Elements of Algebra

CHAP. III. OF ALGEBRA. 203 Sum, or property to be divided. | Order of the children. | Portion of each. | Differences. --- | --- | --- | --- $z$ | 1st | $x = 100 + \frac{z-100}{10}$ | $z-x$ | 2d | $x = 200 + \frac{z-x-200}{10}$ | $100 - \frac{x-100}{10} = 0$ $z-2x$ | 3d | $x = 300 + \frac{z-2x-300}{10}$ | $100 - \frac{x-100}{10} = 0$ $z-3x$ | 4th | $x = 400 + \frac{z-3x-400}{10}$ | $100 - \frac{x-100}{10} = 0$ $z-4x$ | 5th | $x = 500 + \frac{z-4x-500}{10}$ | $100 - \frac{x-100}{10} = 0$ $z-5x$ | 6th | $x = 600 + \frac{z-5x-600}{10}$ | and so on. We have inserted, in the last column, the differences which we obtain by subtracting each portion from that which follows; but all the portions being equal, each of the differences must be $= 0$. As it happens also, that all these differences are expressed exactly alike, it will be sufficient to make one of them equal to nothing, and we shall have the equation $100 - \frac{x-100}{10} = 0$. Here, multiplying by 10, we have $1000 - x - 100 = 0$, or $900 - x = 0$; and, consequently, $x = 900$. We know now, therefore, that the share of each child was 900; so that taking any one of the equations of the third column, the first for example, it becomes, by substituting the value of $x$, $900 = 100 + \frac{z-100}{10}$, whence we immediately obtain the value of $z$; for we have $9000 = 1000 + z - 100$, or $9000 = 900 + z$; therefore $z = 8100$; and consequently $\frac{z}{x} = 9$. So that the number of children was 9; the fortune left by the father was 8100 pounds; and the share of each child was 900 pounds. QUESTIONS FOR PRACTICE. 1. To find a number, to which if there be added a half, a third, and a fourth of itself, the sum will be 50. Ans. 24.