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ALGEBRAICAL
PROBLEMS,
PRODUCING
SIMPLE aw(i QUADRATIC EQUATIONS,
WITH THEIR SOLUTIONS.
DESIGNED AS
AN INTRODUCTION TO THE HIGHER
BRANCHES OF ANALYTICS.
BY THE
REV. M. BLAND, A. M.
U
FELLOW OF ST. JOHN's COLLEGE, CAMBRIDGE
UHITBESI
CAMBRIDGE;
Printed hy J. Smithy Printer to the University:
AND SOLD BY J. DEIGHTON, CAMBRIDGE; LONGMAN, HURST, REES, ORME,
AND BROWN, PATERNOSTER-ROW, AND LUNN, SOHO SQUARE,
LONDON; AND PARKER, OXFORD.
1812.
^2^Jt)
ABVEHTISEMENT.
The following pages contain a collection of
Algebraical Problems designed solely to point out
the various methods employed by Analysts in the
Solution of Equations. They are arranged in the
usual manner : 1 . Simple Equations ; 3. Pure Quad-
ratics, and others which may be solved without
completing the square; and 3. Adfected Quadratics.
Many books have been consulted; and, as utility is
the sole object of this Publication, wherever a proper
Example occured, it has been taken without hesitation,
or altered to suit the pui'pose. At the head of each
Section are given the common Rules; and the whole
concludes with a Collection of Problems without
Solutions, for the Exercise of the Learner,
CONTENTS,
-^ ^ Page
Definitions, &c 1
Solution of Simple Equations, involving only one unknown
Quantity m r. ....6
Solution of Simpk Equations, involving two unknown Quan-
tities 17
Solution of Simple Equations, involving three unknown
Quantities ; •*. 36
V Solution of Pure Quadratics, &c 42
Solution of Adfected Quadratics, involving only one un-
known Quantity..... ...59
Solution of Adfected Quadratics, involving two unknown
Quantities .....90
Solution of Problems producing Simple Equations, involving
only one unknown Quantity .' 124
Solution of Problems producing Simple Equations, involving
two unknown Quantities. 167
Solution of Problems producing Pure Equations 188
Solution of Problems producing Adfected Quadratics ..206
Solution of Problems in Arithmetical and Geometrical pro-
gressions .....244
Praxis.. ..•.•.!##. «.••.*.*. ...M**tiMt....M269
fc-.«i
ERRATA.
p. 112, line 6. for ^j/ + 2,' read Vjr+ 2.
P. 208, line 21. for - 42, read — 52,
;♦.
ALGEBRAICAL PROBLEMS.
SECT. I.
DEFINITIONS.
(1.) An Equation is a proposition, which declares
the equality of two quantities, expressed algebraically.
This is done by connecting these quantities, by the
sign ( = ) ; thus, x- 4 = 6 — x is an equation expressing
the equality of the quantities x — A and 6-x, Also
^ — 5 = O is an equation which asserts that :c — 5 is equal
to nothing, and therefore that the positive part of the
expression is equal to the negative part.
(2.) A Simple Equation is one which, when cleared of
fractions and surds, contains only the first power of the
unknown quantity.
(3.) A Quadratic Equation, or an equation of two
dimensions, is one into which the square of the unknown
quantity enters, with or without the simple power.
(4.) A Cubic Equation, or an equation of three
dimensions, is one into which the cube of the unknown
quantity enters, with or without the simple and quadratic
powers.
(5.) In general, the index of the highest power of
the unknown quantity denotes the number of dimensions
of the equation.
y
2 Definitions,
(6.) A Pure Quadratic is one into which only the
square of the unknown quantity enters.
(7.) An Adfected Quadratic is one which mvolves the
square of the unknown quantity, and also the simple
power and known quantities.
Thus, aj?* + & = is a pure quadratic,
and ax*-\-hx+c=0 is an adfected quadratic.
(8.) The Resolution of Equations is the determining,
from some quantities given, the values of others which
are unknown, so that these latter may answer certain
conditions proposed.
(9.) And these values are called Roots of the
Equation.
(10.) Known quantities are usually expressed by the
/firsjt letters of the alphabet, a, h, c, &c. ; and unknown
quantities by the last, v, x, ?/, &c. And this must b«
always understood, unless the contrary be expressed.
AXIOMS.
(11.) If equal quantities be added to equal quantities,
the sums will be equal.
(12.) If equal quantities be taken from equal quan-
tities^ the remainders will be equal.
(13.) If equal quantities be multiplied by the same
or equal quantities, the products will be equal.
(14.) If equal quantities be divided by the same or
equal quantities, the quotients will be equal.
(15.) If the same quantity be added to and sub-
tracted from another, the value of the latter will not be
altered.
(16.) If a quantity be both multiplied and divided
by another, its value will not be altered.
Reduction of Equations, - 3
(17-) Any quantity may be transposed from one side
of an equation to the other^ by changing its sign :
Because, in this transposition, the same quantity is
merely subtracted from each side of the equation ; and
(12) if equals be taken from equals, the remainders are
equal.
Thus, 5j:— 7 = 2x+2, if 2x be taken from each side,
607 '-2J? - 7 = 2, or 3a? - 7 = 2 j and if — 7 be subtracted,
or (which is the same thing) if +7 be added to each
side, 3x=2 + 7 = 9,
Also, if 0? - a -f ^ = c — 3a?, then, by subtracting
-a + 6 — 3a? from each side, we have a? + 3a? = a — ^-f-c.
Cor, 1 . Hence, if the signs of all the terms on each
side of an equation be changed, the two sides still remain
equal ; because in this change every term is transposed.
Cor. 2. Hence, when the known and unknowh
quantities are connected in an equation by the signs +
or — , they may be separated by transposing the known
quantities to one side, and the unknown to the other.
Cor. 3. Hence also, if any quantity be found on
both sides of an equation, it may be taken away from
each; thus, if 07+?/ = 5-1-2/, thena?=:5. If a-6 = c4-
<;? — ft, then az=:c + d.
(18.) If every term on each side of an equation be
multiplied by the same quantity, the results will be
equal :
Because in multiplying evfery term on each side by
any quantity, the value of the wliole side is multiplied
by that quantity; and (13) if equals be multiplied by
the same quantity, the products will be equal.
Thus, if 0? = 5 -fa, then 6x = 30 + 6a, by multiplying
every term by 6.
4 Reduction of Equations.
Cor. 1. Hence an equation, of which any part is
fractional, may be reduced to an equation expressed in
integers, by multiplying every term by the denominator
of the fraction. If there be more fractions than one in
the given equation, it may be so reduced by multiplying
every term either by the product of the denominators,
or by a common multiple of them ; and if the least
common multiple be used, the equation will be in its
lowest terms.
XXX
Thus, if -H-- + - = 13; if every term be multiplied
by 12, which is the least common multiple of 2, 3, 4 ;
6.r + 4x+3x=156.
Cor. 2. Hence also, *if every term on both sides
have a common multiplier or divisor, that common
multiplier or divisor may be taken away ;
Thus, if -\ = , then also bx -^ a
4 4 4 '-
+ 6 = 4^-7;
, . ax + ab ad 4ax ., u- i • u
Also, it = j , then, multiplying by
— , x-\'b = d-{-4x,
u
(19.) If each side of an equation be raised to the
same power, the results are equal ;
Thus, if :j7 = 6, x^ = 36; if 0? + a = y—b, then
a?* + 2a.r + a^ = z/' - 2ft 2/ + &' ;
And if the same roots be extracted on each side, the
results are equal :
Thus, if ^^ = 49, xz=7;. if 0:^+207+1=2/*- 2/ + ^,
then a? + l=^ — §;
Reduction of Equations. 5
For (13 and 14) when equal quantities on each side
of an equation are multiplied or divided by equal quan-
tities, the results will be equal.
Cor. Hence, if that side of the equation which
contains the unknown quantity be a perfect square, cube,
or other power, by extracting the square root, cube root,
&c. of both sides, the equation will be reduced to one of
lower dimensions :
Thus, if ar*4-8j7 + 16 = 36, x^4=z6,
if ^3^30^*4-30^+1=27, a?+l=3, L
if 0^ + 2^3+0:*= 100, o?* + o?=10.
(20.) Any equation may be cleared of a single radical
quantity by transposing all the other terms to the con-
trary side, and raising each side to the power denomi-
nated by the surd. If there are more than one surd,
the operation must be repeated.
Thus, if ^/o?'+7+ 0: = 7 ,
then (17) by transposition, ^x^-\r7 — 7 — ^\
and (19) by squaring each side, o:*+7=49— 14o:+jr*,
which is free from surds.
Also, if N/cr'' + >/o?'+21- 1 -X,
then (17) by transposition, \/o^* + >/a7* + 21=a;+l,'
and ( 19) by squaring each side, or*,+ >y/o;^ + 21 = 0?*+ 207+ 1 ;
.-. (17. Cor. 3.) ^x'+2\=2x-\-\,
and (19) by squaring eacli side, o?* + 21 =4o:'+4o:+ 1,
which is free from surds.
(21.) Any proportion may be converted into an
equation ; for the product of the extremes is equal to
the product of the means.
, ct c
Let a : :: c: d, by the nature of proportion r = j i
.-. (Cor. 18.) ad=:bc.
6 Solution of Simple Equations
(22.) Examples in which the preceding Rules are
applied^ in the Solution of Equations,
fi 1. Given 4x+36 = 5j: + 345 to find the value of x.
(17) By transposition, 36 — 34 = 5^?— 4^,
and therefore 2 = x.
>C 2. Given x-7=- + -, to find the value of x.
5 3
Here 15, the product of 3 and 5, being their least
common multiple, every term must be multiplied by it
(18. Cor. 1.), and 15a?- 105=3a? + 5a:;
.*. (17) by transposition, 15a?— 3j;— 5j:= 105,
or 7"^= 1^5 5
^ and therefore (18. Cor. 2.) x^ =15.
7
'7^ 3. Given 3aX'-4ab = 2ax-6ac, to find the value
of X in terms of b and c,
(18. Cor. 2.) dividing every term by a, Sx- 4b = 2x-^ 6c;
.' . ( 1 7 ) by transposition, 3a? — 2 r^. 17-3X 4x>2 ^ ^ , 7a: + 14
V 16. Given -^-g — g— =5-60?+^ — ^ »
to find the value of x.
(18. Cor. 1.) multiplying both sides of the equation
by 3x5 = 15,
51 -9J:-20a?- 10 = 75 -90i? + 35o?-f 70 ;
.-. (I7)by trans", 90x-35j?- 20j?-.9jp = 75+70+10- 51,
or 260; = 104;
/-. N 104
.-. (18. Cor. 2.) a:= — r- = 4.
^ ^ 26
K 17. Given a: _- + 4=~^~ _ + __,
to find the value of x,
(18. Cor. 1.) multiplying both sides of the equation
by 2x 5 X 7=70,
7OX - 4207 4- 42 + 280 = 700 - 350? - 60o;+ 80 + 56o? - 56;
.-. (17) by transposition,
70o?+35o:+6oo: — 42r-56o? = 700 + 80--56-280-42j,
or 67x = 402 ;
.-. (18. Cor. 2.) ^= IH = 6.
'I
involving one unknoivn Quantity, 11
t ,o r^' 4X-21 , „, 57-3^ , 5ZC-96
^ 18. Given ^ 4-3I+ ^ - =241 j^
— II a?, to find the value of x,
(18. Cor. 1.) multiplying by 36, the least common
multiple of 4, 9, and 12,
iScT- 84 +135 +5 13 -27^ = 8676 -15^ + 288 -396a:;
.'. (17) by transposition,
l6a? + 15a?+396x- 27^^ = 8676 + 288 + 84 - 135 - 513,
or 400j?=s8400;
.-. (18. Cor. 2.) x=z = 2K
^ ^ 400
,^ g^ 6x+18 ^5 ll-3a: ^ ^. 13-3^
19. Given -f^ 4. -_^ =.50.-48--^
-, to find the value of x,
18 '
' (18. Cor. 1.) multiplying by 36 x 13, the least
common multiple of the denominators,
2162:4-648^2262- 143+390; = 2340a?-22464-
507-h39a:-546 + 52x;
.-. (17) by transposition, 648 + 22464 + 507 +546-
2262- 143 =2340a?+39x + 52a?- 21607-390;,
or 21760 = 2176X;
.-.(18. Cor. 2.) il^ =10=0:.
^ ^ 217b
20. Given ' = -, to find the value
21 4o;-ll 3'
of or.
Multiplying both sides of the equation by 21,
^^_i.i^ 21.r4-l68 ^
.-. (17. Cor. 3.) 16= :jj— n— ■
12 Solution of Simple Equations
/. (18. Cor. 1.) 64a:-.176 = 21a?+l68;
.-. (17) by transposition, 64.r— 21a;= 1684-1765
or 43x=:344;
.-. (18. Cor. 2.Y ^=~ = 8.
43
21. Given -y + ^L__. = ___, to find the
value of X,
Multiplying both sides of the equation by 9,
.-. (18. Cor. 1.) 21a;-39 = 10j:H-5;
.-. (17) by transposition, 21a:- 10^ = 39 + 5,
or 110?= 44;
44
.\ (18. Cor. 2.) 0?= — =4.
X
22, Given 12-a: : - :: 4 : 1, to find the value
2
of X,
(21) Since the product of the extremes is equal to
the product of the means,
X
12— a7 = 4x - = 2a?;
2
/. (17) by transposition, 12 = 2a: + a? = 3a:,
and (18. Cor. 2.) ^ =4 = 0?.
o
23. Given .^ii^ : ^^—^ :: 7 : 4, to find the
2 4 ' '
value of X,
involving one unknown Quantity* 13
/«,\ ^•^ + 4 ^ 18-07 .
(21) — ^ X 4 = -^— X r,
126-70;
or 10o:+8= ^;
4
.-. (18. Cor. 2.) 40jr + 32 = 126-70:;
.-. (17) by transposition, 40o:+7x = 126 — 33,
or 470: = 94 ;
Q4
24. Given a/4jpTT6 = 12, to find the value of x.
{19) squaring both sides of the equation, 407+ l6= 144;
.-. (17) by transposition, 407 = 144 — 16=128;
1 28
.-. (18. Cor. 2.) 07= = 32.
25. Given >^ 2o? -1- 3 + 4 = 7, to find the value of x,.
(17) by transposition, i^2o: + 3 = 7 - 4=3 ; ^"^
/. (19) cubing both sides of the equation, 2o?-|-3 = 27;
.-. (17) by transposition, 2o7 = 27 — 3 = 24,
24
and (18. Cor. 2.) 07= — = 12.
^2
26. Given ^\2-\-x = 2+>>/a7, to find the valu^
of 07.
(19) squaring both sides of the equation,
12 4-07 = 4 4- 4>y/ 07 +07; ■^<--"'"'
.', (17. Cor. 3.) 8 = 4^/1^5
and (18. Cor. 2.) 2 = ^^;
.-. (19) 4 = 07.
27. Given ,^x + AO = 10 ->>/^ to find the value
of 07.
H Solution of Simple Equations
(19) squaring both sides of the equation,
a?4-40=100— 20>>/x + a?;
.-. (17. Cor. 3.) 2oV^=100-40 = 6o,
and (18. Cor. 2.) >>/r=3;
A (19) x = 9.
28. Given ^ x-^lij — S- \/ x, to find the value
of X,
(19) squaring both sides of the equation,
X-'\6=:64—\6a^/1c-\-Xj
.-. (17. Cor. 3.) l6»/^=64+l6 = 80;
.-. (18. Cor. 2.) >v/7=5,
> and (19) x = 25.
29. Given >/^-^=x/^-2, to iind the value
of X.
(19) squaring both sides of the equation,
07 — 24 = a: - 4>y/zc+ 4 ;
.-. (17. Cor. 3.) 4.yjc=24 + 4 = 28 ;
.-. (18. Cor. 2.) »/^=7^
and (19) 07 = 49.
30. Given a^ bX\/x-^2 = ,^fbx + 2.
(19) squaring both sides of the equation,
5o: + 10 = 5a? + 4v^'57+ 4 ;
/. (17. Cor. 3.) lO-'4 = 6 = 4^Jx;
.'. (18. Cor. 2.) ^= ^=V^,
and (19) ^ =50?;
.-. (18. Cor. 2.) ^ =x.
^ ' 20
involving one unknown Quantity » 15
-^ 31. Given — -p=r = "^^ , to find the value of x.
Multiplying both sides of the equation by /^/T,
X
X- aj?=s= - = 1,
X
I- a,x=\y
and (18. Cor. 2.) x— , .
^ ' \ -a
^. V^+28 v^+38 ^ ^ ■ ,
32. Given ~7= = —y^ — ;; , to find the value
^x + 4 ^x + 6
of a?.
(18. Cor. 1.) multiplying both sides of the equation
by (>/r+4) X (^^^"+6),
a? + 34^^l68=a: + 42>/r+152;
.'. (17. Cor. 3.) taking (x 4-34,^7+ 152) from each
side of the equation^
l6 = 8>y/7;
- .-. (18. Cor. 2.) 2 = v^J
.-. (19) 4=0?.
33. Given -^7=^: =1 -j-^LlilL-, to find the
value of a?.
Since 3j7- 1 =(^37+1) x (Vslr-iy;
3^-1 , —
>v/3jp+1
/~3x — 1 ^*
and (17. Cor. 3.) taking ^— from each side.
16 Solution of Simple Equations
3x-l _
— 1 ;
2
.-. (18. Cor. 1.) >/3l-l=2;
.*. (17) l>y transposition, >/3F=2H-l=3;
.*. (19) squaring both sides, 3,r = 9,
(18. Cor. 2.) x= ? =3.
^ ^ 3
34. Given x^s/a" ^x^b"- -^ x^ ~ a, to find the
value of :r.
-f *
(17) by transposition, x -\- a = s/ a'-^-x^ b" + x'^ ;
••• (i9) squaring both sides, :r'+2a:c+a* = a» + ^>/F+F;
.-. (17. Cor. 3.) a;* + 2flj? = x^'F+^;
.-. (18. Cor. 2.) x+2a=:^'FT^\
and (19) squaring both sides, a7' + 4ff j:+4a» = 5'+a?»;
/. (17. Cor. 3.) 4aj7-f4a^=^>»;
.-. (17) by transposition, 4aa7 = ^*- 4a*;
.-. (18. Cor. 2.) a:= ^Izif!
Sb. Given >/2+^ 4- V^= " >-^ — : , to find the
V 2+0?
value of :c.
(18. Cor. 1.) 2+1? + ^ 27+^=4;
.-. (17) by transposition, ^^207+^^ = 4- 2 -^ = 2 "^,
and (19) squaring both sides, 2^ + a:* = 4-4a? + a?*-'
.*. (17. Cor. 3.) 607 = 4;
.-. (18. Cor. 2.) 07=1 = ?.
63
36. Given v/m:I + V^= "7=. to find the
value of a;.
involving one unknown Quantity* 17
(18. Cor. 1.) 5+^ + V5i^f^=15;
/. (17) by transposition, ^ bx +x^=.\b — 5 — a?= 10— ar,
and (19) squaring both sides, 5j? + ^*=100 — SOJ^+a?"*;
.-. (17. Cor. 3.) 25^=100;
.-. (18. Cor. 2.) x= — = 4.
^ ^ 25
37. Given x + ^JTF^f^ = ,^ ■ , to find the
value of 37. ^ *
(18. Cor. 1.) x^ a'-i-x' + a' + a?' ^ 2a' ;
• . (17) by transposition, x^oa- ^x" = 2q' ^^ a'' — jj* = a' - «%
and (19) squaring both sides, ^^.a^+aj^zsa'^— gaV+JP*,
or aV+a?^ = a^-2aV+^;
.-. (17. Cor. 3.) 3aV = a4;
and (18. Cor. 2.) x^^ ^ = ^*;
3a^
••• ^^^> ^=;^
18 * Solution of Simple Equations
Sect. II.
0)1 the Solution of Simple Equations which involve
move than one unknown Quantity,
(23.) If the equations involve several unknown
quantities, and definite values of these are required,
there must necessarily be as many independent equa-
tions as there are unknown quantities. In which case,
the values will be found by exterminating all the
unknown quantities except one ; and this may be done
by either of the three following methods :
1. By equalizing the coefficients of the same unknown
quantity in the several equations.
2. By substitution.
3. By equating different values of the same unknown
quantity.
1. Of exterminating an unknown quantity by the
first method in equations where two unknown quan-
tities are concerned.
If the coefficient of either unknown quantity in one
equation be contained a certain number of times exactly
in the coefficient of the same unknown quantity in the
other, multiply the former equation by that number,
then add it to, or subtract it from, the other equation,
according as the signs are different or the same, and an
equation arises, in which only one unknown quantity
is found.
involving tivo. unknown Quantities. 19
^ A Here the coefficient of a; in
and 4^ + 0?= 10^
the second equation is contained 4 times exactly in
the first;, multiplying therefore the second equation
by 4, -and subtracting the first from it,
4X'\- l6i/ = 64,
and 4x + i/ =34 ;
15^ =30, and t/=2.
Having thus obtained a value of one of the unknown
quantities, the Other may be determined by substituting
JD either equation the value of the quantity found, and
thus reducing the equation to one which contains only
the other unknown quantity. Thus, from the second
of the preceding equations, 07=16 — 4^ =16-8 = 8.
The values of jo and 3/ might be found in a similar
manner, by multiplying the first equation by 4, and
subtracting the second from it.
But if neither of the coefficients be a measure of
the coefficient of the same unknown quantity in the
other equation, multiply the first equation by the co-
efficient of one of the unknown quantities in the second
equation, and the second equation by the coefficient of
the same unknown quantity in the first. If the signs
of the unknown quantity be alike in both, subtract
one equation from the other; if unlike, add them
together, and an equation arises in which only one
unknown quantity is found.
Thus, if 2x + 3y = 23) - .. ., ,. . .
J ^ ^ > In this case neither of the
and 5a?— 23/ = 10 >
coefficients is a measure of the coefficient of the same
unknown quantity in the other equation ; and therefore,
multiplying the first equation by 2, and the second
by 3,
20 Solution of Simple Equations
and l5x—6i/=30;
.\ by addition, igx = 76, and a? = 4;
whence^ as before, 3i/ =: 23 - 2a: = 23 — 8 = 1 5,
and 3/ = 5.
The values of x and i/ might also be obtained, by ^
multiplying the first equation by 5, and the second by
2, and then subtracting the second from the first.
2. By substitution.
Find the value of one of the unknown quantities,
in terms of the other and known quantities, in the more
simple of the two equations ; and substitute this value
instead of the quantity itself in the other equation ;
thus an equation is obtained in which there is only
one unknown quantity.
Thus in the first of the preceding examples ; from
the second equation, x=l6-^ 47/ ; substit^ting therefore
this value of x in the first equation,
4 . l6 — 4y+3/ = 34,
or 64-i6t/-}-i/=34;
.'.by transposition, (64--34 = )30 = 153/,
and therefore 2 = i/ ;
whence, as before, x = S.
Here a value of x might have been obtained from
the second equation, and substituted for it in the first ;
whence an equation would have arisen, involving only i/ ;
the value of which being found, that of x also might be
determined, as before, by substitution.
Or a value of ?/ might be determined from either
equation, and substituted in the other; froni which
would arise an equation involving only x, the value of
invdving tnvo unhncum Quantities. 21
which might be found ; and therefore the value of y
also might be obtained by substitution. ' -
Again/ in the second example; from the first equation
is obtained
2 x = 23 — 33/ ; and therefore x = -^ ;
substituting therefore this value in the second equation,
^ 23 -3y ^
or 115 — 15^-4i/ = 20;
.*. by transposition, 115 — 20 = 15t/+4y, ^
or 95 = 19?/;
.-. 5=y,
23 -3v 23-15 8
and xss ^ = — = - =4.
2 2 2
Here also a value of x might be obtained from the
second equation, and substituted in the first, which
would give an equation involving only y ; or a value
of 1/ might be obtained from either equation, which
substituted in the other would give an equation involving
only X ; the value of which might therefore be found,
and consequently that of i/ might also be determined.
3. By equating different values of the same unknown
quantity.
From each equation find the value of the same
unknown quantity in terms of the other and known
quantities ; then, by equating the values so found, an
equation arises containing only one unknown quantity.
Thus in the first of the preceding examples ; from
the first equation, 2/=34— 4j?,
and from the second equation,
4^= 16 — X j and therefore ^ = —: — ;
22 Solution of Simple Equations
/. =34-41-;
4
consequently, l6-a?=136-l6jp;
. .*. by transposition, 1 6^7 — a' = 136 - 1 6,
or 15 J? =120;
.-. a? = 8,
and v=34-4.r = 34 — 32 = 2, as before.
In this case also, two values of x are deducible from th^
two equations, which would give an equation involving
y only ; and the value of y being determined, that of
X might also be found.
Again, in the second of the preceding examples ;
23 3?/
from the first equation, x = —^ ^ ,
and from the second, x = ^ ;
5
10 + 2?/ __ 23—31/
•** r^ — T^'
and 20 + 42^=115 - 15?/;
by transposition, 4y + 1 5y = 1 1 5 - 20,
. ' * or 193/ = 95;
/. y = 5, and a? = 4, as before.
Here again two values of y might have been found,
which would have given an equation involving only x ;
and from the solution of this new equation, a value
of X, and therefore of y, might be found.
Examples.
and 3x + 7y
1. Given 5zr + 4i/ = 58> ^ . , ^, , r
_/.>-, to find th^ values of x
and y.
involving two unknown Quantities. £S
Multiplying the second equation by 5, and the
first by 3,
15j? + 35i/=335,
and l5x-\-l2i/=:l74;
. • . by subtraction^ 23 1/ = 1 6 1 ,
and . 3/ = 7s
whence, 5t=:58-43( = 58 — 28 = 30,
and therefore x = 6.
If the second equation had been multiplied by 4, and
subtracted from the first when multiplied by 7> ^^
equation would have arisen, involving only x, the value
of which might be determined, and thence, by sub-
stitution, the value of 3/.
Second Method.
Prom the second equation, 3x = 67 — 7^ h
,.x^ 3 •
Substituting this value of x in the first equation,
and 335 -355/ + 122/= 174;
.*. by transposition, 335 — 174 = 353/ — 1 23/,
or l6lsB23y;
whence, as before, the value of x may be found. In
the same manner, a value of x might be found from
the first equation, which substituted in the second,
would give an equation involving only 3/. Or a value
of y ' might be obtained from either equation, and sub-
stituted for it in the other j whence an equation would
arise involving only x, the value of which might be
found, and therefore that of 3/ also determined.
S'Jf Solution of Simple Equatiotu
Third Method.
From the first equation, 5-r= 58 - 4i/ ;
58 -4y
From the second, x = - * ( 2 ! ;
o
58 --4y _ 67 - 7v
and 174- 123^ = 335 -35e/;
by transposition, 351/— 12i/=335 - 174, .
or 233/= 161 ;
whence, as before, a: = 6. - '
In this case, two values of y might be deduced from
the two equations ; and from equating these, there would
arise an equation involving x only ; whose value being
found, that of y also might be determined by sub-
stitution,
2. Given \\x-{-3y^lOO%.
and Ax-^7y=.A/]^ '^ ^"^ ^^^ ^^^"^^ ^^
X and y.
Multiplying the first equation by 7, and the second
by 3,
7Tx-\-2\y = f(Hy^
and \2X'-1\y~ 12 1
/. by addition, 89a; =712,
and 0? = 8 ;
whence 3t/ = 100 — lla?= 100 -88 = 12 ;
/. i/ = 4.
involving two unknown Quantities. 25
3, Given f ^-f =7
and
M =4
find the values of x and y.
(18. Cor. 1.) clearing the equations of fractions, by
multiplying each by 6,
3a? -f 2^ = 42,
and 2a? + 3^ = 48;
and as the coefficients in this case are not aliquot parts,
multiplying the first by 3, and the second by 2;
/. 9x4-63^=126,
and 4x + 6^ = 963
/. by subtraction, bx =30,
and J? = 6 ;
whence 21/ = 42 - 3x = 42 — 1 8 = 24,
and V = 12.
f + 7^ = 99")
>, to
1 + 7^.-51)
4./ Given ^ . .^ ^^^
find the values of
and
X and y,
(18. Cor. 1.) multiplying each equation by 7,
.-. ^ + 497/= 693,
and 4Qx^y= 357;
.'. by addition, 50J7-}-50y=: 1050,
and therefore x-^-y = 21 ;
but since -x + ^Qy = 693,
—^
subtracting the upper equation from the lower,
48^ = 672 ;
.-.,^ = 14,
whence j? = 21 -y = 21 — 14=7.
E
Q6 Svkfion of Simple. Equatiotu
find the values
5. G- ^ + ^
iven ^~^+ 8^= 31>
> , to
^ V + 5 f
and •^-^— +10x=192V
4 -^
of X and i/.
Clearing the first equation of fractions,
.•. by transposition, x + 24y = 91.
Clearing the second equation of fractions,
^-f 54-4ajD = 768;
/. by transposition, 40x+y^T63.
Multiplying the first equation by 40, and subtracting
the second from it,
^ 40j: + 960y = 3640;
AOX'\-y = 763 ;
.-. d^9y =2877,
and ;y = 3 ;
.-. a;=91-24y = 9l-72 = 19.
6. Given -^— ^
to find the values
-^ + 14=18^
of X and ?/.
2tl? '*^ 2/
By transposition, — =4, from the first equation,
an d therefore 2x-y = S,
2i/"4— zr
Also, -^^- — =3, from the second equation,
and therefore 23/4-,r=s9,
involving two unknown Quantities* S7
which, multiplied by 2, gives 2a?+4i/fa:i8 ;
but ax-y^i 8;
/. by subtraction, 52/= 10,
and therefore 3/ = 2,
whence a: = 9-2^=r9-'4 = 5.
7. Given ?^+^ + f = .
find the values
and^y
V^ j.co
of X and ^.
Clearing the first equation effractions, 2«r + 3i/4-2a?==48,
or 4j? + 3y = 48;
and clearing the second of fractions, Ty'-3x—2y:= 22,
or 5y — 3 , to find
From the second equation, (18. Cor. 1.) multiplied
by 13,
8:r-30+% = 41 -6j:4-3 ;
,\ by transposition, 14^^ + % = 74,
and 7a?+ 3^ = 37.
But from the first equation, 5t/ = 3x -\- 3,
or 5t^ — 3x=:3,
Multiplying this equation by 7, 352/— 2lJ?=i 21,
and the former by 3, 95/-{-2lj7= 1 11 ;
/. by addition, 44^ = 133 ;
.-. 3/ = 3;
by
/. x+I= -—• =5, and therefore \r = 4.
10. Given -g j~ = '^-j— /
. ,or * to find
J 2V + 4 2X'hy x+ 131 ^
the values of x and 3/.
(18. Cor. 1.) multiplying the first equation by 60,
12j?-24-200 + 30i?=15j/- 160;
and by transposition, 3 2 j?- 15^ = 74.
Also (18. Cor. 1.) multiplying the second equation
by 24,
l6i/ + 32-6^-3i/ = 6a? + 78i
involving two unknown Quantities, 29
/. by transposition, 131/- 12x = 46.
Now the coefficients of x have ahquot parts ; multi-
plying therefore this by 8, and the preceding by 3,
1042/ -96j7 = 368,
and 96.r— 45z/ = 222;
/. by addition, bQy =590; .'.3/= 10,
and 32a7=152/ + 74 = 150+74 = 224;
.-. J7=7.
ril¥Gia23/-^±^=7+^^^^
t-J^ ■ . W^" o o \l , V . to find the
arid 4x -^ =24i
3 ^
?
values of x and y,
(18. Cor. 1.) from the first equation,
40y-5J7-15 = 140+12a7-83/;
/. by transposition, 48z/— 17^ = 155,
and from the second equation,
2 4a; - 1 6 4- 2y = 1 4 7 - 607 - 3 ;
.-. by transposition, 30j?-f-2^ = 160.
Multiplying this by 24, 483/ + 720j7 = 3840 ;
but 48t/- 17^:= 155 ;
.-. by subtraction, 737^ =3685,
and 0? = 5,
and 23/ = l6o-30J? = l6o- 150=10;
/. yz=zb.
12. Given ^-?iz£z:i« l±^-f^ , ,
18 36 3 Try, tofind
and X : 33/ :: 4
the values of x and 3/.
3 6 L to
30 Solution of Simple Equations
Reducing the first equation to lower terms,
and therefore (18. Cor. 1.) multiplying by 18,
22/- 407 + 1 = 18 — 24 — 6^4- 3a: -3?/ ;
.'. by transposition, 7 =7^ — 1 1?^.
But from the second equation, 7^ = 1 ^y •
Substituting therefore this value in the preceding
equation,
12z/
and therefore x = — ^ = 12
7
,^ g^' 3y — 2+x , . 3
-13. Given x—^ ^^-— = 1 H ^
11 33 S, to
, 33: + 2y y-5 _ na?+152 3?/ + 1
6* 4 12 2
find the values of x and ^.
(18. Cor. 1.) multiplying the first equation by 33,
33j?-92/ + 6-3j? = 33 + 15a?+ ^j
4v
by transposition, 15i?— 9i/ = 27+ -~;
.-. 45j:-27^ = 81 + 4y,
^nd 45j?-31e^ = 81.
(18. Cor. 1.) multiplying the second equation by 12,
6^4-43/-3y + 15 = lla7 + 152- 18^^-6;
.-. by transposition, 1 93/ — 5a; =» 1 3 1 .
Multiplying this by 9, 1715^ - 45a? = 1 179 ;
but 45a;-31y= 81;
.*. by addition, 14O3/ =1260;
involving two unknown Quantities i 31
and 5^=193^-131 = 171-131=40;
.-. ^ = 8.
^^ ^. 80 + 31? ^^1 4^+3jy-8
14. Given — --- — = 18^ ~
15 ^
' and 102^ H = 55
+ 10a?V
find
the values of oc and ^.
(18, Cor. 1,) multiplying the first equation by 105,
the least common niultiple of 3, 7, and 15,
560 + 2 1 j: = 1 925 - 60a: - 452/-+ 1 20 ;
.*. by transposition, 81a: + 45y= 1485 ;
and dividing by 9, gx+ 5i/=l65,
Now from second equation, 50^ -{- 6x - 35 = 275 + 50^;
.'. by tra^isposition, 50^ — 44a; = 310;
and dividing by 2, 252/- 22^7=155 ;
but multiplying the equation
found above, by 5.
> 25 V + 45x = 825;
/., by subtraction.
S7x =670,
and a; =10.
Now
55^= 165 -90:= 165-
.-. t/=15.
90 =
75;
^ni/4
5x-f22/ 3y-12+8Ji
6 .. 5
• = 4-
15+207-
3
4y
J 7a?+13~52/ . ^ 3:i? + 2v-l6
and -^ 7- ^ +^ = 22/ ^
4 -^ 3
\
to find the values of x and 3/.
(18. ^or. 1.) multiplying the first equation by 30,
the least common multiple of 3, 5, and 6,
302/ + 25^+10j/-182/ + 72~48a?=120-150-20:r + 402/;
32 Solution of Simple Eqt4atious
whence, by transposition, 102 = 183/4- 3a^;
and dividing by, 3, 34= 6y-\-x,
Clearing the second equation of fractions,
2IX + 3.9- \by + \2x=z24y- I2JC-83/ + 64;
and by transposition, 45a:— 31y= 25.
Multiplying the former by 45, 45a? + 2701/ = 1 530 ;
/. by subtraction, 301?/ =1505,
and 3/ = 5 ;
whence x = 34-6y = 34 — 30 = 4.
16. Given 1 -f 7?— ^■"^—
o
and : 5x
9
3 12 f
3 -^
to find the values of x and j/,
(18. Cor. 1.) multiplying the first equation by 12,
the least common multiple of 3, 6, and 12,
12 4-50 4- 105/ - 28j?4-24 = 120 - 30?+ 10 - 7,^ ;
.-. by transposition,^ 173/- 25j: = 44,
and (21) from the second equation,
96 -8x , 144-V
9 3
.-. 96 -8j? = 45x- 42-31/;
and by transposition, 138 = 53a; - 3^.
Multiplying this equation by 17, and the one found
above by 3,
5I3/- 75a?= 132,
and -5l2/+901a? = 2346;
.-. by addition, 826a7 =2478,
and a: = 3. ^
Now 3i/ = 53a?- 138 = 159- 138 = 21 ;
.•.^ = 7:
involving two unknown Quantities, 33
17. Given i-f+^=§-l
and -
X
y y i
y X 2J
^ ^ ^ \o find the values
of X and y.
Reducing the first equation to lower terms.
X y y
..44
.'. by transposition, - — - = — l ;
X y
2 4 3
and from second equation, by trans% - - + - = - ;
2 1"
.-. by addition, -7 = -; »
4=x,
A A
and - = -+1 = 2;
y X
/. 2y = 4, and y =
18. Given X -^ =20 ^
23 -JT
V-3
and i/+ -^ =30-
values of x and y.
Multiplying the first equation by 2,
41/ — 2x
2X '-^^ =40-59 + 2x;
23 — 0?
4iJ'-2x
/. by transposition, 19= -^^^ ,
and 437''l9x = 4y — 2X'y
and by transposition, 437 = 17'^ + 43/'
F
to find the
34? Solution of Simple Equations
Also from second equation, 3y + -^ = 90 — 73 + 3y,
and (17. Cor. 3.) ^=17;
••* 3^-9 = 1707-306;
by transposition, 297 = 1 70: - 3^ ;
but 437 = l7a? + 4t/;
/. by subtraction^ 140 = 7y,
and 20 =y,
and 17^ = 297 +3// = 297+6o=357;
.*. a? = 21.
19. Given 8x- 1^±^ = 1^5111}^^
-^^^ 3jr-23/+l J
the values of j? and z/.
Multiplying the first equation by 5 + 22/,
, , n .^ ,^^ 80 + 300^+ 3 2v+120a?y
/. by trans" 40a? + 107 = — -^ — ^-^ ^ ;
•^ _ 33/ - 1
and multiplying by 3i/ - 1,
120.13/- 40j?4-321i/-l07 = 80H-300:Cri-323/+120a?2/;
.•. (17. Cor. 3.) 289y-340:c=187.
And from the second equation,
27a?*- 12i/*+ 15^7+ 22/ + 2 = 27^* -12i/" + 38 J
.*. by transposition, lbx-{'2i/ = 36 ;
\Vhence, the coefficients of x having aliquot parts,
multiplying the first equation by 3, and the second
by 68,
involving two unknown Quantities, 35
8673/-i020a?= 561,
and 1363^+1020^ = 2448;
by addition, 1003 2/ =300^,
and 3/ = 3,
and 15:c = 36-2y=36-6 = 30;
7+~
20. Given 3 5L=5_5£±9
5 3y
107 /
, 4+153, ^^^"T-
and y r; ^ = ;
^ 6x-2 2X+5 ^
I
to find the
values of x and y.,
(18. Cor. 1.) multiplying the first equation by 15y^
.•. 45^ — 2 1 2/ - 6^ = 75y - 25^ - 45 ;
and by transposition, blt/-l9x = 4b.
Multiplying the second equation by 2j?-f5,
2xy+5y- »^ + ^0+^^0^Jf+7^y ^^^y_ M.
.-. (17. Cor. 3.) 53,+i27=!£±£^g2^±7£|^j
and multiplying by 6a? — 2,
321^ 107
30a?3/-102/+ ^ =:8a?+20 + 30:i?3/ + 753/;
.-. (17. Cor. 3.) ^^^^"^^7 =8a?+853/ + 20,
and 321^- 107 = 32a?+3403/ + 80;
and by transposition, -187 = 340«/ - 2892".
The coefficients of y in this case having ajiquot parts ;
multiplying the first by 20, and the last by 3,
3d Solution of Simple Equations
10202/- 3800:= 9OO,
and 1020;y-867x= — 561 ;
.". by subtraction, 487^ = 146l,
and 57 = 3 ;
consequently, 5 1^^ = 45 + iga: = 45 + 57 = 102
.-. y=2.
(24.) If there be three unknown quantities, their
values may be found from three independent equations.
For from two of the equations, a third, which
involves only two of the unknown quantities, may be
deduced by the preceding rules; and from the re-
maining equation, and one of the others, another which
contains the same two unknown quantities. Having
therefore two equations, which involve only two un-
known quantities, these may be determined ; and, by
substituting their values in any of the original equations,
that of the third quantity will be obtained. In some
particular equations, two unknown quantities may be
exterminated at once.
Examples.
1. Given ^+3/ + ;2 = 31
^+3^ + ^ = 31^
a?-f3/-;s = 25 V , to
find the values of
Xy y and %,
Adding the first and third equations, 2^ = 40 ;
involving three unknown Quantities. 37
Subtracting the second from the first, 2« = 6 ;
.*. 2; = 3;
and subtracting the third fi*om the second, 3t/ = l6 j
/. 2/= 8.
2. Given x-\^ i/ + z =29
x-\- 1/ + z = 29^
x-^-^iz + Sz-Gal, to
2 3 4 J
find the values of
and z.
Subtracting the first equation from the second,
l/+2z = 33.
(18. Cor. 1.) multiplying the third equation by 12,
the least common multiple of 2, 3, and 4,
6x + 4i/ + 3z=l20;
multiplying the first equation by 6, 6x + 61/ + 6z=i 1^4;
by subtraction, 2^ + 32 = 54 ;
but 2i/.+42= 66;
,\ by subtraction, «= 12,
and 3/ = 33 — 22 = 33-24 = 9;
also a: = 29-2^~2 = 29-9~ 12 = 8.
In like manner, had the first equation been multi-
plied by 2, and jsubtracted from the second, an equation
would have resulted, involving only x and z ; and had
it been multiplied by 4, and subtracted from the third
when cleared of fractions, another equation would have
been obtained, involvmg also x and z ; whence, by the
preceding rules, the values of x and z would be found,
and consequently the value of t/ also, by substitution.
Or if the first equation be multiplied by 3, and the
second subtracted from it, an equation would arise
involving only x and 7/ ; and if the first, when multi-
plied by 3, be subtracted from the third when cleared
38 Solution of Simple Equations
of fractions, another would arise involving only x and
y ; w^hence the values of x and y might be determined.
And hence the third, that of «, might be found.
Second Method,
"From the first equation, a? = 29 - y — sj ;
.-. substituting this value of x in the second equation,
29— ^ — 2 + 2i/+3;s = 62 ;
.*. by transposition, 3/=33 — 2sr.
Also substituting, in the third equation, the value of
X found from the first,
22^^ + f + !=.«,
/. (18. Cor. 1.) l74-62/-6^ + 4i/-f32=12o,
and by transposition, 54=3z-\-2y;
in which, substituting the value of y found above,
54z=3z+66'-4z;
.'. by transposition, 2 = 12;
whence t/ = 33 — 22 = 33 - 24 = 9,
and a? = 29-^-2 = 29-9— 12 = 8.
It may be observed, that there will be the same
variety of solution, as in th^ last case, according as
«, y, or z, is exterminated. v
Third Method.
From the first equation, j? = 29 — 3/ — z^
and from the second, x=z62 — 2y-3z;
.*. 29-3/ — 2 = 62 — 23/ — 3;s,
and by transposition, 3/ =33 - 2z.
Again, from the third equation, a? = 20— -^ — -i
3 52
2v z
... 29-^-2 = 20- -f - r;
involving three unknown Quantities, 39
z y
and by transposition, 9 — « ~ 3 *
^ Sz
••• 27- — =3/;
whence 27 r: 33 — 2;2; ;
' 2
.'. by transposition, - = 6, ^
and 2;= 12;
whence 3/ = 9, and j: = 8, as before.
The same obsei'vation appHes to this solution, as did
to the last.
3. Given — -^ - -y^ -^ = 5 4- z
10 15 5
, 9^+51/ - 2Z 2x 4-.V - 3g _ 7,y 4-z + 3 1
^"12 4 ' 11 61
1 5VH-32; 2j? + 3z/ — ^ 3x + 2y+7
''''^4, ^^^—+2^=3/-! + -^3:^
to find the values of x, y, and 2;.
Multiplying the first equation by 30, the least
common multiple of 5, 10, and 15,
1 207 + 93/ + 3;s-r- 4^ - 4;2 4- 2x - 2 = 150+6^ - 62-30 ;
.-. by transposition, 8jr4-5i/+52; = 122.
Again, multiplying the second equation by 132,
the least common multiple of 4, 6, 11, 12,
99x -i- 553/ - 22Z - 66x - 331/ 4- 992 = 843/ + 1 22; 4- 36+ 22 ^
.-. by transposition, 33.r- 623/4-652 = 58.
Again, multiplying the third equation by 12, the
least common multiple of 12, 6, 4,
153/+9«-2a:-33/4-2; + 24;2;=123/- 12 4- 6a? 4- 4^+1^^
40 Solution of Simple Equations
.•. by transposition, .Sx-^Ay — 3Az = — 2 ;
but from the first equation, 8.r + by -\- 5^= 1 22 ;
.'. by subtraction, y-\-3^z = 124.
Also the third equation being divided by 2, -
4X'\-2y- \Tz=z — l,
Multiplying this by 33, and the second by 4^
132^+ 66y-b6\z^ - 33,
and 132i?~2482/+2602;= 232;
/. by subtraction, 314?/— 8212= -265;
but 3 142/ + 12246^ = 38936
(by multiplying the equation found above by 314) ;
.-. by subtraction, 13067;s = 39201,
and therefore ;s = 3 ;
w^hence 3/ = 124 — 392 = 124- 117=7,
and 4j:=17z-22/-1=61 -14- 1=36;
.*. a: = 9.
(25.) If there be four unknown quantities, their
values may be found from four independent equations.
For from the four given equations, by the preceding
rules, three may be deduced which involve only three
unknown quantities, the values of which may be found
by the last Article ; and hence the fourth may be found,
by substituting in any of the four given equations, the
values of the three quantities determined.
If there be n unknown quantities and n independent
equations, the values of those quantities may be found
in a similar manner. For from the n given equations,
w- 1 may be deduced, involving only n-\ unknown
quantities; and from these /^— 1, n-2 may be ob-
tained, involving only w — 2 unknovt'n quantities ; and
involving three unknown Quantities. 41
SO on, till only one equation remains, involving one
unknown quantity ; which being found, the values of
all the rest may be determined by substitution.
(26.) If there be more unknown quantities than
independent equations, some of these quantities cannot
be found except in terms of the others ; and by assuming
values of these others, we may obtain an infinite number
of corresponding values of the former quantities, which
will satisfy the conditions proposed.
But if there be fewer unknown quantities than
independent equations, the values of the unknown
quantities may be found from the different equations;
and if these values be the same, some of the equations
are unnecessary; if different, the equations are in-
congruous.
