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已知方程組{3x+y=12 有正整數解(a為整數),求a的值.4x+ay=2 y=12-3x 有正整數解 x=1,y=9 x=2,y=6 x=3,y=3 當x=1,y=9時 4+9a=2無整數解,捨去 當x=2,y=6時 8+6a=2 解得a=-1 當x=3,y=3時 12+3a=2無整數解,捨去 綜上,a=-1 希望對樓主有所幫助,
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