... Ars.A:i=:5, or x:3= if -008 8 15. d, 9. 9.~t~-n~a 2-9 01 g x +!/~-~a..............?. —, r.5..... 1o a Q ifs f(y K)} 2Ar is. x —6, or —. 12.:' —?/~:=l-20 8 } ooopo~rexa1............ or 16' y. —x2'$ yt 3..../-4............. y=2, or......6 17,.i-'9 ) Lx.......Ans.x-', or,-0-1 y x = I v2n..... —....y-. or 3 22Ans' x=7 —Anso wq2 02 1 _ii~3 ~( y=) O OB3, or 2.,jp ~aI^~~~ t.4:'j...... A... c o.....-. 17~~~~~ s 115< 20 1 alf9~ — 19ns., or — I:o x. - yj............1 y4, oiIn solhivg p qesion.8, let =-w, anidl z, the question thou becomes si.:cff=tar to the 9th. I question find the val { oft'from the second equation, as if t w-er a smile unko n. qnut PR -Lgy11 PRDUIN EQUATIONS=:OF TH E oI EGr. It ONVAPItWO)N( TWOI UNKNO'4 WN QUWANTITIES)'bTe sum of two umbers is 19, -n the n sum of their in.t s'52 what aro.the numbers'? A. 2. -The, dirence. of two numbers iy 3 d e dill''or Of heIr sa g9; rqnird the nu aiberst 1As. and 5.,2~"~~ ~~. —, —~ th quso t, he~u C@N'.IL4.Ii. YINC~'PW LE NOV ~UArPP. ~g,~ w~X — 1. Th4e su of tw nmbs is 10 and th suma of thh q'a 52; wha~t~ arte ther.S numet trsd o -.Aeqems. 4 aendG 2,~orjs "" nl. ls L li ~,to,.........IUI O~f o'. J_..'. difrnce o f' two ntlumbers is 3,, andtPosur o tl'e-dlfere ~jl~nee of their~ sciuareslc g.~9 ceuie the n.umbers 3, n. At;diEnso S and 5 EQUATIONS gOF TUHE SECON)D DEGREE. 221 3'. ItI is required to divide the number 25: into t:'wo such parts, tthatt the sum of their square roots shall be 7. Auns. 1. a-ndl 9. 4. The product of a certain nuumber, consisting of two places, by t(.he sum of its digits, is 160, and if it be divided by 4 times the digit'in unit's place, the quotient is 4; required the number. AnSo 32. 5, The difference between two numbers, multiplied by the greater, — _16, but by the less, -12; required the nutmbers.Ans. 8 and 6. 6. Divide 10 into two such parts, that 1 thteir produet shall excteed'their difference by 2',. Ans. 6 and 4. 7. The sum of two numbers is 1 l9, and the sum of t-heir cubes is 370'; irequired the numbers. Ans. 3 and 7. 8. The difference of two numbers is 2, anid the difference of their cubes is 98; required the Ln-umbers. Ans. 5 and 3. 9, The sum of 6 times the greater of twvo numbers, and 5 t{imes the less, is 50, and their product is 20; rueuired the numbers. Ans. s 5 and 4. 10. If a certain number, consisting of two places, is divided by the product of ites digits, the quotient will be 2, and. if.~27 is added to it, the digits will be invertedt; required sthe anh.e -' A.nn 36, I 1. Find three such quiantities, that the quotients arisn:g:from dividinig the products of every two of themtn, by the ole remailning, arc a., b, and c. Ans. ~Vct= b', and::E/bc:, 12. The sumn of two numbers is 9, and'the sum of'their ctubes is 21 times as great as their sum; required the.numbters, Ans. 4 ant]. 5. 13. There are t w o numabers, the sunm of whose squares exceeds twice their product, by 4, and the difference of their squares exceeds half their product, by 4; required the numbers. AnSo 6 and S. 14. The fobre wheel of a carriage makes 6 revoluti.ons more'than the hind wheel, in going 120 yards; but if the circum-ference of ea;c{h'Xwheezs l, is increased 1 yard, it will 1make only 4 revolutions, mo.re 0 than. the hind wheel, in the same distance; required the circumf-t rence of each. wheel. Ans. 4 atd 5 yds. i.5[' Two pe rsons, A and B, depa: ifrom. the samine place, and'trav.el in the same direction; A starts 2 hours before B, and after tra:V0Ieling 30 miles, B overt'akes A; but had each of'them traveled i:half a mile more per hour, B would have traveled 42 miles before ovrtakin. A. At what rate did.they travel? Ans. A. 2' and tli $ iltes per ]hour 222 RAY'S ALGEBRA, PART FIRST. 16. A and B started at the same time, from two different points, toward each other; when they met on the road, it appeared that A had traveled 30 miles more than B. It also appeared, that it would take A 4 days to travel the road that B had come, and B 9 days to travel the road that A had come. Find the distance of A from B, when they set out. Ans. 150 miles. CHAPTER VIII. PROGRESSIONS AND PROPORTION. ARITHMETICAL PROGRESSION. ART. 220.-A series, is a collection of quantities or numbers, connected together by the signs + or -, and in which any one term may be derived from those which precede it, by a rule, which is called the law of the series. Thus, 1+3+5+7+9+, &c., 2+6+18+154+, &c., are series; in the former of which, any term may be derived from that which precedes it, by adding 2; and in the latter, any term may be found by multiplying the preceding term by 3. ART. 221.-An Arithmetical Progression is a series of quantities which increase or decrease, by a common difference. Thus, the numbers 1, 3, 5, 7, 9, &c., form an increasing arithmetical progression, in which the common difference is 2. The numbers 30, 27, 24, 21, &c., form a decreasing arithmetical progression, in which the common difference is 3. R E M A R K.-An arithmetical progression is termed, by some writers, an equidifferent series, or a progression by differences. Again, a, a+d, a-+2d, a+3d, a+4d, &c., is an increasing arithmetical progression, whose first term is a, and common difference d. And if d be negative, it becomes a, a-d, a-2d, a-3d, a-4d, &c., which is a decreasing arithmetical progression, whose first term is a, and common difference d. ART. 222.