Question 1127709

{{{ abs(2y/(y^2 -1))  = 2}}} 

{{{ 2abs(y/(y^2 -1))  = 2}}} 

{{{ abs(y/(y^2 -1))  = 2/2}}} 

{{{ abs(y/(y^2 -1))  = 1}}} 

solutions: since abs value of the {{{ y/(y^2 -1) }}} and  {{{ -(y/(y^2 -1)) }}} is same, we have

{{{ y/(y^2 -1)  = 1}}} or {{{ -y/(y^2 -1)  = 1}}}


1. 
{{{ y/(y^2 -1)  = 1}}}

{{{ y  = y^2 -1}}}

{{{  y^2-y -1=0}}}

{{{(y - 1/2)^2 - 1/4-1 = 0}}}

{{{  (y-1/2)^2-5/4=0}}}

{{{  (y-1/2)^2=5/4}}}

{{{  (y-1/2)=sqrt(5/4)}}}

{{{  y=1/2+ sqrt(5)/2}}} 

{{{  y=(1/2)(1+ sqrt(5))}}} or {{{  y=(1/2)(1- sqrt(5))}}}