Question 859792
Let {{{A=x/y}}}
{{{A+1/A=17/4}}}
{{{A^2+1=(17/4)A}}}
{{{4A2+4=17A}}}
{{{4A^2-17A+4=0}}}
{{{(4A-4)(A-1)=0}}}
Two solutions:
{{{4A-1=0}}}
{{{4A=1}}}
{{{A=1/4}}}
{{{x/y=1/4}}}
and
{{{A-1=0}}}
{{{A=1}}}
{{{x/y=1}}}