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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 21 Jan 2022, 05:21
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Statement 1:
|x + 3| = 4x – 3
Because |x + 3| is never negative, we can say 4x - 3 ≥ 0
x ≥ 3/4
Therefore, x > 0
Statement 1 is sufficient.

Statement 2:
|x – 3| = |2x – 3|

In cases where Mod = Mod, I prefer seeing them as magnitudes or distances rather than equations. It is simpler.
So, the statement says that the distance between x and 3 is the same as the distance between 2x and 3.
There can be 2 scenarios

(1) ---(0)----(x)-----(3)-----(2x)
(2)------(0=x=2x)------3

Because the exact values of x don't matter here, considering the mods as distance rather than equations is way faster.
Also, solving by this method gives you only the valid values of x -> 2 and 0.

Statement 2 is insufficient.

Therefore, A is the answer.
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 06 Feb 2023, 05:42
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Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.
Show more


Hi Bunuel please advise an absolute value is always supposed to be positive, irrespective of number inside the absolute expression, so while solving inequalities when do we assume that expression in inequality can have two value. For example in statement one you said that value on left hand side is an absolute value hence positive but in second statement you have discussed two cases for both values on right and left

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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 06 Feb 2023, 06:45
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Bunuel
Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel please advise an absolute value is always supposed to be positive, irrespective of number inside the absolute expression, so while solving inequalities when do we assume that expression in inequality can have two value. For example in statement one you said that value on left hand side is an absolute value hence positive but in second statement you have discussed two cases for both values on right and left

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Sorry but I don't understand your question. Please elaborate. Thank you!
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 06 Feb 2023, 07:09
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Bunuel
Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel please advise an absolute value is always supposed to be positive, irrespective of number inside the absolute expression, so while solving inequalities when do we assume that expression in inequality can have two value. For example in statement one you said that value on left hand side is an absolute value hence positive but in second statement you have discussed two cases for both values on right and left

Posted from my mobile device
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So value i was under expression that |x|=4 means x can have two values +4 and -4

|-4|= 4 only


But |x+1| will always have positive value for example if |4+1|=5 or |-4+1|=3

That is also what you said in statement one the that |x+3| will always have positive value


Now in statement below

|x – 3| = |2x – 3|

|2x-3| implies we can have two values of |2x-3|for +x and -x.

Let's say x is +-2 then two values are 1 and -7

This means |x-3| = 1 or -7 but then absolute value should always be positive so how can |x-3| be = -7
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 06 Feb 2023, 07:28
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Bunuel
Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.


Hi Bunuel please advise an absolute value is always supposed to be positive, irrespective of number inside the absolute expression, so while solving inequalities when do we assume that expression in inequality can have two value. For example in statement one you said that value on left hand side is an absolute value hence positive but in second statement you have discussed two cases for both values on right and left

Posted from my mobile device


So value i was under expression that |x|=4 means x can have two values +4 and -4

But |x+1| will always have positive value for example if |4+1|=5 or |-4+1|=3

That is also what you said in statement one the that |x+3| will always have positive value


Now in statement below

|x – 3| = |2x – 3|

|2x-3| implies we can have two values of |2x-3|for +x and -x.

Let's say x is +-2 then two values are 1 and -7

This means |x-3| = 1 or -7 but then absolute value should always be positive so how can |x-3| be = -7
Show more

Absolute value is ALWAYS non-negative. But the expression in the modulus can be positive, negative or 0. That is:

    \(|a| \geq 0\) but a itself can be positive, negative or 0.

For example, as you correctly wrote, |x| = 4 means that x can be 4 or -4 but |x| is ALWAYS positive or 0. |x| = -4 has no solutions for x.

Now, about |x – 3| = |2x – 3|. Again, I'm not sure what you are doing there but |x – 3| or |2x – 3| can NEVER be negative!

    If x = 2, then |2x – 3| = |4 - 3| = |1| = 1.
    If x = -2, then |2x – 3| = |-4 - 3| = |-7| = 7.

There are several ways to solve |x – 3| = |2x – 3|. One is shown in the solution you quote another is here but none of them says that |x – 3| or |2x – 3| can be negative because that would be incorrect.
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| Updated on: 10 Feb 2023, 01:29
Originally posted by puneetfitness on 10 Feb 2023, 01:19.
Last edited by puneetfitness on 10 Feb 2023, 01:29, edited 1 time in total.
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Bunuel
Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.
Show more

[url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Bunuel]Bunuel

Hi Bunuel lCan you please advise where i am going wrong
I tried it doing this way

|x+3| = 4x -3
Critical value x = -3

This implies we will test for x < -3 and x > -3

If x < -3

Then -(x + 3) = -4x -3
=>5x = 0 => x = 0
not possible x is not < -3

If x > -3 then x can be -2,-1, 0, <0

If x greater then zero then
x+3= 4x-3
X=2 which is correct x >-3

But if x is -2 or -1then we get

X+3= -4x-3

3x=-6
X=-1 x greater than -3
Now we have two values for x
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 10 Feb 2023, 01:25
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Bunuel
Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

Bunuel
Can you please advise where i am going wrong
I tried it doing this way
|x+3| = 4x -3
|x+3| = x + 3 for x + 3 >= 0 ---> x > -3.
Solving for x,
x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

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All is correct there. You got that x = 2, which is correct, and you can answer the question whether x is greater than 0. Good job!
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 10 Feb 2023, 01:34
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Bunuel
Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

