alabraical Problems
46
ARTES
1837
SCIENTIA
LIBRARY
VERITAS
OF THE
UNIVERSITY OF MICHIGAN
PLURIBUS
TUEBOR
SI-QUÆRIS-PENINSULAM-AMŒNAM`
CIRCUMSPICE
QA
.35
A823
HE YOUNG
1757
N PALY ST's
EXERCISE.
BEING.
A Choice Collection of a Hundred AL-
GEBRAICAL PROBLEMS, exhibiting all
that is curious in Simple and Quadratic Ad-
fected Equations.
The Whole Illuſtrated
With various Methods of COMPUTATION,
as well by plain Numbers and Geometry, as by
Algebra; and calculated to the Capacity of Young
Beginners.
By SAMUEL
ASH BY,
Teacher of the MATHEMATICS.
LONDON:
Printed by C. Jephson, in West-Smithfield,
For JOHN WILCOX, at Virgil's Head, oppofite
the New Church in the Strand.
M, DCC, XXXVII
QA
35
.A823.
•
ΤΟ ΤΗΕ
MATHEMATICAL
SCHOOLS
OF
Great Britain,
AND AL L
YOUNG STUDENTS
I N
ALGEBRA,
This fmall TREATISE is moft humbly
Infcrib'd by the AUTHOR.
+
に​な
​THE
PREFACE
EETING with the fol-
1
?
}
M
lowing Propofitions, pub-
lifhed by a certain Latin
Author, well known to the
World for his eminent Skill
in ALGEBRA, but without their Solu-
tions, and finding them a curious Collection,
was prevailed upon by my own Inclination
to draw out all their Anfwers. Which ha-
ving finished, and fhewn to fome of my Ac-
quaintance, very well skilled in Mathema-
tical Learning, they thought it a Pitty
Such
}
The PREFACE
Such a Variety of useful Propofitions should
lie hid in Oblivion, for that they would be
of great Service, and very much inftruct
and exercife the Young ANALYST, and
prepare him for higher Speculations in Al-
gebraical Inquiries, if they were made
public. I therefore complied to their Pro-
pofal, and it was thought most proper to
print them by themselves, without the Ele-
ments of ALGEBRA, because whoever in-
tends or propoſes to learn ALGEBRA, are
generally furnished with Ward's Introdu-
ction, or at leaſt ought to be, that by being
the ſmaller, the Book may come within the
Compass of every ones Circumftance to pur
chase it.
AS the chief Defign of this Treatiſe is
to exercise the Young ANALYST in Sim-
ple and Quadratic Equations, and Suppo-
fing him but barely acquainted with the first
Principles of ALGEBRA, I have traced
every Operation Step by Step, that thereby.
nothing
*
The PREFACE.
nothing might seem obfcure to the Learner,
and likewife have explained whatever I
thought was not fufficiently evident, leaſt he
Should spend his Time too much in Ufeless
Cogitations.
AND notwithstanding it is adapted to
the Capacity of Young Beginners, I doubt
not but the more skillful ANALYST will
find fomething in it worth his Obferua_
tion. For to many of the lineal Problems I
have not only given their Algebraical In-
veftigations, but also the Geometrical Con-
ftruction and Demonftration; and the Nu-
merical Solutions to many others. And to
Prob. xxxII. I have annexed three Ex-
amples, wherein is fhewn the Geometrical
Reafon of ambiguous Equations, with the
original Derivation of the Method of com-
pleating the Square, together with a ſhort
Specimen of Quadratic Equations, in which
is proved, Algebraically, that the third Cafe
of Quadratic Adfected Equations, is not
univerfally
嗣​。
The PREFACE.
univerfally ambiguous, as Algebraifts gene-
rally take it to be, but only fo in particular;
with many others too numerous here to men-
tion; which are beft fpecified by the Ex-
amples themselves, to which I must refer
the curious Reader.
1:
1
A
2
i
་
A
HUNDRED
Algebraical Problems.
T
HE principal Uſe of the Ana-
lytic Art, is to bring Mathe-
matical Problems to Equati-
ons, and to exhibit thofe Equa-
tions in the moſt fimple Terms
that can be. And the Rules
required to bring out Equations to fuch Pro-
blems, are chiefly thefe following, viz. Addi-
tion, Subſtraction, Multiplication, and Divifion;
Involution and Evolution. But if the Learner
underſtand the firſt four Rules abovementioned,
fo as to manage Equations by them; he will
be fufficiently qualified to anfwer many knotty
and curious Queftions without the help of In-
volution or Evolution; as he will find by the
I, II, III, IV, V, and many other of the fol-
lowing Operations.
B
WHEN
[2]
->
WHEN the Learner has been fome time exer-
cifed in fuch Problems as naturally produce
fimple Equations, he may then proceed to the
Principals of Involution and Evolution, and
learn to compleat the Square, and then the
Calculations to thofe Problems that genuinely
bring out Adfected Quadratic Equations, af-
ter they are ſtated, will be very eaſy to him,
fuch as the LXXXIII, LXXXIV, LXXXV,
and the like.
+
By this Method aforegoing (if ftrictly ob-
ferved) all thofe who have a Defire to learn
this miſterious Art, and are unacquainted with
the Elements thereof, will find Algebra not fo
difficult to attain as perhaps they may ima-
gine: And when the young Analyft has pro-
ceeded fo far in this kind of Arithmetic as
the above Rules will lead him, he will eafily
percieve of himſelf what more is requifite to
make him a compleat Algebraiſt. Which that
The may attain to with more Certainty, let me
adviſe him not to aim to learn a great deal in
a little Time, but to learn a little well; for
it is not the Multiplicity of Propofitions, but
the Method they are delivered in, that ought
to be confidered.
HAVING thus laid down a brief, but plain,
Method for ſpeedily attaining to the Solutions
of Problems producing Simple and Adfected
Quadratic Equations; I will now proceed to
the
[ 3 ]
the following Hundred Queſtions and An-
fwers; but firft fhall premiſe theſe three Parti-
culars, viz. That in all Queſtions whatſoever
theſe three Things are chiefly to be obſerved;
first, The Data or Things given; fecondly, The
Quæfita or Things fought; and thirdly and
laftly, The Conditions or Form of a Queſtion;
which three Diftinctions are fully illuftrated
in the following Operation.
PROBLEM I.
To find a Number, which being multiplied
by 3, fubftracting 5 from the Product,
and the Remainder divided by 2, if the
Number fought be added to the Quotient,
that the Sum may be 40.
FOR a perfect Idea of this, or any
Queſtion propofed, divide it into theſe three
Particulars; first, The Data, which in this
Problem are thefe Numbers 3, 5, 2, and 40;
fecond, The Quafita, which is the Number to
be found; and third, The Condition or Form
of the Queſtion, which in this Problem runs
thus. Which being multiplied by 3, fub-
ftracting 5 from the Product, and the Re-
mainder divided by 2, if the Number fought
be added to the Quotient, that the Sum may
be 40.
B. 2
Now
[4]
Now to folve this Queftion, for the Num-
ber fought put any Symbole or Letter, as for
inftance x; which by the Condition of the Pro-
blem, muſt firſt be multiplied by 3, and 3*
will be the Product; fecondly, Subftract 5 from
that Product, and the Remainder will be
3x-5; thirdly, Divide the faid Remainder by
2, and the Quotient will be 35; fourthly,
2
and lastly, Add the Number fought, viz. x to
this Quotient
40.
3x-5
2
and the Sum is
"
3x-5
2
+x; which Sum by the Queftion is equal to
And thus the Queſtion is tranflated out.
of the Verbal into the Symbolical Expreffion,
and therefore is brought to this
Equation 3-5
1 ×
I +40. And to find the
2
Value of x proceed thus,
23x-5+2x=80
2 +535x=85
354-17 the Number which was to
be found, or the Value of x required.
W. W. D.
FROM this Operation aforegoing the
Learner may obferve, that to the Solution
of Queſtions, which only regard Numbers,
or the abſtracted Relations of Quantities,
there is ſcarce any thing elfe required,
than
[ 5 ]
than that the Problem be tranflated out of the
English, or any other Tongue it is propofed in,
into the Algebraical Language, that is, into
Characters fit to denote our Conceptions of the
Relations of Quantities. For when a Queſtion
is expreffed Analytically, the Calculation or
Form of the Reduction, for the moft part, is
feldom any thing more, than the Condition of
the Problem repeated backward; and thus the
Condition or Form of a Queſtion, in a man-
ner, may ferve for a Rule to inveſtigate the
unknown Quantity by.
BUT the moſt commodious and univerfal
Solutions of Algebraical Problems are gene
rally performed by Symboles or Letters repre-
ſenting the Quantities given as well as thofe
required. Thus, put 3-4, 5-6, 2=c, and
40= d; and fuppofing x fignifies the Number
fought as before; then inftead of this Nume-
3x-5
ral Equation
x=40, we fhall have this
2
Literal one
ах
axi-b
x=d, from whence the
C
Quantity x will be found
zx-b
Thus, 1
+x=d.
1 2
I
xc{2\ax b + cx=cd.
x = b+cx=cd.
+b/3/ax+ cx=cd+b
cd + b
3 ÷a+c/4/2
at
41%
a + c
1
the Value of x required.
B 3
BY
[ 6 ]
..
Tilly
•
By this Method of Computation it appears,
that not only the unknown Quantity is inve-
ftigated, but a general Theorem or Rule is dif-
covered for folving all Queſtions of the fame
Nature; for the Letters a, b, c, and d, may
be put for any other Numbers than what they
here repreſent.
atc
Therefore this Equation
Ex is a general Theorem for finding the
Number fought to all Queftions having the
fame Data and Quæfita as that above.
N. B. THAT the affumed Quantities in all
Operations are eſteemed the very fame as tho'
really known: For any of the known Quan-
tities may be brought to one fide of the Equa-
tion, and made equal to all the reft, by the
fame Principals of Algebra as the unknown Quan-
tities are calculated. To explain which, let any
of theſe known Quantities a, b, c, in this Equa-
ax-b
tion +x=d, as b for inftance, be cleared
C
and brought to one fide the Equation, and all
the reft to the other, The Equation being
thus
ax-b
Stated,
I
+x=d. Work as follows,
C
W ~ H
2
3
I
xc 2 ax-bcx=cd
+b3/cd+b=ax+cx
-cd|4lb=ax+cx-cd.
But
[ 7 ]
But if it had been required to have brought
out c, then,
-cx 5/cd-cx=
5\cd-cx=ax-b
2
ax-b
5
-d-x 6c=
d-x
Or a, Then,
5
+37ax=cd-cx+b
7
x 8a=
cd-cx+b
X
Or 2+b-cx 9ax=cd|b-cx
9
cd+b-cx
x10 a:
X
Hence it is, that Algebraifts fuppofe thofe
things given, which notwithſtanding are the
fole Objects of their Inquiry, until by Re-
duction they become equal to fome certain
Pofition of the given Quantities; and then the
true Value of fuch feigned Numbers are dif.
covered.
FARTHER; to ftrengthen the young Ana-
lyft's Idea in Algebraical Computation, he
muſt underſtand, that the Method of affuming
Letters for the unknown Quantities, is no
more than gueffing the Value of fuch Quanti-
ties at randum. So in the foregoing Propo-
fition, inſtead of making uſe of x, I may fup-
pofe the Quantity fought to be any Number at
Pleaſure, and the Queſtion being Stated with
that Number and thofe given, according to
the
[ 8 ]
the Condition or Form of the Problem, the
true Quantity required will then come out by
the fame Laws of Algebra, as tho' the Number
fought was repreſented by any Letter in the
Alphabet, or other Symbole. Thus, for the
Number to be found put 9, and by the Con-
dition of the Queftion, there will be had this
numeral
!
Equation 1
13×9-5
+9=40.
2
I
2
X223×9-52X9=80
+533×9+2×g=85
85
33249= 17. Here the Cal-
3x2
culus is the very fame, and brings out the fame
Conclufion, as either of the foregoing Methods,
and likewife fhews the Suppofition to be falfe;
for inſtead of 9 I fhould have gueſt 17.
Whence we may conclude, that the ultimate
Defign of Algebraical Operations, is no more.
than to diſcover the Truth from falfe Suppo-
fitions.
MOREOVER feeing my Suppofition was
too little, I may fupply that defect with x, and
fo call the Number fought 9+x; and having
exprefs'd the Problem Algebraically, there will
be obtain'd this
Equation
[ 9 ]
5
Equation 127+35+9+x=40.
2
1 x2 2/27+3x-5+18+2x=80
2+ 35x=40
40
3548. Wherefore 9+8=17 the
Number which was to be found, the fame as
before.
FROM what is here fpecified the Artiſt may
obferve, that let the affumed Number be what
it will, there will ftill Refult the fame Conclu-
fion to the Queſtion propoſed; fo that it mat-
ters not whether the whole or a part of the
Quantity, or fome other Number, be made
the Scope of our Inquiry, fince they all bring
out the true Anſwer to the Problem.
BUT let the unknown Quantity be what it
will, the Algebraiſt muſt accomodate it, ſo as
to bring out the moft fimple Equation poffible.
And fince from the various Species of Pro-
blems different Equations arife, fo thoſe where-
in only one Power of the unknown Quantity
is contained, as in thefe following,
Jax + cx=cd+b
I.
Viz. 2,5x²+16x²=16a XVI.
Ix²=2ab.
3
x²=ab², &c.
Thefe Num-
bers fhew the
XXXVI. each Equation
Operation
XXXVII. is taken from.
are
[ 10 ]
are called Simple Equations.
And thofe
wherein different Powers of the Quantity fought
are contained, as for Example,
x²+ax=-ab.
2
LXIX.
bx²-cx²-zacx-ac. LXXVI.
X -cx²=ab, &c. LXXVII.
are called Compound or Adfected Equations.
3.
2
As there arife different Powers of the un-
known Quantity in Algebraical Operations, fo
there are different Forms of both Simple and
Compound Equations; for ax-ex-cd, is
called a pure Simple Equation, 25x²-16x²
=16a or x²=2ab a Simple Quadratic, and
x³-ab² a Simple Cubic, &c. And in Com-
pound Equations, thofe wherein the unknown
Quantity is found of no more than two different
Powers, whofe Indices are double to each other,
are called Adfected, and is either Quadratic,
or Biquadratic, &c. according to the heigheſt
Dimentions of the Quantity fought. That is,
x²+ax=ab or bx²-cx2acx-a³c, is called a
Quadratic Adfected Equation, but x*--cx²
ab is called a Biquadratic Adfected, &c.
HAVING thus defined the different kinds of
Equations contained in the following Pages, I
will now fhew the Learner how the faid Equa-
tions are brought to a Solution. And firſt
then, for Simple Equations take this general
RULE.
[ n ]
RULE.
-
Exit fuch a Root of the Equation, whoſe Index
Shall be the Power of the unknown Quantity,
and it is done.
As per Example, In this Equation x=2ab,
becauſe the unknown Quantity x is fquared, I
extract the Square Root, and there comes out
zab. So alfo in this Equation x³-ab², by
extracting the Cube Root we have x=³√ab²,
and ſo on.
3.
I SHALL omit the Rule for folving Qua-
dratic Adfected Equations at preſent, until the
Learner has been fome Time excercifed in
Simple Equations; becauſe the Learning of
Rules without having an immediate Ufe for
them, very often ferves more to perplex than
inftruct the Artift; for tho' a Learner be ne-
ver fo expert in the Elements of Algebra, yet
when he comes to the Application of theſe
Principals in the Solution of Propofitions (ex-
cept he is Mafter of a good genious and ftrong
Memory) he will find fuch a Crowd of confu-
fed Ideas in his Mind, that if he is not helped
by the Inftruction of a Maſter or fome other Ac--
quaintance, he will be ready to cry out ne plusul-
tra. But by proceeding according to the Methods
laid down in the foregoing Pages, and obfer-
ving well the following Operations, he will in
a great
I
[ 12 ]
a great Meaſure, be freed from fuch Inconve-
niences, and find, that Algebra is no more
difficult to attain, than common Arithmetic.
For both are founded on the fame Principles,
and aim at the fame End. And differ from
each other only in this particular, that Vulgar
Arithmetic begins it's Computation from the
Quantities given, and fo proceeds forewards to
thofe that are required, whereas Algebraic Com-
putations proceed in a retrograde Order, viz.
from the Quantities fought, as if they were
"given, to the Quantities given, as if they
were fought, to the End that we may fome
way or other come to a Conclufion or Equa-
tion, from which we may bring out the Quan-
tity fought. Which the Reader may obferve
from the Solution to the tenth Queſtion, and
ſeveral others in the proceding Pages.
Note, The Signs made Ufe of in the fol-
lowing Pages, being the very fame as thofe
in Ward's Introduction to Mathematics; the
Reader is referr'd there for their Explanati-
ons.
PROBLEM
4
[ 13 ]
PROBLEM II.
To find a Number, which being multiplied
by 12, and 48 added to the Product,
as much may be produc'd, as if the fame
Number fought were multiplied by 18.
PUT 12a, 48=b, 18c; and let the
Number fought be called x. Then by the
Queſtion we ſhall
have 1ax+bcx
ax 2 cx—ax=b
I
b
48
2
-a3x==
8. The Num-
ca 18-12
ber required.
>
PROBLEM III.
To find a Number, to which if 11 be added,
and 7 fubftracted from the fame Num-
ber, (viz. the first) the Sum of the Ad-
dition may be double the Remainder.
PUT 11=α, 7=b, and call the Number
fought ; then by the Queftion it will ſtand
thus
I
2
1x+a=2x-2b
+2b2x+a+2b=2x
x3x=a+2b=11+7x2=25. The
Number which was to be found.
C
PROBLEM
[ 14 ]
PROBLEM IV.
To find a Number, to which if it's double
(treble, quadruple, &c.) be added, the
Square of the fame Number may be pro-
duc'd.
CALL the Number fought x, to which,
by the Queſtion, add it's double, viz. 2x, and
the Sum is x-2x or 3x; which Sum, per
Query, is equal the Square of x, that is, x².
Ergo 3x=x²; this Equation divided by x,
gives x 3. But the Queftion fays, if treble
the Number fought, viz. x, be added to itſelf,
the Sum will alfo be equal the Square of x;
therefore x3x, or 4x=x², and dividing this
Equation by x, we ſhall have x=4; and again,
if we add to it's Quadruple, the Sun will like-
wife be equal to x², confequently x+4x=x²,
that is, 5x=x²; which being divided by x, the
Queſtion will give x=5, &c.
2
Q; E, I.
PROBLEM
:
[ 15 ]
:
,
PROBLEM V.
To find a Number, which if added to itſelf,
and the Sum multiplied by the fame, and
the fame Number ftill fubftracted from the
Product: And lastly, the Remainder di-
vided by the fame, that it may produce
13.
FOR the Number fought put x, which
added to itſelf, the Sum is 2x, this Sum mul-
tiplied by, the Product is 2x²; from this
Product fubftract x, and the Remainder is
2x²-x: And laftly, divide this Remainder
by x, and the Quotient is
2x²-X
X
but becauſe
x is found in both the Terms of the Numerator,
the Quotient will be 2x-1, which, by the
Queftion, is equal 13-a. Therefore,
I
2
12x-1=a
1230
+12/2x=a+1
at1_13|1=7.
2
2. E. I.
2
:
C 2
PROBLEM
..
[ 16 ]
PROBLEM VI.
To divide the Number 16 into 2 Parts, je
10
that the Square of the greater Part may
exceed the Square of the lefs by 32.
y
To folve this Propofition, according to
Ward's Method, put the greater Part, and
the leſs; and the Square of the greater Part x
is ², and the Square of the leffer is y², the
Difference of theſe two Squares is my²,
which is equal to 32 by the Queftion.
x
Therefore
I
30 2
2x
}
x+y=16
-y²=32$
- y² === 32
3y=16
4y²-256-32x²
2 ty² 5x² = 32 + j²
532 6 y²=x
4
7
=6
32
7256-32x+x=x-32
8256-32x32
8 +32 9256=32x-32
9 +321032x=288
10 3211x
288
32
2
Note, This Method of Exter-
minating unknown Quantities,
is called Comparative Algebra.
9, the greater Part,
which taken from 16, the Remainder is 7,
the leffer Part.
Q. E. I.
BUT the great and incomparible Ma-
thematician, Sir Ifaac Newton, has, in his Uni-
verfal
[ 17 ]
verfal Arithmetic, Page 68, moſt wifely noted,
That the Solutions of Queftions are (for the
moft part) fo much the more expedite and ar-
tificial, by how fewer unknown Quantities you
have at firft. That is, in thofe Queftions
where two, three, or more unknown Quan-
tities are required to be found, by the Affump-
tion of one unknown Quantity only, the reſt,
many times, will be exprefs'd by fuch and
fuch Laws or Properties of Algebra and Geo-
metry, as the Query happens to fall under:
With which Expreffions, the Queſtion being
ftated, and the unknown or affum'd Quantity
being inveſtigated, the other unknown Numbers
will be had from the Expreffions themſelves.
THE better to conceive this Method of
Refolution, the Algebraift is defir'd to obferve
the following
AXIOM S.
1. IF from the Sum of any two Quantities
be taken either of them, the Remainder will
give the other Quantity.
2. THE Difference of any two Quantities.
added to the lefs, gives the greater Quantity;
and fubftracted from the greater Quantity,
gives the leffer; or the whole is equal to all
the Parts taken together.
3. THE Product of any two Quantities
divided by either of them, the Quotient arifing
will give the other Quantity.
C 3
4.
[ 18 ]
4. THE the Quotient of any two Quan-
tities, being multiplied by the leffer, the Pro-
duct is the greater Quantity.
N. B. That what is here faid of fimple Quantities,
the fame may be understood of Squares, Cubes,
and Biquadratic Quantities, &c.
Now, by a right Application of thefe
Axioms, many Queftions in Algebra, where 2,
3, 4, &c. Quantities are requir'd to be found,
may be expeditiouſly ſolv'd, by the help of one
unknown Quantity only. Thus :
PUT 16-a, 32=b, call the greater Part x,
as before, then (by Axiom 1.) the leffer Part
will be a-x; now, the Square of the greater
Part is a², and the Square of the leffer Part is
a²-2axx², which taken from x², the Square
of the greater Part, the Remainder is 2ax—a²,
which (by the Queſtion) is equal to b.
Therefore, 12ax-aª—b
+a²22ax=b+a²
b+a² b
2
I
a
2 ÷2a/3x= 20
32
16
2a
2 2X16
2
the greater Part, the fame as before, whence
the leffer Part is 7.
2. E. I.
By comparing this Method with that afore-
going, we fhall find the very fame Conclufion
is produc'd with quarter the Trouble; for, in
that
[ 19 ]
that Operation, as there are two unknown
Quantities, x and y, concern'd, there muſt be
one of them, as y, exterminated before the
other can be found; which Trouble of Exter-
mination is entirely avoided by this Method
of Subftitution.
ས ས ལ 1:“、·༦ ག ༥་ ་
PROBLEM VII.
To divide the Number 36 into two Parts,
Jo that if 12 be added to the first, and
6 to the fecond, the former may be the
double of the latter.
PUT 360, 12, 6=c, and call the firft
Part x, then the fecond Part (by Axiom 1.)
will be a-x; now, if to the first Fart x be
added b, the Sum is xb, and to the fecond
Part a-x be added c, the Sum is a-x-c,
which is half the Sum a+b, per Query.
Therefore,[1+1=2a-2x+20
1 +2x23x+6=2a-2c
2
—633x=2a2c—b
3 ÷ 34x=2a + 2c-b_2×36-12x612
334x
3
3
24, the firſt or greateſt Part ;
which being known, the leffer Part is 36-24
=12; and theſe two Numbers will answer the
Conditions of the Queſtion.
2. E. I.
PROBLEM
[ 20 ]
PROBLEM VIII.
