Question 994117
.
{{{system(x^2 - 2y^2 = 7,
xy = 3)}}}.


Express y from the second equation,  y = {{{3/x}}},  and substitute it into the first equation.  You will get single equation for the unknown  x:


{{{x^2}}} - {{{2(3/x)^2}}} = {{{7}}}.


Multiply both sides by  {{{x^2}}}.  You will get


{{{x^4 - 7x^2 - 18}}} = {{{0}}}.


Factor:


{{{(x^2+2)*(x^2-9)}}} = {{{0}}}.


The roots are  {{{x^2}}} = -{{{2}}}  and  {{{x^2}}} = {{{9}}}.


The real roots are  x = +/- 3.


Correspondingly  y = +/- 1.


<U>Answer</U>. &nbsp;The pairs &nbsp;(x,y) = (3,1) &nbsp;and &nbsp;(x,y) = (-3,-1) &nbsp;are real solutions.