How to solve this simultaneous equation of 3 variables. (1) x + y + z = a + b + c (2) a x + b y + c z = a 2 + b 2 + c 2 (3) a x 2 + b y 2 + c z 2 = a 3 + b 3 + c 3

Question

How to solve this simultaneous equation of  3  variables.                    (1)                    x        +        y        +        z                            =                    a        +        b        +        c                            (2)                    a        x        +        b        y        +        c        z                            =                              a                  2                +                  b                  2                +                  c                  2                                    (3)                    a                  x                  2                +        b                  y                  2                +        c                  z                  2                                    =                              a                  3                +                  b                  3                +                  c                  3

oedfeuonbk203 · Accepted Answer

By inspection we see that   (  x  ,  y  ,  z  )  =  (  a  ,  b  ,  c  ) is a solution of the given system                    (0)                                                        Eq. 1                            x              +              y              +              z              =              a              +              b              +              c                                                          Eq. 2                            a              x              +              b              y              +              c              z              =                              a                              2                            +                              b                              2                            +                              c                              2                                                                        Eq. 3                            a                              x                              2                            +              b                              y                              2                            +              c                              z                              2                            =                              a                              3                            +                              b                              3                            +                              c                              3                            .                                          The other solution can be found as follows. Solve Eq. 1 for   z. Multiply original Eq. 1 by   a, subtract Eq. 2 and solve for   z. This results in                    (1)                                                        z              =              a              +              b              +              c              −              x              −              y              ,                                                              (2)                                                        z              =                                                b                  −                  a                                                  a                  −                  c                                            y              +                                                a                  b                  +                  a                  c                  −                                      b                                      2                                    −                                      c                                      2                                                                    a                  −                  c                                            .                                          Equate the right hand sides of (1) and (2)                    (3)                                                                                          b                  −                  a                                                  a                  −                  c                                            y              +                                                a                  b                  +                  a                  c                  −                                      b                                      2                                    −                                      c                                      2                                                                    a                  −                  c                                            =              a              +              b              +              c              −              x              −              y              ,                                          and solve for   x                    (4)                    x        =                              c            −            b                                a            −            c                          y        +                              −            a            c            +                          b                          2                        +                          a                          2                        −            b            c                                a            −            c                          .            Substitute   x  ,  z in (0), Eq. 3, using (4) and (2)                    (5)                    a                              (                                          c                −                b                                            a                −                c                                      y            +                                          −                a                c                +                                  b                                  2                                +                                  a                                  2                                −                b                c                                            a                −                c                                      )                            2                +        b                  y                  2                +        c                              (                                          b                −                a                                            a                −                c                                      y            +                                          a                b                +                a                c                −                                  b                                  2                                −                                  c                                  2                                                            a                −                c                                      )                            2                =                  a                  3                +                  b                  3                +                  c                  3                .            Solving for   y we get the solution   y  =  b and the solution                    (6)                    y        =                  B          D                ,            where  A  =      a      3    c  +      a      3    b  −  2      a      2    b  c  −      a      2        b      2    −      a      2        c      2    +  2  a      c      3    −  a  b      c      2    +  2  a      b      3    −  a  c      b      2    −  2  b      c      3    −  2      b      3    c  +  4      c      2        b      2    C  =  2      a      3    c  −  2      a      3    b  +  4      a      2        b      2    −      a      2        c      2    −      a      2    b  c  −  a  c      b      2    −  2  a      b      3    +  a      c      3    −  2  a  b      c      2    +  b      c      3    +  2      b      3    c  −      c      2        b      2    .So the two solutions of (0) are:  (  x  ,  y  ,  z  )  =  (  a  ,  b  ,  c  )    and     (  x  ,  y  ,  z  )  =      (          A      D        ,          B      D        ,          C      D        )    .Eq. (5) is equivalent to      (    c          b          2        +          c          2        b    +    a          c          2        +    c          a          2        +    a          b          2        −    6    a    c    b    +    b          a          2        )    (  y  −  b  )      (    y    −          B      D        )    =  0.

How to solve this simultaneous equation of 3 variables. <mtable columnalign="right center lef

Answered question

Answer & Explanation

New Questions in Algebra I