-If we take an arithmetical series, of which tbe first term is a, and common difference d, we have 1st term --...... a 2d term =lst term +d=a+d 3d term =2d term +d=-a-2d 4th term =3d term +-d=a-+3d, and so on. ARITHMETICAL PROGRESSION. 223 Hence, the cobfficient of d in any term, is less by unity, than the number of that term in the series; therefore,,the nth term -a+(n-1 )d. If we designate the nth term by 1, we have Il-.a+(n —l)d. Hence, the RULE, FOR FINDING ANY TERM OF AN INCREASING ARITHMETICAL SERIES. Multiply the common dijfrence by the number of terms less one, and add the product to the first term; the sum will be the required term. If the series is decreasing, then d is minus, and the formula is 1=a —(n-1)d. This gives the RULE, FOR FINDING ANY.TERM OF A DECREASING ARITHMETICAL SERIES. Multiply the common difference by the number of terms less one, and subtract the productfrom the first term; the remainder will be the required term. EX AM P L ES. 1. The first term of an increasing arithmetical series is 3, and common difference 5; required the 8th term. Here 1, or 8th term =3+(8-1)5=3+35-38. Ans. 2. The first term of a decreasing arithmetical series is 50, and common difference 4; required the 10th term. Here 1, or 10th term =50-(10-1)3-=50-27=23. Ans. In the following examples, a denotes the first term, and d the common difference of an arithmetical series; d being plus when the series is increasing, and minus when it is decreasing. 3. a=3, and d=5; required the 6th term..... Ans. 28. 4. a=20, and d —4; required the 15th term.... Ans. 76. 5. a-=7, and d=-; required the 16th term.... Ans. 10-. 6. a=2', and d-3; required the 100th term.. Ans. 35'. 7. a=0, and d=-; required the 11th term.. Ans. 5. 8. a —30, and d= —2; required the 8th term.. Ans. 16. 9. a —-4, and d-=3; required the 5th term.... Ans. 8. 10. a —10, and d=-2; required the 6th term. Ans. -20. 11. If a body falls during 20 seconds, descending 16-;1 feet the first second, 481 feet the next, and so on, how far will it fall the twentieth second? Ans. 627~ feet. R E VI E w.-220. What is a series? Give examples. 221. What is an arithmetical progression? Give an example of an increasing series. Of a decreasing series? 222. What is the rule for finding the last term of an increasing arithmetical series? Of a decreasing arithmetical series? Explain the reason of these rules. '24 R]A.TY'S ALGEBRA, PARliT FIRST. A.RT. 223o —Givetn, tJhe first term a, the comnlon difference 4, and the numntber of terms ni, to find s, the suml of the s —eries. If we take an arithnetical series of which tihe first'term is 3, common difference 2, and number of terms 5, it may be written in the followingC formns: 3, 3+2, 3~+4, 3+6, 3~ -8, 11 —-2, 11-4, 11 —6, 11 —It is oviorus, that the sura of all the terms in either of 1tese lines, will represent the sumn of the series; that is, s-_ — ( 3 -2)+( 3+4)+t 3-i-6)-i-( 3-+i 8) And s I — (11 2) -i-(l-4)-+(11 —6 —)+( —.8) Adding, 2s.-14 1. - 14 -1 14 + 14 - -14X the nuniber of terms" -4Xhs 5 —70 Whence,.s= —- Of 70e —35. Now, let t — the last term, then writing t.e series both in a direct and inverted order, o:,+ (a+ d)++(ca-~ 2d) + (a+ —3c)+ -. o 1 And,== ( -t-d) >-(-2U )+( -— 3d) a By eadding the corresponding terms, we have 2's -(zVa) + (-I a.) (- a)-) + ~a) * -=(zd-sr) taes:( n as xMz am any times as there are tterms (n') in the series. Hieneee, 2 s —(+ ) asE-t/-_i-a: — ( a ) t. T-his formula gives the following FOn, FIN'D4Ni' TriE SUnI OF AN ARn tn IMiETItCrAI SEiRIES. 3t!fu/ti/ y ha/if the s uat of the two exytrees, by the ntarinber of tf'ern'r.. Froim the preceding, it appears, that the suaot of the 0eae;e.xr es is equal to t oe mmn,/f any other twvo terrms ep.i'E// dnieda ntJi emo t/e 1R,. ite r —Since / -...-.-(n. —t.)drc, if we s:bstittte this in the pXlace of X in the forienlt c=sbc(zoe — (, t b,(coules = - This -give. the fbllowring R.llse, for dfinding t1e~ sum of an aritlhetiea.l series:.'o the double t te. d he tp ec te oet oft the. tc1 ier1 of tews eiss oe, bl f the com-aon ct.t d reeee, esdt tt ht-o1ilyl the ctci f ibyhef thle eizcbery of t0)'Jis Rtz vt x xw. — 223. Wha. t is the rnle or fir findig the stam of an arith. mleti — ca3! eries? Expolsin timhe ieas.. of the rule. ARITtT iETlICAL PROG-RESSION. 2 2 EX AiMPLE S 1. Find the sum of an arithmetical series, of which the firsterim is 3, last term 17, and number of terms 8. ( - ) 8 —80. Anso 2, Find the sum of an arithmetical series, whose:first term is 1, last twerm 12, an.d number of term1s 12. Ans. 78. 3, Fin the sam of an arithmetical series, w hose first term is 0, cormmon diferenc 1I, and number of terms 2.0. Ans. 1 90. 4. Find the sum of an arithmetical scries, whose first term is 3, common difference 2, and numnber of terms 21. Ans. 483. 