Question

数学函数~

若方程组3x+2y=k+1,4x+3y=k-1的解x,y同号，求整数k的值。

Answer 1

3x+2y = k+1 (1)
4x+3y = k-1 (2)
4(1)- 3(2)
-y = 4(k-1) - 3(k-1)
= k -1
y = -(k-1)
from (1)
3x-2(k-1) = k+1
x = (3k-1)/3
x,y同号，
xy > 0
-(k-1)(3k-1)/3 > 0
=>(k-1)(3k-1) < 0
=> 1/3 <k < 1

Answer 2

x=k+5
y=-k-7
xy>0
-7<k<-5
k=-6