Question 1182047

1. 
{{{x + y = 3}}}....eq.1
{{{x^2 + y^2 = 2}}}.....eq.2
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{{{x + y = 3}}}....eq.1, solve for {{{x}}}

{{{x = 3-y}}}....eq.1a

go to

{{{x^2 + y^2 = 2}}}.....eq.2, substitute {{{x}}}

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

{{{9-6y+y^2 + y^2 = 2}}}

{{{2y^2-6y+9-2  =0}}}

{{{2y^2-6y+7  =0}}}

{{{2y^2-6y+7  =0}}}

using quadratic formula we get

{{{y = (1/2 )(3 + i *sqrt(5))}}}

{{{y = (1/2 )(3 - i* sqrt(5))}}}

go to

{{{x = 3-y}}}....eq.1a, substitute {{{y}}}

{{{x = 3-(1/2 )(3 + i *sqrt(5))}}}=>{{{x = 3/2 - (i*sqrt(5))/2}}}

{{{x = 3-(1/2 )(3 - i *sqrt(5))}}}=>{{{x = 3/2 + (i*sqrt(5))/2}}}