It is not the same with the values of x and y, found in the preceding article, for they are not susceptible of being reduced to a more simple expression. It follows, from what I have just said, that when we meet with an expression which becomes, it is proper, before pronouncing upon its value, to see if the numerator and denominator have not a common factor, which becoming nothing, renders the two terms at the same time equal to zero, and which being suppressed, the true value of the proposed expression is obtained. There are, notwithstanding, some cases which elude this method, but the limits of this work will only allow me to note the analytical fact. It belongs properly to the differential calculus, to give the general processes for finding the true value of quantities, which become . 71. It is very evident, from what has been said, that algebraic solutions either answer perfectly to the conditions of a problem, when it is possible, or they indicate a modification to be made in the enunciation, when the things given imply contradictions that cannot be reconciled; or lastly, they make known an absolute impossibility, when there is no method of resolving with the same things given, a question analogous in a particular sense to the one proposed. 72. It may be remarked, that in the different solutions of the preceding question, the changing of the signs of the unknown quantities x and y, corresponds to a change in the direction of the journeys represented by the unknown quantities. When the unknown quantity y was counted from B towards A, it had in the equation x + y = a, the sign+, and it takes the sign for the second case, when the motion is in the opposite direction from B towards C, art. 65, since we had for the first equation a result which differs 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 x= bc d If we suppose neither to owe the other any thing, we must make d = 0, and then the equation would be 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 7y from 74 francs, there will remain 74-7y 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 8 per day. Putting the two quantities equal to each other, we form the equation if x represent the course AR 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 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. We suppose that the courier, who starts from B, sets off a number 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 BC c d 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 ACAB BC= a c d. - 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 and the distance passed over by the courier from the point A would be AR, while that passed over by the other courier would be CRAR — AC; 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 BC represented by cd, becomes greater than a, which represents AB. 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., there would result from it c d = c = 8mls., d = 60h, 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 x 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 dam, or what remains of BC 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 cd a, 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 e 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, 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. We deduce from the first multiplying this value by d, in order to substitute it in the place See note at the end of the Elements of Algebra. Alg. |