·已知集合a={(x,y)│2x+y=1}集合b={(x,y)│x+2y=5}则a∩b
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![·已知集合a={(x,y)│2x+y=1}集合b={(x,y)│x+2y=5}则a∩b](/uploads/image/z/9302342-14-2.jpg?t=%C2%B7%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88a%3D%7B%EF%BC%88x%2Cy%EF%BC%89%E2%94%822x%2By%3D1%7D%E9%9B%86%E5%90%88b%3D%7B%28x%2Cy%29%E2%94%82x%2B2y%3D5%7D%E5%88%99a%E2%88%A9b)
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·已知集合a={(x,y)│2x+y=1}集合b={(x,y)│x+2y=5}则a∩b
·已知集合a={(x,y)│2x+y=1}集合b={(x,y)│x+2y=5}则a∩b
·已知集合a={(x,y)│2x+y=1}集合b={(x,y)│x+2y=5}则a∩b
用中括号把2x+y=1和x+2y=5并列连起来成为一个方程组
解得X=-1,Y=3
所以a∩b={(-1,3)}
2x+y=1;
x+2y=5
解得:x=-1 y=3
所以a∩b={(x,y)│x=-1,y=3}