Solve for reals x , y ∈ R given system of two non-linear equations. 5 x ( 1 + 1 x 2 + y 2 ) = 12 5 y ( 1 − 1 x 2 + y 2 ) = 4 I got 6 x − 1 + 2 y − 1 = 5Substitute x − 1 = x 1 and same for y and got a four degree equation. Is there a short and elegant method for this?

Question

Solve for reals   x  ,  y  ∈      R   given system of two non-linear equations.                    5        x                  (          1          +                      1                                          x                2                            +                              y                2                                              )                                    =        12                            5        y                  (          1          −                      1                                          x                2                            +                              y                2                                              )                                    =        4            I got  6      x          −      1        +  2      y          −      1        =  5Substitute       x          −      1        =      x    1   and same for   y and got a four degree equation. Is there a short and elegant method for this?

kuthiwenihca · Accepted Answer

The two given equations imply the equation  2  x  +  6  y  −  5  x  y  =  0.Since   x  ,  y are different from zero, this is exactly the equation you have found. The rest is easy now. The equation implies that   5  x  −  6  ≠  0, so that   y  =            2      x              5      x      −      6      . Substituting this into the given equations we obtain  (  5      x    2    −  12  x  +  9  )  (  5  x  −  2  )  (  x  −  2  )  =  0.The quadratic polynomial has no real roots, so that we obtain   (  x  ,  y  )  =  (      2    5    ,            −      1        5    ) and   (  x  ,  y  )  =  (  2  ,  1  )There might be a more elegant solution, but this solution gives also another insight - the polynomial   2  x  +  6  y  −  5  x  y is an   S-polynomial which appears from the Buchberger algorithm for polynomial equations.

charmbraqdy · Accepted Answer

From yor equation we obtain   x  =            6      y              5      y      −      2       plugging this in the first equation, simplifying and factorizing we get   −      (    y    −    1    )        (    5        y    +    1    )        (    5                      y                    2              −    4        y    +    4    )    =  0 from here you can compute all solutions.

Solve for reals x,y in R given system of two non-linear equations. 5x(1+1/(x^2+y^2))=12, 5y(1-1/(x^2+y^2))=4

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