Question

Solve the following system?

`x^2-y^2=ln (y/x), 2x^2-3yx+2y^2=9`

Answer 1

See below.

Explanation:

Making the change of variables

`((x),(y)) =1/sqrt2 ((1,1),(-1,1))((X),(Y))`

we get at

#{(2XY=ln((Y-X)/(Y+X))), (7X^2+Y^2=18):}#

or

`{(Y/X+tanh(XY)=0),(7X^2+Y^2=18):}`

but

`Y/X+tanh(XY)=0 rArr Y = 0 forall X ne 0`

so the solution gives

`X = 0` and `Y = pm sqrt(18)`

or in the old coordinates

`{x,y} = { -3,-3}` and `{x,y} = {3,3}`