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

2 Answers
Jul 16, 2018

Answer:

#x<1/2# or #1/2< x <=11/7# or #x >=3#

Explanation:

At first observe that #xne 2# and #xne 1/2# by crossmultiplication we get

#5|x-2|>=|2x-1|# no we distinguish several cases:

a) #x>2# then we get

#5x-10>=2x-1# and we get #x>=3#

b)#1/2<x<2# and we get

#10-5x>=2x-1# and we get #x<11/7#

#1/2<x<=11/7#

in our last case we have

c)#x<1/2#

and we get

#10-5x>=1-2x#

so #x<1/2# and we get the interval like above.

Jul 17, 2018

Answer:

The solution is #x in (-oo,1/2) uu(1/2,11/7]uu[3,+oo)#

Explanation:

You cannot cross multiply when you have an inequality

The inequality is

#|5/(2x-1)|>=|1/(x-2)|#

#<=>#, #5/|2x-1|>=1/|(x-2)|#

#<=>#, #5/|2x-1|-1/|(x-2)|>=0#

There are #2# points to consider

#2x-1=0#, #=>#, #x=1/2# and

#x-2=0#, #=>#, #x=2#

There are #3# intervals to consider

#I_1=(-oo, 1/2)# and #I_2=(1/2, 2)# and #I_3=(2,+oo)#

In the first interval #I_1#, we have

#5/(-2x+1)-1/(-x+2)>=0#

#(5(2-x)-1(1-2x))/((1-2x)(2-x))>=0#

#(10-5x-1+2x)/((1-2x)(2-x))>=0#

#(9-3x)/((1-2x)(2-x))>=0#

#(3(3-x))/((1-2x)(2-x))>=0#

Let #f(x)=(3(3-x))/((1-2x)(2-x))#

Solving this equation with a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##1/2##color(white)(aaaaaa)##2##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##1-2x##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaa)##-##color(white)(aaa)##-##color(white)(aaa)##-#

#color(white)(aaaa)##2-x##color(white)(aaaa)##+##color(white)(aaaa)####color(white)(aaaa)##+##color(white)(aa)##||##color(white)(a)##-##color(white)(aaa)##-#

#color(white)(aaaa)##3-x##color(white)(aaaa)##+##color(white)(aaaa)####color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(a)##0##color(white)(a)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(a)##||##color(white)(a)##+##color(white)(a)##0##color(white)(a)##-#

Therefore,

In the interval #I_1#, #f(x)>=0#,when #x in (-oo,1/2)#

In the second interval #I_2=(1/2,2)#

#5/(2x-1)-1/(-x+2)>=0#

#<=>#, #(5(2-x)-1(2x-1))/((2x-1)(2-x))>=0#

#<=>#, #(10-5x-2x+1)/((2x-1)(2-x))>=0#

#<=>#, #(11-7x)/((2x-1)(2-x))>=0#

Let #g(x)=(11-7x)/((2x-1)(2-x))#

Solving this inequality with a sign chart,

#g(x)>=0# when #x in (1/2,11/7]#

In the third interval #I_2=(2, +oo)#

#5/(2x-1)-1/(x-2)>=0#

#<=>#, #(5(x-2)-1(2x-1))/((2x-1)(x-2))>=0#

#<=>#, #((5x-10-2x+1))/((2x-1)(x-2))>=0#

#<=>#, #((3x-9))/((2x-1)(x-2))>=0#

#<=>#, #(3(x-3))/((2x-1)(x-2))>=0#

#h(x)=(3(x-3))/((2x-1)(x-2))#

Solving this inequality with a sign chart,

#h(x)>=0# when #x in [3,+oo)#

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