Elements of Algebra

CHAP. VI. ADDITIONS. 549 and expunging $x$ and $y$ from these three equations, we should have $(ak - bh)u^2 - (ck - dh)t^2 = (ad - cb)s^2$; so that, making $\frac{u}{t} = z$, the difficulty would be reduced to resolving the simple equality, $\frac{ak-bh}{ad-cb}z^2 - \frac{ck-dh}{ad-cb} = 0$, which is evidently a case of our general method. Having found the value of $z$, we shall have $u = tz$, and the first two equations will give $x = \frac{d-bz^2}{ad-cb}t^2, y = \frac{az^2-c}{ad-cb}t^2$. But if the given triple equality contained only one variable quantity, we should then again have an equality with the unknown quantity rising to the fourth degree. In fact, it is evident that this case may be deduced from the preceding, by making $y = 1$; so that we must have $\frac{az^2-c}{ad-cb}t^2 = 1$; and, consequently, $\frac{az^2-c}{ad-cb} = 0$. Now, calling $f$ one of the values of $z$, which satisfy the above equality, and, in order to abridge, making $\frac{ak-bh}{ad-cb} = e$, we shall have, in general, (Art. 57.) $z = \frac{fm^2-2gm+ef}{m^2-e}$. Then, substituting this value of $z$ in the last equality, and multiplying the whole of it by the square of $m^2-e$, we shall have, $\frac{a(fm^2-2gm+ef)^2-c(m^2-e)^2}{ad-cb} = 0$, where the unknown quantity, $m$, evidently rises to the fourth power.