Question 1129186

{{{x^2 + y^2= 6}}}........eq.1
{{{xy= 1}}} ........eq.2
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{{{xy= 1}}} ........eq.2=> {{{x=1/y}}}

substitute in eq.1


{{{(1/y)^2 + y^2= 6}}}........eq.1

{{{1/y^2 + y^2= 6}}}...multiply by {{{y^2}}}

{{{1 + y^4= 6y^2}}}

{{{ y^4- 6y^2+1=0}}}...factor

{{{(y^2 - 2 y - 1) (y^2 + 2 y - 1)=0}}}


solutions: there will be four   


if {{{(y^2 - 2 y - 1) =0}}}


{{{y = (-(-2) +- sqrt( (-2)^2-4*1*(-1) ))/(2*1) }}}

{{{y = (2 +- sqrt( 4+4))/2 }}}

{{{y = (2 +- sqrt( 2*4))/2 }}}

{{{y = (2 +- 2sqrt( 2))/2 }}}...simplify

{{{y = (1 +- sqrt( 2)) }}}


so, {{{highlight(y [1]= 1 + sqrt( 2)) }}} or {{{highlight(y[2] = 1 - sqrt( 2)) )}}}



if{{{ (y^2 + 2 y - 1)=0}}}


{{{y = (-2 +- sqrt( 2^2-4*1*(-1) ))/(2*1) }}}

{{{y = (-2 +- sqrt( 4+4 ))/2 }}}

{{{y = (-2 +- 2sqrt(2 ))/2 }}}

{{{y = (-1 +- sqrt(2 )) }}}


{{{highlight(y[3] = -1 + sqrt(2 ) )}}}  or {{{highlight(y[4] = -1 - sqrt(2 ) )}}}




now find {{{x}}} using all 4 solutions for {{{y}}}


if {{{y [1]= 1 + sqrt( 2) }}}, since {{{x=1/y}}} we have 


{{{x= 1/(1 + sqrt( 2))= (1 - sqrt( 2))/((1 + sqrt( 2))(1 - sqrt( 2)))=(1 - sqrt( 2))/(1^2 - (sqrt( 2))^2)=(1 - sqrt( 2))/(1 - 2)=(1 - sqrt( 2))/-1}}}
{{{highlight(x[1]=sqrt(2) - 1) )}}}


if {{{y [2]= 1 - sqrt( 2) }}}

{{{x= 1/(1 - sqrt( 2))= (1 + sqrt( 2))/((1 + sqrt( 2))(1 - sqrt( 2)))=(1 + sqrt( 2))/(1^2 - (sqrt( 2))^2)=(1 +sqrt( 2))/(1 - 2)=(1+sqrt( 2))/-1}}}
{{{highlight(x[2]= - 1-sqrt(2)) }}}


if {{{y[3] = -1 + sqrt(2 )= sqrt(2 )-1}}}

{{{x= 1/(sqrt( 2)-1)=(sqrt( 2)+1)/((sqrt( 2))^2-1^2)=(sqrt( 2)+1)/(2-1)}}}

{{{highlight(x[3]=sqrt( 2)+1)}}}



if {{{y[4] = -1 - sqrt(2 ) }}}

{{{x= 1/(-1 - sqrt(2 ))=(-1 + sqrt(2 ))/(((-1)^2 - (sqrt(2 ))^2))=(-1 + sqrt(2 ))/(1-2)=(-1 + sqrt(2 ))/-1=-(-1 + sqrt(2 ))}}}

{{{highlight(x[4]=1-sqrt(2 ))}}}