[tex][ \frac{x+1}{3}]=\frac{x-1}{2}[/tex]
[tex] [ \frac{x+1}{3}] \in \mathbb{Z} \Rightarrow \frac{x-1}{2} \in \mathbb{Z}[/tex]
[tex] [ \frac{x+1}{3}] \leq \frac{x+1}{3} \leq [\frac{x+1}{3}]+1\\ \\ \frac{x-1}{2} \leq \frac{x+1}{3} \leq \frac{x-1}{2}+1[/tex]
[tex]\frac{x-1}{2} \leq \frac{x+1}{3} \leq \frac{x-1+2}{2} \ | \ \cdot 6\\ \\ 3(x-1) \leq 2(x+1) \leq 3(x+1)\\ \\ 3x-3 \leq 2x+2 \leq 3x+3 \ | \ -3x\\ \\ -3 \leq -x+2 \leq 3 \ | \ -2 \\ \\ -5 \leq -x \leq 1 \ | \cdot (-1)\\ \\ 5 \geq x \geq -1[/tex]
[tex]Acum \ vom \ aduce \ acel \ x \ simplut \ la \ \frac{x-1}{2}\\ \\ 5 \geq x \geq -1 \ | \ -1\\ \\ 4 \geq x-1 \geq -2 \ | \ :2\\ \\ 2 \geq \frac{x-1}{2} \geq -1\\ \\ Cum \ \frac{x-1}{2} \in \mathbb{Z} \Rightarrow \frac{x-1}{2} \in \{-1, \ 0, \ 1, \ 2 \}[/tex]
Rezolvi ecuatiile de mai jos si scoti solutiile:
(x-1)/2 = -1
(x-1)/2 = 0
(x-1)/2 = 1
(x-1)/2 = 2
