THE LIBRARY
OF
THE UNIVERSITY
OF CALIFORNIA
LOS ANGELES
GIFT OB
John S.Prell
KEY
TO
WITH
MANY ADDITIONAL EXAMPLES, ILLUSTRATING
THE ALGEBKAIC ANALYSIS.
BY CHAKLES DAVIES, LL.D.,
AUTHOR OF A FULL COURSE OF MATHEMATICS.
S. P R ELL
Lwil & Mechanical Engineer
SAN THAN CISCO, CAL.
A. S. BARNES & COMPANY,
HEW TOEK, CHICAGO AND NEW ORLEANS,
1876.
DAYIES' COURSE OF MATHEMATICS,
I3ST THREE ~E* A. E, T S -
I. COMMON SCHOOL COURSE.
DAVTES' PRIMARY ARITHMETIC. The fundamental principles displayed in Object
Lessons.
DAVIES' INTELLECTUAL ARITHMETIC. Referring all operations to the unit 1 as the only
tangible basis for logical development.
DAVIES' ELEMENTS OP WRITTEN ARITHMETIC. A practical introduction to the whole
subject. Theory subordinated to Practice.
DAVIES' PRACTICAL ARITHMETIC, A combination of Theory and Practice, clear, exact,
brief, and comprehensive.
II. ACADEMIC COURSE.
DAVIES' UNIVERSITT ARITHMETIC. Treating the subject exhaustively as a science, in a
logical series of connected propositions.
DAVTES' ELEMENTARY ALGEBRA. A connecting link, conducting the pupil easily from
arithmetical processes to abstract analysis.
DAVIES' UNIVERSITY ALGEBRA. For institutions desiring a more complete but not the
fullest course in pure Algebra.
DAVIES' PRACTICAL MATHEMATICS. The science practically applied to the useful art?,
as Drawing, Architecture, Surveying, Mechanics, etc.
DAVIES' ELEMENTARY GEOMETRY. The important principles in simple form, but with
all the exactness of rigorous reasoning.
DAVTES' ELEMENTS OP SURVEYING. Re-written in 1870. A simple and full presentation
for Instruction and Practice.
III. COLLEGIATE COURSE.
DAVIES' BOURDON'S ALGEBRA. Embracing Sturm's Theorem, and a most exhaustive
course. Re-written, in 1873.
DAVIES' UNIVERSITY ALGEBRA. A shorter coarse than Bourdon, for Institutions hav-
ing less time to give the subject.
DAVTES' LEGENDRE'S GEOMETRY. A standard work in this country and in Europe.
DAVIES' ANALYTICAL GEOMETRY. A full course of Analysis, embracing the applications
to surfaces of the second order.
DAVTES' DIFFERENTIAL AND INTEGRAL CALCULUS, on the basis of Continuous Quantity
and Consecutive Differences.
DAVIES' ANALYTICAL GEOMETRY AND CALCULUS. The shorter treatises, combined in
one volume.
DAVTES' DESCRIPTIVE GEOMETRY. With application to Spherical Trigonometry, Spher-
ical Projections, and Warped Surfaces.
DAVTES' SHADES, SHADOWS, AND PERSPECTIVE. A succinct exposition of the mathe-
matical principles involved.
DAVTES' NATURE AND UTILITY OF MATHEMATICS, Logically considered.
DAVTES AND PECK'S MATHEMATICAL DICTIONARY, or Cyclopedia of Mathematics.
Entered according to Act of Congress, in the year 1873, by
CHARLES DAVIES,
In the Office of the Librarian of Congress, at Washington.
PREFACE.
A WIDE difference of opinion is known to exist among teachers
in regard to the value of a Key to any mathematical work, and
it is perhaps yet undecided whether a Key is a help or a
hindrance.
If a Key is designed to supersede the necessity of investiga-
tion and labor on the part of the teacher; to present to his
mind every combination of thought which ought to be suggested
by a problem, and to permit him to float sluggishly along the
current of ideas developed by the author, it would certainly do
great harm, and should be excluded from every school.
If, on the contrary, a Key is so constructed as to suggest
ideas, both in regard to particular questions and general science,
which the Text-book might not impart; if it develops jpnethods
of solution too particular or too elaborate to find a place in the
text ; if it is mainly designed to lessen the mechanical labor of
teaching, rather than the labor of study and investigation ; it
may, in the hands of a good teacher, prove a valuable auxiliary.
The KEY TO BOURDON is intended to answer, precisely, this
733414
rr PREFACE.
end. The principles developed in the text are explained and
illustrated by means of numerous examples, and these are all
wrought in the Key by methods which accord with and make
evident the principles themselves. The Key, therefore, not only
explains the various questions, but is a commentary on the text
itself.
Nothing is more gratifying to an ambitious teacher than to
push forward the investigations of his pupils beyond the limits
of the text book. To aid him in an undertaking so useful to
himself and to them, an Appendix has been added, containing a
copious collection of Practical Examples. Many of the solutions
are quite curious and instructive ; and taken in connection with
those embraced in the Text, form a full and complete system of
Algebraic Analysis.
The many letters which. I have received from Teachers and
Pupils, in regard to the best solutions of new questions, have
suggested the desirableness of furnishing, in the present work,
those which have been most approved. They are a collection
of problems that have special values, and their solutions may be
studied with great profit by every one seeking mathematical
knowledge.
FISHKILL LANDING,
July
, LANDING,)
', 1873. f
INTRODUCTION,
ALGEBRA.
1. ON an analysis of the subject of Algebra, we Aig*b.
think it will appear that the subject itself presents no Difficult!.*
serious difficulties, and that most of the embarrassment
which is experienced by the pupil in gaining a knowl-
edge of its principles, as well as in their applications,
arises from not attending sufficiently to the language Langum*.
or signs of the thoughts which are combined in the
reasonings. At the hazard, therefore, of being a little
diffuse, I shall begin with the very elements of the
algebraic language, and explain, with much minute-
ness, the exact signification of the characters that stand Charge
which repre-
for the quantities which are the subjects of the analy- sent quantity
sis ; and also of those signs which indicate the several Sign*.
operations to be performed on the quantities.
2. The quantities which are the subjects of the n
algebraic analysis may be divided into two classes: HowdiTid A
those which are known or given, and those which are
unknown or sought. The known are uniformily repre- ^
sented by the first letters of the alphabet, a, b, c, d,
&c. ; and the unknown by the final letters, a-, y, z,
v, w, &c.
6 INTRODUCTION.
^rLseVor Quantity is susceptible of being increased, di-
diminished. minisbed, and measured ; and there are six operations
Five opera- which can be performed upon a quantity that will
give results differing from the quantity itself, viz.:
First 1st. To add it to itself or to some other quantity ;
second 2d. To subtract some other quantity from it;
Third. 3d. To multiply it by a number;
4th. To divide it;
Fourth.
5th. To raise it to any power;
Fifth.
6th. To extract a root of it.
The cases in which the multiplier or divisor is 1,
are of course excepted; as also the case where a
root is to be extracted of 1.
signs. 4 The six signs which denote these operations
Elements are ^ OQ we jj k nown to be repeated here. These, with
Algebraic t ne gigns of equality and inequality, the letters of the
language.
alphabet and the figures which are employed, make up
lu words and fa e elements of the algebraic language. The words
phrases :
and phrases of the algebraic, like those of every
HO.W inter- other language, are to be taken in connection with
preted.
each other, and are not to be interpreted as separate
and isolated symbols.
5. The symbols of quantity are designed to repre-
sent quantity in general, whether abstract or concrete,
General. whether known or unknown ; and the signs which in-
dicate the operations to be performed on the quanti-
Exampies. ties are to be interpreted in a sense equally general.
When the sign plus is written, it indicates that the
SiB minus.* quality before which it is placed is to be added to
some other quantity : and the sign minus implies the
INTRODUCTION. 7
existence of a minuend, from which the subtrahend is
to be taken. One thing should be observed in regard s ' ns h T
effect on the
to the signs which indicate the operations that are to nature of
be performed on quantities, viz. : they do not at all
affect or change the nature of the quantity before or
after which they are written, but merely indicate what
is to be done with t//e quantify. In Algebra, for ex- Example*:
In Algebr*.
ample, the minus sign merely indicates that the quan-
tity before which it is written is to be subtracted from
some other quantity ; and in Analytical Geometry, that in Analytical
the line before which it falls is estimated in a contrary
direction from that in which it would have been reck-
oned, had it had the sign plus ; but in neither case is
the nature of the quantity itself different from what
it would have been had the sign been plus.
The interpretation of the language of Algebra is interpretation
of the
the first thing to which the attention of a pupil should language-.
be directed ; and he should be drilled on the meaning
and import of the symbols, until their significations
and uses are as familiar as the sounds and combina- ****
tions of the letters of the alphabet.
6. Beginning with the elements of the language, Elements
explained
let any number or quantity be designated by the letter
a, and let it be required to add this letter to itself
and find the result or sum. The addition will be
expressed by
a -f a = the sum.
But how is the sum to be expressed? By simply Signification
regarding a as one a, or la, and then observing that
one a and one a, make tico a's or 2a : hence,
INTRODUCTION.
a + a = 2a ;
and thus we place a figure before a letter to indicate
how many times it is taken. Such figure is called a
Co-efEcient
panct: 7. The product of several numbers is indicated
by the sign of multiplication, or by simply writing the
letters which represent the numbers by the side of
each other. Thus,
4ow indicated ttXbXCXdxf, OT abcdf,
indicates the continued product of a, i, c, d, and /,
rctor. and each letter is called a factor of the product :
hence, a factor of a product is one of the multipliers
which produce it. Any figure, as 5, written before a
product, as
5abcdf,
Co-efficient of is the co-efficient of the product, and shows that the
product is taken 5 times.
actors : g^ jf t jj e num bers represented by a, b, c, d, and
wht the _/, were equal to each other, they would each be
product
become*. represented by a single letter a, and the product
would then become
How
that is, we indicate the product of several equal fac-
tors by simply writing the letter once and placing a
figure above and a Itole at the right >f it, to indicate
INTRODUCTION. 9
how many times it is taken as a factor. The figure Exponent:
BO written is called an exponent. Hence, an exponent where writua.
is a simple form of language to point out how many
equal factors are employed.
9. The division of one quantity by another is indi- Diriion:
cated by simply writing the divisor below the dividend, hoir
n /> expreised
after the manner ot a traction ; by placing it on the
right of the dividend with a horizontal line and two
dots between them ; or by placing it on the right with
a vertical line between them : thus either form of
expression :
a .
, b a, or 6 | a, Three fonnit
indicates the division of b by a.
10. The extraction of a root is indicated by the Roots :
sign -y/. This sign, when used by itself indicates the howindicaud
lowest root, viz., the square root. If any other root
is to be extracted, as the third, fourth, fifth, &c., the index;
figure marking the degree of the root is written above where writtm
and at the left of the sign ; as,
^/~cube root, ^/""fourth root, &c.
The figure so written, is called the Index of the root. Language fa
We have thus given the very simple and general prtitL
language by which we indicate each of the five
operations that may be performed on an algebraic
quantity, and every process in Algebra involves one or
other of these operations.
10 INTRODUCTION.
MINUS SIGN.
language": U- The algebraic symbols are divided into two
classes entirely distinct from each other viz,, the
kow diyidd. letters that are used to designate the quantities which
are the subjects of the science, and the signs which
are employed to indicate certain operations to be per-
Aigebraic formed on those quantities. We have seen that all
processes :
the algebraic processes are comprised under addition,
thei: number, subtraction, multiplication, division, and the extraction
Do not change of roots ; and it is plain, that the nature of a quan-
6 uantuiee ^^ * s no ' ^ a ^ cnan g e( l by prefixing to it the sign
which indicates either of these operations. The quan-
tity denoted by the letter a, for example, is the same,
in every respect, whatever sign may be prefixed to it ;
that is, whether it be added to another quantity, sub-
tracted from it, whether multiplied or divided by any
number, or whether we extract the square or cube or
Algebraic an 7 other root of it. The algebraic signs, therefore,
Slgns: must be regarded merely as indicating opera/tons to
how regarded.
be performed on quantity, and not as affecting the
nature of the quantities to which they may be prefixed.
Plus and "\y e saVj indeed, that quantities are plus and minus,
Minni.
but this is an abbreviated language to express that
they are to be added or subtracted.
Principles of j. In Algebra, as in Arithmetic and Geometry
the science-
From -what all the principles of the science are deduced from tm
definitions and axioms ; and the rules for performing
the operations are but directions framed in conformity
Example. to such principles. Having, for example, fixed b}
definition, the power of thr minus sign, viz., that an;
INTRODUCTION. 11
quantity before which it is written, shall be regarded
as to be subtracted from another quantity, we wish to What *' e wiak
to discover
discover the process of performing that subtraction, so
as to deduce therefrom a general formula, from which
we can frame a rule applicable to all similar cases.
SUBTRACTION.
13. Let it be required, for example, to subtract Subtraction.
from I) the difference between a and c.
b
Now, having written the letters, with Process.
a c
their proper signs, the language of Al-
gebra expresses that it is the difference only between
a and c, which is to be taken from b ; and if this dif- Difference,
ference were known, we could make the subtraction at
once. But the nature and generality of the algebraic
symbols, enable us to indicate operations, merely, and Operations
indicated.
we cannot in general make reductions until we come
to the final result. In what general way, therefore,
can we indicate the true difference ?
If we indicate the subtraction of a
from b, we have b a ; but then we
b a
b a -\- c
Final formula
have taken away too much from b by
the number of units in c; for it was not a, but the dif-
ference between a and c that was to be subtracted
from b. Having taken away too much, the remainder
is too small by c: hence, if c be added, the true re-
mainder will be expressed by b a + c.
Now, by analyzing this result, we see that the sign Analysis of
of every term of the subtrahend has been changed ;
and what has been si own with respect to these quan-
12
INTRODUCTION.
Generalize tities is equally true of all others standing in the same
tiott.
relation : hence, we have the following general rule
for the subtraction of algebraic quantities :
Change the sign of every term of the subtrahend, or
Rnl, conceive it to be changed, and then unite the quantities
as in addition.
Multiplica-
tion.
Signification
of the
language,
MULTIPLICATION.
14. Let us now consider the case of multiplication,
and let it be required to multiply a b by c. The
algebraic language expresses that the
difference between a and b is to be
taken as many times as there are
units in c. If we knew this differ-
a-b
ac be
Process.
ence, we could at once perform the multiplication.
But by what general process is it to be performed
without finding that difference ? If we take a, c times,
the product will be ac ; but as it was only the differ-
ence between a and 6, that was to be multiplied by c,
this product ac will be too great by b taken c times ;
that is, the true product will be expressed by ac be:
hence, we see, that,
If a quantity having a plus sign be multiplied by
another quantity having also a plus sign, the sign of
the product will be plus ; and if a quantity having a
minus sign be multiplied by a quantity having a plus
sign, the sign of the product will be minus.
ejierai case : 15. Let us now take the most general case, viz.,
that in which it is required to multipy a b by c - d.
It nature.
Principle for
the signs.
INTRODUCTION.
13
Let us again observe that the algebraic language
denotes that a b is to be ta-
ken as many times as there
are units in c d ; and if these
two differences were known,
their product would at once
form the product required.
First : let us take a b as
a b
c-d
etc be
ad -f- bd
ac be ad + bd
Iti form.
First step.
many times as there are units in c ; this product,
from what has already been shown, is equal to etc be.
But since the multiplier is not c, but c d, it follows
that this product is too large by a b taken d times ;
that is, by ad bd : hence, the first product dimin- Second step:
ished by this last, will give the true product But, by
the rule for subtraction, this difference is found by HOW taken.
changing the signs of the subtrahend, and then uniting
all the terms as in addition : hence, the true product
is expressed by ac be ad + bd.
By analyzing this result, and employing an abbre- Analysis of
the remit.
viated language, we have the following general prin-
ciple to which the signs conform in multiplication, viz. :
Plus multiplied by plus gives plus in the product ;
plus multiplied by minus gives minus ; minus mul-
tiplied by plus gives minus ; and minus multiplied by
minus gives plus in the product.
General
Principle
16. The remark is often made by pupils that the
above reasoning appears very satisfactory so long as
the quantities are presented under the above form ;
but why will 3 multiplied by ' give plus bd?
Remark.
Particulu
cast.
14 INTRODUCTION.
How can the product of two negative quantities stand'
mg alone be plus ?
Minus sign. In the first place, the minus sign being prefixed to
> and d, shows that in an algebraic sense they do not
to interpre- Titand by themselves, but are connected with other quan-
*V n ' tities ; and if they are not so connected, the minus
sign makes no difference ; for, it in no case affects the
quantity, but merely points out a connection with other
quantities. Besides, the product determined above,
being independent of any particular value attributed
Foim *f the * the letters a, b, c, and d, must be of such a form as
product : ^ Q ^ e ^ rue or a jj values : and hence for the case in
must be true
lot quantitiei which a and c are each equal to zero. Making this
of any value
supposition, the product reduces to the form of -4- bd.
signs in The rules for the signs in division are readily deduced
from the definition of division, and the principles al-
ready laid down.
ZERO AND INFINITY.
Zero and 17. The terms zero and infinity have given rise to
infinity. muc h discussion, and been regarded as presenting diffi-
culties not easily removed. It may not be easy to
frame a form of language that shall convey to a mind,
idas not but little versed in mathematical science, the precise
ideas which these terms are designed to express ; but
we are unwilling to suppose that the ideas themselves
are beyond the grasp of an ordinary intellect. The
terms are used to designate the two limits of Space
and Number.
18. Assuming any two points in space, and joining
INTRODUCTION.
15
them by a straight line, the distance between the points
will be truly indicated by the length of this line, and
this length may be expressed numerically by the num-
ber of times which the line contains a known unit. If
now, the points are made to approach each other, the illustration,
showing th*
length of the line will diminish as the points come meaning of
nearer and nearer together, until at length, when the *
two points become one, the length of the line will
disappear, having attained its limit, which is called
iero. If, on the contrary, the points recede from each
other, the length of the line joining them will con-
tinually increase ; but so long as the length of the illustration,
,. , . f , ,. showing the
line can be expressed in terms of a known unit of me amngof
measure, it is not infinite. But, if we suppose the
Infinity.
points removed, so that any known unit of measure
would occupy no appreciable portion of the line, then
the length of the line is said to be Infinite.
19. Assuming one as the unit of number, and ad-
mitting the self-evident truth that it may be increased
or diminished, we shall have no difficulty in under-
standing the import of the terms zero and infinity, The term*
. Zero and In-
as applied to number. For, if we suppose the unit fi n ;ty applied
one to be continually diminished, by division or other- 1
wise, the fractional units thus arising will be less and niu*tratio.
less, and in proportion as we continue the divisions,
thfy will continue to diminish. Now, the limit or
boundary to which these very small fractions approach,
is called Zero, or nothing. So long as the fractional Zeio:
number forms an appreciable part of one, it is not
zero, but a finite fraction ; and the term zero is only
16 INTRODUCTION.
applicable to that which forms no appreciable part of
the standard.
illustration. If, on the other hand, we suppose a number to be
continually increased, the relation of this number to the
unit will be constantly changing. So long as the num-
ber can be expressed in terms of the unit one, it is
infinity ; finite, and not infinite ; but when the unit one forms
no appreciable part of the number, the term infinite
is used to express that state of value, or rather, that
limit of value.
The terms, 20. The terms zero and infinity are therefore em-
ployed to designate the limits to which decreasing and
employed.
increasing quantities may be made to approach nearer
A limiu. than any assignable quantity ; but these limits cannot
be compared, in respect to magnitude, with any known
standard, so as to give a finite ratio.
