whence, 19z (10 x) + 3800 = 9 (10 + z) (10 ).
Reducing, z 2 19a; = 290 ;
19 , ,/i ;~36i 19 39
hence, = yy290 + = i
.-. a;'=29, a;"=-10.
24. Given ^=^ = -3 + -;
a; + <> *
whence, x (x* 5x) = (a? 9)a? + * + 3.
Reducing,
hence,
or
25. Given
whence, 2s 2 = 3 (x 2 x) + 2 (z 2 2z + 1) ;
72 7 , .749 2
hence, o^--^--, or ^-ij/---;
.-. a;' = 2 and " = -.
26 Given g + 2 s-2_5.
a;_2 s + a-8'
whence, 6 (a? + 4x + 4) 6 (z 3 4a; + 4) = 5 (& 4);
48
therefore, a? - a; = 4 ;
KEY TO DAVIES' BOUEDON.
, , . , , 24 , ,/576~~ 2426
and hy the rule, x=y -^+4= ; ;
5 iiiD O
.*. a;' = 10 and s"= f.
5
~ x x 6 a; 12 5
27. Given --: = - ;
whence, 6(z 2 12z+36) 6(z 2 24z + 144) =5(z 2 18a,-+ 72)
6
162 1008
.Reducing, a; 2 -- x = -- ;
_ 81 ./6561 5040 _ 81 39
= 1/ " " : ~~
oo n-
28. Given
25
' = 24 and a?" = ^
5
a; + 1 13
x + l^ x 6 '
whence, 6a; 2 + 6 (a? + 2z + 1) = 13 (a; 2 + x) ;
1 1 +5
hence, x = v 6^ = 5 ;
/& /*
.-. '= 2 and &"= 3.
1 23
29. Given - = ^ ;
whence, 5 ( + 2) 10 (x 2) = 3 (x* 4) ;
5 , 4 /~ ~25 5 23,
hence, = - ^ |/ 14 + ^ = g ;
.-. a?'=3 and x"= y.
, . 4 5 12
30. Given = + , , = , .. ;
whence, 4(a; 2 +5z+6) + 5(^+4a;+3)=12(a?+3a;+2);
27.
hence, x = ^
.. x'= 3 and x"= 5.
162.] EQUATIONS OF THE SECOND DEGREE. 83
i'ROBLEMS GIVING RISE TO EQUATIONS OF THE SECOND DEGREE
4. A grazier bought as many sheep as cost him 60, and after
reserving 15 out of the number, he sold the remainder for 54, and
gained 2s. a head on those he sold : how many did he buy 1
Let x denote the number purchased :
and x 15, the number sold ;
then will denote the number of shillings paid for 1 sheep,
9
and - the number of shillings received for each.
a: 15
From the conditions of the problem,
1200 _ 1080
x ~ z^l5 " '
clearing of fractions, &c.,
x 3 - + 45* = 9000 :
whence, by the rule,
_ 45 195
22'
hence. x = 75, and x 120,
the positive value only, corresponds to the required solution.
5. A merchant bought cloth for which he paid 33 15*., which he
sold again at 2 8*. per piece, and gained by the bargain as much
as one piece cost him : how many pieces did he buy ?
Let x denote the number of pieces purchased :
675
then will, denote he number of shillings paid for each,
fi4 KEY TO DAVIKS' BOURDON. [163.
and 48# the number of shillings for which he sold the whole.
From the conditions of the problem,
48* -675=^-
x
then, by clearing of fractions, &c.,
whence, by the rule,
225 225
, , , /* - . _ i
16 16
_225^ /
X ~ 32 " V
225 50625 225 255
~T6~ 1024 : ~ "32" " "32"
480
using the positive value only, x = = 15.
HI
6. What number is that, which being divided by the product of
its digits, the quotient will be 3 ; and if 18 be added to it, the order
of its digits will be reversed 1
Let x denote the first digit,
and, y " second "
ehen will Wx -f y denote the number.
From the conditions of the problem,
10* + y _ o
o,
xy
10a:-f-y+ 18 = 1
whence, by reduction,
finding the value of x in terms Df y from the second, and sub-
sti tnting in the first, we have.
163.] EQUATIONS OF THE SECOND DEGREE. 85
10y 20 + y = 3y 2 6y ;
whence, by transposing, &c.,
17 , 20
- y2 = - ; and,
by the rule,
17 / 20 , 289 , 17.
-6-* V ""a + S6 =+ 1T
taking the positive sign, y = 4 ;
whence, a, = 2, and the number is 24.
7. Find a number such that if you subtract it from 10, and mul
tiply the remainder by the number itself, the product will be 21.
Let x denote the number :
from the conditions of the problem,
(10 - *) x = 21 ; or, x z 10* = - 21 j
by the rule,
x = 5 y - 21 + 25 = 5 db 2;
whence, x =. 7, and x = 3.
8. Two persons, A and B, departed from different places at the
same time, and travelled towards each other. On meeting, it ap-
peared that A had travelled 18 miles more than B; and that A
could have performed B's journey in 15|- days, but B would have
been 28 days in performing A's journey. How far did each travel 1
Let x denote the number of miles B travelled ;
x + 18 " " " A "
" " A " in one day,
KEY TO DAVIES' BOTJKDON. [163.
denote the number of miles B travelled in one day ,
_ _l_ I Q
" " days A
15}
tx -f 18\
V28-;
B
from the conditions of the problem,
x+ 18 _ x_ x 2 H- 36s -f 324 _ x*
~>T ~* ~28 ~I5
clearing of fractions, and reducing,
2 324 _ 2916
'^T a ~'
By the rule,
162 /2916 26244
: :t ~~ ~~'
162 216
nence, using the upper sign,
378
* = - = 54.
9. A company at a tavern had 8 15s. to pay for their reckoning ;
but before the bill was settled, two of them left the room, and then
those whc remained had 10s. apiece more to pay than before: how
many were there in the company 1
Let x denote the number in the company.
163.] EQUATIONS OF THE SFCOND DEGREE. 87
175
Then will denote the number of shillings each should pay ;
" paid;
from the conditions of the problem,
17C -|7C
I/O I/O
* 2 *~ '" '
clearing of fractions,
175* 175* + 350 = 10* 2 - 20*;
whence, * 2 2* = 35.
By the rule,
x = 1 y36= 1 6;
using the upper sign, x = 7.
10. What two numbers are those whose difference is 15, and o*
which the cube of the lesser is equal to half their product?
Let x denote the smaller number ;
then will x + 15 " greater "
from the conditions of the problem,
*3 = \ (3.2 + 15*), or, x* = 1 (x + 15) ;
1 15
x
By the rule,
whence, x z -x = .
z
4- a -V~2~^16-4- I -T
using the upper sign, x = 3 ; hence, * -f 15 = 18.
11, Two partners, A and B, gained $140 in trade: A's money
88 KEY TO DAVIES* BOUKDON. [163.
was 3 months in trade, and his gain was $60 less than his stock : B's
money was $50 more than A 's, and was in trade 5 months : what
was A's stock?
Let x denote the number of dollars in A's stock ;
x + 50 " " B's "
x 60 " A's total gain ;
x 60
As gain per month ;
3
x 60
" A's " on 1 dollar ;
50) B's "
( - J (x + 50) 5 B's total gain.
\ OiP /
From the conditions of the problem,
x 60 + (* ~ 60) (x + 50) 5 = 140 ;
clearing of fractions, and reducing,
~~4~ a
By the rule,
105625 _ 325 475
64 ~ ~8~ IT
whence, x = 100.
12. Two persons, A and B, start from two different points, and
travel toward each other.. When they meet, it appears that A has
travelled 30 miles more than B. It also appears that it will take A
4 days to travel the road that B had come, and B 9 days to travel
175.] EQUATIONS OF THE SECOND DEGREE. 89
the road which A had come. What was their distance apart when
they set out?
Let x denote the number of miles B travelled ;
then will ar + 30 " " " A
2
A travels per day ;
x + 30
- days A "
(a:
V
-f 30
B
From the conditions,
x _ x_+ 30 x 2 _ x 2 + 60* -f- 900
r ' T~~ ~9"
whence, by reduction,
* 2 48* = 720
and by the rule,
x = 24 -y/^20 + 576 = 24 36 ;
taking the upper sign, x = 60, and x + 30 = 90 ;
hence, the distance is 150 miles.
EXAMPLES INVOLVING RADICALS OF THE SECOND DEGREE.
CL I CL^ 1 ' (ifi X
3. Given - + \ / - - = T ,
x V x z b
to find the values of x.
90 KEY TO DAVIES' BOURDON. [175.
Multiplying botn members by bz, and transposing,
b y/a 2 x 2 =i x 2 - ab ;
squaring both members,
cancelling 6 2 a 2 , dividing both merr.bers by a 2 and transposing,
x* = 2ab-b* .-. x=
4. Given \ /_^__? + 2
Multiplying both members by
z -f a
a;
multiplying both members by a:, and transposing,
2-y/aa; = b 2 x x a = (b 2 l)x a ;
squaring both members,
4ax = (b* - 26 2 + 1) x z - 2a (6 2 - 1) x + a 2 ;
transposing and reducing,
-,2
rX =
5* _ 26 2 -j- 1 5* 26 2
whence,
/ a 2 2 (6*
1 d: V ~ 6* - 26 2 + 1 + (6* -
- 26 2 -f 1 : ~ 6* - 26 2 + 1 (6* - 26 2 -f 1 ) 2 '
or,
~ b* 2b z + 1 ~ 6* - 26 2 + 1 '
now, b* - 2b z -f 1 = (b - 1) 2 (6 + I) 2 .
175.] EQUATIONS OF THE SECOND DEGREE.
Hence, taking the upper sign and reducing,
91
_
=
_q(64-l) 2 _ a(b + I) 2 a
= ~ ~
(62 _ t )2 - (b- l) 2 (6 4- I) 2 ~ (b - I) 2 '
and taking the lower sign and reducing,
_ a(b - I) 2 _ g(b 1)2 q
~ (52 _ !)2 - (b _ 1)2(6 4. 1)2 - (j + i)a'
or, uniting the two values in a single expression,
a
x
(b =F i) 2
a -J q 2 a; 2
5. Given, . ; = 0, to find x.
a 4- -y/ a 2 a; 2
Clearing of fractions, transposing, &c.,
squaring both members,
2 (i _ j)2 - (6 4. 1)2 ( a 2 _ 3.2) .
transposing and reducing,
2a -^b
(6 4- I) 2
x
6. Given,
a 7ia
, to find x.
Multiplying both terms of the first member by */x
and then dividing both members by a,
I n 2
2x a
clearing of fractions, &c.,
a: a
a,
92
KEY TO DAVTES BOUKDON.
[175.
(1 2n 2 )x (1 n 2 )a = 2w 2 -y/a; 2 ax ;
squaring both members, transposing and reducing,
x * _ aiLrJ^a, = -.L=^l*.
1 r 4-/i^ 1 - ^yj.2
whence by the rule,
^ ~ 1 ]J 9 " \/ 1 XI 9 ' 7l
1 4/i" 2 V 1 4^ (1 -
_
1
2n 3 a
- 4n 2 4n z ~ 1 4n*
Taking the upper sign, and dividing both terms of the fraction by
1 +2n,
2n + n a )a _ (1 - n) 2 q
_
2n
Taking the lower sign, and dividing both terms by 1 2n,
+ 2ra + ft 2 ) _ (1 + n) 2 a
_
+2n
a (I n} 2
taking the two values together, x = ^ -- =-*
n- V "" ' X \ "V a x I X t. c. j
7. Given v = + =\/ T , to find x.
IX
Multiplying both members by ^fx
/a + x -f- i/o a; = ~~= >
V*
squaring ooth members,
175.] EQUATION* OF THE SECOND DEGREE. 93
squaring both members,
cancelling 4o 2 in the two members, and dividing both by x*,
_ x 2 4a
= P~~b'
clearing of fractions, &c.,
z 2 = 4ab 46 2 . . x = 2 Jab &T
a x - , c ,
8. Given ? to find x.
a -+ x
Clearing of fractions, transposing and factoring,
^/2ax + x 2 = (b 1) (a + x).
