Question 178060
Assume you mean
{{{2/(y^2-1)}}} + 3 + {{{1/(y+1)}}}
:
(y^2-1) is the difference of squares, factor
{{{2/((y-1)(y+1))}}} + 3 + {{{1/(y+1)}}}
:
common denominator would be (y-1)(y+1)
{{{(2 + 3(y-1)(y+1) + (y-1))/((y+1)(y-1))}}}
:
{{{(2 + 3(y^2-1) + (y-1))/((y+1)(y-1))}}}
:
{{{(2 + 3y^2 - 3 + y - 1)/((y+1)(y-1))}}}
:
combine like terms
{{{(3y^2 + y - 2)/((y+1)(y-1))}}}
:
numerator will factor:
{{{((3y-2)(y+1))/((y+1)(y-1))}}}
:
cancel (y+1)'s
{{{((3y-2))/((y-1))}}}