Question

关于x,y的方程组x+3y=2k+1；2x-y=k+3的解中,x,y的值都是正数,求k的取值范围.

Answer 1

x+3y=2k+1 (1)

2x-y=k+3 (2)

(1)+(2)*3

7x=5k+10

x=(5k+10)/7

y=2x-k-3=(3k-1)/7

x>0,y>0

(5k+10)/7>0

(3k-1)/7>0

k>-2

k>1/3

所以k>1/3