Question

Помогите с алгеброй, пожалуйста.
Решите неравенства:

Answer 1

$1)\frac{1}{3}\leq\frac{x}{6}+0,5\leq\frac{3}{4} \\\\\frac{1}{3}*12\leq\frac{x}{6}*12+0,5*12\leq\frac{3}{4}*12\\\\4\leq2x+6\leq9\\\\4-6\leq2x+6-6\leq9-6\\\\-2\leq 2x\leq3\\\\-1\leq x\leq1,5\\\\\boxed{x\in[-1 \ ; \ 1,5]}\\\\\\2)x^{2}\geq -3\\\\\boxed{x\in(-\infty \ ; \ +\infty)}$