# How do you solve 3 / |2x-1| >=4 ?

Feb 14, 2017

1/2 < x" le 7/8

#### Explanation:

$0 < | 2 x - 1 | \setminus \le \frac{3}{4}$

Test for its positive and negative ranges

$2 x - 1 \setminus \ge \setminus 0$

Add one to both the sides = $2 x - 1 + 1 \le 0 + 1$

= $2 x \setminus \ge 1$

=$\frac{\text{2x}}{2} \setminus \ge \frac{1}{2}$

= $x \setminus \ge \frac{1}{2}$

2x-1\ge \0.......... "for"x\ge \frac{1}{2},\therefore"for"\:x\ge \frac{1}{2}\quad \le|2x-1|="2x-1"

$2 x - 1 < 0$

$2 x - 1 + 1 < 0 + 1$

$2 x < 1$

Divide both sides by 2
$\frac{\text{2x}}{2} < 1$
$x < \frac{1}{2}$

$2 x - 1 < 0 \text{for" x <1/2 therefore "for} x < \frac{1}{2} | 2 x - 1 | = - \left(2 x - 1\right)$

Evaluate the expression in the following ranges:

$x < \frac{1}{2} , x \setminus \ge \frac{1}{2}$

For x < 1/2

Replace $| 2 x - 1 | \text{with} - \left(2 x - 1\right)$

$\frac{3}{-} \left(2 x - 1\right) \setminus \ge 4 : \frac{1}{8} / \le x < \frac{1}{2}$

= 3/-(2x-1) \ge 4

Simplify 3/-(2x-1)

(apply the fraction rule $\frac{a}{-} b = - \frac{a}{b} = - \frac{3}{2 x - 1}$

Remove parentheses (a) = a

= $- \frac{3}{\text{2x-1}}$
= $- \frac{3}{\text{2x-1}} \setminus \ge 4$

Subtract 4 from both the side

$- \frac{3}{\text{2x-1}} - 4 \setminus \ge 4 - 4$

Refine
$- \frac{3}{\text{2x-1}} - 4 \setminus \ge 0$

calculate $- \frac{3}{\text{2x-1}} - \frac{4}{1}$

Find the LCD = $1 \cdot \left(2 x - 1\right) = 2 x - 1$

= $- \frac{3}{\text{2x-1" - "4(2x-1)"/"2x-1}}$

= $\text{-3-4(2x-1)"/"2x-1}$
= $\text{-8x+7"/"2x-1} \setminus \ge 0$

Compute the signs of 7- 8x
7-8x is 0 if x = 7/8
therefore 7-8x is negative if x > 7/8
7-8x is positive for x < 7/8

Compute the signs of 2x-1
2x-1 is 0 for x = 0.5
2x-1 = negative for x < 0.5
2x-1 is positive for x > 0.5

$\text{7-8x = p , if x <1/2 , p = +}$
$\text{7-8x = p , if x = 1/2, p = +}$
$\text{7-8x = p , if 1/2 < x < 7/8 , p= +}$
$\text{7-8x = p , if x = 7/8 , p = 0}$
$\text{7-8x = p , if x > 7/8 ,p = -}$

$\text{2x-1 = c ,if p , if x <1/2 , c = -}$
$\text{2x-1 = c ,if p , if x = 1/2 , c = 0}$
$\text{2x-1 = c ,if p , if 7/8>x >1/2 , c = +}$
$\text{2x-1 = c ,if p , if x = 7/8 , c = +}$
$\text{2x-1 = c ,if p , if x > 7/8 , c = +}$

$\text{-8x + 7/2x-1 = f , if x <1/2 , f = -}$
$\text{-8x + 7/2x-1 = f , if x = 1/2 , f = undefined}$
$\text{-8x + 7/2x-1 = f , if x = 7/8>x >1/2, f = +}$
$\text{-8x + 7/2x - 1 =f , if x = 7/8 , f = 0 }$
$\text{-8x + 7/2x - 1 =f , if x = 7/8 , f = -}$

So the ranges that satisfy the required condition /le 0 = 1/2 < x" le 7/8