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(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(3x-2\geq{0}\) --> \(x\geq{\frac{2}{3}}\) --> \(x\) positive, so \(|x|=x\) --> \(x=3x-2\) --> \(x=1\). Sufficient

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(3x-2\geq{0}\) --> \(x\geq{\frac{2}{3}}\) --> \(x\) positive, so \(|x|=x\) --> \(x=3x-2\) --> \(x=1\). Sufficient

Answer: B.

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me. Generally in modulus questions we do as below: (1) |x| < 2 I agree with u. -2<x<2 (2) |x| = 3x – 2 In this case i did a. x is positive so x = 3x - 2 , upon solving x = 1 a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2. I am confused !

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(3x-2\geq{0}\) --> \(x\geq{\frac{2}{3}}\) --> \(x\) positive, so \(|x|=x\) --> \(x=3x-2\) --> \(x=1\). Sufficient

Answer: B.

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me. Generally in modulus questions we do as below: (1) |x| < 2 I agree with u. -2<x<2 (2) |x| = 3x – 2 In this case i did a. x is positive so x = 3x - 2 , upon solving x = 1 a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2. I am confused !

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm doing is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: \(3x-2\geq{0}\) --> \(x\geq{\frac{2}{3}}\) --> \(x\) positive, so \(|x|=x\) and no need to consider \(|x|=-x\) --> \(x=3x-2\) --> \(x=1\).

I marked E but i was wrong, my reasoning was as follows:

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0 x=3x-2 x=1

when x<0 -x=3x-2 ( |x|=-x if x is -ve) x=1/2

again 2 values not sufficient. Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong. Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.
_________________

Whatever one does in life is a repetition of what one has done several times in one's life! If my post was worth it, then i deserve kudos

Last edited by conty911 on 27 Aug 2012, 11:20, edited 2 times in total.

Sorry, if this question has already been posted earlier, i tried searching the forum but it never came up.

What is x?

(1) |x| < 2

(2) |x| = 3x – 2

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0 x=3x-2 x=1

when x<0 -x=3x-2 ( |x|=-x if x is -ve) x=1/2

again 2 values not sufficient. Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong. Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.

Merging similar topics. Please refer to the solution above.
_________________

Sorry, if this question has already been posted earlier, i tried searching the forum but it never came up.

What is x?

(1) |x| < 2

(2) |x| = 3x – 2

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0 x=3x-2 x=1

when x<0 -x=3x-2 ( |x|=-x if x is -ve) x=1/2

again 2 values not sufficient. Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong. Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.

Merging similar topics. Please refer to the solution above.

Thanks for merging the topic . By the way the post is tagged in the wrong section , it should be in DS rather than PS.section?
_________________

Whatever one does in life is a repetition of what one has done several times in one's life! If my post was worth it, then i deserve kudos

Re: What is x? (1) |x| < 2 (2) |x| = 3x – 2 [#permalink]

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11 Jul 2013, 00:51

1

This post received KUDOS

Think of statement 2 this way:

If we consider x to be negative, then -x=3x-2 which translates to x=1/2. Now when we test this situation 1/2=3*1/2-2 the outcome is =3/2-2= -1/2. The RHS is not equal to the LHS. Hence x has to be 1, because when we test x=3x-2 we get x=1. and the RHS and LHS are equal to each other

I solved equation 2 and got two values. Hence I thought 2 was insufficient. However, mgmat says "When dealing with absolute values that contain variables, you should always check that the solution is valid".They solve the equation, and then test the value once again in the equation and one the solutions is now invalid. Why is such a test required? Usually when you solve a equation, what you get is the SOLUTION right? Can someone please clarify why one solution will be 'invalid' in case of 'variables with absolute values'? I am unable to visualize that.

I solved equation 2 and got two values. Hence I thought 2 was insufficient. However, mgmat says "When dealing with absolute values that contain variables, you should always check that the solution is valid".They solve the equation, and then test the value once again in the equation and one the solutions is now invalid. Why is such a test required? Usually when you solve a equation, what you get is the SOLUTION right? Can someone please clarify why one solution will be 'invalid' in case of 'variables with absolute values'? I am unable to visualize that.

Merging similar topics. Please refer to the solution above.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(3x-2\geq{0}\) --> \(x\geq{\frac{2}{3}}\) --> \(x\) positive, so \(|x|=x\) --> \(x=3x-2\) --> \(x=1\). Sufficient

Answer: B.

Hope it's clear.

Hi Bunuel, I have a question. (2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

Thanks in advance
_________________

--------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(3x-2\geq{0}\) --> \(x\geq{\frac{2}{3}}\) --> \(x\) positive, so \(|x|=x\) --> \(x=3x-2\) --> \(x=1\). Sufficient

Answer: B.

Hope it's clear.

Hi Bunuel, I have a question. (2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

Thanks in advance

We never considered negative value for |x|.... we considered negative value for just x...... By definition, mod never has a negative value....

When absolute values are given I guess it's better to go for number substitution.

1) |x| < 2 implies x can be either 1 or -1. 0 won't come here as it is neither positive nor negative and there is no need to take it here. 2) |x| = 3x - 2 is possible only when x = 1.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm ding is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm ding is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

I dont think so. You need to follow the following 2 step procedure for ALL absolute value questions:

Treat |x| as following:

1. Let x \(\geq\)0 ----> |x| = x ---> x=3x-2 ---> x=1 and as x=1 satisfies x \(\geq\)0, this is a valid answer.

2. Let |x| < 0 ----> |x| = -x ----> -x=3x+2 ---> x = 1/2 and as x= 1/2 does not satisfy x<0, this solution is rejected. Thus we only 1 value, x = 1 and thus combining the 2 statements is sufficient. Thus C is the correct answer.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

If you consider |x| as positive (or >0) and you get x <0 , then this solution is not possible as you have 2 conflicting scenarios.

HI Enge2012 I completely agree what you have mentioned, and that is way we normally solve mod equation, but if you see the Bunuel solution, he consider |x| as positive first and put equation >= 0 through which he gets +ve X and gain he solve the equation. Now my concern is over above equation what if we get -ve value of X example instead of 3x-2>=0 if we have equation as 3x +2 >=0 now we get x as negative. hope i am cleared with doubt.

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