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ALGEBRAICAL PROBLEMS,
PRODUCING
SIMPLE AND QUADRATIC EQUATIONS,
WiTH
THEIR SOLUTIONS ;
DESIGNED AS
‘
AN INTRODUCTION TO THE HIGHER BRANCHES OF ANALYTICS:
TO WHICH IS ADDED,
AN APPENDIX,
CONTAINING A COLLECTION OF PROBLEMS ON THE NATURE AND
SOLUTION OF EQUATIONS OF HIGHER DIMENSIONS.
BY
MILES BLAND, D.D. FERS. & FS.A.
LATE FELLCW AND TUTOR OF ST. JOHN’S COLLEGE, CAMBRIDGE,
NINTH EDITION,
WITH CONSIDERABLE ADDITIONS.
>
i
LONDON:
WHITTAKER AND CO., AVE MARIA LANE.
1849.
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ADVERTISEMENT.
Tue following pages, of which. ezght large editions have
been favourably received by the public, contain a col-
lection of Algebraical Problems, designed to point out
the various methods employed by Analysts in the So-
lution of Equations. They were originally drawn up for
the use of the younger Students of St. John’s College,
in performance of the duties attached to the office of
Sadlerian Algebra Lecturer; and were printed with the
approbation of the late Very Rev. Dr. Wood, who had
himself at one time designed a similar publication, but
from more important occupations had not found leisure
to collect materials for the work. The Examples are
arranged in the usual manner: 1. Simple Equations;
2. Pure Quadratics, and others which may be solved
without completing the square; and 3. Adfected Quad-
ratics. Utility being the sole object of this Publication,
wherever a proper Example occurred, it has been taken
without hesitation, or altered to suit the purpose. Many
have been selected from the questions which for a very
long period have been proposed annually to the Freshmen
in the College Examinations at St. John’s: and several
successive Editions of the Work have been benefited by
vl ADVERTISEMENT.
the contributions of friends. At the head of each Sec-
tion are given the common Rules; and the whole con-
cludes with a Collection of Problems without Solutions,
for the Kpercise of the Learner.
To the Sixth Edition was added an Appendix, con-
taining a Collection of Problems in Arithmetical, Geo-
metrical, and Harmonical Progressions; and another on
the nature of Equations, and the solution of those of
higher dimensions. |
And the Ninth Edition has been increased by an
additional Section on the Solution of Indeterminate
Equations and Problems, with a corresponding portion
of the Praxis.
ome
ALGEBRAICAL PROBLEMS.
LONDON:
GILBERT & RIVINGTON, PRINTERS,
ST. JOHN’S SQUARE.
CONTENTS.
Definitions, &c.
Solution of Simple Equations, involving only one unknown
Quantity
Solution of Simple Equations, involving two unknown Quan-
tities
Solution of Simple Equations, involving three or more unknown
Quantities .
Solution of Pure Quadratics, &c. . . . ..
Solution of Adfected Quadratics, involving only one unknown
COATES Ne oR Sa PR «ta aah) ("Lhe felts is uaeee oda.
Solution of Adfected Quadratics, involving two unknown Quan-
TIS. "pS PRED Se pas tee aise
Solution of Problems producing Simple Equations, involving
Ori vOne MU KTOWI QUAN reels cot. clu oy ARE a ne
Solution of Problems producing Simple Equations, involving two
unknown Quantities: (2... 4.
Solution of Problems producing Pure Equations
Solution of Problems producing Adfected Quadratics .
Solution of Problems in Arithmetical and Geometrical progres-
SIONS Mt | tee ye se ee ed ae
Solution of Indeterminate Equations and Problems . .
EA ReRRCOMMME) Gh) 3Y = 35
.. the digits are 2, 3, and 4, and the number = 234.
3. The sum of £1. 7s. was to be raised by subscription by
three persons 4, B, and C; the sums to be subscribed
by them respectively forming an arithmetical progression.
But C dying before the money was paid, the whole fell
to A and B; and C’s share was raised between them in
the proportion of 3:2, when it appeared that the whole
sum subscribed by 4 was to the whole sum subscribed
by B:: 4:5. Required the original subscriptions of 4,
B, and C.
Let v — y, x, v + y, be the respective subscriptions of 4,
B, and C;
UHCI 3 2e== 97 ean ec.
Now 5 : 2:: (C’s share =) 9 + y : the part paid by B
2
-° (9 - Y)s
and5:3::9+y: the part paid by dA ==. (9 + y),
-and consequently, 4 paid upon the whole 9 — y + =. (9+ y)
_72—2y
i — 5 3
2 63 + 2y
also B paid upon the whole 9 + a 9O+ty)= romper és
hence, 72 — 2y 3 63 + 2y 334: 5,
and (Alg. 179, 3.) 135 : 63 + 2y 2:9: 5,
and (4/g. 179, 8.) 15 > 63 + 2y 23225;
*. (21) 63 -+ 24 = 75;
by transposition, 2y = 12,
and: 4i== 6:5
.. the sums to be subscribed originally were 3, 9, and 15
' shillings.
248 Examples of the Solution of Problems
4. Four numbers are in arithmetical progression. The sum
of their squares is equal to 276, and the sum of the num-
bers themselves is equal to 32. What are the numbers?
Let 27 = the common difference,
and # + 3y
e+y
aed,
x — 3y
be the numbers ;
then their sum = 42 = 32,
and) Saw ies
also the sum of their squares = 4x° + 20y? = 276, in which
substituting the value of x found above,
256 + 20y? = 276;
by transposition, 20y’ = 20;
Sita
and 7 —-h Is
hence the numbers are 11, 9, 7, 5.
5. The sum of the squares of the extremes of four numbers
in arithmetical progression is 200, and the sum of the
squares of the means is 136. What are the numbers?
Supposing as before, 7 + 3y, x + y, 2 —y, and xv — 34,
to be the numbers ;
then 2v° + 18y’? = 200,
and’ 2a7'- 97/7 136;
.. by subtraction, 16y’? = 64,
and 4y = +8;
~YHt?;
whence 2” = 68 — y’ = 68 — 4 = 64,
WAY OR) eam.
and .*. the numbers are + 14, + 10, + 6, + 2.
6. The sum of the first and second of four numbers in geo-
metrical progression is 15, and the sum of the third and —
fourth is 60. Required the numbers.
Let x, vy, xy’, xy’, be the numbers ;
in Arithmetical and Geometrical Progressions. 249
.~e@+ry=15,
and vy’? + xy*® = 60,
Olly. (2 -+-2y) — 60;
OLiioy: — 6G;
oa iat
BNO te ee,
and (71+ 247.—=) 37—=.153
| @ = 5,
and the numbers are 5, 10, 20, 40.
7. The sum of four numbers in geometrical progression is
8.
equal to the common ratio + 1; and the first term = 7
Required the numbers.
Let # = the common ratio;
.. the numbers are om le a a
Wie ad VEER WALLY:
l+e7+e42? (14+ 2").(1 + 2)
17 - 17 ‘
1+ 2°
17
lao ee,
and 1o:== 2s
andi+a=
and 1 =
9
~m4= 2,
Tee 4 tom Ge
and the numbers are —, —, —, —.
1 ie Nes ES Hay)
A regiment of militia was just sufficient to form an equi-
lateral wedge. It was afterwards doubled by the supple-
mentary, but was still found to want 385 men to complete
a square containing 5 more men in a side, than in a side of
the wedge. How many did the regiment at first contain ?
Let 2 = the number of men in a side of the wedge;
”. (Alg. 192.) (@ +1). = the number of men in the wedge;
mae
250 Examples of the Solution of Problems
“. (@ + 1).2 + 385 = (@ + 5)’,
or 2? + xv + 385 = 2& + 102 + 255
*, by transposition, 360 = 92,
and 40 —\2"+
*, the number of men = 820.
9. After A, who travelled at the rate of 4 miles an hour, had
been set out two hours and three-quarters, B set out
to overtake him, and in order thereto went four miles
and a half the first hour, four and three-quarters the
second, five the third; and so on, gaining a quarter of
a mile every hour. In how many hours would he over-
take A?
Let # = the number of hours;
*, (Alg. 192.) (0 + (@ — 1) 1) x - = the whole number of miles
he travelled ;
but 11 + 4a = the whole number 4 travelled;
1 x
, (0++@-n).%=u +42,
2
or ov += Sos ae
4 ATH 7
Tomes a
and by transposition, — — + — = 225
; Eh as bs eign 9 361
completing the square, — + — +. — = 22 + — = —_;
. P 6 seen RB So Sey, +76 16 ’
1
extracting the root, - + = = +=
es Wego
gs 2 4 2 |
. © = 8,o0r— 115
hence in 8 hours he would overtake him. — 11 not answering
the conditions of the problem.
10. The base of a right- angled triangle is 6, and the sides are
in arithmetical progression ; it is required to find the om
two sides.
in Arithmetical and Geometrical Progressions. 251
Let 6 — x, 6, and 6 + 2 be the sides;
then 36 — 124 + x2 + 36 = 36 + 12@ 4+ 2”
(Eucl. B. I. p. 48.)
by transposition, 24a = 36,
3
and 7 = —;
2
.. the sides are - Lars. and 22,
rae 2
But if 6 be the first term of the progression ; let,
6,6 + 2, 6 + 22 be the sides;
then 36 + 24% + 4a” = 36 + 36 + low + 2”;
by transposition, 32? + 12% = 36,
or 2° + 42 = 123
completing the square, v7? + 4% + 4= 16;
extracting the root, 7 +2=+4,
and # = 2, or — 6,
and the sides are 6, 8, and 10; or 0, 6, and — 6.
The problem is not properly restricted; the algebraical ex-
pression, in this instance, is more precise than the language in
which the problem is stated.
11. A and B set out from London at the same time, to go
round the world (23661 miles), one going Kast, the other
West. A goes one mile the first day, two the second,
and so on. B goes 20 miles a day. In how many days
will they meet; and how many miles will be travelled
by each?
Let # = the number of days;
L
then (Alg. 192.) (w@ + 1) aa the number of miles 4 goes,
and 20” = the number B goes;
2
e+e
ot + com = 29661,
and 2 + 41” = 47322;
252 Examples of the Solution of Problems
: : 41\? 1681
completing the square, 2? + 41” + 5) 41322. ae
190969
iain
. 41 437
*, extracting the root, v + Pe
*, #@ = 198, or — 239;
“. they travel 198 days; A goes 19701, and B 3960 miles.
12. A traveller sets out for a certain place, and travels one
mile the first day, two the second, and so on. In 5 days
afterwards another sets out, and travels 12 miles a day.
How long and how far must he travel to overtake the
first ?
Let # = the number of days ;
then # + 5 = the number the first travels,
and .*, (Alg. 192.) (vw + 6). “te = the distance he travels,
and 124 = the distance the second travels ;
2+ 5
er (x + 6) a = 122,
and 2 + 112% + 30 = 242;
.. by transposition, v*? — 13% = — 30;
: 16 16 4
completing the square, 7 — 13%” + a = “ —30= =;
: 13
extracting the root, # — mr = + Z
Un Pipe) 2a ae
.. they are together at the end of 3, and 10 days after the
second sets out; and 36 and 120 miles is the distance travelled. _
13. A and B, 165 miles distant from each other, set out with
a design to meet; A travels one mile the first day, two the
second, three the third, and so on; B travels 20 miles the
first day, 18 the second, 16 the third, and so on. How soon |
will they meet ?
in Arithmetical and Geometrical Progressions. 253
Let x = the number of days required ;
Menli+2+3+... . + @ = (1 + #).< = the number
of miles A travelled,
x
and 20+ 18+ . - . . + 20-2” +2= (42 — 24).5 =
the number B travelled ;
L 2 L
*, (42 —2av).—-+ (1+ 2).—= (43 — 2).-— = 165,
2 2 2
Or @ — 432 = — 3303
a 184 184 52
completing the square, v? — 43v + a = at — 330 = =;
3
extracting the root, 7 — “ = = ;
.. & = 10, Or 33.
| Hence it appears that they meet in 10 days. On the loth
day B travels 2 miles, and the next day he rests; the following
| day he returns 2 miles; the succeeding day 4, and so on, in-
creasing two miles every day; and on the 33d day he again
comes up with 4, who has been travelling forward, every day’s
journey being one mile longer than that of the preceding day.
14. There are four numbers in arithmetical progression whose
continual product is 1680, and common difference is 4.
Required the numbers.
Let # + 6,# + 2, # — 2, and w — 6, be the numbers;
then (#@ — 36). (xv? — 4) = 1680,
or 2° — 402” + 144 = 1680;
| .. by transposition, v* — 402° = 1536;
completing the square, 2* — 402° + 400 = 1936,
extracting the root, 7? — 20 = + 44;
ao) Ge 1045 OF —_ 245
and # = + 8, or + 24/ (— 6),
jand .. the numbers are + 14, + 10, + 6, + 2; the two other
values of 2 being impossible.
254: Examples of the Solution of Problems
15. The product of five numbers in arithmetical progression is —
945, and their sum is 25. Required the numbers.
Let 7 + 2y, 2+ Y, 2,0 — y, @ — 2y, be the numbers ;
then 64 = 25, and .. 7 = 53
. (a — y’) . (2 — 4y’) = 945, or dividing by # = 5,
(u? — y"). (aw — 4y") = 1889,
or x — 5a°y’ + 4y* = 189,
and 4y* — 125y” + 625 = 1893
by transposition, 4y* — 125y* = — eg
125 15625 8649
completing the square, ay —125y? + (—) = 67 0 =e
125 3
extracting the root, 2y’ — mare ee + 2
also, #@
and the numbers are 9, 7, 5, 3, 1.
16. A Gentleman divided £210 among three servants, in
geometrical progression; the first had £90 more than ‘hey
last. How much had each?
Let vy’, vy, 2 = the number of pounds each had;
then vy’ = &% + 90,
and # + wy + vy’ = 210;
or 2@ + vy + 90 = 210;
by transposition, 27 + wy = 120.
Now from the first equation, 7 = 7 we 7
and from the last, 7 = ites 53
E20 tae 100 Se
~ y + 2 = y’ 25 1’
4 3
in Arithmetical and Geometrical Progressions. 255
4y°—4=3y4+ 6;
by transposition, 4y° — 3y = 10;
. 16
completing the square, 4y’? — 3y + ws E19) poe pa
16 16 16
3 13
extracting the root, 2y — lea Pike AIS
5
and 2y = 4, OFS»
5
and ich YY = 2, On 2
120
whence 7 = == 30; or 160;
Yr 2
and the sums are 120, 60, and 36 pounds.
17. The sum of three numbers in geometrical progression is
35; and the mean term is to the difference of the extremes
as 2to 3. Required the numbers.
Let x, vy, and wy’, be the three numbers ;
woe Pb vy + 47y” = 36;
and vy: vy’ — #3: 23 3,
OF Ys) YAS BAe 35
a Gr
ei Ue Tie waive
oe 3
by transposition, y’? — Y= 13
ats erp Sl Sg be pee
completing the square, y Abin IK Pea ATOR Sc mehENe
: 3 5
extracting the FOO Yeo £73
] } ae?
“. ¥ = 2% or — 5, which last does not answer the conditions ;
.(@ + 27444 =) 72 = 35;
oe V= 55
and the numbers are 5, 10, and 20.
256 Examples of the Solution of Problems
1s. There are three numbers in geometrical progression, the
greatest of which exceeds the least by 15. Also the differ-
ence of the squares of the greatest and least is to the sum
of the squares of all the three numbers as 5: 7. Required
the numbers.
Let x, vy, xy’, be the numbers;
then vy’ — # = 15,
and a? y* — a? ; x y* + ay? + 2° 35557,
ory —liyi ty tlii5i73
SY tl eee el,
by”
2
and yw#—1= +5;
by transposition, y* — -. y= '5:
25 25 121
i MP Uke See 25 é
completing the square, sid Aen hammer
ke 2 5 11
extracting the root, y? — i= + Fee
“. y’ = 4, or — *, which last is impossible,
11th a) et
.. from the first equation, (47 — v7 =) 34 = 15,
and # = 5;
.. the numbers are 5, 10, and 20.
19. The sum of three numbers in geometrical progression is
33, and the product of the mean and the sum of the ex-
tremes is 30. Required the numbers.
Let the numbers be ? a, and ry;
aL
SD aaes hae tl
x
and ( +2y) sah 308
.. by transposition, 13 — x =F +ay= =
in Arithmetical and Geometrical Progressions. 257
and 1342 — x” = 30,
or & — 134% = — 30,
“4h6 16 4
completing the square, #? — 134 + rae = —30=— :
13
extracting the root, # — Tone = c,
ande,”. @ == 10, OY 3:
3
tea ss GO eas Salsa 10>
or 3 + 3y° = 1043
by transposition, 3y’ — loy = — 3,
Or ye ay els
2 ail a 3
completing the square, y’? — -- Y + = = = —-1= ~;
: 5 4
extracting the root, y — Hie Gare
1
Aas Yy = 35 or 3?
and the numbers are 1, 3, 9.
If the other value of x be taken, the corresponding values of y
are impossible.
20. There are three numbers in arithmetical progression, and
the square of the first added to the product of the other
two is 16; the square of the second added to the product
of the other two is 14. What are the numbers?
Let « — y, x, x + y, be the numbers ;
then 227— avy +y’ = 16,
and 2”? — y? = 143
.. by subtraction, 2y? — vy = 2,
and by addition, 42° — vy = 30,
or 2y°= 2+ 2y,
and 427 = 30+ @y;3
.. by multiplication, sz’y? = 60 + 327y + 2°*y’;
S
258 Examples of the Solution of Problems
by transposition, 77°y’ — 32v”y = 60,
32 60
or ay’? — — .ry=—;3
je
completing the square, 777? ste L Ba sae at » 2 ae
p £ q 9 VY 7 y 49s 40
16 26
extracting the root, vy — ia = a3
' 10
ee DT ie ore
{oy = 24+ ry = 8,
and y? = 4;
yor),
and 42” = 30 + #y = 36;
aie POD te Be
PLOY ia mai
.. the numbers are 1, 3, 53 or —5, —3, —1. The other
value of wy was introduced in the operation, and does not
answer the conditions of the question.
21. The sum of four whole numbers in arithmetical progres-
: : : . eae
sion is 20, and the sum of their reciprocals is ay Re-
quired the numbers.
Let 7 — 3y,2—y, 2% + y, 2 + 3y, be the numbers;
then 4% = 20,
ORs
: 1 1 1 1 25
ey NEN 5 V—y + v+y TE Sy oe
4g" — 200y" OB
oY. {7 =
a —=102°y? + 9y* 24’
*, 25 x (9y* — 2504” + 625) = 24 x (500 — 100y’),
or 9y* — 250y° + 625 = 24 x (20 —4y’);
by transposition, 9y* — 154y? = — 145,
completing the square, 9y* — 154y? + —— ata == ih aoe
in Arithmetical and Geometrical Progressions. 259
extracting the root, 3y? — 2 = =,
145
and 3y? = 3, or rang!
.y =1,
= sSCAdE
and y = + 1, or —Y——
and .*. the numbers are 2, 4, 6, 8.
22. There is a number consisting of 3 digits, the first of
which is to the second as the second to the third; the
number itself is to the sum of its digits as 124 to 7; and
if 594 be added to it, the digits will be inverted. Required
the number.
Let the digits be represented by 2, vy, xy’;
then 1007 + lovy + ay :x+ay4+ uy’ 22 124:7,
orl00+ loy+yr>:ltyty 2124273
div’.99 + 9y:1+y+y i: 117: 7;
or Ik -pyei at y ty’? ts 13273
*. 13y? + 13y +13 =7y +773
by transposition, 134? ; 6y = 64,
a
or y’ Le pre Soars —
84]
completing the square, y’ AC eet (2 ve es + aye 169
extracting the root, Rs =+— cae
= 2, or rae
oe Y=2; 13¢
also 100% + lowy + wy’? + 594 = 100%y’ + lowy +2;
by transposition, 99” + 594 = 99ay’,
OF @4-16 = wy = 423
.. by transposition, 6 = 32,
and oi as
.. the digits are 2, 4, 8, and the number is 248.
s2
260
23.
Examples of the Solution of Problems
There are five whole numbers, the three first of which
are in geometric progression; the three last in arithmetic
progression, the second number being the second dif
ference. The sum of the four last = 40, and the product
of the second and last = 64. Required the numbers.
Let x = the first,
and y = the common ratio of the three first ;
.. the numbers are a, vy, vy’, vy’ + vy, vy’ + 2@Yy;3
spy? + 42y =140,
and x*y* + 22’y" = 64.
Multiplying the first equation by xy, and the second by 3,.
