ARTES
LIBRARY
1837
SCIENTIA
VERITAS
OF THE
UNIVERSITY OF MICHIGAN
E PLURIBUS UNUM
TUEBUR
SI-QUAERIS-PENINSULAM AMOENAM
CIRCUMSPICE
QA
35
~H227
1772
•
THE
ELEMENTS
OF
ALGEBRA.
/
1
THE
ELEMENTS
O F
ALGEBRA
IN A
New and Eafy Method;
WITH THEIR
USE and APPLICATION,
IN THE
Solution of a great Variety of Arithmetical
and Geometrical Queſtions;
By general and univerfal RULES.
To which is prefixed an
INTRODUCTION,
CONTAINING
A Succinct HISTORY of this SCIENCE.
By NATHANIEL HAMMOND,
Of the BANK.
The FOURTH EDITION, Corrected.
LONDON,
Printed for the AUTHOR:
And Sold by E. and C. DILLY, in the Poultry; and
W. DOMVILLE, under the Royal Exchange.
M DCC LXXII.
1
vi
The PRE FACE.
t
There are two Things abfolutely neceffary, to
make the Acquifition of any Science as eafy as its
Nature will admit. First, the Difpofition of the
Work, fo that the Rules be clear and diftinct;
and then the Illuftration of thefe Rules by a fuf-
ficient Number of proper and pertinent Examples.
And tho' the excellent Elements of the judicious
EUCLID are of a different Nature, yet in this I
have induſtriouſly ſtudied to imitate him, that I
propoſe no new Rule or Article in this Science of
Investigation, till it become neceffary to carry the
Learner to a further Degree of Knowledge.
Science may be compared to a highly finiſhed
Pile of Building, all the Parts of which being dif-
pofed in the moſt exact Symmetry, they muſt affect
our Perception, and gratify our internal Senfation.
with a more exquifite Pleaſure, than if viewed in a
feparate State: For in fuch a State, to all but the
Learned, they would appear broken and incon-
nected Materials of a mighty Structure, which the
Mind wanting Power to conceive, could enjoy no
Satisfaction in the Contemplation of fuch a Train
of imperfect and confufed Ideas. But, when thus
exhibited in their true Proportion, it will be eaſy,
even for the youngeſt Scholar, to gain a perfect
Notion of each, and, as he advances, a gradual
Comprehenfion of the Beauty reſulting from their
Connection, and how they mutually affift and
ornament each other.
In teaching the abſtract Sciences, Examples have
a ftrong as well as natural Tendency to illuftrate the
Precepts concerning our abſtract Reafonings, efpe-
cially in this Science of Investigation: In the Syn-
thetic Method, or Method by Demonftration, one
Example
The PREFA CE. -
vii
Example or Propofition is fufficient, for a Number
of the fame Propofitions is only a Repetition of the
fame fucceffive Train of Ideas; whereas, in the
Analytic Method, or Method by Investigation, the
fame Conclufion is gained, tho' there may be a great
Variety in the Connection of the Reaſoning, by the
different Difpofition of the Ideas, notwithſtanding
they are directed by the fame general Rule.
And as the principal Difficulty in this Science,
is acquiring the Knowledge of folving of Que-
ftions, I have given a great Variety of theſe
in refpect to Numbers and Geometry, and their
Solutions I choſe to give in the moſt particular,
diſtinct, and plain Manner; and for which the
Reader will find full and explicit Directions. As
it is prepofterous and abfurd, to expect a Perfon
to be a Critic in any Language as foon as he has
paſſed thro' his Grammar; fo, I cannot help think-
ing it wrong to expect a Learner ſhould ſee the
Reaſon of particular elegant Methods of Solution,
before he has practifed a general and univerfal
Method; but after the general Rules are become
eafy and familiar, the Learner may then apply him-
felf to the particular Methods. And I know of no
Work that has illuftrated and exemplified the ge-
neral and univerfal Rules in fo copious a Manner,
as will be found in the following Sheets.
In the Arithmetical Operations, the Decimal
Fractions are continued to two Places only, theſe
being fufficient to fhew the Reader, that if the
Queſtion admits not of an exact Anfwer, he is
yet
near the Truth, and may profecute the Anſwer to
any required Degree of Exactneſs. I have avoided
all tedious Numerical Calculations, as they have no
Tendency
viii
The PREFACE.
Tendency to increaſe the Learner's Knowledge in
Algebra.
The Rules of Vulgar Fractions in Algebra are
omitted, being generally very perplexing to Learners;
but as I have given fufficient Directions how they
are managed whenever they occur in the Solution
of any Queſtion, the Reader will find no Difficulty
in reducing an Equation with Fractional Quan-
tities.
It is neceffary my Reader fhould underſtand
Vulgar and Decimal Fractions in commón Arith-
metick, and the Extraction of the Square Root,
and then I know no Reaſon why a Perfon may not
make himſelf a perfect Maſter of the following
Work, excepting the Geometrical Queſtions, which
he may omit, and proceed to thoſe which require
no Skill in Geometry; for thro' the whole, where
it was neceffary, I have given the fame Directions,
as if I was actually teaching a Scholar.
THE
X
THE
INTRODUCTION.
A
S all Arts have their Beginning, rude and
weak, and reach Perfection by Degrees, fo
that, which is the Subject of the following
Sheets, has been cultivated by fo many illu-
ſtrious Men in our own, as well as in foreign Nations,
that it cannot but appear a natural Introduction to this
Treatife, if we digeft the Hiftory of its Rife and Pro-
grefs into a fuccinct Difcourfe; the rather, becauſe
Books of this Sort are now become very numerous in
ours, as well as in other Languages, and therefore, it is
the more neceffary to record the Names of fuch as have
eminently improved fo uſeful a Branch of Knowledge.
The Word Algebra is certainly derived from the Ara-
bic, but there have been ſome Miſtakes as to its Mean-
ing. When it was firſt introduced in Europe, it was un-
derſtood to be the Invention of the famous Philofopher
Geber; and therefore Michael Stifelius calls it fometimes
Regula Algebra, and fometimes Regula Gebri, whence it
is plain he underſtood by it no more than the Rule of
Geber, or, as we ufually exprefs it, Geber's Rule. But
when we became better acquainted with Arabic Learning,
this Derivation appeared ill founded: In that Language,
this Art is called Al-gjábr W'al-mokábala, which is lite-
rally, the Art of Refolution and Equation. Hence it
is plain, we had the Word Algebra from the Arabic
Name of the Art, and not from the pretended Inventor.
But it may not be amifs to obferve, that the Arabic Name
contains a Definition, or is rather an emphatic Decla-
b
ration
X
The INTRODUCTION.
ration of the Nature and End of this Science; for the
Arabic Verb jábara fignifies to refet, and is properly
uſed in reſpect to Dislocations, and the Verb hábala, im-
plies oppofing, or comparing; and how applicable this
is to what we call Algebra, the Reader, when he is
thoroughly acquainted with this Book, will eafily under-
ftand. As it became better known to the Europeans, it
received different Names; the Italians ftiled it Ars magna,
in their own Language l'Arte Magjore, oppofing to it
common Arithmetick, as the leffer or minor Art. It was
alſo called Regula Cofa, the Rule of Cofs, for an odd
Reaſon: The Italians make uſe of the Word Cofa, to
fignify what we call the Root, and from thence, this
Kind of Learning being derived to us from them, the
Root, the Square, and the Cube, were called Coffick Num-
bers, and this Science the Rule of Cofs. I fhould not
have dwelt fo long on fo dry a Subject, but that it is ab-
folutely neceffary for the underſtanding what follows.
It is a Point ſtill difputed, whether the Invention of
Algebra ought to be afcribed to the Oriental Philofo-
phers, or to the Greeks; but it is a Thing certain, that
we received it from the Moors, who had it from the
Arabians, who own themſelves indebted for it to the Per-
fians and Indians; and yet, which is ftrange enough, the
Perfians refer the Invention to the Greeks, and particu-
larly to Aristotle. Yet, notwithſtanding this, it muſt be
allowed that the Algebra taught us by the Arabians dif-
fers very much from that contained in the Works of
Diophantus, the eldeſt Greek Author on this Art, which
is now extant, and which was difcovered and publiſhed
long after the Algebra taught by the Arabians had been
ftudied and improved in the Weft. But all thefe Dif-
ficulties, which have given fome great Men fo much,
Trouble, may be eaſily furmounted, if we fuppoſe that
the Invention was originally taken from the Greeks, and
new modelled by the Arabians, in the fame Manner as
we know that common Arithmetick was; for this, which
is at leaſt extremely probable, makes the whole plain and
clear,
The INTRODUCTION.
xi
clear, and leaves us at liberty to purſue the Progreſs of
this Art from the firſt printed Treatifes about it.
Lucas Paciolus, a Franciſcan Friar, commonly known
by the Name of Lucas de Burgo Sandli Sepulchri, publifh-
ed at Venice, under the Title of, A Compleat Treatife of
Arithmetick and Geometry, Proportions and Equations, the
firſt Book at preſent extant on this Subject. It was printed
fo early as 1494, and is a very correct Treatife. He af-
cribes the Invention of Algebra to the Arabians, uſes
their Method, and treats very clearly of Quadratic
Equations. After him, feveral Authors wrote on the
fame Subject in Italy, and in Germany; but ftill the Art
advanced little 'till the famous Jerom Carden printed, at
Nuremberg in 1545, in Folio, a Treatife with this Title,
Artis magnæ, five de Regulis Algebraicis Liber unus; and
foon after a fmaller Piece, with the Title of Sermo de
Plus&Minus, wherein were contained Rules for refolv
ing Cubic Equations, which have fince been called Car-
dan's Rules, though they were not invented by him, but,
as himſelf owns, by Scipio Ferreus of Bononia, and Tar-
talea. The next celebrated Writer was a French Monk,
whoſe Name was Baeton, better known to the Learned by
his Latin Appellation of Buteo; he publiſhed in 1559 his
Logistica, in which there was a Treatife of Algebra which
gained him great Reputation: Yet his Excellency lay in
a clear and copious Manner of writing, nor does it ap-
pear that he added any thing to what had been already
diſcovered, except fome Corrections as to Tartalea's
Method of managing Cubic Equations.
Hitherto nothing was known in Europe of the Greek
Analyſis, but in 1575 Xilander publiſhed Diophantus, or
at leaſt a Part of his Works, which are ftill remaining;
and this quickly changed the Face of Things, for it pre-
fently appeared that his was a nearer and more eaſy ivie-
thod, and withal opened a Path to much greater Difco-
veries, which was the Reafon that fucceeding Algebraifts
quitted the Terms made ufe of by Arabic Writers, and
followed his. The Time in which Diophantus flouriſhed
b 2
is
xii
The INTRODUCTION.
1
is not thoroughly fettled. Voffius thinks he lived in the
fecond Century, but others place him in the fourth.
His Works were known to the Arabians, and tranſlated
by them; nay, it is faid, that they have ftill thoſe ſeven
Books of his Arithmetick, which are loft to us. The
famous Arabian Hiftorian Abul Pharaijus, whofe Works
were publiſhed by the learned Pocock, not only mentions
him, but afcribes to him the Invention of Algebra; but
in this he is to be underſtood, as writing according to
the Lights he had; for tho' it be true, that Diophantus
Alexandrinus is the oldeft Author we have which treats
expreſsly of the Analytic Art, yet the Footſteps thereof
are viſible in much older Writers. Theo, who is thought
to have explained the five firſt Propofitions of the thir-
teenth Book of Euclid in the Analytic Way, gives the
Honour of this Invention to Plato; and indeed, it feems
very agreeable to his Genius, and Method of Reafoning
on Mathematical Subjects. By the Junction of both
Lights, and a proper Connection of the Arabic Method
of Inveſtigation with the Greek Terms, which were ſhorter
and eaſier, Algebra quickly became a much more uſeful,
as well as confiderable Science, than it was before.
In our own Country, the firſt Writer upon Algebra
that we know of was Dr. Robert Record, a Phyſician,
who diftinguiſhed himſelf in the Reign of Queen Mary,
by his Skill in the Mathematicks. He firſt publiſhed a
Treatife of Arithmetick, which continued the Standard
in that Branch of Knowledge for many Years, and in
1557 he fent abroad a fecond Part, under the Title of
Cos Ingenii, or the Whetstone of Wit, which is a Treatife
of Algebra; the Word Cos alluding to Coffick Numbers,
or the Rule of Cos, by which Name, as we have before
fhewn, this Art was known abroad. This Treatife is
really a great Curiofity, confidering the Time in which
it was publiſhed, and together with his other Works,
muſt give us a high Idea of this Man's Induſtry and Ap-
plication, whofe Memory notwithſtanding is almoft buried
in Oblivion. But, notwithſtanding the early Publication
of
The
INTRODUCTION.
xiii
of this piece, and that fome English Gentlemen had in
their Travels acquired fome Knowledge of this Kind, as
appears by a Spanish Treatife of Algebra, publiſhed by
Pedro Nunnez in 1567, yet it continued to be fo little
cultivated in England, that John Dee, in his Mathema-
tical Preface prefixed to Sir Henry Billingfley's Tranſlation
of Euclid, printed at London in 1570, fpeaks of it in
very high Terms, and as a Myſtery fcarce heard of by
the Studious in the Mathematicks here. It is however
plain, from fome of his Annotations on Euclid, that he
was tolerably verſed therein, and was even acquainted with
the Manner of applying it to Geometry. In 1579 Leo-
nard Digges, a great Mathematician for thofe Times,
printed a Treatife of Algebra in his Stratioticos; after
which it came to be better known and more ftudied, to
which contributed not a little, the Improvements made
by the Author I am next to mention.
Francis Viete, better known by his Latin Name of
Francifcus Vieta, was a Native of Poitou, in France, and
Maſter of Requeſts to Queen Margaret, firft Wife to
King Henry IV. His Affection to the Mathematicks, and
eſpecially to this Part of it, was ſo ſtrong, that he fre-
quently paffed three whole Days and Nights in his Study
without eating, drinking, or fleeping, except a Nod now
and then upon his Elbow *. He, about the Year 1590,
publiſhed a Treatife of Algebra in quite a new Method,
and by a judicious Mixture of the Greek and Arabian
Rules, with fome Improvements of his own, introduced
that Mode of Calculation which is ftill in Ufe, under the
Title of Specious Arithmetick. Before his Time, only
unknown Quantities were marked by Letters, but fuch
as were known were fet down in Figures according to
the uſual Notation: He made uſe of Letters for both,
only with this Diſtinction, that the known Quantities he
repreſented by Confonants, and the unknown by Vowels.
By this Contrivance he greatly extended the Science, and
which was more, fhewed its Capacity of being farther
extended. For, whereas former Algebraifts had confined
*Thuan, Hift. A. D. 1603.
their
Xiv
The INTRODUCTION.
י
their Inveſtigations to the particular Queſtions propofed
to them, he by this Means produced Theorems capable
of refolving all Demands of a like Nature, inſtead of
particular Solutions. The learned Dr. Wallis has ac-
counted very clearly for the new Title which Vieta gave
to his Algebra. The Romans had a Method of ſtating
Law Queſtions under general Names, fuch as Titus and
Sempronius, Caius and Mevius, whence we derive our
Way of ufing A, B, C, D, on fuch Occafions, which
Method of ſtating the Civilians ftile Species, in Oppo-
fition to the ſtating of real Cafes by true Names. Vieta
having made a Change of the fame Nature in Algebra,
and being, as we obferved before, a Lawyer by Profeffion,
he borrowed from that Science this Title of his new In-
vention, which was received with univerfal Applauſe.
We have likewife many of his Works, under the Name
of Apollonius Gallus, which he affumed on Account of
his first attempting to restore the Works of Apollonius
Pergaus. His Genius was fo extenfive, and his Pene-
tration fo great, that it enabled him to apply his Mathe-
matical Knowledge to moft Subjects; of which we have
a particular Inftance, in his decyphering the Letters
which paffed between the Court of Spain, and the Fac-
tion of the League in France, notwithstanding above
five hundred different Characters were made uſe of in
them. About the fame Time flouriſhed Raphael Bom-
belli, an Italian, who publifhed at Florence a Treatife of
Algebra, wherein he firſt taught how to reduce a biqua-
dratic Equation to two Quadratics, by the Help of a
Cubic.
Our own Countryman, Mr. William Oughtred, was the
next great Improver of Algebra. Building, however, on
what Vieta had already performed. He introduced fuch a
Conciſeneſs, and withal fo plain and perfpicuous a Me-
thod of inveſtigating Geometrical Problems, as acquired
him immortal Reputation. His Clavis Mathematica, or
Key of the Mathematicks, was firft publiſhed in 1631, and is
perhaps the cloſeſt and moſt compendious Syftem hitherto
extant.
The INTRODUCTION.
XV
extant. In this Work he contented himſelf with the Solu-
tion of quadratic Equations, referving thoſe of higher
Powers for another Work, which was his Exegefis Nume-
rofa, which in later Editions is joined to his Clavis. In
both Pieces there were abundance of Additions and Im-
provements, and the Doctrine of Proportions more fully
and clearly ſtated than hitherto it had been; but the
greateſt Excellency in Mr. Oughtred's Book, was his Ap-
plication of the Analytic Method to Geometry, which
he did in a Variety of Cafes, and enabled his Difciples to
proceed ſtill farther than himſelf had done. By Profeffion
he was a Clergyman, and Rector of Albury in Surry,
where he gave himſelf up entirely to his Studies, and to
the Converſation of a very few Friends; he lived to the
Age of Fourfcore and Seven, and died then of Joy, on
May 1, 1660, at hearing the Houfe of Commons had
voted the King's Return. Some have cenfured his
Clavis as too fhort and obfcure, and fo indeed it might
prove for fuch as were altogether unacquainted with theſe
Studies, for whofe Ufe it is plain enough he never de-
figned it; but where Perfons are acquainted with the
Elements of Geometry and Algebra, and have that Saga-
city and Attention which is neceffary to make any con-
fiderable Progrefs in this Sort of Learning, Mr. Oughtred's
Key will be ſtill found a very uſeful Book, and its Style
the moſt perfect in its Kind that has ever been uſed.
Contemporary with him was Mr. Thomas Harriot, an
excellent Mathematician, and who made ſtill greater Im-
provements in this Science. He is placed after Oughtred,
tho' he died long before him, becauſe his Book was not
publiſhed till ſome Time after the firft Edition of Oughtred's
Clavis. It was then printed in a thin Folio by the Care
of Mr. Walter Warner, under the Title of Artis Analytica
Praxis ad Equationes Algebraicas nová, expeditâ, & ge-
nerali Methodo, refolvendas, Tractatus pofthumus, &c. i. e.
A Treatife of the Analytic Art, containing a new, ex-
peditious, and general Method of refolving Equations,
a pofthumous Tract, by the late learned Mr. Thomas Har-
riot.
xvi
The INTRODUCTION.
riot. The Publisher, Mr. Warner, prefixed a Preface of
his own, containing a very judicious, tho' very concife,
Repreſentation of the feveral Parts of Algebra, their
Nature and Dependance on each other, the Extent and
Uſefulneſs of this Art, and the Progreſs thereof to that
Time. In Mr. Harriot's Book, Algebra takes a new
Form, and from him alone it met with more Improve-
ment than from all who had ftudied, or at leaſt all who
had written upon it, before him. He was indeed one
of the greateſt Men this Nation ever produced, and
great Pity it was, that this Work of his did not appear.
in his Life-time, or that his other Pieces, which were of
infinite Value, ſhould be buried in Oblivion. The true
Caufe of the former feems to have been his Courſe of
Life; he was a Dependant on the Earl of Northumberland
and Sir Walter Raleigh, and afterwards upon Sir Thomas
Aylesbury, to whom, if I am rightly informed, he left
many of his Writings, and, as I hinted, the Reaſon of
his not publiſhing them in his Life-time, feems to have
been his Deference for his Benefactors. Happy had it
been, if the reft of the Mathematical Works he left had
been fent abroad (as in his Preface he ſeemed to promiſe
they ſhould) by the intelligent Editor of this excellent
Work.
It is divided into two Parts; and the Author begins his
Improvements by removing every thing that was ufe-
lefs, fuperfluous, or inelegant in former Methods; thus
inſtead of Capitals, he introduced fmall Letters; inſtead
of the Terms, Squares, Cubes, Surfolids, &c. and their
Contractions, he brought in the Powers themſelves,
which made the Operations much more eaſy, natural, and
perfpicuous than they were before. Having thus efta-
bliſhed a plain and accurate Notation, he proceeds to a
Multitude of new Diſcoveries, of which, to the Number
of twenty-three, the Reader may find a full, diſtinct, and
very judicious Account, in the celebrated Treatiſe of Dr.
Wallis. From this admirable Piece of Mr. Harriot's,
Des Cartes took all the Improvements he pretended to
make,
The INTRODUCTION.
xvii
make, as the Doctor juftly obferves, and of which I fhall
furniſh the Reader with fome concife, and I think con-
cluſive Proofs. First, It appears from all the Accounts
we have of the Life of Des Cartes, that he was here in
England when Harriot's Book was publiſhed, which be-
ing written in Latin, in a Branch of Learning about
which that great Man was then very fedulous, it is eaſy
to conceive that he was one of its moft early Perufers;
Secondly, It is certain that he did not publiſh any thing on
this Subject before that Year; Thirdly, His Treatife of
Geometry, wherein thefe new Improvements firſt ap-
peared, was printed in French in 1637 without his Name,
which in all Probability was to try what Opinion the
World would have of them, and whether any of the
French Mathematicians could difcern whence they were
taken; Fourthly, Though he fuffered the two firſt Parts
of his Book to be published in Latin, with his Name, in
1644; yet the third Part, relating to Geometry, did not
appear till 1649, when it was published by Francis Van
Schooten. Theſe are probable Reaſons only, but then,
Fifthly, He follows Harriot diftinctly in Nineteen feveral
Diſcoveries; which that they ſhould be made in the fame
Method and Manner, (except a few Miſtakes) without
confulting Mr. Harriot, is altogether incredible, and was
fo held to be even by his own Countrymen, when,
thro' the Information of the Honourable Mr. Cavendish,
they were made acquainted with Mr. Harriot's Book;
Sixthly, There are fome little Changes, particularly in the
Marks made ufe of by Des Cartes, and which were never
followed by any body, that plainly intimate he only
introduced them, in order to difguife his Method;
Seventhly, It appears that Des Cartes himfelf was ac-
quainted with the Charge brought against him upon this
Head, and yet he never thought fit to juftify himſelf,
nor did ever fo much as declare that he had not feen
the Book he was faid to have copied. On the whole there-
fore, there is all the Reaſon in the World to believe, that
the Honour due to the great Improvement of this Science,
C
which
xviii The INTRODUCTION.
which fitted it for all that it has received fince, from
Foreigners or Engliſhmen, belongs to our Author Harriot,
and not to Des Cartes, who only accommodated thefe
Diſcoveries to Geometrical Subjects.
After him Dr. John Pell, who was Refident for the
Commonwealth of England in Switzerland, publiſhed ſome
new Diſcoveries. The Method he took of doing it was
this, he recommended to Mr. Thomas Brancker a Treatife
of Algebra written in the German Language by Rhonius,
which when he had tranflated, the Doctor reviſed, altered
and added to it. In this Piece there are a great many
curious Things relating eſpecially to Diophantine Algebra,
but delivered very obfcurely, infomuch, that the learned
Dr. Wallis feems to be in doubt, whether himſelf had
reached Dr. Pell's true Meaning. Yet, to this Gentleman,
who wrote in fo perplexed a Way, we ftand indebted for
the Invention of the Regiſter; a Method of great Uſe,
eſpecially to Beginners, the Practice of which was what
chiefly recommended Kerfey's Algebra, and which is con-
ftantly and judiciously preferved throughout the follow-
ing Treatife. It is very likely, that the Darkneſs com-
plained of in Dr. Pell's Writings might be owing to
his Circumftances as well as Temper, for he was a very
bad Economiſt, not through any Vice or Extravagancy,
but by a Neglect of his private Affairs, and ſpending all
his Time in Study.
As for the Rules of John Van Hudde, Mr. Merry,
Erafmus Bartholine, Mr. Hugens, and others, I do not
take Notice of them, becauſe in reality they are no more
than Improvements on, or Deductions from, Harriot.
The fame Thing may be ſaid of what has been written by
Meff. Farmat, de Billy, Fernicle, and other French Mathe-
maticians, who only propoſed Problems for other People
to reſolve, and referved their own Methods of Solution
as impenetrable Secrets: A Practice, which, however it
might intitle them to the Admiration of the Age in which
they lived, can give them no juft Claim to the Praife of
Pofterity; fince if we reap any Benefit from their Diſco-
veries,
The INTRODUCTION.
xix
veries, it is indirectly, and in a Manner againſt their
Intentions.
Dr. Wallis himſelf has alſo made fome very confiderable
Improvements in this Science, eſpecially in refpect to im-
poffible Roots in fuperior Equations; and what he left un-
perfected has been fupplied by the ingenious Mr. Abraham
De Moivre, whofe accurate Performance on that Subject
has been lately publiſhed, in the Algebra of Dr. Saunderfon.
In 1655 Dr. Wallis publiſhed his Arithmetica Infinitorum,
in which he ſquared a Series of Curves, and ſhewed that
if this Series could be interpolated in the middle Spaces,
the Interpolation would give the Quadrature of the Circle.
This Treatife fell into the Hands of the ingenious Sir Ifaac
then Mr. Newton, in the Year 1664, when that Gentle-
man was about Two and Twenty; and he by a Sagacity
peculiar to himſelf, and which can never be enough ad-
mired, derived from this Hint his celebrated Method of
Infinite or Converging Series. In 1665, he computed
the Area of the Hyperbola by this Series to Fifty-two
Figures, which having communicated to Dr. Barrow, he
prevented Mr. Nicholas Mercator's running away with the
Reputation of this Difcovery, who in 1668 publiſhed the
Quadrature of the Hyperbola by an infinite Series. This
was received with univerfal Applauſe, and yet Mr. New-
ton far exceeded him; fince, without ftopping at the Hy-
perbola, he extended this Method by general Forms to
all Sorts of Curves, even ſuch as are Mechanical, to their
Quadratures, Rectifications, and Centers of Gravity, to
the Solids formed by their Rotations, and to the Super-
ficies of thoſe Solids; fo that fuppofing their Determina-
tions to be poſſible, this Series ftopped at a certain Point,
or at leaſt their Sums were given by ftated Rules.
the abfolute Determinations were impoffible, they could
yet be infinitely approximated, as he likewife fhewed,
and which, as a French Writer juftly obferves, is the hap-
pieſt and moſt refined Contrivance for ſupplying the De-
fects of human Knowledge, that Man's Imagination could
poffibly invent. It is alfo certain, that he attained his
Invention
C 2
But if
XX
The INTRODUCTION.
Invention of Fluxions by that Time he was Four and
Twenty, but his Modefty was fo great, that he forbore to
publiſh his Diſcovery, which was the fole Reaſon that the
Honour of it was ever difputed with him.
In 1707, he firſt publiſhed a Syſtem of Algebra under
the Title of Univerfal Arithmetick, and in 1722 gave
another Edition of it, wherein are contained all his Im-
provements in that Art,
From the Rules by him laid down, ftill farther Lights
were ftruck out by fucceeding Mathematicians, fuch as
Dr. Edmund Halley, who publiſhed in the Philofophical
Tranfactions, a Method of finding the Roots of Equations
without any previous Reduction, and the Conftruction of
Equations of the 3d and 4th Power, by the Help of a
Circle and Parabola. Mr. J. Colfon, who obliged the
learned World with a univerfal Refolution, Geometrical
and Mechanical, of Cubic and Biquadratic Equations.
Mr. Colin Mac Laurin, in his Treatiſe of impoffible Roots,
and many others too long to be enumerated here.
But after all theſe Diſcoveries and Improvements, there
has still been a general Complaint, that hitherto we have
had no Book of Algebra plain enough to inftruct fuch as
are inclined to ſtudy this Science without farther Aſſiſt-
ance, or who live in Places where it is not to be had. To
obviate this Objection, the following Treatife was drawn
up, which will be found to contain a clear and copious
Syftem of Algebra, delivered in ſo eaſy and natural a
Method, and with ſuch Perfpicuity and Condefcenfion to
the Feebleness of the Underſtanding, when firſt applied
to this kind of Study, that I felicitate myſelf on having,
prevailed upon its Author to make it publick, as I am
perfuaded it will be of general Ufe, in preventing young
People from being difcouraged at their firft Entrance into
Algebra, which has hitherto hindered Numbers from
cultivating their Inclinations to the Mathematicks.
THE
[ xxi ]
THE
CONTENTS.
I
NTRODUCTION, containing the History of the
Origin and Progrefs of this Science.
Signs, their Signification.
Co-efficients, are the Numbers prefixed to any Quantity.
Quantities, fimple and compound.
Page ix.
I
3
ibid.
4
Axioms, on which Algebra is founded.
Addition, Cafe 1. When the fame Quantities have like Signs, exem-
plified in the adding of fimple and compound Quantities. 5
Cafe 2. When the fame Quantities have unlike Signs,
exemplified in the adding of fimple and compound Quan-
tities.
8
Cafe 3. When the Quantities are different, exemplified
in the adding of ſimple and compound Quantities. 12
13
Examples, where these three Cafes are promiscuously used.
Subſtraction, performed by one general Rule, and exemplified in
the Subtraction of the fame fimple Quantities, with
like and unlike Signs.
15
18
Further exemplified in the Subftraction of the fame
compound Quantities, with like and unlike Signs. 16
Further exemplified in the Subſtraction of unlike
Quantities, fimple and compound.
Multiplication, Cafe 1. When the Signs of the Quantities are
alike, and have no Co-efficients, exemplified in
multiplying ſimple and compound Quantities. ibid.
Cafe 2. When the Signs of the Quantities are alike,
and have Co-efficients, exemplified in multiplying
Simple and compound Quantities.
23
Cafe 3. When the Signs of the Quantities are un-
like, exemplified in multiplying ſimple and compound
Quantities, with or without Co-efficients. 26
Of Algebraic Quantities, by a pure or abfolute
Number,
Examples, wherein all theſe Cafes are promiscuouſly uſed.
29
ibid.
Diviſion,
xxii
CONTENT S.
The
Divifion, Cafe 1. When there are no Co-efficients prefixed, and
the Signs of the Quantities are alike, exemplified in
dividing fimple Quantities by Simple Quantities, and
compound Quantities by fimple Quantities.
31
Cafe 2. When there are no Co-efficients prefixed, and
the Signs of the Quantities are unlike, exemplified in
dividing fimple Quantities by fimple Quantities, and
compound Quantities by fimple Quantities.
36
Cafe 3. When there are Co-efficients prefixed, and the
Signs of the Quantities are either alike or unlike, exem-
plified by Examples as in the two foregoing Cafes. 37
Cafe 4.
When no Quantities in the Divifor are like
thofe in the Dividend, exemplified in dividing fimple and
compound Quantities.
40
When all the Quantities in the Divifor are like thoſe in
the Dividend.
43
Of compound Quantities by compound Quantities. 183
Involution, of fimple Quantities, with or without a Co-efficient. 44
Of compound Quantities, with or without Co-effi-
cients.
Evolution, of fimple Quantities.
Of compound Quantities.
45
47
50
Surd Quantities, Addition of, Cafe 1. When the Surds are alike. 56
-Caſe 2. When the Surds are unlike. 58
Surd Quantities, Subftraction of, Cafe 1. When they are alike. 59
Surd Quantities, Multiplication of,
Cafe 2. When they are unlike. 61
Cafe 1. When there are no ra-
tional Quantities prefixed. 64
Cafe 2. When there are ratio-
nal Quantities prefixed. 65
Surd Quantities, Divifion of, Cafe 1. When there are no rational
Quantities prefixed.
68
Cafe 2. When there are rational
Surd Quantities, Involution of, Cafe 1. When there are no rational
Quantities prefixed.
Quantities prefixed.
70
73
Cafe 2. When there are rational
Quantities prefixed.
74
Surd and rational Quantities con-
nected by the Signs + and -. 77
Equations, what is meant by them.
The different Methods of reducing them,
Questions which produce fimple Equations,
by Addition and Subſtraction.
I
79
exemplified in
and folved.
80
Negative
The CONTENT S.
xxiii
Negative Numbers, when greater than pofitive Numbers, how
fubftracted.
89
Questions, which produce fimple Equations, and folved by Multi-
plication.
90
Which produce fimple Equations, and folved by Di-
viſion.
102
Which produce fimple Equations, and folved by Invo-
lution.
114
Which produce fimple Equations, and folved by Evo-
lution.
122
Wherein all theſe ſeveral Methods of Reduction are
promiscuously uſed.
127
Where the unknown Quantity is in more Terms than
one.
134
Where any Quantity is in every Term of the Equa-
tion.
138
139
The Manner of Regiſtering the Steps in any Algebraic Operation,
explained.
Queftions, which contain two Equations, and two unknown
Quantities, and produce fimple Equations, the un-
known Quantities being only reprefented in Species;
the Manner of folving them.
142
Of the fame Kind, where both the known and unknown
Quantities are reprefented in Species.
152
Which produce Quadratic Equations, the Manner of
folving them explained.
168
Which produce Quadratic Equations, and the first
Power of the unknown Quantity is in more Terms
than one.
Subftitution, its Ufe.
174
ibid.
Queſtions, which produce Quadratic Equations, and the Square
of the unknown Quantity has a Co-efficient.
Quadratic Equations, have three Forms.
178
183
Quadratic Equations, have two Roots or Values, the Manner of
finding them in the three Forms of Qua-
dratic Equations.
ibid.
Quadratic Equations ambiguous, the Manner of expreffing their
Roots.
Queſtions, which produce thofe ambiguous Equations.
Which produce Biquadratic Equations.
194
198
209
Which produce ambiguous Biquadratic Equations. 215
Which contain three Equations, and three unknown
Quantities, the Manner of folving them.
219
Series
xxiv
CONTENT S.
The
Series Converging, their Nature explained in the Refolution of
Adfected Equations, Cafe 1. when the affumed
Root is less than the true Root,
230
Cafe 2. When the affumed Root is greater
than the true Root.
235
Equations, the Manner of folving them, when the unknown
Quantity is to feveral Powers in one Equation, and
but to the first Power in the other Equation. 243
Equations Adfected, have as many Roots as is the higheft Dimen-
fion of their unknown Quantity; that fome
of thefe Roots are affirmative, fome nega-
tive, and fome impoffible.
The Manner of finding thefe Roots.
248
249
Equations, the Manner of refolving them, when the unknown
Quantities are to feveral Powers in both Equations. 257
Queſtions, a Mifcellany of, anfwerea by the general Rules already
explained.
260
Queſtions, of three Equations and three unknown Quantities, when
one Equation has all three unknown Quantities, and the
other two have only two of the unknown Quantities. 265
Which contain four Equations and four unknown Quan-
tities, the Manner of folving them.
296
An Inftance how elegant a Question may be folved, by a judicious
Choice of the unknown Quantities, to exprefs the Conditions of
the Question.
302
The Method of expreffing the Power of any Quantity, by placing
a Figure over it.
304
To know if a Question is limited, or admits of feveral An-
fwers.
306
The Method of raifing Theorems for extracting the Roots of high
Powers, exemplified in raiſing Theorems for the Cube Root.
Cafe 1. When the affumed Root is less than the true Root. 307
Cafe 2. When the affumed Root is more than the true Root. 311
An Example where both Cafes are uſed.
313
Further exemplified in raifing Theorems for extracting the Biqua-
drate Root.
Cafe 1. When the affumed Root is less than the true Root.
Cafe 2. When the affumed Root is more than the true Root.
The Method of turning Equations into Analogies.
The Foundation of tranfpofing Quantities explained.
The Demonftration of the Rules ufed in Subftraction.
317
319
323
324
ibid.
Multiplication.
Divifion.
326
327
ALGEBRA.
[ 1 ]
ALGEBRA.
H
AVING given the Reader an Hiftorical Account of
this Science in the Introduction, we are now to explain
the Signs and Characters ufed by Analytic Writers, and
mention thoſe Axioms or Self-evident Principles of Truth and
Certainty, which are the Foundations of this celebrated Science.
+
Signs. Names.
+ } { Plus or more.
Significations.
The Sign of Addition; as 8 + 4, is 8 is
to be added to 4, and m+ n fignifies the
Number repreſented by m, is to be added
to the Number reprefented by n; again,
2+3+5, fignifies they are all to be added
into one Sum, and b+m+d fignifies that
the Numbers reprefented by b, m, and d
are to be added into one Sum.
The Sign of Subtraction; as 5-2, is
5 lefs by 2, or 2 is to be ſubſtracted from
5, and a-b is a lefs b, or the Number
repreſented by b is to be ſubſtracted from
-} { Minus or lefs. the Number reprefented by a; and 9-2
x}{
-3, is that from 9 there is to be fub-
ſtracted 2, and from the Remainder 3 is
to be fubfracted.
The Sign of Multiplication; as 5×7, is
5 is to be multiplied by 7, and axb, is
the Number reprefented by a, is to be
} { Into or with.< multiplied by the Number reprefented by
b; and 7 × 3 × 2, is that 7, 3, and 2, are
to be multiplied together, which Product
is 42.
B
The
ALGEBRA.
-}{ By.
=}{ Equal.
} { So is.
}{ Involution.
The Sign of Diviſion; as 8÷4, that is
8 is to be divided by 4, and xy, that
is, the Number reprefented by x, is to be
divided by the Number reprefented by y;
or fometimes they are placed like Vulgar
8
Fractions thus, that is, 8 is to be di-
4
vided by 4, and
y
that is, the Num-
ber repreſented by x, is to be divided by
the Number repreſented by y.
The Sign of Equality or Equation; thus
99, that is, 9 is equal to 9, and 2+3
=5, that is, 2 added to 3, is equal to 5:
Again, mny, that is, the Number
reprefented by m is equal to the Number
reprefented by n, added to the Number
reprefented by y; and y-xa+b; that'
is, the Number reprefented by y being
leffened by the Number reprefented by x,
the Remainder is equal to the Number
reprefented by a, added to the Number
reprefented by b.
The Sign of Proportion, or what is
commonly called the Rule of Three, and
is placed between the two middle Num-
ber thus, 35:6 10, that is, as 3 is to
5, fo is 6 to 10; and a;b::c: d, that is,
as the Number repreſented by a is to the
Number repreſented by b, fo is the Num-
ber reprefented by to the Number re-
prefented by d.
The Sign of Involution, or raifing any
Number or Quantity to the Square, Cube,
or any other Power; and the Heighth of the
Involution is generally expreffed by the
Number after the Sign thus, 72, is 7
is to be involved to the Square or fecond
Power; and 3, is 7 is to be involved
7
or raifed to the Cube or third Power; and
a 2, is a is to be involved to the Square
or fecond Power.
The
ነ
} { Evelution.
W
}{
~} {
ALGEBRA.
3
The Sign of Evolution, or the extract-
ing of Roots; and the Root that is taken
is likewife expreffed by the Figure that
follows the Sign, thus 9w 2, is the Square
Root of 9 is to be extracted, and 27 u 3,
is the Cube Root of 27 is to be extracted,
and a a un 2, is the Square Root of a a is
to be extracted.
3
f The Sign of Irrationality, or of a Surd.
Root; that is, the Number or Quantity
has not ſuch a Root as is required to be
extracted; thus the Square Root of 2
will be expreffed thus 2, and the
Irrationality, Square Root of 5 thus/5, and the Cube
Root of 4 thus 4, the little Figure
ftanding over the Sign being 3, fhews it to
be the Cube Root; again, ✔✓15 is the
Cube Root of 15, and where there is no
fuch Figure over the Sign, it fignifies the
Square Root only.
BR
or a Surd
Root.
3
Now before we go farther, it will be neceffary to inform the
Reader, that where any Number is joined to a Quantity, it
thews how many Times that Quantity is taken; thus, 4 a is
four times a, or the Number reprefented by a is to be taken
four times; and 7 m is feven times m, and if y was to be taken
Jeven times, it may be expreffed thus 7 y.
Theſe Numbers are called Co-efficients, or Fellow-Factors, as
they multiply the Quantity; and if any Quantity is without a
Co-efficient, then it is always implied that Unity, or 1, is the
Co-efficient of that Quantity; thus a is the fame as 1, and y
the fame as 1y; for when the Co-efficient is only Unity, or 1,
it is generally omitted.
Quantities that are expreffed or reprefented by fingle Letters,
or feveral joined together like a Word, as a, b, ab, anz, 7 y 2,
are called fimple or fingle Quantities.
But when theſe are connected by the Signs-or, as ab,
am-d, dn+az, they are called compound Quantities.
And fometimes Quantities are fet down in the Manner of
a a+b
Vulgar Fractions, thus,
b
n
72
x y
B 2
The
4
ALGEBRA.
}
The Sign that connects the Quantities belongs to that which
follows the Sign, thus, a + b, where the Sign + belongs to the
Quantity b; again, ac+d, the Sign belongs to the
Quantity c, and, the Sign to the Quantity d.
As to thofe fingle Quantities which have no Sign before them,
it is always underſtood they have the Sign+; thus a is the fame
as+a, and m is the fame as m; and therefore if fingle
Quantities are to have the Sign +, it is commonly omitted, as
they are uſually fet down without any Sign; but the Sign
never omitted, but always placed before the Quantity to which
it belongs.
is
And in Compound Quantities, if the firft or leading Quantity
has no Sign, then it is always underſtood to have the Sign +,
thus, abis the fame as + a + b, and a b is the fame as
;
ab; therefore in Compound Quantities, if the firft or
leading Quantity is to have the Sign, it is generally omitted;
but in thefe Compound Quantities, as well as in Simple Quan-
tities, the Signis never omitted, but always placed before the
Quantity to which it belongs.
Letters fet or joined together like a Word fignifies the Pro-
duct or Rectangle of thefe Letters, thus, a b is the Product
of a multiplied by b, and d n y is the Product of d, n, and y,
multiplied together.
The Operations in Algebra are founded on thefe Axioms.
AXIOM 1.
If equal Quantities are added to equal Quantities, the Sum
of thefe Quantities will be equal.
AXIOM 2.
If equal Quantities are taken or fubftracted from equal Quan-
tities, the Quantities remaining will be equal.
AXIOM 3.
If equal Quantities are multiplied by equal Quantities, their
Products will be equal.
Α ΧΙΟ ΜΑ
ADDITION.
5
AXIOM 4.
If equal Quantities are divided by equal Quantities, their
Quotients will be equal.
AXIOM 5.
If there are feveral Quantities that are equal to one and the
fame Thing, thoſe Quantities are equal one to another.
The Reader having premiſed theſe Things, and underſtanding
what the Signs are intended to exprefs, he may proceed to the
Rules of the Science; and if at first he meets with fome little
Difficulties about the Signs and Co-efficients, I would recom-
mend him to read the foregoing Pages again; and if that and
another Effay or two does not remove the Difficulties of any
particular Example, then to omit that and proceed to the next,
in which perhaps he may fucceed, and that may cauſe the Diffi-
culty in the other to vaniſh.
ADDITION,
In which there are three Cafes.
(1.) Cafe 1. Signs are both affirmative, or both negative,
W HEN the Quantities are alike, and their
add the Co-efficients or prefixt Numbers together, and to their
Sum join the Quantities, prefixing to them the Sign they have
in the Example.
Exam. I.
Exam. 2.
Exam. 3.
Exam. 4.
To
Add
2 a
5 m
4 y
3 a
2 m
·3%
2 %
6%
7 m
7y
Sum 5a
-8%
Exam. 1. The Co-efficients are 2 and 3, which added toge-
ther make 5, to which je ning a the Quantity, it is 5 d, and no
Sign
ALGEBRA.
Sign being prefixt to either 2 or 3a, the affirmative Sign is
underſtood as prefixt to both; hence 5 a, or-5 a is the Sum
required.
Exam. 2. The Co-efficients are 5 and 2, which being added
make 7, to which joining m, it is 7 m, the Sum required; for
the Signs of 5 m and 2 m are both affirmative, by what was faid
in the laſt Example.
Exam. 3. The Co-efficients are 4 and 3, which being added
make 7, to which joining y it becomes 7y; but as 4y and 3y
have both the Sign- before them, therefore prefix the Sign-
to 7y, and then-7 y is the Sum required.
Exam. 4. The Co-efficients are 2 and 6, which added make
8, to which joining z, it becomes 8 z, and prefixing the Sign
for the Reaſon in the laſt Example, we have
required.
8 z, the Sum
Exam. 5.
Exam. 6.
Exam. 7.
Exam. 8.
To
15 my
14 az x
4ady
1 6 y m d
Add
7 my
2aZx
3ady
1 2 y m d
Sum
22my
16 az x
7ady
28 ymd
Exam. 5. The Sum of the Co-efficients 15 and 7 is 22, to
which joining my, it is 22 my, the Sum required; for 15 my and
7 my have both the affirmative Sign, there being no Sign
prefixt.
Exam. 6. The Sum of the Co-efficients 14 and 2 is 16, to
which joining a zx, it is 16 a zx, to which prefixing the Sign
, as both the Quantities to be added have that Sign, then is
16 a z x the Sum required.
Exam. 7. The Sum of the Co-efficients 4 and 3 is 7, to which
joining a dy, it is 7 a dy, and both the Quantities having the
affirmative Sign, therefore 7 a dy is the Sum required.
Exam. 8. The Sum of the Co-efficients 16 and 12 is 28, to
which joining y md, it is 28 y m d, to which prefixing the Sign
as both the Quantities to be added have that Sign, then is
28 y md the Sum required.
و
83
Exam. 9.
Exam. 10.
Ian
Exam. II.
21 dy
Exam. 12.
da
To
Add
2 my
2an
dy
da
3 my
Sum
5 my
зап
224by
244
Exam.
ADDITΙΟ Ν.
7
Exam. 11. The Co-efficients are 21 and 1, for there being
no Co-efficient prefixt to dy, Unity, or I, is always understood
in fuch Cafes to be the Co-efficient; hence the Sum is 22 dy.
Exam. 12. There being no Co-efficient prefixt to either of
the Quantities, Unity, or 1, is the Co-efficient to each; and I
being added to I makes 2, to which joining da, it is 2 da, to
which preĥxing the negative Sign, we have 2 da, the Sum
required.
(2.) If there are two or more Quantities connected by the
Signsor, and are alike to two or more Quantities con-
nected by the Signsor, they are added as in the former
Examples, only taking due Care that the Quantities which com-
pofe their Sum are connected with their proper Signs, according
to the Rule, as in the following Examples.
Exam. 13.
To
2a+76
Add
3426
Sum
5a+96
Exam. 15.
Exam. 14.
6ma+5y
2 ma + 3y.
8ma + 8 y
21ma-2yd
3ma 3rd
24ma + 530
d
+
Exam. 13. Is 2 a 4-7 b to be added to 3 a 2 b. The
Quantities being difpofed as in the Example, it follows from
former Examples that 2 a being added to 3 a makes 5 a, and 7 5
addeď to 2b makes 9b; but as 76 and 26 have both the affe-
mative Sign, to 5 a connect 9b with the Sign +; hence
5a9b is the Sum required.
Exam. 14. Is 6 ma 5 y to be added to 2 ma +3y. Now
by the former Examples 6 m a being added to 2 ma is 8 ma, and
5y being added to 3y is 8y; but as 5y and 3y have both the
affirmative Sign, to 8 a connect 8y with the Sign+; fo will
& ma + 8y be the Sum required.
m
Exam. 15. Is 21 ma 2yd to be added to 3 ma + 3y d.
Now by the former Examples 21 m a being added to 3 ma, the
Sum is 24 ma; and 2yd being added to 3yd, the Sum is 5y d.
But as 2 yd and 3yd have both the affirmative Sign, therefore
connecting 24 ma end 5yd with the Sign +, we have 24 m a
454, the Sum required.
Exam. 16.
то
Add
7 da 15 m
2 da
9 ma
Exam. 17.
14 1 d
4. 12
3 m a
Sum
J
19 m
3 nd
1 2 m a — 17 nd
Exam, 18.
-2mn+15yd
-4mn 4 y d
6 m n + 19 yd
Exam.
8
ALGEBRA.
Exam. 16. Is- 7 da
15 m to be added to 2 da
-4 m.
Now 7 da added to 2 d a is 9 da; but as both theſe Quantities
have the Sign, prefix the negative Sign to 9 d a, and then it
is - 9 da. Again, 15 m added to 4 m is 19 m; and both theſe
Quantities having likewife the negative Sign, prefix it to 19 m;
whence the Sum required is 9 d a
19m.
Exam. 17. Is 9 Ma 14 nd to be added to 3 ma
3nd.
Now 9 m a added to 3 m a, is 12 ma; and both theſe Quantities
having the Sign +, place down 12ma as in the Example:
Then 14 n d added to 3n d, is 17 nd; but both theſe Quantities
having the Sign-, place the Sign-before 17 nd, and the
Sum required is 12 ma —
17nd.
Exam. 18. Is 2mn + 15yd to be added to
4mn +4yd.
Now 2 m n added to 4 m n, is 6 mn; but both theſe Quantities
having the negative Sign, prefix the Sign to 6 mn, and then
it is 6 mn.
And 15 yd added to 4 yd, is 19 yd; and both
thefe Quantities having the affirmative Sign, prefix the Sign +
to 19yd; hence the Sum is — 6 m n † 19 y d.
Exam. 19.
Exam. 20.
Exam. 21.
To
9yd-7a
Add
2yd- a
14 y d + 15 a
2 yd +
-14y+d
a
Sum
11yd-8a
16yd + 16 a
y + d
-151+2d
When you come to add
Exam. 19.
7 a to a, there
being no Co-efficient prefixt to a, Unity, or 1, is always in fuch
Cafes the Co-efficient; and then by what has been already taught,
-7 a being added to
09
the Sum is -8 a, as in the Example.
Exam. 20. And when 1, a, is to be added to a, the Sum is
I
Ja,
15},
for the fame Reafon 16 a.
Exam. 21. And-14 y being added to y, the Sum is
and d being added to d, for the fame Reafon the Sum is 2 d, or
+ 2 d.
(3) Cafe 2. When the Quantities are alike, but the Signs
are one affirmative, and the other negative, fubftract the leffer
Co-efficient from the greater, to the Remainder join the Quan-
tity, and prefix to it the Sign of the greateſt Co efficient.
It is of no Signification whether the Quantity that has the
greateft Co-efficient ftands above or below.
2
Exam.
ADDITIO N.
9
£
Exam. I.
Exam. 2.
Exam. 3.
Exam. 4,
Το
5 a
16 m
21 ad
14 m z
Add
2 a
12 m
-7ad
5 m z
Sum
3 a
4 m
14ad
9 m z
Exam. I.
The Co-efficient 2 fubftracted from
5 leaves 3,
to which joining a it is 3 a, but the Sign of 5 the greatest
Co-efficient is affirmative, therefore 3 a or + 3 a is the Sum
required.
Exam. 2.
The Co-efficient 12 fubftracted from 16 leaves
4, to which joining m it is 4 m, but the Sign of 16 the greateſt
Co-efficient is affirmative, therefore 4 m or + 4 m is the Sum
required.
Exam. 3. The Co-efficient 7 fubftracted from 21 leaves.
14, to which joining a d it is 14 a d, but the Sign of 21 the
greateſt Co-efficient is affirmative, hence 14ador + 14 ad is the
Sum required.
Exam. 4. The Co-efficient 5 fubftracted from 14 leaves.
y, to which joining m z it is 9 m z, but the Sign of 14 the
greatest Co-efficient is affirmative, hence 9 mz or + 9 m z is the
Sum required,
Exam. 5.
Exam. 6.
Exam. 7.
Exam. 8.
Το
14 m
9y
5ལ
Add
7 m
2 y
9Z
א א
9 a m
14 am
Sum
7 in
7y
42
5 a 112
Exam. 5.
The Co-efficient 7 fubftracted from 14 leaves
7, to which joining it is 72, but the Sign of 14 the greateſt
Co-efficient being-, prefix that Sign to 7 m, then is 7 m the
Sum required.
Exam. 6. The Co-efficient 2 fubftracted from 9, there
remains 7, to which joining y it is 7 y, but the Sign of 9 the
greatelt Co-efficient being, prefix that Sign to 7y, and we
have 7y, the Sum required.
Exam. 7.
The Co-efficient 5 fubftracted from 9 leaves
4, to which joining z it is 4 %, but the Sign of 9 the greatest
Co-efficient being negative, prefix the Sign to 4%, and we
bave-42, the Sum required.
C
Exam,
ΙΟ
ALGEBRA.
Exam. 8. The Co-efficient 9 fubtracted from 14 leaves.
5, to which joining a m it is 5 am, but the Sign of 14 the great-
eft Co-efficient being negative, prefix the Sign to 5 am, and
we have — 5 am, the Sum required.
Exam. 9.
To I am
Exam. 10.
Exam. II.
Exam. 12.
8 a d
14ym
ay
Add
a m
gad
16 y m
May
Sum
6 am
ad
2 y m
6ay
—
Exam. 9.
The Co-efficient of am being Unity, or 1,
which fubftracted from 7 leaves 6, to which joining a m it is
6 am, prefixing to it the Sign of 7, the greateſt Co-efficient,
we have 6 am or + 6 am the Sum required.
Exam. 10. The Co-efficient 8 fubftracted from 9 leaves
1, to which joining a d we have I ad or ad, which having
already the Sign of 9, the greateft Co-efficient, hence ad is the
Sum required.
Exam. 12. The Co-efficient of a y being Unity, or 1, which
fubftracted from 7 leaves 6, to which joining ay it is 6 ay,
which having the fame Sign with 7, the greateſt Co-efficient,
6 ay is the Sum required.
4. And if there are feveral Quantities connected by the different
Signs of and, to be added to feveral Quantities connected
by the different Signs of and, the Quantities being alike,
are added as in the ſecond Article, only taking Care to prefix the
Signs, according to the Directions in the first and third Articles.
Exam. 14.
Exam. 15.
To
Add
Sum
Exam. 13.
14a+7 m
3 m
15 my —
7 my + 12a Z
8 My-
1492
17 ay + 8am
3 ay
5 am
2AZ
8 a
6a+ 4m
14 ay 3am
Exam. 13. Is 14 a +7 m to be added to — 8 a a - 3m. Now
by the Rule at Art. 3. the Difference between the Co-efficients 14
and 8 is 6, to which joining a it is 6 a, but 14 the greateſt Co-
efficient having the affirmative Sign, hence 6 a is the Sum of 14 a
added to
8 a.
And the Difference between 7 and 3 the Co-
efficients of m being 4, to which joining m it is 4 m, but as 7
the greateſt Co-efficient has the affirmative Sign, therefore to
6 a connect 4 m with the Sign+, fo is 6a+ 4m the Sum required,
Exam,
ADDITΙΟΝ.
II
Exam. 14. Where
15 m y 14 az is to be added to
7 my + 12a z. Now the Difference between 15 and 7 the two
Co-efficients of my is 8, to which joining my it is 8 my, but as
15 the greateſt Co-efficient hath the negative Sign, therefore pre-
fix the Sign to 8 my, and it is 8 my: And the Difference
between 14 and 12 the two Co-efficients of a z being 2, to
which joining a z it is 2 a z, but as 14 the greateſt Co-efficient
has the negative Sign, therefore to
8 my
connect 2 a z with the
Sign, fo is 8 my-2 az the Sum required.
Exam. 15.
The Difference between 17 and 3 the two Co-
efficients of ay is 14, to which joining ay it is 14 ay, but as 17
the greateſt Co-efficient has the affirmative Sign, therefore place
down 14 ay or † 14 ay. And the Difference between 8 and 5
the two Co-efficients of a m is 3, to which joining a m it is
3 am, but as 8 the greateſt Co-efficient has the affirmative Sign,
therefore prefix the Sign - to 3 a m, ſo is 14 a y + 3 am the
Sum required.
Exam. 16.
7a+167
4 m
Exam. 17.
Exam. 18.
Το
mn - 15+ 78
Add за
7y IIP
8y- 4p
7am-167
11 am + 18 y
Sum- 4a+12 11
4am j- 2†
Exam. 16. By Art. 3. the Difference between 7 and 3 the
two Co-efficients of a is 4, to which joining a it is 4 a, but as 7
the greateſt Co-efficient has the negative Sign, therefore prefix the
Sign
to 4 a, and it is 40. And the Difference between
16 and 4 the two Co efficients of m is 12, to which joining m it
is 12 m, but 16 the greateſt Co-efficient having the affirmative
Sign, prefix the Sign + to 12 m, fo is 4 a 12 m the Sum
required.
Exam. 17. By Art. 3. the Difference between 15 and 7 is 8,
to which joining y it is 8 y, but 15 the greateft Co-efficient
having the negative Sign, prefix the Sign-to 8 y, and it is 8 y.
8y,
And the Difference between 7 and 11 the two Co-efficients of p
is 4, to which joining p it is 4 p, but as II the greateſt Co-
efficient has the negative Sign, therefore prefix the Sign
4p, and it is 4p, fo is 8 y4p the Sum required.
to
Exam. 18. By Art. 3. the Difference between 7 and II is 4,
to which joining a m it is 4 am, but as 11 the greateſt Co-
efficient has the negative Sign, therefore prefix the Sign to
44 m, and it is
And the Difference between 16
C 2
of 22 7/2.
and
IZ
ALGEBRA.
and 18 is 2, to which joining y it is 2 y, but as 18 the greateſt
Co-efficient has the affirmative Sign, therefore prefix the Sign +
to 2 y, fo is 4 am+ 2y the Sum required.
Exam. 19.
Exam: 20.
To
14 my
ma
Add
3 my
+ 4ma
- 5 yd + 13%
vd-
༡༤
Sum
11my +3 ma
८२
·4yd +122
Exam. 19. The Co-efficient of
Exam, 21.
— 14dy + 5mp
dy
m D
·13dy + 4111?
ma is 1, which being by
Art. 3. fubſtracted from 4 leaves 3, to which joining m a it is
3 ma, as in the Anfwer, and by the fame Method in
Exam. 20. If 59t is added to y d or 1 yd, the Sum is
4 yd; and likewiſe in
Exam. 21. If-14 dy is added to dy or I dy, the Sum is
13 dy.
13
5. If the Quantities are alike and the Co-efficients are equal,
but the Signs are one affirmative, and the other negative, theſe
being added together deftroy each other, or the Sum of them is
a Cypher or nothing.
Exam. 1.
Exam. 2.
Exam. 3.
Exam. 4
7 a
5 y
14 m
5 ya
5 y
14 m
5 ya
O
о
о
О
Το
Add - 7a
Sum
Exam. 1. By Art. 3. the Signs being unlike the Co efficients
are to be ſubſtracted, but 7 taken from 7 leaves o, and if to this
we join a it is o a, or no times a, that is, the Quantity a is to
be taken no times or not at all, which is the fame as nothing:
So in the fourth Example, if 5 is fubftracted from 5, there
remains o, or nothing, to which if we join y a, we then have no
times y a, or nothing.
(6.) Cafe 3. When the Quantities are unlike, that is, the
Letters are different, then fet them down one after the other,
with the fame Co-efficients and Signs they have in the Example,
and this is the Sum required.
And they may be fet in any Order, that is, any Quantity
may be fet firft, in the middle or laft, it being not material
how they are ranged, fo as they are but connected with their
proper Signs.
Exam.
ADDITION.
13
1
Exam. 1.
Exam. 2.
Το
2 a
3 m
Exam. 3.
a + d
5 a
2 y
2a-3d
3m+5a
a+d+2y
Add 3 d
Sum
Exam. 1. The Quantities or Letters being unlike, I place
down 2a, and becauſe 3d has the Sign+, therefore after the
2 a put + 3d, fo is 2a+3d the Sum required.
Exam. 2. Having put down the 3 m, after that put +5 a the
other Quantity with its Sign, fo is 3 m + 5 a the Sum required.
Exam. 3. Having put down a, after that put + d, and after
that2y, fo is a+d+ 2y the Sum required.
Exam. 4.
To
2 a 7 m
Add
34+52
Sum 2a —7m+31 +5%
Exam. 5.
2a+15
Z
7 d
2a+15+2
7 d
7 m, after
Exam. 4. Begin and place down 2 4, after that
that + 3y, and after that +5%, fo is 2a — 7 m +31 +5%
the Sum required.
Exam. 5. Begin and place down 2 a, after that +15, after
that + z, and after that 7d, fo is 2 a + 15 + z −7 d the
Sum required.
To
7m+157
Add
4 a + m n
Sum
7 m + 151 - 4 a + 11 12
Add
2 a — 8 d
Sum 16 +7111
2 a
8 d
To 16 +7m
15 m + 7 a
8 y
26
15m+7a+83—2 &
14m — 159
a - 7
14 M-
m 15y+a-7
Examples wherein all the foregoing Cafes are promifcuouſly uſed.
Exam. 2.
Exam. I.
To
7a15d+m
Add
5a+18d
Sum
12a + 3d-1- m
21 X
12 m + 51
8a+ 7m
II
за
5 m
21x+5y
Exam. 1. 7 a added to 5 a makes 12 a, by Art. 1. and -15 d
added to 18d makes 3d, by Art. 3. and there being no Quan-
tity like m, that muſt be placed by itſelf, by Art. 6. and connect-
ing
14
ALGEBRA.
ing theſe Quantities with their proper Signs we have 12 a 3d
+m, the Sum required.
Exam. 2.
7 m added to
8 a added to 11 a makes 3a, by Art. 3. and
12 m is 5 m, by the fame, but 21 x and 5 y
being different, place them down one after another as at Art. 6.
21x+5y the Sum required.
fo is 3 a 5 m
5 am.
Exam. 3.
Exam. 4.
Το
15a+14m- - 16
II Am
7 y d + m n
Add
7a-
14m+y
2 y d — 7 a
Sum
8a — 16+ y
Exam. 3.
is — 8 a,
14 m added to
6 am − 9 y d + m n −7α
15 a added to 7 a
14 m is nothing or o,
8 a, by Art. 3. and
by Art. 5. therefore
take no Notice of thofe Quantities in the Sum, and 16 and
y being different Quantities fet them down by Art. 6. fo is
16+y the Sum required.
— 8 a
2 yd is
Exam. 4. 1 I am added to ·5 am is 6 am, by Art. 3. and
7 y d added to
9yd, by Art. 1. But m n and
7 a being different Quantities fet them down by Art. 6. and
6 am —— 9 y d + mn — 7 a is the Sum required.
To
Add
Sum
Exam. 5.
4 a 17y+15 ap
2ap+ 34-2y
за
13 ap + 7a 19y
Exam. 5.
Exam. 6.
II + 8 m
−7m+15+4a
4 a
2 ap added to 15 ap is 13 ap,
mi + 4
by Art. 3. and
3a added to 4a is 7 a, by Art. 1. and 2 y added to
17y is
19y, by Art. 1. hence 13 ap +7a-19 y is the Sum required.
Exam. 6. 7 m added to 8 m, the Sum is m, by Art. 3:
II, the Sum is 4, by Art. 3. and 4 a added to
4a, the Sum is o, or nothing, by Art. 5. hence m +4 is
the Sum required.
15 added to
In theſe two Examples the fame Quantities are not ſet under
one another, to fhew the Learner that however they are placed,
if the Quantities are alike, they are to be added as if they food
one under the other.
The more perfectly Addition is underſtood, the eaſier it will
render the Work of Subſtraction.
SUBSTRACTION,
[ 15 ]
1
SUBSTRACTION,
7.
I
S performed by one general Rule; change all the Signs of
thofe Quantities which are to be fubftracted, or ſuppoſe
them in the Mind to be changed, then add theſe Quantities to
the others, according to the feveral Rules of Addition, which
will be the Difference or Remainder required.
I would adviſe the Learner to take out the Examples, and
put down thofe Quantities which are to be fubftracted with
contrary Signs, to thoſe they have in the Examples; that is,
making thoſe affirmative which are negative, and thoſe negative
which are affirmative, and then proceed as directed in the
general Rule.
From
Exam. I.
5 a
Exam. 2.
7 m
Exam. 3.
Exam. 4.
5 y
8 Z
Subftract
за
2 11
2 y
4 Z
Remains
2 a
5 m
3y
4 Z
Exam. 1. Here 3 a the Quantity to be fubftracted has the Sign
+, which being made or fuppofed to be made, then by the
general Rule, 5 a is to be added to 3a, the Sum of which is
2a, by Art. 3. and this is the Remainder required.
Exam. 2. In the fame Manner 2 m being fuppofed to have
the Sign -prefixed to it, then by the general Rule, 7 m is
to be added to 2 m, the Sum of which is 5 m, by Art. 3. and
this is the Remainder required.
Exam. 3 And if we fuppofe 2 y to be 2y, or + 2y, then
by the general Rule, 5y added to + 21, the Sum is
by Art. 3. and this is the Remainder required.
Exam. 4. If we ſuppoſe 4≈ to be 4 %, or + 4 %,
by the general Rule, if — 8 ≈ is added to 4 z, the Sum is
by Art. 3. and this is the Remainder required.
31,
then
4 Z,
From
Subftract
Exam. 5.
14 m n
Exam. 6.
7yd
Exam. 7.
Exam.8.
2. m 11
Remains
16 m n
+ 5vd
12yd
5 3*
+31x
4 ay
3 ay
-8yx
7ay
Exam. 5. The Sign of 2 m n being, if we fuppofe it +,
then by the general Rule, 14 m n added to 2 m n, the Sum is
*6mn, by Art, 1. the Remainder required,
2
Exam.
16
ALGEBRA.
{
Exam. 6. If we fuppofe 5yd to be-5yd, then by the
general Rule, 7 y d added to 5 d, the Sum is
12 yd,
3yx, then by the
3yx, the Sum is 8 yx,
by Art. 1. the Remainder required.
Exam. 7. By fuppofing 3yx to be
general Rule,
5 yx added to
by Art. 1. the Remainder required.
Exam. 8. And if we fuppofe
by the general Rule, 4 ay added to
Art 1. the Remainder required.
3ay to be 3 ay, then
3 ay, the Sum is 7 ay, by
From
Subſtract
5 am
a m
Remains
6 am
ay
5ay
-7ad
+ ad
5yd
y d
4ay
-8 ad
4yd
The Truth of Subftraction may be proved as in common
Arithmetic, by adding the Remainder to the Quantity which is
ſubſtracted, and if their Sum is the fame as that from which the
Quantity was fubftracted, the Work is true, otherwife it is
erroneous.
Thus in the four laft Examples 6 am added to-am, the
Sum is 5 am.
5 ay, the Sum is — a y.
7 ad.
And 4 ay added to
And 8 a d added to a d, the Sum is
And 4 y d added to yd, the Sum is 5yd. And in the fame
Manner may the other Examples be proved.
8. If two or more Quantities connected by the Signs + or
are to be ſubſtracted from other like Quantities connected
by the Signsor, it is done in the fame Manner, only
taking due Care to connect the remaining Quantities with their
proper Signs, as was done in the Addition of compound Qantities.
Exam. II.
-5xy-
2 a m
From
Take
Exam. 9.
12a +76
Exam. 10.
7ma+57
3a+26
6ma + 4y
3zy - 4 am
Remains
ma+y
—
8 z y — 6 u ki
ga+56
Exam. 9. By fuppofing 3 a to be-3 a, then 3 a added to
12 a the Sum is 9 a, by Art. 3. and again, fuppoling 2b to
be 2 b, then-2 b added to 7b the Sum is 5b by the fame,
and connecting thefe Quantities we have 9 a + 5 b, the Re-
mainder required.
G
Exam. 10. 6 m a being fuppofed negative, or to be
6 m a, then — 6 m a added to 7 ma the Sum is m a, and 4 y
being
SUBSTRACTION.
17
3 zy, and adding
being fuppofed to be 45, then-4y added to 5 y the Sum is y,
hence may is the Remainder required.
Exam. 11. 3zy being fuppofed to be
this to
5 zy the Sum is Ezy, by Art. 1.
ſuppoſed to be
4 am, by adding that to
by Art. 1. is 6 am, hence
823
and 4 am being
2 am the Sum
6 am is the Remainder
required.
Exam. 12.
Exam. 13.
Exem. 14.
From
14a-5 y
Take
за
5 y
Remains 17 a
4 mn — y
7 m n + 2y
3 m n + 3 y d
d
d
2a+m
5 a
7
~3a+m+7
Exam. 12. The
3 a being fuppofed by the general Rule to
be 3a, and adding that to 14 a the Sum is 17 a, by Art. 1. and
5y being fuppoſed to be 5y, if we add 5y to 51,
the Sum is a Cypher, or nothing, by Art. 5. hence 17 a is the
Remainder required.
the
Exam. 13. The-3 m n being fuppofed to be 3 m n, then by
adding 3 m n to-7mn the Sum is 4 mn,
3 y d being fuppofed to be
3y d, and adding
the Sum is -yd, by Art. 3. hence 4 m n
mainder required.
by Art. 3. and
3yd to 2 y d
yd is the Re-
5 a, if that is
Exam. 14. The 5 a being fuppofed to be
added to 2 a the Sum is 3a, by Art. 3. but the m and 7
being different Quantities, fet them down by Art. 6. only take
particular Care to change the Sign of 7, according to the general
Rule for Subſtraction, then will — 3a + m + 7 be the Re-
mainder required.
From
Exam. 15.
amy
Exam. 16.
