I convert them by the first of the rules cited into the following; 5 X 7 X 2x 3 X5 X 7' 3 X 7 X 4x 3 X5 X 7' 3 X 5 X 5 x 3 X5 X 7 Since, for converting the whole numbers 4 and 12 into fractions, nothing more is necessary than to multiply them by the common denominator of the fractious, namely, 3 x 5 x 7; we have The denominator may now be cancelled, since by doing it we only multiply all the parts of the equation by this denominator, (Arith. 54), which does not destroy the equality of the members. It will become then or 5 × 7 × 2x + 3 × 5 × 7 X 4 3 X 5 X 5 x. an equation without a denominator from which we deduce the value of x by the preceding rules. It is evident from inspection, as also from the mere application of the arithmetical rules referred to, that in the above operation the numerators of each fraction must be multiplied by the product of the denominators of all the others, the whole numbers by the product of all the denominators; then no account need be taken of the common denominators of the fractions thus obtained. The equation 70 x + 420 84 x + 1260-75 x, becomes successively 70x+75x 420, The same process is applicable to literal equations, it being observed, that it is necessary only to indicate the multiplications, which are actually performed when numbers are concerned. Let there be, for example, the equation we deduce from it απ b e dx eh xax-beh xcbhx dx + be x fg, a result which may be more simply expressed by placing the factors of each product one after the other without any sign between them, according to the method given in article 7; and by arranging the letters in alphabetical order, they are more easily read, it then becomes a ehx-bceh bdhx + befg, from which is deduced and a eh x-b dh x = befg+bceh, x= befg+beeh 14. Although no general and exact rule can be given for forming the equation of any question whatever; there is notwithstanding, a precept of extensive use, which cannot fail to lead to the proposed object. It is this, To indicate by the aid of algebraic signs upon the known quantities represented either by numbers or letters, and upon the unknown quantities represented always by letters, the same reasonings and the same operations, which it would have been necessary to perform in order to verify the values of the unknown quantities, had they been known. In making use of this precept, it is necessary, in the first place, to determine with care what are the operations which are contained in the enunciation of the question, either directly or by implication; but this is the very thing which constitutes the difficulty of putting a question into an equation. The following examples are intended to illustrate the above precept. I have taken the two first from among the questions which are solved by arithmetic, in order to show the advantage of the algebraic method. 1. Let there be two fountains, the first of which running for 24h. fills a certain vessel, and the second fills the same vessel by running 33h. what time will be employed by both the fountains running together in filling the vessel? If the time were given we should verify it by calculating the quantities of water discharged by each fountain, and adding them together we should be certain, that they would be equal to the whole content of the vessel. To form the equation we denote the unknown time by x, and we indicate upon a the operations implied by the question; but in order to render the solution independent of the given num bers, and at the same time to abridge the expression where fractions are concerned, we will represent them also by letters, a being written instead of 24h. and b instead of 33h. This being supposed, by putting the capacity of the vessel equal to unity, it is evident, that, The first fountain, which will fill it in a number of hours denoted by a, will discharge into it in one hour a quantity of water expressed by the fraction, and that consequently, in a 1 α a number of hours it will furnish the quantity X or -• (Arith. 53). 1 The second fountain, which will fill the same vessel in a number of hours described by b, will discharge into it in one hour a quantity of water expressed by the fraction, and consequently in a number of hours, it will furnish the quantity xX b' or b 1 The whole quantity of water then furnished by the two fountains, will be and this quantity must be equal to the content of the vessel, which was considered as unity; we have then the equation Divide the product of the numbers, which denote the times employed by each fountain in filling the vessel, by the sum of these numbers; the quotient expresses the time required by both the fountains running together to fill the vessel. Applying this rule to the particular case under consideration, we have whence x = }} = 1 2. Let a be a number to be divided into three parts, having among themselves the same ratios as the given numbers m, n, and p. It is evident that the verification of the question would be as follows; denoting the 1st part by x, we have m: nx: the 2d part = m: px: the 3d part = nx m (Arith. 116.) the three parts added together must make the number to be divided. We have then the equation By reducing all the terms to the denominator m, it becomes m x + n x + px = am ; This result is nothing more nor less than an algebraic expression of the rule of Fellowship, (Arith. 124); for by regarding the numbers m, n, p, as denoting the stocks of several persons trading in company, m+n+p is the whole stock, a the gain to be divided, and the equation x= ma m + n + p shows that a share is obtained by multiplying the corresponding stock into the whole gain, and dividing the product by the sum of the stocks; which reduced to a proportion, becomes the whole stock: a particular stock :: the whole gain: to the particular gain. 15. To form an equation from the following question, requires an attention to some things, which have not yet been considered. A fisherman, to encourage his son, promises him 5 cents for each throw of the net in which he shall take any fish, but the son, on the other hand, is to remit to the father 3 cents for each unsuccessful throw. After 12 throws the father and the son settle their account, and the former is found to owe the latter 28 cents. What was the number of successful throws of the net? If we represent this number by a the number of unsuccessful ones will be 12. *; and if these numbers were given, we should verify them by multiplying 5 cents by the first, to obtain what the father was bound to pay the son, and 3 cents by the second, to find what the son engaged to return to the father. The first number ought to exceed the second by 28 cents, which the father owed the son. We have for the first number x times 5 cents, or 5 x. With respect to the second, there is some difficulty. How are we to obtain the product of 3 by 12 — x? If instead of r we had a given number, we should first perform the subtraction indicated, and then multiply 3 by the remainder; but this cannot be done at present, and we must endeavour to perform the multiplication before the subtraction, or at least, to give the expression an entire algebraic form, similar to that of equations that are readily solved. With a little attention we shall see, that by taking 12 times the number three, we repeat the number 3 so many times too much, as there are units in the number x, by which we ought first to have diminished the multiplier 12, so that the true product will be 36 diminished by 3 taken x times or 3 x, This conclusion may be verified by giving to x a numerical value. If, for example, x were equal to 8, we should have 3 to be taken 12 times 8 times, and, if we neglect - 8 times, we should make the result 8 times the number 3 too much; the true product then will be This result agrees with that which would arise from first subtracting 8 from 12; for then - 12 8 4, and 3 x 4 = 12. This being admitted, since the money due from the father to the son is expressed by 5x, and that which the son owes the father by 36 3 x, the second number must be subtracted from the first in order to obtain the remainder 28; but here is another difficulty; how shall we subtract 363x from 5x, without having first subtracted 3 x from 36? We shall avoid this difficulty by observing, that if we neglect the term 34, and subtract from 5x the entire number 36, we shall have taken necessarily 3x too much, since it is only what |
