Question

How do you solve `2x^2-3y^2+4=0` and `xy=24`?

Answer 1

#{x = pm sqrt[sqrt[865]-1], y =pm sqrt( 2/3 (1 + sqrt[865]))}#

Explanation:

Calling `x_2=x^2` and `y_2=y^2` we have
the equivalent system

# {(2 x_2-3y_2+4=0), (x_2 y_2 = 24^2) :}#

Solving for `x_2,y_2` we obtain the only feasible solution

`{x2 = sqrt[865]-1, y2 = 2/3 (1 + sqrt[865])}`

then

#{x = pm sqrt[sqrt[865]-1], y =pm sqrt( 2/3 (1 + sqrt[865]))}#