A
42
Sect. Ill
On the Solution of Pure Quadratics, and others which
may he solved without completing the Square.
(27.) When the terms of an equation involve the
square of the unknown quantity -only, the value of the
square will be found by the preceding articles ; and
extracting the root on each side of the equation, the
^ unknown quantity itself will be determined.
In the same way any pure equation may be solved ;
for the power of the unknown quantity standing alone
on one side of the equation, the known quantities being
transposed to the other, the simple unknown quantity
will be determined by extracting the root.
And by the same process, any equation containing
the powers of a function of the unknown quantity, or
containing the powers of two unknown quantities, may
frequently be reduced to lower dimensions.
Examples.
\ 1. Given j:'— 17 =f'130- 2:r*, to find the values
of X,
By transposition, 3j?*=147;
.-. ^' = 49,
and X =:=t7'*
* The square root of a quantity may be either -\- ot — , and
consequently all quadratic equations admit of two- solutions. Thus,
_j-7 X -f 7, and — 7 x — 7, are both equal to 49 ; and both, when
substituted for x in the original equation, answer the condition*
required.
Solution of Pure Quadratics^ ^c. ' 43
2. Given x-^-y : a? :: 5 : 3/ ^ ^ j ^i i
^ ^ r 5 to find the values
and xy = b S
of X and y.
Since x+y : x :: 5 : 3;
/. {Wood's Alg, 180.) ^ : .r :: 2 : 3;
- 2.!!r
.-. (21) 3y = 2x, and 3/=-J-
Substitute this value in the second equation,
2a?' ^
.*. — = o,
3
and a?* — 9 5
therefore, extracting the square root, x=^ ±3,
jLX
w^hence v = — = + 2.
jf 3 -
r 3. Given j: + y : x-y \\3 : l) ^ /: j i.u i
' 1 , , ^ r 5 to find the values
and x^-y =bb >
of J? and y.
From the first equation, {Alg. 182.)
2x : 2 V :: 4 : 2 ;
/. {Alg, 184.) X : y :: 2 : li
^ and x=2y.
Substituting this value of x in the second equation,
.-. 81/3-. I/' = 56,
or 7i/3 = 56;
and y=z 2,
whence a? = 21/= 4.
^4. Given x-y \ x :: 5 : 6# ^ , ,
> , to find the values
and a7y* = 384
of X and ?/.
■■\
44 Solution of Pure Quadratics y ^c,
(By Jig. 177 & 181.) X : y :: 6 : 1 ;
.-. (21) x=z6y.
Substituting this value of x in the second equation^
and 6y=384;
.-. r/3= 64,
whence y= 4,
and x=:6y= 24.
-^ ^
5. Given a7 + y : a: :: 7 : 5/ ., , ,
^ * ?- . to find the value
and j:i/ -f 2/* = 1 26 3
of X and 3/.
(^/^. 180.) y I X V. 2 \ b\
.'. (21) 2x = byy
and « =
■ 2 •
•<,
Substituting this value for
2- +^ =
X in the second
126;
equation,
or 7/ =
.-. y- =
and y =
252,
252;
36,
±6;
■
.'. x= f =
±15.
' :: 64
: 1
|, to
6, Given ^+i/l : x-y]
and xy = 63
values of a? and y.
find the
(^/^. 188.) x+y : ,
,'. {Alg. 182.) 207 :
and {Jig. 184.) a? :
.-. (21) 7
and x =
»-«/ ::
2y ::
y ::
7 '
8 :
9 :
9 :
1;
7,
7;
'Without completing the Sguare* ' 45
Substituting this for x in the second equation,
f = 63;
and v = ± 7 ;
7. Given x* + a?v = 12 ^ , , , /»
, „ > , to find the values ot x
and
X* -1-0??/ = 12")
3/Va:?/ = 243'
and 3/.
Adding the two equations together,
0?"^ + 2^3/ +3/* = 36;
.*. extracting the square root, x + 7/= ±6
Now x'' + xi/ = x ,x-\-i/z=z ±6x;
.'. ±6x=zi2,
and x= ±2;
and therefore ^= ±6 + 2= +4.
8. Given x +y =* ^ , to find the values of x
and J?* — 3/* = ^/*j and y.
Since x" -if^x^-y ,x — y^s .x-^y \
,\ s,x-y=d\
but x+y = s;
.'.by addition, 2x =s-\ = — .,
s s
and 0?= — - — ,
2*
46 Solution of Pure Quadratics, ^c.
d^ s^ — d^
and by subtraction, 2//=^ — — = ;
s' - d'
^ 2s
9. Given nA + v/^=5? ^^ ^^^^ ^^^ ^^^^^^ ^^
and sf x — sj y = \j
X and ^.
Adding the two equations, lijlc== 6 ;
.'. va7= 3,
and a; = 27,
and subtracting the equations, 2\J y= A ;
and 3/= 8.
f
13
10. Given a?' + ?/* = # ,c j ^u 1 r
find the values of
•^ a;-3/f , to f
6 ( X
and
-y
and ^.
From the first equation subtracting twice the second,
.-. X— ^) =1, and j?-2/ = l;
.-. x*+i/' = 13,
and 2xy =12;
. by addition, x^ + Ixy +3/* = 25,
and j?+y= ±5 ;
but x—y= 1 ;
, . by addition, 2x = 6', or — 4,
and x = 3, or - 2,
and by subtraction, 2?/ = 4, or —6;
/. 3/ = 2, or -3.
tuithout completing the Square, 47
' 11. Given x^-xy=:4Sy
and xy
X and y.
-^y y{ ^^ ^^^ ^^^ V2S\M^% of
(x^y = ) 3y = (^=^=)l2; ,'.y=.4;
Dividing the first equation by a:, x-y= — ^ 5
X
3x
and the second by y, a? — ?/= — ;
48;/ __ 3a?
•• V ~ 3/ ' .
/. 48^/* = 3a?%
and l6y = x* ;
consequently, ±42/= a?;
and first, suppose -jr 4y = x;
'48y 48y
But if J? = — 4^,
12
A .48
and 0-' = — 4?/= — .
o
'T 12. Given -^^ = 48/
a:«^ » V 3 to find the values of x
, xi/ I and 1/.
and -*~ = 241 ^
Dividing the first equation by the second,
(^X V X
-^^ =)3/^ = 2;
.-.^ = 4;
48 Solution of Pure Quadratics ^ ^c.
Ax —
whence, from the second equation, — ;= z=z4^x = 24;
and x=zS6,
At X* 4-'3a? — 7
^13. Given =1, to find the valued of x.
X
Clearing the equation of fractions,
1 8
X
18
.*. by transposition, x'+2j? = 9+ — ,
X
2 9
or cT .^ + 2=9 . 1+ - = - . 0^ + 2 J
0?= -
0^
and a:* = 9 ;
.•. 0?= +3,
'^ 14. Given x^ + V^=:13- ^ , ,
^ , i L > , to nnd the values of
and
x- + t/- = 13)
x^ ']ry^= bS
V and i/.
Squaring the second equation, x'^ + 2x^y^ 4- 1/^ = 25 ;
but x^ 4.3^^=13 ;
/. by subtraction, 2a?3^» = 12.
Subtracting this from first equation, x^-'2x^y^ +2/^=1 ;
.*. extracting the root, a?t— 3^t= + 1 ;
but x^ 4-y^ = 5 ;
.\ by addition, 2a?t=:6^ or 4,
without completing the Square, 49
and x^— 3, or 2;
whence x =27, or 8 ;
but by subtraction, 2y^ = 4, or 6,
and 3/^ = 2, or 3 ;
.-. 1/ = 8, or 27,
V15. Given xy* +v =21 1 ^ , , , n
"^ "^ t , to find the values of
and xy + 2/* = 333i
X and y.
Squaring the first equation, ^y + 2a??/^ +3/* = 441 ;
but a?y +2/* = 333;
/. by subtraction, 2xy^ = 108 ;
Subtracting this from the second equation,
oc^y^ - 2xy^ +3/* = 225 ;
extracting the root, xy*" ^y= ± 15 ;
I but xy''-\-y=z 21;
'. by subtraction, 2yzz6, or 36,
and 3/ = 3, or 18 ;
but by addition, 2xy'' = 36, or 6;
.*. 0^1/*=. 18, or 3,
18 3
and therefore x:± — , or
9 ' 324'
= ^' "^T58- .
v^l6. Given x'y+xy^ -ISOl ^ , , , p
^ , >, to find the values of
and x^ + 3/^ = 189)
X and y.
Adding 3 times the first equation to the second,
x^ + 3x'y + 3xy^ +3/' = 7^9 i
whence, extracting the cube root, X'{-y=:^9,
H
^0 Solution of Pure Quadratics^ ^e,
, Now x^y-^-xy^ = x +3/ . 0:2^ = 1 80 ;
/. by substitution, 90:2/ =180, -
and xy^20.
But a?* + 3a?y+/ = 81,
and 4xy =80;
.-. by subtraction, x"-'2xy^y^=: 1 ;
.*. extracting the square root, x—y^ ±1,
and x-^-y^^ 9>
' ,1, -
.'. by addition, 2x=: 10, or 8;
.-. a?= 5, or 4;
and by subtraction, 2y=. 8, or 10 ;
.•. 1/= 4, or 5.
^ 17. Given ^ + V^-f3/ = 19 ? ^^ find the values
and x*+^2/ +/=133 5
of a? and y.
Dividing the second equation by the first,
x-Jxy-^ry=^ 7;
biit x-^-^xy^y^l^'y
,\ by addition, 2j? + 2?/ = 26
and 0?+ «/ = 13
and by subtraction, 2/^xy = \2
.-. ^/^ = 6,
and 072/ = 36.
Now from the second equation, x* +073/+^*= 133 ;
and from the last, Sxy = 108 ;
.-. by subtraction, a,* — 2x3/ +3/-= 25;
witJiout completing the Square, 51
and extracting the square root, ^ - 3^ = ± 5 ;
but J? +3/= 13;
.-. by addition, 2x = 18, or 8,
and X = 9, or 4 ;
but by subtraction, 2i/= 8, or 18,
and ?/= 4, or 9.
>ir" 18. Given ■ /. = 5, to find the values
of X.
Since the value of a fraction is not altered if the
numerator and denominator be multiplied by the same
quantity, {Wood's Alg, 89.) Multiply therefore by
and "E^^L^h.
Extracting the root, "~^"°~^ ' = ± -/^
and a- ^ ct — x^^ ±^/b ,x%
.*. by transposition, a + \/T * x = >^a^^^^^ ;
and squaring both sides, a" + 2a^/t. x+bx'^^a^-x^ ;
/. by transposition, {b'^l),x'= ± 2a^s/T. x;
and therefore j? = ± — ^~- .
l+b
19. Given 3^ — ^=y2_^/ ^^ ^^^^ ^^^ ^^1^^^^
o,.^ -.2 . ^ ^ C of « and V.
and ^ + 0? = 4 3 .
Reducing the first equation, ^'^V^-^'"^^ =3/ • («/ - J
52 Solution of Pure Quadratics, tsfc,
3x
and therefore (18. Cor. 2.) — =y,
/ and 3x—y^,
Substituting this value in the second equation, 4x = 4 ;
.'. j?=l,
and y* = 3a?=3;
20. Given x^^ysf^^ 9? ^o find the values
and y^ +X\/ xy=.\Sy
of X and y.
The first equation is x^ y. (a?^+y^) = 9,
and the second, y^ x {x^ 4-y^) = 18.
Dividing the second by the first, 21 = 2 ;
x'^
J- - -L
.'. y^ =2x»,
and y =4x.
Substituting this value in the first equation,
oc" -\-Ax^~Aj?=2,
or 9a?* =9;
consequently, 2/ = 4x = ±4.
^'+2y" ^ -^
21. Given 7=r- = x-\-2xy f ^ ^ , ,,
^j/ \ , to find the
and x—ly"*^ =256-x^y J
values of x and y.
From the first equation, x'^ + 2y^ = x^y + 2xy^ ;
.'. by transposition, x^ — .r^yy = 2j?y^ — 2^%
or ^,(a:-v^) = 2?r.(^-\/y);
.-. ^=:2y^,
I
ivithout completing the Square, 53
and x^^=2y''',
adding these equals to the second equation,
and therefore .r= ± l6 ;
.-. 22/^= ±16,
and ^ = ± 8 ;
.-. ^t=±*2,
and 3/ = 4.
t2. Given x'y-^-oo^y^SO'^
^1.1 5 >' *^
and - + - = - \ X
X y o J
find the values of
and y.
From the second equation, x+y=. -^ ;
but from the first, xy
30
xi-y'
Substituting this value in the other equation,
5 30 25
^^"6 x+y x^y'
.*. x-\-y\ =25;
and extracting the root, x+y= ± 5.
Let J? +3/ =4-5; then ^3^=4-6;
whence x"-\-2xy-\-y'^=i2b,
and Axy = 24 ;
'. by subtraction, a?*— 2a?3/+^^ = l,
and 4? — «/= ± 1 ;
but a?+3^= 5 ;
/. by addition, 2x= 6, or 4; and j? = 3, or 2;
by subtraction, 2^= 4, or 6; and ^ = 2, or 3.
,54 Solution of Pure Qjuadratics, is'c.
But if 00+7/= —5, then ori/= -6;
whence x^-{-2x7/-^i/''= 25,
and 4X7/ = — 24 ;
.'. by subtraction, .r'- 2J7?/-f ?/*= 4^
and extracting the root, j? — 1/= +7,-
but X 4-y = - 6 ;
.-. by addition, 2a: = 2, or -12; and x=l, or -6;
by subtraction^ 2?/ = - 1 2, or 2 ; and y = — 6, or 1 .
i. 23. Given o;^ v + a:y= 6) ,> , , ,
-^ 1 , o o , r ^ to find the values of
and xY+xY = l^}
X and y.
Dividing the second equation by the first, xt/ = 2.
But X^7/-\-X^^=X+7/ ,xi/=z6 ;
^ and a? +^ = 3 ;
whence x^ + 2xy+y^ = Q -,
but 4^:3/ r= 8 ;
.'. by subtraction, x^ — 2x7/ ■^y'^ = 1 ;
and extracting the root, zr — ?/= + i •
also x-\-y=3 ;
.*. by addition, 2x=4^ or 2;
.'• a? = 2, or 1 5
and by subtraction, 2y = 2, or 4;
••, 3^=xl^ or 2.
24. Given ^-^3 : jr^' ,, 6, :ll, to find the
and xy=z320 y
values of a? and y.
without completing the Squarii 55
Since x^-^y^ : x^ -^ 3x^y ->r Sxy'^ -^ y^ :: 6l : l;
{Alg. 180.) Sx^y-Sxy"" : x-^y]^ :: 6o : I,
or 3xyxx-y : x—y\ :: 6o : 1 ;
.S (^/^. 184.) 960 : x-y] :: 60 : 1, dividing
the first and second terms by a? — y 5
also lo : x^y] :: 1 : 1, dividing the first and
third ternis by 60 ;
.'. x—y\ = lo,
and x-y= ±4.
But since ^' — 2x^+3/*= 1 6,
and 4xy =1280;
.*. by addition, x'^ + 1xy-\'y' = 1 296 ;
.*. extracting the root, x+y— ± 3G ;
but x — y= ± 4 ;
.'. by addition, 2x = ± 40, or + 32,
and X =+20, or ± 16;
but by subtraction, 2y =±32, or ±40,
and y = ± 16, or ±20.
25. Given (x' +y') x (x+y) = 2336} , . . ,,
values of x and
y-
From the first equation, x^-\-xy+xy^'\-y^=s2336';
and from the second, oc^ — x'^y — xy'^ +y^ = 576 ;
.•. by subtraction, 2x'^y + 2xy^ =1760;
adding this to the first equation,
X' + 3x'y + 3:r7/* -i-y' = 4096 ;
.*. extracting the cube root, x+y^lG,
56 Solution of Pure Quadratics, ksfe.
and 2j7«/.^ + y = 176o,
or 16 X 2xy=: 1760 ;
.*. XT/ = 65,
Now x'+2a?2/+3/' = *256,
and 4x1/ = 220 ;
.-. by subtraction, x^^2xi/ + i/'^= 36,
and therefore x—y= + 6 ;
but x + 7/=: 16;
,', by addition, 2a? =22, or 10,
and ^r = 1 1 5 or 5 ;
but by subtraction, 2i/= 10, or 22 ;
/. ^= 5, or 11.
26, Given x^ +y^=x-\-y .xyl . ^ j ., ,
^ o ^ ^ ^> , to find the values
and xy-^xy =Axy 3
of X and y.
Dividing the second equation by xy^ ^+!/ = 4 ;
/. x'-{-3x"y'\'3xy^+y'=:64.
But from the first equation, x^— x'y— xy'^-\-y^= 0;
.-. by subtraction, 4x^y+Axy'^ = Q4i
.*. x-\'y .xyzz 16,
and xy = 4.
But x'+2xy-\'y'=\6,
and 4xy =l6;
.-. by subtraction, x^ - 2xy + 2/^ = O ;
and extracting the root, j?— 3^ = ;
but X'\-y=z4'^
by addition, 2a? = 4 ;
without completing the Square, 57
and x=2;
but by subtraction, 2i/ = 4 ;
.-. i/ = 2.
to find
1 5X2/
27. Given (a?*+/) x (x+t/) = -j^
and (^ -y) . (^' + «/*) = ^^
the. values of x and y.
Dividing the second by the first,
(^^+/).(x-y)=^.
Again, dividing the first equation by this last,
.•- x+y = 3x-3y\
.*. by transposition, 4^ = 2a?,
and 2y = a?;
whence (ar* +3^*) ,(x-y)=5y*xy= ^ *
2
or ^ = V.
and therefore a? = 2y = 2.
28. Given (j?*-i^')x(^-3/) =3^^ ? ^^ ^^^
and (a:* - y^) X (a?* - «/*) = 45a?y y
the values of a? and y.
Dividing the second equation by the first,
{x^'hy")x{x'{-y) = l5xi/,
or x^ + x^^y 4- XT/"- + y 3 _ 1 5^ y .
but from the first, x^ — x'^y — xy"" + 2/' =^ 3xy ;
. by addition, 2x^ +2y^=il8xy,
and x^ +y^ = 9.^3/ ;
I
SS Solution of Pure Quadratict^ Is^c,
but by subtraction, 2xy + 2xy^ = 1 2xy ;
.•. dividing by 2x^, x -\-y = 6 ;
whence ^'4-3j?'i/+3j?y*4-i/' = 2l6;
but oc^ +y' = 9^y;
by subtraction, 3x^y + Sxy"" = 2 1 6 - Qxy^
or 3.a: + ?/.cri/=18a:y = 2l6 — 9^^;
.*. 27xy = 2l6,
and xy=^S.
Now x*4-2a^^H-y*=365
and 4j:y =32;
.*. by subtraction, a?* — 2x2/ + «/* = 4 ;
and extracting the root, x~-y— ±2;
but J? + 2/= 6;
.*. by addition, 2x = 8, or 4
.-, X = 4, or 2
and by subtraction, 2y = 4, or 8
/. y = 2, or 4,
59
Sect. IV
Solution of Adfected Quadratics, involving only one
unknown Quantity.
(28.) Let the terms be arranged on one side of
the equation, according to the dimensions of the unknown
quantity, beginning with the highest; and (17) the
known quantities be transposed to the other side ; then,
if the square of the unknown quantity has any coef-
ficient, either positive or negative, let all the terms
be divided by this coefficient (13). If the square of
half the coefficient of the second term be now added
(11) to both sides of the equation*, that side which in-
volves the unknown quantity will become a complete
square; and (19) extracting the square root on both
sides of the equation, a simple equation will be obtained,
from which the values of the unknown quantity may
be determined.
* This is called completing the square ; and that a complete square
is thus obtained may be easily proved.
Let x^±^2ax be the proposed quantity on one side, when the terms
are arranged according to the form prescribed above ; and suppose
r/^=. the quantity requisite to complete the square. Novv the square
of x±^d-=.x'^±^2dx-\-d'^, vehere it is evident that four times the
product of the extreme terms is equal to the square of the middle
term; and therefore, in order that a?^ + 2«^+j/* may be a square,
^x'^f' must be equal to ^a^x"^ -, therefore j/*=:tt*= the square of half
the coefiicient of the middle term.
60 Sgiution of Adfected Qttadratics^
It may be observed, that all equations may be solved
as quadratics, by completing the squares, in which there
are two terms involving the unknown quantity or any
function of it, and the index of one is double |:hat of
the other. Thus, x^ -{-px^z=q, o?^" —px^^rj, x^ + x"' =:a,
x^''-\-ax^ = o, x^+px+q] + x^ -\-px + q=:r, are of the
same form as quadratics, and the value of the unknown
quantity may be determined in the same manner.
Many equations also, in which more than one unknown
quantity are involved, may in a similar manner be
reduced to lower dimensions by completing the square,
as jr*-f-2/'] +7? . ^ -hy^ = r. Instances of this kind occur
in the following
Examples.
1. Given x*-f 4j:=140, to find the values of x.
Completing the square, x* + 4x + 4 = 144;
and extracting the root, ^+2 = ±12;
whence, by transposition, a: =10, or — 14.
Given x* - 6j? + 8 = 80, to find the values of x.
By transposition,- a?* ~ 6a? = 72 5
and completing the square, a^ — 6x-{-g = Sl ;
,\ extracting the root, x^3= ± 9 ;
/'^ whence a? = 1 2, or - 6.
3. Given j:*- 8a?+10 = 19, to find the values of ^.
By transposition, a;" — 8 j? ±= 9 ;
and completing the square, x^ — 8x-\-l6=i25;
.*. extracting the root, a? - 4 = + 5,
and J? = 9^ or - !•
4. Given ^* - a; - 40 = 170, to find the values of x.
{K 2.
involving only one unknown Quantity. 61
By transposition, T?^ — x = 2iO\
and completing the square, a^—x-\- -=210+ 7="-^>
1 29
/. extracting the root, x- - = ± — ,
and J? = 15, or — 14.
1^/ 5. Given 3x*+2.r — 9 = 76, to find the values of a?.
By transposition and division, x^ + -j?= — ;
I .• .1 , 2 1 85 , 1 256
and completing the square, oc -{- ■- x-\- -=—+- = — — ;
1 16
.*. extracting the root, ^+ - = ± -r- ;
whence j: = 5, or — .
6. Given 3j?* — 9^? - 4 = 80, to find the values of .r.
By transposition and division, x^ — 3x= — =28 ;
9 Q 121
. completing the square, x" - 3.- 8-j? 2a?-ll 07-2 ^ n J 4.u^
V/ 10. Given = — ^— , to find the
values of 0?.
by transposition, 26 — 4^ =
involving only one unknown Quantity. 6$
1 2A? — 66
Multiplying by 6, 24-307 — = :i? — 3 ;
12.r~ 66
x-3 '
6j? — 33
and (18. Cor. 2.) 13-2x= ^;
.\ 19JC-39-2ar» = 6i?-33;
/. changing signs, and transposing,
2^'-13j?=-6,
2
and completing the square,
13 11
.\ extracting the root, x -• = ± — ;
.'. a: = o, or -.
. '^ ,, i->- . Sx — S ^ 3x-6 ^ n 1 ^u
I 11. Given 5x =2x+ — - — , to find the '
values of X. ,
(18. Cor. 1.) 10i?*-36:i? + 6 = 4a?"-12j? + 3a??-15a? + 18;
.'. by transposition, 3a?* — 9^=12,
and a?*-3j: = 4;
9 9 25
•. completing the square, a?* — 3^?+- =4+ - = — ;
T 4 4 4
3 5
and extractmg the root, x— - = ± -,
and j? = 4, or —1.
^12. Given 1^ - i^^:^;^ =? 3, to find the values
X 4x
of X,
64 Solution of Adjected Quadratics y'
(18. Cor. 1.) 64x-100+9x=12^*;
whence, .by transposition, 12x* - 73x= — 100,
and x"" X = -- ;
12 12
and completing the square,
. » 73 ,73]* _ 5329 _ 100 _ 629
^ " 12*'^"^24l " 576 12 ""576'
73 23
.'. extracting the root, x- "~2 ~ — "ol *
25
and J? = 4, or — .
V
13. Given 3x ^ = 29, to find the values
X
Here 3x* - 1 69 + 3a; = 29j: ;
;•. by transposition, 3j7'- 26a: = l69,
. , 26 169
and x^ - —X = -^ ;
therefore, completing the square,
13 2
and extracting the root, x- — = ± - 13 ;
13
.'. J? =13, or — -- .
2x^ 4x 3
14. Given 16 = -— 4- 7 r? to find the values
3 5 '5
of
X.
3 Sx 5 7
Multiplying every term by '-, 24 -a?*= — -f — :
2 9
.-. (17. Cor. 1.) and by transposition,
, 6x 57 63
involving only one unknown Quantity. 65
and completing the square,
^^+6£ 9.^63 _9^^ 324.
5 ^ 25 5 25 25 '
Q 18-
.*. extracting the root, x + ^ = + —
5 5
and x = 3/'or — — . \
15.- Given 12. - 1±^ ^ ^; to find the values ^
of X,
22 x^
(18. Cor. 1.) \0x-l4-^2x =
.-. (17. Cor. 1.) ~^-12a^=-14/
*j '
and X ^ = ;
11 11'
.•. completing the square,
11 11' 121 11 /ir
and extracting: the root, x ^ = H ;
6 ' 11 - 11'
21
.-. a? = 3. or — .
' 11
3 J? "^ 4 1' — 2
16. Given + 1 =10 , to find the
X — A 2
values of x.
Clearing the equation of fractions,
6a?-8-f-2j?-8 = 20j:-80-a?* + 6a?-8;
.-. by transposition, , a?*— 18j7= - 72 ;
and completing the square, x*— 1 8a: + 81=81-72 =9;
whence, extracting the root, x -9 = ± 3 ;
and therefore j?=12, or 6.
K
C6 Sclution of Adjected Quadratics^
/ 17. Given—- i — - = —Zl -1, to find the
values of x.
63 — Qj;
Multiplying by 9, 3a?-hl2 |-=4^4.7-9j
u ^ •-.• , . 63 -9a?
,•. by transposition, 14 — x= ^-,
and 17j?-x* — 42 = 63-9^7;
.-.by transposition, and changing signs, ^ — 260?= - 105;
completing the square, x^ — 26x + 169 = 169 - 105 = 64 ;
.'. exti'acting the root, a?— 13 = ± 8 ;
.'. a; = 21, or 5.
^- « 3a - 10 6a?* -40 ^ ^ , ,i
18. Given 3.T =2+ — —, to find the
9 — 2a? 2a?— 1 '
vatlies of a?.
Multiplying by 2a;— 1,
n a o 6a?*- 23a?+ 10 ^ ,
6a?'-3JC ^ — =4a?-2+6a?*-40,
^ . 6a^^*-23a:4-10
or 7^H — ::: =42 ;
' 9- 2a? '
.-. 63a?- 14a?" + 6a?*-23a?+10=:378-84a?;
by transposition, 1 2Ax - 8a?* = 368,
and x^ a? = - 46 ;
.*. completing the square.
^, 31 ^ . 961 961 ,a
^ T^+ 16 " 16 ^^ =
225
16 •
. .. .1 . 31
15
extractinsj the root, a?— — -- = ± ^
» 4 4
23
and therefore .r= — , or 4.
2
involving only one unknown Quantity. 6*7
'9- ^^'^ j^ + err. = tS ' ^-^ «"^ ^'^^
values of x.
35+70? 55^-HlJ?^
Multiplying by b+x, x+ ^_^^ = TT^TT"'
and multiplying by 11 j? — 8,
b- 4x
or 329x4-77x'-280 ^gg^^
o - 4j?
J- -J- u \ 47x4-lla?'~40 ^^ ^ .
or, dividing by 7, g^^^^ =9;*?'
.-. 47a? + llo?*-40 = 54a?-36j?';
by transposition, 47^^^ — 7d? = 40 ;
a 7 40
•••^-T7"=77^
and completing the square,
»_ Jl J. Z]' - -i?- -4- — = I^
^ 47 "^ 94I ■" 8336 "^ 47 8336'
7 87
and extracting the root, x 7 ~ — q2*
40
20. Given — ^ = -^5 to find the values
X x+2 a?+l
of
X.
Dividing every term by 9, ^ - j:^ = ^:j-j-;
.-. (18. Cor. 1.) 10x' + 30a:+20 — 3j?^-3a7=r0^* + 2O^
.-. by transposition and (17. Cor. 1.) 3a?*-7.r = 20,
and x^- ^ o? = — ;
3 o
68 Solution of Adfected Quadratics,
, . , 7 49 20 49 289
completing the square, '^' - 3 -^ + ^ = 3- + ^ = -^ »
,and extracting the root, x- U = ± -^ ;
.-. x = 4, or - -.
21. Given --i + -r-^^ — = -§- , to find the
values of x.
1 19
Multiplying every term by x, j— ^ + ^q-j = g ;
.-. (18. Cor. 1.) 8j[:+32 + 8j;-24 = 9^' + 9a:- 108;
.*. by transposition and (17. Cor. 1.) Qx^^7x= 11 6,
and X* - " X = ;
9 9
completing the square,
.__7 49 ^ 116 49 ^ 4225
^^ 9 ^ 324 9 324 324 '
7 65
.-. extracting the root, x- 75- = ± 7^ ;
lo 1 o
29
.-. x = 4, or .
22. Given — - — 7; =x — 3, to find the values
of X.
(18. Cor. l.j J?'- 10a?'4-l =^^-9^*4-270?- 27;
/. by transposition and (17-' Cor. 1.) x* + 27a? = 28;
and completing the square,
2^7 2Q
.-. extracting the root, x-\ — - = ± -^ .
and x = l, or -28.
involving only one unknown Quantity. 69
23. Given h =2:^, to find the values
T—x X 10^
of X,
Clearing the equation of fractions,
20x"- 140x4-490 = 2030? -29^^
.*. by transposition, 490?* — 343a? = —490,
and o:* — 7j?= - lOj
40 49 Q
.-. completing the square, x^-Tx-{- ~ = — — 10= - ;
7 3 '
and extracting thcroot, x- - = ± - ;
.'. o? = 5, or 2.
_,. 7-120? . X Sx+llO , ^
24. Given -, — = —zz — = -;=r— , to find the
oc^ sjoc ^x'
values ©f 0?.
Clearing the equation of fractions,
7- 12o? = a?'-8o?- llOi
/. 117 = o?*+4o?;
and completing the square, 1 2 1 = 0?^ 4- 4o? + 4 ;
.*. extracting the root, + 1 1 =0? 4- 2 ;
and therefore a? = 9, or - 13.
25. Given ^x+b x Ayo?+12 = 12, to find the
values of x.
Squaring both sides, o?+5 .o?H-12 = 144,
or 0?'+ 170? 4-60 =144;
. • . by transposition, o?^ + 1 70? == 84 ;
and. completing the square,
TtI 28Q 625
70 Solution of Adfected Quadratics^
I 17 ' 25
extracting the root, x-\ — ^ =.+ — .
2 2
and J? = 4, or —21.
26. Given ^a^^a^ = x—h, to find the values
of X.
Cubing each side, x^ - a? —x^ - Shx^ -{• 3b^x — h^ ;
. • . by transposition, 3bx^-3¥x = a^- b^,
a? -b^
and x^-bx= — ^I""'
.\ completing the square,
. ^ b . / Aa?> - b^
extracting the root, «^ - 5 = ± V — , ^ # - ?
and^=5±\/-Y5^.
^. sJax-^2 A — sJ X ^ , ,
27. Given 7=r = 7=- , to find the values
4+V^ V^
of J^.
Clearing the equation of fractions, 2x 4- 2^1c^ 16 — a? ; '
/. by transposition, 3a?+2>/^=l6,
2 ,^ 16
and x-{- -^/x= -^ ;
completing the square, x+ XaJx +-=-:^+- = -;-;
^ 9~ 3
and extracting the root, j^JIc-^ - = ± r ;
.-. >,Jlc = 2^
8
or ^3;
64
and x = 4, or — .
9
involving only one unknown Quantity, 71
28. Given <^/^ — 2x/a?- j? = o^ to find the values
of X.
Dividing by .yj^ vre find x- 2 - ^^^"=0 ;
.*. by transposition, a; — /y/^=2;
and completing the square, x- ^s/^+ 4 ^^"^ 4 ~ 2*
>— 1 3
extracting the root, ^a?— - = ± - ;
.'. js/x = 2, or — 1,
and x = 4, or 1.
29. Given ^^^ + sj 3c^ = 6s/x, to find the values
of X,
Dividing by ;y/x^ a?*-t-a? = 6',
, . 1 « 1 ^ 1 25
/• completing the square, x'-\-x-^ - =b+' = — ;
and extracting the root, x-\- - = + - ,
2 « 2
and ^ = 2, or -3.
X \ j^f X '
Given - = 22^ -f —-- , to find the values \y
30.
of X,
2 — 133 .
Multiplying by 2, and transposing, x— -z^ x^ — ^;'
and completing the square,
2 ._ 1 133 ,1 400
,-> 1 20
/. extracting the root, >/j?- r — ± 3 >
72 Solution of Adjected Quadratics^
1 / — ^y
and ^/x = 7, or - — ;
19
361
.*. a^ = 4Q, or
9
5 1
31. Given =0. to find the values
x-5 20 '
of ^.
Clearing the equation of fractions,
12/s/F— 40 — a? + 5=0;
.*. (17. Cor. 1.) x-l2^=-35y
and completing the square, x- l2^x + 36 = 36- 35 = 1 •,
.*. extracting the root, /v/^-*6= ± 1 ;
.*. ^x = 7, or 5,
and ^ = 49, or 25.
32. Given x^ +x^=:'i56, to find the values of x.
Completing the square, x^' -\-x^+ - =756+ -=— -- ;
4 44'
and extracting the root, x^ + - = ± — s
.-. x^ = 27, or -28,
and a^ = 3, or ^ — 28;
••. 07 = 243, or ^^8^.
33. Given j?3_^4^5g^ to fin^j the values of x. .
Completmg the square, x'-x^-^- - =56-i- - = — r-;
.'. extracting the root, x^ = ± -— , 1
2-2
and J7^ = 8, or -7;
\
involvitig only one unknown Quantity, 73
.*. x^ = 2, or 1/ ~ 7,
and x = A, ov ~T ^.
/'
34. Given 3^:^ + ^^ = 3104, to find the values of jc.
Dividing by 3, o?^ + - . .r^ = ;
and completing the square,
si .1 _ 3104 1 _ 3/249
^"""^3^""^ 36- " 3 "^ 3B " 36 '
/. extracting the root, x^-\- p^=z ± --7- ;
5 97
whence jj«^ = 32, or - — ,
3
and j?^=2, or ^±} ;
.*. x=:64, or — ^
3
35. Given ax^ + ^j?^ = c, to find the values of x,
3 b s c
Dividinjj by a, x'^+ -,x^= -;
and completing the square,
&" + 4ac
4a^ '
/. extracting the root, x^ + — —
Jia
Jh^ + 4ac
2a
9
, 3 ±x/&^ + 4ac
and jr^ = — —
2a
,^hen^^ r- ±V^^ + 4ac.
-&
•T
2«
.74 Solution of Adfected Quadraticsy
36. Given 3^^ — = - 592, to find the valuQS
2
of X,
8
(17. Cor. 1.) ~ -3*- = 592;
J *u r 8 6 . 1184
and therefore x'— -j?^= .
5 5
Completing the square,
5 25 5 25 25
3 77
/. extracting the root, x^ — - = ± -7-?
D
.-. x^= 16, or --,
5
74
and a: = 8, or
5
37. Given af^2ax^ = b, to find the values of x.
Completing the square, x** — 2a ^^ + «* = a* + ^ ;
.-. extracting the root, a;'^ -a= ±^a' + 0,
n A'
and x^ =^a±s/ a- + b\
^2
21
318. Given ^2x + l + 2^*' = . , to find
^ x/2a^-i-l
the values of x.
(18. Cor. 1.) 2x4-1 +2>y/2^*+J?=21 ;
. • . by transposition, 2>/ 2x* + x = 20 — 2x ,*
and therefore .>/ 2:i?* -j- a = 1 - a? ;
involving only one unknown Quantity. 75
,\ squaring both sides, 2j?' + a?=100- 20.r+.X'* ;
and transposing, a?* -f 2 1 J? = 1 00 ;
completing the square,
21
cr*4-21^+ —
2i
441 841
= 100 + ~— = -— -
4 4
21 29
.*. extractuig the root, a?H = ± — ;
.-. x = 4, or — 25.
39. Given
/ ^ /-; 2yFT6olM^9^T540 + 89
V«^ + oO + Va? +9= ,
>v/.r + 6o+V^^ + 9
to find the values of x.
Clearing the equation of fractions,
^ 4- 60 + X' + 9 + 2>>/cr3 ^- 60:r^ + 9.r + 540 =
2^x' + 60a?^-h9i^+ 540 + 89 ;
.*. by transposition, a7^+a^=r20;
1 181
and completing the square, x'-i-x^ - =20 4- - = — >
1 Q
.'. cxtractmg the root, x -i- - = ± - ,
and j?=4, or — 5. ^
40. Given
123+41>y/'J 20>yr4-4a? 20?"
b^-x 3-sJx {bs/x~x).{:^'-^x)
to find the values of x.
This equation is
4 1 . (3 + ^) 4 . (5 V^4- x) 2X''
b^yV^x S-sJx {bjlc- x),{3'-ssj'x)'
and therefore clearing it of fractions,
41 (9 - ^) = 4 . (25cr - x") - 2x\
7t> Solution of Adjected Quadratics^
or 36'9-4Lr--:^100x- 4a?*-2.i*;
.'. by transposition, 6j?^— 141x~ - 369,
and X' a; = —
2 2 '
completing the square,
^ ^K^ _ 2209 _ lf3 _ ^1225,
I . 47 35
.-. extracting the root, x = + — ,
and X— — , or 3.
41. Given a? + 5 = /y/ j? -|- 5 + 6, to find the values
of X.
By transposition, a; -f 6 — v^o? -|- 5 = 6 ;
and therefore completing the square, ^
, -_ 1 5
extractmg the root, ^ x-\-b- 2~ - o'
.-. V J? 4-5 =3, or -2;
and squaring both sides, x + 5=9, or 4 ;
whence j; = 4, or — 1 .
42. Given x + l6 - 7^/a? + l6* = 10 - 4>v/^TT6,
to find the values of x.
By transposition, 2?+ \6- 3>y/a;+ l6= 10 ;
and completing the square,
?Tl6-3V^FrT6 + ? =10+ ^ = ^;
. _ 4 4 4
3 7
.*. extracting the root, ^/^-flo- - = ± - ,
involving only one unknown Qtiantity, 711
and ^a-f-l6' = 5, or — 2 ;
whence .t+16=i25j or 4;
and v/2x*-f3a?4-9 = 6;
and completing the square,
2x*+3cr + 9-5./2FT3Ff94- —=6+— = -f;
4 4 4
^ . 5 7
/. extracting the root, \/ 2x'' -\-3x + g = ± - ;
aud >v/2j?*4-3a? + 9 = 6, or - 1 ;
suppose the value to he 6,
then 2a:^+3^-i-9=36,
^8.3 27
and ^'+ -.0?= Y»
/. completing the square,
^+ s^'^'ie 2 ^ 16 ~ 16"
and extracting the root, ^+ - = + — ,
9
or a? = 3, or ^ - .
2
But if - 1 be taken, A^'ooF^fSxTd = - 1 i
.-. 2a?'+3a? + 9=l,
3
and J?' H — .a: = — 4 ;
completing the square,
a 3 9 9 -55
and extracting the root, a?+ - = ^=^-^- ;
J -3 + x/-55
and X = =^-^^^ ,
4
involving only one unknown Quantity, 79
46. Given VZ±£±6^ ^'^ Jl ,
^'^ V^* 4-^ + 6
to find the values of oc.
(18. Cor. 1.) ,r* + ^ + 6 = 54-4V^^^ + a; + 6 + 6;
.-. by transposition, a^H^ + 6 + 4v^a?*+v/~43
51. Given x^ + 2x .x+4 = 2^x-j-4j to find the
values o( X,
involving only one unkmnvn Quantity, 83
By subtraction, x~ -\-1x ,x-\-A— —x-\'^ \
.*. dividing by 07 + 2, a?.^ + 4=-l;
.-. X^-^-Ax^ -1;
and completing the square^ x'^^ 4a: +4 = 4— 1=3;
extracting the root, a? + 2 = ft v 3 ;
.-. x^ -2±x/^-
33£i
52. Given ■ ^^ + d^^^UL = ^11, to find
sfbx^-x'' 25.r X
the values of x.
Multiplying every term by 'IbsJ bx^ -xf",
.-. 849 + 5 - ^^ = 850x/5 - x"^',
and by transposition, 5 - x' — 850>/5 -x^= — 849 ;
completing the square.
5 -ar-850V^5T?+ 425]*= 180625-849= 17977^;
and extracting the root, sj b — o?^ — 425 = ± 424 s
• whence ^ b — x^ — %AQ^ or 1,
^ and 5 -a?'' = 720801, or 1,
and x'^ -720796, or 4;
.-. X- ±\/- 720796, or ±2.
53. Given 9x + >/T^P^+'3Sr3'= 1 5a?' - 4, to find
the values of x.
By transposition, 9a? + 4 + 1x\J ^x + 4 = 1 5a;* ;
and completing the square,
^ 9a?+4 + 2a.\/9«^ + 4+^' = 1 6a:' ;
and extracting the root, ^^x + 4 +a? = ± 4a: ;
.*. -y/9a7 + 4 = 3a?, or —5a:,
and 9^+4 = 90?% or 250?*; supposing the former ;
^4 Solution of Adfected Quadratics^
then^ by transposition, 9^:* - 9x = 4 ;
111 Q 9 25
and completing the square^ 9^'*- 9«3^+- =4+ - = — i
4 4 4
3 5
.'. extracting the root, 3j?— - = ± - ;
and 3 J? = 4, or - 1 ;
whence j? = - , or — - .
3' 3
But if 9x + 4 = 25ar»,
then, by transposition, 2507* - 9a? = 4 ;
and completing the square,
"qI* 81 481
and extracting the root, 5 jr — -- = ± ^ ;
50
12 4- 8 J?^
54. Given x= -, to find the values of x,
X — b
(18. Cor. 1.) ^'-5^ = 12 + 8xt,
or 0?^ — 4J?=12^-8a?t4-a7;
and completing the square, ^*-4:i: + 4 = l64- 8x^ + x ;
extracting the root, a?-2:=± (4+a?^)^ and fir^t
taking the positive value ;
then, by transposition, a? — x»=6;
1 . 1 i 1 ^ 1 25
completing the square, x — x^-^- - —Ky-\-- = — ;
, X 1 5
extracting the root, a;^ — ;: = ± ~ »
4 4
2 - 2
involving only one unknown Quantity, 85
.-. ^t=3^ or -2,
and a? = 9, or 4.
But if the negative value be used, a?-2= — 4 — ^^;
.'. by transposition '.r + .x't=: —2;
11 -~ 7
and completing the square, x-\-x^--\- - — 2= — ^s
4 4 4
extracting the root, J7^+ - = -'^ ~ ;
-i + yzry
• • M/ — —————— ^—^ »
2 '
and .= zi±N^.
2
55. Given ^ + i^ -49 =9 + ~, to find the
4 x^ X
values of x.
Adding -^ to each side, in order to complete the
X
square,
then -^ 49+ -T=9.+ - + -J
4 0? X x^
extracting the root, - = ±(3H — ),
J^ X V X
and first taking the positive value ;
^ X 8
.*. by transposition, 3= -;
2i X
and therefore *" = — ;
7 7
completing the square,
X' ^4.-9._l£+9. ill.
~ 7 49 ~ 7 ^ 49 49'
8C Soluthn of AdfecUd Quadratics^
extracting the root^ a- - =^± — ,
and JT = 2, or - - .
7x 7 1
But if the negative value be used, — - - = -3— -;
^ ^ 2 X X
. . 7j? 6
.*. by transposition, h3= -,
2 X
and x^-\ = — ;
7 7
.*: completing the square,
7 49 7 49 49 _
and extracting the root, a: + - = ^^-^ — ;
-3 + >v/93
* 4i>^ OC^ 1 7<2?'
56. GivctT^ 1 17a; = 8, to find the values
of .r.
17.2?^
Multiplying by ;2, x^ -f- ~ 34j? = 16^
17^^
.*. by transposition, ^ -^ * - =34a; + 16 ;
and completing the square.
^ 17^?^ . 17^ 17^ o. ,^
2 4 ! ^ "*
\*7 X /\7 X \
extracting the root, x'^-\ = ± y~ — 1-4 ) ;
first let the positive value be taken ;
then, by transposition, ^* = 4,
and ^= + 2.
\
■ involving only one unknown Quantity. 87
But if the negative value be taken,
\7x
and by transposition, x* 4 — — = — 4 ;
2
completing the square,
2^4
389 _ 225
' TT IF'
and extracting the root, x+ -^ — ± - — ;
4 4
/. x= - 8, or .
' 2
t>^ r«- «.. a 841 17 232 1
57* Given 270?* + -L = . + 5,
' ' 30?^ ^3 3x 3x^ ^ '
to find the values of x.
Multiplying every term by 3,
/. by transposition, 81 a;* + 17 + -- = + - — + 15.
X X^ ' X
Adding unity to each side, in order to complete
the square ;
^ X^ X
and extracting the root, 9x+ - = + {~ +4^ .
X \ X '
Let the positive value be taken ;
28
then, by transposition, 90? — 4 = — ;
and therefore 9^*- 4:» = 28 ;
, ^- ,, 4 4 256
completuig the square, 9x*-4a: + - = - 4-28= —^\
88 Solution of Adfeded Quadratics^
2 16
extracting the root, 3a7— - = ± — ;
o o
.-. 3^ = 6, or -— ,
and a; = 2, or .
9
But if the negative value be taken, Qx" + 4a: = - 30 ;
and completing the square,
4 4 -266
2 ±x/-266
extracting the root, 3a? + - = —
-2 + V''^^^266
.-.30:=—-^^ ,
, - 2 + V - 266
and X— ^-^ .
9
' 58. Given x" 5 +«'- -,
X^\ X
values of x.
= — , to find the
a
By transposition, ^'- d "" ^ = """'" i^i '
and squaring both sides,
^ a* a?* 20^ "7 ?!'' , a*
X* a" a X* 0?*
2j?' „ ai^ . 0?*
and by transposition, a?* .a?' r + — - a' = o
or a?* x^-a^l + t-==0;
extracting the root
> 0?-^^-— =0;
involving only one unknown Quantity, 89
j^ a^ — a^
by t^^nsposition^ x = — ,
and ax^^J af^—'O^,
Squaring both sides, a V = j?'* — a* ;
.*. by transposition, ^-aV = a^;
and completing the square,
j:^- aV -f-— ^a'^^- — = — ;
4 4 4
extracting the root, x* - - = ^=^^^^- — '~
1±V^
2
and ^=±a\/ii^^
^ = -^^.«%
90
Sect. V.