5, Find the stu vn of an aSlri'thmeltical series, whose first terrm is 10, comm:aon difference -3, and number of terms 10. A. -t. In -this case, the sum1 of the negative termas exceeds that of the pos tIve. ARn., 2StA. —The cquatioiins lz —ar ( —-l)d and:s-K(at-l), firnish'the means of solving this gencral problem: nzovwinog any three qf the fitve r.tati-es au, d, n, 1I, s, zwhi/ch, ent er in2toen a tritinetiee2ta series, to d term,i'dne the other two. This question furnishes ten problems, t:he solution of which pmre sents no dificulty; for we have always two equations, to determinei thet two unknown quantities, and the equations to be solv ed, are either those of the first or second degree, 1 Let i t be recquired to find a in terms of t, an, and d From the first fomimulia, by transposing, w leave taz.d:-it —. l )d That is, th1e fi~rst tleBrm f an inacreasingt arithnmcztical series is equacg to the test feorm, dini?,inizl?.:d by the prod.cn'tct Of the con m'2n, diftl'rnce into,,the usaibter of lte -ets less oie. From the sane formula, by t.lransposing a,S and dividing b.y n,1-, we find cd.-: That is is, ant ny arith}netfictl series, thgle con'mnon d:Jif`irence. is egruat to the difference (f the' extremes, divided by the nunc222er ojf' teris less one, Examples, illustrating -these principles, will be foun.d. in the col leetion at the close of this subjct. Itv:rL -—,2214. What are the fundamental equations of artlhmetic.al progression, aend to what general problem do they give rise? To'wh3t is the first term of an increasing arithmetical C series equal? To wbhat is the common. difference of an arithmethical series equal,? f326 RtaY'S ALGEBRAI PART TItRST. ApTo -225 3y imeans of the preceding principle, we are enabled to solve the fobllowing problem. Two numobers, a and b, being given, to insert a nunmber, m, of arithnetieal means betwroeen them; that is, so that the numbers inserted, shall form, with the two given num bers, an arithmetical series., Regardinog a and b as the first and last terms of an increasing arithmietical series, if we insert om terms between them, we shall have a series consisting of.m2 terms. But, by the preceding principle, the common difference of this series will be equal to the difference of the extremes divided by the number of termois less b —a. b-a one; that is, d —b b- therefore, the commoan deifi rence wilZ be equal to the dff'erence of the tl wo itnumbers, divided by the nuntber of inean6sp plus one. Let it be required to insert five arithmetical mneans between 3 and 15. Itere d- -i- 2; hence the series is 3, 5, 7, 9, 11, 13, 15. It is evident, that if we insert the same number of mneans between the consecutive terms of an arithmetical series, thle result will formn e, new progression. Thus, if we insert 3 terms between the consecutive terms of the progression, 1, 9, 17, &c., the new series will be 1, 3, 5, 7 y, 9, 11, 13, 15, 17, and so on. E XAMPLES.g 1. Find the sum of the natural series of numbers 1, 2, 3, 4,, carried to 1000 termas. Ans. 500500. 2Q Required the last term, and the sumn of the series of odd numbers 1, 3, 5, 7,... continued to 101 terms. Ans. 2J01 and 10201, 3S. I-tow many times does a common clock stri]he in a week? Ans. 109)2. 4. Find the nth ternm, and the sum of t terms-s of the natural series of numbers 1, 2, 3, 4.... A i..it, and 2}n(n-i-l). 5. Find the nth term, and the sum of n terms, of the series of odd numbhers 1 3, 5, 7. Ans. 2.n-1, and 2a.. 6.'The filrst antd last terms of an arithmetical series are 2 and 29, and tXhe comnmon difference is 3; required the number of termms and the sum of the series. A ns. 0 and 155 7. The first a.nd last terims of a deereasing arithmretical series are 10 and 6, and the number of terms 9; required the commron difference, and the sem of the series. An.s, t ad 2t. ARITTME TI CAL.PROGRESSI N0 2'27 8. The first term of a decreasing arithmetical series is ].0, the number of terms 10, and the sut of the series 85; required. the last term and the comminon difference. Ans. 7 and - 9. RequVired the series obtained from inserting four arithmeticat means between each of the two terms of the series 1, 16, 31, &e. Ans. 1, 4, V, 10, 13, 16, &ee. 10. The sum of an arithnaetical progression is 72, the first terra is 24, and the common difference is — 4; required the number of' terms. Ans. 9 or 4. In finding the vMue of a in this question, it is required. to solve the equation n' -3n —— 36, which has two roots, 9 and 4. These give rise to the two following series, in botlh of -lwhich the sum is 72. First series, 24, 20, 16, 12, 8, 4, 0, -4, — 8 Second series, 24, 20, 16, 12. I 1..A man bought a farmzl, paying for the first acre 1 dollar, or the second 2 dollars, for tie third 3 dollars, (and so on;'when he came to settle, he had to pay 12880 dollars; how hmany acres did the farm contain, and what was the average price per acre? Ans. 160 acres, at n80{ per acre. 12. If a. person, A, start fromn a certain place, traveling La mile8 the frst day, 2ca the second, 2a the third, and so on; lanid at the end of 4 days, B start after him fronm the same place, traveling uniformly 9a miles a dcay; when will B overtake A? Let xz the number of days required; then the disnrtce tra-veled by A in x days =a1-2a-a3a, &e., to z termse,:t-cx(r 1); and the distance traveled by B in (x —4) dalys =9a.(x-4). Whence ax(x:-+-1) -— a(x-r 4). Froim which xzS, or 9. Hence, B overtakes A at the end of S dIays; and since, on the ninth day, A tiSaives 9a1 miles, which is B's uniform rate, they'will be together at the end of the ninth day. This is an instance of the precision with which the solution of an equation points out the circumstances of a problem. 