Question

若方程组｛3x+y=k+1.x+3y=3的解x，y，且2<k<4,则x-y的取值范围是____

Answer 1

方程组｛3x+y=k+1.x+3y=3相减得
2x-2y=k-2
x-y=1/2*(k-2)
因2<k<4
所以0<k-2<2
0<1/2*(k-2)<1
因此0<x-y<1

Answer 2

—3<X-y<-1
过程，x=3-3y则
x-y=3-4y
3（3-3y）+y=k+1
k=8-8y
2<8-8y<4
左右同时减5
-3<3-4y<-1

Answer 3

解 ： 令 x- y = a(3x+y) + b(x+3y) = (3a+b)x + (a+3b)y
所以 3a+b = 1 a+3b = -1 得 a=1/2 b= -1/2
所以 x-y = 1/2(3x+y) - 1/2(x+3y) = 1/2(k+1) - 3/2 = k/2 - 1
由 2<k<4 得 0 < x-y < 1