jCE: 90 C^HT%.
./'.'. . . ♦
DOMATED BY
m. ^. mqe & m'» ^§\mmmai ^tvu^.
Elementary Algebra,
J. HAMBLIN SMITH, M.A.,
OF GONYILLE AND CAIUS COLLEGE, AND LATE LECTUBEB
AT ST. PETEE's college. CAMBEIDQB.
WITH APPENDIX BY
ALFRED BAKER, B.A..
MATH. TUTOR CNIV. COL. TORONTO.
Sth CANADIAN COFYKIGHT EDITION.
NEW BEVISED EDITION.
Authorized by the Education Department, Ontario.
Authorized by tlie Council of Public Ittstruction. Quebec
Recotnmended by the Senate of the Univ. of Halifax.
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TOKONTO:
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Entered according to thf Act o/ the Parliament of the Dominion of Canada,
in the year one thousand eight hundred and seventyseven, by Ada*
MiLXER & Co. , in the Office of the Minister of Agriculture.
PREFACE
The design of this Treatise is to explain all that is
commonly included in a First Part of Algebra. In the
arrangement of the Chapters I have followed the advice
of experienced Teachers. I have carefully abstained from
making extracts from books in common use. The only
work to which I am indebted for any material assistance
is the Algebra of the late Dean Peacock, which I took as
the model for the commencement of my Treatise. The
Examples, progressive and easy, have been selected from
University and College Examination Papers and from
old English, French, and German works. Much care has
been taken to secure accuracy in the Answers, but in a
collection of more than 2.300 Examples it is to be feared
that some errors have yet to be detected. I shall be
grateful for having my attention called to them.
I have published a book of Miscellaneous Exercises
adapted to this work and arranged in a progressive order
so as to supply constant practice for the student.
I have to express my thanks for the encouragement
and advice received by me from many correspondents ;
and a special acknowledgment is due from me to Mr. E.
J. Gross of Gonville and Caius College, to whom I am
ndebted for assistance in many parts of this work.
Tlie Treatise on Algebra by .^Ir. E. J. Gross is a
continuation of this work, and is in some important
points supplementary to it.
J. HAMBLIXSJ^IITH.
Cambridge, 1871.
Digitized by tlie Internet Arcliive
in 2009 witli funding from
Ontario Council of University Libraries
http://www.archive.org/details/elementaryalwestOOsmit
CONTENTS.
CHAP. PAGE
I. Addition and Subtraction i
II. jMultiplication 17
III. Involution 29
IV. Division 33
V. On the Resolution of Expressions into Factors . 43
VI. On Simple Equations 57
VII. Problems leading to Simple Equations . . 6i
VIII. On the Method of finding the Highest Common
Factor 67
IX. Fractions . 76
X. The Lowest Common Multiple .... 88
XI. On Addition and Subtraction of Fractions . 94
XII. On Fractional Equations 105
XIII. Problems in Fractional Equations . . .114
XIV. On Miscellaneous Fractions 126
XV. Simultaneous Equations of the First Degree . 142
XVI. Problems resulting in Simultaneous Equations 154
XVII. On Square Root • . . .163
XVIII. On Cube Root 169
XIX. QUADR.A.HC Equations 174
XX. On Simultaneous Equations INVOLVING Q'"*nKATics 186
XXI. On Problems resulting in Quadratic Equations . 192
XXII. Indeterminate Equations 196
XXIIi. The Theory of Indices 201
XXIV. On Surds 213
XXV. Oy Equations involving Surds . . . .229
CONTENTS.
CHAP. PAGE
XXVI. On the Roots of Equations .... 234
\XVII. On Ratio 243
XXVIII. On Proportion 248
XXIX. On Variation 258
XXX. On Arithmetical Progression .... 264
XXXI. Om Geometrical Progression ' . . . . 273
XXXII. On Harmonical Progression .... 282
XXXIII. Permutations 28"
XXXIV. Combinations 291
XXXV. The Binomial Theorem. Positive Integrai
Index 296
XXXVI. The Binomial Theorem. Fractional and
Negative Indices 307
XXXVII. Scales of Notation 316
XXXVIII. On Logarithms 328
Appendix 344
\nswers . • . . . 345
1
ELEMENTARY ALGEBRA.
I. ADDITION AND SUBTRACTION.
1. Algebra is the science whicli teaches the use of sym
bols to denote numbers and the operations to which numbers
may be subjected.
2. The symbols employed in Algebra to denote numbers
are, in addition to those of Arithmetic, the letters of some
alphabet.
Thus a, b, c x, y, z : a, )3, y : a', b', c' read
a dash, b dash, c dash : a^, b^, c^ read a o«e,
b one, c one are used as symbols to denote numbers.
3. The number o?ie, or unity, is taken as the foundation of
all numbers, and all other numbers are derived from it by the
process of addition.
Thus two is defined to be the number that results from
adding one to one ;
three is defined to be the number that results from
adding one to two ;
four is defined to be the number that results from
adding one to three ;
and so on.
4. The symbol +, read plus, is used to denote the opera
tion of Addition.
Thus 1 + 1 symbolizes that which is denoted by 2,
2 + 1 3,
and a + b stands for the result obtained by adding b to a.
5. The symbol = stands for the words " is e(iual to," or
•' the result is."
[SA.] . ^
ADDITION AND SUBTRACTION.
Thus the definitions given in Art. 3 may be presented in an
algebraical form thus :
1 + 1=2,
2 + 1 = 3,
3 + 1=4.
6. Since
2 = 1 + 1, M'here unity is written twice^
3 = 2 + 1 = 1 + 1 + 1, where unity is written thrte times,
4 = 3 + 1 = 1 + 1 + 1 + 1 pur times,
it follows that
a = l + l + l +1 + 1 with iinity written a times,
6 = 1 + 1 + 1 +1 + 1 with unity written h times.
7. The process of addition in Arithmetic can be presented
in a shorter form by the use of the sign + . Tlius if we have
to add 14, 17, and 23 together we can represent tlie process
thus :
14 + 17 + 23 = 54.
8. When several numbers are added together, it is indiffe
rent in what order the numbers are taken. Thus if 14, 17, and
23 be added together, tlieir sum will be the same in whatever
order they be set down in the common arithmetical process :
14 14 17 17 23 23
17 23 14 23 14 17
23 17 23 14 17 14
54 54 54 54 54 54
So also in Algebra, when any number of symbols are added
together, the result will be tlie same in whatever order the
symbols succeed each other. Thus if we have to add together
the numbers symbolized by a and b, the result is represented
by a + 6, and this result is the same number as that which is
represented by b + a.
Similarly the result obtained by adding together a, b, c
might be expressed algebraically by
a + b + c, or o + c + 6, or b + a + c, or b + c + a, or c + o + 6,
or c + b + a.
9. When a number denoted by a is added to itself tlie
result is represented algebraically by a + a, This result is for
ADDITIO.y AXD SCBTRACTTO.V.
the sake of brevity represented by 2a, the figure prefixed to
the symbol expressing the number of times the number
denoted by a is repeated.
Similarly a + a + a is represented by 3a.
Hence it follows that
2a + a will be represented by 3a,
3a + a by 4a.
10. The symbol — , read minus, is used to denote the ope
ration of Subtraction.
Thus the operation of subtracting 15 from 26 and its con
nection with the result may be briefly expressed thus ;
2615 = 11.
11. The result of subtracting the number h from the num
ber a is represented by
ah.
Again a — h — c stands for the number obtained by taking c
from a — b.
Also a — b — c — d stands for the number obtained by taking
d from a — b — c.
Since we cannot take away a greater number from a smaller,
the expression a — b, where a and b represent numbers, can
denote a possible result only when a is not less than b.
So also the expression ab — c can d'enote a possible result
only when the number obtained by taking b from a is not
less than c.
12. A combination of symliols is termed an algebraical
expression.
The parts of an expression which are connected by the
symbols of operation + and — are called Terms.
Compound expressions are those which have more than one
term.
Thus ab + c — d is a compound expression ■\nade up of four
terms.
When a compound expression contains
hvo terms it is called a Binomial,
three Trinomial,
four or more Multinomial.
ADDITION AND SUBTRACTION.
Terms which are ju'eceded by the symbol + are called posi
tive terms. Terms which are preceded by the symb.>l — are
called negative terms. When no symbol precede.s a t<'rui the
symbol + is understood.
Thus in the expression n h + cd + e f
a, c, e are called positive term.s,
b, d,f negative
The symbols of operation + and — are usually called posi
tive and negative Signs.
13. If the number 6 be added to the number 13, and if
be taken from the result, the final result will plainly be 13.
So also if a number b be added to a number a, and if b he
taken from the result, the final result will be a : that is,
a + bb = a.
Since the operations of addition and subtraction when per
formed by the same number neutralize each other, we conclude
that we may obliterate the same symbol when it presents itself
as a positive term and also as a negative term in the s;ime ex
pression.
Thus aa = 0,
and aa + b = b.
14. If we have to add the numbers 54, 17, and 2? we may
first add 17 and 23, and* add their sum 40 to the number 54,
thus obtaining the final result 94. This process may be repre
sented algebraically by enclosing 17 and 23 in a Bracket
( ), thus :
54 + (l7 + 23) = 54 + 40 = 94.
15. If we have to subtract from 54 the .sum of 17 and 23.
the process may be represented algebraically thus :
54  (17 H 23) = 54  40 = 14.
16. If we have to add to 54 the difference betweon 23 ar'A
17, the process may be represented algebraically thu!<:
54 4 (23 17) =54} 6 = 60.
17. If we have to subtract from 54 the difference between
23 and 17, the process may be represented algebraically ihus :
54(2317) = 546 = 48.
ADDITION AND SUBTRACTION.
18. The use of brackt . is so frequent in Algebra, that
the rales for their removal and introduction must be carefully
considered.
We shall first tree^t of the removal of brackets in cases
where symbols supply the places of numbers corresponding to
the arithmetical examples considered iVts. 14, 15, 16, 17.
Cd«e I. To add to a the sum of b and c.
3 is expressed thus : a + {b + c).
a. irst add b to a, tlie result will be
a + b.
This result is too small, for we have to add to a a numVir
/•eater than b, and greater by c. Hence our final result wili
oe obtained by adding c to a + 6, and it will be
a + b + c.
Case II. To take from a the sum of b and»c.
This is expressed thus : a — {b + c).
First take b from a, the result will be
a — b.
'i his result is too large, for we have to take from a a number
greater than b, and greater by c. Hence our final result will
be obtained by taking c from a — b, and it will be
a — b — c.
Case III. To add to a the difference between b and c.
This is expressed thus : a + {b — c).
First add b to a, the result will be
a + b.
This result is too large, for we have to add to a a number
less than b, and less by c. Hence our final result will be ob
tJiined by taking c from a + b, and it will be
a + b — c.
Case IV. To take from a the difference between b and c.
This is expressed thus : a — {b~ c).
First take b from a, the result will be
a — b.
This result is too small, for we have to take from a a num
ber less than b, and less by c. Hence our final result will be
obtained by adding c to a — b, and it will be
a~b + c.
ADDITION AND SUBTRACTION.
Note. We assume that a, b, c represent such numbers that
in Case II. a is not less than the sum of b and c, in Case III.
b is not less than c, and in Case IV. b is not less than c, and a
is not less than b.
19. Colkcting the results obtained in Art. 18, we have
a + {b + c) = a + b + c,
a — (b + c) — a — b — c,
a+ {b — c) = a + bc,
a — {b — c) = a — b + c.
From which we obtain the following rules for the removal of
a bracket.
Rule I. "Wlien a bracket is preceded by the sign +,
remove the bracket and leave the signs of the terms in it
imchmiged.
Rule II. When a bracket is preceded by the sign — ,
remove the bracket and change the sign of each term in it.
These rules apply to cases in which any number of terms
are included in the bracket.
Thus
a + b + {cd + e /) = a + b + cd + e f,
and
a + b {cd\e—f) = a + bc + d — e+f.
20. The rules given in the preceding Article for the rt
moval of brackets turuish corresponding rules for the intro
uuction of l)rackets.
Thus if we enclose two or more terms of an expression in a
bracket,
T. The sign of each term remains the same if + pre
cedes the bracket :
II. The sign of each term is changed if — precedes the
bracket.
Ex. ab + cd + ef=ab + {c — d) + {ef\
ahtcd + e/=a{bc){d e+f).
ADDITION AND SUB TRA C TTON.
21. We may now proceed to give rules for the Addition
and Subtraction of algebraical expressions.
Suppose we have to aM. to the expression a + b — c the ex
pression d — e +f.
The Sum =a + hc + {de+f)
= a + bc + de+f (by Art. 19, Rule I.).
Also, if we have to subtract from the expression a + b — c the
expression d~e +f.
The Difference =a + bc{de +/)
= a + bcd + ef{hj Art. 19, Rule II.).
We might arrange tlie expressions in each case under each
other as in Arithmetic : thus
To a + b — c From a + b — c
Add de+f Take de+f
Sum a + b — c + d — e +f Difference a + b — c — d + e —f
and then the rules may be thus stated.
I. In Addition attach the lower line to the uyper with the
signs of both lines unchanged.
II. In Subtraction attach the lower line to the upper with
the signs of the lower line changed, the signs of the upper line
bein" unchanged.
The following are examples.
(1) Toa + 6 + 9
Add abQ
Sum a + b + Q + abQ
and this sum =a + a + 66 + 9 — 6
= 2& + 3.
For it has been shown, Art. 9, that a + a = 2a,
and, Art. 13, that 66 = 0.
(2) From a + 6 + 9
Takea66
Remainder a + h + 9 — a + h + Q
and this remainder = 26 + 15.
8 ADDITION AND SUBTRACTION.
22. We have worked out the examples in Art. 21 at full
leni.fth, hut ill practice they may he ahhreviated, by combining
the symbols or digits by a mental process, thus
Toc + (Z + 10 rromc + (^ + 10
Addcd7 Takecrf7
Sum 2c +3 ilemainder 2d + 17
23. We have said that
instead of a + a we write 2a,
a + a + a 3a,
and so on.
The digit thus prefixed to a symbol is called the coefficient
of the term in which it appears.
24. Since 3a = a + a + a,
and 5a = a + a + a + a + a,
Sa + 5a = a + a + a + a + a + a + a + a
= 8a.
Terms which have the same symbol, whatever their coefli
cients may be, are called like terms : those which have diffe
rent symbols are called unlike terms.
Like terms, when positive, may be combined into one by
adding' their coefficients together and subjoining the common
symbol : thus
2a; + 5ic = 7x,
Zy + by + 8y = 16y.
25. If a term appears without a coefficient, unity is to be
taken as its coefficient.
Thus .*■ + 5x = 6a;.
26. Negative terms, when like, may be combined into one
t(;rm with a negative sign prefixed to it by adding the coeffi
cients and subjoining to the result the common symbol.
Thus 2x3i/5i/ = 2x8i/,
lor 2x~'3yby = 2x — (Sy + 5y)
= 2x8y.
So again 3xyiyGij = 3.f lly.
ADDITIOPT AMD SUBTRACTION:
27. If an expression contain two or more like terms, some
being positive and others negative, we mual first collect all the
positive terms into one positive term, then all the negative
terms into one negative term, and finally combine the two
remaining terms into one by the following process. Subtract
the smaller coefficient from the greater, and set down the
remainder with the sign of the greater prefixed and the com
mon symbol attached to it.
Ex. 8a;3a; = 5x,
7x — 4x + 5a; — 3a; = 1 2aj — 7a; = 5x,
a26 + 564& = a + 5666 = a6.
28. The rules for the combination of any number of like
terms into one single term enable us to extend the application
of the rules for Addition and Subtraction in A]g>3bra, and we
proceed to give some Examples.
ADDITION.
(1) a 26 + 3c (2) 5a + 763c4(£
3a 46 5c 6a76 + 9c + 4ci
4a662c 11a +6c
The terms containing 6 and d in Ex. (2} destroying one another.
(3) 7x5)/+ 43 (4) 6m13?i + 5p
x + 'iyWz 8m+ n — ^
Zx~ y{ bz m— n— p
bx~'iy— z m+ 2n + bp
IGx — ly 3z 16m 1 In
SUBTRACTION.
(1) 5a 36+ 6c (2; 3a + 76 So
2a + 56 4c 3a 76+ 4c
(3)
3a 86 + 10c
5a 66 + 2c
2a66 + 2c
(5)
3a
3a; + 7y + 122!
by 2z
146135
(4)
x — y + z
xyz
2z
(6)
7xl9y14z
6x24j/+ 9a
Zx + 2y + 14z x+ by2Zz
ib ADDITION AND SUBTRACTION.
29. We have placed the expressions in the examples given
in the preceding Article under each other, as in Arithmetic,
for the sake of clearness, but the same o]ierations might be ex
hibited by means of signs and brackets, thus Examples (2) of
each rule might have been worked thus, in Addition,
5a + 76  3c  4rf + (6a  76 + 9c + 4fO
= 5a + 763c4(i+ 6a  76 + 9c + 4i
= lla + 6c;
and, in Subtraction,
3a + 768c(3a76 + 4c)
= 3a + 768c3a + 764c
= 146 12c.
Examples.— i.
Simplify the following expressions, by combining like sym
bols in each.
I. 3a + 46 + 5c + 2a + 36 + 7c. 2. 4a + 56 + 6c 3a 26 4c.
3. 6a 36 4c 4a + 56 + 6c.
4. 8a  56 + 3c  7a  26 + 6c  3a + 96  7c + 10a.
5. 5a;3a + fe + 7 + 263x4a9.
, 6. a — 6 — c + 6 + cd + f7a.
7. 5a + 1063c + 263a + 2c2a + 4c.
EXAMPLES.— ii. ADDITION.
Add together
I. a + xandaa;. 2. a + 2x and a + 3a;.
3. a  2x and 2a  x. 4. 3x + 1y and 5x  2i/.
5. a + 36 + 5c and 3a  26  3c.
a26 + 3c and a + 263c. 7. 1 +a;^ and 3x + i/.
2x  3i/ + 4a, hx^y 2;:, and 6,x + 9)/  82.
2a + 6  3x, 3a  26 + x, a + 6  5x, and 4a  76 + 6x.
Examples.— iii. SUBTRACTION.
I . From a + 6 take a  6.
2 3x + i/ 2x — 1/.
3 2a + 3c + 4(i a2c + 3i.
4. X + 2/ + 3 xyz.
ADDITION A ND SUB TRA CTION'. 1 1
5 . From m — n + r take m — n — r.
6 a + b + c a — b — c.
7 3a + 46 + 5c 2a + 7b + 6c.
8 3x + 5ij4z 3x + 2y5z.
30. AVe have given examples ol' the use of a bracket. The
methods of denoting a bracket are various ; thv.s, besides the
marks ( ), the marks [ ], or j j, are often employed. Some
times a mark called "The Vinculum" is drawn over the symbols
which are to be connected, thus «  6 + c is used to rejiresent the
same expression as that represented by a — (b + c).
Often the brackets are made to enclose one another, thus
a[b+\c~{def)\].
In removing the brackets from an expression of this kind it
is best to commence with the innermost, and to iemove the
brackets one by one, the outermost last of all.
Thu8
a[b+\c(def)\]
= a[b+\c{de+f)\]
= a[b+ \c~d + ef\]
= a[b + cd + ef]
= a — b — c + d — e +/.
Again
5x(3x7) l42.c(6x3)f
= 5x3x + 7 42ic6x + 3j
= 5x3.o + 74 + 2x + 6a;3
= 10x.
Examples.— iv. beackets.
Simplify the following expressions, combining all like quan
tities in each.
1. a + 6 + (3a26).
2. a + b{a~3b).
3. 3a + 56 6c (2a + 46 2c).
4. a + 6 — c — (a — 6 — c).
5. 14x(5x9) j43,r(2x3}/.
6. 4x {3x(2xxa){.
7. 15x j7x + (3x + a^){.
12 ADDITION AND SUBTRACTlOh.
8. a.[6+ja(6 + a)[].
9. 6« + [4a j86(2a + 46)22&}76][76 h{8a
(36 + 4a) + 86}+ 6a].
10. 6[6(a + 6)j6(6^6)j].
11. 2(^(6a6) }c(5a + 26)(a36)}.
12. 2^ ja(2a[3a(4a[5a(6ax)J)])}.
13. 25a 196 [36 1 4a (56 6c)}].
31. We liave liitherto supposed tlie syml'ols in every ex
pression iised for illustration to represent syich numbers that
1^'A expressions symbolize results whicli wou Id be arithmetic
j,lly possible.
Thus a — 6 symbolizes a possible result, so long as a is not
less than 6.
If, for instance, a stands for 10 and 6 for 6,
a — 6 will stand for 4.
But if a stands for 6 and 6 for 10,
a — 6 denotes no possible result, because we cannot
take the number 10 from the number 6.
But though there can be no such a thini, as a negative
number, we can conceive the real existence of a negative
quantity.
To explain this we must consider
I. What we mean by Quantity.
II. How Quantities are measured.
32. A Quantity is anything which may be regarded as
being made up of parts like the whole.
Thus a distance is a quantity, because we may regard it as
made up of parts each of themselves a distance.
Again a sum of money is a quardity, because we mav regard
it as made uji of parts like the whole.
33. To measure any quantity we fix upon some known
quantity of the same kind for our standard, or unit, and then
any quantity of that kind is measured by saying how many
times it contains this unit, and this number of times is called
the measure of the quantity.
ADDITION AND SUBTRACTION.
For example, to measure any distaiice a]ohg a road we fix
lip. in a known distance, such as a mile, and express all distances
by saying how many times they contain this unit. Thus 16 is
the measure of a distance containing 16 miles.
Again, to measure a man's income we take one pound as our
unit, and thus if we said (as we often do say) that a i .an's in
con le is 50(T a year, we should mean 500 times the unit, that is,
£5<'0. Unless we knew what the unit was, to say that a man's
inc. me was 500 would convey no definite meaning : all we
shoald know would be that, whatever our unit was, a pound, a
dollar, or a franc, the man's income would be 500 times that
unit, that is, £500, 500 dollars, or 500 francs.
IN.B. Since the unit contains itself once, its measure is
unity, and hence its name.
;'4. Now we can conceive a quantity to be such that wheji
piu to another quantity of the same kind it will entirely or in
p:ii't neutralize its eff"ect.
Thus, if I walk 4 miles towards a certain object and then
rt turn along the same road 2 miles, I may say that the latter
distance is such a quantity that it neutralizes part of my first
j.iurney, so far as regards my position with respect to the point
from which I started.
Again, if I gain £500 in trade and then lose £400, I may
say that the latter sum is such a quantity that it neutralizes
liart of my first gain.
If I gain £500 and then lose £700, 1 may say that the latter
sum is such a quantity that it neutralizes all my first gain, and
not only that, but also a quantity of which the absolute value
Is £200 remains in readiness to neutralize some future gain.
llegardii:g this £200 by itself we call it a quantity which will
have a subtradive effect on subsequent profits.
Now, since Algebra is intended to deal with such questions
in a general way, and to teach us how to put quantities, alike
• iT opposite in their effect, together, a convention is adopted,
I ounded on the additive or subtractive effect of the quantities
in question, and stated thus :
"To the quantities to be added prefix the sign +, and to
the quantities to be subtracted prefix the sign — , and then
>vrite down all the quantities involved in such a question con
nected with these siKUS,"
14 A DDITION A ND SUB TRA C TTON.^
Thus, suppose a man to trade ibr 4 years, and to gain a
pounds the tirst year, to k)se 6 pounds the secoii<l year, tn ga!ii
c pounds the third year, and to lose ti pounds the iuurth year.
The additive quantities are here a and c, which we are to
write +a and +c.
The suhtractive (juantities are here h ;:nd d, whidi we are to
write — h and — d,
:. Eesult of trading — ■\a — h\c — d.
35. Let us next take the case in which the gain for tlie
first year is a pounds, and the loss lor eacli ot three subsequent
years is a pounds.
Eesult of trading = +aaaa
=  2a.
■ Thus we arrive at an isolated quantity of a subtractivi
nature.
Arithmetically we interpret this result as a loss of £2a.
Algebraically we call the result a negative quantity.
When once we have admitted the possibility of the inde
pendent existence of such quantities as this Ave may extend the
application of the rules for Addition and Subtraction, for
I. A negative quantity may stand by itself, and we may
then add it to or take it from some other quantity or expres
sion.
II. A negative quantity may stand first in an expression
which we may have to add to or subtract from any other
expression.
The Rules for Addition and Subtraction given in Art. 21
will be applicable to these expressions, as in the following
Examples.
ADDITION.
(1) 5rt  7ft = — 'la.
(2) 4a366af76=2rti46.
(3) To 4a To 5n3S
Add 3a Add 2a 26
Sum a Sum 3a 56
ADD/T/OX AND SUBTRACTION: 15
(4) 6ffl56 4c + 6 (5) Ixby + Qz
5a + 7612c17 ISx + 'jybz
 a 86 + 19c + 4  ZxS]j+ z
66+ 3c 7 l4x42/ + 5»
SUBTRACTION.
(1) Fiom X
Take ^^y
Remainder x + y
or we might represent the operation thus,
(2) a + 6(o + 6) = a + 6 + a6 = 2a.
(3) —a b{ab)= aba + h= 2a.
(4) 3a+ 46 7c + 10
5a 9^+ 8c +19
8a + 13615c 9
(5) xy{;3x\5x(4:y + 7x)i]
= x — y~ [3x — \ —5x + 4y — Ix {■ ]
=x — y — [3a; + ox  4?/ + lx\
=x — y — Zx — bx + 4y~7x
= 14x + 3i/.
(6) 7a + 56+ 9c12i
36 12c 8d+ 6e
7a+ 86 + 21C 4d6e
In this example we have deviated from our previous prac
tice of placing like terms under each other. This arrange
ment is useful to facilitate the calciiiation. but is not absolutelv
necessary ; for the terms which are alike can be combined
independently of it.
* NoTE. — The meanhig of Subtraction is liere extm !eil so that
the result in Art. 18, Case iv. may he true when b is less than C.
I6 ADDITION AND SUBTRACTION.
Examples. — v.
(I.) ADDITION.
Add togethe'"
1 . 6a + 76,  2a 45, and 3a  5&.
 5a + 66  7c,  2a + 136 + 9c, and 7a  296 + 4r.
2;e — 3?/ + 42,  5x + 4?/ — 72, and  8x — 9y — 32.
4. — rt + 6 — c + (/, a — 26  3c + d, — 56 + 4c, and — 5c + d.
5. a + 6  c + 7,  2a  36  4c + 9, and 3a + 26 + 5c  '**
6. 5x  3a  46, 6?/  2«, 3a — 2i/, and 00  7a;.
7. a + 6 — c, c — a + 6, 26 — c + 3a, and 4a — 3c.
7a  36  5c + 9*/, 26  3c  5(7, and  \il + 15c.
— 12a;  5j/ + 42, 3a; + 2?/  32, and 9a;  3i/ + 2.
9
(2.) SUBTRACTION.
From a + 6 take —a — b.
From a — 6 take  6 + c.
From a — 6 + c take — a + 6 — c.
From 6x  81/ + 3 take  2x + 9?/  2.
From 5a  126 + 17c take  2a + 46  3c.
From 2a + 6  3x take 46 — 3a + 5a;.
From a + 6  c take 3c  26 + 4a.
From a + 6 + c — 7 take 8  c — 6 + a.
From l2x — 'Mjz take 4y  52 + x.
From 8a  56 + 7c take 2c  46 + 2a.
From 9p4q + Zr take bq'6p+r
II. MULTIPLICATION.
36. The operation of findins .'lie sum of a numbers each
equal to h is called Multiplication.
The number a is called the Multiplier.
h Multiijlicand.
This Sum is called tbe Product of the multiplication of h
by a.
This Product is represented in Algebra by three distinct
symbols :
I. By writing the sjTubols side by side, with no sign
between them, thus, ab ;
II. By placing a small dot between the symbols, thus, a.h;
III. By placing the sign x between tlie symbols, thus,
axh ; and all these are read thus, " a into h" or " a times 6."
In Arithmetic we chiefly use the third way of expressing a
Product, for we cannot symbolize the product of 5 into 7 by
57, which means the sum of fifty and seven, nor can we well
represent it by 5.7, because it might be confounded with the
notation used for decimal fractions, as 5 7.
37. In Arithmetic
2x7 stands for the same as 7 + 7.
3x4 4 + 4 + 4.
In Algebra
ab stands for ine same as + 6 + 6+ ... with 6 written
a times.
{a + Vjc stands for the same as c + c + 1 ... with c written
a + 6 times.
[s.A.] B
<8 MULTIPLICATION.
38. To shew that 3 times 4 = 4 tivies 3.
3 times 4= 4 + 4^4
= 1rll + 1 )
'. I.
+1+1+1^1
C
1^1 )
4 times ii= 3 + 3 + 3 + 3
=1^1+1 \
^'^'^' II.
+1+1+1 I
+1+1+1 )
Now the results obtained from I. and II. must be the same,
for the horizontal colunin^: of one are identical with the verti
cal columns of the other.
39. To prove that ah = ha.
ah means that the sum of a numbers each equal to h is to
be taken.
.'. db= 5+6+ with h written a times
= h
+ h
+
to or lines
= 1 + 1 + 1 + \oh terms \
+ 1 + 1 + 1 + to 6 temis f J
+ \
to a lines. )
Again,
ha= a + a + with a written 6 times
= a
+ a
+
to h lines
— 1 f 1 + 1 + to rt terms •\
+ 1 + 1 + 1 + to « terms r y,
to h lines 
MULTIPLICATION. 19
Now the results obtained from I. and II. must be tlie same,
for the horizontal columns of one are clearly the same as the
vertical columns of the other.
40. Since the expressions ah and ha are the same in mean
ing, we may regard either a or h as the multiplier in forming
«#the product of a and 6, and so we may read ah in two ways :
(1) a into 6,
(2) a multiplied by h.
41. The expressions ahc, ach, bac, hca, cah, cba are all the
same in meaning, denoting that the three numbers symbolized
by a, h, and c are to be multiplied together. It is, however,
generally desirable that the alphabetical order of .the letters
representing a product shoula be observed.
42. Each of the numbers a, h, c is called a Factoji of the
product abc.
43. When a number expressed in figures is one of the
factors of a product it always stands first in the product.
Thus the product of the factors x, y, z and 9 i^ represented
by 9xyz.
44. Any one or more of the factors that make up a product
is called the Coefficient of the other factors.
Thus in the expression 2ax, 2a is called the coefficient of x.
45. When a factor a is repeated twice the product would
be represented, in accordance with Art. 36, by aa ; wlien tJiree
times, by aaa. In such cases these products are, for the sake
of brevity, expressed by writing the symbol with a number
placed above it on the right, expressing the number of times the
symbol is repeated ; thus
instead of aa we write a^
aaa a^
aaaa a*
These expressions a, a^, a* are called the second, third,
fourth Powers of a.
The number placed over a symbol to express the power of
the symbol is called the Index or Exponent.
a^ is generally called the square of a.
a? the cube of a.
20 MULTIPLICATION.
46. The product of a^ and o? = a^x o?
= aax aaa = aaaaa = a^.
Thus the index of the resulting power is the sum of the
indices of the two factors.
Similarly a* xa^ = aaaa x aaaaaa
= aaaaaaaaaa = o.^** = a*+*. «
If one of the factors be a symbol without an index, we may
assume it to have an index \ that is
Examples in multiplying powers of the same symbol are
(1) axa'^ — a^~^' = a^.
(2) 7a3 X 5a^ = 7 X 5 X a3 X a' = 35aW = 35aio.
(3) a3 X a« X ^9 = a^+^^ = a^^
(4) x^y X xy^ = x.y.x.y = x~.x.y.y^ = x^"*"^. 1/^+2 _ ^SyS^
(5) a% X a¥ x a^V = a^+i+s. ji+s+r = ^8. jjU
Examples.— vi.
Multiply
I. X into 3y. 2. 3x into 4y. 3, 3xy into 4xy.
4, 3a6c into ac. 5. a^ into a*. 6. a" into a.
7. 3a"5 into 4a.^62_ g, y^^c into Sa^tc^. 9. 15a6''c^ by 12a%.
10. 7aV by 4a6c^. 11. a^.by 3rt^ 12. 4a36a; by 5a6''?/.
13. 19x^2/3 by 4x2/ V. ^4 17a6^2 by 3k?/. 15. 6^y^z^ hy 8x^yz\
16. 3a6cby4axi/. 17. a^i'c by 8a''6''c. 18. 9m^7ip by m%^2.
19. ayz by 6xz^. 20. lla%x by 3ft^"6^*m.
47. The rules for the addition and subtraction of powers
are similar to those laid down in Chap. I. for simple quantities.
Thus the sum of the second and third powers of x is repre
sented by
x^ + x^,
and the remainder after taking the fourth power of y from the
fifth power of y is represented by
and these expressions cannot be abridged.
MULTIPLICATION. it
But when we have to add or subtract the same powers of
the same quantities the terms may be combined into one :
thus
Sy^ + 5y^ + 7y^ = 15?/',
8x*bx* = 3x\
9y^3y^2if = 4i/.
Again, whenever two or more terms are entirely the same
with respect to the symbols they contain, their sum may be 
abridged.
Thus ad + ad = 2ad,
3ab — 2ab = ab,
5a%^ + 6a%^  ga^fes = 2^363,
1a?x— \Oa?x— 12ax= — 15ax.
48. From the multiplication of simple expressions we pass
on to the case in which one of the quantities whose product is
to be found is a compound expression.
To shew that (a + b) c = ac + be.
{a + b) c = c + c + c+ ... with c written a + b times,
= {c + c + c+ ... with c written a times)
+ {c + c + c ... with c ^\Titten b times),
= ac + be.
49. To sheiv that (a — b) c = ae — be.
(a — b)c = c + c + e+ ... with c written a — b times,
= {c + c + c+ ... with c written a times)
— (c + e + c... with c written b times),
= ac — be.
Note. We assume that a is greater than b.
50. Similarly it may be shewn that
(a + b + c) d = ad + bd + cd,
(ab — c) d = ad — bd — ed,
and hence we obtain the following general rule for finding the
product of a single symbol and an expression consisting of two
or more terms.
" Multiply each of the terms by the single symbol, and con
nect the terms of the result by the signs of the several terms
of the eompound expression."
i± MUL T I PLICA TtON.
Examples. — vii.
Multiply
1. a + 6chya. 7. 8//!. + 9m?i + lOn^ by mn.
2. a + 36  4c by 2a. 8. ^a? + 4a't6  ZaW + 40^63 by 2a 6.
3. a^ + 3a^ + 4a by a. 9. y?\j^ — a;?/^ + xy — 7 by a;^/.
4. Ba^^ 5a2  6a + 7 by 3a2. i o. m^  3 )»% ■\'imv?v? by w.
5. a2  2a6 + V by ah. 11. ISa^?, _ 6a262 + oah^ by 12a263
6. o' — 3a62 + J3 by 3a6. 1 2. 13.c^  17a;"'?/ + 5xj/2 — y^ by 8x3.
51. We next proceed to the case in which both multiplier
and nuiltiplicand are comjwmid expressions.
First to nnilti]ily a + b into c + d.
Eepresent c + d by x.
Then (a + b){c + d) = (a + b)x
= ax + bx, by Art. 48,
= a(c + d) + b{c + d)
= ac + ad + hc + hd, by Art. 48.
The same result is obtained by the following process :
c + d
a + b
ac + ad
+ bc + bd
ac + ad + bc + bd
which may be thus described :
Write a + b considered as the multiplier under c + d con
sidered as the multiplicand, as in common Arithmetic. Then
•multiply each term of tlie multiplicand by a, and set down the
result. Next multiply each term of the multiplicand by b, and
set down the result under the result obtained before. The
sum of the two results will be the product required.
Note. The second result is shifted one place to the right.
The object of this will be seen in Art. 56.
Mi 'L TIPL ICA TIOM. 23
52. Next, to multiply a + 6 into c — i.
Represent c — d by x.
Then (a + 6)(cd) = (rt + ?))x
= ax + hx
= a((:  (Z) + ?)(c  rZ)
= ac — a*^ + 6c — M, by Art. 49.
From a comparison of this result with the factors from
which it is produced it appears that if we regard the terms of
the multiplicand c — (Z as independent quantities, and call them
+ cand —d, tlie effect of multiplying the positive terms +a
and +b into the positive term +c is to produce two positive
terms + ac and + he, whereas the effect of multiplying the
positive terms +a and +b into the negative term —d is to
produce tivo negative terms —ad and —hd.
The same result is obtained hy the following process :
c — d
a + b
ac — ad
+ bcbd
ac — ad + bc — bd
This process may be described in a similar manner to that
in Art. 51, it being assumed that a positive term multiplied
into a negative term gives a negative result.
Similarly we may shew that a — b into c + d gives
ac + ad — be — bd.
53. Next to multiply a — b into c — d.
Represent c — d by x.
Then (ab){cd) = {ab)x
= ax — bx
= a{c — d) — b{c — d)
= {ac  ad)  {be  hd), by Art. 49,
— ac — ad — bc + bd.
When we compare this result with the factors from which
it is produced, we see that
The product of the positive term a into the positive
term c is the positive term ac.
±4 MULTIPLICATION.
The product of the positive term a into the negative
term — d is the negative term — ad.
The product of the negative term — h into the positive
term c is the negative term — he.
The product of the negative term — h into the negative
term  rf is the positive term 6(7.
The multiplication of c  d by a — h may be written thus :
cd.
a~b
ac — ad
 be + bd
ac — adbc + bd
54. The results obtained in the preceding Article enable us
to state what is called the Rule of Signs in Multiplication,
which is
"The product of tivo positive terms or of two negative tervm
is positive : the product of tivo terms, one of which is positive aiul
the other negative, is negative."
55. The following more concise proof may now be given of
the Rule of Signs.
To shew that (a — b){c — d) = ac — ad — be + bd.
First, {a  h)M= M +M^M+ ... with M written ab times,
= {M + M + M {■ ... with M written a times)
(M+i¥ + M + ... with M written ft times),
= aMbM.
Next, let M= cd.
Then aM= a (cd)
= {cd)a Art. 39.
= ca — da. Art. 49.
Similarly, bM=cbdb.
.". (a — b)(c — d) = {ca — da) — (cb — db).
Now to subtract (cb — db) from (ca — da), if we take away cb
we take away db too mucli, and we must therefore add dh u>
the result,
.". we get ca  da — cb + db,
which is the same as acadbc + bd. Art. 33.
MUL TI PLICA T/ ON. 2$
So it appears that in multiplying {a h) {c d) we must
multiply each term in one factor by each term in the other
and prefix the sign according to this law : —
When the factors viultiplied have like signs prefix +, when
unlike — to theprodkct.
This is the Rule of Signs
56. We shall now give some examples in ill'istration of the
principles laid down in tlie last five Articles.
Examples in Multiplication wwked out.
(1) Multiply ic + 5 by a; + 7. (2) Multiply x  5 by x + 7.
x+ 5 xb
x+ 7 35 + 7
x^ + 5x x^ — ox
+ 7x + 35 +7x35
a;2 + 12x + 35 x + 2a;35
The reason for shifting the second result one place to the
right is that it enables us generally to place like terms under
each other.
(3) Multiply X + 5 by X  7. (4) Multiply x  5 by x  7.
x + 5 X 5
x7 X 7
x^ + 5x x2_ 53.
7x35  7x + 35
x2_2x35 x'''12x + 35
(5) Multiply x2 + ?/2 by x'i  y'^. (6) Multiply 3ax  5by by 7ax  2by.
7? + if 3ax  ■ hhy
x^if lax 2hy
X* + xh/ 210^x2  Soabxy
xV2/*  dabxy + \Oby^
X*  y* 2lax:  Alahxi^ + lOlj^'^
26 MUL TI PLICA TION.
57. The process in the multiplication of factors, one or
both of which contains more than two terms, is similar to the
processes which we have been describing, as may be seen from
the following examples :
Multiply
'
(1) x^ + .Tt/ + 1/2 by a; — 2/.
(2) a^ + 6a + 9hj a^~6a + 9.
x^ + a;?/ + 2/2
a^ + 6a +9
xy
a^6a +9
01? + X^T/ + XI/2
a* + 6a^ + 9«2
— x^i/  xy^ — y^ '
6a336(i254a
x^y^
+ 9a2 + 54a + 81
a*18a2 + 81
(3) Multiply 3x2 + ^^y _ ^2 by Zx^4xy + y\
3X2+ 43;^ _ y2
3X2 ^y.y ^ y2
9x« + 12x3y 3xy
 12x^1/  \Qxy + 4x?/^
+ 3x!/2 + 4x1/3 _ yi
(4) To find the continued product of x + 3, x + 4, and
x + 6.
To effect this we must muUi]ily x f 3 by x + 4, and then
inltiplv the result by ,(; + 6.
x+ 3
x+ 4
x2+ 3x
+ 4x 4 I fj
x2+ 7x I 12
x+ 6
x''+ 7x2 + 12x
+ 6x2 + 42x + 72
x3 + 13x2 + 54x + 72
Note. Tlie numliers 13 and 54 are called the coefficients of
x2 and X in the expression x^" ISx^f 54x+72, in accordance
with Art. 44.
MUL riPL re A T!0\ :
27
(5) Find the continued product of x + a, a + 6, and z + c.
x^ + ax + bx + ab
x + c
7? + aa;2 + Ix^ + ahx
+ ex + acx + bcx + abc
a? + {a + b + c)x^ + (ab + ac + bc)x + abc
Note. The coefficients of x^ and x in the expression just
obtained are a + b + c and ab + ac + be respectively.
When a coefficient is expressed in letters, as in this example,
it is called a literal coefficient.
Examples. — viii.
Multiply
I. X + 3 by X .'.). 2. a; + 15 by X — 7. 3. x  12 by x + 10
4. X — 8byx — 7. 5. a — 3 by a — 5. 6. y — 6hyy + lS.
7. x24byx2 + 5. 8. x26x + 9 by x26x + 5.
9. X + 5x  3 by x^  5x  3. 10. a^  3a + 2 by a^  3a^ + 2.
II. x^ — x + 1 by X + X— 1. 12. 3:^ + xy + y'^ hy X — xy + y\
13. x^ + xy + y^hyx — y. 14. a  x^ by a* + a%^ + x*.
i^
16
17
18
19
20,
21
22
23
24.
25
26,
x^ — 3x + 3x  1 by x + 3x + 1.
x^ + 3x^y + 9jy' + 21y^ by x — 3?/.
a^ + 2a26 + 4«l + 86^ by a  2b.
SftS + 4a^b + ■lab'^ + ¥ by 2a  b.
cr  2a26 + Za¥ + A¥ by a  2ab  Zb\
a^ + Za"b  2a6'2 + 36^ by a^ + 2a6  3&2.
a — 2ax + 4x2 ]jy ^2 _l 2rta; + Ax\
9rt + 3ax + X 1 ly 9a"  3ax + x^.
X* — 2ax^ + 4a !)}■ r^ f 2ax2 + 4a.
a f 6^ I c^  06  rf c  lie 1 ly a i 6 + c.
x^ + 4xt/ I 5 J/ by x"  Zj?d  2xy^ + 3y\
ab + cd + ac + bd bv ab • cd  ac — bd.
Find the continued product ot the following expression
27. X  a, X i a, x^ t a"*, X* + a*
28. xa, xi&, » — c,
28 ML Y, TIPL ICA TION.
29. 1  a;, 1 + a;, 1 + a;2, 1 + x*.
30. % — y,x + y, x^ — xy + y^, x^ + ocy + y^.
31. a — a;, a + x, a^ + x^, a^ + sc*, a^ + x*.
Find the coefficient of x in the following expansions :
32. (x5)(x6)(x+7). 33. (x + 8)(x + 3)(x2).
34. (x  2) (x  3) (:c 4 4). 35. (xa") (x6)(xc).
36. (x2 + 3x2)(x23x + 2)(x^5).
37. (X2X + 1)(X2 + X1)(X*X2+1).
38. (x  mx + 1) (x^ — mx — 1) (x* — m'x — 1).
58. Our proof of the Rule of Signs in Art? 55 is founded
oil the supposition that a is greater than b and c is greater
tlian d.
To include cases in which the multiplier is an isolated nega
tive quantity we must extend our definition of Multiplication.
For the definition given in Art. 36 does not cover this case,
since we cannot say that c shall be taken — d times.
We give then the following definition. " The operation of
]\[ultiplication is such that the product of the factors a — b and
cvf? tfill be equivalent to ac — ad — bc + bd, whatever may be the
values of a, b, c, rf." .
Now since
(a — b)(c — d) = ac — ad — bc + bd,
make a = and d = Q.
Then (06) (c0) = x c0 x 06c + 6 xO.
or —bxc= —be.
Similarly it may be shewn that
— bx —d= +bd.
Examples. — ix.
Multiply
I . a by — b. 2. a"^ by — a^. 3. a% by  ab.
4. ~4a6by — Safe^. 5. 5x^by— 6x?/2. 6. a'^ — ab + Hhy —a.
7. 3a3 + 4^2 — 5a by — 2a2. 8. —a^ — a — ahy—a—\.
9. 3x"^/ — bxx/ + Ay'^ by — 2x — Zy.
— iSiri^ — 6mn + In"^ by — m + n.
13r17r45 by r3.
Tx^  8x%! — 92 by  x  s.
— X* + x^y — x?y by —y — x.
— y^ — xy — xy — j^ by —x — y.
III. THVOI.TJTION.
59. To this part of Algebra belongs the process called
Involution. This is the operation of multiplying a quan
tity by itself any number of times.
The power to which the quantity is raised is expressed by
the number of times the quantity has been employed as a
factor in the operation.
Tlius, as has been already stated in Art. 45,
a^ is called the second power of a,
a? is called the third power of a.
60. When we have to raise negative quantities to certain
powers we symbolize the operation liy putting the quantity in
a bracket with the number denoting the incfex (Art. 45) jdaced
over the bracket on the riht hand.
Thus ( — of denotes tlie third power of — a,
( — 2.c)* denotes the fourth jjower of — 2x.
61. The signs of all even powers of a negative quantity
will be ])ositive, and the signs of the odd powers will be
negative.
Thus (a)2 = (_a)x(a) = a2,
(^af = {a).{a) {a) = a.{ a)= a^.
62. To raise a simple quantity to any power we multiplv
the index of the quantity by the number denoting the power
to which it is to be raised, and prefix the proper sign.
Thus the square of a^ is a^,
the cube of a^ is a",
the cube of  x^yz^ is  x^yh^.
36 INVOLUTIOX.
63. AVe form the second, third and fourth powers of a + 6
in the following manner :
a + 6
a + 6
a^ + ab
+ «?) +6^
(a + 5)^ = a? + 3n6 + 3a6'' + W
a +6
a4 + 3a36 + 3a26'^ + rt63
(a + by = ft* + 4a36 + 6a%^ + 4a¥+"b\
Here observe tlie following laws :
I. The indices of (i decrease \>y unity in each term.
TI. The indices of b increase by unit}' in each term.
III. The numerical coefficient of the second term is always
the same as the index of the power to which tiie
binomial is raised.
64. We form the second, third and fourth powers of a  6
in the following manner ;
ab
ah
a^ ab
ab +62
(a6)2 = ^2T2a6 + 62
<7  *:
a^  2a~b + ab''
 a^b + 2ahb^
(aby = a^:)al>^ Sali^P
a b
a*  'Sa% + Cab  ab^
 a^b + :Uib  Sab^ + 6*
(a  bf = a*  4a'6 + 6(t6  iab^ + b*.
INVOLUTION. %\
Now observe that the powers of a  6 do not differ from the
powers of a + 6 except that the terms, in which the odd, powers
of 6, as 6', }?, occur have the sign  prefixed.
Hence if any power of a + h be given we can write the
corresponding power of a,  6 : thus
since (« + hf = a* + Sa'i?) + 1 (daW + 1 Oai^ + 5a6* + 6^,
(a  If = a^  5a*6 + X^aW  lOa'^i^ + 5^54 _ y,^
65. Since (a + 6)2 = a2 + 62 + 2a6 and (a  6)2 = a2 + 52 _ 2a&,
it appears that the square of a binomial is formed by the
following process :
" To the sum of the squares of each term add twice the
product of the terms."
Thus (a; + yf = x + if + 2xi/,
(x5)2 = :c2 + 2510x,
(2a;  7i/)2 = U + 49i/  28xi/.
66. To form the square of a trinomial :
a + 6 + c
a + 6 + c
a? + 2a& + ¥ + 2ac + 26c + c.
Arranging this result thus a' + b'^ + c' + 2ab + 2ac + 2bc, we set
that it is composed of two sets of quantities :
I. The squares of the quantities a, b, c.
II. The double products of a, b, c taken two and two.
Now, if we form the square oi abc, we get
abc
abc
a^ab ac
ab + ¥ + bc
ac + bc + c^
a'^2ab + ¥ 2ac + 2bc + c\
The law of formation is the same as before, for we have
32 INVOLVTION.
I. The sqTiares of the quantities.
II. The doul)le products of a,  6,  c taken two by two :
the sign of each result being + or  , according as
the signs of the algebraical quantities composing it
are like or unlike.
67. The same law holds good for expressions containing
more than three terms, thus
(a + 6 + c + (0^ = a2 + 52 + c2 + (i2
+ lah + 2ac + 2arf + 26c + 26(Z + 2cd,
(a& + c(^)2 = a2 + 62 + c2 + ,;2
 '2ah + 2ac  2a(i  26c + 2M  led.
And generally, the square of an expression containing 2, 3,
4 or more terms will be formed l^y the following proci.s :
" To the sum of the squares of each term add twice the
product of each term into each of the terms that follow it."
Examples. — x.
Form the square of each of the following expressions :
I. xva. 2. va. 3. a^ + 2. 4. a; 3. 5. x^ + i/'.
6. x^y\ 7. n' + R 8. a^W. 9. X + 1/ + 2. 10. xy + z.
II. m + wp?. 12. a'" + 2x3. 13. 3?Qx + l.
1 4. 2x2 _ 7 J. + ()_ 1 5 _ yi + if zK 16. X*  ix^y + y*.
17. a^ + P + c^. 1 8. x'^y^z^. 19. x + 2y3z.
20. X  '2y'^ + 5z^.
Expand the following expressions :
21. (x + cf)^. 22. {xaf. 23. (x + 1)^. 24. (x1)^
25. (x + 2)l 26. (rt262)^. 27. {a + b + c)\ 28. (a6c)3.
29. (»i + ?i).(7n  ?i). 30. {in+ny.{m' — n^).
68. An algebraical product is said to be of 2, 3 dimen
sions, when tiie sum of the indices of the quantities composing
the product is 2, 3
Thus ab is an expression of 2 dimensions,
aWc is an exjiression of 5 dimensions.
DIVISTOISr. 33
69. An algebraical expression is called homogeneous when
each of its terms is of the same dimensions.
Thus x'^ + xy + y is homogeneous, for each term is of 2 dimen
sions.
Also 3x^ + 4xi/ + 5i/^ is homo<:eneous, for each term is of 3
dimensions, the numerical coefficients not affecting the dimen
sions of each term.
70. An expression is said to be arranged according to
powers of some letter, when the indices of that letter occur in
the order of their magnitudes, either increasing or decreasing.
Thus the expression a^ + a^x + ax + y? is arranged according
to descending powers of a, and ascending powers of x.
71. One expression is said to l)P of a higher order than
another wlien tlie former contains a higher power of some dis
tinguishing letter than the other.
Thus a^ + ax + rtc + x^ is said to be of a higher order than
a^ + ax + x^, with reference to the index of a.
rr. DIVISION.
72. Division is the ]i!oclss liy which, when a product is
given and we know one ot the factors, ihe other factor is deter
mined.
The product is, vith reference to this process, called the
Dividend.
The given factor is called the Divisor.
The factor which has to be found is called the Quotient.
73. The operation of Division is denoted by the sign =.
Thus ab^a signifies that ab is to be divided by a.
The same operation is denoted by writing the dividend
owr the divisor with a line drawn between them, thus — .
a
In this chapter we shall treat only of cases in which the
dividend contains the divisor an exact number of times.
[S.A.] Q
34 DIVISIOJV.
Case I.
74. When the dividend and divisor are each included in
a single term, we can usually tell by inspection the factors of
which each is composed. The quotient will in this case be
represented by the factors which remain in the dividend, wlien
those factors which are common to the dividend and the di
visor have been removed from the dividend.
Thus X"^*'
Sa^ Zaa .
— = = 6a,
a a
a^ aaaaa „
a' aaa
Thus, when one power of a number is divided by a smaller
power of the same number, the quotient is that power of the
number whose index is the difference between the indices of the
dividend and the divisor.
Thus —=a}'^~' = a\
15a362 _ „,
'3ao
75. The quotient is ^initT/ when the dividend and the
divisor are equal.
Thus ^ = 1; "■^'^^l;
and this will liold true wuen the dividend and the di\isor are
compound quantities.
Thus ■ — r=l; ^r— S=l.
Examples.— xi.
Divide
1. .x" by x'. 2. x^^ by x. 3. xhj"^ by xy.
4. x?'y^^ "hy xyh. 5. 24«6c by 4a?). 6. 72o6'c^ by 9a6c.
7. 256«3ir(;9by USahc^. 8. 1331»i'"»'V'' b' llmn^p\
^. QOa^xif by bxy. ip. 9Ga'6'c3 by 126c,
DIVlStOM. 35
Case II.
76. If the divisor be a single term, while the diviJeml
contains two or more terms, the quotient will be found by
dividing each term of the dividend separately by the divisor
and connecting the results with their proper signs.
m, ax + ix 
Thus = a + 6,
aV + a^x^ + ftx „ „
ax
12xy+16a;y8xy2
4x1/'
2 — 3x^1/2 4 4a;y _ 2.
Examples.— xii.
Divide
1 . x^ + 2x2 + a; by X. 4. m/)x* + m^px^ 4 m^p^ by m'p.
2. if y* + y^y^ by i/^. 5.1 6a^xy  28a^x^ + 4a''x^ by 4a'^x.
3. 8a3 + 16a26 + 24a62by8a. 6. 72xY36xy 18xY by 9x2^/.
7. 81m*ft''  54m^?i^ + 277/!,^/i2p by 3m2?i2.
8. 12xY8xy4xy by 4x3.
9. 169a*6  1 1 7a^62 + 91^25 ,y 13^2^
• 10. 36l65c3 + 2286M13363c5by 1962c.
77. Admitting the possibility of the independent existence
of a term affected with tlie sil^mi  , we can extend the Exam
ples in Arts. 74 — 76, by taking the first term of the dividend
or the divisor, or both, negative. In such cases Ave apply the
Rule of Signs in Multiplication to form a Eule of Signs in
Division.
Thus since —axb=ah,\ve conclude that ^[— = a,
7 7 —ab
ax b=ab, .^ ^a,
—axb = ab, — =a;
and hence the rules
I. When the dividend and the divisor have the shme
sign the quotient is positive.
II. When the dividend and the divisor have different
signs the quotient is negative.
36 D/VIS/OI\r.
78. The followinj; Examples illustrate the conclusions just
obtained :
(I) '^^^.. (3) ^''=9.,.
(5) _ . = lP + ahah + a^.
(6) _4^^2   •' =3a;V4xy + 2.
Examples.— xiii.
Divide
1. 72rt6by9a6. 6.  a V — a2x2 _ ^a; by — ax.
2.  60«8 by  4a3. 7.  34ft3 + 51 ^2 _ i7„_^.2 i,y j „
3 84a;Vby4ry. 8. d,a?h^'2A(eh'^ + 2,2a~¥hy AaW.
4.  ISm^^i^ by Zvm. 9.  144i';3+ 108.721/ 96.r?/2 ^^^ ^^x.
5.  128ft36'c by  86c. 10. ¥xh  Wx'z^  hhfz^ by  ¥z\
Case III,
79. The third case of the operation of Division is that in
which the divisor and the dividend contain more terms than
one. The operation is conducted in the Ibllowing way :
Arrange the divisor and dividend according to the
powers of some one symbol, and ])l>ice them in the
same line as in the process of Long Division in
Arithmetic.
Divide the first term of the dividend by the first term
of the divisor.
Set down the result as the first term of the quotient.
Multiply all the terms of the divisor by the first term
• of the quotient.
Subtract the resulting product from the dividend. If
there be a remainder, consider it as a new dividend,
and proceed as before.
DIVISION. 37
The process will best be understood by a careful study of
the following Examples :
(1) Divide a^ + 2«6 + 6" by a + 6. (2) Divide a?  2ab + b by ab.
a + b)a^ + 2ab + b''{a + h ab) a^  2ab + 6^ (a  6
a" + ab a^ — ab
ab + b' ab + ¥
ab + b" ab + b'^
(3) Divide a;^  y^ by x^ — y^.
x^y^Jx'^y'^^x^hx^y^ + y* ■ ,
a;6 _ a;4y2
a V  T}y'^
xY  y^
xy*  y^
(4)
Divide x^  Aax^ + Aa^j^  o" by x"^  a^.
CC2
 0?) x^  4ax* + 4a*x2  a^ (a;*  3a^x^ + a*
x"  a V
'3ah^ + 4a*x^~a^
3a2x* + 3a%2
aV  a°
(5) Divide 3xt/ + x3 + 1/^1 by i/ + xl.
Arranging the divisor and dividend by descending powers
of x,
x + yl)x^ + 3xy + >/'  i. i^x xy + x + y^ + y+1
a^ + xy  X
xy + x^ + Smj + y^ 1
x^yxy'^ + xy
x^ + xy"^ + 2xy + y^l
x^ + xyx
xy^ + xy + x + y^l •
xy^ + y^y^
xy + x + y^1
Xy + y2y »
x + y 1
x + y1
38 DIVISION.
80. We must now direct the attention of the student to
two points of great importance in Division.
I. The dividend and divisor must be arranged accord
ing to the order of the powers of one of the symbols
involved in them. This order may be ascending or
descending. In the Examples given above we have
taken the descending order, and in the Examples
worked out in the next Article we shall take an
ascending order of arrangement.
II. In each remainder the terms must be arranged in.
the same order, ascending or descending, as that in
which the dividend is arranged at first.
\ °
81. To divide (1) 1 x* by a;3 + x2 + a; + l,
arrange the dividend and divisor by ascending powers of x,
thus :
1+x + x^ + x^
xx^x^x^
xx^a^x^
(2) 48x2 + 6  35x5 + 58x*  70x3 _ 333; by Gx^  5x + 2  Tx^,
arrange the dividend uud divisor by ascending powers of x,
thus :
25x + 6x2  7^3^ 6  23x + 48x2 _ 70^3 + sgx*  35x5 (^3 _ 4^ + 5^2
615x+18x221x3
 SxI 30x2 49x3 + 58x*
 8x + 20x2 24x3 + 28x*
10x2 _ 25x3 + 30x*  35x5
10x2 _ 25^3 + 30x*  35x*
Examples. — xiv.
Divide
1. x2+15x + 50by x+10. 5. x3+ 13x2 454x + 72 by xi6.
2. x2  17x + 70 by x  7. 6. x3 + x^  x  1 by x + 1.
3. x2 + X  12 by X  3. 7. x3 + 2x + 2x 4 1 by x + 1.
4. x2 + 13xl12byx + l. 8. x^  5x3 + 7x2 + 6x + 1 by x2 + 3x + 1.
9. X*  4x3 ^ 2x + 4x + 1 by x2  2x  1.
10. x*4x3 + 6x24x+l by x22x+l.
^1
Dins 10 y. 39
n. a;*x2 + 2xl by a;2 + xl. 12. a;*4x2 + 8x+ 16 l>y x + 2.
13. x^\ ixy + 3xi/ + 12)/^ by x + 4y.
14. a* + 4a36 + 6a262 + 4^53 + 54 ^y ^ + 6.
15. <{5  5«*6 + 10a362  I0a263 + 5^54 _ ^s ijy ^ _ 5,
16. x*  12x' + 50x2  84x + 45 by x2  6x + 9.
17. «■•  4a*b + 4a362 + 4^(2^,3 _ 17^54 _ i265 by a^  2ah  3//.
1 8. 4a2xi  Ua^x^ + lSa*x  Gu^x + a^ by 2(tx2  3a2x + a\
1 9. X*  a;^ + 2x  1 by x2 + X  1.
20. X* + rtx2  2a^ by x2 + 2a2. 23. x^  y^ by x  1/.
21. x2  rSxy  '30y'' by x  15j/. 24. a  b + ihc  c2bya  6 + c.
22. x'' + (/' by X + y. 25. h  'ilr + 36^  i* by &  1.
26. tr  62 _ c2 + ^2 _ 2(afi  6c) ])y a + 6  c  rf.
27. x^ + ?/■' + ^3  2xyz by x + ?/ + z. 28. x^^ + ;/^" by x^ + yK
29. ^2 4.^2 4. 223}  2^2 + 72? _ 3}2 by y q + 3r,
30. a^ + a«62 + a«6* + u¥' + 6« by a* + a% + «62 + a¥ + 6^
31. x^ + x^y~ + x^y^ + xy^' + y^ by x^  x^y + x'y^  xy^ + y*.
32. 4x5  x3 + 4,^. by 2x2 + 3^. ^ 2. 33. a''  243 by a 3.
34. Z;io  k by P  1. 35 a;^  5x2 _ 46.5  40 by x + 4.
36. 48x3  76ax2  6iax + I05a^ by 2x  3a.
37. ISx*  45x' + 82x2 _ 673 + 40 by 3x2 _ 4^ ^ ^
38. 16x*  72a2x2 + 81a* by 2x  3a.
39. Six*  256a* by 3x + 4a. 41. x^ + 2ax2  a2x  2a^ by x2  a'K
40. 2«^ + 3a262a62_363bya262. 42. a* a262_ i264bya2 + 352
43. X*  9x2 _ Q.j.y _ y2 by x2 + 3x + y.
44. X*  6x^y + 9xhf  Ay^ by x2  3xy + 2y\
45. X*  8ly* by X  Sy. 47. 81a*  166* by 3a + 26.
46. a*  166* by a  26. 48. 16x*  81y* by 2x + Sy.
49. 3a2 + 8a6 + 46^ + lOac + 86c + 3c2 by a + 26 4 3c.
50. a* + 4a2x2 + 16x* by a + 2ax + 4x^.
51. X* + x2j/2 + y* hy X'  xy + y^. ,
52. 256x* + 16x2!/2 + y* by 16x2 + 4xy + y\
53. x^ + x*7/  x^2/2 + x^  2x2/2 + y^'by xi^ + xy.
40 DIVISION.
54. ax^ + Sa^x^  <id?x  2a* by a;  a. 55. a^  ^ byx + a.
56. 2x2 + a;?/  3!/^  4i/z  X3  2;^ by 2x + 3y+ z.
57. 9a; + Sa;* + 14x'' + 2 by 1 + 5x + x^.
58. 12  38x + 82x2  1 12x3 + 106x*  TOx^ by Tx^  5x + 3.
59. x^ + 1/^ by X*  x^y + x^y^  xy^ + y*.
60. (ax^ + h\j'^  (a?h + x'^y'^) by ax + by + ab + xy.
61. a& (x + 1/2) + xj/(a2 + 62) by ax + by.
62. X* + (262  a^)x2 + ¥ by x2 + ax + b\
82. The process may in smne cases be shortened by the use
of brackets, as in the following Exaui2:)le.
x + 6^x^ + (a + 6 + c) x2 + (a6 + ac + 6c) x + a6c(x2 + (a + c) x + ac
x^ + 6x2
(a + c) x2 + (a6 + ac + bc) x
(a + c) x2 + (a6 + 6t) x
acx + abc
acx + abc
x — l)x^ mx* + na^ — ?ix2 + mx — 1 (:'•■*  (m  1) x^
x^x* (//i.7il)x2(?n,l)x+l.
 (m  1) X* 4 nx^
(ml)x* + (ml) x3
— (rn — n—l)x^ — nx^
{mn1) a^+{vinl) X
(m 1) x2 + mx
 (m  1) x2 + (in 1) X
x1
x1
Examples.— XV.
Divide
1. X*  (a2 _ 6 _ c) x2  (6  c) ax + be by x2  ax + c.
2. y^(l + m + n) y^ + {hn + In + mn) y  Imn by yn.
3. x^  {m  c) x'* + {n  cm + d)oi^ +
{r + en  dm) x^ + {cr + dn) x + dr by x^  mx + ')ix + r.
4. X* 4 (5 + a) x3  (4  5a + 6) x2  (4a 4 56) x 4 46 by x2 4 5x  4.
5. x*(a464c4rf) x3 4(a64ac4a<i46c46(i4cd) x^
 (a6c 4 a6(/ 4 acd + bed) x 4 abed by x*  (a 4 c) x 4 ac.
division: 41
83. Tlie following Exainples in Division are of great
importance.
Divisor.
Dividend.
Quotient.
x + y
x22/2
xy
xy
x^y^
x + y
x + y
x^ + y^
x^  xy + y^
xy
oc^y^
x^ + xy + y^
84. Again, if vre an'ange two series of binomials consisting
respectively of the sum and the ditference of ascending powers
of x and y, thus
x + y, X" + y, y? + y^, xf^ + ?/■*, x'' + y^, a;" + 7/'', and so on,
xy, X y, sr  y^, x*  y^, af"  y^, x'^  1/", and so on,
x + y will divide the odd terms in the upper line,
and the even in the lower
xy will divide all the terms in the lower,
but none in the upper.
Or we may put it thus :
If n stand for any whole number,
x^ + jj" is divisible by x + y Avhen n is odd,
by xy never ;
x^y" is divisible hy x + y when n is even,
hy xy alv.jys.
Also, it is to be observed that when the divisor is a;y all
the terms of the quotient are positive, and when the divisor is
x + y, the terms of the quotient are alternately positive and
negative.
x^ — v^
Thus^^ — ^ = a^ + x'^y + xy^ + y^,
xy
xJ + 'lf~
^ = x^ afy + xhf  oi?y^ + x^y^  xy^ + y^,
J =x^ xhf + x^y^  x^y^ + xy*  y^.
45 Tilvisroy:
85. These properties may bf easily remembered by taking
the four simplest cases, thus, x + j/, xy, x + y, x^yp, of
which
the first is divisible by x + y,
second xy,
third neither,
fourtli both.
Again, since these properties are true for all values of x and
y, suppose y = \, then we shall have
x^\ , x^l
a;+l X  1
3? + l , , x3l ,
 = xx+l, '=x + x+l.
X+l X I
Also
x^ + l , , ., ,
; = X*  X/ + .X  X + 1,
xi 1
x^ 1
= x^ + X* + x^ + X + X + 1.
X 1
Examples. — xvi.
"Without going through the process of Division write down
the quotients in the following cases :
1. When the divisor is m + n, and the dividends are
respectively
m^  n^, m^ + n^,'ni^ + n^, m^  n^, m^ + ?i^.
2. When the divisor is m  n, and the dividends are
respectively
vi^  n~, m^  n^, m*  7i*, m^  n^, vi' — iiJ.
3. AVhen the divisor is a+1, and the dividends are
respectively
(r1, r»^+ 1, a^ + l, a" + l, a*l.
4. When the divisor is y\, and the dividends are
respectively
2/21, !/5 1,2/51,7/7 _ 1^2/9 _1.
V. ON THE RESOLUTION OF EXPRES
SIONS INTO FACTORS.
86. We shall discuss in this Chapter an operation which
is the opposite of that which we call Multiplication. In Mul
tiplication we determine the product of two given factors : in
the operation of which we have now to treat the product u
given and the factors have to be found.
87. For the resolution, as it is called, of a product into its
component factors no rule can be given which shall be applic
able to all cases, but it is not difficult to explain the process
in certain simple cases. We shall take these cases separately.
88. Case I. The simplest case tor resolution is that in
which all the terms of an expression have one common factor.
This factor can be seen by inspection in most cases, and there
fore the other factor may be at once determined.
Thus a^ + ah = a(a + b),
2a3 + 4a2 + Sa = 2a {a? + 2a + 4),
23?y  1 %xhf + bAxij = 9xy (x  2xy + 6).
EXAMPLPS.— xvii.
#
Resolve into factors :
1. 5a;215x. 5. a^ax^ + hx'^ + cx.
2. 3rc" + 18x26a;. 6. 3afy^2lxY + ^'^x^y*
3. 49yUy + 7. 7. 54a%'i + 108a%'*  24Sa^b\
4. 4x^y\2xy2 + Sxy\ 8. 45a;"(/i^  90^5^7  360xV^.
44 RESOL UTTOJSr INTO FA CTORS.
89. Case 41. The next case in point of simplicity is that
in which four terms can be so arranged, that the first two have
a common factor and the last two have a common factor.
Thus
a;^ + ax + 6a; + a6 = {^ + ax) + (ix + a6)
= .r (.r 4 a) 4 & (x + a)
= (x + 6) (x + a).
Again
ac  ad  be + bd = (ac  ad)  (he  hd)
= a{cd)h{cd)
= (ah) (cd).
Examples. — xvi ii.
Resolve into factors :
1 . x^ axbx + ab. 5. ahx^  axy + hry  y\
2. ab + ax — hx — x"^. 6. abx — ahy + cdx — cdy.
3. bc + hy cy  y^. 7. cdx^ + dmxy  cnxy  m n y'.
4. hm + mn + ab + an. 8. abcxb^dxacdy + bd'y.
90. Before reading the Articles that follow the student is
advised to turn back to Art. 56, and to observe the manner in
which the operation of multiplying a binomial by a binomial
produces a trinomial in the Examples there given. He will
then be prepared to expect that in certain cases a trinomial
can be resolved into two binomial factors, examples of which we
shall now give.
91. Case III. To find the factors of
x + 7x+12.
Our object is to find two numbers whose product is 12,
and whose sum is 7.
These will evidently be 4 ai^d 3,
* .. x^ + 7x + 12 = (x 4 4) (x + 3).
Again, to find the factors of
x2 + 56x + 662.
Our object is t4 find two numbers whose product is 65*,
and whose sum is 56.
These mil clearly be 36 and 26,
.•. X + 56x + 66 = (x + 36) (x + 26).
ffF.SOLUTION INTO FACTORS.
Examples.— xix.
Resolve into factors :
a:2+llx + 30.
X+ 17x + 60.
2/2+13^ + 12.
i/ + 2l!/ + 110.
?7i2 + zbvi + 300.
m? + 23»i + 102.
a2 + 9«6 + 862,
a; + 13ma; + 36??;^.
9. j/2 + 19?!?/ + 48n2.
10. ^ + 2^ + 1002)2.
1 1 . a* + 5x + 6.
12. u:'' + 4x3 + 3.
13. a:2i/2 + 18a:i/ + 32.
14. xh/\1x^y'+l2.
i;. m'o + 10?rt=+16.
93.
16. ?i' + 27?i2+ 14032.
Case IV. To find the factors of
a;— 9x + 20.
Our object is to find two negative terms whose product is 20,
aud whose sum is — 9.
These? will clearly he  5 and  4,
.. x2  9.C + 20 = (,.;  5) (x  4).
Exam ples. — xx.
Resolve into factors :
x7x+ 10.
X  29x + 190.
2/2  237/ + 132.
y  30y + 200.
7r43?i+.460.
6. 'n^57n + 56.
7. 3:^7x^ + 12.
8. a26227a6 + 26.
9. Mc''ll62c3 + 30.
10. xyhl3xijz + 22.
92.
Case V. To find the factors of
x2 + 5x  84.
Our object is to find two terms, one positive and one negative,
w liose product is  84, and whose sum is 5.
These are clearly 12 and  7,
.. x2 + 5x  8^= (x + 12) (x  7).
46
RESOLUTION INTO FACTORS.
Examples.
— xxi.
Resolve into factors :
I. x^^'lxm.
6. ■
62 + 256150.
2. a:2 + 12a;45.
7
a;8 + 3ar*4.
3. rt2+n(j_i2.
8.
a;V + 3xi/154.
4. a2+13a140.
9
7/i'0+15rr65 100.
5. 6 +13?^ 300.
10.
7(2 +17,1 390.
94. Case VI. To find the factors of
;«  3.C  28.
Our object is to find two terms, one positive and one negative,
whose product is  28, and whose sum is  3.
These will clearly be 4 and  7,
.. a;23a;28 = (2; + 4)(j;7).
Examples.— xxii.
Resolve into factors :
I.
■j?hx 66.
2.
x^  Ix  18.
3
ni^  9??i  36.
4
?i2_ 11,^60.
5
1/13J/14.
2 152 100.
.aio _ 9.,.5 _ 10.
cd'24ctZ180.
in^'n  mhi  2.
95. The results of the four jnvcvding articles may be tims
stated in general terms : a trinomial of one of the forms
X" + ax + h, x^  ax + h, x + ax  b, xax h,
nuiy be resolved into two simple factore, when b can be re
solved into two factors, such that their sum, in the fiist two
iorms, or tlieir difference, in the last two forms, is equal to a.
96. We shall now give a set of Miscellaneous Examples oh
the U'solution into factors of expressions which come under
one or other of the cases already explained.
RESOLUTION INTO FACTORS. 47
Examples. — xxiii.
Kesolve into factor.s :
1. a;15x + 36. 8. a;" + TOx + ?ia; + mu..
2. a;^ + 4.o45. ^ 9. 1/^ 41/^ + 3.
3. a^W'  1 Ga6  36. i o. a;^  «&x  cxy + a6c.
4. a:^ 3?;.'x* IOjh^. II. a;2 4. (^fj _ j^ ^. _ g5_
5. T/^ + 1/^90. 12. a;  (c  rf) X  c(i.
6. x*a:110. 13. ab^  bd + cd  abc.
7. ar^ + 3«x + 4ff2x. 14. 4^^  28^!/ + 48i/2. .
97. We have said, Art. 45, that when a number is mnlti
plied by itself the result is called the Square of the number,
and that the figure 2 placed over a number on the right hand
indicates that the number is multiplied by itsell'.
Thus a^ is called the square of a,
(x  yf is called the square oi xy.
Tlie Square Root of a given number is that number
whose square is equal to the given number.
Thus the square root of 49 is 7, because the square of 7
is 49.
So also the square root of a^ is a, because the square of a is
a^ : and the square root of {x  y) is xy, because the square
of xy is (x yy.
The symbol ^1 placed before a number denotes that the
square root of that number is to be taken : thus ,j2b is read
" the square root of 25."
Note. The square root of a positive quantity may be either
positive or negative. For
since a multiplied by a gives as a result a',
and  a multiplied by  a gives as a result a^,
it follows, from our definition of a Square Root, that either a
or  a may be regarded as the square root of a^.
But throughout this chapter we shall take only the positive
value of the square root.
'48 RESOLUTION INTO FACTORS.
. 98. We may now take the case of Trinomials which are
verfect squares, which are really included in the cases dis
cussed in Arts. 91, 92, but which, from the importance they
nssume in a later part of our suliject, demand a s^eparate con
.ii deration.
99. Case VII. To find the factors of
Seeking for the factors according to the hints given in Art
9i, we find them to be a; + 6 and x+ 6.
That is a;2 + 12x + 36 = (x + 6)2.
Examples. — xxiv.
Resolve into factors :
1. a;2+18a; + 81. 6. a;* + 14a;2 + 49.
2. a;2 + 26a; + 169. 7. a;2+ 10xi/ + 25t/2.
3. a;2 + 34x4289. 8. tn!^ + \(5mhi + QAn*.
4. y' + 'iy+l. 9. x« + 24.13 + 144.
5. 22 + 2002+10000. 10. x?/2 + 162x«/ + 6561.
100. Case VIII. To find the factors of
.r^12x + 36.
Seeking for the factors according to the hints given in Art.
92, we find them to be x  6 and x  6.
That is, x2  12x + 36 = (x  6)2.
Examples.— XXV.
