Help me state the solution set?

The definition of |x| is: |x| = x if x≥0 and |x| = -x if x<0

Therefore you always have two cases:

1) if 3x + 2 ≥0 then 3x + 2 > 4, or
... if x ≥ -2/3 then x> 2/3 ... is (2/3 ≥ -2/3) ? ,,, yes, then this is one solution.

2) if 3x + 2 <0 then -3x - 2 > 4, or
... if x<-2/3 then x< -2 ... is (-2/3<-2) ?... no, then this is not a solution

So the only solution is: x>2/3

First, get x via itself so: x <= 8. this suggests that x is comparable to all numbers decrease than or equivalent to eight. era notation: ( or ) skill decrease than/extra advantageous than. [ or ] skill decrease than or equivalent to/ extra advantageous than or equivalent to. SO: x = (- infinity sign, 8]. you already know the infinity sign precise: the parent 8. because of the fact infinity isn't a particular variety, we use parenthesis. because of the fact 8 is a particular variety, we use ]. desire that helps

|3x + 2| > 4

3x + 2 > 4 or -(3x + 2) > 4
3x + 2 > 4 or 3x + 2 < -4
3x > 2 or 3x < -6
x > 2/3 or x < -2

The solution of the problem consists of two inequalities, namely x > 2/3 and x < -2.

Try this way.
Consider |3x+2| < 4

|3x+2| < 4 means
-4 < 3x+2 < 4
Isolate x: subtract 2 throughout
-6 < 3x < 2
divide by 3 throughout
-2 < x < 2/3

But this is the solution of |3x+2| < 4
Therefore, the solution of |3x+2| > 4 is:
x < -2
x > 2/3

Using interval notation, the solution set is:
(-∞ , -2) U (2/3, ∞)