1 5 y d + 20
Subſtract + am+y
Remains 2 a m
- 3rd
16
18 y d +36
Exam. 17.
a
14d +7-
-d+7-8 a
15d+7 a
The Truth of theſe Operations is proved in the fame Man-
ner as in Subtraction of fimple Quantities, by adding the
Remainder to the Quantity which is fubftracted, and obferving
if that Sum is the fame, and has the fame Signs, with those
Quantities from which the Subftraction was made.
Thus,
am, by
we find that
amy, the
Exam. 15.2 a m added to a mi, the Sum is
Art. 3. to which connecting y with the Sign +,
by adding 2 am to amy, the Sum is
Quantity from which the Subftraction was made.
D
Exam.
18
ALGEBRA.
Exam. 16. If 18 y d is added to 3 y d the Sum is 15 yd,
by Art. 3. and 36 added to 16 the Sum is 20 by the fame,
hence the Sum of 18 yd + 36 added to-3yd-16 is 15yd+20,
the Quantity from which the Subſtraction was made.
Exam. 17. By adding 15 d to d the Sum is 14 d, by
Art. 3. and by adding 7 a to 8 a the Sum is a, by the
fame, to which putting down the 7, there being no Quantity
to be added to that, hence 15 d+7 a added tod + 7-8 a
the Sum is 14d7a, the Quantity from which the Sub-
fraction was made.
But if the Quantities to be ſubſtracted are unlike thoſe from
which the Subtraction is to be made, fet down theſe with the
fame Signs and Co-efficients they have in the Example, after
which place the Quantities to be fubftracted with their Co-
efficients, but change their Signs.
Exam. 18.
From
Take
2 a
d
Exam. 19.
Exam. 20.
34
2 a
5 m
2 y
Remains 20- d
34 - 20
5 m + 2y
Exam. 18. Having put down 2 a, after which put-d, the
Quantity to be fubftracted being +d, and 2 ad is the Re-
mainder required.
Exam. 19. Having put down 3y, to that connect 2 a with
the Sign, fo is 3 y 2 a the Remainder required. The
20th Example is done in the fame Manner.
And if compound Quantities are to be ſubſtracted from com-
pound Quantities, but unlike, fet down all the Quantities one
after the other, but change the Signs of thofe Quantities which
are to be fubftracted, as in thefe Examples.
From
Take
2a +5m
3y 2 d
Remains 2a + 5 m — 3.3 + 2 d
-4d+2p
5a+37
4 d +2p+ 5a — 3 y
Having wrote down 2a +5 m, to that connect 3y with the
Sign, it being + in the Example, to which connect 2 d
with the Sign+, it being in the Example.
In the other Example, having wrote down-4 d + 2 p, to
this connect 5 a with the Sign+, it being in the Example,
to which conne& 3y with the Sign, it being in the
Example.
MULTI-
[ 19 ]
MULTIPLICATION,
In which there are three Cafes.
(9.) Cafe 1. WH
HEN the Signs of the Quantities to be
multiplied, are both affirmative, or both
negative, ſet or join the Letters together, and to them prefix
the Sign, which will be the Product required.
Exam. I.
Multiply
Exam. 2.
y
By
Product
a
d
d a
m
my
Exam. 3.
Z
Exam. 4.
da
x
a
da x
a z
Exam. 1. Having joined the Letters da, and each of them
having the affirmative Sign, therefore, by the Rule, da, or
+da, is the Product required.
Exam. 2. Having joined the Letters m and y, and each of
them having the affirmative Sign, therefore, by the Rule, my,
or my, is the Product required.
Exam. 3. Having joined the Quantities z and a, and each of
them having the fame Sign, therefore, by the Rule, a z, or
-† a z, is the Product required.
Exam. 4. Having joined the Quantities d a and x, and both
having the fame Sign, therefore dax, or + da x, is the Pro-
duct required.
Multiply a
Exam. 8.
Exam. 5.
Exam. 6.
Exam. 7.
a m
a m
y
By
a
Product
a a
d
am d
d p
d py
an
aman
Exam. 5. Having joined a a, and both the Quantities being
affirmative, therefore aa is the Product required.
Exam. 6. Having joined the Quantities a m and d, and both
having the Sign-, hence a m d is the Product required.
Exam. 7. Having joined the Quantities y and d p, and be-
cauſe both have the fame Sign, therefore dpy is the Product
required.
Exam. 8. Having joined the Quantities a m and an, and
both having the fame Sign, therefore am an is the Product
required.
D 2
10. If
20
ALGEBRA.
10. If the Multiplicand confifts of two or more Quantities
connected by the Signs + or, then the Multiplier must be
multiplied into each of thoſe Quantities, prefixing to each par-
ticular Multiplication its proper Sign, which will give the Pro-
duct. Thus,
Exam. 9.
Multiply a+ď
By
m
Produc
m a -\- m d
Exam. II.
Exam. 10.
x + y
m x
72
d
р
d z + dy
p m x + p n
Exam. 9. If we multiply a by m, the Product is m a, by
Art. 9. and multiplying d by m, the Product is md, by
Art. 9. but as m and d have both the affirmative Sign, therefore
prefix the Sign + before md, and mamd is the Product
required.
Exam. 10. Multiplying z by d, the Product is dz, and
multiplying y by d, the Product is dy, by Art. 9. but as d and
have both the Sign+, prefixing that Sign before dy we have
dzdy, the Product required.
y
Exam. 11. Multiplying-mx by-p, the Product is pm x,
by Art. 9. and for the fame Reafon n multiplied by
P, the
Product is pn, then connecting pm and pn with the Sign +,
we have pin x + pn, the Product required.
Exam. 12.
Exam. 13.
Exam. 14.
а
ad-1-%
Multiply
By
a
-y
У
ZY
y
d
ady + y z
Product dm j dy
n
ax + xzy
Exam. 12. Multiplying -m byd, the Product is m d, by
what was faid at Example 11, and multiplying-d by―y,
the Product is for the fame Reafon dy, and connecting dm and
dy with the Sign +, we have d m+dy, the Product required.
Exam. 13. Multiplying-a by-x, we have a x for the
Product, as in the last Example, and from multiplying —zy
by-x, we have for the fame Reafon x zy for this Product,
then connecting ax and xzy with the Sign, we have
ax+xzy, the Product required.
Exam. 14. Multiplying ad by y, the Product is a dy, and
multiplying z by y, the Product is y z; but as the Signs of y
and z are both alike, therefore prefixing the Sign
have a dy + yz, the Product required.
to yz, we
II.
It
MULTIPLICATION.
2I
II.
11. It may be proper to caution the Learner, that in
Multiplication it is quite indifferent which Letter he places firſt,
or laft, for to multiply am by d, the Product is a m d, or
m da, or d ma, or a dm, or any of the different Pofitions
in which three Letters can be placed; this will be more
compleatly and fully underſtood when we come to apply the
Science to the Solution of Problems: But, that the Learner may
form fome Idea of the Truth of this, fuppofe we were to multi-
ply 3, 5, and 7 together, the Product will be the ſame in what-
ever Order theſe three Numbers are multiplied. Thus,
3
5
15
7
105
35
ཋ ཋ མ
105
3
7
2 I
3
5
105
This Obfervation I adviſe the Learner to fix in his Mind, to
prevent concluding he has done any of the following Examples
erroneouſly, by happening to place the Letters different from
what they are in the Book.
12. But if the Multiplier and Multiplicand confifts of two or
more Quantities, then begin and multiply the Multiplicand by
any one Quantity in the Multiplier, according to the Directions
in Art. 9. and 1o. after that multiply the Multiplicand by
another Quantity in the Multiplier, and put this Product under
the other, and continue doing this till the Multiplicand has been
multiplied by every Quantity in the Multiplier; then under theſe
Products draw a Line, and add them together by the feveral
Cafes of Addition, and this will be the Product required.
Exam. I.
Multiply a+b
By
m -- n
mame the Product of a+b multiplied by m, by
Art. 10.
nanb the Product of a + b, multiplied by ", by
the fame.
mamb + naab the Sum by Art. 6. the Pro-
duct required.
Multiply
42
ALGEBRA.
Multiply my
By
at d
am+ay the Product of my multiplied by a, by
Art. 10.
mdyd the Product of my multiplied by d, by
the fame.
am+ay + m dyd the Sum by Art. 6. the Pro-
duct required.
Multiply
By
a
m
· d
Z
mamd the Product of a d multiplied by
m, by Art. 10.
az+dz the Product of. a - d multiplied by — ≈,
by the fame.
ma+md + a z+dz the Sum by Art. 6. the Product
required.
Multiply a+b
a + b
By
aa+ab the Product of a + b multiplied by a, by
Art. 10.
ab+bb the Product of a + b multiplied by b, by the
fame.
a a +2ab+bb the Product of a + b multiplied by
a+b.
Now in the Addition of the laft Example I obferve there
is a bin each of the two Lines, and there being no Co-efficient
prefixt, Unity or I being then always understood to be the Co-
efficient, hence I ab added to I ab is 2 ab; the other Quantities
a a and b b are fet down as in the former Examples, therefore
a a +2ab+bb is the Product required.
And in fuch Additions as thefe I recommend it to the
Learner, before he begins to add, to examine the feveral Quan-
tities, and fee if the Letters in any two of them are alike, and if
they are to collect them into one Sum, according to Art. I and
3; remembering, that though the Letters which compoſe the
two Quantities are not in the fame Order in each; yet if they
are but the fame Letters, and no more in one than there is in
the other Quantity, they are the fame, and may be added by
Art. 1 and 3.
I
The
MULTIPLICATION.
23
The four following Examples are for the Exercife of the
Learner.
Multiply a + b
By
aty
a+y
m + y
am + mb
a y + by
Product am+mb+ay + by
Multiply y+m12
By
y + m
y y + y m
y m+mm
Product yy + 2y m + mm
a a + a y
a y + y y
aa+2ay+"y
a
a
d
d
a a + a d
a a + d d
aa + 2 ad + d d
13. Cafe 2. If there are Co-efficients or prefixt Numbers,
then multiply the Numbers as in common Arithmetic, and to
their Products join the Products of the Quantities found by the
laft Caſe, Art. 9.
Exam. I.
Exam. 2.
Exam. 3.
Exam. 4.
Multiply 2 a
5 m
By
3 m
3 y
-7ad
3 m
29
a
Product 6 am
15 my
21 adm
2ay
Exam. 1. The Product of the Co-efficients 2 and 3 is 6: the
Product of a multiplied by m is a m, the Signs being alike, joining
theſe together it is 6 am, the Product required.
Exam. 2. The Product of 5 by 3 is 15: the Product of m
by y is my, for the Signs are alike, and joining theſe together it
is 15 my, the Product required.
Exam. 3. The Product of 7 by 3 is 21: the Product of a d
by mis adm, the Signs being alike, and joining the 21 and
adm it is 21 a dm, the Product required.
Exam. 4. The Product of 2, and I the Co-efficient of a, is
2, to which joining ay, the Product of a and y, it is 2 ay, the
Product required, for the Signs of 2y and a are alike, being
both negative.
Multiply
By
Product
7 am
2 d
14 amd
6 dz
2 a
35 p
·31
2 d z
ď
12dza
Q Y Y P
2 d d z
14. And
24
ALGEBRA.
14. And if there are two or more Quantities with Co-effi
cients connected by the Signs + or, to be multiplied by any
Quantity and its Co-efficient, they are multiplied as in the laft
Article, only connecting the feveral particular Products together
with their proper Signs, as was done at Art. 10.
Exam. I.
Multiply 2a+3b
By
3m
Product 6am +9 bm
Exam. 3.
Exam. 2.
37+5d
2 y 2 A
5m
32
15 y m + 25 d m
6yz +6 za
Exam. 1. Multiplying 2 a by 3 m the Product is 6 am, by
Art. 13. and then multiplying 3 m by 3 b the Product is 9 b m,
by Art. 13. to which prefixing the affirmative Sign, as the
Signs of 36 and 3 m are alike, and 6 am 9 bm is the Product
required.
Exam. 2. Multiplying 3y by 5 m the Product is 15 y m, by
Article 13. then multiplying 5 m by 5d the Product is
25 dm, to which prefixing the affirmative Sign, as the Signs of
5d and 5m are alike, and 15 y m + 25 dm is the Product
required.
—
2a by
Exam. 3. Multiplying 2y by 3 z the Product is 6 y z,
by Art. 9 and 13. Again, multiplying
3% the
Product is 6 a z, for the Signs of 2 a and 3% are alike, and con-
necting 6yz and 6 za with the Sign+, we have 6 y z + 6 za,
the Product required.
Exam. 4.
Exam. 5.
2 d
3m
4 a
Multiply 3 m + 2y
6a
By
Product 18 ma 12ya
+
Exam. 6.
2 A
8 da + 12 ma
31-7 my
бya-j- 14amy
Exam. 4. Multiplying 3 m by 6 a the Product is 18 m a, and
multiplying 2y by 6a the Product is 12 ya, and placing the
Sign + before 12ya, because the Signs of 6 a and 2 y are
alike, we have 18 ma 12ya, the Product required.
+
Exam. 5. Multiplying
2d by 4a the Product is 8 da,
for the Signs of 2 d and 4 a are alike, being both negative,
therefore 8 da or + 8 da is the Product of thefe Quantities.
Now multiplying-4a by 3m the Product is 12 ma, to
which prefixing the affirmative Sign, as 3m and 4 a have the
fame Sign, both being negative, and we have 8 da + 12 ma,
the Product required.
3
Exam.
MULTIPLICATION.
25
Exam. 6. The Product of
3y by 2 a is 6 y a, or + 6 y a,
2 a is 14 amy, or + 14 amy,
and the Product of -7 my by
hence, for the Reafon in the laft Example, 6 y a
the Product required.
Multiply 3 m 2 d
By
4 a
Product 1 2 m a + ŏ d a
14 a my is
·4d-5m
2 Z ·3y
4 a
26
8za + 12a y
8bd +10mb
15. And if there are two or more Quantities with Co-
efficients connected by the Signs + or —, to be multiplied by
two or more Quantities with Co-efficients conne&ed in the
fame Manner, the Quantities are to be multiplied as at Art. 12.
taking due Care to multiply the Co-efficients, as has been taught
Art. 14. Thus,
Multiply 2a+36
3a + 5m
By
6aa9ab the Product of 2 a
3a, by Art. 14.
3 b multiplied by
10 ma15bm the Product of 2 a +36 multiplied by
5 m, by the fame.
baa+gab† 10 m a + 15 v m
the Product re-
quired, being the Sum of the two particular Products, which are
added together by Art. 6.
Multiply 3 m +59
By
zazn
6 am
10ay the Product of 3m+5y multiplied
by 2 a, by Art. 14.
9 m n + 15 yn the Product of 3m +53 multiplied
6 am
by 32, bv Art. 14.
10ay9mn+15yn the Product required,
being the Sum of the two Products, which are added together
by Art. 6.
Multiply 2a+36
By
2a+26
4aa6 ab the Product of 2 a
2a, by Art. 14.
36 multiplied by
4ab + 6 bb the Product of 2a + 36 multiplied by
2b, by the fame.
4aa+1ab6bb the Produ& required. In this
E
Addition
26
ALGEBRA.
Addition the Reader is to obferve that in one Line there is 6 ab,
and in the other Line there is 4 ab, which two Quantities
added together the Sum is 1o a b, by Art. 1. but the 4a a and
6bb being different Quantities, they are fet down by Art. 6.
hence the Product of 2a+3b multiplied by 2a + 2b, is
4aa+10ab+6bb.
Multiply 3 a +76
2a +5 n
6aa14b a
By
1 5 a n + 35 b n
за
3a+26
a + 4 b
3aa-j-2ba
12ba + 8b b
Product 6aa+14ba+15an+35bn
3aa4146a+ 8 b b
16. Cafe 3. When the Signs of the two Quantities to be
multiplied are one affirmative and the other negative, then
multiply the Quantities as before directed, but to their Product
prefix, or the negative Sign.
Exam. I. Exam. 2.
Exam. 3.
Exam. 4.
d m
Multiply-b
a m
a
Z
d
y
By
a
dm z
Product
-ba
a d
amy
Exam. I. The Product of b by a is ba, for Multiplication is
only joining the Letters, but as the Sign of bis —, and that
fo is ba the
of a is, therefore to b a prefix the Sign-,
Product required.
Exam. 2, The Product of a by d is a d, but as the Signs of
a and d are different, therefore prefix the Sign to ad, and
a d is the Product required.
to a my,
Exam. 3. The Product of am by y is a my, but as the Signs
of a mand y are different, therefore prefix the Sign
fo is amy the Product required.
Exam. 4. The Product of dm by z is dmz, but as the
Signs of dm and ≈ are different, therefore prefix the Sign
to dmz, fo is dmz the Product required,
This Operation being the fame as at Art. 9. taking Care
to make the Sign of the Product—, I fhall only fubjoin the
following Examples for the Exercife of the Learner.
Multiply
MULTIPLICATION.
27
Multiply
By
Product
x m
-ym
ут
ax
ma
y
d
x my
ym d
dy
-axdy
-p
map
17. And if two or more Quantities with Co-efficients are to
be multiplied into any one Quantity with a different Sign, they
are multiplied as at Art. 14. taking Care of the Signs arifing in
the Product, according to Art. 9. and 16.
Multiply
By
Exam. I.
3a — 2 ż
Exam. 2.
Exam. 3.
274-5d
5 m
2 }
3m
-3 a
3 d
Product
9am-6zm-6ay—15ad
-15md-6dy
Exam. 1. Multiplying 3a by 3 m, the Product by Art. 13.
is gam,
but as 3 m has the Sign + prefixed to it, and 3a has
the Sign prefixed to it, therefore to the Product 9 am prefix
the Sign, by Art. 16. Again, 2 z multiplied by 3 m the
Product is 6 zm, but as 2x has the Sign to it, and 3m the
Sign, therefore to 6 z m prefix the Sign, by Art. 16. and
— 9am-5zm is the Product required.
Exam. 2. Multiplying 2y by 3a, the Product is 6 ay, but
as the Sign of 2y is +, and that of 3 a is, therefore to 6 ay
prefix the Sign, by Art. 16. Then multiplying 5d by 3a
the Product is 15 ad, but as the Sign of 5 d is +, and that of
3a is, therefore prefix the Sign to 15 ad by Art. 16.
and 6ay 15 ad is the Product required.
ཀ
Exam. 3. Multiplying 5 m by 3d, the Product is 15 md,
but as the Sign of 5 m is and that of 3d is, therefore
prefix the Sign-to 15 md. Again, multiplying 2y by 3d,
the Product is 6yd, but becauſe the Signs of 2y and 3d are
different, therefore prefix the Sign-to 6 dy, and
6 dy is the Product required.
Examples for the Exercife of the Learner.
15 md-
Multiply
By
2a-3b
3m+7y
42
2 d
Product 8az
1262
6md-14dy
57-3b
3 m
-15ym ·9bm
18. And if two or more Quantities with Co-efficients are to
be multiplied by two or more Quantities with Co-efficients, if
their Signs are unlike, yet they are multiplied as at Art. 15.
taking due Care of the Signs of the Product, by Art. 9. and 16.
E. 3
Multiply
28
ALGEBRA.
17.
Multiply
за
2 m
By
4 b + 6 y
-12ab-86m the Product of
3a-2m multiplied
by 4 b, by Art.
18ay-12ym the Product of.
2 m multi-
12ab 8 bm
18 ay
12ym the Product re-
34-
plied by 6y, by Art. 17.
quired, being the Sum of the two particular Products, which are
added by Art. 6.
Multiply 51+3m
By
-7d-3a
-35yd-2x d m the Product of 5y + 3m multiplied
by 7 d, by Art. 17.
-15ay-9am the Product of 5y+3m multiplied
by 3a, by Art. 17.
35 y d
21 dm 15 ay
9am the Product re-
quired, being the Sum of the two particular Products, which are
added by Art. 6.
Multiply
2a+36
By
za — 3 b
4aa-bab the Product of 2 a
3b multiplied
by 2a, by Art. 17.
-6ab9bb the Product of 2a+3b multiplied
by 3b, by Art. 17.
— 4 aa—12ab — 966 the Product required: For
in this Addition the Reader may obferve that there is 6 a b
in each of the two particular Products, which being added toge-
ther by Art. 1. make- 12 ab, but 4 a a and 9 bb being
different Quantities, they must be placed feparate from one
another. There are Examples of this Kind at Art. 15.
19. It may be for the Learner's Advantage to be put in Mind,
that if any Algebraic Quantities are to be multiplied by a pure
Number, that then this Number is to be multiplied into every
one of the Co-efficients of the other Quantities, in all Refpects
as before, and to each particular Product ſet or join that Quan-
tity whoſe Co-efficient was multiplied. Thus,
Exam.
MULTIPLICATION.
29
Exam. 3.
Exam. I.
Exam. 2.
Multiply 2a+36
By 6
-3m-4d
4 d +37
7
9
Product 12a + 18b
21 m — 28 d
36d+27x
Exam. 1. Multiplying 6 by 2 the Product is 12, to which
joining a it is 12 a, then multiplying 3 by 6 it is 18, to which
joining b it is 18 b, and becauſe 6 and 36 have both the Sign +,
therefore by Art. 9. prefix the Sign
the Product required.
3 m is
,
to 18 b, fo is 12 a † 186
and that of 7 is +, there-
to 21 m, and it is
Exam. 2. Multiplying 3 by 7 it is 21, to which joining m it
is 21 m, but as the Sign of
fore by Art. 16. prefix the Sign
Again, multiplying 4 by 7 it is
21 m.
28, to which joining d it is
28 d, but as the Signs of 4 d and 7 are likewiſe unlike, there-
fore to 28 d prefix the Sign
and 21 m 28 d is the
Product required.
Exam. 3. Multiplying 4 by 9 the Product is 36, to which
joining d it is 36 d, and becauſe the Signs of 4d and 9 are alike,
therefore it will be 36 d, or + 36 d; and multiplying 9 by
3 y the Product will be 27 y, to which must be prefixt the Sign
+, becauſe 3y and 9 have the fame Sign, fo is 36 d + 27 j
the Product required.
Examples wherein all the three Cafes of Multiplication are
promifcuouſly uſed.
Multiply 2a-3b
By
5m+2y
IQ ma 15 mb
4 a y — 6 y b
Product 10m a
15mb +4 ay — 6 y &
The 2a being multiplied by 5 m the Product is 10 ma, by
Art. 13. and 3 b being multiplied by the fame 5 m, the
Product is 15 in b, by Art. 16. and 17.
6 y b,
And 2 a being multiplied by 2y the Product is 4 a y, by
Art. 13. and 3 multiplied by 2y the Product is
by Art. 16. and 17.
Now draw the Line and begin to add them, and becauſe the
Quantities are all different, they are added by Art. 6. and there-
for the Product will be 10 ma-15 mb + 4 ay — 6 y b.
2
Multiply
30
ALGEBRA
Multiply
By
-7m+2a
3 y
4 n
21 my--
6 ay
28 m n
8 na
Product
The
21 my +бay 28 m n — 8 na
7 m multiplied by 3 y the Product is
21 my, by
Art. 16 and 17. and 2 a multiplied by 3y the Product is 6 a y,
by Art. 13.
And -7 m multiplied into
Art. 13. and
4n the Product is 28 m n, by
4 n multiplied by 2 a the Product is 8 nag
4n
by Art. 16 and 17.
Now begin the Addition, and becauſe the Quantities are
all different, they are added
21 my + 6 ay + 28 m n —
2a+36
:36
Multiply
By
2 a
4 a a + 6 a b
6 a b
96
b b
by Art. 6. and the Product is
8 na.
Product
4aa-9bb
966
Multiplying 2 a by 2 a the Product is 4 a a, and multiplying
3b by 24 the Product is 6 ab.
6 a b
s
And multiplying 2 a by-3b the Product is
becauſe the Signs of the two Quantities are unlike, and for the
fame Reaſon the Product of 36 by 3b is
9bb.
Now begin the Addition, and I obferve in the firft Line there
is + 6 ab or 6 ab, but in the fecond Line there is
— ɓ a by
now becauſe the Co-efficients are equal, the Quantities alike,
but the Signs being contrary, therefore by Art. 5. thefe Quanti-
ties will deftroy one another, then putting down the 4a a and
-9bb, by Art. 6. we have 4 aa-9 bb, the Product required.
Multiply
By
7 m + 4 a
3a+5
21ma- 12 aa the Product of 7 m † 4 a multi-
plied by
3a.
35m -- 20 a the Product of 7 m -- 4 a multi-
plied by 5, by Art. 19.
Product
21 Md --re 12aa + 35 m -† 20 a
Multiply
DIVISION.
31
Multiply 5a + b
By
2a+3b
10aa2ab the Product of 5 a + b multiplied by 2 a.
15ab3bb the Product of 5 a + b multiplied by 3b.
Product 10aa + 17 ab + 3b b.
DIVISION,
In which there are four Cafes.
20. Cafe 1.
WHE
HEN the Signs of the Quantities to be
divided are both affirmative, or both ne-
gative, reject all thofe Quantities in the Dividend and Divifor
that are alike, and fet down the Remainder, prefixing to it the
Sign+, which will be the Quotient required.
Divide
Exam. I.
a b
Exam. 2.
Exam. 3.
Exam. 4.
By
a
d m
d
In n
ap
712
-Þ
Quotient
b.
12
a
Jil
Exam. 1. Becaufe a is in the Dividend and Diviſor, eject
it, and b being only left, it is the Quotient fought, and is to
have the Sign, becauſe the Signs of ab and a are alike.
Exam. 2. Becaufe d is in the Dividend and Divifor, reject
it, and there being only m left, it is the Quotient fought,
which muſt have the Sign+, becaufe the Signs of d m and d
are alike.
Exam. 3. Becaufe m is in the Dividend and Divifor, reject
it, and a being only left, I write it down for the Quotient
fought, which must have the Sign +, becauſe m n and m have
the fame Sign.
Exam 4. Becaufe p is in the Dividend and Divifor, I reject
it, and place down a, the Quantity left, for the Quotient
fought, which mult have the Sign-, for the Signs of ap and
pare alike.
Exam.
į
32
ALGEBRA.
Exam. 5.
Exam. 6.
Exam. 7.
Exam. 8.
Divide
amd
By
-apy
ру
mda
my z
ma
Z
d
a
d
y m
Quotient
a m
Exam. 5. Becauſe a m is in the Dividend and Diviſor, reject
it, and place down d for the Quotient, which must have the
affirmative Sign, for the Signs of a m d and a m are alike.
Exam. 6. Becaufe py is in the Dividend and Divifor, I
reject it, and place down a, the remaining Quantity, for the
Quotient, which must have the affirmative Sign, for the Sign's
of a py and py are alike.
Exam. 7. Becauſe ma is in the Dividend and Divifor, I
reject it, and place down d for the Quotient, which must have
the Sign+, becauſe the Signs of m da and ma are alike.
Enam. 8. Becauſe is in the Dividend and Divifor, I
reject it, and place down y m, or my, which is the fame thing,
for the Quotient fought, and must have the Sign +, becaufe
the Signs of my z and z are alike.
Divide apz
By
az
Quotient
P
m nd
md
a b c
abdy
C
ay
n
ab
b d
!
The Truth of thefe Operations in Divifion may be proved
like thofe in Arithmetic, for the Quotient and Divifor being
multiplied, the Product will be the Dividend if the Work is
true; thus in the fecond Example of the laft four, by multiplying
the Quotient into-md the Divifor, the Product is md n
or mnd, to which must be prefixt the Sign, by Art. 16. be-
caufe the Signs of md and n are unlike, hence the Product with
its Sign is mnd, the given Dividend.
And in the last Example, if we multiply bd the Quotient by
ay the Divifor, the Product is bday, or abdy, which is the
fame thing, by Art. 11. and this Quantity muft have the affir-
mative Sign, by Art. 9. for the Signs of d and ay are alike,
hence abdy, or abdy, is the Product with its Sign, the fame
as the given Dividend: And fo of any other Example.
+
21. But if all the Quantities in the Divifor are not to be found
in the Dividend, then you must only reject thofe Quantities in
the
DIVISION.
33
the Dividend and Divifor that are alike, placing down the re-
maining Quantities of the Dividend, and under them thoſe of
the Divifor that are not rejected by this Rule; which will
be the Quotient fought, and fland like a Vulgar Fraction in
common Arithmetic.
Exam. I.
Exam. 2.
Exam. 3.
Exam. 4.
Divide
amb
m n dz
By
ay
m na
-dayp
dpz
p n q r
pad
Quotient
mb
dz
ay
n q r
y
a
Z
ad
Exam. 1. Becauſe a is in the Dividend and Divifor, reject
it, and place down mb the remaining Part of the Dividend,
under which drawing a Line, and place y the remaining Part of
721 b
y
the Divifor, fo will be the Quotient fought, which must
have the Sign+, by Art. 20. as the Signs of the Quantities to
be divided are alike.
Exam. 2. Becauſe mn is in the Dividend and Diviſor, reject
it, and place down dz the remaining Part of the Dividend,
under which drawing a Line, and place a the remaining Part of
d ૪.
the Divifor, fo is the Quotient required, and it muſt have
Q
the Sign+, by Art. 20. as the Signs of the Quantities to be
divided are alike.
Exam. 3. Becaufe dp is in both Dividend and Diviſor, reject
it, and write down ay the remaining Part of the Dividend,
under which place z the remaining Part of the Divifor, as in the
two former Examples, fo is,
or + Y, the Quotient re-
a y
quired, for the Signs of the two Quantities to be divided are
alike.
Z
༤
Exam. 4. Becaufe p is in both Dividend and Divifor, reject
it, and write down n qr the remaining Part of the Dividend,
under which place ad the remaining Part of the Divifor, and
nar is the Quotient required, which will be affirmative by
Art. 20. becauſe the Signs of pnqr and pad are alike.
ad
F
Divide
7:
34
ALGEBRA,
Divide
a p q n
ad z
mn d
- y z d b
By
anm
ap
ma
-yza
Quotient
P q
d z
n d
d b
m
P
a
Q
Thefe Operations are proved as at Art. 20. by multiplying
the Quotient by the Divifor; for in the laſt Example the Quotient
db
is which is a Fraction: the Divifor isy z a, which by the
-yza,
a
Rule of Vulgar Fractions in common Arithmetic is made this im-
proper Fraction --
Ι.
>
yza
then the two Fractions to be multiplied
d b
are
and
y za
a
, multiplying the Numerators we have
I
dbyza for the new Numerator, and multiplying the Denominators
we have a for the Denominator, hence the Product is this Fraction
d b
d by z a
a
the affirmative Sign, therefore by Art. 16. prefix the Sign
but as
>
a
y za
I
has the negative Sign, and has
to
d by z a
and it is
the Product with its true Sign:
a
d by z a
a
>
But in this Fraction as d by za is to be divided by a, reject-
ing a both in Dividend and Divifor by Art. 20. we have-dbyz
or-yzdb, the fame with the Dividend in the given Example;
in like Manner may the others be proved.
22. And if there are two or more Quantities connected by
the Signs or to be divided by any fingle Quantity, every
+
Quantity in the Dividend muſt be divided by the Divifor, fetting
down the particular Quotients, as at Art, 20. which must be
connected by the Sign + when the Signs of the Quantities to
be divided are both alike.
Exam. 3
Exam. I.
Divide
By
a b + am
Exam. 2.
m dJ m z
-da
d p q
d
m
a
d + z
a + p q
Quotient b+ m
Exam. 1. Dividing ab by a the Quotient is b, by Art. 20.
and dividing am by a the Quotient is m, by the fame Art, but
29
DIVISION.
35
as am and a have both the affirmative Sign, therefore to m pre-
fix the Sign +, fo is b+m the Quotient required.
Exam. 2. md being divided by m the Quotient is d, by Art.
20. and dividing m z by m the Quotient is z, to which prefix-
ing the Sign, as m'z and m have both the fame Sign, we have
dz for the Quotient required.
Exam. 3. da being divided by d the Quotient is a, and be-
caufe da and a have both the negative Sign, or the Signs are
alike, therefore a muſt have the Sign+, whence it is a or a,
and dividing-dpq byd the Quotient is pq, to which
muſt be prefixed the Sign+, for the Signs of dpq and d are
alike, hence we have a +pq for the Quotient required.
Divide
By
Exam. 4.
abam
Exam. 5.
b m + b n
a
b
Quotient
b + m
m + n
Exam. 6.
-zypzya
Zy
pte
Exam. 4. Dividing
Dividing ab by -a the Quotient is b, by
Art. 20. and it muſt be + b or b, as the Signs of a b and a are
alike: then dividing -am by a the Quotient is m, by Art.
20. becauſe the Signs of am and a are alike, then connecting
band m with the Sign+, and b+m is the Quotient required.
Exam. 5. Dividing bm by b the Quotient is m, by Art. 20.
and dividing bn by b the Quotient is n, and as bn and b have
the fame Sign, therefore prefix the Sign+to n, fo is m+n
the Quotient required.
Exam. 6. Dividing-zy p byzy the Quotient is p, by
Art. 20. and dividing-zy a by-zy the Quotient is a, to
which prefixing the Sign+, for the Signs of zya and zy are
alike, we have pa the Quotient required.
Divide
By
Quotient
- d n Z zad am + a d
dz
n+a
a
m + d
dy - d z
d
Y+z
The Truth of theſe Operations are proved by multiplying the
Quotient by the Divifor; if that Product agrees with the Divi-
dend in its Quantities and Signs the Work is true, otherwiſe
Now in the laft Example the Quotient y + z, and the
Diviford, being multiplied together by Art. 14. they pro-
duce dyd, the given Dividend.
not.
F 2
23. Cafe
36
ALGEBRA.
23. Cafe 2. When the Signs of the Quantities to be divided
are one affirmative and the other negative, find the Quotient of
the Quantities as before; but to them prefix the negative Sign,
or Sign
Exam. 1.
Exam. 2.
Exam. 3.
Exam. 4.
Divide
By
a m
a
m n p
m
ayx
-ay
d m b
d b
Quotient
m
np
א
Z
m
Exam. 1. Becauſe a is in both Dividend and Divifor, rejest
it, and place down m the remaining Part of the Dividend, but
as the Signs of am and a are different, therefore to m prefix
the Sign, and it will bem, the Quotient required.
Exam. 2. Becauſe m is in the Dividend and Divifor, rejest
it, and place down np the remaining Part of the Dividend,
but as the Signs of mnp and m are different, therefore to rp
prefix the Sign, and it will be np, the Quotient re-
quired.
Exam. 3. Becauſe ay is in the Dividend and Divifor,
reject it, and place down z the remaining Part of the
Dividend, but as the Signs of ayz and ay are different,
prefix the Sign to z, and it will be z, the Quotient
required.
Exam. 4. Becaufe db is in the Dividend and Divifor,
reject it, and place down m the remaining Part of the Divi-
dend; but the Signs of the Quantities that are divided being
different, to m prefix the Sign, and it will be-m, the
Quotient required.
Divide
By
m n p
mn p
dyp
m
771 72
dy
da b
d b
a
P
Quotient np
+
24. And if there are two or more Quantities connected by
the Signs or, to be divided by any firgle Quantity, the
Operation is the fame as at Art. 22. only taking due Care,
when the Signs of the Quantities to be divided are different,
to prefix the Sign- before thole Quotients.
Exam.
DIVISI O N.
37
Divide
By
Quotient
Exam. I.
Exam. 3.
Exam. 2.
m n
md
ad + ab
dnz-dzy
a
dz
db
n-y,
m
n-d
Exam. 1. Dividing
mn by m the Quotient is n, by
Art. 20. but as the Signs of m n and m are different, therefore
And
by Art. 23. I prefix the Sign to n, and it is—n.
dividing ―md by m, the Quotient is d; but as the Signs of
m d and m are different, therefore by Art. 23. prefix the Sign-
tod, hencend is the Quotient required.
Exam. 2. Dividing ad by a the Quotient is d, to which
prefix the Sign-, by Art. 23. which makes it-d: then
dividing a b bya, the Quotient is b; but as the Signs of ab
and a are different, therefore by Art. 23. prefix the Sign
b, and d—b is the Quotient required.
to
Exam. 3. Dividing-dnz by dz the Quotient isn, by
Art. 20. and 23. and for the fame Reafon dividing-dzy by
z, the Quotient isy, which placing aftern, we have
ny, the Quotient required.
Divide
By
maz + mzd
M Z
dab-dby
d b
az x
azb
az
ad
a-y
X
-b
Quotient
The Truth of theſe Operations are likewife proved from
multiplying the Quotient by the Divifor, and if it agrees with
the Dividend in its Quantities and Signs, the Work is true,
otherwiſe not.
25. Cafe 3. But when there are Co-efficients joined to the
Quantities, divide the Co-efficients as in common Arithmetic;
and to their Quotients join the Quotient of the Quantities found
by the foregoing Directions; but cautiously remember, that if
the Signs of the Quantities that are divided are alike, the Quo-
tient must have the affirmative Sign, as at Art. 20. but if the
Signs of the Quantities that are divided are unlike, then the
Quotient must have the Sign-prefixt to it, by Art. 23.
Exam. 4.
Exam. I.
Divide
By
16am
2Z
24
Quotient 8 m
47
Exam. 2.
83 Z
Exam. 3.
24
d m
18ma
-6 d
6 a
4 m
3 m
Exam
38
ALGEBRA.
Exam. 1. Dividing 16 by 2 the Quotient is 8, and am di-
vided by a the Quotient is m, joining 8 to the m it is 8 m, and
as 16 am and 2 a have the fame Sign, hence by Art. 20. the
Sign+muſt be prefixt to 8 m, therefore the Quotient is +
8 m or 8 m.
Exam. 2. Dividing 8 by 2 the Quotient is 4, and dividing
yz by z the Quotient is y, joining the 4 to the y it is 4y;
but as 8 yz and 2 z have the ſame Sign, therefore by Art. 20.
prefix the Sign + to 4 y, hence + 4y or 4y is the Quotient
required.
Exam. 3. Dividing 24 by 6 the Quotient is 4, and dividing
dm by d the Quotient is m, joining 4 to the m it is 4 m; but
as 24 dm and 6d have the fame Sign, prefix the Sign + to 4 m,
hence + 4 m or 4 m is the Quotient required.
Exam. 4. Dividing 18 by 6 the Quotient is 3, and dividing
ma by a the Quotient is m, joining 3 to the m it is 3 m, and
as 18 ma and 6a have the fame' Sign, therefore by Art. 20.
the Quotient is + 3 m or 3 m.
Exam. 5.
Exam. 6.
Exam. 7.
Exam. 8.
Divide
1597
8 d m
28yz
12 da
By
за
4 d
7 y
за
Quotient
5y
2 m
4 Z
4 d
Exam. 5. Dividing 15 by 3 the Quotient is 5, and dividing
ay by a the Quotient is y, joining 5 to the y it is 5y, but as
the Signs of 15 ay and 3a are different, therefore by Art. 23.
prefix the Sign to 5, and 5y is the Quotient re-
quired.
Exam. 6. Dividing 8 by 4 the Quotient is 2, and dividing
dm by d the Quotient is m, joining the 2 and m it is 2m; but
as 8 dm and 4 d have the fame Sign, prefix the Sign + to 2 m,
by Art. 20. and + 2 m or 2 m is the Quotient required.
Exam. 7. Dividing 28 by 7 the Quotient is 4, and dividing
y z by y the Quotient is z, joining the 4 and z it is 4 ; but as
28 y z and 7y have different Signs, therefore by Art. 23. prefix
the Sign-to 4 %, fo will-4 z be the Quotient required.
Exam. 8. Dividing 12 by 3 the Quotient is 4, and dividing
da by a the Quotient is d, joining the 4 and d it is 4d; but as
the Signs of 12 da and 3 a are different, therefore by Art. 23.
prefix the Sign to 4d, and then- 4d is the Quotient
required.
-
!
Divide
DIVISION.
39
Divide
By
32 am
18 dza
8 m
9 a
22ym n
Il y n
16a%
8 z
Quotient
4a
2dz
2 m
2 a
26. And if there are two or more Quantities connected toge-
ther with Co-efficients, to be divided by any fingle Quantity and
its Co-efficient, the Operation is ftill performed in the fame
Manner, connecting the particular Quotients as at Art. 22, and
24. ftill carefully remembering that when the Quantities that
are divided have like Signs, whether affirmative or nega-
tive, the Quotient muſt have the affirmative Sign; but if the
Signs of the Quantities that are divided are unlike, then the
Quotient muſt have the Sign-prefixt to it.
Divide
By
Exam. 1.
4am+12ad
2 a
Quotient 2m+6 d
Exam. 2.
-16 my 24mz
4 m
43-6z
Exam. 3.
28 dn-21 db
7 d
412-36
Exam. 1. Dividing 4 am by 2a the Quotient is 2 m, by
Art. 25. and dividing 12 ad by 2 a the Quotient is 6d, and
becauſe the Signs of 2a and 12 ad are alike, prefix the Sign +
to 6d, and we have 2 m+6d, the Quotient required.
Exam. 2. Dividing-16 my by-4 m the Quotient is 47,
by Art. 25. for the Signs of 16 my and 4 m are alike, and
dividing 24 mz by 4m the Quotient is 6 z, for 6 z muft
have the negative Sign prefixt to it, the Signs of 24 m 2 and
4m being unlike; hence 4y6z is the Quotient required.
Exam. 3. Dividing 28 dn by 7d the Quotient is 4n, or
+4n, for the Signs of 28 dn and 7 d are alike: and dividing
21 db by 7d the Quotient is 3b, for 36 muſt have the
negative Sign prefixt to it, as the Signs of 21 db and 7 d are
unlike, hence 47-3b is the Quotient required.
Divide 16 pa—28 pd 24nm+36mz 16 zu-4zd
By
Quotient
4 P
4a+7d
4 m
On - 9 z
2Z
8u-2d
The Truth of thefe Operations are proved likewife from mul-
tiplying the Quotient by the Divifor, for if the Work is true,
the Product will agree with the Dividend in its Quantities and
Signs: In the last Example the Divifor is 2, and the Quotient
is 8 u 2d, now if we
Multiply
40
ALGEBRA.
%
By
Multiply 8 uzd
2 Z
16zu 4
zd the Product is the fame as
the given Dividend, and fo may the other Examples be proved.
27. Cafe 4. But when any Quantities in the Dividend are not
the fame with thoſe in the Divifor, then place down the Dividend
with its Signs and Co-efficients, under which drawing a Line,
and after the Manner of Vulgar Fractions place the Divifor with
its Signs and Co-efficients, and this will be the Quotient
required.
Exam. I.
Exam. 2.
Exam. 3.
Exam. 4.
Divide
b
a m
3 my
2 dy
d
Z
b
By
Quotient
oralo
a m
3 my
2 dy
ď
b
Exam. 1. Becauſe b and a are different Quantities, place
down the Dividend b, under which draw a Line, and place the
Divifor a, fo is the Quotient required.
Exam. 2.
b
a
Becauſe am and d are different Quantities, place
down am the Dividend, draw a Line under it, and place the
a
Divifor d, ſo is 4m the Quotient required.
d
Exam. 3. Becauſe 3 my and z are different Quantities, place
down 3 my
the Dividend, under it draw a Line, and place z the
Divifor, fo is 3my the Quotient required.
Z
Exam. 4. Becauſe 2dy and b are different Quantities, place
down 2 dy the Dividend, draw a Line under it, and place b the
Divifor, and
2 dy
6
b
is the Quotient required.
Divide
By
2 ma
5dz
2y
21 ma
5 d
8 y d
32
Quotient
2
3y
2 m a
31
5 dz
Zy
21 ma
5 d
i
8 y d
3%
Divide
DIVISION.
41
Divide
By
Quotient
+
7 y
m a
7 d
3 mb c
m Z
y d
24d
7 d
7YZ
M Z
3 m b c
y d
24 d
73%
ma
7 y
28. When two or more Quantities connected by the Signs
or are to be divided by any fingle Quantity, and the
Quantities in the Dividend are different from thoſe in the
Divifor, then having fet down all the Quantities in the Di-
vidend with their Signs and Co-efficients, draw a Line under
them all, under which place the Divifor as before, and this
will be the Quotient required.
Exam. I.
Exam. 2.
Exam. 3.
Divide
By
2a+38
7 y 2 m
15%-7da
5 m
3n
4y
Quotient
2a+36
5 m
7y- 2m
3 n
15%-7da
4 y
i
Exam. 1. Becaufe 2 a+3b the Dividend and 5 m the Divifor
are different, therefore place down 2a+3b, under which draw
a Line, and place the Divifor 5 m, ſo is
2a+36
5m
the Quotient
required.
Exam. 2. Becauſe 7y 2 m the Dividend and 37 the Di-
viſor are different, therefore place down y — 2 m, under which
draw a Line, and place the Divifor 3 m, fo is 72 m the
Quotient required.
3n
7 da the
Exam. 3. Becauſe 15 -7 da the Dividend and 4y the
Divifor are different, therefore place down 15 %-
Dividend, under which draw a Line, and place 4 y the Divifor,
15%-7 da
the Quotient required.
fo is
4 y
Divide
4ma-
за
7 db — 5xx
19m 15p
By
༨༤
3V
7 y
Quotient 4ma-3
32
7 db — 5xx
19 m
15 P
5%
3Y
G
73
Divide
42
ALGEBRA.
Divide
3dx-5b
7ym-3 dn
By
2 Y
5 u
3dx-5b
7ym-3 dn
Quot ent
2y
Su
25dp+532-17 118
70
251p+5y≈ 17 m
7 a
وخت
29. If there are two or more Quantities connected by
the Signsor –
to be divided by two or more Quantities
connected by the Signs + or, but the Quantities in the
Dividend are different from thofe in the Divifor, it is only
placing down the Dividend as before, under which drawing a
Line, and place in like Manner the Divifor, and this will be
the Quotient required.
Exam. 3.
14m+5% - II ș
2 d
Exam. I.
Exam. 2.
Divide 2a + m
5-7 d
By
5d+3y
3a+2m
3 V
20+ m
Quotient
5y- 7.d
→ 14m² + 5%
I I X
5d+3y
3a+2m
37-
2 d
Exam. 1. Becauſe the Quantities in the Dividend and Divifor
are unlike, therefore place down 2a+m the Dividend with its
Co-efficients and Signs, under which draw a Line, and place
2a + m
5d+ 3y the Divifor, fo is
the Quotient required.
Exam. 2. Becauſe 5y
3a+2m the Divifor,
Dividend, under which
Divifor, fo is 5-7d
3a + 2 m
Exam. 3, Becauſe
5d+39
7 d the Dividend is different from
therefore place down 5y
7d the
draw a Line, and place 3 a + 2 m the
the Quotient required.
14m +5% -II x the Dividend is
different rom 3y-2d the Divifor, therefore place down
14 in + 5% 11x the Dividend,
and place 3y-2 d the Divifor, and
is the Quotient required.
Divide 4.in
༢* + 22
By
Quotient
4: m
d
2h
draw a Line under it,
14 m +5% II X
3y-2d
5 y
21 pm +19 Z Y
14yz-9dx
-3 m
J
m-5 y
3x+22
5
217 m + 19 zy
dx
J
- 512
g d x
3 111 4 5 n
Divide
DIVISION.
43
Divide
By
-4a+5m-3d
72-81 y
за
4a+3x-5*
— 7 d + 11m
Quotient
-411-5m 3 d 1 a +37-5 ×
72-87
7a+ 11m
2a+37
-52-712
5རྨ
2a+37
-52-77
30. It may be juft obferved to the Learner, that when any
Quantity is divided by itfelf, or the Dividend and Divifor are
alike, that then the Quotient will be Unity, or 1. And if the
Signs of the Quantities to be divided are alike, the Quotient
muft have the Sign +; but if the Signs of the Quantities to
be divided are unlike, then to the Quotient, or 1, prefix the
Sign-.
Divide
By
2
z a b
20 b
14 m n
a
14 m n
5 dz
5dx
+7"
7 y
Quotient
I
I
I
I
For by Art. 25. if we divide the Co-efficients, the Quotient
will be Unity, or I; then, by Art. 20. rejecting all thoſe
Quantities that are alike, both in the Dividend and Divifor,
the Quantities all vanifh, and there will be none to be joined to
the Unity, or 1; whence, in fuch Cafes as thefe, Unity, or 1,
is the Quotient required.
It may be further obferved, that if an abfolute or pure
Number is the Divifor, the Co-efficients in the Dividend, if
there are more than one, muſt be divided by the Divifor, and
to each of thefe Quotients join the refpective Quantities of the
Dividend, as at Art. 26.
Divide
24ma + 18yz
16za+24ym
+ 141d +35%
By
6
8
7
Quotient 4ma +33%
2za+3ym
2 y d — 5%
But if the Divifor will not exactly divide the Co-efficients of
the Dividend, then place the Dividend and Divifor in the Man-
ner of Vulgar Fractions, as in the foregoing Articles.
The Method of dividing compound Quantities by one another,
where the Operation is continued as in common Arithmetic,
being generally perplexing to Learners, it will be explained in
the Method of folving Quadratic Equations, this Divifion not
being neceffary in the prefent Defign before we come to that
Part of the Work.
INVOLUTION.
G 2
[ 44 ]
INVOLUTION.
HIS is only raifing of Powers from any given
31. T Root, and therefore is performed by Multiplication -
:
For the Quantity which is given being multiplied by itfelf will
be the Square of that Quantity, that Product being multiplied
by the given Quantity, this Product will be the Cube of that
Quantity, and that Product multiplied again by the given
Quantity, will be the fourth Power of that Quantity; and ſo
on as in common Arithmetick.
To find the Cube of a
To find the Cube of b
Q
b
The Square of a
a a
a
The Square of b
bb.
b
The Cube of a aa a
To find the Cube of
Now 2 y multiplied by 2y the Į
Product will be by Art. 13.
And 4yy multiplied by 2 y, the
Product will be by Art. 13.
To find the Cube of
The Cube of b bb b
23
23
4yy the Square of 2 y
2 y
}
}
8yyy the Cube of 2 y
33
3%
3 Z
א א
Now 32 multiplied by 3%, the 9zz the Square of 3 ≈
Product will be by Art. 13.
}
32
z
And 9zz multiplied by 32, the 27zzz the Cube of 3 ≈
Product will be by Art. 13.
To
INVOLUTION.
45
To find the 4th Power of
Now -2x multiplied
by-2x, the Product
is by Art. 9. and 13.
And 4 xx multiplied by
2x, the Product
is by Art. 13. and 16.
And-8 xxx multiplied
by-2x, the Product
is by Art. 9. and 13.
2 x
2 X
4xx the Square of — 2 x
2x
- 8 x x x the Cube of — 2 x
2 x
16xxxx the 4th Power of -2x.
In like Manner any other fingle Quantity may be raiſed to
any required Power, and if in the given Quantity there are more
Letters than one, it is done in the fame Manner.
To find the 4th Power of 2 ab
2ab
4 aabb the Square of 2 a b
2ab
8 a a ab bb the Cube of 2 ab
2ab
16 aa aabbbb the 4th Power of 2 a b.
"",
32.
If there are two or more Quantities connected by the
Signs+or, to be raiſed to any given Power, it is ftill
performed by common Multiplication. Two Quantities when
connected by the Sign+, is commonly called a Binomial.
To raiſe the Binomial,
or a+b to the third Power or Cube.
a + b
aa+ab the Prod. of a b multip. by a, by Art. 10.
abbb the Prod. of a+b multip. by b, by Art. 10.
a a + 2 a b + bb the Sum of theſe two Products, or
a + b
the Square of a + b.
aaa+2aab+abb the Product of a a + 2ab+b b
multiplied by a, by Art. 10.
aab 2abbbbb the Product of a a +2ab+bb
multiplied by b, by Art. 10.
aaa+3aab+3abbbbb the Sum of thefe two Pro-
ducts, or the Cube of a + b.
When
46
ALGEBRA.
!
When two Quantities are connected by the Sign-, it is.
commonly called a Refidual.
To raiſe the Reſidual,
or xy to the third Power or Cube.
x-y
xxxy the Product of x
ху
xyyy the Product of
*
y multiplied by x.
-y multiplied by-y.
X X -------- 2xy+yy the Sum of theſe two Products, of
the Square of x —y.
x -y
xxx-2xxy+xyy the Product of xx-2xy+"y
multiplied by x.
-xxy+2xyyyyy the Product of xx-2xy+yy
multiplied by Y.
xxx−3xxy+3xyy-yyy the Sum of thefe two Pro-
ducts, or the Cube of x 3.
And if theſe compound Quantities have Co-efficients, the
Work ftill proceeds as at Art. 18.
To raiſe the Binomial,
or 2a+3b to the third Power.
2a+36
4aa6 a b the Product of 2 a +36 multiplied by 2 a.
6ab9bb the Product of 2 a +36 multiplied by 3 b.
4aa+12ab+9bb the Sum of thefe two Products,
or the Square of 2 a + 3 b.
2a+36
8aaa+24aab + 18 a b b the Product of 4 a a + 12 a b
12 a ab +36 a b b 27 bbb
+9bb multiplied by 2 a.
the Product of 4 a a + 12 ab
+9bb multiplied by 3 b.
8aaa+36aab +54abb +27bbb the Sum of thefe two
Products, or the Cube of 2 a +36.
To
EVOLUTION.
47
To raiſe 3 m + 2y to the third Power.
3m + 2x
To raiſe a
9mm + 6 MY
6 m ++vy
9mm + 12my +4yy the Square of 3 m + 2y
3 m + 2y
27 mmm + 36 m m y + 1 2 m ŷ y
18 m my + 24 myy + 8 yyy
27mmm + 54mmy +36 myy-+8yyy the Cube of
a
2b to the third Power.
2 b
3m+2y.
a a
·2ab
-2ab+ 4b b
2ab+4bb
4ab4bb the Square of a — 2 b
аа
a
26
a a a
4aab + 4 a b b
2 a ab + 8 a b b — 8 b b b
a a a—6aab +12 a b b −8 b b b the Cube of 4-2 b.
In this Example I have placed the fame Quantities under
each other, for the more commodious adding them, though this
is not neceffary, and is a Knowledge the Learner will acquire
from his own Obfervation.
EVOLUTION.
83· TH
33.THIS is the Extraction of Roots, and therefore op-
pofte to Involution, and as Equations in which the
unknown Quantity rifes above the Square are generally adfected,
and refolved by the Method of Converging Series, we fhall con-
fider the Square Root only; and give fuch Directions that the
Learner may generally know, whether the Square Root of fuch
Quantities as commonly occur in the Solution of Queſtions can
be extracted or not.
Now to many Times as any Letter is repeated, fo high is
the Power of that Letter faid to be. Thus, a is a to the first
Power: a a is a to the fecond Power or Square: and a a a a is a
to the fourth Power, &c. as in Involution.
2
And
48
ALGEBRA.
:
And to extract the Root of any fimple Quantity, confider how
many Times the Letter is repeated, or how high the Power
of it is, and if it appears to be the fecond, third, fourth, or
any other Power, divide that Figure which expreffes the Heighth
of the Power by 2, and if it does not divide it exactly, it is a
Surd Quantity, and has no Square Root; but if it divides it
exactly, fet down the Quantity whofe Root you are extracting
as many Times as the Quotient of the above Divifion directs,
and that will be the Square Root required.
Exam. 1. Exam. 2. Exam. 3.
To extract the Square Root of a a
The Square Root is
b b b b
b b b b b b
b b
bb b
a
Exam 1. Here a is repeated twice, or to the fecond Power;
now dividing 2 by 2 the Quotient is 1, therefore fetting down a
once, or a, it is the Square Root required.
Exam. 2. Here b is repeated four times, or to the fourth
Power; now dividing 4 by 2 the Quotient is 2, therefore ſetting
down b twice, or b b, it is the Square Root required.
Exam. 3. Here b is repeated fix times, or to the fixth Power;
now dividing 6 by 2 the Quotient is 3, therefore ſetting down b
three times, or bbb, it is the Square Root required.
The Truth of thefe Operations are proved by Multipli-
cation, for if the Work is right, the Square Root being multi-
plied by itſelf will produce the Quantity from which the Root
was extracted. Thus in Example 2,
The Square Root is
Which being multiplied by itſelf
The Product is the given Square
And fo of any other Example.
b b
b b
b b b b
Exam. 4.
Exam. 5.
d d d d d d
d d d
To extract the Square Root of a aa a
The Square Root is
a a
Exam. 4. Here a is repeated four times, or to the fourth
Power; now dividing 4 by 2 the Quotient is 2, which ſhows
that a muſt be repeated twice, that is, aa is the Square Root
required.
Exam.
EVOLUTION.
49
Exam. 5. Here d is repeated fix times, or to the fixth
Power; now dividing 6 by 2 the Quotient is 3, which fhows
that d muſt be repeated three times, and confequently ddd
is the Square Root required.
And if the Quantity, whofe Root is to be extracted, has dif-
ferent Letters, then confider if the Number of Times each
Letter is repeated can be divided by 2 without any Remainder,
and if they can, fet down each Letter fo many Times as the
Quotient of the reſpective Divifion directs, and joining them,
this will be the Square Root required; but if the Number of
Times any one Letter is repeated cannot be divided by 2, then
the whole Quantity has no Square Root.
Exam. I.
To extract the Square Root of a abb bb
The Square Root is
a.
!
abb
Exam. 2. Exam. 3.
aaaadddd mmpp
aadd
mp
Exam. 1. Here a is repeated twice, and 2 being divided by 2
the Quotient is 1, which fhows a muſt be taken only once, or
Now b is repeated four Times, or to the fourth Power, and
4 being divided by 2 the Quotient is 2, which ſhows b muſt be
repeated twice, or bb, now joining a to bb, and a bb is the
Square Root required.
Exam. 2. Here a is repeated to the fourth Power, and dividing
4 by 2 the Quotient is 2, which ſhows that a muſt be repeated
twice, that is, it must be a a: Again, dis repeated to the fourth
Power, and dividing 4 by 2, the Quotient is 2, which fhows
d muſt be repeated to the fecond Power, or dd. Now joining
a a to dd, we have a add for the Square Root required.
By the fame Method of reaſoning we fhall find in Example 3,
that the Square Root of mmpp is mp.
But when it is found that the given Quantity has not fuch a
Root as is required, then the Square Root of it is expreffed by
prefixing this Sign before it.
Exam. I. Exam. 2.
Exam. 3.
Required the Square Root of a
The Square Root is
✓ a
bb b
Vbbb
ddddd
✓ddddd
Exam. 1. Becauſe a is only repeated once, and as we cannot
divide by 2, and have the Quotient a whole Number, therefore
I conclude a is a Surd Quantity, and accordingly, to express the
Square
H
50
ALGEBRA.
!
Square Root of a, prefix the Sign to it, ſo is a the
Square Root required.
Exam. 2. Here b is repeated three times, and becauſe 3
cannot be divided by 2, and have no Remainder, therefore I
conclude bbb is a Surd Quantity, and to exprefs the Square
Root of it, prefix the Sign to it, fo is bbb the Square
Root required.
Exam. 3. Here d being repeated five times, and as we cannot
divide 5 by 2, and have no Remainder, therefore I conclude that
ddddd is a Surd Quantity, and to express the Square Root of
it, prefix the Sign✔✅to it, ſo is✔ ddddd the Square Root
required.
34. But to extract the Square Root of compound Quantities,
or thoſe connected by the Signs + or, obferve,
First, There must be three Quantities to make it a Square,
for ab multiplied by itfelf, or fquared, the Product is
a a + 2 a b + bb, by Art. 32. whence if there are only two
Quantities it is a Surd. I take no Notice of any greater
Number of Quantities than three, which may compofe a
Square, as they ſeldom occur in any Operation.
Secondly, Whether theſe three Quantities have two dif-
ferent Letters only; there may be Cafes in which there are
more than two different Letters in theſe three Quantities, but
as they feldom happen, I chooſe not to perplex the Learner
with them.
Thirdly, If two of theſe three Quantities are pure Powers
of thoſe two Letters; that is, in the Square of a + b there is
a a and b b, pure Powers of the Quantities a and b.
Fourthly, Whether both thefe pure Powers of the two dif
ferent Letters have the Sign + before them.
Fifthly, If the third of the above three Quantities is twice
the Product of the Square Root of the two pure Powers of the
two different Letters, that is, the Square of a + b being
aa+2ab+b b, the Quantity 2 a b is twice the Product of
the Square Root of a a and b b; and this Quantity may have
either the Sign + or -.
Now if the given Quantity, whofe Root is to be extracted,
anſwers theſe Particulars, its Square Root may be extracted thus.
Sixthly, Extract the Square Root of the two pure Powers of
the two different Letters, according to the Directions at Art. 33.
Seventhly, If the Quantity mentioned at the fifth Particular
has the negative Sign, connect the two Roots mentioned in the
laft Particular with the Sign, and it will be the Square Root
required.
Eighthly a
EVOLUTION.
51
Eighthly, But if the Quantity mentioned at the fifth Particu-
lar has the Sign+, then connect thoſe two Roots with the
Sign, and this will be the Square Root required.
Now let it be required to extract the Square Root of a a
+2ab+bb.
Here are three Quantities by the first Particular.
They have likewife two different Letters, viz. a and b, by
the fecond Particular.
Two of theſe Quantities, viz. aa and bb, are pure Powers
of the two Letters a and b, by the third Particular.
And both theſe pure Powers, viz. aa and bb, have the Sign
+, by the fourth Particular.
Now ſuppoſe we neglect the Conſideration of the fifth Parti-
cular, and attempt the Extraction of the Root by the fixth
Particular.
Then the Square Root of a a, is by Art. 33.
And the Square Root of bb, is by the fame
a
b
And now the third Quantity 2 a b being twice the Product
of the Roots a and b, and having the Sign,
Therefore by the eighth Particular, I connect a and b with
the Sign+, then it is
a+b
Hence I fuppofe ab to be the Square Root of aa+2ab+bb.
But to prove the Truth of the Operation, multiply the Root
by itſelf, and if the Product agrees with the given Quantity, in
its Quantities, Signs, and Co-efficients, the Work is right; if
not the Work is either erroneous, or has no Square Root, and
is a Surd Quantity.
The Root of the laft Example}
was fuppofed to be
} a+b
Which multiplied by itſelf
a + b
aa+ab
ab+bb
aa+2ab+b b
The Product is the given Quantity, which proves that a+b
is the Square Root of a a +2ab+bb.
Required the Square Root of a a + 2 za+zz.
Here are three Quantities by the first Particular.
They have likewife two different Letters, a and z, by the
Second Particular.
Two of thefe Quantities, viz. aa and zz, are pure Powers.
of a and, by the third Particular, whofe Square Roots are
a and z.
And both thefe Powers have the Sign+by the fourth
Particular.
H 2
Now
52
ALGEBRA.
1
Now the third Quantity 22 a is twice the Product of a and
2, the Square Roots of the two pure Powers a a and z z.
Then to extract the Square Root of a a +2za+zz by
the fixth Particular.
The Square Root of a a by Art. 33. is
The Square Root of zz by the fame is
Z
Becauſe the third Quantity 2 a z has the Sign+, therefore by
the eighth Particular, connect a and z with the Sign +, and
az is the Square Root required.
*
To try if the Square Root is
a+z
Multiply it by itſelf
a+z
a a + a z
az+zz
+
aa+2ax+2%
The Product a a +2az+zz, agreeing with the given Quan-
tity, in the Quantities, Signs, and Co-efficients, it appears
that az is the Square Root required.
To extract the Square root of mm
2mp + pp.
Here are three Quantities by the first Particular.
They have likewife two different Letters m and p, by the
fecond Particular.
Two of theſe three Quantities, viz. mm and pp are pure
Powers of m and p, by the third Particular.
And both thefe Powers have the Sign, by the fourth
Particular.
Likewife the third Quantity — 2 mp is twice the Product of
m and p, the Square Roots of the two pure Powers m m and p p.
Then according to the fixth Particular, the Square Root of
mm is
By the fame, the Square Root of pp is
m
p
But as the third Quantity 2 mp has the Sign, therefore by
the Seventh Particular connect m and p with the Sign—, and
mp is the Square Root required.
To try if the Square Root is
Multiply it by itself
m
m-p
P
mp
ገዜ ጎ
mp + pp
m m
2mp +pp
The Product mm - 2 mp+pp, agreeing in every thing with
the given Quantity, it proves mp is the Square Root required.
By
EVOLUTION.
53
By the fame Method of reafoning it will be found that the
Square Root of zz + 2xx + xx, is z +x.
And that the Square Root of aa-2 addd, is a-d.
And that the Square Root of xx 2xmmm, isx —m.
And if it was required to extract the Square Root of a a+ba
b b
+ in extracting the Root of the Fractional Quantity
4
extract the Root of the Numerator for a new Numerator, and
the Root of the Denominator for a new Denominator.
Here the two pure Powers are a a and
But the Square Root of a a is
And the Square Root of
bb
is
4
And connecting theſe we have
The Square Root required.
b b
4
Q
b
2
a +
2
h
To prove the Truth of this Operation, multiply a +
by itſelf,
at
a +
aa+
b
2
2
ab
2
a b b b
IN
2
+
2
4
bb
4
aa+ab+
a multiplied by a the Product is a a, and multiplied by a
b
2
a
I
is ab, (for making a an improper Fraction as in common
2
>
Arithmetic, and multiplying the two Numerators a and b for
a new Numerator, and the
new Denominator, we have
b b
duces
4
ab
2
ab
two Denominators 1 and 2 for a
a b
b
b
4) and multiplied by pro-
2
2
by the fame Rule; and in the Products the F..ions
and having the fame Denominator, adding them accord-
2
ing
54
ALGEBRA.
ing to the Rule for Addition of Vulgar Fractions in Arith
metic, the Sum is
296
""
2
but rejecting the 2 by the Rule for
Divifion of Algebra the Sum is a b.
Therefore when any one of the Quantities appears in a Frac-
tional Manner, we must extract the Square Root of both the
Numerator and Denominator, placing the Square Root of the
Numerator for a new Numerator, and the Square Root of the
Denominator for a new Denominator, and try the Work as
before.
But if we cannot extract the Square Root of both the Nume-
rator and Denominator, then we conclude the given Quantity
to be a Surd.
Now by this Reaſoning we ſhall find the Square Root of
*x + xa + 22, to be x + 21.
a a
4
a
2
And that the Square Root of mm - my+2, is m y
4
2
Suppoſe it was required to extract the Square Root of
xx+2x n − n n.
Here are three Quantities by the first Particular.
They have likewiſe two different Letters, x and n, by the
fecond Particular.
Two of theſe three Quantities, viz. xx and nn, are pure
Powers of x and n.
But both theſe Powers have not the Sign +, for it isn 11,
therefore by the fourth Particular, I conclude that the given
Quantity xx+2xnnn is a Surd Quantity, and its Square
Root cannot be extracted any otherwife than by prefixing the
Signto it, as in Art. 33. Thus, √xx + 2 x n — n n`is,
or expreffes the Square Root of xx + 2 x n − n n.
Let it be required to extract the Square Root of a a +5ab+bb.
Here are three given Quantities by the firſt Particular.
They have likewife two different Letters, a and b, by the
Second Particular.
Two of theſe Quantities, viz. aa and bb, are pure Powers
of a and b.
And both thefe Powers have, the Sign - by the fourth Par-
ticular.
But then the third Quantity 5 ab is not twice the Product of
the Square Roots of a a and bb; for their Roots being a and b,
if they are multiplied the Product is a b, and that being multi-
plied by 2 it is 2 ab: Whereas the third Quantity in the given
Example
EVOLUTION.
55
Example is 5 ab. Hence, I conclude that a a +5 ab + b b is
a Surd Quantity, and to exprefs its Square Root I prefix to it
the Sign, fo will ✓aa + 5 ab + b b be the Square Root of
aa+5ab+bb.
And if it was required to extract the Square Root of
b b
aa+2ab+ it will be found a Surd Quantity, it being
5
impoffible to extract the Square Root of 5, therefore prefix the
Sign to aa+2ab+
Square Root required.
bb
b b
>
and thena a+2ab+ is the
5
5
For the fame Reaſon the Square Root of xx+2xa+
is xx+2xa+
Square Root of 3.
a a
3
o a
3
it being impoffible to extract the
When the Radical Sign or is to be prefixt to the Whole
of any compound Quantity, draw the Top of the Sign over all
thoſe Quantities, which fhews that they are all included under
that Sign; for if the Sign was not to be drawn over all of them,
it may be thought the Square Root of that Quantity was
only to be extracted which ſtands next the Radical Sign.
To extract the Square Root of aaaa+2aab + bb.
Here are three given Quantities by the first Particular.
They have likewiſe two different Letters, a and b, by the
fecond Particular.
Two of thefe Quantities, viz. aa aa and bb, are pure Powers
of a and b, by the third Particular.
And both thefe Powers have the Sign +, by the fourth
Particular.
And the third Quantity 2 a ab is twice the Product of the
Square Roots of aa aa and b b.
For by the fixth Particular, the Square Root of a aa a is a a
And by the fame, the Square Root of bb is
b
And as the third Term 2 a ab in the given Quantity has the
Sign, by the eighth Particular connect a a and b, the two
Roots of aa aa and bb, with the Sign+, fo is a ab the
Square Root of a aaa+za ab + b ba
2
!
}
To
56
ALGEBRA.
