-2|3j| - 8 < -20
Isolate the absolute value term by adding +8 to both
sides:
-2|3j| - 8 < -20
+ 8 + 8
-------------------
-2|3j| < -12
Divide both sides by -2, to leave just the absolute value
on the left. Here we must reverse the inequality < to >
because we divided by a negative number. [When dividing
by a positive number we never reverse the inequality but
when dividing by a negative number, as we are doing here,
we always reverse it.]
-2|3j| -12
-------- > -----
-2 -2
Simplifying,
|3j| > 6
We can take a positive factor outside an absolute value
so we can take out 3
3|j| > 6
Divide both sides by 3. We DO NOT reverse the inequality
this time because we are dividing both sides by a
positive number, 3.
|j| > 6
This means that j can be any number which is 6 or more
units away from 0 on the number line. j can be either
6 or more units away from 0 on the lower side or 6 or
more units away from 0 on the upper side of 0. So the
graph looks like this
<==========]-----------------------[===========>
-10 -8 -6 -4 -2 0 2 4 6 8 10 12
j < -6 OR j > 6
and the interval notation for the solution is
(-¥, -6] È [6, ¥)
Edwin
