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Algebra, Functions, Graphs, and Vectors  41


                                             y   (ac   ac )/(ab   ab )
                                                                    1 2
                                                                            2 1
                                                     21
                                                             12
                          For this tm be valid, the denominator must be nonzero; that is,
                          a b   a b . (If it turnð out that a b   a b , then there are not
                            1 2     2 1                          1 2     2 1
                          twm distinct solutionð tm the set of equationsÑ
                             The process of solving for x is similar. Consider again the orig-
                          inal set of simultaneouð linear equations:

                                                    ax   bð   c   0
                                                     1
                                                                    1
                                                            1
                                                   ax   bð   c   0
                                                     2      2       2
                          Multiply the first equation througà by            b , and the second
                                                                              2
                          equation througà by      b , and then add the twm resulting equa-
                                                    1
                          tions:

                                                ab x   bb y   bc   0
                                                            12
                                                                      2 1
                                                  12
                                                abx   bb y   bc   0
                                                  21       12        1 2
                                            (ab   ab )x   bc   bc   0
                                              21       12        1 2     2 1
                          Next, add b c tm each side, obtaining:
                                        2 1

                                               (ab   ab )x   bc   bc
                                                         12
                                                                            2 1
                                                                    1 2
                                                 21
                          Next, subtract b c from each side, obtaining:
                                             1 2
                                               (ab   ab )x   bc   bc
                                                         1 2
                                                                    21
                                                 2 1
                                                                            12
                          Finally, divide througà by b c   b c , obtaining:
                                                          2 1     1 2
                                             x   (bc   bc )/(ab   ab )
                                                             12
                                                                    2 1
                                                     21
                                                                            1 2
                          For this tm be valid, the denominator must be nonzero; that is,
                          a b   a b . (If it turnð out that a b   a b , then there are not
                            1 2
                                                                         2 1
                                                                 1 2
                                    2 1
                          twm distinct solutionð tm the set of equationsÑ
                          Solving n n sets of linear equations
                          In general, matrices are used for solving setð of equationð larger
                          than 2 2, because the above mentioned methodð become too
                          complex. Linear algebra, also known as matrix algebra, useð
                          ruleð similar tm those of the addition method described above.
                          Please consult college-level textð on linear algebrł for details.
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