Решите неравенство: 3) 1/4-у/3 больше или равно 1/3 - у 6) х/6+1/2>х-1/3

Решите задачу:

$\frac{1}{4} - \frac{y}{3} \geq \frac{1}{3} -y \\ y- \frac{y}{3} \geq \frac{1}{3} -\frac{1}{4} \\ \frac{2y}{3} \geq \frac{4}{12} -\frac{3}{12} \\ \frac{2y}{3} \geq \frac{1}{12} \\ y \geq \frac{1}{12}: \frac{2}{3} \\ y \geq \frac{1}{12} * \frac{3}{2} \\ y \geq \frac{1}{8}$

$\frac{x}{6}+ \frac{1}{2}\ \textgreater \ x- \frac{1}{3} \\ x- \frac{x}{6}\ \textless \ \frac{1}{3}+ \frac{1}{2} \\ \frac{5x}{6} \ \textless \ \frac{5}{6} \\ x\ \textless \ 1$