Question 1178359
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Our starting equations are


    x^2 + y^2 = 2      (1)

    y = {{{1/x}}}             (2)


From equation (2),  

    xy = 1             (3)



So, I will multiply equation (3) by 2 (both sides) and then subtract it from equation (1).  I will get then


    x^2 - 2xy + y^2 = 0,

or

    (x-y)^2 = 0.


It means  x = y,  and then from equation (1) I have


    2x^2 = 2,  x^2 = 1, x = +/- 1.


Thus the two solutions are  

    x = y = 1   and/or   x = y = -1.
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Solved.