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
xy=2
{{x^2=3+y^2}}
2 answers
<<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 y = 2/x and sub into the second to get
no change elsewhere, just a typo
should have said:
if so, change the first into y = 2/x and sub into the second to get
no change elsewhere, just a typo