Question 1021116
{{{3x^2+4xy+5y^2=8 }}}     (Equation A)

{{{ x^2+3xy+2y^2=0}}}       (Equation B)

The key lies in factoring Eqn. B:  {{{ x^2+3xy+2y^2=0}}}<==> (x+y)(x+2y) = 0.
We discuss 2 cases:
(i) When x+y = 0, or y = -x.  Substitute this into Eqn.A
==> {{{3x^2-4x^2+5x^2 = 8}}}==> {{{4x^2 = 8}}} ==> {{{x^2 = 2}}}
==> {{{x =sqrt(2)}}} or {{{x = - sqrt(2)}}} 
==> the solutions ({{{sqrt(2)}}}, {{{-sqrt(2)}}}) and ({{{-sqrt(2)}}}, {{{sqrt(2)}}}).

(ii)  The case of x+2y = 0, or x = -2y, is handled in a similar manner.
After doing so we get the pair of solutions  ({{{-4sqrt(2)/3}}}, {{{2sqrt(2)/3}}}) and ({{{4sqrt(2)/3}}}, {{{-2sqrt(2)/3}}}).

Therefore the solutions are ({{{sqrt(2)}}}, {{{-sqrt(2)}}}), ({{{-sqrt(2)}}}, {{{sqrt(2)}}}), ({{{-4sqrt(2)/3}}}, {{{2sqrt(2)/3}}}), and ({{{4sqrt(2)/3}}}, {{{-2sqrt(2)/3}}}).