If |-3x + 1| < 7, then which of the following represents all possible

Case 1 : (-3x + 1) is positive .

( -3x + 1) <7
-3x < 6
-x < 2
x> -2.

Case 2: ( -3x +1 ) is negative :

- (-3x + 1) <7
3x -1 <7
x < 8/3.

Combining both cases :

-2 <x < 8/3.

Option B fits .


The best answer is B.

Hi chetan2u,

While solving the above, i tried to get the solutions with the following

-7<-3x+1<7

is this wrong?

Thanks

This is also correct..
-7<-3x+1<7
-7-1<-3x<7-1
-8<-3x<6
8>3x>6
8/3>x>-2

Case 1:

| (-3x + 1) <7

- 3x + 1 < 7

x> -2

Case 2:

|–3x + 1| < 7

3x -1 <7

x < 8/3


-2 <x<8/3

The best answer is B.

Another method is to use the distance concept of absolute values.

\(|-3x + 1| < 7\)

\(3|-x + 1/3| < 7\)

\(|x - 1/3| < 7/3\)

-2 < x< 8/3

Answer (B)

When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:
Rule #1: If |something| < k, then –k < something < k
Rule #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive

From rule #1, we can write: -7 < -3x + 1 < 7
Subtract 1 from all sides to get: -8 < -3x < 6
Divide all sides by -3 to get: 8/3 > x > -2 [since we divided by a NEGATIVE value, we had to REVERSE the inequality symbols]
Rewrite as: -2 < x < 8/3

Answer: B

|-3x+1| < 7
|3x-1| < 7

a) x < 1/3

-(3x-1) < 7
3x-1 > -7
3x > -6
x > -2

b) x>=1/3

3x-1 < 7
3x-1 < 7
3x < 8
x < 8/3

Thus, -2<x<8/3 (B)

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