Resolve into factors :
I. x28x + 16. 2. x228x + 196. 3. x236x + 324.
4. 2/2 _ 40j/ + 400. 5 . c2  lOUs + 2500. 6. .i"*  22x2 + 1 2 1 .
7. x2  30x1/ + 225?/2 8. 7H^32?/i.2„2 + 256«*.
n. x« 38x3 + 361.
RESOLUTION INTO FACTORS. 49
101. Case IX. We now proceed to the most important
case of Resolution into Factors, namely, that in which the ex
pression to be resolved can be put in the form of two squares
with a negative sign between them.
Since m^  n^ = (m + n) (m  n),
we can express the difference between the squares of tw<i
quantities by the product of two factors, determined by tlie
following method :
Take the square root of the first quantity, and thasquaie
root of the second quantity.
The sum of the results will form the first factor.
The difference of the results will form the second factoi'.
For example, let a^  Ir be the given expression.
Tlie square root of a is a.
The square root of h is h.
The sum of the results is a + 6.
The difference of the results is a — b.
The factors will therefore be a + 6 and a  6,
that is, a^¥ = {a + b) (a  b).
102. The same method holds good with resjDect to com
pound quantities.
Thus, let a^  (6  c) be the given expression.
The square root of the first term is a.
The square root of the second term is 6  c.
The sum of the results is a + bc.
The difference of the results is a — b + c.
.. a^ {bcy = {a + bc){ab + c).
Again, let (a  by  (c  d) be the given expression.
The square root of the first term is a  6.
The square root of the second term is c  d.
The sum of tlie results is ab + cd.
The difference of the results is abc + d.
:. {ab/ {cdf=(ab + cd^ (abc + d).
[6.A.] • D
50 kESOLUTION INTO FACTORS.
103. The terms of an expression may often be arranged
30 as to form two squares with the negative sign between
them, and then tlie expression can be resolved into factors.
Thus a2 + 52_c2_f^2 + 2a6 + 2cd
= a2 4.2a6 + 62_c2 + 2cdd2
= {o? + 2a6 + 62)  (c2  2cd + d?)
= (a + 6)2(cd)2
= (a + 6 + c  c?) (a + 6  c + (Z).
Examples. — xxvi.
Resolve into two or more factors :
I. x^y^. 2. x29. 3. 4a;225.
4. a* a;*. 5. x^,!. 6. a;''_i.
7. x^l. 8. m'^lQ. 9. 36!/24922.
10. Slxhf  121a262. J I. (a _ 5)2 _ ^2, ■ 12. x^  (m  nf.
13. (a + 6)  (c + (Z)2. 24. 2a;2/a;2i/2^1_
14. {x + yf{xyf. 25. x^2yzyz^.
15. x22xt/ + i/2!2. 26. a24629c2 412ic.
16. (a^6)2(m + Ji,)2. 27. a* 1661
17. a2_2ac + c262_26ji_(^2_ ^g. l49c2.
18. 2bcb^c^ + d\ 29. a2 + 62_c2_rf2_2a5_2crf
19. 2xy + x'^\y'^z^. 30. a  6 + ^2  ^2 _ 2ac + 266?.
20. 2mn  m2  %2 ^ ^2 + 52 _ 2^6. 3 1 . Sa'^.r'^  27ax,
2 1 . (ax + byy  1. 32. a''6''  c*.
22. (ax + %/  (ax  6?/)2. 33. (5x  2)  (x  4)2.
23. la262 + 2a6. 34. {7x + 4yy{2x + 3y)l
35. (753)2 (247)2.
104. Case X. Since
=x2«a: + a2, and ^ = x2 + ax + a2 (Art. 83),
X ~r d X — CV
we know the following important fj^^jts ,•
RESOLUTION INTO FACTORS. 51
(1) The suTfi of the cuhes of two numbers is divisible by
the swm of the numbers :
(2) The difference, between the^ cvhes of two numbers is
divisible by the difference between the numbers.
Hence we may resolve into i'actors expressions in the form
of the sum or difierence of the cubes of two numbers.
Thus ic3 + 2 7 = :<? + 33= (./ + 3) (x  3x + 9)
3/3_64 = 2/343=(2/4)0/ + 42/+16).
Examples.— xxvii.
Express in factors the following expressions :
I. a3 + 63. 1. a? ^W. 3. a3_8. 4. a;3 + 343.
5. 63 _ 125. 6. x3 + 64;/3. 7. a3216. 8. '6y? + Tif.
9. 64a3 100063. 10. 729x3 + 512t/3.
Express vcijour factors each of the following expressions :
II. x^i/^ 12. x^1. 13. 0*^64. 14. 729 j/S.
105. Before we proceed to describe other processes in
Algebra, we shall give a series of examples in illustration of
the principles already laid down.
The student will find it of advantage to work every example
in the following series, and to accustom himself to read and to
explain with facility those examples, in which illustrations are
given of what may be called tlu shorthand method of expressing
Arithmetical calculations by the symbols of Algebra.
Examples. — xxviii.
1 . Express the sum of a and b.
2. Interpret the expression ab + c.
3. How do you express the double of x ?
4. By how much is a greater than 5 ?
5. If a; be a whole number, what is the number next
above it ?
6. Write five numbers in order of magnitude, so that x
phall be the tliird of the five,
RESOLUTION INTO FACTORS.
7. If a be multiplied into zero, what is the result ?
8. If zero be divided by x, what is the result ?
9. What is the sum of a + a + a . . . written d times ?
10. if the product be ac and the multiplier «, what is th«
iiiulli[ilicand?
1 1 . What number taken from % gives 1/ as a remainder ?
12. J. is a; years old, and B is y years old ; how old was A
when B was born ?
13. A man works every day on weekdays for x weeks in
llie year, and during the remaining weeks in the year he does
not work at all. During how many days does he rest ?
14. There are x boats in a race. Five are bumped. How
many row over the course ?
15. A merchant begins trading with a capital of x pounds.
He gains a pounds each year. How much capital has he at
the end of 5 years 1
•' _ «
16. A and B sit down to play at cards. A has x shillings
and B )j shillings at first. A wins 5 shillings. How much has
each wiieu they cease to play ?
17. There are 5 brothers in a family. The age of the eldest
is X years. Each brother is 2 years younger than the one next
above him in age. How old is the youngest ?
18. I travel x hours at the rate of y miles an hour. How
many miles do I travel ?
19. From a rod 12 inches long I cut off x inches, and then
I cut off y inches of the remainder. How many inches are
left ?
20. If n men can dig a piece of ground in q hours, how
many hours will one man take to dig it ?
21. By how much does 25 exceed xl
22. By how much does y exceed 25 ?
23. If a ]iroduct has 2m repeated 8 times as'a factor, how
do you express the product ?
24. By how much does a + 2h exceed a2bl
25. A girl is X years of age, how old was she 5 ye.irs since ?
RESOL UTTON INTO FA C TO RS. 53
26. A boy is y years of age, how old will he he 7 years
hence 1
27. Express the difference between the squares of two
numbers.
28. Express the product arising from the multiplication of
the snui of two numbers into the difference Ijetween the same
numbers.
What value of x will make 8a; equal to 16 ?
What value of x will make 28a; equal to 56 ?
X
What value of x will make ^ equal to 4 ?
What value of x will make a; + 2 equal to 9 ?
What value of x will make a;  7 ecpial to 16 ?
What value of x will make a; + 9 equal to 34 ?
What value of x will make a. 8 equal to 92 ]
Examples. — xxix.
Explain the operations symbolized in the following expres
sions :
1. + 6. 2. (I IP. 3. 4(t2 + fc3_ 4_ 4(f(2^^2)_
5. a2_26 + 3c. 6. a + mKhc. 7. ((( + ?h)(6 c). 8. s]^.
9. ^x + t/. 10. a + 2(3c). II. (a + 2)(3c),
12. . ,. 13. ^^^ ^ . 14. — ■ — ^ .
4ab ■" xy ^ ^f^ + y
Examples. — xxx.
If a stands for 6, b for 5, x for 4, and y for 3, find the val ue
of the following expressions :
I. a + xb~y. 2. a + ybx. 3. 3a + 4yb2x.
4. 3(a + 6)  2(x  y). 5 . (a + x){b y). 6. 2a + 3 r y).
7. (2a + 3)(x + y). 8. 2a + Zx + y. 9. ^^ti'.
10. abx. II. ab{x + y). 12. ay{b\xf.
54 * kEsdLUTION INTO FACTOkS.
13
ah{x y)^.
14.
v/56.
1 6.
isfW
17.
{J^+b)\
19.
^2axy.
20.
a^ + ¥ + y
15 Jy^
18. J5bx.
21. 3a + (2xi/)2.
22. «(&i/)f !a(a;i/)f. 24. 3(« + 5 ?/)H4(/i4x)*.
23. (a62/)2 + (aa; + 2/)2. 25. 3 (tt  6)^ + (4x  1/2)2.
EXAMPLES. — XXXi.
1. Find the value of
Sabc a^ + P + c^, when a = 3, /> = 2, c=l.
2. Find the value of
3? + y^ z^ + 3xyz, when ic = 3, y = 2, g = 5.
3. Subtract «' + c^ from (a + c).
4. Subtract (x  1/) from x + y'^.
5. Find the coefficient of x in the expression
{a + byx{a + bxf.
6. Find the continued product of
2x  VI, 2x + n, x + 2m, x  2n.
7. Divide
acr^ + {be + ad) r^ + {bd + ae)r + be by ar + h;
and test your result by putting
a = b = c = d = e=l, and r = 10.
8. Obtain the product of the four factors
(a + h + c), (b + ca), {c + a b), {o. + bc).
What does this become when c is zero ; when 6 + c = a;
when a = b = c'{
9. Find the value of
(a + 6)(6 + c) {c + d)(d + a){a + c) (bd), '
where b is equal to d.
10. Find the value of
3a + (26  c2) + ! (;2  (2a + 5b) ! + { 3c  (2a + 36)2,
wh(;i a = 0, 6 = 2, c = 4.
RESOLUTION INTO FACTORS.
11. If a = l. // = 2, c = 3, d=A, shew that the numerical
values are ec^ual of
j(Z(c& + a)f{(fZ + c)(6 + o)i,
and of d^  (c^ + 6) + a + 2 (6c  arf).
12. Bracket together "the different powers of x in the follow
in g expressions :
(a) ax^ + bx'^ + ex + dx. .
(/?) ax^  h:i?  ex"  dx} + 2x2.
(7) A'j?  ax^  3x  hi?  5x — ex.
(S) {n + x;'(hxy.
(e) {mx + qx. + 1 )■'  (?!x2 + qx+l )2.
13. Multiply the three factors xa, xb, xc together,
and arrange the product according to descending powers of x.
14. Find the continued product of (x + a) {x + b){x + c).
15. Find the cube of a + h + c; thence without further
multiplication the cubes of a + 'jc; b + c a; c + ab; and
subtract the sum of these three c\\ >e< from the first.
16. Find the product of (3a + 2b) (3a + 2c  3b). and test the
result by making a = 1, 6=:c = 3.
17. Find the continued product of
ax, a + Xj a + x', a'' + a;'*, a* + x^.
1 8. Subtract (b  a) (c  d) from {a  6) (c  d).
What is the value of the result when a = 26 and tZ = 2c ?
19. Add together J) + y) (a + x), xy, ax  by, and a(x + y).
20. What vaiue of x will make the difference between
(x + 1) (x + 2) and (x  1) (x  2) equal to 54 ?
21. Add together ax by, x y, x(x  ?/), and (a  x) {h  y).
22. ^T:iat value of x will make the difference between
(2x + 4) (3x + 4) and (3x  2) (2x  8) equal to 96 ?
23. Add together
2mx  3ny, x + y, 4(m + n)(x y), and mx + vy.
24. Prove that
{x + y + z)^ + x'^ + y^ + z = {x + ij)'^ + {y + z)^ + {x + zy.
56 RESOUmON INTO FACTORS.
25. Find the product of (2a + 36) (2a + 3c  20;, and test the
result by making a = \, 6 = 4, c = 2.
26. If a, b, c, d, e ... denote 9, 7, 5. 3, I, find the values of
ab  cd .,  . 62 _ ^2
___; (6c«f?)(Wc.); ^^; andrt«c^
::/. Find the value of
3a6c  a^ + Ir + c^ when a = 0, 6 = 2, c = l.
;3. Find the value of
.^ , , 2«62 • c^ , , I, 1 n
3a H ^:, Avhen a = 4, 6 = 1, c = 2.
29. Find the value of
(a6c)2 + (6«r)' + (ca6)2'when a=l, 6 = 2, c = 3.
30. Find the value ol'
(rt + 6  c)2 + (a  6 + c)2 + (6+ cfl)2 when a=;l, 6 = 2, c=4.
31. Find the value of
(a + 6)2 + (6 + c)' + (e + a)2 when a=  1, 6 = 2, c= 3.
32. Shew that if the sum of any two nnmhers divide the
ditt'erence of their squares, the quotient is equal to the differ
ence of the two numbers.
33. Shew that the product of the sum and difference of anv
t\Mi numbers is equal to the difference of their squares.
34. Shew that the square of the sum of any two consecu
ti\e integers is always greater by one than four times their
jjriMluct.
35. Shew that the square of the sum of any two consecutive
even whole numbers is four times the square of the odd number
between them. .
36. If the number 2 be divided into any two parts, the
ditlerence of their squares will always be equal to twice liie
difference of the parts.
37. If the number 50 be divided into any two parts, tli
difference of their squares will always be equal to 50 timdi tli.
difference of the parts.
38. If a number n be di^dded into any two parts, the
difference of their squares will always be equal to n times the
difference of the parts.
ON SIMPLE EQUA TIONS. 57
39. If tw'o numbers differ by a imit, their product, together
with the sum of their squares, is equal to the difference of the
cubes of the numbers.
40. Shew tliat the sum of the cubes of any three consecu
tive whole numbers is divisible by three times the middle
number.
VI. ON SIMPLE EQUATIONS.
106. An Equation is a statement that two expressions
are equal.
107. An Identical Equation is a statement that two ex
pressions are equal for all numerical values that can be given
to the letters involved in them, provided that the same value
be given to the same letter in every jiart of the eciuation.
Thus, 0<; + a)2=a;2!2ax + a2
is an Identical Equation.
108. An Equation of Condition is a statement that two
expressions are equal for some particular numerical value or
values that can be given to the letters involved.
Thus, a;+l = 6
is an Equation of Condition, the only number which x can
represent consistently with this equation being 5.
It is of such equations tliat we have to treat.
109. The Root of an Equation is that number which, wlien
])at in the place of the unknown quantity, makes both sides of
the equation identical.
110. The Solution of an Equation is the process of find
ing what number an unknown letter must stand for that the
eipiation may be true : in other words, it is the method of
fuuling the Eoot.
The letters that stand for imknown numbers are usually
X, y, z, but the student must observe tliat any letter may
stand for an unknown number.
111. A Simple Equation is one which contains the
first jiower only of an unknown quantity. This is also called
^n Equation of the First Decree.
58 ON SIMPLE EQUATIONS.
112. The following Axioms form the grounANVork of the
solution of all equations.
Ax. I. If equal quantities be added to equal quantities,
the sums will be equal.
Thus, if a = &,
Ax. II. If equal quantities be taken from eaual quantities,
the remainders will be equal.
Thus, if x = y, ^
xz = y z.
Ax. III. If equal quantities be multiplied b^ equal quan
tities, the products will be equal.
Thus, it a = h.
Ax. IV. If equal quantities be divided uy equal quantities,
the quotients will be equal.
Thus, if xy = xz,
y=z.
113. On Axioms I. and II. is founded a process of great
ntilitv in the solution of equations, called The Traksposition
OF Terms from one side of the equation "c the other, which
may be tlius stated :
" Any term of an equation may be transferred from one side
of the equation to the other if its sign be changed."
For let xa = h.
Then, bv Ax. I., if we add a to both udes, the sides remain
equal :
therefore xa + a = b + a,
that is, x = b + a.
Again, let x + c = (l. •
Then, by Ax. II., if we subtract c liom ^u,•ih. side^ the sides
remain equal :
therefore iC + c~c = d~c,
l.uat is, x=dc,
ON SIMPLE EQUATIONS. 59
114. We may change all the signs of each side of an equa
tion without altering the equalit}'.
Thus, if ax — hc,
xa = cb.
115. We may change the position of the two sides of the
e(nation, leaving the signs unchanged.
Thus the equation a  b = x  c, may be written thus,
X c = a b.
116. We may now proceed to our first rule tor the solution
of a Simple Equation.
Rule I. Transpose the known terms to the right hand side
(>f the equation and the unknown terms to the other, and com
I'ine all the terms on each side as far as possible.
Then divide both sides of the equation by the coefficient of
the unknown quantity.
This rule we shall now illustrate by examples, in which x
stands for the unknown quantity.
Ex. 1. To solve the equation,
5x  6 = 3x + 2.
Transposing the terms, we get
5x  3x = 2 + 6.
Combining like terms, we get
2x = 8.
Dividing both sides of this equation by 2. we get
x = 4,
and the value of x is determined.
Kx. 2. To solve the equation,
7x + 4 = 25x  32.
Transposing the terms, we get
7x~25x= 324.
Combining like terms, we get
18x=36.
Changing the signs on each side, we get
18x = 3t).
Dividing both sides V)y 18, we get
x = 2,
and the value of x is determined.
6o
OX SIMPLE EQUATIONS.
Ex. 3. To solve the equation,
2a;  3a: + 120 = 4j;  6x+ 132.
that is,
2x  3x  4x + 6a; = 13z  liv/,
or,
• 8x7x=I2,
therefore,
x=12.
U.X.
4.
To solve the equation,
3a; + 58(13a;) = 0,
that is,
3x + 5104 + 8x==0.
or.
3a; + 8x=1045, '
or.
llx=99,
therefore,
a;=9.
Ex.
5.
To
f.olve the equation,
()a;2(43x) = 73(17 ■<,
that is,
6a; 8 + 6x= 7 51 + :).£,
or,
or,
6x46:c3a;=751+8,
12x3a;=1551,
or,
9x= 36,
therefore.
x= 4.
EXAMPLES.— :?cxxii.
9. 268x = 801435.
10. 1333x = x83.
11. 133x = 5x o.
12. 127 + 9x= 12x4100.
13. 155x=64x.
14. 3./ 22 = 7xJ6.
15. 8 + 4x=12.c16.
1 6. 5.r  (3x  7) = 4x  ^6x  3o).
17. 6x  2(9  4x) + 3 (5x  7'i = lOx  (4 + 16x) + 35.
18. 9x3(5x6) + 30 = O
19. 12x  5 (9x + 3) + 6(7  8x) + 783 = 0.
20. X  7(4x  11) = 14(x  5)  19(8  x^  € :.
21. ^T + 7)(x3) = (x5)(x15).
1. 7x + 5 = 5x+ll.
2. ]2x + 7 = 8x + 15. ,
3. 236.c+425 = 97x + 564
4. 5x  7 = 3x + 7.
5. 12x9 = 8xl,
6. 124x+19 = 112x + 43.
7. 18 2^=27 5x.
8. 1257x=14512.''.
PROBLEMS LEADING TO SIMPLE EQUATIONS. 6i
22. (x8)(x + 12)=(ic+l)(a;6).
23. {x  2)(7  x) + (x  5) (x + 3)  2(x  1) + 12 =^ 0.
24. (2a!  7) (x + 5) = (9  2.r) (4  x) + 229.
25. (7  6x) (3  2x) = (4.C  3) (3j;  2).
26. 14  a;  5 (x  3) (x + 2) + (5  x) (4  Sx'i = 45.r  76.
27. (x + 5)2(4x)=21x.
28. 5(x  2)2 + 7(x  3)2 = (3x  7)(4.t  19, + 42.
29. (3x  17)2 + (4x  25)2 _ (5_^ _ 09)2 = ] .
30. (x + 5) (x  9) + (x + 10)(x  8) = (2x + 3) (x  7)  1 13.
VII. PROBLEMS LEADING TO SIMPLE
EQUATIONS.
117. When we have a question to resolve by means 01
Algebra, we represent the number sought by an unknown
symbol, and then consider in what manner the conditions of
the question enable us to assert thot tv:o exjjressiotis are equal.
Thus we obtain an equation, and by resolving it we determine
the value of the number sought.
The wliole difficulty connected with the solution of Alge
braical Problems lies in the determination from the conditions
of the question of tiro different expressions having the same
numerical value.
To explain this let us take the following Problem :
Find a number sucli that if 15 be added to it, twice the sum
will be equal to 44.
Let X represent the number.
Then x + 15 will represent the number increased by 15,
fu I 2(x + 15) will represent twice the sum.
But 44 will represent twice the sum,
therefore 2 (x + 15) = 44.
Hence 2x430 = 44,
tliatis, *2x=14,
or, x=7,
and therefore the number sought is 7.
62 PROBLEMS LEADING TO SIMPLE EQUATIONS.
118. We shall now give a series of Easy Problems, in
which the conditions by which an equality between two expres
sions can be asserted may be readily seen. The student should
be thorouffhly familiar with the Exanijdes in set xxviii, the use
of which he will now find.
We shall insert some notes to explain the method of repre
senting quantities by algebraic symbols in cases where some
difficulty may arise.
Examples. — xxxiii.
1. To the double of a certain number I add 14 and obtain
as a result 154. What is the number ?
2. To four times a certain number I add 16 and obtain as
a result 188. What is the number ?
3. By adding 46 to a certain number I obtain as a result a
number three times as large as the original number. Find the
original number.
4. One number is three times as large as another. If I
take tlie smaller from 16 and the greater i'rom 30, the remaiii
deis are equal. What are the numbers %
;. Divide the number 92 into four parts, such that the first
is greater than the second by 10. greater than the third by 18,
and greater than the fourth by 24.
6. Tlie sum of two numbers is 20, and if three times the'
smaller number be added to five times the greater, the sum is
84. What are the numbers ?
7. Tlie joint ages of a father and his son are 80 years. If
the nge of the son were douliled he would be 10 years older
than his father. What is the age of each?
8. A man has six sons, each 4 years older than the one
next to Jiim. The eldest is three times as old as the youngest.
Wiiat is the age of each ?
9. Add .£24 to a certain sum, and the amount ^dll be as
much above ^80 as the sum is below ^80. What is the sum \
10. Thirty yards of cloth and lorty yards of silk together
cost £66, and the silk is twice as valuable as the cloth. Find
the cost of a vard of each.
PROBLEMS LEADING TO SIMPLE EQUATIONS. 63
11. Find the number, the double of which being added to
24 the result is as much above 80 as the number itself is below
100.
12. The sum of ^500 is divided between A, B, C and D.
A and B have together ^280, A and G X260, A and D ^'220.
How much does each receive ?
13. In a company of 266 persons, composed of men, women,
and children, there are twice as many men as there are women,
and twice as many women as there are children. How many
are there of each ?
14. Divide i'1520 between A, B and C, so that A has ^100
less than B, and B i;'270 less than C.
15. Find two numbers, differing by 8, such that four time?
the less may exceed twice the greater by 10.
16. A and B began to play with equal sums. A won £0,
and then three times ^I's money was equal to eleven times B'a
money. What had each at first ?
17. A is 58 years older than B, and ^'s age is as much
above 60 as B's age is below 50. Find the age of each.
18. yl is 34 years older than B, and A is as much above 50
as B is below 40. Find the age of each.
19. A man leaves his property, amounting to J7500, to be
divided between his wife, his two sons and his three daughters,
as follows : a son is to have twice as much as a daughter, and
the wife .£500 more than all the five children together. How
much did each get ?
20. A vessel containing some water was filled up by pour
ing in 42 gallons, and there was then in the vessel 7 times as
much as at first. How many gallons did the vessel hold 1
21. Three persons. A, B, C, have .£76. B has .£10 more
ihan A, and C has as much as A and B together. How much
lias each ?
22. Wliat two numbers are those whose difference is 14,
and their sum 48 ?
23. A and B play at cards. A has £72 and B has £52
when they begin. When they cease playing, A has three times
as much as B. How much did A win ?
64 PROBLEMS LEADING TO SIMPLE EQUATIONS.
Note I. If we have to express algebraically two parts into
which a yiven number, suppose 50, is divided, and we repre
sent one of the parts by x, the other will be represented by
:.()  X.
Ex. Divide 50 into two such parts that the double of one
\rAxt may be three times as great us the other part.
Let X represent one of the parts.
Then 50  x will represent the other part.
Now the double of the first part will be represented bv
2x, and three times the second part will be represented by
3 (50  x).
Hence 2a; = 3 (50 x),
or, 2x=1503x,
or, 5a; = 150;
.. x = 30.
Hence the parts are 30 and 20.
24. Divide 84 into two such parts tliat three times one part
may be equal to four times the other.
25. Divide 90 into two such parts that four times one part
may lie equal to five times the other.
26. Divide CO into two such parts that one part is greater
tlian llie other by 24.
27. Divide 84 into two such parts that one part is less than
t'.ie (idler by 36.
28. Diviile 20 intn two sucli parts that if three times one
1 art be added to five times the other part the sum may be S4.
Note 1 1. When we have to compare the ages of two per
sons at one time and also some years alter or before, we must
lie caretul to remember that hoih will be so many years older
or younger.
Thus if X be the age of .4 at the present time, and 2* be
the age of B at the present time,
The age of .4 5 years hence will be a; + 5,
an<l the age of B 5 years hence \\ iil be 2j + 5.
PROBLEMS LEADING TO SIMPLE EQUATIONS. 65
Ex. ^ is 5 times as old as B. and 5 years hence A will
only be three times as old as B. What are the ages of A and
B at the present time ?
Let X represent the age of B.
Then bx will represent the age of A.
Now a; + 5 will represent £'s age 5 years hence,
and 6x + 5 will represent ^'s age 5 years hence.
Hence 5x + 5 = 3 (x + 5),
or 5x + 5 = 3x+15,
or 2x=10;
.'. x = 5.
Hence A is 25 and 5 is 5 years old.
29. A is twice as old as B, and 22 years ago he was tliree
times as old as B. What is yl's age ?
30. A father is 30 ; his son is 6 years old. In how many
years will the age of the father be just twice that of the son \
31. A\% twice as old as B, and 20 years since he was three
times as old. What is £'s age ?
32. A is three times as old as B, and 19 years hence he will
be only twice as old as B. What is the age of each ?
33. A man has three nephews. His age is 50, and the
joint ages of the nephews are 42. How long will it be before
tlie joint ages of the nephews will be ec^ual to the age of the
uncle \
Note III. In problems involving weights and measures,
after assuming a symbol to represent one of the unknown
quantities, ^ve must be careful to express the other quantities
in the same terms. Thus, if x represent a number of pence, all
the sums involved in the problem 7nust be reduced to pence.
Ex. A sum of money consists of fourpenny pieces and si.x
pences, and it amounts to £1. IBs. 8d. The iiuniber of coins
is 78. How many are there of each sort ?
[s.A.] S
66 PROBLEMS LEADING TO SIMPLE EQUATIONS.
Let X be the number of l'ouT2)enny pieces.
Then Ajy is their value in fence.
Also 78 — X is the number of sixpences.
And 6 (78 — X) is their value in 'pence.
Also £\.. 16s. 8(Z. is eqtiivalent to 440 pence.
Hence 4a; + 6 (78  a) = 440,
or 4a; + 468 Gx = 440,
from which we find x= 14.
Hence there are 14 fourpenny pieces,
and 64 sixpences.
34. A bill of ^100 was paid with guineas and halfcrowns,
and 48 more hulfciowus than guineas were used. How many
of eacli were paid ?
35. A person paid a bill of £3. 14s. w'ith shillings and
hallcrowns, and gave 41 pieces of money altogether. How
many of each were paid ?
36. A man has a sum of money amounting to £11. 13s. 4d.,
consisting only of shillings and fourpenny pieces. He has in
all 300 pieces of money. How many has he of each sort ?
37. A bill of .£50 is paid with sovereigns and moidores of
27 shillings each, and 3 more sovereigns than moidores are
given. How many of each are used ?
38. A sum of money amounting to £42. 8s. is made up of
shillings and halfcrowns, and there are six times as many
halfcrowns as there are shilling. How many are there of
each sort ?
39. I have £5. 1 1*. 3(/. in sovereigns, shillings and pence.
I have twice as many shillings and three times as many pence
as I have sovereigns. How manv have I of each sort J
YIII. ON THE METHOD OF FINDING
THE HIGHEST COMMON FACTOR.
119. An expression is said to be a Factor of another
expression wiieii the latter is divisible by the former.
Thus 3a is a factor of 12a,
5xy of lox^y\
120. An expression is said to be a Common Factor of two
or more other expressions, when each of the latter is divisil)le
by the former.
Thus 3a is a common factor of 12a and 15a,
3xy of Ibx^y^ and 2\x^y^,
4z of 82, 12^2 and I6z^
121. The Highest Common Factor of two or more expres
sions is the expression of highest dimensions by which each of
the former is divisible.
Thus 6a2 is the Highest Common Factor of 12a2 and 18a^,
bx^y of 10x^1/, 15x^2/2
and 25x*2/^
Note. That which we call the Highest Common Factor is
named by others the Greatest Common Measure or the Highest
Common Divisor. Our reasons lor rejecting these names will
be given at the end of the chapter.
122. The words Highest Common Factor are abbreviated
thus, H.C.F.
123. To take a simple example in Arithmetic, it will
readily be admitted that the highest number which will
divide 12, 18, and 30 is 6.
Now, 12 = 2x3x2,
18 = 2x3x3,
30=2x3x5.
68 MET HOP Oh h IN DING THE
Having thus reduced the numbers to their sirajylht factor?,
it appears that we may determine the Highest Common Factoi
in the i'ollowing way.
Set down tlie factors of one of the numbers in any order.
Place beneath theiii tlie factors of the second number, in
!uch order tliat fact(jrs like, any of those of the first number shall
stand under those factors.
Do the same for the third number.
Then the number of vertical columns in which the numbers
are alike Avill be the number of factors in the h.c.f., and if
we multiply the figures at the head of those columns together
the result will be the h.c.f. required.
Thus in the example given above two vertical columns are
alike, and therefore there are two factors in the h.c.f.
And the numbers 2 and 3 which stand at the heads of
those columns being multiplied together will give the H.C.F.
of 12, 18, ana 30.
124. Ex. 1. To find the h.c.f. of aWx and a%^X'.
aWx = aaa .bb .x,
aWx^ = aa . bbb . xx ;
:. 'E.c.F.=aabbx
= a'b^x.
Ex. 2. To find the h.c.f. of 34a26»c* and 51a%*c*.
Ma%^c* = 2 X 17 xaa . bhbbbb . cccc,
5la%*c = 3 X 17 X aaa . bbbb . cc ;
.*. B..c.F. = 17 aabbbbcc
= 17a'6V.
Examples.— xxxiv.
Find the Highest Common Factor of
\. a*h iuu\ a'b^. 3. 14x_ir' and 24.r^.
3. a^yh and x'yz. 4. ibmny and mmhip*.
HIGHEST COMAfOA' FACTOR. 69
5. 18rt7)'c'd and Z^a^hcd?. 8. 17 pq~, 'i4p^q and 5\p^q^.
6. «3?), arb^ and a^6*. 9. Sx^j/Sg*, Ux^z^ and 20x*3/V.
7. 4a7), lOac and 305c. 10. 3(teY, QOx^ and 120x3t/4
125. The student must he urged to commit to memory the
following Table of furms which can or cannot be resolved into
factors. Where a blank occurs after the sign = it signifies
tliat the form on the left hand cannot be resolved into simpler
factors.
x^y^ = {x + y){xtj) x^l=(x+l)(xl)
x^ + y^= a: + 1 =
x^ — y^ = {x — y){x'^ + xy + y^) x?l = {x—l)(x^ + x + l)
3^ + y^ = (x + y){xxy + y^ a^+ l = (x+ 1) (a;^ — x + l)
a^y* = {x'^ + y^){x'^y^) x*l = {x^+l: {x^1)
x* + y^= a;* + 1 =
x^+2xy + y' = {x + yy x'^h2x+l={x+iy
x^2xy + y^ = {xyf x2x+l = {xiy
05^ + 'Ax^y + Sxy^ + y^ = {x + yY x? +.3x2 + 3x + 1 = (x + 1)3
X?  3x2?/ + 33.^^2 _yZ^(^^_ yy x^  3x2 + 3x  1 = (x  1)3
The lefthand side of the table gives the general forms, the
righthand side the particular cases in which 1/= 1.
126. Ex. To find the h.c.f. of x^l, x22xH, and
a;2 + 2x3.
X2I=(xI)(x+1),
x22x + l = (xl)(xl),
x2i2a;3 = (xl)(xl3),
.". H.C.F. =X1.
Examples. — xxxv.
1. a2  52 and a^  b\ 4. a^ + a^ and (a I x)'.
2. a — h and a* — b*. 5. 9x2 _ i ^j^^j ^^x + 1)2.
3. a2_x2 and (a — x)2. 6. 1 25a^ and (1 — bay.
7. x2  1/2, (x + yY and x2 4 Zxy + 2y'.
8. x2 — y^, x^ — y^ and x2 — Ixy + 6?/2.
9. x2 — 1, x3 — 1 and x2 + x  2.
10. 1 — a"^, \ + a^ and a2 + 5^ 1 4.
METHOD OF FINDING THE
127. In large numbers the factors cannot often be deter
mined by inspection, and if we have to find the h.c.f. of two
such numbers we have recourse to the following Arithmetical
Rule :
" Divide the greater of the two numbers by the less, and the
divisor by the remainder, repeating the process until no rr
mainder is left : the last divisor is the h.c.f. required."
Thus, to find the h.c.f. of 689 and 1573.
689; 1573(2
1378
T95;689(3
585
Tb4; 195(1
104
9i; 104(1
91
13; 91 (7
91
.. 13 is the H.i F. of 689 and 1573.
Examples.— xxxvi.
Find the h.c.f. of
1. 6906 and 10359. 4. 126025 and 40115.
2. 1908 and 2736. 5. 1581227 and 16758766.
3. 49608 and 169416. 6. 35175 and 236845.
128. The Arithmetical Rule is founded on the following
)peration in Algebra, which is called the Proof of the Rule foi
finding the Highest Common Factor of two expressions.
Let a and h be two expressions, arranged according to de
scending powers of some common letter, of which a is not of
lower dimensions than h.
Let h divide a with p as quotient and remainder c,
c h g A.
d c r with no remainder.
HIGHEST COMMOX JACTOR. 71
The form of the operMtiun may be shewn thus :
fb
d) c (r
rd
Then we can shew
!. That rf is a common factor of a and 6.
II. That any other common factor of a and 6 is a factor of
rf, and that therefore d is the Higliest Common Factor
of a and b.
For (I.) to shew that rf is a factor of a and b :
b = qc + d
= qrd + d
= {qr + I) d, and .'. d is a factor ot b ;
and a=j6 + c
—P (3<^ + d) + c
= pqc+pd + c
=pqrd+pd + rd
= {pqr+p + r) d, and .•. d is a factor of a.
And (11.) to shew that any common factor of a and b is a
factor of d.
Let 8 be any common Factor of a and b, such that
a = viS and b = n8.
Then we can shew that 8 is a factor of d.
For d = bqc
= b~q(apb)
= b  qa + pqb
=n8  qm8 + pqnS
= {nqm +pqn) 8, and .". 8 is a factor of d.
Now no expression higher than d can be a factor of d ;
:. d is the Highest Common Factor of a and b.
72 METHOD OF FINDING THE
1 29. Ex. To find tlie h.c.f. of x' + 2a; + 1 and
x"' + 2x + 2a ^ I.
a;2 + 2x+l_;a;3 + 2x2 + 2x4l(x
x3 + 2x2 + X
x+l^x2 + 2x+ 1 (,x+ 1
x^ + x
x+1
x+ 1
Hence x+1 being the last divisor is the h.c.f. required.
130. In tlie algelnaical process four devices are frecjuently
useful. Tliese we shall now state, and exemplify each iu the
next Article.
I. If the sign of the first term of a remainder be negative,
we may change the signs of all the terms.
II. If a remainder contain a factor which is clearly not a
common factor of the given expressions it may l)e
removed.
III. We may'nmliiply or divide either of the given expres
sions by any number which does not introduce or
remove a common factor.
IV. If the given expressions have a common factor which
can be seen by inspection, we may remove it from
both, and find the Highest Common Factor of the
parts which remain. If we inultiply this result by
the ejected factor, we shall obtain the Highest Com
mon Factor of the given expressions.
131. Ex. I. To find the h.c.f. of 2x2  x  1 and
6x2 4x 2.
2x«xi;6x24x2(3
6x2 3x 3
 x + 1
HIGHEST COMMON FACTOk. ^%
Change tlie signs of the remainder, and it becomes x— 1.
• a!i;2!fc2x_ 1(2x41
2x2 _ 2x
x1
x1
The H.c.F. required is x — 1.
Ex. 1 1 . To find the h.c.f. of x^ + 3x + 2 and x^ + 5x + 6.
x2 + 3x42;x2 + 5xl6(l
x2 + 3x + 2
2x + 4
Divide the remainder by 2, and it becomes x + 2.
x + 2;x2 + 3x + 2(x+l
x2 + 2x
x + 2
x + 2
The H.c.F. required is x + 2.
Ex. III. TofindtheH.c.F.of 12x2 4.xland 15x"+8x + l.
Multiply 15x2 + 8x + l
by 4
12x2 + a;  1> 60x2 + 32a; + 4 (5
60x2+ 5a;_5
27x + 9
Divide the remainder by 9, and the result is Zxkl.
3x+i;i2x2 + xl(4xl
12x2+ 4x
^x^
3xl
The H.c.F. is therefore 3x + l.
Ex. IV. To find the h.c.f, of x^ — 5x2 + 6x and
xKte2 + 21x.
Remove and reserve the factor x, which is cotuiiKJii tu both
expressions.
U METHOD OF FINDING TfTK
Then we have remaining x^ — 5x + 6 and x — lOx + 21.
The H.c.F. of these expressions is x — 3.
The H.c.F. of the original expressions is therefoie x^ — 3x
Examples.— xxxvii.
Find the h.c.f. of the following expressions :
1. x2 + 7x + 12 and x2 + 9x + 20.
2. x2 + 12x + 20 and x^ + 14x + 40.
3. x2  17x + 70 and x2 13x442.
4. x2 + 5x84andx2 + 21x+108.
5. X + X— 12 and x^ — 2x3.
6. x^ + 5x2/ + ^y^ ^^^ ^^ + ^^V + 9j/^
7. x^  6x?/ + 8j/ and x^ — Sxy + 16?/^.
8. x2  13x1/  30?/2 and x^  18xt/ + 45i/2.
9. x^ — y^ and x — Ixy + 1/.
10. x^ + \f and x^ + 3xi/ + 3x?/2 + 1/^.
11. X* — 1/* and X — 2x1/ + ^2.
12. x^ + 1/^ and x^ + y^.
13. X* — 2/* and 3? + 2x1/ 4 y"^.
14. a^ _ 52 ^ 26c  c and a 4 2a?> 4 6  2ac  26c 4 c.
1 5. 12x2 4 Ixy 4 2/" and 28.C 4 3x!/ — y.
1 6. 6x 4 xxj  1/2 and 39x2 _ 22x1/ 4 3 j/*.
17. 1 5x2 — 8x1/ 4 1/2 and 40x2 — 3x1/ — 1/2.
18. x*5x3 45x2l and x*4x34x2 4x4l.
19. X* + 4x2 + 1 6 an, x5 + X*  2x3 4 1 7^.2 _ 1 Ox 4 20.
20. X* 4 x2(/2 4 1/* and x* 4 2x''i/ 4 3x2)/2 4 2xi/' 4 y*.
21. x"  Gx^ 4 9x2  4 ^mi x^4x^2x*43x2x2.
ftlGHEST COMMON FACTOR. 75
22. 1 5a* + lOa^ft + ^aW + Gai^  36* and Ga^ + IQa^i + 8a62 _ 563.
23. ISa:^  Hcc^?/ + 24a;y2  Ti/^ and 27x3 + SSx^j/  20x1/2 + gi/^.
24. 21x2  83xy  27x + 22!/2 + 99;,' and 12x2 _ 353.^ _ ^y.
33t/2 + 227/
25. 3a312a2a26 + 10rt6262and %a^ \~ia?h^M}PW.
26. 1 8a3 _ I8a2x + 6ax''  Gx^ and 60a2  75ax + 15x2.
27. 21x326x2 + 8xand 6x2x2.
28. 6x* + 29n2x2 + 9rt* and Sx^  15ax2 + a^x  ou\
29. x^ + x^i/2 + x^y + T/3 and x* — ?/*.
30. 2x3 + 1032 + 14a; + 6 and x^ + x2 + 7x + 39.
3 1 . 45a3a; + 3a2x2  9ax3 + 6x* and 1 8a2x  Sx\
132. It is sometimes easier to find the h.c.f. hj reversing
the order in which the expressions are given.
Thus to find the h.c.f. of 21x2 + 38x + 5 and 129x2 + 221x+ 10
the easier course is to reverse the expressions, so that thev
stand thus, 5 + 38x + 21x2 and 10 + 22 lx+ 129x2, and jjjgj^ j^
proceed by the ordinary process. The h.c.f. is 3x + 5. Other
examples are
(1) 187x3  84x2 + 31x  6 and 253x3  14x2 ^ 29x  12,
(2) 371^/3 + 262/250?/ + 3 and 469i/ + 7oi/  103?/  21,
of which the h.c.f. are respectively llx — 3 and 7?/ + 3.
133. If the Highest Common Factor of three expressions
a, b, c be reciuired, find first the h.c.f. of a and b. If d be the
h.c.f. of a and b, then the h.c.f. of d and c will be the h.c.f.
of a, b, e.
i«%. Ex. To find the h.c.f. of
x'3 + 7x2  X  7^ ^.3 ^. 5^.2 _ 2; _ 5^ and x2  2x + 1.
The H.C.F. of x3 + 7x2  X  7 ^nd x3 + 5x2  x  5 will be found
to be x2 — 1.
The H.C.F. of x2l and x22x+l will be found to be
x1.
Pence x— 1 is the h.Cf. of the three expressiona.
76 FRACTIONS.
Examples. — xxxviii.
Find the Highest Common Factor of
1. a;2 + 5x + 6, x2 + 7x+10, and a;2 + i^^20
2. x3 + 4x25, a;33x + 2, andx3 + 4x28x + 3.
3. 2x2 + xl, x2 + 5x + 4, and x^il.
4 V^'fV^^, 3l/22i/l, and 1/3 2/2 + 2/ 1.
5. 3? 4x + 9x  10, x^ + 2x2 _ 3x + 20, ami
x^* + Sx'"  v»x » S.").
6. x3 _ 7a,2 + i6x  12, 3x3  143.2 + igj.^ ^^^,1
5x310x + 7x14,
7. ■j/3 — 5?/2 + ii^_ 15j y3_y2^3y^.5 aj^Q
2i/372/+ieT/ 15.
XoTE. We use the name Highest Common acrm n..«Tead
of Greatest Common Measure or Highest Common Divisor lor the
following reasons :
(1) We liave used the word " Measure '• in An. a
different .*ense, that is, to denote the number of tiW^~ any
quantity contains the ^lnit of measurement
(2) Divisor does not necessarily imply a quanntv wuich
is contained in another an exact number of times. '1 nus in
performing the operation of dividing 333 tiy 13, we can 13
divisor, but we do not mean that 333 coutains 13 au eicact
number of times.
IX. FRACTIONt
135. A QDANTITY a is called an Exact i)ivisoK 01 m civwn
tity b, when h contains a an exact number :)i mnes.
A quantity a is called a Multiple of a (juanniy 0, M^acu a
contains b an exact number of times.
■FJHACTIOXS. 77
136. Hithorio we have treated of quantities wliicli coutniii
the unit of »»^«^«iiremeut in each case an exact ninube: of
times.
We have p'^w to treat of quantities which contain some exact
divisor of a pnmary unit an exact number of times.
137. We must first explain what we mean by a primary
unit.
We said in Art. 33 that to measure any quantity we take a
known standard or unit of the same kind. Our choice as to
the quantity to be taken as the unit is at first unrestricted, but
when once made we must adliere to it, or at least we must
give distinct notice of any change which we make with re.spect
to it. To such a unit we give the name of Primary Unit.
138. Next, to explain what we mean by an exact divisor of
a primary unit.
Keeping our Primary Unit as our main standard of mea
surement, we may conceive it to be divided into a number of
parts of equal magnitude, any one of which we may take as a
Subordinate Unit.
Thus we may take a pound as the unit by which we mea
sure sums of money, and retaining this steadily as the primary
unit, we may still conceive it to be subdivided into 20 equal
parts. We call each of the subordinate units in this case a
shilling, and we say that one of these equal subordinate units is
onetwentieth part of the primary unit, that is, of a pound.
These subordinate units, then, are exact divisors of the
primary unit.
139. Keei^ing the primary unit still clearly in view, we
represent one of the subordinate units by the followinf' nota
tion.
We agree to represent the words onethird, onefifth, and
onetwentieth by the symbols ^, , — , and we say that if
the Primary Unit be divided into three equal parts,  will
represent one of these parts.
78 FRACTIONS.
If we have to represent two of these subordinate units, we
2 3
do so by the symbol  ; if three, by the symbol  ; if four, by
o o "
4
the symbol , and so on. And, generally, if the Primary Unit
be divided into h equal parts, we represent a of those parts by
the symbol ■ .
140. The symbol t we call the Fraction Symbol, or, more
briefly, a Fraction. The number helow the line is called the
Denominator, because it denominates the number of equal
parts into which the Primary Unit is divided. The numbir
above the line is called the Numerator, because it enumerates
how many of these equal parts, or Subordinate Units, are
taken.
141. The term number may be correctly applied to Frac
tions, since they are measured by units, but w'e must be
careful to observe the following distinction :
An Integer or Whole Number is a multiple of the Primary
Unit.
A Fractional Number is a multiple of the Subordinate
Unit.
142. The Denominator of a Fraction shews what multiple
the Primary Unit is of the Subordinate Unit.
The Numerator of a Fraction shews what multiple the
Fraction is of the Subordinate Unit.
143. The Numerator and Denominator of a fraction are
called the Terms of the Fraction.
144. Having thus explained the nature of Fractions, we
next proceed to treat of the operations to which they are sub
jected in Algebra.
145. Def. If the quantity x be divided into b equal parts,
and a of those parts be taken, the result is said to be the
fraction , of x.
Jfxhe the unit, this is called tlie fraction j.
FRACTIONS. 79
146. If the unit be divided into b equal parts,
y will represent one of the parts.
r two
T three
And generally,
T will represent a of the parts.
147. Next let us suppose that each of the b parts is sub
divided into c equal parts : then the unit has been divided
into be equal parts, and
T will represent one of the subdivisions.
=— two
DC
And generally,
a
— a
be
148. To shew that r = t
be b
Let the unit be divided into b equal parts.
Then j will represent a of these parts (1).
Next let each of the b parts be subdivided into c equal
parts.
Then the primary unit has been divided into be equal parts,
and J— will represent ae of these subdivisions (2).
Now one of the parts in (1) is equal to c of the subdivisions
in (2),
.'. a parts are equal to ac subdivisions ;
, a ac
"'b^'k'
8o FRACTIONS.
Cor. We draw Irom this proof two inferences :
I. If tlie numerator and denominator of a fraction be
multiplied by llie same number, the vahie of the frac
tion is not altered.
II. If the numerator and denominator of a fraction be
divided by the same number, the value of the fraction
is not altered.
149. To make the important Theorem established in the
preceding Article more clear, we shall give the following proof
that K = o7x, ^y taking a straight line as the unit of length.
I I I I I I I I I I M I I I I I I I I I
A E D F B C
Let the line AG be divided into 5 equal parts.
Then, if B be the point of division nearest to C,
AB is I of AC. (1).
Next, let each of the parts be subdivided into 4 equal parts
Then AG contains 20 of these subdivisions,
and AB 16
:. ABi^^^oiAJ. (2).
Comparing (1) and (2), we conclude that
4^1^
5~20"
150. From the Theorem established in Art. 148 we derive
the following rule for reducing a fraction to its lowest terms :
Find the Highest Covimon Factor of the numerator and denomi
nator and divide both by it. The res^ilting fraction vnll be
one equivalent to the original fraction expressed tJi the simplest
terms.
FRACTIONS. 8i
151. When the numerator and denominator each consist of
a single term the h.c.f. may be determined by inspection, or
we may proceed as in the following Example : ,
To reduce the fraction , ^ „,, ., to its lowest terms,
10a^6c* _ 2 X 5 X aaabbcccc
12a6V^ 2 X 6 X aabbbcc '
We may then remove factors common to the numerator and
denominator, and we shall have remaining — — j :
" 6x0
.'. the required result will be ^^
152. Two cases are especially to be noticed.
(1) If every one of the factors of the numerator be removed,
the number 1 (being always a factor of every algebraical
expression) will still remain to form a numerator.
3a'C Zaac 1
Thus
I2ah'^ 3 X 4 X aaacc 4ac'
(2) If every one of the factors of the denominator be removed,
the result will be a whole number.
„, I2ah 3 X 4 X oMacc
Thus .^ , = — ^ = 4ac.
3ac o X aac
This is, in fact, a case of exact division, such as we have
explained in Art. 74.
Examples. — xxxix.
Reduce to equivalent fractions in their simplest terms the
following fractions :
4a2
12a3'
8x3
^' 36x2
IQx^yh^
45x^2*'
7o567c«
5' 21a36V
6. tT
3abc
blayz
8xYz^
•iiayz^'
9 6a^y»Z^'
[8.A.]
F
§2 FRACTIONS.
2\0mVp a? 14m*x
lO. . II. . 12. ■•
42m%2jp2" • d\ab' 21m^p 7mx
xy Aax + 2x^ mi + w
3x2/2 — 5x2^z" 8ax^ — 2x2' 3 abc + bcy'
4a^x + 6ah j 12 aF  6ah „ c^4a^
' ■ 8x218?/' ^^^ 86C2C ■ ^ ' c2 + 4ac + 4a2
3x* + 3x2/2 ^ labhfi  7abY
^9' 5x* + 5x27/' ^'^ 14a%x»  14a%2/2'
lOxlOy 5x9 4, 45t^2
"°' 4x2 8x1/ + 4?/^ ^5' lOcx" + 90crfx2'
ax + by , 10a2 + 20a6+106*
^ 26.
7a2x2  7by^' ' 5a^ + 5a%
6ab + Scd 4x2 _ g^y ^ 4^2
27a262x  48c2cZV ^" 48(xi/)2 '
xyxyz  3mx + 5?ix2
2«3 — 2rt32' ■ 3to?/ + 5)!X1/'
153. We shall now give a set of Examples, some of which
may be worked by Resolution into Factors. In others the
H.C.F. of the numerator and denominator must be found by
the usual process. As an example of the latter sort let us
take the following :
To reduce the fraction „, — ^„ — .so ^, to its lowest terms.
•' 2x39x238x + 21
Proceeding by the usual rule for finding the H.C.F. of the
numerator and denominator we find it to be x  7.
Now if we divide x^ — 4x2 — 19x— 14 by x — 7, the result is
x2 + 3a; + 2, and if we divide 2x^9x238x421 byx7, the
result is 2x2 + 5x — 3.
x2 + 3x + 2
Hence the fraction „ (. . "q i^ equivalent to the proposed
fraction and is in its lowest terms.
Examples. xl.
a2+7a + 10 ^ x29x + 20 x« 2x3
a2 + 5a + 6' '' x27x + 12" ^' x^\0x^2l'
FRACTIONS. 83
x218xj/ + 452/^ x^ + x^ + l x^ + 2x3?/^ + 2/'
x^8x?/105i/2' 5 a; + x + l* ' a;^i/6
''■ x3 + 2x2"3x + 20' ^"^^ m37m + 6 '
„ x^5x+ llx — 15_ a^ + 1
a^_x2 + 3x+5 ^' a3 + 2a + 2a+r
a:38x2 + 21x18 , 3ax213ax414a
^" 3x316x'^ + 21x ■ ^ ■ 7x3 17x2 + 6x •
x37x2 + 1 6x12 14x234x + 12
'°" 3x3 14x2 +T6x • 17 9ax23'9ax + 42a'
x* + x3j/ + X1/3  1/^ 10a 24a2 + 14^3
x* — x^?/ x]/'* — ?/'' * 15 — 24a + 3a2 + 6a3"
a3 4 4a  5 2a63 + a6  8a6 + 5a
^^' a33a + 2* ^^' 7631262 + 56 *
63 + 46256 a3_3^2 + 3^_2
3x2 + 2xl a^a%) x^3x^ + 4x2
x3 + x2xl "' a + a12' "^ x3x22x + 2
(x + y + g)2 + ( g  j/)2 + (x  g)2 + (y  a;)2
X2 + ?/2 + ^2
2x*  x3  9x2 + 1 3x  5_ 1 5fj3 ^ab2b'^
^5" 7x319x2 + 17x5 ' 3^" 9a2 + 3a6 262'
16x*53x + 45x + 6 x27x + 10
8x^30x3 + 31x2^1 2" ^■^" 2x2  X  6 "
4x2  1 2_ax + 9a2 x3 + 3x2 + 4,^ + 1 2
^'^' 8x327a3 • 35 x3T4x2T4x + 3*
6.< r^23x2 + 16x 3 x*x2_2x + 2
6x317x2+llx2' ^ 2x3 x^l"'
x36x 2 + nx 6 x3  2x2  1 5x + 36
^9" x32x2x + 2 ■ ^^' 3x2^x^ 15 '
7n3 + m2 + m — 3 3x3 + x25x + 21
■^ ' '»i3 _• Q^v.2 _i_ P^«v» 1^ Q* 3 '
3)762 + 5m + 3' :)"• 6x3 + 29x2 + 26x2r
x^ + 5x* — x2 — 5x X* — x3  4x2 — X + 1
^'' x4 + 3x3x"3 ' ^^: 4x33x28xl "
a2  62  26c  c2 a37a2+I6(x12
?^* a2 + 2a6 + 62?" ^O 3^3 ::T4a2 + i6a '
FRACTIONS.
154. The fraction t is said to be a proper fraction, ■when a
is less than h.
The fraction t is said to be an improper fraction, when a is
greater tlian h.
155. A whole number x may be written as a fractional
number by writing 1 beneath it as a denominator, thus .
156. To prove that 5 of j = r3
a ocL
\
Divide the unit into bd parts.
^^'^n'^^d = 6«^^ (Art. 148)
= r of be of these parts (Art. 147)
= T of 6c of these parts (Art. 148)
= ac of these parts (Art. 147).
But yj = ac of these parts;
a ^ c _ac
•'■bd~bd'
This is an important Theorem, for from it is derived the
Rule for what is called Multiplication of Fractions. We
extend the meaning of the sign x and define , x t (which
according to our definition in Art. 36 has no meaning) to mean
r of „ and we conclude that y >< i = t^. which in words trives
b d b d bd ^
us this rule — " Take the product of the numerators to form
the numerator of the resulting fraction, and the product of the
denominators to form the denominator."
The same rule holds good for the multiplication of three or
more fractions,
FRACTIONS. 85
157. To shew that r^7= t •
a be
The quotient, x, of r divided by 5 is such a number that x
multiplied by the divisor 3 will give as a result the dividend t
. arc _ a
•• li~b''
d p xc d . a
..  of r =  of  ;
c a c
xcd _ ad
" cd be '
ad
■' ^=k
Hence we obtain a rule for what is called Division of
Fractions.
_,. a c ad
Since rr j = T~)
d be
a c _a d
b^d'h^'c'
Hence we reduce the process of division to that of multiph'
cation by inverting the divisor.
158. The following are examples of the Multiplication an 1
Division of Fractions.
2x o _ ^■'^ 3a _ 6ax 2x
I. 3„2>^'^"3„2>^l = 3^ = 
3x_^ _3x^3a_3x 1 _ .3x _ x
^' 26* *~26 • T~26 ^3a~6a6~2a6*
4a^ 3c _ 3 X 4 X ac _ 2a
^' 9c^ ^ 2a~ 2 X 9x ac ~ 3c'
14x2^ '7x_ 14x2 9.v_ 9xl4xa:2? / 2x
^ 27y^'' 9y''27y^^7x~7x27xxy^~^'
2a 96 5c _ 2a x 96 x 5c _3
5' 36 '^ TOc ^ 4a ~ 36 x~10or4a " 4*
86 PR ACTIONS.
x^ — 4x x^ + 7.x_ x(x — 4) x(x + 7)
x^ + 7x^ x4 x(x + l) x — 4
_x(x4)x{x + 7)_
~x2(x + 7)(x4)~
a'i _ 52 ^ 4(a2_ ab) _ «2 ?)2 a2 ^j,
'' a^ + 2ab + b ' a + ab ~ a^ + 2ab + b"^ 4{a'^  ab)
_{a + b){a — b) a{a + b)
~{a + b){a + b) 4a{ab)
_(a + b){ab)a{a + b) _1
~{a + b)\a + b)4a(ab)'~4'
Examples.— xli.
Simplify the following expressions :
3x 7x 3a 26 4x^ 3x
4y^9y' ^' 4b^3a' ■ ^' df^^V'
80253 I5xy2 Q^y2^ 20a%k 2a 46 5f
45x2?/ ^ 24a42 5 ioa^^c'' mxijz' 56 "" 3c ^ 6a"
Sx^y 5yh I2xz Ici^b* 20cM 4ac
4x^ ^ 6x2/ ^ 20x^" 5"c2d3 ^ 42^463 "" shd'
9vihi^ hifiq 24x2w2 25A;3m2 "On^q 3pm
^ ^ ^ X ^ TO X — X — ^
82?3g3 2x2/ 90mn" ' 14712^2 'jop'm 4k^n
Examples. — xlii.
Reduce to simple fractions in their lowest terms :
a 6 a262 x2 + x2 ■t 2  13x + 42
a2 + a6 a2  a 6' ^' x^  7x x2 ^ gx
x^ + 4x 4x2j^l2x x2llx + 30 x23x
X2"^X ^ 3x2Tl 2X 5 • 22 _ y^ + y ^ a;2 _ 53;"
x2 + 3x + 2 x27x + 12 , x24 x2_25
O. r, i X ./^
5x + 6 x2 + x ■ ■ x^ + S^u x^ + 2x'
gi _ 4 a + 3 a29a + 20 a2  7 a
'' a^ — ba + 4 a  1 Oa + 21 «'  5a'
6276 + 6 62+106 + 24 6^  862
8 62:^364 "" 62  1 46 + 48 "" 62 + 66 •
t^R ACTIONS. 87
■y^ xy 27/2 ^x^xy
II.
13
x^  3xy + 2y^ '" x^ + xy {x  yy^
{a + by~c^ c2  (g  b)
d'  {h  cY"" c^  {a + hf
{x  m) — n X  (n  nt)^
(x  n)  m^ X  {m  n)'^'
( a + hy(c + dY {ahf{d cf
(a + cy {b + df ""[a cf {d bf
X? — 2xy + y z^ x + yz
x^ + 2xy + y^z xy + z'
Examples.— xliii.
Simplify tlie following expressions :
2a
36
X
~5c"
Aa ,
rix '
rSab.
bx .
2.
2 l£M^^^ . 8x*^_^2x3
14s ■ 7z' ■^' loab^ ' SOoi^
^ 2^;  2 >  1 • 5x
11 11
X'  a.c + 2 ■ X  r ^' x  17x + 30 ■ X  15'
158. We are now able to justify the use of the Fraction
Symbol as one of the Division Symbols in Art. 73, that is,
we can shew that j is a proj^er representation of the quotient
resulting from the division of a by b.
For let X be this quotient.
Then, by the definition of a quotient. Art. 72,
b xx = a.
But, from the nature of fractions,
, a
X y = a;
a
:.r=x.
THE LOWEST CO^TmO^ :■ i'.^.
159. Hire we may state an important : neorciu, whicn «>'
shall require in the next chapter.
If ad = be' to shew that , = ,.
b a
Since ad = bc,
ad
be
bd~
'bd
a
c
'b~
'cC
X. THE LOWEST COMMOri iviuiTIPLE.
160. An expression is a Common Multh uk of two or
more other expressions when the former is esLnunv divisible by
each of the hitter.
Thus 24x^ is a common multiple of 6, 8x^ and 12a^.
161. The Lowest Common Multiple of two or more
expressions is the expression of loivest dimenbi^ns which is
exactly divisible by each of then, ,
Thus ISx* is the Lowest Couimou iuunipie of 6j;*, ^x^,
and 3x.
The words Lowest Common Multiple are abbreviated
into L.c.M.
162. Two numbers are said to be prime to each other
which have no common factor but unity.
Thus 2 and 3 are prime to each other.
163. If a and b be prime to each other the fraction
is in its lowest terms.
Hence if a and b be prime to each other, uud i=^, «J>tl
if m be the h.c.f. of c and d,
^ 1 1. ^
o = — and = —.
7/1 m
THE L O IVES T COMMOX MUL TIPL E. Sg
164. In finding the Lowest Common Multiple of two or
more 'expressions, each consisting of a single term, we may
proceed as in Arithmetic, thus :
<1) To hnd the l.c.m. of ^a?x and 18ax'^,
2
4a%,
18ax3
a
2a\
9acc3
X
2a\
9x3
2a\
9x2
L.C.M. = 2 X a X .7 X 2a2 X 9x2 = 36^83*^
(2) To find the l.c.m. of ab, ac, be,
a
ab,
ac,
6c
b
b,
c,
be
c
1,
c,
c
1,
1,
1
L.C.M. = a X 6 X c = a6c.
(3) To find the l.c.m. of 12«2c, 146c2 and SGoJ*,
2 12a2c, 14&c2, 36a¥
6
a
6a\
ac,
7bc%
' nc\
18a62
~3a62
h
ac,
lbc\
362
c
ac.
7c2,
36
a.
7c,
36
L.C.M. = 2 X 6 X a X 6 X c X « X 7c X 36 = ^biaWc^.
Examples
Find the L.C.M. of
I. 4a^x and 6<(x.
Zxhj and 12.i:y.
xliv.
4a36 and 8*262.
ax, a"x and rt2x2.
2ax, 4ax2 and x^.
6. ab, ac and 6'c^
7. a'^x, a^y and xy.
8. blaH^, 34ax3 and ax*.
9. 52)'q, lOq^r and 20pqr.
I p. 18ax2, 72ai/2 and 12x?/.
90
THE LOWEST COMMON MULTIPLE.
165. The method of finding tlie l.c.m., given in tlie pre
ceding article, may be extended to the case of compound
expressions, when one or more of their factors can be readily
determinea. Thus we may take the following Examples :
(1) To find the l.c.m. of ax, a^ — x^, and a^ + ax,
a — x, a^ — x^, a^ + ax
1, a + x, a^ + ax
a — x
a +x
1, 1, a
L.C.M. = (« — x){a + x)a = (a^ — .r^) ar=a^ — ax^.
(2) To find the l.c.m. of .t^ i, x*l, and 4x''4.r:»,
a;2l I x^~l,x*l,4x^4x*
I 1, x^+l, 4x*
L.C.M. = (a;2  1) (x2 + 1) 43;^ = (x*  1) 4x* = 4x8 _ 43.4,
166. The student who is familiar with the methods of
resolving simple expressions into factors, especially those given
la Art. 125, may obtain the L.C.M. of such expressions by a
process which may be best explained by the following Ex
amples :
Ex. 1. To find the l.c.m. of ax^ and a^x\
a^  x^ = («  x) (a + x),
o3  a;3 = (rt — x) (a^ + ax + x'^
Now the l.c.m. must contain in itself each of the factors in
each of these products, and no others.
.•. L.C.M. is (a  x) {a + x) (a + ax + x"^,
the factor ax occurring once in each product, and therefore
once onlv in the L.c.Jr.
Ex. 2. To find the l.c.m. of
a~ — b, a^ — 2ab + b'', and a^ 2ab + 1
a252 = (rt + 6)(a6),
a^2ab + b^=(ab){ab),
a^ + 2ab + b^ = la + b)(a + b);
t.C.M. is (a + b){a h) (a  b) {a + b).
THE LOWEST COMMOX MULTIPLE. 91
the factor a — h occurrinrj txoice in one of the products, and a + 6
occurring twice, in another of the products, and therefore each
of these factors must occur lunce in the l.c.m.
Examples. — xlv.
Find the L.C.M. of the following expressions :
1. x^ and ax + x^. 10. x^  1, a;^ + 1 and x^  1.
2. a.2 — 1 and a;2 — X. ii. x^x, x^— 1 and x^ + 1.
3. a^ — 52 and a^ + aJ, 12. x^ 1, x^x and x^ 1.
4. 2xl and 4xl. 13. 2a + 1, 4a^ 1 and 8a^ + l,
5. a + 6 and a^^W". 14. x + ?/ and 2x2 + 2x?/.
6. x+ 1, X 1 and x^— 1. 15. (a + 6) and a2_52_
7. x+ l,x^— 1 andx2 + x+ 1. 16. a + 6, a 6 and a^ — 62_
8. x+1, x2+l andx^+l. 17. 4(1 +x), 4(1 x)and 2(1 x^).
9. X— 1, x^ 1 and x^— 1. 18. x— 1, x + x + 1 and x^— 1.
19. (a — 6) (a — c) and (a — c) (6  c).
20. (^x + l)(x + 2), (x + 2)(x + 3) and (x+l)(x + 3).
21. x^  ?/^, (x + 1/) 2 and (x ijf
22. (a + 3) (a + 1), (a + 3) (a  1) and a^  1.
23. '3?{x — 'ijy, x{x^ — y) and x + y.
24. (x+l)(x+3), (x + 2)(x + 3)(x + 4) and (x + l)(x + 2).
25. x^ — y"^, 2{x — yY and 12 (x^ + i/^).
26. 6(x2 + x?/), ^(xyy) and 10(x2i/2).
167. The chief use of the rule for iinding the l.c.m. is for
the reduction of fractions to common denominators, and in the
simple examples, which we shall have to put before the student
in a subsequent chapter, the rules which we have already given
will be found generally sufficient. But as we may have to find
the L.C.M. of two or more expressions in which the elementary
factors cannot be determined by inspection, we must now pro
ceed to discuss a Rule for finding the l.c.m. of tv, o expressions
which is applicable to every case.
92 THE LOWEST COMMON MULTIPLE.
168. The rule for finding the l.c.m. of two expressions o
and h is this.
Find d the higliest common factor of a and 6.
Then the l.c.m. of a and i = , x h,
a
b
or. = 3 X o. ♦
a
In words, the l.c.m. of two expressions is found by the fol
lowing process :
Divide one of the expressions by (he h.c.f. and multiply the
quotient by the other expression. The result is the L.C.M.
The proof of this rule we shall now give.
169. To find the l.c.m. of two algebraical expressions.
Let a and h be the two algebraical expressions.
Let d be their h.c.f.,
X the required L.C.M.
Now since x is a multiple of a and 6, we may say that
X = ma, X = 7i6 ;
.". ma = nb ;
fii b / J . , v
.. =  (Art. 159).
n a
Now since x is the Loxcest Common Multiple of a and h.
m and n can have no common factor ;
;. the fraction ~ must be in its lowest terms ;
n
:. m = T and n = , lArt. 163).
d d
 Hence, since x = ma,
b
x = ,xa.
d
Also, since x—nb,
" J,
x = ,x 0.
a
THE LOWEST COMMON MULTIPLE. 93
170. Ex. Find the l.c.m. of x2  13x + 42 and x^  19x+ 84.
First we find the h.c.f. of the two expressions to be x — 7.
„, (x213x + 42)x(x219x + 84)
Then l.c.m. = ^ ' \ '.
x1
Now each of the factors composing the numerator is divisible
by X — 7.
Divide x — 13x + 42 by x — 7,'and the quotient is x  6.
Hence l.c.m. = (x  6) (x^  19x + 84) = x^  25x2 _,. iggx  504.
Examples. — xlvi.
Find the l.c.m. of the following expressions :
1 . X + 5x + 6 and .x + 6x + 8.
2. ft' a20 and n^ + a 12.
3. x^ + 3x + 2 and x + 4x + 3.
4. x2+llx + 30 andx2+12x + 35.
5. x29x22 andx213x422.
6. 2x2 + 3x + 1 and x^  x  2.
7. x^ + x^y + xy + y^ and x*  y*.
8. x^  8x + 15 and x + 2x 15.
9. 21x2 _ 2Gx + 8 and 7x''  4x''*  21x + 12.
10. x^ + x^y + x?/2 4 y^ and x^  xij + xy^  y^.
11. a^ + 2ab  ab^  2P and a^  2a'^b  ab + 2¥.
171. To find the l.c.m. of three expressions, denoted by
a, b, c, we find m the l.c.m. of a and b, and then find M the
L.C.M. of m and c. M is the l.c.m. of a, b and c.
The proof of this rule may be thus stated :
Every common multiple of a and 6 is a multiple of m,
and every multiple of m is a multiple of a and b,
therefore every common multiple of m and c is a common
multiple of a, b and c,
and every common multiple of a, b and c is a common
multiple of m and c,
and therefore the L.C.M. of m and c is the l.c.m. of a, b
and c.
94 OM ADDITION AND SUBTRACTION
Examples. — xlvii.
Find the l.c.m. of the following expressions :
x  3,/; + 2, x  4x + 3 and a;^  5x + 4.
x + 5x + 4, a + 4x + 3 and x + 7x+ 12.
X  9x + 20, x^  1 2x + 35 and x^  1 Ix + 28.
4. 6x2  X  2, 21x2  17x4 2 and 14x2 + 5^  i.
5. x^  1, x + 2x  3 and 6x2 _ 3; _ 2.
x3  27, x2  15x + 36 and x^  3x2 _ ^x + 6.
XL ON ADDITION AND SUBTRACTION
OF FRACTIONS.
172. Having established the Rules for finding the Lowest
Common Mi;ltiple of given expressions, we may now proceed
to treat of the method by which Fractions are combined by
the processes of Addition and Subtraction.
173. Two Fractions may be replaced by two equivalent
fractions with a Common Denominator by the following
rule :
Find the l.c.m. of the denominators of the given fractions.
Divide the'L.c.M. by the Denominator of each fraction.
Multiply the first Numerator by the first Quotient.
Multiply the second Numerator by the second Quotient.
The two Products Avill be the Numerators of the equivalent
fractions whose common denominator is the L.C.M. of the
original denominators.
The same rule holds for three, four, or more fractions.
174. Ex. 1. Reduce to equivalent fractions with the
lowest common denominator,
2x + 5 , 4x7
3 and ^.
of FRACTIONS. 9$
Denominators 3, 4.
Lowest Common Multiple 12.
Quotients 4, 3.
New Numerators 8a; + 20, 12x21.
8a; + 20 12x21
Equivalent Fractions
12 ' 12
Ex. 2. Reduce to equivalent fractions with the loweet
common denominator,
56 + 4c 6a 2c 3a 56
a6 ' ac '' be '
Denominators a6, ac, be.
Lowest Common ^Multiple abc.
Quotients c, b, a.
New Numerators 56c + 4c', 6ab  26c, 3a  5a6.
„ , ^ „ . 56c + 4c' 6a6  26c 3a  5a6
lljqiuvaient 1" Tactions — , , = , ; .
a6c abc abc
Examples. — xlviii.
%
Reduce to equivalent fractions with the lowest common
denominator :
3x , 4x ^ a b , d^ab
1. —and • 6. —  and r^.
4 5 a'^6 a6''
3x7 , 4x9 „ 3 ,3
2. — TT and ^ —. 7. ^ and .
6 18 1rX 1z
2x4y , SxSy „ 2 ,2
3.  ■■ ^ and ^rp^. 8. , — and , .
•^ Sx'' lOx 12/ 1+2/''
4a + 56 , 3a — 46 5 1 6
4. — rv and  — . 9. and ^ „
la oa ^ 1 — X 1  x''
4a — 5c 1 3a — 2c a , 6
and ^ ^ ., . 10.  and
5ac 12ac ' ' c c(6 + a!)'
II. , ,.—,, .and
(a6)(6c) (a6)(ac)'
12. r^ TT^ , and
a6(a — 6)(a — c) ac (a  c) (6 — c)'
o6 ON ADD/TW.V AND SUBTRACTION
,.Tr m 1 *.!, 4. * c ad + bc
175. To sliew that v + ?= — I'l" ■
Suppose the unit to be divided into bd equal parts.
7
Then jj will represent ad of these parts,
6c
Id
Now ^ = g, by Art. 148,
, c 6c
and J = r^.
d bd
Hence t + . will represent ad + be of the parts.
But — i," will represent aa + be of the pares.
bd
_, „ a c ad + be
.nerefore^ + ^=^^.
By a similar process it may be shewn that
a e _ ad — be
h~d^~~bdr'
,„„ _,. a c ad + bo
176. Since J 4^=^,
our Rule for Addition of Fractions will run thus :
"Reduce the fractions to equivalent fractions having the
Lowest Common Denominator. Then add the Numerators of
the equivalent fractions and place the result as the Numerator
of a fraction, whose Denominator is the Common Denominator
of the equivalent fractions.
The fraction will be equal to the sum of the original frac
tions."
The beginner should, however, generally take two fractions
at a time, and then combine a third with the resulting fraction,
as will be shewn in subsequent Examples.
. , . a e ad — be
Also, since ^5=^^,
the Rule for Subtracting one fraction from another will be,
OF FRACTIONS. gy
•' Reduce the fractions to eijuivalent fractions having the
Lowest Common Denominator. Then suhtract the Numerator
of the second of the equivalent fractions from the Numerator
of the first of the equivalent fractions, and place the result as
the Numerator of a fractijjn, Avhose Denominator is the Common
Denominator of the equivalent fractions. This fraction will be
equal to the difference of the original fractions."
These rules we shall illustrate by examples of various degrees
of difficulty.
Note. When a negative sign precedes a fraction, it is best
to place the numerator of that fraction in a bracket, before
combin'ing it with the numerators of other fractions.
177. Ex. 1. To simplify
4x  3/ 3x + 7/ _ 5a;  2)/ 9x + 2y
7 ^ ^ ~I4 2l " "*" ~42~'
Lowest Common Multiple of denominators is 42.
Multiplying the numerators by 6, 3, 2, 1 respectively,
24xa % ^ii^l y_ _ lOx4?/ 9x + i'y
~~ 42 "*" 42 42"~^'''~42~
24x  1 8;/ + 9x + 2 1 7/  ( 1 (\c  4y) + 9x + 2y
~ 42
_2 4x18?/ + 9x + 21?/10x + 47/ + 9x + 2;/
42
_ 32x 4 9 ;/
42~'' »
Tv o 'p • ^•r 2x+l 4x + 2 1
Ex. 2. To simplify — h =.
^ 3x ox 7
Lowest Common Multiple of denominators is 105x.
Multiplying the numerators by 35, 21, 15x, respectively.
70.C + 35 _ 84x + 42 1 5x
lU5x l()5x 105x
_ 70x 4 35  (84x + 42) + 15a;
~ lOSx
70x43584x42H5x x7
^ 105x ~T05x'
98 ON ADDITION AND SUBTRACTION
Examples.— xlix.
4x + 7 3a; 4 3a 46 lah + c 13a 4c
I. ^— +— ,v— • 2. ^ + .,— .
5 15 / .J 12
4x  3i/ 3a; + 7i/ bx — 2?/ 9x + ly
3 ~y~"'^~l4 21^'^~42~^"
3a;2i/ 5x7y 8x + 2y
^' ~5x '*' 10a; "^ ~ 25"* cv
4a;2  7i/2 3a;  81/ 52 j/
5' "3x2" + ^6x~ "'' ~~[2~'
. 4a2 + 56_2 3a^26 7^2a
W^^ 56 "^ 9 •
4x + 5 3x  7 9
^' ""3 5x~''"l2x2"
5a + 26 _ 4c  36 6a6  76c
3c 2a 14ac
2a + 5c 4oc  Zc^ bac  2c^
3X1/4 51/2 + 7 6x211
10. 5—5 ' 6 T  \
jc^' xy* x^y
a  6 4a  56 3a  76
a?h a^bc b^c^
178. Ex. To simplify
a6 a+b
a + b ab'
L.C.M . of denominators is a2  62.