Let the Line A B (of 70 Parts) be di-
vided any how in C, [fo that AC may
be
42, BC 28] it is required to divide
the fame Line again in another Point :
For Example, in D, jo that the Rect-
angle A D C may be equal to the Square
D B. Let the Segment CD be enquired,
[which being obtained, AD, BD will
be known.]
PUT 70-a, 42=b, 28=c, and let x re-
preſent the Segment CD, and AD (by Axiom
I and 2.) will be bx, DB-c-x; now mul-
tiplying AD by DC, or their Equivolents
bx by x, the Product is bxx², which (by
the Queſtion) is equal the Square of DB, or
it's Equivolent cx, which is c-2cx+x².
(See Fig. 1.)
Therefore,¡¡¡bx+x²=c²—2¢x¯†x²·
I -x² 2 bx=c²-2cx
+2cx/3 bx+2cx=c²
13
22
3 ÷ 6+204x=
c² 28×28
b+20 42+56
8, the length
of CD, which being known, gives AD=50,
and DB=20.
2. E. I.
The
[ 21 ]
De a
亨
​I
2
The fame otherwiſe.
LET X=AD, whence a-x=DB, and
CD=x-b, as is evident by the Line AB;
and multiplying x-b by x, the Product is
x`—bx, which is equal to the Square of a—x
or to a¹-2ax+x² by the Queſtion.
1
Therefore, 1x-bx—a²—2axx²
2
2
-x² 2 bx=a²-zax
-x^2 -
+2ax32ax-bx=a²
3 →2a-14x=
a²
70x70
24-b 2×7042
hence 50—42=8=CD, as before.
Geometrically thus.
50=AD,
2. E. I.
ON the given Segment CB, (by the 46
Prop. of Euclid's first Book) defcribe the Square
cbBC, and draw the Diagonal Cb; this done,
compleat the Parallelogram ab AB, and draw
it's Diagonal aB, which cuts the other Diago-
nal Cb in the Point e. Now, through the
Point e, and Parallel to bB, or cC, draw the
Line D, and it cuts the Segment CB in D,
the Point requir'd, (See Fig. 2.)
Demonftration.
DRAW through the Point e the Line trs,
parallel to the given Side AB; then (per Cor.
to
[ 22 ]
[
to Prop. 4. of Euclid's fecond Book) the Rect-
angles reDC and dbSe are Squares, therefore
eD is CD, and bS-eS-DB, confequently
□bs=☐es=DB; but the Square of bS or
eS is equal the Square dbSe, therefore the
Squares dbSe and DB are equal; and (by
Prop. 43. of Euclid's firft Book) the Square dbSe
is equal to the Parallelogram teAD,” which is
equal the Rectangle of AD into De, which is
equal to AD into CD, for eD and CD are equal;
therefore the Rectangle of AD into CD is-
the Square of DB.
2. E. D.
OR by conftructing Equation the third, of
Operation the firft, aforegoing, will likewife
give the Point D Geometrically. But the
Limits of this Book being fo fmall, obliges
me to refer the young Analyft to J. C. Stur-
mius's Mathefis Enucleata, or to Harris's Alge-
bra, where he will meet with ample Satif-
faction in this kind of Geometry.
PROBLEM IX.
Let the Line EF be divided any how in G,
[fo that EG may be 6, GF 4] it is re-
quired to produce this right Line EF,
[for Example unto H] fo that the Rect-
angle EHF may be equal to the Square
GH; the Length of FH is required.
PUT 10=a, 4b, and x-FH; then
(by Axiom 2.) EH-a-tx, and GH=b‍fx;
multiply
參
​[ 23 ]
詈
​2
"
multiply ax by x, and the Product is ax+x
and the Square of btw is b²+2bx+x², which
(by the Queftion) is equal to ax-x².
2
2
(See Fig. 3.)
Therefore, axx² =b²+2bx+x²
2
1
|2|ax = b²+2bx
·2bx = b²
2
I
2
-2bx 3 ax
4x4
÷a- -264x=
8; and
a-
-26 10—2×4
thus much muſt the Line E F be produced, to
fulfill the Conditions of the Queſtion.
Or thus.
PUT X EH the whole compounded Line,
and (by Axiom 2.) x-a-FH, and GH=x
a+b; the Rectangle of x-a into x, is
x²-xa, and the Square of x-ab is x²-2ax
—2ab+a²+2bxb, which is equal to x-ax
by the Queſtion.
Therefore, 1x²-ax=x²—2ax—2ab+a²+²
2
b.x+b²
2ax-2bx—a²—2ab+b²
I
b²
2
2a--26 3x=a
18, from which
a2b
take 10, and there remains 8=FH, as before.
Note. In the Margin, at the fecond Step +
fignifies Tranfpofition, or the Rectification of
the first Step by Addition and Subftra&tion.
Or
[ 24 ]
Or likewife thus.
LET AGH, and (by Axiom 2.) FH
=x-b, and EH=x-b+a; which Terms
being ſtated as the Queſtion requires, will
ſtand
1 x 2 = x²+ax—2bx—ab+b²
——2|2bx—ax—b² —ab
thus,
I
b² -ab b
2÷2b-a 3 x = 2b—a
ab
12, from
2 4b-2a
which take 4, and the Remainder is 8=F H,
as in the foregoing Operations.
b² —ab
b
BECAUSE this Equation x= 2b-a 2
ab
is ſomewhat uncommon, and per-
4b-2a
haps may non-plus the Learner, tho' to the
more ſkilful in Algebra it is moft evident, it will
(I prefume) not be improper to give the Reader
its
Explanation.
b2-ab
IN this Fraction
ſeeing a is greater
2b-a,
than b, it is plain, by the Laws of Addition,
that the Sum of bab, the Numerator, is a
negative
یکھو
[ 25 ]
26
way of
a is a
negative Quantity; and by the fame
Reaſoning, the Denominator
negative Quantity alfo: But a negative
Quantity divided by a negative Quantity,
by the Laws of Divifion, the Quotient will be
Affirmative; therefore if you write 10 for a,
and 4 for b, the faid Fraction will become
4×4-10×4
-24
that is,
=12=x. And fo
2X4-10
2
b
ab
likewife
when thrown into
2 4b-2 a
4
IOX 4
Numbers will ftand thus
2
4X4 2 X 10
40
2
=2+10=12.
Which was to
4
be explained.
PROBLEM X.
A General difpofing his Army into a Square
Battle, finds he has 284 Soldiers over
and above; but increafing each fide with
one Soldier, he wants 25 Soldiers to fill
up the Square: How many Soldiers had
be?
PUT 2842, 25-b, and let the Number
of Soldiers placed in Rank and File be called
x; which being fquar'd is x2, which is lefs
than the Number of Soldiers in the Army by
D
284
[26]
284 or a (per Query.) But increasing each fide
with one Soldier, that is, adding one to x the
Sum is x1, whofe Square is x²+2x+1,
which (by the Queftion) is greater than the
whole Army by 25 or b, confequently taking
b from x²+2x+1 the Remainder is x2+2x
+1-b the whole Army, and Adding a to x²,
the Sum is likewife equal the whole Army,
that is, the Sum of a and x is x²+a.
X
-b
2
2
Therefore|1|x²+a=x²+2x+1−b.
x²/2a=2x+1-
+ 32 x
32 x = a+b=1
I
1 2
3
응치
​a+b1_284—25—1154,
2
2
the Soldiers in Rank and File, which being
fquar'd, and 284 added to it, that is, add
284 to 23716 the Square of 154, and the Sum
is 24000 the Number of Soldiers requir'd.
The fame otherwife.
PUT the whole Army, then x-a (by
the Queſtion) is a Square Number, whofe
Square-Root is √x-a the fide of the Square
equal the Number of Soldiers in Rank and
File, to which add 1, and the Sum is x-a.
-a.
1, the Square of which is x—-a+2√x
+1, and becauſe this Square exceeds the Num-
ber of Soldiers by b or 25, fubtract b from it,
and the Remainder is x-a-2x-a.+1—b,
which
[ 27 ]
which (per Query.) is equal the whole Army.
Therefore 1 x-a-2√x—a.†1—b—x.
I
2
土
​2'2√x—a=a+b—1—308=c
23.4x-4a=c²
3 +4a44x=c²-4a
c²+4a_308×308 +4×284
4 45
4
4
24000. The fame as above.
Numerically.
THIS numerical Solution depends on the
Fourth Propofition of the fecond Book of Euclid.
For feeing the Number of Soldiers exceeds the
leffer Square by 284, and wants 25 to fill up
the greater Square, confequently 284 +25
=309 is equal the double Rectangle of the fide
of the leffer Square into 1 more the Square of
1 (by the above faid Propofition.) But becaufe
the Square of 1 is but 1, therefore 308 is the
double Rectangle of the fide of the leffer
Square into i; whence dividing 308 by 2 the
Quotient gives 154 the fide of the leffer Square
as before, which fquar'd produces 23716, to
which add 284 the Sum is 24000 the Number
of Soldiers in the Army.
Q. E. J.
D 2
PROBLEM
[ 28 ]
PROBLEM XI.
A certain Captain fends out of his Sal-
10, and there remain'd ÷ + 15.
diers
How many Soldiers had he?
PUT 10a, 15b, and x the Num-
ber of Soldiers; then the Number of Sol-
diers fent out are -a, and thoſe remaining
NIX
3
+b, which together are equal to x.
Therefore+a++b=x.
I
2
3
x3 2x+3a+32x²+3b=3x
×232x+6a+3x+6b=5x+6a+6b
|| = 6x.
35x4x=6a+6b=6×10+6×15—150,
the Number of Soldiers.
Or thus.
To avoid Fractions put 6x
of Soldiers, of which is 2x,
Q. E. J.
the Number
and is 3x;
2
whence, by the Queftion, we ſhall have this
|1/2x+a+3x+b=6x.Y
2/5x+a+b=6x,
Equation,
Or,
2 -
-5x
D 3
3}x=a+b=10+15=25, which
multiplied
[ 29 ]
multiplied by 6 gives 150, the fame as a
bove.
SEEING
as,
Numerically.
of any Quantity is the fame.
the fame as
that (by the Queſtion)
15 together makes
ber of Soldiers; it is
5
of the whole; and
+-
or +10, and or
25 the Num-
evident that 25, the
Sum of 10 and 15, is Part of the whole
Company. Therefore 25×6=150 the Num-
ber of Soldiers required.
W.W.D.
PROBLEM XII.
There is an Army to which if you add,
1
}
44
, and of itſelf, and take away 5000,
the Sum total will be 100000. What
was the Number of the Army?:
To avoid Fractions put 24x the Num-
ber of the Army, for 24 is a Multiple of
2, 3, and 4, the Denominators of thefe Fra--
ctions,, and; Now of 24x is 12x,
is 8x, and is 6x, whence (by the Que-
ftion)
I
2
D 3
we
[ 30 ]
we have 124x+12x+8x+6x-5000
Or,
= 100000.
250x5000 = 100000.
2 +5000 3 50x105000
3÷5041
105000
x=
=2100 which mul-
50
tiplied by 24 gives 50400, the Number of
the Army.
2. E. J.
PROBLEM XIII.
In the Rectangle ABCD, the Difference of
the greater Side AB, and of the leffer
Side BC, is 12, but the Difference of
the Squares of the Sides is 1680. What
are the Sides of the Rectangle ABCD ?
PUT AB-BC—12—2, AB²—BC²-1680
b, and x-BC the leffer Side; and (per
Ax. 2.) the greater Side AB=x+a, whofe
Square is x²+2axa², and the Square of the
leffer Side is x²; the Difference of theſe two
Squares is 2ax+a, which (per Query) is
equal b.
Therefore 1/2 axa²=b.
2
a² 22ax-ba³,
(See Fig. 4.)
1
I
2293
b
a
1680
12
=64—BC,
24
2 24
2
hence AB 64+12=76, the Sides required.
PROB,
[ 31 ]
PROBLEM XIV.
The Length DE of the Rectangle DEFG,
is twice the Breadth EF; and the Sum
of the Squares of the Length and Breadth
is ten times the Sum of the two Sides DE,
EF. What are the Sides of the Rectangle
DEFG?
}
PUT the Breadth EF, and the Length
DE will be 2x; the Sum of the Length
and Breadth is 2x+x or 3x, and the Sum
of their Square is 4x²x² or 5x², which (by
the Queſtion) is equal 10 times the Sum of the
Sides or 3*.
(See Fig. 5.)
2
Thefore 15x30x.
2
1 ÷÷5x|2|x=6=EF, the leffer Side, con-
fequently the greater Side DE is 12, the Side
of the Rectangle, as was required.
If the Learner is not yet acquainted with
the Doctrine of Proportion, let him peruſe the
feventh Chapter of the fecond Part of Ward's
Introduction to Mathematics; or the fifth Book
of Euclid's Elements of Geometry, and then he
will the eafierly underſtand the following So-
lution, and feveral others where the Buſineſs
of Proportion is required.
PROB.
[ 32 ]
PROBLEM XV.
To find two Numbers in the Proportion of
2 to 3, whofe Product, if they be mul-
tiplied by one another, shall be
54.
PUT 54=a, x the leffer Number, and
as 2:3
3x
2
the greater Number; the
is
3x2
,
2
which (by the
Product of x and 3*
2
Queſtion) is equal to a.
Therefore t
3 x
2
a.
2
2
I
X-
3
alm
24
2
3
2 a
2×54–6, the leffer
3
3
=
Number.
Q. E. J.
2 w 23x=
Number, and as 2:3::6:9 the greater
LEAST the young Algebraift ſhould not
readily percieve the Reafon of multiplying
3x3
this Equation
alm
2
a, at Step the firſt, by
to bring out of the Fraction, which to
the
[ 33 ]
the Profecient will be moft confpicuous, it is
thought proper to give him the following.
EXPLANATION.
3x
2
THAT multiplying -a
a by brings
24
2
3
out x²= at Step the ſecond, will appear
3
very plain if we retain the whole Multiplica-
2
tion; thus,
3x² 2
2×3x²
2a
X
For fince
2 3
3×2
3.
2×3
the Co-efficient of x
is
and becauſe the
,
3×2
Numerator and Denominator are the fame in
all Refpects, they confequently deſtroy each
other, therefore x²-
2a
3
HENCE We have the following general
Method for freeing any litteral Quantity of a
Fractional Co-efficient. Multiply the Quan-
tity by it's Co-efficient inverted, and it is
done.
Note, What I mean by an inverted Fraction
3
is this,
2
when inverted becomes
2
3
FROM hence alſo appears the Reaſon of
multiplying a Fractional Quantity by it's De-
nominator,
[ 34 ]
nominator, by only taking away the Deno-
minator. For
3x-5
(at Step 1. Operation 1.
2
of the firſt Propofition) multiplied by 2 it's
Denominator, that is,
3x=-5x2 = 3x-5×2
2
2
becomes 3x-5, at Step 2; becauſe 2 is found
both in the Numerator and Dénominator, and
fo confequently vaniſhes.
Numerically.
SEEING that 54 is the Rectangle or Pro-
duct of the two Quantities required to be found,
which Numbers having the fame Ratio to one
another as 2 to 3: Therefore multiply 2 and
3 together, and the Product is 6.ˆ Ñow be-
caufe (by the twentieth Pröpofition of the ſixth
Book of Euclid's Elements) all fimilar Planes.
are in a Duplicate Ratio of their like Sides,
fay, as 6: is to 4 (= the Square of 2 the leaft
Ratio) fo is 54 to 36, whofe fquare
Root is 6 the leffer Number; and divide
54 by 6, and the Quotient is 9= the greater
Number.
W. W. D.
+
:
+
PROBLEM
I 35 ]
ས
PROBLEM XVI,
To find two Numbers whofe Ratio is to one
another as 4 to 5: and the Sum of the
Squares of both, is 2624.
PUT x
5x
X
the leffer Number, and as 4: 5
:x: = the greater; the Square of the
4
leffer Number x is x², and the Square of the
greater Number is
25x2
16
>
and the Sum of theſe
Squares is
25x2
, which by the Queſtion
I
16
is equal 2624= a.
Therefore
XIC 225x216x²=16a
16
Іба
2 ÷ 25+163x2 =
25+16
3
พ
16a
2516
16×2624
25+16
F32
$
the leffer Number, which being known, fay,
as 45: 32: 40 the greater Number.
Q, E. J.
Or
[ 361
Or thus.
TAKE 2 the Square of the leffer
Number, from a, and the Remainder is
àx² the Square of the greater Part (per
Note following the Axioms), and the Square
of the Ratios is 16 and 25; now (by the
twenty fecond Propofition of Euclid's fixth
Book), fay,
as
|1|16:25:: x²: a-x2: And (by
the XVIth of the farne Book)
we have 216a-16x²=25x²
2
2 +16x² 3+1x²=16a, the very fame with
Step 2, above, only contracted.
16a
น
3
414x
4I
, I ба
16×2624
4
w
4/5/20
=32, &c.
41
41
Numerically.
THE Squares of the Ratios added toge-
ther is 41, and the Sum of the Squares of the
Numbers required being given 2624, fay
by Proportion, as 41: 16:: 2624: 1024
the Square of the leffer Part, whofe Root is
32=the leffer Number, from 2624 take 1024,
and the Remainder is 1600, whofe Square
Root is 40 the greater Number. W.W.R.
Note,
[ 37 ]
!
Note, This Numerical Rule is founded on
the twenty fecond Propofition of the fixth, and
the eighteenth of the fifth Books of Euclid's
Elements of Geometry.
PROBLEM XVII.
To find the fide of a Square, whofe Area
is to the Sum of the fides in a given
Ratio, as 45 to 12.
PUT 45=a, 12b, and x = the fide of
the Square; then 4x is the Sum of the fides,
and x² the Area of the Square, which are to
each other as a to b, by the Propofition.
2
Therefore 1a: b:: x² : 4%.
1 Ergo 2 bx²=4ax
2
÷
4a
bx 3 x = 40=4x+5=15 the fide of
b
the Square required.
12
N. B. The Word ergo in the Margin at the
fecond Step, fignifies the fame, as tho' I had
quoted the fixteenth Propofition of Euclid's
fixth Book for the Reafon of this Equation
bx=4ax; as is done at Step 2. Operation 2.
of the foregoing Problem.
E
Numerically
[ 38 ]
Numerically.
ALL Squares are fimilar Planes; therefore
divide the Ratio of the Perimiter or Sum of
the fides by 4, and with that Quotient divide
the Ratio of the Area; this laft Quotient will
give the fide of the Square required. Thus,
the Quotient of 12 divided by 4 is 3, and
45 divided by 3 quotes 15 the fide of the
Square as above.
THE Reaſon of this Arithmetical Solution,
is this, that that Number, that expreffes the
Ratio of any Quantity, whether a Line, Su-
perficies, or Solid, hath the fame Properties
as that Quantity, of which it is the Ratio.
PROBLEM XVIII.
To find the fide of a Cube, whofe Superficies
is to the Solidity in a given Ratio, as 6
to II.
PUT 6a, 11b, and x = the fide of the
Cube; then the Solidity will be x³, and the
Superficies 6x2. Now by the Queſtion we ſhall
have \ra: b: 6x²: x³.
Ia
Ergo 2 ax³-6bx²
I
2 ax²
2
༔}
6b
6XII
X
= 11 the fide of
a
6
the Cube required.
Numerically.
[ 39 ]
!
Numerically.
ALL fimilar Solids are to each other as the
Cubes of their like fides (by the thirty third
Propofition of Euclid's eleventh Book), whence
divide the Ratio of the Solidity of the Cube,
by part of the Ratio of the Superficies,
and the Quotient is the Side of the Cube re-
quired. Thus, part of 6 is 1, andȚI
ह
is e fide of the Cube as above. Or
II
ſuppoſe the Proportion of the Superficies and
Solidity had been as 243 to 2916, the fide of
the Cube would then be 72; for a fixth part
of 243 is 40.5, and 2916 divided by 40.5 the
Quotient is 72.
W. W. D..
PROBLEM XIX.
A certain Man hires a Labourer, on this.
Condition, that for every Day he work'd
he ſhould receive 12 Pence, but for
every Day he was idle, he should be
muleted 8 Fence. When 390 Days were
past, neither of them were indebted to one
another. How many Days did he work;
and how many was he idle?
8=c,
x
PUT 390a, 12b, 8 = c, and the
Days he was idle; then (per Ax. 1.) the Days
E 2
he
[ 40 ]
he work'd will be a-x. Now feeing the
8 Pences difcounted for the Days he play'd
was equal the 12 Pences he was to receive for
every Day he work'd, we fhall
I
have 11cx=ab-bx. For c:b:: a—x: x.
+ bx2 bx + cx=ab
叶
​ab
2 ÷ 0 + 4 3 x = 6 +
390 X 12
128
=234
234 the
Days he were idle, which taken from 390
leaves 156 the Days he work'd,
Numerically.
As by the Queſtion 8 times the Number
of Days the Labourer was idle, are equal 12
times the Number of Days he work'd; there
is nothing more required, then to divide 390.
Days into two fuch parts, that 8 times the one
fhall be equal 12 times the other, and this Di-
vifition is done Reciprocally (by virtue of the
fourteenth Propofition of Euclid's fixth Book.)
Thus: As 12+8 (=20): 12:: 390:234
the Days idle; and as 12-8 (= 20): 8
:: 390: 156 = the Days he work'd, the fame
as before, for 156×12=1872=234×8.
W.W.D.
PROBLEM.
[ 41 ]
PROBLEM XX.
A certain Gentleman hires a Servant, and
promifes him 24 Pounds yearly Wages,
together with a Cloak. At 8 Months
End the Servant obtains leave to go a-
way, and instead of his Wages receives
a Cloak 13 Pounds. How much did
the Cloak coft?
To folve this Problem put 12, the Months
in a Year, a, and 8 Months he ferved b,
let the Price of the Cloak be called x; like-
wife put 24c, and 13 d.
his yearly Wages, and x+d
received at 8 Months End (by
therefore fay,
as
[1c+x: a :: x+d : b.
1 Ergo 2 axad=bcbx
-bx 3 ax-bx+ad=bc
・ad 4 ax-bx=bc-ad
2
3
CA
-bx
Then ctx is
the Wages he
the Queftion),..
bc-ad____ 8×24—12×13
4a-b
x = a
a-b
=9
12-8
Q. E. J.
Pounds the Price of the Cloak.
Numerically.
2
To folve this Propofition Arithmetically,
Reaſon thus; 8 Months is or of a Year;
therefore
E 3
[ 42 ]
爹
​2
herefore at the End of 8 Months, at the rate
of 24 Pounds and a Cloak per Year, the Ser-
vant muft receive of 24 Pounds, together
with of the Cloak for 8 Months Service;
but becauſe he received the whole Cloak +13
Pounds, inftead of of his Year's Wages,
which is 16 Pounds and of the Cloak, there-
fore take 13 from 16, and the Remainder is
3 Pounds, equivalent to part of the Cloak,
confequently the Cloak coft 9 Pounds.
W.W.D.
PROBLEM XXI.
A Perfon being ask'd how old he was, an-
fwered, If I quadruple of my Years,
and add of them 50 to the Product
the Sum will be ſo much above 100, as
the Number of my Years is now below
100.
;
To anſwer this or any Queſtion of the like
Nature, the Learner ought to be acquainted
with Arithmetical Progreffion; with which if
he be unacquainted, confult the fixth Chapter of
the first or fecond Part of Ward's Introduction
to the Mathematics.
To avoid Fraction, put 6x = his Age,
of which 4x, and the Quadruple of 4x
16x, to 16x add 3x (= his Age) +50,
and
[ 43 ]
;
and the Sum is 19x+50, which (by the Que-
ftion) as much exceeds 100, as 6x (= his
Age) wants of 100. Whence 6x, 100, and
r9x-50 are a Series of Terms in Arithme-
tical Progreffion, and will ſtand
21
Thus 16.x. 100. 19x50.
Ergo
225x+50=200 (per Lemma 1.
chap. vi. Part I. or Sect. 1. chap.
vi. Part II of Ward's Intro-
duction, &c.
50325x=150
3254x=6, therefore 6x=36 his Age
as was required.