Squaring both members,
2ax + x 2 = (6 2 -26 + 1) (a 2 + 2a* + * 2 ) ; or,
2a* + z* = (6 2 26) (a 2 + 2a* + * 2 ) + a 2 + 2a -f-
whence, by reduction,
/I _ 26 +
-
whence, by the rule,
=
-
26
a 1 -
taking the upper sign, x =
<
a(\
taking the lower sign, x = -- J
whence, a? =
KEY TO DAVIES' BOtTRDON. [175-178.
2d Solution.
Make x -f a = y ; whence, 2ax -f z 2 = y 2 a*
substituting in the equation, and clearing of fractions,
y -f yy a 2 = iy ;
transposing, &c.,
V> 2 a 2 = (5 1) y ;
squaring both members, and cancelling,
- a 2 = (i 2 - 26) y 2 ;
whence, solving with respect to y,
a
j """ **-
substituting for y its value, &c.,
x = a
, ,.
whence, as before,
^/26 - & 2
r ^a(l--
-V/26 - 6 2
TBINOMIAL EQUATIONS
6. Given x* (2bc + 4a 2 ) ar 2 = 6 2 c 2 ; to
By the rule,
4o* = e + 2u 2 2a v f
7. Given 2 - 7 y^ = 99 ; or, 2z - 99 = 7 ^/i
Squaring both members
4* 2 396 + 9801 = 49a? ;
178-181] EQUATIONS OF THE SECOND DEGREE. 95
transposing and reducing,
445 9801
* = -
Ahcnce, by the rule,
445 / 9801 , 198025
ff _ L __ f __. -^. * / __ __ I _ _ , _
* ~~~ <" ~^ \ / /* j J
445 203
or, x = ^ :
648
taking the upper sign, x = - = 81
o
242 121
lower sign, x = r- = r
8 Given, ^ bz* + C -x* = 0, to find*;
transposing and reducing.
whence, or = 1
reducing, * = J
V
2W
EXAMPLES OF REDUCTION OF EXPRESSIONS OF THE FORM OF
Vb.
4. Reduce to its simplest form, \7^ ~
a = 28, 5 = 300. c = 22.
KEY TO DAVIES BOUEDON. [181.
Applying the formula and considering only the upper sign,
v/
28-f 10 1/3 =
5. Reduce to its simplest form, */l + 4^3.
o=l, b - 48, c = 7 ;
applying the formula, &c.,
+ 4-v/- 3 = 2
6. Reduce to its simplest form,
_ 6 2 _ ^J bc _
a = ic, i = 46 2 (6c - 6 2 ), c = b (c - 2b) ;
applying the formula to the 1st radical,
+ 26 -vA - * 8 =
applying the formula to the 2d radical
subtracting the second result from the first,
6 ytc b 2 bc 2b -foe b 2 = 2A.
7 Reduce to its simplest form,
\/ab -r 4c 2 d 2 2
a = aft -f 4c 2 ' P = ZVU-&
or, p = % and p = ~.
4
Using, p = -, we have from (3),
x'= + 8 and x"= 8; whence, y'= -f 2 and y"= 2.
9.
Making
9 j s2 + 4^ + 4^ = 256 (1)
(3y z a^ = 39 (2)
x 1 -\- kpx* + 4j0 2 .r 2 = 256 (3)
3p*x* x* = 39 (4)
From (3) and (4),
1 -(- 4p
Equating,
256 39
ufi&06
768P 2 - 256 = 39 + 156jo + 156J0 3 ,
13 295
and P 2 ^rP=7^r;
KEY TO DAVTES' BOURDON. [184
hence,
78 432 5 59
p -- -eia-' or ^ = 6 and * = -ioa'
Using the first value, we have from (6),
a;2 = 36; . 3'= 6 and x"= 6;
and by substitution,
y'= 5 and y"= 5.
Using the second value, we find,
a' =102 and x" =102;
and by substitution,
y'= 59 and y"=. 59.
10 J6(a? + ^) = 13ajy ...... (1)
(z2 2/2 = 20 ........ (2)
Making
6(1 +^ 2 )a; 2 =: 13^ 2 ...... (3)
(1-^2)^ = 20 ....... (4)
From (3), we have,
6 + 6p 2 = I3p.
Reducing,
_ 13 . /169 _ 144 _ 13 5
'* ^ ~ 12 ^ 144 ~~ 144 ~ 12 '
or
^ = H and p = -.
188.] EQUATIONS OF THE SECOND DEGREE. 101
Using the second value of p, we have from (5),
on
a? = T --=4 = 36, or x'= 6 and x"= 6 ;
1 f
and by substitution, y' 4 and y"= 4.
EQUATIONS OF A HIGHER DEGREE THAN THE FIRST, INVOLVING
MORE THAN ONE UNKNOWN QUANTITY.
( x 2 u + xv 2 = 6 ... (1) )
15. Given, < f- to find x and y.
tzy + zy = i2.. . ( 2 )>
Dividing (2) by (1), member by member,
2
xy = 2, or a; =
Substituting this value of x in (1) and reducing,
* + * = ;
clearing of fractions and reducing,
y 2 - By = - 2 :
whence, y = | 1 /- 2 + i = i I;
or, y = 2, and y = 1 ;
whence a? = 1, and a; = 2.
16. Given, J *' + * + y = 18 ~ y2 ' ' ' 0) It o find , and y.
( xy 6 (2) )
Multiplying both members of (2) by 2 and adding the resulting
equation to (1), member to member,
z' -f 2*y + y 2 + * \-y = 30,
or, (x + y) 2 + x + y = 30 ;
102 KEY TO DAVIES' BOURDON. [189.
whence, by the rule,
whence, x -\- y = 5, and x + y = 6.
Taking the first value of x + y and substituting in it, for y its
/
value derived from (2),
x
x + = 5;
x
clearing of fractions and reducing,
x z 5x = 6 ;
. 5 / _ , 25 5 ^ 1
whence, a . = _y/_6 + = -- ;
ur, x = 3, and # = 2 ; whence, y = 2, and y = 3.
Taking the second value of x + y and proceeding as before,
clearing of fractions, &c.,
a; 2 + 6z = 6 ;
whence, x 3 / 6 + 9 = 3
and by substitution, y = 3
PROBLEMS GIVING RISE TO EQUATIONS OF A HIGHER DEGREE
THAN THE FIRST CONTAINING MORE THAN ONE UNKNOWN
QUANTITY.
2. To find four numbers, such that the sum of the first and fourth
shall be equal to 2s., the sum of the second and third equal to 2/,
the sum of their squares equal to 4c 2 , and the product of the first
%nd fourth equal to the product of the second and third.
189-191.] EQUATIONS OF THE SECOND DEGREE. 103
Assuming the equations,
+ 2 = 2* (1)
x+y = W .... (2)
w 2 -(- z 2 + y 2 + z 2 = 4c 2 . . (3)
uz = xy (4)
Multiplying both members of (4) by 2, and subtracting from (3),
member from member,
u 2 %uz -f- z 2 + a; 2 -f 2ary + y 2 = 4c 2 ;
or, ( z) 2 + (x 4- y) 2 = 4c 2 .
Substituting for x + y its value 2s' and transposing,
(u - z) 2 = 4c 2 - 4s' 2 ,
or, u z -v/4c 2 4s' 2 .
Combining with (1),
,/4c 2 - 4*' 2
and, z s v - ~ - s ^& s' 2 ;
reversing the order of the members of (4), and proceeding as before,
we find in like manner.
and,
4. The sum of the squares of two numbers is expressed by a,
and the difference of their squares by b : what are the numbers?
Let x and y denote the numbers.
From the conditions of the problem,
J04: KEY TO DAVIEs' BOURDON. [191.
x 2 + y 2 = a - (1)
z*-y* = b . (2).
By adding, member to member,
by subtracting,
5. What three numbers are they, which, multiplied two and two.
and each product divided by the third number, give the quotients
, 6, c ?
Let #, y and 2, denote the numbers
From the conditions of the problem,
- = a or, xy =z 02 . (1)
P
= b or, yz = bx (2)
#2
= c or, xz = cy (3).
Multiplying (1), (2) and (3) together, member by member,
x z y z z 2 = abcxyz ;
dividing both memoers by xyz,
xyz = abc (4) ;
substituting in (4) the value of xy taken from (1). and dividing both
members by a,
2? = be . . z =
Substituting the value of yz and dividing by b,
191.] EQUATIONS OF THE SECOND DEGREE. 105
x 2 = ac . ' . x = Joe.
Substituting the value of xz and dividing both members by c,
y z = ab .'. y
6. The sum of two numbers is 8, and the sum of their cubes is
152 ; what are the numbers ?
Let x and y denote the numbers.
t
From the conditions,
x + y = 8 ..... (1)
z3 + y3 = i52 ..... (2);
cubing both members of (1),
x 3 + 3z 2 y + 3xy* + y 3 = 512 (3) ;
subtracting (2) from (3), member from member, and dividing both
members by 3,
x z y + xy z = 120 (4) ;
-
substituting the value of a; taken from (1),
(64 - 16y + y 2 ) y + (8 - y) y z = 120,
or, reducing, y 2 Sy 15 ;
whence, y = 4 -y/ 15 + 16 = 4 1 . . y = 5, y = 3;
whence, from (1) a; = 3, a? = 5.
7. Find two numbers, whose difference added to the difference ol
their squares is 150, and whose sum added to the 3um of theii
squares, is 330.
Let x and y denote the numbers.
From the conditions of the problem,
106 KEY TO DAVIES' BOURDON. [191.
a?_y2 + *-y=150 (1),
' y *^ y \ J t
adding member to member, and reducing,
x 2 + x = 240 ;
whence, x = - -/240 -f ^ = ^ y J
or, considering only the positive solution, .
x = 15;
whence, from (1), by substitution,
8. There are two numbers whose difference is 15, and half their
product is equal to the cube of the lesser number : what are the
numbers 1
Let x and y denote the numbers.
From the conditions of the problem,
*-.y=15 (1),
f - (2);
qp
substituting in (1) and dividing both members by 2,
2 _y _ 15.
y 2~ 2 '
whence,
15 J_ _ 1 11
considering only the positive solution,
y = 3 ; whence, from (1), x = 18.
191-192.] EQUATIONS OF THE SECOND DEGBEE. 107
9. What two numbers are those whose sum multiplied by the
greater, is equal to 77 ; and whose difference, multiplied by the
lesser, is equal to 12?
Let x and y denote the numbers.
From the conditions,
(x + y)x = n, or z*+'zy^W . . . (1) ;
(x-y)y=l2, or xy - y* = 12 . . . (2);
make x = py ; whence,
Tt t or, ya = -JL- . . . (3),
(P -I)y 2 =12, or, y* = j . . . (4);
equating the second members and reducing,
2 _65 _77 i
P l'2 P ~ 12'
65 / 77 , 4225 65 23
whence, p = ^/ - -+ = .
taking the upper sign,
/Qi* O
substituting k (4),
whence, x = -y/2~;
i. i. , 42 21
taking the lower sign, p = ;
= 514 . . (2),
multiplying both members -of (1) by 2, adding and subtracting the
resulting equation to and from (2), member by member,
ac* -f 2xy + 7,2 = 1024 . . (3)
*2-2zy + y 2 = 4 . . . (4);
extracting the square root of both members,
x + y = 32,
*- y = 2,
whence, x = 17 ; y = 15.
13. There is a number expressed by two digits, which, when
divided by the sum of the digits, gives a quotient greater by 2 than
the first digit ; but if the digits be inverted, and the resulting num-
ber be divided by a number greater by 1 than the sum of the
digits, the quotient will exceed the former quotient by 2 : what is
the number?
