Question

xy=2
{{x^2=3+y^2}}

Answer 1

are you solving for the intersection of the two hyperbolas?

if so, change the first into y = x/2 and sub into the second to get

x^4 - 4x^2 - 12 = 0
(x^2 - 6)(x^2 + 2) = 0
x^2 - 6 or x^2 = -2

x = ±√6 or x = ±√2 i (but we probably only want real intersections)

so y = ±√6 /2

Answer 2

<<if so, change the first into y = x/2 and sub into the second to get >>

should have said:

if so, change the first into <b>y = 2/x</b> and sub into the second to get

no change elsewhere, just a typo