Why limits? 21. It may, perhaps, appear somewhat paradoxical,
that zero and infinity should be defined as " the limits
of number and space" when they are in themselves
not measurable. But a limit is that " which sets bounds
Definition of to, or circumscribes ;" and as all finite space and finite
number (and such only are implied by the terms Space
pace a an( j Dumber), are contained between zero and infinity,
we employ these terms to designate the limits of Num-
ber and Space.
OF THE EQUATION.
Subject of 22. The subject of equations is divided into two
.quatious : tg The fa. consists in finding the equation ; that
ho\r divided. r
First part: is, in the process of expressing the relations existing
INTRODUCTION.
17
between the quantities considered, by means of the
algebraic symbols and formula. This is called the
Statement of the proposition. The second is purely statement.
Second part.
deductive, and consists, in Algebra, in what is called
the solution of the equation, or finding the value of Solution,
the unknown quantity ; and in the other branches of
anah-sis, it consists in the discussion of the equation ; ]
' an equati
that is, in the drawing out from the equation every
proposition which it is capable of expressing.
23. Making the statement, or finding the equation, Statement:
is merely analyzing the problem, and expressing its what it u.
elements and their relations in the language of analy-
sis. It is, in truth, collating the facts, noting their
bearing and connection, and inferring some general
law or principle which leads to the formation of an
equation.
The condition of equality between two quantities Equality of
two quanti-
is expressed by the sign of equality, which is placed
between them. The quantity on the left of the sign
of equality is called the first member, and that on lst member,
the right, the second member of the equation. Hence, 2d member -
an equation is merely a proposition expressed aJge- Proposition,
braically, in which equality is predicated of one quan-
tity as compared with another. It is the great formula
of Algebra.
ties:
Hour ex
pressed.
24. Every quantity is either abstract or concrete :
hence, an equation, which is a general formula for
expressing equality, must be either abstract or con-
crete.
Abstract.
Conci etc.
18
INTRODUCTION.
Abstract
q nation.
Concrete
equation.
Five opera-
tions may be
y#rformed.
A ti em*.
First.
Second.
ThM.
An abstract equation expresses merely the relation
of equality between two abstract quantities : thus,
a -\- b =: x,
is an abstract equation, if no unit of value be assigned
to either member ; for, until that be done the abstract
unit one is understood, and the formula merely ex-
presses that the sum of a and b is equal to x, and is
true, equally, of all quantities.
But if we assign a concrete unit of value, that is,
say that a and b shall each denote so many pounds
weight, or so many feet or yards of length, x will be
of the same denomination, and the equation will be-
come concrete or denominate.
25. We have seen that there are five operations
which may be performed on an algebraic quantity
(Art. 3). We assume, as an axiom, that if the same
operation, under either of these processes, be performed
on both members of an equation, the equality of the.
members will not be changed. Hence, we have the
five following
AXIOMS.
1. If equal quantities be added to both members
of an equation, the equality of the members will not
be destroyed.
2. If equal quantities be subtracted from both mem-
bers of an equation, the equality will not be destroyed.
3. If both members of an equation be multiplied by
the same number, the equality will not be destroj ed
INTRODUCTION. 19
4. If both members of an equation be divided by Fourth,
the same number, the equality will not be destroyed.
5. If both members of an equation be raised to Fifth,
the same power, the equality of the members will
not be destroyed.
6. If the same root of both members of an equa- sixth,
tion be extracted, the equality of the members will
not be destroyed.
Every operation performed on an equation will Use of
fall under one or other of these axioms, and they
afford the means of solving all equations which ad-
mit of solution.
26 t Two quantities are said to be equal, when Equality de-
fined,
each contains the same unit an equal number of
times. Hence, the term equal applies to measures,
and has the same signification in Arithmetic, in E( i ual to *&
parts.
Algebra, and in Geometry. If, in Geometry, two
figures can be applied to each other, so as to coin-
cide or fill the same space, they are said to be
equal in all their parts.
27t We have thus pointed out some of the marked
characteristics of Algebra. In Algebra, the quan- Classes of
quantities in
tities, about which the science is conversant, are Algebra,
divided, as has been already remarked, into known
and unknown, and the connections between them,
expressed by the equation, afford the means of
tracing out further relations, and of finding the
values of the unknown quantities in terms of the
known.
20 INTRODUCTION.
SUGGESTIONS FOR THOSE WHO TEACH ALGEBRA.
utters are but j_ ^ careful to exp i ain that ^ i etters employed.
mere gymtnls * *,
are the mere symbols of quantity. That of and in them-
selves, they have no meaning or signification whatever,
but are used merely as the signs or representatives
of such quantities as they may be employed to denote.
Signs indicate 2. Be careful to explain that the signs which are
operations.
used are employed merely for the purpose of indicating
the five operations which may be performed on quan-
tity ; and that they indicate operations merely, without
at all affecting the nature of the quantities before which
they are placed.
Letter* and 3 Explain that the letters and signs are the ele-
uiji'* elements
of language, ments of the algebraic language, and that the language
itself arises from the combination of these elements.
Algebraic 4 Explain that the finding of an algebraic formula
formula
is but the translation of certain ideas, first expressed
in our common language, into the language of Algebra;
its interpreta- anf ] that the interpretation of an algebraic formula is
tion.
merely translating its various significations into common
language.
Language. 5. Let the language of Algebra be carefully studied,
so that its construction and significations may be clearly
apprehended.
Co-efficient. 6. Let the difference between a co-efficient and an
; x oncnt exponent be carefully noted, and the office of each often
explained ; and illustrate frequently the signification of
the language by attributing numerical values to letlers
in various algebraic expressions.
7. Point out often the characteristics of similar and
INTRODUCTION. 21
dissimilar quantities, and explain which may be incor- similar''
quantities'
porated and which cannot.
8. Explain the power of the minus sign, as shown Minus si e
in the four ground rules, but very particularly as it is
illustrated in subtraction and multiplication.
9. Point out and illustrate the correspondence be- Arithmetic
and Algebra
tween the four ground rules of Arithmetic and Alge- compared.
bra ; and impress the fact, that their differences, where-
ever they appear, arise merely from * differences in
notation and language : the principles which govern
the operations being the same in both.
10. Explain with great minuteness and particularity, Equation.
all the characteristic properties of the equation ; the Its properties-
manner of forming it ; the different kinds of quantity
which enter into its composition ; its examination or
discussion ; and the different methods of elimination.
11. In the equation of the second degree, be careful Equation oi
the second
to dwell on the four forms which embrace all the cases, degree.
and illustrate by many examples that every equation
of the second degree may be reduced to one or other
of them. Explain very particularly the meaning of its form.
the term root ; and then show, why every equation of i te roo t.
the first degree has one, and every equation of the
second degree two. Dwell on the properties of these
roots in the equation of the second degree. Show why
their sum, in all the forms, is equal to the co-efficient Thir*nm.
of the second term, taken with a contrary sign ; and
why their product is equal to the absolute term with a Their
contrary sign. Explain when and why the roots are
imaginary.
22 INTRODUCTION.
General 12. In fine, remember that every operation and rule
is based on a principle of science, and that an intelli-
gible reason may be given for it. Find that reason,
and impress it on the mind of your pupil in plain and
should be simple language, and by familiar and appropriate illus-
trations. You will thus impress right habits of inves-
tigation and study, and he will grow in knowledge.
The broad field of analytical investigation will be
opened to his intellectual vision, and he will have
made the first steps in that sublime science which dis-
They lead to covers the laws of nature in their most secret hiding-'
ifnerai lawSl places, and follows them, as they reach out, in omnipo-
tent power, to control the motions of matter through
the entire regions of occupied space.
(See Davies' Nature and Utility of Mathematics,
Article Algsbra).
KEY
EQUATIONS OF THE FIRST DEGREE.
~. 5x 4x 10 7 13z
1. Given ____ 13 = --..
VERIFICATION.
5 x 11.1 4 x 11.1 7 13 x 11.1
12 3 "86
Multiply by 24, least common multiple,
10x11.1 32 x 11.1 312 = 21 52 x 11.1; that is,
556.2= 556.2.
2. Given x -f 18 = 3z 5, to find x.
Transposing and reducing,
- 2ar = - 23 ;
dividing both members by 2,
x= 11J.
3. Given 6 - 2x + 10 = 20 3* 2, to find *.
Transposing and reducing,
x = 2.
4. Given ar "^"H z "^o * = H> to nn ^ #
2 o
Multiplying both members by 6, and reducing,
11* = 66;
whence, x = 6.
24 KEY TO DAVIES' BOURDON. [91.
5. Given 2z - x + 1 = 5x 2, to find x.
tt
Multiplying beth members by 2, transposing and reducing,
Tx = -6;
whence, x = -.
6. Given Box + - - 3 = bx a, to find a.
Multiplying by 2, transposing and reducing,
Qax 2bx = 6 3a ;
factoring the first member of the equation, we have
(6a 26) x 6 3a ;
6 3a
whence, a; =
6a - 26'
7. Given ? i + | = 20 - ^^, to find *,
SI 8 /*
Multiplying both members by 6,
Bx - 9 + 2* = 120 - 3 x + 57 ;
transposing and reducing,
8* = 186 ; . . = 23.
a: + 3 a; a; 5
3. Given h = 4 , to find x.
Multiplying both members by 12.
Gx + 18 + 4x = 48 - 3ar 4- 15 ;
transposing and reducing,
13* = 45 .. . t
EQUATIONS OF THE FIRST DEGREE. 25
ax b a bx bar a
9. Given -- h g = -^ -- g , to find x.
Multiplying both members by 12,
Sax Sb+4a = Gbx 4bx + 4a;
transposing, reducing and factoring,
36
(3a 26) x = 36, . . x
Sax %bx
10. Given --- ; -- 4 =/, to find x.
c a
Multiplying both members by cd,
3adx 2bcx 4cd =fcd ;
transposing, reducing and factoring,
3a
8aa; b 36 c
11. Given - --- s = 4 6, to find x.
i '-~
Multiplying both members by 14,
IQax - 26 - 216 + 7c = 56 - 146 ;
transposing and reducing,
56 + ft
16a
12. Given - 1- - = , to find
Multiplying both members by 30,
Qx - Wx + 20 + 15 x = 130 ;
transposing and reducing,
lla; = 110; . x = 10.
KEY TO DAVIES' BOURDON. [91-92.
.,-,. X X . X X
13. Given
T ^ -j =/, to find x.
abed
Multiplying both members by abed, and factoring,
(bed acd + abd abc) x =abcdf . . x =
bed acd -j- abd abc
ij n- 8* 5 . 4* -2 , , . ,
14. Given a; 1 = x + 1, to find x.
lo 1 1 .
Multiplying both members by 143,
143* - 33* + 55 + 52x 26 = 143* + 143 ;
transposing and reducing,
19* = 114; . *s_-6.
15. Given y - y - ^^ = - 12f|, to find *.
Multiplying both members by 315,
45x 280* - 63* -|- 189 = 3983 ;
transposing and reducing,
-298* =-4172; .-. * = 14.
4* - 2 3* - 1 t
16. Given 2* = - , to find *.
O ii
Multiplying both members by 10,
20* 8*+ 4 = 15ar 5;
transposing and reducing,
3* = - 9 ; . . x = 3.
92.] EQUATIONS OF THE FIRST DEGREE. 27
17. Given Sx -\ = x + a, to find a?.
o
Multiplying both members by 3,
9x + bx d = Sx -f 3a ;
transposing, reducing and factoring,
O_ I J
(6 -f b) x = 3a -f d ; . . x =
18 . Given
a o a -4- b b
to find x.
We see that the least common multiple of the several
fractions of the two members of this equation is,
a*b b 5 .
Hence, multiplying both members of the equation by
a?b b 3 , and performing all the indicated operations, we
shall have,
a?bx + 2ab*x + t>*z aW 2ab* 5 4 3a s b + 3i 3 = 4a 2 ^
- 5ab 3 + J 4 Zatbx + Wx + a 4 a?bx tf& + IPx ;
then, by transposing,
b s x + 2a 2 Ja; aj 3 ^ + a?bx b*x = 4a 2 5 2
J4
factoring, we have,
2J (2a 2 + a5 J 2 ) a; = a 4 + 3a 3 b
dividing by the coefficient of x,
3a s b
2b (2 2 + ab -
'28 KEY TO DAVIES 5 BOUKDON. [92.
19. Given, x = 3x \ (4 x) + ^ .
iii O
Clearing of fractions, and dropping parenthesis,
6x = ISx 12 + 3x + 2.
Transposing and reducing,
-I5x= 10;
2
o .....
3x 7 25 4z 5a; 14
20. Given, -fr- + ^-~ = -^ .
Clearing of fractions,
27a; 63 + 125 20z = 75a; 210.
Transposing and reducing,
_ 682; =272;
x = 4.
2z + 5 40 x Wx 427
21. Given, -^ + -_ - J ^-.
Clearing of fractions,
304a; + 760 + 9880 247z = 1040z 44408.
Transposing and reducing,
983z = 55048 ;
x = 56.
00 ~. x x 5 /2 11 \
22. Given, + 5 = 2;-^ + ij .
Clearing of fractions,
Ux 7x + 35 + 385 = 77z 2 77.
Transposing and reducing,
71z= 497;
a; =7.
a; 1 2 a; + 3 z + 4
23. Given, -^- + -^-=-- + -^ + 1.
Clearing of fractions,
6x 6 + 4z 8 = 3x + 9 + 2.r + 8 + 12.
92.] EQUATIONS OF THE FIRST DEGREE. 29
Transposing and reducing,
5z = 43 ;
= 8f
X ~ l X ~ 2 X ~ 5 a? ~ 6
24. Given,
Performing the indicated subtraction in both members,
1 1
( x _ 2) (z 3) ~~ (x 6) (z 7) '
Clearing of fractions, and performing indicated operations,
Transposing and reducing,
8x = 36 ;
# = 4i.
25. Given, * + aj - - (* + 5) (x- 3) + = 0.
Performing indicated operations,
s? + x-.15-tf-2 x + 15 + ^ = 0,
Transposing and reducing,
x =12; .-. 3 = 12.
6z + 7 2rc 2 2x + 1
26. Gwen,
Clearing of fractions,
42^ + I3x 42 30z + 30 = 42a^ 15a; 18.
Transposing and reducing,
2x = 6 ;
x = 3.
/* i Lriven, ^^ ~
x 2 x 4 x 6 x 8
Performing the indicated subtractions,
2 _ 2
(x 2) (x 4) ~ (x 6) (x 8) '
3 KEY TO DAVIES' BOURDON. [92.
Dividing by 2, and clearing of fractions,
x 2 14z + 48 = tf 6x + 8.
Transposing and reducing,
Sx= 40;
x = 5.
28. Given, (x + I) 2 = (5 + x) x 2.
Performing indicated operations,
z 3 + 2x + 1 = 5x + a? 2.
Transposing and reducing,
3z = 3;
3 = 1.
29. Given, 5-^-=- + -^- = g-^ .
2x 5 x 3 3^ 1
Clearing of fractions,
62 20a; + 6 + 6z 2 llx + 5 = 12o 66a + 90.
Transposing and reducing,
_79
~'
30. Given, - + * a
a b a b -\- a*
Factoring,
jl , 1 > a
/ ^ I , \ __ -__j__
(o d a)~J + a*
Reducing,
b \_ g
aCJ -)/(& + )'
_
~
fa)'
. / n\ i /
81. Given, -|
93-97.] EQUATIONS OF THE FIRST DEGEEE. 31
Performing indicated operations,
1 1
X ~ a
Clearing of fractions, and reducing,
= 8a ;
32. Given, 1.2* - ' lSx ~ >05 = .4* + 8.9.
.5
Clearing of fractions,
.6* .18* + .05 = 3n + 4.45.
Transposing, and reducing,
a; = 20.
_ AK
33. Given, 4.8z - ' = 1.6* + 8.9.
Clearing of fractions,
2.4z .72* + .05 = 0.8z + 4.45.
Transposing and reducing,
.882; = 4.40;
* = 5.
STATEMENT AND SOLUTION OP PROBLEMS.
8. Divide $1000 between A, B, and C, so that A shall have
72 more than B, and C $100 more than A.
Let x denote the number of dollars in B's share.
Then will * + 72 " " A's "
and * + 72 + 100 " " " C's "
From the conditions of the problem,
32 KEY TO DAVIES' BOUKDON. [37.
x + x + 72 + x + 172 = 1000; or, 3 x = 756 , . . x -= 252,
or, A's share is $324, B's share $252 and C's share $424.
9. A and B play together at cards. A sits down with $84 and B
with $48. Each loses and wins in turn, when it appears that A has
five times as much as B. How much did A win?
Let x denote the number of dollars that A wins.
Then will 84 + x denote what A has at last,
and 48 x what B has at last ;
from the conditions of the problem,
84 + * = 5 (48 x) ; or, 84 + x = 240 - 5x ;
whence, ? = 26 ; or, A wins $26.
10. A person dying, leaves half of his property to his wife, one
sixth to each of two daughters, one twelfth to a servant, and the
remaining $GOO to the poor : what was the amount of his property ?
Let x denote the whole number of dollars in the property.
Then will ^ " " " " " in the wife's share.
it
- " " " each daughter's ''
6
and -^ " " " " " the servant'? "
1
from the conditions of the problem,
multiplying both members by 12, transposing and reducing,
- x = - 7200 ; or, x = 7200.
97.] EQUATIONS OF THE FIKST DEGREE. 33
1 1 . A father leaves his property, amounting to $2520, to four
sons, A, B. C and D. C is to have $360, B as much as C and D
together, and A twice as much as B less $1000: how much do A,
B and D receive ?
Let x denote the number of dollars that D receives
Then will x + 360 " " " " B "
and 2* + 720 -1000 " " " A "
from the conditions of the problem,
360 -(- x -f x + 360 -f 2x + 720 - 1000 = 2520;
transposing and reducing,
4x = 2080 ; . . x = 520
or, D's share is $520 ; B's share $880, and A's share $760.
12. An estate of $7500 is to be divided between a widow, two
sons, and three daughters, so that each son shall receive twice as
much as each daughter, and the widow herself $500 more than all
the children : what was her share, and what the share of each child ?
Let x denote the number of dollars in each daughter's share ;
Then will 2x " " " " son's "
and 4x + 3z + 500" " " the widow's "
from the conditions of the problem,
4x + 3x + 4* + 3z -|- 500 = 7500;
transposing and reducing,
Uas = 7000 ; . x = 500.
Daughters' share $500; son's share $1000 ; widow's share $4000.
13. A company of 180 persons consists of men, women and
34: KEY TO DAVIEs' BOURDON. [97-98.
children. The men are 8 more in number than the women, and the
children 20 more than the men and women together : how many of
each sort in the company 1
Let x denote the number of women ;
Then will x + 8 " " men ;
and x + x + 8 + 20 " children.
From the conditions of the problem,
x + x + 8 + x + x + 8 + ZO = lSQ;
transposing and reducing,
4x= 144; . . x = 36.
36 women, 44 men and 100 children.
14. A father divides $2000 among five sons, so that each elder
should receive $40 more than his next younger brother : what is
the share of the youngest ?
Let x denote the number of dollars in the youngest's share.
Then will x 4- 40 " " " " second's "
x + 80 " " " " third's "
x -f 120 " " " fourth's "
x + 160 " " " fifth's "
From the conditions of the problem,
5* + 400 = 2000 ;
transposing and reducing,
5z=1600; .-. 2 = 320.