110 KEY TO DA VIES' BOURDON. [192.
Let x and y denote the digits ; then will
lOx + y lenote the number.
From the conditions,
clearing of fractions, and reducing,
Sx y - x z - xy = (3),
6y 4z x z xy = 4 (4) ;
by subtraction,
12*-f4
fy-12* = 4; .-. y = ;
substituting in (3),
12* + 4 12^ + 4_
0>c 7 ~
clearing of fractions and reducing,
40 4
-* = -
_20 / 4_ 400 _ 20 18
"19 V 19 + 361 ~ 19 19*
Taking the upper sign, gives x = 2 ; whence, y = 4.
14. A regiment, in garrison, consisting of a certain number of
companies, receives orders to send 216 men on duty, each company
to furnish an equal number. Before the order was executed, three
of the companies were sent on another service, and it was then
found that each company that remained would have to send 12 men
additional, in order to make up the Complement, 216. How-mam
192.] EQUATIONS OF THE SECOND DEGREE. Ill
companies were in the regiment, and what number of men did each
of the remaining companies send ?
Let x denote the number of companies, and
y " " " each should send ; then,
y + 12 will denote the number sent by each.
From the conditions of the problem,
*y = 216 (1),
(*-3) (y+12)=216 . (2);
Performing operations, subtracting and reducing,
4x y = 12. . . y = 4a; 12 ;
substituting in (1), 4z 2 - 12* = 216, or ** 3z = 54 ;
whence,
taking the upper sign,
x = 9 ; hence, y = 24, and y + 12 = 36.
15. Find three numbers such, that their sum shall be 14, the
sum of their squares equal to 84, and the product of the first and
third equal to the square of the second.
Let x, y and z denote the numbers.
From the conditions of the problem,
x + y + z = 14 . (1),
* 2 + y 2 + * 2 = 84 . (2),
yz = y 2 (3).
Multiplying both members of (3), by 2, adding to (2) and -educing,
KK1 TO DAVIES* BOURDON. [192.
from(l), x + z = U-y (5);
equating the second members of (4) and (5) and squaring,
84 + y 2 - 196 28y + y 2 ; . . y = 4.
Substituting in (1) and (3),
x + z = 10 . . . . (6)
16
xz = 16 . .
Substituting in (6) and reducing,
2 2 _ 10* = - 16
2 = 5 ,./9~= 5 =fc 3,
V *M
= 8; a = 2,
and by substitution, x = 2 ; a; = 8.
16. It is required to find a number, expressed by three digits,
such, that the sum of the squares of the digits shall be 104; the-
square of the middle digit to exceed twice the product of the other
two by 4 ; and if 594 be subtracted from the number, the remainder
will be expressed by the same figures, but with the extreme digits
reversed.
Let a?, y and z denote the digits ;
then, lOOx + lOy + * will denote the number.
From the conditions of the problem,
z 2 + y 2 + * 2 = 104 . . . (1)
y z Zxz = 4 . . . (2)
100* + 10y + z - 594 =1002 + lOy 4- x (3) ;
193.] EQUATIONS OF THE SECOND DEGREE. 113
subtracting (2) from (1), member from member,
x 2 -i 2xx + z 2 - 100 . . x + z 10 ;
reducing (3) x z = 6 ;
hence, x = 8, and z = 2.
By substitution, y = 6, and the number is 862.
17. A person has three kinds of goods which together cost
$230^. A pound of each article costs as many fa dollars as there
are pounds in that article : he has one-third more of the second than
of the first, and 3^ times as much of the third as of the second .
How many pounds has he of each article ?
Let ar, y and z denote the number of pounds of each article.
From the conditions of the problem,
or x " + * + * = 6525 1
4 16
y = \* .'. 2/ 2 = y 2 .-. (2)
2 = -y .-. z=-2%, and, z 2 = z 2 ... (3) ;
substituting in (1) and reducing,
* 2 = 225 .'. . x = 15,
substituting in (2) and (3),
y = 20 ; z = 70.
18. Two merchants each sold the same kind of stuff: the second
sold 3 yards more of it than the first, and together, they received
35 dollars. The first said to the second, " I would have received 24
dollars for your stuff." The other replied, "And I would have
114 KKY TO DAVTES' BOURDON. [193.
received 12^ dollars for yours." How many yards did each of
them sell?
Let x and y denote the number of yards sold by each.
24
Then will denote the price the first received per yard,
25
and Q- will denote the price the second received per yard.
From the conditions of the problem,
* + 3 = y,
+ . = 35, or, 4S* 2 + 25y 2 = ?0zy :
substituting in the second equation the value of y taken from the
first,
48* 2 + 25 (x z + Qx + 9) = 70* 2 + 210* ;
reducing, x 2 - 20x = 75 ;
whence, x = 10 -y/25^ = 10 5,
/
or, ar = 15 ; x = 5 ;
substituting, y = 18 ; y = 8.
19. A widow possessed 13000 dollars, which she divided into
two parts, and placed them at interest, in such a manner, that the
incomes from them were equal. If she had put out the first portion
at the same rate as the second, she would have drawn for this part
360 dollars interest ; and if she had placed the second out at the
same rate as the first, she would have drawn for it 490 d( liars
interest. What were the two rates of interest ?
Let x and y denote the rates per cent.
Let z denote the 1st portion ; then will 13000 z denote the 2d.
224.] EXTRACTION OF ROOTS. 115
P roni the conditions of the problem,
xz (13000 z\ y
or, = 1300<*-* (1),
j^ = 360, or, 2 y = 36000 . . . (2),
p^-? = 490, or, 13000* -zx = 49000 (3).
Substituting in (1) the values of zy and zx taken from (2) and (3)
and reducing, we find, x = y + 1.
Substituting this value of x and the value of z taken from (2) in
(1), and reducing, we find
72 36
v -- v :
y 13 y 13'
36 /36 1296 36 42
whence, y -.= - ^ - + = __; .-. y = 6 ;
by substitution,
* = 7, and = 6000, 13000 - 2 = 7000.
ADDITION AND SUBTRACTION OP RADICALS.
2. 3 A/4a 2 :
3. 2 A/45 = 6 A/5; 3A/5, .'. ^4/w. 9A/57
4. 3a V* 2c A/*, .'. Ans. (3a 2c) V*-
5. 3 Vi2 2 V2a ;
3A/4^ = 3A/2a; .'. Ans. V%
6. A/243 = 9 A/3; A/27 = 3\/3; A/48 = 4 A/3. -4ws. 16 A/3.
116
KEY TO DAVIES' BOURDON. [224-225.
7.
5 A/720 5 =
= 56
. (13a - 56) VSai
8.
9.
10.
5.
6.
V6 4
16a 4 = 2a yb + 2a, and
+ 2,
; .'. Ans.
MULTIPLICATION OF RADICALS.
X 3 = 1 /64 x
* /r 3 /r i2 /r /T 12 /"
V 2 X V 3 =: V 8 X Vil =! V 648 ; ' ' ^ "
6 /2 10
- VI- 4
^; 3 10"= 3/100; .-. ^ns. 6 337500
10
8.
. 048000.
7/4 42 /4096 3/1 42/i
O ./ t / . 4/_ / . 14 /ft
'VI V 729 ' V2~ V 10384' V
. ' . Ans.
10. The product by the rule for multiplication is
28 . ft . n . 10 43 . H 43 . 13
42 /2
226-227.] TRANSFORMATION OF RADICALS.
9w 2 3m
117
11.
12. The product equals
3m " 3m
.-. Ans. = Va? + P.
DIVISION OF RADICALS.
2. 2 V3 x V4 = 2 v^729 x 1 4 c~ m multiplied by
-- + -=-; 4 5 = 1; m + w = w m;
hence, 6a~ 4 c- m x
4. (q^) multiplied by ^a*.
3341 3
v * v 2
3 X 3~9' 3 X ~3'
hence,
239.] TKANSFOBMATION OF RADICALS.
1 1 1_ J L . 1 1 !_?.
3 X 3 X 3 ~ 27 ' o + o + o o 5
(q 6
6. at divided by a~*.
\ 3 !
hence, -<,*= _
j 3_17.
S i V 4l~~3 4~12'
, 11
hence, at ^- at = a 1 ^.
7. a* divided by .
3_4_ 1
4 5~ 120 J
hence, a* -j- a^ = a~^V.
8. a x i* divided by a~ zb
2_/l\_ 2 1__9_. 3_7_ _1.
5~\2/~~ 5 + 2~10' 4 8~ 8'
hence, a*b* -$- a~%b = flifrJ .
9. 32a^J 6 c^ divided by
= ; -
hence, 32aii0
10. 64a 9 jVt divided by
64-32 = 2; 9-(-9)=18; |-
hence, 64a 9 jV^ -r- 32a-5~ *c~* = 4a 18 i 5 .
11. VS x Vji x
hence, x x
12. Reduce ? , x to its s i mp i es t terms.
120 KEY TO DA VIES' BOUEDON. [239.
Cancelling the
3. Transform the equation
a? _ Q X Z + y x _ 10 = o
into one whose second term shall be wanting.
Here we make the roots of the resulting equation greater
p
than those of the given equation by = 2 ; that is,
o
we make them less than those of the given equation by + 2.
OPERATION.
l_6 + 7- 10 [2
+ 2_8 2
1st quotient, 1 4 1, 12 1st rem.
+ 2-4
2d quotient, 1 2, 5 2d rem.
+ 2
3d quotient, 1, 3d rem.
132 KEY TO DAVIES' BOITKDON. [349.
Hence, the transformed equation is
ys _ 5y _ 12 o.
4. Transform the equation,
a^ + gz 8 3 + 4 =
into one whose second term shall be wanting.
Here we make the roots of the resulting equation greater
p
than those of the given equation by + = 3 ; that is, we
o
make them greater than those of the given equation by 3.
Therefore, the synthetical divisor is 3.
OPERATION.
1 + 9 i + 4|_ 3
3 18 + 57
1st quotient, 1 + 6 19,+ 61 1st rem.
1 3 9
2d quotient, 1 + 3, 28 3d rem.
-3
3d quotient, 1, + 3d rem.
Hence, the transformed equation is
y& _ %sy + 61 = 0.
5. Transform the equation
& _ 8z3 + 7s 2 + 3x + 4 =
into one whose second term shall be wanting.
Here we make the roots of the resulting equation greater
p
than those of the given equation by = 2 ; that is,
wo make them less than those of the given equation by + 2 ;
hence, the synthetical divisor is + 2.
249-259.]
EQUAL ROOTS.
133
OPEBATION.
1-8+ 7+ 3+ 4 [2
+ 2 12 10 14
1st quotient, 1 6 5 - 7, 10 1st rem.
+ 2 8-26
2d quotient, 1 4 13, 33 2d rem.
+ 2- 4
3d quotient, 1 2, 17 3d rem.
+ 2
4th quotient, 1,+ 4th rem.
.-. y 1 17y 2 33# ]
EQUAL ROOTS.
4. What are the equal factors of the equation
8s -f 36 = 0?
The first derived polynomial is
7* - 42s 5 + 50.r 4 + 88z 3 - 129* 2 - 70* + 48,
and the common divisor between it and the first member of the
given equation, is
The equation
x* 3z 3 + Sx 2 + 7x -f- 6 = 0,
cannot be solved directly, but by applying to it the method of equal
roots ; that is, by seeking for a common divisor between its first
member and its derived polynomial,
4z 3 - 9z 2 - Ox + 7,
we find such divisor to be x -f- 1 ; hence, x -\- 1 is twice a factor of
the first derived polynomial, and three times a factor of the fir?t
member of the given equation (Art. 271).
134 KEY TO DAVIES' BOURDON. [359.
Dividing,
6 = 0, by (a +!)* = * 2 + 2ar + 1,
we have, x 2 bx + 6,
which being placed equal to 0, gives the two roots
x =. 2 and x = 3,
and the two factors, x 2 and x 3.
