Elements of Algebra

CHAP. IV. OF ALGEBRA. 213 Suppose that the first has $x$ pistoles, and that the second has $y$ pistoles. Then we shall first have, $x + \frac{2}{3}y = 29$; and also, $y + \frac{3}{4}x = 29$. The first equation gives $x = 29 - \frac{2}{3}y$, and the second $x = \frac{116-4y}{3}$; so that $29 - \frac{2}{3}y = \frac{116-4y}{3}$. From which equation, we obtain $y = 14\frac{1}{2}$; Therefore $x = 19\frac{1}{3}$. Hence the first person had $19\frac{1}{3}$ pistoles, and the second had $14\frac{1}{2}$ pistoles. 619. Question 6. Three brothers bought a vineyard for a hundred guineas. The youngest says, that he could pay for it alone, if the second gave him half the money which he had; the second says, that if the eldest would give him only the third of his money, he could pay for the vineyard singly; lastly, the eldest asks only a fourth part of the money of the youngest, to pay for the vineyard himself. How much money had each? Suppose the first had $x$ guineas; the second, $y$ guineas; the third, $z$ guineas; we shall then have the three following equations: $x + \frac{1}{2}y = 100$; $y + \frac{1}{3}z = 100$; $z + \frac{1}{4}x = 100$; two of which only give the value of $x$, namely, 1st $x = 100 - \frac{1}{2}y$, 3d $x = 400 - 4z$. So that we have the equation, $100 - \frac{1}{2}y = 400 - 4z$, or $4z - \frac{1}{2}y = 300$, which must be combined with the second, in order to determine $y$ and $z$. Now, the second equation was, $y + \frac{1}{3}z = 100$: we therefore deduce from it $y = 100 - \frac{1}{3}z$; and the equation found last being $4z - \frac{1}{2}y = 300$, we have $y = 8z - 600$. The final equation, therefore, becomes $100 - \frac{1}{3}z = 8z - 600$; so that $8\frac{1}{3}z = 700$, or $\frac{25}{3}z = 700$, and $z = 84$. Consequently, $y = 100 - 28 = 72$, and $x = 64$. The youngest therefore had 64 guineas, the second had 72 guineas, and the eldest had 84 guineas.