13. A sets out 3 hours and 20 minutes )efore B, and travels at the rate of 6 niles an hour; in how smany hours will B overtake A, if Ie travel 5 miles th2e first hour, 6 the second, 7 the third, arnd so on? tAns 8 hours. 14. Two -travelers, A and B, set out fron. the same place, at the same tuine.. travels at the constant rate of 3 mniles an houtr, but B's rate of traveling, is 4 muiles the first ho-r, 3'the second, 3 the -third, and so on, in the same series; in how umany hours will A overtatke B? Ans. 5 ours,. a wv. -2.25. }low do you insert i arithmetical meuans beteyeei t;o given ltu.boers? 9528 ]RAY'YS ALGEBRA, LART FIRSTP ARTa, S2til —A. Geometrical Progression is a series of terms, each of which is derived froni the preceding, by multiplying it by a constant quantity, termed thoe ratio. Thus, 1, 2, 4, 8, 16, &c,.is an increasing geometrical series, whose colmmon ratio]. is 2. Also, 54, 18, 6, 2, &e., is a decrerasing /geometrical series, whose commr.1 on ratio is:, Generally, a., ar, art', as, e.,, is a geomeu-trical progression, whose common. ratio is r, and which is an increasetng or decretasing series, according as r is greater, or less than 1, It is obvious, that the commn on ratio in any series, will'be ascertained by dividi.ng any terrm of the series, by that whfich iummere diately precedes it. R E X A it x-.-. - A geometic-lal pt)rogreson is termed, by somel writers, an eqei-rcrcioca/? series, or a series of coJ,/in.ed,rqeort/tioxces, or a,prorerssioaT by qteoie/.sts. AaT. 2721o —To find th'e last termn of the series, Let a denote the first termn, r the conmnon ratio, I the nth term, and s the suln of n' teriis, tLhen, the respective terims of the series will be 1, 2, 83, 4, 5..a 3, ii 1 s —2 s,2.i $' a, ar, a t, a5 c... a r-", ar"' 3 a-2'r.t That is, the exponent of si in the second terIxm is I, in" c' I/itrd terim, 2in the,.fo ri"i terune 3, and so on; hence, the neth ternm. of the series wiill be, l —ar1,'-;'that is,.Atyg festrn ofp7 a g/eoneetric series- is egq, al Ie rodetset At/lse first fternsi, by ite raIio raised to a power, -whose eptonen.t is one less tfialn the nusueIber if lerais. EX1APIPLES. 1. Find the 5thl ter. of the geometric progression, whose, first term is 4, and commion ratio 3. 3ic —43X 3X3SX —-- 8, and 81 X4- -324, fleo fifth termn 2. Find the 6th term of thfe progression 2, 8, 32, &c. Ans. 2048. 3. (Given. the Ist teirm 1, and ratio 9, to.find the 7th termc Ans. 64. 4. Given the est term 4, and ratio 3, to find the 10Oth term. Ans. 78732. R, s::' r w. —-226. Wthat is a Geo~metrical 7Progression? G -xive examples of an increasing, and of a decreasing geometrical series. IHowr may thic common ratio in any geometrical series 3be found? 227. iow is any Sterl'of a geometrical series:found? Expliain te principle of this rIle. GE}OMETRICAL PRtOGRESSIO.N 229 5. Find the 9th termn of the series, 2, 1.0, 50, &c. A. 7125 Qo 6, Given th.e 1st term 8, and ratio, o find. the 15th temn, Ans. 04~'g 7, A man purchased 9 horses, agreeing to pay for the whole what the last would cost, at 2 dollars for the first,; for the second, &e.;' wlhat wvas the average price of each? Ans, $14 458..\r. P.-26r —To find the sum of all the termis of the series. If we multiply any geometrical series by the ratio tlhe result ~will be a new series, of which every term except thie last, will lhave a corresponding term in the first series, Thus, let a, ar', art, ara3, &e., be any geoometrical series, and, its sumni1, -then s=a-Lar-t,-'2-+-ar. s. c o a.-...... r Multiplying this equation by r, we hiave rs +a2+aA a4., o - < -.ar' o e The terrns of the two series are identical, except the figrs term, of the first series, and the last term of the second series. If, then, we subtract the.first equation froma the second, all the remaniing terms of the series will disappear, and we shall have Or ('-1 )s=we(.-.l) Ience, e -— =' ) Since t:=?e....., we have r 1..=as' Theref'oc, F a t tI:enee, tle RULE, vFOR VrNDINGoT rutH'sc or, A c oMusurexoUA SE.SES. ]1fttihpty the east te?a by the r'atio, fi'om tire pro'octet sebtract lute first terin, iand divlde the renmtainder by the rratio less one. ~ X A iMIIP L ES. 1, Find the sum of 10 terms of the progression 2, 6, 18, 54, &e. The last term ==23 —\X3=2X1963 — 39366G S. — _ 18098 — 59048. Ans. ~.-. 31 —1 2. Find the sums of 7 ternms of the progression 1, 2, 4S, &. Arise 127o 3, Find the sumz of 10 terms of the progression 4, 12, 36{, &c. Ansi st 1t 86096. 4. Find the stmn of 9 terms of the series 5, 20, 80, h&..As.s 4690305 5, 3Find the snumt of 8 terms of the series, whose first ter-m is r 6-'-, atntd ratio a *A G-0 -' 4 230 RAY"S ALGEB3RA, PRT FIRST. 6. Find t.he suam of 8- -20+- 50+, &c., to 7 tortrsue. A 32449 7. Find the sum of 3+41-+6"J-, &e, to 5 terms. A. 9- t9 11t xrs a Rc.-If the ratio r is less than 1, the progression is decreasing', and the last term' ias less than a. In order that both terms of the fiaetion shall hbe positive, the signs of the terms must be changed, and w-e haver s —'. The suml of the series when the progression is deereasing,. —i' is, therefore, found by the same rule, as when it is increasing, except that the product of the last term by the ratio, is to be subtracted. from the first term, and the ratio subtraeted fronm unity, instead of sxobtracting unity fr'om the ratio. S. Find the suam of 15' terms of the series 8, 4, 2, 1, &c. Anls. > o 4 5. 