Therefore, (a: 2) and (# 3), each enters twice as a factor of the
given equation ; hence, the factors are
(x - 2) 2 (x 3) 2 (x + I) 3 . Ans.
5. What are the equal factors of the equation
33** + 27z - 9 = ?
The first derived polynomial is
7* 6 - 18* 5 + 45z 4 - 76z 3 + 81* 2 - 66* + 27,
and the common divisor between it and the first member of the
given equation, is
x* 2x 3 + 4z 2 Qx + 3.
The equation
x* 2x 3 + 4z 2 Gar + 3,
cannot be solved directly, but by applying to it the method of equal
roots, as in the last example, we find the derived polynomial to be
4x 3 6x 2 + 80; 6
and the common divisor to be x 1 ; hence, (x 1) enters twice
as :i factor into the derived polynomial, and three times as a factor
into the first member of the given equation.
Dividing
x t 2a; f 4z 2 6* + 3 by (x I) 2 = a; 2 2x + 1,
359-375-384.] SMALLEST LIMIT IN ENTIRE ROOTS. 135
we have for a quotient x z + 3 ;
hence, (z 2 + 3) enters twice as a factor into the first mem-
ber of the given equation ; hence the factors are
(x - I) 3 (x + 3)3.
SUPERIOR LIMIT IN ENTIRE ROOTS.
2 What is the superior limit of the positive roots in the equation
x s _ 3^4 _ Q X 3 _ 25z 2 + 4z 39 = ?
Recollecting that if we use x for x' (Art. 285), we have
X = x 5 3z* 8z 3 25z 2 + 4z 39
T = 5z 4 - 12z 3 - 24z 2 - 50* + 4,
Z = 20x 3 - 36a; 2 - 48* - 50,
V= 60s 2 - 72z - 48,
TF= 120* - 72.
r=120
The least whole number that will render all these derived polyno-
mials positive is 6 ; hence, 6 is the superior limit.
3. What is the superior limit of the positive roots in the equation
x 5 5z 4 13* 3 17x 2 - 69 = 0.
The process of solution is the same as in the last example, and the
limit is found to be 7.
COMMENSURABLE BOOTS.
2. What are the entire roots of the equation
x* 5x 3 + 25* 21 = ?
The divisors of the last term are + 1, I> +3, 3, -|- 7, 7
+ 21, and -21: L = 22; - L"= - 4.
KEY TO DAVIES' BOURDON. [384.
+ 21, +7, -f-3, +1, -1, -3, -
- 1, - 3, -7, - 21, + 21, + 7, +
+ 24, 4- 22, 4- 18, 4- 4, 4- 46, + 32, +
4- 3, 4- G, 4- 4, -46,
4- 2, 4- 4, + 46,
3, - 1, + 41,
- 1, - 1; -41,
therefore, -}- 3 and -f 1 are the two entire roots. Dividing the
first member of the equation by the product of the factors
(x - 3) (x - 1) = x> - 4x + 3,
we have x 2 y = 0.
NOTE. In the 4th line we add the co-efficient of x*, which is 0, and then
divide by the divisors, and thus obtain the 6th line.
3. What are the entire roots of the equation
15z 5 - 19* 4 + 6a; 3 + 15* 2 - I9x + 6 = ?
+ 6, + 3, + 2, 4- 1, - 1, - 2, - 3, - 6,
4- 1, 4- 2, 4- 3, 4- 6, 6, -3, 2, - 1,
-18, -17, -16, -13. -25, -22, -21, -20,
- 3, - 8, -13, t-25, + 11, + 7,
+ 12, + 7, + 2, + 40, + 26, + 22,
4-2, 4-2, -40, -13,
+ 8, +8, -34, - 7,
+ 8, 4-34,
-11, 4-15,
-11, -15,
4- 15:
hcuoe, there is but one entire root, which is -- 1
384-392.] NUMBER OF REAL ROOTS. 137
4. What are the entire roots of the equation
9z + 30s 5 + 22Z 4 + 10s 8 + 17s 2 20z + 4 = ?
This is worked like the preceding example, giving the en-
tire root, 2. Then dividing the equation by x + 2, we
find a new one, which has a root, 2.
NUMBER OF REAL ROOTS.
3. What is the number of real roots of the equation
a?_5z2_|_ 8z 1=0?
By finding the expressions which indicate, by their change of
sign, the existence of real roots (Art. 293 and Example 1), we
have
X= z 3 5y?+Sz I
X,= 3Z 2 IQx + 8
X 8 = 2x 31
X 8 = 2295
x = oo gives 1 2 variations,
x = + oo gives + + H 1 variation ;
hence, there is one real and two imaginary roots (Art. 293).
For x = 0, we have 1 2 variations,
for x = 1, " + H 1 variation ;
hence, the real root lies between and + 1.
4. Find the number, places, and limits of the real roots of
z* Sz 3 + 14s 2 + Ix 8 = 0.
For solution, see Example 3, page 141 of Key.
5. Find the number, places, and limits of the real roots of
the equation
a? _ 23x 24 = 0.
KEY TO DAVIES' BOURDON. [393-396.
X = y? 23x 24
X 1= 3z 2 23
X 2 = 23z + 36
X 3 = 8279.
For x oo, we have, H 1-, 3 variations.
For x= + oo, we have, + + + +, no variations.
Hence, there are 3 real roots, which are easily placed.
6. In the sixth example, we have,
X = y? + f z 2 2x 5
X 1= 3z2 + 3x 2
X 2 = llx + 28
X 3 = - 1186.
Hence, there is but one real root, and that lies between
the limits 1 and 2.
CUBIC EQUATIONS.
1. What are the roots of the equation
X 3 _ 6z 2 + 3* = 18 . . (I) 1 ?
Transforming so as to make the second term disappear,
z 3 - 9* -28 = (2)
^= 9 q = 28 ;
substituting in Cardan's formula, and reducing,
* = 4.
But the roots of the given equation are greater than those of equa-
tion (2) by 2 ; hence x = 6 4
396-405.] CUBIC EQUATIONS. 139
Transposing 18 in equation (1), and dividing both numbers by
x 6, we find
a: 2 + 3 = . ' . * = v / 3;
hence the three roots are 6, -y/ 3, ^/ 3.
2. What are the roots of the equation
X 3 _ 9^2 + 28z = 30 .... (1)?
Transforming, we find
x 3 + x (2) .'. x=0 and a; = v '~ 1.
But the roots of (1) are greater than those of (2) by 8;
hence the roots of (1) are, 3, 3 + T/ 1 and 3 y' 1.
3. z 3 7z + 14 = 20 (1)
Transforming (Bourdon, Art. 266), we have,
7 344 7 344
3*~T~ ^' 6 ' ^ = ~~3' 9 ' = ~T7~ 4
g
Substituting in Cardan's formula, we have ^ , the real root.
o
7
But the roots of (1) are greater by - than those of (2) ; hence,
B
in (1) x = 5.
Transposing and dividing by x 5, we have,
a;2 _ 2x + 4 = ;
hence, the required roots are, 5, 1 + V3, and 1 V &
HORNER'S METHOD OF SOLVING NUMERICAL EQUATIONS.
1. 3? + z 2 + x 100 = 0.
By Sturm's Rule, we find
140
KEY TO DA VIES 5 BOURDON.
[405.
X r= X 3 + X Z + X - 100,
Z l = 3* 2 -f 2* + 1
X z = - 4x + 899
X, = - 2409336.
For cc = oo , 1--| ,
for * = + CD , -f H ,
hence, there is but one real root.
Tor x = 4, . + + .-,
for x = 5, + + + _ ,
hence, the real root lies between 4 and 5.
2. a* - 12* 2 4- 12* - 3 = 3.
By Sturm's Rule,
X =x* - 12* 2 + 12* -3,
JTj = 4* 3 - 24* + 12, or x 3 - Qx + 3,
2 variations,
1 variation ;
2 variations,
1 variation ;
X, = 13*- 9,
X 4 = 20.
For x oo , + -
for x = + oo , H \-
hence, there are 4 real roots.
For x = 4 H f-
for * = 3 \-
for * = h +
for x = -f 1 =F
for .T = + 2 h
for * = + 3 + + 4-
4 variations,
variation ;
4 variations,
3 variations,
3 variations,
1 variation,
1 variation,
variation ;
406.] HORNER'S METHOD. 141
hence, one of the roots lies between 4 and 3, two between
and 1, and the remaining root lies between 2 and 3.
3. x* - 8* 3 + 14* 2 + 4z - 8 = 0.
By Sturm's Eule,
X = x* Sx 3 + 14* 2 + 4* 8,
X I = 4* 3 24* 2 + 28* + 4, or x ? 6* 2 -f 7* + 1,
X 2 = 5z 2 - 17* -j- 6,
X 3 = 76* - 103,
X< = 45475.
For x = co , H 1 ( 4 variations,
for = + 00, + + + + + variation ;
hence, the equation has 4 real roots.
For * = 1 + h 1- 4 variations.
for*= h-l h 3 variations,
for orrr-fl + + f- 2 variations,
for x = + 2 H 1 H 2 variations,
for z=:-f3 + + 1 variation,
fora;= 5 H + + 1 variation,
fora;= 6 + + + + + variation ;
hence, one root is between 1 and 0, one between and 1, one
between 2 and 3, and one between 5 and 6.
4. x 5 I0x 3 + Qx + 1 0.
By Sturm's Rule,
X = x 5 - 10* 3 + 6* + 1,
X I - 5* 4 - 30* 2 + 6,
X 2 = 20* 3 - 24* - 5,
X 3 = 96* 2 - 5* - 24,
X 4 = 43651* + 10920,
X,, = 32335636224.
142
KEY TO DAVIES' BOURDON.
[406.
For x = ~ oo 1__. I- __'_!_
for ar=+ao -f. + _}. _j_ _j_ _j_
hence, the equation has 5 real roots.
For x = 4
for
for
for
for
for
for
hence,
ar=-3
x= I
x =
* = + !
x- +3
x =. + 4
one roc
i i i
+ 4- - + ~ +
4- 4- ~ +
+ + -- + +
4- 4- 4- +
+ + + + + +
one root lies between 4 and
two roots lie between 1 and
one root lies between and
one root lies between 3 and
5 variations,
variation ;
5 variations,
4 variations,
4 variations,
2 variations,
1 variation,
1 variation ;
variation ;
-3,
0,
4-1,
4
APPENDIX.
GENERAL SOLUTh,* OF TWO SIMULTANEOUS EQUATIONS OF THE
FIRST DEGREE.
1. Take the equations,
ax + by c . . . (1),
a'z+b'y = c' ... (2);
multiply both members of (1) by b' and of (2) by 6, then sub-
tracting and factoring, we find
(aV - a'b) x^b'c- be' ;
b'c - be'
ab' - a'b
(3).
r vi ' ac' a'c
In like manner, y = 77 (4).
ab ab
By means of formulas (3) and (4) any two simultaneous equations
of the forms (1) and (2) may be solved.
Thus, 4x -f By = 31,
3* + 2y = 22 :
by comparison with (1) and (2),
a = 4, 6 = 3, c = 31, a' = 3, b' = 2, c' = 22;
by substitution in (3) and (4),
62 - 66 88 - 93
144 APPENDIX.
EXAMPLES.
( -+ y - =2 )
1. Given J 3 4 I to find x and y.
I 3x + 4y = 25 )
By comparison with (1) and (2),
a = , 6=1, c = 2,
a' = 3, ^ = 4, c' = 25 ;
by substitution in (3) and (4),
X = *L--J% = 8> y = 8 ^~ 6 = 4 .
** ~~ 4 *3 ~~ 4
2. Given 4 f- to find a; and y :
(- 5z + 16y = 124)
by comparison,
a = 11, b = 5, c = 1,
a' --= - 5, b' = 16, c' = 124 ;
oy substitution,
- 16 + 620 _ 1364 5
176-25 ' 176 - 25 ~
GENERAL SOLUTION OF THREE SIMULTANEOUS EQUATIONS OF
THE FIRST DEGREE.