15. A purse of $'2850 is to be divided among three persons, A,
P> and C ; A's share is to be T 6 T of B's share, and C is to havf $300
more than A and B together : what s each one's share ?
98.] EQUATIONS OF THE FIRST DEGEEE.
Let x denote the number of dollars in B's share.
Then will ^ " " " " A's "
R X
and x + -j + 300 " C's "
From the conditions of the problem,
* + fl + * + if +300 = 2850;
clearing of fractions, transposing and reducing,
34z = 28050 ; . . x = 825 ;
hence, B's $825 ; A's $450 ; and C's $1575.
16. Two pedestrians start from the same point; the first steps
twice as far as the second, but the second makes five steps while tho
first makes but one. At the end of a certain time they are 300 feet
apart. Now. allowing each of the longer paces to be 3 feet, how for
will each have travelled 1
Let x denote the number of feet travelled by the first.
Then will " " steps " "
8
5x
" " " second.
3
and 11 x , or ^ " feet " "
O O
I
From the conditions of the problem,
1^ - x = 300 ;
o
clearing of fractions, transposing and reducing,
15a:
9z = 1800 ; . . y = 200 ard r = 500.
36 KEY TO DAVIES' BOURDON.
17. Two carpenters, 24 journeymen, and 8 apprentices, received
at the end of a certain time $144. The carpenters received $1 per
day. each journeyman half a dollar, and each apprentice 25 cents :
Slow many days were they employed 1
Let x denote the number of days.
Then will x " dollars due each carpenter. -
x
journeyman.
2
x
and - apprentices ;
from the conditions of the problem,
reducing,
16* = 144 ; . . r = 9.
18. A capitalist receives a yearly income of $2940 ; four fiths of
his money bears an interest of 4 per cent., and the remainder of 5
per cent. : how much has he at interest ?
Let x denote the number of dollars at interest.
4 4
Then will X -r-r^ denote the interest of 1st parcel.
O 100
From the conditions of the problem,
4x 4 x 5
"5" X TOO + 5 X T00 =
clearing of fractions, and reducing,
21* = 1470000 ; . . x = 70000,
98.] EQUATIONS OF THE FIRST DEGREE. 37
19. A cistern containing 60 gallons of water has three unequal
cocks for discharging it ; the largest will empty it in one hour, the
second in two hours, and the third in three : in what time will the
cistern be emptied if they all run together ?
Let x denote the required number of minutes.
Then since the first emits 1 gallon per minute, the second \ of a
gallon per minute, and the third of a gallon,
x will denote the number of gallons emitted by the 1st.
ti U (( (t (C O(]
2
U (( ( U it 0,1
3
From the conditions of the problem,
clearing of fractions and reducing,
11s = 360 .-. x = 32 1 r w.
20. In a certain orchard are apple-trees, -J peach-trees, \ plum-
trees, 120 cherry-trees, and 80 pear-trees : how many trees in the
orchard ?
Let x denote the whole number of trees.
(K
Then will - " " apple-trees.
" " " " peaeh-trees.
I " " " "plum-trees.
From the conditions of the problem,
KEY TO DAVIES' BOURDON. [99.
clearing of fractions, transposing and reducing,
x = - 2400 . . x = 2400.
21. A farmer being asked how many sheep he had, answered that
he had them in five fields; in the 1st he had i, in the 2d , in the.
3d , in the 4th fa, and in the 5th 450 : how many had he ?
Let x denote the whole number of sheep :
Then will - A " " " " in 1st field.
4
x
u u
noted by j-, and the time required to walk back will be de-
/v*
noted by -. From the conditions of the problem, we have,
c
XX
T -f- - = a. Hence,
c
(0 + c}x = abc\ .'. x = j- -- .
6 + c J
35. A can do a piece of work in one-half the time that B
can ; and B can do it in two-thirds the time that C can ;
all together can do it in 6 days. How many days would it
take each to do it singly?
Let x denote the number that it will take A to do it.
Then will %x denote the number of days that it will take B
to do it ; and 3x will denote the number of days it will take
C to do it. Consequently, A can do a part denoted by -
v
in one day, B can do a part denoted by ^-, C can do a
/iX
part denoted by -$- in the same time, and all together can
oX
do a part denoted by - in one day. Hence, from the con-
ditions of the problem,
= or
.-. - = -L or, a: = 11, 2x = 22, and 3z = 33.
x 11
106.] EQUATIONS OF TIIE FIRST DEGREE.
SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE.
(2*4- 3y = 16)
1. Given { > to find x and y.
(3* 2y = 11 )
Multiply both members of the first by 2, and of the second by 3 j
+ Gy = 32 )
whence, by addition, member to member, we have,
13* = 65 ; . . x = 5, also, y = 2.
2. Given
2f 3y_ 9_
"5" H " T ~ 20"
3s 2y_ 61
T + "5" ~ 120 J
i. to find x and y.
Clearing of fractions, and then multiplying both members of the first
by 16, and of the second by 5,
128* + 240y =
450* + 240y
whence, by subtracting, member from member.
= 144)
= 305)
322* = 161 ;
3. Given
x = -, also, by substitution, y = -
* 3
| + 7* = 51
to find x and y.
Multiplying the first by 343 and the second by 7 ;
49* f 2401y - 33957 )
y f- 49* = 357 \
4r6 KEY TO DA VIES* BOURDON. [106-107
by subtraction,
24(% = 33600 ; . . y = 14 ; also, by substitution, x = 7.
4. Given
; to find ar and y.
5 ' 3 4
Clearing of fractions and transposing,
2* - y = 80
47-18y-2100;
multiplying both members of the first by 18, and subtracting the
result from the second, member from member,
11 x 660 ; . . x = 60 ; by substitution, y = 40.
5. Given *
x + y + 2 = 29
* + 2y + 3z = 62
(1)
(2)
(3)
* to find x, y and *.
Combining (1) and (2),
y + 2z = 33 .... (4);
combining (1) and (3)
2y + 3z = 54 . (5) ;
combining (4) and (5)
2 = 12; by successive substitutions, x = 8. y = 9.
pZ* + 4y - 3* = 22 t - (I)'
6. Given -I 4* 2y -f 5z = 18 (2) [; to find r -i and 2.
[& 4- ly - z = 63 (
Combining (1) and (2).
107.] EQUATIONS OF THE FIRST DEGREE.
10y- 112 = 26 . . . (4);
combining (1) and (3),
5y - Sz = 3 . . . (5) ;
combining (4) and (5),
5z = 20 ; . ' . 2 = 4.
By successive substitutions, x = 3, y = 7.
7. Given
+I+I-*
>. ; to find x, y and .
7 i eT ' a
.45 o
Clearing of fractions,
6x + 3y + 22 = 192 . . . (1) ;
20* + 15y + 122 = 900 . . . (2) ;
15x + 12y + 102 = 720 . . . (3) ;
combining (1) and (2),
16x + 3y = 252 . . . (4) ;
combining (1) and (3),
15ar + 3y = 240 . . . (5);
combining (4) and (5),
x = 12;
by successive substitutions, y = 20, z = 30.
( 7z - 22 + 3tt = 17 . . . (1) "
: y-2z+ t=l\ . . . (2)
3. Given ^ 5y 3x - "2u 8 . . . (3) I ; to find x, y,
\y 3?/ -f 2^ = 9 . . . (4) 2, t and .
3z + 8w = 33 . . . (5)
48 KEY TO DAVIES* BOUEDON. 107.
Combining (2) and (4),
4y 4z + 3w = 13 . . . (6) ;
combining (1) and (3),
35 y _ 62 - 5w = 107 . . . (7) ;
combining (5) and (6),
12y -1- 41w = 171 . . . (8);
combining (5) and (7),
35y + ll = 173 . . . (9);
combining (8) and (9),
1303^ = 3909; .-. =3;
by successive substitutions, a: = 2, y = 4, 2 = 3, / = 1.
C3x + 2y 4:z = 15 (1)
9. Given, J 5x 3y + 2z = 28 (2)
[ 3y + z x = 24 (3)
Combining (1) and (3),
ll# + 8z = 87 .,(4)
Combining (2) and (3),
12y + 22z = 148. ...... (5)
Combining (4) and (5),
73y = 365; .-. y = 5.
Substituting and reducing,
* = 7, and 2 = 4.
10. Given,
x y
M= ()
x z
-L . J. o ,-.
108.]
EQUATIONS OF THE FIRST DEGREE.
Adding (1), (2), and (3), member to member, and divid-
ing by 2,
- - --- (4)
Subtracting (1), (2), and (3), successively, from (4),
J. ^t
2J 4 O
i_i
13 4
- + - = - (1)
x y z
11. Given,
1 1_4
Making - = x', - = y', and - = z' t
x y z
2x' + y' = 3z' (4)
Bz' 2y' = 2 (5)
x' + z' = - (6)
Combining (4) and (5),
4z' + 3z' = Gz' + 2,
or, 4z' 3z' = 2 (7)
Combining (6) and (7),
, 6 7
H rp p\ /> ^^ AT* 1*
4 "-t \J y . * t/ p, J Ul ^ *O ^ p
Substituting and reducing,
, _ 10 _ 21
-21' r> -10'
2 7
and, y = = , or, y = T .
50
KEY TO DAVIES' BOUKDON. [108.
1 _ 6z x 9
4~ = 1T~2 + 5 ' ' * '*
12. Given, l + ir = 2/+!j ( 2 )
~^~ * _ JL I 1 _ I y_ t<\\
7 14 "*" 6 ~~ 21 "*" 3 '
Clearing of fractions and transposing,
15y + 10# 242 = 41 (4)
12# + 15x + lGz= 10 (5)
14y + 18z 7z = 13 (6)
Combining (4) and (5),
115z 162 = 214 (7)
Combining (4) and (6),
410s - 441z = 379 (8)
Combining (7) and (8),
8831z = 8831 ;
Substituting and reducing,
x = 2, and y =. 3.
13. Given, ^ + - = 1 (2)
c
Adding (1), (2), and (3), and dividing by 2,
x v z 3
108.]
EQUATIONS OF THE FIRST DEGEEE.
51
Subtracting (1), (2), and (3), successively, from (4),
z 1 c
a
;; .-. * = -.
7x-3y = I (1)
llz-7w=l (2)
14. Given,
4z - 7y = 1 (3)
^ L9z 3w = l (4)
Combining (1) and (4),
21w 57y = 12 (5)
Combining (2) and (3),
28w 77y = 7 (6)
Combining (5) and (6),
3^ = 27; .-. y = 9.
Substituting and reducing,
x 4, z = 16, and u = 25.
T 1 (1)
15. Given,
Clearing of fractions and transposing,
llx - My = - 30 (3)
y = 6, and x = 12.
52 KEY TO DAVIEB' BOUEDON. [108-111.
^_l._ i - ._..
10 15 9 "~ 12 18
-
3~12 16 10
Clearing of fractions, transposing, and reducing,
39z 2y = 80 (3)
U5x + ty = 22G (4)
Combining (3) and (4),
193z = 386 ;
x = %, and y = 1.
PROBLEMS GIVING RISE TO SIMULTANEOUS EQUATIONS OF
THE FIRST DEGREE.
5. What two numbers are they, whose sum is 33 and whose
iifference is 7?
Let x denote the first, and y the second.
From the conditions,
* 4- y 33
*-y= 7;
whence, by combination,
x = 20, y = 13.
t>. Divide the number 75 into two such parts, that thret times fa
greater may exceed seven times the less by 15.
Let x dei.ote the greater, and y the less
From the conditions of the problem,
Ill, 112.] EQUATIONS OF THE FIRST DEGREE. 53
x -f- y = 75
3x - 7y = 15 ;
by combination, lOy = 210 ; .-. y = 21 ; also, x = 54.
7. In a mixture of wine and cider, ^ of the whole plus 25 gallons
was wine, and % part minus 5 gallons, was cider : how many gallons
were there of each ?
Let x denote the number of gallons of wine ;
and y " cider;
then will x -f y " mixture.
From the conditions,
x -4- v
J-^ + 25 = x
3
clearing of fractions, transposing and reducing,
y x = 50
2y + x 15
by combination,
y = 35 ; and x = 85
8. A bill of 120 was paid in guineas and moidores, and the
number of pieces of both >orts that were used was just 100; if the
guineas were estimated at 21s., and the moidores at 27s., how many
were there of each 1
Let x denote the number of moidores ;
and y " guineas ;
then, shce 120 = 2400s., we have, from the conditions,
54 KEY TO DAVIES' BOURDON. [112.
x + y = 100
27* + 21y = 2400 ;
by combination, % = 300 ; . . y = 50 ; also, x = 50.
9. Two travellers set out at the same time from London and
York, whose distance apart is 150 mile* ; they travel toward each
other; one of them goes 8 miles a day, and the other 7; in what
time will they meet?
Let x denote the number of miles travelled by the first ;
V
X
then will
o
and y -
From the conditions,
days
x + y = 150 ;
x__y_
8~7'
second ;
first;
second ;
whence, by combination,
x = 80 and
Uio,
the number of days.
10. At a certain election, 375 persons voted for two candidates ;
and the candidate chosen had a majority of 91 ; how many voted for
each?
Let x denote the number of votes received by the first ;
y " " " " " second;
from the conditions of the problem,
x + y = 375
x = y+ 91;
by combination, x 233, y = 142.
112.] EQUATIONS OF THE FIRST DEGKEE. 55
11. A's age is double B's, and B's is triple C's, and the sum of
all their ages is 140 : what is the age of each ?
Let x denote the age of A ;
y B;
z " " C;
from the conditions of the problem,
*= 2y (1);
y= 3z . . . . (2);
x + y + z = 140 . (3) ;
from (1) and ('2), x = Gz ;
substituting, y = 3z, and x = 62, in (3), and reducing,
102 = 140 ; . . z = 14, x = 84, y 42.
12. A person bought a chaise, horse and harness, for 60; the
norse came to twice the price of the harness, and the chaise to twice
the price of the horse and harness: what did he give for each?
Let x denote the number of pounds paid for the harness;
y u u u (( horse;
z " " " " " c'haise;
from the conditions of the problem,
y = 2x (1)
Z = 2(x + y) . . (2)
x + y + z = GQ (3)
from (2) and (1) z = Gx;
substituting z = Gx and y = 2x in (3)
Jte = 60, . ' . x = 6^, also, by substitution, y =. 13, z = 40;
hence, the price of the chaise was 40; of the horse 136*. Sd. ;
smd that of the hamss 6 13s. 4d.
56 KEY TO DAVIES' BOURDON. [112.
13. A person has two horses, and a saddle worth 50 ; now, if
the saddle be put on the back of the first horse, it will make his
value double that of the second ; but if it be put on the back of the
second, it will make his value triple that of the first : what is the
value of each horse ?
Let x denote the number of pounds the 1st horse is worth j
y " " ' " 2d " "
from the conditions of the problem,
x + 50 = 2y
y + 50 = 3x ;
whence, by combination,
x = 30 y = 40.
14. Two persons, A and B, have each the same income. A saves
of his yearly; but B, by spending 50 per annum more than A,
at the end of 4 years finds himself 100 in debt ; what is the income
of each ?
Let x denote the number of pounds in the income of A ;
y "...' B;
by the conditions of the problem, these are equal ; one only will be
used. Then will
4
-x denote what A spends per year :
5
-x + 50" " B " "
5
from the conditions of the problem,
4x+ 100;
whence, performing indicated operations, transposing and reducing,
4 Z = TO ~~ Wo = T
113-114.] EQUATIONS OF THE FIRST DEGREE. 61
21. A laborer can do a certain work expressed by a. in a time
expressed by b ; a second laborer, the work c in a time '.g the values of x, y and z, we have,
~
v = ms
"
1 m 1 m I m
r ns r ns
1 n 1 n 1
(13).
Had we represented the polynomial x + y + z by s, the alge-
braic work would be slightly diminished, but the preceding method
has been followed in order to show more clearly the process of solu-
tion.
30. Find the values of the estates of six persons, A, B, C, D, E,
F, from the following conditions : 1st. The sum of the estates of A
and B is equal to a ; that of C and D is equal to b ; and that of E
and F is equal to c. 2d. The estate of A is worth m times that of
C ; the estate of D is worth n times that of E, and the estate of F
is worth p times that of B.
1st Solution.
Let x denote the value of A's estate ;
a x " " " B's "
y ' " ; < C's "
b y " " " D's "
z " " " E's '*
c _ g F'S
from the conditions of the problem,
70 KEY TO DAVIES' BOURDON. [115.
* = my
b y = nz
c z = p (a x)
or,
x my = (1)
y + nz = b . (2)
px z = ap c (3)
combining (1) and (2),
x -f- muz = bm .... (4) ;
combining (3) and (4), and finding the value of 2,
bmp + c ap
z = ; , . which denote by P :
mnp + 1
combining (3) and (4), and finding the value of a?,
amnp + bm cmn
x = , which denote by O ;
mnp H- 1
substituting the last value in (1), and finding the value of y,
anp + b en ,.,, ,, D
y = * which denote by R :
mnp + 1
whence,
A's estate is equal to Q,
B's " " " a Q,
C's " " R,
D's " " b R,
E's " " P,
F-s " " c - P.
We might have combined (2) and (3), eliminating 2, and then from
this resulting equation, taken with (1), have found the value of a
and y.
2d. Solution
By a single unknown quantity,
Let x denote the value of C's estate ;
'hen will b - x " " D's
127.] INEQUALITIES. 71
mz denote the value of A's "
a mx " " B's "
p(a-mx) " " F's "
c p( a -mx)" " E's
and from the remaining condition of the problem,
b x = n[c p(a mx)]
whence, by the rule for solving equations of the first degree,
b -f 19, to find the smallest limit of x.
If we add 6 to both numbers of the inequality, we have
5z > 19 + 6, or, 5z> 25;
dividing both numbers by 5, we have
ar>5.
14
2. Given, 3x + x 30 > 10, to find the least limit of*.
m
Reducing, 3x+1x- 30 > 10, or, 10z 30 > 10 ;
adding 30 to both members of the inequality, and dividing by 10,
sre have,
z>4.
3. Given, J - + - f -^ > , to find the least 'imit of ST.
D O <6 <* '-
72 KEY TO DAVIES BOURDON. [127-149.
Multiplying both members by 6, we have,
x 2x -f 3z + 39 > 51 ;
reducing, subtracting 39 from both members, and dividing by 2,
we have,
x >6.
ax cfl
4. Given, - + bx ab > to find the least limit of x.
5 5
Multiplying both members by 5, we have
ax -f 5fce Sab > a 2 ;
adding -f 5ab to both members, and dividing by the co- efficients of
x, we have,
(a + 56) x > a (a + 56) ; or, x > a.
&C J2
5. Given, ax + a& < , to find the largest limit of x.
Multiplying both members by 7, adding 7a&, and dividing by
the co-efficients of x, we have,
(b 7a) x < b (b 7a) or, a < i.
REDUCTION OP RADICALS.
7
1. - . Multiply both terms by 3 + \f5.
3 A/5
y /K
2. Jl V 7=. Multiply ooth terms by VH V3.
3. Multiply both terms by 5\/l3 + 6\/5.
5A/I3-6-N/5
150.] REDUCTION OF RADICALS. 73
- t ^ terms
(5 - A/5) (1 + A/3)
5 -{- A/5 and 1 - A/3; this S 17
(3 + A/3) (1 - A/3) (3 + A/5) (5 + A/5) (A/5 -2).
