Given that |2x + 1| < 3x - 2 and we need to find the range for all possible values of x

Let's solve the problem using two methods

Method 1: Substitution

We will values in each option choice and plug in the question and check if it satisfies the question or not. ( Idea is to take such values which can prove the question wrong)

(A) -3 < x < 0.2

Lets take x = 0 (which falls in this range of -3 < x < 0.2) and substitute in the equation |2x + 1| < 3x - 2
=> |2*0 + 1| < 3*0 - 2
=> | 1| < -2
=> 1 < -2 which is FALSE

(B) x > -3

We can again take x = 0 to prove this one FALSE

(C) x < -3

Lets take x = -4 (which falls in this range of x < -3) and substitute in the equation |2x + 1| < 3x - 2
=> |2*-4 + 1| < 3*-4 - 2
=> | -7| < -14
=> 7 < -14 which is FALSE

(D) 0.2 < x < 3

Lets take x = 1 (which falls in this range of 0.2 < x < 3) and substitute in the equation |2x + 1| < 3x - 2
=> |2*1 + 1| < 3*1 - 2
=> | 3 | < 1
=> 3 < 1 which is FALSE

(E) x > 3

Lets take x = 4 (which falls in this range of x > 3) and substitute in the equation |2x + 1| < 3x - 2
=> |2*4 + 1| < 3*4 - 2
=> | 9 | < 10
=> 9 < 10 which is TRUE

So, Answer will be E

Method 2: Algebra

Now, we know that |A| < B can be opened as (Watch this video to know about the Basics of Absolute Value)
A < B for A ≥ 0 and
-A < B for A < 0

=> |2x + 1| < 3x - 2 can be written as

Case 1: 2x + 1 ≥ 0 or x ≥ \(\frac{-1}{2}\)
=> 2x + 1 < 3x - 2
=> 3x-2x > 1 + 2
=> x > 3
And the condition was x ≥ \(\frac{-1}{2}\) and x > 3 satisfies this
=> x > 3 is a solution

Case 2: 2x + 1 < 0 or x < \(\frac{-1}{2}\)
=> -(2x + 1) < 3x - 2
=> 3x+2x > 2-1
=> 5x > 1
=> x > \(\frac{1}{5}\)
And the condition was x < \(\frac{-1}{2}\), but x > \(\frac{1}{5}\) DOES NOT satisfy this
=> x > \(\frac{1}{5}\) is NOT a solution

So, Answer will be E
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems