用加减法解方程组:{(x+y)/6+(x-y)/10=3 (x+y)/6+(y-x)/10=-1

Question

如题所述

Answer 1

(x+y)/6+(x-y)/10=3 ①
(x+y)/6+(y-x)/10=-1 ②
①可变为：(x+y)/6-（y-x)/10=3 ③
②+③=(x+y)/3=2
得x+y=6 ④
②-③=（y-x）/5=-4
得y-x=-20 ⑤
④+⑤=2y=-14
得 y=-7
④-⑤=2x=26
得x=13