已知－1＜a+b＜3且2＜a－b＜4，求2a+3b的取值范围.
－
＜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＜
.
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