(5 _ V5) (5 + A/5) (1 + A/3) (1 - A/3)
which, after performing the operations indicated, and reducing,
becomes,
- 40 A/15 80 A/3 + 80 \/3 + 32 A/15 _ 1 rr=
20z 2 ~5 V
, r ,, . , . / -
5. - - Multiplying both terms by \a-\-x
V a + x A/a a;
A/ ^j we have,
2a + 2 Aa 2
(7 -2 A/5) + 2 + A/45
6. -i - _1 = - . Multiplying both terms by
2 v 3 A/7
2 A/3 + A/7, the denominator becomes 5. For the numerator,
performing the multiplications and reductions, we have,
7 _ 2 V5~~+ 2 + A/45
2A/3+ A/7
U ^3 4 A/15 + 4 A/3 + 2 A/135
+ 7 A/7 2 A/35 + 2 A/7 + A/315.
But, 2 A/135 = 6 A/15; and A/315 = 3A/35;
hence, the product reduces to
18A/3 + 2A/15+ A/35 + 9A/7.
74 KEY TO DAVIES' BOURDON. [150-158.
But,
18 A/3 = 9 x 2 \/3; and 2 t _ ^..
5 5 >
whence by the rule,
_39 / 28
~ 10 V "r 5 " r7T7r *~ n ~
1521 _ 39 31.
100~~10 10'
v, j 8 4
hence, a; = 7, and- a; = = -.
10 o
12. Given ax -f- b = - a; 2 + -a?, to find the values of a;.
o b a
Clearing of fractions, &c.,
/*2 _, *T A
ti, tt/ I/ .
a
whence, by the rule,
a 2 - 6 A a* - 2a 2 5 + 6 2 a 2 - i a 2 +
:VA^
2a "V 4a 2 2a 2a
hence, a; = a, 2; =
8
a n f% / )** *7* n I /i o**^ r>2
y . *>* u -f- a x c;
1 "-J C-iivAn - /* I ^- _r .__ . /> i
c 2 c 2 ~ c H 2 <*'
to find the values of x.
159.] EQUATIONS OF THE SECOND DEGREE. 77
Clearing of fractions, &c.,
26 a 2 6 2
ffi __ _ /) _^^______
-
whence, by the rule,
_& A^=
~C V C 2
4- 1 - -
~~ *" ^ c c
b + a b a
hemce, x = - , and x =
c c
14. Given mx 2 + mn = 2m-y/n x + nx z , to find the values of x.
Transposing and reducing,
x z
m n m n
whence, by the rule,
m-\fn I mn m z n m-Jn n-\fm
_ y -4 A / ^_ _ _i_ -- T -^ * _ .
~ m n V m n (m n 2 m n m n '
(m n)
m/n n-vw
aence, z =
~V n
since, m n = (-y/m + -y/n) (-y/w -yA)^ (-^
also,
m y^T ny'm _ \/rnn(\/m -y/n)
V"
2rf Solution.
mx z -f- wiw = 2m -y/n"- ar + na;2 j
transposing the first term of the second member, we have,
78 KEY TO DAVIES' BOURDON. [159.
ma 2 2m -\fn x + mn = nx 2 ;
observing that the first member is the square of a binomial whose
terms are -y/m x ^/m y^ we have,
y#T x -y/m y/n = y"n x ;
-y/m - x q= -\/n ' x -y/m -y/jT
-y/n) x = y'my'n : hence,
\/mn \/nm
^ -- - and, x = L --
-y/m y n ym + yn
6a2 b*x ab -
1 5. Given ato 2 -- - -) -- =
c 2 c c 2
whence, 19z (10 x) + 3800 = 9 (10 + z) (10 ).
Reducing, z 2 19a; = 290 ;
19 , ,/i ;~36i 19 39
hence, = yy290 + = i
.-. a;'=29, a;"=-10.
24. Given ^=^ = -3 + -;
a; + <> *
whence, x (x* 5x) = (a? 9)a? + * + 3.
Reducing,
hence,
or
25. Given
whence, 2s 2 = 3 (x 2 x) + 2 (z 2 2z + 1) ;
72 7 , .749 2
hence, o^--^--, or ^-ij/---;
.-. a;' = 2 and " = -.
26 Given g + 2 s-2_5.
a;_2 s + a-8'
whence, 6 (a? + 4x + 4) 6 (z 3 4a; + 4) = 5 (& 4);
48
therefore, a? - a; = 4 ;
KEY TO DAVIES' BOUEDON.
, , . , , 24 , ,/576~~ 2426
and hy the rule, x=y -^+4= ; ;
5 iiiD O
.*. a;' = 10 and s"= f.
5
~ x x 6 a; 12 5
27. Given --: = - ;
whence, 6(z 2 12z+36) 6(z 2 24z + 144) =5(z 2 18a,-+ 72)
6
162 1008
.Reducing, a; 2 -- x = -- ;
_ 81 ./6561 5040 _ 81 39
= 1/ " " : ~~
oo n-
28. Given
25
' = 24 and a?" = ^
5
a; + 1 13
x + l^ x 6 '
whence, 6a; 2 + 6 (a? + 2z + 1) = 13 (a; 2 + x) ;
1 1 +5
hence, x = v 6^ = 5 ;
/& /*
.-. '= 2 and &"= 3.
1 23
29. Given - = ^ ;
whence, 5 ( + 2) 10 (x 2) = 3 (x* 4) ;
5 , 4 /~ ~25 5 23,
hence, = - ^ |/ 14 + ^ = g ;
.-. a?'=3 and x"= y.
, . 4 5 12
30. Given = + , , = , .. ;
whence, 4(a; 2 +5z+6) + 5(^+4a;+3)=12(a?+3a;+2);
27.
hence, x = ^
.. x'= 3 and x"= 5.
162.] EQUATIONS OF THE SECOND DEGREE. 83
i'ROBLEMS GIVING RISE TO EQUATIONS OF THE SECOND DEGREE
4. A grazier bought as many sheep as cost him 60, and after
reserving 15 out of the number, he sold the remainder for 54, and
gained 2s. a head on those he sold : how many did he buy 1
Let x denote the number purchased :
and x 15, the number sold ;
then will denote the number of shillings paid for 1 sheep,
9
and - the number of shillings received for each.
a: 15
From the conditions of the problem,
1200 _ 1080
x ~ z^l5 " '
clearing of fractions, &c.,
x 3 - + 45* = 9000 :
whence, by the rule,
_ 45 195
22'
hence. x = 75, and x 120,
the positive value only, corresponds to the required solution.
5. A merchant bought cloth for which he paid 33 15*., which he
sold again at 2 8*. per piece, and gained by the bargain as much
as one piece cost him : how many pieces did he buy ?
Let x denote the number of pieces purchased :
675
then will, denote he number of shillings paid for each,
fi4 KEY TO DAVIKS' BOURDON. [163.
and 48# the number of shillings for which he sold the whole.
From the conditions of the problem,
48* -675=^-
x
then, by clearing of fractions, &c.,
whence, by the rule,
225 225
, , , /* - . _ i
16 16
_225^ /
X ~ 32 " V
225 50625 225 255
~T6~ 1024 : ~ "32" " "32"
480
using the positive value only, x = = 15.
HI
6. What number is that, which being divided by the product of
its digits, the quotient will be 3 ; and if 18 be added to it, the order
of its digits will be reversed 1
Let x denote the first digit,
and, y " second "
ehen will Wx -f y denote the number.
From the conditions of the problem,
10* + y _ o
o,
xy
10a:-f-y+ 18 = 1
whence, by reduction,
finding the value of x in terms Df y from the second, and sub-
sti tnting in the first, we have.
163.] EQUATIONS OF THE SECOND DEGREE. 85
10y 20 + y = 3y 2 6y ;
whence, by transposing, &c.,
17 , 20
- y2 = - ; and,
by the rule,
17 / 20 , 289 , 17.
-6-* V ""a + S6 =+ 1T
taking the positive sign, y = 4 ;
whence, a, = 2, and the number is 24.
7. Find a number such that if you subtract it from 10, and mul
tiply the remainder by the number itself, the product will be 21.
Let x denote the number :
from the conditions of the problem,
(10 - *) x = 21 ; or, x z 10* = - 21 j
by the rule,
x = 5 y - 21 + 25 = 5 db 2;
whence, x =. 7, and x = 3.
8. Two persons, A and B, departed from different places at the
same time, and travelled towards each other. On meeting, it ap-
peared that A had travelled 18 miles more than B; and that A
could have performed B's journey in 15|- days, but B would have
been 28 days in performing A's journey. How far did each travel 1
Let x denote the number of miles B travelled ;
x + 18 " " " A "
" " A " in one day,
KEY TO DAVIES' BOTJKDON. [163.
denote the number of miles B travelled in one day ,
_ _l_ I Q
" " days A
15}
tx -f 18\
V28-;
B
from the conditions of the problem,
x+ 18 _ x_ x 2 H- 36s -f 324 _ x*
~>T ~* ~28 ~I5
clearing of fractions, and reducing,
2 324 _ 2916
'^T a ~'
By the rule,
162 /2916 26244
: :t ~~ ~~'
162 216
nence, using the upper sign,
378
* = - = 54.
9. A company at a tavern had 8 15s. to pay for their reckoning ;
but before the bill was settled, two of them left the room, and then
those whc remained had 10s. apiece more to pay than before: how
many were there in the company 1
Let x denote the number in the company.
163.] EQUATIONS OF THE SFCOND DEGREE. 87
175
Then will denote the number of shillings each should pay ;
" paid;
from the conditions of the problem,
17C -|7C
I/O I/O
* 2 *~ '" '
clearing of fractions,
175* 175* + 350 = 10* 2 - 20*;
whence, * 2 2* = 35.
By the rule,
x = 1 y36= 1 6;
using the upper sign, x = 7.
10. What two numbers are those whose difference is 15, and o*
which the cube of the lesser is equal to half their product?
Let x denote the smaller number ;
then will x + 15 " greater "
from the conditions of the problem,
*3 = \ (3.2 + 15*), or, x* = 1 (x + 15) ;
1 15
x
By the rule,
whence, x z -x = .
z
4- a -V~2~^16-4- I -T
using the upper sign, x = 3 ; hence, * -f 15 = 18.
11, Two partners, A and B, gained $140 in trade: A's money
88 KEY TO DAVIES* BOUKDON. [163.
was 3 months in trade, and his gain was $60 less than his stock : B's
money was $50 more than A 's, and was in trade 5 months : what
was A's stock?
Let x denote the number of dollars in A's stock ;
x + 50 " " B's "
x 60 " A's total gain ;
x 60
As gain per month ;
3
x 60
" A's " on 1 dollar ;
50) B's "
( - J (x + 50) 5 B's total gain.
\ OiP /
From the conditions of the problem,
x 60 + (* ~ 60) (x + 50) 5 = 140 ;
clearing of fractions, and reducing,
~~4~ a
By the rule,
105625 _ 325 475
64 ~ ~8~ IT
whence, x = 100.
12. Two persons, A and B, start from two different points, and
travel toward each other.. When they meet, it appears that A has
travelled 30 miles more than B. It also appears that it will take A
4 days to travel the road that B had come, and B 9 days to travel
175.] EQUATIONS OF THE SECOND DEGREE. 89
the road which A had come. What was their distance apart when
they set out?
Let x denote the number of miles B travelled ;
then will ar + 30 " " " A
2
A travels per day ;
x + 30
- days A "
(a:
V
-f 30
B
From the conditions,
x _ x_+ 30 x 2 _ x 2 + 60* -f- 900
r ' T~~ ~9"
whence, by reduction,
* 2 48* = 720
and by the rule,
x = 24 -y/^20 + 576 = 24 36 ;
taking the upper sign, x = 60, and x + 30 = 90 ;
hence, the distance is 150 miles.
EXAMPLES INVOLVING RADICALS OF THE SECOND DEGREE.
CL I CL^ 1 ' (ifi X
3. Given - + \ / - - = T ,
x V x z b
to find the values of x.
90 KEY TO DAVIES' BOURDON. [175.
Multiplying botn members by bz, and transposing,
b y/a 2 x 2 =i x 2 - ab ;
squaring both members,
cancelling 6 2 a 2 , dividing both merr.bers by a 2 and transposing,
x* = 2ab-b* .-. x=
4. Given \ /_^__? + 2
Multiplying both members by
z -f a
a;
multiplying both members by a:, and transposing,
2-y/aa; = b 2 x x a = (b 2 l)x a ;
squaring both members,
4ax = (b* - 26 2 + 1) x z - 2a (6 2 - 1) x + a 2 ;
transposing and reducing,
-,2
rX =
5* _ 26 2 -j- 1 5* 26 2
whence,
/ a 2 2 (6*
1 d: V ~ 6* - 26 2 + 1 + (6* -
- 26 2 -f 1 : ~ 6* - 26 2 + 1 (6* - 26 2 -f 1 ) 2 '
or,
~ b* 2b z + 1 ~ 6* - 26 2 + 1 '
now, b* - 2b z -f 1 = (b - 1) 2 (6 + I) 2 .
175.] EQUATIONS OF THE SECOND DEGREE.
Hence, taking the upper sign and reducing,
91
_
=
_q(64-l) 2 _ a(b + I) 2 a
= ~ ~
(62 _ t )2 - (b- l) 2 (6 4- I) 2 ~ (b - I) 2 '
and taking the lower sign and reducing,
_ a(b - I) 2 _ g(b 1)2 q
~ (52 _ !)2 - (b _ 1)2(6 4. 1)2 - (j + i)a'
or, uniting the two values in a single expression,
a
x
(b =F i) 2
a -J q 2 a; 2
5. Given, . ; = 0, to find x.
a 4- -y/ a 2 a; 2
Clearing of fractions, transposing, &c.,
squaring both members,
2 (i _ j)2 - (6 4. 1)2 ( a 2 _ 3.2) .
transposing and reducing,
2a -^b
(6 4- I) 2
x
6. Given,
a 7ia
, to find x.
Multiplying both terms of the first member by */x
and then dividing both members by a,
I n 2
2x a
clearing of fractions, &c.,
a: a
a,
92
KEY TO DAVTES BOUKDON.
[175.
(1 2n 2 )x (1 n 2 )a = 2w 2 -y/a; 2 ax ;
squaring both members, transposing and reducing,
x * _ aiLrJ^a, = -.L=^l*.
1 r 4-/i^ 1 - ^yj.2
whence by the rule,
^ ~ 1 ]J 9 " \/ 1 XI 9 ' 7l
1 4/i" 2 V 1 4^ (1 -
_
1
2n 3 a
- 4n 2 4n z ~ 1 4n*
Taking the upper sign, and dividing both terms of the fraction by
1 +2n,
2n + n a )a _ (1 - n) 2 q
_
2n
Taking the lower sign, and dividing both terms by 1 2n,
+ 2ra + ft 2 ) _ (1 + n) 2 a
_
+2n
a (I n} 2
taking the two values together, x = ^ -- =-*
n- V "" ' X \ "V a x I X t. c. j
7. Given v = + =\/ T , to find x.
IX
Multiplying both members by ^fx
/a + x -f- i/o a; = ~~= >
V*
squaring ooth members,
175.] EQUATION* OF THE SECOND DEGREE. 93
squaring both members,
cancelling 4o 2 in the two members, and dividing both by x*,
_ x 2 4a
= P~~b'
clearing of fractions, &c.,
z 2 = 4ab 46 2 . . x = 2 Jab &T
a x - , c ,
8. Given ? to find x.
a -+ x
Clearing of fractions, transposing and factoring,
^/2ax + x 2 = (b 1) (a + x).
Squaring both members,
2ax + x 2 = (6 2 -26 + 1) (a 2 + 2a* + * 2 ) ; or,
2a* + z* = (6 2 26) (a 2 + 2a* + * 2 ) + a 2 + 2a -f-
whence, by reduction,
/I _ 26 +
-
whence, by the rule,
=
-
26
a 1 -
taking the upper sign, x =
<
a(\
taking the lower sign, x = -- J
whence, a? =
KEY TO DAVIES' BOtTRDON. [175-178.
2d Solution.
Make x -f a = y ; whence, 2ax -f z 2 = y 2 a*
substituting in the equation, and clearing of fractions,
y -f yy a 2 = iy ;
transposing, &c.,
V> 2 a 2 = (5 1) y ;
squaring both members, and cancelling,
- a 2 = (i 2 - 26) y 2 ;
whence, solving with respect to y,
a
j """ **-
substituting for y its value, &c.,
x = a
, ,.
whence, as before,
^/26 - & 2
r ^a(l--
-V/26 - 6 2
TBINOMIAL EQUATIONS
6. Given x* (2bc + 4a 2 ) ar 2 = 6 2 c 2 ; to
By the rule,
4o* = e + 2u 2 2a v f
7. Given 2 - 7 y^ = 99 ; or, 2z - 99 = 7 ^/i
Squaring both members
4* 2 396 + 9801 = 49a? ;
178-181] EQUATIONS OF THE SECOND DEGREE. 95
transposing and reducing,
445 9801
* = -
Ahcnce, by the rule,
445 / 9801 , 198025
ff _ L __ f __. -^. * / __ __ I _ _ , _
* ~~~ <" ~^ \ / /* j J
445 203
or, x = ^ :
648
taking the upper sign, x = - = 81
o
242 121
lower sign, x = r- = r
8 Given, ^ bz* + C -x* = 0, to find*;
transposing and reducing.
whence, or = 1
reducing, * = J
V
2W
EXAMPLES OF REDUCTION OF EXPRESSIONS OF THE FORM OF
Vb.
4. Reduce to its simplest form, \7^ ~
a = 28, 5 = 300. c = 22.
KEY TO DAVIES BOUEDON. [181.
Applying the formula and considering only the upper sign,
v/
28-f 10 1/3 =
5. Reduce to its simplest form, */l + 4^3.
o=l, b - 48, c = 7 ;
applying the formula, &c.,
+ 4-v/- 3 = 2
6. Reduce to its simplest form,
_ 6 2 _ ^J bc _
a = ic, i = 46 2 (6c - 6 2 ), c = b (c - 2b) ;
applying the formula to the 1st radical,
+ 26 -vA - * 8 =
applying the formula to the 2d radical
subtracting the second result from the first,
6 ytc b 2 bc 2b -foe b 2 = 2A.
7 Reduce to its simplest form,
\/ab -r 4c 2 d 2 2
a = aft -f 4c 2 ' P = ZVU-&
or, p = % and p = ~.
4
Using, p = -, we have from (3),
x'= + 8 and x"= 8; whence, y'= -f 2 and y"= 2.
9.
Making
9 j s2 + 4^ + 4^ = 256 (1)
(3y z a^ = 39 (2)
x 1 -\- kpx* + 4j0 2 .r 2 = 256 (3)
3p*x* x* = 39 (4)
From (3) and (4),
1 -(- 4p
Equating,
256 39
ufi&06
768P 2 - 256 = 39 + 156jo + 156J0 3 ,
13 295
and P 2 ^rP=7^r;
KEY TO DAVTES' BOURDON. [184
hence,
78 432 5 59
p -- -eia-' or ^ = 6 and * = -ioa'
Using the first value, we have from (6),
a;2 = 36; . 3'= 6 and x"= 6;
and by substitution,
y'= 5 and y"= 5.
Using the second value, we find,
a' =102 and x" =102;
and by substitution,
y'= 59 and y"=. 59.
10 J6(a? + ^) = 13ajy ...... (1)
(z2 2/2 = 20 ........ (2)
Making
6(1 +^ 2 )a; 2 =: 13^ 2 ...... (3)
(1-^2)^ = 20 ....... (4)
From (3), we have,
6 + 6p 2 = I3p.
Reducing,
_ 13 . /169 _ 144 _ 13 5
'* ^ ~ 12 ^ 144 ~~ 144 ~ 12 '
or
^ = H and p = -.