9. Find the sum of 6 terms of the series 6, 4-, 34, &c. Ans. l9 13w 1.. a-as"r A.uRT. 229-Thhe form3ula s- C, by separating thie namerteGr into two parts, may be placed under the form a a?" Now, g when r is less than 1, it must be a proper firactiort, whicl may be repese nted by;P. te' (-( S) i Since p is less than q, thlse higher the power to which the frlaction is ra.ised, the less will be the nxnmerator compared with'the denominator; that is, the less will bhe te value of the fraction; therefbire, when X becomes very large, the value of o r"21 will be very sm all; and, when?, beoomes iirfnsitely great, the value of -;, or,'-, will be tin finitely sa mall, t hat is, 0. But, when the numerator of a fr'action is zero, its vaue is 0. This reduces the value of:, to a- H[ence, wches the stnumber of tersms of a decreasing geon metrical series'is tifihite, the last term.t is zero, ca;nd the sumn is equtal to th.e first enain divided by one smiS ts the ratio. R B vi w. —228. What is the rule for finging the sum of the termns of a geometrical series? Explain the reason of this rule. TWhen t}he series is dererasing, how must the formula, expressing the suml be'written, so that both terms of the fraction may be positive? 229, What is the rule for:fuding th e sum of a decreasing geometrical series, when the naunber of terms is infinite? Bxplain the reason of this rule. GEOLMETRICAL PROGRESSION. 231 1. Find the sumn of the infinite series I — +,: c. a I s Here a-zl, r — a4nd An 2. Find the sum of the infinite series 1+-K-K+m+, &c. Ains. 2. 3. Find the sum of the infinite series 9-+6+4-}-, &c. A. 27. 4, Find the suni of the infinite series i —— I — t-4 -+, &c. Ans. 5. Find the sum of the infinite series 1+- -- -~-, &c. 2X tSo -- 6. Flnd the saun of the infsite geometrical progsession a —-- -s &.,1 in which the ratio is- ns - 70 If a body moves 10 feet the first second, 5 the next second,'22 the next, and so on, continually, how many feet would it muove over? Ans. 20. AeRt, @. —The two equations, a —ar' —- anud s —-- --- r nish this general problem-: knotwing three of thae le quantlities a, r,,1~, an s, of a geometr'icat. progression, to defermine the fte other two. This problem embraces ten different questions, as in ar'ith meatical progression. Some of the cases, however, involve tihe extracetion of high roots, -the application of logarithms, and the solution of higher equations than ha ve been treated of in the prev ceding pages. The following is one of the mLost simple and. useful of these cases, Having given the first and last terms, and the number of terms of a geometrical progression, to find. the ratio. Here i orrT ", e __1= Hene rinsj (* )v 1. The frst aznd last terms of a geomsetrical series, aire 3 atnd 48, and the number of terms 5; reqtired the intemrediate tierms. Here i-48, a-=3, n-1l5-l-1-4 ience, rs'44s - 6, and 2_ v 64, and ry'4-2 2. In a geomnetrical series of three terms, the first andd last terms are 4 and 06; required the middle term. Aus. 8. In a geometrical progression, containing three terms, the middle term is called a lean proportional between tble other two. 3. Find a niean proportional between 8 and 32. Atns. 16. 4. The first and last ternas of a geometrical seres are 2 and 1 62, and the: number of terms 5; required the ratio. Ans 3.s 232 tRAY'S ALGEBRA, A. RT FIRST. BATI-Ot AND PROPO.GRTI'ON. ART.. *231.-Two quantities of.the saume kind, my be conm pared in two wa5,VS: lst. By finding how ezmuci the one exceeds the othler. 2d.o By findinog hosw.many tinzes tilhe one contains the other. If we compare the numbers 2 and 6, by the first method, we say tla.t 2 is 4 less than 6, or that 6 is 4 greater thanu 2. If we comlpare 2 and 6, by the second method, w5e ay that 6 is equal -to /tree tiz'es 2, or that 2 is oane iti:rd of 6, Th-is mnethod of compasrison gives rise to proportion. 2A.RT 22. —R.a-ti'aio is the quotient which arises from dividing one quantity by ianother of the sa.nme kind. Thus, the ratio of 2 to 6 is 3; the ratio of a to ma is8 i, ti_; tfXn i N -18at, Iln oetparing two numbers or qu(tntities by their quotient' the number expressing the ratio which t.he first heara;Xs; t3 o tihe seeond, will depend on whichl is made the sttandaerd of comopia1triso2n. Thus, in. comparing 2 anud 6, if we make 2 the utit of ireasuare, or stan.dard, we i:tjd, tlhat 6 is three Lc nes the e standard, If we makeo 6 the unit of meas-tnse, or stanlldard, we find, that 2 is oze thi rd of the standard. In finding the ratio of one nltmbher to anothelr, the French ma thematicians a'ways nakte th.eofi'rs~t of the two 1numlsh)bers tho st-landaird of com.parison; while the Singlilsh omake the last named the standard, Thus, the French sa.5y the. ratio of,-2 to 6 is 3; while the English say it is &s The French met, hod is now generally used, in the United States, though, in a few works,, the other is still retilneod, 2d. In order that; two quantities, nany. be co'pared, or nhave a, ratio to o teah other, it is evidently necessary that, they shotld bhe of the state kind, so that one tny be some s, parti of, or some Lnumber of times the other. Thus, 2 yards has a ratio to 6 yarts, because the latter is tlircee timsses the forimer; but 2 yarnds has no ratio t-o 6 doltars, since the one can not be said to be, eitl.her greater, less, or any annmber of thmes the other. ART. e233v, —' When two numbers, as 2 and 6, are compared, thefi first is called t.he anteceden, and the second the cons'eqt,.ae.t An antecedent and consequent, when spoken of as onze, are called a cont?.pt,s When spoken of as two, thtey are called the ter6ms of the ratio. Thus, 2 and 6 together, form a couplet, of which 2 is the -first ters12m, andd 6 the second.,ART..234. —RatiO is expressed in two ways..st; In the ionrm of a fraction, of which the antecetent is the destoz.inator, antd the con2seqsttent the stue.'eri tor..Thus, the ratio of 2 to 6, is expressed by 3; the ratio of 3 to 12, by 7, Yoe ~Rl mvX ws.