[tex] 2\cdot{\frac{{(3x+4)}}{2x+5} \in N=\ \textgreater \ \frac{6x+8}{2x+5} \in N=\ \textgreater \ \\
\frac{3(2x+5)-7}{2x+5}\in N=\ \textgreater \ 3-\frac{7}{2x+5}\in N [/tex]
[tex]\frac{7}{2x+5}\in N =\ \textgreater \ (2x+5)\in\{\pm1,\pm7\}\\
2x+5=1=\ \textgreater \ x=-2\\
2x+5=-1=\ \textgreater \ x=-3\\
2x+1=7=\ \textgreater \ x=3\\
2x+1=-7=\ \textgreater \ x=-4\\
Din\ cele\ 4\ numere\ doar \ numerele\ -3,3 si -4 \ verifica\ datele\ problemei.[/tex]
