Question 603245
Someone has probably solved similar problems and developed an easy general procedure that I've never heard of. Maybe it would help to realize that one of the equations represents an ellipse and the other a hyperbola, and to figure out their axes of symmetry
I'll just try something and see how it works.
If you were to subtract the second equation from the first one, you would have
{{{( x^2+3xy+y^2)-( x^2-xy+y^2)=27-6}}} --> {{{x^2+3xy+y^2-x^2+xy-y^2=21}}} --> {{{4xy=21}}} <---> {{{xy=21/4}}}
(That's the equation of a hyperbola that has the x-axis and the y-axis as asymptotes, and tells me that x and y are both positive or both negative).
If you multiply the second equation times 3 and add the first, you would have
{{{3*(x^2-xy+y^2)+ x^2+3xy+y^2=3*6+27}}} --> {{{3x^2-3xy+3y^2+ x^2+3xy+y^2=18+27}}} --> {{{4x^2+4y^2=45}}} <---> {{{x^2+y^2=45/4}}}
(That's the equation of a circle, centered at the origin,
with radius {{{3sqrt(5)/2=sqrt(45)/2}}} ).
At this point I see two ways forward.
 
I can build expressions for {{{(x+y)^2}}} and {{{(x-y)^2}}} , solve them for {{{x+y}}} and {{{x-y}}} and end up with easy systems of linear equations:
{{{(x+y)^2=(x^2+y^2)+2*xy=45/4+42/4=87/4}}} tells me {{{x+y=sqrt(87)/2}}} or {{{x+y=-sqrt(87)/2}}}
{{{(x-y)^2=(x^2+y^2)-2*xy=45/4-42/4=3/4}}} tells me {{{x-y=sqrt(3)/2}}} or {{{x-y=-sqrt(3)/2}}}
That seems to give too many combinations, but from the 4 sytems of linear equations
{{{system(x+y=sqrt(87)/2, x-y=sqrt(3)/2)}}},  {{{system(x+y=-sqrt(87)/2, x-y=-sqrt(3)/2)}}},  {{{system(x+y=-sqrt(87)/2, x-y=sqrt(3)/2)}}}, and {{{system(x+y=sqrt(87)/2,  x-y=-sqrt(3)/2)}}}, you easily get the answers:
{{{system(x=(sqrt(87)+sqrt(3))/4, y=(sqrt(87)-sqrt(3))/4)}}},  {{{system(x=-(sqrt(87)+sqrt(3))/4, y=-(sqrt(87)-sqrt(3))/4)}}},  {{{system(x=-(sqrt(87)-sqrt(3))/4, y=-(sqrt(87)+sqrt(3))/4)}}}, and {{{system(x=(sqrt(87)-sqrt(3))/4, y=(sqrt(87)+sqrt(3))/4)}}} .
 
Alternatively, I can work part of the previous option up to {{{x+y=sqrt(87)/2}}} or {{{x+y=-sqrt(87)/2}}}
and then use that along with {{{xy=21/4}}}
to make two quadratic equations that will give me the answers.
We know that the quadratic equation {{{(z-x)(z-y)=0}}} (z is the variable here),
with solutions {{{x}}} and {{{y}}} , can also be written as
{{{z^2-(x+y)z+xy=0}}}
We make one of those quadratic equations with {{{x+y=sqrt(87)/2}}} , along with {{{xy=21/4}}} .
We get {{{z^2-(sqrt(87)/2)z+sqrt(3)/3=0}}} , which gives us two solutions for {{{z}}}.
{{{z=(sqrt(87) +- sqrt(3))/4}}} , both positive.
We can make {{{x=(sqrt(87) + sqrt(3))/4}}}  and {{{y=(sqrt(87) - sqrt(3))/4}}}  or
{{{x=(sqrt(87) - sqrt(3))/4}}}  and {{{y=(sqrt(87) + sqrt(3))/4}}} .
With {{{x+y=-sqrt(87)/2}}} and {{{xy=21/4}}} , we make
{{{z^2+(sqrt(87)/2)z+sqrt(3)/3=0}}} , which gives us two negative solutions
{{{z=(-sqrt(87) +- sqrt(3))/4}}} .
We can make {{{x=(-sqrt(87)+sqrt(3))/4=-(sqrt(87) - sqrt(3))/4}}} and {{{y=(-sqrt(87)-sqrt(3))/4=-(sqrt(87)+sqrt(3))/4}}} , or
{{{x=(-sqrt(87)-sqrt(3))/4=-(sqrt(87) + sqrt(3))/4}}} and {{{y=(-sqrt(87)+sqrt(3))/4=-(sqrt(87) - sqrt(3))/4}}} .