Question 1134167
{{{x^2+y^2=4}}}....eq.1
{{{x+y=1}}}.......eq.2
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{{{x+y=1}}}.......eq.2...solve for {{{y}}}

{{{y=1-x}}}.......substitute in eq.1


{{{x^2+(1-x)^2=4}}}....eq.1...solve for {{{x}}}

{{{x^2+1-2x+x^2=4}}}

{{{2x^2-2x +1-4=0}}}

{{{2x^2-2x -3=0}}}....use quadratic formula


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


{{{x = (-(-2) +- sqrt( (-2)^2-4*2*(-3) ))/(2*2) }}} 

{{{x = (2 +- sqrt( 4+24 ))/4 }}} 

{{{x = (2 +- sqrt( 7*4 ))/4 }}} 

{{{x = (2 +- 2sqrt( 7 ))/4 }}} 

{{{x = (1 +- sqrt( 7 ))/2 }}}

solutions:

{{{x = (1 + sqrt( 7 ))/2 }}}

{{{x = (1 - sqrt( 7 ))/2 }}}


go to

{{{y=1-x}}}.......eq.2...plug in {{{x = (1 + sqrt( 7 ))/2 }}}


{{{y=1-(1 + sqrt( 7 ))/2}}}


{{{y=2/2-1/2 + sqrt( 7 )/2}}}

{{{y=1/2 + sqrt( 7 )/2}}}

{{{y=(1 + sqrt( 7 ))/2}}}



{{{y=1-x}}}.......eq.2...plug in {{{x = (1 - sqrt( 7 ))/2 }}}


{{{y=1-(1 - sqrt( 7 ))/2}}}


{{{y=2/2-1/2 - sqrt( 7 )/2}}}

{{{y=1/2 - sqrt( 7 )/2}}}

{{{y=(1 - sqrt( 7 ))/2}}}


ordered pairs are:

({{{ 1/2 - sqrt(7)/2}}}, {{{ 1/2 + sqrt(7)/2}}}) ≈({{{ -0.8}}}, {{{ 1.8}}})

({{{ 1/2 +sqrt(7)/2}}}, {{{ 1/2 - sqrt(7)/2}}})  ≈ ({{{  1.8}}}, {{{ -0.8}}})



{{{drawing ( 600, 600, -10, 10, -10, 10,

circle(1/2 - sqrt(7)/2,1/2 + sqrt(7)/2,.12),
circle(1/2 + sqrt(7)/2,1/2 - sqrt(7)/2,.12),

graph( 600, 600, -10, 10, -10, 10, 1-x, sqrt(4-x^2),-sqrt(4-x^2))) }}}