# How do you solve  | 5 / (2x-1) | >= | 1 / (x - 2) |?

Oct 18, 2017

See below.

#### Explanation:

For $x \ne 2$ and $x \ne \frac{1}{2}$

$| \frac{5}{2 x - 1} | \ge | \frac{1}{x - 2} | \Rightarrow 5 \left\mid x - 2 \right\mid \ge \left\mid 2 x - 1 \right\mid$ or equivalently

$5 \sqrt{{\left(x - 2\right)}^{2}} - \sqrt{{\left(2 x - 1\right)}^{2}} = {\epsilon}^{2} > 0$ and squaring both sides

${5}^{2} {\left(x - 2\right)}^{2} - 10 \sqrt{{\left(x - 2\right)}^{2} {\left(2 x - 1\right)}^{2}} + {\left(2 x - 1\right)}^{2} = {\epsilon}^{4}$ and again

${\left({5}^{2} {\left(x - 2\right)}^{2} + {\left(2 x - 1\right)}^{2} - {\epsilon}^{4}\right)}^{2} = {10}^{2} {\left(x - 2\right)}^{2} {\left(2 x - 1\right)}^{2}$

or

$\left(11 + {\epsilon}^{2} - 7 x\right) \left(9 + {\epsilon}^{2} - 3 x\right) \left({\epsilon}^{2} + 3 x - 9\right) \left({\epsilon}^{2} + 7 x - 11\right) = 0$

and then the solution sets

$\left\{\begin{matrix}11 - 7 x \le 0 \\ 9 - 3 x \le 0 \\ 3 x - 9 \le 0 \\ 7 x - 11 \le 0\end{matrix}\right.$

giving the solution set $x = \left\{\frac{11}{7} , 3\right\}$