Elements of Algebra

CHAP. III. OF ALGEBRA. 319 given numbers, and it is required to find integers for $x$ and $y$ that are not known. If we first separate $y$, we shall have $y = \frac{ax+c}{mx-b}$; and removing $x$ from the numerator, by multiplying both sides by $m$, we have $my = \frac{max+mc}{mx-b} = a + \frac{mc+ab}{mx-b}$, by division. We have here a fraction whose numerator is a known number, and whose denominator must be a divisor of that number; let us therefore represent the numerator by a product of two factors, as $fg$ (which may often be done in several ways) and see if one of these factors may be compared with $mx-b$, so that $mx-b=f$. Now, for this purpose, since $x = \frac{f+b}{m}$, $f+b$ must be divisible by $m$; and hence it follows, that out of the factors of $mc+ab$, we can employ only those which are of such a nature, that, by adding $b$ to them, the sums will be divisible by $m$. We shall illustrate this by an example. Let the equation be $5xy=2x+3y+18$. Here, we have $y = \frac{2x+18}{5x-3}$, and $5y = \frac{10x+90}{5x-3} = 2 + \frac{96}{5x-3}$; it is therefore required to find those divisors of 96 which, added to 3, will give sums divisible by 5. Now, if we consider all the divisors of 96, which are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96, it is evident that only these three of them, viz. 2, 12, 32, will answer this condition. Therefore, 1. If $5x-3=2$, we obtain $5y=50$, and consequently $x=1$, and $y=10$. 2. If $5x-3=12$, we obtain $5y=10$, and consequently $x=3$, and $y=2$. 3. If $5x-3=32$, we obtain $5y=5$, and consequently $x=7$, and $y=1$. 35. As in this general solution we have $my-a = \frac{mc+ab}{mx-b}$, it will be proper to observe, that if a number, contained in the formula $mc+ab$, have a divisor of the form $mx-b$, the quotient in that case must necessarily be contained in the formula $my-a$: we may therefore express the number $mc+ab$ by a product, such as $(mx-b) \times (my-a)$. For