Solve the following system of equations:
(1) $\quad x / y+y / x=6,6 \dot{5} 3846 \dot{1}$,
(2) $\quad 1 / x+1 / y=0,5 \dot{7} 6923 \dot{0}$.

(The numbers on the right side are repeating decimals. Read the solution to problem 1302, then convert the numbers on the right side to common fractions.)
I. The right-hand side segmented decimal fractions in the solution of problem 1302 are converted into the ratios of two natural numbers $A / B$ and $C / D$, similar to the procedure seen in 1 1. The seventh decimal place of the first segmented decimal fraction marks the beginning of the infinite repetition of the 538461 segment. This is a consequence of the fact that when determining the digits of the $A: B$ ratio up to 6.6538461 (and already subtracting the partial product of 1 with $B$ from the partial dividend), the same remainder $R$ reappeared as it did after determining the digits of the ratio up to 6.6, except that in the latter case, the local value of $R$ is $1 / 10$, while in the case of repetition, it is $1 / 10000000 = 1 / 10^{7}$. Thus, the two remainder divisions yield the following expressions for the next representation of $A$:

$$
A = B \cdot 6.6 + R / 10
$$

$$
\text{or} \quad A = B \cdot 6.6538461 + R / 10^{7}
$$

Subtracting the first equality from the second, multiplied by $10^{6}$, gives the desired form:

$\left(10^{6}-1\right) A = B(6653846.1 - 6.6)$,

$$
\frac{A}{B} = \frac{66538461 - 66}{10 \cdot (10^{6} - 1)} = \frac{66538395}{9999990} = \frac{173}{26}
$$

In the simplification, we used the fact that the denominator can be easily factored:

$$
999999 = 999 \cdot 1001, \quad 999 = 27 \cdot 37, \quad \text{and} \quad 1001 = 7 \cdot 11 \cdot 13
$$

Thus, the first segmented decimal fraction can be converted to the common fraction 173/26, which can be easily verified.

Similarly, the common fraction form of the right-hand side of equation (2) is

$$
\frac{C}{D} = \frac{5769230 - 5}{9999990} = \frac{15}{26}
$$

II. Multiply both equations by $x y$, and add $2 x y$ to the new first equation, so that the left side of the final form is equal to the square of the left side of the new second equation:

$$
\begin{gathered}
x^{2} + y^{2} = \frac{173}{26} x y \\
x + y = \frac{15}{26} x y \\
x^{2} + 2 x y + y^{2} = (x + y)^{2} = \left(\frac{173}{26} + 2\right) x y = \frac{225}{26} x y
\end{gathered}
$$

Thus, dividing (1b) by (2a) and then the square of (2a) by (1b), and rearranging, we get

$$
x + y = 15, \quad x y = 26
$$

(We used the fact that $x \neq 0$ and $y \neq 0$, otherwise the system of equations would have no meaning.) Therefore, $x$ and $y$ are equal to the roots of the equation

$$
z^{2} - 15 z + 26 = 0
$$

which are 13 and 2, in any order:

$$
x_{1} = 13, \quad y_{1} = 2, \quad \text{and} \quad x_{2} = y_{1}, \quad y_{2} = x_{1}
$$

The value pair indeed satisfies the system of equations.

Steiner György (Budapest, Radnóti M. gyak. g. III. o. t.) Baranyai Zsolt (Budapest, Fazekas M. gyak. g. III. o. t.)

Remarks. 1. By introducing a new variable $u$ for $x / y$, after converting the infinite decimal fractions to common fractions, we arrive at the system of equations

$$
u + \frac{1}{u} = \frac{173}{26}, \quad \frac{1}{y}\left(\frac{1}{u} + 1\right) = \frac{15}{26}, \quad x = y u
$$

From the first, $u$ is 13/2 and 2/13, from the second, $y$ is 2 and 13, and from the third, $x$ is 13 and 2.

Varga Gabriella (Szombathely, Savaria g. II. o. t.)

2. Most of the solutions chose the method of converting the segmented decimal fractions to common fractions, which, as mentioned in the second remark of problem 1302, proceeds in a way not clarified in the high school curriculum, performing subtraction with numbers written with infinitely many digits as if they were written with finitely many digits. Those who converted the fractions in the manner described above or mentioned this issue received 1 bonus point.[^0]

[^0]: ${ }^{1}$ The editorial board recommended studying the solution of problem 1302 when this exercise was assigned; see K. M. L. 30 (1965) 18. o.