Solving systems of equations in 3 variables x + y + z = 0 x y + y z + x z = c − c 2 x y z = 3 c 2 where, c is some fixed non-zero constant.

Question

Solving systems of equations in 3 variables  x  +  y  +  z  =  0  x  y  +  y  z  +  x  z  =  c  −      c    2    x  y  z  =  3      c    2  where,   c is some fixed non-zero constant.

Jerry Kidd · Accepted Answer

Write your system in the form  x  y  −  (  x  +  y      )    2    =  c  −      c    2  and  −  x  y  (  x  +  y  )  =  3      c    2  Substituting  x  +  y  =  a  x  +  y  =  a  x  y  =  bso you will get  b  −      a    2    =  c  −      c    2  and  −  a  b  =  3      c    2  Can you proceed? Eliminating   b you will get for   a:  0  =      a    3    +  a  (  c  −      c    2    )  +  3      c    2

Alissa Hutchinson · Accepted Answer

Let   z  =  −  x  −  y. Substitute this into the second and third equation. Multiplying the second by y and subtracting the third we obtain  2  x      y    2    +  c  (  c  y  +  3  c  −  y  )  =  0.For   y  =  0 we have   x  =  y  =  c  =  0. Otherwise we can substitute   x  =  −  c  (  c  y  +  3  c  −  y  )  /  (  2      y    2    ). This yields only one polynomial equation in y and c, namely,      c    3    (  −      y    2    −  6  y  −  9  )  +  2      c    2    y  (  y  +  3  )  +  c      y    2    (  2      y    2    −  6  y  −  1  )  −  2      y    4    =  0.I doesn&#039;t get easier than that. But it can be solved over the complex numbers. One particular solution is   c  =  1 and   2      y    3    +  3  =  0,   z  =  3  /      y    2  ,   x  =  −  3  /  (  2      y    2    ). However, it is not correct that a non-linear system in n variables with n equations has a unique solution.

Solving systems of equations in 3 variables x + y + z = 0 x y +

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