Solution ofAdfected Quadratics^ involving two unknown
Quantities.
(29-) If the equations involve two unknown quan-
tities, they may, by the preceding rules, be reduced to
one containing only one of the unknown quanties^ the
values of which may b^ found by Art. 28. ; whence,
by substitution, the values of the other may also be
determined. In many cases, however, it may be con-
venient to solve the equations first, considering one of
the quantities as known ; when the rules for extermi-
nating unknown quantities (23) may be more easily
applied.
Examples.
1. Given x—y=\b^
^ XV V 5 to find the values of x and y,
and -^ =2/^ V
From the second equation, x = 2y^ j
Substituting this in the first, 2y^^y=i 15 ;
^_ 1 _ 15
' ' if 2 ^ "" 2 '
and completing the square,
,_ 1 , JL _ 1^ . ± « 1^^
2-^^ 16' ■" 2 ' 16' "" l6 '
Solution of Adfected Quadratics ^ ^c. ^1
.'. extracting the root, ?/-- = +—,
and 3^ = 3, or — -;
25
whence x = Sz/* = 18, or —- .
2. Given ^^^"^^ =3 f , to find the values of x
^ t and y,
and 92/ -9^ =18^ "^
From the second equation, y — a?=2;
and therefore ^ = j? + 2 ;
but fronci the first, 10^+y = 3^^.
Substituting in this the value of y found above,
10jp + ^ + 2 = 3a?. (a:4-2),
or ll^ + 2 = 3a?V6^;
.*. by transposition, 3a?*-3j? = 2,
5 2
and (18) x" - -^= -;
/. completing the square,
''~3'''^36 ~ 36 "^ 3 "" 36"
5 7 *
/. extractmg the root, .r- ^ = + «,
and a? = 2, or - -;
u^hence w = ^+ 2 = 4, or - .
. 3
3. Given x-^y : a: — v :: 13 : 57 . ,
^ ^ >- , to find the
and i/*+.r = 25
values of x and y.
92 Solution of Adftcted Quadratics,
{Alg. 182.) 2x : 2y ::' 18 : 8
.-. {Alg. 184.) X : y :: 9 -4;
/. (21) 4a: = 9y, and Jp=s-^.
Substituting this value in the second equation,
^•+^=25;
completing the square, ^
^+4 +64 ""64+^^" 64 '
9 41
/. extracting the root, 3/ + f = ± -^ 5
8 o
25
whence ^ = 4, or — -- ;
9y 235
.■.x=-f=2, or --^.
4. Given 4j^2/ = 96 — cT'v*) « , ,
-^ "^ ^ ^- find the values of
|, to
and x-{-y^6
X and y.
From the first equation, x^ip" + 4xy = 96 ;
/. completing the square, a?*!/* + 4xy + 4 ~ 100 ;
and extracting the root, ^1/ + 2 = ± 10 ;
.-. xy = S, or — 12.
Now, squaring the second equation,
a;* -h 2:^3/ + 2/* = 36;
but 4xy =32, or -48;
.*. by subtraction, x"- — 2xy-^y^ = 4, or 84;
whence^ extracting the root, x—y= ±2, or ±\fS4\
but a7+y= 6 J
.•, by addition, 2a? =8, or 4, or 6±2/>/21 ;
invohing Pwo unknown Quantities. 93
whence a? = 4^ or 2, or 3±^/21;
and by subtraction, 23/ = 4, or 8, or 6 + 2>/21;
/. y = 2, or 4, or 3 + ^^21.
5. Given x^+^ + z,= ,8-y7 ^^ ^^^^j^^ ^^^^^^
and xy = o 3
of /'3;
and by subtraction, 23/ = 4, or 6, or —6 + 2^/3";
- . /. y = 2, or 3, or. -3+^/3.
6. Given j?*-f 2a?v+«^* + 2j?= 120- 2^7 ^ . ,
> , to find the
and a?^— 3/*=s8 )
values of x and v.
^yft*
94 Solution of Adfected Quadratics^
By transposition, x-^yl * + 2 . {x-^-y) = 1 20 ;
.-. completing the square, x-^y\^-\-2 . (^+y) + 1 = 121 ;
/, extracting the root, (^+3/) + l = ±11,
and X +3^= 10, or - 12; and first let Vr -f y = 10 ;
Q
from the second equation, a? - 3/ = - ;
.*. by subtraction, 2y = 10— -;
and by transposition, y^ -^ by= - 4 ;
25 25 9
.•. completing the square, y-5i/4-— = — -4=-
.• .1 5 3
and extractmg the root, y — - = + - ;
.-. y = 4, or 1 ;
and ^=10— 3/ =6, or 9.
But if x-\-y- -12,
and 07 ~ V = - 5
8
then 2v = — 12 ,
^ y
and ^' + 6y = — 4 ;
completing the square, y*+6t/ +9 = 9 — 4 = 5;
extracting the root, ^+3 = ± ,>J^\
.•.i/=^3 + V^,
and j?=: - 12-1^= -■9 + >s/5^
7. Given a?* + v*-^-y=78r .. , ,
' "^ ^ ' c X. fin^j the values
and cT^ +
of X and ^.
involving two unknown Quantities. 95
Since ^*+3/*~ (^+3/) = 7^ i
and from the second, ^xy + 2 . (jc +3/) = 78 ;
.'. by addition, ^^4-2^1/4-3/'^ +^+y = 156 ;
and completing the square,
, 1 25
extractnig the root, x-\-y-\- - = + — 5
•'• -3^+3/= 12, or - 13 ; supposing the former;
then a?2/ = 39~(a^+3/)=39--12 = 27,
and a?*4-y' = 78 + (^+3/) = 78 + 12 = 90i
but 2xy = 54;
.• . by subtraction, a?* — 2a? 1/ +3/* = 3 6 ;
and extracting the root, x — y =z ±Q -^
but x-\-y=: 12 ;
.*. by addition, 2j? = 18, or 6,
and J? = 9, or 3 ;
and by subtraction, 2y = 6, or 18,
and 3/ = 3, or 9.
But if a?+3/= - 13,
then j?3/ = 39 + 13 = 52,
and a?'' +3/' = 78 -13= 65 ;
but 2xy =104;
.'. by subtraction, x^-2xy-^y^= —39 ;
and extracting the root, x-^y— ± V - 39 ;
but x-\-y= - 13 ;
.-. by addition, 2x = — 13 ± >/ —39,
, -13 + ^/t-39
and X =- -^
96 Solution of Adfect$d QuadraticSy
but by subtraction, 2y=z — 13 + ^/ - Sg ;
8. Given x^y^-Txy" - 943 = 765)
and xy — y:=^\2 )
values of i? and y.
to find the
From the first equation, by transposition,
x^y^—Jxy^'^lflO;
and completing the square,
...,4 ^^... .49 ^ , 49 6889
xy*- 7^3/*+ -^ =1710 H — i- =
4 ' ■ 4 4
7 83
extracting the root, xy* - - = + — ;
/. j73/* = 45, or -38.
Multiplying the second equation by y, xy*—y^=il2y.
Substituting in this the value of xy^ found above,
45 - 3/* = 1 2y, in one case ;
and by transposition, i/*+12y = 45 ;
completing the square, ^*+l 23/ +36 = 45+36 = 81 ;
extracting the root, y +6= ±g,
and 2/ = 3, or — 15;
whence j? = -r- = 5, or - ;
^ 5
and in the other case, — 3 8 - ^* = 1 2^ ;
whence y^+l2y= -38 ;
completing the square, y^ + l 23^+36 = 36-38=— 2;
extracting the root, y + 6 = ± x/-~2;
_ /. 3^=-6±V^
-38 -38 -19
and x =
y^ '6A^\2j ■-'2. 17T6>v/-2*
involving two unknown Quantities. 9?
9. Given x-^-JsjT^ +y - s/lc + \/y^o\ ^^ ^^^
and s/ x+s/y'^^ 3 ^
the values of x and y.
Completing ^he square in the first equation,
^ . 1 1
and extracting the root, ^Jx — s/y - - = ± ^ »
.'. s/^-s/y=l, or 0;
but from the second equation, >>/^4- ^y = 5 ;
.•. by addition, 2vjc = 6, or 5^
and >>/j7 = 3, or - ;
25
.-. 0^ = 9, or ;
4
but by subtraction, l^fy^^ 4, or 5,
or >\/2/ = 2, or - ;
V7=2,
25
and t^ = 4, or —
10. Given — + -f = -_./, to find the values of
V V Q V
, "^ ^ i X ^xA y,
and x-~y=z2 )
Completing the square in the first equation,
x" 4X . 85 121
X 11
and extracting the root, - + 2 = ± — ;
X 5 17
9S Scltition of Adfected Quadratics^
and a:= — , or — r^ \ supposing the former;
by
then, from the second equation, -— — 1^=2,
or 21/ = 6,
and y=:3 ;
and if the second value be taken, — -^ 3/ = 2
or — 20y = 6 ;
-3
and .= ^1^=1^.
11. Given \/lI + \/S=2)
^+^ 3a;. >, to find the
and xy — ^x-^y) =54 3
values of a? and y.
(18. Cor. 1.) and by transposition,
3a?- 2^'3x,A^x+y-^ {^+y) =0;
.-. extracting the root, >/3j?— >y/^4^=0 ;
by transposition, ^Sx^a^x +y,
and squaring both sides, 3x = x-\- y,
and .*. 2x=y;
substituting this value in the second equation, 2a*— 3j7 = 54,
or JT^ .x = 27:
2 ' '
whence completing the square,
8 3 9 _ 9 _ 441
""" 2^^ 16"^'"^16-"T6 '
involving two unknown Quantities. $B
3 21
extracting the root, x = + —^
and x = 6y or ;
whence 3/ = 2j:=12, or -9.
12. Given\r^-2x'v+V* = 49, 7
> . to find
and x' - 2x^y^'\-y'' - x* -i-y* = 20)
the values of x and y.
Completing the square in the second equation,
81
I Q
extracting the root, x*-^^ — ~ - o»
/. a:*— y = 5, or -4;
but extracting the root)
of the first equation, y ^ '"—.'*
.-. by subtraction, ^^—3/ =±2, or - 12, or 11, or — 3,
Taking the first value, and completing the square,
1 1 9
1 13
extractmg the root, 3/ — - = + - ,
and ^/ = 2, or - 1 ;
/. x= ±sJl-\-y= ±3, or ±>y67
Taking the second value, y* ~ y = — 1 2 ;
11 -47
completing the square, 3/^—^+2 ~;i~^^^ "It — '
extracting the root, y - - = — ^^-^ ?
190 Solution of Adftctcd Quadratics,
and y= ^ ;
whence 'x=. ± */y-7= ± V 1±>/-Z:1I - 7
_v ^ ,
Taking the third value, y* — y= 1 T;
1 1 45
completing the square, y*-y4- - =11+- = — ;
4 '4 4
extracting the root, y— - = iiL •
1 ± 3 J 5
'••y= ;; 5
and
= ±V7-+^=±V 7+ — -^^ = ±V — =^^^!^ —
Taking the fourth value, ^ - ^ = — 3 ;
completing the square, y' — «/+ - = - -3= — — ;
4 4 4
1 iV-ll
extractmg the root, y - - = — ^^^-- — -,
, l±x/-ll
and v= ^^:;^
^ 0±x/^nT ~_^^/ -i3±V:rn
13. Given j?3/4-J?y = 12,*) , to find the values of
and X +a?i/^ = L8 3 ^ and y,
12
From the first equation, a?= — r--- — r ,
^ 3/ -(1+3/)
involving two unknown Quantities, 101
18
and from the second, x = -— ; — r ;
1+3/3
whence — r-r-- — - = --; — r, and dividing by 7— -
y y-y + l
/. 2^*- 23/ + 2 = 33/,
and by transposition, 23/'- 63/= - 2,
5
or 3/'- -.3^= -Is
5 25 25 Q
completing the square, y*'~o*^"^Tfi ~ Tfi"^^"'TS
• .1- . ^ ^^
extractmg the root, y — ^ ~ — I '
1
.-. 3/ = 2, or -;
hence x= — ; — j=2, or 16.
14. Given ^-^=3-^ , ^^ ^^^ ^^^^ ^^^^^^ ^^
4-0? =y-3/^3 '
and
X and 3/,
Adding the two equations together, 4 - a?^ = 3 -y^;
.-. by transposition, i/^—x^-l,
and ^ = 0::^— iP.
Substituting this value in the first equation,
/. a?-x^ = 3 — a? + 2o?^— 1 ;
.*. by transposition, 2^ — 3a?^ = 2,
3 *
and 0? ,x^=l ;
3 1 9 9 25
completing the square, *-5*^+7g = ^ + rg ^TS'
'02 Solution of Adjected, Quadratics^
.3^ 5
extracting the root, j?^ — - = ± -,
and J?'' = 2, or »
' 2
/. j? = 4, or -,
and y=^x^- l] =1^ or ?.
4
15. Given a?" + 1. y = xi/ + 126} ^^2,^1 1
^ ^ > , to find the values
and a?'-M.i/ = j;^y*-7443
of a? and y.
Since quantities w^hich are equal to the same, are
equal to each other,
x-y-744 = xy+ 126;
.'. by transposition, ^y — j:i/ = 870 ;
completing the squai-e, x^i/* — xy + - = 870 -|- - = ;
1 5Q
extracting the root, xy — = ± — ,
and xy=:30, or - 29 ; let the former value be taken,
30
then fi-om the first equation (j7* + 1). — = 156;
X
„ 156^ '26x
2d <2?
and by transposition, x^ — = - l ;
completing the square, x^ *x-\ — ~ = — ^— 1 = ;
13 12
extractmg the root, x — — = ± "t J
whence a? = 5, or -;
o
involving two unknown Quantities. 103
30 ^
and /. 2/ = -- = o, or 150.
29
In the second case, j?' + 1 x ^= - i?9 + 126 = 97,
X
andx^+l = -^;
by transposition, ^' + -r^ = — 1 ;
9409 _6045
3364 3364'
_29
completing the square, ^ + 9 "^ + rjS
97 . J604b
extracting the root, x-{- —^ ± — ;
. ^-.-97±n/^045.
58
__ 29 _ 1682
" "^"^ " ~ ~ 97 + >^/S045 *
16. Given x + y + ^V^=l2 7 ^^ ^^^ ^^^ ^^1^^^
anda;^ + t/3 =189 3'
of X and ;^.
Completing the square in the first equation,
, 1 1 49
/ 1 7
extractmg the root, V-^+^Z+o^ ±o»
.'. V:c+3/ = 3, or - 4,
and .r +3/ = 9, or 16;
.-. a3 + 3:c'3/+3a;y 4-^ = 729, or 4096;
/. by subtraction, 3a?*^H-3a^i^' = 540, or 3907 ;
3907
/. (x+3/). xy-\%0^ or — - — ;
101 Solution of Adfected Quadratics^
.'. ^xy^ 180 in one case, and l6j"y = — — in the other,
whence in the first case xy^20.
Nowx*+2^z/ + «/*=81 ;
but Axy = 80 ;
.'. by subtraction, x^ -'2xy'\-y*^\
and extracting the root, x—y= ± 1
but, x+y= 9
.•. by addition, 2x = lO, or 8,
and X =5, or 4;
but by subtraction, 2^ = 8, or 10;
and y = 4, or 5.
Now in the second case, l6xy = — —
3907
and since ar* + 2a? 2/+^* = 256,
A . 3907
and 4xy =="T2~
835
/. by^ subtraction, x"^- 2xy+y^= -"—-;
* and extracting the root, a:— y= ±\/ ■,
but x+y=l6;
,\ by addition, 2x = l6±\/ — ttt- *
1 . /-835
and X- 8±jV — 3—
but by subtraction, 23/ = 1 6 + \/ — -- — ,
3, = 8 + ^\/z
. _ _ - w - 835
and
involving two unknown Quantities. 105
17. Given ^'H-//' + x-3/= 132 1
^ and {j^ +f) . {X -3/) = 1 220 I ^ *^ ^^'^ ^^^ ^^^^^'
of X and y.
From the first equation 07*+^''= 132 - {x—y)-,
1220
and from the second j?*+v* = ;
u / X 1220
whence 132 - (x — y) = ;
and, .-. 132.(^-1^) -x- yX = 1220;
and (17. Cor. 1.) x^yY - 132. {x—y) = - 1220;
completing the square,
x-yf'-l32.(X'-y) +6H1' = 4356- 1220 = 313^;
extracting the root, x—y -66= ±56;
and a?~^ = 10, or 122, supposing the former,
x^-^y^ =122;
but a?* — 2^1/4- y*= 100;
.'. by subtraction, 2xy= 22,
and since x*'\- i/*=122;
.-.by addition, x^ + 2xy +3/' = 144,
and extracting ,the root, X'^y= ± 12 j
but x—y = 10;
.'. by addition, 2^ = 22, or - 2,
and a?= 11, or — 1 ;
by subtraction^ 2y = 2, or — 22,
and y^'l, or — 11.
But if X —2/ =122,
then ^'4-/ = 10,
and x^-2xy-\-y^ = T22\";
.*. by subtraction, 2xy = 10 - 122Y ;
but a?* + 2/*=10;
by addition, a;' + 2*3/+y = 20-Tl22l';
p
106*- Soiution of Adfrcted Quadratics^
and extracting the root, x -\-y = ± V 20 — 122)* ;
but X'-t/=l22;
'. by addition, 2j; = 122±2\/6 — Si]*)
and x= 6l± ^ -3716;
and by subtraction, 2y = — 1 22 ± 2 V — 3? 16;
.-. 3/=: -61 ±^-3716.
- .V -.Y I ^^ g^^ ^j^^ values of x and y .
and 8^^
From the first equation, x^ = 2y^^
and, /. -ar^=y^;
substituting this value in the second equation,
and iSx^'-x^sz^S;
or (17. Cor. 1.) 0?^- l6xi= - 28 ;
completing the square, x^-l 6x^+64 = 64 — 28 = 36 ;
and extracting the root, a?j — 8 = ± 6 ;
.-. a?i=14, or 2,
and x = T4\^ or 8;
but ^4 = 2'^ ^ = 98, or 2;
•'• y = 98l%or4.
19. Given 0.^+3^^ = 3^ 7 ^ ^^ ^^^ ^^^ ^^^^^ ^^
and a?» 4-^3 = j? ^
a? and y.
Squaring the second equation, .*. x + 2X'i/>'{'t/^=:3C*;
but x^ 4-^T=:3^^
.• . by subtraction, x - ar^ + 2^:^?/^ = a?' — 3« ;
but from the second equation, z/* =« — op^.
involving two unknown Quantities, iOT
Let this value be substituted in the preceding
equation^
then ^-o^^+Sj::^— 2a? = a?*-3a?;
and by transposition, 2a? =s a?* - a?^ ;
and dividing by a?, 2 = a? — a?^ ;
i 1 1 9
completing the square, a?--j?8+-=2H — = -;
1 3
extracting the root, a?^- - = ± - ,
2 2
and j?t=:3j or — 1 ;
.'. a? = 4, or 1,
and 3^4 = 0? — a?^ = 2;
.-. y = 8.
20. Given x^xr^^ ^^'^K'^^ +4, , ^ , .,
ar^ > , to find the
s
and y-^-xy =y^ + 4^
values of x and y.
From the first equation, a?^ + a? — 4a?^=2/*+y+2,
and from the second, x^y-\-3.
Substituting this value for x in the former,
ar^ +3^ -f 3 - Aoc^=y^ -{-y + 2,
and by transposition, x^ - 4x^ =y^ — 1 .
But since a? = i/4-3i /. a?- 4=t/— 1,
by which equation let the preceding one be divided ;
.-. xi=y + l ;
squaring both sides of this equation, x^y^ + 23^ + 1.
Equating therefore the two values of j?,
/ + 23/ + l=2/+3;
,'. by transposition, y+^ = 2;
119
completing the square, y^+y + T = ^'*"4~4»
108 Solution of Adfected Quadratics ^
1 3
extracting the root, y + - = ± - ;
.-. 3/=l, or - 2-,
whence ^7^=^+1=2, or —1,
and therefore a? = 4, or l .
21. Given ^V-^+? = 6^—^-)
y^ X y 4 X* V, ^ to find the values
and x-^y=z2 j
of X and y.
By transposi tion. — +-^4--+^=~;
'' ^ y^ x^ y X 4
.-. adding 2 to each side. "L ^. 2 + -^-H-*?^ f?= — ;
^ x^ y X A
completing the square, - -f - +r- + -^ + -^ — ;•
^ ^ V^^ x/ 4 4
•3? 1/ 1 6
extracting the root, -+'^ + - = ± -;
y X ^ 2
X . y 5 7
.-. x'+y' = ^, or - ^^;
now from the second equation squared,
x'-^y' = f2xy+4;
.-. 2^y + 4 = ^, or - ^;
whence by multiplication and transposition,
xy =r 8, or - Ji '
and since j?"^ — 2xy-^y^ = 4,
. « 32
and 4j;y =32, or - — ;
12
.% by addition, x' + 2 jC2/ +y* — 36, or -y ;
involving two unknown Quantities. 109
2 /s
and extracting the root, oo-{-^= ±6, or + y>^ ;
^^ 1 1
but x-i/= 2 ;
2>/^
.-.by addition, 2a? = 8, or —4, or 2±-^,r;>,
and ^ = 4, or — 2, or 1 + --^== ;
2 /s
.*. by subtraction, 2?/ = 4, or —8, or —2 + —7==;
/. t/ = 2, or -4, or -1± ^^.
22. Given 2a? -i- 2/ = 26 — 7^ 2x-^y 4- 4 -^
fi ^■^ + N^ --. 2^ i ^-^ ~ n/^ >> to find
the values of x and y.
Adding 4 to each side of the first equation,
2a?+«/4-4 =30- 7^/20? +3/ + 4 ; .
.". by transposition, 2j? + 2/ + 4 + 7^2x+y + 4 = 30 j
completing the square,
(2a?H-3/ + 4) +7V2a?+3/ + 4+^=r304- T = ^J
——————— *7 1 '^
extracting the root, ^2X'\-y-\-4-\- - = ± — ;
2 2
.-. ^2a?+3/ + 4:;=3, or -10,
and 2a?+y + 4=9, or 100;
.-. 2a? + 1/ =5, or 96.
Multiplying every term of the §econd equation by
16 2a? + x/^ . , .
—r • 7=- -r X J
2£v+x/.y 2x-^Jy
^x-^y* 2x-s/y
1^ ' ^x-^y
10 Solution of AdfecUd Quadratics^
,\ by transposition, ^^-^1 — tt . ~!~=1;
completing the square.
2j?+>/y r jg 2x-f-v9 64 64 ._ 289.
2x-^r ""^^ ' 2x-ljy ^ 225 ~ "*" 225 "" 225*
2j7H->s/y 8 17
extracting the root^ ^ - TT = ± TT >
^ 2x-Jy 15 - 15
2x-^AJ'y 5 3
2x-^y 3' 5
Let the former vaiue be taken, then
6j?+3^«/= \0x - b*J~yl
and by transposition, 8v^=4r,
and %^y = xi
but if the second value be taken, /£. = — -r ;
3a? - s/y
/. 10J?+5Ayy= -6a:+3/v/3(^
and 16 J? = - 2^y',
or 8 a: =' — /^^y^
Now 3i:+i/ = 5, or 96; supposing the former^
and taking the first value of ^x^A^J'y^
3^ + 4^7=^;
completing the square, y+4v'^^+4=:9;
extracting the root, ^y^+2 = ± 3 ;
••• \/^=l, or -5,
and 3/=l, or 25;
but M^2^'y:^2, or >-:.10.
involving two unknown Quantities, ill
Again, taking the value, 3a?= — - ^^
r
completing the square,
.•• extracting the root, ^ - - = ^v^^^
8 8
and^=i±^,
" -^ 32
and ^=-5^/3^ =
64
Now takingjhe equation 2^+3^ = 96, and the first
value 2j: = 4^y;
then 3/H-4v^^=96;
completing the square, 3^+4>>/^+4 = 100,
and /y^+2= ±10;
/. ^Jy;=S, or —12;
and .•. 2/ = 64, or 144;
whence ^ = 2x/t/ = l6, or —24.
Again, taking the value ^^--^sfv'^
then y- -x/p"=96;
completing the square.
„ ^ /- 1 1 «c ■ 1 6145
112 Solution of Adjected Qtiadrattcsy
extracting the root, ^/^-^ i = -^^^^^
8 8
and V^= -^=~ ;
^ 3073 + >v/6l45
^ 32
1 y- -1+^6145
23. Given x/^+x/^ • s/^—'^^ •' \/^+2 :
to find the values of x and ;y.
From the first equation,
A,/^ : -^/^ :: Ayx+3 : v^+ 1 ;
and from the second,
,\ by transposition, 3/ + s,/!/^ \/^!^ = ^^-^\/^9
but mJ y -\-sJ xy= a? + 3^/5^;
.-. by addition, y-{-'^sfy =4a^ + 4^/^;
completing the square, «/ + 2^y + 1 = 4a?4-4v^ j? + 1 ,
and extracting the root, \fy -V 1 = ± {2sJ x-\-\) j
and .*. \i the positive value be taken, sj y = 2sJ x ;
/. ^+3^/a? = 2>v/a7 + 2a:,
and x — ^x\
and ^ = 1 ;
involving tivo unknown Quantities, 113
and 3/ = 4.
But if the negative value be taken,
^4- 1 = -2^7-1;
by transposition, ^ y = — 2^Jx - 2 ;
and if this value be substituted in the first equation,
-.^JC-S : -3»/F— 2 :: >/r+2 : 1,
or ,^/T+2 : 3^^+ 2 :: y/^^2 : 1;
and since the first term in this proportion is equal to
the third, the second will be equal to the fourth ;
.-. 3^/7+2 = 1 ;
by transposition, 3.^/^= — 1 ;
and j?= - ;
9
whence a/^= - ^ • (\/7+ ^^ ~ "" 3 '
and y= — .
X^ 42 "\
24. Given v -f v ~ = — / ^ n a .\ i £•
^ ^ j: X f , to hnd the values of
J a?* X 54t ^ and y.
and -- + — 7=. = —V
3 2^2/ yJ
Completing the square in the first equation,
./7 .1 42 ^ 1 169
^ ^ X 4x X 4x 4x '
and extractmg the root, ^/y + —^ — ^^ ;
IH ^Mon of Adjected Quadratics,
r ^ -7
36 49
.-. ?/= — or -^.
Again, from the second equation,
2^ - y '
completing the square,
^^^ 2^,j "^ 163^ - // "^ 163/ 163/ '
3 51
and extracting the root, x+ :; — 7==^ ~ ± - — 7= :
13 - 27
— 7=» , or — 7= .
.But v'7=;;7^, or-^;
12 -27 ,~
whence, x= —7=, or — 7==2>^x,
or ^, or -^, or -^ ;
/— -12 -9 27
,% ^ x = 2, or --—, or -j-, or ^
144 81 729
.. ^ = 4, or— , or ^-^, or^,'
A *^^ 49 ^ 49 784 49x196
and 2/=- — , or -^=9, or -^, or V— ^ or '^^^ ■
^ 0? ' a? ^^ 4 ^ 81 ^ 729
25. Given -^ + — ^ =20^ ^^^-I—i ^ n . ^.
y s/y 3^ > > to find the
and JJ + 8=4;y
values of x and y.
l±f)
3^ > > to
involving two unknown Quantities. 115
From the first equation, by transposition,
y
completing the square,
y-^ "^ Vy"^^^ T.+n/ 2^ = 20;
X ,—
+ (f+V5) + i-o+!=H,
extracting the root, - +\/iH- - = ± - ;
••• ^-l-\/y = 4, or -5,
and X -\-y^ = 4y, or - 52/.
Let the former be taken.
Now from the second equation, i? + 8 = 4?/ ;
.'. by subtraction, ^t— g^o,
or y' = 8 ;
.-. y =4;
/. J? = 43/-.8 = l6-8 = 8.
But if X •k-'ip = - 5^,
and .r-f-8 = 42/;
.'. by subtraction, 8—^^= 9^;
.*. by transposition, 8 - 8 1/ = i/ +^^,
or 8.(l-3/)=:y.(l+3/i)
Dividing by (l +3/^), .-. 8(1 -y^) =y ;
.*. by transposition, y + 8^^= 8 ;
completing the square, y -|- Sy^^ 16 = 24,
and extracting the root, ?/^ + 4 = ± 2\/6 ;
/. y^ss — 4±2/>/6^
116 Solution of Adjected Quadratics y
and ?/ = 40+l6V6;
.-. x = 4j/-8 = 152T64x/6^
26. Given 8^ + 233/ = 2a- + 2^ . ^^ ^^^ ^^^
and 343/ + 6a;''-5/ = 13j?3/4-243
values of x and y.
From 'the second equation,
6j?* - 13 J?y = 6j/' - 341/ + 24,
J , 13 5 ^ 34
and x^- -^.a:y= g^*- ^3/ +,4 ;
. completing the square,
.13 169 , 169 . 5 , 34 , .
6 •^^144'^ 144*^ 6-^ 6-^^
289 , 3.4 ,^
extracting the root, x— — ^= ± (^—..y- 2^ ;
5
and first, taking the positive value, x— - 3/- 2.
Let this value of x be substituted in the first equation,
.-. 2qy-i6 + 23i/= 1^^-753/^+603/- l6 + 2t/3i
233^3
/. by transposition, - — "Jby^ + 1 73/ = O,
1 J- J- 1 133v , 300 68
and dividmer bv^ ^, y y-\ =0;
^ ^ 4 ' ^ 133 -^ 133
, 300 68
by transposition, 3/ - 7^3/= - ^;33 ?
completing the square,
300 1501" _ 22500 _^ _68^ _ 13456
^'■" 133'^"*" 1331 " 133]* 133 133I* '
involving two unknown Quantities, 117
extracting the root, 3/- — = + — ;
34
/.^/=2, or— ;
.-, a?= -^—2=3, or
-171
133
But if the negative value be taken,
1
Let this value be substituted in the first equation,
/. by transposition, -^ + ^/= —3/, ^
and dividing by ^ , i/* + ^^ 3/= -^ j
completing the square,
^ ^ 13"^ 26*|. 26! ^52 2B]» '
* ^- ^1 4. . 9 +V 10026*
extracting the root, 3/ + -^ = - ~^ ;
- ~9±\/ 10026' _ -9±3 »/lll4
... y_ ^ _ ^^_
, . 1 55Ty/TTl4
and 07 = 2 V = ^^7; .
3^ 26
27. Given - - S^x' - ^xy^=.^y - l6xy
y '
and 5^ = 4 + 25y
the values of x and y.
From the first equation,
x-^Sx-^y ^X'-f)y* = gy^— iGxy"" ;
to find
118 Solution of AdftcUd. Quadratics, .
/. by transposition,
(•^ - 93/*) - 8^Ty ^jo - 9/- + l6a?3/* = ;,
extracting the root, \/a: — gy* — 4^ty = o,
and ^ X — ^y^ = 4x^y,
or a? — 9y = iS-ry* ;
and J? = (9 + l6x), y^ ;
^
r =
9+l6a?*
5.27 — 4
But from the second eqHation,^* = — ;
5a? — 4 ^ X
and 80a:*- 19a: -36 = 25a?;
/. by transposition, 80x*- 44a? = 36,
J a 22 18
and x^ .0? = — -;
40 40
completing the square.
4 22 ,11
40 40
121 1_8_ 841
1600 40 ""l boo'
extracting the root, x = + -^
40 - 40
/. a?=l, or ~^;
^ 5X-4 1 1
28. Given
find the values
and - -
of X and y.
l6x — y^ = 6y^x^ 1
V involving two unknown Quantities. 119
From the first equation, by transposition,
completing the square, 250? =3/^ 4- Qy^tx^ + 9<^ ;
extracting the root, + 5^ =x z/t 4- 3^:^ ;
2x^f ov-Sx^—y'^,
and therefore Ax, or 64x=y^.
*c* X 12
Now from the second equation, 3= = — ;
, .. ,, a:* ^ , 1 12 1 49
completmt? the square, j=t H = — h = _£. ;
y sHf "^^ x^^x" 40?^'
0?* 1 7
extracting the root, —7= = ± — ;
sfy ^^ 2x'
a?' 4 3
s/ y X X
x^
and —7= = 4, or -3.
Now fJy — Ax, or 64^;
which, substituted in the last equation, gives,
- , or^ = 4,or-3;
X a? / — --
••• 2.or- = ±2, or ±V-3;
whence j?=: ±4, or ± 16, or + 2 />/ - 3 , or + 8.^/ — 3,
and 3/ =256, or 2561', or - I92, or -3x64lV
29. Given ^'-64 = 8x^y 7 ^ ^^ fi„a the values of
and ^ — 4 = 21/20:2 3
* and y.
120 - Solution of Adfected Quadratics ,
From the first equation, by transposition,
1/^ — 8 J?2y = 64 ;
completing the square, y^ - Sx^y + 1 6x = 1 6a? + 64 ;
extracting the root, y -^ 4x2 = + 4 s/a?-h4,
and y = 4j?2 ± 4 ^y a; + 4.
Also from the second equation, y'^2y^x^ = 4;
completing the square, y - 2yia:^ + a? = a? + 4 ;
extracting the root, 3/2 — a?2= + >>/x4-4 j
.*. 42/^ = 4xi ± 4 />/x-r4 =^, from the last equation ;
.*. 4 = i/2, and l6=^;
, I/ — 4 12 2
And from the second equation, x^=z- — ——-_- = -;
^ ' 2«/4 8 3
4
30. Given \/5^x-j-5^+ ^^zlO-y/x ^
SLnd,^ + y/y*=:275 5 '
find the values of x and y.
From the first equation,
^JxT^/y'\-s/^. \/ ^-{-^^ 10 ;
completing the square,
4 4 4
extracting ^he root, V^V^+Vv+^ = + ^^,
and sH/T+J^^z^'E, or - 2^5";
••• \/x + ^y = 5, or 20, supposing the former ;
to
involving two unknown Quantiths. 1^1
.'. by involution,
x^ + 5xy^ H- 10:c% -I- 10^y^+ 5a?4i/* -^i/^ = 3125;
but x^ +2^7= 275 ;
.'. by subtraction,'
5j7*t/5+ 10j^3/ + 10x3/^ + 5i?%' = 2850
or 5.372^^ (x^ + 2xi/i + 2X21/ +3/^) = 2850
and x^yi (j?^4-2xy^4-2a?%+y^) = 570
but x^T/i {x^ + 3x^^ + 3x it/ 4- 3/^) = 1 25xii/i.
.*. by subtraction, xii/'^x {xi/^+xii/) = 125x%3-570;
or xit/i X X2y4 X (xi + t/i) = I25xii/i - 570 ;
or 5x3/= 125x2^2 — 570 ;
.•. xt/— 25X2i/i= — 114 ;
completing the square.
o^ i'l 25
xi/ - 25x2^2 4- —
625 ,. . 169
I I 25 13
extracting the root, x'^y^ - -~ = ± -— • ;
.'. X2y2 = i9^ or 6.
but x + 2x^2 +3^ = 25,
and 4x%2 =24, or J6;
by subtraction, x - 2x23/2 4: 2/ = 1^ or --51.
extracting the root, x2-y2 = -ti, or ±,y — 51 ;
but X2 +3/2 = 5 .
.•. by addition, 2xi = 6, or 4, or 5+^^-51 ;
/. X2 =r 3, or 2, or li-^sLZii ,
R
122 Solution of Adftcted Quadratics^ Is^c,
J ^ , - 13 + V-Sl
and j: = 9, or 4, or — ^; - > ;
by subtraction, Sya = 4, or 6, or 5 + s/ -*'»l ;
1 « o 5+x^-5l
.-. y2 = 2, or 3, or . T^ ;
.•.y=4,or9,or^il±>/IIi;«
2
* The other case, where ^ x.V V'^=20, is solved in the saiT»«
manner.
12S
Sect. VL
On the Solution of Problems which involve Simple
Equations,
(30.) The solution of a problem, or method of dis^
covering by analysis quantities which will answer its
several conditions, is performed by assuming algebraic
symbols to represent the quantities sought, and by de-
ducing equations from the application of these, in the
samfe manner as if they were known quantities, to the
conditions of the problem. The independent equations
derived from this process, if the conditions be properly
limited, will equal in number the unknown quantities
assumed ; and from' the solution of these several equa-
tions by the rules already given (23. 27. 29), the values
of the algebraic symbols will be determined. - Whether
these values are correct, may be determined synthetically,
by applying them instead of their respective symbols to
the several conditions of the problem.
If the conditions of the problem are not properly
limited, that is, are not sufficient in number, or not
sufficiently independent of each other, the resulting equa-
tions will either exceed in number the unknown quan-
tities, and will therefore some of them be identical or
inconsistent, or will be fewer in number than the un-
known quantities, and consequently will admit of an
indefinite number of answers.
124 Examples of the Solution of Problems
In many cases, instead of assuming a symbol to
represent each of the required quantities, it is convenient
to assume one only, and from the conditions of the
problem to deduce expressions for the others in terms of
that one and known quantities. And as the number of
conditions ought to be one more than the number of
quantities thus expressed, there will remain one to be
stated in an equation ; from which the value of the un-
known quantity may be determined (22. 27. 28) : and
this being substituted in the other expressions, their
value also may be discovered.
Examples of the Solution of Prohlems producing simple
Equations involving only one unknown quantity,
1. What number is that, to the double of which if 18 be added,
the sum will be 82 ?
Let j? = the number required.
Then by the problem, 2a: + 18 = 82
.'. by transposition, 2 a: =64
and cr = 32 *
2. What number is that, to the double of which if 44 be added,
the sum is equal to four times the required number ?
Let - - - to the fourth.
/. X + 2X'{-3X-^4X=:60,
or I0x = 60^
,', J? = 6,
and .'. he gave 6, 12, 18, 24 pence respectively to them.
\9,(f Examples of the Solution of Problems
6. A Bookseller sold 10 books at a ceirtain price; afid after-
wards 15 more at the same rate. Now at the latter time he
received 35 shillings more than at the former. What did
he receive for each book ?
Let X = the price of a book.
Then 10x= price of the first set,
and 15 x= price of the second set.
but by problem 150?= l Ox* +35^
.'. by transposition 5 a? = 35,
and a?=7.
7. A Gentleman dying bequeathed a legacy of £.140 to three
servants. A was to have twice as much as B ; and B three
times as much as C. What were their respective shares?
Let j?=5: Cs share,
.'. 3x= JB's share,
and 6j?= A'^ share;
whence (6a:-|-3x + a? = ) 10a?=140,
.". .r=14,
^ .-.received ^84^ B, £A2 and C £A4.
8. Four Merchants entered into a speculation, for which they
subscribed £.4755 ; of which B paid three times as much
as ^ ; C paid as much as A and B ; and D paid as much as
C and B. What did each pay ?
Let x=: number of pounds A paid ;
.*. 3x= number B paid,
4x= number Cpaid,
and 7^= number Z> paid;
.*. (x-f 3x4-4 J7-I-7 <3C = ) 15a? = 4755,
and .*. cr = 317,
.-. they contributed 317, 951, 1268, and 2219 pounds
respectively.
producing Simple Equations. 127
9« A Draper bought three pieces of cloth, which together mea-
sured 159 yards. The second piece was 15 yards longer
than the first, and the third 24 yards longer than the second.
What was the length of each ?
Let X = the number of yards in the first piece, *
.*. x+15 = the number in the second,
and x-j-39= the number in the third,
.\ j? + a;-i-15-|-a?+39=159,
and by transposition 3 a? = 105,
.-. x=35.
/. the lengths are 35, 50, and f4 yards respectively.
10. A cask which held 146 gallons, was filled with a mixture of
brandy, wine, and water. In it there were 15 gallons of
wine more than there were of brandy, and as much water a*
both wine and brandy. What quantity was there of each ?
Let X = the number of gallons of brandy,
/. X H- 1 5 = number of gallons of wine,
and 2x+l5= number of gallons of water,
,\'X + x-{'l5-\-2X'\-l5 = l46,
.'• by transposition, 4x=ll6,
and 07 = 29.
.*. there were 29, 44, and 73 gallons respectively of
brandy,* wine, and water.
11. A person employed 4 workmen ; to the first of whom he
gave 2 shillings more than to the second; to the second 3
shillings more than to the third ; and to the third 4 shillings
more than to the fourth. Their wages amounted to 32 shil-
lings. What did each receive ?
Let x= the sum received by the fourth,
,', x-{-4= - -- - third,
x='/ + - - - - second,
and x+9= - ♦ - - ^irst.
1^ Examples of the Solution of Problems
and by transposition, 4 a: =12,
.-. x = 3.
/. they received 12, 10, 7, and 3 shillings respectively,
12. A Father taking his 4 sons to school divided a certain sum
amongst them. Now the third had 9 shillings more than
the youngest; the second 12 shillings more than the third;
and the eldest 18 shillings more than the second ; and the
whole sum was 6 shillings more than 7 times the sum which
the youngest received. How much had each ?
Suppose the youngest received x shillings,
then the third received x+ 9 -^ - -
the second --a7-f21-- -
and the eldest - - .r-f39 - - -
/. ^+x + 9+a? + 21+a?+39=7^+6,
.*. by transposition, 63 = 3 j?,
and .'. 21 =jC.
consequently they received 21, 30, 42, and 6o shilhngs
respectively.
13. A sum of money was to be divided amongst six poor persons ;
the second received 10c?. the third 14^. the foufth 9.5d. the
fifth 28 and differ-
ence 17 ?
Let X = the less,
.'.x-h 17= the greater,
and .*. a?-hx + 17 = 59,
by transposition, 2j? = 42,
and x = 21 the less,
.'. the greater =38.
16. What number is that, the treble of which increased by 1^,
shall as much exceed 54 as that treble is below 144 ?
Let X = the number.
.-. 3j;+12 — 54 = 144-3 J? by supposition.
/. by transposition, 6a: =186,
and J? = 31.
s
130 ExafMtpiet of the Solution of Prohletm
17. Two persons began to play with equal sums of money : the
first lost 14 shillings, the other won 24 shillings ; and then
the second had twice as many shillings as the first. What
sum had each at first f
Let 07= the sum;
Then x-\4
and
'. by problem 2a;--28=a: + 24
.-. a: = 52.
> =s the sums each had after playing;
18. At a certain election 9^3 men voted, and the candidate
chosen had a majority of Q5, How many voted for each ?
Let x= the number of votes the unsuccessful candidate
had;
,\ x4.63= the number the successful one had.
.-. J? -f^^- 65 =943;
by transposition, 2 j? = 8 7 8,
and J? = 439.
.*. the numbers were 439 ^^'^ ^04.
19. Two Robbers after plundering a house found that they had
35 guineas between them ; and that if one of them had had
4- guineas more, he should have had twice as many as the
other. How many had each ?
Let X = the number one had,
.*. 35— a? = the number the other had,
and 35— J? -1-4 = 2 07.
by transposition, 39=3^:,
and \3^x.
.'. they had 13 and 22 guineas respectively.
producing Simple Equations. ISl
ao. A Mercer having cut 19 yards from each of three equal
pieces of silk, and 17 from another of the same length, found
that the remnants taken together were 142 yards. What
was the length of each piece ?
Let x=. the length,
/. a?— 19= the length of each of the 3 remnants,
and ^ - 17 = the length of the other,
then 3 . (^-19)+^- 17 = 142,
or 3a?-57 +4?- 17 = 143.
by transposition, 4j?=216,
••. ar = 54.
21. A Farmer has two flocks of sheep, each containing the same
number. From one of these he sells 39, and from the other
93 ; and finds just twice as many remaining in one as in the
other. How many did each flock originally contain ?
Let x= the number required.
Then ar— 39 and x — Q3 are the numbers remaining;
.-. x-39=2j?- 186;
and by transposition, 147= a:.
S2. Bought 12 yards of cloth for J[,.\0, 145. For part of it I
gave 19 shillings a yard, and for the rest 17 shillings a yard.
How many yards of each were bought ?
Let x=i the number of yards at 19^. per yard;
A 12^— J? = the number at l^Js,
and 19^= the price of the former,
and 17.(12 — J?) = the price of the latter.
/. 19ar + 204-17x = 214.
and by transposition, 2 a? = 10,
and, a? = 5.
A there were 5 yards at I9 shillings, and 7 ^^ ^7 shil-
lings.
1S2 'Examples ^of ike Solution of Problems
25. Divide the number 197 into two such parts that four times
the greater may exceed five times the less by 50.
Let X = the less,
and .•. 197-3:= the greater.
Then 788-4^=5x4-50;
and by transposition, 738 = 9x.
and 82 =x;
•'• the greater =115.
S4. A Courier, who travels 60 miles a day, had been dispatched
five days, when a second was sent to overtake him ; in order
to which, he must go 75 miles a day : In what time will he
overtake the former.
Let x= the number of days the second courier
then x-fS = the number the first travels ; [travels.
.*. 75^= the number of rniles the second travels,
and 60 (x + 5)= the number the first travels.
But by the supposition they both travelled the same
pumber of miles ;
.-. 75x=6ox+300;
by transposition, 1 5 x = 300,
and x = 20.
25. After A had lost 10 guineas to B, he wanted only
8 guineas in order to have as much money as B : and to-
gether they had 60 guineas. What money had each at first ?
Let X = the number of guineas ^ had ;
.*. 60- x= the number - - - ^ had.
Then after playing A had 'x — 10, and B had 70 — x;
.-. a?-104-8 = 70-a:.
by transposition, 2x = 72,
and x = 36.
.*. they had 36 and 24 guineas respectively.
producing Simple liquations, 133
26. A and B began trade with equal stocks. In the first year
A tripled his stock and had £>%! to spare ; B doubled his
stock f^nd had £.\53 to spare. Now the amount of both
their gains was five times the stock of either. What was
that stock ?
Let a?= the stock;
then 3 0? -I- 27 = A^ stock at the end of the year,
.-. 2X+2T = his gain,
and 2 j?-f 1 53 = ^'s stock at the end of the year ;
.*. jr + 153= 5*s gain.
.-. 5j? = 2x+27+x+153.
by transposition 2 a; = 1 80,
and x = ^0.
0,11. Two workmen A and 5 were employed together for 50 days,
at 5 shillings per day each. A spent sixpence a day less
than B did, and at the end of the fifty days he found he had
saved twice as much as B, and the expence of two days
over. What did each spend per day ?
