[tex]\displaystyle Notam~ \frac{3x-1}{5}=k \in \mathbb{Z}~ \Rightarrow x= \frac{5k+1}{3}. \\ \\ Astfel,~ecuatia~devine: \\ \\ \left[ \dfrac{7 \cdot \dfrac{5k+1}{3}+1}{3}\right]=k \Leftrightarrow \left[ \frac{35k+10}{9}\right]=k. \\ \\ Deci~ k \leq\frac{35k+10}{9}\ \textless \ k+1 \Leftrightarrow 0 \leq 26k+10\ \textless \ 9 \Rightarrow \\ \\ \Rightarrow k \in \left[ -\frac{5}{13};- \frac{1} {26}\right).~Dar~k \in \mathbb{Z},~contradictie!~ \\ \\ Prin~urmare~ecuatia~nu~admite ~ solutii~reale.[/tex]
