Solve the following equations:

$$
\begin{aligned}
& \frac{1.618+\frac{1}{x}}{1.618-\frac{1}{x}}=1.618+\frac{1}{1.618} \\
& \frac{0.618+\frac{1}{x}}{0.618-\frac{1}{x}}=0.618+\frac{1}{0.618}
\end{aligned}
$$

For 2nd graders, substitute the values 1.618 and 0.618 with $(\sqrt{5}+1) / 2$ and $(\sqrt{5}-1) / 2$, respectively.
(1) By replacing 1.618 with 0.618 in all four places, we get (2). Therefore, instead of the two equations, it is advisable to solve the following equation containing the parameter $c$ in place of the numbers 1.618 and 0.618, and then substitute 1.618 or 0.618 for $c$ in the result:

$$
\frac{c+\frac{1}{x}}{c-\frac{1}{x}}=c+\frac{1}{c}
$$

The values $x=0$ and $x=1 / c$ make the left side undefined; thus, they are not roots of the equation.

By adding 1 to both sides, the unknown appears only in the denominator. At this point, it is advisable to take the reciprocal of both sides:

$$
\begin{gathered}
\frac{c+\frac{1}{x}}{c-\frac{1}{x}}+1=\frac{2 c}{c-\frac{1}{x}}=c-\frac{1}{c}+1=\frac{c^{2}+c+1}{c} \\
\quad \frac{c-\frac{1}{x}}{2 c}=\frac{1}{2}-\frac{1}{2 c x}=\frac{c}{c^{2}+c+1}
\end{gathered}
$$

By rearranging and taking the reciprocals again,

$$
\begin{gathered}
\frac{1}{2 c x}=\frac{1}{2}-\frac{c}{c^{2}+c+1}=\frac{c^{2}-c+1}{2\left(c^{2}+c+1\right)}, \text { finally } \\
x=\frac{c^{2}+c+1}{c\left(c^{2}-c+1\right)}=\frac{1}{c}\left(1+\frac{2 c}{c^{2}-c+1}\right)=\frac{1}{c}+\frac{2}{c^{2}-c+1}
\end{gathered}
$$

By substituting $c_{1}=1.618$ and $c_{2}=0.618$, we can reach the goal with fewer calculations if we use the fact that $c_{1}-c_{2}=1$, or $c_{1}-1=c_{2}$, and $c_{2}+1=c_{1}$. Thus, the denominator of the second term of the $x_{1}$ obtained with $c_{1}$ is

$$
\begin{gathered}
\left(c_{1}^{2}-c_{1}\right)+1=c_{1}\left(c_{1}-1\right)+1=c_{1} c_{2}+1, \quad \text { and thus } \\
x_{1}=\frac{1}{c_{1}}+\frac{2}{c_{1} c_{2}+1}
\end{gathered}
$$

Similarly, for $x_{2}$ obtained with $c_{2}$,

$$
\begin{gathered}
c_{2}^{2}-c_{2}+1=\left(c_{2}^{2}+c_{2}\right)-2 c_{2}+1=c_{1} c_{2}+1-2 c_{2}, \quad \text { and thus } \\
x_{2}=\frac{1}{c_{2}}+\frac{2}{c_{1} c_{2}+1-2 c_{2}}
\end{gathered}
$$

Now, $c_{1} c_{2}=0.999$ 924. If we want to calculate $x_{1}$ and $x_{2}$ with three decimal places (thousandths) accuracy, we can round $c_{1} c_{2}$ to 1. This means that we can write $c_{2}$ for $1 / c_{1}$ with sufficient accuracy, and $c_{1}$ for $1 / c_{2}$. Thus,

$$
\begin{aligned}
& x_{1} \approx c_{2}+\frac{2}{1+1}=c_{2}+1=c_{1}=1.618, \quad \text { and } \\
& x_{2} \approx c_{1}+\frac{2}{2-c_{2}}=c_{1}+\frac{1}{1-c_{2}} \approx 1.618+2.618=4.236
\end{aligned}
$$

Thus, the solutions of (1) and (2) with three decimal accuracy are

$$
x_{1} \approx 1.618, \quad \text { or } \quad x_{2} \approx 4.236
$$

With the given square root expressions for $c_{1}$ and $c_{2}$, we have $c_{1} c_{2}=1$, so we only need to calculate the second term of $x_{2}$:

$$
\begin{gathered}
\frac{1}{1-c_{2}}=\frac{1}{\frac{3}{2}-\frac{\sqrt{5}}{2}}=\frac{2(3+\sqrt{5})}{9-5}=\frac{3}{2}+\frac{\sqrt{5}}{2}, \quad \text { thus } \\
x_{2}=\left(\frac{\sqrt{5}}{2}+\frac{1}{2}\right)+\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)=\sqrt{5}+2, \quad \text { and } \quad x_{1}=\frac{\sqrt{5}+1}{2}
\end{gathered}
$$

Gergely Kovács (Győr, Czuczor G. g. I. o. t.) Ilona Friss (Budapest, Radnóti M. g. II. o. t.)