Let x= what A spent per day (in pence) ;
/. 60— a? = what he saved per day,
and 54 — ^= what B saved per day.
and .-. 3000 — 50 J? = 5400 -100 x + 2 J?,
by transposition, 48j7 = 2400,
and J? — 50;
.*. A spent 50 pence, and B 56 pence a day.
€8. A and B begai) to trade with equal sums of money. In
the first year A gained 40 pounds and B lost 40; but in the
second A lost one-third of what he then had, and B gained
a sum less by 40 pounds than twice the sum that A had lost;
when it appeared that B had twice as much money as A.
What money did each begin with i* ^
134 Examples of the Solution of FrohUms
Let X = the number of pounds each had at first,
then a: + 40 = the sum A had after the first year,
and a; - 40 = the sum B had,
also ^ . ( J7 + 40) = the sum A had after the 2** year,
and :c - 40 + ^ (i^ + 40) - 40 = the sum B had ;
/. ^. (x + 40) =x-40+4.. (j?+40) -40,
and^. (j? + 40) =a: — 80;
.-. 2j:-f80 = 3a?--240,
and by transposition, 3 20 = J?.
29. Divide the number 68 into two such parts, that the differ-
ence between the greater and 84 may equal three times the
difference between the Jess and 40.
Let x^ the less,
then 68 — a:= the greater ;
.-. 84- (68 -a:) =3. (40-t)
or l6+a? =120 -3a?.
by transposition, 4x=104,
and jr = 26;
and .-. the greater =42.
30. A and B being at play severally cut packs of cards so as to
take o^more than they left. Now it happened that A cut
off twice as many as B left, and B cut off seven times as
many as A left. How were the cards cut by each ?
Suppose A cut off 2 x cards,
then 62 — 2j?= the number he left,
and jc= the number B left;
.*. 62 — :c= the number he cut off;
, whence 52 -07=364— 14 a^;
by transposition, 1 3 a? = 3 1 2,
and a?=24 ;
.'. A cut off 48, and B cut off 28 cards.
producing Simple Equations, 135
31. What number is that whose one-third part exceeds its one
fourth part by 16 ?
Let 12j? = the number;
or x=\Qi
and .'. the number =12x16 = 192.
S2. Upon measuring the corn produced by a field, being 48
quarters; it appeared that it yielded only one-third part
more than was sown.. How much was that ?
Let 307= the number of quarters sown,
then 3 J? +0^ = 48,
or 4 J? = 48,
and 0? = 1 2 ;
.% the quantity sown was 36 quarters.
33. A Farmer sold 96 loads of hay to two persons. To the
first half, and to the second one fourth of what his stack
contained. How many loads did that stack contain ?
Let 4 0? = the number of loads,
then 2 X = the number the first bought,
and x= the number the second had.
.•. {2x-\-x = ) 3x = 96,
and x = 32.
whence, the stack contained 128 loads.
34<. A Gentleman bequeathed £.210 to two servants ; to one he
left half as much as to the other. What did he leave to
each ?
Let 2x= the sum one received ;
,\ xzz the sum left to the other.
13(5 Examples of the Solution of Problems
,', (2x-^x = ) 3a: = 210,
, and x = fO;
r, they had 140 and 70 pounds respectively.
S5. A prize of ;^.2329 was divided between t;wo persons A and
B, whose shares therein were in proportion of 5 to 12.
What was the share of each ?
Let 5x= A% share,
then 1 2 a? = ^s share ;
.-. (5j?+12j? = ) 17i = 2329,
and x=137 ;
.-. their shares were 685 and 1644 pounds respectively.
36. A sum of money is to be shared between two persons A
and B, so that as often as A receives 9 pounds, B takes 4.
Now it happens that A receives 15 pounds more than B'
What are their respective shares ?
Since for every £,^ that A receives, B receives ^.4,
Let ^x= the whole sum A receives;
.-. 4 X = the whole sum B receives.
/, 9a'=4a? + 15 ;
and by transposition, 5j:=15,
/. j7=3 ;
.*. ^ receives ;{^.27, andi5, jC.12
S7. A Gentleman gave to 3 persons 98 pounds.' The second
received five-eighths of the sum given to the first, and the
third one-fifth of what the second had. What did each
receive
Let 8x= the number of pounds the first received^
.'. 5x=i- - - - - second - -
-It d r— _ . .. _ . third - -
producing Simple Equations. 137
.-. {8x-i-5x + 0C=) 14J7 = 98,
and x = 7;
.*. they received 56^ 35, and 7 pounds, respectively.
38. A person bought two casks of beer, one of which held
exactly three times as much as the other. From each of
these he drew four gallons, and then found that there were
four times as many gallons remaining in the larger, as in the
other. How many were there in each at first?
Let 3 x= the number of gallons in the larger;
and .'. d?= the number in the smaller;
.*. 4 . (j?-4)= 3a?-.4 ;
by transposition, j?=12 ;
.-. they held 36 and 12 gallons, respectively.
39. A man at a party at cards betted three shillings to two
upon every deal. After twenty deals he won five shillings.
How many deals did he win ?
Let X = the number of deals he won ;
.*. 20 — 0?= the number he lost;
.-. 2.r= the money won,
and 3 . (20 - x) = the money lost ;
whence 2 a? - 3 . (20 — x) = 5 ;
.*. by transposition, 5^ = 65,
and x= 13.
40. What two numbers are as 2 to 3 ; to each of which if 4 be
added, the sums will be as 5 to 7 E
Let 2 X and 3 ^ be the numbers ;
.-. 2a?4-4:3^ + 4::5 : 7,
and (21) 14 J? + 28 = 15 a-|-20 ;
by transposition, S = x ;
and .*. the numbers are l6 and 24.
138 Examples of th Solutkn of ^rabiems
41. A sum of money was divided between two persons A and B,
so that the share of A wap to that of ^ as 5 to 3 ; and
exceeded five-ninths of the whole sum by 50 pounds. What
was the share of each person ?
Let 5 i:'= ^'s share ;
.'. 3jr= Rs share,
and 8 57= the whole sum ;
.'. bx= \. 8^ + 50,
or ^=s f. J?-f 10,
aad 9a? == 8x + 90;
.*. by transposition^ .r = 90,
and the sums were 45a and 2J0 pounds.
42. Being sent to market to buy a certain quantity of meat, I
found that if I bought beef, which was then 4 pence a
pound, I should lay out all the money I was entrusted with;
but if I bought mutton which was then threepence halfpenny
a pound, 1 should have two shillings left. How much meat
was sent for?
Let 2x= the number of pounds ;
.\ 8 j: = the price of 2 a? lbs. of beef,
and7x= the price of 2 j? lbs. of mutton,
and 8j? = 7a;4-24;
.-. j:=24;
whence 48 lbs. were sent for.
43. A fish was caught, whose tail weighed 9lbs. ; his head
weighed as much as his tail, and half his body ; and his
body weighed as much as his heaB and tail. What did the
fish weigh ?
Let 2 ^ = the number of lbs. the body weighed ;
then 9+^=;;; the vmght of the tail :
ffroducing SlffvpU Equations. 1 S9
by transposition, 18=0?;
/. the fish weighed 36 -f- 27 4-9 = 72 lbs.
44. The joint stock of two partners whose particular shares
differed by 40 pounds was to the share of the lesser as 14 to
5. Required the shares.
Suppose 14 07= the joint stock;
/. bx= the less,
and 9cr= the greater;
.-. 9a? = 5a?-f 40 ;
by transposition, Ax = 40,
and A?2=10;
/. the shares are 90 and 50 pounds, respectively.
45. A Bankrupt owed to two Creditors 140 pounds; the differ-
ence of the debts was to the greater as 4 to 9. What were
the debts I
Let 4 J? = the difference of the debts ;
and »•. 9'^= the greater,
and 5 a? = the less ;
/. (9T+5:r = ) 14^=140,
and 57= 10 ;
.-. the debts are 90 and 60 pounds.
46. A Gentleman employed two labourers at different times,
one for 3 shillings, and the other for 5 shillings a day. Nov/
the number of days added together was 40 ; and they each
received the same sum. How many days was each em-
ployed ?
Let X = the number of days the second was em-
ployed ;
140 ' Examples of the Solution of Probleftu
.*. 40-07= the number the first was employed ;
.*. 5.r= the sum received by the second,
and 3 . (40— x) = tlie sum received by the first ;
.-. 5 J? =3 . (40-^);
by transposition, 8 j: = 1 20,
and\r=15;
^\ the second was employed 15^ and the first 25 days
47' Some persons agreed to give sixpence each to a waterman
for carrying them from London to Gravesend ; but with this
condition, that for every other person taken in by the way,
three pence should be abated in their joint fare. Now the
waterman took in three more than a fourth part of the num-
ber of the first passengers, in tonsiddration of which he
took of them but five pence each. How many persons were
there at first ?
Let 4 X = the number of passengers at first ;
then j7+3= the number taken in,
and 3 j: + 9 = the sum deducted from their joint fare ;
.-. 24j?- (3j: + 9) =20:p;
by transposition, ^ = 9;
consequently there were 3^ passengers.
48. In a mixture of wine and cyder, half of the whole + 25
gallons was wine, and one-third of the whole — 5 gallons
was cyder. How many gallons were there of each ?
Let 6a? = the number of gallons in all ;
.*. 3a: + 25 = the number of gallons of wine,
and 2 J? — 5 = the number of gallons of cyder ;
.'. 6^ = 3a?+25+2^-5;
by transposition, j? = 20 ;
consequently there were 85 gallons of wine, and 35 of cyder.
producing Simple Equations, 141
49. -A and B engaged in trade, A with ;f .240, and B with £'9Q.
A lost twice as much as B ; and upon settling their accounts
it appeared that A had three times as much remaining as B,
How much did each lose ?
Let X— what B lost ;
.% 96-- j?= what he had remaining;
then 2.1= what A lost,
and 240 — 2 x= what he had remaining; .
.-. 240-2x = 3 . (96— J?)
by. transposition, j:=48;
/. J lost ^.96, and B lost ^.48.
50. Four places are situated in the order of the four letters
Af By C, D. The distance from ^ to D is 34 miles, the
distance from A to B : distance from C to D :: Q : 3, and
one-fourth of the distance from A to B added to half the
distance from C to D is three times the distance from B to
C. What are the respective distances?
Let 2x= the distance from A to B ;
/. 3.r= the distance from C to D.
and C^+ ^-^=) 2x^3. BC:
.-. BC==lx,
and r 2a:-|-3.rH -..:c =^ --^=34;
/, - = 2, and .r = 6 ;
3
whence AB^\2, BC=^4, and C/>=: 18,
142 Examples of the Solution of Problems
51 A Field of wheat and oats which contained 20 acres was put
out to a labourer to reap for six guineas, the wheat at 7
shillings an acre, and the oats at 5 shillings. Now the
labourer falling ill, leaped only the wheat. How much
money ought he to receive according to the bargain ?
Let X = the number of acres of wheat ;
then 20-~x= the number of acres of oats ;
and 7 07= the price of reaping the wheat (in shill*.)
and 100— 5 J? = the price of reaping the oats ;
.-. 7X+100-5 j:=126;
by transposition 2 a? =26,
and 0?= 13 ;
.•. he ought to receive ^.4. \\s.
52. A General having lost a battle found that he had only half
his army + 3600 men left, fit for action ; one-eighth of his
men + 600 being wounded, and the rest, which were one-
ftfth of the whole army, either slain, taken prisoners, or
missing. Of how many men did his army consist ?
Let x= the number required ;
,-. - 4.3600= the number fit for service ; *
2 ^
— 1-600= the number wounded^
o
X
and - = the number missing ;
5
/. jr = ^4-36oo + ?+6oo+f ;
2 o 5
S zr X
by transposition, -^ "" 7 =4200,
and (I6a?-8a:=) 7«3^=l68000,
and x = 24000.
producing Simple Equations, 143
53. Three men, A, B, and C entered into partnership; A paid
in as much as B and one-third of C ; 5 paid in as much
as C and one-third of A ; and C paid in o£'.10 and one-third
oi A. What did each man contribute to the stock ?
Let 3 j: = the sum A contributed ;
/. 10+07= - - C - - - -
and 10 + 2x= - - B - - --
10-f-^
.'. 3a?=10 + 2a; +
3
2x- 10
by transposition, --- = lo -|- ---,
and 337 = 40;
.'. J7 = 20.
and the sums contributed were ^.6o, ^^.56, and ^30, by
A, B, C, respectively.
54. It is required to divide the number 91 into two such parts
that the greater being divided by their difference, the quotient
may be ?•
Let X = the greater ;
.". 91 -x= the less,
and = 7 ;
.'. x=\4x - 637 ;
by transposition, 637 = 13 0? ;
and .-. 49=^0?;
.•. the parts are 49 and 42.
55. From each of 16 coins an artist filed the worth of half a
crown, and then offered them in payment for their original
value : but being detected, the pieces were found to be really
worth no more than 8 guineas. What was their original
value ?
144 • Examples of the Solution of Problems
Let X = the number of sixpences each was worth ;
.'. x^ 5 =^ the number each was worth after fiUng;
.-. l6.(j7— 5) =336.
by transposition, 1 6a: = 41 6,
and jr = 26 = 13 shiUings.
56. A and B made a joint stock of c£.833, which, after a suc-
cessful speculation, produced a clear gain of £.\5S, Of this
B had £A5 more ihan-^; What did each person contribute
to the stock ? ,
Let x= the sum brought in by JB ;
then 833 : .r :: 153 : JB*s gain=^
/. A's gain = — -45,
^ 49
and — +'— -45 = 153
49 49
18 iT
by transposition, = 198 ;
*^
.-. J?= j^ = 49x 11=539,
whence, B brought in £.b3^ and A £.294,
57. Sold a c|uaTility of tobacco for 19 shillings, part at 1 shilling
a pound, and the rest at 15 pence. Now the first part was
to the latter :: 4. : |. How much *was sold of each ?
Since |- : ^ :: 9 = 8,
Let 9 «3? = the number of lbs. of the former ;
.*. Sx~ the number of lbs. of the latter ;
.*. 90?= the number of shillings the first aold for,
and 8a?x4^=10a?= the number of shillings the second
sold for.
producing Simple Equations, 145
.-. (10^ + 9^=) 1907=19,
and ^= 1 ;
there were at 9 lbs. at 1 shilling, and 8 lbs. at 15 pence.
58. A Gentleman gave in chanty ^.4?6 ; a part thereof in equal
portions to 5 poor inen, and the rest in e^ual portions to
7 poor women. Now a man and a woman had between
them £.Q, What was given to the men, and what to the
women ?
Let 6.r= the number of pounds the men received ;
.'. 46— 5 J7= the number the women received ;
.'. ^= the sum one man received,
and 8 — :r = the sum one woman received ;
.-. 56 - 7a? = 46 — bX',
by transposition, 2 j?= 1 0,
and a? = 5 ;
/. the men received ^.25, and the women ;^.21.
59' Suppose that for every 10 sheep a farmer kept, he should
plough an acre of land, and be allowed one acre of pasture
for every 4 sheep. How many sheep may that person keep
who farms 700 acres ?
Let xzz the number of sheep required ;
X
then 10 : J? :: 1 : thle number of acres ploughed = •-->
10^
X
and 4 : a: :: 1 : the number of acres of pasture = - ;
X X ^
and (2a: + 5a?=) 7a: = 20x700;
.-. a: = 20X 100 = 2000,
u
146 Examples of the Solution of Problems
60. A person being asked the hour, answered that it was be-
tween five and six; and the hour and
.*. the number of apples is 75, and pears 25.
77* A person has two sorts of wine, one worth 20 pence a quart,
-and the other 12 pence; from which he would mix a quart
to be worth 14 pence. How much of each must he take ?
Let X = the quantity of the first, the whole quart
being represented by unity ;
.-. 1 — a?= the quantity of the second ;
also 20 x=: the value of the first,
and 12— 12a? = the value of. the second;
.-. 20 0? -{-12- 12 ^
afterwards 2a?-j-4.
1 2 T 4- 24
Having lost \ of this, he had f . (2 x + 4) = — ^- ,
, 12J? + 24 '^
and .-. = 12,
and \2x-\-2A = 7x 12^
by transposition, 1 2 :c = 5 x 1 2,
and .r = 5 ;
.'. he had at first 20 shillings.
94. A Trader maintained himself for 3 years at the expence
of £.50 a year; and in each of those years, augmented
that part of his stock which was not so expended by J
thereof. At the end of the third year his original stock
was doubled. What was that stock ?
Let X = the number of pounds required ;
producing Simple Equations, 16S
then a? — 50 = the sum not expended ; and with this
he traded ;
.'. — - — = his gain the first year,
4
and - . (.r — 50) = the sum he had at the end of the first
year;
4X-200 ^^ 4a? - 350 . u ^ J J -^u
— 50 = =the sum he traded with
3 3
the second year ;
4 4j:-350 l6a?-140O
= the sum he had at the
3.3 9
end of the second year ;
J 16j?-1400 ^ i6j?'-.1850 , u ^ J J
and 50 =r = the sum he traded
9 9
with the third year ;
' ' 5 * Q ~ ^^^ ^""^ ^® ^^^ ^^ ^^^ ^^^ ^^ ^^
third year ;
, 4 l6a?-1850
whence - . -, — — == 2 x,
and 32 .r - 37:00 = 27 a? ;
by transposition, bx = 3700,
and \r = 740.
95. A merchant buys a cask of brandy for £M, and sells
a quantity exceeding three-fourths of the whole by 2
gallons at a profit of ^.9.5 per cent. He afterwards sells
the remainder at such a price as to clear ^.60 per cent, by
the whole transaction ; and, had he sold the whole quantity
at the latter price, he wouM have gained /'.175 per cent.
Required the number of gallons contained in the cask.
Let 4 0? = the number of gallons ;
164 Examples of the Solution of Froblems
,12
then — = the original price per gallon (in pounds,)
12 15
and 100 : 125 :: — : the first price of sale =— ,
OC X
12 33
and lOQ : 1^75 :: — : the latter price of sale == — ;
X X
.\ (3a:-f 2). — + (^--2) . 48= whole gam = 30
X X X
of?
hence 100 : 6o :: 48 : 30 — ^—,
X
6
or 5 ; 3 :: 8 : 5 ; '
30
.-. 24 = 25 ;
, ■ . . 30
by transposition, — = 1 ,
X
and .•. 07 = 30,
and the number required = 4 a? = 1 20 gallons.
96. Water flows uniformly into a cistern, capable of containing
720 gallons, through a pipe ; and at the same time is
discharged by a pump, worked by three men, who take
four strokes in a minute; but this not being sufficient, the
cistern becomes full in 6 hours ; they therefore now put in
another pump, of such power that the quantity discharged at
one stroke by this pump is to the quantity discharged at one
stroke by the former :: 2 : 3; but being obliged to detach one
of their number to work the pump, the former pump makes
only 10 strokes in 3 minutes, and the latter 5 strokes in 2
minutes ; by which means the cistern is emptied in 12 hours.
How much water was discharged by each pump at one stroke ?
and how much flowed in through the pipe in one minute ?
Let 3o? = the number of gallons discharged by the first
pump at one stroke ;
/. 2x = the number discharged by the second.
producing Simple Equations. 165
and 12a: = the quantity discharged by the first in one
second^ when 3 men woiked ;
/. 6x60xl2^ = the quantity discharged in six hours ;
/. 6 X 6ox 12 0? -1-720=720 . (6x + l) = the quantity
introduced through the pipe in that time.
Now when the additional pump is worked,
lOo? = quantity discharged by the first in one minute ;
and 5 X = the quantity discharged by the second in one
minute ;
.*. 15 0? X 12 X 60 = the whole quantity discharged in
12 hours;
.-. 15a?x720 = 720X2.(6a:+l)-f720;
or 15 ^=12d? + 3 ;
by transposition, 3x = 3,
and x=l ;
.•. the first discharged 3 gallons, and the second 2, at one
stroke ; and the quantity introduced by the pipe in one
. ^ 720 X (607+1) ^ ^ ,, ,,
mmute= .. ^ ^ = 2x7=14 gallons.
6 X 60 ^
97. A poor man with a wife and seven children, found during
a scarcity that he could only earn sufficient to procure
I of a white loaf of bread per day for each of his family,
himself included. He therefore applied to the parish officers
for assistance, by whom being allowed a dafly sum = ^ his
earnings, and mixed bread being made by order of Par-
liament, which was cheaper than white in the proportion of
4 to 5, he was now enabled to. procure J of a mixed loaf
per day for each of the family (himself still included) and
had Is, J-rd. over. Required the sum allowed him by
the parish".
Let 07 = the price of a white loaf (in pence ;)
.* . -- = what he earned,
4
166 Examples of the Solution of Problems y tsfc, ,
and ^ = what the parish allowed him ;
8
4x
also -— = the price of a mixed loaf ;
\j
I2x 39 _ ^9^ , 9^ _\ 2J X
and 96^+780 = 135x;
by transposition, 7^0 = 39^7,
and 20 = 0?; '
whence it appears that he earned 45 pence, and had
22^d. allowed him by the parish.
Sect. VII.
Examples of the Solution of Problems producing Simple
Equations involving two unknown Quantities,
]. Afteb a had won four shillings of ^, he'had only half as
many shillings as B had left. But had B won six shillings
of Af then he would have had three times as many as A
would have had left. How many had each ?
Let X =. the number of shillings A had,
and y — the number jB had j
then ?/-4 = 2^ + 8,
and ?y 4- 6 = 3 0? — 1 8 ;
.'. by subtraction 10 = a; — 26,
and by transposition, 36 = x,
and 3/ - 4 = 80 ; .•.«/=: 84 ;/. ^ had 36 and B 84.
A person bought a quantity of brandy and rum for
£.\^, 4s. and gave for the brandy 9 shillings, and rum 6
shillings ^er bottle. He found however that he could have
bought as many bottles of rum as he now had of brandy,
and as many of brandy as he now had of rum for £A. iSs.
less. How much was bought ?
Let X = the number of bottles of brandy,
and y = the number of bottles of rum ;
then 907 -f 63/ = 384^
and 6.r + 93/ = 351 ;
.-. by addition, 15^ +153/= 735,
168 Examples of the Solution of Problenu
And X + y = 49.
But since 3x + 2y =128,
and 2x + 2y — 98 ;
.'. by subtraction, x = 30,
and 3/ = 49 - X = 19;
.-. he bought 30 bottles of brandy, and 19 of rum.
3. What fraction is that, to the numerator of which if 4 be added,
the value is one-half, but if 7 be added to the denominator,
its value is one-fifth ?
X ' , .
Let - = the fraction required;
y
then, = JL and /. 2^ + 8 = v ;
also = 4-, and .*. b x -=.11 4- 7 :
by subtraction, 3 1? — 8 =75
by transposition, 3 a? = 15,
and X = h\
.-. y = 2^ + 8 = 18,
5
and the fraction is -— .
18
4. Find tv/o numbers, the greater of which shall be to the less
as their sum to 42, and as their difference to 6. ,
Let X and y = the numbers ;
then, X \ y :: x + y : 42,
and X : y :\ X -- y '. 6.
But ratios which are equal to the same tatio are equal to
each other ;
producing Simple Equations. l69
.'. a? 4- ^ : 42 :: a? ~ y : 6,
B.W. X -\- y I X -^ y :: (42 : 6 :: ) 7 : 1 ;
/. {Alg:\Q2,) 2x : 2y \y 8 : 6,
and X : t/ :: 4 ; 3 ;
• •
, X =
3
3
and
4
: 3 ::
y
3
: 6;
•*•
3
8,
and
and X
=
±y
3
=
32;
\ the numbers are 32 and 24,
5. What two numbers are those, whose diiFerence, sum, and
product, are as the numbers % 3, and 5, respectively ?
Let X and y = the numbers ;
then X -^ y : X -{- y V, 2 ', 3\
.'. 2x : 2y :: 5 : 1,
or X : y :: 5 : 1 j
, also X -{- y : xy ;: 3 : 5,
or 6y : xy :: 3 < 5 ;
/. 2 : :c :: 1 : 5,
and .*. X = 10,
and 1/ = 1 = 2;
.•. the numbers are 10 and 2.
6. A and B playing at bowls, says A to j6, If you will give me
a guinea, I will bet you half a crown to eighteen pence on
z
170 Examples of the Solution of Problenus
each game, and will, play 3^ games together. B won his
guinea back again, and £.1. 175. besides. How many
games did each win ?
Let X = the number of games A won,
and y = the number B won ;
.-. y + ^ = 36,
and 3i/ + 3^ = 108 ;
but 5y — 3 a? = ll6;
.r. by addition, 8y = 224,
and y = 28;
.•. J? = 36 — 3/ = 8;
.*. A won 8, and jB 28 games.
7. A person exchanged 12 bushels of wheat for 8 bushels of
barley, and £.2. 1^5 ; offering at the same time to sell a
certain quantity of wheat for an equal quantity of barley,
and £.3. 15s. in money, or for £.10. in money. Required
the prices of the wheat and barley per bushel.
Let X = the price of wheat per bushel, in shillings,
and y = the price of barley ;
then ■~— = the number of bushels in the second offer ;
X
/. \2x == 8 2/ -1- 56,
, 200
a»d X 3/ = 125 ;
.-. s y = b X,
and .". 12jf = bx +56;
by transposition, T x = 56,
and 0? = 8 ;
.-. 3/ = ^;
. . the prices of wheat and barley per bushel were 8 and
5 shillings^ respectively.
producing Single Equations^ 17 i
8. A vintner sold at.one time ^0 dozen of port wine, and SO of
sherry, and for the whole received £.120; and at another
time sold 30 dozen of port, and 25 of sherry at the same
prices as before, and for the whole received £.140. What
was the price of a dozen of each sort of wine ?
Let X = the price of a dozen of port,
and ^=------of sherry ;
.*. 20 J? + SOy = 120, or 2a? + 3y = 12,
and 30 .r + 25^/ = 140, or 6a? + 6y = 28.
Multiplying the first equation by 3 ;
6a? + 93^ = 36,
but 6a? + 5 2/ = 28;
.•• by subtraction, 43/ = 8,
and 3/ = 2;
whence 2a? = 12 — 3y = 12 — 6 = 6,
and a? = 3 ;
.-. the prices of port and sherry per dozen were ^.3 and
jf.2, respectively.
^~^^*^~9 *■*"»■**■*
A and B severally cut packs of cards, so as to cut off less
than they left. Now the number of cards left by A added
to the number cut off by JB make 50 ; also the number q£
cards left by both exceed the number cut off, by 64. How
many did each cut off?
Let a? = the number cut off by A\
\ 52 — a? = the number left by him.
Let 7/ = the number cut off hy B ;
. 52 — ^ =K the number left by him;
.*. X + 3/ = the whole number cut ofi^
and 104 — {x + y) = the whole number left,
whence 104 — 2. (a? + ^) = 64 ;
by transposition, 2 . (a? H- y) = 40,
and a: -f y = 30.
172 Examples of the Solution of Problems
Now 52 - X + 3/ = 50 ;
.•.by transposition, a: — 3/ = 2,
but X + y — 20
,\ by addition 2x = 22,
and X = 11;
by subtraction, 2y = 18,
and y = 9 ;
.-. ^cut off 11, and B 9.
10. A countryman, being employed by a poulterer to drive a
flock of geese and turkeys to London, in order to distinguish
his own from any he might meet on the road, pulled 3
feathers out of the tail of each turkey, and one out of the
tail of each goose, and upon counting them, found that the
number of turkeys* feathers exceeded twice those of the
geese by 15. Having bought 10 geese and sold 15 turkeys
by the way, he was surprised to find, as he drove them into
the poulterer's yard, that the number of geese exceeded the
number of turkeys in the proportion of 7 to 3. Required
the number of each at first.
Let X = the number of turkeys ;
,\ 3 X = the number of feathers from their tails ;
let y = the number of geese; and .-. of the feathers,
and 3 a? — 2^ = 15.
Also 3/ + 10:a;— l5::7-3;
.'. 3y -{- 30 = 7x - 105;
by transposition, 7a;-3^=;135,
and 14 a; - 63/ = 270 ;
but from the first equation 9x — 61/ = 45;
.*. by subtraction, 5 a? = 225,
and a? = 45;
.-. 2t^ = 3 a? - 15 = 135 - 15 = 120,
and 3/ = 60 ;
.•. tbere were 45 turkeys, and 60 geese.
prodiucing Simple Equations, 173
11. A farmer with 28 bushels of barley at 2s. ^d. a bushel, would
mix rye at 3 shillings per bushel, and wheat at 4? shillings per
bushel, so that the whole mixture may consist of 100 bushels,
and be worth 3s. 4c?. per bushel. How many bushels of rye,
and how many of wheat, must he mix with the barley ?
Let X = the number of bushels of rye,
and y = the number of wheat ;
then the value of the barley = 1 96 (four-pences),
of the wheat — \2y,
of the rye = 9 j? ;
.*. 196 + 9x + I2y = 1000;
by transposition, Qx + 12^ = 804,
and 3x + ^y = 268.
Now a? + y + 28 = 100,
and by transposition, a: 4- y = 72 ;
,\3x 4-3 2/ = 216,
but 3 X +4y = 268;
,'. by subtraction, y = 52,
and a: = 72 - y = 20.
Hence he must mix 20 bushels of rye, and 52 of wheat.
12. A and B speculate with different sums ; A gains £\50, B
loses £.50, and now As stock is to J3's as 3 to 2. But had
A lost £,SOf and B gained £.100, then A's stock would have
been to B's as 5 to 9- What was the stock of each ?
Let X = A's stock,
and y = i5's ;
then X + 150 : ^ - 50 :: 3 : 2;
.-. 2X + 300 = 32/ - 150,
and by transposition, 3 y — 2 x = 450 ; ^
also X - 50 \ y -{■ 100 :: 5:9;
.-. 90? -450 = 5y + 500;
by transposition, Qx — 5y = 9^0;
174 Examples of the faitttion of Problems
multiplying this equation by 3, and that found above by 6,
27a? - \by = 2850,
and \by - 10a? = 2250;
.-. by addition, 170^ = 5100,
and X — 300 ;
.-. 33/ = 2x + 450 =1050,
and y = 350 ;
.-. As was ;f .300, and Rs £.350.
13. A merchant having mixed a certain number of gallons
of brandy and water, found that if he had mixed 6 gallons
more of each, he would have put into the mixture 7 gallons
of brandy for every 6 of water ; but if he had mixed 6 less of
each, he would have put in 6 gallons of brandy for every
5 of water. How many of each did he mix \
Let X = the number of gallons of brandy,
and y = the Humber of gallons of water ;
then X -{- 6 : y -{- 6 V. T \ 6y
.-. {Alg. 181.) X + 6 : J? - y :: 7 : 1,
but X — 6 : y — 6 :: 6 : 5,
and .*. a: — 2/ : X — 6 :; 1 : 6,
and since x -^ G i x — y ix T : \ \
.', €X OBqualiy j: + 6 : a? — 6 :: 7 *• ^>
and 2 a? : 12 :: 13 : 1,
ovx'. 6 :: 13 : 1 ;
.-. X = 78 ;
whence 84 : y + 6 :: 7 i 6>
or 12 : y + 6 :: 1:6;
.-. 3/ + 6 = 72,
and y ^ 661
.'. he mixed 78 gallons of brandy with 66 of water.
14. A person had a bag of money worth ^^.93 ; but a servant
having robbed him of one-sixth of his moidores, and three-
producing Simple Equation f. 175
fifths of his guiaeas, left hkn only .£.54. 15s. How many
moidores and guineas had he at first.
Let X — the number of guineas,
and y = the number of moidores ;
then -Tt^ + -- = 305,
o 5
and /. 225 3/ + 84 X = lOgSO -,
also 7 J? + 9 2/ = 620 ;
.-. IO83/4-84 J? = 7440 ;
but since 225 ?y + 84 :c = IO95 ;
/. by subtraction, 1173/== 3510,
and 3/ = 30 ;
also 7 a? = 620 — 9 3/ = 620 -270=350 5
.-. a; = 50 ;
.-. he had 50 guineas, and 30 moidores.
15. A vintner bought 6 dozen of port wine and 3 dozen of
white for 12 guineas ; but the price of each afterwards
falling a shilling per bottle, he had 20 bottles of port, and
3 dozen and 8 bottles of white more, for the same sum.
What was the price of each at first ?
Let Of = the price of the port 1 ^^^ ^^^^^^ ^j^ shillings,)
2^=5 * - - - white >
then 72 a? + 36 3/ = 252 =92 0^ + 803/ -172,
and .-. by transposition, 20 .r^- 44 3/=! 72,
or 5:r4- 11 3/ = 43.
Now, since 72^+363/ = 252;
.-. 20^+3/ =7>
and 22 a?+ll3/ = 77>
but 5^7+113/ = 43;
,*. by subtraction, 17'^= 34,
and a; = 2 ;
whence y = 7 — 2a? = 7-"4=3;
176 Examples of the Solution of Problems
/. the price of port was 3*. and of white ^s, per bottle.
16. A rectangular bowling-green having been measured, it
was observed, that if it were 5 feet broader, and 4 feet
longer, it would contain II6 feet more: but if it were 4
feet broader, and 5 feet longer, it would contain US feet
more. Required the length and breadth.
Let X = the number of feet in length,
and y = the number of feet in breadth ;
then {x + A .y-\-5J=)xi/ + 5x + 4i/ + 20 = ll6-f ji^,
and .*. 5 x+4i/ = 96 ;
also {x + 5 , 2/ + ^ = )xi/'\'4x +53/ + 20 = 113 + Xi/ ;
/. 4x + 5t/ = 93 ;
multiplying the former equation by 4, and the latter by 5,
20X -\- 161/ zz 384,
and 20 X -\- 251/ =z 465 ;
by subtraction, 9 2/ = 81,
and2/ = 9;
.-. 4x=93-52/ = 93-45=48,
and 0? = 12 ;
.'.the length was 12 and the breadth 9 feet.
17. Find two numbers in the proportion of 5 to 7, to which
two other required numbers in the proportion of 3 to 5
being repectively added, the sums shall be in the proportion
of 9 to 13; and the difference of those sums= 16.
Let 5 a: and 7 ^ = the two first numbers,
and 3 3/ and 5 y = the others ;
then 5 07 + 3 2^ : T x-^-by :: 9 : 13;
.'. bx-\-3y : 2x-\'2y ;: 9 : 4 ;
ov ^ x-\-3 y '. X + y ::9:2,
and 10 x-{-6y = 9 x±9y ;
by transposition, x = Sy;
producing Simple Equations. 177
but 2 x-\-2y = 16;
.-. (6^ + 2y =) 83/ = 16;
.-. 3/ = 2,
and a: = 6 ;
whence, the two first numbers are 30 and 42 ; the
two others, 6 and 10.
18. A merchant finds that if he mixes sherry and brandy in
quantities which are in the proportion of 2 to 1, he can
sell the mixture at 78 shillings a dozen; but if the pro-
portion be as 7 to 2, he must sell it at 79 shillings a
dozen. Required the price of each liquor.
Let * = the price of the sherry ? ^^^^ .
y = the price of the brandy 3
then 2 a: + 2^ = 3 X 78 = 234,
and 7/^^+2^ = 9x79 = 711;
but the first equation ) 4^^^ ^4685
being multiplied by 2, 3
.-. by subtraction, 3 ^ = 243,
and 0^ = 81 ;
whence, ^ =234 - 2 J? = 234 - 162 = 72 ;
.*. the price of the sherry was 81.?., and of brandy 72.?.
19. A corn factor, mixes wheat-flour, which costs him 10
shillings a bushel, with barley-flour, which costs him 4
shillings a bushel, in such proportion, as to gain 43|./?er cent,
by selling the mixture at 1 1 shillings a bushel. Required
the proportion.
Let the proportion be a? : ^ ;
then 1 X + 4 3/ = the cost of x-^-y bushels,
and II x -h 1 1 y = the selling price ;
.'. x-^T y^ihe gain ;
AA
i
178 Examples of the Solution of Problems
whence^
10a:+43/ : oc+71/ :: 100 : 43| :: 400 t 175 :: 16 : 7,
and 5^ + 22/ \ x-\-7 y :: 8 : 7;
.-. 35x+14^ = 8a? + 563/;
by transposition, 27 x = A2yy
and 90; = 14^ ;
.-. x\yv,lA: 9,
and .-. he must mix 14 bushels of wheat-flour with 9 of
barley.
20. A number consisting of 2 digits when divided by 4,
gives a certain quotient and a remainder of 3 ; when
divided by 9 gives another quotient and a remainder of 8.
Now the value of the digit on the left hand is equal the
quotient which was got when the number was divided by
9; and the other digit is equal ^i^th of the quotient got
when the number was divided by 4. Required the number.
Let X and y = the digits in order ;
then 10 x -\- y = ^the number,
andH£-±^ = ^ + ?;
9 9
.-. lOo: + 3/ = 9^ + 8;
by transposition, x ■{■ y = S ^
, 10a: + y 3 . ^
also J~^ = J + 173^;
.-. 1007 + 3/ = 3 -h 68y;
by transposition, \0 x - 67 y =1 S -y
but from the preceding equation \0 x ->[- 10 y = SO ;
,\ by subtraction, 77 y = 77,
Q,nd y = 1 ;
... X = 8 -1/ = 7,
and the number required is 71,
producing Simple Equations. 179
SI. A man and his wife could drink a barrel of beer in 15 days.
After drinking together 6 days, the woman alone drank the
remaipder in 30 days. In what time would either alone
drink a barrel ? •
Let X = the number of days in which the man could
drink it,
and ^ =the number in which the woman could drink it ;
then = 1,
X y
and h -- + — =1, ^
X y y '
6 36
or - -f — = 1 ;
^ y
hence from the first equation — | — = t; ,
^ X y 15
* from the last - + - = ^ ;
X y Ki
■'■ ^y subtraction, |=g-l^=^=i^.
and .•.?/ = 50;
i "■ 15 ^ — 15 50 150'
and .-. X = 21-^;
.-.the man would drink it in 21-^ days^ and the woman
in 50 days.
22. A purse holds 19 crowns and -6 guineas. Now 4> crowns
and 5 guineas fill —• of it. How many will it hold of each ?
Let 07 = the number of crowns,
and y = the number of guineas ;
180 Examples of the Solution of Frohlems
4
then XI 1 :: 4 : the space occupied by 4 crowns = - .
30
* . . 5
In the same way, the space occupied hy 5 guineas = - ;
4 5 _ 17
•** i "^ «/ ~ b*3 '
X y
The first equation being multipHed by 6, and the second
by^?,
24 30 _ 34
X y " 21 '
and ^ + — = 5 ;
^ y
.-. by subtraction, — = 6 - — -; L- .
•^ j: 21 21
.-. X = 21,
, 5 17 4 5
""^ ^ = 6i - 2T = 65 '
/. 3^ = 63 i ;
/. the purse would hold 21 crowns, or Q2i guineas.
fi3. Some smugglers discovered a cave, which would exactly hold
the cargo of their boat, viz. 13 bales of cotton, and 33 casks
of rum. Whilst they were unloading, a custom-house cutter
coming in sight, .they sailed away with 9 casks and 5 bales,
leaving the cave two-thirds full. How many bales or casks
would it hold ?
Let X = the number of bales,
and y = the number of casks ;
,, 13 33
then — •\- — =1,
X y
producing Simple Equations. 181
and - + - = -.
X y S
Multiplying the first equation by 3, and the second
by 11,
then 19 + 99 ^ 3
X y
and 5^ + 99 ^ li^
X y 3
^,*. by subtraction, — = - ;
.-. 2x = 48,
and X = 24;
^,9 1 5 1 5 3 1
consequently -= - — -= — = — = -.
^ -^^3 ^3 24 24 8^
and y = r^i
and /. tbe cave would hold 24 bales, or 72 casks.
24. Round two wheels, whose circumferences are as 5 to 3, two
ropes are wrapped, whose difference exceeds the difference
of the circumferences by 280 yards. Now the larger rope
apphed to the larger wheel wraps round it a certain number
of times, greater by 12 than the smaller round the smaller
wheel ; and if the larger wheel turns round 3 times as quick
as the other, the ropes will be discharged at the same time-
Required the lengths of the ropes and the circumferences
of the wheels.
Let 5 X and 3 r = the circumferences of the wheels
(in yards) ;
then 2x4- 280 = the difference of the ropes,
and (I5x:3a?:;)5:l:: the length of the longer string :
the length of the shorter.
Let /. 5y and y = the lengths of the ropes (in yards) ;
then 4y = 2x -h 280,
182 Examples of the Solution of Problems
and --^ = ^ + 12 ;
5 vT 3a:
or by transposition, —2. = 12;
/. 4y =72^,
and .•. 72 a; = 2 x + 280 ;
by transposition, *jOx = 280,
and J7 = 4 i
.-. 3/ = 72 ;
/. the circumferences of the wheels are 20 and 12 yards,
and the lengths of the strings 36o and 72 yards.
25. Three guineas were to be raised on two estates, to be charged
proportionably to their values. Of this sum, A^s estate
which was 4- acres more than JB*s, but wor^e by 2 shillings
an acre, paid £.1. 15s. But had A possessed 6 acres more,
and B*s land been worth 3 shillings an acre less, it would
"have paid £.2. 55. Required the values of the estates.
Let X = the number of acres B had ;
then ^ + 4 = the number A had.
Let 1/ =. the value of an acre of ^'s land j
,\y + 2 = the value of an acre of JS's,
and {x^ 4),y : x.(^ + 2) :: 35 : 28 :: 5 : 4 ;
.-. 4xy 4" 163/ = bxy 4- \0x,
and xy = l6y — 10 a?.
Again, {x + 10) .y : x.(y - l) :: 45 : 18 :: 5 : 2;
.*. 2xy -^ 20y = 5xy — 5x,
and by transposition, 3 xy = 20 y + 5 x,
whence 483/ — 30 .r = 20y + 5^;
by transposition, 28 y =. 35 y,
and 4y = 5x,
which value substituted in the first equation, gives
xy = 20j? — 10J7 = lOo:;
and ,". y = 10,
producing Simple Equations, 183
whence x = -^ = S ;
5
.-. the value of ^'s estate = {x -\- 4).y = 120 = £,6;
and the value of 5's = « . (i^ + 2) = 96 = £A. l6s.
26. A coach set out from Cambridge to London with a certain
number of passengers, 4 more being on the outside than
within. Seven outside passengers could travel at 2 shillings
less expence than 4 inside. The fare of the whole amounted
to £.9. But at the end of half the journey, it took up 3
more outside and one more inside passengers'^ in consequence
of which the fare of the whole became increased in the pro-
portion of 17 to 15. Required the /lumber of passengers,
and the fare of the inside ahd outside.
Let X = the number of inside 7
.*. x -f 4 = the number of outside) ^
and y = the fare of an outside passenger ;
.., LK = the fare of an inside passenger,
and 7£L±ll 4-1/ .{x + 4) = 180.
Also — ^ + ■ = the fare of the passengers taken up
halt way =• — -^ ;
o •
O
ori9^Jli : 12 :: 2;1, .
8
+ 2
24,
and 193^ + 2 => 192;
I9.y + 2
8
184 Examples of the Solution of Problems
.-. by transposition, Ipy = 190,
and y = 10 ;
.-. from the first equation, + 10 . (a: + 4) = 180;
by transposition, 28 x = 140,
and a? = 5 ;
.*. there were 5 inside, and 9 outside passengers,
and the fares were 18 and 10 shilUngs, respectively.
27. In one of the corners of a rectangular garden there is a fish-
pond, whose area is one-ninth part of the whole garden ; the
periphery of the garden exceeding that of the fish-pond by
200 yards. Also if the greater side be increased by 3 yards,
and the other by 5 yards, the garden will be enlarged by
64f5 square yards. The fish-pond is a rectangle about the
same diameter with the garden. Required the periphery
of the garden, and the length of each side.
Let X = the length of the lesser side,
and 1/ = the length of the greater ;
X TJ
.*. - and "I = the lengths of the lesser and greater sides
of the fish-pond, (EucL B. 6. Prop. 24.)
Also 2.,(x + 1/) = the periphery of the garden ;
^and — ' ; ^' = the periphery of the fish-pond;
2^
.-. 2 . (a; + «/) - 3 . (^ -F t/) = 300, ^ .
' . or - . (1? + 1/) = 100;
.*. X + 3/ = 150.
Also (1/ -{■ 3) , (x + 5) IT xy + 645 ;
.'.by transposition, 3x + 5t/ s^ 630,
producing Simple Equations. 185
but from the former equation, 3 0? -f- 3y = 450 ;
.•. by subtraction, 2 2/= 180,
and 3/ = 90 ;
.'. X = 150 — ?/ = 60;
and •. the periphery =^300 yards, and the sides are 6o
and 90 yards.
28. A and B each bought of .300 into the stocks, A into the
three per cents.y and B into the four s^ These stocks were at
such a price that B received one pound interest more than
A> When afterwards each of the stocks rose 10 per cent.,
they sold out their money, and A found himself .£.10 richer
than B. Required the prices of the stocks.
Let X = the price of the three per cents.,
and 1/ = the price of the four's ;
900
.*. ar : 300 :: 3 : -^s interest =
and 1/ : 300 :: 4 : Rs interest =
X
1200
900 1200
" ^ " y '
Again, x\ x 4- 10 :: 300 : what A received when selling
300. (r 4- 10)
1 • xi_ T> ., 300.(2/4- 10)
and m the same way B received ^— '- ;
^ y
300. {x -V 10) _ 300. (y + 16)
X ~ 2/
, 300 ^ 300 '
and 30 + = 30 -f + 1
X y
300 _ 300
B B
+ 10,
3S6 Examples of the Solution of Problems
and h 4 = ( - — + i = ) from the first
900 , /900 . , _ \ 1200
y
equation ;
.', by transposition, 4 = ,
and 3/ = 75 ;
300
.'. - — = 6,
•I?
and consequently, a? = 60 ;
.-. the prices of the stocks were 60 and J 5, per cent.
29. ;f '500 was to be lent out at simple interest in two separate
sums, the smaller at 2 per cent, more than the other.
The interest of the greater sum was afterwards increased,
and that of the smaller sum diminished by 1 per cent
By this, the interest of the whole was augmented by
one-fourth of the former value. But if the interest of the
greater sum had been so increased, without any diminution
of the less, the interest of the whole would have been
increased one-third. What were the sums and the mteper
cent, of each ?