3a°y P42 y? = 4074,
and 3a°y° + 6a’y? = 192;
by subtraction, 2a’°y’? = 192 — 40ry;
by transposition, 22°y? + 4ory = 192,
or vy’ + 20"%y = 96;
completing the square, xy’ + 20a”y + 100 = 1963
extracting the root, vy + 10=+14;
*, vy = 4, or — 24.
Now from the first equation, vy. (3y + 4) = 40,
or 4.(3y + 4) = 403
°. 3yY +4= 105
by transposition, 3y = 6,
and y == 2;
{. also ime,
and the numbers are 2, 4, 8, 12, 16.
There are two casks A and B; of which, A the greater
holds 312 gallons. Into A a certain quantity of wine is
put, and B is filled with water; then water is conveyed
out of B into 4 in the following manner. First, a number
of gallons is taken, which is less by two than the square
root of the number of gallons in A, then a quantity less
than the former by two gallons, and so on. Now when B
is in this manner exactly emptied, A is exactly full: and it
in Arithmetical and Geometrical Progressions. 26]
is known that 8 gallons were taken out of B at one time,
after which the quantity left in B was 12 gallons. Re-
quired the number of gallons of wine in A.
Since the quantities taken out of B are in a decreasing
progression, whose common difference is 2, and one term of
this progression is 8, therefore the next terms are 6, 4, 2, the
sum of which is = 12; and therefore the quantity last drawn
out is 2 gallons. Let 2? =the number of gallons of wine in
A, then 2 — 2, x —4, &c. are the numbers of gallons drawn
each successive time; and the number of terms is evidently
= — ip and therefore the whole quantity drawn from 8B is
pG-)-2-5
[~]
x x
e°@ a + pF a oe SIA
4 2 :
5 L j
or ins, es ch Leg
4 2
i ~ 22 1248
Lv — —r% = —;
5 5
: 2 1 1248 1 6241
completing the square, 2? — —# + — = — + — = —_ ;
P 8 d 5 BL 25 5 25 25 ”
: 1
extracting the root, 2 — gic a
and # = 16, or — * which last will not
answer the conditions; therefore #? = 256 =the number re-
quired.
25. The diagonals of 4 squares are in an increasing geome-
trical progression, and the product of the squares of the
diagonals of the extremes is to the product of the dia-
arals of the means as a side of the third is to the square
root of the common ratio divided by 44/2. Required
the diagonal of the third square, and the common ratio,
supposing their difference equal to 45.
262 Examples of the Solution of Problems
Let ? x, xy, and xy’ = the diagonals ;
then since the diagonal : a side :: 2213
the side of the third = pan
2
and ex. wf ive x HADES
Y V2 4/2”
oo By tit ae /y 213
. ay = 42 SY;
and wy? = 4.
Now y — vy = 45,
or y — 4y2 = 45;
completing the square, y — 4y? + 4= 49;
extracting the root, y2 —-2=+7;
“, y? = 9, or — 5, which last does not agree with the con-
ditions; and .. y = 81;
4 4
whence # = — =-;
wo Seay
and .*. the diagonal of the third square = 36.
26. Two persons, 4 and B, traded together. 4 gained every
year £3 more than the preceding year, and the last year he
gained £17. His whole gain was £57. 8B in the first four
years gained £52, and if what 4 put into the common
stock be added to what B gained the second year, the sum
will be £13. How many years did they remain in trade,
and what were their original stocks ?
Since 4’s annual gains are in an increasing arithmetical —
progression, whose common difference is 3, the last term 17, |
and sum 57, if n = the number of terms, then (Alg. 192),
fai—(n—1).3$.- = 57,
or 372 — 3n” = 114;
in Arithmetical and Geometrical Progressions. 263
completing the square, n’ — = n + 1S 20 ACU ee ee.
extracting the root, n — = == ob
19
*. nm = 6, or ia
hence they remained 6 years in trade, and consequently A’s gain
the first year was £2, and his gain in four years was £26.
(Let ..°. # = A’s stock,
and 26 ; 52: 2: B’s stock = 22,
and A’s gain the second year being £5,
Gael te ater ROA el °
Pb oe woe Cad BE
anda 3
*, A’s stock was £3, and B’s £6.
27. A pyramidical pile of cannon-balls, the base of which
was an equilateral triangle, was all used in an engagement,
except the three lowest layers, and 4 balls of the next
layer; these were afterwards formed into a pile with a
rectangular base, having as many balls in one side of
the lowest layer, as there were in the side of the lowest
layer of the pyramidical pile, and 4 in the adjacent side.
What was the number of balls; and what the number of
layers in each pile when complete?
Let x = the number of balls in a side of the lowest layer ;
Pi De aot = the number of balls in that layer,
and (7 — 1) = = the number in the next,
L—1 :
and (7 — 2). Pee the number in the third ;
, 3a°— 34742
264: Solution of Problems.
Now since there were only 4 balls in one side of the second
pile, there can only be four layers, which will contain 4a,
3.(%— 1), 2.(# — 2), and w& — 3 balls respectively ;
32° — 34 +%
*-———___—__ + 4= 10% — 10,
2
or 327 — 34 +2+8= 20% — 205
by transposition, 32° — 234 = — 30,
ee
or 2? —- —xr =— 10;
3
: 23 23\? 529 169
completing the square, 77 — —.w# (22) = — —10= — ;
ater a 3 ae 36 36”
23 13
extracting the root, v — eens ra re
5 . . e
Ande — oer = which last cannot answer the conditions of
the problem. Hence there were 6 layers in the first pile,
and they contained 1, 3, 6, 10, 15, 21 balls, respectively;
.. the whole number of balls in the first pile was 56, and in
the second 50.
(32.) In the preceding solutions it may be observed, that,
in many instances, values of the unknown quantities are de-
duced, which do not agree with the conditions of the pro-
blems. This is always the case when the roots of the equations
are negative; and the circumstance arises from that peculiar
quality of an algebraic expression, by which it is denominated
either positive or negative. The product of two or any even
number of such quantities, whether all of them are positive or
all negative, will only be affected with a positive sign: thus
the quantity vy will represent the product of + #@ x + y, or of
—«#x —y; and a’, of +ax +4, or of —a x —4a; con-
sequently, in the reduction of such quantities to their con-
stituent factors by the rules of division or evolution, these
factors may be considered either as all positive or all negative.
But in common language, in which the conditions of a problem
are expressed, quantity or number is from its very nature what
in Algebra is meant by the term positive, 7. e. it increases any
Solution of Problems. 265
homogeneous quantity to which it is added, and diminishes any
one from which it is subtracted. Hence it may be understood,
why, when quadratic equations are formed to express the con-
ditions of a problem, the resulting roots may exceed in number
what appear to be required as answers to the problem, and why
such as are negative cannot be applied to its conditions.
These roots or values, however, though inapplicable in their
present shape, will, if assumed as positive, become correct
answers to the problem under a different modification of the
conditions. In the equations thence deduced, these former
negative values will appear as positive roots, and the former
positive values as negative roots. Thus, if Prob. 12, page 200,
be transformed into the following, “A detachment from an
“army was marching in regular column with 5 fewer in depth
“than in front; but upon the enemy coming in sight the front
was increased till it became = 845 — the original front; and
“by this movement the detachment was drawn up in 5 lines.
© Required the number of men;” from the solution of this
problem the number is found to be 3900, answering to the num-.
ber which would be found from using the negative value of 2
in the original problem; and the equation for determining
this (7 — 542 = 4225 — 52) differs from the other only in the
sign of x.
|
(
In Prob. 19, page 204, v is found to be equal to + 4, where
the negative value shows, that if the trading vessel had turned
out of its first course in a direction contrary to CE, or on the
opposite side of the line AC, it would have been taken after
' sailing 4 miles in that direction.
In Prob. 1, page 212, the negative value of 2 is found
to be — 130. But if the problem be modified so as to become
“A merchant sold a quantity of brandy, by which he lost
“£29 more than the prime cost, and found that his loss
“was as much per cent. as the brandy cost him. What was
“that price?” the equation for determining the price, is
2
q00 = 29 + 2: which is deduced from the equation to the
266 Solution of Problems.
original problem by changing the sign of x, the positive value
of which is in this case 130.
Also, in Prob. 4, page 213, the negative value of 2 is
— =. Now if the problem were, “ Bought two sorts of linen,
“for the finer of which I gave 6 crowns more than for the
“other. An ell of the finer cost as many shillings as there
“were ells of the finer. Also 28 ells of the coarser (which
‘was the whole quantity) sold at such a price, that 8 ells
“cost as many shillings as one ell of the finer. How many —
“ells were there of the finer; and what was the value of
“each piece?’ an equation arises differing from the equa-
tion to the original problem only in the sign of x, and whose
o,° » ad
positive root is =; whence there were 73 ells of the finer at
7s. 6d. per ell, the whole price of which was therefore
£2. 16s. 3d., and the whole price of the coarser was £1. 6s. 3d. —
And in the very same manner, all the other problems may be
transformed.
(33.) The same reasoning will apply to the case in which all —
the roots of the resulting equation are negative. None of its |
values can in this case be applied to satisfy the conditions of the |
problem; but if the conditions are properly modified, equations
may be deduced, of which these values rendered positive will
become roots, and will satisfy such conditions.
(34.) The same observation holds, if the resulting values be
the square roots of negative quantities, with this exception, that
such roots can never be applied to satisfy the conditions of the
problem under any modification whatever.
SECTION XI.
Examples of the Solution of Equations, where the number of
unknown Quantities exceeds the number of Equations.
(35.) Ir has before been observed, that when the number of
unknown quantities exceeds the number of equations, some of
these quantities cannot be found except in terms of the others;
the values of which, being undetermined, may be assumed at
pleasure; thus admitting a number of answers that will be in-
definite. A problem thus not properly limited is called an in-
_ determinate problem. In such, however, it is not unusual to
annex the condition that the values of the numbers sought
‘should be positive integers, or at least rational; or by other
limitations to lessen the number of answers. In the different
_kinds of these indeterminate problems, different expedients will
be made use of; and different artifices be found appropriate to
questions differently circumstanced. These, however, are best
| learned by practice.
(36.) In the case of a simple equation expressing the re-
‘lation of two unknown quantities, whose corresponding integral
) values are required, the common rule is, to divide the whole
| equation by the less coefficient, and to assume that part of the
‘quotient, which is in a fractional form, equal to a whole num-
‘ber. A new simple equation is thus obtained, in which a repe-
tition of the process takes place. And this is similarly continued
with each new equation, till the coefficient of one of the quan-
tities becomes unity, and that of the other a whole number.
An integral value of the former may then be obtained by sub-
stituting zero or any whole number for the other; and then
from the preceding equations, integral values of the quantities
proposed may be ascertained.
268 Examples of the Solution of Indeterminate Equations.
EXAMPLES.
1. Having given 2z + 3y = 35, to find the corresponding
positive integral values of x and y.
Dividing by 2 the least coefficient of the unknown quantities,
35 — 3 —1
ge Pt yy — 2 ;
y—1
2
Assume
= whole number = m,
Y= 2M+4+ 15
whence # = 16 — 3m;
and here 3m must be less than 16, or m not greater than 5; the
question .*, will admit of 6 answers.
Letun = 0; ‘then vr 16%and 44,
m=1,; v= 13 Y = 3;
m= 2, v= 10 y= 5,
nis; ign #6 y =7,
m= 4, v=4 Y¥ = 9,
m= 5, P= y= 11.
2. Given 72 + 11y= 100, to find the corresponding positive
integral values of # and y. :
Dividing by 7 the least coefficient of the unknown quantities,
4y —2
L=14—-Y— a ;
Assume -t— * = whole number = m;
OT Ae oe
ah dgaee tele shai
; m+1
Again, assume
= whole number = 7; |
. m=2n— 1,
and y =7n— 3,
£=19 —11n; |
where must be less than 2, and can .*, only = 1; :
in which case v = 8 andy = 4,
Examples of the Solution of Indeterminate Equations. 269
3. Given 9% + 13y = 200, to find the corresponding positive
integral values of # and y.
Dividing by 9,
y ne = 22—-y— ot.
Assume are = whole number = m;
“. 2y¥=9m +1,
and y = 4m + an
Again, assume an = whole number = n;
.m=2n— 1,
and y = 9n — 4;
oe & = 28 — 132.
_And since 9 must be greater than 4, and 13” less than 28;
.. 2 must not exceed 2, nor be less than 1.
elm le then dg — loan 3.5,
n= 2, v= 2. y = 14.
4, Given 1142 + 13y = 190, to find corresponding positive
integral values of # and y.
Dividing by 11,
190 — 13y 2y — 3
Ce oe ;
ll ee 11
2y — 3
Assume < = whole number = m;
m+ 1
yom + 1+ ——.
m+
Again, assume = whole number = 7;
then m= 2n—1,
and y = 11n — 4,
Uy 290 — 1371s
270 Examples of the Solution of Indeterminate Equations.
22) . .
where 7 cannot exceed 737 & it cannot exceed 1, and must be
greater than 0.
Tpeten ta hee ae 9,0 ee
Ifin =0; 7 — 37; andy ==:
if” = 2, 7 = — 4, and y — 18, &e.
5. Given 13” + 16y = 97, find corresponding positive in-
tegral values of # and y.
Dividing by 13,
SADArs LY yenayee 3y — 6
13 er
I mie’:
13
v
Assume = whole number = m;
ve YIM + 2;
and 2 = 5 — 16m,
where it is evident that m cannot be = 1, to have a positive
value of x.
Let — 0, then — Sanday ee
It =, tien ¢ — — LF, dud 7—15,-
and so integral negative values may be obtained.
6. Given 17@ + 23y = 183, to find corresponding positive
integral values of # and y. |
Dividing by 17,
pet Sermo el (0 eee ees
17 Ly,
647 —13
= whole number = m;
“. 6Y = 17m + 13,
and y = 3m 4+2—"—.
arnt!
m
Assume = whole number = n;
“ m=6n+1,
and y =17n + 5,
L=4— 23025
Examples of the Solution of Indeterminate Equations. 271
where » = 0 is the only value which will give positive integer
values of # and y; viz.2=4andy=5.
7. Given 19% +5y= 119, to find corresponding integral
values of # and y.
Dividing by 5,
1g 1 92 aad
= + = 24 47 — ——.,
y 5 5
L—1
Assume = whole number = m;
=5M +1,
and y = 20 — 19m;
where 19m must be less than 20, and .*. m cannot exceed 1.
Letm= 0, .*. a= 1, and ¥ = 20,
m=1, .. &©=6, y= 1,
which are the only integral values.
8. Given 46% + 3y = 3668, to find corresponding positive
integral values of 2 and y.
Dividing by 3,
L—2
Yaar hee a iat
| L—2
Assume = whole number = m;
°. 7=3m + 2,
and y = 1192 — 46m;
1192
where m must be less than —, and cannot .*, exceed 25.
Let m= 0,:then 2 = 2, and: y= 1192;
m =, Bs 5, = 1146,
Mm = 2, L = 8, y = 1100,
and so on, by assuming m equal to every number up to 25, we
‘obtain pairs of integral positive values of x and y, the former
‘increasing by 3, and the latter decreasing by 46.
)
9. Given 3% =8y— 16; find the least corresponding
positive integral values of x and y.
272 Examples of the Solution of Indeterminate Equations.
Here » = SY" = 3y—5 —Y
Assume = whole number = m,
y+]
3
.y =3m—1,
and v = sm — 8,
where m must be greater than 1.
Assume m = 2, then # = 8,
and y = 53
the least numbers which will satisfy the conditions.
If m = 3, x = 16, y = 83 and other values of m will give
corresponding values of # and y.
10. Given 77 — 9y = 29; find the least corresponding
positive integral values of # and y.
Here w= TU aa ty + ais.
Assume aE. = whole number = m;
° 2y=7m—1,
m—1
and y = 3m + :
m—1
Let ane ae whole number = m;
._m=an+ i,
and y = 7n + 3,
L= 9M + 8.
Let m=0, then x =8, and y=3; which are the least
whole numbers which satisfy the equation.
By assuming other values of m, corresponding values om x
and y may be obtained.
» |
11. Given 92—7y=6; find the least corresponding positive
integral values of # and y.
97 — 6 27 +1
=wf—1+ .
Here y =
Examples of the Solution of Indeterminate Equations. 273
27+ 1
Assume
= whole number = m;
7m—1 m— 1
= = 3m + = ee
1
= whole number = n;
.m=2n+15
whence # = 7n + 3,
and y = 10n + 3.
If n=0, x =3, and y=3, which are the lowest integral
values. Others may be obtained by assuming different values
of n.
12, Given 11a — 17y =5; find the least corresponding in-
tegral values of 2 and y.
5 Pin pga
Here x = “¥** — Ysa;
1]
ei
—1
Assume % f= whole number = m;
ete al Layee
and #7 =17m + 2.
And if we assume m = 0, # = 2, and y = 1, which are the
Teast numbers. Other values may be obtained by assuming
different values of n.
13. Given 14v4 =4y +73; determine whether positive in-
tegral values of # and y can be found.
-_ 2 1
Here y = oo = Leg ye Sgt, Cane,
4 4
1
| Assume at
= whole number = m;
ve 22 — 4m eae 15
which is impossible; .*. no integers can be found to answer the
sonditions.
14, Given 19% = 14y — 11; find the least corresponding
dositive integral values of x and y.
dd
274 Examples of the Solution of Indeterminate Equations.
197 + 11 5@ +11
Leese ee pale a pg
Assume
he +11
TH whole number. = ™;
14m — 11 m+ 1
ee es TR me .
5 5
Ter ae 1
= whole number = 7;
.m= 5N—13
whence # = 14n — 5,
and y = 19n — 6;
where to obtain positive integers, m cannot be = 0.
Let n = 1, then # = 9, and y = 13, the least values.
15. Given 23” — 9y = 929; find the least corresponding
positive integral values of # and y.
23% — 929 5” — 2
Here y = Wargo Poet cee oa :
5k — 2
Assume = whole number = m;
m—2
7. £= 2m — —.
5
m—2
Let zp whole number = 7;
°. M=5n + 23
whence # = 9n + 4,
and y = 23n — 93;
where any value of 7 less than 5 will give y negative.
Let .*. m = 5, then 2 = 49, and y = 22, the least values.
16. Given 54 + 7y + 112 = 224; find all the positive in-
tegral values of x, y, and z, which satisfy the equation.
24 — 7y —
awe a 1a 2; ok aha sell revere :
5 i)
Examples of the Solution of Indeterminate Equations. 275
= whole number = m;
then z == 5m — 2y — 1,
and 7=47 + 3y— 11m.
If then we assume different values for m and y, correspond-
ing values of # and z may be obtained.
meretnen 7% — 1, and y—1; .. # = 39,2=—23 any other
values of y will give z = 0, or negative.
et m ==, and 7 == 1, then 2 = 28, and z= 7,
y = 2, e&= 3), z=5,
y = 3, & = 34, = 3,
y = 4, v2 = 37; z<=l.
Other values of y will give z negative.
It is evident that m must be greater than te
AS Y
ld iete 7
and with this limitation other values may be found.
and less than
17. Given 17x + 19y + 212 = 400; find all the positive in-
tegral values of #, y, and z which satisfy the equation.
ey ead 2 4z2—
eres bili py, Spas el ei a
17 17
2y+42—9
Assume = whole number = m.
l
ae Yy — 8m rea | 22 + 4 + ant
Let mo = whole number = 7;
“ M=S=N—15
whence y = 17n — 4 — 22,
and # = 28 — 19n + 2.
Substituting for z and n different values, we obtain integral
values of # and y.
r 2
276 Examples of the Solution of Indeterminate Equations.
Ifn=1, andz=1, r= 10, y= 11,
— 9, 711, fa Os
2=3, f= 12,y7= 7, &e.
z2=6, ©=15, y= 15
8 + 2
y)
and m cannot exceed , nor be less than 1.
find the corresponding po-
sitive integral values of 2,
y, and 2.
18, Given 2 + 2y + peered
da + sy + 62 = 47)
From the first equation, 27 + 4y + 62 = 40,
but 4@ + 5y + 62 = 47;
*, by subtraction, 27 + ¥ =
and y = 7 — 223
whence 32 = 20 — 14+ 42 —®@ =6 + 32,
and 2z=2-+ @#.
Let .. @ =1, then y = 5, and z= 3,
Tei; ij 12; fa
v= 3; y=1; 2=535
but if 2 be assumed greater than 3, y becomes negative.
19. Required the positive integral solutions of the equation
vy +20 + 3y = 42.
Here (vw + 3).y = 42 — 2a,
pices al ee, SA
e+3 L+3
: - v + 31s a divisor of 48.
