Suppressing the factor a, common to the numerator and denom are applied in the same manner as those, which we before found for literal equations, with only one unknown quantity; we substitute in the place of the letters, the particular numbers in the example selected. We shall obtain the results in art. 56, by making 77. The values of x and y are adapted not only to the proposed question; they extend also to all those, which lead to two equations of the first degree with two unknown quantities, since it is evident, that these equations are necessarily comprehended in the formulas, A で the expression x = B signifies, that x is equal to the quotient of the A fraction divided by B, and the expression a = で the quotient arising from A divided by the fraction denote by the expression x = and lastly, we It will be perceived by these remarks, that it is necessary to place the bars according to the result, which we propose to express. ax+by=c, provided the letters a, b, d, e, denote the whole of the given quantities, by which the unknown quantities x and y are respectively multiplied, and the letters c and f the whole of the known terms, transposed to the second member. Of the resolution of any given number of equations of the first degree, 1 containing an equal number of unknown quantities. 78. WHEN an equation has as many distinct conditions, as it contains unknown quantities, each of these conditions furnishes an equation, in which it often happens, that the unknown quantities are involved with others, as we have seen already in the problems with two unknown quantities; but if these unknown quantities are only of the first degree, according to the method adopted in the preceding articles, we take in one of the equations the value of one of the unknown quantities, as if all the rest were known, and substitute this value in all the other equations, which will then contain only the other unknown quantities. This operation, by which we exterminate one of the unknown quantities, is called elimination. In this way, if we have three equations with three unknown quantities, we deduce from them two equations with only two unknown quantities, which are to be treated as above; and having obtained the values of the two last unknown quantities, we substitute them in the expression for the value of the first unknown quantity. If we have four equations with four unknown quantities, we deduce from them, in the first place, three equations with three unknown quantities, which are to be treated in the manner just described; having found the value of the three unknown quantities, we substitute them in the expression for the value of the first, and so on. Sce an example of a question, which contains three unknown quantities and three equations. 79. A person buys separately three loads of grain; the first, which contained 30 measures of rye, 20 of barley, and 10 of wheat, cost 250 francs; The second, which contained 15 measures of rye, 6 of barley, and 12 of wheat, cost 138 francs; 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 x be the price of a measure of rye, y that of a measure of barley, ༧ that of a measure of wheat. To fulfil the first condition we observe, that 30 measures of rye are worth 30 x, 20 measures of barley are worth 20 y, 10 measures of wheat are worth 10 z; and as the whole must make 230 francs, we have the equation 30 x 20 y +10%=230. For the second condition we have 15 measures of rye worth 15 x, The proposed question then will be brought into three equations; 30+20y+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+x=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 or without a divisor, as follows. This value being substituted for z in the second and third equations, they become 5x+2y+92-12x-8 y=46, 10x+5y+92-12x8y=75; and reducing the first member of each, we find 92-7x 6 y = 46, 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 184-4 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 = we obtain It appears then, that the price of the rye per measure was 4 fr. that of the barley that of the wheat S, 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 3 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; and by substituting this value, the second and third equations become Making the denominators to disappear, we have 19628y63x+6y-10%=6, 25236 y 81+20 y 22=8; reducing the first member of each, we obtain |