THIS Queſtion may alſo be ſtated other-
wife. Thus, for the Difference between 100
and 6x being the fame as betwen 1950 and
and 100 (per Query), that is 100-6x, and
19x50-100 are equal.
I
Therefore 1100-6x=19x+50-100.
+6x225x+50-100=100
2-50-100 325x=150
3
25 14x=6, &c. the fame as before.
PROB-
[ 44 ]
PROBLEM XXII.
One being asked what Hour of the Day it
was, anfwered, The Day at this Time
is 16 Hours long; if now of the Hours
past be added to of the Remainder, you
will have the Hour defir'd, reckoning
from Sun-rifing.
x=
PUT 16a, the Hours paft, and
(per Ax. 1.) a-x is the Hours remaining;
2a- 2x
of which is
and the half of x is-,
,
3
2
X
24
2X
whofe Sum, viz. -+
2
Hours paft (by the Queſtion).
1 x
Therefore 1 +
2
is equal the
3
24
2x
=x.
3
I
X2 2 +
4a-4x
=2x
3
2
x333x+4a-4x=6x
3-3x+4x47x=4a
4 ÷ 75x=
4a
4× 16
9 the Hour
7
7
of an Hour paft One
of the Day, which is
o' Clock, reckoning from Sun-rife; and may
be
[ 45 ]
2
་
2
be proved thus,
I 12
7
the whole Day, or
the Hours remain-
6448
16 Hours)
7
7
2
ing, of which is =
96
64
and
=
is half
ΣΙ
14
96
the Hours paft: but theſe two Fractions
64
27
and are to each other, for as 96: 21
14
::64:14. Ergo 96×14=64×21.
W.W.P.
Numerically.
4
I
2
BECAUSE the Hours paft, and of
thofe remaining in one Sum,. are equal the
faid Hours paft; it is plain that of the Hours
remaining and the Hours paſt are equal:
Therefore of the Hours of the Day remain-
ing is equal the prefent Time of the Day or
Hours paft, confequently the Hours of the Day
paſt and thoſe remaining, are to one another
in the Ratio to, or as 1 to 1. Now the
Sum of theſe Ratios is, and to divide 16
into two parts that ſhall be to each other
4
उ
as to,
}{\
44
:: 16:
9
Τ
or 94.
61.
the fame as by the foregoing Operation.
W.W.D.
PROB.
[ 46 ]
PROBLEM XXIII.
From Noremberg to Rome are 140 Miles:
A Traveller Jets out at the fame Time
from each of the two Cities, one goes 8
Miles a Day, the other 6. In how many
Days from their first fetting cut will they
meet one another, and how
many Miles
will each of them go?
PUT x the Number of Days they travel
before they meet; then 8x is the Number of
Miles he goes that travels 8. Miles a Day, and
6x is the Number of Miles the other goes that
travels 6 Miles; and the Sum of 8x and 6x is
14: which, by the Queſtion, is equal 140
Miles, the Diſtance between Noremberg and
Rome. Therefore 14*140, which Equation
divide by 14, we ſhall have x = 10 the Num-
ber of Days each travelled before they met,.
whence one travelled 80 Miles and the other
60.
BUT the Number of Days they travelled
before they met each other may be found by
Divifion only; for feeing they both travel the
fame Number of Days, and that each Day's
Journey of both is 14 Miles. Therefore di-
vide 140 by 14 and Quotient gives 10 the
Number of Days as before. Moreover, be-
caufe
[ 47 ]
caufe every Day's Journey is the fame, there-
fore the Place where they meet, divides the
Diſtance between Noremberg and Rome, into
two fuch parts, that are to one another, as the
Number of Miles each travels a Day. That
is, as 8 to 6, or as 4 to 3.
8
80.
Whence, as 8 +6 (= 14):
:: 140:
6
60.
the fame as above.
W.W.D.
******
PROBLEM XXIV.
A certain Meffenger goes 6 Miles every
Day: 8 Days after, another follows
him, and he goes 10 Miles a Day. In
how many Days will he come up to the
first?
PUT 6a, 10=b, 8=c, and x the
Days he travelled that goes 10 Miles a Day,
and the Number of Days the other travelled
will be c-x: Now the Number of Days
each travelled, multiplied by the Number of
Miles each Meffenger goes every Day, gives
the Miles both travels feparately, which are
equal, becauſe they both began their Journey
from the fame Place, and ended it at their over-
taking each other; whence ac ax is the Miles
the firſt Meffenger travelled, and by the Num-
ber
[ 48 ]
ber of Miles he travelled that goes 10 Miles
a Day.
Therefore 1bx ac tax.
I
ax2 bx
ax = ac
ac
6x8
2 ÷b—a|3x=
ba
12, the Days
IO -
6
he travelled that goes 10 Miles a Day, and in
To many Days he came up to the firit Meffen-
Q. E. J.
ger.
Or thus
THE firft Meffenger had gone 48 Miles be-
fore the ſecond Meffenger fet out; but becaufe
the fecond Meffenger gains 4 Miles every Day
of the firft; therefore 48 divided by 4, the
Quotient is 12 the Number of the Days as be-
fore.
W.W.D.
PRÒ B LE M XXV.
A certain Meſſenger goes 6 Miles a Day:
And after he has gone 56 Miles, another
follows him who goes 8 Miles a Day. In
how many Days will he come up to him?
THE foregoing Solution being rightly un-
derſtood, it will be needlefs to give this an
Algebraical Inveſtigation; and indeed the
Arith-
[ 49 ]
い
​Arithmetical Solution is the very fame as that
aforegoing, for divide 56 by 2 (8-6) and
the Quotient is 28, the Number of Days in
which the ſecond Meffenger came up to the
firft.
Q. E. J.
If the Learner defire to ſee more of thefe
kind of Problems, let him perufe the fifth
Propofition (Page 72.) of the Arithmetical
Queſtions, in Sir Ifaac Newton's Univerfal Arith-
metic, with it's Cafes and Examples.
PROBLEM XXVI.
One bought 3 Books, whofe Prices were in
Proportion as 12, 5, 1: If the Price of
the first be doubled, of the fecond trebled,
of the third quadrupled; the Sum of theſe
Products will as much exceed 10 Crowns,
as the Sum of the Prices of the greatest
and middle is below 5. How much did
the faid Book coft ?.
PUT x the Price of the firft Book, and
12
5x
as 12 x: 5:
the Price of the fecond,
X
the Price
I 2
and again, as 12 : x : : 1 :
of the third: Now double the Price of the firſt
F
Book
[ 50 ]
Book is 2x, treble the fecond —
druple the Price of the third is
of theſe Products is 2x+
5x
and qua-
4
༩
; the Sum
3
5x
X
which as
4
3
5x
(the
1 2
much exceeds 10 Crowns, as x+
Sum of the greateſt and middle Prices) is be-
low 5. (per Query.)
Therefore 12x+5x+10=5x-5
1 X 12243*
2+17x360x
4 3
120=60 17x
12060
3+120460x=180
5x
12
4605x=3, the Number of Crowns the
4
firſt Book coſt = 15 Shillings, and, as 12 : 15
56 Shillings, the Price of the fecond,
and again, as 12: 15 :: 1: 1 Shilling,
the Price of the third Book.
I
I
Q. E. J.
PROBLEM
:
[ 51 ]
PROBLEM XXVII.
:
Suppose the Number 50 were to be divided
into two Parts, ſo that the greater Part
being divided by 7, and the less multiplied
by 3, the Sum of this Product and the
former Quotient may make the fame
Number propofed, which was 50.
a
PUT 50, 7=b, 3c and the
greater Part; then the lefs (per Ax. 1.) will be
*
a x: again, is a 7th part of x, and ac
b
cx is three times a x; the Sum of this
X
+ac-cx, which
Quotient and Product is ~ +ac
b
(by the Queſtion) is equal a.
Therefore
I +ac cx = a.
I
2
b
xb2x+abcbcx= ab
+ 3 bcx-x=abc — ab
+3/bcx
abc ab
3 bc
14 x
141**
=35, the greater
bc I
part,
confequently the leffer is 15. QE. J.
F 2
PRO-
[ 52 ]
PROBLEM XXVIII.
Let the Number 20 be divided into two
Parts, fo that the Square of the less part,
being taken out of the Square of the
greater, may leave the very Number pro-
pofed, which was 20 (or may leave the
double, treble, &c. of the Number pro-
pofed.)
PUT 20 = a, x= the leffer part, and (per
Ax. 1.) the greater part will bea-x; the
Square of the greater part is a2-2ax+x²,
and the Square of the leffer x², which taken
out of the Square of the greater part, leaves
a²-2ax, this Remainder (by the Propofition)
is equal to a, or.2a, or 3a, &c.
Therefore/1a²
2ax=a.
I 1 +•2ax2a +2ax=a²
2
1
a32ax=a²
a
a
a-
I
20-I
=9.5
24
2
2
32a4x=
the leffer part, and 20-9.5= 10.5 the
greater.
Q. E. J.
AND becauſe a²-2ax may be equal 2a,
we ſhall have this new
Equation
་
[53]
Equation/1a²
+12/2ax
2ax=2a.
a²
I
20
2
A-2A
a
22a3x=
-19 the leffer
=9
2a
2
part, and the greater is 20-9=11.
AND again, a-2ax being made equal
34, we fhall
2ax=3a.
+22ax = a²
have
Iaz
I
за
a²
за
a
3_20 3
2 24 3x=.
8.5
20
2 2 2 2
the leffer part, and 20-8.5=11.5 is the
greater, and fo on.
W.W.D.
PROBLEM XXIX.
If a Man gains 30 Crowns a Week; how
much muſt be ſpend a Week to have 500
Crowns, together with the Expence of 4
Weeks remaining at the Year's End?
SUPPOSING the Year to confift of 52
Weeks, he will gain 52×301560 Crowns
every Year: Put x = his weekly Expence, and
52x Crowns is what he lays out yearly, which
taken from 1560 the Remainder is 1560
-52x, which, by the Queftion, is equal to
500 + 4*.
Therefore
F 3
[ 54 ]
Therefore 11560-52x= 500 +4.
256x=1060
I
2 ÷56
A 응
​w
1060
56
= 18, the Crowns he
ſpends per Week, equal to 4 l. 7 s. 1 d. §.
Q. E. J.
PROBLEM XXX.
A Labourer, after 40 Weeks in which he
had been at work, lays up 28 Crowns
the Pay of three Weeks; and finds
that he had expended 36 Crowns -- the
Pay of eleven Weeks. What Pay did
he receive a Week?
PUT x his weekly Pay, and 40x is the
Number of Crowns he received for 40 Weeks
Labour; in which Time he laid up 28-3x
Crowns, and expended 36+11x Crowns,
which two Sums together make 8x+64
Crowns, equal to 40x.
Therefore 118x+64-40%.
-8x232x=64
I
2323x
64
2. Whence he recieved
32
2 Crowns or 10 Shillings a Week. W.W.R.
•
PRO-
[55]
:
PROBLEM XXXI.
In the Rectangle ABC, is given the Bafis
AB=9 and the Difference of the other
Sides, that is, the Segment BD, = 3.
Required, the Sides AC, BC?
LET 9=a, 3=b, and x = CA. and be-
caufe CA and CD are equal (See Fig. 6.) BC
=b+x. Now (by the 47th Propofition of
Euclid's firft Book) BC=ACAB, but
BC²= b²+2bx+x², AC-x, and AB=a2.
x²-—a².
Therefore 12+2bx+x²= x² -† a².
2
x² 2b² = 2bx = a²
26²+2bx
2
a² — b²
2
I
2
-b232bx
a²
b2
3
÷ 264x=
26
-9x9—3x3
2X3
= 12,
the Side CA, and 12315 the Length
of the Side CB.
Or thus.
Q. E. J.
THE fame Geometrical Problems may be
ſolved by different Principals of Geometry, as
well as Arithmetical Queftions by different
Laws of Arithmetic. For by drawing the
Circle ADE from C as Center, with the Ra-
dius CA, and continuing the Side BC to E;
then
[ 56 ]
then will BE be =b+2x, and (by the 36th
Propofition of Euclid's third Book) it will be,
as,
IBE: BA:: BA: BD. Or rather
BEX BD = BA².
That is, 262x: a: a: b.
+2x:a:
2 Ergo 362 2bx = a²
3 · b² | 4|2bx = a² — b².
4265x
a²
!
2
·b² 9×9—3×3
=12
26
2×3
the Side CA, &c. the fame as before.
FROM the Solutions to the foregoing Pro-
blem, the Analyft may obferve, that in an-
fwering Queſtions of this Nature, how well he
ought to be acquainted with the Principals of
Geometry. For when a Queſtion of this kind
is propofed, it will be impoffible for him to
give an Anſwer to it, unless he thoroughly
underſtands the Nature of the Figure the Quere
is propofed in, and the different Properties
that may arife from the Application of new
Lines not firft concerned in the Diagram. For
as Propofitions relating to abſtract Quantities,
are folved by the Laws of Arithmetic; fo we
muſt have recourſe to the Laws and Properties
of Geometrical Figures, to anſwer Queſtions
in Geometry. And how the faid Properties
may be employed in bringing Geometrical
Problems to Equations, the moſt illuftrious
Geometer Sir ISAAC NEWTON, in his Uni-
verfal
[ 57 ]
፡
verfal Arithmetic, from Page 86 to 101, hath
largely ſpecified by a Variety of Examples;
which ought to be dilligently and moſt atten-
tively peruſed, and well underſtood by every
one that defigns to be a Proficient in this
Branch of Mathematical Learning.
THE thirty fixth Propofition of Euclid's
third Book being well understood, it will be
eaſy to inveſtigate the Side CA independent of
Algebra, by reafoning thus, feeing BEX BD
BA'.
And becaufe BD and BA are given,
BE will be known alfo; and taking BD from
BE, the Remainder DE will be given likewife;
but (by the Fig.) DE is 2CA. Therefore
DE
CA==DC: Whence, dividing 81
2
(=BA²) by 3 (=BD) the Quotient gives BE
27, from which take 3, and there remains.
DE=24=2CA, confequently CA=12. Ha-
ving thus found CA, add to it BD=3, and
the Sum is BC= 15 the other Side of the
Triangle ABC. Which was required.
Geometrically.
BECAUSE the Side CA, and the Segnient
BD are together in one Sum equal to the Side
BC, and that the Angle BAC is a right one`
by the Propofition. Therefore draw Ab
(BD) perpendicular to BA, and join Bb
with
[ 58 ]
I
with a right Line (See Fig. 7.) which Bifect in
a, at this Point a erect the Perpendicular aC
until it concurs with bA produced to C, laftly,.
draw the Line BC, and BAC is the Triangle
that was to be conftructed.
Demonftration.
THE Angles at a being right, and the Ba-
fis of the Triangles BaC and baC, viz. Ba
and ba, and alfo Ca common to both, the
Sides CB and Cb (by the fourth Propofition of
Euclid's firſt Book) are equal, and ſo (by Defi-
nition 21 of the fame Book) the Triangle BCb is
Ifofceles. With CA Radius, defcribe the Arch
AD, ſo ſhall BD be equal bA. Therefore, &c.
2. E. D.
PROBLEM XXXII.
In the Rectangle Triangle ABC, is given
the Bafis AB = 5, and the Sum of the
other Sides AC+BC-25. Required,
the Sides AC, BC feverally?
PUT 25=a, 5=b, and x=CA; then
(per Ax.1.) CB=a-x. And becauſe the
Angle CAB is right, we have (by the forty
feventh Propofition of Euclid's firft Book)
BC2 CA+BA; but BC2 a² 2ax
=
CA²= x², and BA²= b².. (See
+x²,
Fig. 8.)
2
Therefore
י
1
7
.
1
I
¿
:.
[ 59 ]
Therefore 1a2 - 2ax + x² = b²+x².
I
2
2
2ax
+32ax = a²
b2
2
b2
a²
- b²
2
a
3
÷2a4x
12=CA,
20
2
2a
W.W.R.
*
and 25 1213 CB.
x=
Or thus.
PUT BC, and (per Fig.) AC will be
-a-x; having thus expreffed every Side
of the Triangle ABC, we fhall have (by the
forty ſeventh Propofition of Euclid's firft Book)
BC2 AC²+ AB².
2
2
That is, 1x2 = a²
I
x20
2+2ax 32ax
3
小
​2
-2ax + x²+b².
zax + b²
2
2
a
a² + b²
a
b2
25 +5x5
2
24
2
+
= 13
2X25
BC, and 25-1312 AC. The fame
as before.
Note, The Sides AC and BC may be found
Numerically, Thus, compleat the Circle ADE,
and (by the thirty fixth Propofition of Euclid's
third Book) the Rectangle of BE into the Seg-
ment BD is to the Square of AB, There-
fore BD is equal to the Square of AB divided
by BE (BC+AC 25.) That is, BD
=5×5
[60]
5×5
25-I
!
= 1, and (per Fig.)
=12=AC,
25
2
hence 25 12 = 13= BC.
The fame Geometrically.
THUS, Draw the Line AB= 5 from any
Scale of equal Parts; at right Angles to which,
draw the Line Ab 25 the given Sum
of the Sides AC and BC. This done join Bb
(See Fig. 9) with a right Line; which divide
into two equal Parts in the Point a; from a
erect the Perpendicular aC, and having drawn
the Line BC, the Triangle ABC is that requi-
red to be conſtructed.
Demonftration.
THE Line aC being perpendicular to Bb
and cuts it into two equal Parts, the Triangle
BCb (by the fifth Propofition of Euclid's firft
Book) is Ifofceles. Ergo BCbC, confequent-
ly ACBD = A+¿C. 2. E. D.
HAVING
[61]
AVING thus far exercised the
young Analyſt in fimple Equations
only, it may not be improper to fhew him
what quadratic adfected Equations are, and
how fuch Equations are brought to a Solu-
tion: And for the better Understanding of
which, it is requifite he should know how
they are first produced, and this he will ea-
fily perceive by the following Examples.
EXAMPLE I.
IF inſtead of what is given and required
to be found in the foregoing Problem, there
had been given the Sum of the Legs forming
the right Angles (See Fig. 8.) Viz. AB+ AČ
= a, and the Hypothenufe or longeft Side
BC= b; to find the faid Legs feperately.
Then for the Leg AB put x, and (by Ax. 1.)
the other Leg AC will be equal a
x; hav-
ing thus exprefs'd every Side of the Triangle
ABC, we ſhall have, by Virtue of the 47th
Propofition of Euclid's firft Book, this
G
Equa-
[ 62 ]
1 +
Equation, BC2=AB² + AC²
That is, 2b2x² + a²
2
± 32x 2ax = b²
2
2ax--x².
a².
Ъ
2
3
-24x³-ax=
2
-C.
2
2
a
a²
સ
4
CO 15x2
ax +
2
C
4
4
а
LO
16:
6x
2
5
w 21
พย
6 +
a
2
2
+1
va
2
4
4
2
Co
C.
Becauſe a² is greater than 62, their Dif-
ference will be a Negative.
Having thus found the Leg AB, the other
Leg AC is given from what is ſhewn above.
SEEING therefore at the fourth Step of the
foregoing Operation, there is produced this
Equation x
ax
x²
b²
2
2
·a²
which becauſe it
is conftituted of two different Powers of the
unknown Quantity, viz. x2 and x, whofe In-
dices are double to each other, is called a qua-
dratic adfected Equation. Which, with re-
ſpect had to the Signs
three Varieties or Cafes, viz.
+mx=n.
and
admits of
Sx²+mx² = n.
I x2
2x2
2
mx=n.
men's Mx
2
=n.
1. &c.
371x
2
x² = 1.
2
4
mx
72.
N. B.
[ 63 1
N. B. That the Forms of the above Equa-
tions may be rendered the more Univerſal, I
put m for any Number that may happen to
be the Co-efficient for the unknown Quantity x,
b2
and n for the abfolute Number, as 2
and the like.
a
2-
>
AND for the Solution of fuch Equations,
the Learner muft obferve, that the Method
took it's firft Original from this Confideration.
That feeing on one Side of the Equation in
every Cafe, there is the Square of x, and the
Rectangle of x drawn into the given Quan-
tity m; or what comes to the fame Purpoſe,
the Square of x, and the double Rectangle of
x drawn into it is evident (from the 4th
2
m
2
;
Propofition of Euclid's fecond Book) that
xmx in the firft Cafe, xmx in the
fecond, and mx -- 2 in the third, wants only
m
mn 2
the Square of
that is
to make the faid
2
4
Quantities compleat Squares; for the Square
IN
2
of x + 2 is = x²+mx+
= x²
+
2
2
772
772
,
ofx
is
4
2
-mx +
M2
m
and of x
is = x²
2
mx
4
2
"
which tho' in every
Refpect is the
4-
G 2
fame
2
M²
[ 64 ]
fame in Form to that next before it, yet it be-
ing in the third Cafe is of a different Quality,
becauſe equal to a negative Quantity, as fhall
be fhewn by and by. Whence proceeds this
general Rule for compleating the Square in
all quadratic adfected Equation whatſoever.
RULE.
ADD the Square of the Co-efficient of
the unknown Quantity, to both Sides the
Equation, and the Square will be compleat.
See the following Operations.
m²
1 | x² + mx + "==n+"
4
2
mx+==n+
2²x² ~-mx+
4
2
2
m
4
2
m
4
m
ทา
2
or
x²+mx²+?
4
-mx²+"
m²
4
2
m
4
ent
4
m
2
M1
4
3, xz
-mx+:
m²
n.
mx²+
m
4 4
-n.
4 4
and fo on.
Hence, by extracting the fquare
Root, we ſhall have the following Equations.
1
↑
1[x
[65]
2
2
m
m
IN
m
I
2
4
2
4
2
IN
2
2
√nt
m
m
m
or x2
2
4
2
772
772
3
n.
X
2
4
212
2
m
2
·n.
4
&c.
1
m
2
And by tranſpoſing to the other Side, every
Equation will give the Value of x.
2
m
m
777
2
m
I
X
x
m
2
2x ==+un+
2
4
2
4
2
m
m
m2
or x 2 =
4
2
4
2
m
2
士
​ገን
.m 2
∙n.
2
∙n.
4
4
Except in thoſe
Equations on the
right Hand, which
muſt have the Root
again extracted.
Thus,
2
m
m
2
4
2
m
m
X
2
4
m²
2
2
tv
4
G 3
·12..
N. B.
[66]
N. B. In the third Cafe, or this Equation
mx-x² = n, before the Square can be com-
pleated, the Signs of every Term in the
Equation muſt firſt be changed, which being
done, will ftand thus x INX
n, and
then by compleating the Square, it will be-
2
2
come x mx +
M²
4
=
m²
2
4
n. as is fpe-
cified in the foregoing Operations.
To what is here faid concerning quadra-
tic adfected Equations, much might be ad-
ded, but I fhall content myſelf by inferting
only theſe Obfervations following. And if
the induftrious Reader defire to fee more of
this Doctrine, he is referred to Ward's Introduc-
tion, Chap. viii. Part II. or to John Parfon's
Clavis Arithmetica, Chap. xv. where he will
meet with Plenty of Examples of this
Kind.
FIRST then, It may be obferved that x,
the unknown Quantity, in the firft Cafe, may
be either greater or lefs than it's Co-efficient,
becauſe all the Terms in the Equation are af-
firmative. But in the fecond Cafe, x is al-
ways greater than it's Co-efficient; becauſe
x
2
mx is equal to an affirmative Quantity. And
in the third Cafe, x is always lefs than it's Co-effi-
cient, becauſe mx and n are both negative. Hence,
in compleating the Square, in the first and fecond
Cafes it matters not whether the Square of half the
Co-
[ 67 ]
KÝ VEZ kan skalita:
m² 2
Co-efficient, that is, be greater or lefs than the
4
abfolute Number n. But in the third Cafe it
muſt always be greater than n, elfe there will
be required to extract the fquare Root of a ne-
gative Quantity, which is impoffible.