Now the divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
Let thenv +3=™m™;
c= Mm -— 3,
and y =
48
and y = eins We
m .. must be greater than 3, else # will be negative; and if less
than 24, y will be negative; if = 24, y will =o. :
Examples of the Solution of Indeterminate Equations. 277
Let m= 4, thenz = 1, and y = 10,
m= 6, L= 3, y= 6,
m= 8, v= 5, Y= 4,
Dial 29. J We
216; Ti emetASP e— wal
If all the integral solutions be required,
Peri — ol, then’ 2 — —"2, and y ="46,
eee. L=—1, = 23.
m= 3, v= 0, y= 14,
Mm = 48, Te AD, y=.
20. Required the positive integral solutions of the equation
lary = 52 + 7y + 15.
Here (12@ —7).y = 5@ + 15,
ie BY +15
Y 12%@—7’
aa aha a 215m
gece 19g Fete for 7
and .°. 122 — 7 must be a divisor of 215.
But the divisors of 215 are 1, 5, 43. And since if m=127—7
: a
aT, such values of # only will answer as when increased
by 7 will be divisible by 12, it is evident that 5 is the only one
‘lof the divisors which can be used ;
.t=1,
and y =4.
21. Required the integral solutions of the equation
ey + xv =2H + 3y + 29.
Here (v@ + 3).y = 27 — 2 + 29,
20 — 2 +29 26
SO Y=) ae tte gaa WEF 1) eas
and .*. 2 — 3 must be a divisor of 26.
het il aij a te 1 tes
278 Examples of the Solution of Indeterminate Equations.
Now the divisors of 26 are 1, 2, 13, 263
and om ela 4, aa}
m= 2,= 5,Y= 7,
m=13, ©= 16, y=— 15,
m = 26, # = 29, y= — 29.
22. Find a number which, divided by 2 and 3, leaves re-
mainders respectively 1 and 2.
Let 2 = the number;
aC l e
then is a whole number;
let it =m;
ee mh Py pote
vw Sara 2 .
But is a whole number ;
. 2m—1
Zc. = whole number = 7;
37 + 1 n+
m= =n + °
2
n-+1
Let amas! i
._nm=2p—1,
andm=n+p=3p—135
whence #7 = 2m+1=6p—1.
Assume p= 1, thena# = 5,
p= 2; L=115
and by assuming different values of p, other numbers are ob- -
tained which answer the conditions.
23. Find the least whole number which, when divided by 3.
and 5, has its respective remainders 1 and 3. |
Let 2 = the number;
&e— 1
= whole number = m;
then
aa L= 3m + 1.
Examples of the Solution of Indeterminate Equations. 279
Rm 3
But ey whole number ;
3m — 2
ap eee oe whole number = n.
5n + 2 on +2
and m= "7 =n + :
Let nt ’ — whole number = OS
~ n= 3p—1,
and m=n+2p=5p—1;
C= 3M +1 = 1p — 2.
Let p = 1, then = 13; the least whole number which will
answer the conditions.
24. Required a number which, divided by 11, leaves a re-
mainder 3, but being divided by 19 leaves a remainder 5.
Let « = the number ;
LX—3
then = whole number = m;
Jit = 1m + 3.
U—5 l1l1m—2
But —— = ——— = whole number = n;
19 19
19n +2 8n + 2
2 PEE SM Eants,
11 li
4n-+ 1
Let :
ll
“42 = llp—1,
1lpy—1 a+1
and n = —f—— = 3p egaant
4 4
is et
Assume / a
fee To I
and 2 = 117 — 3,
m= 19" — 5;
whence vw = 2097 — 52.
280 Hxamples of the Solution of Indeterminate Equations.
Let 7 = 1, then # = 157, the least number which can answer
the conditions. Other values may be obtained by assuming
Peat ROPER fed eh
25. Find the least whole number which, divided by 19,
leaves a remainder 7, and divided by 28 will leave a remainder 13.
Let # = the number;
pal |
then = whole number = m,
and w= 19m + 7.
Saar ees 19m—6
Also = whole number = n;
*, 19m = 28n + 6,
and m=n + valaat es
19
3 2
Let “ = whole number = p;
19p — —2
then as ee by ies Sasa
3 3
p—2
Assume = whole number = 7;
©
then p = 3r + 2,
n= 197 + 12,
m = 287 + 18,
and # = 5327 + 349.
Assume 7 = 0; then x = 349, the least number that will
answer the conditions. Other numbers, however, may be de-
termined by assuming different values of 7.
26. A certain number, when divided by 5 and 4, leaves a |
remainder 1; but when divided by 3, leaves no remainder.
Determine the number.
Let v = the number;
oO)
then = whole number = m;
. = 56m, +1.
Examples of the Solution of Indeterminate Equations. 281
ee | 5m m
Also apy aicassial aa m+ Vs whole number.
m
Let 7A whole number = n;
eee tt
and 7 = 20n +1.
20n +1 i—1
—- = 7rn— ne
But - = whole number =
n—
Assume
thenvn= 37 --- 1,
m= 12p + 4,
and # = 60p + 21.
If then p be assumed = 0, # = 21, the least whole number.
[EE i ele
and so on for every number which may be assumed for p.
27. Find a number which, divided by 3, 4, 5, respectively,
‘shall leave 2, 3, 4 for remainders.
Let # = the number;
“2
then ae whole number = m;
° fom 3m + 2.
2-3 3m—1
Also or. aa ae whole number = 7;
4m + ] ie ae
m= -———— =n
3 ST;
n+1
and ~ = whole number = p;
| “. N= 3p — 1;
and m= 4p — 1,
whence # = 12p — 1.
L—4 12p—5 2p
But —— = ——— = whole number = 2p —1 + =,
: ot = whole number, and £ = whole number = 7;
282 Examples of the Solution of Indeterminate Equations.
Ie or,
and 7 = 12p —1=60r—1.
Assume 7 = 1, then 7 = 59;
r= 2, & = 1195
a= 2p Fy WAN RL u ii
28. Find a number less than 400 which is a multiple of 7,
and upon being divided by 2, 3, 4, 5, 6, always leaves 1 for a
remainder.
Let # = the number;
aL—
1
then = whole number = m;
a ee oe
L—) 2m ,
and es whole number = mig let it =n;
m= ee ‘he &
ee — 9 — 2°
Assume — = 7p;
noma 207,
mM == 37,
and v7 = 6p + 1.
Rap Me at Pa Fe whole number ;
4 4 2
<7 . = whole number = g,
and 7) ==:
eae) 7 ae a
Again << = ae = whole number = 2q + “f.
Let2 = ee
5
BET Bny
and # = 607 + 1.
G—1 607
Also Pre ee 107 = whole number.
Examples of the Solution of Indeterminate Equations. 283
By assuming .*, 7 = 1, 2, 3, 4, 5, the values of x will be found
to be 61, 121, 181, 241, 301, 361, which are all less than 400. But
301 is the only multiple of 7; the only number .*, which answers
the conditions.
29. Divide 25 into two parts, one of which may be divisible
_ by 2, and the other by 3.
Let 24 = one part,
and 3y = the other;
then 22 + 3y = 25,
25 —= 32 uw—l
and @ = ——"# = 12 —y— 4,
y—1
Assume % ak whole number = m ;
oe y == 277 + l,
and #@ = 11 — 3m.
And since 11 — 3m is a positive number, 3m must be less
than 11, and .*. m less than 4.
mem = 0, then y = 1, and = 11, .*, the parts are 22 and 3;
pa I, Buia, v= 8, .. the parts are 16 and 9;
m = 2, jo", %=5, .. the parts are 10 and 15;
m = 3, jem e e%=2, .°, the parts are 4 and 21.
And these are the only divisions which can be made.
30. How many ways are there of paying £7 with crowns
and seven-shilling pieces ?
| Let 2 = the number of crowns, ‘i required to pay
and y = the number of seven-shilling pieces, the sum.
| then 52 + 7y = 140,
2
and « = 28 — y — ~4.
ca)
Let e ce i i
2 Y= 5m,
) and wv = 28 — 7m.
28
And as m cannot be = Ree 4, there can be only 3 answers,
or 3 different ways of payment.
284 Examples of the Solution of Indeterminate Equations.
Suppose m= 1, then #= 21, and y =5;
Mm = 2, 24a oie
m = 3, L=7,; y = 15.
31. In how many ways may £80 be paid in sovereigns and
guineas ?
Let # = the number of aire required to pay
y = the number of guineas, the sum.
then 20% + 21y = 80 x 20 = 1600,
and #@ = 80 —y—<.
But i. = whole number = m;
fai = 120 ans
and # = 80 — 21m.
And since 80 — 21m must be a positive whole number, .. m
80 |
must be less than ore and cannot .*. exceed 3. There are .*
only three ways of payment.
Suppose m.= 1, then # = 59, and y = 20;
M.= 2, L = 38, ii 40%
m = 3, &= 17, y = 60.
32. Can £100 be paid in guineas and moidores?
If it can, let x and y be the numbers of each.
then 214 + 27y = 2000,
2000 ee UE akin 6y — 5
21 paca? ye
and #7 =
But “y—* = whole number = m;
(es
oy =3m +
3m 5
Assume t:
which is impossible; and .*. the payment cannot be made.
Examples of the Solution of Indeterminate Equations. 285
33. In how many different ways is it possible to exchange
11 bullocks which are worth £12 each, for sheep which are
worth £2. 5s. each, and pigs which are worth 12s. each?
Let x and y represent the numbers of sheep and pigs which
must be given in exchange ;
then 452 + 12y = 11 xX 240,
or l5v@+ 4y =11 X 80 = 880;
v
age Dee iar titer: 3
L
Assume ae
. &®=4mM,
and y = 220 — 15m;
_ where it is evident that m cannot exceed 14, nor be less than 1;
there are .*. 14 different ways of exchange.
Let m= 1, then 2 4, anda = 905;
M2, aks. Te
and so on, the values of x increasing by 4, and those of y
decreasing by 15.
34. A company of men and women club for the payment
of a reckoning; each man pays 25s., and each woman 16s.; and
it is found that all the women together pay one shilling more
than the men. How many men and women were there?
Let x = the number of men,
and y = the number of women ;
then 16y = 25% + 1,
ov + L
and y = @
y TE Ek 16
Gb
Let ae = whole number = m;
<3 9v = 16™ rma 1,
21 + 1
9 .
and # = 2m —
= whole number = 7;
9m-+i1
| Assume = eoroe
286 Hxamples of the Solution of Indeterminate Equations.
then 2m = 9n — 1,
n—1
and m =4n +
n—1
2
then n = 2p + 1,
m=9p + 45
whence @ = 16p + 7,
and y = 25p + 11.
Let p = 0, then # = 7, and y = 11; the least numbers which
answer the conditions. Other values of x and y may be ob-
tained by assuming different values for p; the number of men
increasing by 16, and the women by 25.
Let
ie
35. A person distributes 4s. 2d. among some beggars, giving
7d. to some, and a shilling each to the rest. How many were
there ? .
Let 2 = the number of those to whom 7d. was given,
y = the number to whom 1s. was given;
then 7% + 12y = 50,
50 — 12y sy — 1
ape eee re
and # = 7—Yyr-
But = = whole number = m;
. by=>7M+4+1,
2m + 1
and y = m + ———_.
2m + 1
And to ae hole number = 7;
n—1
mM =2n +
But “—" = whole number aie
9
. m=2p4+1,
and m = 5p + 23
whence y= m+ 2n=7p + 3,
and #7 =7—y—m=2—12p.
Examples of the Solution of Indeterminate Equations. 287
And as 12y must be less than 2, no whole number substi-
tuted for it will answer the conditions: but if p = 0, 2 = 2, and
y = 2, the numbers required.
36. A wishes to pay a debt of £1. 12s., but has only half-
crowns in his pocket, while B has only fourpenny pieces. How
may they settle the matter most simply between them ?
Suppose 4 to pay # half-crowns, and receive y fourpenny
pieces ;
| then 30% — 4y = 32 X 12,
or 152 — 2y = 192;
v
Bi Ick rm OO ree
e
Let >= m;
“Tr
+ =a,
and y = 15m — 96;
whence m must be greater than 6.
Let m=7; .. v= 14, and y = 9; the smallest number of
coins which will answer the conditions.
37. It is required to divide 24 into three such parts that if
the first be multiplied by 36, the second by 24, and the third by
8, the sum of these products may be 516.
Let x, y, and z be the three parts;
thenv +y + 2= 24,
and 364% + 24y + 8z = 516,
or 9% + 6y + 22 = 129;
but 22 + 2y +22 = 48;
.. by subtraction 7” + 4y = 81,
and y = 20 — 2% + ar
Let
+1
. {Pe + 4b eee eal, to find the
2° |3 3 2° |\@
J
value of 2.
ANS. & = 3.
10— 32
: XB — 12% 2— 62x aa 4
Given ——=2 — ——_—- = x — ———______, to find the
2 13 39
value of x.
ANS. @ = 11.
sere Ce 2 ;
Given ~~ pe src eek le i —_ ed ets aire
33 yh B4 |
value of 2.
ANS. = 2.
2
Given ~ as cute Coen —, to find the value of #.
4e°7—1 1+2”~— 4’
ANS. 2 =>— =
4
hacia ba = — 3b + (d + 6). x, to find the
be a fi :
value of 2.
(d? — ab). bcf
ANS. 2 = 5 Ghee — abcd
300
39.
40.
41.
42.
43.
44.
45.
46.
Simple Equations involving only
Given = a “ + oy — g =/h, to find the value of z.
nk aie ae
Given oe, — dc = bx — ac, to find the value of 2.
Ans. @ = * ee Ce g
eee 7
Given = — —=*= +o 4 to find the
Os
ANS. &@ = 6.
9v+20 4¥%—12
36° 52 — aA
ANS. @ = 8.
Given + , to find the value of 2.
20% + 36 52+20 42
: 86 .
Given ei fa eon te + on to find the
of x.
ANS. 2 = 4.
Given 10% + 17 me 120 + 2 — Neri to find the
18 132 — 16 9
of x.
ANS. @ = 4.
18H — 19 LIP 2le Soe 15
Given ——_—___ = , to find the
28 6@ + 14 14
of x.
ANS ei.
‘ 20% + 84 13” — 2 nh q
Given 22s ertag Bee +-= {LED soe to find the
9 17@7—32 3 12 36 ee
value of 2.
ANS. @ = 4.
value
47.
48.
50.
51.
52.
13.
one unknown Quantity. 301
7@ +6 +42 vf 118 #&—3
Given + —- = — — —.,, to find the
232 — 6 4 21 42
value of x.
ANS. & = 4.
11
, 6 — 52 — 22 1 32
Given SPREE I 56 ae aA bay SE
15 14.(# — 1) 21 6 105
to find the value of 2.
ANS. 2 = 4.
97 — 3 37 — ]
: x 3 4 ey 1 2
een ree ee ee toe indaethe
2 £—1 9 3@7—2
2
value of @.
13
ANS. 2 = ae
Given ee =ac+ a to find the value of z.
ANS. & =
c
: cam dam
Given Fis fae aa a fae , to find the value of 2.
ad—ce
ANS. 2 = cf—bd’
Given -_ Fare : era +2 = = k, to find the value of z.
adfh + bcfh + bdeh + bdfg
NS bdfhk
Given (a+ 2).(6+2)—a.6+0 =< +2", to find
the value of z.
ANS. 2 = -
302
54.
55.
56.
576
58.
59.
60.
61.
62.
63.
Simple Equations involving only
Given we : —— > ++ 14: 5, to find the value of x.
UN ieee
Given ee! : aa —2”7::5:4, to find the value
of x.
ANS. s2.8.
Given 16% + 5: Tr °$ 362 + 10: 1, to find the value
of x.
ANS. 2 = 5.
42 +3.
6x2 — 43 ©
of 2.
ANS. & = 8.
Given 1:32” +19: 3x2 —19, to find the value
Ee en 10z” — 18
Given 52 = eae
ch Fale) 27 +3
, to find the value
of x.
ANS. &@ = 3.
Given (/ (107 + 35) — 1 = 4, to find the value of 2.
ANS. &@ = 9.
Given \/ (97 — 4) + 6 = 8, to find the value of 2.
IANS eat Ae
Given \/ (v + 16) = 2+ 4/2, to find the value of 2.
ANS "9.
Given (/ (2 — 32) = 16 — 4/2, to find the value of z.
ANS. # = 81.
Given \/ (4a + 21) =2°/@ + 1, to find the value of 2.
ANS, & = 255
64.
65.
66.
67.
69.
(70.
one unknown Quantity. 303
Given ay/ (bu —c) =d./ (ex + fx —g), to find the
value of 2.
ac—d'yg
@b—d’.(e+f)
ANS. &2 =
Given «/ (a + ¢ ee ak: ) to find the value of x.
(2 + b)
ANS. dealt WAC ian
Given / ( (a + 2) = A/a? + 5ax@ + 6’, to find the value of zx.
Se ee ail)
3a
Given a + 6.\/ (x + d) =c, to find the value of z.
C—a m
ANS. @ = (
“aed 18+ ox
Gi , to find th ] f
Oy rare Tak 40 o fin e value of 2.
ANS. 2 = 4,
Given va + at 3? to find the value of z.
Va2— fb
2
mane
a—b
Vor—2 4 f/6"—9
Ser+2 4/6r +6
ANS. & = 6.
Given ~ to find the value of z.
Given “29 — 1, = ¥ 8* => to find the value of z.
/5t + 3 2
ANS. & = 3.
304: Simple Equations involving
y A,
72. Given ing wi nian ad 2, to find the value of 2.
3 64—3% 3 #-—2
14
ANS. #2 = —.
13
812° — 9 a 3 2n7'°—) 57 —aee
- 0 ares 2 @
73. Given Tk WW abe EMS 2
; 2 (3@—1).(@ +3)
to find the value of 2.
ANS. & = 10.
74. Given \/{1+2/(2’ + 12)? =1+ 4, to find the value of a,
NN Gees
2 )
75. Given So (ca +d’) + a = ex, to find the value
of a.
ON ier ey.
Pee cli be a’ d
2abce
— 36
» Gi — 9) = ———_.,, to find th faa
76. Given /2+4/(4 — 9) 7 ey) o find the value of #.
ANS. & = 25.
77. Given = ff (v? + 39% + 374) — f/f (wv + 20% + 51)§
ae & + 22
a7 ic % *) to find the value of 2.
ANS. & = 78.
Il. Semple Equations involving two unknown Quantities.
1 GIVEN@ + 1y= 53, |
andy + 3x = 27,{
fe = 8,
ANS.
NS ‘te S
to find the values of # and y.
we
ot ge + 10, |
o
two unknown Quantities. 305
Given 47 + oy = ey) to find the values of x and y.
and sv — l3y = 9,
[xv = 6,
ANS.
ly = 3.
Given = + 2 = 6, |
to find the values of x and y.
and — + a= 52; |
ie fe wee
ly = 16
Given = + 8Y = 194,
to find the values of x and y.
and 7 ew —= 131,
ANs. Ley
ly = 24.
Taped
eae dbs to find the values of wv
3Y — 5 and y.
and +22 — 8 =72,
ANs. ee
y= 5.
| to find the values of w and y.
ANs. le
ly = 10.
306 Simple Equations involving
|
7+o 27 y _
° Gi eo ke as eS a
d nesne t: 4 34 — °> | to find the values of #
5 —: |
and -4 fy “ta 18 — 52, one )
|
ANS. t, we :
WS
syt47 _ | _ oy +33 |
ale oe 14 ’ | to find the values
Bom AY a Deel Y Sale of x and y.
er i a eee eo
s. Giveney+1—
and y — 3 —
ANS.
: 15 — &@ 7H+11 |
9. Given 4a 4+ ——— = caer ne |
- 4 etmoene Neos eae the values
Bey ee tice | of x and y. |
ANS. Eons |
ly = 4.
|
10. Given 4 — al +17 =5y + ae to find the ©
22—6y 5x 21 sy +5 values
= =i |
d Pees SR GAGn ee ice Ts xz and y, |
ANS. ie ‘
ly
3
20 PY | 90 = BY AD), AU SY
2 esr eee 16
11. Given wee Bde aT i A es | to find the
5
values of
| wand y.
ANS. se He
yea
15.
' 12.
114.
two unknown Quantities. 307
ek
b+y 344 2° | to find the values of 2 and y.
and av + 2by =¢, |
_ 26°— 6a’ +c
5 3a :
_ 37 —h +e
oer 36
Given
&
ANS.
Given
12) +--6 BUS UE emai 3Y —2&
fat sess = 39 — ——~ — t "|
and 32 + 4: 2y —3:: 5:3, |
to find the values of 2 and y,
Ogee
ly =9.