Multiplying the numerators by a6 anri a +6 respectively,
we get
a'^  2ab + b a + lab + 6^
a262 "^""^a— ftl
_a2 2a6 + 62 + a^ + 2ab + b
a2  62
_2a2^+^62
~ fi2_j2
or FRACTIONS. 99
EXAMPLES.
1.
1 1
a;  6 a; + 5'
1 1
^' x7 x3'
3
1 1
1 +X lX
?±1 _ ?J"_1
xy x + y'
1 2
5' lX 1X2*
6.
a (ad  be) x
c c{c + dx)
XX o
7. + . 8.
X+y xy
1
X
— +
X
(xyy
2 3a j^
"* x + a {x + ay
2a
1 + 1 .
(a + x) 2a (a  a)
179. Ex. 1.
To simplify
3 5 6
1 + y 11/ 1 + 1/2"
Taking the first two fractions
3 5
1+y l~y
1
33?/ b + by
~1/ ' lt/2
8 + 2y ^
17/'
we can now combine with this result the third of the original
fractions, and we have
3 5
1 + 7/^17/ 1
6

+ 7/2
8 + 27/ 6
17/2 1+7/2
^ 8 + 27/ + 87/2 ^27/3 _ 6  6?/2
~ ' 1jr* "l^p
_ 8 + 27/ + 87 /2 + 27/ 3  6 + 6y2
1^7/*
_ 27/3+147/2.27/ + 2
1y*
too ON ADDITION AND SUBTRACTION
Ex. 2. To simplify
2 2 2
(a6)(6c)"'"(a6)(ca)^(6c)(ca)'
I..C.M. of first two (lenominatois being (a  6) (6  c) (c  a)
_ 2c 2a '26  2c 2
~ (a  6y (6  cH^ a) ^ (ab) (6"^H7T) ■*■ (6  c) (c  a)
262CT _ 2
~ (^Ujlh~=^cj~{^^ ^ (6^ r) (c  «)•
L.C.M. of the two denominators being (a h) {b c) (c  a)
262ffl + 2a26
{a b)(b c) (c a) (a b) (6  c) (c  a)
=0.
Examples.— li.
1 Jl_ 2a _J^ 1 26 4¥
'• Hft''"la''"la2" "^^ «6 a + b a'^ + b^ a*^b*'
1 1 2a; X y x^
1x 1 + x l+X' y x + y x' + xy
X x^ X ^ x + Z x4 x + 5
^ xl x1 x3
a;2 x3 x4
8 ^ _^_ _ 5a^
X  a (x  a)"^ (x  a)3*
1 1 3
^' xl x + 2 (x+l)(x + 2)'
1 3
lo.
(r + 1 ) (x + 2) (x + 1) (x + 2) (x + 3)*
X X
 1 "*'x^'T'^x+r
1 1
£2.
(d + c) (a + d) (tt + c) (a +• e)'
a6 6c ca
'^" (ftTcncTo) (c + a) (aT6) (a + 6) (fTc)*
OF FRACTIONS.
X — a x — b {a  b)'
xb xa (x  a){xb)'
x + y 2:c xyx^
'■ y x + y y{''y')'
1 6.
17
a + 6 6 + c c + ii
(6 c){c a) (c a) (a b) (a — b){b c)'
x 2xy
x^ + xy + y^ x^ — y^'
i8 2,2^2 ^( abr + {bcy + {caf ^
ab bc c — a {a — b){b — c){ca)
a + b 2a ab  a^
'9 'b^~^+b^a^b^¥'
1 1 1
^°' {n+l){n + 2) ()i+l)(?H2)(?i + 3) (?i + l)(7i + 3)*
a^ — be b  ac c^ — ab
2 1 1 +
(a + 6)(a + c) {b + a){b\c) {c + b){c + a)
„ .,• O'b , — ab . ►,«
180. bmce X~"' ^ ~T'~^' ^'^' '
ab _—ab
'b~~^T'
Fioiu this we learn that we may change the sign of the
(lenoiiiinat(ir of a traction it we also change the sign of the
numerator.
Hence if the numerator or denominator, or both, be expres
sions Avith more than one term, we may change ibe sign of
every term in the denominator ii we also change the sign of
every term in the numerator
„ a — b —(ab)
cd {ca)
~  r+i '
or, writing the terms of the new i'raction so that the positive
terms may stand first,
_b — a
dc'
I02 ON ADDITION AND SCB TRACTION
181. tX. To simplify — ^ .
Changing the signs of the numerator and tleiiominator of tlu^
second fraction,
X (a + x)  bax + x
ax a—x
_a!}c + x^( — 5ax + x) ax + x^ + bax x^ _ 6ax
a — x a — x ax'
182. Again, since —ab= the product of a and h,
and «6= tlie proiUict of +a and h,
the sign of a product will he cliani,'ed hy changing the sic;ns of
one of the factors composing the product.
Hence (a — b)(b c) will giv. a set of terms, .
and (6 a) {b c) will give the same set of terms icith dif
ferent signs '
This may be seen hy actual multiplication :
(a  h) {b c) — (ibacb^ + he,
(b a) (b  c)= — ah + ac + Ir — he.
Consequently if we have a fraction
1^
(a  h) {h  c)'
and we change the factor nh into ha, Ave shall in effect
change the sign of every term of the expression which Avould
result from the multiplication of (a  b) into {b  c).
Now we may change the signs of the denominator if we also
change the signs of the numerator (Art. 180) ;
1 J
" (a  6) (6  c) ~ (ba) [b c)'
If we change the signs of two factors in a ilenominator. the
sign of the numerator will remain unaltered, thus
1 1
iab)(bc)~(hn)(c'by
I
I
OF FRA C TICNS. io3
183. £x. Simplify
1 1
(a6)(6c) {ha){ac) (ca){cb)'
First change the signs of the factor (ba) in the second
fraction, changing also the sign of the numerator ; and change
the sigTis of the factor (c  a) in the third fraction, changing
also the sign of the numerator,
, . 1 1 1
the result is , rrr, r + ; Tx / \ ~ 7 w i\
(a b) {b c) (a  o) (a  c) [a c) [c b)
Next, change the signs of the factor (c  b) in the third,
changing also the sign of the numerator,
.11. 1  1 1
the result is ^ tttt ^ + . jv? ;  7 ttt — ;.
(a b) [0 c) [a o) {a c) {a  c) (6  c)
L.c.M. of the three denominators is (a b) {b c) {a  c),
_ ac b+c ab
~{ab){b c) (a  c) (a b) {a c) (b c) (a b)(a c) (6  c)
acb>rc {a b)
(a b)(b c) {a  c) (a  h) (6 c){fl c) "
.0.
Examples.— lii.
X xy 3 + 2x2 — 3a; 16a! — a;^
' xy yx' ' 2 — x 2fx x^ — 4'
jc x x^ 114
^ x+l~lx'''x2l* ^' 61/ + 6 ~ 2?/  2 "•■ 3^7?"
5
1 2 1
(m2)(m3) (ml)(3m) (m  1) (to  2)"
^ (a6)(x + 6)"*"(6a)(x + a)* ''" a262 a^  fe3 "•" a^ + F
1 1 1
°' 4(l+x) 4(xl) 2(l + x2)'
1 ^ 1 • , 1 '
1 _.__„ 1
a(ab)(a — c) b{ba)ibc) cica)(cb)'
I04 ADDITION AND SUBTRACTION OF FRACTIONS.
184. Ex. To simplify
1 1
a;2llx + 30 a;12x + 35'
Here the denominators may Le exprcssi.il in lacinrs, and "Wo
have
1 1
(x — 5)(x — G) (re — 5) (x — 7)"
The L.C.M. of the denominators is (x — 5) (x — 6) (x — 7), and
we have
X— 7 X— 6
+ 
(a;5)(x6)(x7) (x5) (xG) (x7)
_ 2x13
"~(x5)(x6)(x7)'
Examples. — liii.
1
: + ^
x''' + 9x + 20 x2 + 12x + 35*
1 1
x213x + 42 x2^15x + 54'
, __i + i
^' x2 + 7x44 x22x143"
1 2x 1
^' x2 + 3x + 2'*'x + 4x + 3"^x^ + 5x + 6*
m 2ni 2?7),?i
15. — h —
^ n m + n {m + nf
1+x 1x 2
l+x + x^ 1— x + x 1+x^Hx**
5 2 7x 7x
^* 3^( 1^^) ~ T+x "^ 3x2T3 " 3x  3'
1 1 J j
8(xl)^4(3x) 8(x5) (lx)(x3)(x5)
X*
Q. 1  X + X  X^ +  — — .
^ 1+a;
XII. ON FRACTIONAL EQUATIONS.
185. We shall explain in this Chapter the method of
solving, first, Equations in which fractional terms occur, and
secondly, Problems leading to such Equations.
186. An Equation involving fractional terms may be
reduced to an equivalent Equation without fractions by mul
tiplying every term of the equation by the Lowest Common
Multiple of the denominators of the fractional terw^.
This process is in accordance with the principle laid down
in Ax. III. page 58 ; for if both sides of an equation be multi
plied by the same expression, the resulting products will, by
that Axiom, be equal to each other.
187. The following examples will illustrate the process of
clearing an Equation of Fractions.
EX. 1. 1 + ^.8.
The L.c.M. of the denominators is 6.
Multiplying both sides by 6, we get
6.5 6x ,„
T+6=^^'
or,
3x + x = 48,
or.
4x = 48;
.. cc = ]2.
Ex. 2.
X x + \ _
2 + 7"'
The L.c.M. of tie denominators is 14.
Multiplying both sides by 14, we get
14x 14X + 14 ,^
— + 14x2a^
(o6 ON FRACTIONAL EQUATrON"^.
or, • .7a; + 2a; + 2 = 14x28,
or, 7a; + 2x14x= 282,
or, 5x=30.
Changing the signs of both sides, we get
5x = 30;
.. a; = 6.
188. The process may be shortened from tlie foUowin.i,'
considerations. If we have to multiply a fraclion by a multi])le
of its denominator, we may first divide the multiplier by the
denominator, and then multiply the numerator by the quotient.
The result will be a whole number.
Thus,  X 12 = a;x 4 = 4x,
^^x56=(xl)x8 = 8x8.
EX. 1. M + 139.
The L.c.M. of the denominators being 12, if we multiply the
numerators of the fractions by 6, 4, and 3 respectively, and the
other side of the equation by 12, we get
6x + 4x + 3x = 468,
or, 13x = 468;
,. x = 36.
Ex, 2. §^ + ^ = ^.
" X 2x 3x 12'
The L.c.M. of the denominators is 12x. Hence, if Ave mul
tiply the numerators by 12, 6, 4, and x respectively, we get
96 90 + 28= 17x,
or, 34=17x,
or, 17x = 34;
/. x=2.
O.y FRA C no A' A L EQ UA TIONS. 107
EXAMPLES.— liv.
,. 1 = 8. . f = ». , ,M = 8.
X X „ .,„ 4x ^ , 2a; 176 4a;
4.  = 3. 5 36g = 8. 6. =^—
2a: , 7x ^ a; + 2 X  1 a  2
o 2x 4.7; a; x_ 3 X
^ "3" + ^^ 5+^ '^ 2 + 3^44
3x , 5x ^ x + 9 2x 3x6 „
9. — + 5 = — + 2. 19. —4^ = —^ — + 3.
'^4 6 ^4(0
7x , 9x „ 17 3x 29llx 28x+14
,0. _5 = 8. 20. _^  = _^_+_^_
5x _ ^. 7a; 2x10 .
II. r8 = 74T7i. • 21. — = — = 0.
9 12* " 7
X
6
X , ,, X 3x + 4 4x51
12. , — 4 = 24.
22.
7 + 47
 = u
23
^3 = lL
X X
24.
12+x ^ 6
X X
25.
1 1 1
4"+10^ + 20^ =
= 40.
13. 56 = 48^.
3x 180 5x _
^4 4 + — 6— = '^
3x , , X  8
^5 T^^ = 2
, X X X 13 , 1 3x _5 ._,!
'6 2 + 3 + 4=12 ^6 24^ + 2=V%
.,3 31 325
^'^^ ""4 x~x 100'
^ „1 18 X ,1 1 32x 2
^^ 22 + 3=¥ + 3 + 10 + 5
X X 5x _ ,2 ^„
^9 3+46 12= ¥^^
7x+2 , 3x 3x + 13 17x
3° 10—12^=^^.
108 ON FRA CTION.A L EQUA T!Oy<!.
189. it must next be observed tbat in clearing an equation
!)[■ fractions, whenever a fraction is precedeii by a negative sign,
u e must place the result obtained by multiplying that nume
: ilor in a hraclcet, alter the removal of the denominator.
For example, we ought to proceed thus : —
Ex. 1. ^±! = ^^Z^_^
5 2 7
Multiply by 70, the l.c.m. of the denonii:)ators, and we get
I4x + 28 = 35.7;  70  (10.c  10),
or 14.c + 28 = 35a;7010x+10,
I'.om which we shall find a; = 8.
Ex.2. 122^_.4J? + 2^1.
bx 'ix
Multiplying by 15a;, the L.C.M. of the denominators, we get
516x(20a; + 10) = 15x,
or 516jj20x10 = 15x,
from which we shall find a;= 1.
Note. It is from want of attention to this way of treating
fractions preceded by a negative sign that beginners make so
many mistakes in the solution of equations.
Examples.— Iv.
x + 2 „, 5x 5a; 9 3x
1. 5x— 2 = 71. 4 TT = 4— 2
3x ,2 „ 5x4 „ l2x
^ ^—3 = '3 5 2.. —=.—— .
52.7; „ 6x8 , x + 2 14 3 4 5x
3. —  + 2 = x —. 6. 2 = 9 ^ .
5x + 3 3  4x x_31 95x
7' ~8~~ 3 ^l~^ "~6~"
„x + 5x2.(; + 9 „x + 2x
^ l—~^^U '° ^='— 8 = 3
x+1 x4_x + 4 x + 5_x + 2 X — 2
GN FRACTIOXAL EQUATIONS. tog
x + 2 x2 'a;! ^ 2x x + 3 ., _.
ID. i; ^— =3x21.
5 2 7' 7
+7_a
11
x + 9 3x6 „ 2x 2i; + 7 9x  8 x11
H — , r = 3. 17. ^
7x31 8 + 15x_7x8
■ 4 26~~~22~"
8x15 llx1 7x + 2
19.
3 7 13 ■
7x + 9 3x+l_9x13 2499X
20. ~^ 7— 4 T4 •
X ^ XX X 10 X „3
190. Literal e(nations are those in wliicii known quantilifs
are represented by letters, usually the first in the alphabet.
The following are examples : —
Elx. 1. To solve the equation
ax + bc = bx + ac.
that is,
ox bx — ac be,
or,
{ab)x = {a — b)c.
therefore,
x = c.
Ex. 2.
To solve the equation
ax + bx c = hx + cx d,
that is.
(i^x + bx b'x cx = cd,
or.
(a^ + b b  c)x = c — d,
therefoie,
cd
" a^ + bb'c
Examples.— Ivi.
1. ax+bx = c. 4. dm  ox = bc ox.
2. 2a — ex = 3c — 56x. 5. abc ax = ax — ab.
3. bc + ax — d = a^bfx. 6. 3acx — 6bcd=l2cdx + abc.
ox FRACTIOXAL EQUATIONS.
7. A; + ?><xckx 4 3A; = ^x + Zahh — li^  ackx.
8. — ac^ + b'^c + obex = abc + cmx — acx + hc — mc.
9. {a + X + b) (a + b  x) = {a + x) {b — x) — ab.*
10. (a — x){a + x) = 2a^ + 2ax — x'.
11. (a2 + a;)2 = x2 + 4a2 + a*.
1 2. (o"  a;) (a + x) = a* + 2ax — x.
axb x + ac m (px + x^) mx^
13. la = . 17. — ^ ' = mqx\ .
•^ c c px p
3abx 1 „ X , c
TA. ax 7; — = ^. lo. — o = , — x.
^ 2 z ad
Aax  26 x'^ — aax2xa
15. 6a 3— = x. 19. fc^— 2, = y
, 6x41 a{x^\) 3 abx 4xac
16. ax = ~^ . 20. , = .
XX c ox ex
ab + x b' x x — b abx
22.
¥ a'^b a^ 6
3ax — 2b ax — a ax 2
36 26
, ax 05 
2"?. am — 6 — ^H = 0.
■^ 6 m
^a263^ 62^ 3a2c _ 3acx _ ^^^ab^x
(a + 6) a(a46) a + 6~ 6 (a + 6) "
ax^ ax ^ a6 , , 1
25. r + a + — = 0. 27. — = 6c + rf+ .
o~cx C X X
, a(d + x) ax _ m(a — x)
26. ^— J ^ = ac + j. 28. c = a+~' ^.
ax a 3a + x
29. (a + x) (6 + x)  ffl (6 + c) = ^ + x'i
ace (a + by.x , „,
30. — T— ^^ ox = ae36x.
 d a
191. In the examples already given the L.C.M. of the
denominators can 'generally In deterniined by inspection.
When compound expressions appear in the denominators, it
is sometimes desirable to collect the fractions into two, one
ON FRA CTTONA L EQUA TIONS 1 1 1
on each side of the eqiiution. When tliis has been done, we
can clear the equation of fractions liy multiplying the nu
merator on the hft by the denominator on the right, and the
numerator on the right by the denominator on the left, and
making the produ ts equal.
For, if ^ = j, it is evident that ad = bc.
' a
F 4x + 5_13x6_2.'c3_
10 7xl4 ~ 5 '
4a: + 5 2x3_13a:6^
■■ To 5~~ 7.C + 4 '
4x + 5  (4a;  6) _ 1 3x  6 ^
' 10 7x + 4 '
ll_13x6^
•■ 10~ 7x + 4 '
.. ll(7x + 4) = 10(l;ix6);
whence we find a; = — r.
06
Examples.— Ivii.
3x + 7 3X15 ,2 5 ^
4XI5
4x +
3"
xl6
X
2x + 5
2x
5'
2x + 7
4x
1
x + 2
2x
1'
5xl
5x
3
2x + 3
2x
3*
1
2
4^ —
1  5x 1  2x
1 1 3
7 — +
8
x1 x+1 X1'
4x + 3 8x+19 7x29
9 18 5x12"
X .'■  ox _ 2
3 ox — 7 3*
3x + 2 2x  4 ^
'' 3^^^^"3="' '"• ^:rr+^T2=^
II. l(x + 3)^(llx) = (x4)l(x3).
(x+_lM2x + 2)_.^_ J, x+_l_^l_
(x3)(x + 6) " ■ ^ x + 1 xl~lx2
(2x + 3)x 1 , 2 8 4.5
•^ 2.( + i 3x ^ 1  X 1 + X 1  x^
ON FRA C TIONA L EQUA TIONS.
, 4_ _3^ iA = _.?
■ x8''2j;16 24 3a;24'
:c*(4a;220a; + 24) , „
a;2  2.C + 4
„ 2rc* + 2x3 23x2 + 31a;
18. ., —r ; = 2X''  4x  3.
X + 3x  4
(A 2\ 1 3x(45x)
192. Equations into which Decimal Fractions enter do not
present any serious difficulty, as may be seen from the follow
ing Examples : —
Elx. 1. To solve, the equation
•5x = 03x + l41.
Turning the decimals into the form of Vulgar Fractious,
we get
5x_2x_ 141
10 ~ loo "^ loo*
Then multiplying both sides by 100, we get
50x = 3x+141;
therefore 47x=141;
therefore x = 3.
Ex. 2. 1 •2x  i^^^ = ix + 89.
■5
First clear the fraction of decimals by multiplying its
numerator and denominator by 100, and we get
i2x— =^ = 4x + 89;
DO
^, , 12x 18x5 4x 89
therefore ^^y ___ = _ + _;
therefore 60x  1 8x + 5 = 20x + 445 ;
therefore 22x = 440;
therefon x = 20.
ON' FRA CTIONAL EQ UA TIOMS. 1 1 3
Examples. — Iviii.
1. •5.c2 = 25x + 2xl.
2. 325X51 Hx — ■75x = 39 4"5x.
3. •125x + 01x=132x+4.
4. S.c + 1 SOS.t + Sx = 22 95  • 1 95x.
5. •2x01x + 005x=ll7.
6. 24x:^^^;:^^ = 8x + 89.
7. 24x 1075 = 25x. 8. •5x + 2 •75a;=4.r. 11.
9. ^^ + 3875 = 4025.
10. 25x ^=^{ i~^)~"^
2 + .'.•/! ^\ . 5x43
8 "'
85 2 J 11X 48x 34x .„__
"• Yx=^4 .^' ■ ''■ ^—^=^^^3.
23x 5x _2x3_x2 ^^7
14. ?i^ + . ■04(x + 9) = 24I2.
•45X75 12 Sx'e
15...5X + ___ = ____.
, , 35x 24 3x .^,
16. 5 ^ „ =3/5x.
X — 2 8
•135X225 36 09x18
17. •15X + =  ^^.
193. To shew that a simple equation can only have one root.
Let x = a be the equation, a form to which all equations of
the first degree may be reduced.
Now suppose a and /3 to be two roots of the equation.
Then, by Art. 109,
a = a,
and /? = «, ,
therefore a = P\
in other Avords, the two supposed roots are identicaL
XS.A.1 H
XIII. PROBLEMS IN FRACTIONAL
EQUATIONS.
194. We shall now give a series of Easy Problems resulting
for the most part in Fractional Equations.
Take the following as an example of the form in which such
Protjlems should be set out by a beginner.
"Find a number such tliat the sum of its third and fourth
parts shall be equal to 7."
Suppose X to represent the number.
Then  will represent the third part of the number,
o
and  will represent the fourth part of the number.
X X
Hence ^ + t "^^'i^^ represent the sum of the two parts.
But 7 will represent the sum of the two parts.
Therefore ^ "^ 4 ^ "^^
Hence 4a; + 3x = 84,
that is, 7x = 84,
that is, a; =12,
and therefore the number sought is 12.
Examples. — lix.
1. What is the number of which the half, the fourth, and
the tilth jiaits added together give as a result 95 ?
2. ^^"llat is the number of which the twelfth, twentieth,
and fortieth parts Ridded together give as a result 38 ?
3. What is the number of which the fourth part exceeds
the Ulth part by 4 1
PR OR f. EMS IX FRACIIOXAL EQUATIONS. 115
4. Wiiat is the iniiuber of wliicli the twentyfifth part
exceeds the thirtytifth jjart by 8 \
5. Divide GO into two such parts that a seventh part of one
may be ec^ual to an eighth part of the other.
6. Divide 50 into two such parts that onefourth of one
parr being added to fivesixths of the other part the sum may
be 40.
7. Divide KK) into two such parts that if a tliird part of tlie
one be subtracted from a fourth part of the otlier the remainder
may be 11.
8. \Yhat is the number which is greater than the sum of its
third, tenth, and twelfth parts by 58 ?
9. "When I have taken away from 33 the fourth, fifth, and
tenth pai'ts of a certain number, the remainder is zero. Wliat
is the number ?
10. What is the number of which the fourth, fifth, and
sixth parts added together exceed the lialf of the number
by 112?
11. If to the sum of the half, the third, the fourth, and the
twelfth parts of a certain number I add 30, the sum is twice as
large as the original number. Find the number.
12. The difference between two numbers is 8, and the
quotient resulting from the division of the greater by the less
is 3. What are the numbers ?
1 3. The seventh part of a man's property is equal to his
whole property diminished by £1626. What is his property ?
14. The difference between two numbers is 504, and the
quotient resulting from the division of the greater by the less
is 15. What are the numbers ?
15. The sum of two numbers is 5760, and their difference
is equal to onethird of the greater. What are the numbers ?
16. To a certain number I add its half, and the result is as
much above 60 as llie number itself is below 65. Find the
number.
ii6 PROBLEMS IN FRACTIONAL EQUATIONS.
17. The difference between two numbers is 20, and one
seventb of the one is equal to ouethird of the other. What
are the numbers ?
18. The sum of two muubers is 31207. On dividing one
by the other the (^uolient is fouud to be 15 and the remainder
1335. What are tlie numbers ?
19. Tlie ages of two brothers amount to 27 yeais. On
dividing the age of the elder by that of ihe younger the quo
tient is 3i. What is the age of each ?
20. Divide 237 into two sucb parts that one is fourfifths of
the other.
21. Divide £1800 between A and B, so that 5's share may
be twosevenths of ^'s share.
22. Divide 46 into two such parts that the sum of the
quotients obtained by dividing one part by 7 and the other by
3 may be equal to 10.
23. Divide the number a into two such ]iarts that the sum
of the quotients obtained bv dividing one part by 7?i and the
other by n may 1*6 equal to h.
24. The sum of two numbers is a, and their difference is h.
Find the numbers.
25. On multiplying a certain number by 4 and dividing
the product by 3, I obtain 24. Wiiat is the number ?
5
26. Divide £864 between A, B, and G, so that A gets —
of what B gets, and C"s share is equal to the sum of the shares
of A and B.
27. A man leaves the half of his property to his wife, a
sixth part to each of his two children, a twelfth part to his
brotlier, and the rest, amounting to £600, to charitable uses.
What was the amount of his property ?
28. Find two numbers, of which the sum is 70, such that
the first divided by the second gives 2 as a quotient and 1 as
a remainder.
29. Find two niimbers of Avliich the difference is 25, such
that the second divided by tlie tiist ^jives 4 as a quotient and
4 as a renjainde:.
PROBLEMS IN FA' ACTIONAL EQUATIONS. W;
30. Divide tlie number ^08 into two parts snch that the
sum of tlie fourtli of the tjreater and the tliird of the less is
less bv 4 tliau four times the difference between the two parts.
31. There are thirteen days between division of term and
the end of the first twothirds of the term. How many days
are there in the term ?
32. Out of a cask of wine of which a fifth part had leaked
away 10 i^allons were drawn, and then the cask was twothirds
full. How much did it hold 1
'^3. The sum of the ages of a f;ither and son is half what it
will be in 25 years : the difference is onethird what the sum
will be in 20 years. Find the respective a.ges.
34. A mother is 70 years old, her daughter is e.xactly half
that age. How many years have passed since the mother was
3J times the age of the daughter ?
35. A is 72. and B is twothirds of that age. How long is
it since A was 5 times as old as B ?
Note I. If a man can do a ]iiece of work in x hours, the
part of the work which he can do in one hour will be repre
sented by .
•' X
Thus if A can reap a field in 12 hours, he will reap in one
hour — of the field.
Ex. A can do a piece of work in 5 days, and B can do it
12 days. I
do the work ]
in 12 days. How long will A ami B working together take to
Let X represent the number of days A and B will take.
Then  will represent the part of the work they do daily
Now  represents the part A does daily,
and Yg represents the pait B does daily.
irS J'ROBLEMS IN FRACTIONAL EQUATIONS.
Hence  +  will represent the part A and B do daily.
.1 1 1 1
Consequently ^4^^ = .
Hence 12x + 5x = 60,
or 17x = 60;
60
•■• ^ = 17
9
That is, they will do the work in 3r— days.
36. A can do a piece of work in 2 days. B can do it in 3
days. In what time will they do it if they work together ?
37. A can do a piece of work in 50 days, B in 60 days,
and G in 75 days. In what time will they do it all working
together ]
38. A and B together finish a work in 12 days ; A and G
in 15 days ; B and G in 20 days. In what time will they
finish it all working together ?
39. A and B can do a piece of work in 4 hours ; A and G
in 3 hours ; B and C in 5= hours. In what time can A do
it alone ?
, 40. A can do a piece of work in 2;^ days, B in 3.^ days,
and G ill ?> days. In what time will they do it all working
together ?
41. A does  of a piece of work in 10 days. He then calls
in B, and they finish the work in 3 days. How long would B
take to do onethird of the work l>y liimself ?
Note II. If a tiip can fill a vessel in x hours, the part of
the vessi'l llllcd 1>\ it in om Innir will be represented by .
Ex, Three taps running separately will fill a vessel in 20,
30, and 40 minutes respectively. In what time will they fill it
when thev all run at the same time \
PROBLEMS IN FRACTIONAL EQUATIONS. 119
Let X represent the number of minutes they will take.
Then  will represent the part of the vessel filled in >
minute.
Now  represents the part filled by the first tap in 1 minute,
1
30
J_
40
second .
third..
1111
Hence 20 + 30 + 40 = ?
or, multiplying both sides by 120ic,
6a; + 4x + 3.x = 120,
that is, 13a; = 120;
120
••• ^=iy
3
Hence they will take 9 ^^ minutes to fill the vessel.
42. A vessel can be filled by two pipes, runnincr separately,
in 3 hours and 4 hours respectively. In what time will it be
filled when both run at the same time ?
43. A vessel may be filled by three different pipes : by the
first in I5 hours, by the second in 3 hours, and by the third
iu 5 hours. In what time will the vessel be filled when all
three pipes are opened at once ?
4i. A bath is filled by a pipe in 40 minutes. It is emptied
by a wastepipe in an hour. In what time will the bath be
full if both pipes are opened at once ?
45. If three pipes fill a vessel in a, 6, c minutes running
separately, in wliat time will the vessel be filled when all three
are opened at once ?
I20 PROBLEMS IN FRACTIONAL EQUATIONS.
46. A vessel containing 755 "allons can be filled bv three
pipes. The first let<? in 12 gallons in Z minutes, the second
15 gallons in 2r minutes, tlie third 17 gallons in 3 minutes :
in what time will the vessel })e filled by the three pipes all
running together?
47. A vessel can be filled in 15 minutes by three pipes,
one of which lets in 10 gallons more and the other 4 gallons
less than the third each iiiiniite. The cistern holds 2400 gallons.
How much comes throug'n each pipe in a minute ?
Note III. In questions involving distance travelled over in
a certain time at a certain rate, it is to be observed that
Distance ^t^.
— .is = linie.
Rate
That is, if I travel 20 miles at the rate of 5 miles an hour,
number of hours I take = ^.
5
Ex. A and B set out, one from Newmarket and the other
from Cambridge, at the same time. The distance between the
towns is 13 miles. A walks 4 miles an hour, and B 3 miles an
hour. Where will they meet ?
Let X represent their distance from Cambridge when they
nu^et.
Then 13 a: will represent their distance from Newmarket.
X
Then  = time in hours that B has been walking.
13
4
X
A
And
since
both have been walking
the
same
time,
X
13
 X
3"
4
»
or
4.x =
= 39
3x,
or
7x =
.'. x =
= 39;
39
" 7'
PROBLEMS IN FRACTIONAL EQUATIONS. 121
4
That is, they meet at a distance of 5 miles from Cam
bridge.
48. A person starts from Ely to walk to Cambridge (wliich
4
is distant 16 miles) at the rate of 4 miles an hour, at the
y
same time that another person leaves Cambridge for Ely
walking at the rate of a mile in 18 minutes. Where will they
meet ?
49. A person walked to the top of a mountain at the rate
of 2 miles an hour, and down the same way at the rate of
o
3^ miles an hour, and was out 5 hours. How far did he walk
altogether ?
50. A man walks a miles in 6 hours. "Write down
(1) The number of miles he will walk in c hours.
(2) The number of hours he will be walking d, miles.
51. A steamer which started from a certain place is fol
lowed after 2 days by another steamer on the same line. The
first goes 244 miles a day, and the second 286 miles a day. In
how many days will the second overtake the first ?
52. A messenger who goes 31 ^ miles in 5 hours is followed
after 8 hours by another who goes 22 miles in 3 hours. When
will the second overtake the first ?
53. Two men set out to walk, one from Cambridge to
London, the other from London to Cambridge, a distance of
60 miles. The Ibrmer walks at the rate of 4 miles, the latter
3
at the rate of 3 miles an hour. At what distance from Cam
4
bridge will they meet ?
54. A sets out and travels at the rate of 7 miles in 5 hours.
Eight hours afterwards B sets out frrnu the same place, and
travels along the same road at the rate of 5 miles in 3 hours
After what time will B overtake A. ?
122 PROBLEMS TN FRACTIONAL EQUATIONS.
Note IV. In problems relatincj to clocks the chief point to
be noticed is that the minutehand moves 12 times as i'ast as
the hourhand.
The following examples should be carefully studied.
Find the time between 3 and 4 o'clock when the hands of a
clock are
(1) Opposite to each other.
(2) At right angles to each other.
(3^ Coincident.
«g3
(1) Let ON represent the position of the rainiitehand in
Fig. I.
OD represents the position of the hoiuluind in Fig. I.
M marks the 12 o'clock point.
T 3 o'clock
The lines OM, OT represent the position of the hands at
3 o'clock.
Now suppose the time to be x minutes past 3,
Then the minutehand has since 3 o'clock moved over the
urc MDN.
And the hourhand has since 3 o'clock moved over the
arc TD.
Hence arc MDN= tvelve times arc TJX
If then we represent MDN by x,
we shall represent TD by .
Also we shall represent MT by 15,
and DX in 30.
PROBLEMS IN FRACTIOiVAL EQUATIONS. T?3
Now MDN = MT ^TD\ UN,
that is, x=15 + — +30,
or 12a; = 180 + x + 360
or llx = 540;
540
.•.x=— .
Hence the time is 49 niimites past 3.
(2) In Fig. II. the description given of the state of the
clock in Fig. I. applies, except that DN will he represented hy
15 instead of 30.
Now suppose the time to he x minutes past 3.
Then since
MDN= MT+TD + DN,
x=15 + ^ + 15.
from which we get
360
8 ^
that is, the time is 32— minutes past 3.
(3) In Fig. III. the hands are both in the position ON.
Now suppose the time to be x minutes past 3.
Then since
MN=MT+TN,
IK ^
^=15 + ^2,
or 12x=180 + x,
180
or x = ,
4
that is, the time is 16 — minutes past 3.
55. At what time are the hands of a watch opposite to
each other,
(1) Between 1 and 2,
(2) Between 4 and 5,
(3) Between 8 and 9 ]
124 PROBLEMS IX FRACTIONAL EQUATIO.VS.
56. At what time are the hands of a vatch at light angles
to each other,
(1) Between 2 and 3.
(2) Between 4 and 5,
(3) Between 7 and 8 \
57. At what time are the liands of a watch together,
(1) Between 3 and 4,
(2) Between 6 and 7,
(3) Between 9 and 10 ?
58. A person buys a certain number of apples at the rate
of five for twopence. He sells half of them at two a j)enny,
and the remaining half at three a penny, and clears a penny
by the transaction. How many does he buy ?
59. A man gives away half a sovereign more than half as
many sovereigns as he has : and again half a sovereign more
than half the sovereigns then remaining to him, and now has
notliing left. How much hud he at first ?
60. ^Miat must be the value of 71 in order that
may be equal to — wlien a is  ?
3u + 69a
61. A body of troops retreating before the enemy, from
which it is at a certain time 25 miles distant, marches 18 miles
a day. The enemy jairsues it at the rate of 23 miles a day,
but is fiist a day later in starting, then after 2 days is forced
to halt for one day to repair a bridge, and this they have to do
again after two days' more marching. After how many days
from the beginning of the retreat will the retreating force be
overtaken ?
62. A person, after ]iaying an incometax of sixpence in the
pound, gave away onetliirteentli of his remaining income, and
had .£540 left. What was his original income ?
63. From a sum of money I take away £bO more than the
half, then from the remainder £.10 more than the filth, then
fiom the seconil remainder ;£20 more than the fourth part :
and it last onlv i;iO remains. W!;at was the original sum '
PROBLEMS IN FRACTIONAL EQUATIONS. 125
64. I bou;4lit a certain number of eggs at 2 a penisy, and
the same nuuiher at 3 a penny. T sold tlieni ut 5 for twopence.
and lost a petiny. How man}' eg;,'S aid I Luy ?
65. A cistern, liolding 1200 gallons, is tilled by 3 pipes
A, B, C in 24 minutes. The pipe A re'Uiires 30 minutes more
than C to fill tlie cistern, and U) gallons le.~s run tl.rough C per
minute than through .4 and B togellier. What time would
each pipe take to till the cistern by itstdf ?
66. A, B, and (' drink a barrel of beer in 24 days. A and
4
B drink „rds of what C does, and B drinks twice as much as A.
o
In what time would each separately drink the cask ]
67. A and B shoot by turns at a tari;et. A puts 7 bullets
out of 12 into the centre, and B puts in 9 out of 1. Between
them they put in 32 bullets. How many shots did each fire?
68. A farmer sold at market 100 head of stock, horses,
oxen, and sheep, selling two o.xen for every horse. He obtained
on the sale £2, 7s. a head, li he sold the horsgs, oxen, and
sheep at the respective prices .£22, £12, lOs., and £1, 10s., how
many horsesi^oxen, and sheep respectivirly did he sell ?
69. In a Euclid paper A gets 160 marks, and i> just passes.
A gets full marks for bookwork, and twice as many marks
for riders as B gets altogether. Also B, sending answers
to all the questions, gets no marks for riders and half marks
for bookwork. Supposing it necessary to get  of full marks
in order to pass, find the number of marks which the paper
carries.
70. It is between 2 and 3 o'clock, but a person looking at
the clock and mistaking the hourhand lor the minutehand,
fancies that the time of day is 55 minutes earlier than the
reality. What is the true time ?
71. An army in a defeat loses onesixth of its number in
killed and wounded, and 4(X)0 prisoners. It is reintbrced by
3000 men, but retreats, losing a fourth of its nundjer in doing
so. There remain 18000 men. What was the original force /
72. The national debt of a country was increased by one
fourth in a time of war. During t\\ enty years of peace widen
ii6 Oy MISCELLANEOUS FRACTrON^.
followed £25,000,000 was paid off, and at the end of that time
the interest! was reduced from 4J to 4 per cent. It was then
found that the interest was the same in amount as before the
war. What was the amount of the debt before the war ?
73. An artesian well supplies a brewery. The consump
tion of water goes on each weekday from 3 a.m. to 6 p.m. at
double the rate at which the water flows into the well. If
the well contained 2250 gallons when the consumption began
on Monday morning, and it was just emptied when the con
sumption ceased in the evening of the next Thursday but one,
what is the rate of the influx of water into the well in gallons
per hour ?
XIV. ON MISCELLANEOUS FRACTIONS.
195. In this Chapter we shall treat of various matters con
nected with Fractions, so as to exhibit the mode of applying
the elementary rules to the simplification of expressions of a
more complicated kind than those which have hitherto been
discussed.
196. Tlie attention of the student must first be directed
to a point in which the notation of Algebra difiers from that of
Arithmetic, namely wktn a whole number and a fraction stand
side by side vdth no sign between them. •
3 3
Thus in Arithmetic 2' stands for the sum of 2 and .
/ 7
But in Algebra x stands for the product of x and ".
So in Algebra 3— — stands for the product of 3 and ;
° c c
. „a + b 2a + Zb
that 18, 3 = —  —
c c
ON MISCELLANEOUS FRACTIONS. 127
Examples. — Ix.
Simplify the following fractions :
1, a + x + 3. 3. ^ + 2—*^.
X ^ X xy
a + ax jxa .a + b ^a'^ — b^
2. • s 2 — , 4. 4 ,~2. — jT,.
x^ X ah a^\¥
197. A fraction of which the Numerator or Denominator
is itself a fraction, is called a Complex Fraction.
y X
Thus , ■% and — are complex fractions.
a a m
b n
A Fraction whose terms are whole numbers is called a
Simple Fraction.
All Complex Fractions may be reduced to Simple Fractions
by the ]nocesses already described. We may take the follow
ing Examples :
a
b_am_a n _an
^ m~b ' n b m bm
n
b___d_/a c\ /m _p\_adbc , mqnp
^~' m p \b dJ \n q/ bd ' nq
n q
_ad — bc nq _ nq (ad  be)
bd mq  np bd {mq — np)'
,„, 1+x ,, . /, 1\ ,, , xt1
(3) _ = (i+x)^(^l + ^j = (l+x)— —
_l+a; X _x(l +x'
1 + 
X
1 "x+1 1+x *
t28 OM MISCELLA.VF.OUS FRACTTOyS.
1 1
^^ X . 1 Vla; 1 + x/ ■ Vlx 1 + x'
1+xl+x cc + x^+l.
1x 1+x
(5)
1_3;2 • l_a;2
_ 2x lx^_ 2j
~ 1  a;2 1 + x=^ ~ 1+ x"^"
3 3 3 3
3,3, 3(1 .c) , 3,ix
3 lx+3 lx+3 4x
1 — X 1  X
3 3 (4  x) _ 12 3x
'^ 4x + 33x ~4^x + 33x~ 74x'
4 — X
Examples.— IxL
Simplify the following expressions :
4 « «
5 X 7/ X • 1 g*
^' 7 ~ 1' ^' xu" ^' r
3— ^ 1+
23 ^^a
0D
2x + „ 1+i
1 XX
a^ x + rtxa 2x
7. — ^ 8. 2I • 9 r
a x^a'' 1rx^
xu x + V
.X , 1 x+y X y
7+1 pj. II. s.
1 x + 1 xy x + y
1 +
X+y xv
ON MISCELLANEOUS FRACTIONS. 129
., 2771  3 + —
1 + a: + X m
■ '^ TT ^4. 2m 1
1 +  I .,
X x' m
a + b _b_ J_ L i.
b a + h ah ac be
a 6 aft
198. Any fraction may Tie split np into a number of trac
tions equal to the number ol tenns iu its numerator. Tiius
a^ + x^ + x + l x^ I!? X 1
X* X* x'* :f^ X*
1111
X X X^ X*
Examples.— Ixii.
Split up into four fractions, cacli in its lowest terms, the
following fractions :
a* + 3a3 + 2a + 5a 9«3  1 2^2 + 6a  3
'• 2a* ■ ^ 108 ■
a^bc + alri + abc^ + feed' 18;?+ 12y^36r2 + 72sg
abed, ' 'Spqrts
x^3x2y + 3x?/?/ 10x3  25x2 + 75x 125 ^
5' x'y ' ■ 1000 ■
199. The quotient obtained by dividing the unit by any
fraction of that unit is called The Reciprocal of that fraction.
Thus , that is, , is the Reciprocal of ?.
a a ^ 6
b
200. "VVe have shewn in Art. 158, that the fraction symbol
r is a proper representative of the Division of a by b. In
r.s.A.] 1
no ox MISCELLANEOUS LR ACTIONS.
Chapter IV. we treated of C3=es of division in which the divisor
is contained an exact number of times in the dividend. We
now proceed to treat of cases in which the divisor is not con
tained exactly in the dividend, and to shew the proper method
of representing the Quotient in such cases.
Suppose we have to divide 1 by \a. We may at once
represent the result by the fraction . But we may
actually perform the operation of division in the following
■way.
\a) 1 (1 +a + a2 + a3t...
\a
a
i3a4
The (^lotient in this case is interminable. We may carry
on the operation to any extent, but an exact and terminable
Quotient we sliall never find. It is clear, liowever, that the
terms of the Quotient are formed by a certain law, and such
a succession of terms is called a Series. If, as in the case
before us, the .scries may be indefinitely extended, it is called
an Infinite Serie.s.
If we wish to express in a concise i^tiu the result of the
operation, we may sto) at any term of the quotient and write
the result in the following way.
_!__ _a_
la~^'la'
1  a 1  a'
1 , ., «^
;; = 1 + a + a + :; ,
\a \a
= 1 + a T (( + a^ + :; ,
I a 1a'
ON MrsCELLANEOUS FRACTIONS.
13'
always bein^ careful to attach to that term of tlie quotient, at
which we intend to stop, the remainder at that point of the
division, placed as the numerator of a fraction of which the
divisor is the denominator.
Examples. — Ixiii.
Carry on each of the following divisions to 5 terms in the
quotient.
1. 2 by \+a. 7.
2. m by m + 2. 8.
3. a  6 by a + 6. 9.
4. a^ + X by a^  x^. 10.
5. ax by a X. 11,
1 by 1 + 2x  2x2.
1 + X by 1  X + x^
1 + h by 1  2&.
x^ — 6^ by X + 6.
a^ by xh.
b bv a + x.
1 2. a^ by (a + x)^
13. If the divisor be xa, the quotient x2ax. and the
remainder 4a^, what is the dividend ?
14. If the divisor be m  5, the quotient m^ + 5m^ + Ibm + 34,
and the remainder 75, what is the dividend ?
201. If we are required to multiply such an expression as
x^ X 1 , X 1
¥ + 3 + 4^^^23'
we may multiply each term of the former by each term of the
latter, and combine the results by the ordinary methods of
addition and subtraction of fractious, thus
a;2 X 1
X 1
2 3
X^ X^ X
4+6 + 8
X^ X
1
6 9
12
«* X
1
t ' 72
LX
t32 Or^ MISCELLAiWEOUS ER ACTIONS.
Or we may first reduce tlie mulliplicaiid and the multiplier
to single I'ractioiis und proceed in the loUowing way :
(2+3 + 4)423)
_ 6.x2 + 4x + 3 3a:2_ 18ar^4x 6
12 ^ 6 ~ 72
"72+72 72~ 4 +72 12
This latter process will be louud the simpler ty a beginner.
Examples.— ixiv.
Multiply
a a \ , a \ 11,11
'• y6 + 3^'^'45 5 ^ + 6^by^p.
, 11, 1 ,111,111
xa;^ X a c ' a c
7. 1 +  + r by 1   + ^.
8. l+a:f.rbylx + xx3.
5^ 37 2 1_1
9' 2x2 + x'3 ^x^"x 2"
10. pr + 5 + 2 by j:r  T  2.
2b2. If we have to divide such an expression as
^ o 3 1
X x^
by X + , we may proceed as in the division of whole numbers,
carefully observing that the order of descending powers of x
is
*^' ^' *' t' X2 ' X3
ON MISCELLANEOUS ER ACTIO. \S. 133
Any isolated digits, uo 1, 2, .j ... will stand between x
, 1'
and .
X
Tims the expression
■! 1 r. O y ^ 5
arranged according to descendinfj powers of x, will stand thus,
5 3 1
a^ + 3x2 + 5x + 4 +  + _ +
The reason for this arrangement will be given in the Chapter
on the Theorv of Indices.
Ex. x + l ]x3 + 3x + ^ + ,l x2 + 2 + 4
x/ a. x^ ^ X'*
x^+ X
2x
3
X
2x
2
+ 
X
1
 +
X
1
1
 +
X
1
X3
Or we may proceed in the follov/ing way, which will be
found simpler by the beginner.
(x3 + 3x + ^ + l3)(x+^)
3 1
)^lx+ ,
x/
x^ f 3x^ + 3x + 1 , x2 + 1
x^^ ■ X
x« + 3x* + 3x2 + 1 3.
x^ x + 1
X* + 2x' +1 X* 2x^ 1 , ^ i
= = —, + ^ +  = x + 2 + ..
X X^ X' X X'
134 ON MISCELLANEOUS FRACTIONS.
Examples.— Ixv.
Divide :
2 1 V, 1 fill,'
1. %'■ — nDya; + . 4. c° — t bv c — 5.
1 1 X V^ X 1/
2. aj„hy a.. 5. 5 + 2 + % by  + ^.
b^ •' b y^ x^ ^ y X
3. m' + 3bym + . 6. 4 + wg + ri bv — ,r 4 rs
a' w'^ X y . X y
7. .^,343' by 4
1/* x* 1/ X y X
_ 3x5 , , 77 , 43 „ 33 „ , a;^
8. r  4x* 4 — x3  —X  ^x 4 27 by —  X 4 3.
a^ ¥, a b 1113,111
9. i:i + ^byT +  10. , + M + ~i — jrDy + rH — •
^ ¥ a? ■' a a^ ¥ c^ abc ■' a b c
203. In dealing with expressions involving Decimal Frac
tions two methods may be adopted, as will be seen from the
following example.
Multiply Ix  21/ by •03x 4 4?/.
We may proceeil thus, applying the Eules for Multiplication.
Addition, and Subtraction ot Decimals.
•\x—2y
•03x 4 Ay
•003x2 Ouexy
404 xy08y»
•003x"2T03l,v0%2'
Or thus,
_ x2y 3x4^40y
~ 10 ^ lOOT"
^ 3x2 43 4x?/80y3
~ 1060
= 003x2 4. .034x2/  083/2.
The latter method will be found the simpler for a be^nner.
ON MISCELLANE O US ERA C TIONS. 1 35
Examples.— ixvi.
Multiply :
(. Ix 3 by 53; +07, 2. •05x + 7by2 3,
3. Sx  2!/ by •4x + ly, 4. 43x + b2y by •()4x  06?/.
5. Find the value of
a^  6^ + c^ + Zabc when a = 03, h=\, and c = 07.
6. Find the value of
01?  3ax2 + 3a^x  n^ when x = '7 and (i = 'OS. ,
204. When any expression E is put in a form of which /is
E
a factor, then ^ is the other factor.
Thus a + h = a{ \
c 7 T , ab\ac¥hc
So (U) + ac^oc = aoc. ,
and a?+2an/ + 2/2 = x2.(?l±^^±^')
EXAMPLES.— Ixvii.
1. Write in factors, one of which is a^x, the series
a^x + a^ocr + a^x^ + a4X'* + . . .
2. Write in factors, one of which is xyz, the expression
xij XZ + yz.
3. Write in factors, one of which is x^, the expression
X + x!/ + y'.
4. Write in factors, one of which is a + 6, the expression
(a + 6)3c(a + 6)2d(a + 6) + e.
136 O.V MISCELLANEOUS FRACTIONS.
205. We s!i;ill now give two examples of a process by
which, when certain Iractiuns are known to be equal, otlier
relations between the q^uautities involved in them may be.
deteruiineJ.
This i)rocess will be found of great use in a later part of
the suliject. and the student is advised to pay particular
attention to it.
(1) If ^= J, shew that
0,
a — b c — ^'
T ^ a
Let r = X
Then 3 = X ;
a
.■. a = \b,
and c = \d.
Now a + 6^X6+^&^6(X + l)_X + l
a'b \bb~bl\l)~\l'
c + d_\d + d d{X+l) X + 1
and
cd Xdd (Z(Xl) XT
TT C'^b , c + fZ  . , , ;^+i
tience — _^ and ~_j being each equal to — — are equal to
one another.
(2) If = = = .shew that m + u + r = 0.
a — 0 c c  a
Let
ab~^'
oc
r
= X,
ca
then m = XaX6.
n = \b Xc,
r='Kc  Xa ;
.•. m + n + r = X(T X^L \6Xc + XcXff = 0.
ON MISCELLANEOUS FRACTIONS. 137
Examples. — Ixviii.
O. C
1. If r = 7 prove the following relations :
. 'lZ^ — —A ^r\ 8a + & _ 8c + d
, . rt _ _c , c26_a6
, N 3ft _ 3c . ll« + 6_13a + 6
^^^ 4a  56 "" 4c  5rZ" ^^^ TlcTrf ~ 13c+"5"
/ ^ "' + ^^_ C' + tf  «2rt6 + 6_c'ff(' + d*
W a2_p^rr"ci2 ^^) a2 + a6 + 62c^ + cf/ + f;2
Tr ^ m 71 , ,
2. 11 r = T = , then i + 7/1} ?i = 0.
a — n — c c — a
^ jfO c e ^, ^ « la + vic + ne
3. ii r = , = 7, T>rove that y = .7 1 >
a J' ' h Ib + md + nf
a+h b+c_c+t
c a
4 11 — 7;— = = . prove that a = 6 = c.
'■ r, ■ i
3 11 1 =r = r> sliew that fJ= ^/ ^ . •*.
6. II T, J. J he in descending order of magnitude, shew
xifi + cle., , a , , e
i_j7T— >is less than ^ and greater tlian ■^.
7. If '=^ shew that ^^A^4x, + 5y,^
2/1 2/1. '^i + y^/i '^2 + 92/a
T4^ <^ 1 ^1 , rtta6 abb
8. Il5 = ^,shewthat^^ = ^^_^,
9. If^ = .%hewthat7«+.^, = I^,.
o a 6a + ob 3cf5(i
138 Oy MTSCELLANEOUS FRACTIOf^S.
lo. If r be a f roper fraction, shew that r is greater
than r, c being a positive quantity
6 +
II. If r bt an improper fraction, shew that t — : is less
than r, c being a positive quantity.
206. We shall now give a series of examples in the svorking
of which most of the processes connected with fractions will
be introduced.
Examples.— ixix.
I. Find the value of Sa^ H y^ when
a = 4, b = ^, c=l.
^ Sin^pMy 7x^12x + 5 '^"'^ a^'H4a45 
3. Simplifv(^t^_^^)^(«_tP + ^n
^ ^ ' \aj9 a+p/ \a^ a+p/
4. Add together
a;2 t/2 2^ ^2 j,2 3.2 j;2 ji y2
4" e'^S' 4""6"*" 8 ^""^ 4"^ 6"^8'
and subtract 2  x + ^ from the result.
5. Find the value of 5 — .,~ „ — :r^ wheii
a=4, 6 = , c = l.
6. Multiply x2 + 3ax  \a^ by 2x2 ax%,
01 xi ^ a^ft „, 36
7. Shew that . tw = « + 26 t r
' (a  6)^ ah
OM MTSCELLANEOUS FRACTIONS. 130
8. Simplify ?^ + ^^ + 4^!.
^ X xy x^ — xy
01 ^, , 60x3 17x2 4x + l , „, ^ 49
Q. Shew tnat , ., . ^ = 12x25h ^.
^ 5x + 9x  2 .j; + 2
„. ,.„ x*9.r'' + 7x2 + 9x8
10. fei«^Pl'fy^4 + 7^3Z9^2_7^V8'
11. Simplify ^^+ j— .
1. J
1 2. Simplify a+'ah + 6^ f a + «6 + ^V^tt )•
13. Multiply together U + 1)\}' + j>)y ~ \
14. Add to!::'ether , , — , , and shew that if their
^ ° a+V h+V c+l
sum be equal to 1, then ahc = a + 6 + + 2.
^. ., X b h^ h b^ ,
15. Divide! 5 +  + , by xa.
a b c .
r^cH 'a\ =0
16. Simplify r , and shew that it is equal
rC + i^aH ^6
a c
to ^^ ' — T^ — if 2s = a + 6 + c.
be
17. Shew that ^ + p+ _ = — ^.
a^x a~x a^Tx^
1 8. Simpllly r + r  2:^ — r,.
^ •'ab a + b a^ + b
_. ,.. 6 a + 6 a2 + 62
19. Simplify^ 2  + 2^^^ZTy
„. ,.,. aab + ¥ (ily^
I40 ON MISCELLANEOUS FRACTIONS.
2
2 1 . Simplify rj — ,Tr, —
(x2_i)' 2x4x + 2 lx2
... (^^V'vlahc^ a>rh>rc
22. Sunpulyn T. — ,., — ^TT^i— •
'^ '' c^  a^ — b' + 2ab b + c — a
23. Simplify /^■.lj\^^^x^^y
^ X X
. /xrt\3 x2« + 6 , a + 6
24. rind the value of I r I ^j, when x = — s"~'
\x6/ x + a26 2
a"  (6  c)2 Zy  (fi  c) c  (a  6)^
, „. ,.„ (x24x)(x24)a
26. Simplify^— ^,^3i.
27. Simplify ^^^^,^^,^
28. Simplify ^ + — 5 rr, + s rrr^ rr^
•^  ar X X (x^ + l)^ x2+l 2(x + l)'
T,,. . , x^ X a a^ , X a
29. Divide ,   + b by .
a'^axx^'ax
30. Simplify 2—__^^^^4^^.}^^.
31. Simplify ^^ " ^ + ^^'^ ^ (^ : '^l:^ ( V ")'^ ^ ^"  ^^\
„ , lx3x2  l+3a2 + 2x3
3^ ^^^^ (32.c7x^)3 ^^""^ (332^^x^v
33. Sim^lih(i±f,it)^(^J'^^).
*'' ^ • \X'y X' + y/ \xy x + y/
34. si.p,ir,Q:>)(!^i).(^OC.^e0
35. Simplify
a(<6 g 4 a6 + 6 /_ 2a3_ _ , \ / , _ ^gft \
OM MISCELLANEOUS FRACTLOXS. 141
36. Simplify
1 _ 1^+
2(a;l) 4(xl) 4(a;+l) (x  1)' (x + 1)"
37. Prove that
1 1
— +
s—ash&c /111 \
4 + ... =s( +r+ + ... )".
\a be I
ab.c a {a  h) {x  a) h{b — a) {x  h) x {x  u) {x  1)'
38. Tf s = a4 ?) + c+ ... to 71 term?, slie\v that
b s
T , — +
a c
39. MuUith(^,^^.>y ^$^/?_.
jy i •■ \ X'  y X + y/ X — yy + {x + ij)
, ax , a — x^
1 + 1 + , r.
,.r « + •'<; rt4X'
4.0. Simpluv — •..
^ , ^ •' ax U — X
a + x a + X'
41 . Divide x^ + 3  s( ^  x ) + 4f x +  ) liy x+ .
42. If s = rt + 6 + c + ...tow terms, shew th;it
sa s — b .«  r ,
+ + +...=711.
5 S S
43. Divide (" — ^  I l.y ( ..^^. + X0
^•^ \x/ x + 2// \x + y X'yy
1^. /I
44. bimphfy ^—^ T
45 If r_a6 =^;j:ri' P^«^'^ that p^^ = a6.i
a c a
46. Simplify
p* + 4p^q + 6pq' + 4p(f + 2^ ^ 1^ + ^P'1 * •^P^ '^ * 'f
pi  4p^q + Gjyq'  4pq^ + q* ' p^  3pq + 3pg  ^
142 SIMULTAXEOUS F.QUATIOXS
48. Simplify
1 1 y(xi/3 + x + a)
y+~
^ z
1 1 1 a: y
^ iniBlifr * " ■'' " ~ '^ ^"^ ~ ^ • ' ^"^ " ^^'
49.
(a  1/) (a  x)2 (a  1/;2 (« _ x)
3
50.
c.:..„i:f.. f'^'^ 3a6c
i'"' 1 1 1 a^ 0c '
T)c ca ah
Simpliiy l{a'^b'^.
a —  —
51
♦ b
XV. SIMULTANEOUS EQUATIONS OF
THE FIRST DEGREE.
207. To determine several unknown quantities we must
have as many independent equations as there are unknown
quantities.
Thus if we had this equation given,
x + y = 6,
we could determine no definite values of x and y, for
a; = 2) x = 4) x = 3)
or other values miglit he given to x and y, consistently with
the equation. In fact we can find as many pairs of values of
X and 1/ as we please, which will satisfy the equation.
OF THE FIRST DEGREE. 143
"We must have a second equation independent of tlie first,
and then we may tind a pair of values of x and y wliicii will
satis/ u both equations.
Thus, if besides the equation x + y = Q, we had anotlier
equation xy = 2, it is evident that the values of x and y
which will satisfy both equations are
x = 4
y = 2
since 4 + 2 = 6, and 42 = 2.
Also, of all the pairs of values of x and y which will satisfy
one of the equations, there is but one pair which will satisfy
the other equation.
We proceed to shew how this pair of values may be found.
208. Let the proposed equations be
2x + 7i/ = 34
5x + 9i/ = 51.
Multiply the first equation by 5 and the second equation by
2, we then get
10x + 35j/ = 170
10.c+18?/ = 102.
The coefficients of x are thus made alike in both equations.
If we now subtract each member of the second equation
from the corresponding member of the first equation, we shaU.
get (Ax. II. page 58 j
35j/18i/=170102,
or 17(/ = 68;
■• 2/ = 4.
We have thus obtained the value of one of the unknown
symbols. The value 01 the other may be found thus :
Take one of the original equations, thus
2x + 7j/ = a4.
Now, since y — 4,7y = 28;
.: 2x + 28 = 34;
.. x = 3.
Hence the pair of values of x and y which satisfy the
ecjuations is 3 and 4.
144 SIMULTANEOUS EQUATIONS
Note. The process of thus obtaining from two or more
equalioiis an equation, Ironi wliicli one ol the unknown quanti
ties has disappeared, is called Elimination.
209. ^Ye worked out tlie steps fully in the example given
in the last article. We shall now work au example in the lorm
in which the process is usually given.
Ex. To solve the equations
5a; + 4)/ = 58.
Multiplying the first equation hy 5 and the second by 3,
15x435^ = 335
15x4 12// = 174.
Subtracting, 23^=161,
and therefore 2/ = 7.
Now, since 3x4 7;/ = 67,
3x449 = 67,
.. 3x = 18,
.. x = 6.
Hence x = Q and y=*7 are the values required.
210. In the examples given in the two preceding articles
we made the coetticients of x alike. Sometimes it is more con"
venient to make the coefficients ol" y alike. Thus if we have
to solve the equations
29x427/ = 64
13x4 2/ = 29,
we leave the first equation as it stands, and multiply the
second equation by 2, thus
29x4 2?/ = 64
2Gx + 2!/ = 58.
Subtracting, 3x = 6,
and therefore x = 2.
Now, since 13x4y = 29,
264?/ = 29,
.. 2/ = 3.
Hence x = 2 and ?/ = 3 are the values required.
OF THE FIRST DEGREE. 145
Examples — Ixx.
I. 2a; + 7i/ = 41 2. 5.<; + 8i/=101 3. 13.r+ l??/^ 189
3a; + 4i/ = 42. 9x + 2y = 95. 2a;+ i/ = 21.
4. 14x + 9?/ = 15G 5. a; + 152/ = 49 6. 15;(; 19?/= 132
7x + 2)/ = 58. 3x+ 7)/ = 71. Sax + 17?/ = 226.
7. 6a; + 4?/ = 236 8. 39.j; + 277/ = 10o 9. 72a; + 14?/ = 330
3a; +15?/ = 573. 52a; + 297/ =133. G3x+ 7?/ = 273.
211. "We .shall now give some examples in which negative
signs occur attached to tlie coelUcieut ol y in one or both of
the equations.
Ex. To solve the equations:
6a; + .^o?/ = l77
8a; 217/ = 33.
Multiply the first equation by 4 and the second by 3.
24x+140?/ = 7()3
24x 63?/ = 99.
Subtracting, 2037/ = G09,
and therefore y = ^
The value of x may then be found.
Examples.— Ixxi.
I. 2x + 7y = 52 2. 7x 47/ = 55 3. x + 7/ = 9G
3x5// = 16. 15x13// = 109. x7/ = 2.
4. 4x+ 9?/ = 79 5. a; + 197/ = 97 6. 29x14?/ = 175
7x17?/ = 40. 7x53?/ = 121. S7.:;5G?,'4y7.
7. 171x213?/ = 642 8. 43x+ 2?/ = 26G 9. 5x + 9?/ = lfi8
114x326?/ = 244. 12x17?/ = 4. 13x27/ = 57.
fs.A.l 
146 . STMUL TANEOUS EQUA TIONS
212. We have hitherto taken examples in ■which the
coetiicients of x are both positive. Let us now take the lolljv .
ing equations :
5x 7y = 6
9y2x = 10.
Change all the signs of tJie second equation, so that we get
5a; — 7i/ == 6
2x9y= 10.
Multiplying by 2 and 5,
I0xUy = l2
10x45j/=50.
Subtracting,
141, f 45?/ = 12 + 50,
or, Sly = 62,
or, y = 2.
The value of x may then be found.
Examples.— ixxii.
I. 4x~7y = 22 2. 9.c5?/ = 52 3. 17x + 3i/ = 57
7y'Sx=l. 8y'3x = 8. 16i/3x = 23.
4. 7y + 3x = 7S 5. 5.c3i/ = 4 6. 3x + 2i/ = 39
19i/7x = 136. 12!/7x=10. 32/2a;=13.
7. 5y2x = 21 8. 9?/7x=13 9. 12a;+ 7y=176
13x4y=l20. 15x7t/ = 9. 32/19a; = 3.
213. In the preceding examples the values of a; and y have
been jiositive. We shall now give some equations in which x
or y or both have negative values.
Ex. To solve the equations:
2x9y = U
3x4y = 7.
Multiplying the equations by 3 and 2 respectiveiy, we get
6x27y = :yi
6x 8}f=14.
OF THE FIRST DEGREE. i^'j
Subtracting,
19*/ = 19,
or, 192,'=19,
or, y=l.
Now since 9y=  9,
2x  9y will be et[uivalent to 2x  (  9) or, 2x + 9.
Hence, from the first equation,
2x + 9 = ll,
.. x = l.
Examples. — Ixxiii
I. 2x + 3i/ = 8 2. 5x~2y = bi 3. 3x5y = ol
3x + 7y = 7. 19a; 3?/ = 180. 2x + 7i/ = 3.
4. 7i/3x=139 5. 4x+ 9?/= 106 6. 2x7i/ = S
2x + by = i)l. 8a; + 172/=198. 4!/9x=19.
7. 17x+122/ = 59 8. 8x + 3y = :i 9. 69i/17x=103
19x 4?/ = 153. 12x + 9(/ = 3. 14.c 13^/= 41.
214. We shall now take the case of Fractional Equations
involving two unknown quantities.
Ex. To solve the equations,
2x^:^=4
5
3y = 9— 3.
First, clearing the equations of fractions, we get
10xy + 3 = 2O
9y = 27x + 2,
from which we obtain,
10xy = 17
x + 9y = 29,
and hence we may find x = 2, i/ = 3.
148 SIMULTANEOUS EQUATIONS
Examples.— Ixxiv.
I. 1 + 1 = 7 2. iOx +  = 210 3. ^ + 7j/ = 251
1 + 1 = 8. 10!/? = 29().  + 7x = 299.
4. —,^ + 5 = 10 5. 7a; + Tr = '413 6. — ^ = 10^
3 2 5 3
^1 + 7 = 9^ 39x = 142/l609 ^^= + 1.
7. x'^^ = 5 10. ^;^ + 8, = 31
4u :^— = 3. =i— +10x = 192.
3 4
8.  + 8 = 12 II. ?^J^ + 3x = 2y6
x + ?/ w 2x'V „^ 1/ + 3 1/X „ ^^
3x5;/ „ 2x + y x2 10x 1/IO
2 o 3 4
x2y/_x 7/ 2>/ + 4_4 x + y+13
S ____ + _ _ ^ .
13. — ^ + 3x = 47/2
■^ 13
5x + 6i/ 3x2v_5, _ .^
5x3 3.7;19 , 37/x
.4. 2 o" = ^^3
2x + y _ 9x  7 _ % t? _ f^ + ^
~2 8 "4 16 ■
4x + 5?/
^5 40" = ^^
2x  1/ „ 1
3^"^' = 2
OF THE FIRST DEGREE. 149
215. We have now to explain the nietliod of solving Literal
Equations involving two xniknowu quantities.
Ex. To solve the equations,
^ ax\by = c
px + q>j = r.
\.
Multiplying the first equation Ly p and the second by a, we
get
apx + bpy = cp
apx + aqxj = ar.
Subtracting, bjnj — aqij = cp — ar,
or, {bp aq)y = cp  ar ;
_ cp  ar
bp — aq'
We might then find x by substituting this value of y in one
of the original equations, but usually the safest cour.se is to
begin afresh ami make the coefficients of y alike in the original
equations, multiplying the first by q and tlie second by b,
which gives
aqx + bqy = cq
bpx + bqy = br.
Subtracting, aqx — bpx = cq br,
or, {aq — bp)x = cqbr;
_ cq br
aqbp'
Examples.— ixxv.
I. 'mx + ny = e, 2. ax + by = c 3. axby = m
px + qy =f. dx  cy =/. ex + ey = n.
4. ex =dy 5. mxny = r 6. x + y = a
x + y = e. m'x + n'y = r xy = b.
7. ax + by = c 8. abx + cdy = 2 o. , = ^;
' ^ •' ^ h + y 3a + x
dx + fy = c. axcy= rj. ax + 2by = d.
156 ^TMtJLTAYF.OTJS F.QUATIOI^S
\o. bcxi2b aj = II. {b + c)(;x + c — b) + a(7j + a) = 2a^
^, a(c^b^) 2h^ , ay (b + cy
^ be c (bc)x a'
(8b2m)bm
12. 3x + 5i/ = — , ij „ —
0  r/r ,
6a;  , t(b + c + vi) my = ■ni^x + (b + 2m) &m.
216. We now proceed to the solution of a particular class
of Simultaneous Equations in which the unknown symbols
appear as the denominators of fractions, of which the following
are examples.
Ex. 1. To solve the equations,
a b
 + =c
X y
m TO J
=(L
X y
Multiplying the first by m and the second by a, we get
am bm
+ ^ =cm
X y
am an ,
X y ~ '
bm an
— H = cm
y y
ad.
Subtracting,
bm + an .
or, =cmad,
y
or, bm + an = (cm  ad) y,
_^bm + an
^ cm ad'
Then the value of x may be found by substituting this value
of y in one of the original etjuations, or by making the terms
containing y alike, as iu the example given ia Art. 215.
6P THE FIRST DEGREE. t^t
ElX. 2. To solve the equations:
X 3y~27
Ax'^y 72"
Multiplying the second equation 1>y 8, we cret
2__5 _i_
a~3i/~27
2 8_11
X y~ 9'
Subtracting,
5 8_ 4 11
•3y y~27 9"
Changing signs,
5 8 11 4
Sy'^y~9 27'
or.
5 + 24 334
dy 27 '
whence we find
y=9,
and then tlie vahie of x may be found by substituting 9 lor y
in one of the original equations. *
Examples. — Ixxvi.
X y
2.
1 2
 +  = a
a; y
3
a b
 +  = c
X y
i + =^ = 20.
X y
3 4 .
 +  = ?).
X y
b a
 +  = d.
X y
a b
 +  = m
X y
5
X y
6.
5 2 „
Sx by
a b
X y
7 1 ^
X y~
6x 102/ ""■
2
ax
3
:5
8.
m n
1 = m + n
nx my
5
ax
2
by"
3.
n
 +
X
TO ., .,
— =TO + ?l.
I.<2 SIMULTANEOUS EQUATIOh'S
217. There are two other methotls of solvin.n Simultaneoua
El [nations of which we have hitherto made no mention, because
they are not generally so convenient and simple as the method
which we have explained. They are
I. The metliod of Substitution.
If we have to solve the equations
a; + 3?/= 7
2x + 4?/ = 12
we may find the value of x in terms of y from the first equa
tion, thus
a; = 73(/,
and substitute this value for x in the second equation, thus
2 (7 37/) + 47/= 12,
from which we find
i/ = l.
We may then find the value of x from one of the original
equations.
II. The method of Comparison.
If we have to solve the equations
5x + 2j/ = 16
7x3!/= 5
we may find the values of x in terras of y from each equation,
thus
x = —  — , from the first equation.
x = — — , from the second equation.
Hence, equatini; these values of x, we get
16^27/ _5+_3y
5 '~ 7" '
an equation involving only one unknown symbol, from which
we obtain
!/ = 3,
and tlun the value of x may be found fr<ini one of the oriinal
e(iuati'>'>s.
OF TFJE FIRST DEGREE. 153
218. If tliere be ihrRe, unknown symbols, their values may
be found from tliree independent eijuations.
For from two of the equations a third, which involves only
tioo of the unknown symbols, may be found.
And from the remaining equation and one of the others
a fourth, containing only the same two uukuown symbols, may
be found.
So from these two equations, which involve only two un
known symbols, tlie value of these symliols may be found, and
by substituting these values in one of the original equations
the value of the third unknown symbol may be found.
Ex. 5x6y + 4z=15
7.'c + 47/3.;=19
2x+ 7/ + 6.v = 46.
Multiplying the first by 7 and the .second by 5, we get
35.<;42i/ + 283=105
3Jx + 20?/15s = 95.
Subtracting,
62)/ + 433 = 10 (1).
Again, multiplying tlie first of the original equations by 2
and the third by 5, we get
10.c12r/ + 83 = 30,
lOx + by + 30z = 2Z0.
Subtracting,  17?/ 222= 200 (2).
Then, from (1) and (2) we have
62>j43z= 10
17(/ + 222 = 200,
from which we can find ij = 4 and s = 6.
Then substituting these values for tj and z in the first equa
tion we find the value of x to be 3.
Examples. — Ixxvil.
1. 5x + 7y 22 = 13 3. bx3y + 2z = 21
8x + 3!/+ 2 = 17 8x y3z= 3
x4?/ + 103 = 23. 2x + 3i/ + 2z = 39.
2. 5x + 3)/6.i; = 4 4. 4x5y + 2z= 6
3x y + 2z = S 2x + 3?/ 2 = 20
x2y + 2z = ± 7x4.v + 32 = 35,
154 PROBLEMS RESULTING TiV
5 x+ y+ z= 6 8. 4x3j/4 2= 9
5x + 47/ + 3;2 = 22 9x + 1/  62 = 1 6
15a; + 10)/ + 62 = 53. x4i/ + 3z= 2.
6. 8x + 47/3.v = 6 9. 12a; + 5?/ 42 = 29
a; + 32/— z = 7 13x 2^452 = 5b
4x52/ + 4z = 8. 17a; ?/ z = 15.
7. x+ y+ 2 = 30 10. yx + z= 5
8a; + 4?/ + 22 = 51) z  ?/  x =  25
27a; + 9;/ + 32 = 64. x + ?/ + 2 = 35.
XVI. PROBLEMS RESULTING IN SIMUL
TANEOUS EQUATIONS.
219. In the Solution of Problems in which we represent
two of the numbers sought by unknown symbols, usually x and
y, we must obtain two independent equations from the condi
tions of the question, and then we may obtain the values of
the two unknown symbols by one of the processes described in
Chapter XV.
Ex. If one of two numbers be multiplied by 3 and the
other by 4, the sum of the products is 43 ; and if the former be
multiplied by 7 and the latter by 3, the difference between the
results is 14. Find the numbers.
Let X and y represent the numbers.
Then 3a; + 4*/ =43,
and 7x — 'iy = 14.
From these equations we have
21x + 28!/ = 301,
21a; 9?/ = 42.
Subtracting, 37j/ = 259.
Therefore J/ = 7.
and tlu'U tlie value of a; may be found to be o.
Hence the numbers are 5 and 7.
A.
SIMULTANEOUS EQUATIONS. 155
Examples.— Ixxviii.
The snm of two numbers is 28, and tlieir difference is 4,
find the numbers.
ill 2. The sum of two numbers is 256, and their difference is
10, find the numbers.
3. Tlie sum of two numbers is 13'5, and tlieir difference is
1, find the numbers.
■6 ^4. Find two numbers such that the sum of 7 times the
"^greater and 5 times the less may be 332, and the product of
their difference into 51 may be 408.
.jjl 5. Seven years ago the age of a father was four times that
•^of his son, and seven years hence the age of the father will be
double that of the son. Find their ages.
^^6. Find three numbers such that the sum of the first and
•second shall be 70, of the first and third 80, and of the secoud
and third 90.
7. Three persons A, B, and G make a joint contribution
which in the whole amounts to ^400. Of this sum B contri
butes twice as much as A and £20 more ; and G as much as A
and B together. What sum did each contribute?
8. If A gives B ten shillings, B will have three times as
much money as A. If B gives A ten shillings, A will have
twice as much money as B. What lias each ?
9. Tlie sum of £760 is divided between A, B, G. The
shares of A and B together exceed the share of G by £240,
and the shares of B and C together exceed the share of A by
£360. Wliut is the share of each ?
^(^ 10. The sum of two numbers divided by 2, gives as a quo
"^tient 24, and the difference between them divided by 2, gives
as a quotient 17. What are the numbers?