SECONDLY, In the third Cafe, after the
2
m²
Square is compleated, if and ʼn happen to
4
2
be, then, becauſe n is negative, x²- mx
+
m²
4
2
IN
2
=0. whofe Root being extracted
112
0,
and therefore x =
confe-
2
quently 2x=m.
THIRDLY, In the Solution of Equations
of the third Form, the unknown Quantity
will fometimes have two affirmative Values,
the one juft, and the other more than juft;
for which Reafon, this Sort of Equations are
generally called ambiguous; but properly
fpeaking they are only particularly fo. For
when the Co-efficient of x, that is m, is lefs
than 2x, then there will be no Ambiguity at
all: So this Ambiguity can only happen
when m is greater than 2x; for if m be equal
to 2x, then, by the laft Obſervation,
the
Quan-
[ 68 ]
tities
2
m
4
and
n will be equal, and confe-
quently deftroy each other.
which, in this Equation x =
To prove
IN
m²
2
土
​n.
4
which is of the third Form, let us affume 2x
for the Co-efficient of x inftead of m, and
fuppofing 2x-r any Co-efficient of x lefs
than 2x, but greater than x; and by ſubſti-
tuting this Co-efficient, viz. 2x - r, for m
in this Equation x =
2x
have x =
+
2
m
2
2
m²
+
n. we ſhall
4
4xr-x²
2
n.
4
1412
And for the Value ofn, form the original
Equation with the unknown Quantity x and
and it's Co-efficient 2x r, or which comes
all to the fame Thing, fubftitute 2xr for m
in this Equation xmx1, and we fhall
have − n == x² + rx; which being ſubſti-
tuted in the abovefaid Equation for
n. then
4x²+4rx+r²
2x-
go
X
2
+
4x
4xr
4
which being contracted will become x=
土
​4
whence it is plain x =
2x·
2
2X
за
2
2
and
R
[ 69 ]
and not
2x
2
r
x
r. . Therefore
2
this Cafe is not ambiguous when the Value of
the Co-efficient of the unknown Quantity, is
fome Number greater than the faid Quantity, but
lefs than it's Double; which was to be proved..
BUT to prove that any Co-efficient of x
greater than 2x will render this Cafe ambigu-
ous; put 2xr for any Co-efficient of x
greater than 2x, which being fubftituted for
ทา
m in this Équation x == 土
​m
4
2
bove ſpecified, it will become x =
n. as a-
2x - Fr
2
H
2
√4x² + 4xr
4x2
4xr + r²
.whence
4
2x + r
tr
2
r
H+
2x +
or x=
4
+1
jo
2
2
2x + r
r
but
2
+=x+r, which is one af-
2
firmative Value of the unknown Quantity:
but too much by r, the Excefs of the Co-ef-
ficient above twice the Quantity fought; and
2x + r J
2
2
x is the other, the true Value
of the Quantity required. Hence proceeds
the Ambiguity which was to be proved.
SCHO-
[70]
SCHOLIU M.
FROM what is proved above, it appears,
that the Difference between the Co-efficient
2
of x and 2x, that is, is always equal to
2
m n. and that ſeeing from the firſt De-
4
2x
y
r
monftration
gives the unknown
2
Quantity too little, and from the fecond,
2 x² + ² + 2/2
2
r too much by r.
Therefore
when this Cafe is not ambiguous, always add
m
n. to 2/22
4
and the Sum will give x the
Quantity required, but when ambiguous,
2
then always ſubſtract/m²
m
n. from
and
4
the Difference will give the Quantity ſought;
thus, in the former x =
in the latter x =
2
2
m
+ √m²
2
n. and
4
n.
But the
4
m
2
Learner muſt know that the Ambiguity of
this Caſe cannot be determined when the Equa-
tion
[71]
נ
1
tion is expreffed literally; and for this Reafon
m
the Quantities
m²
2
and - n. are always
4
connected together with a doubtful Sign.
2
m
Thus, x =
n. See the third Cafe
2
4
$
៩
aforegoing.
FOURTHLY, In every quadratic adfected
Equation, or indeed all compound Equations
whatſoever, are as many Roots or Values of
the unknown Quantity, either affirmative or
negative, as the Index of the higheſt Power
of the Quantity fought contains Units; and
when one of thofe Roots is diſcovered, every
Root befides is 'found by dividing the whole
Equation by that Root continually, or in
quadratic adfected Equations, having found
one Root, fubftract it from the Co-efficient
of their refpective Equations, and the Re-
mainder will give the other Root.
one
FIFTHLY and Laftly, In the firſt and ſe-
cond Cafes of quadratic Equations,
Root will be affirmative and the other a ne-
gative Quantity; but in the third Cafe, both
Roots will be affirmative, tho' but one of
them will anſwer the Queftion, which muſt
be taken according to the Conditions thereof;
and in the other two Cafes, or indeed all
Equa-
[72]
Equations whatfoever, the affirmative Quan-
tity always folves the Problem, for the nega
tive Quantities are of no other Ufe than to
help to form the original Equation.
2
THUS much I prefume is fufficient for
what may be required in quadratic adfected
Equations; from whence it appears, that this E-
quation x ax = c at the fourth Step in
the Calculus to the foregoing Example, is of
the third Form, and therefore from what hath
been fhewn above relating to this Sort of
Equations, it is evident x =
a
2
土
​2
4
Со
FROM the 47th of Euclid's first Book
it will eaſily appear that the Legs of a right
angled Triangle are equally related to the
other Side, or Hypothenufe. So that if in-
ftead of putting ≈ for AB (See Fig. 8.) I
fuppofe it = AC; then (per Ax. 1.) a-x
- AB, which Expreffion is the very fame as
before, tho'x repreſents a different Part of
a; and being ſtated according to the above-
faid Propofition of Euclid, will ſtand
1
Thus,
[ 73 ]
Thus,
I
IX
2
|x² + a
2
a
22x
2
2
zax+x²= b²
62
2
2ax
b2
a²
2
2
a²
2
2
2
3x
ax=
3
C4x2
a
4
w 25x
2
a
5+
७
a
2
+
ax +
2
2
4
а
a²
4
2
2
2
a²
C.
Co
C
4
2
4
Seeing
a² is greater than b
it is evi-
dent their Difference will be nega-
tive, or
1
I
ነ
+
Now by comparing this Operation with
that aforegoing, we fhall find, that both the
Form of the Calculus and Conclufion is the fame
in each, notwithſtanding different Quantities
are the Scope of our Enquiry. And hence it is
that fome Geometrical Problems are ambigu-
ous, and when fuch occur, the great and in-
comparable Mathematician, Sir Ifaac Newton,
adviſes, that when there happens to be ſuch an
Affinity or Similitude of the Relations of two
Terms to the other Terms of the Queſtion,
that you ſhould be obliged, in making Ufe
of either of them, to bring out Equations ex-
actly alike; or that both, if they are made
Ufe of together, fhould bring out the fame
Dimentions, and the fame Form (only except-
H
ing
[74]
ing perhaps the Signs and -) in the final
Equation (which will be eafily feen) then it
will be the beſt Way to make uſe of neither
of them, but in their Room to chufe fome
third which ſhall bear a like Relation to both,
as fuppofe the half Sum, or half Difference,
or perhaps a mean Proportional, or any other
Quantity relating to both indifferently and
without alike.
Page 126.
See the Univerſal Arithmetic
THAT is, in this Example feeing the Simi-
litude of the Relations of the two Terms or
Legs AB and AC to the other Term or Hy-
pothenufe BC, is fuch, that making uſe of
either of them, brings out Equations exactly
alike; therefore for a more commodious Elec-
tion of the Terms for the Calculus, by the
foregoing Advice, chufe fome third Term,
which, becauſe the Sum of Legs are given, let
be their Difference, for which put x, and by
the proceeding Lemma, the greater Leg
a
X
will be, and the leffer Leg
both which being fquared will
A
x
;
2
2
become
3
[75]
become
I
2
༡ས་.
a
4.
a
ax
+~+
2
4
whofe Sum
2
ax
X
+
4
2
4
+
X
2
b².
2
+
x²
2
=2b2
4
3 +5
is
X 2
a
w 2
3
2
4
15/06
2
2
22b2a2, a Simple Quadratic.
a. AB AC,
√262
=
or AC-AB, according as AB or AC re-
preſents the greater or lefs Leg.
HENCE by computing after this Method,
it appears, that not only the final Equation is
reduced to a fimple Quadratic, but the Am-
biguity of the Propofition is likewife avoided;
and therefore in fome Queſtions it is more
commodious to feek not the Quantities that
are required, but fome others from whence
they may be found; as the Reader will find
if he obſerves ſome of the foregoing and fol-
lowing Operations.
BUT to render the Ambiguity or Nature
of this Example more confpicuous, take the
following Geometrical
H 2
CON
[76]
CONSTRUCTION.
DRAW the Line BD (See Fig. 10.) equal to
the Sum of the Legs AB and AC, in which
take the Point e any where at Pleaſure; from
which Point erect the Perpendicular eb equal
to the Segment eD; from D, by the End of
the Perpendicular eb, draw the Line Dbf of
any convenient Length. This done, with
the Hypothenufe BC as Radius, and from B
the Center defcribe the Arch gcC, and from
the Point C where the faid Arch cuts the
Line Dbf, let fall the Perpendicular CA;
and the Triangle required to be conftructed
is ABC.
DEMONSTRATION.
THE Lines be and CA being both drawn
from the fame Line Dbf perpendicular to the
Line BD, it is evident the Triangles be D and
CBD are fimilar, and therefore eD: eb: :
AD AC. but eD and eb are equal by
Conftruction, confequently AD is equal to
AC. Whence it is plain AB AC is
ABAD = BD.
Q. E. D.
COROL-
!
[ 77 ]
COROLLARY.
FROM the foregoing Conſtruction, ſeeing
the Arch Ccg cuts the Line Dbf in another
Point c, it will be eafy to demonftrate Geome-
trically the Reaſon of the Ambiguity in this
Example. Thus, let fall the Perpendicular
ca, and with a right Line join the Points B
and c.
Then by fimilar Triangles be: eD
:: ca: aD; but be and eD are equal by
Conſtruction, therefore ca is equal to a D,
confequently Baca Ba + aD = BA
+CA; whence the Triangles ABC and aBc
are equal in all Refpects, for BA is ca, and
CA. Now fince from each of the
Points of Interfection C, c, the fame Trian-
gle is produced, it is therefore indifferent
which of the faid Points the Triangle is con-
ftructed from, fince they both bear alike Re-
lation to the Line BD.
Ba=
Ba
2. E. D.
EXAMPLE II.
OR if inftead of the Sum of the Legs,
as in the laft Example, there had been
given their Difference a, and the Hypothe-
nuſe bas before, to find the Legs. Then
by the common Method of Computation
H 3
put
1
[ 78 ]
put
&
the greater Leg, and (per Ax. 2.)
the leffer Leg will bex - a. And by
the 47th Propofition of Euclid's first Book,
we shall have BC AB+ AC² (See
Fig. 9.) That is,
I
2
2
=
1116² = x² — 2ax+a2+x².
I
2
a² 22 x
22x2
223x
3 Co
41
४
4 w 25 x
a
5 + 26
N
2ax = b²
a²
b2
a2
2
ax
2
a²
262
4
2
212
ax +
4
262
4
a
262
a²
2
壮
​2
4
the greater Leg, which was to be found, whence
the leffer Leg will be known. But by putting
x= the leffer Leg, then the greater Leg
will be xa, and by the abovefaid Propo-
fition of Eucild, we fhall have this
Equa-
[79]
2
Equation [122ax + a² + x² = b²
I
a² 22x² + 2ax =
2 ÷ 23x² + ax =
W.
2
b2
a²
2
b2
22
2
262
a²
4
CO 4x² + ax +
w 25x +
3
4
a
5
26x =
a
=
a 2
4
√262
4
a
2
2
a
262
a
+
2
4
SEEING that in the foregoing Operations.
their final Equations are the very fame in
both, except that in the former is affirma-
a
2
tive, which in the latter is a negative Quan-
tity; the third Term therefore to be made
ufe of for the better Election of the Terms
for the Calculation, muſt be the Sum of the
Legs, becauſe their Difference is given, for
which put ; then (per Lemma following) the
x
greater Leg will be+, and the leffer
४ ल
x
a
2
>
2
2
2
and the Sum of their Squares
*?
[ 80 ]
ર
2
+
gure.
2
which is equal to b2 by the Fi-
Confequently I
| N
2
2
+
a²
2
2
= b².
I
X
a2=2b2
2
a2
3 =262
2/2x² +
2
a
3 พ 2 41 x = √26² a. AB+ AC.
=
AND thus, by Help of the half Sum
and half Difference, I have brought out the
refulting Equation in a fimple Quadratic
Form, and freed of Ambiguity: tho' I
might have omitted this Expreffion, for no
fimple Equation, of what Kind foever, can
be ambiguous.
GEOMETRICALLY.
DRAW the Line BD equal the given Diffe-
rence of AB and AC (See Fig. 11) to which
add the Line De of any convenient Length;
from e erect the Perpendicular be equal to De,
and draw the Line Dbf of any Length at
Pleafure; then from B fet off the Hypothe-
nufe BC to the faid Line in C, and let fall.
the Perpendicular CA, fo is ABC the Trian-
gle which was required.
DEMON
[ 81 ]
7
}
DEMONSTRATION.
BECAUSE the Lines AC and eb are both
perpendicular to the Line De, &c. the Tri-
angles DAC and Deb are fimilar, whence
DA AC. but De is eb by
Conſtruction, therefore DA is AC. confe-
quently AB DA=BD.
:
De eb
-
:
Q; E. D.
FARTHER, to fhew the Utility of this Me-
thod of Sir Ifaac. Newton's, take another
Example; and the better to conceive the Rea-
fon of exprefling the half Sum and half Dif-
ference of any two Quantities, obſerve the
following
LEMM A.
IF from half the Sum of any two Quan-
tities be taken half their Difference, the Re-
mainder will give the leffer Quantity; but if
added to the half Sum, gives the greater
Quantity.
DEMONSTRATION.
X
PUT the greater Quantity, and y =
the lefs; and make their Sum xys,
and Difference x-y=d. The half Sum will
be
[82]
be
x + y
and half Difference
2
2'
x
y
2
d
Bla
x + y
Now
x + y
d
2
2
2
2
=y. and
x + y
+
x y
d
+
x.
2
2
2
2
EXAMPLE III.
Q. E. D.
IN the right angled Triangle ABC (See
Fig. 12.) there is given the Hypothenuſe
AC = a, and BD, the Perpendicular let
fall from the right Angle to the Hypothenufe
AC, b. To find the Legs AB and BC;
and the Segments of the Hypothenufe AD
and DC?
To folve this Example by the common
Method put x = one of the Segments, as
ſuppoſe the greater, and (per Ax. 1.) the leffer
will be = a
X.. Now, by the eighth
Propofition of Euclid's fixth Book, the Tri-
angles ADB and BDC are fimilar.
Whence
[83]
{
!
2
Ergo,
3 ax
3 +
4 CO
5 ww 2
Whence
That is,
| AD: BD
::
2x : b :
2 x : b:: b: a
BD DC.
x.
}
SAn Equation of 2
the third Form.
}
x² = b²
b2{
2
4x
ax=
b2
a²
2
a
5x
2
ax +
62
4
4
6
x
2/2
a
11
2
b2.
6 +
012
a
X
-
4
2
a
b², = AD or
2
4
1
}
DC, which foever is the longeft. Having thus
found the Segments of the Hypothenufe, the
Legs AB and BC will be found by Virtue
of the 47th Propofition of Euclid's firft Book;
for in the Triangles ADB and BDC there
is given the Legs AD, DB, and DC.
BUT if x be put for the leffer Segment,
then ax will be the greater, and feeing
the Expreffion is the fame in both Suppofi-
tions, that is, ax reprefent indifferently
the greater and lefs Segment, it is evident
this Example ambiguous, and therefore
fince the Sum of the Segments are given,
repreſent their Difference, and (per Lem-
let
ma)
[ 84 ]
ma) the greater Segment will be
a
x
a
012
+212
and the leſs ——-*. And ſo by the above
2
2'
cited Propofition of Euclid, we ſhall have this
Equation, IAD: BD:: BD: DC.
That is,
2
2 Ergo,
3
X
4
4
士
​5
พ
2
w
45
212
a
а
2
4
2
2
a
+
:b: :b:
2
2
λ 2
= b²
2
4
x 2 --
4.62
5,x² = a² 4b2
1
2
N! X
6 x = √ a² = 462. the Difference
of the Segments of the Hypothenuſe AC, which
а
x
a
X
being ſubſtituted for x in
2
2
+ and
2
2
a
2
.b².
2
the greater Segment will be equal to to
and the leffer =
A
2
4.
a
4
-b². which exactly
agrees with the 7th Step of the laſt Operation.
Hence it appears, that by fubftituting the Va-
lue of x found by this Method of Computa-
tion, for x in the Quantities expreffed by the
half. Sum and half Difference, gives the very
fame
[85]
|
>
Theorems for finding thoſe Quantities, as tho'
they were inveſtigated by the common Me-
thods of Algebra.
GEOMETRICALLY.
ON AC draw the Semicircle AbBC, draw
alfo the Line ac parallel to AC at the Dif-
tance of the given Perpendicular BD, and
where the faid Line ac cuts the Semicircle, as
in the Point B or b, let fall the Perpendicular
BD or bd; this done, draw the Lines AB and
BC, or Ab and bС. and ABC or AbC is the
Triangle required. The Truth of this Con-
ſtruction is evident from the Figure itſelf,
and therefore needs no Demonſtration.
HAVING thus fhewn what Quadratic ad-
fected Equations are, and how they are
brought to a Solution;
Solution; with fome ufeful
Animadvertions on every Cafe, wherein, I
prefume, is fpecified all that is requifite for
finding that Root of an Equation that truly
anfwers the Queftion: I will now conclude
this Synopfis, and fo proceed to give the So-
lutions to the following Problems.
I
PROBLEM
[86]
PROBLEM XXXIII.
Suppofe two Towers, AB 180 Feet high,
and CD 240, at the Distance AC 360
Feet. A Ladder is to be fet upon the
Line AC, at fome Point, fuppofe in E,
of fuch a Length, as from thence it
may reach the Top of both the Towers.
We require the Point E in the Line of
Distance, as alfo the Length of the
Ladder EB, ED?
=
PUT 180 a, 240 b, 360c, and
x = AE (See Fig. 13.) and (per Ax. 1.) c X
EC; but becaufe BE is equal to DE, by
the 47th Propofition of Euclid's firft Book,
we fhall have AB² + AE² = EC² + DC².
which in
Species will
itand thus,
I
2
2
I a
+x²=c²—2cx-|x²+b².
४
2
2cx +
2 a² = c²
+2cx32cx + a² = c² +
2
X
2
3
a.2
- 20 5
420x c² + b²
c²~—b² a²
b2
b 2
a2
2
=215=AE,
20.
which taken from 360 leaves 145 =
which gives the Point E required. Whence
EC,
the
[ 87 ]
2
the Length of the Ladder will be found to
contain 280.4 Feet.
W. W. F.
Conftruction Geometrically.
JOIN the Points B and D with a right
Line, which divide into two equal Parts in
F, from which Point raiſe the Perpendicular
FE, and where it cuts the Line AC, as in the
Point E, there the Ladder muſt be fet to
reach to the Tops of both Towers.
For BF is
= FD, and FE common to both, therefore
the Triangle BED is Ifofceles, and confequently
BE is equal to ED.
2. E. I. and D.
PROBLEM XXXIV.
In the Triangle ABC, the feveral Sides
AB = 13, AC
AC = 14, BC = 15 are
given; and the Perpendicular BD being
drawn. Requir'd the Segments of the
Bafis AD, ÓC?
PUT 13a, 146, 15=c, and x =
the Segment BD, then (per Ax. 1.) cx is
equal the other Segment DC; and becauſe the
Triangles ADB and ADC are right angled,
and the Perpendicular AD common to both,
I 2
we
[88]
we ſhall have (by Virtue of the 47th Propo-
fition of Euclid's firft Book, (See Fig. 14.) this
a
2
Equation 1AB-BD-AC-CD', which in
Species is 242x² = b² — c² + 2cx — x².
+ x² 3 a² = b² — c² + 2cx
3-b² + c² 4/2cx = a² = b² + c²
2
2 }
2
4 ÷ 205x=
205
b² -+- c²
20
=6.6=BD,and
156.6 8.4 the other Segment DC.
2. E. I
OR, by the Application of the 36th Pro-
pofition of the third Book of Euclid's Ele-
ments, the Segments DB and DC may
be found by Proportion, thus as BC AB
+AC:: AC-AB: DC- DB, that is, as
15: 13 + 14 (= 27.): : 14 13(1.)
:(DC- DB.) Now the Half of
added to half 15 gives 8.4 DC the
greater Segment, and fubftracting half 7 from
the Half of 15, the Remainder gives 6.6
DB the leffer Segment, the fame as before.
W. W. D.
27
=
SCHOLIU M.
2
I
FROM the foregoing Operation it may be
obferved from the third Step, that by tranf-
pofing c² to the other Side the Equation, we
fhall
[89]
fhall have a² + c
2
= b² - 2cx. that is
+
AB² + BC² = AC² + 2 x BC × BD. an
Equation that exactly agrees with the 13th
Propofition of the fecond Book of Euclid;
whence the Learner is to take Notice, that by
carefully looking into the Calculations of Al-
gebraical Problems, many curious and uſeful
Theorems are often difcover'd, and fometimes
new Properties of the Figure the Calculus
belongs to, is brought to Light; and to this
Property of Analytical Operations, both
Arithmetic and Geometry chiefly owe their
greateſt Improvements.
PROBLEM XXXV.
In the obtufe angle Triangle DEF, the feve-
ral Sides are given, viz. DE 11, EF 13,
DF 20;
and the Perpendicular FG,
being let fall upon the Bafis produced.
Required the Prolongation of the Ba-
fis EG.
PUT 20 = a, 13 b, 11 = c, and ✰
= EG the Prolongation of the Bafis DE;
then (per Ax. 2.) c + x = DG, now becauſe
the Triangles DGF and EGF are right an-
gled, and the Perpendicular FG common to
both, it is evident (by the 47th Propofition
of Euclid's firſt Book) that DF2
2
= EF² EG2. that is,
I 3
DG 2
in
[ 90 ]
1 2
I
in Species [1a2
+ x²/2a²
+
-
3
204x=
(See Fig. 15.)
2
c² - 20.x-x² = b² — x²
2cx = b²
2
± 32cx = a² — c² — b²
2
2
b= = a²
20
C²
2
=5=EG.
2. E. I.
SCHOLIU M.
AT the fecond Step by tranfpofing - c²
- 2cx to the other Side of the Equation, it
will be a² = b²+c22cx, the fame with
the 12th Propofition of Euclid's fecond Book
of Geometry.
GEOMETRICALLY.
UPON the Side DF draw the Semicircle
DGF, and produce out the Side DE till it
cuts the faid Semicircle in the Point G, then
the Segment EG will be the Prolongation
required.
DEMONSTRATION.
DRAW the Line FG, and (by the 31ft
Propofition of Euclid's third Book) the Angle
DGF is a right Angle, confequently the Line
FG is perpendicular to the Side DE produc'd
to
[91]
to G. Therefore EG is the Prolongation
fought.
Q, E. D.
PROBLEM XXXVI.
In the Rectangle ABCD, is given the
Difference between the Length AB and
the Diagonal BD, that is DE = 2;
and likewife the Difference between the
Breadth ÁD and the Diagonal BD,
that is, FB 9. Required the Sides
of the Rectangle AB, AD?
PUT 9a, 2b, and x = AD=DF;
then DB = a+x, and AB = EB = a + x
b. Now the Triangle ABD being right
angled, there will be had (by the 47th Pro-
pofition of Euclid's firft Book) this
(See Fig. 16.)
Equation, | 1| DB² = AB² + AD².