° 1 —— — i
a, SEE A eo a Rd Adee 8
2 6 3
Y+7 . 3y—8
and fam ee ees
to find the values of x and y.
ANS. Ne ome
ly =4.
Given’? +y:4¢+Yy:24:7, to find the
x
en) 2 + — values of
and. ————— = i xz and y.
Given Sood tS ee Yi FSS
10 15 5
and Yt 5%@—8 _ @ty _ 7e@ +6
12 4 11
to find the values of x and y.
ANs. ie fa
Yor oe
x2
308 Simple Equations involving
4y—17 +2 15 —3H7_ 12y +11
i7. (Given 13% + = : :
lo@ + 7y + 28
“SEN pray?
ue ow eS 12 oh rOy iia AY So te oe
4 5 8 15
to find the values of x and y.
ANS. | ee q
Beye sR: |
|
— 2b). |
18, Given 32 + A ek Urea |
a? — 6’, |
and ee — 2 +(atb+c).by=e + (a+ 2b) ab,
to find the values of x and y. )
20 + Yy 74 + 6a 411 yi 3 WY)
—_——— + URS Se ee I
18 6 .
sat 3y+2, 9y +6 |
; Ranger
to find the values of # and y.
ANS. ae
ly =4.
19. Given
and. lise
32 —5y 247—8yY—9
1
90. Given — mae
12 prs
and = + 24 1}: 4 —F— 24 1238 2 3h
to find the values of # and y.
Auteur
ly = 4.
21.
22.
23.
24.
25.
two unknown Quantities. 809
Given (v7 + 5). (y +7) = (w+ 1). (y—9) + 112, }
and 27 + 10 = 3y + 1, i
to find the values of 2 and y.
fe =3,
ANS.
ys
Given 6x + 9 3U-T5Y = 34 + hpi | to find the
4 42—6 2
Hp it toesy oes values of
of Mehvsae eae ;
an -e Ley : 4 +4 5 {| x and y
ANS. oe
ly=9
Peenidy — sap — <9 4 137 _ to E8 | to find the
27 — 6Y
values of
21—4y 184+ 13
ad3 et ee es OF T dy.
an “ana : 21, wand y
= 7
ANS. :
ie att
ae 2 :
Given 167 + 6y—1= ebilae EER ay to find the
8H —3y+2
ee, values of
and eS ee ee wand y.
20+ 2 +3 3@ + 24 —]
ANS. [ee
ly = 5.
oe ely ly
2
G; 2 __ l6u*+12V7y—8x+5y +28 |
Mel an a ee igo ee
Bertie t oe es 3y pos ey te 108 |
40+ 6y +3
to find the values of x and y.
ANS. (as,
ly =2.
310 Simple Equations involving
A — 247"
26. Given 34+ 6y+1= SR SOS \ toe mae
2N—4yY +3
ata | Bye values of
and by i heir asic hss DL ak, xv and y.
4y—1 3y —4
ANS. [eases
ae
40° —y.(@ + 3y)
2—Yy+a4
and (2 + 7).(y—2) +3=2xey —(y—1).(@ + ));
to find the values of x and y.
27, Given 3y + 11= + 31 — 4a, |
Ans; doin ©?
oa 3.
y 4 7}
+5 -y—2 [
af 3 | Aone tn
28 ey he mAs orale tented ai (x y) |
10
mit
ANS.
Peps
I= 2
y lle
oa — +6
’ &—6 4
20. Cie SEE a Oe SN a
7y 24 6 42 56Y
13 86 14
and 12” — 15y + —-: 1oy — 8@ + — 3:93 —9@ : 6a@——,
to find the values of # and y.
ANS. . te
malt
eS ”
two unknown Quantities. 3ll
72 3y +6 347 —2
Me Wp evaHIE NI 9 — By +,
30. Given — : a ts ae
| 5 8 16”
32 2y as
and —+—~+2):-—#+4-::10i:11
ee ait “2 SG a cls
to find the values of x and y.
ANs. [ares
Jee
; 40 —2 Se Nis a an 4s '5 x 1
1, py SEERA Sl all, Badass Sue IAD PE
3 7 4 5 rf
| gy US ee 1
a :y—2 55 oS ae ere ee ee
and 27—y+15:y—2"+4 MEL pe oe eo on Gee
to find the values of w and y.
ao jv = 18,
ly = 24.
ay — EF 2 “eu + low +13
32. ities acm Gk a
RM ey ah gyi tae
to find the values of # and y.
ANS. - ma
y = 2
3. Given «/y — \/ (y — 2) =4/ (20 — z),] to find the values
and 4/ (y — #):4/(20 — 2) ::3:2,f ofwandy.
ren fe =e
Ly = 25.
312 Simple Equations involving
34. Given? + y= a4, |
v + 2= 0,7 to find the values of x, y, and z.
yYytez=e, |
[eat @+ 0-0)
Ans.jy =4.(@—6+4 0),
aero aie tina
35. Given” —y— z= 6, | |
3Y —e@—-Zz= a to find the values of x, y, and z. |
72 —yY— v= 4,
ease
ANS. 4 ¥ = 21,
zg 12.
36. Giveny+ y— 2=8,
2% —- y+ 3z= a to find the values of x, y, and 2,
42+ 3Y — 2% = 17,
see
ANS. (Y¥ = 5,
eee
S7. Given = + —= 2,
yey aS
1 1 3
ia to find the values of w, y, and z.
1 1 7
yt ea
38.
39.
40.
41,
42.
three unknown Quantities. 313
Given @ + 2y + 3z= 17,
y + 2z + 3x = 13, ; to find the values of 2, y, and z.
1 ul
(eaee 1,
ANS. Iwo
“= 4,
Givenv— y+ z= a
8“ —4y + 2z = 50, ; to find the values of x, y, and_z.
We%—oy+3z2= All
ANS. i = %
ie = 364.
Given 30 — y+ 7z=15,
5x2 + 3y — 22 = 16,; to find the values of a, y, and z.
7@+4y —52= 11,
aa 4,
ANS. yY = 2,
<= 5.
Given 2v + 4y — 32 = 22,
4@ — 2y + 52 = 18, ‘ to find the values of @, y, and z.
On -- 7Y — Gee
fe 35
IXNAS 4 = 7;
Nevah
Given 327 + 2y — 2 = 20,}
2v+3y+t6z= yy find the values of x, y, and z.
e— yt 6z= 41,
(aes
.
-_—
ANS. °
314 Simple Equations involving three unknown Quantities.
43. Given 7@ + l2y + 42 = 128,
32+ 3Y + 72 = 60,
6x2 + y+ 52 = 68,
to find the values of x, y,
and z. |
Peet
ANS, Yy —= 5, |
ie = 3. :
44, Given 62” + 3y — 42 = 22, :
4x — y + 6z = 20, ; to find the values of 2, y, and z.
52 + af 6Zz2= wel
ya
bara:
ANS. heise
\2 = 2.
[
45. Given 11a — 1oy = —2 =, :
to find the values of 2,
V+Z—2yY z2—-y-!) :
ye pe eae y, and 2.
3V =Y+Z2+7,
ae ]
ANS. 4y = 1), |
lee |
sou aera ip ee, ; |
46 CNL oe pd ig nh ots ABE
x Zz .
—+%—== 23, to find the values of #, y, and 2.
je EN ct |
Fas ahs a
[Sey 420s
ANS. 5 ¥ = 60,
panies
2
3.
or
e
6.
iy
Pure Quadratics, &c. 315
III. Pure Quadratics and others which may be solved
without completing the Square.
GIVEN 32” — 4 = 28 + 2’, to find the values of z.
ANS, & = 43
Gj jess
wen’? +ysy: ey to find the values of # and y.
cha ee g,
ANS jem &
“ly =+3.
Given @—yiyii4:5,|
and a? + 47’ Cala:
[est
ANS. i
ie ah,
to find the values of # and y.
Given2 +yi:ev—yiia:b,)
and ay = c’,|
v=e./ (4+),
ANS.
to find the values of w and y.
Given a’ + yi 2? —yiti7: A to find the values of x
and x#y’? = 45, and y.
ANS Feary
ys
Gi — 2Y = 6
Od “td +] to find the values of x and y.
and ry — y? = 18,{
hep el to aaen
Sani kE
Givneg+y:v2—yil: 4,| to find the values of x
and ay = 21,f andy.
fx = 7, or — 3,
ANS.
ly = 3, or — 7.
316 Pure Quadratics and others which may be solved
gs. Given az + bay =c',|
And @.— YoU Lt Mes Bl
Ge es ghee
n— m
Le BO Crees,
9. Given #* + y*°: 2 — y° 2: 559: 127,] to find the values of
. and a’y = 294, # andy.
to find the values of # and y.
ANS.
ANS. Pea
ly = 6.
10, Given a — wy: xy —y’ ::3:7,) to find the values of 2
and ry’ = eh and y. {
ANS. [ae
eee |
1. Given /@ + /y: Se — VY 3: 4: 1,) to find the values
and 2 — y = 16,f of v and y.
[uv = 25,
ly =9.
12. Given x — Sy = —
and «/@ 4- ES ah to find the values of w and y.
[vi 625, |
ly = 16.
ANS.
ANS.
13, Given — i Ay 8 1] to find the values of
and «/w7y =15,/ andy.
{v = 25, or 9,
ANS.
ly = 9, Or 25.
14. Given #* — x’; xy — xy’ +: 7: 2,) to find the values of #
andw+y=6,{ andy.
Neca fe =4, or 2,
ly = 2, or 4.
without completing the Square.
; Weeks ws)
15. Given - +-= 7
y to find the values of x and y.
2
and — = 2
x 9
eee [@/==16, or 3;
16. Given z‘ — y‘ = 369,
ave : y ’ 95] to find the values of wv and y.
and 2” — 7’ = 9, f
tae Mi e= =
ly=a4.
17, Given a’ — 7° = 56,)
|
16 > to find the values of # and y.
andw—y=-, |
LY
oo [v= 4, or — 2,
ly-=2,0r— 4.
; eat hE
18. Ses Sara mesa) aie to
the values of 2.
ANS. @ =
i
2
19. Gi . OS aby
meee ey a9 Die al to find the values of x and y.
and vy? + Y= 14, |
2
le = 5 or2/2
ANS. - { > ee
20. Given Viz+V7y = 6]
and # + y = 72,
== O40 bes
ANS. J eke
ly = 8, or 64.
317
find
318
21.
22.
23.
A
26.
Pure Quadratics and others which may be solved
Given 42? + a Ag ah 16Y, |
2 ey to find the values of # and y. |
iL ober ene |
ANS. frat st |
ial |
Given pada aN Seg + HOE tied ie, ax, to find th |
1 2 + /(2—2") e— fee): 3 |
values of 2.
Ans. o = ./ (41),
|
|
|
|
| |
Given VAC ar = 4, to find the values of 2.
ab
ANS. 2 = + Vb +1)"
Given \/ (}2 + 2) —V/ ($e —2) =o (w@ +3) —VS/(@-3),
to find the values of z. '
Ans. ®=2+5. |
Given 3o/f (49-2) +3 SA Y—2) =A (Ry — 2), 4
and § / (2° — 6y) + of (y* — 92) 3 of (#* — by) 2219 1)
to find the values of # and y.
|
:
ANS. [a= 7,
y = 8.
|
i ¥ 4 lues of
. 8 va |
ST Ng) ALN a
40 y 4° x 9 Ys |
pipette |
ANS :
28.
29.
32.
Tiv
iar,
without completing the Square. 319
Gi 3+ y3 = 20
aia 7s i! to find the values of # and y.
and 23 + ys = 6,
Jukes (vous, or + \/3,
' Ly = 32, or 1024.
Given a* + 2a’y’ + y§ = 1296 —4ry. (@ + ry 4+ ¥’),|
and #7 — y= 4, J
to find the values of # and y.
eee {v= 5, or — 1,
= 1,0r— 5.
Given (a? + 1) . (vi — 1)? =2. (# + 1), to find the values
of 2.
2b 2
ANS. @ = (CS) :
Given Va—/ (4-2) = a, to find the value of z.
Jat /(a—2)
2
ANS. & = a ;
(a +1)
Given Va aaa at =4,] to find the values of x
and fa: Syii Vy i 4 |
f 625
Ans. ! 16”
y = 25
ieee
Given —— —= -
2 \ to find the values of # and y..
and wy — xy’ = 16,
jen se=4, or — 2,
ly = 2, or — 4.
3820 Pure Quadratics and others which may be solved
V (4a +1) + V4e I to find the value of a. ;
J/ (42 +1) — i Peas th
33. Given
4
ANS. % =-.
9
fick Bat vou! Rae ae
at+ae—/(2ax + 2’)
(6 = 1) |
TGA
|
:
34. Given to find the values of a.
ANS. @=+t
1 1
35. Given (S a 2) +- i 2) ates 4a a 2 , to find the
27 — 3 94 +3 13 42
values of 2.
aii
ANS. & = —.
14
es » ind
36. Given V(@—y) +3 @+N=FE— | of
and 2° + y?: wy 3: 34:: 15, wand y.
ANS. te ae : |
y= 3.
37. Given 2‘y’? — 2°y' = 216
, : an >| to find the values of # and y.
and zy — ary’ =6, J 8
{v= 3, or — 2,
ANS. :
ly = 2, or — 3
38. Given 2 + @q/ (vy’) = 208,)
and y* + y ¢/ (ay) = 1033,
ANS ee = 8
ly = 27.
- to find the values of # and y.
39. Given #3 + viyi+ y= 1009, | to find the values of @
and 2° + ay3 + y° = 582193, | and y.
ee) fz == §1,forelG;
ly = 16, or-8t.
40.
41.
a
42.
lisa
without completing the Square. 3821
Given 2 + y’ + vy. (v7 + y) = 68,] to find the values of
and 2° + y*°—3a°=12+3y,f wandy.
pen {v= 4, or 2,
ly = 2, or 4.
Given vy . (v7 + y) = 84,
pnd 23? . (2* +4") 2600, | to find the values of # and y.
we {w= 4, or 3,
"ly = 3, or 4.
ete YY,
etven —— r a
Y to find the values of w and y.
and = = ba pious
a
ANS. {z sii?
ly = 3.
43. Given / e + 3) —/ <— 3) — (2): to find the
values of x.
Ans. =+9//2.
4. Given f/x —SYy=V2. (fat Vy)» | to find thevalues
and (v7 + y)?=2.(@—y)’,{ of # andy.
ANS. a
ly = (3-202).
Y
alm — n)? + g-imn
Given fees iano *, to find the value of 2.
ze 1
A Sat Use
a ] \(m +2)?
ANS. & =[— a :
at —=1
$22 Adfected Quadratics involving only
46. Given y +30/y. \./(a+ b2) —V7y). ve + ta) aa
30/y. weal a— bx )
3 Fata a— bx
Cera rua / ( )s
to find the values of w and y.
_a—1
ANS. Fs hou
y = §0/(2a +1) + 12°.
\ f
and AT
IV. alerted Quadratics involving only one unknown
Quantity.
a. Given v? + 4v = 140, to find the values of 2.
ANS. C= 10, or — 14.
9, Given 2 — 6x + 8 = 80, to find the values of 2.
ANS. x = 12, or ae 0.
3. Given a — 10% + 17 = 1, to find the values of 2.
ANS) ot 55,0102;
4. Given 2 — x — 40 = 170, to find the values of 2.
ANS. # = 15, or — 14.
5. Given 32? — 92 — 4 = 80, to find the values of wz.
ANS. & = 7, or — 4.
6. Given 72? — 21” + 13 = 293, to find the values of z.
ANS. & = 8, or — 5.
. ee eae
7. Given Py ee Te 154, to find the values of x.
5
ANS. 2'= 9. or —
one unknown Quantity. 323
| , 22° L
gs. Given ie +3i= s + 8, to find the values of z.
9
ANS. @ = 3, or — a
io, Given? + 4+ ae = 13, to find the values of z.
ANS. @ = 4;0r' — 2.
. 36 — @
10. Given 47 — ae tee find the values of z.
3
ANS. av = 12, or — re
me Given 16 —
5—
— Bo enh + 3a, to find the values of x.
Lv
6
ANS, = 35 or 5
“+3 16— 22
12. Given + ——— = 51, to find the values of z.
2 20 — 5
6
ANS, # = 5, or so
10
; : 40
13. Given 14 + 4% — —— = 3@ + oe to find the values
of x.
ANS. & = 9, or 28.
+4 —v 42
14, Given —— = Meh wists
— 1, to find the values of z.
ee Sl i a
net 21 OF oe
; 15 — 2 12—3@ 23x@ + 60
15. Given —— — ~~ = 7x — Ef to find the
4 40 — 5 fi
values of 2.
229
ANS. # = 3, or —.
148
324:
16.
17.
18.
19.
20.
21.
22.
Adfected Quadratics involving only
: e+ill 9o+42
Given Te eae ss ae 7, to find the values of z.
1
ANS. # = 3, or — 3
; Lk 42 — 3 34” — 16
Given DEEN NO Nino erica es + —, to find the values
9 42 + 3
Ole
1
ANS. i = 6, or ms WH “.
Given ———— = saath, to find the values of 2.
+60 3H%—5
ANS. # = 14, or — 10.
, 32 — 42 — 10
Given EAS Sagres ame 31, to find the values of 2.
YU+5
10
ANS. & = 7, or — ae
Given — ——~ = 21, to find the values of 2.
vL—1 22
4
: 8a 20
Given — 6 = —, to find the values of 2.
L+2 30
2
ANS. #& = 10, or — i
: 40 27
Given z a aa 13, to find the values of #.
15
13
23.
24.
25.
| 26.
27-
one unknown Quantity. 325
5X2— 12 3H — 24 (a 34
Given ———— + ——— = 9 — ——_,, to find the values
42 —12 15
Uline.
4
184
; 2” — 5
Given + — = 81, to find the values of 2.
—4 L— 3
0
ANS. # = 6, or —.
: x LX
Given eel tied 1, to find the values of 2.
10— 2% 25—34 :
WANS? @ = 8, 0r 1322.
Pe: 42 —5 32 — v + 23
«yd SR al kite, to find the values of x.
YH 302 +7 132
154
ANS. C= ee 25 or Pe) ae
45
: 8x" + 16 122 — 11
Given 2x7 + 18 — Peas eerie 27 — ——_——., to find the
42 +7 22 —3
values of 2.
ANS, 2== 8) Ors,
rye ax’ “+ 20 20
Given — ae ac naa pears to find the values
C+9 20%+18 2
of x.
ANS. & = 4, or — 2,
Given —— eas + cabal Sabeae to find the values of x.
L+6 27 + 4 3H +4
2
ANS. & = 8, or — A
326
30.
31.
32.
33.
34.
35.
36.
37.
Adfected Quadratics involving only
Given ———— eee eges Lara to find the values of 2.
ov +3 5H@+18 5x
4
ANS. & = 6, or — 2.
: 82 —1 4
Coven ee ie a ee etme to find the values of 2,
9+5”0 2+4” +412 |
137
; 8 32
Given “ & , to find the values of 2.
5—-B@ 4-2 #42
ANS.@ == 12, DF sh
13
, 2" — 1 — x
Given ee kash al + z to find the values of z.
3-2, Wael 2 :
it
Z 3 6 11
ee he ee oe CO TIN the Values Ones
ee én — at ee+oe 5a" i
ANS, # = 3, or 7
40° +7H 5e—2' 4x?
Given —_—_ = , to find the values of 2.
19 3+2 9
ANS, & = 3, or — mi
10
“ton +8
Given A Ree nr = 2 + x + 8, to find the values of 2.
ve +e—6
14
ANS. @ = 4, or — ry
V+ 12 &
L+-12
ANS. # = 3, or — 15.
Given 5 =, to find the values of 2.
38.
39.
40.
41.
42.
| 43.
44,
45,
one unknown Quantity. O27
Given \/ (42 + 5) x «/ (7@ + 1) = 30, to find the values
of x.
ANS. Pe oe
28
VE +9 _ V9 — 37
Given = , to find the values of z.
re 9— Cie
ANS, # = 25, or Sida
400
Given ~ aoe <4 to find the values of 2.
L— Vi i 4
—3+f/—7
;
PONS. e == 4, OF 1, OF
2—/ (ex +1)
B+ / (e+ 1)
8
ANS.” = 8, or rer
Given = =, to find the values of #.
; 32 —1 2 -
Given 5 . ——__~ + —— = 3/2, to find the yalues
+5fu Va ’
of 2.
1
ANS. 2==¢1,-0% 3
Given (/2* — a = 32, to find the values of wz.
ANS. # = 4, or (— 5)3.
Given 23 + 7v3 = 44, to find the values of 2.
ANS. # = + 8, or + (— 11)8.
