Ответ:

решение смотри на фотографии

[tex]\displaystyle\bf\\\left \{ {{(x+2)(x-6)\leq (x+2)(x+1)+4} \atop {2(6x-1)\geq 7(2x-4)}} \right. \\\\\\\left \{ {{x^{2} -6x+2x-12\leq x^{2} +x+2x+2+4} \atop {12x-2\geq 14x-28}} \right. \\\\\\\left \{ {{x^{2} -4x-12\leq x^{2} +3x+6} \atop {12x-14x\geq -28+2}} \right. \\\\\\\left \{ {{x^{2} -4x-x^{2} -3x\leq 6+12} \atop {-2x\geq -26}} \right. \\\\\\\left \{ {{-7x\leq 18} \atop {x\leq 13}} \right. \\\\\\\left \{ {{x\geq -2\dfrac{4}{7} } \atop {x\leq 13}} \right.[/tex]

[tex]\displaystyle\bf\\Otvet \ : \ x\in\Big[-2\frac{4}{7} \ ; \ 13\Big][/tex]