Solve the following system of inequalities: $\frac{2x+1}{7x-1} > 5$ and $\frac{x+7}{x-8} > 2$.

(NONE) For the first inequality: $\frac{2x+1}{7x-1} > 5$
$\Rightarrow \frac{2x+1}{7x-1} - 5 > 0$
$\Rightarrow \frac{2x+1 - 35x + 5}{7x-1} > 0$
$\Rightarrow \frac{-33x + 6}{7x-1} > 0$
$\Rightarrow \frac{33x - 6}{7x-1} < 0$
$\Rightarrow \frac{x - 2/11}{x - 1/7} < 0$
Thus,$x \in (1/7, 2/11)$.
For the second inequality: $\frac{x+7}{x-8} > 2$
$\Rightarrow \frac{x+7}{x-8} - 2 > 0$
$\Rightarrow \frac{x+7 - 2x + 16}{x-8} > 0$
$\Rightarrow \frac{-x + 23}{x-8} > 0$
$\Rightarrow \frac{x - 23}{x-8} < 0$
Thus,$x \in (8, 23)$.
The intersection of the intervals $(1/7, 2/11)$ and $(8, 23)$ is empty.
Therefore,there is no solution for the given system of inequalities.