Page 36 - Bahan Ajar Kalkulus Lanjut
P. 36

 1  =  3

                     5    =  1
                    2x    2z
                    10z = 2x

                        2x
                    z =
                        10
                          x
                    z =                                  ) 6 (
                          5


                    Selanjutnya substitusi (5) dan (6) ke (4) diperoleh:

               x  2  + y 2    + z  2  − 30 = 0

                      2x      x  2
                          2
               x  2  +  (−  ) +     − 30 = 0
                          5   5 
                     4x 2     x 2  
               x  2  +  (  ) +       −  30 =  0
                                
                     25       25 

                    5x 2
               x  2  +    −  30 =  0
                    25

               6x  2  −  30 =  0
               25
               6x  2  =  30
               25
               6x 2  = 150

               x  2  =  25
               x  =   5

























                                                              32
   31   32   33   34   35   36   37   38   39   40   41