^^ II. Fiml two numbers such that when the greater is
divided by the less the quotient is 4 and the remainder 3, and
when the sum of the two numbers is increased by 38 and the
result divided by the greater of the two numbers, the quotient
is 2 and the remainder 2.
/A 12. Divide tlie number 144 into three such parts, that
(/'when the first is diviiled by the second the quotient is 3 and
the remainder 2, and when the third is divided by the sum
of the other two parts, the quotient is 2 and the remainder 6.
IS6 PROBLEMS RESULTING IN
H
13. A and B buy a horse for £120. A can pay for it if B
will advance half the money he has in his pocket. B can pay
for it if A will advance twothirds of the money he lias in his
pocket. How much has each ?
.^14. "How old are you?" said a son to his father. The
father replied, "Twelve years hence you will be as old as 1 was
twelve years ap;o, and I shall be three times as old as you were
twelve years ago." Find the age of each.
1,^5. Eequired two numbers such that three times the
greater exceeds twice the less by 10, and twice the greater
together with three times the less is 24.
^7'
li6. The sum of the ages of a father and son is half what it
"^vill be in 25 years. The difference is onethird what the sum
will be in 20 years. Find their ages.
, / ' 17. If I divide the smaller of two numbers by the greater, •
/ the quotient is '21 and the remainder "OIjT. H" I divide the
greater lunuber by the smaller, the quotient is 4 and the
remainder '742. Find the numbers.
18. The cost of 6 barrels of beer and 10 of porter is £51 ;
the cost of 3 barrels of beer and 7 of porter is £32, 2s. How
much beer can be bought for £30?
19J The cost of 7 lbs. of tea and 5 lbs. of coffee is £1, 9s. 4il. :
the cost of 4 lbs. of tea and 9 lbs. of coffee is £1, 7s. : what is
the cost of 1 lb. of each ?
20. The cost of 12 horses and 14 cows is £3S0 : the cost of
5 horses and 3 cows is £130 : what is the cost of a horse and a
cow respectivel}' 1
21. The cost of 8 yards of silk and 19 yards of cloth is
£18, 4s. 2d.: the cost of 20 yards of silk and IG yards of doth,
each of the same quality as the former, is £25, 16s. Sti. How
much does a yard of each cost ?
22. Ten men and six women earn £18, 18s. in 6 days, and
four men and eight women earn £(>, Cs. in 3 days. What are
the earnings of a man and a woman daily ?
1 1)23. A farmer bought 100 acres of laud for £4220, part at
^^£37 an acre and part at £45 an acre. How many acres had
lie of each kind?
SIMULTAi^EOUS EQUATIONS. 157
Note I. A number consisting of two digits may be repre
sented algebraically by lOx + y, where x and y represent the
significant digits.
For consider such a number as 76. Here the significant
digits are 7 and 6, of which the former has in consequence of
ils position a local value ten times as gre'it as its natural
value, and the number represented by 76 is equivalent to ten
times 7, increased by 6.
So also a number of which x and y are the significant digits
will be represented Ijy ten times x, increased by y.
If the digits composing a number lOx + y be inverted, the
resulting number will be lOy + x. Thus if we invert the digits
composing the number 76, we get 67, that is, ten times 6, in
creased by 7.
If a number be represented by lOx + y, the sum of the
digits will be represented by x + y.
A number consisting of three digits may be represented
algebraically by
100x+ lOy + z.
Ex, The sum of the digits composing a certain number is
5, and if 9 be added to the number the digits will be inverted.
Find the number.
Let lOx + y represent the number.
Then x + y will represent the sum of the digits,
and lOy + x will represent the number with the digits inverted.
Then our equations will be
x + y = 5,
10x + y + 9 = l0y + x,
from which we may find x = 2 and ?/ = 3 ;
.". 23 is the number required.
^'^ 24. The sum of two digits composing a number is 8, and if
36 be added to the number the digits will be inverted. Find
the number.
jvi^25. The sum of the two digits composing a number is 10,
and if 54 be added to the number the digits will be inverted.
What is the number ?
1S8 PROBLEMS RESUL TING IN
26. The sum of the digits of a munber less than 100 is 9,
and if 9 be added to the number the digits will be inverted.
What is the nundjer?
27. The sum of the two digits composing a number is 6,
and if the number be divided by the sum of the digits the
quotient is 4. '\Vliat is the number ?
28. The sum of the two digits composing a number is 9,
and if the number be divided bv the sum of the digits the
quotient is 5. What is the number ?
29. If I divide a certain number by the sum of the two
digits of wliich it is composed the quotient is 7. If I invert
the order of the digits and then divide the resulting nund)er
dinnnished by 12 by tlie difference of the digits of the original
number the quotient is 9. What is the number ?
A 30. If I divide a certain number by the sum of its two
digits the quotient is 6 and the remainder 3. If I invert the
digits and divide the resulting number by the sum of the digits
the quotient is 4 and the remainder 9. Find the number.
31. If I divide a certain number by the sum of its two
digits diminished by 2 tlie quotient is 5 and the remainder 1.
If I invert the digits and divide the resulting number by the
sum of the digits increased by 2 the quotient is 5 and the re
mainder 8. Find the number.
^i
32. Two digits which form a number change places on the
addition of 9, and the sum of these two numbers is 33. Find
the numbers.
33. A number consisting of three digits, the absolute value
of eacli digit being the same, is" 37 times the square of any
digit. Find the number.
34. Of the three digits composing a number the second is
double of the third : the sum of the first and third is 9 : the
sum of all the digits is 17. Find the number.
.1 (35. A number is composed of three digits. The sum of the
digits is 21 : the sum of tlie fust and second is greater than the
third by 3; and if 198 be added to the number the digits will
be inverted. Find the number.
SIMUL TANEOUS EQ UA TIONS. I J9
Note II. A fraction of which the terms are unkno\vn may
be represented by .
Elx. A certain fraction becomes ^ when 7 is added to its
denominator, and 2 when 13 is added to its numerator. Find
the fraction.
Let  represent the fraction
a; + 13_
are the equations ; from which we may find a; = 9 and i/=ll.
9
That is, the fraction is yy.
36. A certain fraction becomes 2 when 7 is added to its
numerator, and 1 when 1 is subtracted from its denominator.
What is the fraction ?
37. Find such a fraction that when 1 is added to its
1
3'
numerator its value becomes , and when 1 is added to the
denominator the value is .
4
38. What fraction is that to the numerator of which if 1 be
1
^2
added the value will be ^ : but if 1 be added to the denominator.
the value will be ;^ ?
39. The numerator of a fraction is made equal to its
denominator by the addition of 1, and is half of the deno
minator increased by 1. Find the fraction.
40. A certain fraction becomes  when 3 is taken from the
numerator and the denominator, and it becomes  when 5
i6o PROBLEMS RESUL TING IN
is added to the numerator and the denominator. "What is the
fraction ?
7
41. A certain fraction hecomes ^ when the denominator is
20
increased hv 4, and — ^ when the numerator is diminished by
15 : determine the fraction.
42. What fraction i.< that to the numerator of which if 1 be
added it becomes , and to the denominator of which if 17 be
added it becomes  ?
o
Note III. In questions relating to money put out at
simple interest we are to observe that
T Principal x Rate x Time
Interest = ,
where Eate means the number of pounds paid for the use of
£100 for one year, and Time means the number of years for
which the money is lent.
43. A man puts out £2000 in two investments. For the first
he gets 5 per cent., for the second 4 per cent, on the sum
in\ested, and by the first investment he has an income of
£10 more than on the second. Find how much he invests in
each case.
44. A sum of money, put out at simple interest, amounted
in 10 months to £5250, and in 18 months to £5450. "What
was the sum and the rate of interest ?
45. A sum of money, put out. at simpie interest, amounted
in 6 years to £52(10, and in 10 years to £6000. Find the sum
and the rate of interest.
Note IV. "When tea, spirits, wine, beer, and such com
modities are mixed, it must be observed that
quantity of ingredients = quantity of mixture,
cost of ingredients = cost of mi.\ture.
Ex. I mix wine which cost 10 shillings a gallon with
another sort which cost 6 shillings a gallon, to make 100
SIMULTANEOUS EQUATIONS. i&i
gallons, which I may sell at 7 shillings a gallon vithout profit
or loss. How much of each do I take ?
Let X represent the number of gallons at 10 shillings a gallon,
and \j 6
Then a; + 2/=100,
and 10x + 6t/ = 700,
are the two equations from which we may find the values of
X and y to be 25 and 75 respectively.
46. A winemerchant has two kinds of wine, the one costs
36 pence a quart, the other 20 pence. How niucli of eacli must
he put in a mixture of 50 quarts, so that the cost price of it
may be 30 pence a quart ?
47. A grocer mixes tea which cost him Is. 2fZ. per lb. with
tea that cost him Is. %d. per lb. He lias 30 lbs. of the mi.vture,
and by selling it at the rate of Is. 8(Z. per lb. he gained as
much as 10 lbs. of the cheaper tea cost him. How many lbs.
of each did he put in the mixture?
Note V. If a man can row at the rate of x miles an hour
in still water, and if he be rowing on a stream that runs at the
rate of 1/ miles an hour, then
X + 1/ will represent his rate down the stream,
X — ?/ wp
48. A crew which can pull at the rate of twelve miles an
hour down the stream, finds that it takes twice as long to come
up a river as to go down. At what rate does the stream How ?
49. A man sculls down a stream, which runs at the rate of
4 miles an' hour, for a certain distance in 1 hour and 40 minutes.
In returning it takes him 4 hours and 15 minutes to arrive at
a point 3 miles short of his starting place. Find the distance
he pulled down the stream, and the rate of his pulling.
50. A dog pursues a hare. The hare gets a start of 50 of
her own leaps. The hare makes six leaps while the dog makes
5, and 7 of the dog's leaps are equal to 9 of the hare's. How
many leajJS will the hare take before she is caught ?
l62 PROBLEMS RESUL TIXG IN
51. A grevhoimd starts in pursuit of a hare, at the distance
of 50 of liis own leaps Irom ber. He makes 3 leaps while the
bare makes 4, and he covers as much ground in two leaps as
the hare does in three. How many leaps does each make
before the hare is caught ?
i;2. I lay out halfacrown in apples and pears, buying the
apples at 4 a penny and the pears at 5 a jtenny. I then sell
half the apples and a third of the pears for thirteen pence,
•which was the price at which I bought them. How many of
each did I buy ?
53. A company at a tavern found, when they came to pay
their reckoning, that if there had been 3 more persons, each
would have paid a shilling less, but had there been 2 less,
each would have paid a shilling more. Find the number of
the company, and each man's share of the reckoning.
54. At a contested election there are two members to be
returned and three candidates. A, B, and C. A obtains 1056
votes, B, 9S7, C, 933. Now 85 voted for B and C, 744 for
B only, 98 ibr C only. How many voted for A and C, for
A and B, and for A only ?
55. A man walks a certain distance : had his rate been
half a mile an hour faster, he would have been H hours less
on the road; and had it been half a mile an hour slower, he
■would have been 2h hours more on the road Find the distance
and rate.
56. A certain crew pull 9 strokes to 8 of a certain other
crew, but 79 of the latter are equal to 90 of the former. Which
is the faster crew ?
Also, if the faster crew start at a distance equivalent to
four of their own strokes behind the other, how many strokes
will they take before they bump them ?
57. A person, sculling in a thick fog, meets one barge and
overtakes another which is going at the same rate as the
former ; shew that if a be the greatest distance to which he
can see, and b, b' the distances that he sculls between the
times of his first seeing and passing the barges,
2^1 I
a h h''
STMUL TA NEOUS EQUA TTONS. 1 63
58. Two trains, 92 feet long and 84 feet long respectively,
are moving with uniform velocities on parallel rails in opposite
directions, and are observed to pass each other in one second
and a half ; but when they are moving in the same direction,
their velocities being the same as before, the faster train is
observed to pass the other in six seconds; find the rate in
miles pei» hour at which each train moves.
59. The forewheel of a carriage makes six revolutions
more than the hindwheel in 120 yards ; but only four revolu
tions more when the circumference of the forewheel is increased
onefourth, and that of the hindwheel onefiith. Find the
circumference of each wheel.
60. A person rows from Cambridge to Ely (a distance of
20 miles) and back again in 10 hours, and fihds he can row
2 miles against the stream in the same time that he rows
3 miles with it. Find the rate of the stream, and the time of
his going and returning.
61. A number consists of 6 digits, of which the last to the
left hand is 1. If this numl>er is altered by removing the 1
and putting it in the unit's place, the new number is three
times as great as the original one. Find the number.
XVII. ON SQUARE ROOT.
220. In Art. 97 we defined the Square Root, and explained
the method of taking the square root of expressions consisting
of a single term.
The square root of a positive quantity may be, as we
explained in Art. 97, either positive or negative.
Thus the square root of 4a' is 2a or  2a, and this ambiguity
is expressed thus,
J4a^=±2a.
In our examples in tnis chapter we shall in all cases regard
the square root of a single term as a positive quantity.
l64 OKT SQUARE ROOT.
221. The sfjuare root of a product may be found by taking
the square root of each factor, and multiplying the roots, so
taken, together.
Thus y/^' = ab,
222. The square root of a fraction may be found by taking
the square root of the numerator and the square root of the
denominator, and making them the numerator and denominator
of a new fraction, thus
V4a^_2a
8lP" 9b
4
96'
2bxy^ _ 5ory^
492'^ ^~72^'
Examples. — Ixxix.
Find the Square Root of each of the following expressions ;
2. Slants. 3. 121mio?ii2,.u.
5. 11289a*b^z^. 6. lG9a^%^c^^.
1 25a^6<'
I.
4xy'^.
4
Ma'^b^^cl
9a2
y
1G62
256x^2
289/"
4a2c** ^* 121x«i/">'
625«2
"• 3246'i*
223. We may now proceed to investigate a Rule for the
extraction of the square root of a compound algebraical
expression.
"We know that the square of a + 6 is a'^ + '2nb + b'^, and there
fore a + 6 is the square root of a + 2ab + b'.
If we can devise an operation by which we can derive a + b
from a^ + 2ab + b', we shall be able to give a rule for tlie
extraction of the square root.
Now the first term of tlie root is the square root of the first
term of the square, i.e. a is the square root of a^.
Hence our rule begins :
"Arrange tlie terms in the order of magnitude of the indices
of one of the quantities involved, then take the square root of the
ON SQUARE ROOT. idg
jirst term and net down tlie result as the first term of the root:
subtract its square from the given expression, and bring down the
remainder :'' thus
d^ + 2ab + b (a
a
2ab + b'^
Now this remainder may he represented thus &(2a + 6^:
hence if we divide iab + b"^ by 2a + b we shall obtain rh the
second term of the I'oot.
Hence our rule proceeds :
'" Double the first term of the root and set 'fowr the result as the
first term of a divisor:'' thus our process up to this point will
stand thus :
a^ + 2ab + b^ [a
a?
2a , 2a6 + &2
Now if we divide 2ab by 2a the eRult is b, and hence we
obtain the second term of the root, and if we add this to 2a
we obtain the full divisor 2a + b.
Hence our rule proceeds thus :
'• Divide the first term of the remainder by this first term of the
divisor, and add the result to the first term of the root and also to
the first term of the divisor:" thus our process up to this point
will stand thus :
a^ + 2ab + b{^a + b
a2
2a+b
2ab + 62
If now we midtiply 2a + 6 by 6 we obtain 2ab + b^, which we
subtract from the first remainder.
Hence our rule proceeds thus :
^'Multiply the divisor by the second term of the root and sub
tract the result from the first remainder :' tiius our process will
stand thus :
i66
ON' SQUARE KOOT.
a2 + 2a6+6%a + fe
o2
2tH6
2a6 + 62
2a6 + 62
If there is now no remainder, the root has been found.
If there he a remainder, consider the two terms of the root
already found as one, and proceed as before.
224. The following examples worked out will make the
process more clear.
(1) o22a6 + 62(^a6
ft2
2a 6 I 2a6 + 62
■ 2a6 + 62
Here the second term of the root, and consequently the
second term of the divisor, will have a negative sign prefixed,
because >, — = o.
2a
(2)
(3)
6p + 42
1016
9p^ + 24pq+l6q(^3p + 4q
9p2
24pq + 16^2
24pq + IQq
25x260x + 36(5x6
25x'
 60z + 36
60X + 36
Next take a case in which the root contains three terma.
a + 2ab + b — 2ac  2bc + c{^a + b — e
a2
2a + 6
2ab + b2ac2bc + c^
2ab + 6"
2a + 26  c
 2ac  26c + c^
 2ac  26c + c*
ON SQUARE ROOT. 167
When we obtained the second remainder, we took the double
of + 6, consiflereJ as a single term, and set down the result as
the first part of the second divisor. We tlien divided the first
term of the remainder, — 2ac, by the first term of' the new
divisor, 2a, and set down the result,  c, attached to the part
of the root already found and also to the new divisor, and then
multiplied the completed divisor by c.
Similarly we may proceed when the root contains 4, 6 or
more terms.
Examples.— Ixxx.
Extract the Square Eoot of the following expressions :
1. 4a + V2ab + 9b\ 6. x^  6x^ + I9x'  30x + 2b.
2. lG¥^24kH^ + 9l^ 7. 9x^+12x3+ 10x2 + 4x+l.
3. ab+l62ab + 65Gl. 8. 4r* 12)3+ 13?Gr+ 1.
4. /38?/3 + 361. 9. 4)i* + 4)i37n2_47i + 4.
5. 9a26V  102a6c + 289. 10. l6x+ 13x212x3 + 4x*
11. x8 4x5 + 1 Ox* 12x3 + 9x2.
1 2. 4y*  12yh + 2oyh^  24yz^ + 16a*.
13. a^ + 4ah + 4¥ + 9c' + 6ac + '[2bc.
1 4. a^ + 2a'6 + 3a^b + 4a"¥ + 3ab* + 2ab^ + W.
15. x84x5 + 6x3 + 8x' + 4x+l.
1 6. 4x* + 8ax3 + 4a2x2 + 1 662x2 + \<oabx + \ 66*.
[7. 9  24x + 58x2  116x3 + 129x*  14.0x5 + ioOx«.
f 8. 1 6a*  4Qa?b + 2ba?b'  80a62x + 646'x2 + 64a26a;.
1 9. 9a*  24a^p^  •^OaH + 1 daf + 40apH + 25^2,
20. 4?/*x2  1 2 y^x^ + 1 7i/2x* \2yx^ + 4x^.
2 1 . 25x*2/2 _ 30x37/3 + 29x2?/*  1 2xif + 4y^.
22. 16x*  24x3?/ + 25x2y2 _ 12x»/3 + 4y\
23. 9a212a6 + 24ac166c + 452 + i6c2.
24. x* + 9x2 + 256x3+10x230x.
25. 25x2 _ 20x2/ + 4 )/2 + 9^2 _ 1 2i/3 + SCtea.
26. 4x2 (a;2 _ ^) + ^^3 (j, _ 2) + y2 (43.2 4. 1).
i68
ON SQUARE ROOT.
225. "When .any fractional terms are in tlie expression of
which we have to liiid the Scjuare Root, we may pmceed as in
the Examples just given, taking care to treat the fractional
terms in accordance witli the rules relatiun to fractionsi
8 16
Thus to find the square root of '•''^ — K^ + o^•
9 81
.,8 16/ 4
X" — .c + ■
9 81
{^\
2x
8 16
9^ + 81
8 16
■9^*8l
Since
8 c_8_2_8 1_4
9^ 9 • 1~9''2~9
8 16
Or we niis^ht reduce x^^z + ^ to a single fraction, whicli
9 oi
would he
81.T72X416
8l '
and then take the square root of each of the terms of the
fraction, with the I'oUowin result :
9x4 , . , . ^, 4 '
—  — , which IS the same as x  .
Examples.— ixxxi.
I. 4a'' + ~  a^b.
lb
9 ^ a'
a 9
4 TT + 2 + T.
5. x*2x^^ + 2xx + 
3. «*2 + 7.
6. X* + 2x3.T+ ,.
4
353 54
7. 4a  12a& + ah" + 9)h  t + tt
' 2 lo
ON CUBE ROOT. '^9
16 32
8. x^ + 8x + 24 + ^ + ,.
X* x^
9 ^ . 16 p ., .^ , „ , ,16 ,
lb y i
1 4 9 4 6 12
a;^ y'' z xy xz yz
71 5 n^ 25 .)».
12. a^J^  Gffkrf + " ■' + 9c'ch + Jq ~.
4x2 j,2 9!/2 • ^ 6?/ 12xw
^^ X 3 X 2^
4??i2 9,i,2 16^ 247i
14. —+^ + 4 + — .
n nf n iii
cfi ^ h' c d? ah 2ac ad be bd cd^
^ 5 ■ ¥ "^ iTi "^ 2^ '*' T ~ 6 "*" Ts" ~ y ~ To "^ T " y *
1 6. 49x4  28x3  1 7x2 + 6x + ?.
4
1 7. 9x*  3ax^ + 66x3 + abx'^ + b^x^.
4
1 8. 9x*  2x3  l^lx'' + 2x + 9.
XVIII. ON CUBE ROOT.
226. The Cube Root of any expression is that expression
whose cube or third power gives the proposed expression.
Thus a is the cube root of a^,
3b is the cube root of 276^.
Tlie cube root of a negative expression will be negative, for
since
(a)3=— ax —ax a=a^,
the cube root of  a^ is —a.
170 ON CUBE ROOT.
So also
 3a; is the cube root of  27:c'',
and — 40^6 is the cube root of  %\a^}?.
The sA^mliol I] is used io denote tJae operation of extracting
the cube root.
Examples. — Ixxxii.
Find the Cube Roots of the following expressions :
I, Sal 2. 27xY 3 125m3n3.
4. 216ai263. 5. 34361^8. 6.  lOOOa^iVa.
7. 1728m2in24. 3. 133100/^13/
227. We now proceed to investigate a Rule for finding the
cube root of a compound algebraical expression.
We know that the cube of a + & is a^ + 3a5 + 3a6 + 5^,
and thereiore a + 6 is the cube root of a'' + 3a6 + 3a62 + ft'.
We observe that the first term of the root is the cube root of
the first term ol the cube.
Hence our rule begins:
"Arrange the terms in the orrler of magnitude of the indices of
one of flic quantities involved, then take the cube root of the first
term and set down the result as the first term of the root: subtract
its cube from the given expression, and bring down the remainder;"
thus
a^ + 3ab + 3ab^ + ¥{^a
a3
3ab + 3ab' + P
Now this remainder may be rejiresented thus,
b {3a + 3ab + b) ;
hence if we divi<1e 3ab + 3ab + P by 3a + 3ab + b^, v:e shall
obtain +b, the second term of the root.
Hence onr rule proceeds :
" Multiply the sqiiare of the first term, of the root by 3, and xet
doivn the result as the firU term of a divisor:" thus our process
up to this point will stand thus :
OiV CUBE ROOT. 171
a3 + 7,orh + 3a& + h^ (a
a
3a2 I 3a26+3a62 + 63
Now if we divide 3a6 by 3a2 the result is 6, and so we
obtain the second term oi the root, and if wc add to 3a2 the
expression 3a6 + 6' we obtain the full divisor 3ti + 3a6 + 6^.
Hence our rule proceeds tlius :
" Divide the first term of the remainder by the first term of the
divisor, and add the result to the first term of the mot. Then take
three times the product of the first and second terms of the root,
and also the square of the second term, and add these results to
the first term of the divisor." Thus our jsrocess up to this point
will stand thus :
a^ + 3a^6 + ZalP' + 6^ (^a + 6
«3
3a2 + 3a6 + 62
Za'h + ZaV' + l^
If we now multiply the divisor by h, we obtain
3a26 + 3a6 + 6^
which we subtract from the first remainder.
Hence our rule proceeds thus :
"Multiply the divisor by the second term of the root, and sub
tract the result from the first remainder :" thus our process will
stand thus :
a3 + 3a26 + 3aZ)2 + &3(,a + 6
a
3a2 + 3a6 + 62
3a 6 + 3a 6 + 1^
Zab + Zalfi + h^
K tliere is now no remainder, the root has been found.
If there be a remainder, consider the two terms of the root
already found as one, and proceed as before.
228. The following Examples may render the process more
clear:
172
ON CUBE ROOT.
Ex. 1.
3a212o + 16
rt3_i2a2 + 48ft64(a4
«3
12a' + 48a 64
12a2 + 48a64
Here observe that the second term of the divisor is formed
thus : r
3 times the product of a and — 4 = 3xax 4= — 12a.
Ex. 2. a;«  6r^ + 15a;* 20x3 + 15x2 6x+l(x22x+l
3x*  6x3 + 4_j2 _ (jj., + X5^.4 _ 20x3 + 15x2  6x + 1
6x'+ 12x^8x3
3x'*12x3
+ 15x26x+l
3x* 12x3 + 15x2 6x + l
3x*  12x3 f 15x2 6x + l
Here the formation of the Urst divisor is similar to that in
the preceding Examples.
The formation of tlie second divisor may be explained thus:
Regarding x2 — 2x as one terra
3 (:c2  2x)2 = 3 (x*  4x3 + 4^2) = 3_^4 _ I2x3 + 12x2
3x(x22x^xl = 3x26x
12 = i
and adding these results we obtain as the second divisor
3x* 12x3 + 15x2 6x + l.
Examples. — Ixxxiii.
Find the Cube Root of each of the following expressions:
1. a^2>d^hvZa\rh^. 2. 8a3 + 12a2 + 6a + 1.
3. o3 + 24a26 + 192a62 + 51263.
4. a3 ^ 3(^2^ + 3rt^2 + 53 + 3„2(. 4. 6a5g ^ 352(5 + 3^^^ + "ihc + cl
5. JC3  3x2)/ + 3j;j^2 _ y^ + 3j.2~ _ g_j.y.. ^ 3j^2~ + 3jv2!i _ 3y~2 + j^^
6. 27x"  54x° + 63x*  44.t3 + 2 1x2  6x + 1.
ON CUBE ROOT. 173
7. 1  3a 4 6a  7a3 + 6a*  3a'' + a".
8. x^  3xi/ + 3x1/2 _ ^3 + 833 + 6x^2  Vixxjz + G?/2 + 12x32 _ i2y;:2_
9. a"  12a^ + 54a*  1 12a3 + 108a2  48a+ 8.
10. 8»^''  367/t= + 66»i*  %Zm^ + 33??i2  9m + 1.
11. x^ + 6x.v + 1 2x.i/2 + 8?/3 _ 3x2  1 2xw2  1 2 v^s + 3a;~2 + ^y^i _^z_
12. %m?  36?7i?i + 547JiJi,2 — 2Tu' — Mmr + 36m?(? — 27'7ir
4 67?ir2 — 9?i7"2 — r'.
1 3. ?7i3 + 3??i2  5 H 5 =.
"^ ?/(." 7?l'*
229. The fourth root of an expression is found by taking
tlie sejuare root of the s(iiare root of the expression.
Thus 4/16aS6* = ^l^a^h" = 2a26.
The sixih root of an expression is found by taking the cube
root of the scj^uare root of the expression.
Thus 4/64ai266 = .^8a663 = 2a26.
Examples.— Ixxxiv.
Find the fourth roots of
1. 16u*96a3x + 216ax22T6ax3 + 81cc*.
2. l + 24a2 + 16a<8a32al
3. 625 + 2000X + 2400x2 + 1280x3 + 256a;^.
Find the sixth roots of
4. a«  ^w'h + 15a*62  ma?\? + ISa^t*  Gai^ + ^6.
5. x6 + 6x5 + 15x' + 20x3+ 15x2 + 6x + l.
6. m"  12771^ + 60771*  160?jr + 2407?i2  192m + 64
XIX. QUADRATIC EQUATIONS.
230. A Quadratic Equation, or an equation of two dimen
sions, is one into which the square of an unknown symbol
enters, without or with the tirst power ol' the symboL
Tims a;2=16
and xi6x = 27
are Quadratic Equations.
231. A Pure Quadratic Equation is one into which the
square of an unknown symbol enters, the fiist power of the
symbol not appearing.
Thus, x=16 is a. pure Quadratic Equation.
232. An Adfected Quadratic Equation is one into which
the square of an unknown symbol enters, and also the liist
power of the symbol.
Thus, x^ + 6x = 21 is an adfected Quadratic Equation.
Pure Quadratic Equations.
233. When the terms of an equation involve the square
of the unknown symbol oiibj, the value of this square is either
given or can be found by tlie piocesses described in Ciiapter
XVII. If we then e.xtract the square root of each side of the
equation, the value of the unknown symbol will be determined
234. The following are examples of the solution of Pure
Quadratic Equations.
QUA DRA TIC EQUA TIONS. I7S
Ex. 1. x^=\%.
Taking the square root of each side
x=±4.
We prefix the sign ± to the number on the righthand side
of the etjuation, for the reason given in Ait. 220.
Every pure quadratic equation will therefore have two roots,
equal in magnitude, but with different signs
Ex.2. 4a;2 + 6 = 22.
Here 4x' = 226,
or 4x^=16, '
or x = 4 ;
.. x=±2.
That is, the values of x which satisfy the equation are 2
and  2.
Ex. 3. '^^ ^^
Here 128 (5x26) = 216 (3x24).
or 640x2768 = 648x2864,
or x2= 12 ;
.. X=±V12.
Examples.— ixxxv.
I. x'^=Qi. 2. x2 = a252_ 3. x2 10000 = 0.
4, x23 = 46. 5. 5x29 = 2x2 + 24. 6. 3ax2=192a5c6.
x2 — 12 x24 o
7.  —  — = — . — . II. mx + n=q.
3 4 ^
8. (500 +x) (500 x) = 233359. 12. x2ax + 6 = ax(x 1)
8112 45 57
9 — =3x. 13. 2^J3 = 4^:r5
rl ■> ,„ r.. /^ OW 42 35
10. 5x18x + 6o = (3x3)2. 14. ^^32 = ^733
176 QUADRA TIC EQUA TTO.VS.
Adfeded Quadratic Equations.
235. Adfectecl Qnailratic Equations are solved by adding
a certain term to both sides of tiie e(.uatiou so as to make the
lefthand side a perl'ect sq^uare.
Having arranged the equation so that the first term on tlie
lefthand side is tlie square of tlie uui<no\vn symbol, and the
second term the one containing the lirst power of tiie unknown,
quantity (the known symbols being on the right of the equa
tion), we add to both ddes of the equation the square of half the
coefficient of the second term. The lefthand side of the equa
tion then becomes a perfect square. If we then take the square
root of both sides of tiie equation, we shall obtain two simple
equations, fiom which the values of the unknown symbol may
be determined.
236. The process in the solution of Adfected Quadratic
Equations will be learnt by tlie examples which we shall give
in this chapter, but before we proceed to them, it is desirable
that the student should be satisfied as to the way in wliich an
expression of the form
x^ + ax
is made a perfect square.
Our rule, as given in the preceding Article, is this : add the
square of half the coefficient of the second term, that is, the
square of 5, that is, ^. We have to shew then that
4
is a perfect square, whatever a may be.
This we may do by actually performing the operation of
extracting the square root of x'^ + ax + —, and obtaining the
result X + A with no remainder.
QUADRATIC EQirATIOiVS. 1 77
237. Let us examine this process by the aid of numerical
coefficients.
Take one or two examples from the perfect squares given
in page 48.
We there have
x^+ 18x+ 81 which is the square of x+ 9,
a;2 + 34x + 289 x+ll.
ic — 8x + 16 X— 4,
a;236u; + 324 cc18.
In all these cases the thinl term is the square of half the
coefficient of x.
For 81= (9)^ = (\^)',
289 = (17)^ = (=^y,
324 = (18)2 = (''y.
238. Now put the question in this shape. What must we
add to X' + ax to make it a perfect square i
Suppose b to represent the quantity to be added.
Then x'^ + ax + 6 is a perfect square.
Now if we perform the operation of extracting the square
root of x + ax + b, our process is
x^ + ax + hi x + 
X
2x + H ax + 6
2
a"
ax + r
4
'T
faA.] M
178 Q VADRA TIC EQUA TlONS.
Hence in order that x^ + ax + 6 may ba a perfect square we
must have
t
4
i?o,
or t=,
(ly
That is, 6 is equivalent to the square of half the coefficient
ofx.
239. Before completing the square we must be careful
(1) That the square of the unknown symbol has no coeffi
cient but unity,
(2) That the square of the unkno^vn symbol has a positive
These points will be more fully considered in Arts. 245 and
246.
240. We shall first take the case in Avhich the coefficient of
the second term is an even number and its sign positive.
Ex. a;l 6x = 40.
Here we make the lefthand side of the equation a perfect
square by the following i)rocess.
Take the coetficient of the second term, that is, 6.
Take the half of this coefficient, that is, 3.
Square the result, which gives 9.
Add 9 to both sides of the equation, and we get
x26x + 9 = 49.
Now taking the square root of both sides, we get
x + 3=±7.
QUADRATIC EQUATIONS. 179
Hence we liave two simple equations,
a; + 3=+7 (1),
and 35 + 3= 7 .'. (2).
From these we find the values of x, thus:
froui (1) x = 73, that is, x = 4,
from (2) x= — 7  3, that is, a;=  10
Thus the roots of tlie equation are 4 and  10.
EXAMPLES. — IXXXVi.
I. x + 6x = 72. 2. a;+12x = 64. 3. a;2 + 14x = 15.
4. x2 + 46x = 96. 5. x+128x = 393. 6. x + 8x65 =
7. x2+18x243 = 0. 8. x'^ + 16x 420 = 0.
241. We next take the case in which tlie coethcient of the
second term is an even number and its sign negative.
Ex. x^8x = 9.
The term to be added to both sides is (872)^, that is, (4),
that is, 16.
Completing the square
x28x+ 16 = 25.
Taking the square root of both sides
z4=±5.
This gives two simple equations,
a;4=+5 (1),
a;4=5 (2),
From (1) x=5+4, .. x = 9;
from (2) x=5 44, .. x=l.
Thus the roots of the e'j;iation are 9 and  1.
I
1 80 Q UADRA TIC EQ UA TIONS.
EXAMPLES. — IXXXVii.
I. a;26a: = 7. 2. xAx = ^. 3. a;220x = 21.
4. a;22x = 63. 5. a;2 12x+ 32 = 0. 6. x214x + 45 = ()
7. x''  234x + 13688 = 0. 8. (x  3) (x  2) = 3 (5x + 14).
9. x(3x17)x(2x + 5) + 120 = 0.
10. (x  5) 4 (x  "7) = X (x  8) + 46.
242. We now take the case in wliich the coefficient of the
second term is an oiii number.
Ex. 1. x27x = 8.
The term to be added to both sides is
Completing the square
, ,. 49 ^ 49
o ^ 49 81
or, x'^  7x + r = r
4 4
Taking the square root of both sides
7 .9
•"2=±2
This gives two simple equations,
7 9
^2=+2 <!>•
7_ 9 ^^
^~2~~2 ^''^•
From (1) ^"^9+2' or!^=9) •■•x = 8;
9 7 —2
from (2) x=   + , or, x = — , .. x= 1.
Thus the roots of the equation are 8 and  1.
QUA DRA TIC EQUA TIONS. l8l
Ex. 2, a;2x = 42.
The coefficient of the second term is 1
The term to be added to both sides is
/. a;'  x + :: = 42 + 
4 4
1 169
or, xx + ^= — ;
1 ^13
2  2
Hence the roots of the equation are 7 and —6.
Examples. — Ixxxviii.
I. a;2+7a; = 30. 2. x2llx=12. 3. x2 + 9x = 43.
4. x213x=140. 5. x2 + x = — . 6. x2x = 72.
7. x2 + 37x = 3690. 8. x2 = 56 + x.
9. x(5x)(2x(x7)10(x6) = 0.
10. (5x21)(7x33)(17x+15)(2x3) = 448.
243. Our next case is that in which the coefficient of the
second term is a fraction 0/ which the numerator is an even
number.
Ex. i?jx = 2\.
5
The term to be added to both sides is
4 4 4
5 2o 2o
„ 4 4 529
r82 Q UADRA TIC EQUA TIONS.
2 ^23
5  5
21
Hence the values of x are 5 and  =.
Examples. — Ixxxix.
„ 2 35 ,4 3 „ 28x 1 ^
I. x2x = g. 2. ^^ + 5^= 25 3. ^''9 + 3 = f^
,83_ ,43 „16 16
4 ^ n^ll = ^ 5 ^^ + 35^ = 7 6. x==y'; = y.
7. x2^x + ^ = 0. 8. a;2_4 ^45_
244. "We now take the case in which the coefficient of the
second term is a fraction u7iose numerator is an odd number.
Ex. ^W^
The term to be added to both sides is
2 7 49_13G 49
''•'' ~3'' + 36~ 3 "^36'
„ 7 49 1681
°' ^3^^36 = 36'
66
17
Hence the values of x are 8 and — ^.
Examples.— xc.
I. X22T=8.
2. x2j2; = 98.
5
3. .x2 + _a. = 39.
4. x^ + ^x=76.
g
5. x2x=16.
6. x^x + 6 = 0.
7. x2— X
34 = 0.
8.
^3 3
/ 4
QUADRATIC EQUATIONS. 183
245. The square of the unknown symbol tnust not be pre
ceded by a negative sign.
Hence, if we have to solve the equation •
6x — X = 9,
we change the sign of every temi, and we get
x26x= 9.
Completing the square
a;26x + 9 = 99,
or x^  6x + 9 = 0.
Hence a;  3 = 0,
or x = 3.
Note. We are not to be surprised at finding only one
vaJue for x. The iuterpretatiuu to be j>laced on such a result
is, that the two roots of the equation are equal in value and'
alike in sign.
24(5. The square of the unknown symbol must have no
coefficient but wiity.
Hence, if we have to solve the equation
5x23x = 2,
we must divide all the terms by 5, and we ^.et
, 3x 2
X" — — = .
o o
2
From which we get x = l and x= — .
247. In solving Quadratic Equations involving literal co
efficicnts of the unkiiown symbol, the same rules will apply as
in tlie cases of numerical coefficients.
Thus,' to solve the equation
?^?2 = 0.
X a
Clearing the equation of fractions, we get
2a2x22ax = 0;
therefore x^2ax= 2a\
or x^ + 2ax = 2a^.
r84 Q UADRA Tld EQUA TIONS.
Completing the sqiiare
X + 2ax + a^ = Sa^,
whence* x + a=^±. ^J'i • a ;
therefore a; =  o + ^3 . a, or x = — a  ^^3 . a.
The following are Examples of Literal Quadratic Equations.
EXAMPLES.— XCi.
7m
I. x2 + 2ax=a2. 2. x^ — 4ax = 7a2. 3. x^ + Zmx = ^ .
, 5n ._372^ _rt2 fe2
4. ^Y^~2~ ''■ (x + a)2 (x«)2*
5. x^ + (al)x = a. 8. adxacx = bcxbd.
6. x'^+ {ah)x = ab. 9. cx{ — y = (a + b)3:?^.
a"x^ 2ax b ^
10. To +2 = 0
^ , Sa^x 6a2 + a6262 JZx
11. abx{ = „ . .
c c^ c
12. (4a2  9cd) X + (4a2c2 + 4abd') x + (ac^ + bd^^ = 0.
248. If both sides of an equation can be divided by the
nnknowTi symbol, di\ide by it, and observe that is in that
case one root of the equation.
Thus in solving the equation
x^2x2 = 3x,
we may divide by x, and reduce the equation to the form
x22x = 3,
from which we get
x = 3 or .r=  1.
Then the three roots of the original equation are 0, 3 and  1.
We shall now give some Miscellaneous Examples of Quad
ratic Equations.
Q UADRA TIC EQUA TIONS. 185
Examples.— xcii.
I. x27a; + 2 = 10. 2. x5x + 3 = 9. 3. a;2llx7 = 5.
4. x213x(j = 8. 5. x' + 7x18 = 0. 6. 4x  "^ = 22.
x3
7. x9x + 20 = 0. 8. 5x3— ~4 = — J,— ■
x3 2
9. xGx14 = 2. 10. —^ —:!? = 2.
^ x^3 2x + 5
4x X 7
x + 7~2xT3'
14 2^^""~3^+~8 = ^ ^5 '^•^  = 26. 16. 2x = 18x40.
4 + 3x 15 — X 7x — 14 „ , o
3x5 ^f__l 7_2x5_3x7
'^' 9x 3x25~3' ^°' 4~^+y""~27~'
4X10 73x 7 , ^,., , ,,
21. ; =. 22. (x3, + 4x = 44.
x + o X 2 ^ ^
x+11 „ 9 + 4x ^ , „ ,11
21. —^='—^—' 24. 6x + x = 2. 25. x^x = ^.
26. x2x = 210. 27. — % +  = 3. 28. ^11=5.
' X 4 1 X 3 3
X 3 x1
= 15.
1
2
x + 2'
3
"5'
10
3 ¥"
14 2x
X
22
" 9"
12
8
32
30.
15x' — 7x = 46.
32.
4x 20  4x
5x X
34
X 7
x + 60 3x5'
— +  = 2
' 5x'4x xf2' •'"' 7x X 10'
JV ^ + 7^ = ,o 36. ;^— ; + — — = 2
37. x+(a + 6)x + rt6 = 0. 38. x2(6a)xa6 = 0.
39. x^  2ax 4 rt' — t = 0. 40. X  (rt a^)x — a^ = 0.
, a 2a „ a + 62
41. x2 + ^x^ = <. 42. x^x + l=0.
XX. ON SIMULTANEOUS EQUATIONS
INVOLVING QUADRATICS.
249. For the solution of Simultaneovis Equations of a de
gree hi^'Iier than the first no lixed rules can be laid down. We
shall ])oiiit out the methods of solution which may be adopted
with advantage in particular cases.
25('. If the simple power of one of the unknown symbols
can be expressed in terms of the other symbol by means of one
of the given equations, the Method of Substitution, explained
in Art. 217, may be employed, thus:
Ex. To solve the equations
x + y = 50
xy = 600. .
From the first equation
x = 50y.
Substitute this value for x in the second equation, and we
get (50  y) . y = 600.
This gives 50/ i/^ = 600.
From which we find the values of y to be 30 and 20.
And we may then find the corresponding values of x to be
20 and 30.
251. But it is better that the student should accustom
himself to work such equations symmetrically, thus :
To solve the equations
x + y = 50 (1),
x!/ = 600 (2),
From ( 1 ) x'^ + 2xy + y = 2500.
From (2) 4x7/ = 2400.
O^ SIMULTANEOUS EOUArrONS. &^c. 187
Subtracting, x^ — 2xy + y = 100,
:. xy=± 10.
Then from this equation and (1) we find
X = 30 or 20 and y = 20 or 30.
Examples. — xciii.
I. x + y — 40 2. x+v = 13 3. a; + i/ = 29
xj/ = 300. xy = m. xy=\ 00.
4. .c — 1/=19 5. xy = 45 6. x?/ = 99
•c?/ = 66. xy = 250. i^l/ = 1 00.
252. To solve the equations
xy=\2 (1),
a;2+!/2 = 74 (2).
From (1) '*x22.r;/ + i/2 = 144 V^)
Subtract this t'roui (2), then
2x1/= 70,
.• 4x1/= 140.
Add this to (3), then
x^ + 2xy + y = A,
:. X + 1/ = ± 2.
Then from this equation and (1) we get
X = 7 or 5 and y= — 5 or — 7.
Examples. — xciv.
I. xy = A 2. xy=l() 3. x«=14
x2 + 2/2 = 4o. x2 + 2/2 = 178. x2 + i/2 = 436.
4. xti/ = 8 5. x + y = l2 * 6. x + i/ = 49
x2 + i/2 = 32. x2 + i/2=104. ' x2^y2=,ie81.
i88 ON SIMUL TANEOUS EQUA TJONS
253. To solve the equations
a^ + 2/3 = 35 (1),
a; + 2/ = 5 (2).
Divide (1) by (2), then we get
x^xy + f^l (3),
From (2) a;2 + 2xi/ f i/^=25 (4),
Subtracting (3) from (4),
3a;2/ = 18,
.". 4^2/ = 24.
Then I'rum this equation and (4) we get
x''^2x?/ + i/2= 1,
:.xy=±\;
andl'rom this etiuation and (2) we find
x = 3 or 2 and ^ = 2 or 3.
Examples.— xcv
I. ftr' + i/3 = 91 2. a;^ + i/ = 341 ' 3. x3 + i/3 = 1008
x + y = l. x^y = \\. x + i/=12.
4. ic3_y3 = 5(J 5, 3.3_^3 = 98 6. x3i/3 = 2T9
xy = % xy = 2. xy = S.
'254, To solve the equations
s+r6 ^'^'
1 1 13 ,,
F + ?=3«' ^')
From (1), by squaring it, we get
1 2 1 25 ,.,,
^ + — + 2=^ (3).
x^ xy y' 3d
From this subtract (2), and we liave
^_'12
xi/ 36 ' .
^_24
xn 36'
IXVOLVING QrADA\4 7/CS. 189
Now subtract this from (3), and avo gi't
x^ xy y'^ 36'
X y  o •
and from this equation anil (1) we find
x = 2 or 3 and i/ = 3 or 2.
Examples. — xcvi.
I.
1 1 9
x'^y~20'
2.
1 1_3
X y~A'
3
1 1 r.
X 1/
1 1 41
1 1 5
x2"^/~16'
^ + 4 = 13.
X 1/
4.
1 1_ 1
X y~\2
5
11=2^
X y 2
6.
11 = 3.
X ?/
I 1 7
x^ 1/ 144'
x^ ?/^ 4"
X^ i/^
255. To solve the equations
x2+3xj/=7 a),
^ + 4y'=18 (2}
If we add the equations we get
x + 4xi/ + 4!/ = 25.
Taking the square root of each side, and taking only the
positive root of the riglithand side into account,
X + 2?/ = 5 ;
.•. x = 5 ly.
Substituting this value for x in (2) we get
(522/)2/ + 4!/2=18,
an equation by which y may be determined.
Note. In some examples we mut subtract the second
equation from the first in order to get a perfect square.
190 ON SIMULTANEOUS EQUATIONS
256. To solve the equations
x^f = i^ (1),
a;2 + xi/ + i/2=p (2).
Dividing (1) by (2) we get xi/ = 2 '3),
squaring, x''±xy\y = \ (4}.
Subtract this from (2), and we have
3^2/ = 9;
.'. 4x?/ = i2.
Adding this to (4), we get a;2 + 2.rj/ + ?/2= 16 ;
.. X + ?/ = ± 4.
Then from this equation and (3) we lind
a; = 3 or — 1 , and ;/ = 1 or — 3.
257. To solve the equations
a;2 + y2 = (;5 ^Y\
01/ = 28 (2,.
Multiplying (2) by 2, we have
! + i/2 = 65)
2a:?/ = 56) '
.. X2 + 2X!/ + J/2=121^_
a;22x2/ + i/= 9) '
.. x + i/=±ll (A),
xxj=± 3 (B).
The equations A and B furnTsh four pairs of siiii])lr
equations,
x + i/=ll, .r + ?/=ll, a + !/=ll, a: + )/=ll,
xi/ = 3, x2/=3, x2/ = 3, x?/=3.
from whicli we find the values of x to be 7, 4, 7 aixl 4.
and the corresponding values of ?/ to be 4, 7. —4 and  7.
258. The aititiee, l\v wliich the solution of the equation.^
eiven in this article is eti'ected, is a])plicable to cases in wliich
the equations are homogeneonx mid <>/ the savie orilcr.
INVOLVING QUADT?ATICS. 19^
To solve the equations
x2 + xy = 15,
Suppose y = mx.
Then x'^ + mx^=l^. from the first eonation.
and mx'^ m'^x^ = z, from the secona equation.
Dividing one of these equations by the other,
x^ + mx^ _15
mx^  m^x~ 2 '
x^n+m) 15
or ^^ — = —
x^ {m  m^) 2 '
1 + m .15
o^ 2= o
m — m^ 2
From tliis equation we can determine the values of m.
2
One of these values is ^, and putting this for m in the
o
2
equation x + 7nx= 15, we get x'^ + x''=l5.
From which we find a;= ±3,
and then we can find y from one of the original equations.
259. The examples which we shall now give are intended
as an exercise on the methods of solution explained in the
four preceding articles.
Examples.— xcvii.
I. a;^ — y^ = 37 2. x + 6x?/ = 144 3. x'^ + xy = 2l0
x + xy + y = 37. 6xy + 36?/ = 432. y + xy = 23 1.
4. a;2 + 7/2 = 68 5. x^ + y^=l52 6. 4x + 9xy=l90.
xy=l(J. x^xy + y=l9. 4x5y = l0.
7. x^ + xy + y = 39 8. x^ + xy = 6Q 9. 3x + 4x!/ = 20.
3y'5xy = 2o. xy — y' = b. 5xy + 2y = l2.
\o. x^xy + y = 7 11. x'^ — xy = 35 12. 3x^ + 4xy + 5ij = 7l.
:ix^ + l3xy + 8y = liy2. xy + y^ = 18. ox + 7y = 29.
i2,.'X^ + y = 212S 14. u;2 + 9xy = 340 15. x^ + y==225
x^xy + y=l24:. 7xy y'^=\7l. xy=]()8.
XXI. ON PROBLEMS RESULTING IN
QUADRATIC EQUATIONS.
260. The method of stating problems resiiltiug in Quad
ratic Equations does not require any general explanation.
Some of the Examples which we shall give involve ane
unknown symbol, others involve tivo.
Ex. 1. What number is that whose square exceeds the
number by 42 ?
Let X represent the number.
Then x^ = x + 42, ^
or, a;'''x = 42;
thereiore x — x +  = —j ;
4 4
whence x  ^ = ± ^.
And we find the values of .'• to be 7 or — 6.
Ex. 2. The sum of two numbers is 14 and the sum of
their squares is 100. Find the numbers.
Let X and y represent the numbers.
Then xf!/ = l'i,
and .x2 4r = i00.
Proceeding as in Art. 252, we find
x = 8 ur 6, y = 6 or 8.
Hence the numbers are 8 and 6.
ON PROBLEMS RESULTING, dj'c. 193
Examples. — xcviii.
*<j \ I. What number is that whose half multiplied by its third
part gives 864?
2. What is the number of which the seventh and eiglith
parts being multiplied together and the product divided bj'
2
3 the quotient is 298^ ?
^ 3. I take a certain number from 94. I then add the
number to 94.
I multiply the two results together, and the result is 8512.
What is the number ?
{.4. What are the numbers whose product is 750 and the
quotient of one by the other 3 ?
5. The sum of the squares of two numbers is 13001, and
the difference of the same squares is 1449. Find the numbers.
* 6. The product of two numbers, one of which is as much
•^ above 21 as the other is below 21, is 377. Find the numbers.
A \ 7. The half, the third, the fourth and the fifth parts of a
^certain number being multiplied together the product is 6750.
Find the number.
8. By what number must 11500 be divided, so that
the quotient may be the same as the divisor, and the re
mainder 51 .'
g. Find a number to which 20 being added, and from
which 10 being subtracted, the square of the first result added
to twice the square of the second result gives 17475.
10. The sum of two numbers is 2G, and the siim of their
squares is 436. Find the numbers.
11. Tlie difference between two numbers is 17, and the
sum of their squares is 325. What are the numbers ?
1 2. What two numbers are they whose product is 255 and
the sum of whose squares is 514 ]
a/ '3 Divide 16 into two parts such that their product
' added to the sum of their squares may be 208.
[S.A.] N
194 ON PROBLEMS RESULTING
\
14. What number added to its square root gives 'as a
result 1:332 ]
3
15. What number exceeds its square root by 48^?
16. What number exceeds its square root by 2550 ?
^ 17. The product of two numbers is 24, and their sum
niultiiilied by their difference is 20. Find the number.*.
18. What two numbers are those whose sum multiplied
(; by the greater is 204, and whose difference multiplied by the
less is 35 ?
f\ 19. What two numbers are those whose I'.ifference is 5
'and their sum multiplied by the greater 228 ?
; 20. Find three consecutive numbers whose product is
V equal to 3 times the middle number.
^ 21. The difference between the .squares of two consecutive
niimbers is 15. Find the numbers.
3 2. The sum of the squares of two consecutive numbers is
481. Find the numbers.
23. The sum of the squares of three consecutive numbers
is 365. Find the numbers. '
Note. If 1 buy x apples for y pence,
 will represent the cost of an apple in pence.
If I buy X sheep for z pounds,
 will represent the cost of a sheep in pounds.
Ex. A boy bought a number of oranges lor 16(f. Had he
bought 4 more for the same money, he would have paid
onethird of a penny less for each orange. How many di<l
he buy ?
Let as represent the number of oranges.
Then — will represent the cost of an orange in pence.
„ 16 16 1
Hence — =  — , + ^,
X a; + 4 3
or 16(3x + 12) = 48x + x2 + 4x,
or x2 + 4.i = 192,
from which we find the values of x to be 12 or — 16.
Therefore he bouglit 12 oranges.
IN Q UA DRA TIC EQCA TIOXS. 19$
24. T buy a number of handkerchiefs for £\l. Had I
bought 3 more for tlie ^^anle money, they would have cost one
shilling each less. How many did I buy '{
25. A dealer bought a number of calves for £80. Had he
bought 4 more for the same money, each calf would liave cost
£\ less. How many did he buy ?
26. A man lx)ught some pieces of cloth for £33. 15s.,
which he sold again for £1. 8.?. the piece, and gaiueil as much
as one piece cost him. What did he give for each piece ?
27. A merchant bought some pieces of silk for £180.
Had he bought 3 pieces more, he would have paid £3 less for
each piece. How many did he buy ?
28. For a journey of 108 miles 6 hours less would have
sufficed had one gone 3 miles an hour faster. How many
miles an hour did one go ?
29. A grazier bought as many sheep* as cost him £60.
Out of these he kept 15, and selling the remainder for £54,
gained 2 shillings a head by them. How many sheep did
be buy ?
30. A cistern can be filled by two pipes running together
in 2 hours, 55 minutes. The larger pipe by itself will fill it
sooner than the smaller by 2 hours. What time will each
pipe take separately to fill it ?
31. The length of a rectangular field exceeds its breadtli
by one yard, and the area contains ten thousand and one
hundred square yards. Find the length of the sides.
32. A certain number consists of two digits. The left
hand digit is double of the righthand digit, and if the digits
be inverted the product of the number thus formed and the
original number is 2268. Find the number.
33. A ladder, whose foot rests in a given position, just
reaches a window on one side of a street, and when turned
about its foot, just reaches a window on the other side. If tlie
two positions of the ladder be at right angles to each other,
and the heights of the windows be 36 and 27 feet respectively,
find the width of the street and the length of the ladder.
tg6 ON PROBLEMS RESULTING, drr.
34. ('lot]), bein<,r wetted, shrinks up  in its length and
o
~ in its width. If the surface of a piece of cloth is di
3
minished by 5 square jards, and the length of the 4 sides
by 4 yards, what was the length and width of the cloth %
35. A certain number, less than 50, consists of two digits
whose difference is 4. If the digits be inverted, the difference
between the squares of the number thus formed and of the
original number is 3960. Find the number.
36. A plantation in rows consists of 10000 trees. If there
had been 20 less rovvs, there would have heen 25 more trees in
a row. How many rows are there ?
37. A colonel wished to form a solid square of his men.
The first time he had 39 men over: the second time he in
creased the side of the square by one man, and then he found
that he wanted 50 men to complete it. How many men were
there in the regiment ?
XXII. INDETERMINATE EQUATIONS.
261. WHEisr tlie number of unknown symbols exceeds that
of the independent equations, the number of simultaneous
values of the symbols will be indefinite. We propose to ex
plain in this Chapter how a certain number of these values
may be found in the case of Simultaneous Equations involving
two unknown quantities.
Ex. To find the integral values of x and y which will
satisfy the equation
3x + 7y=lO.
Here 3a;=107/;
.. x=32j/ + ^^.
Now if X and y are integers, — must also be an integer.
INDE TERMINA TE EQUA TIOXS. 197
1 —1/
Let — ;— = in, then 1 — ^ = 3?7i ;
.". i/ = 1 — 3m,
and a; = 3 — 2i/ + m = 3 — 2 + 6m + ?^^ = 1 + 77w ;
or the general solution of the equation in whole numbers is
x = l + 1 m and y = \ — 3?7i,
where rii may be 0, 1, 2 or any integer, positive or
negative.
If m = 0, x= 1, 1/= 1 ;
if m=l, x= 8, 3/= 2;
if m = 2, x= 15, 1/= 5;
and so on , from which it appears that the only positive inte
gral values of x and y which satisfy the equation are 1 and 1 .
262. It is next to be observed that it is desirable to divide
both sides of the equation by the smaller of the two coefficients
of the unknown symbols.
Ex. To find integral solutions of the equation
lx>rby = Z\.
Here by = ^\lx:
1  2x
1 — 2e
Let — ^ = m, an integer.
Then 1 2x = 5m, whence 2x=l 5fli;
1 — m
2 2"^
T . 1 — m . ,
Let =n, an integer.
Then 1 m = 2n, whence m = l 2n.
Hence x = n27?i = ?i — 2 + 4n = 5n — 2 ;
y = 6x + )n = 65?il2 + l2«, = 97».
Now if n = (K x= 2, 2/=: 9;
if n = l.x= 3,2/= 2;
if n=2, X— 8,ys:~5.
and 80 on.
198 INDE TERMLVA TE EQUA TIO.VS.
263. In how many ways can a person pay a bill of £13
with crowns and guineas?
Let X and y denote the number of crowns and guineas.
Then 5a; + 21?/ =260;
.. 5a; = 260211/;
x = 524v.
^ 5
Let ^ = m, an integer.
Then y = 5m,
and x = 524y — m = 52 21m.
If 771 = 0, 2 = 52, y= 0;
m=l, x = 31, y= 5;
m = 2, a;=10, y = 10;
and higher values of m will give negative values of x.
Thus the number of ways is three.
264. To find a number which when divided by 7 and 5
will give remainders 2 and 3 respectively.
Let X be the number.
x'2
Then — „— =an integer, suppose m;
I
and =an integer, suppose n.
Then x — 7ni + 2 and x = 5?i + 3;
.. 7?rt + 2 = 5?i + 3;
2ml
:. 5n = 7m  1, whence n = m + 
5
Let — ' — =p, an integer.
Then 2m = 5p+l, whence m=2p »— g— v
Let ~n— — 9.} *^ integer.
Then j9 = 2gl,
m = 2) + (7 = 4(72 + 7 = 5g2,
x=7m + 2 = 357 12.
INDE TERM IS' A TE EQ L 'A T/ONS. I99
Henct if
» q = 0,x=l'2;
if
Q = \,x= 23;
if
q=2,x= 58;
and so on.
Examples. — xcix.
Find positive integral solutions of
I. 5x + 72/ = 29. 2. 7x+192/ = 92.
3. 13x+19!/=ll70. 4. 3x + 5i/ = 26.
5. \Axby = l. 6. llx+15?/ = 1031.
7. llx + 7i/ = 308. 8. 4x19?/ = 23.
9. 20x9!/ = 683. 10. 3x + 77/ = 383.
II. 27x + 4i/ = 54. 12. 7x + 9^ = 653.
13. Find two fractions with denoiuinators 7 and 9 and
their sum ^~.
DO
14. Find two proper fractions with denominators 11 and
82
13 and their difference 77:.
14.3
15. In how many ways can a debt of £\. 9s. be paid in
florins and halfcrowns ?
16. In how many ways can £20 be paid in halfguineas
an(f halfcrowns ?
17. What number divided by 5 gives a remainder 2 and
by 9 a remainder 3 ?
18. In how many different ways may £11. 15a. be paid in.
guineais and crowns ?
19. In how many different ways may £4. lis. Qd. be paid
with halfguineas and lialfcrowns .'
20. Shew that 323x 527?/ =1000 cannot be satisfied by
integral values of x and y.
INDETERMINATE EQUATIONS.
21. A farmer buys oxen, sheep, and hens. The whole
number bought was 100, and the whole price .£100. If the
oxen cost .£5, the sheep ;£1, and the hens Is. each, how many
of each had he? Of how many solutions does this Problem
admit ?
22. A owes B 4s. lOd.; if A has only sixpences in his
pocket and B only fourpenny pieces, how can they best settle
the matter ?
23. A person has £12. 4s. in halfcrowns, florins, and shil
lings ; the number of halfcrowns and florins together is four
times the number of shillings, and the number of coins is the
greatest possible. Find the number of coins of each kind.
24. In how many ways can the sum of £h be paid in
exactly 50 coins, consisting of halfcrowns, florins, and four
penny pieces \
25. A owes B a shilling. A has onlj' sovereigns, and B has
only dollars worth 4s. 3d. each. How can A most easily pay Bl
26. Divide 25 into two parts such that one of them is
/ divisible by 2 and the other by 3.
27. In how many ways can I pay a debt of £1. 9s. with
crowns and florins ?
:b
28. Divide 100 into two parts such that one is a multiple
of 7 and tlie other of 11.
29. The sum of two numbers is 100. The first divided by
3 5 gives 2 as a remainder, and if we divide the second by 7 the
remainder is 4. Find the numbers *
^ 30. Find a number less than 400 which is a multii^li^ <•» 7^
'f and which when divided by 'z, o, *, 5. 6. gives as a iciiiiuuder
in each case 1.
XXIII. THE THEORY OF INDICES.
265. The number placed over a symbol to express tlie
power of the symbol is called the Index.
Up to this point our indices have in all cases been Positive
Whole Numbers.
We have now to treat of Fractional and Negative indices ;
and to put this part of the subject in a clearer light, we shall
commence from the elementary principles laid down in Arts.
45, 46.
266. First, we must carefully observe the following results :
(a3)2=a6.
For a^ X a^ = a . a . a . a . a = a^,
and (a^y = a^.a^ = a.a.a.a.a .a=a^.
These are examples of the Two Rules which govern all
combinations of Indices. The general proof of these Rules we
shall now proceed to give.
267. Def. "When m is a positive integer,
a" means a, a. a with a written m times as a factor.
268. There are two rules for the combination o^ndices.
Rule I. a'"xa" = a''^.
Rule II. {'(*")••=«.
269. To prove RvLE 1.
a"^ = a.a .a to m factors,
a''a^a.a to /i factors.
402 THE THEORY OF INDICES.
Therefore
^" X a" = (rt . o . a to m factors) x (a . a . a to ?i. factors)
= a .a. a io ^m + n) factors.
= a'"+", by the Definition.
To prove Rule 11.
(a'")" = a'' .oT' .a" tc n lactors,
= (a.a. a m m taeiors) (a . a . a ... to m factors) . . .
repeated n times,
= a.a .a to mn factors,
= 0"", by the Definition.
270. We have deduced immediatehj from the Definition
that when m and n are positive integers a" x a'' = a'*+ . When
m and n are not positive integers, tlie Definition has no mean
ing. We therefore extend the Definition by saying that a" and
a", whatever m and n may lie, shall be such Uiat a" x a' = a"*+",
and we shall now proceed to shew what meanings we assign to
a" in consequence of this definition, in the following cases.
p
271. Case I. To find the meaiiing of a', p and q being
positive integers.
? p p,p
a''xa'> = a'' ',
P P T fj.? ? ?+?+?
«» X «' X o» = a» 1 xa'' = a'' » »;
and by continuii^g this process,
xa'x to (/ lactors = a» « «
But by the nature of the symbol 4/
i^a^ >^ ^a'' X to q factor8 = a';
p p
:. a^xa'' X to q factors = ^/a' x ^o' x . . . to ^ factora ;
p
THE THEORY OF INDICES. 203
272. Case II. To find the meaning of a~\ s beirig a po.si.
tive number, ivhole or fractional.
We must first find the meaning of a".
We have
Now
a
"xa"
.•. </,"
— i.
a'
X a~'
= 1;
:. a~'
_ 1
~ a''
273. Thus the interpretation of a*" has been deduced Irom
Rule I. It remains to be proved that this interpretation
agrees with Rule II. This we shall do by shewing that Rule
II. follows from Rule I., whatever m and n ma\' be.
274. To shew that {a"')" = a'"" for all values of m and ?i.
(1) Let n be a positive integer : then, whatever m may be,
(ft"*)" = a" . rt"* . o"" to n factors
™mf"ii"»+ ... to n t«rm3
(2) Let n be a positive fraction, and equal to ~,p andV
being positive integers ; then, whatever be the value of m,
^  ^ + ^+...109 terms
(a"*)' X («")« X to 5' factors = (a")' '
= «"•", by (1).
But a' xa' x to 5' factors = a ' '
that is, Ca''y = a'^.
204 THE THEOR V OF INDICES.
(3) Let n— —s., s being a positive number, whole or frac
tional : then, whatever m may be,
(«") =  — _ bv Art. 272,
(a*")" ^
= ;j, by (1) and (2) of this Article ;
that is, (0°*)"= ^— 
275. We shall now orive some examples of the mode in
which the Theorems established in the preceding articles are
applied to particular cases. We shall commence with exam
ples of the combination of the indices of two .single terms.
276. Since x'" x x" = x'''+",
(1) x" X af— = yf^' = X*.
(2) X' X X = x'+i.
(4) ft^—'.fc" ''xa"'".6''".c
= a'""+""'.6"''+^".C
= 1.1. c
= c.
Since (x"')" = x"",
(1) (.c6)3 = x6^3 = a.ij.
(2) (x^)^ = x*'^=.o'.
(3) {a^^ = a"'^=^(iK
278. Since x' = 4/x%
(1) x^= Vr*.
(2) x^=^i2;
THE THEORY OF INDICES. 205
Note. When Examples are given of actual numbers raised
to fractional powers, they may often be put in a form more fit
for easy solution, thus :
(1) 144^ = (144)3 = (V141)'=12'=1728.
(2) 125^ = (125^)''i = ( 4/125)''^ = 5 = 25.
279. Since (ic"*)" = a;™",
(1) j(a;"')"j'' = (x""')'' = x""'P.
(2) {(«"*)"}'■ = («'""/ = «"•"'.
(3) I (x")" i*" = (x'"") " = X"""'.
280. Since x" = — ,
X"
we may replace an expression raised to a negative power by
the reciprocal (Art. 199) of the expression raised to tlic. same
positive power : thus
(1) ai = . (2) a^= \. (3) a~^= \.
Examples. — c.
(1) Express with fractional indices :
1 . ^x5 4 4/x2 + ( Jxy. 3. 4/^^ A ( ^'af + a J^.
(2) Express with negative indices so as to remove all p<n\ er.s
trom the denominators :
1 a 6^ 3 o(? 5x^ X
X x^ x'^ X* '^ 42/3'' Tt/s* yz
x^ 3x 4 ocy I z
y^ y^ y*' ^^^ bx^y'^ x?y^'
(3) Express with negative indices s< as to remove all powers
from the numerators ;
206 THE THEORY OF INDICES.
1 x X? x* 4o6 3f( l]x
t t f M^ ih^^ ^/(T'ofi
(4) Express with rootsymbols and positive indices :
i o » I
2 12 X^SXX*
y^ yi 3yi •
2 i f
I o X X X
2. x"3 + (/~S + 23. 4 "X "^ ir^ "*" ~^'
281. Since x"^x" = — = x'" .x~" = x"",
X"
(1) X«=X^ = X«3 = X*.
(2) x3HxS = x38 = X5 = 1
(3) x"^x"" = x"''""'^x"''^" = x".
(4) a'ra'^ = a'<^' = a'»^=a=— .
(5) x*^x^ = x5 ^ = x*
(6) X^^X^ = X^"6=X^"^ = X~^ = X~^=^
x^
282. Ex. Multiply a^ a' + a'\ by a' + 1,
a^ — a^ + a'l
o'+l
a*'  a*' + a*'a'
ay1
EXAMPLES.— Ci.
Multiply
1. x'' + x'y' + T/*' by .r*'  x'j/' + y*.
2. a'" + 3a^y' + 9at/*" + 27y'" by a"  Sy".
3. x** 2ax*' + 4a« by x*^ + 2ax='' + 4a\
THE THEORY OF INDICES 207
a"* + 5" + c' by a"'  6" + c".
a" + 6"  2c' by Sa"*  6 + c''.
x*" — x"i/" + 2/" by x^" + ar''^/" + 1/**.
ap*+p _ 5?° 4. cp by a''"' + 6*'' + c^"*.
Form the square of ar''' + x' + 1.
Form the square of x^ — x^ + 1.
283. Ex. Divide, x*"  1 by x"  1.
x"  1) x*"  1 {x'" + x'*" + X' f 1
x^p .
1
a?p.
x^
7?"
1
T?"
xJf
xf
1
af
1
Examples.— cii.
Divide
I . X*" — y*"* by X"'  y". 3, x*'  y'' by x'  y.
10
x"* + 1/'" by x" + y\ 4. o"'' + 6'°' by a^*" + 6'
x'"  243 by x"  3.
a*" + 4a"'x^" + 16x*" by a*" + 2a"'x" + 4x^"
Qx" + 3x*'' + Ux'" + 2 by 1 + Sx' + x^''.
14&*"'c'"  ISi'^c'"*  56''" + 4b"'c'"' by 6»~ + 6"c^'"  26^'"c"
Find the square root of
a*" + Ga'"" + IQa*"" + 20a^ + 15a'» + Ga" + 1.
Find the square root of
2o8 THE THEOR V OF INDrCES
I
Fractional Indices.
284. Ex. Multiply J  ah^ + b^ by J + 6*.
Jah^ + b^
a^ + b^
a  a%* + 0*6*
+ ah^ah^ + b
a +b
Examples. — ciii. '
Multiply
1. x^2x^+lhj x^l.
2. 2/* + 2/^ + ?/^+ 1 by J/* 1.
3* a*  x^ by a^ t a^x^ + x^.
4. a^ + b^ + c^  a*b^  a^c^  b^c^ by a* +b^ + c^.
5. 5x^ + 2x^y^ + 3x^1/2 + 7y^ by 2xi  ??/*.
4 31 sa 12 4, a 1
6. to"' + TO'7i' + 7*i,^?i^ + 7?i"?i'^ + "" by TO"  n",
7. m^  2dhn^ + 4d by m^ + 2dhn> + ld~.
8. 8 J + 4ah^ + 5ah^ + 96^ by 2«^  36*.
Foim the square of each of tlie following expressions :
9. x^ + a^. 10. x^a^. II. x'^ + y^'.
12. a + ti 13. x22x* + 3. 14. 2x' + 3x'+4.
11;. x^y^ + zK 16. x* + 2i/Ja*
THE THEoky OF INDICES.
2c 9
285. Ex. Bividt abby ija i/b.
1 1
Putting a^ for ^'a, and b^ for i/b, we jjroceed thus ;
J b^)ab{J + ah^ + aihi + },i
3 1
aa^b'^
ah^b
ah^a^b^
ah^b
a^b^ah^
ah^  h
ah^b
EXAMPLES.— Civ.
Divide
1. xyhy x~y^
2. a — bhy o^ + 6*
3. xy \)y x^ y'
4. a + b by a^ + b^
1 I
5. x + yhyx^+y''
1 1
6. m — n by ?)i^ — 7i''.
.s _ a,/i.
7.x Sly by x^  3y^.
8. 81a166by 3«5^2Ai.
9. ax by a;^ +a~.
1
10. 1*1 — 243 by m" — 3.
1 1
11. a;+17x + 70 by a:2 + 7.
12. x^ + x^  12 bv X* — 3.
13. 63 _ 3t 5 + 36 _ 5I by b^  1.
14. x + y+z 3.'?:3y3;.'3 ^y x* + 1/* + 2^.
g 1 1
15. X  5x3  46x3  40 by x^ + 4.
1 .!_ 1 1 i 1
16. m + m^n^ + n by m2 m*ri* + ?i2,
17. ^  4^)* + 6p2 •_ 4pi + 1 by ^  2pi + 1.
18. 2x + x^y^ ~3y 4i/^3^  xh^  2 by 2x^ + 3^/2 + z^.
4 31 2g la 4.
19. x + 1/ by x"^ x^y^ +x^y" x'^y^ +y".
Sio THE THEORY*OF INDICES.
Negative Indices.
28(5. Ex. Multiply x~^ + x~y~'^ + x~'^y~^ + y~^ by x"' — y~^.
x^^ + x~~y~^ + a:~^T/~^ + y~^
 x~^i/~^ — x~^y~^ — x~'i/~^  y"
x"*i/~*
Examples. — cv.
Multiply
I. «i + 61 by (ii  61. 2. x3 + 62 by xr^  62.
3. x^ + x + xi + x^ by xxi. 4. X I4X2 by x2+ l+x~2.
5 . a2 + 62 by a2  62. 6. a"!  61 + fi by ai + fti + ci.
7.1 + «6i + a62 by 1  a6i + a'^b^.
8. a262 + 2 + a262 by a262  2  a'b.
9. 4x3 + 3x2 + 2xi + 1 by x2  xi + 1.
r o. ^x2 + 3xi  1 by 2x2  x~i  J.
2 3 ^ 2
287. Ex. Divide x^ + l+x' by xl+ x~\
Xl+XlJ x2+l +X2 (^X+1+X~l
X2  X + 1
X + X2
X  1 + x^
lxi + x3
lXl + X2
Note. The order of the powers of a is
a', a^, (1(1, a", a~^, a~', a'^'..
u serii^s which may be written thus
3 2 1 1 1 1
a a" a**
THE THEOR Y OF INDICES.
EXAMPLES.— CVi.
Divide
I. a;2  X" by *; t i ' j. > 6~^bya — 6~^
3. 771^ + ?i~^ by 7?i + 7^~*. 4. c^  tZ~^ by c  d~^.
5. x^~^ + 2 + x~y'^ by x?/~i + x~^y.
6. a* + a'^ts + h* by a^ _ ^i^i + 52.
7. x^y~^  x~^y^ — Sxy~^ + '3x~^y by xy~^  x~hj.
^ 3x5 . . 77x3 43X2 33xi „,
g. — 4x4g J+27
a;2
by 7:;^ — x~'^ + 3.
^ 2
g. a^6~^ + «~^6^ by «6~^ + rt"'6.
10. a~^ + 6"^ + c~^  3a~^6~^c'i by a~^ + ft^ ^c~^
288. To shew that (rt5)'' = a". J",
(at)" = a6.a6. a?)... to 71 factors
= (a . a . a . . . to n iactors) x (6 . 6 . 6 . . . to ti factors}
= a" . 6".
We shall now give a series of Examples to introduce the
various forms of combination of indices explained in this
Chapter.
Examples. — cvii.
1 . Divi de x^  4x!/ + Ax^y + Ay^ by x* + 2x~y^ + 2y.
_i_ _i_
2. Simplify )(x»"*)3.(x6)'j3». 3. Simplify (.r^o* . xi«^)^l
j _i i_U
,. ) 1 1 x + a xa
4. Simpliiv < TT — 5 — ^ 2 r: — 3
^ ^  jx''a^ x^ + a'' x + a
\ a
THE THEORY OF INDICES.
5. Multiply x2 + 4xi  1 by isr'^  2xr^  \.
01 it
af^' i""* x*~^
6. Simplify ' — ^ . 7. Divide x^  2/"" by x" + ?/".
8. Multijjly (a^ + 6^)' by a^  1^.
9. Divide a — 6 by 4^rt  4/^. 10. Prove that (a^)" = (a").
11. If a"'" = (a'")", find 7?i iu terms of n.
1 2. Simplify cc''+*+' . a;*+'~' . x'*~'+' . x*^^.
13. Simplify(^j^(^,^) . 14. Divide 4^' by —.
15. Simplify [j (a") }^][ j (a"')" I"']
1 6. Multiply iC + 1" 2c" by 2a"'  36.
17. Multiply a'""})"" by a"^6'>"c.
18. Shew that +<^^")^:("^^^^ = ^^^.
19. Multiply x^ + x^ + 1 by x^  x^ + 1
and their product by x^  x^ + 1.
