The solution of inequality with absolute value in the first grade of Higher Education Given p = {a | A-1 | ≤ 2}, t = {a | A-B | ≤ 2}, if P ∩ t ≠ empty set, the value range of B is obtained This problem I calculate - 1 ≤ a ≤ 3, later want to ask how to do |ax+3|0,A={x||x-a|

The solution of inequality with absolute value in the first grade of Higher Education Given p = {a | A-1 | ≤ 2}, t = {a | A-B | ≤ 2}, if P ∩ t ≠ empty set, the value range of B is obtained This problem I calculate - 1 ≤ a ≤ 3, later want to ask how to do |ax+3|0,A={x||x-a|

1. From | A-B | less than or equal to 2, it can be seen that a is less than or equal to 2 + B is greater than or equal to B-2. From the intuitive figure of the number axis, it can be seen that in order to meet the meaning of the question, there must be 2 + b greater than or equal to - 1 or B-2 less than or equal to 3
2. I fainted. It's the algorithm of the first inequality in the first topic. I just want to discuss the positive and negative of A
3. Classification discussion, discuss the size of a and quarter
4. In silence