Absolute value question?

Because 4 < x + 3 < 6 says that x + 3 is a positive number which is less than 6. Since x + 3 is a positive number, then |x + 3| and x + 3 are the same thing.

Since x + 3 < 6 and x + 3 is positive, then |x + 3| < 6.

Those aren't equivalent inequalities though. While it is true that the first inequality implies the second one (any x that satisfies the first one satisfies the second one), it isn't true the other way.

For instance, let x = -8. Then |x +3| = |-5| = 5, which is certainly less than 6. So |x + 3| < 6.

But x + 3 = -5 which is definitely not between 4 and 6. So it's not true that 4 < x + 3 < 6.

If 4 < x + 3 < 6, then

|x + 3 - 5| < 1

|x - 2| < 1

Verify:

|x - 2| < 1

-1 < x - 2 < 1

-1 + 5 < x - 2 + 5 < 1 + 5

4 < x + 3 < 6

4 < x + 3 < 6

1 < x < 3

thus I x + 3 I < 6

|x+3|<6

i) x+3<6

x<3

ii) -(x+3)<6

-x-3<6 |x+3| < 6?

-x<9

x>-9

-9<x<3

-6<x+3<6

It's like saying that if x>1 then x>-9