That is, 2 a2+2ax+x²= x²+a² +
2
3
3x2
2ax-2ab+x²
2bx 2abb²
2
2bx+b².
C4x² - 2 bx +b² = 2ab
4 ww 25x -b=√zab.
5 + 6/6x = √2ab. -+ b = √2 × 9×2.
+6.
+2=8= AD, the Breadth of the Rect-
angle,
[ 92 ]
angle, which being known, the Length AB
will be found = 15.
Q, E. I.
OR put xEF; then (by the Figure)
a + x = AB, b+x= AD,
+x
x
and
AD,
and a +6
BD, which being ſtated as above di-
rected, it will
ftand thus a²+2ab
+2a
2
+2b
x+b² +²x²
a² +
+ 20
十
​+- x²
2
X
2
b² -|- 26
+
2
X
I
士
​४
2
2ab
2 WJ 2
131x
=√2ab. =√2×9×2.=6=EF,
whence the Side AB = 15, AD = 8, and
BD = 17.
W. W. R.
SINCE the End and Defign of Algebraical
Solutions is to bring out the moſt elegant and
fimple Equations poffible; and becauſe ſome
Ways are much neater than others; where-
fore (fays the great Sir Isaac Newton, in his
Univerfal Arithmetic, Page 98) if the Method
you take from your firft Thoughts for fol-
ving a Problem, be but ill accommodated to
Computation, you muſt again confider the
Relation of the Lines, until you have hit on
a Way as fit and elegant as poffible. For
thofe
[ 93 ]
X
21
<
thoſe Ways that offer themſelves at firſt
Sight, may often create fufficient Trouble if
they are made Ufe of. Thus, it would not
have been more difficult to have fallen upon
the following Method than upon one of the
precedent ones.
= DB,
For having put x =
b, and AD = x
X
whence, by the Property of the Figure or
Triangle ABD, this Equation will be
then AB
a;
obtained 1x2=x²-2bx+b²---x²-2ax+a³.
+19
I
2
3
W
2x2
2bx
24x
a 2
-62
x²-2bx2ax + b²+2ba+a²=2ab
3/x
241 x
a-b
√zab.
4 +a+65x==a+b+√2ab.=17=DB, &c.
HENCE, by computing after this Rate, I
have fallen into larger and more perplexed
Algebraical Terms, than either of the former
Methods.
BUT of all theſe three Ways of Computa-
tion, the fecond Operation is much more fim-
ple than either of the other two; for in thofe
Calculations their refulting Equations come out
in an adfected quadratic Form, whereas this
produces a fimple Quadratic only. From
which may be drawn the following
SCHO-
[ 94 ]
1.
SCHOLIU M.
2
SEEING, at the fecond Step, there comes
out this Equation x² = 2ab, which turned
into an Analogy, will be 2a:x:x:b, or
2b:x::
2b: x: xsathat is, 2FB: EF:: EF: DE,
or 2 DE: EF : : EF : FB, whence x or EF is
a mean Proportional between 2a and b, or
2b and a; or between 2 FB and DE, or
2 DE and FB. Hence it will not be difficult to
conftruct this Problem Geometrically, which
I ſhall leave to exerciſe the ingenious A-
nalyſt.
PROBLEM XXXVII.
3
In a Rectangle DEFG, the right Line
DK is drawn from the Ang e D to the
oppofite Side, cutting the Diagonal EG
at right Angles in H: And there is given
the Segment HK 2, and HE
= 16. Required the Sides of the Rect-
angle?
To folve this Problem put 16 = a, 2 =b,
and x = GH; and becauſe the Triangles
KGD and GDE are right angled, and DK
and GE cut each other at right Angles in H
by the Queſtion. Therefore by fimilar
Tri-
[ 95 ]
*
Triangle {
That is,
(See Fig. 17)
1 KH: GH::GH: HD.
2 GH: HD:: HD: HE. S
3b:x:x:
4x:
X
b
Ergo, 5 ax =
2
X
4
b2
X
xb² 6 ab² x=
4
5
7
w 38 8 x=
x 7 x
3
ab 2
*
b
2
2
X x2
b
: a.
=³√ab². =√16 × 2 × 2. =4.
having thus found GH = 4, DH will be
given 8; whence (by the 47th Propoſition
of Euclid's firſt Book) the Length PE will be
= 18.02; and the Breadth DG = 8.75,
the Sides of the Rectangle.
W. W. F.
OBSERVATION.
THIS Problem, if rightly confider'd, is
no other than the finding two mean Propor-
tionals between two given Quantities; for fee-
ing HK HG :: HG: HD:: HD: HE;
it is evident the Lines HG and HD, are two
mean Proportionals between the given Lines
HK and HE. Put the firſt Mean, and,
x
from
**
[ 96 ]
from the Laws of Geometrical Progreffion,
202
the ſecond will be =
; having thus obtain-
a
ed the two Means in Species, multiply them
203
together, and their Product will be which
a
is equal to the Rectangle of a into b; that is,
X3
a
=ab. This Equation multiplied by a, be-
3
2
comes x = a²b, and extracting the Cube
Root x = √ab. the fame as before. But if
it be required to find three mean Propor-
tionals between a and b ; and putting x =
Mean, the ſecond will be
X
x²
a
X
firft
and the third
or into itſelf, and
a
2
= multiply x by 2, or
2
a23
which is a x b; that
a
there will be had
X4
is,
а
a===
x²
a
ab, which multiplied by a², x²=a³b,
and extracting the fourth Root,
Root, or the
fquare Root of the fquare Root,
we fhall have
x = 4√a'b, or x = ✔✔ab.. Or if it were
required to find four mean Proportionals be-
tween a and b, then if x repreſents the firſt
Mean,
[ 97 ]
20
2
Mean, the ſecond will be, the third
the fourth
4
23
205
3
a
is
a
3
x
a
2 >
and
and the Form of the Equation
ab; which reduced, and the Root
3
Nab.
√a*b.
extracted, gives x
Now from
thefe Equations x³ a²b, x4 a'b, and
a¹b, it appears, that x, the firſt mean
Proportional, will always be raiſed to a
Power, whofe Index is the Number of Means
fought more 1, and a, the firft Term of the
Series, will be raiſed to that Power whofe In-
dex is the Number of Means required.
Hence, putting n any Number of mean
Proportionals between a and b, we fhall have
this univerfal Theorem for folving all Quef-
N+I
aions of this Nature, viz. √a^b. but to ef
fect the fame Logrithmetically, put : a
1:a =
Logrithm of a, and 7: b = Logrithm of b,
1:
and we ſhall have this Logrithmetical Theo-
n x l : a
nxl:a+1:6
rem, viz.
n +
>
I
either of which
Theorems gives the firft mean Proportional of
any Geometrical Series whatfoever. But the
latter, for it's Facility and Expedition in Cal-
culation, is far preferable to the former.
K
BECAUSE
[ 98 ]
"BECAUSE moft of our Writers of Arith-
metic have only fhewn how to extract the
quare and cube Roots of Numbers, it may
not be improper, in this Place, to give the
Reader a Rule by which all Roots may be
extracted univerfally, taken from Sir Ifaac
Newton's Arithmetic, Page 31.
RULE.
EVERY third Figure beginning from Unity
is first of all to be pointed, if the Root to be
extracted be a Cubic one; or every fifth, if it
be a Quadrato-Cubic, or of the fifth Power,
&c. and then fuch a Figure is to be writ in
the Quotient, whoſe greatest Power (i. e. whoſe
Cube, if it be a Cubic Power, or whofe Qua-
drato-Cube, if it be the fifth Power, &c.)
fhall either be equal to the Figure or Figures
before the first Point, or the next lefs; and then
having fubftracted that Power, the next Figure
will be found by dividing the Remainder aug-
mented by the next Figure of the Refolvend, by
the next greatest Power of the Quotient, multi-
plied by the Index of the Power to be extracted,
that is, by the triple Square of the Quotient,
if the Root be a Cubic one; or by the Quin-
tiple Biquadrate, i. e. five Times the Biqua-
drate if the Root be of the fifth Power, &c.
And having again fubftracted the greatest Power
of
[ 99 ]
of the whole Quotient from the first Refolvend,
the third Figure will be found by dividing that
Remainder augmented by the next Figure of the
Refolvend, by the next greatest Power of the
whole Quotient multiplied by the Index of the
Power to be extracted; and fo on in infi-
nitum.
2
NOTE. When the Biquadratic Root is to
be extracted, you may extract twice the fquare
Root, for the fquare Root of x is x², and
the fquare Root of x² is x. And when the
Root of the fixth Power is to be extracted,
you may first extract the fquare Root, and
then the Cube Root out of that ſquare Root, for
the fquare Root of x is x, and the Cube
Root of x³ is x.. And the fame is to be obſerved
in other Roots, whofe Indexes are not prime
Numbers.
6
PROBLEM XXXVIII.
Let there be a Circle, whofe Diameter is
AB, with another lefs Circle whofe Dia-
meter AC, touches within in A: and
from the Center of the greater Circle
D, draw the Radius DE at right An-
gles to AB, cutting the Periphery of the
leffer Circle in F. Now there is given
K 2
BC
[100]
5..
BC (the Difference of the Diameters)
9, with the Segment EF
Required the Diameters AB, AC of the
Said Circles?
BECAUSE the Circles AEB and AFC touch
each other within, the Center of the leffer Cir-
cle (by the 11th Propofition of Euclid's third
Book) will be in the Diameter of the great
Circle drawn from the Point of Contact.
Wherefore putting x = DB DE DA
the Radius of the greater Circle, 9a, and
5b; then (per Figure) DE=x-b, and
DC=X a; now (by the 13th Propofition
of Euclid's fixth Book) it
2 ♡
will be
(See Fig. 18.)
ADXDC=DF2. which exprefs'd
2bx+b².
2
in Species 22 — ax = x² —
x23
ax=2bx+b²
+2bx42bx ax = b²
b² 2
3
5 X5
42b-a5x=
=25
26
a
2X5-9
DB. Confequently AB the Diameter of
the greater Circle is 50, which being known,
the Diameter AC of the leffer Circle will be
=41 (509.)
2 E 1.
Or Thus.
FROM this evident Principle, that the Dif
ference of the Halves of any two Quantities
is
[101]
is equal to the half Difference of the Quantities
themſelves; it follows, that PD, the Diffe-
rence between the two Radii AD and AP, is
a
equal the half Difference of the Diameters
2
AB and AC. Whence putting x AP the
Radius of the leffer Circle, then AD- DE
1!
= DB = ≈
B
十
​DC
५
DE
a
a
2 and DF
2
FE =x+2 - b; which, by
2
the aboveſaid Propofition of Euclid, will be
brought to an Equation in the very fame
Manner as Step the fecond of the preceding
Operation ſpecifies.
N}
Thus,
ADXDC=DF. which expreffed
in Species, will become
2x
2
a²
4
a
a
2
(=x+-xx- =x
+ax-2bx+
±32bx-ax=
2
·ab+b².
ab+b²,
a
4
a²
2
2
·ab-fb²
2
3÷2b-a4x=-
2
b2
26
a
26
a
2
Q 2
+
20.5= AP, which being known, the Dia-
K 3
meter
[102]
I
meter AC will be twice 20.5, that is,
and 419
as before.
41.
50 the Diameter AB the fame
W. W. D.
PROBLEM XXXIX.
Two Companions have got a Parcel of
Guineas; fays A to B, if you will give
me one of your Guineas, I shall have as
many as you will have left. Nay, re-
plies B, if you will give me one of your
Guineas, I fhall have twice as many as
you will have left. How many Guineas
had each of them?
PUT x A's Number of Guineas, and
the Number of Guineas B had will be x
+2; and by the Condition of the Queſtion,
we fhall obtain this.
Equation, 1x+3=2x-2.
I
x23=x
2
2 + 23 x = 5, hence A had 5 Guineas
and B had 7.
Q, E. I.
PROBLEM
1
*
[ 103 ]
PROBLEM XL.
A certain Perfon bought two Horfes, with
the Trappings, which coft 100 Pounds;
which Trappings if laid on the first
Horfe A, both Horfes will be of equal
Value: but if the Trappings be laid on
the other Horfe, he will be double the
Value of the firft. How much did the
Said Horfes coft?
To folve this Problem put 100 a, the
Value of the fecond Horfe and Trappings
together, and (by Ax. 1.) the first Horfe will
bea-x; but by the Queftion, double
the Value of the firft Horfe is equal the Price
of the fecond Horſe and Trappings together;
whence this Query will be reduced to the fol-
lowing Equation, which being stated will
ftand
Thus, Ix = 2a.
I + 2x2 3x = 2a
2 ÷
313
༡༩
2X.
2a
2 X 100
= 663 Pounds,
3
3
and 100 6612 = 33 Pounds the Price of
the firſt Horſe; from 66 take 33 and the
Remainder is 33, which divided by 2 the
Quo-
[ 104 ]
Quotient is 16
Pounds the Price of the
Trappings, and the Sum of 33 and 163 is
Pounds what the ſecond Horſe coſt.
50
NUMERICALLY
To folve this Problem by Numbers, rea-
fon thus; fince the Trappings laid on the firſt
Horſe makes both the Horſes of equal Value,
and becauſe both the Horfes and Trappings
together coft 100 Pounds, the fecond Horfe
confequently muft coft 50; but (by the Quef
tion) if the Trappings be laid on the other
Horſe, he will be double the Value of the
firft, therefore the firft Horſe coſt of 100
which is 33 Pounds; now take 33 from
50, or 50+ 33 from 100, the Remainder
will be 163 Pounds, the Price of the Trap-
pings.
W. W. D.
PROBLEM XLI.
A Vintner has two Sorts of Wine, viz. A
and B: which if mixed in equal Parts,
a Flaggon of mixed will coft 15 Pence;
but if they be mixed in fefqui-alter Pro-
portion, as if you should take two Flag-
gons of A as often as you take three of
B, a Flaggon will coft 14 Pence. Re-
quired the Price of each Wine fingly?
SEEING
[105]
جام شود
;
SEEING by the former Part of the Quef-
tion, a Flaggon of Wine of equal Mixture
coft 15 Pence, two Flaggons of the fame
Mixture will coft 30 Pence; and becauſe, by
the latter Part of the Queftion, two Flaggons
of the Wine A mix'd with three Flaggons of
the Wine B coft fourteen Pence a Flaggon, it
is evident that five Flaggons of this Wine
comes to 70 Pence. Put 30= a, 70=b,
and the Price of a Flaggon of Wine A,
and a Flaggon of the Wine B will coft a-x
Pence; and fince the Sum of the Prices of
two Flaggons of A or 2x, and three Flag-
gons of B or 3a
34 3x, added together, is
equal to 70 Pence, or b,
x
Therefore 12x
1/2x + 3a
3a
3x=b.
I 3x22x + 3a
+
3a =
b
b+3x
2
b
33x
за
b + 2x
3 2x 4x=3a — 6 = 3 × 30~70=20
Pence, which taken from 30 leaves 10 Pence,
the Price of a Flaggon of the Wine B.
W. W. R.
I HAVE hitherto given the Arithmetical So-
lution to many of the foregoing Problems,
and fhall obferve the fame Order in fome of
the following Propofitions alfo; purely to
fhew the young Analyft, that fome Queſtions
that
[106]
that are propoſed in Algebra, may, by ma-
ture Confideration, be folved by Arithmetical
Principles only.
HENCE, in this Queftion, it will be no
hard Tafk to find the Price of a Flaggon of
each Sort of Wine independent of Algebra,
and that too, even by the moft fimple Laws
of Arithmetic. Thus, becauſe the Wine when
mix'd with equal Quantities of each Sort coft
15 Pence a Flaggon, it is evident that two
Flaggons of A mixed with two Flaggons
of B will coft 60 Pence; and when mix'd
in fefqui-alter Proportion, it is plain that five
Flaggons of fuch a Mixture, at the Rate of
14 Pence the Flaggon, will amount to 70
Pence: now, fubftracting 60 from 70, that
is, fubftracting the Price of two Flaggons of
A mix'd with two Flaggons of B, from the
Price of two Flaggons of A-mix'd with three
Flaggons of B, the Remainder is ten Pence.
the Price of a Flaggon of the Wine B,
whence a Flaggon of the other Sort of Wine
A will coft twenty Pence. The Prices re-
quired. Thus you fee, that by a right Way
of thinking, fome Things that may feem dif-
ficult to Algebraic Computation, may be
fòlved by the eafieft Rules in Arithmetic.
PROB-
I 107 ]
PROBLEM XLII.
A Son afk'd his Father how old he was;
his Father anfwer'd him thus; If you
take away 5 from my Years, and di-
vide the Remainder by 8, the Quotient
will be of your Age; but if you add
2 to your Age, and multiply the Whole
by 3,
and then fubftract 7 from the
Product, you will have the Number of
Years of my Age. What was the Age
of the Father and the Son?
PUT the Father's Age, and (by the
X
Queſtion) the Son's Age wil be
or
8
5
× 3
315 ; but by adding 2 to the Son's
Age, and multiplying that Sum by 3, and 7
fubftracted from the Product; this laſt
Remainder is equal the Father's Age,
or x.
Therefore,
I
2 ♡?
3
19** 53
8
x.
+8 829x ·53=8x
8x+53
5339x8x + 53
8x4x53 Years, the Father's Age.
Hence
[ 108 ]
t
Hence the Son's Age will be found to be 18,
the Anfwer which was required.
PROBLEM XLIII.
To find out two Numbers, to the Sum whereof
if you add 6, the Whole fhall be double
the greater; and if you fubftract 2 from
their Difference, the Remainder will be
half of the least.
X
PUT the greater Number, and y the
leffer; which being ſtated according to the
Conditions of the Queftion, will ftand
9x+y+6 = 2x.
Thus,
I
X
2
X 2
1216
32x
2 =
212
2
I-
5
X
2y4=y
4y=x-6. Subftitute x-6
for y in the third Step, and
58 X 6
+66x14 the greater Number,
and by the fourth Step, the leffer will be found
8. Which was to be fought.
=
NOTE. This Method of exterminating
Quantities by fubftituting their Adequates for
the
}
[ 109 ]
the Quantities themſelves in fome other Equa
tion, is called fubftitutive Algebra. And if
the Learner defires to fee more of this Kind
of Calculation, let him read Sir Isaac Newton's
Univerfal Arithmetic from Page 60 to 66;
where he will meet with the beſt Inftructions
of this Kind.
PROBLEM XLIV.
To find two Numbers, the Product whereof
is 240, and the Triple of the greater
divided by the less is 5.
PUT 240 a, 5 = b, x = the greater
Number, and (per Ax. 3.) the leffer Number
a
will be = The Triple of the greater
Number is 3x, which divided by the leffer
the Quotient is
is equal to b.
Whence,
I
3
a
I
2
2
3x²
'
a
3x2
♡
x
a
2
which, by the Queſtion,
= b.
ab
1
2
W 2
3 x
18/
11
√240 X 5:
3
L
=20
20
the
[110]
the greater Number, and the leffer will
240
be
12.
20
Q. E. I.
PROBLEM XLV.
Two Men have a Mind to purchase a
Houſe rated at 1200 Pounds; fays A to
B, if you will give me of your Mo-
ney, I can purchase the House alone,
but fays B to A, if you will give me
of yours, I shall be able to purchase the
Houfe. How much Money had each of
them?
PUT 1200a, x= A's Money, and of
B's Money will be ax, to which add ½
itfelf, and the Sum is
за
3x
2
the Money that
½
B had; now, by the Queſtion, of A's-Mo-
ney added to B's Money, that Sum will pur-
chafe the Houfe, whence it is plain
that
1
~
♡
I
X 4
2 +
3 ÷
3*
+
4
ба
3x
за
2
3x
= a.
3x=4.a
=20
24 1200 X 2
4x=
3
3
=800 Pounds
A's
[ 111 ]
A's Money, hence B had 600. for of 600
of 800 added to 600,
added to 800, or
either Sum will be 1200.
W. W. F.
PROBLEM XLVI.
Some young Men and Maids had a Rec-
koning of 37 Crowns to pay for a Treat,
and this was their Conditions, that eve-
ry young Man ſhould pay three Crowns,
and every Maid two. Now, if there
had been as many young Men as there
were Maids, obferving the fame Condi-
tions, the Reckoning would have come to
four Crowns less than it did. How ma-
ny young Men and Maids were there?
3.x
PUT 37 = a, x = the Number of young
Men, and the Number of Crowns they paid
is 3x, whence, according to the firft Condi-
tion of the Queſtion, the Maids paid a
Crowns, and becauſe they paid two Crowns
a piece, the Number of Maids will be expref-
fed by this Fraction
a
2
3*
Now, by the fe-
a
3*
cond Condition, there being fuppos'd 2
young Men, and x Maids, and the Men pay-
L 2..
ing
༈
[112]
ing three Crowns each, and the Maids two,
the Number of Crowns the Men paid will be
за
2
9%
and the Maids 2x Crowns, the
Sum of which, viz.
за 9x
+2x, by the
2
Queftion, is equal to a — 4.
Therefore, I
за
9 x
2
+2x= a -4°
.I
× 223α
за
5x=2a
- 8
2
+35:
35x = a +8
a + s
37 +8
3
4.
9. the
5
5
Number of young Men, hence the Number
of Maids will be found to be 5. for 9 × 3
+5×2 = 37, and 5x3 + 9×2 = 33
2. E. Ï.
37 - 4.
37
PROBLEM XLVII.
A General, who had fought a Battle,
upon reviewing his Army, whofe Foot
was thrice the Number of his Horfe,
finds that before the Battle 120 of
his Foot had deferted, and of his Horfe
+120, befides of his whole Army
were fent into Garriſons (reckoning the
Sick
[ 113 ]
1
Sick and Wounded) and of his Army
remained; the reft, who were wanting,
being either flain or taken Prifoners;
now, if you add 3000 to the Number of
the Slain, the Sum will be equal to half
the Foot he had at the Beginning. What
were the Numbers of each?
For the Number of Horfe put x; then 3x
is equal the Number of the Foot, and 4x will
be the whole Army; the Number of the Foot
120, and of the Horfe
deferted
3x
12
120, and likewife 4 Part of the whole
at 120,
20
now the Sum
Army fent into Garriſon is x;
of the Horfe and Foot together with
the whole Army is
Part of de
Ser
3x
X
+
+x, or
13x tod
;
12
20
ΙΟ
I2X
8
8
12X
142
or
8
5
to which add of the whole Army, viz.
and the Sum will be +
13x
10
which Sum taken from 4x (= the
my) the Remainder will be 4x
whole Ar-
14x
or
5
6x
equal the Number of the Slain or taken
5
Pri
1
[114]
A
Prifoners.
Whence, by the Queſtion,
6х
اق
5
3000 is equal to half the Foot he had at
the Beginning.
6x+15000 =
6x
Therefore,
+ 3000 =
3*
5
2
15x
IF X5 2
2
2
3-
4
x2312x + 30000 = 15×
12x43x= 30000 ·
3
5x 10000 the Number of the
Horfe, and the Number of the Foot will be
found to be 30000, whence there was 12000
flain or taken Prifoners, and the whole Army
he had at first was 40000 Men.
W. W. F.
PROBLEM XLVIII.
To divide 100 twice into two Parts, fo
that the major Part of the first Divi-
fion may be treble the minor Part of the
fecond Divifion; and the major Part of
Second may be double the minor Part of
the first..
PUT
[ 115 ]
$
x
PUT 100= a, the major Part of the
first Divifion, and (per Ax. 1.) the minor Part
will be a x; and, by the Queſtion, the
major Part of the fecond Divifion will be
=20
= --
2x, and minor Part = 7. Now the
3°
Sum of the two Parts of the fecond Divifion
(per Query) is equal to a, whence this Prob-
lem will be reduced to the following E-
quation,
Viz. I 2a 2x+ =a.