188.] EQUATIONS OF THE SECOND DEGREE. 101
Using the second value of p, we have from (5),
on
a? = T --=4 = 36, or x'= 6 and x"= 6 ;
1 f
and by substitution, y' 4 and y"= 4.
EQUATIONS OF A HIGHER DEGREE THAN THE FIRST, INVOLVING
MORE THAN ONE UNKNOWN QUANTITY.
( x 2 u + xv 2 = 6 ... (1) )
15. Given, < f- to find x and y.
tzy + zy = i2.. . ( 2 )>
Dividing (2) by (1), member by member,
2
xy = 2, or a; =
Substituting this value of x in (1) and reducing,
* + * = ;
clearing of fractions and reducing,
y 2 - By = - 2 :
whence, y = | 1 /- 2 + i = i I;
or, y = 2, and y = 1 ;
whence a? = 1, and a; = 2.
16. Given, J *' + * + y = 18 ~ y2 ' ' ' 0) It o find , and y.
( xy 6 (2) )
Multiplying both members of (2) by 2 and adding the resulting
equation to (1), member to member,
z' -f 2*y + y 2 + * \-y = 30,
or, (x + y) 2 + x + y = 30 ;
102 KEY TO DAVIES' BOURDON. [189.
whence, by the rule,
whence, x -\- y = 5, and x + y = 6.
Taking the first value of x + y and substituting in it, for y its
/
value derived from (2),
x
x + = 5;
x
clearing of fractions and reducing,
x z 5x = 6 ;
. 5 / _ , 25 5 ^ 1
whence, a . = _y/_6 + = -- ;
ur, x = 3, and # = 2 ; whence, y = 2, and y = 3.
Taking the second value of x + y and proceeding as before,
clearing of fractions, &c.,
a; 2 + 6z = 6 ;
whence, x 3 / 6 + 9 = 3
and by substitution, y = 3
PROBLEMS GIVING RISE TO EQUATIONS OF A HIGHER DEGREE
THAN THE FIRST CONTAINING MORE THAN ONE UNKNOWN
QUANTITY.
2. To find four numbers, such that the sum of the first and fourth
shall be equal to 2s., the sum of the second and third equal to 2/,
the sum of their squares equal to 4c 2 , and the product of the first
%nd fourth equal to the product of the second and third.
189-191.] EQUATIONS OF THE SECOND DEGREE. 103
Assuming the equations,
+ 2 = 2* (1)
x+y = W .... (2)
w 2 -(- z 2 + y 2 + z 2 = 4c 2 . . (3)
uz = xy (4)
Multiplying both members of (4) by 2, and subtracting from (3),
member from member,
u 2 %uz -f- z 2 + a; 2 -f 2ary + y 2 = 4c 2 ;
or, ( z) 2 + (x 4- y) 2 = 4c 2 .
Substituting for x + y its value 2s' and transposing,
(u - z) 2 = 4c 2 - 4s' 2 ,
or, u z -v/4c 2 4s' 2 .
Combining with (1),
,/4c 2 - 4*' 2
and, z s v - ~ - s ^& s' 2 ;
reversing the order of the members of (4), and proceeding as before,
we find in like manner.
and,
4. The sum of the squares of two numbers is expressed by a,
and the difference of their squares by b : what are the numbers?
Let x and y denote the numbers.
From the conditions of the problem,
J04: KEY TO DAVIEs' BOURDON. [191.
x 2 + y 2 = a - (1)
z*-y* = b . (2).
By adding, member to member,
by subtracting,
5. What three numbers are they, which, multiplied two and two.
and each product divided by the third number, give the quotients
, 6, c ?
Let #, y and 2, denote the numbers
From the conditions of the problem,
- = a or, xy =z 02 . (1)
P
= b or, yz = bx (2)
#2
= c or, xz = cy (3).
Multiplying (1), (2) and (3) together, member by member,
x z y z z 2 = abcxyz ;
dividing both memoers by xyz,
xyz = abc (4) ;
substituting in (4) the value of xy taken from (1). and dividing both
members by a,
2? = be . . z =
Substituting the value of yz and dividing by b,
191.] EQUATIONS OF THE SECOND DEGREE. 105
x 2 = ac . ' . x = Joe.
Substituting the value of xz and dividing both members by c,
y z = ab .'. y
6. The sum of two numbers is 8, and the sum of their cubes is
152 ; what are the numbers ?
Let x and y denote the numbers.
t
From the conditions,
x + y = 8 ..... (1)
z3 + y3 = i52 ..... (2);
cubing both members of (1),
x 3 + 3z 2 y + 3xy* + y 3 = 512 (3) ;
subtracting (2) from (3), member from member, and dividing both
members by 3,
x z y + xy z = 120 (4) ;
-
substituting the value of a; taken from (1),
(64 - 16y + y 2 ) y + (8 - y) y z = 120,
or, reducing, y 2 Sy 15 ;
whence, y = 4 -y/ 15 + 16 = 4 1 . . y = 5, y = 3;
whence, from (1) a; = 3, a? = 5.
7. Find two numbers, whose difference added to the difference ol
their squares is 150, and whose sum added to the 3um of theii
squares, is 330.
Let x and y denote the numbers.
From the conditions of the problem,
106 KEY TO DAVIES' BOURDON. [191.
a?_y2 + *-y=150 (1),
' y *^ y \ J t
adding member to member, and reducing,
x 2 + x = 240 ;
whence, x = - -/240 -f ^ = ^ y J
or, considering only the positive solution, .
x = 15;
whence, from (1), by substitution,
8. There are two numbers whose difference is 15, and half their
product is equal to the cube of the lesser number : what are the
numbers 1
Let x and y denote the numbers.
From the conditions of the problem,
*-.y=15 (1),
f - (2);
qp
substituting in (1) and dividing both members by 2,
2 _y _ 15.
y 2~ 2 '
whence,
15 J_ _ 1 11
considering only the positive solution,
y = 3 ; whence, from (1), x = 18.
191-192.] EQUATIONS OF THE SECOND DEGBEE. 107
9. What two numbers are those whose sum multiplied by the
greater, is equal to 77 ; and whose difference, multiplied by the
lesser, is equal to 12?
Let x and y denote the numbers.
From the conditions,
(x + y)x = n, or z*+'zy^W . . . (1) ;
(x-y)y=l2, or xy - y* = 12 . . . (2);
make x = py ; whence,
Tt t or, ya = -JL- . . . (3),
(P -I)y 2 =12, or, y* = j . . . (4);
equating the second members and reducing,
2 _65 _77 i
P l'2 P ~ 12'
65 / 77 , 4225 65 23
whence, p = ^/ - -+ = .
taking the upper sign,
/Qi* O
substituting k (4),
whence, x = -y/2~;
i. i. , 42 21
taking the lower sign, p = ;
= 514 . . (2),
multiplying both members -of (1) by 2, adding and subtracting the
resulting equation to and from (2), member by member,
ac* -f 2xy + 7,2 = 1024 . . (3)
*2-2zy + y 2 = 4 . . . (4);
extracting the square root of both members,
x + y = 32,
*- y = 2,
whence, x = 17 ; y = 15.
13. There is a number expressed by two digits, which, when
divided by the sum of the digits, gives a quotient greater by 2 than
the first digit ; but if the digits be inverted, and the resulting num-
ber be divided by a number greater by 1 than the sum of the
digits, the quotient will exceed the former quotient by 2 : what is
the number?
110 KEY TO DA VIES' BOURDON. [192.
Let x and y denote the digits ; then will
lOx + y lenote the number.
From the conditions,
clearing of fractions, and reducing,
Sx y - x z - xy = (3),
6y 4z x z xy = 4 (4) ;
by subtraction,
12*-f4
fy-12* = 4; .-. y = ;
substituting in (3),
12* + 4 12^ + 4_
0>c 7 ~
clearing of fractions and reducing,
40 4
-* = -
_20 / 4_ 400 _ 20 18
"19 V 19 + 361 ~ 19 19*
Taking the upper sign, gives x = 2 ; whence, y = 4.
14. A regiment, in garrison, consisting of a certain number of
companies, receives orders to send 216 men on duty, each company
to furnish an equal number. Before the order was executed, three
of the companies were sent on another service, and it was then
found that each company that remained would have to send 12 men
additional, in order to make up the Complement, 216. How-mam
192.] EQUATIONS OF THE SECOND DEGREE. Ill
companies were in the regiment, and what number of men did each
of the remaining companies send ?
Let x denote the number of companies, and
y " " " each should send ; then,
y + 12 will denote the number sent by each.
From the conditions of the problem,
*y = 216 (1),
(*-3) (y+12)=216 . (2);
Performing operations, subtracting and reducing,
4x y = 12. . . y = 4a; 12 ;
substituting in (1), 4z 2 - 12* = 216, or ** 3z = 54 ;
whence,
taking the upper sign,
x = 9 ; hence, y = 24, and y + 12 = 36.
15. Find three numbers such, that their sum shall be 14, the
sum of their squares equal to 84, and the product of the first and
third equal to the square of the second.
Let x, y and z denote the numbers.
From the conditions of the problem,
x + y + z = 14 . (1),
* 2 + y 2 + * 2 = 84 . (2),
yz = y 2 (3).
Multiplying both members of (3), by 2, adding to (2) and -educing,
KK1 TO DAVIES* BOURDON. [192.
from(l), x + z = U-y (5);
equating the second members of (4) and (5) and squaring,
84 + y 2 - 196 28y + y 2 ; . . y = 4.
Substituting in (1) and (3),
x + z = 10 . . . . (6)
16
xz = 16 . .
Substituting in (6) and reducing,
2 2 _ 10* = - 16
2 = 5 ,./9~= 5 =fc 3,
V *M
= 8; a = 2,
and by substitution, x = 2 ; a; = 8.
16. It is required to find a number, expressed by three digits,
such, that the sum of the squares of the digits shall be 104; the-
square of the middle digit to exceed twice the product of the other
two by 4 ; and if 594 be subtracted from the number, the remainder
will be expressed by the same figures, but with the extreme digits
reversed.
Let a?, y and z denote the digits ;
then, lOOx + lOy + * will denote the number.
From the conditions of the problem,
z 2 + y 2 + * 2 = 104 . . . (1)
y z Zxz = 4 . . . (2)
100* + 10y + z - 594 =1002 + lOy 4- x (3) ;
193.] EQUATIONS OF THE SECOND DEGREE. 113
subtracting (2) from (1), member from member,
x 2 -i 2xx + z 2 - 100 . . x + z 10 ;
reducing (3) x z = 6 ;
hence, x = 8, and z = 2.
By substitution, y = 6, and the number is 862.
17. A person has three kinds of goods which together cost
$230^. A pound of each article costs as many fa dollars as there
are pounds in that article : he has one-third more of the second than
of the first, and 3^ times as much of the third as of the second .
How many pounds has he of each article ?
Let ar, y and z denote the number of pounds of each article.
From the conditions of the problem,
or x " + * + * = 6525 1
4 16
y = \* .'. 2/ 2 = y 2 .-. (2)
2 = -y .-. z=-2%, and, z 2 = z 2 ... (3) ;
substituting in (1) and reducing,
* 2 = 225 .'. . x = 15,
substituting in (2) and (3),
y = 20 ; z = 70.
18. Two merchants each sold the same kind of stuff: the second
sold 3 yards more of it than the first, and together, they received
35 dollars. The first said to the second, " I would have received 24
dollars for your stuff." The other replied, "And I would have
114 KKY TO DAVTES' BOURDON. [193.
received 12^ dollars for yours." How many yards did each of
them sell?
Let x and y denote the number of yards sold by each.
24
Then will denote the price the first received per yard,
25
and Q- will denote the price the second received per yard.
From the conditions of the problem,
* + 3 = y,
+ . = 35, or, 4S* 2 + 25y 2 = ?0zy :
substituting in the second equation the value of y taken from the
first,
48* 2 + 25 (x z + Qx + 9) = 70* 2 + 210* ;
reducing, x 2 - 20x = 75 ;
whence, x = 10 -y/25^ = 10 5,
/
or, ar = 15 ; x = 5 ;
substituting, y = 18 ; y = 8.
19. A widow possessed 13000 dollars, which she divided into
two parts, and placed them at interest, in such a manner, that the
incomes from them were equal. If she had put out the first portion
at the same rate as the second, she would have drawn for this part
360 dollars interest ; and if she had placed the second out at the
same rate as the first, she would have drawn for it 490 d( liars
interest. What were the two rates of interest ?
Let x and y denote the rates per cent.
Let z denote the 1st portion ; then will 13000 z denote the 2d.
224.] EXTRACTION OF ROOTS. 115
P roni the conditions of the problem,
xz (13000 z\ y
or, = 1300<*-* (1),
j^ = 360, or, 2 y = 36000 . . . (2),
p^-? = 490, or, 13000* -zx = 49000 (3).
Substituting in (1) the values of zy and zx taken from (2) and (3)
and reducing, we find, x = y + 1.
Substituting this value of x and the value of z taken from (2) in
(1), and reducing, we find
72 36
v -- v :
y 13 y 13'
36 /36 1296 36 42
whence, y -.= - ^ - + = __; .-. y = 6 ;
by substitution,
* = 7, and = 6000, 13000 - 2 = 7000.
ADDITION AND SUBTRACTION OP RADICALS.
2. 3 A/4a 2 :
3. 2 A/45 = 6 A/5; 3A/5, .'. ^4/w. 9A/57
4. 3a V* 2c A/*, .'. Ans. (3a 2c) V*-
5. 3 Vi2 2 V2a ;
3A/4^ = 3A/2a; .'. Ans. V%
6. A/243 = 9 A/3; A/27 = 3\/3; A/48 = 4 A/3. -4ws. 16 A/3.
116
KEY TO DAVIES' BOURDON. [224-225.
7.
5 A/720 5 =
= 56
. (13a - 56) VSai
8.
9.
10.
5.
6.
V6 4
16a 4 = 2a yb + 2a, and
+ 2,
; .'. Ans.
MULTIPLICATION OF RADICALS.
X 3 = 1 /64 x
* /r 3 /r i2 /r /T 12 /"
V 2 X V 3 =: V 8 X Vil =! V 648 ; ' ' ^ "
6 /2 10
- VI- 4
^; 3 10"= 3/100; .-. ^ns. 6 337500
10
8.
. 048000.
7/4 42 /4096 3/1 42/i
O ./ t / . 4/_ / . 14 /ft
'VI V 729 ' V2~ V 10384' V
. ' . Ans.
10. The product by the rule for multiplication is
28 . ft . n . 10 43 . H 43 . 13
42 /2
226-227.] TRANSFORMATION OF RADICALS.
9w 2 3m
117
11.
12. The product equals
3m " 3m
.-. Ans. = Va? + P.
DIVISION OF RADICALS.
2. 2 V3 x V4 = 2 v^729 x 1 4 c~ m multiplied by
-- + -=-; 4 5 = 1; m + w = w m;
hence, 6a~ 4 c- m x
4. (q^) multiplied by ^a*.
3341 3
v * v 2
3 X 3~9' 3 X ~3'
hence,
239.] TKANSFOBMATION OF RADICALS.
1 1 1_ J L . 1 1 !_?.
3 X 3 X 3 ~ 27 ' o + o + o o 5
(q 6
6. at divided by a~*.
\ 3 !
hence, -<,*= _
j 3_17.
S i V 4l~~3 4~12'
, 11
hence, at ^- at = a 1 ^.
7. a* divided by .
3_4_ 1
4 5~ 120 J
hence, a* -j- a^ = a~^V.
8. a x i* divided by a~ zb
2_/l\_ 2 1__9_. 3_7_ _1.
5~\2/~~ 5 + 2~10' 4 8~ 8'
hence, a*b* -$- a~%b = flifrJ .
9. 32a^J 6 c^ divided by
= ; -
hence, 32aii0
10. 64a 9 jVt divided by
64-32 = 2; 9-(-9)=18; |-
hence, 64a 9 jV^ -r- 32a-5~ *c~* = 4a 18 i 5 .
11. VS x Vji x
hence, x x
12. Reduce ? , x to its s i mp i es t terms.
120 KEY TO DA VIES' BOUEDON. [239.
Cancelling the
3. Transform the equation
a? _ Q X Z + y x _ 10 = o
into one whose second term shall be wanting.
Here we make the roots of the resulting equation greater
p
than those of the given equation by = 2 ; that is,
o
we make them less than those of the given equation by + 2.
OPERATION.
l_6 + 7- 10 [2
+ 2_8 2
1st quotient, 1 4 1, 12 1st rem.
+ 2-4
2d quotient, 1 2, 5 2d rem.
+ 2
3d quotient, 1, 3d rem.
132 KEY TO DAVIES' BOITKDON. [349.
Hence, the transformed equation is
ys _ 5y _ 12 o.
4. Transform the equation,
a^ + gz 8 3 + 4 =
into one whose second term shall be wanting.
Here we make the roots of the resulting equation greater
p
than those of the given equation by + = 3 ; that is, we
o
make them greater than those of the given equation by 3.
Therefore, the synthetical divisor is 3.
OPERATION.
1 + 9 i + 4|_ 3
3 18 + 57
1st quotient, 1 + 6 19,+ 61 1st rem.
1 3 9
2d quotient, 1 + 3, 28 3d rem.
-3
3d quotient, 1, + 3d rem.
Hence, the transformed equation is
y& _ %sy + 61 = 0.
5. Transform the equation
& _ 8z3 + 7s 2 + 3x + 4 =
into one whose second term shall be wanting.
Here we make the roots of the resulting equation greater
p
than those of the given equation by = 2 ; that is,
wo make them less than those of the given equation by + 2 ;
hence, the synthetical divisor is + 2.
249-259.]
EQUAL ROOTS.
133
OPEBATION.
1-8+ 7+ 3+ 4 [2
+ 2 12 10 14
1st quotient, 1 6 5 - 7, 10 1st rem.
+ 2 8-26
2d quotient, 1 4 13, 33 2d rem.
+ 2- 4
3d quotient, 1 2, 17 3d rem.
+ 2
4th quotient, 1,+ 4th rem.
.-. y 1 17y 2 33# ]
EQUAL ROOTS.
4. What are the equal factors of the equation
8s -f 36 = 0?
The first derived polynomial is
7* - 42s 5 + 50.r 4 + 88z 3 - 129* 2 - 70* + 48,
and the common divisor between it and the first member of the
given equation, is
The equation
x* 3z 3 + Sx 2 + 7x -f- 6 = 0,
cannot be solved directly, but by applying to it the method of equal
roots ; that is, by seeking for a common divisor between its first
member and its derived polynomial,
4z 3 - 9z 2 - Ox + 7,
we find such divisor to be x -f- 1 ; hence, x -\- 1 is twice a factor of
the first derived polynomial, and three times a factor of the fir?t
member of the given equation (Art. 271).
134 KEY TO DAVIES' BOURDON. [359.
Dividing,
6 = 0, by (a +!)* = * 2 + 2ar + 1,
we have, x 2 bx + 6,
which being placed equal to 0, gives the two roots
x =. 2 and x = 3,
and the two factors, x 2 and x 3.
Therefore, (a: 2) and (# 3), each enters twice as a factor of the
given equation ; hence, the factors are
(x - 2) 2 (x 3) 2 (x + I) 3 . Ans.
5. What are the equal factors of the equation
33** + 27z - 9 = ?
The first derived polynomial is
7* 6 - 18* 5 + 45z 4 - 76z 3 + 81* 2 - 66* + 27,
and the common divisor between it and the first member of the
given equation, is
x* 2x 3 + 4z 2 Qx + 3.