Let X =? the less sum ;
.*. 500 — jT = the greater ;
let ^ + 1 3= the interest of the less ;
.\ y — 1 = the interest of the greater,
.x.{y^\) , (500-x).(y-l) 0? , ^ ^ ^,
and — f^^ ^ + -^ / ^-^ . — ' = —. H- $ V - 5 =the
100 ^ 100 50 -^
former interest,
*°^ Tm "^ "^ — ' z=z by =: the second interest ;
b / X \
and .•. 4y ^ -^ -^ by -r- b;
X
by transposition, y = 5 — 77:^
producing Simple Equations. 187
Again, the third interest =±li^^±i} + (^OQ-^-y ^
* 100 ^ 100
3x ^ ^ Sx
or h 15 1/ = • h 20 2/ — 30 ;
100 ^ '^ 100 ^ -^ •
by transposition, (20-^ = ) 20--^ = 53/ = 25-^
from the former equation ;
• • 20 ^^
and X = 100;
also y + 1 =6— —=4;
•^ 50
.\ the sums were 100 and 400 pounds,
and the rates of interest ^.4 and ^.2^ respectively*
Sect. VIII.
Examples ojf the Solution of Problems producing pure
Equations. '
1. What two numbers are those, whose sum is to the greater
as 10 to 7; and whose sum multiphed by the less pro-
duces 270 ? ,
Let I Ox = their Bum ;
.'. Jx = 'the greater number,
and 3 J? = the less -,
whence 30 j?" = 270,
and a?* = 9 J
r, X = ± 3,
and the numbers are ± 21 and ± 9-
S. Th^re are two numbers in the proportion of 4 to 5, the
diflfcrence of whose squares is S 1 . What are those numbers ?
Let 4 X and 5 x = the numbers ;
then (25 J?* - 16a:' = ) 9a?' = 81;
.-. X' = 9.
and J? = ± 3,
and the numbers are 4- 12 and =fc 15,
3. What two numbers aje those, whose difference is to the
greater as 2 to 9, and the difference of whose squares is 128 ?
^Examples of the Solution of TrdblemSy Is^c. 189
Let 2 a? = their difference ;
.-. 9 ^ = the greater,
and 7 ^ = the less ;
.*. (81 x' - 49.1?^ =) 32 a:* = 128,
and J?- = 4;
.-. a: = ± 2,
and the numbers are +18 and +14.
4. A mercer bought a piece of silk for of.l6. 45.; and the
number of shillings which he paid for a yard was to the
number of yards as 4? : 9. How many yatds did he buy,
and what was the price of a yard ?
Let 4 ^ = the number of shilHngs he paid for a yard ;
.-. 9^7 = the number of yards,
and 36 a:* = (the price of the vsrhole =) 324;
.-. a;' = 9, .
and /. a? = + 3 ;
consequently there were 27 yards, at 12^. per yard.
4. It is required to divide the number 18 into two such parts,
that the squaees.of those parts may be in the proportion of
25 to 16.
Let X = the greater part ;
then 18 — X == the less;
/. x"- : 18 - xX :: 25 : 16,
and (^/g-. 188.) a? : 18 — of :: 5 : 4 ;
.-. X : 18 :: 5 : 9',
and .a? : 2 :: 5 : 1 ;
.'. X = 10,
and.the parts are 10 and B.
190 Examples of the Solution of Problems
i
5. Find three numbers in the proportion of i, -, and|; the sum
of whose squares is 724,
Reducing the fractions to a common denominator,
the required numbers will evidently be in the proportion
of 6, 8, and 9 ;
let /. 6x, Sx, and 9 ^, represent the numbers ;
then (36a?' + ^4^^ + 810?' =) 181 x" = 724 ;
.-. x" = 4,
and X =: ± 2y
and consequently, the numbers are ±12, ± l6,
and-f 18.
6. It is required to divide the number 14 into two such parts
that the quotient of the greater part divided by the less, may
be to the quotient of the less divided by the greater as
16 : 9.
Let X = the greater part ;
/. 14 ~ X = the less,
X 14 - a: ;^
: 9;
or x^ : 14 -xl* :: l6
: 9;
.-. {Alg. 188.) a: : 14 -a? :: 4
and X : 14 :: 4:7;
: 3,
.'. X : 2 :: 4 : 1,
and a? = 8;
the parts are 8 and 6.
7 What two numbers are those whose difference is to the
less as 4 to S ; and their product multiplied by the less is
equal to 504 ?
Let 4 a? = the difference ;
then 3 a? = the less.
producing pure Equations, 191
and 7 ^ = the greater ;
whence 63 j? = 504,
or x' = 8 ;
.-. a? = 2 ;
and the numbers are 14 and 6.
8. What two numbers are as 5 to 4, the sum of whose cubes
is 5103?
Let 5 X and 4 x = the numbers ;
.-. (I25a?3 + 64^ =) 189x3 =r 5103, '
and ar* = 27 ;
.-. X = 3,
and the numbers are 15 and 12.
A number of boys set out to rob aa orchard, each carrying
as many bags as there were boys in all, and each bag capable
of containing 4 times as many apples as there were boys.
They filled their bags, and found the number of apples was
2916. How many boys were there?
Let X = the number of boys ;
then jf = the number of bags,
and 4 a?' = the number of apples ;
.-. Ax' = 2916,
and a?3 = 729 j
.-. ^ = 9;
,*. there were 9 boys.
10. A person bought for one crown as many pounds of sugar,
as were equal to half the number of crowns he laid outi
In selling the sugar he received for every 25lbs. as many
192 Examples of the Solution of Problems
crawns as the whole had cost hi pa ; and he received on the
whole 20 crowns. How many crowna did he lay out, and
what did he give for a pound ?
Let X = the number of crowns he laid out ;
JC
.'. - = the number of lbs. for one crown,
and — = the number of lbs. in all,
2
and -- = the selling price of one lb. ;
X" X
2 25
and x"^ = 1000;
whence .r = 1 ;
.'. he laid out 10 crowns, and gave one shilling for a lb.
11. A detachment from an army was marching in regular
column with 5 men more in depth than in front ; but upon
the enemy coming in sight, the front was increased by 845
men ; and by this movement the detachment was drawn up
in five lines. Required the number of men.
Let X = the number in front ;
,'. gc jf 5 = the number in depth,
and jf + 5x = 5x4- 4225,
.-. X* = 4225,
and X ==. ± 65 ;
.-. the number of men= 5 x 4-4225 =4550, the negative
value not answering the conditions of the problem.
IC. A number of shillings were placed at equal distances on a
table so as to form the sides of an equilateral triangle ; then
from the middle of each side a number* of shillings, equal to
producing pure Eqttatkns. 1^3
the square root of the number in the side were taken, and
placed upon the corner shilling opposite to that side ; it then
appeared that the number on each sid^ Was to the number
previously upon it, as 5 to 4. Required the number of
shillings on one side ^t first.
Let a?* = the number;
then x"* + ^ : ^* ;: 5 : 4,
and * : a;* :: 1 : 4;
.*. X' = AXy
and J? = 4;
whence a?* = l6 = the number required.
13. A certain sum of money is divided every week among the
resident members of a corporation. It happened one week
that the number resident was the square root of the number
of pounds to be divided. Two men however, coming into
residence the week after, diminished the dividend of each
of the former individuals d^^.l. 6*. 8c?, What was the sum
to be divided ?
Let x^ = the number of pounds;
then X = the number of men resident, and also = the
sum each received. ,
Hence, x — - =
3 X + 2 '
or x* + 3^-3 « « ;
12 8
by transposition, - a? ,= - ,
o o
and X = 4 ;
.'. a'* = 16 = the sum required,
14. Two partners A and B dividing their gain (of.SO.), JB took
£J10 ; A^ money continued in trade 4 months, and if the
cc
i94f Examples ^the Sokiion ofTrohlems
number 50 be divided by A^s money, the quotient will give
the number of months that B's money, which was c£,100,
continued in trade. What was A% money, and how long
did ^'s money continue in trade ?
Suppose As money was x pounds ;
60
.•. — =the number of months -B's money was in trade^
and since B gained <£.20, A gained ^£.40 j
.'. Ax \ :: 2 : 1 ;
X
2500 , ,
or X : :: 1 : 1 ;
X
.-. x^ = 2500,
and a? = ± 50; ,
.•. As money was dC.50, and Rs money was one month
in trade.
15. Two workmen A and B were engaged to work for a certain
number of days at different rates. At the end of the time,
^who had played 4 of those days, had 75 shillings to receive ;
but B who played 7 of those days, received only 48 shillings.
Now had B only played 4 days, and A played 7 days, they
would have received exactly alike. For how many days were
they engaged ; how many did. each work, and what had
each per day ?
Let X = the number of days for which they were
engaged ;
.'. :r — 4 = the number A worked,
and a? - 7 = ^^e tiumber B worked,
and —. — = the number of shillings A received per day,
:c— 4
and = the number of shillings B received per day ;
x-7
producing pure Equaiions, 195
75 . (3? - 7) _ 48 , (j? - 4)
0?— 4 "" X — 7 '
and 25 .. x - 7I* = l6 . x - 4\' ;
/. 6 . (a: - 7) =: ± 4 . (a? -. 4) ;
17
.-. X = 19, or -^ ;
and .*. they were engaged to work 19 days,
and A worked 15, and J5 12 days,
and A received 5 shillings, and B 4 shillings per day.
16. Two travellers A and B set out to meet each other, A leav-
ing the town C at the same time that B left i>. They
travelled the direct road CD, and on meeting it appeared
that A had travelled 18 miles more than B; and that ^ could
have gone B^s journey in 154 days, but B would have been
28 days in performing A*s journey. What was the dis-
tance between C and D,
Let 0?= the number of miles A has travelled ;
.*. a; — 18 = the number B has travelled,
and X - 18 : a: :: 15| : the number of days J travelled
63 X
•"~4.(a? - 18) '
also X : X " 18 :: 28 ; the number of days B travelled
_ 28. (j? - 18) ^
— ———————— ;
X »
28 .(x - 18) _ 63 a?
•'• ~ "" 4 . (a? - 18) '
or 16. a? - is]"" = 9a;*;
/. 4 .(a? - 18) = ± 3a?,
and X = 72, or 10^ ;
whence A travelled 72, and B 54 miles ;
and /. the whole distance CD 126 miles.
19^ ExampUs of the Solution of Problems
n, A and B lay out some money on speculation. A disposes
of his bargain for £.\\y and gains as m\xc\v per cent, as B
lays out ; ^'s gain is of .36, and it appears that A gains four
times as mxxch per cent, as j5. Required the capital of each.
Let 4 X == -B's capital, and .*. A's gain per cent. ;
then X = B'% gain per cent._, -
and 100 : 4\r :: a: : 36;
.-. Ax' = 36 X 100,
a^d a?* = 9 x 100 ;
/. a? = ± 30,
and .-. -B's capital = 120,
and 220 : 100 :: 11 : A% capital = ~- = 5.
18. The Captain of a privateer descrying a trading vessel 7
miles ahead, sailed 20 miles in direct pursuit of her, and then
observing the trader steering in a direction perpendicular to
her former course, changed his own course so as to overtake
her without making another tack. On comparing their
reckonings it was found, that the privateer had run at the
rate of 10 knots in an hour, and the trading vessel at the
r^te of 8 knots in the; same time. Required the distance
sailed by the privateer.
Let A, B, he the original places of the privateer
and trader, E the point of concourse, D the place where
A B DC
E
the captain changed his course, CE being perpendicular
to^C. AB = 7, AB = 20.
Now (10 : 8 ::) 5 : 4 :; the velocity of the privateer
: Ihe velocity of the trader
•.: AD : BC :: 20 : BC;
producing pure Equations » 197
,. 5C=H0jLi= 16;
5
/. DC = 16 ^ 52> = i6 - 13 = 3,
and DE : CE :i ^ : 4. Let CE = x;
.-. VimF : :c :: 5 : 4,
and 9 + x" : x" :: 25 : l6;
/. 9 : a?* ::* 9 : l6 ;
/. x^ = l6^ and ^ = d= 4,
and .-. DE = 5, and AD + DJ5; = 25.
19. A vintner draws a certain quantity of wine out of a full
vessel that holds ^^BQ gallons ; and then filling the vessel
with water, draws oiF the same quantity of liquor as before,
and so on, for four draughts, when tliere were only 81
gallons of pure wine left. How much wine did he draw
each time?
Let X = the number of gallons drawn the first time ;
.•. 256 — J? = the quantity of wine lefl,
and 256 : 256. - x 11 x i the quantity of wine drawn
, , . X , (256 - x)
the second time = .. ;
^^^ X . (256 - x) 256 -TT) ^,
., 256 -. a: ,56> = 256 ' = '^'
quantity left after the second draught. ,
. a? = the quantity drawn
ha the same way, — ^.
^ the third time,
255* — x]
^^^ ^ - = the quantity left,
2561
198 Examples of the Solution of Problemi
256 - x\ 256 - x^
and — ^i(- . X and ,3 . " = the quantities drawn
^^^ ' 2561
and left the fourth time;
256 - A
whence .-,3 — = 81,
256)
and 256 - X = isel"" x 3 = 64 X 3 = 192;
.*. by transposition, 64 = x,
and the quantities drawn off each time were 64, 48, 36,
and 27 gallons.
201 What two numbers are those, whose difference multiplied
by the greater produces 40, and by the less 15 ?
Let X = the greater,
and y = the less ;
.'. x^ ^ xy ^ 40,
and xy - y * = 15;
,*. by subtraction, or* - 2a^ + ^* = 25,
and X - y = ± 5 ;
.'. from the first equation ± b x =■ 40,
and X = ± 8,
and from the second equation ± by ^ 15,
and 3/ = ± 3 ;
.'. the numbers are + 8, and ± 3.
ei. What two numbers are those whose difference multiplicd>
by the less produces 42, and by their sum 133 ?
Let X = the greater,
and y = the less j
••• {^ - y) ^y ^ 42,
and {x -- y) . (x + y) =z 133 ;
producing pure Equations, 199
by subtraction, {x - y).x =zQ\, subtracting the first
equation from this, (^ - ^) . (^ - 3/) = 49 ;
or X - y\ = 49 ;
.-. ^ - 3/ = ± 7.
whence + 7 ^ = 42,
and y = ± 6,
and ^ = ± 7 4- y = =t 13 ;
.-. tlie numbers are ± 13, and + 6.
22. In a mixture of rum and brandy, the difference between
the quantities of each is to the quantity of brandy as 100
is to the number of gallons of rum; and the same difference
is to the quantity of rum as 4 to the number of gallons of
brandy. How many gallons are there of each ?
Let X = the number of gallons of rum,
and 1/ = the number of gallons of brandy ;
.'. X " 1/ i y :: 100 : x,
and X \ X — y :: y : 4;
.*. ex oequaliy x i y \i 2tt y : x,
and 07" = 25 1/*;
.'. X = =*= byy the negative value not answering the
conditions of the problem.
Now from the second proportion by \Ay \\ y : 4 ;
or 5 : 1 :: ^ : 1 ;
.-. y—b,
and X = 25 ;
/. there are 25 gallons of rum, and 5 of brandy.
23. What two numbers are those whose difference being mul-
tiplied by the greater, and the product divided by the less,
quotes 24 ; but if their dilKerencebe multiplied by the less,
and the product divided by the greater, the quotient is 6?
wo Example J of the Solution of Problems
Let X = the greater,
and y = the less ;
X
then {x - y) .- = ± 24,
and (a: - 3/) . ^ = 6 ;
X
x"
dividing the first equation by the second, j-^ = 4 ;
or a? = ± 2y,
und .'. in the first case, {x -i^ = )3/= 12, and ^ = 24 ;
but if J? = — 23/; - 3 2^ X — 2=z24;
,-. y = 4, and a: = — 8.
>^*^^s»^S^^s»sA^s».#^
24' It is required to find two numbers such, that the product of
the greater and square root of the less may be equal to 48,
and the product of the less and square root of the greater
may be 36.
Let ^* and ^* be the two numbers j
.\ x*y = 48, and xy^ = 36;
48 , , 36
X ^ ^ ' y ^
4 3
or - = - ;
^ y .
?iX
.-. y = - ;
3 x^
whence =48,
and :r3 = 64 ; .
.'. X — 4,
and consequently, y =: 3 ;
/. the numbers are 16, and 9.
producing pure Equations, ^01
25. Find two numbers such that the square of the greater mul-
tiplied by the less may be equal to 448, and the square of
the less multiplied by the greater may be 392.
Let X = the greater and y == the less ;
then x'y^: 448, and xy"" = 392 ;
8 7
OK - = - ;
X y
Sy '
and consequently, -—• =s 392,
or -^ r= 49;
" .-. y^ = 343, and y — 7 \
.\ a: = 8 ;
.% the numbers are 8, and 7*
26. A and ^ carried 100 eggs between them to market, and
each received the same sum. If A had carried as many as
B he would have received 18 pence for them, and if B had
taken only as many as A, he would have received only 8
pence. How many had each ?
Let X = the number Al had,
and y = the number B had ;
18
then — = the price of one egg of A's (in pence);^
Q
and - = the price of one of B's ;
X
18 a? _ 8y
" y "^ X '
V D
S102 Examples ef the Solution of Problems ^
and 9 a:* = 4i/*;
.'. 3jj = ± 2«/, the negative value of which will not
answer the conditions of the problem.
3 X
Now (a? + y = ) a: H = 100;
Ji
/. (2j? + 3a: =) 5 j? = 200,
and a? = 40 ;
and /. y = 6o. /
27. What two numbers are those, which being both multiplied
by 27 the first product is a square, and the second the root
of that square: but being both multiplied by 3, the first
product is a cube, and the second the root of that cube.
Let X and y be the numbers ;
then j2Tx
=
27 y.
and .*. X
=
3//;
}
also 4/3*?
■^^
3y,
and /. X
=
9y';
whence 93/^
=
27y\
and y
=
3;
r. X = 27y^
=
243 ;
,\ the numbers
are
: 343,
and 3.
28. It is required to find the three sides of a right-angled tri-
angle from the following data. The number of square feet
. ia the area is equal to the number of feet in thehypothenuse
4. the sum in the other two sides ; and the square described
upon the hypothenuse is less than the square described upon
a line equal in length to the two sides, by half the product of
numbers representing the base and area.
Let X = the number of feet in the altitude,
and y = the number in the base ;
producing pure Equations, ^03
/. >v/x* + y^ := the number in the hypothenuse, {EucL
B.l. p. 47.)
and -^ = the area 5
2
also ^' H- 1/' = ( a? H- y Y - 1^:^ = ) J:"-l-2a;3/+3/*- |xy*;
.'. by transposition^ J j?y* = 2 a;^,
and 3/ = 8 ;
hence from the first equation, 4 j? = \/a?^ + 64 4- -^ + 8,
and by transposition, 3 x — 8 =^J x^ '\- QAy
.-. 9^'^ - 48 a: + 64 = j;' + 64,
and 8 a?* = 48 a? ;
.*. a? = 6;
whence the hypothenuse = ^64 + 36 = 10;
/. the sides are 6, 8, and 10 feet, respectively.
29. A farmer lias 2 cubical stacks of hay. The side of one is
3 yards longer than the side of the other ; and 'the difference
of their contents is 1 1 7 solid yards. Required the side of
each.
Let X = the side of the greater,
and y = the side of the less ;
.-. X' •- f ^ 117,
and X " y = 3 ;
cubing the latter equation, x^-3 x^-^S xy'^ — y^ = 27 ;
but x^ -y = 1175
.-. by subtraction, Sx'y — 3xy^ = 90,
and xy , {x ^ y) = 30,
or 3 ^^ = 30 ;
.% xy =10,
'IMH Examples if the Solution of Problems
Now a?* - 2j:y + 3/* = 9,
and Axy =40
.'. by addition, a?* -\- 2xy + y' = 49,
and a? 4- 1^ = ± 7 5
but X — y -=:■ 3 ;
.". by addition, 2 a: = 10, or - 4 ;
.-. J? = 5, or — 3,
and by subtraction, 2y = 4, or — 10;
.-. y = 2, or - 5,
and the sides of the stacks are 5, and 2 yards, respectively.
30. When a parish was inclosed, the allotment of on6 of the pro-
prietors consisted of two pieces of ground ; one of which
was in the form of a right-angled triangle ; the other was a
rectangle, one of the sides of which was equal to the hypo-
thenuse of the triangle, the other, to half the greater side ;
but wishing to have his land in one piece, he exchanged
his allotments for a square piece of ground of equal area, one
side of which equalled the greater of the sides of the triangle
which contained the right-angle. By this exchange he found
that he had saved ten poles of railing. What are the re-
spective areas of the triangle and rectangle ; and what is the
length of each of their sides ?
Let 2 07 = the greater side of the triangle,
and y = the less ;
r, ^ Ax^ -k-yt" = the hypothenuse; and also the greater
side of the rectangle,
and X = the less side of the rectangle ;
.*. xy= the area of the triangle,
and Xs/^ ^^ + ^* = the area of the rectangle ;
.'. 4x^ = xy + x^ 4x^ ^y"" ,
or 4 X - y — s/^^'+^^i
also 8«4-10 = 2 J7+y + >/4a?^ 4- «/*+ 2 a: 4-2^^4 a?* -f^%
producing pure Equations. 205
OX Ax + 10 =3/ + 3^T^M-l/^; in which equation sub-
stituting the vaUie of /y/4a;* + y^ found above;
.-. 4x + 10 = 2/ H- 3 {Ax - ^) = 12x — 2t/;
.*. by transposition, 2^ = 8,r — 10,
and y ^ 4x — b;
,\ from the first equation, 5 = \/ 4 x^ -+• 4 a? — 5l%
and 25 = 4 X* + l6 a?* - 40 j? + 25 ;
by transposition, 40 a: = 20 a?* ;
.*. 2=0?,
and i/ = 4a?-5=3;
.". the sides of the triangle are 3, 4, and 5 ; the sides of
the rectangle are 2, and 5 ; and the areas of the triangle
and rectangle are 6, and 10, respectively.
Sect. IX
Examples of the Solution of Problems producing
Adfected Quadratic Equations.
1. A Merchant sold a quantity of brandy for <£.39, and
gained as much per cent, as the brandy cost him. What
was the price of the br^dy ?
Let X = the price of the brandy ;
X*
then 100 I X :: X I the eain = ,
^ 100 '
or x'' = 3900 - lOOx;
by transposition, a?* + 100^ = 390O,
completing the square, x'' + 100j? + 5o1* = 3900 + 2600
= 6400;
extracting the root, a? + 50= ±80;
.-. X = 30, or - 130;
.-. the price was ^.30.
2. There are two numbers whose difference is 9, and their sum
multiplied by the greater produces 266» What are those
/ numbers ?
Let a? = the greater;
.-. a? — 9 = the less,
and x.{2x - g) = 266;
, 9 _ 266
Examples of the Solution of Problems y ^c. 207
- . , ^ 9 81 266 81 2209
completing the square, -^ 7^-^ + rr. = ~5~ ■*" TS "^ TfT* '
9 47
extracting the root, a? — - = i T '
.'. a; = 14, or -;
^ 2
37
X - 9 = 5, or - —
and both values answer the conditions of the problem-
S, It is required to find two numbers, the first of which may
be to the second as the second is to 16 ; and the sum of the
squares of the numbers may^be equal to 225.
Let X — the first number ;
.-. A^ iGx ^ the second,
and a?* + 16 X = 225 ;
completing the square, x^ -\-l6x + 64 = 225 -f- 64 = 289 5
extracting the root, j? + 8 = + 17;
/. X = 9, or — 25 ;
but as this latter value of x makes the second number
an impossible quantity, 9 is the only value of x which
answers the conditions, and therefore the numbers are
9 and 12.
Bought two sorts of linen for 6 crowns. 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) were
at such a price that 8 ells cost as many shillings as 1 ell
of the finer. How many ells were, there of the finei*,
and what was the value of each price?
Let X = the number of ells of the finer ;
.-, x^ = the price of the finer (in shillings,)
208 EKomples ofth Solution of Problems
7 X
and 8 : 28 :: a: : the price of the coarser = — ;
7 X
.-. X* 4- Y = 3^5
7 49 49 529
completing the square, a:^ + -a7 + -^=30 +rb= TF* 5
7 23
extracting the root, ^ -f - = d: — ,
and J? = 4, or — — ,
' 2 '
and .-. the price of the finer = 16 shilHngs, and of the
coarser =14 shillings.
5. Two partners^ and B gained of. 18 by trade, ^s money
was in trade 12 months, and he received for his principal
and gain £M. Also B*s money, which was of.SO, was in
trade 16 months. What money did A put into trade?
Let X = the number of pounds he put in ;
/. 26 — X = the number he gained,
and 12cr + 16 x 30 : 12x :; 18 : 26 - x;
,\ X -{- 40 : X :: 18 : 26 — x,
and 18j? =. 1040 - 14 x - x' ;
by transposition, xf* + 32j? = 1040;
completing the square, x* + 32 j? + 16]* = 1040+ 256
= 1296;
extracting the root, ^ + l6 = +36;
.-. a? = 20, or — ^ — r2.
and consequently A put £,20 into trade.
6. A person bought some sheep for o£.72 ; and found that
if he had bought 6 more for the same money, he would
have paid £.1 less for each. How many did he buy, and
what was the price of each ?
pricing Adfected Quadratic Equations, ^0
Let X r= the number of sheep bought ;
72 .
then — = the price of one (in pounds^)
and -A = the price of one, if he had bought 6 more;
72 . , 72
X
Ftr^ + ^ = ^'
/. 72^ + a?* + 6 a? = 72 J? + 432;
.*. oj* + 6 J? = 432 i
completing the square, a?^ + 6 a? + 9" = 441 ;
extracting the root, a: + 3 = ± 21,
and X = 18, or - 24,
72
and ,•. he bought 18, and the price of one = j-
=s 4 pounds.
?• The plate of a looking-glass is 18 inches by \% and is to be
framed with a frame of equal width, whose area is to be equal
to that of the glass. Required the width of the frame.
The area of the glass = 12 x 18 = 2l6.
Let X = the width of the frame (in inches ;)
then the areaof the frame = (18 +2a?).(l2 +2a;)-.2l6,
and .-. (18 + 2 x) . (12 + 2 a;) ~ 2l6 = 2l6,
or 4a?* + 60a7 t= 216,
and a?* + 15 a? = 54;
15
completing the square, a;' 4- 15 a? + *r
441
225
= 54 -h — -
4
15 21
extracting the root, a? + — = ± -g" 5
.'. a? = 3, or — 18,
and /. the width must be 3 inches.
E E
210 ExampUs of the Solution of ProUems
8. There are two square buildings, that are paved with stones, a
foot square each. The side of one building exceeds that of
the other by 12 feet, and both their pavements taken toge-
ther contain 2120 stones. What are the lengths of them
separately ?
Let J? and x + 1 2 = the number of feet in the isides of
each;
••. jp* and a?-Hl2l* = the number of stones in the squares,
and a:* + J?* + 24a: + 144 = 2120;
by transposition, 2:c*+ 24 J7 = 1976,
or 0?* + 12 a: = 988 ;
completing the square, a:* + 12j: + 36 = 988+36= 1024 ;
extracting the root, a: + 6 = ± 32 ;^
.-. 0? = 26, or - 38,
whence the lengths are 26, and 38 feet, respectively.
9. A labourer dug two trenches, one of which was 6 yards
longer than the otlier, for £a1. 16s. and the digging of
each of them cost as many shillings per yard as there were
. yards io its length. What was the length of each?
Let X and x + 6 = the number of yards in each ;
.'. x' + X +6\" = 356 shillings,
or 2 a* -f 12x + 36 = 356;
by transposition, 2 x^ 4- 12i: = 320;
-or a?* -h ^x = 160 ;
completing the square, x* + 60" -|- 9 = 169 ;
extracting the root, x +3 = ±13,
and X = 10, or — 16; N
.'. the lengths were 10, and 16 vards.
10. A company at a tavern had «£.8. 155. to pay ; but before
the bill was paid, two of them sneaked off, when those who
remained had each 10 shillings more to pay. How many
were in the company at first \
producing Adjected Quadratic Equations. 211
Let X = the number;
175
then — ^ = the number of shilHngs each had to pay at first,
175
and - — - =r the number each had to pay, after two had
sneaked off;
X— 2 X \X — 2 X/
and .-. 10. J?, (j? - 2) = 175 X 2,
or a?* — 2x =r 35 ;
completing the square, a?' — 2a:+l=36f
extracting the root, a? — 1 = ± 6,
and .'.a; = 7> or — 5 ;
consequently there were 7 at first.
11. A grazier bought as many sheep as cost him £,60 ; out of
which he reserved 15, and sold the remainder for £.54i,
gaining 2 shillings a head by them. How many sheep did
he buy, and what was the price of each ?
Let X = the number ;
.'. -^ = the price of each in pounds.
and
<'-")-(j + n;)-"^
or {x — 15) . (600 4- x) = 540 2?,
and x^ + 585 J? — 90OO = 540^;
by transposition, a:* + 45 a? = 900O ;
n2
45 I
completing the square, a;* + 45x-!---- = 9000 +
2025 38025
212 Examples of the Zohitton of TrohUms
45 195
extracting the root, a: + — = ± -r^ ,
and X = 7^? or - 120 ;
and .'. the number bought was 75,
4
and the price = - ^. = 16 shillings.
12. j4 and J? set out from two towns which were at the distance
of 247 miles, and travelled the direct road till they met. A
went 9 miles a day ; and the number of days, at the end of
which they met, was greater by S than the number of milei
which B went in a day. How many miles did each go ?
Let X = the number of days they travelled ;
,\ Qx = the number of miles A went,
and 247 - 9 X = the number B went, ,
247 — Q X
and = the number B went per day;
and X* — 3x = 247 — 9a?;
by transposition, x^ + 6 x = 247 ;
completing the square, a?' + 6 j? + 9 = 256,
extracting the root, x -{- 3 = ± 16,
and X = 13, or — 19,
and .•. j4 went II7, and B 130 miles.
13. A person bought two pieces of cloth of different sorts ;
whereof the finer cost 4 shillings a yard more than the other;
for the finer he paid c£*. 18; but the coarser, which exceeded
the finer in length by 2 yards, cost only <£.16. How many
yards were there in each piece, and what was the price of
a yard of each ?
producing Adfeded Quadratic Equations, 2 1 3
Let X = the number of yards of the finer ;
.*. a? -f 2 = the number of yards of the coarser,
18
and — «. = the price of a yard of the finer (in pounds) ;
also = the price of a yard of the coarser ;
18 _ l6 1
" X " X -\' 2 S'
and 90 jr + 180 = SOo? + a;* 4- 2 ar ;
by transposition, x* -- 8 0? = 1 80 ;
completing the square, x'^ — ^x + l6= 1 80 + l6 = l 96;
extracting the root, a? — 4 = ± 14 ;
/. X = 18, or - 10,
consequently, there were 18 yards of the finer, and 20
of the coarser; and the prices were d£»l, and 1 6 shillings,
respectively.
\^, A set out from C towards D, and travelled 7 miles a day.
After he bad gone 32 miles, B set out from D towards C,
and went every day-rrth of the whole journey ; and after he
had travelled as many days as he went miles in one day, he
met A, Required the distance of the places C and Z).
Suppose the distance was x miles ;
X
.'. — = the number of miles B travelled per day ; and
also = the number of days he travelled before he met A,
x"^ *7 X
by transposition, ^ — = - 33 ;
X^ 1 Q /p
completing the square^ -^ |-36 = 36-32=s4;
214? Examples of the Solution of Problems
X
extracting the root, —• - 6 = ± 2 ;
X •
.-. — = 8, or 4,
19
and X = 152, or "jQ, both whicli values answer the con-
ditions of the problem. The disiauce therefore of Cfrom
D was 152, or ^6 miles.
15. A and B sold 130 ells of silk, (of which 4-0 ells were vfs,
and 90 jS's) for 42 crowns. Now ^ sold for a crown one-
third of an ell more than B did. How many ells did each
'• sell for a crown.
Let X = the number B sold ^
.-.« + != the number ^ «,ld ( '" ' «""""'
5
QO
and j: : go :: 1 : the price of 90 ells = — ,
X
1 , . r 11 120
and X + - : 40 :: I : the price of 40 elis = - — -— -
3 ax "X" 1
X 3 J7 +. 1 '
or 7 = — +
15 20
com
X 3a? + 1 '
whence, 21 a:' + 7x = 45 a: 4- 15 + 20x;
by transposition, 21 r* — 58j: = 15,
, ^ 58 15
p,e.„,.,.e^„.....-£5...|-.L5+-=-t
extracting the root, a? — -^ = ±
29 34
21
9
j?rodticw^ A dfected Quadratic Equations, 215
5
.-. X = 3, or - — ;
whence, B sold 3 ells, and\^ 3 J, for a crown.
16, Three merchants ^, B, and C, made a joint stock, by which
ihey gained a sum less than that stock by c£.80. A'% share
of the gain was £.00 ; and his contribution to the stock v/as
•i*. 17 more than ^'s. Also B and C together contributed
£,325. How much did each contribute ?
Let X = the number of pounds that j4 contributed ;
,\x- 17 = the number that B contributed,
and 325 - (x- 17) = 342 - a? = the number that Q
contributed ;
•. 325 -f X = the whole stock,
and 3 25 -f x — 80 = 245 -f j? = the whole gain;
.-. 325 4- ^ : ^ :: 245 -f 0? : 60,
and a?* + 245 x = 6ox + 1950O;
by transposition, a?' + 185 x = 19500 ;
= 19500
^\ ^ 185
completing the square, rr* -f 185 ^ H
2i
?>A12b 112225
185 335
extractmg the root, x -{ — ~- = ± -— ,
2 2
and 0? = 7^> <^r - 260 ;
.*. the stocks of A^ B and Cwere 75, 58, and 267 pounds,
respectively.
17. The joint stock of 2 partners A and B was £A\Q, A\
money was in trade 9 months, and B\Q months : when they
shared stock and gain, A received «£.228, and B £.152.
What was each man's stock ?
2l6 Examples of the Solat'ion of Problems
Let X = A'^ stock ;
.*. 228 — jr = his gain ;
also 4l6 - x = jB's stock,
iv+B^f^*^ and X — 164 = his gain ;
-X^-^[X'Uu)^ iu. and .•. 64 = the whole gain,
and 9 J? -f 6. (4l6 ~ x) :Qx :: 64 : 228 - x;
or 3x -h 2.(41(5 - x):3x :: 64: 228 - a?;
.-. 192x = (.r+ 832) .(228 -^) = 189696 -604 x-.a?»;
by transposition, j?* + 796 a? = 1 89696;
completing the square, o?^ + 796 x 4- 398]' = 1 89696
+ 158404 = 348100;
extracting the root, x + 398 = ± 590,
w and X = 192, or — 988;
.-. the stocks were c£.193, and i;.224.
18, A body of men were formed into a hollow square, three deep,
when it was observed that with the addition of 25 to their
number, a solid square might be formed, of which the
number of men in each side would be greater by 22 than the
square root of the number of men in each side of the hollow
square. Required the number of men in the hollow square.
Let X = the number of men in a side of the hollow
square ;
.'. X* — X — 61* = the whole number of men,
7 and x^ - X — 6\^ + 25 = ^M-22l%
or 12 a: - 36 + 25 = a? + 44a:'^ + 484 i
.*. by transposition, 1 1 a? — 44 a?^ = 495 ;
or j; — 4 a?i = 45 ;
completing the square, ^ — 4 a:J + 4 =: 49 ;
extracting the root, j?2 — 2 = Hh 7 i
.-. xi = 9, or — 5,
producing Adfected Quadratic Equations • SI 7
and X = 81, or 25,
and .'. the whole number = 936.
'^■^ ^ ■f^**'**'^^**^*
19« A mercer bought a number of pieces of t^'^o different kinds
of silk for ,£.92. 35. There were as many pieces bought of
each kind, and as many shillings paid per yard for them as a
piece of that kind contained yards. Now 2 pieces, one of
each kind, together measured 19 yards. How many yards
were there in each t
Let a: = the number of yards in one piece ; and .-. =the
number of pieces, and also the number of shilHngs 'per
yard;
.-. 19 - a? = the number in the other,
and x^, and 19 - o^ = the whole prices of each kind ;
.-. a?' + 19-^3 ^ 1843^
or 57 a;' - 1083 jr + 6859 = 1843;
by transposition, 57 a;* - 1083 a? = - 50l6;
or x^ — 19a: = - 88 ;
= 2^-ss4
completing the square^ a?* — 19 a? + —
2i
IQ 3
extracting the root, x ^ = + - ;
.*. a: = 11, or 8;
.•. 35 — a: = 8, or U,
both which values answer the conditions of the problem ;
.*. there were 11 yards in one, and 8 in the other.
20. A square court-yard has a rectangular gravel-walk round it.
The side of the court wants 2 yards of being ^ times the
breadth of the gravel-walk ; and the number of square yarda
in the walk exceeds the number of yards in the periphery
of the court by 92. Required the area of th« court.
F F
218 Examples of the Solution of Problems
Let X = the breadth of the walk (in yards,)
.*. 6 a? — 2 = the side of the court,
and Ax — 2 = the side of the interior square ;
.•. 6 a? — 2]* — 4x - 2f = the area of the walk,
and 20 1?- — 8 a? — 92 = 4 X (6 a? - 2 ;)
by transposition, 20a?* — 32a? = 84;
, 8 21
.'. X a? = — ;
5 5 '
completing the square, x'--a?+ — = — + ^^ = —
4 11
extracting the root, a? — - = ± — ;
.-. ^ = 3, or - ?,
and 6 a? - 2f =T6|' = 206, the area required.
a I. A merchant bought 54 gallons of Coniac brandy, and a
certain quanlity of British. For the former he gave half
as many shillings />er gallon as there were gallons of British,
and for the latter 4 shillings per gallon less. He sold the
mixture at 10 shillings per gallon, and lost £,28- l6s. by
his bargain. Required the price of the Coniac, and the
number of gallons of British.
Let 2 0? = the number of gallons of British ;
.*. 0? = the number of shillings one gallon of
Coniac cost,
and 54 a: = the price of all the Coniac;
also a? — 4 = the number of shillings one gallon of
British cost,
and 2 J?* — 8 a? = the price of all the brandy ;
.-. 2 a?* - 8 a? 4- 54a? = 10 . 54 + 2 a? + 576;
by transposition, 2a?^ + 26 a? = 11 16,
or o:^ 4- 13 a? = 558 ;
j?roducmg Ad/ected Quadratic Equations. 219
completing the square, x'^ + 13 a: -f — | = 558 +
169 2401
, extracting the root, x -{ =» + — ,
and 0? = 18, or — 31 ;
,\ he bought 36 gallons of British ; the Coniac cost 18
shilHngs per gallon,
and .*. the whole price = «£.48. 12^.
22. During the time that the shadow on a sun-dial, which shews
true time, moves from one o'clock to five, a clock, which
is too fast a certain number of hours and minutes, strikes
a number of strokes = that number of hours and minutes,
and it is observed that the number of minutes is less by
41 than the square of the number which the clock strikes
at the last time of striking. The clock does not strike
twelve during the time. How much is it too fast?
Let X = the number of hours too fast ;
then the clock strikes, {x + 2) + (x + 3) + {x + 4)
+ (a; + 5) times = 4 07 + 14,
and the number of minutes = o?"^ + lOa? -f 25 — 41 =
07* + 10 0? - 16;
.-. X + x^ + lOx " 16 =z 4x + 14;
by transposition, o?"* -f 7 ^ = 30 ;
tI^ 49 169
completing the square, x^ + 7x i-^\ = 30 + — = j
■* [ 4^ 4
7 13 ' -
extracting the root, x + -- = ± — ;
J/ 2
.'. 0? = 3, or — 10,
and the number of minutes = 23 ;
.*. the clock is too fast 3 hours and 23 minutes. -
220 Examples of the Solution of Problems
23. A vinter sold 7 dozen of sherry and 12 dozen of claret
for £,50. He sold 3 dozen more of sherry foro£*.10 than
he did of claret for £.Q, Required the price of each.
Let X = the price of a dozen of sherry (in pounds ;)
.'. X : 10 :: 1 : the number of dozens of sherry for
.£.10, = ^,
X
and 3 == ^ — = the number of dozens of
X X
claret for ^.6;
10 - 3x
; 1 :: 6 : the price of a dozen of claret
6x
10 - 3 J?'
72 X
and 70x — 21x* + 72x = 500 — 150 .r;
by transposition, 292 r — 2\ x^ = 500,
292 500
or ^-.— .x = ^_;
completmg the square, or — i^.x + — --| = ,. .^
__ 10816
21»
146 104
extractmg the root, x — = ± -— ,
and a: = 2, or ;
* 21
/. the price of a dozen of sherry was ^.2, and the price
of a dozen of claret = = £,,3.
10 — 30?
24. A and B hired a pasture into which A put 4 horses, and
B as many as cost him 18 shillings a week. Afterwards
producing Adfected Quadratic Equations. 22 1
B put in two additional horses, and found that he must pay
20 shillings a week. At what rate was the pasture hired ?
Let X = the number of J5's horses at first ;
18
then — = the pay of each per week (in shillings ;)
72
/. — = what A paid,
72
and — -f 18^= the price of the pasture ;
also X 4- 6 = th6 whole number of horses in the
second case ;
72
.-. 0? + 6 : a?+ 2 :: — + 18 : 20 :: 72 + 18 a? : 20 x;
X
.-. 20 X* + 120 a? = 18 x^ + 108 a? + 144 ;
by transposition, 2 x^ + \2x = 144,
or a;' 4- 6 07 = 72 ;
completing the square, x"" 4-6a?-}-9= 72 + 9 = 81;
extracting the root, x + 3 =:> ±9^
and 07 = 6, or — 12;
72
•. B had 6 horses in the pasture at first, and [-18
X
= 30 shiUings, per week, was the price of the pasture.
25. An upholsterer has two square carpets divided into square
yards by the lines of the pattern. Now he observes, that if
he subtracts from the number of squares in the smaller carpet,
the number of yards in the side of the other, the square
of the remainder will exceed the difference of the number
of squares in the smaller carpet, and the number of yards
in its side, by 88. Also the difference of the lengths of
the sides of the carpets is 6 feet. Required the size of
each carpet.
Let X = the number of yards in a side of the less ;
.*. 07 + 2 = the number in a side of the greater.
222 Examples of the So/ution of Problems
and a:*— a?- 2® = a;' — ^4-88=^*-.^- 2 +90;
by transposition, .r* - x— 2|* — (x'' — J? - 2) = 90 ;
completing the square, a?**— ^ — 2)* — (x* — a; — 2) + ^
I . 361
extracting the root, x*--a?— 2—-=±-~,
and J? * — a: = 12, or — 7, the former of which only will
give a possible value of x; and .-.
completing the square, x^ — x -{• - = 12+-=-^;
1 7
extractmg the root, x = + - ,
2 2
and J? = 4, or - 3 ;
consequently the carpets contain 16 and 36 square yards,
respectively.
26. A man playing at hazard won at the first throw, as much
money as he had in his pocket ; at the second throw, he won
5 shillings more than the square root of w^hat he then had ;
at the third thiow, he won the square of all he then had ;
and then he had £.\IQ, l6s. What had he at first ?
Let X = the number of shillings he had at first ; /
. .'. 2j?= the number he had after the first throw,
^2 x-\-5 = the number won the second throw,
and 2 X + Ay2x + 5 = the number he had after the
second throw ;
also 2x + ^2x -{- bY = the number won the third
throw ;
.-. 2x + ^2x-{-5'(' + (2X + ^2x + 5) = 2256';
compK the square, 2 J7-|->>/2x + 5l' + (2 a: + »/2i + 5) +.
producing Adfected Qitadratic Equations, ^€3
= 2256+1 = 9-!^;
4 4 '
2 05
extracting the root, 2a?+^2a; + 5 + -=+ •,
.-. 2j? + >/2T= 42, or - 53, the latter of which
gives impossible values of .r, and .*. 2aj + /^2x = A2,
to answer the conditions of the problem ;
. 1 1 169
completing the square, 2X'\-^ 2x + -=42+- = — - ;
. 1 13
extracting the root, v2x + -=± ;
2 2
.-. sj 2x =6, or - 7,
and 2 a? = 36, or 49 ;
49
.-. J? = 18, or ~ ;
and consequently he began with 18 shilHngs.
27. What number is that, which being divided by the product
of its two digits, the quotient is 2, and if 27 be added to it,
the digits will be inverted ?
Let X and y be the digits ;
/. \0x •\- y ^=- the number,
and ^ = 2 ;
/. 10 J? +2/ = 2 xy;
also lOcT + «/ + 27 = 10?/ 4-^;
by transposition, 9 .3? + 27 = 9 3/,
or ^ + 3 = 3/ ;
which value of 3/ being substituted in the first equation.
\0x :{- X + 3 — 2x ,{x + 3),
or 1 1 X -f 3 = 2 0?" -f 6 1' ,•
by transposition, 2 x"^ — b x — 3.
5.3
5-^ = i^
€24 Examples of the Solution of Problems
w .1 ^ 5 . 25 3 , 25 49
completing the square, x^ --j?4- -^ = n"'"7S~7S'
. , 5 7
extracting the root, x— - = ± -: >
and X = 3, or — , the first of which only answers the
conditions ; .•. ^ = 6, and the number is 36.
58. There are three numbers, the difference of whose differences
is 8; their sum is 41; and the sum of their squares 699.
What are the numbers ?
Let X = the second number,
and 2/ = the difference of the second and tiie least ;
.*. X — J/, X, and x + y + S are the numbers,
and their sum =3j?-|-8=41j
by transposition, 3 x = 33,
and a: = 11 ;
.-.11 - ^V + 121 + 19 + ^V = 699,
or 603 + 163/ + 2.y* = 699;
by transposition, 2i/* + l6^ = 96,
and y" + Sy =48;
completing the square, 3/* + 83/ + 16 = 64 ;
extracting the root, 3/4- 4 = + 8,
and 3/ = 4, or - 12, both which values answer the
conditions ; and the numbers are 7^ H^ ^^^^ 23.
29. There ^re three numbers, the difference of whose differences
is 5 ; their sum is 44 ; and continual product is 1950. What
are the numbers ? ,
Let X = the second number,
and y =s the difference of the second and the least;
producing Adfected Quadratic Equations. 225
.•. the numbers are oc - y, x, x + y + 5,
and {x - y + X -{-' X +3/ + 5 =) 3 x -j- b =44;
by transposition, 3x = 39,
and J? = 13;
.-. (13 - 3/). 13. (18 +3^) = 1950,
and (13 - ^).(18 +3/) = 150,
or 234 - 53/ - i/* = 150;
r.y' + 53/ = 84;
, . , 25 25 361
completing the square, y* + 5 ^ H = 84 -\ = ~T"'
1 5 19
extractmg the root, 3/ -f - = ± — ^
and y ^ 7) ov — 12, both which values answer the con-
ditions ; /. the numbers are 6, 13, and 25.
30. There were two rows of counters, of which the upper row
exceeded the lower by one. A certain number having been
taken from the upper row, and as matiy as then remained
from the lower row, it was found that the square of the
number remaining in the lower row, added to the square root
of that number, is equal to 72 divided by the excess of the
number taken from the upper row above unity. Required
the number of counters taken from the upper row.
