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直線y=3x+p與x軸的交點是: 以y=0代入,得:x=-p/3 即:直線y=3x+p與x軸的交點是(-p/3,0) 這個點也在直線y=-2x+q上,代入,得: 0=-(2p/3)+q 3q-2p=0 -----------------【這個就是p與q的關係式】
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