Solve simultaneously: { 6 ( x + y ) = 5 x y 21 ( y + z ) = 10 y z 14 ( z + x ) = 9 z x Obviously one of the solutions is ( 0 , 0 , 0 ) but I'm more interested in the other .I actually solved this question , but using hard way.I expanded each equation then got one variable in terms of other. For example after expanding first equation I get , 6 x + 6 y = 5 x yand then 5 x y − 6 x = 6 ySo, x ( 5 y − 6 ) = 6 ythen, x = 6 y / ( 5 y − 6 )Similarly I found the value of x from the third equation ( in terms of z), equated both and got z in terms of y, substituted that in equation two and finally got the values. The solution is ( 2 , 3 , 7 ). You might have guessed that this is a very tedious method.

Question

Solve simultaneously:      {                            6          (          x          +          y          )                          =                          5          x          y                                      21          (          y          +          z          )                          =                          10          y          z                                      14          (          z          +          x          )                          =                          9          z          x                        Obviously one of the solutions is   (  0  ,  0  ,  0  ) but I&#039;m more interested in the other .I actually solved this question , but using hard way.I expanded each equation then got one variable in terms of other. For example after expanding first equation I get ,  6  x  +  6  y  =  5  x  yand then  5  x  y  −  6  x  =  6  ySo,  x  (  5  y  −  6  )  =  6  ythen,  x  =  6  y  /  (  5  y  −  6  )Similarly I found the value of   x from the third equation ( in terms of   z), equated both and got   z in terms of   y, substituted that in equation two and finally got the values. The solution is   (  2  ,  3  ,  7  ). You might have guessed that this is a very tedious method.

ran1suel23 · Accepted Answer

You can divide the equations, to get                              1          x                +                  1          y                                    =                  5          6                =                  35          42                                              1          y                +                  1          z                                    =                  10          21                =                  20          42                                              1          x                +                  1          z                                    =                  9          14                =                  27          42                    Subtracting the second equation from the first yields      1    x    −      1    z    =      15    42  and adding that to the last      2    x    =      42    42    =  1  ⇒  x  =  2.Subtracting from the last yields      2    z    =      12    42    ⇒  z  =  7.Inserting   x  =  2 for example into the first equation yields   y  =  3.

Lesly Weiss · Accepted Answer

If all you want to do is get a quick, quick answer and you&#039;re confident that there is only one answer, with integers, try this.   6  |  5  x  y, so   6  |  x  y. Similarly   21  |  y  z and   14  |  x  z.There may be many ways to distribute  2  ,  3  ,  7 among   x  ,  y  ,  z, but the most &quot;natural&quot; to make these three properties all true is:   2  |  x,   3  |  y,   7  |  z. As it happens, that&#039;s already a solution, but if it weren&#039;t I&#039;d try   x  =  2      x    ′    ,  y  =  3      y    ′    ,  z  =  7      z    ′  ; this will at least allow you to cancel the   6  ,  21  ,  14 on the LHS&#039;s.

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