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Subtracting this Equation from the fourth, and transposing,

2(x+y)=676; hence

(8) x+y=338.

Substituting this value of x+y in the third Equation, and uniting similar terms,

24336

√xy=156; whence x=

y

Substituting this value of x in the eighth Equation,

24336

+y=338; whence y=104.

Y

Since x=

24336
9
y

we have x

31. Three merchants made a joint stock, by which they gained a sum less than that stock by $80. A's share of the gain was $60, and his contribution to the stock was $17 more than B's; also B and C together contributed $325. How much did each contribute?

Let x represent A's contribution, y B's, and z C's; then (x+y+z)-80 represents their gain.

=234.

Now, the whole stock is to A's contribution, as the whole gain to A's gain.

Therefore, x+y+z:x::(x+y+z−80):60.

From the conditions of the problem, we shall also have

x-y=17;

y+z=325.

Converting the proportion into an Equation,

60(x+y+z)=x(x+y+z—80).

Substituting for y+z in this Equation its value from the second Equation,

60(x+325)=x(x+245).

This Equation will give x=$75; and we shall now, from the first Equation, easily find y=$58; and substituting this value in the second Equation, we find z= = $267.

32. Of three numbers in geometrical progression, the greatest exceeds the least by 15, and the difference of the squares of the greatest and the least is to the surn of the squares of the three numbers as 5 to 7. What are the numbers?

Let x represent the least term, and y the ratio of the progression;

then

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2y+-5y2 = 12.

From this Equation, y=2; and substituting this value for y in the first Equation, we find x=5, the least term.

Then xy=5×2=10, the mean term; and xy2=5× 4= 20, the greatest term.

5(x2+x2 y2+x2y1)

33. Two persons set out from different places and traveled towards each other. On meeting, it appeared that A had traveled 24 miles more than B, and that A could have gone B's journey in 8 days, while B would have been 18 days in performing A's journey. What distance was traveled by each?

y

Let x represent the number of miles A traveled, and y the number of miles B traveled.

The first Equation will then be

x=y+24.

Now, since A could have gone B's journey in 8 days,

Then x

represents the number of miles A went per day.

8

And since B could have gone A's journey in 18 days,

x

represents the number of miles B went per day.

18

x

Y÷ or
18'

У 8x

or represents the number of days A traveled; and 8' y

"

18y

-, represents the number of days B traveled.

X

Since both traveled the same number of days, we have

8x

18y

x

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Y

The two Equations will give x=72, and y=48 miles.

34. The joint stock of two partners was $416. A's money was in the business 9 months and B's 6 months. When they shared stock and gain, the first received $228, and the second $252. What was each man's amount of stock?

Let x represent A's capital, and y B's capital; then we have for the first Equation,

x+y=416.

Now, 228-x represents A's gain; and

252-y represents B's gain.

And since the shares of profit are to each other in the compound ratio of capital and time (131), we have

228-x: 252-y:: 9x: 6y.

Converting this proportion into an Equation, uniting similar terms, and dividing both sides by 3,

456y+xy=756x.

From the first Equation,

y=416—x.

Substituting this value of y in the second Equation, and reducing, x2+796x=189696;

from which x=$192; and 416—192=$224, is the value of y.

35. The sum of $700 was divided among four persons, A, B, C, and D, whose shares were in geometrical progression; and the difference between the greatest and the least was to the difference between the two means as 37 to 12. What were the several shares?

Let x represent A's share, and y the ratio of the progression; then xy, xy2, and xy3, respectively, represent B's, C's and D's shares. By the conditions of the problem,

x+xy+xy2+xy3=700;
xy3-x: xy2-xy:: 37: 12.

This proportion converted into an Equation becomes

12(xy3—x)=37(xy2 —xy).

Dividing both sides of this Equation by xy-x, 12(y2+y+1)=37y ;

from which x=$108.

Hence, xy=108 ×

4

from which y= 3

Substituting this value of y for y in the first Equation, and adding similar terms,

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EXERCISES

2

(1)2 =

On Affected Cubic and Biquadratic Equations.

1. Find the values of x in the Equation

x3-6x2+11x=6.

$192; and xy3

The divisors of the known term 6 are 1, 2, 3, 6; and it will be found that the Equation

x36x2+11x-6=0

is divisible by X- −1, x—2, and x-3; hence 1, 2, and 3, are the roots of the given Equation, (254).

2. Find the values of x in the Equation x39x2+26x— 24.

Or, having one of the roots as 1, the other two may be found from the quadratic Equation

–5+6=0,

which results from dividing the second Equation by x-1.

The divisors of the absolute term 24, are 1, 2, 3, 4, 6, 12, and 24; and we shall find that the Equation

x39x2+26x-24=0,

is divisible by x-2, x-3, and x--4; hence 2, 3, and 4, are the roots of the given Equation.

3. Find the values of x in the Equation

x3-3x2-6x=-8.

The divisors of 8 are 1, 2, 4, and 8; and the Equation

23-3x2-6x+8=0,

is divisible by x-1, x+2, and x- -4; then 1, -2, and 4, are the required values of x.

4. Find the values of x in the Equation

2x36x28x=-24.

Dividing both sides of this Equation by 2, we have

x3-3x2-4x=-12.

The divisors of 12, are 1, 2, 3, 4, 6, and 12; and the Equation

x3-3x2-4x+12=0,

is divisible by x-2, x+2, and x- 3; hence 2, 2, and 3, are the roots of the given Equation.

5, Find the values of x in the Equation

x+2x3-13x2-14x+24=0.

The divisors of the absolute term 24, are 1, 2, 3, 4, 6, 12, and 24; and the Equation is divisible by x2-1, x+2, x−3, and x+4; then 1, -2, 3, and 4, are the required values of x.

Or, having found one of the roots, as 1, the other three may be found from the cubic Equation

x3-3x2-10x-24=0,

which results from dividing the given Equation by x-1.

6. Find an approximate value of x in the Equation x2+10x2+5x260.

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