Question

关于x,y的方程组2x-y=3a+1,x+2y=4-2a的解x,y都是正数,求a的最小整数值

Answer 1

2x-y=3a+1 (1)
x+2y=4-2a (2)

(1)×2+(2)
5x=4a+6
x=(4a+6)/5>0
4a+6>0
a>-3/2

y=2x-3a+1>0
2x>3a-1
所以2(4a+6)/5>3a-1
8a+12>15a-5
a<17/7
所以-3/2<a<17/7
所以a的最小整数值是a=-1

Answer 2

2x-y=3a+1 ①
x+2y=4-2a ②

①×2+②，得：5x=4a+6>0，a>-3/2

②×2-①，得：5y=7-7a>0，a<1

所以-3/2<a<1
那么a的最小整数值为-1

Answer 3

2x-y=3a+1(1)
x+2y=4-2a(2)
(2)*2 2x+4y=8-4a(3)
(3)-(1) 5y=8-4a-3a-1
5y=7-7a
y=7(1-a)/5
4x-2y=6a+2(4)
(2)+(4) 5x=4a+6
x=2(2a+3)/5
x>0,y>0
2a+3>0
a>-3/2,1-a>0,a<1
-3/2<a<1
a=-1

Answer 4

2x-y=3a+1
y=2x-3a-1>0
x>(3a+1)/2

x+2y=4-2a
5x-8a-6>0
x>(8a-6)/5
(3a+1)/2=(8a-6)/5
15a-16a=-12-5
a=17

Answer 5

x=(4a+6)/5
y=(7-7a)/5
x>0 y>0
a=-1