Question

СРОЧНО ПОЖАЛУЙСТА ПОМОГИТЕ!!!!ДАЮ 35 БАЛЛОВ!!!ОЧЕНЬ НУЖНО

Answer 1

$\left\{\begin{array}{ccc}\dfrac{2x+1}{7}-\dfrac{3x+1}{5}\leq-1 \\\dfrac{1-3x}{2}-\dfrac{x+6}{3}\leq4\end{array}\right \\\\\\\left\{\begin{array}{ccc}\dfrac{2x+1}{7}\cdot35-\dfrac{3x+1}{5}\cdot35\leq-1\cdot35 \\\dfrac{1-3x}{2}\cdot6-\dfrac{x+6}{3}\cdot6\leq4\cdot6\end{array}\right \\\\\\\left\{\begin{array}{ccc}5\cdot(2x+1)-7\cdot(3x+1)\leq-35 \\3(1-3x)-2(x+6)\leq24 \end{array}\right \\\\\\\left\{\begin{array}{ccc}10x+5-21x-7\leq-35 \\3-9x-2x-12\leq24 \end{array}\right$

$\left\{\begin{array}{ccc}-11x-2\leq-35 \\-11x-9\leq24 \end{array}\right\\\\\\\left\{\begin{array}{ccc}-11x\leq-33 \\-11x\leq33 \end{array}\right \\\\\\\left\{\begin{array}{ccc}x\geq3 \\x\geq-3 \end{array}\right \ \Rightarrow \ x\geq3\\\\\\Otvet:\boxed{x\in[3 \ ; \ +\infty)}$

Answer 2

Ответ:

$[3;+\infty)$

Пошаговое объяснение:

$\left \{ {{\frac{2x+1}{7}-\frac{3x+1}{5}\leq -1}|*35 \atop {\frac{1-3x}{2}-\frac{x+6}{3}\leq 4}|*6} \right.=>\left \{ {{5(2x+1)-7(3x+1)\leq -35} \atop {3(1-3x)-2(x+6)\leq 24}} \right.=>\\\\\\=>\left \{ {{10x+5-21x-7\leq -35} \atop {3-9x-2x-12\leq 24}} \right.=>\left \{ {{-11x-2\leq -35} \atop {-11x-9\leq 24}} \right.=>\left \{ {{-11x\leq -33} \atop {-11x\leq 33}} \right.=>\\\\\\=>\left \{ {{x\geq 3} \atop {x\geq -3}} \right.=>x\geq 3\\\\\\x\in[3;+\infty)$