Given 4x3 + # = 39, to find the values of #.
is \*
ANS. & = 729, or (o>) :
328
46.
47.
48.
49.
50.
51.
52.
53.
Adfected Quadratics involving only
Given 32° + 424° = 3321, to find the values of 2.
sos
ANS. @ = 3, or — 41.
Given -- +2= ae to find the values of x.
2
ANS. @ = 4, or ee
4
Given x3 + ane se 3 + x8, to find the values of z.
Lv
ANS. # = 4, or (— 7)3.
6/are ane many a
Given yf = ee =. = “v=, to find the values of 2.
27
ANS, e= 1, or — nS
Given 32" 2" — etal = 4, to find the values of 2.
Vx"
ie aS ( i es
; = {8)2n — — }2n,
NS. & = (8)2n, or =)
mS — 3 cs ee
aVa—ah _ W+3Ve— 37 1, oa the value
Given a
2+ 2 2/i—3
of x.
ANS, # = 4, OF. <5.
Given 2%. (v7 + a’): = 22". (4 + 2a) + a’. (x — a), to
find the values of x.
a
ANS. & => or — a.
Given adx — acxz’ = bcx — bd, to find the values of 2,
87.
|
54.
55.
56.
58,
159,
60.
one unknown Quantity. 329
2 rn2
Given < — ° +> as = 0, to find the values of 2.
ANS. 2 = — a) bsy@—0)
a
Given 9a‘ b*x’? — 6a°b’x = 0’, to find the values of z.
a’? 4
ANS. “= NaS VAN a
3a 6
Given (a + 0) .2° = cx a S55 to find the values of x.
eee ck + se)
.(@ + 4)
Given 3 // (112 — 87) = 19 + / (34% + 7), to find the
values of 2.
ANS. & = 6, or ie)
625
Given (/ (22 +7) + (34 — 18) = / (7% + 1), to find
the values of 2.
ANS. & = 9, or — =.
Given 7. / 5) — Je + 45 = Gv (ioe +56),
to find the values of 2.
14568980
ANS. & = 20, or sarah
Given
16—46/r 88+ 33/@ 2? — 5x” +11
SEE ee * @—3V2). (+ Vay
to find the values of z.
PENA tae 935,01 (7.
330
61.
62.
63.
64.
65.
66.
67.
68.
69.
Adfected Quadratics involving only
Given
54-0 a _ 230 — Vv, "2 —3u +4
L+24/z 6+ f/x (v7 +2/x) x (6+ Sa)
to find the values of 2.
32
ANS. & = 5, or — —.
15
Given a2 + /e7i:7—-Srir3Vet+e: 24/2, to find the
values of 2.
ANS. # = 9, or 4.
Given 2 + 11 + 4/ (x + 11) = 42, to find the values of 2.
Ans. v = +5, or + V/ 38.
Given (7 — 5)° — 3. (v — 5)3 = 40, to find the values of a.
ANS. & = 9, or (— 5)3 + 5. |
Given 2 + 4/(# + 6) = 2+ 34/(wx + 6), to find the values
of x.
ANS. # = 10, or — 2.
Given (x + 5)? — 42” = 160, to find the values of a.
ANS. @ = + 3, or + «/ (— 15).
Given x? — 72 + 4/ (x —7x + 18) = 24, to find the values
of x.
, ak
ANS. £2 = 9, or — 2, oh tse
Given 9” — 427 + / (42° — 97 + 11) = 5, to find the
values of 2.
oe te
ANS. £& = 2, or be or a
Given a + 4/ (5% + x’) = 42 — 52, to find the values of a.
—54 221
one unknown Quantity. 331
2 J (#@ +2) _ 17
came ‘2 regan Choe + CACeeyy. to find the
values of x.
. Given
3
ANS. # = 6, or wary
a L 4 21
Given ee a. Se a ae to find the values of xv.
a
ANS. # = 12, or — 3, or ae
eet to Se eG
Given oe on — = aa to find the values of 2.
ANS. # = 7, or — 3.
- Given 22 pimp ® erases ——., to find the values of x.
3H—5 3H +5 a
pores Bey
FSi hg NT OF ee
“v+@r—4
Va
, to find the values of 2.
Givne + /rt+e=
ANS. @ = 4, OF. 1.
2
_ ; i eee , to find the values of 2.
w—4 258
39
; 2
Given (« + =) +e=42— to find the values of 2.
—7+AY 17.
2
WAM Mer 45 OF 2, OF
332
ree
78.
79.
80.
8l.
83.
Adfected Quadratics involving only
Given v7 + 4—2 a) 52. = ——, to find the values
of x.
ANS. Qpeeet tot Olat Aa Lys
Given iia + J («" ~ 2) =, to find the
values of x.
Given \/ {(v — 1). (# — 2)} +/{(@ — 3). (#-—a)} = V2,
to find the values of 2.
TNS xe ay OLE.
Given z*. ( + =) — (3° + x) =70, to find the values —
of 2.
aca, sa
ANS. #@ = 3, or — an or alls Vas |
3 6
Given 2? — = +] es a ae to find the values of 2.
naan Ge a
ANS. # = 4, or — 8, or ate (on),
Given Wa ra 92") a P ee sal find the values
of 2.
ANS, = 55, or a Y (15661)
Given : ;
e@+ir—s 2+27—8s 27—13%—8
find the values of 2.
DNS) imeet fOr als
84.
85.
t
86.
b
87.
88.
39.
}
0.
ide
}
one unknown Quantity. 333
Given 3. {(v — 1)’ — v7}? + 24 = 341 + 2. (x — 1)’, to find
the values of 2.
34/3 +4/(— 109)
24/3 ;
Given 2 — 243 + ox — \/x = 6, to find the values of z.
—5+4/(—11)
mayer ett
ANS. & = 5, or — 23 or
PONS ata 45,0Fc1, OF
132°
Given 2* + — 39x = 81, to find the values of x.
—13+ — 155
PUN Ssrae oct. S108 en alee Ure
Given 2? — 24% + 4 = 24/ (x — 1), to find the values of x.
Ans, = 4+ 4/6, or + \/—2.
2
f= aay aie
: e+s 4046 Fi
Given - _ _ are as to find the values of x.
ANS. & = 4, or — 23 or —1 + 4/ (— 3).
Given /x — iets ae to find the values of z.
v / L—2
+ a
PUN Sy 1G, Ofe1,. 08 tee
Given 7 — anil at Sy + LM kes find the values of x.
AT oat ast Vie
—3FV—7
ANS ae 9, OF 4, 0F = REY
Given 42* + = 44° + 33, to find the values of @.
3 Bitar — 43) .
ANS. 2 = 2, or — 53 Fass ten
304 Adfected Quadratics involving only
g2. Given @.(/# + 1)? = 102. (@ + eres)
[ 27
ANS. <
640
4.
two unknown Quantities. 307
oy Eta ig fae crag ry
5v 5 3
and “24 ¥ 22 PY + 2,
to find the values of wv and y.
3
xv = 6, or — —,
266
ANS. J
| Bee gee oTBE
Mee ick hear
Given 2’? — 7’? =’, | to find the values
(w+y+ 6)? + (w@—y +t by =20',f of v and y.
—b+// (2a — BB + 20’)
Oe Naa re eae an eae CR Sear weirs es a
ahs c i: = / (207 — 8? + 20’)
rs : fy uF oy
eae) (geen a ee)
Given 2’ + y:27—yi::9:7, Ie find the values of z
andl +a? ;y+4::5y +7: 3y, and y.
jpatoorti/(-2),
14
y = 2, or — —.
19
ANS.
Given x’ + 22°y = 441 — xy’, | to find the values of x
and vy = 3+ 2, Wai hop
v= 3, or — 7, or —2+4/(— 17),
ANS. 5 $4/ (— 17)
ren he
|\y sr ore, or
Given x’ + 4y’ = 256 —4xy, ) to find the values of «
and 3y? — 2? = 39, J and y.
fv =+6, or + 102,
ANS.
; ay == 5, or 59;
338 Adfected Quadratics involving
9. Given (@ + y)? — 3y = 28 + 3a, | to find the values of #
and 27y + 3% = 35, j andy.
— 5 +4/ (— 255)
4 ]
vi
eae ie or
| 7 Ue (= 255)
4
y == 2, oF =, OF
ANS.
10. Given (27 — 4y)’ + @ — 2y = 5,| to find the values of #
and 2? — y’ = 8, Jo gandiy.
ly.
ANS.
ee 7
Y — 1, or 3° :
to find the values |
a
|
We Given 3 @—) =1 4+ GQryy |
of # and
and J@+ytVe-n=s | ut
te 13
eee
ANS.
5
ly Tig"
; 2 Ay — ., \ to find the
12. Given a+ 10%7+y=119—2 Sy x (x + 2), | a: of
and # + 2y = 13, 2 andi
—6o9+ ,
es = 5, ree or ie Ree VAN
2 4
ANaH 121 = 4/ (241) ‘
ly = 4, or - or Se ee ;
2a" 39
eh Given Ld gy al |
y" = y j 49° \ to find the values of x and y. |
and a + y’ = 65,
ANSWER,
= 4, 0r +
44/(— 65). opp (= 450. 30 3410)
G 5 as
14.
16.
two unknown Quantities. 309
Givenew +y+/(e + y) =6,) to find the values of wx
and xz? + y* = 10, f and y.
9+ /—é61
9 b
-
v= 3, o0ril13; or
ANS. v
+ — 61
etl Or as or NT
Given 2 + 44/ (@’ + 3y +5) = 55 — 3y,| to find the values
and 64 — 7y = 16, J of wand y.
— = ce a 5
a phe (389 y
—7o+ 5
ly = 2, or — ca or bad ACE )
7 49
he = 5, or
ANS. .
Given # + 34 + y = 73 — 2xy,| to find the values of #
and y’ + 3y + @= 44, f andy.
Ae (@ = 4, or 16; or — 12 + 1/58,
‘ly = 8, or—73 or—1 4 58.
Pee Vee M7 ) to find the
wy) Y — 4o/(@ + y)’ values of
ae J wand y.
lie 6, Obi; An elie IOP
ANS. oe
LY = 2, or 1, or
re SnkVAMoePB ETE
8
Given y — y? = 16 — cea to find the values of # and y.
and 28 —-y=2 + 422,
58|"
ANS.
16,.0F res
yp ee: 289°
Z2
340
19.
20.
21.
22.
23.
Adfected Quadratics involving
. 4 4 —_ br
Given 2 + y' = 975] to find the values of # and y.
andw+y=
5+ — 151 |
fee aor as or ==
a se (= 151)
2
y = 2, 0r 35 OF
a gl ee to find the
Su JC )+ Mga —2y % values of |
and 2? BA rea ony | x and y.
= 6, or 3,
ANS. 3
(ee or —.
Z
|
Given v +4\/e+4y=21 $3Vy ta ay "Plo find the
and (/a + /y =6, J
values of x and y.
= 25
ANS.
| 169
Gi 2 ; "y) = (% — 4).
iven 37 + 2/ (wy’ + 92°y) =(@— 4) -¥%| to find ti
and 62 +yiyii@+-5:3 J !
values of # and y.
[v=
A
NS. fy Cee
|
to find the values of
Givenv+y=5, 1]
= 455, x and y.
aa
and (a + y’) x (#@ + y’)
|
)
two unknown Quantities. 341
24. Givenz + y — J()-,;
and #’ + y’? = 41,
ae or 3.4/8,
piano 4, rot of 2,
Y
: Salle ay
| 95, eke a of Sten 136 = — Hg Te | to find the values of #
ay to find the values
ey of #@ and y.
ANS,
andvy+4=14—y, | ANS
Pee caae ors £5 / (— 8),
: ANS. -
ly =4, or6; ors¢5./(—©
Ms. Given @t¥_ 2-9 _ 4
\?. eet y ree to find the values of
ery (252) 4 wand y.
OS (@ YY
3 5
& = 3, or —3 or, or
ANS
35 1 135
HT EE 05) A emer ecenceer al DT ok ea ibe
| J = Kage ha leet
ae. Given (6 0/2 + 6\/y) +2 p20 Vy |
| and # — y = 12,
to find the values of x and y.
59536
ite = 16, or ——,
8l
—
| 28. I mn =
" een — 438 = 12ary” >| to find the values ee and y.
| andy’ =12+2xy, |{
Ro fewer
y = 6.
Adfected Quadratics involving
2 +>
29. Given an ae pe Deere & + ey | to find the values of
y y z ee.
| wand y.
and 4y’ — vy = 2,
~
50
faahahe my er
ANS. |
= ] or = a
Y > 5
Given /iQit a? tye} tia ty t=4)
and (4 — 2”)? = 18 — 4y’, |
to find the values of # and y.
Oe Ol / 105
30.
—
2
AL ce RV ey ee to find the
31. Given = =
o 2 40
© A a he tay values of
and yy — fay’ = G? x and y.
196 289
puke ks OY =, OF eS ok &
Y = 4; 9 ) 9g’ 3°
D ABN/AM Era a— Jf (ev —y) |
e Gi , — 2 9
ee J (2 — y’) Caiaves VAD e a)
and oat) ene ee
to find the values of wv and y.
ANS. ;
two unknown Quantities.
343
xv —15y—14 :
33. Given 5y + Vv ; J ) = “ — 36, to find the
5 eae e EN cea y values of
erage. (F+5)-% w and y.
ANSWER,
+ 5 5+
ie or — aL ore vases: or 4 WD
5 20 4
i = are
es br i oan / iy maces 3 4/ (3849)
12 120 8
; e+ #) eee a yx ( 42 )
34, Given /( as togicg- 5 Gea
and Acai Cir Bees 2 <= 7 1;
Va —/(w@—y—1)
to find the values of w and y.
(amt or #3 or es Ge a
ANS, .
os a —
ly = 2, or — 1: ee
35. Given aa" + yb" = 2 (ax)? : (by)2, to find the values
. and wy = ab, of w and y.
2n S m—n genta,
“= bmtn : La 2 a AeA ae a4 OE aa n
ANS A eee
0 mH Pant 2s
\a 20 +b of (an ay puny bmn
weet Yt Ve y') 9 .
30. = —,
36 eee fy fe 4") ay (7+ y),
and (@’ + yy +ev—y=2x.(xv? + y) + 506,
to find the values of # and y.
me as
[a= sor Hat anes) 20g)
6¢ gins ——' 1209
ly =, or— 25 or va ).
J
ANS.
344, Adfected Quadratics involving
37+ Given 4 = JE+ ev. Ji “i”? | to find the values
ie e of w# and y.
and ——- — ——= = s
16 625 625
@ = 4, or —, or ——-, or
144”
ANS. :
| ("4 (= Ne
y = 64, or — or
; L 61
38. Given J2 + /f= —= +1, | to find the values of #
y Vay
yin ees and y.
and /a'y + / yx = 78, dl
371 spi
ae or 163 or — ——_ +347;
12
ANS.
371 re
ake SR or eae.
39. Given @ + 4/(3y? — 11 + 2v@) =7 + 2y — to find the
e+y values of |
and 3y—“e+7)= : 4
J (3y ae 2 anda
ANS. fe ai
=
40. Given # + y*=1+ 2%y + 3a°y*,| to find the values of |
anda®+y=oaye+oy+a+1,f wvandy. |
v= 2,0r—1 |
Ans. | ; ° |
y=1.
: 2 ss a Pee
41. Given v’y—4=4ahy a | to find the values of # andy.
Li— 3 = why}. (43 — y3),
{v=1],
ANS. ly = gt
t
L a4,
145.
two unknown Quantities
| 42. Given 5—2\/(y +2) = 2
4
Bde ata t/ha
y Y
to find the values of 2 and y
—{Va-—3VyP
B)
{e~=4,
ANS
ly =}.
EN Bae ACh a VALE ed
43. Given ae Pes Ga) cae
and —.y‘ =y’4 — 1, |
to find the values of z and y
5 85
‘ [erg Oleg?
NS
5929
lyase, rt /(-3 + ame )
Given (7 — 2 y—Vfxy.(y —1) =27? —2, to find the
Pee MALY — 13 values of
ae Dae ry —138 ° wand y.
eae eae
ANS. ee ees
ie Flee
)=a2-y, ronindatie
] f
ma Ve +9) 3 _ 28-3 Hay, values o
22 J (@ ty)
ae xv and y.
345
346 Adfected Quadratics involving
46. Given 2’? —y’?=3, |
and (af +y'+a7y.@—y)y+e—-y= 328, {
to find the values of x and y.
— 2+2 — 13 ;
yg —— aw Pp (8) Bm /—1,0r# SAAR 4
ANS. - $2 /
Woe por 82 ona, on ey eee ).
47. Given y= 2’. (ay — Oz),
and 2’? = ax — by,
fe =a-+ br,
sort ly =ar-+ br’;
} to find the values of # and Yo |
where r =p V/p’?—1,
a—b+fae+20b+58'
and p = A;
=y.(l— ay"), | to find the values
3 + 22” —- 4x"
48, Given ——_,.——_
x —1
and (2”’ — 1). (2y’ — 1) =3, |
crt
NS.
ane he or + /—1; or & 44/ (5+ V 33). :
)
of x and y. ;
U4
49. Given —# +4—40y° =140—y’. J (a -
and at — 2. (2 4 ise Mad Sid,
LS) y y
to find the values of # and y.
two unknown Quantities. 34:7
50. ee, y + 9) Ve. to find the
x y LY
a Jj A values of
Y es PAN igeTS: re
and vemiares AE | xv and y.
ANS aoe:
La == 95.
Bee Ay typi Aleit oy es
even y < Vu ’ | to find the values
and a (y +1) =36.(y° +),
jendev ep roy
ANS. —
lyati(re/®), or 2. (1
Given (#° + 1).y =(y? + 1).2",
and (y° + 1).27=9.(# +1).y°, J
ANS.
53.
e=at// (a? +1), wherea=+
Given
J
and /at/is.(y—VW2)—4t=
to find the values of # and y.
ANSWER,
= 4 or = or — or yah sty
’ 9° Q5° 3 — 3°
5 ys 13
y = 3, or 55 or =, or — 1; Nene yay See
a?
Bl aR A ital Eatin en
vx
of # and y.
|
1. to find the values
of # and y.
VA (3 TY 1),
y= b+ (b° —1), where b = 34/3.4/ {3+ V3}.
3
—-
J
y +1;
: 788 + 24 / 644
2
25
37/644,
or ea
5
348 Problems producing Simple Equations,
1
54. Given —.(#?—1) + 4. (223—1) = (yi + £3)
3 3
3
oe,
3
Tee oat 133 1 2 Yi
GN
3 y Y3 36 | ¥f3 V3 U3
1S
to find the values of # and y.
ANS. ) 2 ie
ly =8.
VI. Problems producing Simple Equations, involving only one
unknown Quantity.
1. Waar number is that, from the treble of which if 18
be subtracted, the remainder is 6? Ans. 8.
2. What number is that, the double of which exceeds four-
fifths of its half by 40? Ans. 25.
3. In fencing the side of a field, whose length was 450
yards, two workmen were employed; one of whom fenced
9 yards, and the other 6 per day. How many days did they
work? Ans. 30.
4. A Mercer bought 4 pieces of silk, which together measured |
50 yards; the second was twice, the third three times, and the.
fourth four times as long as the first. What were the respective |
lengths of the pieces ?
Ans. 5, 10, 15, 20 yards.
and afterwards 17 bushels at the same rate; and at the second
time received 36 shillings more than at the first. What was the
price of a bushel?
|
5. A Farmer sold 13 bushels of barley at a certain price |
Ans. 9 shillings.
involving only one unknown Quantity. 349
6. A person bought 198 gallons of beer, which exactly filled
4 casks; the first held twice as much as the second, the second
twice as much as the third, and the third three times as much
asthe fourth. How many gallons did each hold?
Ans. 108, 54, 27, and 9 gallons.
7. A Silversmith has 3 pieces of metal, which together weigh
-48 ounces. The second weighs 12 ounces more than the first,
and the third 9 ounces more than the second. What are their
respective weights?
Ans. 5, 17, and 26 ounces.
gs. A Vintner fills a cask, containing 96 gallons, with a mix-
ture of brandy, wine, and water. There are 20 gallons of water
more than of brandy, and 17 more of wine than of water. How
many are there of each?
Ans. 13 gallons of brandy, 33 of water, and 50 of wine.
9. A Gentleman buys 4 horses; for the second of which
he gives £12 more than for the first; for the third £6 more
than for the second; and for the fourth £2 more than for
the third. The sum paid for all was £230. How much did
each cost ?
Ans. 45, 57, 63, and 65 pounds.
