Question 107658
{{{(2y-3)/(y^2-1) – (4-y)/(1-y^2)}}}

{{{(2y-3)/(y^2-1) – (4-y)/(1-y^2)}}}….{{{1-y^2)}}} we can write as 
{{{-(-1+y^2)}}} or {{{-(y^2 -1)}}} 
{{{(2y-3)/(y^2-1)}}} – {{{(4-y)/(1-y^2)}}}



{{{(2y-3)/(y^2-1)}}} – {{{(4-y)/(-(y^2-1))}}}

wher ...{{{(y^2-1)}}}... is common denominator

{{{(8y- 2y^2 -12 + 3y)/(y^2-1)}}}...{{{(y^2-1)}}} factors to {{{(y-1)(y+1)}}}

{{{(- 2y^2 + 11y -12 )/(y-1)(y+1)}}}



now we can factor the quadratic {{{(- 2y^2 + 11y -12 )}}}; it factors to  {{{(-2y +3)(y-4)}}}

then we have

{{{(-2y +3)(y-4)/(y-1)(y+1)}}}