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Bunuel
Which of the following inequalities is equivalent to \(|2x-|x||<3\)?

A. \(0<x<2\)

B. \(0<x<3\)

C. \(-1<x<3\)

D. \(0<x<1\)

E. \(-3<x<1\)

Consideering value of 'X' +ve or -ve two solution set will be there.
First lets consider X>0,
Then |2X-|X||<3
|2X-X|<3
|X|<3
X<3 (as X is+ve)
Thus first solution 0<X<3.......1
Sexond condition X<0,
Then |2X-|X||<3
|2X-(-X)|<3
|3X|<3
-3X<3 (as X is -ve)
X>-1
Thus second solution becomes
-1<X<0............2
Merging 1 & 2 we will get
-1<X<3.
Thus answer C.

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Asked: Which of the following inequalities is equal to |2x-|x||<3?

|2x-|x||<3
-3 <2x - |x| < 3

Case 1: x>=0
-3 < 2x - x < 3
-3 < x < 3
Since x>0
0 <= x < 3

Case 2: x<0
-3 < 2x + x < 3
-3 < 3x < 3
-1 < x < 1
Since x<0
-1 < x < 0

Combining above 2:
-1 < x < 3
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Making graph of \(y1=-|x|\) and \(y2=2x\) , then adding both gives \(f(x)=y1+y2\)

Check interval when \(f(x)<3\) gives \(-1<x<3\)
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chetan2u , I have gone through you post on Absolute modulus. However, I am finding difficulty to how to approach this question. Could you please explain in your way or which would be the best way to solve this problem?
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Bunuel
Which of the following inequalities is equivalent to \(|2x-|x||<3\)?

A. \(0<x<2\)

B. \(0<x<3\)

C. \(-1<x<3\)

D. \(0<x<1\)

E. \(-3<x<1\)

SHSAHOO

Here, I will not use any method but make use of options.
I’ll choose 2 first, so |2*2-|2||=2<3…true. Discard A, D and E
Next choose 0, so 0<3…true. Discard B
Answer: C
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Can someone please explain why we get to merge the two solutions?

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Cuttlefish
Which of the following inequalities is equivalent to \(|2x-|x||<3\)?

A. \(0<x<2\)

B. \(0<x<3\)

C. \(-1<x<3\)

D. \(0<x<1\)

E. \(-3<x<1\)­

Can someone please explain why we get to merge the two solutions?

Posted from my mobile device
­When solving, we get two ranges for which the given inequality holds: -1 < x < 0 and 0 ≤ x < 3. So, -1 < x < 0 or 0 ≤ x < 3. However, notice that these two solution sets represent a single continuous set -1 < x < 3. So, it's more convenient to write it this way. Also, this is how the correct answer is represented.

Hope it's clear.­
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Bunuel

Cuttlefish
Which of the following inequalities is equivalent to \(|2x-|x||<3\)?

A. \(0<x<2\)

B. \(0<x<3\)

C. \(-1<x<3\)

D. \(0<x<1\)

E. \(-3<x<1\)­

Can someone please explain why we get to merge the two solutions?

Posted from my mobile device
­When solving, we get two ranges for which the given inequality holds: -1 < x < 0 and 0 ≤ x < 3. So, -1 < x < 0 or 0 ≤ x < 3. However, notice that these two solution sets represent a single continuous set -1 < x < 3. So, it's more convenient to write it this way. Also, this is how the correct answer is represented.

Hope it's clear.

Apologies, I thought I understood but the reality is I still have some lingering questions. How come 0 gets to be eliminated? If we combine the two solution ranges, then it's - 1 < x < 0>= x <3
I just don't get how x can be both non-negative and negative at the same time.­
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Cuttlefish

Bunuel

Cuttlefish
Which of the following inequalities is equivalent to \(|2x-|x||<3\)?

A. \(0<x<2\)

B. \(0<x<3\)

C. \(-1<x<3\)

D. \(0<x<1\)

E. \(-3<x<1\)­

Can someone please explain why we get to merge the two solutions?

Posted from my mobile device
­When solving, we get two ranges for which the given inequality holds: -1 < x < 0 and 0 ≤ x < 3. So, -1 < x < 0 or 0 ≤ x < 3. However, notice that these two solution sets represent a single continuous set -1 < x < 3. So, it's more convenient to write it this way. Also, this is how the correct answer is represented.

Hope it's clear.

Apologies, I thought I understood but the reality is I still have some lingering questions. How come 0 gets to be eliminated? If we combine the two solution ranges, then it's - 1 < x < 0>= x <3
I just don't get how x can be both non-negative and negative at the same time.­
­
-1 < x < 0 implies:

--------(-1)----(0)------------(3)­

0 ≤ x < 3 implies:

--------(-1)----(0)------------(3)­

Now, combine the two ranges to get -1 < x < 3:

--------(-1)----(0)------------(3)­

­Hope it's clear.­
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We need to solve the inequality:
\(∣2x−∣x∣∣<3\)

Step 1: Consider Absolute Value Definition
We define cases based on xxx:
  • Case 1: \(x≥0\)
    Since\( ∣x∣=x\), the expression simplifies to: \(∣2x−x∣=∣x∣<3\)
    which results in: \(−3<x<3\)
  • Case 2:\( x<0\)
    Since \(∣x∣=−x\), the expression simplifies to:
    \(∣2x−(−x)∣=∣3x∣<3\)
    which results in: \(−1<x<1\)

Step 2: Combine the Two Cases
From the first case: \(−3<x<3\)
From the second case: \(−1<x<1\)
The valid range satisfying both cases is: \(−1<x<3\)

Step 3: Identify the Correct Answer
Looking at the answer choices, the correct answer is: C.\(−1<x<3\)
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