\section*{Problem 4 - 061234}

Determine all and only those real numbers \(x\) that satisfy the equation

\[
\left[\frac{5+6 x}{8}\right]=\frac{15 x-7}{5}
\]

where \([a]\) denotes the greatest integer not greater than \(a\); for example, \(\left[\frac{13}{2}\right]=6,[-6.5]=-7\) and \([6]=6\).
Assume there exists a real number \(x_{0}\) such that

\[
\left[\frac{5+6 x_{0}}{8}\right]=\frac{15 x_{0}-7}{5}
\]

holds. Then,

\[
\frac{15 x_{0}-7}{5} \leq \frac{5+6 x_{0}}{8}<\frac{15 x_{0}-7}{5}+1
\]

This implies, by multiplying by 4:

\[
12 x_{0}-\frac{28}{5} \leq \frac{5}{2}+3 x_{0}<12 x_{0}-\frac{8}{5} \quad \text { hence } \quad \frac{41}{10}<9 x_{0} \leq \frac{81}{10}
\]

and further, after dividing by 3 and subtracting \(\frac{7}{5}\):

\[
-\frac{1}{30}<\frac{15 x_{0}-7}{5} \leq \frac{13}{10}
\]

Since \(\frac{15 x_{0}-7}{5}\) is an integer, it follows that either \(\frac{15 x_{0}-7}{5}=0\) or \(\frac{15 x_{0}-7}{5}=1\) holds. This yields \(x_{0}=\frac{7}{15}\) or \(x_{0}=\frac{4}{5}\).

By substituting these values into the original equation, one can show that these two values are solutions to the original equation. Therefore, the equation has the two solutions \(x=\frac{7}{15}\) and \(x=\frac{4}{5}\) and no others.

\section*{Adapted from \([5]\)}