Question 320155
Use a substitution, let {{{u=y^(-1)}}}.
{{{u^2=y^(-2)}}}
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{{{y^(-2)-3y^(-1)+4=0}}}
{{{u^2-3u+4=0}}}
{{{(u-3)(u-1)=0}}}
Two solutions:
{{{u-3=0}}}
{{{u=3}}}
{{{y^(-1)=3}}}
{{{highlight(y=1/3)}}}
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{{{u-1=0}}}
{{{u=1}}}
{{{y^(-1)=1}}}
{{{highlight(y=1)}}}