To prove which put down the
fuppofed Square Root
Which multiplied by itfelf
}
aa + b
aa + b
a a a a + a a b
a ab - bb
a a a a + 2 a a b + b b
The Product aa aa +2 a ab+bb, agreeing with the given
Quantity in every Particular, proves the Square Root to be as
above.
To extract the Square Root of yyyy-2yyxxx.
Here the given Quantities agree with the first five Parti-
culars as before.
X
By the fixth Particular I find the Square Root of yyyy is yy
By the fame, that the Square Root of x x is
But as the third Term 2yyx in the given Quantity has the
Sign, therefore by the feventh Particular I connect yy and
the two Roots with the Sign-, and fay, or fuppofe y y to
be the Square Root required.
To prove which put down?
the fuppofed Root
Which multiply by itſelf
The Product, agreeing with}
the given Quantity
yyx
yy-x
уух
y y y y
•yyx + xx
—— x
yyyy — zyy x + xx
By the fame Method we fhall find the Square Root of
n n n n + 2 n nd+dd to be nn+d.
And that the Square Root of xxxx + 2 xxyy +yyyy is
xx + yy.
And we fhall find that dddd+ 3 ddy+yy is a Surd Quantity,
and its Square Root must be expreffed by prefixing the Radical
Sign to it, thus dddd+3d dy yÿ.
We fhall likewife find that p p p p + 2 p pytyy is a Surd
Quantity, and to extract its Square Root, is only to prefix to it
the Radical Sign, thus ✔PPP p + 2 p p y + xy.
x
Of
[57]
OF SURD QUANTITIES.
TH
HESE are fuch Quantities whofe Roots cannot be
exactly extracted, and as they arife in the Reſolution of
Algebraic Queſtions, we fhall explain fo much of them only,
as is neceffary to the prefent Defign.
Addition of Surd Quantities, in which there are
two Cafes.
35. Cafe 1. When the Quantities under the Radical Signs are
alike, add the rational Quantities, or thoſe which are without
the Radical Sign together, by the Rules of Addition at Art.
1, 2, 3, 4, 5, 6, and to this join the Surd Quantities, and
this will be the Sum required.
And if there appears to be no rational Quantities without the
Radical Sign, then Unity, or 1, is always fuppofed to be the
rational Quantity.
Exam. 4.
Exam. I.
Exam. 2.
To vam
2 √ dy
Exam. 3.
6 m√d + a
5 y ✓ dm + x
Add Jam
4 m / d + a
y / dm + z
10m√d + a
Sum 2 am
✔dy
3√ dy
6 y ✓ dm + z
Exam. 1. There being no rational Quantities, therefore
Unity, or 1, is the rational Quantity to each. Now I added
to i makes 2, to which joining the Surd✔✔am, we have
2 am, the Sum required.
Exam. 2. The rational Quantities being 2 and 1, their Sum
is 3, to which joining✓dy we have 3 √ dy, the Sum required.
Exam. 3. The rational Quantities are 6 m and 4 m, which
being added make 10 m, to which joining the Sud ✓d + a
we have 10 mdm + a, the Sum required.
The
+a,
Exam. 4. The rational Quantities are 5 y
and J, which
being added make 6 y, to which joining the Surd ✔d m + z
we have 6 y✔✅dm+z, the Sum required.
Exam. 5.
To 13yd ✓2 -
x
Exam. 6.
15% √ da + p
Exam. 7.
-7m√ da
Add 5yd/z-x
- 3 z₁/da+p
2 m/ da
12 z √ da + p
Sum 18 ydzo mi to
-y
9m/da-y
Exam. 5. The rational Quantities 13yd and 5 yd being
added make 18 y d, to which joining the Surd Quantity √ z — x
we have 18 y dx, the Sum required.
I
Exam,
58
ALGEBRA.
Exam. 6. The rational Quantities 15 % and
3% being
added, their Sum by Art. 3. is 12z, to which joining the Surd
✔da+p we have 12 z✓da+p, the Sum required.
Exam. 7. The rational Quantities - 7 m and — 2 m being
added make—9 m, to which joining day, we have
To
-9 m✓ da—y, the Sum required.
23√ma+m
Add —31 √/ma+m
15m√ da—z. p
7 m√ da—zp
zp
16 dp√ 14+ p
-12dp√ 14+P
Sum 5% √ma+m
To 49√zd — za
Add 3y v zd— za
Sum jy √ zd Za
8m√ da-zp
51 √mp + x
4y ✓ mp + x
y ✓ mp + x
4 dp✓ 14+e
7 Z
z✓ma-
8zma-
Z ✓ ma
36. Cafe 2. When the Letters under the Radical Sign are
different, then place them down one after the other with the
fame Signs they have in the Queftion, in the Manner as at
Art. 6. and this will be the Sum required.
Exam. I.
Exam. 2.
Exam. 3.
To √ a
ν
√b+m
mda+y
Add / b
Suma+√
√d+y
m/z
√ b+m: + √d ty
√d+y
m√da+y: +m✓✓ Z
vz
Exam. 1. The Letters under the radical Signs being different
put downa, then becaufeb has the Sign+, therefore after
a put, after which put b, and a+b, is the
Sum required.
· Exam. 2. The Letters under the radical Signs being different
put down/b+m: after which place two Dots to fhow that
Surd goes no farther, then becauſe dy has the Sign +,
therefore after the Quantity b+m: put +, and after that
the Surd/d+y, and we have √b+m: + √d+y, the
Sum required.
Exam. 3. The Letters under the radical Signs being different
put down mda+y: and becauſe the Quantity m✅✔✅z has
the Sign+, therefore after m↓ da+y: put the Sign+, after
which put the Quantity mz, and we have mda+y:
+m√, the Sum required.
Exam.
Of S U R D QUANTITIE S.
59
To
Exam. 4.
✔da
Add -zm
Sum yda- z ✓ m
Exam. 6.
Exam. 5.
5 √ damy
2m ✓ zm
— 5 √ da—y: — 2 m
-
2 m ✓ z m
2 m √ bz + n
39 √ dz.
b
2m ✓ bx+n: +39 ✓ dz
b
Exam. 4. The Letters under the radical Signs being different
da, and becauſe zm has the Sign
put down y
there-
fore after y✔da put the Sign, and after that the Quantity
Z m, and y✔ da z✔m is the Sum required.
Exam. 5. The Letters under the radical Signs being different
put down
Sign, therefore after-5da-y: put the Sign—, and
after that the Quantity 2 m ↓ xm, and
and — 5 √ da
5√day:
5 √ da y and becauſe 2 m√√ zm has the
2 m√ zm is the Sum required.
✓
Exam. 6. Becauſe the Letters under the radical Signs are
different I put down - 2 m bz+n, but 3 y√dz — b
having the Sign+, therefore after 2 mbz+n: put the
Sign, and after that the Quantity 3y dz-b: and
2 m √ bz+n: + 3y √ d z − b is the Sum required.
✔
To -5
Add
7 ✓ m
√
da
Sum-5✓da +1 √m
To −33√p+r
Add m/d
m✓ b ma
3√3p+9
m✓ bma + 3 √ y p + q
14 m✓ da +p%
7 √x+p
Sum-3p+r:+m√d 14m √da+pz: +7√x+p
To -51✓dp-7
Add 79 √zm+a
Sum-5✓ dp-x: +7x
x: +71 / 2 m + a
I 2
Subftraction
7
60
ALGEBRA.
:
Subftraction of Surd Quantities, in which there are
two Cafes.
37. Cafe 1. When the Letters under the Radical Signs are
alike, fubftract the rational Quantities from the rational Quan-
tities by Art. 7. and to the Difference join the Surd Quantities,
which will be the Remainder required.
Exam. I.
Exam. 2.
Exam. 7.
Exam. 4.
From
5√da 5m/mz 14y/d+z 21pm/db-r
Subſtract 3 da 2m/mz_31√d+z 19pm、/db-r
Remains 2da 3m/mz 11√d+z
2pm ✓ db r
Exam. 1. The rational Quantities are 5 and 3, fubftracting
3 from 5 there remains 2, to which joining the Surdd a we
have 2 da, the Remainder required.
Exam. 2. The rational Quantities are 5 m and 2 m, fubftract-
ing 2 m from 5 m there remains 3 m, to which joining the Surd
m z we have 3 mm z, the Remainder required.
Exam. 3. The rational Quantities are 14 y and 3y, ſubſtract-
ing 3y from 14y there remains 11y, to which joining the Surd
✔d+z we have 11yd+z, the Remainder required.
Exam. 4. The rational Quantities are 21 pm, and 19 pm,
ſubſtracting 19 pm from 21 pm, there remains 2 pm, to which
joining the Surddbr we have 2 pm db-r, the
Remainder required.
✓
Exam. 7.
From
Exam. 5.
17 db a
Exam. 6.
~51 √d + a
Subftract-4 dba
3y / d + a
5 m√d + ab
6m/d+ab
Remains 21 d✓ ba
— 8 y√d + a
m ✓ d + a b
4 d:
Exam. 5. The rational Quantities are 17 d and
Now to fubftract-4 d from 17 d, by the Rule for Subftraction
at Art. 7. change the Sign of -4 d, or fuppofe it to be changed,
then-4 d becomes + 4 d or 4 d; then by Art. 7. if we add
17 d to 4 d it is 21 d, which is the Remainder that arifes by
fubftracting 4d from 17d; now to this 21d join the Surd
ba, and 21dba is the Remainder required.
Exam. 6. To fubftract the rational Quantity 3y from
5y, we muſt by Art. 7. change or fuppole the Sign of 3y
ta
Of SURD QUANTITIES.
61
to be changed, which will make it 3y: then by the fame
Art. 3y added to
5y, it is
8y, which is the Remainder
that arifes from fubftracting the rational Quantities, therefore
to this 8 y join the Surd Quantity/d+a, and-8yd + a
is the Remainder required.
Exam. 7. Here the rational Quantities are · 5 m and — 6 m,
and by the Rule for Subftraction Art. 7. if we fuppofe the Sign
of 6 m to be changed, it becomes + 6 m or 6 m, and then
adding 5 m to 6 m it is m, the Remainder arifing from fub-
ftracting the rational Quantities; and if to this m we join the
Surd d+ab we have m✓d+ab, the Remainder required.
Exam. 8.
21m √d + a
Exam. 9.
− 9 d√ m n + p.
Exam. 10.
12yd-an
31₁/d-an
From
Subſtract 9m / d + a
Remains 12 m✓d + a
-2 d₁/mn + p
— 7 d√ m n +P
151✓d—an
Exam. 11.
Exam. 12.
From
4 a √ m
14p √ dy
Subſtract
2 am — p
-3px/d — v
Remains
6 a√m
-P
17 p✓d-y
Exam. 14.
From Ja vap
Subſtract 2 a/ap
21 √ ap
a x
9/ap
ap — ax
Remains 5 a✓ ap
12 √ ap
a x
21 √ da-
Exam. 15.
Exam. II.
-5a√x+y
зал
3a./x+7
8a / x + 3
Exam. 16.
14 √ da
Z
W/da- Z
Z
The Truth of thefe Operations are proved as in Subtraction
of common Numbers. Thus at Example 1. the Remainder is
2da, and the Quantity fubftracted was 3 da, now if
we add thefe together by Art. 35. the Sum is 5 da, which
being the fame Quantity from which 3d a was ſubſtracted,
it proves the Work to be true.
Again, at Example 6. the Remainder is
83d+a: the
Quantity ſubſtracted was 3ydu: Now by Art. 35. if
to-8yd+awe add 3d+a, the Sum is-5ya+a,
which being the Quantity from which 31 √ +† a was ſub-
stracted, it proves the Work to be true.
3
38. Cafe
02
ALGEBRA.
!
38. Cafe 3. When the Letters under the radical Signs are
different, fet them down one after the other, as at Art. 36.
but in fetting them down take Care to change the Signs of thofe
Quantities that are to be ſubſtracted, by Art. 7. and this will
be the Remainder required.
Exam. 1.
From 2 da
Subſtract 3✓m
Exam. 2.
Exam. 3.
2m✓ dp
51 √ a
3 ✓ z
- 3 √b
Remains 2da 3 √m 2m ✓ dp y√ z
—
51√a+3√b
Exam. 1. The Letters under the radical Signs being different
place down 2✓da, and becauſe 3m the Quantity to be
fubftracted has the Sign+, therefore after 2 da place
the Sign and after that the Quantity 3 m, and
2✓ da―3√m is the Remainder required.
Exam. 2. Becauſe the Letters under the radical Signs are
different put down 2 m dp, and becauſe y✓✓z the Quantity
to be fubftracted has the Sign +, therefore after 2 m dp
put the Sign, and after that yz, and 2 mdp — y √ z
is the Remainder required.
Exam. 3. Becauſe the Letters under the radical Signs are
different put down 5 ya, but as 3b the Quantity to
be fubftracted has the Sign-, therefore after 5ya put the
Sign+, and after that 3b, and 59 √ a + 3b is the
Remainder required.
Exam. 5.
Exam. 4.
From mda+p
Subſtract 2 a
-53√a
d√ b
Remains mda +p: -2 √ a
− 5 y√a + d v b
Exam. 6.
From
5m √ a
Subftract
Remains 5 ma + z √ p + q
Exam. 4. Becauſe the Letters under the radical Signs are
different put down mda+p, but as 2✓✓a the Quantity to
be fubftracted has the Sign +, therefore after m/ da +p put
the Sign, and after that 2a, and m✓da+p: −2 √ a
is the Remainder required.
Exam.
Of SURD QUANTITIES.
63
Exam. 5. Becaufe the Letters under the radical Signs are
different put down-5ya, but as-db the Quantity to
be ſubſtracted has the Sign therefore after 5 y√ a put the
Sign, and after that db, and 5ya+db is the
Remainder required.
Exam. 6. Becauſe the Letters under the radical Signs are
different put down 5 ma, but asxp+q has the Sign
before it, therefore after 5 ma put the Sign, and after
that ✔✅p+9, and 5 m✓a+z✔p + q is the Remainder
required.
From 5 √ atp
Subſtract my
Remains 5 ✔a+p: - m√y
From 3 u√d + p
Subftract 2 n % -y
Remains 3u✓d+p: -2n√√x-y
From
-5√P+%
Subſtract 3n √ m
Remains
ท
m√ p
yda-p
m√p+ y√da-p
-5n√da
3√m
-5n√da+3√m
14 da
5+z: -3n√/m 14√da: -7 P+?
The Truth of thefe Operations are proved in the fame Man-
ner as in the laft Article, by adding the Remainder to the
Quantity that was fubftracted; and if their Sum makes the
Quantity from which the other was taken, the Work is true,
if not, there is a Miſtake.
Thus at Example 1. the Remainder is
To which if we add the Quantity }
The Sum is 2da, the fame in
the given Example. For in this
2✓ dan 3 √ 12
3√m
2 da
Addition, adding + 3m to -3m, the Co-efficients and
Quantities being the fame and the Signs contrary, they deſtroy
one another, or the Sum is nothing, by Art. 5.
Again at Example 5. the Remainder is
To which if we add the Quantity }
fubftracted
The Sum is 57a, the fame
as in the given Example. For here
53 √ a + d√b
d ✓ b
-5√a
db being added to d ✅✔✅ b or + d ✓b, they deſtroy one
another
64
ALGEBRA.
another as in the laſt Inſtance. In like Manner the Reader
may prove any of the other Examples.
Multiplication of Surd Quantities, in which there
are two Cafes.
39. Cafe 1. When there are no rational Quantities but Unity
joined to the Surd Quantities, then multiply the Surd Quantities,
as in Multiplication of rational Quantities, but to their Product
prefix the radical Sign.
Exam.4.
Exam. I.
Exam. 2.
Exam. 3.
Multiply a
By ✓ m
✓ m n
✓ d
Lo
√py
✓ xx
Va
Product
am
✓ m n d
√py z
Exam. 1. Multiplying a by m, the Product is a m, to which
prefixing the Sign, we have✅✔✅ am the Product required.
Exam. 2. Multiplying m n by d, the Product is mnd, to
which prefixing the Sign, we have mn d, the Product
required.
Exam. 3. Multiplying py by z, the Product is pyz, to
which prefixing the Sign, we have ✔py z, the Product
required.
Exam. 4. Multiplying zx by a, the Product is zx a, to
which prefixing the Sign, we havez xa, the Product
required.
Exam. 5.
Multiply
pa
By
Z
Product ✓paz
Exam. 7.
Multiply
a + b
By
}
Exam. 6.
/x
Vay
x
Exam. 8.
M N - Z
✓ m n
Va
Product ✓ay + y b
✓ mna
A Z
Exam. 7. Multiplying a + b by y, the Product is a y+yb,
by Art. 10. to which prefixing the Sign, and drawing it
over all the Quantities, we haveay+yb, the Product
Jequired.
Exam.
Of SURD QUANTITIES.
65
Exam. 8. Multiplying_m n z by a the Product is mna
az, by Art. 10 and 16. to which prefixing the radical Sign
as in the laſt Example, we have✔mna➡az the Product
required.
Exam. 9.
Èxam. 10.
Exam. II.
Multiply ✔ap + z
Vax-
ap
√d-y
By
✔m
Product ✓ap y + y z
Vazm
ap m
✓ dp - py
Exam. 9. Multiplying ap +z by y, the Product is a py+yz,
to which prefixing the radical Sign, we have yapy + y z the
Product required.
Exam. 10. Multiplying azap by m, the Produ&t is a z m
a pm, to which prefixing the radical Sign, we have
azm-apm the Product required.
Exam. 11. Multiplying d-y by p, the Product is dp-py,
to which prefixing the radical Sign, we have
Product required.
dp-py the
√ap-z
✓ d
Multiply
ab
By Va -P
✓zy
i/d+ y
Product ✔a ab — a b p
✓ zy d + zyy
✓ apd — dz
Multiply
By
ap + z
↓ m
day
Nd-z
√ m
Ja-py
apm + zm
√ ayd—ayz
V ma mpy
Product
40. Cafe 2. When there are rational Quantities joined to the
Surds, then multiply the rational Quantities together as in Mul-
tiplication of rational Quantities, after which multiply the Surd
Quantities together by the laft Article, and joining thefe two
Products, this will be the Product required.
If there are no rational Quantities prefixt, then Unity, or 1,
is always fuppofed to be the rational Quantity.
Multiply am
Exam. I.
Exam. 2.
Exam. 3.
Exam. 4.
By
dy
ap✓ %0
2 √ a
ap
3 √ m n
р
✓ mp
3 m / d
*
Product admy
Zap ✓ za
a
33 m np ✓ mp d
K
Exam.
66
ALGEBRA.
Exam. 1. Multiplying the rational Quantities a and d, the
Product is ad, and multiplying the Surds ✔✅m by✔y, the
Product is my by Art. 39. joining this to the rational
Quantity a d, we have admy, the Product required.
Exam. 2. Multiplying the rational Quantities a p by 2, the
Product is 2 ap, and multiplying the Surds z by a, the
Product is za by Art. 39. joining theſe, and 2 ap √ za is
the Product required.
•
Exam. 3. Multiplying the rational Quantities 3 and a, the
Product is 3 a, and multiplying the Surds mn by p, the
Product is mn p, by Art. 39. joining theſe we have
3amnp, the Product required.
Exam. 4. Multiplying the rational Quantities y and I, (for I
is the rational Quantity of mp, there being no rational Quan-
tity prefixt) the Product is y, and multiplying the Surds ✓mp
byd, the Product is mp d by Art. 39. and joining theſe
we have y✔mpd, the Product required.
Exam. 5.
Exam. 6.
Exam. 7.
Exam. 8.
Multiply am✓ P
I √ p q
m√2p
2a√3%
By z ✓ d
d ✓ z
43√y
3d√48
bad ✓ 12 zy
Product amz✓pd y d ✓ pqz 4 my 2py
Exam. 5. The Product of the rational Quantities is a mz,
and the Product of the Surds is pd, theſe being joined we have
amz√ pd, the Product required.
Exam. 6. The Product of the rational Quantities is y d, and
the Product of the Surds is pqz, thefe being joined we
have y d√p qz, the Product required.
Exam. 7. The Product of the rational Quantities is 4 m y,
and the Product of the Surds is 2 py, theſe being joined we
have 4 my 2py, the Product required.
Exam. 8. The Product of the rational Quantities is 6 a d,
and the Product of the Surds is 12 zy, thefe being joined we
have 6 ad 12 zy, the Product required.
✓
Exam. 9. Exam. 10.
By
Multiply y✔p
av
Product ya✓ PL
Exam. II.
Exam. 12.
✓ m n
a√y
2√dx
3√2%
5 √73
a mny
✓
2√ dxz
2√d.
15 √ 1432
Exam.
Of SURD QUANTITIES.
67
1
Multiply maty
By
Exam. 13.
Exam. 14.
Exam. 15.
d✓ m
p Z
ap
y √ d
a√ ap + z
x/y
ax√apy +yz
Product maap +py dy✓md—pzd
Exam. 13. Multiplying the rational Quantities m and a, the
Product is ma, and multiplying a +y by p, the Product is a p
+py, but prefixing to this the Sign, becauſe they are
Surds, we have ap+py for the Product of the Surds, which
joining to ma the Product of the rational Quantities, we have
ma√ ap +py, the Product required.
Exam. 14. Multiplying the rational Quantities d and y, the
Product is dy, and multiplying the Surdsm-p z by ✓ d,
the Product is/md-pzd, which being joined to dy, the
Product of the rational Quantities, we have dy ymd-pzd,
the Product required.
Exam. 15. The Product of the rational Quantities is a x, and
the Product of the Surds is ✓apy+yz, thefe being joined
we have a xapy+yz, the Product required.
Exam. 16.
Multiply ampy + d
By y/2
Product a my p y z + z d
Multiply m✔p d
By
Exam. 18.
ad-a
Product ma✓ p d d - pd a
į
Exam. 17.
2✓ am
avp
y
2a√amp-py
Exam. 16. The rational Quantities am and y being multi-
plied, the Product is a my, and py+d being multiplied by z,
the Product is pyz+zd; but before it prefix the radical Sign,
becauſe theſe Quantities are Surds, then it is p y z + zd,
joining this to the Product a my, we have a my ✅py z + zd,
the Product required.
K 2
Exam.
68
ALGE BRA.
Exam. 17. The Product of the rational Quantities 2 and a,
is 2 a, and the Product of the Surds is
amp-py: joining
theſe we have 2 a √ amp-py, the Product required.
Exam. 18. The Product of the rational Quantities m and a,
is ma, and pd multiplied into da, is pdd-pda, to which
prefix the radical Sign, becauſe theſe are Surds, and this becomes
√pdd-pda, now joining it to ma, we have ma√ pdḍ—pda,
the Product required.
Multiply a✔p—y
By 2 b✓ m
Product 2 ay bpm-my
Multiply 2a√ 33 +2
By
b d
Product 2 ab 3 y d + d z
3√m n
25a
√n + b
a√2
6√5ma-5na a √ zn+zb
3ma
зум Vpa A -- A Z
Multiply 5✔y
X
By
39√26
a
Product 15a2by2bx
Divifion of Surd Quantities, in which there are
two Cafes.
41. Cafe 1. When there are no rational Quantities joined
with the Surd Quantities, reject all thofe Quantities in the
Dividend and Divifor that are alike, as at Art. 20. and ſet
down the Remainder, to which prefix the radical Sign, and
this will be the Quotient fought.
Exam. I.
Exam. 2.
Exam. 3.
Exam. 4.
By
Divide ✓ m n
Im
✓ ma
abd
abd
Ja
ab
Va
Quotient
n
V m
V d
✓bd
Exam. 1. Becauſe m is in both Dividend and Divifor,
reject it, and place down n the remaining Part of the Dividend,
to which prefixing the radical Sign, and
required.
n is the Quotient
Exam.
Of SURD QUANTITIES.
69
Exam. 2. Becauſe a is in both Dividend and Divifor, reject
it, and place down m the remaining Part of the Dividend,
to which prefixing the radical Sign, we have ✅m, the Quotient
required.
Exam. 3. Becauſe a b is in both Dividend and Divifor, reject
it, and place down d the remaining Part of the Dividend, to
which prefixing the radical Sign, we have √d, the Quotient
required.
Exam. 4. Becauſe a is in both Dividend and Divifor, reject
it, and place down b d the remaining Part of the Dividend, to
which prefixing the radical Sign, we have ✔ bd, the Quotient
required.
Exam. 5.
Exam, 6.
Exam. 7.
Exam. 8.
Divide
may
√ b z d
√ bzd
Vypa
By
لو که
V b d
√ zd
√yp
md
√2
No
Na
Quotient
Exam. 5. Becauſe y is in both Dividend and Divifor, reject
it, and place down md with the Sign✔ before it, and md
is the Quotient required,
Exam. 6. Becauſe b d is in both Dividend and Divifor, reject
it, and place down z with the Sign✔ before it, and is
the Quotient required.
Exam. 7. Becaufe zd is in both Dividend and Divifor, reject
it, and place down b with the Sign ✓ before it, and we have
b, the Quotient required.
Exam. 8. Becauſe yp is in both Dividend and Diviſor, rejec
it, and place down a with the Sign before it, and a is
the Quotient required.
Exam. 9.
Exam. 10.
Exam. II.
Exam. 12.
Divide max
By
✓
x
√ndy
√ny
Jayz
abd
Va
Quotient ma
✓ y z
Vad
V b
Exam. 13.
Exam. 14.
Divide
am+a p
By
a
✓ p y − p n
P
Quotient m + p
M
Exam. 15.
b d — b 112
√bd
✓ b
d
712
Exam.
70
ALGEBRA.
Exam. 13. If we divide am+ap by a, the Quotient is
m+p by Art. 22. but becauſe they are Surds, prefix the Sign
√ to m+p, and m+p is the Quotient required.
Exam. 14. Dividing pypn by p, the Quotient is yx,
by Art. 22 and 24. to which prefixing the Sign, we have
√yn, the Quotient required.
Exam. 15. Dividing b dbm by b, the Quotient is dm,
by Art. 22 and 24. to which prefixing the Sign, we have
d-m, the Quotient required.
Divide
✓
bn + ba
mx
md
nz
np
By
Im
√ n
Quotient n + a
✓
x-d.
Divide x —
ལ
ZY
Vad tay
✓ b d — b m
By
Va
Quotient ✓
y
√d +x
vd-
m
The Truth of thefe Operations are proved by multiplying the
Quotient by the Divifor, for if that produces the Dividend, the
Work is true, otherwiſe it is erroneous. Thus in Example 2,
Page 68. the Divifor is a, and the Quotient ism, which
being multiplied by Art. 39. the Product is ma, the given
Dividend.
And at Example 6, Page 69. the Divifor is bd, and the
Quotient is, which being multiplied by Art. 39. the Product
is bzd, the given Dividend.
And at Example 13, the Divifor isa, and the Quotient is
/mtp, which being multiplied by Art. 39. the Product is
am+ap, the given Dividend; in the fame Manner may
any of the other Examples be proved.
42. Cafe 2. When there are rational Quantities joined with
the Surds, divide the rational Quantities by the rational Quan-
tities, by the Rules in Divifion of rational Quantities; and to
their Quotient, join the Quotient of the Surds found by the laſt
Article, which will be the Quotient required.
Divide ay mn
Exam. I.
✓
By
/m
Quotient yn
I
Exam. 2.
Exam. 3.
y d Vaz
“
Z
¿ m + y z
+yz
mz
Exam. 4.
ma ayn
avay
m ✓ n
Exam,
Of SURD QUANTITIES.
71
Exam. 1. Dividing the rational Quantities ay by a, the
Quotient is y by Art. 20. and dividing mn bym, the
✔mn
Quotient is n by Art. 41. now joining y to√n, we have
✔n, the Quotient required.
Exam. 2. Dividing the rational Quantities, bm by m, the
Quotient is b by Art. 20. and dividing y z by ✓z, the
Quotient is by Art 41. now joining bandy, we have
by, the Quotient required.
Exam. 3. Dividing the rational Quantities yd by 7, the
Quotient is d by Art. 20. and dividing a z by ✔a, the
Quotient is ✔z by Art. 41. now joining d and ✔✅z, we
have dz, the Quotient required.
Exam. 4. Dividing the rational Quantities ma by a, the
Quotient is m by Art. 20. and dividinga y n by
Quotient is n by Art. 41. now joining m and
have mn, the Quotient required.
ay, the
n, we
Exam. 5.
Exam. 6.
Divide ayn
By
aym
n√xy
Quotient n√n
m√ a
Exam. 5. Dividing the
Exam. 7.
Exam. 8.
mn xay xav
✓ m n
nd
dzan
an
z/an
d √ p
rational Quantities ay n by a y, the
Quotient is n by Art. 20. and dividing ✔✅mn by ✓m, the
Quotient is n by Art. 41. now joining n and, we have
nn, the Quotient required.
Exam. 6. Dividing the rational Quantities mn by n, the
Quotient is m by Art. 20. and dividing
We
xay by ✔y, the
Quotient is a by Art. 41. now joining m anda,
have ma, the Quotient required.
Exam. 7. Dividing the rational Quantities x a by a, the
Quotient is, and dividing nd byn, the Quotient is
✔d by Art. 41. now joining and d, we have,
the Quotient required.
Exam. 8. Dividing the rational Quantities dz by z, the
Quotient is d, and dividing ✓ anp by on, the Quotient is
✓ by Art. 41. now joining d´and
p
the Quotient required.
Divide 4 m n ✓ ab
my ✓ az
p, we have d✔ po
✔p, p,
Exam. 9.
Exam. 10.
Exam. II.
Exam. 12.
dnx y
8an ra
2 ma
J Z
72
Quotient 2 no
Q
a
44√r
2 nd
By
Exam.
72
ALGEBRA.
!
Exam. 13.
Exam. 14.
Divide
mx √ p q
4an
rd
By
x√ p
a ✓ d
Exam. 15.
xz✓ my p
z my
Exam. 16.
rm✓ dz
r✓ d
Quotient m✔✔q
4 n√ r
X √ P
m ✓ Z
Exam. 17.
Exam. 18.
Exam. 19.
Divide
By
m✓ a
mn√ ap + ax y p√zd+zm day ✓ym + yr
day
Quotient n✔P + x
y √ d + m
y ✓ m + r
Exam. 17. Dividing the rational Quantities mn by m, the
Quotient is n by Art. 20. and dividing ✔ap + ax by ✅a,
we have p+x by Art. 41. and joining ʼn and √p + x,
we have np+x, the Quotient required.
Exam. 18. Dividing the rational Quantities y p by p, the
Quotient is y by Art. 20. and dividing ✓zd + z m by √ 2,
the Quotient is d+m by Art. 41. joining y and ✔d+m,
we have y✓d+m, the Quotient required.
Exam. 19. Dividing the rational Quantities day by da,
the Quotient is y by Art. 20. and dividing✔y m + y r by
✔y, the Quotient is mr by Art. 41. joining y and
√m+r, we have y✓m+r, the Quotient required. The
following Examples are done in the fame Manner.
་
Divide 4 an✓dy+dn an √ pz-pb 6bdy ✓pm + pd
By
2a/d
Quotient 2ny + n
a
3bd ✓p
Divide
pn/dxdb
Ey n d
Quotient p√x
23m+d
12bapy—px
3a/p
4b√y-x
anx ✓ pd - pm
ax/p
n ✓d — 1
The Truth of thefe Operations are proved likewife from
multiplying the Quotient by the Divifor, and if that Product
makes the Dividend, the Work is true, if not, there is a
Miftake. Thus in
Exam.
7
73
Of SURD QUANTITIE S.
Exam. 1. The Quotient is yn, and the Divifor is a √ m;
now multiplying y✔✔n by am, by Art. 40. first multiply
the rational Quantities y and a, this Product is a y, and multi-
plying ✔n by ✔m, this Product is √ mn, and joining this to
ay we have ay✔mn the Product, which being the fame as the
given Dividend, proves the Work to be true.
And at Exam. 5. the Divifor is aym, the Quotient is
n✔n, now multiplying a y✔✅m by nn, according to Art.
40. we firſt multiply the rational Quantities ay by n, and this
Product is ayn; then multiplying ✔m byn, this Product
is √mn, and joining this to a yn, the Product is ayn√↓ mn,
which being the fame with the given Dividend, the Work
is true.
And at Exam. 17. the Divifor is ma, and the Quotient is
n√p+x, and multiplying theſe by Art. 40. we firft multiply
the rational Quantities m and n together, and this Product is m n,
then multiplyinga by✓p+x, this Product is ap +ax,
which being joined to m n, the Product is m n √ ap + ax, the
fame as the given Dividend; and fo may any of the other
Examples be proved.
Involution of Surd Quantities.
43. Cafe 1. When there are no rational Quantities joined
with the Surds, it is only fetting the Quantities down without
their radical Sign, which raiſes the given Root as high as is the
Index of the radical Sign.
-
Exam. 1. Exam. 2. Exam. 3. Exam. 4.
Raife to the Square✓ a
or fecond Power
The Square
√m n
√na
a
m n
na
b
This being no more, according to the Rule, but to ſet down
the Quantities without their radical Sign, it is fo eaſy as not to
want any farther Explanation.
The Reaſon on which the Operation is founded, is, that any
Quantity or Number being multiplied by itſelf, will produce the
Square of that Quantity or Number, thus 2 x 24, whence
4 is the Square of 2, and a xaaa, which is the Square of a,
and fo of any other Quantity. Now fuppofing the Square Root
of a was to be extracted, which by Art. 33. isa. But
L
23
74
ALGEBRA.
as ✔a is the Root, and a was the Square from which that Root
was extracted, hence✔a multiplied intoa, muſt produce a,
by what has been juft faid: Now a multiplied bya, is
✔aa by Art. 39. and asa a fignifies the Square Root of a a,
which is a by Art. 33. it follows, that to involve any Surd that
has no rational Quantities joined with it, is only to fet down
the Quantities without their radical Sign.
To find the Square or
fecond Power of
The Square
}√ax
√nd
V pr
ax
nd
pr
א
z
א
And if there are feveral Quantities connected by the Signs +
or, and are all under the radical Sign, they are involved in
the fame Manner.
Raife to the fecond}
Power or Square
The Square
Raife to the fecond
vato
a+b
✔an
·d
a+b
an-d
√p+ny
p+ ny
}
The Square
Raife to the 2d
}
Power or Square ✔pd—n √dz+zy √pm—nd
pd-n dz+zy pm-nd
Power or Square✔a+y—d ✔am—n+db √pz +z x—x d
The Square
aty-d am-n+db pz+zx-xd
44. Cafe 2. When there are rational Quantities joined with
the Surds, then involve the rational Quantities as high as is the
Index of the Surd, and multiply thefe involved Quantities into
the Surd Quantities, after the radical Sign is taken away.
Raife to the Square
Exam. I.
am
Exam. 2.
Exam. 3. Exam. 4.
b √ n z
The Square
aam
b b n z
d √ y
ddy
zz✓b
Zzzz b
Ex. 1. The rational Quantity a ſquared is by Art. 31.
The Surd Quantity/m being put down
without the radical Sign is
n}
a a
m
Thefe being multiplied, the Product is the Square required aam
1
Exam.
Of SURD QUANTITIES.
75€
b b
nz
Ex. 2. The rational Quantity b ſquared is by Art, 31.
The Surd Quantity/nz without the radical Sign is
Theſe multiplied, the Product is the Square required bbnz
Exam. 3. The rational Quantity d ſquared is
The Surd Quantity ✔y without the radical Sign is
Theſe multiplied, the Product is the Square required
Exam. 4. The rational Quantity zz fquared is
The Surd Quantity without the radical Sign is
Theſe multiplied, the Product is the Square required
dd
Y
ddy
ZZZZ
b
zzzzb
Exam. 5.
Exam. 6.
Exam. 7.
Exam. 8.
Raife to the Square an✔p
dz ✓ y x
p√xy
aannp
d d z z y x
pp xy
da/%
dda az
The Square
Exam. 5. The rational Quantity an fquared is
The Surd Quantity ✔p without the radical Sign is
Theſe multiplied, the Product is the Square required
aan n
P
aannp
d d z z
yx
Exam. 6. The rational Quantity dz fquared is
The Surd Quantity✔y x without the radical Sign is
Theſe multiplied, the Product is the Square required d d z zyx
Exam. 7. The rational Quantity p fquared is
The Surd Quantity/xy without the radical Sign is
Thefe multiplied, the Product is the Square required ppxy
Exam. 8. The rational Quantity da fquared is
The Surd Quantity without the radical Sign is
Theſe multiplied, the Product is the Square required
pp
xy
dda a
aa
2
ddaaz
Raiſe to the Square
The Square
m✓pz
m n ✓ d
a √rd
py√m
mm pz m m n n d
aard
Рруут
Raife to the Square
xpd
xn√ a
xn√ a
z√ px
az✓ d
The Square
xxp d
xx n na
zzpx
a a z zd
And if there are more Quantities than one under the radical
Sign, connected with the Signs + or then after the rational
Quantities are involved, or raiſed as high as is the Index of the
(
L 2
2
Surd;
76.
ALGEBRA.
Surd; place thefe under the radical Quantities, without their
Sign, then multiply them by the Rule of Multiplication at
Art. 10. &c. and this will be the Square required.
Exam. 3.
Exam. I.
Raiſe to the Square a✓m+y
Exam. 2.
b √ d + z
m✓ z-
The Square is
aam+aay
bbd + b b z
MM Z-M MX
Exam. 1. The Surd Quantity✔m + y
without the radical Sign is
The rational Quantity a ſquared is
Theſe being multiplied according to Art. 10.
the Product is the Square required
Exam. 2. The Surd Quantity ✔d +
without the radical Sign is
The rational Quantity b fquared is
א
2
m+y
a a
}
aam+a ay
d+ z
b b
Theſe multiplied, the Prod. is the Square required bb d + b b z
Exam. 3. The Surd Quantity ✔z—x
*}
without the radical Sign is
The rational Quantity m fquared is
Thefe multiplied, the Product is the
Z-X
m m
Square required
Raife to the Square z√ a + n
The Square
e}
11 m Z-
m m x
Exam. 4.
Exam. 5.
Exam. 6.
x√b-d
zza+zzn
xxb-xxd
d√ x + y
ddz + d dy
"}
a + n
}
Exam. 4. The Surd Quantity ✔a + n
without the radical Sign is
The rational Quantity z fquared is
Theſe multiplied, the Product is the ?
Square required
Exam. 5. The Surd Quantity ✓b-d}
without the radical Sign is
The rational Quantity. fquared is
Theſe multiplied, the Product is the }
Square required
ZZ
z za + z z n
b-d
x x
x d
xx b - xx
Exam.
Of SURD QUANTITIES.
77
Exam. 6. The Surd Quantity ✔z+
without the radical Sign is
x + 1}
+"}
x+y
The rational Quantity d fquared is
dd
Theſe multiplied, the Product is the
Square required
e}
d d z + ddy
Raife to the Square y✔a- n n
n √ at d
d √ p-z
The Square
yya — yy n
nna+nnd
ddp-d dz
Raife to the Square
e✔p-
d√e + y
Z zyn - Y
The Square
eep - eer
d de + ddy
ZZN- ZZY
45. Cafe 3. But if there are rational Quantities connected
with the Surd Quantities by the Signs+or, they are in-
volved in the fame Manner as compound Quantities, at Art. 32.
carefully obſerving the Directions concerning the Multipli-
cation of Surd Quantities, at Art. 40. and their Involution at
Art. 43.
To raiſe to the Square or ſecond Power
Putting down again the fame Quantity
Now multiply a +✔b by a, and a multi-
plied by a, the Product is a a, and ✔✅ b multi-
plied by a, the Product is ab, by Art. 40.
therefore a+b multiplied by a is
Then multiply a +✔✅b, byb, and a mul-
tiplied by✓b, the Prod. is a✓✓b by Art. 40.
and multip. by b, the Prod. is b, by Art.
43. whence a+b multiplied by✔✔b is
The Sum is a a +2a√b+b: for a ✅ b
a + √ b
a + √ b
a a + a √ b
a √ b + b
added to a√ bis 2a√b, by Art. 35. whence {aa+2a√b+b
the Square of a + √ b is
To raife to the Square or fecond Power
Putting down again the fame Quantity
The Product from multiplying d+✓ byd,
by what is mentioned in the laft Example is S
z
The Product from multiplying d + ✓ x by }
✔z, by what is mentioned in the laft Example is }
d + √ z
d + √ ≈
dd+d√≈
d√x+x
The Sum added as in the laft Example is }dd+2 d√z+%
the Square of d + √x
To
78
ALGEBRA.
x a
Va
'x - Ja
a
To raiſe to the Square or ſecond Power
Putting down again the fame Quantity
The Prod. from multiplying x-✔a by x, for
multiplied by x, the Prod. is xx, and V
multiplied by x, the Prod. is xa, by Art. *** √α
&iva, by Art. 5
40. the Signs of a and x being different
The Product from multiplying x-✓ a by
✔a, for x multiplied by-a, the Product
isa, the Signs being unlike, but — ✔ a
multiplied by-✔a, the Signs being alike, the
Product is a a or a by Art. 39 and 43.
Their Sum is the Square of x-√ a
-x√a+a
X
**−2× √a+
To raife to the Square or fecond Power
Putting down again the fame Quantity
The Product from multiplying by y, } yy
from what is mentioned in the laſt Example is
y ✓√x
a
y = √ x
— y √ x
The Product from multiplying by − y √ x + x
-x, from what is faid in the laſt Example is
y√x }
Their Sum is the Square of y-x-3y-21 √x + x
y√*
To raiſe to the Square or fecond Power
Putting down the fame Quantity
Multiplying b+xa by b we have
Multiplying b+ √xa byxa we have
The Sum being the Square of b +✔xa,
To raiſe to the Square or fecond Power
Putting down the fame Quantity
Multiplying m+dz by m we have
Multiplying m + d z by ✓✓ d z we have
The Sum being the Square of m+✔dz,
To raiſe to the Square or fecond Power
Putting down the fame Quantity
b+ √xa
b + √x a
bb + b√x a
b √xa + xa
bb+2b√xa+xa
m+ ✓ dz
m+ / dz
mm \ m ✓ dz
m✓ dz + dz
mm+2m√dz+dz
The Square of z - ✓ dn
Z Z -
Z- ✓ d n
Z ✓ dn
ZZ. -x√
z ✓ d n
: / d n + d n
༧/
2 z √✓ d n + d n
2
Το
Of SURD QUANTITIES.
79
To raiſe to the Square or fecond Power
Putting down again the fame Quantity
The Square of p-√y%
p-√yz
1- √yz
PP-P√yz
-p√yz+y%
pp 2p ✓ y z + y z
Of E QUATIONS.
H
AVING thus copiouſly explained all the Rules neceffary
to be known, in order to the Solution of Queftions, we
come now to their Ufe and Application in the Reduction of
Equations, or the Method by which Problems are folved, and
Queſtions answered.
When any Problem or Queftion is propofed to be anſwered
Algebraically, for the feveral Numbers that are in the Queſtion
we generally put Letters, reprefenting likewiſe the Numbers
which are to be found by Letters, and for Diftinction Sake uſe
the Vowels for the unknown Numbers, or thoſe that are to be
found, and Confonants for thofe that are known, or given.
Then we begin to exprefs all the Conditions of the Queftion,
by ranging and connecting the Letters, by Help of the fore-
going Signs, in fuch a Manner that they fhall reprefent all the
Circumftances of the Queftion, this being only to tranflate the
Queſtion from Engliſh into Algebra.
Thus if the Propofition, that 6 being added to 5, the Sum is
equal to 11, was to be expreffed in Algebra.
Now ſuppoſe b = 6, d = 5, m = 11.
Then the above Propofition will be expreffed thus, b+dm.
And when any Letters or Numbers are fo connected, that
between any of them there appears this Sign =, it is called an
Equation, for the Sign = fignifies Equality or Equation, and
in the due ordering and managing thefe Equations confifts the
whole of the Analytic Science, or Algebra.
Equations confift of Quantities or Letters, fome known, and
others unknown, and the grand Work is fo to manage the Equa-
tions, that exprefs what is given in the Queftion, by the Rules
of Certainty and Science, that all the known Quantities may
at laft be found on one Side of the Equation, and the unknown
Quantity by itſelf on the other of the Equation: For when
this is done, the Equation is brought to a Solution, and the
Queſtion is anſwered.
And
80
ALGEBRA.
And that Part of Algebra which teaches how to manage thefe
Quantities, fo as to carry all the known Quantities on one Side,
leaving the unknown Quantity by itſelf on the other Side of
the Equation, is called the Reduction of Equations, which is done
by Addition, Subftraction, Multiplication, Divifion, Involution,
and Evolution, according as the Cafe requires.
To reduce an Equation by Addition, or Subſtraction.
46.
WE
HEN any known Quantities are on the fame Side
of the Equation with the unknown Quantity, and
connected by the Signs + or, to reduce fuch an Equation is
only to tranſpoſe or place the known Quantities on the other Side
of the Equation, or Sign of Equality, prefixing to them their
contrary Sign, that is, thofe Quantities which have the Sign +,
after they are tranfpofed must have the Sign —, and thoſe which
have the Sign must have the Sign+.
Queſtion 1. To find that Number to which 6 being added, and
fubftracting 15 from this Sum, the Remainder may be equal to 11.
1 a
2a+b
Now fuppofe a = the Number fought, b = 6, d= 15,m=11.
Then I am to find a Number, which call
To which 6 or b being added, it is by Art. 6.
From which Sum 15 or dis to be fub-
ftracted, that is, to ab connect d by
the Sign —, then it is
Which a+b-d is to be equal to II or
m, that is
Now to reduce this Equation, or to an-
fwer the Queftion, I obſerve d, a known
Quantity, is on the fame Side of the
Equation with the unknown Quantity
a, therefore tranfpofe d, that is, write
down the remaining Part of that Side of
the Equation without d, and place it on
the other Side with the Sign +, it hav-
ing before the Sign then we have
Again b is a known Quantity on the fame
Side of the Equation with a, then by
taking it away from that Side of the
Equation, and placing it on the other Side
with the contrary Sign, or, we have
}
3a+b-d
4a+b―d=m
5a+b=m+d
6a=m+db
Here
To reduce an Equation, &c.
81
Here the Queſtion is folved, for the unknown Number or
Quantity a, is equal to the Number repreſented by m, added to
the Number reprefented by d, from which Sum fubftracting the
Number repreſented by b.
II repreſented by m
15 reprefented by d
26 Sum of the Numbers reprefented by m and d
6 repreſented by b, to be ſubſtracted
Remains 20 which is a, or the Number fought.
And that this is the Number required, is thus proved, from
the Conditions of the Queftion.
I fay the Number fought is
For if to this is added
The Sum is
From which fubftracting
There remains as the Queſtion required
20
6
26
15
II
Queſtion 2. A Man being aſked how many Shillings he had,
faid, if you add 15 to their Number, and then fubftract 20 from
that Sum, and then add 19 to the Remainder, I ſhall have 64
Shillings. How many Shillings had he?
Let a
the Number of Shillings fought, b = 15, d = 20,
m = 19, n = 64.
Then, A Man had a certain Number
of Shillings, which call
To which 15 orb being added we have,
by Art. 6.
From which Sum taking away 20 or d,
that is, connect d by the Sign-
To which adding 19 or m we have, by
Art. 6.
Which a+b-d+m is to be equal
to 64 or n, hence
Now to reduce this Equation, or an-
ſwer the Queſtion: I begin with
tranſpoſing m a known Quantity, by
putting down the remaining Part of
that Side of the Equation, and placing
m on the other Side with the con-
trary Sign, which gives
M
}
a
:}
2
a+b
3a+b-d
4a+b-d+m
5/a+b=d+m=n
6a+b―d=n
And
82
ALGEBRA.
And to tranſpoſe d another known
Quantity, put down the remaining
Part of that Side of the Equation, and
d on the other Side with a contrary
Sign, whence we have
And lastly, by tranfpofing b, that is,
placing it on the other Side of the
Equation with a contrary Sign, we
have
7a+b=n—m+d
8|a=n―m+d—b
That is, if from the Number repreſented by n we ſubſtract
that reprefented by m, and to the Remainder add the Number
reprefented by d, and from this Sum fubftract the Number re-
preſented by b, the Remainder will be the Number fought.
64 repreſented by n
19 repreſented by m, to be fubftracted
45 or n— m
20 repreſented by d, to be added
65 or n m + d
15 reprefented by b, to be fubftracted
50 the Number fought or a; and therefore the Man
had 50s. at firft, which is thus proved, from the Conditions of
the Queſtion.
I fay he had at firft
}
For if to them you
add
And from that Sum fubftract
And then add to the Remainder
It makes what the Queſtion requires
50s.
15
65
20
45
19
64
Queſtion 3. A Countryman afked another how many Eggs he
had, Why, fays he, if you fubftract 15 from their Number, and
then add 21 to thofe that are left, and fubftract 7 from that Sum,
but if you add 19 to what is then left I shall have 43 Eggs.
How many Eggs had he?
Let
To reduce an Equation, &c.
83
Let a = the Number of Eggs, b = 15, d = 21, m =7,
n = 19, p = 43•
Now the Countryman had a Num-
ber of Eggs, which call
From which 15 or b being fub-
ftracted, or connecting b by the
Sign- we have
To which a
I a
2a.
b
3 a
b + d
S
4a
b + d― m
a b, if we add 21 or
d, we have by Art. 6.
From which Sum fubftracting 7,1
or connecting m by the Sign-}
To which adding 19 or ", we have
by Art. 6.
And this a- b+d—mn is to
mn is to
be equal to 43 or p, hence
Now to reduce this Equation, or
anſwer the Queftion, I begin
with tranſpoſing n, by putting
}|
5a-b+d—m + n
6 a−b+d¬m+n=p
down the remaining Part of that 7a-bd-mp-n
Side of the Equation, and n on
the other Side with its contrary
Sign, then
Now tranfpofe m, by putting down
the remaining Part of that Side.
of the Equation, and m on the
other Side with its contrary Sign,
and we have
Then tranfpofe d, by putting down
the remaining Part of that Side
of the Equation, and d on the
other Side with its contrary Sign,
then
Laftly, tranfpofe b, by putting
down the remaining Part of
that Side of the Equation, and
b on the other Side with its con-
trary Sign, and it is
8a-b+dp―n-m
9a-bp-nm-d
15a=p―n + m―d + b
Hence a, the unknown Quantity or Number of Eggs, is
equal to the Number repreſented by p, fubftracting from it the
Number repreſented by n, adding to this Remainder the Number
repreſented by m, fubft:acting from this Sum the Number
M 2
repreſented
F
84
ALGEBRA.
4.
reprefented by d, and adding to the Remainder the Number
repreſented by b.
43 repreſented by p
19 repreſented by n, to be ſubſtracted
24 or p. n
7 repreſented by m, to be added
31 orp-n+ m
21 repreſented by d, to be ſubſtracted
10 or p―n+m
d
15 repreſented by b, to be added
25 the Number fought or a; and therefore the Man
had 25 Eggs, which is thus proved, from the Conditions of the
Queſtion.
I fay the Man had
For if from them you fubftract
And to the Remainder add
And from this Sum fubftract
And to the Remainder add
It makes what the Queſtion requires
25 Eggs
15
IO
21
31
7
24
19
43
Queſtion 4. To find that Number to which 19 being added, if
from that Sum we fubftract 50, and add 7 to the Remainder,
and fubftract 60 from this Sum, and by adding 6 to that Re-
mainder, this Sum may be 22.
Let a
Number fought, b19, d = 50, m = 7, n=60,
p6, g 22,
Now I am to find a Number, }
which call
}|
I a
To which 19 or 3 being added, }
we have by Art. 6.
From which Sum fubtracting 50
ord, that is, connecting d by
the Sign, and it is
2a+b
3a+b-d
And
To Reduce an Equation, &c.
85
And to this Remainder adding 7 } | 4a+b=d+m
or m, we have by Art. 6.
From this Sum fubftracting 60
or n, that is, connecting n
by the Sign-
5a+b=d+m-n
And to this Remainder adding 66a+b=d+m-n+p
orp, we have
And this a+b-d+m―n+er
is to be equal to 22 or g, hence
Now to anſwer the Queftion,
tranſpoſe p, by putting down
the remaining Part of that
Side of the Equation, and p
on the other Side with its con-
trary Sign, hence
Then tranfpofe n, by putting
down the remaining Part of
that Side of the Equation,
and n on the other Side with
its contrary Sign, then
Then tranſpoſe m, by putting
down the remaining Part of
7a+b―d+m-n+p=g
8a+batman-g-
9{a+b=d+m=g−p+n
that Side of the Equation, roa+b—d=8—p+n—m
and m on the other Side with
its contrary Sign, whence
Then tranfpofe d, by putting
down the remaining Part of
that Side of the Equation,
and d on the other Side with
its contrary Sign, and
Laftly, tranfpoſe b, by putting
down the remaining Part of
11\a+b=8—p+n—m+d.
that Side of the Equation, |12|a=8—p+n—m+d—b
and b on the other Side with
its contrary Sign, we have
Hence a, the unknown Number, is equal to the Number
reprefented by g, ſubſtracting from it the Number repreſented
by p, adding to the Remainder the Number reprefented by n,
fubtracting from this Sum the Number reprefented by m, adding
to the Remainder the Number reprefented by d, and fubftracting
from this Sum the Number reprefented by b.
I
22 the
86
ALGEBRA.
22 the Number repreſented by g
6 the Number reprefented by p, fubftract
16 or g-p
60 the Number reprefented by n, add
76 org―p+n
7 the Number repreſented by m, ſubſtra&t
69 or g
-p+n-
m
50 the Number reprefented by d, add
119 org-p+ n
p + n − m + d
19 the Number reprefented by b, fubſtract
100 the Number fought or a, which is thus proved, from
the Conditions of the Queftion.
I fay the Number fought was
For if to that you add
100
19
119
50
And from the Sum fubftract
And to the Remainder add
And from the Sum fubftract
69
7
76
60
And add to the Remainder
It makes what the Queftion requires
16
6
22
The Directions to the two following Queſtions are not quite
fo copious, that the Judgment of the Learner may be a little
more exerciſed.
Queſtion 5. A Number of Men were walking on a Bowling-
Green, one Man aſked another how many there were, the other
replied, if you fubftract 7 from their Number, and add 15 to
the Remainder, and fubftract 9 from the Sum, and add 56 to
the Remainder, and fubftract 2 from that Sum, this will leave
ICO. To find the Number of Men on the Bowling-Green.
Let a = Number of Men on the Bowling-Green, = 7,
d = 15, 8=9, m = 56, n = 2, p = 100,
I am
To reduce an Equation, &c.
87
I am to find the Number of
Men on the Bowling-Green,
which call
}
I
a
From which 7 or
or b being
fubftracted, which is only
to connect b by the Sign-
To which Remainder adding 2
15 or d, we have by Art. 6.
From which Sum fubftracting
9 or g, or connecting g by
the Sign, and we have
To this Remainder adding m
or 56, by Art. 6.
From which Sum fubftracting
2 or n, that is, connecting
n by the Sign —, it is
Which ab + d − g + m
b+d—g+m — n
is to be equal to 100 or p,
hence
2 a
3a-b+d
3 a
4a-b+d-g
5\a—b+d―g+m
6a-b+d-g+m—n
7a=b+d—g+m-n=p
8a—b+d—g+m=p+n
9a ·b+d—g=p+n—m
10ab+d=p+n—m+g
-b=p+n⋅
n―m+g.
12\a=p+n¬m+g—d+b
100 is the Number reprefented by p
By tranſpofing n we have
By tranfpofing m we have
By tranfpofing g we have
By tranfpofing d we have
By tranfpofing b we have
m+g-d
1
2 or n, to be added
102 or p + n
56 or m, to be ſubſtracted
46 or p+n
112
9 or g, to be added
55 or p + n
-m+.
m + g
15 or d, to be fubftracted
40 or p+n m + g
7 or b, to be added
་
d
47 the Number fought or a, for a=p+n- m+g
-d+b.
Now to prove 47 was the Number of Men that were on
the Bowling-Green, let us try if it will anſwer the Conditions
of the Queſtion.
I fay
88
ALGEBRA.
I fay the Number of Men were
For if from them you fubftract
And add to the Remainder
And from the Sum fubftract
And add to the Remainder
{
And from the Sum fubftract
It makes what the Queſtion requires
་
47
40
15
55
9
46
56
102
2
100
Queſtion 6. A Perfon required another to tell him how many
Shillings he had, by saying that if to their Number was added 5,
and from this Sum fubftracting 3, and adding 16 to the Re-
mainder, and from that Sum fubftracting 50, and adding 54 to
the Remainder, he should then have 43 Shillings.
How many
Shillings had he?
Let a the Number of Shillings fought, b = 5, d = 3,
m = 16, n=50, p = 54, 9=43•
The Perfon had a certain
Number of Shillings, which
call
To which 5 or b being added, Į
we have
I a
2a+b
From which Sum fubftracting} 3
3 or d, we have
To which Remainder addingĮ
16 or m, we have
From which Sum ſubſtracting
50 or n, we have
To which Remainder adding
54.or p, we have
Which abd + m
n + p
is to be equal to 43 or 9,
hence
The Queftion being now ex-
preffed in Algebra, by tranf
pofing p, we have
3a+b-d
4a+b=d+m
·
5a+b
5a+b=d+m — n
6a+b
6a+b¬d+m-n+p
7a+b-d+m-n+p=q
8 a+b=d+m—n=q— $
By
To reduce an Equation, &c.
89
By tranfpofing n we have
By tranfpofing m we have
By tranfpofing d we have
9a+b=d+mqp+n
10 a+b―d=q − p + n
十九​一
​11a+b=q—p+n―m + d
·m
Laftly, by tranfpofing b we have 12│a=q-p+nm+d-b
43 is the Number repreſented by 9
-54 from which fubftracting 54 or p, there remains - II
-II or g-p, fee below.*
50 adding 50 or ʼn to
39 or q p + n
11, the Sum is 39
16 from which fubftracting 16 or m
23 or q p+n-m
3 to which adding 3 or d
26 or q p+n―m + d
5 from which fubftracting 5 or b
21 hence 21 is the Number fought; which is thus proved:
I ſay the Perfon had
For if to them you add
21 Shillings
And from the Sum fubftract
5
26
3
And to the Remainder add
And from the Sum fubftract
There remains a negative or
And if to this Remainder we add
It makes what the Queftion requires
23
16
39
50
-I I
54
43
When a negative Number is to be fubftracted from an
affirmative Number, and the negative Number is greateft, as
in this Cafe, it is only to take the Difference of the two
Numbers, and place the Sign before it; and if the next
Number to be added is affirmative, and greater than the nega-
tive Remainder, then it is only fubftracting the negative Re-
mainder from the affirmative Number which is to be added,
and this will be the Sum.
N
If
go
ALGEBRA.
If the Learner finds any Difficulty in conceiving this, he
may collect all the affirmative Numbers into one Sum, and all
the negative Numbers into another; and ſubſtracting the Sum
of the Negatives from the Sum of the Affirmatives, the Re-
mainder is the Anfwer to the Queſtion.
In the last Queſtion,
1
or Numbers are
The affirmative Quantities}
9 = 43
n
= 50
d
3
Sum of the negative Numbers
The negative Quantities }
or Numbers are
96
75
-p=
21a as before
p = -54
m =
b =
- 16
5
75.
To reduce an Equation by Multiplication.
47. In the laft Article, the unknown Quantity was connected
with the known Quantities by the Signs + or - only, but it
may happen that the unknown Quantity may be divided by
fome known Quantity; in this Cafe, multiply every Part or all
the Terms of the Equation by that known Quantity; and the Part
of the Equation containing the unknown Quantity will be
then multiplied and divided by the fame Quantity, take down
this Equation, rejecting the known Quantity from that Part
of the Equation where it both multiplies and divides the un-
known Quantity, by Art. 20. it being in both Dividend
and Divifor After this Equation is fet down, if there are any
other Quantities connected with the unknown one by the Signs
+ or, tranfpofe them to the other Side of the Equation as
in the laſt Article, by which Method we fhall have all the
known Quantities on one Side of the Equation, and the un-
known one by itſelf on the other Side, which is the Solution of
the Queſtion.
Queſtion
To reduce an Equation, &c.
91
?
Queſtion 7. A Gamefter challenging another to play for as many
Guineas as he had in his Hand, the other required to know how
many there were, he replied, if you divide their Number by 5, and
add 19 to the Quotient, I shall then have 23 Guineas in my
Hand. How many Guineas had he?
Let a the Number of Guineas fought, b = 5, d = 19,
m = 23.
Then the Gamefter had a certain Num-}
ber of Guineas, which call
Which being divided by 5 or b, the Quo-
tient is by Art. 27.
a
To which Quotient if we add 19 or d,
b
we have by Art. 6.
a
And this — + d, is to be equal to 23 or
b
m, therefore we have
The Queſtion being expreffed in Algebra
a
by the Equation dm, in which
b
the unknown Quantity a being divided
by b; now by the Rule, multiply
every Part or Quantity in the Equa-
tion by b, and in this Multiplication,
multiply only the Numerator a of the
a
Quantity by b, according to the Rule
b
of Vulgar Fractions in Arithmetic, and
we have
Becauſe bis in both Dividend
a
}
a
2
b
3
+d
b
a
+d=m
b
5
ab
+bd=bm
b
and
ab
Divifor of the Quantity
hence by
ab
the Rule, rejecting b from
only, and ≥ 6 a + b d = b m
b
placing down the remaining Part a, and
all the other Parts of the Equation,
without any Alteration, we have
Tranfpofing bd by the laſt Article, it
it }
being a known Quantity, then
N 2
3719
7a=bm-bd
Here
92
ALGEBRA.
Here the Queftion is anſwered, for a the unknown Quan-
tity is equal to the Product of the two Numbers repreſented by
b and m, fubftracting from it the Product of the two Numbers
repreſented by b and d.
The Number reprefented by b is 5, the Number re-
prefented by m is 23, which two Numbers being multi-
plied is bm or
The Number reprefented by b is 5, the Number re-
prefented by d is 19, which two Numbers being multi-
plied is b d or
}
115
}
95
Subftracting b d from bm, that is, 95 from 115, leaves }
b m - b d or
20
Which is the Number fought, or the Guineas the Gamefter
had, and is proved from the Conditions of the Queſtion, thus,
I fay the Gamefter had
For if that Number is divided by 5, the Quotient is
But if to this 4 we add
It makes what the Queftion requires
20 Guineas
4
19
23
Queſtion 8. To find that Number which being divided by 15,
if to the Quotient we add 27, and fubftract 13 from the Sum,
the Remainder will be 18,
Þ
Let a the Number fought, b = 15, d = 77, m = 131
= 18.
Now I am to find a Number,
which call
氨
​Which being divided by 15 or b,
we have by Art. 27.
To the Quotient or
b=15, d=27,
}|
I a
a
ܩܙ
ܨ
b
2
a
if we
b
add 27 or d,
we have by
by S
3
b
010
+ d
Art. 6.
From this Sum if we fubftract
13 or m, that is, connect m
by the Sign, it is
a
Whichd-m is by the
b
Quction to be equal to 18
or p, hence we have
a
10
+d-
712
a
5
+d—m=p
b
The
To reduce an Equation, &c.
93
The Queftion being now ex--
preffed in Algebra by this E-
a
quation +d—m=p,
m=p, and
b
the unknown Quantity a being
divided by b, multiply every
Part of the Equation by b as in
the laft Queftion, and then we
have
Becauſe bis in both Dividend-
and Divifor of the Quantity
ab
b
7 reject 6 from this Quan-
b
,
6
a b
+ db — m b = p b
b
tity only as in the laft Queftion, 7a + db → mb = pb
placing down a and the remain-
ing Quantities in the Equation
without any Alteration, then
we have
Becauſe mb is a known Quan-
tity, tranfpofe it by the Direc-
tions in the laft Article, and
we have
Becauſe db is a known Quan-
tity, tranfpofe it by the fame.
Directions, and we have -
8a + db = p b + mb
pb
9a
p b + m b - d b
Now a the unknown Quantity being by itfelf on one Side of
the Equation, the Queftion is folved; for a, the unknown Quan-
tity, is equal to the Product of the two Numbers reprefented by
p and b, added to the Product of the two Numbers reprefented
by m and b, fubftracting from this Sum the Product of the
Numbers reprefented by the Letters d and b.
The Number repreſented by p is 18, the Number re-
prefented by bis 15, the Product of theſe two Numbers
is pb or
The Number reprefented by m is 13, the Number re-
prefented by bis 15, the Product of thefe two Numbers
is mb or
The Sum is pb+ mb or
CA
}
270
195
405
The
94
ALGE BRA.
The Number reprefented by d is 27, the Number
reprefented by bis 15, the Product of thefe two Num-
bers is 405, which being fubftracted from the Sum of
the other two Products
Leaves pb-m b
Therefore 60 is
d b or a
405
60
equal to a, or 60 is the Number fought,
which is thus proved from the Queſtion.
I fay the Number fought is
For if that is divided by 15 the Quotient is
To which Quotient, or 4, if we add
The Sum is
And if from this Sum we fubftract
There remains what the Queſtion requires
60
4
27
31
13
18
Queſtion 9. A Man being asked how many Shillings he had,
replied, if you divide the Number I have by 25, and ſubſtract 3
from the Quotient, and then add 51 to this Remainder, and from
the Sum fubftracting 40, I shall have 12 Shillings left. How
many Shillings had he?
Let a the Number of Shillings the Man had, b = 25,
d = 3, m = 51, p = 40, Z=12..
Now the Man had a certain
Number of Shillings, which
call
Which being divided by 25 or
b, we have by Art. 27.
a
From the Quotient or if
b
we fubftract 3 or d, that is,
connecting d by the Sign
To the Remainder adding 51 orm,
we have by Art. 6.
From which fubftracting 40 or
p, that is, connecting p by
the Sign
a
Which
b
we have
dmp is by
the Queſtion, to be equal to
12 or 2, hence we have
}
I a
2
alo
a
3
b
다
​a
b
d
d + m
d+m-p
a
5
a
d+mp=2
I
The
To reduce an Equation, &c.
95
The Queſtion being now ex-
preffed in Algebra, and the
unknown Quantity a being
divided by b, multiply every
Quantity in the Equation by
b, as in the two laft Queftions,
then we have
And rejecting b out of the Quan-
ab
tity only, becauſe it is
b
in both Dividend and Divifor,
and fetting down the reft as in
the two laft Queftions, we
have
Becauſe pb is a known Quantity,
tranfpofe it by Art. 46. and
we have
Becauſe mb is a known Quan-
tity, tranfpofe it in like Man-
ner, then we have
Becauſe db is a known Quan-
tity, by tranfpofing it we
have
;
ав
7 →db+mb — pb zb
b
8a-db+mb-pbxb
9a-db+mb=zb+pb
10 adb zb + pb — mb
11a=zb+pb―mb+db
Now it appears the unknown Quantity, or a, is equal to
the Product of the two Numbers reprefented by z and b, added
to the Product of the two Numbers reprefented by p and b,
fubftracting from this Sum the Product of the two Numbers.
repreſented by m and b, and adding to this Remainder the Pro-
duct of the two Numbers reprefented by d and b.
The Number reprefented by z is 12, and that by bĮ
is 25, the Product of these two is z bor
The Number reprefented by p is 40, and that by b
"}
}
300
is 25, the Product of theſe two is pb or
The Sum is z b + p b or
1000
1300
The Number repreſented by m is 51, and that by b} 1275
is 25, the Product of theſe two is m
b
or
Which being fubftracted from the Sum of the other
two, leaves z b + p
b
m bor
The Number reprefented by d is 3, and that by b
is 25, the Product of theſe two is db or
Which added to the laft Remainder, the Sum is
zb + pb11bdb or a
25
75
100
Whence
96
ALGEBRA.
Whence the unknown Quantity a, or the Number of
Shillings the Man had is 100, which is thus proved, from the
Conditions of the Queſtion.
I fay the Man had
For if that Number is divided by 25, the Quotient is
From which Quotient if we ſubſtract
Remains
To which adding
The Sum is
From which fubftracting
There remains what the Queſtion requires
Shillings.
100
| | | |
12
Queſtion 10. A Country Servant, who underfood Algebra,
being afked by his Mafter how many Cows there were in the Field,
replied, if you add 13 to their Number, and divide that Sum
by 8, and then add 19 to the Quotient, and fubftract 11 from
that Sum, there will be 12 Cows left. How many Cows were
there?
Let a the Number of Cows, b=13, d=8, m=19,
p = 11, x = 12.
Now there were in the Field a
certain Number of Cows, which
call
b
To which 13 or 6 being added,
we have by Art. 6.
ab
}
a
}
Which a + b being divided by 3
8 ord, we have by Art. 28.
To which if we add 19 or m, we
have by Art. 6.
From which if we fubftract II
or p, we have by connecting
p with the Sign-
Which a+b+m—p is by the
d
Queſtion to be equal to 12 or
*, hence we have
2a+b
a+b
d
}
a + b
+ m
d
a + b
5
+m-p
d
6
d
a+b+m-p="
Becauſe
To reduce an Equation, &c.
