Find all solution in R for the following system of equations: { x + 3 x − y x 2 + y 2 = 3 y – x + 3 y x 2 + y 2 = 0

Question

Find all solution in   R for the following system of equations:      {                            x          +                                    3              x              −              y                                                      x                2                            +                              y                2                                              =          3                                      y          –                                    x              +              3              y                                                      x                2                            +                              y                2                                              =          0

tatoas9f · Accepted Answer

Complexify. Let   z  =  x  +  i  y. Then for       x    2    +      y    2    ≠  0:                                          (          x          +                                    3              x              −              y                                                      x                2                            +                              y                2                                              )                +        i                  (          y          −                                    x              +              3              y                                                      x                2                            +                              y                2                                              )                                    =        (        3        +        i        0        )                                                  ⟺                            (        x        +        i        y        )        +                              (            3            −            i            )            (            x            −            i            y            )                                              x              2                        +                          y              2                                                          =        3                                                  ⟺                            z        +                              3            −            i                    z                                    =        3                                                  ⟺                                      z          2                −        3        z        +        (        3        −        i        )                            =        0                                                  ⟺                                                  (            z            −                          3              2                        )                    2                                    =        i        −                  3          4                                                          ⟺                            z                            =                              3            ±                          4              i              −              3                                2                .            Now,   (  1  +  2  i      )    2    =  −  3  +  4  i, so the solutions are  z  =            3      ±      (      1      +      2      i      )        2    ,either   z  =  2  +  i or   z  =  1  −  i, which, in real form, yields   (  x  ,  y  )  =  (  2  ,  1  ) or   (  x  ,  y  )  =  (  1  ,  −  1  ).

Aidyn Cox · Accepted Answer

Great solution and answer. I didn&#039;t know that complexifying will provide a real solution. Thanks again.

Find all solution in <mi mathvariant="double-struck">R for the following system of equations:

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