10. A poor man had 6 children, the eldest of which could
earn 7d. a week more than the second; the second sd. more
‘than the third; the third 6d. more than the fourth; the fourth
4d. more than the fifth; and the fifth sd. more than the
‘youngest. They altogether earned los. 10d. a week. How
‘much could each earn a week?
ANS. 38, 31, 23, 17, 13, and 8 pence per week.
11. An express set out to travel 240 miles in 4 days,
but in consequence of the badness of the roads, he found
\that he must go 5 miles the second day, 9 the third, and 14 the
fourth day, less than the first. How many miles must he travel
each day?
Ans. 67, 62, 58, and 53 miles.
550 Problems producing Simple Equations,
12. There are 5 towns, in the order of the letters, 4, B, C,
D, E. From Ato Eis so miles. The distance between B and
C is 10 miles more, between C and D is 15 miles less, and
between D and E 17 miles more than the distance between 4
and B. What are the respective distances ?
Ans. From A to B 17; from B to C27; from Cto D2;
and from D to E 34 miles.
13. A gentleman gave 27 shillings to two poor persons; but
he gave 5 shillings more to one than to the other. What did he
give to each?
Ans. 11, and 16 shillings.
14. What number is that, the treble of which is as much
above 40, as its half is below 51?
Ans. 26.
15. Two workmen received the same sum for their labour;
but if one had received 15 shillings more, and the other 9 shil-
lings less, then one would have had just three times as much as
the other. What did they receive?
Ans. 21 shillings each.
16. ‘Two merchants entered into a speculation, by which
one gained £54 more than the other. The whole gain was
£49 less than three times the gain of the less. What were the
gains ?
Ans. £103, and £157.
17. The perimeter of a triangle is 75 feet, and the base 1s
11 feet longer than one of the sides, and 16 feet longer than the
other. Required their respective lengths.
Ans. 34, 23, and 18 feet.
is. A company settling their reckoning at a tavern, pay
8 shillmgs each; but observe, that if there had been 4 more,
they should only have paid 7 shillings each. How many were
there ? .
Ans. 28.
involving only one unknown Quantity. 351
19. Divide the number 46 into two such parts, that one of
them being divided by 7 and the other by 3, the quotients may
together be equal to 1o.
Ans. 28 and 18.
20. A certain sum is to be raised upon two estates, one of
which pays 19 shillings less than the other; and if 5 shillings
be added to treble the less payment, it will be equal to twice the
‘greater. What are the sums paid?
Ans. 33, and 52 shillings.
21. Having bought a certain quantity of brandy at 19 shil-
lings a gallon, and a quantity of rum exceeding that of the
brandy by 9 gallons at 15 shillings a gallon, I find that I paid
one shilling more for the brandy than for the rum. How many
gallons were there of each?
Ans. 34 of brandy, and 43 of rum.
/ 22. ‘Two persons, 4 and 3B, have each an annual income of
£400. A spends every year £40 more than B, and at the end of
4 years, the amount of their savings is equal to one year’s
‘income of either. What does each spend annually?
Ans. £370, and £330, respectively.
23. 75 Ye) s
ii eae
2. Given 5” + 8y = 153, find corresponding positive in-
| tegral values of w and y.
eee fv=2, y= 1, fv ==, 135 7 == 11,
la = 21, y = 6, [aoe 5, y = 16.
3. Given 52 + 21y = 2000, find corresponding positive
integral values of x and y.
[az = 3795 Y= 5,
lw = 358, y = 10;
Frith 17 other corresponding pairs, the values of x decreasing
by 21, and those of y increasing by 5.
ANS.
| 4, Given 82 + 3y = 17, find corresponding positive integral
y values of x and y.
i ANS 0194, yin 3:
5. Given 10# + 17y = 71, find corresponding positive in-
‘tegral values of @ and y.
=o
ANS. ee ;
' ly =3.
6. Given 112+ 7y = 108, find corresponding positive in-
‘tegral values of # and y.
ANS. # = 6, y= 6.
pd2
404 Indeterminate Equations and Problems.
7. Given 11% + 15y = 1031, find corresponding positive
integral values of # and y.
AS se = /OlgHnG 7/12,
lw =76, and y/== 13>
with five other corresponding pairs, the values of # decreasing
by 15, and those of y increasing by 11.
gs. Given 132 + 7y = 141, find corresponding positive in-
tegral values of # and y.
ANS. i rae
Y =
9. Given 13% + 14y = 200, find corresponding positive in-
tegral values of # and y.
5R = 0)
ANs.| ;
MS ky
10. Given 172 + 7y = 310, find corresponding positive in-—
tegral values of # and y. ;
fe= 3, [v= 10, [v= 17,
ANS.
ay ==137,0 |) 120, sy ee
11. Given 27” + 16y = 1600, find corresponding positive
integral values of w and y.
ROT ee. 16, ea 32, {@ = 48,
Y = 73; y = 46, | = 19.
12. Given 71” + 17y = 1005, find corresponding positive
integral values of # and y.
ANS age?
“ly= 9
13. Given 99% + 19y = 1900; find corresponding positive
integral values of # and y. |
[v= 19,
ANS. <
ly = 1.
|
Indeterminate Equations and Problems. 405
14. Given 52—7y =3; find the least corresponding po-
sitive integral values of w and y.
ANS on
NS.
yaar
15. Given 7e—12y = 19; find the least corresponding
positive integral values of x and y.
16. Given 11v —18y = 63; find the least corresponding
positive integral values of # and y.
fa=o,
ANS.
ly = 2.
17. Given 132 — 17y = 54; find the least corresponding
_ positive integral values of @ and y.
1s. Given 19% — 117y = 113; find the least corresponding
positive integral values of x and y.
milf
ANS. é 4
Ue De
19. Given 14% —5y = 7; find the least corresponding
positive integral values of # and y.
(v= 3,
ANS. ly ae
_ 20. Given 177 —7y = 13 find the least corresponding
positive integral values of x and y.
{v= 55
ly = 12.
21. Given 242 = 13y + 16; find the least corresponding
ANS.
positive integral values of w and y.
|
fv=5,
ANS. iN ale
406 Indeterminate Equations and Problems.
22. Given 3% + 7y + 172 = 100; find all the positive in-
tegral values of x, y, and z, which eitisty the equation. |
{Gama Vi 16,
PWS EAL he ono pas Ieee <=
ies pl eee ae 5
'
23. Given lov + lly + 122 = 300; find all the positive —
integral values of x, y, and z, which athe the equation. ht
fate = 20, V=21, v= 22, :
TANS2 t/t, 0) = nD ae tome 4
js Tey pire i OE Sabet
24. Given 1747 + 23y + 3z = 200; find all the positive in- is
tegral values of wv, y, and z, which ade the equation.
Cor 3, @©= 2, %7= 1 &%= 6,
ANS Sa aoe Te att 9 eh 3, 0 7/— e
zou, 2tbah See ioe 25,
25. Given 20% + 15y + 6z = 171; find all the positive in-
tegral values of 2, y, and z, which cbt the equation.
op SaP 3a ae G5
ANS. {y= 1. soya Fee:
z2=16 2=>6,
26. Given 35” + 43y + 55y = 4000; find all the positive:
integral values of #, y, and z, which satisfy the equation. %
Li 105, pee ob ree
A \"= 5.8/0 yak SU See
z= 2 <= 3. <= 1,
27. 6a + 7y + 4z = 122,)to find the corresponding positiv e
lle +sy—6z= | integral values of a, y, and z. ;
ANS see = 0506 8s, eae ‘
(i
Vy the corresponding
28. Given 37+ 5y + 72= 560,
ositive integral values
ox + 25y + 492 = 2920, P g §
of x, y, and z.
fv=15, y=82, 2=15,
=50, y=40, &£= 30.
ANS.
Indeterminate Equations and Problems. 407
29. Required the positive integral solutions of the equation
2ey+L+_y=195.
pice fve= 8 #=11,
; Ly = li. y= 8.
20. Required the positive integral solutions of the equation
sry —4y + 37> 14.
ren a 3 and 2,
mk Bnd 4¢
31. Required the positive integral solutions of the equation
bey = 2H 4+ 3y 4+ 18.
(v= 5, [ve=3, [x=7,
ANS.
me ly =10. ly =2. ly=1.
32. Required the positive integral solutions of the equation
72y — 5x2 = 3y + 39.
| ee fe = 1, 3, 5, 21.
(y== Tl, 3, 2, Ie
33. Required the integral solutions of the equation
2Vy —3a°+y=1.
ANS. deipris?
ly = 4.
34. Find the least whole number which divided by 3 and 7
leaves remainders 1 and 2.
ANS. 16.
35. Find a number which divided by 6 leaves a remainder
2, and divided by 13 leaves a remainder 3.
Ans. The least number is 68.
36. Find the least whole number which divided by 17 shall
Jeave a remainder 7, but being divided by 26 the remainder
shall be 13.
ANS. 143.
408 Indeterminate Equations and Problems.
37. Find the least whole number which being divided by
28 will leave a remainder 17, and being divided by 19 will leave
a remainder 13.
ANS. 241.
38. What number is that which divided by 23 gives a re-
mainder 22, and being divided by 37 gives a remainder 36?
ANS. 850.
39. Find a number which divided by 39 leaves a remainder
16, and divided by 56 leaves a remainder 27.
ANS. 1147.
40. Find two fractions whose denominators shall be 7 and
‘ l
9, and their sum ~.
4 3
ANS. ae} and oe
if 9
41. Find a number which divided by 2, 3, 5, respectively,
leaves as remainders 1, 2, 3.
ANS. 23, 53, 113, &c.
42, Find a number which divided by 4, 5, 6, respectively,
shall leave 3, 3, and 5 for remainders.
ANS. 23, 83, 143, &c.
43. Find a number such that when divided by 11 there
shall be a remainder 3; when divided by 19 there shall be a
remainder 5; and when divided by 29, a remainder 10.
Ans. The least number is 4128.
44, Find the least number which being divided by 28, 19,
and 15, leaves remainders 13, 2, and 7.
ANS. 97.
45. Find the least number which divided by 6, 5, 4, 3, and 2,
respectively, shall leave 5, 4, 3, 2, and 1, respectively, remaining.
ANS. 59.
Indeterminate Equations and Problems. 409
46. Find the least number which divided by 3, 5, 7, and 2,
shall leave remainders 2, 4, 6, and 0, respectively.
ANS. 104.
47. Find the least whole number which divided by 16, 17,
18, 19, and 20, shall leave 6, 7, 8, 9, and 10 remainders re-
spectively.
ANS. 232550.
48. Find the least whole number which being divided by
the nine digits respectively, shall leave no remainder.
ANS. 2520.
49. Divide 100 into two such parts that the one may be
divisible by 7 and the other by 11.
ANS. 56 and 44.
50. Divide 100 into two such parts, that dividing the first
by 5 there may remain 2, and dividing the second by 7 the
remainder may be 4.
82 and 18,
ANS. i" and 53,
12 and 8s.
51. In how many different ways may £1000 be paid in
crowns and guineas? —
Ans. 190 different ways.
52, In how many different ways can £43. los. 6d. be paid
with half-guineas and sovereigns ?
Ans. In 3 different ways;
paying 81 half-guineas and 1 sovereign
3) 41 29 93 22 39
29 1 39 3) 43 9?
53. In how many different ways is it possible to pay a bill
of £351 with guineas and pieces of 27s. each ?
Ans. In 36 different ways ;
_ paying 333 guineas and one piece of 27s., and diminishing the
_ former number by 9, and increasing the latter by 7.
4.10 Indeterminate Equations and Problems.
54. In how many different ways is it possible to pay £20
in half-guineas and half-crowns ?
Ans. 7 different ways.
55. In how many ways can £100 be paid in guineas and
pistoles (17s. each) ?
Ans. There are 6 different ways of payment ;
paying 7 guineas and 109 pistoles,
99 «24 9 §8 9
increasing each time the number of guineas by 17, and diminish-
ing the number of pistoles by 21.
56. A Gentleman has to pay £1000, and has only two sorts
of coins, guineas valued at 21s. 6d. and Louis d’ors at 17s. each.
How many different ways may the payment be made with these
coins ?
Ans. 27 different ways ;
paying 32 of the former and 1136 of the latter, and increasing
the former by 24 and diminishing the latter by 43, in each case.
57. If I have 9 half-guineas and 6 half-crowns in my purse,
how may I pay a debt of £4. 11s. 6d.?
Ans. By s half-guineas and 3 half-crowns.
58. In how many ways can an equivalent for 13 dollars, at
3s. each, be given in English crowns and seven-shilling pieces ?
Ans. Only one, viz. 5 crowns and 2 seven-shilling
pieces.
59. A company of men and women spend £50. The men
pay each 19s., and each woman 13s. How many men and |
women are there?
Ans. 2 and 74; or 15 and 55;
or 28 and 36; or 41 and 17.
60. A Farmer laid out the sum of £1770 in purchasing —
horses and oxen. He paid £31 for each horse, and £21 for —
each ox. How many horses and oxen did he buy?
Ans. 9 horses and 71 oxen,
OF'S0 a eee
OF 81 3" eA eee
ea
Indeterminate Equations and Problems. 411
61. ‘Two women have together 100 eggs. One says to the
other, When I count mine by eights, there is an overplus of 7.
The second replies, If I count mine by tens, I find the same
overplus of 7, How many eggs had each?
Ans. The first had 63, and the second 37;
or the first had 23, and the second 77.
62, What quantities of tobacco at 16d. and lod. per pound,
may be mixed with 50 pounds of tobacco at sd. per pound, so
that the whole may be worth a shilling per pound?
Ans. 51 lbs. of the former, and 2 lbs. of the latter;
52 lbs. 4 lbs. “
and so on, increasing the former by 1, and the latter by 2.
63. A has moidores only; B only crowns. How can B
pay A £1. 12s. most easily ?
Ans. By paying 28 crowns, and receiving 4 moidores.
64. A having only one-pound notes, owes 12s. to B, who
has only seven-shilling pieces. What is the least number
which 4 and B must respectively interchange, so that the debt
may be discharged ?
Ans. He must give 2 notes, and receive 4 seven-shilling
pieces.
65. A person buys some horses and oxen. He pays £31
for a horse, and £20 for each ox; and he finds that the oxen
cost him £7 more than the horses. How many horses and
oxen did he buy?
Ans. The least numbers are 3 and 5: and other answers
may be obtained by adding 20 to the former, and 31
to the latter continually.
66. Find two numbers, such that their product added to
their sum may be 79.
ANS. 1 and 39; 3 and 19; 4 and 15; and 9 and 7.
67. Find three integers, such that if the first be multiplied
by 3, the second by 5, and the third by 7, the sum of the pro-
ducts may be 560. But if the first be multiplied by 9, the
second by 25, and the third by 49, the sum of the products
may be 2920.
ANS. 15, 82, and 15;
or 50, 40, and 30.
412 Indeterminate Equations and Problems.
6s. It is required to buy 20 fowls for 20 shillings; viz.
geese at 4s., quails at 6d., and pigeons at 3d. each.
ANS. 3 geese, 15 quails, and 2 pigeons.
69. Thirty persons, men, women, and children, spend 50s.
in a tavern; the share of each man is 3s., that of a woman 2s.,
and that of a child 1s. How many persons were there of each
class ?
ANS. 1 man, 18 women, and 11 children; and there are
s other answers, the number of men and children
increasing by 1, and the women decreasing by 2.
70. and
87. Determine three numbers, such that their sum shall be
a square, and the sum of their squares a perfect fourth power.
ANS. (119)?, 2. (119). (180), and 2. (180), whose sum is
(349)*, and the sum of their squares (281)‘.
ss. Determine three square numbers which shall be in
arithmetical progression, but the sum of their square roots shall
be a cube.
Aws. (169), (845)’, and (1183).
APPENDIX I.
I. Problems in Arithmetical Progression.
1. DeTermineE the esth term of the series 13, 122,
123, &c.
2. Having given the first and last terms of an arithmetic
progression, and their common difference; determine their
number of terms.
3. Having given the first and last terms of an arithmetic
progression, and the number of terms; determine the progres-
sion, in general; and in the particular case where the first term
is 100, the last 1, and the number of terms 19.
4. Find the series in arithmetic progression having 29 terms,
of which the first 1s 3, and the last 17.
5. Having given the first term of an arithmetic series = 2,
the number of terms = 7, and the common difference = 1; find
the last term.
6. Show that +, +, and — 4, are in arithmetic progression ;
and find the 11th term of the series.
7. Show that +5, 1, 4, are quantities in arithmetic pro-
gression; and find the sum of s terms of the series of which
they are the first three terms.
8. In an arithmetic progression, it is observed that the
‘fifth and ninth terms are 13 and 25: what is the 7th term?
9. The sum of the two first terms of an arithmetic pro-
gression is 4, and the 5th term is 9; find the series.
10. If three quantities are in an increasing arithmetic pro-
416
APPENDIX.
gression; show that the second will have to the first a greater
ratio than the third to the second.
ine
12,
Find the sums of the following series:
1+34+5+7 + &c. to n terms.
1+5+9+ 13+ &c. to n terms.
1+4+7+410+ &c. to 12 terms.
5+7+9+ 11+ &c. to 50 terms.
2+ 24+ 22 +3 + &c. to 13 terms.
ban vc < + &c. to 16 terms.
4. 8
eee to 12 terms.
> + 1 + 12 + &c. to 8 terms.
—9—7—5 — &c. to 20 terms.
—5—3-—1, &c. to 8 terms.
11+8s+5+ &c. to 8 terms.
l
5
—~—]1——— &c. to 29 terms.
2 2
44 43
15 + oc + o + &c. to 16 terms.
= 4+ <4 ot &c. to 19 terms.
a Pate = 4+ = -— + &c. to 8 terms.
16." 15
Pim Mey Min oe aaa Resear
n n
na—b+(n—1).a+ (n—2).a+6+4 &c.tonterms.
(a+ 27)? + (a+ 2’)+ (a—2z)’ roe to n terms.
Ce) He aa
Gis Oni Sat
a+b a+b
lad
2
Show that + &c. to n terms, is
een at
ih ees mn
APPENDIX. AlL7
13. Show that the sum of the (m— n)™ and (m + n)™
terms of any arithmetical progression will be equal to twice
the m™ term.
14, It is required to divide (a) into n parts proportional
to the numbers 1}, 2, 3, &c.
15. Having given the first and last terms, and the sum of
an arithmetic series; determine the common difference, and
apply it to the case where the first term is = 1, the last = 50,
and the sum = 204.
16. Having given the first term = 1, the number of terms
=n, and the sum = s, of an arithmetic series; determine the
common difference.
17. Having given the n™ term of an arithmetic series, and
also the sum of m terms; determine the series.
18. If the first term of an arithmetic series be = 1, and
the common difference = m, the sum of n terms of the series
is}. {mn? — (m —2).n.
19. How many terms of the series — 7 — 5 — 3 —, &c.
‘amount to 9200?
20. If the first term of an arithmetic series be = 1, the
‘common difference = 4, and the sum = 120; determine the
‘number of terms.
_ 21. Ifthe first term of an arithmetic series be = 31, the
‘common difference = 14, and the sum = 22; determine the
number of terms.
22. The first term of an arithmetic series is 3, the common
idifference is 4, and the sum of 7 terms is 1081; find », and ex-
plain the double answer.
23. If the first term of an arithmetic series be = 11, the
common difference = — 5, and the sum = 5; determine the
number of terms.
24, There are five numbers, of which the first two are 21,
Ee
A418 APPENDIX.
3-2;, and each number exceeds the preceding one by the same
fraction ; find the numbers and the sum of them.
25. The sum of an arithmetic series is = 1455, the first
term = 5, and the number of terms = 30; determine the com-
mon difference.
26. The sum of an arithmetic series 1s 91, the common
difference 2, and the last term 19; find the number of terms.
27. The sum of a certain number of terms of the series
21 +19 +17 + &c. is 120; find the last term and the number
of terms.
28. If a be the first term, 4 the second, 7 the last term
of an arithmetic series, the sum = ; = ; decoded
29. In the expression s = {2a + (n—1).d}.4n; if n be
negative, point out the form of the series which satisfies that
condition; and in finding », show when the values, one, both,
or neither are congruent values supplying interpretable so-
lutions. Determine the series in the case of a =7, d= 2,
5 40:
30. If abe the first and 7 the last term of an arithmetic
series, in which the common difference is 8, and the number of
n. (1? — a’)
2.(n—1)8
31. The sum of » terms of any increasing arithmetic series,
whose common difference is equal to the least term, will be
terms 2; then will the sum of the series =
equal to the sum of (m + 1) magnitudes, each of which is half |
the greatest term of the progression.
32. Ifthe number of terms of an arithmetic series be odd;
show that the sum of the series is equal to the middle term
multiplied by the number of terms. i
n—
?
od. The n™ term of an arithmetic progression is :
and the sum of n terms is — . (32 + 1); find the series.