97
Becauſe a, the unknown Term,
is Part of the Fraction
a+b
d
where the Divifor is d, there-
fore multiplying every Quan-7
tity in the Numerator of
the Fraction by d, by the
Rule of Vulgar Fractions in
Arithmetick, and the other
Quantities as before, then
Becauſe d is in every Term of
the Dividend and Divifor of
the Fraction ad + b d
ject d, from
d
"> re-
ad + b d only,
d
by Art. 22. and 24, and ſet
down all the reft as before,
then
Now begin to tranfpofe pd, it 7
being a known Quantity,
then we have
Becauſe md is a known Quan-
tity, therefore tranfpofe it,
and we have
Becauſe b is a known Quan-
tity, therefore tranfpofe it,
and we have
ad+bd
d
+m d − p d = x d
8 a+b+md—pd = xd
9a+b+md=xd + pd
10 a+b=xd+pd—md
11a=xd+pd-md-b
By this it appears that a, the unknown Quantity, is equal to
the Product of the two Numbers reprefented by x and d, added
to the Product of the two Numbers reprefented by p and d,
ſubſtracting from this Sum the Product of the two Numbers
repreſented by m and d, fubftracting ftill from this Remainder
the Number repreſented by b.
The Product of the two Numbers reprefented by x
and d is xd, or
*}
} 96
?}
The Product of the two Numbers reprefented by p} 88
and d is pd, or
Their Sum is x dp d, or
184.
The
98
ALGEBRA.
The Product of the two Numbers reprefented by m
and dis m d, or
Which fubftracted from the Sum of the other two
Products, there remains x d + p d
m d
From which fubftracting the Number reprefented by b
}
152
32
13
The Remainder is xd+pd-md-b, which is equal} 19
to a, or the Number fought
And that 19 Cows were in the Field, is thus proved from
the Conditions of the Queftion.
I fay the Number of Cows were
For if to them we add
The Sum is
Which divided by 8, the Quotient is
To which Quotient if we add
The Sum is
From which Sum ſubſtracting
11
There remains what the Queſtion requires
I
19
13
32
4
19
23
II
12
Queſtion 11. Two young Gentlemen were difputing how many
Men were at a public Diverfion, but not agreeing, they referred
it to a third Perfon, who, being ſkilled in Algebra, instead of a
direct Answer, replied, that if you ſubſtract 115 from their
Number, and divide the Remainder by 50, and add 39 to that
Quotient, from which Sum fubftracting 16, and adding 68 to
the Remainder, this laft Sum will be equal to 101.
How many
Men were there?
Let a the Number of Men fought, b=115, ¢=50,
d=39, n=16, p = 68, x = 101.
There were a certain
Number of Men, which
call
From which 115, or b, be-
ing fubftracted, we have S
Which Remainder of a—b,
being divided by 50, or
c, we have by Art. 28.
To which Quotient if we
add 39, ord, we have
24
I a
b
a
b
C
3
4
a
C
+ d
From
To reduce an Equation, &c.
99
From this Sum if we
fubftract 16 or n,
have
a
C
we
}
b
5
+d-n
a
b
6
S
C
To which Remainder if we
add 68 orp, we have
a b
Which +d-n+p
C
is, by the Queſtion, to
be equal to 101 or x,
hence
Becauſe a, the unknown
Quantity, is Part of
a-b, which being di-
C
vided by c, therefore
multiplying by c, as
in the last Queftion, we
have
Becauſe c is in every
Term
of the Dividend and Di-
ca-cb
vifor of
C
7
8
a
+d-n+p
b
+d=n+p=x
C
ca
cb
+cd-en+cp=cx
C
reject c,
9a-b+cd-on + cp=cx
as in the last Queſtion,
and fet down all the
reft as before, and we
have
Now tranfpofe cp, it be-
ing a known Quantity,
then it is
Tranſpoſe en, it being a
10|a—b c d—cn=cx — cp
1
known Quantity, and we
have
IIa
ΙΙ
⋅ b + c d = c x=cp+cn
12/4
=cx-cp+c n
c p + c n − c d
Tranfpofe cd, it being a
known Quantity, and we
have
And tranfpofing b, it being
a known Quantity, we
have
13a=cx-cp+cn-cd+b
Hence it appears that a, the unknown Quantity, is equal to
the Product of the two Numbers reprefented by c and x, fubftract-
ing from it the Product of the two Numbers reprefented by c and f,
O 2
adding
100
ALGEBRA.
adding to that Remainder the Product of the two Numbers
repreſented by and n, fubftracting from this Sum the Product
of the two Numbers repreſented by c and d, and adding to this
Remainder the Number repreſented by b.
c
The Product of the two Numbers reprefented by 5050
and x is cx, or
The Product of the two Numbers reprefented by
and p is cp, or
The Remainder is cxcp, or
''}
3400
1-650
yc} 800
The Product of the two Numbers repreſented by c
and n is cn, or
Which added to the laft Remainder, the Sum is
1 x cp+ cn, or
The Product of the two Numbers reprefented by c
and d is ed, or 1950, which being ſubſtracted
The Remainder is cx-
cp+cn
ed, or
Adding the Number reprefented by b, or
} 2450
}
1950
500
115
The Sum is ex-cp+cn-ed+b, which is equal } 615
to a, the Number fought
And that there were 615 Men is proved from the Condi-
tions of the Queſtion.
I ſay there were
For if from them we ſubſtract
Remains
Which being divided by 50, the Quotient is
To which adding
The Sum is
From which ſubſtracting
The Remainder is
To which adding
Men.
615
115
500
ΙΟ
39
49
16
33
68
101
There remains what the Queſtion requires
Queſtion 12. There is a certain Number to which 9 being added,
and dividing this Sum by 5, if from this Quotient we fubftract 6,
and add 101 to the Remainder, from that Sum fubftracting 10,
there remains 97. What is the Number?
Let
To reduce an Equation, &c.
ΙΟΣ
Let a = the Number fought, b = 9, c = 5, d = 6, m = 101,
† = 10, x = 97•
Now I am to find a certain}
Number, which call
To which 9, or b, being
added, we have by Art. 6. S
This being divided by 5,7
we have by Ar-
or c,
ticle 28.
From which fubftracting 6, }
or d, we have
To which adding 101, or
m, we have by Art. 6.
From this fubftracting 10,
or p, that is, connecting
p
by the Sign, it is
a+b
Which
C
d+m-p
is to be equal to 97, or
x, hence
The Queſtion being thus
expreffed in Algebra, be-
gin and multiply by c,
for the Reaſon in the laft
Queſtion, then we have
Rejecting from
C
ca + c b
C
and ſetting down the reſt
as before, then
Now tranfpofing cp, it
being a known Quantity,
we have
Tranfpofing cm, it being
a known Quantity, we
have
Tranſpoſing cd, it being
a known Quantity, we
have
Laftly, tranfpofing b, it
being a known Quantity,
we have
2
Ia
ف.
2a+b
a+b
3
C
a+b
4
}|5
a + b
d + m
C
6
a + b
d+m-p
C
}
с
a+b
7
·d+m-p=x
C
8
ca + c b
-cd+cm-cp=cx
C
9/a+b―cd+cm=cp=cx
Ica+b=cd+cm=cx+cp
11a+b—cd=cx+cp=cm
12a+b=cx+cp-cm+cd
I
13a=cx+cp-cm+cd-f
ہو
The
102
ALGEBRA.
The Algebraick Operation being finiſhed, the Numerical Work
is thus.
The Product of the two Numbers reprefented by c and 485
x is cx, or
}
The Product of the two Numbers repreſented by c and } 50
pis cp, or
The Sum is cx + cp, or
535
The Product of the two Numbers reprefented by cand} 505
m is cm, or
Subftracting, the Remainder is cx+cpcm, or
The Product of the two Numbers repreſented by c and
d is cd, or
Adding, the Sum is cx+cp-cm + cd, or
The Number reprefented by b is 9, fubftracting
30
}
30
The Remainder is cx+cp cm+cd-b, which is
is }
equal to a, or the Number fought
And is thus proved from the Conditions of the Queſtion.
I ſay the Number fought is
For if to this we add
The Sum is
Which being divided by 5, the Quotient is
From which fubftracting
The Remainder is
To which adding
The Sum is
From which ſubſtracting
51
9.
60
I 2
6
6
ΙΟΙ
107
60
9
51
There remains what the Queſtion requires
To reduce an Equation by Divifion.
IO
97
48. In the laft Article the unknown Quantity was divided by
a known Quantity, in the Equation that aroſe from the Con-
ditions of the Queftion; in this Article the unknown Quantity
will be multiplied into a known Quantity, in the Equation that
arifes from the Conditions of the Queſtion; when this happens,
divide every Quantity on both Sides of the Equation, by the fame
known Quantity into which the unknown Quantity is multiplied,
then
To reduce an Equation, &c.
103
then you will find the unknown Quantity to be multiplied and
divided by the fame Quantity; now place down this Equation,
rejecting only the Letter from that Quantity, where it multiplies
and divides the unknown Quantity, as in the laft Article; then
tranfpofe the Quantities as before, but if there are none to be
tranfpofed the Queftion is folved.
If any Quantities are connected with the unknown one by
the Signs + or it will be moft convenient for the Learner
to tranſpoſe them before he begins to divide by the Rule juſt
given.
Queſtion 13. A Perfon required another to tell him how many
Shillings he had, by faying that if their Number was multiplied by
13, and if from that Product was fubftracted 25, he should then
bave 170 Shillings. How many Shillings had he?
Let a the Number of Shillings the Perfon had, b = 13,
d = 25, m = 170.
A Perfon had a certain Number of
Shillings, which call
Which multiplied by 13, or b, we
have by Art. 9.
From the Product, or ba, if we fub.
ſtract 25, or d, we have
Which Remainder bad is by the
Queſtion to be equal to 170, or m,
hence
Becaufe d is on the fame Side of the
Equation with the unknown Quan-
tity, and connected by the Sign-,
therefore tranfpofe d, then
There being no more Quantities to be
tranfpofed, and the unknown Quan-
tity being multiplied by b, therefore
divide both Sides of the Equation by
b. Now ba divided by b gives
ba
b
by Art. 27. and m 4 d divided by
b, gives m+d by Art. 28. therefore
we have
b
a
}
26 a
$3
ba – d
4ba-
ba-dy
n
5ba=m+d
b a
6
b
m + d
b
Becaufe
104
ALGEBRA.
Becauſe is in both Dividend and
ba
Diviſor of the Quantity, reject b
by Art. 20. and putting down the
other Quantities without any Altera-
tion as in the foregoing Queftion, we
have
m + d
70
b
From hence it appears that a, the unknown Quantity, is
equal to the Sum of the two Numbers reprefented by m and d,
divided by the Number reprefented by b.
The Number repreſented by m is
The Number reprefented by dis
178
25
195
The Sum is m+d, or
And dividing 195, or m+d, by 13, or b, the Quotient is
m + d
b
or 15, which is a, or the Number fought.
The Truth of which is thus proved from the Conditions of
the Queſtion.
I ſay the Perfon had
For if that is multiplied by
[]
1
15 Shillings
13
The Product is
From which fubftracting
There remains what the Queftion requires
45
15
195
25
170
Queſtion 14. A Butcher feeing a Drover going to Market with
a Number of Sheep, afked how many there were; the Drover
anfwered, if you multiply their Number by 9, and fubftract 157
from that Product, and add 168 to the Remainder, I fhall then
have 2000 Sheep. How many Sheep had he ?
遍
​Let a the Number of Sheep, b=9, d= 157, m=168,
P = 2000.
Then
To reduce an Equation, &c.
105
}
་
The Drover had a certain Number)
a
Number}
Which multiplied by 9, or b, we
have by Art. 9.
From the Product fubftracting 157, or
d, that is, connecting d by the
Sign, we have
To which Remainder adding 168, or
m, we have by Art. 6.
This ba-d+m is by the Queſtion
to be equal to 2000, or p, hence
we have
Now according to the Rule begin with
tranfpofing m, and we have
Then tranfpofing d, we have
The Quantities being all tranſpoſed]
that were connected by the Signs +
or, and the unknown Quantity
being multiplied by b, therefore by
the Rule divide both Sides of the
Equation by b, but ba divided by b,
ba
gives and pmd divided
b'
by b, gives 1m+d
by Art. 28.
b
hence we have
ba
I
1 la
}
2 ba
3ba-d
} | 4
ba-d+m
5 ba―d+m=p
6ba―d=p—m
7 ba = p → m + d
ba P
m + d
8
b
b
Rejecting b from the Quantity
b
becauſe it is in both Dividend and
Divifor, and placing down the re-
maining Parts of the Equation with-
out any Alteration as before, we have
9|a=
m + d
b
The Algebraic Work is now finished, for the unknown
Quantity a is on one Side of the Equation by itſelf, and it
appears to be equal to the Number reprefented by p, fubftract-
ing from it the Number reprefented by m, adding to this Re-
mainder the Number reprefented by d, and dividing this Sum
by the Number repreſented by b.
P
The
1
106
ALGEBRA.
The Number reprefented by pis
2000
From which ſubſtracting the Number repreſented by 168
m, which is
There remains p―m, or
To which adding the Number repreſented by d
The Sum is p―m+d, or
1832
157
1989
And dividing this 1989, or p-m+d, by 9, the Num-
ber repreſented by b, the Quotient is Pm+d or 221, which
b
is a, or the Number of Sheep the Drover had; and is proved by
the Conditions of the Queftion thus.
I fay the Number of Sheep were
For that being multiplied by
The Product is
From which ſubſtracting
There remains
To which adding
The Sum is what the Queſtion requires
221
9
1989
157
1832
168
2000
Queſtion 15. A Man being asked what he gave for his Horſe,
answered, if you multiply the Number of Pounds I gave by 5, and
then add 15 to the Product, and from that Sum fubftract 50,
and to the Remainder adding 25, from which Sum fubftracting
14, this Remainder will be equal to 81. What did he give for
his Horſe?
Let a what he gave for the Horfe, b5, d= 15,
c = 50, p = 25, m = 14, x=81,
Let the Pounds which the Perfon
gave for the Horſe be called }
Which multiplied by 5, or b, we
have by Art. 9.
To which Product if we add 1
15, or d, we have by Art. 6.
From this Sum fubftracting 50,
I a
2 ba
3 ba+d
or c, that is, connecting c by 4 ba+d-c
the Sign, we have
Το
To reduce an Equation, &c.
107
To which Remainder adding 25; } | 5 | 6a+d=c+p
or p, we have
From which Sum fubftracting} 6 ba+d=c+p.
14, or m, we have
Which ba+d-c+p—m
is by the Queſtion to be
equal to 81, or x, hence we
have
Now tranſpoſing m, we have
And tranfpofing p, we have
And tranfpofing c, we have
And tranfpofing d, we have
The Quantities connected by
the Signs+ or -, being
now all tranſpoſed, I obferve
the unknown Quantity to
be multiplied by b, there-
fore divide every Term on
both Sides of the Equa-
tion by b. Now dividing
ba by b, it is
ba
and
dividing x+m—p+c — d by
b, we have x+m-p+c-d
b.
as in the foregoing Queſtions,
hence we have
Rejecting b from the Quantity
ba
b'
becauſe it is in both
Dividend and Divifor, and
placing down the reſt of the
Equation without any Al-
teration as before, and we
have
7\ba+d-c+p — m = x
8 ba+d=c+ p = x+m
9 ba+d-c=x+m-p
10 ba+d=x+m-p+c
11ba=x+m-p+c-d
12
ba
x+m-p+c-d
b
b
I
13 a
x+m-p+c-d
b
That is, a, the unknown Quantity, is equal to the Number
repreſented by x, added to the Number reprefented by m, fub-
ftracting from their Sum the Number reprefented by p, adding to
this Remainder the Number reprefented by c, fubftracting from
this Sum the Number reprefented by d, and dividing this Re-
mainder by the Number repreſented by b..
P 2
Now
108
ALGEBRA.
i
Now x is
To which adding m, or
81
14
The Sum is x + m, or
From which fubftracting p, or
95
There remains x + m—p, or
To which adding c, or
The Sum is x + m − p + c, or
From which ſubſtracting d, or
There remains +m-p+c-d, or
x
25
70
50
120
15
105
Now dividing this 105, or x+m-pc-d, by b, or 5,
the Quotient is
x + m− p + c― d
,
or 21, which is equal to a,
b
or Number of Pounds the Horfe coft.
Which is proved from the Conditions of the Queſtion,
thus,
I fay the Horſe coſt
For if that is multiplied by
The Product is
To which adding
The Sum is
From which fubftracting
There remains
To which adding
The Sum is
From which ſubſtracting
There remains what the Queftion requires
21 Pounds
5
105
15
12 2/2 = | 27
81
Queſtion 16. There is a certain Number which being multiplied
by 7, if from this Product we fubftract 21, and to the Remain-
der add 11, and from this Sum fubftract 23, and add to
the Remainder 33, this last Sum will be 210. What is the
Number?
Let a the Number fought, b=7, d = 21, x = 11,
<= 23, p = 33, r 210.
Now
i
{
1
To reduce an Equation, &c. 109
Now there is a certain Number
fought, which call
Which multiplied by 7, or b,
we have by Art. 9.
From which ſubſtracting 21, or
d, that is, connecting d by
the Sign-, we have
To this adding 11, or x, we
have by Art. 6.
}|
I a
2 ba
3
ba-d
4
ba
d + x
by
5 ba-d+x-C
6 ba-d+x―cte
From which fubftracting 23, or
c, that is, connecting
the Sign, we have
we}/6/ba.
To which adding 33, or p, we
have by Art. 6.
And this bad+x-c+p
is by the Queftion to be
equal to 210, or r, hence
we have
The Queſtion being now ex-
preffed in Algebra, begin the
Solution by tranſpoſing p, and
then we have
Tranfpofing c we have
Tranfpofing x we have
Tranfpofing d we have
All the Quantities being now
tranſpoſed that were connect-
ed by the Signs + or
and the unknown Quantity
being multiplied by b, di-y
vide every Term, or both
Sides of the Equation by b,
as in the laft Example, and
we have
Now reject bout of the Quan-
ba
tity becauſe it is in both
Dividend and Divifor, and
fetting down the remaining.
Parts of the Equation, as
in the laſt Queſtion, and we
have
7ba-d+x-c+p=r
8ba-d+x-c=r-p
9 ba➡d+x=r-ptc
10ba-dr-p+c-x
11bar-p+ c = x + d
I2
ba
b
r-p+c−x+d
b
13 a = " — p + c = x+d
b
To
110
ALGEBRA.
To find what a is in Numbers.
The Number reprefented by r, is
From which fubftracting the Number reprefented
by p, which is
There remains r-p, or
1.
To which adding the Number repreſented by c
The Sum is r-p+c, or
210
}
33
177
23
From which ſubſtracting the Number repreſented by x
There remains rpcx, or
To which adding the Number repreſented by d
The Sum is rp + c −x+d, or
200
II
189
2I
210
or 7, the Quotient is
And dividing this 210 by b, or 7,
r-p+c-x+d
, or 30, which is
b
fought, and is thus proved.
I fay the Number fought is
For if this is multiplied by
The Product is
From which ſubſtracting
There remains
To which adding
The Sum is
From which ſubſtracting
There remains
To which adding
equal to a, or the Number
The Sum is what the Queftion requires
30
7
210
21
189
I I
200
23
177
33
210
Queſtion 17. A Gamefter challenged another to play with him
for as many Guineas as were in his Hand; but being asked how
many they were, anfwered, if you multiply their Number by 10,
and ſubſtract 100 from the Product, and to the Remainder
add 55, and from the Sum fubftract 31, and adding to this
Remainder 115, I shall then have 539 Guineas. How many
had he?
b—10, c
Let a the Number of Guineas fought, b= 10, 100,
d=55, m = 31, 115, p= 539.
Then
To reduce an Equation, &c.
Then a Gamefter had a certain.
Number of Guineas, which
call
a
Which being multiplied by 10,2 ba
or b, we have by Art. 9.
From which fubftracting 100, } 3 bac
or, we have
To which adding 55, or d, we
have by Art. 6.
|
}
4 ba— c + d
5 ba-c+d— m
* }
6 | ba− c +d—m+x
From this fubftracting 31, or
m, we have
To which adding 115, or x,
we have
This by the Queftion is to be
equal to 539, or p, hence we
have
Then by tranfpöfing & we have
Tranfpofing m it is
Tranfpofing d we have
And tranfpofing c
Now divide by b, as before di-}
rected, and we have
ba
And rejecting b from and]
7ba-c+d―m+x=p
8 ba-c+d—m
-m=p-x
c+d=p−x+m
9ba-c+d
10 ba c = p −x+m — d
II b⋅a= Р x+m―d+c
p−x +m
ba
d + c
12
b
b
placing down the reſt as be-
13a=
p — x + m
·d + c
b
fore, then
b
The Queftion being now folved in Algebra, we are to find
what a is equal to in Numbers.
Now p is equal to
From which fubftracting x, or
There remains p—x, or
To this adding m, or
The Sum is p -x+m, or
From this fubftracting d, or
There remains p−x+m-d, or
To this adding c, or
The Sum is p-x+md + 6, or
2
539
115
424
31
455
55
400
100
500
But
I 12
ALGEBRA.
!
But dividing this 500 by b, which is 10, the Quotient is 50,
the Number of Guineas the Gamefter had; and is thus proved
from the Conditions of the Queſtion.
I fay the Gamefter had
For if that Number is multiplied by
The Product is
From which fubftracting
There remains
To which adding
The Sum is
From which fubftracting
There remains
To which adding
The Sum is what the Queftion requires
1
50 Guineas
ΙΟ
500
100
400
55
455
31
424
115
539
Queſtion 18. A Perſon being aſked how many Hours it was
past Noon, replied, if you multiply the Number of Hours paft
Noon by 7, and fubftract 5 from the Product, and to the Re-
mainder add 9, and from the Sum ſubſtract 3, and to this Re-
mainder adding 4, this Sum will be equal to 12.
How many
Hours was it past Noon, or what of the Clock was it?
Let a the Number of Hours it was paft Noon, or the
Number fought, m = 7, p = 5, d = 9, c = 3,
x = 12.
b = 4,
Then there is a certain Number
of Hours paft Noon, which
call
This multiplied by 7, or m, we
have by Art. 9.
From which fubftracting 5, or
p, we have
To this adding 9, or d, we have
by Art. 6.
From which fubftracting 3, or
c, that is, connecting
the Sign, we have
To which adding 4, or b,
have by Art. 6.
a
2m a
3ma-p
4 ma
p+ d
by
5ma-p+d-c
we ?
} [ 6 ]ma − p + d − c + b
p+d=c+b
Which
To reduce an Equation, &c.
113
Which by the Queſtion is to be Į
}|7 na-p+d-c+b=x
equal to 12, or x, hence
Now tranfpofe b, and we have
Tranfpofing c, then
Tranfpofing d, and
Tranfpofing p, we have
Now dividing by m, as in the
former Queſtions, and we
have
Rejecting m from the Quantity
8na-p+dc=x-b
9 ma➡p+d=x− b + c
10 map=x-b+c-d
I
ma
ma
m
I2
,
ma as before, and we have
13a=
m
b+c
d+p
x b + c d + p
m
x − b + c − d + p
m
The Algebraic Work being finiſhed, we may find what a is
in Numbers thus.
Now x is equal to
From which fubftracting b, of
There remains xb, or
To which adding c, or
I2
4
8
U
The Sum is x- b+c, or
From which fubftracting d, or
There remains x ·b + c −d, or
To which adding p, or
The Sum is x-b+c-d+p, or
3
I I
9
2
5
7
And dividing this by m, or 7, the Quotient is 1, which
is equal to a, or the Number of Hours it was paſt Noon,
hence it was I of the Clock in the Afternoon.
Which is thus proved from the Conditions of the Queſtion.
I fay the Number of Hours paft Noon were
For if that is multiplied by
The Product is
From which fubftracting
There remains
To which adding
The Sum is
From which fubftracting
There remains
To which adding
The Sum is what the Queſtion requires
Q
a!
7
7
5
2
9
I I
3
8
4
J 2
TA
114
ALGEBRA.
To reduce an Equation by Involution.
49. Hitherto there has been no Equation in which the un-
known Quantity has had the radical Sign prefixt before it, or
has been connected with known Quantities under the radi-
cal Sign; but as this is a Cafe which frequently happens, we
are now to explain the Manner, how fuch Equations are
managed.
If any Part of an Equation is a furd Quantity, but the un-
known Quantity is not under the radical Sign, then there is
no Occafion to clear this Equation of its Surds; but if the
unknown Quantity is under the radical Sign, then the Equation
muſt be cleared of its Surds.
And when there is a given Equation where the unknown
Quantity is under the radical Sign, and there are known Quan-
tities without the radical Sign on that Side of the Equation, and
connected by the Signs+or, tranſpoſe all thofe Quantities
which are without the radical Sign, to the other Side of the
Equation; then raiſe both Sides of the Equation to the Square,
if the radical Sign expreffes the Square Root, or to the Cube, if
the radical Sign expreffes the Cube Root, and fo on; by which
Means the Equation will be cleared of its Surds.
After this, if there are no known Quantities on the ſame Side
of the Equation with the unknown one, the Queſtion is folved;
but if there are ftill known Quantities on the fame Side of the
Equation with the unknown Quantity, the Equation is to be
reduced by fome of the Methods before explained, at Art. 46,
47, 48.
The Square Root is expreffed by this Sign, and the Cube
3.
Root by the fame Sign with a 3 on the Top, thus ✓; and if
any Root is taken befides the Square Root, the Figure over the
Sign fhews what Root it is; but when it is only the Square
Root, then there is generally no Figure over the Sign.
Queſtion 19. Two Gentlemen were talking of the Number of
Acres there were in a Park, the Park-Keeper being prefent, and
difpofed to how his Learning, told them, that if they extracted
the Square Root of the Number of Acres in the Park, from which
Square Root fubftracting 5, this Remainder will be equal to 50.
How
many. Acres were there in the Park?
Let
A
To reduce an Equation, &c.
115:
d
Let a the Number of Acres in the Park, b = 5,
= 50.
of } |
I
a
Now there were a certain Number of
Acres in the Park, which call
The Square Root of which, by Art. }
33. is
From which fubftracting 5, or b, that
is, connecting b by the Sign-, it is
Whicha:-b by the Queftion?
is equal to 50, or d, hence
The Queſtion being now expreffed
in Algebra, and obferving that b
is not under the radical Sign, there-
fore tranſpoſe b, then
Now all the Quantities being tranf-
pofed, which were not under the
radical Sign, fquare both Sides
of the Equation, as the radical
Sign expreffes the Square Root.
But the Square of a is a, by
Art. 43. and the Square of d+b
is d d + 2 db + bb, by Art. 32.
and making the fe equal to one
another, for the Square of equal
Quantities or Numbers must be
equal, and we have
2✓ a
3√a: —b
4√a: b = d
-
5√a=d+b
6a=dd +2db + b b
Hence it appears that a, the unknown Quantity, is equal to
the Square of the Number reprefented by d, added to twice the
Product of the two Numbers reprefented by d and b, and this
Sum added to the Square of the Number reprefented by b.
dd, or 2500
The Square of the Number reprefented by d is dd, or
The Product of the two Numbers reprefented by d
and bis db, or 250, and twice that Product is
2 db, or
The Sum is d d + 2 db, or
The Square of the Number repreſented by b is bb, or
500
3000
25
The Sum is dd + 2 db +bb, or 3025, which is 3025
a, or the Number fought
Q 2
Hence,
116
ALGEBRA,
200
Hence, I fay, there were 3025 Acres in the Park, which is
thus proved, from the Conditions of the Queſtion.
The Number of Acres in the Park were
Now the Square Root of that Number is
From which ſubſtracting
There remains what the Queſtion requires
3025
55.
ๆๆๆ
50
Queſtion 20. A Perfon, who had been fortunate at Gaming,
was aſked how many Guineas he had won, to which he answered,
that if the Square Root of their Number was extracted, from
which Root fubftracting 7, he should then have 16 Guineas,
What Number of Guineas did he win?
Let a = the Number of Guineas won, b=7, d= 16.
Now a Perfon won a Number of
Guineas, which call
The ſquare Root of which by Art.}
33. is
From which ſubſtracting 7, or b, wel
have
Which
ab by the Queſtion is
to be equal to 16, ord, hence
Now becauſe b is not under the
radical Sign, therefore tranſpoſe b,
then
All the Quantities not under the ra-
dical Sign being now tranfpofed,
in order to clear the Equation of
the Surd, raife both Sides of the
Equation to the Square or fecond
Power. But the Square of a is
a, by Art. 43. and the Square of
d + biş d d + 2 db + b b, by
Art. 32. and making thefe two
equal to one another, for the Square
of equal Quantities or Numbers
muſt be equal, and we have
I a
2 √ a
} 3√ a
a - b
4√ a
4√a―b=d
5√ a = d + å
6a=dd2d b - b b
That is to fay, the unknown Quantity, or a, is equal to the
Square of the Number reprefented by d, added to twice the Pro-
duct of the two Numbers reprefented by d and b, to which Sum
add the Square of the Number reprefented by b.
The
To reduce an Equation, &c.
117
The Square of the Number reprefented by d is dd, or
The Product of the two Numbers reprefented by d and
bis db, or 112, and twice that Product is 2 db, or
The Sum is dd2 db, or
The Square of the Number reprefented by b is bb, or
}
256
224
480
49
The Sum is dd +2db+bb, or 529, which is equal} 529
to a, or the Number fought
Therefore the Perfon won 529 Guineas; and is thus proved,
from the Conditions of the Queſtion.
I ſay the Number of Guinéas he won was
For the Square Root of that Number is
And if from that Square Root we ſubſtract
There remains what the Queſtion requires
}
529
ཀླག་
16
Queſtion 21. A Gentleman having fold his Eftate, an imper-
tinent illiterate Perfon aſked him what he had fold it for, why,
Sir, replied he, if you extract the Square Root of the Number of
Guineas for which I fold it, and add 17 to that Number, this
Sum will be equal to 317. How many Guineas had the Gentle-
man for his Eftate?
Let a the Number of Guineas for which the Eſtate was
fold, b17, d = 317.
Now the Eftate was fold for al
Number of Guineas, which call
The fquare Root of which by Art. Į
33. is
To which 17, or b, being added, we l
have
Which ab by the Queftion is
to be equal to 317, ord, hence
The Queſtion being now exprefled
in Algebra, and becaufe b is
not under the radical Sign,
therefore tranfpoſe b, and we
have
Now fquare both Sides of the Equa
tion, and make them equal to one
another, for the Reafons mentioned
in the two laft Queftions, and we
have
a
2/a
3√a+6
4√ √ a+b=
+
ď
5✓a=d-b
6a=dd-2dbbb
From
118
ALGEBRA.
From hence we know that a, the unknown Quantity, is
equal to the Square of the Number reprefented by d, fubftracting
from it twice the Product of the Numbers repreſented by d and
b, and adding to the Remainder the Square of the Number re-
prefented by b.
The Square of the Number reprefented by d is 100489
dd, or
The Product of the two Numbers reprefented by
}
b and d is db, or 5389, and twice that Product is 10778.
2 db, or
or
Į
Which ſubſtracted, the Remainder is dd2 db, 89711
The Square of the Number repreſented by b is $}
bb, or
—
which}
289
The Sum is d d — 2 db +bb, or 90000, which 90000
is equal to a, and is the Number fought
And that the Eſtate was fold for 90000 Guineas, is thus
proved from the Conditions of the Queſtion.
I ſay the Eſtate was fold for
For the fquare Root of that is
To which if we add
The Sum is what the Queftion requires.
90000 Guineas
300
17
317
Queſtion 22. A young Gentleman, when he came of Age, aſked
his Guardian the annual Rent of the Eftate his Father left him,
to which he was answered, that if he extracted the fquare Root of
the Number of Pounds for which the Estate was rented, and to
this Root if he added 27, it would be equal to 100 Pounds.
What was the annual Rent of the Estate?
Let a = the Rent of the Eftate, m 27, x = 100.
Now the Rent of the Eftate was
The fquare Root of which by Art.}
33. is
To which 27, or m, being added, we
have
Which by the Queſtion is to be equal
to 100, or x, hence
I
a
2 Va
} 3√a+m
4a+m
= =X
*
The
To reduce an Equation, &c.
f1g
The Queſtion being now expreffed
in Algebra, begin by tranfpofing
m, for the Reafons mentioned in
the former Queftions, and then
we have
Now fquaring both Sides of the E-
quation, to take away the radical
Sign, as was done in the foregoing
Queftions, and then we have
5 √ a = x
6a=xx―2xm+mm
And there being no more Quantities to be tranfpoſed, the
Queftion is folved; for we may find the Value of a in Num-
bers from the Algebraic Work, thus:
The Square of the Number reprefented by x is xx, or
The Product of the two Numbers repreſented by
the Letters x and m is x m, or 2700, and twice that
Product is 2 xm, or
Which fubftracted leaves x x - 2 x m, or
The Square of the Number repreſented by m is mm, or
The Sum is 5329, or xx
equal to a, or the Number fought
2 xm + mm, which is
10000
5400
4600
729
s}
} 5329
And that the annual Rent of the Eftate was 5329 Pounds,
is proved from the Conditions of the Queftion.
I ſay the annual Rent of the Eftate was
For the fquare Root of that is
To which there being added
The Sum is what the Queftion requires
5329 Pounds
73
27
:
100
Queſtion 23. To find that Number to which 1290 being added,
if the fquare Root of this Sum is extracted, from which Root fub-
Atracting 29, the Remainder may be 71.
Let a the Number fought, b 1290, d 29, *=71.
There is a Number fought, which
I call
To which 1290, or b, being added,
we have
I
a
2a +6
#
The fquare Root of which Sum by 3a +
ать
From
Art. 34 is
I
120
ALGEBRA.
From which ſubſtracting 29, or 2 | 4
d, we have
"}
Which by the Queftion is equal 5
to 71, or x, hence
Now begin the Solution with
tranfpofing d, it not being
under the radical Sign, and
then
All the Quantities on one Side-
of the Equation being now
under the radical Sign, to
take away that, as the un-
known Quantity is under
it, fquare both Sides of the
Equation as before. Now the
Square of ✔a + b is a + b,
by Art. 43. and the Square
of x + d is x x + 2 x d + dd,
by Art. 32. and as the Squares
of equal Numbers, or Quanti-
ties, muſt be equal to one ano-
ther, hence
a+b:~ď
✓a+b:~d=*
/a+ b = x+ď
7a+b=xx + 2 x d + d d
Now tranfpofe b, it being a 8a=x+2xd+dá — ↓
known Quantity, and then
From whence we may find the Value of a in Numbers.
The Square of the Number reprefented by x is xx, or 5041
The Product of the two Numbers repreſented by x
and d is xd, or 2059, and twice that Product is
2 x d, or
The Sum is x x + 2 x d, or
The Square of the Number reprefented by dis dd, or
The Sum is xx + 2 x d + dd, or
From which fubftracting the Number reprefented by b
}
4118
9159
840
10000
1290
There remains 8710, or xx + 2 x d + dd—b, } 8710
which is a, or the Number fought
And is thus proved from the Conditions of the Queſtion.
I fay
To reduce an Equation, &c.
121
*
I fay the Number fought is
For if to that we add
The Sum is
The fquare Root of which is
8710
1290
10000
100
From which fubftracting
There remains what the Queſtion requires
t
29
71
Queſtion 24. A Perfon being asked his Age, replied, that if
from my Age you fubftract 11, and extract the fquare Root of the
Remainder, to which Root adding 13, this Sum will be equal to
20. What was the Age of the Perfon?
Let a the Number of Years, or Age of the Perfon,
b = 11, m = 13, d= 20.
Now the Age of the Perfon is
From which if we fubftract 11, 2
or b, we have
I
a
2a-b
The ſquare Root of which by 3
Art. 34. is
To which adding 13, or m, we
have
Which by the Queftion is equal
to 20, or d; hence we have
The Queſtion being thus ex-
preffed in Algebra, and m not
being under the radical Sign,
therefore tranfpoſe m; then
Now fquare both Sides of the
Equation, to clear it of the
Surd, as in the former Que-
ftions. But the Square of
✔a—b, is a-b, by Art.
43. and the Square of d-m
by Art. 32. is dd
+mm; then as the Squares of
equal Quantities are equal, we
have
2 d m
And by tranfpofing b we have
3√ √
a
-b
4√ a
b:+m
5√
a
:+m=d
6
a b = d — Ma
7a-b=dd-2dm+mm
8\a=dd-2dm+mm+b
By which we find what a is in Numbers. Thus,
R
The
122
ALGEBRA.
4}
}
The Square of the Number reprefented by d is dd, or
The Product of the two Numbers reprefented by d
and m is dm, or 260, and twice that Product is 2 dm, or S
Which 520 fubftracted from 400, leaves d d- 2 dm,
or 120, (fee the Numerical Work in Queſtion 6.)
The Square of the Number repreſented by m is m m, or
Which 169 added to-120, makes d d-2 dm + mm,
er + 49, (fee the Numerical Work in Queſtion 6.)
To which adding the Number reprefented by b
The Sum is 60, which I fay ise, or the Age of the
Perfon
400
520
-120
169
49
11
© } 60
And is proved from the Conditions of the Queſtion, thus:
I ſay the Perſon was
For if from that you fubftract
There remains
60 Years old
II
49
The ſquare Root of which is
To which adding
7
13
1
The Sum is what the Queſtion requires
20
To reduce an Equation by Evolution.
50. This is done by the Extraction of Roots, for if after all
the known Quantities have been carried to the other Side of the
Equation from the unknown Quantity, it appears that one
Side of the Equation is the Square, Cube, or any Power of the
unknown Quantity, then extract fuch Root of both Sides
of the Equation as will deprefs or lower this Power of the un-
known Quantity to the firft Power; that is, if one Side of the
Equation is the Square of the unknown Quantity, then the
Square Root muſt be extracted; and if it is the Cube of the un-
known Quantity, then the Cube Root must be extracted, and
fo on, which depreffing the unknown Quantity to the firft
Power, the Queftion is anſwered.
Queſtion 25. What is that Number, if to the Square of which
there is 51 added, the Sum may be 100?
Leta the Number fought, b = 51, m = 100.
Now
To reduce an Equation,
123
c.
Now there is a Number fought, which}
I call
The Square of which by Art. 31. is
To which 51, or b, being added, we Į
have
a
2aa
3
a. a + b
4 aa+b=m
}|5
And this a a+b is by the Queſtion to Į
be equal to 100, or m; hence
The Queſtion being expreffed in Al-
gebra, begin and tranfpofe b; then
The known Quantities being now all
on one side of the Equation, and the
other Side being a a, or the Square
of a; therefore by the Rule extract
the fquare Root of both Sides of the
Equation. Now the fquare Root of
aa is a, by Art. 33. and the Square
Root of mb ism-b, by Art.
34. and as the fquare Root of equal
Quantities muſt be equal, therefore
-m-b
5a
6a= √ m
✓
Hence a, or the Number fought, is equal to the Number
reprefented by m, fubftracting from it the Number reprefented
by b, and extracting the fquare Root of the Remainder.
The Number repreſented by m, is
From which fubftracting b, or
There remains m b, or
The fquare Root of which ism-b, or 7, and
b, or 7, and }
is equal to a, or the Number fought
And is thus proved:
I fay the Number ſought is
The Square of which is
To which adding
The Sum is what the Queftion requires
¿}
།
100
51
49
7
7
49
51
100
Queſtion 26. A Merchant had gained fo many Pounds, that if
from the Square of their Number is fubftracted 101, and to the
Remainder add 500, this Sum will be 3000 Pounds. What had
the Merchant gained?
R 2
Let
124
ALGEBRA.
Let a the Gain of the Merchant, b= 101, m 500,
p3000.
Then a Merchant had gained a certain
Number of Pounds, called
The Square of which is by Art. 31.
From which fubftracting 101, or b, we
have
To which adding 500, or m, we have
This, by the Queſtion, is to be equal
to 3000, or p; hence
By tranfpofing m we have
By tranfpofing b it is
I a
2 a a
3aa-
·b
4aa b+m
5aab+m=p
6aa—b=p—m
7aa=pm + b
27 aap
By extracting the ſquare Roots, as at the
fixth Step of the laft Example; then 8a=p-m+b
} √p—m+b
That is, a is equal to the Number reprefented by p, fub-
ſtracting from it the Number repreſented by m, and adding to
this Remainder the Number reprefented by b, and extracting
the fquare Root of the Sum.
The Number reprefented by pis
From which fubftracting m, or
There remains p
m, or
3000
500
2500
ΙΟΙ
2601
"}
51
To which adding b, or
The Sum is p m + b, or
The ſquare Root of which is ✔ p −m + b, or
51, equal to a, the Number fought
And is thus proved, from the Conditions of the Queſtion.
I ſay the Merchant gained
For the Square of that is
From which fubftracting
There remains
To which adding:
The Sum is what the Queſtion requires
51 Pounds
2601
ΙΟΙ
2500
500
3000
Queſtion 27. If to the Square of the Number of Miles a Per-
fon had travelled there is added 97, fubftracting from the Sum
251, and adding to the Remainder 160, this Sum will be 10006.
How many Miles had he travelled?
Let
To reduce an Equation, &c.
125
Let a the Number of Miles he had travelled, b = 97,
M = 251, x = 160, z=10006.
Then a Perfon had travelled a cer-
tain Number of Miles, called
The Square of which is by Arti-
cle 31.
To which adding 97, or b, it is
}|
a
2 a a
From which ſubſtracting 251, or 4 aa+b-m
m, gives
To which adding 160, or x, it is
Which by the Queftion is to be?
equal to 10006, or z, whence
By tranfpofing it is
By tranfpofing m we have
By tranfpofing b then
By extracting the fquare Root, as
at the eighth Step of the laft
Example, we have
3aa+b
5aa+b
6aa+b
m + x = z
-m+x
7aa+b m = 2
8}a a + b = % x + m
9aa=z— x + m
IO a ✓ Z- x+m-
b
That is, from the Number repreſented by z, fubftract the
Number repreſented by x, to the Remainder add the Number
repreſented by m, from which Sum fubftract the Number repre-
fented by b, extract the fquare Root of the Remainder, and it
will be the Number fought.
The Number reprefented by z is
From which fubftracting x, or
There remains z — x, or
To which adding m, or
The Sum is z −x+m, or
From which ſubſtracting b, or
There remains zx+m
The fquare Root of which, or
the Number fought
-b, or
A
10006
160
:
9846
251
10097
97
10000
−x+m—b, is
is }
100
PROOF.
126
ALGEB R A.
ļ
PROOF.
I fay the Perfon had travelled
For the Square of that is
To which adding
The Sum is
From which fubftracting
There remains
To which adding
The Sum is what the Queſtion requires.
100 Miles
10000
97
10097
251
9846
160
10006
Queſtion 28. A General, upon numbering his Army, found,
that if from the Square of the Number of Men in his Army, there
was fubftracted 3196, and to the Remainder adding 2721, from
which Sum fubfiracting 1711, there would remain 99997814.
To find the Number of Men in the Army?
Let a the Number of Men in the Army, b = 3196,
m =2721, *= 1711, Z =
2 99997814.
The Number of Men in the Army}
was
The Square of which is by Ar-
ticle 31.
From which fubftracting 3196,
or b, it is
To which adding 2721, or m,
gives
I
a
2
a a
$3 aa-
b
} | 4
4aa
b + m
From which fubfracting 1711,5 aab+ m − x
or x, we have
aa-
6 [aa-b+m-x=Z
Which by the Queſtion is equal to 6
99997814, or z; hence
First, by tranfpofing x
By tranfpofing m
By tranfpofing b
By extracting the fquare Root, as
in the former Examples
By Numbers thus:
7aa-b+m = x + x
8aa-b=
b=x+x-
m
9aa = x+x- m + b
as }|10|4 =
10 a = √x + x — m + b
z I
is
To reduce an Equation, &c.
127
z is in Numbers
To which adding ×, or
The Sum is z + x, or
From which fubftracting m, or
There remains zx-
m, or
To which adding b, or
The Sum is x + x − m + b, or
The fquare Root of which is a, or the Number
fought
Which is thus proved:
I fay the Number of Men in the Army were
For the Square of that is
From which fubftracting
There remains
To which adding
The Sum is
From which fubftracting
There remains what the Queftion requires
99997814
1711
99999525
2721
99996804
3196
100000000
}
10000
100000000
10000
3196
99996804
2721
99999525
1711
99997814
51. Theſe being the particular Methods by which Equations
are reduced, or Queftions anſwered, we fhall now add fome
Examples where all theſe Methods are promifcuouſly uſed.
Queſtion 29. A Merchant broke for fo many Pounds, that if
their Number was multiplied by 4, and the Product divided by 6,
and extracting the Square Root of the Quotient, from which ſub-
fracting 60, there remains 40. What was the Sum for which
the Merchant broke?
Let a the Number of Pounds fought, 4, d=6,
m = 60, p
60, p = 40.
Then the Merchant broke for al
Number of Pounds, called
Which multiplied by 4, or b, we?
have
2}
I G
}
216a
ba
3
This divided by 6, or d, we
have
The
128
ALGEBRA.
The fquare Root of which } | 4
From this fubftracting 60,5
or m, we have
Which by the Queſtion is
equal to 40, or p, hence
Becauſe m is not under the
radical Sign, therefore
tranſpoſe it, by Art. 49.
Now fquaring both Sides of
the Equation by Art. 49. S
And multiplying by
Art. 47. then
Rejecting d from
>
ba
d
ba
th
d
6
ba
m
d
་
8
of }
y } |
9
d, by
dba
d
and putting down the
other Quantities without
any Alteration, as at
Art. 47. we have
Dividing by b, by Art. 48.}
then
ba
Rejecting b from and
b
"
putting down the other
Quantities without
without any
Alteration, as at Art.
47, or 48, we have
In Numbers thus:
dpp 9600
=
+2 dpm = 28800
+ dmm=21600
bd
√²a=p+m
ba
| 0/2 = pp+2pm+mm
d
dba = dpp+2dpm+dmm
d
10 badpp +2dpm+dmm
ba
dpp +2dpm+dmm
II
b
b
124=
a
dpp +2dpm + dmm
b
Sum 60000 or d p p + 2 d pm + dm m
Now dividing 60oco, or dpp +2 dpm + dmm, by 4, or b,
dpp +2dpm+dmm
or 6000, divided by 4= 15000,
we have
b
which is equal to a, or the Number of Pounds for which the
Merchant broke,
PROOF.
To reduce an Equation, &c.
129
PROOF.
15000
4
6)60000
10000 (100 the fquare Root of 10000
60
40 as the Queſtion requires.
Queſtion 30. A Gentleman having bought a Houfe, and being
difpofed to try the Knowledge of his Son in Algebra, told him, if
the Number of Pounds the Houfe coft was divided by 8, and that
Quotient multiplied by 50, and extracting the fquare Root of
the Product, to which adding 10, this Sum would be 60 Pounds.
What did the Houſe coſt?
Let a the Price of the Houfe, b 8, d= 50, m = 10,
p = 60.
Now the Price
the Price of the}
Houfe is
Which divided by 8, or b,}
it is
This multiplied by 50, or
d, we have
The fquare Root of which
is, by Art. 33.
I
a
2
b
}
da
3
b
da
4
b
da
To which adding 10, or m
5
+m
Б
Now fquaring both Sides of
This, by the Queſtion, is
equal to 60, or p, hence S
The Queftion being now
expreffed in Algebra, and
m not being under the
radical Sign, tranſpoſe it
by Art. 49. then
the Equation, by Art. 49.
And multiplying by b, by l
da
6
+m=p
b
da
7
p-
m
Ъ
da
8
=pp-2pm + mm
b
b da
9
bpp -2bpm + b mm
Art. 47.
1 6
S
Rejecting
130
ALGEBRA.
Rejecting b from
bda
and
b
putting down the reſt as
at the twelfth Step of
the laſt Queſtion, then
Dividing by d, by Art. 48.}
10 da=bpp~ 2 bp m + b m m
d
da
_bpp — 2bpml + b mm
II
then
d
da
Rejecting d from and
putting down the reſt as
124
bpp — 2 bpm + b m m
d
at Art. 47, or 48. and
In Numbers:
~2bpm=
9600
bpp = 28800
19200
+bmm = 8co
d=5|0)2000|0
4004, the Number of Pounds the
Houſe coft.
PROOF.
8)400
50
50
2500 (50 the Square Root of 2500
ΙΟ
60 as the Queſtion requires.
CONSECTAR Y.
If the Reader compares the eighth, ninth, and tenth Steps of
the laſt Work, he will find that to multiply any Fraction by its
Denominator, or any Dividend by its Divifor, is only to reject
the Denominator, or Divifor, from that Quantity, and multiply,
it into all the other Quantities; thus, the Equation at the eighth
=pp-2pm + mm, which being multiplied by
Step is da
b
its
To reduce an Equation, &c.
131
its Denominator b, we have at the tenth Step dabpp
— 2 b p m + b mm; the ninth Step, or bda = b pp - 2 bpm
b
+bmm, being only a more particular Illuftration of the
Work.
And by comparing the tenth, eleventh, and twelfth Steps of
the fame Work, it appears, that to divide any Quantity, by
any Letter in that Quantity, is only to reject that Letter from
the Quantity, and placing it as a Divifor to the other Quan-
tities; thus, at the tenth Step, the Equation is da = bpp -
2 bpm + b mm, which being divided by d, gives at the twelfth
b p p − 2 bpm + b m m
Step a =
; the eleventh Step or
d
da
d
b p p — 2 bpm + b mm, being only a more particular Illuftration
d
of the Work.
Therefore we ſhall for the future leave out ſuch Steps as the
ninth and eleventh : I did not chooſe to do it before, my Deſign
being to make this curious Science as eafy as poffible.
Queſtion 31. A Running-Footman being ſent of an Errand,
was told, that if he ſquared the Number of Miles he was to run,
and multiplied the Square by 4, and divided the Product by 40,
to this Quotient adding 500, from which Sum ſubſtracting 1400,
and extracting the fquare Root of the Remainder, it would be 10.
How
many Miles was the Footman to run?
Let a the Number of Miles the Footman was to run,
b = 4, d=40, m
=40, m=500, x = 1400, p = 10.
The Number of Miles the Foot-
man was to run let be
Which being fquared is by
Art. 31.
}
a
2 aa
This being multiplied by 4, 3 ba
or b, we have
This being divided by 40, or d,}
it is
To which adding 500, or m,
gives
S 2
baa
4
d
baa
5
十九
​From
132
ALGEBRA.
From which fubftracting 1400, } 6
or x, we have
The fquare Root of which is
by Art. 34.
४
baa
+ m
d
}|7
baa
+m− x
d
baa
+m· x = Þ
Which by the Queftion, is equal 8
to 10, or p, therefore
Now ſquare both Sides of the
Equation, by Art. 49. and
s
baa
d
9 +m x = pp
d
é
baa
By tranfpofing x, we have
}
ΙΟ
+m=pp+x
d
ba a
I I
d
By tranſpoſing m, we have
By multiplying by d by the
Confectary, Page 130.
=pp+x-m
12|baa=dpp+ dx - dm
And dividing by b by the Con-}|13|a a=
Setary, Page 130.
Now extracting the ſquare Root, }|14|4=
by Art. 50.
In Numbers:
dpp4000
+dx = 56000
dm=
60000
20000
b = 4) 40000
dpp + d x
dx ~
b
d m
d p p + d x - dm
b
10000 (100 the ſquare Root of 10000, hence
the Footman was to run 100 Miles.
4
a a
40
PROOF.
+509—1400 = 10
I have not drawn out the Proof of the laft Queftion into Par-
ticulars, but only expreffed it at once; that is, four times the
Square of a (which is found to be 100) being divided by 40, if
to this Quotient we add 500, and from this Sum fubftract 1400,
And
the fquare Root of this Remainder will be equal to 10.
now I fhall exprefs all the Conditions of the Queſtion at the firſt,
Equation,
To reduce an Equation, &c.
133
Equation, that the Learner may form fome little Judgment in
what Manner to fhorten his Work; and if he conceives how the
Proof of the laft Queftion is expreffed, it will eafily lead him to
the Knowledge of expreffing the Conditions of the Queſtion, or
raife fuch Equations as arife from the Queftion without parti-
cularizing every Circumftance. But if the Learner finds any
Difficulty in this, he may proceed as before.
Queſtion 32. A Gentleman who had been at the Gaming-
Tables, and lofing, fome of his Acquaintance laughing at him for
his Folly, afked how much he had left; to which he answered, if
you Square the Number of Pounds I have loſt, and divide that
by 4, multiplying this Quotient by 10, to which Product add
3900, then extracting the fquare Root of this Sum, from which
fubftracting 80, the Remainder will be equal to 90. How much
had he loft?
Let a = the Number of Pounds loft, b = 4, d = 10,
m = 3900, p = 80, z = 90.
Then by the Queſtion
By tranfpofing p, it
not being under the
radical Sign, by Art.
49. we have
By fquaring both Sides
of the Equation, by
Art. 49. then
By tranfpofing m, it is
Multiplying by b by
the Confectary, Page
130.
Dividing by d by the
ſame
Extracting the fquare 7
Root, by Art. 50.
daa
+m:-p=%
b
daa
2
+m=x+p
b
da a
3
+m=2x+2xp+pp
b
da a
4
=xx+2xp + pp-m
b
5daa=bzz+2bz p+b pp− b m
6|aa=
b x x + 2 b xp + b p p − b m
7a=
d
bxx + 2b xp + bpp-bm
d
In Numbers:
dz z
134
ALGEBRA.
i
bzz=32400
2bxp=57600
bpp=25600
115600
bm 15600
d=10) 10000|0
10000 (100 = a, the Number of Pounds loft.
I
O
PROOF.
10 aa
a a
+3900:- 80 90
4
To reduce an Equation when the unknown Quantity
is in feveral Terms.
52. When the unknown Quantity is in more Terms than
one, bring all thofe Terms which have the unknown Quantity
to one side of the Equation, taking Care that the greateſt Co-
efficient of the unknown Quantity has at laft the affirmative
Sign, and carrying all the Quantities that are known on the
other Side of the Equation; then divide both Sides of the
Equation by all the Co-efficients of the unknown Quantity, con-
nected with the fame Signs of and, as they then happen
+
to have, which will reduce the Equation, as in the following
Examples.
If the unknown Quantity fhould be in more than two
Terms, tranfpofe thofe Terms in fuch a Manner, that the Sum
of the pofitive Co-efficients of the unknown Quantity may
exceed the Sum of the negative Co-efficients of the unknown
Quantity, and then divide as before directed.
Queſtion 33. There is a certain Number which being multiplied
by 10, if this Product is divided by 2, to this Quotient adding 19,
and fubftracting 99 from that Sum, the Remainder will be equal
to the Number fought.
=
Let a the Number fought, b = 10, d = 2, m = 19,
x = 99.
2
By
To reduce an Equation, &c.
135
ba
I +m-z=a
d
By the Queſtion
By tranfpofing z
By tranfpofing m
ba
2
+m=a+z
d
ba
d
3 =a+z
a+z-m
dm
5 ba-da-dz-dm
By multiplying by d by the 4 bada+dz.
Confectary, Page 130.
Becauſe d is less than b, tranſ-
poſe da, that both the Terms
which have the unknown
Quantity, may be on the fame
Side of the Equation, then
And dividing according to the
Rule by b-d, the two Co-
efficients of a, we have
6a=
d z — d m
b-d
Number fought.
= 20 the
1
10 a
2
PROOF.
+ 1999 a
The Divifion at the fifth and fixth Steps, viz. that bada,
divided by bd, fhould leave only a, may perhaps a little
perplex the Learner; and if it does, I advife him to examine
Art. 10. where he may obferve, that in multiplying any com-
pound Quantity by a fingle Letter, that Letter goes into every
Term of the Product, therefore the Multiplier is not fo many
Times that Letter as the Number of Terms are in which that
Letter is found, but only that fingle Letter multiplied fuc-
ceffively into all the other Quantities; hence, if this Product
is to be divided by all thofe Quantities, the Quotient will be
the fingle Letter, and not fo many Times that Letter as the
Number of Terms are in which it is found. See farther the
Proof of the Queſtion 38. and Art. 22.
1
Queſtion 34. A Gentleman bought an Eſtate for so many Pounds,
that if they were multiplied by 4, and this Product divided by 5,
from which Quotient fubftracting 600, and adding to the Re-
mainder 6 Times what the Eftate coft, this Sum will be equal te
6200 Pounds. How much did the Eftate coft?
Let
136
ALGEBRA.
Let a the Number of Pounds the Eſtate coft, b= 4,
d = 5, m = 600, p=6, x = 6200.
Then by the Queſtion
By tranfpofing m, we have
Multiplying by d by the Con-
fectary, Page 130
Dividing by b+dp, the Co-
efficients of a, as in the laſt
Queſtion, and we have
-
ba
I
m+pa = x
d
ba
2
+pa=x+m
d
3 ba+dpa=dx+dm
4
a =
Therefore the Eſtate coft 1000 Pounds.
PROOF.
d x + d m
b+dp
= 1000
4 a
600+6a= 6200
5
Queſtion 35. A Perfon had a certain Number of Shillings,
which multiplied by 4, this Product being divided by 11, to the
Quotient adding 90, and from this Sum taking away 30, the
Square Root of this Remainder will be equal to the fquare Root of
the Number of Shillings fought, after being diminiſhed by 10.
Let a the Number of Shillings fought, b = 4, d=11,
*=90, p = 30, % = 10.
Then by the Queſtion
1b a + x = p = √ a=x
ba
d
}
ba
+x➡p=a
d
Becauſe there is no Quantity on
each Side of the Equation but
what is under the radical Sign,
therefore ſquare both Sides of
the Equation, by Art. 49.
Multiplying by d by the Con-
fectary, Page 130.
Becaufe d, one Co-efficient of a,
is greater than b, the other
Co-efficient of a, tranſpoſe
ba, then
2
א
Z
}|3|ba+dx—dp=da— dz
4 dx-dp=da-dz-ba
Tranſpoſing
To reduce an Equation, &c.
137
Tranpofing dz
Dividing by db, the two-
Co-efficients of a,
Queſtion 33. Step 6. we
have
as at
wes
=
|5|dz + dx dp da—ba
61a=
dz + d x = dp = 110
d-b
(the Number fought.
If the Learner chooses to have the unknown Quantity
on the left Side of the Equation, he might have put the 5th Step
thus, da-badz + dx-dp, this being only to change
the Sides of the Equation, not to alter their Value.
PROOF.
If a 110, then
4 a
I I
+90-30=√α-10
a
Queſtion 36. A Running-Footman forward to fhow his Learn-
ing, being in Company, faid, if the Number of Miles he had run
was multiplied by 7, to which Product adding 550, and fubfiract-
ing 20 from that Sum, and dividing the Remainder by 10, the
fquare Root of the Quotient will be the fame, as if you added
14 Miles to thofe he had run, and extracted the fquare Root of
that Sum.
Let a the Number of Miles he had run, b
m = 20, p = 10, × = 14.
Then by the Queſtion
Quan-
There being no
tity without the radi-
cal Sign, therefore fquare
both Sides of the Equa-
tion as at the fecond Step
of the laſt Queſtion
Multiplying by p by the
Confectary, Page 130.
Becauſe p, one Co-efficient
of a, is greater than b,
other Co-efficient of
therefore tranſpoſe b a
By tranfpofing px
the
a,
Dividing by p-b, the two
Co-efficients of a, as at
Queſtion 33. Step. 6. we
have
7, d=550,
= 7,
I
ba + d
P
m
=√a+x
ba+d- m
=a+x
P
3bad-m=p a + p x
2
} 3 ba+a
T
4d-m=pa+px-ba
5d-m-px-pa-ba
a
m px
130 the
(Number of Miles required,
PROOF,
138
ALGEBRA.
1
PROOF.
7a+550— 20
ΙΟ
= √ a + 14
To reduce an Equation when the fame Quantity, either known
or unknown, is in every Term of the Equation.
53. In any Algebraic Operation, if the fame Quantity, either
known or unknown, is in every Term of any Equation, then
divide every Term of the Equation by that Quantity, which will
reduce the Equation to more fimple Terms, as in the following
Queftions.
Queſtion 37. To find a Number which multiplied by 4, and
the Product added to the fame Number multiplied by 56, and
divided by 7, this Sum will be equal to the Square of the Number
fought.
Let a = the Number fought, b = 4, d= 56, m = 7.
Then by the Queſtion
}
da
Iba +
- аа
772
}
2mba+da = maa
Multiplying by m by the?
Confectary, Page 130.
Becauſe a is in every Term
of the Equation, divide
by a, then
Dividing by m, the Co-
efficient of a, by the Con-
fectary, Page 130. and
3mb+d=ma
4/a
4α==
mb + d
12 the Num-
m
(ber fought.
PROOF.
56 a
4a+
=aa
7
Queſtion 38. There are two Towns at fuch a Distance, that if
the Number of Miles between them is multiplied by 79, and this
Product added to their Distance, the Square Root of this Sum will
be equal to the Distance of the two Towns multiplied by 2.
Lot
To reduce an Equation, &c.
139
Let a the Diftance of the Towns, b = 79, m = 2.
Then by the Queftion
There being no rational Quantities on
the fame Side of the Equation where
the radical Sign is, fquare both Sides
of the Equation, and
Dividing by a, it being in every Term
of the Equation, and
I√bata = ma
2 ba+a=mmaa
36+1=mma
Dividing by mm, the Co-efficient of a{4 a =
b+ I
M M
Hence the Diſtance between the two Towns is 20 Miles.
PROOF.
√79a+a= 2 a
=20
If the Reader does not eafily conceive that dividing b a + a,
or ba+1 a at the fecond Step, by a, gives b + 1, as at the
third Step, I advife him to confider what is faid at Queſtion
33; to which may be added, that b + 1 × a = ba+a, whereas
b + 1 × 2 a = 2 b a + 2 a, a Product very different from
batia
baa. Or it may be explained thus,
=b+ 1,
Q
the a being rejected by Art. 22 and 26.
The Manner of registering the Steps of an Algebraic
Operation explained.
54. Having explained to the young Analyft the different
Methods of managing Equations, to fave the Trouble of uſing
fo many Words, I fhall now fhow him the Method of register-
ing the Steps, introduced by the ingenious Dr. John Pell.
To register the Steps of an Analytic Operation is only to ex-
preſs in the Margent of the Work by Symbols, inſtead of Words,
what has been done; and to render it as eafy as may be to
the Learner, we ſhall refume the Work of one of the former
Queſtions, and exprefs by Words what is done in one Column,
in another Column exprefs the fame Thing by Symbols, or
Characters, and in the third Column place the Work itſelf,
that by comparing the Operation with the Obfervations that
follow it, the Reader may the more easily underſtand the
Manner of regiſtering the Steps.
T 2
Queſtion
140
ALGEBRA.
Queſtion 39.
A Running-Footman being fent of an Errand,
was told, that if he fquared the Number of Miles he was to run,
and multiplied it by 4, and divided the Product by 40, to this
Quotient adding 500, from which Sum fubftracting 1400, and
extracting the fquare Root of the Remainder, it would be 10,
How many Miles was the Footman to run? (this is Queſt. 31.)
Let a =
the Number of Miles the Footman was to run,
b = 4, d = 40, m = 500, x = 1400, p = 10.
Then, by the Queſtion, we have
the fame Equation as at the
eighth Step, Queſtion 31.
Squaring both Sides of
the Equation, or in-
volving them to the
fecond Power, by
Art. 49.
By tranfpofing x at the
fecond Equation
By tranfpofing m at the
third Equation
Multiplying the fourth?
Equation by d
Regifter
baa
I
+m-
d
11
b a a
12 2
+m-x=pp
d
baa
2+x3
3-m
baa
4
4x d
4 x d
5
÷ 6
b
6
a
app+dx—dm
6 un 2
7a= √
b
Dividing the fifth Equa-} 5
Extracting the fquare
Root of the fixth E-
quation, by Art. 50. S
30 ac² + m = p p + x
d
= = p p + x — m
pp+xm
d
baad p p + dx — d m
аа -
d p p + d x - d m
b
For another Inſtance let us take Queſtion 33.
=10
Queſtion 40. There is a certain Number, which being multiplied
by To, if this Product is divided by 2, to this Quotient adding 19,
and ſubſtracting 99 from that Sum, the Remainder will be equal
to the Number fought.
Let a the Number fought, b = 10, d= 2, m = 19,
2 = 99.
Then
To reduce an Equation, &c.
141
Iba
Then by the Queſtion
I
+m—z=a
Regiſter
d
By tranfpofing z from the
firft Equation
I 1+%
By tranfpofing m from the
fecond Equation
-
Į
2-m
Multiplying the third E- 3xd
quation by d
By tranfpofing da from
the fourth Equation
Dividing the fifth Equa-
tion by b―d, the two
Co-efficients of a, by
Art. 52.
ba
2 +m=a+z
d
ba
3=a+xm
d
4 bada+dz-dm
4-da 5 ba-da— dz—dm
dz-dm
5÷b-d6a=
=20
b-d
From theſe two Examples we may obferve, that to register
any Operation, is only to put down the Figure which ftands in
the Column against that Equation, from which we intend
to raiſe the next Equation, and after that the Sign of either
Addition, Subſtraction, Multiplication, Divifion, Involution and
Evolution, according as the Cafe requires, and after this the
Quantity which fuffers the Alteration.
Thus at Queſtion 39, the firft Equation being raiſed or in-
volved to the fecond Power produces the fecond Equation, there-
fore, I fay in the Regifter 12, that is, the firſt Equation
involved to the fecond Power gives the fecond Equation, and
in the fame Operation.
Becauſe the fourth Equation is produced from the third, by
tranſpoſing m with the Sign, therefore in the Register I fay
3-m, that is, the third Equation-m, produces the fourth
Equation. And,
As the fifth Equation is produced from the fourth by multi-
plying by d, therefore I fay in the Regifter 4 × d, that is, the
fourth Equation multiplied by d, produces the fifth Equation.
And,
As the fixth Equation is produced from the fifth by dividing
by b, therefore, I fay in the Register 5b, that is, the fifth
Equation divided by b, produces the fixth Equation. And,
As the feventh Equation is produced from the fixth by ex-
tracting the fquare Root, I fay in the Regifter 6 w 2, that is,
the fixth Equation having the Square Root extracted, produces
the feventh Equation.
2
Whence,
142
ALGE BRA.
Whence, as I faid above, to regiſter any Operation, is only
to put down whether it is the firft, fecond, third, fourth, or any
other Equation, which fuffers the Alteration, and from which
the new Equation is raiſed; and after that Figure to exprets in
Characters, or Signs, the Alteration that is then made to gain
the new Equation.
The Method of refolving Questions that
contain two Equations, and two un-
known Quantities.
55.T'
HE foregoing Questions requiring only one unknown
Number to be found, their Conditions were all
expreffed in one Equation, which Equation being reduced by
the Rules already delivered, the Queftion was anſwered.
But if the Queſtion requires two unknown Quantities to be
found, then there are generally raiſed two Equations from the
Queftion, each of them including both the unknown Quan-
tities; which may be reſolved by this
RULE.
Find what the fame unknown Quantity is equal to in each
of the two Equations, which arife from the Conditions of the
Queftion, then make theſe two Equations equal to one another,
and in this Equation there will be but one unknown Quantity,
confequently if this Equation is reduced by the Rules already
given at Art. 46 to 53. we ſhall find what this unknown
Quantity is.
And to find the Value of an unknown Quantity in any
Equation, is only to find what it is equal to, therefore all the
other Quantities, whether known or unknown, muſt be carried
to the other Side of the Equation by the Directions at Art. 46
to 53. and then it will appear to what this unknown Quantity
is equal, as this makes one Side of the Equation, the other Side
of the Equation being known Quantities, with the other un-
known Number or Quantity fought.
Finding
The Method of refolving Questions, &c.
143
Finding the Value of the fame unknown Quantity in each
of the given Equations, and making thefe two Equations equal
to one another, which clears the Work of that unknown
Quantity whofe Value was found, is called exterminating an
unknown Quantity.
Queſtion 41. To find two Numbers, if the greater is added
to the leffer, the Sum is 262.
But if from the greater you ſubſtract the leffer, the Remainder
is 144.
=
Let a the greater Number, and the leffer Number
fought, b = 262, x = 144.
Now the Sum of the two Num-
bers in Algebra is a + e, which
is equal to 262, or b, hence
we have
And the leffer Number being
ſubſtracted from the greater is
ae, which is equal to 144,
or x, hence we have
144, S
Ia+e=b
By the
Queſtion
2 a
The Conditions of the Queftion being now expreffed, there
appears in the above two Equations, two unknown Quantities.
a and e, therefore according to the Rule find what a is equal to
in the first Equation, by tranfpofing e.
I -
Е
31α=
Now find what a is equal to
in the fecond Equation, by
tranfpofing e
2 + e
4a=x+e
Therefore make the third and fourth Equations equal to one
another, for they are both equal to the fame Quantity a, which
exterminates that unknown Quantity: this Step is registered by
placing the 3 and 4 with a Point between them as in the Work,
which expreffes that the fifth Equation is from comparing the
third and ourth Equation together
3·4 151x+e=bme
The
1
144
ALGEBRA.
The unknown Quantity e being on both Sides of the Equa-
tion, bring it on one Side of the Equation, by Art. 52.
5 + e| 6 | x + 2 e = b
6. x 7 2e=b
7÷28 e=
b- x
2
x
Here it appears that e, or the leffer Number fought, is equal
to b, or 262, ſubftracting from it x, or 144, and dividing the
Remainder by 2.
