20 б. Помогите прошу!!! Решите систему!!! Лучше подробно всё!!!

$\left \{ {{ \frac{x-5}{6} \leq \frac{3x-1}{4}} \atop { \frac{x+2}{3}\ \textgreater \ \frac{x+3}{5}}} \right. \\ \left \{ {{ \frac{2(x-5)-3(3x-1)}{12} } \leq 0 \atop { \frac{5(x+2)-3(x+3)}{15}}\ \textgreater \ 0} \right. \\ \left \{ {{ \frac{2x-10-9x+3}{12}} \leq 0 \atop { \frac{5x+10-3x-9}{15}}\ \textgreater \ 0} \right. \\ \left \{ {{ \frac{-7x-7}{12} \leq 0 } \atop { \frac{2x+1}{15} }\ \textgreater \ 0} \right. \\ \left \{ {{-7x-7 \leq 0} \atop {2x+1\ \textgreater \ 0}} \right. \\ \left \{ {{-7x \leq 7} \atop {2x\ \textgreater \ -1}} \right. \\$
$\left \{ {{x \geq -1} \atop {x\ \textgreater \ -0,5}} \right. \\$
x∈(-0,5;+∞).
Ответ: (-0,5;+∞).