[url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Bunuel]Bunuel

Hi Bunuel lCan you please advise where i am going wrong
I tried it doing this way

|x+3| = 4x -3
Critical value x = -3

This implies we will test for x < -3 and x > -3

If x < -3

Then -(x + 3) = -4x -3
=>5x = 0 => x = 0
not possible x is not < -3

If x > -3 then x can be -2,-1, 0, <0

If x greater then zero then
x+3= 4x-3
X=2 which is correct x >-3

But if x is -2 or -1then we get

X+3= -4x-3

3x=-6
X=-1 x greater than -3
Now we have two values for x
Show more

You got that x = 2. How it can be -2 or any other value? Only x= 2 satisfies |x+3| = 4x -3.
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 10 Feb 2023, 01:41
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Bunuel
Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

[url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Bunuel]Bunuel

Hi Bunuel lCan you please advise where i am going wrong
I tried it doing this way

|x+3| = 4x -3
Critical value x = -3

This implies we will test for x < -3 and x > -3

If x < -3

Then -(x + 3) = -4x -3
=>5x = 0 => x = 0
not possible x is not < -3

If x > -3 then x can be -2,-1, 0, <0

If x greater then zero then
x+3= 4x-3
X=2 which is correct x >-3

But if x is -2 or -1then we get

X+3= -4x-3

3x=-6
X=-1 x greater than -3
Now we have two values for x

You got that x = 2. How it can be -2 or any other value? Only x= 2 satisfies |x+3| = 4x -3.
Show more

My apologies but in such case where we get x>-3 then which values are we suppose to test for x only positive or negative as well

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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 10 Feb 2023, 01:46
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My apologies but in such case where we get x>-3 then which values are we suppose to test for x only positive or negative as well

Posted from my mobile device
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Your question is not clear. What do you mean "test"? For what ?

|x + 3| = 4x – 3

When x <= -3, |x + 3| = -(x + 3), and |x + 3| = 4x – 3 becomes -(x + 3) = 4x – 3, which gives x = 0. Discard this solution because x = 2 is NOT in the range we consider (x > -3).

When x > -3, |x + 3| = x + 3, and |x + 3| = 4x – 3 becomes x + 3 = 4x – 3, which gives x = 2. This solution is OK since x = 2 is in the range we consider (x > -3).

So, |x + 3| = 4x – 3 ha only one solution x = 2. DONE.

What else do you want to test here?
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 10 Feb 2023, 01:58
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puneetfitness


My apologies but in such case where we get x>-3 then which values are we suppose to test for x only positive or negative as well

Posted from my mobile device

Your question is not clear. What do you mean "test"? For what ?

|x + 3| = 4x – 3

When x <= -3, |x + 3| = -(x + 3), and |x + 3| = 4x – 3 becomes -(x + 3) = 4x – 3, which gives x = 0. Discard this solution because x = 2 is NOT in the range we consider (x > -3).

When x > -3, |x + 3| = x + 3, and |x + 3| = 4x – 3 becomes x + 3 = 4x – 3, which gives x = 2. This solution is OK since x = 2 is in the range we consider (x > -3).

So, |x + 3| = 4x – 3 ha only one solution x = 2. DONE.

What else do you want to test here?
Show more


When now here is what I am getting confused at when I assume x is greater than -3

Let say x= -2 the then on left hand side we get x+3 equal to positive

But on right hand side of equation which is not in modulus 4x-3 will 4x continue to be 4x or it will become -4x because we had assume x to be -2
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 10 Feb 2023, 02:15
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puneetfitness


My apologies but in such case where we get x>-3 then which values are we suppose to test for x only positive or negative as well

Posted from my mobile device

Your question is not clear. What do you mean "test"? For what ?

|x + 3| = 4x – 3

When x <= -3, |x + 3| = -(x + 3), and |x + 3| = 4x – 3 becomes -(x + 3) = 4x – 3, which gives x = 0. Discard this solution because x = 2 is NOT in the range we consider (x > -3).

When x > -3, |x + 3| = x + 3, and |x + 3| = 4x – 3 becomes x + 3 = 4x – 3, which gives x = 2. This solution is OK since x = 2 is in the range we consider (x > -3).

So, |x + 3| = 4x – 3 ha only one solution x = 2. DONE.

What else do you want to test here?


When now here is what I am getting confused at when I assume x is greater than -3

Let say x= -2 the then on left hand side we get x+3 equal to positive

But on right hand side of equation which is not in modulus 4x-3 will 4x continue to be 4x or it will become -4x because we had assume x to be -2
Show more

When x is negative in some expression it's not the reason to replace it with -x. Say we have x = -1. Do you change it and write -x = -1? This does not make sense. If x is negative then x there stands for that negative number.
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 26 Dec 2024, 01:51
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Hi Bunuel,
I have understood why (1) is sufficient
For (2), I have doubts- when both the sides of equality have mods, how shall we interpret and open them?
I did it this way that I opened one side with classic approach

1) x-3>=0 ==> x>=3
x-3=2x-3 ==> x=0 (invalid since x>=3 is the range)

2) x-3<0 ==> x<3
x-3=-(2x-3) ==> x=2 (valid)
(2 is sufficient)
Answer shall be D

But i think i am missing on the aspect about 2x-3 that it can also open as positive negative both.
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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| 26 Dec 2024, 02:41
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Hi Bunuel,
I have understood why (1) is sufficient
For (2), I have doubts- when both the sides of equality have mods, how shall we interpret and open them?
I did it this way that I opened one side with classic approach

1) x-3>=0 ==> x>=3
x-3=2x-3 ==> x=0 (invalid since x>=3 is the range)

2) x-3<0 ==> x<3
x-3=-(2x-3) ==> x=2 (valid)
(2 is sufficient)
Answer shall be D

But i think i am missing on the aspect about 2x-3 that it can also open as positive negative both.
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1. Please review and study solutions provided on the previous pages carefully.


2. Brush-up fundamentals:

Absolute Value



For more check Ultimate GMAT Quantitative Megathread




3. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.­
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