H6. Solve the pair of simultaneous equations
$$
\begin{aligned}
x^{2}+3 y & =10 \quad \text { and } \\
3+y & =\frac{10}{x} .
\end{aligned}
$$
Solution
First, let us number the two given equations, so that it is easy to refer to them.
$$
\begin{array}{r}
x^{2}+3 y=10 \\
3+y=\frac{10}{x}
\end{array}
$$

It is possible to eliminate one of the two unknowns by substituting from equation (2) into equation (1), but this leads to a cubic equation. We present another method that avoids this.

By subtracting $x \times$ equation (2) from equation (1), we get
$$
x^{2}+3 y-3 x-x y=0
$$
so that
$$
(x-3)(x-y)=0 \text {. }
$$

Hence either $x=3$ or $x=y$. We deal with each of these two cases separately.
$$
x=3
$$

Using equation (1), say, we obtain $y=\frac{1}{3}$.
$$
\boldsymbol{x}=\boldsymbol{y}
$$

Using equation (1) we obtain
$$
x^{2}+3 x=10
$$
so that
$$
x^{2}+3 x-10=0 \text {. }
$$

Hence
$$
(x-2)(x+5)=0,
$$
and therefore either $x=2$ or $x=-5$. When $x=2$ then $y=2$; when $x=-5$ then
$$
y=-5 \text {. }
$$

By checking in the two given equations, we find that all three solutions are valid. Thus there are three solutions of the simultaneous equations, namely
$$
\begin{aligned}
\text { either } x & =-5 \text { and } y=-5, \\
\text { or } x & =2 \text { and } y=2, \\
\text { or } x & =3 \text { and } y=\frac{1}{3} .
\end{aligned}
$$