In x & lt-2,-2 <; = x < = 3, x>3 seeks the solution in three intervals.

1. Suppose x

Then x+2

Original formula:

-x-2-x+3=5

-2x=4

x=-2

Give up the contradiction between hypothesis and hypothesis.

2. Hypothesis -2

Then x+2 >; =0，x-3 & lt； =0

Original formula:

x+2-x+3=5

5=5

It's an identity

So when -2

3. Suppose x> III

Then x+2 >; 0，x-3 & gt； 0

Original formula:

x+2+x-3=5

2x=6

x=3

refuse

To sum up, when-2 < = x <; When =3, the original equation holds.

In fact, it is assumed that the content in each absolute value is positive or negative after removing the absolute value.

But one situation has been abandoned.

can try