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Mustu, for (1) |x + 1| = 2|x – 1| when looking at X<0 case, how come you multiply only right-hand side by negative, but not both sides? thanks

We have to consider 4 cases totally..

1) + + 2) + - 3) - + 4) - -

If u analyse more closely, then u will find that case 1 and 4 ( - and - cancels out) are the same and cases 2 and 3 are the same( + and - 0r - and + are the same). Hope this is clear.

Regards, Mustu

Could you post all the case in this question and reason why are these cases ?I have a lower hand in Inequalities ;Therefore , I can't figure out what are referring about .

In this question, you were going to add a link to read up more about the approach. Seems like you forgot to add the link. Can you please send me the link. Thank you.

In this question, you were going to add a link to read up more about the approach. Seems like you forgot to add the link. Can you please send me the link. Thank you.

This is fine as long as you know why you are doing this:

4LEX wrote:

Am I doing this right:

Statement 1:

(1) |x + 1| = 2|x – 1|

(x+1)=2(x-1) (When both the terms are positive, x > 1) x+1=2x-2 3=x (NO) (Valid value for x since 3 >1) & (x+1)=-2(x-1) (When -1 < x < 1, (x+1) is positive but (x-1) is negative so you are put a negative sign here) x+1=-2x+2 3x=1 x=1/3 (YES) (Valid value since -1 < 1/3 < 1)

There would be another case x < -1. In that case both the terms will be negative. -(x+1)=-2(x-1) giving x = 3 (Not a valid value since 3 is not less than -1) I am assuming that you saw the two negatives will get canceled out and give x = 3 which will not be valid so you skipped this step. In some questions, you could get a valid value here. So you have only 2 values for x (3 and 1/3).

Insufficient B,C,orE

Statement 2

(2) |x – 3| > 0

(x-3)>0 x>3 (NO)

-(x-3)>0 -x+3>0 -x>-3 x<3 (MAYBE)

Not sufficient

C or E

Combined:

Statement 1: x=1/3,3 Statement 2: x <> 3

Since x CANNOT equal 3, x = 1/3

Since |1/3| < 1, both statements are sufficient to answer the prompt.

From statement 1, we are able to get two values of x; they are \(x=3\) and \(x=1/3\). Two values of x, hence insufficient. From statement 2, all we know is that the distance from x is more than 0 or it indirectly implies that x is not 0. Not enough information. Hence insufficient.

On combining these two statements, we come to know that x cannot be 3 and x=1/3. Since \(1/3\) < 1, hence \(|x|<1\). +1C.

Please do add the OA while posting the questions.
_________________

This is my approach: Is |x|<1? 1st start from statement 2, cause it is easier, |x – 3| > 0 just tell us x is note equal to 3, so it is insufficient to solve the target question 2nd for statement 2: |x + 1| = 2|x – 1| we have to separate the condition to x<-1, -1<x<1, x>1 that is , |x|<1 and |x|> 1 to do further thinking 1) when |x|<1, we could know we will get the solution in this range after solving equation, thus get the answer "YES" for question |x|<1 2) when |x|>1, we could know we will get the same answer in x<-1 and x>1 condition, and we could assure the answer is "NO" for target question so based on above, statement 2 is insufficient to solve the target question

we only left option C and E now. To test whether statements together will help to solve target question, we could use the denied solution x=3 in statement 2 to statement 1 to see whether it is one of the two solutions of equation.

If it is one of the solutions, then statement 2 will help to reduce the two solutions to one, thus, support the target question. We could feel free to choose option C If it is not one of the solution, then statement 2 will not help to reduce the number of solutions, thus, we could feel free to choose option E.

Let us test now. LS 3+1|=4 RS:2*|3-1|=4, we could know x=3 is one of the two answers. Thus we could choose C

Dear faceharshit, I'm happy to respond. I dare say, this problem is a little bit harder than what the GMAT will ask of you.

Statement #1: |x+1| = 2|x-1| If we are given |P| = |Q|, this means: P = Q OR P = -Q. Notice that the word "or" is not a piece of garnish there: rather, it is an essential piece of mathematical equipment.

|x + 1| = 2|x - 1|

Case I (x + 1) = 2(x - 1) x + 1 = 2x - 2 x = 3

Case II (x + 1) = -2(x - 1) x + 1 = -2x + 2 3x = 1 x = 1/3

This, from statement #1, we have x = 3 or x = 1/3. With this, we do not have sufficient information to answer the prompt question. This statement, by itself, is insufficient.

Statement #2: |x-3| > 0 Forget about everything we did in statement #1. Here, x could equal 10, in which case |x| is not less than 1, or x could equal 0, in which cases |x| is less than 1. We can pick different values that satisfy |x-3| > 0, x = 10 and x = 0, that give two different answers to the prompt question. Therefore, we do not have sufficient information to answer the prompt question. This statement, by itself, is insufficient.

Combined: #1 gives us x = 3 or x = 1/3 The value x = 3 does not satisfy the second statement, so we reject that value. The value x = 1/3 is only value that satisfies both statements, and with this, |x| < 1.

Combined, the statements are sufficient. Answer = (C)

'Is |x| < 1' implies 'Is distance of x from 0 less than 1?' i.e. does x lie within -1 and 1 (excluding the points -1 and 1)?

1. |x+1| = 2|x-1|

This tells you that distance of x from -1 is twice the distance of x from 1. There are two values of x for which this is possible:

Attachment:

Ques3.jpg [ 8.77 KiB | Viewed 760 times ]

The red line is twice the length of the blue line in both the cases. For the first case, x lies somewhere between 0 and 1 but for the second case, x lies at 3. Hence we can't answer whether x will lie between -1 and 1 from this statement alone.

2. |x-3| > 0 This tells us that x is a point whose distance from 3 is more than 0. That means it is not at 3 but on its left or right. This statement alone doesn't tell us whether x lies between -1 and 1.

Both statements together: Stmnt 1 tells us that x lies between -1 and 1 or at 3. Stmnt 2 tells us that x doesn't lie at 3. Then there is only one option left: x must lie between -1 and 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).

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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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