Given that |-3x + 1| < 7 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. -2 < x

Lets take x = 5 (which falls in this range of -2 < x ) and substitute in the equation |-3x + 1| < 7
=> |-3*5 + 1| < 7
=> |-14| < 7
=> 14 < 7 which is FALSE

B. -2 < x < \(\frac{8}{3}\)

Lets take x = 0 (which falls in this range of -2 < x < \(\frac{8}{3}\) ) and substitute in the equation |-3x + 1| < 7
=> |-3*0 + 1| < 7
=> |1| < 7
=> 1 < 7 which is TRUE
In test, we don't need to solve further. But I am solving to completed the solution.

C. \(-2 \leq x \leq \frac{8}{3}\)

Lets take x = 5 (which falls in this range of \(-2 \leq x \leq \frac{8}{3}\)) and substitute in the equation |-3x + 1| < 7
=> |-3*\(\frac{8}{3}\) + 1| < 7
=> |-8 + 1| < 7
=> |-7| < 7
=> 7 < 7 which is FALSE

D. \(x < -2\) or \(x > \frac{8}{3}\)

Lets take x = -3 (which falls in this range of \(x < -2\) or \(x > \frac{8}{3}\)) and substitute in the equation |-3x + 1| < 7
=> |-3*-3 + 1| < 7
=> |-9 + 1| < 7
=> |-8| < 7
=> 8 < 7 which is FALSE

E. \(x \leq -2\) or \(x \geq \frac{8}{3}\)

We can again take x = -3 to prove this one FALSE

So, Answer will be B

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

=> |-3x + 1| < 7 can be written as

Case 1: -3x + 1 ≥ 0 or x ≤ \(\frac{1}{3}\)
=> |-3x + 1| = -3x + 1
=> -3x + 1 < 7
=> 3x > 1 - 7
=> 3x > -6
=> x > \(\frac{-6}{3}\)
=> x > -2
And the condition was x ≤ \(\frac{1}{3}\), so answer will be the range common in x ≤ \(\frac{1}{3}\) and x > -2
=> -2 < x ≤ \(\frac{1}{3}\) is the solution



Case 2: -3x + 1 < 0 or x > \(\frac{1}{3}\)
=> |-3x + 1| = -(-3x + 1) = 3x - 1
=> 3x - 1 < 7
=> 3x < 7 + 1
=> 3x < 8
=> x < \(\frac{8}{3}\)
And the condition was x > \(\frac{1}{3}\), so answer will be the range common in x > \(\frac{1}{3}\) and x < \(\frac{8}{3}\)
=> \(\frac{1}{3}\) < x < \(\frac{8}{3}\) is the solution



Final answer will be a combination of the two answers
-2 < x ≤ \(\frac{1}{3}\) and \(\frac{1}{3}\) < x < \(\frac{8}{3}\)
=> -2 < x < \(\frac{8}{3}\)

So, Answer will be B
Hope it helps!

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