用加减法解二元一次方程：5/(x+1)+4/(y-2)=2①;7/(x+1)+3/(2-y)=13/20②
问题描述：
用加减法解二元一次方程：5/(x+1)+4/(y-2)=2①;7/(x+1)+3/(2-y)=13/20②
答
5/(x+1)+4/(y-2)=2 (1)
7/(x+1)+3/(2-y)=13/20
所以7/(x+1)-3/(y-2)=13/20 (2)
(1)*3+(2)*4
15/(x+1)+12/(y-2)+28/(x+1)+12/(y-2)=6+13/5
43/(x+1)=43/5
x+1=5,x=4
代入(1)
5/5+4/(y-2)=2,4/(y-2)=-1,y-2=-4,y=-2
所以x=4,y=-2