When any Equation is divided by an abfolute Number, as
the feventh Equation is divided by 2, place them in the Regiſter
as uſual, but draw a Line over the 2 to diſtinguiſh that it is an
abfolute Number by which you divide, and not by the ſecond
Equation in the Work.
Now b=262
144
2) 118
59e, the leffer of the two Numbers fought.
It being now known what e is in Numbers, we may find a
by the third or fourth Equation, and by the third Equation we
have a — b—e.
But b 262
e
59
203 = a, the greater of the two Numbers fought.
Whence 203 and 59 are the two Numbers required in the
Queſtion, and is thus proved from their Conditions.
The greater Number is 203
The leffer Number is
59
262 Sum
203
59
144 Remains.
Queſtion 42. Two Men difcourfing of their Money, founa
that if the Number of Shillings each had were added together, the
Sum would be 38.
But if from him that had the greater Number of Shillings,
there be fubftracted twice the Number of Shillings the other Perfon
had, there would remain 5. How many had each Man?
Let
The Method of refolving Questions, &c. 145
J
Let a
the greater Number of Shillings, e the leffer
Number of Shillings, b= 38, x=5.
And becauſe the Sum of their
Shillings, or a+e, was 38, or
b, hence
And twice the leffer Number
being taken from the greater,
or a
*, hence
2 e, was equal to 5, or
Now to find the Value of a in
the firſt Equation, tranſpoſe e.
I
a
+e=b
By the
Queftion
21á·
Ze=x
I
e
3a=b
a = be
And to find the Value of a in the
fecond Equation, tranſpoſe 2 e.
2+2e4a= x + 2e
Make the third and fourth Equations equal to one another,
becauſe they are both equal to the fame Quantity a, and regifter
it as directed in the laſt Queſtion; and this exterminates the
unknown Quantity a.
3.4151x+2ebc
The unknown Quantity e being on both Sides of the Equa
tion, bring it on one Side of the Equation, by Art. §2.
5 + e 6 x + 3 e = b
6
x73e=b
b- X
b.
3
7 ÷ 38 e =
Hence the Queſtion is anfwered, for b 38
5
3 33
I e, the leffer
11 = e,
Number of Shillings.
And as e is now known, we may find what a is by the third
or fourth Equation; taking the fourth Equation, we have
U
་ ་ ་
146
ALGEBRA.
* = 5
2e22
27=a, the greater Number of Shillings.
PROOF.
The greater Num-27
ber of Shillings
The Leffer Num-}
ber of Shillings
II
Sum 38
27
Twice the leffer
Number of Shillings 22
Remains 5
Queſtion 43. Two Men laying a Wager concerning the Num-
ber of Sheep in two Droves, as they could not decide it, appealed
to a third Perfon, who told them, that if 31 was added to the
Number of Sheep in the greatest Drove, that Sum would be equal
to twice the Number of Sheep in the leaft Drove.
But if they added 44 to the Number of Sheep in the least
Drove, that Sum would be as many as were in the greateft Drove,
and defired they would now find the Number of Sheep in each
Drove.
Let a the Number of Sheep in the greateſt Drove, e the
Number of Sheep in the leaft Drove, x = 31, d=44.
Now the Number of Sheep in
the greateſt Drove being added
to 31, is equal to twice the
Number of Sheep in the leffer
Drove, hence
Ia+x=2e\
1
By the
Queftion
2e + d
I - x 3α== 2 e
а
2e-
***
And the Number of Sheep in the
leaft Drove when added to
44, being equal to the Num-
ber of Sheep in the greateſt
Drove, we have
Now by the third Equation, a is equal to 2 e-x, and by the
fecond Equation, a is equal to ed, therefore make theſe
Equations equal to one another, for they are both equal to the
fame Quantity a, which exterminates a, as before.
2.3
The Method of refolving Questions, &c.
147
d44
x=31
2.3
4- e 5
5+x
4 2e-x=e+d
e
6
* d
e=d+x
75e, the Number of Sheep in the leaft Drove.
Then having found e, we may find a by the ſecond Equation.
e = 75
d = 44
1194, the Number of Sheep in the greateſt Drove.
PROOF.
119
317
150 which is twice the Number of Sheep in the leaſt Drove.
75
44
119 which is the Number of Sheep in the greateſt Drove.
Queſtion 44. Two Gentlemen who had fold their Eftates, by
comparing what each Eftate was fold for, found, that twice the
Sum of what both the Estates was fold for was 11468 Pounds:
And if what the leaft Eftate was fold for, be ſubſtracted from
what the greatest Eftate was fold for, there will remain 1408
Pounds. For how much was each Eftate fold?
Let a
for, e
the Number of Pounds the greateſt Eftate was fold
the Number of Pounds the leaſt Eſtate was fold for,
b=11468, x=1408,
By the firft Condition
By the fecond Condition.
Find the Value of a, in the
firft Equation.
I
7 2
2a+20=6
a
e = x
Į - 2 e 3
2a=b-ze
32
4
a →
b-20
~
U 2
Now
148
ALGEBRA.
Now find the Value of a, in the fecond Equation,
2+e1510= x + e
Make the fourth and fifth Equations equal to one another,
becauſe they are both equal to the fame Quantity a, and
therefore must be equal to one another, by which a will be
exterminated.
?
b-ze * *Here we have
2
4 · 5
6
x+e=
6 x 2
7
7+ 2e
8
8
2X 9
2x+2e=b2e
2x+4e=b
4e=b 2x
b.
2 x
94 IO e
4
only to find what
e is by the Rules
already delivered,
at Art. 46 to 53.
2163, the Pounds for
which the leaft Eftate was fold; and e being now known, then
11
By the fifth Step | II | a=x+e=3571, the Pounds for
which the greateſt Eftate was fold.
PROOF.
Now if a=3571, and e = 2163, then 2a +2 e = 11468,
and a- 1408.
=j
Queſtion 45. Two Gamefters, A and B, found, that if twice
the Number of Pounds won by A was added to what had been
won by B, the Sum was 48 Pounds:
And if what had been won by ▲ was added to three times what
bad been own by B, the Sum was 39 Pounds. What was the
Sum won by each Gamefter ?
=
=
Let a the Pounds won by A, the Pounds won by B,
b = 48, x = 39.
By the firſt Condi-}
tion
By the fecond Con- 1
dition
I 2a+e=b
2 a+3e=x
Find
The Method of refolving Questions, &c.
149
Find the Value of a, from the firſt Equation.
2a=b — e
b- e
I
e 3
32 4
a =
2
Now find the Value of a, from the ſecond Equation.
23e5a=x -3e
Make the fourth and fifth Equations equal to one another,
becauſe they are each equal to the fame Quantity, which Equa-
tion will exterminate a.
2
=*=3e*
b—e=2x-6e
b+5e=2x
b
e
4.5
6
6 × 2
7
7+6e
8
8-b 9
9-5 IO
e=
Then from the
fifth Equation S
4I
5e = 2x — b
2 x -b
5
* Here we have
only to find e by
the Rules already
delivered, at Art.
46 to 53.
= 6 Pounds, won by B.
a=x—3e = 21 Pounds, won by A.
PROOF.
2a+e=48
a3e=39
Queſtion 46. What are those two Numbers, that twice the
greater being added to three times the leffer, the Sum is 29:
And three times the greater being ſubſtracted from five times
the leffer, the Remainder is 4.
e
Let a = the greater Number, the leffer Number,
b = 29, m = 4.
By the firft Condi-}
tion
}
I
12
} |
By the fecond Con- Į
dition
2a+3e=b
2 | 56-3a=m
Find
!
1
150
ALGEBRA.
Find the Value of a, from the firſt Equation.
I
3 e
3
2 a
b- 3e
3÷2 4
b — ze
a
2
Now find the Value of a, from the fecond Equation; tranf-
pofe 3a, becauſe it has the negative Sign.
2 +3a
5
5e=m+3a
5 — m
6
5e—m = 3 a
Or
7
3a=5e- m
7+3 8
5e-m
3
Make the fourth and eighth Equations equal to one another,
for they are each equal to the fame Quantity a, and this un-
known Quantity will be exterminated.
4.
8
5 e-m
9
b-3e
3
10e-
2
2 m
9x210
3e
3
10 X 3
II
II
10 e
10 e 2 m =
38 — де
11 + 9 e
12
19e —
36
12
2m 1319e3b+2 m
3b+2m
13 19 14e =
5, the leffer Number.
19
By the fourth?
Equation
15
b-ze
=
7, the greater Number..
2
PROOF.
2a+3e29
5e3a= 4
Queſtion 47. Two Travellers, A and B, meeting on the Road,
found, that if the Number of Miles travelled by A was divided
by five, adding to this Quotient three times the Number of Miles
travelled by B, the Sum was 249:
2
But
The Method of refolving Questions, &c. 151
But if twice the Number of Miles travelled by A were added
to four times the Number of Miles travelled by B, the Sum was
Miles had each travelled?
540.
How many
Let the Number of Miles travelled by A, e the
Number of Miles travelled by B, x = 249, ≈ = 540.
By the firft Con- }
dition
By the fecond Con-
dition
I
+3e=x
5
22a + 4e = z
I × 5 3a +15e=5x
3—154 a = 5 x — 15 e
The Value of a being now found by the firſt Equation, find
its Value from the fecond Equation.
24e52a = z
526a=
—
46
-4e
2
Now make the fourth and fixth Equations equal to one ano-
ther as before, which exterminates a.
4
6
7
א
-48
=5*
=5x=15e
2
7x28x
4e=10x- 30 e
8 +30e9% + 26e = 10x
9 ≈1026e=10x-%
10 X
Z
10
÷
26 11e =
Then by the 4th
Equation
I I
12a = 5 x
·12a = 5 x — 15 e = 120,
26
75, the Miles tra-
(velled by B.
the Miles
(travelled by A.
PROOF.
5
+3e=249
2a + 4e = 540
The
152
ALGEBRA.
The Learner being now a little converfant with theſe Kind
of Queſtions, let the laft be repeated, and put Letters for all
the Numbers both known and unknown, and if he finds any
Difficulty in folving it, by comparing the two Operations, the
former may in fome Manner explain this; and to illuftrate it
the more, I have placed the Equations in the laft Work, againſt
their correſpondent Equations in the next Operation.
Queſtion 48. Two Travellers, A and B, meeting on the Road,
found, that if the Number of Miles travelled by A was divided
by 5, and adding to the Quotient 3 times the Miles travelled by B,
the Sum was 249:
But the Miles travelled by A being multiplied by 2, and added
to 4 times the Miles travelled by B, the Sum was 540. How
many Miles had each travelled?
Let a the Number of Miles travelled by A, e the
Number of Miles travelled by B, × = 249, ≈ = 540, as before,
but now put d 5, m = 3, 9 = 2, p = 4.
By the firft Con-}
dition
}\
By the fecond Con-}
dition
3-
a
a
I +me=x, that is, ²+3e=x
d
5
29a+pe=x, that is, 2a +4e=z
1 x d 3a+dme = dx, that is, a +15e=5x
-dme 4 la=dx dme, that is, a=5x-15e
Having found the Value of a from the first Equation, find
its Value from the fecond Equation.
2- pe 1 5 19 a
519a = z z-pe,
5- 9 6 a =
>
pe, that is, 2 a - Z
x-pe that is, a =
4 e
Z
4 e
9
2
Now make the fourth and fixth Equations equal to one ano-
ther, for they are both equal to the fame Quantity a, which
exterminates that unknown Quantity.
Z
•
6
4
-pe
Z
7
dx
dme, that is,
40
9
2
.5x- 15e
7x98x-pe=dqx-dme q, that is, z4e
= 10X 30e
8 + d meq
The Method of refolving Questions, &c.
153
8+dmeq 9dmeg+z-pe=dqx, that is, 26 e
+ z = 10x
9-211dmeg-pe=dqx-z,
= 10%- Z.
that is, 26 e
The unknown Quantity e being in two Terms, therefore
divide by both the Co-efficients of e,
as at Art. 52.
10÷d m
9
-pl:
dq x— Z
€ 75, that is, e=
d mq - p
IO X
26
by B.
א
Z
=75, the Miles travelled
And it being found that e is 75, we may find a by the fourth
or fixth Equation to be 120.
And now for the future we ſhall put Letters for the Numbers
that are known, as well as for thoſe that are unknown.
Queſtion 49. There are two Armies ready to engage; if the
Number of Soldiers in both Armies are added together, and that
Sum multiplied by 4, the Product is 84440:
But if the Number of Men in the greatest Army be multiplied
by 2, and added to the Product of the Number of Men in the
leffer Army multiplied by 3, the Sum is 52219. To find the
Number of Men in each Army?
Let a
the Number of Men in the greateſt Army, ethe
Number of Men in the leffer Army, d= 4, m = 84440, z = 2,
x=
3, b
52219.
By the firft Con-}
dition
By the fecond
Condition
fecond}
|: 1
I
|da+de=
e-m
2
| -
za+xe=b
Find the Value of a, in the first Equation.
I
de
3
da = m
-de
771
de
3÷d
4
a=
d
Now find the Value of a from the fecond Equation,
{
X
2-e
154
ALGEBRA.
2 −x e
5-
za=b-x e
b
xe
5÷2 6
a =
Z
Make the fourth and fixth Equations equal to one another
to exterminate a.
m de
b-xe
4.6 7
d
Z
d b - dx c
7xd 8
m
de =
Z
8xx 9
Z m
z de db-dxe
Now in this Equation e being on both Sides, find which of
its Co-efficients dx or zd is the greateft. zd is 8, but dx is
12, therefore tranfpofe dxe, that the unknown Quantity,
with the greateſt Co-efficient, may have the affirmative Sign,
as at Art. 52.
9+ dxe 10
dxe + zm
zde = db
II
Z m II
d
xe
zde = db
Z m
db- Z m
11 ÷ d x − z d 12
e=
= 9999
dx - zd
b
xe
quation
By the fixth E-}
a
13
Z
=III, the Number
of Men in the greateſt
Army.
Dividing the eleventh Equation by dx-zd, the two Co-
efficients of e, as at Art. 52. gives the twelfth Equation.
PROOF.
4a+4e84440
2a+3e= 52219
Queſtion 50. A Gentleman bought a Pair of Horfes for his
Coach, his Son having learnt Algebra, the Father propofed for
him to determine the Price of each Horfe from faying,
That if the Pounds both Horfes coft were multiplied by 4,
and this Product divided by 8, the Quotient was 20:
But
The Method of refolving Queſtions, &c. 155
But if the Pounds the best Horſe coft were multiplied by 3, and
this Product added to 5 times the Pounds the worst Horje coft,
the Sum was 158 Pounds. Now what was the Price of each
Horje?
་
Let a the Pounds the beſt Horfe coft, e the Pounds
the worſt Horſe coſt, b=4, d= 8, m = 20, p = 3, x = 5,
2 = 158.
By the first Con-1
dition
By the fecondĮ
Condition
-}|
ba+be
I
=m
d
S
2 pat xe 2
ba
dm - be
1 x d 3 ba+be=dm
3-be 4
2.
4÷b5a=
dm - be
6
хе pa = z
6÷p7a=
b
Z- xe
P
x e
To exterminate a
d m
be
Z- xe
5.78
b
P
d m
pbe
8xP9
Z-xe
b
9xb10pdm
pb e = b z — b x e
pbe = b z
pd m
10+ bxe 11 b xe + p d m
pdm 12 b x e · p b e = b z
pd m
bz
12÷bx-pb13
e-
b x
рь
By the ſeventh
Equation
Z
xe
14a=
א
Р
PROOF.
19 Pounds, the Price
of the woft Horſe.
21 Pounds, the Price of
the beſt Horſe.
4a+4e
=.20.
8
3a +5e=158.
Queſtion 51. Two young Gentlemen, who had ftudiet Numbers,
not agreeing about their Age, referred the Difpute to their Father,
who fmiling told them, that if the Age of the eldest was divided
X 2
by
156
ALGEBRA.
by 2, to which Quotient adding 4 times the Age of the youngest,
and extracting the fquare Root of this Sum, it will be 10:
But if the Age of the eldest was multiplied by 3, and added
to the Age of the youngest multiplied by 5, this Sum will be 201.
To find the Age of each Perfon?
Let a
b
= 2, d
the Age of the elder, e = the Age of the younger,
m = 10, p
= 4,
Z
= 4, m = 10, p = 3, 2 = 5, r = 201.
By the firft Con-
dition
By the fecond Į
Condition
I
+de = m
2
pa+ze = r
I
Becauſe in the firft Equation, a the unknown Quantity, is
under the radical Sign, therefore fquare both Sides of the
Equation, as at Art. 49. The 12 in the Register fignifies
that the firſt Equation being involved or raiſed to the fecond
Power or Square makes the third Equation, for is the Sign
of Involution.
I &
a
23 +de=mm
3-de4
Ъ
a
m m
de
b
b de
4xb5a= bm m
2 - ze 6 pa=r—ze
j Ze
6÷p7a=
P
5.78
Now to exterminate a
r - Z l
8 xp9r
P
ze = p b mm
9 + pb de 10 pbde+r Ze
ΙΟ -r11pbd e
11÷pbd-212e=
= bm m
b de
pbde
pbm m
Ze
p b m m
p b m m
p b d -
Z
By the feventh}
Ze
13a=
=21, the Age of the
(youngeſt.
= 32, the Age of the
P
(eldeſt.
PROOF.
The Method of refolving Questions, &c. 157
PROOF.
√²+40= 10.
4e
2
3a+5e=201.
Queſtion 52. Two Tradefmen, A and B, comparing their
Gains, found, that if the Pounds gained by A were multiplied by
2, to which adding 3 times the Pounds gained by B, the fquare
Root of this Sum was 11 Pounds :
But if 6 times the Pounds gained by B, were added to the
Quotient of the Pounds gained by A divided by 10, this Sum
was 47 Pounds. To find the Gains of each Tradefman?
Let a the Pounds gained by A, e
the Pounds gained
by B, b=2, d= 3, n
3, n = 11, p = 6, z = 10, x = 47.
By the firft Con-
dition
I
√ ba+de = n
By the ſecond
Condition
a
2
pet
Z
In the first Equation the unknown Quantity a being under
the radical Sign, fquare both Sides of the Equation as in the
laft Queftion.
1 2 3 ba+denn
4÷b5a=
2 X Z
zpe+a=xx
3
3-
de 4 ba = nn
de
n n
de
b
6
6—zpe
5.7 8
zpe
=2x
zx - z pe
b
=
zpe 7a=ZX
n n de
8xb9nn de bzxbxpe
9+bzpe 10 bzpe+nn-de = b z x
10 — n n ] 1 1 } b x pe — de e = b xx nn
II -
158
ALGEBRA.
11-bzp-d 12
By the feventh 13
|
Step
!
b zx
n n
e=
bap
Z
d
= 7 Pounds gained
(by B.
(by A.
a = zx —zpe = 50 Pounds gained
PROOF.
√2a+3e=11.
be+
a
= 47•
ΙΟ
Queftion 53. Two Perfons, A and B, owe fuch a Sum of
Money, that if the Pounds A owes are divided by 5, to which
Quotient adding 4 times the Pounds В owes, and extract the
fquare Root of this Sum, it will be 6 Pounds:
But if from 3 times the Pounds A owes, is fubftracted 50
times the Pounds В owes, and extract the ſquare Root of this
Remainder, it will be 10 Pounds. What did each Perfon owe?
Let a
the Pounds A owes, e the Pounds B owes,
3, x = 50, 2 = 10.
m = 5, n = 4, d = 6, p = 3, x
√²+ne=d
By the firft Con-
dition
I
m
2
a-
By the fecond Į
Condition
"}
To find the Value of a in the first Equation, raiſe it to the
fecond Power as in the laſt Queſtion.
a.
1 2 3
+ne = d d
m
a
3 - ne
4
=dd
d dne
m
4 x m
5
a = md d -mne
To find the Value of a in the fecond Equation, raiſe it to
the ſecond Power as before.
2
22
The Method of refolving Questions, &c.
159
22 6
6+xe 7
pase=22
pa=zz+x e
8 a=
+
22 xe
P
Now make the fifth and eighth Equations equal to one ano-
ther to exterminate a.
5.8
9
224 xe = m d d — m n e
P
9xp 10 2x+xe=pmdd-pmne
10+ pmne 11 pm ne+zz+xe=pmd d
zx12pmne + xe=pmdd
p m d d Z Z
I I
12 ÷ pm n + 13e =
Z Z
4 Pounds, the Debt
(of B.
pm n + x
Then by the
eighth Step
}
22+xe
14a=
100 Pounds, the Debt
P
(of A.
PROOF.
a
12+4e=6.
5
√3a-50e=10.
Queſtion 54. Two Men, A and B, going to Market with Eggs,
if the Number of Eggs that A had were multiplied by 6, to which
adding 100, and dividing the Sum by the Number of Eggs that
B had, the Quotient is 16:
And if from 9 times the Number of Eggs A had, is fub-
ftracted 4 times the Number of Eggs B had, there remains 350.
How many Eggs had each Perfon?
Let a the Number of Eggs A had,
Eggs B had, d = 6, m = 100, p = 16,
the Number of
b = 9, x = 4,
z = 35.2.
da + m
a+m
I
៩
By the Queftion.
e=2
2 ba
Ixe 3 da+m=pe
3 → m
160
ALGEBRA.
3- m
4
d
5 a
ре
d
4 da = pe—m
a=pe-
m
2 +
2 - xe6ba=2xe
-
6÷÷ ÷b
x + x e
a
b
Make the fifth and feventh Equations equal to one another
to exterminate a.
1
5.718
d
m
8 x d 9pe - m
b
dz - dxe
b
9xb10bpe-bm=dz + dxe
10- dxe11bpe-dxe
b m = d z
11+bm 12 bpe-dxe=dz +bm
dz + b m
1 2 → bp — d x 13 e
·bpdx 13e =
bp
dx
=25, the Number
(of Eggs B had.
Step
By the feventh }
Z + xe
14a=
= 50,
b
50, the Number of
(Eggs A had.
PROOF.
6a+ Ico
9 a
e
= 16.
4 e = 350.
Queſtion 55. Two Perfons, A and B, lofing at the Gaming-
Table, were asked how much they lost, to which A replied, that
if the Number of Pounds I left be multiplied by 3. and add
100 to the Product, if this Sum is divided by the Number of
Pounds B loft, the Square Root of this Quotient will be 10
Pounds:
But if the Pounds B loft be multiplied by 250, from which
Product fubftracting 600, and dividing the Remainder by the
Pounds A loft, the fquare Root of this Quotient will be 2 Pounds.
How much had each Perfon loft ?
Let
The Method of refolving Questions, &c.
161.
Let a the Pounds A loft, e = the Pounds B loft, d= 3,
m = 100, n=10, x = 250, z=600, b = 2.
*
da + m
I
=n ท
e
By the Queſtion.
xe
Z
2
= b
a
To find the Value of a in the firſt Equation, raiſe it to the
fecond Power by Art. 49.
da+
da + m
e
= n n
nn
damenn
1 & 2
3
3 x e
4
4-
m
5
m
5
| 6
en n
m
a =
d
da = en n
To find the Value of a in the fecond Equation, raiſe it to
the ſecond Power by Art. 49.
xe
א
22 7
Z
= b b
a
уха 8
хе- z = abb
xe
8 b b
9
- Z
b b
= a
Make the fixth and ninth Equations equal to one another,
to exterminate a.
6.9 10
en n
m
xe.
Z
d
bb
10 x d
d xe — dz
II
ennm=
b b
I I x b b
12
b b n n e — b b m = d x e — dz
Becauſe dx, one Co-efficient of e, is greater than bbnn, the
other Co-efficient of e, therefore tranfpofe b b nne, by Art. 52.
Y
12
162
ALGEBRA.
14 + dz
15÷d x-b bnn
16
bbm
12-bbnne | 13 | bbm dxe-dz- b b n né
Or 14 dxe-dz-bbnne
15 dxe-bbnne-dz-bb m
dz-bbm
d x - b b n n
4, the Pounds B
(loft.
By the ninth Step 17
x e
2
a
100, the Pounds A
b b
PROOF.
3a+100
= 10.
e
250e600
a
=2.
1
(loft.
1
Queſtion 56. In the right-angled Triangle ABC, there is
given the Bafe AB=4, and the Difference between the Hypo-
thenufe AC and Perpendicular BC= 2. To find the Hypo-
thenufe AC and Perpendicular BC?
Let AC = a, BC = e, AB = b = 4, m = 2.
C
Having put Letters for the three
Sides of the Triangle, and amongſt
theſe there being two unknown
Quantities a and e, therefore we
muft raiſe two Equations either from
the Properties of the Figure, or from
the Conditions of the Queſtion.
And in the Solution of Geome-
trical Queſtions, I would recom-
mend it to the Learner, that after
all the Parts of the Figure which
are neceffary to the Solution of the Queftion are expreffed by
Letters, to obferve how many of them' are unknown, for gene-
rally fo many different Equations are raiſed from the Properties.
of the Figure, or the Conditions of the Queftion; afterwards
the Wark is regulated by the Rules already given.
A
Now from the Property of the Figure, the Square of the
Hypothenufe AC, or aa, is equal to the Square of the Bafe
AB, or bb, added to the Square of the Perpendicular BC,
or ee, by 47 e 1.
That
The Method of refolving Questions, &c.
163
That is I aa = bb + ee from the Property of
the Figure by 47 e 1.
Becauſe by the Queftion, the Difference between the Hypothe-
nufe AC, or a, and Perpendicular BC, or e, is = 2, orm.
Hence
em by the Conditions of the
amen
121ª Queſtion.
Having raiſed the two Equations, proceed as in the former
Examples, that is, firft find the Value of a in the firft Equation,
by the Extraction of Roots, as at Art. 50..
|3|a=
I w 2 | 3 |α = √bb + ee
Now find the Value of a, in the fecond Equation.
2+e|4|a= m + e
Make the third and fourth Equations equal to one another,
to exterminate a.
3·4151m+e=√bb + ce
Becauſe è the unknown Quantity is under the radical Sign,
and there being no other Quantities on that Side of the
Equation, but what are under the radical Sign, therefore fquare
both Sides of the Equation, as at Art. 49.
52 6
mm + 2 mee + e − b b + ce
mm + 2me=
2 me = b b
6-ee 7
7 —mm 8
b b
8÷2m 9
e =
- mm
2 m
b b
m m
=3, the Perpendicu-
(lar BC.
By the fourth Step 10
10a=m+e=5, the Hypothenuſe AC.
To prove theſe are the three Sides of a right-angled Tri-
angle, fquare the Hypothenufe 5, and that will be equal to the
Square of the Bafe 4, added to the Square of the Perpendicular
33
3 for this is the celebrated Property of the right-angled
Triangle to have the Square of the Hypothenufe equal to the
Sum of the Squares of the Bafe and Perpendicular.
Y 2
Queſtion
164
ALGEBRA.
:
A
Then I
C
Queſtion 57. In the right-angled
Triangle ABC, given the Perpen-
dicular BC= 3, and the Difference
between the Hypothenufe AC, and
Baſe AB = 1. To find the Hypo-
thenufe AC, and Bafe BA?
=
=
Let AC a, BC 3b,
B AB e, x = 1.
AB=e,
=
aa=bb+ee, by the Property of the
Figure, as in the laft Queftion.
ex by the Queſtion..
And 2
a
There being as many Equations raiſed from the Property of
the Figure, and the Conditions of the Queftion, as there are
unknown Quantities, the Work proceeds upon the fame general
Rules, thus
6
I w 2
2+ e
•
4
3
5 & 2
7-
ee
X X
82*
4
3+ 5O Z∞
6
716
a = √ bb + e e
a = x + e
te
x+e=√bb+ce
9
e
By the fourth Step 10
A
D
tee
xx+2xe + ee=bb + ee
xx+2xe=bb
2xebb XX
bb - x x
2 x
= 4, the Bafe A B.
a=x+e=5, the Hypothenufe AC.
୯
,<
Queſtion 58. In the right-angled
Triangle ABC, there is given the
Hypothenufe AC = 5, the Bafe AB
= 4, and the Perpendicular BC=3,
to find the Perpendicular BD, let
fall from the Angle B, upon the Hy-
pothenufe AC.
Let ACb5, AB=m=4,
BBC 3, DC-a, AD=e.
= x
The
The Method of refolving Questions, &c. 165
The Queſtion requiring that we find BD, if we find CD
we can anſwer the Queftion, for the Triangle BDC being a
right-angled Triangle, BD being perpendicular to A C, con-
fequently BC being known, and by finding DC, we fhall
afterwards eafily find DB, by the common Property of the
Triangle.
It is exactly the fame, if we find AD, for the Triangle
ADB is right-angled, and A B is given by the Queftion.
Now BD being a Perpendicular common to the two Tri-
angles ABD, and BDC, let BDp, then from the right-
angled Triangle ABD, we have mm eepp, and by the
right-angled Triangle CBD, we have xx-a app, from the
fame Reaſoning as in the two laſt Queſtions.
1
Confequently I
m2 m
4-
- e e
and xx
ee
ee—xx — aa, for both m m -----
a a, are equal to the fame
Quantity pp, and therefore equal to
one another.
And 2a+e=
a+eb, that is, AD+DC-AC
by the Figure.
To find the Value of a in the firft Equa-
tion.
aa+mm-ee = xx
aa + mm = x x + ee
6│a = √xx + e e
m m
m m
Itaa
3 + ee
3
4
m m
5
aa=xx+ee
5 US
WJJ 2
2
e 7a = b
b-
Now find the Value of a in the fecond
Equation.
e
6.7 8√ xx tee mm = b
8 29xx Tee -
0
9-
eel IO XX
10 + 2 be
mm = bb
bb-2be Tee
- b b — 2 be
mm = b b
mm
bb
b b + mm
1 1/2 be + xx —-
11+ mm 12|2be + xx=
I 2 - -xx13
I 2 be l b + m m
xx
1326 14 e
b b + m m —xx
e =
= 3.2=AD.
26
Having found AD to be 3.2 it will be eafy to find DB
by what was faid above. Thus,
3
4
16 the
166
ALGEBRA.
16 the Square of AB.
10.24 the Square of AD.
5.76 (2.4 = DB, the Perpendicular required.
4
44) 176
176
A
D
B
Queſtion 59. In the ob-
lique Triangle ADB, there
is given the Side AB = 15,
the Side BD=12, and the
Side AD=6, to find the
Perpendicular BC falling
without the Triangle from the
Angle B, on the Side AD,
continued.
This Queſtion will be
anſwered from finding DC,
for the Triangle BCD be-
ing right-angled, and DB
being known from finding
DC, we may then find
BC from the common Pro-
perty of the Triangle DB C, as in the laſt Queſtion.
=
Let A B b 15, AD=m=6, DB = x 12,
= =
DC a, then ACAD+DC=m+a, BC = e.
=
Becauſe the Triangle ABC is right-angled, therefore if from
the Square of AB, or bb, we fubftract the Square of AC, or
mm + 2ma+aa, the Remainder is equal to the Square of
CB, or e e.
Therefore | 1 | bb — m m
2 ma
a a = el.
Becauſe the Triangle DBC is right-angled, by the fame
Reafoning we have
Again | 2 | xx
a a = e e.
And
The Method of refolving Questions, &c. 167
!
And as the first and fecond Equations are each
ee, there-
fore make them equal to one another, which exterminates every
Power of in thofe Equations.
I. 2 3 **
a a = bbm m
m m 2 ma
b b - m m
3+ aa 4 x x = b b
* bb.
4+2ma 52 ma+xx=b b
5-xx 62m a = b b —mm
xx
2ma-aa
b b
62m 7a=
mm-
x x
3.75= DC.
2 m
And from hence we may find BC as was faid above, thus
BD, or x = 12
12
144
DC, or a 3.75
3.75
1875
2625
1125
14.0625
144.
14.0625
129.9375 (11.39 = BC, the Perpendicular required.
I
21) 29
21
223) 893
669
2269) 22475
20421
2054
Of
1
[168]
1
56.
Of Quadratic EQUATIONS.
W
HEN all the known Quantities are on one Side of
the Equation, and thofe Quantities only on the other
Side which have fome Power of the unknown Quantity; then
if the unknown Quantity appears to be to the fecond Power or
Square in one Term, and to the firft Power only in another.
Term; or if in one Term, its Power or Heighth is double its
Power or Heighth in another Term, and there is no other Power
of the unknown Quantity in the Equation, thefe Equations are
called Quadratic, as in the following Queſtions.
Queſtion 60. Two Men had fuch a Number of Shillings, that
the leffer being fubftracted from the greater, there remains 10:
But the Number of Shillings one Man had multiplied by the
Number of Shillings the other Man had, the Product is 75. To
find each Man's Number of Shillings ?
Let a
had,
=
the greater Number of Shillings one of the Men
the leffer Number of Shillings the other Man had,
b = 10, m = 75•
Then
And
Ι
I te
3
a e = m
a =
b+e
m
1 2
I la—e=b
}
By the Queſtion,
2 e
4
a =
e
3.45
5 × e
6
e + b =
m
e
ee+be=m
From comparing the fixth Equation with what is ſaid above,
it appears to be Quadratic, for one Quantity is e e, or e to the
fecond Power, and in the other Quantity it is only e, or e to
the first Power.
!
And
Of Quadratic EQUATION s.
169
And to refolve this Equation, take b the Co-efficient of e
to the firſt Power, and divide it by 2, the Quotient is
fquare or multiply by itſelf, and the Product is
to both Sides of the Equation, thus
The
b b
b
גי
which
2
b b
which add
4
600 | 7 | ee + be + 60 = m + b
ㅣ ​기
​4
4
b
in the Regiſter fignifies, that the fixth Step is made
a Square at the feventh Step, or the Square is compleated.
b b
Now if we compare the Side of the Equation ee+be+ —,
4
with fome of the Examples at Art. 34. we ſhall find it to be
a rational Quantity, or a Square, therefore extract the ſquare
Root of both Sides of the Equation :
b
b b
7 uu 2
8
et
✓m+.
2
4
b
b b
b
8-
9
e = √m +
2
4
2
In Numbers.
m = 75
b b
25
4
100 the ſquare Root of which is 10
b
5=
2
5e, the Number
of Shillings one of the Men had.
M
Then by the fourth Step a =
15, the Number of
.
Shillings the other Man had.
Z
PROOF.
1
170
ALGEBRA.
PROOF.
a-e = IO
a e = 75
Queſtion 61. There are two Numbers, if the Square of the
leffer is taken from the greater, there remains 36:
But the greater being added to 6 times the leffer, the Sum is
148. What are the two Numbers ?
= -
Let a the greater Number, e the leffer Number,
b = 36, m = 6, x = 148.
Then I
a
And 2
I+ee
me 4
-ee- - b
a+me = x
}By the Queſtion.
3
a
b tee
a = x-me
b+ee = x → me
2
3.4 5
5-b
6
e e = x
6 + me
7
me
b
ee+me = x ·b
The unknown Quantities being brought on one Side of
the Equation, the Equation appears to be Quadratic, by
Art. 56.
Now the Co-efficient of the firft Power of
e is m,
m
m m
which divided by 2 is
',
this ſquared is
and adding
2
4
m m
to both Sides of the Equation as in the laſt Queſtion,
4
we have
70미
​m m
m m
8 ee + me +
X ----
- b +
4
4
The 7 fignifies that the feventh Equation is made a com-
pleat Square, at the eighth Step.
And extracting the Roots of both Sides of the Equation, as
in the last Queftion,
ແ
8 w 2
Of Quadratic EQUATION s.
171
m
8 w 2
9
e+
2
=√x-
.b+
m m
4
m
9
2/2
m m
m
IO e=
X
6 +
= 8,
4
2
By the fourth Step 11
a=x
a
(the leffer Number.
-me=100, the greater Num-
PROOF.
ee = 36
a + 6e = 148
(ber.
Queſtion 62. In the Parallelogram ABCD, if from the
longeft Side AB multiplied by 3, is fubftracted the Square of
the shorteft Side BC, the Remainder will be 5:
But if the longest Side A B is added to 4 times the shorteft Side
BC, the Sum is 30. To find the Sides of the Parallelogram A B,
and BC?
A
D
B
C
AB, BC = e, d
Let a AB, BC:
= e, d = 3, m = 5, % = 4, x = 30.
I
1 2
da-ee m
a + ze=x
da = m + e e
te
I tee
3÷d
3
4
a
5
a
Z - Z e
m + e e
X
d
Z 3
Ze
} By the Queſtion.
4.5
172
ALGEBRA.
4.5
m + ee
6
d
73
8
6 x d
7-m
8 + dze 9
ze
m+ee = dx-dze
-dze- m
e e = d x
ee+dze=dx-m
Now the Equation appears to be Quadratic by Art. 56.
and the Co-efficient of e is dz, which divided by 2, is
dz
2
this ſquared is ddzz, which added to both Sides of
4
the Equation, as in the two laſt Examples, we have
9 0 0 | 10 | ee+ds
10/ee+dze+
g co
d d z z
d d z z
=dx-m+
4
4
And extracting the Roots of both Sides of the Equation, as
in the two laft Queſtions,
II
dz
d d z z
10 w 2 I I
e+
= √ dx
-m+
2
4
d d z z
dz
dz
2
I2
e = √dx.
m+
= 5
4
2
(= BC.
From the fifth Step 13
a = X — % € = 10 — A B.
PROOF.
3a-ee=5
a + 4e = 30
Queſtion 63. Two Gentlemen having had their Parks ſurveyed,
bad loft the Account, but remembered, that if the Number of
Acres in A's Park was added to the Number of Acres in B's
Park, the Sum was 110:
But if the Number of Acres in B's Park was multiplied by
80, from which Product fubftracting the Square of the Number of
Acres in A's Park, there remained 400. How many Acres was
there in each Park?
Let
Of Quadratic EQUATIONS.
173
Let a the Number of Acres in A's Park, e the Num-
ber of Acres in B's Park, b≈ 110, m
=110, m = 80, x = 400.
I
a + e = b
2
me
a a
x
}By the Queftion.
I
3
2 + a a
4
e = b → a
me = x + a a
x+aa
4 m
5
.
m
x+aa
3.5
6
-ba
m
6 x m
78
8 + ma
9
x + aa = m b
a a = m b ma
ma
X
a a + ma = m b X
Here the Equation appears Quadratic, and compleating the
Square as in the former Examples, we have
900
öl
10 | aa+ma+
m m
m m
= m b
m b − x +
4
4
And extracting the fquare Roots of both Sides of the Equa-
tion, as in the former Examples,
10 w 2
I I
m
2
m m
at √ mb
x+
4
m m
m
II
m
2
From the third}
Step
}
12
| 13
a = √ mb x +
= 60,
4
2
= b
(the Number of Acres in A's Park.
a=50, the Number of Acres
(in B's Park.
PROOF.
ate=110
80e-aa=400
The
174
ALGEBRA.
The Manner of fubftituting one Quantity for feveral
others explained.
57. But if, after the Work is prepared for having the Square
compleated, it appears that the firft Power of the unknown
Quantity is in more Terms than one, it will be more conve-
nient to fubftitute fome other Letter, for the Co-efficients of the
firft Power of the unknown Quantity, as in the following
Examples.
Queſtion 64. A Gentleman propofed to give his two Sons, A
and B, each an Eftate, on the Condition, they could tell him what
were their Rents, by knowing, that if the Square of the Rent
of the Eftate he intended to give A was added to the fame Rent
multiplied by 7, and the Sum added to the Rent of the Eftate he
intended to give B, when multiplied by 4, this Sum would be
4220 Pounds:
But if the Sum of the Rents of the two Eftates was divided by
10, the Quotient was II Pounds. What was the Rent of each
Eftate?
Let a
=
the Rent of the Eſtate A was to have, e the
Rent of the Eſtate B was to have, b = 7, m = 4, d = 4220,
p = 10, x = 11.
aa+ba + me = d
2
ate
P
By the Queſtion.
X
Thefe being the two Equations which arife from the Que-
ftion, and becauſe the Terms are more fimple that have the un-
known Quantity e, than thofe that have the unknown Quantity
a, it may be more convenient to find the Value of e, in each
of the two given Equations. This Caution the Learner may
obſerve for the future, to find the Value of that unknown
Quantity whofe Terms are the moft fimple in the given Equa-
tions; and thoſe may be taken for the more fimple, whofe Powers
are the loweſt in both the Equations that arife from the Que-
ftion; thus, if one of the unknown Quantities is only to the first
Power in both the given Equations, when the other unknown.
Quantity is to the fecond Power in one of the given Equations,
the Terms of the former may be faid to be more fimple, and
there-
2
Of Quadratic EQUATIONS.
175
therefore beſt to find the Value of that unknown Quantity:
The Reader will find this Method obferved in the following
Queſtions, and comparing their Work with what is faid may
make this Direction more intelligible.
I bal 31aa+me = d — ba
3-aa4me ==
d-ba
a a
d - ba
a a
4 ÷ m
5
e=
m
2 xp6a+e = px
6- a e
7 =p·x
5.78px
a
d — ba
a a
a
m
8 xm 9 mpx - ma=d-ba-a a
9aa10aa mpx
10 + ba 11
aa+
mad — ba
a a + ba+mp x ma-d
11 - mp x
mp x[12]aa+ba ma=d.
d—mp x
Here the Equation appears to be Quadratic, and the first
Power of a is in two Terms, viz. ba and ma, the two Co-
efficients being b and m, and connected by the Sign
But b and m, being known Quantities, therefore b m
4 = 3, now ſubſtitute, or put z = 3, or z = b
the laſt Equation is,
7
m, then
By Subftitution [13] aa+za = d — mpx; for by Sub-
ftitution zaba ma, and therefore in the room of b a — ma,
we uſe only z a. Now taking z for the Co-efficient of a, and
compleating the Square as before,
22
Z Z
£ 3 C
14 aa+za+
d-
mp x +
4
4
ZZ
14 un 215a+
2
√d—mp x +
a = √✓ d― mpx +
d-
15-2 16la
By the feventh }
Step
ZZ
4
4
א
Z
60,
2
the Rent of the Eftate which A was
to have.
e = p x
a = 50,
50, the Rent of the
(Eftate which B was to have.
PROOF.
176
ALGEBRA.
PROOF.
aa+7a+4e4220.
ate
=II.
ΙΟ
for after the
mz, then to have
It may be juft obferved to the Learner, that the Method of
Subſtitution is only to fave Trouble and Labour,
twelfth Step, if we had not ſubſtituted b
compleated the Square, we muſt have divided b
Co-efficients of a by 2, the Quotient of which is
ſquared is
bb — 2 bm + mm
·4
b
m the two
m
which
2
and this muſt have been added to
both Sides of the Equation, whereas by fubftituting b-m — %,
the Quantity to be added on both Sides of the Equation is
only 22.
Z Z
4
Queftion 65. A Draper bought a Parcel of Linen, and a
Parcel of Woollen Cloth, if the Square of the Pounds he gave for
the Linen Cloth be divided by 4, and to this Quotient there is
added the Pounds each Sort coft, the Sum is 1000 Pounds:
But if the Pounds the Linen coft is added to the Quotient of
the Pounds the Woollen caft, divided by 8, the Sum is 65 Pounds.
How much was given for-each Sort ?
Let a
-
the Pounds the Linen, coft, e=the Pounds the
Woollen coſt, b = 4, d = 1000, m² 8, x =
= 65.
a a
I
+a+e=d
b
e
By the Queftion.
2a +
X
m
a a
b
e = d
d
+e=d-a
ä a
b
3-
a3
a a
b
4
2 m
Of Quadratic EQUATION s.
177
2x m 5 ma+e=mx
5-ma
61e
e = m x
m a
4.67mx
7+
a a
b
8
3+99
ma = d — a
a a
b
аа
+mx
mad
d — a
b
a a
a
+a+mx
m a
d
b
a a
9-mx10
6
+a—ma=d.
m a = d — m x
10xb11aa+ba-bma-bd-bmx
=
Here the Equation appears Quadratic, and the firſt Power of
the unknown Quantity a, has two Co-efficients b and b m, both
which are known, but b-b m = 4 — 32
=432=- -28, therefore
as 28 is a negative Quantity, fubftitute z= 28, or
— z = b — m, then the laft Equation becomes,
ba-bm a
By Subftitution | 12| aa—ża= b d — bmx, for ba—
is a negative Quantity, bm being greater than b: And com-/
pleating the Square as before,
A
ZZ
c
12 C13|aa-
а
zat
4
2
for
X
2
2 | 2
Z Z
bd-bmx+
4
ชช
=+
by Art. 9.
4
And extracting the fquare Root as in the former Queftions,
2
Z
13 w 214 a
✓bd―bmx+
ชช
1
4
a
= √ bd ÷ bmx +
ZZ
:+
=60
2
(Pounds, the Linen coft. -
སྙིང་མཚན་
14 +
By the fixth E-
quation,
2
}
4
16 emx ma
= = 40 Pounds, the
(Woollen coft,
$
A a
PROOF
178
ALGEBRA.
PROOF.
Ra
+a+e=1000.
4
a+ = 65.
8
To refolve a Quadratic Equation when the Square of
the unknown Quantity has a Co-efficient.
58. But if the Square of the unknown Quantity has any
Co-efficient befides Unity, or 1, then before you begin to com-
pleat the Square, divide every Term in the Equation by that
Co-efficient, after which compleat the Square, and proceed as
before.
Queſtion 66. To find two Numbers, that the Square of the
greater being multiplied by 4, if this Product is added to 3 times
the leffer, the Sum ſhall be 1606:
But if 5 times the greater is added to 6 times the leffer, the
Sum fhall be 112.
Let a
b
= 4,
the greater Number,
d = 3, m = 1606, p = 5,
12
*
the leffer Number,
x = 6, z = 112.
}By the Queftion.
baa
baa + dem
patxe
2
I-baa
3
3÷d
de=m
m baa
4
e =
d
2-pa
ра
5
5-x
6
e=x-pa
z-pa
e =
x
א
4.6
7
Z
X
ра
m
-baa
==
d
x m
xba a
7 xx
8
1
Z ---
pa
d
8 x d
Of Quadratic EQUATIONS.
179
8 x d
9+xba a
9
IO
IO- dz II
dz — dpa = x m -x baa
xbaa+dzdpa=xm
x ba a dp a = xm
dz
The Equation appearing to be Quadratic, and all the known
Quantities, except thofe which contain the unknown one, being
on one side of the Equation, and the higheſt Power of the
unknown Quantity having a Co-efficient, divide by that Co-
efficient.
11÷x6|12| aadpa-xm-dz
I
|
x b
=
x b
To avoid the Trouble of dividing, the Co-efficient of a,
by 2, and fquaring the Quotient, and adding it to both Sides
of the Equation to compleat the Square, as in the former
Queſtions, fubftitute
dp
xb
.625 then,
xmed z
xb
rr
x m
d z
By Subftitution 13 aa-ra=
со 14 aa-rat -
1300
4
x b
+
Now extracting the ſquare Root as in the laſt Queſtion,
r
14 w 2
xm ----
15
a
15 +
12
16
By the fixth Step | 17
| | 6
2
a = √
e -
x m
*6
z - pa
x
dz
x b
+
dz
+
rr
4
4
:+
rr
4
r
= 20,
2
(the greater Number.
2, the leffer Number.
PROOF.
4ª a + 3 e = 1606.
5a + 6e = 112.
A a 2
Queſtion
180
ALGE BRA.
Queſtion 67. Two Gamefters, A and B, lofing at the Gaming-
Tables, upon comparing their Loffes, found, that if the Square of
the Pounds A loft was multiplied by 5, and this Product added to
6 times the Pounds B loft, the Sum was 548 Pounds :
But if the Pounds A loft was multiplied by 3, and to this Pro-
duct adding the Pounds B loft multiplied by 2, the Sum was 46
Pounds. To find the Lofs of each?
Let a the Pounds A loft, e = the Pounds E loft, x = 5,
m = 6, d=548, b = 3, ≈ = 2, r = 46.
I -
I
z
1\xaa+me=d}
2ba+ze= r
xaa 3 me = d — x ⋅a a
3 m 4e =
d. хаа
m
ba5ze=r-
r - ba
By the Queſtion.
2
ba
526e
Z
4.67
ba
d — x a a
Z
m
7 xm8
r m
·mba
- d — xa a
Z
8 x x 9rm — mba = z d
zd- ----- Z X A a
9+zxaa 10 zxaa +rm m ba
1Q rm II z za a mba - zd
a = zd
r m
The Equation being Quadratic, and all thofe Terms which
contain any Power of a being on one Side of the Equation,
divide by the Co-efficient of its higheſt Power.
mb a
z d - r m
II - Zx 12 a a
Zx
Subſtitute
Ey Subftitution 13a a
·pa=
Z X
m b
Z X
z d — r m
14 aa—pa+
p a + p p
13 14 aa
4
Z X
z d — r m
ZX
- 1.8
+pp
4
14 w 2
Of Quadratic EQUATIONS.
181
P
z d
d — r m
PP
14 w 2 15
215a
+
2
Z X
4
15 +
p16a=
z d
r m
рр
+
2
4
r - b a
By the fixth Step 17 e =
Z
Z X
(Pounds, the Sum loft by A.
8 Pounds, the Sum
(loft by B.
: + 1/2
= 10
PROOF.
5aa+6e548
3a+2e = 46
Queſtion 68. Two Brothers, A and B,
trying each other's
Skill in Algebra, fays the eldest Brother, the Sum of our Ages
is 45:
But, fays the youngest, if they are multiplied together, the Pro-
duct is 500. What is the Age of each of them?
Let a = the Age of the eldeft, e = the Age of the youngeſt,
s = 45, p = 500.
I
I
1 2
a+e=s
P
a e
se
By the Queſtion.
e31a
2e4a=
3.45
P
e
P
e
5x e6p=se — e e
Becauſe the Square of the unknown Quantity has the Sign,
therefore tranfpofe it, that the higheſt Power of the unknown
Quantity may have the affirmative Sign.
6 +ee 7 lee+p=se
7 p8ee = se
semp
8 — se gee - se
se
910ee
р
ss
Į
SS
set
4 4
-p
IQ w. 2
180
ALGEBRA.
Queſtion 67. Two Gamefters, A and B, lofing at the Gaming-
Tables, upon comparing their Loffes, found, that if the Square of
the Pounds A loft was multiplied by 5, and this Product added to
6 times the Pounds B loft, the Sum was 548 Pounds :
But if the Pounds A loft was multiplied by 3, and to this Pro-
duct adding the Pounds B loft multiplied by 2, the Sum was 46
Pounds. To find the Lofs of each ?
Let a the Pounds A loft, e = the Pounds E loft, x = 5,
m = 6, d=548, b = 3, ≈ = 2, r = 46.
I
xaa + me = d} By the Queſtion.
2ba+ze = r
I — xaa 3|me=d- xa a
2
3 m 4e =
ba5ze=r-
5-26e
d - xa a
711
ba
ba
Z
4.67
f
ba
d — x a a
ช
m
r m
mba
7xm8
= d
d-xa a
Z
8 x x 9rm - mba zd- zxa a
9+zxaa 10 zxaa+rm—mba = z d
1 Q — r m \ 1 1 | 2 x a a mb a = z d r m
The Equation being Quadratic, and all thofe Terms which
contain any Power of a being on one Side of the Equation,
divide by the Co-efficient of its higheft Power.
mb a
z d — r m
II zx 12 a a
ZX
Z X
Subſtitute - p =
Ey Subftitution 13a apa=
136 □ 14 aa-pa
m b
1.8
Z X
zd-
Z X
=zd=rm + pp
pat pp
4
zx
4
14 w 2
Of Quadratic EQUATIONS.
181
พ.
14 w 2 15 a
Р
z d
d—r m
+
PP
2
Z X
4
z d — r m
+
PP:t
+ 1/2
= 10
Z X
4
2
15 + 216a=
2
By the fixth Step 17 e =
Z
b a
ba
(Pounds, the Sum loft by A.
= 8 Pounds, the Sum
(loft by B.
す
​PROOF.
5aa+6e=548
3a+2e= 46
Queſtion 68. Two Brothers, A and B, trying each other's
Skill in Algebra, fays the eldest Brother, the Sum of our
is 45:
Ages
But, fays the youngest, if they are multiplied together, the Pro-
duct is 500. What is the Age of each of them?
Let a =
the Age of the eldeft, e= the Age of the youngeſt,
s = 45, p = 500.
ate=s
P
e3a=s
I
2
a e
By the Queftion.
I
e
Р
2
e4a
e
р
3.45
S
-e
e
5 x e
6
p =se- e e
Becauſe the Square of the unknown Quantity has the Sign-
therefore tranfpofe it, that the higheſt Power of the unknown
Quantity may have the affirmative Sign.
· 6 + ee 7 lee + p = se
7- p8ee = se -p
8 se gee
se
SS
9c10ee-
set
1
+15
10 w. 2
182
ALGEBRA.
SS
10 w 2
I I
2
4
ss
11 +
12 e= +
2
2
4
P 25, the Age
(of the youngeſt.
By the third Step 13
a=s—e=20, the Age of the eldeſt.
This Anſwer to the Queſtion contains an Abfurdity, for e that
is put for the Age of the youngeſt Brother is 25, when a that is
put for the Age of the eldest Brother is only 20.
The two Roots of Quadratic Equations explained.
59. And now we fhall explain to the young Analyſt, that in
every Quadratic Equation, the unknown Quantity has two
Values or Roots, fometimes one is affirmative, and the other
negative, and fometimes both are affirmative.
There are three Forms of Quadratic Equations.
The firft is the fixth Step of Queſtion 60, where we have
ee + be = m.
et
And of this Form are the Equations at Queſtion 61, Step 7.
Queſtion 62, Step 9. Queſtion 63, Step 9. Queſtion 64,
Step 12.
The fecond Form is the twelfth Step of Queftion 65, where
we have a a ~za=bd - bmx.
And of this Form are the Equations at Queſtion 66, Step 11.
Queſtion 67, Step 13.
The Difference between these two Forms of Quadratic Equa-
tions, is only in the loweſt Power of the unknown Quantity
having the Sign + or, for in the firſt Form it has the Sign
+, it being be, but in the fecond Form it has the Sign
for
it is za.
And if the loweſt Power of the unknown Quan-
tity has feveral Co-efficients connected by the Signs + or -,
as at Queſtion 64, Step 12. Queſtion 65, Step 11. Then if
the Sum of the pofitive or affirmative Co-efficients exceeds the
Sum of the negative Co-efficients, the Equation is of the firſt
Form: But, on the contrary, if the Sum of the negative Co-
efficients exceeds the Sum of the pofitive or affirmative Co-
efficients, then the Equation is of the fecond Form.
2
But
Of Quadratic EQUATIONS.
183
{
But the third Form is the ninth. Step of the laſt Queſtion,
where we have ees e = p, which differs from the other
two Forms of Quadratic Equations, in this, that if the Side of
the Equation, which is known, confifts but of one Quantity, as
in the preſent Cafe, it has the Sign; and if that Side of the
Equation confifts of ſeveral known Quantities connected by the
Signsor, that then the Sum of the negative Quantities
is always greater than the Sum of the affirmative Quantities;
but in the first and fecond Form, if there is but one known
Quantity, which compofes that Side of the Equation, it will
always have the affirmative Sign; and if there are ſeveral known
Quantities connected by the Signs + or, that then the
Sum of the affirmative will always exceed the Sum of the
negative Quantities.
Now of the two Values or Roots of a in the first and fecond
Form of Quadratic Equations, one is affirmative, and the other
negative; and as the negative Value in thefe Equations does
not come out in the Operation without a Miſtake in the Work,
therefore theſe two Forms of Quadratic Equations give the
true Numbers required.
But the two Values or Roots of a in the third Form are both
affirmative, and the Anfwer fometimes giving one, and ſome-
times the other Root, and it being doubtful in many Cafes
which of theſe two Values of a will anſwer the Conditions of
the Queſtion; this Form of Quadratic Equations is therefore
called the Ambiguous Form.
Before we fhow the Reafon of theſe two Values or Roots of
the unknown Quantity in Quadratic Equations, and how from
having found one Number, or Value, the Learner may find the
other Number; we fhall explain the Divifion in Algebra, where
the Quotient confifts of feveral Quantities connected by the
Signs and
60. The Nature of Divifion explained, when the Quotient confifts
of feveral Quantities connected by the Signs + or -.
To render this the eafier to the Learner, let us refume
Example 1, Article 22, where we are to divide a b + am by
a, which being placed as ufual in common Arithmetic, thus,
Now
{
ང
در
184
ALGEBRA.
Now the Number of times a may be
ab
had in abis b, that is, b is the Quotient | a) a b+am (b+m
of a b divided by a; place b in the
Quotient, multiply it by a, and place
the Product ab as in common Diviſion,
and ſubſtracting it from a b -- am the
Dividend, there remains a m; then find
how many times a will go in a m, and
it is m, that is, m is the Quotient of a m
divided by a, and becauſe the Signs of
the Divifor a, and Dividend a m are
alike, therefore it must be + m, which
being placed in the Quotient, and mul-
tiplied by a, the Product is a m, which
placing under a m, and fubftracting it
from am, there remains o.
Hence the Quotient is b + m.
To divide x x + x m + xa b by x.
x) x x + x m + xa b (x + m+a b
xx
xm + xab
+xa
x m
x a b
xa b
О
+am
+am
Here dividing xx by x, the Quotient is x, which placed
in the Quotient, and multiplied by the Divifor x, and placing
the Product xx under the Dividend, from which fubftracting
it, there remains x m + x a b.
Then dividing xm by x, the Quotient is m, orm, for
the Signs of xm and x are alike, put m in the Quotient, by
which multiply the Divifor x, and put the Product xm under
xm + xa b, and ſubſtracting, there remains x a b.
Then dividing xab by x, the Quotient is a b, or +ab,
for the Signs of x a b and x are alike, put +ab in the Quotient,
by which multiply the Divifor x, and put the Product x ab,
under xab, and fubftracting, there remains o, hence the Quo-
tient is +m+ab.
X
To
Of Quadratic EQUATION Ss.
185
To divide x x + 2 x a + a a by x + a.
x + a) xx + 2 x a + aa (x + a
xx+
ха
xa + a a
xa + a a
О
Dividing xx by x, the Quotient is x, by which multiplying
the Divifor x + a, the Product is x x + xa, which placed under
the Dividend, and ſubſtracted, there remains x a + aa.
Then dividing xa by x, the Quotient is a, or + a, for the
Signs of xa and a are alike, puta in the Quotient, multi-
plying it by the Divifor x+a, the Product is x a + a a, which
put under the Remainder xa+aa, and fubítracting, there
remains o, hence the Quotient is x + a.
To divide a abb by a + b.
a+b) aa—bb (a - b
a a + a b
·ab bb
a b — bb
O
Dividing aa by a, the Quotient is a, and multiplying the
Divifor by a, gives a a+ab, this fubftracted from the Divi-
dend leaves ab-bb; for here the Quantity a b, which is
to be fubftracted, is, by the Rule for Subítraction, to have its
Sign changed, and then added, hence +ab becomes in the
Remainder ab.
Then dividing-ab by a, the Quotient is b, for the
Signs of ab and a are now unlike; multiplying the Divifor
ab, byb, and fubftracting the Product-ab-bb, from
the Remainder ab — bb, there remains o, hence the Quo-
tient is a — b.
To divide a a a
a —x) a a a
a a a
3
xxx by a X.
za a x + za xx
3ªax +3axx - xxx (aa
a a x
2aax + 3a x x
+3
2aax+2ax x
-
2ax+xx
*
a x x -
X X X
a x x
B b
O
X X X
In
186
ALGEBRA.
In thefe Divifions we may at Pleafure take any Term in the
Dividend we have a Mind to uſe firft, and find how many
times any Term in the Divifor can be had in it, and when the
Divifor is multiplied by the Quotient Quantity, we fubftract it
from the whole Dividend, that is, take any Term in the Pro-
duct, from any Term in the Dividend, without regarding
whether they ftand immediately over one another or no.
And to diſcover how many times any one Quantity can be
had in another, we are only to confider into what Quantities we
muſt multiply that Term in the Divifor, to make it the fame
with the Term in the Dividend, at which we aſk the Queftion.
Or, it is no more than to find the Quotient, which ariſes from
dividing that particular Quantity in the Dividend, by the Quan-
tity in the Divifor, which is done by the Rules in Divifion.
Let us take the laft Example, and change the Poſition of the
Quantities:
−x+a) −xxx+aaa+3axx — zaax (xx+aa-2ax
X X X
+ axx
aaa+2axx
за
a a x
aax
aaa
2A X X —
2axx
2aax
-2ûax
O
where we have the fame Quotient as before.
The Truth of theſe Operations is proved as in Divifion of
common Numbers, for if the Work is true, the Quotient being
multiplied by the Divifor, the Product will be the given Divi-
dend; thus in the laſt Example,
xx+aa-2ax is the Quotient.
X
Ļa
~xxx―aax+ 2 axx the Product from multiplying xx+a a
a x x + aa a
X X X
2 ax, by- XC.
2 aax the Product from multiplying xx +
2 ax, by a.
a a
3aax+3axx+aaa the fame with the given Di-
vidend, for though they do not ftand in the fame Pofition as
in the Example, yet as the Quantities in each Term are
alike, and they have the fame Co-efficients, and connected by
the fame Signs, their whole Value, or Amount, muſt be the
fame.
The
Of Quadratic EQUATION s.
187
The Manner of finding the two Roots, or Values, of
the unknown Quantity in Quadratic Equations.
61. Now to find the other Value of a, in the Ambiguous
Quadratic Equation, Queftion 68.
Take the Work at the Step immediately before you begin to
compleat the Square, which is at the ninth Step, where the Equa-
eese = -Þ
tion is
Make this Equation equal to nothing, }ee-se+p=0
by tranſpoſing p
Then put it in Numbers, and it is
By the Work we found
ee - 45e +500 = 0
Make this Equation equal to nothing, by tranf-
pofing the 25, thus,
5000
e = 25
-}c - 25
25 =
= 0
Then divide e e 45 e + 500 by e
25, thus,
e -
8 = 25)ee - 45 e + 500 (e — 20
ee - 25 e
20e +500
20e + 500
O
Hence the Quotient is e-20, but as the Dividend is nothing,
for e e
45e+500 = o as above; and as the Diviſor e - 25
is nothing, for e-250 as above, it follows that the Quotient
must be nothing, or equal to nothing, that is, e 200; then
tranfpofing 20, we have e20, which is the other Value of e,
in this Quadratic Ambiguous Equation; therefore, I fay the
youngest Brother was but 20 Years of Age.
And upon this Value of e, if we take the third Step of the
Work to the Queſtion, that is, a = s e, we fhall find a = 25,
whence the eldest Brother was 25 Years of Age, and thefe are
the true Ages of the two Brothers; for their Ages anfwer the
Conditions of the Queſtion, and it is a poffible Cafe, whereas
though the other Numbers anſwered the Conditions of the
Queſtion, yet it was impoffible for the youngest Brother to be
25, when the eldest was but 20 Years old.
B b 2
From
188
ALGEBRA.
From the Work of the Queſtion we found
But now we have found
The Sum of theſe two Values of e, is
e = 25
e
20
45
But obferving where we put this Quadratic Equation into
Numbers, and made it equal to nothing, we ſhall find the
Co-efficient of the firſt Power of e to be
the two Values of e is + 45, as above,
Quadratic Equations, Algebraifts give us this
45, but the Sum of
and concerning theſe
SCHOLIU M.
62. That in Quadratic Equations the Sum of both the Roots,
r Values, of the unknown Quantity, is equal to the Co-efficient
of the lowest Power of the unknown Quantity, at the Step im-
mediately preceding the compleating the Square, but will have the
contrary Sign; that is, if the Co-efficient of the loweft Power
of the unknown Quantity has the Sign +, the Sum of both
the Roots will be the fame as the Co-efficient, but will have
the Sign-
And if the Co-efficient of the loweſt Power of the unknown
Quantity has the Sign, then the Sum of both the Roots,
or Values, will be the fame as the Co-efficient, but will have
the Sign +.
Therefore having found any one Root, the other is eaſily
found.
63. To find the other Value of the unknown Quantity in the first
Form of Quadratic Equations, or where the Co-efficient of the
Lowest Power of the unknown Quantity has the Sign +, it is done
by adding the Value of the unknown Quantity found from the
Operation, to the Co-efficient of its loweft Power, and to their
Sum prefix the Sign-.
Thus at Queſtion 60, Step 6, the Co-efficient of e, }
is b, or
To which adding the Value of e, as found by that
Operation
The Sum is
}
ΙΟ
}
5
15
And
Of Quadratic EQUATIONS.
189
And prefixing to this 15 the Sign, and this is the other
Value of e, that is, e =— 15, which is an imaginary Value
of e, it being abfurd for a poſitive Quantity to be equal to a
negative one.
However, we ſhall find this imaginary Value of e, if we pro-
ceed by Divifion according to the Directions at Art. 61.
For the fixth Step, Queſtion 60, is that which immediate-
ly precedes the compleating the Square, where the Equation
ee + be = m
is
Which is in Numbers
-
ee +10e=75
Tranſpoſe 75, to make the Equation} +10-75=0
equal to nothing
By the Work we found.
e = 5
Tranſpoſe 5, to make this Equation equal to nothing e― 5 = 0
Then dividing ee + 10 e-75 by e-5.
e −5) ee + 10e — 75 (e + 15
e e
- 5e
15e-75
15e-75
I
Hence the Quotient is e+ 15,
but as the Dividend is equal
to nothing, for ee + 10e−75 = 0, and as the Divifor e 5
is equal to nothing, for e 5=0, as above, confequently the
Quotient must be equal to nothing, that is, e + 15 = 0, by
tranfpofing the 15, we have e-15, as before.
For another Example of this Kind take Queftion 61, where
the ſeventh Step is that which immediately precedes the com-
pleating the Square, the Equation being ee+me = x — b,
which being put in Numbers is
e e + 6 e = 112
By tranfpofing 112, to make the Equa- Į
tion equal to nothing, we have
By the Work it was found
} ee+6e-112=0
°}
Tranfpofing 8, to make the Equation equal to Į
nothing, we have
And dividing to find the other Root of e, as before,
:
e=8
e
&= a
3
190
ALGEBRA.
e—8)ee+6e− 112 (e + 14
e e
8 e
14 e
14 e
- 112
II2
Hence the Quotient is e - 14, which for the fame Reafon as
before, it is e+ 14 = 0,
o,
Value of e.
hence e
14, for the other
And this Value of e will be found by the Rule. Art. 62.
Thus at Queſtion 61, Step 7, the Co-efficient of e
is m, or
} 6
the} 8
To which adding the Value of e, found at the
Operation
The Sum is
14
Then by the Rule prefixing the Sign to 14, we have
-14 for the other, or imaginary Value of e, the fame as
before.
But if we add theſe two Values of e together, we ſhall find
their Sum anſwer to the Scholium, Art. 62.
The firft Value of e is
The fecond Value of is
.
8
14
6
Hence their Sum is the fame with the Co-efficient of e, but
has the contrary Sign.
If the Reader has a Mind to profecute this Speculation, he
may try Queſtion 62, Step 9. Queſtion 63, Step 9. Queſtion
64, Step 12, or 13, which are Equations of this firft Form,
as well as fome that follow them.
To find the other Value or Root of the unknown Quan-
tity in the Second Form of Quadratic Equations.
64. The ſecond Form of Quadratic Equations, is when the
Co efficient of the loweſt Power of the unknown Quantity has
the Sign; in this Cafe ſubſtract the Co-efficient of the lowest
Power, fuppofing it affirmative, of the unknown Quantity in the
given Equition, at the Step immediately preceding the compleating
[
I
the
Of Quadratic EQUATIONS.
191
he Square, from the Value of the unknown Quantity found by the
Work, to the Remainder prefix the Sign, and it will be the other
Value of the unknown Quantity. Or place down the Co-efficient
with its Sign, to which add the Value of the unknown Quantity
found by the Work, and to this Sum prefix the Sign —, and it will
be the other Value, or Root of the unknown Quantity.
疲
​An Equation of this fecond Form is Step 12, Queſtion 65,
bmx.
where we have a a— za = b d — b m x.
Here the Co-efficient of a, is-z, or
28
And the Value of a found in that Equation is
+60
The Sum is 32, but to it prefix the Sign, and
32, the other Value of a, which is imaginary, S
32
it is
as it has the Sign-.
And if we proceed by Divifion according to the Directions at
Art. 61. we fhall find this imaginary Value of a.
Thus if we take the twelfth Step of Queftion 65, which
immediately precedes compleating the Square, we have this
Equation
Which being put in Numbers is
Tranfpofing 1920 to make the
Equation equal to nothing
By the Work it was found
a a
a a
aa-
za = b d bmx
<
- 28 a = 1920
28 a — 1920 — 0
a=60
°}
a
60=0
Tranſpoſe 60 to make the Equation equal to Į
nothing
And dividing to find the other Root of a, as before,
a — 60) a a — 28 a
1920 (a +32
a a
600
32 a
1920
32 a
1920
О
.འ
Hence the Quotient is a + 32, which becauſe the Dividend
and Divifor are each equal to nothing, confequently the Quotient
muſt be equal to nothing, hence
By tranfpofing 32, we have
imaginary Value of a, as before.
a +32=0
a
32, the fame
And
192
ALGEBRA.
And if we add thefe two Values of a together, we fhall find
their Sum agree with the Scholium, Art. 62.
