Question 1077333
all pairs of real numbers ({{{x}}},{{{y}}}) such that{{{ x + y = 6}}} and {{{x^2 + y^2 = 28}}}

{{{x + y = 6 }}}...solve for {{{x}}}
{{{x = 6 -y}}}

substitute in {{{x^2 + y^2 = 28}}}

{{{(6-y)^2 + y^2 = 28}}}.....solve for {{{y}}}

{{{36-12y+y^2 + y^2 = 28}}}

{{{36-12y+2y^2  -28=0}}}

{{{2y^2-12y+36  -28=0}}}....both sides divide by {{{2}}}

{{{y^2-6y+18  -14=0}}}

{{{y^2-6y+4=0}}}.......complete square

{{{(y^2-6y+b^2)-b^2+4=0}}}

{{{(y^2-6y+3^2)-3^2+4=0}}}

{{{(y-3)^2-9+4=0}}}

{{{(y - 3)^2 - 5 = 0}}}

{{{(y - 3)^2 = 5 }}}

{{{(y - 3) = sqrt(5 )}}}

solutions:

{{{y=3+sqrt(5 )}}} or {{{y=3-sqrt(5 ) }}}


now find {{{x}}}

{{{x = 6 -y}}}=>{{{x = 6 -(3+sqrt(5 ))=6-3-sqrt(5 )}}}=>{{{x=3-sqrt(5 )}}}
or
{{{x = 6 -y}}}=>{{{x = 6 -(3-sqrt(5 ))=6-3+sqrt(5 )}}}=>{{{x=3+sqrt(5 )}}}


so, ({{{x}}},{{{y}}}) pairs are:

({{{3+sqrt(5 )}}},{{{3-sqrt(5 )}}})

and

({{{3-sqrt(5 )}}},{{{3+sqrt(5 )}}})