Elements of Algebra

CHAP. IV. OF ALGEBRA. 211 615. If there were more than three unknown quantities to determine, and as many equations to resolve, we should proceed in the same manner; but the calculations would often prove very tedious. It is proper, therefore, to remark, that, in each particular case, means may always be discovered of greatly facilitating the solution; which consist in introducing into the calculation, beside the principal unknown quantities, a new unknown quantity arbitrarily assumed, such as, for example, the sum of all the rest; and when a person is a little accustomed to such calculations, he easily perceives what is most proper to be done*. The following examples may serve to facilitate the application of these artifices. 616. Question 4. Three persons, A, B, and c, play together; and, in the first game, a loses to each of the other two, as much money as each of them has. In the next game, B loses to each of the other two, as much money as they then had. Lastly, in the third game, A and B gain each, from c, as much money as they had before. On leaving off, they find that each has an equal sum, namely, 24 guineas. Required, with how much money each sat down to play? Suppose that the stake of the first person was $x$, that of the second $y$, and that of the third $z$: also, let us make the sum of all the stakes, or $x + y + z, = s$. Now, A losing in the first game as much money as the other two have, he loses $s - x$ (for he himself having had $x$, the two others must have had $s - x$); therefore there will remain to him $2x - s$; also B will have $2y$, and c will have $2z$. So that, after the first game, each will have as follows: $A = 2x - s$, $B = 2y$, and $c = 2z$. In the second game, B, who has now $2y$, loses as much money as the other two have, that is to say, $s - 2y$; so that he has left $4y - s$. With regard to the other two, they will each have double what they had; so that after the second game, the three persons have as follows: $A = 4x - 2s$, $B = 4y - s$, and $c = 4z$. In the third game, c, who has now $4z$, is the loser; he loses to A, $4x - 2s$, and to B, $4y - s$; consequently, after this game, they will have: * M. Cramer has given, at the end of his Introduction to the Analysis of Curve Lines, a very excellent rule for determining immediately, and without the necessity of passing through the ordinary operations, the value of the unknown quantities of such equations, to any number. F. T. P 2