APPENDIX. 419
34, The first term is n? — n+ 1, and the common difference
is 2; prove that the sum of » terms is n°; thence show that
P=1, 2=>3+4+5, 2=7+9+11, &c. If the first term be
n”-!—n-+ 1, the sum of n terms is n”; and thence show that
=i, 2=7-+9, 3° = 25 + 27+ 29 + &e.
35. Thesum of n terms of an arithmetic series is pn + gn’,
whatever be the value of 2; find the m™ term.
36. In an arithmetic series the first term and the common
difference are the same, and the sum of the series is always
equal to the number of the terms + the square of that number;
determine the series.
37. If from any square number (n’) there be subtracted
the sum of an arithmetic progression beginning from unity,
having a common difference unity, and continued to as many
terms as there are units in the root of the number (n); the
remainder will be the sum of the progression continued to
(2 — 1) terms.
388. The sum of an even number of terms of any arith-
metic series whose common difference is equal to the least term,
will be four times the sum of half that number of terms di-
minished by half the last term; the first term bemg the same
in each case.
39. The latter half of 2” terms of an arithmetical series is
equal to trd of the sum of 3” terms of the same series.
40. Prove that 1, 3, 5, 7, &c. is the only arithmetic pro-
gression beginning from 1, in which the sum of the first half
of any even number of terms bears to the sum of the second
half the same constant ratio; and determine that ratio.
41. The sum of » terms of the series 1, 3, 5,7, &c. is to
the sum of (nm — 1) terms of the series 2, 4, 6, &c. :: mi n—13
required a proof.
42, The difference between the sums of m and n terms of
an arithmetic progression: the sum of (m +) terms :: m—n
:m+n.
Ee2
4.20 APPENDIX.
43. ‘The two first terms of an arithmetic progression being
together = 18, and the three next = 12; how many terms,
beginning with the first, must be taken to make 28? and explain
the reason of the double solution.
44, The first two terms of an arithmetic series being to-
gether = 18, and the third term = 12; how many must be taken
to make 78?
45. The first and third terms of an arithmetic series being
21 and 14 respectively; find the second term and the sum of 19
terms.
46. The (n+ 1)" term of an arithmetic progression is
ma—nb
a—b
47. How many terms of the series 1, 3, 5, 7, &c. must
be added together to produce the (2m) power of a given
quantity 7?
; required the sum of the series to (27 + 1) terms.
48. Find » terms of the indefinite series 3, 5, 7, &c. whose
sum may be the (m)' power of n.
49. Inthe two series 2, 5, 8, &c. and 3, 7, 11, &c., each
continued to 100 terms; find how many terms are identical.
50. Having given (a) and (0), the (m)™ and (n)™ terms of
an arithmetic series; determine the value of the (#)™ term.
51. Having given as before; determine the sum of (p)
terms of the series.
52. Having given (a) and (6), the (m)™ and (n)™ terms of
an arithmetic series, and its last term (a + 0); determine the
first term, common difference, and number of terms.
53. The sum of m terms of an arithmetical series is m, and
the sum of n terms is m; show that the sum of (m+) terms is
— (m+n), and the sum of (m—n) terms is (m—n) . ( + va ‘
54, If the m™ term be n, and the n term m; of how many
]
|
APPENDIX. 421
terms will the sum be 3 (m+) .(m+n— 1); and what will be
the last of them.
55. In an arithmetic series, if the (m + n)™ term = p, and
the (m—n)™ term = g; show that the (m)™ term =3.(p+ q),
and the (n)" term = p — (p—q). —.
56. If 2, y, and z be respectively the p, g, and 7** terms
of an arithmetic progression; show that (p — g).z+ (r—p).y
+(qg—7r).v%=0.
57. There are two series in arithmetic progression, the
sums of which to n terms are :: 13 —7n:3n+ 13; prove that
their first terms are as 3 : 2, and their second terms as — 4: 5.
58. If S, S’, S” be the sums of three arithmetic series,
= the first term of each, and the respective differences be
1, 2,3; prove that S + 8” = 2S’.
59. There is a series of m quantities in arithmetic pro-
gression whose first term is a, and common difference d; show
that if there be taken v terms of the first, n — 1 of the second,
&c. to the nv inclusive, the sum will be
n.(n + 1)
ites - {3a + (n— 1). ad}.
60. If S,, S,, S,.... are the sums of m terms of different
arithmetic series having the same first term, and common dif-
ferences 1, 2, 3.... respectively; show that S,, S,, &c. are in
arithmetic progression ; and when that first term is 1, prove that
mn’. (n + 3)
Seer. +S = ;
61. If there be (p) arithmetical progressions, each begin-
‘ning from unity, whose common differences are 1, 2, 3...
show that the sum of their (n)™ terms is
= 3-t(v—1)-p' + (n+ 1) - ph.
62. Ifaandd are respectively the first term and common
difference of an arithmetic series, S, the sum of terms, S_, , ,
4.22 APPENDIX.
the sum of (n+ 1) terms,&c. prove that S,+ 8,,,+85,,, + &e.
a
= —1).2.—— — 2).(7— 1).n.———.
to m terms = (37 — 1). erste 2).(~—1).n ae
ONDA S54.) S53 Uy emac rs S, be the sums of (p) arithmetic
progressions continued to » terms, and their first. terms be
1, 2, 3, 4, &c. and their common differences 1, 3, 5, 7, &c.;
show that S,, + S, + S, + ..0.. + 8 = 3. (np + 1) np.
Ode LEO os iy eet S, are (p) arithmetic series each con-
tinued to m terms, whose first terms are the first (py) even
numbers, and common differences the first (p) odd numbers
respectively ; show that
m.(n + 1
(n+ Dy
65, 1858), 85, 94. +--: S,,, be the sums of n terms of 2” arith- .
metical progressions, whose first terms are the same, and com- —
mon differences d, 2d, 3d....2nd; then will
(S, + S,.....+ 8,,) —(S, + Sot... + 8,,_,) = 3. (n—1).d.
66. If S_ denote generally the sum of m terms of any
arithmetic progression; prove that
/ (Src |
S 9= 8,2.
n@
—S.n.(n— 2).
67. Find three arithmetic means between 1 and 11; and —
seven between 1 and — 4; and fifteen between 3 and 47. |
68. If between all the terms of an arithmetic progression f
the same number of arithmetic means be inserted; show that —
‘
ae
the new series will still form an arithmetic progression. d
ie
69. Insert 6 arithmetic means between 4 and 2; and find —
their sum. 4
70. Insert ” arithmetic means between a and 8, and apply i
the general expression to insert 3 terms between 5 and — a5
find the sum of the series for 20 terms. %
APPENDIX. 423
71. There are ” arithmetic means between 3 and 17, and
the last is 3 times as great as the first; find the number of
means.
72. There are n arithmetic means between 1 and 31, and
the 7th mean is to the (n—1)™:: 5:9; prove that the number
is 14,
73. Between aand 4, a being less than 4, insert m means such
that a, — a, a, — G,,...... 6 — a, form an arithmetic progression
whose common difference is d; and find the limits between
which the value of d must lie.
74. The sum of n arithmetic means between 1 and 19 is to
the sum of the first (n — 2) of them:: 5:3; determine the
means.
75. If S be the sum of an arithmetic progression, a, J, ¢, d,
&c. to n terms; determine the sum of the series
Sta, St(a+bd),St(a+b+0), &.
76. In any arithmetic progression of which a is the first
term and 2a the common difference; prove that the number of
terms which must be taken to make a sum S, 1s ih 8, S being
a
assumed such that is any square number, but no other.
77. Determine the sum of » terms of the triangular num-
bers 1, 3, 6, 10, 15, &c. the terms of which series are 1, 1 + 2,
1+2+ 3, &c. the successive sums of 1, 2, 3, &c.
78. Determine the sum of terms of the pyramidal num-
bers, 1, 4, 10, 20, 35, &c. the successive sums of 1, 3, 6, 10,
15, &e.
79. Find the sum of n terms of the series 1’, 2’, 3”, &c.
80. Solve the equation (v7 — 1) + 2.(¥— 2) + 3 (w — 3) to
6 terms = 14.
81. Having given the first term and common difference
424, APPENDIX.
of an arithmetic progression; find the sum of m terms and the
sum of their squares.
82. Having given the sum of (2”) quantities in arithmetic
progression, and the sum of their squares; determine the quan-
tities themselves.
83. In the series 1, 2, 3, 4..... 100, determine the sum of the
numbers which are not squares.
84. Prove that the sum of the series 1’ + 3’ + 5’, &c. to
nm terms = “ . (42? — 1). And determine the sum of the series
a’ + (a+ 6b)? + (a + 26)’ + &c. to n terms.
85. The sum of the series 0, 1, 2, 3, &c. continued to an
unknown number of terms, being = 1225; determine the sum
of their squares.
86. The sum of 9 terms of the series n’ + (n + 1)’
+ (n + 2)? + &c. = 501; determine the value of n.
87. Determine the sum of 10 square numbers, whose roots
are in an arithmetic progression, the least term of which is = 3,
and the common difference = 2.
88. The square of any number of digits less than ten,
each of which is unity, will, when reckoned from either end,
form the same arithmetic series whose common difference is —
unity, and greatest term the number of digits in the root.
Required a proof.
89. A man is employed ina certain manufacture, where the
quantity of work which he produces in the 1st, 2nd, srd,....
days are 1, 3, 5, 7,..... For the first day’s work he receives
a shilling, and afterwards in proportion to his work. If a new
workman is added every day under the same conditions, how
much money will have been paid to the men after n days?
90. A gentleman owed to each of two persons, A and Bb,
an equal sum of money, which he discharged as follows: to 4
he paid £8 the first payment, £12 the second, £16 the third, ~
Poa oe
APPENDIX. 425
and so continued increasing £4 each payment. Now B at his
first payment received but £1, the second £4, the third £9,
increasing according to the square of the number of payments.
Determine what he owed each person, and the number of pay-
ments required to discharge the debt.
91. Compare the sum of the numbers 1, 2, 3, 4, &c. with
the sum of their cubes.
92. Find the sum of » cube numbers, whose roots are, in
arithmetic progression, the least term of which is a, and com-
mon difference d.
93. Determine the arithmetic progression, the number of
whose terms is 11, their sum 220, and the sum of their cubes
147400.
94, Prove that when 1 is indefinitely great,
a’ + (a+ 6)’ + (a+ 26)" + &.tonterms JB
mn’ +1 wa r+ if
95. Find the sum of (m) terms of a series of polygonal
numbers, which numbers are formed by assuming any arith-
metic series that has its first term 1, and difference a whole
number, and by making generally the (m)' polygonal number
equal to the sum of (n) terms of the arithmetical series.
96. If the first term of an arithmetic series be (a), the
last term (J), the common difference (d); and S,, S,, S,......
S_, be the sums of the 1st, end, ard,....... (m — 1) powers
10min
of the terms; prove that (7+ d)”—a"™=mdS,,_,+m. ;
m—-1m— 2
Ag? Same ec
3
oS, 5 Mt:
OF GeLeb. d,- @ AT, @:—s27 cand. 0,0. 7.0 + 7+ ot 75 t ke. an inf.
63. Given =, : the first two terms of an arithmetic pro-
gression; find the sum of 15 terms. And if the same quantities
be the first two terms of a geometrical progression; find the
sum of 15 terms.
64. If the terms of an arithmetic progression, a, a+7,
a+a2r, &c. be multiplied by the corresponding terms of a
geometric progression, 0, bd, bd’, &c. of the same number of
terms; determine the sum of the resulting series.
65. Determine the sum of 7 + 5 + &c. an arithmetic
series of 12 terms; also of — : ab Z — &c. a geometric series
of 5 terms: and also of the infinite series, whose terms are
the products of the corresponding terms of these series con-
tinued in inf.
66. There are two infinite geometric progressions, each
beginning from 1, whose sums are o and o'; prove that the
APPENDIX. 437
”
sum of the series formed by multiplying their corresponding
1
Ooo
terms 1s ———, :
oto-—1
6%. If o, represent the sum of a geometric progression
continued in inf., «, the sum of the squares of the terms, o, the
sum of their cubes, &c.; then will
l a. ,
day lh SS A Es — Hi
om oO» 4 asa 1 a+r
68. IFS =1 + 24+ 2424 &e. ining
99 RE 8
3 5 i A
As) —— 7 y.
and S, = 1 tee, 5 + Ke. tn inf:
Prove: (hats S665 0505509751,
69. If 8 =1 4+ + &e. ining.
1
1 1
S,=1—--+ —— KC eceeeee
y jie
1 ] 1
S,=l1+— 57+ 7+—......
s=lt+stats
prove that S, x S, = S,.
70. Determine the sum of the series
ar + 3ar? + 6ar + loar* + &e.
which arises from multiplying the terms of a geometric pro-
gression by the corresponding terms of a series of triangular
numbers.
71. Find the sums of the following series:
1+ 227 + 327+ &c. to n terms.
2 3 4 ex:
1+-—-+—+-—>+ &c. in inf.
Lie edad ee 4
2a" 80 i Ag joke.
a Sat ig home: + &c. in inf.
a a a’
438 APPENDIX.
1.2.04+2.3.a°+3.4.a° + &c. in inf. and to n terms.
1.2.3.0+2.3.4.2°+3.4.5.a° + &c. in inf. and to n terms.
1.2.3.4.” + 2.3.4.5.a" + 3.4.5.6.a"% + &c. ininf. & ton terms.
1.2+ 32? + 52° + 72" + &e. in inf.
1.27 +52? 4+ 94° + &c. in inf.
1.2.2 +3.4.47 + 5.6.2 + &c. in inf.
1
1
o
22.3.0 +4.5.6.2° +7.8.9.2° + &c. in inf.
.3.04+2.4.27°+3.5.2° + &e. in anf.
5.27 +3.6.2° + &c. in in.
6.07 +5.8.x° + &c. in inf.
1.3.0+4.6.a°+7.9.x2° + &c. in inf.
5.0? + 5.7.0 + &e. in inf.
72. Prove that the sum of m terms of the series 1° + 3° + 5°
+ 7° + &c. is a hexagonal number whose root is 7’.
73. Prove that the sum of the reciprocals of the n powers
of the odd numbers is to the sum of the reciprocals of the same
powers of the even numbers :: 1 ; 2”—1.
- 3242424 ...4.%
74. Reduce to its lowest terms ———————_._,
1—2+4+4— &c.
Lie COD se mitts
P) ES) 1 — oe, tee Soe
O27 Jarek
‘ A t—444-—&c.
III. Problems in Harmonic Progression.
1. Kxpxuarn the nature of harmonic progression; and
continue in both directions the series 2, 3, 6.
2. Continue the harmonic progression....3, 4, 6....up-
wards and downwards. How far can it be continued either
way?
3. Continue to 3 terms each way the series 1, 14, 12.
APPENDIX. 439
4. Ifaand 0 be the first two terms of a series in harmonic
progression, continue the series to three more terms, and find
the n‘ term.
~~
5. Prove that the reciprocals of quantities in harmonic
progression are in arithmetic progression.
6. In any harmonic progression, the product of the first
two terms is to the product of any two adjacent terms as the
difference between the two first is to the difference between the
- two others.
7. In any harmonic progression, the difference between the
first two terms is to the difference between any two others as
the second term diminished by (m) times the difference between
the first and second is to the last; where » =the number of
terms between the first and last.
8. In any harmonic progression, the second term dimi-
nished by (mn) times the difference between the first and second
is to the last as the product of the two first is to the product of
the two last; n as before.
9. Any term of an harmonic progression is equal to the
product of the first two terms divided by the difference between
[the second] and [z times the difference between the first and
second |.
10. Thesum of any two terms of an harmonic progression is
greater than twice the intermediate mean term; and this excess
is greater, as they are the more remote.
11. Find a fourth harmonic proportional to 6, 8, 12.
: tb bon Mm
12. Find the sum of n terms of the series —, —, —,
nm mn
3n — 2m
a Se a7 oo * &e.
mn
13. If the two extremes and the number of terms in an
harmonic progression be known; the intervening series may be
found.
44.0 APPENDIX.
14. Insert two harmonic means between 2 and 4; two
between 6 and 24; four between 2 and 12; six between 1 and
20; six between 3 and -°,; and seven between 10 and 12.
15. Find two numbers whose difference is 8, and the har-
monic mean between them 1+.
16. Insert 2 harmonic means between a and 6; and if a,
be the first harmonic mean, prove that
Te Oee Ce ey es
e ee e ] e
17. The difference between two numbers ts 18, and 4 times
the geometric mean is equal to 5 times the harmonic mean;
find the numbers.
18. Prove that a geometric mean between two quantities is
a mean proportional between an arithmetic and harmonic mean
between the same two quantities.
19. If (a) be an arithmetic, (6) a geometric, and (c) an har-
monic mean between two quantities; show that @ is greater
than 6, and J greater than ec.
20. The difference of the arithmetic and harmonic means
between two numbers is 1+; find the numbers, one being 4
times the other.
21. The arithmetic mean between two .numbers exceeds
the geometric by 13, and the geometric exceeds the harmonic
by 12; what are the numbers ?
22. If the arithmetic mean between two quantities # and y
be m times the harmonic mean; then will
Tey Wy a he ey he ee
23. If the geometric mean between two quantities x and y,
be to the harmonic as 1 : 2; show that
eiyiitY/i—-w):1-Sf—n).
24. Ifm harmonic means be inserted between a@ and pa;
prove that the ratio of the first : the last is = to the ratio of
m+pr:mp+i.
ei ete eae
APPENDIX. 441
25. Ify be an harmonic mean between # and z, and # and
z be respectively the arithmetic and geometric means between a
and 4, show that
2.(a + b)
Y= er
{OOF
Ua) Aa)
26. Having given (a) the sum of three numbers in har-
monic progression, and (d) their continual product; determine
the numbers.
27. There are three numbers in harmonic progression; if
1 be subtracted from the first, the progression becomes geo-
metric; and if 4 be subtracted from the third, it becomes
arithmetic. What are the numbers?
28. From each of three quantities in harmonic progression,
what quantity must be subtracted that the three results may be
in geometric progression ?
29. There are four numbers, the first three of which are
in arithmetic and the last three in harmonic progression; prove
that the first has to the second the same ratio which the third
has to the fourth.
30. The sum of three consecutive terms of an harmonic
progression, whose first term is 3, is = 14,;; determine the
progression, and continue it both ways.
31. If S and s be the sums of two infinite series, the
common ratios of whose terms are R and r respectively; then
S, s, R, r are in harmonical progression, the form of each series
being (r, 7’, r*, &c.) and r and R fractional.
82. Given M and N, the m and nv terms of an harmonic
progression; find the first and second terms.
83. Having given the two first terms of an harmonic pro-
gression; determine the (m)'" term.
34. Having given the (m)™ and (z)™ terms of an harmonic
progression ; determine the (m + nm)" term.
442 APPENDIX.
35. Ifa, b, ¢ be the p*, g™, 7 terms of an harmonic pro-
gression; then will (p — g).ab+(r—p).ac+ (q—r).bc=0.
36. Ifa™ = bY = c* = &c., and a, b, c, &c. be in geometric
progression ; then will 2, y, z, &c. be in harmonic progression.
37. Compare the lengths of the sides of a right-angled tri-
angle, when the squares described upon them are in harmonic
progression.
APPENDIX ILI.
1. Form the equation, whose roots are
Pi ABE oe WY Air (eo bones
2. Also whose roots are + «/— 2, 3, and 4.
3. Also whose roots are 1 + 4/—2, and 2+ (/—3.
4. Form the biquadratic equation, two of whose roots are
1+ fa’, and — /—28,
Also if two of the roots be 4/3, and — «/—5.
Cr
6. Determine the equation whose roots are
is al apa
aad ( io 54 rs , and — a.
7. Form the equation, of which the roots are the different
values of a@ + 4 b.
8. Given that an equation has one root; show that it will
have as many roots as it has dimensions.
9. If any coefficient in an equation be changed; prove that
all the roots will be changed.
an
APPENDIX. 443
10. Ifa be a root of the equation
oP arth ise. + Px’ + Q2?+ Rx + S=0,
hey R,
andif> + R=R,— + Qa Q,, &e.3
h S r
show that a R,, Q,, &c. are integers.
1l. Ifa, 3, y, &c. be the roots of the equation
ES) EN 1 me ea Qe +8 R= 03
show that
pQ—nP
rit
OA ae Lalkeehl egatie ceatieat erage ae
Pa day teal yp
12. The roots of the equation
id oat 1 da ec — Qv + R=0, being a, B, y, &c.;
show that
2 2 2 2 2 2
Se ULE tingid eet Als gt yarn ag abet
Boy BN oy a B
= (p29). —P.