2. Take the equations,
ax + by + cz = d (1),
a'x + Vy + c't = d' . (2),
~n~, i irr., i ..//_ j// /o\
ax-}-by + cz-=a (6).
From (1) and (2) we obtain, by eliminating e,
(So, - ca'} x + (c'b - cb') y = c'd -cd / - - (4).
ADDITIONAL EXAMPLES. 145
In like manner, from (1) and (3),
(c"a - ca") x + (c"b - cb") y = c"d - cd" (5; ;
combining (4) and (5) and eliminating y, we find
_ (c"b - cb") (c'd - cd 1 ) - (c'b - cb') (c"d - cd")
~ (c'a ca' ) (c"b -cb") - (c"a- ca") (c'b - cb')
In like manner,
_ (c'a - cap (c"d - cd") - (c"a- ca") (c'd - cd')
y - (c'a _ ca') (c"b _ cb") - (c"a- ca") (c'b - cb') ***
_ (a"6 - ab") (afd - ad') - (a'b ab') (a"d ad")
~ (c'a - ca') (b"a -ba") - (c"a- a"c) (b'a - ba' )
Formulas (6), (7) and (8) enable us to solve all groups of simul
taneous equations of the form of (1), (2) and (3). Thus,
2x + Sy + 4z = 29,
3* + 2y + 50 = 32,
4x + 3y + 2 = 25 :
Vy comparison with (1), (2) and (3),
a =2, b =3, c =4, d =29,
a' =3, b' = 2, c' =5, ab '~ a ' b
a
These values of x and y correspond to those already deduced by
previous methods.
As an example, let it be required to find the values of x and y
from the equations
7z + 3y = 33 (2) ;
multiplying both members, of (1) by n,
3nx ny = 5ra (3) ;
adding (2) and (3), member to member,
(3% + 7) x + (3 n) y 5n + 33 (4) :
1 st. Assume 3 n = ; . . n =: 3 ;
substituting in (4), and reducing,
15 + 33 _ Q
: 9 + 7
2d.' Assume 3n + 7 = 0;
ADDITIONAL EXAMPLES. 140
substituting in (4), and reducing,
= 4.
S+ 3
EXAMPLES.
1. Given { } to find x and y.
V, J
Multiplying both members of the first equation by n and adding to the
second, member by member,
making n = - and reducing,
m
making n = - and reducing,
2 Given
y-2
/> __,_-. ^
a? 7 -o
Q
4y =3
to find x and y.
Reducing and transposing,
7*-y = 33 (1),
12y - x = 19 (2) :
multiplying by n, and adding and factoring,
9
150 APPEltDIX.
(In 1) x (n 12) y = 33n + 19 :
making n = - , we find y = 2 ;
making n = 12, we find x = 5.
T+? = s
3. Given -^ ^ to find x and y.
~jr~ ~7 =
Multiplying by n, adding and factoring,
(9n + 14) - + (6% - 6) - = 36/i + 10 ;
14 1 1
making n = , we have - = o ; . . y = ->
y
making n = 1, we have - = 2; . . x = -
MISCELLANEOUS GROUPS OF SIMULTANEOUS EQUATIONS OF THE
FIRST DEGREE.
82; 5y 9
] . Given < > to find a: and y.
+ - -
i, 5a; 3y 4
Combining and eliminating -,
tG
OK Of Z^ T9 '
itii) Jf/ y to is/
1 DO i 20
whence, - = 57 ? -. 1 r= IT
ADDITIONAL EXAMPLES.
Conr/Dining and eliminating -
y
\ _ JLU -
9 25/ x ~ 27 12 '
1 65
whence, x = 192'
= 5
2. Given { i. to find x and y ;
I r*. *
c 1
5 J
Clearing of fractions and reducing,
* + 6y= 15 (1),
2z 5y = 4 . (2) :
combining and eliminating ar,
Yly = 34 ; . . y 2 ;
substituting in (1), ar = 3.
y-2
A.
ry ~" ~
3. Given \ \ to find a? and y.
3
Clearing of fractions and reducing,
7a;-y = 33 (1),
12y-ar=19 . . (2);
combining and eliminating ar,
83y = 166; .-. y = 2;
substituting in (1), x = 5.
152
APPENDIX.
4. Given
5 * 6 4 + 2y = 24
to find x and y.
Clearing of fractions and reducing,
5z -f 12y = 148,
25* 2y = 182 ;
combining and eliminating y,
155* = 1240 ; . . x - 8 ;
substituting, we find y = 9.
5. Given
6
to find x, y and c.
Clearing of fractions and reducing,
3x + 2y+ s = 72 . . . (1),
x + 3y + 20 = 48 . (2),
3*+ 2* = 60 ... (3);
combining (1) and (2), eliminating y,
11* -2= 120 . (4);
combining (3) and (4), eliminating 2,
25* = 300 ; . . * = 12 ;
by substitution in (3), 2 = 12,
" " (1), y=12.
ADDITIONAL EXAMPLES.
153
6. Given
Clearing of fractions,
21*+14y + 60 = 924
10s -f 6y+152 =
x+ 2y+ 2 = 11
combining (1) and (3), eliminating 2,
to find or, y an< <.
(1),
(2),
(3);
(4);
(5) ;
combining (2) and (3), eliminating 2,
5* + 24y = 780
combining (4) and (5), eliminating ar,
70y = 2100 ; . . y = 30 ;
from (5), x - 12 ; from (3), z = 42.
f 2* - 3y + 22 = 13 (1)
2v * = 15 ... (2)
2y+ = 7 ... (3)
I 5y + 3v = 32 . . . (4) J
Combining (2) and (4), eliminating v,
lOy + 3x = 19 (5) ;
combining (5) and (1), eliminating a?,
29y - 62 = - 1 . . . (6) ;
7. Given
t( find x, y, t
and v.
154
APPENDIX.
combining (6) and (3), eliminating z,
41y = 41; .'. y = 1 :
by successive substitutions, z = 5, x = 3, v = 9.
8. Given
3 5 19 4
z + z~24 + y
24
y 2 ~ 8
Transposing, reducing, &c.,
x y
a; y z 24
(1),
(2),
(3):
combining (1) and (3), also (2) and (3), eliminating -,
10- +43- = (4),
x /24
8 ^- 11 y == 24 ' ' ' (5)J
combining and eliminating -,
x
2B *l = i&> ' p = l? or y=12 >
substituting in (5), a; = 6 ; whence ; 2 = 8.
f 10+ 6y 4a; _ 4
9. Given <( f- to find x and y.
126+ 8a;-17y _ 35
100 12ar+"7y ~ 13 J
ADDITIONAL EXAMPLES. 155
Clearing of fractions and reducing,
- 2* + 3y = - 1 ... (1),
262* - 233y = 931 (2).
Combining and eliminating x,
160y = 800; .'. y = 5;
by substitution in (1), x = 8.
r ax + by = c 2 \
10. Given < a(a + x)_ > to find x and y.
(W+7)- )
Clearing of fractions and reducing,
ax -f- by = c 2 - (1), '
ax by ^ - a 2 (2) ;
>y addition,
52 _i_ r 2 a 2
2aa; = 6 2 c 2 -a 2 .-. x =
2a
by subtraction,
2y = c 2 -f- a 2 6 2 ; . . y = ^-
MISCELLANEOUS EXAMPLES OF EQUATIONS OF THE FIRST, SECOND
AND HIGHER DEGREES, CONTAINING BUT ONE UNKNOWN
QUANTITY.
1. Given 3* 2 4 = 28 + * 2 , to find x :
transposing and reducing,
x 2 =5 16 ; . . x = 4
2 . Given ^--5 - *? = 117 - fo, to find x;
o o
1 56 APPENDIX.
Clearing of fractions, transposing and reducing,
x z = 25 ; . . * = 5.
3. Given a; 2 + ab = 5a: 2 , to find x ;
transposing and reducing,
* 2 = V ' ' ' * = 2
4. Given
+7 a? -7
;o find a?.
a; 2 7* a; 2 f 7a; a; 2 73'
Clearing of fractions,
x* 4- 14a: 3 - 24a; 2 - 1022a: 3577 a;* + 14a; 3 4- 24a: 2 10222
4- 3577 = 7a: 3 343a: ;
transposing and reducing,
21a; 3 = 1701ar, or z 2 = 81 ; . '. a? = 9.
/*-2 . /* + 2
5. Given W , 2 + V _ 2 ~ ^> to fin< ^ * '
multiplying both members by <^x 4- 2,
-/a; 2
multiplying both members by -y/a; 2.
x 24-a;4-2 = 4 -\/x z 4, or a; = 2 y^a; 2 4 ;
squaring both members,
16 16 4 /
x = 4a; 2 16, or a; 2 = = X 3 ; . ' . a; = - -/3.
6. Given
x 4- y^Sa; 4- 10 = 8, .o find x ;
ADDITIONAL EXAMPLES.
Transposing and squaring both members,
5x + 10 = 64 IGz +
whence, x z 21a; = 54 ;
21 / 441 21 15
by the rule, x = \/54 + : = j
. . a; = 3, * = 18.
7. Given 5 \f* + 7 ^ _ 10 8, to find x :
make \f& = y ; whence, ^/z* = y 2 ;
substituting and reducing,
7 108
7 /108 49 ' 7 4?
whence, y= - JQ * \/5 ' f 100 = ~TO 10 ;
27
. . y = 4 and y = ;
5
27\ 3
from which, x=^y 3 =S and a;=y / y^= \/| ^-
8. Given 3z 2 + 10z = 57, to find x.
By division,
+ - x = 19 ; whence,
o
_5 14
/. x = 3 and a: = 6^
9. Given (a; - 1) (x 2) = 1, to find x.
Performing indicated operations and reducing,
o;3 Qj. 1 .
*' * J^/ ^~* X
158 APPENDIX.
115
10. Given - z 2 - x = -, to find a:.
Dividing both members by -, or multiplying by 2,
m
2 5
x z - x = - ; whence,
_ 5 1 1 7
-
.-. x =11, *=-
11. Given *LZli?_i* a, to find *.
8 x x 2
Clearing of fractions,
2z 2 14z + 20 (5x + 24 x 2 ) = 20s 32
, . 39 28
reducing, a; 2 - x = -- ; whence,
u
39 / 28 , 1521 39 31
X r \/ h , =
10 V 5 100 10
4
12. Given __ - __ = , to find A
Clearing of fractions,
35 (x + 3 a; + 1) = & -\- 2a; 3 ;
reducing, x z + ( 2x = 143 ; whence,
x - 1 -v/l4T=: - 1 12;
ADDITIONAL EXAMPLES. 150
.-. * = 11, *= -13.
24
13. Given x H -- - = 3r 4, to find x.
x 1
Clearing of fract ons,
x 2 x + 24 = 3* 2 3x 4x + 4;
reducing, x 2 3x = 10 ; whence,
3, / in ,9 37
^-^10 + - = ;
. . x = 5, * = 2.
a; x + 1 13
14. Given - + - - = -3-, to find ar.
a; + 1 a; 6
Clearing of fractions,
6z 2 + Qx z + 12a; + 6 = 13* 2 + 13ar ;
reducing, x 2 + x = 6 ; whence,
- 1 5
- = ^-;
. . a: = 2, ar = 3.
a; 4
15. Given -- = x 8, to find a;
Since a; 4 = (^fx 2) (y^+ 2),
we have, by performing indicated operations,
y/x 2 = x 8, or -y/aT= ar 6 ;
squaring both merr bers,
x = x z l2a; + 36 j
or, x* 1 3x = 36 ; whence,
13 f~ ~l69 13 5
= -^-36 + - = -^-;
160
APPENDIX.
. . x 9, x =s 4.
16. Given
Reducing,
* = -
17* 2 + 19* -
* 2 + 19 *
- 1848 =
1848
0, to find 5
whence,
19 355
r 17
17 '
19 /I 848
361
34 ~V 17
1156
34
17. Given
x 9 jf, and a; = 11.