The Value of a found by the Operation, Queſtion 65, is 60
The Value of a now found is
Their Sum is 28, or + 28, the fame Number as the Į
Co-efficient of a, but with a contrary Sign
32
}
28
Another Equation of this fecond Form is Queſtion 67,
Step 13, where the Equation is aa - pa=
z d
r m
Z X
Which being put into Numbers is a a — 1.8 a—82
Tranfpofing 82, to make the Equa-aa-1.8a-82=0
tion equal to nothing
By the Work it was found
a = 10
° } a
10 =
Tranfpofing 10 to make the Equation equal to
nothing
And dividing to find the other Value of a, as before,
a—10) aa— 1.8a-82 (a8.2
a a
10 a
8.2 a
82
8.2 a
82
O
Hence the Quotient is a +8.2 which must be equal to nothing,
for the Dividend and Divifor are each equal to nothing: but if
a + 8 2 = 0,
= —
By tranfpofing 8.2 we have a 8.2 which is the other
Value of a, and it is imaginary, becauſe it has the Sign—.
The fame imaginary Value of a may be found by Art. 64,
thus,
The Co-efficient of a, is
The Value of a found by the Queſtion, is
The Sum is
– 1.8
IO.
8.2
Now to this 8.2 prefix the Sign, and we have 8.2 for
the imaginary Value of a, the fame as before.
And
Of Quadratic EQUATIONS.
193
And if theſe two Values of a are added together, their Sum
wili agree with the Scholium, Art. 62.
The firft Value of a, is
The fecond Value of a, is
Sum
10.
8.2
But the Co-efficient of a, is
1.8
1.8
65. But in the ambiguous, or third Form of
Quadratic Equations,
If the Value of the unknown Quantity found by the Operation,
is fubftracted from the Co-efficient of its lowest Power, at the
Step immediately before the Square is compleated, the Co-efficient
being fuppofed affirmative, the Remainder is its other Value.
At Queſtion 68, Step 9, the Co-efficient of e is s, or 45
The Value of e, found by the Operation, is
The Remainder is the other Value of e
25
20
And it is this fecond Value of e that is the true Anſwer to
the Queſtion, as was obſerved Page 187; and here the
Learner may again obſerve, that both the Values in this Cafe.
are affirmative, which makes this be called the ambiguous Caſe,
but in the other two preceding Cafes, or in the four former
Examples, the other Value of the unknown Quantity was
negative, which is only an imaginary Value, it being impoffible
for an affirmative, or pofitive Quantity, which the Queſtion
requires, to be a negative, or equal to a negative Quantity.
But we may find the other Value of e, in this ambiguous Cafe,
by Diviſion, as in the former Inſtances, thus,
The Equation, Queſtion 68, Step 9,
immediately before the Square was com-
pleated, is
Which being put in Numbers, is
Tranfpofing 500, to make the
Equation equal to nothing
By the Work it was found
e e
ee — 45e=.
se =
P
500
}
ee - 45e + 500 =
e = 25
25 = 0
And
Tranfpofing 25, to make the Equation equal
to nothing
С с
194
ALGEBRA.
">
And dividing to find the other Value, or Root of e, as
before,
e
e — 25) ee — 45 € + 500 (e — 20
ee — 25 ė
20e +500
20e + 500
Hence the Quotient is e
O
20, which must be equal to nothing,
for the Reafon in the former Cafes, but if
Tranſpofing 20, we have
other Value of e, the fame as before.
e — 20
e = 20 the
And in this ambiguous Caſe, if we add the two Values of e
together, we fhall find them agree with what is ſaid at the
Scholium, Art. 62.
The firſt Value of e, is
The fecond Value of
e, is
But the Co-efficient of
в
is
· 45.
25
20
45
The Manner of expreffing the two Roots of an
ambiguous Quadratic Equation explained.
66. Now to explain the ufual Manner in which Algebraifts
express the Value of the unknown Quantity, in the ambiguous
Quadratic Equation; let us reſume the Solution of Queſtion 68,
at the eighth Step, where there is this
Equation
I
e e = s e
P
I - se
2
2 CD
3
e e
3 w 2
4 +
4
e
2
e e - se =
-set
S
4
SS
4
4
p
ss
5
e
P
2
2
4
That
?
Of Quadratic EQUATION s.
195
That is, prefix both the Signs + and -, to the Quantity under
the radical Sign, for that being added to, or the rational Quan-
S
2
tities on that Side of the Equation, gives one of the Values of e,
but if it is ſubſtracted from, or the rational Quantities on that
2
Side of the Equation, then it gives the other Value of e, thus,
s = 45
s = 45
225
180
4) 2025=5S
506.25=
SS
4
—p = 500.
6.25 ==—p (2.5 =
4
45) 225
225
4
4
Then to find the two Values or Roots of e.
S
=22.5
2
p = 2.5
4
25. one of the Values of e.
N/
=22.5
p = -2.5
4
20. the other Value of e, which two Values
ofe are the fame as was found at Art. 61.
Cc 2
And
196
·
ALGEBRA.
And this is the common Method in which Algebraifts fet
down, or exprefs the Value of the unknown Quantity, in the
ambiguous Quadratic Equation.
The Reafon of Quadratic Equations having two different
Values of the fame unknown Quantity, is becauſe the fame
Quadratic Equation can be formed from two different Suppofi-
tions, or Values of the unknown Quantity, or fuppofing the
fame unknown Quantity to be equal to two different Numbers.
For let us reſume the Equation e e se
p, or e e
=500, in this ambiguous Equation we found the firſt Value
of e to be 25, by making e equal to 25, we have
I ✪ 2
Multiplying the firſt
Equation by 45, the
Co-efficient of e, in the
given Equation
I
1 2
e = 25
ee = 625
-45e=1125
3
2 +3
4
e e
45e
45e=— 500, the fame
with the given Quadratic
Equation.
And if we take the other Value of e, viz, 20, we can form
the given Equation, for
Let
I 2
IX-45 the Co-7
efficient of e, in the
given Equation
1 2
3
2+3
4
e e
I ė 20
ee
400
45e=— 900
45e= 500, the fame
with the given Quadratic
Equation.
Likewife if we take the firft Form of Quadratic Equations,
viz. ee+be=m, or ee + 10 e 75, fee Queſtion 60, Step 6.
Now the two Values of e in this Equation we found to be 5,
and 15, and from either of thefe Values of e, we can form
the given Quadratic Equation,
Suppofe
1 & 2
27
I
e = 5
2
ee = 25
I X
Of Quadratic EQUATIONS. 197
IX 10 the Co-effi-
cient of e, in the given
3
10e = 50
Equation
2 +3
1
Again, ſuppoſe
1 & 2
4 ee+10e=75, the fame with
I
12
I X IO as above
3
2+3 4
the given Quadratic Equation.
e — — 15
ee 225, for
15 X -
15
=225, the Signs being
alike.
10e= 150
ee + 10 e = 75, the fame E-
quation as before.
bmx, or aa-
28 a
And if we take the fecond Form of Quadratic Equations, viz.
aa-za- b d — bmx,
1920, ſee Queſtion
65, Step 12. The two Values of a in this Equation we found
to be 60, and 32, from either of which we can form the
given Equation, for
Suppofe | I
I 2 2
a = 60
aa = 3600
I X 28 the Co-
efficient of a, in the
given Equation
3
28a=-
1680
2+3
4
a a
28 a
1920, the fame
1
I X
Again, if
with the given Equation.
32
aa = 1024, for
= +1024.
I
a
I Ⓒ 2
2
}
-28a
3
2 + 3
4
a a
28 the Co-
efficient of a, as above S
32 X
32
896, for-28x-32
= + 896.
-
28 a
1920, the fame
with the given Equation.
From this the Learner may obferve, that making the unknown
Quantity equal to either of its Values, and raiſing this Equation to
the Square, and adding it to the former Equation, after it has been
multiplied by the Co-efficient of the lowest Power of the unknown
Quantity in the Quadratic Equation, this Sum will be the given
Quadratic Equation.
Queſtion
198
ALGE B R A.
Queſtion 69. Two Men, A and B, difcourfing of their Shil-
lings, A, who had the greatest Number, faid, if my Number of
Shillings is divided by your's, and this Quotient is added to your
Number of Shillings, the Sum will be 15:
How many'
But if the Sum of both our Shillings is multiplied by 4, and
this Product divided by 10, the Quotient will be 22.
Shillings had each Perfon?
Let a = the Number of Shillings A had, or the greateſt
Number, the Number of Shillings B had, or leffer Number,
s = 15, m = 4, n = 10, d= 22.
a
Then I
+e=s
e
And 2
ma+me
By the Queſtion.
=d
22
I Xe
3
3 — e e
4
a =
2 x n
5
5- me
6
a+ee = se
se-ee
ma+med n
ma dn—me
d n
me
6÷m 7
a
m
d n
me
4.7
8
=se-ee
ве
m
8 x m
9
d n
memse
mee
9+ mee
ΙΟ
mee + d n
+di memse
IO - dn
II
mee
memse-
d n
dn
11
mse
12
mee
-me
mse=
Here the Equation appears to be Quadratic, and of the ambi-
guous Kind; becauſe dn, the known Side of the Equation, has
the negative Sign. Then by Art. 58, dividing by m, the Co-
efficient of ee,
12 m
|
| 13 | cc-e-
·e ŝe =
d n
m
For m be-
ing in every Term on one Side of the Equation, dividing that
Part of the Equation by m, is only to caft away m, out of every
Term of that Side of the Equation, and to divide the other
Side
-
Of Quadratic EQUATION s.
199
Side of the Equation is only to place m as a Denominator to it.
The Equation being now prepared for compleating the Square,
and the firft Power of e being in two Terms, viz. e
whofe Co-efficients are
Therefore
I, and
Subſtitute
S,
se,
I-s, then by
dn
Subftitution 14
e e
Ze
སྦུ
Z Z
14 CO
15
ее ка
·ze+
4
4
ZZ
dn
m
15 w 2 16
17 -
2 | 2
Z
d n
Z
א
16+
2
e
א
Z
土
​2
ZZ
4
Z Z
4
m
d n
= 8± 3,
m
(that is, e is either 5, or 11.
And if e is 5, we fhall find a = 50, by the fourth Step,
which two Numbers of Shillings anſwer the Conditions of the
Queſtion; or, if we ſuppoſe e = 11, then by the fourth Step
we ſhall find a = 44, which two Numbers will likewife anfwer
the Conditions of the Queftion: But fometimes one of the
Numbers, or Roots, of thefe ambiguous Equations, will not
anſwer all the Conditions of the Queſtion, as at Queſtion 74,
and then the other Root muſt be found.
Queſtion 10. Two Merchants, A and B, had gained in Trade,
but A, who gained the most, found, that if the Square of the
Pounds he gained was multiplied by 2, and the Product added to
8 times the Pounds B gained, if this Sum was divided by 4, the
Quotient was 816 Pounds:
But if 3 times the Pounds A gained, was added to 10 times
the Pounds B gained, and this Sum divided by 40, the Quotient
was 5 Pounds. How many Pounds had each Man gained?
Put a the Pounds gained by A,
=
by B, x = 2, m = 8, p = 4, d
=
e=
e the Pounds gained
816, b = 3, % = 10,
=
r
40, n=5.
IX A 2
200
ALGEBRA.
t
I
2
Ixp
3
3 − xa a
4
xaa + me
P
ba+ze
r
= d
By the Queſtion.
n
xaa+me = pd
me = pd
хаа
4÷m
5
p d
xa a
e=
2 X r
6
m
ba+ze = r n
6-ba
7
ze = r n
ba
r n
ba
8
e
Z
r n
ba
5.8
pd-
xa a
9
ช
m
9 x z
10 irn ----
— ba
ba=
zpd― zxa a
m
10 x m
II
mrn—mba
mba = z pd — z x a a
Tranſpoſe zxa a, that the higheft Power of the unknown
Quantity may have the Sign +.
11 + zxa a 12
I
12 mr n 13
=
zxaa+mrn—mba z pd
z xa a — mb a = z p d — m r n
mr n
The Equation now appears to be Quadratic, but to know if
it is ambiguous, find which Quantity is greateſt zpd, or mrn,
but z pd is 32640, and m r n is only 1600, hence zp d
= 32640-1600 31040, which being an affirmative Num-
But be-
ber, the Equation is not ambiguous, by Art. 59.
cauſe the Square of the unknown Quantity has a Co-efficient,
therefore, by Art. 58,
mb a
z p d — m r n
13÷2x 14
a a
Z X
Zx
m b
Subſtitute
S
then by
ZX
Subftitution 15
a a
sa=
15
col
16
ㅁ
​| aa-sa+
2
z pd — mr n
Z
SS
4
zx
zpd.
ZZ
mr n
+
55
4
16 ws
Of Quadratic EQUATIONS.
201
55
+ +
S
zpd.
p d — m r n
16 w 2
17
a
+
2
ZX
4
S
17+
18
a = √
z pd
mr n
2
Z X
2
By the eighth Step 19
e =
4
(=40, the Pounds gained by A.
r n − b a
Z
8, the Pounds gained
(by B.
Queſtion 71. A Father, by his Will, left his two Sons, A and
B, fuch a Portion, whereof A had the greatest Fortune; that if
the Square of the Number of Pounds he was to have, be multiplied
by 2, and to this Product there is added the Number of Pounds B
was to have multiplied by 35, the Sum was 6400 Pounds:
But if the Number of Pounds A was to have, be multiplied by
20, and this Product added to the Number of Pounds B was to
have multiplied by 15, the Sum was 1600 Pounds. To find the
Fortune of each?
Let a
:
the Fortune of A, e the Fortune of B, x = 2,
= 20, z=15, r = 1600.
m = 35, d=
d = 6400, b = 20, z =
72
xaa+me = d
ba + ze = r
at
me = d — xa a
e =
I
2
I
хаа
3
d
x a a
3 ÷ m
4
m
2
b q
5
ze = r
.ba
r
ba
6
e=
Z
4.6
7
7 x Z
8
r
-
8 xm |
|
9
} By the Queſtion.
Tranſpoſe z xa a,
affirmative.
r—ba
Z
ba =
d
zd.
m
хаа
m
z xa a
| mr — mba — zd
mba — z d— z x a a
that the higheſt Power of a may be
zxaa + mr
9 + zxa a
10
mba = z d
ΙΟ
77 1º
II
Z Xa a
mba = z
d.
m r
D d
The
202
ALGEBRA.
The Equation now appears to be Quadratic, and to know if
it is ambiguous, find what zd and mr are in Numbers. But
zd-
mr = 96000 56000 40000, a pofitive Quantity,
whence the Equation is not ambiguous by Art. 59. And be-
cauſe the Square of the unknown Quantity has a Co-efficient,
therefore by Art. 58.
mb a
ba
zd—m
mr
12
a a
ZX
Z X
mb
Subſtitute
then by
ZX
Subſtitution
13
a a
sa=
1360
14 a a
sa+
4
ค
S
z d
14 w 2
15
a
이
​S=
zd - mr
ZX
zd-
z d—m r
zx
+
SS
+
1
4
mr
SS
2,
ZX
4
་
z d — m r
SS
15 +
16 a=
+
:+
2 |
Z X
4
2
=49.9999, &c. becauſe of the Imperfection of the Decimal
Fraction; the true Number being 50, from which by
r—ba
The fixth Step 17 e-
= 40.
Z
Queſtion 72. What are those two Numbers, the Quotient
of the greater divided by 5, and added to the leffer, the Sum
may be 12:
But the Product of the two Numbers divided by 4, the Quotient
is 40?
a =
Put the greater Number,
m = 5, p = 12, d= 4, * = 40.
e =
the leffer Number,
.
I X m
3-me
2 x d
I
| = = = +e=p
a e
By the Queſtion.
= x
2
d
3
a + me = mp
4
a = mp
mp
5
a e=dx
me
5 →
Of Quadratic EQUATIONS.
203
i
dx
=mp―me
dx
5 →
e 6
a
4.6
7
e
7 x e
8
8 + mee
9
9 -d x
ΙΟ
ΙΟ
mpe
I I
dx = m pe — me e
mee+dx=mpe
mee=mpedx
mee—mpe-dx
Here the Equation not only appears Quadratic, but Ambiguous,
for dx the known Side of the Equation is negative. Now
by Art. 58.
dx
m
II - m
12
e e
•pe
12 CO
13
ee-p
ee - pet
pp - pp
dx
4
4
m
PP
dx
13 w 2
14
11
2
14 +
€
15
2
2
4
m
PP
d x
=6±2,
m
4
that is, e is either 8, or 4: But if
e = 8, then by
dx
The fixth Step 16
a =
= 20.
e
Or if e 4, then
40, either of which anſwers the Queſtion.
Queſtion 73. Two young Gentlemen having been at the Gaming-
Tables, and being asked by their Friend what they loft, which being
afbamed to own, A faid, if the Number of Pounds I loft is divided
by 4, and this added to the Number of Pounds B loft divided by 2,
the Sum is 9 Pounds:
But if the Product of the Number of Pounds we both loft is
divided by 10, and extracting the Square Root of this Quotient,
it will be 4. How much did each Perfon lofe?
the Number
10: as the
Let a the Number of Pounds loft by A, e
of Pounds loft by B, b = 4, d = 2, m = 9, p
Number 4 is in the firft Part of the Queſtion, and it being
again repeated, there is no Occafion for any new Letter.
D d 2
j
204
ALGEBRA.
I
2
a
b
+
a e
e
d
b
m
By the Queſtion.
I
པ
d
3
a
b
m
3 x b 4a= bm
225
5 x p
ae
Р
= b b
6 a e = p b b
6÷÷÷e7a=
e
d
be
d
p b b
e
p b b
ЪЪ
be
4.78
b m
e
d
bee
8 x e 9
d
p b b = b m e
9 x d 10dpbb dbme-bee
Dividing by b, by Art. 53.
10b11dpbdme-ee
To have the higheft Power of e
mative, tranfpofe e e.
affir-
13-dme 14
14ee
11 +ee12ee+dpb=dme
12-dpb13eed me
Here the Equation appears quadratic,
and ambiguous.
dp b
d me
dp b
d d m m
d d m m
14c15ee
dme +
- dp b
4
4
d m
'd d m m
15 w 216e-
= √
dp b
2
4
16+
d m
2
d m
'd d m m
17
土​v
· dpb = 9±1,
2
4
that is, e is either 8, or 10, whence by the fourth, or feventh
Steps, we ſhall find a
20, or 16.
Queſtion 74. In the right-angled Triangle ABC, there is
given the Hypothenuſe AC = 10, and the Sum of the Baſe AB
and Perpendicular BC= 14. To find the Bafe AB and Per-
pendicular BC? See Figure, Page 206.
Let
Of Quadratic EQUATIONS.
205
Let A Cb 10, AB+BCd 14, AB = a; and
becauſe A B+ BC=d, therefore from this fubftracting A B,
or a, we have BC=d-
a.
Having expreffed all the Sides of the Figure in Symbols, and
there being but one unknown Quantity, we are only to raiſe
one Equation from the Property of the Figure; and the Triangle
ABC being right-angled, we have by 47 e 1 the Square of the
Hypothenuſe AC, or bb, equal to the Square of the Bafe AB,
or a a, added to the Square of the Perpendicular BC, or dd
— 2 d a + a a, that is,
I b b = d d — 2 da + 2 aa
In Symbols | | 362
dd
I
2
2 aa
z da = b b
d d
Here the Equation appears quadratic, and becaufe dd is
greater than bb, it is likewife ambiguous, for bbdd100
196 96 a negative Quantity; but as the Square of the
unknown Quantity has a Co-efficient, therefore divide by it
by Art. 58.
bb-dd
2-2 3
aa — da =
2
d d
d d
b b - d d
3Co 4
aa-da+
+
4
4
2
d
d d
b b
d d
4 w 2
5
a
+
2
4
2
d
5+ 6
d
dd
b b - d d
a = +
+
=7±1
2
2
4
2
from whence the Bafe A B may be either 8, or 6; fuppofing
the Bafe 8, then becauſe by the Queftion, the Sum of the Bafe
and Perpendicular is 14, the Perpendicular B C will be 6; but
if we ſuppoſe the Baſe to be 6, then from the fame Reaſoning
the Perpendicular B C will be 8.
And the Queftion not limiting which is longeft, either the
Baſe AB, or Perpendicular B C, we may take either 6, or 8,
for the Length of the Bafe AB, for either will anfwer the Con-
ditions of the Queftion.
But if the Queſtion had faid that the Bafe AB, is longer
than the Perpendicular BC, then we must take a =
d d b b — d d
+ v
+
4
2
d
2
=8, by which we fhall find the Perpen-
di ular
206
ALGEBRA.
dicular BC= 6; for if we take a =
d_dd +
bb-dd
4
2
2
=6, then we ſhall find the Perpendicular BC= 8, which can-
not be, becauſe the Queſtion is fuppofed to determine the Baſe
AB, to be longer than the Perpendicular B C.
A
Queſtion 75. In the right-angled
Triangle ABC, given the Hypothe-
nuſe AC = 10, the Perpendicular
BC, being shorter than the Bafe
AB, by ſubſtracting the Perpendi-
cular BC from the Bafe AB, and
multiplying the Difference by 20,
and dividing this Product by 8, the
Quotient is 5. What is the Length
B of the Bafe AB, and Perpendicu
lar BC?
Let AC = b = 10, AB = a, BC = e, d = 20, m =
% = 5.
I
3 w 2
I
2
8,
aa+ee = bb by the Property of the
Figure, as in the laſt Queſtion.
da-de
m
a a = b b
a = √ bb
√bb.
=z by the Queſtion.
z
e e
-ее
de = zm
da = zm + de
ее
3
19
4
2 x m
5+ de
5
da
6
6 ÷ d! 7
zm + de
a =
d
zm + de
4.7 8
d
e e
Squaring both Sides of the Equation, becauſe the unknown
Quantity is under the radical Sign.
802
9
9 × d d
10 + ddee
zzmm + 2 zmde + ddee
d d
=bb
bb-ee
10 zzmm+2zmde+ddee-ddbb-ddee
zzmm + 2 zmde +2ddee bbd d
I I
II
Z Z M M
I 2
2 ddee + 2zmde bbdd
Z ZM M
2
Dividing
Of Quadratic EQUATIONS.
207
Dividing by the Co-efficient of ee by
Art. 58.
ee +
12-2 dd 13 |
13/0
That is 14
ee +
2 zm de
2 d d
Z me
d
==
b b d d — z z M M
2 dd
b b d d
dd-
Z Z M M
for
2 d d
Z me
>
rejecting 2 d, as
d
N
2 zmde
2 d d
=
in Divifion.
Hence the Equation is quadratic, but it cannot be ambiguous,
becauſe both the Quantities e e + zme having the Sign+, the
d
whole Side of the Equation must be affirmative, and confequently
the other Side of the Equation must be alfo affirmative, other-
wife an affirmative Quantity would be equal to a negative Quan-
tity, which is abfurd. Now,
Z m
Subſtitute x =
=2, by Art. 57-
d
then by
Subftitution 15
1500 16
+
b b d d — z z M M
xx
2 d d
b b d d
Z Z M M
e
4
2 d d
x x
4
b b d d
16 w Z
17
e+
+1/
Z Z M M
x x
+
2
2 d d
4
17
이
​*
b b d d -
Z Z M M
XX
18
e-
+
2
6
2 dd
BC.
4
| *
2
Then by Step 7th 19
zm + de
= 8 = AB.
d
The fame Question done otherwife.
Let A Cb 10, A B = a, then by 47 e 1, BC
= ✓bbaa; fuppofe d= 20, m = 8, x=5, as before.
Now
208
ALGEBRA.'
Now all the Sides of the Triangle being expreffed in Symbols,
and there being only one unknown Quantity, there is but one
Equation required, which may be raiſed from the Conditions of
the Queftion, and thefe I fhall particularly exprefs to prevent
any Difficulty to the Learner.
orb
Now | 1 | a, is the Baſe AB, which is longer than
the Perpendicular BC, or bba a, therefore connecting
√ bb—a a to a by the Sign
| |
We have 2 abba a for the Difference
between the Bafe and Perpendicular, which is to be multiplied
by 20, or d, then
We have 3 da d√ bb — a a
But this Product is to be divided by 8, or m, then
m (is to be equal to 5, or z.
d ✓ bb. a a
da
d / b b на
We have
4
>
and this Quotient
d a
Whence
5
=z, by the
(Queſtion.
m
Becauſe the unknown Quantity is divided by m, therefore by
Art. 47.
5xm6|da-db b―a a z m
=
Becauſe the unknown Quantity is multiplied by d, therefore
by Art. 48.
Z M
6
7
a
√bb.
a a =
d
Now tranſpoſe the Surd, becauſe it has the Sign-, the higheſt
Power of the unknown Quantity being Part of it.
Z m
7+ √bb-aa
аа
8
a a
a =
d
Z m
Z m
8-
a a = a
d
9
d
There being no Quantities on the fame Side of the Equation
with the Surd, raiſe both Sides of the Equation to the ſecond
Power to take away the radical Sign.
9 & 2
Of Quadratic EQUATION Ss.
209
92
IO
b b
- аа-аа
a a = a a
2 zma
d
ZZ M M
Z Z m m
+
d d
2 z ma
10+ a a
2 aa
+
bb
= b b
d
dd
Z Z M M
2 z ma
Z Z M M
I I
12
2 a a
=
bb.
d d
d
dd
zma
b b
Z Z M M
12 2
13
a a
d
2
2 dd
b b
Z Z m m
Here the Equation is quadratic, but becauſe
2
2 d d
= 50
2 = 48, a pofitive Quantity, it is not ambiguous.
Z m
Now by Art. 57, ſubſtitute
x =
X
2.
d
b b
Z Z m m
Then 14
a a
xa=
2
xx
**
14 O
15
aa-
1+
2 dd
4
+
b b
2
15 w 2 16 a
-xa+
Z Z M M
2dd
४ |
2
४ ति
16 +
17
a=
2
2
(8
8:
4
xx
4
**
4
+
+
Z Z M M
bb
2
2 d d
b b
Z Z M M
2
2 d d
the Bafe AB, as before.
Hence in the right-angled Triangle ABC, becauſe we have
given the Hypothenufe AC, which is 10, and having now
found the Bafe AB to be 8, therefore the Perpendicular B C
✓ ICO 646.
Queſtion 76. Two Merchants, A and B, becoming Bankrupts,
owe fuch Sums of Money, that if from the Number of Pounds A
owes, we fubftract the Square of the Number of Pounds B owes,
there remains 1900 Pounds :
But if the Square of the Number of Pounds B owes, is multi-
plied by the Number of Pounds A owes, the Product is 81000002
Pounds. To find the Debt of each Merchant ?
E e
Let
*
210
ALGEBRA.
Let a = the Money owed by A, e the Money owed by
1900, d = 81000000.
B, b
I
а
e e = b
2
a e e = d
By the Queſtion.
I
1 + ee
3
a = b + e e
2-ee
4
3.4 5
5 x ce 6
a =
d
e e
ee + b =
d
ee
e e e e + be e =
d
Perhaps this Equation may appear new to the young Analyſt,
but by turning to Art. 56. he will find it to be a Quadratic
Equation, for the unknown Quantity is only in two Terms, and
in one of them its Power or Height is double its Power or
Height in the other, for it is e e e e and e e, therefore take b
the Co-efficient of e e, the loweft Power of e in the prefent Cafe;
divide it by 2, fquare the Quotient, and add it to both Sides
of the Equation, as before, thus,
b b
b b
6c□ 7
e e e e + be e+
d +
4
4
Extracting the fquare Root as ufual,
b
b b
7 w 2
8
ee+
d+
2
4
b
2
Tranfpofing becauſe it is a known
Quantity,
b
b b
b
8 -
9
e e = √ d +
2
4
2
Now extracting the fquare Root to
deprefs e to the firft Power.
9 ws 2
IO
e
b b
b
90 Pounds,
4
2
(the Money B owed.
bb
4
To extract the Square Root of the Quantity✔d+
b
2
, is only to place again the radical Sign before the fame
•
Quantity,
Of Quadratic EQUATIONS.
211
Quantity, drawing it over the radical Sign already there, and
the other Quantities without that Sign, if there are any; for
though theſe were not included in the firft Root, yet as they
were afterwards tranfpofed to that Side of the Equation, and the
Root is again required to be taken, they will now be included
under the radical Sign of this ſecond Extraction.
Becauſe of the two radical Signs, I fhall fet down the Nume-
rical Work, to make the Operation the plainer.
d = 81000000
b b
902500
4
b b
bb
81902550 = d +
(9050 =
d +
4
4
81
b
·950=
1
1805) 9025
9025
О
8100 = √d +
and the fquare Root of 8100
is 90 = √ √ d +
2
b b
b
4
2
= 10000 Pounds, the Money
The fame Queſtion anſwered by exterminating the unknown
Quantity e.
b b
b
4
2
Then by the third Step a
bee
owed by A.
I
a-ee b
:
S
By the Queſtion, as
before.
b tee
I
3-
e e
b
2
3
4
2 → a
5
a e e = d
a =
e e
6+
-6
e e
ее-
a-
d
a
Make the fourth and fifth Equations equal to one another,
for each is equal to e e.
4.5
6
a
b
6 x a
7
a a
d
a
b a = d
E e 2
J
760
212
ALGEBRA.
氰
​700
8
a a
b b
bat =d+
4
b b
b
4
bb
8 u, 2
a
9
2
b.
b b
9+
|
IO
a = √d +
+ = 10000
2
4
e=
4
b
2
(Pounds, as before.
And by the fourth Step e-a-b, or by the fifth Step,
d
a
= 90 Pounds, as before.
From hence the Learner may obſerve, there are different
Methods of anfwering the fame Queftion, and that fome are
more elegant than others, as they give the Anſwer in more
fimple or lefs complicated Terms: And in this Part of the
Science he is to exerciſe himſelf according to his own Prudence
and Judgment, and fome Meaſure in Proportion as he under-
ftands and conceives the general and univerfal Methods by
which Queſtions are anfwered; it being only my Defign to
illuftrate theſe by pertinent Examples, with fuch Solutions as
arife in an obvious Manner from the Directions, that the
Learner may acquire fome general Idea of the Nature and
Excellency of Algebra.
Queſtion 75. Two Running Footmen, A and B, meeting on the
Road, found, if the Number of Miles A had run was multiplied
by 5, and fubftracting from this Product the Square of the Miles
run by B, there remained 100:
But if the Square of the Miles run by B, was multiplied by the
Number of Miles run by A, and the Product multiplied by 2, this
Product was 80000. How many Miles had each Perſon run?
Let a
the Number of Miles run by A, e = the Number
of Miles run by B, m = 5, x = 100, d
5, % = 100, d = 2, b = 80000.
1
mae e = x
2
daee b
} By the Queſtion.
I + ee
3÷m
3
ma = x+ee
4
a =
x + ee
b
m
2 ÷ dee
5 a =
dee
4.5
Of Quadratic EQUATIONS.
213
6
x+ee
b
4.5
m
dee
m b
6 x m
7
· ee+x=-
dee
7x dee
8
deeee + x dee — mb
Here the Equation appears to be of the fame Kind with the
laft, that is quadratic, but not ambiguous. Now by Art. 58.
9 ÷ d
m b
9
e e e e + xe e =
d
And compleating the Square as in the
laft Queſtion,
xx
m b
xx
10 D
IO
eeee + xee+
+
4
d
4
x
m b
11 un 2
I I
ee+
2
d
+
xx
4
५
m b
**
x
12
12
e e =
+
X
d
m b
4
xx
2
13 U
1 3 ບນ 2
13 e-
+
* |
d
4
2
2) 400000 = m b
200000 =
mb
d
2500
x x
4
202500 (450
m b
d
+
xx
4
16
50=
1
2
85) 425
425
m b
xx
x
400 (20 =
d
+
the
4
2
(Number of Miles run by B.
Then by the fourth Step a =
x tee
100, the Number
in
of Miles run by A.
This
214
ALGE BRA.
This Queftion, as the laſt, may be refolved in a more fimple
Manner, if we exterminate the unknown Quantity e inftead of
a, thus,
I
1 2 3
ma — ee = x 1 By the Queſtion, as
da e e b
ma = x + ee
}
before
I tee
3 - X
4
e e ma
b
2 ÷ da
5
ee =
da
b
4.5
6
m a
da
6 x da
7
d m a a
d x a = b
Dividing by dm the Co-efficient of a a, by Art. 58.
xa
b
dx a
x a
7 ÷d m 8
a a
-
for
m dm
d m
m
x
the d being rejected as in Divifion.
Now the Co-efficient of a being divided by 2,
m
is
X
2 m
For making 2 an improper Fraction by the Rule in common
Arithmetic, is 2
I
: But by the Rule for Diviſion of Vulgar
X
Fractions in Arithmetic 2) (-2/
xx
is
4 m m
I
m
m
X
Now fquaring is
2 m
and adding this to both Sides of the Equation, the
Square is compleated, by Art. 56.
4 m m
= √
b
+
d m
x x
x a
x x
b
8c0
9
a a
+
m
x
9 w 2
IO
a
x
b
10+
I I
a=
2 m
d m
2 m
+
+
d m
X X
4 m m
4 mm
:+
x
2 m
مل میشود
4 m m
=100
(as before.
20, as
Then by the fourth or fifth Step we fhall find e—20,
before.
Queſtion
Of Quadratic EQUATION S. 215
Queſtion 78. It is required to find two fuch Numbers, that
the greater being added to the Square of the leffer, the Sum
may be 19:
But if the greater is multiplied into the Square of the leſſer, the
Product may be 90.
=
Let a the greater Number, e = the leffer Number,
s = 19, p = 90.
a + e e
a =
By the Queſtion.
2
a e e
P
I - ее
3
a = s
ее
P
2 ee
4
P
3.4
5
ee
5 x e e
6
6 + eeee
6+e
c་
e e
-e e
p = see
e e e e
Tranfpofe e e e e to make it affirmative.
eeee + p = see
7
see
8
e e e e
-
8
P
9
e e e e
see+p=0
see = - p
Here the Equation not only appears quadratic, the Powers.
of the unknown Quantity e, being the fame as in the two
laft Queſtions, but it is likewife ambiguous, for that Side of
the Equation which is known is negative, viz.
ㅁ ​10
e e e e -seet
p.
SS
S S
P
4
4
S
IO w 2
I I
e e
2
N|
S
S
12
e e
2
+1
12
土
​4
55
4
p the Equation
(being ambiguous, as above.
5 S
12 w 2
13 e =
-P
2
4
That is, by reafon of the Ambiguity of the Equation, it may
S
bee=
+
p, ore =
2
4
2
4
Let
2
216
ALGEBRA.
A
1
Let us fuppofe e =
今
​کیا
2
+
4
55
90.25
4
--90.
=-p
√.25=.5=✓
9.5=
10. =
P
4
2
+
-p
2
4
10) 3.162 neareſt =
9
61) 100
61
626) 3900
3756
6322) 14400
12644
✓
SS
p the Value
2
4
(of e.
Then by the third Step a = s -ee
9, if we take 10 for
the Square of e, the fquare Root of 10 being equal to e.
By trying theſe Numbers according to the Conditions of the
Queſtion, we have
a+ee = 19
aee
90 taking 10 for the Square of e, as above.
But becauſe the Value of e is a Fraction which does not ter-
minate, and therefore its exact Value cannot be found, let us
try the other Root, viz. e =
S
SS
Р
2
4
SS
90.25
go.
4
- P
√.25 = .5=
4
2
こ
​44
Of Quadratic EQUATIONS.
217
S
= 9.5
2
S S
4
S
2
2
4
p = -.5
-9. extracting the fquare Root of 9. we have
4
p3, for the other Value of e exact.
Then by the third Step as - ee = 10.
And trying theſe two Numbers by the Conditions of the
Queſtion, we have
a + ee = 19 2 As the Queſtion requires, whence the two
a e e = 90 Numbers are 10 and 3.
I have been particular in the Arithmetical Work of this
Queftion, that the Learner may fee the Method of finding both
the Values of the unknown Quantity, in any ambiguous quadratic
Equation, when the unknown Quantity is to the fourth Power.
But in this Queſtion, if we exterminate e inſtead of a, we
ſhall have a more fimple Solution.
I
2
a tee=s
a e e = p
}
By the Queſtion as
before.
I
а
3
2 a
4
e e = S
ee =
P
a
a
4.3
5
a
5 x a
6
6 + aa
7
7
Р
8
8-
sa
9
a a
-P
p = sa
a a
a at p = sa
aa-sa
sa
P
Here the Equation appears quadratis and ambiguous, as before.
~sat
9 | 10 | α =
a a
SS
s s
4.
Ff
4
?
I w
218
ALGEBRA.
SS
10 ເນ 2
II
a
2
4
S
II+ 12 a =
2 |
55
P
2
4
Let us firſt fuppofe the Root to be
4) 361 = ss
90.25
P
2
4
90.
.25=
4
whence .5V.
P
4
S
Then
= 95
2
SS
p =
-0.5
4
p, but the fquare Root of .25 is .5
9.a, for one of the Roots of the am-
biguous Equation, and from this Root, or Value of a, we ſhall
from the third, or fourth Step, find,
fquare Root of 10, as before; but this
whoſe Root cannot be exactly extracted,
Root, or Value of a, then we have a =
that e is equal to the
being a furd Number,
therefore find the other
S
55
p.
2
4
S
= 9.5
2
+
+v
p = .5
4
10.a, the other Root of the ambiguous
Equation; then by the third, or fourth Step, we ſhall find e to
be equal to the fquare Root of 9, which is 3; and theſe two
Numbers 10, and 3, anfwer the Conditions of the Queftion.
It may not be improper in this Place to add, that if the
Learner meets with any Queftions, where the Anfwers come
out
Of Quadratic EQUATIONS.
219
out in Decimal Fractions, he is not from thence to conclude
they are not the true Anſwers, as theſe are very frequent and
common: But if the Equation is ambiguous, it will be proper
to find the other Root, which may be free from Fractions;
and if this Root anfwers the Conditions of the Queftion, he
has then found the Anfwer compleat: But if the Queſtion.
will not admit of fuch an Anſfwer, he can then only approach
to the true Anfwer in continuing his Fractions at Pleafure;
but hitherto I have endeavoured to avoid thefe Circumftances,
as they only fatigue the Learner, and perplex his Mind, inſtead
of increafing his Judgment, or advancing his Knowledge in
this Science.
66. The Method of refolving Questions,
that contain three Equations, and three
unknown Quantities.
F! IND the Value of one of the unknown Quantities, in one of
the given Equations :
For the fame unknown Quantity in the other two Equations,
write, or put this Value, which exterminates that unknown
Quantity from thofe two Equations; and reduces the Question
to two Equations, and two unknown Quantities, which may be
refolved as the foregoing Questions, by Art. 55. that is,
Find the Value of one of thefe two unknown Quantities, in
each of thoſe two Equations, and making theſe two Equations
equal to one another, exterminates another unknown Quantity,
for this laft Equation will have only one unknown Quantity,
which being reduced by the Directions already given, will give
the Value of that unknown Quantity in Numbers, from which
it will be eaſy to determine the Value of the other two.
To help the Learner in his Choice which to exterminate, if
one of the three, unknown Quantities is not multiplied, or
divided by either of the other two, but theſe are multiplied, or
divided by one another, then it will be eaſieſt to find the Value
of that unknown Quantity, which is not multiplied, or divided
by the others,
F f 2
Or
220
ALGEBRA,
}
Or if one of the unknown Quantities fhould be to the first
Power only in all the three given Equations, and the other two
are raiſed to ſome higher Power, then it may be eaſieſt to exter-
minate the unknown Quantity, which is to the firft Power
only.
--
And if all the three unknown Quantities are only to the first
Power, and none of them are multiplied or divided by one ano-
ther, then if one of them has no Co-efficient but Unity, and
the other two have Co-efficients, it may be eaſieſt to exter-
minate that unknown Quantity, whofe Co-efficient is Unity.
Thefe Directions may be of Ufe to the Learner, in affifting
his Choice which unknown Quantity to exterminate, and a little
Care and Attention will help his Judgment in this Part of the
Science; I fhall only juft mention, that if any particular Diffi-
culties arife from the exterminating one unknown Quantity, it
may not be improper to make an Effay how the Work will
proceed, from exterminating fome other unknown Quantity.
Queſtion 79. There are three Numbers whofe Sum is 18:
The first being added to three times the fecond, from which Sum
fubftracting twice the third, the Remainder is 9:
But if the first is added to four times the third, from which
Sum fubftracting twice the fecond, the Remainder is 21. What
are the three Numbers ?
Let a
the first Number, e = the fecond Number, y = the
third Number, b = 18, m =
I
4 2
9, p=21.
a+e+y=b
a+ 3e − 23
2y=m
2e=p
3
a + 4y — 2 e
I
y
4
a + e = b − y
4 e
5
a = b—y—e
By the Queſtion.
Having found the Value of a in the firft Equation, in the
room of a in the fecond and third Equations, put its Value
bye, thus.
2.5 6 b-ye÷ze-2y=m
3.5 7
6 contracted 8
7 contracted
9
b-y-e+47-2ep
b
·31+ 2e=m
b+39-3e=&
Here the Queftion
is reduced to two E-
quations, and two un-
known Quantities,
for a is exterminated.
Now
The Method of refolving Questions, &c. 221
Now find the Value of either y, or e, in each of theſe two
laft Equations, and 3 y being in each Equation, find what that
is equal to.
8+39
m
1016+2e = m + 3y
3y =6+ 2e
6+3y=1+3e
10
m ΙΙ
9+3e
12
12 6 13 3y = p + 3e-b
- b
•
II 13 | 14 |P + 3 e
b = b + 2e.
+2e- m
Here we have an Equation with one unknown Quantity only,
which is reduced in the common Manner, thus,
15pte
ptemb
6p+e= 2 b
e = 2 b
26 772
b
m
کو
m
14
2 e
15 + b
16
16 - p
17
p = 6, then
13÷3
18
P + 3e
b
y
7, and
3
By the fifth Step
19
a = b — y — e
y-e=5.
Hence the three Numbers fought are 5. 6. and 7.
PROOF.
a+e+ y = 18
a+3e2y = 9
=
a + 4 y 2 e 21.
Queſtion 80. Three Men, A, B, C, difcourfing of their Shil-
lings, found, that if twice A's Shillings was added to B's Shillings,
and from that Sum fubftracting C's Shillings, there remains 15:
And if B's Shillings was added to three times C's Shillings, and
from that Sum fubftracting A's Shillings, there remains 31:
But if fix times A's Shillings was added to four times C's Shil-
lings, and this Sum added to B's Shillings, the Sum was 97.
How many Shillings had each Perfon?
Let a the Number of Shillings of A, e thofe of B, and
y thofe of C, b = 15, d = 31, m97.
=
I
23
2a+e—y=b
e37a=d
e | 3y — a =
6a+ 4y + e = m
te
By the Queftion.
Becaufe
i
222
ALGE B R A.
Becaufe has no Co-efficient but Unity, begin with finding
the Value of e, as being the moſt ſimple.
za+e=b+ y
I+y
4 — 2 a 5
4
e = b + y
2 a
Now in the ſecond and third Equations, in the room of e,
put its Value, or by — 24, as in the laſt Queſtion.
ty
2 5 6 b+y
•
3.5
2a+37 a - d
7 b + y −2a + 6a+ 4y = m
Here the Queftion is reduced to two Equations, and two
unknown Quantities, e being exterminated, and therefore pro-
ceeding, as in the laſt Queſtion,
6 contracted
8
7 contracted
9
14
b+4y-3a=d
4a+5y+b = m
1
Now find the Value of either of the unknown Quantities in
both thefe Equations: To find the Value of y,
8 + 3a101b+4y=d+3 a
10-b II 4 y = d + 3a — b
y =
9-b13 4a+5y=m
d+3a-b
II-4 12
4
m- b
13-4a
14 51 = m
·b — 4 a
m
b
4 a
145 15
15y =
5
12. 15
16
d + 3a — b
m
b
4 a
4
5
Here we have an Equation with only the unknown Quantity a
16 × 4
17 d + 3 a
a−b
4 m
b =
4 b — 16 a
5
17 X 5
18
18 | =
5d+15a-5b — 4 m — 4 b — 16 a
18+ 16 a 19 5 d +31a—5b = 4 m
1956 20
5d+31a=4m+b
31a=4m+b - 5 d
4m+b— 5d
20
-5d
21
21 31 22 a =
46
= 8, then
By
31
The Method of refolving Queſtions, &c. 223
By the 12th Step 23y=d+3a-b
= 10, and
4
By the fifth Step | 24 e=b+y=2a=9
PROOF.
2a+e-y=15
e+ 3y a = 31
6a+4y+e = 97
I have done theſe two Queftions without putting Letters for
the given Numbers, it being more eafy and familiar; but now
to do the laft univerfally, let us put Letters for the Numbers
2. 3. 6 and 4 which are given in the Queſtion, and comparing
the former Operation with the following, may render it more
eafy ;
but if the Learner finds this too perplexing, he may
neglect it, and proceed to the next.
•
Let a e and y be the three unknown Numbers as before,
and x = 2, ≈= 3, s = 6, p
Z 3, s=6, p = 4, then,
I
1xa+e-y=b
e+zy—a=d
By the Queſtion.
3
sa+py+em.
2
Becauſe e has no fpecious Co-efficient in either of the given
Equations, find the Value of e.
I ty
4
4
x a
5
xa+e=b+y
e=b+y-xa
Now in the fecond and third Equations, in the room of e put
its Value, or by — xa,
b + y
xa+zy — a = d
2.5 6
3.5 7 sa + p y + b + y — x a = m
Here the Queftion is reduced to two Equations, and two
unknown Quantities, e being exterminated; but becauſe of the
Specious Co-efficients we cannot contract them as before: Now
find the Value of y, in both thefe Equations.
1
2
6 + a
224
ALGEBRA.
xa+zy = d + a
6 + a81b + y
8 + xa9b + y + zy = d + a + x a
9—b10xy+y=d+a+xa
10÷2+111y=
d + a + x a
b
b
the Co-efficients of y being
7 + x a 12s a +py+b+y= m + x a
xa12s
13 s a + py + " m + x a-
12
-b
13 — sa 14py + y = m + xa b-sa
14+1+115y
m+xa-b
P+I
II. 1516
d+a+xa—b
Z + 1
sa
(= + 1
the Co-efficients of y being
m+xa b- sa
(p+ I
an Equation
(with only the unknown Quantiy a.
16x2+1 17 d+a+xa−b—zm+zxa—zb—zsa+m+xa—b – sa
16xz+117d+a+xa—b—
17 xp + 1 18 pd + pa + px a− p b + d + a + xa—b=zm
+zxa zb- zsa+m+x a b-sa
The Learner may think theſe Multiplications diſcouraging,
though perhaps they are not fo perplexing as he may imagine,
for at the ſeventeenth Step where m + xa-b sa is x z + 1,
put down the Product of it by z firft, which is z m + z xa
zb. zsa, after which he need only write m + xa-b—sa,
the next Part of the Multiplier being Unity; or, if it had been
another Letter, it had been no more than repeating the Multi-
plicand, with the multiplying Letter joined to each of its Quan-
tities, placing them one after another, taking due Care of the
Signs by the Rules for Multiplication.
In the fame Manner he will find the eighteenth Step multi-
plied, and a little Attention will familiarize the Operation; but
if there is any Difficulty in multiplying theſe compound Quan-
tities, the Learner may fet them down one under the other, and
multiply them in the ufual Manner.
18-xa191p d+pa+pxa—pb+d+a−b=zm+zxa
z b
zsa+m
b-
sa
19 +420 pd+pa+pxa-pb+d+a=zm+zxa-zb
4/20pd
zsa +m sa
Now tranſpoſe all the unknown Quantities to one Side of
the Equation, and all the known ones to the other Side of the
Equation.
20-
The Method of refolving Questions, &c. 225
20
Zx a
21 | pa+px
pd+pa + px a—pb+d+a—zza
Z m
Z
b
zsa+m- s a
A
21+zsa 22 pd+pa+pxa—pb+d+a+zsa
Z xa = Z M
xb+msa
22 +sa 23 pd+pa+pxa-pb+d+a+zsa
zxa+sa=zm
zb + m
p d
23-pd|24|pa+pxa-pb+d+a+zsa-zxa
+sa=zm
zb+m
24 + pb | 251pa+pxa+d+a+zsa-zxa+sa
Z m z b + m
p d + p b
d 26 pa+pxa + a + zs a zxatsa
=Z M z b + m
25-
26
27
a
- p d + p b — d
zm―zb+m—p d + p b - d
p+px+1+25-2x+5
= 8,
the Divifor is the Co-efficients of a,
connected by their Signs.
d+a+xa
b
Then by 11th Step
28
y
=IQ
2 + I
And by 5th Step
29
e = b + y − x a = 9
Queſtion 81. There are three Travellers, A, B, C, who have
traveled in all 62 Miles :
But if the Miles A travelled is multiplied by 2, and added
to the Miles B travelled multiplied by 3, this Sum is equal to the
Miles C travelled multiplied by 17:
And if 4 times the Miles C travelled, is added to the Miles
B travelled multiplied by 2, this Sum is equal to the Miles tra-
velled by A. To find the Miles each travelled?
Let a
by B, y
the Miles travelled by A, e the Miles travelled
the Miles travelled by C; p = 62, b = 2, d = 3,
m = 17, x=4, and 2 being in the Queftion before, put no
new Letter for it.
I
123
a+e+y=P
bat de e = my By the Que lion.
x y + be
a
Becauſe a feems to be in as fimple Terms as any in the three
given Equations, and having its Value already by the third Equa-
tion, therefore for a in the first and fecond Equation write its
Value xybe at the third Equation, which exterminates a.
G G
1.3
226
ALGEBRA,
I. 3 4 x y +be+ e + y = p
2·35bxy+bbe + de=my
Here the Queftion is reduced to two Equations, and two une
known Quantities, then proceed, as before, to find the Value
of either y or e, in each of theſe Equations, as ſuppoſe y.
4
be e16 xy+e+y=p — be
6 — e7xy+y=p — be e
7x+1/8 y=
p-be. e
x+1
the Co-efficients of y being x + 1,
5 bx9my-bxy=bbe + de
bbe + de
9÷m―lx10y=
m — b x
bbe de
the Co-efficients of y being m- -bx.
p-be-e
an Equation with only
mb x * + I (one unknown Quantity.
mp_mbe
8. 1011
11 xm-xb12bbe+de=
+bbxe + bxe
me—pbx+bbxe+bxe
x + 1
12 x x + 113 xbbe+xde+bbe+de=mp—mbe-me-p b x
13-xbbe 14
xde+bbe+de=mp―mbe-me-p b x + bx e
Now bring the Terms that have the unknown.
Quantity, to one Side of the Equation.
14+mbe 15x de+bbe+de+mbe-mp-me-pbx+b
15+ me 16x de+bbe+de+mbe+me=mp — p b x + bx e
16-bxe17xde+bbe+de+mbe+me-bxe=mp-pbx
1718e=
m p − p b x
xe.
=9, the
xd+bb + d +mb+m-b.x
Divifor is the Co-efficients of e, connected by
their Signs.
By Step 190 = 7
8, or 10.}/19/1
=1
By Step 3.20 a = 46
201a
Queſtion 82. Three Men, A, B, C, diſcaurfing of their Shil-
lings, found, that A's Shillings added to C's Shillings, the Sum
was double B's Shillings:
And A's Shillings added to three times B's Shillings, from which
Sum fubftracting C's Shillings, there remained 13 Shillings:
But if A's Shillings was added to the Product of B's and C's
Shillings, the Sum was 34. How many Shillings had each
Perfon?
Let
The Method of refolving Questions, &c.
227
Let a
A's Shillings, e B's Shillings, y C's Shillings'
b = 13, d = 34.
123 +
4
e=
a + y = 2 e
a+3e
y = b
}
a + ey = d
a = 2 e -y
2e−y+3e—y=b
2.4
3·4
5
6
2e −y+ey=d
5 contracted
75e— 23 =
= b
By the Queſtion.
Here the Queſtion is
reduced to two Equations
and two unknown Quan-
tities, a being extermi
nated.
Now find the Value of e, or y, in the fixth and ſeventh
Equations, fuppoſe e.
6+9 8
2e+ey=d+y
d + y
8+ 2+ 3
9
e
2+ y
7 +23
10
5e=b+2y
b+2y
10 ÷ 5
I I
e
5
¿ + 2y
9.1!
12
5.
12 x 2 + y
12x2+
13
13x5
14
d+, an Equation with
2+ only one unknown
Quantity,
2b +41 + by + 2xy=d+y
5
26 + 4y + by + 2y y=5d+5y
+23
Now bring all the Quantities that have y, to one Side of the
Equation.
15 2b −y + by + 2y y = 5 d
162yy+by-y=5d-26
145
15-26 16
Here the Equation appears quadratic, the unknown Quantity
being to the ſecond and firſt Power only, but is not ambiguous,
5d being greater than 26; then by Art. 58, divide by the
Co-efficient of yy.
by y
16 ÷ 2 | 17 | 99 + 6 = =
2
G g 2
5d-2b
2
The
228
ALGEBRA.
The Work being now prepared for compleating the Square,
becauſe the Co-efficient of
b
I
y is
to avoid the Trouble of
2
dividing this Fraction by 2, and fquaring the Quotient, ſub-
itute by Art. 57. x=
b. I
= 6.
2
Then 18
5d-2b
yy + xy=
2
18c0
x
x x
19 | yy + xy + x = = 5 d=2b + **
19 w 2 20 " +
4
5d - 26
2
+
4
xx
4
20
2
2
મ
x
5 d
d26
x x
x
|
21
y = √
+
= 6,
2
2
4
2
(C's Shillings.
S
By the 9th, or ?
11th Steps
By the 4th Step 23
22
|
e=5, B's Shillings.
a 4, A's Shillings.
=
Queſtion 83. Three young Gentlemen, A, B, C, having been
at the Gaming-Tables, from comparing their Loſſes, found, that if
from twice the Pounds A loft, was fubftracted the Pounds B loft,
there remained the Pounds C loft:
And that the Pounds A loft, added to the Pounds B loft, and
this Sum added to twice the Pounds C loft, the Sum was 19
Pounds:
But if to the Product of A's and C's Lofs, there is added B's
Lofs, the Sum is 26 Pounds. How much did each Perfon lofe?
Let a =
b = 26.
A's Lofs, e = B's Lofs, y C's Lofs, d
2a-e=y
| | 2+2 + 2 =d}By the Queftion.
2
a
e y
3 ay +e=b
19,
Becauſe e ſeems to be in the moſt fimple Terms, therefore
find its Value.
te 4
از
2 a = y + e
+ + 3 / 3 / ==
4
5
e = 2 a y
2.5
The Method of refolving Questions, &c.
229
2.5
The Queſtion is here
Equa-
6a+2a−y+2y=dreduced to two
tions and two unknown
3.5
7
ay+2a-y=b
6 contracted
8
3a+y=d
Quantities, for e is ex-
terminated.
:
7 — 2 a
9
9÷a
Find the Value of a, or y, in the feventh and eighth Equations.
a y — y = b — 2 a
b — 2 a
I
IO
y =
a — I
8
8 — 3 a
I I
y =
d
d — 3 a
b
2 a
IO. II
I 2
= d
за
1
12 X a I
13
a- I
b— 2a = da — za a―d+za
3a
Now bring all the Quantities that have a on one Side of the
Equation, obferving to have the higheft Power of a affirmative.
3⋅aab5a=da-d
13+3aa
14
3aa+b―2a=da―d+3a
14- за
15— da
15
16
16 — b
17
3aab5a-da-d
3a a — 5 a— da = - d — b
Here the Equation appears both quadratic and ambiguous, for
the unknown Quantity is to the ſecond and firft Power only,
and it is ambiguous, becaufed-b the Side of the Equation
which is known, is negative; dividing by the Co-efficient of
aa, as in the laſt Queſtion,
—
d — b
17+ 3 | 18 | aa—50 — da = = =
| 2 3
17÷3
3
The Work being now prepared for compleating the Square,
fubftitute
x=
- 5 — d
3
in the laſt Example.
Then 19
18 c
8 the Co-efficients of a, as
ala
d - b
аа
xa=
3
| 20 |
Xx
d
b
4
3
aa-xa+
+
**
4
19 w
230
ALGEBRA.
મ
19 w 2
21
a
20+
2
| N
22
a=
2
2
*
d - b
3
+
4
d - b xx
+
4
For the Practice of the Learner, let us
Then by the tenth, or eleventh Steps
And by the fifth Step
4,
3
(+ 1 = 3, or 5.
fuppofe a = 3
e
y = 10
:6-10
=-4, which is an Impoffibility, that e an affirmative Quan-
tity, can be equal to a negative 4.
Now let us fuppofe
Then by the tenth, or eleventh Steps
And by the fifth Step
Then 2 a
e = y
a + e + 2y = 19
ayte = 26
a = 5
y = 4
e = 6
And theſe three Numbers anfwering the Conditions of the
Queſtion, are the true Numbers fought; from hence the young
Analyft may obferve, that in quadratic ambiguous Equations, if
one of the Roots of the unknown Quantity does not anſwer the
Conditions of the Queſtion, he ſhould find the other Root, and
try that, before he concludes his Work erroneous.
I fhall now fhow the Learner the excellent Method of
refolving all Equations, be their Powers never fo high, by the
univerfal Method of Converging Series.
67. The Refolution of Adfected Equations,
by the univerſal Method of Converging
Series.
Ex. 1.
CASE I.
UPPOSE there was given aaa+ a = 9282, to
SUP
find a
Then fuppofe, or imagine a to be
Confequently the Cube of a, or a a a, is
20
8000
Theſe being added together, becauſe it is a aata 802.0
in the given Equation, the Sum is
2
Hence
Of Adfected Equations, &c.
231
Hence a muſt be more than 20, for if that had been the true
Root, the Cube of 20 added to its firft Power, or 20, muſt
have been equal to 9282 the given Number, for theſe are the
fame Powers of a as in the given Equation; but that Sum being
only 8020, which being less than 9282, the Value of a muſt
be more than 20. To find how much that is?
Let r = 20, and for what 20 wants of the true Number or
Root, pute:
Then will r+ea, or the true Root of the Equation,
hence by determining what is, we find the Number that is to
be added to r or 20, which Sum will be the Root of the given
adfected Equation, to do which, put down,
1|r+e=a
Now raife this Equation to the third Power, becauſe we have
a a a in the given Equation.
I
13 | 2 | rrr + 3rre + 3ree + eeeaa a
Add the first and fecond Equations together, becauſe in the
given Equation it is a a ata.
1 + 2 3
it is
3.45
4
5 in Numbers
6
That is
7 — 8020.
8
-60 9
920.016+.
tel
10
For the Reafon of adding the
Quotient Figure, or 1, to the
Divifor, fee Article 68, in the
next Page,
rrr+3rre+3ree+eee+r+e= aaa+a
But from the given Equation
aaa + a = 9282
rrr + 3rre+gree +eee+r+e=9282
each Equation being equal to a aata.
Putting this Equation into Numbers,
and rejecting the Powers of e above e e.
8000+1200e +60ee +20+e = 9282
8020 +1201e+ 60 ee 9282
+60
1201 e + 60 e e = 1262
Divide by the Co-efficient of e e, and
we have,
20.016e+ee=21.0333 ad infinitum.
Dividing by 20.016+e, that is, by the
Co-efficient of e plus e, and we have,
2.1.03333
в
20.016 +e
In Numbers thus:
20.016) 21.0333 (1 = e
I
21016
21.016
17 the Remainder
(being very finalļ reject it.
By
232
ALGEBRA.
By this it appears that e1, that is, I is to be added to the
firft fuppofed Number 20, which Sum is to be the Value of a,
or the Root of the given adfected Equation.
We affumed r = 20
And found e = İ
r + e = 21 = a
To try whether 21 is the true Root, raiſe it to the ſeveral
Powers of a in the given Equation.
a = 21
a = 21
21
42
a a = 44!
a
21
441
882
1
a a a = 9261
a
21
Then à a a + a =
+ a = 9282, which being the fame with the
Number in the given Equation, it appears that a = 21.
68. By reviewing the Operation, the Learner may obſerve,
Firft, That we fuppofed a Number for the true Root, which
upon Trial was found less than the true Root.
Secondly, For that Deficiency or Want, we put e,
other Letter.
or any
Thirdly, By connecting r the Number firſt ſuppoſed to be
the Root, with e by the Sign+, we have r+e for the true Root,
r being a known Quantity, and e the unknown Quantity.
Fourthly, We raife re to the feveral Powers of the un-
known Quantity, that are in the given adfected Equation.
Fifthly, Then we add thefe feveral Equations together, re-
jecting all the Powers of e, or of the unknown Quantity above
the Square, for in the given Equation all the Powers of the
unknown Quantity have the Sign +, but when any of theſe
have the Sign, then their refpective Equations must be
fubftracted, as at Step 5, Example 3, Page 238.
Sixthly, By thefe Means we have an Equation in the Terms
of r and e, equal to the given Equation.
Seventhly,
Of Adfected Equations, &c.
233
Seventhly, This Equation is put into Numbers, r being a
known Quantity, and the leſs abſolute Number is tranſpoſed to
the Side of the Equation of the greater abfolute Number, and
fubftracted from it.
Eighthly, After this, the Equation is divided by the Co-
efficient of the Square of e, or the unknown Quantity.
Ninthly, This laft Equation is divided by the Co-efficient of
e plus e, which leaves e on one Side of the Equation by itſelf.
Tenthly, In the Arithmetical Work, becauſe the laſt Divifor
confiſts of a Number plus e, therefore, as the Quotient Figure
is found, it is added to the Diviſor to make it compleat; and
if the numerical Operation had been continued to more Places
of Figures in the Quotient, then the Quotient Figure muſt be
twice added, once when it is found, and once at the next Step
in the Divifion, as in the next Page.
Eleventhly, The Quotient thus found being the Value of e,
or the unknown Quantity, it is added to the Number firft fup-
poſed to be the Root of the Equation, which is repreſented by r,
and this Sum is fuppofed to be the Root required.
This Operation to find e is the fame as the common Method
of finding the unknown Quantity, till we come to the tenth
Step, where the unknown Quantity making Part of the Diviſor,
it is carried to the other Side of the Equation, and the Divifor
being a known Number +e, the Quotient as it is found is
added to the Diviſor, to make it compleat, as before-mentioned.
But if this Number fhould not be the true Root, the Opera-
tion must be repeated, making the Number thus found =r,
and at the ſecond Operation, the Work in any common Cafe
will be ſufficiently exact: And from the Repetition of the Ope-
ration, whereby we approach nearer and nearer to the true Root,
this Method is called the Method of Converging Series, or of
Approximation.
Example 2. Suppoſe a aa+a a + a = 42997850, to find a.
Suppoſe a to be
Then the Cube of a, or a a a is
And the Square of a, or a a is
Theſe being added, becauſe it is aaa+aat a
in the given Equation, the Sum is
300
27000000
90000
} 27999399
+aata}
But 42997850, the given Number, is greater than 27995300,
therefore the Root must be more than 300.
Hh
Now
234
ALGEBRA.
Now let r 300, and e
= what 300 wants of the true
Root.
Then
13
I
1 2
}
I Ⓒ 2
1+ 2+ 3
3
4
But
5
4.5
6
6 in Numbers
7
That is 8
8-27090300 9
9÷901 ΙΟ
r+e=a
rrr+3rre + 3ree
+eee = aaa
rr+are+ee aa
According to Parti-
cular 4, Art. 68.
r+e+rrr+3rre+ 3ree+rr+ 2re + ea
= aaa+aa+a, by Particulars 5
and 6, Art. 68.
aaa+aa+a= 42997850, from the
given Equation.
r+e+rrr+3rre+3ree+rr+2re-
+ee = 42997859
13300.334+e
II
e
300+e+27000000+270000e+9ocee
+90000+60ce +ee = 42997850
27090300+270601e+901ee4299785a
270601 e +901ee=15907550
300.334 e +ee = 17655.43 from
Particular 8, Art. 68.
17655.43, from Particular 9,
300.334+e
(Art. 68.
300.334) 17655.43 (50.34e, the firft Figure being
5.
Divifor 350.334 1751670
50.3
1387300
1201902
Divifor 400.634
•34
1853980
Divifor 400.974
1603896
£
250084
in the Place of Tens, place the
5 under the Place of Tens in
the Divifor; and this the
Reader is to obſerve, to place
the Quotient Figures he adds
to the Diviſor, under thofe of
the fame Denomination.
द
e
* = 300
50.34
r+e=·350-34a, hence I fuppofe the Root of the given
Equation is 350.34 but to try it, raife 350.34 to the feveral
Powers of a in the given Eqnation.
1
@=
Of Adfected Equations, &c.
235
a = 350.34
a = 350.34
140136
105102
1751700
105102
aa 122738.1156
a
350.34
4909524624
3682143468
61369057800
3682143468
a a a = 43000071.419304
a a= 122738.1156
аа
a =
•
350.34
Then a aa aa + a = 43123159.874904 which being
greater than 42997850, the Root cannot be fo much as 350.34.
and this leads us to explain the Method of finding the true Root,
when the Number affumed is greater than the Root required, or,
CASE 2.
Let us take the laſt Example, viz. aaa+aa+a=42997850.
And ſuppoſe the Root to be 350.34 which we know is too
much by the laft Operation:
Now put the Number to be fubftracted from 350.34
fuppofing r = 350.34 and to r connecting e by the Sign -,
we have,
rea, or the true Root
I
x 3
2
rrr-3rre+zree-eee
- ваа
1 & 2 3 rr2re + ee=aa
By Parti-
cular 4.
Art. 68.
5
Now collect theſe three Equations by Art. 68, Particulars
and 6, and rejecting e e e, we have,
I
1 + 2 + 3 4
But
4.5 6
r—e+rrr—3rre+3ree+rr-2re
+ee=aaa+aa+a
5|aaa+aa+a=42997850, from the
given Equation.
r_e+rrr—3rre+3ṛee+rr-are
+ee=42997850
Hh 2
6 in
236
ALGEBRA.
6 in Num.
7
That is
8-
9
350.34➡e+43000071.42-368214.3468 é
ee+122738.1156-700.68 e
+1051.02
+ee = 42997850
8 | 43123159.8756-368916.0268e+1052.02ee
= 42997850
Now tranfpofe 42997850, it being leſs than
43123159.8765
125309.8756-368916 0268e+1052.02 e e
=0, for one Side of the Equation being
fubftracted from the other Side, the Re-
mainder muſt be nothing, as both Sides of
the Equation are equal. Now tranſpoſe
the feveral Quantities which contain e to
the other Side of the Equation.
125309.8756+1052.02ee368916.0268
368916.0268e-1052.02ee125309.8756
Here dividing by the Co-efficient of ee as
before,
12 | 350.673eee119.1135
13
But now divide by the Co-efficient of e
minus e.
e =
119.1135
350.673-e
9 +
ΙΟ
10
II
12-
In Numbers:
350.673) 119.1135 (.34
•3
Divifor 350.373 1051119
- ·34
1400160
Divifor 350.033 1400132
28
Having thus determined to be .34 it must now be ſubſtracted
from r, becauſe it was affumed r —
— — the true Root.
But 7 was fuppofed.
, which we have found =
Hencer
350.34
.34
350. a, the Root of the giyen
adfected
Of Adfected Equations, &c.
237
adfected Equation, which is proved by raifing 350. to the feveral
Powers of a in the given Equation,
Thus, a a a = 42875000
ва
== D
122500
350
Confequently a a a+aa+a=42997850 which being the
fame Number as in the given Equation, it fhows that a is
exactly equal to 350.
In the above Operation, at the thirteenth Step, the Learner may
obferve, that the Divifor is 350.673-e, therefore here, as the
Quotient Figure is found, we fubftract it from the Part of the
Divifor 350.673 to have the Divifor compleat, which is likewife
done twice, once before the Divifion is made at that Figure,
and once afterwards: But in the firſt Cafe, when r is affumed
too little, then the Quotient Figure is added, as at Particular 10,
Art. 68. the Sign then being contrary to what it is now.
The Learner may further obferve, that by this fecond Ope-
ration, we have found the true Root, whereas by the firſt Ope-
ration it was .34 too much, and therefore if the true Root does
not come out at the firft Operation, make a ſecond Operation,
fuppofing the Number found at the firft Operation to be r, and
call it re, or r -e, for the true Root as the Occafion requires,
that is, as the Number at the firft Operation is either greater or
leffer than the true Root; which fecond Operation will give the
true Root very near, and near enough for any common Cafe,
though if the Arithmetical Divifions were continued, as they
will not terminate, do not give the true Root exactly, as in
the Divifion of thofe Decimal Fractions which never termi-
nate; in ſuch Divifions we leave off when the Quotient is to
a fufficient Degree of Exactnefs, fo the fame is done here
when we are near enough the Truth; and in common Cafes,
two or three Places of Decimal Fractions are fufficient, and
according as they happen the true Root is fometimes found;
and in general, continue the Divifion, at the fecond Opera-
tion, to as many Places of Decimal Fractions as are in the
Number found in the first Operation: And after the Number
found at the fecond Operation is added to, or fubftracted from
the Number found at the first Operation, if there is a very
ſmall Fraction you may reject it; but if the Fraction fhould
be very near an Unit, then take 1 for it, which add to the
Integers, and try whether the whole Number thus found is not
the true Root. In Arithmetical Queſtions, whofe Anſwers are
often
238
ALGEBRA.
often in whole Numbers, this Caution may help the Learner
to chufe the true Root exactly.
