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設大船有x只,小船有y只.由題意可得方程組 x+y=16 4x+3y=60 解得x=12,y=4 即大船有12只,小船有4只 以上為每只船恰好坐滿時的解法,若不一定恰好坐滿,則為不等式組 x+y=16① 4x+3y>=60② 由①得x=16-y,代入②可得y
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