Question

求教一道关于二元一次方程取值的问题！！！望各位高人指点

当方程组kx-y=5,2x+3ky=7的解满足x＞0，y＜0,k的范围应是？

Answer 1

kx-y=5;
2x+3ky=7;
解出:
x=(15k+7)/(3k^2+2);
y=(7k-10)/(2k^2+2);
则:15k+7>0;
7k-10<0;
所以-7/15<k<10/7

Answer 2

3k²x-3ky=15k
2x+3ky=7
(3k²+2)x=15k+7
x=(15k+7)/(3k²+2)>0
k>-7/15
y=kx-5
=(15k²+7k)/(3k²+2)-5
=(7k-10)/(3k²+2)<0
k<10/7
∴-7/15<k<10/7