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

1 Answer
Feb 14, 2017

Answer:

#1/2 < x" le 7/8#

Explanation:

#0< |2x-1|\le 3/4#

Test for its positive and negative ranges

#2x-1\ge \0#

Add one to both the sides = #2x -1+1 le0+1 #

= #2x \ge 1#

=#"2x"/2 \ge 1/2#

= #x \ge 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"#

#2x-1 < 0#

add 1 to both sides

#2x-1+1<0+1#

#2x<1#

Divide both sides by 2
#"2x"/2<1#
#x<1/2#

#2x -1<0 "for" x <1/2 therefore "for" x < 1 /2 |2x-1| = -(2x - 1)#

Evaluate the expression in the following ranges:

#x <1/2, x\ge1/2#

For x < 1/2

Replace #|2x-1| "with" - ( 2x -1) #

#3/-(2x-1)\ge4 : 1/8 /le x < 1/2#

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

Simplify 3/-(2x-1)

(apply the fraction rule #a/-b = -a/b = -3/(2x-1)#

Remove parentheses (a) = a

= #-3/"2x-1"#
= #-3/"2x-1" \ge 4#

Subtract 4 from both the side

#-3/"2x-1" -4\ge 4-4#

Refine
#-3/"2x-1" -4 \ge 0#

calculate #-3/"2x-1" -4/1#

Find the LCD = #1 * (2x-1) = 2x - 1#

= #-3/"2x-1" - "4(2x-1)"/"2x-1"#

= #"-3-4(2x-1)"/"2x-1"#
= #"-8x+7"/"2x-1"\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

#"7-8x = p , if x <1/2 , p = +"#
#"7-8x = p , if x = 1/2, p = +"#
#"7-8x = p , if 1/2 < x < 7/8 , p= +"#
#"7-8x = p , if x = 7/8 , p = 0"#
#"7-8x = p , if x > 7/8 ,p = -"#

#"2x-1 = c ,if p , if x <1/2 , c = -"#
#"2x-1 = c ,if p , if x = 1/2 , c = 0"#
#"2x-1 = c ,if p , if 7/8>x >1/2 , c = +"#
#"2x-1 = c ,if p , if x = 7/8 , c = +"#
#"2x-1 = c ,if p , if x > 7/8 , c = +"#

#"-8x + 7/2x-1 = f , if x <1/2 , f = -"#
#"-8x + 7/2x-1 = f , if x = 1/2 , f = undefined"#
#"-8x + 7/2x-1 = f , if x = 7/8>x >1/2, f = +"#
#"-8x + 7/2x - 1 =f , if x = 7/8 , f = 0 "#
#"-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#