Page 91 - Elementary Algebra Exercise Book I
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ELEMENTARY ALGEBRA EXERCISE BOOK I                                             inequAlities




               3.7 Solve the inequality  lg(x − x − 6) < lg(2 − 3x).
                                            2
               Solution: The inequality holds if and only if

                                                  2
                                                 x − x − 6 > 0
                                                     2 − 3x> 0
                                                  2
                                                 x − x − 6 < 2 − 3x
               ⇒

                                                   x< −2 or x> 3
                                                          x<  2
                                                              3
                                                       −4 <x< 2
               ⇒−4 <x< −2, which is the solution of the original inequality.


               3.8     a, b  are real numbers and  a + b =2, show  a + b ≤ 2.
                                                   3
                                                        3
                                                                                                                                  2
                                                                 3
                                                                                                                3
                                                                                                     3
                                                                                               2
                                                                             3
                                                                                                           3
               Proof: Suppose  a + b> 2, then   b> 2 − a ⇒ b > (2 − a) =8 − 12a +6a − a ⇒ a + b > 8 − 12a +6a =
                                                                3
                                                                                                    3
                                                                                               2
                                                                                                          3
                                                                                                                                 2
                                                                                                               3
                                                                            3
                                                 b> 2 − a ⇒ b > (2 − a) =8 − 12a +6a − a ⇒                  a + b > 8 − 12a +6a =
                                                 2
                                                                            2
                                              6a − 12a +6+2 = 6(a − 1) +2 > 2
 b> 2 − a ⇒ b > (2 − a) =8 − 12a +6a − a ⇒ a + b > 8 − 12a +6a = 6a − 12a +6+2 = 6(a − 1) +2 > 2 , a contradiction to
                                                                           2
 3
 3
                      3
                 3
           3
     2
                                                2
                                        2


 2
 2
 6a − 12a +6+2 = 6(a − 1) +2 > 2  a + b =2. Hence,  a + b ≤ 2.
                     3
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