College Algebra, Chapter 1, 1.7, Section 1.7, Problem 48

Solve the inequality $\displaystyle \frac{1}{|2x-3|} \leq 5$. Express the answer using interval notation.


$\begin{equation} \begin{aligned} \frac{1}{|2x-3|} &\leq 5\\ \\ \frac{1}{5} &\leq |2x -3| && \text{Divide by 5 and multiply } |2x-3|\text{ on each side} \end{aligned} \end{equation}$

We have,


$\begin{equation} \begin{aligned} 2x -3 &\geq \frac{1}{5} && \text{and}& -(2x-3) &\geq \frac{1}{5} && \text{Divide each side by -1}\\ \\ 2x -3 &\geq \frac{1}{5} && \text{and}& 2x -3 &\leq - \frac{1}{5} && \text{Add 3}\\ \\ 2x &\geq \frac{9}{5} && \text{and} & 2x &\leq \frac{7}{5} && \text{Divide by 2}\\ \\ x &\geq \frac{9}{10} && \text{and} & x &\leq \frac{7}{10} \end{aligned} \end{equation}$


The solution set is $\displaystyle \left(-\infty, \frac{7}{10} \right) \bigcup \left( \frac{9}{10}, \infty \right)$