20. Multiply a"  6a"— ^ x + ca"— ^ a;2 by a" + 6ti"' x  ca''h^.
2 1 . Divide x^*"*^'  i/2«<^ii by x*^*"" + !/«<'".
22. Simplify j (a")" "•i'»+i.
23. Multiply x^"" + x'^yf" + x'y^ + y"'' hy af—y'.
24. Write down the values of 625^ and 12~^.
25. Multiply .••'"•'"•  2/'— 1)" by x"  y".
26. xM u] tiply x^ + 3.C  1 by x^  :>x"i.
XXIV. ON SURDS.
289. All numbers which we cannot exactly determine,
because they are not multiples of a Primary or Subordinate
Unit, are called SurdS.
290. We shall confine our attention to those Surds which
originate in the Extraction of roots where the results cannot
be exhibited as whole or fractional numbers.
For example, if we perform the operation of extracting the
square root of 2. we obtain 14142..., and though we may
carry on the process to any required extent, we shall never be
able to stop at any particular point and to say that we have
found the exact number which is equivalent to the Square
Root of 2.
291. We can approximate to the real value of a surd by
finding two numbers between which it lies, differing from each
other by a fraction as small as we please.
Thus, since V2 = 14 142
14 15 1
a/2 lies between :— and —, which differ by :r ;
10 10 ■' 10
also between — and — , which differ by tt^k,
100 100' ■^ 100
also between ^ and ^^^, ,vhich differ.by ^.
And, generally, if we find the square root of 2 to n places
of decimals, we shall find two numbers Ix'twec^; wliich ^2 lies.
differinLT ironi eucli other by the fraction ,^ .
214 ON^ SURDS.
292. Next, we can alwaj's find a fraction differing from the
real value of a surd by less than any assigned quantity.
For example, suppose it required to find a fraction differ
ing from ^'2 by less than ^o
Now 2(12)''^, that is 288, lies between (16) and (17)2,
.'. 2 lies between ( t;^) and (rs) ;
.•. ^2 lies between ^ and j^ ;
.. J2 differs from r— by less than r^.
12 ^ 12
293. Surds, though they cannot be expressed by whole or
fractional numbers, are nevertlieless nuinlxrsof which we mav
form an approximate idea, and we may make three assertions
respecting them.
(1) Surds may be compared so far as asserting that one is
greater or less than another. Thus ^^^3 is clearly greater than
^'2, and 4^9 is greater than ^fS.
(2) Surds may be multiples of other surds : thus 2 ^^2 is
the double of J2.
(3) Surds, when multiplied together, may produce as a
result a whole or fractional number: thus
V2x ^2 = 2,
294. The symbols ^^a, ^a, ^/a, i^a, in cases where the
second, third, fcurth, and n*^ roots respectively of a ainnot be
exhibited as wliole or fractional numbers, will represent surds
of the second, third, fourth, and Jt"" order.
These symbols we may, in accordance with tlie principles
laid down in Chapter XXIII., replace by a*, a^, a*, a".
ON SURDS. 215
295. Surds of the same order are those for which the root
symljol or surdiudex is the same.
1
Tliiis ^a, 3 >Ji^h), 4 i^l(inn), r^ are surds of the same order.
Like surds are those in which the same rootsymbol or surd
index appears over the same quantity.
Thus 2 sja, 3 Ja, 4a^ are like surds.
296. A whole or fractional number may be expressed in
the form of a surd, by raising the number to the power denoted
by the order of the surd, and placing the result under the
symbol of evolution that corresponds to the surdindex.
Thus 0= Mja\
b ' Ib^
297. Surds of different orders may be transformed into surds
of the same order by reducing the surdindices to fractions
with the same denominator.
Thus we may transform ^fx and ^y into surds of the same
order, for
and. ^y = y^^y^ = ^l/y\
and thus both surds are transformed into surds of the twelfth
order.
Examples.— cviii.
TransforTn into Surds of the same order :
I. Va;and ^y. 2. 4/4 and ^2. 3. ^(18) and 4/(50).
4. 'J^2 und ;i/2. 5. ^/rt and ;;/6. 6. 4^(a + 6) and i^{ab).
298. If a whole or fractional number be multiplied into a
surd, the product will be represented by plaqing.the multiplier
and the multiplicand side by side with no sign, or with a dot
(.) between them.
Thus the product of 3 and ^f2 is represented by 3 ^^2,
of 4 and 5 v'2 by 20^2,
of rr and Jc by a ^/c.
2i6 ON SURDS.
299. Like surds may he combined by the ordinary pro
cesses of addition and subtraction, that is, by adding the
coefficients of the surd and placing the result as a coefficient
of the surd.
Thus ,v/« + «/« = 2 V**?
X Jc ^/c = (.C  1) i^C
300. We now proceed to prove a Theorem of great ini
]>ortance, which may be thus stated.
The root of any expression is the saw,e as the product of the
roots of the separate factors of the expression, that is
sj(ah) = ^la . ^h,
^{xyz)=^x.^y. »/z,
;:/(pqr)= ;'2). s^q.^fr.
We have in fact to shew from the Theory of Indices that
1 11
(aby =0" . h".
Now \(ahy>r = (abf = ab,
11 11 '^ 1
and Irt". ?)" j" = (a")". (6")" = rt". 6" = rt.6;
^ 111
.". (ab)" =rt". b".
301. We can eometimes reduce an expression in the form
of a surd to an equivalent expression with a whole or frac
tional niimber as one factor.
Thus v'("2) = V(fi X 2) = ^/CM) . ^/2 = 6 ^/2,
4/(128) = ^(64 X 2) = ^(64) . ^72 = 4 ^f'2,
!j{a'x) = a^a" . Zfr = a . ^J/x.
O.V SURDS.
Examples.— cix.
Reduce to equivalent expressions with a whole or fractional
number as one factor :
I. V(24). 2. ^/(50). 3. V(4a3).
4. s'{l2fiaH^). 5. v/(32?/s3). 6. ^/(lOOOa).
7. V(720c2). 8. 7.V(396x) 9. 18.J(^x3).
T. II' \^('t^ + 2ax + ax).
12. V(a^2x2/ + XJ/2). 13. v'(5Oa2_iO0a645O&).
14. V(63c*?/42cy + 7y3). 15. 4/(54rt662),
16. 4/(1 60xV). 17. 4/(108m9ni»y.
18. 4/(1372ai65i6). ig. 4/(3;* + 3x31/ + 3xV ^ xr)
2a 4/(rt*3a36 + 3a262a&3).
302. An expression containing two factors, one a surd, the
other a whole or fractional number, as 3 »J2, a ^x, may be
transformed into a complete surd.
Thus 3 v'2 = (32)i V2 = V9 . V2 = ^/(18),
a^fx = {a^)K ^x= 4/a3. ^x= ^{a^x).
Examples.— ex.
Reduce to complete Surds :
I. 4V3. 2. 3^/7
4 24/6. 5. 3/^''^
7. 4«V(3x).
9. (»^+^)^G~3
3. 54/9.
6. 3 V«
^■•W(£>
'«+«(,7y
\x + v/ ' \x  2xv + v''
2i8 ON SURDS.
303. Surds may be compared by transforming them into
surds of the same order. Tlius if it be required to determine
whether s/^ be greater or less than 4^3, we proceed thus :
V2 = 2^ = 26= 4/23= ^8,
4/3 = 3^ = 3^=4/32=4/9.
And since 4^9 is greater than ,^8,
^3 is greater than ^'2.
Examples. — cxi.
Arrange in order of magnitude the lullowing Surds :
1. J3 and 4/4. 6. 2 ^87 and 3 ^33.
2. VlO and 4/15. 7. 2 4/22, 3 4'7 and 4 V2.
3. 2 V3 and 3 ^'2. 8. 3 ^/19, 5 4/I8 and 3 4/82.
4. ^J'l^rld^{~). 9 2 4^14, 5 4/2 and 3 4'3.
5. 3^7 and 4^3. 10. ^ ^72,  ^3 and ^ v'i.
304. The following are examples in the application of the
rules of Addition, Subtraction, Multiplication, and Division to
Surds of the same order.
1. Find the sum of ^'18, ^a28, ami ^'32.
v/(18)4 v/(128)+ V(32)= ^'(9x2)+ ^/(64x2)+ v'(16 x 2)
= 3V2 + 8s'2^4^'2
= 15 ^'2.
2. From 3 ^/(75) take 4 ^/(12).
3 ^/(75)  4 x/(12) = 3 x/(25 x 3)  4 ^'f4 x 3)
= 3.5.^'34.::. x'3
= 15 ^3  8 J3
= 7^3.
ON SURDS.
3. Multiply v/R ^y V(12).
^/8x v/(12)= v/(8xl2)
= V(96)
= V(16 X 6)
= 4^6.
4. Divide ^/32 by ^18.
x /(32) _ >v/(16x 2) ^ 4^2 ^ 4
V(18) V(9x2) 3V2 3"
Examples. — cxii.
Simplify
1. V(27)+ 2^(48) + 3^(108). 11. ^6 x */8,
2. 3^(1000) +4^(50) + 12^(288). 12. ^(14) x ^(20).
3. a VC^^a;) + & sj(t»^x) + c sj{c^x). 1 3. ^/(50) x V(200).
4. ^(128) + 4/(686) + 4/(16). 14. 4/(3rt26) X 4/(9a6''!).
5. 7 4/(54) +3 4/(16) +4/(432). 15. 4/(12a6) x 4/(8a^6S).
6. V(96) V(54). 16. ^/(12) V3.
7. V(243)V(48). 17. x/(18)vio'V
8. 12 ^(72) 3^(128). 18. 4/(rt^)^ 4/(«?').
9. 5 4/(16) 2 4/(54). 19. 4/(a36)^ 4/(a63).
10. 7 4/(81)  3 4/(1029). 20. V(^2 + ^,3y) ^ ^/^^ + 2a;2y + a;^)/^).
305. We now proceed to treat of the Multiplication of
Compound Surds, an operation which will be frequently ?e
quired in a later part of the subject.
The Student must bear in mind the two following Rules ;
Rule I. sjax Jb= ^/(ab),
Rule II. ^ax ^a = a,
which will be true for all values of a and b.
ON SURDS.
EXAMPLES.— cxiii.
Multiply
s]x by ^y.
V(3;2/)i>y Vy
6 ,^x by 3 sjx.
7V(* + l)by 8V('X+1)
lO^a^by 9V(a;l).
9 \'35 ^y  'J^
10. V(^l) by  sl{^.l).
11. 3 ^/.c by  4 ^x.
12.  2 ^a by  3 ,^a.
13 \/(a;7)by  ^x.
14. 2 V(a; + 7) by 3 ^a:
15. 4Vra2l)by 2^(a2l).
V(3x) by ^/(4x) . 16. 2 V(a^  2a + 3) by  3 ^{a^  2a + 0).
306. The following Examples will illustnite tlie wny of
proceeding in forming the products of Compound Surds.
Ex. 1 . 1 o multiply ^x + 3 by ^a: + 2.
^/x + 3
Vx + 2
« + 3^x
+ 2v'a; + 6
X + 5 v'a; + 6
Ex ?. To multiply A^x + Zjy by 4 ^r  3 ^hj.
AJx + ZsJy
4 Vx  3 y/j/
16x + 12v'(x?/)
12V(xj/)9?/
16x  9?/
Ex. 3. To form tlu' scpiare ofV(a;7) ^^x.
V(x7) Vx
^(x7)Va;
x7  ^/(x27x)
 ^/(x27x)+a;
2x72\/(x''2'7x)
ON SURDS. at
Ex AMPLES.— CXiv.
Multiply
I. ^x + 7 hj ,^'o: + 2. 2. v'^5 by Vx+3.
3. J(a + 9) + 3 by ^f{a + 9)  3.
4. V(a4)7byV(a4) + 7.
5. S^x1 hj i^x + 4.
6. 2^/{xi>) + 4hy3J{x5)6.
7. ^(6 + x) + ^fx by ^/(6 + x)  v'x.
8. V(3;;+l)+ v/(2xl)byV3x V(2a;1).
9. s^a + J{a  x) by .Jx  J(a  x).
I o. V (3 + x) + Jx by ^/(3 + x).
11. sjx+ ijy+ ijz hy Jx Jy+ Jz.
12. Ja+ J(a — x)+ Jx hy Ja J(ax)+ Jx.
Form the squares of the following expressions :
13. 21+ ^/(x29). 17. 2V^3.
14. J{x + Z)+ J(:x + 8). 18. J{x + y) J{xy).
15. JX+ J{xA). 19. Jx.J{x+l)J{xl).
1 6. J{x  6) + v'a;. 20. ^/(.c + 1) + V^ . V(^'  1 • •
307. We may now extend the Theorem explaineil in
Art. 101. We there shewed how to resolve expression^ df
the form
a26«
into factors, restricting our observations to the case of perfect
squares.
The Theorem extends to the difference between any tivo
quantities.
Thus
ab={Ja+ Jh){Ja Jb).
»
x^y = {x+ Jy) {x Jy).
lx==(l+ Jx) (1 Jx\
222 ON SURDS.
308. Hence we can always find a multiplier which will
fVie tVoiii surds an expression of any of the /oitr forms
I. a+ s/b or 2. Ja+ Jb,
3. a s,fh or 4. Ju Jb.
j.'Oi since the first laid third of these expressions give
as a product a'^~b, which is free from surds, and since the
second and fourth give as a product ab, which is free from
surds, it follows that the required multiplier may be in all
cases found.
Ex. 1. To find the multiplier which will free from surds
each of the following expressions:
I. 5+V3. 2. ^6+^5. 3. 2 ^o. 4. x/7 ^'2.
The multipliers will be ^
I. 5^3. 2. V6V5. 3. 2+^5. 4. V7+V2.
The products will be
I. 253. 2. 65. 3. 45. 4. 72.
That is, 22, 1, 1, and 5.
ct
Ex. 2. To reduce the fraction ^_ ^^ to an equivalent.
fraction with a denominator free from surds.
Multiply both terms of the fraction by 6+ ,^c, and it be
comes
ab + atjc
b^c '
which is in the required form.
Examples.— cxv.
Express in factors :
I. cd.
2. c2d.
3
cd^.
4. 11/.
5. 1Sx\
6.
5m \. ■
7. 4a23x.
8. 98?i.
9
11«16.
10. p'^  4r.
1 1 . jj  83^
12.
rt*"  b\
ON SVRDS. 223
Reduce the following fractions to equivalent fractinus Avitli
denominators free from surds,
,, 1 „ N^L , 4 + 3V2
16 ^ 17 V3 2V2
V«+Vx V(™'+*1) V('h21)
^" Va  V*' ■ \/(»i2 + 1) + ^{m^  i)
* 1 Vrc' ~^' a~ s/(a2l)
V(a + x) + v'(a  a;) ^ a+ sj{a^x)
^{a + x)— sjia — x)' ~ ' a sj{a?  x^)'
309. The squares of all numbers, negative as well as posi
tive, are positive.
Since there is no assignable niamber the square of which
would l)e a negative quantity, we conclude that an expression
which appears under the form sfi  'i^) represents an impossible
quantity.
310. All impossilile square roots may be reduced to one
common form, thus
V(«2)=Vlax(l)f=>2.N/(l) = a.V(l)
^(a;)=VI^ x(l)\=Jx .^'■1).
Where, since a and sjx are possible numbers, the whole
impossibility of the expressions is reduced to the appearance of
^(  1) as a factor.
311. Def. By ,^/(l) we understand an expression which
ivhen multiplied l»y itself produces  1.
Therefore
}n/(i)P=U'(i)!v^(i)=(i)v^(i)= V(i),
*U/(i)l*=U'(i)!MV(i)P=(i)(i)=i,
«Vd so on.
:524 ON SURDS.
Examples.— cxvi.
^lultiply, oLservini,' tluit
^  ax ^1 h=  ^fab.
1. 4+ ^/(:3)l_.y4 V(3).
2. V32V(2)l,y ^'3 + 2 ^/(  2).
3. 4V(2)2V2l.y^^/(2)3V2.
4. V(2)+ V(3j+ x/(4)by V(2) V(3) ^/(4).
5. 3 V(  «) + x^(  b) by 4 V(  «)  2 v/(  6).
6. a + s,f(  a) by a  ^f(  a).
7. a^{a) + b^'{h) by a .yA;  a)  6 V(  &)■
8. a+/5v'(l) bya/ix/(l). "
9. 1 V(le') by 1+^/(162).
I o. t''^'" + e"''^ '" by e""^ '''  e"^^'".
312. "We sliall now gi\ e a few Miscellaneous Examples to
illustrate the principles explained in this Chapter.
Examples. — cxvii.
1. bnnphly ^^^^^J^^^A
2. Prove that 1+ ^(1)^+11 v/(l)j2 = 0.
4. Prove that 11+ ^f{\)'r \1 ^{l)\^= ^'(16).
; . D i \ i 1 1 e .r^ + (I M ly x + ^/2ax + a^.
6. Divide 7a'' 4?c' by m— ^^f2mn + n.
7. Siniplil'y ,^f {x^ + 2xy + xy) + ^' {x^  2xy + xy^).
8. Simplify , „ , yr, ami verify by puttniL:
„ . ;) and i = 4.
ox sunDS. 225
9. l^iiul the square of « >» /r  sf{cd).
10. Find the square of aV'^ — j^r
11. Siniplit'y
12. Smipuiy — — — ^ i.
x/(l^'')
„. ,.,. icl ( a;l \x )
13. Simphtv { — , r + T > ,
14. Form the square of . /( \v + •' )  /i / ( !  3 ).
15. Form the square of i^(^x + a)  sj{x — a).
16. :\Iiiltiply J/(a^'»"6''"'+V"') by xy(a''&"^*c— ^0
17. Raise to the 5"" power —\ — a^l{ 1).
1 8. Simplify 4/(81 )  ^l{ 512)+ 4/(192).
19. Simplify ^^y( 33 ).
20. Simplify ~~„ j '4/(32:'a;^  GSjjV + 441^^^: ._ 1029^52) j .
X— I
21. Simplify 2('h \)^ (  _, 1 5— , \
•'  ^^V 2/i*6/i3 + 6?t^2?i/
22. Simplify 2(?i  1) ^(63) + \ v/(112)  ^'(^j!^
2
../!l75(nl)2c^!xA_2 /(^;j
23. Wliat is the difference between
s/jl7 v/OW)!x VI17+ v/(33)J
and 4/ ! ( i") + ^/( 1 :!! )) I X 4/ ) f;5  V( 129)1
[S.A.]
226 ON SURDS.
313. We have now to treat of the method of finding the
Square Root of a Binomial Surd, that is, of an expression of
one of the following forms :
m+ s]n^ m— /y/n,
where m stands for a whole or fractional number, and tjn for
a surd of the second order.
314. We have first to prove two Theorems.
Theorem I. If Ja = m+ ^n, m must fee zero.
Squaring both sides,
a='mP'\ 2m ^n + n ;
.". 2ni ^11 = a — m — n ;
, a — Tn? — n
that is, V*i, a surd, is c(iual to a whole or fractional number,
which is impossible.
Hence the assumed equality can never hold unless in =0, in
which case ijn= s,hi.
Theorem II. 7/"fe+ ^'a=^m^ Jn, then must fe=7?i, and
For, if not, let b — m, + x.
Then m + x+ ^a=m+ ^/n,
or x+ ^fa= x^i ;
which, by Theorem I., is inipossiblii unless a; = 0, in which case
h = vi and ^'(f= i>^^n.
315. To find the Square Boot of : + ^fb.
Assume V(*+ V^)= >/•*+ \'v
Then a+ ,Jb = x + 2 VCr.V) + y ;
••• x + y = a (1).
2n'('."V v7, ^i),
froni which we have to tiud x auu ij.
ON SURDS. 227
Now from (1) »2 ^ 2x1/ + 1/^ = a,
and from (2) 4xy = h ;
.•. x"2x2/+ (/ = a2 — 6;
Also, x + y = a.
From these equations we find
and y
2 " i2 '
Similarly we may show that
^'(»  ^'») = ^l " "4"" *' }  ^^^^f^l .
316. The practical use of this method will be more clearly
seen from the following example.
Find the Square Root of 18 + 2 VC^T).
Assume V{ 18 + 2 ^(77)  = V« + Vy.
Then 18 + 2 V(77) = a; + 2 ^(xy) + y ■
.. a; + 2/ = 18 )
2V(a^) = 2V(77)r
Hence x^ + 2xy + 1/ = 324 )
4aj/ = 308J"'
.'. x^  2xy + y"^ = \^ ;
:.xy=±A;
also, x + y=l8.
Hence a = ll or 7, and y = l or 11.
That is, the square root required is ^^(11)+ ^^7.
228 ON SURDS.
Examples. — cxviii.
Find the square roots of tlie following Binomial Surds:
I. 10 + 2^/(2^. 2. 16^2^(55). 3 92^(14).
4. 9442V5 3 132^/(30). 6. 3812^(10).
7. 144V6. 8. 10312^/(11). 9. 7.^>  12 ^/(21).
10. 8712v'(42). IT. 3_^v/(10). 12. .5712^/(1.5).
317. It is often easy to determine the square roots of
expressions such as those given iu the preceding set ot
Examples hxj insjiedion.
Take for instance the expression 18 + 2 \/(77).
What we want is to find two numbers wliose sum is 18 and
whose product is 77 : these are evidently 11 and 7.
Then 18 + 2 V(7V) = 11 + 7 + 2 ^(11 x 7)
= U/(ll)+^/7p.
That is v/(ll)+ \'' is the .s(iuare root of 18 + 2 ^/(77).
To effect this resolution by inspection it is necessary that the
coefficient of the surd should be 2, and this we can always ensure.
For example, if the proposed expression be 4+ /v/(15), we
proceed thus :
8 + 2V(15) 5 + 3 + 2^(5x3)
4+ V(15) =
2
V2
~\ J2 J '
:. — 75^ is the square root of 4+ \/(15).
Again, to find the Square Root of 28  10 is/3.
2810^/3 = 282^/(75)
= 25 + 32v/(2.)'x.3)
= :5 V3)2;
:. 5  ^3 's f '16 sipuire root required.
XXV. ON EQUATIONS INVOLVING SURDS.
318. Any equation may be cleared of a single surd, by
transposing all the other terms to the contrary side of the
equation, and then raising each side to the power correspond
ing to the order of the surd.
The process will be explained by the following Examples.
Ex. 1. ^'.r = 4.
Raising both sides to the second power,
a; =16.
Ex. 2. 4/x = 3.
Raising both sides to the third jjower,
a; = 27.
Ex. 3. Via;2 + 7)x=l.
Transposing the second term,
J(a;2 + 7) = r+a;.
Raising both sides to the second power,
X + 7 = 1 + 2x + a;2,
.. x = 3.
Examples.— cxix.
I. Jx = 1. 2. v/c = 9. 3. x^ = b.
4. 4/a; = 2. 5. x = Z. 6. 4/x = 4.
7. v/(x + 9) = 6. 8. ,./(x7)7, 9. V(.r15) = 8.
10. (x9)^=12. II. ^(4x16) = 2, 12. 2()3Va; = 9.
230 ON EQUATIONS INVOLVING SURDS.
13. 4/(2a; + 3) + 4 = 7. 17. ^/(4x2 + 5x2) = 2x + l.
14. h\CsJx = a. 18. x/(9x212a;51) + 3 = 3x.
15. V0^"9) + x = 9. 19. v^(^''"^ + '^)«=K
16. ^(x^ 11) = x 1. 20. ^'iLoy? — '^inxVii)hx = m.
319. When ^iro surds are involved in an equation, one at
least may be made to disappear Ly disposing the tenns in
such a way, tliat one of the surds stands by itself on one side
of the equation, and then raising each side to the power cor
responding to the order of the surd. If a surd be still left, il
can be made to stand by itself, and removed by raising each
side to a certain power.
Ex. 1. ^(x16)+ v'c = 8.
Transposing the second term, we get '
^/(x16) = 8 ^Ix.
Then, squaring both sides (Art. 306),
3;16 = 6416V« + a;;
therefore 1 6 ^/.c = 6 i + 1 6,
or 16Va: = 80,
or /y/x = 5 ;
x = 25.
Ex. 2. V(^  5) + sK^ + T) = 6.
Transposing the second term,
V(c5) = 6 ^'Crr';).
Squaring both sides, x  5 = 36  1 2 sj{x + 7) + x + 7 ,
therefore 12 ^'(x + 7) = 36 + x + 7x + 5.
or 12V(x + 7) = 48,
or V(x + 7) — 4.
Squaring both sides, x + 7 = 1 6 ;
therefore x = 9
r V EQUA TIONS LWOLVIXG SURDS. 23 1
Examples. — cxx.
1. v''(16 + x)+ Jx = 8. 6. 1+ v/(3a; + l)= x/(4x + 4).
2. ^f{.C\(3) = 6 s,tx. 7. l~ ^J{l■6x) = •ls/(^■'•c)■
2,. s/{x + 15) + ^.0= 15. 8. a  ^/(x  a) = ^x.
4. ^'{x 21)= ^'x  1. 9. V^ + v/(x  7?l) = y.
5. v'(cl) = 3 v/(u; + 4). 10. V(x1)+ ^'(:c4)3 = 0.
320. When surds appear in the denominiitors of fractions
in equations, tlie equations may be cleared of fractional terms
by the process described in Art. 186, care being taken to
follow the Laws of Combiualiou of Surd Factors given in
Art. 305.
Examples.— cxxi.
36 28
2. Vx+,/(.21) = ^^. 4. V(x15)+V^ = ^i*^^.
9a
^I{ax) + h^ ba jjx+l6 _ s'' + S2
'' x + 6 ~h ^{ax)' 9 ^/x + :r~^Zr+12'
o /I , / N /I / \ 4+ ,^/x v/:c8 ./;/; 4
8. (1 + Vx) (2  Va;) = — ^ . 10. \_ ,. = >, — .
321. The following are examples of Surd Equations result
ing in quadratics.
Ex.1. 2^x^^^'5.
r'learing the equation of fractions, 2a; + 2 = 5 ^jz.
232 ON EQUA TIONS INVOL VING SURDS.
Squaring both sides, we get 4x2 48x + 4=25x;
whence we find re = 4 or .
4
Ex.2. V('' + 9) = 2V^3.
Squaring both sides, a; + 9 = 4x  1 2 ^/x + 9 ;
therefore \1 sjx = 3a;,
or A: ,Jx=x.
Squaring both sides, 16x = x2.
Divide by x, and we get 16 = a;.
Hence tlie values of x which satisfy the equation are 16
and (Art. 248).
Ex.3. v/(2x+l)+2^x = ^^^j^.
Clearing the equation of fractions,
2a; + l + 2v'(2x2 + x) = 21;
therefore 2 ^(2x2 + x) = 20  2x,
or V(2x2 + ;i;) = 10x.
Squaring both sides, 2x"^ + x = 1 00  20x + x*,
whence x = 4 or 25
322. We sliall now give a set of examples of Surd Equa
tions some of which are reducible to Simple and others to
Quadratic Equations.
Examples.— cxxii.
I. 4x  12 ^/x = 16. 4. V(6x 11)= V(249  fix^).
2." 4514Vx=x. 5. >/(6x) = 2 ^/(2xl).
3. 3V(7 + 2.c2) = 5^/(4x3). 6. x2 ^',43x) + 12 = 0.
7. v/(2x + 7) + V(3x 18)= v\7x + 1).
8. 2 V(204  5x) = 20  ^'(3x  68).
ON EQUATIONS INVOLVING SURDS. 233
9. Vx4 = ^^. 14. V(x + 4)+ V(2x1) = 6.
10. V:c+ll=^^?^. 15. V(13x1) v/(2xl) = 5.
>y X — 11
11. V(.c + 5). V(a; + 12) = 12. 16. V(7x+1) V(3x+1) = 2.
1 2. V(a; + 3) + V(a; + 8) = 5 ^x. 17. VCl + x) + V^ = 3.
525
13 v'(25 + x)4 V(25x) = 8. 18. v/x+ V(a! + 9975)=7=.
20. V(x2l) + 6 = ^^^
. 21. V(('^«)" + 2«/) + 6S=a;a+i.
22. Vl(^ + «)' + 2aft + 6J=6aa;.
23. V(x + 4) V^=J(,r + ).
a; — 1 5
24 ^;/^I^=^ + 4 26. V(a; + 4)+ V(a; + 5) = 9.
V(a;4)
25 . V(4 + .r)  v'3 = ^x. 27. ^fx + ^{x  4) = j^
28. x2 = 21+ ^(^29).
29. V(50 + a;) V(50x)=2,
30. V(2xr4) J(+6) = l.
31. V^3 + .r)+^/x= ^
V(3 + x)'
1 _J ^1^ ^
3^ V(a; + 1) ■*" \./{x ~i)~ ^/{x'  !)•
3x r ■^f(4x — x^
XXYI. ON THE ROOTS OF EQUATIONS.
323. We have already proved that a Simple E(iuation can
have only one root (Art. 193) : Ave have now to prove that a
Quadratic Equation can have only two roots.
324. We must first call attention to the following fact :
If m7i = 0, either m = 0, or n = 0.
Thus there is an ambiguitv : but if we know that m cannot
be equal to 0, then we know for certain that n = 0, and if we
know that w cannot be equal to 0, then we know for certiiin
that m = 0.
Further, if lmn = 0, then either 1 = 0, or 7?i = 0, or n = 0, and
so on for any number of factors.
Ex. 1 . Solve the equation (x  3) (x + 4) = 0.
Here we must have
x3 = 0, or x + 4 = 0,
that is, X = 3, or X — — 4.
Ex. 2. (x  3a) (5x  26) = 0.
Here \m must have
x3a = 0, or 5x — 26=0,
, . 26
that IS, «=3a, or x = — .
o
OM THE ROOTS OF EQUATIONS. 23$
Examples. — cxxiii.
I. (a;2)(a;5)=0. 2. (x3) (x + 7) = 0. 3. (a; + 9)(x + 2)=0,
4. (x5a)(a;6?*) = 0. 6. (19x227) (14a; + 83)=0.
5. (2a; + 7)(3.c5) = 7. (5x4m)(6x lln) = 0.
8. (a;2 + hax + Sa^) (x^  Tax + 1 2a2) = 0.
9. (x^  4) (x  2«x + cfi) = 0.
10. X (x^  5x) = 0.
11. (acre  2ffi + 6) {bcx + 3a  6) = 0.
12. (ex  (f ) (ex  e) = 0.
325. The general form of a quadratic equation is
ax^ + bx + c = 0.
Hence aix^ + x + ) = 0.
\ a a/
Now a cannot =0,
.. x^ + x +  = 0.
a a
... b e
Wnting x> for  and q for , we may take the following
as the type of a quadratic equation of which the coeflBcient of
the first term is unity,
x'^irfx + q — O.
326. To show that a quadratic equation has only two roots.
Let x^ +px + 5' = he the equation.
Suppose it to have three different roots, a, b, c.
Then a'^ + ap + q = (1),
¥+bp + q = i..(2),
c2 + cp + q = (3).
Subtracting (2) from (1),
a^b^+(ab)p = 0,
or, {ab){a + b\p) = 0.
236 ON THE ROOTS OF EQUATIONS.
Now ab does not equal 0, since a and 6 are not alike,
:. a + h+p = (4).
Again, subtracting (3) from (1),
a^ — c^ + (a — c) p = 0,
or, {a — c){a + c+p)=0.
Now a — c does not equal 0, since a and c are not alike,
.. a + c+p = (5).
Then subtracting (5) from (4), we get
6c = 0, and therefore h = c.
Hence tliere are not more than hvo distinct roots.
327. We now procetd to show the relations existing be
tween the Roots of a quadratic equation and the Coefficients
of the terms of tlie equation.
328. x'^^px + q=0
is tlie general form of a qtiadratic equation, in which the co
efficient of the first term is unity.
Heni'e x'^+px= —q
x'^ + 'px+^^=^q,
Now if a and /? be the roots of the equation,
«=iV('i') '"•
^iV(t') <''•
Adding (1) and (2), we '.y\
a + j3= p (.3).
v
ON THE ROOTS OF EQUATIONS.
Multiplying (1) and (2), we get
or a/3=^^+2,
or ay8 = (2 (4^.
From (3) we learn that tiiM, sum of the roots is equal to the
coefficient of the second term with its sign changed.
From (4) we leam that the product of the roots is equal to
the last term.
329. The equation x'^ + px + q = has its roots real and
different, real and equal, or impossible and different, according
as 'p is > = or < Aq.
For the roots are
2
iV(?')'""
and _2 _/(»?. A „,rti4£!zil).
iV(?'>
First, let p~ be greater than Aq, then >J{p^  Aq) is a possible
quantity, and the roots are different in value and Ijoth real.
Next, let2'^ = 4g', then each of the roots is equal to the real
quantity ^.
Lastly, let ^^ be less than Aq, then \f{p — Aq) is an impos
sible quantity and the roots are different and both impossible .
Examples.— cxxiv.
I. If the equations
ax + bx + c = 0, and a'x^ + h'x + c' = 0,
have respectively two roots, one of which is the reciprocal of
the other, prove that
(aa'  cc')^ = {aV  he') {a'b  b'c).
238 ON THE ROOTS OF EQCA
2. If a, /? be the roots of the equation ax + 6a; + c = (), prove
that
.) no &^ — 2ac
' a
3. If a, ^ be the roots of tlie equation ay? + 6x + c = 0, prove
that
ac'j? i {2ac ~ b'^) x + ac = ac \^~ r,)\^— )
4. Prove that, if tlie roots of the equation ax + bx + c = i) be
equal, nx + bx + c is a perfect square witli respect to x.
5. If a, y8 represent the two roots of the equation
x^{l + a) a; + ^(l + a + «") =0,
show that a + /3' = a.
33O. If a and /3 be the roots of the equation x^+px + q=Oy
th en x'^ + jjx + 3 = (a;  a) (x  yS).
For since ^=  (a + /?) and q = afB,
3? + px + q = x ~ {a + (B) X + a/B
= {x~a){x(3).
Hence we may form a quadratic equation of which the roots
are given.
Ex. 1. Form the equation whose roots are 4 and 5.
Here xa = x — 4andx/3 = x5;
.•. the equation is (x  4) (x — 5) = ;
or, x9x + 20 = 0.
Ex. 2. Form the equation whose roots are ^ and  3.
2
.111(1 r— /?=)•! 3 ■
Here x  a = x  and x  ^ = x + 3 ;
th«
equation is f x  ^ j (x + 3) = ;
er, (2xl)(x + 3)=0;
or, £3: + 5x3 = 0.
UN THE ROOTS OF EQUATIONS. 239
Examples.— cxxv.
Form the equations whose roots are
I. 5 and 6. 2. 4 and 5. 3. 2 and 7.
12 5
4. 2 and. 5. 7aud 6. v/3 and  ^3.
7. m + 7i and. m — n. 8.  and . 9. 7^ and.
a jd pa
331. Any expression containing x is said to be a Function
of X. An expression containing any symbol x is said to Ll; a
positive integral function of a; when all the powers of x con
tained in it liave positive integral indices.
3 1
For example, bx^ + 2r^ + ^x* + j~x^ + 3 is a positive integral
■ 1 •
function of :r, but Qx^ + Scc^ + 1 and 5a"  2x~^ + 3x + 1 are
1
not, because the first contains x^, of which the index is not
integral, and the second contains a;"^, of which the index is not
positive.
332. The expression 5x^ + 40;' + 2 is said to be the expres
sion corresponding to the equation 5x^ + 4x^ + 2 = 0, and the
latter is the equation corresponding to the former.
333. If a be a root of an equation, then xa is a factor
of tlie corresponding expression, provided the equation and
expression contain only positive integral powers of x. This
principle is useful in resolving such an expression into factors.
We have already proved it to be true in the case of a quadratic
equation. The general proof of it is not suitable for the stage
at which tlie learner is now supposed to be arrived, but we
■will illustrate it by some Examples.
240 ox TUF. ROOTS OF EQUATTOA^S.
Ex. 1 . Rrsul ve 2oc2  5x + 3 into factors.
If we solve the equation 2x^5.15 + 3 = 0. we shall find thai
its roots five 1 and .
Now divide 2/5.r + 3 by x1 ; the quotient is 2j63
that is o(.,: I);
.'. the L;i\'eii eA])ression = 2 (a;  1) ( x  ^ I.
Ex. 2. Eesolve 2x^ + a;— ll.f 10 jnto factor.?.
By trial we find that this expre.ssion vanishes if we put
x=  1 ; tliat is, — 1 is a root of the e(jiiation
Sx^ + x^ll.t 10 = 0.
Divide the expression l>y x4 1 : the quotient is Sx^x 10 ;
.'. the expre.ssion = (2x  x  10) (x + 1)
= 2(x^5)(. + l).
We must now resol\e x  b into factor.s, by solving the
corresponding equation x^ ~' — b=0.
The roots of this equation are  2 and g;
.. 2x3 + x2  1 Ix 10 = 2 (.»• + 2) (x  ^) (x + 1)
= (x + 2)(2./;5)(x + l).
Examples.— cxxvi.
Resolve into simijle factors the following expressions :
I. .t311x2 + 36x36. 2. x^7.c2 + l4x8.
3. x"*  5.12  4(i.  40. 4_ 4x3 + 6.12 + X1.
5. 6.r3+ll.»;29x14. 6. 3?^y^ ^^Zxyz.
7. a^P~c^2ahc. 8. 3x3x223x + 21.
9. 2.f3  5x2 _ i7.r + 20. ■ 10. 15.13 + 41.r2 + 5.r  21.
ON THE ROOTS OF RQClATlONS. 241
334. Tf we can find one root of such an equation as
2a;3 + a;2llx10 = 0,
we can find all the roots.
One root of the equation is  1 ;
.. (x + l)(2x2x10) = 0;
.. x+l = 0, or2a;2a;10 = 0;
.. x=  1, or —2, or .
Similarly, if we can find one root of an equation involving
the 4"" power of x, we can derive from it an equation involving
the 3'* and lower powers of x, from which we may find the other
roots. And if again we can find one root of this, the other
two roots can be found from a (quadratic equation.
335. Any equation into which an unknown symbol or ex
pression enters in two terms onl3', having its index in one of
the terms double of its index in the other, may be solved as a
([uadratic equation.
Ex. Solve the equation x^ — Qx^ = l.
Regarding x^ as the quantity to be obtained by the solution
III the equation, we get
therefore x^3=±4; •
therefore x^=7, or x^= — 1.
Hence x= ^'7 or x= ^ 1^
and one value of ^^  1 is  1.
336. In some cases by adding a certain quantity lo both
sides of an equation we can bring it into a form capable of
solution, thus, to solve the equation
x2 + 5.>; + 4 = 5 ^'(x^ + 5,/; + 28),
add 24 to each side.
Then x^ + 5x + 28 = 5 s'{x'^ + 5x + 28) + 24 ;
or, a;2 + 5a; + 285 V(a'"^ + 5x + 28) = 24.
This is now in the form of a quadratic e([uation, the un
known quantity being ^f{x + 5x + 28), and completing the
square we have
fsAl Q
242 0\ THE kOOTS OF EQUATIONS.
95 191
.. V(a;2 + 5.r + 28)=±^;
whence »J{x^ + 5x + 28) = 8 or  3 ;
.. .r'' + 5x + 28 = 64 or 9;
from which we may find four values of x, viz. 4,  9, ani
5+ V(51)
Examples.— cxxvii.
Find roots of the following equations :
I. x* 12x2=13. 2. x6+14x3 + 24=0.
3. x»t 22x^ + 21=0. 4. x'" + 3x"' = 4.
5. x3X» = ^. 6. ^=20;=.
7. x2 + 3xi = ^. 8. x'"'x'' = 20.
9. x22x + 6(x22x + 5)2 = ll.
10. x^x + S V(2x25x + 6) = — ^ — .
11. x22V(3x22ax + 4) + 4 = *(x + ^ + l).
12. ax + 'i >J{xax + a'^=x^ + '2,a.
337. Every equation has as many roots as it has dimen
sions, and no more. This we have proved in the case of
simple and quadratic equations (Arts. 193, 323). The general
proof is not suited to this work, but Ave may illustrate it by
the following Examples.
Ex. 1. To solve the equation x^ 1=0.
One root is clearly 1.
Dividing b} x — 1, we obtain x + x + 1 = 0, of which the roots
1+ V3^^.. lx/3
are ^r^'— and ^ .
2 «
ON RA no. 243
Hence the three roots are 1, ~ and ^^ .
Ex. 2. To solve tht equation x^\=0.
Two of the roots are evidently + 1 and  1.
Hence, dividing by (x l)(x + 1), that is by a;^ 1, we obtain
a;2 + 1 = 0, of which the roots are v^— 1 and  v''— I
Hence the /our roots are 1,  1, ^^ 1, and — \'— 1.
The equation x^ — 6x^ = 7 will in lii<;e luauner have six
roots, for it may be reduced, as in Art. 335, to two cuT)ic
equations, x^  7 = and x^ + 1 = 0,
each of which has three roots, which may be found as in
Ex. 1.
XXVII. ON RATIO.
338. If a and B stand for two unequal quantities of the
same kind, we may consider their inequality in two ways. We
may ask.
(1) By ichat quantity one is greater than the other ?
The answer to this is made by stating the difference be
tween the two quantities. Now since quantities are represented
in Algebra by their measures (Art. 33), if a and b be the
measures of A and B, the difference between A and B is
represented algeljraically by ab.
(2) By how many times one is greater than tlie other?
The answer to this question is made by stating the number
of times the one contains the other.
Note. The quantities must be of the same kind. We can
not compare inches with hours, nor lines with surfaces.
339. The second method of comparing ^4 and B is called
finding the Eatio of A to B, and we give the following ileti
nition.
Def. Eatio is the relation which one quantity bears to
another of the same kind with respect to the number of t,ime:
the one contains the other.
244 ON RATIO.
340. The ratio of A io B is expressed thus, A : P>.
A and B are called the Terms of the ratio.
A is called the Antecedent and B the Conskquent.
341. Now since quantities are represented in Algebra hy
their measures, we must represent the ratio between two
(quantities by tlie ratio Ijetween their measures. Our next
step then must be to sliovv how to estimate tlie ratio between
two numbers. This ratio is determined by finding how many
times one contains the other, that is, by obtaining the quotient
resulting from the division of one by the other. If a and 6,
then, be any two numbers, the fraction j will express the ratio
of a to b. (Art 136.)
342. Thus if a and b be the measures of A and B respec
tively, the ratio of A to JB is represented algebraically by the
fraction r.
343. If a or b or both are surd numbers, the fraction ^
may also be a surd, and its approximate value can be found In'
Art. 291. Suppose this value to be ' , where m and n are
whole numbers : then we sliould say that the ratio A : B is
aj (proximately re])resented by — .
344. Ratius may be compared witli each other, by com
paring ihe fractions by wliich they arc denoted.
Thus the ratios 3 : 4 ami 4 : 5 may be compared by com
3 4
paring the fractions  and .
These are equivalent to — and ^ resi)ectlvely ; and since
gx is greater than 7,—, the ratio 4 : 5 is greater than the
latio 3:4
ON RATIO. Ms
Examples. — cxxviii.
1. Place in order of magnitude the ratios 2 : 3, 6 : 7. 7 : 9.
2. Compare the ratios x + 3y : x + 2y and x + 2ij : x + //.
3. Compare the ratios x5y : x — 4y and xZy : x  ly.
4. What number must be added to each of the terms of th«
latio a : h. that it may become the ratio c : d?
5. The sum of the squares of the Antecedent and Conse
quent of a Eatio is 181, and tlie product of the Antecedent
and Consequent is 90. What is the ratio?
345. A ratio of greater inequality is one whose antecedent
is greater than its consequent.
A ratio of less inequality is one whose antecedent is less than
its consequent.
This is the same as saying a ratio of greater inequality is
represented by an Improper Fraction, and a ratio of less in
equality by a Proper Fraction.
346. A Ratio of greater inequality is diminished by adding
the same nuwher to both its terms.
Thus if 1 be added to both terms of the ratio 5 : 2 it becomes
6 : 3, which is less than the former ratio, since g, that is, 2, is
less than .
And, in general, if x be added to both terms of the ratio
a : h, where a is greater than 6, we may compare the twu
ratios thus,
ratio a + x : 6 + a is less than ratio a : b,
if i be less than y,
b + x V
.„ ' ab+bx . ■, ,, ab + ax
it ,i5 — r be less than = — ;— ,
62 + bx V + bx^
if ab + bx be less than ab + ax,
if 6x be less than ax,
if b be less than a.
Now b is less than o ;
:. a +x .b + x \?, less than a : h.
246 CyV l^A no.
347. We may observe that Art. 346 is iiieiely a repetition
of that which we proposed as an Example at the end of the
chapter on Miscellaneous Fractions. There is not indeed any
necessity for us to vvearj' the reader with examples on Ratio:
for since we exjiress a ratio by a fraction, nearly all that we
mi.t,'ht have had to say about Ratios has been anticipated in
our remarks on Fractions.
348. The student may, however, work tlie following Theo
rems as Examples.
(1) If fl : 6 be a ratio of greater inequality, and x a positive
quantity, the ratio a — o:: b — x is greater than the ratio a : b.
(2) If (/ : h he a ratio of less inequality, and x a positive
quantity, llie ratio a + x : b +x is greater than the ratio a : b.
(3) If a : i be a ratio of less inequality, and x a positive
quantity, the ratio a — x: b — x is less than the ratio a : b.
349. In some cases we may from a single equation involv
ing two unknown symbols determine the ratio between the
two symbols. In other words we may be ahle to determine the
relative values of the two symbols, though we cannot determine
their absolute values.
Thus from the equation 4x = 3?/,
X 3
we get  = .
y 4
A^ain, from the equation 3x2 = 2?/,
■we"et'., = ^; and therefore =\t.
" 2/ 3 y x/.3
Examples. — cxxix.
Find the ratio of x to y from the following equations :
1. 9.1 = 6)/. 2. ax = by. 3. axby = cx + dy.
4. x42a?/ = 5?/. 5. a;2 12,r)/= 13;r. 6. x' + mxy = n^y.
7. Find two numbers in the ratio of 3 : 4. of \\ hich the
sum is to the sum of their scjuares :: 7 : 50.
8. Two numbers are in the ratio of 6 : 7, and when 12 is
addid to each ihe resulting numbers are in the ratio 1 I 12 : 13.
Find the nimibers.
OK RATIO. 247
9. The sum ol' two iiujuLers is 100, and the nunriieis are
in the ratio of 7 : 13. Find them.
10. The ditt'erence of the squares of two numbers is 48,
and the sum of the nvimber^ is to the difierence of the num
bers in the ratio 12:1. Find the numbers.
11. If 5 gold coins and 4 silver ones are worth as much as
3 gold coins and 12 silver ones, find the ratio of the value of a
gold coin to that of a sih'er one.
12. If 8 gold coins and 9 silver ones are Avorth as much as
6 gold coins and 19 silver ones, find the ratio of the Aaliie of a
silver coin to that of a srold one.
350. Ratios are compounded by multipljing together the
fractions by wliich they are denoted.
Thus the ratio compounded of a : 6 and c : fZ is ac : hd.
Examples. — cxxx.
Write the ratios compounded of the ratios
1. 2:3 and 4:5.
2. 3 : 7, 14 : 9 and 4 : 3.
3. a; — y : x^ + y^ and x  xy + y : x + y.
4. a^ — b^ + 2bc  c^ : a^  6  2hc  c^ and a + brC : a + h  r.
5. m^ + n^ : vi^  n^ and m — n : m + n.
6. x^ + 5x + 6 : y'' — ly + 12, and y^  3)/ : x + 3x.
351. The ratio a^ : b'^ is called the Duplicate Ratio of a ; 6.
Thus 100 : 64 is the duplicate ratio of 10 : 8,
and 36a;2 : 2oy^ is the duplicate ratio of 6x : by.
The ratio a^ : ¥ is called the Triplicate Ratio of a : &.
Thus 64 : 27 is the triplicate ratio of 4 : 3,
.and 343x^ : 1331)/^ is the triplicnte ratio of fx : lly.
248 ox PROPORTION.
352. The definition of Ratio given in Euclid is the sanu! jf^
in Algebra, and so also is the expression for the ratio that one
quantity bears to another, that is, A : B. But Euclid cannot
employ fractions, and hence he cannot represent the value of a
ratio as we do in Altjebra.
XXVIII ON PROPORTION.
353. Proportion consists in the equality of two ratios.
The algebraic test of Proportion is tlud the two fractions
representing the ratios must he equal.
Thus the ratio a : b will be equal to the ratio c : d,
and the/o?(? numbers a, b, c, d are in such a case said to be in
proportion. ^
354. If the ratios a ■ b and c ; d form a proportion, we
express the fact thus :
a : b = c : d.
This is the clearest manner of expressing the equality of the
ratios a : b and c : d, but there is another way of expressing
the same fact, thus
a : b :: c : d,
which is read thus,
a is to 6 as c is to d.
The two terms a and d are called the Extremes.
, b and c the Means.
355. When four numbers are in proportion,
product of extremes = product of means.
Let a, b, c, d lie in jiroportion.
ON PROPORTION. 249
Multiplying both sides of the equation by M, we get
ad = he.
Conversely, if ad = bc we can show that a : b=c ', d.
For since ad = be,
dividing both sides by bd, we get
ad_hc
hrVd'
that is, h^ d' ^'^' " '•^ = '^ '• ^'
356. liad = bc,
Dividing by cd, we get  = j, ife a : c = b : d;
d c
Dividing by ab, we get r = , it^ '^ : 6 = c : a ;
Dividing by ac, we get  = , i.e. d : c = b .a.
357. From this it follows that if any 4 numbers be so
related that the product of two is equal to the product of the
other two, we can express the 4 numbers in the form of a pro
portion.
The factors of one of the j^roducts must form the extremes.
The factors of the other product must form the means.
358. Three, quantities are said to be in Continued Pro
portion when the ratio of the first to the second is equal to
the ratio of the second to the third.
Thus a, b, c are in continued projjortion if
a : b = b : c.
% The quantitj' b is called a Mean Proportional lietween
a and c.
25© ON proportion:
Four quantities are said to be in Continued Proportion
when the ratios of the first to the second, of the second to
tlie third, and of the tliird to the fourth are all equal.
Tlius a, b, c, d are in continued proportion when
a : b = b : c = c : d.
359. We showed in Art. 20.5 the process by wliicli when
two or more fractions are known to be equal, otlier relation?
between the numbers involved in them may be determined
That process is of course applicable to Examples in Ratio and
Proportion, as we shall now show by particular instances.
Ex. 1. li a : b = c : d, prove that
a^ + b'^ : a^ ¥ = <:' + d^:c^ d^.
Smce a : o = c : d, t= j.
a
Let r=X. ThenT = \;
d
:. a = \b, and c = \d.
Now
and
ft^ + 6' _ X^b + ¥ _ />(\^+l) _ X2+J
c'^ + d^ _ \\l + d'^ _ d (X^+l)_X^ + l
a^ + b'^_c + d
^t^b^~^~d^''
Hence
ti^at is, a2 + 6« : a2 _ 52 = c2 + ^^2 . ^2 _ ^2^
Ex. 2. If « ; fe :: c : d, prove that
a:c:: ^{a* + ¥): i/{c* + d^),
LetJ = X. Then^ = X;
d '
,: a = \b, and c = \d.
ON PROPORTION. 251
a _yb _h
c y^d d'
i/ic'+d^) ~ ^(Md* +'d') ■" ;^d^. */^\* 41) ^¥~'d
Hence
that is, a:c:: ^{a^ + b*) : ij{c* + d^^ .
Ex. 3. ] f a : 6 = c : f? = e : /, prove that each of these ratios
is equal to the ratio a + c + e: b + d +f.
Let   = \,  = X, ^ = X.
Then a = \b, c = Xrf, e = X/.
^ a + c + e 'Kb + \d + \f_H b + d+f) _^
°^'^ b + 'd+f~~b + d+f " 'b + d+f
TT a + cre a c e
^"^^^ b^drrb=d=f'
that is, a + c + e : b + d +f— a : b = c : d = e :/.
Ex. 4. If a, b, c are in continued jiroportion, show that
a"^ + b'^ : b'^ + c^ ~ a : c.
Let ~ = X. Then = X.
c
Hence a='\b and 6 = Xc:.
a^ + b'^ _Xb + b_b\ \^+l )_b(\^' + l)_b'^ _ac_a
b'^ + c'^ ~ T + C' ~ Tc^^Tc ~ f^X'4l)~ c''^ ~ c^ ~ c"
Ex. 5. If Uxi + b : 15t + d=l2a + b: 12c + d, ])rove that
a :b = c : d.
Since 15a + 6 : 15c + c?=12a + ?* : 12c + ci,
and since product of extremes = product of means.
252 ON PROPORTION.
(15CH6) (12c + i) = (15c + d) (12a +6),
or, 180ac+ 126c + 15ad + 6d = 180ac + 12a<Z+ 156c + 6rZ,
or, 126c + Ybad = 12ad + 156c,
or, 3ad = 36c,
or, atZ = 6c.
Whence, by Art. 355, a : b = c : d.
Additional Examples will be found in page 137, to which
we may add the following.
EXAMPLES. — CXXXi.
1 . li a : b = c : d, show that a + b: a = c + d :e.
2. U a : b = c : d. show tliat a^  1 : b^ = c  d"^ : d^.
"?. It a, : Oi = a2 : 6,, show that — ^ , ==.
4. If a : 6 :: c : c?, show that
3a + ab + 26'' : 3a  26^ : : 3o + cd + 2d^ : 3c2  2d.
5. If ffl : 6 = c : rf, show that
ft2 + 3ab + ¥ : c^ + 3tY/ + d = 2«6 + 362 . 2cd + 3d.
6. Ifa:6 = c:rf = e :/ then a : b — mc — ne : mdnf.
7. If — a, —6, any parts of a, b, be taken from a and 6
n n
respectively, show that a, b, anil the remainders form a propor
tion.
8. If a : 6 = c : d = e :/, show that
ac : bd = la^ + mc + ne : lb + ind + nf.
9. If (/, : 6i = aj : 63 = 03 : 63, show that
(/, 4^,2 + ^^2 . 5^2 ^5^2 + 5^2 .. „^. . i,i^
av PROPORTION. 253
10. If ai : 6i = rt2 : 62 = a3 : hi, show that
a^a.2^ + a^Og + 03(11 : h^.^ + 6063 + 6361 = a^ : 6,2.
rpa2a6 + 62 ccd + c?2 ,,,.,, a c a d
11. It „ — , , „ = .,— — — „, snow that either r = 3 or t = •
12. If a2 + 6 : rt^  /* = c2 + c?2 . cii  rf2^ phow that
a: b — c : (^.
13. If rt : 6 = c : (Z, show that
(rt  c) (a^  r') (6  d) (¥ d^y
14. If rtj : h^ — a., : /*.,, show that
On the Geometrical Treatment of Proportion.
360. The definition of Proportion (viz. the equality of
ratios) is the same in Euclid as in Algebra. (Eucl. Book v.
Def. 6 and 8.)
But the ways of testing whether two ratios are equal are
quite different in Euclid and in Algebra.
The algebraic test is, as we have said, that the two fractions
representing the ratios must be equal.
Euclid's test is given in Book v. Def. 5, where it stands
thus :
" The first of four magnitudes is said to have the same ratio
to the second which the third has to the fourth, when any
equimultiples whatsoever of the first and third being taKen
and any equimultiples whatsoever of the second and fourth :
" If the multiple of the first be less than that of the second,
the multiple of the third is also less than that of the fourth :
" If the multiple of the first be equal to that of the second,
the multiple of the third is also equal to that of the fourth :
or.
254 ON PROPORTION.
" If the multiple of the first be greater than that of the
second, the multiple of the third is also greater than that of
the fourth."
We shall now show, first, how to deduce Euclid's test of the
equality of ratios from the algebraic test, and secondly, how to.
deduce the algebraic test from that employed by Euclid.
361. I. To show that if quantities be proportional accord
ing to the algebraical test they will also be proportionai
according to the geometrical test.
If a, 6, c, d be proportional according to the algebraical
test,
a _c
Multiply each side by — , and we get
ma _mc
nb ruT
Now, from the nature of fractions,
if ma be less than nb, mc will also be less than nd, and
if ma be equal to nb, mc will also be equal to nd, and
if ma be greater than nb, mc will also be greater than nd.
Since then of the four quantities a, b, c, d equimultiples have
been taken of the first and third, and equimultiples of the
second and fourth, and it appears that when the multiple of
the first is greater than, equal to, or less than the multiple of
the second, the multiple of the third is also greater than,
equal to, or less than tlie multiple of the fourth, it follows that
a, b, c, d are proportionals according to the geometrical test.
362. II. To dediioe the algebraic test of proportionality
from that given by Euclid.
Let a, h, c, d be proportional according to Euclid.
Then if s is not equal to 3,
let , be equal to , (1).
EXAMPLES ON RA TIO. ±%^
Take to and n such that
via, is greater than nh,
but less than n (i + x) (2).
Then, by Euclid's definition,
TOC is greater than nd (3).
But since, by il), 77,— ^ = —7?
and, by (2), wa is less than i!(6 + x),
it follows that 7/i.c is less than nd (4).
The results (3) and (4) therefore contradict each other.
Hence (1) cannot be true.
Therefore 7 is equal to ^.
We shall conclude this chapter with a mixed collection of
Examples on Ratio and Proportion.
EXAMPLES. — CXXXii.
1. VL ah .hc •..h : c, show that i is a mean proportional
between a and c.
2. If a : 6 : : c : rf, show that
a^^W : ^\ = c'' + d^:^.
a+b c+d
and a : b :: ^/{ma* + nc*) : i/(^mb^+ iid*).
3. li a : b :: c : d, prove that
ma — nb _ mc  nd
ma + nb mc + nd'
4. If ba + 2b': 7a + 36 : : 56 + 3c : 76 + 3c,
6 is a mean proportional between a and c.
5. If 4 quantities be proportional, and the first be the
greatest, the fourth is the least.
If a + 6, TO 4 n, mn,a — b be four such quantiti««j show that
h is greater than n.
25^ k^AMPLES ON RA TlO.
6. Solve the equation
x\ : a;2 = 2x + l : x + 2.
7. If — , — = — 5—, show that the ratios a : b and c : d are
b a
also equal.
8. In a mile race hetween a bicycle and a tricycle, their
rates were proportional to 5 and 4. The tricycle had halfa
niinute start, but was beaten by 176 yards. Find the rates of
each.
9. li a : b :: c : d and a is the greatest of the four quanti
ties, show that a + d is greater than b~ + c^.
01 .1, .plOa + fe 12a + 6 , , ,
10. bhow that it vt; i=rEi 3) i^hen a : :: c : a.
lOc + d 12c + d'
11. U X : y :: Z : 2 and x : 25 : : 24 : 1/, find x and y.
12. If a, b, c be in continued proportion, then
(1) a : a + b :: ab : ac;
(2) (a2 + f^) (62 + C) = {ab + bcf.
13. If a : : : c : a, show that — j— = — t— ;
and hence solve the equation
ah — bc — dx_a — h — c
bc + dx b + c '
14. If a, b, c are in continued proportion, show that
a y nib : a  mb :: b + vie : b  mc.
15. li a : h :: ."> : 4, find the value ot the ratio
1 3
16. The sides of a triangle are as 2 : 3 : 4, and the peri
2 4 '^
meter is 205 yards: tiiid the sides.
17. The sides of a triangle are as 3 : 4 : 5, and the peri
meter is 480 y^rds : find tlie sides.
AND PROPORTION. 257
1 8. Assuming a + 6 :^ + 9 '■'p — i • ab, prove that the sum
of the greatest and least terras of any proportion is greater than
the sum of the other two.
'^'1 19. A waterman rows 30 miles and back in 12 hours, ■md
he finds that he can row 5 miles with the stream in the same
time as 3 against it. Find the rate of the stream.
A,^ 20. There are three equal vessels A, B, C ; the first con
tains water, the second brandy, the third brandy and water.
If the contents of B and G be put together, it is found that the
mixture is nine times as strong as if the contents of A and G
had been put together. Find the ratio of the brandy to the
water in the vessel G.
21. A factor buys a certain quantity of wheat which he
sells again so as to gain 5 per cent, on his outlay, and thus
clears £16. Had he sold it at a gain of 5s. a quarter lie would
have cleared as many pounds as each quarter cost shillings.
How many quarters did he buy, and what did each quarter
cost him ?
22. A man buys a horse and sells it for £144, gaining as
much per cent, as the horse cost him. What was the price of
the horse 1
23. I buy goods and sell them again for £96, gaining as
much per cent, as the goods cost. "What is the cost price ?
24. A man bought some sheep and sold them again for £24,
gaining as much per cent, as the sheup cost him. What did he
give for them ?
^^ 25. A certain crew, who row 40 strokes per minute, start
'at a distance equivalent to four of their own strokes behind
another crew, who row 45 strokes to the minute. In 8 minutes
the former succeed in bumping the latter. Find the ratio
between the lengths of the strokes of the two boats.
26. The time which an express train takes to travel a
journey of 180 miles is to that taken by an ordinary train a»s
9 : 14. The ordinary train loses as much time from stoppnges
as it would take to travel 30 miles without stoppini;. The
express train only loses half as mucli time as the o:hfi in this
25S OA' VARIATION.
manner, and it also travels 15 miles an hour quicker. Sup
posing the rates of travelling uniform, what are they in miles
per hour ]
, 27. An article is sold at a loss of as much per cent, a? it
/lis, worth in pounds. Show that it cannot be sold for more
than ^25.
XXIX. ON VARIATION.
363. If a sum of money is put out at interest at 5 per cent,
the principal is 20 times as great as the annual interest, what
ever the sum may be.
Hence if x be the principal, and y the interest,
x = 2()i/.
Now if we change x we must change w in ike same propor
tion, for so long as tlie rate of interest remains the same, x
will always be 20 times . as great as y, and hence if a: be
doubled or trebled, y will also be doubled or trebled.
This is an instance of what is called Direct Vari.\tion,
of which we may give the I'ullowing definition.
Def. One quantity y is said to vary directly as another
quantity x, wlien y depends on x in such a manner tiiat any
increase or decrease made in the value of x produces a propor
tional increase or decrease in the value of 1/.
364. If x = my, where m is a constant quantity, that is. a
quantity which is not altLrcd by any change in the values of j,
and y,
y will vary directly as x.
For any increase made in the value of x must produce u
proportional increase in the value of y. Thus if x be doubled,
y must also be doubled, to prtserve the e(uality of x and my,
since m cannot be changed.
ON VAI^IATION. 259
365. Suppose a man can reap an acre of corn in a day.
Then 10 men can reap 60 acres in 6 days,
and 20 men can reap 60 acres in 3 dayss
So that to do the same amount of work if we double the
number of men we must halve the number of days.
This is an instance of what is called Inverse Variation,
of which we may give the following definition.
Def. One quantity y is said to vary inversely as another
quantity x, when y depends on x in such a manner that any
increase or decrease made iu the value of x produces a propor
tional decrease or increase in the value of y.
366. If a; = — , where m is constant,
y
y will vary inversely as x.
For any increase made in the value of x must produce a pro
portional DECREASE in the value of y. Thus if x be doul)led,
y must be halved, to preserve the equality of x and — .
■ For 2x = = — .
y y
2
367. If 1 man can reap 1 acre in 1 day,
5 men can reap 20 acres in 4 days,
and 10 men can reap 80 acres in 8 days.
That is, the number of acres reaped will depend on the
product of the number of men into the numl^er of days.
This is an example oi joint variation, of which we may give
the following definition.
Def. One quantity x is said to vary jointly as tw^o others
;/ and j, when any change made in x produces a proportional
change in the product of y and z.
368. One quantity x is said to vary directly as y and
inverseh' as z when x varies as .
z
2bo O.y \ AklAT.'OS'.
369. Theorem, ll x varies as y when % is constant, and
«s z when '/ is eonstiiiit, then wlien y and z are both variable,
% varies as yz.
Let x — tth. yz.
Then we have to show thnt in is constant.
Now when z is constant,
X varies as y ;
.". mz is constant.
Now z cannot involve y, since z is constant when y changes,
and therefore m cannot involve y.
Similarly it may be shown that m cannot involve z ;
:. m is constant,
and X varies as yz.
370. The symbol oc is used to express variation; thus xocy
stands for the words x varies as y.
371. Variation is only an abbreviated form of expressing
proportion.
Thus when we say that x varies as y, we mean that x bears
to y the same ratio that any given value of x beiirs to the
corresponding value of y, or
x : J/ = a given value of x : the corresponding value of y.
And similarly for the other kinds of variation, as will be
.seen from our examples.
Ex. 1. If xoc y and i/oc,t, to show that xocj.
Let x=my, and y = iiz.
Then substituting this value of y in the first equation.
x = m}U ;
•Old therefore, since mn is constant,
• OCS.
ON VARIATION. 2C1
Ex. 2. If a;cci/ and xocz, then will xcc ^/(i/js).
Let x = mi/, and x = ns.
Then x^ = mnyz\
;. x= v/(mTO) . V(l/«)
Now J(mn) is constant ;
.. re ex: VM
Ex. 3. If y vary as x, and when x=l, i/ = 2, what wiU be
the value of y when x = 2 ?
Here 1/ : x= a given value of y : corresponding value ot x;
:.y:x = 2:l:
.. y = 2x.
Hence, when x = 2, y = 4.
Ex. 4. If A vary inversely as B, and when A = 2, B=12,
what will B become Avhen A =91
Here yl : j, = a given value of A
corresponding
value
ofS'
A
1
12"
9
12"
 ^ .
2
*^
2
1
\2'
1
B
24
9
8
~3'
1
Hence, when ^ = 9,
whence
Ex. 5. If A vary jointly as B and G, and when yl = 6, J5 = 6
and (7= 15. find the value of A when 5= 10 and C=3.
Here
A : BG= a given value of A : corresponding value of BC\
:. A :BC='6 : 6x15;
.. 90.4 =6B0.
262 Uy iARIAlJOW
Henee, when J5= 10 and C=3,
90^ = 6 X 10 X 3 ;
••"*90^
Ex. 6. If z vary as x directly and y inversely, and if when
2 = 2, x = 3 and i/ = 4, what is the value of z when x=15 and
TT a; . ■ 1 r corresponding value of x
Here % :  =*'a civen value of z : ^ — ^.^ = jr ;
y " corresponding value oi y
X
\ z :
y
:!.
. 32
■■ 4
_2j;
nd j/ =
8,
32
4
30
8'
.'. «=
120 ^
^T4 = ^
Examples. — cxxxlii,
II
1. If ^oc^ and Boc — then will ^ocC.
2. If .loc^then Avill^oc^.
3. U Acr.B and Coc D then will ^ C<x 5D.
4. If xccj/, and when x = 7, J/ = 5, find the value of z whew
y = 12.
5. If xec , and when x» lO, y = t, find the value of y when
:c = 4.
ON VARIATION. 263
6. \i xazyz, and when x = l, i/ = 2, a = 3, find the value of y
when X = 4 and ^ = 2.
7. If xoc^, and when x = 6, ]/ = 4, and 2 = 3, find the value
of X when 1/ = 5 and a = 7.
8. If 3x + 5?/ oc 5a; + 3y, and when a; = 2, 1/ = 5, find the value
. X
of .
y
9. If ^cci> and B^ozC^, express how J. varies in respect
of a
10. If z vaiy conjointly as x and y, and 2=4 when x=l
and 2/ = 2, what will be the value of x when s = 30 and y = Si
11. If ^ocB, and when A is 8, 5 is 12; express A in
terms of B.
12. If tlie square of x vary as the cube of y, and x = 3 when
7/ = 4, find the equation between x and y.
13. If the square of x vary inversely as the cube of y, and
« = 2 when i/ = 3, find the equation between x and y.
14. If the cube of x vary as the square of y and x = 3 when
i/ = 2, find the equation between x and 3/.
I?. If XOC5J and i/oc, show that xoz.
^ z' y
16. Show that in triangles of equal area the altitudes vary
iiurersely as the bases.
17. Show that in parallelograms of equal area the altitudes
vary inversely as the bases.
18. H y=p + q + r, where p is invariable, q varies as x, and
r varies as x^, find the relation between y and x, supposing
that when a;=l, y = 6; when x = 2, y=ll ; and when x = 3,
2/ = 18.
19. The volume of a pyramid varies jointly as the area of
its base and its altitude. A pyramid* the base of which is 9
264 ON ARITHMETICAL PROGRESSION.
feet square and the height of which is 10 feet, is found to con
tain 10 cubic yards. What must be the height of a pyramid
upon a base 3 feet square in order that it may contain 2 cubic
yards ?
20. The amount of glass in a window, the panes of which
are in every respect equal, varies as the number, length, and
breadth of the panes jointly. Show that if their number varies
as the square of tlieir breadth inversely, and their length varies
as their breadth inversely, the whole area of glass varies as the
square of the length of the panes.
XXX. ON ARITHMETICAL PROGRESSION.
372. An Arithmetical Progression is a series of
numbers wliich increase or decrease hxj a constant difference.
Thus, tlie following series are Arithmetical Progressions:
2, 4, 6, 8, 10;
9, 7, 5, 3," 1.
The Constant Difference being 2 in the first series and  2
in the second.
373. In Algebra we express an Arithmetical Progression
thus : taking a to represent the first term and d to represent
the constant ditt'erence, we shall have as a series of numbers in
Arithmetical Progression
o, a + d, a + Id, a + ;j(/,
and so on.
We observe that the terms of the series differ only in the
coefficient of d, and that each coefficient of d is always less by 1
than the number of the term in which that particular coefficient
stands. Thus
the coefficient of d in the 3rd term is 2,
in the 4th 3,
? in the 5th 4.
OM ARITHMETICAL PROGRESSION. 265
Consequently the coefficient of d in the m"" term will be
Therefore the v!^ term of the series will be a i (n  1; d..
374. If the series be
a, a + d, « + 2(i,
'anil % the last term, the term next before z will clearly be 2  d.
and the term next before it will be s  2d, and so on.
Hence, the series written backwards will be
2, z  fZ, 3  2rf, a + 2d, a + rf, a.
375. To find the s^im of a series of numbers in Arithvietical
Progression.
Let a denote the iirst term.
... d the constant difference.
... z the last term.
... n the number of terms.
... s the sum of the 7i terms.
Then s = a+(a + d) + (rt + 2(f)+ +{z~2d) + (zd) +z.
Also s = z + {zd)+ {z2d)+ +(« + 2d) + (a + rf) + a,
the series in the second case being the same as in the lirst, Vmt
written in the reverse order.
Therefore, by adding the two series together, we get
•2s={a + z) + (a + z) + (a + z)+ + {a + z) + (a + z) + (a + z) ;
and since on the righthand side of this equation we have a
series of n numbers each equal to a + z, we get
2s = n{a + z)',
This result may be put in another form, because in the
place of z we may put a + {n—l)d, by Article 373.
Hence s = ~\a + a + {n i.)d\,
thatio, ={2a + (nl)di.
/
266 ON ARITHM£:TICAL PROGRESSION.
376. We have now obtained the following results :
a(=a + (Ml)rf (A),
«=(« + ^') (B),
< = )2« + (nl)cf( fC).
From one or more of these equations we have in Examples
to determine the values of a, d, n, s or z. We shall now jjro
(•ee<l to j^ive instances of such Examples.
Ex. 1. Find the LAST TERM of the series
7, 10, 13, ...... to 20 terms.
Taking the equation z = a+ {n — \)i,
for a put 7 and for n put 20, and we get
« = 7 + (20l)i,
or, !J! = 7+19d.
Now d is always found by taking the first term from ike second^
and in this case,
r^=107 = 3;
.. 2 = 7 + 19x3 = 7 + 57 = 64.
Ex. 2. Find the last term of th  series
12, 8, 4, to 11 term.=^.
In the equation z = a^ {n—\)d,
]iut a = 12 and n=ll.
Then z=\<i + \Od.
Now d = 812=4.
Hence, a = 12 40 =28.
Examples. — cxxxiv.
Find the last term of each of the following siiries •
1. 2, 5, 8 to 17 terms.
2. 4, 8. 12 to 50 terms.
On arithmetical PROGRESSIOiV. Z'oJ
3
^ 29 15 . .^ ^
7,r,Tr to 16 terms.
' 4 2
4.
1 5
^,—1,  to 23 terms.
r..
5 11 . n.
6' 2' 6 to 12 terms.
6.
12, 8, 4 to 14 terms.
7
3, 5, 13 to 16 terms.
8.
wln2n3
, , to ?i terms.
n n n
9
(x ^yf^x^ + y, {xijf to n terms.
lO.
a fe 4a 36 7a — 56
., ^, — ^— to 71 terms.
a + 6' a + 6' a + 6
377. Ex. 1. Find the sum of the series
3, 5, 7 to 12 terms.
In the equation s = {2a+ («  1) (l\
put 3 for a and 12 for n, and we get
19
Now d = 5  3 = 2, and so
s = ^{6 + 22; =6x28 = 168.
Ex. 2. Find the sum of the series
10, 7, 4 to 10 terms.
s = !2a + (»l)d{:
put 10 for a and 10 for n, then
,=12j20 + 9d(.
26S ON ARITHMETICAL PROGRESSION.
Now rf«=7  10=  3, and therefore
a = l?)2<v27!=5xf 7)= 35.
EXAMPLES. — CXXXV .
Find the sum of the following series :
1. 1, 2. 3 to 100 tdins.
2. 2, 4, () to 50 t(Mm3.
3. 3, 7, 11 to 20 terms.
4. , , 7 to 15 terms.
4' 2 4
5. 9, 7, 5 to 12 term
6. . ,  to 17 terms.
6' 2' 6
7. 1, 2, 3 TO n terms.
8. 1, 4, 7 to ?i terms.
g. 1, 8, 15 TO n terms.
n  1 ?!  2 ?i  3 . ,
10. , , to 7? terms.
n n n
378. Ex. What is tlie Constant Differekci!; when the
first term is 24 and the tenth teiiii is  12?
Takintf tlie equation (A),
z = a + (n  l)d,
ajid re^'ardinij tlie tenth as the last term, we get
12 = 24 + (10l)rf.
or  36 = Off,
whence v,m obtain d= — 4.
ON ARITHMETICAL PROGRESSION. t j
Examples. — cxxxvi.
What is the Constant Difference in the following cases %
I. When the first term is lOo and the twentieth is — 14.
2 c fiftyfirst is  x.
3 —= fortyninth is 5.
3 3
4 — ^ twentyfifth is 21.
5 10 sixth is 20.
6 150 ninety first is 0.
379. Ex. What is the First Term when
the 4()th term is 28 and the 43rd term is 32 ?
Taking equation (A),
2 = a + (n l)c?,
and regarding the last term to be the 40th, we get
28 = a + 39(7 (1).
Again, regarding the last term to be the 43rd, we get
32 = af42tf (:)
From equations (1) and (2) we may find the value of a to
be 24.
Examples. — cxxxvii.
I. What is the first term when
(i) The 59th terra is 70 and the 66th term is 84;
(2) The 20th term is 93  356 and the 21st is 98  376 ;
(3) The second term is ^ and the 55th is 58 ;
(4) The second term is 4 and the 87th ir;  SO ?
fijro ON ARITHMETICAL PROGRESSION.
2. The Sinn of the 3rd and 8th terms of a series is 31, and
the sum of the 5th and 10th terms is 43. Find the sum of
10 terms.
3. The sum of the 1st and 3rd terms of a series is 0, and
the sum of the 2nd and 7th terms is 40. Find the sum of
7 terms.
4. If 24 and 33 Ije the fourth and fifth terms of a series,
what is the 100th term ]
5. Of how many terms does an Arithmetical Progression
consist, whose difference is 3, fir.st term 5 and last term 302 ?
