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Rezolvati in \( \mathbb{R} \) sistemul \( \begin{cases}
\left[\frac{x+3}{2}\right]=\frac{y-2}{3}\\
\left[\frac{y-1}{2}\right]=\frac{x+2}{2}\end{cases} \).

Din ecuatia a doua \( \frac{x+2}{2}\in\mathbb{Z} \) de unde \( x=2k \) \( k\in\mathbb{Z} \)

Inlocuind in prima \( \left[\frac{2k+3}{2}\right]=\frac{y-2}{3} \) de unde \( y=3k+5 \) si atunci ecuatia a doua devine

\( \left[\frac{3k}{2}\right]=k-1 \) deci

\( k-1\le \frac{3k}{2}< k \)

Asadar \( k\in\{-2,-1\} \) deci solutiile sunt \( (-4,-1);(-2,2) \)