-2 31., In how many ways, may two quantities of'the s ame kind be comp-ared? Comupare tho numbelrs 2, and 6 by the first method. Bly the second, 232. Wha-t is ratio? Give an illustration. 233, When two numbiers arc ecomparedt, what is the first called? The seondr? Give a.,n example.e IRATIO AND PROPO RTION. t233 2d, By ptacing twro points (:) be'tweenb the terms of the ratios Thus, the ratio of 2 to 6, is-written 2:e; the ratio of 3 to 8, 3 8, 6,c ART. 23. -5 —ThJ1e ratio of two.quantitiies, may be either at, whole number, a common fracti.on, or an interiainate decimal, Thus, the ratio of 2 to 6 is 2, or 3. The ratio of 1 0 to 4 is Q, or 2.236'1The ratio of 2 to vS /5 is, 2 r. 18W'e see, from this, that the ratio of two quantit.ies can not always be expressed exactly, except Sby symbols; but, by taking a sufficient number of decirnal places, it nmay be found to any required degree of exactness. AnrT. 236,g-Sincee the ratio of two numn-bers is expressed by a fraction, of which t.he antecedent is the denominator, and the consectuent the numerator, it follows, that whatever is true with regard to a fraction, is true with regard to the terms of a ratio. Hence, 1st. T.b mulzti-ply tie confsequeent, or to dividte the antecedent of a -ratio by a.cy,iunber,'17amtipties the ratio by teil ntmbin r. (Articles 122, 125.) Thus, the ratio of 4 to 12, is 3. The ratio of 4- to 12X5, is 3X5. The ratio of 4- 2 to 12, is 6, which is equal to 3X2. 2d. lb divide the conaseq.ettent, or to m7uttply f/the aznteceden/ of a'ratio by a nyna, umber, divides the,ratio by that nzumber, (Articles 123, 124.) Thus, t~he ratio of 3 to 24, is 8. The ratio of 3 to 24 —~2, is 4, which is equal to 8.-2. The ratio of 3X2 to 24, is 4, which is equal to 8-. 2. 3d. To multiply, olr divide, both, the antecedent and consequtelnt lf? a 2ratio by anyi.?tnumber, does 2not alter the r"atio. (Artieles 126, 127.) Thus, the ratio of 6 to 18, is 3. The ratio of 6X2 to 18/2, is 3. Trhe ratio of 6~-2 to 18-~-2, is 3. A.r. 2273 — When the two numbe-rs are equal, the ratio is said to be a ratio of eqtuality. When the second number is greater than Br:,svsX-w. -23.4:. When are the antecedent and consequent of a ratio ca.lled a coupletd? When the ternms of a ratio? By whiat two imethods is ratio expressed? Give an example. 235. Wihatt forms -nay the ratio of two quantities havelr? 236. flow is a raltio afitaceted hby Inultiplying th.e Consequent, or dividing tlhe antecedent? Why?.ow ris a- ratio affected by dividing the conseque'nt. or multiplying the antecedent?'Why? low i s a ratio affected, by either multiplying or dividing both ant codent and cons.equenat by tie sam numtber? Why? ~34 RAKYS AL1GEBRAa PART FIRST. the first, the ratio is said to be a ratio of greater inequt.ality, and when it is less, the ratio is said to be a ratio of less inequalitty. Thus, the ratio of 4 to 4, is a ratio of equality. The ratio of 4 to 8, is a ratio of greater ineqnuity. The ratio of 4 to 2, is a ratio of less inequality. W'e see, from this, that a ratio of equality may be expressed by 1; a ratio of greater inequality, by a number greater than 1; and a ratio of less inequality, by a number less than 1. ART. 238 —-When the corresponding terms of two or more ratios are multiplied together, the ratios are said to be compoundced, and the result is termed a copouwnd,ratio. Thus, t1he ratio 130, compounded with the ratio;, is ) In tbhis case, 3 multiplied by 5, is said to have to 10 multiplied by 6, the ratio compounaded of the ratios of 3 to 10 and 5 to 6. AnRT 239. —Ra.tios may be compared with each other, by reducing the fractions which represent them, to a con.mmoot denoninator. Thus, to ascertain whether the ratio of 2 to 5 is greater than the ratio of 3 to 8, we have the two fractions, i and -, which being reduced to a common denominator, are and a-f; and, since the first is less than the second, we infer, that the ratio of K2 to 5 is less than the ratio of 3 to S. PROPOR TI ON. ART..4O.-Proportion is an equality of ratios. Thus, if a, b, c, b d d are four quantities, such that b s equal to, the a, 3, c, d form a proportion, and we say that a is to b, as c is to d; or, that a has the same ratio to b, that c has to d. Proportion is written in two ways. Ist. By placing the double colon between the ratios. Thus, 2d. By placing the sign of equality between them. Thus, a b —-c: d. The first method is the one generally used. From fthe preceding definition, it follows, that when four quantities re in proportion, the second divided by the frst., gives the same qeotient as the fourth divided by the third, This is the test of the proportionality of four quantities. Thus, if 3, 6, 5, 10 are Ravr:w~..