If |2x + 1|

Question

If |2x + 1| < 3x - 2, then which of the following represents all possible values of x ? (A) -3 < x < 0.2 (B) x > -3 (C) x < -3 (D) 0.2 < x < 3 (E) x > 3

GMATinsight · Answer

Given, |2x + 1| < 3x - 2
Since |2x + 1| is ALWAYS NON-NEGATIVE (Modulus of any function is always non-negative) therefore, 3x - 2 must ALWAYS BE POSITIVE

i.e. 3x - 2 > 0
i.e. x > 2/3

Hence, Only  Options E fits

For, |2x + 1| = 3x - 2
±(2x + 1) = 3x - 2

CAse-1: (2x + 1) = 3x - 2 i.e x = 3

CAse-2: -(2x + 1) = 3x - 2 i.e x = 1/5 (NOT ACCEPTABLE as x must be greater than 2/3)

i.e. x Must be greater than 3

Answer: Option E

BrentGMATPrepNow · Answer

STRATEGY: Students who fail to recognize the differences between high school math and GMAT math will attempt to solve the given inequality and then look for their answer among the answer choices (here's an article about high school math vs GMAT math: https://gmatclub.com/forum/gmat-math-vs ... 88210.html). The much faster (and less prone to errors) GMAT math approach is to test values. Let's first test whether x = 0 is a solution to the given inequality by plugging it in to get: |2(0) + 1| < 3(0) - 2 Simplify to get: |1| < -2, which simplifies further to get 1 < -2 Since the resulting inequality is NOT TRUE, we know that x = 0 is NOT a solution, which means we can eliminate any answer choices that says x = 0 IS a solution. So, we can eliminate A and B Now let's test another value. How about x = 10 . Plug it in to get: |2(10) + 1| < 3(10) - 2 Simplify to get: |21| < 28, which simplifies further to get 21 < 28 Since the resulting inequality is TRUE, we know that x = 10 IS a solution, which means we can eliminate any answer choices that says x = 10 is NOT a solution. So, we can eliminate C and D By the process of elimination, the correct answer is E

BrushMyQuant · Answer

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

ashKing12 · Answer

Is this a correct way to solve?

we know  lxl < a  means  -a < x < a  which is same as saying  -a < X AND X < a

So -3x + 2 < 2x +1 < 3x -2

-3x +2 < 2x +1 AND 2x +1 < 3x - 2

1/5 < X AND 3 < X

AND means Intersection so intersection of both gives 3 < X (E)

If |2x + 1| < 3x - 2, then which of the following represents all possi

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