Question

已知－1＜a+b＜3且2＜a－b＜4，求2a+3b的取值范圍.

Answer 1

－＜2a+3b＜

∵a+b，a－b的范圍已知，
∴要求2a+3b的取值范圍，
只需將2a+3b用已知量a+b，a－b表示出來(lái).
可設(shè)2a+3b=x（a+b）+y（a－b），用待定系數(shù)法求出x、y.
設(shè)2a+3b=x（a+b）+y（a－b），
∴解得
∴－＜（a+b）＜，
－2＜－（a－b）＜－1.
∴－＜（a+b）－（a－b）＜，
即－＜2a+3b＜.
錯(cuò)解：解此題常見(jiàn)錯(cuò)誤是：－1＜a+b＜3，                                                            ①
2＜a－b＜4.                                                                                                           ②
①+②得1＜2a＜7.                                                                                                 ③
由②得－4＜b－a＜－2.                                                                                         ④
①+④得－5＜2b＜1，∴－＜3b＜.                                                                ⑤
③+⑤得－＜2a+3b＜.