Question

已知－1＜a+b＜3且2＜a－b＜4，求2a+3b的取值范围.

Answer 1

－＜2a+3b＜

∵a+b，a－b的范围已知，
∴要求2a+3b的取值范围，
只需将2a+3b用已知量a+b，a－b表示出来.
可设2a+3b=x（a+b）+y（a－b），用待定系数法求出x、y.
设2a+3b=x（a+b）+y（a－b），
∴解得
∴－＜（a+b）＜，
－2＜－（a－b）＜－1.
∴－＜（a+b）－（a－b）＜，
即－＜2a+3b＜.
错解：解此题常见错误是：－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＜.