Then multiply the last number thus found by the absolute quantity (b) in the numerator of the given fraction, and divide the product by the denominator; so shall the remainder be the true value of x, required; provided the number of terms in the upper line be even, and the sign of b negative, or, if that number be odd and the sign of b affirmative; but, if the number of terms be even, and the sign of b affirmative, or vice versa, then the difference between the said remainder and the denominator of the fraction will be the true answer. In the general method here laid down a is supposed less than c, and that these two numbers are prime to each other: for, were they to admit a common measure, whereby b is not divisible, the thing would be impossible, that is, no integer could be assigned for x, so as to give the value of ax ± b c an integer: the reason of which, as well as of the lemma itself, will be explained a little further on here it will be proper to put down an example or two, to illustrate the use of what has been already delivered. Examp. Let the given quantity be 87x-50 256. These four examples comprehend all the different cases that can happen with regard to the restrictions specified in the latter part of the rule: I shall now show the use thereof in the resolution of problems. PROBLEM I. To find the least whole number, which, divided by 17, shall leave a remainder of 7; but being divided by 26, the remainder shall be 13. Let x be the quotient, by 17, when 7 remains, or, which is the same, let 17x+7 express the number sought; then, since this number, when 13 is subtracted from it, is divisible by 26, it is manifest that 17x+7-13 26 or 17x-6 must be a whole number: whence, by proceeding according to the lemma, x will be found 8; and consequently 17x+7=143, the number required. See the operation. = * PROBLEM II. Supposing 9x+13y=2000, it is required to find all the possible values of x and y in whole positive numbers. By transposing 13y, and dividing the whole equation by 9, we have x = = 222 2-4y ; 9 which, as x is a whole positive number, by the question, must also be a whole positive number, and so likewise 4y-2 9 ; from which the least value of y, in whole numbers, will come out = 5; and consequently the corresponding value of x=215. From whence the rest of the answers, which are 16 in number, will be found, by adding 9, continually, to the last value of y, and subtracting 13 from that of x, as in the annexed table, which exhibits all the possible answers in whole numbers, -213 | 2021 18 x=215|202189|176|163|150|137|124|111|98|85 72 59 46 33 20 7 y= 5 14 23 32 41 50 59 68 77 86 95 104 113 122 131 140 149 In the same manner, the least value of y, and the greatest of x, being found, in any other case, the rest of the answers will be obtained, by only adding the coefficient of x, in the given equation, to the last value of y, continually, and subtracting the coefficient of y from the corresponding value of x. Hence it follows, that, if the greatest value of x be divided by the coefficient of y, the remainder will be the least value of x, and that the quotient +1 will give the number of all the answers. But it is to be observed, that the equations here spoken of, are such, wherein the said coefficients are prime to each other; if this should not be the case, let the equation given be, first of all, reduced to one of this form, by dividing by the greatest common measure. PROBLEM III. To find how many different ways it is possible to pay 100l. in guineas and pistoles, only; reckoning guineas at 21 shillings each, and pistoles at 17. Let x represent the number of guineas, and y that of the pistoles; then the number of shillings in the guineas being 21x, and in the pistoles, 17y, we shall therefore 2000-17y have 21x+17y=2000, and consequently x= 21 =95+ 5—179; which being a whole number, by the 21 question, it is manifest that = teger: now the least value of y, in whole numbers, to answer this condition, will be found =4, and the expression itself = 3; the corresponding, or greatest value of x being 92; which being divided by 17, the coefficient of y (according to the preceding note) the quotient comes out 5, and the remainder 7: therefore the least value of x is 7, and the number of answers (=5+1)=6; and these are as follow, To determine whether it be possible to pay 100l. in guineas and moidores only; the former being reckoned at 21 shillings each, and the latter at 27. Here, by proceeding as in the last question, we have least terms, and the numbers 6 and 21, at the same time, admitting of a common measure, a solution in whole numbers (by the note to the preceding lemma) is impossible. The reason of which depends on these two considera-" tions; that, whatsoever number is divisible by a given number, must be divisible also by all the divisors of it; and that any quantity which exactly measures the whole and one part of another, must do the like by the remaining part. Thus, in the present case, the quantity 6y-5, to have the result a whole number, ought to be divisible by 21, and therefore divisible by 3, likewise (which is, here, a common measure of a and c :) but 6y, the former part of 6y5, is divisible by 3, therefore the latter part -5 ought also to be divisible by 3; which is not the case, and shows the thing proposed to be impossible. PROBLEM V. A butcher bought a certain number of sheep and oxen, for which he paid 100l.; for the sheep he paid 17 shillings a-piece, and for the oxen, one with another, he paid 7 pounds a-piece; it is required to find how many he had of each sort. Let x be the number of sheep, and y that of the oxen; then, the conditions of the question being expressed in algebraic terms, we shall have this equation, viz. 17x 2000 140y +140y =2000; and consequently x = 17 which being a whole number, must therefore be a whole number likewise: whence, by proceeding as above, we find y = 7, and x = 60; and this is the only answer the question will admit of; for the greatest value of x cannot in this case be divided by the coefficient of y, that is, 140 cannot be had in |