How to solve | 3 x + 6 | &gt; 15 ?

Question

How to solve             |      3      x      +      6      |        &amp;gt;    15  ?

Denisse Bray · Accepted Answer

The definition of the absolute value is:

| a | = a 888888

if and only if

88888 a \geq 0

| a | = - a .888

if and only if

888888 a < 0

As a result, we must address both:

3 x + 6 > 15

and

- (3 x + 6) > 15

I

3 x + 6 > 15

3 x > 9

x > 3

II

- (3 x + 6) > 15

3 x + 6 < - 15

3 x < - 21

x < - 7

This results in:

- \infty < x < - 7

and

3 < x < \infty

The following can be written as a union of intervals:

(- \infty, - 7) \cup (3, \infty)

How to solve | 3 x + 6 | > 15 ?