The equation
x* 2x 3 + 4z 2 Gar + 3,
cannot be solved directly, but by applying to it the method of equal
roots, as in the last example, we find the derived polynomial to be
4x 3 6x 2 + 80; 6
and the common divisor to be x 1 ; hence, (x 1) enters twice
as :i factor into the derived polynomial, and three times as a factor
into the first member of the given equation.
Dividing
x t 2a; f 4z 2 6* + 3 by (x I) 2 = a; 2 2x + 1,
359-375-384.] SMALLEST LIMIT IN ENTIRE ROOTS. 135
we have for a quotient x z + 3 ;
hence, (z 2 + 3) enters twice as a factor into the first mem-
ber of the given equation ; hence the factors are
(x - I) 3 (x + 3)3.
SUPERIOR LIMIT IN ENTIRE ROOTS.
2 What is the superior limit of the positive roots in the equation
x s _ 3^4 _ Q X 3 _ 25z 2 + 4z 39 = ?
Recollecting that if we use x for x' (Art. 285), we have
X = x 5 3z* 8z 3 25z 2 + 4z 39
T = 5z 4 - 12z 3 - 24z 2 - 50* + 4,
Z = 20x 3 - 36a; 2 - 48* - 50,
V= 60s 2 - 72z - 48,
TF= 120* - 72.
r=120
The least whole number that will render all these derived polyno-
mials positive is 6 ; hence, 6 is the superior limit.
3. What is the superior limit of the positive roots in the equation
x 5 5z 4 13* 3 17x 2 - 69 = 0.
The process of solution is the same as in the last example, and the
limit is found to be 7.
COMMENSURABLE BOOTS.
2. What are the entire roots of the equation
x* 5x 3 + 25* 21 = ?
The divisors of the last term are + 1, I> +3, 3, -|- 7, 7
+ 21, and -21: L = 22; - L"= - 4.
KEY TO DAVIES' BOURDON. [384.
+ 21, +7, -f-3, +1, -1, -3, -
- 1, - 3, -7, - 21, + 21, + 7, +
+ 24, 4- 22, 4- 18, 4- 4, 4- 46, + 32, +
4- 3, 4- G, 4- 4, -46,
4- 2, 4- 4, + 46,
3, - 1, + 41,
- 1, - 1; -41,
therefore, -}- 3 and -f 1 are the two entire roots. Dividing the
first member of the equation by the product of the factors
(x - 3) (x - 1) = x> - 4x + 3,
we have x 2 y = 0.
NOTE. In the 4th line we add the co-efficient of x*, which is 0, and then
divide by the divisors, and thus obtain the 6th line.
3. What are the entire roots of the equation
15z 5 - 19* 4 + 6a; 3 + 15* 2 - I9x + 6 = ?
+ 6, + 3, + 2, 4- 1, - 1, - 2, - 3, - 6,
4- 1, 4- 2, 4- 3, 4- 6, 6, -3, 2, - 1,
-18, -17, -16, -13. -25, -22, -21, -20,
- 3, - 8, -13, t-25, + 11, + 7,
+ 12, + 7, + 2, + 40, + 26, + 22,
4-2, 4-2, -40, -13,
+ 8, +8, -34, - 7,
+ 8, 4-34,
-11, 4-15,
-11, -15,
4- 15:
hcuoe, there is but one entire root, which is -- 1
384-392.] NUMBER OF REAL ROOTS. 137
4. What are the entire roots of the equation
9z + 30s 5 + 22Z 4 + 10s 8 + 17s 2 20z + 4 = ?
This is worked like the preceding example, giving the en-
tire root, 2. Then dividing the equation by x + 2, we
find a new one, which has a root, 2.
NUMBER OF REAL ROOTS.
3. What is the number of real roots of the equation
a?_5z2_|_ 8z 1=0?
By finding the expressions which indicate, by their change of
sign, the existence of real roots (Art. 293 and Example 1), we
have
X= z 3 5y?+Sz I
X,= 3Z 2 IQx + 8
X 8 = 2x 31
X 8 = 2295
x = oo gives 1 2 variations,
x = + oo gives + + H 1 variation ;
hence, there is one real and two imaginary roots (Art. 293).
For x = 0, we have 1 2 variations,
for x = 1, " + H 1 variation ;
hence, the real root lies between and + 1.
4. Find the number, places, and limits of the real roots of
z* Sz 3 + 14s 2 + Ix 8 = 0.
For solution, see Example 3, page 141 of Key.
5. Find the number, places, and limits of the real roots of
the equation
a? _ 23x 24 = 0.
KEY TO DAVIES' BOURDON. [393-396.
X = y? 23x 24
X 1= 3z 2 23
X 2 = 23z + 36
X 3 = 8279.
For x oo, we have, H 1-, 3 variations.
For x= + oo, we have, + + + +, no variations.
Hence, there are 3 real roots, which are easily placed.
6. In the sixth example, we have,
X = y? + f z 2 2x 5
X 1= 3z2 + 3x 2
X 2 = llx + 28
X 3 = - 1186.
Hence, there is but one real root, and that lies between
the limits 1 and 2.
CUBIC EQUATIONS.
1. What are the roots of the equation
X 3 _ 6z 2 + 3* = 18 . . (I) 1 ?
Transforming so as to make the second term disappear,
z 3 - 9* -28 = (2)
^= 9 q = 28 ;
substituting in Cardan's formula, and reducing,
* = 4.
But the roots of the given equation are greater than those of equa-
tion (2) by 2 ; hence x = 6 4
396-405.] CUBIC EQUATIONS. 139
Transposing 18 in equation (1), and dividing both numbers by
x 6, we find
a: 2 + 3 = . ' . * = v / 3;
hence the three roots are 6, -y/ 3, ^/ 3.
2. What are the roots of the equation
X 3 _ 9^2 + 28z = 30 .... (1)?
Transforming, we find
x 3 + x (2) .'. x=0 and a; = v '~ 1.
But the roots of (1) are greater than those of (2) by 8;
hence the roots of (1) are, 3, 3 + T/ 1 and 3 y' 1.
3. z 3 7z + 14 = 20 (1)
Transforming (Bourdon, Art. 266), we have,
7 344 7 344
3*~T~ ^' 6 ' ^ = ~~3' 9 ' = ~T7~ 4
g
Substituting in Cardan's formula, we have ^ , the real root.
o
7
But the roots of (1) are greater by - than those of (2) ; hence,
B
in (1) x = 5.
Transposing and dividing by x 5, we have,
a;2 _ 2x + 4 = ;
hence, the required roots are, 5, 1 + V3, and 1 V &
HORNER'S METHOD OF SOLVING NUMERICAL EQUATIONS.
1. 3? + z 2 + x 100 = 0.
By Sturm's Rule, we find
140
KEY TO DA VIES 5 BOURDON.
[405.
X r= X 3 + X Z + X - 100,
Z l = 3* 2 -f 2* + 1
X z = - 4x + 899
X, = - 2409336.
For cc = oo , 1--| ,
for * = + CD , -f H ,
hence, there is but one real root.
Tor x = 4, . + + .-,
for x = 5, + + + _ ,
hence, the real root lies between 4 and 5.
2. a* - 12* 2 4- 12* - 3 = 3.
By Sturm's Rule,
X =x* - 12* 2 + 12* -3,
JTj = 4* 3 - 24* + 12, or x 3 - Qx + 3,
2 variations,
1 variation ;
2 variations,
1 variation ;
X, = 13*- 9,
X 4 = 20.
For x oo , + -
for x = + oo , H \-
hence, there are 4 real roots.
For x = 4 H f-
for * = 3 \-
for * = h +
for x = -f 1 =F
for .T = + 2 h
for * = + 3 + + 4-
4 variations,
variation ;
4 variations,
3 variations,
3 variations,
1 variation,
1 variation,
variation ;
406.] HORNER'S METHOD. 141
hence, one of the roots lies between 4 and 3, two between
and 1, and the remaining root lies between 2 and 3.
3. x* - 8* 3 + 14* 2 + 4z - 8 = 0.
By Sturm's Eule,
X = x* Sx 3 + 14* 2 + 4* 8,
X I = 4* 3 24* 2 + 28* + 4, or x ? 6* 2 -f 7* + 1,
X 2 = 5z 2 - 17* -j- 6,
X 3 = 76* - 103,
X< = 45475.
For x = co , H 1 ( 4 variations,
for = + 00, + + + + + variation ;
hence, the equation has 4 real roots.
For * = 1 + h 1- 4 variations.
for*= h-l h 3 variations,
for orrr-fl + + f- 2 variations,
for x = + 2 H 1 H 2 variations,
for z=:-f3 + + 1 variation,
fora;= 5 H + + 1 variation,
fora;= 6 + + + + + variation ;
hence, one root is between 1 and 0, one between and 1, one
between 2 and 3, and one between 5 and 6.
4. x 5 I0x 3 + Qx + 1 0.
By Sturm's Rule,
X = x 5 - 10* 3 + 6* + 1,
X I - 5* 4 - 30* 2 + 6,
X 2 = 20* 3 - 24* - 5,
X 3 = 96* 2 - 5* - 24,
X 4 = 43651* + 10920,
X,, = 32335636224.
142
KEY TO DAVIES' BOURDON.
[406.
For x = ~ oo 1__. I- __'_!_
for ar=+ao -f. + _}. _j_ _j_ _j_
hence, the equation has 5 real roots.
For x = 4
for
for
for
for
for
for
hence,
ar=-3
x= I
x =
* = + !
x- +3
x =. + 4
one roc
i i i
+ 4- - + ~ +
4- 4- ~ +
+ + -- + +
4- 4- 4- +
+ + + + + +
one root lies between 4 and
two roots lie between 1 and
one root lies between and
one root lies between 3 and
5 variations,
variation ;
5 variations,
4 variations,
4 variations,
2 variations,
1 variation,
1 variation ;
variation ;
-3,
0,
4-1,
4
APPENDIX.
GENERAL SOLUTh,* OF TWO SIMULTANEOUS EQUATIONS OF THE
FIRST DEGREE.
1. Take the equations,
ax + by c . . . (1),
a'z+b'y = c' ... (2);
multiply both members of (1) by b' and of (2) by 6, then sub-
tracting and factoring, we find
(aV - a'b) x^b'c- be' ;
b'c - be'
ab' - a'b
(3).
r vi ' ac' a'c
In like manner, y = 77 (4).
ab ab
By means of formulas (3) and (4) any two simultaneous equations
of the forms (1) and (2) may be solved.
Thus, 4x -f By = 31,
3* + 2y = 22 :
by comparison with (1) and (2),
a = 4, 6 = 3, c = 31, a' = 3, b' = 2, c' = 22;
by substitution in (3) and (4),
62 - 66 88 - 93
144 APPENDIX.
EXAMPLES.
( -+ y - =2 )
1. Given J 3 4 I to find x and y.
I 3x + 4y = 25 )
By comparison with (1) and (2),
a = , 6=1, c = 2,
a' = 3, ^ = 4, c' = 25 ;
by substitution in (3) and (4),
X = *L--J% = 8> y = 8 ^~ 6 = 4 .
** ~~ 4 *3 ~~ 4
2. Given 4 f- to find a; and y :
(- 5z + 16y = 124)
by comparison,
a = 11, b = 5, c = 1,
a' --= - 5, b' = 16, c' = 124 ;
oy substitution,
- 16 + 620 _ 1364 5
176-25 ' 176 - 25 ~
GENERAL SOLUTION OF THREE SIMULTANEOUS EQUATIONS OF
THE FIRST DEGREE.
2. Take the equations,
ax + by + cz = d (1),
a'x + Vy + c't = d' . (2),
~n~, i irr., i ..//_ j// /o\
ax-}-by + cz-=a (6).
From (1) and (2) we obtain, by eliminating e,
(So, - ca'} x + (c'b - cb') y = c'd -cd / - - (4).
ADDITIONAL EXAMPLES. 145
In like manner, from (1) and (3),
(c"a - ca") x + (c"b - cb") y = c"d - cd" (5; ;
combining (4) and (5) and eliminating y, we find
_ (c"b - cb") (c'd - cd 1 ) - (c'b - cb') (c"d - cd")
~ (c'a ca' ) (c"b -cb") - (c"a- ca") (c'b - cb')
In like manner,
_ (c'a - cap (c"d - cd") - (c"a- ca") (c'd - cd')
y - (c'a _ ca') (c"b _ cb") - (c"a- ca") (c'b - cb') ***
_ (a"6 - ab") (afd - ad') - (a'b ab') (a"d ad")
~ (c'a - ca') (b"a -ba") - (c"a- a"c) (b'a - ba' )
Formulas (6), (7) and (8) enable us to solve all groups of simul
taneous equations of the form of (1), (2) and (3). Thus,
2x + Sy + 4z = 29,
3* + 2y + 50 = 32,
4x + 3y + 2 = 25 :
Vy comparison with (1), (2) and (3),
a =2, b =3, c =4, d =29,
a' =3, b' = 2, c' =5, ab '~ a ' b
a
These values of x and y correspond to those already deduced by
previous methods.
As an example, let it be required to find the values of x and y
from the equations
7z + 3y = 33 (2) ;
multiplying both members, of (1) by n,
3nx ny = 5ra (3) ;
adding (2) and (3), member to member,
(3% + 7) x + (3 n) y 5n + 33 (4) :
1 st. Assume 3 n = ; . . n =: 3 ;
substituting in (4), and reducing,
15 + 33 _ Q
: 9 + 7
2d.' Assume 3n + 7 = 0;
ADDITIONAL EXAMPLES. 140
substituting in (4), and reducing,
= 4.
S+ 3
EXAMPLES.
1. Given { } to find x and y.
V, J
Multiplying both members of the first equation by n and adding to the
second, member by member,
making n = - and reducing,
m
making n = - and reducing,
2 Given
y-2
/> __,_-. ^
a? 7 -o
Q
4y =3
to find x and y.
Reducing and transposing,
7*-y = 33 (1),
12y - x = 19 (2) :
multiplying by n, and adding and factoring,
9
150 APPEltDIX.
(In 1) x (n 12) y = 33n + 19 :
making n = - , we find y = 2 ;
making n = 12, we find x = 5.
T+? = s
3. Given -^ ^ to find x and y.
~jr~ ~7 =
Multiplying by n, adding and factoring,
(9n + 14) - + (6% - 6) - = 36/i + 10 ;
14 1 1
making n = , we have - = o ; . . y = ->
y
making n = 1, we have - = 2; . . x = -
MISCELLANEOUS GROUPS OF SIMULTANEOUS EQUATIONS OF THE
FIRST DEGREE.
82; 5y 9
] . Given < > to find a: and y.
+ - -
i, 5a; 3y 4
Combining and eliminating -,
tG
OK Of Z^ T9 '
itii) Jf/ y to is/
1 DO i 20
whence, - = 57 ? -. 1 r= IT
ADDITIONAL EXAMPLES.
Conr/Dining and eliminating -
y
\ _ JLU -
9 25/ x ~ 27 12 '
1 65
whence, x = 192'
= 5
2. Given { i. to find x and y ;
I r*. *
c 1
5 J
Clearing of fractions and reducing,
* + 6y= 15 (1),
2z 5y = 4 . (2) :
combining and eliminating ar,
Yly = 34 ; . . y 2 ;
substituting in (1), ar = 3.
y-2
A.
ry ~" ~
3. Given \ \ to find a? and y.
3
Clearing of fractions and reducing,
7a;-y = 33 (1),
12y-ar=19 . . (2);
combining and eliminating ar,
83y = 166; .-. y = 2;
substituting in (1), x = 5.
152
APPENDIX.
4. Given
5 * 6 4 + 2y = 24
to find x and y.
Clearing of fractions and reducing,
5z -f 12y = 148,
25* 2y = 182 ;
combining and eliminating y,
155* = 1240 ; . . x - 8 ;
substituting, we find y = 9.
5. Given
6
to find x, y and c.
Clearing of fractions and reducing,
3x + 2y+ s = 72 . . . (1),
x + 3y + 20 = 48 . (2),
3*+ 2* = 60 ... (3);
combining (1) and (2), eliminating y,
11* -2= 120 . (4);
combining (3) and (4), eliminating 2,
25* = 300 ; . . * = 12 ;
by substitution in (3), 2 = 12,
" " (1), y=12.
ADDITIONAL EXAMPLES.
153
6. Given
Clearing of fractions,
21*+14y + 60 = 924
10s -f 6y+152 =
x+ 2y+ 2 = 11
combining (1) and (3), eliminating 2,
to find or, y an< <.
(1),
(2),
(3);
(4);
(5) ;
combining (2) and (3), eliminating 2,
5* + 24y = 780
combining (4) and (5), eliminating ar,
70y = 2100 ; . . y = 30 ;
from (5), x - 12 ; from (3), z = 42.
f 2* - 3y + 22 = 13 (1)
2v * = 15 ... (2)
2y+ = 7 ... (3)
I 5y + 3v = 32 . . . (4) J
Combining (2) and (4), eliminating v,
lOy + 3x = 19 (5) ;
combining (5) and (1), eliminating a?,
29y - 62 = - 1 . . . (6) ;
7. Given
t( find x, y, t
and v.
154
APPENDIX.
combining (6) and (3), eliminating z,
41y = 41; .'. y = 1 :
by successive substitutions, z = 5, x = 3, v = 9.
8. Given
3 5 19 4
z + z~24 + y
24
y 2 ~ 8
Transposing, reducing, &c.,
x y
a; y z 24
(1),
(2),
(3):
combining (1) and (3), also (2) and (3), eliminating -,
10- +43- = (4),
x /24
8 ^- 11 y == 24 ' ' ' (5)J
combining and eliminating -,
x
2B *l = i&> ' p = l? or y=12 >
substituting in (5), a; = 6 ; whence ; 2 = 8.
f 10+ 6y 4a; _ 4
9. Given <( f- to find x and y.
126+ 8a;-17y _ 35
100 12ar+"7y ~ 13 J
ADDITIONAL EXAMPLES. 155
Clearing of fractions and reducing,
- 2* + 3y = - 1 ... (1),
262* - 233y = 931 (2).
Combining and eliminating x,
160y = 800; .'. y = 5;
by substitution in (1), x = 8.
r ax + by = c 2 \
10. Given < a(a + x)_ > to find x and y.
(W+7)- )
Clearing of fractions and reducing,
ax -f- by = c 2 - (1), '
ax by ^ - a 2 (2) ;
>y addition,
52 _i_ r 2 a 2
2aa; = 6 2 c 2 -a 2 .-. x =
2a
by subtraction,
2y = c 2 -f- a 2 6 2 ; . . y = ^-
MISCELLANEOUS EXAMPLES OF EQUATIONS OF THE FIRST, SECOND
AND HIGHER DEGREES, CONTAINING BUT ONE UNKNOWN
QUANTITY.
1. Given 3* 2 4 = 28 + * 2 , to find x :
transposing and reducing,
x 2 =5 16 ; . . x = 4
2 . Given ^--5 - *? = 117 - fo, to find x;
o o
1 56 APPENDIX.
Clearing of fractions, transposing and reducing,
x z = 25 ; . . * = 5.
3. Given a; 2 + ab = 5a: 2 , to find x ;
transposing and reducing,
* 2 = V ' ' ' * = 2
4. Given
+7 a? -7
;o find a?.
a; 2 7* a; 2 f 7a; a; 2 73'
Clearing of fractions,
x* 4- 14a: 3 - 24a; 2 - 1022a: 3577 a;* + 14a; 3 4- 24a: 2 10222
4- 3577 = 7a: 3 343a: ;
transposing and reducing,
21a; 3 = 1701ar, or z 2 = 81 ; . '. a? = 9.
