Question 155556
{{{1+2/(b-1)=2/(b^2-b)}}}
{{{((b-1)+2)/(b-1)=2/(b^2-b)}}}
{{{(b+1)/(b-1)=2/b*(b-1)}}}
{{{b+1 = 2/b}}}
{{{b^2+b -2=0}}}
{{{(b+2)(b-1) = 0}}}
so b = 1 or b = -2,

But if b= 1, then the term {{{2/(b-1)}}} = 2/0 and it is not defined
Hence, b=-2