Question 353523
{{{x^2 + 3*x^2=27}}}  (1)
{{{xy = 6}}}         (2)
From equation (2), {{{y = 6/x}}}.
Substituting into (1),
{{{x^2 +3*(6/x)^2=27}}},
{{{x^2 + 108/x^2=27}}},
Multiplying both sides by {{{x^2}}}, we get
{{{x^4 +108 = 27*x^2}}}, or
{{{x^4 - 27*x^2 + 108 = 0}}}.  Treating the equation as a quadratic equation in {{{x^2}}}, and applying the quadratic formula, we get 
{{{x^2 = (27 +- sqrt(27^2-4*1*108))/2}}},
{{{x^2 = (27+-sqrt(297))/2}}}.  Hence 
{{{x = sqrt((27+-sqrt(297))/2)}}}, or 
{{{x = -sqrt((27+-sqrt(297))/2)}}}. The values that we get from the first x-value are {{{x = 4.703}}} or {{{2.2098}}}.  from the second x-value, we get the negatives:{{{x = -4.703}}} or {{{-2.2098}}}.  
The corresponding y-values are (obtained from equation (2)) 1.2758, 2.7152, 
-1.2758, -2.7152.