L

1

duced, we have 2 s* + 12 s2x2+2x=b, or or x*+6s*x=b-s;

1

by completing the square, x2+6s2x2+98*=b+8s; by

1 2

2

1

evolution, x2 +382 = ±√b+8s*, x2=-382±√b+8s",

1

2

and x=±√-3s* ±√6+8s*=1; whence 8+x=4, and s x=2.

32. Given the sum 10, and the sum of the fifth powers 17050, of two numbers, to find them?

Let 2s=10, b=17050, 2x=the difference of the numbers required; then s+x=the greater, and s-x=the less, and by proceeding as in problems 30 and 31, we have 2 s +20 83x2 + 10 sx*

b

4 s

=b, whence x== √之十 --s2=2, 8+x=7, and s-x=3.

10 s

5

33. Given the product p, and the sum of the nth powers s, of two numbers, to find them?

Let x and y represent the numbers, then by the problem xTM + y=s, and xy=p; from the second equation y=- ; this value sub

stituted for y in the first, gives

+P -=s, or x2+p"=sx", or

x-sx=-p"; hence, completing the square, x2-sx" +

-p"; whence r".

2

S

82

4

34. Given the product p, and the difference of the nth powers d, of two numbers, to find them?

Let x and y be the numbers, then xy=p, and x-y"=d; whence by proceeding as in the foregoing problem, x=

d+ √d2 + 4 p°, and y=p+ √

2

d+ √d2+4p"

2

d+d2+4 p

2

35. Required the values of x and y in the following equations, viz. x2 x y3=2y2, and 123 √x−y=22 ?

Let u=3√x, z= √y, then w=x, and z2=y; the given equations become uz3=2z, and 12 u-z=22; divide the last but

one by 2 zt, and z=2; this equation added to the preceding, gives

12 u=22+ -, or u2-24 u=-44; this equation resolved, gives

u=2, z=

2
и

2

z=(=)2, x= x=(u3=) 8, and y=(z2=) 4.

36. If 18 oxen in 5 weeks can eat 6 acres of grass, and 45 oxen in 9 weeks eat 21 acres of the same, how many must there be to eat 38 acres in 19 weeks, the grass being allowed to grow uniformly ?

Let a=18, b=5, c=6, d=45, m=21, n=9, r=38, s=19, 1=the quantity eaten by an ox in a week, w=the quantity on an acre at first, z=the weekly increase on an acre after the first 5 weeks, x=the number of oxen required, p=(n-b=) 4, t= (s-b=) 14; then will rw=the grass on r acres at first, and rtz= the increase on r acres in tweeks; the sum of these, by the problem, equals the quantity x oxen ate in s weeks, that is, sx=rw+rtz; again, mw=the grass on m acres at first, and mpz=the increase of the same in p weeks; the sum of these two equals what d oxen ate in n weeks, that is, mw+mpz=dn; also cw=(the grass on c acres at first,=)the quantity a oxen can eat in b weeks, that is, cu=ab,

ab

whence w==; to mp times the first equation, add rt times the second, and mpsx + mrtw + mprtz=dnrt+mpru + mprtz, or mpsx =dnrt+mprw-mrtw; for win this equation, substitute its equal

ab
and the equation becomes mpsx=dnrt+
인

cmpsx=cdnrt+abmpr-abmrt; whence x=

=

abmpr abmrt

C

,

cdnt + abmp-abmt cmps

cdnt+abmxp-t 34020+1890 x-10
cmps
9576

answer.

Xr=

×38=60, the

OF

X

37. A waterman, who can row 11 miles an hour with the tide, and 2 miles an hour against it, rows 5 miles up a river and back

again in 3 hours; now supposing the tide to run uniformly the same way, required its velocity *?

Let m=11, n=2, p=5, r=3, v=the velocity required, and x=the time he rowed with the tide, then will r-x=the time he rowed against it; whence (x :p :: 1 hour :)P=his velocity with the tide, and (r-x:p :: 1 hour:) P=his velocity against

P

x

tide; now since the tide assists him=v when he goes with it, it must evidently retard him=v when he goes against it; whence 2v=the difference of his velocity with, and against tide, .. P=2v, or v= : now because his velocity with, is to his velocity against, tide, as m ton; so his time of rowing with, is to his time of rowing against, tide, as n to m, since the time is inversely as the velocity; whereforex:r-x::n : m, *.x=

6

13

p
p
2x 2r-2x

nr

m+n 7 13 6 13

of an hour the time he rowed with tide, and r-x=2 hours=the time he rowed against it; for x substitute its value

in the equation above derived, and it becomes v=(P

12

=)

P 2x 2-2x 12 65 65 3510 19 =4 miles per hour=the 792 44

13-P-2-13-12 66

velocity of the tide.

38. The ages of five persons, A, B, C, D, and E, are such, that the sum of the first four is 95, that of the three first and last 97, that of the two first and two last 103, that of the first and three last 106, and that of the four last 107; required the age of each?

Let a=95, b=97, c=103, d=106, e=107, s=the sum of all their ages, and let x, y, z, v, w, be put for their ages respectively; then will s-w=a, s-v=b, s-z=c, s-y=d, and s-x

Velocity (from the Latin velor, swift,) is that affection of motion, whereby a moving body passes over a certain space in a certain time; or in common language, it is the degree of swiftness with which a body moves: it is likewise named celerity, from the Latin celer, swift or nimble,

=e; add these five equations together, and the sum is (58x-y-z-v-w=5s-s=) 48=a+b+c+d+e; whence s=

a+b+c+d+e

4

; now if this value be substituted for s in the five preceding equations, we shall thence obtain the required numbers, viz. w=32, v=30, z=24, y=21, and x=20, being the ages of E, D, C, B, and A, respectively.

39. To find a point in the straight line which joins two luminaries, or in the line produced, which is equally enlightened by

both '.

Let a their distance apart, x=the distance of the least of them from the required point, then a+x=the distance of the other: let the quantity of light emitted by the first in a given time be to that emitted by the second in the same time, as m ton; then

be the ratio of the effects they produce, supposing

n

a+x

m

equal: but these effects are by hypothesis equal; whence x

a2+2ax+x2 -ma2, x2+

2

*. ma2+2 amx+mx2 ==nx2, or m-n.x* +2 amx=

2 am m-n

am

m-n

x+

2

x=

am m--n ma2

=the distance required.

m-n

=±√

x2+

am

m-n

40. The weight w, and the specific gravity of a mixture, and the specific gravities a and b, of the two simples which compose it, being given, to find the quantity of each ?

• A luminary, (from the Latin lumen, light,) is a body that gives light, as the sun, moon, a planet, star, &c.

The double sign serves both cases, viz. a +x when the point required is beyond the smaller luminary, and a-r when it is between them; also in the answer, the upper sign - applies to the first case, and the lower sign + to the second.

The gravity of a body, (from the Latin gravis, heavy,) is its weight,

Let x=the weight of the simple, whose specific gravity is the

greatest, then w-x=the weight of the other.

41. Suppose two bodies, A and B, to move in opposite direc tions towards the same point with given velocities, the distance of the places from whence they set out, and the difference of the times in which they begin to move, being likewise given, thence to determine the point where they meet?

Let d=the distance from A to B at the time of setting out, x=A's distance from the point of meeting, then d-x=B's distance from the point of meeting; let t=the difference between the times of their beginning to move, and suppose A moves through the space a in the time n, and B through the space b in the time m, then (a: n::x:)-=the time of A's motion, and (b: m::d-x:)

d-x.m

b

nx

a

=the time of B's motion; whence by the problem,

nx

a

bt+dm
bn+am -ama.