Let X + 1 = the number in the upper row ;
/. X = the number in the lower,
y -f 1 = the number taken from the upper row ;
,'. X — y — the remainder in the upper row,
and y = the remainder in the lower row ;
-'./+ sfy = —y
and 3/' .+ 2r = 72 ;
1 1 289
completing the square, y^^y^ +2 =72 + -== —7— 5
G G
226 Examples of the Solution of Problems
,1 17
extracting the root, ^/"^ + - = ± '5' »
.-. 1^ = 8, or - 9,
and y^ = 64, or 81";
,-. y = 4, or f^^, the latter of which is excluded by
the nature of the question.
31, A grocer sold 80 pounds of mace, and 100 pounds of cloves
for ^.65 ; but he sold 60 pounds more of cloves for c£.20,
than he did of mace for <£. 10. What was the price of a
pound of each ?
Let X = the price of a lb. of mace (in pounds),
and y = the price of a lb. of cloves ;
then X : lO :: 1 : the number of lbs. of mace for ^.10 = — .
X
20
In the same way — =thenumberoflbs. of cloves for ^.20;
20 ^^ , 10
.-. — = 60 + — ,
2/ ^
2 ^ 1 6x + I
or - = 6 + - = .
y XX
. 2X
and .-. y = ^ — — .
^ 6x -\- I
Again, 80 a? + 100 y = 65,
or 16 X + 20y = 13;
n 40 J?
and g6x^ + 560? = 78 j? + 13;
by transposition, 960?* — 22 j? ^ 13,
producing Adjected Quadratic "Equations, 227
completing the square, ^' - 2£ . a: + ^* = =y, +
13 _ 1369
extracting the root, a? — —r. = + -— :
1 13
and .•.!? = -, or - -—, which last does
not answer the conditions ; and 3/ = - .
/. the price of a pound of mace is 10 shillings, and of
a pound of cloves, is 5 shillings.
S2. A and B engage to reap a field for £A. IO5. ; and as
A alone could reap it in 9 days, they promise to complete
it in 5 days. They found however that they were obliged
to call in C, an inferior workman, to assist them for the
two last days, in consequence of which B received Z^, 9d,
less than he otherwise would have done. In what time
could B or C, alone reap the fiejd ?
Let X = the number of days in which B could reap the field,
and t/ = the number iu which C could reap it ;
then - + - : ^ :: 90 : the number of shillings jB would
l) X X
have received =
9+ ^'
5 450
and - X 90 = = the humber he did receive :
XX
8IO' 450 _ ^, 15
— 3J = ---,
g -i- X ^x ^ 4
54 30 1
or — = - ;
9 + « ^ 4
223 JSf(mp^s ofthi Solut'm of Problems
.-. 2lGx - 1080 - \20x = 90? + .r';
by transposition, x' - 8? j? = — 1080;
7569 7569
completing the square, x' — 871: + -- — ^ = -
3249
1080 =
4
87 57
extracting the root, x — -i- =r ± -—-,
2 2
and a; » 72, or 15.
» ,552
Let X = 15, then - -h -? + - = 1 ;
9 . 15 3^
u 4. V 2 , >5 1 1
by transposition, - = 1----=-;
.'. y = 18 the number of days in
which C could reap the field. The other value of x
is excluded by the nature of the question.
33. ThroT?ping out the three court cards from a suit of spades,
and placing the remainder in two heaps, I find the supi
of the pips iu the smaller heap is to the sum in the greater,
as the number of cards in the greater heap is to the number
of cards in the smaller. But if I add the seventh card
to the smaller heap, the diflference of the number of pips
in the two heaps is equal the square of the number of
cards in the smaller. Required the number of pips and
cards in each.
The whole number of cards =10,
and the whole number of pips = 55.
If .-. X = the number of cards in the larger heap;
\0 - X = the number in the smaller,
and if 1/ = the number of pips in the smaller ;
bo — 1/ := the number in the larger ;
.\ y '. bb — y :: x : 10 — J7,
and {Alg, 179.) y : 55 :: j? : 10;
. jjroducing Adjected Quadratic EquatiofU, 2S1D
or y : 1 1 :: 0? : 2 ;
.'. 2 1/ = II X.
Again, after the change, ^ + 7 = the number of pips
in the smaller heap,
and 55 — 3/ — 7 = 48 — 1^ = the number of pips in
the larger heap,
and .*. their difference = 23/ - 41 = 11 - ^''5
and by substituting for 2 1/ its value, 1 1 a? — 41 = 11 —il
=r 121 — 22j? + x\
and .-. by transposition, ^* - 33 J? = — l62;
' _ 1089 _
"~ 4
33
completing the square, x* — 33 a; -i
2
162 = -T-i
4
33 21
extracting the root, x — "-^ = ± — ,
2 2
and J? = 6, or 27 ; but 27 being inconsistent with the
nature of the problem, the number of cards in the larger
heap = 6, and .*. the number in the smaller heap = 4 ;
ll.r
and the number of pips in the smaller = *~- = 33,
and .-. the number in the greater heap =22.
34. The fore- wheel of a carriage makes 6 revolutions more
than the hind-wheel in going 120 yards; but if the
periphery of each wheel be increased one yard, \i will
make only 4 revolutions more than the hind-whjpel in the
same space. Required the circumference of each.
Let X = the number of yards in the circumference of
the larger,
and y = the number in the circumference of the less;
^, 120 120 ^
then = D,
« y
250 Examples of the Solution of Prohlems
or 20?/ = 20 a; — xy,
,\ by transposition, xy = 20 x — 20 y.
. . 120 120
Again, = -^ 4,
^ ' a: 4- 1 3^ 4- 1
or 30 . (^ + 1) = (x + 1) . (29 - y,)
or 30y + 30 = 29 x + 29 - 0:3/ - i/;
by transposition, o'y = 29 a? — 1 — 31y,
and .-. 29a: - 1 - Sly r= 20x - 20y;
by transposition, 90?= 11^ + 1,
11 y + 1
or a: = — i^ ;
9
I u ..i ..• 1 1 V^ + y 220 V 4- 20
.-. by substitution, — '-^ ^ = sS: — L-^ — 20y,
y . y
or 11^' + .V = 220?/ + 20 — 180y = AOy + 20;
, _ 39 __ 20
y FT '^ " 11 •
,3
1521
— :« +
completing the square, ?/* — -^ . ^ + -^
extracting the root, y = ± -^ ;
5
.*. y = 4, or -r — ;
•^ ' 11
/. the number of yards in the circumference of the less
= 4, and the number in the circumference of the greater
35. On the late jubilee, a gentleman treated his tenantry at the
following rate. He allowed for each poor child a certain num-
ber of sixpences, for eacH poor woman sixpence more, and
for each poor man sixpence still in addition. The number
producing Adjected Quadratic Equations. 231
of women was one-fourth greater than the number of men ;
the number of children was equal to twice the square of the
difference between the numbers of men and women ; and the
whole expence was c£.8. 2s. But had each child been
allowed as much as each woman, the expence on their
account added to nine times the difference of what the men
and women cost, would have been £A. IS*. Required the
number of men, women, and children, and the allotment to
each.
Let 4 J? = the number of men ;
.'. 5x = the number of women,
and 2 j?*= the number of children.
Let y = the number of sixpences each child had ;
.". y-\-\ = the number each woman had,
and y-\-2 = the number each man had ;
.-. 2x*^ + Qxy +13 0? = 324;
B\so2x^y +2x^'\-9xy- 27 x = 196;
.*. by subtraction, 2 o?* — 40 x = - 128,
and 0?* — 20x =z —64;
completing the square, o?^— 20 cT + 100= 100— 64 = 36 ;
extracting the root, x — 10 = ± 6^
and 0? = 16, or 4, the former of which will not answer
the conditions of the problem ; .-. the number of men
was 16, of women 20, and of children 32.
Also 32 y + 36 y + 32 = 324,
or 68 y = 272;
,\ y =4;
.'. each child had 2 shillings, each woman 2.9. 6d, and
each man 3s,
36. A and B were going to market, the first with cucumbers,
and the second with three times as many e^g^ ; and they
find that if B gave all his eggs for the cucumbers, A would
lose 10 pence, according to the rate at which they were then
^32 Examples of the Solution of Problems
selling. A therefore reserves two-fifths of his cucumbers ; by
which B would lose sixpence, according to the same rate.
But By selling the cucumbers at sixpence apiece, gains
upon the whole the price of 6 eggs. Required the number
of eggs and cucumbers, and their price.
Let X = the number of cucumbers,
and y = the price of one ;
.*. 3 J? = the number of eggs,
and '^ = the price of one egg ;
also - xy ^ xy — l6,
or 3 xy = 5 xy — 80 ;
.\ 2xy = 80,
and xy = 40;
also|.6x-(xi/-10)= ''-^"V'"^ '
X
18 ■
or— -J?* - 30 X = 60,
5
-and 9x* - 75x = 150;
'7^Y 5625
completing the square, 9.r* - 75 a: + -^ = —^ +150
__ 11025
' 36* ' . . '
75 105
extractmg the root, 3 x j^ = i "IT' >
and 3 x = 30, or - 5, the latter of which is excluded
by the nature of the problem ; ,\ x = 10,
A 40 ^
and V = — = 4 .
Hence the number of eggs was 30, and of cucumbers 10 ;
/. the price of a cucumber was 4 pence, and of an egg
x%f - 10
3x
1 penny.
producing Adjected Quadratic Equations', 233
37. A person bought a certain number of larks and sparrows
for 6 shillings. He gave as many pence jaer dozen for larks
as there were sparrows, and as many pence per score for
sparrows as there were larks. If he had bought 10 more
of each, (the price of larks remaining the same) and had given
as much per dozen for sparrows as he gave per score for
larks, they would have cost £a. Ss, 5d. Required the
number of each.
Let X = the number of larks, and .••• = the number
of pence per score for sparrows,
1/ = the number of sparrows^ and .*. = the
number of pence per dozen for larks ;
and oc^ = 15 X 36 = 540.
Again, if ^ + lo = the number of larks,
and 1/ 4- 10 = the number of sparrows;
then the price of the larks = 2/ x = -^ ^ ,
^ I 12 12 '
= 10.51+^,
12 '
and the price per dozen of sparrows = — x 20 =--^ ;
.*. the price 01 the sparrows = -— TlT^y
J 54 -f i/ 5. (y* 4- lOy)
and (54 -f 1/) 30 + 5 . (3/^4- 10 ?/) = 36 x 305,
or/ + 163/4- 324 = 36 X 61 = 2196 i
by transposition, 3/* -f- l6t/ = 18/2 ;
completing the square, 3/* + 1 63/ 4- 64 = 1 8^2 -f 6*4 = 1 ^36 ;
extracting the root, 2/ + 8 = ± 44 ;
.'. y = 36, or — 52, the latter of which will not answer
the conditions of the problem,
H H
23* Examples of the Solution of Problems
, 15 X 36
and X = =15;
.'. he bought 15 larks, and 36 sparrows.
38. A poulterer bought a certain number of ducks and 18 turkeys
for c£.5. 105. ; each turkey costing within one shilling as
much as three ducks. He afterwards bought as many ducks
and 5 over, and 20 turkeys, giving one shilling 9. piece more
for each duck and turkey than before ; and finds that the
value of his former purchase is to the value of the present
one :: 2 : 3. Required the number of ducks, and the
" prices of the ducks and turkeys at the first purchase*
Let X = the number of ducks required,
and 1/ = the price of a duck ;
.*. 3y - 1 = the price of a turkey,
and XT/ 4- 54^ — 18 = 110,
or xi/ + 54^ = 128.
Now at the second purchase, x4-5 =the number of ducks,
?/ -f 1, and 3 7/ = the price of a duck and turkey, re-
spectively;
.-. 110 : xt/ + X + 65 2/ + 5 :: 2:3,'
or 55 : x;^ + x +^ 65y + 5 :: 1 : 3 ;
.\ xi/ + X + 651/ + b = 165,
and XT/ -{- X + 652/ = 160 ;'
but xt/ 4- 541/ = 128;
.'. by subtraction, x + 11^/= 32,
or X = 32 — 1 1 y, which being substituted in the first
equation,
323/ — 111/* + 543/ = 128,.
or 111/* - 863/ = - 128,
. , 86 128
and/--. 3^= - — ,
producing Adjected Quadratic Equations,
^^S
1849
121
43
= +
completing the square, y*
128 _ 441
11 "" TTT*
extracting the root, y -
and y = — , or 2, the former which makes op
negative, and .*. the price of a duck is 2 shilHngs, and
the price of a turkey = 5 shi Rings. Also the number
of ducks = 32 - 11^ = 10.
21
11^
39. There are three towns A, B, and C ; the road from B to
A forming a right angle with that from B to C Now
a person has to go from B to A, but after travelling a
certain distance towards A, he crosses over by the nearest
way to the road which leads from C to A, and when on this
road he is 3 miles from A and 7 from C. He then proceeds
to A, and when arrived there he finds that he has gone
a distance, equal to one^fourth of the distance from B to
C, more than he would have done, had he gone the direct
road from B to A» Required the distance of B from A
and C.
and since the shortest path from a given point to a given
straight line is a perpendicular draWn from that point.
^36 Examples of the Solution of Problems
draw ED perpendicular to AC; E being the point
where he leaves the road BA ; .*. D is the point where
he enters the road CA; ,\ CD = 7, and DA =.3.
By similar As BA : CA :: DA : AE,
or y
and BA
or y
10
BC
X
30
3 : ^^ = ^A
Z)^ : DE,
DE =
3x
y
3x
.-. — + 3 = — + ^,
or 3 . (a? + » = 30 + - xy,
Stod 2 xy - 24 . (a: + 3/) = — 240 ;
but X* + y = 100;
.•. by addition, o?"^ 4- 2 xy + y' - 24 . (j? + «/) = — 140;
completing the square, x -\- y] * - 24 , (x + y) -{- 144 =
144 - 140 = 4;
extracting the root, j;+y-12=±2;
/. X 4- y = 14, or 10, the former of which only
answers the conditions, .*. x^ + 2 xy + y"^ = 196,
but 3f^ +y= 100;
.'. by subtraction, 2xy =96,
?nd nT* - 2xy +y^ =si 4y
whence x - y :=z ± 2,
but X -^y = 14 ;
.'. by addition, 2 a: =1 6, or 12,
and by subtraction, 2?/ = 12, or 16;
/. J? = 8, or 6,
and y = 6, or 8 ;
the distance of 5 from C is 8, or 6 miles, and from
^is 6, or 8 miles.
producing Adfeded Quadratic Equations, 237
40. In a garden is a square bowling-green, a side of which
is 30 yards, and near to it is a rectangular grass-plot.
The number of square yards in the area of the grass-plot
192
is a mean proportional between ■-—-, and the number of
square yards contained in the grass-plot and bowling-green
together. Also the number of square yards contained in
the square described on the diameter of the grass-plot is
a mean proportional between 10, and the number of square
yards contained in the aforesaid square increased by the
number contained in the bowling-green. Required the
area and sides of the grass-plot.
Let X and y = the number of yards in the sides ;
.*. xy '■= the area,
and -— : xy :: xy i xy + 900;
a » 192 , 1 72800
.-. x^y* = -~- .xy + -i-- 'y
-^ 79 ^ 79
, ^ ... , ^ 192 172800
by transposition, x^y^ - _. ^3^ = ;
w .u 1 % 192 , 951' 1 72800 , qS
completing the square, x'y^ - —xy+^\ =-L_.4.|.
13660416
^ w •
extracting the root, xy — — — ± ~— • ;
/ J/ / y
/. xy = 48, or — -, the former of which only
answers the conditions of the problem.
Again, 10 : a?* + 2/^ :: J?" + y- : X^ + 3/* + 900 ;
.-. x' + fY = 10 . (a;^ + /) + 9000 ;
by transposition, a?* + y^"" - 10 . (a?'' + 3/=*) = 900O;
completing the square, x* -h y^t — 10 . (.r* -j- y^) -}-
25 = 9025 ;
extracting the root, x^ + «/* -^ 5 = 9^ »
238 Examples of the Solution of Problems
. • . J7* + «^* = 1 00, or - 90, the latter of which will not
answer the conditions ;
but, 2 a?3/ = 96 ;
.-. by addition, a?'-|-2^y+3/* = 196, and a? + 1/ = ± 14;
by subtraction, x^-^^xy +3/' = 4, and a: — t/ = ± 2 ;
/. by addition, 2 a; = ± 16, or + 12 ;
/. ^ = ± 8, or ± 6 ;
by^ subtraction, 2y= + 12, or±l6;
/. y = + 6, or + 8;
but the positive values will only answer the conditions,
and the area is 48 square yar(Js.
41. A rectangular vat, 3 feet deep, when filled to the depth of
2 feet, holds less than when completely filled, by a number
of cubic feet equal to 24, together with half the number
of feet in the perimeter of the base. It is also observed,
that the length of a pole, which reaches from one of the
corners of the top to the opposite corner of the bottom of
the vat, is equal to one-eighth of the number of feet in the
square inscribed on the diagonal of the bottom. Required
the dimensions of the vat.
Let X and y be the number of feet in the sides of the
base ;
then {^ xy - 2 xy =) xy = 24 -^ x + y,
and V 9 + ^' + 3/* = § • (^' + ^') 5
r, x" -h y* - S s/ 9 + ^" +T^ = 0.
and (9 + oc^ + y") - 8 V 9 + >r^ -f ^^ = 9 ;
completing the square, (9 + x"" + y'*) - 6 ^^ 9 + x^ -{• y'^
+ 16 ;= 25 ;
extracting the root, \/ 9 + a?* -f y — 4 = +5,
and ^ 9 +x^ + f = 9v or - 1 ;
.-. 9 + 0?' +y\=. 81, or 1;
producing Adfecied Quadratic Equations, 239
by transposition, j?^ +3^' = 72, or — 8, the latter of which
is impossible ; . ,
but 2 0? 1/ = 48 + 2 . (a? -f 3/) ;
.-. by addition, jr* + 2j?3/ + 3/* = 120-}-2 . (.r+y) ;
by transposition, x -^ yX — 2,(x +3/) = 120 ;
completing the square, x + 2/|* — 2 , (x -{- y) ■{• 1 = 121;
extracting the root, ^4-y— 1 = ±11,
and i: + 3/ = 12, or- 10, the latter of which is im-
possible ;
.-. x^ + 2 a?3/ + 3/* r= 144 .
but 4X7/ = 144 ;
.', by subtraction, :c' — 2 jri/ + 3/* = O,
and X — 1/ = O;
but X ■{- 7/ = 12;
.'. by addition, 2x =z 12,
arid a? = 6 ;
/. 3/ = J? =6, and the base is a square whose side is
6 feet.
42. A person bought two cubical stacks of hay for c£.41, each
of which cost as many shillings per solid yard as there were
yards in a side of the other, and the greater stood on more
ground than the less by 9 square yards. What was the
price of each ?
Let X = the number of yards in a side of the larger,
• and 3/ = the number in a side of the less ;
then x^ and 3/' = the number of soUd yards in the stacks,
and a:*, and 3/^ = the number of square yards in their
bases ;
.'. ^' - / == 9,
and x^y + y^x = 820;
^ " xy
G40 Examples of the Solution of Problems
xY
but^ - 2x^if + y = 81 ;
8201*
■. by subtraction, Ax'^y'' = -^ — 81,
X y
or X
,\ 820]' 81a:'y"
by transposition, 07* y-H-— .x'^*= — r = 4l6|' = 168100;
completing the square, x^y^+ — . x^y''+~\ = 168100
+
81 „ _ 81
8
6561 _ 10764961
64 "" 64
81 3281
extracting the root, j;' y* + -- = + — -— ;
\ 8 8
,«. X* i^* = 400, or , the latter of which is impossible 5
r. xy ^ ± 20, the positive value only answering the
conditions of the problem ;
J , <. 820 820
and X* + y^ = — - = = 41 ;
^ xy 20
but ^xy = 40;
.-. by addition, x^ + 2xy'\-y^ = 81, and x+y= +9,
and by subtraction, x*'-2xy -\-y*=l; .\ x- y=: ± 1 ;
/. by addition, 2x = ± 10, or + 8,
and X = ± 5, or ± 4 ;
by subtraction, 2y = ± 8, or + 10,
and y = ± 4, or + 5 ;
.•. the prices were <£.25^ and £.l6.
43. A and B put out different sums to interest, amounting
together to of .200. jB's rate of interest was £A, per cent.
producing Adjected Quadratic Equations. 241
more than A^, At the end of 5 years, JB's accumulated
simple interest wanted but £a to be double of -^'5. At the
end of 10 years, A*^ principal and interest was to jB's as
5 : 8. Required the separate sums put out by each, and
the ratejBer cent.
Let 4 J? == ^s money (in pounds) ;
.*. 4 . (50 — ^) = J5's money;
y = A^ rate of interest ;
.-. 3^ -f 1 = ^'s rate ;
••. "^ = -^'s interest after 5 years,
and ^ ^^ \y + ) _ ^.g interest afterd years ;
. (50-J^).Cy+ 1) . , _ . ocy
5 "^ ^ ^' 5 '
or 501^ — ^3^ 4- 50 - J? + 20 = 2X1/ ;
.-. 3^3^ = 503/ + JO - X ... . [1].
Again, after 10 years, A"s capital and interest = 4 :p
5*5'
and 5's = 2 . (50-^)>(.y+ H) .
5
. g IQ^ 4- X2/ ^ ^ (50 - X),(l/ + 11) .. 5 .g.
5 ' 5
.-. 80^ -f 8j?,y = 250y - 5X1/ + 2750 - 55 a:;
by transposition, 13 a? 3^ = 2501/ + 2/50 - 135 a? . [2]
but from [l] 15 xi/ = 2502/ + 350 - 5x;
.". by subtraction, 2x1/ = 130 a? — 2400 . . . [3]
multiplying [l] by 13, and [2] by 3 ;
/. 650i/-\-SlO- 13 0? = (390^3/ = )750;i/ + 8250 - 405a?;
by transposition, IOO3/ = 392 a? - 7^^40,
or 50;^ = 196a? - 3670, which being multiplied by x,
50 xy = 196 a?^- 3670a? i but [3] being multiplied
by 25,
I I
242 'Examples of the Solution of Problems
50X7/ = 3250 J? - 60000;
.-. 196 x» - 3670 X = 325007 - 60000 ;
by transposition, 196 a?* — 6920.C = ~'60000;
173ol*
completing the square, 196cr* — 692OX + —^ —
= 2J92900 ^ ^ 5J900
49 49 '
,, 1730 250
extracting the root, 14 a? — = ±*— z— »
1500
.*. I4a? = 28O5 or -— — ;
.-. 0? = 20;
/. ^'s money =4j? = 80 pounds, and 5's=120 pounds,
, 196 X 20-3670
and V = -^ = 5 ;
^ 50
•. A's rate of interest was 5 per cent., and ZTs 6 per cent.
44. When the price of brandy was three times the price of
British Spirit, a merchant made two mixtures of brandy and
British spirit, and the prices per gallon were in the ratio of
9 to 10. He afterwards mixed twice as much brandy with
the same quantity of British spirit in each case, and the
relative price was the same as before. Required the ratio
of the quantities mixed.
Suppose at first x gallons of British spirit were mixed
with one gallon of brandy, in one case ; and 1/ gallons of
British spirit with one gallon of brandy in the other case ;
then
-a: -4-1 : 1 i: x + 3 : / ^1 , ^- . r ^,
X -^if the relative prices or the
tirst mixtures per gallon ;
and i/+l:l ••3/ + 3:jJf^
, X -h S y + 3
hence, : -^ , :: 9:10.
prc^ucing Adjected Quadratic Equations. 243
' . X + 6 y + 6
In the same manner — ■ — - : :: 9 : 10,
X + 2 7/ -h 2
and ,'. XT/ + X + 3^ + S : xy+3x + ^ + 3 :: 9 : 10,
diV*". x^ + X. + 3 1/ + 3 : 2a; - 2?/ :: 9 : 1 ;
.-. X2/+ X + 3i/ + 3 = 18a? - 18?/;
^ by transposition, xy — 17 x -{- 21 i/ + 3 =0j
also 0??/+ 2 I? +63/ + 12 : xy + 6a7^-2i/+12 :: 9 : 10;
div**'. X1/ + 2X + 63/ + 12 : 4 07 — 4 2/ :: 9 : 1 ;
.\ a?t/ +2a:-h 6;y + 12 = 36 x - 36y;
by transposition, xy — 34.x •\- A2y •\- 12 = 0;
but a?3/ — 17a: + 21 2/4- 3 =0;
••. by subtraction, 17 a? — 21 1/ — 9 = O ;
also 2 073/— 34a? + A2y + 6 = 0,
and a?2/ — 34a?+ 42t/ 4- 12= 0;
.-. by subtraction^ ^^ — 6=0,
or xy •==: 6.
Now X = — -^ ;
and 21/ + 93/ = 102,
, , 3 34
or 3/' + y;y = -yr;
3 Q 34 9 961
completing the square, y^ + jV+j^^ = y + Tge"^ 196 '
3 31
extracting the root, 3/ + ^ = ± 74 >
17
.-. y = 2, or- y ;
.-. 0? = 3;
/.the first mixtures were in the ratioof 3 to 1, and 2 to 1 ;
and the second in the ratio of 3 to 2, and of equality.
Sect. X,
Examples of the Solution of Problems in Arithmetical
and Geometrical Progressions.
1. A PERSON bought 7 books, the particular prices of which
(in shillings) were in arithmetical progression. The price of
the next above the cheapest was 8 shillings, and the price of
the dearest 23 shillings. What was the price of each book ?
Let X = the price of the cheapest,
and ^ = the common difference;
then X + 1/ = the price of the second = 8,
and X +6i/= the price of the dearest = 23 ;
.-. by subtraction, 5 i/ = 15,
and y = 3 i
.-. x = 8-y=8— 3=5,
and .-. the prices are 5, 8, 11, 14, 17, 20, 23 shillings,
respectively.
A number consists of 3 digits, which are in arithmetical pro-
gression ; and this number divided by the sum of its digits
is equal to 26; but if 198 be added to it, the digits will be
inverted. Required the number.
Let 07 - 3/ J
^ r represent the digits ;
Examples of the Solution of Problems^ ^c. 245
then the nijmber will be 100 . (^ — «/) + 1007+^+3^ =
llljc - 993/;
. Ilia? - 99.2/ ^ ^Q
or 37^ - S3y =• 2Qx^
and /.by transposition, 1 1 a? = 33 y,
and .r = 3 1/.
Again, Ilia?- 99^+ 198 = 100 . (x+3/)4.l0x+(a:-3^)
= 111 J? 4- 99y;
.-. by transposition, 1983/ = 198,
and 3^ = 1 ;
.-.. 0? = 3y = 3 ;
.'. the digits are 2, 3, and 4, and the number = 234.
5. The sum of £.1. 7«. was to be raised by subscription by three
persons Ay B, and C ; the sums to be subscribed by them re-
spectively 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 y^ was
to the whole sum subscribed hy B :: 4 : 5. Required the
original subscription of ji, B, and C.
Let X " y^ Xy x+y, be the respective subscriptions of
A, J5,and C;
then 3 J? = 27 ; and .\ x =^ g.
Now 5 : 2 :: (Cs share =5)9+3/: the part paid by B
= f . (9 + 3^),
3
and 5 : 3» :: 9 + y : the part paid hy A ^ 5 * (^ "*" ^)»
«46 EKampIes of the Solution tf ProHetns
3
and consequently, A paid upon the whole 9 — 3/ + r •
also -B paid upon the whole 9 + - (9+3/)= 7-^ ;
hence, 72 ~ 23/ : 63 4- 23/ :: 4 : 5,
' and {Alg, 179.) 135 : 63 + 23/ :; 9 : 5,
and {Alg. 186.) 15 : 63 + 23/ :: 1 : 5 ;
/. {i\) 63 +2.2/ = 75;
by transposition , 2 y = 12,
and 3/ = 6 ;
.-. the sums to be subscribed originally were 3, 9, and
15 shillings.
Four numbers are in arithmetical progression. The sum of
their squares is equal to S76, and the sum of the numbers
themselves is equal to 32. What are the numbers ?
Let 2y = the common difference,
and X + 33/'
the numbers ;
^ -^ ^ I be
X - y {
X - 3vJ
X - 3y.
then their sum = Ax = 32,
and .'. x = S',
also the sum of their squares = 4x^ + 20^" = 276, in
which substituting the value of x found above,
256 + 203/- =276;
by transposition, 20y^ = 20 ;
.-. f = 1;
and 3/ = ± 1 ;
hence the numbers are 11, 9, 7> 5»
in Arithmetical and Geometrical Progressions, 247
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, x ^Sy, x -^-y^x-y^ and x -- 3y,
to be the numbers ;
then 2x^ + 18?/* = 200,
and 2 0?* -f 23/* = 136;
.-. by subtraction, l6^' = 64,
and 4y = ± 8 ;
.-. ^ = It 2;
whence x' = 68 - 3/' = 68 — 4 = 64,
and x = + 8,
and /. the numbers are ± 14, ±10, +6, db 2.
6. The sum of the first and second of four numbers in geome-
trical progression is 15, and the sum oi the third and fourth
is 60. Required the numbers.
Let Xf xy, xy^, xy^, be the numbers ;
.*. 07 + j^y = 15,
and xy"^ + xy^ = 60,
or y"- . {x ■{■ xy) = 60,
, or Iby'^ = 60 ;
.-. / = 4,
, - and 3/ = ± 2,
and (a?4-2a7=)3o?=15;
.'. 07 = 5,
and the numbers are 5, 1 0, 20, 40.
7. The sum of four numbers in geometrical progression is
equal to the common ratio + 1 ; and the first term = ^7*
Required the numbers.
Let X = the common ratio ;
24S Examples of the Solution of Problemt
,, , la? X* x^
/. the numbers are — , — . — , — .
\r 17 17' 17
and 1-f a: = ^-^^+^ + ^=:(^+^\(^+"\
, , 1 + a?'
andl=-^;
.-. 17 ~ 1 + ^*,
and 16 = x"
.-. ± 4 = J7,
and the numbers are -~ , --., — , — .
\f \f 17' 17
8. A regiment of militia was just sufficient to form an
equilateral wedge. It was afterwards doubled by the
supplementary, 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 X = the number of men in a side of the wedge ;
X
.•. {Alg. 212.) (x+l).- = the number of men in the
wedge ;
.-. (a? + 1) . .r + 385 = «? 4- .5?,
or x" ^ X -\- 385 = a?* + 10a? + 25 ;
by transposition 36o = 9 x,
and 40 = a: ;
. the number of men = 820.
After y^ 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 evejy
hour. In how many hours would he overtake A \
in Arithmetical and Geometrical Progressions, 249
Let X = the number of hours ;
/. {Alg, 212.)( 9 + {^ - 1) ^) X f = the whole number
of miles he travelled ;
but 11 + 4 a? = the whole number A travelled ;
.-. (9+ \{x-^ 1)).| = 11 +407,
or 9 07 + - - f - 22 + 8 0?,
4 4
^ S X
and by transposition^ — H = 22 ;
4 4
X^ 3x ^ Q 361
completing the square, ^ + _ + ^ = 22+^ = 'Y6''
, 0? 3 IQ
extractmg the root, - +^- = + -^;
0? ■ 11
.*. - = 4, or — — ;
2 ' 2 *
.-. 0? = 8, or - 11 ;
hence in 8 hours he would overtake him. — 1 1 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
other t3?yo sides.
Let 6 — X, 6, and 6 -^ x be the sides ;
then 36 - 12 0? + 0?' + 36 3= 36 + 12a; + x*
{EucL B. L p. 48.)
by transposition, 24 a? = 36,
and 07 = - ;
2
.•. the sides are -, — , and -^^
2 2- 2
^SO EKamples of the Solution of Problems
But if 6 be the first term of the progression ; let,
6, 6 + a:, 6 + 2j?be the sides ;
then 36 + '24x + 4jr* = 36 -f 3<) + 12a? + x" -,
by transposition, 3 x^ + \2x = 36,
\ or a?* + 4 07 = 1 2 ;
completing the square, cir* + 4 a? + 4 = l6;
extracting the root, x + 2 = =ir 4,
and X ^ 2, or — 6,
and the sides are 6, 8, and 10 ; or O, 6, and — 6.
The problem is not properly restricted ; the algebraical
expression, in this instance, is more precise than the
language in which the problem is stated.
W. A and B set out fiom London at the same time, to go
round the world (23661 miles) one going £ast, 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 X = the number of days ;
X
then {Alg, 212.) (x-fl).- = the number of miles
ji
A goes,
and 20 J? = the number B goes;
... 'L±l +20 0:^ 23661,
and r' +^41 07 = 47322;
ID*
completing the square, a?* + 41a? H \ = 47322 -f-
i68i _ 190969
~l 4 '
.-. extracting the root, x -\ = + — -\
/. X = 198, or - 239;
/.they travel 1 98 days; ^goes 19701, and J5 396o miles.
in Arithmetical and Geometrical Progressions, Q51
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 after-
wards another sets out, and travels 12 miles a day. How
long and how far must he travel to overtake the first ?
Let X = the number of days j
then X + 5 = the number the first travels,
and .-. {Alg, 212.) {x + 6). ^ "^ ' - = the distance he
travels,
and 1 2 X = the distance the second travels ;
.-. (x + 6). ^ = 120?,
2
and ^* + llo? -f 30 = 24^i
.'. by transposition, o?^ — 13 .r = - 30;
completing the square, x'^ - 13aj+ "7^~"7^~^^~"5r'
13 7
extracting the root, x ^ ;= ± - ,
and X = 3, or 10.
.'. they are together at the end of 3, w^-lO days after
the second sets out ; and 3 6 and 120 miles is the distance
travelled.
13. AandB, l65 miles distant from each other, set out with
a design to meet ; A travels 4g|«; mile the first day, two the
second, three the third, and so on ; B travels 20 miles the
first day, 18 the second, l6 the third), aiid so on ; How soon
will they meet ? \
Let X = the number of days required ;
L +2+3+ . .
of miles A travelled,
then 1+2+3+ .... +.r= (l?l-a?>'. ^ = the number
re.
9^9, ExampUs of the SoJufum of Problems
ai>d 20+18+ .... +20- 2JP + ?= (42 - 2a:).- =r
2
the number B travelled ;
.-. (43 - ^) . I ^ 165,
or x' - 43j: = - 330;
1€4Q 1^4Q
completing the square, x^ - 43j: H ^= -— 330
_ 529
~ 4 '
43 23
extracting the root, x — — = ± — ;
2 2
.'. j: 5? 10, or 33.
Hence it appears that they meet in 10 days. On
the 10th day B travels 2 miles, and the next day he
rests *, the following day he returns 2 miles ; the suc-
ceeding day 4, and so on, increasing 2 miles every day ;
and on the 33d day he again comes up with A, 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 X + 6, J? + 2, X — 2, and x — 6, be the numbers j
then (a?* - 36) . (x* - 4) = l680,
or a?* - AOx" + 144= 168O5
.*. by transposition, .r^ — 40 a?^ = 1536;
completing the square, x^ — 40 a?'' + 400 = 1936,
extracting the root, j?* - 20 = + 44 ;
.'. 0?* = 64, or - 24,
and X = + 8, or + 2\/ - 6,
in Arithmetical and Geometrical Progressions. 555
and .'. the numbers are ± 14, ± 10, ± 6, ± 2 ; the two
other values o( x beiug impossible.
15. The product of five numbers in arithmetical progression
is 945, and their sum is 25. Required the numbers.
Lret X + 2^? ^ + 1/, X, X — 1/y X ^ 2 7/, he the numbers;
then 5 X = 25, and .\ x = 5 ;
also, X, (a?'- 3/*) . (a?* - 4^*) = 945, or dividing by x=:5,
{x^ ^ y^).{x^ - 4^) = 189,
or a;* - 5 x^y"' -¥ Ay^ = I89,
and Ay' - 125^* + 625 = 189;
by transposition, -43/* — 125 j/* = — 436;
completmg the square, Ay^ ~ 125 y^ + - — j = — -^
lo
extracting the root, 2y* = + -— ;
109
.-. 2/ = -^, orSi
.•.^* = l^,or4;
.•.3^ =± ^^^^ or ± 2,
and the numbers are 9, 7> 5, 3, 1.
16. A gentleman divided £.2\0 among three servants, in geo-
metrical progression ; the first had £.90 more than the last.
How much had each ?
Let xy% xyy x = the number of pounds each had;
then xy'^ zsi.x + 90,
atid x+xy + xy^== 210;
254 EMomplts of the Solution of Problems
or 2 X + xy -{■ QO = 210 ;
by transposition, 2 x + ^ ^ = 120.
90
Now from the first equation, x
and from the last, x =
V-1'
120
3^ + 2'
1 20 90
" y + 3 3/' - 1 •
4 _ 3
.-. 4/ - 4 = 31/ + 6;
by transposition, 4^' — Sy = lOj
9 9 169
completing the square, 4y^-3^ + -7j=104-Yg = -rrg- ;
3 13
extracting the root, 2y — - «= ± "j" j
5
and 2^3/ = 4, or- - ;
5
and /. y = 2, or - -;
whence x = — — - = 30, or 160,
and the sums are 120, 60, and 30 pounds. «
17, The sum of three numbers in geometrical progression is
35 ; and the mean term is to the difference of the extremes
as 2 to 3. Required the numbers.
X
Let - , X, and xy^ be the three numbers;
y
y
in Arithmeiical and Geometrical Progressions, 255
. and 0? : xy :: 3 : 3,
or \ : y :: 2 : 3 :
2
r. 2^'— - =3,
^ y
and?/' - 1 = - .y;
3
by transposition, ^ — - . y = 1 ;
completing the square, /-2^'^l6'^^"*"T6'^T6'
. , 3 5
extracting the root, y - ~ = ± - >
.-. y = 2, or — - , which last does not answer the con-
ditions ; '
/. X = 10,
and the numbers are 5, 10, and 20.
18. 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, xr/y xy"* be the numbers ;
then xy"^ — a^ = 15,
and x'^y^ - j?* : x'y' + a?>* + a^* :: 5 : 7,
or 3/'^ - 1 : y -H 3/' + 1 :: 5 : 7 ;
S56 Examp/es of the Solution of Problems
.-. y^ - 1 : 5^' f 3 :: 5 : 2,
and 3/^ - 1 = -^ + 5 ;
5
by transposition, y* - - . ^^ = 6 ;
5 25 26 121
completing the square, 5^* - ^ * "^^ "^ 16 ^^ "*"r6~"T6" '
5 11
extracting the root, y* — - = + — ;
3
.•, y* = 4, or , which last is impossible,
and y = ± 2 ;
/. from the first equation, (4a?-a?=)3j?=:15,
and ^ = 5 ;
.-. the numbers are 5, 10, and 20.
19. The sum of three numbers in geometrical progression is IS,
and the product of the mean and the sum of the extremes
is 30. Required the numbers.
X
- Let the numbers be - . x. and xy ;
y
, X
then - + j: + ^3/ = 13^
and (- + 0:3/ J . J? = 30 ;
X 30
/. by transposition, 13 — j; = - -^ xy — — ,
and 13 ^ — X* = 30,
or ^ - 13j: = - 30;
169 169 49
completing the square, x* - 13 J7+ -— < = -~- - 30 = --- ;
13 7
extractmg the root, x — -- = ± o »
tn Arithmetical and Geometrical Progressions. Q57
and /. ^ s= 10, or 3.
Ifx =5 3 ; then - + 3y = 13 - 3 = 10,
W
or 3 + 3y = \0y\
, by transposition, 31/* — \0y = - 3,
. o 10
or ^* - -^ -y- - i ;
completing the square, y* — ^^.y4 ^=— — iss— ;
5 4
extracting the root, ^ - 7 == ± - ;
3 3 ♦
,\y = 3, or ji
and the numbers are 1, 3, 9^
If the other value pf x b^ taken, the Qorrespouding values
gf y are impossible.
20. There are three numbers ia arithmetical progression, and
the square of the first added to the prodi^ct pf the other two
is 16; the square of the second added to the product pf the
other two \% 14. What are the numbers t
Letx—y, Xy x+y, be the numbers;
then 2x* " xy + ^ = ]l6^
and 2 a:' - / = 14;
/. by subtraction, 2y* — xy = 2,
and by addition, 4# — xy — 30,
or 2^* = 2 + ^y,
and 4 a?' ==30+ x y i
.-. by multiplication, Sx^y"" = 60+ 32xy -^x^yU
by transposition, 7^''y'^ ^ ^^xy = ^^>
LL
258 Examples of the Solution of Froblems
or x^y^ ^— .ccy^ —',
completing the square, xy^— — . xy + ^ = —
7 ^9 7
256 __ 676
■*■ 49 ""79 '
16 26
extracting the root, xy— — - = ± — ;
.-. jr^^ = 6, or ~ y ;
.-. 23/' = 2 + J?3/ = 8,
and y* = 4 ;
.-. y =r ± 2,
and 4 J?* = 30 + j:?/ = 36;
/. 2 a: = ± 6,
and a; = ± 3 ;
.•. the numbers are 1, 3, 5 ; or — 5, — 3, — 1. The
other value of xy was introduced in the operation, and
does not answer the conditions of the question.
SI. The sum of four whole numbers in arithmetical progression
25
£4
25
is 20, and the sum of their reciprocals is ~ . Required the
numbers.
Let x — Sy, X —3/, x -\- y, x-^3y, be the numbers ;
then 4 0? = 20,
or a? = 5.
.1,1 1 , 1 25
Again, h + -f — . „ = — ,
° ^x-Sy x-^y X -^ y x + 3y 24'
4x^ -^ 20xy^ 25
or = — ;
^ — lOo^^y* -f 9j/* 24
.'. 25 X (9^^ - 250/ + 625) ^ 24 k (500- IOO3/*),
in AritJtmettcal and Geometrical Progretstons, 259
or 93^ - 250 y* + 625 = 24 X (20 - 4i/') ;
by transposition, Qy^ - 154i/^ = —145;
completing the square, 93/^ - 1542^* + -222 -. ^^29
• -145 =i^;.
9
77 68
extracting the root, 3 2/*-^=+ ,
3 o
145
and 3^* = 3, or — ;
, , 145
.-. 1/* = 1, or—,
, , , + a/ 145
and 3/ = ± 1, or -=-^^^ ^,
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 a?, xy, xy* -,
then 100a? + lOxy + xy'* : x + xy + xy'^ :: 124 : 7,
or 100 + lOy + 3/* : 1 + ^ + 3/' :: 124 : 7;
div''^ 99 + 93/ : 1 + 2/ + y - 117 : 7,
or 11 + y : I -{■ y + y' :i 13 : 7;
.'. 13/ + 13 2^ + 13 = 73/4-77;
"by transposition, 13 3/* + 6y = 64,
, . 6 64
or 3/^ + ^3. 3/ = T3 5
QGO EMmpIfi of the SeitaioH tf Problems
completing the square, ^» + ^.i^+±. p=|| +
9 841
13]* " 169'
extracting the root, 3^ + ^ = ± ?| ;
« 32
.-. y = 3, or-:^^;
also 100a: -f lOjry + a?^* 4- 594 = 100^' + ^Oxy + x-,
by transposition, 99
5
and j: = 6, or - , which last cannot answer the condi-
tions 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.
(31.) In the preceding solutions it may be observed,
that, in many instances, values of the unknown quantities
are deduced, which do not agree with the conditions of
the problems. This is always the case when the roots
of the equations are negative ; and the circumstance
arises from that peculiar quality of an algebraic expres-
sion, by which it is denom.inated 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
M M
266 Solution of Problems,
the quantity xy will represent the product of 4-^ x -f 2/,
or of- X y — y\ and a*, of + a x + a, or of- a x - a :
consequently 5 in the reduction of such quantities to their
constituent 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 arc expressed, quantity or number
is from its very nature what in Algebra is meant by the
term positive, i. e. it increases any 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 modifi-
cation of the conditions. In the equations thence de*
duced, these former negative values will appear as positive
roots, and the former positive values as negative roots.
Thus, if Prob. 11, page I92, 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 in-
*' creased 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 number which would be found from
using the negative value of x in the original problem ;
and the equation for determining this {x^ — 5 x = 4225
— 5 x) differs from the other only in the sign of x.
Solution of Prcb/ents. ^67
In Prob. 18, page 196, x is found to be equal to ±4,
where the negative value shews 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 206, the negative value of a? 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 £.3^ 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 deter-
mining the price, is — = 39 + *> which is deduced
from the equation to the original problem by changing
the sign of x, the positive value of which is in this
case 130.
' Also, in Prob. 4, page 207, the negative value of x is
15
— , Now if the problem were '^ Bought two sorts of
*^ linen, for the finer of which I gave 6 crowns ?nore 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 elk 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 diflfering
from the equation to the original problem only in the
. 15
sign of X, and whose positive root is — ; whence there
ji
were 7i ^'^ ^^ ^^ fi^^^r at ^s, 6d. per ell, the whole
price of which was therefore £.2, l6.s. 3c?., and the whole
price of the coarser was c£.l. ^s, 3d, And in the very
same manner, all the other problems may be transformed.
268 Soluticft t>f Problems.
(32.) 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 ai'e
properly modified, equations may be deduced, of which
these values rendered positive will become roots, and will
satisfy such conditions.
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 condi-
tions of the problem under any modification whatever.
Sect. XL
PRAXIS.
I. Simple Equations involving only one unknown
Quantity,
T. Given 36 - 4 ^ = 17 ^ + 15* to find the value of x.
Answer a: = 1. .
X
2. Given 2a?+7 - - =28 - 2a?, to find the value of a?.
Ans. a; = 6.
3. Given x^ + 1707=370:- 4 o?% to find the value of x.
Ans. a? =a 4.
V OC X x
4. Given g - 4 + ^^ =? 3 - 2 "*" *^' *^ ^^^ ^^®
value of X,
Ans. 07 = 12.
X + \ 3 0? — 4
5. Given — -— + 1 = — r — , to find the value of or.
3 5
Ans. 0? = 8.
6. Given ^^^j^ + So: = 20 + 1^ . "" ^ , to find the
value of X.
Ans. 0: =: 3.
270 Simple Equatiotis involving only
I. Given iL±i + 8i; = i^-^^ + 36, to find the'
«3 3
value of a:. Ans. a: = 4.
8. Given ■^- + 4 = + ■ .. ' — , to find
2 4 o
the value of x> Ans. j? = 3.
^. 2i; - 3 . 18 — a? llx-3 5 ^ .
9. Given--^^^ + — ^— = - ^ -^,tofind
the value of x. Ans. a? = 6.
3j?4-4 19+^
10. Given r — 3 = 35 - 5 a?, to find the
value of X,
Ans. a? = 8.
