设a={(x,y)|2x-y=1},b={(x,y)|2x+y=0},求a∩b
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![设a={(x,y)|2x-y=1},b={(x,y)|2x+y=0},求a∩b](/uploads/image/z/11449892-20-2.jpg?t=%E8%AE%BEa%3D%EF%BD%9B%EF%BC%88x%2Cy%EF%BC%89%7C2x-y%3D1%EF%BD%9D%2Cb%3D%EF%BD%9B%EF%BC%88x%2Cy%EF%BC%89%7C2x%EF%BC%8By%3D0%EF%BD%9D%2C%E6%B1%82a%E2%88%A9b)
设a={(x,y)|2x-y=1},b={(x,y)|2x+y=0},求a∩b
设a={(x,y)|2x-y=1},b={(x,y)|2x+y=0},求a∩b
设a={(x,y)|2x-y=1},b={(x,y)|2x+y=0},求a∩b
就是求:
2x-y=1
和2x+y=0的交点
解得
x=1/4
y=-1/2
从而
a∩b={(1/4,-1/2)}