5|x + 3| - 2 < 18
5|x + 3| < 20
|x + 3| < 4
Now, remembering that absolute value is the distance from 0 and 4 and -4 are equidistant from 0, x + 3 has to be less than 4 or greater than -4, since 3 and -3 are also equidistant from 0 and therefore have the same absolute value.
x + 3 < 4
x < 1
x + 3 > -4
x > -7
Since those both fall in the same area of the number line...
-7 < x < 1, AKA x > -7 and x < 1