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268. The negative value of y makes known there

y fore a rectification in the enunciation of the problem ; since that, instead of adding 7y to 12c in the first equation, and 5y to 8x in the second, y being considered a positive or an absolute number, we must subtract them in order to have the sum given for the common wages of these three persons; or what is the same thing, if, in place of considering the money attributed to the wife and son as a gain, we would regard it as an expense made by them to the charge of the workman ; then we must subtract this money from what the man would have gained alone, and there would be no contradiction in the equations, since they would become

60-7y=46, and 40-5y=30; from either of which we would derive y=2; and we should therefore conclude that if the workman gained 59. per day, his wife and son's expense is 28., which can be otherwise verified thus :

For 12 days work, he receives 5X12 or 60s.; the expense of his wife and son for 7 days, is 2x7 or 14s; and there remain 46 shillings.

Again, he receives for 8 days work 5 X 8 or 40s., the expense of his wife and son during 5 days is. 2x5 or 108.; therefore his clear gain is 30 shillings.

269. It is very evident that, in place of the enunciation of (Prob. 9), we must substitute the following: in order that the problem proposed may be possible, with the above given quantities :

A labourer working for a gentleman during 12 days, having had with him, the first 7 days, his wife and son, who occasion an expense to him, received 46 shillings ; he has wrought, afterwards, for 8 other days, on 5 of which he had with him his wife and son, whose expenses he must still defray, and he received 30 shillings. Required the salary of the workman per day, and also, tre expense of his wife and son in the same time.

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Designating by x the daily wages of the workman, and by y the expense of his wife and son, for the same time; the equations of the problem shall be

12x – 7y=46, and 8x ---5y=30; which, being resolved, will give

x=5s, and y=2s. 270. Although negative values do not answer the enunciation of a concrete question, as has been observed (Art. 199), yet they satisfy the equations of the problem, as may be readily verified, by substituting 5 for e, and -2 for y, in the equations (Art. 267), since they would then become identically equal.

Prob. 10. Two pipes, the water flowing in each uniformly, filled a cistero containing 330 gallons, the one running during 5 hours, and the other during 4; the two same pipes, the first running during two hours, and the second three, filled another cistern containing 195 gallons. The discharge of each pipe is required.

Let x represent the discharge of the first in an hour; y that of the second in the same time.

And in order to have a general solution, put a=5, b=4, c=330, a'=2,b'=3, c=195; then by the conditions of the problem we shall have these two equations,

axt-by=c, and a'x+by=c'; which, being resolved as in (Art. 209), will give b'c-bc'

ac'a'c ab'a'b Now, by restoring the values of a, b, c,&c., we have

990--780 210


15 8 Thus, the first pipe discharges 30 gallons per hour, and the second 45.

and y=ab' -a'b

= 30;

and y=


271. Let us now suppose that the first pipe running during 3 hours, and the second during 7, filled a cistern containing 190 gallons ; that afterwards, the first

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running 4 hours, and the second 6, filled a cistern containing 120 gallons.

In this case, a=3, b=7, c=190, a=4; b=6, c'= 120; and, consequently, b'c-bc'=1140-840=300, ab' -a'b=18—28=-10, ac'-a'c=360–760= 100, which will give x=-30, and y=40.

In order to understand the meaning of these results, we must return again to the conditions of the problem, or what amounts to the same, we must try how these values of x and y satisfy the equations of the problem:

Thus, if we substitute - 30 for x, and 40 for y, in the equations 3x +7y=190 and 4x+6y=120, resulting from the above problem, we find first, that 3x=90, and 7y=280, consequently, 3x+7y=-90+280, which in effect is equal to 190.

In like

430 +6y is found to be -- 120+240, which is equal to 120.

Having, therefore, discovered how the values 30 and +40 of x and y answer the equations 3x + 7y =190 and 4x+6y=120, we perceive at the same time how they would answer the conditions of the problem ; for since the use that has been made of the quantities 3x and, 4x, which express the quantities of water discharged by the first pipe in the first and second operation, was to subtract them from 7y and from 6y, which express the quantities furnished in the same operations by the second pipe. The first pipe must be considered in this case as depriving the cisterns of water instead of furnishing any, as it did in the preceding problem, and as it was supposed in expressing the conditions of this problem.


272. Hence, in almost every question solved after a general manner, we may always conclude that when the value of the unknown quantity becomes negative, the quantity expressed by it, should be considered as being of an opposite kind from what it was supposed in expressing the conditions of the problem.

What has been said with respect to unknown quantities, is equally applicable to known quantities, that is, when a general solution is applied to any particular case, if any of the given quantities, a, b, c, &c. in the problem, are negative.

273. Let it be proposed, for example, to find what should be, in the foregoing problem, the discharges of two pipes, that the first furnishing water during 3 hours, and the second 4, may fill a cistern containing 320 gallons, and that the second pipe afterwards furnishing water during 6 hours, whilst the first discharges it during 3 hours, may fill a cistern containing 180 gallons.

We have only to put in the general solution (Art. 270), a=3, b=4, c=320, a'=-3, b=6, d=180. and there will result x=40, and y=50.

From whence it appears that the discharge of the first pipe is 40 gallons per hour, either to carry away the water as in the second operation, or to furnish it as in the first, and the discharge of the second, 50 gallons an hour, which it furnishes in both operations.

Prob. 11. A certain sum of money put out to interest, amounts in 8 months to 2971, 12s. ; and in 15 months its amount is 3061 at simple interest. What is the sum and the rate per cent ?

Ans. 2881, at 5 per cent, Prob. 12. 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, Prob. 13. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes; but the denominator being doubled, and the numerator increased by 2, the value becomes

Ans. .

Prob. 14. A Farmer parting with his stock, sells to one person 9 horses and 7 cows for 300 dollars; 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 dollars, and of a horse 24 dollars.

Prob. 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 3. After they became 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. Prob. 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. Prob. 17. A sets out express from C towards D, and three hours afterwards B sets out from D towards C, travelling 2 miles an hour more than A. When they meet it appears that the distances they have travelled are in the proportion of 13 to 15; but had A travelled five hours less, and B had gone 2 miles an hour more, they would have been in the proportion of 2 : 5. 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.

Prob. 18. A Farmer hires a farm for 2452 per annum, the arable land being valued at 21 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. 93 acres of arable, and 35 of pasture. Prob. 19. A and B playing at backgammon, A bet

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