taking the value of y in the first of these equations, we find By means of this value, the second equation becomes being cleared of the denominator 34, it is changed into 34 × 252 +56 × 73 % - 56 × 190- 34 x 103% = 34 × 8, or into The reduction of the first member of this result gives By going back with the value of z to that of y, we have The prices then were 2 fr. for carrying a small vase, 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 Su-2y= 2, 2x+3y=39, 5 x 7% = 11, containing the unknown quantities u, x, y and z. With a little attention we see, that by taking the value of x : 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 ed in this state a seventh part, and five years more, when he had a son, whom he survived four years, and who attained only to half the age of his father, what was the age of Diophantus when he died? Answer, 84 years. 3. A merchant drew, every year, upon the stock he had in trade, the sum of 1000 francs for the expense of his family; still his property increased every year, by a third part of what remained after this deduction, and at the end of three years it was doubled; how much had he at the beginning of the first year? Answer, 14800 francs. 4. A merchant has two kinds of tea, the first at 14 francs a pound, the second at 18 francs; how much ought he to take of each to make up a chest of 100 pounds, which should be worth 1680 francs ? Answer, 30 pounds of the first and 70 of the second. 5. A person filled, in 12 minutes, a vessel containing 39 gallons, with water, by means of two fountains, which were made to run in succession, and one discharged 4 gallons per minute and the other 3, how long did each fountain run ? Answer, the first 3 minutes, and the second 9. 6. At noon the hour and minute hands of a watch are together, at what point of the dial will they next be in conjunction ? Answer, at 1 hour 5 minutes and Obs. This problem refers itself to that of art. 65. 7. A man, meeting some beggars, wishes to give them 25 cents each, but finds upon counting his money, that he wants 10 cents in order to do it; he then gives them only 20 cents each, and has 25 cents left; how much money had he, and what was the number of beggars? Answer, 7. 8. Three brothers purchased an estate for 50000 francs, and the first wanted, in order to complete his part of the payment, half of the property of the second; the second would have paid his share with the help of a third of what the first owned, and the third required to make the same payment, in addition to what he had, a fourth part of what the first possessed; what was the amount of each one's property? Answer, the first had 50000 francs, the second 40000, and the third 42500. 9. Three players after a game count their money, one had lost, the other two had gained each as much as he had brought to the play; after the second game, one of the players, who had gained before, lost, and the two others gained each a sum equal to what he had at the beginning of this second game; at the third game, the player, who had gained till now, lost with each of the others a sum equal to that, which each possessed at the beginning of this last game; they then separated, each having 120 francs; how much had they each when they commenced playing? Answer, he, who lost at the first game, had 195 francs, General formulas for the resolution of equations of the first degree. 83. To obviate the inconvenience referred to in the beginning of the last article, we shall represent all the coefficients of the same unknown quantity by the same letter, but distinguish them by one or more accents, according to the number of equations. General equations with two unknown quantities are written thus ; ax + by = c a' x + b'y=c. The coefficients of the unknown quantity a are both represented by a, those of y by b; but from the accent, which is placed over the letters in the second equation, it may be seen, that they are not considered as having the same value, as the corresponding ones in the first. Thus a' is a quantity different from a, b'a quantity different from b. If there are three equations, they are expressed; a x+by+cz=d a' x + by + c z = d' a" x + b′′y + c" z = d′′. All the coefficients of the unknown quantity ∞ are designated by the letter a, those of y by b, those of z by c; but the several letters are distinguished by different accents, which show, that they denote different quantities. Thus a, a', a", are three different quantities. The same may be said of b, b', b", &c. Following this method, if we have four unknown quantities, and four equations, we may write them thus ; a x+by+c z+d u=e a′′ x + b′′ y + c′′ ≈ + d′′ 84. To avoid fractions, and simplify the calculation, we may vary the process of elimination in the following manner. Let there be the equations ax + by = c ax + by = c; it is evident, that if one of the unknown quantities, x, for example, has the same coefficient in the two equations, we have only to subtract one of these equations from the other, in order to make this unknown quantity to disappear. This may be seen at once in the equations 11y-9y=27-15, or 2y= 12, or y = 6. It is evident, that the coefficients of x may be immediately made equal in the equations ax + by = c a' x + b' y = c' by multiplying the two members of the first by a', the coefficient of x in the second, and the two members of the second by a, the coefficient of x in the first; we thus obtain, a a x + a' by = a' c a a'x + a b'ya c'. = Then subtracting the first of these from the second, the unknown quantity a disappears; and we have (a b'a'b) ya c'—a'c, an equation, which contains only the unknown quantity y; from this we may deduce, The method, we have just employed, may always be applied to equations of the first degree, to exterminate any one of the unknown quantities. By exterminating, in the same manner, the unknown quantity y, we may find the value of x. If we apply this process to three equations, containing x, y and %, we may first exterminate ar from the first and second, then from the first and third; we thus obtain two equations, which contain only y and x, from which we may exterminate y. When this calculation is performed, the equation containing ~, |