若直线y=-1/2x+m与直线y=-2x+2m+3的交点在第四象限,m为整数,则m为

y=-1/2x+m
x + 2y = 2m (1)
y=-2x+2m+3
2x +y = 2m + 3 (2)
(1) - (2) * 2
==> -3x = 2m - 2 * (2m + 3)
==> -3x = 2m - 4m - 6
==> -3x = -2m - 6
==> x = 2(m+3)/3
(2) - (1) * 2
==> -3y = 2m+3 - 4m
==> -3y = -2m + 3
==> y = (2m -3) / 3
交点在第四象限
x > 0 y < 0
==> 2(m +3)/3 > 0
(2m - 3) / 3 < 0
==> m > -3
m < 3/2
==> -3 < m < 3/2
m为整数
所以m= -2,-1,0,1

若直线y=-(1/2)x+m与直线y=-2x+2m+3的交点在第四象限，m为整数，则m为
令-(1/2)x+m=-2x+2m+3，得(3/2)x=m+3，故x=2(m+3)/3，
y=-2[2(m+3)/3]+2m+3=(2m-3)/3；
已知交点在第四象限，故有：
2(m+3)/3>0..............(1)
(2m-3)/3<0...........

联立
y=-1/2x+m
y=-2x+2m+3
解得
x=(2m+6)/3
y=(2m-3)/3
因为交点在第四象限
所以
(2m+6)/3>0
(2m-3)/3<0
解得
-3又因为m为整数
所以m可以为-2,-1,0,1