<PRE><B>Question 1072925</B>
{{{ 1/(2x+2) + 5/(x^2-1) = 1/(x-1)}}}
Factor the denominators
{{{ 1/(2(x+1)) + 5/((x-1)(x+1)) = 1/(x-1)}}}
Multiply by the common denominators 2(x+1)(x-1), cancel and we have
(x-1) + 2(5) = 2(x+1)
x - 1 + 10 = 2x + 2
9 - 2 = 2x - x
7 = x
:
:
Check in the original equation, replace x=7
{{{ 1/(14+2) + 5/(49-1) = 1/(7-1)}}}
{{{1/16 + 5/48 = 1/6}}}
{{{3/48 + 5/48 = 8/48}}}</PRE>