Let x , y , z &gt; 0 and such { x x y − z 2 = − 1 7 y y z − x 2 = 2 5 z z x − y 2 = − 3 show that: x + y + z = 6

Question

Let   x  ,  y  ,  z  &amp;gt;  0 and such      {                                                      x                              x                y                −                                  z                  2                                                              =          −                                    1              7                                                                                      y                              y                z                −                                  x                  2                                                              =                                    2              5                                                                                      z                              z                x                −                                  y                  2                                                              =          −          3                        show that:  x  +  y  +  z  =  6

Jackson Trevino · Accepted Answer

Ones first impression is that this is a set of three quadratic equations. So if we would proceed by eliminating two of the three variables, we would get a higher order equation that presumably can not be solved. However, in this case the functions are such that we can side-step these complications. With a nice trick we can work with linear equationsю  x  y  −      z    2    =  P  x  y  z  −      x    2    =  Q  y  z  x  −      y    2    =  R  zSubtract these equations from each other (e.g. the second from the first). This yields:  (  x  −  z  )  S  =  P  x  −  Q  y  (  y  −  x  )  S  =  Q  y  −  R  z  (  z  −  y  )  S  =  R  z  −  P  xwhere we have defined   S  =  x  +  y  +  z. We now proceed as if   S an adjustable parameter, and solve the set of   3  +  1 linear equations (of which one can be omitted, since it depends on the other two). The final result is:  x  =  (      S    2    −  Q  S  +  Q  R  )  C  y  =  (      S    2    −  R  S  +  R  P  )  C  z  =  (      S    2    −  P  S  +  P  Q  )  Cwhere   C  =      S          3              S        2            −      (      P      +      Q      +      R      )      S      +      Q      R      +      R      P      +      P      Q      Inserting   P  =  −  7  ,  Q  =  5      /    2  ,  R  =  −  1      /    3, we find all sets of values   (  x  ,  y  ,  z  ) in terms of   S. In principle   S can range from   −  ∞ to   +  ∞. But if we are only interested in positive solutions, than we must restrict   S to values equal to or larger than   5      /    4  +  (  1      /    12  )      345  . This comes down to   S    ⟹    2.797848. There is one solution in terms of natural numbers:   x  =  1  ,  y  =  2  ,  z  =  3.

Let x,y,z>0 and suc {x/xy-z^2=-1/7, y/yz-x^2=2/5, z/zx-y^2=-3 show that: x+y+z=6

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