Elements of Algebra

304 ELEMENTS PART II. 1. $u = 1$ gives the number of horses $x = 9$, and that of oxen $y = 71$; wherefore the former cost 279 crowns, and the latter 1491; in all 1770 crowns. 2. $u = 2$ gives $x = 30$, and $y = 40$; so that the horses cost 930 crowns, and the oxen 840 crowns, which together make 1770 crowns. 3. $u = 3$ gives the number of the horses $x = 51$, and that of the oxen $y = 9$; the former cost 1581 crowns, and the latter 189 crowns; which together make 1770 crowns. 10. The questions which we have hitherto considered lead all to an equation of the form $ax + by = c$, in which $a$, $b$, and $c$, represent integer and positive numbers, and in which the values of $x$ and $y$ must likewise be integer and positive. Now, if $b$ is negative, and the equation has the form $ax - by = c$, we have questions of quite a different kind, admitting of an infinite number of answers, which we shall treat of before we conclude the present chapter. The simplest questions of this sort are such as the following. Required two numbers, whose difference may be $6$. If, in this case, we make the less number $x$, and the greater $y$, we must have $y - x = 6$, and $y = 6 + x$. Now, nothing prevents us from substituting, instead of $x$, all the integer numbers possible, and whatever number we assume, $y$ will always be greater by $6$. Let us, for example, make $x = 100$, we have $y = 106$; it is evident, therefore, that an infinite number of answers are possible. 11. Next follow questions, in which $c = 0$, that is to say, in which $ax$ must simply be equal to $by$. Let there be required, for example, a number divisible both by $5$ and by $7$. If we write $N$ for that number, we shall first have $N = 5x$, since $N$ must be divisible by $5$; and farther, we shall have $N = 7y$, because the number must also be divisible by $7$. We shall therefore have $5x = 7y$, and $x = \frac{7y}{5}$. Now, since $7$ cannot be divided by $5$, $y$ must be divisible by $5$: let us therefore make $y = 5z$, and we have $x = 7z$; so that the number sought $(N)$ will be $= 35z$; and as we may take for $z$, any integer number whatever, it is evident that we can assign for $N$ an infinite number of values; such as $35, 70, 105, 140, 175, 210, \&c.$ If, beside the above condition, it were also required that the number $N$ be divisible by $9$, we should first have $N = 35z$, as before, and should farther make $N = 9u$. In this man-