a result which does not differ from that given in the article cited; but it should be observed, that the journey y being made up of multiples of the space c passed over in an hour by the courier from B, and this space having the same direction as the space y, ought to be supposed to have the same sign, and consequently to take the sign —, when — is applied to y; we have accordingly A simple change of sign then is sufficient to comprehend the second case of the question in the first, and it is thus that algebra gives at the same time the solution of several analogous questions. We have a striking example of this in the problem of art. 15. It is here supposed that the father owed the son a sum d; if we would resolve the question on the contrary hypothesis, that is, by supposing that the son owed the father the sum d, it would be sufficient to change the sign of d in the value of x, and we If we suppose neither to owe the other any thing, we must make d = 0, and then the equation would be d= Nothing can be easier than to verify the two solutions by putting anew the problem into an equation for each of the cases, which we have enunciated. 73. It was only to preserve an analogy between the problems 56 and 64, that I have employed two unknown quantities in the second. Each may be resolved with only one unknown quantity; for when we say that the labourer received 74 francs for 12 days' work performed by himself and 7 days' work by his wife and son, it follows that, if we call y the daily wages of the woman and son, and take 7 y from 74 francs, there will remain 74-7 y for the 12 days' labour of the man; from which we 74-7y infer that he earned 12 per day. By a similar calculation for the 8 days' service, we find that he earned 50-5 y per day. Putting the two quantities equal to each other we form the equation if x represent the course A R of the courier from A, BR = a—x would be that of the courier who set off from B towards A. These two distances being passed over in the same time by the couriers whose rate of travelling per hour in miles is denoted by the numbers b and c respectively, we have The difference between the solutions, which I have now given and those of articles 56 and 64, consists merely in this, that we have formed and resolved the first equation by the assistance of ordinary language, without employing algebraic characters, and it is manifest, that the further we carry this, the less will remain to be effected by the other. 74. We sometimes add to the problem of art. 64 a circumstance, which does not render it more difficult. A R C B We suppose that the courier, who starts from B, sets off a num ber d of hours before the other, who goes from A. It is evident that this amounts only to a change of the point of departure of the first, for if he travelled a number c of miles per hour, he would pass over the space B C cd in d hours, and would be at the point C, when the other courier set off from A; so that the interval of the points of departure would be = By writing then a-cd in the place of a in the equation of the preceding article, we have If the couriers proceeded in the same direction, the interval of A B C R the points of departure would be AC AB+BC=a+cd; = and the distance passed over by the courier from the point Л would be A R, while that passed over by the other courier would be CR AR-AC; we have then = 75. Enunciated in this manner the problem presents a case, in which the interpretation of the negative value found for x is attended with some difficulty; it is when the couriers being supposed to proceed in opposite directions, we give to the number d a value such, that the space B C represented by c d becomes greater than a, which represents A B. C R A B Now the courier from the point B arrives at C on the other side of A at the moment, when the courier from A sets off towards B ; there is then an absurdity in supposing that the two couriers can thus come together. If we should take, for example, a=400mls., b=12mls., c= gmls., d=60b, there would result from it cd=480mls., thus the point C would be 80mls. on the other side of A, with respect to the point B; but we find, Thus the coming together of the couriers takes place in a point R, 48mls on the other side of the point A, but between A and C ; although it seems that the courier from B, being supposed to continue his journey beyond the point C, can be overtaken by the other courier only after he has passed this point. To understand the question resolved in this sense, we may substitute in the place of a the negative member-m, and the or by changing the signs in the two members, We see that the distance passed over by the courier from the = point B, is c d -a m, or what remains of B C after AB and AR are subtracted, that is CR, and that AC cd-a. This is just what would take place if the second courier had started immediately from the point C, where he is at the departure of the first; but as they travel in opposite directions, they must necessarily meet between A and C. Thus, this case is similar to the first of those of art. 74, where it is sufficient to change a cd into cda, in order to obtain the value, which m has according to the above equation.* * 76. The problem of art. 56, taken in its most enlarged sense, may be enunciated as follows; A labourer having passed a number a of days in a family, and having with him his wife and son during a number b of days, received a sum c; he lived afterward in the same family a number d of days; he had with him this time his wife and son during a number c of days, and he received a sum f; we inquire what he earned per day, and what was allowed per day to his wife and son. Let x represent constantly the daily wages of the labourer, and Y that of his wife and son; for the number a of days he has ax, and for the number b of days his wife and son have by, so that. ax+by = c; for the number d of days, he has dx, and for the number e of days his wife and son have ey, thus, dx+ey=f. These are the general equations of the question. multiplying this value by d, in order to substitute it in the place * See note at the end of the Elements of Algebra. a result which does not differ from that given in the article cited; but it should be observed, that the journey y being made up of multiples of the space c passed over in an hour by the courier from B, and this space having the same direction as the space y, ought to be supposed to have the same sign, and consequently to take the sign, when is applied to y; we have accordingly A simple change of sign then is sufficient to comprehend the second case of the question in the first, and it is thus that algebra gives at the same time the solution of several analogous questions. We have a striking example of this in the problem of art. 15. It is here supposed that the father owed the son a sum d; if we would resolve the question on the contrary hypothesis, that is, by supposing that the son owed the father the sum d, it would be sufficient to change the sign of d in the value of x, and we have If we suppose neither to owe the other any thing, we must make d0, and then the equation would be x= bc a+b Nothing can be easier than to verify the two solutions by putting anew the problem into an equation for each of the cases, which we have enunciated. 73. It was only to preserve an analogy between the problems 56 and 64, that I have employed two unknown quantities in the second. Each may be resolved with only one unknown quantity; for when we say that the labourer received 74 francs for 12 days' work performed by himself and 7 days' work by his wife and son, it follows that, if we call y the daily wages of the woman and son, and take 7 y from 74 francs, there will remain 74-7 y for the 12 days' labour of the man; from which we 74-7 y infer that he earned per day. 12 By a similar calculation for the 8 days' service, we find that he earned 50-5 y per day. Putting the two quantities equal to cach other we form the equation |