X326a-5x=3a
35x=3a
35%
X
3
I
за
2 ±
за
за 3X100
354 x
=60 the major
5
5
Part of the first Diviſion, which being
known, the minor Part will be
=
100 60
40; hence the major Part of the ſecond
Divifion will be found to be 100
=80, and the minor Part = 100
60
3
60×2
80 (or
) 20, the Divifions which were
quired.
re-
t
PROB-
[116]
/
PROBLEM XLIX.
To divide 30 twice into two Parts, fo that
the major Part of the first Divifion
with the minor of the fcond may be 333
and the Sum of the minor Parts fub-
Atracted from the Sum of the major,
may leave 14 remaining.
PUT 30= a, 14 b, 33 = c, x
x== the
major Part of the firſt Divifion, and (per
Ax. 1.) the minor Part will be ax; but
in the ſecond Divifion the minor Part will be
= cx, and the major Part = a - c
c+x.
Having thus exprefs'd the Parts of both Di-
vifions, the Sum of the major Parts is 2x+a
-c, from which take the Sum of the minor,
viz. a + c = 2x, and (by the Queftion) the
Remainder is equal to b; whence there will
be obtained the following Equation.
Viz.
114x20=b.
I
+2c24x=b+2c
b+2c
2
X
14+2×33—20
4
4
which
the major Part of the first Diviſion,
taken from 30 there remains 10
nor; and again take 20 from 33,
the mi-
the Re-
mainder
[
17 ]
mainder, viz. 13 gives the minor Part of the
Second Divifion, which Part taken from 30
leaves 17 the major, which was to be di-
vided.
17=
PROBLEM L.
A Man, his Wife, and his Son's Ages make
up 96 Years, ſo that the Husband's and
and Son's Years together make the Wife's
+15; but the Wife's and the Son's
make the Husband's 2. What was
the Age of each?
FOR the Hufband's Age put x, for the
Wife's put y, and for the Son's put z.
Then S
I
4 +
5
I
ENN & W
7
+y
8÷2
2
1x+y+2=96.
2x+2=y+15. by the Queftion.
3y┼x=x+2.
4x=96—x—2
52x=94
6x=47, Years, the Huſband's Age.
7y=96-y-15
8 2y=81
9y=40, the Wife's Age.
6+9 10x+y=87 /
1—10 (11| 2=8 the Son's Age.
2. E. I.
The
[ 118 ]
~
&
The fame otherwife.
x
THUS, put the Huſband's Age as be-
fore, and (per Ax. 1.) the Wife's and Son's
Ages together will be 96 -x; which, by the
Queſtion, is equal to the Hufband's Age
2. Whence we have this
Equation, 1196x=x+2.
I
2
22x94
131x = = 47 Years, the Hufband's
2131x
Age. Hence 96 - 47 is equal to the Wife's
and Son's Ages together, that is, 49. Now
let x repreſent the Wife's Age, and the Son's
Years will be 49 xthen, by the former
Part of the Queſtion, this Problem will be
reduced to the following
Equation, 1147 +49
+22x = 81
I
x+15.
2 2/3/x = 40 Years, the Wife's Age,
and the Son's Age will be 96 47
428. the fame as before.
W. W. D.
PROB-
[ 119 ]
:
#
3
PROBLEM LI.
Three Merchants from three different Fairs
meet together at an Inn, where they
reckon up their Gains, and find them
the Sum of 780 Crowns. Moreover, if
you add the Gain of the first and fe-
cond, and fubftract the Gain of the
third from the Sum, there remains the
Gain of the first +82 Crowns; but if
you add the Gain of the fecond and
third, and from the Sum fubftract the
Gain of the first, there remains the Gain
of the third 43 Crowns. What was
the Gain of each?
each
To anſwer this Query, let x, y, and z re-
preſent each Merchant's Gain, then ftating the
Queſtion according to the Conditions thereof,
it will
Stand thus.
I
3
Ixy
+y+x=780.
2xty-z
=x+82.
+xxz-43.
342x823Z
2|5|22—698-
5
[120]
7
8
x
-349- 2'
ſubſtitute 349-
for z in the fourth
Step, and 72x=474+
812 ml a
3*
18
2
=474
x
*=474×- =
x
2
2
= 316 Crowns,
3
which being known, the third Merchant's Gain
will be found at the fixth Step to be 191 Crowns,
and the Sum of thefe two Merchant's Gains,
viz. 316 + 191 (= 507) ſubſtracted from
780 the Sum of all their Gains, gives 273
Crowns the fecond Merchant's Gain, which
was to be found.
NOTE, This Problem may be folved by
the Help of one unknown Quantity only,
which the Learner may try at his Leifure;
but that Method of Computation is too ab-
ftruce for young Beginners, in Propofitions
that are pretty much compounded, as this is.
And for this Reafon I have affumed Letters
for every Quantity fought, in fome Queſtions,
purpoſely to render the Calculus more intelli-
gible to the young Analyſt.
BROB-
1
[ 121 ]
#
{
บ
+
PROBLEM LII.
Three Perfons, A, B, C, owe a certain Sum
of Money, fo that A and B together owe
210 Crowns; B and C 290, and C and
A 400. What did each of them owe?
By obferving the Form of this Problem,
each Perfon's particular Debt may be readily
found by one unknown Quantity only. Thus,
put 2104, 290=b, 400=c, x=the Num-
ber of Crowns A owes, and (by Ax. 1.) a-x
will be Number of Crowns B owes, which
taken from b, gives b-ax the Number
of Crowns C owes; now the Sum of x and b
a+x is b-a+2x, which, by the Que-
ftion is equal to c, and will ſtand
ba+2x=c.
=b+ a
Thus,
I
1-b+ a22x=c-
c—b+a_400—290+210
2
}
2
2
160, the Crowns A owes, which taken from
210 leaves 50, the Crowns B owes, and 50
fubftracted from 290 there remains 240 the
Number of Crowns C owes, which was to be
found.
M
PROB-
[122]
1
PROBLEM LIII.
To find three Numbers, fo that the first and
half of the Remainder, the fecond and
of the Remainder, and the third and
of the Remainder, may always make 34.
FOR the Numbers fought put x, y, z. And
y + z
IX
=a=34.
2
x
per Query,
2
ルート
​+x
=a=34.
3
3x+
-ty
a=34•
4
I
2
хх
2
42xty-tx= 2a
207
×3 531 + x + z = za
X3
64x+x+y=4a.
3
X4
×42
4- 2X
Z7y
=20
2X
z, fubftitute
this for y in the
8.6a 5x = 2x=3a. Or 5x
92a
22
за
+2x=za
x+32=4a.
+32
5&6Steps
9
+x 1032 =
10
311Z
=2a
十七
​2a + x
Or
-
X
2ax which fubſtitute
3
for z in the
8 Step,
V
[ 123 ]
8 Step, 12
4a + 17 x ===
- за.
3
12
×3 13 40
17x=9a
5a
5×34
141715x=
10, the firſt
17. 17
1417x
13 4a 14 17x=5a
Number, whence the fecond Number will be
22, and the third 26.
Q. E. J.
PROBLEM LIV.
Let a Square be divided into 9 Small
Squares: We are to find and difpofe the
Numbers through the feveral fmall Areas,
jo that the Sum of every three, taken
either laterally or diagonally, may be al-
ways 15.
THIS Queſtion, or any other of the like
Nature, is only to be folv'd Mechanically, ex-
cept in thoſe Problems where the Numbers of
fmall Squares are odd, as in this we are about to
anfwer; then the middle Numbers will be
found Analytically, but the Pofition of the Reſt
Mechanically. Thus let the fmall Squares be
repreſented by the Letters a, b, c, &c. as the
Fig. Specifies;
ab
C
elf
gbm
M 2
A
Then
أم
[124]
Then by the
Queftion,
1\a+e+m=15
2b+e+b=15
3/ +e+g=15
1+2+3 4a+b+c+3ẹ + m + b + g
45. Again,
5a+b+c=15
per Query,m+b+8=15
5
4
8
+6 677a+b+c+m+b+8=30.
+1小
​7 8
зе
15
÷ 3
9e5, the middle Number.
W. W. F.
Now to find the Pofitions of the other
Numbers Work Mechanically.
Thus, fet
down all the Numbers in their progreffive Or-
der, and they will ſtand
I 2 3
thus,
4 5
6
; and having
7
8
B
drawn the Square
ABCD fo as 1,
fall.
3, 7, 9, may
without,
1/2 3
A 4 5 6 c
78/9
D
Then
[ 125 ]
Then place I between 8 and 6,9 between 4 and
2, 3 between 4 and 8, and 7 between 2 and 6;
and there will come
2 7
out 9 5
I
: The Anfwer
4 3 8
which was required.
F
If the curious Reader defire to be more ac-
quainted with the Principles of Magic Squares
let him perufe Ronayne's Algebra.
LV. THEOREM.
Let any Numbers whatſoever be given, if
you fubftract every lefs Number from
that which is the next greateſt: I say,
that the Sum of thofe Differences is
equal to the Difference of the greatest
and leaft Numbers.
DEMONSTRATION.
LET the Lines fa, hb, Kc, md, and oe,
(See Fig. 19.) reprefent any Series of Num-
bers, all drawn from the fame right Line ae,
and Parallel to each other; and having drawn
the pricked Lines mr, Kl, hi, and fg, all pa-
rallel
M 3
[126]
rallel to ae, it is evident that oemd on, md
md=on,
Kc=nl, Kc—bb=li, bb — fa=ig, confe-
quently the Sum of theſe Differences, viz. on
+nl+li+ig is equal the Difference between
the greater Number or Line ce and the Num-
ber or Line fa, that is, oe- fa= on +nl
+litig.
PROBLEM
2. E. I.
LVI.
To find a Number, which being multiplied
by 6, and the Product fubftracted from
the Square of the Number to be found,
the Remainder will be 280.
x =
2
PUT 280 a, 6b, and the Number
to be found, whofe Square is x², and multi-
plied by b, the Product is bx, which Product
fubftracted from x2, the Remainder is equal
to a, by the Queſtion.
Therefore, 1x2
bx = a.
b²
b2
I CD 2'x²
bx +
4
4
b
bz
2
w 2 3x
vat
2
4
b
b
b² 6
6x6
3 +
4
trat~
+√280+
2
2
4 2
4
20, the Number which was to be found.
The
}
[127]
The fame otherwiſe.
2
THUS, let x + 3 be the Number fought,
whofe Square is x²+6x+9, and Product
when multiplied by 6 is 6x+18; whence
by the Queſtion
we have 1x2+6x96x-18=280.
That is, 2x²-9280.
2
2
+9;3x² = 289
3 w 214x=17, and 17 +3=20, the fame
as before.
W. W. F.
AND thus, by affuming a compound Quan-
tity for the Number to be found, the refulting
Equation of the above Calculus is reduced to a
fimple Quadratic, which the Learner is di-
fired to obferve.
PROBLEM. LVII.
To find a Number, which being multiplied
by 8, and the Product added to the
Square of the Number to be found, the
Sum will be 660.
PUT 660-a,8=b,and x=the Number fought,
whofe Square is x2, and multiplied by b, the
Product
[ 128 ]
Product is bx; whence by the Queſtion, we
fhall
have, 1x2 bx=a.
+
I CO2x² + bx +
23x+
b
2
w
2
b
b
3
4x
2
2
а
b2
2
a+
4
4
b²
4
b2
= 22 the
4
trat
Number which was required.
Or thus.
X
=
2
PUT 4 the Number to be found,
the Square of which is x²- 8x16, and 8
times x4 is 8x32; whence (per Query)
there will come out this,
Equation, 1x28x168x-32-660
That is,
2x²-16660
2 +16 3x²=676
ws
3
us 2 14x=26, and 26
=26, and 26 - 4
22 the
fame as above.
୧
W. W. F.
PROBLEM
[ 129 ]
PROBLEM LVIII.
To divide 140 into two Parts, ſo that the
Product of thofe Parts may the Square
of 56, that is 3136.
PUT 140=a, 3136=b, x= one of the
Parts, and (by Ax. 1.) a-x will be the other
Part; which Parts multiplied together, the
Product will be ax-x, which Product, by
the Queſtion, is equal to b.
That is, I'ax-x² — b.
1 + 2x²
I
士
​2
2 CO 3x²
3 w
IN
2
131x
14'x
a
ax +
12
ا{
ax=-
-b
a²
b
4
Ą
b.
a
140
b. =
2
2
4
140 X 140
+√
3136. =112, having thus
4
4 + 2/5/20
H
4
a
2
found one of the Parts, ſubſtract it from 140
and the Remainder is 28 the other Part.
2. E. F.
Or
[ 130 ]
Or thus.
PUT the Difference of the two Parts, and (per
a
X
Lemma) the greater Part will be, and the
leffer Part
faid Parts
a
2
2
a²
2
4
4
2
; and the Rectangle of the
or
a
2
Queſtion, is equal to b..
That is,
,
which, by the
4
Q2
2
b..
I
X
× 7/2/22
4
X
2
46
2.3x²
2
=a2
46
3
w 24x=√ a²-4b.=√140×140—4×3136.
84. the Difference of the two Parts requir'd,
which being known, the greater Part will be
140
84
+
2
2
84
2
112, and the leffer Part
=28, the fame as before.
140
2.
PRQ B-
-
LU
[ 131 ]
}
}
4.
PROBLEM LIX.
Let 969 Soldiers be drawn up into an ob-
long Battle, fo that the Difference of the
greater and lefs Sides is 40. Required
the Number of the Soldiers of each Rank
in Length and Breadth ?
NOTE, The Import of this Problem
is no other than to find two Numbers whofe
Product is 969, and Difference 40. Put 969
=a, 40=b, x the Number of Soldiers in
Breadth, and the Number of Soldiers in Length
(by Ax. 2.) will be xb, which multiplied
by the Breadth, the Product is x2 + bx; and
by the
Queſtion, 12+ bx = a.
Queſtion, | 1 | 2
I CO2x² + bx +
2
b2
b²
a
+
4
4
b
2
2
พว
b
3
+
vat
2
4
b² 2
40
3
2
4jx
λ=
trat
2
4
2
40 X 40
4
+√969 +
Soldiers in Breath, and 401757, the
=17; the Number of
Soldiers in Length.
W.W.R.
Or
[ 132 ]
X
Or thus.
{
PUT the Sum of the Soldiers in Length
and breadth, and (per Lemma) the Number
of Soldiers in Length will be
Breadth
b
2 2
X
b
and in
2
2
which Length and Breadth
multiplied by each other, the Product is
bz
4
"
4
and this Product, by the Queſtion, is
equal to a.
That is, 1
I
X
x
4
ลง
2
2
a.
} 2
= 4a
2
X
2
4
2
2 +b² 3 = 4a+b²
3
พ
24 x = √ 4a+b² = √ 4×969+40×40=74.
the Number of Solders in Length and Breadth,
whence the Length will be =
74+
40
57,
2
2
74 40
and the Breadth
17. the fame as
2
2
before.
PROB-
[ 133 ]
1
PROBLEM LX.
Again, let 480 Soldiers be drawn up into
an oblong Battle, fo that the Sum of the
greater and lefs Sides is 52. Required
the Number of the Soldiers of each Rank
in Length and Breadth?
PUT 480=a, 52=b, x=one of the Sides
of the oblong Battle, and (by Ax. 1.) the other
Side is b-x; which Sides multiplied together
the Product is bxx2, which Product, by
the Queſtion, is equal to a, hence the Problem
is reduced to this
Equation. 1bx-x²-a.
2
2 X · bx = — a
I
I 2
b2
2 CO
CD 3x
x² - bx +
4
b
3
w24 x
2
b
b
2
4 +
5x =
+
2
2
4
b
2
4
b2
4
A
a
a. = 40, the
Number of Soldiers in Length, and the Num-
ber in Breadth will be 52-4012, the
Ranks which was required.
N
The
[ 134 ]
PUT
The fame otherwife.
the Difference of the Ranks in
Length and Breadth, then (per Lemma) the
b
X
Length will be
+
+
2
2
and Breadth
b
X
>
4
2
2
h2
4
whoſe Product when multiplied together is
>
4
which, by the Queſtion, is equal to a.
That is, I
x426
2
4
2
2
4
= a.
x² = 4a
x = b²
13*
4a
3
ww 24 x = √b²
—4a.√52×52-4×480.
พ
28, the Difference of the Ranks, &c. whence
the Length of the Oblong will be
=40 Soldiers, and the Breadth
2
52
28
+
2
52
28
= 12.
2
2
W. W. R.
PROBLEM.
[ 135
J
PROBLEM LXI.
In the Square ABCD is given the Diffe-
rence of the Diagonal and the Side,
that is EC=6. Required the Side of
the Square?
FOR AC the Side of the Square put x, and'
(by Ax. 2.) the Diagonal BC 6+x; and,
by virtue of the forty feventh Propofition of
the firſt Book of Euclid, we fhall (See Fig. 20.)
have,
1x 2 + 12.x +36=2x².
I-x²-12x2x
2
4
CO 3 x
3
4
Side of the Square.
12x=36
12x+36=
w 24x-6=√72. = 6√2.
+65x=6+6√2.14.48, the
Q. E. I.
BUT by confidering that all Square Plains
are fimilar Figures, this Problem will be eaſily
folved Numerically. Thus, if the Side of a
Square be 1, the Diagonal of that Square will
be 1.414, and confequently the Difference
between the Side and Diagonal will be .414
+. Now by Proportion it will be, as .414
16:14.49 the fide of the Square. And
Univerfally, by dividing the Difference be-
tween the Diagonal and Side of any Square,
N 2..
by
[136]
3
by .414, the Quotient will give the Side of
the Square required.
GEOMETRICALLY
By the forty fixth Propofition of Euclid's firft
Book, make a Square abCd (See Fig. 21.) of
any Magnitude at pleafure, and draw the Di-
agonal bC, then with bd Radius defcribe the
Arch de, and join the Point d and e with a
right Line; this done, on the Diagonal bC, fet
off EC=6, and draw the Line ED Parallel
to ed until it meets with Cd produced to D.
And CD will be the Side of the Square re-
quired to be found.
DEMONSTRATION.
A
BECAUSE the Lines ed and ED are Parallel
to each other, the Triangles eCd and ECD (by
the fecond Propofition of Euclid's fixth Book)
have their Sides Propotional, whence it will
be. eC:EC:: Cd: CD, therefore &c.
Q. E. D.
PROB-
[137]:
:
PROBLEM LXII.
The Rectangle EK is added to the Square
DF (being of the fame Heighth ;) whoſe
Breadth EL is given=2, and also the
Area of the whole compound Rectangle
DK, =60. Required the Side of the
Square?
PUT 60= a, 2=b, and x=ED the Side
of the Square; then the Area of the ſaid
Square will be x², and bx the Area of the
Rectangle EK, which Areas add together
their Sum is x² + bx the Area of the whole
compounded Rectangle DK; whence comes
(See Fig. 22.)
out this
2
Equation, 1x2 bx=a.
rx
+
b2
2
:
I
CQ 2x² + bx +
a
+
4
4
h
b=
2 w z 13x
=√a+
2
4
b
b²
2
3
4x
+wa+
= 6.8г
2
2
4
=ED, the Side of the Square which was to
be found.
N 3
Or
}
[ 138 ]
Or thus.
PUT x=LD+LK, and (per Lemma)·
b
X
x
b
LD
+
and
>
LK; whence,
2
2
2
2
b
the Product is
2.
2
multiplying+by
x2
b2
"
4 4.
which, by the Queſtion, is equal to
the compound Rectangle DK.
Therefore, I
Ii
2.
1x 2
b2
a.
4
4
ba
44.
X
N
x2
41 2
22 +62 3x² = 4a+b²
3
zw2/4/x = √4a+b². = √ 4×60 +4..
15.62 =LD+LK, which being known,
15.62 2
there will be given LD
2
2
it 8.81.
and LK
15.62
2
= 6.81 = ED, the
2
2.
fame as before.
PROB-
139]
PROBLEM LXIII.
A Man buys fome Ells of Cloth for 70
Crowns; and finds, that if he had 4
Ells more, he had then bought every Ell
2 Crowns cheaper. How many Ells did
be buy?
LET 70a, 4b, and 2 c.
And for
the Number of Ells of Cloth put x, and the
a
Price of Ell will be Crowns: But fuppo-
<
x
fing he had had x-b Ells for a Crown, then
a
I Ell at that rate will coft
x+
which is 2 or c Crowns less than
Crowns,
C
Now,
X
fince the Difference of theſe two Quotients is
2 or c Crowns, add that Difference to the leffer
a
Quotient
x + b²
to the end they may become
equal, and there will be had this
a
Equation, 1
xt b
to
1 x x + b² a + cx + cb =
2
a
X
a + ab
**
× x 3 ax + cx² + cbx = ax + ab
3 = ax14/cx² + cbx=ab
4 ÷c
ཊ» ༥ ན༨ལབ* ག&བྷཡོ
[140]
+
4
÷ c5 x² + bx
==
ab
С
5
СО
6
w 2/7x+
2
2
ca6, x² + bx +
b2
ab
C
ab
4
+
b²
4
C
b
81x
7
2
2
b²
4
!
ab
b²
=
+. 10 the
C
4
Number of Ells he bought for 70 Crowns,
which coft him
7o
7
Crowns an Ell, for had
ΙΟ
70
he had 4 Ells more, then
=5 Crowns
10+ 4
would be the Price of an Ell, which is 2 lefs
than 7, the Anſwer which was required.
PROBLEM LXIV.
A fet of boon Companions dining at an Inn
the Reckoning in all came to 175 Shil-
lings: But, before the Bill was paid.
off, two of them flunk away, and then
the Club of those that remained came to
How many
10 Shillings a Man more.
were there in Company?
PUT 175=a, 10=b, and for the Num-
ber of boon Companions put x; then, had
not
[141]
.
"
not 2 of them flunk away, each muſt have paid
12 Shillings towards the Reckoning, but becauſe
X
thofe that remained cleared the Score, each
paid
X
a
Shillings a-piece, which by the
2
Queſtion is 10 or
Shillings more than
a
Now to bring theſe two Quotients to an Equa-
lity add 10 or b Shillings to the leffer, and
we ſhall have the following
Equation, I +b=
, I XX-22
2
W3
ax 20
X
X
a
+ bx
2b a
x x 3 ax2a + bx² -2bx = ax
2
+4bx² - 2bx=2a
46x2
2 a
4
-65x2
2x=
5 CQ6
CO6x22x
2X
b
2x+1=
b
24
+I
2a
6 ww 217 x I =
b
+1.
24
2x175 +1.
2x175+1
IO
=7 the Number in Company, which was re-
quired. And had not two of them flunk off,.
each
[ 142 ]
175
each would have paid = 25 Shillings,
7
but becauſe 5 paid the Reckoning, they will
175
pay
5
35 Shillings a-piece, which is 10
Shillings more than 25.
PROBLEM LXV.
To divide the Number 21 into two Parts,
So that if the greater be divided by the
lefs, and again the lefs by the greater,
and then the first Quotient being multi-
plied by 4, and the latter by 25, the
Number produced may be equal.
LET 214, 4b, 25=c, and xthe
greater part, and y
2
the lefs; then by the
1x + y = a: 7
ty=
1bx
y
cy
x;
which fubftituted
in the fecond Step
Queſtion.
I
x 3 y = n
bx
ac
CX
for y, gives 4
a
X
*
bx 2
4
xx5
ас
CX
a
X
Ly
-
x6bx²=a²c — acx acxcx²
—
424
!
6
[143]
1
i
L
2
+7cx² - bx
6
7
-
cb8x2
CO 10x
8 in Numbers 9x2
9
IO
I I
w 21 IX
2
=
2ACX-
arc
часк
a²c
C
b
cb
50x=- 525
50x625=100
25=√ 100. = 10
+25/12x=25+10=15,
the greater
part, and 21-15=6 the leffer, the parts re-
quired to be found.
PROBLEM LXVI.
Let the Line AB be divided in C, fo
that AC may be 8, and CD 6: We
are to divide the fame Line AB in D,
So that the Rectangle under AD and
DC
may be equal to the Rectangle un-
der AC and CB, or to the Product
from 8 and 6, which is 48. Required
the Segment CD?
