Question 681944
{{{x+y=1}}}...............1

{{{x^2+xy-y^2=-1}}}........2
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{{{x+y=1}}}...............1...solve for {{{x}}}

{{{x=1-y}}}.......substitute in 2
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{{{x^2+xy-y^2=-1}}}........2

{{{(1-y)^2+(1-y)y-y^2=-1}}}.........solve for {{{y}}}

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

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

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

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

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

{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

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

{{{y = (-1 +- sqrt( 1+8 ))/2 }}} 

{{{y = (-1 +- sqrt(9))/2 }}} 

{{{y = (-1 +- 3)/2 }}} 

solutions:

{{{y = (-1 +3)/2 }}} 

{{{y = 2/2 }}} 

{{{y = 1 }}}

and 

{{{y = (-1 -3)/2 }}} 

{{{y = -4/2 }}} 

{{{y = -2 }}}


now find {{{x}}}


if {{{y = 1 }}}:

{{{x=1-y}}}

{{{x=1-1}}}

{{{x=0}}}


if {{{y = -2 }}}:

{{{x=1-(-2)}}}

{{{x=1+2}}}

{{{x=3}}}


so, solution pairs are:

1. {{{y = 1 }}} and {{{x=0}}}

2. {{{y = -2 }}} and {{{x=3}}}