一題絕對值不等式

若︱ax+1︱≦b 的解為－2≦x≦6 則數對(a，b)=____
有人給的解題技巧是
1.取中點 2.同減中點 3.等量乘除
想請問如何運用 有點看不懂
Sol
6－2=4
4/2=2
－-2<=x<=6
－2－2<=x－2<=6－2
－4<=x－2<=4
｜x－2｜<=4
｜－x+2｜<=4
｜－x/2+1｜<=2
So
a=－1/2，b=2
(a，b)=(－1/2，2)

| ax + 1 | ≤ b

-b ≤ ax + 1 ≤ b

-b - 1 ≤ ax ≤ b - 1

若 a > 0
(-b - 1)/a ≤ x ≤ (b - 1)/a

考慮
(-b - 1)/a = -2
(b - 1)/a = 6

(b - 1)/(-b - 1) = -3
(b - 1)/(b + 1) = 3
b - 1 = 3b + 3
2b = -4
b = -2
a = -1/2　不合

2015-07-07 22:40:03 補充：
若 a < 0
(b - 1)/a ≤ x ≤ (-b - 1)/a

考慮
(-b - 1)/a = 6
(b - 1)/a = -2

(b - 1)/(-b - 1) = -1/3
(b - 1)/(b + 1) = 1/3
3b - 3 = b + 1
2b = 4
b = 2
a = -1/2

(a, b) = (-1/2, 2)