transposing to one member all the terms involving x, it be comes -3a2x+8abx+3b2x=2a2c+5abc+b2c-12ab+8ab*+4b3, from which we deduce x= 2a2c+5abc + b2 c— 12 a b + 8 a b2 + 4 b3 of questions having two unknown quantities, and of negative quantities. 55. THE questions, which we have hitherto considered, involve only one unknown quantity, by means of which, with the known quantities, are expressed all the conditions of the question. It is often more convenient, in some questions, to employ two unknown quantities, but then there must be, either expressed or implied, two conditions, in order to form two equations, without which the two unknown quantities cannot be determined at the same time. The question in art. 3, especially as it is enunciated in art. 4, presents itself naturally with two unknown quantities, that is, with both the numbers sought. Indeed if we denote the least by x, the greatest by y, their sum by a, their difference by b, we have by the enunciation of the question, x + y = a Each of these two quantities being considered by itself, we can determine one of the unknown quantities. If we take the second, for example, we deduce the value of y, which is y=b+x, a value which seems at first to teach us nothing with regard to what we are seeking, since it contains the quantity x, which is not given; but if instead of the unknown quantity y in the first equation, we put this, its equivalent; the equation, containing now only one unknown quantity x, will give the value of x by the process already taught. We have in fact by this substitution .b and putting this value of x in the expression for y=b+x=b+ Y, we have then for the two unknown numbers the same expressions as in art. 3. - It is easy to see indeed, that the above solution does not differ essentially from that of art. 3; only I have supposed and resolved the second equation y — xb, which I contented myself with enunciating in common language in the article cited; and from it I deduced, without algebraic calculation, that the greater number was x + b. 56. I take another question. A labourer having worked for a person 12 days, and having with him, during the 7 first days, his wife and son, received 74 francs; he worked afterward with the same person 8 days more, during 5 of which, he had with him his wife and son, and he received at this time 50 francs; how much did he earn per day himself, and hors much did his wife and son earn ? Let x be the daily wages of the man, y that of his wife and son ; 12 days' work of the man will amount to 12 x, 7 days' work of his wife and son 7y; we have then by the first statement of the question, 12x + 7y=74; 8 days' work of the man will give and 5 days' work of his wife and son we have then by the second statement 8x+5y =50. 8x, 5y; Proceeding as in the preceding question, we take the value of y in the first equation, which is and substitute this value in the second, multiplying it by 5, the coefficient, and it becomes an equation, which contains only the unknown quantity x. By reducing it we have transposing 4 x to the second member, and 350 to the first, we obtain 370-350=4x 20=4x x = x 5= x. Knowing x, which we have just found equal to 5, if we place this value in the formula 74-12x y= 7 the second member will be determined, for we have y= 74-12 X 5 74-60 14 thus y=2. The man then earned 5 francs per day, while his wife and son earned only 2. No 57. The reader has perhaps observed, that in resolving the above equation 370-4 x 350, I have transposed 4x to the second member. I have proceeded thus to avoid a slight difficulty, that would otherwise have occured, and, which I will now explain. By leaving 4 x in the first member, and transposing 370 to the second, we have and reducing the second according to the rule in art. 19, there will result from it 4x=- 20. But as we have avoided, in the preceding article, the sign --, which affects the quantity 4 x, by transposing this quantity to the other member; and as in like manner the quantity 350 - 370 becomes by transposition 370-350; and since a quantity, by being thus transferred from one member to the other changes the sign (10), it is evident that we may come to the same result by simply changing the sign of each of the quantities +350-370, which gives 4 x, or .which is the same as 4x= 350+370, 4x= 370-350 370-350=4x. We might also change the signs after reduction, and the equation becomes, as above, -4x= · 20 4 x = 20. It follows from this, that we may transpose indifferently to one member or to the other, all the terms involving the unknown quantity, observing merely to change the signs of the two members in the result, when the unknown quantity has the sign —. 58. Having undertaken, by means of letters, a general solution of the problem of art. 56, I will now examine a particular case. I suppose that the first sum received by the labourer to be 46 francs, and the second 30, the other circumstances remaining as before; the equations of the question will then be multiplying this value by 5, in order to substitute it in the place of 5 y, in the second, we have and the signs being changed agreeably to what has just been remarked, 4 x = 20, If we substitute this value instead of x in the expression for y, it will become solated quantity 14? We understand its import, when there are two quantities separated from each other by the sign, and 1 when the quantity to be subtracted is less than that from which it is to be taken; but how can we subtract a quantity when it is not connected with another in the member where it is found? To clear up this difficulty it is best to go back to the equations, which express the conditions of the question; for the nearer we approach to the enunciation, the closer shall we bring together the circumstances which have given rise to the present uncertainty. I resume the equation 12x+7y=46, I put in the place of x its value 5, and it becomes 60+7y=46. This equation by mere inspection presents an absurdity. It is impossible to make the number 46 by adding any thing to the number 60, which exceeds it already. I take also the second equation 8x+5y = 30, and putting 5 in the place of x, I find 40+5y=30; the same absurdity as before, since the number 30 is to be formed by adding something to the number 40. Now the quantities 12 x or 60 in the first equation, 8 x or 40 in the second, represent what the labourer earned by his own work; the quantities 7 y and 5 y stand for the earnings of his wife and son, while the numbers 46 and 30 express the sum given as the common wages of the three; we must see then at once in what consists the absurdity. According to the question, the labourer earned more by himself than he did by the assistance of his wife and son; it is impossible then to consider what is allowed to the woman and son, as augmenting the pay of the labourer. But if, instead of counting the allowance made to the two latter persons as positive, we regard it as a charge placed to the account of the labourer, then it would be necessary to deduct it from his wages; and the equations would no longer inve contradiction, as they would become 60—7 y = 46, 40-5 y = 30; we deduce from the onc as well as from the other |