LET 8 a, 6b, 48 = c, and for the
Segment CD (See Fig.
Segment AD will be
Queſtion we fhall
23.) put x, and the
ax. And by the
have
[ 144 ]
have 1px²+ax=c.
1 C= 2x² + ax + ²² = c + 2 =
2
a
4
trot
2
a²
8
∞ol a
4
2
=4=CD the Segment re-
2
ww 23x +
a
+;
2
a
3
4
2
8 × 8
+ √48 +
quired.
4
** ***
PROBLEM. LXVII.
Let there be a Rectangular Garden ABCD,
the Length of which AB is thrice the
Breadth AD: And reckoning 18 Per-
ches from B towards A, that is BE,
and drawing EF parallel to AD, let
the Area of the remaining Rectangle
ED be given 120 fquare Perches.
What was the Length and Breadth of
the ſaid Garden?
=
PUT 120=a, 18=b, and x= AE; then
(per Fig.24.) AB= b+x, and (by Ax. 3.) AD
a
AB.
A B.
Now fince by the Queftion 3AD
Therefore,
[ 145 1
*
I
2
со 3.
за
Therefore, 16+x=34
X
xx2x² + bx = za
2
ca! 3! x² + bx +
за
b2
3a+
4
4
b
b 2
3
w 24x 十
​3a-+--
2
4
b²
2
4
5x =
+√3a+
= 12.
2.
2
4
Hence the Length AB 18+12=30, and the
120
Breadth AD=
30
=10=
2. E. J.
12
3
१
!
PROBLEM LXVIII.
Let 600 Soldiers be difpofed into an oblong
Battle; which the Colonel willing to
make broader, finds that if he takes away
10 Ranks from the Length, he fhall aug-
ment the Breadth with two Ranks. What
was the Number of his Soldiers through
every Rank in Length and Breadth?
PUT 600a, 10b and 2c; and for
the Number of Soldiers in Length put x, then
(per Ax. 3.) the Soldiers in Breadth will be`
a
; x
and by the Queftion -b multiplied
X
by
[146]
ན
a
by the Product will be equal the given
Number of Soldiers a; hence the Prolem will
be reduced to the following
Equation, acx
I
12
× x2ax + cx²
X
2
ab
bca.
X
ab —Ùcx = ax
box
+ 3 cx²-bcx=ab
·
ab
2
3
94/x
bx=
C
4
CD52
15/20
bx +
5
2
4-
ab
C
4
2 ab b2
+
b =
+
C
4
b
b
Gb
2
6 ÷
+
60, and fo
2
C
4
many Soldiers the Colonel placed in the longeſt
Rank or Length, which being known, the
Number of Soldiers placed in the leffer Rank
600
or Breadth will be
= 10; for 60 ΙΟ
60
(= 50) multiplied by 102 (12) the Pro-
duct is the fame as 10 times 60,
that is, 600.
Q. E. F.
P.ROB-
K
[147]
PROBLEM LXIX.
A Man buys a Horfe, which he fells again
for 56 Crowns, and gains as many
Crowns in 100 as the Horfe coft him?
How much did he give for the Horfe?
PUT 100=a, 56b, and x = the Num-
of Crowns the Horſe coſt him; then (per Ax.
) he will gain b-x Crowns by felling the
Horſe again. Now to bring this Queftion
to an Equation, fay by Proportion, Thus,
asa : : : : bx: Therefore by multi-
plying the Means together, and Extremes,
there will arife this
*
I
Equation, 1x2 — ab — ax.¨¯
1 +ax 2x² + ax=ab.
аг
a
2 CO 3x² + ax + 2 == ab + ==
4
4
a
3 w24x
x +
= √ ab +²²²
2
a
2
4
z
a
4
5x=
21
2
+ √ ab + = = 40
a
4
Crowns, and ſo much he gave for the Horſe,
which was required to be found.
L
PROB-:-
[ 148 ]
PROBLEM LXX.
A certain Linnen-Draper buys two Sorts
of Linnen for 30 Crowns, one finer, the
other coarfer. An Ell of the fineſt coſt
as many Crowns as he had Ells: And alfo
28 Ells of the coarfeft at fuch a Price,
that 8 Ells coft as many Crowns as one
Ell of the fineft. How many Ells of the
finest Linnen did he buy, and what Price
did he give for them both?
To folve this Problem, put x = the Num-
ber of Ells of fine Linnen; and fince one Ell
coft as many Crowns as he had Ells of that
fort, confequently x Ells coft x2 Crowns; and
by the other Condition of the Queſtion, 28
Ells of coarfe Linnen will amount to
28x
8
or
3.5x Crowns, whence x²+3.5x is the Price
of both Sorts of Linnen, which, by the Que-
ftion, is to 30 Crowns.
Therefore, 13.5x=30.
I
2
C□ 2x² + 3.5 +3.0625=33.0625
W 23x1.75=5.75
31.754x=4, the Number of Ells of
fine Linnen, as was to be found, for which he
gave
[ 149 ]
7
gave 16 Crowns, and the 28 Ells of coarfe
Linnen coft him 14 Crowns. Q, E. F.
A
PROBLEM LXXI.
In a certain Rectangular Garden, the
Length of which AD is 22 Perches,
and the Breadth AB is 10, the Walk
DG is to be made in a Situation parallel
to the Sides of the Figure, ſo that the
Area of the faid Walk or Gnomon DG
may be equal to the remaining Rectangle
FC, or that the Gnomon DG may
be half of the whole Figure ABCD
propofed. Required the Breadth of the
faid Gnomon DE, BG?
10=b,
PUT 22a, 10 b, and for the Breadth
of the Walk DE (See Fig. 25.) puts then
GB=b-x, and the Area of the Gnomon
GD is axbxx2; which, by the Queſtion,
is equal to half the given Parallelogram AC,
Therefore, ax+bx — x²
土
​ab
2
I
+ 2x²-ax-bx
-bx=
ab
2
2
I
CO 3-
ab. b
a
4
2
4
0 3
3
?
[ 150 ]
a
b
3
พ
w 24
√ a² + b²
2
.9.29
4
4+
a+b
2
15/2
x=6±9.29=3.29 Perches,
the Breadth of the Gnomon.
W.W.R
PROBLEM LXXII.
Of three proportional Numbers there is the
middle Term givin = 12, and the Diffe-
rence of the Extremes 10. Required
the Extremes?
=
PUT 12 =0, 10 = b, x
=
the leffer Ex-
treme, and (per Ax. 2.) the greater will be
+b; and by the Laws of Geometrical Pro-
portion we ſhall have this
Analogy, a :: a : x+b.
1 Ergo, 2 x2 bx = a²
+
b²
2
2
2
со
C□ 3x² + bx +
= a²
+
!
4
4
62
3
214
14/x+
a²+
2
4
b
b
2
4
2
15
2
2
+√a²+
leffer Extreme, and the greater will be 8-10
18, for 8: 12 :: 12: 18, which was to be
found.
BUT
b².
8, the
4
[ 151 ]
BUT becauſe there is given the Difference
of the Extremes, let x reprefent their Sum, and
X
(per Lemma) the greater Extreme will be +:
and the leffer
8/28
x
b
2
2
Ђ
2
b2
And from this
:a::a:
X
Ъ
2 2
b
2 2
we fhall have this
= 26=the
Analogy, 1
Equation, 2
2
4
2
4
2
a²
× 43x² - b² = 4a².
3 + b² 4²x² = 4a² + b²
4
2
ww 215 x = √ 4a²+b². 26 the Sum of
the Extremes, which being found, the greater
26
Extreme will be + =18, and the leffer
26
ΙΟ
2
2
2
ΙΟ
2
=8, the very fame as by the fore-
going Method.
PROBLEM
[ 152 ]
*
PROBLEM LXXIII.
Of three proportional Numbers there is
given the Sum of the first and fecond
= 10, and the Difference of the fecond
and third = 24. Required the feveral
Numbers?
LET 10a, 24b, and for the ſecond
Number put ; then (per Ax. 1.) ax the
firſt, and the third is bx. And by the
=
Property of Geometrical Proportion it will
be, as Ila x:x :: x : b+x.
Ergo, 2 ab-bx+ax — x² = x²
32x² + bx ax = ab
bx ax ab
2
3÷24
४
4/x² +
x²+
b.
a
make
2
2
2
ab
Then,
2
α,
2
CO
+
4
2
+1
5x² + cx
5 C☐ 6x²+cx+
ab
+
C
2
4
6
w27x
+
2
2
ab
7
012
+
=8, the fe-
2
2
4
2
8x=-
cond Number in the Series, whence the firft
and the third = 24 +8
will be 10 8
= 32; for as 2:38:32.
W.W.R.
PRO-
T
:
[ 153 ]
PROBLEM LXXIV.
Of four proportional Numbers there is given
the third = 12, alfo the Sum of the first
and fecond = 8; befides the fecond Num-
ber being fubtracted from it's Square,
the Remainder is to be the fourth.
quired the faid Numbers?
Re-
To folve this Problem, put 12a, 8b,
and for the fecond Number put x; then (per
Ax. 1.) the firft Number will be b-x,
and by the Queſtion, x2-x will be the fourth
Number. And from the following Analogy,
viz. b
x:x:: @: x²
4: -x, there will ariſe, by
multiplying the Means and Extremes, this
I
3 in
2 3 4 5 6
4
Equation, 11x2
I
2X
13.x
Numbers, 4x
CO
- X3
bx
+x²=
= ax.
2
3
bx 2
ax
bx
X
bx
2
a_b
9x= 20
9x+20.25.25
w z 6'x
6x-4.5=√.25.=.5
+4.5.7x=4.5.5=5.
Having
thus found the fecond Number, the first
will be 853, and the fourth =25
=
-520; for as 3: 5: 12: 20, the Num-
bers which was required.
PRO-
[154]
(
PROBLEM LXXV.
Of four Numbers in continued Proportion
there is given the Sum of the Means
=24, and likewife the Sum of the
Extremes = 56. Required the faid
Numbers (fuppofing that the first is the
least of all?)
PUT 24
2a, 56=b, and 2x = the Dif
ference of the fecond and third Terms; then
(per Lemma) a-x will be equal the fecond
Term, and the third will be a-x; and by-
this Analogy, as a-xatxa + x
a²+2ax + x²
a-x
gain, as
a
+x:
2ax+x
a²
2
a + x
2
the fourth Term. A-
x : a
x : : a
Hence there
= the firſt.
will arife the following Equa-
tion, I
|a²——2ax+x² + a²+2ax+x³=b.
a+x
a
X
1 × a+ x 2 a² —2ax+x² +ª³+3a²x+3ax²+x²,
x
=ab + bx
—3a²x+3ax²-x³
2
2
2
2
a-
X
2xa-x
²+3a² x+3ax²+x²}=
}=a²
=a² b———bx² -
2
3
[ 155 1
}
A
3
6
*
ويم
6.
thus,
÷ =
2
2
2
+46ax² + bx²=a²b— 2a³, this
Equation in Numbers will ſtand
5128x2 = 4608
1286x2=36
WJ
⇒
:
:
~ 2|7|x = √ 36. = 6, which being found,
the fecond Term will be 12—6—6, and
the third = 12+6=18; and as 18: 6::6
2 the firft Term. Again as 6 18 18
54 the fourth, or having found the first
Term, the fourth will be = 56—2.54, for
as 26: 18: 54.
PROBLEM
:
W.W. R.
LXXVI.
Two Country-women, A and B, carry 100
Eggs together to Market, in the Jale of
them, one took as much Money as the other:
but A (who had the largest, and confe-
quently the best Eggs) Jays to B, had I
carried as many Eggs as you, I ſhould
have had 18 Pence for them; B replies,
if I had brought as many Eggs as you,
I should have had but 8 Pence for them.
How many Eggs had each?
To anſwer this Queſtion Univerfally, put
100=a, 18=b, 8=c, and a the Number of
Eggs A carried to Market, and (by Ax. 1.)the
Number of Eggs B carried will be a-x.
Now fince the Eggs A carried to Market were
fold at fuch a Price, that had they been equal
in
[ 156 1
in Number to thofe B carried, he would
have had b Pence for them. Therefore as
bx
a-x x: b:: x:
a-x
the Pence A fold her
Eggs for and again, had B had the Number
of Eggs A carried to Market, fhe could
have made but c Pence of them.
Therefore
ас CX
x: c :: a X:
the Number of Pence
X
B received for her Eggs. Hence, by the
Queftion, there will be had this,
bx
ac-cx
Equation,
a-
X
X
a²c-acx-acx+cx²
1 xa-x2 bx =
2 3
xx 3 bx²= a²c
2
X
acx acx + cx²
+4 bx² - cx² + 2acx= a²c
2
4÷b-c5 x² +
6 x²+
5
CO 6x2
СО
2 ACK
banc
240x
a²c
bc
+
a2c2
6
ac
b2
2
b² = 2b c + c ²
a²bc
2bc + c²²
a²bc
zbc +
ws 27:
b-
b²
2
C
а
ас
b-c
18'x
b — c
b
ac
✔ bc.
a
✔ bc. = 40,
+
b-
C
the
1
[ 157 ]
I
$
1
}
the Number of Eggs A carried to Market, and
the Number B carried will be 100-40 60.
40=60.
2. E. J.
PROBLEM LXXVII.
Two Country-men, A and B, fell their Corn
at different Prices: A fells 20 Bufhels;
and B received for one Bushel as many
Crowns as he fold Bufhels: A perceives
that if he had fold as many Bushels as
B received Crowns, he should then have
received 252 Crowns ; but both together
received 176 Crowns. How many Bushels
did B fell, and what Price had A?
PUT 204, 252 = b, and 176=c, and
x= the Number of Bufhels of Corn B fold,
for which, by the Queftion, he received x
Crowns: And fuppofing A had fold x² Buſhels
of Corn for b Crowns, at the rate he fold a
Bufhels for. Therefore for a Bufhels he will
ab
2
receive Crowns; which two Sums are equal
-to c.
x
P
That
[ 158 ]
I
That is, 1x2
I
H
2
3
ab
x
2
C.
Xx2 2x²+ab=cx²
Xx 2 x
+3
H
CD
}
다
​2
CX -ab
cx² +
C
4.
aù
...4
2
C
2
4.
พ
wJ 25
x
12
ab.
4.
5+6x
2
2
áb.
2
4
6: wJ 217
ab.. 6 the -
2
4.
+
5
Bufhels of Corn B fold, for which he received
36 Crowns, and the Price A had for his 20
Bufhels will be found, by what is fhewn above,
to be 140 Crowns, for as 36: 252:20:
: 140.
>
PROBLEM LXXVIII.
Two Merchants fell 21 Ells of Cloth: The
first fells 1 Ell for as many Crowns, as
is of the Number of Ells that the ſe-
cond bad; and the fecond fells 1 Ell for
as many Crowns, as is of the Number
of the Ells that the first had. The Sale
being
[159]
7
being over, they had taken 48 Crowns "in
all: How many Ells did each fell, and
at what Price?
x
To folve this Queftion put 21 = a, 48=b,
and the Number of Ells the firft Merchant
fold; then (per Ax. x.) ax will be the Num-
ber of Ells the fecond Merchant fold, and be-
cauſe the firſt Merchant fold one Ell for as ma-
ny Crowns as is of the Number of Ells that
the fecond had. Therefore as 1:
ax
5
2
5
X
:: X
X the Crowns the firft Merchant
*
fold his Cloth for; and again, fince the fecond
Merchant fold one Ell for as many Crowns
as is of the Number of Ells that the firft
had.
ร
x
Therefore as 1: :: a - X:
ax x
the Number of Crown's the fecond Merchant
-föld his Cloth for.
ax
A
N-
Confequently I
+
5
3
1 X
2
H
3/8x2
and 312 8ax8x215b
8ax=156
15b
3
84x
axi
ী
P2
1
=b.
= b.
[ 160 ]
4
со
2
5x
ax +
а
a 2
??
2
a
2
5
u 2 6 :
2
4
a
6 +
4
2
a²
15b
8
4
2100
156
8
15b
-7
2
2
+1
土
​<= 15,
4
8
the Number of Ells the firft Merchant fold,
and the fecond fold 21 156 Ells. Ha-
ving thus found the Number of Ells of Cloth
each fold, fay as 1 ::: 15: 18 the Number
of Crowns the first Merchant received for his
Cloth, and as 1 ::: 6:30 Crowns, the
Price the fecond Merchant fold his Cloth at,
for 18+30=48. Which was required.
I
PROBLEM LXXIX.
Two Merchants have a parcel of Silk; the
firft 40 Ells, the fecond 90: The firſt
Jells for a Crown of an Ell more than
the fecond: When the Sale was over,
they had taken between them 42 Crowns.
How many Ells did each of them fell
for a Crown?
FOR the Number of Ells the firft Mer-
chant fold for I Crown put x+, and the
fecond
[ 161 ]
T
fecond will fell x Ells; then as x
Ells: 1
40
Crown: 40 Ells:
Crowns, and Ells
x
x + 1/
90
: :
1 Crown: 90 Ells: Crowns; which
X.
two Sums together are equal to 42 Crowns ;
whence there will be had the following
Equation,
Viz.
7
40
x + //
IXX+240 +
90
+ =42.
x
1
90x +30=42x+14
x
xx3130x+30=42x²+14x
3130x442x²-116x=30
2 3 4
425x2
2
2.76x=.7143
5: CO 6x -2.76x+1.9044-2.6187
6. w 27 x 1.38=√ 2.6187. ·
7.+1.388x=2.998, or rather 3, the
Number of Ells the fecond Merchant fold for
1 Crown, whence the firft Merchant muſt
fell 3 Ells for the fame Money.
2.E.J..
P 3
:
PROBLEM
B
[[162]
1
PROBLEM LXXX.
To find a Number, to the quadruple of
which if you add 91, the whole fhall be
to the Square of the Number fought,
as 3 to 4.
To folve this Queftion Univerfally, put 91
a, and for the Ratios 3 and 4 put m and n,
likewife for the Number fought put x, then
it's Quadruple-will be 4x, and Square=2;
hence by the Queftion we have this
Analogy, 14x+a: 262 :: m : N.
1 Ergo, 24nx + an = mx².
·4nx3 mx²
2
4nx= an
3
-34x²
2
4xx an
m
m
4
CD
2
4Nx
4n
m
m
2n
- 5
w 2 6
m
472
2
・amn + 4nº
amn+4n
m2
2
m2
2n
2n
6+
7x=
amn + 4n²
m
m
M2
=13.99
or rather 14, the Number which was to be
found.
;PROB-
i
I
[ 163 ]
1
PROBLEM LXXXI.
:
To find a Number, from the double of which
if you fubtract 12, the Square of the
Remainder lefs I, will be nine times the
Number fought.
PUT 12a, 9=b, and for the Number
fought put x, then the double of x leſs a is 2x
-a, whofe Square is 4x-4x+a², from
which take 1, and the Remainder, by the
Queftion, is equal to b times x.
That is, 14x2
I
24x
4ax+a
4ax bx=
i = bx.
bx
a²+ I
a² + 1
2
3/*
ax
4
4
b
Subſtitute 2m for
a
4
a² + I
Ι
and n for
4
And
4
2mx
2
12.
4
5
CO5x
w 2
6
2mx+ m² = m²
M². - N
x = m = √ m² n.
7\x = m + √m²
n.
11, the
Number required. For the Square of 22 — 12
(10) is 100, and 100-1 (=99) is=9×11.
PRO B-
[ 164 ]
PROBLEM LXXXII.
To divide the Number 19 into two Parts,
So that the Sum of the Squares of the
Parts will be 193.
PUT 19a, 193b, and x = one of the
Parts; then (by Ax. 1.) the other Part will
be a-x; and the Sum of the Squares of the
faid Parts is a22ax+2x2, which by the
Queſtion is equal to b, hence we.
2ax+2x²=b.
2ax = b — a²
b- a²
have
I
a²
2
I
• a² 22x
2
2
2
2
x
ax=
3
3 CO 4x
2
ba
a
2b-a
2
4
2
4
4
a
2b
a
}
4
w25
5 x
2
4
a
5 + - 26
2
2
6x= +
the greater Part, and 19 127 the leffer,
which was to be divided.
2b-
a²
12, equal
4
OF
[165]
Or thus.
PUT X the Difference of the two Parts, then
a
(by the Lemma) is the greater Part,
2
2
a
and
will be the lefs; and the Sum of
2
2.
2
Squares of the faid Parts will be
а
+
2
2
*
which by the Queftion is equal to b.
2
а
That is,
+
2
I
X
十七
​a²
2
3x
2
3
X 2
2
2
=b.
× Σ2 a² + x² = 2b
=2b
ws24x=√2b
2
a
a². 5.
Having
thus found the Difference, the Parts are given
from what is fhewn above.
PROBLEM LXXXIII.
To divide 7 into two Parts, fo that the
Difference of the Squares, which are
made from the treble of the lefs Part,
and the double of the greater, may
be 17.
PUT 7a, 17 b. and call the greater
Part x; then (per Ax. 1) the leffer will be
[166]
;
→ .
a; the treble of the leffer Part is 34
- 3x, and double the greater 2x; which parts
being ſquared becomes 9a2 18ax +9x²
and 4x2, and 4x2 taken from 9a2
-9x2 is equal to ga² 18ax+5x2, which
by the Queftion, is equal to b. Hence we
have this
2
Equation, 119a2 -18ax+5x²=b.
1- 9a² 25x2
18ax=b
18ax
2
дач
18ax
b
25
53x2
9a2
5
5
18ax
3
CO l
· 5
4
w 25 x
+324a²
100
· 18a =√√√206 +144a²
ΙΟ
20b+144a²
100
189
·IO
5+ 6
which is the greater Part fought, whence the
leffer will be
7-43, for 3×39, and
9 times 9 is 81: Again 2 x 48, and 8 times
& is equal 64; take 64 from 81, and the Re-
mainder is 17.
2. E. 7. and D.
100
18a
+
20b+1440
=·4,
ΙΟ
100
PROB
[167]
PROBLEM LXXXIV.
A Man buys a Piece of Linnen, and by
felling it again, he gains 12 Crowns
of what he bought it for: And finds by
this Means that he had gained as much
for 100 Crowns as the Linnen cost him.
What price was the Linnen bought and
jold at ?
PUT 124, 100 b, and 10x the..
Number of Crowns the Linnen coft him; then
a-x is what he gained by felling it again.
And from this
|
:
}
Analogy 1b 10x :: 10x : a
have 2abbx 100x 2.
-x, we fhall
2 + bx 3100x2 bx-ab, and becauſe b hap-
pens to be equal to 100, we fhall
4x²+x=a
have
CD
4 CQ 5 x²+x+25=a+.25:
w26x+5=√ a¯†.25.
5
7'x •5+√a+.25.=3, which
multiplied by 10, the Product is 30, the Num-
ber of Crowns the Linnen coft him, and 12
30
10.
=9 Crowns, is what he gained by felling
it
1
[168]
it again, confequently the Linnen was fold for
39 Crowns; the Anfwer required.
PROBLEM LXXXV.
;
A Man buys 18 Ells of Cloth of different
forts and colour, fuppofe red and black
what he bought of each coft 40 Crowns:
And he pays for every Ell of red Cloth
1 Crown more than for the black. How
many Ells of each fort did he buy?
PUT 40a, 18 b, and x-
= b, and the Num-
ber of Ells of black Cloth; then (by Ax. 1.)
the Number of Ells of red will be bx.
And, fince he gave a Crowns for each fort,
a
1 Ell of black Cloth will coft - Crowns,
a
b-x
X
Crowns will be the Price of 1 Ell
and
a
of red; but, by the Queſtion,
exceeds
b
x
a
a
18
by Crown, therefore add 1 to
, and there
will arife this
Equation,
[ 169 ]
Equation, I
X
a
bx
ax
ax
3 abbx
1
xx2a+ x
b
X
2xb-x
χ - ax
3
and
+19
5
6
7
w 2
4x²+2ax-bx=ab. Put 2m=2a-b,
-bx=ab.Put2m=2a–b,
5/x²+2mx= ab
6x² + 2mx+
+2mx
2mx+ m² = ab + m²
7x+m=√⋅ab +m².
m 8x=- m + √ ab + m² = 10,
the Number of Ells of black Cloth, and 18
108, the Number of Ells of red.