6. Supposing that a body falls through a space of 16^^ feet
in the first second of its fall, and in each succeeding second
32 feet more than in the next preceding one, how fir will a
body fall in 20 seconds?
7. What debt can be discharged in a year by weekly pay
ments in arithmetical progression ; the first pajTiient being 1
shilling and the last ^5. 3*'. ?
8. Find the 41st term and the sum oi 4i lernis in each of
the following series :
(1) 0,4,13
(2) 4a2, 0, 40.2
(3) 1 + a, 5 + 3.r, 9 + 5x
(4) 4 1'4
V.5.) 4> 20
9. To how many terms do the following series extend, and
what is the sum of all the terms ?
(1) 1002 10,2.
(2) 0, :: ,186.
ON ARITHMETICAL PROGRESSION. 271
(3) 22X, Sa; 723je.
/ X 1 1
(4) 2' 4 ^
(5) m\ 137(1 m), 135>a m).
(6) a; + 254, x + 2, x2.
380. To insert 3 arithmetic niMyis between 2 and j.O.
The number of terms will be 5.
Taking the equation z = a + {n i) d,
we have 10 = 2 + (bl)d.
Whence 8 = 'id; :. d=2.
Hence the series will be
. 2. 4, 6, 8, 10.
Examples. — cxxxviii.
1. Insert 4 arithmetic means between 3 and 18.
2. Insert 5 arithmetic means between 2 and —2.
2
■*. Insert 3 arithmetic means between 3 and .
^ 3
4. Insert 4 arithmetic means between  and .
381. To insert 3 arithmetic menus between a and b.
The number of tenu^ in the series will be 5. since the;,
are to be 3 terms in addition to the iirst term a and the last
term b.
Taking the equation 2 = a + (n — iy a,
we have to find d, having giveii
a, z = b and n — b.
272 UN ARl'l HAfETICAL PROGRESSION.
Hence h = a\(^\)d.
or, 4<i=6a, .". d=— p .
Hence the series will be
6 — a h — a 3(6 — a) .
«, a + —4—, « + 2"' '^ + ~ 4 — "' °'
that IS, a, ^p, — g— , —J—, 6.
Examples. — cxxxix.
1. Insert 3 arithmetic means between rfi and n.
2. Insert 4 arithmetic means between m + 1 and in\.
3. Insert 4 arithmetic means between 11^ and ?! + 1.
4. Insert 3 arithmetic means between x^ + y and a;'^ — y'.
382. We shall now give the general form of the proposition
" To iiisert m arithmetic means beticeen a aiid b."
The number of terms in the series will be 7?i + 2
Then taking the equation z = a + (n'\)d,
we have in this case b = a + {m + 2  1 ) f/,
or, b=a+{m + l)d.
Hence d= ,,
m + V
and the form of the series will be
, 26  2a , b — a .
m+1 m+V '
bmb + ia bm + a ,
m+1 ' m + l' '
a
ba
' m+1'
26 2a
a + — ,
m+1 '
that
is,
am + b
avi  a + 26
'^' »r+T'
m + 1 '
XXXI. ON GEOMETRICAL PROGRESSION.
383. A Geometrical Progression is a series of mnuliers
which increase or decrease by a constant factor.
Thus the following series are Geometrical Progressions.
2, 4, 8. 16, 32, 64;
12 3 2 A ^•
^^' "*' 4' 16' 64'
_1 ^ __! Jl
2' 16' 128' 1024'
The Constant Factors being 2 in the tirst series,  in tin
4
second, and —  in the third.
8
Note. That which we shall call the Constant Factor is
usually called the Common Ratio.
384. In Aljj;eV)ra we express a Geometrical Progression
thus : taking a to represent the jfirst term and / to represent
the Constant Factor, we shall have as a series of numbers in
(ieonii'trical Progression
a, of, af'^, af^, and so on.
We observe that the terms of the series differ only in the
index of/, and that each index of/ is always less by 1 than the
number of the term in which that particular index stands.
Thus the index of/ in the 3rd term is 2,
in the 4th 3,
in the 5th 4
Consequently the index of/ in the nth term will be n  1.
Therefore the ?ith term of the series will be a/"~'.
[s.A,] 8
274 ON GEOMETRICAL PkOGKESSlON.
Hence if z be the last term,
385. If the series contain 7! terms, a being the first term
and / tlie Constant Factor,
the last term will be a/""',
the last term but one will be a/"~*,
the last term but two will be a/*~*.
Now a/"' x/=a/'i x/i = o/'i+' = a/",
a/"^ X /= «/^ X /' = a/"'+' = a/\
a/^s X /= a/'=' X /I = a/"='+' = a/*.
386. We may now proceed to Jind the tmm of a teries of
numbers in Geometrical rrogression.
Let a denote the first term,
/ the constant factor,
71 the number of terms,
s the sum of the n terms.
Then s = a + af+ af+...+ af^ + of'' + afK
Now multiply both sides of this equation by/, then
fs = af+ af^ + af+ ... + af"^ + «/"' + af".
Hence, subtracting the first equation from the second,
fss=^af"a.
■•• «(/!)=« (/"I);
•■' fi •
Note. The proposition just proved presents a difficult}' to
a beginner, which we shall endeavour to explain. When we
multiply the series of ?! terms
a + af+afi + af'^ + af^ + c^f^
OiV GEOMETRICAL PROGRESSION. 275
by/, we shall obtain another series
af+af + af + + a/'' + «/» + a/",
which also contains n terms.
Though we cannot fill up the gap in each series completely,
we see that the terms in the two series must be the same,
except the first term in the former series, and the last term in
the latter. Hence, when we subtract, all the terms will dis
appear except these two.
387. From the formulae :
2 = a/"' (A),
.."^' (B,
prove the following :
(a) sJj^. (y) a=fz{f\)s.
f
s — a
z'
388. Ex. Find the last term of the series
3, 6, 12 to 9 terms.
The Constant Factor is , that is, 2.
In the formula
3 = a/— »,
putting 3 for a, 2 for/, and 9 for n, we get
3 = 3x25 = 3x256 = 768.
Examples.— cxl.
Filid the last term of the following series
1. 1, 2, 4 to 7 terms.
2. 4, 12, 36 to 10 terms.
3. 5, 20, 80 tu 9 terms.
276 ON GEOMETRICAL PROGRESSION.
4. 8, 4, 2 to 15 terms.
5. 2, 6, 18 to 9 terms.
6 ^' 1^' 4. to 11 term..
2 1 1 ^ n,
7 3' 3' 6 to 7 term*.
389. Ex. Find thf sum of the series
3
2
3
6, 3, ^ to 8 terms.
Generally, s= — 7— j —
anti here a = 6,/=^, « = 8,
2 2
6__ 6_
256 256 _ 766
" _1 1 ~~"6T*
2 2
EXAMPLES.— CXli.
Find the .sum of the following series :
1. 2, 4, 8 to 15 terms.
2. 1 , 3, 9 to 6 terms.
3. a, ax^, ax* to 13 terms.
4.. a, , ., to 5) terms
,r ./
0^x, a X, — ; — to 7 terms.
' a + x
osr Geometrical pkockESSioM. 277
6. 2, 6, 18 to n terms.
7. 7, 14, 28 to ?i terms.
8. 5, 10, 20 to 8 terms.
2 1 1 ^ r, ^
9 3J 35 g to 7 terms.
390. To find the sum of an Infinite Series in Geometrical
Progression, when the Constant Factor is a proper fraction.
If/ be a proper fraction and n very large,
/" is a very small number.
Hence if the number of terms be infinite, f" is so small that
we may neglect it in the exjiression
,_«(/" 1)
/I, '
and we get
a
"I/'
391. Ex.1. Find the sum of the series 5 + 1 + 7 + to
infinity.
Here /=1 = !'
4
_ 0^ 3 16_1
•'•*~i7~7~3~T~^3
^4
3 2 8
Ex. 2. Sum to infinity the series g ~ o + 07 ~
Here /=^=!;
3 3
a 2 V 2 27
/ 4\ , 4 26"
(9) ^+
9
27^ ox GEOMETRICAL PROGRESSION,
Examples. — cxlii.
Find the sum of tht I'olldwiug infiniie series;
I. 1, i \ 9 4^ 2*. ..
2' 4'
2. 1. . ~ lo. 2z^,  SSa;^
i in
3 3, ,  II. o, 6,
o Z t
2 11 11
4 o. o> 5, 12.
3' 3' 6' 10' 10^'
13. X, I/, .
3 1
)• 4' 4'
11 ^ 86
2" ~3 ''^' 100' 10000
•7 8, I, 15. 54444 _
3'
8. l, 5, 16. 83636,
392. To inst'ti 3 geometric means heticeen 10 and 160.
Taking the equation z^af"^^,
we put 10 for a, 160 for z, an.l 5 for ?i, and we obtain
160=10./'':
.. 16=/*.
Now 16 = 2x2x2x2 = 2*;
• 2* =/*.
Hence /=i;. and the serie.s will be
10, 20, 40, 80, 160.
ON GEOMETRICAL PROGRESSlOX. if^
Examples.— cxliii.
1. Insert 3 geometric means between 3 and 243.
2. Insert 4 geometric means between 1 and 1024.
3. Insert 3 geometric means between 1 and 16.
4. Insert 4 geometric means between  and — .
393. To insert m geometric riieans between a (ind b.
The number of terms in the series will be m + 2.
In the formula z = af'''''^,
putting b for z, and to + 2 for n, we get
or, 6 = f^"'+l;
•••' ~a'
or, /=~t:.
Hence the series will be,
1 _i_ _}_ 1
rt, a X — p , a X
6^— r, 6^— T, t,
that is,
II J 1
a, (rr . &)+', (ft"i.62)m+i^ ^ (a^S—y+i, (; . Z/")"^!, />.
394. AVe shall now give some mixed Examples ou Aiitl:
raetical and Geometrical Progression.
Examples. — cxliv.
I. Sum the following series :
(i) 8 + 15 + 22+ tol2terms.
(2) 116 + 108 + 100+ to 10 terms.
28o ON GEOMETRICAL PROGRE.SSTOM.
(3) 3 + 2'^12"^ to infinity.
'4) 2   + —  to infinity.
4 oz
1 2 11
(5) 2~3~y ^"^"^ terms.
112
(6) 9~o+q— to 6 terms.
1 5
(7) g1^ to 29 terms.
(8) s + l + l?+ toSterms.
(9) 3 + 9 + 27"^ ••••■■• to infinity.
, , 3 14 Ol i ^/^i .
(10) V — T7^r to 10 terms.
5 10 lo
('0 /v/? v'6 + 2V(l''') to 8 terms.
V 5
, , 7 7 35 ^  i
(12) ^ + s — r+ to .5 terms.
o 2 4
2. If the continued product of 5 terms in Geometrical
Progression be 32, show that the middle term is 2.
3. If a, h, c are in arithmetic jirogression, and a, ?/, (• :i;v
.1 • 1 *i 4. ^ <* + <'
in geometrical jirogressioii, show that 17 = 5 — 77 — r.
4. Show that the arithmetical mean between a and h i
i:reater than tlie geometrical mean.
5. The sum of the first three terms of an arithmetic series
is 12, and the si.xth term is 12 also. Find the sum of the first
6 terms.
6. What is necessary that «, 6, c may be in geometric pro
grescsion ?
ON GEOMETRICAL PROGRESSION. 281
7. If 271, X and ^r are in cfeometric protrression, what is x?
8. If 2n, ?/ and — are in aritlimetic progression, what is 1/?
9. The sum of a geometric progression whose firet term is
1, const nit factor 3, and number of terras 4, is equal to the sum
of an arithmetic progression, whose tirst term is 4 and constant
difference 4 ; how many terms are there in the arithmetic pro
gression?
10. The tirst (7 + ?i) natural numbers when added together
make 153. Find n.
11. Prove that the sum of any number of terms of the
series 1, 3, 5, is the square of the number of terms.
12. If the sum of a series of 5 terms in arithmetic progres
sion be 95, show that the middle term is 19.
13. There is an arithmetical progression whose first term is
1 4
3„, the constant difference is 1,;. and tlie sum of the terms is
22. Required the number of terras.
14. The 3 digits of a certain number are in arithmetical
progression ; if the number be divided liy the sum of the digits
in the units' and tens' place, the quotient is 107. If 396 be
subtracted from the number, its digits will be inverted.
Required the number.
15. If the {'pVfjf' term of a geometric progression be'm,
and the {p — qf" term be n, show that the 2^"" term is >^f{mn).
16. The ditt'erence between two numbers is 48, and tlie
arithmetic mean exceeds tlie geometric by 18. Find the
numbers.
17. Place three aiithmctic means between 1 and 11.
18. The first term of an increasing arithmetic series is "034,
the constant difference •0004, and the sum 2748. Find the
number of terms.
19. Place nine arithmetic jueans between 1 ami  1,
282 ON HARMONIC AL PROGRESSION.
20. Prove that every term of the series 1, 2, 4, is
greater by unity than the sum of all that precede it.
21. Show that if a series of my) tenns forming a geometrical
progression whose constant factor is r be divided'into sets of p
consecutive terms, the sums of the sets will foim a geometrical
progression whose constant factor is r'.
22. Find five numbers in arithmetical progression, such
that their sum is 55, and the sum of their squares 765.
23. In a geometrical progression of 5 terms the difference
of the extremes is to the difference of the 2nd and 4th terms
as 10 to 3, and the sum of the 2nd and 4tli terms equals twice
the product of the 1st and 2nd. Find the series.
24. Show that the amounts of a sum of money put out at
Compound Interest form a series in geometrical progression.
25. A certain number consists of three digits in geometrical
progression. The sum of the digits is 13, and if 792 lie added
to the number, the digits will be inverted. Find the number.
26. Tlie population of a county increases iu 4 years from
10000 to 146 il ; what is the rate of increase ?
XXXII. ON HARMONICAL PROGRESSION.
395. A Harmonicai Progression is a series of numbeis
of which the reciprocals form an Arithmetical Progression.
Tims the series of numl)ers «, h, c, <l, is a Harmon ical
• .•1 .1111. ..,.,
Progression, 11 the series , r> > "7? ^s an Anthmetical
a bed
Progression.
If a, b, e be in Harmonicai Progression, b is called the
Harmonicai Mean between </ and c.
Note, There is no way of finding a general expression for
the sum of a Harmonicai Series, but manv problems with
OA' HARMONICAL PROGRESSTON. 283
reference to sncb a series maybe sobbed by inverting tbe terms
and treating the reciprocals as an Arithmetical Series.
396. J/a,,b, c he in Harmonical Progression, to show that
a : c :: a — b : b — c.
Since , ,) ' are in Arithmetical Protrression,
c 6 5 a'
bc a—b
DC ao
ab ab
a a — b
or  = T^ — .
c b — c
397. To insert m harmonic means between a and b.
First to insert m arithmetic means between  and t
a h
Proceeding as in Art 357, we have
a '
or a = 6 + (m + l).a6(£
, ab
ab (?«.+ 1)
Hence the aritlimetic series will be
11^ ab 1 , 2 (ffl  6) 1 m{ab ) 1
a' rt'a6(ni+l)' a ab{m+iy a a6(m+l)' b'
1 6?M + a 6//1 + 2ab am + b 1
a' a6(»irl)' ab{m+l)' ah m + iyb'
Therefore the Harmonic Series is
ab(m + \) abjm+^l) ab{m + l)
' 6m r a ' hm + 2ab' am + b '
284 ON HARMONICAL PROGRESSION.
398. Given a and h the first two terms of a series in Har
monical Progression, to find the n*'' term.
, T are the first two terms of an Arithmetical Series of
o
which the common difference is t — .
6 a
The w"' term of this Arithmetical Series is
1 (n — 1) (o  6) _ 5 + Tia  a — n6 + ft
a ah ah
*
(Tta  g)  inh  2b) _ ( n  1) a {n2) h
ah ~ ah
.'. the n** term of the Harmonical Series is
(rr^)a(n2)6"
399. Let a and c be any two numbeis,
6 the Harmonical Mean between tliem.
1111
Then t — = i>
b a c
2 a + c
or T= ;
ac
,_ 2ac
~a + c'
400. The following results should be remembered.
Arithmetical Mean between a and c = —^ — .
Geometrical Mean between a and c= ^ac.
2ac
Harmonical Mean between a and c = — — .
, a + G
ON HARMONICAL PROGRESSION. 285
Hence if we denote the Means by the letters A, G, H
respectively,
A X 11=—— X
= ac
that is, (? is a mean proportional between A and H.
401. To show that A, G, H are in descending order of
magnitude.
Since ( ^'a  aJc) must be a positive quantity.
( V«  */c)^ is greater than 0,
or a — 2 ,Jac + c greater than 0,
or a + c greater than 2 i^fac,
a + c . , —
or — greater than ^ac ;
that is, A is greater than G.
Also, since a + c is greater than 2 ^ac,
Jac (a + c) is greater than 2ae ;
,— . , 2ac
:. Jac IS greater than — ; — ;
^ * a + c
i.e. G is greater than H.
Examples.— cxlv.
I. Insert two harmonic means between 6 and 24.
2 four 2 and 3.
3 three  and .
4 foiir and—.
286 ON HARMONICAL PROGkESSlON.
5. Insert five harmonic means between — 1 and 2~^.
6 five ^and.
7 SIX 3 and — .
8 n 2x and By.
9. Tlie sum of three terms of a harmonical series is Yg> *"*!
the first term is  : find the series, and continue it both ways.
10. The arithmetical mean between two numbers exceeds
the geometrical by 13, and tlie geometrical exceeds the har
monical by 12. What are the numbers?
11. There are four numbers a, 6, c, d, the first three in
arithmetical, the last three in harmonical progression ; show
that (t : 6 = c : rf.
12. If X is the harmonic mean between m and n, show that
_1_ _1_ = J_ 1
xtii xn m n
13. The sum of three terms of a harmonic series is 11, and
the sum of their squares is 49 ; find the numbers.
14. If X, y, z be the //"", 5"*, and r* terms of a h.p., show
that {rq)y + {P r)xz + {q p) xy = 0.
15. If the H.M. between each pair of the numbers, a, b, c
he in a. P., then b', a, c'^ will be in H.P. : and if the h.m. be in
H.P., b, a, c will be in H.P.
16. Show that ^ — ~+ =4, >7, or >10, according as
c — c — a
c is the A., G. or H. mean between a and b.
XXXIIi. PERMUTATIONS.
402. The different arrangements m respect of order of suc
cession wliich can be made of a given number of things are
called Permutations.
Thus if from a box of letters I select two, P and Q, I can
make two permutations of tliem, placing P first on the left and
then on the right of Q, thus :
P, Q and Q, P.
If I now take three letters, P, Q and R, I can make six per
mutations of them, thus :
P, Q, B ; P, R, Q, two in which P stands first.
Q,P,R; Q,R,P, Q
R,P,Q; R,Q,P, R
403. In tlie Examples just given all the things in each case
are taken together ; but we may be required to find how many
permutations can be made out of a number of things, when a
certain number only of them are taken at a tinie.
Thus the permutations that can be formed out of the letters
P, Q, and R taken tivo at a time are six in number, thus:
P,Q; P,R; Q,P; Q,R; R,P; R, Q.
404. To find the nuvdicr of jJermutations of n different things
taken t at a time.
Let a,h, c, d ... stand for n difi'erent things.
First to find the number of permutations of the n things
taken two at a time. ,
If a be placed before each of the other things 6, c, d ... of
which the number is n— 1, we shall have n—\ permutations
in which a stands first, thus
ah, ac, ad,
2^8 PERMUTA TIONS.
If I be placed before each of the other thiiifjs, a, c, d ... we
shall have «  1 permutations in which b stands first, thus :
ba, be, bd,
Similarly there will be n 1 permutations in which c stands
first: and so of the rest. In this way we get every possible
permutation of the 71 things taken two at a time.
Hence there will be n . (n  1) permutations of n things taken
two at a time.
Next to find the number of permutations of the n things
taken three at a time.
Leaving a out, we can form (n 1) . (n — 2) permutations of
the remaining (n  1) things taken tivo at a time, and if we
place a before each of these permutations we shall have
(« 1) . (?i 2) permutations of the n things taken three at a
time in which a stands first.
Similarly there will be (n  1) . (n — 2) permutations of the
n things taken three at a time in which b stands first : and so
for the rest.
Hence the whole number of permutations of the n things
taken three at a time will be n.(nl). {n2), the factors of
the formula decreasing each by 1, and the figure in the last facto?
being 1 les^s than the niuiiber taken at a tinu.
We now assume that the formula holds good for the number
of permutations of n things taken r—1 at a time, and we shall
proceed to show that it will hold good for the number of per
mutations of n things taken r at a time.
The number of permutations of the n things taken r—1 at
a time w iU be
n.{nl).(n2) [„ } (r 1)  I [],
tliat is ?i..(?il). («.2) (nr + 2).
'Leaving a out we can form {n  1) . (n  2) («  1 — r + 2)
permutations of the (nl) remaining things takrn r — 1 at a
time.
Putting a before each of these, we shall have
(nl). {n2) (nr+l)
periiuitatiniis of the n things taken r at a time in which a
stands fir>l.
PERMUTA TIONS. 289
So again we shall have (to — l).(n — 2) (?ir + l) per
mutations of the n things taken r at a time in whicli h stands
first ; and so on.
Hence the whole numtier of permutations of the n things
taken r at a time will be
n.(«l).(?i2) (7ir+l).
If then the formula holds good when the n things are taken
r 1 at a time, it ■will hold 'good when they are taken r at a
time.
But we have shown it to hold when they are taken 3 at a
time ; hence it will hold when they are taken 4 at a time, and
so on : therefore it is true for all integral values of r*
405. If the 71 things be taken all together, r = n, and the
formula gives
n. (n— 1) . (?i2) (n — nl 1) ;
that is, n.(nl).(7i2) 1
as the number of permutations that can be formed of n dif
ferent things taken all together.
For brevity the formula
TO. (71 1). (712) 1,
which is the same as 1.2.3 to,
is written 1 77. This symbol is called /a ciorwZ n.
Similarly \r is put for 1 . 2. 3 r ;
[r1 for 1.2.3 {r\\
Ohs. i 7i = n . 1 71  1 = n . (ti — 1) . ?! — 2 = &c.
406. To find the numbei of jpermutations of n things taken all
together ivhen certain of the things are alike.
Let the n things be represented by the letters a, b, c, d
and suppose that a recurs p times,
b q times,
c r times,
and so on.
* Another proof of this Theorem may be seen in Art. 475.
£s.A.l ^
290 PERMUTA TIONS.
Let P represent the whole number of permutations.
Then if all the p letters a were changed into f other letters,
different from each other and from all the rest of the n letters,
the places of these p letters in any om permutation could now
be interchanged, each interchange giving rise to a new permu
tation, and thus from each single permutation we could form
1.2 p permutations in all, and the whole nutnber of per
mutations would be (1 . 2 ...^) P, that is [p . P.
Similarly if in addition the g letters h were changed into 5
letters different from each other and from all the rest of the 7i
letters, the whole number of permutations would be
k.l^.P;
and if the r letters c were also similarly changed, the whole
number of permutations would be
ind so on, if more were alike.
But when the^, g, and r, &c., letters have thus been changed,
we shall have n letters all different, and the number of permu
tations that can be formed of them is \ n (Art. 405).
Hence P .\p . \q .\r = ?i ;
\p.\q. [r
Ex AMPLES. — CXivi.
1. How many permutations can be formed out of 12 things
taken 2 at a time ?
2. How many permutations can be formed out of 16 things
taken 3 at a time ?
3. How many permutations can be formed out of 20 things
taken 4 at a time 1
\ 4. How many changes can be rung with 5 bells out of 8 ?
5. How many permutations can be made of the letters in
the word Examination taken all together \
y,^. In how many ways can 8 men be placed side by side ?
CO MB IN A no MS. igt
7. In how many ways can 10 men be placed side by side ?
8. Three flags are required to make a signal. How many
signals can be given by 20 flags of 5 different colours, there
being 4 of each colour ?
9. How many different permutations can be formed out of
the letters in Algebra taken all together ?
I o. The number of things : number of permutations of the
things taken 3 at a time = 1 : 20. How many things are there?
11. The number of permutations of in things taken 3 at a
time : the number of permutations of j?i + 2 things taken 3 at
a time = 1:5. Find m.
12. In the permutations of a, b, c, d, e, f, g taken all
together, find how many begin with cd.
13. Find the number of permutations of the letters of the
product a^b^c* written at full length.
14. Find the number of permutations that can be formed
out of the letters in each of the following words : Conceit,
Talavera, Calcutta, Proposition, Mississippi.
XXXIV. COMBINATIONS.
407. The Combinations of a number of things are the
diflerent collections that can be formed out of them by taking
a certain number at a time, without regard to the order in
which the things stand in each collection.
Thus the comliinations of a, b, c, d taken tuo at a time are
ab, ac, ad, be, bd, cd.
Here from each combination we could make tico permuta
tions : thus ab, ba ; ac, ca ; and so on : for ab, ha are the same
combination, and so are ac, ca. <
Similarly the combinations of a, b, c, d taken three at a time
are abc, abd, acd, bed.
Here from each combination we could make six permuta
lions ; thus abc, acb, bac, bca, cab, cba : and so on.
igi COMBIXATIONS.
And, generally, in accordance with Art. 405, any combina
tion of n things niuy he made into 1 , 2 . 3 ... n permutations.
408. To fuul tJie number of combinations of n different things
taken x at a time.
Let C, denote the nnmher of combinations required.
Since each conibinatinn contains r things it can be made
into I r permntations (Art. 405) ;
.•. the Avliole number of permutations = : r . (7,.
But also (from Art. 404) the wliole number of permutations
of n tilings taken r at a time
— n{n—\) (nr + 1);
., I r . C, = n (?i  1) (?i  r + 1) ;
. ^ _ n{n\) (?tr + l)
409. To show that the number of combinations of n things
taken t at a time is tlis same as the number taken n — r at a
time.
_, n. (n 1) (7ir+l)
^'" 1.2.3 r '
and c n,(nl) \ninHl\
1.2.3 (nr)
_ n.(nl) (r+1)
~ 1.2.3 {nr) '
Hence
C, _ n.(nl) (nr+l) 1.2. 3 (nr )
C^~ 1.2.3 r ^n.{nl) (r+1)
n.(?il) (nr+l). (nr) 3.2.1
"^ 1.2.3 r. (r+1) (nl).n
\n
= 1.
That is. O.'O,^
COMBINATIONS. 293
410. Making r=], 2, 3 r 1. r, r+ 1 in order,
, _ p _ " ''^ — 1 /I _ 'I 711 712
^ ^^71.011) (»r+2)
1.2 (rl)
„ n.(nX) (7ir + 2). Oi r+1)
1 .2 (r1). '•
71 . (n — 1) (7? — r + 1 ) . (71 — r)
1.2 r.(r+l)
c;.=i.
Hence the general expression for the factor connecting Cv,
one of the set of numbers Cj, Cj, C^i C',, with C^i,
that which stands next before it, is , that is,
^^^7^r + l
r
With regard to this factor , we observe
r
(1) It is always positive, because 71 + 1 is greater than r.
(2) Its value continually decreases, for
7!  r + 1 71+1
r 1'
which decreases as r increases.
11 J ^ 1^
(3) Though continually decreases, yet for several
•»
successive values of r it is greater than unity, and therefore
each of the corresponding terms is greater than the preceding.
(4) When r is such that '^— is less than unity the cor
responding term is less than the preceding.
294 . COMB IN A T/OXS.
71 — 7* * 1
(5) If 11 and r be such that '■ — = 1, C, and C^, are a
pair of equal terms, each greater than any preceding or suLse
quent term.
Hence up to a certain term (or pair of terms) tlie terms in
crease, and after that decrease : this term (or pair of terms) is
the greatest of the series, and it is the object of the next Article
to determine what value of r gives this greatest term fur )iair
of terms).
411. To find the value of t for which the number of combina
tions of n thincjs taken r together is the greatest.
n.(nl ) ( n rH2)
^r,  jf_2 (ri)
^ _ n. (n1) fw7 + 2 ) (nr + 1)
' 1.2 [r[) * r
„ _n.(7il) (nr+l) nr
^^^ 1.2 r r+l ■
Hence, if 0, denote the number of combinations required,
C C
j^ and ^ must neither of them be less than 1.
a nr+l
But Jt— = y
Cr r+l
and rr~ = — •
C'^i nr
vt r+l. T* + l.
Hence is not less than 1 and is not less than 1,
r n — r
or, n — r+ 1 is not less than r and r+l not less than n r,
or, n + 1 is not less than 2r and 2r not less than n — l;
:. 2r is not greater than ?! + 1 and not le.ss than n—l.
Hence 2r can have only three values, 7i — 1, n, n + 1.
Now 2r must be an even number, and therefore
(1) If n be odd, ?!  1 and 7i + 1 being both even numbers,
2r may be equal to 7i  1 or ?» + 1 ;
COMBTNA rroNs. 295
n— 1 w+ 1
(2) If n be even, n\ and n + 1 being both odd numbers,
2r can only be equal to n ;
n
■■' = 2
Ex. 1. Of eight things how many must be taken together
that the number of combinations may be the greatest pos
sible ?
Here « = 8, an even number, therefore the number to be
taken is 4, which will give = —  —  —  or 70 combinations.
1x2x3x4
KXi 2. If tlie number of things be 9, then the numlier
9 _ 1 9 I 1
to be taken is — v— or — g— , that is 4 or 5, which will givf
respectively
9x8x7x6
1x2x3x4
9x8x7x6x5
, or 126 combinations, and
or 126 combinations.
1x2x3x4x5
Examples. — cxlvii.
/^ I. Out of 100 soldiers how many different parties of 4 can
be chosen ]
( 2. How many combinations can be made of 6 things taken
' 5 at a time /
A 3. Of the combinations of the first 10 letters of the alphabet
/ 'taken 5 together, in how many will a occur ?
^ /\ 4. How many words can be formed, consisting of 3 cnn
sonants and one vowel, in a language containing 19 consonants
and 5 vowels ?
5. The number of combinations of n things taken 4 at a
time : the number taken 2 at a time =15 : 2. Find n.
6. The number of combinations of n things, taken 5 at
296 COMBINA TIONS.
3
a time, is 3_ times the number of combinations taken 3 at a
time. Find n.
. 7. Out of 17 consouants and 5 vowels, how many words
\ r can be formed, each containing 2 vowels and 3 consonants ?
. Q 8. Out of 12 consonants and 5 vowels how many words can
be formed, each containing 6 consonants und 3 vowels ?
9. The number of permutations of n things, 3 at a time, is
6 times the number of combinations, 4 at a time. Find n.
10. How many different sums may be formed with a guinea,
a halfguinea, a crown, a halfcrown, a shilling, and a sixpence ?
; ^ II. At a game of cards, 3 being dealt to each person, any
one can have 425 times as many hands as there are cards in
the pack. How many cards are there ?
I 12. There are 12 soldiers and 16 sailors. How many dif
/ ferent parties of 6 can be made, each party consisting of S
soldiers and 3 sailors ?
//,
13. On how many nights can a different patrol of 5 men be
drfiughte<l from a corps of 36 ? On how many of these would
any one man be taken \
XXXV. THE BINOMIAL THEOREM.
POSITIVE INTEGRAL INDEX.
412. The Binomial Theorem, first explained by
Newton, is a method of raising a binomial expression to any
])ower without going through the process of actual multipli
cation.
413. To investigate the Binomial Theorem for a Positive
Integral Index.
THE BINOMIAL THEOREM. 29?
By actual multiplication we can show that
(x + aj (z + a,) = x^ + («! + a^) X + Oja,
(x + ai) (x + Oa) (x + Og) =x3 + (a^ + Oj + 03) x^
(x + aj) (x 4 a^) (x + 03) (x + a^) = X* + (oj +02 + 013 + 04) i? .
+ (ttitta + ajCTj + c^a^ + a^a^ + a.ja4 + 0304) x*
+ (cfi<*2% + ffliffl2«4 + dfl'^di + a2a3a4) x + a^ajCtja^.
In these results we observe the following laws :
I. Each product is composed of a descending series of
powers of x. The index of x in the first term is the same as
the number of factors, and the indices of x decrease by unity
in each succeeding tenu.
II. The number of terms is greater by 1 than the number
of factors.
III. The coefficient of the _^rs< term is unity.
of the second the sum of a^, a.^, tij ...
of the third the sum of the products of
%, rtj, rt3 ... taken two at a time.
of the/o?t?;/i the sum of the products of
Oj, «2, ^3 ... taken three at a time.
And the last term is the product of all the quantities
«1, «2> «3
Suppose now this law to hold for 7i — 1 factors, so that
(x + tti) (x + aj) (x + «3) (x + a„_i)
= x"i + S'l . rc"2 + S^ . x"^ + ,^8 . x"^+ + S„_i,
where .S\ = a^ + aj + 03 + . . . + a„_i,
that is, the sum of aj, a.,, 03 ... a„_i,
Sf=aja.2 + a^a.^ + a^Oj + . . . + aia„_i + a„a„_i + ...
that is, the sum of the products of dj, a^, a^ ... a.»_i,
taken two at a time.
298 THE BT.VOMIAL THEOREM.
S3 = a^a^ai + aiCiM^ + . . . + a^aM^^i + aia^a„_y + ...
that is, the sum of the products of Oj, aj...a^„
taken three at a time,
that is, the product of a^, a^, 0.3 ... dni
Now multiply both sides hy x + a„.
Then
{x + ai)(x + a.,) ... (.T + a„_i) {x + a„)
=x" + Si X"' + ,S', X" + S3 x"^ + ...
+ a„ x"~^ + a„Sj x"'^ + a„S.^ x"^^ + . . . + a,S„ _i
=x'' + {Si + a„) x"i + (S3 + a„Si) x"'"^
+ {S3 + a„S.,) x"^ +... + a„S„_i.
Now Si + a„ = ai + a„ + a3 + ... +a„_i + a„
that is, the sum of Oj, a.^, 03... a„,
/Sj + a„Si = S^, + rr„ (f?! + Oj + . . . + a._i),
that is, the sum of the products of a^, aj,..a,„
taken two at a time,
Sg + a„S.2 = S3 + a„ {aia.2 + a^a^ +...),
that is, the sum of the products of a^, aj...a,,
taken three at a time,
that is, the product of Oi, a,, ^3 ... o,.
If then the law holds good for nl factors, it will hold good
for n factors : and as we have shown that it holds good up to 4
factors it will hold for 5 factors : and hence for 6 factors : and
so on for any number.
THE BINOMIAL THEOREM. ig^
Now let each of the n quantities a^, a^, a^... a„he equal to
a, and let us write our result thus :
{x + a^) {x + a.^) ...{x + a„) ^x' + Ai . x"~* + ^2 . x"'+ ... +A^.
The lefthand side becomes
{x + a) {x + a)...{x + a) to n factors, that is, {x + a)'.
And on the righthand side
Ai = a + a + a+ ...to n terms = ?;«,
A^ = a^ + a^ + a^+ ...to as many terms as are equal to the
number of combinations of n things taken two at a time, that
. n .(nl)
. _ n.{nl)
.. A^ ^^ .a,
A3 = a^ + a^ + a^+ ...to as many terms as are equal to the
number of combinations of n things taken three at a time, that
. n.{nl) .{n2)
1.2.3
_^. (nl). (n2 )
^' 17273 •"''
A„ = a . a . a ...to n factors = a".
Hence we obtain as our final result
/ N„   1 n . (n  I') „ . ,
{x + a)" = x" t Tiaa;""'  ^^ — ^■' ax*^
n.(n\) . (n2) , .^
1.2.3 r...rt*
414. Ex. Expand (x + a)6.
Here the number of terms will be seven, and we have
^6.5.4.3 ,2,6.5.4.3.2 , .
+ 1727^74 "^^17273:476 " ^^^"
— x^ + Qaufi + 15aV I 20aV i 15a*x*i 6a^x + afi.
300 THE BINOMIAL THEOREM.
Note. The coefficients of terms equidistant from the end
and from the beginning are the same. The general proof of
this will be given in Art. 420.
Hence in the Example just given when the coefficients of
font terms had been found those of the other three might have
beeu written down at once.
Examples.— cxlviii.
Expand the following expressions :
I. (a + x)*. 2. (6 + c)8. 3. (a + 6)^
4. (x + i/)8. 5. (5 + 4a)*. 6. {a^^hcf.
415. Since
1 n . («  1) , ,
(,c + a) " = a:" + naT^"'^ + ^ — „— ' . aV'* 4 ... + a",
if we put x= 1, we shall have
(1 +a)" = l +na + — Y~a~~ ^ "^ ••• +<* •
416. Every binomial may be reduced to such a form that
the part to be expanded may have 1 for its first term.
Thus since x + a = x(l+Y
(x + a) = x"(l+^);
and we may then expand (l +  j and multiply each term of
the result by x".
Ex. Expand (2.c + 3y)\
(2x + 32/)6=(2x)''.(l+y
, 54.3.2 /3i/y /3yy
"^1.2.3.4\2x/ ■^V2x/ I
THE BINOMIAL THEOREM. ^oi
= 32x5 + 240x*y + 720x3i/2 + loSOxy + SlOxi/^ + 243?/.
417. The expansion of (x — a)" will be precisely the same as
that of {x + a)", except that the sign of terms in which the odd
powers of a enter, that is the second, fourth, sixth, and other
even terms, will be negative.
Thus
[X  a)" = X" — 7utx"~' + — ^ —  — . ah:"'*
TO ■ (w  1) . (?i  2)
1.2.3
for (x — a)'=\x + (a)\'
^x' + nia) x'' + ^_lC^_rLl) ( _ afx'' + &c.
Ex. Expand (a  c)*.
/ ^^ ^ r4 5432 5.4.3,, 5.4.3.2^ ^
(a  cf = a»  5a*c  j— g ~ r~2"~3 "^ 12 3 4 ~
= a^  5a*c + 10u\'2  lOa^c^ + 5ac*  c^
Examples. — cxlix.
Expand the following expressions :
I. (ax)«. 2. (bcy. 3. (2x3j/)».
4. (l2x)5. 5. (lx)io. 6. (a^by.
418. A trinomial, as a + b + c, may be raised to any power
by the Binomial Theorem, if we regard two terms as one, thus :
+^4^*(«»)''*^
302 THE BINOMIAL THEOREM.
Ex. Expand (l+x + a;2)3.
(l+x + x)^ = (l+x)3 + ;3(l+a;)'''.a;2 + ^(l+x).x* + x«
= (1 + 3ic + 3a;''^ + x3) + 3 (1 + 2x + a;2) o?
+ 3(l+u,).c' + x«
= 1 + 3a; + 3a;2 + ic^ + 3:c2 + 6x3 + 3x^ + 3a;'»
+ 3XS + .7''
= 1 + 3x46x2 + 7x3 + 6x* + 3xS + x«,
Examples.— cl.
Expand the following expressions :
I. (ft + 26c)'. 2. (l2x + 3x2)3. 3. (x3_a.2 + j.^3_
4. (3x^ + 2x« + l)3. 5. ^x + 1). 6. (a^ + 6^c^);
419. To jind the r"" or general term of the expansion of
(z + a)".
We have to determine three things to enable us to write
down the r"" term of the expansion of (x + a)".
1. The index of x in that term.
2. The index of a in that term.
3. The coefficient of that term.
Now the index of x, decreasing by 1 in each term, is in the
r* term ?i — 7+ 1 ; and the index of a, increasing by 1 in each
term, is in the r"" term r— 1.
For example, in the tliird term
the index of x is n — 3 + 1, that is, n2 :
the index of a is 3  1, that is, 2.
m assigning its proper coefficient to the ;•"' term we have to
determine tlie last factor in the denominator and also in the
numerator of the fraction
n.{nl).{n2).(nZ)
1.2.3.4
TffE BINOMIAL THEOREM. 303
Now the last factor of the denominator is less by 1 than the
number of the term to which it belongs. Tlius in the 3'* term
the last factor of the denominator is 2, and in the ?•"■ term the
last factor of the denominator is r— 1.
The last factor of the numerator is formed by subtracting
from 7i«the number of the term to which it belongs and adding
2 to the result.
Thus in the S"* term the last factor of the numerator is
713 + 2, that is ?i— 1 ;
in the 4* 71 4 + 2, that is 9i2 ;
and so in the j"" ?i  r + 2.
Observe also that the factors of the numerator decrease by
unity, and the factors of tlie denominator increase by unity, so
that the coetticient of the r"" term is
n.(nl). {n  2) {n  r + 2)
1.2.3 (r1) ~"'
Collecting our results, we write the r* term of the expansion
of (x + a)" thus :
n.(7il).(n2) inr + 2) '
1.2. 3 (r1) •" ••" •
Obs. The index of a is the same as the last factor in the
denominator. The sum of the indices of a and x is n.
Find
Examples. — cli.
The 8*'term of (1+x)".
The 5'" term of (a^  ft^)".
The 4'" tf rm of («  6)i«>.
The 9"" term of (2a6cd)».
The middle term of (a — 6)^^.
The middle term of (a^ + h°)^.
The two middle terms of (a  by^.
The two middle terms of (a + x)^.
304 THE BINOMIAL THEOREM.
9. Show that the coefficient of the middle term of
1.3.5 (4nl)
(a + as)*" is 2"" x
1.2 .3 2n
10. Show that the coefficient of the middle term of
(a + xr is 2^ X ^2!^) (^" + ^) (^" 1) (^^^ 1)
1.2
420. To &how that the coefficient of the r"" temi from the
btyianing of the expansion of (x + a)" is identical with the coeffi
cient of the r"^ term from the end.
Since the number of terms in the expansion is n+ 1, there
are n+l—r terms before the r"" term from the end, and there
fore the r^'term from the end is the (n — r + 2)'^ term from the
beginning.
Thus in the expansion of (x + a)*, that is,
X* + 5ax* + lOa^x^ + lOa^x^ + 5a*x + a°,
the 3rd term from the end is the (5  3 + 2)"', that is the 4"" term
from the beginning.
Now if we denote the coefficient of the r*^ term by (7„
and the coefficient of tbe (?) r + 2)"' term by C«_^2,
we have
n.{nl) (nr + 2)
C,=
C'.Hl —
1.2 (r1)
_ n.(nl) {n(nr + 2) + 2{ ,
1.2 (717 + 21)
n. (n 1) r
1 .2 (nr+l)
Hence
C, n.(nl) (nr + 2) 1. 2 (nr+l)
CZ^,~ 1.2 (rl) "" n.(n^) r
n.(Til) (n r + 2 ).(n r+ 1) 2.1
 1.2 (r 1) .r (n  1). n
In , . , ,
= 4^:= = 1, which proves the proposition,
n
THE BINOMIAL THEOREM 305
421. To find the greatest term in the expansion of (x + a)% n
being a positive integer.
Tlie r* term of the expansion {x + a)' is
n.{nl) {nr + 2) ,
1.2 (r1)
The (r + 1)"" term of the expansion (x + a)" is
n.(nl) (nr + 2). jnr + l)
1.2 (rl).r
Hence it follows that. we obtain the (r + l)* term by multi
plying the r"" term by
n  r + I a
r ' x'
When this multiplier is first less than 1, the r"" term is the
greatest in the expansion.
Now .  is first less than 1
r X
when nara + a is first less than rx,
or na + a first less than rx + ra,
or r (x + a) first greater than a {n + 1),
r first greater than — ^^ .
or
x + a
re \, 1 *. <^ ('"' + ^) i.u nr+1 a . , ,,
If r be equal to , then . = 1, and the
x+a « r X
(r + l)"" term is er^ual to the r*, and each is greater than anv
other term.
Ex. Find the greatest term in the expansion of {4 + ay,
when « = „.
Here ±±^^£11.)^^^^^.^^.
X + a ^3 11 11 ^'
^ + 2 T
The first whole number greater than 2^ is 3, therefore th«
greatest term of the expansion is the .3rd.
[s.A.j 17
2o6 THE BINOMTAL THEOREM.
422. To find the sum of all the coefficients in the expansion
of{l+x)\
a /^ \, 1 n . (71— 1) „
Since (1 +x)" = l +71X+ —  — ^— V +
n.(?z — 1) ,
<— ^g V' + ruf^ + af
putting x = l, we get
_. , n.(ni) n.(nl)
2"=l + n + — 1 2~ '*' — 1 "2" '
or, 2" = the siiiu of all the coefficients.
423. To shon: that the sum of the coefficients of the odd term
in the expansion of (1 + a;)" is equal to the smn of the coeffi^yients
of the even terms.
Since
,, , , 7i,(nl) , n. (n l).(n — 2) ,
(l+x)=l + T?x + — j^'^ + 123 ^"^
putting x=  1, we get
(11) _ln + j2~ j23 +
or.
,„^ .,.(,,.(.., ^ }
= sum of coefficients of odd terms  sum of co
efficients of even terms ;
.". sum of coefficients of odd terms = sum of coefficients of
even terms.
Hence, by the preceding Article,
2"
sum of coefficients of odd terms = — = 2»~^;
2"
sum of coefficients of even terms = g = 2"~*.
XXXVI. THE BINOMIAL THEOREM.
FRACTIONAL AND NEGATIVE INDICES.
42L We have shown that when m is a positive integer,
N_ T m.(m\) ,
(l+x)" = Hmz+ l^r — ^ X+
We have now to show that this equation holds good when
. . , . 3 . .
TO is a positive fraction, as , a negative integer, as  3, or a
3
negative fraction, as  .
We shall give the proof devised by Euler.
425. If m be a positive integer we know that
,, ,_ , m.(m.— 1), m . Cm1) . lm2) ,
(l+a;)"=ir7nj;+ p^ — 3^ + ^ , ^ g ^x^+
Let us agree to represent a series of the form
m . (m — 1) „
1+77!X+ p2 ^ + •'
by the symbol /(m\ irhatever the valw of m maii be.
Then we know that when m is a positive integer
(l+x)"'=/(m) ;
and we have to show that, also, when m is fractional or
negative
■ ■' (l+.r;;=/(»0.
o J/ \ ,  m.(ml) ,
Sine* f{m) = l + 'mx\ ; ■ z+
/(??) = 1 + nx + — ~^ — X+
3o8 THE BIAOMTAL THEOREM.
If we multiply together the two series, we shall obtain an.
expression of the form
1 + aic + &x2 + cx^ + dx*+
that is, a series of ascending powers of x in which the coeffi
cients a, 6, c are formed by various combinations of
m and n.
To determine the mode in which a and h are formed, let ua
commence the multiplication of the two series and continue it
as far as terms involving a;^, thus
,, , _ TO . (m  1)
/(m) = l+mx4 — ^^—^ — x'+
/(n) = l +«a; + f^"
f{m) xf(n) = l+mz+ — pg"" ■*"
+ nx + mnx^+
n.(nl) ,
^ 1.2
/ s (m.(m — 1)
l + (m + n).i +  ^g^
n . (u — 1) , ,
1.2 (
Comparing this product with the assumed expression
l+ax + bx + cz^ + dx* +
we see that a = m + n,
. , to.(toI) n.(n—l)
and b = — i"^ +mn + — ] o
m^ m + 2mn + 7i — n
"" T72
(m + n) . (m + n— 1)
" 172 •
P/^ACIYOXAL AND NEGATIVE INDICES. 309
Similarly we could show hy actual viultiplication that
(m + n) . (m + n — l). {m + n2)
*" 1T273 '
,_ (m + 7i) . (m + n — 1) . (m + n — 2) . (m + n  3)
TTaTsTi •
Thus we might determine the successive coefficients to any
extent, but we may ascertain the law of their formation by the
following considerations.
Tlnj forms of the coefficients, that is, the way in which m
and n are involved in them, do not depend in any way on the
values of m and n, but will be precisely the same whether m
and 71 be positive integers or any numbers whatsoever.
If then we can determine the law of their formation when
m and n are positive integers, we shall know the law of their
formation for all values of m and n.
Now when m and n are positive integern,
/(m) = (l+x)",
/(u) = (l+x)";
" /(™) x/(w) = (1 + x)" X (1 + x)
= (Ha;)"'+
, , , (m + n) .{m + n \) ,
= l + {m + n)x + ^ Ya — V+ ...
=/(m + n).
Hence we conclude that whatever he the values of m and n
f(m)x f{n)=f{m + n).
Hence f{m + n+p)=f{m).f{n+p)
=f(m).fin).f{p),
and »o generally
/{m + n+p + ...)=f{m).f{n)./{p)...
3 to THE BLVOM/AL 'IHEOREM.
Xuw let m = n=p= ... =j, h ami k being positive integ(;rs,
then
^h h h
^fh h h ^ , \
/i^^ + j + ^+.. tot terms j
=/a)./(J)./©...to. facto.
h (h \
, h k\k V,
k 1.2
which proves the theorem for a positive fractional index.
Again, since f{m).f{n)=f{m + n) for all values of m and n,
let 71= VI, then
/(m)./(m) =/(«!?»)
=/(0).
, . , mJml) .,
Now the series l + mx+ ^—h ^'+'
becomes 1 when vi = 0, that is,/(0) = l ;
.. /(m)./(770=l;
•■■•^(™^=/(7^=(TT^=^^ + ^>'^'
.. (l+x)=/(7n)
■■ / \  771 (  771  1) 2 ,
= l+(77l)x+ j—g X2+ ...
which proves the theorem for a negative index, integral or
fractional.
FRACTIONAL AND NEGATIVE INDICES. 3:1
426. Ex. Expand (a + x)2 to four tenna.
.a'' .J? ...
1
1.2.3
3
111 4. T ft 5
z z o
= a2+ g + — ^
2a 2 8a2 16«
Or we might pmceed thus, as is explained in Art. 416.
a0 .= ia')a^) ^
1 /I
= aMl + ^ ^^
2 a 1.2 o^
= a^ '1+ „H , ... ^
( 2a 8a2 16a3" j
i , X X2 X2
2a2 8a^ IGa^
1.2.3
Examples. — clii.
Expand the following expressions :
I. (1 + x)2 to five terms. 7. (1 x)^ to five terms.
2. (1 +a)^ to four terms.
3. (a f x)^ to five terras.
4 (1 + 2x)2 to five terms.
5. ( a + —^ H to four terms
1 i i
6. (o^ ^x*;' to four terms.
8. (1  a^)^ to four terms.
9. (1 — 3x)^ to four terms.
10. (x — ^y^to four terms.
1 1. (1 — x)* to four terms.
12. ( o" "" g ) ^^ three terms.
312 THE BINOMIAL THEOREM.
427. To exjxnid (1 +x)~'.
(1 +x) = l + («). x + ^1^^"^^ X*
n.(nl).{n2)
1.2.3
= 1 na; + j2~x ^ ^ ^ .x+
the terms being alternately positive and negative.
Ex. Expand (1 +x)~^ to five terms.
/, , x, , o 3.4, 3.4.5, 3.4.5.6 .
= 1  S.c + 6x2 _ 10x3 4 15x*  ...
428. To expand (1  a;)—.
_ n{ nl)(n2) .
1.2.3
n.(n + l) „ n. (?i + l)(n + 2) ,
the tenns being all positive.
Ex. Expand (1 x)~^ to five terms.
„ ,, , „ 3.4, 3.4.5, 3.4.5.6,
(lx)3=l+3x + — .2+^_^_^^+____.^+...
= l + 3x + 6x2 + 10x3+ 15x*+ ___
Examples.— cliii.
Expand
1. (1 +a)~2 to five terms. 4. ( 1  5) to five terms.
2. (1  3x)i to five terms. 5. (a22.r)~^ to five term*.
3. ( 1 — T ) to four terms. 6. (a^ — x^)""" to lour tenus.
THE BINOMIAL THEOREM. 313
429. To ex'pand{\+x)'n,
11 \ n /
n
2 "*"
_l(_l_l)(_I_2)
1.2.3
x'^^ ...
n 2?i^ fan*
Examples. — cliv.
Expand
I. (1 + x2)~2 to five terms. 4. (1 + 2a;)~2 to five terms.
3. (1 — c'^)~2 to five terma. 5. (ft^ + x*)'^ to four terms.
2 _i
3 (a^ + is*) ^ to four terms. 6. (o' + x^) ' to four terms.
430. Observations on the general expression for the term involving
X' in the exjjayisions (1 + x)" and (1  x)".
The general expression for the term involving x', that is the
(r+ l)"" term, in the expansion of (1 +x)'' is
n.(nl)...(7ir+l)
1.2 r ' •'■'■
From this we must deduce the form in all cases.
Thus the (r+ l)* term of the expansion of (1 x)" ifl found
liy changing x into ( — x), and therefore it is
n.(nl)...(7ir+l)
r72~;:~r ^""^^
*"' V^^ 1.2 r '
314 THE BINOMIAL THEOREI.I.
If n be negative and = — m, the (r+ 1)* term of the expan
sion of (1 +x)" is
( — m)(m— 1)... m — r + 1) ,
1.2 r *'
(1)'. j?n..(m+l)...(m + rl)jx'.
°'"' 1727::::^.:::::::^^
If n be negative and = m, the (r+ 1)* term of the expan
sion of (1 + x)" is
(l). )m.(m+l )...(m + rl) { _
1.2 r "^ ^'
m. (m+ 1) ... Cwi + r 1) ,
1.2 ....r •^•
Examples. — civ.
Fiuil the r"' terms of the following expansions :
1. (l+ay. 2. (lx)i2. 3. {ax)\ 4. (5x + 2!/)».
5. (1+x)^ 6. (l3x)^. 7. (la;)".
9. (l2x)~^. 10. (a2x2)"5.
1 1. Find the (r + 1)"" term of (1 x)^.
'* S. (a + x)^'.
12. Find the (r+ 1)* term of (1 4s;) .
13. Find the (r + 1)'*" term of (1 + x)''.
14. Show that the coefficient of x'+* in (1 +x)''+^ is the sum
of the coefficients of x' and^ j;"^* in (1 +x)",
I ;. What is tlie fom th term of ( « — )  ?
• 16. "What is the tiftli term of (a^i')^ ?
17. Wliat is the ninth term of (a2 + 2x) ?
18. What is the tenth term of (a + 6)"" ?
ig. What IS the seventh term of ya^b)" \
THE BINOMIAL THEOREM. 315
431. The following are examples of the application of the
Binomial Theorem to the approximation to roots of numbers.
(1) To approximate to the square root of 104.
V104 = ^/(lOO + 4) = 10 ( 1 + ~f
=ioi+i.i+tk:'J.(Ay
( 2 100^ 1.2 VlOO/
^2V2 /V2 /.(J \%
ia')a ^)
= 10l4A__?__ + _J . 1
( 100 10000 1000000 [
= 1019804 nearly.
(2) To approximate to the fifth root of 2.
4/2 = (1 + 1)^
= 1 + 1 + 1. l(ll) + l.l.Cll)(l2U...
5^2 5\5 / 6 5 \5 / \5 /
^j^l_^ _3 21_
5 25 "^250 2500"^'"
= 1 + ^ + nearly
25 2500
= 11236 nearly.
(3) To approximate to the cube root of 25.
4/25=4/(272) = 3il.U.
Here we take the cube next ahove 25, so as to make the
second term of the binomial as small as possible, and then
proceed as before.
Examples.— clvi.
Approximate to the following roots :
I. ;/31. 2. ^108. 3. 4^260. 4. 4^31.
XXXVII. SCALED OF NOTATION.
432. The sjinbols employed in our common system of
Arithmetical Notation are the nine digits and zero. These
digits when written consecutively acquire local values from
their positions ■s\ith respect to the place of imits, the value of
every digit increasing tenfold as we advance towards the left
hand, and hence the number ten is called the Radix ot the
Scale.
If we agree to represent the number ten by the letter t, a
number, expressed according to the conventions of Arithmetical
Notation by 3245, would assume the form
3<3 + 2?2 + 4« + 5
if expressed according to the conventions of Algebra.
433. Let us now suppose that some other number, as^re,
is the radix of a scale of notation, then a number expressed in
this scale arithmetically by 2341*will, if five be represented by
/, assume the form
2/3 + 3/2r4/+l
if expressed algebraically.
And, generally, if r be the radix of a .scale of notation, a
number expressed arithmetically in that scale by 6789 will,
when expressed algebraically, since the value of each digit
increases rfold as we advance towards the left hand, be repre
sented by
67' + 7r* + 8r + 9.
434. The number which denotes the radix of any scale will
be represented in that scale by 10.
Thus in the scale whose radix is five, the number fire will
be represented by 10,
Scales of xo ta tion. 3 17
In the same scale seven, being equal to five + two, ■will
therefore be represented by 12.
Hence the series of natural numbers as far as tv:entyj\,vt will
be represented in the scale whose radix is five thus :
1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22. 23, 24, 30, 31,
32, 33, 34, 40, 41, 42. 43, 44, 100.
435. In the scale whose radix is eleven we shall require
a new symbol to express the number ten, for in that scale the
number eleven is represented by 10. If we agree to express
ten in this scale by the symbol t, the series of natural numbers
as far as twentythree will be represented in this scale thus :
1, 2, 3, 4, 5, 6, 7, 8, 9, t, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, It, 20, 21
436. In the scale whose radix is tioelve we shall require
another new symbol to express the number eleven. If we
agree to express this number by the symbol e, the natural
numbers from nine to thirteen wdll be represented in the scale
whose radix is twelve thus :
9, t, e, 10, 11.
Again, the natural numbers from twenty to twentyfive will
be represented thus :
IS, 19, If, le. 20, 21.
437. The scale of notation of which the radix is two, is
called the Binary Scale.
The names given to the scales, up to tliat of which the
radix is twelve, are Ternary, Quaternary, Quinarj, Senarv,
Septenary, Octonary, Nonary, Denary, Undenary and Duo
denary.
438. To perform the operations of Addition, Subtraction,
Multiplication, and Division in a scale of notation whose index
is r, we proceed in the same way as we do for numbers ex
pressed in the common scale, with this difference onlv, that r
must be used where ten would be used in the common scale :
which will be understood better by the following examples.
3 1 8 SCALlLS OF NOTA TION.
Ex. 1. Find the sum of 4325 and 5234 in the senary scale.
4325
5234
the sum 14003
which is ohtained by adding the numbers in vertical lines,
carrying 1 fur every six contained in the several results, and
set' ing do%vu tlie excesses above it.
Thus 4 units and 5 units make nine units, that is, six units
together with 3 units, so we set down 3 and carry 1 to the
next column.
Ex. 2. Find the difference between 62345 and 53466 in
the septenarv scale.
62345
53466
the difference = 5546
which is obtained by the following process. We cannot take
six units from five units, we therefore add seven units to the
five units, making 12 units, and take six units from twelve
units, and then we add 1 to the lower figure in the second
column, and so ou.
Ex. 3. Mul iply 2471 by 358 in the duodenary scale.
247 1
358
17088
e < e 5
7 193
8333 18
Ex. 4. Divide 367286 by 8 in the nonary scale.
8 ; 367286
~42033
The following is the process. We ask how often 8 is contained
in 36, which in the nonary scale represents thirtythree units ;
the answer is 4 and 1 over. We then ask how often 8 is con
tained in 17, which in the nonary scale represents sixteen units;
the answer is 2 nnd no remainder. And so for the other digits.
SCALES OF NOTAtiON.
M$
Ex. 5. Divide 1184323 hy 589 in the duodenary scale.
589; 1184323 ('2483
e5fi
22f3
KeO
3e32
39i!0
1523
1523
Ex. 6. Extract the square root of 10534521 in Vhe senary
scale.
10534521 ( 2345
4
43
253
213
504
4045
3224
5125
42121
42121
I
2
3
4
5
6
7
8
9
lo
scale.
Examples. — clvii.
Add 23561, 42513, 645325 in the septenary scale.
Add ,3074852, 4635628, 1247653 in the nonary scale.
Subtract 267862 i'roni 358423 in the. nonary scale.
Subtract 124321 from 211010 in the quinary scale.
Multiply 57264 by 675 in the octonary scale.
Multiply 1456 by 6541 in the septenary scale.
Divide 243012 by 5 in the senary scale.
Divide 3756025 by 6 in the octonary scale.
Extract the square root of 25400544 in the senary scale.
Extract the square root of 56898(1 in the duodenary
^io SCALES 01' NorATiohr.
439. To transform a given integral number from one scale to
another.
Let N be the given, integer expressed in the first scale,
r the radix of the irw scale in which the number is to
be expressed,
a, b, c in,}:!, q tlie digits, n + l in number, expressing
the number in the Jiew scale ;
so that the number in the new scale will be expressed thus :
ar" + br"""^ + cr"~ + + mf + jn + q.
We have now from the equation
J\r= ar" + 6r"~^ + cr""2 ^ ^ ^,ij2 ^ pj. ^ q^
to determine the values of a, 6, c m, p, q.
Divide N by r, the remainder is q. Let A be the quotient :
Uien
A = ar"^ + br" + cr"^^ + +mr+p.
Divide A by r, the remainder is p. Let B be the quotient ;
then
B = ar''^ + br"^ + cr"^+ +m.
Hence the
iirst digit to the right of the number expressed in the
new scale is q, the first remainder ;
second p, the second remainder ;
third m, the third remainder ;
and thus all the digits may be determined.
Ex. 1. Transform 235791 from the common scale to the
scale whose radix is 6.
6
235791
6
39298 remainder 3
6
6549 remainder 4
6
1091 remainder 3
6
181 remainder 5
6
30 remainder 1
6
5 remainder
I remainder 5
The number required is therefore 5015343.
SCALES OF NOTATIOiY.
32i
The digits by which a number can be expressed in a scale
whose radix is r will bel, 2, 3 t 1, because these, with 0,
are tiie only remainders which can arise from a division in
which the divisor is r.
Ex. 2. Express 3598 in the scale whose radix is 12.
12
12
12
12
3598
299 remainder t
24 remainder e
2 remainder
remainder 2
.•. the number required is l^&i.
440. The method of transforming a given integer from one
scale to another is of course applicable to cases in which both
scales are other than the common scale. We must, however,
be careful to perform the operation of division in accordance
with the principles explained in Art. 438, Ex. 4.
Ex. Transform 142532 from the scale whose radix is 6 to
llie scale whose radix is 5.
5
142532
5
20330
remainder 2
5
2303
remainder 3
5
300
remainder 3
5
33
remainder 3
5
4
remainder 1
remainder 4
The required number is therefore 413332.
Examples. — clviii.
Express
1. 1828 in the septenary scale.
2. 1820 in the senary scale.
3. 43751 in the dnodenarv scal(».
rs.A.i
SCALES OF NOTATIOK.
4. 3700 in the quinary scale.
5. 7631 in the binary scale.
6. 215855 in the duodenaxy scale.
7. 790158 in the septenary scale.
Transform
8. 34002 from the quinary to the quaternary scale.
9. 8978 from the undenary to the duodenary scale.
10. 3256 from the septenary to the duodenary scale.
1 1. 37704 from the nonary to the octonary scale.
12. 5056 from the septenary to the quaternary scale.
13. 654321 from the duodenary to the septenary scale.
14. 2304 from the quinary to the undenary scale.
441. In any scale the positive integral powers of the num
bei which denotes the radix of the scale are expressed by
10,100, 1000
Thus twentyfive, which is the sqiiare of five, is e.xpressed in
the scale whose radix is live by 100: one hundred and twenty
five will be expressed by 1000, and so on.
Generally, the ?;"' power of the number denoting the radix
in any scale is exjjressed by 1 followed by n cyphers.
The highest number that can be expressed byj:^ digits in a
scale whose radix is r is expressed by ;•''  1.
Thus the highest nuni\)er that can be exjjressed by 4 digits
in the scale whose radix is five is
10^  1, or 10000  1, that is 4444.
The least number that can be expressed by 2' digits in a
scale whose radix is r is expressed by j^^
Thus the least numlier that can be expressed by 4 digits in
the scale whose radix is five is
JO*' or 103, tiij^t is 1000,
SCALES OF NO TA TION. 323
442. In a scale whose radix is r, the sum of the digits of
an integer divided by (»■ 1) will leave the same remainder as
the integer leaA^es when divided by r — 1.
Let iV be the number, and suppose
Then
JV=a(r''l) + 5(r''il) + c(r"2l)4 ... +m(r2 l)+p(r 1)
+ jrt + 6 + c+ +m+j:> + 7{.
Now all the expressions r"  1, r""^ — 1 r 1, r— 1 are
divisible by r  1 ;
N . ^ a + b + c+ m + p + q
.. ;= an integer + :; —  ;
r— 1 ° r— 1
which proves the proposition, for since the quotients differ by
an integer, their fractional parts must be the same, that is, the
remaindt^ra after division are the same.
Note. From this proposition is derived the test of the
accuracy of the result of Multiplication in Arithmetic by cast
ing out the nines.
For let A = Qm + a,
and B = ^n + h ',
then AB=Q{Qmn + an + 6m) + ah ;
that is, AB and ah wlien divided by 9 will leave the same
remainder.
Radical Fraction?.
443. As the local value of each digit in a scale whose radix
is r increases 7'fold as we advance from right to left, so does
the local value of each decrease in the same proportion as we
advance from left to right.
If then we affix a line of digits to the right of the units'
place, each one of these having from its laosition a value
oner"" part of the value it would have if it Avere one place
further to the left, we shall have on the right hand of the
units' place a series of Fractions of which the denominators
324 SCALES Of NOTAlIOAr.
are successively r, r'', r^, , while the immerators may be
any numbers between r— 1 and zero. These are called
Eadical Fractions.
In our common sy.^tem of notation the word Radical is
replaced by Decimal, because ten is the radix of the scale.
Now adopting the ordinary sj'stem of notation, and markins^
the place of units by putting a dot ' to the right of it, we have
the following results :
2464789 = 2 X 10^4 4x 1046 + A + _^"^^ + __^_9^.
In the denary scale
2464789 = 2:
in the quinary scale
3244213 = 3x 10^+2. 10444A + _L+_l_ + J_,
remembering that in this scale 10 stands ioifive and not for teii
(Ait. 434).
444. To shoio that in any scale a radical fraction is a proper
fraction.
Suppose the fraction to contain n digits, a, b, c
Then, since r  1 is the highest value that eacli of the digits
can have,
 + 5 + ... is not greater than (r 1)^+ t, + ... to n terms)
r r^ \r r /
than(rl) 'y~
not greater
1
r
( r" — \ ")
not greater tlian (r  1) ■; — I ;
(?'\rl)j
not greater than ;
not greater than I  —
r"
SCALES OF iVOTA TTON: 325
Hence the criveu Iraction is less than 1, and is therefore a
proper fraction.
445. To transform a fraction expressed in a given scale into
a radical fraction in any other scale.
Lut F be tlie given fraction expressed in the first scale,
r the radix of the new scale in which the fraction is to
be expressed,
a, b, c.the digits expressing the fraction in the nev/
scale, so that
r r r^
from which eqnation the values of a, 6, c.are to be deter
mined.
Multiplying both sides of the equation by r,
TP be
r r*
b c
Now  + ^+ ••• is a proper fraction by Art. 444.
Hence the integral part of Fr will =a, the first digit of the
new fraction, and the fractional part of Fr will
b c
=  + ,+ ...
r r
Giving to this fractional part of Fr the symbol F^^ we have
Alultiplying both sides of the equation by r,
F,r = b + + ...
r
_ Hence the integral part of F^r=^b, the second digit of the new
Taction, and thus, by a similar process, all the digits of the
lew fraction may be found.
\26 SCALES OF NOTATION.
3
Ex. 1. Express = as a radical fraction in the quinary
Kcale :
7 7 7'
15^5
5 , 25 ., 4
7 / 7
4 . 20 ^ 6
i?x5 = — = 2 + ,
7 / 7'
6 , 30 , 2
7 7 7'
2 , 10 , 3
^xo = =l+;
therefore fraction is •203241 recurring.
Ex. 2. Express •84375 in the octonary scale :
•84375
8
675000
8
6^00000
The
fraction required
is •66.
Ex.
3.
Transform ■
42765 from the
nonary
to the
senary
scale.
•
•42765
6
278133
6
5 •23820
6
155430
6
365800
TTie fraction required is •2513...
SCALES OF NOTATION.
327
Ex. 4. Transfonn 6l24i275 from the duodenary to the
quaternary scale :
•^275
4
34«58
4
175 i8
4
25e68
4
le(28
Number required is 102232303121
4
el24
4
2937
 remainder
4
834
remainder 3
4
20e
 remainder 2
4
62
 remainder 3
4
16
 remainder 2
4
4
 remainder 2
4
1
 remainder
0
 remainder 1
Examples. — clix.
25
1. Express :^^ in the senary scale.
3
2. Express — in the septenary scale.
3. Express 23' 125 in. the nonary scale.
4. Express 1820"3375 in the senarj' scale.
5. In what scale is 17486 written 212542 ?
6. In what scale is 511173 written 1746305 ?
7. Show that a number in the Common scale is divisible
(1) by 3 if the sum of its digits is divisible by 3.
(2) by 4 if the last two digits be divisible by 4.
(3) by 8 if the last three digits be divisible by 8,
(4) by 5 if the number ends with 5 or 0.
32^ ON LOGARITHMS.
(5) by 11 if the difference between the sum of the digit*
in the odd places and the sum of those in the even
places be divisible by 11.
8. If iV be a number in the scale whose radix is r, and n
be the number resultinsr when the digits of N are reversed,
show that iV 7i is divisible by r 1.
XXXVIII. ON LOGARITHMS.
446. Def. The Logarithm of a number to a given base
is the index of the power to which the base must be raised to
give the number.
'^ Thus if m = a', x is called the logarithiu of m to the base a.
For instance, if the base of a system of Logarithms be 2,
3 is the logarithm of the number 8,
because 8 = 2^:
and if the base be 5, then
3 is the logarithm of the number 125,
because 125 = 5^
447. The logarithm of a number in to the base a is written
thus, logaWi ; and so, if vi = a',
X = log^m.
Hence it follows that m = a'"^"".
448. Since 1 = 0", the logarithm of unity to any base is
zero.
Since a = a}, the logarithm of the base of any .system
i3 unity.
449. We now proceed to describe th;it wliish is called the
Common System of logarithms.
The ba,8e of the system is 10.
ON LOGARITHMS.
%n
By a system of logarithms to tlie base 10, we mean a succes
sion of values of x whicli satisfy the equation
771=10"
for all positire values of m, integial or fractional.
Such a system is formed by the series of logarithms of
the natural numbers from 1 to 100000, which constitute the
logarithms registered iu oiir ordinary tables, and which are
therefore called tabular logarithms.
450. Now
1 = 100,
10 = 101,
100 = 102,
1000 = lO'',
and so on.
Hence the logarithm of
1 is 0,
of 10 is 1,
of 100 is 2,
of 1000 is 3.
and so on.
Hence for all numbers between 1 and 10 the logarithm is a
decimal less than 1,
between 10 and 100 the logarithm is a decimal between 1
and 2,
between 100 and 1000 a decimal between 2
and 3, and so on.
451. The logarithms of the natural numbers from 1 to 12
stand thus in the tables :
No.
Log
1
00000000
2
03010300
3
04771213
4
06020600
5
06989700
6
07731513
No.
Log
7
08450980
8
09030900
9
09542425
10
10000000
11
10413927
12
10791812
The logarithms are calculated to seven places of decimals
33<5 ON LOGARITHMS.
452. The integral parts of the logarithms of numbers
higher than 10 are called the characteristics of those logarithms,
and the decimal parts of the logarithms are called the mantisscB.
Thus ■ 1 is the characteristic,
•0791812 the mantissa,
of the logarithm of 12.
453. The logarithms for 100 and the numbers that succeed
it (and in some tables those that jirecede lOOj have no charac
teristic prefixed, becfiuse it can be supplied by the reader, l)eing
2 for all numbers between 100 and 1000, 3 for all between
1000 and 10000, and so on. Thus in the Tables we shall
tind
No.
Log
100
0000000
101
0043214
102
0086002
103
0128372
104
0170333
105
0211893
which we read thus :
the logarithm of 100 is 2,
of 101 is 20043214.
of 102 is 20086002; and so on.
454. Logarithms are of great use in making arithmetical
computations more easy, for by means of a Table of Logarithms
the operation
of j\Iultiplication is changed into that of Addition,
. . . Division Subtraction,
... Involution Multiplication,
...Evolution Division,
as we shall show in the next four Articles.
455. The logarithm of a product is equal tc. the sum of the
logarithms of its factors.
Oy LOGARITHMS. 33»
Let
m = a',
and
n = a".
Then
mn = a"'^' ;
••• log,
mn = x + y
= log„m + ^O'^ji.
Hence it follows that
log^mnp = log^TO + \og^n + log^^'j
and similarly it may be shown that the Theorem holds good
for any number of factors.
Thus the operation of Multiplication is changed into that of
Addition.
Suppose, for instance, we want to find the product of 246
and 357, we add the logarithms of the factors, and the sum is
the logarithm of the product : thus
log 246 = 23909351
log 357 = 25526682
their sum = 49436033
whicli is the logarithm of 87822, the product required.
Note. We do not write logio246, for so long as we are
treating of logarithms to the particular base 10, we may omit
the suffix.
456. 77ie logarithm of a quotient is equal to the logarithm of
the dividend diminished by the logarithm of the divisor.
Let m = a",
and n = a^.
Then ^a'";
n
, m
.: los„——x — y
= log^7)i — log„n..
Thus the operation of Division is changed into that of Sub
traction.
532 ON LOGARITHMS.
If, for example, we are required to divide 371 "49 by 52376,
we proceed thus,
log 371 49 = 25699471
lo" 52376 = 17191323
their difference = '8508148
which is the logarithm of 7092752, the quotient required.
457. The, logarithm, of any jwiver of a number is equal to the
product of the logarithm of the number and the index denoting the
poiver.
Let
771 = a".
Then
TO' = a";
.•. logjn' = rx
= r . lo" m.
Thus the operation of Involution is changed into Multipli
cation.
Suppose, for instance, we have to find the fourth power of
13, we nauy proceed thus,
log 13 = 11139434
4
44557736
which is the logarithm of 28561, the number required.
458. The logarithm of any root of a number is equal to the
quotient arising from the division of the logarithm of the number
by the nurnher denoting the root.
Let
m = a*.
Then
1 X
m' = a" ;
1 ' a;
_1
r '
log^TO.
Thus the operation of Evoluiion is changed into Division.
ON LOGARITIJMS. y^.
If, for example, we have to find the fifth root of 16807 we
proceed thus, '
5 _42254902, the log of 16807
•8450980
which is the logarithm of 7, the root required.
459 The common system of Logarithms has this advantae
overall otliers for numerical calculations, that its base is th"
same as tlie radix of the common scale of notation.
Hence it is that the same mantissa serves for all numl.ers
which have tlie same significant digits and difler only in the
position of the place of units relatively to those digits.
For, since log 60 = log 10 + log 6=1+ log 6,
log 600 = log 100 + log6 = 2 + log6,'
log 6000 = log 1000 + l(3g 6 = 3 + log 6,
t is clear that if we know the logarithm of any number as 6
ve also J.novv tlie logarithms of the numbers resulting' from
nultiplymg that number by the powers of 10.
So again, if we know that
log 17692 is 247783,
TO also know that
log 17692 is 1247783,
log 17692 is 2247783,
log 17692 is 3247783,
log 17692 is 4247783,
log 176920 is 5247783.
.1^ unit^^'' ™"'^ ""'^'^ *^'^' ""^ '^' logarithm., ot numbers less
Since \ = \Q\
•^=^o=l«"^
334 ON LOG/.RITRMS
the logarifhm of a numl^er
between land "1 lies between and 1,
between land '01 land 2,
between •()! and '001 —2 and 3,
and so on.
Hence the logarithms of all numbers less than unity are
negative.