II. Given 4a? '—- =7 - , to find
6 4
the value of x.
Ans. a: = 2.
-,. 21 -3x 4a: + 6 ^ 5a?-fl
12. Given g — = 6 - — j — , to
find the value of x.
Ans. a: = 3.
^. 8af + 4 7^ + 3 32 - 4a: 5a?+l3
13. Given -j-j — = —J- j^,
to find the value of ar.
Ans, a? = 5.
^. 27 - 9a? 5a? + 2 6l 2x 4- 5
14. G,ven:c + — j g- = ^ 5—
^ to find the value of a?.
12 '
Ans. X = 5,
ont unhiown Quantiti^. 271
15. Given 1 = DC -{ -— ,tofind
11 13. 2
the value of x*
Ans. j; = 9,
1.6. Given p— — ^H ^—^5; — ,toJfind
the value of x,
Ans. X = 9-
.», i- « * 4*^ - ^ 2x + 11 7-8J? ^ £ ,
17. Given 2x- — — — = -— - — -— — , to find
11. O f
the value of i*.
Ans. 07 = 7-
18. Given -^^. 5— =—8 T^ ^ '^
find the value of x,
Ans. o; =;: 8.
^, ^ Sx+lb 7 - 6x 35 9^7+11
19. Given4- -5 L_^=^+^^__^,
to find the value of x.
Ans. 0? = 3.
^. 707-43 , ,,^ 13 + 50? ^^^ 3o?-12
20. Giveni^ j^ +14| = 276 ~
^ i — 13 07^ to find the value of x,
o
Ans. X = 19.
13 1^
9
^. , 1 3 07 — 13 12 + 7o7 ^ ,
21. Given 407+ —- - r^ —^—=707—33
9 + 507 no?-- 17 ^ n .,, 1 t
■■ — • '. , — ,JL ^ to find the value of 0?.
Ans» 0? = 15.
272 Simple Equatiom involving onlij
3 8 16 ^'*
16-lO.r 4^ 4- 37 ^ ^ , . , . _
+ 7^; -z — - , to find the value of 07.
11 7
" , Ans. ^ = 17. ,
23. Criven — = 4. » to find the
36 5x - 9 4
value of X,
^ 4a? + 17 = 2 ^T + 1, to find the
value of X,
Ans. >r = 16.
?<37. Given ^^-^ = ^ +i^, to find the value
V^ + 3 >/F+ 25
of :c. ^
Ans. 0? = 9.
38. Given \--- = ^,. ^, ^, to hnd the
V 6^ + 2 4 V 6.r + 6
value of X,
Ans. a; = 6.
N N ' •
274 Simple Equations involving
^ 5 J? — 9 isJ b X 3
39. Given ■ ; ■ • — 1 = ^ , to find the
value of X.
i Ans. j: = 6.
40. Given v 1 + a; ^ a?* + 12 = 1 + / 16 + x = / ^^ > to find the
value of X.
Ans. X = 9.
II. JSimple Equations involving two unknown Quantities:
Xl. Given x + Ib^ = 36,1 ^ , ,, , ^ .
^ ^ , , o ^ n/^ I to find the values of x and y.
^ and y 4- 3 X ==: 20, ) "^
%. Given 4x + 9i^ = 47,7 ^ r,, , - ,
> to find the values of x and y.
and 8x- 13t/ = 1,) -^
5? Given g + 4 = ^^ /
N to find the values of x and y.
two unknown Quantities, 275
^. Given | + sy =± 194 1
y ^ Mo find the values of x and y.
and I +8^= 131,\
Aks. '^ = ^^'
't ^. 3^ -. 1
5. Given -
and -
4 ■•■ ^^ ~ ^ = ^' / to find the
y - 2 . ^ ■■ - ^„ I of ar and «
24.
- Q 1
values
-^ . 8« ■
6. Given ga; H ^= 70,,
the values of x
An6
+ -5-= 70,1 ^^ g^^^j
and 7y- '-1^ = 44 '"^^•.
Ans. f = ^'
|y = 10.
I- 7. Given ^
5
and
^ —5 4-^ = 3^-5, 1 to find the
51/ -7 4a?-3 _ f values of a?
- ^ g «- 18 - 6a?, ^ and y.
2
Ans
(a? = 3,
8. Given .+ l^^-^ = 7-93LtL^
7 j^ — >1 to find the
Ans. f = ^'
27^ Simple Equations involving
9. Given 4oc -\ =21^'" '
707 -h 11 ■\
+ ^ +*— Yg— ^/tofind the
2x±v_ _^2y + 4i values of
— a? 4- 2 <, J 0? and y.
Ans. f = "'
and 3?/
-^ 5
^. 3^+5v 4x-\-7
10. Given 0? ^-:^ + i7 = 5v-f ■ . :
2^ ~ 6i/
11. Given
So::
5£-7_^+l 8v + 5r values of
~~n 5 IT' J ^^^^y-
— ?-+-^-r — = 4+ 2»l to find the
„ _ „ > values of x
+.V 9£— 7_3.y+9 4X+5.V I .
and 3
Ans.
bx-Z , 3?/-j; ^ . 3a'- 19
ven ~
Or a.
and
16
C.r = O
Ans.
^. 7^+6,43/-9 o 13-a? 3.y-a?
12. Given -^ + ^=3.: T^"^^
and 3x4-4 : 21^-3 :: 5 : 3,
to find the values of x and y,
Ans
\y = 9.
,3. Given 5_£±13_8^-3x-^^^_^r£-|^^
,,d If : "-^%4X :: 4 : 21,
to find the values of a? and y.
Ans. }^ = ''
(^ = 4.
two unhtown Quantities. ^77
14. Given ^5 ___=5+-3-,
and -^— ^^ 3^ = -Tr-'
to find the values of x and 3^,
15. Given i,, + ^MJi}l±£.li:-^^l2y^
12 4 3
12a; + 7?/ + 28
g' '
. 9a;+18 12+5V— 6a; \bx — 3y - i
and '^^ — — —
ya; + V — 10
15 '
to find the values of x and y.
Ans
Ca; = 2,
16. Given —J— + 4^_g = 3:f + ^— ,
to find the values of x and y.
Ans. -J
\y = 9-
^- . o.. 4.y + 13a; 12a; + 8
17. Given 4X-344. - ^j _^y = 3 ,
, „ , 21 - 4y 18a; +• 13
and ix^-^j—^ = — g 2i,
to find tlie values of x and y.
5.
A^s. {^^
278 Simple EquAtions involving two unknown Quatiiitiet*
24 4- n *~
18. Given4x + 3v + L _ l6^^-hl2^-8^ + 5t/-h28
20? + 1 40? - 2 *
J « . „ 8ar — ISv'* + 108
and 2o: + 4 = 3v+ — r-Tr- — ,
to find the values of x and y^
Ans. \' = ^'
\y = 2.
tf+6« 3y + 6 _ 3x-2
19. Given J l-_i 1£_ = 5 - il
and if + ^ + 2} : ? - ^ + 1 •
23^2 3 6
8 16'
lOi : II,
to find the values of x and y.
Ans. f = "^
^3^ = 3.
6^+2 3o?w-31 . ^^ , ,^
12i^ ^-L— 1- hlOo?+13
;20. Given 4 + fL— =v+ ii ,
11 "^ 3or '
170
J 2a: 30? — 6 ^^^ + -3^
and --- - ~ = r. ,
3 1/ + 7 61/ + 27
to find the values of x and y.
Ans. [^ - T^
\y = 2.
III. Pwrc Quadratics and others which may he solved
without completing the square.
M^ 1. Given 5o?* — 16 = 20 + 0?% to find the values of x,
Ans. 0? = ± 3.
•>i
2. Given x + i/:i/ :: 3 : 1,^^ ^^^ ^j^^ ^^^^^^^ of a; and 3^.
and xi/= S^}
Ans.
(x = ± 4
Jj^ = ± 2.
3. Given ^ - ^ : ^ :: 3 : 5- / ^ ^ J , , ^ ,
, , "^ *^^ - ^>to find the values ot a? and v»
and 07* + 43^* = 164,5
Ans. f = ± «'
ly = ± 5.
■^. Given V+y' : ar* -?/':: 25 : 7 J to find the values of
and x^' = 36,3 X and^.
"75. Given X* - a^V = 48, ^
^ , , >to find the values of x and y.
and a?^- ^'= 12,) "^
Aks. f = ± «'
ty = ± 2.
-^6. Given a;+y : a;*-y' :: l : 3,^to find the values of
and xy = 10, ^ of a; and y.
Ans. }" = ^' «•• - =^'
C^ = 2, or — 5.
/^'^. Given x^^^^ . x^-^y^ :: 234 : 109,|to find the values
and xy'^ = 175,i ofj^andy.
Ans. s '
ly = 5.
>Y8. Given x' - xy \ xy -- y^ \x b : 3y | to find the values of
and x'^y = 75,) ^ and y»
Ans. < '
ly = 3.
1^ 280
280 Pure Quadrafks and others whkh may he solved ^
. Given x^ -3/' : x'y-xif :: 3 : 1,1 to find the values
y and X + 1/ =" 6,3 of x and y.
I3/ = 3.
-f 10. Given v'^^ x/J: sfx - >/^:: 5 : 1'"^ to find the
and X
values of x and v.
fy :: 5 : 1,7
Ans. \ ^ '
ly ='4.
X 11. Given A^^- .y^= 3,7 to find the values of x
and v^3F + .C^^ = 7 J and y,
. ix = 625,
^- {3/ = 16.
y 12. Given a:-3/ : x/^-n/^:: 7 : 1,7^^ ^"^ the value?^
^ and x/^ = 124 of 0. and 3^.-
A^^' I3, = 9, or 16.
T 13. Given -+- = -,#
•^ ^ ^ \ to find the values of x and y.
and — = -j\
xy 9 y
ex ^ 6, or 3,
^^^- U=:3, or 6.
y. 14. Given a?*-3^ = 65,7 ^^^^^^^^^^^^^^^j.^^^^^^
and X* — ^' = 5,^
Ans
Co; = ± 3,
t/^ = ± 2.
X 15. Given o:^ - 3/' = 56,y
/ 16 y to find the values of x and y,
and X -3^ =^,^
Ans.
X = 4, or - 2,
.3/ =2, or - 4.
without ^otnpleting the Square, 2S1
' ID. Given ,- - , = •^,
1 - V 1 - ^' 1 + V 1 - ^' -^
to find the values of 07.
Ans. :r = + - .
- 2
[ T 17. Given x'y -^t y^ = ll6,| ^
> 1 I . > to find the values of ^ and y,
\ and 0:7/2 4- ^ = 1 4^ 3
Ans,
y = 4, or 10.
^ l^Given Xj x \ Ipif = 6,7 to find the values of x
\ and 0? +3/ = 72,3 and y.
"^ ra: = 64, or 8,
. ^ ^'''' i^ = 8, or ^A,
^ \ 19. Given 4x' + - = ~ + 103/,/to find the values of O
and ^' + 3^=55 J ^^ndj/. / ^ ^ (^
\
Ans. <
ly = 10.
1 . 1
^('-20. Given . t ,. = «a?,
to find the values of x.
Ans. * = ± \/l±-l
^-p 21. Given o?^ + y^ = 20,7
V' J » . » /. / to find the values of a; and y.
\ and x^ + y^ = ,6, 3
i Ans. {^ = ± 8. or ± ^"i,
(1/ = 32, or 1024.
o o
284 Twre QuadratUs and others nuhich mmf he solved
^It^ 22. Oiven 'a?* + 2a?V + 3^ = 1^96 - 4^2^ . (^H^+y*)?
/ and X — 7/ = 4,
to find the values of x and y.
Ca: = 5, or - 1,
Ans. ? , -
^y = 1, or — 5.
V' 23. Given v^ + "^ JL =A = 4,1 to find the values of
and Vj • Vi/ '*'- \/ y ' 4J .
(•625
f Given 1 - i = - 1
2 . iven y " J^ 4'>to find the values of x and ?/.
and x^y-xy''^ 16J
625
Ans.
*"'• iv-^'
ven ^
_. >v/ 4x + 1 4- J 4x ^ Ito find the value *i
25. Given ^ . ^^7= = 9w c ^
Ans. a? = -.
^26. Given a:*t/^ -- x^y^ = 2l6,7 to find the values of
and xy - a?i/* = 6, \ x and y.
Ans. Xy^l
*'\(^' :27. Given JC* ■\-xlJ xy"^ =• 208, 7 to find the values of '
(7^ and y'^-^ylj'^'^ 1053,5 a: and 3/.
Ans. \y ^ ^ 2j,^
29
without completing tJte Square. 283
26. Given a^^j^y^+y^ = IOO9/) to find the values
and x^ + x^y'^+y^ = 582X93^) of x and y,
(3^ = 16, or 81.
Given J?* + ^' + x y . (a? + 3^) = 68^
and .r' 4- y' - 3a?* = 12 + 32/% 3 ^^ ^*^^ ^^®
values of x and y.
Ans. j^ = 4^ <^2,
C^ = 2, or 4.
30. Given xy . {x + y) x=: 84, :)to find the values of
and xy . {a?» + ^^') = 3600, 5 X and ^.
^y = 3, or 4. .
>C3i. Given :^±£JL±J^ = >. 1
'^ 0? + y " f to find the values of x
and ^'"^-y+.y' = o C ^^^ y-
y
x^-^ xy H- .y^ -^
^ "+ y f
Ans. J^ = ± ^>
^2/ =3.
IV. Adjected Quadratics involving only one unknown
Quantity,
1. Given x-' + Sx^ 33, to find the values of x,
Ans. X = 3, or - ll.
2. Given a?» -f. 6a? + 4 = 59, to find the values of x,
Ans. 0? = 5, or - 11.
3. Given X' - ioa? +17 = 1, to find the values of x.
Ans. a? = 8, or 2.
4 Given a?* - 0? + 3 :^ 45, to ^d the values of .t,
Ans. 0?= 7, or;— 6,
284 Adjected Quadratics involvings only
5. Given 5 a?' — 407 + 3 = 169, ^o find the values of 07.
Ans. 0? = 6, or — — .
5
6. Given 7o?*-21o: + 13 =293, to find the values
of X,
Ans. a: =i 8, or - 5.
^fc/ tx^ 4 X
r- 7. Given — + -r 19 = 154-, to find the values of 0?.
3 o
Ans. 07 = 9, or i.
5
7o7 — 8
8. Given x + 4 + = 13, to find the values of or.
X
Ans. 07 = 4, pr — . 2.
35 — 3 07
9. Given 60? + * = 44, to find the values of 0:.
. X
Ans. 07 = 7j ^"^ "~ S*
7 -L /r Q I 40?
10. Given 14 + 4o7 - ■ = 307+^-—^ — , to find
X "" f ^
the values of 07.
Ans. 07 = 9, or 28.
^. 307 + 4 30 — 207 7-^ — 14 ^ £ J ,,
11. Given — -— ^- = — -r — , to find the
5 07 — o 10
values of 07.
Ans. 07 = 36, or 12.
-^. 15-07 12-307 ^ 2307 + 60 ^
12. Given — =1x = , to
4 407 — 5 7
find the values of 07.
229
Ans. X = 3, or - ^35"
one unhnowH Quantity, 285
13. Given l±li + §-illf = 7 to find the values
of ^. ,
Ans. j? = 3. or .
2
14. Given ■ H : — = 3 H — . to find
9 4x4-3 ^ 18 '
the values of x.
Ans. j? =s 6. or - -i-.
^ 4
15. Given 1^4 . ■ ^ =3}, to find the values
jc a; + 5
of X,
Ans. a; = 7, or — —
^ ^. 5jc — 12 3a: - 24 7j; ~ 34 ^
16. Given — ^_ + ^--^ = 9 - .___., to
find the values of o:.
477
Ans. a? = 12, or — ^.
184
. ^. 4a; -. 5 3i—7 9a: + 23 ^ a j ^u
17. Given P-f. = ^— to find the
' a: 3a: 4-7 13a: ^
values of a:.
Ans. a: = 2, or
45
18. Given 2a: 4- 18 - ^^^±^ « 27 - ''^''''\
4.X'\-7 ' 2a^^ -3 '
to find the values of x.
Ans. X r: 8, or 5.
286 Adjected Quadratics involving only
19. Given —^ + ' — r-7 = ^ — r-7* ^^ fi"" ^^^^
^ 07 + 620? + 4 3x + 4
values of 0:,
o
Ans. 0: = 8, or - - .
^^ n- 4 307 + 6 3o? + 5 ^ ^ , ,
20. Given - — --- + - — -—--r = — , to find the
207 + 3 507+18 507^
values of x.
Ans. 07 = 6, or .
2
«, r^« 8 .807-17 407 + 3 . ^,..
21. Given ---—- h ^ . ' = - — —-;, to find the
9 + 507 2 + 407 207+ 12'
values of 07.
283
Ans. 07 = 3, or -
137'
12 8 32
22. Given ; h = — —~, to find the values
5-07 4—07 07+2'
of 07,
" . 58
Ans. op = 2, or — .
' 13
23. Given ^ + - . ■ = ~ ^ to find the
007 — 07' 07*+207 bX
values of 07.
Ans. 07 = 3, or p .
- ^. 4,x;' + 7o: . 5o7 - A-" 4o7* ^ ^ , ,
24. Given— ^ + ^^-j^ = -^^ ^^ fi^^ ^^^
values of 0?.
Ans. 07=3^ ^^ "" To *
one unknown Quantity, 287
^^' ^"^^^^ "^ ' ilt^i ' = ^* + ^ + 8, to find the values
of X.
14
Ans. ^ = 4, or —
' 3
26. Given '^-^ + _£_=, 5 ^ ^ to find the Values
of ^.
^^.^ Ans. X = 3, or - 15.
2r. Given ^Ax + 5 x vT^TT = 30, to find th^
values o{ X,
Ans. o: = 5, or — ~^ .
28
_ / 4
28. Given ^^^ = ^^ "" ^^ , to find the values
of ^.
Ans. X = 25, or ^^^^
40O
29. Given ^j?» --7^ = ^ J?, to find the values of x.
Ans. a? = 4, or - 5l 3".
30. Given x^ + 7# = 44, to find the values of x.
Ans. 07 = ± 8, or ± - n]"^.
31. Given 4^4 + ^^ = 39, to find the values of x,
Ans. .X =729, or ^ .
4 i
288 Adjected Quadratics involving only
>( 32. Given - + 2 = if , to find the values of x.
■* x^
Ans. j? = 4, or V 3 .
4
33. Given 3 x- ^^ ^=4, to find the values of x.
8 2«
Ans. x =~8l^" , or - — r' •
34. Given
8-3^^"" 4 + >>/i (8-3.yx).(4+v'i)'
to find the values of x, ,
Ans. X = 93, or 7.
35. Given
54 - ^^x -. 23j:-46^^ 7^ ~ 3a? -f 4
a: + 2v/i 6 + ^r {a; + 2V^)x(6 + V^)'
to find the values of x.
32
Ans. X = 5, or ----- .
15
36. Given x + a^Ic: x-a^x:i Sy/x-^-G : 2v^ to find
the values of a?. V
Ans. a: = 9, or 4. . ''^
. \
37. Given a?' + 11 + ^af" + 11 =42, to find the values
~ of a?.
Ans. a? = + 5, or + ^38.
38. Given a?~5)'-3.a7 - 5)-i- = 40, to find the values
of X.
Ans. a: =s 9, or - 5)^ + 5
one unknown Quantity, 289
7^ 39. Given x+ ^7+^' = 2 + Ss/VT^, to find the
values of x,
Ans. X = 10, OF - 2.
40. Given PT5I* ~ 4 ^^ = 160, to find the values of :r.
' Ans. X = ± 3, or + ^ _ 15.
41. Given ^^ - 7a: + V'.i?^-. 7^+18= 24, to find the
values of x,
Ans. a? = 9, or - 2, or ^^tJjT^ ^
2
4i?^ Given 907-4.^*+ x/4a7^-9^+ii = 5, to find the
values of ^.
Ans. :r = 2, ori,or9±N/"^.
4' 8
43. Given x« + ./s^r + *' = 42 - 5x, to find the
values of x.
Ans. a: = 4, or - 9, or ^_lWMI.
44. Given —2 + ^-^^ ^ ^ ly _ . ^ ,
;>,i ^ 5 ~ ' /-~--^ > to find
J? + 2|« -^ 4v^j?-f 2
the values of x.
o
Ans. X =z 6, or — - .
2
45. Given j;^+ -^^=---, to find the values
of X.
Ans. X = 12, or- 3, or jg^+x/^"
PP
£90 Adjected Quadratics involving only
^^. 3a? +5 3a?— 5 25 ^ a a ^\
^6- <^'^^° 3731 + 3-¥Tl = ^TtS ' ^'^ «"^ ^''^
values of jr
,
Ans. a? = ± 9.
Givena?+>sy
_ X ^r X 4 ^ , ,
47-
a? + 2 = 1— 9 ^"^ iiiiu tiic
\J X
values of x.
Ans. a? = 4, or 1.
48.
Given _^, + / , - „^/\, to find the
a;* — 4 * a?' - 4 25 a?* '
values of a?.
Ans.x = ± 3, or ± V Y^ .
49.
Given a? + -
' 8
+a? = 42 - - , to find the values of x,
X
— 7 + / 17
Ans. a? = 4, or 2, or ^^^ —
/x 4- 4 3
50. Given a? + 4 - 2 V/ = - — 7 > ^^ ^^^ the
'^ X — 4 , a?— 4
values of a:.
Ans. a? = ± 5, or ± s/Tj.
^' ^ 5x 2537* 6*4 ^ ^ , ,,
51. Given a?* h 15 = — ^ r , to find the
2 16 a?*
values of a?.
Ans. a? = 4, or — 8, or ^ .
52. Given ^H + V^^flu^ = M, to find the
V^a?^-9a?* 7^ 2a?'
values of x,
Ans. a? = ± 5, or + ~ — .
one unknown Quantiti/,' 291
53. Given 3 .^-1>- J +2^ = 341+2. x-i\\ to find
the values of x.
Ans. x = 5,ov-2; or V3±7~[^
54. Given x* + — 39^ = 81, to find the values
of X.
Ans. ^ = + 3, or -"l^ + V— 155
- ' 6
55. Given 4x^ + | = 4V + 33, to find the values
of jr.
Ans. ^ = 2, or - - ; or ^ ± ^^^ .
2 4
56. Given 7^*-.6^i (a: -i2)=r24- 14^4-15^7^, to
find the values of x.
Ans
. a? = 16, or 1; or ±V-11-1
57. Given4a7+l> + 4a:i(4a? + l) = 19l2-(lOa:+3^),
to find the values of j?.
Ans. :c = 9, or 1? ; or -9Q + n/~181
4 8
3 ^
68. Given 8 a;*- 13 = -— + V 6 o:^ + 52 ^', to find the
values of 07.
Ans. ^=2,or-i^;or^-±^^lH£i.
8' 128
^9 j AdfecUd Quadratics involving
39. Given 4«* + 21.r + SirVZ-^* ~ 5j? = 207 r>
to find the values of x.
A.s..=3. 6r^^^or -^^^^%^^^l ^.
60. Given f^JlJV^ = 2^ - ?£-Z-^4^, to find the
2x -V* 2x + aJx
values of x.
Ans. X := 4.
V. Adfected Quadratics involving two unknown
Quantities.
1. Given X + 43/ = 14^) to find the values of x
and y + 4 a: = 2y -M ij and r/.
d3^ = 15,
X = — 46, or 2,
or 3.
^. 2x + 7v 51 4- 2x-\
4x "^ 10 ' f to find the values
andl£^=3^-2,) ^^^ ^"^^•
56
^- ^ 640
or .
351
3/ = 4,
^. 4xy + 2v-l , 4?/ + 3x + 1 17-x
3. Given -y—^ 1=^^-5 —■■
7 5
to find the values of x and y,
^"''- i , 144^
4453
5450
tivo unknown Quantities, 293
o«^ I _. :» . / , 5 o I to find the
and 1 + a?* : 2/ + 4 :: 52/ + 7 : 3z/r, j
values of x and y.
'^ = 2, or - -.
i
5. Given ^» + 2j?3^, = 441 - oe^y%i to find the values
and Ty = 3 + ^, 3 .' of ^ and y.
507 = 3, or- 7, or ~2± v/ - 17,
4 1 It ^"1777
7 -2±^-i7 :
6. Given a7* + 4^=:2l6- 4a^y,|to find the values of
and 4y- a?^ = 64,i ^ and y.
Ans. i^ = ±^^
7. Given x •\- y]"^ -3^ = 28 + 3ci?,7to find the values
and 20^ + 3 X = 36, J of x and 3^.
Ans < 2 ■*
■ >y = 3, or|,or nAl^^liSa.
V ^ 4
8. Given ^* + lOx 4- y = 1 19 - 2 Vy X (^ + 5),
and X -\- 2y ^ \3,
to find the vakies of j? and y.
\x =5, or -^
\J ^4, or ?
y _ - 69 + ^/241
4
121 + >»./"241
= 5, or-, or ^j ,
Ans.
or
8
294 Adjected Quadratics involving
Q. Given ~H = 4^,(10 find the values of x
J o . « ck \ and y.
and. a?- + t/* = 65, ) ^ ,
Answer,
. ^ . 4>w/— 65 \/-450 + 30x/3410
^= ±4, or ±--^~; ; or 4- ^ ,
7 7
65 15 + ^3410
3/ = 7, or-y ; or -^ ■
10. Given a? + y + 'J x ->r y = 6,1 to find the values of
and a?* J- y = 10, 5 -^ and y.
9± x/'"^^!
Ans.
f.=3, orl;orMj^^,
(3^=1, or 3; or5i^/Hi\
2
11. Given :c* + 5 V '^* + 2 y + 3 = 47 — 23/,
and 6 J? - 53/ = 9,
to find the values of x and y,
32 - 6 + V 2551
r , 32 -6 +
la; = 4, or g- ; or
I ^ 237 — 81 +
(3^ = 3, or -~f; or -;
V 2551
25 ' 25
12. Given x" '\- 3x -\- y = 73 - 20:3/0 to find the values
and y^ ^ sy ^ X t= 44,5 of a? and y.
Ans4^ = 4, or 16; or - 14 ± sf^,
(y= 5, or -7;
or - 1 ± ^ 58,
13. Given . . V3 +— -= . i —
and X = ^
9 + 3^- 119
^O/ = v^, Ul o, ur
Ans.
r 17 ^to find the
" 4y/ X -\-y, > values of x
x = 3/*-|-2,) andy.
two unknown Quantities, 295
14. Given 3/ - ir = l6 - jr,^ to find the values of x
and 28 -. 3/ = 07 -f 4ar^lS . and 7/,
c . ^'
\x = 4, or — ,
/ V= 16, or ^ — .
V*^ ' 289
1 5 Given ^ -4- v^ = QT i
, ^ ^' ^ t to find the values of x and 2/.
and X + 1/ =5^3 '^
i
''[to
Ans. ^ _ 2 •
y = 2, or3;ori±4Zin,
,6. Given \/'-£^ + \/~^^ = 2,,, « ,
^ 207 ^ 307 - 2y '^to find
and a?" - 18 = 07 . (4 2/ - 9).
the values of x and 2/.
To? = 6, or 3,
Ans. i o ^
^y = 3. or-.
17. Given a? jf 4 >v/T + 4 2/ = 2 1 + 8 ^/'^ + 4 ^"r^,
and aJ X •{■ ^ y = 6,
to find the values of 0? and y.
lx= 25^ or — ,
Ans. < ^
18. Given a; +3^ = 5, i to find the values
and (x'+ y') X (a;'+ y') = 455, i of a; and y.
Ans. -j 2-2^ 3_'
(y=3.or3;or!-+i\/-i2?.
Adfeciid Quadratics invQlving
19.Givena;+« + \/i-±J^= -JL), . ,,, ,
T-^T- V ^ __ ^ ^_yy\to find the values
and *^ + y- = 41,j ^'f * *"^ y-
^« = + 5, or ± 3V - ,
Ans. 7 ^
^y = ± 4, or ± V -^ •
20.- Given — , + ^ =^36 2xvJto find the valuej
and ^ H- 4 = 14 -
Ans.
• 1/, / to find the va
i of ^ and y.
\x =6, or 4; or 5 ±5 V - — . ^
(?/ =4, or6; or5T5 V - jj.
21. Given ^^-^-::-^ = 4^,")to find the values of
and 2^.
^^^^'^TT^-rrf'^i'^ofii
v/EI + 1 = _±_ C ^
and
' ^ ^ 9^x-y'J
45
135
64
22. Given sf'^^lJTTTJj ^ 2 \/^ = 9 - 2 V^^
f o 45 3
^^ l. = 3,or-;orj^,or^^,
j ^ 135 1
:r ' "* 8 ' '''* 64
and a? — 2/ = 1 2.
to find the values of x and y,
784
I a; = lo, or
Ans.
Qx = .16. or — ,
{y = 4, or — .
81
23. Given y*— 432 = liJxj/^^tofind the values of x
andj/" = 12 + 2xy,S and y.
Ans
(X = 2,
two unknown Quantities, 297
24. Given —
4 4 + ,v _ 8 + 4.y , I2y ^^ 1
— - ^ -^ __ ^ r " 1 » r
y y X X* r
and ^y' — xy = X, J
to find the
values of
jjc and y.
G = 2,
\y = 1.
= 2, or - -~ ,
Ans. '
50
3
5
or- -
25. Given \/ 1 + ^]=* + ^* + Vl - 4' -Ty" = 4,
and 4 - ^'^' = 18 - 4/,
to find the values of x and y.
^x = ± 1, or ± a/io,
Ans. ^ 3 / — r
^3^ = ± - , or ± 3\/ - i .
26. Given^+-^i^-y- -^f " ^ - ^ = 2 £.,
J? — ^ X + ^ >y/a? + X + y 40
and y* - J-^ = 1^,
to find the values of a: and y.
196 289
\x = y, or
Ans.
#7/ = 4- or — , _.
9 9
i- ^ 196 289 ,^
I J? = 9, or -^ , or — , or 16,
) _ 14 68 4
(y = 4, or __,or-~,or^-.
27. Given ^, -^^ = 4J y , ' . ,
and J?. (j7+ y)= 52 — ^ x"" -\-xy -{-4y
to find the values of x and i/.
10
r lu
t^=±5,or±— -,
(y = + 4, or ± -4= •
aa
298 Adfected Quadratics involving two unknown Quantiti£S.
28.
2 a?* X 1
;.Given^V- +J V- • V^. = 6.
X ^ y 2 ^ 1/ ^ x^
and »- -
3/ 3^3^ 3 '
to find the values of x and i/.
l6 625 625
-^ . or -TT-, or ,
9 ' 64 ' 144'
^^'- 1 . 256 6^' ::6i^^
ia;= 4,
y=64,
= 04, or ^ — , or -- — , or
en
9 ' 32' ' 48 I
29. Given x^ — y^ ^ 3,
and j;* + yl* + a?y . a?' -/T 4- a?* - / = 328,
to find the values of x and y.
C /— T . \/'d + 2j- 13
kx=±2, or±2V - 1, or± V - ^ }
1.3/= ±1, or4:2V -Ijor ± V =—21 .
2
30. Given ^y' - y^ + V4y--l6 7x=^,
and y^ + Vs . (i/ - V^ - 4 = y + 1,
to find the values of x and y.
4 4 4_,^./ 13 788±2x/644
x = 4,or-^,or-,or-- + l6V " -J . or ^^ ,
^ 7 i/"*^ 13 7 ±n/ 644
3/ = 3,or-,or-l,ori,orl±2V -Y'"^"" 25 '
VI. Problems producing Simple Equations^ involving
only one unknown Quantity.
1. What 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.
Problems producing Simple Equations y ^e, 299
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 ?
' Am, 5, 10, 15, 20 yards.
5. A farmer sold 13 bushels of barley at a certain
price ; and afterwards 1 ^ 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 ?
Ans. 9 shillings.
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 as the 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.
8. A vintner fills a cask, containing 96 gallons, with
a mixture 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.
500 Problems producing Simple Equations^
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
c£.2 more than for the third. The sum paid for all was
o£.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 jd, a week more than the second ; the second
Sd, more than the third ; the third 6rf. more than the
fourth ; the fourth 4d. more than the fifth ; and the fiflh
bd, more than the youngest. They altogether earned
10*. 10^. a week. How much could each earn a week ?
Ans. 38, 31, 23, \7, 13, and 8 pence per
week.
1 1 . 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.
12. There are 5 towns, in the order of the letters A, B,
C D, E, From ^ to ^ is 80 miles. The distance
between B and C is 10 miles more, between C and D
is 15 miles less, and between Z) and E \7 miles more
than the distance between A and B, What are the re-
spective distances ?
Ans. from ^ to ^ 17 ; from B to C 27 ; from
CtoD2y 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.
involving only one unknown Quantity. 30 1
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 shillings 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 i^.5 4 more than the other. The whole
gain was ^£.49 less tlian three times the gain of the less.
What were the gains }
Ans. c£.103, and £.157.
17. The perimeter of a triangle is Tb feet, and the
base is 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.
1 8. A company settling their reckoning at a tavern, pay
8 shillings each; but observe,. that if there had been 4
more, they should only have paid 7 shillings each. How
many w^re there ?
Ans, 28.
19. 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.
20. Having bought a certain quantity of brandy at 19
shillings a gallon, and a quantity of rum exceeding that
302 Problems producing Simple Equations,
of the brandy by 9 gallons, at 15 shillings a gallon^ I find
that I paid 1 shilling more for the brandy than for the
rum. How many gallons were there of each.?
Ans, 34 of brandy, and 43 of rum.
21 . Two persons, A and 5, have each an annual income
of ^.400. A spends every year £,40 more than jB, 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.
22. A draper sold two pieces of cloth, by one of which
he lost £.6 more than by the other ; and his whole loss
was ^.5 less than treble the less loss. What were the
losses sustained by each piece ?
Ans. £.11, and £.17.
23. A person engaged to reap a field of corn for 5
shillings an acre, but leaving 6 acres not reaped, he re-
ceived £2. lOs. Of how many acres did the field
consist ?
Ans. 16.
24. In a naval engagement, the number of ships taken
was 7 more, and the number burnt 2 fewer, than the
number sunk. Fifteen escaped, and the fleet consisted
of 8 times the number sunk. Of how many did the
fleet consist ?
Ans. 32.
25. A farmer hires a farm for £.175. 15^. a year; part
of which he is not allowed to plough : he gives £.2 per
acre for the^ arable, and for the rest, which was 5 acres
less, he gives £l. bs. per acre. How m^ny acres were
arable, and how many not ?
Ans. 56 acres arable, 5 1 not ?
involving only one unknown Quantity. 303
26. A cistern is filled in twenty minutes by three
pipes, one of which conveys 10 gallons more, and the other
5 gallons less, than the third, per minute. The cistern
holds 820 gallons. How much flows through each pipe
in a minute?
Ans, 22, 7, and 12 gallons.
27. A and B began to play ; A with exactly - of the Sum
which B had. After winning ^.10, he found that they
had each the same sum. What had each at first ?
Am. A had of .16, and B £.36.
28. A person has a certain number of horses at livery
stables, and three times as many at grass. He keeps
15 in constant employment-, and his whole, number is
seven times the number in the stables. Required the
whole number.
Ans, 35.
29. Two men at the distance of 150 miles set out to
meet each other, one goes 3 miles in the time the other
goes 7* What part of the distance does each travel ?
Am. One 45, and the other 105 miies.
30. A and B begin trade, A with triple the stock
of J?. They each gain c£-50, which makes their stocks
in the proportion of 7 to 3. What were their original
stocks ?
Am. As was X'.300, and Es £.100.
31. A charitable person distributed £,.b. \As. amongst
some poor women and children, giving to each woman
6 shillings, and to each child two; and the number of
women was to the number of children as 4 :-7 How
many were relieved ? "^7i it
Ans. 12 women, and 21 children.
^S04 Problems producing Simple Equations,
32, There are two numbers in the proportion of- to - ,
which being increased respectively by 6 and 5, are in the
2 1
proportion of - to - . Required the numbers.
Ans, 30 and 40.
33 . A farmer has a stack of hay, from which he sells
a quantity which is to the quantity remaining in the pro-
portion of 4 to 5. He then uses 15 loads, and finds that
he has a quantity left which is to the quantity sold as 1
to 2. How many loads did the stack at first contain ?
Ans. 45.
34. There are 3 pieces of cloth, whose lengths are in
the proportion of 3, 5, and 7 ; and 6 yards being cut off
from each, the whole quantity is diminished in the pro-
portion of 20 to 17. Required the length of each piece
at first.
Ans. 24, 40, and 56 yards.
35. The number of days that 4 workmen were em-
ployed were severally as the numbers 4, 5, 6, 7 > their
daily wages were the same, viz. 3 shillings, and the sum
received by the first and second was 36 shillings less than
that received by the third and fourth. How much did
each receive ?
Ans. 36, 45, 54, and 63 shillings.
36. From two casks of equal size are drawn quantities,
which are in the proportion of 6 to 7 ; and it appears that
if 16 gallons less had been drawn from that which is now
the emptier, only half as much would have been drawn
from it as from the other. How many gallons were
drawn frc^ each ?
Ans. 24 and 28.
invohirlg only One unkfiown Quantity, 305
37. On the inclosure of a parish, a proprietor had for his
allotment two pieces of land, which were of the form of
rectangular parallelograms. The longer sides of the
parallelograms were in the ratio of 6 to 11, and the
adjacent sides of the less as 3 to 2. The periphery of
the less was 135 yards more than the longer side of the
greater. Required the sides of the less, and the longer
side of the greater.
Ans. The sides of the less were 90 and 60 -, and the
longer side of the greater was l65 yards.
38. Two persons A and 5 travelling, each with ^£.80,
meet with robbers, who take fi-om A twice as much as
from B, and ^.5 over, and leave A within c£.13 half as
much as B, How much is taken from each ?
Ans, 69, and 32 pounds.
39. A person distributes forty shillings amongst 50
people ; giving to some nine-pence each, and to the rest
fifteen pence. How many were there of each ?
Ans, 45, and 5.
40. A person put out a certain sum to interest for 6|
years, at 5 per cent, simple interest, and found that if he
had put out the same sum for 12 years and 9 months at
4 per cent, he would have received ^.185 more. What
was the sum put out ?
Ans, iJ.lOOO.
'41. Two persons, ^and jB, were partners, -i^'s money
remained in the firm 6 years, and his gain was one-fourth
of his principal ; and JB's money, which was j£.50 less than
A'Sy had been in the firm 9 years, when they dissolved
partnership, and it appeared that if B had gained £.6, 5s.
less, his gain and principal would have been to A's gain
RR
$06 Problems producing Simple Equatkm,
and principal as 4 to 5. What was the principal of
each ?
J71S, 1\200, and ^M 50.
42. A merchant bought wheat at the rate of £,3. 10*. for
5 bushels. He afterwards bought some inferior, which
was in quantity to the former as 3 to 4, at the rate of
£,3, 1 2s. for 8 bushels J and sold the whole for 10
shillings a bushel ; in consequence of which, he lost
£.7, l6s, by the bargain. How much of each did
he buy ?
Ans, 48 bushels of the better, and 36 of the worse.
43. Three persons. A, B, and C, spent equal sums at a
tavern. C having no money, the reckoning was paid by
A and B. When C came to reimburse them, he paid
4 times as much to A as to 5, and observed, that if B
had paid 3 shillings more of his reckoning, their demands
would have been equal. Required the sum each spent,
and the respective parts of Os reckoning that A and B
paid.
Ans, Each spent 10 shillings ; ^paid 8, and B 2.
44. A farmer had 2 flocks of sheep, one of which
contained 40, and the other was sold for ^6.30; but one
sheep of the latter was worth four of the other; and the
value of the first flock was only £a more than the price
of eight sheep of the second. How many sheep did the
second flock contain, and what was the value of a sheep
of each ?
Ans. The number was 15, and the prices £.2,
and 10 shilHngs.
45. A and B playing at billiards, A bet 5 shillings to
4 on every game, and found that after a certain number
of games he had won 10 shillings. Had B won ono
game more, the number won by him would have been
involving onhf one unhtown Quantity, 307
to the number won by A as 3 to 4. How many did
each win ?
Ans, ^won 20, and 5 14.
46. A besieged garrison had such a quantity of bread,
as would, if distributed to each at 10 ounces a day, last
6 weeks; but having lost 1200 men in a sally, the
governor was enabled to encrease the allowance to 12
ounces per day for 8 weeks. Required the number of
men at first in the garrison.
Ans. 3200.
47. A composition of copper and tin containing 100
cubic inches weighed 505 ounces. How many ounces
of each metal did it contain, supposing a cubic inch of
copper to weigh 5| ounces, and a cubic inch of tin to
weigh 4 1 ounces.
Ans, 420 of copper, and 85 of tin.
48. There are two towns, A and J5, which are 131
miles distant from each other. A coach sets out from A
at 6 o'clock in the morning, and travels at the rate of 4
miles an hour without intermission, in the direct road
towards B. At 2 o'clock in the afternoon of the same day
a coach sets out from B to go to A, and goes at the rate
of 5 miles an hour constantly. Where will they meet ?
Ans. 7^ miles from A, and 55 from B.
49. Out of a certain sum, a man paid his creditors ,£.96 ;
half of the remainder he lent his friend ; he then spent
one-fifth of what now remained ; and after all these deduc-
tions had one-tenth of his money left. How much had
he at first ?
Ans. <£.128.
50. A and B, in order to keep up the price of copper
308 Problems producing Simple Equations y
to £,S6 per ton, agree for a certain time to sell all the
copper they mise, jointly ; yet so that each shall be paid
proportionably to the quantity he raises. Now the whole
quantity raised in the stipulated time was 235 tons;
and J receives £.4214 more than B. Required the
quantity raised by each.
^ns, 142 tons by J, and 93 by B.
51. Bought two pieces of linen, one of which wanted
12 yards of being four times as long as the other. The
longer cost 5 shillings, and the shorter 4 shillings a yard ;
23 yards being cut off from the longer, and 5 from the
shorter, and the remainders being sold for 1 shilling
a yard more than they cost, I received £.7, 2s, How
many yards of each were there ?
Ans, 40, and 13.
52. A courier, passing through a certain place A,
travels at the rate of 13 miles in 2 hours; 12 hours
afterwards another passes through the same place, tra-
velling the same road, at the rate of 26 miles in 3 hours.
How long, and how far must he travel before he over-
takes the first.
Arts, 36 hours, and 312 miles.
53. As ^ and B were going to school, J^ first shot an
arrow in the direction in which they were going, which
B took up and shot forward ; and so on alternately till the
arrow had passed exactly from one mile- stone to another ;
when it appeared that A had shot the arrow 8 times,
and B 7 times. . Some" time afterwards, A and B were
on the opposite banks of a river, the breadth of which
they wished to ascertain ; A first shot the arrow across
the river, and it flew 13 yards beyond the bank on
which B stood ; B then took it up, and from the place
where it had fallen, shot it back across the river : it now
involving only one unknown Quantity. 309
fell 92. yards beyond the bank upon which A stood*
Required the breadth of the river.
Ans. 100 yards.
.54. Three merchants, A, B, and C, enter into a specu-
lation; B subscribes oC. 1 more than four-fifths of what
A floes; and C c£.30 more than half of what 5 does. A's
gain is two- fifths of his subscription, and B's is
^.148. What are the respective sums subscribed, and
the whole gain ?
Ans. The sums subscribed are 450, 370, and
215 pounds; and the whole gain is ^.414.
55^ There are two places, 154 miles distant, from
which two persons set out at the same time to meet, one
travellino; at the rate of 3 miles in two hours, and the
other at the rate of 5 miles in four hours. How long,
and how far did each travel before they met.
Am. 56 hours ; and 84, and 70 miles.
bS. A sets out from a certain place, and travels at
the rate of 7 miles in ^ve hours ; and eight hours after-
wards B sets out from the same place, and travels the
same road at the rate of 5 miles in three hours. How
long, and horn far must A travel before he is overtaken
hy B.
Ans. 50 hours, and 70 miles.
57. A man lent out a certain sum to interest at ^.8
per cent, per annum. He suffered this to accumulate at
simple interest for 12 years; and then putting out the
principal and interest at the same rate found that the
present annual interest exceeded the former by c£.38. 8s,
Jlequired the sum put out each time.
Ans. o£.500 the first time, and £.^80 the second.
3 10 Priiblems producing Simple Equations,
58. A waterman finds by experience that he can with
the advantage of a common tide row down a river from^
to B which is 18 miles in an hour and a half, and that
to return from B to A against an equal tide, though
he rows back along the shore, where the stream is only
three-fifths as strong as in the middle, takes him just two
hours and a quarter. It is required from hence to find at
what rate per hour the tide runs in the middle where it
is strongest.
Ans, At the rate of two miles and a half per hour.
59- The ingredients of a loaf of bread are rice, flour,
and water, and the weight of the whole is 15lbs. The
weight of the rice augmented by 5lbs. is two- thirds of the
weight of the flour, and the weight of the water is one-
fifth of the weight of the flour and rice together. Re-
quired the weight of each.
Ans, Rice 2 lbs., flour 10^ lbs., water 2} lbs.
60. Suppose two fingers of a watch (a) and (b) were
together on Sunday noon at 1 2 o'clock, and that the motion
of each was such that (a) moved round the horarycircle
in one hour, and b in l^ hour. When wilbthey be toge-
ther again for the first time ?
Ans. 6 1 hours. ^
61. A draper bought a piece of cloth i^o£'.69, from
which he cut off* 1 1 yards. He then met with another
piece of equal goodness for which he gave .£.21, and
found that if it had been one yard longer, its length
would have been to the length of the remainder of the
first as 2 to 3. How many yards were there in each
piece, and what was the price of a yard ?
Ans. 23 in the first, and 7 yards in the second ;
and the price £.3 per yard.
62. Divide the number igs into 5 such parts, that the
involving only one unknown Quantity. ^^ 1 1
first increased by one, the second increased by two, the
third diminished by three, the fourth multiphed by four,
and the fifth divided by five, may be all equal,
Ans. 23, 22, 27, 6, and 120, are the numbers.
63. A and B set out from two places, C and 7>, at the
same time, towards E ; the road from C to E being
through D, A travels 7 miles an hour, and at that rate
of travelling would have overtaken B 5 miles before he
got to E ; but after arriving at Z>, he travels 6^ miles an
hour, in consequence of which he overtakes 5 just as he
enters E, Supposing B to travel 5 nliles an hour, what
are the distances between C, Z), and E,
Arts, From C toD 14 miles, and from D to E 40.
