I'm not sure if I did these correct. I got (No Solution) for both.

>Absolute value inequalities<

7|4a-6|<42

second inequality

3|z+4|-13 < - 1

7|4a-6| < 42

|4a-6| < 42/7

|4a-6| < 6

remove the absolute value brackets by solving for both + / -

4a-6 < 6

4a < 12

a < 12/4

a < 3

now negative inside

- (4a-6) < 6

-4a + 6 < 6

-4a < 0

equality sign flips direction when multiply or divide by a negative.

a > 0

So "a " has to be both greater than zero and less than 3.

0 < a < 3