-237. What is a ratio of equality? Of greater inequality? Of less inequaity? Give examples. 238$. When are 1wo or more ratios saaid to be compounded? Give an example. 239. ow may ratios be 0om1pared to each other? 240. Wtht is proportion? Give an example. H;ow are four qatntiets in proportion written? Give exareples. RATIO AND) PROPORTION, 235 tohe four terms of a true proportion, so that 3: 6:: 5: 10, we must have 0s10 If these fractions are equal to each other, the proportion'is true; if they are not equal to each other, it is false. Thus, let it be required to find whiether 3: 8: -: 5 The first ratio is s the second is 5- o and lo therefore, 3, 8, 2, 5 are not proportional quantities. 1 E M KAs sx.-The words ratio and proportion, in common llanguage, are,sometines confounded with each other. Thns, two quantities are said to be in the proportion of 3 to 4, instead of, in the ratio of 3 to 4. A ratio slbsists between two quantities, a proportion only between Jour. ]It requires tszo equal ratios to form a proportion. ARTs. 924i —-In the proportion a: bt: c d, each of the quantities a, b, c, d, is called a term. The first and last terms are called the extremaes, the second and third, the means. AnT. 242-Of four proportional quantities, the first and third are called antecedents, a-nd the second and fourth, conseque'nts (Art. 233); and the last is said to be a fourth proportional to the other fthree, taken in their order. AnT. 243r.- Three quantities are in proportion, when the first has the saime ratio to the second, th1at the second has to the third. In this case, the emiddle term is called at mean. proportional between the other two. Thus, if we ihave the proportion a:::: e then b is called a 2zean proportional between a and c, and e is called a thirdl proportional to a and b. ART. 244-PROPOsRTION I.-J/l every proportion, Ih.e iproduci' f the means'is equtal to the product of the extremes. Let a b:: c:d. Then, since this is a true proportion, the quotient of the second divided by tlhe first, is equal to the quotient of the fourth divided by the third, Therefore, we must have b d a C Multiplying both sides of this equality by ae, to clear it of frace abe ade tiens, we have- Or, zad, Illustration by numbers. 3: 6:: 5: 0, and 6XS= 3XI0o;R vi x w.-.240. Give examples of a true and false proportion. What is a test of the proportionality of four quantities? 241, What are the first and last terms of a proportion called? The second and third termns? 242. What are the first and third terms of a proportion called? The second and fourth? 243. When are three quantities in proportion? fGire an example What is the second clterm called? The third? 236 RASY S ALGEBRA, PART FIRST. Fromn the equation bc=aatd, we have d=bc, c ad b a and be cc b e d$ fromn which we see, -that if any three terms of a proportion asro given, the fourth may be yreadily found. The first three terms of a proportion, are atc, bd, and acey; w hat is the fourth? Ans. bdxy. gt x AR a. —— tls proposition furnishes a more convenient, tet of the proportionality of four quantities, than the method given in Article 240. Thus, to ascertain whether 3: 8: 2: 5, it is merely necessary to coinpare the product of the meaens atnd the extremes and, sincea 3X5.is:no equal to 8X2, we infer that the proportion isfalse. A.nT. 6T2i3$6 —PoProsIrTItoN I.-i-Cnver'sely, if lie proadlt qf two qulantitties is equccc t~o the prod uci oj two others, ivwo oft thers mzay be nmde e Jd'he cea's, ar'd icte otherw two tice extreeaces of a proporlion. Leot v gbc-ad& Dividing each of tfhese equals by ac, we hrave be ad b d......;Or, That is, a: b. c - d. Iilustration. 5SX 4X 1O, and 4 - 5: S - 10. Art, 24. —Paorosir oiOT It 1 —- / thr"eec qpuantiiies cae i'i con ieath iced uroportion, t/,e jo dccl qc/' lthe extr3'ces is eCTcal t to the stqupcre qf ice Mea'ts. If a:b: b c Then, by Art. 244, ac —bb=zb'2 I. follows, from Art. 245, that the converse of this proposition. is also true. Thus, if ac=-&2, Then, a: b::b e c. That is, if the product of thie first and third of tao qutantilies, is equat to btlie sqt~uar'e of a secortd, tce'iest is tlo the second, as lihe secfond is to the thirdo llustration. If 4: 6: 6: 9, then 4X9 —62-=36 If 2X&8=i6, then 2: 1/16 /i6 8 Or 2:4::4:8. AnT. 24-.-PRoP'o SITION IV'. —,yTffour.quZantities, care inZ propor2 hiot, tihe/y wilt be i'n propor"tion by ALTERNu TIoN; tl.ct'htis, thejirst till isc,;e ithe saqne ratio to lthe third, hatl the s'econd hais to thle jbtcr'7h. Let a: ab c: c'. This g odesa — 6 Midtlplyying both ides by c, -.t.ct cc RATIO NAD PRO PORTION. 237 Dividing both sides by b, a —c Thatis, a: c: b d. IlXlustration~ 2 7: ~: ], 6 and 21 - 2: 7e 21 APa.T. 24- -— PR OPOSTION'V.