Question 350047
Substitute, let {{{u=(x-1)/x}}}
Then,
{{{u^2-u-2=0}}}
{{{(u-2)(u+1)=0}}}
Two solutions:
{{{u-2=0}}}
{{{u=2}}}
{{{(x-1)/x=2}}}
{{{x-1=2x}}}
{{{highlight(x=-1)}}}
.
.
.
{{{u+1=0}}}
{{{u=-1}}}
{{{(x-1)/x=-1}}}
{{{x-1=-x}}}
{{{2x=1}}}
{{{highlight(x=1/2)}}}