{ 1 − 4 x ( x 2 + y 2 ) = 0 1 − 4 y ( x 2 + y 2 ) = 0 and I tried to do it by subtracting the first equation from the second, yielding ( 4 x − 4 y ) ( x 2 + y 2 ) = 0. Clearly this is satisfied when x = y, which gives ( x , y ) = ( 1 2 , 1 2 ), or when x 2 + y 2 = 0, which gives ( x , y ) = ( 0 , 0 ).

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{                            1          −          4          x          (                      x            2                    +                      y            2                    )          =          0                                      1          −          4          y          (                      x            2                    +                      y            2                    )          =          0                        and I tried to do it by subtracting the first equation from the second, yielding   (  4  x  −  4  y  )  (      x    2    +      y    2    )  =  0. Clearly this is satisfied when   x  =  y, which gives   (  x  ,  y  )  =  (      1    2    ,      1    2    ), or when       x    2    +      y    2    =  0, which gives   (  x  ,  y  )  =  (  0  ,  0  ).

Xavier Vargas · Accepted Answer

You didn&#039;t go wrong. Subtracting the equations will preserve any existing solution, but may add others, which aren&#039;t solutions to the original equations (as here). So you need to feed the possible answers back into the original equations, as you have done to check that they are valid.

ht1o4qgqdy · Accepted Answer

{                            1          −          4          x          (                      x            2                    +                      y            2                    )          =          0                                      1          −          4          y          (                      x            2                    +                      y            2                    )          =          0                        Let   S the set of solutions.You can summarise the logic steps:1. If   (  x  ,  y  ) is a solution:  4  x  (      x    2    +      y    2    )  =  4  y  (      x    2    +      y    2    )    ⟹        x    2    +      y    2    =  0   or   x  =  y      ⟹    (  x  ,  y  )  =  (  0  ,  0  )   or     x  =  y  =      1          4      (              x        2            +              y        2            )        =      1          8              x        2                ⟹    x  =  y  =      1          8              1              /        2              =      1    2  You have proved   S  ⊂  {  (  1  /  2  ,  1  /  2  )  ,  (  0  ,  0  )  }2. If   x  =  (  1  /  2  ,  1  /  2  ) the equation is true. If   x  =  (  0  ,  0  ) the equation is false. You have proved   S  =  {  (  1  /  2  ,  1  /  2  )  }.

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