Question

ПОМОГИТЕ СРОЧНО!!!
з роз'ясненням
Даю 40 балов​

Answer 1

$\displaystyle\bf\\\left \{ {{\dfrac{x}{3} > \dfrac{x+1}{4} } \atop {3(x+1)+5\geq 4(x-3)+15}} \right. \\\\\\\left \{ {{\dfrac{x}{3} \cdot 12 > \dfrac{x+1}{4} } \cdot 12\atop {3x+3+5\geq 4x-12+15}} \right. \\\\\\\left \{ {{4x > 3x+3} \atop {3x+8\geq 4x+3}} \right. \\\\\\\left \{ {{4x-3x > 3} \atop {3x-4x\geq 3-8}} \right.\\\\\\\left \{ {{x > 3} \atop {-x\geq -5}} \right.\\\\\\\left \{ {{x > 3} \atop {x\leq 5}} \right. \\\\\\Otvet \ : \ x\in\Big(3 \ ; \ 5\Big]$

Целые решения системы неравенств  :  4  и  5