1 5
- a; 2 + - x = 27, to find ar.
O xi
Multiplying both members by 3,
15
+ x = 81 ; whence,
. 225 15 39
15
. . x = 6, and x =
18. Given
12
-, to find r.
Transposing and reducing,
28 a; 2
s_ 4 - *-4'
squaring both members and clearing of fractions,
x* 16 = 784 56z 2 + ** ;
reducing, ** 57* 2 = 800 ;
28
ADDITIONAL EXAMPLES. 161
by Rule. Art. 124,
/57~ l~ , 3249 /57 7
x = db y db y 800 + J = iW-^-i-;
hence r x = 5, and x = 4-y/2.
2* + 9 , 4* - 3 , 3a; 16
19. Given = 3 H , to find x.
9 4x + 3 18
Clearing of fractions,
16* 2 + 84* + 54 + 72* - 54 = 216* -(- 162 + 12* 2 - 55* - 48 ;
5 114
reducing, * 2 - * = - ; whence,
114 548
f 64" ~8 '
. . * = 6, and * = 4
20. Given a 3 + * + 2 -/* 2 -f a: + 4 = 20, to find *.
Making z 2 -{- x = y, and reducing,
2 /y + 4 = 20 - y ;
squanng both members,
4y + 16 = 400 40y + y 2 ;
reduc'ng, y 2 44y = 384 ; whence,
y = 22 V 384 + 484 = 22 10 ;
.-. y = 32, and y = 12 :
taking the first value : * y and substituting in the equation,
x z + x y,
x 2 + x = 32 ; whence.
162
APPENDIX.
taking the second value of y,
x 2 + x = 12 ; whence,
21. Giveh \/ x
transposing,
squaring both members,
= x, to find x.
whence, by reduction,
x* - x + 1 = 2 V^
Placing x 2 x = y
(1),
squaring both members,
y z + 2y + 1 = 4y ; whence,
y 2 -2y=-l; .'. y = 1 V- 1 + 1, or y=
substituting in (1), a; 2 a: = 1 ;
22. Given a; 2 6z = 6a + 28, to find x :
transposing, a? 1 2x = 28 ; whence,
x = 6 -y/28 + 36"= 8 ; . . x = 14, = 2. *
ADDITIONAL EXAMPLES. 163
23. Given x** 2z 3n -f x* 6 = 0, to find :
making, x* = y ; whence,
y* - 2y 3 + y - 6 = ;
causing the second term to disappear (Arts. 263 and 313),
3 91
^ ~ 2 Z = Itf '
id 1U
By the rule for solving trinomial equations (Art. 124).
/3 10
also.
1 . 1
.-. 2==t--v/13, and z = -
24. Given
x* -4- 0.x 3 I 8
-^_ = ^2 + ,,. + 8 to find Xm
x 2 + x 6
Clearing of fractions,
x* -f 2x 3 +
2x 3
2* 48 ;
, 2 56
reducing, x 2 + - ar = ; whence,
O D
. . x = 4, and a; = 4J.
2
25, Given * a 1 = 2 4- -, to find x.
x
Reducing, a 3 3a - 2 = (1) ;
164: APPENDIX.
comparing with x 3 -\- px -f- q =. 0,
P=-Z, ? = -2;
by Cardan's formula,
* = 3 v/r+\A = 2:
dividing both members of (1) by x 2,
x z + 2a: -\- 1 = ; whence,
*= -1,
and the two roots are each equal to 1 ; hence, the three roots
re> +2, 1, and 1.
26. Given 2 e n\ ,
1. Given { } to find x and y.
< a; + y = 12 . , . (2) i
Make a; = v + w, and y v w.
From (2), we have (v + w) + (y 70) = 12 ; . . v = 6.
From (1), we have, (v + w) 3 + (v w) 3 = 18 (v 2 w 2 ) ;
or, reducing, v 3 + 3vw> 2 = 9 (v z w 2 ) ;
substituting the value of v, wi have
216 + 18u 2 = 9 (36 - w 2 ), or 27w 2 = 108; .'. w=2,
hence,
x = v + w = 62 = 8 al 4 = v w=6^2=4 and 8.
(z2 + y-^53 . . . (1))
2. Given { f to find x and y.
( - = 14 ... (2))
Multiplying both meipW.rs of (2) by 2, and adding and subtracting,
we have
x 2 - + 2a;y ' 2 = 81 ; . ' . x + y = 9,
x i _ 9 /f/f : c yZ _ 25 ; . . x y = 5 ;
hence, rz > 1 , and 7 y = + 2, and 7.
U + y* = 82 . . . (1))
3. Given { \ to find ar and y.
U +y = 4 . . . (2))
Raising both numbers of (2) to the 4th power, adding to (1), mem
her to member, and dividing by 2,
x* -f 2z 3 y + 3zV + 2ary 3 + y* = 169 ;
ADDITIONAL EXAMPLES. 169
extracting the square root of both members,
* 2 + xy + y 2 = 13 . . . (3);
squaring both members of (1),
ar + 2xy + y 2 = 16 . . . (4) ;
subtracting (3) from (4), member from member,
xy = 3, or 3xy 9 . . (5) ;
subtracting from (3), member from member,
z 2 2zy + y 2 = 4 ;
whence, x y = 2 . . . (6).
Combining (2) and (6),
x 3, and 1 ; y = 1, and 3.
( 5* + 3y = 19 . . (1) )
4. Given < > to find x and y
('7x z -2y z =lO . . (2)}
From (1), we find
_ 19 - 3y 2 _ 361 - 114y + 9y 2
~5 ; ~25~
Substituting in (2), and reducing,
y
2 _798_ 2277
~ 13 ~~ 13 '
399^ X-2277 , 159201 399 360
whence, y = ^ + _ = ir ~
759
hence, y = , and y = 3 ;
by substitution, x = -, and a; = 2.
18
170 APPENDIX.
(x + 4y = 14 . . (1))
5. GrtVb.1 { > to find x and y.
(y + 4a; = 2y + 11 (2))
From (1), = 14 4y, or 4* = 56 16y ;
subtracting ana reducing,
y 2 - 18y = - 45 ;
whence, y = 9 ^/ 45 + 81 = 9 6 = 15 and 3 ;
hence, x = 2 and 46.
z 2 + 4y 2 = 256 - 4zy . . (1)
6. Given { > to. find x and y.
3y 2 - a; 2 = 39 ... (2)
)
>
)
Transposing in (1), and extracting the square root of both members,
* + 2y=16; .'. a; =16 2y;
or, x 1 = 256 ^ 64y + 4y 2 ;
substituting in (2), and reducing,
y 2 =p 64y = - 295 ;
whence, y = 32 -y/ - 295 + 1024 = 32 27 ;
. . y = 59 and 5 ;
substituting, x = 102 and 6.
7. Given ? f to find a; and y.
(a; 2 + 2:y =84 . . (2))
Subtracting (1) from (2), member from member,
y 2 + xy = 60 . . . (3) ;
adding (3) to (2), member to member,
x 2 + 2xy + y 2 = 144 ; . '. x -f y = 12 . . . (4).
ADDITIONAL EXAMPLES. 171
Dividing (1) by (4), member by member,
x-y= 2;
hence, x . 7, y = db 5.
(x-y = 4 . . . (1))
8. Given { r to find ar
( xy = 45 . . . (2) )
From (1), x = y + 4, which, in (2), gives
y 2 -f4y = 45; .'. y = 2 -/45 + 4 = 2 7 ;
.'. y = + 5, and 9;
and by substitution, x = + 9, and 4.
fary + *y2 = 12 . . . (1) )
9. Given \ t to find x and y.
( x + xy 3 = 18 . . . (2) )
Dividing (2) by (1), member by member, and reducing,
1-fy 3 3 1 y + y z 3
,. . = = or
y ~2'
g
by reduction, y 2 - y = 1 ;
v
5 / , , 25 5 3
whence, y = - i / 1 +T^ = T= t T;
4V 1644
1
. . y = 2, and y = ;
by substitution, a: = 2, and or = 16.
1 1
X "y = fa\ 1
j. x y A * '~4
Clearing (1) of fractions,
9z 2 + 36a;y = 85y 2 (3).
From (2), x = 2 + y ; .' . s 2 = 4 + 4y + y 3 ;
substituting in (3), and reducing,
27 9
y2 ~io y = io ;
27 /9 , 729
wnpnfp v -t- * / -4- ~~
y ~ 20 V 10 ^ 400 ~
. . y = 3 and y = -
17
by substitution, x = 5 and a; =
27 db 33
ADDITIONAL EXAMPLES. 173
(f! + ^+* + *: = ?Z..m)
12. Given J y* z 2 y * 4 v ' I to find x and y.
( *-y = 2 (2) J
X II
Make - + - = z (3) ; whence, by squaring,
hence, from (1), by substitution and reduction,
35 1 735 , 1 - 1 6
* 2 + z = T ; .-. , = -- v / T + - = __;
5 7
or, z = - and --;
substituting the positive value of z in (3), and clearing of fractions,
2 _L_ 2 . //I \
From (2), * = y + 2 ; . ' . z 2 = y 2 + 4y + 4,
and zy = y 2 -f 2y ;
substituting in (4) and reducing,
hence, y = 2 and y = 4 ;
by substitution, x = 4 and y = 2.
r x z + y 2 -f 2 2 = 84 . (1) >j
13. Given -{ * + y + z = .14 (2) I to find ar, y.
I 22 = y 2 (3) J and 2 '
Substituting in (1) and (2) the value of y from (3), and reducing
z 2 4- 2xz + z 2 = 84 -f tz or a; + z t/84 -f a* (4)
From (2) and (3), * + 2 = 14-^7 (5)
174: APPENDIX.
Equating the second members,
14 - ^/xz -v/84 + xz -,
squaring both membeis,
196 - 28 V** 4- = 84 + xz ;
.educing, -y/arz = 4, or 2 = 16 (6) ;
hence, from (5), x + = 10 (7) ;
1 A
aobstituting in (7) for z its value , and reducing,
sc
a 2 - 10*= - 16;
* = 5 ^/ 16 + 25 = 5 3 ;
or. a; = 8, a; = 2 :
by substitution, 2 = 2, 2 = 8, and y 4.
f * + y + -v/# + y = 12 . (1
14. Given { ; 2 + y = 41..(
From (1), by transposition,
V*+y = 12 - (x + y) ;
squaring both members,
* + y = 144 - 24 (x + y) + (* + J/) 2 ;
reducing, (a; + y) 2 25 ( + y) = 144 ;
25 ~ 1 ^ 5 " 25 7
whence, x -f y = 16, or x + y = 9.
Tlie first value does not satisfy (1), unless the radical have the
negative sign ; adopting, therefore, the second value, from which
x = 9 y, or x 2 = 81 18y + y 2 ,
ADDITIONAL EXAMPLES.
which in (2) gives, after reduction,
y 2 9y = 20 ; whence,
T = Hr' ; .'. y = 5 and y = 4;
by substitution, x = 4 and x = 5.
l X 3-y3~U1[ . . . (1)1
15. Given { f to find x and y.
(x -y = 3 ... (2))
Cubing both members of (2), subtracting from (1), member from
member, and dividing both members by 3, we have
x z y -xy z = 30, or (r. y) xy = 30 . (3) ;
dividing (3) by (2), member by member,
xy = W; .'. y = ;
X
substituting in (2), and reducing,
x z - 3* = 10 ;
3 I 93?
whence, x = - ^ 10 + - = - -;
. . x = 5, x = - 2 ;
by substitution, y = 2, and y = 5.
(x* + r*ffi= 208 (1))
16. Given 1 [ to find a? and y,
- (2) }
These equations may be written,
4 2. 444
x*+x*y*= 208, or x s (x s +y s )= 208 . (3),
^ + ^=1053, yVhar^lOSS ... (4)
176 APPENDIX.
Dividing (3) by (4), member by member,
aJ _ 208 _ 16
~| ~~ 1053 ~ 1
y
extracting the 4th root of both members,
I ' '
substituting in (3),
64 . 16
= 208 ; whence,
208
y2 = 208, or y = 729 ; ' y = 27,
and by substitution, x = 8.