W. W. R.
PROBLEM LXXXVI.
A Man buys 120 Pounds of Pepper, and
as many of Ginger: and received for a
Crown one Pound of Ginger more than
of Pepper. So that the whole Price of
the Pepper came to 6 Crowns more than
the Price of the Ginger. How many
Pounds of each did he buy for a Crown ?
the Number
PUT 120a, 6b, and
of Pounds of Pepper he bought for 1 Crown,
and for 1 Crown he had +1 Pounds of
Ginger. And by this Analogy, as x Pounds
е
of
[ 170 ]
Y
of Pepper
a
:- Crowns
X
1 Crown: : a Pounds of Pepper
the Price of the Pepper; and a-
gain, as x Pounds of Ginger : 1 Crown
:: a Pounds of Ginger:
a
x + 1 I
Crowns, the
Price of Ginger. But becauſe the whole Price
of the Pepper came to b Crowns more than the
Price of Ginger, to the end therefore that they
may become equal. add b to
a
>
and the
X
Sum is
x + I
+b, or
x +
a+bx+b
I
M
I
we ſhall
have
a + bx + b
a
I
x + I
b
whence
2
3
4.
5
1 x x + 2a + bx + b = ax fa
xx√3 ax + bx² + bx=ax+a
I 4 bx² + bx = a
2
十七
​= 65 x² + x = 1/
b
CO | 6x²++.25 = 1/25
b
6
UN 217
x+.5= √ / +.25.
.5
Co
8
४
·5 +
+25=49
[ 171 ]
the Pounds of Pepper bought for 1 Crown, and
4+15 is the Number of Pounds of Gins
ger bought for the fame Money.
Q. E. J.
PROBLEM LXXXVII.
A Man buys 80 Pounds of Pepper and 36:
Pounds of Saffron, fo that for 8 Crowns
he had 14 Pounds of Pepper more than
he had of Saffron for 26 Crowns, and
what he laid out amounted to 188
Crowns. How many Pound of Pepper
had he for 8 Crowns, and how many of
Saffron for 26?
To folve this Queſtion, put x = the Num-
ber of Pounds of Saffron bought for 26 Crowns,
and the Number of Pounds of Pepper bought
for 8 Crowns will be 14. Hence as x
Pounds of Saffron : 26 Crowns :: 36 Pounds
the Price of 36
936
of Saffron : Crowns
↑
Pounds of Saffron. And again, as x+14
Pounds of Pepper : 8 Crowns: : 80 Pounds
640
of Pepper:
Crowns
N+ 14
the Price of 80
Founds of Fepper. But the 80 Pounds of Pep-
per and 36 Pounds of Saffron together coft
188 Crowns by the Queſtion.
Q 2,
Therefore,
[172]
{
•
៩
་
+.
1936
640
Therefore, I
+
=188.
5 x + 14
640x
I
xx2936+
= 188%
x+14
2xx+143 936x+13104+640x=188x2
3
+2632x.
+4188x21056x= 13104
41885 x²+5.6x=69.7
+
5
6
7
CO
2
C6x² +5.6.x +7.8477.54
w27x2.8=√77.54.8.8
-2.8,81x=6, the Pounds of Saffron
bought for 26 Crowns, and for 8 Crowns he
had 20 Pounds of Pepper, which will be ea-
fily proved from the Analogies above.
Q. E. J.
PROBLEM LXXXVIII.
A and B between them owe 174 Pounds,
A pays 8 Pounds a Day, and В pays
the first Day 1 Pound, the fecond 2, the
third 3, and fo on.
In how many Days
will they clear the Debt, and how much
did each of them owe?
PUT 174a, 8b, and for the Number
of Days they required to pay the Debt in put
x; then the Number of Pounds A muſt pay
of
[173]
of the Debt will be bx, and, by the Laws of
Arithmetical Progreffion, multiply the Num-
ber of Terms leſs 1, viz. x— 1, by the com-
mon Difference of the Series, which is 1, and
the Product is x1 equal the Difference be-
tween the two Extremes, to which add 1, and
the Sum is x the laft Term. Now the last
Term more the firſt is = x 1, which Sum -
x-1,
=
x-
multiplied by ~ (=the Number of Terms).
2
the Product will be
x²
2
2
+, the Sum of all
the Series = the Number of Pounds B owes,
which added to bx the Sum will be = a,
That is, I
I
3
4
Fix² + x
+x
2
+ bx = a.
× 2'2x² + x + 2bx=2a. Put 1+26=2m,
And 3x2
x² + 2mx=2a
CD4x2
СО
4x² + 2 mx + m² = 2a+m²
ww 25x+m=√ 2a+m².
พบ
5 — m'6 x = — m + √2a + m² = 12,
the Days in which A and B paid the Debt,
of which, by what is fhewn above, A owed
96 Pounds, and B 78, the Anſwer required.
Notwithſtanding the foregoing Method of
Computation produces an Adfected Quadratic
Q3
Equation,
[174]
174-8
165
155-8 3 = 11
Equation, the Number of Days of Payment:
may be found by Subtraction only. Thus,
8+ I 9 165
8+2=10155
23
144
144 8
4=12
132
4
132
8
5
13
119 5
119
105
90
74-
со со сосо
8
14
105
8
—
7=15
8 = 16
90 7
61=
74
57
+ 9=
9=17
8+10= 18
57 9
39 10
39-
8 --- 11
19
2011
20
8 3 +12
12 20
=2015
012
THE greateſt Number in the right Hand
Column being 12, give the Number of Days
of Payment, the fame as before; which be-
ing known, the reſt is obvious..
PROBLEM LXXXIX.
A certain Man intends to travel as many
Days as he has Crowns: It happens that
every following Day of his fourney be
had as many Crowns as had the Day
before, befides two Crowns over and
above: and when he came to his four-
ney's end he finds he had in all 45 Crowns.
How many Crowns had he at first?
THIS
[ 175 ]
THIS Queſtion imports no more than to
find a Series of Numbers in Arithmetical Pro-
portion whofe Sum is given 45, and common
Difference 2; fuch, that the firſt Term and
Number of all the Terms fhall be equal.
the Number of
To effect which put
Crowns he had in his Pocket when he firſt ſet-
out, or Number of Days he Travel'd, and
(by Corollary 2. Chap. 6. Part I, of Ward's
Introduction) we fhall have x 1X2 2x
2 the Difference between the two Extremes,
to which add the firft Term, viz. x, and the
Sum is 3x2 the laft Term. Now the Sum
of the firft and laft Terms,
multiplied by the
(=
2
the Product will be 2x2
=
viz. 4x — 2,
Number of Terms.)
x the Sum of all
the Series, which by the Queftion is equal to
45 Crowns, hence there comes out this
Equation, 12x2 x = 45•
I
22x²
2
४/०
<= 22.5
32 ++=22.5+:
to 22.5十六
​2.
CO 3
3
2
3
w24x-1=√22.5+18=44
4 +5x=5, the Number of Crowns
he had at first, for 5+7+9+1+13
=45.
2. E. J.
PROBLEM
[176]
PROBLEM XC.
A certain Traveller goes 9 Miles a Day,
three Days after another follows him,
who the first Day travels 4 Miles, the
Second 5, the third 6, and so on, gaining
a Mile every Day. In what Time will
he overtake the former.
X
PUT the Number of Days he Tra--
veled that goes 9 Miles a Day, and the Num-
ber of Days the other Traveled will be x-3;
hence he that Travels 9 Miles a Day, Travel-
ed in all 9x Miles, and the other, becauſe he
makes his Journey in Arithmetical Proportion,.
whofe firft Term of the Series is 4, and com-
mon Difference 1, and likewife the Number
of Terms x 3, by the Quotation in the
foregoing Solution, will Travel
x²-x
2
2
12
Miles, which must confequently be equal to
9x, whence arifes this
*
2
Equation, I
x²+x-
12
9x.
2
I
X 2 2
2x2x-12=18x
2
3
4
2
+ 3x² = 17x=12
4x
2
·17x+72.25=84.25
WJ 2 5x 8.5=√84.25.9.17
x=17.67, the Number of Days
5+8.5/6x=
he
[ 177 ]
he travelled that goes 9 Miles a Days, whence
the other Traveller will overtake him at the
end of 14.67 Days, the Anfwer required.
PROBLEM XCI.
Two Travellers fet out at the fame Time
from two Cities, the one from A, and
the other from B, which are 78 Miles
diftant one from another; one of them goes
6 Miles every Day; and the other 2
Miles the first Day, 2 and a half the fe-
cond, three the third, and fo on, adding
half a Mile to every Day's Journey. In
what Time will they meet with one
another?
In order to folve this Problem, reduce the
diſtance between the two Cities into half Miles
which will be 156; and he that Travels
6 Miles a Day will go every Day 12 half
Miles, and the other will go 4 half Miles the
firſt Day, 5 the fecond, 6 the third, &c.
increaſing a Mile every Day 'till he meets
with the former. Now fince they both fet out
from the two Cities at the fame Time, and
proceed towards one another, they will each
make his Journey in the fame Number of
Days, which Number call x, then he that
Travels 12 half Miles a Day will defcribe the
ſpace
1
[178]
ſpace 12x half Miles in x Days, and the other,
fince the Number of half Miles he Travels
every Day is a Rank of Numbers in Arithme-
tical Progreffion, whofe firft Term is given
= 4,
and common Difference 1, and the
Number of Terms=x, will make his Journey
x²+7x
Σ
2
half Miles in the fame Time, which
added to 12x, the Sum will be equal to the
diſtance between the two Cities; whence the
Problem will be reduced to this
Equation, I
|x² +
2+7x
12x156.
2
I
2 со
3/2.2
+31x240.25 = 552.25
x22x231x312:
3 w 24 x 15.5= √ 552.25.=23.5
4-15.55x8, the Number of Days they
muſt travel before they meet.
Or thus.
W.W.R.
ADD together fucceffively the Number of
Miles they both Travel every Day, until the
Sum amounts to 78 Miles the Diſtance between
the two Cities A and B; and the Number of
fuch Additions will give the Anfwer required.
fee the Operation
Days.
[ 179 ]
7
5
1
Days.
216
3456
6
+2
ナナト
​2.5 +8
+3+ 16.5
+3.5+ 25.5
+4+ 35
+4.5 +45
6+5+55.5
6 +5.5 + 66.5
8.0
16.5
25.5
35.0
45.0
55.5
66.5
78.0
Miles.
Seeing in the left Hand Column the Num-
ber of Additions are 8, they therefore met
each other after eight Days Travel. the fame
as before.
PROBLEM XCII.
Again, Two Travellers fet out at the fame
time from two Cities, the one from A,
and the other from B, which are 120
Miles diftant from one another; the firſt
goes 5 Miles a Day, and the other 3
Miles less than the Number of Days in
which they meet. When will they meet?
X
PUT the Number of Days in which
they meet, and the Number of Miles the fe-
cond Traveller goes every Day will be x-3.
And they will meet one another after the firſt
has gone 5x Miles, and the other x 3x
Miles.
2
Whence,
[ 180 ]
Whence, 1x
2
2
Or, 2x²+2x= 120
CO
3x+5x=120.
3
CQ 3x² + 2x+1=121
· 214
ພ 4x+1 = √
4
✓ 12-1. = II
-5x= 10 Days, the Time required.
15100
PROBLEM XCII.
A Poft fets out from A towards B, who
travels 8 Miles a Day: After he had
gone 27 Miles, another fets out from B
to meet him, who goes every Day of
the whole Journey, or Distance of the
Places A and B, and meets the firſt Poft
after fo many Days as is of the faid
Diftance. Required the Distance of A
and B?
FOR the Diſtance between the two Places A
and B put 20x, and the fecond Poft will go
every Day x Miles; and becauſe he met the
firft after he had travelled as many Days as
was an x Part of the whole Diſtance, he will
make his Journey x2 Miles in x Days; and
the firſt Poft will make his Journey to the
Place where they meet 8x+27 Miles.
Whence,
[181]
Whence, | 1'x²+8x+27=20x.
+ 2
I
2
CO
3
4
४ ४ ४
2
2
12x27
3 ~12x+36=9
w 24x
-6=3
x=9.confequently 20x=180Miles,
the Diſtance between A and B. Which was
required.
PROBLEM XCIV.
Two Merchants A and B go Partners, B
brings 420 Crowns, and A receives out
of the Gains 52 Crowns, and the Sum of
both their Shares is 854 Crowns.
54 Crowns. How
much did A bring, and how much did B
receive out of the Gains?
To anſwer this Problem put 420a, 52
=b, 854c, and for the Stock A put in put x;
then as x (A's Stock): b (what he received
out of the Gains) :: a ( the Number of
Crowns B brought in):
ab
B's Gain; and by
adding both the Merchants Shares together, the
ab
Sum, viz. x++a+b, will be equal to
X
c, hence we have the following
R
Equation,
[ 182 ]
Equation,
I
2
ab
x+=+a+b=c.
xx 2x²+ab+ax+bx=cx
² + ax + bx-cx=-ab
+
3
x²
2
3 in Numb. 4 X -382x=-21840
4
over A
5
6
со 5
+
W
2382x36481=14641
26x-191=√14641.=121
+1917x=312 the Numb. of Crowns.
A put
into the Stock. And as 312: 52: 420:
70 the Number of the Crowns B received
out of the Gain.
PROBLEM
W. W. R.
XCV.
A Son asks his Father how old he was? his
Father replied thus; If you take 4 from
my Age, the Remainer will be thrice the
Number of your Years: But if you take
1 from your Age, half the Remainer will
be the Square Root of my Age. Required
the Age of the Father and Son?
PUT the Father's Age, and the Son's
X
Age will be =
X
-4
Years, now divide
x-4
3
3
X
4 I
which,
6 2
1 by 2 and the Quotient is
و
by the Queſtion, is equal to the fquare Root
of x.
Whence
[ 183 ]
4 I
2
x28x+16
x.
+4
I
\3_8x +16_ *1=,
Whence, I
6
I
O 2
2
+14
са
นย
5
23 +
4
36
6
4
-8x+16-6x+24+9=36x
50x=-49
·50x625=576
6~25=√576.=24
6 +2517x=49 Years the Father's Age,
and the Son's Age, by what is fhewn above, will
be found to be 15 Years.
IV. W. R.
PROBLEM XCVI.
To find two Numbers, the Sum of whose
Squares may be 317, and the Product,
if they may be multiplied by one ano-
ther, 154.
PUT 317a, 154b, and
Number; then (per Ax. 3.)
leffer, and by the
h2
Queftion, x²+==a.
I
2
1
2
X
XX 2x +b²-ax
士
​=3/204
ax
2b?
-b?
R 2
w
the greater
b
will be the
3.
#the
[184]
3
外
​со
ww 25
a
x
2
ax²+
a²
a
2 4
2
a
4
4
5
12
-b².
2
4
6
w27 x = v
-b².. = 14,
= 14, the
2
4
greater Number, and the leffer will be 154
14-
11; for the Square of 14 is 196, of a in
121, and 196121=317. QE. I. and D.
The fame otherwife.
x=Sum
THUS, put } of the two Numbers.
y=Diff.
then (per Lemma.) the greater Number will
be+2, and the leffer=
•
leffer=X_Y which being-
2 2
ftated according to the Condition of the Que-
ftion, will ftand
Thus
Y_—b.
4
xy
ドーナ
​4
2
2
xy, y²
4 2
4.
2 Con-
[185]
2 Y
2 Contracted 3+
2
a
2
X2
I
X 24
=26
2
2
3+4
5
ת!
5x²=a+2b
ww 26x=√a+26=25. which being
known, the Numbers required
will be found from what is
fhown above.
But if for the leffer Number be put x, and
xy for the greater. Then by the
Queſtion,{ | 1 | x²y=b.
I
2 x2
3
3x
for x², gives 4
4
5
6
xy.
C
+x²y²=a.
y
}
which fub. in the 24 Step
b by²
+ =a
لو
لا
5b+by² = ay
2
ay
b
ay
b
a
+
46
2
2
22
462
i
I
7
[186]
a
2
462
a²
7
ربية
3 28
26
a
8 +
a
26/9/9 =
2 b
462
2
I.
.24
1.1.27 the
Value of y being thus found,
known from this Equation
x will be
b
· (See Step
y
the third) for the Root, being extracted gives
}
11.01 or rather 11, the leffer Num-
y
ber, and the greater Number will be xy
= 11 X 1.27 = 13.97 or rather 14. the fame
as by the foregoing Operations.
II
This Method of Computation I never met
with in any Author; nor will I prefume to
fay it is my own Invention; for it was firft
communicated to me by my moſt ingenious
and worthy Friend Mr Samuel Waters. But
how he came by it, or whether it was his own
Contrivance, I am at a lofs to determine,
PRO
[187]
1
PROBLEM XCVII.
To find two Numbers, the Product of which
may be 108, and the Difference of the
the Squares 63.
PUT 108 a, 63=b, and call the greater
Number x; then (per Ax. 3.) the leffer Num-
ber will be, and fince the Difference of the
"
Squares of the faid Numbers is equal to be, we
fhall have this
Equation,
I
2
2
I
a²
2
X
2
=b.
X
4
x x² 2 x
±3x4
СО
++
4,8
w 215 x
४
a² =
=bx2
bx² = a²
-bx²
b
2
22!2
2
[]
2
b²
十
​4.
2
a² +
=√a² +
2
a
√e² +
4
2
b =
4.
2
3
4
5
+
16
4
b2
6
w 2
2
12, the
2
4
108
greater Number, and
12
gives 9 the leffer.
Q. E. J.
The
-
Ъ
[ 188 ]
The fame otherwiſe.
x= leffer
THUS, put greater} Number, and the
xygreater
Difference of their Squares will be x2y2-2,
and Product x²y.
I x² y = a.
2 x2 y22 by the Queſtion.
Whence,
2y2—13
x2
2
y²
by
y 2
b
I
4
a
I
er
by =
азг
a
2
by
لا
I
which fubftitute
for x in the ift
Step, and
2
4× y² — I
응
​5
5 ± &÷61
6
CO7Y
w28y
h
2
b
b
a
2
by
+
402
a
20
162
4a2
2
7
8
+ 9 =
24
20
+
4a2
bz
4a2
ti
+ 1. = 1.03
+1. = 1.31 +.
Having thus found y, x may be known from
any of the three firft Steps, which the Learner
may find at his Leifure.
PRO-
1
[ 189 ]
PROBLEM
XCVIII.
Two Farmers fell two forts of Corn : A fells
6 Bushels; B receives in all for his 20
Crowns: Now, fays B to A, if we add
the Number of my Bushels to the Number
of your Crowns, the Sum will be 28. Says
A to B, and if I add the Square of my
Crowns to the Square of your Bufhels, the
Sum will be 424. How many Bushels
did B fell, and how many Crowns did A
receive?
2
LET 6=a, 20 b, 28c, 424 d, and
for the Number of Buſhels B fold put x; then
c-x will be the Number of A's Crowns,
whofe Square is c² -2cxx², and the Square
of B's Bufhels will be x2; which Squares added
together, their Sum is c²-
2cx2x2 which,
by the Queſtion, is equal d; whence arifes this
2
Equa-
[ 190.]
Equation,! 1 /c2
C
I
2
2
- 2cx + 2x²=d.
2/2x²-2cx=d.c²
23x²-cx=
CO
3
CD
2
4/2
dc²
2
2
-180. For c
is greater than d.
-cx+
zd.
4
4.
WW2
5x
2
zd-
2
c²
2
с
dc²
4
2
d.
2
- C
4.
C
3
2
2
ー​ト
​+
ር
2
4
2
4.
L
6x
5
+3
210
11
له ان
2
4
2d-c 2
C
土
​4
له أن
18 the Number of Bufhels B fold, which
taken from 28, will leave 10 the Number
of Crowns A received.
W.W.R.
•
PRO-
3
[ 191 ]
PROBLEM CXIX.
To find two Numbers, the first of which
+2 multiplied into the fecond—3, may
produce 110: and on the contrary, the
firſt-3, multiplied by the fecond + 2,
may produce 80.
FOR the first Number put x, and y the fe-
cond, then x + 2 multiplied into y3,
and x2 multiplied into y + 2 produces the
following
Equations, 2x+2x-3y-6-80.
S|1|xy=3x+25-6=110.
1 +3x+63 xy+2y=116+3x
2 =2x64xy-3y=86-2x
M
3
÷ +2
116+3x
51
x+2
86—2x
4x36y=
x-3
116+3x
86—2x
5=6
7
xx+28116+3x=
8
x+2 X-3
XX-39116x-3x²-348-9x= 86x
- 2x²+172-4x
86x-2x²-172-4x
X3
[192]
9
IO
I I
I 2
13
土
​± 105x² + 25x=520%
2
÷ 511x²+5x=104
C☐ 12x²+5x+6.25=110.25
213x+2.5=√110.25.=10.5
-2.514x10.5
2.58, the firſt
Number, and, by the 5th or 6th Step, the
Second will be found to be 14: for 8+2×14
-3=110. and 8-3×14+2=80.
Or thus,
2. E. I. and D.
LET x repreſent the firft Number as above,
IIO
and is equal the fecond lefs 3. therefore
x+2
IIO 3 will be the fecond Number, to
x+2
which add 2, and the Sum is
Sum multiplied by x-
110x-330
x-2
110
x+2
+5,which
3 the Product will be
+5 15, and the faid Product,
by the Queſtion, is equal to 80, hence is ob-
tained the following
Equation,
[ 193 ]
110x-330
Equation, I
+5x15=8o.
x+2
2
3456
I XX-22 110x-330+5x²- 5× — 30
2
=80x160
+ 35x²+25x=520
5 4x²+5x=104
52757-6.25—IIO.25
CO
wΣ 6x+2.5=√110.25.10.5
ber, and
before,
2
•257x=10.5-2.5-8 the first Num.
+2+3=14 the fecond, the fame as
HAVING now finiſhed thefe XCIX Queſtions
and Anfwers further to excite the Genius of
the young Analyft, I fhall conclude thefe
few Pages with the following Problem, but
leave the Solution to imploy his leiſure Hours.
ro.
S
PRO-
3
[134]
+
PROBLEM C.
Suppoſe a Cask holds 81 Gallons of
Wine when full, out of which a
certain Quantity is exhausted, and
then the Cask is filled up again with Wa-
ter; the fame Quantity being again
drawn out as at first, and the Cask
again filled up with Water, and ſo on
four Times (always filling the Cask with
Water after every Evacuation) there is at
laft found 16 Gallons of Wine left in
the Cask befides Water. The Question is,
what Quantity of Wine was drawn out
each Time?
see found" voli page
235
*
$
2
FINI S.
Fig. 1 A
a
CP B
c d
b
Fig. 8
Fig. 12
a
B
C D
A
C D B
Fig. 13
D
G F
H
F
B
A
Fig.4
В
B
Fig. 2 t
Fig.3 E
Fig. 4
b
d A
c
D
Fig. 22 I
L
C
F K
E
C C
D
B
Fig. 23
CD B
D
C
Fig. 9
a
Fig. 15
A
Fig.16
D
E
e
F
E
3
D
E
G
D
F
B
Fig.18
Fig.24
A
E
B
Fig.5
G
Fig.17
D
H
E
C
Fig. 6
Fig.10
Fig. 19 nm
F
PA
F
D
F
D
Fig. 25
A
It-
K
B
A
h
B
A e ag
C
B
Fig. 7
D
Fig.
e d сБа
ΑΓ
Fig. 2.1
G
BA
а
B
la
Fig.20
B
E
B
D.
A e.
e
ЈЕ
C
Place this at y End of Bock
UNIVERSITY OF MICHIGAN
3 9015 06360 9369
4
735