Elements of Algebra

210 ELEMENTS SECT. IV. two equations between $z$ and $y$, to be resolved by the preceding rule. 614. Suppose we had found the three following equations: $3x + 5y - 4z = 25$, $5x - 2y + 3z = 46$, $3y + 5z - x = 62$. If we deduce from each the value of $x$, we shall have from the 1st $x = \frac{25-5y+4z}{3}$, 2d $x = \frac{46+2y-3z}{5}$, 3d $x = 3y + 5z - 62$. Comparing these three values together, and first the third with the first, we have $3y + 5z - 62 = \frac{25-5y+4z}{3}$; multiplying by 3, gives $9y + 15z - 186 = 25 - 5y + 4z$; so that $9y + 15z = 211 - 5y + 4z$, and $14y + 11z = 211$. Comparing also the third with the second, we have $3y + 5z - 62 = \frac{46+2y-3z}{5}$; or $46 + 2y - 3z = 15y + 25z - 310$, which, when reduced, becomes $356 = 13y + 28z$. We shall now deduce, from these two new equations, the value of $y$: 1st $14y + 11z = 211$; or $14y = 211 - 11z$, and $y = \frac{211-11z}{14}$. 2d $13y + 28z = 356$; or $13y = 356 - 28z$, and $y = \frac{356-28z}{13}$. These two values form the new equation $\frac{211-11z}{14} = \frac{356-28z}{13}$, whence, $2743 - 143z = 4984 - 392z$, or $249z = 2241$, and $z = 9$. This value being substituted in one of the two equations of $y$ and $z$, we find $y = 8$; and, lastly, a similar substitution in one of the three values of $x$, will give $x = 7$.