\frac{4x+9}{2}-\frac{x-1}{3} } \atop {\frac{5x-2}{3}\leq \frac{2x+13}{2}-\frac{x+2}{3}}} \right. \left \{ {{16x+2>12x+27-2x+2} \atop {10x-4\leq 6x+39-2x-4}} \right. \left \{ {{6x>27} \atop {6x\leq 39}} \right. \left \{ {{x>4,5} \atop {x\leq 6,5}} \right. \\\\\left \{ {{-(4,5)+++++++++>x} \atop {++++++++[6,5]-->x}} \right. x\in (4,5;6,5]" alt="\sf \displaystyle \left \{ {{\frac{8x+1}{3}>\frac{4x+9}{2}-\frac{x-1}{3} } \atop {\frac{5x-2}{3}\leq \frac{2x+13}{2}-\frac{x+2}{3}}} \right. \left \{ {{16x+2>12x+27-2x+2} \atop {10x-4\leq 6x+39-2x-4}} \right. \left \{ {{6x>27} \atop {6x\leq 39}} \right. \left \{ {{x>4,5} \atop {x\leq 6,5}} \right. \\\\\left \{ {{-(4,5)+++++++++>x} \atop {++++++++[6,5]-->x}} \right. x\in (4,5;6,5]" align="absmiddle" class="latex-formula">
ответ: ![(4,5;6,5]](https://tex.z-dn.net/?f=%284%2C5%3B6%2C5%5D "(4,5;6,5]")