Question 1160786
 {{{3/(4x+3) <  2/3  < 2/(x+5)}}}
 

if {{{a < u< b}}} then {{{a<u}}} and{{{ u < b}}}


 {{{3/(4x+3) <  2/3}}} and {{{2/3  < 2/(x+5)}}}


first solve


{{{3/(4x+3) <  2/3}}}.........subtract {{{2/3}}} from both sides

{{{3/(4x+3) -2/3<  2/3-2/3}}}

{{{3/(4x+3) -2/3< 0}}}......common denominator is {{{3(4x+3)}}}

{{{9/3(4x+3) -2(4x+3)/3(4x+3)< 0}}}

{{{(9 -2(4x+3))/3(4x+3)< 0}}}

{{{(9 -8x-6)/3(4x+3)< 0}}}

{{{(3 -8x)/3(4x+3)< 0}}} ..........find solutions of both  numerator and denominator

{{{(3 -8x)< 0}}}

{{{highlight(x>3/8)}}}

or

{{{3(4x+3)< 0}}}
{{{(4x+3)< 0}}}
{{{4x< -3}}}
{{{highlight(x< -3/4)}}}



and, now solve  {{{2/3  < 2/(x+5)}}}

 {{{0  < 2/(x+5)-2/3}}}

{{{0  < 6/3(x+5)-2(x+5)/3(x+5)}}}

{{{0  < (6-2(x+5))/3(x+5)}}}
{{{0  < (6-2x-10)/3(x+5)}}}

{{{0  < (-2x-4)/3(x+5)}}}

solutions:

{{{0  < -2x-4}}}
{{{2x  <-4}}}
{{{highlight(x  <-2)}}}

or

{{{0  <3(x+5)}}}
{{{0  <x+5}}}
{{{-5 <x}}}
{{{highlight(x>-5)}}}



so,  the intervals that satisfy the required condition  < {{{0}}} are:

{{{highlight(x>3/8)}}} or {{{highlight(x< -3/4)}}}

and 

{{{highlight(-5<x<-2)}}}

Merge overlapping intervals which is your solution:  

{{{highlight(red(-5<x<-2))}}}