Solve the inequality problem that involves modulus

[chwala]

Homework Statement

See attached

Relevant Equations

modulus

This is the question * consider the highlighted question only *with its solution shown (from textbook);

My approach is as follows (alternative method),
##|\frac {5}{2x-3}| ##< ## 1##
Let, ##\frac {5}{2x-3} ##⋅##\frac {5}{2x-3}##=##1##
→##x^2-3x-4=0##
##(x+1)(x-4)=0##
it follows that the critical values are ##x_1=-1## and ##x_2=4##, which will give the correct solution once we confirm/establish the valid region in the given inequality.
Is their another method on this type of questions? ... regards

 

[anuttarasammyak]

[tex]5<|2x-3|[/tex]
[tex]5/2<|x-3/2|[/tex]
[tex]x-3/2<-5/2, \ 5/2<x-3/2[/tex]
[tex]x<-1, \ 4<x[/tex]

 

[chwala]

[tex]5<|2x-3|[/tex]
[tex]5/2<|x-3/2|[/tex]
[tex]x-3/2<-5/2, \ 5/2<x-3/2[/tex]
[tex]x<-1, \ 4<x[/tex]

Nice one, I will write this in my notebook ...good way to simplify ...##ax...##

 

[Mark44]

##5<|2x-3|##
##5/2<|x-3/2|##
##x-3/2<-5/2, \ 5/2<x-3/2##
##x<-1, \ 4<x##

The use of commas in the last two lines implies "and", which can't happen here, as it's impossible for x to be both less than -1 and larger than 4.
Here's how I would write this:
##|2x - 3| > 5##
##2x - 3 > 5 \text{ or } 2x - 3 < -5##
##2x > 8 \text{ or } 2x < -2##
##x > 4 \text{ or } x < -1##

 

[chwala]

Thanks Mark...a learning point for me too...

The use of commas in the last two lines implies "and", which can't happen here, as it's impossible for x to be both less than -1 and larger than 4.
Here's how I would write this:
##|2x - 3| > 5##
##2x - 3 > 5 \text{ or } 2x - 3 < -5##
##2x > 8 \text{ or } 2x < -2##
##x > 4 \text{ or } x < -1##