I wonder whether the system of equations and inequations below have a solution. If there are solutions, what are they? A numerical solution is also desired. { c 1 1 − x 1 + c 2 1 − x 2 + c 3 1 − x 3 = 0 c 1 1 − x 4 + c 2 1 − x 5 + c 3 1 − x 6 = 0 c 1 ln ⁡ x 1 1 − x 1 + c 2 ln ⁡ x 2 1 − x 2 + c 3 ln ⁡ x 3 1 − x 3 = 0 c 1 ln ⁡ x 4 1 − x 4 + c 2 ln ⁡ x 5 1 − x 5 + c 3 ln ⁡ x 6 1 − x 6 = 0 x 1 ( 1 − x 1 ) x 4 ( 1 − x 4 ) = x 2 ( 1 − x 2 ) x 5 ( 1 − x 5 ) = x 3 ( 1 − x 3 ) x 6 ( 1 − x 6 ) &gt; 1 where c 1 , c 2 , c 3 are constants, and x i ∈ ( 0 , 1 ) , i = 1 , 2 , 3 , 4 , 5 , 6

Question

I wonder whether the system of equations and inequations below have a solution. If there are solutions, what are they? A numerical solution is also desired.      {                                                      c              1                                      1              −                              x                1                                              +                                    c              2                                      1              −                              x                2                                              +                                    c              3                                      1              −                              x                3                                              =          0                                                                c              1                                      1              −                              x                4                                              +                                    c              2                                      1              −                              x                5                                              +                                    c              3                                      1              −                              x                6                                              =          0                                                  c            1                    ln          ⁡                                                    x                1                                            1                −                                  x                  1                                                              +                      c            2                    ln          ⁡                                                    x                2                                            1                −                                  x                  2                                                              +                      c            3                    ln          ⁡                                                    x                3                                            1                −                                  x                  3                                                              =          0                                                  c            1                    ln          ⁡                                                    x                4                                            1                −                                  x                  4                                                              +                      c            2                    ln          ⁡                                                    x                5                                            1                −                                  x                  5                                                              +                      c            3                    ln          ⁡                                                    x                6                                            1                −                                  x                  6                                                              =          0                                                                                x                1                            (              1              −                              x                1                            )                                                      x                4                            (              1              −                              x                4                            )                                =                                                    x                2                            (              1              −                              x                2                            )                                                      x                5                            (              1              −                              x                5                            )                                =                                                    x                3                            (              1              −                              x                3                            )                                                      x                6                            (              1              −                              x                6                            )                                &amp;gt;          1                        where       c    1    ,      c    2    ,      c    3   are constants, and       x    i    ∈  (  0  ,  1  )  ,  i  =  1  ,  2  ,  3  ,  4  ,  5  ,  6

haingear8v · Accepted Answer

A simple set of solutions are any   (      c          1        ,      c          2        ,      c          3        ) such that       c          1        +      c          2        +      c          3        =  0,       x          1        =      x          2        =      x          3        =      1    2  ,       x          4        =      x          5        =      x          6        &amp;lt;      1    2

I wonder whether the system of equations and inequations below have a solution. If there are solutio

Answered question

Answer & Explanation

New Questions in Pre-Algebra