Page 36 - MODUL 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 
                                      
                                +
                     x  2    +  ( 4x 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
























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