MISCELLANEOUS PROBLEMS.
1. A courier starts from a place and travels at the rate of 4 miles
per hour ; a second courier starts after him, an hour and a half
later, and travels at the rate of 5 miles per hour : in how long a
time will the second overtake the first, and how far will he travel ?
Let x denote the number of hours travelled by 2d courier :
then will x + 1% " " " " 1st "
5x " " " miles " 2d "
and 4(x -f H) " " " " " 1st ;t
From the conditions of the problem,
bx = 4 (x + li) ; . . x = 6 and 5x = 30.
2. A person buys 4 houses for $8000 ; for the second he gave
half as much again as for the first ; for the third, half as much again
as for the second ; and for the fourth, as much as for the first and
third together : what does he give for each 1
ADDITIONAL EXAMPLES. 177
Let x denote the amount paid for 1st house : then will
ar + 1 u er hour slower, he would have been 6 hours longer in completing
ihe journey : how many miles did he travel per hour ?
Let x denote the number of miles travelled per hour. Then will
105
denote the number of hours.
x
From the conditions,
105 105
= + 6, or 105* = 105* - 210 + 6x 2 - 12* ;
x 2 x
reducing, x 2 2* = 35 ;
.-. x = 1 -v/35 + 1 = 1 6 ; .'. x = 7.
14. The continued product of four consecutive numbers is 3024 :
what are the numbers ?
Let x denote the least number.
! .From the conditions of the problem.
x(x + 1) (* + 2) (x + 3) = 3024,
or x* + 6x 3 + II* 2 + Gx 3024 = 0.
A superior limit of the real positive roots is 9 (Art. 279). Ne-
glecting the divisor 1, and all negative divisors, we may proceed by
the rule (Art. 285), as follows :
9, 8, 7, 6, 4, 3, 2,
- 336, 378, - 432, 504, - 756, - 1008, - 1512,
- 330, - 372, - 426, 498, - 750, - 1002, - 1506,
-, - 83, -., - 334, - 753,
., .'72, .., _ 323, 742,
, , , 12, , , 371,
ft 0,'K
*, ", ", U, , , OUt,
', ", , 1, ", , ",
" "> "> "> "> "t "
Hence, 6 is the required value of x, and the numbers a- e 6, 7, 8
and 1).
ADDITIONAL EXAMPLES. 183
15. Two couriers start at the same instant for a point 39 miles
distant ; the second travels a quarter of a mile per hour faster than
the first, and reaches the point one hour ahead of him : at what
rates do they travel 1
Let x denote the number of miles per hour of first courier.
39
Then will denote the number of hours he travels.
x .
From the conditions,
reducing,
or * =
16. The fore-wheels of a wagon are 5 \ feet, and the hind-wheels
7 feet in circumference ; after a certain journey, it is found that the
fore-wheels have made 2000 revolutions more than the hind-wheels ^
how far did the wagon travel ?
Let x denote the number of feet.
From the conditions of the problem,
1197
multiplying both members by -5^-,
BSB
_ 2394000 _ 598500
7 W* ^ x 32 8 '
57 x 42 x = 598500,
V
15 a; = 598500,
x= 39900.
184 APPENDIX.
17. A wine merchant has 2 kinds of wine ; the one costs 9 shil*
lings per gallon, and the other 5. He wishes to mix them together
in such quantities that he may have 50 gallons of the mixture, and
so that each gallon of the mixture shall cost 8 shillings.
Let x and y denote the number of gallons of each, respectively.
From the conditions,
x+ y = 50 . . .- . (1),
9x + 5y = S(x + y) . . (2) ;
substituting for x + y its value in (2),
9* + 5y = 400 . . . . (3);
combining (1) and (3),
4y = 50-, .'. y = 12$, and x = 37.
18. A owes $1200 and B, $2500, but neither has enough to pay
his debts. Says A to B, " Lend me the eighth part of your fortune,
and I can pay my debts." Says B to A, " Lend me the ninth part
of your fortune, and I can pay mine :" what fortune had each 1
Let x and y denote the number of dollars in the fortunes of A
and B.
From the conditions of the problem,
x + | = 1200, or 8x + y = 9600,
o
y + ^ = 2500, or x -f 9y = 22500 ;
y
combining and eliminating x,
71y = 170400; .-. y = 2400, x = 900.
19. A person has two kinds of goods, 8 pounds of the first, and
9 of the second, cost together $18,46; 20 pounds of the first, arrd
16 of the second, cost together $36,40: how much does each cost
per pound ?
ADDITIONAL EXAMPLES. 185
Let x and y denote the cost of a pound of each in cents.
From the conditions of the problem,
Sx+ 9y=1846,
20* + 16y = 3640 ;
combining and eliminating x,
13y=1950: .'. y = 150, and x = 62.
20. What fraction is that to the numerator of which if 1 be
added the result will be , but if 1 be added to the denominator the
result will be J ?
Let x denote the numerator, and y the denominator.
From the conditions of the problem,
x+ 1 1
V
x 1
or 3x + 3 = y,
or 4x = I + y ;
i+y 4
hence, by combination, * = 4 and y = 15. Ans. y^.
21. A shepherd was plundered by three parties of soldiers. The
first party took -J- of his flock and of a sheep ; the second took
of what remained and ^ of a sheep ; the third took -J of what then
remained and ^ of a sheep, which left him but 25 sheep : how many
had he at first ?
Let x denote the number of sheep. Then, after being plundered
by the 1st party, he would have
O_ 1
sheep ;
4
after being plundered by the 2d party, he would have
3x I /3x 1 . 1\ x I
1 /
\
12
186
APPENDIX.
after being plundered by the 3d party, he would have
x 1 /x 1 l\ Z 3
1\ x
27 =
2
from the conditions of the problem,
- = 25, or x 3 = 100; .-. x 103.
22. What two numbers are those whose product is 63, and the
square of whose sum is equal to 64 times the square of their dif
ference 1
Let x and y denote the two numbers.
From the conditions of the problem,
xy - 63 (1),
(x + yY = 64 (x - y) 2 . . (2);
extracting the square root of both members of (2),
x -{- y = 8 (x ?/), or 7# = 9y ; . ' . x = 2-y ;
substituting in (1), f y 2 = 63 ;
. . y 2 49 and y = 7, also x = 9.
23. The sum of two numbers multiplied by the greater gives
209 ; their sum multiplied by their difference gives 57 : what are
the two numbers ?
Let x and y denote the numbers.
From the conditions of the problem,
(x + y) x = 209, or x 2 4 zy= 209 . . (1),
(z + y)(x-y)= 57, or a* - y* = 57 . . (2);
subtracting (2) from (1), member from member,
adding (3) and (1), member to member,
x 2 + 2xy 4- v 2 = 361 ; . . x -4- y = ] 9 ;
ADDITIONAL EXAMPLES. 187
209
hence, from (1), = -- = 11; also, y = 8.
24. Three numbers are in arithmetical progression ; their sura is
15, and the sum of their cubes is 495 : what are the numbers 1
Let x, y and z denote the numbers,
From the conditions of the problem,
y x = z y . . (1)
x -hy +z = 15 . . (2)
x 3 + y 3 + z 3 = 495 . . (3) ;
from (1), 2y = z + or, which in (2), gives y = 5 ;
substituting in (2) and (3),
z + x = 10 . . (4)
3 -f x 3 = 370 . . (5) ;
dividing (5) by (4), member by member,
2 2 - zx '+ x* = 37 . . (6) ;
squaring both members of (4),
z 2 + 2zx + x 2 - = 100 . . (7) ;
combining (6) and (7),
21
'6zx = 63, or * zx = 21 ; . * . z = ;
X
substituting in (4),
21
* + = 10, or x z 10* = - 21 ;
.-. x = 5 -/ 21 + 25 = 5 2 ; hence, x = 7. or 3.
25. Divide the number 16 into two parts such that 25 times the
square of the first shall be equal to 9 times the square of the
second.
188 APPENDIX.
Let x denote one part ; then will 16 x denote the other.
From the conditions,
25* 2 = 9 (256 - 32* + * 2 ) = 2304 - 288* + 9* 2 ; .
reducing, x z + 18* = 144 ;
.-. x = 9 -/144 + 81 = 9 15, or * = 6,
since the negative value does not satisfy the problem understood in
the numerical sense.
26. There are two numbers such that the greater multiplied bj>
the square root of the less is 18, and the less multiplied by the
square root of the greater is 12 : what are the numbers ?
Let * and y denote the numbers.
From the conditions of the problem,
y^=lS . . (1)
*V7=12 . . (2);
multiplying (1) by (2), member by member,
y^r)3"=216; .'. xy - 36 . . (3);
adding (1) and (2), member to member,
(V^+ V7) V*y = 30 . . (4),
or -\/x + yV = 5 ;
squaring both members,
* + y + 2V*7=25 . . (5),
or * + y = 13 (6);
combining (3) and (6),
x = 9. y = 4.
ADDITIONAL EXAMPLES. 189
27. What two numbers are those the square of the greater of
which being multiplied by the lesser gives 147, and the square
of the lesser being multiplied by the greater gives 63 1
Let x and y denote the numbers.
From the conditions of the problem,
*2y = 147 (1)
xy*= 63 . . (2);
multiplying (1) and (2), member by member,
*V = 9261 . . (3),
or xy= 21 . . (4);
dividing (2) by (4), member by member,
y = 3 ; in like manner, x = 7.
This method of solution might be applied to the equations of the
preceding example.
.
28. There are two numbers whose difference is 2, and the product
of their cubes is 42875 : what are the numbers ?
Let x and y denote the numbers.
From the conditions of the problem,
*-y = 2 . . (1)
X 3 y 3 _ 42875 . . (2) ;
extracting the cube root of both members of (2),
35
ary = 35; .-. y = ;
substituting and reducing,
** - 2* = 35,
x 1 -/S5 + 1 = 1 6;
. . x = 7, and 5, y = 5, and 7.
190 APPENDIX.
29. A sets out from C towards D, and travels 8 miles each day ;
after he had gone 27 miles, B sets out from D towards C, and goes
each day ^ of the whole distance from D to C ; after he had
travelled as many days as he goes miles in each day, he met A
what is the distance from D to C?
Let x de lote the number of miles from D to C.
X
Then, will denote the number of miles B travels per day,
\j
also the number of days that he travels ;
a; 2
hence, --^-r denotes the number of miles travelled by B,
27 + Sx " " " " " A.
"From the conditions of the problem,
* 2 _L 07 _L Sx _
lOO 4 " "^20"
clearing'of fractions and reducing,
x 2 - - 240* = - 10800 ;
.-. x - 120 -/- 10800 + 14400 = 120 60 ;
whence, x = 60, x = 180.
30. There are three numbers ; the difference of the differences of
the 1st and 2d, and 2d and 3d, is 4 ; their sum is 40, and their con
tinued product is 1764 : what are the numbers ?
Let #, y and z denote the numbers.
From the conditions of the problem,
(*-y)-(y-*)= 4 . . (l)
x + y + z= 40 . . (2)
. . (3);
ADDITIONAL EXAMPLES. 191
combining (1) and (2), eliminating x and 2,
3y = 36; .-. y = 12;
substituting in (2) and (3),
x + z= 28 . . (4)
xz = 147 . . (5);
combining (4) and (5),
x = 7, or 21 ; y = 21, or 7.
31. There are three numbers in arithmetical progression : the
sum of their squares is 93, and if the first be multiplied by 3, the
second by 4, and the third by 5, the sum of the products will be 66 :
what are the numbers ?
Let x denote the first number, and y their common difference.