13. Ifa — pa"-! + qv"? — &. + W= A, and abe any
root of the equation 2” — px"~! + gz”—~?— &c. + W=0, prove
that x — a is a divisor of the expression
x” — px-1+ ga"? — Kt W.
14. Take away the second term of the following equa-
tions:
a’ — 92° + 2067 — 34 = 0.
v—3e@ +47 —5=0.
a+ 242° — 124° +47 —30=0.
wt+sa4t+a2’?’—r—10=0.
mw no =
a So. CHa.
15. Take away the third term of the following equations :
1 v2 — 62" + 97 — 20=0.
9. #—427+57—-2=0.
44,4, APPENDIX.
16. Prove that the third term of the equation
v— pe +qx—7r=0,
cannot be taken away by the common method, if p’ be less
than 3g. Show how it may be taken away in this case.
17. In an equation of m dimensions, show that the second
and third terms may be taken away by the same transformation,
when the square of the sum of the roots is to the sum of their
Squares (7: 1.
18. Exterminate the last term but one of an equation of
five dimensions by the solution of a simple equation.
ip das Mt De
m i oD
into one whose coefficients shall be integral.
19. Transform the equation 2 —
20. Transform the equation y* — 2py’ — 33p’y + 4p” = 0,
into one whose coefficients shall be numerical.
21. Transform the equation
a” — at pan} + gx? — a?ra"-> 4+ &c. = 0,
into one whose coefficients are rational.
22. Transform the following equations into others whose
terms shall be alternately positive and negative :
: SR
1 eee Tes MM
2 2 + a — 192° + lla + 30=0.
23. Transform the equation a’ + x2 — 10% + 4 = 0, Into
one whose roots shall be greater by 4 than the roots of the given
equation.
24. Transform the equation 2* — 42° + 6a? — 12 = 0, into
one whose roots shall be greater by 5 than the roots of the given
equation.
25. Transform the equation 2 — 6x? + 9% — 12 = 0, into
one whose roots shall be less by 6 than the roots of the given
equation.
APPENDIX. 445
26. Transform the equation 32° — 1247 + 15” — 21 =0,
into one whose roots shall be treble the roots of the given
equation.
27. Transform the equation v* — 2”? — 3% + 4=0, into
one whose roots shall be one-eighth of the roots of the given
equation.
28. Ifa, b,c, &c. be the roots of the equation
wen — per} + Caen mw &e. = 0;
transform it into one whose roots are ma, mb, mc, &c.
29. If the roots of the equation 2 — px’ + ga —r=o0,
be a, 0, c; transform it into another whose roots shall be
atbat+cbo+e.
30. Transform the equation #* — 40x + 39 = 0, into one
whose roots shall be the sum of every two roots of the original
equation.
3l. Transform the equation 2° — px’ + gx — r = 0, whose
roots are a, 6, c, into another whose roots are
1 1 1
a+b ate b+e°
32. Ifthe roots of the equation
v—px+gqxu—r=o0, bea, b,c;
determine the equation whose roots are ab, ac, bc.
30. Transform the equation 2 — pa’ + qx —r=0, into
one whose roots shall be mean proportionals between the roots
of the equation, and a given quantity (m).
34, Ifthe roots of the equation 2 — px + ¢ =0,beaandd;
determine the equation, of which — /a, and — /0 are roots.
35. Ifthe roots of the equation
xz’ —px+gq=o0, bea and b;
determine the equation whose roots are an arithmetic, a geo-
metric, and an harmonic mean between a and 0.
44.6 APPENDIX.
86. Transform the equation 2° — 22° +27 —4=0, into
one whose roots are the squares of the roots of the original
equation.
87. Transform the equation 2° — px’? + qx — r = 0, whose
roots are a, J, c, into one whose roots are a’, 0’, c’.
38. Transform the equation 2 — px’ + gv —r = 0, whose
; oe)
roots are a, 5, c, into one whose roots are GP BP? oF
39. Transform the equation x* — px’ + gx — r = 0, whose
roots are a, 6, c, into one whose roots are
@7+0,e70+0¢,0 +c’.
40. ‘Transform the equation 2° — pa’ + gx — r= 0, whose
roots are a, 6, c, into one whose roots are
1 4 1 1 + 1 1 it. 1
a b?? a ce? & C
41, ‘Transform the equation 2 — 6x’ + 114 — 6 =0, into
one whose roots are ageless :
G0” Ft ¢ b? + ¢c
42. Transform the equation a* + a + #’?+4#+1=0, into
one whose roots shall be the squares of the roots of the given
equation; and show from the roots themselves that the trans-
formation is correct.
43. ‘Transform the equation 2° — px’ + qx — r = 0, whose
roots are a, 0, c, into one whose roots are
a b a Cc b Cc
(5 + Z)> (G+ 3)» ana (3+ 5):
44, Transform the equation 2° — pa’ + gx — r = 0, whose
roots are a, b, c, into one whose roots are
c b a
at+b—cCat+e—Vb+e—a
APPENDIX, 44,7
45. Transform the equation
v— px +qr%—r=od,
whose roots are a, b, c, into another whose roots are
atb+tabat+ec+ac,o6+c+4 be.
46. Transform an equation into one whose roots shall
be the squares of the differences of the roots of the original
equation; and show by means of this transformation how the
number of impossible roots in an equation of five dimensions
may be detected.
47. Transform the equation #*—pa"—1+ qa"~?— &c.=0,
into one whose roots are the reciprocals of every (x — 1) roots
of the original equation.
48. If the roots of the equation # — px’ + qv —1r = 0, be
a, b,c; transform it into one whose roots are a’, b°, c’.
49. If 2 — 22° + 1=0; deduce the equation of which the
roots are the cubes of the roots of the original equation.
50. Transform the equation x" — pa”-1+ gxu"—-*— &c.=0,
into one whose mm" term shall be a given quantity.
51. Determine the roots of the equation
at — 40/2 + 62° 4— 44/8 +2=0.
52. Solve the equation x#° — 4@° — 3% + 12 =0, one root
of which is of the form / a.
53. One root of the equation z* — 6a + 6% + 8 = 0, being
1+ / 3; find all the roots.
54. One root of the equation 2 — 11” + 37” — 35 =0,
being 3 + 4/2; find all the roots.
55. One root of the equation z* + #2 — sx2° — 167 —8 = 0,
being 1 — 53 find all the roots.
AA8 APPENDIX.
56. Solve the following equations, two of whose roots are
equal : ‘
lL. #@—72? +164 —12=0.
2 v+8a2?> + 207+ 16=0.
3. @—5a°+8r7—4=0.
4, @—52°—8H +48=0.
5. PP — 2 — 82 + 12 = 0:
6 pO pa neste © F
2 16
ri Pe pee eae
7 9261
57. The equation 32° — 102° + 15% + 8 = 0, has three equal
roots; determine them.
58. Solve the equation a — 144° + 614° — 84¥ + 36 =0,
whose roots are of the form a, a, 0, 0.
59. Solve the equation
av — 132° + 672° — 1712” + 2162 — 108 = 0,
whose roots are of the form a, a, a, 0, 0.
60. The equation a°—2x°+6z‘— sx’ + 124°—87+8=0,
has equal roots; determine them.
61. Solve the equation a + pa’ + qa’? + r# + s =0, which
has two pairs of equal roots.
62. Solve the following equations, which have two roots of
the form + a, — a,
40° — 3227 —xv+8=0.
wv — 50° — 54" + 454% — 36=0.
v + 32° — 72 — 27H —18=0.
v+a—11H? +07 4+ 18 =0.
m0 te
63. Solve the equation
a — 102#* + 292° — 10x? — 6227 + 60 = 0,
two of its roots being 3 and 4/2.
64, The equation 2 — 15z* + 66a — 80 = 0, has two roots
whose sum is 133 find all the roots.
APPENDIX. 44,9
65. The equation 2 — 45a’ — 40x + 84 = 0, has two roots
whose difference is 3; determine all the roots.
66. The roots of the equation 2° — 152° + 66% — 80 = 0,
have a common difference; determine them.
67. In the equation #° — 6x? + 11a —6=0, one root is’
double another; determine all the roots.
68. The product of two roots of the equation
av + a — 62”? — 80% + 1200 = 0, is 30;
determine all the roots.
69. Determine the roots of the equation
xv — 17x"? + 94” — 168 = 0,
two of them being in the proportion of 2 : 3.
70. In the equation x — 102 + 27” — 18 = 0, the greatest
root is double of the second, and the second treble of the third;
determine all the roots.
71. One root of the equation # —52*°—#+5=0 1s 5;
determine all the roots.
72. The equation x° — zy aa ly — 1 =0, has two roots of
the form 4a, =; determine all the roots.
73. The roots of the equation
6a* — 432° + 1072” — 1084 + 36 = 0,
are of the form a, 5, : and ; ; determine them.
74, Determine the roots of the equation
a — 102° + 352” — 50a + 24 = 0,
they being of the forma +i1,a—1,6+1,0—1.
Gg
450 APPENDIX.
75. Solve the following equations, whose roots are in
arithmetical progression :
lL &@—62? —447 + 24=0.
2. #@ — 9x”? +234 —16=0.
3. @—62°> + 11% —6=0.
4. #2 —327+67+8=0.
xv’ — 102° + 35a? — 50” + 24=0.
Or
°
6. vt —sa’°+ 142° 4+ 8a@— 15=0.
Pol et ae — 1 her” 1 921+ 18& 0.
76. The roots of the equation # — px"—! + ga”~? — &e.
= 0 being in arithmetical progression ; prove that the least root is
pes lara fa am tT) usa)
NAt ni— 1 iF
” 7
and the common difference
2 {(m_— 1) .3p' — 6g].
a4 n> —1 li
77. Solve the following equations, whose roots are in
geometrical progression :
lL #—72’ + 14%7—8=0.
2 @ — 1327 + 39% —27 =
3. &@ — 142? + 562 — 64 =0.
4, x — 26a" + 156% — 216 = 0.
5. & — px? + gx—r=o.
6 @+pert+qe +re+s=0.
78. Ifthe roots of the equation
a” — par} + Ob ae oe &e.= —_—
be in geometrical progression; having ae p = 15, q = 703m
find n, 7, &c.
APPENDIX. za Ae |
79. Solve the following equations, whose roots are in har-
monical progression :
l wv — 112? + 367 — 36=0.
2. «#— 132° + 547 —72=0.
3 Do Es
Shoat ie ae eS er 0;
2 6
4. 82 —62?—3H7+1=0.
WL a ee aes Sena a eet — a 4
6 ax’ — ba’? —c#+1=0.
80. -In the common cubic 2° — px? + gv — r =0, if the
roots are in harmonic progression, and p, q, 7, integer numbers,
then 7 is the square of the greatest root. Apply this to solve
the equation 2° — 23a° + 135” — 225 = 0. |
81. If the roots of the equation 2° — pa’? + qv~—r=0,
be in harmonic progression; show that
» py—3r p— spar + or
tre ms
2+ ¢ 0,
Lv
contains the greatest and least.
82. Ifthe roots of the equation
Deep og oe ea + Qa’? — Pe +L=o0,
be in harmonic progression, then will the greatest and least be
nf (n+1).L
J (n+ 1).P = 9/ $3. (w= 1)?. P= Gn. (w= 1). QLY
J/(m+1).P+46/33.(n— 1)’. P? —6n.(n—1).QL2
83. Solve the equation x* — 31a’ + 300% — 900 = 0, whose
roots are successive triangular numbers.
84. Explain the method of finding the equal roots of
equations, and apply it to the equations
lL #@—92?+4%4 12=0.
2. — 132" + 672° — 1712" + 216% — 108 = 0.
Gge2
452 - APPENDIX.
85. Having given the equation
22° — 12”° + 19”? — 6H +9=0,
determine whether it has equal roots.
86. Ifthe equation v + ga — rx2’® — t =0 has two roots
equal to each other; prove that one of them will be a root of
; Sha. 47
. 2q
drat ye eS, ee Se 4
the quadratic x#* + = 2+ oF FE:
87. Show that if an equation have two equal roots, and the
terms be multiplied by the terms of an arithmetic progression,
the result will = o.
88. If an equation have (nm) equal roots, the equation
formed by multiplying the terms by the terms of an arithmetic
progression has (z — 1) of them.
89. Having given
xv — px + gx — r = 0, whose roots are a, 8, c3)
and w — p'v’ + g'v — r' = 0, whose roots are a, b,c’; J
find c, and c'.
90. One root being common to the two equations
xv — 9x + 267 — 24 = 0,
and 2 — 7a? +7x# + 15=0,
find the remaining roots of each.
91. Determine all the roots of the two equations which
have one common root
v— 32° + 1le~—9=0,
e— 52? +1lw7—7=0.
92. Solve the equation 2° — 1 = 0; and show that its roots —
are of the form a, 6, 67, |
93. The roots of the equation x2 + px’? + 1=0, must be
of the form a, 8, - 33 exhibit them in that form.
APPENDIX. 453
94. Solve the following recurring equations:
1 2t—3a° + 24° — 347 +1=0.
2. 2 + 6ax* — 200°2" — 6x4 + a =0.
4 53 2 5 vol
us ee a ae aS eae
4. wtil =o.
5. av? — 21at + 372° — 372° + 2127 —1=0.
4 15 37 ay 15
Goa ae a age a:
T 2 2 2
Pe Ane 100 e190 e 4a eo a Se 0.
8 a ae ns — 0
95. Reduce
9 8 7 6 5 4 3 2 —*
e—pxe+qer—ra4+sae—sae+re’—qe+pr—1=0,
to an equation of four dimensions.
96. Exhibit the quadratic factors of the equation
wm +t1i=o.
97. Show that when m is a prime number, the roots both
real and imaginary of the equation 2” + 1 = 0, are different
powers of any one of its roots.
98. In any recurring equation # — pa"! + &c. = 0,
whose roots are a, 6, c, &c.; prove that
2 6? 2 2 ae —
ata ta tat ke. = (p'— 29+ Vn). (p'—29—- V2).
99. The roots of the cubic equation 2° — gv” + 7 = 0, are
g° y?
real when = exceeds is
100. The solution of the cubic equation 2 + gv +r=0
is dependent on the solution of the equation #* — 1 =0.
454: APPENDIX.
101. Having given, 1, a, (3, the cube roots of 1,
i ce Yea a
and A ; Ng, of) f?
and B = is Seth: & = aN
L 277 |
+
prove synthetically that 4 + B, ad + PB, and B4 + aB are
the three roots of the equation.
102. If two roots of the cubic equation
a —qe+r=0 beat bY —3, and a—bY/—3;
By 3
then will — = + Wea = (6 — a)’.
103. Explain in what case, and for what reason, Cardan’s
formula for the solution of a cubic equation does not enable us
to determine the roots.
104. Determine whether Cardan’s rule is applicable to the
solution of the equation 2° — 23747 — 884 = 0.
105. Show that Cardan’s rule for the solution of a cubic
equation is applicable when all the roots are possible, and two
of them equal; and by means of it, find the roots of the
equation x + 6a? — 32=0.
106. Solve the following equations by Cardan’s rule:
e—o9v—l4=0.
xv — 62% —40=0.
xv — 9x + 28 =0.
v+32° +9" —13=0.
xv—6xr° + 347— 18 =0.
a’ — i2n”? + 57% — 94 =0.
e+ 6x2? +20” + 15=0.
av — 122° + 367 —7=0.
a" — Hem + ge" —r=0.
107. Find by the doctrine of permutations, the roots of the
equation v — ga+r=o.
APPENDIX. 455
108. Assuming the quadratic factors in Des Cartes’s solu-
tion of a biquadratic to be 2? + aw + b=0,and 2 —axr+c=0;
find the reducing equation in (0) or (ce): and show that it may
be depressed to a cubic.
109. If a@ be a root of Des Cartes’s reducing cubic, the
four roots of the equation z + ga’ +rxv+s=0,
are vB A (iss Ut to ye 87 2),
2 2
qY
110. Prove that Des Cartes’s solution of a biquadratic
succeeds when all the roots are possible, and two of them
equal; and apply it to solve the equation
e—6e+ 82°? +6%—9=0.
lll. Find the roots of the following equations by Des
Cartes’s method:
lL #@—427° — 8x + 32=0.
vt — 327 — 4% —3=0.
aot — 62° + 527 +27 —10=0.
e+ o7°—~ 72" — 8x + 12 = 0.
me OG wv
a en, Oe
112. Give Euler’s solution of a biquadratic; and show that
the cubic involved in that method has its roots four times less
than the roots of the cubic in Des Cartes’s.
113. Solve the equation 2 = 12¥ + 5, by the method
attributed to Waring. One root of the reducing cubic is 2.
114. Ifa+ BW/—1 bea root of the equation
e+pet+ge+re+s=0,
two roots are
i{- (p+ 2a) 4 pay -2(¢ B+ em) |.
456 APPENDIX.
115. Prove that if (uw) the last term of any equation be
resolved into prime factors a, 9, y, so that w =a” 2” y’, then
the number of divisors of uw will = (m + 1). (mn +1).(p +t 1).
116. Find by the method of divisors the roots of the
following equations:
lL #—62? +54 +12=0.
v— on? + 227 — 24=0.
av — 62° — 16x + 21=0.
wv’ — 42° — 84 + 32 =0.
av’ + a — 2927 — 97 + 180 = 0.
32° — 262? + 34% — 12 =0.
82° — 2627 +1la7+10=0.
sx — 452” + 734 — 30 = 0.
CON Oo KF W Wb
Perr ae eee en ee ee
117. In the method of divisors, show how the number of
substitutions may be lessened :—and in the equation
av — xv — 162? + 55” — 75 =0,
determine whether 3, 5, and — 5 are roots.
118. Apply the rule for quadratic divisors to the equation
xt — 172° + 88x — 172% + 112 =0.
119. Solve by the method of divisors, the equation
62° + 532° — 952" — 2h7 +42 =0.
120. It is always possible to find those roots of numeral
equations which are whole numbers or rational fractions with-
out the aid of formule of approximation.
121. Ifa be an approximate root of the equation
e+ px’ + qx =r, so that a + pa’+qa=r,
a.(r—7,)
TOV Ty 20° + pa
r or r, being used according as a is greater or less than 1.
Approximate by this formula to the value of # in the equation
xv — 20 = 5,
prove that 7 =a@+
5 very nearly ;
APPENDIX. 457
122. Three given quantities (¢ + 2), (a@+2)+h,(a+z2) +h,
approximations to the root @ of an equation, being substituted
for the unknown quantity, give results n, n + 6, n + 0’; show
that z will be very nearly found from the equation
2. (hd’ —h’d) + 2. (WV s’ —h*8) + nhh’. (h’ —h) =0.
123. Approximate to a root of the following equations :
lL #@—x7—50=0.
2. v—2727—5=0.
3. 2+ 27 —30=0.
4. e@+a’+r2=90.
5 av —6H +1=—0.
6 #@—20°+37—4=0.
ret ae = 3.
8. wv —12% + i= 0.
9. 22° — 162° + 402° — 307 ++1=0.
o. {ot eras
2Vy—y =2.
a: , Nik ae YE
y —x“v=6.
12. ar Meee
|e +y=s.
124. In the equation 2 + 92° + 4% = 80, approximate to
the value of x by means of a series of converging fractions.
125. Express the roots of 2° —7x# + 7=0, by continued
fractions ; and determine the accuracy of the approximation of
any converging fraction deduced from these.
126. If a be an approximate value of w in any equation,
and J, c be the results when a is substituted for z in the ori-
ginal and in the limiting equation; then will
=a-— a near!
= y-
458 APPENDIX.
127. Determine the number of positive and negative roots
in the equation #* + 42* — 192° — 34v” + 60x + 36 =0, of which
all the roots are real. :
128. Determine the same in the equation
x — 5a — 152° + 852" — 26H — 120 = 0.
129. Prove without resolving the equation into factors,
that if two numbers (a) and (4) when substituted for the un-
known quantity in the equation v — px"! + ga"—?— &c.=0,
give results affected with contrary signs, there is at least one
real root between (a) and (0).
130. If P represent the sum of the positive terms in any
equation, and a series of quantities be successively substituted
for the unknown quantity; prove that the successive increments
of P may be made less than any assignable quantity.
131. If P and WN be the greatest positive and negative
coefficients in the equation
eee 12. sd eee Pa ites: ee TA ata Re 0;
then a superior limit to the roots is N + 1; and an inferior
limit is ea or oe according as u 18 positive or negative.
132. If M2"-™ be the first negative term of the equation
B+ 0087 cn — Ma” — &e.= 0,
and if P be the greatest negative coefficient, then1 +