/*-2 . /* + 2
5. Given W , 2 + V _ 2 ~ ^> to fin< ^ * '
multiplying both members by <^x 4- 2,
-/a; 2
multiplying both members by -y/a; 2.
x 24-a;4-2 = 4 -\/x z 4, or a; = 2 y^a; 2 4 ;
squaring both members,
16 16 4 /
x = 4a; 2 16, or a; 2 = = X 3 ; . ' . a; = - -/3.
6. Given
x 4- y^Sa; 4- 10 = 8, .o find x ;
ADDITIONAL EXAMPLES.
Transposing and squaring both members,
5x + 10 = 64 IGz +
whence, x z 21a; = 54 ;
21 / 441 21 15
by the rule, x = \/54 + : = j
. . a; = 3, * = 18.
7. Given 5 \f* + 7 ^ _ 10 8, to find x :
make \f& = y ; whence, ^/z* = y 2 ;
substituting and reducing,
7 108
7 /108 49 ' 7 4?
whence, y= - JQ * \/5 ' f 100 = ~TO 10 ;
27
. . y = 4 and y = ;
5
27\ 3
from which, x=^y 3 =S and a;=y / y^= \/| ^-
8. Given 3z 2 + 10z = 57, to find x.
By division,
+ - x = 19 ; whence,
o
_5 14
/. x = 3 and a: = 6^
9. Given (a; - 1) (x 2) = 1, to find x.
Performing indicated operations and reducing,
o;3 Qj. 1 .
*' * J^/ ^~* X
158 APPENDIX.
115
10. Given - z 2 - x = -, to find a:.
Dividing both members by -, or multiplying by 2,
m
2 5
x z - x = - ; whence,
_ 5 1 1 7
-
.-. x =11, *=-
11. Given *LZli?_i* a, to find *.
8 x x 2
Clearing of fractions,
2z 2 14z + 20 (5x + 24 x 2 ) = 20s 32
, . 39 28
reducing, a; 2 - x = -- ; whence,
u
39 / 28 , 1521 39 31
X r \/ h , =
10 V 5 100 10
4
12. Given __ - __ = , to find A
Clearing of fractions,
35 (x + 3 a; + 1) = & -\- 2a; 3 ;
reducing, x z + ( 2x = 143 ; whence,
x - 1 -v/l4T=: - 1 12;
ADDITIONAL EXAMPLES. 150
.-. * = 11, *= -13.
24
13. Given x H -- - = 3r 4, to find x.
x 1
Clearing of fract ons,
x 2 x + 24 = 3* 2 3x 4x + 4;
reducing, x 2 3x = 10 ; whence,
3, / in ,9 37
^-^10 + - = ;
. . x = 5, * = 2.
a; x + 1 13
14. Given - + - - = -3-, to find ar.
a; + 1 a; 6
Clearing of fractions,
6z 2 + Qx z + 12a; + 6 = 13* 2 + 13ar ;
reducing, x 2 + x = 6 ; whence,
- 1 5
- = ^-;
. . a: = 2, ar = 3.
a; 4
15. Given -- = x 8, to find a;
Since a; 4 = (^fx 2) (y^+ 2),
we have, by performing indicated operations,
y/x 2 = x 8, or -y/aT= ar 6 ;
squaring both merr bers,
x = x z l2a; + 36 j
or, x* 1 3x = 36 ; whence,
13 f~ ~l69 13 5
= -^-36 + - = -^-;
160
APPENDIX.
. . x 9, x =s 4.
16. Given
Reducing,
* = -
17* 2 + 19* -
* 2 + 19 *
- 1848 =
1848
0, to find 5
whence,
19 355
r 17
17 '
19 /I 848
361
34 ~V 17
1156
34
17. Given
x 9 jf, and a; = 11.
1 5
- a; 2 + - x = 27, to find ar.
O xi
Multiplying both members by 3,
15
+ x = 81 ; whence,
. 225 15 39
15
. . x = 6, and x =
18. Given
12
-, to find r.
Transposing and reducing,
28 a; 2
s_ 4 - *-4'
squaring both members and clearing of fractions,
x* 16 = 784 56z 2 + ** ;
reducing, ** 57* 2 = 800 ;
28
ADDITIONAL EXAMPLES. 161
by Rule. Art. 124,
/57~ l~ , 3249 /57 7
x = db y db y 800 + J = iW-^-i-;
hence r x = 5, and x = 4-y/2.
2* + 9 , 4* - 3 , 3a; 16
19. Given = 3 H , to find x.
9 4x + 3 18
Clearing of fractions,
16* 2 + 84* + 54 + 72* - 54 = 216* -(- 162 + 12* 2 - 55* - 48 ;
5 114
reducing, * 2 - * = - ; whence,
114 548
f 64" ~8 '
. . * = 6, and * = 4
20. Given a 3 + * + 2 -/* 2 -f a: + 4 = 20, to find *.
Making z 2 -{- x = y, and reducing,
2 /y + 4 = 20 - y ;
squanng both members,
4y + 16 = 400 40y + y 2 ;
reduc'ng, y 2 44y = 384 ; whence,
y = 22 V 384 + 484 = 22 10 ;
.-. y = 32, and y = 12 :
taking the first value : * y and substituting in the equation,
x z + x y,
x 2 + x = 32 ; whence.
162
APPENDIX.
taking the second value of y,
x 2 + x = 12 ; whence,
21. Giveh \/ x
transposing,
squaring both members,
= x, to find x.
whence, by reduction,
x* - x + 1 = 2 V^
Placing x 2 x = y
(1),
squaring both members,
y z + 2y + 1 = 4y ; whence,
y 2 -2y=-l; .'. y = 1 V- 1 + 1, or y=
substituting in (1), a; 2 a: = 1 ;
22. Given a; 2 6z = 6a + 28, to find x :
transposing, a? 1 2x = 28 ; whence,
x = 6 -y/28 + 36"= 8 ; . . x = 14, = 2. *
ADDITIONAL EXAMPLES. 163
23. Given x** 2z 3n -f x* 6 = 0, to find :
making, x* = y ; whence,
y* - 2y 3 + y - 6 = ;
causing the second term to disappear (Arts. 263 and 313),
3 91
^ ~ 2 Z = Itf '
id 1U
By the rule for solving trinomial equations (Art. 124).
/3 10
also.
1 . 1
.-. 2==t--v/13, and z = -
24. Given
x* -4- 0.x 3 I 8
-^_ = ^2 + ,,. + 8 to find Xm
x 2 + x 6
Clearing of fractions,
x* -f 2x 3 +
2x 3
2* 48 ;
, 2 56
reducing, x 2 + - ar = ; whence,
O D
. . x = 4, and a; = 4J.
2
25, Given * a 1 = 2 4- -, to find x.
x
Reducing, a 3 3a - 2 = (1) ;
164: APPENDIX.
comparing with x 3 -\- px -f- q =. 0,
P=-Z, ? = -2;
by Cardan's formula,
* = 3 v/r+\A = 2:
dividing both members of (1) by x 2,
x z + 2a: -\- 1 = ; whence,
*= -1,
and the two roots are each equal to 1 ; hence, the three roots
re> +2, 1, and 1.
26. Given 2 e n\ ,
1. Given { } to find x and y.
< a; + y = 12 . , . (2) i
Make a; = v + w, and y v w.
From (2), we have (v + w) + (y 70) = 12 ; . . v = 6.
From (1), we have, (v + w) 3 + (v w) 3 = 18 (v 2 w 2 ) ;
or, reducing, v 3 + 3vw> 2 = 9 (v z w 2 ) ;
substituting the value of v, wi have
216 + 18u 2 = 9 (36 - w 2 ), or 27w 2 = 108; .'. w=2,
hence,
x = v + w = 62 = 8 al 4 = v w=6^2=4 and 8.
(z2 + y-^53 . . . (1))
2. Given { f to find x and y.
( - = 14 ... (2))
Multiplying both meipW.rs of (2) by 2, and adding and subtracting,
we have
x 2 - + 2a;y ' 2 = 81 ; . ' . x + y = 9,
x i _ 9 /f/f : c yZ _ 25 ; . . x y = 5 ;
hence, rz > 1 , and 7 y = + 2, and 7.
U + y* = 82 . . . (1))
3. Given { \ to find ar and y.
U +y = 4 . . . (2))
Raising both numbers of (2) to the 4th power, adding to (1), mem
her to member, and dividing by 2,
x* -f 2z 3 y + 3zV + 2ary 3 + y* = 169 ;
ADDITIONAL EXAMPLES. 169
extracting the square root of both members,
* 2 + xy + y 2 = 13 . . . (3);
squaring both members of (1),
ar + 2xy + y 2 = 16 . . . (4) ;
subtracting (3) from (4), member from member,
xy = 3, or 3xy 9 . . (5) ;
subtracting from (3), member from member,
z 2 2zy + y 2 = 4 ;
whence, x y = 2 . . . (6).
Combining (2) and (6),
x 3, and 1 ; y = 1, and 3.
( 5* + 3y = 19 . . (1) )
4. Given < > to find x and y
('7x z -2y z =lO . . (2)}
From (1), we find
_ 19 - 3y 2 _ 361 - 114y + 9y 2
~5 ; ~25~
Substituting in (2), and reducing,
y
2 _798_ 2277
~ 13 ~~ 13 '
399^ X-2277 , 159201 399 360
whence, y = ^ + _ = ir ~
759
hence, y = , and y = 3 ;
by substitution, x = -, and a; = 2.
18
170 APPENDIX.
(x + 4y = 14 . . (1))
5. GrtVb.1 { > to find x and y.
(y + 4a; = 2y + 11 (2))
From (1), = 14 4y, or 4* = 56 16y ;
subtracting ana reducing,
y 2 - 18y = - 45 ;
whence, y = 9 ^/ 45 + 81 = 9 6 = 15 and 3 ;
hence, x = 2 and 46.
z 2 + 4y 2 = 256 - 4zy . . (1)
6. Given { > to. find x and y.
3y 2 - a; 2 = 39 ... (2)
)
>
)
Transposing in (1), and extracting the square root of both members,
* + 2y=16; .'. a; =16 2y;
or, x 1 = 256 ^ 64y + 4y 2 ;
substituting in (2), and reducing,
y 2 =p 64y = - 295 ;
whence, y = 32 -y/ - 295 + 1024 = 32 27 ;
. . y = 59 and 5 ;
substituting, x = 102 and 6.
7. Given ? f to find a; and y.
(a; 2 + 2:y =84 . . (2))
Subtracting (1) from (2), member from member,
y 2 + xy = 60 . . . (3) ;
adding (3) to (2), member to member,
x 2 + 2xy + y 2 = 144 ; . '. x -f y = 12 . . . (4).
ADDITIONAL EXAMPLES. 171
Dividing (1) by (4), member by member,
x-y= 2;
hence, x . 7, y = db 5.
(x-y = 4 . . . (1))
8. Given { r to find ar
( xy = 45 . . . (2) )
From (1), x = y + 4, which, in (2), gives
y 2 -f4y = 45; .'. y = 2 -/45 + 4 = 2 7 ;
.'. y = + 5, and 9;
and by substitution, x = + 9, and 4.
fary + *y2 = 12 . . . (1) )
9. Given \ t to find x and y.
( x + xy 3 = 18 . . . (2) )
Dividing (2) by (1), member by member, and reducing,
1-fy 3 3 1 y + y z 3
,. . = = or
y ~2'
g
by reduction, y 2 - y = 1 ;
v
5 / , , 25 5 3
whence, y = - i / 1 +T^ = T= t T;
4V 1644
1
. . y = 2, and y = ;
by substitution, a: = 2, and or = 16.
1 1
X "y = fa\ 1
j. x y A * '~4
Clearing (1) of fractions,
9z 2 + 36a;y = 85y 2 (3).
From (2), x = 2 + y ; .' . s 2 = 4 + 4y + y 3 ;
substituting in (3), and reducing,
27 9
y2 ~io y = io ;
27 /9 , 729
wnpnfp v -t- * / -4- ~~
y ~ 20 V 10 ^ 400 ~
. . y = 3 and y = -
17
by substitution, x = 5 and a; =
27 db 33
ADDITIONAL EXAMPLES. 173
(f! + ^+* + *: = ?Z..m)
12. Given J y* z 2 y * 4 v ' I to find x and y.
( *-y = 2 (2) J
X II
Make - + - = z (3) ; whence, by squaring,
hence, from (1), by substitution and reduction,
35 1 735 , 1 - 1 6
* 2 + z = T ; .-. , = -- v / T + - = __;
5 7
or, z = - and --;
substituting the positive value of z in (3), and clearing of fractions,
2 _L_ 2 . //I \
From (2), * = y + 2 ; . ' . z 2 = y 2 + 4y + 4,
and zy = y 2 -f 2y ;
substituting in (4) and reducing,
hence, y = 2 and y = 4 ;
by substitution, x = 4 and y = 2.
r x z + y 2 -f 2 2 = 84 . (1) >j
13. Given -{ * + y + z = .14 (2) I to find ar, y.
I 22 = y 2 (3) J and 2 '
Substituting in (1) and (2) the value of y from (3), and reducing
z 2 4- 2xz + z 2 = 84 -f tz or a; + z t/84 -f a* (4)
From (2) and (3), * + 2 = 14-^7 (5)
174: APPENDIX.
Equating the second members,
14 - ^/xz -v/84 + xz -,
squaring both membeis,
196 - 28 V** 4- = 84 + xz ;
.educing, -y/arz = 4, or 2 = 16 (6) ;
hence, from (5), x + = 10 (7) ;
1 A
aobstituting in (7) for z its value , and reducing,
sc
a 2 - 10*= - 16;
* = 5 ^/ 16 + 25 = 5 3 ;
or. a; = 8, a; = 2 :
by substitution, 2 = 2, 2 = 8, and y 4.
f * + y + -v/# + y = 12 . (1
14. Given { ; 2 + y = 41..(
From (1), by transposition,
V*+y = 12 - (x + y) ;
squaring both members,
* + y = 144 - 24 (x + y) + (* + J/) 2 ;
reducing, (a; + y) 2 25 ( + y) = 144 ;
25 ~ 1 ^ 5 " 25 7
whence, x -f y = 16, or x + y = 9.
Tlie first value does not satisfy (1), unless the radical have the
negative sign ; adopting, therefore, the second value, from which
x = 9 y, or x 2 = 81 18y + y 2 ,
ADDITIONAL EXAMPLES.
which in (2) gives, after reduction,
y 2 9y = 20 ; whence,
T = Hr' ; .'. y = 5 and y = 4;
by substitution, x = 4 and x = 5.
l X 3-y3~U1[ . . . (1)1
15. Given { f to find x and y.
(x -y = 3 ... (2))
Cubing both members of (2), subtracting from (1), member from
member, and dividing both members by 3, we have
x z y -xy z = 30, or (r. y) xy = 30 . (3) ;
dividing (3) by (2), member by member,
xy = W; .'. y = ;
X
substituting in (2), and reducing,
x z - 3* = 10 ;
3 I 93?
whence, x = - ^ 10 + - = - -;
. . x = 5, x = - 2 ;
by substitution, y = 2, and y = 5.
(x* + r*ffi= 208 (1))
16. Given 1 [ to find a? and y,
- (2) }
These equations may be written,
4 2. 444
x*+x*y*= 208, or x s (x s +y s )= 208 . (3),
^ + ^=1053, yVhar^lOSS ... (4)
176 APPENDIX.
Dividing (3) by (4), member by member,
aJ _ 208 _ 16
~| ~~ 1053 ~ 1
y
extracting the 4th root of both members,
I ' '
substituting in (3),
64 . 16
= 208 ; whence,
208
y2 = 208, or y = 729 ; ' y = 27,
and by substitution, x = 8.
MISCELLANEOUS PROBLEMS.
1. A courier starts from a place and travels at the rate of 4 miles
per hour ; a second courier starts after him, an hour and a half
later, and travels at the rate of 5 miles per hour : in how long a
time will the second overtake the first, and how far will he travel ?
Let x denote the number of hours travelled by 2d courier :
then will x + 1% " " " " 1st "
5x " " " miles " 2d "
and 4(x -f H) " " " " " 1st ;t
From the conditions of the problem,
bx = 4 (x + li) ; . . x = 6 and 5x = 30.
2. A person buys 4 houses for $8000 ; for the second he gave
half as much again as for the first ; for the third, half as much again
as for the second ; and for the fourth, as much as for the first and
third together : what does he give for each 1
ADDITIONAL EXAMPLES. 177
Let x denote the amount paid for 1st house : then will
ar + 1 u er hour slower, he would have been 6 hours longer in completing
ihe journey : how many miles did he travel per hour ?
Let x denote the number of miles travelled per hour. Then will
105
denote the number of hours.
x
From the conditions,
105 105
= + 6, or 105* = 105* - 210 + 6x 2 - 12* ;
x 2 x
reducing, x 2 2* = 35 ;
.-. x = 1 -v/35 + 1 = 1 6 ; .'. x = 7.
14. The continued product of four consecutive numbers is 3024 :
what are the numbers ?
Let x denote the least number.
! .From the conditions of the problem.
x(x + 1) (* + 2) (x + 3) = 3024,
or x* + 6x 3 + II* 2 + Gx 3024 = 0.
A superior limit of the real positive roots is 9 (Art. 279). Ne-
glecting the divisor 1, and all negative divisors, we may proceed by
the rule (Art. 285), as follows :
9, 8, 7, 6, 4, 3, 2,
- 336, 378, - 432, 504, - 756, - 1008, - 1512,
- 330, - 372, - 426, 498, - 750, - 1002, - 1506,
-, - 83, -., - 334, - 753,
., .'72, .., _ 323, 742,
, , , 12, , , 371,
ft 0,'K
*, ", ", U, , , OUt,
', ", , 1, ", , ",
" "> "> "> "> "t "
Hence, 6 is the required value of x, and the numbers a- e 6, 7, 8
and 1).
ADDITIONAL EXAMPLES. 183
15. Two couriers start at the same instant for a point 39 miles
distant ; the second travels a quarter of a mile per hour faster than
the first, and reaches the point one hour ahead of him : at what
rates do they travel 1
Let x denote the number of miles per hour of first courier.
39
Then will denote the number of hours he travels.
x .
From the conditions,
reducing,
or * =
16. The fore-wheels of a wagon are 5 \ feet, and the hind-wheels
7 feet in circumference ; after a certain journey, it is found that the
fore-wheels have made 2000 revolutions more than the hind-wheels ^
how far did the wagon travel ?
Let x denote the number of feet.
From the conditions of the problem,
1197
multiplying both members by -5^-,
BSB
_ 2394000 _ 598500
7 W* ^ x 32 8 '
57 x 42 x = 598500,
V
15 a; = 598500,
x= 39900.
184 APPENDIX.
17. A wine merchant has 2 kinds of wine ; the one costs 9 shil*
lings per gallon, and the other 5. He wishes to mix them together
in such quantities that he may have 50 gallons of the mixture, and
so that each gallon of the mixture shall cost 8 shillings.
Let x and y denote the number of gallons of each, respectively.
From the conditions,
x+ y = 50 . . .- . (1),
9x + 5y = S(x + y) . . (2) ;
substituting for x + y its value in (2),
9* + 5y = 400 . . . . (3);
combining (1) and (3),
4y = 50-, .'. y = 12$, and x = 37.
18. A owes $1200 and B, $2500, but neither has enough to pay
his debts. Says A to B, " Lend me the eighth part of your fortune,
and I can pay my debts." Says B to A, " Lend me the ninth part
of your fortune, and I can pay mine :" what fortune had each 1
Let x and y denote the number of dollars in the fortunes of A
and B.
