Given the inequality:

5 - 3 l 2t -3 l < -16

First, we will solve the way we solve any equation.

We need to isolate the absolute values on the left side.

Let us subtract 5 from both sides.

==> -3 l 2t -3 l < -16-5

==> -3 l 2t -3 l < -21

Now we will divide by -3 and reverse the inequality.

==> l 2t -3 l > -21/-3

==> l 2t -3 l > 7

Now we have two cases:

Case(1):

(2t -3 ) > 7

==> 2t > 10

==> t > 5

==> t = (5, inf)..............(1)

Case(2):

-(2t-3) > 7

==> -2t +3 > 7

==> -2t > 4

==> t < -2

==> t= (-inf, -2)............(2)

From (1) and (2) we conclude that:

**t= (-inf, -2) U (5, inf)**

**OR:**

**t= R- [ -2, 5]**

We have to solve the inequality 5 - 3*l2t-3l < -16

5 - 3*l2t - 3l < -16

=> - 3*l2t - 3l < -16 - 5

=> - 3*l2t - 3l < -21

=> 3*l2t - 3l > 21

=> l2t - 3l > 7

=> 2t - 3 > 7 and 2t - 3 < -7

=> 2t > 10 and 2t < -4

=> t > 5 and t < -2

**Therefore t should be greater than 5 or less than -2.**

To find for what values of t does the inequality hold?

5-3l 2t-3l < -16.

We subtract add 16 to both sides and then add 3|2t-3| to both sides:

5+16 < = 3|2t-3|

21 = < = 3|2t-3|.

We divide by 3:

21/3 < = |2t-3|.

7<= |2t-3|.

-7 <= 2t- 3 < =7.

We add +3 :

-7+3 <= 2t < 7+3.

-4 <= 2t <=10.

We divide by 2:

-2 < = t < = 5.

So t belongs to the closed interval (-2, 5).