Elements of Algebra

CHAP. IV. OF ALGEBRA. 209 Suppose the mule's load to be $x$ hundred weight, and that of the ass to be $y$ hundred weight. If the mule gives one hundred weight to the ass, the one will have $y+1$, and there will remain for the other $x-1$; and since, in this case, the ass is loaded twice as much as the mule, we have $y+1=2x-2$. Farther, if the ass gives a hundred weight to the mule, the latter has $x+1$, and the ass retains $y-1$; but the burden of the former being now three times that of the latter, we have $x+1=3y-3$. Consequently our two equations will be, $y+1=2x-2$, and $x+1=3y-3$. From the first, $x=\frac{y+3}{2}$, and the second gives $x=3y-4$; whence we have the new equation $\frac{y+3}{2}=3y-4$, which gives $y=\frac{11}{5}$; this also determines the value of $x$, which becomes $2\frac{3}{5}$. The mule therefore carried $2\frac{3}{5}$ hundred weight, and the ass $2\frac{1}{5}$ hundred weight. 613. When there are three unknown numbers, and as many equations; as, for example, $x+y-z=8$, $x+z-y=9$, $y+z-x=10$; we begin, as before, by deducing a value of $x$ from each, and have, from the 1st $x=8+z-y$; 2d $x=9+y-z$; 3d $x=y+z-10$. Comparing the first of these values with the second, and after that with the third, we have the following equations: $8+z-y=9+y-z$, $8+z-y=y+z-10$. Now, the first gives $2z-2y=1$, and, by the second, $2y=18$, or $y=9$; if therefore we substitute this value of $y$ in $2z-2y=1$, we have $2z-18=1$, or $2z=19$, so that $z=9\frac{1}{2}$; it remains, therefore, only to determine $x$, which is easily found $=8\frac{1}{2}$. Here it happens, that the letter $z$ vanishes in the last equation, and that the value of $y$ is found immediately; but if this had not been the case, we should have had P