From the conditions of the problem,
* 2 + (* + 2/) 2 + (* + 2y) 2 = 93 . . (1)
3* + 4 (x + y) + 5 (x + 2y) = 66 . . (2) ;
performing indicated operations and reducing,
3z 2 -f 5y 2 + Gxy = 93 . . (3)
12* -f I4y = 66, or 6* + 7y = 33 . . (4).
33 Qx
From (4), y = ;
1089 396* -f- 36z 2 33* 6* 2
.'. y*-- - & - i and *y= ^.,
substituting in (3) and reducing,
^ _ 198 290
" "25^ * ~ " "25 '
192 APPENDIX.
- , , 296 , 9801 99 49
whence, x = - x / - + = _;
148
Taking the second value of x, we find y = 3, and the numbers
are 2, 5 and 8.
The problem supposes the numbers entire, therefore the 1st value
of x is not used.
32. There are three numbers in arithmetical progression whose
sum is 9, and the sum of their fourth powers is 353 ; what are the
numbers ?
Let a;, y and z denote the numbers.
From the conditions of the problem,
2y = x + z . . (1)
x + y + z = 9 . . (2)
z* + y* 4- s 4 = 353 . . (3).
From (1) and (2) we find y = 3;
substituting in (2) and (3),
x + z = 6 . . (4)
z* + z* = 272 . . (5) ;
raising both members of (4) to the 4th power,
a;4 + 4 X 3 Z _|_ Q X Z Z Z _j_ 4 XZ 3 + s * _ J296 . . (6) ;
adding equations (5) and (6), member to member, and dividing by 2
Z* + 2* 3 z 4- 3z 2 2 2 + 2z2 3 -f 2* = 784 . . (7);
extracting the square root of both members,
2 2 -f xz + z 2 = 28 . . (8);
ADDITIONAL EXAMPLES. 193
squaring both members of (4),
z 2 + 2xz + 2 = 36 . . (9);
from (8) and (9) we find
xz = 8 . . (10);
from (4) and OO) we get
x = 2, or 4 ; s = 4, or 2 :
hence, the numbers are 2, 3 and 4.
33. How many terms of the arithmetical progression 1, 3, 5, 7,
&c., must be added together to produce the 6th power of 12 ?
The 6th power of 12 is 2985984.
From Art. 175 we have the formula,
d-2a
w =
Ir. the present case, o = 1, d = 2, and S = 2985984 ;
-V/16 X 2985984 , WM
substituting, n = = 1728.
34. The sum of 6 numbers in arithmetical progression is 48 ; the
product of the common difference by the least term is equal to the
number of terms : what are the terms of the progression ?
Let x denote the 1st term, and y the common difference.
From the conditions of the problem,
Qx+ 15y = 48, *y = 6; . . y=^;
substituting and reducing,
a* - 8* = - 15 ;
.-. x = 4 / 15 + 16 = 4 1, or * = 5, * = 3;
whence, y = f > y = 2 :
194: APPENDIX.
hence, the series is 3.5.7.9.11.13,
or 5 . 6 . 7f . 8 . 9f . 11.
35. What is the sum of 10 square numbers whose square roots
are in arithmetical progression the least term of which is 3, and the
common difference 2?
Let x denote the sum.
The progression of roots is
3.5.7.9.11.13.15.17.19.21,
and the series of squares,
9 . 25 . 49 . 81 . 121 . 169 . 225 . 289 . 361 . 441.
1st order of diffs, 16, 24, 32, 40, &c.,
2d order of diffs, 8, 8, 8, &c.,
3d order of diffs, 0, 0, &c.
From Art. 210, making
S' = x, a = 9, n = 10, ^ = 16, d z = 8, d 3 = 0, &c.
x = 90 + 45 x 16 + 120 x 8 --= 1770.
36. Three numbers are in geometrical progression whose sum i
95, and the sum of their squares is 3325: what are the numbers?
Let a;, y and z denote the numbers.
From the conditions of the problem,
y* = a* . . (1)
x z + y + z 2 = 3325 . . (2)
x + y + z = 95 . . (3) ;
combining (1) and (2),
x* + 2xz \- 2 2 = 3325 + xz , . (4) ;
ADDITIONAL EXAMPLES. 19
combining (1) and (3),
x -f v ^xz + z = 95 . . (5) ;
from (4) and (5),
x + z = ^3325 + xz
x + 2 = 95
hence, -y/3325 + xz = 95
squaring both members,
3325 + xz = 9025 190
taking the 1st value of y, we find
x = 4, s = 25.
39. A, B and C purchase coffee, sugar and tea at the same prices ;
A pays $11,62-| for 7 pounds of coffee, 3 pounds of sugar, and 2
pounds of tea ; B pays $16,25 for 9 pounds of coffee, 7 pounds of
sugar, and 3 pounds of tea; C pays $12,25 for 2 pounds of coffee,
5 pounds of sugar, and 4 pounds of tea : what is the price of a
pound of each 1
Let x, y and z denote the number of cents that the coffee, sugar
and tea cost, respectively.
From the conditions of the problem,
^+ fy + 21z = 1162$ . . (1)
9*+ 7y + 3z= 1625 . (2)
2 x + 5y -f 4 z = 1225 . . (3) ;
ADDITIONAL EXAMPLES. 197
clearing (1) and (3) of fractions,
30* + 12y + 90 = 4650 (4)
4 ar+lly-f 8z -2450 (5).
From (2) and (4),
3* 9y = 225, or x 3y = 75 (6) ;
from (2) and (5),
QQx + 23y = 5650 . (7);
from (6) and (7), y = 50;
by substitution, x = 75, z = 200.
40. Divide 100 into 2 such parts that the sum of their square
roots shall be 14.
Let x denote the first part.
From the conditions of the problem,
y/ar-f -v/100 x = 14;
squaring both members and reducing,
V/l 00* - a? = 48 ;
squaring both members and reducing,
* 2 -100r= -2304;
.-. x = 50 -/- 2304 -I- 2500 = 50 14,
x = 64, and 36.
41. In a certain company there were three times as many gentle-
men as ladies ; but afterwards 8 gentlemen with their wives went
away, and there then remained five times as many gentlemen as
ladies : how many gentlemen, and how many ladies were there
originally ?
12
198 APPENDIX.
Let 3z denote the number of gentlemen ; then will x denote the
number of ladies.
From the conditions of the problem,
3x 8 = 5 (x 8) ;
. . x = 1 6, and 3x = 48.
42. Find two quantities such that their sum, product, and the
difference of their squares, shall all be equal to each other.
Let x and y denote the quantities.
From the conditions of the problem.
x +y -xy . . (1)
x z y 2 - xy (2) ;
by division of (2) by (1), we have
x y = 1, or a; = y -f- 1 ;
substituting in (1),
2y + 1 = y z + y, -or y 2 y = 1 ;
. 1 /T~T
whence,
hence, x =
43. A bought 120 pounds of pepper, and as many pounds of
ginger, and had one pound of ginger more for a dollar than of
pepper ; the whole price of the pepper exceeded that of the gingei
by 6 dollars : how many pounds of pepper, and how many of
ginger had he for a dollar ?
Let x denote the number of pounds of pepper for a dollar.
ADDITIONAL EXAMPLES. 199
From the conditions of the problem,
The negative value does not conform to the conditions of the
special problem.
44. Divide the number 36 into 3 such parts that the second shall
exceed the first by 4, and that the sum of their squares shall be
equal to 464.
Let #, y and z denote the parts.
From the conditions of the problem,
x + y + z = 36 . (1)
y-x = 4 . . (2)
x i + y 2 - + z 2 = 464 (3) ;
from (1), z 2 + 2xy + y 2 = 1296 - T2z + z 2 ' (4) ;
from (2), z 2 2xy + y 2 = 16 ...... (5) ;
adding (4) and (5), member to member.
2* 2 + 2y 2 = 1312 72z + z 2 (6) ;
from (3), 2* 2 + 2y 2 = 928 - 2z 2 . . . (7) ;
equating the second members and reducing,
2 2 - 24* = - 129 ;
.-. z = 12 -V/-128 + 144 = 12 4;
hence, z 16, z = 8 ;
substituting the first value in (1),
* + y = 20 (8);
200 APPENDIX.
from (2) and (8), y = 12 and x = 8.
.45. A gentleman divided a sum of money among 4 persons, so
that what the first received was ^ that received by the other three ,
what the second received was that received by the other three ;
what the third received was -j that received* by the other three, and
it was found that the share of the first exceeded that of the last by
$14 : what did each receive, and what was the whole sum divided ?
Let x, y, z and w denote the number of dollars that each received.
From the conditions of the problem,
2# = y + z -f w ' (1)
3y = x + s + w (2)
42 = x + y + w - (3)
x w = 14 (4) ;
from (2) and (3),
x + w 3y z
x -f- w = 4z y ; whence, 3y z =. 4z y,
01 4y = 5s, z A y . . (5) ;
from (4), w = x 14 (6) ;
substituting the values of w and z in (1) and (2),
2x = y + f y + x 14
3y x -\- %y -\- x 14; whence, by reduction,
5* 9y = 70
10*-lly= 70;
, *. x = 40, y = 30 ; and by substitution, z = 24, to = 26.
. 46. A woman bought a certain number of eggs at 2 for a penny,
and as many more at 3 for a penny, but on selling them at the rate
ADDITIONAL EXAMPLES. 201
of 5 for 2 pence, she lost 4 pence by the bargain ; how many did
she buy ?
X
Let x denote the number at each price. Then will + 5-
i o
denote the number of pence paid, and - - will denote the
number of pence received.
From the conditions of the problem,
x x %(x + x\
^ + 3 = V 5 ' + 4 ; reducing, x = 120.
47. Two travellers set out together and travel in the same direc-
tion ; the first goes 28 miles the first day, 26 the second day, 24 the
third day, and so on, travelling 2 miles less each day ; the second
travels uniformly at the rate of 20 miles a day : in how many days
will they be together again 1
Let x denote the required number of days. The distance
travelled by the first in x days is
[(Art. 176), since a = 28, d = 2, and n = a;], denoted by
\x [56 - (x I) 2 ], or 29* a: 2 ;
and the distance travelled by the second is denoted by 20x:
hence, we hare
29a; x z = 20ar, or x = 9.
48. A farmer sold to one man 30 bushels of wheat and 40 of
barley, for which he received 270 shillings. To a second man he
gold 50 bushels of wheat and 30 of barley, at the same prices, and
received for them 340 shillings : what was the price of each ?
Let x denote the number of shillings for 1 bushel of wheat,
and y " " " " " " barley.
202 APPENDIX.
From the conditions of the problem,
30* + 40y = 270 (1)
50a; + 30y = 340 . (2) ;
whence, HOy = 330, or y = 3; hence, x = 5.
49. There are two numbers whose difference is 15, and half their
product is equal to the cube of the lesser number : what are the
numbers ?
Let x and y denote the numbers ;
from the conditions of the problem,
x y = 15
xy = 2y 3 or, x = 2y 2 ;
substituting and reducing,
-- v = :
2 y 2 '
1 5 1 111
5 25
hence, y = 3, and - ; also, x = 18, and
50. A merchant has two barrels and a certain number of gallons
of wine in each. In order to have an equal quantity in each, he
drew as much out of the first cask into the second as it already
contained ; then again he drew as much out of the second into the
first as it then contained : and lastly, he drew again as much from
the first into the second as it then contained, when he found that
there was 16 gallons in each cask : how many gallons did each
originally contain ?
ADDITIONAL EXAMPLES. 203
Let x denote the n imber of gallons in the first cask, and y the
number in the second ;
x y will denote the quantity in the first cask after the first drawing,
and 2y the quantity in the second cask ; after the second drawing,
2y (x y) or 3y x will denote the quantity in the second,
and 2 2y the quantity in the first cask ; after the third drawing,
2or 2y (3y x} or 3x 5y will denote the quantity in th
first cask, and Qy 2x the quantity in the second.
From the conditions of the problem,
3* 5y = 16
6y 2z = 16.
By combination,
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