"We do not require a separate table for these logarithms, for
w^e can deduce them from the logarithms of numbers greater
Lhan unity by the following process :
log6 =log j^ =log6loglO =log6l,
log06 =logi^ =log6loglOO =log62,
log 006 = log ^^ = log 6  log 1000 = log 6  3.
Now the logarithm of G is •7781513.
Hence
log6 =  1 + 7781513, which is written 17781513,
log 06 =  2 + 7781513, which is written 27781513,
log 006=  3 + 7781513, which is Avritten 37781513, i
the characteristics only being negative and the mantissse
positive. I
461. Thus the same mantissce serve for the logarithms of
all numbers, whether greater or less than unity, which have the
same significant digits, and differ only in tlie position of the
place of units relatively to those digits.
It is best to regard the Table as a register of the logarithms
of numbers which have one significant digit before the decimal
point.
ox LOGARITHMS. 335
No. ! Log
For instance, when we read in the tables 144  1583625, we
interpret the entry thus
log 144 is 1583625.
We then obtain the following rules for the characteristic to
be attached in each case.
I. If the decimal point be shifted one, two, three ...n
places to the right, prclix as a characteristic 1, 2, 3 ... n.
II. If the decimal point be shifted one, two, tlu'ee...TO
places to the left, prefix as a characteristic f, 2, 3 ... w.
Thus log 144 is 1583625,
.. log 144 is 11583625,
log 144 is 21583625,
log 1440 is 31583625,
log 144 is 11583625,
log 0144 is 21583625,
log 00144 is 31583625.
462. In calculations with negative characteristics we follow
lae rules of algebra. Thus,
(1) If we have to add the logarithms 364628 and 242367,
M3 first add the mantissoe, and the result is 106995, and then
add the characteristics, and this result is 1.
The final result is 1 + 106995, that is, 06995.
(2) To subtract 56249372 from 32456973, we may arrange
the numbers thus,
 3 + 2456973
5 + 6249372
1 + 6207601
the 1 carried on from the last sul)traction in the decimal places
changing — 5 into — 4, and then — 4 subtracted from  3 giving
1 as a result.
Heuce the resulting logarithm is 16207601.
336 ON LOGARITHMS.
(3) To multiply 37482569 Ly 5.
374825C9
5
127412845
the 3 carried on from tlie last multiplication of the decimal
l>laces being added to — 15, and thus giving — 12 as a result.
(4) To divide 142456736 Ly 4.
Increase the negative characteristic so that it may be exactly
divisible by 4, making a proper compensation, thus,
142456736 = 16 i 22456736.
142456736 16 + 22456736 
Then ^ = ^ =4 + 5614184;
and so the result is 45614184,
Examples. — clx.
1. Add 31651553, 47505855, 66879746, 26150026.
2. Add 46843785, 56650657, 38905196, 34675284.
3. Add 25324716, 36650657, 58905196, 3156215.
4. From 2483269 take 3742891.
5. From 2352678 take 5 4286 19.
6. From 5349162 take 3624329.
7. Multiply 24596721 by 3.
8. Multi])ly 74296S3 by 6.
9. Multiply 92843617 by 7.
10. Divide 63725409 by 3.
1 1. Divide 14432962 by 6.
1 2. Divide 453627188 by 9.
463. "We shall now explain how a system of logarithms
calculated to a base a may be transformed into another system
of which the base is 6.
ON LOGARITHMS.
Let m be a number of which the logarithm in the first
system is x and in the second y.
Then m = a*',
and 771=6".
Hence h^ = a',
■•^=iog.^;
■ X logj) '
\
■yio^y^
Hence if we multiply the loyiirithm of any number in the
system of which the base is a by = we shall obtain the
logarithm of the same number in the system of which the 'base
is h.
This constant mulliiilier . — , is called The Modulus of the
log„o ■'
system of which the base is b with reference to the system of
which the base is a.
464. The common system of logarithms is used in all
numerical calculations, but there is another system, which we
must notice, emj^loyed by the discoverer of logarithms, Napier,
and hence called The Napierian System.
The base of this system, denoted by the symbol e, is the
number which is the sum of the series
of which sum the first eight digits are 27182818.
465. Our common logarithms are formed from the Loga.'
rithms of the Napierian System by multiplying each of tha
[s.A.j Y
338 ON LOGARITH.\fS.
■latter by a common multi])lier culled Tlie Modulus of the
Couimon System
Tliis modulus is, in accordance with the conclusion of
Art. 463, i— ^.
log, 10
That is, if I and iV V)e the logarithms of the same number in
the common and Napierian systems respectively.
Now log, 10 is 230258509 ;
1 . ]
or 43429448,
■ ■ log, 10 230258509
and so the modulus of tlie common system is 43429448.
466. To prove that log,6 x log(,rt = 1.
Let ft; = log A
Then 6 = a';
..  = log,a.
X
Thus loga6 X logjft = .; x 
= 1.
467. The following are simple examples of the method ot
applying the princi})les explained in this Chapter.
Ex. 1. Given log 2 = 3()l()3()0, log 3 = 4771213 and
log 7 = 8450980, find log 42.
Since 42 = 2x3x7
log 42 = log 2 + log 3 + log 7
= 3010300 4 4771213 + 8450980
=r 16232493.
ON LOGARITHMS. 339
Ex. 2. Giveu log 2 = 3010300 and log 3 = 4771213, find
the logarithms of 64, 81 and 96.
log 64:* log 26 = 6 log 2
log 2 = 3010300
6
log 64 = 18061800
log 81 = log 3* = 4 log 3
log 3 = 4771213
4
log 81 = 19084852
log 96 = log (32 X 3) = log 32 + log 3,
and log 32 = log 2^ = 5 log 2;
.. log 96 = 5 log 2 + log 3 = 15051500 + 4771213 = 19822713.
Ex. 3. Given log 5 = 6989700, find the logarithm of
^(625).
log (625)^ = i log 625 = ^ log ~g = J (]og625log 100)
= ^(Iog5''2) = i(4log52)
= i (27958800 2) = 1136657.
Examples.— clxi.
1. Given log 2 = 3010300, find log 128, log 125 and
log 2500.
2. Given log 2 = 3010300 and log 7 = 8450980, find the
logarithms of 50, 005 and 196.
3. Given log 2 = 3010300, and log 3 = 4771213, find the
logarithms of 6, 27, 54 and 576.
4. Given log 2 = 3010300, log 3 = 4771213, log 7 = 8450980,
find log 60, log 03, log 105, and log 0000432.
340 ON LOGARITHMS.
5. Given log 2 = SOinS'OO, log 18 = 1 2552725 and
log 21 = 13222193, find log 00075 and log 315.
6. Given log 5 = 6989700, find the logarithms of 2, 064,
J,
and (5,0; •
7. Given log 2 = 3010300, find the logarithms of 5, 125,
8. What are the logarithms of 01, 1 and 100 to the base
10? What to the base or?
9. What is the characteristic of log 1593, (1) to base 10,
(2) to base 12 ?
10. Given —^ = 8, and x = 3i/, find x and y.
11. Given log 4 = 6020600, log 104 = 0170333 :
(a) Find the logarithms of 2, 25, 832, (625)"~.
(6) How many digits are there in the integral part ot
(104)0000?
12. Given log 25 = 13979400, log r03 = 0128372 :
(a) Find the logaritlims of 5, 4, 5r5, (•064)"'~.
(6) How many digits are there in the integral part of
(103)000?
13. Having given log 3 = 4771213, log 7= 8450980,
log 11 = 10413927:
find the logarithms of 7623,  ^ and ^^r^.
14. Solve the equations :
(i> 4096'=^. (4) a?)=<;.
(2) (:4y = 625. (5) a^.\^^^c^\
(^) a^.]f = m. (6) a'lr =&~\
ON LOGARITHMS. 34!
468. We have explained in Arts. 459 — 461 the advantages
of the Common System of Logarithms, wliich may be stated in
a more general form thus :
Let A be any sequence of figures (such as 235916), having
one, digit in the integral part.
Then any niiniber iV having the same sequence of figures
(such as 235916 or 00235916) is of the form A x 10", where n
is an integer, positive or negative.
Therefore logjjxY= logjo( J. x 10") = log^,^ + n.
Now A lies between lO*' and 10\ and therefore log ^ lies
between and 1, and is therefore a proper fraction.
But logjjiV and logjo.4 differ only by the integer 71 ;
.". logjp^4 is the fractional part of logu,iV.
Hence the logarithyns of all numbers having THE same
SEQUENCE OF FIGURES have the same mantissa.
Therefore one register serves for the m.antissa of logarithms of all
such numbers. This renders the tables more comprehensive.
Again, considering all numbers which have the same
sequence of figures, the number containing t'co digits in the
integral part =10. J., and therefore tlie characteristic of its
logarithm is 1.
Similarly the niimber containing m digits in the integral
part = 10". A, and therefore the characteristic of its logarithm
is m.
Also numbers which have no digit in the integral part and
one cypher after the decimal point are equal to A . 10~' and
A . 10~^ respectively, and therefore the characteristics of their
logarithms are  1 and — 2 respectively.
Similarly the number having m cyphers following the decimal
point = ^ . 10<™+";
.'. the characteristic of its logarithia is ~{m + 1).
Hence we see that the characteristics of the logarithms of all
nuvibers can be determined b)j inspection and therefore need not be
itj'istered. This renders the tables less bulky.
34:2 ON LOGARITHMS.
469. The method of using Tables of Logarithms does
not fall within the scope of this treatise, but an account of
it may be found in the Author's work on Elementaky
Trigonometry.
470. We proceed to give a short explanation of the way
in which Logarithms are applied to the .solution of questions
relating to Compound Interest.
471. Suppose r to represent the interest on .£1 for a year,
then the interest on P pounds for a j^ear is represented by
Fr, and the amount of P pounds for a year is represented,
by P + Pr.
472. To find the amount of a (jiven sum for any time at
conifpound interest.
Let P be the original principal,
r the interest on £\ for a year,
n the number of years.
Then if P^, P„, P.^...P„ be the amounts at (he end ol
1, 2, 3 . . . n years,
Pi = P +Pr = P (1 + r,
P2 = Pi + PirP,(l+r)=P(l+7f
P3 = P„ + P.,? = P., (1 + 7) = P (1 + if,
P,. = P(l+r)".
473. Now suppose P„, P and r to be given : then by the aid
of Logarithms we can find n, for
logP„ = log !P(l + r)"
= log P + nlog(l+r) ;
_ log 7'„l()gP
log(i+r)
I
ON LOGARITHMS. . 343
474. If the interest be payable at intervals other than a
year, the fornmla P^ = P(1 +r)" is applicable to the solution of
tlie question, it being observed that /• represents the interest
on £\ for the perio'l on wliich the interest is calculated, half
yearly, quarterly, or for a*iy other period, and n represents the
number of such periods.
For example, to find the interest on P pounds for 4 years
at compound interest, reckoned quarterly, at 5 per cezit. per
annum.
Here r=l of A = l^ = .0i25,
n = 4x4 = 16;
.. P„ = P(1 + 0125)16.
Examples.— clxii.
N.B. — The Logarithms required may l)e found from the
extracts from the Tables given in pages 329, 330.
1. In how many years will a sum of money double itself
at 4 per cent, compound interest ?
2. In Iiow many years will a sum of money double itself
at 3 per cent, compound interest \
3. In how many years will a sum of money double itself
at 10 per cent, compound interest ?
4. In how many years will a sum of money treble itself
at 5 per cent, compound interest ?
5. If £F at compound interest, rate ?•, double itself in n
years, and at rate 2r in m years : show that in : n is greater
than 1 : 2.
6. In how many years will £1000 amount to £1800 at
5 per cent, compound interest ?
7. In how mnny years will £P double itself at 6 per cent,
per ann. compound interest payable halfyearly 1
APPENDIX.
475. The following is another method of proving the prin
cipal theorem in Permutations, to which reference is made in
the note on page 289.
To prove that the number of pernjfiitatioHs of n things taken r at
a time is n . (n  1) (n — r + 1).
Let there be n things a, h, c, d
If n things be taken 1 at a time, the number of permutations
is of course n.
Now take any one of them, as a, then n  1 are left, and
any one of these may be put after a to form a permutation,
2 at a time, in which a stands first: and hence since there are
n things which may begin and each of these n may have n  1
put after it, there are altogether n (n — 1) permutations of n
things taken 2 at a time.
Take any one of these, as ab, then there are n2 left, and
any one of these may be put after ab, to form a permutation,
3 at a time, in which ab stands first : and hence since there
are n{n — 1) things which may begin, and each of these n{n  1)
may have n2 put after it, there are altogether n(n — 1) (ti  2)
permutations of n things taken 3 at a time.
If we take any one of these as abc, there are ?i  3 left, and
so the number of permutations of n things taken 4 at a time is
n.(nl){n2){n3).
So we see that to find the number of permutations, taken
r at a time, we must multiply the nvimber of permutations,
taken r— 1 at a time, by the niimber formed by subtracting
r— 1 from n, since this will be the number of endings any one
of these permutations may have.
Hence the number of permi;tations of n things taken 5 at a
time is
n(nl)(7j2) (n3) x (n4), orn(n 1) (h 2) (n3) (n4);
and since each time we multiply by an additional factor the
number of factors is equal to the number of things taken at a
time, it follows that the number of permutations of n thinga
taken r at a time is the product of the factors
n.(nl)(n2) (nr+1).
A :^ S W E R s.
i.
(Page 10.)
I.
5a+ 76f 12c.
"7
a + 3b + 2c.
3
2a + 26 + 2a
4
6a + 2h + 2c.
5
'2x7a + 3b2.
6.
0.
7
126 + 3c.
ii. (Page 10.)
I. 2a. 2. 2a + 5A 3. 3a — 3x. 4. Sx + Sr/.
5. 4a + 6 + 2c. 6. 2ti. 7. 4. S. 13xy6z.
9. 10a— 76 X.
iii. (Page 10.)
I. 26. 2. a: + 2]/. 3. a + 5c + d. 4. 2y{2z.
5. 2r. 6. 26 + 2c. 7. o36c. 8. By + z.
iv. (Page 11.)
I. 4a 6, 2. 46. 3. a 46 4c. 4. 26.
5. 14x + 2. 6. 2x + a. 7. 6x — a. 8. a.
9. 2a 6. 10. 2a. 11. c. 12. x + 3o.
13. 29a 276 + 6c.
^r. (Page 16.)
Addition.
I. 7a26. 2. 106 + 6c. 3. llxSy6z.
4. 665c + 3d. 5. 2a. 6. 2x2a + b + 4y.
7. 7a + 46  4c. 8. 7a — 6 + 7c. 9. — 6?/ + £<;.
[S.A.] ^*
34&
AxYSlV£RS.
I.
4
7
lo.
Subtraction.
2a + 26.
8xl7j/ + 5.
 3a + 36 — 4c.
6rt  6 + 5c.
2. a  c. 3. 2a  26 + 2c.
5. 7a 166 + 20c. 6. 5a368x.
8. 26 + 2c 15. g. llx7i/ + 45;.
II. 12^95 + 2r.
I.
2xy.
5
a?.
9
180a^65c*.
13
76x*i/%3.
16.
12a6can/.
19.
ahx^yh^.
Vi. (Page 20.)
2. \2xy. 3. 12a;2i/2.
6. a*. 7. 12^561
10. 28a"6r'". 11. Ba'
14. 51a6*c2/2;.
17. 8a"6V.
20. 33a206i6m2x.
4. 3a26c2
8. 35a66c*.
12. 20a*b^xy.
15. 48x8t/i<'2«.
18. ^mhi^p^.
Vii. (Page 22.)
I. a2 + o6ac. 2. 2rt + 6a6 — Sac. 3. a^ + Za^ + Aa^.
4. 9a515a'*18a^ + 21a2. 5. a'j _ 2a252 + ^js^
6. 3a569a*63 + 3a264. 7. 8/)i% + 9m2ji2+ lOmiiA
8. 18a66 + 8a5626a*63 + 8a36*. 9. a; Y  ary + x^  7xi/.
to. m^n  2mn^ + Smn^ — 71'*. 11. 1 44a^6*  72a'*6^ + 6()a^6^
1 2. 104a;*i/  136.c'!/2 + 4t)x)/''  ^xyK
I.
:c + 12x + 27.
4
x^ 15.r + 56.
7.
jc* + a;  20.
9
.T*31x2 + 9.
II.
.v;*  .X + 2x 
14.
a«  .r«.
16,
a;*81y*,
viii. (Page 27.)
2. x2 + 8x  105. 3. X'  2x  1 20.
5. «2 — 8a+15. 6. i/' + 7j/ — 7S.
8 . X*  1 2.c3 + 50x2 _ 84x + 45.
10. a" — 3a^ — 3a* + 1 3a'  6a — Ga + 4.
12. .x* + .>;2v2 + (/■». 13. x^y^
15. a^5.i;3 + 5x 1.
17. a* 166*. 18. 16a* 6*.
ANSWERS. 347
19. a^  Aa*h + 4aW + Aa^h^  llah*  1265,
20. a= + ba*h + aW  lOa1? + 12a6^  2}y>.
21. a* + 4ax + J6j;*. 22. Sla^ + Qa^x^ + x*.
23. x8 + 4ax*+16a*, 24. a^ + 6^ + c3 _ 3a6c.
25. x^ + x*y  9x3?/2 _ 20x2^/3 + 2x7/< + 15?/*.
26. a^fi + cd  rtc  6cZ. 27. s?  a\
28. x^  ax + 6.';  cx^ — abx + acx  hex + abc.
29. 1 c*. 30. x^ — y^. 31. a^Sx^^ 32. 47.
33. 2. 34. 14. 35. ab + ac + hc. 36. 60.
37. 2. 38. m^.
ix. (Page 28.)
I. a%. 2. a*. 3. _a363_ ^ l^aW.
5. 30xV. 6. a3 + a26a62. 7, 6a58a*+ lOal
8. a* + 2a3 + 2a2 + «. 9.  6x3y + x^y^ + y^j/S _ 1 9y4_
lo. 5m3 + 7/i,2/i 137H7i2 + 77i3. II.  IS/'^  22?2 + 96r + 135.
12.  7X* + X^Z + 8x2^2 + 9x^2 + 923
13. x« + xy. 14. x* + 2x3?/ +2x22/2 + 2X2/3 + 1/4.
X. (Page 32.)
I. x'^ + ^ax + a^. 2. x22ax + a2. 3. a: + 4x + 4.
4. x2_gj.^.9_ g_ x* + 2.r2y2 + ,y. 6 x42x2?/2 + y4.
7. a6 + 2a363^56_ g_ a62rt363 + 56
9. x^ + j'2 + ^2 + 2x?/ + 2x2 + 2yz.
10. x2 + 1/2 + ^2 _ 2x?/ + 2x3  2yz.
11. m2 + n2 + 2) + 72 + 2 m n  2mp 27nr~2iip2nr+ Jj r.
12. x* + 4x32x212x + 9. 13. X*  12x3 + 50x2 84x + 4y.
14. 4x*  28x3 + 85x2 126x + 81.
15. x* + i/ + ^+2xY2xz^2y^'^,
34^ ANSWERS.
1 6. cc8  8x«2/2 + 1 8x*i/*  8a;2i/8 + f,
17. a6 + 66 + c« + 2a3i3 + 2a3c3 + 26V.
18. x^ + 2/8 + 2^  2x^2/3  2x^»^ + 2yh^.
19. x^ + 4t/2 + Qz^ + 4x1/  6x2 — 1 '2,yz.
20. X* + 4?/* + 252*  4x^2/2 + 10x2g2 _ 2O2/V.
21. x^ + 3ax2 + Sa^x + a^. 22. x^  Sax^ + Sa'x  o*.
23. x3 + 3x2 + 3x+l. 24. x33x2 + 3xl.
25. x3 + 6x2+12x + 8. 26. a63a<62 + 3a26*6fi.
27. a' + 3a26 + 2,a¥ + 6^ . c' + 3a«c + 6a6c + Zhh + 3ac2 + 36c2.
28. a3  3a26 + ZaV^ h^c^ Zah + Gahc  Wc + Sat  Zhc.
29. m*  2?>i?i2 + n*. 30. m* + 2m^n  2mn^  n*.
xl. (Page 34.)
I. a^. 2. x*. 3. x^?/. 4. x*y^. 5. 66c. 6. 8c*.
7. 16a266c8. 8. 121m«ri«2>^ 9 12a3xy*. 10. 8a*6c^.
xii. (Page 35.)
I. x' + 2x + l. 2. 2/' 1/2 + 2/!. 3. a* + 2rt6 + 36.
4. X* + m2?x + m^p*. 5. Aay Ix + x^. 6. 8x^1/^ — 4.r!/2 _ 2y.
7. 27m%*18m%'* + 97ny. 8. 3xy  2x!/^  y*.
9. 13u269a62 + 76. 10. 196V + 12&V 76c*.
Xiii. (Page 36.)
I. 8. 2. 15a^ 3. 21x't/'.
4. 6m2rj. 5. 16a^6. 6. axJrax + l.
7. 2a2 + 3ax*. 8. 2 + 6a=6  8a*66.
9. — 1 2x2 4. 9_^. j^ _ 8y2_ 10.  x^ + i^x V + fry*.
Xiv. (Page 38.)
I. x + 5. 2. x10. 3. x + 4. 4. x+12.
^. x2+7x+12. 6. a;l. 7 x^ + x+l.
ANSIVERS. 349
8. x33x2 + 33;+l. 9. X2X1. 10. x^'ix+l.
II. x^x + l. 12. x32x2 + 8. 13. x + 3y'^.
[4. a^ + ^a^ + Zah' + y^. 15. a*  4a35 + Ga^fcs _ 4a{,3 + j4
16. x26x + 5. 17. a^ — ^a~h + Zah^ + A¥.
1 8. 2rtx^  3ax + a'. 19. x^  x + 1 . 20. x^ — a^.
21. x + 2i/. 22. X*  x^y + x^i/^  X1/3 + 2/*.
23. x^ + x*2/ + x'ff^ + xy' + xj/* + y^. 24. « + 6 — c.
25. 6 + 2&61 26. a6 + cd.
27. x^ — xy — x.: + y^ — i/z+2^ 28. x*' — x^2/ + x^!/*  x^j/" + y*.
29. 2J + 29r. 30. a*  0^6 + a'6'^  a6^ + 6*.
31. X* + x% + X"2/2 + xt/' + 7/*. 32. 2x'  Sx''^ + 2x.
33. a^ + 3a3 + 9a2 + 27a + 81. 34. ^7 + ^* + ^.
35. x29x10. 36. 24x^2ax35a2.
37. 6x27x + 8. 38. 8x3+12ax218a2x27a».
39. 27x3  36ax2 4 48a2x  64a3. 40. 2a + 36.
41. x + 2a. 42. a^Alfi. 43. x'^3xy.
44. x3xy2y. 45. x^ + Sx^y + 9xi/2 + 27i/3.
.46. a^ + 2a% + 4ab^ + 8¥ 47. 27a318a26+ I2a628R
48. 8x312x2i/ + 18x?/2_27i/». 49. 3« + 26 + c.
50. a22ax + 4x2. 51. x^ + xy + y" 52. I6x4xy + y'.
53. x^ + xyy. 54. flx2 + 4«x + 2flA 55. ax.
56. xyz. 57. 3XX + 2. 58. 46x + 8x''10x'.
59. x + y. 60. ax + byabxy. 61. bx + ay.
62. x^  ax + 6.
XV. (Page 40.)
I. x2 + ax + 6. 2. 2/2  (^ + to) 1/ + Zm. 3. «;'4cx + (/.
4. x^ + axb. 5. x2  (6 + 0?) X + 6ci.
xvi. (Page 42.)
I. m~n, m^ — mn + v?, m* — in?n + ni^n  mn^ + n*,
vv"  mhi + &c., m*  m'n + <Ssc.
556 ANSWERS.
2. 'm^n,w?^ mn + n, w? + mhi + &c., m* + mhi + &c.,
m® + m*n + &c.
3. a  I, a^  a + I, a*  a^ + &c., a^a^ + &c., a^  a^ + &c.
4. y + l,y'^ + y+l, y* + y^ + &.c., y^ + y^ + &c., y^ + y'' + &,c.
xvii. (Page 43.)
I. 5a; (x 3). 2. 3x{x^ + Gx2). 3. 7(7i/22i/ + 1).
4. 4a;y (a;2  3x1/ + 2?/2), 5. a;(x^ — ax^ + 6x + c).
6. 3xY (x^i/  7x + V). 7. 27a%^{2 + 4a%'^9a^b3).
8. 45xy(xV2x8i/).
xviii. (Page 44.)
I. (xa)(x6). 2. (ax)(6!x). 3. (by)(c + y).
4. (a + m) (6 + n). 5. (ax + y) (bx  y). 6. (a6 + cd) (x  j/).
7. {ex + my) {dx  nyy 8. (ac  bd) (bx  dy).
xix. (Page 45.)
I. (x + 5)(x + 6). 2. (x + 5)(x + 12). 3. {y + U){y + l).
4. (!/ + ll)(i/+10). 5. (»i + 20)(?n + 15). 6. (m + 6)(m + 17).
7. (a. + 86) (a + 6). 8. (x + 4?)i)(x + 9m). 9. {y + 3n)(y + l6n).
10. (s; + 4^j) (2 + 25;?). II. (x^ + 2) (x2 + 3).
12. (x^+l)(x3 + 3), 13. {xy + 2){xy+l6).
14. (xY + 3) (xy + 4), 15. (m5 + 8)(7rt5 + 2).
16. {n + 20q){n + 7q).
XX.
(Page 45.)
V (x5)(x2).
2. (x19)(x10).
3. {yU)iyl2).
4. iy20)(y10).
5. (n 23) (71 20).
6. (7i56)(jil).
7. (.x34)(x33).
8. (ab  26} {ab I).
9. (6'c»5)(6V6). 10. (xi/~ll)(xy«2).
Al^SlVEHS. 35»
xxi. (Page 46.)
I. (a; 4 12) (a; 5). 2. (x + 15)(x3). 3. (a+12)(al).
4. (a + 20) (a 7). 5. (& + 25) (6  12). 6. (6 + 30) (6 5).
7. (x* + 4)(x*l). 8. (x!/+14)(x2/ll).
9. (m5 + 20)(m55). 10. (7i + 30) (ji 13).
xxii. (Page 46.)
I. (xll)(a; + 6). 2. (x9)(.r; + 2). 3. (m 12) (/?!, + 3).
4. (7i15)(n + 4). 5. (2/14)(i/ + l). 6. (3 20) (2 + 5).
7. (x5_i0)(x5 + i). 8. (cd30)(cd + 6).
9. (m%  2) (m% + 1). 10. (;>Y  12) (i^V + '^)'
xxiii. (Page 47.)
I.
(x3)(x12).
2, (x + 9)(a;5).
3
(a618)(a6 + 2).
4. (x*  5m) (x* + 2m).
5
(l/3+10)(j/39).
6. (x2+10)(x2ll).
7
z (.r^ + Zax + 4a2).
8. (x + to) (x + n).
9
(2/33)(r/3l).
10. (xy — ab) (xc).
II.
{x + a) (x  6).
12. (x  c) (x + d).
13
(a6  d) (6  c).
14. 4.(x47/)(x32/).
xxiv.
(Page 48.)
I.
(x + 9)2. 2. (x +
13)2.
3. (x + 17)2. 4. (2/ + 1)2.
5
(2+100)2. 6. (X2 +
■7)2.
7. (x + 52/)2. 8. (m2 + 87*2)2,
9. (x3 + 12)2. 10. (X7/ + 81)2.
XXV. (Page 48.)
I. (x4)2. 2. (x14)2. 3. (x18)2 4. (7/ 20)2.
5. (350)2. 6. (X211)2. 7. (x157/)2. 8. (77^2  1 67*2) ».
9. (it' 19)2.
3S2 ANSH'EH^.
xxvi. (Page 50.)
1. {x + y){xy). 2. (x + 3)(x3). 3. (2x + 5) (2x  5).
4. (a2 + x2)(a'x^). 5. (a; + l)(a;l). 6. (x3 + 1) (x^ 1).
7. (:c* + 1) (x*  1 ). 8. (m2 + 4) {m^  4).
9. (61/ + Tz) (6?/  72). 10. (9xr/ + lla6) (9xi/lla6).
II. {ah + c) {ahc). 12. (x + mn) (xm + n).
13. (a + b + c + cO (« + ^<^'~^) H 2xx2y.
15. (xi/ + z)(xi/z).
16. {ah + m + n) {ahmn).
17. (fflc + 6 + (0 (^c^c^) 18. (a + 6c) (a6 + c).
19. (:c + t/ + z) (x + ya). 20. (a6 + mn) (a6m + n).
21. {ax + h]i+l){ax + hy\). 22. 2axx2by.
23. (Ha6) (la+'O 24. (l+xi/)(lx + 2/).
25. (X + 2/ + 2) (X1/2). 26, (a + 26 3c) (a 26 + 3c).
27. (rt2 + 46)(rt246). 28. (1 + 7c) (1  7c).
29. {ab + c + d){abc d). 30. (a + 6  c  rf) (a  6  c + d).
31. 3ax(ax + 3)(ax3). 32. (a^t^ + c*) (a^t^  c*).
33. 12(xl)(2x + l). 34. {9x + ly){5x + y).
35. 1000x506.
xxvii. (Page 51.)
I . ((( + /;; (cr nh + b"^). 2. (a  b) (a« + a6 + i^).
3. («  2) (rt2 + 2a + 4).
4. (x + 7) (.r7x + 49).
5. (65) {b + 56 + 25). 6. (x + 4?/) (x^  ixy + 16?/2).
7. (a6)(rt2 + 6rt + 36). 8. (2x + 3^) (4x2  6xj/ + 9i/»).
9. (4a  106) (Ifia^ + 40a6 + 10062).
' 10. (9x + Sy) (8 lx2  72xy + 64 j/2).
II. {x + y) {jc  xy + y) {x  y) {x + xy + y)
ANSWERS. 353
20. n^.
21. 25 z.
5. x5.
26. 1/ + 7.
29. 2.
30. 2.
34. 5.
35. 10.
12. (x+l)(x2x + l)(xl)(a;2 + x+l).
13. (a + 2) (a2  2a + 4) (a  2) (a2 + 2a + 4).
14. (3 + 2/)(9~3y + 2/2)(32/)(9 + 37/ + 7/2).
xxviii. (Page 51.)
I. a + 6. 2. Take 6 from a and add c to the result.
3. 22/., 4. a 5. 5. x + l. 6. x— 2, x1, x, x+l, x + 2.
7. 0. 8. 0. 9. da. 10. c. II. xi/. 12. xy.
13. 365 6x. 14. x10. 15. x + 5a.
16. A has X + 5 shillings, B has 1/  5 shillings.
17. x8. 18. xy. 19. 12X2/.
22. y — 25. 23. 256r/i*. 24. 4b.
27. x2_^2 28^ (x + 2/)(x?/).
31. 28. 32. 7. 33. 23.
XXiX. (Page 53.)
1. To a add b.
2. From the square of a take the square of h.
3. To four times the square of a add the cube of b.
4. Take four times the sum of the squares of a and b.
5. From the square of a take twice b, and add to the result
three times c.
6. To a add the product of m and b, and take c from the
result.
7. To a add m. From b take c. Multiply the results
together.
8. Take the square root of the cube of x.
9. Take the square root of the sum of the squares of x and y.
10. Add to a twice the excess of 3 above c.
1 1. Multiply the sum of a and 2 by the excess of 3 ab^ve c.
[S.A.] g
354 ANSWERS.
I.
2.
2.
0.
3
17.
4
31.
7
105.
8.
27.
9
14.
10.
120.
•3
30.
14.
5.
15
3.
16.
4.
12. Divide the sum of the squares of a and h by four times
the product of a and h.
13. From the square of x subtract the square of y, and take
the square root of the result. Then divide tliis result
by the e.xcess of x above y.
14. To the square of % add the square of ?/, and take the
square root of the result. Then divide this result by
the square root of the sum of x and y.
XXX. (Page 53.)
5. 20. 6. 33.
II. 210. 12. 1458.
17. 49. 18. 10.
19. 12. 20. 4. 21. 43. 22. 20. 23. 29. 24. 41536. 25. 52.
xxxi. (Page 64.)
I. 0. 2. 0. 3. 2ac. 4. Ixy. 5. a^^h".
6. 4x* + (6m  6?i) x'  (4m ^ + 9??in + 4?r) x'
+ (6™^?i — 6m?i) X + 4m^n^.
7. cr^ + dr + e. 8.  a*  6*  c* + 2*262 + 2tt2c2 + 2ic2.
When c = 0, this becohies  a*  6* + 2*262. When
6 + c = «, the product becomes 0. When a = h = c, it
becomes 3a*. 9. 0. 10. 34.
12. (a) (a + 6)x2+(c + rf)x. (/S) (a6)x3(c + (Z2)x2.
(7) (4a)x3(3 + ?))x2(5 + c)x. (5) a^  62 + (2a + 26) x.
(e) (7)1.2 _ ^2^ a^s ^ ^271! 2 — 2?((;) x^ + (2wi — 2?i) x2.
1 3. .x^ _ ((,, 4. 6 + c) x2 + (((6 + rtc + 6c) X  a6c.
14. x^ + (« + 6 + c) x2 + (rt6 + ac + 6c) X + a6c.
15. (a + 6 + c)3 = a3 + 3a6 + 3rt62 + 63 + c3 + 3rt2c
+ 6a6c + 36'c + Zac^ + 36c2.
(d + 6 _ c)3 = a^ + 3a26 + 3<»6' + 6^  c^  3rt2c
 ^a\)c  36c + 3(a2 + Zhc",
ANSCVERS.
(6 + ca)3=a3 + 3a263rt62 + 63 + c3 + 3a2c
6a&c + 362c3ac2 + 35c2.
(c + a  6)3 = «3 _ 3f^25 + 2aV' h'^ + c^ + ^ah
 6rt6c + 362c + 3rtc2  36c^.
The sum of the hi?t three subtracted from the first gives
24a6c.
1 6. 9a2 + 6ac3«6 + 46'662. 17. a^^x^^.
1 8. 2ac  26c — 2«fZ + 26c/. The value of the result is — 26c.
19. a6 + a:i/ + (6+ l+2a)a; + (2a6 1)2/.
20. 9. 21. 06 + 2: + (a 6+1) a; (a + 6 + 1)7/.
22. 2. 23. (7m + 4?i + l)a;+ (1 6>i — 477i)?/.
25. 4a2 + 6ac + 2a6 + 96c662. 26. 3; 128; 3; 118.
27. 9. 28. 44. 29. 20. 30. 35. 31. 18.
xxxii.
(Page 60.)
I.
3.
2. 2.
3 1
4
7. 5 2.
6. 2.
7
3.
8.
4.
9. 9.
10.
A
?!s. 54.
II.
2.
12. 9.
13 9.
14. 7.
15. 3.
16.
7
17
2.
18. 8.
19. 10.
20. 6.
21. 4.
22.
lit.
23
3.
24. 15.
25. 1.
26. 2.
27. 3.
28.
4.
29.
6.
30. 1.
xxxiii
(Pa^e 62.)
I. 70. 2. 43. 3.' 23. 4. 7,21. 5. 36,26,18.1:2.
6. 12, 8. 7. 50, 30. 8. 10, 14, 18, 22, 26, 30. 9. .iC
10. 12 shillings, 24 shillings. 11. 52.
12. A has £130, B il50, C jElSO, D £90.
13. 152 men, 76 women, 38 children. 14, £350, £450, £720.
15. 21, 13. 16. £8. 15s. 17. 84, 26. 18. 62, 28.
19. The wife £4000, each son, £1000, each daughter £5no.
20. 49 gallons. 21. £14. £24, £38. 22.31,17
356 AX.sirhRS.
23. £21. 24. 48, 36. 25. 50, 40. 26. 42, 18.
27. 60, 24. 2.S. 8, 12. 29. 88. 30. 18. 31. 4a
32. 57, 19. 32, 4. 34 SO. 128.. 35. 19, 22.
36. 200, 100. 37. 23, 20. 38. 53. 318. 39. 5, 10, 15.
xxxiv. (Page 68.)
I. a%. 2. xyz. 3. 2xy. 4. 15m2?ijs. 5. 18a&c(f.
6. a2j2_ 7_ 2. 8. 172)2. 9 4a;2j/222. 10. SOxV
XXXV. (Page G9.)
I. ah. 2. a'^fcl 3. a — x. 4. a + x. 5. 3x + l.
6, l5a. J. x + y. 8. xy. 9. x1. 10. 1+a.
XXXVi.' (Page 70.)
I. 3453, 2. 36. 3. 936. 4. 355. 5. 23. 6. 2345.
xxxvii. (Page 74.)
I. x + 4. 2. x+10. 3. x7. 4. x + 12.
5. x3. 6. x + 2y. 7. x4!/. 8. xl5y.
g. xy. 10. x + y. 11. xy. 12. x + y.
13. x + y. 14. a + 6 c. 15. ix + y. 16. 3x!/.
17. bxy. 18. x* + x^ 4X + X + 1. 19. x2x + 4.
20. x^ + xy + y"^. 21. x^ + x" — x1. 22. 3a^ + 2a66.
23. Zx — y. 24. 3xlli/. 25. 3a6.
26. 3(ax). 27. 3x2. 25. 3x2 + al
29. x2 + 2/l 30. x + 3. 31. (3a + 2x)a.
xxxviii. (Page 76.)
I. xf2. . 2. x1. 3. x + 1. 4. y1.
5. x22x + 5. 6. x2. 7. J/ 2!/ + 6.
ANSWEfiS. 357
xxxix. (Page 81.)
J^ 2a; _56 2x2
aW'c^ , 4xy 3y _ 5hc
5" ~3~' 36c* 7* 2aa* * 4a^'
4 5 m a 2??ix
3x^2/^' .P ' " a + 6* ■ Sm^p — x
1 2a + x jr^ o2
3?/ 5x3* 4ax — x' *' 6c' ' 2x3y'
3ab .  c2a 3
17. sr . 18. ^. 19. .
' 2bc + c c + 2a ^5
5 1 2
20. ^ — . 21. s srr 22.
2x2y* ' 7ax7by' ' 9abx—12cdx'
xy 62 1 , 2a + 26
23. .r^. 24. ^.. 25. TT 26. ;;— .
1  X
7. 12 28. .
Xl. (Page 82.)
a + 5 x5 a + l
a + 3* * x3' '■ 037*
. ^JL^y x^x+l 6 — +^'
^ x + 7y ^ oi?y^
x2 X3 x2_5a; + 6
^* x+"4* ■ x + r ^" 3x2 7x ■
x2  5X + 6 x2 + XT/  y'
3x2 _ 8x ' * x2  xy  2/2*
a2 + 5a + 5 6^ + 56 m2 + 4m
'"• a2 + a2* ^^' 6^ + 65* ^^ m2 + m6'
a2a + l 3ax7a 14x6
"5* a2 + a+l' ^ ■ 7x2 3x' ^7 g^x  21a'
g 10fl 14a2 2a62 + .3a65a
' ' 159a6a2' '9" 762^I"56 ^*
358 ANSWERS.
a^a+l 3xl ab
20. , „ ;,. 21. „ ,, 22. 5.
a;  2x + 2  2x + 'ixb
23. ^^32 24. 3. 25. ^^_5 •
4x + 9x + l 2x3a „ x3
2x3x2" ^'^* 4x2 4.6ax + 9a2 ^^ x^'
m — 1 x^ + 5x
2Q. T 30 5 3'
6a + 26
x2 + 4 , X3 + X2
^5' X + X+1' ^ 2x^ + 2x + r ^'^'
x^2x + 3 x32x^2x + l
^ 2x2 + 5x3" ^^' 4x27xl " '^°* 3a28a
x + 3 ■
x5
2j; + 3'
a:2 + xi2
3x + 5 ■
a2_5a + 6
Xli. (Page 86.)
I.
\2f
5. ax.
1
^ 2
2x'
2 3y3
^•1
3
7 8
bkm?
10. j
4pq
4
by
9ax'
8.
Sa^c'
9d2'
4
3mnxi/
423g2
xlii. (Page 86.)
ah 4 (x 4 2) (x  4)
~W ^' 3' ^' x(x2) •
(xl)(x 6) x6 g (x2)(x5)
x''' * ^* X — 3' * a^ '
1 01 n V c a+b
7. 1. 8. 0. 9. — ^— . 10. f.
x—y ca —0
x — m + n  x — yz*
II. . 12. 1. 13. ~ .
x + mn "' x + y + z
JJVSIV£J?S.
359
xliii.
(Page 87.)
lOae
^ 2bx
3
3. ^. 4.
4
36?ix"
3
5 4
, hx
^ 4i'
5x
7 I4
^ x2
9
1
x2'
Xliv. (Page 89.)
I. 12a3x2. 2. 12x2?/2. 3. Sa^ftz 4. a^x*.
5. 4ax3. 6. aW(^. 7. a3x22/2. 8. 102a2xi
9. 20p222r. 10. l^ax^y^.
Xlv. (Page 91.)
I. x2^(j^a.)_ 2. x^x. 3. a{a'^}p).
4. 4x21. 5. a3 + 63. 6. x2l.
7. (x5l)(x + l), 8. (x2 + l)(x3+l).
9. (X + I)(x31). 10. X*l.
II. x(x3l) (xHl). 12. X (x + 1) (x^  1).
13. (2al)(8a3 + l). 14. 2.(;2 + 2x2/.
15. (a + 6)2 (a 6). 16. a2_ft2_
17. 4(lx2). 18. x3l.
19. (a  6) (a  c) (6  c). 20. (x + 1) (x + 2) (x + 3').
21. (x + 2,')2(xi/)2. 22. (a + 3)(a"''l).
23. x2(x2i/). 24. (x + l)(x + 2)(x + 3)(x + 4).
25. 12(x?/)2(x3 + 2/3). 26. 120x1/ (x2 1/2).
Xlvi. (Page 93.)
I. (x+2)(x + 3)(x + 4). 2. (a5)(a + 4)(a3).
3^ (x+l)(xH2)(x + 3). 4 (x + 5)(x + 6)(x + 7).
5. (xll)(x + 2)(x2). 6. (2,: + 1) (x+1) (x2).
36o ANSWERS.
7. (x2 + y)(x + y)(x2 + i/2)(x2/). 8. (x5)(x3)(x + 5).
9. (7z4)(3x2)(x23). 10. (a;2 + j/2)(x + i/)(x2/).
II. (a2 62) (a + 26) (a 26).
xlvii. (Page 94.)
I. (a;2)(a;l)(x3)(a;4). 2. (a; + 4) (x + 1) (a; + 3).
3. (a; 4) (a; 5) (a; 7). 4. (3x  2) (2x + 1) (7x  1).
5. (x+l)(xl)(a; + 3)(3x2)(2x+l).
6. (x3)(x2 + 3x + 9)(x12)(x'2).
xlviii. (Page 95.)
15x 16x 9x21 4x9
^' W "20"* ^' 18 ' ~T8~'
4x8?/ 3x2 Bxy 20a + 256 Q>a'%ah
lOx ' 10x2 • 4 iOrt^~' 10a2 •
48a60ac 15a 10c , ahW a^a%
3  3x 3 + 3a;
2j^2!/2 22y2
1X" 1X2' • l_2/4' i_y
5 + 5x 6 a6 + ax b
lO.
lx2' lx2' • c(6 + x)' c{b + x)'
a—c b—c
{a'^'b)(h~c)(acy (ab) (bc) (ac)'
c{bc) 6( a6)
o6c(a6) (ac) (6 — c)' abc (a  6) (o  c) (6  c}'
xlix. (Page 98.)
15X+17 71a 206 56c 32x + 9tf
I . 2. . 1. .
15 84 ^ 42
16x» + 55 x'+ 4x1/  55!/ , 27x2  2x2;/ _ ^ gjy _ 28^2
^ "■ 50x ■ '■ li:.'
AI^SWERS. 361
ISOffl^ + 54ffl6 + 3316^  20a62 S Ox^ + 64 x2 + 84a; + 4 5
9062 • 7 gQ^2
35rt2 + 23a6 + 2l6c42c 2 Aa?c  Zac^  3ac + 7c»
2 lac ■ ^" ac^
lly28xY4xy7x'
3a*  7a^b + 4a%c  5ahh + ahc^  ¥c*
aWc^
1. (Page 99.)
2xl
4 ^ 2
(x6)(x + 5y
 (x7)(x3y •"■ (i+x)(ixy
4
4x1/
 1 y. a + bx
1+x" 'c + dx'
(x + i/)(x3/y
7
2x2
2xi/ 2x4 5a
(xt/)2 9 (^^^jr
{x + y){xyy
10.
1
(a + x) (ax)"
U. (Page 100.)
I.
2
4x 2x 86'
1x*" ^" 1x*" "^^ a8~l>
5
x + y
y
, 3x3 + 20x2 _ 32a; _ 235
• (x + 4)(x3)(x+7)"
7.
3x3 24x2 + 60x
46 3x22ax6a
(x2)(x3)('x
4) ^ (xa)3 •
9
6
X
(xl)(x + 2)(x+l)' — (x+l)(x + 2)(x + 3)"
1 1
3x2
x2r
ed
■ {a + c){a + d){a + ey ^^' '
14
2. 15.
y . 16. 0. 17. ";^^^
362 ANSWERS.
i8. 0. 19 A^, 20. 0. 21. 0.
a + 6
lii. (Page 103.)
?/ 1 3x2 y + 6
^ ^ 3(1^
^' x1/' ^" 2 + « ^* x'r "^ 3(12/2)
5. 0. 6. , r^ rr. 7
^ (x + a) (x + 6) c* </
1 2 1
1x* ^ (x2)(y8;) aoc
liii. (Page 110.)
2x + ll 2(x8)
(x + 4)(x + 5)(x + 7)* ■ (x6)(x7)(x9)'
2x  1 7 2 7/1^ + 4m2n + m?!^
4 rr^ 5
■'■ (x4)(x + ll)(x13)' ^ x + 3" ^" n(m + 7i)2
,  Ilx3x2 + 25xl „  1
6. 0. 7. ;r7^ ,, . 8. 0. 9. :, ■.
' 3(lx*) ^ l + x
liV. (Page 107.)
I.
16.
2.
12.
3
15.
4.
28.
5
63.
6.
24.
7
60.
8.
45.
9
36.
10.
120.
II.
72.
12.
96.
13
64.
14.
12.
15
28.
16.
1.
17
8.
18.
9.
19.
7.
20.
4.
21.
5.
22.
1.
23
1.
24.
3
2'
25.
100.
26.
24.
27.
2
28.
6.
29.
24.
30.
4.
IV. (Page 108.)
16. 2. 5. 3. \. 4. 1 S 8.
AA'SU'EI^S.
363
^•4
12. 12.
18. 9.
7. 9.
13 8.
19. 9.
8. 2.
9. 11.
14. 7. 15. 9.
20. 9. 21. 10,
10. 6.
16. 7.
II. 2.
17. 7.
14
25.
c
a + 6'
6c — rfm
a — 5
3a&2Jfe3
4acl
a
10. 2
3a +1
„ ahd + ac
18. — J — T.
aa + a
22. 1.^
Ivi. (Page 109.)
3c 2a
^" 56^T'
6 (a + c)
^ 1 + a
15
2.
18a + 2&
4a + 3 ■
19. 61.
23. 6m.
6c
c26"
o a (m  3c + 3o)
c a + m
26.
29.
12. 0.
16. ^, — .
0^6 — bc + d
6bd + ah
3a^l2cl
{a + hf
b — a'
13
_6_
ar
f
21.
2a^
F"i
3a36c + 2a%^ + «6^
^"^ 63 + 3a3cT3a26c + 2a2p
c
ac
T'
27.
30
a6l
6c + d'
a'e (c — d)
Xofi'+Wjd'
I.
2.
6.
1
7*
I.
9.
16. 12.
Ivii. ^Tage 111.)
2. 15.
7 5
2'
3 1
4
13"
8. 6. 9. 7.
12. 19. 13. 1. 14. 4.
1
2" *^ 8*
17. 2. 18. \. 19. i
7
5 To
10. 6.
15. 
20. 3.
35
364 AmiVERS.
Iviii. (Pacre 11:1)
4f)Q
I. 20. 2. 3. 3. 40. 4. ~. 5. 60.
J ^ 46 ^
6. 10. 7. 5. 8. 20. q. 3. 10. ^.
II. 8. 12. 100. 13. 0. 14. 1. 15. 5.
16. . 17. 5.
liX. (Page 114.)
I.
100.
2.
240. 3. 80.
4. 700.
5. 28,32.
6.
A.\
7. 24, 76,
8. 120.
9, 60.
10.
960.
II. 36. 12.
12,4.
13. £1897.
14.
540, 36.
15. 3456, 2304.
16. 50.
17. 35, 15.
18.
29340, 1867
19. 21, 6.
20.
IO5I, 13l
21. X has £1400, B has £400. 22. 28, 18.
m (nb  a) n (mb a) a + b a — b
23. — !^ ■% 5^ . 24. ^r, 5. 25. 18.
■^ nm mn ^2' 2 ^
26. £135, £297, £432. 27. £7200. 28. 47, 23.
29. 7,32. 30. 112,96. 31. 78. 32. 75 gallons.
33. 40, 10. 34. 20. 35. 42 years. 36. 1^ days.
37. 20 days. 38. 10 days. 39. 6 hours. 40. I53 days.
41.
4 days.
42. 1.:, hours.
43 48'.
44.
2 hours.
45
abc
, , minutes.
ab + ac + oc
46. 48.
47. 51;r, 6I.3, 47.J gallons. 48. 9_ miles from Ely.
000 i
ANSWERS. 365
, , 1 ac Id ^13
49. 14 miles. 50. J, — . 51. 11—.
30
52. 42 hours. 53. 30. miles. 54. 50 houi's.
55. (1) 38^ past 1. (2) 54^ past 4. (3j 10 past'S.
56. (1) 27 past 2. (2) 5^ and also 38— past 4.
9' 6'
(3) Slj past 7, and also 54— past 7.
57. (1) 16^ past 3. (2) 32^ past 6. (3) 49^ past 9.
58. 60. 59. £3. 60. ^. 61. ISidays
62. .£600. 63. ^£275. 64. 60.
65. 90', 72', eC. 66. 126, 63, 56 days. 67. 24
68. 2, 4, 94. 69. 200. 70. 2*, 5—.
71. 30000. 72. X200000000. ' 73. 50.
Ix. (Page 127.)
z* + ax + 3a
I. .
X
a2 + 3ax2x'
^" x{xyy
2a3 + 6a26 + 6a62 + 26»
"^ (o6)(a2 + 62) •
Ixi. (Page 128.)
813X
xy V y 4 j^_^_
x^ + ox'^ + l
5 2a;2x3 + r
, x^x + l a^ + a + 1
D  — . • 7 •
X a
3«6
ANSWERS.
8.
1
X. Q. . lO
X
X. II, 
2x]/
a(a2 + 2a6 + 262)
(a + 6)2
14. «tL
I 1
^3
'^' c(a6c)'
Ixii.
(Page 129.)
I.
13 15
2.
fi 6 c (i
1 +  + J + .
a c a a
3
^_3 + ?_y.
y'^ y X 3?'
4
i2~¥"^18~3(j'
5
6p 4} 12r 24s
grs jjrs fqs fqr
6.
100 40^^40 !5
Ixiii. (Page 131.)
1. 22a + 2a2_2a3 + 2a*
, 2 4 8 16
2. 1 + — 3 + —4
m m" 771"* m*
, 26 262 265 26*
' a a a"^ a*
2x2 2x* 2x« 2x»
x2 x^ X* X^
5. X + — +2 + 3 + 4
' a a^ a^ a*
^ 6 6x 6x' hs? 6x*
a a^ a^ a* a°
7. 1  2x + 6x2 16x3 + 44x*
8. l + 2x + x2x32x*
9. 1+36 + 66= + 126' + 246*
, ^ ,, 263 26*
10. x6.c + o^ + ^
X X^
ANSWERS. 367
a2 a26 0^52 a^ a^
X x^ x^ x* x^ '
, 2x 3x2 4a;3 5a4
12. 1 — + ..  , +—r....
n. n n9 n*
a a a'" a*
13. x^3ax + 2a^x + 4a\ 14. m<  lOm^  41to  95.
Ixiv. (Page 132.)
J x^ x2 23x 1 , a^ _49a^ la 1
9^4'''l20'''20' ^' 20 "600''" 60" 15
3
^*^ 4.
x*+l + 
X*
5 ^
6.
12 11
a^ ac b'^ c^'
7 l+a2 + a^
«■ '4'f
x«
'64
9
5 7 107 5 7
X*"^2'i3i2x2 + 6x'^6 ^°
¥ a* 0*^ ^•
Ixv. (Page 134.)
1 ,1 , m 1
I. «. 2. a + T. 3. m2 + 
X b n n^
, c' c2 c 1 XV
^ a d^ d^ d^ •' y X
6. ^ + 7 + 75. 7. 02 + ^2. 8. x55x2+x + 9.
62 ""^o2 ^° a2 ab ac"*" 62 6c "^"?
Ixvi. (Page 135.)
I. 05x2 •143x 021, 2. •01x2+l25x21.
3. 12x2 + 13x2/ 141/2. 4 172x2 05x2/ 3 12?/2.
5. 0. 6. 300763.
368 ANSWERS.
Ixvii. (Page 135.)
/, «o «, „ a4 , \
1. OiXl 1+— a: + — x2 + — x3+ ... I.
^ aj ttj ttj /
2. a;i/2( + ). 3. x2(l+^ + ^).
" \z y xf \ X xV
4 (a + 6) I (a + 6)2c(a + 6)d + ^l.
Ixix. (Page 138.)
2x2 + 3x5 ,a2 + 5(j_i4 2«}7
I. 46. 2. — = r — and — — r — . 3. v  ,
7x5 a + 9 ^ a^+f'^
37x271/21922 11
4 24 • 5 9
, 60x* + 42ax']07a2x2+10aSx + 14o*
6. j2 •
„ x' + xhi + 2w^ X  8 x2 a
8 tY^ Sr lO —,5 II ^i 4 12. :; — ,
x{T/ — y^) X + 8 1  X* 16
„ 1 ah + ac + hc + 2 a + 26 + 2c + 3
^' ~F a6c + a6 + ac + ic + a + 6 + c + 1'
1 6 62 52 8a252 6(a2 + 6»)
15. 2 5 18. 4 ,4. 19. 75 — roi.
■' a ttx a^x ax a* — o* ^ a {a^  6^)
a^ + ¥ 1 a + b c
22.
^ (a6)2.(a2 + 6')" 2(x + l)2 "" a6 + c
A 1 ^ (x4)(x + 2)2
23. X. 24. 0. 25. 1. 26. ^^ .
27.
X
,2
29 ^2 + i 30. 1.
W X'
2 + 5x + 17x2ll.r '21x*
3^ (3^2x'^7x?V*
28.
X3(x2 4
31
^1
3.
33
r' + y^
34.
2.
i
ANSIVERS.
369
35
2a6 . r.
r 36. 0.
^9 :^,(;i + 2/2)
X
40, ,
41.
x2 + 3x + 3 + ,.
X X
(.^ + 2/2)2
^^ x« + 3/*
44 1
46.
^ + ^ A7
1
48. 1.
pq ^'' (x2 +
l)(a;3 + l)
49
2a2  ax  ay. 50.
a + 6 + c
(a363)2.
Ixx. (Page lib.)
I. x = 10 2. x=9 3. x = 8
2/ = 3. y = 7. i/ = 5.
4. x = 6 5. x=19 6. x = 5
l/ = 8. i/ = 2. 2/ = 3.
7. x = 16 8. x=2 9. x = 4
y = 35. y=l. i/ = 3.
Ixxi. (Page 145.)
I.
x=12
2. x = 9
3. x = 49
4. x = 13
y = A.
2/ = 2.
2/ = 47.
t/ = 3.
5
x = 40
6. x = 7
7. x = 5
8. x = 6
!/ = 3.
l/ = 2.
i/=l.
y = 4.
9
x = 7
2/ = 17.
Ixxii.
(Page 146.)
I.
x = 23
2. x = 8
3. x = 3
4. x = 5
2/ = 10.
2/ = 4.
!/ = 2.
t/ = 9.
5
x = 2
6. x = 7
7. x=12
8. x = 2
t/ = 2.
y = 9.
y=9.
2/ = 3.
9
x = 3
/ = 20.
rs.A.]
sa
370
1
ANSWERS.
Ixxiii.
(Page
147.)
I.
x = 7
2.
x=9
3
x = 12
4
x= 2
2/= 2.
J/=3.
1/=3.
i/ = 19.
5
x= 5
6.
x=3
7.
x = 7
8.
1
^ = 2
2/ = 14.
y=2.
3/= 5.
■^ 3
9
x=2
y=i.
Ixxiv.
(Page
148.)
I.
x = 6
2.
x = 20
3
x = 42
4
x = 10
2/ = 12.
2/ = 30.
2/ = 35.
y = 5.
5
x = 9
6.
x = 4
7
x = 5
•
8.
x = 40.
2/= 140.
2/ = 9.
y = 2.
2/ = 60.
9
13
x=12
2/ = 6.
x = 6
lO.
14.
x=19
1/ = 3.
x=19^
II.
15.
x = 6
2/ = 12.
1
12.
3201
^~ 708
278
^=59
y= 17
y=^'
5*
Ixxv. (Page 149.)
eg nf ce + bf em + bn
x^ * 2 X= '^ "^ x =
mq — np ' bd + ae ^' ae + bc
_ mf— ep _cdaf _ an — cm
mq  np' ^~ bd + ae ^ ae + bc'
de n'r + n/ , a + b
x = — r^ c. x= — , 7 6. x = — ^—
c + d ■' mn +mn 2
_ ce _ to/ — mV _ a — 6
^~c + d' ^~mn' + m'n' ^~ 2~ "
c(/6c) „ 1 2b^6a^ + d
x= ^;; , / 8. x = T 9. x= 5
a/6(i aft ^ 3a
_c(acrf) _J_ 3a'fe^ + rf
^~ afbd '''cd y~ 36 •
ANSWERS. 371
a ofi bm
10. x = T II. x = i — 12. x = r
be b + e b — m
_a + 2b _ ¥c^ _ bm
^ c ■ ^~~a ■ ^~b + m'
Lxxvi. (Page 151.)
_ 1 _ ^_ _ 6  g''
'• ^~2 ^ ''~b2a 3 "'bdac
1 2 _ ¥a^
^4 '^~S^^ y~bca<r
2a 61 , 1
26 61 1
103' ^ 5'
x =  8. x = 
a n
1 1
Ixxvii. (Page 153.)
I. x=\
2.
x = 2
y = 2
y = 2
2 = 3.
2 = 2.
5. x = l
6.
x = l
2/ = 2
2/ = 4
2 = 3.
2 = 6.
9. x = 2
10.
x = 20
t/ = 9
2/ = 10
3. x = 4 4. x = 5
2/ = 5 y = 6
2=8. 2 = 8.
2 8. x = 5
7 ^ = 3 , = 6
y=7 2!=7.
2 = 36^.
2=10.
Ixxviii. (Page 15'.)
I. 16, 12. 2. 133, 123. 3. 725, 625.
4. 31, 23, 5. 35, 14. 6. 30, 40, 50,
372 ANSWERS.
7. £60, £140, £200. 8. 22s., 26s. 9. £200, £300, £260.
10. 41, 7. II. 47, 11. 12. 35, 11, 98. 13. £90, £60
14.60,36. 15.6,4. 16.40,10. 17.503,1072
18. 10 barrels. 19. ^s.,\s. Sd. 20. £20, £10.
21. 15s. \Qd., 12s. 6(f. 22. 4s. 6d., 3s. 23. 35, 65
24. 26. 25. 28. 26. 45. 27. 24. 28. 45.
29. 84. 30. 75. 31. 36. 32. 12. 33. 333.
34. 584. 35. 759. 36. I 37. A 38. I
2 7 35 19
39.3 40.^9. 41.41. 42.40
43. £1000. 44. £5000, 6 per cent. 45. £4000, 5 per cent.
46. 31^, 18 47. 20, U). 48. 3 miles an hour.
49. 20 miles, 8 miles an hour. 50. 700. 51. 450,600.
52. 72, 60. 53. 12, 5s. 54. 750, 158, 148.
55. 15 and 2 miles. 56. The second, 320 strokes. 58. 50,30.
5
59. 4 yd. and 5 yd. 60. , 6, 4 miles an hour respectively.
61. 142857.
IxxiX. (Page 164.)
I. '2rt]. 2. 9af6*. 3. IIw^hV. 4, Sa?lf>c.
5. 267a26x». 6. ISa'S^c^. 7. ~. 8. — V
^ ' Ah 2a?
5a63 16x6 25a
^ llxV '° 171/2 ^' "1S6
Ixxx. (Page 167.)
I. 2a + 36. 2. 4fr^3P. 3. a6 + 81. 4. 1/819.
5. 3a6c17. 6. xZx + b. 7. 3a; + 2x + l.
^N^IVERS. 373
8. 2r23r+l. 9. 2u2 + 7i2. 10. l3x + 2x2.
II. x32x2 + 3x. 12. 2^232/3 + 42. 13. a + 26 + 3c.
14. a^ + arb + aki' + h\ 15. x^2x — 2xl.
16. 2j;2 + 2ax + 46. 17. 3  4x + 7x  lOx^.
18. 4a2_5o6 + 86x. 19. Za^Aap^bt.
20. 2i/2a;  3yx2 + 2x^. 2 1 . 5x^2/  3x?/2 + 2 y^.
22. 4x2  3x2/ + 2i/2. 23. 3a 26 + 4c. 24. x^ — Zx + b.
25 . 5x  2i/ + 3^. 26. 2x2  y + i/2.
IXXXi. (Page 168.)
3 a  1
5. x2_a;+ 6. x2 + x.
8. x2 + 4 + ^. 9. i^a^x + ia.
x' 6 4
II. 6m +^.
n 5
2x 3i/ z
13. — ^ + .
"^ z z X
a b e _d
^5 3"4''"5 2"
« o ftX ,
17. 3x2 — + OX.
Ixxxii. (Page 170.)
I.
4
4
a b
b^a
7
2a36 + ^.
4
10.
1 2 3
+ .
X y z
12.
ab  3cd + Y
14.
2m 3»
16.
7x22x
18.
3x213.
I. 2a.
2. 3x2t/2. •
3
 binn.
4
 6o<6.
5. 7V>c<'.
6. lOafc^c*.
7
 I'lm'n^.
8.
llo'6«.
%1A
AJ^SWMRS.
Ixxxiii. (Page 172.)
I.
ah.
2
!. 2a + 1. 3. a + 86.
4. a + 6 r c.
5
xy + z.
6. 3x22x+l. 7.
1  a + o2
8.
xy + 2z.
9. a24a + 2. 10.
2m23m+l.
II.
x + 2y — z.
12. 2m3;ir. 13.
IXXXiv. (Page 173.)
1
TO 4 1 .
TO
I.
2a  3x.
2. 1  2a.
3. 5 + 4x.
4
ah.
5. x+1.
IXXXV. (Page 175.)
6. TO  2.
I.
±8.
2. ±ah. 3. ±100.
4 ±7.
5
± v'(ll).
6. ±8a2cl 7. ±6.
8. ±129
9
±52.
10. ±4. II. ± J('
7?l /
12.
W(.
I>
13. ± x/6. 14. ±2v^2
Ixxxvi. (Page 179.)
I.
6, 12.
2. 4, 1(5. 3. 1, 15.
4. 2, 48.
5
3, 131.
6. 5, 13. 7. 9, 27.
Ixxxvii. (Page 180.)
8. 14, 30.
I.
7,1.
2.
5,1. 3. 21, 1.
4. 9,  7.
5
8,4.
6.
9, 5. 7. 118, 116.
8. 10±2v'34.
9
12, 10.
lO.
14,2.
Ixxxviii. (Page 181.)
I.
3, 10.
,^ ■■ 7 25
2. 12, 1. 3 2' y
4. 20, 7.
5
1 5
4' 4"
6. 9, 8.
7. 45, 82.
8.
8, 7.
9. 4. 15.
TO. 290, 1.
ANSWERS. 37S
Ixxxix. (Page 182.)
I.
7
3'
5
3'
2.
1 3
5' 5"
3
3.^
4
1,
3
11*
5
3 5
5' f
6.
^.1
7.
8,
2
3"
8.
xc.
(Page 182.)
I.
3,
8
3'
2.
u.,.
3 ^. ¥■
4
8,
19
2*
5
^.^
^ ^. 1
7.
8,
17
4"
8.
7 3
2' 14"
xci. (Page 184.)
I. a±V2a 2. 2rt±^/ll.a. 3. 2'~"2*
, ^ , cfi + ah a^ — ah
5. 1, a. 6. 6, a. 7. — •
•^ ab ' a + b
c+ J{c'^ + 4ac) c ^ (c^ + 4ac)
^" 2 (a + 6) ' 2 (oTi) '
4
Zn,
n
~2'
8
d
h
c'
a
0.
62
¥
ac'
ac
2a b 3a + 26
II. , ,
ac be
ac^ + bd^ ac + bd^
2a + 3d Vc' ~2a^3f?v'c'
XCii. (Page 185.)
I. 8, 1. 2. 6, 1. 3. 12, 1. 4. 14, 1.
9
4'
5. 2,  9. 6. 6, 5. 7 5, 4. 8. 4,  1. 9. 8,  2.
376 ANSWERS.
lo. 3, ^. II. 7,^. 12. 12, 1. 13. 14, 1.
14. \\ 15 13, y. 16. 5,4. 17. 36,12.
o ^ c 25 5 „ 10' „ 10
18.6,2. 19.18,3. 2°7,y. 31. .,y.
^ r o 1 12 2 1
22.7,5. 23.3,2. 24.2,3. "5.3.,.
26. 15,14. 27. 2, . 28. 3, ^. 29. 2,.
o 23 o 14 .5 o 21
30. 2, . 31. 3, — . 32. 4, „. 33. 3
15' ^ ■ ' 3' ^ ' 3' •'■^" ' ir
58
13"
58
34. 14,10. 35. 2,. 36. 5,2. 37. a, 6. 38. a,h.
, , o , a 2a. ft 6
39. a + 6, a 6. 40. a, a'. 41. ^, y. 42. p .
xciii. (Page 187.)
I. a; = 30 or 10 2. a; = 9or4 3. 2; = 25 or 4
y = 10 or 30. y = 4 or 9. j/ = 4 or 25.
4. 2 = 22 or 3. 5. x = 50 or  5 6. a;=100 or  ]
j/ = 3or22. i/ = 5or50. i/=lorl()0
XCiv. (Page 187.)
I. x = 6or2 2. a;==13or3 3. x = 20or6
y = 2 or — 6. i/ = 3or — 13. y = Gor20.
4. x = 4 5. a;=10or2 6. x = 40 or 9
y = 4. 1/ = 2 or 10. j/ = 9or40.
XCV. (Page 188.)
I. a; = 4 or 3 2. a: = 5 or 6 3. a = 10 or 2
i/ = 3or4. J/ = 6 or 5. t/ = 2 or 10.
4. a;==4or2 5.a; = 5or3. 6. x=7or4
y = 2 or — 4, 1/ = 3 or  5. y = 4 or  7.
i
AA'SIVERS. 377
XCVi. (Page 189.)
1 . 2 = 5 or 4 2. a; = 4 or 2
y ==4 or 5. y = 2 or 4.
1
4 a; = 3 5. x = ^
y = 4. y = 2. y =
xcvii. (Page 191.)
3
1 1
x = — or —
3 2
1 1
^ = 2°^ 3
6.
1
I
I.
j:=^i or
3
2.
x=±6
3
x=±10
y — 'i 01
4.
2/= ±3.
y=±n.
4
x=±8
5
2 = 5 or
3
6.
95
x = 5or^
33
!/ = 2ory.
y=±2.
y = 3 or
5.
7
x=±2
y=±5.
8.
x=6
y=5.
9
z=±2
2/=±l.
10.
x=±2
II.
2=±7
12.
11
2=3 or 
!/ = 2or.
y=±3.
i/=±2.
'3
a = 10 or
12
14,
x = 4 or
85
"8"
19
8"
15
2=±9 or ±12
?/ = 12 or
10.
t/ = 9 or
y=±l2 or ±9
xcviii. (Page 193.)
I. 72. 2. 224. 3. 18. 4. 50, 15, 5. 85, 76.
6. 29, 13. 7. 30. 8. 107. 9. 75. 10. 20, G.
II. 18,1. 12. 17,15. 13. 12,4. 14. 1296. 15. 56^
16. 2601. 17. 6, 4. 18. 12, 5. 19. 12, 7. 20. 1, 2, 3.
21. 7,8. 22. 15,16. 23. 10,11,12. 24. 12. 25. 16.
26. £2, 5s. 27. 12. 28. 6. 29. 75. 30. 5 and 7 liours.
31. 101 yds. and 100 yds. 32. 63. 33. 63 It., 45 It.
34. 16 yds., 2 yds. 35. 37. 36. 100. ^j. 1975,
378 ANSWERS.
XCix. (Page 199.)
1.35 = 3 2. x = b 3. 05 = 90, 71, 52. ..down to 14
i/ = 2. 2/ = 3. / = 0,13,26 upto52.
4. a: = 7, 2 5. x = 3,8,13... 6. x = 91, 76, 61 ...down to 1.
y=\,A. i/ = 7,21,35... i/ = 2,13,24 up to 68.
7. a; = 0, 7, 14,21,28 8. a; = 20,39... 9. a; = 40,49...
i/ = 44,33, 22, 11,0. i/ = 3,7... i/=13,,33...
10. a = 4, ll...uptol23 II. x = 2 12. x = 92,83....2
1/ = 53, 50... down to 2. y = 0. y=l, 8... 71.
4 3 8 2
13. i^ and. 14. yyand— . 15. 3ways, viz. 12,7,2; 2,6,10.
16.7. 17.12,57,102... 18.3. 19. '2.
21. 19 oxen, 1 sheep and 80 hens. There is but one other
solution, that is, in the case where he bought no oxen,
and no hens, and 100 sheep.
22. A gives 5 11 sixpences, and B gives A 2 fourpenny pieces.
23. 2, 106, 27. 24. 3.
25. A gives 6 sovereigns and receives 28 dollars.
26. 22, 3 ; 16, 9; 10, 15; 4, 21. 27. 5. 28. 56, 44.
29. 82, 18 ; 47, 53 ; 12, 88. 30. 301.
C. (Page 205.)
^27 2 2 3 jf
(1) I. x"2+x^ + x2. 2. x^y + x^y^+x'y.
4 5fi. 121312
3. o5 + a=i+a2. 4. x^yz^ + a^yH + a''yz' .
(2) I. x' + ax^+b^x' + 3x*. 2. x^y* + 3xy* + 4y*.
x^y~h~^ 5xhrh~^ , ,
3. ^^ +7 — +ar*2"'
4. ^^ + ^g' + x''y*z.
11 1_ ^ 1_ _1_ J^
ANSWERS. 379
4 3 1
a ^b~c^ X ^y
1 1 1
(4) :. 24/.^.34/(x,^).i,. 2. ^^^ + i
ci. (Page 206.)
I. x*'' + x'^''y^'' + y*''. 2. a*"81i/*".
3. x*'' + 4aV + 16a'», 4. tr" + Sa^c'  6=" + c^
5. 2a " + 2a'"6"  4a"'c''  a'"b  b"+^ + 26c' + crc + b"c^  2c'^".
6. x^' + x""'"". 2/""'"" — ^''j/"" 7/"'"""+"'. 7. x*'' + ar"y'' + y*\
8. a='^  a''''' b"^ + a^^^ c" + a''°+'' . S^^*  6 + S''" c" + a''^+'' c'"
 b^^c^" + c.
9. x*' + 2x«'' + 3x2'' + 2x''Hl. 10. x*''2x'"' + 3x''2x''+l.
cii. (Page 207.)
1. x^" + ar'^y'" + x'"if"' + y^"'.
2. x^" — x'"!/" + x^'y"  x^'y^" + y*".
3 . r'' + x^'i/' + x^'y^ + x'y^' + x'y*' + ]f'.
4. a'^"  a"6^ + a*''6*»  a^''&*'' + h^.
5. x'* + Sx** + Qx" + 27x'' + 8] .
6. a"  2a"'x" + 4x". 7. 2x'' + 3x*^
8. 46"'c"'552"'.
9. a'"' + 3«="' + 3a'" + l. lo. a"' + 6'' + c',
38o ANSWERS.
ciii. (Page 208.)
I. x3a;^ + 3x^l. 2. y\. 3. a^x^.
4. a + h + cZaH^c^. 5, 10a;lla;*i/* + 5xi?/*21?/.
4 12
6. vw — w. 7. m^ + 4d^m3 + 16rf.
8. 16a + 8a"^6^ + \Qah^ + 18a^6^  2Aah^  12a^6^  15a^ b?
276.
fi 1 2 a 1 1 A
y. . +2a^x^ + a^. 10. x32a3x3+a3.
4 3 2 A '^ i
II. x^ + 2x^y^ +y^. 12. a + 2a65' + 62^
13. x4x* + 10x212x* + 9.
1 4. 4x* + 1 2x' + 25x'^ + 24x'^ + 16.
15. x^  2x^2/^ + 2x^z* + 2/* — 22/*2* + 3*.
16. x2 + 4x%4  2x*a* + 41/2  Ay^z^ + a^^
Civ. (Page 209.)
I. x^ + j/* 2. a^ — 6^, 3. yfl {T^y* + y'\
2112 ^ZlZS^XSik
^. a^a^h^ + o^. 5. x^ x^y \x^y^ x^y ^y.
6. m 8 + m* w* + m2 71,3 4. m* «, 2 ^ jnByj,^ + n^.
7. X* + 3x^1/^ + 9x*i/2 + 27t/^.
8. 27a^ + 18a^6i + 12ai6^ + 86^. 9. a^xi
10. ?>i"+3m5 + 9m5 + 27m5 + 81.
11. x2 + 10. 12. x^ + 4. 13. h + 2h^h^.
a 11 1152 11
14. x^ x^y^ —x^z^ +y^ + z^ —y^z^.
15. x39x310. 16. m^ + 1111^11^ + 11^.
11 111 11
J7. 'p~2p^ + \. 18. x (/;:. 19. x'^+y^.
ANSIVERS. 381
CV. (Page 210.)
I. a2&2 2. x«6*. 3. a;4x4.
4. iC' + l+oj*. 5. a'*_j4_ 5_ a2 + 2aici62 + c2.
7. l + a262 + a*64. 8. a^J"*  a*6*  4a262 _ 4.
9. 4a;5x* + 3a;~' + 2a;2 + x^Vl.
cvi. (Page 211.)
I. xx^. 2. a + 6~^ 3. m2mni + w,2.
4. c* + c3cii + c2i2 + cd3 + d4. 5, xyi + xY
6. a'^ + a'^h^ + h~\ 7. a;2?/2 _ 2 + a;2,^2_
8. x35x2 + lxi + 9. 9. a262l + a262
I o. a2  aifti  aici + 62  &ici + c2.
cvii. (Page 211.)
2 11 y?*+I2
I. x*2x%2 + 2y. 2. X '"' •
108+18a
3. X 3i2 . A
2a
(x*a*)^'
^ , 22 _, 421 _, 10 . 1
5. 7x*+ gx 3^^a; «yxi + . 6. x".
7. x"i/". 8. a2 + 2a26'52a^656^
9. a^ + a3 63+63. u. 771 = ™"^'. 12. x^+2H2=,
13. x^*. 14. I6a'^. 15. a'^'p.
16. 20^"" + 2a" 6''  4(( c"3a"63/>''+i + 66c". 17. c.
382 AJ\rslVE/?S.
19. x^ + x^hl.
