Объясните мне по какой системе решается номер 5))

Решите задачу:

$\left \{ {{3(x-1+\frac{4-3x}{4}-(1-2(x-1-\frac{x+2}{5})))\leq5x-7,} \atop {\frac{x^2+8x+7}{x^2+2x+1}-1\leq0;}} \right. \\ 60(x-1+\frac{4-3x}{4}-(1-2(x-1-\frac{x+2}{5})))\leq100x-140, \\ 60x-60+15(4-3x)-60(1-2(x-1-\frac{x+2}{5}))\leq100x-140, \\ 60x-60+60-45x-60+120(x-1-\frac{x+2}{5})\leq100x-140, \\ 15x-60+120x-120-24(x+2)\leq100x-140, \\ 135x-180-24x-48\leq100x-140, \\ 111x-100x\leq228-140, \\ 11x \leq 88, \\ x \leq 8; \\$
$\frac{(x+7)(x+1)-(x+1)^2}{(x+1)^2}\leq0, \\ x+1\neq0, x\neq-1, \\ \frac{(x+1)(x+7-x-1)}{(x+1)^2}\leq0, \\ \frac{6(x+1)}{(x+1)^2}\leq0, \\ \frac{1}{x+1}\leq0, \\ x+1\ \textless \ 0, \\ x\ \textless \ -1; \\ \left \{ {{x\leq8,} \atop {x\ \textless \ -1;}} \right. \\ x\ \textless \ -1, \\ x\in(\infty;-1).$