M4. What are the solutions of the simultaneous equations:
$$
\begin{aligned}
3 x^{2}+x y-2 y^{2} & =-5 \\
x^{2}+2 x y+y^{2} & =1 ?
\end{aligned}
$$
Solution
For convenience, we number the equations, as follows:
$$
\begin{aligned}
3 x^{2}+x y-2 y^{2} & =-5 \\
x^{2}+2 x y+y^{2} & =1
\end{aligned}
$$

By adding equation (1) to $5 \times$ equation (2), we obtain
$$
8 x^{2}+11 x y+3 y^{2}=0 \text {, }
$$
that is,
$$
(8 x+3 y)(x+y)=0 .
$$

Thus either $8 x+3 y=0$ or $x+y=0$.
In the first case, from equation (2) we obtain $25 x^{2}=9$, so that $x= \pm \frac{3}{5}$. Because $8 x+3 y=0$, if $x=\frac{3}{5}$, then $y=-\frac{8}{5}$, and if $x=-\frac{3}{5}$, then $y=\frac{8}{5}$.

Checking, we see that each of these is also a solution to equation (1).
In the second case, from equation (2) we get $0=1$, so this is impossible.
Hence the solutions of the given simultaneous equations are $x=\frac{3}{5}, y=-\frac{8}{5}$; and $x=-\frac{3}{5}, y=\frac{8}{5}$.

Alternative
We may factorise the left-hand side of each of the given equations as follows:
$$
\begin{aligned}
(3 x-2 y)(x+y) & =-5 ; \\
(x+y)^{2} & =1 .
\end{aligned}
$$

From equation (4) it follows that $x+y= \pm 1$.
When $x+y=1$, from equation (3) we get $3 x-2 y=-5$. Solving these two linear simultaneous equations, we obtain $x=-\frac{3}{5}$ and $y=\frac{8}{5}$.

Similarly, when $x+y=-1$, we obtain $x=\frac{3}{5}$ and $y=-\frac{8}{5}$.
Checking, we see that each of these is also a solution to the original equations.
Hence the solutions of the given simultaneous equations are $x=\frac{3}{5}, y=-\frac{8}{5}$; and $x=-\frac{3}{5}, y=\frac{8}{5}$.