1. Solve the pair of simultaneous equations
$$
\begin{aligned}
x^{2}-2 x y & =1, \\
5 x^{2}-2 x y+2 y^{2} & =5 .
\end{aligned}
$$
SolUtion
From the first equation, we have $x^{2}=1+2 x y$
Substituting this into the second equation, we obtain $5(1+2 x y)-2 x y+2 y^{2}=5$, which simplifies to $y^{2}+4 x y=0$

Factorising, we have $y(y+4 x)=0$, and so $y=0$ or $y=-4 x$
If $y=0$ we have, from the first equation, $x= \pm 1$
If $y=-4 x$ we have, from the first equation, $x= \pm \frac{1}{3}$
So there are four solutions, namely
$$
(x, y)=(1,0),(-1,0),\left(\frac{1}{3},-\frac{4}{3}\right),\left(-\frac{1}{3}, \frac{4}{3}\right)
$$