The Reaſon why this Method does not abfolutely give the true
Root is the arbitrary rejecting all the Powers of e above e e.
Example 3. Admit a a aa aa ≈ 46526760, to find a.
Now fuppofe a 400.
Then a a a is
And aa is 160000, which must be fubftracted}
becauſe it is a a in the given Equation
To which adding a, or 400, it being + a in
in }
the given Equation
Hence a a a
aata is
64000000
160000
63840000
400
63840400
\
Which exceeding 46526760 the Number in the given Equa-
tion, a muſt be less than 400.
Then let r = 400, e = the Number that 400 is too much,
which being the fecond Cafe, Page 235.
Hence
1 & 3
1 & 2
123
j j
e
a
3rre+3ree — e e e — a a a
2re+ee=aa
Becauſe in the given Equation the Quantities a aa and a are
affirmative, therefore add the firſt and ſecond Equations together.
1 + 2 | 4 | r—e+rri―3ṛre+3ree-eee=aaa+a
Becaufe in the given Equation it is a a, therefore fubftract
the third Equation from the fourth, or Sum of the first and fecond
Equations. And here the Reader is to obferve, that if in the
given adfected Equation, any Powers of the unknown Quantity
have the Sign, the Equation which arifes from involving
-e to fuch Powers, is to be fubftracted instead of being added.
2
4-3 5
But 6
5.6
7
r—e+rrr—3rre+3ree-eee-rr
+2ṛe- e e = a a a a a-t- a
a a a—a a + a = 46526760, by the
given Equation.
r_e+rrr-3rre+3ree-eeerr
+are-ee = 46526760
Putting
Of Adfected Equations, &c.
239
t
Putting this Equation in Numbers, and rejecting all the
Powers of e above e e.
7 in Numbers 8
£ contracted | 9
10
400e + 64000000480000e+
I 200 e e
46526760
160000 + 800e — ee=
63840400 + 1199 e e 479201 e =
46526760
Tranſpoſe 46526760 it being less than
63840400
17313640 +1199 ee-479201 e = 0
Now tranſpoſe the Quantities that have
e, to the other Side of the Equation.
479201 e 17313640 +1199 ee
9-
10 +
I I
II
[2
479201 e
12-
小
​=
1199ee 17313640
Dividing by the Co-efficient of ee,
13 399.66 e-ee = 14440.06
14
|
Now dividing by the Co-efficient of e
minus e.
14440 06
399.66 — e
In Numbers thus:
399.66) 14440.06 (40,16 =e.
40.
Divifor 359.66 143864
40.I
53660
Divifor 319.56
1
31956
.16
217040
Divifor 319,40
191640
25490
Now 400
40.16
r
e
e = 359.84
a, and to try if this is the true Root,
raiſe it to the feveral Powers of a, in the given Equation.
240
ALGEBRA.
$
5
a = 359.84
a = 359.84
143936
287872
323856
179920
107952
a a = 129484.8256
a=
359.84
5179393024
10358786048
11653634304
6474241280
3884544768
a a a=46593819.643904
aa ==
129484.8256
Remains 46464334.818304
+a=
359.84
Sum, or aaa-aa+a=46464694.658304 which being
leſs than 46526760 the Number in the given Equation, the
Root or a muſt be more than 359.84.
Therefore, for a fecond Operation, fuppofe r = 359.84 and
what it wants of the true Root, then it being r+e=a,
it is now the first Cafe, Page 230.
Therefore
I
I
3
2
1 & 2
3
r+e=a
r r r + z r r e + 3ree + e e e — a a a
rr+2rete e = a a
Add the first and fecond Equation toge-
ther, becauſe in the given Equation
it is a ta a a.
I + 2 4 rrr+3rre+3ree+eee+r+e=aaa+a
From this Equation ſubſtract the third
Equation, becauſe it isa a in the
given Equation.
rrr+3rre+3ree+eee+rters
·2re - ee = a ɑ ɑ —
aa + a
6 | aaaaa+a = 46526760 by the
4 - 3
5
But
5.6
7
given Equation.
xrr+3rre+зree+eee+r+e-
J2 reee = 46526760
·rr
Put
Of Adfected Equations, &c.
241
7 in Numbers
8 contracted
9-
Put this Equation in Numbers, and re-
ject the Powers of e, above ee.
846593819.644 + 388454 • 4768 e +
1079.52ee359.84+e―129484.8256
-719.68ece=46526760
946464694.6584 +387735-7968e+
1078.52ee46526760
10387735.7968 e +1078.52 e e =
62065.3416
10 11
Dividing by the Co-efficient of ee.
359.5 e tee = 57.547
Now dividing by the Co-efficient of
plus e.
II
I 2
57.547
In Numbers thus ;
359.5)
359.5+
57.547 (.16 = &.
+
.I
Divifor 359.6
3596
.16
215870
Divifor 359.76 215856
14
The Reader will obferve that in this Divifion, I have taken at
once two Figures from the Dividend, viz. 70, becauſe in adding
the .16 to the Divifor, the Number of Places there is increaſed-
by one, therefore I take one Figure more from the Dividend
than is uſual; which is recommended to the Reader's Attention,
as he may again meet with the fame Cafe.
Now r
te
359.84
.16
+ + e = 360.00 a, which will be found to be the true
Root, by involving it to the feveral Powers of a in the given
Equation, and adding or fubftracting them according as thoſe
Powers of a are there connected by the Signs + or —.
It may be proper to inform the Learner, that the nearer
the Number is taken to the true Root, the nearer the Opera-
tion will come to the Truth, and therefore after he has tried the
firſt Suppoſition, if he thinks he can make a fecond Suppofition
nearer the Truth, it will be right to do it, which perhaps
I i
}
may
242
ALGEBRA.
may bring out the Root fo near at the firft Operation, that
it may fave him the Trouble of making a fecond Operation.
Thus,
If a a a + aa + a=4942070, to find a.
Suppoſe a to be 100.
Then the Cube of a, or a a a is
And the Square of a, or a a is
And a is
1000000
ICOOO
100
The Number in the given Equation is
ΙΟΙΟΙΟΟ
4942070
If we ſuppoſe a = 100, then the Sum of its
feveral Powers are
}
ΙΟΙΟΙΟΟ
Difference wanting
3931970
Now let us make a fecond Suppofition thus,
If a 200,
Then the Cube of a, or a a a is
8000000
And the Square of a, or a a is
40000
And a is
200
Sum of the feveral Powers of a, if a is 200
The Number in the given Equation
8040200
4942070
3098130
J
Difference over
For a third Suppofition, ſuppoſe it 160 and try with that,
and if it be less than just, it muſt be r
160 and r+e=
a; if 160 be too much or more than just, then it muſt be
во
When there are two Suppofitions made, one being more
than juſt, and the other less than just, it may be convenient to
make a third Suppofition between the two, and proceed by
Cafe 1 or 2, according as the fuppofed Number is more or less
than just.
Having explained the Method of refolving adfected Equations,
we proceed to fuch Queftions as produce thefe Equations.
The
[243]
The Manner of folving Questions,
Questions, when
the unknown Quantity has feveral Powers
in one Equation, and only the first
Power in the other Equation.
69. the unknowe
HEN the unknown Quantities are to the first and
W
fecond Power in one Equation, and but to the firſt
Power in the other Equation, find the Value of that unknown.
Quantity, in the Equation where its Terms are the more fimple;
raife this Equation, or Value of the unknown Quantity, to the
feveral Powers of the unknown Quantity in the other Equation;
then in that Equation for the feveral Powers of the unknown
Quantity, write, or put theſe Values, which exterminates that
unknown Quantity, leaving an Equation with only one un-
known Quantity, which may be refolved by fome of the
Methods already explained.
Queſtion 84. There are two Numbers, if the Square of the
greater is divided by the leffer, to this Quotient adding the greater,
from which Sum fubftracting the Square of the leffer, the Remain-
der is 100:
And the Sum of the two Numbers is 50. What are the Num-
bers fought?
Let a the greater, e = the leffer Number, m = 100,
P = 50.
I
2
a a
e
+a-ee=m
a+e=p
By the Queſtion.
In the first Equation both the unknown Quantities are to the
firft and fecond Power; but in the fecond Equation they are
only to the firft Power; therefore, according to the Directions,
find the Value of a or e, in the fecond Equation.
2 el 3│a=pe
Becauſe a is to the fecond Power in the firft Equation, raiſe
the third Equation to the fecond Power.
Ii2
3G
244
ALGEBRA.
324aa=pp-2 petee
Now for a a and a in the firſt Equation, write their reſpec-
tive Values, pp 2 pe+ee, and pe, found by the third
and fourth Equations, then we have,
1.3.4
5 x e
That is
7 +eee
8 + pe
9 in Numbers
5
6
879
9
IO
That is
II
pp-2pe tee + p- e — e e = m,
PP
e
an Equation from which a is exter-
minated, and contains only the un-
known Quantity e.
e e em e
2рetee+pe—ee — eee -
pp - pe
·e e e
= me
pp-pe=me+eee
pp=eee+me+pe
e
eee + 100 et 50 e = 2500
e e e ÷ 150 e = 2500
Here the Equation appears to be adfected, and to refolve it,
let us fuppofe e = 9.
Then eee = 729
And 150e
1350
2079 which being less than 2500, therefore e
muſt be more than 9.
Then let r9, and y
e, then by Cafe 1, Art. 67,
I ✪ 3
I
7 2
what 9 wants of the true Value of
we have,
r + y = e
r r r + зrry + 3ryy=eee, rejecting
the Powers of y above yy.
Becauſe in the given Equation e is multiplied by 150, there-
fore multiply the first Step by 150.
1X 150 | 3 | 150 r + 150 y ≈ 150 e
Now add the fecond and third Equations together, becaufe
the like Powers of e in the adfected Equation, are connected by
the Sign +.
2+3 4
But
5
rrr +3rry+3ryy +150r + 150y
=eee + 150 e
eee + 150 e=2500, by the given
Equation.
4.5
Of folving Equations, &c.
245
6 rrr + 3rry+зryy +150r +1509
4.5
6 in Numbers
7
7 contracted
8
8
2079
9
9÷27 ΙΟ
=2500
729+243 +27 39 + 1350 + 1509
= 2500
2079 +3933 +27 yy 2500
393y+27y=421
Dividing by the Co-efficient of yy.
14.55y + y = 15.59
Now dividing by the Co-efficient of y
plus y
· 10 ÷ 14.55 + "
II
小
​y =
15.59
14.55+y
Operation 14.55) 15.59 (1.=y
I.
15.55
Divifor 15.55
4 Remainder neglected.
r 9
y = I
r+y=10=e, which being involved and tried will be
found to be the true Root: Hence 10 is the leffer Number fought.
Then by the third Step of the Work to the Queſtion a =
e=40, the greater Number fought.
= p
In the Divifion for finding y, the Learner may obſerve, that
as the two next Figures in the Quotient will be Cyphers, and in
the Places of Fractions, and the third Figure being of ſo ſmall
a Value, I proceed no further in the Divifion, but leave it as in
the Work, and fo happen to find the true Value of e.
Queſtion 85. Two Men, A and B, have fuch a Number of
Pounds, that the Pounds A has, divided by the Pounds B has,
and from this Quotient ſubſtracting three times the Square of B's
Pounds, and to the Remainder adding the Square of A's Pounds,
the Sum is 27:
But if from the Pounds A has, there is fubftracted the Pounds
B has, the Remainder is 5. How many Pounds had each Man?
the Money of A, e the Money of B, d=27,
Put a
= = 5.
I
a
e
3ee+aad
|: 1 : - 3re + = d } By the Quefion.
2 A
In
246
ALGEBRA.
In the firft Equation a and e being to the firſt and ſecond
Power, and to the first Power only in the fecond Equation,
therefore by the Directions find the Value of a, or e, in the
fecond Equation, fuppofe we find the Value of a.
2+e|3|a=x+e
Raife this to the fecond Power, becauſe a is to the fecond
Power in the firft Equation.
32 | 4 | aa = xx+2xe + ee
Now for a and a a in the firft Equation, write their refpective
Values, xe, and xx+2xe +ee.
I.3.4 5
5 x e
That is
7 in Numbers
8 + 2eee
6-10ee
6
7
8
9
.00
IO -
e
1 I
II — 25 e
25e112
x + e
e
-3ee+xx+2xe+eed, here
a is exterminated, for the Equation
contains only the unknown Quantity e.
x+e—zeee+xxe+2exe+eee = de
x+e−2eee+xxe+2xee=de
5+e—2eee+25e+10ee = 27 e
5+e+25e+10ee=2eee +27 e
5+e+25e=2eee-10ee27e
5+25e=2eee 10ee +26e
10ee+e=5
2 e e e
To refolve this Equation, ſuppoſe e = 6.
Then 2 eee = 432
10ee = ==
-360
72
+e=6
78 which being greater than 5, the Num-
ber in the given Equation, hence e cannot be fo much as 6,
therefore,
Let r6, and y what 6 is too much, then by Cafe 2,
Page 235.
I ✪ 3
3 | 2
ggg
- y = e
3 rry + зry y = e e e re-
jecting the Powers of y above yy.
Becauſe
Of folving Equations, &c.
247
Becauſe in the given Equation e e e is multiplied by 2, there-
fore multiply the laſt Equation by 2.
2x2 | 3 | 2rrr—6rry +6 ryy2eee
Now raiſe r — ye to the fecond Power, after which multi- -
ply it by 10, becauſe it is 10 ee in the given adfected Equation.
I G 2
4 X 10
4
5
1
rr2ryyyee
Iorr
20ry + 10yy 10 еe
Then add or fubftract the Equations that are equal to 2 ee e,
10ee and e, according as thofe Quantities have the Signs +
or, in the given adfected Equation.
у
3- 5+ I 6 2 r r r — 5rry 6 ryy—10rr+20ry
10gy+ry = 2eee - 10 eete·
Zeee 10eee5, by the given
Equation.
But
6.7
7
པ་
8
8 in Numbers 9
2rrr-6rry+6ryy-10rr+20ry
—10yyry=5
432-216y+36 ÿy — 360 + 1209
360+
10yy+6y=5
9 contracted 1078-97y+26yy=5
Tranſpoſe 5 it being less than 78
10 — 5 I I
II + 977 12
12-26уy 13
13 ÷ 26 | 14 |
14 3.73-y 15 y
73-97y+ 26yyo, for one Side
of the Equation fubftracted from the
other, the Remainder muſt be no-
thing, both Sides of the Equation
being equal to one another.
73+26yy= 97 y
97 y — 26 y 773
Divide by the Co-efficient of yy.
3.731 — y y =
3.7313 2.807.
Now divide by the Co-efficient of y
minus y.
2.807
3.73-y
Operation 3.73) 2.807 (1.=y
2.73.
I.
Divifor 2.73
7 Remainder neglected.
r = 6
મ
248
ALGEBRA.
}
=
*
= 6 by Suppofition,
I
r—y=5=e, which being involved and tried it will be
found to be the true Root, hence B had 5 Pounds.
Then by the third Step of the Work to the Queſtion ax
+e=10 Pounds, the Money A had.
70. The Numerical Method of refolving adfected Equations
being explained, we shall now show the Learner, that every ad-
fected Equation has as many Roots, either real or imaginary, as
are the higheft Dimensions of its unknown Quantity.
For in any Equation where the higheft Power of the un-
known Quantity is the Biquadratic, or fourth Power, then
there may be four Values of the unknown Quantity; if it is
only to the third Power, then there may be thiee Values of the
unknown Quantity, and fo on: But there cannot be more
Roots or Values of the unknown Quantity than there are Di-
menfions in the Equation.
Theſe Roots are fometimes affirmative, and fometimes nega-
tive, and fome Roots are impoffible. The Reader obferving how
Quadratic Equations were compounded and generated, may
better understand the Nature of theſe Roots. Thus,
Suppofe a = 1, then a-
2 = 0.
1=0, again ſuppoſe a = 2, then
Now multiply theſe two together
= 1
a
2
a a
a
2a+2=
7
}
O
An Equation of two Dimenſions, which a a-3a+2=
has two Roots, viz. 1 and 2
Again, let a 3, then
From multiplying theſe together, we have an
Equation of three Dimensions, and which has
3 Roots, viz. I. 2 and 3.
Laftly, fuppofe a=-5, then
An Equation of 4 Dimenſions, and
which has 4 Roots, viz. 1.2.3.
and
a a a a
a 3
a a a 3aa2a=
3aa+ga 6
}aaa—baa+11a
7-6=
a+5=0
6a=0
·6 aaa+11 aa
+5aaa-30aa+55a-300
-5, and fo of any other Power, a
aaaa aaa-19a a +49a-30 = 0
Theſe feveral Multiplications must all be
Multiplicand and Multiplier are each o.
a—30
becauſe the
ง
2
By
Of folving Equations, &c.
249
By the fame Method that we found the two Roots in Qua-
dratic Equations, we may find the Roots of thefe Equations.
For fuppofe we had this Equation aa aa — a aa— 19aa49a
30: o given, which being refolved by the Method of Con-
verging Series, we fhall find a 1, whence I is one of the
Roots of the given adfected Equation; now tranfpoſe 1 to make
it a-1 = 0, take the given Equation, which being equal to
nothing, and dividing it by a 1, the Quotient mult be equal
to nothing, thus,
a—1—0) aaaa— aaa—19 aa+49a—300 (aaa—19a+30=0
aaaaaaa
19 aa+49a―30
19 aa +19a
30a-30
300-30
0
Here we find the Quotient to be a a a
19a+30=0, and
folving this Equation by the Method of Converging Series, we
fhall find a 3, for another of the Roots of the given adfected
Equation.
Then a 30) a aa—19a+30=0(aa+3a—10=0
a a a
заа
Заа
19 a +30
заа
- 9 a
10a +30
10a +30
O
Hence we have got this Quadratic Equation aa+3a-10 = 0,
whence a a+3a= 10, the two Roots of which are 2 and —- 5,
the two remaining Roots of the given adfected Equation; in
the fame Manner all the poffible Roots of any other Equation
are determined.
And to give the Learner an Inſtance where ſome of the Roots
of an Equation are impoffible :
Suppofe a a a4 aa + 4a — 16 = o, by tranfpofing 16 and
refolving the Equation by the Method of Converging Series,
we fhall find a = 4: Then tranſpoſing 4 to make it a 4 = 0,
and making the given Equation equal to nothing, and dividing
thus,
K k
4
250
ALGEBRA.
a — 4 = 0) a a a—4aa+4 a−16 = 0 (a a + 4 = 0
a a a
4 a a
4 a
16
4 a
16
0.
Becauſe the Dividend and Divifor are both equal to nothing,
therefore the Quotient must be equal to nothing; but if a a +4
o, then aa=— 4 an Equation which has no real or poſſible
Root in Nature, it being impoffible to generate or produce a
negative Square, for minus multiplied into minus, as well as plus
multiplied into plus, makes the Product affirmative, or plus.
Queſtion 86. Three Merchants, A, B, and C, found the
Pounds A and B had gained, were equal to twice the Pounds C
had gained:
But if the Pounds A gained were added to twice the Pounds B
gained, and this Sum added to the Pounds C gained, it made
19 Pounds
And the Sum of the Squares of each Perfon's Gain was equal
to 77 Pounds. How much did each Perfon gain?
Let a
the Gain of A, e = the Gain of B, y = the Gain
of C, m = 19, p = 77.
123
I
e
4
42 5
a + c = 2 y
a + 2e+y=m
By the
aa+ee + y y = p S Queſtion.
a = 2y e. Raife this Equation to
the fecond Power, it being aa at
the third Equation.
aa=4yy4ye + ee
Now for a, and a a in the fecond and third Equations write.
their respective Values, viz. 2y-e, and 4 yy-4ye+ce.
2 4 6
3.57
3y+e=m
5yy-4ye+2ee = p.
Here the Queftion is reduced to two Equations and two
unknown Quantities, for a is exterminated, therefore in the
fixth Equation, find the Value of e, or y, and raiſe it to the
fecond Power, for thoſe Quantities are to the fecond Power in
the ſeventh Equation.
6-37
Of folving Equations, &c.
251
6
31
8
em 3Y
8 0 2
9
ee = mm
6 my+9yy Multiply this
Equation by 2, becauſe it is 2 ee in
the feventh Equation.
9 x 2
ΙΟ
2 ee = 2 m m 12 my
18yy
Now in the ſeventh Equation for e and 2 e e write their Values
at the eighth and tenth Steps.
7.8.10 II
12 my
5yy—4ym + 12yy+2mm
+ 18 y y = p, an Equation with
only the unknown Quantity y.
354y-16my + 2mm = p
11 contracted
I 2
12- 2 m m
13 | 35yy
16 my = p.
p-2 mm, here the
Equation appears quadratic, and it is
likewife ambiguous, for 2 mm
greater than p.
1335 14 yy-
1415YY
16 my
p — 2mm
==
35
35
16 my
+
256 mm
256 m m
35
4900
4900
+
p — 2 mm
35
16 m
The Co-efficient of y is
which being divided by 2, or
35
2
by the Rule in common Arithmetic for Divifion of Vul-
I
16m
gar Fractions, the Quotient is
the Square of which is
70
256 mm
4900
16 m
256 mm
p - 2mm
15 vw 2
16y
+
70
4900
35
16+
16 m
70
16 m
256 mm
P
m m
17|1=
土
​+
70
4900
35
By the 8th Step
By the 4th Step
= 4.9999 or 3.68.57
But 4.9999 is the Number, the Anſwer
being 5. then if y5
18
e - in
19
a = 2 y
Kk 2
31 = 4
e = 6
Queſtion
252
ALGEBRA.
Queſtion 87. A, B, and C, having been at the Gaming-Table,
found the Pounds A loft added to the Pounds C loft was equal to
twice the Pounds В loft:
But the Pounds A loft added to the Pounds B loft, and this
added to twice the Pounds C loft, the Sum was 22 Pounds:
And the Product of what A and B loft, being added to three
times the Product of what B and C loft, the Sum was 120
Pounds. How much did each lofe?
Let a
=
y=the
the Sum A loft, e the Sum B loft, y the Sum
C loft, d= 22, n = 120.
M
7
I
23 +
a + y = 2 e
a+e+2y=dBy the Queſtion.
ae + 3ey = n
3 e + y = d
I
-y
4
a = 2 e -y
2.4
5
3.4
6
2 ee e y + zey = n
ey
By the fifth and fixth Steps, the Queftion is reduced to two
Equations, and two unknown Quantities, and becauſe y is only
to the first Power in both Equations, find the Value of y in each
of them.
5- 3 e 7
6 — 2ee
8
y = d―ze
2e y = n
82e 9 y
n
n- 2 ee
2ee
n
2 e
2ee
7.9
= d
10
d — 3 e
2 e
10 X 2 e
ΙΙ
n
11 + 6e e
12
12
72
13
4 e e
2 de
13-2de
14
2 ee — 2 de — 6ee
4ee+n=2 de
n
Here the Equation is quadratic and
ee
ambiguous.
de
14÷4
15
15 CO
16
e e
n
2
4
de
dd
d d
n:
+
for the
2
16
16
4
d
Co-efficient of e is
divided by 2 as in the laſt Queſtion,
the
which being
2
Of folving Equations, &c.
253
the Quotient is, the Square of
4
d d
which is
16
d
d d
n
16 w 2
17
e
4
16
4
d
d
dd
n
17 +
18
土
​= 5.5 ± .5
4
4
16
4
(
Then by Step 7th | 19 | y=d3e4
And by Step 4th | 20 a = 2e — y =
= 8
6, or 5, if e = 6,
But if e 5, then by the feventh Step yd3e=7,
=
and by the fourth Step a 2 e 3 = 3•
Queſtion 88. There are two Numbers, the Sum of their Squares
being added to their Sum, is 338:
And their Product is 156.
What are the Numbers ?
Let a and e be the two Numbers fought, b = 338, m = 156.
aa+ee+a+e=b
} By the Queſtion.
Then
I
2
a e = m
2- e
3
a
32
4
I. 3.4
5
a a
m m
m
e
m m
ee
teet
ee
m
e
mm + eeee +
5 x e e
6
e e m
But as
e
Hence 7
+eb, an Equation
having the unknown
Quantity e only.
e e m
+ eee = bee
e
em, the e being rejected
(by Art. 20.
mm + eeee+em+eee=bee
There being only the known Quantity mm, tranfpoſe the
others fo that m m may be at laft affirmative; and this is to be
2
obferved,
254
ALGE BRA.
obferved, that in tranfpofing the Quantities in theſe adfected
Equations, the Side of the Equation which is known may at
laft be affirmative.
7
e e e e
8
8 — e e e
9.
9-em
ΙΟ
Or
II
11 in Numbers
12
Now fuppofee 10.
mm+em + e e e = be e
mm + em = bee e e e e
mm bee
ееее
e e e e eee + bee
eee
e e e e
e e e
me
mem m,
it being the common Method to
place thefe Equations, according to
the higheſt Power of the unknown
Quantity.
-eeee-cee +338 ee-156e=24336.
Then
e e e e
10000
e e e
1000
I 1000
+338 e e 33800
156e=
22800
1560
-ee ee-eee +338 ee-156e=21240 which being less than
24336 the Number in the given Equation, therefore muſt be
more than 10.
Let r = 10, and put y = what it wants of being the true
Root.
Then
I √r + y = e
rty
14 2
1 & 3
3
1 & 2
4
r r r r + 4 r r ry+6rryy=eeee, all
the Powers of y above y y being rejected.
rrr + 3rry + 3ryyeee, rejecting
all the Powers of y abovey y.
rr+2ry+yy=ee
yy.
Becauſe in the given Equation it is 338 e e, therefore multiply
the fourth Equation by 338.
4× 338 | 5 | 338 rr + 676 ry +338 y y = 338 e e
Becauſe
Of folving Equations, &c.
255
Becauſe in the given Equation it is 156 e, therefore multiply
the firſt Equation by 156.
I 156 | 6 | 156r+ 156y = 156 e
Now the ſecond, third, fifth, and fixth Equations being equal
to the feveral Powers of e, and multiplied by the fame Co-
efficients as in the given Equation, add or fubftract them accord-
ing to the Signs thofe Powers have in that Equation.
—2—3+5-6
7
But 8
7.8 9
9 in Numbers
IO
10 contracted
ΙΙ·
-rrrr-4rrry-6rryy-rrr-3rry
-3ryy-338rr+676ry+338yy
156y=
- 156r
+338ee156e
-
ееее - сес
— eeeeeee +338ee—156e=24336
by the given Equation.
rrrr 4rrry бrryy — rrr — 3rry
-3ryy+338 rr+676 ry+338yy
-156156y=24336
10000— 4000 y — 600 y y —
-300 y 30yy + 33800 +6760y
+33847-1560-156324336
I I
21240
12
21240 + 2304y292yy
2304 1 292yy=3096
1000
24336
Now divide by the Co-efficient of yy.
12 292 13 7.89y-yy10.6
And dividing by the Co-efficient of y
minus y,
10.6
137.89—3 14 y =
7.89-y
Operation 7.89) 10.60 (1.7 y
I.
6.89
Divifor 6.89
1.7
3710
3633
Divifor 5.19
77
r = 10 by Suppofition.
+ y = 1.7
r+y=11.7 which being involved and tried, it will be found
too little; therefore for a fecond Operation,
Suppoſer = 11.7 and y what it wants of the true Root.
Then
256
ALGEBRA.
Then
I
1 4
2
I & 3
3
I ✪ 2
4
r + y = e
rty
r r r r + 4 r r r y+6rryyeeee, the
Powers of y above y y being rejected.
r r r + 3 r r y + 3 ry y = q e e, the
Powers of y above y y being rejected.
rr + 2ry + y y
e e
Becauſe in the given Equation it is 338 e e, therefore multiply
the laſt Equation by 338.
4 × 338 | 5 | 338 r r +676 r y + 338 y y = 338e e
Becauſe in the given Equation it is 156 e, therefore multiply
the firſt Equation by 156.
1 X 156 | 6 | 156r+156 y = 156 e
Now add or fubftract the Equations that are equal to e e e e,
e e e, 338 e e and 156 e, according to the Signs thoſe Quantities.
have in the given adfected Equation.
—2—3+5—6
7
But 8
7.8
9
9 in Numbers
IO
-rrrr-4rrry-6 rryy-rrr-3rry
−3ryy +338 rr+676ry + 338 vy
- 156 r
156y
+338ee156e
-сеее
eee
-eeeeeee +338ee156e24336,
by the given Equation.
-rrrr -
4rrry 6rryy gagago 3rry
−3ryy+338rr+676ry+338 y y
-156r156y=24336
-18738.8721-6406.452y821.3433
-1601.613410.67y35. Iyy
+ 46268.82 + 7909.2 y + 338 y y
1825.2156y=24336
24103.1349 +936.078y — 518.44 vy
=24336
10 contracted
I I
I 1
12
936.078y — 518.44 y y
518.44 y y
13
12 13
131.805 y
14 y
YY.
232.8651
Dividing by the Co-efficient of yy.
1.805y-yy.4491.
Now dividing by the Co-efficient of y
minus y, that is, by 1.805-y.
4491
1.805-y
Operation
Of folving Equations, &c.
257
Operation 1.805) .4491 (.297 =y.
.2
Divifor 1.605 3210
.29
12810
Divifor 1.315
11835
97
9750
Divifor 1.218
8526
1224
+"
r = 11.7 by Suppofition,
.297
ry is 11.997 e, which is fomething too little, the true
Value being 12. but this may inform the Learner of the Nature
of folving theſe high adfected Equations, every Operation ap-
proaching nearer and nearer to the true Root, from whence it
may be found to any affignable Degree of Exactnefs.
m
And having found e to be 12, then by the third Step of the
Work to the Queſtion, we have a = =13, the other Num-
ber fought.
¿
71. The Method of refolving Equations when the unknown
Quantity is to feveral Powers in both Equations.
When both the unknown Quantities are to the firt and
fecond Power in both Equations, find the Value of the Square
of the unknown Quantity in each Equation, and make theſe
two Equations equal to one another; which Equation will have
the firft Power only of the unknown Quantity, its Square being
exterminated by that Equation.
Then find the Value of the firft Power of the unknown
Quantity in this laſt Equation, which raiſe to the ſecond Power;
and in either of the two given Equations in which it may be,
moſt conveniently done, for this unknown Quantity and its
feveral Powers, write their reſpective Values, which will give
an Equation with only one unknown Quantity, and is to be
reduced by the Rules already explained.
LI
Queftion
258
ALGEBRA.
Queſtion 89. To find two Numbers, the Sum of whofe Squares
is equal to the leffer multiplied by 20:
And the Square of the leſſer being added to their Product, the
Sum is 16.
Let a
the greater Number, e = the leffer Number,
m = 20, d≈ 16.
I
2
aa+ee=me }By the Queſtion.
ee+ae = d S
Begin to exterminate e according to
the Directions, that is, find the Value
of ee in both the given Equations.
e em e
e e = d — a e
a a
a a = d — a e, here ee is ex-
terminated, now find the Value of e.
me +ae a a = d
me+aed + a a
d + a a
i-a a
3
2- - a e
4
3.4 5
me
5 + ae
6
6 + a a
7
7÷m+a
8
8 & 2
9
се
le=
e=
m + a
Raife this Value of e to the ſecond Power.
d d + 2 da a + a a a a
mm + 2 ma + aa
Now in the firft Equation for ee and e, write their refpective
Values at the eighth and ninth Steps.
1.9.8 10
aat
d d + zda a + a a a a
ata
mm + 2 m a + a a
md + maa
m + a
an Equation clear of 4,
having only the unknown Quantity a.
To clear this Equation of the Fractions, obferve that m m
+ 2 mata a is the Square of ma, the former arifing
from the Involution of the latter by the eighth and ninth
Steps, and in the Multiplication of Fractions, it being the
fame thing to divide the Divifor, as to multiply the Dividend,
dd+2 daa+aa a a
to multiply
by ma, we only change
>
mm + 2 mataa
the Divifor to ma, that being the Quotient of mm + 2 m a
+a a divided by ma, the reft of the Multiplication is the
fame as ufual.
TO X
Of folving Equations, &c.
259
dd + 2 daa+aaaa
m+a
mmaa+maaa+maaa+qa q q + d d
+2 daataaaammd+mma a
+mda+ma a a
10xm + a II
maa+aaa+
= md + ma a
11 x m + a
12
12. -maaa
13
14 — mma a
14-mda
mmaa+maaa+aaaa+ḍd + 2 d q a
+aaaa=mmd+mmaa+m da
14 maaa+aaaa+dd+2 daataaaa
15
15-dd 16
16 in Numbers 17
172 18
=mmd+mda
maaa + 2 aaaa + d d + 2 ḍaa -
mda = mm d
2 aa aa + maaa + 2 da a---- mda
= m m d — d d
| 2aaaa+20aaa + 32aa — 320a = 6144
Becauſe the Co-efficient of a aaa will
divide the other Co-efficients without
any Remainder, divide by it.
aaaa+10 aaa+1бa a—160 a—3072
aaaa+10aaa+16a a—160a=3072
Which Equation being refolved by the Method of Converging
Series, we fhall find a 6, or nearly to it, 6 being the true
Root, from whence by the eighth Step = 2.
Queftion 90. There are two Numbers, if the greater is added
to its Square, and from this Sum we fubftract the Square of the
leffer, the Remainder is 94:
But the Square of the leffer, being added to the leffer, this Sum
is equal to twice the greater.
Let a = the greater Number, the leffer Number, m≈94.
I
2
a+aa-eem
ee + e = 2 a
"} By the Queftion.
Begin with finding the Value of ee in
each Equation.
aa+a=mtee
I tee
3- m 4
3
aa+a
mee
2
e
5
ee = 2 a
24
4.5
6+e
6
aata- m2 a
-e, here e e is ex-
terminated, now find the Value of e,
78
e+aa + a
?
eta a
ma
L12
m = 2 a
8 + m
260
ALGEBRA.
8+m
9
9 — aa
ΤΟ
10 G 2
II
e+aa = a + m
e = a + m a a
Raiſe this Value of e to the fecond Power.
ee=aa+2am+m m−2 a a a—2 m a a
+aaaa
Now for e e and e in the fecond Equation, write their refpec-
tive Values, found at the tenth and eleventh Steps.
•
2. II ΙΟ 12
aa+2am+mm ---- 2aaa
+a+m—a a = 2 a
2maa + aaaa
12 in Numbers | 13
13 188a8836-2 a aa—188aaa aa a
+a+94 = 2 a
13 contracted | 14 | 187a+8930—2aaa—188aa+aaaa=0
Tranfpofe the feveral Powers of a, that
8930 the known Part of the Equa-
tion may be affirmative.
14
188aa
a aa a¦ 15 -aaaa—187a+8930 — 2aaa·
15 + 2a a a 16
−aaaa+2aaa=187a+8930—188aa
16 + 188 aa | 17 |—aaaa+2aaa+188aa = 187a+8930
17 — 187 a | 18 | −aaaa+2aaa+188aa—187a=8930
Which Equation being refolved, we fhall find a = 10, or
nearly to it, ro being the true Root.
Then by the tenth Step ea+m- ----a a = 4.
We ſhall now proceed to the Solution of ſeveral Geometrical
Problems upon the fame general Principles, and if the Learner is
not fufficiently acquainted with the Elements of Geometry, to
difcover how the Equations are formed from the Properties of
the Figure, he may omit theſe Queftions, and proceed to the
others which require no Knowledge in Geometry.
Queſtion 91. In the oblique Triangle ABC, given the Difference
between the Sides AC and BC-8, and the Difference between
the Segments of the Bafe AE and EB 10, and the Perpendicular
CE, let fall from the vertical Angle C, upon the Bafe A B ≈ 16.
To find the Sides AC, CB, and Baſe A B?
Upon C as a Center, with the Radius CB, draw the Circle
GBFD, and continue A C to G.
Hence CB
CD, as Radius of the fame Circle, whence
AD is the Difference of the Sides, or the Difference between
AC and CB=8.
And
Of folving Equations, &c.
261
And BEEF, as FB is bifected at E by the 3.c.3
hence AF is the Difference of the Segments of the Bafe, or
the Difference between A E and EB = 10.
C
D
A
F
E
B
Let AD d
8, and DC
Let AFb
10, and FE
+2a.
Let CE == 16.
p
CBe, then AC=d+e.
EB-a, hence AB-b
Having two unknown Quantities, a and e, and no Equation
from the Conditions of the Quefton, we muſt raiſe two Equa-
tions from the Properties of the Figure.
Now the Lines AG and AB are drawn from within a Circle,
and touch without at the Point A, therefore by 37.e.3 AG
× AD = AB × AF, all which Lines are expreffed in Sym-
bols, except AG, but CG = CD=e, and AC
e, and AC = d + e,
therefore AG d + 2e, hence we have in Symbols,
I
d + 2 ex d = b + 2 a × b the ſhort
Lines over the Quantities, fignifies
that they are both to be multiplied by
the Quantity which follows the Sign
of Multiplication.
But the Triangle CEB is right-angled,
therefore by
pp + a a = e e
dd2 debb + 2 ba
47.e. I
2
From the firft
3
3-dd
4
2
de=bb+2b a — d d
Now find the Value of either a or e,
in the third Equation.
But as we ſhall have Occafion to fquare this Equation, for
when the Value of e is found, that Equation must be raiſed to the
fecond
262
ALGEBRA.
fecond Power, it being ee in the fecond Equation; and bb
-dd being a known Quantity, to avoid Trouble, fubftitute
x=bbdd.
Then 5
2 de=x+2 ba
5÷2d
6
x+2ba
e =
2 d
Raife this to the fecond Power, becauſe
it is ee in the fecond Equation.
xx+4xba+4bba a
62
7
ee =
4dd
xx + 4xba + 4 bha a
2.7
8
=pp taa,
4 d d
(an Equation
clear of e.
8 x 4 d d 9
9-4dda a
10-
xx+4xba+4bbaa=4ddpp+4dda a
Bring all the Powers of a to one Side
of the Equation.
10 4bbaa-4 ddaa+4xba+xx=4ddpp
4bbaa-4ddaa+4xba4ddpp-xx
II
Here the Equation appears quadratic, the unknown Quantity
being only to the firft and ſecond Power; but as the Square of
the unknown Quantity has Co-efficients, therefore by Article 58,
divide the Equation by 4bb-4dd, the Co-efficients of a a.
2/aa + 4 x ba
11÷4bb-4dd | 12
|12|
4bb-4dd
4 d d p p
x x
4bb-4dd
The Work being now prepared for compleating the Square, and
the Co-efficient of a being a Fraction, to fave the Trouble of
dividing it by 2, and fquaring the Quotient according to Art. 57,
4xb
ſubſtitute y
y =
4 b b - 4 d ď
aa+ya+12 = 4 dd pp - xx
Then 13
aa+ya=
4ddbp-xx
466
4 d d
1300
14
14 w 2
15
-
4
2
15-16
2
a =
4 4 tb-4dd
d d p p − x x
466
+
yy
4
+
yy
4 d d
4
x x
yy.
+
↑
2
4 d d p p
4bb-4dd
= 16.7.
{
Then
Of folving Equations, &c.
263
x + 2 ba
Then by Step 6th
17 e=
Hence 18
19
2 d
=23.12
AC=d+e 31.12
CB=e=23.12
20 AB = b + 2a = 43·4
Queſtion 92. In the oblique Triangle ABC, there is given
the Sum of the Sides AC and BC-8, and the Difference of the
Segments of the Bafe AE and BE 2, with the Perpendi-
cular CE, let fall from the vertical Angle at C upon the Baſe
AB 1. To find the Sides AC, BC, and Bafe AB?
=
Upon C as a Center with
CB=CD,
the Radius CB, draw the
Circle GBFD, and con-
tinue AC to G.
Then CG
being all Radii of the
fame Circle, whence AG
is the Sum of the Sides,
or AC + CB = 8.
And FEEB, for F BA
is bifected at E, by the
3. e. 3,
3, whence AF is
G
C
D
F
E
B
the Difference of the Segments of the Bafe, or the Difference
between A E and BE
2.
The Conftruction of this Figure being the fame as the laft,
we can raiſe the fame two Equations from the Figure, but in-
ftead of AD being given, we have AG given. Let AG, or
AC+CB=s8, and DC-CG a, whence D'G 2a,
then AG-DG
DGAD:
Put AF = d=2, and FE
let CE=pi.
- 2 a.
EB, then A B=d+26,
Now as in the laſt Queſtion, becauſe the Lines AG and A B
are drawn from the Circumference within the Circle, and touch
at the Point A without the Circle, hence by 37.4.3 AGXAD
ABX AF, that is,
in Symbols
I
That is
2
I
by 47 . e
SX S 2a=d+ 2 ex d
ss—25a= d d + 2 de
3pp+ee = aa the Triangle CEB be-
ing right-angled, and DC=CB=a.
2
Here
264
ALGEBRA.
Here the Queftion is expreffed by two Equations, and two
unknown Quantities.
2+25a
4 d d
4
ss= d d \
2 de + 2 sa
5
ss - d d
d d =
2 de + 2 sa
5- 2 de
d d - 2 de
25a=SS
Subſtitute as before xss-
dd, for
when the Value of a is found, it must
be raiſed to the fecond Power, to com-
pare it with a a in the third Equation.
25a=X 2 de
Then 7
725
8
a=
X 2 de
25
Raife this Equation to the fecond Power,
becauſe it is a a in the third Equation.
x x 4 xde +4ddee
455
xx-4x de+ 4d dee
802; 9
a a =
*
3.910
pp + ee=
10 X 4SS II
ΙΙ
455
4sspp + 4ssee=xx-4xde+4ddee
Now bring all the Powers of e to one
Side of the Equation.
11-4ddee 12 4ssee-4ddee+4s.spp=xx-4x de
12 + 4 x de 13 4ssee-4ddee+4x de+4s spp = xx
13-4sspp | 14 | 4ssee-4ddee+4xde=xx4sspp
Here the Equation appears quadratic, but is not ambiguous,
for xx is greater than 4 s spp. Dividing by the Co-efficient of
ee, by Art. 58.
| 15 | ee+
14-455—4 dd | 15
4 x de
455-4dd
**
=
·4sspp
455-4dd
The Work being prepared for compleating the Square, fub-
Atitute y =
455
4x d
4dd
x d
ss-dd
Then 16
**
eet ye
-4sspp
455-4dd
1617 ee+ye+
yy
xx-4sspp
4
455-4dd
4
17 ມ 3
ແມ
พ
Of folving Equations, &c.
265
17 ա 2
w
18
e + 2
x x
·4sspp
+
yy
2
455
4 d d
4
18
212
19 |
4 s spp
+
yy.
y
45 S
4
d d
4
2
Then by Step 8th
20
21
3.03=BC.
Hence
& =
=2.86
x 2 de
25
AC=sa = 4.97
And 22 BA=d+ 2e=7.72
Queſtion 93. In the right-angled Triangle ABC, there is
given the Area of the Triangle equal to 24, and the Sum of the
Hypothenufe AC and Perpendicular BC equal to 16. To
find the Sides of the Triangle?
y, AB=a,
Let AC
BC=e, s 24, d 16.
Here being three un-
known Quantities, the re
must be raiſed three E-
quations from the Que-
ftion, and the Properties
of the Figure.
Now as the Triangle A
ABC is right-angled,
therefore,
By 47. e. I
I
2
3
B
aa+ee="y } By the Queſtion.
}By
y + e = d
a e
2
= 5,
from the Rule for finding the
Area of the Triangle, for the
Product of the Bafe and Perpendicular of any Triangle being
divided by 2, the Quotient is the Area.
The firſt Equation has all the three unknown Quantities,
but the other two have only two of them. Now if we take
that Quantity which is in all the three Equations, and find the
Value of it in one of them, and in its Room write that Value
in the other two Equations, the Queftion will be then reduced
to two Equations, and two unknown Quantities; thus in the
fecond Equation find the Value of e.
M m
2 - Y
256
ALGEBRA.
e = d
d - y
But as it is ee in the firft Equation,
therefore,
2 — j | 4
4 0 2
5
I. 5
6
ad
3.4 7
2
ee=dd-2 dyyy
aa+dd — 2dy+yy = yy
a ỳ
- S
Two Equations
with only two un-
known Quantities.
Find the Value of y, in each of theſe
Equations.
aa+dd — 2 dy = o, for yy being taken
away by the Subftraction, that Side
of the Equation is nothing.
2 dy = aa + d d
y =
a a + d d
2 d
= 2 s
a day
a d = 2 s +ay
ad
6-yy
8
9 + 2 dy
9
9÷2d
IO
7 X 2
I I
II tay
12
12
25
13
à y = a d
25
13-a 14 y
ad
25
y =
a
a a + d d
ad - 2 s
10. 14 15
2 d
a
15 x a
16
a a a + d d a
= ad
ad — 2 3
2 d
16 x 2 d
17
17 in Numbers
18
18
512 a
19
aaa + d d a2dda- 4 ds
a a a + 256 a = 512 a
a a a 256a= 1536
1536
Here the Equation is adfected, therefore tranfpofe the Quan-
tity fo that the Side of the Equation which is known may have
the affirmative Sign.
19256 a
20 1536 21
+
21 aaa |
aaa + 1536
20
a a a =
ааа
256 a
=
1536
256 a
22 |
= 1536
aaa + 256 a =
Which Equation being refolved by the Method of Converging
Sries, we fhall find a 8 nearly, for 8 is the true Root; from
whence the other Sides of the Triangle are eafily determined.
Queſtion 94. In the right-angled Triangle ABC, there is
drawn E parallel to the Perpendicular BC, given the Per-
pencicular BC = 24, the Segment of the Hypothenuſe EC = 15,
and
Of folving Equations, &c.
267
and the Segment of the Bafe AG = 20. To find the Hypothenuse
AC and Baſe AB?
Draw ED parallel to A B.
Let BC c = 24, EC
= n = 15, A G = b = 20,
G® B = E D = a, then A B
=b+a, AE=e, then A C
= n.+ e.
n.te.
Here being two unknown
Quantities, we muſt raiſe two
Equations.
Now the Triangles AGE
and EDC are fimilar,
hence,
•
e 6
by 4
in Symbols 2
whence
by 47. e. I
123
4
A
E
D
G
B
AG: AE::ED: EC
b:e::a: n
a e = bn, for Quantities that are in
continual Proportion, the Product of
the Extreams and Means are equal.
bb + 2 ba+aa+cc=nn+2ne+ee,
the Triangle ABC being right-
angled, and as theſe two laft Equa-
tions contain the Queftion, therefore
f n
a =
e
b b n n
3 ÷ e
5
526
aa =
e e
4.5.67
2 b b n
bbt
b b n n
+
e
ee
+cc=nn+2ne
(+ ee
b b n n
7xe 8
bbe + 2 b b n +
e
+cce=nne
(+2nee+eee
8 x e 9
9 in Numbers 10
bbee+2bbne+bbnn+ccee=nnee
+2neee + e e e e
400 ee + $2000e + 90000 + 576 e e
=225ee +30 e e e + e e e e
976 ee + 12000 e + 90000 = 225 e s
+30eee + e e e e
751ee+12000e+90000 = 30eee + ceee
+30eee-751ee12000e +90000
eeee +30eee—751ce — 12.000€
that is
11
12
eeee
12000
14
M m 3
II - 225e e
12751 ee 13
13 — 12000 e
9000Q
Which
268
1
ALGEBRA.
=25
Which Equation being refolved by the Method of Converging
Series, we fhall find e 25 nearly, for 25 is the true Root.
Then by the fifth Step a = 12, whence AC = n+e=
and AB = b + a = 32.
40,
Queſtion 95. In the Triangle ABC, given the Baſe BC
42.5 and the Angle at B = 49° : 45′ and the Angle at
C= 42° 30′ to find the Perpendicular AD, let fall from the
vertical Angle at A upon the Bafe BC.
A
B D
The Triangle ADB is right-angled,
and the Angle ABD being given, all
the Angles of the Triangle ABD are
known, therefore by plain Trigonome-
try, we can find the Ratio between the
Sides BD, and AD, though we do not
C know the Length of either of them, for
as the Sine of the Angle BAD, is to the
Logarithm of any Number affumed for the Side BD, fo is the
Sine of the Angle at B to a fourth Number, which is the Lo-
garithm of the proportional Number for the Side A D.
Therefore affuming Unity, or 1, for the Side BD, we have
As the Sine of the Angle at A
Is to the Log. of the Side B D
So is the Sine of the Angle at B - 49° 45′
To the Log. of the Side A D
40° 15'
9.810316
I.
0.0
9 882657
9.882657
9.810316
1.18
.072341
Hence the Sides AD and BD are to one another, as 1.18
is to I.
42.5 AD-a, m≈ 1.18 and p = 1.
Now let BC6 = 42.5 AD
Conſequently,
pa
I m: p::a:
| 1 | mp3
BD, that is, as
m the Numbers which
exprefs the Proportion of AD and D B, are to one another, fo
is the true Length of AD to the true Length of B D.
By the fame Reafoning, in the Triangle ADC, becauſe all
the Angles are known, therefore the Ratio of the Sides AD
and DC are known, and afluming A D to be 1. and pro-
AD
ceeding by Trigonometry as before, we ſhall find the propor-
tional Number for CD to be 1.1
Now
Of folving Equations, &c.
269
Now putting d = 1.1 and p= 1, as before, we have
p:d::a:
| 2 | pidica
da
Р
= DC, by the fame
Reaſoning as at the
first Step.
And as we have now expreffed in Symbols the two Parts
BD and DC of the Baſe BC,
=b, that is, BD+DC=BC.
Hence
pa
da
3
+
m
Р
m da
3 x m
4
pat
= m b
mb
P
4 x p
5
ppa+m damb p
5÷pp + mb
6
m b p
a =
=21.82 = AD.
pp + md
Queſtion 96. In the right-
angled Triangle ABC, there is
given the Sum of the Sides equal
to 12, and the Area equal to 6.
To find the Hypothenuſe AC?
Lets
AC = y.
12, b = 6, BC = a,
Then by the Property of the A
Triangle, AB√yyaa.
B
Hence I a + y + √ ♡ Y
a a - s by the
(Queftion.
2
a
012
yy—a a = b, from the Rule for
(finding the Area of the Triangle.
Now becauſe there is the fame Surd in both Equations, find
what the Surd is equal to in the fecond Equation, and write its
Value for the Surd in the firft Equation.
2 →
a
1
2
3√yyaa
b
2b for b =
b
and
a
I
26, by the Rule for Divi-
a
2
a
fion of Fractions in common Arith-
metic.
1.3
270
ALGEBRA.
1
B
1. 3
4
4 x a
5
I
602
679
7
7+
8
9.
IO
5-
9 × 2
8.10
II + 2 ay
12 - 2 SA
13+ss
13-25
A
b
a + 3 + 2/6 =
y
a
=S
aa+ay+2b=sa, now the Queſtion
is contained in the firft and fifth
Equations.
✔yy. aa =s -
yy
a
y
yy — aa—ss — 2sa — 25y+2ay+aa+yy
2aa=2sa+2 sy — 2 ay — s s
a a = sa ay 26
2 aα = 2 sa-
2 ay.
4 b
II 2sa-2ay-4b2sa+2sy-2ay-ss
12 2sa-4b=2sa+2sy—ss
1346=25y-ss
b
14 2 syss
ss - 46
15 y=
124°, the Angle CAD
SS
C
D
2 $
46
5 = AC.
Queſtion 97.
In the Tri-
angle ABC, there is given
the Sum of the Sides BC+BA
+ AC = 85, the Area =
200, and the Angle at A=
124°. To find the Sides of the
Triangle?
=
Lets 85, b=200, AC
=a, becauſe the Angle BAC
56°, and CD being a Perpen-
dicular let fall on A B continued, all the Angles of the Triangle
ACD are known, conſequently the Ratio between AC and
CD is known, for affuming CD to be Unity, or I, then in
the Triangle A CD by Trigonometry,
As the Sine of the Angle CAD
Is to the Log. of the Side CD
So is the Radius
9.918574
-
56° : 00'
I.
0.000000.
90°:00
10.000000
10.000000
9.918574
1.21
0.081426
To the Log. of the Side A C
Hence we know the Sides AC and CD are as 1.21 to 1.
2
Calling
Of folving Equations, &c.
271
Calling m = 1.21 d= 1, therefore,
I
da
m:d::a: = CD, as at the firft
M
or fecond Steps, Queſtion 95.
Now B A being confidered as the Bafe of the Triangle B AC,
and CD as its Perpendicular, hence by the Rule for finding the
Area of the Triangle, BA x DC 2b, that is in
Symbols 2
da
26 =
× BA.
m
da
2 b m
2-
3
da
m
2 b m
d a
= BA, for 2 b or
2 b
I
da
m
by the Rule for Divifion
of Vulgar Fractions.
d da a
And
4
AD = √ aa—
m m
for the Tri-
angle ACD is right-angled, where
a = A C, and
+ v
a a
da
-= CD.
m
d d a a
d d a a
= BA +
(AD=BD
i da a
m m
3 +45
2b m
da
mm
52
6
4 b b m m
+
4 b m
a a
d d a a
d a
+aa
1 & 2
d d a a
7
J
m m
6+7
4 b b m m
4
b
772
d d a a
8
+
a a
d d a a
da
m m
2
m m
2
- BD, or BD
(fquared
= CD, or CD fquared.
+ aa = CB, or CB fquared.
Having now got an Expreffion equal to the Square of CB,
we muſt endeavour to find another Expreffion for CB from
fome other Data.
Now the Sum of all the Sides is given, that is,
9 | B
272
ALGEBRA.
9 | BC + AC+ AB = s
2 b m
But =
10 AC a,
AC-a,
and AB
=
da
2 bm
II | BC + a +
S
da
by the
third Step.
9.10
2 b m
2 b m
II
12
S
da
da
2 b m
12
a
13
S
da
©
13 214
4 s b m
sa
da
= BC + a
a = B C
25a+.
2
+aa=BC, or BC fquared.
4 b ma
+
da
4 bbm m
d da a
4shm
4
8.14 15
SS
-25a+
bma 14 bbm m
+
da
+aa =
d daa
da
da
4bbmm 4bm
+
aa
dd a a
ddaa
m m
+aa
▲ b b m m
16
4 s b m
25a+
15
d d a a
d a
4 b m a
da
+ aa
4 bm
d d a a
-
Vaa-
+ aa
da
m m
4 s t m
4
b m
16 — a a
aa 17
55
2sa+
a
d a
d a
4 b m
d da a
a a
da
m m
Here the Learner may obſerve that the unknown Quantity is
under the radical Sign, and therefore as fuch Equations are
generally ſquared to take away the Surd, the fame is to be done
here; but as it is a a in all the Quantities under the radical Sign,
we can extract the fquare Root of a a, and join it to the rational-
Quantity, leaving the remaining Part of the Surd under the
radical Sign, thus 4bm
4b m
VI
d d
m m
da
>
a a
i d a a
4 b m
a
m m
da
I
dd
m m
whence the feventeenth Equation becomes
18 | s s
Of folving Equations, &c.
273
18
· — 4 sb m
4
b m a
2sa+
da
d a
4 bm
d d
I
,
by which Means
d
m m
we have faved the Trouble of fquar-
ing the feventeenth Step.
18 x d
19 ssda
4 sbm - 2 d saa + 4 bma =
4 bm a I
dd
m m
The Equation being now cleared of its Fractions, it appears
quadratic, for the Powers of a are only to the firft and fecond
Power, and the Surd is Part of one of the Co-efficients of a.
19 20 2dsaa+4bma✓1
H72/12
·:-4bma-ssda
(=-45bm
dd
I
d d
4 bmayı
4bma-.
-ss da
m2 772
202ds
21
aa+
2ds
4 s b m
2 b m
2 d s
d
dd
Now
4 bm √ 1
2 d s
being a known Quantity, and
45, put x =
4 bm —
m m
45.
2 b m
then 22
аа
x a =
d
x x
XX
2 b m
22
23
a a
xat
4
4
d
X
Xx
W
23 w 2
24
a
=
2
4
2 b m
d
X
X
x x
2b m
24 +
25
a =
the Equa-
2
2
4
d
Whence we shall find a =
2b m
d a
tion being ambiguous by the 22d Step.
And by the third Equation
27.21 = AC.
And by the thirteenth Equations
N n
17.78 BA.
2 b m
2
40.01 BC.
da
This
274
ALGEBRA.
This being the moſt difficult Solution we have yet had, a
Review or Summary Account of the Operation may not be
uſeleſs to the Learner, in giving him fome Idea how to begin
and form a Judgment in fuch Cafes.
Now becauſe the Angles of the Triangle ACD are known,
we have the Ratio of the Sides given, whence affuming CD as
known, I find a proportional Number for A C, and from thence
I can exprefs CD and AC in Symbols, and CD being con-
fidered as the Perpendicular to the Triangle B A C, of which the
Baſe is B A, from the Rule for finding the Area of a Triangle,
I obtain an Expreffion for BA; and I exprefs AD in Symbols,
from knowing AC and CD, and adding AD to BA, I have
Expreffions for BD and DC, each of which being ſquared, their
Sum is equal to the Square of BC.
Then from fome other Data I find an Expreffion for BC,
and becauſe the Sum of the Sides is given, and having Expref-
fions for the two Sides BA and AC, it is eaſy to find an
Expreffion for BC, as at the thirteenth Step, which being
fquared, is made equal to the former Square of BC, and the
Equation is reduced, as in the Work.
This and feveral other Queftions are taken from Sir ISAAC
NEWTON, the perpetual and everlaſting Honour, Ornament,
and Glory of our Nation; and I have only endeavoured to
accommodate his Solutions to the Learner, in explaining them
in a more copious Manner.
Queſtion 98. In the Triangle ABC, there is given the Altitude
CD = 7。 the Baſe A B = 10, and the Sum of the Sides BC+
AC=23. To find the Sides of the Triangle?
B
A
C
Becauſe the three Sides of
the Triangle BAC will be
eafily expreffed in Symbols,
and the Triangle BCD being
right-angled, we fhall eafily
find to what BD is equal.
Again, as the Triangle ACD
Dis right-angled, and the Sides
AC and CD are known in
Symbols, therefore A D is known in Symbols.
Now if from BD before found in Symbols, we fubſtract B A,
there remains another Value for AD, which being made equal
to the former, we have an Equation, which is. fufficient, if we
uſe but one unknown Quantity.
And
Of folving Equations, &c.
275
And as here will be a new Method of expreffing the Quan-
tities fought, 1 refer the Reader to Queſtion 41, where at Step 8.
he will find, that in any two Numbers, if their Difference is
the
fubftracted from their Sum, and the Remainder divided by 2,
Quotient will be equal to the Jeffer; and at the fame Question,
if he exterminates e and finds a, or the greater Number, it will
be equal to the Sum and Difference of the two Numbers added
together and divided by 2.
Therefore put x = CD
CD=7, b
7, b=BA = 10, c = half the
Sum of the Sides BC+AC=11.5 and a = half their Dif-
ference, then the greater Side or BC=c+a, and the leffer
Side or AC = c — a, now in the Triangle BCD
√cc+2ca + a a − x x
BD, for
BC=c+a, and CD—x
And in the Triangle A CD,
2 √cc-2ca+aa—xx=AD, for
AC c-a, and CD=x
by 47 .e. I
I
by 47. e. I
but 3
BA=b
1-3
4
− × × : — b − B D
2.4
5
-b
50 2
6
CC
2ca + aa
we have
7
7 ±
8
802
9
√ic+2ca+aa-xx:
BAAD
√cc-2ca+aa-xx=√cc+2ca+aa—xx
xx=cc+2ca+aa-
xx
2b √cc+2ca+au—xx:+bb
By tranfpofing the Quantities which
deftroy one another
·4ca=·
·2b₁/cc+2ca+aa—ax: +bb
2b ✓cc+2ca+aa — x x = b b + 4 ca
4bbcc + 8bbca + 4bbaa4bbx x = b b b b
+8bbca 16cca a
9-8bbca 10 4bbcc +4bbaa — 4bbxx = bbbb16ccaa
16ccaa4bbcc+4bbaa-4bbxx - bbbb
4bbaa4bbcc 4 b b xx - bb bb
|
1 0 - b b b b
1 I
I I
4 b b a a
12
It ccaa
a a
1216cc-4bb13
4 bbc c - b b b b — 4 b b x x
b b x x
16cc
4cc - bb
466
b
See Page 276.
b b x x
b b
4
b b
13 w 2
a =
14
4
4 cc
b b
x x
b
3.69
4
400
b
bb
Nn 2
Whence
276
ALGEB R A.
BC, and c- a = 7.81 AC.
perplexed to fee the Contractions at
Steps, they may be illuftrated thus
b b b b 4 b b xx
16cc- 4bb
Whence c + a = 15.19
If the Learner ſhould be
the thirteenth and fourteenth
4 b b c c - b b b b - 4 b b x x 4 b b c c -
16cc-4bb
16cc-4bb
for it is the fame Thing whether the Quantities that compoſe the
Numerator, are placed fucceffively one after another like one
continued Fraction, or placed feparately and diftinctly, like dif-
ferent Fractions, the Quantities that compofe the Denominator
being placed under each diftinct Numerator.
But 16 cc 4 b b ) 4 b b c c — b b b
bib b
4
4 b b c c - b b b b
O
The Quotient Quantity is b b, and as the Co-efficients of the
Divifor are reſpectively four Times more than thofe of the Di-
vidend, therefore under the Quotient Quantity b b place 4, and
is the Quotient exact.
b b
4
And this Fraction _ 4 b b xx
16cc-4bb
b b x x
4cc—bb,
for it is only
dividing the Co-efficients by 4, therefore the Contractions are
as at the thirteenth Step.
The Contractions at the fourteenth Step, arife from its being
bb in all the Terms under the radical Sign, for it is only placing
the fquare Root of bb without the radical Sign, by which
I b b
4
means
b b
4
or ✔
is b✔
I
4
Queſtion 99. In the Triangle BCA, there is given the Baſe
AB 6, the Sum of the Sides AC+ BC= 18, and the ver-
AB=6,
tical Angle at C 30° 00', To find the Sides AC and BC.
:
Let fall the Perpendicular A D, and in the Triangle ACD,
becauſe the Angle at € is given, all the Angles of that Triangle
are known, and therefore the Ratio of the Sides is known, by
which means we can get an Expreffion for CD.
And
Of folving Equations, &c.
277
And becaufe AD is a Perpendicular that falls within the
Triangle, and the Angle at C is acute, therefore by 13. e. 2,
BC fquared added to AC fquared, is equal to B A fquared added.
to the Product of 2 BCxCD, from whence we ſhall have ano-
ther Expreffion for C D. Now if we can exprefs the Sides of
the Triangle with one unknown Quantity, this Equation be-
tween the two Values of CD will be fufficient.
D
*
B
A.
In the Triangle ACD, becauſe the Angle at C is known,
and AD being a Perpendicular to CB, all the Angles of the
Triangle ACD are known, therefore affuming CD=1, by
Trigonometry,
60° : 00′ -9.93753′
As the Sine of the Angle CAD
Is to the Log. of the Side CD
So is the Sine of the Angle CDA - 90 : 00
I.
To the Log, of the Side A C
1.15
0.000000
10.000000
10.000000
9.937531
0.062469
Hence we know that as 1.15 is to 1, fo is AC to CD.
Then let AB=6=x, half the Sum of the Sides AC
BC
=9b, and half their Differencea, then as in
the laſt Queſtion, the greater Side or BC = b + a, and the
leffer Side or AC = b —a, d = 1.15 n = 1.
Becauſe AC is to CD, as 1.15 is to 1.
1
Therefore
278
ALGEBRA.
1
Therefore
I
d:n::b
a:
nb.
d
na
- CD
by 13. e. 2
2
2 -
3
2 bb + 2 aa
bb+2ba+aa+bb−2 ba+aa××
+2b+2axCD
xx=2b+2 a × CD
2 b b + 2 aa
x x
3÷26+2a
4
=CD
26+2a
The fhort Line over the two Quantities 2 b+2a in the
ſecond and third Equations, fignifies they are both to the mul-
tiplied into CD, otherwife there would be no Diftinction whe-
ther CD is to be multiplied into 2 a only, or all the Quantities
on that Side of the Equation.
Now make an Equation between the two Values of CD
found at the firſt and fourth Equations.
2b6+2aa-xx
I. 4 5
26+2a
5 x 26+2a
6
6 x d
7 +2 na a
8 + d xx
9 - 2 db b
78
9
10
=3
nb - na
d
2bb2aa—xx—2nbb-2bna+2bna—znaa
d
2dbb +2daa-dxx-2 nbb-2 naa
2 na a+ 2 db b + 2 da a¬d x x = 2 n b b
znaa+2 db b+2daa=2 nbb+dxx
2naa+2 daa-2 nbb+dxx-2 db b
2 n b b + d x x — 2 d b b
10 ÷ 2n + 2 d
I I
aa =
dx
2 n + 2 d
II w 2
12
2nbb+dxx — 2 db b
= 1.99
2n+2 d
=
Whence BC= b+a= 10.99 and AC ba = 7.01
Queſtion 100. In the Fish Pond ABCD, there is given the
Side AD = 36, DC = 35, CB = 40, and A B
38; the
Angle at A113°, the Angle at B= 60°, the Angle at C 100°,
and the Angle at D=87°, and the Fish Pond is to be furrounded
with an Area of 700, and every where of the fame Breadth. To
find the Breadth of the Walk?
Suppoſing the Walk to be drawn round the Pond as in the
Figure, let fall the Perpendiculars AK, BL, BM, CN, CO,
DP, DQ and A I, by which the Walk is divided into
Four Parallelograms AKLB, BMNC, COPD, DQIA,
and
Of folving Equations, &c.
279
and into four Trapezia AIEK, BLFM, CNGO and
DPHQ, and the Area of theſe four Parallellograms and four
Trapezia is equal to the given Area of the Walk.
F
M
N
G
10
L
B
C
A
D
P
K
E I
Q
H
Let the Breadth of the Walk be a, and the Sum of the Sides
AD+DC+CB+BA= 143 b, then the Area of the
four Parallellograms will be ba. Let x = 700.
Draw AE, BF, CG and DH, becauſe the Triangles
AIE and AK E are equal, therefore the Angle A EK and
A EI are equal, and each of theſe Angles are equal to half the
Angle at A which is 113°, hence the Angle A EI is 56° : 30´.
Then in the Triangle A EI all the Angles are known, and
confequently from plain Trigonometry, we can find the Ratio
of the Sides EI and I A, for affuming EI to be Unity, or I,
we have
As the Sine of the Angle E AI
Is to the Log. of the Side EI
So is the Sine of the Angle A EI -
33° : 30′ - 9.741889
Ι.Ο
0.000000
56° : 30'
9.921107
9.921107
9.741889
To the Log. of the Side AI
1.51
C.179218
Hence
280
ALGEBRA.
Then let d
Hence de::a:
d
Hence we know that AI is to EI as 1.51 is to 1.
1.51 and e I.
e a
=EI, which being the Baſe of the
Triangle EIA,
Hence
2
elå
e a
a
ea a
X
2
2 d (Triangle EIA, and
the Area of the
2 X 2
3
e a a
d
the Area of the Trapezium
(EIAK.
Now in the Trapezium BLF M, becauſe the Angle B is 60°,
we have the Angle LFB = 30°, for the fame Reaſons as before;
whence in the Triangle LF B, if we affume BL to be Unity,
or 1, we fhall find the proportional Number for FL to be
1.73 hence as I is to 1.73 fo is BL to LF, let f= 1.73
A
Then
4
ef::a:
af = LF
с
a
4 X.
5
afx
a
aaf
-the Area of the
2
e
2
5×2
6
a a f
e
2 e (Triangle BLF.
the Area of the Trapezium
(BL FM.
;
Again, in the Trapezium CNGO, becauſe the Angle at C
is 1000, the Angle CGN is 50°, for the fame Reafon as before
and affuming Unity for NG, we fhall find 1.19 to be the pro-
portional Number for NC, whence we know that CN and
NG are as 1.19 is to 1, let g
1.19
Then 7 g
e:
e a
a:
NG, which being
g
100
the Bafe of the Triangle CNG,
e a
a
ea a
Hence
8
X
2
28
e a a
8 x 2
9
مه
= the Area of the
(Triangle CNG.
the Area of the Trapezium
(CNGO.
Laftly, in the Trapezium DPHQ, becaufe the Angle D
is 87, the Angle DHP is 43° 30', for the fame Reafon as
before;
Of folving Equations, &c.
281
before; and affuming Unity for DP we fhall find 1.07 to be
the proportional Number for PH, hence we know that as 1 is
to 1.07 fo is DP to PH, let s = 1.07
Then ΙΟ e:s::a:
the Area of the Tri-
as
PA, then as before
e
a
IO X
12 1
II
as
X
a
aas
e
2
2 e
(angle DPH, hence
a as
II X 2
12
e
the Area of the Trapezium
But it was before found that
(DPHQ;
13 bathe Area of the four Parallello-
(grams.
