Question

已知方程组x+2y=3 ax+2y=2k+1 且0＜y-x＜1 则k的取值范围是 要过程

Answer 1

先由第一格已知条件得：x=3-2y
带入0<y-x<1得1<y<4/3;
由ax+2y=2k+1解得：k=(3a-1)/2 - (a-1)y
讨论a>1,a=1,a<1三种情况下K的范围；
a>1:(a+5)/6 < k < (a+1)/2;
a=1:k=1;
a<1;(a+1)/2 < k < (a+5)/6;

Answer 2

由两个方程可解：
x=(2k-2)/(a-1)
y=3/2-(k-1)/(a-1)
y-x=3/2-(3k-3)/(a-1)
∵0<y-x<1
∴(3k-3)/(a-1)<3/2
(3k-3)/(a-1)>1/2
解得（a+5）/6<k<(a+1)/2