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

Jul 16, 2018

$x < \frac{1}{2}$ or $\frac{1}{2} < x \le \frac{11}{7}$ or $x \ge 3$

#### Explanation:

At first observe that $x \ne 2$ and $x \ne \frac{1}{2}$ by crossmultiplication we get

$5 | x - 2 | \ge | 2 x - 1 |$ no we distinguish several cases:

a) $x > 2$ then we get

$5 x - 10 \ge 2 x - 1$ and we get $x \ge 3$

b)$\frac{1}{2} < x < 2$ and we get

$10 - 5 x \ge 2 x - 1$ and we get $x < \frac{11}{7}$

$\frac{1}{2} < x \le \frac{11}{7}$

in our last case we have

c)$x < \frac{1}{2}$

and we get

$10 - 5 x \ge 1 - 2 x$

so $x < \frac{1}{2}$ and we get the interval like above.

Jul 17, 2018

The solution is $x \in \left(- \infty , \frac{1}{2}\right) \cup \left(\frac{1}{2} , \frac{11}{7}\right] \cup \left[3 , + \infty\right)$

#### Explanation:

You cannot cross multiply when you have an inequality

The inequality is

$| \frac{5}{2 x - 1} | \ge | \frac{1}{x - 2} |$

$\iff$, $\frac{5}{|} 2 x - 1 | \ge \frac{1}{|} \left(x - 2\right) |$

$\iff$, $\frac{5}{|} 2 x - 1 | - \frac{1}{|} \left(x - 2\right) | \ge 0$

There are $2$ points to consider

$2 x - 1 = 0$, $\implies$, $x = \frac{1}{2}$ and

$x - 2 = 0$, $\implies$, $x = 2$

There are $3$ intervals to consider

${I}_{1} = \left(- \infty , \frac{1}{2}\right)$ and ${I}_{2} = \left(\frac{1}{2} , 2\right)$ and ${I}_{3} = \left(2 , + \infty\right)$

In the first interval ${I}_{1}$, we have

$\frac{5}{- 2 x + 1} - \frac{1}{- x + 2} \ge 0$

$\frac{5 \left(2 - x\right) - 1 \left(1 - 2 x\right)}{\left(1 - 2 x\right) \left(2 - x\right)} \ge 0$

$\frac{10 - 5 x - 1 + 2 x}{\left(1 - 2 x\right) \left(2 - x\right)} \ge 0$

$\frac{9 - 3 x}{\left(1 - 2 x\right) \left(2 - x\right)} \ge 0$

$\frac{3 \left(3 - x\right)}{\left(1 - 2 x\right) \left(2 - x\right)} \ge 0$

Let $f \left(x\right) = \frac{3 \left(3 - x\right)}{\left(1 - 2 x\right) \left(2 - x\right)}$

Solving this equation with a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$\frac{1}{2}$$\textcolor{w h i t e}{a a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$1 - 2 x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$2 - x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(white)(aaaa)+$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$3 - x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(white)(aaaa)+$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$

Therefore,

In the interval ${I}_{1}$, $f \left(x\right) \ge 0$,when $x \in \left(- \infty , \frac{1}{2}\right)$

In the second interval ${I}_{2} = \left(\frac{1}{2} , 2\right)$

$\frac{5}{2 x - 1} - \frac{1}{- x + 2} \ge 0$

$\iff$, $\frac{5 \left(2 - x\right) - 1 \left(2 x - 1\right)}{\left(2 x - 1\right) \left(2 - x\right)} \ge 0$

$\iff$, $\frac{10 - 5 x - 2 x + 1}{\left(2 x - 1\right) \left(2 - x\right)} \ge 0$

$\iff$, $\frac{11 - 7 x}{\left(2 x - 1\right) \left(2 - x\right)} \ge 0$

Let $g \left(x\right) = \frac{11 - 7 x}{\left(2 x - 1\right) \left(2 - x\right)}$

Solving this inequality with a sign chart,

$g \left(x\right) \ge 0$ when $x \in \left(\frac{1}{2} , \frac{11}{7}\right]$

In the third interval ${I}_{2} = \left(2 , + \infty\right)$

$\frac{5}{2 x - 1} - \frac{1}{x - 2} \ge 0$

$\iff$, $\frac{5 \left(x - 2\right) - 1 \left(2 x - 1\right)}{\left(2 x - 1\right) \left(x - 2\right)} \ge 0$

$\iff$, $\frac{\left(5 x - 10 - 2 x + 1\right)}{\left(2 x - 1\right) \left(x - 2\right)} \ge 0$

$\iff$, $\frac{\left(3 x - 9\right)}{\left(2 x - 1\right) \left(x - 2\right)} \ge 0$

$\iff$, $\frac{3 \left(x - 3\right)}{\left(2 x - 1\right) \left(x - 2\right)} \ge 0$

$h \left(x\right) = \frac{3 \left(x - 3\right)}{\left(2 x - 1\right) \left(x - 2\right)}$

Solving this inequality with a sign chart,

$h \left(x\right) \ge 0$ when $x \in \left[3 , + \infty\right)$

graph{5/(|2x-1|)-1/(|x-2|) [-14.24, 14.24, -7.12, 7.12]}