From the conditions of the problem,
x + | = 1200, or 8x + y = 9600,
o
y + ^ = 2500, or x -f 9y = 22500 ;
y
combining and eliminating x,
71y = 170400; .-. y = 2400, x = 900.
19. A person has two kinds of goods, 8 pounds of the first, and
9 of the second, cost together $18,46; 20 pounds of the first, arrd
16 of the second, cost together $36,40: how much does each cost
per pound ?
ADDITIONAL EXAMPLES. 185
Let x and y denote the cost of a pound of each in cents.
From the conditions of the problem,
Sx+ 9y=1846,
20* + 16y = 3640 ;
combining and eliminating x,
13y=1950: .'. y = 150, and x = 62.
20. What fraction is that to the numerator of which if 1 be
added the result will be , but if 1 be added to the denominator the
result will be J ?
Let x denote the numerator, and y the denominator.
From the conditions of the problem,
x+ 1 1
V
x 1
or 3x + 3 = y,
or 4x = I + y ;
i+y 4
hence, by combination, * = 4 and y = 15. Ans. y^.
21. A shepherd was plundered by three parties of soldiers. The
first party took -J- of his flock and of a sheep ; the second took
of what remained and ^ of a sheep ; the third took -J of what then
remained and ^ of a sheep, which left him but 25 sheep : how many
had he at first ?
Let x denote the number of sheep. Then, after being plundered
by the 1st party, he would have
O_ 1
sheep ;
4
after being plundered by the 2d party, he would have
3x I /3x 1 . 1\ x I
1 /
\
12
186
APPENDIX.
after being plundered by the 3d party, he would have
x 1 /x 1 l\ Z 3
1\ x
27 =
2
from the conditions of the problem,
- = 25, or x 3 = 100; .-. x 103.
22. What two numbers are those whose product is 63, and the
square of whose sum is equal to 64 times the square of their dif
ference 1
Let x and y denote the two numbers.
From the conditions of the problem,
xy - 63 (1),
(x + yY = 64 (x - y) 2 . . (2);
extracting the square root of both members of (2),
x -{- y = 8 (x ?/), or 7# = 9y ; . ' . x = 2-y ;
substituting in (1), f y 2 = 63 ;
. . y 2 49 and y = 7, also x = 9.
23. The sum of two numbers multiplied by the greater gives
209 ; their sum multiplied by their difference gives 57 : what are
the two numbers ?
Let x and y denote the numbers.
From the conditions of the problem,
(x + y) x = 209, or x 2 4 zy= 209 . . (1),
(z + y)(x-y)= 57, or a* - y* = 57 . . (2);
subtracting (2) from (1), member from member,
adding (3) and (1), member to member,
x 2 + 2xy 4- v 2 = 361 ; . . x -4- y = ] 9 ;
ADDITIONAL EXAMPLES. 187
209
hence, from (1), = -- = 11; also, y = 8.
24. Three numbers are in arithmetical progression ; their sura is
15, and the sum of their cubes is 495 : what are the numbers 1
Let x, y and z denote the numbers,
From the conditions of the problem,
y x = z y . . (1)
x -hy +z = 15 . . (2)
x 3 + y 3 + z 3 = 495 . . (3) ;
from (1), 2y = z + or, which in (2), gives y = 5 ;
substituting in (2) and (3),
z + x = 10 . . (4)
3 -f x 3 = 370 . . (5) ;
dividing (5) by (4), member by member,
2 2 - zx '+ x* = 37 . . (6) ;
squaring both members of (4),
z 2 + 2zx + x 2 - = 100 . . (7) ;
combining (6) and (7),
21
'6zx = 63, or * zx = 21 ; . * . z = ;
X
substituting in (4),
21
* + = 10, or x z 10* = - 21 ;
.-. x = 5 -/ 21 + 25 = 5 2 ; hence, x = 7. or 3.
25. Divide the number 16 into two parts such that 25 times the
square of the first shall be equal to 9 times the square of the
second.
188 APPENDIX.
Let x denote one part ; then will 16 x denote the other.
From the conditions,
25* 2 = 9 (256 - 32* + * 2 ) = 2304 - 288* + 9* 2 ; .
reducing, x z + 18* = 144 ;
.-. x = 9 -/144 + 81 = 9 15, or * = 6,
since the negative value does not satisfy the problem understood in
the numerical sense.
26. There are two numbers such that the greater multiplied bj>
the square root of the less is 18, and the less multiplied by the
square root of the greater is 12 : what are the numbers ?
Let * and y denote the numbers.
From the conditions of the problem,
y^=lS . . (1)
*V7=12 . . (2);
multiplying (1) by (2), member by member,
y^r)3"=216; .'. xy - 36 . . (3);
adding (1) and (2), member to member,
(V^+ V7) V*y = 30 . . (4),
or -\/x + yV = 5 ;
squaring both members,
* + y + 2V*7=25 . . (5),
or * + y = 13 (6);
combining (3) and (6),
x = 9. y = 4.
ADDITIONAL EXAMPLES. 189
27. What two numbers are those the square of the greater of
which being multiplied by the lesser gives 147, and the square
of the lesser being multiplied by the greater gives 63 1
Let x and y denote the numbers.
From the conditions of the problem,
*2y = 147 (1)
xy*= 63 . . (2);
multiplying (1) and (2), member by member,
*V = 9261 . . (3),
or xy= 21 . . (4);
dividing (2) by (4), member by member,
y = 3 ; in like manner, x = 7.
This method of solution might be applied to the equations of the
preceding example.
.
28. There are two numbers whose difference is 2, and the product
of their cubes is 42875 : what are the numbers ?
Let x and y denote the numbers.
From the conditions of the problem,
*-y = 2 . . (1)
X 3 y 3 _ 42875 . . (2) ;
extracting the cube root of both members of (2),
35
ary = 35; .-. y = ;
substituting and reducing,
** - 2* = 35,
x 1 -/S5 + 1 = 1 6;
. . x = 7, and 5, y = 5, and 7.
190 APPENDIX.
29. A sets out from C towards D, and travels 8 miles each day ;
after he had gone 27 miles, B sets out from D towards C, and goes
each day ^ of the whole distance from D to C ; after he had
travelled as many days as he goes miles in each day, he met A
what is the distance from D to C?
Let x de lote the number of miles from D to C.
X
Then, will denote the number of miles B travels per day,
\j
also the number of days that he travels ;
a; 2
hence, --^-r denotes the number of miles travelled by B,
27 + Sx " " " " " A.
"From the conditions of the problem,
* 2 _L 07 _L Sx _
lOO 4 " "^20"
clearing'of fractions and reducing,
x 2 - - 240* = - 10800 ;
.-. x - 120 -/- 10800 + 14400 = 120 60 ;
whence, x = 60, x = 180.
30. There are three numbers ; the difference of the differences of
the 1st and 2d, and 2d and 3d, is 4 ; their sum is 40, and their con
tinued product is 1764 : what are the numbers ?
Let #, y and z denote the numbers.
From the conditions of the problem,
(*-y)-(y-*)= 4 . . (l)
x + y + z= 40 . . (2)
. . (3);
ADDITIONAL EXAMPLES. 191
combining (1) and (2), eliminating x and 2,
3y = 36; .-. y = 12;
substituting in (2) and (3),
x + z= 28 . . (4)
xz = 147 . . (5);
combining (4) and (5),
x = 7, or 21 ; y = 21, or 7.
31. There are three numbers in arithmetical progression : the
sum of their squares is 93, and if the first be multiplied by 3, the
second by 4, and the third by 5, the sum of the products will be 66 :
what are the numbers ?
Let x denote the first number, and y their common difference.
From the conditions of the problem,
* 2 + (* + 2/) 2 + (* + 2y) 2 = 93 . . (1)
3* + 4 (x + y) + 5 (x + 2y) = 66 . . (2) ;
performing indicated operations and reducing,
3z 2 -f 5y 2 + Gxy = 93 . . (3)
12* -f I4y = 66, or 6* + 7y = 33 . . (4).
33 Qx
From (4), y = ;
1089 396* -f- 36z 2 33* 6* 2
.'. y*-- - & - i and *y= ^.,
substituting in (3) and reducing,
^ _ 198 290
" "25^ * ~ " "25 '
192 APPENDIX.
- , , 296 , 9801 99 49
whence, x = - x / - + = _;
148
Taking the second value of x, we find y = 3, and the numbers
are 2, 5 and 8.
The problem supposes the numbers entire, therefore the 1st value
of x is not used.
32. There are three numbers in arithmetical progression whose
sum is 9, and the sum of their fourth powers is 353 ; what are the
numbers ?
Let a;, y and z denote the numbers.
From the conditions of the problem,
2y = x + z . . (1)
x + y + z = 9 . . (2)
z* + y* 4- s 4 = 353 . . (3).
From (1) and (2) we find y = 3;
substituting in (2) and (3),
x + z = 6 . . (4)
z* + z* = 272 . . (5) ;
raising both members of (4) to the 4th power,
a;4 + 4 X 3 Z _|_ Q X Z Z Z _j_ 4 XZ 3 + s * _ J296 . . (6) ;
adding equations (5) and (6), member to member, and dividing by 2
Z* + 2* 3 z 4- 3z 2 2 2 + 2z2 3 -f 2* = 784 . . (7);
extracting the square root of both members,
2 2 -f xz + z 2 = 28 . . (8);
ADDITIONAL EXAMPLES. 193
squaring both members of (4),
z 2 + 2xz + 2 = 36 . . (9);
from (8) and (9) we find
xz = 8 . . (10);
from (4) and OO) we get
x = 2, or 4 ; s = 4, or 2 :
hence, the numbers are 2, 3 and 4.
33. How many terms of the arithmetical progression 1, 3, 5, 7,
&c., must be added together to produce the 6th power of 12 ?
The 6th power of 12 is 2985984.
From Art. 175 we have the formula,
d-2a
w =
Ir. the present case, o = 1, d = 2, and S = 2985984 ;
-V/16 X 2985984 , WM
substituting, n = = 1728.
34. The sum of 6 numbers in arithmetical progression is 48 ; the
product of the common difference by the least term is equal to the
number of terms : what are the terms of the progression ?
Let x denote the 1st term, and y the common difference.
From the conditions of the problem,
Qx+ 15y = 48, *y = 6; . . y=^;
substituting and reducing,
a* - 8* = - 15 ;
.-. x = 4 / 15 + 16 = 4 1, or * = 5, * = 3;
whence, y = f > y = 2 :
194: APPENDIX.
hence, the series is 3.5.7.9.11.13,
or 5 . 6 . 7f . 8 . 9f . 11.
35. What is the sum of 10 square numbers whose square roots
are in arithmetical progression the least term of which is 3, and the
common difference 2?
Let x denote the sum.
The progression of roots is
3.5.7.9.11.13.15.17.19.21,
and the series of squares,
9 . 25 . 49 . 81 . 121 . 169 . 225 . 289 . 361 . 441.
1st order of diffs, 16, 24, 32, 40, &c.,
2d order of diffs, 8, 8, 8, &c.,
3d order of diffs, 0, 0, &c.
From Art. 210, making
S' = x, a = 9, n = 10, ^ = 16, d z = 8, d 3 = 0, &c.
x = 90 + 45 x 16 + 120 x 8 --= 1770.
36. Three numbers are in geometrical progression whose sum i
95, and the sum of their squares is 3325: what are the numbers?
Let a;, y and z denote the numbers.
From the conditions of the problem,
y* = a* . . (1)
x z + y + z 2 = 3325 . . (2)
x + y + z = 95 . . (3) ;
combining (1) and (2),
x* + 2xz \- 2 2 = 3325 + xz , . (4) ;
ADDITIONAL EXAMPLES. 19
combining (1) and (3),
x -f v ^xz + z = 95 . . (5) ;
from (4) and (5),
x + z = ^3325 + xz
x + 2 = 95
hence, -y/3325 + xz = 95
squaring both members,
3325 + xz = 9025 190
taking the 1st value of y, we find
x = 4, s = 25.
39. A, B and C purchase coffee, sugar and tea at the same prices ;
A pays $11,62-| for 7 pounds of coffee, 3 pounds of sugar, and 2
pounds of tea ; B pays $16,25 for 9 pounds of coffee, 7 pounds of
sugar, and 3 pounds of tea; C pays $12,25 for 2 pounds of coffee,
5 pounds of sugar, and 4 pounds of tea : what is the price of a
pound of each 1
Let x, y and z denote the number of cents that the coffee, sugar
and tea cost, respectively.
From the conditions of the problem,
^+ fy + 21z = 1162$ . . (1)
9*+ 7y + 3z= 1625 . (2)
2 x + 5y -f 4 z = 1225 . . (3) ;
ADDITIONAL EXAMPLES. 197
clearing (1) and (3) of fractions,
30* + 12y + 90 = 4650 (4)
4 ar+lly-f 8z -2450 (5).
From (2) and (4),
3* 9y = 225, or x 3y = 75 (6) ;
from (2) and (5),
QQx + 23y = 5650 . (7);
from (6) and (7), y = 50;
by substitution, x = 75, z = 200.
40. Divide 100 into 2 such parts that the sum of their square
roots shall be 14.
Let x denote the first part.
From the conditions of the problem,
y/ar-f -v/100 x = 14;
squaring both members and reducing,
V/l 00* - a? = 48 ;
squaring both members and reducing,
* 2 -100r= -2304;
.-. x = 50 -/- 2304 -I- 2500 = 50 14,
x = 64, and 36.
41. In a certain company there were three times as many gentle-
men as ladies ; but afterwards 8 gentlemen with their wives went
away, and there then remained five times as many gentlemen as
ladies : how many gentlemen, and how many ladies were there
originally ?
12
198 APPENDIX.
Let 3z denote the number of gentlemen ; then will x denote the
number of ladies.
From the conditions of the problem,
3x 8 = 5 (x 8) ;
. . x = 1 6, and 3x = 48.
42. Find two quantities such that their sum, product, and the
difference of their squares, shall all be equal to each other.
Let x and y denote the quantities.
From the conditions of the problem.
x +y -xy . . (1)
x z y 2 - xy (2) ;
by division of (2) by (1), we have
x y = 1, or a; = y -f- 1 ;
substituting in (1),
2y + 1 = y z + y, -or y 2 y = 1 ;
. 1 /T~T
whence,
hence, x =
43. A bought 120 pounds of pepper, and as many pounds of
ginger, and had one pound of ginger more for a dollar than of
pepper ; the whole price of the pepper exceeded that of the gingei
by 6 dollars : how many pounds of pepper, and how many of
ginger had he for a dollar ?
Let x denote the number of pounds of pepper for a dollar.
ADDITIONAL EXAMPLES. 199
From the conditions of the problem,
The negative value does not conform to the conditions of the
special problem.
44. Divide the number 36 into 3 such parts that the second shall
exceed the first by 4, and that the sum of their squares shall be
equal to 464.
Let #, y and z denote the parts.
From the conditions of the problem,
x + y + z = 36 . (1)
y-x = 4 . . (2)
x i + y 2 - + z 2 = 464 (3) ;
from (1), z 2 + 2xy + y 2 = 1296 - T2z + z 2 ' (4) ;
from (2), z 2 2xy + y 2 = 16 ...... (5) ;
adding (4) and (5), member to member.
2* 2 + 2y 2 = 1312 72z + z 2 (6) ;
from (3), 2* 2 + 2y 2 = 928 - 2z 2 . . . (7) ;
equating the second members and reducing,
2 2 - 24* = - 129 ;
.-. z = 12 -V/-128 + 144 = 12 4;
hence, z 16, z = 8 ;
substituting the first value in (1),
* + y = 20 (8);
200 APPENDIX.
from (2) and (8), y = 12 and x = 8.
.45. A gentleman divided a sum of money among 4 persons, so
that what the first received was ^ that received by the other three ,
what the second received was that received by the other three ;
what the third received was -j that received* by the other three, and
it was found that the share of the first exceeded that of the last by
$14 : what did each receive, and what was the whole sum divided ?
Let x, y, z and w denote the number of dollars that each received.
From the conditions of the problem,
2# = y + z -f w ' (1)
3y = x + s + w (2)
42 = x + y + w - (3)
x w = 14 (4) ;
from (2) and (3),
x + w 3y z
x -f- w = 4z y ; whence, 3y z =. 4z y,
01 4y = 5s, z A y . . (5) ;
from (4), w = x 14 (6) ;
substituting the values of w and z in (1) and (2),
2x = y + f y + x 14
3y x -\- %y -\- x 14; whence, by reduction,
5* 9y = 70
10*-lly= 70;
, *. x = 40, y = 30 ; and by substitution, z = 24, to = 26.
. 46. A woman bought a certain number of eggs at 2 for a penny,
and as many more at 3 for a penny, but on selling them at the rate
ADDITIONAL EXAMPLES. 201
of 5 for 2 pence, she lost 4 pence by the bargain ; how many did
she buy ?
X
Let x denote the number at each price. Then will + 5-
i o
denote the number of pence paid, and - - will denote the
number of pence received.
From the conditions of the problem,
x x %(x + x\
^ + 3 = V 5 ' + 4 ; reducing, x = 120.
47. Two travellers set out together and travel in the same direc-
tion ; the first goes 28 miles the first day, 26 the second day, 24 the
third day, and so on, travelling 2 miles less each day ; the second
travels uniformly at the rate of 20 miles a day : in how many days
will they be together again 1
Let x denote the required number of days. The distance
travelled by the first in x days is
[(Art. 176), since a = 28, d = 2, and n = a;], denoted by
\x [56 - (x I) 2 ], or 29* a: 2 ;
and the distance travelled by the second is denoted by 20x:
hence, we hare
29a; x z = 20ar, or x = 9.
48. A farmer sold to one man 30 bushels of wheat and 40 of
barley, for which he received 270 shillings. To a second man he
gold 50 bushels of wheat and 30 of barley, at the same prices, and
received for them 340 shillings : what was the price of each ?
Let x denote the number of shillings for 1 bushel of wheat,
and y " " " " " " barley.
202 APPENDIX.
From the conditions of the problem,
30* + 40y = 270 (1)
50a; + 30y = 340 . (2) ;
whence, HOy = 330, or y = 3; hence, x = 5.
49. There are two numbers whose difference is 15, and half their
product is equal to the cube of the lesser number : what are the
numbers ?
Let x and y denote the numbers ;
from the conditions of the problem,
x y = 15
xy = 2y 3 or, x = 2y 2 ;
substituting and reducing,
-- v = :
2 y 2 '
1 5 1 111
5 25
hence, y = 3, and - ; also, x = 18, and
50. A merchant has two barrels and a certain number of gallons
of wine in each. In order to have an equal quantity in each, he
drew as much out of the first cask into the second as it already
contained ; then again he drew as much out of the second into the
first as it then contained : and lastly, he drew again as much from
the first into the second as it then contained, when he found that
there was 16 gallons in each cask : how many gallons did each
originally contain ?
ADDITIONAL EXAMPLES. 203
Let x denote the n imber of gallons in the first cask, and y the
number in the second ;
x y will denote the quantity in the first cask after the first drawing,
and 2y the quantity in the second cask ; after the second drawing,
2y (x y) or 3y x will denote the quantity in the second,
and 2 2y the quantity in the first cask ; after the third drawing,
2or 2y (3y x} or 3x 5y will denote the quantity in th
first cask, and Qy 2x the quantity in the second.
From the conditions of the problem,
3* 5y = 16
6y 2z = 16.
By combination,
UNIVERSITY OF CALIFORNIA LIBRARY
Los Angeles
This book is DUE on the last date stamped below.
MAR i 8 1955
Form L9-100m-9,'52(A3105)444
JTRE LIBRARY
Library
A 000210489 1
W
<
UL72