The terms of the preceding fractions were simple quantities, but if we had fractions, the terms of which were polynomials, we should have to perform, by the rules given for compound quantities, the operations indicated upon simple quantities; it is thus that we have 54. Understanding what precedes, we can resolve an equation of the first degree, however complicated. If we have, for example, the equation we begin by making the denominators to disappear, indicating only the operations; it becomes then (a+b)(x−c) (3a+b)+4b(a−-b)(3a+b)=2x(a−-b)(3a+b)-—ac(a—b); performing the multiplications, we have 3a2x+4abx+b2x-3a2c-4abc-b2c+12a2b—Sab2 — 4b3 — 6a2x-4 abx-2 b2 x-a2c+ubc; transposing to one member all the terms involving x, it becomes -3a2x+8abx+3b2x=2a2c+5abc+b3c-12a2b+8ab2+4b3, from which we deduce x= 2 a2c+5abc + b2 c 12 a2 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, their difference by b, we have, by the enunciation of the question, Each of these two equations 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. and putting this value of x in the expression for 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 a + b. 56. I take another question. A labourer having worked for a person 12 days, and having with him, during the first 7 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 how much did his wife and son earn? Let x be the daily wages of the man, that of his wife and son; 12 days' work of the man will amount to 12x, 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 y 8x, 5y; Proceeding as in the preceding question, we take the value of 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. reducing it, we have By transposing 4 x to the second member, and 350 to the first, we obtain Knowing, which we have just found equal to 5, if we place this value in the formula the second member will be determined, for we have The man then earned 5 francs per day, while his wife and son earned only 2. 57. The reader has perhaps observed, that in resolving the above equation 3704x350, I have transposed 4x to the second member. I have proceeded thus to avoid a slight difficulty, that would otherwise have occurred, 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 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 350370, which gives + - 4 x, We might also change the signs after reduction, and the equation 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 I suppose 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 case. 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, x = 20 = 5. If we substitute this value instead of x in the expression for y, it will become Now how are we to interpret the sign, which affects the insulated quantity 14? We understand its import, when there are two quantities separated from each other by the sign, and |