已知关于x,y的方程组2x+y=1-m,x+2y=3的未知数x,y满足x-y>0,求m的取值范围

设
x-y
=a(2x+y)+b(x+2y)
=2ax+ay+bx+2by
=(2a+b)x+(a+2b)y
则有
2a+b=1 1
a+2b=-1 2
1式+2式得
a+b=0 3
1式-3式得
a=1
2式-3式得
b=-1
所以
x-y
=(2x+y)-(x+2y)
=1-m-3>0
m