Question 963208

{{{xy = -1}}}
{{{x + y = -6}}}
--------------------------solve the system of equations:
 
from
{{{x + y = -6}}}=>{{{x=-6-y}}}

go to {{{xy = -1}}}, substitute {{{x}}}:

{{{(-6-y)y = -1}}}....solve for {{{y}}}

{{{-6y-y^2 = -1}}}

{{{0=y^2+6y -1}}}..............use quadratic formula


{{{y = (-6 +- sqrt( 6^2-4*1*(-1) ))/(2*1) }}}

{{{y = (-6 +- sqrt( 36+4 ))/2 }}}

{{{y = (-6 +- sqrt(40 ))/2 }}}

{{{y = (-6 +- sqrt(4*10 ))/2 }}}

{{{y = (-cross(6)3 +- cross(2)sqrt(10 ))/cross(2) }}}

{{{y = (-3 +- sqrt(10 )) }}}

solutions:

{{{y = -3 + sqrt(10 ) }}} or {{{y = -3 - sqrt(10 ) }}}

go back to {{{x=-6-y}}}, plug in {{{y}}}

{{{x=-6-(-3 + sqrt(10 ))}}}

{{{x=-6+3 - sqrt(10 )}}}

{{{x=-3- sqrt(10 )}}}

or

{{{x=-6-(-3 - sqrt(10 ))}}}

{{{x=-6+3 +sqrt(10 )}}}

{{{x=-3+sqrt(10 )}}}

so, solution sets are:

{{{x=-3- sqrt(10 )}}} and {{{y = -3 + sqrt(10 ) }}} 

or

{{{x=-3+sqrt(10 )}}} and {{{y = -3 - sqrt(10 ) }}}