Now collect the Area of the Trapezia and Parallelograms
into one Sum, and make them equal to the given Area of the
Walk.
d
e
of
+
= x= 700
f
e a a
a as
+
+ba
g
e
S
в
+ + + the Co-
d e
g e
efficients of aa = 4.302
paa+ba =
ba
ваа
a a
3+6+9+12+13 | 14
+
fubftitute
151P =
14. 15
16
x
16 ÷
Р
17
a a +
ba
Ba
P
b b
b b
x
17 c 0
18
aat
+
+
P
4PP
4PP
이
​b
bb
*
18 w 2
19
at
+
+
2 p
4pp
P
b
b b
X
b
19
2,0 a =
+
4.35
2p
4 pp
P
2P
the Breadth of the Walk.
Becauſe the Angles and Sides of the Fish Pond are given,
the Figure may be drawn; but for the Eafe of the Numerical
Calculation I have chofe fuch Numbers, as will not exactly
agree with a Geometrical Figure.
Queſtion 101. In the right-angled Triangle ABC, given the
Perimeter or Sum of the Sides AC + CB+AB 24, and the
Perpendicular
O o
1
7
282
ALGEBRA.
Perpendicular CD = 4.75 let fall from the Right-angle at C.
upon the Hypothenufe A B. To find the Sides of the Triangle?
A.
D
B
C
Let CD64.75 AB+BC
+CA==24, A Ba, then the
Sum of two of the Sides, or AC
+ CB x a, and as at Que-
ſtion 98 let the Difference of
y
the ſame two Sides AC and CB.
And becauſe the greater Number or
Leg is equal to the Sum and Dif-
ference of the two Numbers or Legs
divided by 2, as in the laſt Queſtion,
x—a+y
a+y, and BC the leffer
therefore AC the greater Leg
Leg
a Y
2
2
,
Having Expreffions for A B, BC, and A C, the three Sides.
of the Triangle ABC, in which there are two unknown Quan-
tities a and y, we must raife two Equations from the Properties
of the Figure, and becauſe the Triangles ABC, BCD are
fimilar, therefore by
AB: BC::AC:CD
x-a-y:: x-a+y:b
2
4.e
e. 6
I
In Symbols
2
a:
2
whence
3
ab=
xx-xa+xy-xa-aa-ay-xy+ay-yy
x x
that is
4
ab =
4
2xa+aa- yy
4
And becauſe the Triangle ACB is
right-angled,
2xy + 2 ayaa + y y
4
xx—2xa+2x3—2ay+aa+yy_aa
XX
2 x a
by 47. e. I
5
+
2 x x
5 contracted.
6
4
4 x a + zaa + 2y y
4
=aa
Hence the two Fquations which contain the Queftion are the
fourth and fixth, and as y is only to the Square in each of them,
find in both the Value of yy. But the fixth Equation becomes
7 | xx
Of folving Equations, &c.
283
Sv
7
7
X 2
8
8 +
9
10
I I
4 X
X 4
10 + 11
9. II
12
a a
13 + 4 ab
14 + 2xa
15 + xx
I 2
| 1
2 x a + a à + yy
2
=aa
xx-2xa+aa+yy = 2a a
yy=aa + 2 x a − x x
4 a b = x x − 2 x a + a a — y y
xa+aa
y y = x x — 2 x a + a a
aa+2xa
13
2 xa-
| 14
-
xx = xx
xxxx÷ 2x a
4 ab
2xa+aa
4 ab
4 ab
X X
x
2 x a
a b
x x
XX
= 2xx
xx
= 10 A B.
2 xa+4ab
15 | 4 xa + 4 a
164xa + 4ab
2 x x
16 ÷ 4x+46 |
17
a=
4x+46
9 w 2
18 y
2x+26
Now a being found, therefore
Vaa + 2xa — x x = 2.
x
a + v
thence 19
AC=
– 8.
2
a
and
20
|BC=
y = 6.
2
Queſtion 102. In the right-angled Triangle ABC, given the
Hypothenufe AB 10, and the Sum of the Sides and Perpendi-
cular CD, that is AC + CB + C D = 18.75 To find the
Sides AC and BC? Vide laft Figure.
x
Let A Bb 10, x = 18.75 CD a, then AC+ CB
a; now putythe Difference between the Legs AC
and CB, then as in the two laft Queftions AC, or the greater
y
x-a+, and the leffer Leg, or C B =
Leg is
2
,
CB=
x
2
-y
Having expreffed the Sides of the Triangle A B C in Symbols,
in which there are two unknown Quantities a and y, we muſt
raiſe two Equations from the Properties of the Figure, and
becauſe ABC is a right-angled Triangle, therefore
by 47. e. I
I
x x
2 x a + 2 x y −2 ay+aa+yy
4
2.xy+2ay+aa+yy = b b
=
xx-2xa·
+
4
002
The
284
ALGEBRA.
The two Triangles ABC and CBD being fimilar, therefore
by 4.e.6| 2 |AB: AC:: CB: CD, that is in Symbols,
36
x-a+y
X
a
لا
: a
2.
2
34 ba-
xx-xa+xy-xa+aa-ay-xyНay-yy
4
Hence the Queſtion is contained in the
first and fourth Equations.
xx−2xa+aa+y y
I contracted 5
b b
2
x x − 2 x a + aa
yy
4 contracted 6|ba =
4
Now in both thefe Equations find the Value of yy, there
being no other Power of y.
བ་
xx-2 xa+aa+yy=2b b
5 × 2
7 ±
8 y y = 2b b
aa+2xa-
x x
6×494b a = xx 2 xa+aa—
9±10 yy=xx — 2x a + a a
8.10
II tea
12
xx
yy
a- 4 b a
II xx-2 xa+aa-4ba2bb — a a
+ 2x a - XX
12 xx-2xa+2aa—4ba2bb+2xa
XX
x x 13-2xa+2aa-4ba2bb+2xa-2xx
13 — 2x a 14
14
215
2aa
4 xa-
a a
4ba2bb— 2 × ×
2 x a 2 b a = b b --- x x
Becauſe xxis greater than bb, there-
fore the Equation is ambiguous.
fubftitute 16 Z =
then 17 aa
ZZ
b b − x x +
17c18aa-za+ = b b
18 w
w 2 19
a
4
2x
26
57.5
za = b b
x x
ZZ
4
2 | ~
√bb
bb -
ZZ
xx +
4
ZZ
Z
19. + 2/2/2 20 a
2
2
Z
· ²± √ b b − x x +
2
4.78
4 (or 52.72
by the 10th Step 21 | y=√√✓xx− 2 xa+aa—4 ba—1.99
Then
Of folving Equations, &c.
285
X
=
Then AC = *="+"=7.98 and CB = a — y — 6.
2
=
2
In the above Equation where a 4.78 or 52.72 the Value
of a muſt be 4.78 for it cannot be 52.72 as the Sum of the
three Quantities is only 18.75.
Queftion 103. In the right-angled Triangle ABC, there is
given the Sum of the Sides AC+ BC= 14, and the Perpendi-
cular CD = 4.75 To find the Sides of the Triangle ? See
Figure, Queſtion 101.
Let AC+BC=x=14, AC-BC-y the Difference
of the Sides, then, as in the preceding Queſtions, the greater
Side or AC = x+, and the leffer Side or BC -
2
Put A Ba and DC-b4.75
X
2
Having expreffed all the Sides of the Triangle ABC in
Symbols, amongst which two are unknown, viz. a and y, we
muft raiſe two Equations from the Figure, then becauſe the
Triangle ABC is right-angled, therefore
by 47. e. 1
I
that is
2
xx+2xy+yy
4
+
xx 2 x y + YY —a a
4
x x + y Y = a a
2
Becauſe the Triangles ABC and CBD
are ſimilar,
3AB: AC::CB: CD
by 4. e. 6
in Symbols 4
a:
x + y
2
x-y:b
2
xx-yy
4.
5
ba =
4
Hence the Queſtion is contained in the fecond and fifth Equa-
tions, and becauſe there are no other Powers of y but y y in either
of thoſe two Equations, find the Value of y y in both Equations.
2 X 2 6
xx + y y = 2 a a
6-
x x
7
уу =2aa
5X4
xx
84ba=xx-yy
8+ yy 9 yy + 4b a = xx
+13 |
9 —
286
ALGEBRA.
y y=xx- 4 ba
9 - 4 b a 10
7.10
II
2 aa
12
13
12-2
13 CO
x x
2aa4ba = 2xx
aa + 2 ba=**
4 ba
$4 aa+2ba+bb = xx+bb
15a+b=√√x x + bb
16
14 ա 2
15-b
7 w 2
x y
a = √xx+bb: b 10.03 or
neglecting the Fraction A B 10.
7|1=√2a a 2 x = 2
x
y
Then AC= * + 1 = 8, and BC = = =
C = ± − 1 = 6.
2
2
The fame Question done in another Manner.
Let AC+BC=x=14, AC=a, then B C xa,
CD = b = 4.75 and becauſe the Triangle A B C is right-
angled, therefore A B = √xx−2xa+2aa.
Here we have Expreffions for all the Sides of the Triangle,
with only one unknown Quantity, and therefore one Equation
will be fufficient. And as the Triangles A B C and CBD are
fimilar, therefore
by 4. e. 6 1
¦
in Symbols
AB: AC::CB: CD
2
√xx
2.* 3
b √ xx
2 xa + 2 aa:a::x a: b
2 xa+2a a = x a — a a
Square both Sides of the Equation, the unknown Quantity
being under the radical Sign.
3 2 4 | bb x x − 2 b b xa +2bbaa=xxa a
0
+1
4 ±
5
2 x aa aa aa a
Ranging the Equation according to the
Powers of the unknown Quantity.
a a a a 2 x aaa + x xa a — 2 b b a a
+2bbxa=bb xx
Tho' the Equation here appears as if adfected, yet it may be
refolved by compleating the fquare, as in Quadratics.
And to give the Learner a clear Idea how this is done, if he
fquares any three Quantities mnz, in the Square he
will find fix Terms, mmnn † zz — 2 m n z m x + 2 n z
2,
three
Of folving Equations, &c.
287
three being pure Powers of the Quantities fquared, and the
other three will be double Rectangles, or Products of thefe
Quantities, and therefore any Expreffion that comes under theſe
Circumftances, may have its ſquare Root extracted.
Now aa aa is the Square of
And xxa a is the Square of
a a
x a
And 2xaaa is the double Rectangle, or Product of theſe Roots.
And 2 bbaa is the double Rectangle, or Product of b b xaa.
And 2 bbxa is the double Rectangle, or Product of bb xx a.
From hence it appears that the above Equation of five Quan-
tities has two of them, a aa a and xxa a, whoſe ſquare Roots
may be taken, and that the other three Quantities are double
Rectangles of thofe two Roots, a a and x a, and a third Quan-
tity bb, therefore multiply this Quantity bb by itſelf, and add
the Product b b b b to both Sides of the Equation, which makes
it a compleat Square, thus,
6
aaaa 2xaaa+xxaa
+ bbbb
2bbaa+2bbxa
bb bb - b b x x
a a — xa — bb = √ b b b b + b b x x
= 6 √bb + ≈ ≈
aa—xa=bb + b ✓ b b + xx
6 w 2
7
7 +66
8
.8c0
X X
xx
9
a a—x at
==+bb+b √bb + xx
√bb+xx
4
4
x
x x
9 w 2
ΙΟ
a
2
+bb+b√bb +xx
4
४
I I
a
2
2
+
XX
4
+bb+b √bb + xx
18.9 -AC, a different Value
of what it had before, for then it
was only 8.
To explain this to the Learner, if he extracts the Square Root
of a a
2xa+xx, he will find it to be a -- x, or x - a,
the double Rectangle, viz. 2 ax having the Sign we are fure
either a, or x muſt be negative; but in this Caſe we are to de-
termine which is to be negative by the Confequences that follow,
for if there follows an Impe ffibility in fuppofing ax to be the
Root, then the Root muſt be x a.
To apply this to the Square before us at the fixth Step, viz.
2 xa a a + x x a a 2 b b a a + 2 b b x a + b b b b .
A A A A
Now
288
ALGEBRA.
…
a a
Now the fquare Root of a a aa is
And the fquare Root of xxaa is
And the ſquare Root of bbbb is
a a
x a
b b
But as 2x a aa the double Rectangle, or Product of a axxa
has the Sign, therefore it must be in the fquare Root either
—xa, orxa—aa; but as an Impoffibility attends putting
it a a
xa, we now put it xa a a, and to determine what
Sign bb muſt have in the Root, now the double Product 2 b b a a
having the Sign therefore it muſt be bb or bb, as
2 b b x-a a produces — 2 b ba a, then taking the
fixth Equation
7 аава 2x aaa + x x a a
2 b b a a
+ 2 b b xa + b b b
b
—
− b b b b
t
b b x x
7 w 2
8
xa—a a + b b
✓ b b b b + b b x x
b ✓ bb + xx
Becauſe a a is negative tranſpoſe it,
a a + b√bb + xx¬xa+bb
aa−x a+b✓ b b + x x
bb =
8 + aa
9
Q
x a
II
a a
10-b6b+xx
ΙΟ
b b
x a = b b − b ✓ b b + xx-
Here the Equation appears quadratic, and becauſe-bbb+xx
is greater than bb, it is likewife ambiguous.
xx
xa + ** = * * + bb — b✅bb+xx
4 4
IICO [2
aa-xa+
X
12 ບຸນ 2 13
a
2
X
13+
X
2
14
a=
土
​√
2
4
x x
x x
4
+bb-b√bb + xx
·
+bb - b ✓ bb + x x
5.89A
=7±1.11=8.11 or 5.89 AC.
Queſtion 104. In the right-angled Triangle ABC, there is
given the Sum of the Legs AC+BC≈ 14, and the Sum of the
Hypothenufe and Perpendicular A B+CD = 14.75 To find
the Sides of the Triangle? See Figure, Queftion 101.
—
Let AC+BC= 14 = x, A B+C D= 14.75b, AC a,
AB=y, then BC=x-a, and CD-b-y.
Having now expreffed the Sides of the Triangle in Symbols,
in which there are two that are unknown, therefore raiſe two
Equations from the Properties of the Figure.
And
Of folving Equations, &c.
289
And becauſe the Triangle ABC is right-angled, therefore by
47 e. 1 I xx−2x a + 2 aayy
| | | |
And becauſe the Triangles A B C and CBD are fimilar,
therefore by
40.6 2
in Symbols
AB: AC:: BC: CD
3
y: a : : x
a:b-y
3 ...
4
by-yy
yy = xa
a a
Now both the unknown Quantities being to the first and
fecond Power, in the fourth Equation, and it being y y only in
the firft Equation, and theſe two Equations containing the Con-
ditions of the Queftion, find the Value of y y in each Equation.
4±
I.5
5
26 7∞
yy by + a a -x a
by
by-
aa—xa = xx
+aa
- x Q
x−−2xa+2a a
6 — a a
7 + xa
x x 2 xa+aa
8
by = xx
xata a
xx-xa+aa
9
Y
b
8÷b
이
​Raife this Equation to the fecond Power, and make it to
the firft Equation, as there it is only yy; whereas in the fourth
Equation, if we were to exterminate y, we muſt uſe the Values
of y and y y.
902
xxxx-2axxx+20arx
2aaax+aaxx+aaaa
10 yy=
b b
XXXX
I. 10❘ II
b b
I l x bb
12
2axxx+2aaxx- 2aaax + aaxx+aaaa
x x 2xa +2aa
xxxx-2axxx+2aaxx-2a a ax+aaxx+aaaa
=xxbb-2bb xa +2bba a
Tranfpofing and ranging the Equation, according to the
higheſt Dimenfions of the unknown Quantity.
12 13
±
±| |
aaaa
2 x aaa+2 a ax x z b b a a − a axx
2a xxx + 2 b b xa+xxxx-xx bb = 0
Tho' the Equation now appears to be adfected, yet the fquare
Root may be compleated, as in the laſt.
To fhow the Learner how this is to be done, if he fquares any
four Quantities, (for the Root of the above Equation will confift
P P
t
290
ALGEBRA.
of ſo many Quantities) he will find ten Terms in the Square,
four of which are pure Powers of the Quantities that were
fquared, and the other fix will be double Rectangles of thoſe
Quantities, of which each particular Root will conftitute a Part
of three of the Rectangles.
Now in the above Equation a a a a, a a x x, x x x x,
Are the Squares of
And the Quantities
a a,
2 x a a a, 2
ax,
xx,
a axx, 2 a x x x,
are the double Rectangles of thoſe Parts, or Roots.
And by examining 2bbaa, 2 bb xa, xx bb, the remaining
Terms in the above Equation, the first two are double Rec-
tangles of bb xa a and b b x ax, but the laſt Term is only a
fingle Rectangle of bb xxx, therefore to compleat the Square
there wants -xx bb, which when added to xx bb, will
make that a double Rectangle of bb x x x, and as we have no
pure Power of bb, which being ſquared is b b b b, hence if we
add b b x x + b b b b to our Equation, we ſhall make it a
Square, therefore
14
2xaaa + aaxx
2bbaaaaxx
-2axxx+2bb xa+xxxx-2xx bb
+ b b b b = b b b b - x x b b
Before we proceed, perhaps the Learner might have obferved.
that xx bb is the Square of xb, and therefore might ſuppoſe
that to be one of the Roots, but then he will find x b to make
a Part only of two of the Rectangles, whereas, if it had been
one of the Roots, it would have made a Part of three of the
Rectangles. Befides, if xb had been one of the Roots, it muſt
have had the Sign + in the Square, by Art. 34.
14 2
| 15 |
aa—ax+xx-b b = ✓✓ b b b b − x x bb
=b√bb-xx
The Manner of extracting the Root is thus, I firſt extract the
fquare Root of a aaa, which is a a, then the fquare Root of
a axx; the next pure Power is a x, and to determine whether
a x must have the Sign + or, obferve the Sign of the double
Rectangle of theſe two Roots, viz. of 2 xa a a, which becauſe
it is, I therefore in the Root make it a x.
The next pure Power is x x x x, whoſe Root is x x, then
obferve the Sign of the double Rectangle of this and one of
the two former Roots, as of 2 a a xx, which being +, and the
Root a a being +, therefore in the Root make it +xx.
The
Of folving Equations, &c.
291
The last pure Power is bb bb, whofe Root is bb, then obſerve
the Sign of the double Rectangle of this and one of the former
Roots, as the laft Root xx, but the double Rectangle of thefe is
2xxbb, which being negative, and the Sign of xx being +,
therefore place the Sign- before b b.
aa− a x + x x = b b + b✓bb-xx
a a — ax = b b
1 5+ b b
16
16. — x x
17
170
18
аа
-ax+
bb - x x + b √ b b
xx
x x
= b b
3 x x
+
4
4
18 ww 2
X
W
19
a
2
xx
b√b b—xx, for XX
4
3xx
4
/ bb - 3 * * + b√bb −xx
√bb_3xx
4
19 + =
x
20 a=
-+√bb-3**
+b √bb-xx
2
2
4
= 18.79= AC, which is impoffible, for AC+BC = 14 by
the Queſtion, confequently A C cannot be 18.79
This impoffible Conclufion is owing to taking the Root of the
Equation at the fifteenth Step, for as
a a x a a produces
aa aa, as well as a a x aa, therefore in the Extraction of fuch
Roots, it is doubtful whether the Root is a a, or a a, let us
now make a new Extraction, and ſuppoſe it to be
14 w 2
2 | 21 |-
aa.
-aa+ax-xx+bbbb bb - xxbb
= b √bb
xx
Having fuppofed the Root of aa aa the first pure Power
to be
aa, I go to the next pure Power, which is a axx,
whoſe Root is a x; but to determine its Sign, obferve the Sign
of the double Rectangle of theſe two Roots, viz. of 2 aaax,
into +
which being, I therefore make it a x,
produces -.
as
The next pure Power is x x x x, whofe Root is xx, then
obferve the Sign of the double Rectangle of this, and either of
the two former Roots, as of a x, now the Sign of 2 a xxx is
therefore in the Root make it xx, for + ax× - xx produces
axxx.
The laft pure Power is bb bb, whofe Root is bb, and obferve
the Sign of the double ReЯangle of this, and either of the other
Roots, as fuppofe the laft, the double Rectangle of theſe two
Pp 2
Roots
292
ALGEBRA.
}
Roots is 2 bb xx, which being —, therefore make it + b b, as
-××× b b gives
xx bb.
Now tranſpoſe a a, it being negative.
21+ aa 22
aa+b✓bb-xx-ax-xx+bb
22-b✓bb-xx 23 bb −
a a = b b + a x-xx-bbb xx
23-ax 24 aa—ax = b b − x x - b bb-xx
— ✓
Here the Equation is quadratic, and becaufe -xx-
bbb-xx is greater than bb, it is therefore ambiguous.
24 c
O
XX
—bb 3xx-b/bb-xx
aa-ax+
ax += = b b
4
4
미
​25/
x
25 w 2
26
2
- * = √
√ b b — 3 * * — b√
√bb_3xx
-
b √ bb - xx
4
26 +
४
3xx
- b √bb — xx
2
4
* = 14
x = 14
56
14
27 | a = = =
a == ± √bb
2
b = 14.75
b = 14.75
7375
10325
XX=
5900
196
1475
196
217.5625=bb
**
21.5625 (4.64 = √ bb - xx
16
86) 556
516
924) 4025
3696
329
14.75
Of folving Equations, &c.
293
14.75=b
4.64 = √bb**
5900
8850
5900
68.4400 b✔bb - xx
147.
3xx
4
215.44 = b√bb ≈ x + 3 x x
− x
4
xx = 196
3
4) 588= 3xx
3**
147=
4
3xx
bbb.
b √ bb - xx
4
217.5625 = bb
-215.44
3 x * -b √ bb — **
4
2.1225 (1.46 neareſt = √bb-
I
24) 112
96
286)-1625
7 ==
2
± 5.46 = √bb-3 **
8.46a= A C.
AC.
or 5.54 = a= A C.
4
bb b
*
But if a 8.46 then by the ninth Step y =
a = 8.46
= 14.
3384
846
ax = 118.44
xata a
b
8.46
294
ALGEBRA.
8.46 = a
8.46 a
5076
3384
6768
71.5716 = a a
196
=xx
267.5716 = a a + x x
118.44 =
a x
b = 14.75) 149.1316 (10.11 = y = A B.
1475
1631
1475
1566
1475
91
Becauſe a AC = 8.46 therefore BC=x-a5.54.
That theſe are the three Sides of a right-angled Triangle may
be tried, by fquaring and adding them, to fee if they agree with
the Property of the Figure.
5.54
5.54
10.II
10.11
8.46
8.46
2216
ΙΟΙΙ
5076
2770
ΙΟΙΙ
3384
2770
ΙΟ Ι ΙΟ
6768
30.6916
102.2121
71.5716
71.5716
102.2632
102.2121
.0511 the Difference which arifes from the Inaccuracy
of the Fractions.
But if the laft Proceſs is too perplexing, the fame Queftion may
be done otherwife, thus,
Let
1
Of folving Equations, &c.
295
Let AC+BC= 14= x, and a the Difference between
AC and BC, whence, as in the former Queftions, the greater
Leg or AC = *+a
and the leffer Leg BC-a,
2
2
Again, put AB+CD≈ 14.75b, and the Difference be-
tween AB and CD=y, then for the Reaſons already men-
¿±³, and CD =
+-y
tioned A B =
2
Y
2
Now becauſe the Triangle AC B is right-angled,
by 47 e. I
I
xx+2xa + a a
+
x x − 2 x a + a a
4
4
b b + 2 by + yy
-
4
Becauſe the Triangles ACB and BCD are fimilar, therefore
by 4 e. 6
2
AB: AC:: BC: CD
in Symbols 3
6+ y
x+a.x
a. b-y
2
2
2
2
bb-yy
X X-A A
or bb
3.
yy=xx
da
4
4
4
x x + a a
+a
from the firft
bb +2 by +yy
5
2
4
The Queſtion being contained in the fourth and fifth Equa-
tions, and there being no other Powers of a but a a in both thoſe
Equations, exterminate that unknown Quantity.
xx+yy—bb
2xx+2aa=bb + 2by+yy
2aa=bb+2 by + y y − 2 x x
bb + 2 by + yy — 2xx
4 + 6
a a=xx + yy
5 X.4
7
8
7
2 x x
82
9
a a
6.9
ΙΟ
xx+yy—bb=
II -yy
2
-
bb+2by+yy→ 2xx
2
*
IO X 2 I I 2xx+2yy2bb bb+2by+yy— 2xx
12 2x+yy2bb bb+2by-2xx
132xx+3y=3bb +2 by — 2xx
14yy=3bb + 2 by — 4 xx
12 +266
13-
2 xx
14-2 by 15 yy - 2by=3bb — 4 ≈≈
I
|
|
Here
296
ALGEBRA.
Here the Equation is quadratic, and fince
than 3 b b it is ambiguous.
1500
16 w 2
4x is
4 x x is greater
16 | yy 2 by + b b = b b + 3 b b — 4 xx
=
4bb-4xx
17y—b=√4bb
—4xx=2 / b b — x x
17+b 18 y = b + 2 b b —xx
y=b±2✓b
= 24.05 or 5.45
14.75±9.3
But y cannot be 24.05 for the Sum of the Legs is only 14.75
therefore y = 5.45
then by Step 6th | 19 | a = √xx+yy— b b — 2.85
b y
Then A B = ' + = 10.1
2
a
2
+ a
10.1 AC=
A C=* ±ª = 8.42 BC =
2
=
5.57 which three Numbers nearly agree with the
Property of the right-angled Triangle, but not exactly, becauſe
of the Imperfection of the Fractions.
The Reader may obferve, that in feveral of the Geometrical
Queſtions, after Letters are put for one or more of the unknown
Quantities, we then get Expreffions for the other Parts of the
Figure from its Properties, and therefore avoid ufing a greater
Number of unknown Quantities, and in general the Solution of
Queſtions are more neat and elegant, the fewer unknown Quan-
tities are uſed in the Work.
The Method of refolving Questions, which
contain four Equations, and four un-
known Quantities.
72. WH
HEN the Queftion contains four Equations, and
there are four unknown Quantities in each Equation;
find the Value of one of the unknown Quantities in one of the
given Equations, and for that unknown Quantity in the other
three Equations write this Value of it, which then reduces the
Question to three Equations, and three unknown Quantities.
Then
Of folving Equations, &c.
297
Then find the Value of one of theſe three unknown Quan-
tities in one of theſe three Equations, and for that unknown
Quantity in the other two Equations write this Value of it,
which reduces the Queftion to two Equations, and two un-
known Quantities.
Then find the Value of one of the unknown Quantities in
each of theſe two Equations, and making theſe Equations equal
to one another, we fhall have an Equation with only one
unknown Quantity, which being reduced, will answer the
Queſtion.
Queſtion 105.
A, B, C, D.
If A's Share was added to twice B's Share, from which Sum
fubftracting twice C's and D's Shares, there remains 650
Pounds:
A Father gave 1000l. to his four Sons
And if from A's Share there is fubftra&ted three times B's
Share, to the Remainder adding twice C's Share, from which
Sum fubftracting five times D's Share, there remains 400 Pounds:
But if to A's Share there is added four times B's Share, from
which Sum fubftracting three times C's Share, and to the Remainder
adding fix times D's Share, the Sum is 1150 Pounds. How much
had each Son?
=
Let a A's Share, e➡ B's Share, y C's Share, u = D'
Share,s=1000, m=650, n= 400, b = 1150.
1234+ no
from the firſt
5.2
5.3
7
5.4
1 8
ate+y+x=5
a + 2e − 2 y — 2 u ⇒ m
a
5u = n
·3e+27-54
a+4e−3y+ 6 u
ste
u - y — e
= m
+ e − 37 — 34 =
4ety
$
6u=1
s+3e-43+5ub
By the
Queftion.
Here the Queſtion is reduced to three Equations, and three
unknown Quantities.
from the fixth
9
9.7
9.8
ΙΟ
8 II
e = m +34 +31
5 ~ 4 17 ---- 1 224 -12y+45+y—6u— n
S
m
s+3m+9u+ 93 — 35—43+54=6
Here the Queſtion is reduced to two Equations, and two
unknown Quantities.
eq
10 con-
298
ALGEBRA.
II contracted
from the twelfth
14y =
from the thirteenth 15 y=
14. 15
16
10 contracted
12 5. s — 4 m — 18 21 ----- y= n
1325+3 m + 14 u+5y=b
55- 4 m
18 น
II
n
6 + 25 — 3 m — 14 u
5
b+25—3m—14u
55—4m-1848-78
5
II
16 x
17
17 ±
1864 ×
u =
1864
19 u น -
116 +225 +33m-154u=255-20171
90u - 5 n
11 b — 35 — 13m+5n
11b-35 13 m + 5 n = 50,
64
the Share of D.
b + 25-
3 m
14 น
y =
= 100,
5
then by Step 15th
20 y
the Share of C.
and by Step 9th 21 e=m+3u+3y=100, the Share of B.
and by Step 5th | 22 a=s—4—3—¿750, the Share of A.
And in the fame Manner may any other Queftion in the like.
Circumftances be anſwered.
I fhall now add a few Queſtions of a different Nature, and
ſuch as are generally firft propofed to Learners, but as they
require a little more Sagacity to exprefs their Conditions, have
hitherto been avoided, imagining the Learner is more per-
plexed to exprefs, or find out the Equations refulting from fuch
Queſtions, than to refolve the Equations; and therefore they
were thought not fo proper at the Beginning of this Work.
Queſtion 106. A Perfon bought two Horſes A and B, which
with the Trappings coft 100 Pounds:
Now if the Trappings were laid on the Horfe A, both Horſes
were of equal Value:
But if the Trappings be laid on the Horfe B, he will be double
the Value of the Horfe A. How much did each Horſe coſt?
Let b = 100, a = the Value of the Horfe B and Trappings,
then b— a = the Value of the Horſe A.
Now becauſe the Horfe B and Trappings are double the
Value of the Horfe A,
hence
Of folving Equations, &c.
299
hence I
1+20 2
2÷313
a = 26-2a by the Queftion.
3α=2b
26
200
a-
-662 Pounds, the
3
3
3
Price of the Horfe B and Trappings.
Confequently 100-661
33 Pounds, the Price of the Horſe A.
3
3
1
But to find what the Trappings coft, and by that Means to
find the Price of the Horfe B, let y the Price of the Trappings.
Now the Trappings taken from the Horfe B, and laid upon
the Horfe A, both Horfes being then of equal Value,
therefore I 33 +9 = 66
+2=66.
2
y
3
3
2
I ty
2
33
3
I
2 — 33
3
29=33
= 33 //
3
3
3÷2
4
y= 162 Pounds, the Price of the
+ 16
3
alm
Confequently 33
Horfe B.
3
{Trappings.
50 Pounds, the Price of the
Queſtion 107. A Labourer in 40 Weeks Labour faved 28
Crowns -the Pay of three Weeks, and found he had spent 36
Crowns the Pay of eleven Weeks. How much did he receive
+
a Week?
Let a = his weekly Pay.
Then he had faved Crowns
And ſpent Crowns
28 3.a
36 +11 a
64+ 80
And as the Sum of theſe two must be equal to what he re-
ceived for his forty Weeks Labour,
1 40 @ = 64+ 8 a
therefore
I 8 a
232
2
32a=64
3
a = 2 Crowns, his weekly Pay.
Q92
Queſtion
300
ALGEBRA.
Queſtion 108. A Servant was hired for 12 Months, for which
he was to have 24 Pounds with a Cloak; when he had ferved 8
Months he has Leave to go away, and instead of his Wages receives
a Cloak and 13 Pounds. How much did the Cloak coſt?
Let a = the Price of the Cloak, b = 12, d = 24, m =
* = 13.
8,
Now d+a is what he did receive for ferving eight Months.
And as the Pay for eight Months was proportional to what
he was to receive for twelve Months, therefore,
I
d+a:b:: x+a: m
When any four Quantities, or Nume
bers, are in Geometrical Propor-
tion, the Product of the Extreams
and Means are equal,
md+ma=b x + b a
bama = md
md - b x
therefore
2 +
3
3÷b —m
md - b x
4
a=
=9 Pounds, the Price
bm
of the Cloak.
Queſtion 109. There is a Footman A, who goes 6 Miles a
Day, and 8 Days after B follows him and goes 10 Miles a Day.
In how many Days will B overtake A?
10, a
Let b = 6, d=8, m = 10, the Number of Days B
travels to overtake A, then as A begun to walk eight Days
before B,
Hence the Number of Days that A travels, is
And the Number of Miles A travels, is
And the Number of Miles B travels, is
d + a
b d + b a
ma
But when B overtakes A, they muft have travelled an equal
Number of Miles.
Therefore
I
I
-ba
2 ma
ma = b d + b a
· b a = b d
b d
2÷m-
b
3
a =
m—b
人
​12, the Number of Days
required, or the Time in which B
will overtake A.
Queſtion
Of folving Equations, &c.
301
Queſtion 110. If a Scribe can in 8 Days write 15 Sheets,
How many fuch Scribes can write 405 Sheets in 9 Days?
Let a = the Number of Scribes, b= 8, d = 15, m = 405,
n = 9.
ds
Then
b:d::n:
I
d the Number of Sheets the
b
dn
and 2
b
Scribe can write in nine Days.
bm
:I::m: = the Number of Scribes
d n
to write the 405 Sheets in nine Days.
b m
3240
hence
3
d n
24, the Number
135 (of Scribes required.
Queſtion 111. A can do a Piece of Work once in 3 Weeks, B
can do it three times in 8 Weeks, and C can do it five times in
12 Weeks. In how long Time can they do it jointly ?
5x
Let a the Time required, b = 1, d= 3, g = 8, n = 5₂
m12, the Number 3 occurring twice, I put only d for it,
Then I
d:b::a:, the Part of the Work
d
that can be done by A in the Time
fought.
and 2
g: d::a:
da, the Part of the Work
&
that can be done by B in the Time
fought.
na
and 3
m:n::a:
the Part of the Work
m
that can be done by C in the Time
fought.
And as theſe three Parts are to be equal to 1, or one Work,
ba da na
therefore 4 + +
= 1,
d
g
m
I
I
whence
a=
5
b
n
I
+
+
+
5
m
3
12
by
302
ALGEBRA
by reducing the Fractions + 3 3 + 2/5/20
minator, and adding
5 9
+ 3 3 +
+ = = 2/
12
3
Whence a=
10100
8
0100
9
3
12
to common Deno-
and abbreviating them, we ſhall find
of a Week, by the Rule for Divifion
(of Vulgar Fractions.
If the Week confifts of 6 Days
8
9) 48 (5 Days
45
3
And the Days confift of 12 Hours
9) 36 (4 Hours, that is, they will pef-
form the Work in five Days
four Hours.
36
+
da
+
na
m
g b a + d d a
I, may be reduced thus;
m
= d
+ 8 d n a = gd
ba
Or the Equation
g
g
4 x d 6
dda
dna
bat
+
g
6x8
7
7 x m
8
mg d
288
8
9
a
m g b + m d d + g d n
324
8
of a Week as above.
9
m
mg ba+m d da + g dna≈m g d
73. Having in this eafy familiar Manner, by general and
univerfal Rules, explained to the Learner the Elements of this
celebrated Science, it may not be improper to raiſe his Curiofity,
and animate him to exerciſe his Judgment in the Choice of
Quantities for the Solution of the fame Queftion, to give an
Inftance how much the Solution of Queftions becomes more neat
and elegant, by a judicious Choice of repreſenting the unknown
Quantities. The Queftion and its Solution is from the ingenious
Mr. JOHN WARD's Young Mathematician's Guide.
Queſtion
Of folving Equations, &c.
૩૦૬
Queſtion 112. A Man playing at Hazard, or Dice, won the
firſt Throw juſt ſo much Money as he had in his Pocket; the fecond
Throw he won the fquare Root of what he then had, and five
Shillings more; the third Throw he won the Square of all he then
had; after which his whole Sum was 1121. 16s. od. What
Money had he when he began to play?
Suppofe
then
I a his firft Sum.
123
2a= his Sum after the first Throw.
3√ √20:+5 his Winnings at the
fecond Throw.
4√20:+2a + 5 = his Sum after
the ſecond Throw.
and
2+3
42
5
4+ 5 6
2a + 40 √2a +10 √2a + 4 a @
+ 20 a + 25 = his Winnings at
the third Throw.
24@+40√2@+11√20+40 a
+302256 Shillings.
Now to avoid thefe furd Quantities, let us make a fecond
Suppofition; thus,
123
Let
then
and
2+3
4
4 2
5
4+ 5
6
20 a = his firſt Sum.
4aa
his Sum after the first Throw.
2a+5= his Winnings at the fecond
Throw.
4aa+20+5
fecond Throw.
his Sum after the
16 aaaa +16 aaa+ 40 aa + 20 a
+4aa +25
the third Throw.
his Winnings at
16 aaaa +16 aaa + 48 a a + 22 a
+30256
But to avoid theſe high Equations, let us make a third Sup-
pofition; thus,
a a
Let I
= his firft Sum.
2
then 2
and 3
2+3 4
a abis Sum after the firft Throw.
a+5= his Winnings at the fecond
Throw.
ao+a+5= his Sum after the ſecond
Throw.
But
304
ALGEBRA.
But as it was the Square of a a+a+ 5 he won at the third
Throw, to avoid the Trouble of fquaring it,
e=aa+a+5
fubftitute 5
then 6
ee
5+6
his Winnings at the third
Throw, confequently,
ус
700
78
8 au 2 9
0.5 ΙΟ
5.10
II
II
12
12 CO
13 un 2
14 -0.5j
whence
ee + e = 2256 Shillings
eete + 0.25 = 2256.25
e +0.5= 47.5
e47. Becauſe at the fifth Step e was
fubftituted for aa+a+5
aa+a+5=47
a a + a=42
13aa+a+0.2542.25
14a +0.5=6.5
1510=6
15
aa
16 = 18 Shillings, the Money he had
2
(when he first began to play.
The Learner will eafily obferve, that the third Solution is
more neat and elegant than either of the other two; tho' I
know of no general Rule that is given for the Choice of the
Quantities to ftate the Queſtion, but it is left to the Judgment
and Sagacity of the Reader, and as fuch Methods must be
attended with particular Difficulties to a Learner, I have avoided
the perplexing him with them; but as he has now a general
Method of folving Equations, he may exerciſe his Judgment at
his own Difcretion, in the Choice of different Quantities to
repreſent the fame Queſtion.
The Method of expreffing the Power of
any Quantity, by placing a Figure
over it.
74. T
HERE is a more compendious Method of expreffing
the high Powers of any Quantity, than writing
them at length, by placing a Figure over the Quantity thus,
3
I
4
a is aaaa, and a is a a a, and a is e, and
2 3
ab is a ab bb,
that is, the Figure that ftands over the Letter ſhows to what
2
Power
Of expreffing the Power of any Quantity. 305
Power that Letter, or Quantity, is involved, which Method of
Notation is generally ufed when the Powers are high. The
Figures placed over the Quantity are called Exponents. The Mind
being a little accuſtomed to this Method of Notation, will as
eafily manage an Algebraic Procefs, when the Powers are ex-
preffed by Exponents, as if they were repeated at length; and for
the further Eafe of the Learner, in this Method of Notation, we
will refume the Solution of Queflion 90, expreffing the Powers
by Exponents, that the Learner may compare both the Opera-
tions together.
I
1 2
I + e² | 3
3- In
4
5
3 to 7∞
6
e
a²+a-
? —es in
e² + e = 2 a
a²
+ a = m +e
a² + a
2 a
a² + a
+a- m
112 = 82
2 a
}to find a and e.
2
a²+a—me = 2 a
2
e
4.5
6+0
7 a
8
a²
8 + m
9
10
e
II
10 & 2
2.II. 10 12
12 in Numbers 13
13 contracted 14
13-a4
15+2a3 16
16188 a
17 - 187 a
m + e = a
a² + c = a + m
e
ė = a + mi n²
ег
аг
a²+2am+m²— 2 a³ — 2 ma² + ai
+2am + m² — 2a³ — 2ma² + at + a
+ m
188
·a²
2 a
a+88362a3188 a² +at+a
+942 a
187a+ 89302a3 188 a² -+- aˆ = 0
152+= 1870 + 89302a3188 a²
—34 + 2a³ = 187a+8930 188 a²
17-a+ + 2a3 + 188a² 187a+8930
18] −24 +2 a³ + 188 a² — 187 a
at
—
8930
In the fame Manner the Learner may attempt the Solution of
any of the other Queftions, expreffing the Powers by Exponents:
One Thing is to be carefully obferved, that the Exponent belongs
only to the Letter which ftands under it, and when it is only
Unity, or 1, it is never fet down, like the Co-efficient when it
is Unity only, it is generally omitted in the Expreffion.
Rr
The
[ 306 ]
if a Question is
of one Anfwer;
that is, admits
The Method of knowing
I'mited, or admits but
75.
or if it is indetermined, that is,
of feveral Anfwers.
TH
HE Queſtion being ſtated, that is, all the Equations
being expreffed which are neceffary for the Solution
of the Question, then if there are more unknown Quantities.
than there are Equations, the Queftion admits of a Variety of
Anſwers, and is therefore unlimited or indetermined, ex. gr.
Suppoſe a +e=40} to find a, e, and y.
Andey 20
Here there are three unknown Quantities, and only two
Equations.
Now e being in both the given Equations, you may ſuppoſe it
any Number under 20, the leaft of the two given Numbers, as
16.
for Example fuppofe e
Then the first Equation is a + 16 = 40.
And the fecond Equation is 16+y=20.
From whence it will be eafy to find a and y, but if e is fup-
poſed any other Number under 20, then there will be found
different Numbers for a and y, and the like of any other Que-
flion, where the Number of unknown Quantities, are more
than the Equations which arife from the Queſtion.
But when the Number of given Equations are just as many
as the unknown Quantities required to be found, then the Que-
fion generally admits but of one Anfwer, for then each of the
Quantities fought hath generally but one fingle Value, thus ag
at Queflion 80, where we have
I
2
a+e+y= b = 18
a+3e-2y
2y=m= 9
31a+4y- 2 e
P = 21
P=
Where a = 5, e =
6, and y = 7.
But
To extract the Cube Root.
307
4
But when the Number of given Equations exceeds the Number
of Quantities fought, they not only limit the Queftion, but often
render it impoffible, as one of the Equations may be inconfiftent
with another; as for Example,
1 late == 162
2 | ae=48
3
| a
e = 22
}
to find a and e.
Now here are three Equations, and but two unknown Quan-
tities, and the first and fecond Equations include a poffible Cafe,
and it may be found what the Numbers are.
And if we take the fecond and third Equations, they like-
wife include a poffible Cafe, for it may be determined what thoſe
Numbers are.
But all three Equations together render the Cafe impoffible,
the first Equation being incompatible with the third, as the Sum
of two Numbers cannot be leſs than their Difference.
;
To raife or invent a Method to extract the
Cube Root.
76. TH
76.HIS is no more than the Method of Converging
Series applied to the Solution of an Equation, one Side
of which is the unknown Quantity, and is a pure Cube, or
raiſed to the third Power only, ex. gr.
Suppoſe a a a≈ 9261, where 9261 is a Cube Number, now
to find what a, or the Number is that being cubed will produce
9261, is to extract the Cube Root of 9261.
By the common Method of diftinguiſhing of how many
Places the Root will confift, by placing a Point over the
Place of Units, and another over every third Figure, the Root
will confift of two Places, therefore fuppofe the Cube Root
to be
20
20
400
20
8000 which being less than
9261 the given Number, the Cube Root of 9261 muſt be more
than 20.
Rr 2
New
308
ALGEBRA.
Now putr
20, and e for what 20 wants of the true Root,
then is r+e = a, or the Cube Root of 9261, and proceed as
in the Method of Converging Series, Cafe 1. Page 230.
If I r+e= a,
Raife this Equation to the third Power, becauſe it is the
Cube Root, which is to be extracted.
1 & 3
but
2
3
2 3
•
4
rrr + 3rre | 3ree+eee≈a a a
aaa 9261 by the Example,
-
rrr + 3rre +3ree+eee=9261
Put this Equation into Numbers, and reject all the Powers of
e above e e, as in the Method of Converging Series.
5
4 in Numbers
5-8000
6÷60 7
8000 + 1200e +60 ee=9261
Becaufe 8000 is less than 9261, tranf
pofe 8000
61200e
7÷2Q+e 8
1200e +60 ee = 1261
Dividing by the Co-efficient of ee
20e +ee = 21.01
Dividing by 20+e, that is, by the Co-
efficient of e plus e, as in the Method
of Converging Series,
21.01
20+e
Operation in Numbers,
20) 21.01 (1 = e
+e=1 I
Divifor 21 21
.01 Remainder rejected.
† 20
+e=
I
r + e = 21 = a,
which being tried will be found to be the
Cube Root of 9261. And by the fame Method may the Cube
Root of any other Number be extracted.
But to fave the Trouble of repeating this Operation, when
any Cube Root is to be extracted, the above Procefs may be
made more general, by not turning the Equation at the fourth
Step into Numbers, and putting any Letter for the given Num-
ber, whofe Cube Roots to be extracted.
Suppofe
To extract the Cube Root.
309
Suppoſe as before a a a = 9261, let b = 9261.
Then a a a= b, to find a, or to extract the Cube Root.
Now make a Suppofition that 20 is the Root, which being
tried as before, it will be found too little. Then put 20 = r,
and becaufe 20 is too little, r in this Cafe is ufually called lefs
than juft; and for what r wants of the true Root put e, whence
re will be the true Root, or equal to a.
Hence
I
1 & 3
2
but
3
2
3 4
r+e = a
Raife this Equation to the third Power
as before.
:a
r r r + 3 r r e +3ree+eee — a a a
aaa = b as above
r r r + 3 r r e +3ree+eee = b
As we know rrr to be less than b, by finding the Cube of 20
was lefs than the given Number, therefore tranfpofe rrr, and
reject the Powers of e above e e.
4 — r r r 5
=
3rre+3ree brrr
Dividing by the Co-efficient of ee,
retee=
53r 6
tee
b
rrr
3r
As there will be another Divifion before the Operation is
finiſhed, to keep the Fraction as fimple as may be, fubftitute
D=
b — r r r
3r
Then
7
re+ce=D
Now dividing by re, that is, the Co-efficient of plus e,
8 |
D
THEOREM 1.
r+e
Operation in Numbers,
ра до ра
b-9261
8000
3r= 60) 1261 (21.01 = D.
120
61
60
100
60
40
r = 20.
310
ALGEBRA.
r = 20) 21.01 = D (1 = e.
+e= I I
Divifor 21 21
.or Remainder neglected.
20
+e= I
r+e = 21 = a, the Cube Root required as before.
Now fuppofe it was required to extract the Cube Root of
132651.
Here according to the Method of pointing, the Root will
confift of two Places, and to make a tolerable near Suppofition
at the firft Trial, the firſt Period being 132, I confider what
whole Number cubed will be the neareſt to 132, and I find it
to be 5, then as the Root confifts of two Places, I fupply the next
Place with a Cypher, and fuppofe the Root to be 50, which
I know is less than the true Root, as the Cube of 5 is leſs
than 132.
Hence, as before, we are to determine what the Number is,
that 50 wants of the true Root of 132651.
Then putting r≈ 50, and e what it wants of the true Root,
and b 132651, we have juſt the fame fubftituted Letters as in
the laft Example; and if the Operation was repeated it will be
exactly the fame, it is therefore needlefs to repeat the Work,
but only obferving the Equation, or Theorem to find e, which
D
is e = -
r -- e
and by Subftitution we have D=
Now b = 132651
-- yr gu
125000
3r=150) 7651 (51.006=D.
750
b
yo yo qo
3r
151
150
1000
900
100
y = 50
To extract the Cube Root.
311
t
r≈ 50) 51.006 = D (1
- &
+e= I
Divifor 51 51
.006 Remainder neglected.
Now it was r = 50
We have found e
I
r+e= 51 the Cube Root of 132651, which
being tried will be found to be true.
And in the fame Manner, the Cube Root of any other Num-
ber may be extracted, without repeating the Algebraic Work,
when the Number affumed for the Root is lefs than the true
Root: But when the Number affumed for the Root is too much,
or more than the true Root, then we proceed as in the following
Example, in the fame Manner as at the fecond Cafe of Converging
Series, Page 235.
Required to extract the Cube Root of 24389, or a ca
=24389.
By the ufual Method of pointing, the Root will confift of two
Figures, the firft Period of the given Number is 24, and the
Cube of 3 being the neareſt of whole Numbers to 24, and fup-
plying the other Place of the Root with a Cypher, I fuppofe 30
to be the Cube Root of 24389, but the Cube of 30 is 27000,
which being more than the given Number, the Cube Root can-
not be fo much as 30.
Therefore let r = 30, which is now too great or more than juſt,
and what 30 is too much call e, then will 7 —
r e = a, or
the true Cube Root required, and calling the given Number
24389=b,
we have
1
1 & 3
2
but 3
2.3 4
e = a
Raife this Equation to the third Power
as before.
·3rre+3res -eee-a a a
a a a = 24389 b, as b is put for the
given Number.
rrr-3rre+3ree-eee=h
Becauſe b is less than rrr, tranſpoſe b and reject the Powers
of e above e e.
+ - " | 51
yo yo yo
I
3rre+3reco, for one
Side of the Equation ſubítracted from
the other Side muft leave o, or nothing.
Then
312
ALGEBRA.
Then tranfpofe all the Powers of e, to the other Side of the
Equation.
3rrerrr-b÷zree
5+3rre 6
6- 3ree 7
3 rre
3ree = rgo qu
b
Dividing by the Co-efficient of ee,
7 ÷ 3"
8
g yg b
re
e e
3r
As there will be another Divifion before the Operation is
finiſhed, to keep the Fraction as fimple as may be, ſubſtitute
G:
3r
b
Then 9 [re-ee = G
|
Now dividing by re, that is, by the Co-efficient of e
minus e,
·| 10 | e=
G
THEOREM 2.
re
- e
Operation,
b
rrr = 27000
24389
3r=90) 2611 (29.01 = G.
180
811
810
100
90
IQ
r= 30) 29.01 = G (1 =
I
Divifor 29 29
.01 Remainder neglected.
ė =
*
30
I
re = 29 = a, which being cubed will be found the true
Cube Root of 24389.
In
To extract the Cube Root.
313
In this Cafe, the Quotient, Figure is fubftracted from the Di-
vifor as it is found, the Divifor at the tenth Step being re,
whereas in Theorem I, p. 309, it was at the eighth Step r+e.
Now as the firſt ſuppoſed Root muſt be too great or too little,
unleſs it happens to be taken exact at the firſt Time, therefore
theſe two Theorems will extract the Cube Root of any Number,
as in the following Example.
Let it be required to extract the Cube Root of 14526.784
From pointing the whole Numbers according to the uſual
Method in common Arithmetic, the Root will confift of two
Places of Integers, the firft Period of the given Number being 14,
the Cube of the whole Number which is nearest to 14 is 2;
and ſupplying the other Place of the Root with a Cypher, I fup-
poſe the Root of the given Number to be 20, which is too little,
or less than just, the Cube of 2 the firſt Figure in the Root being
lefs than 14, the firft Period in the given Number.
e
Then putting b=14526.784 r = 20, and what 20 wants
of the true Root, we proceed as at Theorem 1, page 309, where
D
e
r+e
and by Subftitution D=
b=14526.784
8000.
-rrr =
b
rrr
3r
3r=60) 6526.784 (108.78 neareſt = D.
60
526
480
467
420
478
r = 20) 108.78 =D (4.44 = b.
+e=4
Divifor 24
96
4.4 1278
Divifor 28.4 1136
.44 14200
Divifor 28.84 11536
2664
Sf
The
314
ALGEBRA.
The Reader will obferve that the Quotient Figure is added
twice to the Divifor to compleat it, in the fame Manner as at
the Method of Converging Series, Page 233.
Now r 20
+e= 4.44
r+e = 24.44 and to try whether this is the true Root of
the given Number cube it.
24.44
24.44
9776
9776
9776
4888
597-3136
24.44
23892544
23892544
23892544
11946272
14598.344384 which being greater than the given Num-
ber, the Root cannot be ſo much as 24.44
e will
To approach ftill nearer to the true Root, make a fecond
Operation, fuppofing the Number laſt found, viz. 24.44 to be r,
and put e for what that Number is too much, then r -
be the true Root, and putting the given Number 14526.784
G
b, we proceed as at Theorem 2, Page 312, where e
rrrb
and by Subfiitution G
2
b
3r
rrr = 14598.344384
14526.784
3r73.32) 71.560384 (.976 G.
65988
55723
51224
43998
r
r
g= 24.
43992
To extract the Cube Root.
315
r = 24.44) .9760 G (.04 = e.
.04
9760
Divifor 24.40
Now r 24.44 by the firft Operation.
e =
.24
r-e= 24.4 which being cubed, will be found the true
Root of 14526.784
Therefore by the fecond Operation the true Root is found.
For a further Variety, let it be again required to extract the
Cube Root of the fame Number 14526.784
But let us fuppofe the Cube Root to be 30, the Cube of
which being 27000, the Root cannot be fo much as 30, then
putting r 30, we fhall have r too great or more than just,
and putting e what it is too much, then re will be the true
Root, and calling the given Number 14526.784 = b, we pro-
ceed as at Theorem 2, Page 312, where e =
and by
Subftitution G =
g gr g
3r
rrr = 27000.
b=14526.784
G
3r90) 12473.216 (138.591 = G.
90
347
270
773.
720
532
450
821
810
116
90
26
Sf 2
go ~ 30
316
ALGEBRA.
r =
30) 138.591 = G (5.7
5
Divifor 25
125
5.7 1359
Divifor 19.3
1351
8
7 = 30.
l =
e =
-5.7
24.3 to try whether this is the Cube Root of
14526.784 cube 24.3
·24.3
2463
729
972
486
599.49
24.3
177147
236196
118098
14348.907 which being lefs than the given Number
14526.784 the Cube Root mult be
more than 24.3
Now for a fecond Operation, and let r 24-3 and what it
wants of the true Root call e, then will re be the true Root,
and ftill calling the given Nnmber 14526.784b, we now
proceed as at Theorem 1, Page 309, where e = ———,
D
r te
>
and by
Subftitution D =
b-rrr
b
3r
To extract the Biquadrate Root.
317
b = 14526.784
go ya y 14348.907
3r72.9) 177.877 (2.44 = D.
1458
3207
2916
2917
2916
I
همه اره
r24.3) 2.44D (.1 = "
+ . I
Divifor 24.4
244
r 24.3 by the firft Operation.
+e=
.I
te= 24.4 the true Root as before.
r+
In the fame Manner may the Cube Root of any other Num-*
ber be extracted, and tho' the true Root may not always be ex-
actly had, yet by repeating the Operation you may approach to it,
within any affignable Degree of Exactneſs, and if a fmall Miftake
happens in the firft, it will be corrected at the fecond Operation.
To extract the Biquadrate, or fourth Root.
TH
HIS Operation proceeds in the fame Manner as in the
Cube Root, only raifing the r+e or re to the
fourth Power, thus,
Required the Biquadrate or fourth Root of 194481, or of
194481.
a a a a
By placing a Point over the Place of Units, and another over
every fourth Figure, we hall find the Root will confift of two
Figures: And the firft Period of the given Number being 19,
now the Biquad are or fourth Power of 2 being 16, which
being the neateit in Integers, and fupplying the other Place of
the Rout with a Cypher, fuppofe 20 to be the Biquadrate
Root of 194481; but the Biquadrate of 20 being only 160000,
the Root mult be more than 20. Now let r 20, and putting
e for what 20 wants of the true Root, then will rea
be the true Root required; and calling the given Number
194481b, then a a aab.
Now
318
ALGEBRA.
Now I r+e=a.
Raife this Equation to the fourth Power, becauſe it is the
Biquadrate Root that is to be extracted.
14 2
But
3
2.3
4
rrrr + 4rrre+6rree = a aa a,
all the Powers of e above ee being
rejected.
aaaab, b being put equal to the
given Number.
r r r r + 4 r r re+6rree = b.
Becauſe rrrr is less than b, tranfpofe
rrrr.
b — r r r r
4rrre+6rree-b
5
Dividing by the Co-efficient of ee,
4 - r r go yo
56 rr
6
2re +
b
yo yo yo yu
tee
e e =
orr
3
As there will be another Divifion before the Operation is
finiſhed, therefore as in the Cube Root, put D =
re
Then | 7|274
27
Now dividing by
3
plus e,
2r
8-
+ e
8
3
+ee = D.
b-
go p q p
6 rr
+e, that is, the Co-efficient of a
2 r
D
+e
THEOREM 1.
3
Operation b = 194481
r yr y
- 160:00
6rr=2400) 3+481 (14.367
(14.367 = D.
2400
10481
9600
8810
7200
16100
14400
17000
2r
To Extract the Biquadrate Root.
319
27
13.33) 14.367 = D (1. = e.
3
+ I.
14.33
1433
37 Remainder neglected.
r = 20
+e=
I
r+e = 21 = a, which being raiſed to the fourth Power,
will be found to be the Biquadrate Root of the given Number.
And if we here take the firft Root too great, or more than
the Truth, the Operation is the fame as raiſing the ſecond Theorem
for the Cube Root.
Suppoſe a a a a = 456976, to find the Biquadrate Root.
The Root being found to confift of two Figures as before,
and the firft Period in the given Number being 45, and the`
Biquadrate of 3 being 81, fupplying the other Place of the
Root with a Cypher, let us fuppofe 30 to be the Root, but the
Biquadrate of 30 is 810000, which being more than 456976,
the Root cannot be fo much as 30.
Then putting r = 30, and e what 30 is too much, we have
rea the Root required; and putting b = 456976, we
then have a aaa=b.
I
Now
1 & 4
ज़ि
But
2
3
a
Raifing this Equation to the fourth
Power as before, and neglecting the
Powers of e above ee.
g gr pr
аааа
2 3 4 rrrr
•
b
4rrre+6rree = aaaa
4 rrre + 6rree b
Becauſe b is less than rrrr therefore tranfpofe b.
4
b
5 r r r r — b — 4 rrre+6rree = 0,
one Side of the Equation being ſuð-
ftracted from the other, must leave
nothing, now tranfpofe the feveral
Powers of e.
5 + 4 rrre
6
Arrre = j jo jo j
b + brree
6.
320
ALGE BRA.
S
G
= r —
b
6-6rree ク ​4 rrre — 6 rree
6 r ree = r r r z
Divide by the Co-efficient of ee.
7÷6rr
2re
8
3
r r r r —
b
-ee
6rr
For the fame Reafon as in the laft Operation, ſubſtitute
rrr r -b
6rr
2re
Then 9
ве
G
3
27
Now divide by -e, that is, by the
3
Co-efficient of e less.
12r
9.
e
10
e=
3
1
Operation,
rrrr810000
—b――456976
2 r
3
G
C.
THEOREM 2.
6rr=5400) 353024 (65.374 = G.
32400
29024
27000
20240
16200
40400
37800
26000
21600
;
4400
20.) 65.374 G (4.08 = ec
27
3
4
Divifor 16. 64
.08 13740
Divifor 15.92 12736
1004
r
= 30
To extract the Biquadrate Root.
321
r = 30
e
- 4.08
r = e = 25.92 = a, and to try if this is the Root,
25.92
25.92
5184
23328
12960
5184
671.8464
671.8464
26873856
40310784
26873856
53747712
6718464
47029248
40310784
raiſe 25.92 to the fourth Power.
451377.58519296 which being less than the given Number
456976, the true Root must be more
than 25.92
Then for a fecond Operation let r
25.92 and for what it
wants of the true Root put e, that now r+e = a, and ſtill
calling the given Number 456976b, this is exactly the fame
Cafe as when we raiſed the firft Theorem, Page 318, for the
Biquadrate Root, whence we have no Occafion to repeat
the Algebraic Work, but to ufe that Theorem, where
b — r r r r
D
·
2 r
ابن
+
te
and by Subſtitution D=
6rr
Tt
b
11
322
ALGEBRA.
rrrr
b =456976.
451377.5852
6rr 4031.0784) 5598.4148 (1.3888 D.
40310784
=
156733640
120932352
358012880
322486272
355266080
322486272
32779808
r
27 = 17.28) 1.3888 = D (.08 = c.
3
+ .08
Divifor 1736 13888
r25.92 by the firft Operation.
= .08
+e=
r+e = 26. = a, which being involved to the fourth
Power, will be found the true Biquadrate Root of 456976.
The Reader will eafily obferve that theſe two Theorems will
extract the Biquadrate Root of any given Number, in the fame
Manner as the two Theorems did for the Cube Root.
In the fame Method may Theorems be raiſed to extract any
Root, it being no more than to fuppofe a Number to be the re-
quired Root, and try whether it is too great or too little; then
calling it re, or rea, or the true Root, as the Occafion
requires, and raife this Equation as high as the Root is to be
extracted, after which the Operation is the fame as before.
Tor
·
[323]
77.
To turn Equations into Analogies.
Sthe
UPPOSE there was given this Proportion a: b : : c : d;
then multiplying Extreams and Means we have this
Equation ad bc, now as we get an Equation from Quantities
in continual Proportion, by multiplying the Extreams and
Means, and making one Product equal to the other. Hence to
turn any Equation into an Analogy, is only the reverſe, by
taking the Quantities that compofe either Side of the Equation,
and making them the two Extreams, and the Quantities that
compofe the other Side of the Equation, and making them the
two Means in the Proportion.
To turn the Equation m dz a into an Analogy.
One Side of the Equation is compofed of the Quantities m
and d.
And the other Side of the Equation is compofed of the Quan-
tities z and a.
Hence placing theſe Quantities according to the Direction,
ve have m: z::a:
d
Ord: z::a: m
Orz: d::m: a
Orz: m :: d: a, &c.
For multiplying the Extreams and Means of either of theſe
Proportions, we fhall ftill have the given Equation dmza.
Again, fuppofe the Equation an bdx, and it is required to
find the Proportion of a to b.
Now one Side of the Equation is compofed of the Quantities
a and n.
And the other Side of the Equation is compoſed of the Quan-
tities b and dx.
But in ranging thefe Quantities, make the Quantities a and
whofe Proportion is required, the first and fecond Terms in
the Proportion, and place the other two Quantities fo, that if
the Extreams and Means were to be multiplied, they will pro-
duce the given Equation, and then we fhall find a: b:: dx: n.
From the Equation d ny bxz, to find the Proportion of
d to b.
By the Directions we ſhall find d: b : : x x : n
Tt 2
From
324
ALGEBRA.
From
is a : d:
= dpx.
an
m
=pxd, to find the Proportion of a to d, which
n
px: for multiplying Extreams and Means
m
To find the Proportion of a to z from
a n
rn
an
b z
here
2
y
X
b
n
a: 2 ::
x y
dny.
To find the Proportion of a b to d, from ab or I ab
Here abd::ny: I, for multiplying Extreams and Means
we have a bdny.
78. I fhall now fhow the Learner, the Certainty of the
Rules on which this Science is founded; this I have purpofely
omitted in the Beginning of the Work, imagining it unrea-
ſonable to expect a Learner to fee the Force of a Demonftra-
tion in Algebra, before he is acquainted with its Characters and
Language.
The Foundation of tranfpofing Quantities.
T
HIS is grounded on the obvious Truth, that every Thing
is equal to itself; that is, mm, and
-y= -y; whence
to tranſpoſe any Quantity, is only to make that Quantity equal
to itſelf, prefixing to it the contrary Sign, and adding it to the
given Equation. Suppoſe there is given
I
a—b+d—mz, to tranfpofe, b, d,
and m.
b = b
4
Now
1 + 2
And
2
3
a+d-m=x+b
d
d
5
a
m=x+b- d
mm
a
1=x+b―d+m
3 +4
Laftly 6
5+6 7
SUBSTRACTION.
I fay to fubftract a negative Quantity from a pofitive, is only
to change the Sign of the negative Quantity, and add it to the
t
pofitive
The Rules demonftrated.
325
pofitive Quantity, and this Sum will be the Remainder required.
That is,
If x
y
23, for
is fubftracted from xy, I fay the Remainder is
Suppofe
I
And
2
x+y=m
x - y = n
Now if it can be proved that 2y is equal to the Difference
between m and n, it follows that to fubftract a negative Quantity
is to change its Sign and add it.
2 + 1 | 3 | x = n + y
Now in the firft Equation for x write ny, for that is
equal to x.
Then
4 — n
4 In + 2y = m
52y=m N.
n. Q. E. D.
I fay further, that to fubftract a negative Quantity from a
negative Quantity, is done by changing the Sign of the Quantity
to be ſubſtracted, and then adding them by the Rules in Addi-
tion, and the Sum will be the Difference required.
Suppoſe
I | x -
2y = m
And 2 x y = n
Now the ſecond Equation being fubftracted from the firſt
according to the Rule, leaves -y=m n; and if it can be
proved that-ym-n, then to fubftract a negative Quantity
from a negative Quantity, is only to change its Sign and add it.
2+3
I
3 4
x = n + y
n + y — 2 y
112
5- 12 5
That is 6
y — 2y = m — n
y = m
n.
Q. E. D.
And that mn is a negative Quantity is evident, for
2y cannot be fo great as xy, they being fuppofed pofitive
Quantities, and therefore m cannot be fo great as 17;
X
con-
fequently
326
ALGEBRA.
fequently m-n is a negative Quantity, and therefore may be
equal to ―y.
And in
MULTIPLICATION,
I fay unlike Signs being multiplied give in the Product;
that is,
аха
·aa.
To prove which, I take for granted the following
LEM M A.
That no Quantities connected by the Sign only, or by
the Sign-only, can be equal to nothing. That is, it cannot
be-a-bo, or a + b = 0, though it may be abo,
or b-
a=0.
Now, if poffible, let-a× a produce a a where the Sign of
the Product is affirmative.
Let
I
772
2
I X 2 3
a = 0
a=a, that is, every Quantity is equal
to itſelf.
ma+aa=oa, by the Suppofition,
that is, maaa is equal to nothing, which is against the
Lemma, therefore a xa cannot produce a a.
But, I fay
- a × a =
aa.
Let I
m
a
о
2
I X 2
3
+a+a, for any Quantity is equal
to itſelf.
and multiplying by
ma — a & = 0 a, that is, ma a a is
equal to nothing, whence ma = aa. Now that ma = aa is
evident, for ma=0, therefore ma,
a we have maa a, confequently
I fay further, that like Signs tho'
in the Product. That is,
duce
not
aa, for
Let
I
m
1 2
о
axa =- aa. Q. E. D.
being multiplied, pro-
a x a produces aa, and
-
—a—a, every negative Quantity
being equal to itſelf.
Now, if poffible, letaxa produce
ma
a a
a a. Then
oa, by the Suppo-
maa a is equal to nothing, which is against
1 × 2 3
fition, that is,
2
the
The Rules demonftrated.
327
the Lemma, therefore ax -a cannot produce a aj but
the Sign must be + or affirmative; which may be further
proved thus,
Let I m a=0
2
I X 2 3
a
a
ma+aa=0a, that is,
is equal to nothing, from whence ma = a a.
mataa
And that ma= a a
is evident, for mac, therefore ma, and multiplying by
@, we have ma=aa. Hence a Xaaa. Q.E.D.
DIVISION.
аа.
As unlike Signs in Multiplication produce in the Product,
I say that,
In Divifion, unlike Signs being divided, give-in the Quo-
tient, that is, if ab-bbo, and both Sides of the Equation
be divided by b, I fay the Quotient will be ab, and not a+b.
Suppoſe unlike Signs to give + in the Quotient.
If I
Let 2
I 2 3
a b — b b = 0
b = b
a+b=
+ b = —, by the Suppoſition, that is
a+b is equal to nothing, which is against the Lemma, there-
fore an Abfurdity follows the Suppofition, that unlike Signs
give in the Quotient; but I fay unlike Signs give in the
Quotient.
abbb=0
>
Let
I
2
b = b
I→ 2
3
a — b = —, that is, a − b is equal to
nothing, whence ab, and that a = b is thus proved.
}
I + b b
4
4
ab
bb
÷b
5
a = b. Q. E. D.
I fay further, that like Signs being divided, though they are
negative, give in the Quotient, that is, a bb b divided by
+
b, the Quotient isa+b, and not-a-b.
If
}
328
ALGEBRA.
W
If like Signs though give in the Quotient; then
— ·
I a b b b = 0
Let
2
b
- b
1÷2 3
a
b ====,
=by
by the Suppofition, that
is,ab is equal to nothing, which is againſt the Lemma,
therefore an Abfurdity follows the Suppofition, that like Signs
though — give in the Quotient.
But, I fay, ab-bb divided by
ab, that is, like Signs though
tient. For,
Let
I - 2
I ab bbo
1 2
—b=—b
-b
b, the Quotient is
give + in the Quo-
3 ―a+b=—-, that is,
- a + b is
equal to nothing, whence ba, and that ba is evident,
thus,
(
I + b b
4 b
4
5
a b = b b
ab. Q. E. D.
both
By this the Learner will fee that like Signs though
in Multiplication and Diviſion, muſt give + in the Product and
Quotient, for an Abfurdity follows the contrary Hypothefis, or
Suppofition, of their producing-in either the Product or
Quotient.
The other Principles of this Science are very obvious, being
the plain Confequences of the Axioms mentioned in the Be-
ginning of the Work.
8-1915
FINI S.
>
1
UNIVERSITY OF MICHIGAN
:
+
3 9015 06531 3515
A
F
;+
A 543635