-To f} Jf tur quanctities are it proporItio, they wzil b oein2 roportion by INV.SINn; thait is, the second will be to the frst as the fo-urth to &the irdid. Let a:b: c:d. By Art. 244, adobc. Dividing both sides by b, ad a- C Dividing both. sides by b-d, That is, b a: d: c. IlIustration, 2 5:: 15, and3 5: 2: 5. A.RFT. 2:.9t —-PRO)POSITI ON VI. —f two SC3 qf' roportions ]have an anteceden- and coensequen.t i-n, the one, equagal to an antecedent atnd conseq.gsett in?, lte oliter, tiQcnhe'e o'7 te rm, s will be proportional. Let a'b' c:d (I.).And a:b: e:f (2.) Then will c e: f For, from Ist proportion,.Ioence, -c.. e 0 ~ This gives, c: d:: e f Illustrat1ion. 3:;5: 6 10 3: 5::: 5 And: 10::9: 15. EI M A. —This propositiio is generally termed equality of rcattio, t is almost setf- evidet,;. AnT..2@,-0-PnorossTIon TII. fJurtt qPuantties ar2 e -ea prioporlion, they wiltl be inz proportion bty COMPOSsi'oIN; that is, the'mst.m of, ihe first and secondt, etill be to the second, as the stun of the thir d and ft:urth, is to the Jb'urth. Let: b::c: d Then will a+-b b:: c+d: d From thae t propoatioan, b=ad, b;y -Art. 244, :~S238 ARA 22Y ALGTEBRA tPART FIRSTT Adding bdn to each, bdtbd, bc4bd= —ad+-bd; Or b(c+d)-=d(a+b). Dividing elch side by c-(bd,) c+d b d By 6+b, an-b- c —— d' This gives, ar4- b: b:: ct+d d. Illustration. 3 4: 6: 8 3+4:4: 4:6+88; Or, 7:4:: 14 S. it E M A RI K.-In a similar manner, it imay be proved, that the sutm of the first and second terms, will be to the first, as the sum of the third and fourth is to the thirdc ART. 2 — PRo.rIoro YTI. -—.f foiur quatzities are in Proportion, they will be iz proportion by Divxsrio,; that is, the differcnce oj' the first and second, will be to the second, as the difference oq the ihird and fibns, is to the jburth. Let a: b:::: d, Then will a —b b: c-d d From the 1st proportion, bcad, by.Art, 244. Subtracting bd fronm each, bd-:bd bc-bcd=adZ bd 5 Or, b(c-(d)=d(a 6-). Dividing each side by c-d, b- d(a ) b d By a —b d This gives, a —b: b:c —d d Illustration. 8: 5:: 16: 10 8 —5~5:: 16 10: 1I0; Or, 3: 5 6 I 0 g a a. A R SC.-In a similar manner, it may be proved, that the difference of the first and second will be to the first, as the difference of the third and'fourth is to the third. ART. 22 —PnorosxITION IX.-Jf four quantities are in propavlion, he sum of the first and second till be to th eir di frence, as the sum of- the third and fourth is to their difference. Let a: b::c:d, (1.) Then will a+qb: a-b cd: cd From the Ist by composition, Art. 250, a+b: b: c-+-d d. By aternatio, a+b: c+d:: b: d, Art 247~ RATIO AND PROPORTION. 239 This gives, at+-b From. the 1st, by division, a-b b:: c9 —d: d, By alternation, a —b: c —::: b: d; c —d d c+dl c —d This gives, - henee a —b heb axb-'a- b' That is, a+b: c+-d: a- b: c-d; Or, by alternation, a+b: a-b: c+d: c —d. Illustration. 5: 3:: 10: 6 5+3 5-3:: 10+6-: 10-6 Or, 82: 16 4. ART. 25 $.-PROPOSITION X.-Iffour quantities are in proportion, like powers, or roots, of those quattities, will also be in proportion, Let a b c: do, Then will a4: b:: c. dd":b d For, since -ba c If we raise each of these equals to the nth power, we have: an Cn That is, a' *. b. c: d% TWhere n may either be a whole number or a fraction. [llustration. 2 3:: 4: 6 2:2 32:: 42: 6 Or 4:9:: 16:36 Also,:':: n'n2a2: Mn2b And.d2 a': Vb:: V/.mi' ~2':'/;2/2 Or a b:: nma s zb. ARTs. 24@ —Pno0roSITION XI. —If two sets of quantities are, ri propor on, the pro&tcts of the conresponding terms will also be in proportion Let a: b: c:cd, (1.) And,sz.,r-,s; (2.) Then will am b::er ds b d n s For, from the I st, ac; and from the 2d, -. —:1ultiplying these equais together, b is d Qs bn di -X-""-X- /or -- a 711 c r amn cr This givs,:n bn:r: r: s6 240 RAY'S A.LGEBRA: PART FIRtST. Illurstra'tion,: 5: 6: 10, 4: 3: 8:6, 1.2: 15 48: S60. ART.. 255~ —- PrIEPOSTION So.- JIIrh any codzsinned prnoposrtion, thagt A, en / otber of poortions?aving e sane ratio, azn?/y osnme antecedenzt is o its con'seqZten, as the sanem of all the antecedendts is to - the sitar of all tihe cotnsequejts. Let a: b c: d: -zt:: n7, &c. Then will a - b:: a-+c-q —n: b{-d+z n; Since a: b ~c g d, we hZve b —ad. Since a: b::'n - n9, i e have bm —:an. Adding ab to each, ab-=abo The seums of these equalities give ab -o+bc b-n2-ab-ad+a;,' Or b(a:-F-ck +ne) a(b'J kFi2-). Di-viding by a+c —n, b a(b.- ) b b -d+-n, Dividing bothi sides by a a a+cF — in'his gives, a: i: a c —L n -. — bt-i-d —n. Illustration, 3 - 4: 6' 8::9. 12 3:4::3 6 99: 4+-8+12 Or~ 3: 4: 18:2,4. R. E Ms A it.-iin the preceding demonstrattions, the proof has generally been made to involvse the definition of proportion, that is, that the four b d quantities, b, c, d, are in propiortion, iwhen. This is regarded as a matter of great importance -to the pupil. If the instructor chooses to dispense with this, as some writers do, several of the de monstrations may be somewhat shortened. There areo several other Propositions in Proportion, that may be easily demonstrated, in a manner similax to the preeedinhg, but they are of 0so little use, as not to be worthy of the pupil's attention. NTO1W P BLIS3ED oD RAY'S ALGEBRA, PART I.-HIIGHEI-I ALGEBRA, RAY'S ALGEB3RA, PART SECOND, for advanced students, contain.s a concise review of the eleamentary principles presented in PxA Rnsti, with -more difficult ex amlies for practice, Also, a full discussion of the higher practical. p.rts of the science, embracing the General Theory of equ-ations with SiruRtI's celebrated theorem illustratedt by examples; tIoaR.ce's. mlethod of resolving ntluerical equations, A&-c., -e. Designed to be a thorough treatise for l-Ie-s S cHooIs and for CoLTrsis. lThe author ias. endeavored to present every sublject in a plain and simple manner, with.nsu-lrou:8 interesting and appropriate illustrations and examples. TIE END