64. A gentleman wishing his two daughters to
receive equal portions when they became of age, be-
queathed to the elder the accumulated insterest of a
certain sum of money, bought at the time of his death
into the 4 per cent, stock at 88 ; and to the younger
the accumulated interest of a sum less than the former
by ^.3500, bought at the same time into the 3 per
cents, at 63. Supposing their ages at the time of their
father's death to have been 17 and 14, what would be
the sum bought into the stocks in each case, and what
would be th^Atune of each ?
A)is, ^^e sums would be c£.7700, and o6'.4200 >
and fortune o£'.1400.
65. Out of a common pack of cards, a certain number,
including the ten of diamonds, was dealt equally amongst
four persons, the dealer turning up the last card, which was
the ten of spades, which he gives himself. Now if twice
the number of cards had been dealt to each, the ten of
spades being turned up by the dealer, and the ten of
diamonds being still dealt out, the chance of the dealer's
S12 Problems pyodtictng Simple Equations f
having the ten of diamonds would be to the chance
ajjainst him as 3 : 10. Required the number of cards
dealt to each the second time.
Ans. 10.
66. Two companies of soldiers consisting of equal num-
bers were sent out under A aftd B, from two liostile camps,
to reconnoitre. Falling in with each other, a skirmish
ensued, in which A lost 50 killed and prisoners, and B
had 20 killed. A however having being reinforced by a
party equal to five-sevenths of the number which B had
remaining 5 and B having been reinforced by a number
greater by 46 than three-fifths of the number which A had
remaining, they renewed the engagement, when A \yas
forced to retire with the additional loss of 30 men.
When the returns were made, B found he had again lost
20 men, but that he had then twice as many men re-
maining as A had. How many had each at first ?
Ans. 90.
6T. a sportsman, who kept an account of the number
of birds which he killed, found that each succeeding
season he wanted 50, in order that the number killed
might bear the proportion of 3 : 2 to the number killed
in the preceding year. In the fourth year^e found that
he had killed I/O fewer than three timljPthe number]
killed in the first year. How many did be kill the first
year?
Ans. 180.
6S, Four men walking abroad found a purse containing
shillings only, out of which every one of them took a
number at a venture. Afterwards comparing their
numbers together, they found that if the first took 25
shillings from the second, it would make his number equal
to what the second bad left. If the second took 30
involving one unknown Quantiit/, SIS
shillings from the third, his money would then be triple
what the third had left. And if the third took 40 shil-
lings from the fourth, his money would then be double of
what the fourth had left. Lastly, the fourth taking 50
shillings from the first, he would then have three times
as much as the first had left, and 5 shillings over. What
had each ?
Ans, 100, 150, 90, and 105 shillings, respectively.
69* Fifteen current guineas should weigh 4 ounces ; but
a parcel of light gold being weighed and counted, was
found to contain 9 more guineas, than was supposed from
the weight ; and a part of the whole, exceeding the half by
10 guineas and a half, was found to be 1-^ oz. deficient
in weight. What was the number of guineas ?
Ans. I89.
70. A merchant bought a quantity of wheat for of .200,
half of which he reserved for his private use. He then
sold 3 bushels more than 4- of the remaining quantity at
such a price as to gain c£.40 per cent. But the price of
wheat having advanced, he sold the remainder at such a
price as to gain £.67 per cent, by what he then sold.
And had the whole been sold at this latter price he would
have gained ^. 160 per cent. How much did he buy^ and
how did he ^ell it ?
Ans, He bought 400 bushels; and sold the first
portion at 14^., and the second at 26s. per
bushel.
Vn. Problems producing Simple Equations , involving
two unknown Quantities.
1. A draper bought two pieces of cloth foro£*.12. 13*.;
one being 8^., 9jxd the other 9^. pqr yard. He sold them
s s
314 ProhUms producing Simple Equations f
each at an advanced price of 2*. per yard, and gained by
the whole £.3, What were the lengths of the pieces?
Ans. 17 yards of the first, and 13 of the second.
3. A bill of £,26, 5s. was paid with half guineas and
crowns, and twice the number of half guineas exceeded
three times the number of crowns by 17. How many
were there of each ?
Ans, 40 half guineas, and 21 crowns.
3. Two labourers, A and B, received £.5. 17^. for their
wages ; A having been employed 15, and and B 14 days ;
and A received for working four days 1 1 s. more than B
did for three days. What were their daily wages ?
Ans, A had 5^., and B 3s, a day.
4. A person had two casks, the larger of which he
filled with ale, and the smaller with cyder. Ale being
half a crown, and cyder 11^. per gallon, he paid £,S, l6s, ;
but had he filled the larger with cyder, and the smaller
with ale, he would have paid i^.ll. 5*. 6d. How many
gallons did each hold ?
Ans, The larger contained' 18, and the smaller 11
gallons.
5. A person expends half a crown in apples and pears,
buying his apples at 4, and his pears at 5 a penny ; and
afterwards accommodates his neighbour with half his
apples and one- third of his peats for 13 pence. How
many did he buy of each ?
Ans, 72 apples, and 60 pears.
6* Two persons, A and B, played cards, each with a
diflferent sum. After a certain number of games, A had
won half as much as he had at first, and found that if he
had 1 5s. more he would have had just three times as
Mvohing two unknown Qimntit'tes, 3 1 5
much as Bt But B afterwards won 10.?. back, and he
had then twice as much as A, What had each at first ?
Am. A had 14, and B 19 shilHngs.
7. A certain sum of money put out to interest, amounts
in 8 months to £.2^T, 12a.; and in 15 months its
amount is ^.306 at simple interest. What is the sum,
and the rate per cent. ?
Am. c£. 288, at 5 /?er cent.
8. A farmer being asked how many quarters of wheat he
had sold in the market, answered if he had sold 8 quar-
ters more, and got 'Js, 'per quarter more than he did, he
should have received £A\. \bs. more than he had: but
if he had sold ^ quarters more at 8*. per quarter more,
he should have had c^.ll. 17^. more. How many
quarters did he sell, and what was the price ?
Am, 13 quarters, at 11^. per quarter.
9. There is a number consisting of two digits, the
second of which is greater than the first ; and if the
number be divided by the sum of its digits, the quotient
is 4 : but if the digits be inverted, and that number
divided by a number greater by 2 than the difference of
the digits, the quotient becomes 14. Required the
number.
Ans. 48.
10. What fraction is that, whose numerator being
doubled, and denominator increased by 7, the value
2
becomes - ; but the denominator being doubled, and the
%j
3
numerator increased by 2, the value becomes - ?
Am. - .
5
516 Problems producing Simple Equations,
1 1 . A farmer parting with his stock, sells to one person
9 horses and 7 cows for i^.300 ; and to another, at the
same prices, 6 horses and 13 cows for the same sum.
What was the price of each ?
Ans, The price of a cow was £.12^ and of a horse
^ i:.24.
12. A farmer hires a farm for ^.245 per ann,, the
arable land being valued at £»2 an acre, and the pasture at
28 shillings : now the number of acres of arable is to half
the excess of the arable above the pasture as 28 : 9-
How many acres were there of each ?
Ans. 98 acres of arable, and 35 of pasture.
13. A person owes a certain sum to two creditors.
4
At one time he pays them «£ .53, giving to one — of
the sum which is due^ and to the other £.3 more than
TV of his debt to him. At a second time he pays them
3
«£*.41, giving to the first ]^ of what was due to him, and
to the other - of what was due to him. What were the
debts ?
Ans, <£.121, and £.36.
14. A and B playing at backgammon, A bet 3^. to 2*.
on every game, and after a certain number of games found
that he had lost 17 shillings. Now had A won 3 more
from B, the number he would then have won, would
have been to the number B would have won as 5 to 4,
How many games did they play ?
Ans. 9.
involving two unknown Quantities, S17
15. A vintner has two casks of wine, from the greater
of which he draws 15 gallons, and from the less 11 ; and
finds the quantities remaining in the proportion of 8 to S*.
After they become half empty, he puts 10 gallons of water
into each, and finds that the quantities of liquor now in
them are as 9 to 5. How many gallons will each hold ?
Ans, The larger 79, and the smaller 35 gallons.
16. A person having laid out a rectangular bowling-
green, observed that if each side had been 4 yards-longer,
the adjacent sides would have been in the ratio of 5 to 4 ;
but if each had been 4 yards shorter, the ratio would have
been 4 to 3. What are the lengths of the sides?
Ans, 36, and 28 yards.
17. At an election for two members of parliament,
three men oflfer themselves as candidates, and all the
electors give single votes. The numbers of voters for the
two successful ones are in the ratio of 9 to 8 ; and if the
first had had 7 more, his majority over the second would
have been to the majority of the second over the third as
12:7- Now if the first and third,had formed a coalition,
and had one more voter, they would each have succeeded
by a majority of 8. How many voted for each ?
Ans, 369, 328, and 300, respectively.
18. Two shepherds, A and B, are intrusted with the
charge of two flocks of sheep. A^s consisting chiefly of
ewes, many of which produced lambs, is at the end of
the year increased by 80 ; but B finds his flock dimi-
nished by 20 ; when their numbers are in the proportion
of 8 to 3. Now had A lost 20 of his sheep, and B had
an increase of 90, the numbers would have been in the
proportion of 7 to 10. What were the numbers ?
Ans. As 160, and B's 110.
313 Problems producing Simple Equations ^
19. Two persons, A and B, can perform a piece of work
in 16 days. They work together for 4 days, when A
being called off, B is left to finish it, which he does in
36 days more. In what time would each do it separately?
Ans. A in 24 days,, and B in 48 days.
20. There is a cistern, into which water is admitted by
three cocks, two of which are of exactly the same dimen-
5
sions. When they are all open, — of the cistern is filled
in four hours ; and if one of the equal cocks be
7
stopped, - of the cistern is filled in ten hours and forty
minutes. In how many hours would each cock fill the
cistern ?
Ans, Each of the equal ones in 32 hours, and the
other in 28.
21. Some hours after a courier had been sent from A
to By which are 147 miles distant, a second was sent,
who wished to overtake him just as he entered B ; in
order to which he found he must perform the journey in
28 hours less than the first did. Now the time in which
the first travels 17 miles added to the time in which the
second travels 56 miles is 13 hours and 40 minutes.
How many miles does each go pet' hour ?
Ans, The first goes 3, and the second 7 miles an
hour.
22. Two loaded waggons were weighed, and their
weights were found to be in the ratio of 4 to 5. Parts of
their loads, which were in the proportion of 6 to 7 being
taken out, their weights were then found to be in the
ratio of 2 to 3 ; and the sum of their weights was then
ten tons. What were the weights at first?
Ans. 16, and 20 tons.
involving two unknown Quantities. 319
23. A gentleman gave away a certain sum in charity
to 14 men and 15 women. Had the sum been less by
12 shillings, and only half the number of men relieved,
the rest being divided amongst the women, each woman
would have received two shillings more than each man
did. But if there had been only 8 women, and the rest
had been divided amongst the men, each man would
have received twice as much as each woman. How
much money was given away ?
yins. 24 guineas.
24. When wheat was 5 shillings a bushel, and rye 3
shillings, a man wanted to fill his sack with a mixture of
rye and wheat for the money he had in his purse. If
he bought 7 bushels of rye, and laid out the rest of his
money in wheat, he would want 2 bushels to fill his sack ;
but if he bought 6 bushels of wheat, and filled his sack
with rye, he would have 6 shillings lefl. How must he
lay out his money, and fill his sack ?
Ans, He must buy 18 bushels of wheat, and 12
bushels of rye.
25. A draper bought two pieces of cloth of different
kinds for oC-37. 4a\ : there were 6 yards of the coarser
more than there were of the finer; and had the coarser
cost 2 shillings a yard more than it did, 6 yards of the
coarser would have cost just as much as 5 yards of the
finer. He afterwards bought 4 yards of the finer, and
1 2 of the coarser at the same prices pe?' yard, and found
their value less than that of the former pieces in the ratio
of 20 : 31. How many yards did he buy the first time,
and what did he give per yard for each ?
j4ns, 9 yards of the finer, and 15 of the coarser ;
and the prices were 36, and 28 shillings per
yanl.
S20 Problems producing Simple Equations,
26. A mercer bought two pieces of silk of different
lengths for ^.50.; the price of two yards of the shorter
was 6s, 8d, more than the price of 3 yards of the longer^
and each piece cost the same sum. He cut off two yards
from each^ and sold the rest for £.53, I2s. Now if he had
sold the whole at that rate, he would have gained ^£.5
by each piece. How many yards did each piece contain?
Ans. 25, and 15 yards.
27. ^ sets out express from C towards D, which is
112 miles distant, and travels a certain number of miles
pel' hour. Three hours afterwards B sets out from D
towards C, and goes 2 miles an hour more than -^, and
when they meet, finds that he has travelled 60 miles.
Now had ^travelled five hours less, and B gone 2 miles
an hour more, he would have travelled 80 miles at their
meeting. How many miles did each go per hour, and
how many hours did they travel before they met ?
Ans. A went 4, and B 6 miles an hour, and they
travelled 10 hours after B set out.
28. A and B engaged to reap a field of corn in 12
days. The times in which they could severally reap
an acre are as 2:3. After some time, finding them-
selves unable to finish it in the stipulated time, they called
in C to help them ; whose rate of working was such, that
had he wrought with them from the beginning, it would
have been finished in 9 days. Also the times in which
he could severally have reaped the field with A alone, and
with B alone, are in the proportion of 7 to 8. When
was C called in?
Ans, After 6 days.
29. Two mixtures are made of brandy and sherry; the
quantities of brandy in each being as 4 to 3 ; and the
difference of the quantities of sherry being greater by
involving two unknown Quantities. 32 1
25 gallons than the difFerfence of the quantities of brandy.
Also, if three times the quantity of brandy had been put
into the first mixture, and twice the quantity into the
second, the quantities of brandy would have been
proportional to the quantities of sherry. But if the
sherry in the second mixture had been mixed with the
brandy in the first, and the sherry in the first with the
brandy in the second, the whole mixtures would then
have been in the ratio of 5 to 6. Required the quan-
tities of brandy and sherry in each mixture.
Ans. The quantities of brandy are 80, and 6o gallons,
and the quantities of sherry are 90, and 45 gallons.
30. Two Spanish muleteers, A and B, were seated under
a tree in order to dine ; and on examining, found their
stock of provisions to consist of 5 small loaves ^of bread,
three of which were As property, and a bottle 'Of wine,
which was '-B's. A stranger, who happened to come
up at the time, was invited to partake of their fare,
which was just sufficient for three persons ; and at parting,
being pleased with their behaviour, he gave them what
Spanish money he had about him, which amounted to
6s, 5f«?., to be equitably shared between them. Now
as many shillings as a loaf cost pence would, with four-
pence more, at the next town have bought six such loaves
and four bottles of the same wine ; and when the money
was divided, B received 1^. lO^d. more than ^. What
was the price of each loaf, and a bottle of wine ?
AnSi A loaf cost 9 pence, and a bottle of wine
1 1 J pence.
VIII. Problems producing pure Equations,
1. Find two numbers, which are in the proportion of
8 to 5, and whose product is equal to SGo.
Am,. ± 24, and ±15.
T T
222 Problems producing pure Equations,
2. There are two numbers, 'whose sum is to their difFer-
rence as 8 to 1^ and the difference of whose squares is
128. What are the numbers ?
Ans, ± 18^ and ± 14.
3. In a court there are two square grass-plots; a side
of one of which is 10 yards longer than the side of the
other ; and their areas are as 25 to 9. What are the
lengths of the sides ?
Ans, 25, and 15 yards.
4. A person bought two pieces of linen, which together
measured 36 )rards. Each of them cost as many
shillings per yard, as there were yards in the piece ; and
their whole prices were in the proportion of 4 to 1.
What w*Te the lengths of the pieces ?
Ans. 24, and 12 yards.
5. There are two numbers, whose sum is to the
less as 5 to 3; and whose difference, multiplied by
the difference of their squares, is 135. Required the
numbers.
Ans, 9, and 6.
6. There are two numbers, which are in the pro-
portion of 3 to 2 ; the difference of whose fourth
powers is to the sum of their cubes as 26 to 7. Required
the numbers.
Ans, 6, and 4.
7. There is a field in the form of a rectangular paral-
lelogram, whose length is to its breadth in the propor-
tion of 6 to 5. A part of this, equal to one- sixth of the
whole, being planted, there remain for ploughing 625
square yards. What are the dimensions of the field ?
Ans. The sides are 30, and 25 yards.
Problems producing pure Equations, S23
8. Some gentlejnen made an excursion ; and every
one took the same sum. Each gentleman had as many,
servants attending him as there were gentlemen ; and
the number of pounds which each had was double the
number of all the servants; and the whole sum of
money taken out was «£.3456. How many gentlemen
were there?
Ans, 12.
9. Divide the number 49 into two such parts, that the
quotient of the greater divided by the less may be to
4
the quotient of the less divided by the greater as —
a
to?.
4
Am, 28, and 21.
10. A detachment of soldiers from a regiment being-
ordered to march on a particular service, each company
furnished four times as many men as there were com-
panies in the regiment ; but these being found ta be
insufficient, each company furnished 3 more men ;
when their number was found to be increased in the
ratio of 17 to 16. How many companies were there in*
the regiment ? .
Ans. 12.
11. A charitable person distributed a certain sum
amongst some poor men and women, the numbers of
whom were in the proportion of 4 to 5. Each man
received one-third of as many shillings as there were
persons relieved ; and each woman received twice as many
shillings as there were women more than men. Now
the men received all together 18^. more than the women.
How many were there of each ?
Ans. 12 men, and 15 women.
324 Froblems producing pure Equations,
13. A gentleman who had a certain number of horses,
kept part of them at livery stables, for which he paid
£A, \0s, per week. The rest he kept at home, and
their number was to the number kept at the livery
stables as 7 to 3. He found that the expence of keeping
5^at home was just equal to that of keeping 4 at the
stables ; and the number of shillings that 1 horse cost
him at home was to the number of horses kept at home
as 6 to 7. How many horses had he ?
Ans. 6 at the livery stables, and 14 at home.
13. A city barge, with chairs for the company and
benches for the rowers, went a summer excursion, with
two bargemen on every bench. The number of gentle-
men on board was equal to the square of the number of
bargemen, and the number of ladies was equal to the
number of gentlemen, twice the number of bargemen,
and one over. Among other provisions, there were a
number of turtles equal to the square root of the
number of ladies; and a number of bottles of wine less
than the cube of the number of turtles by 36 1. The
turtles in dressing consumed a great quantity of wine,
and the party having staid out till the turtles were all
eaten, and the wine all gone, it was computed, that sup-
posing them all to have consumed an equal quantity,
(viz. gentlemen, ladies, bargemen, and turtles,) each indi-
vidual would have consumed as many bottles as there
were benches in the barge. Required the number of
turtles.
Ans. 19.
14. From two towns, C and.Z), two travellers, -^and By
set out to meet each other ; and it appeared that when
3
they met, B had gone 35 miles more than - of the dis-
ProbUms producing pure Equations, 325
tance that A had travelled ; but from their, rate of tra-
velling, A expects to reach Cm 20 hours, and 50 minutes ;
and B to reach Z) in 30 hours. Required the distance
of C from D.
Ans. 275 miles.
1 5 . A farmer bought two flocks of sheep, the first of
which contained 1 8 fewer than the second. If he had
given for the first flock as many pounds as there were
sheep in the second, and for the second as many pounds
as there were sheep in the first, then the price of 6
sheep of the first flock would have been to the price of
7 sheep of the second in the proportion of 7 to 6. Re-
quired the numbers in each flock.
Am, 108, and 1 26.
16. A poulterer bought a number of ducks and turkeys,
the number of ducks exceeding the number of turkeys by
8. For each duck he gave half as many shillings as
there were turkeys, and for each turkey half as many shil-
lings as there were ducks. He afterwards bought another
small flock of turkeys containing 4 fewer than the
number of turkeys h^ bought before ; and having given
for each of them as many shillings as there were turkeys
in the flock, he found, that if his former purchase had cost
16 shillings more, it would have cost exactly four times
as much as the present one. How many ducks and
turkeys did he buy at first ?
Ans. 12 turkeys, and 20 ducks.
17. Two men, A and B, enter into partnership with
stocks which are in the proportion of 9 to 8 ; and after
trading one year, A finds his share of their gain to amount
to one-third of his stock. They continued to trade for as
many years as are equal to three-fourths of the number
r>f pounds which B contributes to the stock, and find
326 Problems producing pure Equations.
their whole gain amount to «£.l666. What did each
contribute to the stock ; and how many years did they
trade ?
Ans. A contributed £.^3, B £,b^ ; and the
number of years is 42.
18. A person wishing to ascertain the area of a certain
quadrilateral field, found that he could determine it the
most readily by dividing it into two portions, one of
which was of the form of a rectangular parallelogram, the
shorter side of which measured 6o yards. The other was
of the form of a right-angled triangle, whose shortest side
was equal to the shorter side of the parallelogram, and the
other side, containing the right-angle, was equal to the
diagonal of the parallelogram ; and the area of the triangle
was to the area of the parallelogram as 5 to 8. What'
was the area of the field ?
Ans. 7800 square yards.
19. A merchant laid out a certain sum upon a specu-
lation, and found at the end of a year that he had gained
^.69. This he added to his stock, and at the end of
another year found that he had gained exactly as much
'per cent, as in the year preceding. Proceding in the
same manner, and each year adding to his stock the
gain of the year preceding, he found at the beginning of
the fifth year that his stock was to the original stock as
81 to 16. What was the sum he first laid out ?
Ans. £a2>%,
20. There is a number consisting of two digits, which
being multiplied by the digit on the left hand, the pro-
duct is 46 ; but if the sum of the digits be multiplied by
the same digit, the product is only 10, Required the
number.
Ans. 23.
Problems producing pure Equations. 32*7
21. Prom two towns, C and D, which were at the
distance of 396 miles, two persons, A and B, set out at
,the same time, and meet each other, after travelling as
many days as are equal to the difference of the number
of miles they travelled per day ; when it appears that A
has travelled 21 6 miles. How many miles did each
travel per day ?
Am, A went 36, and B 30.
22. There are two numbers, whose sum iS' to the
greater as 40 is to the less, and whose sum is to the less
as 90 is to the greater. What are the numbers ?
Am, 36, and 24.
23. It is required to find two numbers such, that the
product of the greater and the cube of the less may be
to the product of the less and cube of the greater as 4 to
9; and the sum of the cubes of the numbers may be 35.
Am. 3, and 2.
24. The paving of two square court-yards costj£.205 ;
a yard of each costing one-fourth of as many shillings as
there were yards in a side of the other. And a side of the
greater and less together measure 41 yards. Required
the length of a side of each.
Ans. 25, and 16 yards.
25. A person bought a number of apples and pears,
amounting together to 80. Now the apples cost twice as
much as the pears : but had he bought as many apples
as he did pears, and as many pears as he did apples, his
apples would have cost lOc?., and his pears 3s. gd. How
many did he buy of each ?
Ans. 60 apples, and 20 pears.
26. A person. exchanged a quantity of brandy for a
828 Problems producing pure Equations,
quantity of rum and ^.11.5^.; the brandy and rum
being each valued at as many shilHngs 'per gallon as there
were gallons of that liquor. Now had the rum been
worth as many shillings per gallon as the brandy was,
the whole value of the rum and brandy would have been
^.56. 5^. How many gallons were there of each ?
Ans. 25 gallons of brandy, and 20 of rum.
27. There are two rectangular vats, the greater of
which contains 20 solid feet more than the other.* Their
capacities are in the ratio of 4 to 5 ; and ^heir bases are
squares, a side of each of which is equal to the depth
of the other. What are the depths ?
Ans. 5 feet, and 4 feet.
28. Bought two "square carpets for ^£.62. 1^. 5 for each
of which 1 paid as many shillings per yard as there
were yards in its side. Now had each of them cost as
many shillings per yard as there were yards in a side of
the other, I should have paid 17*. less. What was the
size of each ?
Ans. One contained 81, and the other 64 square
yards.
29. The number of men in both fronts of two columns
of troops, A and B, when each consisted of as many ranks
as it had men in front, was 84 ; but when the columns
changed grovmd, and A was drawn up with the front B
had, and B with the front A had, the number of ranks
in both columns was 91. Required the number of men
in each column.
Ans. 2404, and I296.
30. A field in the form of a rectangular parallelogram
was planted with trees placed at such distances as to have
four on every square yard. The expence of planting
S.\'
Problems producing pure Equations . 329
was such, that every 40 trees cost one-third of as many
shillings as there were yards in the diagonal of the paral-
lelogram. But had they been planted at such a price as
that every hundred should have cost as many shillings as
there were yards in the shorter side of the parallelogram,
the expence would have been less by ^£.224. Now a square
described upon the diagonal of the parallelogram would
8
be equal to - of the square described on the less side,
together with, the square described on a line which is
equal to the difference of the sides. Required the dimen-
>ions of the parallelogram.
Am. The longer side is 80, and the shorter 6o
yards.
•«-«v#**><-*v#>^«><
IX. Problems producing Adfected Quadratics.
1. What two numbers are those, whose sum is 19, and
yi'hose difFerence multiplied by tHe greater is 60 ?
Ans, 12, and 7»
2. If^ the square of a certain number be taken from 40/
arid the sqiiare root of this difference be increased by 10,
and the sum multiplied by 2, and the product divided by
the number itself, the quotient will be 4. Required the
number.
Ans. 6.
3. There is a field in the form of a rectangular paral-
lelogram, whose length exceeds the breadth by l6 yards ;
and it contains 96O square yards. Required the length
and breadth.
Ans. 40, and 24 yards,
uu
3S0 Troblems producing AdfecUd Quadratics,
. 4. A person being asked his age answered, if you add
the square root of it to half of it, and subtract 12, there
will remain nothing. Required his age.
Alls. 16.
5. Two casks of ale were bought for £,2. IBs, one of
which contained 5 gallons more than the other, and the
price pe7' gallon was 2 shillings less than one-third of the
number of gallons in the less. Required the number of
gallons in each, and the price pet' gallon.
Ans. The numbers were 1 2, and 1 7> and the price
per gallon 2 shillings.
6. From two places at the distance of 320 miles, two
persons, A and B, set out at the same time to meet each
other. A travelled 8 miles a day more than B, and the
number of days in which they met was equal to half the
number of miles B went in a day. How many miles
did each travel per day, and how far did each travel ?
Ans, A went 24, and B l6 miles per day ; A went
128, and B 192 miles.A .>e«^ >^i^^ - /^^/ikJd^
7. The difference between the hypothenuse and base
of a right-angled triangle is = 6, and the difference
between the hypothenuse and the perpendicular is = 3.
What are the sides ?
Ans, 9, 3, and 6. to. 4^v^^..^i,=^ /F^^ .ft
8. In a parcel which contains 24 coins of silver and
copper, each silver coin is worth as many pence as there
are copper coins, and each copper coin is worth as many
pence as there are silver coins, and the whole is worth
18 shillings. How many are there of each ?
Ans. 6 of one, and 18 of the other.
ProbUms producing Adfected Quadratics- 331
9. A farmer received ^£.7- 4*. for a certain quantity of
wheat, and an equal sum at a price less by 1*. 6^. 'per
bushel for a quantity of barley, which exceeded the
quantity of wheat by 16 bushels. How many bushels
were there of each ?
Am. 32 bushels of wheat, and 48 of barley.
10. Two messengers, A and 5, were dispatched at the
same time to a place 90 miles distant ; the former of
whom riding one mile an hour more than the other,
arrived at the end of his journey an hour before him.
At what rate did each travel per hour ?
Arts, A went 10, and B 9 miles per hour.
1 1 . Bought a number of books, consisting of folios,
quartos, and octavos, for «£.96. 12*. Fourteen foHos
(which was the whole number) cost 3 times as much as
all the quartos ; and one quarto cost as many shillings as
there were quartos. The number of octavos was 32,
and their value was such, that 4 of them cost as much as
one quarto. Required the value of each, and the number
of quartos.
Ans, There were 21 quartos, each foho cost 4^
guineas, each quarto one guinea, and each
octavo 5*. Zd,
12. A man travelled 105 miles, and then found that
if he had not travelled so fast by 2 miles an hour, he
should have been 6 hours longer in performing the same
journey. How many miles did he go per hour ?
Ans, 7 miles.
13. Bought two flocks of sheep for ^.65. 13*., one,
containing 5 more than the other. Each sheep cost as
many shillings as there were sheep in the flock. Re-
quired the numbers in each flock.
Ans. 23, and 28.
^52 Problems j)rQduc'tng Adjected Quadratics.
14. A regiment of soldiers, consisting of 1066 men, is
formejd into two squares, one of which has four men niore
in a side than the other. What number of men are in
a side of each of the squares ?
Ans, 21, and 25.
15. What number is that, to which if 24 be added,
and the square root of the sum extracted, this root shall
be less than the original quantity by 18.
Ans. 25 .
16. After takmg the kings, queens, and knaves out of
a pack of cards, the rest were divided into three heaps.
The number of pips contained in the second heap was
found to be 4 times the square of the number in the first
heap ; and had the third heap contained 5 more pips
than it did, the number in it would have been exactly
half of what the first and second heap contained. Re-
quired the number of pips in each heap.
Ans. 6, 144, and 70.
17. A tailor bought a piece of cloth for <£.147, from
which he cut off 12 yards for his own use, and sold the
remainder for ^.120. 5*. gaining 5 shillings /?er yard.
How many yards were there, and what did it cost him
-per yard ?
Am, 49 yards, at ^£.3 'per yard.
18. A regiment of foot was ordered to send 2l6 mei;i
on garrison duty, each company being to furnish an
equal number ; but before the detachment marched, 3 of
the companies were sent on another service, when it was
found that each company that remained was obliged to
furnish 12 additional men, in order to make up the com-
plement 216. How many companies were there in the
Problems producing Adjected Quadratics- 353
regiment, and \vhat number of men was each company
ordered to send at first ?
Ans. There were 9. com panics; and each waste
send 24 men.
19. A poulterer bought 15 ducks and 12 turkeys for
five guineas. He had two ducks more for 18 shilHngs
than he had of turkeys for 20 shilUngs. What was the
price of each?
Ans, The price of a duck was 3s,, and of a turkey 5^.
20. Two men, A and 5, entered into a speculation, to
which B subscribed ^.15 more than A. After 4 months,
C was admitted, who added £.bO to the stock •, and at
the end of 12 months from Cs admission they found they
had gained £.15^; when A withdrawing received for
principal and gain <£.88. What did he originally
subscribe ?
Ans, oC.40.
21. A wall was built round a: rectangular court to a
certain height. Now the length of one side of the court
was two yards less than 8 times the height of the wall,
and the length of the adjacent side was 5 yards less than
6 times the height of the wall ; and the number of square
yards in the court was greater than the number in the
wall by 178. Required the dimensions of the court, and
the height of the wall.
Am\ The sides were 30, and 19, and the height
4 yards.
22. A ship containing 74 sailors, and a certain number
of soldiers besides officers, took a prize. The sailors re-
ceived each one-third as many pounds as there were sol-
diers, and the soldiers received £,3 a piece less, and
S34f Problems producing Adfected Quadratics,
£,76S fell to the share of the officers. Had the officers
however received nothing, the soldiers and sailors might
have received half as many pounds per man^ as there were
soldiers. How many soldiers were there, and how much
did each receive ?
Ajis, There were 36 soldiers, each soldier received
^.9, and each sailor X.12.
23. A poulterer going to market to buy turkeys, met
with four flocks. In the second were 6 more than three
times the square root of double the number in the first.
The third contained three times as many as the first and
second ; and the fourth contained 6 more than the square
of one-third of the number in the third ; and the whole
number was 1938. How many were there in each
flock ?
Am, The numbers were 18, 24, 126, 1770, re-
spectively.
24. A body of men are just sufficient to form a hollow
equilateral wedge, three deep ; and if 597 be taken away,
the remainder will form a hollow square, four deep, the
front of which contains one man more than the square
root of the number contained in a front of the wedge.
What is the number of men ?
Ans, 693.
25. Two men, ^ and B, undertake to perform a piece
of work in four days, for which they are to receive a
certain number of shillings ; but after some time, finding
that they shall not be able to finish it in the time pro-
posed, they call in C to assist them, and upon an equi-
table division of the money, C receives a sum equal to
the square root of the whole number of shillings ; but
had they been obliged to call in C to their assistance
Prohleins producing Adfected Quadratics. 335
1 + - of a day sooner, his share of the money would
have been two-fifths more. How long did C work, and
what did he receive ?
Ans, He worked 2 days, and received 5 shillings.
26. A cask, whose content is 20 gallons, is filled with
brandy, a certain quantity of which is then drawn off
into another cask of equal size ; this last cask is then filled
with water ; after which the first cask is filled with
the mixture, and it appears, that if 6|- gallons of the
mixture be drawn oflf from the first into the second cask,
there will be equal quantities of brandy in each. Re-
quired the quantity of brandy first drawn oflf.
Ans. 10 gallons.
27' There are three numbers, the diflTerence of whose
differences is 5 ; their sum is 20 ; and their continua}
product 130. Required the numbers.
Ans, 2, 5, and 13.
28. There are three numbers, the difference of whose
differences is 3 ; their sum is 21 ; and the sum of the
squares of the greatest and least is 137. Required the
numbers.
Ans. 4, 6, 11. '
29. There is a number consisting of 2 digits, which
when divided by the sum of its digits gives a quotient
greater by 2 than the first digit. But if the digits be in-
verted, and then divided by a number greater by unity
than the sum of the digits, the quotient is greater by 2
than the preceding quotient. Required the number.
Ans. 24.
30. A certain sum was to be raised on three estates
belonging to A, B, and C, at the rate of one shilling per
acre. Now the number of acres A and B had, were as
336 Problems producing Adfected Quadratics.
3 to 7 ; and if the number of acres in the whole were
divided by one-third of the product of the numbers in the
3
first and third, the quotient would be ^ • Also the sum
paid by A and C was 36 shillings less than the sum of
three times the money paid by C, and - of the money
paid by B, Of how many acres did each estate consist ;
and what was the whole sum to be raised ?
Arts. A had l^, B 28, and C 30 acres ; and the
sum was ^.3. 10^.
31. A butcher bought a certain number of calves and
sheep, and for each of the former gave as many shillings as
there were sheep, and for each of the latter one-fourth as
much. Now had he given 4 shillings more for each of the
former, and 2 shillings more for each of the latter, he
would have paid seven pounds more. But had a sheep
cost as much as a calf, he would have expended <£.56. 8.y.
How many did he buy of each ; and what were their
prices ?
Arts. 23 calves, and 24 sheep ; and their prices were
24, and 6 shillings, respectively.
32. Two persons, A and B, comparing their wages,
observe that if ^ had received per day in addition to
what he does receive, a sum equal to one-fourth of what B
received per week, and had worked as many days ay B
received shillings ^er day, he would have received £.2. 8s, J
and had B received 2 shillings a day more than ^ did,"
and worked for a number of days equal to half the
number of shillings he received per week, he would have
received £.4, ISs. What were their daily wages ?
Am. As 5 shillings, and Rs 4. ^
Problems producing Adfected Quadratics. 837
33. There are three towns in the order of the letters,
A, B, C. The difference betwen the distances from A
to B -and from Bto C is greater by four miles than the
distance from B to D, Also the number of miles
between B and D is equal to two-thirds of the number
between A and C And the number between A and B
is to the number between C and D as seven times the
number between B and C : 26. Required the respective
distanc*es.
Ans. AB = 42, BC = 6, CD = 26 miles.
34. A person bought a quantity of cloth of two sorts
for £,7- 185. For every yard of the better sort he gave
as many shillings as he had yards in all ; and for every
yard of the worse as many shillings as there were yards
of the better sort more than of the worse. And the whole
price of the better sort was to the whole price of the
worse as 72 to 7' How many yards had he of each ?
Am\ 7 y^^ds of the better, and 9 of the worse.
35. A farmer sold a certain number of bushels of
barley, and ten bushels of wheat for £,7, Ip-^* Now
each bushel of wheat cost within 3 shillings as much as
two bushels of barley. He afterwards sold as many
bushels of barley and four more, and fifteen bushels of
wheat, and received two shillmgs pe?' bushel more for his
wheat and barley than he did before ; when he found
that if he had received ^£.1. 4*. more, he should just
have received twice as much as he did before. How
many bushels of barley did he sell the first time ; and
what were the prices per bushel of the wheat and barley?
Ans, 7 bushels of barley ; and the prices of wheat
and barley were 1 1*. and 7«s. per bushel.
36. In digging among some ruins the workmen found
9 urns, together containing 60 gold coins ; the second
XX
338 Problems producing Adfccted Quadratics.
and eighth containing 8 and 4 respectively. They
secreted a certain number of these, greater than the
number they left; which being afterwards recovered,
it was found that the number of urns secreted was to the
number left as the number of coins secreted was. to the num-
ber remaining. Now if instead of taking the second urn
they had carried off the eighth, then the number of coins
taken away would have been to the number remaining as
the square of the number of urns secreted to the difference
between that square and 20 times the number of urns
remaining. Required the numbers of urns and coins
secreted.
Ans, 6 urns, and 40 coins.
37. Two men, A and 5, set out from the same place
to travel. A goes in 6 days twice as many miles as 3
goes in 5 daysj but does not arrive at the end of his
journey till 5 days after B has arrived at the end of his,
when he finds that he has travelled 259 miles more than B,
But had B gone 2 miles jper day more than he did, and"
A stopped 6 days sooner, A would then have gone only
37 miles more than jB. How many miles did each travel'
^er day, and how many days did they travel ?
Ans, A travelled 11 days, and 35 miles 'per day,
B travelled 6 days, and 21 miles 'per day.
38. Bacchus caught Silenus asleep by the side of a full
cask, and seized the opportunity of drinking, which he
continued for two -thirds of the time that Silenus would
have taken to empty the whole cask. After that Silenus
awoke, and drank what Bacchus had left. Had they
drunk both together, it would have been emptied two
hours sooner, and Bacchus would have drunk only half
what he left Silenus. Required the time in which they
would empty the cask separately.
Am, Silenus in 3 hours, and Bacchus in 6.
Problems producing Adjected Quadratics. 339
39. Two persons, A and 5, comparing the distances
they have travelled, found that the square of the number
of miles which A usually walked per hour, exceeded the
square of the number which B usually walked by 5 ; and
that if to the square of the product of those numbers there
be added the square of the sum of their fourth powers,
augmented by the product of the square of the difference
of their squares into the square of the product of the
numbers themselves, the aggregate amount would be
10345. How many miles did each walk j»er hour ?
Ans, A walked 3, and B 2 miles.
40. Two plantations, one of an oblong, and the other
of a square form, contain the same number of trees, and
they have one fence common to both, viz. that which
bounds the end of the oblong one. Upon every pole in the
square are planted as many trees as there are poles in the
square, and upon every pole in the oblong four times as
many trees as there are poles in the breadth, besides 144
in the hedges. Also the area of the oblong wants 6
poles to be to the area of the square as 3. to 3. Re-
quired the number of trees.
Am. 1296.
41. There are two sorts of metal, each being a mixture
of gold and silver, but in different proportions. Two
coins from these metals of the same weight are to each
other in value as li to IJ ; but if to the same quan-
tities of silver as before in each mixture double the
former quantities of gold had been added, the values of.
two coins from them of equal weights would have been
to each other as 7 to 1 1 . Determine the proportion df
gold to silver in each mixture, the values of equal weights
of gold and silver being as 13 to 1.
Am. The proportion of gold to silver is 1 19 in
the first mixture, and 1 : 4 in the second.
340 FrobUms producing Adjected Qtiadratics.
42. A mason has two cubical pieces of white marble
of exactly the same size, and two cubical equal pieces of
black, larger than the other. The number of solid yards
in the four pieces is 9 more than 1 1 times the number of
yards in a side of a white one, together with 1 2 times the
number in a side of a black one. He afterwards finds
another block, the length of which is 2 yards longer than
a side of one of the white pieces, and the width 4
times the length of a side of the black one ; and this
when laid on its largest side occupies a space greater by
3 yards than the difference between 4 times the space
occupied by a black, and 3 times the space occupied by
a white one. Required the dimensions of the blocks.
Ans, The side of a white block is 1 yard, and of a
black one 3 yards ; the length of the other is
3 yards, and the width 12 yards.
X. Problems in Arithmetical and Geom^rical Pro-
gressions.
1. There are three numbers in arithmetical pro-
gression, whose sum is 21; and the sum of the first and
second is to the sum of the second and third as 3 to 4.
Required the numbers »
Ans> 5, 7, 9.
2. There are six towns in the order of the letters, A^ B,
C, D, E, F, whose distances from each other are in an
increasing arithmetical progression. The distance from
^ to C is 16 miles, and from Cto ^ is 24 miles. Re-
quired their respective distances.
Ans, From ^ to JB is 7, from J5 to C 9, from C
to Z) 11, from D to E 13, and from E to F
15 miles.
ProMems in Arithmetical and Geometrical Progressions. S41
3. A person makes a mixture of 51 gallons, consisting
of brandy, rum, and water, the quantities of which are
in arithmetical progression. The number of gallons of
brandy and rum together is to the number of gallons of
rum and water together as 8 to 9. Required the quan-
tities of each.
Ans, 15 gallons of brandy, 17 of rum, and 19 of
water.
4. During a scarcity, a person wished to make a
mixture of 24 bushels, consisting of wheat, oats, and
barley, the quantities of each forming an increasing arith-
metical progression. Not being able however to procure
any barley, he mixed additional quantities of wheat and
oats in the proportion of 2 to 3, so as to complete his 24
bushels, when he found the whole quantities of wheat and
oats to be in the proportion of 5 to 7. How many bushels
of each did he originally intend to mix ?
Ans, 6 of wheat, 8 of oats, and 10 of barley.
5. The difference between the first and second of four
numbers in geometrical progression is 36, and the dif-
ference between the third and fourth is 4. What are
the numbers? Ans, 54, 18,6, and 2.
. 6. A person employed three workmen, whose daily
wages were in arithmetical progression. The number
of days they worked was equal to the number of shillings
that the second received jp^r day. The whole amount of
their wages was seven guineas, and the best workman
received 28 shillings more than the worst. What were
their daily wages ?
Ans, 5, 7 3 2i^d 9 shillings.
7 . There are three numbers in geometrical progression ;
the sum of the first and second of which is 9, and the sum
of the first and third is 1 5 . Required the numbers,
^w;^. 3, 6,-9,11
342 Problems in Arithmetical
8. There are three numbers in geometrical pro-
gression, whose sum is 14; and the sum of the first
and second is to the sum of the second and third' as 1
to 2. Required the numbers.
Ans. 2, 4, 8.
9. There are three numbers in geometrical progression,
whose continued product is 64, and the sum of their
cubes is 584. Required the numbers.
Ans. 2, 4, 8.
10. There are four numbers in geometrical progression,
the second of which is less than the fourth by 24 ; and
the sum of the extremes is to the sum of the means as
7 to 3. Required the numbers.
Ans. 1, 3, 9, 27.
11. From two towns which were 168 miles distant,
two persons, A and J5, set out to meet each other ; A
went 3 miles the first day, 5 the next, 7 the third, and
so on ; B went 4 miles the first day,' 6 the next, and
so on. In how many days did they meet ?
Ans, 8.
12. A traveller set out from a certain place, and
went 1 mile the first day, 3 the second, 5 the next,
and so on, going every day 2 miles more than he
had gone the preceding day. After he has been gone
3 days, a second sets out, and travels 1 2 miles the first
day, 13 the second, and so on. In how many days will
the second overtake the first ?
Ans. In 2, and 9 days.
13. A person has two pieces of ground, one of
which is in the form of an equilateral triangle, and the
and Geometrical Progressions. . 343
other of a rectangular parallelogram, one side of which
is equal to a side of the triangle, and the other side is
8 yards less. These he plants with trees at the distance
of two yards from each other, and finds that there are
3 more on the rectangle than on the triangle. What
are the lengths of the sides ?
Arts, A side of the triangle is 20 yards, and the
sides of the parallelogram are 20 and 12 yards.
14. There are four numbers in arithmetical pro-
gression, whose sum is 28 ; and their continual product
is 583. Required the numbers.
Ans. 1, 3, 9, 13.
15. There are four numbers in arithmetical pro-
gression ; the sum of the squares of the first and second
is 34 ; and the sum of the squares of the third and
fourth is 130. Required the numbers.
Am, 3, 5, 7, 9-
16. The sum of d£.700 was divided among four
persons, whose shares were in geometrical progression ;
and the difference between the greatest and least was to
the diflference between the means as 3J to 12. What
were their respective shares ?
Am. £.108, £.IAA, £AQ2, £.256.
17. Five persons undertake to reap a field of 87 acres.
The five terms of an arithmetical progression, whose
sum is 20, will express the times in which they can
severally reap an acre^ and they all together can finish
the undertaking in 60 days. In how many days can
each separately reap an acre.
Am. 2, 3, 4, 5, 6 days.
344? Problems in Arithmetical
18. Out of a vessel containing 24 gallons of puii^
spirit, a vintner dr^w oflT at three successive times a
certain number of gallons, which formed an increasing
arithmetical progression, in which the difference between
the squares of the extremes was equal to l6 times, the
mean, and filled up the vessel with Vvater afler each
draught, till he found what he last drew off, reduced
it to one-sixth of its original strength. Required the
number of gallons of pure spirit drawn off each time.
Ans. 12, 8, 3-^.
19. The Fly starts 10 miles before the Telegraph;
but the Fly coachman having made an appointment
with the driver of the Telegraph, walks his horses so as
to be overtaken at the end of the second mile. Now it
is observed, that the nutnber of revolutions made in a
given time by the hinder wheel of the Fly, its fore wheel,
and the hinder wheel of the Telegraph increase in arith-
metical progression, and that the circunrferences of these
wheels, viz. of the fore wheel of the Fly, its hinder
wheel, and the hinder wheel of the Telegraph increase
in a geometrical progression, whose common ratio is the
same as the common difference of the arithmetical pro-
gression. It is required to find the ratio that the wheels
bear to each other.
♦
Ans. 1, 2, 4 are their proportional lengths.
20. A company of merchants fitted out a privateer,
each subscribing i;'.100. The captain subscribed no-
thing, but was entitled to a of. 100 share at the end
of every certain number of months. In the course
of 25 months he captured three prizes, which were
in geometrical progression, the middle term being one-
fourth of the cost of the equipment^ the common ratio
and Geometrical 'Progressions. 345
the number of months which Entitled the captain to
his 06" 100 share/ and tlieir sum £.\3Tb more than
the cost of the equipment. After deducting d.S^b
for prize money to the crew, the captain's" share of the
remainder amounted to one-fifth of that of the com-
pany. Required the nilmher of merchants^ and the
captain's J)ay.
Am,: The number of merchants was 25, ^nd
the captain was entitled to a ^.100 share
*- at the end of every 5 months.
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