摘要：22． 解:(1)设购买乙种电冰箱台.则购买甲种电冰箱台. 丙种电冰箱台.根据题意.列不等式:·························································· 1分 ．······················································· 3分 解这个不等式.得．·························································································· 4分 至少购进乙种电冰箱14台．······················································································ 5分 (2)根据题意.得．·············································································· 6分 解这个不等式.得．·························································································· 7分 由(1)知． ． 又为正整数. ．··········································································································· 8分 所以.有三种购买方案: 方案一:甲种电冰箱为28台.乙种电冰箱为14台.丙种电冰箱为38台, 方案二:甲种电冰箱为30台.乙种电冰箱为15台.丙种电冰箱为35台, 方案三:甲种电冰箱为32台.乙种电冰箱为16台.丙种电冰箱为32台．·················· 10分

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