The system of equations x 2 + y 2 − x − 2 y = 0 and x + 2 y = c ( 1. ) x 2 + y 2 − x − 2 y = 0 ( 2. ) x + 2 y = cSolving for y in ( 2. ) gives y = ( c − x ) / 2Is there a way to simplify equation ( 1. )?Because at the end I arrive at c 2 − 2 x − c = 0and can't proceed. Need to get typical form of square equation A x 2 + B x + C = 0.The solution for c is 0 or 5.For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?(g) x 2 + y 2 − x − 2 y = 0 , x + 2 y = c, Ans. c = 0 or 5.The process is to solve for y, then to substitute that into second equation. From that we get A , B and C. Delta being B 2 − 4 A C we can get parameter.So y = ( c − x ) / 2 x 2 + y 2 − x − 2 y = 0 x 2 + ( ( c − x ) / 2 ) 2 − x − 2 ∗ ( ( c − x ) / 2 ) = 0and i got lost...

Question

The system of equations       x    2    +      y    2    −  x  −  2  y  =  0 and   x  +  2  y  =  c  (  1.  )        x    2    +      y    2    −  x  −  2  y  =  0    (  2.  )    x  +  2  y  =  cSolving for   y in   (  2.  ) gives  y  =  (  c  −  x  )  /  2Is there a way to simplify equation   (  1.  )?Because at the end I arrive at      c    2    −  2  x  −  c  =  0and can&#039;t proceed. Need to get typical form of square equation   A      x    2    +  B  x  +  C  =  0.The solution for c is   0 or   5.For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?(g)         x    2    +      y    2    −  x  −  2  y  =  0  ,    x  +  2  y  =  c, Ans.   c  =  0 or   5.The process is to solve for   y, then to substitute that into second equation. From that we get   A  ,  B and   C. Delta being       B    2    −  4  A  C we can get parameter.So  y  =  (  c  −  x  )  /  2      x    2    +      y    2    −  x  −  2  y  =  0      x    2    +  (  (  c  −  x  )  /  2      )    2    −  x  −  2  ∗  (  (  c  −  x  )  /  2  )  =  0and i got lost...

lydalaszq · Accepted Answer

If you substitute correctly, you will get the equation  5          x      2    −  2    c    x  +        c      2    −  4    c  =  0If you solve this, you will get   x  =      1    5    (  c  ±  2      c    (    5    −    c    )    ). The other value is given by   y  =      1    2    (  c  −  x  ), but you don&#039;t need this to answer your question.If the two sets of solutions have the same values, then we must have   c  (  5  −  c  )  =  0, hence   c  =  0 or   c  =  5.Addendum: To get the above equation, replace   y in       x    2    +      y    2    −  x  −  2  y  =  0 by   y  =      1    2    (  c  −  x  ) (from the second equation). That is, the equation                              x          2                +                  1          4                (        c        −        x                  )          2                −        x        −        (        c        −        x        )                            =                              1          4                (        4                  x          2                +        (        c        −        x                  )          2                −        4        c        )                                          =                              1          4                (        5                  x          2                −        2        c        x        +                  c          2                −        4        c        )            Then the solutions are (ignoring the       1    4  , of course):                    x                            =                              1          10                          (          2          c          ±                      4                          c              2                        −            20            (                          c              2                        −            4            c            )                    )                                                  =                              1          10                          (          2          c          ±                      16            c            (            5            −            c            )                    )                                                  =                              1          5                          (          c          ±          2                      c            (            5            −            c            )                    )

Bruno Pittman · Accepted Answer

Great expert answer!

The system of equations x 2 </msup> + y 2 </msup> −

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