20. a"^ + 2a'"+»* . bcu^  a""^^ bx  a"^^ c V.
21. x'^'"i/^'^". 22. a"~^ 23. x^'tf".
~^' ''' 144 ^^' ic'"''xV~''"c""""""i/" + r"
26. x + Zx^2x^7x^ + 2x~^.
cviii. (Page 215.)
I. 4'x3, 4'j/'. 2. '4^(1024), jys.
3. 4/(5832), 4/(2500). 4. •";'2", "■;/2'. 5. ';/a', 'V^/h".
6. 4/(a2 + 2a6 + 6'0, 4/(a33a26 + 3a5263).
cix. (Page 217.)
I. 2v'6. 2. 5^'2. 3. 2a Ja. 4. oad ^(5d).
5. 4zV(2y2). 6. 10x/(10a). 7. 12c ^'5.
V5x „ ,a
II. (a + x).,Ja. 12. {xy)i^lx. • 13. 5(a6).^'2.
14 (3c2t/).V(7l/). 15 3u^4'(26).
16. 2x?/2 . 4/(20a»7\ 17. 3m3„3^/(4„).
18. va^fe^ 4/(46). 19. (x + y).^x. 20. (ot).4/a.
ex. (Page 217.)
I. V(48). 2. V(63). 3 4^(1125). 4 v.96).
5. I~. 6. V(9a). 7. >v'(48a^x). 8. ^'{Zah .
AmWERS. 383
CXi. (Page 218.)
The numbers are here arranged in order, the highest on the
left hand.
I. x^3, 4/4. 2. J\0, 4/15. 3. 3^/2,2^/3.
6. 2 ^/87, 3 s^33. 7. 3 4'7, 4 ^'2, 2;'22.
8. 5 4/I8, 3 ^'19. 3 s"^:^. 9 5 ^'2, 2 ^^14, 3 4/3.
10. x'2,^,^3,i,'4.
cxii. (Page 219.)
I. 29 ^'3. 2. 30 ^ao+ 164^/2. 3. {a^ + h^ + c^) ^'x.
4. 134/2. 5. 33 4'2. 6. V6. 7. 5^'3.
8. 48^2. 9. 44/2. 10. 0. II. 4v'3.
12. 2^'(70). 13. 100. 14. 3a6. 15. 2a6 4/(126).
,6. 2. ,7. I .8. 4/? ,9. J
V.
X
20
.+X1/
cxiii. (Page 220.)
I. ^{xy). 2. s'{xyy^). 3. x + !/. 4. s'{v^y^).
5. ISx. 6. 56(.f+l). 7. 90v'(rx). 8. 2x^3.
9. X. 10. 1x. II. 12x. 12. 6rt.
1 3.  s'{^  7x). 1 4. 6 v^(x2 + 7x). 15. 8 (a2  1 ).
16. 6a2+12a18.
CXiv. (Page 221.)
1. x + 9^'.r+14. 2. x2^/x15. 3. a.
4. a53. 5. 3x + 5^/x28. 6. 6.v54. 7. 6.
8. V(9x^ + 3x) + ^'(6x2  3x)  ^'(6x  x  1 )  2x + 1 .
3^4 AmWE}?S.
s/iax) + sj{ax — x^)  js/{a^ ax)a + x.
3 + X+ ^{Sx + x"^). 11. xy + z + 2s,/xz.
2x + 2j{ax). 13. 4Z2 + 42j{x^9)+x^.
2x+ll+2V(a;2+lla; + 24). 15. 2x  4 + 2 ^(a;'''  4x).
2a;6 + 2V(a;26x). 17. 4x + 9U^x.
2x2s/{xy). 19. x + 2xl2y/{x?x).
x2 + l + 2V(a^x).
cxv. (Page 222.)
I. {^c+ ,^d)(Jc ^fd). 2. {c+ ^d){c ^d).
3. ( s'c + d){Jcd). 4. (1 + Jy) (1  Vy).
5. {\+ ^•i.o:){\~ >JZ.x). 6. (V5.m + l)(V5.ml).
7. 2a+ v/(3x)n2a V(3x){. 8. 3 + 2V(2n)} j32v/(2n){.
9. U'(ll).« + 4nV(ll).«4i. 10. {p + 2^r){p2Jr).
II. (v'p+ V3.2)(ViJ V3.2). 12. {a" + 6^'na"6^}.
«W6^ oW^. 15. 24+17^2.
■^ a b a — o
16. 2+^2. 17. 3 + 2^3. 18. 32^/2.
a + x + 2 yj{ax) 1 + x + 2 ^x
IQ. • 20. ^ .
^ ax 1x
0+ V(a2x2)
22. TO v'(w*l)
2a2x2 42aV(a2x2)
23. 2a2  1 + 2a V(a2  1). 24
CXVi. (Page 224.)
I. 19. 2. 11. 3 826v/(l). 4 5+4^/3.
5. 2h\2 ^'{ah)\2a. 6. a + a. 7. i^a'.
I
AA'SW^/^S. 385
cxvii. (Page 224.)
I.
x+y ^ x+y
SJixy) 3. 2^{xy)'
5. x^ v'2.ax + a.
6.
vi^+ ^f2.vin + n^. 7. 2x i^x.
2« v''^  26 Va
0.  , .
ab
9
ah J „ fd
^+cd2ac^^. 10.
'^'^ + aU
II.
xl
'3 ,,.•
14. l^\/{^^y ^5. 2x2V(x2a2).
16. a'^¥c. 17. l45a2(2a2) + a(10a2 a4_5)^(_l)
18. 8 + 74/3. 19. 4^/(3cx). 20. x^^CS^?'").
21. 4/(47i0. 22. (9?i10).^7. 23. 0.
t cxviii. (Page 228.)
"l. v/7+V3. 2.^/11+^5. 3. ^/7v/2. 4.73^5.
5. v/10 V3. 6. 2^53^2. 7. 2^3 v/2. 8. 3^112.
9. 3 V"  2 v/3. 10. 3 V7  2 v/6. 11. h ^10  2). 12. 3 ^/5  2 ^3.
I
cxix. (Page 229.)
•I. 49, 2. 81. 3. 25. 4. 8. 5. 27. 6. 256.
7. 27. 8. 56. 9. 79. 10. 153. 11. 6. 12. 36.
13. 12. 14 '^ — 2^' 15 5 16. 6.
17. 3. 18. 10. 19. _. 20. ^3^
.rs,A.i 2 p
:86 ANSWERS.
cxx. (Page 231.)
I. 9. 2. 25. 3. 49. 4. 121. 5. 1^.
7.0,8. 8. (2) 9. (^).
6. 8,0.
10. 5.
cxxi. (Page 231.)
I. 26. 2. 25. 3. 9. 4. 64, 5 ^^
o
6. ^ 7. a. 8. \ or 0. 9. 64. 10. 100.
5 4
cxxii. (Page 232.)
I. 16, 1. 2. 81, 25. 3. 3, 2^. 4. 10,  13.
5. 5.? 6. 4.32. 7.9,3? 8. 28.1f
9. 49. 10. 729. II. 4, 21. 12. 1 or—. 13. ±24.
145 25
14. 5 or 221, 15. 5or— r. 16. 5 or 0. 17. i^. 18. 25.
^ ^ 121 ' 36
19. ±9^/2. 20. ± ^/65 or ± *y5. 21. 2a.
22. 2a. 23. gOJ^^e '^ 4 "=5 1^.
26. ^1^. 27. ^. 28. ±5 or ±3 ^'2. 29. ±14.
1,30. 6ory. 31. 1. 32. ^. 2>2,. 2or0. 34. or ^j^.
cxxiii. (Page 23.'i.)
I. 2,5. 2. 3, 7. 3. 9,2. 4. ba,Gh.
_7 5 A ?^ _§^ ^ Yk^l
5 2' 3' l9"' 14" ^' 5 ' 6 ■
A.VSlVERS. 387
8. 2a,  3a and 3a, 4a, 9. ±2, a. 10. 0,5.
2a  & &  3a d e
, — r — . 12. , .
ac be c c
CXXV. (Page 239.)
I. a;2llx + 30 = 0. 2. .x^ + x 20 = 0. 3. x2 + 9x + 14 = 0.
4. 6x27x + 2 = 0. 5. 9x2 58a; 35 = 0. 5 a;23 = 0.
7. x^ — 2mx + mn = 0. 8. x^ i^c + — ^ = 0.
a/3 a/3
2 a^  /3
a/3 •
cxxvi. (Page 240.)
I. (x2)(x3)(x6). 2. (xl)(x2)(x4).
3. (xio)(x+i)(x+4). 4.4(x+i)(x+l::^)(x+^:!^).
5. (x + 2)(x+l)(6x7).
6. (x + 1/ + 2) (.(;2 + y^ + z^ — xy — xz yz).
7. (afec) (a2 + 6^ + c2 + a6 + ac6c),
8. (xl)(x + 3)(3x7). 9. (xl)(x4)(2x + 5).
10. (x+1) (3x + 7)(5x3).
cxxvii. (Page 242.)
I. Vl3orVl. 2. 4/ 2 or 4/ 12. 3. ^V _ 1 or 4/ ?1 .
4. lorV4. S.^aor^. 6 25.,,.;
1 L
7 ::9F^7 ^ (D'*^^ (D 9 1 or 1 ±2 VI
„ 1 5± V1329
10. 3 or —  or ~ .
3 4
A.VSWERS.
„ a + 6 a±2J(a23o)
11. a + 2, or  — , or  ^\ \
o o
1 2. 0, or a, or ^' — '.
cxxviii. (Page 245.)
I. 6 : 7, 7 : 9, 2 : 3. 2. The second is the greater.
3. The second is the greater.
4. ^iz^, 5. I0:9or9:10.
«
CXXix. (Page 246.)
I. 2:3. 2. h:a. 3. 6 + (Z:ac. 4. ±^61:1.
5. 13 : 1, or,  1 : 1. 6. ± s,l{rn^ + 4m^  in : 2. 7. 6, 8.
8. 12,14. 9. 35,65. 10. 13,11. 11. 4:1. 12. 1:5.
CXXX. (Page 247.)
8 8 xv ah+c
1. TT 2. pr. 3. . 4. ,
15 9 ^ x + y ^ aoc
m^  mn + n^ , (x + 2) y
^' !«?■■¥ wn + n^' ' (y — 4)x'
cxxxii. (Page 255.)
6. X = 4 or 0. 8. 440 yds. and 352 yds. per minute.
II. x = 30, 2/ = 20.
62
^3 T
9
15 41.
16. 50, 75 and 80 yards.
17
120, 160, 200 yards.
19. I5 miles per hour.
20. 1 : 7.
21. 160 quarters, ^2. 22. £80. 23. £60.
24. £20, 25. 90:79. 26. 45 miles and 30 miles.
ANSWEI^S. 389
cxxxiii. (Page 262.)
4. 16. 5. 5. 6. 12. 7. 3^. 8. I
9. Aoz(P. 10. 5. II. A=Ib. 12. 64x2 = 91/3.
13. x2 = — 3. 14. 4x3=271/2. 18. i/ = 3 + 2x + x2. 19. 18ft.
CXXXiv. (Page 266.)
I. 50. 2. 200. 3. lo. 4. 32.^
5. 2 6. 40. 7. 117. 8. 0.
9,20/ o\ 3an  26?^  2a + 6
9. x^ + y^2{n'2)xy. 10. r .
CXXXV. (Page 268.)
I. 5050. 2. 2550. 3. 820. 4. 30.
5. 24. 6. 34. 7. ^^^^l 8. ?^^^l
b 2 2
7n2  5» w  1
10. 2.
CXXXvi. (Page 269.)
I. 6.
5. 2.
2.
X
"25"
1
3 8
4
7
8*
6.
4
cxxxvii. (Page 269.)
1. (I) 46. (2) 362. (3) ?. (4) 44.
2. 155. 3, 112. 4. 888. 5. 100.
390 ANSWERS.
6. 6433. 7. £135. 4s.
8. (i) 355,7175. (2)  ISGa^, 3116«
(3) 161 + 81a;. 3321 + 1681X. (4) 119^, 2357^.
(5) 8^, I74I.
9. (i) 126, 63252. (2) 25, 2250.
(3) 45, 15705X. (4) 99, 1163^.
4
(5) 71, 4899(1 m). (6) 65, 65x + 8190.
cxxxviii. (Page 271.)
I. 6, 9, 12, 15. 2. \\, , 0,  1^.
„_5^ 5 1 ^ 1^ ? 11
CXXXiX. (Page 272.)
3m + TO ?n. + n m + 3?i
4 '
2 ' 4 ■
5m 4 3
5mHl 5m — 1
5to3
2.
5 '
5 ' 5 '
5 •
6n2 + l
5n2 + 2 5?i2 + 3
5n2 + 4
J
5 '
5 ' 5 '
5 ■
4
2x2 + 1/2
9 '
^2 2x2^2
CXl. (Page 275.)
I. 64. 2. 78732. 3. 327680. 4. J.
^ ^ 2048
5. 13122. 6. 16384. 7. — .
' ' 96
ANSWERS. 391
CXli. (Page 276.)
I.
65534
2. 364.
3 ;^i
4
x^ {x
■1)
1)'
(ax)jl
5" (a + u;)». (1
(^ + ^)^t 6. 31,
— ax)
7
7(2»
■!)•
8. 425.
CXlii. (Page
43
9 96
278.)
I.
2.
4
^ 3
27
3 8
^ 1 5. ll.
6.
•3.
7
4
3.4
^1 16x5
9 ^^3 ^° 8x2+1
II.
a2
aV
12
1
• 9 ^3.
x2 86
x + y '■^ 99"
15
49
90"
. 46
^6 55
cxliii. (Page 279.)
I. 9, 27, 81.
2. A,
, 16, 64;
,256.
3 2,4,
3 9 27
81
^ 4' 8' 16
32'
cxliv.
(Page
279.)
I. (i) 558.
(2) 800.
/ ^ 18
(3) T
(4) f
(5) 2
(^) 486
(7)
1189
2 ■
(8) 13^.
(9) 1
(10) 84.
(II) 
9999 V3
(Vio + i).V5'
, , 3157
5. 42.
6. ac=b'^.
7
±1.
8. n+i.
471
392
ANSWERS.
9
4.
10,
. 10.
13
4.
14. 642.
1 6.
49, 1.
17
3 6,
4
18. 60.
19.
4 3
5' 5'
2
5'
1
5'
0,
1
5'
2
6'
3
~5'
4
"5"
22.
3, 7,
11,
15,
19.
23
5,
15,
45, 135, 405.
25.
139.
26.
10
per
cent.
cxlv. (Page 285.)
1111 „ „ , 2
^
6' 9'
12'
15*
>•
^, ^, ^, L,
3*
6.
3 3
4' 2'
0°,
3 3
2' 4'
7
6 3 6 3
5' 4' 11' 7'
6
17'
3
10'
8.
6a5?/ (n + 1)
3?i2/ + 2a; '
6x1/ (% + l)
3?ii/ + Ax — 'iy
»
6x1/ (n + l)
■■' 2nx + Sy '
9
1
4'
1
2'
1 1 1
'*' 2' 4' 6'
5
^"^31'
5 5 15
24' 17' 2' 3'
5
4*
10.
104,
234.
cxlvi.
13
(Page
2, 3, 6.
290.)
I.
132.
2. 3360.
3
116280.
4
6720.
5.
Ill
8'
6.
40320. 7.
3628800. 8. 125.
9
2520.
10.
6.
II. 4.
12.
120.
13
1260.
14. 2520, 6720, 5040, 1663200, 34650.
cxlvii. (Page 295.)
I. 3921225. 2. 6. 3. 126. 4. Ii628a
5. 12. 6. 12. 7. 816000. S. 3353011200.
9. 7. 10. 63. II. 62. 12. 123200. 13. ;376992 ; 52360
ANSWERS. 393
cxlviii. (Page 300.)
a* + 4^x + %d?x^ + Aax^ + x*.
66 + 66=c + 156^c2 + m?(? + 1552c* + Gtc^ + c«.
a" + 7a66 + 21a562 + 35^453 + 35^35* + 'iXaPlP + TaS^ + h\
7? + 8x^y + 283fiy^ + 563^y^ + '70x*y* + 563^y^ + 28x2y«
+ 8xy'{y\
625 + 2000a + 2400a2 + 1280a3 i 256a*.
cxlix. (Page 301.)
1 . a^  Ga^x + 15a'*x2  20a V + 1 5a2x*  6ax^ + x^.
2. V  Wc + 216^02  356*c3 + 356V  216V + Ihc^  c\
3. 32x5 _ 240x*i/ + 720x31/2  lO^Oxy"^ + 810xi/*  243i/5.
4. 1  1 Ox + 40x2 _ 80x3 + SOx*  32x5.
5. l10x + 45x2 120x3 + 210x* 252x5 + 210x6 120x7
+ 45x8 10x9 + xi».
6. a24 _ 8a2i62 + 28ai86*  56tti566 + 70ai268  56a96^o
+ 28a66i2_8o36u + 5i6^
9
Cl. (Page 302.)
1 . a3 + 6a26  3a«c + 1 2o62 _ 1 2a6c + 3ac2 + 863 _ 1 262c + 66c2  c^.
2. 1  6x + 21x2  443^ + 63x*  54x5 + 27x6.
3. x93x« + 6x"7x6 + 6x53x* + x3.
4. 27x454x^ + 63x3 + 44x2 + 21x3+6x^ + 1.
5. x3+3x25+?3,.
' X x^
6i o^ + 6^  c^ + Zah^ + 3a«6^  3a^ci  36^ci + 2a^c^
i 1 111
+ 365c^6a*Z)*c*.
594 ANSWERS.
Cli. (Page 303.)
r. 330a;7. 2. ^^ha}%^. 3.  IGlTOOa^^^s^
4. 192192a666c8d8. 5. 12870a86s.
6. TOa^ii 7. 92378ai069and92378a96i«.
8, I7l6a7a;6 aud iheaSx^.
Clii. (Page 311.)
1 1 2 _L 3_ ^ 4
I. , l+^aJgX +^ga; ^28**
, la a^ 4a^
i X a;2 5J.3 loa;*
3. «^ +  — 5+ — I n
3aS 9a5 Sla^ 243a^
4. 1 + X  X2 + X^  X*.
I 1 1 _& „ 5 a ,
•^ 6 54
1 4 _i 1 2  i 4
6. «^+5«"^*25*"^'125'
x^ X* x^ 5x^
7 2~8~16~ l28"
_ , 7 „ 14 , 14 
8. l3a^ + ^«*,ia«.
9x 27x2 135
9 ■^T~~32' 128'^
10. a=^^y + 6^+5i^
ANSWERS. 395
,5 __5^ , 35
6'^ 72"^ 1£96
/2\2 2 /3\i _i 3/3U 4 ,
Cliii. (Page 312.)
1 . 1  2a + 3a2  4a'' + 5rt^ 2. 1 + 3x + Qx^ + Tls? + 81:c*
5 „ 5 , , Zr? x^ 5x^
5 . a''" + 10a'' c + 60a'*x + 280a'«./;3 + 1 1 20a' V.
, 1 , 6x^ 21.x^ 56x
a'^ I I a^
Cliv. (Page 313.)
, x2 3a;* 5a:6 35»8
2 8 16 128
3x2 15^ 353.6 315^
^ ^^ 2""^ 8 "*" 16 "^"128 •
2 7 98
3. X ^X Z +^^X Z —X . .
j_ 3x2_5x3 35^ . 1 «' 3x* 5x6
"^^ . 2 2 ■*■ 8 • '• a 2^38,j5i6^^r
, 1 x^^ 2x6 143.9
■ a"3a5"''9a^~81ai<>
Clv. (Page 3 U.)
I. 7^;9;) ,^1. 3. (i).12^L^(M::r) ,
1^ ;••(»•!) ^ ^ 1.2...(rl) • •
3. (_i)^i. 8.7...( 10r) .,
396
ANSJFEJiS.
4
i.V;;.y_iy • (5^)" • (22/)^ 5 (  1)^ »• • x^
6.
r.(r+l).(7 + 2) „ 1.3.5...(2r3) /xX^.
6 •^^''^ • 7. i.2.3...(rl)W
8.
1.2.5...(3r7) / x\\\
1.2.3...(71) "V 3a/ •
9
7.9.11...(2r + 3)
1.2.3... (r1) •* •
lO.
a"2 3.7.11... (4r 5) /^Y^^^
4'^* 1.2. 3. ..(r1) *\cJ •
II.
(r+l)(r+2) 1.3.5...(2rl)
2 •''• ' 1.2.3..:r ^^ ^•
13
1.3.5...(2rl) 5 1
1.2.3...r •^^''^' '5. le^i^.
1 6.
3 .,„ 429 xi«
128 •«^' »7 128a^
1 8.
1.2 9 •'' •^•
19.
(1  5?7i) (1  4m) (1m) l«
1.2 6m6 •"
clvi. (Page 315.)
I. 314137.... 2. 195204....
3. 304084.... 4. 198734....
Clvii. (Page 319.)
I. 1045032. 2. 10070344. 3. 80451.
4. 31134. 5. 51117344. 6. 14332216.
7. 31450 and remainder 2, 8. 522256 and reinainder 1.
9. 4112. 10. 2437.
ANSWERS.
397
clviii. (Page 321.)
I. 5221. 2. 12232. 3. 2139e. 4. 104300.
5. 1110111001111. 6. atee. 7 6500145.
8. 211021. 9. 6^12. 10. 814. 11. 61415.
12. 123130. 13. 16430335. 14. 27^
I. 41.
4. 1223220052.
Clix. (Page 327.)
2. 162355043.
5. Senary.
3. 251.
6. Octonary.
I. 12187180.
4. 4740378.
7. 53790163.
10. 21241803.
Clx. (Page 336.)
2. 77074922.
5. 2924059.
8. 40578098.
II. 3738827.
3. 24036784.
6. 3724833.
9. 629905319.
12. 161514132
Clxi. (Page 339.)
1. 21072100 ; 20969100 ; 33979400.
2. 16989700; 36989700; 22922560.
3. 7781513 ; 14313639 ; 17323939 ; 27604226.
4. 17781513; 24771213; 0211893; 56354839.
5. 48750613; 14983106.
6. 3010300; 28061800; 2916000.
7. 6989700; r0969100; 33910733.
8. 2, 0, 2 : 1, 0, 1.
9. (I) 3.
(2) 2.
9
10. a: = 5,i/ = ;
ANSWERS.
11. (a) 3010300; 1397940C; 19201233; 19979588. (6)103.
12. (a) 6989700; G020600; 17118072; 19880618.
(6) 8.
13. 38821260; 14093694; 37455326.
14. (i) x=r {2)x = 2. (3)a; =
(4)a; =
(5)x =
(6)a; =
6' ^ / • V J/ Yog a + log 6"
los c
log a + 2 log h'
4 lo" 6 + log c
2 log c + log /)  3 log a*
log c J
log ft + ??i lo^ 6 + 3 log c
clxii. (Page 343.)
I. 176 years. 2. 234 years.
3. 7 2725 years nearly. 4. 225 years nearly.
6. 12 years nearly. 7. 1 1724 years,
APPENDIX.
The following papers are from those set at the llatrinulatioL
Examinations of Toronto, Victoria, and McGiil LTniversi
t^es. and at the Examinations for Second Class Provincial
Certificates for Ontario.
UNIVERSITY OF TORONTO.
Junior Matric., 1872 Pass.
1. Multiply ^x'lxy + y^hjlid' + lxyy'.
Divide a*  816* by a ± 36 and {x + af  {y  by
by X + a — y + b.
2. What qnantity subtracted from x^ + px + q ''ril]
make the remainder exactly divisible by a; — a .?
Shew that
(a + b + cf {a+ b + c) (a' + b'' + c' ab be  ca)
 3abc = 3 (a + 6) (b + c) {c + a).
3. Solve the following equations :
(a)^{2x3) + Hi^x7) = Ux^).
4a; — 7 3a; — 5
(^)i.^r=Tn^r=r2=20
i<^)
X — 1 ^x — 2
1111
4 X — 5 X — 6
2/+I y x + 2 11
(d) « + 2=l' 3^5 =18
4. In a certain constituency are 1,300 voters,
ind two candidates, A and B. A is elected Vjy a
tt APPENDIX.
certain majority. But the election having been de
clared void, in the second contest {A and B being
again the candidates), B is elected by a majority of
10 more than A's majority in the first election ; find
the number of votes polled for each in the second
election ; having given that, the number of votes
polled ior B in the first case : number polled in the
second case : • 43 • 44.
Junior Matric, 1872. Pass and Honor.
1. Multiply a; + y + 2*  2y^ zi + 2zi a^  2ar4 yi by
X + y + zi + 2yi zi — 2zi x^ — 2x^ yi, and
divide a' + 86' + 27 c^—18abc by a* + 45» + 9 c'—
2ab — Sac — 66c.
2. Investigivte a rule for finding the H. C. D. of
two algebraical expressions.
If X + c be the II. C. D. of af + px + q, and x* f
p' X + q', show that
{qq'Yp {qq) (pp) + 9 (ppY^^^
3. Shew how to find the square root of a binomial,
one oi' whose terms is rational and the other a quad
ratic surd. What is the condition that the result may
be more simple than the indicated square root of the
given binomial 1 Does the reasoning apply if one of
the terms is imaginai y 1 Show that *y/ — 4m' = ^m
+ ^ m.
4. Shew how to solve the quadratic aquation aa^ 4
6a; + c = 0, and discuss the results of giving difierent
values to the coethcients.
If the roots of the above equation be as p to 9
, 6» {p + qY
show that — = •
ac
APPENDIX. iB
6. Solve the equations
xy +y»10 = 0.
(c)
a:' + 6ic + 2 £c'+6x+6 a:*+6a + 4
x* + 6a; + 8
a* + 6 a; +10*
* •
(i) 6 x'  5 af*  38 ic*  5 aj + 6 = 0.
6. Shew how to find the sum of w terms of a geometnp
series. What is meant by the sum of an infinite
series ? When can such a series be said to hav? •
sum %
Sum to infinity the series 1 j 2r + 3 r* ( Ac.
and find the series of which the sum of n terms i&
aF — .
a — \
7. Find Che condition that the equations
ax\hy — cz — ^.
a, a; + 6, 3/ — Ci 2 = 0.
e«, a; + 6, y — c, » = 0.
may be satisfied by the same values of x, y, z.
8. A number of persons were engaged to do a })Iece
of work which would have occupied them m hours if
they had commenced at the same time ; instead of
doing so, they commenced at equu! intervals, and then
continued to work till the whole was finished, tne
payments being proportional to the work done by
each ; the first comer received r times as much as the
last : find the time occupied.
APPENDIX.
Junior Matric, 1872. Honor.
1. There are three towns, A, B, and C ; the road
fi'om B to A forming a right angle with that from B
to C. A person travels a certain distance from B
towards A, and then crosses by the nearest way to the
road leading from C to A, and finds himself three
miles from A and seven from C. Arriving at ^, he
finds he has gone farther by onefourth of the distance
from B to C than he would have done had he not left
the duect road.* Requiied the distance of B from A
and C.
2. If ay \r h x jx + a^ _bz + cy ^ ^^^^ ^^
c h a
ay*
a h e
3. Solve the equations x* — yz = a*, y' — zxb*, «* —
a:y = c\
4. If a, h, and c be positive quantities, shew that
a« (6+c) + 6» (c + «) + c« (a + 6) > %abc.
5. Find the values of x and y from the equations
o 5?/ + 3
2y + ^ = 1,
x '
£c* + 5x + 2/ (y  1) = 24.
6. A steamer made the trip from St. John to Boston
via Yarmouth in 33 hours ; on her return she made
two miles an hour le.ss between Boston and Yarmouth,
but resumed her former sjieed between the latter place
and St. John, thereby making the entire return pas
sage in 11 of the time she would have required had
her diminished speed lasted throughout ; had she
made her usual time between Boston and Yarmouth,
and two miles an hour less between Yarmouth and
APPENDIX. T
St. John, her return trip would have been made in
iJ of the time she would have taken had the whole
of her return trip been made at the diminished rate.
Find the distance between St. John and Yarmouth
and between the latter place and Boston.
Junior Matric, Honor. 1
Senior Matric, Pass. )
1. Solve the following equations :
, . i x^ 2a;?/ + 2?/" ^ xj/
(a) .... I
1874.
a? + xy + tf = 63.
4a;— ?)xy = 171.
ZyA:xy= 150.
1 1 1
a? xy y^
1 1 1 o
— . + — r„+ .= 133.
X* u^y^ y"
Am i find one solution of the equations
{d) ..
(&)
y* — x*^ 68.
a^ + sf x = y.
2. Find a number whose cube exceeds six times the
next greater number by three.
3. Explain the meaning of the terms Highest com
mon measure and Lowest common multiple as applied
to algebraical quantities, and prove the rule for finding
the Highest common measure of two quantities.
4. Reduce to their lowest terms the following
fractions :
\a) .
(6)
i Sar* + i^^^^x — 10^ '
x' + lOx' + 35a; + 50a; + 24
x' + IBa;^ + irj,/2 + 342.^ + 360
▼1 APPENDIX.
5. Find the sum of n terms of the series — , \, —
\, (fee, and the ccth term of the series
a; + 1 2 2> — x
6. Find the relations between the roots and co
efficients of the equation ax* +;yx + ^ = 0.
Solve the equation
a;* + 6a;'+10x« + 3a;=110.
7. A cask contains 15 gallons of a mixture of wine
and water, which is poured into a second cask con
taininif wine and water in the proportion of two of the
former to one of the latter, and in the resulting mixture
the wine and water are found to be equal. Had the
quantity in the second cask originally been only one
half of what it was, the resulting mixture would have
been in the proportion of seven of wine to eight of
water. Find the quantity in the second cask.
8. What rate per cent, per annum, payable half
yearly, is equivalent to ten per cent, per annum, pay
able yearly.
9. A is engaged to do a piece of work and is tn
receive $3 for every day he works, but is to forfeit
one dollar for the first day he is al)sent, two for the
second, three for the third, and so on. Sixteen davs
elnps • l)efore he finishes the work and he receives §26.
Find the number of days he is absent.
Cliange the enunciation of this problem so as to
apply to the negative solution.
Junior .Vatric, 1876. Pass.
1. Explain the use of negative and fractional in
dices in Algebra.
Multiply, A by i/'«'' and the product bv V"
APPENDIX. vii
Simplify , writing the factors all in one
line.
2. Multiply together a* + ax ■¥:!?, a + x, a*ctx + 3^,
ax, and divide the product by a^  ar*.
3. Divide 1 by 1  2x + x^ to six terms, and give
the remainder. Also divide 27a;''6x^ + ^ by Sa;^ +
2x + J.
a, ^ n a»«
■> + •
4.
Multiply
a +6
by a
+ 6 .
6.
Solve the
equations :
(!)•
3a; + 4 7a;  3
5 2
a;16
4 •
(2).
( X (y + z) =
■ly{z + x) =
{z{x + y)
.24,
.45,
= 49.
^ior Matric
, 1876.
Hnvn
1. An oarsman finds that during the firet half of
the time of rowing over any course he rows at the
rate of five miles an hour, and during the second
half, at the rate of four and a half miles. His course
is up and down a stream which flows at the rate ol
three milus an ho\ir, and he finds that by going down
the streani first, and up afterwards, it takes him one
hour lojiu:er to go over the course than by going first
up and then down. Find the length of the course.
2. Shew that if a^ 6^ tr" be in ^.P., then wilJ h  <■,
« + rt, a + 6 be in II.P.
Also, if a, 6, c be in A. P., then will
he 7 ca ah
a + , b + , cr 
^ c c + a a^ It
be in U.P.
APPENDIX.
3. If s = ffl + 6 + c, then
y/{as + be) (bs + ac) (cs + ah) = (sa) (s — b) (s  c)
4. If a, + fta + +a„^ —, then
(6  «,)*+ + (s  a„)^ = a,2 + a/+ +a''«.
5. If the fraction  —  —  , when reduced to a re
2n + 1
petend, contaijis 2n figures, shew how to infer the last
n digits after obtaining the fiist n.
Find the value of Jy by dividing to 8 digits,
6. Solve the equations
X — y + zS,
xv + xz = '2 + 1/z,
Junior Matric, 1876. Honor.
1. Shew that the method of finding the square
loot of a number is analagous to that of finding the
square root of an algebraic quantity.
Fencing of given length is placed in the form of
a rectangle, so as to inchnle the greatest possible area,
which is found to be 10 acies. The shape of the
field is then altered, but still remains a rectangle, and
it is found that with 162 yards more fencing, the
same area as before may be enclosed. Find the sidea
of the latter rectangle.
2. Prove the rule for finding the Lowest C^ommon
Multiple of two compound algebraic quantities.
Find the L.O.M. of a»  6=» + c' + 3a6c and d(b^c)
^{c + a)^^ (« + b)+abc.
3. If a, p be the roots of the equation 3^+px + g =
0, shew that the equ;ition may be thrown into the
form (x — a) (x f3)  0.
APPENDIX. Lx
3 + v/2 is a root of the equation aj* — 5a;' + 2a^ + a;
f 7 = : find the other roots.
4. (1) Shew how to extract the square root of a
binomial, one of whose terms is rational,
and the other a quadratic surd.
(2) Find a factor which will rationalize x^ — y^.
5. a, b are the first two terms of an H. P., what is
the nth term ?
JI a,l, che in. H. P., shew that
h^ci  cf = 2c{b  a)^+ 2a"(c  by.
6. A and B are to race from M to 'N and back. A
moves at the rate of 10 miles an hour, and gets a start
of 20 minutes. On A's returning from N, he meets
B moving towards it, and one mile from it ; but A is
oveiiaken by B when one mile from J\I. Find the
distance from ]\I to N.
7. Solve the equations
(1). ar' + 82a^+lla;+U.
X 51
(2).
V JC 12 a:y
Second Class Certificates, 1873.
1. Multiply +_+iby f+l.
b a •' b a
„ ^, , a' 3ab + 26 a^  lab + 1 26'
2. Shew that —. — „,
0.26 a ob
;an be reduced to the form 36,
APPENDIX.
3, Reduce to its lowest terms the fraction,
. b^ 1
"^ "*■ 12 + 9
^ 1
4. (a) Prove that a^  ?/" is divisible hj xy with
ut remainder, when m is any positive integer.
ih) Is there a remainder when a;"" 100 i>i
^ided by a;  1 ] If so, write it down.
0. Given ax + by = 1,
,xy 1
and  + , = ^'
a ah
Find the difference between x and y.
6, Given 3  n^^zM _ j'^ t! 
^x\) 3(x+i) "•
Find X in terms of wi,
^. X 2 ^ 7a; + 16
/. Given  =5. Find the value of ;= n*'
y 3 73^ + 24
<. Given = 1,
xy X ¥y
, 6 10
and 3. Find x and v.
X ~y x\y ^
9. There is a number of two digits. )^y inverting
:.Iie digits we obtain a number which is less by 8 than
hree times the original number ; but if we increase
bach of the digits of the original number by unity,
and invert the digits thus augmented, a number is
obtained which exceeds the original number by 29.
Find the number.
10. A student takes a certain number of minutes
to walk from his residence to the Normal School.
Were the distance ^th of a mile greater, he would
need to incKMs. his pace (number of miles j^er hour)
APPENDIX.
by ^ of a mile in the hour, in order to reach the
school in the same time. Find how much he would
have to diminish his pace in order still to reach the
school in exactly the same time, if the distance were
■^^ of a mile less than it is.
Second Class Certificates, 1875.
1. Find the continued product of the expressions,
a + b + c, c + a6, b + ca, a + bc.
a' + a^b a{ab) 2ab
2. Simplify ^,^~ y  ^^^  ^^,
3. Find the Lowest Common Multiple of 3a^  2a;  I
and ^a?'2.x'^x+\.
4. Find the value of x from the equation, ax —
a* — '6bx 6bx — 5a^ bx + 4 i
 — ab^ bx+ — K — — f
a 2a 4
5. Solve the simultaneous equations,
05 V
jrt.
c d
 +— =
x y
n.
6. In the immediately preceding question, if a
['U})il should say that, when nb — md, and be ^ ad, the
values of x and y obtained in the ordinary method,
have the form f, and that he does not know Iio'a io
interpret such a result, what would you reply ?
7. Two travellers set out on a journey, one with
%^ 00, the other with $48 ; they meet with robbers,
who take from the first twire as much as they take
from the second ; and wliat remains with the first is
5 times that which remains witii the second. How
i;aMir'> mor^ey did each traveller lose '?
APPENDIX.
8. A and B labor together on a piece of work for
two days ; and tben B finishes the work by himself
in 8 days ; but A, with half of the assistance that B
coukl render, would have finished the work in 6 days.
In what time  could each of them do the whole work
alone %
9. P and Q are travelling along the same road in
the same direction. At noon P, who goes at the rate
of j?i miles an hour, is at a point A ; while Q, who
goes at the rate of n miles in the hour, is at a point
B, two miles in advance of A. When are they to
gether \
Has the answer a meaning when m — n is nega
tive 1 Has it a meaning when.m = ?i? If so,
state what inter^jretation it must receive in these
cases.
10. P is a number of two digits, x being the left
hand digit and y the right. By inverting the digits,
the number Q is obtained. Prove that 11 (a; + y)
(P— Q) = 9(a;— 2/)(P + Q).
Second Class Certificates, 1876.
1. Divide (1 + m) o^ — {m ¥n) xy {x — y) — {n — 1) y'
by ar^— iy + 2/«.
Shew that {a + a^U + hY—{a — aiM + 1)^ is ex
actly divisible by 2ai6i.
2. Resolve into factors x* + layii (z* — t^) — y*,
11^ {h — c) + b^{G — a) 4 c{a — h), »ind 25a;* +
bx^ — X — 1,
3. If x^ hpci? k q.r + r is exactly di\'isible by a:' +
mx + n. then nq — io^ = 7in.
4. Prov»^ that if m be a common measure of p and
APPENDIX. xiii
q, it will also measure the difierence of any multiples
of p and q.
Find the G. C. M. oi x*—^x^ ^ {(i—\)x^ + yx—
q and x* — qoi? + {p — l)a^ + qx — p and of 1 +
x^ + x¥^ and 2x + 1x^ + 3a,* + 3a;^'
5. Prove the rule for multiplication of fractions.
a* — (y — zY 2/* — (2 — cc)^ «^ — {x — ?/)'
Simplify
(2/ + zf—zc' (z + a;)2— y* (x + yf
a?
and 5— Ts, — ^ — r2 +
6. Wliat is the distinction between an identity and
au equation ? If a; — a = y + b, prove x — b = y + o.
Solve the equations (2 ^x) (7/1 — 3) = — 4 — '2ii'x,
16a;— 13 40.x— 43 32a^30 20:^24^
7. What are simultaneous equations ? Explain why
there m\xst be given as many independent equations
as there aie unkno'vn quantities involved. If there
is a gieatei number of squations than unknown o'^.^.n
tities, what is the inlerence '(
Eliminate x anci v irom xhe eor.atioiib ax ^ by
= c, ax + bv = c , ax + U'y = <~.
8. Solve the equations —
( 1 ) y/n + x^ 'Wn — X = m
(2) 3a; + w + s=i;j
3'y + 2 + a;= lo
'iz ^ x + y  17
9. A j)erson has two kintls of foreign money ; it
takes a pieces of the first kind to make one £, and b
pieces of the second kind : he is ofieied one £. for c
pieces, Low many pieces of each kirul must he take 1
rfr APPENDIX.
10. A person starts to walk to a railway station
four and arhalf miles olF, intending to arrive at a
certain time ; but afiur walking a mile and aliaif he
is detained twenty minutes, in consequence of which
he is obliged to walk a mile and ahalf an hour faster
in order to reach the station at the appointed time.
Find at what pace he staited.
11. {a) If y = ^ then will ^j^, = ^,.
(6) Find by Homer's method of division the
value of
a*+ 290a;'+ 279ar'—2892j:*—586a>— 312 when
a; = —289.
(«) Shew without actual multiplication that
(a + 6 + cf — {a + 6 ^ c) (a* — ah + 6'— 6c + c*— ac)
McGILL UNIVEE.SITY.
First Year Exhibitions, 1873.
1. The clifFerence between the first and second oi
four numbers in geometrical progression is 12, and
the difference between the 3rd and 4th is 300 ; find
them.
2. Find two numbers whose difference is 8, and
t])e harmonica! mean between them 1^.
3. Prove the general formula for finding the sum
of an arithmetical series.
4. The diflerences between the hypotenuse and the
two sides of a rightangled triangle are 3 and 6
fe.s]»ectively ; find the sides.
6. Solve the equations
ce^ + 2/^ = 25 , x + y=l;
X a; + 1 13
a; + 1 X ~ 6 '
x + y + 2; = 0, x + 2/ = st; x7)=y \ z
03+4 3x + 8
+11= .
3a3 + 5 2a3 + 3
6. A cistern can be filled by two pipes in 24' and
D '' respectively, and emptied by a third in 20' ; in
what time would it be tilled, if all three were running
together.
7. Shew that
aj^ ^}?  ^ (a + b + c) (a + bc)
^^ 2ah ~ 2ab
tvi APPENDIX.
8. Prove the rule for finding the gi'eatest common
measure of two quantities.
First Year Exhibitions. 1874.
1. The sum of 15 terms of an arithmetic series is
600, and the common ditierence is 5 ; find the first
term.
2. Find the last term and the sum to 7 terms of
the series
14+16&C.
3. Find the arithmetical, geometric, and harmonic
means between 3 and 1^.
4. The difierence between the hypotenuse and each
of the two sides of a rightangled triangle is 3 and G
respectively ; find the sides.
5. The sum of the two digits of a certain number
is six times their difierence, and the number itself
exceeds six times their sum by 3 ; find it.
6. Solve the equations : —
X y = l; «'?/'= 19
3.C  7 4a; 10 .^,
X + x + b ^^'
x\{u1) = b\ 4?/i (a;+10) = 3.
232:c+l 8^5 .^
7. A man could reap a field by himself in 20 hours,
but with his son's help for 6 hours, he could do it in
16 hours ; how long would the son be in reaping tho
lield by himself?
8. Find the value in its simplest form of
x + y 2a; x' y  x^ ^
y ~ x + y 3^yy^
tvii APPENDIX.
9. Find the greatest common measure of
Sar* f 3a:^  15a; + 9 and 3a;^ + 3a;'  21a;* — 9^.
First Year Exhibitions, 1876.
1. Solve the equations
12a
\a + a; + /rt — X = =,— ^,
\ \ O V ff + a;
■X y X y X y
 ^  = 1 _ ; _ f _= 1 +_.
a b cab c
2. Reduce to its siuipiest form the expression ; —
7 V54 + 3 VFe + ^' 2  5 4/128.
3. Find the greatest common measure of
2x'h^^ — Sx + 5 and 7x' — 1 2x+5.
4. Simplifjfc
5. A nuoaoer consists uf two digits, of which the
lett is twice the right, and tlie sum of the digits is
oneseventh of the number itself. Find the n;imber.
6. Solve the following : —
X y X z 1/ z
_ + 1^ + 1,  +  =2,  +  =3;
a b 'I r. be
1 1
X y
7. Find the sum o\' n terms of the series 1, 3; 5,
7, &c.
(a.) Shew that the reinprocals of the first four
terms, and also of any consecutive four terms, are ir»
harmonical proportion.
tjNIVERSITY OP VICTORIA COLLEGE.
Matriculation, 1873.
1. What is the " dimetision " of a term ? WLtn k
ail expression said to be " homogeneous " ?
2. Remove the bxackets from, and simplify the
following expression : —
{•la — Zc + id)  \M ~ {m ^ Za)\ + 5a — {— 4
__j)_ ^3a — {4a — 5rf — 4).
3. P'.ove the " E.ule ot Signs" in ISIultiplication
4. INLultiply a — hjx + .
ax
5. Di vide ax^ + bx'' * cx\ dhy x — ♦•.
6. L^ivide 1 by 1 + a;.
7. ^'''nd the Greatest C mmon Measure of 6a* —
aV — \'Ik and ya''  \'2a^a^ — 6a.r  Sx*.
tt, i'rom 3a — 2c —  ~ _ subtract; 2a — x —
x^ — 1
a — X
(1+1^42^
9. Civ en ■: ' ,> to find a; and v.
I'l l>ivide tne iiui.i )er a into four such paits that
t!if M^coud shall SAceed the first by m, the thiid shall
exii.( 1 tilt; >^oooad oy n, ana the fourth shall exceed
the third hv p.
) I. .V .sum o^ moi^e^ pat out at siiople intersd
APPENDIX. xU
amouxits in m months to a dollars, and in n monthi
to h dollars. Required the sum and rate per cent.
12. Given a' + a6  5x•^ to find the values of x.
13. Divide the number 49 into two sucn parts that
the quutient ot the greater divided by the less niay
be to the quotient of the less divided by the greater,
as I to .
14. L>ivide the number 100 into two such parts that
thcii product ma^ be equal to the dilierence of their
squares.
j" ar* 4 a;?/ — 56, ]
15. Given I >tofind\aluesofa;aud2/.
l«y+2/60j
16. A farmer bought a numbei of sheep for $80,
and if he had bought four more for the same money,
he would have paid %\ less for eu,ch. How many did
he buy 1
Matriculation, 1874.
1. Find the Gieatest Common Pleasure of 26^ —
iO'ib'' + 9>a% and 9^4^ — 2>a¥ + Zd^U" — 9rt '/?. rti.! de
moiistrate the rale.
a* + 7^ n o* — (UB
2. Add tc'ether a — x + , 3a — ,
=^ a + x a + x
„ 3a* — 2af , . a + x
2x — , and — 4a — ;.
a — X a — XT
3. Divide + by — ='
1+a; 1 — X 1 — X i+x
and reduce.
4. Given I (x — a) — lo {2x — 3b) — ^ (a — x)
= 10a 4 116 to find x.
5. A sum of mone/ was divided ainoii!:^ tin ee per
sons, A, B, aiid C, at> follows : tJie slian of 1
©.,.rewded 4 of the sha.es of h and C >\v §1J0: th^
APPENDIX.
share of B, f of the shares of A and V)y 810 ;
and the share of C,  of the shares of A and B by
^120. What was each person's share?
6. Given  ^3 ^ ]f^_'^^_4^ . 12  ^^ ^^^ * ^'^'^ 2/
7. Shew that a quadratic equation of one unknown
quantity cannot hav^e more than two loots.
8. Given ~ .— — ; to find the vahie of x.
4 + V a; >/x
9. The e is a stank of hay whose lenfth is to its
breadth ab 5 to 4, and whose height is to its Vueadth
as 7 to 8. It is wotth as mai y cents per mibic foot
as it is feet in breflcith; and the whole is worth at
that rate 224 times as many cents as bhere art square
teet on the bottom. Fjnd the cdu^ensions of the stack.
y/ocy + b \
10 Given ^ /" V to find x and y.
J^ = y/xy — 4 I
x>ry /
11. In attempting to arrange a number of counteis
in the form of a square it was found there wer«t se\ en
over, and when the side of the square was incie:ised
oy one, there was a deficiency of 8 to complete the
square. Find the number of countere.
12. Reduce to its simplest form
g' — (6 — cf ^ b^ — {c — ay _^ c» — {(i^b)*
(a + c)« — 6» (a + by — c» (6 + c)' — a"'
13. A and B ciin do a piece of work in 12 days;
in hew many days could each do it alone, if it would
lake A 1 days longer than B 1
y w I ^ ^^
1 4. Given ) x— y = A ) x, y, z,
\ X* + 1/ ^z' + w' = 62^
APPENDIX. larf
15. Find the last term, and the sum of 50 terms,
of the series 2, 4, 6, 8, ifec.
( M '
IG. Writo down the expansion of {x —  >
17. How many u..ut»ir)nc swains may be rung on
tea different bells, supposing all the combinations to
produce diffeent notA" '
ANSWERS
Junior Matric l87'J. I'af^s.
{x + ay + {x + a) {yb) + {ybY. 2. a^^ap + q
3. (a), 1^; {b), U; (c), 4^; (J), ^ ^. 4. 640, 660.
Junior Maine. , 1872. jPcws a/w? Honor,
1. I si + (.rJ  yi) I ' I 2i _ (xJ — .vi) I ' =
I «j _ (a^ , .,*)» I '; ,, + 2b + 3c. 2. We have
c* — ;?ci 5 = aiid c* — ^'c + g' = 0, fron?
vhicli to elimDate c.
4. If /3 be one root,   _ fi ^1 + Y '' = i8'^
aad, eliminating j, — = ^ — 2Z .
ac pg
6. (a), 4, — 7, ^(— 3± v/277) ; (6), 3, 2, ; — 3,— 3
— _, L; i,  ~ (c),3
^_ /2. ((f), Divide tL rough bv x^ and put y for
jc+ , and :.y^ — 2 for a:^+ — , then y =
_ or —  and a; = 3. i, — 4 or — 3.
3 2 ' *
ANSWERS. xxUi
6. ; >a 4a Ja f ... V
«• rrr
Junior Mdtric, 1872. Honor.
L 8 ami 6 miles. 2. Each of the first set ol'
fractions may be shewn equal to
X y
labc « or 2ahc h or 2ahc
b' + c'a* c' + a^'b''
s
, which are therefore equaL
a' + b'c^ ^
3. Multiplying the equations successively by y, z, a
and 2, X, y, we obtain c^x + a^y + h^z = 0,
b'x + c'l/ + a'z  ; thence —. — j^, = . . ^ . , =
a —o'c —car
and X —
4. «* + 6^>2a6,..c(a» + 6^)>2rt6c, &c.
5. 3,0; 2, 5; 3, 6; 8, 1. 6. 00 and 2tO mla
Junior Matric, Uonor. \ .„_
Senior Matric, Pass.
1. (a), From first x— 'ly or .//, and then solutions are
3, 3; 3,— 3^y2r, v/21; V2l,— V2f:
(6).>e(41drs/7"69^H37±/7Gy). (c). i.i;
i.i; ii;i>i ('0, 4,18. 2. 3.
, , , 33x«+61a;+ 10 ,,, a;^ + 3x + 2
^ '' a: + 2 ^ T^ + llaj + 30
ixiv ANSWERS.
_ X (3— x)
6. a;  2 and a; + 5 are factors, and roots are, 2, — 5,
H3±^/35): 7. 7^ gals.
8. 4.88 percent. 9. I days.
He receives $3 every day the work continues ;
he returns nothing the first day he is idle,
$1 the second, and so on, and the number of
days he works is IG.
Junior Matric, 1876. Pas$.
1. a' ; a""'' 6""' c~V 2. a«  cc» ; a'+^,
3. 1 + 2a; + 3ar^ 4 4a^ + 5a;* + Gar* + ; rem. 7a:^
Gx'. 90.^ — 6a;+l.
4. a^+^ab)^' '' ^ (ab) + b'^'.
5. (1), 2. (2), 2, 5, 7; or 2,5,7.
JiMiior Matric., 1876. Honor.
1. 35 mis. 2. (2), These quantities are in H. P. if
,&c., are in A. P., i.e., if a, 6, c
ah + ac + be
are in .4. P.
5. It may be shewn that the remainder at the nth
decimal place is 2n ; hence if the nth digit be
increased by unity, and the whole subtracted
from 1, the remainder is the remaining part
nf the period.
6. a = 4,a; = 2or3^=3or2;al,« = 2*^ro;
y2.7l0.
ANSWERS. xxf
Junior Matric, 1876. Honor,
1. 121 and 400 yards.
2. (a — h + c) {ah + 6c f ca) (a* + 6* + c* + rtb + be — ca)
3. Iri4,tionaI roots go in pairs/. 3 — i/iT is a rooi ,
and other roots a"e ^ (—1 zbyZIg).
S 13 2 la.
4. X + le'y^ + ic_y^ + .ry f xiy^ + j/^.
•"»■ i— 7 ^T7 A\ ^ ^ "'^^•
0+ (?^ — 1) (a — b)
1. (1), Plainly x + 2 divides both sides, and roots
are— 2,24 /f. {2), x= 3, >/ ^ i cv I ; x =
— 3, y — 4 or — i.
Second Class Certificates, 1873.
^ Kb^a)^=b^^^a^
2. {ab){a\b) = 2,b.
5. ('/. — t*) (a; — y) = ; .*. if a be not = 6, x y — ;
it' a — b,x — y may have any value.
4 31 4to, , , . , , ,
6. , , o '« • n, provided x be not =  23 ;
1 i//i 1 ) "* ^ ' '
then fraction becomes § and is indetei minate.
o 1 1
xy ' a; + / ^ ' ' ^
9. 13 10. ^ of a mile per hour.
«xvi ANSWERS.
Second
Class Certificates,
IS
7f
1.
2{a'h'
+ hh + c^a*) 
(a^ + b' + c').
2 \
a f o
3.
(3x +
\){\x^
2^
3x+l).
4.
2a(26^  5)
4a  o6
5.
he
— ad
^6c
a^
7?ic — na
6. X and y are indeterminate : there Ls but one
equation. 7. 8SS, .$44. 8. 1 4 Jays, 1 1 g Jays.
2
9 ^i^ hrs. m — n negative means that thev
m — n °
2
were together his. before noon, m — n,
n — n
they are neve»^' together.
10. Each side equals 99(x^ — y).
Second Class Certificates, 187C.
1. {\+m)x{ln)ij. 2. (x + y)'(.ry); {ah)
{bc) {ca); (S.yrl) (5ar' + xr 1).
3. Let the other factor be x¥ a; multijjly and eijuato
coetiicients ; eliminating a, nq — 'n  na; other
condition in pn — mn — r. 4. x— 1; 1 ^ t».
g { x + yz) {xy + z) {>/ + z x) . _1__
[x + y + zf a  b
6. §; 1.
7. a'ih'c  be) + b''[^ac' — dr) + c'{a'h — ah) = 0.
8. (1,) Cube, and 3(« + x)i (>ix)i (n») = m*  ^jj,
q a (c — 6) 6 (a — c)
a — 6 a — i
10. 3 miles an hour.
AA.i'^l^'ER^* xxvii
11. (a), See §359. {b), 2,000. (c), Snnstitute suc
cessively — b, —c, —a for a, b, c, in tho left
hand side, and it appears that a + b, b + c,
c + a are factors, and /. expression is of form
N{a + b) (6 + c) {c + a); putting abc 1 ,
we get iV'= 3.
First Year Exhibit/ions, 1873.
1.3,15,75,375. 2. 9 and 1, or V^ and H 4.9,12.
5. (a), 4, 3; 3,4. (6),2,3. (c), 4,5, 6. {d),\
6.40'. l.J^l±^Ll^ = .
2ab
Firxf. Year Exhibitions, 1874.
1. 5. 2. {—if; 3277. 3. 2^; 1\; 2^^.
4. 9, 12. 5. 75.
6. (a),3,2;— 2,— 3. (6), 7 or— If (c),5,3. {d),U.
7. 30 hours. 8. JL. 9. 3(a; + 3).
x + y
First Year Exhibitions, 1876.
11111
1
1
4 3a 6 c a b
c
b ' b ' \ 1 1 ' 1 1
,1
a« ^^ 7 «=" 6"*
c*
2.
— 12 '^/2. 3. X — 1. 4. m.
5.
21, 42, 63, or 81. 6. o, b, 2c; 1, 1
ixviii ANSWERS.
Matriculation, 1873.
2. \\a — 3c — 5(Z + m. 4. — ax.
5. ax' + (ar + 6) aj  (a/ ' + br + c) f
ar' + 6a^ Jfcr + d
X — r
C. l—x¥ x'— x^ ^ .. .. 7. 3ft' + 4a*.
8. (^t — .r) {x'—2)^ g 144, 216.
x^ — 1
10. \ {a — 2>i)i — "In — ^j), lire.
^^ vib~na 1200 (a — 6)
?u — ?i mh — na
12. drJv'^ 13. 28, 21.
U. 50 {VI —1), 50 (3/5).
15. x= zfc 10, ^= =F10; x = ±W2, y ^ ^ 3»/2
16. 16.
Matriculation, 1874.
1 1 o 4a* + a'x — 2rt;c' + ar* • .
1. a— 6. 2. ;^ 3w 1.
cc* — a^
4. _ 5a — 36. 5. 600, 480, 360.
6. 2, 4 ; 4, 2. 8. 4 or 9f
9.20,16,14ft. 10. 40, 10; 10, 40. 11.56.
12. 1. ' 13. 30 and 20 days.
14. 6, 2, 41, U,or2, — 6,— 1^, — 4.
15. 100, 2550.
16. x^ — 7a^ + 21af'— 35.r + 35a:' — Sla:"* + 7«»
— x\ 17. 1023.
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English Grammar
the same subjects are presented with much greater fulness, aiiu carried to
a more advanced and ditlicult stage. The work contains ample materials
for the requirements of Competitive Examinalions reai^hmg at least the
standard of the Matriculation Examination of the University of London
The Shorter English Grammar.
is intended for learners who have but a limited amount of time at their dis
posal for English studies ; but the experience of schools in which it has
been the only English Grammar used, has shown that, when well mastered,
this work also is sullicient for the London JIatriculation Examination.
(lE. J. (Sage S: €o'e. lctD Cbucatioual gMorks.
THE BEST ELEMENTARY AND GRAMMAR COMPOSITION.
Revised Ed. of Miller's Language Lessons.
Now in iiidcstiuctible iron biiidini^. Sixth edition; 2C0th thousand, with
Exaniiaation Papers for admission to High Schools. Adapted as an Intro
ductory Text Book to Mason's Grammar.
(a TIIOROIGII EXAMIKA7I0S GIVEN).
St. Thomas, Nov. 30th, 1878.
To the Trosident and Members of tlie County of Elj^in Teacher's .Associa
tion : — In adcordance with a motion pas.sed at tlie last regular mectin<j of
the Association, appointing the undersigned a Committee to consider the
respective merits of different English Grammars,with a view to suggest the
most suitable one for Public Schools, wc beg leave to report, that, after ful"
ly comjiaiing the various editions that have been recommended, we believe
that " Miller's Swinton's Language Lessons " is l)est .adapted to the wants
of junior pupils, and we would urge its authorization on the Goverimient,
and its introduction into our Public Schools.
Signed, A. F. Bitler, Co. Inspector, .1. McLean, Town Inspector.
J. Millar, M. A., Head M.aster St. Thomas High School.
A. Steele, M. A., " Orangeville High SchooL
N. Campbell, " Co. of Elgin Model School.
It was moved and seconded that the report be received and adopted.—
Carried unanimously.
i^" TO AVOID MISTAKES, ASK FOR
REVISED EDITION MILLER'S SWINTON'S.
PROOFS OF THE SlTEiUORITT C)F MILLERS EDITION'
Miller's Swinton's is authorized by the Education Department for use in
the Schools of Ontario.
Only Edition adopted by the Protestant Board of Education of Montreal'
and used in many of the principal Schools of the Province of Queijcc.
Only Edition used in the Schools of Newfoundland.
Only Edition adopted by the Su,ii of Education for the Schools of Manitoba
Miller's Revised Swinton's is used in ninetenths of the principal Schools
of Ontario.
Only Edition prepared as an introductory Book to Mason's Granmiar
both haying the same Definitions.
J. ^agc ^ €00. S^W ©bucational cMorks. I
EXAMINATION SERIES.
Canadian History.
Bv James L. Hughes, Inspector of Public Schools, Toronto.
Price, 25 Cents.
HISTORY TAUGHT BY TOPICAL METHOD.
A PRIMER IN CANADIAN HISTORY, FOR SCHOOLS AND STUDENTS PREPARING FOR
EXAMINATIONS.
1. The historj is di\idcd into periods in accordance with the great na
tional changes that have taken place.
2. The history of each period is given topically initead,of in chronolog
ical order.
3. Examination questions are given at the end of each chapter.
4. Examination papers, selected from the official examinations of the
different provinces, are given in the Appendix.
5. Student's review outlines, to enable a student to thoroughly test his
own progress, are inserted at the end of each chapter.
6. Special attention is paid to the educational, social and commercial
progress of the country.
7. Constitutional growth is treated in a brief but comprehensive exer
cise.
SS" By the aid of this work students can prepare and review for exam
inations in Canadian History more quickly than by the use of any other
work.
Epoch Primer of English History.
By Rev M. Creighton, M. A., Late Fellow and Tutor of Merton College,
Oxford.
Authorized by the Education Department for use in Public Schools,
and foi admission to the High Schools of Ontario.
Its adaptability to Public School use over all other School Histories will
be shown by the fact that —
In a brief compass of one hundred and eighty pages it covers all the
work required for pupils preparing for entrance to High Schools.
The price is less than onehalf that of the other authorized histories.
In using the other Histories, pupils are compelled to read nearly thice
times as much in order to secure the same results.
Creighton's Epoch Primer has been adopted by the Toronto School
Board, and many of the principal Public Sclxjul* in Ontario.
J. QSagc ■& €o's Jlcto CSbunttional (iilorks.
Authorized for use in the Schools of Ontario.
The Epoch Primer of English History.
By Rev. M. Creigiito.n, il. A., Late Fellow and Tutor of Mcrton College,
OMord.
Sixth Edition,    P rice, 30 Cents,
Most thorough. Aberdeen Joirsal.
This volume, taken with the eight small volumes cojitaining the ac
counts of the different epochs, presents what may be regarded as the most
thorough course of elementary English History ever published.
What was needed. Toro.sto Daily Gloeb.
It is just such a manual as i^s needed by public school pupils who are
going up for a IJiigh School ju.tc.
Used in separate schoo 3. M. Stafford, PRiPiST.
We are using this History m our Convent and .Separate Schools in Lind
say.
Very concise. Hamilton TrMEs.
A very concise little book that should be used in the Schools. In its
pages will be found incidents of English History from A. D. 43 to 1870, in"
t^rcstitig alike to young and old.
A favorite. London Advertiser.
The book will prove a favorite with teachers preparing pupils for the
entrance examinations to the High Schools.
Very attractive. Rritisii Wiiio, Kingston.
This little book, of one hundred and fort}' pages, presents history in a
very attractive shape.
Wisely arranged. Canada Pre.sbvterian.
The epochs chosen for the division of English History are well marked
—no mere artificial milestones, arbitrarily erected by the author, but reaj
natural landmarks, consisting of great and important events or remarkable
changes.
Interesting. YARMorTn Tribune, Nova Scotu.
With a perfect freedom from all looseness of style the interest is sn 'tH
sustained throughout the narrati\c that those who commence t^.
will tiiul it difficult to leave off with its perusal incomplete. •''
Comprehensive. Litkrart World.
The special value of this historical outline is that it gives the reader a
comprehonsjve view of the course of memorable events and epochs,
M. 3. (gage ^ Co'e. ^cto Cbitrational cHorks. ;
THE BEST ELEMENTARY TEXTBOOK OF THE YEAR. '
Gage's Practical Speller. ,
A MANUAL OF SPELLING AND DICTATION. 1
Price, 30 Cents. '
Sixty copies ordered. Molxt Forest .Advocate, i
After careful inspect on we unhesitatingly pronounce it the best spell '
ing book ever in use in our public schools. The Practical Speller secures ■
an easy access to its contents by the very systematic arrangements of the
wDrds in topical classes ; a permanent impression on the memory bj' the
frequent review of difficult words ; and a saving of time and eflort by the
selection of only such words as are difficult and of connnon occurrence
Mr. Reid, H. S. Master heartily reconunends the work, and ordered some
sixty copies. It is a book that should be on every business man's table as
well as in the school room.
Is a necessity. Presb. Witness, Halifa.x.
We have already had repeated occasion to speak highly of the Educa
tional Series of which this book is one. The "Speller" is a necessity ; and
we have seen no book which we can recommend more heartily than the one
before us.
Good print. Bowmasville Observer.
The " Practical Speller " is a credit to the publishers in its general get 1
up, classification of subjects, and clearness of treatment. The child wh«
uses this hook will not have damaged eyesight through bad print. I
o I
What it is. Strathrov Age.
It is a series of graded lessons, lontaining the words in general use,
with abbreviations, etc. ; words of similar pronunciation and different spell
ing a collection of the most difficult words in the language, and a number
of literary selections which may be used for dictation lessons, and commit"
ted to memorj' by the pupiis
B\«ery teacher should introduce it. Canadian Statesman.
It is an improvement on the old spelling book. Every teacher should 
rntroduce it into his classes 
The best yet seen. Colchester Scn, Nova Scotia.
Itis away ahead of any"speller"that we have heretofore seen. Our public I
schools want a good spelling book. The publication Ijcfore us is the best j
1 je ha^ e \ et seen.
J. ©age & QTo.s' Jlciu (giincattoniH SBorkg.
Gage's Practical Speller.
A new Manual of Siielling and Dictation. PriO«, 30 Cents
Pl:il.MI.MS.NT Fi^ATuan
The book is divided into five parts as follows :
PART 1
Coiitaitu the words in common use in laily life together with ahbrevia
tions, foinn, etc. if a boy has to leave school early, he should at least
know how to spell the uurds of couuuoii occurrence in connection with his
business.
PART fl.
Gives words liable to be spelled incorrectly beiauM the same sounds are
spelled in various ways in them.
PART III.
Contains words pronounced alike but spelled differently with diticre'it
meanings.
PART IT.
Contains • larj;e collection of the most difKcult words in common use,
and is intended to supply material for a general review, and for spelling
matches and tests.
PART V.
Contains literary selections which are Intended to be memorized and re'
cited as well as used for dictation lesson* and lessons in mtiraU.
DICTATION LESSONS.
All the lessons are gultable for dictation lessons on the slate or in dicta
tion book.
RRVIBWa.
These will be found throughout the book.
An excellent compendium. Alex.McRae.Prin. Jr.ad'ii.Dujby.N.S.
I rcgard it as a necessity and an excellent com] cmiium of the subject
of which it treats. Its natural and judicious anaiiement wi.^1 accords
with its title. I'uinls instructed in its principles, umier the care of dilijrent
teachers, cannot fail to become correct spellers. It ^reat \alue will, doubt
less, secure for it a wide circulation. I liave seen no liook on the subject
which I can more cordially recommend than "Tliu Practical Speller."
Supply a want long felt. J'lm J"hj,^t,n, l.P.S., Belleville.
Tne hints for teaching spelling are exiellent. ! bave shown it to a num
ber of experienced teachers, and they all think it is the best and most prac
tical work on spellini^ and dictation ever presented to tlif public. It will
supply a want long felt by teachers.
Admirably adapted. Colin rr. liosene, l.P.S., Wol/vHU, A', s.
The arr».njren\ent an. I :,'rading of the different classes of woixls I reirard
as excellent. Mucli benefit must arise from cnnnnittiii!,' to memory the
"Literary Selections." The work is admirably adapted to our public
schools, and 1 shall recommend it as the best I ha\ e seen.
cE. J. (Sage ^ QLo's. <^eU) (gbucational eMorks
THE BEST ELEMENTARY TEXTBOOK OF THE YEAR.
: Gage's Practical Speller.
I A MANUAL OF SPELLING AND DICTATION.
' Price, 30 Cents. !
: Sixty copies ordered. Mount Forest Advocate. ■
I After careful inspect on we unhesitatingly pronounce it the best spell j
I ing book ever in use in our public schools. The Practical Speller secures :
I an easy access to its contents by the very systematic arrangements of the
! words in topical classes ; a permanent impression on the memory by the
( frequetit review of difficult words ; and a sa\i!ig of time and effort by the
1 selection of only such words as are difficult and of common occurrence 
I Mr. Reid, H. S, Master heartily recommends the work, and ordered some 
I sixty copies. It is a book that should be on every business man's table as 1
i well as in the school room.
Is a necessity. Presb. Witxes.s, Halifax.
We ha\e already had repeated occasion to speak highly of the Educa
tional Series of which this book is one. The "Speller" is a necessity ; and 1
we have seen no book which we can recommend more heartily than the one '
liefore us. !
o • (
Good print. Bow.maxville Ob.server. ]
The " Practical Speller" is a credit to the publishers in its general get !
".p, classification of subjects, and clearness of treatment. The child whe
uses this book will not have damaged eyesight through bad print. i
What it is. Strathrot Age.
It is a series of graded lessons, containing the words in general use, i
■vith abbreviations, etc. ; words of similar proiumciation and different spell
ing a collection of the most difficult words in the language, and a number
of literary selections which may be used for dictation lessons, and commit"
ted to memory by the pupils. 
Every teacher should introduce it. Caxadian Statesman. }
It is an improvement on the old spelling book. Every teacher should 1
introduce it into his classes
The best yet seen. Colchester St7<, Nova Scotia.
It is away ahead of any"speller"that we have heretofore seen. Our public
schools want a good spelling book. The publicaticTt \efore us is the best
we have vet seen.
8E. S ®«9^ * ^'^ <^^to (Ebucational SRorke.
WORKS FOR TEACHERS AND STUDENTS, BY JAS. L. HUGHES.
Examination Primer in Canadian History.
On the Topical Method. By Jas. L. HruiiK.s, iti^iwctor of Schools, To.
rooto. A Primer for Students preparing for Examination. Price, 25c
Mistakes in Teaching.
By Jaj. Lauoiilin IIooiies. Second edition. Price, 50c.
AIK>PTKS B7 8TATR UNIVRRaiTT Of IOWA, AS A.S BI.EMKNTART WORg FOR 081
OF TKACH8R8.
This work discuascf) in a terse manner over one hundred of the mistakes
commonly made by untrained or inexiierienced Teachcis. It is Jesi;;fied to
warn young Teachers of the errors they are liable to make, and to help the
oideT raembiii!^ of the profession to discard whatever methods or habits may
be preventing their higher success.
The niistAkes are arranged under the following head* :
1. Mistakes in Management. 2. Mistakes in Discipliiie. 1. Mistakes in
Methods. 4. Mistakes in Manner.
How to Secure and Retain Attention.
By Ja8. LAuanLiN Huoiibs. Price. 25 Cente.
Comprising Kinds of Attention. CharActcristlcs of Positive .\ttention!
CSiaracteristics of The Teacher, flow to Control a Class. Developing Men
t«l Activity. Cultivation of the Senses.
rFrom The School an'd U.vtvkrsitt Maqazlsb, Loyrios, Eno.)
"ReiMete «ith valuable hints and practiail suggestions which are evident
ly the result of wide experience in the scholastic profession "
Manual of Drill and Calisthenics for use in
Schools.
By J. L. HoonBS, Public .School Inspector, Toronto, Graduate of Militar\
School, H. M. 2dth Regiment. Price. 40 Cents.
The work contains : Ttie Squad Drill prescribed for Public Schools In On
tario, with full and explicit directions for teaching it. Free Gymnastio Ex
ercises, carefully selected from the best Gennan and .American sj" .uns,
and arranged in proper classes. Gennan Caliethenic Exercises, as *»ught
by the late Colonel Goodwin in Toronto .N'omial .'^ohl>ol, and in Enicland.
Several of the l>est Kinderirarten Games, atid a few choice Exercise Songs.
The instructions throughout the book are divested, as far as possible, of
Jnneoessarj technicalities.
"A most valuable book for every teacher, i«rtlcii!arly In oonntry placer
It emliraccs all that a school teacher should teach his pupils on this subject.
Any teacher can use the e&ry drill lessons, and by doing so he will be con
ferring a benefit on his country."— C. Kadcliffr PRAU.tALT, Ucy "if Mrct
Life Guards, Orill Instractor .\ona»l axl Model School*. Toronto
KIRKLAND k SCOTT'S EXAMINATION PAPERS.
Suitalsle for Intermediate Examinations.
BEPRINTBD FROM
GAGE'S SCHOOL EXAMINER
AND STUDENTS ASSISTANT FOR 1881.
COMPILED BT
Thohab Eibeland, M.A., Science Master, Normal School, Mid
Wm. Scott, B. a., Head Master, Provincial Model School.
PRICE,  SO CKNTS.
Thit vohnne contains papers on ArUhmetie, Euclid, Geography,
Algebra, Bookkeevitui, History, Statics and Hydrostaiies, English
liiUrtHure, French (July, 18S0), Chemistry, English Orammar.
FROM THE PREFACE.
In reapoDM to the desire of a larxt: numt>er of Teachers, we reprint
the KxuninatioD Papers suitable foi the Annual Intermediate Ex
fcmination, which have appeared in the number*, for 1881, ot Gage's
"School Kxaiiiiner and Student's Assistant."
The steadily tncreastng circulation of this monthly magazine, and
the numerous letters received testifying to the great value of the
questions in the vmHoos cubjeots reouired for the Examinations,
plainly indicate that sach k periodical U a mo«t oaeful aid to both
MAcher and student.
The almost exhsastire nature of the questions on each subject
bringv the student Into clot^ acquaintance with every needful point;
and the drill experienced in thinlving and working out the answers is
of Incalculable practical benefit to those who wish to exo«l at written
examiDatloua.
When we state that the editors of this department ot tiM SeJwol
Examiner are Messrs. T. Sirkland, M.A., and W. Scott, B.A., we con
sider tiiat It is a suihcient guarantee for the excellenos and appro
priateness of the work, 04 these gentlemen have eam«d a wide reputa
tion as specialists in science and hterature.
lo consequence ot numerous applications (or the PrwKsh Paper given
at the Intermediate Elxamination, 1880, w« reprodoo* it in this book.
Hints and Answers to the Above, 50 Cents.
W. J. OAGE <& CO..
t. (OTBK,
Vm. J. 6iigc & doQ. Ilclu €butational
eMorks.
The Canada Schooi Journal
HAS RKXEIVED AN HONORABLE MENTIOM AT PARIS EXHIBITION, 1878
Adopted by nearly every County in Canada.
Reconniiended by the Ministe of Education, Ontario.
Recoiiiniended by the Council of Public Instruction, Quebec.
Recomniendfd by Chief Supt of Education, New Brunswick.
Rcccniniendcd by Chief Supt. of Education, Nova Scotia.
Reconinicnded liy Chief Supt. of Education, British Columbia.
Recommended by Chief Supt. of Education, Manitoba.
IT IS EDITED BY
A Committee of some of the Leading Educationists in Ontario, assisted
by able Provircial Editors in the Provinces of Quebec, Nova Scotia; New
Brunswick, Prnce Edward Island, Manitoba, and British Columbia, thus
having each s«H;tion of the UomiiiioD fully represented.
CONTAINS TWENTYFOUR PAGES OP READING MATTER.
Live Editorials; Contributions on important Educationa! topics; Selec
tions Readings for the School Room ; and Notes and News from eacb Pro
vince. I
PiiACTicAi, Department will always contain useful hints on methods of
teaching different subjects.
MAniBMATicAL Departmbnt gives solutions to difBcult problems also on
Examination Papers
Official Department contains sueh regulations as may be Issued trom
time to time
Sulscription. 81 00 per annum, strictly in advance.
Read TUB F:i!,i,"u ■ i i'tter frov John Oreenleaf Whittibb, thb Fa
ucis AMI.RH'AN I'ul.l
I b"^ve also rt'cei\til a .No of the •' Canada School Journal,' which seems
to me the brightest and most readable of Educational ilagannes I am very
truly thy friend, John Greenleaf Whittier.
A Club of I.Ojo Subscribers from Nova Scotia.
(Copy) Edccation Office, Halifax, N S . Nov. 17, 1878.
Messrs. Adam Miller & Co., Toronto, Ont
Dear Sirs,— In order to meei, the wishes of our teachers in various parts j
of the Province, and to secure for them the adxantajje ot vcur excellent
periodical, i hereby subscribe in their behalf for one thousaii'' (1,000) ccpits >
at club rates mentioned in .\our recent esteemed fa\or subscription! vil,
begin with January issue, and lists will be forwarded to your office in a li»
days. Yours truly,
David Allison, Chief Supt. of Education.
Address. W. J. GAGE & CO., Toronto, Canada.
i