Answer
$\displaystyle \left\{x\ |\ \frac{9}{35} \leq x \lt \frac{2}{7} \right\}$ or $\displaystyle \left[\frac{9}{35}, \ \frac{2}{7} \right)$
Work Step by Step
Comparing reciprocals,
$\displaystyle \frac{1}{2-7x}\geq\frac{5}{1}\quad\Leftrightarrow\quad\frac{2-7x}{1}\leq\frac{1}{5}$
... and, we have the restriction that the denominator
of $\displaystyle \frac{1}{2-7x}$ must be positive (not zero): $\left[\begin{array}{l}
2-7x \gt 0\\
-7x \gt -2\\
x \lt 2/7
\end{array}\right]$
$ 2-7x\displaystyle \leq\frac{1}{5}\qquad$ ... add $-2$
$-7x\displaystyle \leq-\frac{9}{5}\qquad$... multiply with $-\displaystyle \frac{1}{7}$
$ x\displaystyle \geq\frac{9}{35}\qquad$... and, the restriction ... $x \lt \displaystyle \frac{2}{7}=\frac{10}{35}$
$ \displaystyle \frac{9}{35} \leq x \lt \frac{2}{7}$
Solution set: $\displaystyle \left\{x\ |\ \frac{9}{35} \leq x \lt \frac{2}{7} \right\}$ or $\displaystyle \left[\frac{9}{35}, \ \frac{2}{7} \right)$
