The third, which contained 10 measures of rye, 5 of barley, and 4 of wheat, cost 75 francs; It is asked, what the rye, barley, and wheat cost each per measure ? Let a be the price of a measure of rye, y ༤ that of a measure of barley, that of a measure of wheat. To fulfil the first condition we observe, that 30 measures of rye are worth 30 x, 10 measures of wheat are worth 10 %; and as the whole must make 230 francs, we have the equation 30x+20 y +10x=230. For the second condition we have 15 measures of rye worth 15 x, 15x+6y+12% = 138. For the third condition we have 10 measures of rye worth 10 x, The proposed question then will be brought into three equations; 30 a 20 y + 10 = 230, 15x+6y+ 12 % = 138, 10x+5y+ 4% = 75. Before proceeding to the resolution, I examine the equations, to see if it is not possible to simplify them by dividing the two members of some one of them by the same number (12), and I find that the two members of the first may be divided by 10, and those of the second by 3. Having performed these divisions I have only to occupy myself with the equations 3x+2y+ = 23, 10x+5y+4%=75. As I can select any one of the unknown quantities in order to deduce its value, I take that of in the first equation, because this unknown quantity having no coefficient, its value will be entire er without a divisor, as follows. 2y. This value being substituted for z in the second and third equations, they become 5x+2y+92-12x8 y=46, 10x+5y+92-12x8 y=75; and reducing the first member of each, we find To proceed with these equations, which contain only two unknown quantities, I take in the first the value of the unknown quantity y, and I obtain and by substituting this value in the second equation, it becomes The denominator 6 may be made to disappear by the usual method, but observing that the denominator is divisible by 3, I can simplify the fraction by multiplying it by 3, agreeably to article 54 of Arithmetic. I have then The denominator 2 being made to disappear, it becomes 1844 x 46+7x=150; the first member being reduced gives Substituting this value in the expression for that of y, I find and by substituting these values in the expression for that of z we obtain It appears then, that the price of the rye per measure was 4 fr. that of the barley that of the wheat 3, 5. This example, while it illustrates the method given in the preceding article, ought to be attended to on account of the abbreviations of calculation, which are performed in it. 80. I proceed now to resolve the following problem. A man, who undertook to transport some porcelain vases of three different sizes, contracted that he would pay as much for each vessel that he broke, as he received for those, which he delivered safe. He had committed to him two small vases, four of a middle size, and nine large ones; he broke the middle sized ones, delivered all the others safe, and received the sum of 28 francs. There were afterwards committed to him seven small vases, three of the middle size, and five large ones; he rendered this time the small and the middle sized ones, but broke the five large ones, and he received only s francs. Lastly he took charge of nine small vases, ten middle sized ones, and eleven large ones; all these last he broke, and received in consequence only 4 francs. It is asked what was paid him for carrying a vase of each size. Let x be the sum paid for carrying a small vase, y that for carrying a middle sized one, that for carrying a large one. It is evident that each sum, which the porter received, is the difference between what was due to him for the vessels delivered safe, and what he had to pay for those which were broken; accordingly the three conditions of the problem furnish respectively the following equations ; 2 x 4y+9x=28, 7x+3y- 5% 3, 9x10y-11%= 4. The first of these equations gives and by substituting this value, the second and third equations become Making the denominators to disappear, we have 19628 y63x+6y-10%=6, reducing the first member of each, we obtain taking the value of y in the first of these equations, we find By means of this value, the second equation becomes 103% = 8; being cleared of the denominator 34, it is changed into 34 x 103% 34 × 8, 8568 +4088 % 10640- 3502 z=272. The reduction of the first member of this result gives 586 % 2072=272, whence we deduce ༧= 2344 or ༧=4. By going back with the value of z to that of y, we have one of a middle size, a large one. The prices then were 2 fr. for carrying a small vase, 3 4. This example is sufficient to show how to proceed in all similar cases. 81. It sometimes happens, that all the unknown quantities do not enter at the same time into all the equations; the method, however, is not changed by this circumstance; it is sufficient, carefully to examine the connexion of the unknown quantities in order to pass from one to the others. Let there be, for example, the four equations 3 u 2y= 2, 2x+3y=39, 5–7%=11, 4y+3x=41, containing the unknown quantities u, x, y and z. With a little attention we see, that by taking the value of a in the second equation, and substituting it in the third, the result containing only y and z, will, by being combined with the fourth equation, give the values of these two quantities; and having the value of y, we obtain those of u and x, by means of the first and second equations. The following is the process; 82. The method now explained is applicable to literal equations as well as to numerical ones; but the multitude of letters, which it is necessary to employ to represent generally the things given, when the number of equations and unknown quantities exceeds two, has led algebraists to seek for a more simple manner of expressing them. I shall treat of this in the following article; but in order to furnish the reader with the means of exercising himself in putting a problem into an equation, and resolving it, I have subjoined a number of questions, and have placed at the end of each the answer that is required. 1. A father, being asked the age of his son, said, if from double the age that he is of now, you subtract triple of what he was six years ago, you have his present age. Answer. The child was 9 years old. 2. Diophantus, the author of the most ancient book on Algebra, that has come down to us, passed a sixth part of his life in infancy, a twelfth part of it in youth; afterward he was married and pass |